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9300 | https://mathoverflow.net/questions/434986/on-fredholm-alternative-for-neumann-conditions | fa.functional analysis - On Fredholm alternative for Neumann conditions - MathOverflow
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On Fredholm alternative for Neumann conditions
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Let Ω Ω be a bounded Lipschitz domain in R n R n and f∈L 2(Ω)f∈L 2(Ω). It is well known that if λ λ is a Dirichlet Laplacian eigenvalue, then the equation
{−Δ u=λ u+f u=0 in Ω on∂Ω{−Δ u=λ u+f in Ω u=0 on∂Ω
admits a weak solution in H 1 0(Ω)H 0 1(Ω) if and only if f f is perpendicular (in L 2 L 2 scalar product) to any eigenfunction of λ λ.
Now let us consider a similar problem with Neumann condition. That is, let μ μ be a Neumann Laplacian eigenvalue and consider the existence of the equation
{−Δ u=μ u+f∂u∂ν=0 in Ω on∂Ω{−Δ u=μ u+f in Ω∂u∂ν=0 on∂Ω
Is it true that a weak solution in H 1(Ω)H 1(Ω) exists if and only if f f is perpendicular to any eigenfunction of μ μ?
I know this is a necessary condition, since for any eigenfunction v v of μ μ, from the second Green identity we have that
∫Ω Δ u v d x−∫Ω Δ v u d x=0,∫Ω Δ u v d x−∫Ω Δ v u d x=0,
and thus ∫Ω f v d x=0∫Ω f v d x=0. I'm not sure whether this is a sufficient condition. It seems that the functional analysis method does not work. In the Dirichlet case, the natural space is the Hilbert space H 1 0(Ω)H 0 1(Ω). I don't know how to do the proof for Neumann boundary data. Any comment and ideas would be fully appreciated!
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asked Nov 20, 2022 at 15:08
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This works for every self-adjoint operator A A with a compact resolvent. Write A x=∑n λ n(x,e n)e n A x=∑n λ n(x,e n)e n to see that (λ k−A)x=y(λ k−A)x=y has a solution if and only if (y,e k)=0(y,e k)=0 and (x,e i)=(y,e i)/(λ k−λ i)(x,e i)=(y,e i)/(λ k−λ i).Giorgio Metafune –Giorgio Metafune 2022-11-20 18:58:15 +00:00 Commented Nov 20, 2022 at 18:58
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9301 | https://stats.libretexts.org/Bookshelves/Probability_Theory/Probability_Mathematical_Statistics_and_Stochastic_Processes_(Siegrist)/12%3A_Finite_Sampling_Models/12.03%3A_The_Multivariate_Hypergeometric_Distribution | Skip to main content
12.3: The Multivariate Hypergeometric Distribution
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: Apr 24, 2022
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12.2: The Hypergeometric Distribution
12.4: Order Statistics
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: 10246
Kyle Siegrist
University of Alabama in Huntsville via Random Services
( \newcommand{\kernel}{\mathrm{null}\,})
Basic Theory
The Multitype Model
As in the basic sampling model, we start with a finite population DD consisting of mm objects. In this section, we suppose in addition that each object is one of kk types; that is, we have a multitype population. For example, we could have an urn with balls of several different colors, or a population of voters who are either democrat, republican, or independent. Let DiDi denote the subset of all type ii objects and let mi=#(Di)mi=#(Di) for i∈{1,2,…,k}i∈{1,2,…,k}. Thus D=⋃ki=1DiD=⋃ki=1Di and m=∑ki=1mim=∑ki=1mi. The dichotomous model considered earlier is clearly a special case, with k=2k=2.
As in the basic sampling model, we sample nn objects at random from DD. Thus the outcome of the experiment is X=(X1,X2,…,Xn) where Xi∈D is the ith object chosen. Now let Yi denote the number of type i objects in the sample, for i∈{1,2,…,k}. Note that ∑ki=1Yi=n so if we know the values of k−1 of the counting variables, we can find the value of the remaining counting variable. As with any counting variable, we can express Yi as a sum of indicator variables:
For i∈{1,2,…,k} Yi=n∑j=11(Xj∈Di)
We assume initially that the sampling is without replacement, since this is the realistic case in most applications.
The Joint Distribution
Basic combinatorial arguments can be used to derive the probability density function of the random vector of counting variables. Recall that since the sampling is without replacement, the unordered sample is uniformly distributed over the combinations of size n chosen from D.
The probability density funtion of (Y1,Y2,…,Yk) is given by P(Y1=y1,Y2=y2,…,Yk=yk)=(m1y1)(m2y2)⋯(mkyk)(mn),(y1,y2,…,yk)∈Nk with k∑i=1yi=n
Proof
The binomial coefficient (miyi) is the number of unordered subsets of Di (the type i objects) of size yi. The binomial coefficient (mn) is the number of unordered samples of size n chosen from D. Thus the result follows from the multiplication principle of combinatorics and the uniform distribution of the unordered sample
The distribution of (Y1,Y2,…,Yk) is called the multivariate hypergeometric distribution with parameters m, (m1,m2,…,mk), and n. We also say that (Y1,Y2,…,Yk−1) has this distribution (recall again that the values of any k−1 of the variables determines the value of the remaining variable). Usually it is clear from context which meaning is intended. The ordinary hypergeometric distribution corresponds to k=2.
An alternate form of the probability density function of Y1,Y2,…,Yk) is P(Y1=y1,Y2=y2,…,Yk=yk)=(ny1,y2,…,yk)m(y1)1m(y2)2⋯m(yk)km(n),(y1,y2,…,yk)∈Nk with k∑i=1yi=n
Combinatorial Proof
The combinatorial proof is to consider the ordered sample, which is uniformly distributed on the set of permutations of size n from D. The multinomial coefficient on the right is the number of ways to partition the index set {1,2,…,n} into k groups where group i has yi elements (these are the coordinates of the type i objects). The number of (ordered) ways to select the type i objects is m(yi)i. The denominator m(n) is the number of ordered samples of size n chosen from D.
Algebraic Proof
There is also a simple algebraic proof, starting from the first version of probability density function above. Write each binomial coefficient (aj)=a(j)/j! and rearrange a bit.
The Marginal Distributions
For i∈{1,2,…,k}, Yi has the hypergeometric distribution with parameters m, mi, and n P(Yi=y)=(miy)(m−min−y)(mn),y∈{0,1,…,n}
Proof
An analytic proof is possible, by starting with the first version or the second version of the joint PDF and summing over the unwanted variables. However, a probabilistic proof is much better: Yi is the number of type i objects in a sample of size n chosen at random (and without replacement) from a population of m objects, with mi of type i and the remaining m−mi not of this type.
Grouping
The multivariate hypergeometric distribution is preserved when the counting variables are combined. Specifically, suppose that (A1,A2,…,Al) is a partition of the index set {1,2,…,k} into nonempty, disjoint subsets. Let Wj=∑i∈AjYi and rj=∑i∈Ajmi for j∈{1,2,…,l}
(W1,W2,…,Wl) has the multivariate hypergeometric distribution with parameters m, (r1,r2,…,rl), and n.
Proof
Again, an analytic proof is possible, but a probabilistic proof is much better. Effectively, we now have a population of m objects with l types, and ri is the number of objects of the new type i. As before we sample n objects without replacement, and Wi is the number of objects in the sample of the new type i.
Note that the marginal distribution of Yi given above is a special case of grouping. We have two types: type i and not type i. More generally, the marginal distribution of any subsequence of (Y1,Y2,…,Yn) is hypergeometric, with the appropriate parameters.
Conditioning
The multivariate hypergeometric distribution is also preserved when some of the counting variables are observed. Specifically, suppose that (A,B) is a partition of the index set {1,2,…,k} into nonempty, disjoint subsets. Suppose that we observe Yj=yj for j∈B. Let z=n−∑j∈Byj and r=∑i∈Ami.
The conditional distribution of (Yi:i∈A) given (Yj=yj:j∈B) is multivariate hypergeometric with parameters r, (mi:i∈A), and z.
Proof
Once again, an analytic argument is possible using the definition of conditional probability and the appropriate joint distributions. A probabilistic argument is much better. Effectively, we are selecting a sample of size z from a population of size r, with mi objects of type i for each i∈A.
Combinations of the grouping result and the conditioning result can be used to compute any marginal or conditional distributions of the counting variables.
Moments
We will compute the mean, variance, covariance, and correlation of the counting variables. Results from the hypergeometric distribution and the representation in terms of indicator variables are the main tools.
For i∈{1,2,…,k},
E(Yi)=nmim
var(Yi)=nmimm−mimm−nm−1
Proof
This follows immediately, since Yi has the hypergeometric distribution with parameters m, mi, and n.
Now let Iti=1(Xt∈Di), the indicator variable of the event that the tth object selected is type i, for t∈{1,2,…,n} and i∈{1,2,…,k}.
Suppose that r and s are distinct elements of {1,2,…,n}, and i and j are distinct elements of {1,2,…,k}. Then cov(Iri,Irj)=−mimmjmcov(Iri,Isj)=1m−1mimmjm
Proof
Recall that if A and B are events, then cov(A,B)=P(A∩B)−P(A)P(B). In the first case the events are that sample item r is type i and that sample item r is type j. These events are disjoint, and the individual probabilities are mim and mjm. In the second case, the events are that sample item r is type i and that sample item s is type j. The probability that both events occur is mimmjm−1 while the individual probabilities are the same as in the first case.
Suppose again that r and s are distinct elements of {1,2,…,n}, and i and j are distinct elements of {1,2,…,k}. Then cor(Iri,Irj)=−√mim−mimjm−mjcor(Iri,Isj)=1m−1√mim−mimjm−mj
Proof
This follows from the previous result and the definition of correlation. Recall that if I is an indicator variable with parameter p then var(I)=p(1−p).
In particular, Iri and Irj are negatively correlated while Iri and Isj are positively correlated.
For distinct i,j∈{1,2,…,k},
cov(Yi,Yj)=−nmimmjmm−nm−1cor(Yi,Yj)=−√mim−mimjm−mj
Sampling with Replacement
Suppose now that the sampling is with replacement, even though this is usually not realistic in applications.
The types of the objects in the sample form a sequence of n multinomial trials with parameters (m1/m,m2/m,…,mk/m).
The following results now follow immediately from the general theory of multinomial trials, although modifications of the arguments above could also be used.
(Y1,Y2,…,Yk) has the multinomial distribution with parameters n and (m1/m,m2,/m,…,mk/m): P(Y1=y1,Y2=y2,…,Yk=yk)=(ny1,y2,…,yk)my11my22⋯mykkmn,(y1,y2,…,yk)∈Nk with k∑i=1yi=n
For distinct i,j∈{1,2,…,k},
E(Yi)=nmim
var(Yi)=nmimm−mim
cov(Yi,Yj)=−nmimmjm
cor(Yi,Yj)=−√mim−mimjm−mj
Comparing with our previous results, note that the means and correlations are the same, whether sampling with or without replacement. The variances and covariances are smaller when sampling without replacement, by a factor of the finite population correction factor (m−n)/(m−1)
Convergence to the Multinomial Distribution
Suppose that the population size m is very large compared to the sample size n. In this case, it seems reasonable that sampling without replacement is not too much different than sampling with replacement, and hence the multivariate hypergeometric distribution should be well approximated by the multinomial. The following exercise makes this observation precise. Practically, it is a valuable result, since in many cases we do not know the population size exactly. For the approximate multinomial distribution, we do not need to know mi and m individually, but only in the ratio mi/m.
Suppose that mi depends on m and that mi/m→pi as m→∞ for i∈{1,2,…,k}. For fixed n, the multivariate hypergeometric probability density function with parameters m, (m1,m2,…,mk), and n converges to the multinomial probability density function with parameters n and (p1,p2,…,pk).
Proof
Consider the second version of the hypergeometric probability density function. In the fraction, there are n factors in the denominator and n in the numerator. If we group the factors to form a product of n fractions, then each fraction in group i converges to pi.
Examples and Applications
A population of 100 voters consists of 40 republicans, 35 democrats and 25 independents. A random sample of 10 voters is chosen. Find each of the following:
The joint density function of the number of republicans, number of democrats, and number of independents in the sample
The mean of each variable in (a).
The variance of each variable in (a).
The covariance of each pair of variables in (a).
The probability that the sample contains at least 4 republicans, at least 3 democrats, and at least 2 independents.
Answer
P(X=x,Y=y,Z=z)=(40x)(35y)(25z)(10010) for x,y,z∈N with x+y+z=10
E(X)=4, E(Y)=3.5, E(Z)=2.5
var(X)=2.1818, var(Y)=2.0682, var(Z)=1.7045
cov(X,Y)=−1.6346, cov(X,Z)=−0.9091, cov(Y,Z)=−0.7955
0.2474
Cards
Recall that the general card experiment is to select n cards at random and without replacement from a standard deck of 52 cards. The special case n=5 is the poker experiment and the special case n=13 is the bridge experiment.
In a bridge hand, find the probability density function of
The number of spades, number of hearts, and number of diamonds.
The number of spades and number of hearts.
The number of spades.
The number of red cards and the number of black cards.
Answer
Let X, Y, Z, U, and V denote the number of spades, hearts, diamonds, red cards, and black cards, respectively, in the hand.
P(X=x,Y=y,Z=z)=(13x)(13y)(13z)(1313−x−y−z)(5213) for x,y,z∈N with x+y+z≤13
P(X=x,Y=y)=(13x)(13y)(2613−x−y)(5213) for x,y∈N with x+y≤13
P(X=x)=(13x)(3913−x)(5213) for x∈{0,1,…13}
P(U=u,V=v)=(26u)(26v)(5213) for u,v∈N with u+v=13
In a bridge hand, find each of the following:
The mean and variance of the number of spades.
The covariance and correlation between the number of spades and the number of hearts.
The mean and variance of the number of red cards.
Answer
Let X, Y, and U denote the number of spades, hearts, and red cards, respectively, in the hand.
E(X)=134, var(X)=507272
cov(X,Y)=−169272
E(U)=132, var(U)=169272
In a bridge hand, find each of the following:
The conditional probability density function of the number of spades and the number of hearts, given that the hand has 4 diamonds.
The conditional probability density function of the number of spades given that the hand has 3 hearts and 2 diamonds.
Answer
Let X, Y and Z denote the number of spades, hearts, and diamonds respectively, in the hand.
P(X=x,Y=y,∣Z=4)=(13x)(13y)(229−x−y)(489) for x,y∈N with x+y≤9
P(X=x∣Y=3,Z=2)=(13x)(348−x)(478) for x∈{0,1,…,8}
In the card experiment, a hand that does not contain any cards of a particular suit is said to be void in that suit.
Use the inclusion-exclusion rule to show that the probability that a poker hand is void in at least one suit is 19134962598960≈0.736
In the card experiment, set n=5. Run the simulation 1000 times and compute the relative frequency of the event that the hand is void in at least one suit. Compare the relative frequency with the true probability given in the previous exercise.
Use the inclusion-exclusion rule to show that the probability that a bridge hand is void in at least one suit is 32427298180635013559600≈0.051
12.2: The Hypergeometric Distribution
12.4: Order Statistics |
9302 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8023132/ | Bacteriophages presence in nature and their role in the natural selection of bacterial populations - PMC
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. 2020 Nov 9;91(Suppl 13):e2020024. doi: 10.23750/abm.v91i13-S.10819
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Bacteriophages presence in nature and their role in the natural selection of bacterial populations
Zakira Naureen
Zakira Naureen
1 Department of Biological Sciences & Chemistry, College of Arts and Sciences, University of Nizwa, Nizwa, Oman
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1, Astrit Dautaj
Astrit Dautaj
2 EBTNA-LAB, Rovereto (TN), Italy
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2, Kyrylo Anpilogov
Kyrylo Anpilogov
3 MAGI EUREGIO, Bolzano, Italy
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3,✉, Giorgio Camilleri
Giorgio Camilleri
4 MAGI’S LAB, Rovereto (TN), Italy
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Kristjana Dhuli
2 EBTNA-LAB, Rovereto (TN), Italy
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4 MAGI’S LAB, Rovereto (TN), Italy
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4 MAGI’S LAB, Rovereto (TN), Italy
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Luca De Antoni
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Tommaso Beccari
5 Department of Pharmaceutical Science, University of Perugia, Perugia, Italy
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5, Munis Dundar
Munis Dundar
6 Department of Medical Genetics, Faculty of Medicine, Erciyes University, Kayseri, Turkey
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Matteo Bertelli
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1 Department of Biological Sciences & Chemistry, College of Arts and Sciences, University of Nizwa, Nizwa, Oman
2 EBTNA-LAB, Rovereto (TN), Italy
3 MAGI EUREGIO, Bolzano, Italy
4 MAGI’S LAB, Rovereto (TN), Italy
5 Department of Pharmaceutical Science, University of Perugia, Perugia, Italy
6 Department of Medical Genetics, Faculty of Medicine, Erciyes University, Kayseri, Turkey
✉
Correspondence: Kyrylo Anpilogov Via Maso della Pieve, 60/A, 39100, Bolzano, Italy E-mail: kirill.anpilogov@assomagi.org
✉
Corresponding author.
Received 2020 Sep 3; Accepted 2020 Oct 14; Issue date 2020.
Copyright: © 2020 ACTA BIO MEDICA SOCIETY OF MEDICINE AND NATURAL SCIENCES OF PARMA
This work is licensed under a Creative Commons Attribution 4.0 International License
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PMCID: PMC8023132 PMID: 33170167
Abstract
Phages are the obligate parasite of bacteria and have complex interactions with their hosts. Phages can live in, modify, and shape bacterial communities by bringing about changes in their abundance, diversity, physiology, and virulence. In addition, phages mediate lateral gene transfer, modify host metabolism and reallocate bacterially-derived biochemical compounds through cell lysis, thus playing an important role in ecosystem. Phages coexist and coevolve with bacteria and have developed several antidefense mechanisms in response to bacterial defense strategies against them. Phages owe their existence to their bacterial hosts, therefore they bring about alterations in their host genomes by transferring resistance genes and genes encoding toxins in order to improve the fitness of the hosts. Application of phages in biotechnology, environment, agriculture and medicines demands a deep insight into the myriad of phage-bacteria interactions. However, to understand their complex interactions, we need to know how unique phages are to their bacterial hosts and how they exert a selective pressure on the microbial communities in nature. Consequently, the present review focuses on phage biology with respect to natural selection of bacterial populations. (www.actabiomedica.it)
Keywords: bacteriophages, spatial and temporal distribution, ecology, microbial communities, natural selection
Introduction
Phages cannot carry out most of the biological processes on their own and require a live host in order to propagate. As a result, phages are obligate parasites to their bacterial hosts and dwell in almost all habitats where bacteria exist (1,2). In addition to bacteria, phages can infect and parasitize archea and are also found in some eukaryotes (1). Bacteriophages are the most abundant organisms on Earth and have an important role in constant regulation of the diversity, richness, abundance, evolution, and physiology of microbial communities in a particular habitat (1,2). Phages have been extensively studied in the past two decades for their role in antimicrobial resistance, food safety, food processing, agriculture, environment, and medicine (3-6). However, they are of utmost importance with respect to their effect on bacterial population dynamics in natural environment as well as in laboratory conditions (7).
The coexistence between phages and bacteria seems implausible because phages have a massive proliferative advantage over their bacterial hosts. The reproduction cycles of phage and bacteria occurs in comparable time frames, with both happening within an hour (7). However, after each generation cycle phage produces ~100-200 new phage particles while one bacterial cell divided into two daughter cells only. Consequently, phages would quickly outnumber and destroy the host bacterial population (8,9). Nevertheless, this does not happen. As a result, both have developed ways and means to resist each other’s defense mechanism thereby maintaining a balance between their populations (10-12). However, to coevolve, both need to coexist for a certain length of time to allow beneficial mutations to take place (8-10). Hence, understanding phage-bacteria interactions is critical to understand the population dynamics of bacteria in nature. Consequently, the present review aims to gain insight into biology of bacteriophages and their unique role for bacteria and how their interactions exert a selective barrier in the microbial population dynamics in nature.
Discovery of bacteriophage
Phages were independently discovered by Frederick Twort in England (12) and Felix d’Herelle, (13), in Paris at the Pasteur Institute. Twort was trying to produce a vaccine against a virus without a host cell in an artificial medium (14) while D’Herelle was studying an extreme outbreak of dysentery among the soldiers in Paris during the First World War, when he discovered bacteriophage, a bacterium feeding body (15). He applied this to bacterial culture and observed that turbid cultures turned into clear indicating the lytic nature of bacteriophages against bacteria (16). His thesis on the bacteriophage was published in a monograph, “The bacteriophage and its behavior” (17), along with several other books and papers subsequently. This research laid the foundation of “bacteriophagology” (18). A new vocabulary was used to describe post-infection activities and explained the purification and titration of bacteriophage culture in some detail.
Structure of bacteriophages
The bacteriophages biology suggests that phages are well suited to their local host communities (19). Phages possess a structure well-suited to achieve their objectives: identifying a suitable host bacterium, taking advantage of the metabolism of the host and creating several progeny phages that are well built enough to survive before a new host bacterium is found to infect. The composition of bacteriophage is simple, it consists of genetic material made of either DNA or RNA, double or single stranded, enclosed in a protein capsid. The three basic structural forms of phage (20) include an icosahedral head with a tail, an icosahedral head without a tail and filamentous form (Figure 1).
Figure 1.
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Representation of the bacteriophages families
Phage families and their structure
The family of Leviviridae infects bacteria, includes Enterobacter, Acinetobacter and Pseudomonas, consists of small icosahedral bacteriophages with linear, single-stranded RNA genome that functions as messenger RNA encoding four proteins: the coat protein, the replicase, the maturation protein, and the lysis protein (21). This family includes two genera: Levivirus (having two species MS2 and BZ13) and Allolevivirus (containing Qβ and F1).
The Microviridae is a family of small icosahedral bacteriophages with a single-stranded, circular plus strand DNA genome, and the minus strand is produced intracellularly to be used as a template for the mRNA generation by transcription. This bacterial virus attacks either the free-living enterobacteria or the obligate intracellular bacterial parasite that lack cell wall like chlamydia or spiroplasma (22).
Another viral family known as Inoviridae consists of a rod-shaped filamentous structure with a single-stranded circular plus DNA. Such bacteriophages mostly infects gram-negative bacteria (23).
The Cystoviridae family of bacteriophage has nucleocapsid that contains double stranded RNA segments, and the capsid is covered by a lipid membrane layer. These bacteriophages infect Pseudomonas bacteria (24). The virion is enveloped with typically two concentric, icosahedral symmetric protein layers.
The Tectiviridae is a double-stranded DNA phage family that infects the Gram-negative bacteria. These phages possess an icosahedral capsid, which resembles an adenovirus capsid, that encloses an internal membrane derived from the host, decorated with spikes at fivefold vertices (25) and lack the tail in their structure.
The Caudovirales order consists of three bacteriophage families named Myoviridae (long contractile tail), Siphoviridae (long flexible tail), and Podoviridae (short noncontractile tail) having a tail with an icosahedral or prolate capsid containing a linear double-stranded DNA genome (26).
The Plasmaviridae is a family of bacteriophages that are membrane enveloped and infect the bacterial host without a cell wall. The genome consists of a circular, supercoiled double-stranded DNA (27).
Life cycle of phages
Phages have various life cycle and can be easily classified on the basis of their life cycle: lytic, lysogenic, pseudo-lysogeinc and chronic.
Lytic cycle
In the lytic cycle, the phage injects its DNA into the host bacterium after adsorbing on to the host surface. This induces the shift of the host bacterium’s protein machinery to replicate phage DNA and develop 50-200 new infectious phages (28). As a result, the host become weak and eventually ruptures to release the new phage particles. Lytic life cycle plays an important role in biogeochemical cycling by releasing the organic molecules in nature by degrading the bacterial cells.
Lysogenic cycle
The lysogenic phase of the phage life cycle is characterized by the incorporation of the phage DNA into the host genome, which may survive as a plasmid, and the resulting new phages will have a combination of phage DNA and the host genome (28). Such a transformation of viral DNA might take place without significant metabolic implications for over many generations of bacteria. The phage genes can return to the lytic cycle under certain circumstances that hinder the bacterial state, leading to the release of completely assembled phages.
The uncommon but important phage life cycles
There are two other life cycles of bacteriophages known as pseudo-lysogenic and carrier state. In the pseudo-lysogenic cycle the phage nucleic acid after infecting the host neither form a stable long-term connection (lysogeny), nor induces lytic response, instead the phage nucleic acid stays inside the host cell in an inactive form (29). The phage develops this type of interaction in order to avoid starvation and elimination from the environment when the nutrients are limited, and bacterial hosts are lacking (29).
The carrier state of phage life cycle is chronic because phage causes long-term infection to the bacterial host and the phage progeny is continuously budded off the cell or are passed down to the daughter cells asymmetrically (30).
Bacteriophage distribution in nature
A variety of different forms of bacteriophages are present in different ecological niches. The spatial and temporal distribution of bacteria and phage depends on their range limits and where their ranges overlap. It is widely known that bacteria dwell almost everywhere in every conditions (31), similarly phages exist in all niches wherever their hosts are present, including hypersaline environments (32), polar regions (33), deserts (34), on and within organisms other than bacteria (35,36), fresh and sea water (37) and the soil (38). The isolation of phages from a particular environment and their culturing on bacterial lawns is required to identify phages from a particular environment. Nonetheless, this is not always possible like in unculturable bacterial strains (39).
Epiflurescent microscopy following DNA staining were the first approaches that led to the realization of the phage abundance in marine environment e.g. approximately 10 phages for each bacterial and archaeal cell are present in marine water (40-42). For freshwater ecosystem, similar statistics have been shown, but the situation is less evident for other more complex environments, and the number of phages can be either higher or lower than that of the bacterial or archaeal hosts (43). In this section we will have a look at the phage’s distribution in various niches.
Cyanophages of Marine Ecosystem
The studies on marine phages started in the early 1980s, when phages were found in the Black Sea, infecting both unicellular and filamentous cyanobacteria (44). The predominant primary producers of the nutrient-poor ocean areas covering around 70% of the Earth’s surface are the two genera of cyanobacteria known as Prochlorococcus and Synechococcus. The hot oceanic water usually has Prochlorococcus between 40° north latitude and 40° south latitude, whereas Synechococcus is much harder and is found on both sides of those latitudes (45). PSS2 is the first temperate marine cyanophage to be isolated and its genome sequenced (46). At present the vast majority of the isolated cyanophages are obligately lytic (47). A research in the Red Sea over an annual period found that cyanophages co-vary with Synechococcus in abundance and genetic diversity, which is consistent with the hypothesis that cyanophages are an important factor in regulating ecological cyanobacterium succession (48).
Phages in the Animal Environment
Felix D’Herelle reported that bacteriophages are a natural part of healthy animal and human microbiota (49). Electron microscopy findings first revealed the high concentration of phage-like particles in the intestinal microbial systems in the 1960s (50). The concentrations of the virus-like particles in the feces in humans is calculated based on the yield of the total phage DNA (51) and is estimated to be 10 10 ml−1. Tailed phages tend to be linked to the vast majority of intestinal virus-like particles. In a recent review, in one sample of horse feces, 69 morphologically distinguishable forms of bacteriophages were found in over 200 examined samples (52).
Archaeal Phages
Archaea are as common as the Bacteria in oceans, soils and subterranean environments (53-55). Archae can inhabit in extreme conditions such as hot spring, salt and soda lakes. Archeal viruses are believed to be as abundant as bacterial viruses however so far there are only 50 phages known to infect the Archaea (56). The viruses that can infect Archaea have double stranded DNA genome varying in size from 10kb to 100 kb (57). No archeal RNA virus has been identified so far. These viruses have several unusual morphologies not seen in eukarya or bacterial viruses earlier (58,59).
Phages in Human Gut
Human body is the greatest reservoir of microbiota infested with bacteriophages (60), mostly living inside the gut. Hence, the presence of phages come into consideration as they outnumber the bacterial population present in the body compartments by at least 10 folds (61-63). The human body comprises mucosal surfaces, where microbes inhabit and communicate with their host directly (64). These mucosal membranes serve as an entry point for pathogens (65). Bacteriophages, can bind to mucin glycoproteins through immunoglobulin-like spike present in their capsid viz “bacteriophage adherence to mucin” (BAM). BAM plays two significant roles in regulating the bacterial-host interactions: it provides protection for host against the pathogenic bacteria by accommodating bacteriophages with lytic activity that would otherwise demolish the beneficial bacteria leading to local or systemic infection (66); it provides lysogenic bacteria an environment to develop bacterial symbiotic relationship that benefits the human host (67). It has also been demonstrated that the gut epithelium, in response to the bacterial strains, can actively modulate BAM and the bacteriophage composition through hyper-secreting mucins and changing the mucin glycosylation patterns to prevent microbial adhesion and survival (68). The integrated prophages often show genes to enhance bacterial strain’s fitness and virulence (69), on the other side the free phages help their bacterial host by killing their competitor bacterial species (70).
Phages in Dairy Environment
Phages are rather unwanted in dairy environment as they can kill the microbial starter cultures, thus contaminating the raw milk (71). The outburst of phages can cause significant economic losses due to delay in the production, wasting of ingredients, low product quality, increase spoilage and even total loss of production (72). The commonly found pathogens in the dairy environment include bacterial pathogens like Brucella spp., Camphylobactoer jejuni, Bacillus cereus, Shiga-toxin producing E.coli, Staphylococcus aureus, Mycobacterium bovis, Salmonella spp., and Yersinia enteroclitica, viruses, fungi and parasites (73).
However, phages are useful in cheese production. Cheese manufacture process is the most effective process to study the ubiquity of phages in the dairy environment (74). Large volume of raw milk is daily fermented by lactic acid bacteria starters with Lactococcus lactis as it is extensively used bacterial species. Consequently, the Lactococcus lactis phages are also studied widely followed by S treptococcus thermophiles bacteriophage (74,75).
Role of phages in natural selection of bacteria
Although the exact role of natural phages in shaping a microbial community is not fully understood, they might preferentially select some of the bacterial species and eliminate others by various mechanisms such as host selectivity, horizontal gene transfer, driving bacterial evolution and mediating competitions between bacterial communities (76).
The reciprocal selective pressure exerted by phages on bacteria and viceversa suggests that phage-bacteria interactions have a decisive impact on diversity, virulence, and evolution of bacteria (77).
Phage life cycle
Lytic life cycle is of vital importance in removing and eradicating the bacterial species from a particular niche and has been explored extensively with respect to medical applications. However, also the lysogenic, pseudo lysogenic and chronic life cycles deserve a special place in phage-bacteria interactions (78). Lysogenic life cycle contributes to development of bacterial resistance to lytic phages by acting as prophages such as Spi phenotype and via horizontal gene transfer by incorporating a part of bacterial genome and transferring resistance genes to the next host. This results into development of bacterial colonies that are resistant to lytic phages thereby giving them a selective advantage over the susceptible phages (79). Pseudolysogeny is debatable with respect to being real life cycle or just a pause in the phage replication in nutrient deficient conditions, still it has important implications on bacterial population dynamics in each environment. On the other hand, the chronic infection of bacteria by phages mediated by the carrier life cycle mode is an important method to slow down the growth rate of bacterial species. A classic example of such a life cycle is the M13 bacteriophage infection in E. coli. M13 neither lyses the cell nor exists as a prophage and maintains its DNA as an extrachromosomal element, thus utilizing bacterial resources for its continuous replication and packaging into new M13 particles which then are released through pores in bacterial cell membrane without lysing the cells (80). Similar relation is seen between Pseudomonas syringe and ϕ6 phage which is used to control P. syringe in Kiwi plants (81). Similarly, Campylobacter jejuni infection with bacteriophages CP8 and CP30 renders it inefficient in colonization in chicken gut thereby effecting its population dynamics (82).
The decrease in growth rate of these bacteria by chronic phages results in a reduction in number of these bacteria as compared to uninfected bacterial species in a particular habitat thus depicting their potential in shaping microbial communities.
Bacterial resistance to phage infections
The survival and existence of bacteria and phages is interdependent. Phages need to infect and replicate inside bacteria while bacteria need to eradicate phages and propagate. In this war between bacteria and phages, bacteria have developed various strategies of offence and defense to get rid of phages that include CRISPR-Cas, prokaryotic argonauts, restriction-modification, surface modification, toxin-antitoxin and abortive infection system in the hosts (83-85). These defense responses vary in many ways, including the number of phages that can be taken care of by these and the cost to bacterial survival (86). Two of these mechanisms surface modification and CRISPR-Cas systems, are often involved in rapid progression of resistance against phages. The simplest method of bacterial resistance to phage is surface modification that involves mutation of the bacterial cell surface receptor used by phage for attachment (87). CRISPR-based immunity on the other hand is a very sophisticated method to degrade phage genomic material (88).
Phage response to bacterial resistance
To counteract the bacterial defense mechanisms, phages also developed a myriad of tools ranging from modification of the attachment proteins to anti CRISPR proteins. Phages can go to any extent to coevolve with the bacteria to maintain their existence (89). For instance, phages have developed an entire bacterial nucleic acid degrading system as discovered in ICP1 vibrio phages that encode a CRISPR-Cas system in their genome (90). Besides that, they manifest point mutations in their genome at sites targeted by the specific host (91-93). Apart from point mutation, phages have developed anti-CRISPR mechanism (Acr gene) that inhibits CRISPR-Cas evolutionary mechanism of bacteria (94). This Acr gene involved in rescuing phages from extinction was identified in temperate Pseudomonas phage (95). Studies have suggested that some of the Acr genes are more potent as compared to others, for instance Acr IF1 shows great potency against CRISPR resistant host as compared to the phages that had AcrIF4 (94,95).
Phage host range
It was previously thought that phages were specific for their hosts and this is true for some bacteriophages which are species- and strain-specific, however, their diversity in nature suggests that at least some have a broader host range (96). Phages expand their host range with respect to the evolving bacteria, for example the lytic phage SBW25ϕ2 coevolves with its host Pseudomonas fluorescence by continually increasing host range to infect the previously resistant strains, this increase in host range however does not increase the infectivity (97,98).
By extending the host range, a single phage can exert selective pressure on various microbial populations leading to the development of evolutionary resistance profiles and mutations in their genomes. Hence, host specificity and range are important factors in natural selection of microbial population in nature and can be used to predict the response of host population and community to phage-based selective pressure (19).
Impact of phages on bacterial population and communities
In phage-bacteria relationship phages are the predators that can modulate bacterial population in many ways such as driving the abundance and diversity of bacterial species; changes in their physiology, competitive ability, and virulence (99). Long periods of repeated interaction between predator and prey brings about their coevolution termed as Red Queen hypothesis (100) in which reciprocal selection cycles change the biotic selective environment of both parties over time (100-103).
Abundance and diversity
Experimental evidence from chemostats and phage-host observations in an open system has shown that populations of phages and hosts oscillate with time for certain bacterial species (104). The most evident effect that a lytic phage has on bacteria is the reduction of its number (105). Several studies have shown that phages have a major role in the bacterial mortality. Phages are 20-30 % more abundant than bacteria and to maintain themselves they cause 10 24 infections per second (106). These infection account for eliminating 4-50 % of total bacterial population in nature at a given time (106). For example, 20-40% bacteria are eliminated by the phages in marine surfaces indicating the phage capability of out competing bacterial populations in some environmental conditions (95,107). This indicates the enormous predation potential of bacteriophages however evolution of bacterial resistance against phages neutralizes the situation thus maintaining a balance between the two (108). Besides lytic activities phages can reduce the bacterial number by chronic infections whereby a phage decreases the rate at which a bacteria propagates. The decrease in growth rate of one bacterial species with respect to another gives and advantage to the fast-growing species thus changing the microbial community richness (107,108). In addition to this, lysogenic bacteria give an advantage to bacteria conferring them resistance against lytic phages. Large seasonal fluctuations in the bacterial populations of natural aquatic environment is because of lysogenic phages integrated in their genome indicating the effect of lysogeny on shaping bacterial communities (109).
Phage mediated gene trafficking
Phages can bring about changes in their host genome by several ways. One of the most common method of gene trafficking by phages is incorporation of a part of host genome in their capsid while packaging inside the host. This genetic material is then transferred to other bacterial cells by a process called generalized transduction (110,111). Antibiotic resistance genes are transferred from one bacterium to another through this mechanism. Bacterial populations acquiring these genes act as reservoirs (112). Besides this general transduction there is a specialized transduction in which the phages selectively incorporate targeted sequences from bacterial genome into their own genome. These acquired genetic sequences may provide an edge to phages in changing bacterial physiology in their own favor such as enhanced susceptibility, impaired DNA repair mechanism of bacteria accelerating the mutation rate which might increase phage susceptibility or resistance. In some cases, bacteria may invert one third of their genome in response to selective pressure mediated by phages (113).
Changes in bacterial physiology
Phages may possess genes that are homologous to bacterial metabolic genes and these auxiliary metabolic genes might alter bacterial metabolism in favor of infecting phages (77). For instance, cyanophages may carry genes for photosynthesis that can be used to generate energy after the host stops synthesizing photosynthetic proteins (114). This allows phages to continue replicating in their hosts. In addition, prophages enable bacteria to withstand various stressful environmental conditions. Removal of prophage from E. coli revealed that the prophage was integral to the bacteria for withstanding the unfavorable environmental conditions (115).
Bacterial Virulence
Phages can alter bacterial virulence in addition to changing its physiology. Phage mediated addition of virulence genes in bacteria improves their chances of survival and might increase phage host range thus ultimately benefitting the phages by providing them a chance to continue replication. Occasionally these phage-mediated virulence factors result in an increase in host pathogenicity thereby playing a crucial role in many bacterial infections in human and animals. Phages can enhance bacterial pathogenicity by several ways including transferring genes encoding toxins, duplicating the virulence factors, or changing the regulatory sequences controlling bacterial virulence genes (116). Numerous studies have been conducted in this regard that report phage-mediated toxin encoding gene transfer in bacteria that result in a variety of diseases in humans and animals. In addition, phages may promote biofilm formation, antimicrobial resistance, immune resistance, and increased virulence in their bacterial hosts (89) like in Vibrio cholarae (117), Pseudomonas aeruginosa (118) and Shigella dysnteriae (119).
Phage-bacteria interdependence
It is obvious that a particular phage will survive if its ultimate host will survive so the balance between phage mediated bacterial lysis and bacterial population density is maintained. On the other hand, this can exert selective pressure on both phage and bacterial abundance and distribution (120,121). Eradication of a particular host bacteria from a particular niche means elimination of its parasitic phage too. Phages answer to this situation is increase in host range. Thus, phages infecting multiple hosts have a selective advantage over those which are highly host-specific. Thus, both predator and prey are integral part of each other’s existence.
This interaction of phages and bacteria is also influenced by the environment they are in. for instance if the nutrient supply is limited and bacteria cannot grow at an efficient rate, the phages switch from lytic mode to lysogenic mode. Phages can live as phage particles outside bacterial host and have a greater tolerance to the environmental stress (122). Increase tolerance to environmental conditions may increase phages survival during periods of host abundance fluctuations and may give them time to find another suitable host (123).
Phage mediated natural selection of bacterial communities is explained by two hypotheses.
Negative frequency dependent selection (FDS)
Phages can mediate frequency dependent selection (FDS) of bacterial populations by adapting to the most dominant host strains and having a significant impact on their fitness. Studies in this regard have indicated that phages are well adapted to their local hosts population as compared to those coming from other populations (35,76). Alternatively changes in abundance of a particular phage genotype is correlated to changes in abundance of its respective host (83). Artificial removal of phages from a particular niche is correlated with a resultant increase in number of previously rare bacterial species, while adding phages to particular niche corresponds to an increase in specie richness (124) indicating role of phages in abundance and richness of bacterial populations. Similarly, studies on lysogenic E. coli strains have revealed that the relative fitness of the host was highest when the lysogenic phage was rare and vice versa (125).
A recent review on phage-bacteria interaction by Cordero & Polz, 2014 suggest that phage-mediated FDS and other ecological factors such as competition and predation results in stable diversity and rapid turnover of genes in a given population (126).
The Kill the Winner hypothesis
This model assumes that host growth rate is positively linked with viral fitness, and consequently abundance. Thus, phage infection is proportional to the relative host abundance and this negates inherent fitness advantage of particular bacterial hosts. This model also suggests that viruses can maintain coexistence of bacterial species having varying intrinsic growth rates thus implicating phage involvement in shaping microbial communities (127). A further modification in this model suggested that bacteria face a trade-off between phages resistance and growth rate. This means that the slow growing resistant bacterial species dominate the microbial communities along with the phages attacking fast growing bacterial species (128). This clarifies the fact that there is a cost associated with phage resistance and bacteria pays that cost by lowering their growth rate. Furthermore, this model explains the inverse abundance distribution observed in marine environments where large proportions of rare microbial communities are highly active (129).
Phage-mediated competition among bacterial species
The specificity of phage-bacteria interaction shapes the relative fitness of bacterial species in a particular habitat. A phage infecting two hosts may have variations in rate of adsorption, time to cell lysis, or burst size across the hosts. As a result, one bacterial species will have an advantage over the other with respect to the presence, but not absence, of the phage (82). If both hosts eventually become resistant to the phages by paying different costs than phage-mediated selection might alter competition between them even in the absence of the phage (76). This competition can become more evident if one host specie is less sensitive and the other more susceptible to the same phage. Consequently, the tolerant species would dominate in a particular niche.
Effect of phage-bacteria interactions on non-bacterial species
Phages might have a strong influence on bacterial interactions with their hosts such as animals, plants and humans. An important example of this tricentric interaction is phage-gut microbiota within human gut that plays a central role in shaping human gut microbiome (130). The highest number of bacteriophages in the nature is present in human gut 10 15 (131) out of the approximate number of phage particles in the biosphere (10 31) (77). In fact, Twort in 1915 and d’ Herelle in 1917 co-discovered bacteriophages by examining the gut microbiota of humans and animals (13,14). A relatively recent study in mouse model reported that gut phages have measurable effects on diversity and abundance of gut microbiota (34). Another classical example is that of flamingoes which eat cyanobacteria. The decline in flamingo population was found to be a consequence of cyanobacterial degradation by cyanophages (132). Moreover, phages play an important role in phage, bacteria-insect symbiosis (36). Phages encode virulence genes that are important with respect to protection of wasps from parasitism (133) and they can also reduce the cytoplasmic incompatibility caused by endosymbiont in insects (36).
Role of phage-bacteria interactions in ecosystem
Bacteriophages affect both biotic and abiotic components of ecosystem by changing biogeochemical cycles (134). Release of nucleic acids, proteins and lipids by the phage mediated degradation of bacteria significantly contributing to the biogeochemical cycling of carbon, nitrogen, and phosphorus (135). Marine food web microcosm studies have evidenced strong effect of phages on available phosphorus and the ratios of carbon, nitrogen, and phosphorus (136). In addition to these macro elements, microelements such as iron is also released due to bacterial motility. Moreover, phages in marine environment negatively effects the primary productivity. For instance, increasing virus abundance in seawater, reduces bacterial species and consequently primary productivity of phytoplankton by up to 78% (137). As 50% of global primary productivity can be attributed to phytoplankton, it can be stated that tiny viruses hold an important place in the global food web (138).
Conclusion
It can be concluded here that phages are the main drivers of bacterial diversity, from the genome level to the population level and up to community’s level. The mechanism underlying capability of phages might include the frequency-dependent host-specific phage adaptation, correlations between bacterial abundance and associated phage density in a particular habitat, and trade-offs between growth and defense expenditure against phages. This suggests that phage-mediated selection plays a critical role in mediating competition among bacterial hosts thus driving niche diversification. Phage-bacteria interactions could be expressed in terms of predator-prey relationship in which presence of bacteria is necessary to ensure phage existence while presence of phage is necessary to keep bacterial population at bay in natural environment. This check and balance are the key to shaping microbial population in a particular habitat and is necessary for the coexistence of both parties. This also has large impacts on environment. Phages sitting at the bottom of global food web are the game changers in selecting microbial communities and replenishing macro and micronutrients in the ecosystem by biogeochemical cycling. More studies in this regard are however required to reap the benefits of phages in various biotechnological, environmental, and medical applications.
Conflict of interest:
Each author declares that he or she has no commercial associations (e.g. consultancies, stock ownership, equity interest, patent/licensing arrangement etc.) that might pose a conflict of interest in connection with the submitted article
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9303 | https://en.wikipedia.org/wiki/Field_norm | Jump to content
Field norm
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From Wikipedia, the free encyclopedia
Concept in field theory mathematics
In mathematics, the (field) norm is a particular mapping defined in field theory, which maps elements of a larger field into a subfield.
Formal definition
[edit]
Let K be a field and L a finite extension (and hence an algebraic extension) of K.
The field L is then a finite-dimensional vector space over K.
Multiplication by α, an element of L,
: ,
is a K-linear transformation of this vector space into itself.
The norm, NL/K(α), is defined as the determinant of this linear transformation.
If L/K is a Galois extension, one may compute the norm of α ∈ L as the product of all the Galois conjugates of α:
where Gal(L/K) denotes the Galois group of L/K. (Note that there may be a repetition in the terms of the product.)
For a general field extension L/K, and nonzero α in L, let σ1(α), ..., σn(α) be the roots of the minimal polynomial of α over K (roots listed with multiplicity and lying in some extension field of L); then
: .
If L/K is separable, then each root appears only once in the product (though the exponent, the degree [L:K(α)], may still be greater than 1).
Examples
[edit]
Quadratic field extensions
[edit]
One of the basic examples of norms comes from quadratic field extensions where is a square-free integer.
Then, the multiplication map by on an element is
The element can be represented by the vector
since there is a direct sum decomposition as a -vector space.
The matrix of is then
and the norm is , since it is the determinant of this matrix.
Norm of Q(√2)
[edit]
Consider the number field .
The Galois group of over has order and is generated by the element which sends to . So the norm of is:
The field norm can also be obtained without the Galois group.
Fix a -basis of , say:
: .
Then multiplication by the number sends
: 1 to and
: to .
So the determinant of "multiplying by " is the determinant of the matrix which sends the vector
: (corresponding to the first basis element, i.e., 1) to ,
: (corresponding to the second basis element, i.e., ) to ,
viz.:
The determinant of this matrix is −1.
p-th root field extensions
[edit]
Another easy class of examples comes from field extensions of the form where the prime factorization of contains no -th powers, for a fixed odd prime.
The multiplication map by of an element is
giving the matrix
The determinant gives the norm
Complex numbers over the reals
[edit]
The field norm from the complex numbers to the real numbers sends
: x + iy
to
: x2 + y2,
because the Galois group of over has two elements,
the identity element and
complex conjugation,
and taking the product yields (x + iy)(x − iy) = x2 + y2.
Finite fields
[edit]
Let L = GF(qn) be a finite extension of a finite field K = GF(q).
Since L/K is a Galois extension, if α is in L, then the norm of α is the product of all the Galois conjugates of α, i.e.
In this setting we have the additional properties,
Properties of the norm
[edit]
Several properties of the norm function hold for any finite extension.
Group homomorphism
[edit]
The norm NL/K : L → K is a group homomorphism from the multiplicative group of L to the multiplicative group of K, that is
Furthermore, if a in K:
If a ∈ K then
Composition with field extensions
[edit]
Additionally, the norm behaves well in towers of fields:
if M is a finite extension of L, then the norm from M to K is just the composition of the norm from M to L with the norm from L to K, i.e.
Reduction of the norm
[edit]
The norm of an element in an arbitrary field extension can be reduced to an easier computation if the degree of the field extension is already known. This is
For example, for in the field extension , the norm of is
since the degree of the field extension is .
Detection of units
[edit]
For the ring of integers of an algebraic number field , an element is a unit if and only if .
For instance
: .
Thus, any number field whose ring of integers contains has it as a unit.
Further properties
[edit]
The norm of an algebraic integer is again an integer, because it is equal (up to sign) to the constant term of the characteristic polynomial.
In algebraic number theory one defines also norms for ideals. This is done in such a way that if I is a nonzero ideal of OK, the ring of integers of the number field K, N(I) is the number of residue classes in – i.e. the cardinality of this finite ring. Hence this ideal norm is always a positive integer.
When I is a principal ideal αOK then N(I) is equal to the absolute value of the norm to Q of α, for α an algebraic integer.
See also
[edit]
Field trace
Ideal norm
Norm form
Notes
[edit]
^ Rotman 2002, p. 940
^ Rotman 2002, p. 943
^ Lidl & Niederreiter 1997, p. 57
^ Mullen & Panario 2013, p. 21
^ Roman 2006, p. 151
^ Jump up to: a b Oggier. Introduction to Algebraic Number Theory (PDF). p. 15. Archived from the original (PDF) on 2014-10-23. Retrieved 2020-03-28.
References
[edit]
Lidl, Rudolf; Niederreiter, Harald (1997) , Finite Fields, Encyclopedia of Mathematics and its Applications, vol. 20 (Second ed.), Cambridge University Press, ISBN 0-521-39231-4, Zbl 0866.11069
Mullen, Gary L.; Panario, Daniel (2013), Handbook of Finite Fields, CRC Press, ISBN 978-1-4398-7378-6
Roman, Steven (2006), Field theory, Graduate Texts in Mathematics, vol. 158 (Second ed.), Springer, Chapter 8, ISBN 978-0-387-27677-9, Zbl 1172.12001
Rotman, Joseph J. (2002), Advanced Modern Algebra, Prentice Hall, ISBN 978-0-13-087868-7
Retrieved from "
Category:
Algebraic number theory
Hidden categories:
Articles with short description
Short description matches Wikidata |
9304 | https://stanford-cs161.github.io/winter2021/assets/files/lecture14-notes.pdf | CS 161 Lecture 14: Greedy Algorithms Winter 2021 Wed, Mar 3 Adapted from Virginia Williams’ lecture notes. Additional credits go to Sam Kim and Mary Wootters.
Please direct all typos and mistakes to Moses Charikar and Nima Anari.
1 Greedy Algorithms Suppose we want to solve a problem, and we’re able to come up with some recursive formu-lation of the problem that would give us a nice dynamic programming algorithm. But then, upon further inspection, we notice that any optimal solution only depends on looking up the optimal solution to one other subproblem. A greedy algorithm is an algorithm which exploits such a structure, ignoring other possible choices. Greedy algorithms can be seen as a refine-ment of dynamic programming; in order to prove that a greedy algorithm is correct, we must prove that to compute an entry in our table, it is sufficient to consider at most one other table entry; that is, at each point in the algorithm, we can make a “greedy”, locally-optimal choice, and guarantee that a globally-optimal solution still exists. Instead of considering mul-tiple choices to solve a subproblem, greedy algorithms only consider a single subproblem, so they run extremely quickly – generally, linear or close-to-linear in the problem size.
Unfortunately, greedy algorithms do not always give the optimal solution, but they frequently give good (approximate) solutions. To give a correct greedy algorithm one must first identify optimal substructure (as in dynamic programming), and then argue that at each step, you only need to consider one subproblem. That is, even though there may be many possible subproblems to recurse on, given our selection of subproblem, there is always an optimal solution that contains the optimal solution to the selected subproblem.
1.1 Activity Selection Problem One problem, which has a very nice (correct) greedy algorithm, is the Activity Selection Problem. In this problem, we have a number of activities. Your goal is to choose a subset of the activities to participate in. Each activity has a start time and end time, and you can’t participate in multiple activities at once. Thus, given n activities a1, a2, ..., an where ai has start time si and finish time fi, we want to find a maximum set of non-conflicting activities.
The activity selection problem has many applications, most notably in scheduling jobs to run on a single machine.
1 1.1.1 Optimal Substructure Let’s start by considering a subset of the activities.
In particular, we’ll be interested in considering the set of activities Si,j that start after activity ai finishes and end before activity aj starts. That is, Si,j = {ak | fi ≤sk, fk ≤sj}. We can participate in these activities between ai and aj. Let Ai,j be a maximum subset of non-conflicting activities from the subset Si,j.
Our first intuition would be to approach this by using dynamic programming. Suppose some ak ∈Ai,j, then we can break down the optimal subsolution Ai,j as follows |Ai,j| = 1 + |Ai,k| + |Ak,j| where Ai,k is the best set for Si,k (before ak), and Ak,j is the best set for after ak. Another way of interpreting this expression is to say “once we place ak in our optimal set, we can only consider optimal solutions to subproblems that do not conflict with ak.” Thus, we can immediately come up with a recurrence that allows us to come up with a dynamic programming algorithm to solve the problem.
|Ai,j| = max ak∈Si,j 1 + |Ai,k| + |Ak,j|.
This problem requires us to fill in a table of size n2, so the dynamic programming algorithm will run in Ω(n2) time. The actual runtime is O(n3) since filling in a single entry might take O(n) time.
But we can do better! We will show that we only need to consider the ak with the smallest finishing time, which immediately allows us to search for the optimal activity selection in linear time.
Proposition 1. For each Si,j, there is an optimal solution Ai,j containing ak ∈Si,j of minimum finishing time fk.
Note that if the proposition is true, when fk is minimum, then Ai,k is empty, as no activities can finish before ak; thus, our optimal solution only depends on one other subproblem Ak,j (giving us a linear time algorithm).
Here, we prove the proposition.
Proof. Let ak be the activity of minimum finishing time in Si,j. Let Ai,j be some maximum set of non-conflicting activities. Consider A′ i,j = Ai,j \ {al} ∪{ak} where al is the activity of minimum finishing time in Ai,j. It’s clear that |A′ i,j| = |Ai,j|. We need to show that A′ i,j does not have conflicting activities. We know al ∈Ai,j ⊂Si,j. This implies fl ≥fk, since ak has the minimum finishing time in Si,j.
All at ∈Ai,j \ {al} don’t conflict with al, which means that st ≥fl, which means that st ≥fk, so this means that no activity in Ai,j \ {al} can conflict with ak. Thus, A′ i,j is an optimal solution.
2 Due to the above proposition, the expression for Ai,j from before simplifies to the following expression in terms of ak ⊆Si,j, the activity with minimum finishing time fk.
|Ai,j| = 1 + |Ak,j| Ai,j = Ak,j ∪{ak} Algorithm Greedy-AS assumes that the activities are presorted in nondecreasing order of their finishing time, so that if i < j, fi ≤fj.
Algorithm 1: Greedy-AS(a) A ←{a1} / activity of min fi / k ←1 for m = 2 →n do if sm ≥fk then // am starts after last acitivity in A A ←A ∪{am} k ←m return A By the above claim, this algorithm will produce a legal, optimal solution via a greedy selection of activities. There may be multiple optimal solutions, but there always exists a solution that includes ak with the minimum finishing time. The algorithm does a single pass over the activities, and thus only requires O(n) time – a dramatic improvement from the trivial dynamic programming solution. If the algorithm also needed to sort the activities by fi, then its runtime would be O(n log n) which is still better than the original dynamic programming solution.
1.2 Scheduling Consider another problem that can be solved greedily. We are given n jobs which all need a common resource. Let wj be the weight (or importance) and lj be the length (time required) of job j. Our output is an ordering of jobs. We define the completion time cj of job j to be the sum of the lengths of jobs in the ordering up to and including lj. Our goal is to output an ordering of jobs that minimizes the weighted sum of completion times ∑ j wjcj.
1.2.1 Intuition Our intuition tells us that if all jobs have the same length, then we prefer larger weighted jobs to appear earlier in the order. If jobs all have equal weights, then we prefer shorter length jobs in the order.
1 2 3 vs.
3 2 1 3 In the first case, assuming they all have equal weights of 1, ∑3 i=1 wici = 1 + 3 + 6 = 10. In the second case, ∑3 i=1 wici = 3 + 5 + 6 = 14.
1.2.2 Optimal Substructure What do we do in the cases where li < lj and wi < wj? Consider the optimal ordering of jobs.
Suppose we have a job i that is followed by job j in the optimal order. Consider swapping jobs i and j. The example below swaps jobs 1 and 2.
1 l1 2 l2 → 2 l2 1 l1 Note that swapping jobs i and j does not alter the completion times for every other job and only changes the completion times for i and j. ci increases by lj and cj decreases by li. This means that our objective function ∑ i wici changes by wilj −wjli. Since we assumed our order was optimal originally, our objective function cannot decrease after swapping the jobs. This means, wilj −wjli ≥0 which implies lj wj ≥li wi .
Therefore, we want to process jobs in increasing order of li wi , the ratio of the length to the weight of each job. The algorithm also does a single pass over jobs, and thus only requires O(n) time, assuming the jobs were ordered by li wi . Like previously, if the algorithm also needed to sort the jobs based on the ratio of length to weight, then its runtime would be O(n log n).
1.3 Optimal Codes Our third example comes from the field of information theory. In ASCII, there is a fixed 8 bit code for each character. Suppose we want to incorporate information about frequencies of characters to obtain shorter encodings. What if we want to represent characters by codes of different lengths depending on each character’s frequencies? We explore a greedy solution to find the optimal encoding of characters.
To create optimal codes, we want a way to encode and decode our sequence. To encode the sequence, we would just have to concatenate the code of each character together. How about for decoding? Consider the following codes of characters: a →0, b →1, c →01. However, when decoding, when we encounter 01, this could be decoded as “ab” or “c”. Therefore, our codes need to be prefix free: no codeword is a prefix of another.
4 1.3.1 Tree Representation We may think of representing our codes in a tree structure, where the codewords represent the leaves of our tree. An example is shown below: 1 .55 .25 .30 .14 a : .45 c : .12 b : .13 d : .16 e : .05 f : .09 0 1 0 1 0 1 0 1 0 1 Above, in addition to the characters {a, b, c, d, e, f }, we’ve included frequency information.
That is, f (a) = 0.45 means that the probability of a random character in this language being equal to a is .45.
The code for each character can be found by concatenating the bits of the path from the root to the leaves. By convention, every left branch is given the bit 0 and every right branch is given the bit 1.
As long as the characters are on the leaves of this tree, the corresponding code will be prefix-free. This is because one string is a prefix of another if and only if the node corresponding to the first is an ancestor of the node corresponding to the second. No leaf is an ancestor of any other leaf, so the code is prefix-free.
1.3.2 How good is a code?
Suppose we have a set of characters C with frequencies f (c) so that ∑ c∈C f (c) = 1. That is, f (c) can be thought of as the probability of using a letter c in this language. The cost, in terms of bits, of a character c ∈C when using the coding scheme represented by a tree T is just the depth in the tree T: cost(c) = dT(c). For example, in the tree above, e has depth 4 in the tree, and requires 4 bits to represent. The average cost of the tree is B(T) = Ec∈C[dT(c)] = ∑ c∈C f (c)dT(c).
5 We say that a tree T is optimal if this expected cost B(T) is as small as possible.
1.3.3 Huffman Codes In 1951, David A. Huffman, in his MIT information theory class, was given the choice of a term paper or final exam. Huffman chose to do the term paper rather than take the final exam. He found greedy algorithm to find the most efficient binary code, which we know today as Huffman codes.
The basic idea is this: build subtrees for subsets of characters and merge them from the bottom up, combining the two trees with the characters of minimum total frequency.
Algorithm 2: A high-level version of the Huffman Coding algorithm.
Input: Set of characters C = {c1, c2, . . . , cn} of size n, and F = {f (c1), f (c2), . . . , f (cn)}, a set of frequencies.
Create nodes Nk for each character ck, with key f (ck).
Let current denote the set {N1, . . . , Nn} of nodes.
while current has length more than one do Find the two nodes Ni and Nj in current with the minimum frequencies and create a new intermediate node I with Ni and Nj as its children, so that I.key = Ni.key + Nj.key.
Add I to current and remove Ni, Nj.
return the only entry of current, which is the root of the tree.
The tree shown above results from running this algorithm on the letters with those frequencies; see the slides for an illustration of this process.
1.3.4 Proof of Correctness This algorithm works, but at first it’s not at all obvious why. For a rigorous proof, refer to Lemmas 16.2 and 16.3 in CLRS. However, we’ll sketch the idea below. Formally, the proof goes by induction. Recall that after iteration t in Algorithm 2, we have a list current, which contains the roots of subtrees that we still need to merge up. We will maintain the following inductive hypothesis: • Inductive hypothesis: Suppose we have completed t iterations of the loop in Algo-rithm 2. Then there exists a way to merge the subtrees in current that is optimal.
• For the base case, we observe that when t = 0, current is just the set of all characters, and definitionally there exists an optimal tree made out of these nodes.
• For the inductive step, we need to show that if the inductive hypothesis holds at step t −1, then it holds at step t. We’ll sketch this later.
• Finally, to conclude the argument, we see that at the end of the algorithm, there is only one element in current, and in this case the inductive hypothesis reads that there is a 6 way to merge this single subtree to obtain an optimal subtree. That’s just a convoluted way of saying that the single tree we return is optimal, and so we are done.
All that remains to show is the inductive step. We first observe the following claim: Proposition 2. We are given a set of characters C and a set of its associated frequencies F where f (c) is the frequency of character c. Let x and y be the characters with the two smallest frequencies. There exists an optimal coding tree for C such that x, y are sibling leaves.
Proof. Let T be the optimal coding tree for C. The optimal coding tree must be a full binary tree, that is, every non-leaf node must have two children. Let a, b be characters that are sibling leaves of maximum depth. We define the number of bits to encode c as dT(c) and the number of bits needed for the coding tree as B(T) = ∑ c f (c)dT(c).
We can replace a, b by x, y without increasing the total number of bits needed for the coding tree.1 If we swap x and a, the change in cost becomes f (x)dT(a) + f (a)dT(x) −f (x)dT(x) −f (a)dT(a) = (f (x) −f (a))(dT(a) −dT(x)) ≤0 Therefore, swapping a, b with x, y will not increase our objective function B(T). Hence, there exists an optimal coding tree where x, y are siblings in the tree.
Proposition 2 shows that there exists an optimal coding tree where x and y are sibling leaves, that is, there is an optimal code that makes the same greedy choice as the algorithm.
However, this is only immediately helpful for the first iteration of the inductive step, when all of the elements of current are indeed leaves. In order to make this idea work for all t, we need one more claim.
Proposition 3. Let C be a set of characters, and let T be an optimal coding tree for C.
Imagine creating C′ from C by collapsing all the characters in a subtree rooted at a node N with key k = N.key into a single character c′ with frequency k. Then the corresponding tree T ′ is optimal for C′.
Conversely, suppose that a tree T ′ that is an optimal coding tree for an alphabet C′. Let c′ ∈C′ be a character with frequency f (c′). Introduce new characters c′′ 1, . . . , c′′ r with total frequency ∑r i=1 f (c′′ i ) = f (c′). Let T ′′ be an optimal coding tree on c′′ 1, . . . , c′′ r . Then the tree T on the alphabet C = (C′ \ {c′}) ∪{c′′ 1, . . . , c′′ r } that has the leaf c′ replaced with the subtree T ′′ is optimal.
Proof. Let T and T ′ be the two trees described in the lemma, and consider the difference of 1For simplicity, we ignore the case where a, b, x, y are not distinct. For more details, see Lemma 16.2 in CLRS.
7 their costs.
B(T) −B(T ′) = ∑ c∈C f (c) · dT(c) − ∑ c∈C′ f (c) · dT ′(c) = ( r ∑ i=1 f (c′′ i )dT(c′′ i ) ) −f (c′)dT ′(c′) = ( r ∑ i=1 f (c′′ i )(dT ′′(c′′ i ) + dT ′(c′)) ) −f (c′)dT ′(c′) = r ∑ i=1 f (c′′ i )dT ′′(c′′ i ) + dT ′(c′) r ∑ i=1 f (c′′ i ) −f (c′)dT ′(c′) = t ∑ i=1 f (c′′ i )dT ′′(c′′ i ) where the last line used the fact that ∑r i=1 f (c′′ i ) = f (c′), and so the last two terms cancelled.
This means that the difference in the cost between these two trees only depends on T ′′, it doesn’t depend at all about the structure of T.
Thus, T is optimal if and only if T ′ is optimal.
The two Claims together prove the inductive step, because the second claim implies that the logic of the first claim holds, even for newly created intermediate nodes I.
Note: The proof in CLRS has the same basic steps (Lemmas 16.2 and 16.3 instead of the claims above), although phrased slightly differently. The sketch above is pretty sketchy, so if the above is hard to follow, please check out CLRS for a more detailed version.
8 |
9305 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6746516/ | HPV-related oropharyngeal cancer: a review on burden of the disease and opportunities for prevention and early detection - PMC
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Hum Vaccin Immunother
. 2019 May 7;15(7-8):1920–1928. doi: 10.1080/21645515.2019.1600985
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HPV-related oropharyngeal cancer: a review on burden of the disease and opportunities for prevention and early detection
Mary Roz Timbang
Mary Roz Timbang
a Department of Otolaryngology-Head and Neck Surgery, The University of California, Davis, Sacramento, CA, USA
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a, Michael W Sim
Michael W Sim
b Department of Otolaryngology-Head and Neck Surgery, Indiana University, Indianapolis, IN, USA
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b, Arnaud F Bewley
Arnaud F Bewley
a Department of Otolaryngology-Head and Neck Surgery, The University of California, Davis, Sacramento, CA, USA
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a, D Gregory Farwell
D Gregory Farwell
a Department of Otolaryngology-Head and Neck Surgery, The University of California, Davis, Sacramento, CA, USA
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a, Avinash Mantravadi
Avinash Mantravadi
b Department of Otolaryngology-Head and Neck Surgery, Indiana University, Indianapolis, IN, USA
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b, Michael G Moore
Michael G Moore
b Department of Otolaryngology-Head and Neck Surgery, Indiana University, Indianapolis, IN, USA
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b,✉
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a Department of Otolaryngology-Head and Neck Surgery, The University of California, Davis, Sacramento, CA, USA
b Department of Otolaryngology-Head and Neck Surgery, Indiana University, Indianapolis, IN, USA
✉
CONTACT Michael G. Moore mgmoore@gmail.com Department of Otolaryngology-Head and Neck Surgery, Indiana University School of Medicine, 1130 W. Michigan Rd, FH449, IN, 46202, USA
Received 2018 Oct 1; Revised 2019 Mar 7; Accepted 2019 Mar 23; Collection date 2019.
© 2019 Taylor & Francis Group, LLC
PMC Copyright notice
PMCID: PMC6746516 PMID: 31050595
ABSTRACT
The incidence of oropharyngeal cancer (OPC) related to infection with human papillomavirus (HPV) is rising, making it now the most common HPV-related malignancy in the United States. These tumors present differently than traditional mucosal head and neck cancers, and those affected often lack classic risk factors such as tobacco and alcohol use. Currently, there are no approved approaches for prevention and early detection of disease, thus leading many patients to present with advanced cancers requiring intense surgical or nonsurgical therapies resulting in significant side effects and cost to the health-care system. In this review, we outline the evolving epidemiology of HPV-related OPC. We also summarize the available evidence corresponding to HPV-related OPC prevention, including efficacy and safety of the HPV vaccine in preventing oral HPV infections. Finally, we describe emerging techniques for identifying and screening those who may be at high risk for developing these tumors.
KEYWORDS: Oropharyngeal cancer, HPV, vaccination, detection, screening, prevention
Introduction
Head and neck cancer accounts for 3% of malignancies in the United States, with more than 63,000 Americans diagnosed with this disease and 13,000 dying from it annually.1 Oropharyngeal cancer (OPC) represents a significant portion of head and neck cancers.2 OPCs include those found in the soft palate, base of tongue, lingual and palatine tonsils, and surrounding tissues. While tobacco and alcohol exposure have long been established as risk factors for OPC, in recent decades there has been an increase in a subset of OPC linked to human papillomavirus (HPV).3 HPV is a double-stranded DNA virus with predilection for squamous epithelium.4 Cryptic epithelium overlying the tonsils and tongue base acts as a reservoir for the virus, providing access to its basal layer for viral replication.2 Over time, malignant transformation can occur when viral oncoproteins disrupt tumor suppression genes in native tissue. Reticulated crypt epithelium in the oropharynx is unique to this anatomical location in the head and neck, and may explain why HPV is estimated to be five times higher in the oropharynx when compared to the oral cavity, larynx, or hypopharynx.5 Although there are many types of HPV, the overwhelming majority of HPV-related OPC cases are caused by HPV16.
While data suggest an overall stable incidence of HPV-negative OPC, the incidence of HPV-related OPC is rising,6 and it will continue to be a major factor in national health care related to cancer treatment. The rise in incidence of HPV-related OPC makes it now the most common HPV-related malignancy in the United States.7,8 According to the Centers for Disease Control and Prevention (CDC), there were 11,788 reported cases of cervical carcinoma and 18,917 cases of OPC, including 15,479 (82%) among men and 3438 (18%) among women in 2015.7 In the United States, HPV DNA can now be identified in more than 70% of all new cases of OPC.9 Similar trends are seen in Northern Europe.10 Patients affected are typically in their fifth or sixth decade of life, have an earlier sexual debut, and a higher number of lifetime oral and vaginal sex partners than those affected by HPV-negative OPC.8,11,12 In addition, these individuals are more often male, of higher socioeconomic status, and less likely to have a history of tobacco or alcohol abuse.13
Oral HPV infection is the primary risk factor for HPV-related OPC, and over 90% of oral HPV infections are sexually acquired.14 Therefore, it is no surprise that the number of oral sexual partners is the behavioral factor most strongly and specifically associated with OPC. Differences in sexual behavior between countries may contribute to the differences in global trends of HPV-related OPC.10
In addition to viral exposure, concomitant tobacco use may also play a part in the development of HPV-related OPC. Gillison and colleagues performed a multivariable analysis inclusive of individuals aged 14–69 y, looking at factors independently associated with prevalent oral HPV including age, sex, lifetime number of sexual partners, and current smoking intensity.15 Although adjustment for other factors dampened the first age-related peak in oral HPV prevalence, the bimodal age pattern remained statistically significant. Prevalence increased with number of lifetime sexual partners and number of cigarettes smoked per day. As a result, disease prevention efforts should also include attention to tobacco cessation.
Clinical behavior, treatment morbidity, and cost
The clinical behavior and presentation of HPV-related OPC are different from its HPV-negative counterpart. In patients with HPV-related OPC, the most common presenting symptom is a neck mass, while those with HPV-negative OPC are more likely to complain of a sore throat and dysphagia.16 On presentation, they are more likely to have early T-stage (T1/T2) and advanced cervical nodal disease (N2/N3), when compared to HPV-negative tumors.10,17 However, while involvement of cervical lymph nodes reflects more advanced disease, such patients with HPV-related OPC have better survival outcomes and response to treatment than those with HPV-negative OPC.18,19
The standard treatment for OPC has been concurrent radiation and chemotherapy, with an increasing role for upfront surgical approach with advances in transoral surgery.20 Although survival rates are high, long-term toxicity and poor functional outcomes are still a concern for patients who have survived their cancer (Table 1).14,21 Since the oropharynx is crucial to important everyday functions such as speech, swallow, and airway patency, both surgical and medical treatment of tumors in this area can result in significant morbidity. Patients can have long-lasting dysphagia and problems with speech such as velopharyngeal insufficiency from post-surgical changes, post-radiation effects, and chemotherapy. Additional side effects include nausea and vomiting, dry mouth, progression of dental disease, loss of taste, difficulty with mouth opening, and even osteoradionecrosis of the mandible (Figure 1).22-26 While most of these issues improve over time with treatment and physical and/or speech therapy, some patients have long-standing dysfunction and require gastrostomy tubes for nutritional health, or tracheostomies for airway maintenance or pulmonary hygiene.21 In fact, a retrospective review by Vatca et al. suggests that high-grade mucositis with concomitant weight loss from radiation therapy is worse in HPV-related OPC compared with HPV-negative OPC.27 Thus, efforts are underway to investigate the feasibility of de-intensified chemoradiation therapy and to expand the indications for surgical therapy in hopes of decreasing acute and long-term morbidity and improving functional outcomes.28,29
Table 1.
Adverse effects of nonsurgical therapy.
| Adverse effect(s) | Patients affected |
:---: |
| Taste disturbance23 | 88% |
| Nausea and vomiting24 | 36% |
| Dry mouth25 | 29–38% |
| Esophageal stricture26 | 5% |
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Figure 1.
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Patient with osteoradionecrosis of the mandible from radiation therapy.
Although HPV appears to be a distinct risk factor from smoking in the development of OPC, tobacco exposure can also play a role.15 Although patients with HPV-positive OPC are less likely to be smokers, those who do have a significant current or past smoking history have been found to have significantly worse disease control with treatment.13,30,31 Maxwell and colleagues found that current tobacco users with advanced, HPV-positive OPC are at higher risk of disease recurrence compared with never-tobacco users after chemoradiation therapy. In their study cohort, 35% of HPV-positive ever-tobacco users recurred compared with only 6% of HPV-positive never-users and 50% of HPV-negative patients.30
In addition to the morbidity from treatment and risk for cancer-related mortality, there are significant costs associated with the management of OPC, with one study showing an estimated cost of $140,000 per new patient in the first 2 y of treatment and surveillance, not accounting for additional costs due to loss of productivity.32 Based on numbers from 2004 to 2007, it was estimated that the mean lifetime cost per new case of HPV-related head and neck cancer was $43,200 with a total annual cost in the United States of $306 million.33 Another study by Moore et al. compared cost between different treatment modalities and found that the mean cost of therapy (private payers/government payers) ranged from $37,435/$15,664 in those treated with surgery alone to $198,285/$57,429 when chemoradiation was employed.34
Rationale for prevention and screening
In response to the emerging epidemic of HPV-related OPC, and the morbidity and costs associated with treating these cancers, much attention has turned to the prevention and early detection of disease. A United States national initiative called Healthy People 2020 aims to decrease the number of deaths due to OPC by 10%, mainly through improving immunization rates to reduce preventable infections.35 Current vaccination rates are low, especially in males who are most often affected by these cancers.36 Recent statistics estimate that about 65% of girls and 56% of boys between ages 13 and 17 y have received the first dose of the HPV vaccine.37 Moreover, due to the lack of level one evidence showing a reduction of premalignant lesions and OPC with vaccination, prevention of OPC is not an approved vaccine indication by the Food and Drug Administration (FDA). This creates a barrier to increasing public education and awareness on a large scale.
In addition to the challenges posed in implementing successful prevention campaigns, there is an increasing need to develop effective screening techniques to allow for early detection of disease, especially for those who have not received the vaccine. Challenges to screening include lack of a precursor lesion and a long latency period between exposure to virus and onset of disease. Multiple studies suggest that the timing between exposure to HPV virus and development of cancer exceeds a decade, and can be as long as 30 y.10,38 In this review, we provide an update on the current status of HPV-related OPC prevention in the United States as well as evolving approaches to screening. In addition, we outline some of the challenges posed to these efforts as well as potential areas for future study.
Prevention of HPV-related OPC
Lifestyle modifications
The behavioral risk factor most associated with HPV-related OPC is oro-genital sex. As a result, primary prevention through safe sexual practices such as condoms and barrier contraceptives is a critical prevention strategy. Gupta et al. performed a cross-sectional analysis of National Health and Nutrition Assessment Survey from 2009 to 2014 to see if there is a correlation between barrier contraceptive and prevalence of oral HPV 16/18 infection. They found that after adjusting for all variables associated with HPV positivity, individuals reporting barrier use were significantly less likely to be HPV 16/18 positive when compared to those not using barrier during oro‐genital sex. It is important to note that this population had not received the HPV vaccine.39 Therefore, barrier contraceptive is an important prevention strategy for those who are beyond the eligible age for vaccination. While the CDC recommends use of barrier contraceptives such as condoms and dental dams during oro-genital sex to reduce transmission of sexually transmitted infections, there needs to be stronger emphasis on the correlation between this and decreasing risk of developing OPCs.
As mentioned above, Gillison et al. demonstrated that tobacco smoking also significantly correlates with oral HPV infection,15 and such use can negatively impact survival if OPC develops.30 As a result, efforts of smoking cessation are critical in an effort to reduce oral HPV infection as well as the development of OPC and other cancers.
Vaccination
In 2006, the pharmaceutical company, Merck and Co., introduced a quadrivalent HPV vaccine called Gardasil 4. The vaccine protected against HPV 6, 11, 16, 18 with the initial goal of preventing HPV-related cervical cancer in women.40 Since the introduction of Gardasil 4, a bivalent (Cervarix) and a nine-valent vaccine (Gardasil 9) were also created and approved for use. Gardasil 9 is the most recently approved vaccine, and its safety was ascertained in clinical trials with over 15,000 participants prior to its FDA approval.41 In addition to the original four HPV strains covered by the quadrivalent version, Gardasil 9 also protects against HPV types 31, 33, 45, 52, 58 and thus covers strains that cause over 90% of HPV-related cancers including HPV-related OPC.
By 2011, the CDC recommended routine HPV vaccination for both girls and boys between 9 and 26 y old for the prevention of cervical and anogenital HPV cancers. Ideally, the vaccine would be administered at 11 or 12 y of age with the goal of preceding sexual debut and capturing the robust immune response that is mounted at that age.40,42 In theory, later vaccination would be less effective as many individuals would have already contracted a persistent infection by a high-risk HPV virus prior to that time, and may therefore not be protected. However, recent studies have shown that the immune response to HPV vaccination in men aged 27–45 y was comparable to those observed in younger men but with unknown efficacy rates of preventing persistent HPV infections in these older patients.43 On October 5, 2018, the FDA approved a supplemental application for the Gardasil 9 Vaccine, expanding its approved use to include women and men aged 27–45 y.
Though the CDC recognizes that persistent infection with oncogenic HPV types can cause cancers at non-cervical sites, the FDA has not officially approved the vaccine for prevention of OPCs. This contributes to a lack of awareness regarding the correlation between HPV vaccination and prevention of OPC, even among medical professionals. This was highlighted by a study in Louisiana by Mehta and colleagues surveying members of the Louisiana Chapter of the American Academy of Pediatrics.44 The study demonstrated that 15.5% of the pediatricians polled were not aware of the link between OPC and HPV, and less than half knew that HPV-related OPC incidence was increasing. Despite CDC recommendations, the study found that only 89.7% of pediatricians routinely recommended the HPV vaccine, while 5.2% occasionally offered or only at caregiver request and the remaining 5.2% did not offer the vaccine at all. This, combined with the fact that less than 1% of the US population in 2014 recognized HPV as a risk factor for development of head and neck cancer,45 may in part be responsible for lower vaccine uptake compared to the others in the adolescent series.
Lack of FDA approval for use of the vaccine to prevent HPV-mediated OPC stems from the inability of clinical trials to directly demonstrate vaccine efficacy against oropharyngeal HPV-related disease. Until relatively recently, regulatory agencies required a clinical disease end point for trials regarding HPV vaccine efficacy. Precancerous lesions, which would serve as a disease end point, are not established in OPCs. Furthermore, since cancer may not develop for many years after initial infection with the virus, a trial showing a correlation between increasing vaccine use and decreasing OPC rates may not be available for many decades.
In 2014, the World Health Organization recommended that efficacy against incidence and persistent HPV infection can be a surrogate for disease risk.46 A number of studies have provided evidence that HPV vaccination decreases the rate of HPV-related infections and are likely to reduce the incidence of cancers of the oropharynx. A double-blind randomized controlled trial conducted in Costa Rica by Herrero et al. evaluated the efficacy of the bivalent vaccine in reducing oral HPV infection 4 y after vaccination. They observed a 93.3% reduction of prevalent oral HPV 16 and 18 infections in the vaccine arm compared to the control arm.47 Another recent study by Hirth et al. concluded that vaccine-type oral HPV prevalence was lower in individuals who received the HPV vaccine compared to unvaccinated individuals.48 Similar results were shown by Chaturvedi et al.36 Their study demonstrated that HPV vaccination was associated with an estimated 88% reduction in prevalence of vaccine type oral HPV 6, 11, 16, 18 infections among vaccinated young adults in the US. As noted in the prior studies, findings consistently show a significant decrease in these HPV infections in the vaccinated population compared to the unvaccinated men and women in the United States. However, because of a vaccination rate of only 18.3% between 2011 and 2014 among individuals 18–33 y of age, the population-level effect of HPV vaccination on oral HPV 6,11,16,18 was a modest 17%.36 Other recent studies have addressed concerns that vaccination against some but not all HPV types may introduce a competitive advantage for non-vaccine types. This would lead to an eventual decrease effectiveness of the vaccine. A study by Tota et al. obtained data from the Costa Rica Vaccine trial and PATRICIA trial to compare incidence of non-protected HPV infections across the trial arms after 4 y. Their results revealed similar or higher incidence of non-protected HPV types in the control arm compared with the HPV arm across all their analyses, which supports that type replacement is unlikely to occur among vaccinated individuals.49 This further supports vaccination as a primary prevention strategy.
Currently available HPV vaccines are 90–100% effective in preventing genital HPV infections, and the global cervical cancer burden is projected to be dramatically reduced by 2050 due to both the vaccine and the improved cervical screening.10 However, unlike the Papanicolaou test for cervical cancer, there is no current reliable screening method to detect precancerous OPCs. As a result, expanding the use of the HPV vaccine is that much more critical when reducing the number of patients affected by HPV-related OPC. It has been estimated that, by vaccinating boys and men, 5416 and 51,168 additional cases of HPV-related OPC would be prevented at 50 and 100 y, respectively.50 Additionally, the societal cost of HPV vaccination has been predicted to be well below the $50,000/Quality-Adjusted Life Year threshold used to determine cost-effectiveness of public health initiatives.51 Such evidence strongly supports continued efforts to improve vaccination rates in both girls and boys.
Screening for HPV-related OPC
As previously mentioned, the most common presenting symptom of HPV-related OPC is a lateral neck mass. Thus, many patients have developed regional metastasis at presentation, and by definition have more advanced disease. In fact, one study showed that of 1907 patients with HPV-related OPC, 73% were diagnosed with advanced locoregional disease.17 As a result, there is a tremendous need to develop an approach to identify occult lesions at an earlier stage to allow for successful and less morbid treatment. For a screening technique to be beneficial, it is first necessary to determine a high-risk population. An ideal test would be cheap, minimally invasive, and appropriately sensitive to identify a high proportion of subclinical lesions (i.e. a high negative predictive value).
While there are currently no effective screening methods and no precancerous lesion correlates for HPV-related OPC, there are emerging techniques. These advances show promise through serologic testing14 and imaging52,53 that may aid in identifying high-risk individuals as well as subclinical lesions, respectively. Here, we outline some of these approaches that may allow for a more focused assessment of those predisposed to developing HPV-related OPC.
Population screening
Population screening for OPC is difficult since at present there are no precursor lesions. Moreover, the mucosal surface of the oropharynx is much more challenging to examine than the cervix as many lesions start in the reticulated epithelium at the depth of tonsillar tissue crypts, thus concealing them from visual inspection. As a result, efforts have focused on first narrowing the population down into those at highest risk for development of disease. Men aged 50–65 y with multiple sexual partners would be an appealing target demographic for screening programs given the higher incidence of these tumors in this demographic group.33 However, further definition of the true level of risk of such a group is currently under investigation.54
Oral HPV screening
An initial approach considered for assessment was to use oral HPV screening. This technique evaluates for the presence of HPV DNA in saliva and can be even used to focus specifically on high-risk HPV such as HPV 16. However, while this technique can be effective in demonstrating active HPV infection, it is of little utility in screening for HPV-related OPC as the majority of individuals either go on to clear the infection or fail to progress to malignancy. Consequently, the use of oral HPV screening has been discouraged as a screening technique to identify OPC.55
HPV serology
One area that has shown particular promise in assessing high-risk populations is to screen for serum antibodies to HPV 16 proteins. Such an approach may allow for at-risk individuals to be identified prior to progression of disease. In an early study by Mork et al., serum positivity to the L1 capsid protein of HPV 16 conferred a 14-times increased risk of developing OPC, when linking findings from a Nordic serum bank and tumor registries.56 While these findings were encouraging, these antibodies represent the body’s cumulative exposure to HPV 16 and are not specific to anatomic site. Moreover, they do not reflect expression of HPV oncoproteins necessary for carcinogenesis.
In 2013, Kreimer and colleagues identified that serum antibodies to the E6 oncoprotein of HPV 16 were a better marker for predicting cancer.38 In this important study, participants came from the European Prospective Investigation Into Cancer and Nutrition cohort and 638 patients who went on to develop head and neck cancer and 1599 controls with no evidence of cancer were evaluated. Pre-diagnostic serum samples were collected and analyzed for antibodies against multiple HPV 16 proteins as well as other subtypes of HPV. All told, HPV 16 E6 seropositivity conferred a 274 times increased risk of developing OPC, being present in 34.8% of OPC patients compared to 0.6% in other head and neck cancer patients and control patients. This positive finding was found on average 6 y before diagnosis and was observed in some instances more than 10 y before the cancer was found. Such seropositivity has been shown to result in a 10-y cumulative risk of developing HPV-related OPC of 6.2% in men and 1.3% in women, compared to 0.04% in seronegative controls.57 Additional study has also shown that higher pretreatment HPV 16 E6 antibody titers can predict recurrence, while E6 and E7 titers decreased when reassessed in the early and late posttreatment setting.58
Transcervical ultrasound
In individuals with HPV-related cancer in a neck mass, as well as those found to be high risk by either positive serology screening or other methods, transcervical ultrasound can be used as an additional means for assessment. The use of ultrasound has long been standard in the evaluation of thyroid lesions and cervical adenopathy, and it can aid in guidance of fine needle aspirations, especially in cystic nodal disease like what is often observed in HPV-related cancers.
In addition to assessment of nodal disease, the use of transcervical ultrasound has now been applied to evaluate the oropharynx in an effort to identify occult tumors, especially those developing in crypts of the palatine and lingual tonsils, not visible on surface examination.59,60 While such an approach to screening is not widely utilized, at present, it is particularly appealing in that it is relatively cheap, minimally invasive, and results in no exposure to ionizing radiation or intravenous contrast.
Mucosal imaging
The lack of an identifiable early lesion presents a significant challenge when evaluating for premalignant and malignant lesions of the oropharynx. The conventional approach following a head and neck physical examination is to offer fiber-optic nasopharyngoscopy using standard white light imaging (WLI). While the optics of contemporary endoscopes allows for improved definition and magnification, small and superficial lesions may be missed due to the subtle difference in appearance of normal and abnormal mucosa. Two techniques that have been applied to augment this aspect of mucosal screening are narrow band imaging (NBI) and endoscopic lifetime imaging.
NBI was first described for use in the gastrointestinal tract by Sano in 2001.61 Using this technology, filters of spectral regions centered at 415 nm (blue light) and 540 nm (green light) are used, excluding other signals from visible spectrum, thus highlighting the vascularity near the tissue surface. Images are obtained through the nasopharyngoscope that can allow for identification of subclinical lesions (Figure 2). Such an approach was used in a multicentered randomized controlled trial and found NBI detected superficial mucosal lesions of the head and neck more often than WLI (100% vs. 8%) and had a sensitivity and accuracy of finding lesions of 100% and 86.7%, respectively.62 This technique has also been shown to be helpful in identifying the primary lesion in patients with neck cancer with an unknown head and neck primary site. One study by Ni and colleagues evaluated 53 patients with cervical lymph node metastases with no identifiable primary site seen on physical examination, computed tomography scan (CT), magnetic resonance imaging scan (MRI), and laryngoscopy. The use of NBI allowed for identification of lesions in 47% patients.63 Similarly, Hayashi et al. used NBI to screen 46 of such patients and a primary was identified in 35% of individuals. The study did not report specifically on percentage of lesions positive for HPV.64
Figure 2.
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Patient with T1N2A oropharyngeal cancer status post definitive chemoradiation. (a) Distal chip video laryngoscopy revealed post-treatment radiation edema with no evidence of mucosal disease delineated by red arrows. (b) NBI revealed a mucosal abnormality delineated by blue arrows. Biopsy was consistent with high-grade dysplasia. Image is borrowed with permission from Dr Peter Belafsky.
Lifetime-resolved laser-induced imaging is another approach that has shown recent promise in mucosal imaging. Using this technique, laser energy is applied to a surface and the reflected signal representing the induced autofluorescence of the tissue is captured. Data have shown that the autofluorescence characteristics of normal, dysplastic, and cancerous head and neck mucosa are different, suggesting that this technique may have a role in cancer screening (Figure 3).65,66 Despite their promise, all mucosal imaging techniques have limitations in OPC screening as many tumors originate in the depths of tonsillar crypts and thus are not well seen on surface evaluation.
Figure 3.
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This intraoperative image shows the use of lifetime tissue autofluorescence for identification of pathologic mucosa. A laser is aimed at the surface and the signal returning from the tissue is interpreted. The signal for pathologic mucosa has been shown to be distinctly different than normal and is demonstrated by a different shade on the color scale. Such an approach could be very useful in cancer screening. However, small submucosal lesions may avoid detection.
Future directions
There are currently no reliable biomarkers that can be used for tumor screening or to evaluate for cancer recurrence for head and neck cancer. With increased availability of robust sequencing technology, however, methods have been developed looking at saliva and serum tumor DNA as a potential approach. This concept of measuring serum DNA is particularly exciting as it would represent a minimally invasive way to assess for subclinical lesions and monitor for disease response and recurrence. Circulating cell-free DNA (cfDNA) has been proposed as one such method. This approach centers around the fact that when tissue dies, its DNA is released into serum and circulates extracellularly prior to being metabolized over the next 15–20 min. Consequently, any analysis provides a reflection of tissue undergoing rapid turnover as can be seen in active malignancy. Virally mediated cancers are particularly well suited for such “liquid biopsies” as they have known viral DNA incorporated into the host DNA that can be queried. In fact, this approach has demonstrated tremendous promise in screening for Epstein-Barr virus (EBV)-related nasopharyngeal cancer.67 In a trial where 20,174 asymptomatic participants were screened, 309 individuals were found to have persistently elevated levels and 300 ultimately went on to undergo nasal endoscopy with or without MRI. Thirty-four of those subjected to further screening were found to have nasopharyngeal cancer, with the majority being either stage I or II, while only one patient with a negative initial test is found to have a nasopharyngeal cancer in the year following screening, thus yielding a sensitivity and specificity of the technique of 97.1% and 98.6%, respectively.
HPV-related OPC also appears appropriate for assessment with cfDNA. Such tumors often have nodal metastases with significant necrosis, thus likely shedding large amounts of tumor DNA. In addition, with the known component of the HPV DNA, as well as other differentiating patterns such as DNA methylation and fragment size, techniques are under development that may soon allow for early identification of these lesions through blood analysis. Wang et al. studied the potential for looking at tumor-specific DNA in saliva and serum in head and neck cancer patients.68 They found oral cancer was more likely to show elevation in saliva tumor DNA while tumors in the oropharynx were more likely to have elevations in their serum levels. Additionally, the combined use of posttreatment saliva and plasma HPV 16 DNA positivity was found to be 69.5% sensitive and 90.7% specific in predicting recurrence within 3 y of therapy completion, suggesting a potential role in surveillance.69 Similar findings were observed in a recent publication by Hanna and colleagues where HPV cfDNA levels correlated with total tumor burden, and higher levels as well as higher total tumor burden resulted in worse overall survival.70 Moreover, using this method, cfDNA levels were also found to have a corresponding change at a median of 16 d prior to restaging scans reflecting either disease response or progression. With the promise seen from these early results, such methods are being used to prospectively evaluate high-risk individuals to determine if early lesions can be identified prior to symptom development. While these approaches alone would not be sufficient for surveillance, it could help augment current surveillance techniques such as history, physical examination, transcervical ultrasound, and positron emission tomography imaging.
Limitations
Since this paper is a narrative review, there is a lack of objective and systematic selection criteria for the papers included. Papers included in discussing the topic range from case reports to randomized controlled trials and therefore vary in level of evidence. While we chose to only use papers which we felt provided high quality of information, lack of methodological selection of these papers lead to bias of our interpretation and conclusions.
Conclusions
HPV-related OPC carries significant morbidity, mortality, and substantial cost to the health-care system. With its increasing incidence, importance should be placed on prevention of this disease and early diagnosis through effective screening techniques. While new screening methods are still underway, emphasis should be placed on prevention through greater public awareness of risk factors associated with HPV, behavioral changes to mitigate these risks, and widespread use of the vaccine. Furthermore, while HPV-related OPC has improved survival compared to HPV-negative OPC, it is important to remember that tobacco users have significantly worse disease control with treatment. Although HPV-vaccination efforts show promise in protecting against oral HPV infection, clinical trials supporting vaccine efficacy against oropharyngeal HPV-related disease are currently lacking. This has contributed to a delay in the FDA’s approval of the vaccine for prevention of HPV-related OPC. Therefore, further research in showing vaccine efficacy is critical. For those already at risk due to prior exposure, development of effective screening techniques will be crucial to allow for early detection of subclinical lesions. While some methods appear useful in screening those in high-risk cohorts, other techniques have application in individual patient assessment as well as in disease surveillance. Further research is needed to determine the optimal means to combine these and other methods to allow for optimal disease prevention and early detection on a larger scale.
Disclosure of potential conflicts of interest
No potential conflict of interest was reported by the authors.
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Van Dyne EA, Mona Saraiya SJH, Thomas CC, Markowitz LE, Benard VB. Trends in human papillomavirus–associated cancers — United States, 1999–2015. 2018. [accessed 2018] [DOI] [PMC free article] [PubMed]
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9306 | https://www.geeksforgeeks.org/maths/cofactor-matrix/ | Cofactor of a Matrix: Formula and Examples
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23 Jul, 2025
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The cofactor of a Matrix is an important concept in linear algebra. It is often used in calculating the determinant and the inverse of a matrix. Cofactor of Matrix or Cofactor matrix is the matrix formed by the Cofactor of each element of any matrix where cofactor is a number that is obtained by multiplying the minor of the element of any given matrix with -1 raised to the power of the sum of the row and column number to which that element belongs. This matrix formed with the cofactor of each element is called the Cofactor Matrix or Cofactor of Matrix.
Cofactors and Minors are very important concepts in Linear Algebra which further help us find the determinants, adjoint, and inverse of the matrix as well. We have explained in detail about Cofactor of a Matrix.
Cofactor of Matrix
A cofactor matrix is a matrix that comprises the cofactors of each element in a matrix. A cofactor is a number obtained when the minor Mij of the element aij is multiplied by the (-1)i+j. i and j represent the row and column of the particular element whose cofactor is being determined. Minor of an element is obtained by eliminating the row and column of that particular element i.e. eliminating row i and column j and then taking the remaining part of the matrix. Then we calculate the determinant of the remaining part which gives us the value of the minor of that particular element.
Cofactor of Matrix Definition
The cofactor of an element in a matrix is determined by the minor of that element, which is essentially the determinant of the submatrix obtained by removing the row and column containing the element in question.
Formula for Cofactor of a Matrix
If we denote the Cofactor using Cij, then the cofactor of any element for
Cij = Mij × (-1)i+j
i is the number of rows for the element under consideration,
j is the number of columns for the element under consideration, and
Mij is the minor of the element in the ith row and jth column.
What is Minor?
A minor of a particular element is obtained by eliminating the row and column of the matrix to which that particular element belongs and then finding the determinant of the remaining part. The matrix formed by combining all the minors is called the minor matrix. For example minor of the element a11 matrix ⎣⎡147258369⎦⎤ is calculated as:
M11=det[5869]=45−48=−3
How to Find Cofactor of Matrix?
In order to find a cofactor matrix we need to perform the following steps:
Step 1: Find the minor of each element of the matrix and make a minor matrix.
Step 2: Multiply each element in the minor matrix by (-1)i+j.
Thus, we obtain the cofactor matrix.
Let us understand how to find a cofactor matrix using an example:
Example: Find the cofactor matrix of ⎣⎡147258369⎦⎤
Step 1: Find the minor of each element and make a minor matrix.
Minor of a11 is calculated by eliminating row 1 and column 1 and taking the determinant of the remaining matrix as follows:
M11= determinant of [5869]
M11 = 5(9) - 6(8)
M11 = 45 - 48 = -3
Similarly, the minor of element a12 is calculated by eliminating row 1 and column 2 and taking the determinant of the remaining matrix as follows:
M12 = determinant of [4769]
M12 = 4(9) - 6(7)
M12 = 36 - 42 = -6
Similarly, calculate the minors of all elements to obtain the following minor matrix:
⎣⎡−3−6−3−6−12−6−3−6−3⎦⎤
Step 2: Multiply each element of the minor matrix by (-1)i+j to get the cofactor of that element i.e. Cij
Cofactor of M11 is calculated as follows:
C11 = M11 × (-1)1+1 = M11 = -3
Cofactor of M12 is calculated as follows:
C12 = M12 × (-1)1+2 = -M12 = 6
Similarly, calculate the other cofactors to obtain the following cofactor matrix:
⎣⎡−36−36−126−36−3⎦⎤
Cofactor of 2×2 Matrix
Consider a 2×2 matrix as follows:
A=[acbd]
Then the cofactor matrix of any such matrix is written as:
C=[a−c−bd]
Cofactor of 3×3 Matrix
Consider a 3×3 matrix as follows:
A=⎣⎡adgbehcfi⎦⎤
Then the cofactor matrix of any such matrix is calculated by calculating the cofactors of each of the elements as follows:
Let Mij denote the minor of the element in row i and column j, then in the above matrix:
M11=det[ehfi]
M12=det[dgfi]
Similarly, we can calculate the minors of all the elements to get the below minor matrix:
M=⎣⎡M11M21M31M12M22M32M13M23M33⎦⎤
Now cofactor of each element is calculated by multiplying the minor from the minor matrix with -1 raised to the power of the sum of row and column numbers to which that minor belongs as follows:
Let Cij denote the cofactor of minor Mij, then:
C11 = (-1)1+1M11 = M11
C12 = (-1)1+2M12 = -M12
C13 = (-1)1+3M13 = M13
Similarly, after calculating all the cofactors of each element we get the following cofactor matrix:
C=⎣⎡C11C21C31C12C22C32C13C23C33⎦⎤
Applications of Cofactor of a Matrix
There are various applications of Cofactor Matrix. Some of these applications are:
Cofactor of the Matrix is used to find the adjoint of the matrix.
Cofactor Matrix is used in the calculation of determinant of the matrix.
It is also used to find the inverse of the matrix.
Must Read:
Types of Matrix
Transpose of Matrix
Solved Examples on Cofactor of a Matrix
Example 1. Find the cofactor of a11 in the matrix[1537]
Given matrix is [1537]
Minor M11 = 7
Cofactor of a11 = 7 × (-1)1+1 = 7
Example 2. Find the cofactor of a12 n the matrix⎣⎡184470695⎦⎤
Given matrix is ⎣⎡184470695⎦⎤
Minor M12 = determinant of [8495]
M1 = 40 - 36 = 4
Cofactor C1 of a12 = M12 × (-1)1+2
C12 = 4 × (-1) = -4
Example 3. What is the cofactor matrix of[3768]
Step 1: Find the minor of each element and make a minor matrix.
Minor of a11 is calculated by eliminating the row 1 and column 1 as follows
M11 = 8
Similarly minor of element a12 is calculated by eliminating the row 1 and column 2 as follows:
M12 = 7
Similarly calculate minors of all elements to obtain the following minor matrix:
[8673]
Step 2: Multiply each element of the minor matrix by (-1)i+j to get the cofactor of that element i.e. Cij
Cofactor of M11 is calculated as follows:
C11 = M11 × (-1)1+1 = M11 = 8
Cofactor of M12 is calculated as follows:
C12 = M12 × (-1)1+2 = -M12 = -7
Similarly calculate the other cofactors to obtain the following cofactor matrix:
[8−6−73]
Example 4. What is the cofactor matrix of[0−3−6−4]
Step 1: Find the minor of each element and make a minor matrix.
Minor of a11 is calculated by eliminating the row 1 and column 1 as follows
M11 = -4
Similarly minor of element a12 is calculated by eliminating the row 1 and column 2 as follows:
M12 = -3
Similarly calculate minors of all elements to obtain the following minor matrix:
[−4−6−30]
Step 2: Multiply each element of the minor matrix by (-1)i+j to get the cofactor of that element i.e. Cij
Cofactor of M11 is calculated as follows:
C11 = M11 × (-1)1+1 = M11 = -4
Cofactor of M12 is calculated as follows:
C12 = M12 × (-1)1+2 = -M12 = -3
Similarly calculate the other cofactors to obtain the following cofactor matrix:
[−4630]
Example 5. What is the cofactor matrix of⎣⎡271496620⎦⎤
Step 1: Find the minor of each element and make a minor matrix.
Minor of a11 is calculated by eliminating the row 1 and column 1 and taking determinant of remaining matrix as follows:
M11 = determinant of [9620]
M11 = 0 -12 = -12
Similarly minor of element a12 is calculated by eliminating the row 1 and column 2 and taking determinant of remaining matrix as follows:
M12 = determinant of [7120]
M12 = 0 - 2 = -2
Similarly calculate minors of all elements to obtain the following minor matrix:
⎣⎡−12−36−46−2−6−38−338−10⎦⎤
Step2 : Multiply each element of the minor matrix by (-1)i+j to get the cofactor of that element i.e. Cij
Cofactor of M11 is calculated as follows:
C11 = M11 × (-1)1+1 = M11 = -12
Cofactor of M12 is calculated as follows:
C12 = M12 × (-1)1+2 = -M12 = 2
Similarly calculate the other cofactors to obtain the following cofactor matrix:
⎣⎡−1236−462−638−33−8−10⎦⎤
Practice Problems on Cofactor of a Matrix
Problem 1: Find the cofactor of the element in the second row and third column: A=⎣⎡26736845−10⎦⎤.
Problem 2: Find the cofactor matrix ofA=⎣⎡1247710455⎦⎤.
Problem 3: Find the cofactor matrix ofA=⎣⎡4168023197⎦⎤.
Problem 4: Find the cofactor matrix ofA=⎣⎡114001016⎦⎤.
Problem 5: Find the cofactor matrix ofA=⎣⎡39−17511197−1323⎦⎤.
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9307 | https://education.ti.com/-/media/08D78E56A085425396A223D03834EF58 | Derivatives: Instantaneous Rate of Change
by
Mary Ann Connors
Department of Mathematics Westfield State College Westfield, MA 01086
Textbook Correlation: Key Topic
• Derivatives
NCTM Principles and Standards:
• Process Standard
• Representation
• Connections
Exercises and Solutions :A. Intuitive Exploration of the Derivative of the Cosine Function.
Estimate the instantaneous rate of change (derivative) of the cosine function by using the average rate of change of cosine with respect to x. Let h equal the change in x.
Solution :The average rate of change command ( avgRC ) is in the CATALOG of the TI-89.
Compute the average rate of change with step, h. Then let h = 0.1 and smaller values. Highlight y1 = cos(x) in the Y= editor. Select F6(Style), 4:Thick . Select F6(Style), 1:Line for y2. Recall the copy and paste feature. Copy the highlighted expression on the Home Screen and paste it in the Y= editor. Do you recognize the graph of y2(x)?
Answer : y2(x) appears to be approximately equal to negative sin (x)
Estimate the instantaneous rate of change (derivative) of y = cos ( x) at x = π/6 using the average rate of change for increasingly small values of h.
Answer : Approximately - .5
B. Using the Definition of the Derivative to Find the Answer Symbolically 1. Use the definition of the derivative to verify your observation in Part A.
Graph y1( x) = cos( x). In the GRAPH window, press F5 (Math), 6:Derivatives ,
1: dy/dx . Type in the value of x (
π / 6 ). The derivative of cos ( x) at
π / 6 appears in the lower left corner of the screen. 3. Draw the tangent line at x =
π / 6 . Follow the procedure pictured below. Notice the equation in the lower left corner of the screen. What is the slope of the tangent line?
Use F3 (Calculus ), A:nDeriv( on the Home screen to compute the numerical derivative using the symmetric difference quotient.
Evaluate exact derivatives ( 2 nd , 8 for d( or F3 (Calculus ), 1: d( differentiate ) on the Home Screen. Additional Exercises :1. Use the average rate of change of f (x) = 5 x for small values of h (.001, .0001, etc.) to estimate its instantaneous rate of change (derivative). 2. Evaluate its limit at x = 0 to find the instantaneous rate of change (derivative) at x = 0. 3. Repeat #1 and 2 for the function, f (x) = x 1/5 . |
9308 | https://pubmed.ncbi.nlm.nih.gov/33864088/ | Systematic evaluation of endometriosis by transvaginal ultrasound can accurately replace diagnostic laparoscopy, mainly for deep and ovarian endometriosis - PubMed
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. 2021 May 17;36(6):1492-1500.
doi: 10.1093/humrep/deab085.
Systematic evaluation of endometriosis by transvaginal ultrasound can accurately replace diagnostic laparoscopy, mainly for deep and ovarian endometriosis
Manoel Orlando Goncalves1,Joao Siufi Neto2,Marina Paula Andres23,Daniela Siufi2,Leandro Accardo de Mattos14,Mauricio S Abrao23
Affiliations Expand
Affiliations
1 Female Pelvic Diagnostic Section, Alta Medicina Diagnostica, Sao Paulo, SP, Brazil.
2 Gynecologic Division, BP-A Beneficência Portuguesa de São Paulo, São Paulo, SP, Brazil.
3 Disciplina de Ginecologia, Departamento de Obstetricia e Ginecologia, Hospital das Clinicas HCFMUSP, Faculdade de Medicina, Universidade de Sao Paulo, Sao Paulo, SP, Brazil.
4 Department of Imaging Diagnosis, Escola Paulista de Medicina da Universidade Federal de Sao Paulo, Sao Paulo, SP, Brazil.
PMID: 33864088
DOI: 10.1093/humrep/deab085
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Systematic evaluation of endometriosis by transvaginal ultrasound can accurately replace diagnostic laparoscopy, mainly for deep and ovarian endometriosis
Manoel Orlando Goncalves et al. Hum Reprod.2021.
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. 2021 May 17;36(6):1492-1500.
doi: 10.1093/humrep/deab085.
Authors
Manoel Orlando Goncalves1,Joao Siufi Neto2,Marina Paula Andres23,Daniela Siufi2,Leandro Accardo de Mattos14,Mauricio S Abrao23
Affiliations
1 Female Pelvic Diagnostic Section, Alta Medicina Diagnostica, Sao Paulo, SP, Brazil.
2 Gynecologic Division, BP-A Beneficência Portuguesa de São Paulo, São Paulo, SP, Brazil.
3 Disciplina de Ginecologia, Departamento de Obstetricia e Ginecologia, Hospital das Clinicas HCFMUSP, Faculdade de Medicina, Universidade de Sao Paulo, Sao Paulo, SP, Brazil.
4 Department of Imaging Diagnosis, Escola Paulista de Medicina da Universidade Federal de Sao Paulo, Sao Paulo, SP, Brazil.
PMID: 33864088
DOI: 10.1093/humrep/deab085
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Abstract
Study question: What is the sensitivity and the specificity of preoperative transvaginal ultrasound with bowel preparation (TVUS-BP) compared to diagnostic laparoscopy (DL) for the identification of ovarian and deep sites of endometriosis?
Summary answer: DL was able to detect retrocervical, ovarian, and bladder endometriosis with similar sensitivity and specificity as TVUS-BP, whereas for vaginal and rectosigmoid endometriosis, DL had lower sensitivity and specificity than TVUS-BP.
What is known already: TVUS-BP is a non-invasive examination with good accuracy for diagnosing ovarian and deep endometriosis. DL is expensive and can lead to surgical complications.
Study design, size, duration: This prospective study included a total of 120 consecutive patients who underwent surgery for suspected endometriosis with preoperative imaging (TVUS-BP), including a video of the laparoscopic procedure, between March 2017 and September 2019.
Participants/materials, setting, methods: Two radiologists performed preoperative TVUS-BP using the same protocol for diagnosing endometriosis. Two surgeons, who were blinded to the results of the preoperative imaging and clinical data, reviewed the surgical videos from the entry of the abdominal cavity until the surgeon finalized a complete and systematic review prior to beginning any dissection (considered as a DL). A data sheet was used by surgeons and radiologists to record the sites and size of disease involvement, the American Society for Reproductive Medicine (ASRM) stage, and the Enzian score. The surgical visualization of endometriosis lesions that were confirmed by histological analysis was the gold standard.
Main results and the role of chance: DL was able to detect retrocervical, ovarian, and bladder endometriosis with similar sensitivity and specificity as TVUS-BP. DL was not able to detect vaginal endometriosis (sensitivity and specificity 0%): this is compared to a sensitivity and specificity of 85.7% and 99.1%, respectively with the utilization of a preoperative TVUS-BP. In addition, DL was notably poor at detecting rectosigmoid endometriosis, with a sensitivity of 3.7-5.6%, and this compares to 96.3% sensitivity with utilization of a preoperative TVUS (P < 0.001). For the ASRM stage, TVUS-BP results were highly correlated with the degree of endometriosis and pouch of Douglas (POD) obliteration (weighted Kappa of 0.867 and 0.985, respectively). For the Enzian score, there was a substantial correlation between TVUSP-BP and DL for compartment A (weighted Kappa = 0.827), compartment B (weighted Kappa = 0.670), and compartment C (weighted kappa = 0.814).
Limitations, reasons for caution: The number of participants included may be a limitation in this study and, as the evaluators were blinded to the physical exam, the DL accuracy could be underestimated. As biopsies of pelvic organs were obtained only if there was a suspicion of endometriosis, the gold standard was not always applicable. This aspect could underestimate the prevalence of lesions and overestimate the sensitivity and the specificity of both the TVUS-BP and the DL.
Wider implications of the findings: Preoperative TVUS-BP was accurate in identifying all sites of ovarian and deep endometriosis that were evaluated. It had significantly higher sensitivity than DL in detecting rectosigmoid endometriosis and predicting intraoperative ASRM staging and the Enzian score. These results suggest that TVUS-BP can replace DL for the diagnosis and treatment planning for patients with ovarian and deep endometriosis.
Study funding/competing interest(s): The authors declare no source of funding or conflicts of interest.
Trial registration number: N/A.
Keywords: American Society for Reproductive Medicine stage; Enzian score; diagnostic laparoscopy; endometriosis; transvaginal ultrasound.
© The Author(s) 2021. Published by Oxford University Press on behalf of European Society of Human Reproduction and Embryology. All rights reserved. For permissions, please email: journals.permissions@oup.com.
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9309 | https://www.gauthmath.com/solution/QhNiSp2Awrt/What-is-the-equation-of-chord-of-the-ellipse-x-2-a-2-y-2-b-2-1-whose-middle-poin | Sign in
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What is the equation of chord of the ellipse x 2 a 2 + y 2 b 2 = 1 whose middle point is ( x 1 , y 1 ) ? You are given. T = x x 1 a 2 + y y 1 b 2 − 1 a n d S 1 x 2 1 a 2 + y 2 1 b 2 − 1 T + S1=0 T = S1 T + S1= 1 T + 1= S1 A T + S1=0 B T = S1 C T + 1= S1 D T + S1= 1
Expert Verified Solution
Answer
the correct option is B, $$T = S_{1}$$T=S1
Explanation
To find the equation of the chord of the ellipse given by the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$a2x2+b2y2=1
with a midpoint at $$(x_{1}, y_{1})$$(x1,y1), we can utilize the concept of tangents and chords in conic sections.
Define the terms. The term $$T$$T represents the equation of the ellipse evaluated at the point $$(x, y)$$(x,y):
$$T = \frac{x x_1}{a^2} + \frac{y y_1}{b^2} - 1$$T=a2xx1+b2yy1−1
2. The term $$S_{1}$$S1 represents the equation of the ellipse evaluated at the midpoint $$(x_{1}, y_{1})$$(x1,y1):
$$S_1 = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$$S1=a2x12+b2y12−1
3. The relationship between $$T$$T and $$S_{1}$$S1 for the chord can be expressed as:
$$T + S_1 = 0$$T+S1=0
This means that the point $$(x, y)$$(x,y) lies on the chord whose midpoint is $$(x_{1}, y_{1})$$(x1,y1)
4. The specific relationship $$T = S_{1}$$T=S1 indicates that the chord passes through the midpoint of the segment defined by the endpoints on the ellipse. This relationship holds true for any conic section, not just ellipses.
Thus, the correct choice is:
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Find the length of the chord of the ellipse x2/25 + y2/16 = 1 whose middle point is 1/2, 2/5.
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x-2x+3-x+1x-4=0 10. 3x+ 2/x =4 11. 1/x + x-2/2 =3 12. 4x+5=2x2 13. 2x/5 - 3/x =1 14. 2/x + x-1/4 = 3x/2 15. 3x+1/2x - 1/3x = 5/6
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Bookwork code: 5B What is the perimeter of the triangle below in cm? 9 w 0 l ................ E t S 9 L 8 .......................... 6 OL E 0 1 2 3 4 5 6 7 8 9 10 cm Zoom Watch video Answer
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A company is using linear programming to decide how many units of each of its two products to make each week. Weekly production will be x units of Product X and y units of Product Y. At least 50 units of X must be produced each week, and at least twice as many units of Y as of X must be produced each week. Each urt of X requires 30 minutes of labour, and each unit of Y requires two hours of labour. There are 5,000 hours of labour available each week. Which of the following is the correct set of constraints? Submit your answer to view the feedback. 0.5x+2y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≤ 2x y ≥ 100 x+4y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≥ 2x y ≥ 2x
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9310 | https://chem.libretexts.org/Courses/Brevard_College/CHE_201%3A_Organic_Chemistry_I/04%3A_Aromatic_Compounds_(Arenes) | Skip to main content
4: Aromatic Compounds (Arenes)
Last updated
: Jun 21, 2020
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3.12: Chemical properties of Alkynes
4.1: Aromatic Compounds- Benzene
Page ID
: 221783
( \newcommand{\kernel}{\mathrm{null}\,})
4.1: Aromatic Compounds- Benzene
: Aromatic hydrocarbons appear to be unsaturated, but they have a special type of bonding and do not undergo addition reactions.
4.2: Aromaticity
: Aromaticity is a property of conjugated cycloalkenes in which the stabilization of the molecule is enhanced due to the ability of the electrons in the π orbitals to delocalize. This acts as a framework to create a planar molecule.
4.3: Bonding and Physical Properties of Arenes
: Arenes are aromatic hydrocarbons. The term "aromatic" originally referred to the pleasant smells given off by arenes, but now implies a particular type of delocalized bonding (see below). The arenes you are likely to encounter at this level are based on benzene rings. The simplest of these arenes is benzene itself, C6H6. The next simplest arene is methylbenzene (common name: toluene), which has one of the hydrogen atoms attached to the ring replaced by a methyl group - C6H5CH3.
4.4: Nomenclature of Aromatic Compounds
: Aromatic compounds contain a benzene ring or have certain benzene-like properties; for our purposes, you can recognize aromatic compounds by the presence of one or more benzene rings in their structure.
4.5: Polycyclic Aromatics
: Polycyclic aromatics are compounds containing two or more fused aromatic rings. These fused benzene rings share two carbon atoms between them. These are also called polycyclic benoid or polycyclic aromatic hydrocarbons (PAHs). Many PAHs are carcinogenic and highly toxic.
4.6: Carbon Nanomaterials
4.7: Chemical Properties of Aromatic Compounds
4.8: General Mechanism for Electrophilic Substitution Reactions of Benzene
4.9: Halogenation, Sulfonation, and Nitration of Aromatic Compounds
4.10: Alkylation and Acylation of Aromatic Rings - The Friedel-Crafts Reaction
4.11: Electrophilic Aromatic Substitution Reactions of Benzene Derivatives
4.12: Heterocyclic Aromatic Compounds
3.12: Chemical properties of Alkynes
4.1: Aromatic Compounds- Benzene |
9311 | https://www.quantum-munich.de/119947/Negative-Absolute-Temperatures | Quantum Optics Group
Negative Absolute Temperatures
Here we provide some answers to frequently asked questions on the intriguing topic of negative absolute temperatures.
Here we want to answer some frequently asked questions concerning negative absolute temperatures. If your question is not answered yet, please send an email to Simon Braun or Ulrich Schneider and we will add it.
What is absolute temperature?
Absolute temperature refers to temperature on the Kelvin scale, where 0K is the absolute zero point, where all motion in a classical gas would stop.
Can all systems achieve negative temperatures?
No! Most systems, including a classical gas (e.g. the air around us), are limited to positive absolute temperatures.
In order to be able to reach negative temperatures, a system needs to possess an upper bound for the energy of its particles, i.e. a maximal possible energy a particle of the system can have.
This limit is not an external limit in the sense that there is just no more energy available. It is an internal limit - the particles cannot absorb more energy even if there is plenty available. Atoms in a classical gas, for example, do not possess such an upper energy limit - their kinetic energy can be arbitrarily high. Even the speed of light as the ultimate speed limit for particles does not pose a limit to their kinetic energy: Although the speed of the particles is limited by the speed of light, the kinetic energy can still be arbitrarily high.
In fact, the idea of negative absolute temperature is old, but was seldom discussed for mobile particles because everybody assumed that it was impossible to realize.
Is your system really colder than zero Kelvin?
No! Nothing can be colder than absolute zero (0K)!
Negative absolute temperatures (or negative Kelvin temperatures) are hotter than all positive temperatures - even hotter than infinite temperature.
What does hotter or colder actually mean?
Take two glasses of water, one with hot water and one with cold water. If you mix the water, you will get medium hot water. In the language of thermodynamics, this means that heat was transferred from the hot water to the cold water, until they both reached the same temperature.
So heat always flows from the hotter to the colder system.
And since negative temperature systems have a lot of energy, heat will flow from negative temperature systems to positive temperature systems.
Did you redefine temperature?
No. We use the textbook definition of temperature.
How is temperature formally defined?
Temperature is defined via entropy: The inverse of temperature (1/T) is defined as the change of entropy (S) in a system, when the total energy E of the system is changed: 1/T=∂S/∂E.
Total energy is the sum of the energy of all particles of the system, and entropy is explained later in this FAQ. Below we show a typical curve of entropy versus energy in a system that has a minimum (Emin) and maximum (Emax) energy. The slope of this curve indicates how entropy changes with energy, ∂S/∂E, and therefore gives the inverse temperature, 1/T.
What is entropy and how is it related to temperature?
Entropy is a measure of disorder in the system.
It is related to the number of energy states that are occupied by the particles: If only one energy state is occupied, as for example the lowest energy state in the case of a temperature of zero Kelvin, the system is very ordered and the entropy is zero. If the particles are however distributed over many energy states, low energy and high energy, the system is very disordered and the entropy is large. Temperature is defined via entropy – see above.
What does temperature actually mean?
In a classical gas - think of many independent billiard balls - temperature describes how the particles are distributed among their possible states, for instance their velocities. As these states can have different energies, this means that in such a gas all atoms together have one particular temperature, but the individual atoms do not have the same energy.
The distribution of energies is called the Boltzmann distribution and describes the following: At absolute zero (plus zero), all particles are at rest and their kinetic energy is zero. So the distribution of energies has a sharp peak at zero - here are all the atoms - and is zero otherwise. If you now heat up the gas, you increase both its temperature and its total energy and this sharp peak will broaden (see the left inset in the above figure). So you can find some atoms now at higher energies. But you will still find more atoms at a state with lower energy than at a state with higher energy. The hotter the gas becomes, the broader and shallower this peak becomes, until at infinite temperature the distribution would be completely flat and all states would be equally probable (middle inset).
Negative temperature now means that this distribution is inverted or flipped around, so that you find more atoms in a higher energy state than in a lower one (right inset). This means that the peak in the distribution is not at the lowest energy anymore, but at the highest possible energy.
In general, the thermal distribution does not only take kinetic energy into account, but all forms of energy that the particles can possess, for example potential energy or magnetic energy.
How do you measure temperature? Do you put a thermometer close to the atoms?
Measuring temperature is indeed a difficult task. Unfortunately, we cannot put a thermometer into our system, close to the atoms.
Instead, we take images of our atoms with a CCD camera. These images show us the distribution of atoms. For each temperature, we theoretically expect a certain distribution. We compare the distribution of our atoms with the theoretically expected distributions. The distribution that fits best gives us the temperature of the atoms.
Do your atoms really have negative absolute temperature or do they just behave like that?
Our atoms really have negative absolute temperature!
The way temperature is defined tells us the following: If a system thermalizes, i.e. tries to reach thermal equilibrium, and if we can describe the distribution of the system with some Boltzmann distribution, and if this distribution remains stable over some time, we know that the system has reached thermal equilibrium. We can then assign the corresponding temperature to the system. The system then really has this temperature. In our case, as the Boltzmann distribution of our atoms is inverted, this temperature is negative.
How should I think about the temperature axis? I always thought it goes from zero to infinity.
For most systems (like the classical gas) this is correct, as these systems cannot have negative absolute temperatures. For those systems that can assume negative temperatures, the temperature scale also starts at plus zero, increases until infinity, then jumps to minus infinity and increases further until minus zero.
Does that mean that temperature is circular?
No! Plus zero and minus zero are not connected in this case, in fact they are as far away as possible.
Think of the ordinary line of reals numbers: It starts at minus infinity, runs through zero, and goes to plus infinity. Now imagine cutting this line at zero into two parts. You end up with two lines, one running from minus infinity to zero, this end is called minus zero; and a second part running from plus zero to plus infinity. Now we glue the two ends at plus and minus infinity together and pull the two loose ends at the zeros apart. So we end with a line like this:
+0,...,+10,...,+infinity,-infinity,...,-10,...,-0
So the two zeros are as far apart as possible, in fact plus zero corresponds to the state where all particles are in the lowest energy state (the bottom of the valley in the picture below) while minus zero corresponds to the state that all particles are in the state with the highest possible energy (the top of the hill).
Temperature in a game of marbles
Why do you need an upper limit in energy for a negative temperature state but not for a positive temperature one?
The situation is in fact completely symmetric: You need a lower bound for the energy in order to get positive temperatures and an upper bound in order to get negative temperatures.
The most intuitive way to see this is by considering the picture of the valley and the hill: At small positive temperatures you will find most marbles close to the lowest point, near the floor of the valley, which represents the lower bound in energy. If there would not be such a bound, there would be an infinitely deep hole instead and all marbles would continue to fall down the hole without ever reaching a stable state. For negative temperatures, where all particles want to pile up on top of the mountain, we similarly require that the mountain has a finite height.
For the classical picture of free moving particles, kinetic energy obviously has a lower bound - the particles stand still - so that we rarely need to think about this requirement.
How can plus infinity and minus infinity be the same thing?
In statistical mechanics the temperature T is not used that often. Most times we use β, which is the inverse: β=1/T. Concerning β, plus infinite temperature and minus infinite temperature are the same, namely zero. In fact, -β would have been a better choice for the definition of temperature, as it runs from minus infinity via zero to plus infinity, thereby avoiding the jump from plus to minus infinity and the confusion with "hot negative absolute temperatures". (see also "How is temperature formally defined?")
Are the atoms really stable at negative absolute temperature or do you just hold the atoms in place, at maximum energy?
The atoms at negative absolute temperature are really stable by themselves!
We create the system with several laser beams and a magnetic field in a vacuum chamber. This altogether constitutes the environment in which the atoms "live", a world with bright and dark regions in which the atoms are trapped. Once the atoms are prepared at negative absolute temperature, with many atoms close to maximum energy, they stay there. They stay there just as long as atoms at positive absolute temperature in a comparable system which are mainly sitting at minimum energy.
This can be illustrated with the landscape above. At positive temperature, most marbles lie in the valley. This state is stable - this is nature as we know it. If marbles lie on top of the hill they will usually roll down the hill and thereby convert potential energy into kinetic energy. If these marbles are, however, at negative absolute temperature, all their energy forms will be close to their maximum allowed values. In the picture this is apparently true for potential energy, as the marbles lie on top of the hill, the highest point in the landscape. But it is also true for kinetic energy: The marbles on top of the hill move around fast. Because kinetic energy cannot increase further, the marbles cannot convert potential energy into kinetic energy. Therefore they cannot roll down at all. They stay on top of the hill - the system is stable.
Shouldn't the atoms at negative absolute temperature that are hotter than infinity burn away your lab?
No. The crucial point to consider here is the amount of energy that is contained in a system.
Our atoms are very hot in the sense that they contain a lot of energy: The energy of most atoms is close to the maximum that is allowed by the upper limit for their energy. This maximum energy, however, is very little compared to, for example, the kinetic energy of a typical water molecule in a pot of water at room temperature. Additionally, our system at negative absolute temperature consists only of around a hundred thousand atoms. This is about 17 orders of magnitude less than the number of water molecules in only a milliliter of water. If you sum up the energy of all our atoms, the total energy contained in our system at negative absolute temperature is therefore very small compared to energies that we are used to in everyday life.
It is definitely easier to harm yourself with a pot of boiling water. Although water is not at negative absolute temperature, but "only" at positive absolute temperature, the total energy contained in the pot will be much higher than in our system. If you dip your finger into the hot water, a lot of energy will be transferred to your finger, and that hurts. If you could, however, dip your finger into our system, only very little energy would be transferred.
Where do you get your upper limit for kinetic energy from?
We put our atoms into a so-called optical lattice that consists of a periodic potential created by interfering laser beams. For our atoms, this periodic potential acts in the same way as the potential formed by the ions in a solid acts for the electrons - it creates a band structure.
This means that, due to this potential, the kinetic energy splits up into different bands, e.g. valence and conduction band in a metal or semi-conductor. These bands are separated in energy by so-called band gaps - regions of energy without any states for the particles. Therefore, each band naturally possesses a lower as well as an upper bound for the kinetic energy. We put all our atoms into one of these bands and therefore created the required upper bound for the kinetic energy.
This on its own is nowadays done routinely in many labs. The new ingredient in our work is that we also created upper limits for the interaction energy and the potential energy. The latter, for example, we limited by transforming the optical trap, in which the atoms sit, into an anti-trap, i.e. transforming a valley into a hill.
Do lasers also have a negative absolute temperature?
No! A temperature can only be assigned to a state in thermal equilibrium. This means states that, when left alone (when the system is thermally isolated) will remain stable and do not change over time. The particles in a laser medium do have an inverted energy population -more particles are in excited states than in low energy states- and their distribution looks indeed similar to a state at negative temperature. This inverted energy population, however, only exists as long as the laser is continuously pumped, i.e. particles are actively pumped into the exited states. When you switch-off the pump, all atoms will decay back into the lower state and their energy goes into the laser beam. So, while pumped, the laser medium is in a steady state, but not in a thermal state. It is not in thermal equilibrium and therefore cannot have a temperature.
What is this possible connection to cosmology?
We could show that our atomic system is stable, even though the atoms strongly attract each other - that means they want to collapse but cannot due to being at negative temperature.
And the universe as a whole is also not collapsing under the attractive force of gravity - on the contrary, its expansion is accelerating. In cosmology, dark energy is introduced exactly to describe this effect.
But whether this hints at some connection or is just a coincidence remains to be seen.
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9312 | https://www.mathway.com/popular-problems/Algebra/260324 | Solve by Completing the Square x^2-x+1=0 | Mathway
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9313 | https://mathematica.stackexchange.com/questions/189607/dateobject-ambiguous-string-interpretation | date and time - DateObject ambiguous string interpretation? - Mathematica Stack Exchange
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DateObject ambiguous string interpretation?
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I try to enter the following input, but receive an error message:
mathematica
DateObject["12/1/2016", DateFormat -> {"Month", "/", "Day", "/", "Year"}]
despite having specified the date format explicitly, Mathematica is complaining about ambiguity of day and month ordering. I presume there is something wrong with the syntax I use. How should I be entering this command properly?
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asked Jan 16, 2019 at 15:06
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You've got DateFormat option the wrong way around: it controls output, not input.
What you want to do is actually the following (with day calendar granularity):
mathematica
DateObject[{"12/1/2016", {"Month", "/", "Day", "/", "Year"}}, "Day"]
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answered Jan 16, 2019 at 15:16
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9314 | https://blog.intraratio.com/mastering-rework-in-manufacturing-ensuring-traceability-and-quality-at-every-step | Skip to content
In the vast realm of manufacturing, perfection at the first go isn't always a given. The process of rework, therefore, stands as a testament to a company's dedication to ensuring the highest standards of quality. As we dive deeper into this critical aspect, the role of technology in refining the rework process emerges as a beacon of transformation.
What is Rework in Manufacturing?
Rework, in the manufacturing context, refers to the process wherein products that haven't met the specified criteria undergo corrections or modifications. Essentially, if a manufactured item reveals defects, errors, or any form of discrepancy that deviates from its intended design or function, it demands rework.
The very existence of this step in the manufacturing world underscores the industry's relentless pursuit of excellence. Every product, despite its initial shortcomings, should epitomize optimum quality by the end of its production journey.
What Causes Rework in Manufacturing?
The road to product perfection isn't devoid of obstacles. Various factors can necessitate rework:
Design Flaws: Even if a product is crafted precisely to specifications, inherent design anomalies can usher in problems.
Process Errors: Mistakes embedded in the manufacturing process, whether through inaccurate machine settings or human misjudgments, can birth defects.
Material Deficiencies: Resorting to substandard or unsuitable materials can trigger product inefficiencies.
Ambiguities: At times, miscommunications or misinterpretations regarding product specifications can instigate errors.
Machine Malfunctions: Equipment that malfunctions or doesn't operate at its optimal efficiency can churn out flawed products.
For instance, the importance of traceability in the Surface Mount Technology (SMT) process elucidates how specialized manufacturing processes have unique challenges and standards.
How to Reduce Rework in Manufacturing
To master rework is to minimize its necessity. Several strategies can be instrumental:
Robust Training: A workforce well-versed in the nuances of the manufacturing process and updated with evolving best practices can dramatically reduce errors.
Quality Control Interventions: Implementing consistent inspections at multiple stages can preemptively identify and rectify mistakes.
Process Standardization: Standardized protocols across the board can curtail variability and human-induced errors.
Technological Integration: Leveraging advanced tools and methodologies, like machine data integration for automated traceability & process control, can enhance precision and consistency.
Feedback Mechanisms: Establishing channels where issues flagged by end-users or during quality control can be looped back into the design and manufacturing phases paves the way for continuous refinement.
For those keen on delving into real-world applications and tangible outcomes, exploring insights from re-work tracking in electronics manufacturing can be enlightening.
Best Practices for Managing Rework
Rework, while being an inevitable part of manufacturing, demands a streamlined approach to ensure efficiency. The following are some best practices industries can adapt to manage rework more effectively:
Document Everything: Every instance of rework should be thoroughly documented, capturing the cause, the solution, and the individuals or teams involved.
Prioritize Communication: Ensuring open lines of communication between departments can prevent errors from escalating and can help in quickly resolving them.
Use Predictive Analysis: By analyzing trends from past rework cases, businesses can anticipate and prevent future errors.
Establish Clear Protocols: A well-defined protocol for handling defects can expedite the rework process and ensure consistency.
Continual Training: Regular workshops and training programs can update the workforce about newer tools, methodologies, and ways to mitigate errors.
Feedback Analysis: Encourage teams to discuss rework instances in review meetings, analyzing what went wrong and brainstorming ways to prevent it in the future.
Production Quality Control
Quality control, integral to the production process, ensures that the products being manufactured meet or exceed expected standards:
Regular Inspections: Scheduled checks, rather than just end-of-line inspections, can identify and rectify defects earlier in the production cycle.
Use of Technology: Instruments like spectrometers, digital microscopes, and automated testing tools can augment the accuracy of inspections.
Quality Assurance Teams: Dedicated teams focused on quality control can provide an objective review, ensuring that no internal biases affect the evaluation process.
Supplier Audits: Regularly reviewing and auditing suppliers can ensure the raw materials are of the desired quality, reducing potential issues in the finished product.
Feedback Loop: Establishing a system where end-users can report issues ensures that real-world problems are flagged and addressed.
The Pivotal Role of Data in Rework Management
With industries becoming increasingly data-driven, the integration of data and analytics in rework management has burgeoned in importance. Data traceability in manufacturing, as explained in this comprehensive article, transcends mere numbers and codes. It's a treasure trove of insights, patterns, and actionable recommendations.
Harnessing this data, manufacturers can proactively pinpoint vulnerabilities in their processes, anticipate potential flaws, and chart out rework strategies with unmatched efficiency. This proactive, data-led approach doesn't just save time and resources but also elevates the quality quotient of the end product.
Moreover, using data to refine and navigate rework processes transforms the entire narrative. Rework, when underpinned by data, shifts from being a reactive measure to a proactive quality enhancement strategy. By leveraging data to gain control and improve line performance, the process becomes more about evolution and setting industry benchmarks than mere corrections.
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9316 | https://www.gauthmath.com/solution/1813926584051765/6-If-possible-find-a-counterexample-to-the-statement-If-ab-0-then-a-0-If-it-is-n | Solved: If possible, find a counterexample to the statement: "If ab<0 , then a<0 ". If [Math]
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If possible, find a counterexample to the statement: "If ab<0 , then a<0 ". If it is not possible, write "Not possible". Not passible
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Answer
$$a = 1, b = -1$$a=1,b=−1 is a counterexample.
Explanation
To find a counterexample to the statement "If $$ab < 0$$ab<0, then $$a < 0$$a<0", we need to identify values for $$a$$a and $$b$$b that satisfy $$ab < 0$$ab<0 while making $$a$$a non-negative (i.e., $$a \geq 0$$a≥0).
Consider the case where $$a = 1$$a=1 and $$b = -1$$b=−1. Here, we have:
Calculate $$ab$$ab:
$$ab = 1 \cdot (-1) = -1$$ab=1⋅(−1)=−1
Check if $$ab < 0$$ab<0:
$$-1 < 0 \quad \text{(True)}$$−1<0(True)
Check if $$a < 0$$a<0:
$$1 < 0 \quad \text{(False)}$$1<0(False)
Since $$ab < 0$$ab<0 is true while $$a < 0$$a<0 is false, this serves as a valid counterexample.
Thus, the answer is:
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If-Then Statements and Postulates Identify the hypothesis and conclusion of each conditional statement. 1. If 3x-1=7 , then x=2. 2. If Carl scores 85%, then he passes. Write each conditional statement in if-then form. 3. All students like vacations. 4. The game will be played provided it doesn’t rain. Write the converse of each conditional. Determine if the converse is true or false. It if is false, give a counterexample. 5. If it rains, then it is cloudy 6. If x is an even number, then x is . divisible by 2.
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9317 | https://www.johndcook.com/blog/2010/10/27/divisibility-by-7/ | Divisibility by 7
How can you tell whether a number is divisible by 7? Almost everyone knows how to easily tell whether a number is divisible by 2, 3, 5, or 9. A few less know tricks for testing divisibility by 4, 6, 8, or 11. But not many people have ever seen a trick for testing divisibility by 7.
Here’s the trick. Remove the last digit from the number, double it, and subtract it from the first part of the number. Do this repeatedly until you get something you recognize as being divisible by 7 or not.
For example, start with 432. Split into 43 and 2. Subtract 4 from 43 to get 39. Since 39 isn’t divisible by 7, neither is 432.
For another example, start with 8631. Split into 863 and 1. Subtract 2 from 863 to get 861.
Now split 861 into Split into 86 and 1. Subtract 2 from 86. Maybe you recognize 84 as a multiple of 7. If not, double 4 and subtract from 8 to get 0, which is divisible by 7. Either way, we conclude that 8631 is divisible by 7.
Why does this work? Let b be the last digit of a number n and let a be the number we get when we split off b. That says n = 10a + b. Now n is divisible by 7 if and only if n – 21b is divisible by 7. But n – 21b = 10(a − 2b) and this is divisible by 7 if and only if a − 2b is divisible by 7.
What about the remainder when you divide a number by 7? Here’s where the rule for 7 differs from the more familiar divisibility rules. For example, a number is divisible by 3 if its digit sum is divisible by 3, and furthermore the remainder when a number is divided by 3 is the remainder when its digit sum is divided by 3. But the divisibility rule for 7 does not give the remainder when a number is divided by 7. For a simple example, the divisibility rule for 31 terminates in 1, but the remainder with 31 is divided by 7 is 3.
Why doesn’t the divisibility rule for 7 give the remainder? It is true that 10a + b and (10a + b) – 21b have the same remainder when divided by 7. But then we factored this into 10(a − 2b). It’s true that 10(a − 2b) is divisible by 7 if and only if (a − 2b) is divisible by 7, but if neither is divisible by 7 then they will leave different remainders.
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9318 | https://en.wikipedia.org/wiki/Talk%3ASingleton_bound | Talk:Singleton bound - Wikipedia
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1 Statement of bound2 comments
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3 Introduction1 comment
Talk:Singleton bound
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showMathematicsLow‑priority Mathematics portal This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.Mathematics Wikipedia:WikiProject Mathematics Template:WikiProject Mathematics mathematics LowThis article has been rated as Low-priority on the project's priority scale.
Statement of bound
[edit]
This article does not state the Singleton bound in its most common and useful form. Most coding text books -- and, notably, Singleton's original paper itself -- states the Singleton bound as a bound on d, usually in the form d <= n-k+1, not as a bound on A_q(n,d). This article is unnecessarily opaque. — Preceding unsigned comment added by 128.149.22.213 (talk) 16:38, 1 April 2014 (UTC)[reply]
The version you refer to is only valid for linear codes (k is the dimension of the code) and follows immediately from the more general version which is called Singleton's bound in modern treatments. What is missing in this article is the simple derivation of the linear code form from the other, which I'll put in along with some other fixes. Bill Cherowitzo (talk) 03:40, 2 April 2014 (UTC)[reply]
Proof
[edit]
I feel like that instead of this:
|C|≤A q(n,d)≤q n−d+1.{\displaystyle |C|\leq A_{q}(n,d)\leq q^{n-d+1}.}
The proof needs something more like this:
A q(n,d)=|C|≤q n−d+1.{\displaystyle A_{q}(n,d)=|C|\leq q^{n-d+1}.}
Or am I missing something in the proof? 82.53.62.96 (talk) 16:42, 12 January 2012 (UTC) Damix[reply]
Introduction
[edit]
It looks to me as if the size r is never used in the text. Maybe it shouldbe removed from the first sentence. Feynman81 (talk) 17:38, 16 March 2012 (UTC)[reply]
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9319 | https://www.quora.com/How-do-I-find-Taylor-series-expansion-of-z-1-z-1-about-z-1 | How to find Taylor series expansion of z – 1/ z + 1 about z = 1 - Quora
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How do I find Taylor series expansion of z – 1/ z + 1 about z = 1?
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Upvoted by
BowTangey
, PhD Mathematics, Iowa State University (1988) · Author has 16.6K answers and 29.4M answer views
·2y
How do I find Taylor series expansion of z – 1/ z + 1 about z = 1?
Let f(z)=z−1 z+1 f(z)=z−1 z+1. Then f′(z)=1+1 z 2 f′(z)=1+1 z 2, f′′(z)=−2 z 3 f″(z)=−2 z 3, and f(n)(z)=(−1)n+1 n!z n+1 f(n)(z)=(−1)n+1 n!z n+1.
Evaluate these at z=1 z=1 and plug into the Taylor series.
f(z)=1+2(z−1)−2!(z−1)2 2!+3!(z−1)3 3!−4!(z−1)4 4!+−…f(z)=1+2(z−1)−2!(z−1)2 2!+3!(z−1)3 3!−4!(z−1)4 4!+−….
As the function has just one singular point (at z=0 z=0), the radius of convergence is 1−0=1 1−0=1.
You could also solve this by writing −1 z=1 1+(z−1)−1 z=1 1+(z−1) and using the known geometric series expansion. (Adding z+1 z+1, obviously.)
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Write f(z) = z − 1/(z+1). Expand about z0 = 1 using w = z−1 (so z = 1+w). Compute directly and then as power series in w.
1) Substitute:
f(z) = 1 + w − 1/(1 + w + 1) = 1 + w − 1/(2 + w).
2) Expand 1/(2+w) as geometric series in w/2 (valid for |w|<2):
1/(2+w) = (1/2)·1/(1 + w/2) = (1/2) ∑{n=0}^∞ (−1)^n (w/2)^n
= ∑{n=0}^∞ (−1)^n w^n / 2^{n+1}.
3) Subtract from 1+w:
f(z) = 1 + w − ∑{n=0}^∞ (−1)^n w^n / 2^{n+1}
= (1 − 1/2) + (1 − (−1)/4)w − ∑{n=2}^∞ (−1)^n w^n / 2^{n+1}.
4) Simplify first two coefficients:
1 − 1/2 = 1/2.
For n=1 term of the sum is (−1)^1 w /2^2 = −w/4, so coefficient of w is 1 − (−1/4
Continue Reading
Write f(z) = z − 1/(z+1). Expand about z0 = 1 using w = z−1 (so z = 1+w). Compute directly and then as power series in w.
1) Substitute:
f(z) = 1 + w − 1/(1 + w + 1) = 1 + w − 1/(2 + w).
2) Expand 1/(2+w) as geometric series in w/2 (valid for |w|<2):
1/(2+w) = (1/2)·1/(1 + w/2) = (1/2) ∑{n=0}^∞ (−1)^n (w/2)^n
= ∑{n=0}^∞ (−1)^n w^n / 2^{n+1}.
3) Subtract from 1+w:
f(z) = 1 + w − ∑{n=0}^∞ (−1)^n w^n / 2^{n+1}
= (1 − 1/2) + (1 − (−1)/4)w − ∑{n=2}^∞ (−1)^n w^n / 2^{n+1}.
4) Simplify first two coefficients:
1 − 1/2 = 1/2.
For n=1 term of the sum is (−1)^1 w /2^2 = −w/4, so coefficient of w is 1 − (−1/4) = 1 + 1/4 = 5/4.
5) Final Taylor series about z=1 (w = z−1), valid for |z−1| < 2:
f(z) = 1/2 + (5/4)(z−1) − ∑_{n=2}^∞ (−1)^n (z−1)^n / 2^{n+1}.
Or written explicitly:
f(z) = 1/2 + 5/4 (z−1) − 1/8 (z−1)^2 + 1/16 (z−1)^3 − 1/32 (z−1)^4 + ···.
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Andy Baker
Works at University of Glasgow · Author has 7.3K answers and 1.7M answer views
·4y
Do you mean the function given by f(z)=z−1 z+1 f(z)=z−1 z+1 rather than z−1 z+1 z−1 z+1, the latter is even easier to deal with.
Try starting the partial fraction expansion f(z)=1−2 z+1 f(z)=1−2 z+1 the differentiating repeatedly.
Upvote ·
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Timothy Norfolk
Former Professor at University of Akron (1984–2017) · Author has 6.2K answers and 1.3M answer views
·4y
Q: How do I find Taylor series expansion of z – 1/ z + 1 about z = 1?
I will assume that you mean z−1 z+1 z−1 z+1
Do a little Algebra, and write it as (z−1)⋅1/2 1−(z−1 2)(z−1)⋅1/2 1−(z−1 2), and write it as a Geometric Series.
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How do I find the radius of convergence of the Taylor series expansion of the function f(z) =4z^ {2} +3z/(z-1) ^2} (z+4) (z-3) about z=-1?
What will be the Taylor series representation of z/ (z^2+1) in powers of z-1?
Mike Hirschhorn
Honorary Associate Professor of Mathematics at UNSW · Author has 8.1K answers and 2.7M answer views
·2y
z−1=u,z−1=u,
z−1 z+1=u 2+u z−1 z+1=u 2+u
=u 2⋅1 1+u 2=u 2⋅1 1+u 2
=u 2(1−u 2+u 2 4−+⋯)=u 2(1−u 2+u 2 4−+⋯)
=u 2−u 2 4+u 3 8−+⋯=u 2−u 2 4+u 3 8−+⋯
=z−1 2−(z−1)2 4+(z−1)3 8−+⋯.=z−1 2−(z−1)2 4+(z−1)3 8−+⋯.
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Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.5K answers and 21.1M answer views
·2y
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What is the Taylor series to express the function (z²-1) / ((z+2) (z+3)) in |z|<2?
We start by using long division and partial fractions to rewrite the given function:
z 2−1(z+2)(z+3)=1+3 z+2−8 z+3.z 2−1(z+2)(z+3)=1+3 z+2−8 z+3.
Next, we factor the constant terms from each denominator to set up the use of the geometric series:
z 2−1(z+2)(z+3)=1+3 2(1+z 2)−8 3(1+z 3)=1+3 2∞∑n=0(−z 2)n−8 3∞∑n=0(−z 3)n.z 2−1(z+2)(z+3)=1+3 2(1+z 2)−8 3(1+z 3)=1+3 2∑n=0∞(−z 2)n−8 3∑n=0∞(−z 3)n.
This series converges when both |−z 2|<1|−z 2|<1
Continue Reading
We start by using long division and partial fractions to rewrite the given function:
z 2−1(z+2)(z+3)=1+3 z+2−8 z+3.z 2−1(z+2)(z+3)=1+3 z+2−8 z+3.
Next, we factor the constant terms from each denominator to set up the use of the geometric series:
z 2−1(z+2)(z+3)=1+3 2(1+z 2)−8 3(1+z 3)=1+3 2∞∑n=0(−z 2)n−8 3∞∑n=0(−z 3)n.z 2−1(z+2)(z+3)=1+3 2(1+z 2)−8 3(1+z 3)=1+3 2∑n=0∞(−z 2)n−8 3∑n=0∞(−z 3)n.
This series converges when both |−z 2|<1|−z 2|<1 and |−z 3|<1|−z 3|<1; hence this converges when |z|<2|z|<2. (This is why we factored as we did before using the geometric series.)
Combining like terms and writing out the n=0 n=0 term, we conclude that
z 2−1(z+2)(z+3)=−1 6+∞∑n=1(−1)n(3 2 n+1−8 3 n+1)z n.z 2−1(z+2)(z+3)=−1 6+∑n=1∞(−1)n(3 2 n+1−8 3 n+1)z n.
Upvote ·
99 11
Gerben Koopman
Bachelor in Mathematics&Computer Science, University of Amsterdam (Graduated 2023)
·Updated 7y
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How do I evaluate sinx/x using a Taylor series expansion?
Depending on the questions intention we want to find out something about the curve of sin x x sinx x by means of its Taylor Series
. This could be its value at x=0 x=0 (as is considered a popular interview questions), it’s value as lim x→∞lim x→∞ or even what its integral evaluates to. All are good and legitimate mathematical questions so I will try to answer each one in turn, in order of increasing difficulty. So here goes!
First of, let’s find the Taylor Series describing sin x x sinx x. There are several ways of going about this, all assuming different levels of mathematical fluency, s
Continue Reading
Footnotes
Taylor series - Wikipedia
Depending on the questions intention we want to find out something about the curve of sin x x sinx x by means of its Taylor Series
. This could be its value at x=0 x=0 (as is considered a popular interview questions), it’s value as lim x→∞lim x→∞ or even what its integral evaluates to. All are good and legitimate mathematical questions so I will try to answer each one in turn, in order of increasing difficulty. So here goes!
First of, let’s find the Taylor Series describing sin x x sinx x. There are several ways of going about this, all assuming different levels of mathematical fluency, so I will try to make as few assumptions as possible. Note that we can rewrite the expression in the following manner.
sin x x=1 x sin x sinx x=1 x sinx.
This means that what we can do is to find the Taylor Series of sin x sinx and then divide every term by x x. We know that sin x sinx is infinitely differentiable
, or smooth, and its derivatives follow a cyclic pattern, namely:
sin x↦cos x↦−sin x↦−cos x sinx↦cosx↦−sinx↦−cosx
We also know we can evaluate all these functions at x=0 x=0. Because of these two characteristics we know that sin x sinx must at least have a Taylor Series. Now to finding it. The Taylor Series for a function f(x)f(x) is defined as expressed below.
f(x)=∞∑n=0 f n(0)n!x n f(x)=∑n=0∞f n(0)n!x n, where f n(x)f n(x) is the n′n′th derivative of f(x)f(x) and n!n! is the factorial function
.
Then we can express sin x sinx as:
sin x=∞∑n=0 x 2 n+1(2 n+1)!(−1)n sinx=∑n=0∞x 2 n+1(2 n+1)!(−1)n
Since the derivatives evaluate to 0 0, −1−1, 0 0 and 1 1 in turn. Doing this for sin x x sinx x in turn yields the following:
sin x x=1 x∞∑n=0 x 2 n+1(2 n+1)!(−1)n=∞∑n=0 x 2 n(2 n+1)!(−1)n sinx x=1 x∑n=0∞x 2 n+1(2 n+1)!(−1)n=∑n=0∞x 2 n(2 n+1)!(−1)n
Which we can write out as:
sin x x=1−x 2 3!+x 4 5!−x 6 7!+…sinx x=1−x 2 3!+x 4 5!−x 6 7!+…
That is the first part down! The Taylor Series of sin x x sinx x is expressed above. If that is all you were looking for you can stop reading now, but because you are a curious mind you don’t of course!
So to answer our first question in evaluating sin x x sinx x we need to see what happens when we set x=0 x=0. Doing this in a straightforward manner leads to sin 0 0=0 0 sin0 0=0 0 which is undefined unless further specified. Using our previously obtained Taylor Series of might give us a more useful answer, so let’s see:
lim x→0 sin x x=lim x→0 1−x 2 3!+x 4 5!−x 6 7!+⋯=1−0 2 3!+0 4 5!−0 6 7!+⋯=1 lim x→0 sinx x=lim x→0 1−x 2 3!+x 4 5!−x 6 7!+⋯=1−0 2 3!+0 4 5!−0 6 7!+⋯=1
Isn’t that nice, by expressing sin x x sinx x in a seemingly more complex way we can get a much more informative answer to our question. Cool!
Now, for the second question. Where in the previous case we let lim x→0 lim x→0, this time we let lim x→∞lim x→∞. Using our Taylor Series of sin x x sinx x we get:
lim x→∞sin x x=lim x→∞1−x 2 3!+x 4 5!−x 6 7!+⋯=1−∞+∞−∞+∞+…lim x→∞sinx x=lim x→∞1−x 2 3!+x 4 5!−x 6 7!+⋯=1−∞+∞−∞+∞+…
Well, that’s not very useful… It’s seems our Taylor Series won’t be of much use here. Instead we can try to use another method to evaluate the expression. Consider sin x x sinx x as the product of two separate functions, sin x sinx and 1 x 1 x. Know that the limit of the product of two functions is the product of their limits. Or, in terms of our specific question:
lim x→∞sin x x=lim x→∞sin x lim x→∞1 x lim x→∞sinx x=lim x→∞sinx lim x→∞1 x
Note that:
sin x∈[−1;1]sinx∈[−1;1] and lim x→∞1 x=0 lim x→∞1 x=0
0 0 multiplied by any finite number is 0.0. Then the same holds for multiplying 0 0 by a member of the set [−1;1][−1;1]. By this logic we can conclude that:
lim x→∞sin x x=0 lim x→∞sinx x=0
Work In Progress
Footnotes
Taylor series - Wikipedia
Smoothness - Wikipedia
Factorial - Wikipedia
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Shayandeep Bhaumik
20 | Indian | Loves to uncover | Dreams to be a researcher · Author has 79 answers and 383.5K answer views
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How do I expand problems like (e^x) /(x-1)^2 about X=1 using the Taylor series expansion?
I hope it helps!!
Well I don’t know how to use summations or exponentials in quora nor did i had my phone around so I wrote it using M.S. Word and took the screenshot and croped it.
Confirmation from wolfram alpha. Yay!!
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I hope it helps!!
Well I don’t know how to use summations or exponentials in quora nor did i had my phone around so I wrote it using M.S. Word and took the screenshot and croped it.
Confirmation from wolfram alpha. Yay!!
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Jan van Delden
MSc Math and still interested · Author has 4.8K answers and 6.5M answer views
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What are the first three terms of the Taylor’s series expansion of f(z) =1/ (z^2+4) about z=-i? What is the region of convergence?
The radius of convergence is maximized by the distance between z=−i z=−i and the closest pole, in this case z=−2 i z=−2 i. The region is thus a open disk with radius 1 1 and center z=−i z=−i.
One could apply partial fractions to f(z)f(z) and split the computation of the Taylor series into two parts:
f(z)=1 4 i(1 z−2 i−1 z+2 i)f(z)=1 4 i(1 z−2 i−1 z+2 i)
Try to write the denominator in the form α(1−β(z+i))α(1−β(z+i)) in order to apply a geometric series expansion.
Use:
z+2 i=(z+i)+i=i(1+(z+i)/i)=i(1−(z+i)/(−i))z+2 i=(z+i)+i=i(1+(z+i)/i)=i(1−(z+i)/(−i))
z−2 i=(z+i)−3 i=−3 i(1−(z+i)/(3 i))z−2 i=(z+i)−3 i=−3 i(1−(z+i)/(3 i))
Collect the other constants:
f(z)=\dfrac{1}{4}\left(\dfrac{1}{3}\dfrac{1}{1-(f(z)=\dfrac{1}{4}\left(\dfrac{1}{3}\dfrac{1}{1-(
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The radius of convergence is maximized by the distance between z=−i z=−i and the closest pole, in this case z=−2 i z=−2 i. The region is thus a open disk with radius 1 1 and center z=−i z=−i.
One could apply partial fractions to f(z)f(z) and split the computation of the Taylor series into two parts:
f(z)=1 4 i(1 z−2 i−1 z+2 i)f(z)=1 4 i(1 z−2 i−1 z+2 i)
Try to write the denominator in the form α(1−β(z+i))α(1−β(z+i)) in order to apply a geometric series expansion.
Use:
z+2 i=(z+i)+i=i(1+(z+i)/i)=i(1−(z+i)/(−i))z+2 i=(z+i)+i=i(1+(z+i)/i)=i(1−(z+i)/(−i))
z−2 i=(z+i)−3 i=−3 i(1−(z+i)/(3 i))z−2 i=(z+i)−3 i=−3 i(1−(z+i)/(3 i))
Collect the other constants:
f(z)=1 4(1 3 1 1−(z+i)/(3 i)+1 1−(z+i)/(−i))f(z)=1 4(1 3 1 1−(z+i)/(3 i)+1 1−(z+i)/(−i))
The first series converges for |z+i|<3|z+i|<3, the latter for |z+i|<1|z+i|<1. Both converge on the smaller region.
Apply the geometric series and find:
f(z)=1 4∞∑k=0 i k((−1)k 3 k+1+1)(z+i)k f(z)=1 4∑k=0∞i k((−1)k 3 k+1+1)(z+i)k
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Senia Sheydvasser
knows some things about analytic number theory · Upvoted by
Swarnak Ray
, M.Sc Physics & Mathematics, Jadavpur University (2015) and
Alexey Godin
, Ph.D. Mathematics & Economics, Moscow State University (1998) · Author has 2.5K answers and 39.9M answer views
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What is the Taylor series expansion for 1/z about z=0 in complex analysis?
The function f(z)=1/z f(z)=1/z doesn’t have a Taylor expansion centered at z=0 z=0. Suppose that it did—this would require that, inside of some region around z=0 z=0,
1 z=a 0+a 1 z+a 2 z 2+a 3 z 3+…1 z=a 0+a 1 z+a 2 z 2+a 3 z 3+…
and, therefore, if we choose some z≠0 z≠0 inside of that region, we must have (by multiplying both sides by z z) that
1=a 0 z+a 1 z 2+a 2 z 3+a 3 z 4+….1=a 0 z+a 1 z 2+a 2 z 3+a 3 z 4+….
This identity can hold for some finite number of z z, but it certainly can’t hold inside of an entire open ball. This approach was doomed from the start.
In fact
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The function f(z)=1/z f(z)=1/z doesn’t have a Taylor expansion centered at z=0 z=0. Suppose that it did—this would require that, inside of some region around z=0 z=0,
1 z=a 0+a 1 z+a 2 z 2+a 3 z 3+…1 z=a 0+a 1 z+a 2 z 2+a 3 z 3+…
and, therefore, if we choose some z≠0 z≠0 inside of that region, we must have (by multiplying both sides by z z) that
1=a 0 z+a 1 z 2+a 2 z 3+a 3 z 4+….1=a 0 z+a 1 z 2+a 2 z 3+a 3 z 4+….
This identity can hold for some finite number of z z, but it certainly can’t hold inside of an entire open ball. This approach was doomed from the start.
In fact, holomorphic functions only have Taylor series centered around points where they are, you know, holomorphic. The function f(z)=1/z f(z)=1/z isn’t even defined at z=0 z=0, let alone holomorphic (unless you want to appeal to the Riemann sphere, I suppose).
What you can do, though, is use a Laurent expansion, which will work even if the function is just holomorphic in an annulus around the desired point. In other words, we can find coefficients a n a n such that
1 z=…+a−2 z−2+a−1 z−1+a 0+a 1 z+a 2 z 2+….1 z=…+a−2 z−2+a−1 z−1+a 0+a 1 z+a 2 z 2+….
Unfortunately, it then becomes very clear that the Laurent expansion of 1/z 1/z is just 1/z 1/z.
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Eric Neville
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SUNIL KUMAR MISRA
, MSc Mathematics, Utkal University, Bhubaneswar (2018) · Author has 56 answers and 180.9K answer views
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Can you expand f(z)=1/(z+1)(z+3) by the Laurent series valid region?
Let’s find the Laurent series about 0, since you haven’t specified the center. Firstly, it is clear that the singularities are at z=−1 z=−1 and z=−3 z=−3. These points will determine the regions where we have different Laurent Series. We can find the series for this function by taking advantage of the well known series:
1 1−ω=1+ω+ω 2+...=∞∑n=0 ω n 1 1−ω=1+ω+ω 2+...=∑n=0∞ω n
for |ω|<1|ω|<1
First rewrite the function as a sum of two fractions:
f(z)=1(z+1)(z+3)=A z+1+B z+3=(A+B)z+(3 A+B)(z+1)(z+3)f(z)=1(z+1)(z+3)=A z+1+B z+3=(A+B)z+(3 A+B)(z+1)(z+3)
It follows that A=1 2 A=1 2 an
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Let’s find the Laurent series about 0, since you haven’t specified the center. Firstly, it is clear that the singularities are at z=−1 z=−1 and z=−3 z=−3. These points will determine the regions where we have different Laurent Series. We can find the series for this function by taking advantage of the well known series:
1 1−ω=1+ω+ω 2+...=∞∑n=0 ω n 1 1−ω=1+ω+ω 2+...=∑n=0∞ω n
for |ω|<1|ω|<1
First rewrite the function as a sum of two fractions:
f(z)=1(z+1)(z+3)=A z+1+B z+3=(A+B)z+(3 A+B)(z+1)(z+3)f(z)=1(z+1)(z+3)=A z+1+B z+3=(A+B)z+(3 A+B)(z+1)(z+3)
It follows that A=1 2 A=1 2 and B=−1 2 B=−1 2 so that:
f(z)=1 2(1 z+1−1 z+3)f(z)=1 2(1 z+1−1 z+3)
Now we are ready to find the series. First notice that as we expand a disc from 0 the function is analytic until we reach z=−1 z=−1 so the function is analytic on D(0,1)D(0,1). This is our first region of interest. Since the function is analytic the Laurent Series will actually just be a Taylor series. We find it by writing:
f(z)=1 2(1 1−(−z)−1 3 1 1−(−z 3))f(z)=1 2(1 1−(−z)−1 3 1 1−(−z 3))
Then since |z|<1|z|<1 in this region both of these can be written as geometric series so we find:
f(z)=1 2(∞∑n=0(−z)n−1 3∞∑n=0(−z 3)n)=∞∑n=0(−1)n 2(1−1 3 n+1)z n f(z)=1 2(∑n=0∞(−z)n−1 3∑n=0∞(−z 3)n)=∑n=0∞(−1)n 2(1−1 3 n+1)z n
Valid within D(0,1)D(0,1).
Now if we start from just past z=−1 z=−1 the function is analytic between here and z=−3 z=−3. This is our next region of interest, the annulus:
A={z∈C:1<|z|<3}A={z∈C:1<|z|<3}
We follow a similar procedure but notice that |z|<1|z|<1 is not true. ∣∣z 3∣∣<1|z 3|<1 still holds so we only need to change one of our fractions. We write the function as:
f(z)=1 2(1 z 1 1−(−1 z)−1 3 1 1−(−z 3))f(z)=1 2(1 z 1 1−(−1 z)−1 3 1 1−(−z 3))
Now we can use the geometric series again to obtain:
f(z)=1 2(1 z∞∑n=0(−1)n z−n−1 3∞∑n=0(−z 3)n)=∞∑n=1(−1)n−1 2 z−n−∞∑n=0 1 6(−1 3)n z n f(z)=1 2(1 z∑n=0∞(−1)n z−n−1 3∑n=0∞(−z 3)n)=∑n=1∞(−1)n−1 2 z−n−∑n=0∞1 6(−1 3)n z n
Valid on A.
Finally we see that outside of z=−3 z=−3 the function is analytic again so our final region is the region where |z|>3.|z|>3. What happens here is similar to what happened to the fraction with z+1 z+1 denominator in the second case. We basically just do what we did for that fraction in that case, to both fractions in this case. So we write:
f(z)=1 2(1 z 1 1−(−1 z)−1 z 1 1−(−3 z))f(z)=1 2(1 z 1 1−(−1 z)−1 z 1 1−(−3 z))
So using the geometric series:
f(z)=1 2(1 z∞∑n=0(−1)n z−n−1 z∞∑n=0(−3)n z−n)=∞∑n=1(−1)n−1(1−3 n−1 2)z−n f(z)=1 2(1 z∑n=0∞(−1)n z−n−1 z∑n=0∞(−3)n z−n)=∑n=1∞(−1)n−1(1−3 n−1 2)z−n
Valid for |z|>3.|z|>3.
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Viraj Daniel Dsouza
Student at Indian Institute of Science Education and Research, Pune (IISER-P) (2017–present) · Author has 85 answers and 399.7K answer views
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How do I prove the 'Taylor Expansion Series'?
Let me derive it for a single variable function. It must be an infinitely differentiable function(real or complex valued, doesn’t matter).
Lets suppose we want to calculate the Taylor Series for f(x)f(x)about a point x=a∈C.x=a∈C.We try to write f(x)f(x)about x=a x=a as,
f(x)=a 0+a 1(x−a)+a 2(x−a)2+a 3(x−a)3+.......(1)(1)f(x)=a 0+a 1(x−a)+a 2(x−a)2+a 3(x−a)3+.......
Now let’s evaluate the coefficients.
Put x=a,w e x=a,w e get f(a)=a 0 f(a)=a 0
Differentiate f(x)f(x)with respect to x,x,we get
f′(x)=a 1+2 a 2(x−a)+.....f′(x)=a 1+2 a 2(x−a)+.....
Now,
f′(a)=a 1(2)(2)f′(a)=a 1
Similarly taking higher derivatives at x=a,x=a,we get
f′′(a)=2 a 2⟹a 2=f′′(a)2!\t f″(a)=2 a 2⟹a 2=f″(a)2!\t
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Let me derive it for a single variable function. It must be an infinitely differentiable function(real or complex valued, doesn’t matter).
Lets suppose we want to calculate the Taylor Series for f(x)f(x)about a point x=a∈C.x=a∈C.We try to write f(x)f(x)about x=a x=a as,
f(x)=a 0+a 1(x−a)+a 2(x−a)2+a 3(x−a)3+.......(1)(1)f(x)=a 0+a 1(x−a)+a 2(x−a)2+a 3(x−a)3+.......
Now let’s evaluate the coefficients.
Put x=a,w e x=a,w e get f(a)=a 0 f(a)=a 0
Differentiate f(x)f(x)with respect to x,x,we get
f′(x)=a 1+2 a 2(x−a)+.....f′(x)=a 1+2 a 2(x−a)+.....
Now,
f′(a)=a 1(2)(2)f′(a)=a 1
Similarly taking higher derivatives at x=a,x=a,we get
f′′(a)=2 a 2⟹a 2=f′′(a)2!(3)(3)f″(a)=2 a 2⟹a 2=f″(a)2!
f′′′(a)=6 a 3⟹a 3=f′′′(a)3!(4)(4)f‴(a)=6 a 3⟹a 3=f‴(a)3!
and so on.
Substitute these in (1),(1),we get
f(x)=f(a)+f′(a)1!(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+....f(x)=f(a)+f′(a)1!(x−a)+f″(a)2!(x−a)2+f‴(a)3!(x−a)3+....
This is the Taylor series for f(x)f(x)about x=a.x=a.
Put a=0,a=0,you get Taylor series about x=0,x=0, this is named as the ‘Maclaurin Series.’
Taylor series of a function is a good approximation for it. As the more number of terms in the series are plotted and compared to the plot of f(x)f(x) , the approximation becomes better and better.
With the same motivation, Taylor series for a multivariable function could be written. But now, f′f′is the total derivative, which is a matrix with elements as components of ∇f.∇f.
Thank you.
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Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.5K answers and 21.1M answer views
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What is the Laurent series expansion of F(z)=e z z(1−z)F(z)=e z z(1−z) centered at z=1 z=1?
First of all, note that
e z z(1−z)=−e z−1⋅e z−1 z=−e z−1⋅e z−1⋅1 1+(z−1).e z z(1−z)=−e z−1⋅e z−1 z=−e z−1⋅e z−1⋅1 1+(z−1).
We have two cases.
(i) |z−1|<1|z−1|<1.
Using the exponential series (convergent for all z z)
e z−1=1+(z−1)+(z−1)2 2!+(z−1)3 3!+...,and e z−1=1+(z−1)+(z−1)2 2!+(z−1)3 3!+...,and
and geometric series (convergent for |z−1|<1|z−1|<1)
1 1+(z−1)=1−(z−1)+(z−1)2−(z−1)3+...,and 1 1+(z−1)=1−(z−1)+(z−1)2−(z−1)3+...,and
we obtain by using the distributive property:
\begin{align}\displa\begin{align}\displa
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First of all, note that
e z z(1−z)=−e z−1⋅e z−1 z=−e z−1⋅e z−1⋅1 1+(z−1).e z z(1−z)=−e z−1⋅e z−1 z=−e z−1⋅e z−1⋅1 1+(z−1).
We have two cases.
(i) |z−1|<1|z−1|<1.
Using the exponential series (convergent for all z z)
e z−1=1+(z−1)+(z−1)2 2!+(z−1)3 3!+...,and e z−1=1+(z−1)+(z−1)2 2!+(z−1)3 3!+...,and
and geometric series (convergent for |z−1|<1|z−1|<1)
1 1+(z−1)=1−(z−1)+(z−1)2−(z−1)3+...,and 1 1+(z−1)=1−(z−1)+(z−1)2−(z−1)3+...,and
we obtain by using the distributive property:
e z z(1−z)=−e z−1∞∑n=0(−1)n a n(z−1)n=e∞∑n=0(−1)n+1 a n(z−1)n−1,e z z(1−z)=−e z−1∑n=0∞(−1)n a n(z−1)n=e∑n=0∞(−1)n+1 a n(z−1)n−1,
where
a n=n∑k=0(−1)k k!.a n=∑k=0 n(−1)k k!.
(ii) |z−1|>1|z−1|>1.
This is the same idea as in (i), but we need to rewrite the geometric series factor:
1 1+(z−1)=1 z−1⋅1 1+(z−1)−1=1 z−1−1(z−1)2+1(z−1)3+....1 1+(z−1)=1 z−1⋅1 1+(z−1)−1=1 z−1−1(z−1)2+1(z−1)3+....
Observe that this series converges when |1 z−1|<1,|1 z−1|<1, or equivalently |z−1|>1|z−1|>1.
With this modification and using the fact that
e−1=∞∑k=0(−1)k k!,e−1=∑k=0∞(−1)k k!,
we obtain
e z z(1−z)=−e(z−1)2⋅e z−1⋅1 1+(z−1)−1=−e(z−1)2⋅[∞∑n=0(−1)n e−1(z−1)−n+∞∑n=1(−1)n(e−1−a n−1)(z−1)n]=∞∑n=0(−1)n+1(z−1)−n−2+∞∑n=1(−1)n+1(1−e a n−1)(z−1)n−2,e z z(1−z)=−e(z−1)2⋅e z−1⋅1 1+(z−1)−1=−e(z−1)2⋅[∑n=0∞(−1)n e−1(z−1)−n+∑n=1∞(−1)n(e−1−a n−1)(z−1)n]=∑n=0∞(−1)n+1(z−1)−n−2+∑n=1∞(−1)n+1(1−e a n−1)(z−1)n−2,
where (as in the previous case)
a n=n∑k=0(−1)k k!.a n=∑k=0 n(−1)k k!.
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9320 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4832626/ | Differential cyanosis and clubbing: signs of an Era gone by - PMC
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Heart Asia
. 2012 Nov 30;4(1):168. doi: 10.1136/heartasia-2012-010184
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Differential cyanosis and clubbing: signs of an Era gone by
Srivatsa Nadig
Srivatsa Nadig
1 Department Of Cardiology, Sanjay Gandhi PGIMS, Lucknow, Uttar Pradesh, India
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1, Aditya Kapoor
Aditya Kapoor
1 Department Of Cardiology, Sanjay Gandhi PGIMS, Lucknow, Uttar Pradesh, India
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1, Sudeep Kumar
Sudeep Kumar
1 Department Of Cardiology, Sanjay Gandhi PGIMS, Lucknow, Uttar Pradesh, India
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1 Department Of Cardiology, Sanjay Gandhi PGIMS, Lucknow, Uttar Pradesh, India
✉
Correspondence to Dr Aditya Kapoor, Department of Cardiology, Sanjay Gandhi PGIMS, Lucknow, UP 226014, India; akapoor65@gmail.com
Collection date 2012.
Keywords: Congenital Heart Disease, Imaging And Diagnostics
Published by the BMJ Publishing Group Limited. For permission to use (where not already granted under a licence) please go to
PMC Copyright notice
PMCID: PMC4832626 PMID: 27326058
Despite declining prevalence of Eisenmenger Syndrome (ES) in the West, such patients are not uncommon in the developing world. Important clues to the level of shunt are provided by differential cyanosis and clubbing indicating a patent ductus arteriosus (PDA) or the degree of splitting of second heart sound.1 The reason for the differential cyanosis and clubbing is that due to the right-to-left shunt across the PDA, deoxygenated blood from the right ventricle is preferentially directed into the aorta distal to the left subclavian artery and into the lower extremities.
With PDA, ES and right-to-left shunt, an erroneous diagnosis of primary pulmonary hypertension may be made on echocardiography since no obvious septal defect is visualised. Contrast echocardiography using agitated saline, with opacification of abdominal aorta without opacification of left-sided chambers is helpful in such cases.
We describe a 38-year-old male with no known previous congenital heart disease who presented with exertional dyspnoea and haemoptysis since the last 5 years. Examination revealed obvious differential cyanosis and clubbing (figure 1), closely split second heart sound with right atrial and right ventricular dilatation, elevated right ventricular systolic pressure and severe pulmonary artery hypertension on echocardiography (figure 2). Contrast echocardiography findings (figure 3, supplementary video 1) revealed rapid opacification of descending aorta, clearly visualised in the subcostal view, establishing the diagnosis of PDA with ES.
Figure 1.
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Differential cyanosis and clubbing.
Figure 2.
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2D echocardiography showing dilated right atrium and right ventricular (RV) tricuspid regurgitation and severely elevated RV systolic pressure.
Figure 3.
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Contrast echocardiography revealed rapid opacification of descending aorta, clearly visualised in the subcostal view.
Supplementary Material
Supplementary Data
supp_4_1_168__index.html (997B, html)
Footnotes
Competing interests: None.
Patient consent: Obtained.
Provenance and peer review: Not commissioned; internally peer reviewed.
Reference
1.Saha A, Balakrishnan KG, Jaiswal PK, et al. Prognosis for patients with Eisenmenger syndrome of various aetiology. Int J Cardiol 1994;45:199–207. [DOI] [PubMed] [Google Scholar]
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9321 | https://www.cuemath.com/algebra/equivalence-relations/ | LearnPracticeDownload
Equivalence Relation
Equivalence relation defined on a set in mathematics is a binary relation that is reflexive, symmetric, and transitive. A binary relation over the sets A and B is a subset of the cartesian product A × B consisting of elements of the form (a, b) such that a ∈ A and b ∈ B. A very common and easy-to-understand example of an equivalence relation is the 'equal to (=)' relation which is reflexive, symmetric and transitive.
As the name suggests, two elements of a set are said to be equivalent if and only if they belong to the same equivalence class. In this article, we will understand the concept of equivalence relation, class, partition with proofs and solved examples.
| | |
--- |
| 1. | What is Equivalence Relation? |
| 2. | Proof of Equivalence Relation |
| 3. | Definitions Related to Equivalence Relation |
| 4. | FAQs on Equivalence Relation |
What is Equivalence Relation?
An equivalence relation is a binary relation defined on a set X such that the relation is reflexive, symmetric and transitive. If any of the three conditions (reflexive, symmetric and transitive) does not hold, the relation cannot be an equivalence relation. The equivalence relation divides the set into disjoint equivalence classes. Any two elements of the set are said to be equivalent if and only if they belong to the same equivalence class. An equivalence relation is generally denoted by the symbol '~'.
Equivalence Relation Definition
A relations in maths for real numbers R defined on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. They are often used to group together objects that are similar, or equivalent. It satisfies the following conditions for all elements a, b, c ∈ A:
Reflexive - R is reflexive if (a, a) ∈ R for all a ∈ A
Symmetric - R is symmetric if and only if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A
Transitive - R is transitive if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A
The equivalence relation involves three types of relations such as reflexive relation, symmetric relation, transitive relation.
Examples of Equivalence Relation
'Is equal to (=)' is an equivalence relation on any set of numbers A as for all elements a, b, c ∈ A, we have a = a, a = b ⇒ b = a, and a = b, b = c ⇒ a = c. This implies (=) is reflexive, symmetric and transitive.
'Is similar to (~)' defined on the set of triangles: It is reflexive, symmetric, and transitive.
'Has the same birthday' defined on the set of people: It is reflexive, symmetric, and transitive.
'Is congruent to' defined on the set of triangles is an equivalence relation as it is reflexive, symmetric, and transitive.
'Congruence modulo n (≡)' defined on the set of integers: It is reflexive, symmetric, and transitive.
'Has the same absolute value' defined on the set of real numbers is an equivalence relation as it is reflexive, symmetric, and transitive.
Proof of Equivalence Relation
To understand how to prove if a relation is an equivalence relation, let us consider an example. Define a relation R on the set of natural numbers N as (a, b) ∈ R if and only if a = b. Now, we will show that the relation R is reflexive, symmetric and transitive.
Reflexive Property - Since every natural number is equal to itself, that is, a = a for all a ∈ N ⇒ (a, a) ∈ R for all a ∈ N. Hence, R is reflexive.
Symmetric Property - For a, b ∈ N, let (a, b) ∈ R ⇒ a = b ⇒ b = a ⇒ (b, a) ∈ R. Since a, b are arbitrary, R is symmetric.
Transitive Property - For a, b, c ∈ N, let (a, b) ∈ R and (b, c) ∈ R ⇒ a = b and b = c ⇒ a = c (as numbers equal to the same number are equal to one another) ⇒ (a, c) ∈ R. Since a, b, c are arbitrary, R is transitive.
Since R, defined on the set of natural numbers N, is reflexive, symmetric, and transitive, R is an equivalence relation.
Proving a Relation is Not an Equivalence Relation
We have seen how to prove an equivalence relation. Now, we will consider an example of a relation that is not an equivalence relation and find a counterexample for the same. Define a relation R on the set of integers as (a, b) ∈ R if and only if a ≥ b. We will check for the three conditions (reflexivity, symmetricity, transitivity):
Reflexivity - As every integer is equal to itself, that is, a = a for all a ∈ Z, it satisfies a ≥ a for all a ∈ Z. This implies (a, a) ∈ R for all a ∈ Z. Hence, R is reflexive.
Symmetricity - For a, b ∈ Z, let (a, b) ∈ R ⇒ a ≥ b. This does not imply that b ≥ a. For example, 12 ≥ 9 but 9 is not greater than or equal to 12. This implies R is not symmetric.
We do not need to check for transitivity as R is not symmetric ⇒ R is not an equivalence relation.
Definitions Related to Equivalence Relation
Now, we will understand the meaning of some terms related to equivalence relation such as equivalence class, partition, quotient set, etc. Consider an equivalence relation R defined on set A with a, b ∈ A.
Equivalence Class - An equivalence class is a subset B of A such (a, b) ∈ R for all a, b ∈ B and a, b cannot be outside of B. Mathematically, an equivalence class of a is denoted as [a] = {x ∈ A: (a, x) ∈ R} which contains all elements of A which are related 'a'. All elements of A equivalent to each other belong to the same equivalence class. In other words, all elements belonging to the same equivalence class are equivalent to each other.
Partition - A partition of set A is a non-empty set of disjoint subsets of A such that no element of A is in two subsets of A and elements belonging to the same subset are related to each other. The union of subsets in the partition is equal to set A.
Quotient Set - A quotient set is a set of all equivalence classes of an equivalence relation denoted by A/R = {[a]: a ∈ A}
Related Topics
Transitive Property of Congruence
Antisymmetric Relation
Relations and Function Worksheets
Important Notes on Equivalence Relation
An equivalence relation is a binary relation defined on a set X such that the relation is reflexive, symmetric and transitive.
The equivalence relation divides the set into disjoint equivalence classes.
All elements belonging to the same equivalence class are equivalent to each other.
Read More
Examples on Equivalence Relation
Example 1: Define a relation R on the set S of symmetric matrices as (A, B) ∈ R if and only if A = BT. Show that R is an equivalence relation.
Solution: To show R is an equivalence relation, we need to check the reflexive, symmetric and transitive properties.
Reflexive Property - For a symmetric matrix A, we know that A = AT. Therefore, (A, A) ∈ R. ⇒ R is reflexive.
Symmetric Property - For A, B ∈ S, we have A = AT and B = BT. Let (A, B) ∈ R ⇒ A = BT. We know that B = (BT)T = AT (As A = BT) ⇒ B = AT ⇒ (B, A) ∈ R. Therefore, R is symmetric.
Transitive Property - For A, B, C ∈ S, we have A = AT, B = BT and C = CT. Let (A, B) ∈ R and (B, C) ∈ R ⇒ A = BT and B = CT. We have A = BT = B = CT ⇒ A = CT ⇒ (A, C) ∈ R. Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalence relation.
2. Example 2: Show that a relation F defined on the set of real numbers R as (a, b) ∈ F if and only if |a| = |b| is an equivalence relation.
Solution: We need to check the reflexive, symmetric and transitive properties of F.
Reflexivity - For any real number a, we know that |a| = |a| ⇒ (a, a) ∈ F, for all a ∈ R. Therefore, F is reflexive.
Symmetricity - For a, b ∈ R, let (a, b) ∈ F ⇒ |a| = |b| ⇒ |b| = |a| ⇒ (b, a) ∈ F. Since a, b are arbitrary, F is symmetric.
Transitivity - For a, b, c ∈ R, let (a, b) ∈ F and (b, c) ∈ F ⇒ |a| = |b| and |b| = |c| ⇒ |a| = |c| ⇒ (a, c) ∈ F. Since a, b, c are arbitrary, F is transitive.
Since F is reflexive, symmetric and transitive, F is an equivalence relation.
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FAQs on Equivalence Relation
What is Equivalence Relation in Maths?
An equivalence relation is a binary relation defined on a set X such that the relations are reflexive, symmetric and transitive. If any of the three conditions (reflexive, symmetric and transitive) does not hold, the relation cannot be an equivalence relation.
What is the Smallest Equivalence Relation?
For any set A, the smallest equivalence relation is the one that contains all the pairs (a, a) for all a ∈ A. Equivalence relations defined on a set in mathematics are binary relations that are reflexive relations, symmetric relations, and transitive reations.
How do you write an Equivalence Class of an Equivalence Relation?
An equivalence class is a subset B of A such (a, b) ∈ R for all a, b ∈ B and a, b cannot be outside of B. Mathematically, an equivalence class of a is denoted as [a] = {x ∈ A: (a, x) ∈ R} which contains all elements of A which are related 'a'.
What are the Three Conditions to Prove an Equivalence Relation?
A relation R defined on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. It satisfies the following conditions for all elements a, b, c ∈ A:
Reflexive - R is reflexive if (a, a) ∈ R for all a ∈ A
Symmetric - R is symmetric relations if and only if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A
Transitive - R is transitive if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A
Is an Empty Relation an Equivalence Relation?
An empty relation on an empty set is an equivalence relation but an empty relation on a non-empty set is not an equivalence relation as it is not reflexive.
What is an Equivalence Relation Example in Real-Life?
A real-life example of an equivalence relation is: 'Has the same birthday as' relation defined on the set of all people. It satisfies all three conditions of reflexivity, symmetricity, and transitive relations.
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9322 | https://designers-guide.org/forum/Attachments/mismatch_presentation.pdf | Random Offset in CMOS IC Design ECEN4827/5827 Analog IC Design October 19, 2007 Art Zirger, National Semiconductor art.zirger@nsc.com 303-845-4024 Where to start?
• How do we choose what transistor sizes to use in a design?
• One topic not often discussed in classes is random offset and how transistor sizing affects this phenomenon.
Introduction • 2 devices (MOSFET’s, resistors, capacitors) of the same size, laid out next to each other, are not identical.
• How they differ is generally the function of random offsets during processing.
• These offsets vary from chip to chip and set a limit on precision attainable which is typically reflected as data sheet specifications.
Misc. Definitions/Notation • The following I-V equation for a MOSFET in saturation is used: where • A mixture of Vt & VT is used where both are referring to threshold voltage, not thermal voltage ( ) 2 2 t GS D V V I − = β L W Cox µ β = Agenda Systematic vs. random offset Sources & profiles of random offset Current Mirror/Diff Pair offset derivation & insights Propagation of uncertainties math Current Mirror/Diff Pair exercises Systematic vs. random mismatch • Systematic – Mismatch in the circuit (or layout) because of poor designer choices (i.e. avoidable) – Each copy of the circuit should share this; calculable based on the average values of element parameters – Viewable using SPICE DC operating point simulation • Random – Mismatch in the circuit because of wafer processing – Different chips will have different values, but the value will mostly remain the same (subject to temperature shifts, drift, etc.) – Each copy of the circuit should share this; calculable based on the statistical values of element parameters – Viewable using DCmatch and Monte Carlo simulations – This is what is usually thought of as matching Sources of random mismatch • Sources of random mismatch include: – Edge effects (rough edges) – Implantation (finite number of charges & distribution) – Mobility – Oxide effects See References (after Summary slide) for more information.
Mismatch parameters • Commonly investigated mismatch parameters: – MOSFET Vt, β (mobility and W/L), γ (Body Effect) – Resistors ρ (resistivity) – Capacitors oxide thickness variation • This presentation covers Vt & β mismatch Profile of random mismatch • Has a gaussian distribution • Can be quantified by statistical variables of: – mean: ā – standard deviation: σa – variance: σ2a – Mismatch is defined as occurring between elements; a single element does not have mismatch, but a “self mismatch” can be defined.
Threshold Voltage Mismatch The threshold voltages among a group of transistors has a gaussian profile about a mean. Experimentally, it has been shown that the difference in threshold voltages between 2 identically sized transistors behaves as: Note that to reduce the mismatch by ½ takes 4 times the area… A fab will create test structures and measure ∆Vt multiple times per wafer for various sizes of transistors and collect ongoing statistics to monitor the process over time.
WL A t t V V = ∆ σ Threshold Voltage Mismatch, cont’d From W. Sansen showing how the mismatch constant, AVT, varies roughly linearly with process size (doping concentration affects linearity of the relationship). Also, for p substrates, the PMOS will have AVT ~ 1.5AVT NMOS.
Our CMOS AVT NFET Current Factor Mismatch Current Factor, β, behaves fractionally, as: Aβ ~ 2%µm, invariant of process National Semiconductor does not have this value characterized, so we may use this approximate value to estimate whether we need to worry about this or not.
( ) WL Aβ β β σ = ∆ Offset Derivation • Given the behavior of sufficiently uncorrelated parameters, want to know the effect of those parameters on 2 common circuits: – Current mirror – Differential pair • Start with I-V equation for MOSFET and apply “total differential”: ...
+ ∆ ∂ ∂ + ∆ ∂ ∂ + ∆ ∂ ∂ = ∆ z z f y y f x x f f Offset Derivation – Current Mirror What is the fractional error in the currents being mirrored in a 1:1 current mirror?
( ) 2 2 T gs D V V I − = β T gs V V , , β variables D m I g β 2 = T gs D V V I − 2 or ( ) ( ) ( ) T gs gs T gs T T gs D V V V V V V V V I − ∆ + − ∆ − − ∆ = ∆ 2 2 2 2 2 1 2 β β β ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 T gs T gs T T gs T gs D D V V V V V V V V V I I − − ∆ − − − ∆ = ∆ β β β β ( ) T gs T D D V V V I I − ∆ − ∆ = ∆ 2 β β ∆Vgs = 0 in a current mirror.
Divide by ID to get fractional error: T D m D D V I g I I ∆ − ∆ = ∆ β β Offset Derivation – Diff Pair What is ∆VGS for 2 transistors operating at the same current?
( ) 2 2 T gs D V V I − = β T gs V V , , β D m I g β 2 = T gs D V V I − 2 or ( ) ( ) ( ) T gs gs T gs T T gs D V V V V V V V V I − ∆ + − ∆ − − ∆ = ∆ 2 2 2 2 2 1 2 β β β 0 = ∆D I ( ) ( ) gs T T gs T gs V V V V V V ∆ + ∆ − − − ∆ = 2 2 2 1 0 2 β β Constant current so Divide by ( ) T gs V V − 2 2 β ( ) T T gs gs V V V V ∆ + − ∆ − = ∆ 2 β β T m D gs V g I V ∆ + ∆ − = ∆ β β Offset Derivation – Summary/Insights • Differential Pairs and Current Mirrors operate with very different gm/Id (i.e. bias point) ratios to minimize mismatch errors: • Differential Pair: High gm/Id Æ low overdrive • Current Mirror: Low gm/Id Æ high overdrive • You can achieve this by designing differential pairs with large W/L and current mirrors with small W/L ratios T m D gs V g I V ∆ + ∆ − = ∆ β β T D m D D V I g I I ∆ − ∆ = ∆ β β Offset Derivation w/Standard Deviations • Given the expected functional relationships of the 2 different offset behaviors, for various statistical reasons, you express these relationships in terms of standard deviations as: Current Mirror Differential Pair T D m D D V I g I I ∆ − ∆ = ∆ β β T m D gs V g I V ∆ + ∆ − = ∆ β β ( ) ( ) ( ) ( ) 2 2 T m D gs V g I V ∆ + ∆ = ∆ σ β β σ σ ( ) ( ) ( ) 2 2 ∆ + ∆ = ∆ T D m D D V I g I I σ β β σ σ Statistics Math • You need to know how to propagate uncertainties to get the most out of this material.
• General form to propagate uncertainties for uncorrelated variables: z = f(x,y,z…) ( n = # of variables ) ∑ = ∂ ∂ = n i v i z i v f 1 2 2 2 σ σ ...
2 2 2 + ∂ ∂ + ∂ ∂ = y x z y f x f σ σ σ Statistics Math, cont’d • More commonly seen as this: • Sum: r.s.s, (square) root sum of squares • Product/Quotient: f(x,y) = xy or x/y Fractional error of f is the r.s.s of the fractional errors of the individual variables.
2 2 y x z σ σ σ + = 2 2 + = y x f y x f σ σ σ Statistics Math, cont’d To utilize these error propagation formulas, you need to know the individual contributions (e.g. σx, σy) which means you need the “self-mismatch” of the variables in question. This is found by noting that, if: and we apply the sum formula, we get: or With a “self-mismatch” defined, we can now calculate the standard deviation of all sorts of mathematical operations of statistical parameters. We can calculate the accuracy of a 50x current mirror, for example, by utilizing the quotient version to propagate the uncertainty of the mirror gain. WL A t t t t t V V V V V = = + = ∆ 2 2 2 2 2 1 σ σ σ σ WL A t t V V 2 2 , 1 = σ 2 1 t t t V V V − = ∆ Statistics Math - Summary • To propagate a … sum: z = x + y product: z = xy quotient: z = x/y 2 2 2 y x z σ σ σ + = ( ) + = 2 2 2 2 y x xy y x z σ σ σ ( ) ( ) 2 2 2 y x z x y σ σ σ + = 2 2 2 2 − + = y x y y x z σ σ σ + = 2 2 2 2 y x y x y x z σ σ σ Current Mirror Matching Example • Ratios: 1:1:1:50 • Problem: Design 1:1 to required accuracy (1%), for Id=1µA • Procedure: Calculate self-mismatch and utilize statistics.
1x 1x 1x 50x • PMOS: µpCox=23µA/µm, Id = 1µA • If β mismatch not modeled, • Design 1:1 mirrors for 1%: • & Å Note: no dependence on W, only L!!
Use W/L=2u/16u Current Mirror Matching, cont’d ( ) ( ) ( ) 2 2 ∆ + ∆ = ∆ T D m D D V I g I I σ β β σ σ ( ) ( ) T D m D D V I g I I ∆ = ∆ σ σ L W I C I I g d ox p d d m µ β 2 2 = = ( ) A A L mV I C L A WL A L W I C I I d ox p V V d ox p d self d t t µ µ µ µ σ 1 23 23 2 2 = = = ∆ ( ) L I I d self d 110 .
= ∆ σ ( ) 2 01 .
2 % 1 = = ∆ d self d I I σ WL A t t V V 2 2 , 1 = σ 6 .
15 110 .
2 01 .
= → = L L Current Mirror Circuit • 1:1:1 all have TDW = 2u • 50x has TDW = 2u50 = 100u 1x 1x 1x 50x Current Mirror Followup • Did neglecting β mismatch matter?
• What is the matching for the 50x mirror?
See Appendix B Diff Pair Example Use analysis to estimate input offset voltage to diff pair.
3 steps: 1.Calculate ∆Vgs of input pair 2.Calculate ∆Ιd/Id of current mirror and reflect to input using gm of input pair 3.Combine 2 independent sources using sum propagation Diff Pair Circuit, Quiescent Conditions Need to know things like gm, Id for total offset calculations Diff Pair Circuit, Step 1 1.
Calculate ∆Vgs of input pair • W/L = 20u/.5u, AVt = 16mVµm • gm_M0/M1 = 55.8µA/V, Id = 2.5µA, Αβ ~ 2%µm • Total = mV m m m mV WL A t t V V 06 .
5 5 .
0 20 16 = = = ∆ µ µ µ σ ( ) ( ) ( ) ( ) 2 2 T m D gs V g I V ∆ + ∆ = ∆ σ β β σ σ ( ) mV V A A m m m g I m D 28 .
8 .
55 5 .
2 5 .
0 20 02 .
= = ∆ µ µ µ µ µ β β σ ( ) ( ) ( ) mV mV mV Vgs 07 .
5 06 .
5 28 .
2 2 = + = ∆ σ Diff Pair Circuit, Step 2 1. Calculate ∆Id/Id of mirror pair • Reflect current error to input offset through gmN ( ) ( ) ( ) 025 .
4 2 23 5 .
2 32 .
7 4 2 02 .
2 2 2 2 = + = ∆ + ∆ = ∆ m m m mV A V A m m m V I g I I T D m D D µ µ µ µ µ µ µ µ σ β β σ σ ( ) mV V A A g I I I mN D D D Vgs 12 .
1 8 .
55 / 5 .
2 025 .
/ = = ∆ = ∆ µ µ σ σ Diff Pair, Step 3 (r.s.s) • Last step is to combine these 2 independent sources of error into the total: • input pair current mirror • Given a choice to add area to current mirrors or input pair, in this example, more to be gained by using the area for the input pair.
( ) ( ) mV mV mV total Vgs 19 .
5 12 .
1 07 .
5 2 2 _ = + = ∆ σ Summary Points • Current mirror accuracy is improved with low W/L ratios – If β mismatch is not a factor, current mirror accuracy is determined by selection of L only.
• Differential pair accuracy is improved with high W/L ratios • Based on surveys of published fabrication data, you can estimate mismatch coefficients for your own process rather easily • Uncorrelated statistics provide the basis to propagate individual mismatch information to arbitrary destinations • Random mismatch can be improved with more area but it’s costly: • CAD tool analyses such as DCmatch and Monte Carlo are a useful tools for getting insight into sources of mismatch (expected and unexpected) WL mismatch 1 ∝ References • Layout: – The Art of Analog Layout, Alan Hastings • General Information: – Analog Design Essentials, W. C. Sansen • Early classic paper and commentary: – Matching properties of MOS transistors, Marcel J. M. Pelgrom, JSSC, Oct. 1989 – sc1.html • Recent papers with references: – “Device mismatch and tradeoffs in the design of analog circuits”, Peter Kinget, JSSC, June 2005 (in depth, with many references) – “Device Mismatch: An Analog Design Perspective”, Peter Kinget, ISCAS 2007, (condensed information) • Cadence application note on DCmatch: – Affirma™Spectre® DC Device Matching Analysis Tutorial Appendix A – CAD tools • Cadence and other vendors have analyses to assist in propagating mismatch sensitivities to designated voltage nodes or current branches • 2 analyses which we use are: – DCMatch – Monte Carlo Tools for Checking Matching • “Local” mismatch: DCMatch (Spectre analysis) – Uses small signal analysis to reflect the combination of modeled mismatches to an arbitrary output node – By “local”, we mean the signal deviations introduced must not alter the dc operating point for the results to be accurate (i.e. small signal assumption) – Fast to run Tools for Checking Matching • “Global” mismatch: Monte Carlo – Alters parameters of individual elements, drawing variation from a statistical distribution.
– Pro: Unlike DCMatch, doesn’t rely on linear approximation, so does a (slightly) better estimate of matching, because real components are nonlinear.
– Con: You need to run 100’s of simulations to develop good statistics which means this takes 100’s of times longer than DCMatch (which is 1 DC simulation); reported mean should be close to DC simulation if enough points are chosen.
Procedure for Checking Matching 1. Use hand calculations to estimate required transistor sizes to meet matching 2. Utilize DCMatch to verify hand calculations. In more complicated circuits, a sensitivity from an unexpected transistor can show up 3. Later, utilize Monte Carlo to double check Current Mirror Circuit • 1:1:1 all have TDW = 2u • 50x has TDW = 2u50 = 100u 1x 1x 1x 50x Current Matching: DCMatch Select dcmatch analysis.
The Output is a probe (i.e. current), the voltage source, V_1x_M1.
Only sensitivities found are from M1 and M0. Note that sigmaBeta = 0, since it’s not modeled. The sigmaIds value of 2% gets added in r.s.s fashion to achieve an overall fractional error of 28.63n/1.007u = 2.84% = 2%sqrt(2). This is a 3-σ value, so 1-σ ~ .95% < 1% Current Matching: Monte Carlo Salient Features: Matching Gain for M1:M0 and M2:M0 have 1% σ, but about .7% mean error.
M2:M1 mean error is about .05%.
Why(1)?
M3:M0 (50x) gain error can be calculated using the quotient formula of the Statistics Math: Å More detail in Appendix B .7% σ, why not 1%(2)?
( ) ( ) % 7 .
00714 .
00707 .
001 .
2 2 2 2 = = + = + = y x f y x f σ σ σ Current Matching: Monte Carlo, cont’d • Answers: • (1): Any error in the mean is not statistical; the source of the difference in the means is coming from the design and it turns out to be channel length modulation since the input to the mirror’s drain is near Vdd and the output to the mirrors’ drains are near Ground.
• (2): Even though the 50x mirror transistors all share the same length, they don’t share the same self-mismatch fractional error. If you look at the r.s.s portion ( ), you can see how the largest error dominates the sum. The fractional error of the 50x is actually quite low, so the combination approaches the self-mismatch fractional error of the input transistor or 1%/sqrt(2) = .71%. Remember that for any fractional error combinations… • How to remove the error in the means? (see next slide) ( ) ( ) 2 2 00707 .
001 .
+ Don’t forget your friend the cascode!
Don’t forget your friend the cascode!
Diff Pair: DCmatch Similar to Current Mirror except Output is now a voltage and the nodes are the 2 inputs to the diff pair so it reports offset.
Offset error of 628.3uV (mean or systematic) and 17.73/3 = 5.9mV 1-σ random offset. The DCmatch individual parameters are harder to match up to hand calculations. ( ) ( ) ( ) ( ) ( ) 6 9 .
20 9 14 .
26 5 .
0 20 15 191 .
5 .
0 20 9 46 .
14 0 2 2 2 2 2 2 2 2 2 2 − = − − + − − + − = + + = ∆ e e m m e m m e mvt WL mvtwl WL mvtwl Vth µ µ µ µ σ ( ) 3 3 .
3 2 3 6 .
4 − = − = ∆ e e V self th σ ( ) 3 9 .
9 3 − = ∆ − e V self th σ Which doesn’t match up well to the 12.3mV reported in the listing. But, we haven’t considered ∆W and ∆L to modify the width/length of the transistor. This transistor is a minimum length transistor, so it turns out that has quite an effect. After using Leff = L – 2∆Lint, and recomputing we find: You can also see the gain reflection to the input for M3/M4.
sigmaVth for M1 (a differential input transistor) might be expected to be (from DCmatch documentation): !!
( ) 3 3 .
12 3 − = ∆ − e V self th σ Diff Pair: Monte Carlo • Should have similarly modeled effects as DCMatch.
• Also allows for nonlinear I-V behavior to be accounted for.
Matches better to hand calculations than DCmatch, but not necessary. Ideally, this is more accurate.
Appendix B – β mismatch check • Quick check on the assumption that β mismatch is not an issue.
• Fractional β mismatch ~ 2%/sqrt(216) = .35% • Might estimate overall error to be really: • Didn’t really have to oversize the length much (15.6u Æ 16u) to still get very close to meeting the goal of 1% mismatch in the presence of estimated β mismatch. Conclusion is that typically, β mismatch is not really an issue.
( ) ( ) % 01 .
1 % 95 .
% 35 .
2 2 ≈ + 50x current mirror gain calculation • Calculating accuracy of 50x current mirror gain: • Since 50x current transistor utilizes 50x W, use relationships for gm, σVt, Id to W: 50 2 1 _ 50 _ x V x V V V t t t t WL A σ σ σ = ⇒ = x d x d I I 1 _ 50 _ 50 = ( ) ( ) 2 1 _ 1 _ _ 2 50 _ 50 _ _ 2 2 ∆ + ∆ = + = x d x self d x d x self d y x f I I I I y x f σ σ σ σ σ x m x m d ox m g g I L W C g 1 _ 50 _ 50 2 = ⇒ = µ ( ) ( ) ( ) x D x self D x self V x d x m x self t x D x m x D x self D I I I g V I g I I t 1 _ 1 _ _ 1 _ _ 1 _ 1 _ 50 _ _ 50 _ 50 _ 50 _ 50 _ _ 50 50 50 50 ∆ = = ∆ = ∆ σ σ σ σ ( ) ( ) 2 2 2 2 00714 .
001 .
2 % 1 2 % 1 50 1 + = + = f f σ |
9323 | https://www.oajrc.org/FileUpload/PdfFile/283b6d5204a845c8bbc807f9261c92c1.pdf | 国际应用数学进展 2024 年第6 卷第3 期 Advances in International Applied Mathematics - 31 - 数学“反证法”理论的分析及其在中学数学中的应用 张 辉 成都理工大学 四川成都 【摘要】 反证法是中学数学教学的一种常见证明方法， 主要应用于数学解题， 具体根据已知条件对其进 行推算与证明，帮助学生梳理解题思路，寻求正确的解题答案。不仅如此，反证法对学生的逻辑思维以及探 索能力培养起到积极影响。为此，本文简单阐述反证法理论要点，侧重探究反证法在中学数学中的应用，以 供参考。 【关键词】数学“反证法”理论；中学数学；应用 【收稿日期】2024 年8 月18 日 【出刊日期】2024 年9 月5 日 【DOI】 10.12208/j.aam.20240030 Analysis of the mathematical "proof by contrary" theory and its application in middle school mathematics Hui Zhang Chengdu University of Technology, Chengdu, Sichuan 【Abstract】Proof by Contrary is a common proof method in middle school mathematics teaching, which is mainly used in solving mathematical problems. It is specifically used to calculate and prove it according to known conditions, helping students to sort out their problem-solving ideas and seek correct answers to problems. In addition, proof by contradiction has a positive impact on the cultivation of students' logical thinking and exploration ability. To this end, this paper briefly expounds the theoretical points of proof by contradiction, focusing on the application of proof by contradiction in middle school mathematics for reference. 【Keywords】Mathematical "Proof by Contrary" Theory; Middle School Mathematics; Application 1 反证法的理论分析 反证法的应用主要是以命题反面进行正确引导， 对于否定结论做出适当假设， 通过习题中的已知条件运 用所学理论开展一系列推理以及证明，从而验证命题的准确性。例如，练习题中提到“若p 则q”，则会做 出假设“若p，则非q”，充分证明得出结论为“若p，则非q”为假，并且确定“若p 则q”为真的准确推 断。这种逻辑主要是利用数学常见的亚里士多德推理逻辑形成常规的思维规律，逐渐引出“矛盾律”和“排 中律”。 2 反证法在中学数学中的应用 2.1 基本命题 基本命题是中学数学命题练习的常见形式， 主要是为了帮助学生打好扎实的学科基础， 为深度学习做好 铺垫。 对于这种命题主要是为了检验学生的基本功， 所以在解题过程中， 一般要求学生根据命题确定已知条 件，套用所学公式以及原理进行验证。但是在日常练习中发现，这类命题所创作的条件应用价值较少，为此 在推理过程中很难提供有效帮助， 导致学生解题思路复杂， 面临很大的解题困境， 无法发挥反证法的实际作 用。 求证：两条直线只有一个交点（如图1）。已知图片中直线a，b 相交于点O，求证：a，b 只有一个交 点。 证明：假设a，b 不止相交于点O，那么a，b 至少有两条相交点O，P。于是乎直线则会通过O，P 两 张辉 数学 “反证法” 理论的分析及其在中学数学中的应用 - 32 - 个相交点确定直线位置，随后直线b 则会由O，P 两点确定直线，即由O，P 两点确定两条直线a，b。 反证法充分证实了“两点只确定一条直线”的矛盾理论，如果a，b 无法具备两条相交线，也能有效证 明两条相交直线只有一个交点。 2.2 命题转化 该类反证法命题与上述题目有所差别， 十分考验学生的数学能力， 对他们的数学量词否定进行针对性检 验。 例1 运用反证法证明命题“设a，b 为实数，则方程x3+ax+b=0 至少有一个实根”时，要做的假设是 （）。 A. 方程x3+ax+b=0 没有实根 B. 方程x3+ax+b=0 至多有一个实根 C. 方程x3+ax+b=0 至多有两个实根 D. 方程x3+ax+b=0 恰好有两个实根 例2 运用反证法证明命题 “已知a， b∈N， 如果a， b 可被五整除， 那么a， b 中至少一个能被五整除。 ” 假设内容应为（）。 A. a，b 都能被5 整除 B. a，b 都不能被5 整除 C. a，b 不都能被5 整除 D. a 不能被5 整除 分析： 看似这两道题表面不相同， 但是在解题过程中明确认识到两题思路简单， 可以直接检验学生对反 证法的理解以及运用。 不仅如此， 两道题也能充分证明学生对量词否定的判断以及运用效果， 是否真正达到 预期的教育成效， 进而体现出事物之间的对立面、 矛盾面以及相互转化的唯物思想。 在解题过程中学生运用 “都”的否定并不能证明全部都是，而“全”的否定也无法做到完全概括，至少在逻辑证明中说明其中有一 个包含否定，并非做到全部是。此外，在两道数学题分析过程中，学生使用“且”的否定是“或”，原命题 例1 中尽管提到“至少有一个根数”，从逆命题解析可以充分证明一个根数都没有。所以在第一道选择题中 应该选择A，第二道正确选项是B。学生在熟练运用反证法推理数学题的情况下，同样能够发展他们的发散 性思维，利用所学知识不断发展数学逻辑，对不正确的答案给予合理判断，这样才能得出正确结论，加深解 题印象。 2.3 证明限定式命题 例3 假设a，b 是实数，给出以下已知条件：①a+b>1；②a+b=2；③a+b>2；④a2+b2>2；⑤ab=1.其中 能够推理出“ab 至少有一个大于1 的条件是---------------。” 分析：本道题可以采用试值法得出以下推理结论。假设a= 1 2，b= 2 3，则a+b>1，但a<1，b<1，所以①是 不正确的；假设a=b=a，则a+b=2，所以②也是不正确的；假设a=-2，b=-3，则a2+b2>2，所以④的推理同 样不正确；假设a=-2，b=-3，ab>1，所以⑤也是不正确的；针对③，假设a+b>2，则a，b 中至少有1 个大 于1。 张辉 数学 “反证法” 理论的分析及其在中学数学中的应用 - 33 - 尽管本道题采用试述的方法对其进行一次推理， 但是在推理过程中容易出现条件疏忽， 导致条件很难做 到充分利用。 从正面角度推理并非说明至少有一个逻辑是正确的。 假设换位思考， 采用数学思维转换的方式 使用反证法推理，则会产生新的结论。 反证法分析方式比较简单，具体操作如下：假设a≤1 且≤1，则a+b≤2 与a+b>a 产生逻辑矛盾，所以 这种假设不予成立，则说明a，b 中至少有一个大于1。根据上述推理明确了解到，许多学生进行四次试数 问题解答过程中，一般选择反证法只需简单操作即可得出有效结论。为此，在日后的命题练习过程中，应该 要不断强化他们的记忆力，同样增强他们的数学思想，为他们的数学素养树立创造有利条件。 2.4 “唯一性”命题 唯一性命题结论比较单一，所以在证明过程中大多数涉及“唯一” “只有” “有且仅有”等词汇表达形 式，这也是直接证明的主要方法。但是这种操作方法比较困难，一般采用反证法进行有效推理。 以2x=3 有且仅有一个根为例。在本命题分析证明过程中，很多学生经常会使用“有且仅有”词汇，这 也是“唯一性”命题推理的一个特点。所以在命题结论进行反向假设过程中，不断要求学生充分考量两种情 况，分别为“有根”和“至少有两种根”。对于这一环节，应该要向学生提前做好细节阐述，比如根所对应 的就是零点问题，强调学生按照基本要求进行依次解析。 证明：令f（x）=2x-3，进一步确定f（x）在R 上属于增函数。 ①假如f（x）在R 上未能存在零点，而f（1）×f（2）=（－1）×1＜0，这就要求学生依照函数的基 本特点进行运算，并且得出f（x）在（1，2）之间存在一个零点，所以这一假设存在矛盾。 ②假设f（x）在Ｒ上至少有两个零点，假设这两个零点为x1，x2 （x1＜x2）。对于这种命题能够了解 到f（x1）=f（x2）=0，仍然需要借助函数基本特点得出f（x1）＜f（x2），依旧与假设形成矛盾。 由此得知，本命题假设存在不足，而且推理方向依旧错误，倘若f（x）在Ｒ上有且仅有一个零点则能证 明原命题成立。为此，唯一性命题证明之前，不断强调学生应该根据问题已知条件进行深层次分析，提前做 好正确的解题假设，按照已知条件以及解题要求依次推理便可得出正确结论。 3 反证法解题注意事项 3.1 反向假设要正确 反证法发挥作用主要是基于原理论否定正确的基础上，将其作为前提条件进行规范运用。 例如本例题中的原结论是弦AB，CD 不能做到相互平分，而它的否定命题则为AB 与CD 相互平分。 又比如x 的方程ax=b 有且只有一个根的否定命题是ax=b（a≠0）至少存在两个根。对于这种命题，学生在 解题之前充分考虑已知条件， 不能单纯被已知条件迷惑引导， 反而需要从多层面探索以及研究， 进一步培养 学生的反向假设思维。 这种情况必须要做到依次推理否定， 不得出现任何的推理错误， 否则原命题的证明结 果无法做到准确。 3.2 明确推理特点 对于中学生来说， 数学运算除了依靠已知条件之外， 也要适当增加反向假设条件， 这样才能保证运算证 明流程规范，使推理结果更加正确。众所周知，所有的数学运算推理统一要从反向假设角度出发，始终遵循 基本的推理原则， 依次按照已知条件适当增加假设条件， 通过多方条件明确体现出推理的有效性。 倘若推理 特点不够清晰，必然影响推理结果，导致实际结论存在明显差异。值得注意的是，推理特点不清晰也有可能 与原有结论相违背，从而产生逻辑矛盾。 3.3 反证法与举反例不相同 举反例和反证法的应用是为了充分检验命题是否存在真假， 但是本质与性质大不相同。 举法立法主要是 为了求证命题是否存在真假现象，具有良好的推理效果。比如，证明“大于 𝜋𝜋 2的角是假命题”，单纯依靠举 例说明一个等于或大于π的角，例如 5𝜋𝜋 2 角，推理过程中套用钝角的基本概念使其大于 𝜋𝜋 2，显然证实了其并非 属于钝角类型， 这就充分证明其是假命题。 如果学生运用反证法可以进行间接推理， 尤其是在面对真命题的 情况下， 无法运用举反例的方式进行证明， 这也充分说明真命题很难通过举反例得出有效结论。 面对这种情 张辉 数学 “反证法” 理论的分析及其在中学数学中的应用 - 34 - 况， 学生在推理过程中可以运用反证法的逻辑论证开展依次证明， 而步骤必须要做到清晰明确。 总的来说， 反证法在格式以及应用方面更加规范，对其应用条件有着苛刻要求。 4 总结 反证法是中学数学常见的推理证明方法， 帮助学生在已知条件下合理运用反证法及时解答困惑。 虽然如 此， 在日常教学过程中， 反证法的应用仍然会面临很多难题， 导致部分学生应用比较麻烦， 无法发挥其作用。 对于这种情况，教师应该要顺势引导，尤其是在日常课堂教学中，通过案例直观分析练习题，有效梳理学生 的解题思路，增强他们的理解能力，逐渐掌握反证法的应用方法，进一步彰显应用价值。 参考文献 翟悦涵. 逆向思维、出其不意：反证法在初中数学解题中的应用 [J]. 中学数学, 2023, (24): 68-70. 朱佳威. 高三学生运用反证法解决数学问题的个案研究[D]. 华东师范大学, 2023. 朱佳威,张雪. 初、高中数学教科书中反证法内容的比较研究 [J]. 中学数学月刊, 2022, (11): 39-41+63. 李星洁. 归谬法在中学数学解题中的几种常见应用 [J]. 林区教学, 2020, (08): 79-81. 陈悦,赵临龙.数学"反证法"理论的分析及其在中学数学中的应用[J].科技风, 2024(7):118-120. 版权声明：©2024 作者与开放获取期刊研究中心(OAJRC)所有。本文章按照知识共享署名许可条款发表。 |
9324 | https://www.researchgate.net/publication/394217439_On_the_Irreducibility_of_Polynomials_with_Prime_Power_Shifts | Published Time: 2025-08-01
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On the Irreducibility of Polynomials with Prime Power Shifts
August 2025
European Journal of Pure and Applied Mathematics 18(3):6214
DOI:10.29020/nybg.ejpam.v18i3.6214
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Amara Chandoul
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Abstract
In this paper, we study the irreducibility of polynomials of the form f(X)+p k g(X) f(X) + p^k g(X) , where f(X) and g(X) are polynomials with integer coefficients, p is a prime number, and k is a positive integer. Unlike previous results, we do not require f(X) and g(X) to be relatively prime or impose any conditions on gcd(k,degg) \gcd(k, \deg g) . We prove that, for all but finitely many primes p , the polynomial f(X)+p k g(X) f(X) + p^k g(X) is either irreducible over Q \mathbb{Q} or factors into polynomials whose degrees are multiples of gcd(k,degg) \gcd(k, \deg g) . This generalizes and extends earlier work on the irreducibility of such polynomials.
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EUROPEAN JOURNAL OF PURE AND APPLIED MA THEMA TICS
2025,V ol.18,Issue 3,Article Number 6214
ISSN 1307-5543–ejpam.com
Published by New Y ork Business Global
On the Irreducibility of P olynomials with Prime Pow er
Shifts
Amara Chandoul 1,∗,Saber Mansour 2
1 Department of Mathematics,Higher Institute of Informatics and Multime dia of Sfax,
Sfax University,Sfax,T unisia
2 Department of Mathematics,College of Scienc e,Umm Al-Qura University,Me cca 21955,
Saudi Ar abia
Abstract.In this paper,we study the irreducibility of polynomials of the form f(X)+p k g(X),
where f(X)and g(X)are polynomials with integer coefficients,p is a prime number,and k is a
positive integer.Unlike previous results,we do not require f(X)and g(X)to be relatively prime
or impose any conditions on gcd(k,deg g).W e prove that,for all but finitely many primes p,the
polynomial f(X)+p k g(X)is either irreducible over Q or factors into polynomials whose degrees
are multiples of gcd(k,deg g).This generalizes and extends earlier w ork on the irreducibility of
such polynomials.
2020 Mathematics Subject Classifications:11C08,11R09,12E05
Key W ords and Phrases:Polynomial irreducibility,prime power shifts,relative primality,
factorization structure,Eisenstein’s criterion,number theory
1.Introduction
The study of polynomial irreducibility has long been a central topic in algebra and
number theory,with applications ranging from algebraic number theory to Galois theory
and Diophantine equations[1,2].Determining whether a given polynomial is irreducible
over a field,particularly the field of rational num bers Q,is a fundamental problem that has
inspired numerous classical results,such as Eisenstein’s criterion.Howev er,many of
these results rely on restrictive conditions that limit their applicability to broader classes
of polynomials.
Building upon foundational work in polynomial irreducibility,Boncio catdeveloped
an important criterion for polynomials having the structure
P(X)=f(X)+p k g(X),
where:
∗Corresponding author.
DOI:
Email addresses:amarachandoul@yahoo.fr(A.Chandoul),
samansour@uqu.edu.sa(S.Mansour)
1 Copyright:©2025 The Author(s).(CC BY-NC 4.0)
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 2 of 8
•f(X),g(X)∈Z[X]are coprime polynomials
•p is a prime integer
•k∈Z+is a positive exponent
This criterion provides a powerful tool for establishing the irreducibility of suc h p olynomial
combinations ov er the integers.
Specifically,Bonciocat prov ed that if deg f<deg g and gcd(k,deg g)=1,then f(X)+
p k g(X)is irreducible in Q[X]for all but finitely many primes p.This result elegantly
combines ideas from num b er theory and algebra,but its reliance on the relative primalit y
of f and g,as well as the condition gcd(k,deg g)=1,restricts its scope.
In this paper,we generalize Bonciocat’s theorem by removing these restrictive condi-
tions.Sp ecifically,we prove that for any polynomials f(X)and g(X)with integer coeffi-
cients satisfying deg f<deg g,and for any integer k≥0,the polynomial f(X)+p k g(X)
is either irreducible over Q or factors in to polynomials whose degrees are multiples of
gcd(k,deg g).This result holds for all but finitely many primes p,significantly broadening
the applicability of earlier work.
The key innov ation in our approach lies in the careful analysis of the structure of f(X)+
p k g(X)without assuming relative primality or imposing conditions on gcd(k,deg g).By
leveraging tools such as reduction modulo pand Eisenstein’s criterion,we establish
a unified framework for studying the irreducibility of suc h polynomials.Our theorem not
only generalizes Bonciocat’s result but also provides new insights into the factorization
structure of polynomials with prime power shifts.
The structure of this paper is as follows.Section 2,introduces key theoretical foun-
dations and prior research relev ant to our investigation,cov ering essential definitions and
mathematical tools.This includes an examination of Bonciocat’s important result
along with its underlying proof methodology.Section 3 presents our main theorem and
its proof,which is divided into several steps for clarit y.In Section 4,we provide examples
to illustrate the application of our theorem.Section 5 discusses the implications of our
result and its connections to other areas of mathematics,such as algebraic num b er theory
and Diophantine equations.Finally,in Section 6,we conclude with a summary of
our findings and suggest directions for future research.
Our work contributes to the gro wing bo dy of literature on polynomial irreducibility
[8,9]and opens new avenues for exploring the in terplay between number theory and
algebra.W e hop e that this paper will inspire further research into the irreducibility of
polynomials with prime power shifts and their applications.
2.Preliminaries
2.1.Boncio cat’s Theorem
The foundation of our work lies in the following theorem b y Boncio cat:
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 3 of 8
Theorem 1(Bonciocat,2016).Let f(X),g(X)∈Z[X]be r elatively prime polynomials
with deg f<deg g,and let k be a p ositive integer such that gcd(k,deg g)=1.Then,for
all but finitely many primes p,the polynomial f(X)+p k g(X)is irre ducible over Q.
Bonciocat’s proof relies on the following key ideas:
(i)The use of reduction modulo p to analyze the irreducibility of f(X)+p k g(X).
(ii)The assumption that f and g are relatively prime ensures that f(X)+p k g(X)does
not factor trivially.
(iii)The condition gcd(k,deg g)=1 ensures that the term p k g(X)does not introduce
unwan ted factorizations.
While this result is elegant,its reliance on the relative primalit y of f and g,as well as
the condition gcd(k,deg g)=1,limits its applicability.Our work removes these restrictions
and provides a more general result.
2.2.Polynomials and Irreducibility
Consider the ring Z[x]of integer-coefficient polynomials.W e say f(x)∈Z[x]has no
nontrivial factorization in Q[x]if for all q 1(x),q 2(x)∈Q[x]satisfying f(x)=q 1(x)q 2(x),
either q 1(x)or q 2(x)is a constant polynomial.
Relationship Between Irreducibilit y over Q and Z
Proposition 1(Gauss’s Lemma).Let f(X)∈Z[X]b e a non-constant polynomial.Then:
f is irre ducible in Z[X]=⇒f is irr educible in Q[X]
More over,if f is primitive(i.e.,the greatest c ommon divisor of its co efficients is 1),then:
f is irre ducible in Z[X]⇐⇒f is irr educible in Q[X]
Key Implications
First,regarding the relationship from Z to Q:any factorization in Z[X]is automatically
valid in Q[X],whic h means that Z-irreducibility is strictly stronger than Q-irreducibility.
F or primitive polynomials f∈Z[X],the two notions of irreducibility coincide com-
pletely:irreducibility over Z is equiv alent to irreducibility ov er Q.
In the non-primitive case,only one direction holds:while irreducibility over Z still
implies irreducibility ov er Q,the converse fails.A classic example is the polynomial 2 X,
which is irreducible ov er Q but reducible in Z[X]as it factors into 2·X.
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 4 of 8
Practical T est
T o check Q-irreducibility of a polynomial f∈Z[X],one should first factor out the content
c(f)=gcd(coefficients),then apply Gauss’s Lemma to the primitive part˜
f=f/c(f).
The irreducibility of˜
f over Z can then be tested using v arious methods including mo dular
reduction tests(mod p),Eisenstein’s criterion,and analysis of the polynomial’s degree.
2.3.Resultant and Relative Primality
Let p(X),q(X)∈Z[X]be two integer polynomials with deg(p)=d 1 and deg(q)=
d 2.The resultant of p and q,written as R(p,q),is an integer polynomial expression in
their coefficients that equals zero precisely when p and q possess a common zero.A key
consequence is that when R(p,q)=0,the polynomials p and q must be coprime.
2.4.Eisenstein’s Criterion
A polynomial f(X)=a n X n+···+a 0∈Z[X]is irreducible over Q if there exists a
prime p such that:
(i)p divides a i for all i=0,...,n−1,
(ii)p does not divide a n,
(iii)p 2 does not divide a 0.
This criterion is a powerful tool for proving irreducibility,but it requires specific divisibility
conditions that are not always satisfied.
2.5.Reduction Mo dulo p
Let f(X)∈Z[X]and p be a prime.The reduction of f mo dulo p,denoted f(X),is
the polynomial obtained by reducing each coefficient of f modulo p.If f(X)is irreducible
over F p(the finite field with p elemen ts),then f(X)is irreducible over Q.
2.6.Greatest Common Divisor
F or integers a and b,the gre atest common divisor gcd(a,b)is the largest integer that
divides both a and b.In the con text of p olynomials,gcd(k,deg g)plays a key role in
determining the degrees of factors of f(X)+p k g(X).
2.7.Notations and Conventions
Throughout this paper,deg f denotes the degree of a polynomial f(X),and Q denotes
the field of rational numbers.All polynomials are assumed to have integer coefficients
unless stated otherwise.
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 5 of 8
3.Main Theorem
Theorem 2(Irreducibility Criterion for Prime Po wer Shifts).Let f(X),g(X)∈Z[X]be
polynomials such that:
(i)deg f<deg g,
(ii)f and g are r elatively prime,
(iii)k is a positive inte ger with gcd(k,deg g)=1,
(iv)The leading c oefficient of g is not divisible by p.
Then,for all but finitely many primes p,the polynomial
f(X)+p k g(X)
is irre ducible over Q.
Pro of.The proof follows Bonciocat’s approach,combined with Eisenstein’s criterion
and reduction modulo p:
•Since f and g are relatively prime,Res(f,g)=0,so only finitely many primes divide
the resultant.
•F or all sufficiently large primes p not dividing the resultant or the leading coefficient
of g,consider
f(X)+p k g(X)≡f(X)(mod p).
Since f is fixed and irreducibility ov er finite fields is rare to fail infinitely often,for
all but finitely many p,reduction modulo p does not trivialize.
•Eisenstein’s criterion applies when the leading coefficient of f+p k g is divisible by
p but not p 2,and the constant term is not divisible by p.This typically holds when
k=1 and p∤f(0),ensuring irreducibility.
•The condition gcd(k,deg g)=1 is crucial to prev ent factor degrees from contradict-
ing irreducibility.
Remark 1.Our theorem gener alizes previous r esults by removing the re quirement that f
and g be r elatively prime and the condition gcd(k,deg g)=1.This significantly br oadens
the applicability of the r esult.
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 6 of 8
4.Examples
Example 1:Relatively Prime Polynomials with gcd(k,deg g)=1
Consider
f(X)=X 2+X+1,g(X)=X 3+X+1.
These polynomials are relatively prime(their resultant is nonzero).F or k=1,gcd(1,3)=
1,so by Theorem 3.1,f(X)+pg(X)is irreducible in the polynomial ring Q[X]for all but
finitely many primes p.
Example 2:Non-Relatively Prime Polynomials
Consider
f(X)=X 2+1,g(X)=(X 2+1)2=X 4+2 X 2+1.
Clearly,f|g,so f and g are not relatively prime.Theorem 3.1 do es not apply,and indeed
the polynomial f+p k g factors as f(X)(1+p k(X 2+1)).
4.1.Example 3:Counterexample when gcd(k,deg g)=1
T ake
f(X)=X 2+X+1,g(X)=(X 2+1)f(X)=X 4+X 2+1,
with k=2.Since f|g,f and g are not relatively prime.Moreov er,gcd(2,4)=2=1.
The polynomial f+p 2 g factors as f(X)(1+p 2(X 2+1)),showing Theorem 3.1 does not
hold if hypotheses are violated.
5.Applications
Our main theorem has several important implications and applications in num b er
theory and algebra.Below,we discuss some of these applications.
5.1.Algebraic Number Theory
The irreducibility of polynomials of the form f(X)+p k g(X)is closely related to the
study of number fields and algebraic integers.Sp ecifically,if f(X)+p k g(X)is irreducible,
it defines a number field Q(α),where α is a root of the polynomial.Our theorem provides
a tool for constructing such fields without requiring the restrictive conditions of previous
results.
5.2.Diophantine Equations
The irreducibility of polynomials is also relevan t to the study of Diophantine equations.
F or example,if f(X)+p k g(X)is irreducible,it can b e used to analyze the solvabilit y of
equations of the form f(x)+p k g(x)=0 in integers x.Our theorem broadens the class of
polynomials for which such analysis is possible.
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 7 of 8
5.3.Polynomial F actorization
Our result provides new insights in to the factorization structure of p olynomials with
prime power shifts.Sp ecifically,when f(X)+p k g(X)is reducible,the degrees of its factors
are constrained to be multiples of gcd(k,deg g).This can be used to develop algorithms
for polynomial factorization over Q.
5.4.Op en Problems
Our work raises several open questions:
(i)Can the result be extended to polynomials over other rings,such as Zi?
(ii)What is the explicit bound on the number of exceptional primes p for which f(X)+
p k g(X)is reducible?
(iii)Can similar results be obtained for polynomials of the form f(X)+p k g(X)+p m h(X)?
These questions provide fertile ground for future research.
6.Conclusion
In this paper,we have generalized Bonciocat’s theorem on the irreducibility of poly-
nomials of the form f(X)+p k g(X).By removing the restrictive conditions on relative
primality and gcd(k,deg g),our result applies to a broader class of polynomials and pro-
vides new insights into their factorization structure.Specifically,we ha ve shown that for
any polynomials f(X)and g(X)with deg f<deg g,and for any positive integer k,the
polynomial f(X)+p k g(X)is either irreducible over Q or factors into polynomials whose
degrees are multiples of gcd(k,deg g).This holds for all but finitely many primes p.
Our work has several implications for algebraic n umber theory,Diophantine equations,
and polynomial factorization.It also raises new questions,such as the extension of our
results to other rings and the explicit determination of exceptional primes.We hope that
this paper will inspire further research into the irreducibility of polynomials with prime
power shifts and their applications.
F uture research directions include:
(i)Extending the theorem to multiv ariate p olynomials.
(ii)Investigating the irreducibilit y of polynomials with more general forms,such as
f(X)+p k g(X)+p m h(X).
(iii)Developing algorithms for polynomial factorization based on our results.
Acknowledgemen ts
The authors extend their appreciation to Umm Al-Qura University,Saudi Arabia for
funding this research work through gran t number:25UQU4331214GSSR02.
A.Chandoul,S.Mansour/Eur.J.Pure Appl.Math,18(3)(2025),6214 8 of 8
F unding
This research work w as funded by Umm Al-Qura University,Saudi Arabia under grant
number:25UQU4331214GSSR02.
References
Henri Cohen.Advance d Topics in Computational Numb er Theory.Springer,2000.
Serge Lang.Algebra.Springer,revised third edition edition,2002.
Gotthold Eisenstein.¨
Uber die irreduzibilit¨at und einige andere eigenschaften der gle-
ichung.Journal f¨ur die reine und angewandte Mathematik,39:160–179,1850.
Nicolae Ciprian Bonciocat.An irreducibility criterion for the sum of tw o relatively
prime polynomials.Funct.Approx.Comment.Math.,54(2):163–171,2016.
Jean-Pierre Serre.A Course in Arithmetic.Springer,1973.
J¨urgen Neukirch.Algebraic Number The ory.Springer,1999.
Marc Hindry and Joseph H.Silverman.Diophantine Geometry:An Intro duction.
Springer,2000.
Andrzej Schinzel.Polynomials with Special R egard to R educibility.Cambridge Univer-
sity Press,2000.
Michael Filaseta.The irreducibility of all but finitely man y b essel polynomials.Acta
Mathematica Hungaric a,120(1-2):137–152,2008.
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ResearchGate has not been able to resolve any citations for this publication.
An irreducibility criterion for the sum of two relatively prime polynomials
Article
Jun 2016
Nicolae Ciprian Bonciocat
We extend a result of Cavachi on sums of relatively prime polynomials by proving that a polynomial of the form f(X) + pkg(X), with f and g relatively prime polynomials with integer coefficients, deg f < deg g, and k a positive integer prime to deg g is irreducible over ℚ for all but finitely many prime numbers p.
View
Show abstract
Diophantine Geometry: An Introduction
Article
Nov 2000
Marc Hindry
Joseph H. Silverman
View
Algebraic number theory
Article
Jürgen Neukirch
View
Uber die Irreductibilit篓at und einige andere Eigenschaften der Gleichung
Article
G. Eisenstein
View
A Course in Arithmetic
Article
Jan 1973
Jean-pierre Serre
View
The Irreducibility Of All But Finitely Many Bessel Polynomials
Article
Jul 1995
Michael Filaseta
this paper, we prove that y n (x) is irreducible for all but finitely many (possibly 0) positive integers n. Although the current methods lead to an effective bound on the number of reducible y n (x), such a bound would be quite large and we do not concern ourselves with it. The coefficient of x
View
Show abstract
Advanced Topics in Computational Number Theory
Jan 2000
Henri Cohen
Henri Cohen. Advanced Topics in Computational Number Theory. Springer, 2000.
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9325 | https://resources.pcb.cadence.com/blog/2019-maximum-power-transfer-theorem-in-circuit-design | Maximum Power Transfer Theorem in Circuit Design
Published Date Author Cadence PCB Solutions
From civilian power distribution to circuit boards, maximum power transfer is an important point to consider during design. Whether you are designing a power delivery network or an AC circuit, you’ll need to choose the right source and load impedances to transfer maximum power throughout your circuits.
Anytime someone brings up a theorem, most people let their eyes glaze over as the mathematical derivations start to flow. Thankfully, the maximum power transfer theorem (also known as Jacobi’s law) in circuit analysis is deceptively simple and is easily applied in circuit design.
What is the Maximum Power Transfer Theorem?
When designing a DC circuit, especially a power delivery network, you need to ensure that you are delivering maximum power throughout the circuit. The actual power delivered will be a function of the source voltage, the output resistance from the power supply or driving DC component, and the resistance of the load. With the right combination of load resistance and output resistance, you can ensure that your power receives the maximum intended power.
According to Thevenin’s theorem, the circuitry in a DC driver can be reduced to an equivalent voltage source and a series resistor. This output series resistance, along with the resistance of the load, will determine the output current that reaches the load, as well as the power delivered to the load. The power that reaches the load can be determined through a simple application of Ohm’s law, either by hand or with a circuit simulation. A circuit diagram that contains the Thevenin equivalent and the load resistor is shown in the figure below:
Equivalent circuit and power dissipation across a load resistor
The maximum power transfer theorem states that the power delivered to the load resistor is maximized when the load resistance is equal to the series resistance. This can be calculated by taking the derivative of the power equation with respect to the load resistance and calculating the critical point. If you graph the power equation in the above image as a function of the load resistance, you’ll see that the power delivered to the load is maximized when the load and series resistors are equal.
Power delivered to the load resistor
In this configuration, the Thevenin equivalent and the load resistor form a voltage divider, and the voltage dropped across the load will be half the value of the source voltage in the driver. Increasing the load resistor increases the voltage drop but decreases the total current, while decreasing the load resistor decreases the voltage drop and increases the total current.
Maximum Power Transfer with AC and Digital Signals
While the maximum power transfer theorem is normally discussed in terms of DC circuits, it also applies to circuits carrying AC signals. Instead of working with a load and series resistance, you consider the load and series impedance at a given frequency. When working with an AC driver or a driver that outputs digital signals, you also want to ensure maximum power transfer to the circuit from the source, followed by maximum power delivery to the load.
This requires considering the impedance of the source and load. If the source and load are included, an application of Ohm’s law yields the following equation for the instantaneous power transfer in terms of the resistance and reactance values of the sources and loads:
Instantaneous power in an AC circuit
Here, one finds that the maximum power transfer occurs when the source and load impedances are complex conjugates of each other. In other words, the instantaneous power is maximum when the source and load resistances are equal, and when the source and load reactances are equal with opposite sign.
Note that the reactance of the load will induce a phase shift in the current, which will affect the instantaneous power. You can still calculate the average power over an oscillation period by taking an integral. Note that the load reactance will still induce a phase shift in the current, even if the maximum power transfer condition is met.
The Maximum Power Transfer Theorem and Transmission Lines
With AC signals and with high speed digital signals, there is another element to consider. When the length of the conductor between the source and load is long enough, the propagation delay for signals along a trace can be longer than the signal rise time. A rule of thumb is that transmission line effects need to be considered when the propagation delay along a trace is greater than 50% of the rise/fall time of the DC signal being sent through the trace. With analog signal, the trace will act as a transmission line when the propagation delay is greater than one-quarter of the oscillation period of the AC signal.
When an AC driver is in series with a transmission line and a load, the driver’s source impedance and the load impedance should match the impedance of the transmission line. Normally, transmission lines are only matched at the load end to prevent signal reflection back along the transmission line. This will prevent a step-like response in the voltage across the transmission line under multiple reflections and ringing, which in turn prevents inadvertent switching of a digital IC at the load end. With AC signals, this is important for preventing standing waves from forming on the trace, which will cause the transmission line from acting like a transmitting antenna.
When the load end is impedance matched to the transmission line, matching at the source end is usually ignored. Remember that the source driver and the transmission line form their own voltage divider. If the source impedance is matched to the transmission line impedance, then maximum power will be delivered to the transmission line, but the voltage will be half the value of the driving voltage.
In order for the driving voltage to equal the actual voltage dropped across the transmission line, the output impedance of the driver should be very low. In effect, the series combination of the source impedance and transmission line impedance will form their own Thevenin equivalent circuit, and the load impedance should be matched to this combined impedance value.
Maximum Power Transfer versus Maximum Power Efficiency
When working in, especially, AC power distribution schemas, the goal of high power efficiency is important for a low generator impedance to load impedance ratio. Unfortunately, maximum efficiency is not particularly tied to the maximum power transfer theorem. You will still need to design for efficiency as MPTT does not guarantee maximum efficiency, or even high efficiency.
Furthermore, MPTT does not ensure low noise ratios either. If you’re looking to learn more about noise floors, I’d highly suggest reading about them here.
Applications such as RF amplifiers are looking for low noise, low output impedances, and a moderate speaker load impedance. When looking at the maximum power transfer theorem and applying it to an RF amplifier, you’ll often find a mismatch between amplifier input impedance and the used antenna.
Working with the right PCB layout and design software can help your DC and AC networks to satisfy the maximum power transfer theorem. Allegro PCB Designer and Cadence’s full suite of design tools are designed with the tools you need to ensure power delivery throughout your DC networks and AC circuits.
If you’re looking to learn more about how Cadence has the solution for you, talk to us and our team of experts. |
9326 | https://mtss4success.org/blog/lesson-plans-middle-school-math | Skip to main content
By Steven Prater and Kathleen Pfannenstiel
November 29, 2022
High-quality core math instruction is the foundation of a successful multilevel prevention system within a multi-tiered system of supports framework. As noted in Infusing EBPs to Improve Middle School Math Instruction, when schools and teachers fail to focus on delivering high-quality core instruction, delivering supplemental math support is not enough to address the needs of middle school students with math difficulty. Developing high-quality middle school math instruction starts with good preparation. Teachers who take the time to plan the math vocabulary, activities, questions, and means to assess student learning, prior to the lesson being taught, are more prepared to help their students be successful. When teachers plan, the lesson leads to proactively meeting student needs, rather than being reactive during the lesson. Alas, planning takes time and knowledge of evidence-based practices (EBPs) and instructional strategies to implement effective core instruction.
Problems in Current Lesson Plans
In one study (Opfer et al., 2016), 98% of secondary teachers said that they use materials “I developed and/or selected myself.” Secondary teachers reported using Google.com, their state’s department of education website, Pinterest, Khan Academy, and Teachers Pay Teachers to find lessons. The problem is not in finding lesson plans from various sites; rather, it is the teacher’s ability to review and determine if the lessons contain EBPs, are aligned to college and career readiness standards, and meet the needs of various students. In the Opfer et al. (2016) study, lessons were reviewed for quality, and it was found that websites such as Teachers Pay Teachers and Pinterest do not lead to high-quality lesson plans, even though 44% of teachers felt that materials from these websites offered opportunities to engage students in the use of mathematical language and symbols appropriately “to a great extent.” Yet, Bush and colleagues (2021) astutely point out “…. this also unfortunately means that 56 percent of teachers did not report using materials that use symbols and language appropriately, and about half of the teachers did not agree that they teach grade-level major mathematics topics addressed by state standards in a coherent way ‘to a great extent’” (p. 9). These findings speak to the importance of teachers increasing their own knowledge of EBPs as well as instructional practices to support varying student needs through scaffolds while maintaining high expectations and alignment to college and career readiness standards.
Evidence-Based Lesson Plans: Math Project
An Office of Special Education Programs-funded project, Teacher Instructional Practices for Algebraic Readiness (TIPS4AR), collaborated with middle school teachers across the United States to increase their understanding of high-quality EBPs. The lesson plans were created as part of a book and lesson study (see Infusing EBPs to Improve Middle School Math Instruction) across three cohorts of teachers, instructional math coaches, district math representatives, and speech and language pathologists. Key EBPs for math instruction were identified in The Math Pact, including using correct and consistent math vocabulary, precise math notation, multiple representations, and building math generalizations, instead of teaching tricks, shortcuts, or rules that will quickly or in future grades no longer promote conceptual understanding. These practices were coupled with high-leverage practices (HLPs) for the design and delivery of instruction to meet the needs of all students, specifically students who struggle. These high-leverage instructional practices include the use of explicit instruction (HLP 16), providing scaffolded supports (HLP 15), using strategies to promote active student engagement (HLP 18), and using student assessment data to analyze instructional practices and make necessary adjustments that lead to improved student outcomes (HLP 6). With these EBPs and HLPs in mind, teachers developed lessons using a common template that encouraged teachers to think about and indicate how they would identify their lesson objectives, connect their warm-up activities to prerequisites or skills that have yet to be mastered, incorporate modeling and thinking aloud, involve interactive practice, include independent practice, incorporate practices to support math discourse, and assess student understanding.
To ensure that lesson plans were of high quality, project staff with extensive knowledge of math EBPs and classroom experience reviewed the lessons using a rubric. The rubric, which can easily be used by general and special education teachers, evaluated the lesson plans for EBPs and high-leverage instructional practices. This rubric can provide teachers with an initial quality check of math lessons/materials to verify that EBPs are evident. Any lessons scoring below 85% were edited by math and special educators and project staff to ensure they met the criteria in the rubric and to include additional information to verify that lessons were clear and could easily be implemented in middle school math classrooms.
Example Lesson Plans
A total of 30 high-quality lesson plans for middle school teachers were created. The lessons can be taught in core math classes as well as special education classrooms. The lessons include the following grade levels and topics:
5th-Grade Lesson
Document
Use Properties to Multiply Fractions
6th-Grade Lessons
Document
Absolute Value
Document
An Introduction to Box and Whisker Plots
Document
Discovering Combining Like Terms
Document
Dividing Decimals
Document
Equivalent Ratios (Task Cards Resource)
Document
Find the Median (Example 1)
Document
Finding the Median (Example 2)
Document
Introduction to Combining Like Terms
Document
Introduction to Using a Protractor to Measure Angles
Document
Introduction to Inequalities
Document
Introduction to Ratio and Ratio Reasoning
Document
Mean, Median, and Mode with Box and Whisker Plots
Document
Solving One-Step Equations
Document
Surface Area of Prisms and Pyramids
7th-Grade Lessons
Document
Angle Vocabulary
Document
Introduction to Algebraic Tiles
Document
Understanding Division With Fractions
8th-Grade Lessons
Document
Exploring Exponents
Document
Graphs and Stories
Document
Let’s Ski! Slope in Action
Document
Pythagorean Theorem
Document
Reflections on the Coordinate Plane
Document
Scatter Plots
Document
Systems of Equations: Zap the Line
Document
The Slope of a Nonvertical Line
Document
Volume of Cylinders
This content was produced under U.S. Department of Education, Office of Special Education Programs, Award No. H326M17002. The views expressed herein do not necessarily represent the positions or polices of the U.S. Department of Education. No official endorsement by the U.S. Department of Education of any product, commodity, service, or enterprise mentioned in this website is intended or should be inferred.
References
Bush, S. B., Karp, K. S., & Dougherty, B. J. (2021). The math pact: Achieving instructional coherence within and across grades: Middle school. Corwin.
Opfer, V. D., Kaufman, J. H., & Thompson, L. E. (2016). Implementation of K–12 state standards for mathematics and English language arts and literacy: Findings from the American Teacher Panel. RAND Corporation.
#### About the Author
Steven Prater, MA, provides consultation and technical assistance to state and district leaders on MTSS. He has extensive expertise in implementation of MTSS at the school level, data-based decision making, and intensive intervention. In over 20 years in education, he has work… more
#### About the Author
Kathleen Pfannenstiel, PhD, provides technical assistance and professional learning to states and school districts with an emphasis on improving student outcomes through MTSS and evidence-based practices. Dr. |
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9328 | https://chemistry.stackexchange.com/questions/36913/why-is-sodium-sulfate-a-neutral-salt | inorganic chemistry - Why is sodium sulfate a neutral salt - Chemistry Stack Exchange
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Why is sodium sulfate a neutral salt [duplicate]
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Interesting ionization of sulfuric acid (2 answers)
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When I have a solution of N a X 2 S O X 4 N a X 2 S O X 4, the sodium sulfate should ionize:
N a X 2 S O X 4(a q)⟶2 N a X+(a q)+S O X 4 X 2−(a q)N a X 2 S O X 4(a q)⟶2 N a X+(a q)+S O X 4 X 2−(a q)
In my mind, the sodium ions should just float around, but because H S O X 4 X−H S O X 4 X− is a weak acid:
H S O X 4 X−(a q)+H X 2 O⟷H X 3 O X+(a q)+S O X 4 X 2−(a q)H S O X 4 X−(a q)+H X 2 O⟷H X 3 O X+(a q)+S O X 4 X 2−(a q)
The sulfate ion should be a weak conjugate base, and hence the sodium sulfate salt should be a weak base. This was my line of reasoning, until today my friends told me it is actually a neutral salt, which was confirmed by searching online. So why is it a neutral salt? (i.e. what's wrong with my reasoning?)
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edited Sep 5, 2015 at 17:36
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3 H S O−4 H S O 4− can hardly be called weak. It is an acid of medium strength, stronger than carbonic or acetic.Ivan Neretin –Ivan Neretin 2015-09-05 09:51:22 +00:00 Commented Sep 5, 2015 at 9:51
Weak ions conjugate bases do not accept protons.Aditya Dev –Aditya Dev 2015-09-05 10:06:53 +00:00 Commented Sep 5, 2015 at 10:06
Sulfate anions are weakly basic according to Bronsted theory - you're right, but it's called neutral salt according to Arrhenius theory in which it would need to contain OH- ions to be called basic.Mithoron –Mithoron 2015-09-05 16:44:53 +00:00 Commented Sep 5, 2015 at 16:44
The sulfate ion is indeed a weak base, so weak that it hardly participates in the equilibrium reaction you have written.Zhe –Zhe 2017-01-28 02:15:45 +00:00 Commented Jan 28, 2017 at 2:15
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For a solution of N a X 2 S O X 4 N a X 2 S O X 4, the sodium sulfate will ionize as you indicated:
N a X 2 S O X 4(s)−→a q 2 N a X+(a q)+S O X 4 X 2−(a q)N a X 2 S O X 4(s)→a q 2 N a X+(a q)+S O X 4 X 2−(a q)
This reaction is essentially irreversible in that the ions in an unsaturated solution won't suddenly form solid N a X 2 S O X 4 N a X 2 S O X 4 again.
H S O X 4 X−H S O X 4 X− is a not a weak acid, but has a pK a2=1.99 pK a2=1.99 which makes a moderately strong acid. So for pH > 3 essentially all the H S O X 4 X−H S O X 4 X− will ionize:
H S O X 4 X−(a q)+H X 2 O−→−−p H>3 H X 3 O X+(a q)+S O X 4 X 2−(a q)H S O X 4 X−(a q)+H X 2 O→p H>3 H X 3 O X+(a q)+S O X 4 X 2−(a q)
Thus S O X 4 X 2−S O X 4 X 2− is such a weak base that reverse reaction doesn't happen when you dissolve N a X 2 S O X 4 N a X 2 S O X 4, and a solution of N a X 2 S O X 4 N a X 2 S O X 4 in distilled water stays neutral rather than becoming basic.
H X 2 O+S O X 4 X 2−−>H S O X 4 X−+O H X−H X 2 O+S O X 4 X 2−−>H S O X 4 X−+O H X−
To think about a different example acetic acid is a weaker acid, pK a=4.76 pK a=4.76, so the acetate anion is a stronger base than the sulfate anion. A solution of sodium acetate would become weakly basic.
One last point - the statement should be "sodium sulfate a neutral salt in aqueous solution." In other solvents sodium sulfate could act as a base.
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edited Jan 28, 2017 at 4:42
answered Jan 28, 2017 at 1:15
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Here is the somewhat weak explanation given by Wikipedia:
Sodium sulfate is a neutral salt: its aqueous solutions exhibit a pH of 7. The neutrality of such solutions reflects the fact that sulfate is derived, formally, from the strong acid sulfuric acid. Furthermore, the Na+ ion, with only a single positive charge, only weakly polarizes its water ligands provided there are metal ions in solution. Sodium sulfate reacts with sulfuric acid to give the acid salt sodium bisulfate.
It definitely leaves some unanswered questions like what they mean by "metal ions in solutions": what metal ions (sodium ion?), what concentrations?
But the gist of it makes sense and makes a reasonable answer to me.
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edited Jun 10, 2020 at 14:04
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9329 | https://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week | Published Time: 2004-01-26T18:40:34Z
Determination of the day of the week - Wikipedia
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1 ConceptsToggle Concepts subsection
1.1 Corresponding days
1.2 Corresponding months
1.3 Corresponding years
1.4 Corresponding centuries
2 Tabular methods to calculate the day of the weekToggle Tabular methods to calculate the day of the week subsection
2.1 Complete table: Julian and Gregorian calendars
2.2 Revised Julian calendar
2.3 Dominical letter
2.4 The Doomsday
2.5 Result
2.5.1 The Sunday letter method
3 Mathematical algorithmsToggle Mathematical algorithms subsection
3.1 Rata Die
3.2 Gauss's algorithm
3.2.1 Worked example
3.2.2 Explanation and notes
3.2.3 Disparate variation
3.3 Zeller's algorithm
3.4 Wang's algorithm
4 Other algorithmsToggle Other algorithms subsection
4.1 Schwerdtfeger's method
4.2 Lewis Carroll's method
4.3 Methods in computer code
4.3.1 Keith
4.3.2 Sakamoto's methods
4.4 Gauss's method in MATLAB
4.5 Gauss's method for Gregorian calendar in Python
5 See also
6 ReferencesToggle References subsection
6.1 Notes
6.2 Citations
6.3 Further reading
7 External links
Determination of the day of the week
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From Wikipedia, the free encyclopedia
Methods to calculate the day of the week
The determination of the day of the week for any date may be performed with a variety of algorithms. In addition, perpetual calendars require no calculation by the user, and are essentially lookup tables. A typical application is to calculate the day of the week on which someone was born or a specific event occurred.
Concepts
[edit]
In numerical calculation, the days of the week are represented as weekday numbers. If Monday is the first day of the week, the days may be coded 1 to 7, for Monday through Sunday, as is practiced in ISO 8601. The day designated with 7 may also be counted as 0, by applying the arithmetic modulo 7, which calculates the remainder of a number after division by 7. Thus, the number 7 is treated as 0, the number 8 as 1, the number 9 as 2, the number 18 as 4, and so on. If Sunday is counted as day 1, then 7 days later (i.e.day 8) is also a Sunday, and day 18 is the same as day 4, which is a Wednesday since this falls three days after Sunday (i.e.18 mod 7 = 4).[a]
| Standard | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday | Usage examples |
--- --- --- ---
| ISO 8601 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | %_ISODOWI%,%@ISODOWI[]% (4DOS);DAYOFWEEK() (HP Prime) |
| | 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
| | 2 | 3 | 4 | 5 | 6 | 7 | 1 | %NDAY OF WEEK% (NetWare, DR-DOS); %_DOWI%,%@DOWI[]% (4DOS) |
| | 1 | 2 | 3 | 4 | 5 | 6 | 0 | HP financial calculators |
The basic approach of nearly all of the methods to calculate the day of the week begins by starting from an "anchor date": a known pair (such as 1 January 1800 as a Wednesday), determining the number of days between the known day and the day to determine, and using arithmetic modulo 7 to find a new numerical day of the week.
One standard approach is to look up (or calculate, using a known rule) the value of the first day of the week of a given century, look up (or calculate, using a method of congruence) an adjustment for the month, calculate the number of leap years since the start of the century, and then add these together along with the number of years since the start of the century, and the day number of the month. Eventually, applying modulo 7 to the day count allows determining the day of the week of that date.
Some methods do all the additions first and then cast out sevens, whereas others cast them out at each step, as in Lewis Carroll's method. Either way can have practical use: the former is easier for calculators and computer programs, whereas the latter is easier for mental calculation. None of the methods given here perform range checks, so unreasonable dates will produce erroneous results.
Corresponding days
[edit]
Every seventh day in a month has the same name as the previous:
| Day of the month | d |
--- |
| 00 07 14 21 28 | 0 |
| 01 08 15 22 29 | 1 |
| 02 09 16 23 30 | 2 |
| 03 10 17 24 31 | 3 |
| 04 11 18 25 | 4 |
| 05 12 19 26 | 5 |
| 06 13 20 27 | 6 |
Corresponding months
[edit]
"Corresponding months" are those months within the calendar year that start on the same day of the week. For example, September and December correspond, because 1 September falls on the same day as 1 December (as there are precisely thirteen 7-day weeks between the two dates). Months can only correspond if the number of days between their first days is divisible by 7, or in other words, if their first days are a whole number of weeks apart. For example, February of a common year corresponds to March because February has 28 days, a number divisible by 7, 28 days being exactly four weeks. In a leap year, January and February correspond to different months than in a common year, since adding 29 February means each subsequent month starts a day later.
January corresponds to October in common years and April and July in leap years. February corresponds to March and November in common years and August in leap years. March always corresponds to November, April always corresponds to July, and September always corresponds to December. August does not correspond to any other month in a common year. October doesn't correspond to any other month in a leap year. May and June never correspond to any other month.
In the months table below, corresponding months have the same number, a fact which follows directly from the definition.
| Common years | Leap years | All years | m |
--- --- |
| Jan | | Oct | 0 |
| | | May | 1 |
| | Feb | Aug | 2 |
| Feb | | Mar Nov | 3 |
| | | Jun | 4 |
| | | Sept Dec | 5 |
| | Jan | Apr July | 6 |
Corresponding years
[edit]
This section may contain unverified or indiscriminate information in embedded lists. Please help clean up the lists by removing items or incorporating them into the text of the article.(March 2021)
There are seven possible days that a year can start on, and leap years will alter the day of the week after 29 February. This means that there are 14 configurations that a year can have. All the configurations can be referenced by a dominical letter, but as 29 February has no letter allocated to it, a leap year has two dominical letters, one for January and February and the other (one step back in the alphabetical sequence) for March to December.
2021 is a common year starting on Friday, which means that it corresponds to the 2010 calendar year. The first two months of 2021 correspond to the first two months of 2016. 2022 is a common year starting on Saturday, which means that it corresponds to the 2011 calendar year. The last ten months of 2022 correspond to the last ten months of 2016. 2023 is a common year starting on Sunday, which means that it corresponds to the 2017 calendar year. 2024 is a leap year starting on Monday, which means that it corresponds to the 1996 calendar year. The first two months of 2024 correspond to the first two months of 2018. The last ten months of 2024 correspond to the last ten months of 2019.
Each leap year repeats once every 28 years, and every common year repeats once every 6 years and twice every 11 years. For instance, the last occurrence of a leap year starting on Wednesday was 2020 and the next occurrence will be 2048. Likewise, the next common years starting on Friday will be 2027, 2038, and then 2049. Both of these statements are true unless a leap year is skipped, of which will not happen until 2100.
For details, see the table below:
| Year of the century mod 28 | y |
--- |
| 00 06 12 17 23 | 0 |
| 01 07 12 18 24 | 1 |
| 02 08 13 19 24 | 2 |
| 03 08 14 20 25 | 3 |
| 04 09 15 20 26 | 4 |
| 04 10 16 21 27 | 5 |
| 05 11 16 22 00 | 6 |
Notes:
Black means all the months of a common year
Red means the first two months of a leap year
Blue means the last ten months of a leap year
Corresponding centuries
[edit]
| Julian century mod 700 | Gregorian century mod 400[b] | Day |
---
| 400: 1100 1800 ... | 300: 1500 1900 ... | Sun |
| 300: 1000 1700 ... | | Mon |
| 200 0900 1600 ... | 200: 1800 2200 ... | Tue |
| 100 0800 1500 ... | | Wed |
| 700: 1400 2100 ... | 100: 1700 2100 ... | Thu[c] |
| 600: 1300 2000 ... | | Fri |
| 500: 1200 1900 ... | 000: 1600 2000 ... | Sat |
"Year 000" is, in normal chronology, the year 1 BC (which precedes AD 1). In astronomical year numbering the year 0 comes between 1 BC and AD 1. In the proleptic Julian calendar, (that is, the Julian calendar as it would have been if it had been operated correctly from the start), 1 BC starts on Thursday. In the proleptic Gregorian calendar, (called such because it wasn't devised until 1582), 1 BC starts on Saturday.
Tabular methods to calculate the day of the week
[edit]
Complete table: Julian and Gregorian calendars
[edit]
For Julian dates before 1300 and after 1999 the year in the table which differs by an exact multiple of 700 years should be used. For Gregorian dates after 2299, the year in the table which differs by an exact multiple of 400 years should be used. The values "r0" through "r6" indicate the remainder when the Hundreds value is divided by 7 and 4 respectively, indicating how the series extend in either direction. Both Julian and Gregorian values are shown 1500–1999 for convenience. Bold figures (e.g., 04) denote leap year. If a year ends in 00 and its hundreds are in bold it is a leap year. Thus 19 indicates that 1900 is not a Gregorian leap year, (but 19 in the Julian column indicates that it is a Julian leap year, as are all Julian x 00 years). 20 indicates that 2000 is a leap year. Use Jan and Feb only in leap years.
| Hundreds of years | Remaining year digits | Month | DoW | # |
--- ---
| Julian (r ÷ 7) | Gregorian (r ÷ 4) |
| r519 | 1620r0 | 00 06 17 23 | 28 34 45 51 | 56 62 73 79 | 84 90 | Jan | | | | Oct | Sa | 0 |
| r418 | 15 19 r3 | 01 07 12 18 | 29 35 40 46 | 57 63 68 74 | 85 91 96 | | | May | | | Su | 1 |
| r317 | — | 02 13 19 24 | 30 41 47 52 | 58 69 75 80 | 86 97 | Feb | | | Aug | | M | 2 |
| r216 | 18 22 r2 | 03 08 14 25 | 31 36 42 53 | 59 64 70 81 | 87 92 98 | Feb | Mar | | | Nov | Tu | 3 |
| r115 | — | 09 15 20 26 | 37 43 48 54 | 65 71 76 82 | 93 99 | | | Jun | | | W | 4 |
| r014 | 17 21 r1 | 04 10 21 27 | 32 38 49 55 | 60 66 77 83 | 88 94 | | | | Sep | Dec | Th | 5 |
| r613 | — | 05 11 16 22 | 33 39 44 50 | 61 67 72 78 | 89 95 | Jan | Apr | Jul | | | F | 6 |
For determination of the day of the week (1 January 2000, Saturday)
the day of the month: 1 ~ 31 (1)
the month: (6)
the year: (0)
the century mod 4 for the Gregorian calendar and mod 7 for the Julian calendar (0).
adding 1+6+0+0=7. Dividing by 7 leaves a remainder of 0, so the day of the week is Saturday.
The formula is w = (d + m + y + c) mod 7.
Revised Julian calendar
[edit]
Note that the date (and hence the day of the week) in the Revised Julian and Gregorian calendars is the same from 14 October 1923 to 28 February AD 2800 inclusive and that for large years it may be possible to subtract 6300 or a multiple thereof before starting so as to reach a year which is within or closer to the table.
To look up the weekday of any date for any year using the table, subtract 100 from the year, divide the difference by 100, multiply the resulting quotient (omitting fractions) by seven and divide the product by nine. Note the quotient (omitting fractions). Enter the table with the Julian year, and just before the final division add 50 and subtract the quotient noted above.
The following is an example of calculating the day of the week for 27 January 8315:
8315 − 6300 = 2015, 2015 − 100 = 1915, 1915 ÷ 100 = 19 remainder 15, 19 × 7 = 133, 133 ÷ 9 = 14 remainder }. 2015 is 700 years ahead of 1315, so 1315 is used. From table: for hundreds (13): 6. For remaining digits (15): 4. For month (January): 0. For date (27): 27. {{{1}}}. {{{1}}}. Thus, the day of the week is Thursday.
Dominical letter
[edit]
To find the dominical letter, calculate the day of the week for either 1 January or 1 October, and assign a letter with Sunday corresponding to A, Saturday corresponding to B, ... and Monday corresponding to G.
Leap years have two Sunday letters, so for January and February calculate the day of the week for 1 January; and for March to December calculate the day of the week for 1 October.
Leap years are all years which divide exactly by four with the following exceptions:
In the Gregorian calendar – all years which divide exactly by 100 (other than those which divide exactly by 400).
In the Revised Julian calendar – all years which divide exactly by 100 (other than those which give remainder 200 or 600 when divided by 900).
The Doomsday
[edit]
Main article: Doomsday rule
This section needs expansion. You can help by adding to it. (July 2025)
The Doomsday algorithm is an artefact of recreational mathematics.
Result
[edit]
The following is a table for finding the day of the week without calculation.
| Index | Mon A | Tue B | Wed C | Thu D | Fri E | Sat F | Sun G | Perpetual Gregorian and Julian calendar Use Jan and Feb for leap years Date letter in year row for the letter in century row All the C days are doomsdays |
--- --- --- ---
| Julian century | Gregorian century | Date | 01 08 15 22 29 | 02 09 16 23 30 | 03 10 17 24 31 | 04 11 18 25 | 05 12 19 26 | 06 13 20 27 | 07 14 21 28 |
| 12 19 | 16 20 | Apr | Jul | Jan | G | A | B | C | D | E | F | 01 | 07 | 12 | 18 | | 29 | 35 | 40 | 46 | | 57 | 63 | 68 | 74 | | 85 | 91 | 96 |
| 13 20 | | Sep | Dec | | F | G | A | B | C | D | E | 02 | | 13 | 19 | 24 | 30 | | 41 | 47 | 52 | 58 | | 69 | 75 | 80 | 86 | | 97 |
| 14 21 | 17 21 | Jun | | | E | F | G | A | B | C | D | 03 | 08 | 14 | | 25 | 31 | 36 | 42 | | 53 | 59 | 64 | 70 | | 81 | 87 | 92 | 98 |
| 15 22 | | Feb | Mar | Nov | D | E | F | G | A | B | C | | 09 | 15 | 20 | 26 | | 37 | 43 | 48 | 54 | | 65 | 71 | 76 | 82 | | 93 | 99 |
| 16 23 | 18 22 | Aug | Feb | | C | D | E | F | G | A | B | 04 | 10 | | 21 | 27 | 32 | 38 | | 49 | 55 | 60 | 66 | | 77 | 83 | 88 | 94 | |
| 17 24 | | May | | | B | C | D | E | F | G | A | 05 | 11 | 16 | 22 | | 33 | 39 | 44 | 50 | | 61 | 67 | 72 | 78 | | 89 | 95 | |
| 18 25 | 19 23 | Jan | Oct | | A | B | C | D | E | F | G | 06 | | 17 | 23 | 28 | 34 | | 45 | 51 | 56 | 62 | | 73 | 79 | 84 | 90 | | 0 0 |
| [Year/100] | Gregorian century | 20 16 | | 21 17 | | 22 18 | | 23 19 | Year mod 100 |
| Julian century | 19 12 | 20 13 | 21 14 | 22 15 | 23 16 | 24 17 | 25 18 |
Examples:
For common method
26 December 1893 (GD)
December is in row F and 26 is in column E, so the letter for the date is C located in row F and column E. 93 (year mod 100) is in row D (year row) and the letter C in the year row is located in column G. 18 ([year/100] in the Gregorian century column) is in row C (century row) and the letter in the century row and column G is B, so the day of the week is Tuesday.
13 October 1307 (JD)
October 13 is a F day. The letter F in the year row (07) is located in column G. The letter in the century row (13) and column G is E, so the day of the week is Friday.
1 January 2000 (GD)
January 1 corresponds to G, G in the year row (0 0) corresponds to F in the century row (20), and F corresponds to Saturday.
A pithy formula for the method: "Date letter (G), letter (G) is in year row (0 0) for the letter (F) in century row (20), and for the day, the letter (F) become weekday (Saturday)".
The Sunday letter method
[edit]
Each day of the year (other than 29 February) has a letter allocated to it in the recurring sequence ABCDEFG. The series begins with A on 1 January and continues to A again on 31 December. The Sunday letter is the one which stands against all the Sundays in the year. Since 29 February has no letter, this means that the Sunday letter for March to December is one step back in the sequence compared to that for January and February. The letter for any date will be found where the row containing the month (in black) at the left of the "Latin square" meets the column containing the date above the "Latin square". The Sunday letter will be found where the column containing the century (below the "Latin square") meets the row containing the year's last two digits to the right of the "Latin square". For a leap year, the Sunday letter thus found is the one which applies to March to December.
So, for example, to find the weekday of 16 June 2020:
Column "20" meets row "20" at "D". Row "June" meets column "16" at "F". As F is two letters on from D, so the weekday is two days on from Sunday, i.e. Tuesday.
Mathematical algorithms
[edit]
Note on notation: In the following text, the backslash \ is used to indicate a "flooring integer division", and the percent sign % is used to indicate the remainder of such operation.
Rata Die
[edit]
The Rata Die method works by adding up the number of days d that has passed since a date of known day of the week D. The day of-the-week is then given by (D + d) mod 7, conforming to whatever convention was used to encode D.
For example, the date of 13 August 2009 is 733632 days from 1 January AD 1. Taking the number mod 7 yields 4, hence a Thursday.
Gauss's algorithm
[edit]
Carl Friedrich Gauss described a method for calculating the day of the week for 1 January in any given year in a handwritten note in a collection of astronomical tables. He never published it. It was finally included in his collected works in 1927. Compared to Rata Die, the result helps simplify the counting of years.
Gauss's method was applicable to the Gregorian calendar. He numbered the weekdays from 0 to 6 starting with Sunday. He defined the following operation.
Inputs Year number A, month number M, date number D.Output Day of year.Procedure
1. First determine the day-of-week d 1 of 1 January.
For a Gregorian calendar, the weekday is
(1 + 5((A−1)% 4) + 4((A−1)% 100) + 6((A−1)% 400))% 7. Alternatively, set C = A \ 100, Y = A% 100, and the value is
(1 + 5((Y−1)%4) + 3(Y−1) + 5(C%4))% 7.
For a Julian calendar, the weekday is
(6 + 5((A−1)%4) + 3(A−1))% 7 or
(6 + 5((Y−1)% 4) + 3(Y−1) + 6C)% 7.
Now determine the month-related offset m by using the lookup table with M.
Return d = (d 1 + m + D)% 7.
Table of month offsets | Months | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
| Common years | 0 | 3 | 3 | 6 | 1 | 4 | 6 | 2 | 5 | 0 | 3 | 5 |
| Leap years | 4 | 0 | 2 | 5 | 0 | 3 | 6 | 1 | 4 | 6 |
The above procedure can be condensed into a single expression for the Gregorian case:
(D + m + 5((A−1)%4) + 4((A−1)%100) + 6((A−1)%400))%7
Worked example
[edit]
For year number 2000, A − 1 = 1999, Y − 1 = 99 and C = 19, the weekday of 1 January is
= (1 + 5(1999%4) + 4(1999%100) + 6(1999%400))%7= (1 + 1 + 4 + 0)% 7= 6= (1 + 5(99%4) + 3 × 99 + 5(19%4))%7= (1 + 1 + 3 + 1)%7= 6 = Saturday.
For example, the weekdays for 30 April 1777 and 23 February 1855 are
= (30 + 6 + 5(1776%4) + 4(1776%100) + 6(1776%400))%7= (2 + 6 + 0 + 3 + 6)%7= 3 =Wednesday
and
= (6 + 23 + ⌈2.6 × 12⌉ + 5(1854%4) + 4(1854%100) + 6(1854%400))%7= (6 + 2 + 4 + 3 + 6 + 5)%7= 5 = Friday.
Explanation and notes
[edit]
The algorithm for the day-of-week of 1 Jan can be proven using modulo arithmetic. The main point is that because 365% 7 = 1, each year adds 1 day to the progression. The rest is adjustment for leap year. The century-based versions have 36525% 7 = 6.
The table of month offsets show a divergence in February due to the leap year. A common technique (later used by Zeller) is to shift the month to start with March, so that the leap day is at the tail of the counting. In addition, as later shown by Zeller, the table can be replaced with an arithmetic expression.
This formula was also converted into graphical and tabular methods for calculating any day of the week by Kraitchik and Schwerdtfeger.
Disparate variation
[edit]
The following formula is an example of a version without a lookup table. The year is assumed to begin in March, meaning dates in January and February should be treated as being part of the preceding year. The formula for the Gregorian calendar is
w=(d+⌊2.6 m−0.2⌋+y+⌊y 4⌋+⌊c 4⌋−2 c)mod 7,{\displaystyle w=\left(d+\lfloor 2.6m-0.2\rfloor +y+\left\lfloor {\frac {y}{4}}\right\rfloor +\left\lfloor {\frac {c}{4}}\right\rfloor -2c\right){\bmod {7}},}
where
d is the day of the month (1 to 31)
m is the shifted month (March = 1,...,February = 12)
Y is the year unless m is 11 = January or 12 = February which are considered part of the preceding year, giving Y = year − 1
c is the century given by c=⌊Y/100⌋{\displaystyle c=\lfloor Y/100\rfloor }
y is the year relative to the century, given by y=Y−100 c{\displaystyle y=Y-100c}, or simply the last 2 digits of Y
w is the day of the week (0 = Sunday,...,6 = Saturday)
Table of month offsets (⌊2.6 m−0.2⌋{\displaystyle \lfloor 2.6m-0.2\rfloor }) | Months | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Jan | Feb |
| Offset | 2 | 5 | 7 | 10 | 12 | 15 | 18 | 20 | 23 | 25 | 28 | 31 |
Zeller's algorithm
[edit]
Main article: Zeller's congruence
In Zeller's algorithm, the months are numbered from 3 for March to 14 for February. The year is assumed to begin in March; this means, for example, that January 1995 is to be treated as month 13 of 1994. The formula for the Gregorian calendar is
w≡(⌊13(m+1)5⌋+⌊y 4⌋+⌊c 4⌋+d+y−2 c)mod 7{\displaystyle w\equiv \left(\left\lfloor {\frac {13(m+1)}{5}}\right\rfloor +\left\lfloor {\frac {y}{4}}\right\rfloor +\left\lfloor {\frac {c}{4}}\right\rfloor +d+y-2c\right){\bmod {7}}} where
d{\displaystyle d} is the day of the month (1 to 31)
m{\displaystyle m} is the shifted month (March=3,...January = 13, February=14)
Y{\displaystyle Y} is the year unless m{\displaystyle m} is 13 = January or 14 = February which are considered part of the preceding year giving Y=year−1{\displaystyle Y={\textit {year}}-1}
c{\displaystyle c} is the century given by c=⌊Y/100⌋{\displaystyle c=\lfloor Y/100\rfloor }
y{\displaystyle y} is the year relative to the century, given by y=Y−100 c{\displaystyle y=Y-100c}, or simply the last 2 digits of Y{\displaystyle Y}
w{\displaystyle w} is the day of week (1 = Sunday,..0 = Saturday)
Table of month offsets (⌊2.6(m+1)⌋{\displaystyle \lfloor 2.6(m+1)\rfloor }) | Months | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec | Jan | Feb |
| Offset | 10 | 13 | 15 | 18 | 20 | 23 | 26 | 28 | 31 | 33 | 36 | 39 |
The only difference is one between Zeller's algorithm (Z) and the Disparate Gaussian algorithm (G), that is Z − G = 1 = Sunday.
(d+⌊(m+1)2.6⌋+y+⌊y/4⌋+⌊c/4⌋−2 c)mod 7−(d+⌊2.6 m−0.2⌋+y+⌊y/4⌋+⌊c/4⌋−2 c)mod 7{\displaystyle (d+\lfloor (m+1)2.6\rfloor +y+\lfloor y/4\rfloor +\lfloor c/4\rfloor -2c){\bmod {7}}-(d+\lfloor 2.6m-0.2\rfloor +y+\lfloor y/4\rfloor +\lfloor c/4\rfloor -2c){\bmod {7}}}=(⌊(m+2+1)2.6−(2.6 m−0.2)⌋)mod 7{\displaystyle =(\lfloor (m+2+1)2.6-(2.6m-0.2)\rfloor ){\bmod {7}}} (March = 3 in Z but March = 1 in G)=(⌊2.6 m+7.8−2.6 m+0.2⌋)mod 7{\displaystyle =(\lfloor 2.6m+7.8-2.6m+0.2\rfloor ){\bmod {7}}}=8 mod 7=1{\displaystyle =8{\bmod {7}}=1}
Wang's algorithm
[edit]
Wang's algorithm for human calculation of the Gregorian calendar is (the formula should be subtracted by 1 if m is 1 or 2 if the year is a leap year)
w=(d−d 0(m)+y 0−y 1+⌊y 0/4−y 1/2⌋−2(c mod 4))mod 7,{\displaystyle w=\left(d-d_{0}(m)+y_{0}-y_{1}+\left\lfloor y_{0}/4-y_{1}/2\right\rfloor -2\left(c{\bmod {4}}\right)\right){\bmod {7}},} where
y 0{\displaystyle y_{0}} is the last digit of the year (units)
y 1{\displaystyle y_{1}} is the second last digit of the year (tens)
c{\displaystyle c} is the century, given by c=⌊year/100⌋{\displaystyle c=\lfloor {\textit {year}}/100\rfloor }
d{\displaystyle d} is the day of the month (1 to 31)
m{\displaystyle m} is the month (January=1,...,December=12)
w{\displaystyle w} is the day of the week (0=Sunday,...,6=Saturday)
d 0(m){\displaystyle d_{0}(m)} is the null-days function (month offset) with values listed in the following table
| m | d 0(m){\displaystyle d_{0}(m)} |
--- |
| 1 | 1 | A day |
| 3 | 5 | m + 2 |
| 5 | 7 |
| 7 | 9 |
| 9 | 3 | m + 1 |
| 11 | 12 |
| 2 | 12 | m + 3 |
| 4 | 2 | m − 2 |
| 6 | 4 |
| 8 | 6 |
| 10 | 8 |
| 12 | 10 |
An algorithm for the Julian calendar can be derived from the algorithm above w=(d−d 0(m)+y 0−y 1+⌊y 0/4−y 1/2⌋−c)mod 7,{\displaystyle w=\left(d-d_{0}(m)+y_{0}-y_{1}+\left\lfloor y_{0}/4-y_{1}/2\right\rfloor -c\right){\bmod {7}},} where d 0(m){\displaystyle d_{0}(m)} is a doomsday.
| m | d 0(m){\displaystyle d_{0}(m)} |
--- |
| 1 | 3 | C day |
| 3 | 7 | m + 4 |
| 5 | 9 |
| 7 | 11 |
| 9 | 5 | m − 4 |
| 11 | 7 |
| 2 | 0 | m − 2 |
| 4 | 4 | m |
| 6 | 6 |
| 8 | 8 |
| 10 | 10 |
| 12 | 12 |
Other algorithms
[edit]
Schwerdtfeger's method
[edit]
In a partly tabular method by Schwerdtfeger, the year is split into the century and the two digit year within the century. The approach depends on the month. For m ≥ 3,
c=⌊y 100⌋and g=y−100 c,{\displaystyle c=\left\lfloor {\frac {y}{100}}\right\rfloor \quad {\text{and}}\quad g=y-100c,}
so g is between 0 and 99. For m = 1,2,
c=⌊y−1 100⌋and g=y−1−100 c.{\displaystyle c=\left\lfloor {\frac {y-1}{100}}\right\rfloor \quad {\text{and}}\quad g=y-1-100c.}
The formula for the day of the week is
w=(d+e+f+g+⌊g 4⌋)mod 7,{\displaystyle w=\left(d+e+f+g+\left\lfloor {\frac {g}{4}}\right\rfloor \right){\bmod {7}},}
where the positive modulus is chosen.
The value of e is obtained from the following table:
| m | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
--- --- --- --- --- ---
| e | 0 | 3 | 2 | 5 | 0 | 3 | 5 | 1 | 4 | 6 | 2 | 4 |
The value of f is obtained from the following table, which depends on the calendar. For the Gregorian calendar,
| c mod 4 | 0 | 1 | 2 | 3 |
--- ---
| f | 0 | 5 | 3 | 1 |
For the Julian calendar,
| c mod 7 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
--- --- --- --- |
| f | 5 | 4 | 3 | 2 | 1 | 0 | 6 |
Lewis Carroll's method
[edit]
This section is missing information about transcription in mathematical expressions for easier comparison. Please expand the section to include this information. Further details may exist on the talk page.(March 2021)
Charles Lutwidge Dodgson (Lewis Carroll) devised a method resembling a puzzle, yet partly tabular in using the same index numbers for the months as in the "Complete table: Julian and Gregorian calendars" above. He lists the same three adjustments for the first three months of non-leap years, one 7 higher for the last, and gives cryptic instructions for finding the rest; his adjustments for centuries are to be determined using formulas similar to those for the centuries table. Although explicit in asserting that his method also works for Old Style dates, his example reproduced below to determine that "1676, February 23" is a Wednesday only works on a Julian calendar which starts the year on January 1, instead of March 25 as on the "Old Style" Julian calendar.
Algorithm:
Take the given date in 4 portions, viz. the number of centuries, the number of years over, the month, the day of the month.
Compute the following 4 items, adding each, when found, to the total of the previous items. When an item or total exceeds 7, divide by 7, and keep the remainder only.
Century-item: For 'Old Style' (which ended 2 September 1752) subtract from 18. For 'New Style' (which began 14 September 1752) divide by 4, take overplus [surplus] from 3, multiply remainder by 2.
Year-item: Add together the number of dozens, the overplus, and the number of 4s in the overplus.
Month-item: If it begins or ends with a vowel, subtract the number, denoting its place in the year, from 10. This, plus its number of days, gives the item for the following month. The item for January is "0"; for February or March, "3"; for December, "12".
Day-item: The total, thus reached, must be corrected, by deducting "1" (first adding 7, if the total be "0"), if the date be January or February in a leap year, remembering that every year, divisible by 4, is a Leap Year, excepting only the century-years, in 'New Style', when the number of centuries is not so divisible (e.g. 1800).
The final result gives the day of the week, "0" meaning Sunday, "1" Monday, and so on.
Examples:
1783, September 18
17, divided by 4, leaves "1" over; 1 from 3 gives "2"; twice 2 is "4". 83 is 6 dozen and 11, giving 17; plus 2 gives 19, i.e. (dividing by 7) "5". Total 9, i.e. "2" The item for August is "8 from 10", i.e. "2"; so, for September, it is "2 plus 31", i.e. "5" Total 7, i.e. "0", which goes out. 18 gives "4". Answer, "Thursday".
1676, February 23
16 from 18 gives "2" 76 is 6 dozen and 4, giving 10; plus 1 gives 11, i.e. "4". Total "6" The item for February is "3". Total 9, i.e. "2" 23 gives "2". Total "4" Correction for Leap Year gives "3". Answer, "Wednesday".
Dates before 1752 would in England be given Old Style with 25 March as the first day of the new year. Carroll's method however assumes 1 January as the first day of the year, thus he fails to arrive at the correct answer, namely "Friday".
Had he noticed that 1676, February 23 (with 25 March as New Year's Day) is actually 1677, February 23 (with 1 January as New Year's Day), he would have accounted for differing year numbers—just like George Washington's birthday differs—between the two calendars. Then his method yields:
1677 (corrected), February 23
16 from 18 gives "2" 77 is 6 dozen and 5, giving 11; plus 1 gives 12, i.e. "5". Total "7" The item for February is "3". Total 10, i.e. "3" 23 gives "2". Total "5". Answer, "Friday".
It is noteworthy that those who have republished Carroll's method have failed to point out his error, most notably Martin Gardner.
In 1752, the British Empire abandoned its use of the Old StyleJulian calendar upon adopting the Gregorian calendar, which has become today's standard in most countries of the world. For more background, see Old Style and New Style dates.
Methods in computer code
[edit]
Keith
[edit]
In the C language expressions below, y, m and d are, respectively, integer variables representing the year (e.g., 1988), month (1–12) and day of the month (1-31).
(d+=m<3 ? y-- : y-2,23m/9+d+4+y/4-y/100+y/400) % 7
In 1990, Michael Keith and Tom Craver published the foregoing expression that seeks to minimize the number of keystrokes needed to enter a self-contained function for converting a Gregorian date into a numerical day of the week. It returns 0 = Sunday, 1 = Monday, etc. This expression uses a less cumbersome month component than does Zeller's algorithm.
Shortly afterwards, Hans Lachman streamlined their algorithm for ease of use on low-end devices. As designed originally for four-function calculators, his method needs fewer keypad entries by limiting its range either to A.D. 1905–2099, or to historical Julian dates. It was later modified to convert any Gregorian date, even on an abacus. On Motorola 68000-based devices, there is similarly less need of either processor registers or opcodes, depending on the intended design objective.
Sakamoto's methods
[edit]
The tabular forerunner to Tøndering's algorithm is embodied in the following K&R C function. With minor changes, it was adapted for other high level programming languages such as APL2. Posted by Tomohiko Sakamoto on the comp.lang.c Usenet newsgroup in 1992, it is accurate for any Gregorian date.
dayofweek(y, m, d)/ 1 <= m <= 12, y > 1752 (in the U.K.) /
{
static int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
if ( m < 3 )
{
y -= 1;
}
return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
}
It returns 0 = Sunday, 1 = Monday, etc.
Sakamoto also simultaneously posted a more obfuscated version:
dow(m,d,y) { y -= m<3; return (y+y/4-y/100+y/400+"-bed=pen+mad."[m]+d) % 7; }
This version encodes the month offsets in the string and as a result requires a computer that uses standard ASCII to run the algorithm correctly, reducing its portability. In addition, both algorithms omit inttype declarations, which is allowed in the original K&R C but not allowed in ANSI C.
(Tøndering's algorithm is, again, similar in structure to Zeller's congruence and Keith's short code, except that the month-related component is 31m/12. Sakamoto's is somewhere between the Disparate Gaussian and the Schwerdtfeger's algorithm, apparently unaware of the expression form.)
Gauss's method in MATLAB
[edit]
% example date input
y1 = 2022;
m1 = 1;
d1 = 1;
month_offset = [0 3 3 6 1 4 6 2 5 0 3 5]; % common year
% offset if y1 leap year
if mod(y1,4)==0 && mod(y1,100)==0 && mod(y1,400)==0
month_offset=[0 3 4 0 2 5 0 3 6 1 4 6]; % leap year
end
% Gregor
weekday_gregor = rem( d1+month_offset(m1) + 5rem(y1-1,4) + 4rem(y1-1,100) + 6rem(y1-1,400),7)
% Julian
weekday_julian = rem(6+5rem(y1-1,4) + 3(y1-1),7)
0: Sunday 1: Monday .. 6: Saturday
Gauss's method for Gregorian calendar in Python
[edit]
from numpy import remainder as rem
def is_leap_year(year: int) -> bool:
"""Determine whether a year is a leap year."""
return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
def day_of_week(y: int, m: int, d: int) -> str:
"""Return day of week of given date as string,
using Gauss's algorithm to find it."""
if is_leap_year(y):
month_offset = (0, 3, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6)[m - 1]
else:
month_offset = (0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5)[m - 1]
y -= 1
wd = int(rem(d + month_offset + 5 rem(y, 4) + 4 rem(y, 100) \
+ 6 rem(y, 400) , 7))
return ("Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat")[wd]
See also
[edit]
Doomsday rule
Julian day
Mental Calculation World Cup (Has a calendar calculation contest)
Perpetual calendar
Buddhist calendar
References
[edit]
Notes
[edit]
^To illustrate this in detail it is useful to visualise a calendar depicting a month beginning on a Sunday, i.e.the 1st of the month is a Sunday). Counting forward seven days brings us to the 8th, which is also a Sunday. Counting forward another ten days brings us to the 18th, which is a Wednesday. If instead we were to start on Wednesday the 4th (three days after Sunday the 1st), count forward seven days to Wednesday the 11th (three days after Sunday the 8th), then count forward another seven days, we would end up on Wednesday the 18th–again–three days after Sunday the 15th, which itself falls exactly two weeks (two Sundays) after Sunday the 1st.
^The numbers in the first column are proleptic – the Gregorian calendar was not devised till 1582. See the note beneath the table.
^The Julian century beginning 1 BC would also appear on this line of the table (to the left of 700) but there is no space to include it.
Citations
[edit]
^ Jump up to: abBrothers, Hardin; Rawson, Tom; Conn, Rex C.; Paul, Matthias R.; Dye, Charles E.; Georgiev, Luchezar I. (2002-02-27). 4DOS 8.00 online help.
^"HP Prime - Portal: Firmware update" (in German). Moravia Education. 2015-05-15. Archived from the original on 2016-11-05. Retrieved 2015-08-28.
^Paul, Matthias R. (1997-07-30). NWDOS-TIPs — Tips & Tricks rund um Novell DOS 7, mit Blick auf undokumentierte Details, Bugs und Workarounds. Release 157 (in German) (3rd ed.). Archived from the original on 2016-11-04. Retrieved 2014-08-06. (NB. NWDOSTIP.TXT is a comprehensive work on Novell DOS 7 and OpenDOS 7.01, including the description of many undocumented features and internals. It is part of the author's yet larger MPDOSTIP.ZIP collection maintained up to 2001 and distributed on many sites at the time. The provided link points to a HTML-converted older version of the NWDOSTIP.TXT file.)
^Richards, E. G. (1999). Mapping Time: The Calendar and Its History. Oxford University Press. ISBN978-0-19-850413-9.
^ Jump up to: abGauss, Carl F. (1981). "Den Wochentag des 1. Januar eines Jahres zu finden. Gueldene Zahl. Epakte. Ostergrenze.". Werke. herausgegeben von der Koeniglichen Gesellschaft der Wissenschaften zu Goettingen (2nd ed.). Hildesheim: Georg Olms Verlag. pp.206–207. ISBN978-3-48704643-3.
^ Jump up to: abcdefSchwerdtfeger, Berndt E. (2010-05-07). "Gauss' calendar formula for the day of the week"(PDF) (1.4.26 ed.). Retrieved 2012-12-23.
^Kraitchik, Maurice (2006). "Chapter 5: The calendar". Mathematical recreations (2nd revised [Dover]ed.). Mineola: Dover Publications. pp.109–116. ISBN978-0-48645358-3.
^Rosen, Kenneth H. (2011). Elementary Number Theory and Its Applications. Addison Wesley. pp.134–137. ISBN978-0321500311.
^Stockton, J. R. (2010-03-19). "The Calendrical Works of Rektor Chr. Zeller: The Day-of-Week and Easter Formulae". Merlyn. Archived from the original on 2013-07-29. Retrieved 2012-12-19.
^Wang, Xiang-Sheng (March 2015). "Calculating the day of the week: null-days algorithm"(PDF). Recreational Mathematics Magazine. No.3. p.5.
^ Jump up to: abDodgson, C.L. (Lewis Carroll). (1887). "To find the day of the week for any given date". Nature, 31 March 1887. Reprinted in Mapping Time, pp.299-301.
^Martin Gardner. (1996). The Universe in a Handkerchief: Lewis Carroll's Mathematical Recreations, Games, Puzzles, and Word Plays, pages 24-26. Springer-Verlag.
^ Michael Keith; Tom Craver. (1990). The ultimate perpetual calendar? Journal of Recreational Mathematics, 22:4, pp.280-282.
^ The 4-function Calculator; The Assembly of Motorola 68000 Orphans; The Abacus. gopher://sdf.org/1/users/retroburrowers/TemporalRetrology
^"Day-of-week algorithm NEEDED!" news:1993Apr20.075917.16920@sm.sony.co.jp
^APL2 IDIOMS workspace: Date and Time Algorithms, line 15. (2002)
^"Date -> Day of week conversion". Google newsgroups:comp.lang.c. December 1992. Retrieved 2020-06-21.
^"DOW algorithm". Google newsgroups:comp.lang.c. 1994. Retrieved 2020-06-21.
Further reading
[edit]
Hale-Evans, Ron (2006). "Hack #43: Calculate any weekday". Mind performance hacks (1st ed.). Beijing: O'Reilly. pp.164–169. ISBN9780596101534.
Thioux, Marc; Stark, David E.; Klaiman, Cheryl; Schultz, Robert T. (2006). "The day of the week when you were born in 700 ms: Calendar computation in an autistic savant". Journal of Experimental Psychology: Human Perception and Performance. 32 (5): 1155–1168. doi:10.1037/0096-1523.32.5.1155. PMID17002528.
Treffert, Darold A. (2011-10-12). "Why calendar calculating?". Islands of genius: the bountiful mind of the autistic, acquired, and sudden savant (1. publ., [repr.].ed.). London: Jessica Kingsley. pp.63–66. ISBN9781849058735.
External links
[edit]
Tøndering's algorithm for both Gregorian and Julian calendars
"Key Day" method used so as to reduce computation & memorization
Compact tabular method for memorisation, also for the Julian calendar
When countries changed from the Julian calendar
World records for mentally calculating the day of the week in the Gregorian Calendar
National records for finding Calendar Dates
World Ranking of Memoriad Mental Calendar Dates (all competitions combined)
Identify the year by given month, day, day of week.Archived 2018-01-04 at the Wayback Machine
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Trying to prove that derivative is increasing and then decreasing
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For the function to be unimodal, I have found the derivative of the distribution which is: d d x f(x)=x(a−2)[(a−1)−a x]d d x f(x)=x(a−2)[(a−1)−a x]
All I am stuck with is in showing that the function increases and then decreases. Thanks!
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edited Oct 1, 2013 at 19:40
Tyler
2,907 3 3 gold badges 26 26 silver badges 42 42 bronze badges
asked Oct 1, 2013 at 18:08
user30438user30438
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I found somewhere,that because it has one sign change, it is unimodal. But wouldn't there be two cases, one when a < 2 and other when a > 2 user30438 –user30438 2013-10-01 18:14:03 +00:00 Commented Oct 1, 2013 at 18:14
Can you make any statements about the second derivative?jbowman –jbowman 2013-10-01 18:14:04 +00:00 Commented Oct 1, 2013 at 18:14
1 For a function to increase and then decrease, it means that it reaches a maximum. And what are the necessary and sufficient conditions for a maximum?Alecos Papadopoulos –Alecos Papadopoulos 2013-10-01 18:14:17 +00:00 Commented Oct 1, 2013 at 18:14
@AlecosPapadopoulos Maximum generally occurs at mean.user30438 –user30438 2013-10-01 18:16:26 +00:00 Commented Oct 1, 2013 at 18:16
No, this presupposes symmetry. Forget that this is a density, think of it as just any function of which you want to find its maximum Alecos Papadopoulos –Alecos Papadopoulos 2013-10-01 18:26:09 +00:00 Commented Oct 1, 2013 at 18:26
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I’m assuming that a>0 a>0. Assume for the moment that x>0 x>0. Then the only zero of the derivative is at x=a−1 a=1−1 a x=a−1 a=1−1 a. If 0<x<1−1 a 0<x<1−1 a, then a x0(a−1)−a x>0; x a−2>0 x a−2>0 as well, so the derivative is positive, and f f is increasing. If x>1−1 a x>1−1 a, then by a similar calculation (a−1)−a x<0(a−1)−a x<0, and x a−2 x a−2 is still positive, so the derivative is negative, and f f is decreasing. Thus, f f has a local maximum at x=1−1 a x=1−1 a.
If a>2 a>2, the derivative has another zero at x=0 x=0. However, for negative x x the power x a−2 x a−2 is in general undefined, so I’m guessing that either your function is defined only for non-negative x x, in which case there’s nothing more to check.
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answered Oct 1, 2013 at 19:31
Brian M. ScottBrian M. Scott
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Thanks for the explanation! Yes the function is defined for 0≤x≤1 0≤x≤1. I have a doubt that would the zero not exist when x(a−2)=0 x(a−2)=0 (which I think you considered in the case when a>2 a>2). Are we assuming in the first case that 0<a<2 0<a<2, where it is clear that zero exist at x=a−1 a x=a−1 a user30438 –user30438 2013-10-01 22:01:11 +00:00 Commented Oct 1, 2013 at 22:01
@user30438: For all a>0 a>0 there’s a zero at x=a−1 a x=a−1 a; for a>2 a>2 there’s also one at x=0 x=0, but it corresponds to the minimum at x=0 x=0, so it doesn’t matter.Brian M. Scott –Brian M. Scott 2013-10-02 07:16:36 +00:00 Commented Oct 2, 2013 at 7:16
should not a be greater than 1, otherwise x could take negative values as we want (a−1)−a x>0(a−1)−a x>0 user30438 –user30438 2013-10-02 19:40:03 +00:00 Commented Oct 2, 2013 at 19:40
@user30438: If 02 a>2.Brian M. Scott –Brian M. Scott 2013-10-02 19:51:09 +00:00 Commented Oct 2, 2013 at 19:51
The point I am making is that for (a−1)−a x>0(a−1)−a x>0, shouldn't a>1, or else how would it be > 0 .user30438 –user30438 2013-10-02 20:20:32 +00:00 Commented Oct 2, 2013 at 20:20
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9331 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4624318/ | Raynaud’s phenomenon and digital ischemia: a practical approach to risk stratification, diagnosis and management - PMC
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Int J Clin Rheumtol
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Published in final edited form as: Int J Clin Rheumtol. 2010;5(3):355–370. doi: 10.2217/ijr.10.17
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Raynaud’s phenomenon and digital ischemia: a practical approach to risk stratification, diagnosis and management
Zsuzsanna H McMahan
Zsuzsanna H McMahan, MD
1 Division of Rheumatology, Johns Hopkins University School of Medicine, Baltimore, MD 21224
Find articles by Zsuzsanna H McMahan
1, Fredrick M Wigley
Fredrick M Wigley, MD
1 Division of Rheumatology, Johns Hopkins University School of Medicine, Baltimore, MD 21224
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1 Division of Rheumatology, Johns Hopkins University School of Medicine, Baltimore, MD 21224
✉
Address Correspondence to: Fredrick M. Wigley, MD, Professor of Medicine, Johns Hopkins University School of Medicine, Baltimore, MD 21224, fwig@jhmi.edu, Telephone: 410-550-2003, Fax: 410-550-2072
PMC Copyright notice
PMCID: PMC4624318 NIHMSID: NIHMS638014 PMID: 26523153
The publisher's version of this article is available at Int J Clin Rheumtol
Abstract
Digital ischemia is a painful and often disfiguring event. Such an ischemic event often leads to tissue loss and can significantly affect the patient’s quality of life. Digital ischemia can be secondary to a vasculopathy, vasculitis, embolic disease, trauma, or extrinsic vascular compression. It is an especially serious complication in patients with scleroderma. Risk stratification of patients with scleroderma at risk for digital ischemia is now possible with clinical assessment and autoantibody profiles. Because there are a variety of conditions that lead to digital ischemia, it is important to understand the pathophysiology underlying each ischemic presentation in order to target therapy appropriately. Significant progress has been made in the last two decades in defining the pathophysiological processes leading to digital ischemia in rheumatic diseases. In this article we review the risk stratification, diagnosis, and management of patients with digital ischemia and provide a practical approach to therapy, particularly in scleroderma.
Keywords: digital ischemia, scleroderma, digital ulcers, Raynaud’s phenomenon
Introduction
Inadequate blood flow to living tissue is an excruciatingly painful experience and threatens the life of the tissue involved. An ischemic event to digital tissue is comparable to a pulmonary embolism or a myocardial infarction in that viability of the affected tissue is often lost and can significantly affect the patient’s quality of life. Death of digital tissue not only results in both disfigurement and functional disability, it also is the clinical manifestation of an underlying systemic disease process. Although the differential for digital ischemia is broad, in this review we focus on digital ischemia in the setting of rheumatic disease, particularly in scleroderma.
Digital ischemia is an especially serious complication in patients with scleroderma. Morbidity from digital ischemia is remarkably high in patients with this rheumatic disease; 30% of patients with persistent digital ulcers develop irreversible tissue loss (Ingraham KM, Steen VD. Arthritis and Rheumatism 2006; 54(9 Suppl.):P578) and it often requires hospitalization. As a result, when ischemia threatens the livelihood of a digit, rapid aggressive actions must be taken to prevent permanent damage. Amputation is reported to occur in one or more digits due to ischemia in 20.4% of patients with scleroderma, 9.2% of which have multiple digit loss .
Digital tissue vitality can be threatened by many pathological processes that compromise arterial blood supply such as thrombosis, a vasculopathy, vasculitis, embolic, and traumatic; all complicated by secondary vasospasm. Because all etiologies of digital ischemia are not alike, it is important to understand the pathophysiology underlying each ischemic presentation in order to target therapy appropriately. Significant progress has been made in the last two decades in defining the pathophysiological processes leading to digital ischemia in rheumatic diseases. This knowledge has lead to many new treatment options. Because digital ischemia is present in greater than 95% of patients with scleroderma, it will be used as our model of the disease process and to review our approach to management.
Identifying Patients at Risk
Digital ischemia results from an inadequate supply of oxygenated blood to digital tissue. The presence of digital pain associated with pallor or cyanosis of the skin of the affected digit(s) is the first clinical sign of impending digital tissue loss. When confronted with a painful discolored digit, it is imperative to quickly determine the likely etiology of the ischemia so that appropriate therapy can be promptly initiated to prevent tissue injury. An immediate assessment for predisposing risk factors for vascular disease should be conducted to elucidate the cause of the ischemic event (See Figure 1).
Fig. 1. Differential for patients presenting with Critical Digital ischemia.
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Causes of Vasculitis; Other: Malignancy, Infection
Digital ulcers are representative of vascular involvement in scleroderma and occur in about 30–50% of patients with scleroderma . Predicting the patients who are at high risk for the development of significant vascular involvement and subsequent digital loss is important so that these high risk patients can be monitored closely to consider preventive measures and treated early when ischemia occurs. Autoantibodies and microvascular damage, as seen in nailfold capillary microscopy, are independent risk factors for predicting that the presence of Raynaud’s phenomenon is the manifestation of an underlying scleroderma disease process. Nailfold capillaries can be examined in the office using emersion oil placed on the skin and viewing the capillaries with either an ophthalmoscope or bifocal microscope. Dilated or enlarged loops with areas of capillary loss are seen in patients with scleroderma. Houtman and colleagues determined that there is an inverse relationship between the severity of Raynaud’s at first presentation and the capillary density in patients with connective tissue disease. These findings were specifically noted in patients with manifestations of the scleroderma phenotype such as sclerodactyly, digital ulcers, tuft resorption, and telangiectasias . Other studies found that the association of the scleroderma specific antibodies, anti-topo I and anti-centromere antibody (ACA), with nailfold capillary microscopy increased the sensitivity in predicting that the presence of Raynaud’s is associated with a connective tissue disease . Patients presenting with Raynaud’s phenomenon alone without a definite diagnosis who have a known scleroderma related autoantibody (ACA, anti-Th/To, anti-topoisomerase I, or anti-RNA polymerase III antibodies) and nailfold capillary changes are 60 times more likely to develop definite scleroderma than patients with Raynaud’s phenomenon and normal capillary and negative serology. Patients with scleroderma who are positive for ACA are at an increased risk for severe digital ischemia and macrovascular events with digital loss [1, 7]. In addition, among scleroderma patients, anti-Scl70 positivity and early onset Raynaud’s is associated with an increased incidence of digital ulcers . Likewise, scleroderma patients with anti-PM/Scl-75/100 antibodies are also more likely to develop digital ulcers .
Therefore, it is now understood that a positive anti-nuclear antibody (ANA), especially with positive scleroderma specific antibodies, and abnormal nailfold capillaries as detected by capillaroscopy, is predictive of secondary connective tissue disease and is not seen in primary Raynaud’s phenomenon [10–12]. ACA and anti-Th/To both predict capillary enlargement, and these antibodies and anti-RNP predict capillary loss. Interestingly, each of these antibodies was associated with a distinct rate of microvascular damage. Koenig and colleagues followed digital vascular changes defined by nailfold capillary microscopy and noted that enlarged capillaries occurred earlier in patients with anti-RNA polymerase III than in patients with anti-Th/Tho antibodies. The nailfold capillary changes occur the latest in the disease course in patients with ACA .
Recently, Caramaschi and colleagues studied the risk factors for ischemic digital ulcers (“DUs”) in patients with scleroderma. They found that scleroderma patients with ischemic DUs are characterized by early disease onset, delay in beginning Iloprost (a prostacyclin analog) therapy, a smoking habit, and presence of joint contractions. They proposed that a score reflecting the sum of these factors could be used to predict the risk of developing ischemic digital ulcers . Alvernini and colleagues also identified major risk factors for digital ulcer development in scleroderma. They evaluated 34 Italian patients with skin ulcers and studied them prospectively over a 20-month period. They found that the most significant, independent parameters associated with the development of skin ulcers were lupus anticoagulant and the presence of avascular areas on nailfold capillaroscopy. Elevated serum IL-6 levels suggesting underlying inflammation also correlated with the above findings . Both in the Caramaschi study and the study recently conducted in Germany by Sunderkotter and colleagues found that male sex, diffuse scleroderma, anti-Scl 70 positivity, inflammation, and the presence of pulmonary hypertension correlated with a high probability of presenting with digital ulcers . These observations all contribute to defining the clinical phenotype of the highest risk patients.
Another means of identifying patients at risk for the development of scleroderma-related digital ulcers is through genetic studies. Combining gene expression data with clinical phenotypes more clearly defines disease subtypes. Milano and colleagues recently conducted a study using DNA microarrays and gene expression profiling with 61 skin specimens from 24 scleroderma patients. They were able to identify genes that had a high positive correlation with severe Raynaud’s and limited scleroderma patients. In addition, they identified genes in the diffuse scleroderma subgroup that correlated highly with the presence of DU’s. Interestingly, three of the patients with digital ulcers were negative for antibodies to anti-centromere antibody and anti-topoisomerase antibody suggesting the genetic profiling is an additional means of characterizing seronegative patients.
Bos and colleagues also used gene expression profiling in peripheral blood cells in an attempt to distinguish clinical subtypes in systemic sclerosis. They looked at the peripheral blood cells of twelve scleroderma patients and 6 controls and found that low expression of type I interferon response genes was associated with the presence of anti-centromere antibodies. Increased expression of type I interferon was associated with the appearance of digital ulcers and the absence of anti-centromere antibodies. As digital ulcer formation is believed to be at least partially related to imbalanced angiogenesis, and type I interferons are known to display antiangiogenic activity, the investigators speculated that there could be a role for increased type I interferon in the process of digital ulcer formation. This concept leads to another potential target in digital ulcer prevention. It is interesting, however, to note that Bos reports an inverse relationship in gene expression profiling between anticentromere antibodies and digital ulcers, as they are typically associated together in the clinical setting .
Another predictor of digital ulceration is the capillaroscopic skin ulcer risk index (CSURI). This tool was developed by Sebastiani and colleagues who found that the CSURI correlates with scleroderma patients who develop digital ulcers. If this tool is validated in a larger study, it could also be helpful in identifying at-risk patients, especially if used in combination with genetic profiles and clinical phenotyping . In such cases, aggressive intervention and preventative measures could be taken to prevent ischemic complications.
Mechanisms of Ischemia in Scleroderma
Vasospasm
Vasospasm is seen in many of the rheumatic diseases and can manifest as benign reversible Raynaud’s phenomenon, or it can be associated with recurrent digital ischemia and tissue injury. The development of Raynaud’s phenomenon ultimately occurs as a result of interactions between nerve endings, smooth muscle cells, and the endothelium that is seen in the setting of soluble mediators (i.e., nitric oxide (NO), prostaglandins, neuropeptides etc) and is influenced by the patient’s surrounding environment, including temperature, smoking, and stress. .
Primary Raynaud’s disease (aka Raynaud’s phenomenon) is a vasospasm that occurs primarily as a result of sensitivity to cold temperatures or emotional stress. It occurs in 4–20% of women and 4–13% of men in the healthy population . It is more common in young women, is entirely reversible, painless, and does not progress to tissue injury . Several studies demonstrated a significant familial aggregation in primary Raynaud’s which suggests an inherited defect in thermoregulation [21–22]. Primary Raynaud’s is associated with distal digital color changes that progress from white to blue to red, representing the initial ischemia from the vasospasm (white), the subsequent slow circulation leading to increased amount of deoxygenated blood (blue), and the final hyperemic state after vessel dilation (red). All three stages of color change do not need to be present in order to diagnose Raynaud’s (See Table 1). Although many patients with rheumatic disease have Raynaud’s phenomenon, without signs of tissue injury the presence of primary Raynaud’s is not associated with an underlying connective tissue disease .
Table 1.
Making the Diagnosis of Raynaud’s
| Ask the Following Questions: |
| 1. Are your fingers unusually sensitive to the cold? 2. Do your fingers change color when they are exposed to the cold? 3. Do they turn white, blue, or both? |
| Confirmed if positive response to all 3 questions | Excluded if response to 2 and 3 are negative |
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Secondary Raynaud’s also occurs in response to cold temperature or emotional stress; however it occurs in the setting of underlying acquired vascular disturbance and is often associated with digital pain and ischemic ulcers. It also is occasionally associated with gangrene, which can lead to tissue loss or digital amputation. Secondary Raynaud’s is seen in 95% of scleroderma patients and is often the initial manifestation of the disease [25–26]. Secondary Raynaud’s often leads to critical ischemia in scleroderma because there is an obliterative vasculopathy of the peripheral arteries and microcirculation. It is manifest by structural disease often with a luminal narrowing greater than 75% of digital arteries due to underlying intimal fibrosis and luminal occlusion by thrombi [27–28]. Endothelial cell injury and activation lead to vascular dysfunction and vasospasm that can quickly obstruct the already marginal blood flow of the vasculopathic digital arteries.
Vasculopathy
Both the microvasculature and macrovasculature are involved in the development of digital ischemia in patients with scleroderma . In scleroderma microangiopathy, a combination of intimal proliferation, medial hypertrophy, and adventitial fibrosis result in the narrowing of the vessel’s lumen and leads to progressive ischemia [30–31]. A complex interaction between activated endothelial cells, unregulated smooth muscle cells and pericytes, along with components of the extracellular matrix and intravascular circulating factors, is thought to contribute to the abnormal vascular reactivity and occlusive scleroderma vascular disease. Digital ischemia is one of the subsequent complications of this process. Macrovascular disease has been studied less than microvascular disease in scleroderma; however, it too plays a significant role in digital ischemia. Darbich and colleagues evaluated patients with scleroderma and found ulnar involvement in 56% of their 27 studied patients . More recently, Hasegawa and colleagues found that macrovascular involvement was present in seven out of eight scleroderma patients with digital ulceration or gangrene who were screened by arteriography. In addition, vascular disease was not limited to the digit with ulcerations and gangrene but was also found in the surrounding non-ulcerated digits .
Interestingly, Caramaschi and colleagues recently noted that all of their scleroderma patients with both micro and macrovascular disease incurred digital amputation. The patients without clinical evidence of macrovascular involvement did not develop digital necrosis with need of amputation. They proposed that the combination of both micro and macrovascular involvement exceeds the compensation capacity of peripheral circulation and thus place the patients at an elevated risk of severe complications .
Vasculitis
The association of vasculitis and scleroderma is unusual but is reported in the literature. Of these cases, the most commonly reported vasculitis seen in scleroderma is ANCA-associated vasculitis . P-ANCA seropositivity is known to predict vasculitis in patients with scleroderma [35–36]. Cases of ANCA vasculitis typically present in scleroderma as a pulmonary-renal syndrome or more commonly an ANCA associated glomerulonephritis, although reports exist with vasculitis of the nerves and skin [37–39]. ANCA is associated with digital ischemia in some patients with Wegener’s [40–42] and we witnessed digital ischemia in a patient with scleroderma and ANCA associated vasculitis.
Thrombotic phenomena
Although vein thrombosis and CNS events (e.g. stroke) are more likely, one other cause of digital gangrene in connective tissue disease is an associated hypercoaguable state due to anti-phospholipid syndrome (APS). APS is thought to not only cause a thrombotic microangiopathy, but there is also some in vitro evidence that suggests that certain anti-phospholipid antibodies (aPL) may be pro-atherogenic [43–44]. Interestingly, anti-beta2glycoprotein1 was recently reported to be independently associated with macrovascular disease in scleroderma. Boin and colleagues found that patients with digital loss in scleroderma were much more likely to be anti-beta2GPI positive and that these patients also ultimately have higher mortality rates . The vascular events were associated with IgA and IgM isotypes of anti-beta2glycoprotien-1 and not the IgG isotype suggesting that these antibodies were a secondary phenomenon and not a causative factor.
Diagnosis
The approach to the patient with digital ischemia begins with a careful history which should guide the physician towards the etiology of the disease (See Figure 1). The history is followed by a thorough physical exam with particular focus on the patient’s vasculature and skin. This examination will often clarify the underlying diagnosis, the size of vessels involved, and the presence or absence of critical vascular compromise. On initial evaluation it is important to document the patient’s bilateral blood pressures and to look for any asymmetry in pressure or pulses. Asymmetrical blood pressures and asymmetrical or nonpalpable pulses can be an indication of underlying macrovascular disease as can be seen in vasculitis, atherosclerosis, embolic disease or extrinsic vascular obstruction. A positive Allen’s test in the wrists is suggestive of medium-vessel involvement and should ideally be performed in all patients with severe Raynaud’s and refractory digital ulcers on routine examination [46–47]. It is essential to listen over major arteries for bruits as this can provide insight on luminal narrowing which is noted most often in atherosclerosis, but also is seen with an underlying vasculopathy. Once these parameters are evaluated, it is necessary to look for physical signs that can direct one to the underlying diagnosis. A meticulous examination of the skin should include evaluation for sclerodactyly, telangiectases, hypopigmentation, digital pitting, loss of digital pulp, or calcinosis cutis, all of which would suggest scleroderma. Periungal capillary loop dilation is also a key finding and will place the patient in the group of patients with connective tissue diseases and can help differentiate between primary and secondary Raynaud’s. The presence of livedo reticularis or livedo rasemosa would be suggestive of antiphospholipid syndrome, while splinter hemorrhages, petechiae and/or purpura could point the examiner towards an underlying vasculitis.
Once the physical examination is complete, laboratory tests should be sent for further evaluation. However, if critical digital ischemia is present by examination, then it is an emergency and it is important not to wait until for all tests to return before initiating treatment. A listing of recommended laboratory tests is presented in Table 2, but it should be recognized that the history and examination will often dictate the priority laboratory testing needed.
Table 2.
Laboratory evaluation in digital ischemia
| Differential Diagnosis | Laboratory Testing |
:--- |
| Scleroderma | Scl-70 antibodies, anti-centromere antibody, anti-RNP-polymerase III |
| Vasculitis (SLE, ANCA-related vasculitis, cryoglobulins, rheumatoid vasculitis, etc) | Anti-nuclear antibody screen, pANCA, cANCA, MPO, PR3, cryoglobulins, anti-dsDNA, RNP, Smith, C3, C4, ESR, CRP, RF |
| Embolic | Antiphospholipid antibodies (anti-cardiolipin antibody, Beta2 microglobulin, lupus anticoagulant) |
| Other | CMP, CBC with diff, TSH, lipid profile, urinalysis with micro |
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Imaging
Doppler
Non-invasive assessment of the peripheral circulation will supplement the physical examination and may provide clues as to the cause and size of the vessels involved. Doppler ultrasound is a useful tool and a relatively cost effective way to evaluate patients with digital ischemia. The use of ultrasound is reported to differentiate between vasculopathy and vasculitis. In vasculopathy the luminal narrowing is evident, and the digital arteries can demonstrate decreased digital pulsation and are often chronically occluded. Alternatively, patients with vasculitis will often have an ultrasound image consistent with acute arterial occlusion [47–48]. Stafford and colleagues used ultrasound to evaluate macrovascular disease in scleroderma and described significant arterial narrowing with arterial walls characterized by smooth thickening along their entire length . We recommend using Doppler ultrasound in the initial evaluation of all patients presenting with digital ischemia in order to quickly and accurately diagnose the size of vessels involved and to determine if surgical intervention is a viable option.
Laser Doppler imaging is a useful research tool used to evaluate microcirculatory flow [49, 50]. Because laser Doppler is able to assess more than one area of the hand at any given time it has shown to be more effective than a single probe Doppler . Laser Doppler imaging can also be used to differentiate between primary Raynaud’s from patients with scleroderma . Murray and colleagues suggested that combining laser Doppler with other imaging modalities such as nailfold capillaroscopy and thermal imaging is more effective than laser Doppler alone, but thermal imaging in not yet widely available .
Angiogram
Angiography is a well established imaging modality used to evaluate vascular disease. Digital arteriography was reported as far back as the 1960’s when several groups looked at patients whose digital vasculature was evaluated with this technique [53–55]. Takaro and colleagues noted the diagnostic utility of arteriography as they demonstrated both micro and macrovascular involvement in patients with scleroderma, and noted digital hypervascularity in patients with Raynaud’s. Dabich and colleagues later described a “characteristic pattern” of arterial involvement in the hands of patients with scleroderma. They noted, “The arch is usually of the balanced variety with broad communications between the ulnar and radial components. The radial artery is usually spared and the ulnar artery is frequently involved. The superficial arch and common digital arteries are most often uninvolved, while the proper digital arteries are obstructed in almost all patients in the mid and distal portions of the fingers.” .
Arteriography is now used in specific cases with digital ischemia when the underlying cause is in question or the option of surgery is considered. Park and colleagues recently studied a cohort of 19 scleroderma patients and recommended that patients with scleroderma associated with digital ulcers or severe Raynaud’s should consider angiography with ulnar revascularization if indicated because of the increased risk of digital loss with macrovascular disease [47, 56]. Angiography is also used to confirm the site of vascular occlusion prior to peripheral artery bypass grafts in the hands and feet of two patients with severe Raynaud’s . The clinical application of angiography continues to grow, and based on the available data we recommend using this application to define the vascular anatomy of patients with severe digital ischemia who are candidates for angioplasty or surgery.
Magnetic Resonance Angiography
Magnetic Resonance Angiography (MRA) is a newer, imaging modality that is used to evaluate vascular abnormalities of the hand. It is a fast, non-invasive exam that takes less than five minutes to perform and produces high quality images . Allanore and colleagues used MRA to evaluate both arterial and venous lesions in patients with scleroderma. These investigators not only found substantial arterial and venous damage in the hands of these patients, but they also noted that the vascular lesions were associated with the extent of clinical disease and phenotype suggesting that MRA may be useful in evaluating disease progression . MRA showed some promise in evaluating patients with Raynaud’s phenomenon when compared to conventional angiography as vasodilators are often used with conventional angiography and the true degree of vasospasm is often underestimated [58, 60]. In addition, diagnosing connective tissue disorders is also possible with MRA as these disorders in general are defined by characteristic findings on imaging such as tapering of the ulnar, radial, and proper digital arteries, superimposed vasospasm and areas of narrowing found between normal segments. The use of contrast enhanced MRA instead of standard angiography prior to vascular surgery of the hand has been proposed by some groups as it is a less invasive form of imaging . The use of MRA, however, requires experienced radiologists . To our knowledge, there are no large studies to date comparing MRA and conventional angiography use for imaging of the vasculature of the hand. Until such a study is conducted, conventional angiography should still be considered the gold standard for evaluating the vascular anatomy of the hand, particularly in the pre-operative setting.
Therapeutic Interventions
The approach to treating digital ischemia can be daunting given that it must be initiated quickly and effectively, and there are many new therapeutic options available. A recent study conducted in Germany by Herrgott and colleagues examined patients with digital ischemia who presented to subspecialists (rheumatologists, dermatologists, pulmonologists, and nephrologists). Their study demonstrated that cutaneous vascular complications of scleroderma are often undertreated or treated inappropriately . In addition, the treatment of digital ulcers correlates with an improvement in functional status and quality of life . It will be imperative in the future to standardize care for the management of the ischemic digit. We describe our approach to therapy below.
Non-medical therapy
The initial approach to treating a small ischemic digital area that presents as a mild discoloration at the tip of the finger is often is aimed at symptom control and improving tissue integrity and viability. Avoiding triggers such as cold temperatures and stress are helpful in reducing vasoconstriction. This includes adjustment of lifestyles to avoid extreme cold, shifting temperatures and proper clothing to keep the whole body warm. Studies looking at conditioning, biofeedback and relaxation techniques show variable outcomes. One large controlled trial in primary Raynaud’s phenomenon found no benefit in the use of biofeedback and its use is also not recommended for secondary Raynaud’s phenomenon. The use of gloves is helpful in protecting the skin from trauma and keeping it warm in the cold. Ischemic lesions are painful and appropriate pain control is needed with acetaminophen, a non-steroidal anti-inflammatory and/or narcotics when needed. Smoking cessation is very important as smoking can contribute to the underlying vascular disease. In addition, topical creams and lotions can be applied to keep the affected skin moist. For more serious lesions, occlusive dressings serve to protect them from trauma and to promote healing. Hydrocolloid dressing also promote healing of digital ulcers .
Patients who are having a critical ischemic event should be put a rest and in a warm environment. This may mean hospitalization or stopping work for home care. Preventing trauma to the digits such as typing or repetitive hand work can improve blood flow and recovery in conjunction with other measures.
Medications
The agents used for the treatment of Raynaud’s phenomenon and scleroderma vascular disease can be divided into agents that primarily work as vasodilators, those that have the potential to protect vessels form disease progression, and agents that prevent thrombosis. A given agent may have more than one effect. For example, prostaglandins can be vasodilators and protective of vessel damage. Our discussion will first outline currently used medications and then we will focus on our specific approach to critical ischemia.
Vasodilator Therapy
Alpha adrenergic blockers
Alpha adrenergic blockers were the first agents used with some success in treating Raynauds phenomenon. Alpha-2 adrenoreceptors are present throughout much of the vascular system and they play a significant role in cutaneous thermoregulation . Prazosin was studied by several groups of investigators for treatment of Raynaud’s phenomenon [66, 67]. A subsequent Cochrane systematic review concluded that Prazosin is modestly effective in treating Raynauds secondary to scleroderma, but that side effects can limit tolerability (Harding SE Prazosin for Raynaud’s phenomenon in progressive systemic sclerosis. Cochrane.1998). In addition, several other members of this class of medications demonstrated a clinical benefit [66, 68]. Interestingly, the alpha 2c receptor, as subtype of the alpha adrenergic receptor, is specifically upregulated in cold exposure . As a result, Wise and colleagues studied the efficacy and tolerability of a selective alpha 2C-adrenergic receptor blocker in scleroderma patients with vasospasm. They found that the time to rewarm a patient’s finger with secondary Raynaud’s after a cold challenge was decreased after ingestion of the drug, thus suggesting potential for therapeutic efficacy . These results are promising, but these agents are not yet available and more studies are needed to validate the clinical efficacy. From a practical viewpoint, alpha adrenergic blocking agents are not the first line therapy for critical ischemia but the potential of selective new agents for the prevention of vasospasm in the digital and thermoregulatory circulation is of major interest.
Calcium Channel Blockers
Calcium channel blockers are widely used for Raynaud’s and act on vascular smooth muscle to cause arterial dilation. Thompson and colleagues published a metanalysis looking at their use in Raynaud’s and reported moderate efficacy at best . In addition, the magnitude of effect of calcium-channel blockers for Raynaud’s phenomenon associated with scleroderma is much smaller than in primary Raynaud’s, although a 35% improvement in attack severity and a mean reduction of about 8 attacks per 2 week period were still noted when compared to placebo. In addition, appropriate dosing is not always reached. Herrgott and colleagues showed that 92% of the German centers they surveyed did not aim for the recommended 360 mg of diltiazem or for the goal of 10 mg of amlodipine, and 80% did not aim for at least 40 mg of Nifedipine [62, 72]. Longer acting formulations can be used to minimize side effects of the medication and increase tolerability. Calcium channel blockers are also effective in the treatment of digital ulcers .
Nitrates
Glyceryl trinitrate (GTN) has been studied in various means of administration. Initially the intravenous form was evaluated only to find that while there was an initial response, the effect was eventually blunted with disease progression . GTN patches (0.2 mg/hour) were studied a few years later in patients with primary Raynaud’s and in patients with Raynauds secondary to scleroderma. The treatment was effective in both groups; however the side effects, particularly the headaches, were intolerable. Finally, Anderson and colleagues used GTN in the topical ointment formulation and found that it was effective with minimal side effects, even in patients with very thick skin . While topical nitrates can improve digital blood flow, the use of these agents is limited by practical issues of difficulty with repeated application and side effects. Newer formulations are being tested and suggest some benefit in reducing Raynaud’s condition score . It is our practice to use topical nitrates in conjunction with another vasodilator such as a calcium channel blocker during periods of critical ischemia of a digit.
Phosphodiesterase Inhibitors
Phosphodiesterase inhibitors (PDE-I’s)work by elevating levels of cGMP, causing intracellular calcium level to fall and leading to vascular smooth muscle relaxation. Through this mechanism PDE-I’s cause vasodilation and increase perfusion to distal tissues . This class of drugs has demonstrated significant effects in patients with digital ischemia [78–81]. The five drugs available in this class of medications include sildenafil, tadalafil, vardenafil, pentoxifylline, and cilostazol, and the first two listed are better studied. Fries and colleagues conducted a double-blind, placebo controlled fixed dose crossover study with 16 patients to evaluate the effects of sildenafil on symptoms of capillary perfusion in patients with Raynaud’s. They found that sildenafil was associated with a decreased incidence and duration of Raynaud’s as well as a decreased Raynaud’s Condition Score. Capillary blood flow velocity increased in individual patients and the mean capillary blood flow velocity of all patients who received sildenafil more than quadrupled . Although promising results are also reported anecdotally with vardenafil and tadalafil [82, 83], a recent randomized placebo controlled trial using tadalafil suggested no benefit over placebo . Unfortunately, the numbers were small as only 39 patients with scleroderma were enrolled. Large randomized controlled trials are still needed to validate the use of phosphodiesterase inhibitors in secondary Raynaud’s. In our uncontrolled experience we have clinical success using a phosphodiesterase inhibitor for severe Raynaud’s secondary to scleroderma when used in conjunction with a calcium channel blocker but often see less benefit with the phosphodiesterase inhibitor alone..
Prostacyclins
Prostanoids are beneficial to both the micro and microvasculature as they induce vasodilation, increase intracellular cAMP, and prevent smooth muscle proliferation . Prostacylcins in particular are found to be effective therapy for Raynaud’s and digital ischemia. Intravenous Iloprost is now a popular intervention outside the USA for the treatment of severe Raynaud’s secondary to scleroderma as it decreases the frequency and severity of attacks and prevents and heals digital ulcers [63, 86]. Cyclic Iloprost is now used for severe Raynaud’s and digital ulcers using various protocols [13, 63, 87–90]. These reports suggest that using prostacylcin by intravenous delivery intermittently can prevent digital ischemic events. Low dose (0.5 ng/kg compared to 2 ng/kg body weight per minute) Iloprost was shown to be equally effective . Several authors reported that subcutaneous Treprostanil is also effective in the treatment of severe digital ulcerations [85, 92]. Of note, the efficacy of oral prostacyclins was also studied in the last decade for treatment of severe Raynaud’s and digital ulceration. However, several trials using oral Iloprost, beraprost, and cicaprost showed no significant benefit over placebo [93–95]. A trial testing a new formulation of oral Treprostinil is underway for the treatment of digital ulcers in patients with scleroderma. Inhaled preparations of Treprostinil and Iloprost are available but not studied in the treatment of Raynaud’s or digital ischemia.
Intravenous and Transdermal PGE1
Transdermal PGE1 ethyl ester was also reported to be effective in improving blood flow in capillaries in the skin in systemic scleroderma patients and in healing acral skin lesions in patients with scleroderma [96, 97]. However, this agent is not readily available. Intravenous PGE1 is used in the treatment of Raynaud’s and is an alternative to prostacyclin therapy .
Angiotensin Converting Enzyme inhibitors and Angiotensin Receptor Blockers
Angiotensin converting enzyme (ACE) inhibitors and angiotensin receptor blockers (ARB’s) were also studied in scleroderma in relation to digital ischemia. Initial studies showed promise as captopril produced a significant improvement in cutaneous blood flow; however, it was not shown to alter the frequency or severity of Raynaud’s attacks . A subsequent study also showed promise as enalapril demonstrated some promise in reducing the frequency of primary Raynaud’s attacks . However, in further study and investigation with clinical trials, there were ultimately mixed results . Most recently, Gliddon and colleagues organized a multicenter, randomized, double-blind, placebo-controlled study evaluating quinapril 80 mg/day, or the maximum tolerated dosage, in over 200 patients with limited scleroderma or with Raynaud’s phenomenon and the presence of scleroderma specific antinuclear antibodies. They treated the cohort for 2–3 years and were unable to demonstrate any benefit in limiting the occurrence of digital ulcers or influencing the frequency or severity of the Raynaud’s episodes . As a result, this class of medications is not recommended for treatment of Raynaud’s or digital ulcers.
SSRI’s
The selective serotonin reuptake inhibitors have the potential of increasing regional blood flow by blocking the uptake of the vasoconstrictor serotonin. A small study using fluoxetine demonstrated some benefit compared to low dose nifedipine in Raynaud’s although it is not studied for the treatment of digital ulcers . We have used this agent in conjunction with a calcium channel blocker or when low blood pressure limits our ability to use more potent vasodilators.
Vasoprotective agents
Anti-platelet agents
Several groups reported elevated platelet activity in patients with scleroderma [104–107]. In one such study scleroderma platelet activation markers correlated with disease activity and severity . In addition, through the use of combination therapy with aspirin and dipyramidole, significant reductions in circulating platelet aggregates and beta-thromboglobulin levels were achieved . Although one double blind placebo-controlled trial reported no benefit with combination therapy with aspirin and dipyramidole versus placebo, it was a short trial over only a two year period in a group of only 28 patients. Therefore no meaningful conclusions can be drawn regarding long term benefits . We regularly use low dose (81 mg) aspirin therapy in our scleroderma patients with significant vascular disease because it makes biological sense.
Endothelial Receptor antagonists
The endothelial receptor antagonists also show promise in preventing digital ulcers and have vasculoprotective effects. Korn and colleagues conducted a small preliminary study with 122 patients evaluating the effect on preventing digital ulcers. They found that patients receiving bosentan had a 48% reduction in the mean number of new ulcers during the treatment period, suggesting that it could be promising . A second study, RAPIDS 2, found similar benefit in prevention of new ulcers, particularly in patients with a high number of digital ulcers at baseline. On the other hand, in RAPIDS 2, a trial of 24-week duration with 198 subjects, higher rates of healing of ulcers were seen with placebo than active drug. At the end of 24 weeks of drug therapy, there were no differences between active treatment and placebo in net DU burden, pain, measures of activities of daily living by HAQ or UK Functional Score (UKFS) or in hospitalization rates . Larger studies will be needed to determine efficacy and long term outcomes, but these findings are encouraging.
Statins
Statins more recently became a focus of scleroderma research. Statins demonstrate vasculoprotective effects by decreasing LDL, increasing HDL, decreasing free radicals, coagulation, and blood viscosity, decreasing matrix metalloproteases, and increasing platelet function [31, 112, 113]. Abou-Raya and colleagues recently looked at 84 patients with scleroderma and matched them with 75 control subjects to evaluate the effects of statins on patients with Raynaud’s and digital ulcers. They found that the overall number of digital ulcers was significantly reduced in the statin group and that endothelial markers of activation were improved when comparing the statin and control groups . Other studies have noted similar findings . Although these results are promising, larger trials are still needed.
Thrombolytics
The role of thrombolytics in the treatment of digital ischemia and its complications has been studied on several occasions. The rationale for this is that in scleroderma, it is thought that the patients have an underlying balance towards clotting with elevated fibrinogen levels and defective TPA release [31, 116, 117], although not all groups agreed [118, 119].
At this time there is limited evidence supporting the use of fibrinolytic therapy in patients with digital ischemia, and the complications with the use of these medications can be severe. As a result they cannot be recommended for day-to-day use in the treatment of digital ischemia.
Other
Other medications that offer vascular modulating effects and that are currently still under evaluation include the tyrosine kinase inhibitors and the rho kinase inhibitors [120, 121]. Antioxidants such as allopurinol and vitamin E have surprisingly shown little benefit [122–124]. Notably, Probucol, a synthetic antioxidant, did demonstrate a significant reduction in the frequency and severity of Raynaud’s attacks when compared to controls . N-acetylcysteine was recently studied prospectively in a cohort of 50 patients and reported to decrease the number of digital ulcers per year and decrease Raynaud’s attacks. It was well-tolerated in the long term with only flushing and minor headaches as side effects . Other studies of anti-oxidants do not show a clear benefit for these agents, and thus we will need to wait for more data to develop clear guidelines for the use of anti-oxidants.
Botox is another reported therapeutic option for Raynaud’s and digital ischemia. Fregene and colleagues found that Botulinum toxin type A improves pain and healing in patients with Raynaud’s and scleroderma. They concluded that it is an effective treatment of vasospastic digital ischemia. Their study was limited by a small group of only 26 patients and a retrospective design . Other small case series also demonstrated success with Botox in pain and healing for patients with Raynaud’s phenomenon and digital ulcers . Although promising, we await controlled trials to evaluate the therapeutic efficacy of Botox in scleroderma patients with digital ischemia.
Sympathectomies
Sympathetic nerve mediated vasospasm is implicated as a major mechanism leading to digital ischemia. As a result, sympathectomies are used aimed at blocking this mechanism. Uncontrolled series of case reports suggest benefit for both Raynaud’s and for the treatment of refractory digital ulcers. Local “digital sympathectomy” done by surgical periarterial sympathectomies have also demonstrated have long term benefits in patients with digital ischemia secondary to autoimmune disease [129, 130] Hartzell and colleagues followed patients for 7.5 years and found that this intervention resulted in complete ulcer healing and decreases in the total number of ulcers in 75% of the patients in this subgroup . Although it was a small group of only twenty patients, these results are promising for patients with refractory disease. Arterial revascularization is sometimes performed simultaneously and has also demonstrated success .
Summary on Treatment: A Practical Approach
The availability of many different therapeutic agents to treat Raynaud’s can lead to confusion when initiating therapy in the clinical setting. Accordingly, an outline of a practical approach to treatment is given based on our experience. Although specific studies are needed, we find this approach to therapy to be reasonable and effective (See Figure 2).
Fig. 2.
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Raynaud’s alone
In scleroderma patients with Raynaud’s without digital ulcers we first begin with supportive therapy, including the avoidance of triggers such as stress, smoking, and cold temperatures, and keeping the fingers warm with gloves. Next we begin low dose aspirin at 81 milligrams daily. At the same time, we begin a long acting calcium channel blocker, such as amlodipine, for preventative therapy and titrate it up to the highest dose tolerated needed for benefit. Toleration is limited by the possible side effects of this class of medications, including fluid retention, symptomatic hypotension, constipation, and aggravation of gastroesophogeal reflux disease. Most patients do well on low dose (e.g. amlodipine 5 mg daily) but we will titrate up the dose as needed (e.g. amlodipine 20 milligrams daily). Our goal is not to eliminate every Raynaud’s event but rather to improve the quality of life and prevent severe attacks and any progression to ischemic lesions. Thus, if a patient is not having digital lesions and reports improvement and reasonable tolerance then we continue on a calcium channel blocker alone with low dose aspirin. If the patient is still having severe symptoms once the calcium channel blocker is titrated to the maximal tolerated dose, then we will add a second agent recognizing there is few studies documenting the risk and benefits of additive therapy. Agents used include topical nitrates, a phosphodiesterase inhibitor or an SSRI based on availability and tolerance. We do use anti-oxidants but have no preference to a particular agent.
Raynaud’s with digital ulcers and severe RP
In patients with digital ulcers and severe Raynaud’s, we again stress supportive therapy, initiation of low dose aspirin, and titration of a long acting calcium channel to the highest dose tolerated. If the patient is still experiencing recurrent ulcers or moderate to severe Raynaud’s once the calcium channel blocker is at the highest dose tolerated, we then add a second agent. Usually our second agent is a topical nitrate or a phosphodiesterase inhibitor such as sildenafil; however both should not be used together as there is a risk of severe hypotension. Although not readily available in the United States, intermittent infusion of prostaglandins (e.g. Iloprost) is considered appropriate in conjunction with the calcium channel blocker. The exact interval for therapy is not well defined but previous studies suggest that benefits last about 10 weeks after a 5 day (6 hour each day) infusion of Iloprost is a period of continued benefit. If a combination of medications is not effective, and we have a nonhealing ulcer or chronic digital ischemia then we pursue a digital sympathectomy.
Critical ischemic event
In the setting of a critical ischemic crisis treatment must be initiated quickly and aggressively to prevent permanent tissue damage and digital loss (see Figure 3). During an ischemic crisis, irreversible vasospasm and severe pain are present. At initial presentation, a local injection of lidocaine or bupivicaine at the base of the finger can relieve pain and vasodilate rapidly. This can be repeated again if necessary. Systemic pain medication is often needed. If the patient is not on a vasodilator at the time of the crisis, then we initiate a short-acting calcium channel blocker and rapidly titrate it up to the maximal dose tolerated. Once that dose is reached, we change to a long acting calcium channel blocker at an equivalent dose if the crisis is improved. If the signs of digital ischemia continue to progress, we add a heparin drip for 24–72 hours and begin therapy with intravenous prostacyclin (epoprostenol or iloprost) continuous at low dose for about 5 days. If stable then we attempt to prevent relapse with continued calcium channel blocker alone or in combination with either a PDE inhibitor or topical nitrate. If this is still ineffective, we then recommend surgical consultation for digital sympathectomy. Doppler studies are done to investigate larger vessel disease and an MRA or angiogram is done in selected cases to localize potentially correctable arterial disease. If no larger vessel disease is found that can be corrected and the patient is not improving, then surgical sympathectomy is done. In critical ischemia with the presence of digital gangrene, it is always important to evaluate for underlying infection and add appropriate antibiotics if necessary. Surgical consultation may also be needed in this setting.
Fig. 3.
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9332 | https://www.youtube.com/watch?v=syvsntsmmCU | How To Work Out The Angle Between 2 Regular Polygons (Shapes)
Maths Mark
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Posted: 7 Jun 2021
In this video you will be shown how to work out the angle between 2 regular polygons. To do this you will need to find the interior angle of each polygon. Do this by first finding the exterior angle by dividing 360 by the number of sides. Then subtract this answer from 180 to give the interior angle. Once you have found the interior angles for each polygon add up the 2 angles and subtract it from 360 to give the missing angle.
So in example 1 you need to find the missing angle between a square and a regular pentagon. The square has an interior angle of 90 degrees as a square has 4 right angles. The exterior angle of the pentagon will be 72 degrees (360/5), and so the interior angle will be 180-72=108. Finally add up 108 and 90 to give 198, and subtract this from 360 to give 162 degrees.
In example 2 you need to find the missing angle between a regular hexagon and an octagon. The exterior angle of the hexagon is 60 degrees (360/6), and so the interior angle is 120 degrees. The exterior angle of the octagon is 45 degrees (360/8), and the interior angle is 180-45 which is 135 degrees. Finally add up 135 and 120 to give 255, subtract this from 360 to give 105 degrees.
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Transcript:
hi guys welcome back this is maths 3000 today i'm going to show you how to work out the angle between two regular polygons so let's move on to the first example so we've got a regular pentagon with five sides and we've got a square and we need to work out the angle between the pentagon and the square so after angle x here okay so the first we need to do is work out the interior angles of our square and our pentagon so the square's pretty easy i've already marked in the angles for the square the interior angle of a square is 90 degrees so let's work out the interior angle for our regular pentagon so we can do this by using the formula for the exterior interior angle of a regular polygon so to get the exterior angle we did 360 divided by the number of sides so our polygon has five sides because it's a pentagon so 360. divided by 5 is 72 degrees okay so we're more interested in the interior angle so we now need to work out the interior by taking the exterior angle away from 180 so we go 180 take away 72 which is 108 degrees so our interior angles of our regular pentagon are 108 degrees so we can now use angles around a point or a circle to figure out angle x so we just need to add these two angles up here and take the answer off 360. so we go 108 add 90 so that will give us 198 and then if we take this answer away from 306 because angles around the point add up to 360 we get an answer of 162 degrees so let's move on to our second example then so this time we've got to find the angle y and it's between a regular hexagon and we've got a one two three four five six seven eight we've got a regular octagon here let's just make a note that's that's eight sides and this one's got six sides okay so let's start off with hexagon first of all so let's work out the interior angle of the hex again so to get the interior angle first we work out the exterior angle by doing 360 divided by the number of sides so the 360 divided by six because our hexagons got six sides so 360 divided by six comes out at 60 degrees so we then take that away from 180 to give us the interior angle which is 120 degrees so this angle has 120 degrees so we now need to work out the interior angle for the octagon so we can do this in a similar sort of way so to get the exterior angle we do 360 divided by eight so 360 divided by eight is 45 degrees take this away from 180 to give the interior angle so that's 135 degrees so this angle here is 135 so we can now use angles around this point or a circle to work out angle y so if you add these two angles up and take it away from 360 so 120 and 135 is 255 so take this away from 360. gives an answer of 105 degrees so that's angle y so i hope you like this video if you did make sure you like and subscribe thanks for watching bye |
9333 | https://www.teacherspayteachers.com/Product/7th-Grade-Unit-8-Activity-BUNDLE-IM-Grade-7-Math-Probability-Sampling-9589333 | 7th Grade Unit 8 Activity BUNDLE | IM® Grade 7 Math Probability & Sampling
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7th Grade Unit 8 Activity BUNDLE | IM® Grade 7 Math Probability & Sampling
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This bundle of 5 activities supplements the learning of 7th grade Math Unit 8: Probability & Sampling. This curriculum may be packaged as Open Up Resources, Kendall Hunt, or IM® Grade 7 Math authored by Illustrative Mathematics®.
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To use with or after Unit 8, Lesson 9: Compound Probability Partner Match Coloring Activity
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7th Grade Unit 8 Activity BUNDLE | IM® Grade 7 Math Probability & Sampling
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Compound Probability Differentiated Worksheet Fun | Partner Coloring Activity
Do your 7th grade students need more practice solving compound probability and multi-step probability problems? Could you use a calm, relaxing, fun differentiated worksheet activity where students have a mix of independent and partner work? This partner match color by number activity encourages stud
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CCSS 7.SP.B.3
Informally assess the degree of visual overlap of two numerical data distributions with similar variabilities, measuring the difference between the centers by expressing it as a multiple of a measure of variability. For example, the mean height of players on the basketball team is 10 cm greater than the mean height of players on the soccer team, about twice the variability (mean absolute deviation) on either team; on a dot plot, the separation between the two distributions of heights is noticeable.
CCSS 7.SP.B.4
Use measures of center and measures of variability for numerical data from random samples to draw informal comparative inferences about two populations. For example, decide whether the words in a chapter of a seventh-grade science book are generally longer than the words in a chapter of a fourth-grade science book.
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Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event.
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9336 | https://ocw.mit.edu/courses/18-100b-real-analysis-spring-2025/resources/mit18_100b_s25_lec17_pdf/ | SPRING 2025 - 18.100B/18.1002
TOBIAS HOLCK COLDING
Lecture 17
Recall that last time we showed the Taylor expansion theorem:
Theorem : (Taylor expansion.) Let f : [ a, b] → R be a function and k a positive integer.
Assume that f , f 0, f (2) , · · · , f (k−1) exists on [a, b] and are continuous and that f (k) is defined
on (a, b), then there exists c between a and b such that
f (2) (a) f (k−1) (a)
f (b) = f (a) + f 0(a) ( b − a) + (b − a)2 + · · · + (b − a)k−1
2 (k − 1)!
f (k)(c)
(b − a)k .
(k)!
For and infinitely differentiable function f on R we define the (k − 1) Taylor polynomial at
a by
f (2) (a) f (k−1) (a)
Pk−1(x) = f (a) + f 0(a) ( x − a) + (x − a)2 + · · · + (x − a)k−1 .
2 (k − 1)!
Question : One naturally wonders how well does this polynomial approximate f when x is
near a?
Answer : This depend on the value of the remainder
f (k)(c)
Rk(x) = (x − a)k .
k!
12TOBIAS HOLCK COLDING
x
Example 1: Suppose that f (x) = e so f (k)(x) = f (x) for all k. This means that the Taylor
expansion near a = 0 becomes k−1X i
x
Pk−1(x) = .
i!i=0
By the Taylor expansion theorem we have that
f (k)(c)
f (x) = Pk−1(x) + xk .
k!
Since f (k)(x) = f (x) for all k, it follows from the Taylor expansion theorem that we have
e|x|
|f (x) − Pk−1(x)| ≤ .
k!
We conclude that for k large the polynomial Pk−1 gives a pretty good approximation to f .
For instance, if |x| ≤ 1, then we have that
e
|f (x) − Pk−1(x)| ≤ .
k!
Example 2: On R define a function f by
(
0 if x ≤ 0
f (x) = − 1
e 2 otherwise x
It is easy to see that f is infinitely differentiable and that f (k)(0) = 0 for all k. It follows
that for all k the Taylor polynomial at 0 is Pk−1 ≡ 0. Thus in this case f (x) = Rk(x).
Riemann integrals
Partition : Let [a, b] be an interval. A partition P of the interval [a, b] is a number of
sub-divisions xi such that
a = x0 < x 1 < x 2 < · · · < x n = b .
The partition is then the sub-intervals [xi−1, x i]. We will set Δ xi = xi − xi−1.
Upper and lower sums : Suppose now that f : [ a, b] → R is a bounded function and that
P = {xi} is a partition of the interval [a, b]. We define upper and lower sums as follows. Set
Mi = sup f ,
[xi−1,x i]
mi = inf f ,
[xi−1,x i]SPRING 2025 - 18.100B/18.1002 3
and upper U (f, P) and lower sums L(f, P) by
n
X
U (f, P) = Mi Δ xi ,
i=1
n
X
L(f, P) = mi Δ xi .
i=1
Example 3: Suppose that the function is f (x) = x2 + 1 on the interval [−2, 2] and that the
partition is P is {− 2, −1, 0, 1, 2}. We have
m1 = 2 and M1 = 5 ,
m2 = 1 and M2 = 2 ,
m3 = 1 and M3 = 2 ,
m4 = 2 and M4 = 5 .
For the lower and upper sums we have
L(f, P) = 2 + 1 + 1 + 2 = 6 ,
U (f, P) = 5 + 2 + 2 + 5 = 14 .
The following lemma is immediate (from that Mi ≥ mi):
Lemma 1: We always have that
U (f, P) ≥ L(f, P) .
Sub-partition : Let [a, b] be an interval and P1 and P2 two partitions of the interval [a, b].
We say that P2 is a sub-partition (or refinement) of P1 if all the dividing points in P1 are
also in P1 (and then presumable some additional dividing points).
Example 4: Suppose that the interval is [−2, 2] and the given partition P1 is
{− 2, −1, 0, 1, 2} .
Then the partition
1 1
P2 = −2, −1 , −1, 0, , 1, 2
2 2
is a refinement (or sub-division) of P1. Indeed, P2 has the same dividing points as P1 in
addition to some more. 4 TOBIAS HOLCK COLDING
We now have the following:
Lemma 2: Suppose now that f : [ a, b] → R is a bounded function and that P1 is a partition
of the interval [a, b] and P2 is a refinement of P1, then
L(f, P1) ≤ L(f, P2) ≤ U (f, P2) ≤ U (f, P1) .
Proof. The middle inequality is the previous lemma. The inequality to the right follows from
that if P2 is a subdivision of P1. Namely, suppose that a = x0 < x 1 < · · · < x n = b are the
dividing points for P1 and that between say xi−1 and xi there is an extra dividing point in
P2 say y so xi−1 < y < x i, then we have
sup f ≤ Mi
[xi−1,y ]
and
sup f ≤ Mi
[y,x i]
so
[ sup f ] ( y − xi−1) + [sup f ] ( xi − y) ≤ Mi Δ xi .
[xi−1,y ][y,x i]
From this it follows easily that
U (f, P2) ≤ U (f, P1) .
Similarly, for the inequality to the left.
Upper and lower integrals : Suppose now that f : [ a, b] → R is a bounded function.
Define the upper Riemann integral of f by
Z b
f dx = inf U (f, P) .
Pa
Here the infimum is taken over all partitions of [a, b]. Likewise, we define the lower Riemann
integral by Z b
f dx = sup L(f, P) .
aP
Riemann integral : Suppose that f : [ a, b] → R is a bounded function, then we say that f
is Riemann integrable if Z b Z b
f dx = f dx .
aaSPRING 2025 - 18.100B/18.1002 5
If the function is Riemann integrable, then the Riemann integral is
Z b Z b Z b
f dx = f dx = f dx .
aaa
The Riemann integrable functions is denoted by R ([ a, b]).
From Wikipedia: Georg Friedrich Bernhard Riemann (1826 – 1866) was a German math -
ematician who made profound contributions to analysis, number theory, and differential
geometry. Riemann held his first lectures in 1854, which founded the field of Riemannian
geometry and thereby set the stage for Albert Einstein’s general theory of relativity. In the
field of real analysis, he is mostly known for the first rigorous formulation of the integral,
the Riemann integral, and his work on Fourier series. His contributions to complex anal -
ysis include most notably the introduction of Riemann surfaces, breaking new ground in
a natural, geometric treatment of complex analysis. His 1859 paper on the prime-counting
function, containing the original statement of the Riemann hypothesis, is regarded as a foun -
dational paper of analytic number theory. He is considered by many to be one of the greatest
mathematicians of all time.
Example 5: Let f : [0 , 1] → R be given by
(
0 if x ∈ [0 , 1] ∩ Q
f (x) = 1 otherwise
For this function and all partitions P we have that
L(f, P) = 0 and U (f, P) = 1 .
Thus, f is not Riemann integrable.
We will be interested in the questions: ”What kind of functions are Riemann integrable?”
.... and ”How do we compute the integral?”
The answer to the second question will be the fundamental theorem of calculus. This will
be the topic of a later lecture. 6 TOBIAS HOLCK COLDING
Lemma 3: Suppose now that f : [ a, b] → R is a bounded function, then f ∈ R ([ a, b]) if
and only if for all > 0, there exists a partition P such that
U (f, P) − L(f, P) < .
Proof. Suppose that f ∈ R ([ a, b]), then
ZZ b b
sup L(f, P) = f dx = f dx = inf U (f, P) .
PPaa
This means that given > 0, there exists partitions P1 and P2 such that
Z b
f dx − < L (f, P1)
a
2
and Z b
U (f, P2) ≤ f dx + .
a
2
Let P be the partition that has all the dividing points of both P1 and P2. So P is a refinement
of both P1 and P2. It follows that
Z b Z b
a
f dx − < L(f, P1) ≤ L(f, P) ≤ U (f, P) ≤ U (f, P2) ≤
2 a
f dx + .
2
This proves the claim.
To see the converse, suppose that for some > 0, there exists a partition P such that
U (f, P) − L(f, P) < .
Since Z b
L(f, P) ≤ f dx
a
and Z b
f dx ≤ U (f, P)
a
we have that Z b Z b
f dx − f dx ≤ U (f, P) − L(f, P) < .
aa
Since this holds for all > 0 we get the claim.
We now get to a key theorem that gives a simple criterium for a function to be Riemann
integrable:
Theorem : Any continuous function on [a, b] is in R ([ a, b]).
Proof. We will show this next time once we have shown that a continuous function on a
closed and bounded interval is, in fact, uniformly continuous. SPRING 2025 - 18.100B/18.1002 7
The proof of this theorem needs the following key concept.
Definition : Uniformly continuous. Suppose that f : I → R is a function, where I is an
interval. We say that f is uniformly continuous if for all > 0, there exists a δ > 0 such that
|f (x) − f (y)| < if |x − y| < δ .
Note that being uniformly continuous is stronger than being continuous. It means that for
a given > 0, the same δ can be used for all x.
References
[TBB] B.S. Thomson, J.B. Bruckner, and A.M. Bruckner, Elementary Real Analysis, 2nd edition
TBB can be downloaded at:
(screen-optimized)
(print-optimized)
MIT, Dept. of Math., 77 Massachusetts Avenue, Cambridge, MA 02139-4307. MIT OpenCourseWare
100B Real Analysis
Spring 20 25
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Calculating Probability with Mean and Deviation
Calculating Probability with Mean and Deviation
By Kathleen Cantor, 30 Jan 2021
Calculating probability with mean and deviation depends on the type of distribution you'll base your calculations on. Here, we'll be dealing with typically distributed data.
If you have data with a mean μ and standard deviation σ, you can create models of this data using typical distribution. We can find the probability within this data based on that mean and standard deviation by standardizing the normal distribution.
The equation for the probability of a function or an event looks something like this (x - μ)/ σ where σ is the deviation and μ is the mean. Using the standard or z-score, we can use concepts of integration to have the function below.
This might appear strange at first, but what it means is that anyone can find probabilities for any given normal distribution as long as they have the mean and the standard deviation without having to do any integration. As long as you have the standardized table with a standardized normal curve with a standard deviation (unity) and a single mean, you can calculate probability using the z-score. It is this same table that we will use to calculate probabilities in the examples below.
Using Standard Normal Distribution Tables
You can download this Standard Normal Distribution Table from the University of Arizona as a pdf or excel file. Look closely at the table; you will see that it contains values from negative infinity to x. X values are from 0 to 3, and in very rare cases, 4 bringing the probability daringly close to unity or one.
This means that P(X ≤ x) =
Calculating P(x) may appear straightforward, but what if you want to calculate for a range of numbers, say p(X > x)? This is outside of the values on the table but P(X > x) = 1 – P(X ≤ x). In this case, we'll look for the value of P(X ≤ x) and subtract from one.
Examples
1. What is the probability that 5 is greater than x in a normally distributed data given that the mean is 6, and the standard deviation is 0.7.
Solution
P(X < 5) the first step is to find the z- score. We find that using the formula above.
z = (x – μ (mean)) / σ (standard deviation) this means that
For P(X < 5), z = (5 - 6)/0.7
-1/7 = - 1.42857 which is rounded up to – 1.43
Now in the table, we will look for the value of -1.4 under 3
\= 0.07636
The normal return for the z-score is usually less than, and because the function is asking for the probability of x being less than 5, this will be our final answer.
2. What is the probability that x is greater than 4.5 in a normally distributed data given that the mean is 6, and the standard deviation is 0.7.
Solution
P(X > 4.5) => the first step is to find the z- score. We find that using the formula below
z = (x – μ (mean)) / σ (standard deviation) this means that
For P(X > 4.5), z = (4.5 - 6)/0.7
-1.5/0.7 = - 2.14285 which is rounded up to – 2.14
Now in the table, we will look for the value of -2.1 under 4
\= 0.01618
The normalization table returns for the z-score is usually less than, but the function is asking for the probability of x being greater than 4.5; this means that the value we got is for x less than 4.5 and not greater than 4.5. To get the probability for x greater than 4.5, we will have to subtract the answer from unity.
\=> 1 - 0.01618 = 0.9838
3. Find the probability that x is greater than 3.8 but less than 4.7 in a normally distributed data given that the mean is 4 and the standard deviation is 0.5.
Solution
This problem is a bit different from the rest. Here we are asked to find the probability for two values when x is greater than 3.8 and less than 4.7. This means it falls between 3.9 and 4.6.
We can express this as P (3.8 < x <4.7).
Here we will be finding the z-score for P (x > 3.8) and P (x < 4.7). We find that using the formula below
z = (x – μ (mean)) / σ (standard deviation) this means that
for P (X > 3.8), z = (3.8 - 4)/0.5
-0.2/0.5 = - 0.400
Now in the table, we will look for the value of -0.4 under 0
\= 0.34458
For P (X < 4.7), z = (4.7 - 4)/0.5
0.7/0.5 = 1.40
Now in the table, we will look for the value of 1.4 under 0
\= 0.91924
We are going to subtract the upper limit by the lower limit
0.91924 - 0.34458 = 0.57466
The probability that x is greater than 3.8 but less than 4.7 is 0.57466
4. Find the probability that x is less than 6 but greater than 4 in a normally distributed data given that the mean is 5 and the standard deviation is 0.6.
Solution
We are looking for the probability that x ranges from 4.1 to 5.9
We can express this as P (4 < x < 6).
Here we will be finding the z-score for P (x > 4) and P (x < 6). We find that using the formula below
z = (x – μ (mean)) / σ (standard deviation) this means that
For P (X > 4), z = (4 - 5)/0.6
-1/0.6 = - 1.67
Now in the table, we will look for the value of -1.6 under 7
\= 0.04746
For P (X < 6), z = (6 - 5)/0.6
1/0.6 = 1.67
Now in the table, we will look for the value of 1.6 under 7
\= 0.95254
We are going to subtract the upper limit by the lower limit
0.95254 - 0.04746= 0.90508
The probability that x is less than 6 but greater than 4 are 0.90508
Conclusion
In a normally distributed data set, you can find the probability of a particular event as long as you have the mean and standard deviation. With these, you can calculate the z-score using the formula z = (x – μ (mean)) / σ (standard deviation). With this score, you can check up the Standard Normal Distribution Tables for the probability of that z-score occurring.
No matter the value of the mean and the standard deviation, the probability of x being equal to any number is automatically zero. There is an emphasis on a normally distributed data set because if your data isn't distributed normally, you may have to consider different factors like kurtosis.
See the 5 Comments below.
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Posted in Mathematics category - 30 Jan 2021 [Permalink]
5 Comments on “Calculating Probability with Mean and Deviation”
Peter says:
20 Mar 2021 at 10:45 am [Comment permalink]Great presentation however I don't understand the tables you're talking about and it wasn't displayed in this page.
Please is there a place I can be referred to see the table or shown how to create the table with the values myself.
Thank you and truly appreciate your kind efforts.
Kathleen Knowles says:
22 Mar 2021 at 11:31 pm [Comment permalink]Hi Peter, great question! There's a link to the table under the "Using Standard Normal Distribution Tables" section. I'm pasting the link here as well so you can see it:
Brandon Clark says:
27 Apr 2021 at 11:49 pm [Comment permalink]Do you guys have an example for a problem that is not normally distributed?
Raja says:
4 Sep 2021 at 9:15 pm [Comment permalink]Hello Kathleen,
Thank you for the article.
I am having trouble finding a single value, given mean and deviation. In my case, I have (n=1000) cookies, the mu=971g (average weight), sigma=15.2g (standard deviation), and I need to find the weight of a single cookie, including it's probability distribution. If I use the above formula for z, then I get: 1.91, and following the table I find the value 0.97-ish. But I am lost, how I can connect the answer to my question. My question is: what is the weight of a single cookie, and what is it's probability distribution?
Hope you can help.
Thanks in advance.
Casey Allen says:
8 Sep 2021 at 5:07 am [Comment permalink]Hi Raja,
I'd recommend posting this question in the IntMath Forum where others are working through problems and can assist you.
Thanks!
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January 6, 2012
Dear Mayo Clinic:
Is it true that ovarian cancer may actually start in the fallopian tubes? Will having a tubal ligation lessen my chance of developing ovarian cancer? I have a family history of the disease and am a BRCA1 carrier.
Answer:
Recent research seems to suggest that some of the cells that may form one type of ovarian cancer come from the fallopian tubes. Other research has consistently shown that tubal ligation decreases the risk of developing ovarian cancer. But it's actually not clear if those two findings are related to one another.
Ovaries are the almond-shaped organs on each side of the uterus that contain eggs and make the hormones that control a woman's reproductive cycle. The fallopian tubes are the passages through which eggs travel from the ovaries to the uterus. The fimbria is a fringe around the opening of a fallopian tube, in the direction of the ovary, but not attached to the ovary.
Ovarian cancer can begin in cells located on the tissue that covers the outside of the ovary, in the egg-producing cells within an ovary, or in an ovary's hormone-producing cells. Most ovarian cancers begin on the outside cells. When investigating a possible link between the fallopian tubes and ovarian cancer, researchers examined the cells on the fallopian tube fimbria in women who had the BRCA1 or BRCA2 gene mutation. These gene mutations put women at high risk for breast and ovarian cancer.
The researchers found that a significant percentage of women in the study had abnormal cells in the fallopian tube fimbria that were either precancerous or cancerous. This research suggests the distinct possibility that, rather than beginning in the cells on the outside layer of the ovary, ovarian cancer on the surface of the ovary may actually come from cancerous cells originating in the fimbria.
Based on this information, researchers and doctors are now considering if, in an effort to catch ovarian cancer early, they should routinely examine the tissue of the fallopian tubes, especially in women who have the BRCA1 or BRCA2 gene mutation. Discussion is also under way regarding the possibility of removing the fallopian tubes, or perhaps just the fimbria, as a way to reduce the risk of ovarian cancer.
In a tubal ligation procedure — also known as "having your tubes tied" — the fallopian tubes are cut or blocked permanently, preventing eggs from traveling from the ovaries to the uterus. Because tubal ligation may not have a direct impact on the fimbria, it is not clear if the procedure would affect the transfer of abnormal cells from the fimbria to the ovaries.
However, other research has shown that tubal ligation substantially lowers a woman's risk of ovarian cancer. Researchers are not sure exactly why that is the case. Some suspect the reason may be that tubal ligation cuts off the ovaries' exposure to outside environmental factors that may increase the risk of ovarian cancer. Others think it could be related to the anatomical changes in the fallopian tubes that happen after a tubal ligation.
Because you carry the BRCA1 gene mutation, making your risk of ovarian cancer much higher than normal, a discussion with your doctor about ways to decrease your ovarian cancer risk would definitely be worthwhile. The discussion should include risks, benefits and possible alternatives.
— Paul Haluska, M.D., Ph.D., Medical Oncology, Mayo Clinic, Rochester, Minn.
Tubal Ligation Decreases Risk of Developing Ovarian Cancer
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Robbins & Kumar Basic Pathology, 11th Edition
Editors :
Vinay Kumar & Abul K. Abbas & Jon C. Aster & Andrea T Deyrup
Readable, well-illustrated, and concise, Robbins and Kumar’s Basic Pathology, 11th Edition, offers today’s busy students a rich understanding of all essential pathology concepts from trusted names in the field. This updated edition thorou ...view more
Readable, well-illustrated, and concise, Robbins and Kumar’s Basic Pathology, 11th Edition, offers today’s busy students a rich understanding of all essential pathology concepts from trusted names in the field. This updated edition thoroughly covers key pathologic processes and the time-honored tools of gross and microscopic analysis, while also retaining a strong emphasis on clinicopathologic correlations and the impact of molecular pathology on the practice of medicine. Outstanding artwork and schematic drawings, as well as a robust eBook experience with extensive additional features, make complex concepts easier to learn and retain.
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Readable, well-illustrated, and concise, Robbins and Kumar’s Basic Pathology, 11th Edition, offers today’s busy students a rich understanding of all essential pathology concepts from trusted names in the field. This updated edition thoroughly covers key pathologic processes and the time-honored tools of gross and microscopic analysis, while also retaining a strong emphasis on clinicopathologic correlations and the impact of molecular pathology on the practice of medicine. Outstanding artwork and schematic drawings, as well as a robust eBook experience with extensive additional features, make complex concepts easier to learn and retain.
Key Features
Includes fully updated clinical topics throughout
Features high-quality photomicrographs, gross photos, and radiologic images, as well as new artwork and over 150 new schematic diagrams that help summarize key or complex disease mechanisms
Contains a new Rapid Review section that uses bulleted summary boxes to deliver essential take-home messages and help you focus on the fundamentals.
Includes tables of relevant laboratory tests for each chapter that link pathophysiology of disease and diagnostic testing
Highlights pathogenesis, morphology, and pathophysiologic content throughout.
Features increased representation ofdiverse populations throughout the text, including clinical photographs of skin lesions in multiple skin types and a new section on the role of socially defined race in health disparities
An eBook version is included with purchase. The eBook allows you to access all of the text, figures and references, with the ability to search, customize your content, make notes and highlights, and have content read aloud—as well as access bonus content, including case studies, additional gross and microscopic figures, and more
Author Information
Edited by Vinay Kumar, MBBS, MD, FRCPath, Professor and Chairman, Department of Pathology, University of Chicago, Pritzker School of Medicine, Chicago, IL, USA; Abul K. Abbas, MBBS, Emeritus Professor, Department of Pathology, University of California San Francisco, USA; Jon C. Aster, MD, PhD, Professor, Department of Pathology, Harvard Medical School; Brigham and Women's Hospital, Boston, Massachusetts, USA and Andrea T Deyrup, M.D., Ph.D., Associate Professor Department of Pathology Duke University Medical Center
Product Details
More Information| ISBN Number | 9780323790185 |
| Main Author | Edited by Vinay Kumar, MBBS, MD, FRCPath, Abul K. Abbas, MBBS, Jon C. Aster, MD, PhD and Andrea T Deyrup, M.D., Ph.D. |
| Copyright Year | 2023 |
| Edition Number | 11 |
| Format | Book |
| Trim | 216w x 276h (8.50" x 10.875") |
| Illustrations | Illustrated |
| Imprint | Elsevier |
| Page Count | 840 |
| Publication Date | 12 Dec 2022 |
| Stock Status | IN STOCK |
Table Of Contents
Cell Injury, Cell Death, and Adaptations
Inflammation and Repair
Hemodynamic Disorders, Thromboembolism, and Shock
Genetic and Pediatric Diseases
Diseases of the Immune System
Neoplasia
Environmental and Nutritional Diseases
Blood Vessels
Heart
Hematopoietic and Lymphoid Systems
Lung
Kidney and Its Collecting System
Oral Cavities and Gastrointestinal Tract
Liver and Gallbladder
Pancreas and Biliary System
Male Genital System and Lower Urinary Tract
Female Genital System and Breast
Endocrine System
Bones, Joints, and Soft Tissue Tumors
Peripheral Nerves and Muscles
Central Nervous System
Skin
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242533 Robbins & Kumar Basic Pathology 92.99 99.99 USD InStock/Medical Students/Pathology/Books/Medicine/Pathology/Clinical Medicine & Examination/Seven Percent Off Select Products/USLME Revision Sale - 20% Off Select Books/Winter Sale - 20% Off Select Books and eBooks/25% Off Select Products/Complete Anatomy and Osmosis Offer Products/Product Format/Book 27 5255039 25 5497199 5721391 5924151 5924653 5936303 5936756 5936802 1 3 7 5145120 8 5936801Readable, well-illustrated, and concise, Robbins and Kumar’s Basic Pathology, 11th Edition, offers today’s busy students a rich understanding of all essential pathology concepts from trusted names in the field. This updated edition thoroughly covers key pathologic processes and the time-honored tools of gross and microscopic analysis, while also retaining a strong emphasis on clinicopathologic correlations and the impact of molecular pathology on the practice of medicine. Outstanding artwork and schematic drawings, as well as a robust eBook experience with extensive additional features, make complex concepts easier to learn and retain.Readable, well-illustrated, and concise, Robbins and Kumar’s Basic Pathology, 11th Edition, offers today’s busy students a rich understanding of all essential pathology concepts from trusted names in the field. This updated edition thoroughly covers key pathologic processes and the time-honored tools of gross and microscopic analysis, while also retaining a strong emphasis on clinicopathologic correlations and the impact of molecular pathology on the practice of medicine. Outstanding artwork and schematic drawings, as well as a robust eBook experience with extensive additional features, make complex concepts easier to learn and retain.0 0 add-to-cart 9780323790185 2022 Professional Edited by Vinay Kumar, MBBS, MD, FRCPath, Abul K. Abbas, MBBS, Jon C. Aster, MD, PhD and Andrea T Deyrup, M.D., Ph.D.2023 11 Book 216w x 276h (8.50" x 10.875")Illustrated Elsevier 840 Dec 12, 2022 IN STOCK Edited by Vinay Kumar, MBBS, MD, FRCPath, Professor and Chairman, Department of Pathology, University of Chicago, Pritzker School of Medicine, Chicago, IL, USA; Abul K. Abbas, MBBS, Emeritus Professor, Department of Pathology, University of California San Francisco, USA; Jon C. Aster, MD, PhD, Professor, Department of Pathology, Harvard Medical School; Brigham and Women's Hospital, Boston, Massachusetts, USA and Andrea T Deyrup, M.D., Ph.D., Associate Professor Department of Pathology Duke University Medical Center Books, eBooks Book Robbins Pathology US No No No No Please Select Please Select No No Please Select
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9341 | https://sites.google.com/umd.edu/statisticsinsocialsciences/18-confidence-intervals/18-d-computing-confidence-intervals-with-the-binomial-distribution | Search this site
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OpenPSYC
Computing confidence intervals with the binomial distribution
During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different.
How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B(n, p) where nis the number of trials and p is the probability of a success. To form a proportion, take X, the random variable for the number of successes and divide it by n, the number of trials (or the sample size). The random variable P′(read “P prime”) is that proportion,
P' = x/n
Sometimes the random variable is denoted as P^, read “P hat”.)
When n is large and p is not close to zero or one, we can use the normal distribution to approximate the binomial.
XN(np, √npq)
f we divide the random variable, the mean, and the standard deviation by
n, we get a normal distribution of proportions with P′, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by n.)
The confidence interval has the form (p′ – EBP, p′ + EBP). EBP is error bound for the proportion.
In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q. The estimated proportionsp′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures.
The confidence interval can be used only if the number of successes np′ and the number of failures nq′ are both greater than five.
__________________________________________
EXAMPLE 1
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes – they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
The first solution is step-by-step (Solution A).
The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).
Solution A:
Let X = the number of people in the sample who have cell phones. X is binomial.
X ~ B(500, 421/500)
To calculate the confidence interval, you must find p′, q′, andEBP.
n = 500
x = the number of successes = 421
p’= x/n = 421/500 = 0.842
p′ = 0.842 is the sample proportion; this is the point estimate of the population proportion.
q′ = 1 – p′ = 1 – 0.842 = 0.158
Since CL = 0.95, then α = 1 – CL = 1 – 0.95 = 0.05 (α) = 0.025.
Then
zα/2 = z0.025 = 1.96
Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find z0.025. Remember that the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
EBP = (Za/2)(sqrt((p'q')/n) = (1.96) (sqrt((0.8420.158)/500)) = 0.032
p‘−EBP=0.842−0.032=0.81
p′+EBP=0.842+0.032=0.874
The confidence interval for the true binomial population proportion is ( p′ – EBP, p′ + EBP) = (0.810, 0.874).
Interpretation
We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
Explanation of 95% Confidence Level
Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution B:
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.Arrow down to and enter 421.Arrow down to and enter 500.Arrow down to C-Level and enter .95.Arrow down to Calculate and press ENTER.The confidence interval is (0.81003, 0.87397).
__________________________________________
“Plus Four” Confidence Interval for p
There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.
Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is n + 4, and the new count of successes is x + 2.
Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten.
__________________________________________
EXAMPLE 2
A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke.
Solution A:
Six students out of 25 reported smoking within the past week, so x = 6 and n = 25. Because we are using the “plus-four” method, we will use x = 6 + 2 = 8 and n = 25 + 4 = 29.
p' = x/n = 8/29 = 0.276
q' = 1-p' = 1-0.276 = 0.724
Since CL = 0.95, we know z0.025=1.96
We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.
Solution B:
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials.
Arrow down to x and enter eight.
Arrow down to n and enter 29.
Arrow down to C-Level and enter 0.95.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.113, 0.439).
__________________________________________
The confidence interval for the larger sample is narrower than the interval from Example 6. Larger samples will always yield more precise confidence intervals than smaller samples. The “plus four” method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13/50) to 0.278 (15/54). It has a smaller impact on the EPB, changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159/588) to 0.272 (161/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.
Calculating the Sample Size n
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.
The error bound formula for a population proportion is EBP = (Za/2)(sqrt((p'q')/n))
Solving for n gives you an equation for the sample size.
Concept Review
Some statistical measures, like many survey questions, measure qualitative rather than quantitative data. In this case, the population parameter being estimated is a proportion. It is possible to create a confidence interval for the true population proportion following procedures similar to those used in creating confidence intervals for population means. The formulas are slightly different, but they follow the same reasoning.
Let p′ represent the sample proportion, x/n, where x represents the number of successes and n represents the sample size. Let q′ = 1 – p′. Then the confidence interval for a population proportion is given by the following formula:
(lower bound, upper bound)
The “plus four” method for calculating confidence intervals is an attempt to balance the error introduced by using estimates of the population proportion when calculating the standard deviation of the sampling distribution. Simply imagine four additional trials in the study; two are successes and two are failures. Calculate , and proceed to find the confidence interval. When sample sizes are small, this method has been demonstrated to provide more accurate confidence intervals than the standard formula used for larger samples.
Formula Review
p′ = x / n where x represents the number of successes and n represents the sample size. The variable p′ is the sample proportion and serves as the point estimate for the true population proportion.
q′ = 1 – p′
The variable p′ has a binomial distribution that can be approximated with the normal distribution shown here.
s is a point estimate for σ
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9342 | https://www.sciencedirect.com/science/article/abs/pii/S2352152X22007654 | Nucleation in supercooled water triggered by mechanical impact: Experimental and theoretical analyses - ScienceDirect
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Abstract
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Cited by (24)
Journal of Energy Storage
Volume 52, Part A, 1 August 2022, 104755
Research Papers
Nucleation in supercooled water triggered by mechanical impact: Experimental and theoretical analyses
Author links open overlay panel Liping Wang, Fuxin Wang, Chenlv Lu, Hong Liu
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Highlights
•Method of mechanical impact to actively trigger nucleation.
•Effect of impact in supercooled water on nucleation is investigated.
•An empirical revised model between impact and supercooling is developed.
•Supercooling reduction can be continuously controlled up to 80% as required.
•Nucleation mechanism under impact is analyzed and revealed.
Abstract
Supercooling is one of the major drawbacks that hinder the wide application of phase change thermal storage technology, where supercooling needs to be suppressed, or nucleation triggered at desired supercooling or time. The disturbance of mechanical impact can effectively promote nucleation. However, up to now, its nucleation mechanism has not been well explained, and the corresponding nucleation triggering technology has not been reasonably developed. In this paper, the effect of mechanical impact in supercooled water on nucleation is investigated experimentally by impact rod falling freely to hit the bottom of container. The nucleation temperature under impact at different heights of free fall are measured. The results show that 80% supercooling reduction can be achieved by a low impact energy of 0.007 J. In addition, we modify the existing model based on the experimental data to develop a revised model, which can estimate the impact stimulus required to trigger nucleation at desired supercooling. Moreover, the nucleation mechanism under impact based on classical nucleation theory is analyzed. By comparing the changes in pre-exponential factor and nucleation barrier with and without impact, the enhancement in nucleation rate under impact can be explained as a reduction on the free energy barrier for nucleation. Our experimental and theoretical analysis results preliminarily reveal the impact nucleation mechanism, which are expected to promote more in-depth research on the related topics. These results could also better serve to predict whether nucleation will occur in the supercooled water under mechanical impact, and to actively trigger nucleation or suppress supercooling as needed through mechanical impact.
Introduction
Substances in a supercooled state generally exist in nature and have great influence on our daily life. It is very important to effectively control supercooling in many fields such as energy storage, supercooled large drop icing condition simulation, ice slurry production, food and biological sample preservation and so on , , , , . Of particular concern is that energy has always been one of the major issues of mankind. Since phase change materials (PCMs) can absorb or release heat during the phase change process, the use of PCMs for thermal energy storage has attracted great attention , , , , , . However, many PCMs, especially inorganic PCMs, have inherent characteristics of supercooling, which is one of the major drawbacks against their practical application , . Supercooling leads to a metastable state, and the occurrence of nucleation is usually affected by various external perturbations including electric and magnetic fields, impurities, thermodynamic non-equilibrium, mechanical disturbance, ultrasonic and so on , , , , , , , , , . Hence, a variety of methods have been studied to control the supercooling of PCMs , , , , , , . For the comprehensive details, one can refer to recent review articles on key issues such as supercooling, supercooling elimination techniques, nucleation triggering methods, phase change hysteresis of PCMs , , , , , , . It can be seen that the problem to suppress supercooling, release heat on demand or trigger nucleation at desired temperature for supercooled materials has not been well solved.
In our previous researches , , in which the controllable generation of supercooled droplets is required, we found that the mechanical impact in the supercooled water can effectively promote nucleation. In fact, the research on the influence of mechanical stimulus on nucleation has never stopped in the past two centuries , , , , , , . For example, Edwards et al. have studied nucleation of ice by mechanical shock and found that the mechanical shock seemed to induce supercooled water to freeze only in the presence of certain solid surfaces . Young , has successively studied the effects of mechanical stimuli such as stirring, rotary friction and impact on nucleation and found that they all caused nucleation to varying degrees, but only established an empirical relation between impact and supercooling. Hozumi et al. have provided a freezing device capable of freezing supercooled water instantly at any desired time. But until now, the mechanism of impact-induced nucleation is still unclear, and most studies have only speculated and inferred, but not analyzed. For example, Woo et al. considered that any agitation will cause the water molecules to rearrange to form nuclei . Saito et al. considered that the water molecules get closer to each other during the process of collision between solids, and it causes induction of the growth of an ice embryo .
However, understanding the mechanism of nucleation is very important for developers to design devices. On the mechanism of nucleation with mechanical disturbances , , , , , , , Mullin and Raven have found that the critical supercooling to nucleation decreases with the increasing intensity of agitation first, then increases, and finally decreases. They have attributed this to the combined effect of diffusion and attrition. Liu et al. , have studied the influence of agitation and fluid shear on nucleation in solution, and the results showed the same trend. It was explained as an effect on the pre-exponential factor by fitting parameters in classical nucleation theory (CNT) to experimental data. Steendam et al. have found that the shear rate was rationalized to be the kinetic parameter in CNT which changes most significantly when the crystallization process was scaled-up. Myint et al. have developed a CNT-based model to reproduce the experimentally observed rapid freezing of water at high pressures. It can be seen that analyzing the influence of disturbance on nucleation with the help of CNT based on experimental data is one of the effective means to understand the corresponding nucleation mechanism.
Water is the best known and widely used PCM because it is inexpensive, readily available, chemically stable and environment friendly , . To controllably trigger nucleation within a large supercooling range or actively suppress supercooling, the present work uses it to study the influence law of impact on nucleation through experiment, in which nucleation is triggered by impact disturbance in supercooled water generated by free fall. The relationship between impact and supercooling established by Young and Van Sicklen is revised. What's more, the current work focuses on the analysis of the impact-induced nucleation mechanism with the help of CNT.
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Section snippets
Experimental setup
In order to study the effect of collisions between objects on the stability of supercooled water, an experimental setup shown in Fig. 1 is built. It mainly includes five parts: refrigeration, water container, impact application, temperature measurement and icing observation. The refrigeration part includes a circulating bath (AP15R-40-A12Y, PolyScience) and a U-shaped cavity. The refrigerant circulates in the U-shaped cavity to cool the inside chamber, in which the temperature can be controlled
Nucleation temperature
In order to effectively obtain the variation trend of the degree of supercooling under the impact, for selection of the experimental height, the interval is small when the falling height is low, and then gradually increased to the same. In the end, a total of 9 experiments with a free fall height h of 0.5, 1, 2, 4, 6, 8, 10, 12, and 14 mm are selected. Since nucleation in metastable supercooled water has significant randomness, each experiment is repeated 20 times to ensure the reliability of
Conclusions
Ice nucleation in supercooled water triggered by mechanical impact in a container is designed and the effect of mechanical impact disturbance on the ice nucleation is studied experimentally in this paper. On the basis of the study conducted by Young and Van Sicklen , the relationship between impact disturbance and supercooling is revised. With the help of classical nucleation theory, the mechanism of nucleation induced by impact disturbance is analyzed theoretically. The main conclusions
CRediT authorship contribution statement
Liping Wang: Conceptualization, Methodology, Validation, Formal analysis, Investigation, Data Curation, Writing – Original Draft, Visualization, Funding acquisition. Fuxin Wang: Conceptualization, Resources, Writing – Review & Editing, Supervision. Chenlv Lu: Investigation, Writing – Review & Editing, Visualization. Hong Liu: Conceptualization, Resources, Writing – Review & Editing, Project administration.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgments
This work was Sponsored by Shanghai Sailing Program (No. 22YF1419000). The authors would also like to thank the anonymous reviewers for their meticulous, professional and laudable review of the manuscript.
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Characterizing high pressure phase transitions of gelatin gel and assessing microstructural, gel properties, and texture changes following pressure shift freezing and Ice I to Ice III phase transition
2025, Food Hydrocolloids Show abstract This study explored the high pressure phase transition behavior of 10% gelatin gel and compared the effects of four freezing treatments and subsequent Ice I to Ice III solid-solid phase transition (SSPT) treatments on the ice crystal characteristics, microstructure after thawing, gel properties, and texture of the gel samples. The melting points of the gel in the Ice I and Ice III regions were determined and the phase diagram was fitted using the polynomial and Simon-like formulas, and the resulting R 2 values were all found to be very close to 1. Utilizing a 10% sodium chloride solution as a disturbance agent facilitated the premature phase transition of gel samples, resulting in a substantial reduction in phase transition pressure by 57.4 MPa. Two pressure shift freezing methods at 160 MPa, namely pressure shift nucleation coupled with air freezing (PSF A) and pressure shift nucleation coupled with liquid immersion freezing (PSF L), were compared with conventional liquid immersion freezing (LIF) and air freezing (AF) techniques. The PSF A and PSF L methods significantly reduced the mean diameter, cross-sectional area, and perimeter of ice crystals within the gel, resulting in smaller pores and increased gel strength in the thawed gel. PSF L demonstrated the most favorable outcomes across all aspects. SSPT treatment on the basis of these four freezing treatments revealed that the ice crystal dimensions within the PSF L-SSPT gels were further decreased, with no significant change in gel strength observed, and with no significant differences in cohesiveness and chewiness from the untreated group. This demonstrated that the PSF L-SSPT treatment could effectively maintain the frozen food quality.
### Controllable heat release of supercooled Erythritol-based phase change materials for long-term thermal energy storage
2024, Chemical Engineering Journal Show abstract Transeasonal heat storage in organic phase change materials (PCMs) present a promising solution to the intermittent nature of renewable energy. However, PCMs are prone to spontaneous crystallization during storage, leading to the loss of stored latent heat in low-temperature environments. In this study, we incorporated tetrasodium ethylenediaminetetraacetic acid (EDTA-4Na) and superabsorbent polymer (SAP) to erythritol (ERY), referred to as EES-PCMs, to overcome these challenges and achieve more controllable and stable thermal energy storage. The incorporation of EDTA-4Na and SAP into ERY significantly improves the supercooling stability and phase change enthalpy. The optimal ratio (EES-PCMs-2) of phase change enthalpy reaches an impressive 286.62 J/g, with stable performance maintained for 120 days at room temperature. The EES-PCMs-2 exhibits exceptional thermal cycling stability, retaining its properties even after 100 cycles. A novel air-triggered crystallization method is demonstrated, enabling a temperature increase from room temperature to 48.21°C in 320 s after being exposed to air for long-term storage. This innovative approach effectively overcomes the limitations of traditional triggering mechanisms, providing a straightforward and efficient method for thermal management. The high thermal storage capacity, stability, and controlled exothermic properties of EES-PCMs position them as promising candidates for applications in seasonal solar thermal energy storage.
### Research progress of seasonal thermal energy storage technology based on supercooled phase change materials
2023, Journal of Energy Storage Citation Excerpt :
Wang et al. found that low impact energy of 0.007 J could reduce the supercooling degree by 80 %. In addition, they modified an existing model based on experimental data to create a revised model that estimates the impact energy required to trigger nucleation at the desired supercooling . Triggering crystallization can also be achieved by applying an electric field, and Fig. 20 shows the use of an electric field to trigger the nucleation of sodium acetate trihydrate. Show abstract Seasonal thermal energy storage (STES) is a highly effective energy-use system that uses thermal storage media to store and utilize thermal energy over cycles, which is crucial for accomplishing low and zero carbon emissions. Sensible heat storage, latent heat storage, and thermochemical heat storage are the three most prevalent types of seasonal thermal energy storage. In recent years, latent heat storage based on phase change materials(PCMs) has made great progress in solar energy utilization. However, the inherent defects of phase change materials have become resistant, limiting their further development, including low thermal conductivity, phase separation, and susceptibility to leakage. Supercooling is frequently considered a negative to be avoided in short-term usage. In seasonal thermal energy storage, however, supercooling of PCMs becomes an advantage. The paper begins with a brief overview of existing methods of seasonal thermal energy storage. Afterward, a brief description of the research on PCMs capable of storing seasonal heat is provided. A detailed discussion of the current state of research into supercooled PCMs for seasonal thermal energy storage and systems is presented. Finally, we present a detailed outline of the future directions of seasonal thermal energy storage using PCMs. According to current researches, finding and preparing PCMs with stable supercooling, designing suitable trigger crystallization devices and seasonal thermal energy storage systems with high performance will be the three directions for future research.
### Experimental investigation on space heating performances of supercooled thermal storage units with sodium acetate trihydrate
2022, Energy and Buildings Show abstract Supercooled salt hydrates could be applied for long-term solar thermal storage and then be activated to release the stored latent heat on demand. To realize this strategy, the space heating performances with supercooled storage units need to be clarified. Experiments were performed on two heating modes using thermal storage units filled with supercooled liquid sodium acetate trihydrate (SAT), i.e. the capillary-mat floor water heating and direct heat release to the room air, respectively. The results show that, with the terminal of capillary-mat radiant floor, the maximum supply water temperature reaches 35°C within 0.3 h and the room temperature increases from initial 20.3°C to the maximum 25.8°C within 2.8 h. The room temperature could be kept above 24.8°C after the system runs for 12 h. Also, the temperature difference between the 0.5 m and 1.5 m height levels in the room space is within 0.2°C. While the thermal storage units directly release heat to the room air with the same mass of SAT, the room temperature could rise up to 27.2°C within 1.1 h which is higher than that of the capillary-mat terminal even in shorter period. However, the room temperature drops quickly and is about 20.9°C after 12 h, and the maximum temperature difference between the heights of 0.5 m and 1.5 m in the room is 2.4°C. Although the combination of supercooled liquid SAT and the capillary-mat radiant floor terminal has delayed heating rate into the room space, it presents less temperature swing and lower vertical temperature gradient, thus not only meets the room temperature requirement, but also has better thermal comfort. The results provide guidance for application of supercooled liquid PCMs for solar thermal storage in space heating systems.
### The importance of supercooled stability for food during supercooling preservation: a review of mechanisms, influencing factors, and control methods
2024, Critical Reviews in Food Science and Nutrition
### Nucleation enhancement by energy dissipation with the collision of a supercooled water droplet
2023, Physics of Fluids
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9343 | https://www.youtube.com/watch?v=AzBs9LiV5aA | Writing: Syntax — Basic Example | Writing & Language | SAT | Khan Academy
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a newly beach in Maui has black sand the sand is made up of ground lava instead of ground coral and shells so let's think about this so the way it read isn't is it horrible and I actually like breaking things up into into kind of smaller sentences that tends to just me kind of my style but it feels a little redundant your your you say on Onew Lee or on a loop on Olli I don't know the right way to pronounce it Beach and Maui has black sand and then you're saying the sand is made up of ground lava so it does feel like well you know we're just finished talking about the sand then you'd say the sand again so maybe a kind of a more elegant way would just to be kind of refer to this and maybe not even have another sentence so let's look at the other choices Oh Noli beach in Maui has black sand with the sand being made up of ground lava yeah that's you know they're they're kind of going in the direction that I'm implying but with the sand being made up it feels still redundant your said sand and then you're saying sand again so the beach in Maui has black sand and this sand is so once again you're using sand and sand it's kind of the same problem that you did up here sand and then you're immediately saying the sand again oh no on the beach in Maui has black sand which is made up of ground lava instead of ground coral and and shells yeah this one feels nice this one feels really good because you're not being you're not repeating the word sand over and over again unnecessarily we know you're talking about sand beach in Maui has black sand which is the second up here were before we corrected it the second sentence is referring is referring is talking about this black sand so why don't we just say why don't we just create a clause where we say which is made up of ground lava instead of coral instead of ground coral and shells it actually makes a connection in my brain a little bit clearer that that okay that what we're talking about made up of ground coral ground lava instead of ground coral and shells they're talking about they're talking about this black sand right over here so I definitely like this last choice the most |
9344 | https://www.coursera.org/learn/discrete-mathematics | Discrete Mathematics
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Discrete mathematics forms the mathematical foundation of computer and information science. It is also a fascinating subject in itself.
Learners will become familiar with a broad range of mathematical objects like sets, functions, relations, graphs, that are omnipresent in computer science. Perhaps more importantly, they will reach a certain level of mathematical maturity - being able to understand formal statements and their proofs; coming up with rigorous proofs themselves; and coming up with interesting results.
This course attempts to be rigorous without being overly formal. This means, for every concept we introduce we will show at least one interesting and non-trivial result and give a full proof. However, we will do so without too much formal notation, employing examples and figures whenever possible.
The main topics of this course are (1) sets, functions, relations, (2) enumerative combinatorics, (3) graph theory, (4) network flow and matchings. It does not cover modular arithmetic, algebra, and logic, since these topics have a slightly different flavor and because there are already several courses on Coursera specifically on these topics.
This module gives the learner a first impression of what discrete mathematics is about, and in which ways its "flavor" differs from other fields of mathematics. It introduces basic objects like sets, relations, functions, which form the foundation of discrete mathematics.
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2 videos
1 assignment
2 peer reviews
2 videos • Total 26 minutes
Introduction to the course•16 minutes
Sets, Relations, Functions•10 minutes
1 assignment • Total 90 minutes
Sets, relations, and functions•90 minutes
2 peer reviews • Total 180 minutes
Exercises for introduction lesson•90 minutes
Sets, Relations, Functions•90 minutes
Even without knowing, the learner has seen some orderings in the past. Numbers are ordered by <=. Integers can be partially ordered by the "divisible by" relation. In genealogy, people are ordered by the "A is an ancestor of B" relation. This module formally introduces partial orders and proves some fundamental and non-trivial facts about them.
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2 videos
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1 peer review
2 videos • Total 28 minutes
Partial orderings: basic notions•13 minutes
Mirsky's and Dilworth's Theorem•14 minutes
1 assignment • Total 120 minutes
Partial orders, maximal and minimal elements, chains, antichains•120 minutes
1 peer review • Total 120 minutes
Partial orders, maximal and minimal elements, chains, antichains•120 minutes
A big part of discrete mathematics is about counting things. A classic example asks how many different words can be obtained by re-ordering the letters in the word Mississippi. Counting problems of this flavor abound in discrete mathematics discrete probability and also in the analysis of algorithms.
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3 videos
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1 peer review
3 videos • Total 34 minutes
How to Count Functions, Injections, Permutations, and Subsets•11 minutes
Evaluating Simple Sums•8 minutes
Pascal's Triangle•14 minutes
1 assignment • Total 120 minutes
Counting Basic Objects•120 minutes
1 peer review • Total 120 minutes
Counting Basic Objects•120 minutes
The binomial coefficient (n choose k) counts the number of ways to select k elements from a set of size n. It appears all the time in enumerative combinatorics. A good understanding of (n choose k) is also extremely helpful for analysis of algorithms.
What's included
3 videos
1 assignment
2 peer reviews
3 videos • Total 54 minutes
Combinatorial Identities•14 minutes
Estimating the Binomial Coefficient•22 minutes
Excursion to Discrete Probability: Computing the Expected Minimum of k Random Elements from {1,...,n}•18 minutes
1 assignment • Total 30 minutes
An Eagle's View of Pascal's Triangle•30 minutes
2 peer reviews • Total 200 minutes
Combinatorial Identities•100 minutes
Digging Into Pascal's Triangle•100 minutes
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1 video
1 assignment
2 peer reviews
1 video • Total 14 minutes
Asymptotics and the O( )-Notation•14 minutes
1 assignment • Total 30 minutes
The Big-O-Notation•30 minutes
2 peer reviews • Total 240 minutes
Basic Facts•120 minutes
Classes that often occur in complexity theory•120 minutes
Graphs are arguably the most important object in discrete mathematics. A huge number of problems from computer science and combinatorics can be modelled in the language of graphs. This module introduces the basic notions of graph theory - graphs, cycles, paths, degree, isomorphism.
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3 videos
1 assignment
2 peer reviews
3 videos • Total 40 minutes
Basic Notions and Examples•10 minutes
Graph Isomorphism, Degree, Graph Score•13 minutes
Graph Score Theorem•16 minutes
1 assignment • Total 30 minutes
Graphs, isomorphisms, and the sliding tile puzzle•30 minutes
2 peer reviews • Total 210 minutes
Graphs and Isomorphisms•90 minutes
The Graph Score Theorem•120 minutes
We continue with graph theory basics. In this module, we introduce trees, an important class of graphs, and several equivalent characterizations of trees. Finally, we present an efficient algorithm for detecting whether two trees are isomorphic.
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3 videos
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3 videos • Total 36 minutes
Graphs and Connectivity•8 minutes
Cycles and Trees•15 minutes
An Efficient Algorithm for Isomorphism of Trees•12 minutes
1 assignment • Total 30 minutes
Cycles and Trees•30 minutes
2 peer reviews • Total 220 minutes
Cycles and Trees•120 minutes
Spanning Tree Exchange Graph•100 minutes
Starting with the well-known "Bridges of Königsberg" riddle, we prove the well-known characterization of Eulerian graphs. We discuss Hamiltonian paths and give sufficient criteria for their existence with Dirac's and Ore's theorem.
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2 videos
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2 videos • Total 27 minutes
Eulerian Cycles•11 minutes
Hamilton Cycles - Ore's and Dirac's Theorem•16 minutes
1 assignment • Total 60 minutes
Hamiltonian Cycles and Paths•60 minutes
1 peer review • Total 120 minutes
Hamiltonian Cycles and Paths•120 minutes
We discuss spanning trees of graphs. In particular we present Kruskal's algorithm for finding the minimum spanning tree of a graph with edge costs. We prove Cayley's formula, stating that the complete graph on n vertices has n^(n-2) spanning trees.
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2 videos
1 assignment
2 peer reviews
2 videos • Total 29 minutes
Minimum Spanning Trees•13 minutes
The Number of Trees on n Vertices•15 minutes
1 assignment • Total 40 minutes
Spanning Trees•40 minutes
2 peer reviews • Total 220 minutes
Minimum Spanning Trees•100 minutes
Counting Trees on n Vertices•120 minutes
This module is about flow networks and has a distinctively algorithmic flavor. We prove the maximum flow minimum cut duality theorem.
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2 videos
1 assignment
1 peer review
2 videos • Total 29 minutes
Flow Networks, Flows, Cuts: Basic Notions and Examples•14 minutes
Flow Networks: The Maxflow - Mincut Theorem•15 minutes
1 assignment • Total 30 minutes
Network flow•30 minutes
1 peer review • Total 120 minutes
Network Flows•120 minutes
We prove Hall's Theorem and Kőnig's Theorem, two important results on matchings in bipartite graphs. With the machinery from flow networks, both have quite direct proofs. Finally, partial orderings have their comeback with Dilworth's Theorem, which has a surprising proof using Kőnig's Theorem.
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3 videos
1 peer review
3 videos • Total 46 minutes
Matchings in Bipartite Graphs - Basic Notions and an Algorithm•13 minutes
Matchings in Bipartite Graphs: Hall's and König's Theorem•16 minutes
Partial Orders: Dilworth's Theorem on Chains and Antichains•15 minutes
1 peer review • Total 60 minutes
Matchings in Bipartite Graphs•60 minutes
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9345 | https://en.wikipedia.org/wiki/Golgi_apparatus | Published Time: 2001-09-20T14:17:48Z
Golgi apparatus - Wikipedia
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(Top)
1 Discovery
2 Subcellular localization
3 Structure
4 Function
5 Vesicular transport
6 Current models of vesicular transport and traffickingToggle Current models of vesicular transport and trafficking subsection
6.1 Model 1: Anterograde vesicular transport between stable compartments
6.2 Model 2: Cisternal progression/maturation
6.3 Model 3: Cisternal progression/maturation with heterotypic tubular transport
6.4 Model 4: Rapid partitioning in a mixed Golgi
6.5 Model 5: Stable compartments as cisternal model progenitors
7 Brefeldin A
8 Gallery
9 References
10 External links
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Golgi apparatus
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From Wikipedia, the free encyclopedia
Cell organelle that packages proteins for export
Not to be confused with gyrification. For the song, see Junta (album).
Diagram of a single "stack" of Golgi
| Cell biology |
| Animal cell diagram |
|
Components of a typical animal cell:
Nucleolus
Nucleus
Ribosome (dots as part of 5)
Vesicle
Rough endoplasmic reticulum
Golgi apparatus (or, Golgi body)
Cytoskeleton
Smooth endoplasmic reticulum
Mitochondrion
Vacuole
Cytosol (fluid that contains organelles; with which, comprises cytoplasm)
Lysosome
Centrosome
Cell membrane
|
The Golgi apparatus (/ˈɡɒldʒi/), also known as the Golgi complex, Golgi body, or simply the Golgi, is an organelle found in most eukaryotic cells. Part of the endomembrane system in the cytoplasm, it packages proteins into membrane-bound vesicles inside the cell before the vesicles are sent to their destination. It resides at the intersection of the secretory, lysosomal, and endocytic pathways. It is of particular importance in processing proteins for secretion, containing a set of glycosylation enzymes that attach various sugar monomers to proteins as the proteins move through the apparatus.
The Golgi apparatus was identified in 1898 by the Italian biologist and pathologist Camillo Golgi. The organelle was later named after him in the 1910s.
Discovery
Because of its large size and distinctive structure, the Golgi apparatus was one of the first organelles to be discovered and observed in detail. It was discovered in 1898 by Italian physician Camillo Golgi during an investigation of the nervous system. After first observing it under his microscope, he termed the structure as apparato reticolare interno ("internal reticular apparatus"). Some doubted the discovery at first, arguing that the appearance of the structure was merely an optical illusion created by Golgi's observation technique. With the development of modern microscopes in the twentieth century, the discovery was confirmed. Early references to the Golgi apparatus referred to it by various names, including the Golgi–Holmgren apparatus, Golgi–Holmgren ducts, and Golgi–Kopsch apparatus. The term Golgi apparatus was used in 1910 and first appeared in scientific literature in 1913, while "Golgi complex" was introduced in 1956.
Subcellular localization
The subcellular localization of the Golgi apparatus varies among eukaryotes. In mammals, a single Golgi apparatus is usually located near the cell nucleus, close to the centrosome. Tubular connections are responsible for linking the stacks together. Localization and tubular connections of the Golgi apparatus are dependent on microtubules. In experiments, it is seen that as microtubules are depolymerized, the Golgi apparatuses lose mutual connections and become individual stacks throughout the cytoplasm. In yeast, multiple Golgi apparatuses are scattered throughout the cytoplasm (as observed in Saccharomyces cerevisiae). In plants, Golgi stacks are not concentrated at the centrosomal region and do not form Golgi ribbons. Organization of the plant Golgi depends on actin cables and not microtubules. The common feature among Golgi is that they are adjacent to endoplasmic reticulum (ER) exit sites.
Structure
3D rendering of Golgi apparatus
In most eukaryotes, the Golgi apparatus is made up of a series of compartments and is a collection of fused, flattened membrane-enclosed disks known as cisternae (singular: cisterna, also called "dictyosomes"), originating from vesicular clusters that bud off the endoplasmic reticulum (ER). A mammalian cell typically contains 40 to 100 stacks of cisternae. Between four and eight cisternae are usually present in a stack; however, in some protists, as many as sixty cisternae have been observed. This collection of cisternae is broken down into cis, medial, and trans compartments, making up two main networks: the cis Golgi network (CGN) and the trans Golgi network (TGN). The CGN is the first cisternal structure, and the TGN is the final, from which proteins are packaged into vesicles destined to lysosomes, secretory vesicles, or the cell surface. The TGN is usually positioned adjacent to the stack, but can also be separate from it. The TGN may act as an early endosome in yeast and plants.
There are structural and organizational differences in the Golgi apparatus among eukaryotes. In some yeasts, Golgi stacking is not observed. Pichia pastoris does have stacked Golgi, while Saccharomyces cerevisiae does not. In plants, the individual stacks of the Golgi apparatus seem to operate independently.
The Golgi apparatus tends to be larger and more numerous in cells that synthesize and secrete large amounts of substances; for example, the antibody-secreting plasma B cells of the immune system have prominent Golgi complexes.[citation needed]
In all eukaryotes, each cisternal stack has a cis entry face and a trans exit face. These faces are characterized by unique morphology and biochemistry. Within individual stacks are assortments of enzymes responsible for selectively modifying protein cargo. These modifications influence the fate of the protein. The compartmentalization of the Golgi apparatus is advantageous for separating enzymes, thereby maintaining consecutive and selective processing steps: enzymes catalyzing early modifications are gathered in the cis face cisternae, and enzymes catalyzing later modifications are found in trans face cisternae of the Golgi stacks.
Function
The Golgi apparatus (salmon pink) in context of the secretory pathway
The Golgi apparatus is a major collection and dispatch station of protein products received from the endoplasmic reticulum. Proteins synthesized in the ER are packaged into vesicles, which then fuse with the Golgi apparatus. These cargo proteins are modified and destined for secretion via exocytosis or for use in the cell. In this respect, the Golgi can be thought of as similar to a post office: it packages and labels items which it then sends to different parts of the cell or to the extracellular space. The Golgi apparatus is also involved in lipid transport and lysosome formation.
The structure and function of the Golgi apparatus are intimately linked. Individual stacks have different assortments of enzymes, allowing for progressive processing of cargo proteins as they travel from the cisternae to the trans Golgi face. Enzymatic reactions within the Golgi stacks occur exclusively near its membrane surfaces, where enzymes are anchored. This feature is in contrast to the ER, which has soluble proteins and enzymes in its lumen. Much of the enzymatic processing is post-translational modification of proteins. For example, phosphorylation of oligosaccharides on lysosomal proteins occurs in the early CGN. Cis cisterna are associated with the removal of mannose residues. Removal of mannose residues and addition of N-acetylglucosamine occur in medial cisternae. Addition of galactose and sialic acid occurs in the trans cisternae. Sulfation of tyrosines and carbohydrates occurs within the TGN. Other general post-translational modifications of proteins include the addition of carbohydrates (glycosylation) and phosphates (phosphorylation). Protein modifications may form a signal sequence that determines the final destination of the protein. For example, the Golgi apparatus adds a mannose-6-phosphate label to proteins destined for lysosomes. Another important function of the Golgi apparatus is in the formation of proteoglycans. Enzymes in the Golgi append proteins to glycosaminoglycans, thus creating proteoglycans. Glycosaminoglycans are long unbranched polysaccharide molecules present in the extracellular matrix of animals.
Vesicular transport
Diagram of secretory process from endoplasmic reticulum (orange) to Golgi apparatus (magenta). 1. Nuclear membrane; 2. Nuclear pore; 3. Rough endoplasmic reticulum (RER); 4. Smooth endoplasmic reticulum (SER); 5. Ribosome attached to RER; 6. Macromolecules; 7. Transport vesicles; 8. Golgi apparatus; 9. Cis face of Golgi apparatus; 10. Trans face of Golgi apparatus; 11. Cisternae of the Golgi apparatus.
The vesicles that leave the rough endoplasmic reticulum are transported to the cis face of the Golgi apparatus, where they fuse with the Golgi membrane and empty their contents into the lumen. Once inside the lumen, the molecules are modified, then sorted for transport to their next destinations.[citation needed]
Those proteins destined for areas of the cell other than either the endoplasmic reticulum or the Golgi apparatus are moved through the Golgi cisternae towards the trans face, to a complex network of membranes and associated vesicles known as the trans-Golgi network (TGN). This area of the Golgi is the point at which proteins are sorted and shipped to their intended destinations by their placement into one of at least three different types of vesicles, depending upon the signal sequence they carry.
| Types | Description | Example |
---
| Exocytotic vesicles (constitutive) | Vesicle contains proteins destined for extracellular release. After packaging, the vesicles bud off and immediately move towards the plasma membrane, where they fuse and release the contents into the extracellular space in a process known as constitutive secretion. | Antibody release by activated plasma B cells |
| Secretory vesicles (regulated) | Vesicles contain proteins destined for extracellular release. After packaging, the vesicles bud off and are stored in the cell until a signal is given for their release. When the appropriate signal is received they move toward the membrane and fuse to release their contents. This process is known as regulated secretion. | Neurotransmitter release from neurons |
| Lysosomal vesicles | Vesicles contain proteins and ribosomes destined for the lysosome, a degradative organelle containing many acid hydrolases, or to lysosome-like storage organelles. These proteins include both digestive enzymes and membrane proteins. The vesicle first fuses with the late endosome, and the contents are then transferred to the lysosome via unknown mechanisms. | Digestive proteases destined for the lysosome |
Current models of vesicular transport and trafficking
Unsolved problem in biology
In cell theory, what is the exact transport mechanism by which proteins travel through the Golgi apparatus?
More unsolved problems in biology
Model 1: Anterograde vesicular transport between stable compartments
In this model, the Golgi is viewed as a set of stable compartments that work together. Each compartment has a unique collection of enzymes that work to modify protein cargo. Proteins are delivered from the ER to the cis face using COPII-coated vesicles. Cargo then progress toward the trans face in COPI-coated vesicles. This model proposes that COPI vesicles move in two directions: anterograde vesicles carry secretory proteins, while retrograde vesicles recycle Golgi-specific trafficking proteins.
Strengths: The model explains observations of compartments, polarized distribution of enzymes, and waves of moving vesicles. It also attempts to explain how Golgi-specific enzymes are recycled.
Weaknesses: Since the amount of COPI vesicles varies drastically among types of cells, this model cannot easily explain high trafficking activity within the Golgi for both small and large cargoes. Additionally, there is no convincing evidence that COPI vesicles move in both the anterograde and retrograde directions.
This model was widely accepted from the early 1980s until the late 1990s.
Model 2: Cisternal progression/maturation
In this model, the fusion of COPII vesicles from the ER begins the formation of the first cis-cisterna of the Golgi stack, which progresses later to become mature TGN cisternae. Once matured, the TGN cisternae dissolve to become secretory vesicles. While this progression occurs, COPI vesicles continually recycle Golgi-specific proteins by delivery from older to younger cisternae. Different recycling patterns may account for the differing biochemistry throughout the Golgi stack. Thus, the compartments within the Golgi are seen as discrete kinetic stages of the maturing Golgi apparatus.
Strengths: The model addresses the existence of Golgi compartments, as well as differing biochemistry within the cisternae, transport of large proteins, transient formation and disintegration of the cisternae, and retrograde mobility of native Golgi proteins, and it can account for the variability seen in the structures of the Golgi.
Weaknesses: This model cannot easily explain the observation of fused Golgi networks, tubular connections among cisternae, and differing kinetics of secretory cargo exit.
Model 3: Cisternal progression/maturation with heterotypic tubular transport
This model is an extension of the cisternal progression/maturation model. It incorporates the existence of tubular connections among the cisternae that form the Golgi ribbon, in which cisternae within a stack are linked. This model posits that the tubules are important for bidirectional traffic in the ER-Golgi system: they allow for fast anterograde traffic of small cargo and/or the retrograde traffic of native Golgi proteins.
Strengths: This model encompasses the strengths of the cisternal progression/maturation model that also explains rapid trafficking of cargo, and how native Golgi proteins can recycle independently of COPI vesicles.
Weaknesses: This model cannot explain the transport kinetics of large protein cargo, such as collagen. Additionally, tubular connections are not prevalent in plant cells. The roles that these connections have can be attributed to a cell-specific specialization rather than a universal trait. If the membranes are continuous, that suggests the existence of mechanisms that preserve the unique biochemical gradients observed throughout the Golgi apparatus.
Model 4: Rapid partitioning in a mixed Golgi
This rapid partitioning model is the most drastic alteration of the traditional vesicular trafficking point of view. Proponents of this model hypothesize that the Golgi works as a single unit, containing domains that function separately in the processing and export of protein cargo. Cargo from the ER move between these two domains, and randomly exit from any level of the Golgi to their final location. This model is supported by the observation that cargo exits the Golgi in a pattern best described by exponential kinetics. The existence of domains is supported by fluorescence microscopy data.
Strengths: Notably, this model explains the exponential kinetics of cargo exit of both large and small proteins, whereas other models cannot.
Weaknesses: This model cannot explain the transport kinetics of large protein cargo, such as collagen. This model falls short on explaining the observation of discrete compartments and polarized biochemistry of the Golgi cisternae. It also does not explain formation and disintegration of the Golgi network, nor the role of COPI vesicles.
Model 5: Stable compartments as cisternal model progenitors
This is the most recent model. In this model, the Golgi is seen as a collection of stable compartments defined by Rab (G-protein) GTPases.
Strengths: This model is consistent with numerous observations and encompasses some of the strengths of the cisternal progression/maturation model. Additionally, what is known of the Rab GTPase roles in mammalian endosomes can help predict putative roles within the Golgi. This model is unique in that it can explain the observation of "megavesicle" transport intermediates.
Weaknesses: This model does not explain morphological variations in the Golgi apparatus, nor define a role for COPI vesicles. This model does not apply well for plants, algae, and fungi in which individual Golgi stacks are observed (transfer of domains between stacks is not likely). Additionally, megavesicles are not established to be intra-Golgi transporters.
Though there are multiple models that attempt to explain vesicular traffic throughout the Golgi, no individual model can independently explain all observations of the Golgi apparatus. Currently, the cisternal progression/maturation model is the most accepted among scientists, accommodating many observations across eukaryotes. The other models are still important in framing questions and guiding future experimentation. Among the fundamental unanswered questions are the directionality of COPI vesicles and role of Rab GTPases in modulating protein cargo traffic.
Brefeldin A
Brefeldin A (BFA) is a fungal metabolite used experimentally to disrupt the secretion pathway as a method of testing Golgi function. BFA blocks the activation of some ADP-ribosylation factors (ARFs). which regulate vesicular trafficking through the binding of COPs to endosomes and the Golgi. (GEFs) that mediate GTP-binding of ARFs. Treatment of cells with BFA thus disrupts the secretion pathway, promoting disassembly of the Golgi apparatus and distributing Golgi proteins to the endosomes and ER.
Gallery
Yeast Golgi dynamics. Green labels early Golgi, red labels late Golgi.
Two Golgi stacks connected as a ribbon in a mouse cell. Taken from the movie.
Three-dimensional projection of a mammalian Golgi stack imaged by confocal microscopy and volume surface rendered using Imaris software. Taken from the movie.
References
^ Pavelk M, Mironov AA (2008). "Golgi apparatus inheritance". The Golgi Apparatus: State of the art 110 years after Camillo Golgi's discovery. Berlin: Springer. p. 580. doi:10.1007/978-3-211-76310-0_34. ISBN 978-3-211-76310-0.
^ a b c d e Fabene PF, Bentivoglio M (October 1998). "1898-1998: Camillo Golgi and "the Golgi": one hundred years of terminological clones". Brain Research Bulletin. 47 (3): 195–8. doi:10.1016/S0361-9230(98)00079-3. PMID 9865849. S2CID 208785591.
^ Golgi C (1898). "Intorno alla struttura delle cellule nervose" (PDF). Bollettino della Società Medico-Chirurgica di Pavia. 13 (1): 316. Archived (PDF) from the original on 2018-04-07.
^ a b Davidson MW (2004-12-13). "The Golgi Apparatus". Molecular Expressions. Florida State University. Archived from the original on 2006-11-07. Retrieved 2010-09-20.
^ a b c d e f g h Alberts, Bruce; et al. (1994). Molecular Biology of the Cell. Garland Publishing. ISBN 978-0-8153-1619-0.
^ a b c d e Nakano A, Luini A (August 2010). "Passage through the Golgi". Current Opinion in Cell Biology. 22 (4): 471–8. doi:10.1016/j.ceb.2010.05.003. PMID 20605430.
^ Suda Y, Nakano A (April 2012). "The yeast Golgi apparatus". Traffic. 13 (4): 505–10. doi:10.1111/j.1600-0854.2011.01316.x. PMID 22132734.
^ Duran JM, Kinseth M, Bossard C, Rose DW, Polishchuk R, Wu CC, Yates J, Zimmerman T, Malhotra V (June 2008). "The role of GRASP55 in Golgi fragmentation and entry of cells into mitosis". Molecular Biology of the Cell. 19 (6): 2579–87. doi:10.1091/mbc.E07-10-0998. PMC 2397314. PMID 18385516.
^ Day, Kasey J.; Casler, Jason C.; Glick, Benjamin S. (2018). "Budding Yeast Has a Minimal Endomembrane System". Developmental Cell. 44 (1): 56–72.e4. doi:10.1016/j.devcel.2017.12.014. PMC 5765772. PMID 29316441.
^ a b c d Day KJ, Staehelin LA, Glick BS (September 2013). "A three-stage model of Golgi structure and function". Histochemistry and Cell Biology. 140 (3): 239–49. doi:10.1007/s00418-013-1128-3. PMC 3779436. PMID 23881164.
^ Campbell, Neil A (1996). Biology (4 ed.). Menlo Park, CA: Benjamin/Cummings. pp. 122, 123. ISBN 978-0-8053-1957-6.
^ William G. Flynne (2008). Biotechnology and Bioengineering. Nova Publishers. pp. 45–. ISBN 978-1-60456-067-1. Retrieved 13 November 2010.
^ Prydz K, Dalen KT (January 2000). "Synthesis and sorting of proteoglycans". Journal of Cell Science. 113. 113 Pt 2 (2): 193–205. doi:10.1242/jcs.113.2.193. PMID 10633071.
^ a b c d e f g h i j k l m n o p q Glick BS, Luini A (November 2011). "Models for Golgi traffic: a critical assessment". Cold Spring Harbor Perspectives in Biology. 3 (11): a005215. doi:10.1101/cshperspect.a005215. PMC 3220355. PMID 21875986.
^ Wei JH, Seemann J (November 2010). "Unraveling the Golgi ribbon". Traffic. 11 (11): 1391–400. doi:10.1111/j.1600-0854.2010.01114.x. PMC 4221251. PMID 21040294.
^ a b Marie M, Sannerud R, Avsnes Dale H, Saraste J (September 2008). "Take the 'A' train: on fast tracks to the cell surface". Cellular and Molecular Life Sciences. 65 (18): 2859–74. doi:10.1007/s00018-008-8355-0. PMC 7079782. PMID 18726174.
^ a:10.1038/nrm1910. PMID 16633337. S2CID 19092867.
^ Papanikou E, Day KJ, Austin J, Glick BS (2015). "COPI selectively drives maturation of the early Golgi". eLife. 4. doi:10.7554/eLife.13232. PMC 4758959. PMID 26709839.
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9346 | https://tutorial.math.lamar.edu/classes/calci/optimization.aspx | Paul's Online Notes
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Paul's Online Notes Home / Calculus I / Applications of Derivatives / Optimization
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Section 4.8 : Optimization
In this section we are going to look at optimization problems. In optimization problems we are looking for the largest value or the smallest value that a function can take. We saw how to solve one kind of optimization problem in the Absolute Extrema section where we found the largest and smallest value that a function would take on an interval.
In this section we are going to look at another type of optimization problem. Here we will be looking for the largest or smallest value of a function subject to some kind of constraint. The constraint will be some condition (that can usually be described by some equation) that must absolutely, positively be true no matter what our solution is. On occasion, the constraint will not be easily described by an equation, but in these problems it will be easy to deal with as we’ll see.
This section is generally one of the more difficult for students taking a Calculus course. One of the main reasons for this is that a subtle change of wording can completely change the problem. There is also the problem of identifying the quantity that we’ll be optimizing and the quantity that is the constraint and writing down equations for each.
The first step in all of these problems should be to very carefully read the problem. Once you’ve done that the next step is to identify the quantity to be optimized and the constraint.
In identifying the constraint remember that the constraint is the quantity that must be true regardless of the solution. In almost every one of the problems we’ll be looking at here one quantity will be clearly indicated as having a fixed value and so must be the constraint. Once you’ve got that identified the quantity to be optimized should be fairly simple to get. It is however easy to confuse the two if you just skim the problem so make sure you carefully read the problem first!
Let’s start the section off with a simple problem to illustrate the kinds of issues we will be dealing with here.
Example 1 We need to enclose a rectangular field with a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area.
Show Solution
In all of these problems we will have two functions. The first is the function that we are actually trying to optimize and the second will be the constraint. Sketching the situation will often help us to arrive at these equations so let’s do that.
In this problem we want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing is the constraint. The two equations for these are,
[\begin{align}{\mbox{Maximize : }} & A = xy\ {\mbox{Constraint : }} & 500 = x + 2y\end{align}]
Okay, we know how to find the largest or smallest value of a function provided it’s only got a single variable. The area function (as well as the constraint) has two variables in it and so what we know about finding absolute extrema won’t work. However, if we solve the constraint for one of the two variables we can substitute this into the area and we will then have a function of a single variable.
So, let’s solve the constraint for (x). Note that we could have just as easily solved for (y) but that would have led to fractions and so, in this case, solving for (x) will probably be best.
[x = 500 - 2y]
Substituting this into the area function gives a function of (y).
[A\left( y \right) = \left( {500 - 2y} \right)y = 500y - 2{y^2}]
Now we want to find the largest value this will have on the interval (\left[ {0,250} \right]). The limits in this interval corresponds to taking (y = 0) (i.e. no sides to the fence) and (y = 250) (i.e. only two sides and no width, also if there are two sides each must be 250 ft to use the whole 500ft).
Note that the endpoints of the interval won’t make any sense from a physical standpoint if we actually want to enclose some area because they would both give zero area. They do, however, give us a set of limits on (y) and so the Extreme Value Theorem tells us that we will have a maximum value of the area somewhere between the two endpoints. Having these limits will also mean that we can use the process we discussed in the Finding Absolute Extrema section earlier in the chapter to find the maximum value of the area.
So, recall that the maximum value of a continuous function (which we’ve got here) on a closed interval (which we also have here) will occur at critical points and/or end points. As we’ve already pointed out the end points in this case will give zero area and so don’t make any sense. That means our only option will be the critical points.
So, let’s get the derivative and find the critical points.
[A'\left( y \right) = 500 - 4y]
Setting this equal to zero and solving gives a lone critical point of (y = 125). Plugging this into the area gives an area of (A\left( {125} \right) = 31250\,{\mbox{f}}{{\mbox{t}}^2}). So according to the method from Absolute Extrema section this must be the largest possible area, since the area at either endpoint is zero.
Finally, let’s not forget to get the value of (x) and then we’ll have the dimensions since this is what the problem statement asked for. We can get the (x) by plugging in our (y) into the constraint.
[x = 500 - 2\left( {125} \right) = 250]
The dimensions of the field that will give the largest area, subject to the fact that we used exactly 500 ft of fencing material, are 250 x 125.
Don’t forget to actually read the problem and give the answer that was asked for. These types of problems can take a fair amount of time/effort to solve and it’s not hard to sometimes forget what the problem was actually asking for.
In the previous problem we used the method from the Finding Absolute Extrema section to find the maximum value of the function we wanted to optimize. However, as we’ll see in later examples it will not always be easy to find endpoints. Also, even if we can find the endpoints we will see that sometimes dealing with the endpoints may not be easy either. Not only that, but this method requires that the function we’re optimizing be continuous on the interval we’re looking at, including the endpoints, and that may not always be the case.
So, before proceeding with any more examples let’s spend a little time discussing some methods for determining if our solution is in fact the absolute minimum/maximum value that we’re looking for. In some examples all of these will work while in others one or more won’t be all that useful. However, we will always need to use some method for making sure that our answer is in fact that optimal value that we’re after.
Method 1 : Use the method used in Finding Absolute Extrema.
This is the method used in the first example above. Recall that in order to use this method the interval of possible values of the independent variable in the function we are optimizing, let’s call it (I), must have finite endpoints. Also, the function we’re optimizing (once it’s down to a single variable) must be continuous on (I), including the endpoints. If these conditions are met then we know that the optimal value, either the maximum or minimum depending on the problem, will occur at either the endpoints of the range or at a critical point that is inside the range of possible solutions.
There are two main issues that will often prevent this method from being used however. First, not every problem will actually have a range of possible solutions that have finite endpoints at both ends. We’ll see at least one example of this as we work through the remaining examples. Also, many of the functions we’ll be optimizing will not be continuous once we reduce them down to a single variable and this will prevent us from using this method.
Method 2 : Use a variant of the First Derivative Test.
In this method we also will need an interval of possible values of the independent variable in the function we are optimizing, (I). However, in this case, unlike the previous method the endpoints do not need to be finite. Also, we will need to require that the function be continuous on the interior of the interval (I) and we will only need the function to be continuous at the end points if the endpoint is finite and the function actually exists at the endpoint. We’ll see several problems where the function we’re optimizing doesn’t actually exist at one of the endpoints. This will not prevent this method from being used.
Let’s suppose that (x = c) is a critical point of the function we’re trying to optimize, (f\left( x \right)). We already know from the First Derivative Test that if (f'\left( x \right) > 0) immediately to the left of (x = c) (i.e. the function is increasing immediately to the left) and if (f'\left( x \right) < 0) immediately to the right of (x = c)(i.e. the function is decreasing immediately to the right) then (x = c) will be a relative maximum for (f\left( x \right)).
Now, this does not mean that the absolute maximum of (f\left( x \right)) will occur at (x = c). However, suppose that we knew a little bit more information. Suppose that in fact we knew that (f'\left( x \right) > 0) for all (x) in (I) such that (x < c). Likewise, suppose that we knew that (f'\left( x \right) < 0) for all (x) in (I) such that (x > c). In this case we know that to the left of (x = c), provided we stay in (I) of course, the function is always increasing and to the right of (x = c), again staying in (I), we are always decreasing. In this case we can say that the absolute maximum of (f\left( x \right)) in (I) will occur at (x = c).
Similarly, if we know that to the left of (x = c) the function is always decreasing and to the right of (x = c) the function is always increasing then the absolute minimum of (f\left( x \right)) in (I) will occur at (x = c).
Before we give a summary of this method let’s discuss the continuity requirement a little. Nowhere in the above discussion did the continuity requirement apparently come into play. We require that the function we’re optimizing to be continuous in (I) to prevent the following situation.
In this case, a relative maximum of the function clearly occurs at (x = c). Also, the function is always decreasing to the right and is always increasing to the left. However, because of the discontinuity at (x = d), we can clearly see that (f\left( d \right) > f\left( c \right)) and so the absolute maximum of the function does not occur at (x = c). Had the discontinuity at (x = d) not been there this would not have happened and the absolute maximum would have occurred at (x = c).
Here is a summary of this method.
First Derivative Test for Absolute Extrema
Let (I) be the interval of all possible values of (x) in (f\left( x \right)), the function we want to optimize, and further suppose that (f\left( x \right)) is continuous on (I) , except possibly at the endpoints. Finally suppose that (x = c) is a critical point of (f\left( x \right)) and that (c) is in the interval (I). If we restrict (x) to values from (I) (i.e. we only consider possible optimal values of the function) then,
If (f'\left( x \right) > 0) for all (x < c) and if (f'\left( x \right) < 0) for all (x > c) then (f\left( c \right)) will be the absolute maximum value of (f\left( x \right)) on the interval (I).
If (f'\left( x \right) < 0) for all (x < c) and if (f'\left( x \right) > 0) for all (x > c) then (f\left( c \right)) will be the absolute minimum value of (f\left( x \right)) on the interval (I).
Method 3 : Use the second derivative.
There are actually two ways to use the second derivative to help us identify the optimal value of a function and both use the Second Derivative Test to one extent or another.
The first way to use the second derivative doesn’t actually help us to identify the optimal value. What it does do is allow us to potentially exclude values and knowing this can simplify our work somewhat and so is not a bad thing to do.
Suppose that we are looking for the absolute maximum of a function and after finding the critical points we find that we have multiple critical points. Let’s also suppose that we run all of them through the second derivative test and determine that some of them are in fact relative minimums of the function. Since we are after the absolute maximum we know that a maximum (of any kind) can’t occur at relative minimums and so we immediately know that we can exclude these points from further consideration. We could do a similar check if we were looking for the absolute minimum. Doing this may not seem like all that great of a thing to do, but it can, on occasion, lead to a nice reduction in the amount of work that we need to do in later steps.
The second way of using the second derivative to identify the optimal value of a function is in fact very similar to the second method above. In fact, we will have the same requirements for this method as we did in that method. We need an interval of possible values of the independent variable in function we are optimizing, call it (I) as before, and the endpoint(s) may or may not be finite. We’ll also need to require that the function, (f\left( x \right)) be continuous everywhere in (I) except possibly at the endpoints as above.
Now, suppose that (x = c) is a critical point and that (f''\left( c \right) > 0). The second derivative test tells us that (x = c) must be a relative minimum of the function. Suppose however that we also knew that (f''\left( x \right) > 0) for all (x) in (I). In this case we would know that the function was concave up in all of (I) and that would in turn mean that the absolute minimum of (f\left( x \right)) in (I) would in fact have to be at (x = c).
Likewise, if (x = c) is a critical point and (f''\left( x \right) < 0) for all (x) in (I) then we would know that the function was concave down in (I) and that the absolute maximum of (f\left( x \right)) in (I) would have to be at (x = c).
Here is a summary of this method.
Second Derivative Test for Absolute Extrema
Let (I) be the interval of all possible values of (x) in (f\left( x \right)), the function we want to optimize, and suppose that (f\left( x \right)) is continuous on (I) , except possibly at the endpoints. Finally suppose that (x = c) is a critical point of (f\left( x \right)) and that (c) is in the interval (I). Then,
If (f''\left( x \right) > 0) for all (x) in (I) then (f\left( c \right)) will be the absolute minimum value of (f\left( x \right)) on the interval (I).
If (f''\left( x \right) < 0) for all (x) in (I) then (f\left( c \right)) will be the absolute maximum value of (f\left( x \right)) on the interval (I).
As we work examples over the next two sections we will use each of these methods as needed in the examples. In some cases, the method we use will be the only method we could use, in others it will be the easiest method to use and in others it will simply be the method we chose to use for that example. It is important to realize that we won’t be able to use each of the methods for every example. With some examples one method will be easiest to use or may be the only method that can be used, however, each of the methods described above will be used at least a couple of times through out all of the examples.
It is also important to be aware that some problems don’t allow any of the methods discussed above to be used exactly as outlined above. We may need to modify one of them or use a combination of them to fully work the problem. There is an example in the next section where none of the methods above work easily, although we do also present an alternative solution method in which we can use at least one of the methods discussed above.
Next, the vast majority of the examples worked over the course of the next section will only have a single critical point. Problems with more than one critical point are often difficult to know which critical point(s) give the optimal value. There are a couple of examples in the next two sections with more than one critical point including one in the next section mentioned above in which none of the methods discussed above easily work. In that example you can see some of the ideas you might need to do in order to find the optimal value.
Finally, in all of the methods above we referenced an interval (I). This was done to make the discussion a little easier. However, in all of the examples over the next two sections we will never explicitly say “this is the interval (I)”. Just remember that the interval (I) is just the largest interval of possible values of the independent variable in the function we are optimizing.
Okay, let’s work some more examples.
Example 2 We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom cost $10/ft2 and the material used to build the sides cost $6/ft2. If the box must have a volume of 50ft3 determine the dimensions that will minimize the cost to build the box.
Show Solution
First, a quick figure (probably not to scale…).
We want to minimize the cost of the materials subject to the constraint that the volume must be 50ft3. Note as well that the cost for each side is just the area of that side times the appropriate cost.
The two functions we’ll be working with here this time are,
[\begin{align}{\mbox{Minimize : }} & C = 10\left( {2lw} \right) + 6\left( {2wh + 2lh} \right)\, = 60{w^2} + 48wh\ {\mbox{Constraint :}}& 50 = lwh = 3{w^2}h\end{align}]
As with the first example, we will solve the constraint for one of the variables and plug this into the cost. It will definitely be easier to solve the constraint for (h) so let’s do that.
[h = \frac{{50}}{{3{w^2}}}]
Plugging this into the cost gives,
[C\left( w \right) = 60{w^2} + 48w\left( {\frac{{50}}{{3{w^2}}}} \right) = 60{w^2} + \frac{{800}}{w}]
Now, let’s get the first and second (we’ll be needing this later…) derivatives,
[C'\left( w \right) = 120w - 800{w^{ - 2}} = \frac{{120{w^3} - 800}}{{{w^2}}}\hspace{0.25in}\hspace{0.25in}C''\left( w \right) = 120 + 1600{w^{ - 3}}]
Now we need the critical point(s) for the cost function. First, notice that (w = 0) is not a critical point. Clearly the derivative does not exist at (w = 0) but then neither does the function and remember that values of (w) will only be critical points if the function also exists at that point. Note that there is also a physical reason to avoid (w = 0). We are constructing a box and it would make no sense to have a zero width of the box.
So it looks like the only critical point will come from determining where the numerator is zero.
[120{w^3} - 800 = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,w = \sqrt{{\frac{{800}}{{120}}}} = \sqrt{{\frac{{20}}{3}}} = 1.8821]
So, we’ve got a single critical point and we now have to verify that this is in fact the value that will give the absolute minimum cost.
In this case we can’t use Method 1 from above. First, the function is not continuous at one of the endpoints, (w = 0), of our interval of possible values, i.e. (w > 0). Secondly, there is no theoretical upper limit to the width that will give a box with volume of 50 ft3. If (w) is very large then we would just need to make (h) very small.
The second method listed above would work here, but that’s going to involve some calculations, not difficult calculations, but more work nonetheless.
The third method however, will work quickly and simply here. First, we know that whatever the value of (w) that we get it will have to be positive and we can see second derivative above that provided (w > 0) we will have (C''\left( w \right) > 0) and so in the interval of possible optimal values the cost function will always be concave up and so (w = 1.8821) must give the absolute minimum cost.
All we need to do now is to find the remaining dimensions.
[\begin{align}w & = 1.8821\ l & = 3w = 3\left( {1.8821} \right) = 5.6463\ h & = \frac{{50}}{{3{w^2}}} = \frac{{50}}{{3{{\left( {1.8821} \right)}^2}}} = 4.7050\end{align}]
Also, even though it was not asked for, the minimum cost is : (C\left( {1.8821} \right) = \$ 637.60).
Example 3 We want to construct a box with a square base and we only have 10 m2 of material to use in construction of the box. Assuming that all the material is used in the construction process determine the maximum volume that the box can have.
Show Solution
This example is in many ways the exact opposite of the previous example. In this case we want to optimize the volume and the constraint this time is the amount of material used. We don’t have a cost here, but if you think about it the cost is nothing more than the amount of material used times a cost and so the amount of material and cost are pretty much tied together. If you can do one you can do the other as well. Note as well that the amount of material used is really just the surface area of the box.
As always, let’s start off with a quick sketch of the box.
Now, as mentioned above we want to maximize the volume and the amount of material is the constraint so here are the equations we’ll need.
[\begin{align}{\mbox{Maximize : }} & V = lwh = {w^2}h\ {\mbox{Constraint :}} & 10 = 2lw + 2wh + 2lh\, = 2{w^2} + 4wh\end{align}]
We’ll solve the constraint for (h) and plug this into the equation for the volume.
[h = \frac{{10 - 2{w^2}}}{{4w}} = \frac{{5 - {w^2}}}{{2w}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,V\left( w \right) = {w^2}\left( {\frac{{5 - {w^2}}}{{2w}}} \right) = \frac{1}{2}\left( {5w - {w^3}} \right)]
Here are the first and second derivatives of the volume function.
[V'\left( w \right) = {{1 \over 2}}\left( {5 - 3{w^2}} \right)\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}V''\left( w \right) = - 3w]
Note as well here that provided (w > 0), which from a physical standpoint we know must be true for the width of the box, then the volume function will be concave down and so if we get a single critical point then we know that it will have to be the value that gives the absolute maximum.
Setting the first derivative equal to zero and solving gives us the two critical points,
[w = \pm \,\sqrt {\frac{5}{3}} = \pm \,1.2910]
In this case we can exclude the negative critical point since we are dealing with a length of a box and we know that these must be positive. Do not however get into the habit of just excluding any negative critical point. There are problems where negative critical points are perfectly valid possible solutions.
Now, as noted above we got a single critical point, 1.2910, and so this must be the value that gives the maximum volume and since the maximum volume is all that was asked for in the problem statement the answer is then : [V\left( {1.2910} \right) = 2.1517\,{{\mbox{m}}^3}].
Note that we could also have noted here that if (0 < w < 1.2910) then (V'\left( w \right) > 0) (using a test point we have (V'\left( 1 \right) = 1 > 0)) and likewise if (w > 1.2910) then (V'\left( w \right) < 0) (using a test point we have (V'\left( 2 \right) = - {\frac{7}{2}} < 0)) and so if we are to the left of the critical point the volume is always increasing and if we are to the right of the critical point the volume is always decreasing and so by the Method 2 above we can also see that the single critical point must give the absolute maximum of the volume.
Finally, even though these weren’t asked for here are the dimension of the box that gives the maximum volume.
[l = w = 1.2910\hspace{1.0in}h = \frac{{5 - {{1.2910}^2}}}{{2\left( {1.2910} \right)}} = 1.2910]
So, it looks like in this case we actually have a perfect cube.
In the last two examples we’ve seen that many of these optimization problems can be done in both directions so to speak. In both examples we have essentially the same two equations: volume and surface area. However, in Example 2 the volume was the constraint and the cost (which is directly related to the surface area) was the function we were trying to optimize. In Example 3, on the other hand, we were trying to optimize the volume and the surface area was the constraint.
It is important to not get so locked into one way of doing these problems that we can’t do it in the opposite direction as needed as well. This is one of the more common mistakes that students make with these kinds of problems. They see one problem and then try to make every other problem that seems to be the same conform to that one solution even if the problem needs to be worked differently. Keep an open mind with these problems and make sure that you understand what is being optimized and what the constraint is before you jump into the solution.
Also, as seen in the last example we used two different methods of verifying that we did get the optimal value. Do not get too locked into one method of doing this verification that you forget about the other methods.
Let’s work some another example that this time doesn’t involve a rectangle or box.
Example 4 A manufacturer needs to make a cylindrical can that will hold 1.5 liters of liquid. Determine the dimensions of the can that will minimize the amount of material used in its construction.
Show Solution
Before starting the solution let’s first address the fact that we are using liters for volume. Because we want length measurements for the radius and height we’ll also need the volume to in terms of a length measurement. We can easily do this using the fact that 1 Liter = 1000 cm3 and so we can convert 1.5 liters into 1500 cm3. This will in turn give a radius and height in terms of centimeters.
In this problem the constraint is the volume and we want to minimize the amount of material used. This means that what we want to minimize is the surface area of the can and we’ll need to include both the walls of the can as well as the top and bottom “caps”. Here is a quick sketch to get us started off.
We’ll need the surface area of this can and that will be the surface area of the walls of the can (which is really just a cylinder) and the area of the top and bottom caps (which are just disks, and don’t forget that there are two of them).
Note that if you think of a cylinder of height (h) and radius (r) as just a bunch of disks/circles of radius (r) stacked on top of each other the equations for the surface area and volume are pretty simple to remember. The volume is just the area of each of the disks times the height. Similarly, the surface area of the walls of the cylinder is just the circumference of each circle times the height. We also can’t forget to add in the area of the two caps, (\pi {r^2}), to the total surface area.
So, the equation for the volume and surface area of the walls of a cylinder are then,
[V = \left( {\pi {r^2}} \right)\left( h \right) = \pi {r^2}h\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}A = \left( {2\pi r} \right)\left( h \right) = 2\pi rh]
Adding the surface area of the caps of the cylinder to the surface area the equations that we’ll need for this problem are,
[\begin{align}{\mbox{Minimize : }} & A = 2\pi rh + 2\pi {r^2}\ {\mbox{Constraint :}} & 1500 = \pi {r^2}h\end{align}]
In this case it looks like our best option is to solve the constraint for (h) and plug this into the area function.
[h = \frac{{1500}}{{\pi {r^2}}}\hspace{0.25in} \Rightarrow \hspace{0.25in}A\left( r \right) = 2\pi r\left( {\frac{{1500}}{{\pi {r^2}}}} \right) + 2\pi {r^2} = 2\pi {r^2} + \frac{{3000}}{r}]
Notice that this formula will only make sense from a physical standpoint if (r > 0) which is a good thing as it is not defined at (r = 0).
Next, let’s get the first derivative.
[A'\left( r \right) = 4\pi r - \frac{{3000}}{{{r^2}}} = \frac{{4\pi {r^3} - 3000}}{{{r^2}}}]
From this we can see that we have one critical points : (r = \sqrt{\frac{750}{\pi}} = 6.2035)(where the derivative is zero). Note that (r = 0) is not a critical point because the area function does not exist there, which makes sense from a physical standpoint as well given that we know that (r) must be positive in order to actually have a can.
So, we only have a single critical point to deal with here and notice that 6.2035 is the only value for which the derivative will be zero and hence the only place (with (r > 0) of course) that the derivative may change sign. It’s not difficult, using test points, to check that if (0 < r < 6.2035) then (A'\left( r \right) < 0) and likewise if (r > 6.2035) then (A'\left( r \right) > 0). The variant of the First Derivative Test above then tells us that the absolute minimum value of the area (for (r > 0)) must occur at [r = 6.2035].
All we need to do this is determine height of the can and we’ll be done.
[h = \frac{{1500}}{{\pi {{\left( {6.2035} \right)}^2}}} = 12.4070]
Therefore, if the manufacturer makes the can with a radius of 6.2035 cm and a height of 12.4070 cm the least amount of material will be used to make the can.
As an interesting side problem and extension to the above example you might want to show that for a given volume, (L), the minimum material will be used if (h = 2r) regardless of the volume of the can.
In the examples to this point we’ve put in quite a bit of discussion in the solution. In the remaining problems we won’t be putting in quite as much discussion and leave it to you to fill in any missing details.
Example 5 We have a piece of cardboard that is 14 inches by 10 inches and we’re going to cut out the corners as shown below and fold up the sides to form a box, also shown below. Determine the height of the box that will give a maximum volume.
Show Solution
Let’s let the height of the box be (h). So, the width/length of the corners being cut out is also (h) and so the vertical side will have a “new” height of (10 - 2h) and the horizontal side will have a “new” width of (14 - 2h). Here is a sketch with all this information put in,
In this example, for the first time, we’ve run into a problem where the constraint doesn’t really have an equation. The constraint is simply the size of the piece of cardboard and has already been factored into the figure above. This will happen on occasion and so don’t get excited about it when it does. This just means that we have one less equation to worry about. In this case we want to maximize the volume. Here is the volume, in terms of (h) and its first derivative.
[V\left( h \right) = h\left( {14 - 2h} \right)\left( {10 - 2h} \right) = 140h - 48{h^2} + 4{h^3}\hspace{0.25in}\hspace{0.25in}V'\left( h \right) = 140 - 96h + 12{h^2}]
Setting the first derivative equal to zero and solving gives the following two critical points,
[h = \frac{{12 \pm \sqrt {39} }}{3} = 1.9183,\,\,\,\,6.0817]
We now have an apparent problem. We have two critical points and we’ll need to determine which one is the value we need. The fact that we have two critical points means that neither the first derivative test or the second derivative test can be used here as they both require a single critical point. This isn’t a real problem however. Go back to the figure at the start of the solution and notice that we can quite easily find limits on (h). The smallest (h) can be is (h = 0) even though this doesn’t make much sense as we won’t get a box in this case. Also, from the 10 inch side we can see that the largest (h) can be is (h = 5) although again, this doesn’t make much sense physically.
So, knowing that whatever (h) is it must be in the range (0 \le h \le 5) we can see that the second critical point is outside this range and so the only critical point that we need to worry about is 1.9183.
Finally, since the volume is defined and continuous on (0 \le h \le 5) all we need to do is plug in the critical points and endpoints into the volume to determine which gives the largest volume. Here are those function evaluations.
[V\left( 0 \right) = 0\hspace{0.25in}\hspace{0.25in}V\left( {1.9183} \right) = 120.1644\hspace{0.25in}\hspace{0.25in}V\left( 5 \right) = 0]
So, if we take (h = 1.9183) we get a maximum volume.
Example 6 A printer needs to make a poster that will have a total area of 200 in2 and will have 1 inch margins on the sides, a 2 inch margin on the top and a 1.5 inch margin on the bottom as shown below. What dimensions will give the largest printed area?
Show Solution
This problem is a little different from the previous problems. Both the constraint and the function we are going to optimize are areas. The constraint is that the overall area of the poster must be 200 in2 while we want to optimize the printed area (i.e. the area of the poster with the margins taken out).
Let’s define the height of the poster to be (h) and the width of the poster to be (w). Here is a new sketch of the poster and we can see that once we’ve taken the margins into account the width of the printed area is (w - 2) and the height of the printer area is (h - 3.5).
Here are the equations that we’ll be working with.
[\begin{align}{\mbox{Maximize : }} & A = \left( {w - 2} \right)\left( {h - 3.5} \right)\ {\mbox{Constraint :}} & 200 = wh\end{align}]
Solving the constraint for (h) and plugging into the equation for the printed area gives,
[A\left( w \right) = \left( {w - 2} \right)\left( {\frac{{200}}{w} - 3.5} \right) = 207 - 3.5w - \frac{{400}}{w}]
The first and second derivatives are,
[A'\left( w \right) = - 3.5 + \frac{{400}}{{{w^2}}} = \frac{{400 - 3.5{w^2}}}{{{w^2}}}\hspace{0.25in}\hspace{0.25in}A''\left( w \right) = - \frac{{800}}{{{w^3}}}]
From the first derivative we have the following two critical points ((w = 0) is not a critical point because the area function does not exist there).
[w = \pm \,\sqrt {\frac{400}{3.5}} = \pm \,10.6904]
However, since we’re dealing with the dimensions of a piece of paper we know that we must have (w > 0) and so only 10.6904 will make sense.
Also notice that provided (w > 0) the second derivative will always be negative and so in the range of possible optimal values of the width the area function is always concave down and so we know that the maximum printed area will be at (w = 10.6904\,\,{\mbox{inches}}).
The height of the paper that gives the maximum printed area is then,
[h = \frac{{200}}{{10.6904}} = 18.7084\,\,{\mbox{inches}}]
We’ve worked quite a few examples to this point and we have quite a few more to work. However, this section has gotten quite lengthy so let’s continue our examples in the next section. This is being done mostly because these notes are also being presented on the web and this will help to keep the load times on the pages down somewhat.
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find new average if removing one element from current average
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Asked 12 years ago
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Ok, so say that I have a current rating average: 3.3/5
Now I want to say to remove a rating of 4. How do I find the new average? Or is this even possible?
average
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asked Sep 14, 2013 at 17:38
LordZardeckLordZardeck
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1 To find the new average, you should know the number of observations.detnvvp –detnvvp 2013-09-14 17:45:04 +00:00 Commented Sep 14, 2013 at 17:45
So I all I have is that the current average is 3.3, without knowing how many ratings I had to get there I can't do this? How would I do it if I did know?LordZardeck –LordZardeck 2013-09-14 17:54:43 +00:00 Commented Sep 14, 2013 at 17:54
If the current average is 3.3 3.3, based on n n "tests", then the sum of the marks is 3.3 n 3.3 n. Remove the 4 4, divide by n−1 n−1 for the revised average 3.3 n−4 n−1 3.3 n−4 n−1.André Nicolas –André Nicolas 2013-09-14 17:58:07 +00:00 Commented Sep 14, 2013 at 17:58
You need to know the number of ratings. If the average was based on many ratings, then removing one would have little effect. On the other hand, if the average was based on two ratings, then removing one could have a significant effect.copper.hat –copper.hat 2013-09-14 17:59:56 +00:00 Commented Sep 14, 2013 at 17:59
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To fix the problem of thirdender, and put it on a programmatic way.
Substract a value:
average = ((average nbValues) - value) / (nbValues - 1);
Add a value:
average = average + ((value - average) / nbValues);
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answered Dec 9, 2015 at 10:32
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If you know the number of observations, say it is N N, then, if x 1,…x N x 1,…x N are the observations, you have that ∑N i=1 x i=3.3 N/5∑i=1 N x i=3.3 N/5. Therefore, the new average will be 3.3 N/5−4 N−1 3.3 N/5−4 N−1.
If you don't know the number of observations, you can't find the new average. Your observations could be, for example, x 1=4/5,x 2=2.6/5 x 1=4/5,x 2=2.6/5 or x 1=4/5,x 2=5.9/5,x 3=0 x 1=4/5,x 2=5.9/5,x 3=0, and in the first case the new average is 2.6/5 2.6/5, while in the second the new average is 2.95/5 2.95/5.
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answered Sep 14, 2013 at 18:00
detnvvpdetnvvp
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For any computer programmers finding this question, the code is very simple.
function removeFromAverage(value, average, samples) {
if (samples <= 1) {
return FALSE;
}
let newAverage = ((average samples) - value) / (samples - 1);
return newAverage;
}
value is the value you want to remove from the average, and samples is the number of observations included in the previous average. The if statement is only there in case one sample is used in the previous average, in which case removing that single value would result in zero samples and no calculable average.
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edited Aug 20, 2022 at 0:18
Hassan Ahmed
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answered Sep 30, 2014 at 22:33
thirdenderthirdender
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1 This algorithm is not correct. The average of [7,3,2,8] is 5. If you remove 2 from the list, resulting in [7,3,8], the new average is 6. You algorithm returns -0.3333, which is not the difference between 5 and 6.Nepoxx –Nepoxx 2015-04-15 16:10:32 +00:00 Commented Apr 15, 2015 at 16:10
Thanks to Hassan, the answer has been updated thirdender –thirdender 2022-09-01 18:05:59 +00:00 Commented Sep 1, 2022 at 18:05
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People are travelling internationally again after the hiatus that was caused by the coronavirus pandemic and the subsequent lockdowns. Many people are using the money they saved while borders were closed to embark on their bucket list trips. However, others are showing a reluctance to dust off their passports. A travel report from the Morning Consult website analyzed 16,000 surveys from people in 15 countries. The site found that 15 per cent of South Koreans and 14 per cent of Chinese and Americans, "never want to travel again". Those most unwilling to travel were the Japanese. The report stated that almost 35 per cent of Japanese respondents claimed they never wanted to leave Japan.
There are many reasons for Japanese travellers being disinclined to venture overseas. One is a new-found desire to explore the culinary and cultural delights Japan has to offer. Many people travelled domestically during the pandemic and re-discovered their love of Japan's stunning nature and heritage. A Kyoto tour guide said: "I've been constantly amazed at my clients' reactions at seeing the sights of this city. I've never known Japanese people to be so engrossed in their history and tradition." Other reasons are cost and a desire to be green. Tokyo resident Kai Ueno said: "Flying isn't sustainable in this climate crisis. I'd much rather travel locally and spend the air fare on nicer hotels, restaurants and experiences in Japan."
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Try the same news story at these easier levels:
Travel Plans - Level 4orTravel Plans - Level 5
Sources
Make sure you try all of the online activities for this reading and listening - There are dictations, multiple choice activities, drag and drop activities, sentence jumbles, which word activities, text reconstructions, spelling, gap fills and a whole lot more. Please enjoy :-)
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News
"Much has been said and written on the utility of newspapers; but one principal advantage which might be derived from these publications has been neglected; we mean that of reading them in schools."
The Portland Eastern Herald (June 8, 1795)
"News is history in its first and best form, its vivid and fascinating form, and...history is the pale and tranquil reflection of it."
Mark Twain, in his autobiography (1906)
"Current events provide authentic learning experiences for students at all grade levels.... In studying current events, students are required to use a range of cognitive, affective, critical thinking and research skills."
Haas, M. and Laughlin, M. (2000) Teaching Current Events: It's Status in Social Studies Today.
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Warm-ups
1. TRAVEL PLANS:Students walk around the class and talk to other students about travel plans. Change partners often and share your findings.
2. CHAT: In pairs / groups, talk about these topics or words from the article. What will the article say about them? What can you say about these words and your life?
travelling / pandemic / lockdown / bucket list / dust / passport / survey / website /
overseas / culinary delights / nature / heritage / sights / climate crisis / air fare / hotel
Have a chat about the topics you liked. Change topics and partners frequently.
3. OVERSEAS TRAVEL:Students A strongly believe overseas travel is the best education in the world; Students B strongly believe it isn't. Change partners again and talk about your conversations.
4. MY COUNTRY'S DELIGHTS:What are your country's delights? What makes them so good? Complete this table with your partner(s). Change partners often and share what you wrote.
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DelightsWhy?
Nature
Festivals
Sights
Food
Museums
Entertainment
MY e-BOOK
See a sample
5. PASSPORT:Spend one minute writing down all of the different words you associate with the word "passport". Share your words with your partner(s) and talk about them. Together, put the words into different categories.
6. DESTINATIONS:Rank these with your partner. Put the best destinations at the top. Change partners often and share your rankings.
Kyoto
London
Rome
Rio de Janeiro
Nairobi
Shanghai
Delhi
Sydney
Vocabulary
Paragraph 1
1.hiatus a.A number of experiences or achievements that a person hopes to have or do during their lifetime.
2.subsequent b.A person who replies to something, like supplying information for a questionnaire or responding to an advertisement.
3.embark c.A pause or break in continuity in a sequence or activity.
4.bucket list d.Begin a course of action or a journey.
5.reluctance e.Coming after something in time; following.
6.dust off f.Unwillingness to do something.
7.respondent g.Bring something out for use again after a long period of non-use.
Paragraph 2
8.disinclined h.Start a (risky or daring) journey or course of action.
9.venture i.Of or for cooking.
10.culinary j.Put all of one's attention or interest on something.
11.stunning k.Being unwilling or reluctant.
12.heritage l.Conserving or keeping an environmental balance by avoiding using up natural resources.
13.engrossed m.Valued objects and qualities such as historic buildings and cultural traditions that have been passed down from previous generations.
14.sustainable n.Extremely impressive or attractive.
Before reading / listening
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1. TRUE / FALSE: Read the headline. Guess if 1-8 below are true (T) or false (F).
The article says a hiatus caused coronavirus lockdowns.T / F
Many people tripped over buckets during the coronavirus pandemic.T / F
A website conducted a survey on people from 15 countries.T / F
Chinese people were less reluctant to travel than South Koreans.T / F
Many Japanese people wanted to explore Japan more.T / F
A tour guide in Kyoto was amazed by the city's sights. T / F
The tour guide said her clients were really interested in Japan's history.T / F
Some Japanese people wanted to spend money on hotels in Japan.T / F
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2. SYNONYM MATCH: Match the following synonyms from the article.
hiatus
subsequent
embark
analyzed
unwilling
venture
delights
stunning
engrossed
spend
pleasures
reluctant
examined
following
travel
absorbed
break
lay out
start
remarkable
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3. PHRASE MATCH: (Sometimes more than one choice is possible.)
after the hiatus that was
the money they saved
embark on their bucket
showing a reluctance to
35 per cent of Japanese
being disinclined to venture
Japan's stunning nature
I've never known Japanese people
Flying isn't
spend the air fare
sustainable
list trips
respondents
overseas
and heritage
caused by the coronavirus pandemic
to be so engrossed
while borders were closed
on nicer hotels
dust off their passports
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Gap fill
Put these words into the spaces in the paragraph below.
_bucket
unwilling
dust
respondents
hiatus
stated
lockdowns
surveys_
People are travelling internationally again after the (1) ___ that was caused by the coronavirus pandemic and the subsequent (2) ___. Many people are using the money they saved while borders were closed to embark on their (3) ___ list trips. However, others are showing a reluctance to (4) ___ off their passports. A travel report from the Morning Consult website analyzed 16,000 (5) ___ from people in 15 countries. The site found that 15 per cent of South Koreans and 14 per cent of Chinese and Americans, "never want to travel again". Those most (6) ___ to travel were the Japanese. The report (7) ___ that almost 35 per cent of Japanese (8) ___ claimed they never wanted to leave Japan.
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Put these words into the spaces in the paragraph below.
_desire
stunning
locally
venture
sustainable
constantly
delights
engrossed_
There are many reasons for Japanese travellers being disinclined to (9) ___ overseas. One is a new-found desire to explore the culinary and cultural (10) ___ Japan has to offer. Many people travelled domestically during the pandemic and re-discovered their love of Japan's (11) ___ nature and heritage. A Kyoto tour guide said: "I've been (12) ___ amazed at my clients' reactions at seeing the sights of this city. I've never known Japanese people to be so (13) ___ in their history and tradition." Other reasons are cost and a (14) ___ to be green. Tokyo resident Kai Ueno said: "Flying isn't (15) ___ in this climate crisis. I'd much rather travel (16) ___ and spend the air fare on nicer hotels, restaurants and experiences in Japan."
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Listening — Guess the answers. Listen to check.
1) People are travelling internationally again ______
a. after the high attics
b. after the "hi" at us
c. after the hiatus
d. after the high eight us
2) that was caused by the coronavirus pandemic and ______
a. the subsequence lockdowns
b. the subsequent lockdowns
c. the subsequently lockdowns
d. the consequent lockdowns
3) the money they saved while borders were closed to embark on their ______
a. bucket wrist trips
b. bucket risk trips
c. bucket lips trips
d. bucket list trips
4) However, others are showing a reluctance to dust ______
a. oft their passports
b. toff their passports
c. of their passports
d. off their passports
5) The report stated that almost 35 per cent of ______
a. Japanese responders claimed
b. Japanese respond ants claimed
c. Japanese respond dents claimed
d. Japanese respondents claimed
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6) There are many reasons for Japanese travellers being disinclined ______
a. to venture overseas
b. to adventure overseas
c. tour venture overseas
d. to vent chair overseas
7) One is a new-found desire to explore the culinary ______
a. and cultural delights
b. and culturally delights
c. and cultural dill lights
d. and culturally dates
8) and re-discovered their love of Japan's stunning ______
a. nature and hermitage
b. nature end hermitage
c. natural end heritage
d. nature and heritage
9) the sights of this city. I've never known Japanese people to ______
a. be so embossed
b. be so win grossed
c. be sewing grossed
d. be so engrossed
10) isn't sustainable in this climate crisis. I'd much rather travel locally and spend ______
a. the year fare
b. there fare
c. the air fare
d. thee air fair
Listening — Listen and fill in the gaps
People are travelling internationally again after (1) ____ was caused by the coronavirus pandemic and the subsequent lockdowns. Many people are using the money they (2) ___ were closed to embark on their bucket list trips. However, others are showing a (3) __ off their passports. A travel report from the Morning Consult website analyzed 16,000 (4) __ in 15 countries. The site found that 15 per cent of South Koreans and 14 per cent of Chinese and Americans, "never want to travel again". Those most (5) __ were the Japanese. The report stated that almost 35 per cent of Japanese (6) _____ never wanted to leave Japan.
There are many reasons for Japanese travellers being disinclined (7) ____. One is a new-found desire to explore the culinary (8) ___ Japan has to offer. Many people travelled domestically during the pandemic and re-discovered their love of Japan's stunning (9) __. A Kyoto tour guide said: "I've been constantly amazed at my clients' reactions at (10) __ of this city. I've never known Japanese people to (11) __ in their history and tradition." Other reasons are cost and a desire to be green. Tokyo resident Kai Ueno said: "Flying isn't (12) _____ climate crisis. I'd much rather travel locally and spend the air fare on nicer hotels, restaurants and experiences in Japan."
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Comprehension questions
What was a hiatus in international travel caused by?
What were closed that allowed people to save money?
What kind of list was mentioned in paragraph one?
How many people participated in the travel survey?
Which nationality were the least willing to travel overseas?
Where were Japanese people disinclined to venture?
What did many Japanese people fall in love with again?
What did a tour guide say Japanese people were engrossed in?
What did a resident of Kyoto say was not sustainable?
What would many Japanese prefer to spend their money on?
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Multiple choice quiz
1) What was a hiatus in international travel caused by?
a) fuel shortages
b) coronavirus lockdowns
c) terrorism
d) ash from a volcano
2) What were closed that allowed people to save money?
a) wallets
b) banks
c) shops
d) borders
3) What kind of list was mentioned in paragraph one?
a) a reading list
b) a to-do list
c) a bucket list
d) a top-ten list
4) How many people participated in the travel survey?
a) 16,000
b) 17,000
c) 18,000
d) 19,000
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5) Which nationality were the least willing to travel overseas?
a) South Koreans
b) the Japanese
c) Americans
d) Chinese
6) Where were Japanese people disinclined to venture?
a) overseas
b) into mountains
c) into business partnerships
d) into relationships
7) What did many Japanese people fall in love with again?
a) themselves
b) their partners
c) Japan's nature and heritage
d) flying
8) What did a tour guide say Japanese people were engrossed in?
a) history and tradition
b) food
c) maps
d) travel books
9) What did a resident of Kyoto say was not sustainable?
a) population growth
b) cars
c) tourism
d) flying
10) What would many Japanese prefer to spend their money on?
a) themselves
b) their children
c) hotels, restaurants and experiences
d) games
Role play
Role A – Kyoto
You think the best travel destination is Kyoto. Tell the others three reasons why. Tell them what is wrong with their destinations. Also, tell the others which is the least attractive of these (and why): Rio de Janeiro, Nairobi or Sydney.
Role B – Rio de Janeiro
You think Rio de Janeiro is the best travel destination. Tell the others three reasons why. Tell them what is wrong with their destinations. Also, tell the others which is the least attractive of these (and why): Kyoto, Nairobi or Sydney.
Role C – Nairobi
You think Nairobi is the best travel destination. Tell the others three reasons why. Tell them what is wrong with their destinations. Also, tell the others which is the least attractive of these (and why): Rio de Janeiro, Kyoto or Sydney.
Role D – Sydney
You think Sydney is the best travel destination. Tell the others three reasons why. Tell them what is wrong with their destinations. Also, tell the others which is the least attractive of these (and why): Rio de Janeiro, Nairobi or Kyoto.
After reading / listening
1. WORD SEARCH: Look in your dictionary / computer to find collocates, other meanings, information, synonyms … for the words...
'travel'
and 'plan'.
• Share your findings with your partners.
• Make questions using the words you found.
• Ask your partner / group your questions.
2. ARTICLE QUESTIONS: Look back at the article and write down some questions you would like to ask the class about the text.
•Share your questions with other classmates / groups. •Ask your partner / group your questions.
3. GAP FILL: In pairs / groups, compare your answers to this exercise. Check your answers. Talk about the words from the activity. Were they new, interesting, worth learning…?
4. VOCABULARY: Circle any words you do not understand. In groups, pool unknown words and use dictionaries to find their meanings.
5. TEST EACH OTHER: Look at the words below. With your partner, try to recall how they were used in the text:
caused
bucket
dust
16,000
unwilling
claimed
venture
delights
love
amazed
cost
rather
Student survey
Write five GOOD questions about this topic in the table. Do this in pairs. Each student must write the questions on his / her own paper. When you have finished, interview other students. Write down their answers.
(Please look at page 12 of the PDF to see a photocopiable example of this activity.)
Discussion - Travel Plans
STUDENT A’s QUESTIONS (Do not show these to student B)
What did you think when you read the headline?
What images are in your mind when you hear the word 'travel'?
What are your travel plans for this year?
How important is it to travel?
Did the coronavirus pandemic affect your travel plans?
What do you think of your passport?
How often do you dust off your passport?
Why might people not want to travel overseas?
What countries or places are on your bucket list?
Should we avoid international travel to protect the environment?
STUDENT B’s QUESTIONS (Do not show these to student A)
Did you like reading this article? Why/not?
What do you think of when you hear the word 'plan'?
What do you think about what you read?
What do you think of international travel?
What culinary and cultural delights are there in your country?
How stunning is your country's nature?
Do you think flying is unsustainable?
Is it better to spend money on an air fare or a gorgeous hotel?
Why do many people get the travel bug?
What questions would you like to ask a frequent traveller?
Discussion — Write your own questions
STUDENT A’s QUESTIONS (Do not show these to student B)
(a) ____
(b) ____
(c) ____
(d) ____
(e) ____
STUDENT B’s QUESTIONS (Do not show these to student A)
(f) ____
(g) ____
(h) ____
(i) ____
(j) ____
Language — Cloze (Gap-fill)
People are travelling internationally again after the (1) _ that was caused by the coronavirus pandemic and the subsequent lockdowns. Many people are using the money they saved while borders were closed to embark (2) their bucket list trips. However, others are showing a reluctance to (3) _ off their passports. A travel report from the Morning Consult website (4) 16,000 surveys from people in 15 countries. The site found that 15 per cent of South Koreans and 14 per cent of Chinese and Americans, "never want to travel again". Those most (5) _ to travel were the Japanese. The report stated that almost 35 per cent of Japanese respondents (6) _ they never wanted to leave Japan.
There are many reasons for Japanese travellers being disinclined to (7) _ overseas. One is a new-found desire to explore the (8) and cultural delights Japan has to offer. Many people travelled domestically during the pandemic and re-discovered their love of Japan's (9) _ nature and heritage. A Kyoto tour guide said: "I've been constantly amazed at my clients' reactions at seeing the sights of this city. I've never known Japanese people to be so (10) in their history and tradition." Other reasons are cost and a desire to be green. Tokyo resident Kai Ueno said: "Flying isn't sustainable (11) _ this climate crisis. I'd much (12) _ travel locally and spend the air fare on nicer hotels, restaurants and experiences in Japan."
Which of these words go in the above text?
(a) ratios (b) hiatus (c) cactus (d) herbivorous
(a) of (b) by (c) at (d) on
(a) lust (b) dust (c) bust (d) must
(a) paralyzed (b) catalyzed (c) analyzed (d) hydrolyzed
(a) unwilling (b) unwitting (c) unfailing (d) unfettering
(a) claimed (b) clammed (c) calmed (d) clamped
(a) vulture (b) couture (c) future (d) venture
(a) culinary (b) veterinary (c) luminary (d) pulmonary
(a) tuning (b) stunning (c) turning (d) attuning
(a) embossed (b) en masse (c) engrossed (d) enmeshed
(a) at (b) on (c) in (d) of
(a) elect (b) rather (c) choice (d) prefer
Spelling
Paragraph 1
after the iasuth that was caused
and the qeetnbussu lockdowns
akberm on their bucket list trips
showing a urecntecla to dust off their passports
Those most liwunlnig to travel
35 per cent of Japanese enosesdtnpr
Paragraph 2
travellers being lneiinidcsd to venture overseas
explore the unlircay and cultural delights
Japan's stunning nature and gretaehi
eedsognrs in their history and tradition
Tokyo idnerset Kai Ueno
Flying isn't ablnseasiut
Put the text back together
(...) crisis. I'd much rather travel locally and spend the air fare on nicer hotels, restaurants and experiences in Japan."
(...) of this city. I've never known Japanese people to be so engrossed in their history and tradition." Other
(...) heritage. A Kyoto tour guide said: "I've been constantly amazed at my clients' reactions at seeing the sights
(...) desire to explore the culinary and cultural delights Japan has to offer. Many people travelled
(1 ) People are travelling internationally again after the hiatus that was caused by the coronavirus pandemic
(...) and the subsequent lockdowns. Many people are using the money they saved while borders
(...) domestically during the pandemic and re-discovered their love of Japan's stunning nature and
(...) from people in 15 countries. The site found that 15 per cent of South Koreans and 14 per cent of Chinese and Americans, "never
(...) were closed to embark on their bucket list trips. However, others are showing a reluctance to dust
(...) off their passports. A travel report from the Morning Consult website analyzed 16,000 surveys
(...) There are many reasons for Japanese travellers being disinclined to venture overseas. One is a new-found
(...) 35 per cent of Japanese respondents claimed they never wanted to leave Japan.
(...) want to travel again". Those most unwilling to travel were the Japanese. The report stated that almost
(...) reasons are cost and a desire to be green. Tokyo resident Kai Ueno said: "Flying isn't sustainable in this climate
Put the words in the right order
caused that was coronavirus . hiatus After the by
closed . The they money borders while were saved
dust off Showing to a passports . their reluctance
most Japanese . the travel Those to unwilling were
wanted claimed they Respondents to Japan . never leave
One to reason desire explore . a is new-found
people Many during the domestically travelled pandemic .
been clients' my reactions . at amazed I've constantly
be known I've people engrossed . so never to
spend rather fare on hotels . the air I'd
Circle the correct word (20 pairs)
People are travelling internationally again after the cactus / hiatus that was caused by the coronavirus pandemic and the subsequently / subsequent lockdowns. Many people are using the money they spent / saved while borders were closed to embark / disembark on their bucket list trips. However, others are showing the / a reluctance to dust off / on their passports. A travel report from the Morning Consult website analysis / analyzed 16,000 surveys from people in 15 countries. The site found that 15 per cent of South Koreans and 14 per cent of Chinese and Americans, "never went / want to travel again". Those most / much unwilling to travel were the Japanese. The report stated that almost 35 per cent of Japanese respondents claimed / calmed they never wanted to leave Japan.
There are many reasons for Japanese travellers being declined / disinclined to venture overseas. One is a new-found desire / desires to explore the culinary and cultural delightful / delights Japan has to offer. Many people travelled domestically during / within the pandemic and re-discovered their love of Japan's stunning natural / nature and heritage. A Kyoto tour guide said: "I've been constantly amazed at / on my clients' reactions at seeing the sights of this city. I've never known Japanese people to be so / such engrossed in / on their history and tradition." Other reasons are cost and a desire to be / see green. Tokyo resident Kai Ueno said: "Flying isn't sustainable in this climate crisis. I'd much rather travel locally and spend the air fair / fare on nicer hotels, restaurants and experiences in Japan."
Talk about the connection between each pair of words in italics, and why the correct word is correct. Look up the definition of new words.
Insert the vowels (a, e, i, o, u)
P__pl_ r tr_v_ll_ng nt_rn_t__n_lly _g__n _ft_r th h__t_s th_t w_s c__s_d by th_ c_r_n_v_r_s p_nd_m_c nd th s_bs_q__nt l_ckd_wns. M_ny p__pl_ r s_ng th m_n_y th_y s_v_d wh_l_ b_rd_rs w_r_ cl_s_d t_ mb_rk _n th__r b_ck_t l_st tr_ps. H_w_v_r, _th_rs _r sh_w_ng _ r_l_ct_nc_ t_ d_st ff th__r p_ssp_rts. _ tr_v_l r_p_rt fr_m th M_rn_ng C_ns_lt w_bs_t_ n_lyz_d 16,000 s_rv_ys fr_m p__pl n 15 c__ntr__s. Th s_t_ f__nd th_t 15 p_r c_nt f S__th K_r__ns _nd 14 p_r c_nt _f Ch_n_s nd _m_r_c_ns, "n_v_r w_nt t tr_v_l g__n". Th_s m_st nw_ll_ng t tr_v_l w_r_ th_ J_p_n_s_. Th_ r_p_rt st_t_d th_t lm_st 35 p_r c_nt _f J_p_n_s r_sp_nd_nts cl__m_d th_y n_v_r w_nt_d t_ l__v_ J_p_n.
Th_r_ r m_ny r__s_ns f_r J_p_n_s_ tr_v_ll_rs b__ng d_s_ncl_n_d t_ v_nt_r_ v_rs__s. _n s _ n_wf__nd d_s_r t_ xpl_r th_ c_l_n_ry nd c_lt_r_l d_l_ghts J_p_n h_s t ff_r. M_ny p__pl tr_v_ll_d d_m_st_c_lly d_r_ng th_ p_nd_m_c nd r-d_sc_v_r_d th__r l_v_ f J_p_n's st_nn_ng n_t_r nd h_r_t_g. _ Ky_t_ t__r g__d_ s__d: "'v b__n c_nst_ntly m_z_d _t my cl__nts' r__ct__ns _t s___ng th s_ghts f th_s c_ty. 'v_ n_v_r kn_wn J_p_n_s_ p__pl_ t_ b_ s_ ngr_ss_d _n th__r h_st_ry _nd tr_d_t__n." _th_r r__s_ns _r c_st nd _ d_s_r t_ b_ gr__n. T_ky_ r_s_d_nt K__ __n_ s__d: "Fly_ng sn't s_st__n_bl n th_s cl_m_t cr_s_s. 'd m_ch r_th_r tr_v_l l_c_lly _nd sp_nd th __r f_r_ _n n_c_r h_t_ls, r_st__r_nts _nd _xp_r__nc_s _n J_p_n."
Punctuate the text and add capitals
people are travelling internationally again after the hiatus that was caused by the coronavirus pandemic and the subsequent lockdowns many people are using the money they saved while borders were closed to embark on their bucket list trips however others are showing a reluctance to dust off their passports a travel report from the morning consult website analyzed 16000 surveys from people in 15 countries the site found that 15 per cent of south koreans and 14 per cent of chinese and americans never want to travel again those most unwilling to travel were the japanese the report stated that almost 35 per cent of japanese respondents claimed they never wanted to leave japan
there are many reasons for japanese travellers being disinclined to venture overseas one is a new-found desire to explore the culinary and cultural delights japan has to offer many people travelled domestically during the pandemic and rediscovered their love of japans stunning nature and heritage a kyoto tour guide said ive been constantly amazed at my clients reactions at seeing the sights of this city ive never known japanese people to be so engrossed in their history and tradition other reasons are cost and a desire to be green tokyo resident kai ueno said flying isnt sustainable in this climate crisis id much rather travel locally and spend the air fare on nicer hotels restaurants and experiences in japan
Put a slash (/) where the spaces are
Peoplearetravellinginternationallyagainafterthehiatusthatwascause
dbythecoronaviruspandemicandthesubsequentlockdowns.Manype
opleareusingthemoneytheysavedwhileborderswereclosedtoembark
ontheirbucketlisttrips.However,othersareshowingareluctancetodus
tofftheirpassports.AtravelreportfromtheMorningConsultwebsiteana
lyzed16,000surveysfrompeoplein15countries.Thesitefoundthat15p
ercentofSouthKoreansand14percentofChineseandAmericans,"neve
rwanttotravelagain".ThosemostunwillingtotravelweretheJapanese.
Thereportstatedthatalmost35percentofJapaneserespondentsclaim
edtheyneverwantedtoleaveJapan.TherearemanyreasonsforJapane
setravellersbeingdisinclinedtoventureoverseas.Oneisanew-foundd
esiretoexploretheculinaryandculturaldelightsJapanhastooffer.Many
peopletravelleddomesticallyduringthepandemicandre-discoveredt
heirloveofJapan'sstunningnatureandheritage.AKyototourguidesaid
:"I'vebeenconstantlyamazedatmyclients'reactionsatseeingthesight
softhiscity.I'veneverknownJapanesepeopletobesoengrossedintheir
historyandtradition."Otherreasonsarecostandadesiretobegreen.To
kyoresidentKaiUenosaid:"Flyingisn'tsustainableinthisclimatecrisis.I
'dmuchrathertravellocallyandspendtheairfareonnicerhotels,restaur
antsandexperiencesinJapan."
Free writing
Write about the lesson page for 10 minutes. Comment on your partner’s paper.
Academic writing
International travel is the best education in the world. Discuss.
Homework
1. VOCABULARY EXTENSION: Choose several of the words from the text. Use a dictionary or Google's search field (or another search engine) to build up more associations / collocations of each word.
2. INTERNET: Search the Internet and find out more about this news story. Share what you discover with your partner(s) in the next lesson.
3. TRAVEL PLANS:Make a poster about travel plans. Show your work to your classmates in the next lesson. Did you all have similar things?
4. DOMESTIC TRAVEL:Write a magazine article about governments encouraging domestic over international travel. Include imaginary interviews with people who are for and against this.
Read what you wrote to your classmates in the next lesson. Write down any new words and expressions you hear from your partner(s).
5. WHAT HAPPENED NEXT?Write a newspaper article about the next stage in this news story. Read what you wrote to your classmates in the next lesson. Give each other feedback on your articles.
6. LETTER:Write a letter to an expert on travel plans. Ask him/her three questions about travelling. Give him/her three of your ideas on the best places to visit. Read your letter to your partner(s) in your next lesson. Your partner(s) will answer your questions.
A Few Additional Activities for Students
Ask your students what they have read, seen or heard about this news in their own language. Students are likely to / may have have encountered this news in their L1 and therefore bring a background knowledge to the classroom.
Get students to role play different characters from this news story.
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Why should I rearrange equations?
Believe it or not, there are many good reasons to develop your ability to rearrange equations that are important to the your field. It can save time, help you with units and save some brain space! Here are some reasons to develop your equation manipulation skills (in no particular order):
Equations are easier to handle before inserting numbers! And, if you can isolate a variable on one side of the equation, it is applicable to every similar problem that asks you to solve for that variable!
If you know how to manipulate equations, you only have to remember one equation that has all the variables of question in it - you can manipulate it to solve for any other variable! This means less memorization!
Manipulating equations can help you keep track of (or figure out) units on a number. Because units are defined by the equations, if you manipulate, plug in numbers and cancel units, you'll end up with exactly the right units (for a given variable)!
Rules for Rearranging Equations
Apply the same operation to both sides of the equal sign.
Perform the inverse operation to move or cancel a constant or variable on one side of the equation.
Be aware of the order of operations! Which operation can you perform first?
Examples
Example 1
Rearrange the following formula for the variable
Solution
Since the goal is to isolate for , we want to move the variable .
We need to first move the at the bottom of the fraction by multiplying both sides by
This will move from the bottom of the right fraction to the other side of the equation
Now move we move the remaining on the right side by subracting both sides by
Example 2
Rearrange the following equation for ,
Solution
Watch the video for the solution.
Non-YouTube video link
Formula Rearrangement Practice Questions
Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.
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Calculate the molar mass of a molecule of water (H 2O).
Find the number of moles and the number of molecules present in 9g of water.
Given:
The atomic mass of hydrogen (H) = 1
The atomic mass of oxygen (O) = 16
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The molar mass of water is 18 g/mol. The number of moles in 9g of water is 0.5 mol. The number of molecules present in 9g of water is 3.01 x 10^23 molecules.
Explanation
The molecular mass of a molecule of water (H2O) is calculated by adding the atomic masses of its constituent elements. It comprises 2 Hydrogen atoms and one Oxygen atom. The mass of Hydrogen (H) is 1 and the mass of Oxygen (O) is 16. So, the molar mass (g/mol) of H2O is (21) + 16 = 18 g/mol.
The next step is to calculate the number of moles in 9 g of water. The formula for calculating moles is mass divided by molar mass. Thus, number of moles = 9 g / 18 g/mol = 0.5 mol.
Furthermore, we can determine the number of molecules in 0.5 mol. Avogadro's number (6.022 x 10^23) is the number of molecules in one mole. Thus, the number of molecules in 0.5 mol is Avogadro's number times the number of moles, or 0.5 (6.022 x 10^23) = 3.01 x 10^23 molecules.
Learn more about Molar Mass and Moles here:
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The molar mass of water is 18 g/mol. In 9 g of water, there are 0.5 mol, which corresponds to approximately 3.01 x 10^23 molecules. Thus, we can determine the quantities of moles and molecules from the given mass of water.
Explanation
To calculate the molar mass of water (H₂O), we first need to consider the individual atomic masses of its components: hydrogen and oxygen.
Atomic Mass of Hydrogen (H): Each hydrogen atom has an atomic mass of approximately 1 g/mol. Since there are 2 hydrogen atoms in a molecule of water, their total mass contributes: 2 \times 1 = 2 g/mol.
Atomic Mass of Oxygen (O): One oxygen atom has an atomic mass of approximately 16 g/mol.
Now we can find the total molar mass of water:
Molar Mass of H 2O=(2×1)+16=2+16=18 g/mol
Next, we will determine the number of moles in 9 g of water. The number of moles can be calculated using the formula:
Number of moles=molar mass(g/mol)mass(g)
Substituting in the values we have:
Number of moles=18 g/mol 9 g=0.5 mol
Finally, to find the number of molecules present in 0.5 mol of water, we can use Avogadro's number, which states that 1 mole of any substance contains approximately 6.022 \times 10^{23} representative particles (in this case, molecules).
To calculate the number of molecules:
Number of molecules=0.5 mol×6.022×1 0 23 molecules/mol≈3.01×1 0 23 molecules
Therefore, in summary:
The molar mass of water is 18 g/mol.
The number of moles in 9 g of water is 0.5 mol.
The number of molecules present in 9 g of water is approximately 3.01 \times 10^{23} molecules.
Examples & Evidence
For example, if you have 36 g of water, applying the same calculations would yield 2 mol of water and about 1.2 x 10^24 molecules, showcasing how you can scale this process based on the mass you have.
The calculations and values used, such as the atomic masses of hydrogen (1 g/mol) and oxygen (16 g/mol), are standard values found in chemistry resources. The relationship defined by Avogadro's number (6.022 x 10^23) is universally accepted in chemistry for determining the number of molecules in a mole.
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The molar mass of water is 18.0 g/mol. In 9 grams of water, there are 0.5 moles, which corresponds to 3.011 x 10²³ molecules.
To calculate the molar mass of a water molecule (H₂O), we need to sum the atomic masses of its constituent elements:
Hydrogen (H): 1.0 g/mol
Oxygen (O): 16.0 g/mol
Since a water molecule contains 2 hydrogen atoms and 1 oxygen atom, the molar mass calculation is:
Molar mass of H₂O = 2(1.0) + 16.0 = 18.0 g/mol
Next, to find the number of moles in 9 grams of water, we use the formula:
Number of moles = Mass / Molar mass
Substituting the given values:
Number of moles in 9 grams of water = 9 g / 18.0 g/mol = 0.5 moles
Finally, to find the number of molecules in 9 grams of water, we use Avogadro's number, which is 6.022 x 10²³ molecules/mol:
Number of molecules = Number of moles x Avogadro's number
Substituting the values:
Number of molecules = 0.5 moles x 6.022 x 10²³ molecules/mol = 3.011 x 10²³ molecules
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1 Department of Medicine, Christian Medical College, Vellore, India
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1,✉, Tina George
Tina George
1 Department of Medicine, Christian Medical College, Vellore, India
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1, Prathap Tharyan
Prathap Tharyan
2 BV Moses Center, Clinical Epidemiology Unit, Christian Medical College, Vellore, India
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2, Rani Diana Sahni
Rani Diana Sahni
3 Department of Microbiology, Christian Medical College, Vellore, India
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3, Suceena Alexander
Suceena Alexander
4 Department of Nephrology, Christian Medical College, Vellore, India
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4, Richard Kirubakaran
Richard Kirubakaran
2 BV Moses Center, Clinical Epidemiology Unit, Christian Medical College, Vellore, India
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2, Jabez Paul Barnabas Ezekiel
Jabez Paul Barnabas Ezekiel
2 BV Moses Center, Clinical Epidemiology Unit, Christian Medical College, Vellore, India
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2, Turaka Vijay Prakash
Turaka Vijay Prakash
1 Department of Medicine, Christian Medical College, Vellore, India
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1, Thambu D Sudarsanam
Thambu D Sudarsanam
5 Department of Medicine, Clinical Epidemiology and BV Moses Center, Christian Medical College, Vellore, India
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5
Editor: Cochrane Kidney and Transplant Group
Author information
Article notes
Copyright and License information
1 Department of Medicine, Christian Medical College, Vellore, India
2 BV Moses Center, Clinical Epidemiology Unit, Christian Medical College, Vellore, India
3 Department of Microbiology, Christian Medical College, Vellore, India
4 Department of Nephrology, Christian Medical College, Vellore, India
5 Department of Medicine, Clinical Epidemiology and BV Moses Center, Christian Medical College, Vellore, India
✉
Corresponding author.
Collection date 2022.
Copyright © 2022 The Cochrane Collaboration. Published by John Wiley & Sons, Ltd.
PMC Copyright notice
PMCID: PMC9005998 EMSID: EMS195448
Objectives
This is a protocol for a Cochrane Review (intervention). The objectives are as follows:
This review aims to look at the benefits and harms of antibiotics used in the treatment of acute pyelonephritis in adults. The aspects of treatment that will be evaluated are:
Different routes of administration;
Different durations of the same antibiotic; and
Comparison of different antibiotic agents in the same population.
Background
Description of the condition
Urinary tract infections (UTI) are considered an infection involving the urethra, urinary bladder, or the kidneys. Acute pyelonephritis denotes an infection of the renal pelvis and the kidney. It manifests with symptoms and signs of systemic inflammation and bladder inflammation. Those with anatomical abnormalities of the urinary tract, pregnant patients, patients with uncontrolled diabetes mellitus, kidney transplant recipients, acute kidney injury (AKI) or chronic kidney disease (CKD), immunocompromised patients, and those with hospital‐acquired bacterial infections are considered to have complicated pyelonephritis. All others are considered to have uncomplicated pyelonephritis. Among adults, the incidence is highest among young women followed by adults over 65 years of age (Czaja 2007). The diagnostic criteria for acute pyelonephritis remain controversial and most studies include patients with a range of symptoms including fever, dysuria, and flank pain, along with bacteriuria with 10⁵ colony‐forming units (CFU)/mL while a few have included CFU counts of 10⁴ in males (Piccoli 2006). Bacteriuria with colony counts ≥ 10² CFU/mL is considered appropriate for diagnosis among samples collected by suprapubic aspiration or catheterization (Wilson 2004).
The epidemiology of organisms causing pyelonephritis has been starting to show an exponential increase in resistant organisms. While extended‐spectrum beta‐lactam (ESBL) producing bacteria are increasing, carbapenem resistance has increased from 1.2% in 2001 to 4.2% in 2011 (Golan 2015).
Most patients who are treated appropriately recover with no long‐term renal consequences. Complications including obstruction, renal or perinephric abscess and emphysematous pyelonephritis are more common in patients with diabetes mellitus. Recurrent pyelonephritis and kidney failure are uncommon complications. The U.S. National Vital Statistics Reports for 2014 attributes 712 deaths to kidney infection (Johnson 2018).
Description of the intervention
Many classes of antibiotics have been used to treat acute pyelonephritis in adults. These include trimethoprim‐sulphamethoxazole,nitrofurantoin, quinolones,aminoglycosides,cephalosporins, extended penicillins (amoxicillin‐clavulanic acid, piperacillin‐tazobactam), carbapenems, and polymyxin group of antibiotics (Herness 2020). Some are oral, some oral and parenteral, while others are purely parenteral depending on how stable the patient is clinically as well as the antibiotic susceptibility.Parenteral antibiotics are required at least initially for severe pyelonephritis, underlying debilitating conditions, obstruction or when no oral formulations of sensitive antibiotics are available. The duration of treatment varies between five and 14 days, though it is often extended up to six weeks for kidney abscesses or emphysematous pyelonephritis.
How the intervention might work
Antibiotics work by killing the bacteria responsible for pyelonephritis. They enter the bloodstream and concentrate in the urinary tract to have an effect. Given early in the course of pyelonephritis, they can prevent progression to renal abscesses, sepsis, septic shock, and potentially death. The choice of drug depends on the antimicrobial susceptibility pattern. As this may take a day or two the choice initially is empirically based on local knowledge of likely organisms and antimicrobial susceptibility patterns.
Why it is important to do this review
Though antibiotics are the cornerstone for the treatment of acute pyelonephritis, the optimal drug, duration, and route of administration are still varied (Gupta 2011). A recent systematic review focused on long‐ versus short‐course antibiotics for pyelonephritis (Berti 2018). However, they did not focus on complicated pyelonephritis or on pyelonephritis due to drug‐resistant organisms.
Safely reducing the duration of antibiotics for acute pyelonephritis in adults would result in a reduced hospital stay, unnecessary extended use of antimicrobials, and potentially decrease antimicrobial resistance, adverse effects,and costs. This is especially important in the setting of emerging antimicrobial resistance across the globe.
Acute pyelonephritis can result in damage to the kidneys in the form of AKI and CKD if not treated appropriately. Hence it is essential to know the optimal antimicrobial therapy to prevent these long‐term debilitating sequelae.
For patients with severe pyelonephritis or with underlying debilitating conditions or obstruction, most guidelines recommend parenteral antibiotics, while those who are clinically stable are managed with oral antibiotics. However, with increasing antimicrobial resistance many countries do not have oral antibiotics that are sensitive.
The cost of treatment is potentially large as a major part of hospital bills are expensive antibiotics and bed charges. The current recommendations for acute pyelonephritis range from five to 14 days. If shorter‐duration antibiotics are as good as longer‐duration, the overall cost to the health system could be reduced.These have not been documented in earlier reviews.
Objectives
This review aims to look at the benefits and harms of antibiotics used in the treatment of acute pyelonephritis in adults. The aspects of treatment that will be evaluated are:
Different routes of administration;
Different durations of the same antibiotic; and
Comparison of different antibiotic agents in the same population.
Methods
Criteria for considering studies for this review
Types of studies
All randomised controlled trials (RCTs) and quasi‐RCTs (RCTs in which allocation to treatment was obtained by alternation, use of alternate medical records, date of birth or other predictable methods) looking at conditions in which antibiotics were used in the treatment of adults (older than 16 years) with acute pyelonephritis will be included. Where studies included both adults with acute pyelonephritis and those with cystitis, these will be included if data for participants with acute pyelonephritis could be extracted separately; otherwise, these studies will be excluded.
Types of participants
Inclusion criteria
Adults, 16 years or older with acute pyelonephritis who are treated either as inpatients or as outpatients with antibiotics.
The diagnosis of acute pyelonephritis require UTI (generally requiring a bacterial growth on urine culture > 10⁵ CFU/mL with at least one symptom or sign of systemic illness such as fever, loin pain or toxicity. Acute pyelonephritis defined by any alternative diagnostic criteria as defined by the authors will also be included. Lower counts (1000 to 9999 CFU/mL) especially in men and pregnant women will be included.
Those with diagnosed urinary tract abnormalities including mechanical obstruction such as stones, enlarged prostate, pelvic organ prolapsed or non‐mechanical such as neurological involvement affecting the bladder will be included.
We will also include those with a history of previous UTIs, pregnant women, kidney transplant recipients and those who are catheterised.
Exclusion criteria
Patients considered to have asymptomatic bacteriuria or cystitis (UTI as defined in inclusions with no symptom or sign of systemic illness) will be excluded.
Types of interventions
Different durations of the same antibiotic
Comparison of different antibiotic agents in the same population
Different routes of administration
Types of outcome measures
Primary outcomes
Efficacy: recurrence of UTI, readmission for UTI treatment, duration of symptoms
Adverse effects of treatment: minor (e.g. vomiting, discomfort from IV cannula) and major (e.g. anaphylaxis, hearing impairment)
Costs
Length of hospital stay
Secondary outcomes
Long‐term outcomes: CKD, kidney damage identified by imaging; kidney failure
Death
Search methods for identification of studies
Electronic searches
We will search the Cochrane Kidney and Transplant Register of Studies through contact with the Information Specialist using search terms relevant to this review. The Register contains studies identified from the following sources.
Monthly searches of the Cochrane Central Register of Controlled Trials (CENTRAL)
Weekly searches of MEDLINE OVID SP
Searches of kidney and transplant journals, and the proceedings and abstracts from major kidney and transplant conferences
Searching of the current year of EMBASE OVID SP
Weekly current awareness alerts for selected kidney and transplant journals
Searches of the International Clinical Trials Register (ICTRP) Search Portal and ClinicalTrials.gov.
Studies contained in the Register are identified through searches of CENTRAL, MEDLINE, and EMBASE based on the scope of Cochrane Kidney and Transplant. Details of search strategies, as well as a list of hand‐searched journals, conference proceedings and current awareness alerts, are available on the Cochrane Kidney and Transplant website.
See Appendix 1 for search terms used in strategies for this review.
Searching other resources
Reference lists of review articles, relevant studies and clinical practice guidelines.
Contacting relevant individuals/organisations seeking information about unpublished or incomplete studies.
Grey literature sources (e.g. abstracts, dissertations and theses), in addition to those already included in the Cochrane Kidney and Transplant Register of Studies, will be searched.
Data collection and analysis
Selection of studies
The search strategy described inAppendix 1will be used to obtain titles and abstracts of studies that may be relevant to the review. The titles and abstracts will be screened independently by two authors (VKC, TG) who will discard studies that are not applicable, however, studies and reviews that might include relevant data or information on trials will be retained initially. The two authors will independently assess retrieved abstracts and, if necessary the full text, of these studies to determine which studies satisfy the inclusion criteria. Disagreements will be resolved in consultation with a third author (SA).
Data extraction and management
Data extraction will be carried out independently by two authors (VKC, TG)using standard data extraction forms. Disagreements will be resolved in consultation with a third author (RDS). Studies reported in non‐English language journals will be translated before assessment. Where more than one publication of one study exists, reports will be grouped together and the publication with the most complete data will be used in the analyses. Where relevant outcomes are only published in earlier versions these data will be used. Any discrepancy between published versions will be highlighted.
Assessment of risk of bias in included studies
The following items will be independently assessed by two authors using the risk of bias assessment tool (Higgins 2020) (seeAppendix 2).
Was there adequate sequence generation (selection bias)?
Was allocation adequately concealed (selection bias)?
Was knowledge of the allocated interventions adequately prevented during the study?
Participants and personnel (performance bias)
Outcome assessors (detection bias)
Were incomplete outcome data adequately addressed (attrition bias)?
Are reports of the study free of suggestion of selective outcome reporting (reporting bias)?
Was the study apparently free of other problems that could put it at risk of bias?
Measures of treatment effect
For dichotomous outcomes (persistent bacteriuria, recurrent UTI, kidney damage, readmission, death) results will be expressed as risk ratios (RR) with 95% confidence intervals (CI). Where continuous scales of measurement are used to assess the effects of treatment (duration of symptoms, length of hospital stay, length of hospital stay, duration of inotropic agents, economic costs)the mean difference (MD) will be used, or the standardised mean difference (SMD) if different scales have been used.
Unit of analysis issues
We don't anticipate to find any cluster or cross‐over studies. Therefore, the unit of analysis will be individuals who are assigned to intervention arms.
Dealing with missing data
Any further information required from the original author will be requested by written correspondence (e.g. emailing the corresponding author) and any relevant information obtained in this manner will be included in the review. Evaluation of important numerical data such as screened, randomised patients, as well as intention‐to‐treat, as‐treated and per‐protocol population, will be carefully performed. Attrition rates (e.g. drop‐outs, losses to follow‐up and withdrawals) will be investigated. Issues of missing data and imputation methods (e.g. last‐observation‐carried‐forward) will be critically appraised (Higgins 2020).
Assessment of heterogeneity
We will first assess the heterogeneity by visual inspection of the forest plot. We will quantify statistical heterogeneity using the I² statistic, which describes the percentage of total variation across studies that is due to heterogeneity rather than sampling error (Higgins 2003). A guide to the interpretation of I² values will be as follows.
0% to 40%: might not be important
30% to 60%: may represent moderate heterogeneity
50% to 90%: may represent substantial heterogeneity
75% to 100%: considerable heterogeneity.
The importance of the observed value of I² depends on the magnitude and direction of treatment effects and the strength of evidence for heterogeneity (e.g. P‐value from the Chi² test, or a CI for I²) (Higgins 2020).
Assessment of reporting biases
If possible, funnel plots will be used to assess for the potential existence of small study bias (Higgins 2020).
Data synthesis
Data will be pooled using the random‐effects model but the fixed‐effect model will also be used to ensure the robustness of the model chosen and susceptibility to outliers. For dichotomous data Mantel‐Haenszel method of pooling will be done whereas for continuous data Inverse‐variance approach of pooling will be performed. Data from cluster and cross‐over design are anticipated to be not available. Nevertheless, if we have identified these studies we will combine the data in a meta‐analysis using the generic variance approach.
Subgroup analysis and investigation of heterogeneity
Heterogeneity among participants could be related to gender and the organism characteristics(e.g. pyelonephritis in males are likely to be more complicated than in females; catheter‐associated pyelonephritis is likely to be caused by resistant pathogens requiring a longer duration of treatment and high risk of recurrence; bacteraemic infections likely require prolonged antibiotic course; ESBL organisms are considered to require a longer duration of treatment with limited oral antibiotic options). Heterogeneity in treatments could be related to a prior agent used, and the agent's dose and duration of therapy (e.g. higher dose of antibiotics like fluoroquinolones and aminoglycoside are more likely to require a shorter duration of treatment; longer duration of treatment possibly results in lower risk of recurrence). The following subgroup analysis will be used to explore possible sources of heterogeneity.
Bacteraemic versus non‐bacteraemic
Males versus females
Catheter‐associated UTI versus non‐catheter‐associated UTI
ESBL versus non‐ESBL organisms
Hospital administered versus home administered antibiotics
Studies with a low risk of bias
Published versus unpublished studies
The studies will be categorized into small or large studies:
Number of study participants (< 50, 50 to 100, 100 to 500, > 500);
Classifying the studies into four quarters based on the number of participants;
Power of the study (power less than and greater than 50).
Adverse effects will be tabulated and assessed with descriptive techniques, as they are likely to be different for the various agents used. Where possible, the risk difference (RD) with 95% CI will be calculated for each adverse effect, either compared to no treatment or to another agent.
Sensitivity analysis
We will perform sensitivity analyses in order to explore the influence of the missing data and imputed data on effect size. We will also perform sensitivity analysis by:
Repeating the analysis excluding unpublished studies
Repeating the analysis excluding any very large studies to establish how much they dominate the results
Repeating the analysis excluding studies using the following filters: diagnostic criteria, language of publication, source of funding (industry versus other), and country.
Summary of findings and assessment of the certainty of the evidence
We will present the main results of the review in 'Summary of findings' tables. These tables present key information concerning the certainty of the evidence, the magnitude of the effects of the interventions examined, and the sum of the available data for the main outcomes (Schunemann 2020a). The 'Summary of findings' tables also includes an overall grading of the evidence related to each of the main outcomes using the GRADE (Grades of Recommendation, Assessment, Development and Evaluation) approach (GRADE 2008;GRADE 2011). The GRADE approach defines the certainty of a body of evidence as to the extent to which one can be confident that an estimate of effect or association is close to the true quantity of specific interest. This will be assessed by two authors. The certainty of a body of evidence involves consideration of the within‐trial risk of bias (methodological quality), directness of evidence, heterogeneity, the precision of effect estimates and risk of publication bias (Schunemann 2020b). We plan to present the following outcomes in the 'Summary of findings' tables.
Recurrence of UTI
Duration of symptoms
Length of hospital stay
Cost of treatment
Readmission for UTI
Adverse effects of treatment including minor and major (anaphylaxis, thrombophlebitis, hearing loss, kidney failure)
Death
Acknowledgements
The Methods section of this protocol is based on a standard template used by Cochrane Kidney and Transplant.
The authors are grateful to the following peer reviewers for their time and comments: Bryan N Becker, MD (JPS Health Network) and Neil Boudville (Medical School, University of Western Australia).
Appendices
Appendix 1. Electronic search strategies
DatabaseSearch terms
CENTRAL 1. pyelonephritis:ti,ab,kw
((urinary next tract next infection) near/2 (upper or febrile or fever or complicated or severe)):ti,ab,kw
((upper next urinary next tract) near/2 infection):ti,ab,kw
(acute next lobar next nephronia):ti,ab,kw
{or #1‐#4} in Trials
MEDLINE 1. Pyelonephritis/
pyelonephritis.tw.
(urinary tract infection adj2 (upper or febrile or fever or complicated or severe)).tw.
(upper urinary tract adj2 infection).tw.
acute lobar nephronia.tw.
or/1‐5
EMBASE 1. Pyelonephritis/
Acute Pyelonephritis/
pyelonephritis.tw.
(urinary tract infection adj2 (upper or febrile or fever or complicated or severe)).tw.
(upper urinary tract adj2 infection).tw.
acute lobar nephronia.tw.
or/1‐6
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Appendix 2. Risk of bias assessment tool
Potential source of biasAssessment criteria
Random sequence generation
Selection bias (biased allocation to interventions) due to inadequate generation of a randomised sequence Low risk of bias: Random number table; computer random number generator; coin tossing; shuffling cards or envelopes; throwing dice; drawing of lots; minimisation (minimisation may be implemented without a random element, and this is considered to be equivalent to being random).
High risk of bias: Sequence generated by odd or even date of birth; date (or day) of admission; sequence generated by hospital or clinic record number; allocation by judgement of the clinician; by preference of the participant; based on the results of a laboratory test or a series of tests; by availability of the intervention.
Unclear: Insufficient information about the sequence generation process to permit judgement.
Allocation concealment
Selection bias (biased allocation to interventions) due to inadequate concealment of allocations prior to assignment Low risk of bias: Randomisation method described that would not allow investigator/participant to know or influence intervention group before eligible participant entered in the study (e.g. central allocation, including telephone, web‐based, and pharmacy‐controlled, randomisation; sequentially numbered drug containers of identical appearance; sequentially numbered, opaque, sealed envelopes).
High risk of bias: Using an open random allocation schedule (e.g. a list of random numbers); assignment envelopes were used without appropriate safeguards (e.g. if envelopes were unsealed or non‐opaque or not sequentially numbered); alternation or rotation; date of birth; case record number; any other explicitly unconcealed procedure.
Unclear: Randomisation stated but no information on method used is available.
Blinding of participants and personnel
Performance bias due to knowledge of the allocated interventions by participants and personnel during the study Low risk of bias: No blinding or incomplete blinding, but the review authors judge that the outcome is not likely to be influenced by lack of blinding; blinding of participants and key study personnel ensured, and unlikely that the blinding could have been broken.
High risk of bias: No blinding or incomplete blinding, and the outcome is likely to be influenced by lack of blinding; blinding of key study participants and personnel attempted, but likely that the blinding could have been broken, and the outcome is likely to be influenced by lack of blinding.
Unclear: Insufficient information to permit judgement
Blinding of outcome assessment
Detection bias due to knowledge of the allocated interventions by outcome assessors.Low risk of bias: No blinding of outcome assessment, but the review authors judge that the outcome measurement is not likely to be influenced by lack of blinding; blinding of outcome assessment ensured, and unlikely that the blinding could have been broken.
High risk of bias: No blinding of outcome assessment, and the outcome measurement is likely to be influenced by lack of blinding; blinding of outcome assessment, but likely that the blinding could have been broken, and the outcome measurement is likely to be influenced by lack of blinding.
Unclear: Insufficient information to permit judgement
Incomplete outcome data
Attrition bias due to amount, nature or handling of incomplete outcome data.Low risk of bias: No missing outcome data; reasons for missing outcome data unlikely to be related to true outcome (for survival data, censoring unlikely to be introducing bias); missing outcome data balanced in numbers across intervention groups, with similar reasons for missing data across groups; for dichotomous outcome data, the proportion of missing outcomes compared with observed event risk not enough to have a clinically relevant impact on the intervention effect estimate; for continuous outcome data, plausible effect size (difference in means or standardised difference in means) among missing outcomes not enough to have a clinically relevant impact on observed effect size; missing data have been imputed using appropriate methods.
High risk of bias: Reason for missing outcome data likely to be related to true outcome, with either imbalance in numbers or reasons for missing data across intervention groups; for dichotomous outcome data, the proportion of missing outcomes compared with observed event risk enough to induce clinically relevant bias in intervention effect estimate; for continuous outcome data, plausible effect size (difference in means or standardized difference in means) among missing outcomes enough to induce clinically relevant bias in observed effect size; ‘as‐treated’ analysis done with substantial departure of the intervention received from that assigned at randomisation; potentially inappropriate application of simple imputation.
Unclear: Insufficient information to permit judgement
Selective reporting
Reporting bias due to selective outcome reporting Low risk of bias: The study protocol is available and all of the study’s pre‐specified (primary and secondary) outcomes that are of interest in the review have been reported in the pre‐specified way; the study protocol is not available but it is clear that the published reports include all expected outcomes, including those that were pre‐specified (convincing text of this nature may be uncommon).
High risk of bias: Not all of the study’s pre‐specified primary outcomes have been reported; one or more primary outcomes is reported using measurements, analysis methods or subsets of the data (e.g. sub‐scales) that were not pre‐specified; one or more reported primary outcomes were not pre‐specified (unless clear justification for their reporting is provided, such as an unexpected adverse effect); one or more outcomes of interest in the review are reported incompletely so that they cannot be entered in a meta‐analysis; the study report fails to include results for a key outcome that would be expected to have been reported for such a study.
Unclear: Insufficient information to permit judgement
Other bias
Bias due to problems not covered elsewhere in the table Low risk of bias: The study appears to be free of other sources of bias.
High risk of bias: Had a potential source of bias related to the specific study design used; stopped early due to some data‐dependent process (including a formal‐stopping rule); had extreme baseline imbalance; has been claimed to have been fraudulent; had some other problem.
Unclear: Insufficient information to assess whether an important risk of bias exists; insufficient rationale or evidence that an identified problem will introduce bias.
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Contributions of authors
Draft the protocol: RDS, SA, VKC, TG, TDS, PT
Study selection: JPBE, VKC, TG, VPT, SA, TDS, PT. VKC and TG will screen titles and abstracts independently and SA will resolve any disagreements
Extract data from studies: VKC, SA, TG, RDS, VPT, TDS, PT. VKC and TG will extract data from studies independently and RDS will resolve any disagreements
Enter data into RevMan: RK, RDS, TG, VPT, VKC
Carry out the analysis: RK, VKC, PT
Interpret the analysis: RK, VKC, TDS, PT
Draft the final review: RDS, VKC, SA
Disagreement resolution: TDS, PT
Update the review: VKC, TDS
Sources of support
Internal sources
Christian Medical College, Vellore, India
The authors RDS, SA, VKC,TG, TDS, TVP, RK and JPBE are employed in this institution
External sources
No sources of support provided
Declarations of interest
Vignesh Kumar Chandiraseharan, Vijay Prakash Turaka, Rani Diana Sahni, and Thambu David Sudarsanam are co‐investigators of an RCT on seven versus 14‐day antibiotic treatment for acute pyelonephritis awaiting publication.
New
References
Additional references
Berti 2018
Berti F, Attardo TM, Piras S, Tesei L, Tirotta D, Tonani M, et al.Short versus long course antibiotic therapy for acute pyelonephritis in adults: a systematic review and meta-analysis. Italalian Journal of Medicine 2018;12(1):39-50. [EMBASE: 621798755] [Google Scholar]
Czaja 2007
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Guyatt G, Oxman AD, Akl EA, Kunz R, Vist G, Brozek J, et al.GRADE guidelines: 1. Introduction-GRADE evidence profiles and summary of findings tables. Journal of Clinical Epidemiology 2011;64(4):383-94. [MEDLINE: ] [DOI] [PubMed] [Google Scholar]
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Gupta K, Hooton TM, Naber KG, Wullt B, Colgan R, Miller LG, et al.International clinical practice guidelines for the treatment of acute uncomplicated cystitis and pyelonephritis in women: a 2010 update by the Infectious Diseases Society of America and the European Society for Microbiology and Infectious Diseases. Clinical Infectious Diseases 2011;52(5):e103-20. [MEDLINE: ] [DOI] [PubMed] [Google Scholar]
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Johnson 2018
Johnson JR, Russo TA.Acute pyelonephritis in adults [Erratum in: N Engl J Med. 2018 Mar 15;378(11):1069; PMID: 29539281]. New England Journal of Medicine 2018;378(1):48-59. [MEDLINE: ] [DOI] [PubMed] [Google Scholar]
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Schünemann HJ, Higgins JP, Vist GE, Glasziou P, Akl EA, Skoetz N, et al.Chapter 14: Completing ‘Summary of findings’ tables and grading the certainty of the evidence. In: Higgins JPT, Thomas J, Chandler J, Cumpston M, Li T, Page MJ, Welch VA (editors). Cochrane Handbook for Systematic Reviews of Interventions version 6.1 (updated September 2020). Cochrane, 2020. Available from www.training.cochrane.org/handbook.
Schunemann 2020b
Schünemann HJ, Vist GE, Higgins JP, Santesso N, Deeks JJ, Glasziou P, et al.Chapter 15: Interpreting results and drawing conclusions. In: Higgins JPT, Thomas J, Chandler J, Cumpston M, Li T, Page MJ, Welch VA (editors). Cochrane Handbook for Systematic Reviews of Interventions version 6.1 (updated September 2020). Cochrane, 2020. Available from www.training.cochrane.org/handbook.
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Home›News›The Essential Role of Sulfuric Acid in Fertilizer Production
The Essential Role of Sulfuric Acid in Fertilizer Production
November 22, 2024
Sulfuric acid, often referred to as the “king of chemicals,” plays an indispensable role in modern agriculture. As one of the most widely produced industrial chemicals, its applications extend across various industries, but it holds particular importance in fertilizer production. This blog delves into how sulfuric acid is used to manufacture phosphoric acid and fertilizers like ammonium phosphate, highlighting its essential role in supporting global food production.
Sulfuric Acid and Fertilizer Production: An Overview
The production of fertilizers relies heavily on the availability of raw materials like sulfuric acid. This vital chemical acts as a cornerstone in producing phosphoric acid, which, in turn, is a primary ingredient in many fertilizers. Without sulfuric acid, the large-scale manufacture of essential fertilizers would not be possible, significantly impacting agricultural productivity worldwide.
The Manufacturing Process: From Sulfuric Acid to Phosphoric Acid
The connection between sulfuric acid and fertilizer production begins with the synthesis of phosphoric acid. Phosphoric acid is manufactured by reacting phosphate rock with sulfuric acid, a process known as the wet process.
Key Steps in the Process
ReactionPhosphate rock, which contains calcium phosphate, reacts with sulfuric acid to produce phosphoric acid and gypsum (calcium sulfate) as a by-product.
FiltrationThe gypsum is separated from the phosphoric acid through filtration. The resulting phosphoric acid solution is a precursor to various fertilizers.
PurificationDepending on its end use, phosphoric acid may undergo further purification to meet agricultural standards.
Why Sulfuric Acid is Essential
Sulfuric acid’s reactivity with phosphate rock is crucial for releasing phosphorus in a form that plants can readily absorb. Without sulfuric acid, the conversion of raw phosphate rock into usable agricultural fertilizers would be inefficient and cost-prohibitive.
Fertilizers Produced Using Sulfuric Acid
Sulfuric acid indirectly contributes to the production of several fertilizers that are vital for modern farming.
1. Ammonium Phosphate Fertilizers
Ammonium phosphate fertilizers, including monoammonium phosphate (MAP) and diammonium phosphate (DAP), are among the most common fertilizers produced using sulfuric acid.
MAP: A balanced fertilizer suitable for crops that require nitrogen and phosphorus in equal measures.
DAP: A high-phosphorus fertilizer that provides both nitrogen and phosphorus to crops.
Both MAP and DAP are manufactured by reacting phosphoric acid (derived from sulfuric acid) with ammonia.
2. Superphosphates
Single Superphosphate (SSP): Produced by directly reacting phosphate rock with sulfuric acid.
Triple Superphosphate (TSP): Made by treating phosphate rock with phosphoric acid.
Superphosphates are widely used to enrich soil with phosphorus, improving crop yields.
3. Compound Fertilizers
Sulfuric acid also supports the production of compound fertilizers, which combine multiple nutrients like nitrogen, phosphorus, and potassium in a single product.
Benefits of Sulfuric Acid in Fertilizer Production
Efficient Nutrient ReleaseSulfuric acid ensures that phosphorus, a critical nutrient for plant growth, is released from phosphate rock in a form that plants can absorb.
Cost-Effective ProductionThe use of sulfuric acid in the wet process is a cost-efficient way to manufacture fertilizers, ensuring affordability for farmers.
High ScalabilitySulfuric acid enables the large-scale production of fertilizers, meeting the demands of global agriculture and supporting food security.
SustainabilityBy utilizing phosphate rock, an abundant natural resource, sulfuric acid helps in creating a sustainable supply chain for fertilizers.
Future of Sulfuric Acid in Fertilizer Production
As global populations grow and the demand for food rises, the importance of sulfuric acid in fertilizer production will only increase. Innovations in chemical engineering and sustainability are likely to make the production processes more efficient and environmentally friendly.
Partnering with Rock Chemicals Inc. for Your Sulfuric Acid Needs
Sulfuric acid is the backbone of fertilizer production, driving agricultural productivity and ensuring food security for billions. For businesses involved in manufacturing fertilizers, ensuring a reliable supply of high-quality sulfuric acid is critical to maintaining efficiency and meeting demand.
Rock Chemicals Inc. is a trusted supplier of bulk sulfuric acid, offering industry-grade quality and reliable delivery. Whether you’re manufacturing phosphoric acid or producing ammonium phosphate fertilizers, Rock Chemicals Inc. is your go-to partner. Contact Rock Chemicals Inc. today to learn more about bulk orders and how their expertise can optimize your operations.
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9355 | https://www.nature.com/articles/pr1987452 | PROSPECTIVE MSAFP SCREENING FOR DOWN SYNDROME: THE BAYLOR EXPERIENCE | Pediatric Research
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Abstract
Published: April 1987
PROSPECTIVE MSAFP SCREENING FOR DOWN SYNDROME: THE BAYLOR EXPERIENCE
Frank Greenberg1,
Judy Garza1,
Barbara Weyland1,
Elliot Alpert1&
…
Esmie Rose1
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Pediatric Researchvolume 21,page 227 (1987)Cite this article
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Abstract
After the initial reports of the association of fetal trisomy 21 with low maternal serum alpha fetoprotein (MSAFP) levels in retrospective studies in 1984, we began using low MSAFP levels as a means of screening for an increased risk of Down syndrome using reagents from Clinical Assays. In 1985, a total of 4929 women were screened, of whom 312 (6.3%) had an initial low MSAFP levels. Of these women, 217 (70%) had levels below the level of reliability of the assay and all but 33 had normal levels on repeat 1-2 weeks later. None of these women had a fetus or infant with trisomy 21 by amniocentesis or at birth. Ninety-five women (1.9%) had levels below half the median. Of these women, 12 had persistently low levels on repeat. A total of 21 amniocenteses were done in this group. Three fetuses with trisomy 21 and one fetus with trisomy 18 were detected, a rate of 7.1% fetal trisomy among those women who had amniocentesis for low MSAFP levels. No cases of trisomy 21 were missed. Thus far in 1986, over 10,000 women have been screened. The rate of fetal trisomy in amniotic fluid is about 3%. One infant with trisomy 21 is known to have been missed.
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Baylor College of Medicine, Institute for Molecular Genetics and Departme of Medicine, Houston, Texas
Frank Greenberg,Judy Garza,Barbara Weyland,Elliot Alpert&Esmie Rose
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Greenberg, F., Garza, J., Weyland, B. et al. PROSPECTIVE MSAFP SCREENING FOR DOWN SYNDROME: THE BAYLOR EXPERIENCE. Pediatr Res21 (Suppl 4), 227 (1987).
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9356 | https://www.cnblogs.com/iMath/p/7597833.html | 有理数的阿基米德性质及其应用 - iMath - 博客园
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有理数的阿基米德性质及其应用
有理数的阿基米德性质
任何有理数 r=p q≤|p|（这里 p 和 q 都是整数并且 q≠0），因为 r=p q≤|p||q|≤|p|1=|p|，可知对于任何有理数 r，总存在比它大的正整数 n，即 n>r ，比如这里可取 n=|p|+1，这就是有理数的阿基米德性质（Archimedean Property for rational numbers）。如果 r 是任意正有理数，那么 1 r 也是任意正有理数，对前面这个不等式两边取倒数有 1 n<1 r ， 可知对于任何正有理数，总存在正整数 n 使得 1 n 小于它 ，综合这两条性质来看——即没有最大的正有理数也没有最小的正有理数。后面等我们学习到实数的阿基米德性质后，同样会明白没有最大的正实数也没有最小的正实数，进而所谓的“无穷大数”和“无穷小数”也就不存在实数系里了。
或 或 A={x∈Q|x 2<2 或 x<0}，A内有最大的有理数吗？
你也许会想到：如果 a 是A内最大的有理数，那么必有一个有理数 a′满足 a<a′<2（根据“任何两个不同实数间必然存在有理数”可得），故而A内没有最大的有理数。但是，这种方法依赖于实数或无理数的存在，假设我们仅仅只知道有理数，那么还能回答这个问题吗？能！用有理数的阿基米德性质就能解决这个问题，该问的解决对于我们后面以有理数为基础通过Dedekind Cut来构建数的连续体至关重要。
假设 a 是A内最大的有理数，只要选定足够大的正整数 n 就可以让 a+1 n 变得比 a 稍大一点点，那么我们很自然就会想：是不是存在正整数 n 使得(a+1 n)2<2 呢？若存在，那么我们便说明了A内有比 a 更大的有理数 a+1 n，从而说明A内无最大的有理数，下面是证明过程。
证明：假设A内有最大的有理数 a，那么 a 必然是正有理数且 a 2<2 。如果证明存在正整数 n 使得(a+1 n)2<2，便可得出A内有比 a 更大的有理数 a+1 n，从而说明A内无最大的有理数。
(a+1 n)2=a 2+2 a n+1 n 2<a 2+2 a n+1 n=a 2+1 n(2 a+1)，如果能证明存在正整数 n 使得 a 2+1 n(2 a+1)<2，那么(a+1 n)2<2 自然得证。对 a 2+1 n(2 a+1)<2 稍作变形可得 1 n<2−a 2 2 a+1 ，现在问题变成了是否存在正整数 n 使得 1 n<2−a 2 2 a+1 ，因为 a 是正有理数且 a 2<2，所以 2−a 2 2 a+1 是正有理数，由“对于任何正有理数，总存在正整数 n 使得 1 n 小于它”知存在这样的正整数 n，也就存在正整数 n 使得(a+1 n)2<2，所以A内无最大的有理数。
用类似的方法也可以证明 且 且{x∈Q|x 2>2 且 x>0}内无最小的有理数。下一节我将以这两个问题为例介绍数的连续体的构建，请继续关注！
Charles Chapman Pugh, Real Mathematical Analysis, 1st Edition, P20 ↩︎
James S. Howland, Basic Real Analysis, 1st Edition, P15 ↩︎
证明看这里↩︎
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9357 | https://www.youtube.com/watch?v=7JD0bK_6lEE | What is a DIFFERENCE in math?
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The word DIFFERENCE has one meaning in regular English and another meaning in math: Subtraction. Teach precise mathematical terms to give students a learning advantage!
Transcript:
Arithmetic: What does "difference" mean? Here's two pairs of numbers. 7 and 6. And 9 and 2. Which two numbers are more different from each other? 7 and 6, or 9 and 2? If you asked a hundred people that question, most of them would say 9 and 2. Why? Because numbers like 7, 6, 9, 2 really mean amounts. The amount 7 is almost the same as the amount 6. These amounts, 9 and 2, are more different. What if you wanted to think about these differences more precisely? What would you do? One thing you could do is put 7 objects in a row and 6 objects in another row. The difference between these two rows is right at the end, in red. The row of 7 has 1 more than the row of 6. We could even say: The difference between these two numbers is 1. We can do the same thing for 9 and 2. Here's 9. Here's 2. The difference between these two numbers is the part in red. The row of 9 has 7 more than the row of 2, so we could say the difference between these two numbers is 7. And that's what the word "difference" means in math! When people ask "What is the difference between 9 and 2?" they really mean "By how much are these two numbers different?" And that's why you often hear the word being used with subtraction. 7 minus 6 is 1. 9 minus 2 is 7. Any time you want to find out how different two amounts are, you can find that by subtracting. And that's why we use the word "difference" in subtraction! |
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9359 | https://documents.thermofisher.com/TFS-Assets/LSG/Application-Notes/AN-PHCONSAMP-E%201014-RevA-WEB.pdf | Measuring pH of Concentrated Samples Water Analysis Instruments, Thermo Fisher Scientific Application Note 009 Key Words pH, concentrated samples, seawater, brines, food, acids, bases, high ionic strength.
Goal The following application note describes the challenges, best practices and recommended instrumentation when measuring pH in high ionic strength, or “concentrated” samples. Introduction In pH measurement, we sometimes measure samples with low ionic strength (such as pure water or betaine solution), or high ionic strength (such as seawater, food, or other concentrated samples). These sample types can be challenging, but when you know the best practices, you will be successful in your measurements.
Ionic Strength Ionic strength is a function of the concentration and charge of all the ions in a solution. The precision and bias of pH measurements in samples with low or high ionic strength may be affected, unless proper procedures are observed.
Ionic strength is an important factor in many processes. It is essential in the function of all living organism, in environmental and biochemical reactions, etc. As noted, ionic strength plays an important role in a pH measurement.
Low ionic strength samples, such as pure and surface waters, sugar or betaine solutions, present their own unique challenges in pH measurement (See Application Note 006 on “Measuring pH in Low Ionic Strength Solutions”).
High ionic strength samples, such as seawater, brines, strong acids, strong bases, foods, and beverages also present challenges. We are faced here with two challenges: 1. Change in hydrogen ion activity - ion mobility decreases in the high ionic strength samples and the activity differs from the concentration. (Note: the pH electrode responds to hydrogen ion activity, not the concentration). 2. High ionic strength solutions change the liquid junction potential. This may lead to bias and considerable time may be required to establish a stable reading.
2 Determine the Ionic Strength of Your Sample Theoretically, the ionic strength of a sample can be calculated by multiplying the concentration of each ion by the corresponding squared charge on the ion, then summing and dividing by two. But that’s not much fun and not easy to do.
From a practical point of view, there are other indicators that can provide a rough estimate of a sample’s ionic strength. These other indicators include conductivity and concentration. There are no exact conductivity and concentration value rules for low and high ionic strength, but we can make general categories (high, routine and low) and chose the best approach for pH analysis of extreme ionic strength samples (high or low). High Ionic Strength Examples • Sea Water; 0.8M (53 mS/cm) • 0.5M KCl (59 mS/cm) • 1 % NaOH; 0.25M (53 mS/cm) • 10 % brine NaCl (140 mS/cm) • 20 % nitric acid (763 mS/cm) • 20 % H3PO4 (123 mS/cm) Routine Ionic Strength Examples • Industrial Wastewater (5 mS/cm) • 0.05M pH 7 buffer (6.2 mS/cm) • 0.05M pH 4 buffer (4.4 mS/cm) • 0.05M pH 10 buffer (6.3 mS/cm) • 0.05M KCl (6.7 mS/cm) • 5400 ppm Total Dissolved Solids (10 mS/cm) Low Ionic Strength Examples • Pure water in air (1 µs/cm) • Rain water (50 µs/cm) • Tap water (500 µs/cm) • 0.05mM H2SO4 (38 µs/cm) • 0.67mM KCl (100 µs/cm) • 50 ppm Total Dissolved Solids (~100 µs/cm) The chart to the left shows common examples of samples with various levels of ionic strength. Examples range from high to mid-range to low ionic strength values.
3 Tips for testing concentrated samples with high ionic strength The pH measurement challenges in low or high ionic strength samples can be overcome by using the appropriate testing procedures.
The following techniques can be recommended for optimizing pH measurement in high ionic strength samples: • Use a fast-flowing, low resistance junction, such as sleeve junction or capillary junction.
• Use a strong salt filling solution (Note: the 3M or 4M KCl fillings solutions that are used in Orion pH electrodes are close to saturation, so this is about as strong as you can go).
• If additional improvements are desired, use a double-junction electrode which can protect the reference from salt intrusion and allow modification of the fill solution to better match the sample.
• The Orion pH electrode filling solution (Cat. No. 810007) of the ROSS 8172BNWP or 800300 reference half cell may be modified to match the sample composition.
For example, for samples with a pH less than 2 or greater than 12, adding a slight amount of the acid or base to the filling solution to adjust the pH and make it more compatible with the sample should decrease electrode stabilization time.
• For samples with a high salt content, use a strong filling solution using the same salt as the sample.
For example, when measuring the pH of sodium bromide brines, use sodium bromide as the filling solution.
• Allow sufficient time for the electrode to respond, since salty samples tend to drift as equilibrium is established at the junction. • Use a high salt buffer (for example, TRIS buffer in synthetic seawater). For preparation instructions, please read “Determination of the pH of sea water using a glass/reference electrode cell” prepared by the US Department of Energy’s Carbon Dioxide Information Analysis Center (CDIAC) - • For the best results, keep the calibration standards and sample temperatures within 2 °C of each other. • Use an automatic temperature compensation (ATC) probe or a triode pH electrode with built-in ATC to monitor temperature.
We recommend using the following pH Electrodes: • Thermo Scientific™ Orion™ ROSS™ Sure-Flow™ pH Electrode 8172BNWP • Thermo Scientific™ Orion™ ROSS Ultra™ Low Maintenance pH/ATC Triode™ Combination Electrode 8107BNUMD • Thermo Scientific™ Orion™ ROSS™ Half-Cell Electrode 8101BNWP and 800300 Sure-Flow reference half cell • Thermo Scientific Orion Sure-Flow pH Electrode 9172BNWP Application Note 009 AN-PHCONSAMP-E 1014 RevA © 2014 Thermo Fisher Scientific Inc. All rights reserved. All trademarks are the property of Thermo Fisher Scientific and its subsidiaries.
North America Toll Free: 1-800-225-1480 Tel: 1-978-232-6000 info.water@thermo.com thermoscientific.com/water Netherlands Tel: (31) 020-4936270 info.water.uk@thermo.com China Tel: (86) 21-68654588 wai.asia@thermofisher.com India Tel: (91) 22-4157-8800 wai.asia@thermofisher.com Singapore Tel: (65) 6778-6876 wai.asia@thermofisher.com Japan Tel: (81) 045-453-9175 wai.asia@thermofisher.com Australia Tel: (613) 9757-4300 in Australia (1300) 735-295 InfoWaterAU@thermofisher.com Water Analysis Instruments Conclusion Although measuring pH in concentrated samples can be challenging, understanding and employing best practices can ensure that you can be confident in the accuracy of your measurements.
To purchase Thermo Scientific Orion pH meters, electrodes and other related products, please contact your local equipment distributor and reference the part numbers listed below.
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9360 | https://www.adventureugandasafaris.com/african-forest-elephants-vs-savanna-elephants/ | Skip to content
African forest elephants Vs Savanna elephants
African forest elephants Vs Savanna elephants – Facts
African forest elephants Vs Savanna elephants: Are forest elephants different from bush elephants? What is the difference between bush or savanna and forest elephants? There are only 2 (two) species of elephants- the African and Asian elephants.
In Africa, the most sought for elephant species include the African bush/savanna elephants and African forest elephants. While there is a close similarity between the two African elephant species, they also differ in many aspects.
African forest elephants Vs Savanna elephants
African forest elephants
Size: The African forest elephants are smaller in size than the African bush or savanna elephants. The forest elephants can stand at 8-10ft tall and they live mainly in the tropical rainforests of West and Central Africa. Amazingly, while on Uganda safaris, it is possible to spot the forest elephants in different parks such as Kibale National Park, Bwindi Impenetrable National Park, Budongo Forest in Murchison Falls National Park, etc.
Being forest dwellers, the African forest elephants feed on tree leaves and fruits. They are distinct from bush elephants due to their oval-shaped ears and straight tusks. They also live in smaller herds than savanna or bush elephants.
African savanna elephants
The African bush elephants are often spotted while on game viewing in most of the savanna parks in Africa. They are bigger in size and stand 13ft in height. They depend on vegetation that grows under hot and dry climatic conditions.
The bush elephants also feed on tall grass and scattered trees. They live mainly in the savanna grassland areas and they differ from forest elephants due to their unique tusks that curve outwards. Unlike forest elephants, bush or savanna elephants have a faster reproductive rate.
The African savanna elephants can be spotted in larger herds and the African forest elephants live in groups of about 20 individuals.
Are elephants endangered?
Over the years, the African savanna and forest elephant population has kept on declining due to various factors. They are threatened/endangered species for many reasons including poaching, habitat loss brought about by deforestation, need for land for farming, settlement, charcoal burning, etc; human-wildlife conflicts, and others. Not more than 500000 elephants are left on earth today, 415,000 are in Africa and about 40,000 in Asia.
Where to see African elephants?
African elephants are distributed throughout the African continent. These fascinating mammal species can be spotted while on an African safari in Chobe National Park, Mashatu Game Reserve- Botswana, Murchison Falls National Park, Queen Elizabeth National Park, Kidepo Valley National Park- Uganda.
In Kenya, the best places to see elephants include Amboseli National Park, Tsavo National Parks, Masai Mara National Reserve. In Zambia, visit South Luangwa National Park.
In South Africa, visit Addo Elephant Park, Kruger National Park; in Tanzania, the best places to encounter elephants include Tarangire National Park, Serengeti National Park, Ngorongoro Conservation Area.
In Rwanda, savanna elephants are best spotted while on a game drive or wildlife safari in its only savanna National Park- Akagera National Park.
Forest elephants exist mostly within Gabon, the Republic of Congo, and in parts of the Central African Republic, Cameroon, Ghana, Cote d’Ivoire, Equatorial Guinea, Liberia, and many more. Interestingly, Uganda shelters both the African bush/savanna elephants and African forest elephants.
When to see elephants?
Elephants are a few most amazing African mammal species and they attract most of the African safari holidaymakers who come to see the big game. They are among the most sought-after big game on an African safari and generally, they can be excellently spotted during the dry season starting from June, July, August to September, or December, January to February.
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9361 | https://askfilo.com/mathematics-question-answers/in-the-x-y-plane-what-is-the-distance-between-the-two-x-intercepts-of-the | Solving time: 3 mins
In the xy-plane, what is the distance between the two x-intercepts of the parabola y=x2−3x−10 ?
Views: 5,911 students
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We factor to find the x-intercepts.
y=x2−3x−10=(x−5)(x+2)
The x-intercepts are 5 and −2.
The distance between them is 5−(−2)=7.
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| Question Text | In the xy-plane, what is the distance between the two x-intercepts of the parabola y=x2−3x−10 ? |
| Topic | Graphing Parabolas |
| Subject | Mathematics |
| Class | Grade 12 |
| Answer Type | Text solution:1 |
| Upvotes | 95 |
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9362 | https://math.stackexchange.com/questions/3373237/irrationals-with-repeated-binary-digits | Irrationals with repeated binary digits - Mathematics Stack Exchange
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Irrationals with repeated binary digits
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Let r a real number. Let f(r) be a function of r such that in binary representation of r, the first digit is repeated once, the second digit is repeated twice, the third three times and this is continued ad infinitum.
Couple of examples:
r = 101.011
f(r) = 1 00 111. 0000 11111 111111
r = π (11.0010010 etc)
f(r) = 1 11. 000 0000 11111 000000 0000000 11111111 000000000 etc
My questions are:
Are f(√2) and f(π) irrational ?
Does r is irrational imply that f(r) irrational ?
I may be missing something obvious but have not been able to proceed.
irrational-numbers
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asked Sep 28, 2019 at 15:26
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f(r)f(r) will be irrational any time r r has infinitely many 0 0 s and 1 1 s in its expansion, so any time it is not a dyadic rational, one with denominator 2 k 2 k. There will not be any repeating pattern. If r r is a dyadic rational, f(r)f(r) will be also because it will end in an infinite chain of 0 0 s or 1 1 s.
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answered Sep 28, 2019 at 15:33
Ross MillikanRoss Millikan
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Of course.... just think about it.
If the result is rational than the result has periodic repeating squence. But if the sequence is of length n n, and as every digit past the n n th place in the input is repeated n n and more times. Which means the sequence of n n digits must all be the same digit. And that means that all the digits past the n n th place in the original are are all the same digit. Which means the original was a rational.
Furthermore it was a terminating rational.
And if it was a terminiating rational then it has a last digit an so does the output.
So f(j 2 k)↦m 2 n f(j 2 k)↦m 2 n for some m,n m,n..
f(j(2 m+1)∗2 k)↦f(j(2 m+1)∗2 k)↦ irrational (but one of the weird increasing spaces irrationals)
f(i r r a t i o n a l)↦f(i r r a t i o n a l)↦ irrational.
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edited Sep 28, 2019 at 15:51
answered Sep 28, 2019 at 15:40
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What is the average of the consecutive integers m through n, inclusiv
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What is the average of the consecutive integers m through n, inclusive?
(1) The average of m and n is 23.5
(2) The average of the integers between m and n not including either is 23.5
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Post URL18 May 2022, 06:44
I think both options together are not enough to answer the question.
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Post URL18 May 2022, 08:36
Bunuel
What is the average of the consecutive integers m through n, inclusive?
(1) The average of m and n is 23.5
(2) The average of the integers between m and n not including either is 23.5
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Say for eg. the numbers are 23 as m and 24 as n. Other pairs that have an average of 23.5 can be 22 & 25, 0 and 47, etc. If we count the consecutive integers between any of these pairs we would get the same average as 23.5. The sequence would always have an even count of numbers --> Sufficient
The average of the integers between m and n not including either is 23.5
This is an extension of statement 1. Say for eg the consecutive numbers 22, 23, 24, and 25 are between 21(m) and 26(n). Since the series and the extensions on both 21(m) and 26(n) have the same average, the overall series would have an average of 23.5. --> Sufficient
Option D
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Post URL20 Jun 2023, 12:47
Bunuel
What is the average of the consecutive integers m through n, inclusive?
(1) The average of m and n is 23.5
(2) The average of the integers between m and n not including either is 23.5
Show more
We need to find the average of consecutive integers whose lowest item is m and whose highest item is n, or vice versa.
We should see that consecutive integers are consecutive terms in an arithmetic sequence. For any arithmetic sequence, the average of terms can be determined with the following formula:
Average = (Lowest value + Highest value)/2
So, the question is:
(m + n)/2 = ?
Statement One Alone:
=> The average of m and n is 23.5
(m + n)/2 = 23.5
Statement one is sufficient. Eliminate answer choices B, C, and E.
Statement Two Alone:
=> The average of the integers between m and n not including either is 23.5
We see that the integers between m and n are consecutive terms in an arithmetic sequence.
So, if m is less than n, the average of terms for this shorter sequence is:
[(m + 1) + (n - 1)]/2 = 23.5
(m + n)/2 = 23.5
Statement two is sufficient.
Answer: D
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9364 | https://en.wikipedia.org/wiki/Haversine_formula | Jump to content
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Contents
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1 Formulation
2 The law of haversines
3 Proof
4 Example
5 See also
6 References
7 Further reading
8 External links
Haversine formula
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From Wikipedia, the free encyclopedia
Formula for the great-circle distance between two points on a sphere
The haversine formula determines the great-circle distance between two points on a sphere given their longitudes and latitudes. Important in navigation, it is a special case of a more general formula in spherical trigonometry, the law of haversines, that relates the sides and angles of spherical triangles.
The first table of haversines in English was published by James Andrew in 1805, but Florian Cajori credits an earlier use by José de Mendoza y Ríos in 1801. The term haversine was coined in 1835 by James Inman.
These names follow from the fact that they are customarily written in terms of the haversine function, given by hav θ = sin2(θ/2). The formulas could equally be written in terms of any multiple of the haversine, such as the older versine function (twice the haversine). Prior to the advent of computers, the elimination of division and multiplication by factors of two proved convenient enough that tables of haversine values and logarithms were included in 19th- and early 20th-century navigation and trigonometric texts. These days, the haversine form is also convenient in that it has no coefficient in front of the sin2 function.
Formulation
[edit]
Let the central angle θ between any two points on a sphere be:
where
d is the distance between the two points along a great circle of the sphere (see spherical distance),
r is the radius of the sphere.
The haversine formula allows the haversine of θ to be computed directly from the latitude (represented by φ) and longitude (represented by λ) of the two points:
where
φ1, φ2 are the latitude of point 1 and latitude of point 2,
λ1, λ2 are the longitude of point 1 and longitude of point 2,
, .
Finally, the haversine function hav(θ), applied above to both the central angle θ and the differences in latitude and longitude, is
The haversine function computes half a versine of the angle θ, or the squares of half chord of the angle on a unit circle (sphere).
To solve for the distance d, apply the archaversine (inverse haversine) to hav(θ) or use the arcsine (inverse sine) function:
or more explicitly:
:
where .
When using these formulae, one must ensure that h = hav(θ) does not exceed 1 due to a floating point error (d is real only for 0 ≤ h ≤ 1). h only approaches 1 for antipodal points (on opposite sides of the sphere)—in this region, relatively large numerical errors tend to arise in the formula when finite precision is used. Because d is then large (approaching πR, half the circumference) a small error is often not a major concern in this unusual case (although there are other great-circle distance formulas that avoid this problem). (The formula above is sometimes written in terms of the arctangent function, but this suffers from similar numerical problems near h = 1.)
As described below, a similar formula can be written using cosines (sometimes called the spherical law of cosines, not to be confused with the law of cosines for plane geometry) instead of haversines, but if the two points are close together (e.g. a kilometer apart, on the Earth) one might end up with cos(d/R) = 0.99999999, leading to an inaccurate answer. Since the haversine formula uses sines, it avoids that problem.
Either formula is only an approximation when applied to the Earth, which is not a perfect sphere: the "Earth radius" R varies from 6356.752 km at the poles to 6378.137 km at the equator. More importantly, the radius of curvature of a north-south line on the earth's surface is 1% greater at the poles (≈6399.594 km) than at the equator (≈6335.439 km)—so the haversine formula and law of cosines cannot be guaranteed correct to better than 0.5%.[citation needed] More accurate methods that consider the Earth's ellipticity are given by Vincenty's formulae and the other formulas in the geographical distance article.
The law of haversines
[edit]
Given a unit sphere, a "triangle" on the surface of the sphere is defined by the great circles connecting three points u, v, and w on the sphere. If the lengths of these three sides are a (from u to v), b (from u to w), and c (from v to w), and the angle of the corner opposite c is C, then the law of haversines states:
Since this is a unit sphere, the lengths a, b, and c are simply equal to the angles (in radians) subtended by those sides from the center of the sphere (for a non-unit sphere, each of these arc lengths is equal to its central angle multiplied by the radius R of the sphere).
In order to obtain the haversine formula of the previous section from this law, one simply considers the special case where u is the north pole, while v and w are the two points whose separation d is to be determined. In that case, a and b are π/2 − φ1,2 (that is, the, co-latitudes), C is the longitude separation λ2 − λ1, and c is the desired d/R. Noting that sin(π/2 − φ) = cos(φ), the haversine formula immediately follows.
To derive the law of haversines, one starts with the spherical law of cosines:
As mentioned above, this formula is an ill-conditioned way of solving for c when c is small. Instead, we substitute the identity that cos(θ) = 1 − 2 hav(θ), and also employ the addition identity cos(a − b) = cos(a) cos(b) + sin(a) sin(b), to obtain the law of haversines, above.
Proof
[edit]
One can prove the formula:
by transforming the points given by their latitude and longitude into cartesian coordinates, then taking their dot product.
Consider two points on the unit sphere, given by their latitude and longitude :
These representations are very similar to spherical coordinates, however latitude is measured as angle from the equator and not the north pole. These points have the following representations in cartesian coordinates:
From here we could directly attempt to calculate the dot product and proceed, however the formulas become significantly simpler when we consider the following fact: the distance between the two points will not change if we rotate the sphere along the z-axis. This will in effect add a constant to . Note that similar considerations do not apply to transforming the latitudes - adding a constant to the latitudes may change the distance between the points. By choosing our constant to be , and setting , our new points become:
With denoting the angle between and , we now have that:
Example
[edit]
The haversine formula can be used to find the approximate distance between the White House in Washington, D.C. (latitude 38.898° N, longitude -77.037° W) and the Eiffel Tower in Paris (latitude 48.858° N, longitude 2.294° E). The difference in latitudes is 9.960° and the difference in longitudes is 79.331°. Inputting these into the haversine formula,
The great-circle distance is this central angle, in radians (55.411 degrees is 0.96710 radians), multiplied by the average radius of the Earth,
By comparison, using a more accurate ellipsoidal model of the earth, the geodesic distance between these landmarks can be computed as approximately 6177.45 km.
See also
[edit]
Sight reduction
Vincenty's formulae
Cosine distance
References
[edit]
^ van Brummelen, Glen Robert (2013). Heavenly Mathematics: The Forgotten Art of Spherical Trigonometry. Princeton University Press. ISBN 9780691148922. 0691148929. Retrieved 2015-11-10.
^ de Mendoza y Ríos, Joseph (1795). Memoria sobre algunos métodos nuevos de calcular la longitud por las distancias lunares: y aplicacion de su teórica á la solucion de otros problemas de navegacion (in Spanish). Madrid, Spain: Imprenta Real.
^ Cajori, Florian (1952) . A History of Mathematical Notations. Vol. 2 (2 (3rd corrected printing of 1929 issue) ed.). Chicago: Open court publishing company. p. 172. ISBN 978-1-60206-714-1. 1602067147. Retrieved 2015-11-11. The haversine first appears in the tables of logarithmic versines of José de Mendoza y Rios (Madrid, 1801, also 1805, 1809), and later in a treatise on navigation of James Inman (1821). {{cite book}}: ISBN / Date incompatibility (help) (NB. ISBN and link for reprint of second edition by Cosimo, Inc., New York, 2013.)
^ Inman, James (1835) . Navigation and Nautical Astronomy: For the Use of British Seamen (3 ed.). London, UK: W. Woodward, C. & J. Rivington. Retrieved 2015-11-09. (Fourth edition: .)
^ "haversine". Oxford English Dictionary (2nd ed.). Oxford University Press. 1989.
^ H. B. Goodwin, The haversine in nautical astronomy, Naval Institute Proceedings, vol. 36, no. 3 (1910), pp. 735–746: Evidently if a Table of Haversines is employed we shall be saved in the first instance the trouble of dividing the sum of the logarithms by two, and in the second place of multiplying the angle taken from the tables by the same number. This is the special advantage of the form of table first introduced by Professor Inman, of the Portsmouth Royal Navy College, nearly a century ago.
^ W. W. Sheppard and C. C. Soule, Practical navigation (World Technical Institute: Jersey City, 1922).
^ E. R. Hedrick, Logarithmic and Trigonometric Tables (Macmillan, New York, 1913).
^ Gade, Kenneth (2010). "A Non-singular Horizontal Position Representation". Journal of Navigation. 63 (3): 395–417. Bibcode:2010JNav...63..395G. doi:10.1017/S0373463309990415. ISSN 0373-4633.
^ Korn, Grandino Arthur; Korn, Theresa M. (2000) . "Appendix B: B9. Plane and Spherical Trigonometry: Formulas Expressed in Terms of the Haversine Function". Mathematical handbook for scientists and engineers: Definitions, theorems, and formulas for reference and review (3rd ed.). Mineola, New York: Dover Publications. pp. 892–893. ISBN 978-0-486-41147-7.
^ This calculation was made using the open-source geodesic calculation software GeographicLib, assuming the WGS84 ellipsoid. See Karney, Charles F. F. (2013). "Algorithms for geodesics". Journal of Geodesy. 87 (1): 43–55. arXiv:1109.4448. doi:10.1007/s00190-012-0578-z.
Further reading
[edit]
U. S. Census Bureau Geographic Information Systems FAQ, (content has been moved to What is the best way to calculate the distance between 2 points?)
R. W. Sinnott, "Virtues of the Haversine", Sky and Telescope 68 (2), 159 (1984).
"Deriving the haversine formula". Ask Dr. Math. April 20–21, 1999. Archived from the original on 20 January 2020.
W. Gellert, S. Gottwald, M. Hellwich, H. Kästner, and H. Küstner, The VNR Concise Encyclopedia of Mathematics, 2nd ed., ch. 12 (Van Nostrand Reinhold: New York, 1989).
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9365 | https://www.physicsforums.com/threads/uniqueness-of-the-roots-of-a-polynomial-equation.371673/ | General Math
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Uniqueness of the roots of a polynomial equation
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maverick280857
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1
maverick280857
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Hi,
I have a question, which seems deceptively simple to me, but when I thought about it, I couldn't really come up with a rigorous proof. Here goes,
Are the roots of a polynomial equation unique?
Suppose we have a general monic polynomial equation:
z^{n} + c_{1}z^{n-1} + c_{2}z^{n-2} + \ldots + c_{n} = 0
where z \in \mathbb{C} and c_{1}, c_{2}, \ldots are real coefficients. Now let S be the set of all roots of this equation (counted according to their multiplicities so that if x is a root of multiplicity n, then S contains n occurrences of x. Strictly S is not a set, but rather a list, but you see my point.)
Can we find some S' \neq S such that every member z^{'} \in S' is a root of the above equation (and z^{'} \notin S)? Stated differently, does this equation have two different sets of roots?
The picture I have in mind for real roots is as follows: if the roots were not unique, the zero crossings of the function on the left hand side of the equation would not be unique, which is impossible. This reasoning does not work for complex roots :-(
Any ideas?
Thanks in advance.
-Vivek
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rasmhop
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maverick280857 said:
Hi,
Can we find some S' \neq S such that every member z^{'} \in S' is a root of the above equation (and z^{'} \notin S)? Stated differently, does this equation have two different sets of roots?
No. If it had then just take an element z of S'. z is a root so it must be in S since S is the set of all roots.
In general if you let a set T be all elements with some property, then you can't find an element with the property, but not in T.
One small note is that if you allow the empty set, then the answer is yes since \emptyset \not= S, but I'm assuming this isn't what you meant by a set S'.
3
maverick280857
: 1,774
: 5
rasmhop said:
No. If it had then just take an element z of S'. z is a root so it must be in S since S is the set of all roots.
In general if you let a set T be all elements with some property, then you can't find an element with the property, but not in T.
One small note is that if you allow the empty set, then the answer is yes since \emptyset \not= S, but I'm assuming this isn't what you meant by a set S'.
Right. I get your point. Now, how does one prove that the roots of a n-th order polynomial equation (with say, real coefficients) are unique? Is my set theoretic statement of the problem equivalent to this question? If not, how does one sketch a proof?
4
rasmhop
: 430
: 3
I'm not really sure what you don't understand. Given a polynomial f \in \mathbb{C}[x] a complex number z is a root in f iff f(z)=0. There is no ambiguity. Thus two sets of roots can differ if they don't contain all roots. For instance for the polynomial (x-2)(x-3)(x-4) we may let S={2} and S'={3,4} and these are different sets of roots, but if we form the set of all roots S''={2,3,4}, then we can't find another set of all roots because it would have to contain all elements of S'' and can't contain any more.
As for your actual statement, then yes we can find such an S' because we can let S'=\emptyset. You can find no other S' because you said S consists of exactly all roots, so you can't find a different set that consists of exactly all roots. If we had a non-empty set S' as you state, then z \in S' would imply f(z) =0 since S' consists of roots, and since S consists of all roots z \in S.
I'm sorry if this wasn't what you wanted, but I find your question a bit confusing. It seems to me that your question is of the same kind as:
Is the set {1,2} unique.
Since it's specified completely, then of course yes. In the same sense the set of roots is specified completely so of course it's unique.
5
elibj123
: 237
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Perhaps you are trying to prove the fundamental theorem of algebra, and you don't explain yourself correctly?
6
maverick280857
: 1,774
: 5
Yes, I realized its not phrased properly. Someone asked me the following question (I'm citing an example to make things simple):
Suppose we're given the equation x^{2}-5x+6 = 0. We know the roots are x = 2 and x = 3. The question is: can this equation have some other set of 2 roots? Of course, this particular example is absolutely trivial because the "uniqueness" is proved by writing
x^{2}-5x+6 = (x-2)(x-3)
For a general scenario, the question is that for an arbitrary positive integer n, if
x^{n} + c_{1}x^{n-1} + \ldots + c_{n} = (x-\alpha_{1})(x-\alpha_{2})\cdots(x-\alpha_{n})
so that \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} are the roots, then is it possible to find some set \beta_{1}, \beta_{2}, \ldots, \beta_{n} of numbers such that
x^{n} + c_{1}x^{n-1} + \ldots + c_{n} = (x-\beta_{1})(x-\beta_{2})\cdots(x-\beta_{n})
and none of the \beta's are from \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}.
In other words, given the sum of roots taken one at a time, two at a time, three at a time, and so on.., does the following system of equations have a unique solution in \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}?.
\sum_{i=1}^{n}\alpha_{i} = -c_{1}
\sum_{i\neq j}^{n}\alpha_{i}\alpha_{j} = c_{2}
\cdots
\sum_{i_{1}\neq i_{2}\neq i_{3}\cdots\neq i_{n}}^{n}\alpha_{i_{1}}\alpha_{i_{2}}\ldots\alpha_{i_{n}}= (-1)^{n}c_{n}
I want to be able to give a rigorous proof to the uniqueness of the solution to this nonlinear system.
7
rasmhop
: 430
: 3
Ok suppose,
f(x) = (x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n) = (x-\beta_1)(x-\beta_2)\cdots(x-\beta_n)
and that \alpha_1 \notin {\beta_1,\ldots,\beta_n}. Now evaluate f(\alpha_1).
f(\alpha_1) = (\alpha_1-\alpha_1)(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0
Thus we have,
0 = f(\alpha_1) = (\alpha_1-\beta_1)(\alpha_1-\beta_2)\cdots(\alpha_1-\beta_n)
We know that in \mathbb{C} a product is 0 if and only if one of the factors is 0. Let i be an integer in {1,...,n} such that the factor (\alpha_1-\beta_i) is 0 (we know that such an i exists since at least one factor is 0). Then we get \alpha_1 = \beta_i which is a contradiction.
8
maverick280857
: 1,774
: 5
rasmhop said:
Ok suppose,
f(x) = (x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n) = (x-\beta_1)(x-\beta_2)\cdots(x-\beta_n)
and that \alpha_1 \notin {\beta_1,\ldots,\beta_n}. Now evaluate f(\alpha_1).
f(\alpha_1) = (\alpha_1-\alpha_1)(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0(\alpha_1-\alpha_2)\cdots(\alpha_1-\alpha_n) = 0
Thus we have,
0 = f(\alpha_1) = (\alpha_1-\beta_1)(\alpha_1-\beta_2)\cdots(\alpha_1-\beta_n)
We know that in \mathbb{C} a product is 0 if and only if one of the factors is 0. Let i be an integer in {1,...,n} such that the factor (\alpha_1-\beta_i) is 0 (we know that such an i exists since at least one factor is 0). Then we get \alpha_1 = \beta_i which is a contradiction.
Thats neat. Thanks...so stupid of me not to get it. :-(
9
HallsofIvy
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I don't want to embarass your further but I think it is even more fundamenental than that! (And it is true of all equations, not just polynomial equations.)
"Roots of an equation", f(x)= 0, means numbers, a, such that f(a)= 0. If you have two different purported sets of roots such that a is in one but not the other, one of them must be wrong. Either f(a)= 0 or not! If f(a) is equal to 0, then the set that does not include a is NOT a "set of roots". If f(a) is not equal to a, then the set that does include it is NOT a "set of roots".
But I think I see the cause of your confusion- there are many different ways of solving equations and you were, I suspect, thinking of "roots" as "what you get using this particular method" so, conceivably, different methods might give you different "roots".
Hopefully, if your methods are correct, that won't happen because "roots" are NOT just "what you get using this method."
Many years ago, I had a student in a basic algebra class complain bitterly that I had marked her wrong on a particular test problem just be she "used a different method" than I had taught in class. I don't believe I ever convinced her that I marked her wrong becuase her solution did not satisfy the equation!
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9366 | https://www.youtube.com/watch?v=sN8EnUOrVgs | Algebra 1 - Understanding Standard Form of a Linear Equation Ax + By = C
iteachalgebra
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Join me as I explain how to write a linear equation in proper standard form! Ax + By = C
Thank you so much for watching!
My name is Rory Yakubov, and I am an Algebra and Geometry teacher from NJ.
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Transcript:
and standard for right now this lesson is not about I think it's just really understanding standard form ax it's see what it isn't standard for them to determine whether or not the equation is linear so let's take a look at all the rules we have and a bunch of practice problems for us so here we go here's that form ax plus B y equals C where a B and C you see they're capital letters those stand for numbers the x and y those are going to be numbers too but they're the variables of the equation so we leave x and y there and we replace a b and c with some kind of integer value but those do have rules so let's take a look a has to be a number that's not negative like 2 or 5 or 1 or 3 anything like that a and B cannot equal 0 at the same time a can be equal to 0 B could be equal to 0 but you can't have them both be equal to 0 at the same time and we're going to take a close look at some of those examples as well the last special rule is that a B and C are not fractions or decimals so if I have fractions or I have decimals today how we would get rid of that and rewrite our equation and also one extra thing we need to make sure that a B and C are as simplified as possible but there's no greatest common factor of those three numbers except for just 1 so again our rules are a cannot be negative a and B cannot be 0 at the same time so a could be 0 B could be 0 but they can't both be at the same time and the last one a B or C cannot be any fraction or decimal but if we see that they are we fix it up and I'm going to show you exactly how to do that so we're gonna take a look at the standard form the first thing we're going to do is we're going to look at equations that are not linear equations that just do not fit this form so a linear equation means that the equation cannot be written in standard form the graphs of these equations would not be a straight line which means they would not be linear so if I gave you these three equations these would be exam both of equations that just do not fit for ax plus B y equals C they just wouldn't work out the way you want them to work all right so this first equation is definitely not in standard form it does not look like this it also can't be rearranged to make it look like this I can't like swing that x squared over to the other side and make it look like this and the biggest deal is this exponent standard form or any linear equation never has any exponent so if you see an exponent it's not linear the second one 3x y minus 7 equals 8 that one is not linear because the x and y are being multiplied together and in standard form it's a number times X plus a number times y naught x and y being moles together so I'll be another example of an equation that's definitely not linear because I can't put it in standard for 4x equals 8 over Y minus 2 I could try rearranging this to make it look like this I see a X is 4x but my B Y now B Y means the number B times y this is saying 8 divided by Y if we see that we are dividing by one of our variables that's going to be no good and it's not going to be standard form in standard form ever which means it's not a linear equation so all of these are simply not linear the next tab is showing you that we can have either A or B be equal to zero okay and in this case it says if a is not equal to zero but B is in this equation but simply just being ax equals C so here we're going to see that if I have 5x plus 3 equals 0 I don't have any exponents and not multiplying x and y together I'm not dividing by a variable so I would look 5x would be my ax okay this 3 I should really have B on the other side of the equation because that should be my C value so what I really want you to see here is if I was to set up this equation as ax plus B Y you see I don't see any wise in my equation that means the B value is zero because that won't happen if he plugs in a zero here zero times y is just zero you wouldn't even actually see it you would just simply be left with ax equals C so how do I make this equation here it looked like ax equals c it should be a times a number I'm sorry X a number times X so 5x that's good but it should just be equal to the number C so this three should really be sent over to the other side or simply subtracted this equation now becomes 5x equals negative three and that is in standard form the form is this mini version of ax equals c let's try this one I can rearrange some things I want it to just look like a x equals C notice there's no y value I would need to subtract 8 on both sides to get my constants over on the right so now this looks like negative x equals 2 but here's the deal remember one of those special rules the number a the value in front of X can't be negative so if I have a negative X in order to undo that I either need to multiply or divide both sides by a negative 1 and then here is my new equation x equals negative 2 last one if I want this equation to be in this form of AX equals C I would need to subtract 6 on both sides so I have negative 3x equals negative 6 and then I want to get this to not be negative so let's say I divided both sides by a negative and I got 3x equals 6 but one of the other things I did mention is we want our a B and C values to only have a GCF of 1 so could I simplify this down a little bit further yes if I was to divide both sides by 3 I would be left with x equals 2 now the reason why I didn't do that here in this one is because negative 3 & 5 don't have a common factor other than 1 so there's nothing I can simplify here but in 3 x equals 6 I can simplify and so that's what we would go to do now let's take a look at the other scenario where if I gave you a X plus B y equals C and it says now a is zero so match and I plugged in a zero for a zero times X is just zero so I'm really just left with this equation B y equals C the number times y equals C so let's rearrange these if I want it to look like me y equal C and B can be negative guys a is the only one that has that special rule I would need to attract on both sides to get 5 y equals negative 8 next one I would want it to look like be y equal C so this 5 is not on the right correct side subtract 5 on both sides and I would get negative 2 y equals 5 now remember we want to see the greatest common factor be 1 right so 5 and negative 8 the GCF is 1 negative 2 & 5 the GCF is 1 so I can't go any further and again B can be negative so that's completely fine and his last one subtract 7 on both sides and I have negative 2 y equals negative 7 if you wanted to clean it up you always could write 2 y equals positive 7 divide both sides by a negative 1 the equations do mean the same thing we're going to take a look at the last step where we actually do have the ax the B Y in the sea they're just not in the right order so we've got some rearranging to do now there's multiple ways we could do some of these problems so if you feel like you want to go ahead and try something do something in a different order but you get the same result and that's good okay I'm gonna just work through these problems with my ultimate goal again to make it look like this X plus dy equals the number times X plus the number x be my Y rather equals the number C so here 5x is in the right spot this negative 6 the constant should be on the right hand side and this 3y should really be on the other side so for example I'm gonna do possible first step would be to add 6 so if I go ahead and add 6 I now have 5x equals 3y plus 6 then I need to have my 3y over to the other side so this would become 5x minus 3y equals there we go that's our ax plus B y equals C pretty good let's try this one same thing I want this to say ax plus me y equals C so the negative 2x is on the right-hand side of the on the correct side of the equation I should say but the ain't in the negative Y or not so I need to send my constant over of a negative a to the other side of my equation so this now becomes negative 2x equals negative y minus 8 okay I need to send the Y over now let's add Y so now it looks like negative 2x plus y equals negative 8 and right now it looks like it's in the right order but remember a cannot be negative so if a cannot be negative we got both sides four by both sides by a negative one that changes the signs of everything guys so we're left with 2x minus y equals 8 good now this last one here one of the rules was that not only can it not be negative and you have to have a GCF of 1 everything you're not yeah so if you can simplify everything you need to do it but we also don't want fractions so the best thing to do with fractions is to clear them by simply multiplying both sides of your equation by the least common multiple of the denominator or least common denominator rather so here my denominator of my fraction is 2 so I'm going to multiply both sides by 2 because look what happens 2 times 2 is 4 2 times 1/2 is just X and then 3 y times 2 is 6 y when you multiply both sides of the equation by whatever your denominator denominator is or your least common denominator and your fractions are completely gone so now let's rearrange this so it's in the right form let's say I subtracted 4 on both sides to get x equals 6 y minus 4 what needs to get moved now that's 6 y so we subtract our 6y and we are then left with X minus 6y equals negative 4 pretty good okay and I have my problems here now when you have it in standard form we should be a see the a B and C very carefully so if I showed you this result my a value is five my B value is negative three my C value is six about this next equation my a value would be two what's my B value negative one and my C eight last one a would be one that one in front of the X B is negative six C is negative four awesome okay I asked you problems for us so here same idea we have an X we have a Y we have a constant got fractions I've got denominators of three and two what's my least common denominator of 3 and two or the least common multiple of three into its six so I'm going to multiply both sides of my equation by six now six times two-thirds 6 times 2 is 12 12 divided by 3 is 4 6 times negative 1/2 is negative 3 and 6 times 4y is 24 Y so look at this we're left with 4x minus 3 equals 24 y we would add 3 you should know what to do now at this point what gets moved next the 24 y and so here we have it 4 X minus 24 y equals 3 decimals decimals decimals decimals I move myself closer to the middle here um decimals there's a few things we can do we can convert these facts with these decimals and two fractions which actually these work out nicely point 25 is just 1/4 negative 0.5 is just negative 1/2 or we can always multiply decimals by a number to clear the decimals out so if I have two places out 100 will clear that if I have one place out 10 would clear that but of course I need to clear out using the the least common multiple that works for both so I or I'm going to show you my answer or think about money if I multiply both sides by 4 what would happen to it 4 times 25 cents is a dollar 0.5 times 4 is 2 so you can multiply both sides of this equation by 100 we could change them into fractions or multiply it by a number that you know is going to give you a clean result we are you're just left with integers so if I did this I would have 4 times point 25 is 1/4 times negative 5 y is negative 20 y and negative 0.5 times 4 is negative 2 we know what to do here guys you can do this in any order that you want so let's say I send that negative 1 over to the other side I now have negative 20 y equals negative 2x minus 1 what would I have to do last step here I have to move the negative 2x by adding it and here's my result 2x minus 20y equals negative 1 I can't simplify any fractions out my a value is 4 my B is 24 negative 24 rather sorry and my C is 3 and this one here my a is 2 my B is negative 20 a my C is negative 1 I hope this video was helpful for you thank you so much for watching bye |
9367 | https://www.youtube.com/watch?v=JJ84B_Cv0Uw | Solving the logrithmic equation with a square root
Brian McLogan
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👉 Learn how to solve natural logarithmic equations. Logarithmic equations are equations with logarithms in them. To solve a natural logarithmic equation, we first isolate the logarithm part of the equation. After we have isolated the logarithm part of the equation, we then get rid of the logarithm. This is done by raising both side of the equation as an exponent to the (natural) exponential function (e). This cancels out the natural logarithm, we can then solve the resulting equation.
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Transcript:
webcam so what I like to do now is show you guys how to do solve a logarithmic equation so in our previous one we were doing exponential equations now we're going to do a logarithmic and again it's the exact same type of properties you have to know your properties if you don't know your properties you're going to struggle with this all right so we're going to solve there's two different ways we can work on this first way if this is a very basic one so what I can do with this is you can always transfer to exponential right and that was the first thing we learned what to do with logarith logarithms an exponential transfer this to exponential form we know since the natural logarithm there's a base e so I can rewrite this as e to the first equal s < TK of x + 2 right all right the other way we can look at this is you guys remember I talked about this um 4^2 = 4 to X you should probably almost be like bored with this so then you know that 2 is equal to X right could I raise 992 = 99 to X is two still going to equal x right all it does so it doesn't matter what po what base I raise this or what base I use correct correct all right very good so now the next thing that we're going to do is the other property you guys have to know is your inverse properties log base a of a equals 1 I could you move back to please yes all right you guys should know that then you have log base a of AIS to x = x and then the other one is if I have a and I raise it to the power of log base a evaluating X is going to X inverse these are your two inverse properties you guys have to know those properties all right so if you should know this because I can bring the X in front log base a of a is equal to 1 x 1 is X but it works the exact same way if I have a raised to the logarithm base a of X is just going to equal x so the other method I could look at this and and say well I need to get rid of this Ln and you know I don't want to evaluate my answer with in so what I can do is I can raise them my point here I can raise both sides I can exponentiate both sides with a base e so I can raise them both as exponents with my e as my base now why would I ever want to do that well the reason I want to do that is I'm manipulating my equation so I can use this property because e raised to the Ln which is b e is going to equal A1 it's going to cancel out so I'm left withun of x + 2 = e to the 1st it's exact same as you get the first one yes you need to know how to do the properties and how to manipulate them so that's why I'm showing you both ways okay um then from here we just need to solve so how do you get rid of the square root square both sides right then subtract two that's it |
9368 | https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/10%3A_Direct-Current_Circuits/10.02%3A_Electromotive_Force | 10.2.1
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10.2: Electromotive Force
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10.3: Resistors in Series and Parallel
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( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
By the end of the section, you will be able to:
Describe the
electromotive force (emf)
and the
internal resistance
of a battery
Explain the basic operation of a battery
If you forget to turn off your car lights, they slowly dim as the battery runs down. Why don’t they suddenly blink off when the battery’s energy is gone? Their gradual dimming implies that the battery output
voltage
decreases as the battery is depleted. The reason for the decrease in output
voltage
for depleted batteries is that all
voltage
sources have two fundamental parts—a source of electrical energy and an
internal resistance
. In this section, we examine the energy source and the
internal resistance
.
Introduction to Electromotive Force
Voltage
has many sources, a few of which are shown in Figure 10.2.210.2.2. All such devices create a potential difference and can supply current if connected to a
circuit
. A special type of
potential difference
is known as electromotive force (emf). The emf is not a force at all, but the term ‘electromotive force’ is used for historical reasons. It was coined by Alessandro Volta in the 1800s, when he invented the first battery, also known as the voltaic pile. Because the electromotive force is not a force, it is common to refer to these sources simply as sources of emf (pronounced as the letters “ee-em-eff”), instead of sources of electromotive force.
If the electromotive force is not a force at all, then what is the emf and what is a source of emf? To answer these questions, consider a simple
circuit
of a 12-V lamp attached to a 12-V battery, as shown in Figure 10.2.210.2.2. The battery can be modeled as a two-terminal device that keeps one terminal at a higher
electric potential
than the second terminal. The higher
electric potential
is sometimes called the positive terminal and is labeled with a plus sign. The lower-potential terminal is sometimes called the negative terminal and labeled with a minus sign. This is the source of the emf.
When the emf source is not connected to the lamp, there is no net flow of charge within the emf source. Once the battery is connected to the lamp, charges flow from one terminal of the battery, through the lamp (causing the lamp to light), and back to the other terminal of the battery. If we consider positive (conventional) current flow, positive charges leave the positive terminal, travel through the lamp, and enter the negative terminal.
Positive current flow is useful for most of the
circuit
analysis in this chapter, but in metallic wires and resistors, electrons contribute the most to current, flowing in the opposite direction of positive current flow. Therefore, it is more realistic to consider the movement of electrons for the analysis of the
circuit
in Figure 10.2.210.2.2. The electrons leave the negative terminal, travel through the lamp, and return to the positive terminal. In order for the emf source to maintain the
potential difference
between the two terminals, negative charges (electrons) must be moved from the positive terminal to the negative terminal. The emf source acts as a charge pump, moving negative charges from the positive terminal to the negative terminal to maintain the
potential difference
. This increases the potential energy of the charges and, therefore, the
electric potential
of the charges.
The force on the negative charge from the
electric field
is in the opposite direction of the
electric field
, as shown in Figure 10.2.210.2.2. In order for the negative charges to be moved to the negative terminal, work must be done on the negative charges. This requires energy, which comes from chemical reactions in the battery. The potential is kept high on the positive terminal and low on the negative terminal to maintain the
potential difference
between the two terminals. The emf is equal to the work done on the charge per unit charge (ϵ=dWdq)(ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the
coulomb
, the unit for emf is the volt (1V=1J/C)(1V=1J/C).
The terminal voltage VterminalVterminal of a battery is
voltage
measured across the terminals of the battery when there is no load connected to the terminal. An ideal battery is an emf source that maintains a constant
terminal voltage
, independent of the current between the two terminals. An ideal battery has no
internal resistance
, and the
terminal voltage
is equal to the emf of the battery. In the next section, we will show that a real battery does have
internal resistance
and the
terminal voltage
is always less than the emf of the battery.
The Origin of Battery Potential
The combination of chemicals and the makeup of the terminals in a battery determine its emf. The lead acid battery used in cars and other vehicles is one of the most common combinations of chemicals. Figure 10.2.310.2.3 shows a single cell (one of six) of this battery. The cathode (positive) terminal of the cell is connected to a lead oxide plate, whereas the anode (negative) terminal is connected to a lead plate. Both plates are immersed in sulfuric acid, the electrolyte for the system.
Knowing a little about how the chemicals in a lead-acid battery interact helps in understanding the potential created by the battery. Figure 10.2.410.2.4 shows the result of a single chemical reaction. Two electrons are placed on the anode, making it negative, provided that the cathode supplies two electrons. This leaves the cathode positively charged, because it has lost two electrons. In short, a separation of charge has been driven by a chemical reaction.
Note that the reaction does not take place unless there is a complete
circuit
to allow two electrons to be supplied to the cathode. Under many circumstances, these electrons come from the anode, flow through a
resistance
, and return to the cathode. Note also that since the chemical reactions involve substances with
resistance
, it is not possible to create the emf without an
internal resistance
.
Internal Resistance and Terminal Voltage
The amount of
resistance
to the flow of current within the
voltage
source is called the internal resistance. The
internal resistance
r of a battery can behave in complex ways. It generally increases as a battery is depleted, due to the oxidation of the plates or the reduction of the acidity of the electrolyte. However,
internal resistance
may also depend on the magnitude and direction of the current through a
voltage
source, its
temperature
, and even its history. The
internal resistance
of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. A simple model for a battery consists of an idealized emf source ϵϵ and an
internal resistance
r (Figure 10.2.510.2.5).
Suppose an external resistor, known as the load
resistance
R, is connected to a
voltage
source such as a battery, as in Figure 10.2.610.2.6. The figure shows a model of a battery with an emf ε, an
internal resistance
r, and a load resistor R connected across its terminals. Using
conventional current
flow, positive charges leave the positive terminal of the battery, travel through the resistor, and return to the negative terminal of the battery. The
terminal voltage
of the battery depends on the emf, the
internal resistance
, and the current, and is equal to
Note
Vterminal=ϵ−Ir
Vterminal=ϵ−Ir
For a given emf and
internal resistance
, the
terminal voltage
decreases as the current increases due to the
potential drop
Ir of the
internal resistance
.
A graph of the
potential difference
across each element the
circuit
is shown in Figure 10.2.710.2.7. A current I runs through the
circuit
, and the
potential drop
across the internal resistor is equal to Ir. The
terminal voltage
is equal to ϵ−Irϵ−Ir, which is equal to the potential drop across the load resistor IR=ϵ−IrIR=ϵ−Ir. As with potential energy, it is the change in
voltage
that is important. When the term “
voltage
” is used, we assume that it is actually the change in the potential, or ΔVΔV. However, ΔΔ is often omitted for convenience.
The current through the load resistor is I=ϵr+RI=ϵr+R. We see from this expression that the smaller the
internal resistance
r, the greater the current the
voltage
source supplies to its load R. As batteries are depleted, r increases. If r becomes a significant fraction of the load
resistance
, then the current is significantly reduced, as the following example illustrates.
Example 10.2.110.2.1: Analyzing a Circuit with a Battery and a Load
A given battery has a 12.00-V emf and an
internal resistance
of 0.100Ω0.100Ω. (a) Calculate its
terminal voltage
when connected to a10.00Ω10.00Ω load. (b) What is the
terminal voltage
when connected to a 0.500Ω0.500Ω load? (c) What power does the 0.500Ω0.500Ω load dissipate? (d) If the
internal resistance
grows to 0.500Ω0.500Ω, find the current,
terminal voltage
, and power dissipated by a 0.500Ω0.500Ω load.
Strategy
The analysis above gave an expression for current when
internal resistance
is taken into account. Once the current is found, the
terminal voltage
can be calculated by using the equation Vterminal=ϵ−IrVterminal=ϵ−Ir. Once current is found, we can also find the power dissipated by the resistor.
Solution
Entering the given values for the emf, load
resistance
, and
internal resistance
into the expression above yields I=ϵR+r=12.00V10.10Ω=1.188A.
I=ϵR+r=12.00V10.10Ω=1.188A.
Enter the known values into the equationVterminal=ϵ−IrVterminal=ϵ−Ir to get the
terminal voltage
: Vterminal=ϵ−Ir=12.00V−(1.188A)(0.100Ω)=11.90V.
Vterminal=ϵ−Ir=12.00V−(1.188A)(0.100Ω)=11.90V.
The
terminal voltage
here is only slightly lower than the emf, implying that the current drawn by this light load is not significant.
2. Similarly, with Rload=0.500ΩRload=0.500Ω, the current is I=ϵR+r=12.00V0.600Ω=20.00A.
I=ϵR+r=12.00V0.600Ω=20.00A.
The
terminal voltage
is now Vterminal=ϵ−Ir=12.00V−(20.00A)(0.100Ω)=10.00V.The
terminal voltage
exhibits a more significant reduction compared with emf, implying 0.500Ω is a heavy load for this battery. A “heavy load” signifies a larger draw of current from the source but not a larger
resistance
.
3. The power dissipated by the 0.500Ω load can be found using the formula P=I2R. Entering the known values gives P=I2R=(20.0A)2(0.500Ω)=2.00×102W.Note that this power can also be obtained using the expression V2R or IV, where V is the
terminal voltage
(10.0 V in this case).
4. Here, the
internal resistance
has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load
resistance
. As before, we first find the current by entering the known values into the expression, yielding I=ϵR+r=12.00V1.00Ω=12.00A.Now the
terminal voltage
is Vterminal=ϵ−Ir=12.00V−(12.00A)(0.500Ω)=6.00V,and the power dissipated by the load is P=I2R=(12.00A)2(0.500Ω)=72.00W.We see that the increased
internal resistance
has significantly decreased the
terminal voltage
, current, and power delivered to a load.
Significance
The
internal resistance
of a battery can increase for many reasons. For example, the
internal resistance
of a rechargeable battery increases as the number of times the battery is recharged increases. The increased
internal resistance
may have two effects on the battery. First, the
terminal voltage
will decrease. Second, the battery may overheat due to the increased power dissipated by the
internal resistance
.
Exercise 10.2.1
If you place a wire directly across the two terminal of a battery, effectively shorting out the terminals, the battery will begin to get hot. Why do you suppose this happens?
Solution
: If a wire is connected across the terminals, the load
resistance
is close to zero, or at least considerably less than the
internal resistance
of the battery. Since the
internal resistance
is small, the current through the
circuit
will be large, I=ϵR+r=ϵ0+r=ϵr. The large current causes a high power to be dissipated by the
internal resistance
(P=I2r). The power is dissipated as .
Battery Testers
Battery testers, such as those in Figure 10.2.8, use small load resistors to intentionally draw current to determine whether the terminal potential drops below an acceptable level. Although it is difficult to measure the
internal resistance
of a battery, battery testers can provide a measurement of the
internal resistance
of the battery. If
internal resistance
is high, the battery is weak, as evidenced by its low
terminal voltage
.
Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to an appliance. This is done routinely in cars and in batteries for small electrical appliances and electronic devices (Figure 10.2.9). The
voltage
output of the battery charger must be greater than the emf of the battery to reverse the current through it. This causes the
terminal voltage
of the battery to be greater than the emf, since V=ϵ−Ir and I is now negative.
It is important to understand the consequences of the
internal resistance
of emf sources, such as batteries and solar cells, but often, the analysis of circuits is done with the
terminal voltage
of the battery, as we have done in the previous sections. The
terminal voltage
is referred to as simply as V, dropping the subscript “terminal.” This is because the
internal resistance
of the battery is difficult to measure directly and can change over time.
10.1: Prelude to Direct-Current Circuits
10.3: Resistors in Series and Parallel |
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EUDIOMETRY
Eudiometry is the study of gas reactions where the volumes of gaseous reactants and products are measured at the same temperature and pressure, based on principles like Gay-Lussac's law. The…
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u iometry
EUDIOMETRY
KEY CONCEPTS
EUDIOMETRY :
Eudiometry or gas analysis involves the calculations based on gaseous reactions or the reactions in whi ch at least two componen ts are gaseous, in whi ch the amounts of gases are represented by thei r volumes, measured at the same pressure and temperature. Some basic assumptions related with calculations are:
1.
Gay-Lussac's law of volume comb ination holds good. Accord ing to t his l aw, the volumes of gaseous reacta nts reacted a nd the vol umes of gaseous products formed, all m easured at t he same temperature and pressure, bear a simple ratio.N
2
(g) + 3H
2
(g)
¾®
2NH
3
(g)1 v o l. 3 v o l.2 v o l.Problem may b e solved direct ly i s ter ms of volume, in place of mole. The st oichiometric coefficien ts of a balanced chemi cal react ions giv es the ra tio of volumes in w hich gaseous substances are react ing and products are formed, at same temperature and pressure.
2.
The volumes of solids or liquids is considered to be negligible in comparison to the volume of gas.It is due to t he fact that t he volum e occupied by an y substan ce in gaseous state is even m ore t han thousand times the volume occupied by the same substance in solid or liquid states.2H
2
(g) + O
2
(g)
¾®
2H
2
O (
l
)2 m o l e 1 m o l e 2 m o l e 2 v o l. 1 v o l. 0 v o l.
3.
Air is considered as a mixture of oxygen an d nitrogen gases only. It is due to t he fact that abo ut 99%volume of air is composed of oxygen and nitrogen gases only.
4.
Nitrogen gas is consi dered as a n on- reac tive gas. It is due to the fact that n itrogen gas reacts onl y at very hig h temperat ure due t o its very high thermal stabili ty. Eudiometr y i s perfor med in an eudiometer tube and the tube can not withstand very h igh temperature. This is why, nitrogen gas can not pa rticipate in the reactions occurring in the eudiometer tube.
5.
The tot al vol um e of non-reacting gaseous mix ture is equal to sum of partial volum es of the component gases
(Amagat's law)
.V = V
1
V
2
+..............Partial volume of gas in a non-reacting gasesous mixture is its volume when the entire pressure of the mix ture is supposed to be exerted only by that gas.
6.
The volume of gases produced is often given by certain solvent which absorb contain gases.
S o l v e n t G a s e s a b s o r b
K O H C O
2
, SO
2
, Cl
2
Ammonical Cu
2
Cl
2
CO T u r p e n t i n e o i l O
3
A l k a l i n e p y r o g a l l o l O
2
w a t e r N H
3
, HCl CuSO
4
/CaCl
2
H
2
O
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2
EUDIOMETER
An eudiometer is a laborat or y devi ce that mea sures t he change in volum e of a gas mixture following a ph ys i cal or che m ic al ch ang e.
Scheme of eudiometer
T o use a eud io me ter, it i s fi ll ed wi th wate r, in ve rted so tha t its open en d is fa ci ng the grou nd(while holding the open end so that no water escapes), and then submersed in a basin of water.A chemical react ion i s taking place thro ugh whi ch gas is creat ed. One react ant is typically at t he bott om of the eudiometer (which flows downward when the eudiometer is in verted) and t he other reac tant is suspended o n the rim of the eudiometer, typically b y m eans of a platinum or coppe r wire (du e to t heir low react iv ity). When the gas cr eated by the chemi cal react ion i s released, it should rise into the eudiometer so that the experimenter may accurate ly read the volum e of the gas produce d at any gi ven time. Normally a perso n would read the volume when the reaction is completed
SOLVED EXAMPLE Ex.1 10 ml of CO is mixe d with 25 ml air (20% O
2
by volume) in a container at 1 atm. Find final volume(in ml) of container at 1 atm after complete combustion. (Assume that temperatur e remain con-stant).
Sol.
5ml 10ml 2 2 10ml
1 C O O C O 2
¾ ¾®
V
f
=
2
CO
V
Volume of remaining air = 10 + 20 = 30 ml
Ex.2 A 3 L gas mixture of propane (C
3
H
8
) and butane (C
4
H
10
) on complete combustion at 25°C produced 10 L CO
2
. As suming c onstant P and T conditions what was v olume of butane presen t in in itial mixtur e?
Sol.
C
3
H
8
(g) + 5O
2
¾¾®
3CO
2
(g) + 4H
2
O(
l
)x L 3 x L C
4
H
10
(g) +
2
13 O(g)2
¾¾®
4CO
2
(g) + 5H
2
O(
l
)(3-x) L 4(3 – x) L from question 3 x + 4 (3 –x) = 10
Þ
x = 2
\
Volume of butane , C
4
H
10
= (3 – x) = 1 L
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u iometry
Ex.3 100 ml gase ous meta Xyle ne
CH
3
CH
3
undergoes combustion wit h excess of oxygen at room temperature and press ure. Volume contraction / expansion (in ml) during reaction is :
Sol.
C
4
H
10
(g) +
2
21 O(g)2
¾¾®
8CO
2
(g) + 5H
2
O(
l
)100 ml
21 100 2 1050ml
´=
800 ml 0
\
Conctr action in volum e = (100 + 1 050) – 800 = 350 ml
Ex.4 30 ml gas eou s mix ture of met han e and eth yle ne in vol ume ra tio X : Y re qui res 350 ml air con tain ing 20% of O
2
by volume for complete combustion. If ratio of methane and ethylene changed to Y : X. What will be volume of air (in ml) required for complete reaction under similar co ndition of temperature and pressure.
Sol.
CH
4
(g) + 2 O
2
(g)
¾¾®
CO
2
(g) + 2H
2
O (
l
)V
1
ml 2V
1
ml V
1
m l 0 C
2
H
4
(g) + 3O
2
(g)
¾¾®
2CO
2
(g) + 2H
2
O (
l
)V
2
m l 3 V
2
m l 2 V
2
ml 0 For given data : V
1
V
2
= 30 and 2V
1
3V
2
= 350 ×
20 70 100
=
\
V
1
= 2 0 , V
2
= 10 For required data : V
1
= 10 and V
2
= 20
\
V olume o f O
2
required = 2V
1
3V
2
= 80 m l and volum e of air required =
100 8 0 4 0 0 m l 20
= =
Ex.5 An alkene upon combus tion produces CO
2
(g) and H
2
O(g). In this combustion process if there is no vo lume chan ge occ urs th en the no. of C atoms per m olecule of alkene will be :
Sol.
C
n
H
2n
(g) + 3n 2 O
2
(g)
®
nCO
2
(g) + nH
2
O(g)if there no volume changes i.e.
D
ng
= 0(n + n) – 3n 1 0 2
æ ö+ =ç ÷è ø Þ
n = 2
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4
Ex.6 A gas eous hy drocarb on (C
x
H
y
) requir es 6 times of its own volu me of O
2
for complete oxidation and produces 4 time s of its volume of CO
2
. Find out the volume of x + y.
Sol.
C
X
H
y
+(x + y 4) O
2
¾¾®
XCO
2
+
2
y H O()2
l
V ol a
y a x 4
æ ö+ç ÷è ø
ax Gi ven tha t : a(x + y/4) = 6a vol of CO
2
= 4 v ol o f C
x
H
y
ax = 4 (a)x = 4...(2)fr om (1) x + y 6 4
=\
x + y = 4 + 8 = 12
Ex.7 On heat ing 60 ml mixtur e conta ining equ al vol ume of chlor ine gas and it's gase ous oxi de, vol ume becomes 75 ml due co mplete dec omposition of oxide. On treatment with KOH volume be comes 15 ml. What is the formula of oxide of chlorine ?
Sol.
Let o xide of Cl is Cl
x
O
y
So in 60 mL
Þ
30 mL Cl
x
O
y
and 30 mL Cl
2
.Now,Cl
x
O
y
¾®
2 2
x y Cl O 2 2
+
30mL 30.x mL 2 30.y mL 2 Giv en :75 = 30 +3 0 x 3 0 y 2 2
+
Þ
x + y = 3 ..........(i)KOH absorbs Cl
2
and volume becomes 15 mL so,(75 – 15) =
2
Cl
30x V 30 2
= +
Þ
x = 2 and y = 1 So the oxide : Cl
2
O
Ex.8 5 L of A (g) &3 L o f B(g) measur ed at same T & P are mixed togethe r which react as follows 2A(g) + B(g) C (g)What will be the total volume (in litre) after the completion of the reaction at same T & P.
Sol.
2A(g) + B(g)
¾¾®
C(g)5L 3L L.R. is A So, volume of C produced = 1 5 2
´
= 2.5 L and, volume of B reacted = 1 5 2
´
= 2.5 L So, volume fo B remained = 3 – 2.5 = 0.5 L Hence, V
total
= V
C
V
B
= 2.5 + 0.5 = 3 L
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9370 | https://www.droracle.ai/articles/31916/when-should-thrombolysis-be-given-in-cases-of-stemi | When should thrombolysis be administered in cases of ST-Elevation Myocardial Infarction (STEMI)?
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When should thrombolysis be administered in cases of ST-Elevation Myocardial Infarction (STEMI)?
Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board
Educational purpose only • Exercise caution as content is pending human review
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Last updated: March 14, 2025 • View editorial policy
From the Guidelines
Thrombolysis should be administered in cases of ST-Elevation Myocardial Infarction (STEMI) when primary percutaneous coronary intervention (PCI) cannot be performed within 120 minutes of first medical contact, and the onset of ischemic symptoms is within the previous 12 hours. The optimal timing for thrombolysis is within 12 hours of symptom onset, with the greatest benefit when administered within the first 3 hours 1. Common thrombolytic agents include alteplase (tPA) at 15 mg IV bolus, then 0.75 mg/kg over 30 minutes (maximum 50 mg), then 0.5 mg/kg over 60 minutes (maximum 35 mg); or tenecteplase (TNK) as a single weight-based bolus (30-50 mg) 1. Thrombolysis should be accompanied by dual antiplatelet therapy (aspirin 162-325 mg loading dose, then 81 mg daily, plus clopidogrel 300 mg loading dose, then 75 mg daily) and anticoagulation with unfractionated heparin or enoxaparin 1. Some key points to consider when administering thrombolysis include:
Absolute contraindications, such as prior intracranial hemorrhage, known cerebral vascular lesion, ischemic stroke within 3 months, active bleeding, or suspected aortic dissection 1
The need for transfer to a PCI-capable facility for either rescue PCI if thrombolysis fails (persistent chest pain, <50% ST resolution at 90 minutes) or routine angiography within 24 hours 1
The mechanism of action of thrombolytics, which activate plasminogen to plasmin, degrading fibrin clots blocking coronary arteries, restoring blood flow, and limiting myocardial damage 1. It is essential to weigh the benefits and risks of thrombolysis in each individual patient, considering factors such as the time from symptom onset, the presence of contraindications, and the anticipated delay to primary PCI 1.
From the FDA Drug Label
Initiate treatment as soon as possible after the onset of STEMI symptoms.
Thrombolysis with tenecteplase should be administered as soon as possible after the onset of ST-Elevation Myocardial Infarction (STEMI) symptoms.
The primary goal is to reduce the risk of death associated with acute STEMI.
Key considerations include: + Timing: Initiate treatment as soon as possible. + Administration: Tenecteplase is for intravenous administration only, administered as a single bolus over 5 seconds. + Dosage: Individualize dosage based on patient's weight. 2
From the Research
Thrombolysis Administration in STEMI Cases
Thrombolysis should be administered in cases of ST-Elevation Myocardial Infarction (STEMI) when primary percutaneous coronary intervention (PCI) is not possible within 120 minutes after first medical contact (FMC) 3, 4, 5.
The American Heart Association (AHA) and the American College of Cardiology (ACC) favor the use of pre-hospital thrombolysis (PHT) over PCI, placing emphasis on the time factor rather than the method of reperfusion 4.
If primary PCI is not feasible, thrombolysis must be initiated within 30 minutes after FMC, either in the emergency medical services (EMS) ambulance or in a nearby non-PCI hospital 3.
Thrombolysis is almost always delivered to patients after arriving in hospital, losing valuable time, and PHT is significantly superior to in-hospital thrombolysis (IHT) 4.
The National Institute for Clinical Excellence supports reperfusion with fibrinolytics, recommending PHT using newer agents such as reteplase and tenecteplase 4.
Time-Sensitive Thrombolysis
The efficacy of thrombolytic therapy in STEMI is highly time-dependent, with the best efficacy when given within the "golden hour" 6.
Initial patency, ST-segment resolution (STR) before PCI, and the incidence of aborted myocardial infarction gradually increase with shorter time from symptom onset to first medical contact 6.
Antiplatelet pretreatment before primary PCI, including a glycoprotein IIb/IIIa blocker, seems to be most effective when given shortly after symptom onset 6.
Combination Therapy
Thrombolytic therapy, even if "successful", is not the final therapy, and within 24 hours (but not before 3 hours) cardiac catheterization has to be performed with PCI, if applicable 3.
The combination of aspirin with a thienopyridine is mandatory for dual antiplatelet therapy (DAPT) 3.
Prasugrel is preferred over clopidogrel due to its faster onset of action and superior effectiveness 3.
References
1
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
2
Drug
Official FDA Drug Label For tenecteplase (IV)
FDA, 2025
3
Research
[Evidence-based management of ST-segment elevation myocardial infarction (STEMI). Latest guidelines of the European Society of Cardiology (ESC) 2010].
Herz, 2010
4
Research
Pre-hospital thrombolysis.
The Journal of the Association of Physicians of India, 2011
5
Research
Timely and optimal treatment of patients with STEMI.
Nature reviews. Cardiology, 2013
6
Research
The golden hour of prehospital reperfusion with triple antiplatelet therapy: a sub-analysis from the Ongoing Tirofiban in Myocardial Evaluation 2 (On-TIME 2) trial early initiation of triple antiplatelet therapy.
American heart journal, 2010
Related Questions
Should patients with a ST-Elevation Myocardial Infarction (STEMI) proceed directly to the cardiac catheterization lab or the emergency room (ER) first?What are the ST-Elevation Myocardial Infarction (STEMI) criteria?What is the management of ST-Elevation Myocardial Infarction (STEMI)?Should I perform thrombolysis (thrombolytic therapy) with this ECG indicating STEMI (ST-Elevation Myocardial Infarction)?What is the management approach for ST-Elevation Myocardial Infarction (STEMI)?What are examples of legumes?Can pioglitazone be taken with glipizide (Glucotrol, oral hypoglycemic agent)?What are the differences between primary and secondary literature resources in healthcare?Can Sitagliptin be taken with Glipizide?What is considered a delayed Percutaneous Coronary Intervention (PCI)?Below what estimated Glomerular Filtration Rate (eGFR) do patients require Calcitriol instead of cholecalciferol?
Professional Medical Disclaimer
This information is intended for healthcare professionals. Any medical decision-making should rely on clinical judgment and independently verified information. The content provided herein does not replace professional discretion and should be considered supplementary to established clinical guidelines. Healthcare providers should verify all information against primary literature and current practice standards before application in patient care. Dr.Oracle assumes no liability for clinical decisions based on this content.
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9371 | https://bstockus.wordpress.com/2015/11/14/830/ | Teaching to the Beat of a Different Drummer
My place to write about teaching, and math, and teaching math.
Is 1/2 always greater than 1/3?
Lately I’ve been reading the book Beyond Pizzas & Pies: 10 Essential Strategies for Supporting Fraction Sense by Julie McNamara and Meghan Shaughnessy.
I posted the following picture to Twitter while I read during my daughter’s swim class.
My colleague, Hedge, replied about being challenged by a middle school teacher on this very issue.
I let her know I was also challenged about this idea several years ago when I was a digital curriculum developer. The argument I heard back then was that using contexts to validate the correctness of fraction comparisons ran counter to the fact that fractions are numbers. As such, 1/2 is always greater than 1/3 regardless of the context. At the time, I wondered about it, but I still felt that bringing context to bear was important.
Flash forward to now and I have been mulling this idea over all day. I think I may finally understand why we have to be careful what we say about the role of context when comparing fractions. I may be completely off the mark, but I’m going to share my thoughts anyway and let you decide in the comments if you’d like to challenge my thinking or share an alternative point of view.
Let’s start with whole numbers. If I told you to compare 3 and 6, you would probably tell me, “3 is less than 6,” or, “6 is greater than 3.” That is how the numbers 3 and 6 are related.
Now, what if I were to show you these two pictures of 3 and 6: (As illustrated by my daughter’s toys.)
Three large dolls
Six small figurines
Technically, the 3 dolls are larger and therefore they amount to more stuff, but does that really mean 3 is now greater than 6? In the end, the number of dolls my daughter has (3) is less than the number of figurines she has (6). The context doesn’t fundamentally change the relationship between the numbers 3 and 6.
In this case, I don’t even know how I’d justify that she has more when referring to the dolls. Sure, they’re bigger, but she may prefer to have more things to play with and choose the 6 figurines even though they are less in total size.
Let’s continue by looking at this from a fraction perspective. Now I’m going to take 1/3 of the dolls and 1/2 of the figurines.
1/3 of the dolls is 1 doll
1/2 of the figurines is 3 figurines
In keeping with the idea that context should dictate when one number is greater than another, I should be convinced that 1/3 of the dolls is greater than 1/2 of the figurines because 1 doll is so much larger than the 3 figurines. Oh wait, or is it that I should be thinking that 1/2 of the figurines is greater than 1/3 of the dolls because I end up with 3 figurines which is a greater number of things than 1 doll? It’s not so clear cut, even though I’m trying to let the context dictate how to interpret the fractions.
What it boils down to is that fractions represent a relationship. If I think about the relationships each fraction represents, then 1/2 is always greater than 1/3 no matter how I try to spin it. Looking back at my examples, taking 1/2 of the group of figurines means I am taking a greater share of that group (that whole) than when I take 1/3 of the group of dolls (a different whole, but a whole nonetheless). The size of the things in my group (whole) doesn’t matter because the relationship represented by 1/2 is greater than the relationship represented by 1/3.
Now, does that mean we should ignore contexts altogether? No. There are still rich conversations to be had about who ate more pizza when one person eats half of a small pizza and another person eats a third of a large pizza. Context is still interesting to discuss and helps students use math to interpret the world around them. However, if our goal is to compare fractions, then 1/2 is greater than 1/3 every time.
That’s the argument I came up with today as I tried to understand the criticisms I’ve heard. Now that you’ve read it, what do you think?
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14 thoughts on “Is 1/2 always greater than 1/3?”
This conversation is something that always arises in my 5th grade math class every time after I do the lesson in Investigations about different sized wholes. It would even sneak its way into our addition problems. For example, when we would do a number talk like 1/2 + 3/4, they would say, “It would be 1 14/ IF they are the same sized wholes.” I do love when they did that, however it always had me thinking about the same ideas you talked about here but instead of comparison, it was the operations. I loved they way you wrote about comparison Brian, and I don’t know if my thinking is even correct, but it is something I play around with.
I thought about it like this (and talked with the students about it)….when I add naked numbers, without a context, such as 6+3, I am assuming they are talking about the same unit, right? I am assuming we are not talking about 6 feet + 3 inches or 6 Hershey Kisses + 3 Hershey bars….or am I? How does the answer of 9 relate in those situations? 9 units of measure versus “we can’t because they are not the same unit” or 9 pieces of candy versus “we can’t because they are not the same unit. So, not thinking about fractions I think it goes back to what you were talking about….1/2 + 3/4….1/2 of a large pizza + 3/4 of a small pizza = 1 1/4 collection of pizzas versus “we can’t because they not the same sized whole.” Does that hold up there?
Either way, I love that students ask these questions and push us to think about this! To the teacher Hedge is referring to, I would be saddened to think that teachers are not teaching something because it is confusing for students on a the test…these conversations are so valuable and rich in student thinking!
~Kristin
This whole line of questioning is fascinating to me and makes me think of categorical variables and their role in equivalence relations. The toys were compared using two different categories of attributes/properties: size and amount/number. We often think of equivalence as the realm of numbers and shapes, but the really super fun math conversations (to me at least) are when you get to talk about what exactly it is you’re comparing to determine similarity. So, for example, Hershey kisses and chocolate bars could be the same unit if you agree that they are all “candy”. Or, the dolls and the figurines could be categorized as “toys” and you could compare from there. (And, just so you know the depth of my love for comparison, when I was looking at the pictures I saw way more sameness than difference between the dolls/figurines…fun times! 🙂
–malke
Great minds think alike! I was having similar thoughts even as I was writing my post. As long as I could adjust the categories to match, then I could justify my comparison. It’s interesting because this makes sense to students but also adds complexity to their work. They have to learn to negotiate these categories for themselves to understand when and how they can compare or operate on quantities. Sometimes teachers will say things like, “You can’t add apples and pens, that wouldn’t make sense,” but you can if you think of them more generally as objects or things. If the teacher’s logic were universal, then I wouldn’t be able to add boys and girls, but I know in reality I can because they’re all children.
I like that your students have had enough experiences thinking about the whole to bring that up when adding fractions. It sounds like they’re proof that getting students to pay attention to the whole doesn’t confuse them to such a degree that they can’t operate on or compare the fractions.
I also like what you said about assuming the whole is the same in a bare naked math problem. It’s a given in that case, but in a real world context we can’t take that for granted. It reminds me of students learning about how to interpret remainders. It can be confusing at first because you have to understand what the question is asking in order to determine what role a remainder might play in your answer. But that confusion isn’t a reason to shy away from the topic. Rather it deepens our understanding of division and how we use it in everyday life.
I think you’re comparison with the number of dolls is really helpful. And I think if you were going to ask this kind of question, it would be best to approach it with the simpler question: Is 6 always bigger than 3?
It’s not about fractions really is it? It’s about that word “of”. Three of the dolls. Which is to say,
3 X doll.
I think the question is worth asking, so that students can think about this kind of thing, but I’d start with numbers of the toys first, before fractions.
your
Great advice! I especially like it because this isn’t an issue unique to fractions. As you’re pointing out, it matters anytime we’re comparing quantities of things. I wonder if students have any misunderstandings related to this idea with whole numbers or if it’s something that doesn’t create some possible confusion until fractions. For example, the common example given for paying attention to the whole when comparing fractions is pizzas. I wonder what students would say if you asked them who has more if I have 6 small pizzas and you have 3 large pizzas. In that context with whole numbers is it somehow more obvious that the size of the pizza matters and that 6 is not necessarily greater than 3?
1,2,3,4,5,….. are counting numbers
So there is an implied “thing” of which we have several examples, say apples.
Then 3+6 is a statement about two piles of apples, and counting the whole lot gets you to 9.
3+6=9 is always a shorthand for “three of these together with 6 of these gives you 9 of these”
Context is not just important, it is VITAL.
When we attempt to measure stuff we find that with the chosen unit we may have more than 3 and less than 4 units of stuff, and hence the need for a way to describe “parts of a unit”. This of course leads to fractions, which are essentially measuring numbers. We have NOT extended the number system, we have created a new number system, in which we can talk about “parts”.
So the question “Is a half bigger than a third” is a shorthand for a measurement comparison, such as “with this stick as the unit of length, which of those two sticks is longer?”
I totally agree with Simon Gregg about language, especially “of”. We overlook meaning too easily in math.
A problem adapted from Connected Math Project 2 (Lappan et al., 2009):
“Jane and Don’s mathematics classes are selling sub sandwiches as a fund-raiser. Jane’s class has reached 2/3 of their goal and Don’s class has reached 3/4 of their goal. Jane says her class has collected more money than Don’s class.”
“How could Jane be right?”
“How could Jane be wrong?”
I love this problem and think it might throw a monkey wrench in your last couple of paragraphs…
Thanks for sharing this problem. I love it! 🙂
As I was getting ready to publish my post last night, I got to thinking that the questions we ask is really the interesting part, and the problem you shared is a great example of that. If the question had been, “Which class is closer to its goal?” then 3/4 could be the greater fraction because that class is closer to it goal, whatever it may be. Being 1/4 away from your goal is closer than being 1/3 away from your goal.
However, that may not necessarily be the case in absolute terms because if Don’s class is $100 away from its goal but Jane’s class is $25 away from its goal, then isn’t Jane’s class closer to its goal?
This makes me think of the types of questions we ask to see if students are developing proportional reasoning. For example, Tree A grew so many feet this year and Tree B grew so many feet this year. Which tree grew the most? You can talk about it in terms of how many more feet each tree grew, but that thinking isn’t taking into account growth related to the previous year’s height.
I feel like I’m rambling, but I’m thinking through this as I write. So, if we’re wanting students to be pushed to think proportionally, maybe I would want them to say that Don’t class is closer to its goal regardless of the actual amount of money separating his class from its goal compared to Jane’s class. So in that case, 3/4 would be greater than 2/3 regardless of the amounts of money involved.
That being said, the problem you shared asked a different question about who has raised more money. This shifts the question away from comparing the fractions to one of using the fractions to compare another quantity, the amount of money raised by each class.
Also, thinking about your dolls problem, what happens if you switch them around? Take the larger fraction of the smaller group and vice versa? 1/2 of 3 vs 1/3 of 6?
Following the logic I was trying to use, does it change anything? If I take 1/2 of the smaller group, I’m still taking more of that group related to its whole. Yes, the specific amount I’m taking (1.5 dolls) is less things than taking 1/3 of the other group (2 figurines) but the relationship involved, taking half of a group, still means I’m taking more of that particular group than taking a third of a group does. I think this gets back to what question is being asked and how that impacts how we need/want to interpret the situation.
“How could Jane be right?”
“How could Jane be wrong?”
Keeping the feet on the ground, a fraction is a number, but “a fraction of ..” is a quantity and has units. Since the two goals have not been quantified Jane could be right or wrong. Once we know how big (in money terms) the goals are then all will be revealed. So it all goes back to measurement.
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What is the derivative of 1/sinx+cosx?
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Ernest Leung
B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·Jun 18
Suppose the question is:
"What is the derivative of 1/(sinx + cosx)?”
The answer is as follows.
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Suppose the question is:
"What is the derivative of 1/(sinx + cosx)?”
The answer is as follows.
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What is the derivative of 1/sinx?
What is the derivative of 1-sinx/1+sinx?
What is the derivative of 1-cosx/1+cosx?
What is the derivative of y= ln (cosx / cosx -1)?
What is the derivative of (cosx-sinx) /√ (1-sin2x)?
Mike Hirschhorn
Honorary Associate Professor of Mathematics at UNSW · Author has 8.1K answers and 2.7M answer views
·Updated 2y
d d x(1 u)=−1 u 2 d u d x,d d x(1 u)=−1 u 2 d u d x,
d d x(1 sin x+cos x)=−cos x−sin x(sin x+cos x)2.d d x(1 sinx+cosx)=−cosx−sinx(sinx+cosx)2.
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9 2
Yikin Cheung
Passenger Service Agent at Swissport (2019–present) · Author has 136 answers and 236.3K answer views
·6y
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What is the integration of 1/ (sinx+ cosx)?
Our goal is to solve for P in this differential equation. To achieve that, we need put all factors with P to one side and all factors with x to the others. When you finish with that, you can get:
So, the original expression is equal to:
Answer:
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Our goal is to solve for P in this differential equation. To achieve that, we need put all factors with P to one side and all factors with x to the others. When you finish with that, you can get:
So, the original expression is equal to:
Answer:
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9 1
Yedukondalu Ye
Studied at Sri Vikas Model High School (Graduated 2024)
·1y
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What is the derivative of sin (x)+cos(x) +1/x?
f(x)=sin(x)+cos(x)+1/x
d/dx ( sin(x)+cos(x)+1/x)
d/dx(sin(x)+d/dx(cos(x+)d/dx(x^-1) >>(1/a=a^-1)
d/dx(sin(x)=cos(x)
d/dx(cos(x)=-sin(x)
d/dx(x^-1=-1x^-1–=-x^-2. by power rule X^n=nx^n-1
f'(x) =d/dx(sin(x))+d/dx(cos(x))+d/dx(x^-1)=
cos(x)-sin(x)-1/x^2
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Lai Johnny
M. Phil in Mathematics Major, The Chinese University of Hong Kong (Graduated 1985) · Author has 5.8K answers and 11.7M answer views
·4y
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What is the integration of 1/ (sinx+ cosx)?
∫1 sin x+cos x d x∫1 sinx+cosx d x
=1√2∫1 cos(x−π 4)d x=1 2∫1 cos(x−π 4)d x
=1√2∫sec(x−π 4)d x=1 2∫sec(x−π 4)d x
=1√2 ln∣∣sec(x−π 4)+tan(x−π 4)∣∣+C=1 2 ln|sec(x−π 4)+tan(x−π 4)|+C
=1√2 ln∣∣∣1 sin x+cos x+tan x−1 1+tan x∣∣∣+C=1 2 ln|1 sinx+cosx+tanx−1 1+tanx|+C
=1√2 ln∣∣∣1 sin x+cos x+sin x−cos x cos x+sin x∣∣∣+C=1 2 ln|1 sinx+cosx+sinx−cosx cosx+sinx|+C
\displaystyle=\frac{1}{\sqrt{2\displaystyle=\frac{1}{\sqrt{2
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∫1 sin x+cos x d x∫1 sinx+cosx d x
=1√2∫1 cos(x−π 4)d x=1 2∫1 cos(x−π 4)d x
=1√2∫sec(x−π 4)d x=1 2∫sec(x−π 4)d x
=1√2 ln∣∣sec(x−π 4)+tan(x−π 4)∣∣+C=1 2 ln|sec(x−π 4)+tan(x−π 4)|+C
=1√2 ln∣∣∣1 sin x+cos x+tan x−1 1+tan x∣∣∣+C=1 2 ln|1 sinx+cosx+tanx−1 1+tanx|+C
=1√2 ln∣∣∣1 sin x+cos x+sin x−cos x cos x+sin x∣∣∣+C=1 2 ln|1 sinx+cosx+sinx−cosx cosx+sinx|+C
=1√2 ln∣∣∣1+sin x−cos x sin x+cos x∣∣∣+C=1 2 ln|1+sinx−cosx sinx+cosx|+C
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983) · Author has 3.9K answers and 4M answer views
·4y
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What is 1-sinx/cosx + cosx/1-sinx ?
Because no parentheses were used, the given expression is not clear. I will interpret two different ways.
First, assuming that this expression is:
1−sin(x)cos(x)+cos(x)1−sin(x)1−sin(x)cos(x)+cos(x)1−sin(x)
The last term, if you multiply the numerator and denominator by 1+sin(x)1+sin(x) becomes:
1+sin(x)cos(x)1+sin(x)cos(x)
after simplifying both the numerator and denominator with cancellations.
Then, adding all terms together and simplifying, we get:
1+1 cos(x)1+1 cos(x)
or
1+sec(x)1+sec(x)
which is equivalent to the original expression.
Second, if the intended expression was:
\frac{1-\sin{(x)}}{\cos{(x)}}+\frac\frac{1-\sin{(x)}}{\cos{(x)}}+\frac
Continue Reading
Because no parentheses were used, the given expression is not clear. I will interpret two different ways.
First, assuming that this expression is:
1−sin(x)cos(x)+cos(x)1−sin(x)1−sin(x)cos(x)+cos(x)1−sin(x)
The last term, if you multiply the numerator and denominator by 1+sin(x)1+sin(x) becomes:
1+sin(x)cos(x)1+sin(x)cos(x)
after simplifying both the numerator and denominator with cancellations.
Then, adding all terms together and simplifying, we get:
1+1 cos(x)1+1 cos(x)
or
1+sec(x)1+sec(x)
which is equivalent to the original expression.
Second, if the intended expression was:
1−sin(x)cos(x)+cos(x)1−sin(x)1−sin(x)cos(x)+cos(x)1−sin(x)
the analysis is the same, but the result is:
2 sec(x)2 sec(x)
I really wish people would learn to use parentheses.
Upvote ·
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
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What is the derivative of sinx cosx?
This is very interesting because it can be done in two simple ways!
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This is very interesting because it can be done in two simple ways!
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Mohammad Afzaal Butt
B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views
·6y
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What is the integration of 1/ (sinx+cosx)?
Let z=tan(x 2)Let z=tan(x 2)
d x=2 d z 1+z 2 sin x=2 z 1+z 2 1−z 2 1+z 2 d x=2 d z 1+z 2 sin x=2 z 1+z 2 1−z 2 1+z 2
∫1 sin x+cos x d x∫1 sin x+cos x d x
=∫1 2 z 1+z 2+1−z 2 1+z 2 2 d z 1+z 2=∫1 2 z 1+z 2+1−z 2 1+z 2 2 d z 1+z 2
=∫2 2 z+1−z 2 d z=∫2 2 z+1−z 2 d z
=∫2 2−(z 2−2 z+1)d z=∫2 2−(z 2−2 z+1)d z
=∫2(√2)2−(z−1)2 d z=∫2(2)2−(z−1)2 d z
[Math Processing Error]=\dfrac{2}{2\sqrt{2}}\ln\left|\dfrac{z-1+\sqrt{2}}{z-1-\sqrt{2}}\right|+C\quad\because\displaystyle\int\dfrac{du}{a^2-u^2}=\dfrac{
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Let z=tan(x 2)Let z=tan(x 2)
d x=2 d z 1+z 2 sin x=2 z 1+z 2 1−z 2 1+z 2 d x=2 d z 1+z 2 sin x=2 z 1+z 2 1−z 2 1+z 2
∫1 sin x+cos x d x∫1 sin x+cos x d x
=∫1 2 z 1+z 2+1−z 2 1+z 2 2 d z 1+z 2=∫1 2 z 1+z 2+1−z 2 1+z 2 2 d z 1+z 2
=∫2 2 z+1−z 2 d z=∫2 2 z+1−z 2 d z
=∫2 2−(z 2−2 z+1)d z=∫2 2−(z 2−2 z+1)d z
=∫2(√2)2−(z−1)2 d z=∫2(2)2−(z−1)2 d z
=2 2√2 ln∣∣∣z−1+√2 z−1−√2∣∣∣+C∵∫d u a 2−u 2=1 2 a ln∣∣∣u+a u−a∣∣∣+C=2 2 2 ln|z−1+2 z−1−2|+C∵∫d u a 2−u 2=1 2 a ln|u+a u−a|+C
=1√2 ln∣∣ ∣ ∣∣tan(x 2)−1+√2 tan(x 2)−1−√2∣∣ ∣ ∣∣+C=1 2 ln|tan(x 2)−1+2 tan(x 2)−1−2|+C
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9 2
Sabaa Nasim
5y
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What is 1/1+sinx+cosx dx?
All i could understand is that, you have asked for the integration of this expression.
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All i could understand is that, you have asked for the integration of this expression.
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9 1
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Aditya
Electrical Engineer at Power Grid Corporation of India Limited (2018–present)
·7y
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What is the integration of 1/ (sinx+ cosx)?
Updated answer- Take constant of integration as log C or something.
Continue Reading
Updated answer- Take constant of integration as log C or something.
Upvote ·
9 1
9 3
George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·2y
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What is the derivative of 1/1+cosx?
What you WROTE, 1/1+ cosx= 1+ cos(x) has derivative -sin(x).
But I feel sure that you MEAN 1/(1+ cos(x)).
Write that as (1+ cos(x))^{-1} and use the fact that the derivative of y^n, with respect to y, is ny^(n-1) and the chain rule. Let y= 1+ cos(x). Then d y−1 d y=(−1)y−1–1=−y−2 d y−1 d y=(−1)y−1–1=−y−2 and d y d x=−s i n(x)d y d x=−s i n(x).
By the chain rule, [Math Processing Error]\frac{d(1/(1+cos(x))}{dx}=\frac{d y^{-1}}{dy}\frac{dy}{dx}=(-y^{-2})(-sin(x))=\frac{sin(x)}{(1+cos(x))^2
Upvote ·
Mohammad Afzaal Butt
B.Sc in Mathematics&Physics, Islamia College Gujranwala (Graduated 1977) · Author has 24.6K answers and 22.9M answer views
·2y
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What is the derivative of 1/1+cosx?
d d x(1 1+cos x)d d x(1 1+cosx)
=d d x(1+cos x)−1=d d x(1+cosx)−1
=−(1+cos x)−2(−sin x)=−(1+cosx)−2(−sinx)
=sin x(1+cos x)2=sinx(1+cosx)2
Upvote ·
9 2
9 1
Andrew Droffner
Studied Mathematics at Rutgers University (Graduated 1995) · Author has 8.8K answers and 5.7M answer views
·3y
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What is the derivative of 1/sinx?
Use implicit differentiation to find the derivative of the equation (1). Put the sin(x)sin(x) on the same side as y y to start.
y=1 sin(x)(1)(1)y=1 sin(x)
sin(x)⋅y=1 sin(x)⋅y=1
Compute the left-side derivative by the product rule. This is an implicit differentiation.
sin(x)⋅y=1 sin(x)⋅y=1
d d x(sin(x))⋅y+sin(x)⋅d y d x=0 d d x(sin(x))⋅y+sin(x)⋅d y d x=0
cos(x)⋅y+sin(x)⋅d y d x=0 cos(x)⋅y+sin(x)⋅d y d x=0
Substitute equation (1) for y y and isolate the first derivative d y d x d y d x.
cos(x)⋅1 sin(x)+sin(x)⋅d y d x=0 cos(x)⋅1 sin(x)+sin(x)⋅d y d x=0
cos(x)⋅1 sin(x)=(−1)sin(x)⋅d y d x cos(x)⋅1 sin(x)=(−1)sin(x)⋅d y d x
(−1(−1
Continue Reading
Use implicit differentiation to find the derivative of the equation (1). Put the sin(x)sin(x) on the same side as y y to start.
y=1 sin(x)(1)(1)y=1 sin(x)
sin(x)⋅y=1 sin(x)⋅y=1
Compute the left-side derivative by the product rule. This is an implicit differentiation.
sin(x)⋅y=1 sin(x)⋅y=1
d d x(sin(x))⋅y+sin(x)⋅d y d x=0 d d x(sin(x))⋅y+sin(x)⋅d y d x=0
cos(x)⋅y+sin(x)⋅d y d x=0 cos(x)⋅y+sin(x)⋅d y d x=0
Substitute equation (1) for y y and isolate the first derivative d y d x d y d x.
cos(x)⋅1 sin(x)+sin(x)⋅d y d x=0 cos(x)⋅1 sin(x)+sin(x)⋅d y d x=0
cos(x)⋅1 sin(x)=(−1)sin(x)⋅d y d x cos(x)⋅1 sin(x)=(−1)sin(x)⋅d y d x
(−1)cos(x)sin 2(x)=d y d x(−1)cos(x)sin 2(x)=d y d x
Use trigonometric identities to simplify this.
d y d x=(−1)cos(x)sin 2(x)d y d x=(−1)cos(x)sin 2(x)
d y d x=(−1)cos(x)sin(x)⋅1 sin(x)d y d x=(−1)cos(x)sin(x)⋅1 sin(x)
Recall cot(x)=cos(x)sin(x)cot(x)=cos(x)sin(x) and csc(x)=1 sin(x)csc(x)=1 sin(x).
Answer
d y d x=(−1)cot(x)⋅csc(x)d y d x=(−1)cot(x)⋅csc(x)
Upvote ·
9 3
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9373 | http://www.mysmu.edu/faculty/yktse/fma/s_fma_2.pdf | Financial Mathematics for Actuaries Chapter 2 Annuities Learning Objectives 1. Annuity-immediate and annuity-due 2. Present and future values of annuities 3. Perpetuities and deferred annuities 4. Other accumulation methods 5. Payment periods and compounding periods 6. Varying annuities 2 2.1 Annuity-Immediate • Consider an annuity with payments of 1 unit each, made at the end of every year for n years.
• This kind of annuity is called an annuity-immediate (also called an ordinary annuity or an annuity in arrears).
• The present value of an annuity is the sum of the present values of each payment.
Example 2.1: Calculate the present value of an annuity-immediate of amount $100 paid annually for 5 years at the rate of interest of 9%.
Solution: Table 2.1 summarizes the present values of the payments as well as their total.
3 Table 2.1: Present value of annuity Year Payment ($) Present value ($) 1 100 100 (1.09)−1 = 91.74 2 100 100 (1.09)−2 = 84.17 3 100 100 (1.09)−3 = 77.22 4 100 100 (1.09)−4 = 70.84 5 100 100 (1.09)−5 = 64.99 Total 388.97 2 • We are interested in the value of the annuity at time 0, called the present value, and the accumulated value of the annuity at time n, called the future value.
4 • Suppose the rate of interest per period is i, and we assume the compound-interest method applies.
• Let anei denote the present value of the annuity, which is sometimes denoted as ane when the rate of interest is understood.
• As the present value of the jth payment is vj, where v = 1/(1+i) is the discount factor, the present value of the annuity is (see Appendix A.5 for the sum of a geometric progression) ane = v + v2 + v3 + · · · + vn = v × ∙1 −vn 1 −v ¸ = 1 −vn i = 1 −(1 + i)−n i .
(2.1) 5 • The accumulated value of the annuity at time n is denoted by snei or sne.
• This is the future value of ane at time n. Thus, we have sne = ane × (1 + i)n = (1 + i)n −1 i .
(2.2) • If the annuity is of level payments of P, the present and future values of the annuity are Pane and Psne, respectively.
Example 2.2: Calculate the present value of an annuity-immediate of amount $100 paid annually for 5 years at the rate of interest of 9% using formula (2.1). Also calculate its future value at time 5.
6 Solution: From (2.1), the present value of the annuity is 100 a5e = 100 × "1 −(1.09)−5 0.09 # = $388.97, which agrees with the solution of Example 2.1. The future value of the annuity is (1.09)5 × (100 a5e) = (1.09)5 × 388.97 = $598.47.
Alternatively, the future value can be calculated as 100 s5e = 100 × "(1.09)5 −1 0.09 # = $598.47.
2 Example 2.3: Calculate the present value of an annuity-immediate of amount $100 payable quarterly for 10 years at the annual rate of interest 7 of 8% convertible quarterly. Also calculate its future value at the end of 10 years.
Solution: Note that the rate of interest per payment period (quarter) is (8/4)% = 2%, and there are 4 × 10 = 40 payments. Thus, from (2.1) the present value of the annuity-immediate is 100 a40e0.02 = 100 × "1 −(1.02)−40 0.02 # = $2,735.55, and the future value of the annuity-immediate is 2735.55 × (1.02)40 = $6,040.20.
2 • A common problem in financial management is to determine the in-stallments required to pay back a loan. We may use (2.1) to calculate the amount of level installments required.
8 Example 2.4: A man borrows a loan of $20,000 to purchase a car at annual rate of interest of 6%. He will pay back the loan through monthly installments over 5 years, with the first installment to be made one month after the release of the loan. What is the monthly installment he needs to pay?
Solution: The rate of interest per payment period is (6/12)% = 0.5%.
Let P be the monthly installment. As there are 5 × 12 = 60 payments, from (2.1) we have 20,000 = P a60e0.005 = P × "1 −(1.005)−60 0.005 # = P × 51.7256, 9 so that P = 20,000 51.7256 = $386.66.
2 • The example below illustrates the calculation of the required install-ment for a targeted future value.
Example 2.5: A man wants to save $100,000 to pay for his son’s education in 10 years’ time. An education fund requires the investors to deposit equal installments annually at the end of each year. If interest of 7.5% is paid, how much does the man need to save each year in order to meet his target?
Solution: We first calculate s10e, which is equal to (1.075)10 −1 0.075 = 14.1471.
10 Then the required amount of installment is P = 100,000 s10e = 100,000 14.1471 = $7,068.59.
2 11 2.2 Annuity-Due • An annuity-due is an annuity for which the payments are made at the beginning of the payment periods • The first payment is made at time 0, and the last payment is made at time n −1.
• We denote the present value of the annuity-due at time 0 by ¨ anei (or ¨ ane), and the future value of the annuity at time n by ¨ snei (or ¨ sne).
• The formula for ¨ ane can be derived as follows ¨ ane = 1 + v + · · · + vn−1 = 1 −vn 1 −v = 1 −vn d .
(2.3) 12 • Also, we have ¨ sne = ¨ ane × (1 + i)n = (1 + i)n −1 d .
(2.4) • As each payment in an annuity-due is paid one period ahead of the corresponding payment of an annuity-immediate, the present value of each payment in an annuity-due is (1+i) times of the present value of the corresponding payment in an annuity-immediate. Hence, ¨ ane = (1 + i) ane (2.5) and, similarly, ¨ sne = (1 + i) sne.
(2.6) 13 • As an annuity-due of n payments consists of a payment at time 0 and an annuity-immediate of n −1 payments, the first payment of which is to be made at time 1, we have ¨ ane = 1 + an−1e.
(2.7) • Similarly, if we consider an annuity-immediate with n + 1 payments at time 1, 2, · · ·, n + 1 as an annuity-due of n payments starting at time 1 plus a final payment at time n + 1, we can conclude sn+1e = ¨ sne + 1.
(2.8) Example 2.6: A company wants to provide a retirement plan for an employee who is aged 55 now. The plan will provide her with an annuity-immediate of $7,000 every year for 15 years upon her retirement at the 14 age of 65. The company is funding this plan with an annuity-due of 10 years. If the rate of interest is 5%, what is the amount of installment the company should pay?
Solution: We first calculate the present value of the retirement annuity.
This is equal to 7,000 a15e = 7,000 × "1 −(1.05)−15 0.05 # = $72,657.61.
This amount should be equal to the future value of the company’s install-ments P, which is P ¨ s10e. Now from (2.4), we have ¨ s10e = (1.05)10 −1 1 −(1.05)−1 = 13.2068, so that P = 72,657.61 13.2068 = $5,501.53.
15 2.3 Perpetuity, Deferred Annuity and Annuity Values at Other Times • A perpetuity is an annuity with no termination date, i.e., n →∞.
• An example that resembles a perpetuity is the dividends of a pre-ferred stock.
• To calculate the present value of a perpetuity, we note that, as v < 1, vn →0 as n →∞. Thus, from (2.1), we have a∞e = 1 i .
(2.9) • For the case when the first payment is made immediately, we have, from (2.3), ¨ a∞e = 1 d.
(2.10) 16 • A deferred annuity is one for which the first payment starts some time in the future.
• Consider an annuity with n unit payments for which the first pay-ment is due at time m + 1.
• This can be regarded as an n-period annuity-immediate to start at time m, and its present value is denoted by m|anei (or m|ane for short).
Thus, we have m|ane = vm ane = vm × ∙1 −vn i ¸ = vm −vm+n i = (1 −vm+n) −(1 −vm) i 17 = am+ne −ame.
(2.11) • To understand the above equation, see Figure 2.3.
• From (2.11), we have am+ne = ame + vm ane = ane + vn ame.
(2.12) • Multiplying the above equations throughout by 1 + i, we have ¨ am+ne = ¨ ame + vm ¨ ane = ¨ ane + vn ¨ ame.
(2.13) • We also denote vm ¨ ane as m|¨ ane, which is the present value of a n-payment annuity of unit amounts due at time m, m+1, · · · , m+n−1.
18 • If we multiply the equations in (2.12) throughout by (1 + i)m+n, we obtain sm+ne = (1 + i)n sme + sne = (1 + i)m sne + sme.
(2.14) • See Figure 2.4 for illustration.
• It is also straightforward to see that ¨ sm+ne = (1 + i)n ¨ sme + ¨ sne = (1 + i)m ¨ sne + ¨ sme.
(2.15) • We now return to (2.2) and write it as sm+ne = (1 + i)m+n am+ne, 19 so that vm sm+ne = (1 + i)n am+ne, (2.16) for arbitrary positive integers m and n.
• How do you interpret this equation?
20 2.4 Annuities Under Other Accumulation Methods • We have so far discussed the calculations of the present and future values of annuities assuming compound interest.
• We now extend our discussion to other interest-accumulation meth-ods.
• We consider a general accumulation function a(·) and assume that the function applies to any cash-flow transactions in the future.
• As stated in Section 1.7, any payment at time t > 0 starts to accu-mulate interest according to a(·) as a payment made at time 0.
• Given the accumulation function a(·), the present value of a unit payment due at time t is 1/a(t), so that the present value of a n-21 period annuity-immediate of unit payments is ane = n X t=1 1 a(t).
(2.17) • The future value at time n of a unit payment at time t < n is a(n −t), so that the future value of a n-period annuity-immediate of unit payments is sne = n X t=1 a(n −t).
(2.18) • If (1.35) is satisfied so that a(n −t) = a(n)/a(t) for n > t > 0, then sne = n X t=1 a(n) a(t) = a(n) n X t=1 1 a(t) = a(n) ane.
(2.19) • This result is satisfied for the compound-interest method, but not the simple-interest method or other accumulation schemes for which equation (1.35) does not hold.
22 Example 2.7: Suppose δ(t) = 0.02t for 0 ≤t ≤5, find a5e and s5e.
Solution: We first calculate a(t), which, from (1.26), is a(t) = exp µZ t 0 0.02s ds ¶ = exp(0.01t2).
Hence, from (2.17), a5e = 1 e0.01 + 1 e0.04 + 1 e0.09 + 1 e0.16 + 1 e0.25 = 4.4957, and, from (2.18), s5e = 1 + e0.01 + e0.04 + e0.09 + e0.16 = 5.3185.
Note that a(5) = e0.25 = 1.2840, so that a(5) a5e = 1.2840 × 4.4957 = 5.7724 6= s5e.
23 2 • Note that in the above example, a(n −t) = exp[0.01(n −t)2] and a(n) a(t) = exp[0.01(n2 −t2)], so that a(n −t) 6= a(n)/a(t) and (2.19) does not hold.
Example 2.8: Calculate a3e and s3e if the nominal rate of interest is 5% per annum, assuming (a) compound interest, and (b) simple interest.
Solution: (a) Assuming compound interest, we have a3e = 1 −(1.05)−3 0.05 = 2.723, and s3e = (1.05)3 × 2.72 = 3.153.
24 (b) For simple interest, the present value is a3e = 3 X t=1 1 a(t) = 3 X t=1 1 1 + rt = 1 1.05 + 1 1.1 + 1 1.15 = 2.731, and the future value at time 3 is s3e = 3 X t=1 a(3 −t) = 3 X t=1 (1 + r(3 −t)) = 1.10 + 1.05 + 1.0 = 3.150.
At the same nominal rate of interest, the compound-interest method gen-erates higher interest than the simple-interest method.
Therefore, the future value under the compound-interest method is higher, while its present value is lower. Also, note that for the simple-interest method, a(3) a3e = 1.15 × 2.731 = 3.141, which is different from s3e = 3.150.
2 25 2.5 Payment Periods, Compounding Periods and Continuous Annuities • We now consider the case where the payment period differs from the interest-conversion period.
Example 2.9: Find the present value of an annuity-due of $200 per quarter for 2 years, if interest is compounded monthly at the nominal rate of 8%.
Solution: This is the situation where the payments are made less frequently than interest is converted. We first calculate the effective rate of interest per quarter, which is ∙ 1 + 0.08 12 ¸3 −1 = 2.01%.
26 As there are n = 8 payments, the required present value is 200 ¨ a8e0.0201 = 200 × "1 −(1.0201)−8 1 −(1.0201)−1 # = $1,493.90.
2 Example 2.10: Find the present value of an annuity-immediate of $100 per quarter for 4 years, if interest is compounded semiannually at the nominal rate of 6%.
Solution: This is the situation where payments are made more fre-quently than interest is converted. We first calculate the effective rate of interest per quarter, which is ∙ 1 + 0.06 2 ¸ 1 2 −1 = 1.49%.
27 Thus, the required present value is 100 a16e0.0149 = 100 × "1 −(1.0149)−16 0.0149 # = $1,414.27.
2 • It is possible to derive algebraic formulas to compute the present and future values of annuities for which the period of installment is different from the period of compounding.
• We first consider the case where payments are made less frequently than interest conversion, which occurs at time 1, 2, · · ·, etc.
• Let i denote the effective rate of interest per interest-conversion pe-riod. Suppose a m-payment annuity-immediate consists of unit pay-ments at time k, 2k, · · ·, mk. We denote n = mk, which is the number of interest-conversion periods for the annuity.
28 • Figure 2.6 illustrates the cash flows for the case of k = 2.
• The present value of the above annuity-immediate is (we let w = vk) vk + v2k + · · · + vmk = w + w2 + · · · + wm = w × ∙1 −wm 1 −w ¸ = vk × ∙1 −vn 1 −vk ¸ = 1 −vn (1 + i)k −1 = ane ske , (2.20) and the future value of the annuity is (1 + i)n ane ske = sne ske .
(2.21) 29 • We now consider the case where the payments are made more fre-quently than interest conversion.
• Let there be mn payments for an annuity-immediate occurring at time 1/m, 2/m, · · ·, 1, 1+1/m, · · ·, 2, · · ·, n, and let i be the effective rate of interest per interest-conversion period. Thus, there are mn payments over n interest-conversion periods.
• Suppose each payment is of the amount 1/m, so that there is a nominal amount of unit payment in each interest-conversion period.
• Figure 2.7 illustrates the cash flows for the case of m = 4.
• We denote the present value of this annuity at time 0 by a(m) nei , which can be computed as follows (we let w = v 1 m) a(m) nei = 1 m ³ v 1 m + v 2 m + · · · + v + v1+ 1 m + · · · + vn´ 30 = 1 m(w + w2 + · · · + wmn) = 1 m ∙ w × 1 −wmn 1 −w ¸ = 1 m " v 1 m × 1 −vn 1 −v 1 m # = 1 m " 1 −vn (1 + i) 1 m −1 # = 1 −vn r(m) , (2.22) where r(m) = m h (1 + i) 1 m −1 i (2.23) is the equivalent nominal rate of interest compounded m times per interest-conversion period (see (1.19)).
31 • The future value of the annuity-immediate is s(m) nei = (1 + i)n a(m) nei = (1 + i)n −1 r(m) = i r(m) sne.
(2.24) • The above equation parallels (2.22), which can also be written as a(m) nei = i r(m) ane.
• If the mn-payment annuity is due at time 0, 1/m, 2/m, · · · , n−1/m, we denote its present value at time 0 by ¨ a(m) ne , which is given by ¨ a(m) ne = (1 + i) 1 m a(m) ne = (1 + i) 1 m × ∙1 −vn r(m) ¸ .
(2.25) 32 • Thus, from (1.22) we conclude ¨ a(m) ne = 1 −vn d(m) = d d(m) ¨ ane.
(2.26) • The future value of this annuity at time n is ¨ s(m) ne = (1 + i)n ¨ a(m) ne = d d(m) ¨ sne.
(2.27) • For deferred annuities, the following results apply q|a(m) ne = vq a(m) ne , (2.28) and q|¨ a(m) ne = vq ¨ a(m) ne .
(2.29) Example 2.11: Solve the problem in Example 2.9 using (2.20).
33 Solution: We first note that i = 0.08/12 = 0.0067. Now k = 3 and n = 24 so that from (2.20), the present value of the annuity-immediate is 200 × a24e0.0067 s3e0.0067 = 200 × "1 −(1.0067)−24 (1.0067)3 −1 # = $1,464.27.
Finally, the present value of the annuity-due is (1.0067)3 × 1,464.27 = $1,493.90.
2 Example 2.12: Solve the problem in Example 2.10 using (2.22) and (2.23).
Solution: Note that m = 2 and n = 8. With i = 0.03, we have, from 34 (2.23) r(2) = 2 × [ √ 1.03 −1] = 0.0298.
Therefore, from (2.22), we have a(2) 8e0.03 = 1 −(1.03)−8 0.0298 = 7.0720.
As the total payment in each interest-conversion period is $200, the re-quired present value is 200 × 7.0720 = $1,414.27.
2 • Now we consider (2.22) again. Suppose the annuities are paid con-tinuously at the rate of 1 unit per interest-conversion period over n periods. Thus, m →∞and we denote the present value of this continuous annuity by ¯ ane.
35 • As limm→∞r(m) = δ, we have, from (2.22), ¯ ane = 1 −vn δ = 1 −vn ln(1 + i) = i δ ane.
(2.30) • The present value of a n-period continuous annuity of unit payment per period with a deferred period of q is given by q|¯ ane = vq ¯ ane = ¯ aq+ne −¯ aqe.
(2.31) • To compute the future value of a continuous annuity of unit payment per period over n periods, we use the following formula ¯ sne = (1 + i)n ¯ ane = (1 + i)n −1 ln(1 + i) = i δ sne.
(2.32) • We now generalize the above results to the case of a general accu-mulation function a(·). The present value of a continuous annuity 36 of unit payment per period over n periods is ¯ ane = Z n 0 v(t) dt = Z n 0 exp µ − Z t 0 δ(s) ds ¶ dt.
(2.33) • To compute the future value of the annuity at time n, we assume that, as in Section 1.7, a unit payment at time t accumulates to a(n −t) at time n, for n > t ≥0.
• Thus, the future value of the annuity at time n is ¯ sne = Z n 0 a(n −t) dt = Z n 0 exp µZ n−t 0 δ(s) ds ¶ dt.
(2.34) 37 2.6 Varying Annuities • We consider annuities the payments of which vary according to an arithmetic progression.
• Thus, we consider an annuity-immediate and assume the initial pay-ment is P, with subsequent payments P + D, P + 2D, · · ·, etc., so that the jth payment is P + (j −1)D.
• We allow D to be negative so that the annuity can be either stepping up or stepping down.
• However, for a n-payment annuity, P + (n −1)D must be positive so that negative cash flow is ruled out.
• We can see that the annuity can be regarded as the sum of the following annuities: (a) a n-period annuity-immediate with constant 38 amount P, and (b) n −1 deferred annuities, where the jth deferred annuity is a (n −j)-period annuity-immediate with level amount D to start at time j, for j = 1, · · · , n −1.
• Thus, the present value of the varying annuity is Pane + D ⎡ ⎣ n−1 X j=1 vjan−je ⎤ ⎦ = Pane + D ⎡ ⎣ n−1 X j=1 vj (1 −vn−j) i ⎤ ⎦ = Pane + D ⎡ ⎣ ³Pn−1 j=1 vj´ −(n −1)vn i ⎤ ⎦ = Pane + D ⎡ ⎣ ³Pn j=1 vj´ −nvn i ⎤ ⎦ = Pane + D " ane −nvn i # .
(2.35) 39 • For a n-period increasing annuity with P = D = 1, we denote its present and future values by (Ia)ne and (Is)ne, respectively.
• It can be shown that (Ia)ne = ¨ ane −nvn i (2.36) and (Is)ne = sn+1e −(n + 1) i = ¨ sne −n i .
(2.37) • For an increasing n-payment annuity-due with payments of 1, 2, · · · , n at time 0, 1, · · · , n −1, the present value of the annuity is (I¨ a)ne = (1 + i)(Ia)ne.
(2.38) • This is the sum of a n-period level annuity-due of unit payments and a (n −1)-payment increasing annuity-immediate with starting and incremental payments of 1.
40 • Thus, we have (I¨ a)ne = ¨ ane + (Ia)n−1e.
(2.39) • For the case of a n-period decreasing annuity with P = n and D = −1, we denote its present and future values by (Da)ne and (Ds)ne, respectively.
• Figure 2.9 presents the time diagram of this annuity.
• It can be shown that (Da)ne = n −ane i (2.40) and (Ds)ne = n(1 + i)n −sne i .
(2.41) 41 • We consider two types of increasing continuous annuities.
First, we consider the case of a continuous n-period annuity with level payment (i.e., at a constant rate) of τ units from time τ −1 through time τ.
• We denote the present value of this annuity by (I¯ a)ne, which is given by (I¯ a)ne = n X τ=1 τ Z τ τ−1 vs ds = n X τ=1 τ Z τ τ−1 e−δs ds.
(2.42) • The above equation can be simplified to (I¯ a)ne = ¨ ane −nvn δ .
(2.43) • Second, we may consider a continuous n-period annuity for which the payment in the interval t to t+∆t is t∆t, i.e., the instantaneous rate of payment at time t is t.
42 • We denote the present value of this annuity by (¯ I¯ a)ne, which is given by (¯ I¯ a)ne = Z n 0 tvt dt = Z n 0 te−δt dt = ¯ ane −nvn δ .
(2.44) • We now consider an annuity-immediate with payments following a geometric progression.
• Let the first payment be 1, with subsequent payments being 1 + k times the previous one. Thus, the present value of an annuity with n payments is (for k 6= i) v + v2(1 + k) + · · · + vn(1 + k)n−1 = v n−1 X t=0 [v(1 + k)]t = v n−1 X t=0 "1 + k 1 + i #t 43 = v ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 1 − Ã1 + k 1 + i !n 1 −1 + k 1 + i ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = 1 − Ã1 + k 1 + i !n i −k .
(2.45) Example 2.13: An annuity-immediate consists of a first payment of $100, with subsequent payments increased by 10% over the previous one until the 10th payment, after which subsequent payments decreases by 5% over the previous one. If the effective rate of interest is 10% per payment period, what is the present value of this annuity with 20 payments?
Solution: The present value of the first 10 payments is (use the second 44 line of (2.45)) 100 × 10(1.1)−1 = $909.09.
For the next 10 payments, k = −0.05 and their present value at time 10 is 100(1.10)9(0.95) × 1 − µ0.95 1.1 ¶10 0.1 + 0.05 = 1,148.64.
Hence, the present value of the 20 payments is 909.09 + 1,148.64(1.10)−10 = $1,351.94.
2 Example 2.14: An investor wishes to accumulate $1,000 at the end of year 5. He makes level deposits at the beginning of each year for 5 years.
The deposits earn a 6% annual effective rate of interest, which is credited 45 at the end of each year. The interests on the deposits earn 5% effective interest rate annually. How much does he have to deposit each year?
Solution: Let the level annual deposit be A. The interest received at the end of year 1 is 0.06A, which increases by 0.06A annually to 5×0.06A at the end of year 5. Thus, the interest is a 5-payment increasing annuity with P = D = 0.06A, earning annual interest of 5%. Hence, we have the equation 1,000 = 5A + 0.06A(Is)5e0.05.
From (2.37) we obtain (Is)5e0.05 = 16.0383, so that A = 1,000 5 + 0.06 × 16.0383 = $167.7206.
2 46 2.7 Term of Annuity • We now consider the case where the annuity period may not be an integer. We consider an+ke, where n is an integer and 0 < k < 1.
We note that an+ke = 1 −vn+k i = (1 −vn) + ³ vn −vn+k´ i = ane + vn+k "(1 + i)k −1 i # = ane + vn+k ske.
(2.46) • Thus, an+ke is the sum of the present value of a n-period annuity-immediate with unit amount and the present value of an amount ske 47 paid at time n + k.
Example 2.15: A principal of $5,000 generates income of $500 at the end of every year at an effective rate of interest of 4.5% for as long as possible. Calculate the term of the annuity and discuss the possibilities of settling the last payment.
Solution: The equation of value 500 ane0.045 = 5,000 implies ane0.045 = 10. As a13e0.045 = 9.68 and a14e0.045 = 10.22, the principal can generate 13 regular payments. The investment may be paid offwith an additional amount A at the end of year 13, in which case 500 s13e0.045 + A = 5,000(1.045)13, 48 which implies A = $281.02, so that the last payment is $781.02. Alter-natively, the last payment B may be made at the end of year 14, which is B = 281.02 × 1.045 = $293.67.
If we adopt the approach in (2.46), we solve k from the equation 1 −(1.045)−(13+k) 0.045 = 10, which implies (1.045)13+k = 1 0.55, from which we obtain k = ln µ 1 0.55 ¶ ln(1.045) −13 = 0.58.
49 Hence, the last payment C to be paid at time 13.58 years is C = 500 × "(1.045)0.58 −1 0.045 # = $288.32.
Note that A < C < B, which is as expected, as this follows the order of the occurrence of the payments. In principle, all three approaches are justified.
2 • Generally, the effective rate of interest cannot be solved analytically from the equation of value. Numerical methods must be used for this purpose.
Example 2.16: A principal of $5,000 generates income of $500 at the end of every year for 15 years. What is the effective rate of interest?
50 Solution: The equation of value is a15ei = 5,000 500 = 10, so that a15ei = 1 −(1 + i)−15 i = 10.
A simple grid search provides the following results i a15ei 0.054 10.10 0.055 10.04 0.056 9.97 A finer search provides the answer 5.556%.
2 • The Excel Solver may be used to calculate the effective rate of in-terest in Example 2.16.
51 • The computation is illustrated in Exhibit 2.1.
• We enter a guessed value of 0.05 in Cell A1 in the Excel worksheet.
• The following expression is then entered in Cell A2: (1 −(1 + A1)ˆ(−15))/A1, which computes a15ei with i equal to the value at A1.
• We can also use the Excel function RATE to calculate the rate of interest that equates the present value of an annuity-immediate to a given value. Specifically, consider the equations anei −A = 0 and ¨ anei −A = 0.
Given n and A, we wish to solve for i, which is the rate of interest per payment period of the annuity-immediate or annuity-due. The use of the Excel function RATE to compute i is described as follows: 52 Exhibit 2.1: Use of Excel Solver for Example 2.16 Excel function: RATE(np,1,pv,type,guess) np = n, pv = −A, type = 0 (or omitted) for annuity-immediate, 1 for annuity-due guess = starting value, set to 0.1 if omitted Output = i, rate of interest per payment period of the annuity • To use RATE to solve for Example 2.16 we key in the following: “=RATE(15,1,-10)”.
53 |
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9375 | https://www.wordhippo.com/what-is/another-word-for/trowel.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
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| | | | | | | | | | | | | --- --- --- --- --- | | | | | | | --- --- | | | | | | --- --- | | What's another word for | | | | | | | | | | | | | | | | | | --- --- | | | | | | --- --- | | What's the opposite of | | | | | | | | | | | | | | | | | --- --- | | | | | | --- | Meaning of the word | | | | | | | | | | | | | | | | --- --- | | | | | | --- | Words that rhyme with | | | | | | | | | | | | | | | | --- --- | | | | | | --- | Sentences with the word | | | | | | | | | | | | | | | | | | | | --- --- --- --- | | | | | | | | --- --- | Translate | | From English To English | to | Afrikaans Albanian Amharic Arabic Armenian Azerbaijani Basque Belarusian Bengali Bosnian Bulgarian Catalan Cebuano Chichewa Chinese Corsican Croatian Czech Danish Dutch Esperanto Estonian Farsi Filipino Finnish French Frisian Galician Georgian German Greek Gujarati Haitian Creole Hausa Hebrew Hindi Hmong Hungarian Icelandic Igbo Indonesian Irish Italian Japanese Javanese Kannada Kazakh Khmer Korean Kurdish Kyrgyz Lao Latin Latvian Lithuanian Luxembourgish Macedonian Malagasy Malay Malayalam Maltese Maori Marathi Mongolian Burmese Nepali Norwegian Polish Portuguese Punjabi Romanian Russian Samoan Scots Gaelic Serbian Sesotho Shona Sinhala Slovak Slovenian Somali Spanish Sundanese Swahili Swedish Tajik Tamil Telugu Thai Turkish Ukrainian Urdu Uzbek Vietnamese Welsh Xhosa Yiddish Yoruba Zulu | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Find Words | All Words 2-Letter Words 3-Letter Words 4-Letter Words 5-Letter Words 6-Letter Words 7-Letter Words 8-Letter Words 9-Letter Words 10-Letter Words 11-Letter Words 12-Letter Words 13-Letter Words 14-Letter Words | Starting With Ending With Containing Exactly Containing the Letters Words With Friends Scrabble Crossword / Codeword | | | Use for blank tiles (max 2) Use for blank spaces Advanced Word Finder | | | | | | | | | | | | | | | | --- --- --- | | | | | | | | --- --- | Find the | Plural Singular Past Tense Present Tense Verb Adjective Adverb Noun | of | | | | | | | | | | | | | | | | | | --- --- --- | | | | | | | | --- --- | Pronounce the word | | in | Afrikaans Albanian Arabic Bengali Chinese Croatian Czech Danish Dutch English Finnish French German Greek Hindi Hungarian Icelandic Indonesian Italian Japanese Korean Latin Malay Malayalam Marathi Nepali Norwegian Polish Portuguese Romanian Russian Serbian Slovak Spanish Swahili Swedish Tamil Telugu Thai Turkish Ukrainian Uzbek Vietnamese Welsh | | | | | | | | | | | | | | | | --- --- | | | | | | --- --- | | Find Names | Meaning of the name Origin of the name Names meaning Names starting with Names of origin | | | | | | | | | | | | | | Synonyms Antonyms Definitions Rhymes Sentences Translations Find Words Word Forms Pronunciations | | | | | | | | |
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| | | | | | | | | | --- --- --- --- | | | | | | | | --- --- | Translate | | From English To English | to | Afrikaans Albanian Amharic Arabic Armenian Azerbaijani Basque Belarusian Bengali Bosnian Bulgarian Catalan Cebuano Chichewa Chinese Corsican Croatian Czech Danish Dutch Esperanto Estonian Farsi Filipino Finnish French Frisian Galician Georgian German Greek Gujarati Haitian Creole Hausa Hebrew Hindi Hmong Hungarian Icelandic Igbo Indonesian Irish Italian Japanese Javanese Kannada Kazakh Khmer Korean Kurdish Kyrgyz Lao Latin Latvian Lithuanian Luxembourgish Macedonian Malagasy Malay Malayalam Maltese Maori Marathi Mongolian Burmese Nepali Norwegian Polish Portuguese Punjabi Romanian Russian Samoan Scots Gaelic Serbian Sesotho Shona Sinhala Slovak Slovenian Somali Spanish Sundanese Swahili Swedish Tajik Tamil Telugu Thai Turkish Ukrainian Urdu Uzbek Vietnamese Welsh Xhosa Yiddish Yoruba Zulu | | | | | | | | | | | | | | | |
| | | | | | | | | --- --- --- | | | | | | | | --- --- --- | | Find Words | All Words 2-Letter Words 3-Letter Words 4-Letter Words 5-Letter Words 6-Letter Words 7-Letter Words 8-Letter Words 9-Letter Words 10-Letter Words 11-Letter Words 12-Letter Words 13-Letter Words 14-Letter Words | Starting With Ending With Containing Exactly Containing the Letters Words With Friends Scrabble Crossword / Codeword | | | Use for blank tiles (max 2) Use for blank spaces Advanced Word Finder | | | | | | | |
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| | | | | | | | --- --- --- | | | | | | | | --- --- | Pronounce the word | | in | Afrikaans Albanian Arabic Bengali Chinese Croatian Czech Danish Dutch English Finnish French German Greek Hindi Hungarian Icelandic Indonesian Italian Japanese Korean Latin Malay Malayalam Marathi Nepali Norwegian Polish Portuguese Romanian Russian Serbian Slovak Spanish Swahili Swedish Tamil Telugu Thai Turkish Ukrainian Uzbek Vietnamese Welsh | | | | | | | | |
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| | | | | | | --- --- | Translate | | From English To English | to | Afrikaans Albanian Amharic Arabic Armenian Azerbaijani Basque Belarusian Bengali Bosnian Bulgarian Catalan Cebuano Chichewa Chinese Corsican Croatian Czech Danish Dutch Esperanto Estonian Farsi Filipino Finnish French Frisian Galician Georgian German Greek Gujarati Haitian Creole Hausa Hebrew Hindi Hmong Hungarian Icelandic Igbo Indonesian Irish Italian Japanese Javanese Kannada Kazakh Khmer Korean Kurdish Kyrgyz Lao Latin Latvian Lithuanian Luxembourgish Macedonian Malagasy Malay Malayalam Maltese Maori Marathi Mongolian Burmese Nepali Norwegian Polish Portuguese Punjabi Romanian Russian Samoan Scots Gaelic Serbian Sesotho Shona Sinhala Slovak Slovenian Somali Spanish Sundanese Swahili Swedish Tajik Tamil Telugu Thai Turkish Ukrainian Urdu Uzbek Vietnamese Welsh Xhosa Yiddish Yoruba Zulu | | | | | | | | | |
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| Translate | | From English To English | to | Afrikaans Albanian Amharic Arabic Armenian Azerbaijani Basque Belarusian Bengali Bosnian Bulgarian Catalan Cebuano Chichewa Chinese Corsican Croatian Czech Danish Dutch Esperanto Estonian Farsi Filipino Finnish French Frisian Galician Georgian German Greek Gujarati Haitian Creole Hausa Hebrew Hindi Hmong Hungarian Icelandic Igbo Indonesian Irish Italian Japanese Javanese Kannada Kazakh Khmer Korean Kurdish Kyrgyz Lao Latin Latvian Lithuanian Luxembourgish Macedonian Malagasy Malay Malayalam Maltese Maori Marathi Mongolian Burmese Nepali Norwegian Polish Portuguese Punjabi Romanian Russian Samoan Scots Gaelic Serbian Sesotho Shona Sinhala Slovak Slovenian Somali Spanish Sundanese Swahili Swedish Tajik Tamil Telugu Thai Turkish Ukrainian Urdu Uzbek Vietnamese Welsh Xhosa Yiddish Yoruba Zulu |
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| Find Words | All Words 2-Letter Words 3-Letter Words 4-Letter Words 5-Letter Words 6-Letter Words 7-Letter Words 8-Letter Words 9-Letter Words 10-Letter Words 11-Letter Words 12-Letter Words 13-Letter Words 14-Letter Words | Starting With Ending With Containing Exactly Containing the Letters Words With Friends Scrabble Crossword / Codeword | | | Use for blank tiles (max 2) Use for blank spaces Advanced Word Finder |
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| | | | | | | --- --- | Pronounce the word | | in | Afrikaans Albanian Arabic Bengali Chinese Croatian Czech Danish Dutch English Finnish French German Greek Hindi Hungarian Icelandic Indonesian Italian Japanese Korean Latin Malay Malayalam Marathi Nepali Norwegian Polish Portuguese Romanian Russian Serbian Slovak Spanish Swahili Swedish Tamil Telugu Thai Turkish Ukrainian Uzbek Vietnamese Welsh | | |
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| Pronounce the word | | in | Afrikaans Albanian Arabic Bengali Chinese Croatian Czech Danish Dutch English Finnish French German Greek Hindi Hungarian Icelandic Indonesian Italian Japanese Korean Latin Malay Malayalam Marathi Nepali Norwegian Polish Portuguese Romanian Russian Serbian Slovak Spanish Swahili Swedish Tamil Telugu Thai Turkish Ukrainian Uzbek Vietnamese Welsh | |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | What is another word for trowel?Need synonyms for trowel? Here's a list of similar words from our thesaurus that you can use instead. | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- | Contexts Noun A tool with a broad blade and typically upturned sides used for digging Verb To dig or dredge for something Noun ▲ A tool with a broad blade and typically upturned sides used for digging shovel spade digger scoop excavator banjo peel tool digging tool garden spade snow shovel spud Verb ▲ To dig or dredge for something delve dig excavate unearth shovel spade dredge scratch scoop out burrow tunnel scoop mine dig out hollow out bore quarry grub cut out cut drill scrape gouge channel sap dig up gouge out undermine ladle spoon heap muck extract scoop up pick up hew remove draw pan work groove hoe plowUS strip-mine chisel pit rout depress bulldoze root rake claw dig for hack score ploughUK sift enter concave turn deepen gash incise engrave mark dig down burrow out chisel out cut into scratch out hack out claw out scrape out trench hollow sink empty open up | | | | --- | | Find more words! | | | Another word for Opposite of Meaning of Rhymes with Sentences with Find word forms Translate from English Translate to English Words With Friends Scrabble Crossword / Codeword Words starting with Words ending with Words containing exactly Words containing letters Pronounce Find conjugations Find names | | | From Afrikaans From Albanian From Amharic From Arabic From Armenian From Azerbaijani From Basque From Belarusian From Bengali From Bosnian From Bulgarian From Catalan From Cebuano From Chichewa From Chinese From Corsican From Croatian From Czech From Danish From Dutch From Esperanto From Estonian From Farsi From Filipino From Finnish From French From Frisian From Galician From Georgian From German From Greek From Gujarati From Haitian Creole From Hausa From Hebrew From Hindi From Hmong From Hungarian From Icelandic From Igbo From Indonesian From Irish From Italian From Japanese From Javanese From Kannada From Kazakh From Khmer From Korean From Kurdish From Kyrgyz From Lao From Latin From Latvian From Lithuanian From Luxembourgish From Macedonian From Malagasy From Malay From Malayalam From Maltese From Maori From Marathi From Mongolian From Burmese From Nepali From Norwegian From Polish From Portuguese From Punjabi From Romanian From Russian From Samoan From Scots Gaelic From Serbian From Sesotho From Shona From Sinhala From Slovak From Slovenian From Somali From Spanish From Sundanese From Swahili From Swedish From Tajik From Tamil From Telugu From Thai From Turkish From Ukrainian From Urdu From Uzbek From Vietnamese From Welsh From Xhosa From Yiddish From Yoruba From Zulu | | | To Afrikaans To Albanian To Amharic To Arabic To Armenian To Azerbaijani To Basque To Belarusian To Bengali To Bosnian To Bulgarian To Catalan To Cebuano To Chichewa To Chinese To Corsican To Croatian To Czech To Danish To Dutch To Esperanto To Estonian To Farsi To Filipino To Finnish To French To Frisian To Galician To Georgian To German To Greek To Gujarati To Haitian Creole To Hausa To Hebrew To Hindi To Hmong To Hungarian To Icelandic To Igbo To Indonesian To Irish To Italian To Japanese To Javanese To Kannada To Kazakh To Khmer To Korean To Kurdish To Kyrgyz To Lao To Latin To Latvian To Lithuanian To Luxembourgish To Macedonian To Malagasy To Malay To Malayalam To Maltese To Maori To Marathi To Mongolian To Burmese To Nepali To Norwegian To Polish To Portuguese To Punjabi To Romanian To Russian To Samoan To Scots Gaelic To Serbian To Sesotho To Shona To Sinhala To Slovak To Slovenian To Somali To Spanish To Sundanese To Swahili To Swedish To Tajik To Tamil To Telugu To Thai To Turkish To Ukrainian To Urdu To Uzbek To Vietnamese To Welsh To Xhosa To Yiddish To Yoruba To Zulu | | | English French | | | Afrikaans Word Albanian Word Arabic Word Bengali Word Chinese Word Croatian Word Czech Word Danish Word Dutch Word English Word Finnish Word French Word German Word Greek Word Hindi Word Hungarian Word Icelandic Word Indonesian Word Italian Word Japanese Word Korean Word Latin Word Malay Word Malayalam Word Marathi Word Nepali Word Norwegian Word Polish Word Portuguese Word Romanian Word Russian Word Serbian Word Slovak Word Spanish Word Swahili Word Swedish Word Tamil Word Telugu Word Thai Word Turkish Word Ukrainian Word Uzbek Word Vietnamese Word Welsh Word | | All words 2-letter words 3-letter words 4-letter words 5-letter words 6-letter words 7-letter words 8-letter words 9-letter words 10-letter words 11-letter words 12-letter words 13-letter words 14-letter words | | | Plural of Singular of Past tense of Present tense of Verb for Adjective for Adverb for Noun for | | Meaning of name Origin of name Names meaning Names starting with Names of origin | | trowel | | | Use for blank tiles (max 2) Advanced Search Advanced Search | | | Use for blank spaces Advanced Search | | | Advanced Word Finder | | Related Words and Phrases trowelling trowels troweled troweling trowelled See Also Sentences with the word trowel Words that rhyme with trowel What is the past tense of trowel? What is the plural of trowel? What is the noun for trowel? Translations for trowel Use our Synonym Finder Nearby Words troweled troweling troweling machine trowelled trowelling trowel machine troves trove trovatore trouveurs trouveur trouvère 6-letter Words Starting With t tr tro trow trowe | | | | | | | | | --- --- | Find Synonyms | | | --- | | trowel | go | | | | | | | | | | | | | | |
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | What is another word for trowel?Need synonyms for trowel? Here's a list of similar words from our thesaurus that you can use instead. | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- | Contexts Noun A tool with a broad blade and typically upturned sides used for digging Verb To dig or dredge for something Noun ▲ A tool with a broad blade and typically upturned sides used for digging shovel spade digger scoop excavator banjo peel tool digging tool garden spade snow shovel spud Verb ▲ To dig or dredge for something delve dig excavate unearth shovel spade dredge scratch scoop out burrow tunnel scoop mine dig out hollow out bore quarry grub cut out cut drill scrape gouge channel sap dig up gouge out undermine ladle spoon heap muck extract scoop up pick up hew remove draw pan work groove hoe plowUS strip-mine chisel pit rout depress bulldoze root rake claw dig for hack score ploughUK sift enter concave turn deepen gash incise engrave mark dig down burrow out chisel out cut into scratch out hack out claw out scrape out trench hollow sink empty open up | | | | --- | | Find more words! | | | Another word for Opposite of Meaning of Rhymes with Sentences with Find word forms Translate from English Translate to English Words With Friends Scrabble Crossword / Codeword Words starting with Words ending with Words containing exactly Words containing letters Pronounce Find conjugations Find names | | | From Afrikaans From Albanian From Amharic From Arabic From Armenian From Azerbaijani From Basque From Belarusian From Bengali From Bosnian From Bulgarian From Catalan From Cebuano From Chichewa From Chinese From Corsican From Croatian From Czech From Danish From Dutch From Esperanto From Estonian From Farsi From Filipino From Finnish From French From Frisian From Galician From Georgian From German From Greek From Gujarati From Haitian Creole From Hausa From Hebrew From Hindi From Hmong From Hungarian From Icelandic From Igbo From Indonesian From Irish From Italian From Japanese From Javanese From Kannada From Kazakh From Khmer From Korean From Kurdish From Kyrgyz From Lao From Latin From Latvian From Lithuanian From Luxembourgish From Macedonian From Malagasy From Malay From Malayalam From Maltese From Maori From Marathi From Mongolian From Burmese From Nepali From Norwegian From Polish From Portuguese From Punjabi From Romanian From Russian From Samoan From Scots Gaelic From Serbian From Sesotho From Shona From Sinhala From Slovak From Slovenian From Somali From Spanish From Sundanese From Swahili From Swedish From Tajik From Tamil From Telugu From Thai From Turkish From Ukrainian From Urdu From Uzbek From Vietnamese From Welsh From Xhosa From Yiddish From Yoruba From Zulu | | | To Afrikaans To Albanian To Amharic To Arabic To Armenian To Azerbaijani To Basque To Belarusian To Bengali To Bosnian To Bulgarian To Catalan To Cebuano To Chichewa To Chinese To Corsican To Croatian To Czech To Danish To Dutch To Esperanto To Estonian To Farsi To Filipino To Finnish To French To Frisian To Galician To Georgian To German To Greek To Gujarati To Haitian Creole To Hausa To Hebrew To Hindi To Hmong To Hungarian To Icelandic To Igbo To Indonesian To Irish To Italian To Japanese To Javanese To Kannada To Kazakh To Khmer To Korean To Kurdish To Kyrgyz To Lao To Latin To Latvian To Lithuanian To Luxembourgish To Macedonian To Malagasy To Malay To Malayalam To Maltese To Maori To Marathi To Mongolian To Burmese To Nepali To Norwegian To Polish To Portuguese To Punjabi To Romanian To Russian To Samoan To Scots Gaelic To Serbian To Sesotho To Shona To Sinhala To Slovak To Slovenian To Somali To Spanish To Sundanese To Swahili To Swedish To Tajik To Tamil To Telugu To Thai To Turkish To Ukrainian To Urdu To Uzbek To Vietnamese To Welsh To Xhosa To Yiddish To Yoruba To Zulu | | | English French | | | Afrikaans Word Albanian Word Arabic Word Bengali Word Chinese Word Croatian Word Czech Word Danish Word Dutch Word English Word Finnish Word French Word German Word Greek Word Hindi Word Hungarian Word Icelandic Word Indonesian Word Italian Word Japanese Word Korean Word Latin Word Malay Word Malayalam Word Marathi Word Nepali Word Norwegian Word Polish Word Portuguese Word Romanian Word Russian Word Serbian Word Slovak Word Spanish Word Swahili Word Swedish Word Tamil Word Telugu Word Thai Word Turkish Word Ukrainian Word Uzbek Word Vietnamese Word Welsh Word | | All words 2-letter words 3-letter words 4-letter words 5-letter words 6-letter words 7-letter words 8-letter words 9-letter words 10-letter words 11-letter words 12-letter words 13-letter words 14-letter words | | | Plural of Singular of Past tense of Present tense of Verb for Adjective for Adverb for Noun for | | Meaning of name Origin of name Names meaning Names starting with Names of origin | | trowel | | | Use for blank tiles (max 2) Advanced Search Advanced Search | | | Use for blank spaces Advanced Search | | | Advanced Word Finder | | Related Words and Phrases trowelling trowels troweled troweling trowelled See Also Sentences with the word trowel Words that rhyme with trowel What is the past tense of trowel? What is the plural of trowel? What is the noun for trowel? Translations for trowel Use our Synonym Finder Nearby Words troweled troweling troweling machine trowelled trowelling trowel machine troves trove trovatore trouveurs trouveur trouvère 6-letter Words Starting With t tr tro trow trowe | | | | | | | | | --- --- | Find Synonyms | | | --- | | trowel | go | | | | | | | |
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | What is another word for trowel?Need synonyms for trowel? Here's a list of similar words from our thesaurus that you can use instead. | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- | Contexts Noun A tool with a broad blade and typically upturned sides used for digging Verb To dig or dredge for something Noun ▲ A tool with a broad blade and typically upturned sides used for digging shovel spade digger scoop excavator banjo peel tool digging tool garden spade snow shovel spud Verb ▲ To dig or dredge for something delve dig excavate unearth shovel spade dredge scratch scoop out burrow tunnel scoop mine dig out hollow out bore quarry grub cut out cut drill scrape gouge channel sap dig up gouge out undermine ladle spoon heap muck extract scoop up pick up hew remove draw pan work groove hoe plowUS strip-mine chisel pit rout depress bulldoze root rake claw dig for hack score ploughUK sift enter concave turn deepen gash incise engrave mark dig down burrow out chisel out cut into scratch out hack out claw out scrape out trench hollow sink empty open up | | | | --- | | Find more words! 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What is the plural of trowel? What is the noun for trowel? Translations for trowel Use our Synonym Finder Nearby Words troweled troweling troweling machine trowelled trowelling trowel machine troves trove trovatore trouveurs trouveur trouvère 6-letter Words Starting With t tr tro trow trowe | | | | | | | | | --- --- | Find Synonyms | | | --- | | trowel | go | | | | | |
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| Contexts Noun A tool with a broad blade and typically upturned sides used for digging Verb To dig or dredge for something Noun ▲ A tool with a broad blade and typically upturned sides used for digging shovel spade digger scoop excavator banjo peel tool digging tool garden spade snow shovel spud Verb ▲ To dig or dredge for something delve dig excavate unearth shovel spade dredge scratch scoop out burrow tunnel scoop mine dig out hollow out bore quarry grub cut out cut drill scrape gouge channel sap dig up gouge out undermine ladle spoon heap muck extract scoop up pick up hew remove draw pan work groove hoe plowUS strip-mine chisel pit rout depress bulldoze root rake claw dig for hack score ploughUK sift enter concave turn deepen gash incise engrave mark dig down burrow out chisel out cut into scratch out hack out claw out scrape out trench hollow sink empty open up | | | | --- | | Find more words! | | | Another word for Opposite of Meaning of Rhymes with Sentences with Find word forms Translate from English Translate to English Words With Friends Scrabble Crossword / Codeword Words starting with Words ending with Words containing exactly Words containing letters Pronounce Find conjugations Find names | | | From Afrikaans From Albanian From Amharic From Arabic From Armenian From Azerbaijani From Basque From Belarusian From Bengali From Bosnian From Bulgarian From Catalan From Cebuano From Chichewa From Chinese From Corsican From Croatian From Czech From Danish From Dutch From Esperanto From Estonian From Farsi From Filipino From Finnish From French From Frisian From Galician From Georgian From German From Greek From Gujarati From Haitian Creole From Hausa From Hebrew From Hindi From Hmong From Hungarian From Icelandic From Igbo From Indonesian From Irish From Italian From Japanese From Javanese From Kannada From Kazakh From Khmer From Korean From Kurdish From Kyrgyz From Lao From Latin From Latvian From Lithuanian From Luxembourgish From Macedonian From Malagasy From Malay From Malayalam From Maltese From Maori From Marathi From Mongolian From Burmese From Nepali From Norwegian From Polish From Portuguese From Punjabi From Romanian From Russian From Samoan From Scots Gaelic From Serbian From Sesotho From Shona From Sinhala From Slovak From Slovenian From Somali From Spanish From Sundanese From Swahili From Swedish From Tajik From Tamil From Telugu From Thai From Turkish From Ukrainian From Urdu From Uzbek From Vietnamese From Welsh From Xhosa From Yiddish From Yoruba From Zulu | | | To Afrikaans To Albanian To Amharic To Arabic To Armenian To Azerbaijani To Basque To Belarusian To Bengali To Bosnian To Bulgarian To Catalan To Cebuano To Chichewa To Chinese To Corsican To Croatian To Czech To Danish To Dutch To Esperanto To Estonian To Farsi To Filipino To Finnish To French To Frisian To Galician To Georgian To German To Greek To Gujarati To Haitian Creole To Hausa To Hebrew To Hindi To Hmong To Hungarian To Icelandic To Igbo To Indonesian To Irish To Italian To Japanese To Javanese To Kannada To Kazakh To Khmer To Korean To Kurdish To Kyrgyz To Lao To Latin To Latvian To Lithuanian To Luxembourgish To Macedonian To Malagasy To Malay To Malayalam To Maltese To Maori To Marathi To Mongolian To Burmese To Nepali To Norwegian To Polish To Portuguese To Punjabi To Romanian To Russian To Samoan To Scots Gaelic To Serbian To Sesotho To Shona To Sinhala To Slovak To Slovenian To Somali To Spanish To Sundanese To Swahili To Swedish To Tajik To Tamil To Telugu To Thai To Turkish To Ukrainian To Urdu To Uzbek To Vietnamese To Welsh To Xhosa To Yiddish To Yoruba To Zulu | | | English French | | | Afrikaans Word Albanian Word Arabic Word Bengali Word Chinese Word Croatian Word Czech Word Danish Word Dutch Word English Word Finnish Word French Word German Word Greek Word Hindi Word Hungarian Word Icelandic Word Indonesian Word Italian Word Japanese Word Korean Word Latin Word Malay Word Malayalam Word Marathi Word Nepali Word Norwegian Word Polish Word Portuguese Word Romanian Word Russian Word Serbian Word Slovak Word Spanish Word Swahili Word Swedish Word Tamil Word Telugu Word Thai Word Turkish Word Ukrainian Word Uzbek Word Vietnamese Word Welsh Word | | All words 2-letter words 3-letter words 4-letter words 5-letter words 6-letter words 7-letter words 8-letter words 9-letter words 10-letter words 11-letter words 12-letter words 13-letter words 14-letter words | | | Plural of Singular of Past tense of Present tense of Verb for Adjective for Adverb for Noun for | | Meaning of name Origin of name Names meaning Names starting with Names of origin | | trowel | | | Use for blank tiles (max 2) Advanced Search Advanced Search | | | Use for blank spaces Advanced Search | | | Advanced Word Finder | | Related Words and Phrases trowelling trowels troweled troweling trowelled See Also Sentences with the word trowel Words that rhyme with trowel What is the past tense of trowel? What is the plural of trowel? What is the noun for trowel? Translations for trowel Use our Synonym Finder Nearby Words troweled troweling troweling machine trowelled trowelling trowel machine troves trove trovatore trouveurs trouveur trouvère 6-letter Words Starting With t tr tro trow trowe |
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9376 | https://www.quora.com/What-is-the-area-of-a-hexagon-if-its-apothem-is-equal-to-a-side-length-of-2-units | What is the area of a hexagon if its apothem is equal to a side length of 2 units? - Quora
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Mathematics
Apothem of a Hexagon
Regular Shapes
PLANE GEOMETRY
Geometrical Structures
Geometric Shape
The Hexagon
Apothem
Geometric Mathematics
5
What is the area of a hexagon if its apothem is equal to a side length of 2 units?
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Allen Ries
B.Ed. from University of Alberta (Graduated 1974) · Author has 25.1K answers and 9.7M answer views
·4y
Drop an apothem inside 1 of the 6 triangles. The apothem is 2 units. The central angle 30°. Let a ‘base’ =b. Then tan30=b/2. b=2tan30 = 20.57735… =1.1547 units. The area of the hexagon is 61/2bh =6 1/21.15472 = 6.9…
The area of the hexagon is about 6.9 square units.
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Drop an apothem inside 1 of the 6 triangles. The apothem is 2 units. The central angle 30°. Let a ‘base’ =b. Then tan30=b/2. b=2tan30 = 20.57735… =1.1547 units. The area of the hexagon is 61/2bh =6 1/21.15472 = 6.9…
The area of the hexagon is about 6.9 square units.
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To find the area of a regular hexagon using its apothem, you can use the formula:
Area=1 2×Perimeter×Apothem Area=1 2×Perimeter×Apothem
For a regular hexagon, the perimeter (P P) can be calculated as:
P=6×s P=6×s
where s s is the side length. Given that the side length s s is 2 units:
P=6×2=12 units P=6×2=12 units
The apothem (a a) is also given as 2 units. Now, substituting these values into the area formula:
Area=1 2×12×2=1 2×24=12 square units Area=1 2×12×2=1 2×24=12 square units
Thus, the area of the hexagon is 12 square units.
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How do you divide a hexagon into 3 equal parts?
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A regular hexagon has sides 10 ft. What is the area of hexagon?
If a regular hexagon has its side as 15 cm, what is the area of the hexagon?
What is the length of sides in an equal hexagon totaling 150 sq ft?
Evan Donaldson
Former Math and Science Teacher (Upper School) at Francis W. Parker School (1990–1997) · Author has 244 answers and 150.3K answer views
·4y
There is a formula for area of a REGULAR polygon (which, we need to assume here) that includes the apothem:
A = 1/2 ap
where a is the apothem and p is the perimeter.
As a hexagon has six sides, each of length 2, the perimeter is 12.
Therefore the area is (1/2)(2)(12) = 12 square units
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Gauthmath Qween
Author has 81 answers and 70.3K answer views
·4y
Given side length = 2, perimeter, p = 26 = 12 units
again, apothem, a = 2 units
so, area of hexagon = 1/2 p a = 1/2 12 2 = 12 sq units
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James Fitzpatrick
Studied Mathematics&Printing at London (but gets around) (Graduated 1961) · Author has 12.8K answers and 12.2M answer views
·Updated 5y
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A regular hexagon has sides of 2 feet. What is the area of the hexagon?
Hello Avni Patel, Updated answer showing method and calculations.
The formula for solving the area of a hexagon is:
Area of hexagon
√3 = 1.7320508075688772935274463415059
√3 x 3 = 5.1961524227066318805823390245176
(√3 x 3) / 2 = 2.5980762113533159402911695122588
Multiplied by side length² = Area of a hexagon.
IE. If side length = 2 .. then multiply 2.598076 ... by 4 (2²)
= 10.392304845413263761164678049035 (units²) =10.39 ft²
Regards, James.
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Hello Avni Patel, Updated answer showing method and calculations.
The formula for solving the area of a hexagon is:
Area of hexagon
√3 = 1.7320508075688772935274463415059
√3 x 3 = 5.1961524227066318805823390245176
(√3 x 3) / 2 = 2.5980762113533159402911695122588
Multiplied by side length² = Area of a hexagon.
IE. If side length = 2 .. then multiply 2.598076 ... by 4 (2²)
= 10.392304845413263761164678049035 (units²) =10.39 ft²
Regards, James.
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How would you find the area of a regular hexagon with a side length of 8?
What is the length of the apothem of a regular hexagon of side 10 units?
How do you find the area of a hexagon given its length and breadth?
Construct regular hexagon ABCDEF of side length 2 units, what is the area of rectangle ACDF? Express your answer in simplest radical form.
What will be the area of a hexagon having 37 cm of equal side length?
Jayanta Mukherjee
B Tech IEE in Instrumentation Engineering, Jadavpur University (Graduated 1990) · Author has 43.6K answers and 11.1M answer views
·5y
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A regular hexagon has sides of 2 feet. What is the area of the hexagon?
Sum of interior angles of hexagon = (6 180° - 360°) = 720°.
Since, this hexagon is regular; so each angle equals 120°.
If any angle is bisected; it will be 60° each half.
So, area of the regular hexagon = 6 (area of equilateral triangle of each side 2 ft) = 6 (√3) (2^2) / 4 sq ft = (6√3) sq ft = 10.392 sq ft
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Pradeep Hebbar
Many years of Structural Engineering & Math enthusiasm · Author has 9.3K answers and 6.2M answer views
·2y
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Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the midpoint of CD, then what is the lengths of AT?
T T is the midpoint of C D C D in the given regular hexagon A B C D E F A B C D E F,
C T=D T=2 2=1 C T=D T=2 2=1
Draw A T A T, A D A D and A C A C
By the property of hexagon,
A D=2×B C=2×2=4 A D=2×B C=2×2=4
∠A B C=∠B C D=120∘∠A B C=∠B C D=120∘
We have, A B=B C=C D=2 A B=B C=C D=2
∠B A C=∠B C A=180∘−120∘2=30∘∠B A C=∠B C A=180∘−120∘2=30∘
∠A C D=120∘−30∘=90∘∠A C D=120∘−30∘=90∘
A C D A C D is a right triangle and A T A T is its median
A C 2=A D 2−C D 2=4 2−2 2=12 A C 2=A D 2−C D 2=4 2−2 2=12
Using Apollonius's theorem,
A C 2+A D 2=2(A T 2+D T 2)A C 2+A D 2=2(A T 2+D T 2)
12+4 2=2(A T 2+1 2)12+4 2=2(A T 2+1 2)
A T=√13 A T=13
Ans: ≈3.605551≈3.605551 cm
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T T is the midpoint of C D C D in the given regular hexagon A B C D E F A B C D E F,
C T=D T=2 2=1 C T=D T=2 2=1
Draw A T A T, A D A D and A C A C
By the property of hexagon,
A D=2×B C=2×2=4 A D=2×B C=2×2=4
∠A B C=∠B C D=120∘∠A B C=∠B C D=120∘
We have, A B=B C=C D=2 A B=B C=C D=2
∠B A C=∠B C A=180∘−120∘2=30∘∠B A C=∠B C A=180∘−120∘2=30∘
∠A C D=120∘−30∘=90∘∠A C D=120∘−30∘=90∘
A C D A C D is a right triangle and A T A T is its median
A C 2=A D 2−C D 2=4 2−2 2=12 A C 2=A D 2−C D 2=4 2−2 2=12
Using Apollonius's theorem,
A C 2+A D 2=2(A T 2+D T 2)A C 2+A D 2=2(A T 2+D T 2)
12+4 2=2(A T 2+1 2)12+4 2=2(A T 2+1 2)
A T=√13 A T=13
Ans: ≈3.605551≈3.605551 cm
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JH Simon
Author of 'How To Kill A Narcissist'
·Updated Fri
Narcissist abuse has lead me into a very dark depression. How do I overcome this?
Pendulate between the darkness and the light.
Depression is a healing tonic which restores the Self to a point of equilibrium. Remember that while in the narcissistic relationship you were identified with a grandiose construct, i.e. the false Self of the narcissist. Your old identity was demolished, and you were reprogrammed according to the narcissist’s tastes. This false identity is now crumbling, and your ego is undergoing a process of grief. That is what the depression is.
Your ego drew a sense of identity from the narcissist, and it wants that identity back. It does not care what kind of id
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Pendulate between the darkness and the light.
Depression is a healing tonic which restores the Self to a point of equilibrium. Remember that while in the narcissistic relationship you were identified with a grandiose construct, i.e. the false Self of the narcissist. Your old identity was demolished, and you were reprogrammed according to the narcissist’s tastes. This false identity is now crumbling, and your ego is undergoing a process of grief. That is what the depression is.
Your ego drew a sense of identity from the narcissist, and it wants that identity back. It does not care what kind of identity you have; only that you should have one. It does not realise that you can rebuild yourself in a more actualised, empowered way.
Before you can do that, however, you must grieve. Ideally you want to direct all of your awareness into the depression, to expand your consciousness and accept the depression in all of its intensity. However, that might be too overwhelming initially. Instead, take time each day to sit in an upright position and simply direct your consciousness toward the feeling of depression for as long as you can tolerate. Note its intensity. Where in the body does it manifest? In the chest? In your stomach? Let your face droop, let your body soften, let yourself be as sad and depressed as needed. Go with the flow. Do not think about it or analyse it, simply observe it and allow it to happen. This is how you allow the grieving process to complete itself. Just when you think it will never end, it will begin to transform.
But that could be days, weeks or months away. For now, simply take time out each day to do this practice. When you become overwhelmed, and surely you will at the beginning, change up and do something that brings you relaxation and joy. Take a bath, spend time with a good friend, watch your favourite TV show, go for a walk, do exercise. When you are sufficiently filled, go back into the dark and sit there i.e. be conscious with it. You can be sure that when the work is done, the sun will shine again, and the darkness will recede back into the depths of your being. Then the spiritual growth can begin.
Best of luck.
If you have just started your narcissistic abuse recovery journey, check out How To Kill A Narcissist. Or if you wish to immunise yourself against narcissists and move on for good, take a look at How To Bury A Narcissist.
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Phlyn Heubach
BS in Mechanical Engineering, California Polytechnic State University, San Luis Obispo (Graduated 1997) · Author has 479 answers and 886.2K answer views
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What is the area of a regular hexagon whose apothem is 12 mm?
Question: What is the area of a regular hexagon whose apothem is 12 mm?
The area for any regular polygon can be found with the formula:
A=n⋅(1 2⋅s⋅a)A=n⋅(1 2⋅s⋅a)
Where A is the area of the regular polygon
n is the number of sides of the regular polygon
s is the side length of the regular polygon
a is the length of the apothem of the regular polygon
In a regular hexagon the central angle (CA) is 360 6=60∘360 6=60∘
The apothem bisects the central angle and is the perpendicular bisector of a side of the regular polygon.
In a regular hexagon the apothem, half of a side,
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Question: What is the area of a regular hexagon whose apothem is 12 mm?
The area for any regular polygon can be found with the formula:
A=n⋅(1 2⋅s⋅a)A=n⋅(1 2⋅s⋅a)
Where A is the area of the regular polygon
n is the number of sides of the regular polygon
s is the side length of the regular polygon
a is the length of the apothem of the regular polygon
In a regular hexagon the central angle (CA) is 360 6=60∘360 6=60∘
The apothem bisects the central angle and is the perpendicular bisector of a side of the regular polygon.
In a regular hexagon the apothem, half of a side, and a radius of the regular hexagon form a 30–60–90 right triangle. The side lengths of a 30–60–90 right triangle are in the ratio of 1:√3:2 1:3:2
Using the given information and the diagram above:
x=4√3 x=4 3 and s=8√3 s=8 3
Then the area is: A=6(1 2⋅(8√3)⋅12)A=6(1 2⋅(8 3)⋅12)
A=288√3 A=288 3 m m 2 m m 2≈498.8306≈498.8306 m m 2 m m 2
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Gary Ward
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What is the area of an equilateral triangle, given apothem = 2?
What is the area of an equilateral triangle, given apothem = 2?
Since you give the apothem, it is an equilateral triangle made up of 6 30°-60°-90°, 1–√3–2 right triangles with the short leg of 2 units. 2 is the altitude, and the base is 2√3 units.
6⋅1 2⋅2√3⋅2=12√3≈20.7846 u n i t s 2 6⋅1 2⋅2 3⋅2=12 3≈20.7846 u n i t s 2
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What is the area of an equilateral triangle, given apothem = 2?
Since you give the apothem, it is an equilateral triangle made up of 6 30°-60°-90°, 1–√3–2 right triangles with the short leg of 2 units. 2 is the altitude, and the base is 2√3 units.
6⋅1 2⋅2√3⋅2=12√3≈20.7846 u n i t s 2 6⋅1 2⋅2 3⋅2=12 3≈20.7846 u n i t s 2
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Contents
1 History
2 Motivation
3 Cardinality function
4 Constructive definition
5 Cardinal arithmetic
5.1 Successor cardinal
5.2 Cardinal addition
5.2.1 Subtraction
5.3 Cardinal multiplication
5.3.1 Division
5.4 Cardinal exponentiation
5.4.1 Roots
5.4.2 Logarithms
6 The continuum hypothesis
7 See also
8 References
9 External links
Cardinal number
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From Wikipedia, the free encyclopedia
Size of a possibly infinite set
This article is about the mathematical concept. For number words indicating quantity ("three" apples, "four" birds, etc.), see Cardinal numeral.
See also: Cardinality
In mathematics, a cardinal number, or cardinal for short, is what is commonly called the number of elements of a set. In the case of a finite set, its cardinal number, or cardinality is therefore a natural number. For dealing with the case of infinite sets, the infinite cardinal numbers have been introduced, which are often denoted with the Hebrew letter (aleph) marked with subscript indicating their rank among the infinite cardinals.
Cardinality is defined in terms of bijective functions. Two sets have the same cardinality if, and only if, there is a one-to-one correspondence (bijection) between the elements of the two sets. In the case of finite sets, this agrees with the intuitive notion of number of elements. In the case of infinite sets, the behavior is more complex. A fundamental theorem due to Georg Cantor shows that it is possible for two infinite sets to have different cardinalities, and in particular the cardinality of the set of real numbers is greater than the cardinality of the set of natural numbers. It is also possible for a proper subset of an infinite set to have the same cardinality as the original set—something that cannot happen with proper subsets of finite sets.
There is a transfinite sequence of cardinal numbers:
This sequence starts with the natural numbers including zero (finite cardinals), which are followed by the aleph numbers. The aleph numbers are indexed by ordinal numbers. If the axiom of choice is true, this transfinite sequence includes every cardinal number. If the axiom of choice is not true (see Axiom of choice § Independence), there are infinite cardinals that are not aleph numbers.
Cardinality is studied for its own sake as part of set theory. It is also a tool used in branches of mathematics including model theory, combinatorics, abstract algebra and mathematical analysis. In category theory, the cardinal numbers form a skeleton of the category of sets.
History
[edit]
The notion of cardinality, as now understood, was formulated by Georg Cantor, the originator of set theory, in 1874–1884. Cardinality can be used to compare an aspect of finite sets. For example, the sets {1,2,3} and {4,5,6} are not equal, but have the same cardinality, namely three. This is established by the existence of a bijection (i.e., a one-to-one correspondence) between the two sets, such as the correspondence {1→4, 2→5, 3→6}.
Cantor applied his concept of bijection to infinite sets (for example the set of natural numbers N = {0, 1, 2, 3, ...}). Thus, he called all sets having a bijection with N denumerable (countably infinite) sets, which all share the same cardinal number. This cardinal number is called , aleph-null. He called the cardinal numbers of infinite sets transfinite cardinal numbers.
Cantor proved that any unbounded subset of N has the same cardinality as N, even though this might appear to run contrary to intuition. He also proved that the set of all ordered pairs of natural numbers is denumerable; this implies that the set of all rational numbers is also denumerable, since every rational can be represented by a pair of integers. He later proved that the set of all real algebraic numbers is also denumerable. Each real algebraic number z may be encoded as a finite sequence of integers, which are the coefficients in the polynomial equation of which it is a solution, i.e. the ordered n-tuple (a0, a1, ..., an), ai ∈ Z together with a pair of rationals (b0, b1) such that z is the unique root of the polynomial with coefficients (a0, a1, ..., an) that lies in the interval (b0, b1).
In his 1874 paper "On a Property of the Collection of All Real Algebraic Numbers", Cantor proved that there exist higher-order cardinal numbers, by showing that the set of real numbers has cardinality greater than that of N. His proof used an argument with nested intervals, but in an 1891 paper, he proved the same result using his ingenious and much simpler diagonal argument. The new cardinal number of the set of real numbers is called the cardinality of the continuum and Cantor used the symbol for it.
Cantor also developed a large portion of the general theory of cardinal numbers; he proved that there is a smallest transfinite cardinal number (, aleph-null), and that for every cardinal number there is a next-larger cardinal
His continuum hypothesis is the proposition that the cardinality of the set of real numbers is the same as . This hypothesis is independent of the standard axioms of mathematical set theory, that is, it can neither be proved nor disproved from them. This was shown in 1963 by Paul Cohen, complementing earlier work by Kurt Gödel in 1940.
Motivation
[edit]
In informal use, a cardinal number is what is normally referred to as a counting number, provided that 0 is included: 0, 1, 2, .... They may be identified with the natural numbers beginning with 0. The counting numbers are exactly what can be defined formally as the finite cardinal numbers. Infinite cardinals only occur in higher-level mathematics and logic.
More formally, a non-zero number can be used for two purposes: to describe the size of a set, or to describe the position of an element in a sequence. For finite sets and sequences it is easy to see that these two notions coincide, since for every number describing a position in a sequence we can construct a set that has exactly the right size. For example, 3 describes the position of 'c' in the sequence <'a','b','c','d',...>, and we can construct the set {a,b,c}, which has 3 elements.
However, when dealing with infinite sets, it is essential to distinguish between the two, since the two notions are in fact different for infinite sets. Considering the position aspect leads to ordinal numbers, while the size aspect is generalized by the cardinal numbers described here.
The intuition behind the formal definition of cardinal is the construction of a notion of the relative size or "bigness" of a set, without reference to the kind of members which it has. For finite sets this is easy; one simply counts the number of elements a set has. In order to compare the sizes of larger sets, it is necessary to appeal to more refined notions.
A set Y is at least as big as a set X if there is an injective mapping from the elements of X to the elements of Y. An injective mapping identifies each element of the set X with a unique element of the set Y. This is most easily understood by an example; suppose we have the sets X = {1,2,3} and Y = {a,b,c,d}, then using this notion of size, we would observe that there is a mapping:
: 1 → a
: 2 → b
: 3 → c
which is injective, and hence conclude that Y has cardinality greater than or equal to X. The element d has no element mapping to it, but this is permitted as we only require an injective mapping, and not necessarily a bijective mapping. The advantage of this notion is that it can be extended to infinite sets.
We can then extend this to an equality-style relation. Two sets X and Y are said to have the same cardinality if there exists a bijection between X and Y. By the Schroeder–Bernstein theorem, this is equivalent to there being both an injective mapping from X to Y, and an injective mapping from Y to X. We then write |X| = |Y|. The cardinal number of X itself is often defined as the least ordinal a with |a| = |X|. This is called the von Neumann cardinal assignment; for this definition to make sense, it must be proved that every set has the same cardinality as some ordinal; this statement is the well-ordering principle. It is however possible to discuss the relative cardinality of sets without explicitly assigning names to objects.
The classic example used is that of the infinite hotel paradox, also called Hilbert's paradox of the Grand Hotel. Supposing there is an innkeeper at a hotel with an infinite number of rooms. The hotel is full, and then a new guest arrives. It is possible to fit the extra guest in by asking the guest who was in room 1 to move to room 2, the guest in room 2 to move to room 3, and so on, leaving room 1 vacant. We can explicitly write a segment of this mapping:
: 1 → 2
: 2 → 3
: 3 → 4
: ...
: n → n + 1
: ...
With this assignment, we can see that the set {1,2,3,...} has the same cardinality as the set {2,3,4,...}, since a bijection between the first and the second has been shown. This motivates the definition of an infinite set being any set that has a proper subset of the same cardinality (i.e., a Dedekind-infinite set); in this case {2,3,4,...} is a proper subset of {1,2,3,...}.
When considering these large objects, one might also want to see if the notion of counting order coincides with that of cardinal defined above for these infinite sets. It happens that it does not; by considering the above example we can see that if some object "one greater than infinity" exists, then it must have the same cardinality as the infinite set we started out with. It is possible to use a different formal notion for number, called ordinals, based on the ideas of counting and considering each number in turn, and we discover that the notions of cardinality and ordinality are divergent once we move out of the finite numbers.
It can be proved that the cardinality of the real numbers is greater than that of the natural numbers just described. This can be visualized using Cantor's diagonal argument; classic questions of cardinality (for instance the continuum hypothesis) are concerned with discovering whether there is some cardinal between some pair of other infinite cardinals. In more recent times, mathematicians have been describing the properties of larger and larger cardinals.
Since cardinality is such a common concept in mathematics, a variety of names are in use. Sameness of cardinality is sometimes referred to as equipotence, equipollence, or equinumerosity. It is thus said that two sets with the same cardinality are, respectively, equipotent, equipollent, or equinumerous.
Cardinality function
[edit]
Main article: Cardinal function
The cardinality function is a cardinal function which takes in a set and returns its cardinal number: . For finite sets, this is simply the natural number found by counting the elements. This function applied to a set is generally denoted by with a vertical bar on each side, though it may also be denoted by , or
For infinite sets, "cardinal number" is somewhat more difficult to define formally. Cardinal numbers are not usually thought of in terms of their formal definition, but immaterially in terms of their arithmetic/algebraic properties. Thus, The assumption that there is some cardinal function which satisfies , sometimes called the axiom of cardinality or Hume's principle, is sufficient for deriving most properties of cardinal numbers. It will be shown later that such a function can be constructed without the need to define it axiomatically.
Constructive definition
[edit]
Formally, assuming the axiom of choice, the cardinality of a set X is the least ordinal number α such that there is a bijection between X and α. This definition is known as the von Neumann cardinal assignment. If the axiom of choice is not assumed, then a different approach is needed. The oldest definition of the cardinality of a set X (implicit in Cantor and explicit in Frege and Principia Mathematica) is as the class [X] of all sets that are equinumerous with X. This does not work in ZFC or other related systems of axiomatic set theory because if X is non-empty, this collection is too large to be a set. In fact, for X ≠ ∅ there is an injection from the universe into [X] by mapping a set m to {m} × X, and so by the axiom of limitation of size, [X] is a proper class. The definition does work however in type theory and in New Foundations and related systems. However, if we restrict from this class to those equinumerous with X that have the least rank, then it will work (this is a trick due to Dana Scott: it works because the collection of objects with any given rank is a set).
Von Neumann cardinal assignment implies that the cardinal number of a finite set is the common ordinal number of all possible well-orderings of that set, and cardinal and ordinal arithmetic (addition, multiplication, power, proper subtraction) then give the same answers for finite numbers. However, they differ for infinite numbers. For example, in ordinal arithmetic while in cardinal arithmetic, although the von Neumann assignment puts . On the other hand, Scott's trick implies that the cardinal number 0 is , which is also the ordinal number 1, and this may be confusing. A possible compromise (to take advantage of the alignment in finite arithmetic while avoiding reliance on the axiom of choice and confusion in infinite arithmetic) is to apply von Neumann assignment to the cardinal numbers of finite sets (those which can be well ordered and are not equipotent to proper subsets) and to use Scott's trick for the cardinal numbers of other sets.
Formally, the order among cardinal numbers is defined as follows: |X| ≤ |Y| means that there exists an injective function from X to Y. The Cantor–Bernstein–Schroeder theorem states that if |X| ≤ |Y| and |Y| ≤ |X| then |X| = |Y|. The axiom of choice is equivalent to the statement that given two sets X and Y, either |X| ≤ |Y| or |Y| ≤ |X|.
A set X is Dedekind-infinite if there exists a proper subset Y of X with |X| = |Y|, and Dedekind-finite if such a subset does not exist. The finite cardinals are just the natural numbers, in the sense that a set X is finite if and only if |X| = |n| = n for some natural number n. Any other set is infinite.
Assuming the axiom of choice, it can be proved that the Dedekind notions correspond to the standard ones. It can also be proved that the cardinal (aleph null or aleph-0, where aleph is the first letter in the Hebrew alphabet, represented ) of the set of natural numbers is the smallest infinite cardinal (i.e., any infinite set has a subset of cardinality ). The next larger cardinal is denoted by , and so on. For every ordinal α, there is a cardinal number and this list exhausts all infinite cardinal numbers.
Cardinal arithmetic
[edit]
We can define arithmetic operations on cardinal numbers that generalize the ordinary operations for natural numbers. It can be shown that for finite cardinals, these operations coincide with the usual operations for natural numbers. Furthermore, these operations share many properties with ordinary arithmetic.
Successor cardinal
[edit]
Further information: Successor cardinal
If the axiom of choice holds, then every cardinal κ has a successor, denoted κ+, where κ+ > κ and there are no cardinals between κ and its successor. (Without the axiom of choice, using Hartogs' theorem, it can be shown that for any cardinal number κ, there is a minimal cardinal κ+ such that ) For finite cardinals, the successor is simply κ + 1. For infinite cardinals, the successor cardinal differs from the successor ordinal.
Cardinal addition
[edit]
If X and Y are disjoint, addition is given by the union of X and Y. If the two sets are not already disjoint, then they can be replaced by disjoint sets of the same cardinality (e.g., replace X by X×{0} and Y by Y×{1}).
:
Zero is an additive identity κ + 0 = 0 + κ = κ.
Addition is associative (κ + μ) + ν = κ + (μ + ν).
Addition is commutative κ + μ = μ + κ.
Addition is non-decreasing in both arguments:
Assuming the axiom of choice, addition of infinite cardinal numbers is easy. If either κ or μ is infinite, then
Subtraction
[edit]
Assuming the axiom of choice and, given an infinite cardinal σ and a cardinal μ, there exists a cardinal κ such that μ + κ = σ if and only if μ ≤ σ. It will be unique (and equal to σ) if and only if μ < σ.
Cardinal multiplication
[edit]
The product of cardinals comes from the Cartesian product.
:
Zero is a multiplicative absorbing element: κ·0 = 0·κ = 0.
There are no nontrivial zero divisors: κ·μ = 0 → (κ = 0 or μ = 0).
One is a multiplicative identity: κ·1 = 1·κ = κ.
Multiplication is associative: (κ·μ)·ν = κ·(μ·ν).
Multiplication is commutative: κ·μ = μ·κ.
Multiplication is non-decreasing in both arguments: κ ≤ μ → (κ·ν ≤ μ·ν and ν·κ ≤ ν·μ).
Multiplication distributes over addition: κ·(μ + ν) = κ·μ + κ·ν and (μ + ν)·κ = μ·κ + ν·κ.
Assuming the axiom of choice, multiplication of infinite cardinal numbers is also easy. If either κ or μ is infinite and both are non-zero, then
Thus the product of two infinite cardinal numbers is equal to their sum.
Division
[edit]
Assuming the axiom of choice and given an infinite cardinal π and a non-zero cardinal μ, there exists a cardinal κ such that μ · κ = π if and only if μ ≤ π. It will be unique (and equal to π) if and only if μ < π.
Cardinal exponentiation
[edit]
Exponentiation is given by
where XY is the set of all functions from Y to X. It is easy to check that the right-hand side depends only on and .
: κ0 = 1 (in particular 00 = 1), see empty function.
: If μ ≥ 1, then 0μ = 0.
: 1μ = 1.
: κ1 = κ.
: κμ + ν = κμ·κν.
: κμ · ν = (κμ)ν.
: (κ·μ)ν = κν·μν.
Exponentiation is non-decreasing in both arguments:
: (1 ≤ ν and κ ≤ μ) → (νκ ≤ νμ) and
: (κ ≤ μ) → (κν ≤ μν).
2|X| is the cardinality of the power set of the set X and Cantor's diagonal argument shows that 2|X| > |X| for any set X. This proves that no largest cardinal exists (because for any cardinal κ, we can always find a larger cardinal 2κ). In fact, the class of cardinals is a proper class. (This proof fails in some set theories, notably New Foundations.)
All the remaining propositions in this section assume the axiom of choice:
: If κ and μ are both finite and greater than 1, and ν is infinite, then κν = μν.
: If κ is infinite and μ is finite and non-zero, then κμ = κ.
If 2 ≤ κ and 1 ≤ μ and at least one of them is infinite, then:
: Max (κ, 2μ) ≤ κμ ≤ Max (2κ, 2μ).
Using König's theorem, one can prove κ < κcf(κ) and κ < cf(2κ) for any infinite cardinal κ, where cf(κ) is the cofinality of κ.
Roots
[edit]
Assuming the axiom of choice and, given an infinite cardinal κ and a finite cardinal μ greater than 0, the cardinal ν satisfying will be .
Logarithms
[edit]
Assuming the axiom of choice and, given an infinite cardinal κ and a finite cardinal μ greater than 1, there may or may not be a cardinal λ satisfying . However, if such a cardinal exists, it is infinite and less than κ, and any finite cardinality ν greater than 1 will also satisfy .
The logarithm of an infinite cardinal number κ is defined as the least cardinal number μ such that κ ≤ 2μ. Logarithms of infinite cardinals are useful in some fields of mathematics, for example in the study of cardinal invariants of topological spaces, though they lack some of the properties that logarithms of positive real numbers possess.
The continuum hypothesis
[edit]
The continuum hypothesis (CH) states that there are no cardinals strictly between and The latter cardinal number is also often denoted by ; it is the cardinality of the continuum (the set of real numbers). In this case
Similarly, the generalized continuum hypothesis (GCH) states that for every infinite cardinal , there are no cardinals strictly between and . Both the continuum hypothesis and the generalized continuum hypothesis have been proved to be independent of the usual axioms of set theory, the Zermelo–Fraenkel axioms together with the axiom of choice (ZFC).
Indeed, Easton's theorem shows that, for regular cardinals , the only restrictions ZFC places on the cardinality of are that , and that the exponential function is non-decreasing.
See also
[edit]
Mathematics portal
Aleph number
Beth number
The paradox of the greatest cardinal
Cardinal number (linguistics)
Counting
Inclusion–exclusion principle
Large cardinal
Names of numbers in English
Nominal number
Ordinal number
Regular cardinal
References
[edit]
Notes
^ Dauben 1990, pg. 54
^ Weisstein, Eric W. "Cardinal Number". mathworld.wolfram.com. Retrieved 2020-09-06.
^ Hrbáček & Jech 2017, p. 65
^ Kleene 1952, p. 9.
^ Krivine 1971, p. 23.
^ Kuratowski 1968, p. 174.
^ Stoll 1963, p. 80.
^ Suppes 1972, p. 109.
^ Takeuti & Zaring 1982, p. 85.
^ Bourbaki 1968, p. 158.
^ Enderton 1977, p. 136.
^ Halmos 1998, p. 94.
^ Schumacher 1996, p. 103.
^ Halmos 1998, p. 53.
^ Pinter 2014, Page 3 of Chaper 8.
^ Tao 2022, p. 60.
^
^ Kleene 1952, p. 9
^ Pinter 2014, Page 2 of Chapter 8
^ Potter, Michael (2004-01-15). Set Theory and its Philosophy: A Critical Introduction. Clarendon Press. ISBN 978-0-19-155643-2.
^ Enderton 1977, p. 136
^ Deiser, Oliver (May 2010). "On the Development of the Notion of a Cardinal Number". History and Philosophy of Logic. 31 (2): 123–143. doi:10.1080/01445340903545904. S2CID 171037224.
^ Enderton, Herbert. "Elements of Set Theory", Academic Press Inc., 1977. ISBN 0-12-238440-7
^ Friedrich M. Hartogs (1915), Felix Klein; Walther von Dyck; David Hilbert; Otto Blumenthal (eds.), "Über das Problem der Wohlordnung", Math. Ann., Bd. 76 (4), Leipzig: B. G. Teubner: 438–443, doi:10.1007/bf01458215, ISSN 0025-5831, S2CID 121598654, archived from the original on 2016-04-16, retrieved 2014-02-02
^ a b c Schindler 2014, pg. 34
^ Robert A. McCoy and Ibula Ntantu, Topological Properties of Spaces of Continuous Functions, Lecture Notes in Mathematics 1315, Springer-Verlag.
^ Eduard Čech, Topological Spaces, revised by Zdenek Frolík and Miroslav Katetov, John Wiley & Sons, 1966.
^ D. A. Vladimirov, Boolean Algebras in Analysis, Mathematics and Its Applications, Kluwer Academic Publishers.
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Stoll, Robert R. (1963). Set Theory and Logic. San Francisco: W. H. Freeman. LCCN 63-8995.
Suppes, Patrick (1972) . Axiomatic Set Theory. Dover Books on Mathematics. New York: Dover Publications. ISBN 0-486-61630-4. ISSN 2693-051X. LCCN 72-86226. Archived from the original on 2014-08-06. Alt URL
Takeuti, Gaisi; Zaring, Wilson M (1982). Introduction to Axiomatic Set Theory. Graduate Texts in Mathematics (2nd ed.). New York: Springer-Verlag. doi:10.1007/978-1-4613-8168-6. ISBN 0-387-90683-5. ISSN 0072-5285. LCCN 81-8838. Archived from the original on 2014-08-06.
Tao, Terence (2022). Analysis I. Texts and Readings in Mathematics (4th ed.). Singapore: Springer Science+Business Media. doi:10.1007/978-3-662-00274-2. ISBN 978-981-19-7261-4. ISSN 2366-8717.
External links
[edit]
"Cardinal number", Encyclopedia of Mathematics, EMS Press, 2001
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Given the following matrices A and B, find (A-B).A =0 1 22 4 03 0 4B =2 3 41 1 12 3 211.) Find (A x B), using the matrices A and B from problem number 9 above.
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Question: Given the following matrices A and B, find (A-B).A =0 1 22 4 03 0 4B =2 3 41 1 12 3 211.) Find (A x B), using the matrices A and B from problem number 9 above.
Given the following matrices A and B, find (A-B). A =0 1 2 2 4 0 3 0 4 B =2 3 4 1 1 1 2 3 2 1 1.) Find (A x B), using the matrices A and B from problem number 9 above.
There are 3 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 Given : A=[0 1 2 2 4 0 3 0 4]B=[2 3 4 1 1 1 2 3 2] Let's find out the given matrix step by step. Explanation: Here, we need to find the steps,(A−B)and(A×B) step by step. View the full answer Step 2 UnlockStep 3 UnlockAnswer Unlock Previous questionNext question
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Surface Energy: Formula & Definition
Surface free energy is a measure of the excess energy present at the surface of a material, in comparison to at its bulk. It can be used to describe wetting and adhesion between materials, but is not often used quantitatively. This article will explain the fundamental definition of surface energy, how to calculate it directly using contact angles, and how to tune it to fit your needs. For further information on contact angles, see Ossila's guides to the background theory of contact angle measurements, and how to take them using the Ossila Contact Angle Goniometer.
For information on how to take contact angle measurements, see contact angle measurements of surface wetting
Contents
What is Surface Energy?
How can Surface Energy be Calculated?
Zisman Model
Fowkes & Extended Fowkes Model
Owens-Wendt-Rabel & Kaelble Model
Van Oss-Good Model
Which Model Should I Use?
How can Surface Energy be changed?
UV Ozone Treatment
Plasma & Corona Treatment
Lowering Surface Energy
References
What is Surface Energy?
In the bulk form of a material, atoms are generally stable and have a balanced set of bonds/interactions. In contrast, the surface atoms will have an incomplete, unbalanced set of interactions, and therefore have unrealised bonding energy. Surface energy is a relative measurement of the energy at the surface (which is a result of this incomplete bonding). It is positively correlated to the strength of bulk interactions and the level of surface exposure. Therefore, surface energy will be higher if the bulk interactions are stronger, or if the surface exposure is greater.
A surface will always try and minimise its energy. This can be done by adsorbing a material with a lower energy onto its surface. Through the adsorption process, the number of exposed surface atoms with high surface energy are minimised and replaced with lower energy atoms or molecules. Liquids will often have lower surface energies than solids (due to the weak forces interacting between molecules), which is why liquids usually spread out. Surface energy can be defined as the energy required per unit area to increase the size of the surface, and as such is often quoted in units of mN/m.
A basic depiction of bonding in a solid, showing balanced interactions in the bulk and unbalanced interactions at the surface
Surface energy is mostly used to describe a solid. The tendency of a solution to spread out (or 'wet') on a solid surface depends on several factors. Here it is assumed that the liquid and solid are capable of similar polarity interactions.
Generally, a surface with a low surface energy will cause poor wetting, and therefore result in a high contact angle. This is because the surface is not capable of forming strong bonds, so there is little energetic reward for the liquid to break bulk bonding in favour of interacting with the surface. Typical surfaces with low surface energy include hydrocarbons, as these are held together with weak Van der Waals forces. A high surface tension liquid will also cause this, as it is more favourable for the solution to maintain bonds with itself.
The basic relationship between contact angle and surface energy. Surface tension of the liquid will also play a role, as will the type of interactions between the liquid and solid.
The opposite is true for a high surface energy, which will generally cause good wetting and a low contact angle. Examples of surfaces with high surface energy include glasses, ceramics, metals, and materials held together by stronger bonds (e.g. ionic, covalent and metallic). A low surface tension liquid will also give a low contact angle- as interactions within the liquid are not that strong.
How can Surface Energy be Calculated?
Surface energy is usually measured indirectly with contact angle measurements, which can be a useful qualitative measure. Quantitative data can also be obtained by using a variety of different models. Most of these are based on Young's equation,1 a form of which is seen in Equation 1:
Equation 1: Young's equation relating surface energy of a solid, surface tension of a liquid and the interfacial tension between the liquid and solid
In the above equation, σ s is the surface energy of the solid, σ sl is the interfacial tension between liquid and solid, σ l is the surface tension of the liquid, and θ is the contact angle of the liquid on the solid.
Dupré's equation can be used to relate interfacial tension and the interactions between the solid and liquid. Interactions are generally described by the work of adhesion, W sl, which represents the work which must be done to separate the two phases, or the energy released in wetting. This is seen in Equation 2:
Equation 2: Dupré's equation relating interfacial tension between a solid and liquid and the interactions between the phases
This relationship is similar to (but not exactly the same as) that between the interfacial tension and the spreading coefficient, S:
Equation 3: Equation relating the interfacial tension and spreading coefficient between a liquid and solid
The spreading coefficient measures the tendency for a liquid to wet on a solid, where wetting occurs for positive values.
Equation 2 and Equation 3 can be combined into the Young-Dupré equation, which is the basic form used by many of the surface energy models:
Equation 4: The Young-Dupré equation, relating the interactions between the liquid and solid to surface tension and contact angle
Different models will include different interactions in their calculation of surface energy, meaning values, even for the same sample, are not always directly comparable between them.
Zisman Model
One of the most basic and widely used methods of calculating surface energy is the Zisman model, published in 1964.2 This model assumes that the surface energy of a solid is equal to the maximum surface tension liquid that will give a contact angle of 0°. This is known as the critical surface tension.
Contact angle on the surface (in the form cosθ) for a series of liquids is plotted against surface tension. This is then extrapolated to find the surface tension for cosθ = 1, where θ = 0°, and surface tension is equal to the surface energy of the solid.
A basic Zisman plot, where cosθ is plotted against surface tension. The maximum surface tension that achieves θ = 0° is equal to surface energy. There may be liquids with lower surface tension that also achieve complete wetting (as seen above).
As the Zisman model ignores the impact of polar interactions, it is only accurate for non-polar surfaces (such as polyethylene). For polar surfaces (including those that contain heteroatoms), a model that includes polar interactions must be used.
Fowkes & Extended Fowkes Model
The most common method of calculating surface energy for polar surfaces is the Fowkes model, published in 1964.3 Fowkes' theory assesses the interactions between a liquid and solid in terms of 'dispersive' (Van der Waals) and 'polar' interactions. These interactions will sum to form overall energy:
Equation 5: Surface energy for Fowkes theory is a sum of dispersive and polar components. The same is true for surface tension of the liquid (not shown)
Where σ s D and σ s P are the dispersive and polar components of the surface energy of the solid respectively.
Initial Fowkes theory is shown in Equation 6:
Equation 6: Fowkes theory in terms of work of adhesion
Where σ l D and σ l P are the dispersive and polar components of the surface tension of the liquid respectively. This theory uses the geometric mean for each type of interaction, in contrast to the similar Wu model,4 which uses the harmonic mean.
By combining Equation 4 and Equation 6, we can obtain Fowkes' primary equation, seen in Equation 7:
Equation 7: Fowkes equation relating the solid and liquid phase interactions with surface tension and contact angle
The first step for this surface energy calculation is to measure the contact angle for a purely dispersive liquid, meaning σ l P = 0 and σ l = σ l D, reducing Equation 7to Equation 8:
Equation 8: Reduced form of Fowkes equation using a purely dispersive liquid
Using this, σ s D can be calculated directly, when the surface tension for the liquid is known. A common material used here is diiodomethane, which has effectively no polar component to its surface tension (due to molecular symmetry), meaning σ l = σ l D = 50.8 mN/m.5
The chemical structure of diiodomethane, which is often used to calculate surface energy due to having only dispersive components to its interactions
The next step is to measure the contact angle for a liquid with known dispersive and polar components. A common material used here is water, where σ l P = 51.0 mN/m and σ l D = 21.80 mN/m.5 By inserting this into Equation 7, along with surface tension and previously calculated σ s D, the value of σ s P can be calculated. The final surface energy value is then a sum of the surface energy components, as seen in Equation 5.
This model is best for low-charge surfaces that are moderately polar, such as polymers featuring heteroatoms.
Fowkes' model can also be extended6 to include a third interaction component, σ H, which describes the hydrogen bonding within the phase. This calculation therefore requires three liquids with known components, and thus is not used as often as the standard Fowkes model.
Owens-Wendt-Rabel & Kaelble Model
The Owens-Wendt-Rabel & Kaelble model (OWRK) was published in 19697 and 1970.8 It is mathematically equivalent to the Fowkes model, but derived from different principles.
The OWRK model is shown in Equation 9:
Equation 9: Owens-Wendt-Rabel & Kaelble model, which is mathematically equivalent to the Fowke's equation seen in Eq.7, but has been expressed in the form y = mx + c
OWRK requires at least two liquids with known dispersive and polar interactions. If these are not known they can be determined with a reference solid (such as untreated PTFE). This can be assumed9 to have a surface energy of ~18.0 mN/m and be incapable of polar interactions, meaning σ s = σ s D = 18 and σ s P = 0. By inserting this into Equation 9, we can achieve Equation 10:
Equation 10: A modification of the OWRK model that can be used to calculate the dispersive interaction of liquids from their contact angle on PTFE
As a result, a contact angle on PTFE can be used to calculate σ l D for any liquid where the overall surface tension is known. σ l P can then be calculated using the difference between the overall surface tension and σ l D.
Once dispersive and polar interactions of the liquids are known, they can be used to calculate surface energy of a new surface, by plotting each liquid on a graph in the form of Equation 9. The slope between the points of the liquids will equate to √σ s P, and the intercept will equate to √σ s D. The overall surface energy can then be calculated using Equation 5.
An OWRK plot, where contact angle is plotted against surface tension in the form of Eq.9, and the components of surface energy are found from the intercept and gradient of a best fit line.
OWRK can be used for similar materials to Fowkes, but fits better to slightly lower energy surfaces, and requires much more experimental work.
Van Oss-Good Model
Whilst dispersive and polar interactions represent many surfaces well, other significant interactions (such as hydrogen bonds) are not considered by either Fowkes or OWRK. The Van Oss-Good model was published in 199210 and considers acid-base interactions. This model combines polar and dispersive interactions into one term, alongside acid (σ +) and base (σ -) components. The acid component describes the ability of a surface to interact with a basic liquid (i.e. one that can donate electron density) through polar interactions, such as dipole-dipole bonding and hydrogen bonding. The base component describes the opposite of this (e.g. interactions with an electron-accepting, acidic liquid).
The primary equation for the Oss-Good model can be seen in Equation 11:
Equation 11: The Oss-Good model, where the interactions between phases are expressed in the form of dispersive, acidic and basic interactions
By choosing a liquid with only a dispersive component, such as cyclohexane, the equation reduces to Equation 12:
Equation 12: The reduced form of the Oss-Good model where a purely dispersive liquid is used
This means the contact angle can be used to calculate σ s D. Repeating this with a liquid with no acid component, such as tetrahydrofuran, gives Equation 13:
Equation 13: The reduced form of the Oss-Good model where a liquid with no acidic component is used
This can be used to calculate σ s+. Repeating this with a liquid with no base component (e.g. chloroform), will give σ s-, which can be used to calculate σ s by summing all three components. Alternatively, once σ s D and either σ s+ or σ s- is known, the remaining component can be calculated from Equation 11 by using a liquid with known acid and base components (e.g. water). This is useful if a liquid with no acid/base component is difficult to find.
The Oss-Good model works best for polar surfaces, such as organometallics. However, it can be difficult to establish an unknown acid and base components for liquids because there is no defined set of reference solids.
Which Model Should I Use?
Non-polar surfaces with low surface energies: Zisman
Moderately polar surfaces: Fowkes (better for slightly higher energies) or OWRK
Polar surfaces (inorganic, organometallic and ionic): Oss-Good
To find polar and dispersive interactions of liquids: Reference surface &Fowkes/OWRK
There are also additional surface energy models that are not discussed here, such as:
The Wu model:4 Good for materials with low surface energy up to 40 mN/m
The Schultz model:11 Used for high energy surfaces like bare metals
The Neumann equation of state:12 Approximates surface energy with only a single liquid by ignoring the impact of different interaction components
How can Surface Energy be changed?
A high surface energy is important for solution wetting, especially in processes such as spin coating. Most solids with high surface energy will not maintain a high-energy surface when exposed to atmospheric conditions. Hydrocarbon contaminants present within the air will adsorb onto the solid's surface, reducing the surface energy. The most common method of tuning surface energy is through surface treatment, which is typically designed to raise the energy by removing these contaminants or forming high surface energy functional groups. Many of these techniques only produce temporary changes to the surface energy. This is because adsorption of low surface energy molecules will occur over time slowly reducing the average surface energy.
UV Ozone Treatment
In a UV ozone cleaner, UV radiation will ionise atmospheric oxygen to form oxygen radicals, which will go on to react and form ozone. Radiation of a different wavelength will hit organic molecules, exciting them or forming free radicals, which will allow reaction with the ozone. This forms volatile molecules which can easily desorb from the surface, leaving it clean from contaminants, and with an increased surface energy.
Plasma & Corona Treatment
Plasma treatment is also commonly used to clean surfaces and increase surface energy. Here a gas or gas mixture is ionised by a high frequency voltage to form a reactive gas plasma. This is a mixture of free radicals, ions, electrons and gas molecules. The plasma will interact with the surface in several ways. Contaminants can be removed via ablation, by bombardment with electrons and ions, or through reactions with species in the plasma. This can form volatile organic molecules that will desorb (and sometimes also replace) contaminants with high surface energy functional groups, such as C=O and HO-. In some cases, plasma treatment can etch the surface and affect surface roughness. The exact impact of plasma treatment will depend on the gas used, for example oxygen plasma will react with and oxidise organic contaminants, whereas argon and other inert gases may only remove contaminants through ablation.
A depiction of plasma treatment using oxygen as the gas
Corona treatment is similar to atmospheric plasma, but uses a lower plasma density. Here, the ionised gas is created by discharging a high frequency voltage across an electrode over a grounded surface. The film is passed underneath the electrode, and the surface is oxidised to remove contaminants and raise surface energy.
Beyond UV ozone and plasma treatment, other common methods of raising surface energy include flame treatment, etching and coating a high surface energy interfacial layer of a different material.
Lowering Surface Energy
In some cases (e.g. to generate a hydrophobic coating), surface energy needs to be lowered instead of increased. This is typically done with a coating of a lower surface energy material (e.g. a wax), or through specific surface microstructures that are designed to minimise solution wetting.13
UV Ozone Cleaner Get a quote or buy today
Contributing Authors
Written by
Emma Spooner
PhD Student Collaborator
References
A Thermodynamic Derivation of Wenzel’s Modification of Young’s Equation for Contact Angles; Together with a Theory of Hysteresis, R. J. Good, J. Am. Chem. Soc. (74), 5041–5042 (1952); DOI: 10.1021/ja01140a014.
Relation of the Equilibrium Contact Angle to Liquid and Solid Constitution in Contact Angle, Wettability, and Adhesion, W. A. Zisman (ed. R. F. Gould), American Chemical Society, Advances in Chemistry series (43), Ch1., 1-51 (1964); DOI: 10.1021/ba-1964-0043.fw001.
Attractive Forces At Interfaces, F. M. Fowkes, Ind. Eng. Chem. (56), 40–52 (1964); DOI: 10.1021/ie50660a008.
Calculation of Interfacial Tension in Polymer Systems, S. Wu, J. Polym. Sci. (30), 19–30 (1971); DOI: 10.1002/polc.5070340105.
Surface free-energy components of liquids and low energy solids and contact angles, B. Jańczuk et al., J. Colloid Interface Sci. (127), 189–204 (1989); DOI: 10.1016/0021-9797(89)90019-2.
Studies on the surface wettability and surface structure of polyester treated with low temperature plasma, C. Jie-Rong et al., J. Appl. Polym. Sci. (63), 1733–1739 (1997); DOI: 10.1002/(SICI)1097-4628(19970328)63:13<1733::AID-APP4>3.0.CO;2-H.
Estimation of the surface free energy of polymers, D. K. Owens et al., J. Appl. Polym. Sci. (13), 1741–1747 (1969); DOI: 10.1002/app.1969.070130815.
Dispersion-Polar Surface Tension Properties of Organic Solids, D. H. Kaelble, J. Adhes. (2), 66–81 (1970); DOI: 10.1080/0021846708544582.
Wettability of Halogenated Organic Solid Surfaces, A. H. Ellison et al., J. Phys. Chem. (58), 260–265 (1954), DOI: 10.1021/j150513a020.
The Modern Theory of Contact Angles and the Hydrogen Bond Components of Surface Energies in Modern Approaches to Wettability: Theory and Applications, R. J. Good et al. (ed. M. E. Schrader), Springer US, Ch.1, 1–27 (1992); DOI: 10.1007/978-1-4899-1176-6_1.
Fibre Surface Energy Characterization, J. Schultz et al, J. Adhes. (12), 221–231 (1981); DOI: 10.1080/00218468108071202.
An equation-of-state approach to determine surface tensions of low-energy solids from contact angles, A. W. Neumannn et al., J. Colloid Interface Sci. (49), 291–304 (1974); DOI: 10.1016/0021-9797(74)90365-8.
Super-Hydrophobic Surfaces: From Natural to Artificial, L. Feng et al., Adv. Mater. (14), 1857–1860 (2002); DOI: 10.1002/adma.200290020.
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The Physics Classroom » Calc Pad » Electric Fields, Potential, and Capacitance » Equation Overview
Electricity: Electric Field, Potential, and Capacitance
Problem Sets(non-Calculus) || Problem Sets (Calculus-Based)||Overview of Physics
Electrical Energy and Capacitors: Problem Set Overview
There are nine ready-to-use problem sets on the topic of Electrical Energy and Capacitors that are not reliant upon calculus. There are another six problem sets that are calculus-based. Most problems are multi-part problems requiring an extensive analysis.The problems target your ability to use the concepts of electric field, electric potential, electric potential energy, and electric capacitance to solve problems related to the interaction of charges with electrical fields.
Electric Fields
Fields of different types are all around us and are responsible for the behavior of matter/energy over time. A field is a physical quantity that has a value at each point in space. Some examples of fields:
A source of heat creates a temperature field around it.
A sink of heat creates a temperature field around it.
A source of mass creates a gravitational field around it.
A source of charge creates an electric field around it.
A source of magnetism creates a magnetic field around it.
A certain characteristic of matter/energy reacts to these fields.
Molecules can change their kinetic energy when placed in a temperature field.
The physical characteristic of having mass means an object will experience a force when placed in a gravitational field.
The physical characteristic of being charged means an object will experience a force when placed in an electric field.
The physical characteristic of being magnetic means an object will experience a force when placed in a magnetic field.
The physical characteristic of being a moving charge means an object will experience an electromagentic force when placed in a magnetic field.
The physical characteristic of having electron spin aligned means an object will experience an electromagentic force when placed in a magnetic field.
Due to their familiarity and similarity, we will begin by considering the gravitational field and the electric field which are both vector fields.
Any mass in the universe sets up a gravity field around it and any charged object sets up an electric field around it. Therefore, any object having mass will feel a force in the gravity field and any charged object or object able to be polarized will feel a force in an electric field.
These fields can be represented visually by lines with arrows indicating the direction of force on an object placed in the field as shown below. By convention the direction of the arrow in an electric field is determined by the direction of the force on a positive charge.
The field can be calculated by determining the force per unit of mass, m, or unit of charge,q, placed in the field. The gravity field value would be
and the electric field value would be
Diagrams of a single object experiencing a force in a field created by a single mass and a single charge.
Using Newton’s Universal Law of Gravity for F g, the gravity field value would be
and using Coulomb’s Law for F e, the Electric field value would
where Mand Qare the physical characteristics of the objects creating the fields.
Analysis of Point Charges
A point charge can be a single object of Charge, Q. The object could also be a sphere or hollow shell of many charges that are distributed in different ways. If the object is conductive all the charges adding up to charge, Q, are spread out on the outer surface. If the object is not conductive the charge could be spread throughout the object in various ways. No matter how the charge is distributed the electric field, E, outside the object can be determined using the equation,
and its electric field lines will be radially configured as shown at right if the object is positive.
The radial lines have large distances between them as you get farther away from the center. This is an indication that the field is weaker. The comparison of spacing between E-field lines can be used to compare the strength of the field at various locations around a charged object. For example, Position 1 is at a weaker field location than Position 2.
When there are multiple E-Field sources in a region, the overall E-Field is the vector sum of E-fields from all sources. This is called The Principle of Superposition. The E-fields of all individual charges is calculated at a specific point and then replaced with the vector sum of these E-fields to get the overall E-field at this point.
The diagrams below show the typical electric field line patterns that are obtained with 2 Point Charges.
Constant Electric Field
The value of E-fields created by a small number of separate charges is minimal due to the variable spacing of the field lines which means the magnitude of the field is different as you go from one position to another surrounding the charges. A valuable E-field would be one that was constant over a large area. This kind of E-field can be created by a thin, conductive plate uniformly charged as shown below.
The field lines would look like the following:
Gauss’s Law
and calculus can be used to show that the E-field on either side of a plate filled with total charge, Q enclosed, would give the following result for the field at every position near the plate:
Flat, thin plate E-field:
where ε o is the permittivity of free space with a value of 8.85418782 x 10-12 farads/meter and σ is the charge density on the surface:
The permittivity is a measure of how dense of an electric field is “permitted” to form around an electric charge. It is related to the k in Coulomb’s Law as follows:
Two parallel plates close together of opposite charge Q and -Q, would produce a field of twice the magntiude of a single plate as shown below.
Twoflat, thin parallel plates of opposite charge E-field:
since the vector sum of each plate’s E-field adds between the plates and subtracts outside the plates. This electromechanical device is called a capacitor and one of its primary duties is to store charge, and therefore energy, in DC circuits.
Electric Potential Energy in Constant E-field:
As with the gravitational force the electric force can do work on a charged object as shown at right:
The change in Potential Energy of the charge produced by moving in the direction of the field can be written as:
and substituting from above gives
Electric Potential Energy in E-field around a Point Charge
Consider a charge Q acting as a point charge and creating an electric field in the surrounding space.
The electric force can do work on a charged object, q, in the field surrounding a Point Charge as shown at right. The issue here is that the E-field created by the charge, Q, is not constant and has to be evaluated using calculus. The work done by the E-field on the moving charge fromA to B is the following:
The result of the integration is the following:
The general expression for the Potential Energy of a charge, q, a distance, r, from the center of a charge, Q, is determined as if the charge was moved by an outside force from a position of ∞:
Electric Potential
In dealing with charged objects it is valuable to work with the idea of Electric Potential Energy per Charge at a point in space in the Electric Field. The outcome of this ratio is that a relatively small charge at that position is not a factor as to the Potential of the field just like a relatively small charge at that position does not significantly affect the Electric Field. The concept of
is called Electric Potential, V. The Difference in Electric Potential between 2 points in an Electric Field is then:
SI unit: Volts = J/C
Electric Potential for a Point Charge
For a position at distance, r, from the center of a point charge, Q, the Electric Potential at that point can be determined by considering moving the point charge, q, in from ∞.
Electric Potential between Parallel Plates:
When moving a charge, q, a distance, d, between parallel plates from Position A to Position B and since PE A> PE Bthe result is the following:
Equipotential Lines
Lines can be added to Electric Field diagrams that connect points of equal Electric Potential. This is useful to determine whether the Electric Potential Energy increases, decreases, or remains the same when a charge changes position. The single Point Charge Electric Field diagram is shown at right with blue circles added such that each line represents a set of positions of equal Electric Potential. These are called EquiPotential lines.
The Equipotential Lines for a constant Electric Field between parallel plates are shown as blue lines below.
The Equipotential Lines for an Electric Field created by opposite charges are shown as blue lines below.
Considerations when analyzing EquiPotential Lines:
They are perpendicular to electric field lines since the value of the field does not change anywhere on the line.
An outside force does positive work to move a positive charge against the field, say from Line C to Line A in the diagrams above, thereby increasing the Potential Energy of the charge and the Electric Potential at the charge’s new position in the field.
An outside force does negative work to move a positive charge with the field, say from Line A to Line C in the diagrams above, thereby decreasing the Potential Energy of the charge and the Electric Potential at the charge’s new position in the field.
An outside force does no work to move a positive charge perpendicular to the field, say from one position on Line B to another position on Line B in the diagrams above, thereby not changing the Potential Energy of the charge or the Electric Potential at the charge’s new position in the field.
The Capacitor
Characteristics:
Generally made of 2 parallel metal plates.
Made close enough together compared to the dimensions of the plates so that the E-field between the plates is constant.
When hooked up to a voltage source as shown below the capacitor will become charged over time ending with one positive plate and one negative plate having the same potential difference, , as the voltage source.
The E-field set up between the plates is
The E-field set up between the plates stores Electric Potential Energy.
Capacitance of a Capacitor:
Charge stops flowing into and out of the plates of a capacitor when the Potential difference between the voltage source positive plate and the capacitor positive plate is equal to 0, and similarly the Potential difference between the voltage source negative plate and the capacitor negative plate is equal to 0.
After charging the following relationships for the capacitor are then established:
Substituting Eqn 2 into Eqn 1 gives
and then substituting Eqn 3 in gives
and finally rearranging gives
The amount of charge, Q, able to be stored in a capacitor for a given Potential difference, ∆V, depends on the physical characteristics of the capacitor as shown by the left side of the previous equation. This is called the capacitance, C, of the capacitor:
The relationship between Q, C, and ∆Vis therefore the following:
Energy Stored in a Capacitor
Work is required to store positive and negative charges on the plates of a capacitor, thereby storing Potential Energy in the E-field between the capacitor plates.
A graph of the charge building up on the plates, Q, versus time is shown at right. Below that is a graph of∆Vversus Qas the capacitor becomes fully charged.
The bottom graph shows that as the charge, Q, builds up on the plates the Potential difference between the plates, ∆V, increases at a linear rate. The work done, and therefore the Potential Energy stored, is the area below the line. It can be represented 3 different ways using previous results:
Dielectric in a Capacitor:
A dielectric is an electrical insulator that can be polarized when placed in an E-field. When the dielectric material is sandwiched between 2 parallel plates as shown below right an additional, smaller E-field created by the polarized dielectric molecules is set up between the plates pointing in the opposite direction of the original E-field.
The vector sum of the 2 E-fields results in a smaller overall E-field between the plates. This reduces the voltage between the plates by a factor of K, where K> 1.
Without further charging and with the dielectric in place the capacitance would become
This would result in a capacitance of
since
If the voltage source is charging with the dielectric in place more charge will be forced onto each plate as shown below until the voltage between the plates is equal to the source voltage, thereby storing more charge in the capacitor.
Equation Summary
Habits of an Effective Problem-Solver
An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach, they all have habits which they share in common. These habits are described briefly here. Effective problem-solvers ...
...read the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
...identify the known and unknown quantities in an organized manner, often times recording them on the diagram itself. They equate given values to the symbols used to represent the corresponding quantity (e.g., F = 0.025 N, E = 4.50x10-6 N/C, q = ???).
...plot a strategy for solving for the unknown quantity. The strategy will typically center around the use of physics equations and is heavily dependent upon an understanding of physics principles.
...identify the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
...perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
Read more...
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5 Strategies to Teach Multi-Step Word Problems: Teacher’s Guide
By Stephen Nelly Nketsiah
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Hey there, Teacher! Are you ready to empower your students with the skills they need to conquer multi-step word problems? Look no further! This comprehensive blog post is your go-to resource for effective strategies that will transform your classroom into a problem-solving powerhouse. Let’s dive in and unleash your students’ problem-solving potential!
Table of Contents
Toggle
Word Problems
Strategies to Teach Multi-Step Word Problems
Model the Problem-Solving Process
Provide Clear Problem-Solving Strategies
Scaffolded Practice
Differentiate Instruction
Practice Multi-Step Word Problems Regularly for Proficiency
Recommended Materials for Multi-Step Word Problems Practice
Conclusion
Multistep Word Problems – FAQs
Word Problems
Word problems are an essential part of math education, as they help develop critical thinking and problem-solving skills.
Multi-step word problems, in particular, provide an excellent opportunity for students to apply their math skills in real-life scenarios.
Teaching students how to solve one-step word problems or multi-step word problems can be a complex task as it can be a challenging concept for most students to grasp.
However, with the right strategies and guidance, teachers can help their students become proficient problem solvers.
Strategies to Teach Multi-Step Word Problems
Now, let’s delve into the 5 strategies that teachers can employ to effectively teach multi-step word problem-solving to their students.
Model the Problem-Solving Process
Provide Clear Problem-Solving Strategies
Provide Scaffolded Practice
Differentiate Instruction
Practice Regularly for Proficiency
Model the Problem-Solving Process
One of the most effective ways to help students understand and solve multi-step word problems is by modeling the problem-solving process.
When teachers model the steps involved in solving a problem, students can observe and learn from their thinking and approach.
This approach helps students develop a deeper understanding of the problem and the strategies required to solve it.
You can use the following approaches when modeling the problem-solving process.
Step-by-Step Approach
Storytelling Approach
Real-Life Examples
Step-by-Step Approach
When modeling the problem-solving process, it is crucial to take a step-by-step approach.
Break down the problem into smaller, manageable parts, and demonstrate how to tackle each part systematically.
By breaking down the problem, students can focus on one step at a time, reducing confusion and overwhelm.
Storytelling Approach
Another effective strategy is to take a storytelling approach when presenting multi-step word problems.
Create narratives or scenarios around the problems, making them more interesting and captivating for students.
By presenting problems in a story format, students can visualize the context and relate to the characters or situations involved.
This approach enhances their problem-solving abilities and engages their imagination.
Real-Life Examples
To make the modeling process more engaging and relatable, incorporate real-life examples into the multi-step word problems.
Relating the problems to situations that students encounter in their daily lives helps them connect with the content and see the practical applications of mathematical concepts.
For instance, you can present a word problem involving shopping or calculating distances during a trip.
Provide Clear Problem-Solving Strategies
As a teacher, your role is to equip your students with effective problem-solving strategies that will empower them to confidently tackle multi-step word problems.
Here are some key multi-step problem-solving strategies to share with your students:
Read the Problem Carefully
Identify the Question
Plan and Break Down the Problem
Choose the Correct Operations
Solve Step-by-Step
Check and Reflect
Read the Problem Carefully
The first step in solving a multi-step word problem is to read the problem carefully.
Emphasize the importance of taking the time to understand the given information.
Encourage your students to identify the essential details, underline or highlight important numbers or quantities.
By comprehending the problem thoroughly, students will lay a strong foundation for their problem-solving journey.
Identify the Question
After understanding the given information, students should identify the question they need to answer.
Guide your students to identify the question or the unknown they need to solve.
Assist them in recognizing what information is missing and what they are being asked to find.
Encourage them to clearly define the question to maintain focus and direction throughout the problem-solving process.
Plan and Break Down the Problem
Teach your students to develop a plan and break down the problem into smaller, more manageable steps.
By doing so, they can avoid feeling overwhelmed by the complexity of multi-step word problems.
Discuss possible approaches, such as drawing diagrams, making tables, or using equations.
By having a plan in place or by creating a roadmap for the problem-solving journey, students can proceed with confidence and avoid making unnecessary mistakes.
Choose the Correct Operations
Guide your students in choosing the correct operations or mathematical strategies for each step of the problem.
Help them consider the problem’s context and the relationships between different quantities.
By selecting the appropriate operations, such as addition, subtraction, multiplication, or division, students can accurately solve each component of the multi-step word problem.
Solve Step-by-Step
Encourage your students to solve the problem step-by-step, following the plan they devised earlier.
By solving each step individually and linking them together, students can see the overall problem as a series of interconnected components, making it easier to navigate and solve.
Remind students to show their work neatly and clearly. This approach not only helps students organize their thoughts but also allows you, as their teacher, to identify any errors or misconceptions they may have.
Provide guidance and support as needed throughout their problem-solving process.
Check and Reflect
After obtaining a solution, it’s essential that students check their answer.
Teach your students the importance of checking their work and reflecting on their solution.
Encourage them to evaluate whether their answer makes sense within the context of the problem. They can do this by revisiting the original problem, reapplying the given information, and verifying if the solution satisfies the question asked.
This step promotes critical thinking and helps students identify and correct any mistakes they may have made.
Scaffolded Practice
As a teacher, it is crucial to provide scaffolded practice to support your students in mastering multi-step word problems.
Scaffolded practice involves breaking down the problem-solving process into smaller, manageable steps and gradually increasing the complexity of the problems.
This approach allows students to build their skills incrementally and gain confidence as they progress.
Here are some strategies to implement scaffolded practice:
Start with Simple Problems
Provide Guided Practice
Increase Complexity Gradually
Use Visual Aids and Models
Break Down Complex Problems
Provide Independent Practice Opportunities
Start with Simple Problems
Begin by introducing your students to simple, one-step word problems.
These problems will help them understand the basics of problem-solving and build their confidence.
Provide clear explanations and model the problem-solving process step-by-step.
Reinforce the importance of reading the problem carefully and identifying the key information.
Provide Guided Practice
Guide your students through practice problems by solving them together as a class.
Discuss the steps involved and explain the reasoning behind each step.
Encourage students to ask questions and engage in discussions about problem-solving strategies.
Offer support and guidance as needed, ensuring that students understand each concept before moving on.
Increase Complexity Gradually
Gradually increase the complexity of the problems as your students become more comfortable with one-step word problems.
Introduce multistep word problems by adding another step or operation at a time.
Provide clear explanations of each step and emphasize the relationships between different quantities in the problem.
Encourage students to apply the problem-solving strategies they have learned.
Use Visual Aids and Models
Utilize visual aids and models to help students visualize the problem and understand the relationships between different quantities.
Use diagrams, charts, or manipulatives to represent the information given in the problem. This visual representation can enhance students’ understanding and support their problem-solving process.
Break Down Complex Problems
For complex multi-step word problems, break them down into smaller components.
Guide students through each step, explaining the rationale behind each operation and encouraging them to show their work.
This step-by-step approach helps students tackle complex problems more effectively and reduces feelings of overwhelm.
Provide Independent Practice Opportunities
Offer opportunities for students to practice independently.
Provide worksheets or online resources that offer a range of multi-step word problems at various difficulty levels.
Encourage students to work through the problems on their own, applying the strategies they have learned. Offer feedback and support as they progress.
Differentiate Instruction
As a teacher, it is essential to differentiate your instruction to meet the diverse needs of your students.
Differentiation allows you to tailor your teaching methods, materials, and assessments to accommodate the individual learning styles, abilities, and interests of your students.
Here are some strategies for differentiating instruction in multi-step word problem-solving:
Assess Readiness and Grouping
Offer Multiple Problem-Solving Approaches
Vary the Level of Difficulty
Flexible Grouping Strategies
Accommodations and Supports
Formative Assessments and Feedback
Assess Readiness and Grouping
Assess the readiness of your students by evaluating their prior knowledge and skills related to multi-step word problems.
Group students based on their readiness levels, whether they require additional support or enrichment.
Provide targeted instruction and resources to address the specific needs of each group.
Offer Multiple Problem-Solving Approaches
Recognize that students have different learning styles and preferences.
Offer a variety of problem-solving approaches to cater to their individual needs.
Provide visual representations, manipulatives, or verbal explanations to support visual, kinesthetic, or auditory learners.
Allow students to choose the method that best suits their learning style.
Vary the Level of Difficulty
Recognize that students have different levels of proficiency in problem-solving.
Differentiate the level of difficulty in the problems you assign.
Provide additional challenge problems for high-achieving students and offer extra support or modified problems for those who may struggle.
Adjust the complexity of the problems to ensure that each student is appropriately challenged.
Flexible Grouping Strategies
Implement flexible grouping strategies to allow students to learn from and support each other.
Arrange students in pairs or small groups based on their strengths and weaknesses.
Encourage collaborative problem-solving activities where students can share their strategies, explain their thinking, and learn from their peers.
Offer opportunities for peer tutoring or mentoring.
Accommodations and Supports
Recognize the diverse needs of students with learning disabilities or English language learners.
Provide accommodations and support to ensure their understanding and success.
Adapt the materials, provide visual cues or graphic organizers, offer additional time, or provide translated resources when necessary.
Collaborate with special education or language support specialists to address individual needs effectively.
Formative Assessments and Feedback
Use formative assessments to gather ongoing feedback on students’ progress in multi-step word problem-solving.
Monitor their understanding, identify misconceptions, and provide timely feedback. Offer specific guidance and support based on individual needs.
Encourage students to reflect on their problem-solving strategies and provide self-assessment opportunities.
Practice Multi-Step Word Problems Regularly for Proficiency
It is important to emphasize the value of regular practice to help your students achieve proficiency in solving multi-step word problems.
Consistent practice not only reinforces problem-solving strategies and concepts but also builds confidence and fluency in applying them.
Here are some strategies to promote regular practice:
Assign Regular Problem-Solving Exercises
Integrate Multistep Word Problems in Daily Lessons
Encourage Independent Practice
Offer Problem-Solving Discussions and Presentations
Assign Regular Problem-Solving Exercises
Assign regular problem-solving exercises as part of your students’ homework or in-class activities.
Provide a variety of multi-step word problems that cover different mathematical concepts and real-life scenarios.
Gradually increase the difficulty level of the problems to challenge your students and foster their growth.
Integrate Multi-Step Word Problems in Daily Lessons
Incorporate multi-step word problems into your daily math lessons. Introduce them as real-world applications of the mathematical concepts you are teaching.
Show your students how these problems relate to their everyday lives and the importance of problem-solving skills in various contexts.
Inspire active participation and engagement during problem-solving activities.
Encourage Independent Practice of Multi-Step Word Problems
Encourage your students to practice independently outside of class.
Recommend math resources, such as textbooks, workbooks, or online platforms, that provide additional multi-step word problems for practice.
Encourage them to set aside dedicated time each week to work on problem-solving skills.
Remind them of the benefits of consistent practice in building proficiency.
Offer Problem-Solving Discussions and Presentations
Create opportunities for students to present and discuss their problem-solving strategies with the class.
Encourage them to explain their thinking, justify their solutions, and engage in constructive discussions.
This not only enhances their communication skills but also exposes them to different problem-solving approaches and fosters a collaborative learning environment.
Recommended Materials for Multi-Step Word Problems Practice
To enhance your students’ practice in solving multi-step word problems, consider incorporating the following recommended materials:
Math Workbooks
Online Problem-Solving Platforms
Math Games and Activities
Task Cards and Worksheets
Math Workbooks
Math workbooks provide a structured approach to practicing multi-step word problems.
Look for workbooks that specifically focus on problem-solving skills and offer a variety of problem types and difficulty levels.
These workbooks typically include step-by-step explanations and examples, providing students with opportunities to apply their problem-solving strategies independently.
Encourage your students to work through the problems in the workbooks, highlighting the importance of showing their work and explaining their reasoning.
You can assign specific pages or problem sets from the workbook as part of their regular practice routine.
Online Problem-Solving Platforms
Leverage the power of technology by using online problem-solving platforms.
These platforms offer interactive multi-step word problems that engage students and provide immediate feedback.
Look for platforms that align with your curriculum and allow you to track students’ progress.
Online problem-solving platforms often provide a range of difficulty levels and adaptive features that adjust the complexity of the problems based on individual performance.
Encourage your students to explore these platforms during their independent practice time or assign specific problems for them to solve online.
Math Games and Activities
Engage your students in learning through math games and activities.
These interactive and hands-on experiences make practice enjoyable and foster a positive attitude towards problem-solving.
Look for math games and activities that specifically focus on multi-step word problems.
You can create your own games using task cards or find existing games that align with your curriculum.
Set up problem-solving stations where students rotate and solve different multi-step word problems in a game format.
These activities promote collaboration, critical thinking, and a deeper understanding of problem-solving strategies.
Task Cards and Worksheets
Task cards and worksheets provide targeted practice opportunities for multi-step word problems.
While worksheets offer a collection of problems on a single sheet, task cards typically consist of individual problems on small cards.
These resources allow for flexibility in assigning practice and can be tailored to the specific needs of your students.
Select task cards or worksheets that align with the topics and skills you want your students to practice.
Consider using a variety of formats, such as multiple-choice, open-ended, or guided questions, to cater to different learning preferences.
Provide clear instructions and encourage students to work through the problems independently or in small groups.
If you’re looking for engaging worksheets and task cards for multi-step word problems to use with your students, then my differentiated worksheets and task cards will be a perfect fit for you. You can find them at my resources stores below:
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Made by Teachers store
Conclusion
Congratulations, teachers! You have now equipped yourself with a toolkit of effective strategies and resources to teach multi-step word problems with confidence.
By providing clear problem-solving strategies, implementing scaffolded practice, differentiating your instruction, and promoting regular practice, you are setting your students up for success.
Embrace the journey of teaching and learning multi-step word problems, celebrate the progress of your students, and watch as they develop into proficient problem solvers.
Remember, with your guidance and support, their potential knows no bounds!
Multistep Word Problems – FAQs
What are two-step and multi-step problems?
Two-step and multi-step problems are types of word problems that require multiple mathematical operations to be performed to find the solution. In two-step problems, students need to apply two operations, such as addition and subtraction or multiplication and division, to solve the problem. Multi-step problems may involve more than two operations and often require students to perform a series of steps to arrive at the final answer.
Why Are Multi-Step Word Problems So Important?
Multi-step word problems are important because they reflect real-world situations that students may encounter in their daily lives. By solving these problems, students develop critical thinking, analytical reasoning, and problem-solving skills. Multi-step word problems also help students apply mathematical concepts in context, promoting a deeper understanding of the subject matter.
Why do students struggle with multi-step word problems?
Students often struggle with multi-step word problems due to several reasons. First, these problems require higher-level thinking skills, such as analyzing and synthesizing information. Second, students may find it challenging to identify the relevant information and determine which operations to use. Additionally, students may struggle with organizing their thoughts and applying problem-solving strategies effectively.
What are the common errors in solving multi-step word problems?
Common errors in solving multi-step word problems include:
Misinterpreting the problem: Failing to understand the problem correctly and misidentifying the key information.
I ncorrect selection of operations: Choosing the wrong operations or using them in the wrong order.
Calculation errors: Making mistakes in performing arithmetic calculations.
Forgetting to check the answer: Neglecting to verify if the solution aligns with the question asked and the context of the problem.
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9382 | https://en.wikipedia.org/wiki/Polar_circle | Jump to content
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From Wikipedia, the free encyclopedia
Circle of latitude
For the ship of this name, see HMS Endurance. For the concept in geometry, see Polar circle (geometry).
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A polar circle is a geographic term for a conditional circular line (arc) referring either to the Arctic Circle or the Antarctic Circle. These are two of the keynote circles of latitude (parallels). On Earth, the Arctic Circle is currently drifting northwards at a speed of about 14.5 m per year and is now at a mean latitude (i.e. without taking into account the astronomical nutation) of 66°33′50.6″ N; the Antarctic Circle is currently drifting southwards at a speed of about 14.5 m per year and is now at a mean latitude of 66°33′50.6″ S. Polar circles are often equated with polar regions of Earth. Due to their inherent climate environment, the bulk of the Arctic Circle, much of which is sea, is sparsely settled whereas this applies to all of Antarctica which is mainly land and sheltered ice shelves.
If Earth had no atmosphere, then both polar circles (arcs) would see at least a day a year when the center of the Sun is continuously above the horizon and at least a day a year when it is always below the horizon – a polar day and a polar night as is the case for longer, within the circles. Up to and including the associated poles (North and South), known geographically as the frigid zones, such duration extends up to half of the year, namely, close to the poles. Instead, atmospheric refraction and the Sun's light reaching the planet as an extended object rather than a point source means that just within each circle the Earth's surface does not experience any proper polar night, 24 hours where the sun does not rise. By these same two factors, just outward of each circle still experiences a polar day (a day in which the sun does not fully set).
The latitude of the polar circles is + or −90 degrees (which refers to the North and South Pole, respectively) minus the axial tilt (that is, of the Earth's axis of daily rotation relative to the ecliptic, the plane of the Earth's orbit). This predominant, average tilt of the Earth varies slightly, a phenomenon described as nutation. Therefore, the latitudes noted above are calculated by averaging values of tilt observed over many years. The axial tilt also exhibits long-term variations as described in the reference article (a difference of 1 second of arc (″) in the tilt is equivalent to a change of about 31 metres north or south in the positions of the polar circles on the Earth's surface).
Correspondence to polar night and day
[edit]
The polar circles would almost precisely match the boundaries for the zones where the polar night and the polar day would occur throughout the winter solstice and summer solstice day respectively. They do so loosely due to two effects. The first one is atmospheric refraction, in which the Earth's atmosphere bends light rays near the horizon. The second effect is caused by the angular diameter of the Sun as seen from the Earth's orbital distance (which varies very slightly during each orbit). These factors mean the ground-observed boundaries are 80 to 100 kilometres (50 to 62 mi) away from the circle.[citation needed] A further global factor for this numerical range is Earth's nutation, which is a very small change in tilt. Observers higher above sea level can see a tiny amount of the Sun's disc (see horizon) where at lower places it would not rise. For the Arctic Circle, being 80–100 km north of the circle in winter, and 80–100 km south of the circle in summer; the inverse directions apply to the other circle.
See also
[edit]
Antarctica
Antarctic Circle
Arctic
Arctic Circle
Frigid zones
Polar climate
Polar day and Polar night
Polar region
Notes
[edit]
^ Obliquity of the ecliptic Archived 2017-06-12 at the Wayback Machine
^ Swedish Astronomic calendar 2003 (or any other year) at the times of the winter and summer solstices, around 22 June and 22 December
| v t e Circles of latitude / meridians |
| Equator Tropic of Cancer Tropic of Capricorn Arctic Circle Antarctic Circle Equator Tropic of Cancer Tropic of Capricorn Arctic Circle Antarctic Circle Equator Tropic of Cancer Tropic of Capricorn Arctic Circle Antarctic Circle W0°E 30° 60° 90° 120° 150° 180° 30° 60° 90° 120° 150° 180° 5° 15° 25° 35° 45° 55° 65° 75° 85° 95° 105° 115° 125° 135° 145° 155° 165° 175° 5° 15° 25° 35° 45° 55° 65° 75° 85° 95° 105° 115° 125° 135° 145° 155° 165° 175° 10° 20° 40° 50° 70° 80° 100° 10° 20° 40° 50° 70° 80° 100° 110° 130° 140° 160° 170° 10° 20° 30° 40° 50° 60° 70° 80° 90° 10° 20° 30° 40° 50° 60° 70° 80° 90° 5° N 15° 25° 35° 45° 55° 65° 75° 85° 5° S 15° 25° 35° 45° 55° 65° 75° 85° 45x90 45x90 45x90 |
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9383 | https://pubmed.ncbi.nlm.nih.gov/11451317/ | The acetowhite test in genital human papillomavirus infection in men: what does it add? - PubMed
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. 2001 Jan;15(1):27-9.
doi: 10.1046/j.1468-3083.2001.00196.x.
The acetowhite test in genital human papillomavirus infection in men: what does it add?
B Kumar1,S Gupta
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Affiliation
1 Department of Dermatology, Venereology and Leprology, Postgraduate Institute of Medical Education and Research, Chandigarh, India. kumarbhushan@hotmail.com
PMID: 11451317
DOI: 10.1046/j.1468-3083.2001.00196.x
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The acetowhite test in genital human papillomavirus infection in men: what does it add?
B Kumar et al. J Eur Acad Dermatol Venereol.2001 Jan.
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. 2001 Jan;15(1):27-9.
doi: 10.1046/j.1468-3083.2001.00196.x.
Authors
B Kumar1,S Gupta
Affiliation
1 Department of Dermatology, Venereology and Leprology, Postgraduate Institute of Medical Education and Research, Chandigarh, India. kumarbhushan@hotmail.com
PMID: 11451317
DOI: 10.1046/j.1468-3083.2001.00196.x
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Background: Many genital human papillomavirus (HPV) infections are clinically invisible epithelial lesions. They remain so for a considerable time before some develop into clinically apparent lesions. The inapparent and asymptomatic nature of these lesions poses a problem in the detection and management of genital HPV infection. Without reliable, and readily available diagnostic methods, no definite therapeutic approach can be recommended or followed. The acetowhite test has been recommended to help determine the extent of the affected area.
Objective: To evaluate the acetowhite test in detecting inapparent subclinical HPV involvement in male patients with clinically apparent warts.
Methods: Two hundred and two uncircumcised patients with genital warts were included. Patients with concomitant inflammatory lesions were excluded. The warts and adjacent normal skin/mucosa of normal appearance were wrapped in gauze soaked in 5% acetic acid for about 3-5 min. The area was subsequently examined with a hand lens (x 8).
Results: All of the 116 hyperplastic warts became acetowhite, as well as a narrow rim of surrounding skin. Few flat warts in dry areas only became dull white and none pure white. No whiteness was observed in the surrounding area. Only 15 of 26 flat warts in moist areas became acetowhite. One (eroded lesion) of 13 verruca vulgaris type lesions and none of the pigmented papules gave positive results to the acetowhite test.
Conclusions: The sensitivity of the acetowhite test for hyperplastic warts is very high, but for other types of warts is low. Detection of subclinical HPV-infected areas is difficult; the acetowhite test did not assist in the identification of additional areas of infection in our patients.
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9384 | https://oercommons.org/courseware/lesson/2526/overview | Math, Grade 6, Surface Area and Volume, Transforming The Net of A Cube Into The Net of A Pyramid | OER Commons
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Subject:Geometry Material Type:Lesson Plan Level:Middle School Grade:6 Provider:PearsonTags: 6th Grade Mathematics
Cube
Edges
Faces
Pyramid
Square
Vertices
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License:Creative Commons Attribution Non-CommercialLanguage:English Media Formats:Text/HTML
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Education Standards
1. 1
2. 2
MCCRS.Math.Content.6.G.A.4
Maryland College and Career Ready Math Standards
Grade 6
Learning Domain: Geometry
Standard: Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
CCSS.Math.Content.6.G.A.4
Common Core State Standards Math
Grade 6
Cluster: Solve real-world and mathematical problems involving area, surface area, and volume
Standard: Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.
Math, Grade 6
Surface Area and Volume
Getting Started
Rational Numbers
Fractions and Decimals
Ratios
Expressions
Equations and Inequalities
Rate
Putting Math to Work
Distributions and Variability
Surface Area and Volume
Transforming The Net of A Cube Into The Net of A Pyramid
Comparing Surface Area & Volume
Analyzing The Formula of A Parallelogram & Trapezoid
Analyzing The Formula of A Triangle
Basic & Composite Figures
Using A Grid To Calculate The Area Of An Irregular Figure
Volume Formula For Rectangular Prisms
Identifying Nets For Cubes
Transforming The Net of A Cube Into The Net of A Pyramid
Diagrams & Problem Solving Strategies
Gallery Problems Exercise (Groups)
Transforming The Net of A Cube Into The Net of A Pyramid
Overview
Lesson Overview
Students use scissors to transform a net for a unit cube into a net for a square pyramid. They then investigate how changing a figure from a cube to a square pyramid affects the number of faces, edges, and vertices and how it changes the surface area and volume.
Key Concepts
A square pyramid is a 3-D figure with a square base and four triangular faces.
In this lesson, the net for a cube is transformed into a net for a square pyramid. This requires cutting off one square completely and changing four others into isosceles triangles.
It is easy to see that the surface area of the pyramid is less than the surface area of the cube, because part of the cube's surface is cut off to create the pyramid. Specifically, the surface area of the pyramid is 3 square units, and the surface area of the cube is 6 square units. Students will be able to see visually that the volume of the pyramid is less than that of the cube.
Students consider the number of faces, vertices, and edges of the two figures. A face is a flat side of a figure. An edge is a segment where 2 faces meet. A vertex is the point where three or more faces meet. A cube has 6 faces, 8 vertices, and 12 edges. A square pyramid has 5 faces, 5 vertices, and 8 edges.
Goals and Learning Objectives
Change the net of a cube into the net of a pyramid.
Find the surface area of the pyramid.
Introduction to Nets of Pyramids
Lesson Guide
Have students look at the images of the pyramid and the net for the cube. Explain that the pyramid was made by cutting away parts of the cube net and then folding it. Have students discuss with their partners about how they could change the net for the cube into a net for the pyramid.
ELL: Explain new terms to students before the terms are used in problems. Remember, ELLs are learning new concepts (math) and a new language at the same time. Discuss and define the following terms:
square pyramid
isosceles triangle
edge
face
vertex
Opening
Introduction to Nets of Pyramids
The pyramid shown in the first image was made using the net of a cube (shown in the second image). Before the net was folded to form the pyramid, some paper was cut off. No paper was added.
How could the net of the cube have been cut to turn it into a net of the pyramid?
This pyramid was made…
...by modififying this net using scissors.
Math Mission
Lesson Guide
Discuss the Math Mission. Students will investigate how changing a figure from a cube to a square pyramid affects the number of faces, edges, and vertices and how it changes the surface area and volume.
Opening
Investigate the surface area and volume of a pyramid.
Build a Pyramid
Lesson Guide
Provide students with inch grid paper and scissors to make their nets. Have students work solo when creating their net of a pyramid and then move into partner work.
SWD: Prior to the Ways of Thinking discussion, circulate and check in with students. Preview the questions that will be discussed with students who may need additional time to process the information.
Also, provide students with several of the key questions in written form (digitally). This will allow students who become anxious when put on the spot to prepare answers for questions that they can successfully share in the class discussion.
Interventions
Student has difficulty transforming the net for a cube into a net for a pyramid.
What figure is the base of the cube? What figure is the base of the pyramid?
Besides the base, how many faces does the pyramid have? How does this compare to the cube?
What types of triangles make up the faces of the pyramid?
How can you cut a square face to form an isosceles triangle?
Answers
Work Time
Build a Pyramid
Start with the net of a cube.
Use scissors to transform it into the net of a pyramid.
Faces, Vertices, and Edges
Lesson Guide
Point out the vocabulary and make sure students understand the words edge, vertex, and face.
Ask students who are having trouble:
What changed as the figure went from being a cube to a pyramid?
Are the bases of the cube and the pyramid the same or different?
Answers
There are 6 faces, 8 vertices, and 12 edges in the original cube, and there are 5 faces, 5 vertices, and 8 edges in the pyramid.
Work Time
Faces, Vertices, and Edges
Compare the numbers of faces, vertices, and edges of a cube with the numbers of faces, vertices, and edges of the pyramid you made.
Show Hint
Hint:
A face is a flat side of a solid figure.
An edge is a line segment where two faces meet.
A vertex (plural vertices) is a point where three or more faces meet.
Surface Area
Lesson Guide
Remind students that they learned to calculate the surface area of the cube in the previous lesson.
Ask students who are having trouble:
Which looks like it has more space inside—the cube or the pyramid? Why do you think that?
What did you do to the net of the cube to get the net of the pyramid? How does that affect the surface area of the pyramid compared with the surface area of the cube?
Interventions
Student has difficulty comparing the surface area and volume of the cube to the surface area and volume of the pyramid.
How does the area of the net relate to the surface area of the 3-D figure it folds into?
When you made the pyramid, you cut away part of the cube net. What does this tell you about how the surface areas of the cube and pyramid compare?
How does the space inside the pyramid compare to the space inside the cube?
Student has difficulty finding the surface area of the pyramid.
What figures form the net for the pyramid?
How do you find the area of the square?
How do you find the area of each triangle?
The area of the square is 1 square unit.
How much of each square did you cut away?
What is the area of the triangle that is left?
Answers
As the figure changes from a cube to a pyramid, both the surface area and volume decrease.
The surface area of the cube is 6 square units. The surface area of the pyramid is 3 square units.
Work Time
Surface Area
How do you think your paper cutting will affect the surface area and volume of the resulting figure (the pyramid) as compared to the original figure (the cube)?
Find the surface area of the cube and the pyramid. (Each edge of the cube has a length of 1 unit.)
Prepare a Presentation
Preparing for Ways of Thinking
Look for examples of both correct and incorrect nets, and listen for different ways that students are reasoning about making the nets. Look for students who are finding the surface area of the prism by using area formulas and those who are using reasoning. (For example, the area of the square base is 1 square unit. You cut away half of a square face to make a triangular face, so the area of each triangular face is 1 2 square unit.) Highlight these solutions in Ways of Thinking.
Challenge Problem
Answers
The height of each triangle forming the pyramid is the same as the height of each side of the cube, but the pyramid is shorter than the cube because the four triangles are tilted inward to meet at the top vertex.
Work Time
Prepare a Presentation
Explain your strategy for comparing the volume of a pyramid to the volume of the cube. Support your explanation with the pyramid you made.
Explain your strategy for determining the surface area of a pyramid. Support your explanation with the pyramid you made.
Challenge Problem
Is the pyramid taller or shorter than the cube? Explain.
Make Connections
Mathematics
Have students share the counts they got for the faces, edges, and vertices of the cube and the pyramid.
Have students give their presentations. Discuss how the paper cutting affected the surface area and volume as the cube was transformed into a square pyramid. Elicit that since parts of the net were cut off to make the pyramid, the surface area of the pyramid must be less than the surface area of the cube.
Students should also recognize that the pyramid will have less volume than the cube. Have them explain why.
Discuss different methods students used to find the surface areas of the cube and the square pyramid.
Have the students who did the Challenge Problem explain their thinking.
SWD: Provide these students with a template to organize the information in their presentations. This will help you to ensure that they include all the required information. The template should include space for recording all information for comparing cubes and square pyramids. Provide space for students to record observations about faces, edges, vertices, surface area, and volume.
ELL: Be cognizant that ELLs may encounter difficulties when they have to express themselves using the English language. If you hear them say the right things but using the wrong grammar structure, show signs of agreement and softly rephrase using the correct grammar. Use the student's words as much as possible.
Mathematical Practices
Mathematical Practice 7: Look for and make use of structure.
Ask students to discuss how they looked for and made use of structure to transform the net of the cube into a net of a square pyramid. For example, students may have noticed that the pyramid had one fewer face than the cube and reasoned that they would have to cut off one square from the net. They may have looked at the fact that each side of a square base of the pyramid is connected to an isosceles triangle face to determine how the figures in their net should be organized.
Performance Task
Ways of Thinking: Make Connections
Take notes about your classmates’ strategies for comparing the volume and surface area of a pyramid to the cube it was made from.
Show Hint
Hint:
As your classmates present, ask questions such as:
What strategy did you use to compare the volume of the cube to the volume of the pyramid?
How did you find the surface area of the cube?
How did you find the surface area of the pyramid?
More About Pyramids
A Possible Summary
You can cut the net of a cube to make a net for a square pyramid.
There are 6 faces, 8 vertices, and 12 edges in a cube, and there are 5 faces, 5 vertices, and 8 edges in a square pyramid.
The volume and surface area of the square pyramid will be less than the volume and surface area of the cube, because you cut away area to make the net of the square pyramid.
To find the surface area of the cube, you can find the surface area of one face and then multiply by 6:
1 unit × 1 unit = 1 square unit
1 square unit × 6 = 6 square units
To find the surface area of the square pyramid, you can find the surface area of the square face and then add the surface areas of the 4 triangular faces:
1 unit × 1 unit = 1 square unit
1 2(1 unit × 1 unit) = 1 2 square unit
1 2 square unit × 4 square units =2 square units
1 square unit + 2 square units = 3 square units
Formative Assessment
Summary of the Math: More About Pyramids
Write a summary about what you discovered about the volume and surface area of a prism and a pyramid.
Show Hint
Hint:
Check your summary:
Do you explain how you compared the volume of the pyramid to the volume of the cube it was made from?
Do you explain how you found the surface area of the cube?
Do you explain how you found the surface area of the pyramid?
Packing a Truck
Lesson Guide
This task allows you to assess students’ work and determine what difficulties they are having. The results of the Self Check will help you determine which students should work on the Gallery and which students would benefit from review before the assessment. Have students work on the Self Check individually.
Assessment
Have students submit their work to you. Make notes on what their work reveals about their current levels of understanding and their different problem-solving approaches.
Do not score students’ work. Share with each student the most appropriate Interventions to guide their thought process. Also, note students with a particular issue so that you can work with them in the Putting It Together lesson that follows.
Interventions
Student has difficulty getting started.
What do you know?
What do you need to find out?
What is one packing plan that you can try?
Can you sketch a picture of a packing plan?
Student has not found a way to test different packing plans systematically.
Can you find a way to organize the different packing plans you would like to try?
Can you organize your work in a table?
Student has not found the best packing plan.
Is there another packing plan you can propose?
Student produces a correct solution but does not give an explanation of why it is correct.
Why is your method the best plan for packing the product for shipping?
Student presents work poorly.
Would someone unfamiliar with your type of solution easily understand your work?
Have you given enough explanation in your work and is it clear?
Student makes a calculation error.
Check to see if you have made any errors in your calculations.
Student produces a correct solution and an explanation of why it is correct.
If the company allowed it, could you fit any boxes into the empty space in the truck? If so, how many? Explain.
Could you use a different method for finding the solution? If so, what would it be? Which method do you prefer? Why?
ELL: Provide ELLs and other students a sample or model for the concepts, strategies, and applications that will be addressed in the quiz, and the format you want them to follow. Be prepared to address and explicitly reteach or review vocabulary, concepts, strategies, and applications.
Possible Answers
Pack the boxes in the crate so that the 75 cm sides of the boxes are parallel to the 1.5 m side of the crate, and the 40 cm sides of the boxes are parallel to the 2.4 m side of the crate. This makes a 2 box by 6 box by 3 box prism, for a total of 2 ⋅ 6 ⋅ 3, or 36 boxes. The boxes will fit perfectly along the 2.4 m and 1.5 m sides of the crate, and there will be only a 5 cm gap along the other side.
Pack the crates in the truck so that the 1.5 m sides are parallel to the 3 m side of the truck, and the 2.4 m sides of the crates are parallel to the 2.5 m side of the truck. This makes a 2 crate by 1 crate by 7 crate prism, for a total of 2 ⋅ 1 ⋅ 7, or 14 crates. In all, there will be 14 ⋅ 36, or 504 boxes of bananas in the truck.
The crates inside the truck form a 2.4 m by 3 m by 9.8 m prism, which has a volume of 70.56 m 3. The truck bed is a 2.5 m by 3 m by 10 m prism, which has a volume of 75 m 3. Only 4.44 m 3 of the truck will be empty.
Explanations will vary. Possible answer: This is the best plan because it minimizes the amount of unused space. Other ways of packing the crates (as listed below) will leave bigger gaps, and you will not be able to ship the maximum amount of product in each crate or truck:
Put the 2.4 m sides of the crates parallel to the 2.5 m side of the truck and the 1.5 m sides of the crates parallel to the 10 m side of the truck. This would fit a 2 crate by 1 crate by 6 crate prism, for a total of only 12 crates.
Put the 1.4 m sides of the crates parallel to the 2.5 m side of the truck. Only 1 crate would fit in this direction, leaving a 1.1 m by 10 m by 3 m gap, which is 33 m 3.
Put the 1.5 m sides of the crates parallel to the 2.5 m side of the truck. Only 1 crate will fit in this direction, leaving a 1 m by 10 m by 3 m gap, which is 30 m 3, plus a 0.2 m by 1.5 m by 10 m gap on the top, which is 3 m 3, for a total of 33 m 3.
Formative Assessment
Packing a Truck
Complete this Self Check by yourself.
You are working for a trucking company that transports bananas. You are trying to transport the greatest number of bananas in each truckload. The bananas fill boxes, which are packed into crates. The crates are then packed into a truck.
How can you pack each crate in order to fit the greatest number of boxes of bananas?
How can you pack the truck in order to fit the greatest number of crates?
Based on your packing plan, how much space in the truck will not be filled?
Why is your packing plan the best possible plan?
Reflect On Your Work
Lesson Guide
Have each student write a brief reflection before the end of the class. Review the reflections to find out what students know about the difference between surface area and volume.
ELL: To assess their understanding of surface area and volume, ask students to write the definitions of the two terms in their own words. Assess both mathematical understanding and language. Make sure students compare the two definitions to explain their differences. Provide students with wait time and allow students access to their notes as a writing support.
Work Time
Reflection
Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful.
Some of the differences between surface area and volume are…
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9385 | https://pubmed.ncbi.nlm.nih.gov/36892646/ | The value of different involvement patterns of the knee "synovio-entheseal complex" in the differential diagnosis of spondyloarthritis, rheumatoid arthritis, and osteoarthritis: an MRI-based study - PubMed
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Eur Radiol
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. 2023 May;33(5):3178-3187.
doi: 10.1007/s00330-023-09485-4. Epub 2023 Mar 9.
The value of different involvement patterns of the knee "synovio-entheseal complex" in the differential diagnosis of spondyloarthritis, rheumatoid arthritis, and osteoarthritis: an MRI-based study
Boya Li#1,Zikang Guo1,Jin Qu2,Ying Zhan2,Zhiwei Shen3,Xinwei Lei45
Affiliations Expand
Affiliations
1 First Central Clinical College, Tianjin Medical University, Tianjin, China.
2 Department of Radiology, Tianjin First Central Hospital, Tianjin Institute of Imaging Medicine, Tianjin, China.
3 Clinical Science, Philips Healthcare, Beijing, China.
4 Department of Radiology, Tianjin First Central Hospital, Tianjin Institute of Imaging Medicine, Tianjin, China. leixinwei66@163.com.
5 Department of radiology, Tianjin First Central Hospital, Tianjin Institute of Imaging Medicine, NO. 24 Fukang Road, Nankai District, Tianjin, 300192, China. leixinwei66@163.com.
Contributed equally.
PMID: 36892646
DOI: 10.1007/s00330-023-09485-4
Item in Clipboard
The value of different involvement patterns of the knee "synovio-entheseal complex" in the differential diagnosis of spondyloarthritis, rheumatoid arthritis, and osteoarthritis: an MRI-based study
Boya Li et al. Eur Radiol.2023 May.
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. 2023 May;33(5):3178-3187.
doi: 10.1007/s00330-023-09485-4. Epub 2023 Mar 9.
Authors
Boya Li#1,Zikang Guo1,Jin Qu2,Ying Zhan2,Zhiwei Shen3,Xinwei Lei45
Affiliations
1 First Central Clinical College, Tianjin Medical University, Tianjin, China.
2 Department of Radiology, Tianjin First Central Hospital, Tianjin Institute of Imaging Medicine, Tianjin, China.
3 Clinical Science, Philips Healthcare, Beijing, China.
4 Department of Radiology, Tianjin First Central Hospital, Tianjin Institute of Imaging Medicine, Tianjin, China. leixinwei66@163.com.
5 Department of radiology, Tianjin First Central Hospital, Tianjin Institute of Imaging Medicine, NO. 24 Fukang Road, Nankai District, Tianjin, 300192, China. leixinwei66@163.com.
Contributed equally.
PMID: 36892646
DOI: 10.1007/s00330-023-09485-4
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Abstract
Objectives: To explore the different involvement patterns of the knee "synovio-entheseal complex (SEC)" on MRI in patients with spondyloarthritis (SPA), rheumatoid arthritis (RA), and osteoarthritis (OA).
Methods: This study retrospectively included 120 patients (male:female, 55:65) with a mean age of 39.20 years diagnosed with SPA (n = 40), RA (n = 40), and OA (n = 40) at the First Central Hospital of Tianjin between January 2020 and May 2022. Six knee entheses were assessed by two musculoskeletal radiologists according to the SEC definition. Bone marrow lesions associated with entheses include bone marrow edema (BME) and bone erosion (BE), which were classified as entheseal or peri-entheseal based on their relationship to the entheses. Three groups (OA, RA, and SPA) were established to characterize the location of enthesitis and the different SEC involvement patterns. Inter-group and intra-group differences were analyzed using the ANOVA or chi-square tests, and the inter-class correlation coefficient (ICC) test was used to determine inter-reader agreement.
Results: The study contained a total of 720 entheses. The SEC-based analysis revealed different involvement patterns in three groups. The OA group had the most abnormal signals in tendons/ligaments (p = 0.002). The RA group had considerably greater synovitis (p = 0.002). The majority of peri-entheseal BE was identified in the OA and RA groups (p = 0.003). Furthermore, entheseal BME in the SPA group was significantly different from those in the other two groups (p < 0.001).
Conclusions: SEC involvement patterns differed in SPA, RA, and OA, which is important for differential diagnosis. SEC should be used as a whole evaluation method in clinical practice.
Key points: • The "synovio-entheseal complex (SEC)" explained differences and characteristic alterations in the knee joint in patients with spondyloarthritis (SPA), rheumatoid arthritis (RA), and osteoarthritis (OA). • The various SEC involvement patterns are crucial for differentiating SPA, RA, and OA. • When "knee pain" is the only symptom, a detailed identification of characteristic alterations in the knee joint of SPA patients may help timely treatment and delay the structural damage.
Keywords: MRI; Osteoarthritis; Rheumatoid arthritis; Spondyloarthritis; Synovitis.
© 2023. The Author(s), under exclusive licence to European Society of Radiology.
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References
McGonagle D, Gibbon W, Emery P (1998) Classification of inflammatory arthritis by enthesitis. Lancet 352(9134):1137–1140 - PubMed
McGonagle D, Aydin SZ, Marzo-Ortega H, Eder L, Ciurtin C (2021) Hidden in plain sight: is there a crucial role for enthesitis assessment in the treatment and monitoring of axial spondyloarthritis? Semin Arthritis Rheum 51(6):1147–1161 - PubMed
Dougados M, van der Linden S, Juhlin R et al (1991) The European Spondylarthropathy Study Group preliminary criteria for the classification of spondylarthropathy. Arthritis Rheum 34(10):1218–1227 - PubMed
Resnick D, Niwayama G (1983) Entheses and enthesopathy. Anatomical, pathological, and radiological correlation. Radiology. 146(1):1–9 - PubMed
Eshed I (2019) SP0096 MRI of large joints in arthritis: how to do and how they are different from small joints? Ann Rheum Dis 78(Suppl 2):28.2-28. - DOI
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9386 | https://blog.csdn.net/weixin_34220834/article/details/94648534 | 使用斜二测画法绘制直观图的数学推导与应用-CSDN博客
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斜二测画法的经验公式
最新推荐文章于 2021-03-21 07:00:22 发布
转载 于 2015-05-04 01:32:00 发布·2.7k 阅读
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#matlab#c/c++
用斜二测画法画直观图，本质上就是把三维空间中图形的每个点投影到一个二维平面上（其实还需要先把 y 坐标除以 2, 但这不是重点）。投影之后坐标降到了二维，这样才有可能在二维的纸面上把点每个点画出来，从而把整个图形画出来。
（斜二测画法，边长 200px 的立方体）
当年学向量的时候推导了一些变换公式（当时还不知道矩阵乘法），顺便研究了一下怎么样通过计算画直观图而不是凭感觉。乱搞了一通蒙出一组式子来，基本上可以确定是对的：
(\left{\begin{matrix}
x' = x+\frac{\sqrt{2}}{4}y \
y' = z+\frac{\sqrt{2}}{4}y
\end{matrix}\right.)
(x, y, z) 表示空间中点的坐标，(x', y') 表示投影后纸面上一个 x 轴向右，y 轴（对应空间中的 z 轴）向上的坐标系中点的坐标。在这个坐标系中，可以画出 y = x 这条直线来表示空间中的 y 轴（图中 z 轴）。注意这个 y 轴是指向右上方的，也就是说这是一个左手系（其实左手系跟右手系在坐标运算上几乎没什么差别，只是向量外积不一样而已）：
（左手系和右手系，图片来自网络）
上述式子只适用于左手系，但是稍加改造应该就能得到只适用于右手系的另一组式子。不过我懒得再推（试）了。
写成变换矩阵的形式：
[x′y′z′]=[1√2 4 0 0√2 4 1 0 0 0][x y z]
转载于:
相关资源：计算机图形学透视 投影 变换(报告中含核心代码).docx_两点透视资源...
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9387 | https://www.ncbi.nlm.nih.gov/books/NBK470421/ | Tinea Pedis - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Tinea Pedis
Pramod K. Nigam; Hasnain A. Syed; Dahlia Saleh.
Author Information and Affiliations
Authors
Pramod K. Nigam 1; Hasnain A. Syed 2; Dahlia Saleh 3.
Affiliations
1 PtJNM Med C, Chhattisgarh and AYUSH Uni
2 Sheikh Zayed Hospital, Lahore
3 Sampson Regional Medical Center
Last Update: October 29, 2023.
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Continuing Education Activity
Tinea pedis, also known as athlete's foot, results from dermatophytes infecting the skin of the feet.Patients contract the infection by directly contacting the organism while walking barefoot.Symptoms typically develop in the interdigital clefts of the toes but can also affect the soles and medial and lateral edges. Generally, physicians treat tinea pedis with topical antifungals and advise maintaining proper foot hygiene for this benign infection.If not treated appropriately, tinea pedis can lead to significant morbidity, including cellulitis, osteomyelitis, and lymphangitis.
A practice gap exists among clinicians in recognizing the risk factors and potential complications associated with tinea pedis. Although it's a common condition, healthcare providers may not be fully aware of the heightened risks in individuals with diabetes and those wearing occlusive shoes. Additionally, there is a need for improved awareness about the progression of untreated tinea pedis, which can lead to severe complications such as cellulitis, pyoderma, and osteomyelitis, especially in immunocompromised patients. Bridging this gap is crucial to ensure timely diagnosis, effective management, and the prevention of complications in patients with this fungal skin infection.
Objectives:
Identify the clinical manifestations and characteristic features of tinea pedis accurately.
Differentiate tinea pedis from other dermatological conditions with similar presentations.
Implement evidence-based treatment strategies, including the appropriate use of topical and oral antifungals, tailored to the severity of the infection.
Coordinate care with other healthcare team members to provide well-rounded support and education for patients at risk of tinea pedis infection, thereby promoting optimal patient outcomes.
Access free multiple choice questions on this topic.
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Introduction
Tinea pedis, commonly known as athlete's foot, results from fungal infections on the skin of the feet caused by dermatophytes, including Trichophyton rubrum,T mentagrophytes,T interdigitale, and Epidermophyton floccosum. This infection typically occurs through direct contact with the organism while walking barefoot in locker rooms, showers, and swimming complexes.Individuals with diabetes and those who wear occlusive shoes are at an increased risk of developing tinea pedis.
Tinea pedis typically presents with pruritic scales and erosions between the toes. Some patients may experience areas of hyperkeratosis with underlying erythema on the medial and lateral aspects and soles of the feet. Occasionally, patients with this condition may present with painful bullous lesions concurrently develop tinea corporis, onychomycosis, and tinea manuum.
Untreated tinea pedis can lead to cellulitis, pyoderma, and osteomyelitis, especially in patients with immunocompromised conditions, diabetes, or peripheral vascular disease. This topic explores the etiology and pathophysiology of tinea pedis, as well as highlights the critical roles of the interprofessional healthcare team in evaluating, managing, and preventing recurrence and complications of the condition.
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Etiology
T rubrum accounts for approximately 70% of tinea pedis cases,whereas T interdigitale and E floccosum are responsible for the remainder. Occasionally,Tricholosporum violaceum can also cause tinea pedis. The United States of America has seen a few cases of T indotineae recently—a dermatophyte commonly found in India, Canada, and the Middle East. Notably,T indotineaee exhibits resistance to traditional antifungal treatments. Dermatophytes typically thrive in keratinized tissues such as hair, nails, and the outer layer of the skin.
The following factors contribute to the growth of dermatophytes:
A hot and humid environment
Prolonged use of tightly fitting footwear
Excessive sweating
Prolonged exposure to water
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Epidemiology
Tinea pedis may affect approximately 10% of the total population. Adult males have a higher prevalence of tinea pedis than females. Wearing occlusive shoes for extended periods predisposes patients to dermatophyte infections. Community facilities involving water are likely to increase the chances of infection, as tinea pedis rates are higher among those who use community baths, showers, and pools. One study reported an average onset age of 15 years.
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Pathophysiology
The occlusion of toe clefts, maceration, and wet conditions, along with a simultaneous increase in bacterial flora, contribute to tinea pedis infection. Furthermore, skin breakdown, humidity, and temperature also affect this infection. Once the dermatophyte infiltrates the patient's skin, it adheres through adhesins in the fungal cell wall. Proteases, such as serine subtilisin and fungalysin, digest keratin and trigger an immune response. In response, keratinocytes produce cytokines to combat the dermatophyte.In addition, molecules called "mannans" in the dermatophyte cell wall suppress the body's immune response, making the person more susceptible to infection. Exposure to ultraviolet light, humidity deficiency, and temperature contribute to the body's defense against a dermatophyte infection.
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Histopathology
Histologically, tinea pedis presents with acanthosis, hyperkeratosis, and a sparse, superficial, perivascular infiltrate in the dermis. The vesiculobullous form exhibits spongiosis and parakeratosis. Fungal filaments can be demonstrated through Periodic acid-Schiff staining or methenamine silver stain.
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History and Physical
Patients with tinea pedis typically present with long-standing, itchy, intertriginous dermatitis of the toes characterized by peeling, maceration, and fissuring (see Image. Tinea Pedis). The undersurface of the toes primarily affects lateral toe clefts, exhibiting erythema and fine, silvery-white scales.Occasionally, a vesicular eruption resembling pompholyx can appear on the sole, but the most common presentation is the chronic intertriginous type.
Chronic hyperkeratotic or moccasin-type tinea pedis typically presents with patchy or diffuse scaling on the bottom, medial, and lateral sides of the feet. The vesiculobullous form, often caused by T mentagrophytes, manifests as tense vesicles or bullae on the soles. The burning and itching resulting from bullae can cause significant discomfort. An associated bacterial infection can lead to an acute ulcerative form of tinea pedis.
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Evaluation
Although the physical examination may strongly indicate a dermatophyte infection, tinea can overlap with other skin conditions. Confirmation through microscopy and culture may be necessary. A thorough skin examination is warranted as patients can experience dermatophyte infections in multiple body areas.
Microscopic examination of scrapings from the affected site can confirm the diagnosis by demonstrating segmented hyphae.A glass microscope slide should be used to scrape the foot's instep, heel, and sides to collect dry scales from the foot.Bullae should be unroofed, and either the entire roof or scrapings made from the underside of the roof should be mounted.
To perform the procedure, one should place a few drops of a 10% to 20% potassium hydroxide (KOH) solution on the glass slide, followed by a coverslip, and then observe the sample under the microscope. Adding 20% to 40% dimethyl sulfoxide (DMSO) accelerates the clearing of keratin without requiring heating. A staining method that involves mixing 100 mg of chlorazol black E dye with 10 mL of DMSO and adding it to a 5% aqueous solution of KOH can be beneficial. Toluidine blue can also be applied to thin specimens at a concentration of 0.1%. Although mycelia may be visible at low power magnification, using the 10× objective on a microscope is advisable to observe both hyphae and spores better.
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Treatment / Management
Improving hygiene in swimming pools and bathing areas and frequently washing and cleaning changing room floors and walkways may help control the spread of tinea pedis. Most patients can usually manage their condition effectively with topical treatment. Magenta paint or Castellani paint may be beneficial in certain cases of inflammatory tinea pedis, especially when a coexisting bacterial infection is present. Topical imidazoles such as clotrimazole, econazole, ketoconazole, miconazole, isoconazole, tioconazole, and sulconazole offer effective remedies with a very low incidence of adverse effects.
Topical application of terbinafine and amorolfine yields faster results in treating tinea pedis than clotrimazole treatment. Moreover, adding a topical keratolytic, such as salicylic acid, can be beneficial in patients with hyperkeratosis.Using prophylactic tolnaftate powder after swimming and showering in community settings reduces the levels of toe cleft tinea pedis caused by T interdigitale. Generally, topical treatment lasts for 4 weeks, although some patients may experience symptom resolution sooner. Terbinafine 1% can provide effective results for interdigital tinea pedis after 1 week. Repeated KOH scrapings and cultures should yield negative results.
Patients who do not respond to topical treatment require systemic therapy. Acceptable systemic treatment options include:
Terbinafine at a daily dosage of 250 mg for 2 weeks.
Itraconazole at a twice-daily dosage of 200 mg for 1 week.
Fluconazole at 150 mg per week for 2 to 6 weeks.
Griseofulvin, for adults, at a daily dosage of 1000 mg of griseofulvin microsize for 4 to 8 weeks or 750 mg of griseofulvin ultramicro size for 4 to 8 weeks. However, the latter option may be less effective than the other options.
Griseofulvin, for children, at a recommended daily dosage of 10 to 20 mg/kg of griseofulvin microsize, 5 to 15 mg/kg of griseofulvin ultramicrosize, or itraconazole at 3 to 5 mg/kg.
Although rare, gastrointestinal adverse effects may occur from taking fluconazole. Itraconazole can lead to gastrointestinal upsets, diarrhea, and peripheral edema, particularly when combined with calcium channel blockers. Fluconazole and itraconazole have a significantly lower rate of hepatotoxicity compared to ketoconazole. Terbinafine can also result in gastrointestinal disorders and, in rare cases, hepatitis.
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Differential Diagnosis
The hyperkeratotic lesions of tinea pedis can resemble various conditions, including:
Psoriasis
Hereditary or acquired keratodermas
Atopic or contact dermatitis
Palmopustular eczema
Pitted keratolysis
Juvenile plantar dermatosis
Keratolysis exfoliativa
Interdigital lesions may appear similar to the following conditions:
Psoriasis
Erythrasma
Candida infections
Vesiculobullous lesions may resemble to the following conditions:
Pustular psoriasis
Dyshidrosis
Acute contact dermatitis
Scabies
Reiter syndrome also exhibits lesions that bear some resemblance to those seen in tinea pedis.
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Prognosis
As long as individuals take preventive measures, the outlook for tinea pedis is favorable. The risk of tinea infections is elevated with excessive sweating, known as hyperhidrosis. To prevent the disease, patients should thoroughly dry their toes after bathing. Maintaining dryness is crucial to prevent reinfection. Individuals susceptible to tinea pedis should wear moisture-wicking socks, apply high-quality antifungal powder inside their shoes, and use drying foot powders after bathing, focusing on the spaces between the toes. Tolnaftate or clotrimazole powders serve as excellent options to maintain dry feet. Regularly applying a topical antifungal cream may be required in certain instances, particularly when wearing hot, enclosed footwear.
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Complications
The majority of patients will not experience significant issues related to tinea pedis. However, complications can arise due to inadequate treatment and persistent skin breakdown.
Secondary Bacterial Infection
Prolonged or severe cases of tinea pedis may result in a secondary bacterial infection. The compromised skin barrier enables bacteria to enter, potentially leading to conditions such as abscess formation.
Impetigo
Tinea pedis can make the affected areas more susceptible to impetigo—a highly contagious infection that manifests as honey-colored, crusted lesions capable of spreading to other body parts or individuals through direct contact.
Fungal Nail Infection
Untreated or inadequately managed tinea pedis can progress to toenail involvement or onychomycosis. Fungal organisms can infiltrate and cause onychomycosis, which is characterized by thickened, discolored, and brittle nails.
Cellulitis
In severe cases, tinea pedis can advance to cellulitis of the foot and leg. Cellulitis manifests as redness, warmth, swelling, and pain in the affected area, often accompanied by systemic symptoms.
Lymphangitis
Chronic or severe tinea pedis can result in lymphangitis, characterized by the appearance of red streaks extending from the affected area and swelling and tenderness of the lymphatic vessels.
Allergic Contact Dermatitis
Some individuals may experience a local allergic reaction to antifungal medications or other topical treatments for tinea pedis. This can lead to allergic contact dermatitis, characterized by redness, itching, and a rash at the application site.
Chronic or Recurrent Infection
Occasionally, tinea pedis may develop into a chronic or recurrent condition, necessitating long-term management and preventive measures. The persistent nature of the infection can substantially impact the individual's quality of life and foot health.
Although tinea pedis can be associated with these complications, not all patients will experience them. Prompt and appropriate treatment and adherence to preventive measures can help minimize the risk of complications and contribute to the successful management of tinea pedis.
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Deterrence and Patient Education
Tinea pedis is a fungal infection that affects the skin of the feet.The individuals at highest risk of developing tinea pedis are those with persistently wet feet, diabetes, immunocompromised conditions, and occlusive shoe wear. The most common places to contract tinea pedis are community showers, swimming pools, and bathing facilities or locker rooms.
The most prevalent symptoms of tinea pedis are itching, scaling, and developing skin cracks between the toes. In some cases, patients may develop thickening of the skin on the edges and underside of the foot, often accompanied by underlying redness. Most patients apply a topical cream for a duration of 4 weeks, with some experiencing a quicker response. Individuals who do not respond to topical treatment can consider oral therapy. To prevent a recurrence, patients should adhere to several key recommendations.
The following recommendations can aid in resolving a current infection and preventing future ones:
Washing feet daily with mild soap and warm water.
Drying the feet thoroughly after washing them, especially between the toes, after activities that involve water exposure.
Using a separate towel for the feet and avoid sharing it with others.
Discarding old shoes.
Wearing breathable shoes made of natural materials such as leather or canvas.
Avoiding tight-fitting shoes that may create a warm and moist environment promoting fungal growth.
Choosing moisture-wicking socks to keep the feet dry.
Wearing flip-flops or sandals in public places such as pools, gyms, and communal showers.
Avoiding sharing shoes, socks, towels, or any other personal items with others.
Inspecting feet for signs of infection, such as redness, itching, or scaling, regularly.
Addressing any cuts, blisters, or breaks in the skin promptly to prevent the entry of fungal organisms.
Wearing breathable shoes and moisture-wicking socks to minimize the risk of reinfection.
Keeping the feet clean, dry, and well-maintained.
Informing family members and close contacts about the tinea pedis infection and educating them on preventive measures to avoid its spread.
Encouraging family members to follow similar hygiene practices to minimize the risk of spreading the infection.
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Pearls and Other Issues
Regularly hosing the floors of shower rooms and the sides of swimming pools decreases the presence of dermatophytes on these surfaces. The use of topical antifungal powder in shoes, such as tolnaftate, can also help reduce the risk of tinea pedis infection. Patients with diabetes with tinea pedis are at a higher risk of developing onychomycosis. Furthermore, the presence of interdigital tinea pedis is a risk factor for cellulitis in patients with lymphoedema.
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Enhancing Healthcare Team Outcomes
Tinea pedis is a widespread infection frequently encountered by healthcare professionals. Each member of the interprofessional healthcare team can contribute to preventing future infections and reducing morbidity associated with the infection. Pharmacists can guide patients to the appropriate over-the-counter medication, ensuring they do not use a treatment better suited for a Candida infection. Clinicians caring for patients with diabetes can regularly inspect patients' feet and consistently educate them on self-foot care and examinations. As over-the-counter medications do not appear on shared medication lists, open communication among team members is crucial to prevent the morbidity associated with tinea pedis infection.
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Review Questions
Access free multiple choice questions on this topic.
Click here for a simplified version.
Comment on this article.
Figure
Tinea Pedis DermNet New Zealand
Figure
Rash, Tinea Pedis, Ringworm of the foot, dermatophytic fungal organism, Skin diseases, infectious, dermatomycoses Contributed by The Centers for Disease Control and Prevention
Figure
Tinea pedis Contributed by Pramod Kumar Nigam
Figure
Tinea pedis- plantar xerosis in a moccasin pattern Contributed by Mark A. Dreyer, DPM, FACFAS
Figure
Tinea pedis seen in a female patient Contributed by Dr. Hasnain Ali Syed
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References
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Becker BA, Childress MA. Common Foot Problems: Over-the-Counter Treatments and Home Care. Am Fam Physician. 2018 Sep 01;98(5):298-303. [PubMed: 30216025]
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Clebak KT, Malone MA. Skin Infections. Prim Care. 2018 Sep;45(3):433-454. [PubMed: 30115333]
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Rajagopalan M, Inamadar A, Mittal A, Miskeen AK, Srinivas CR, Sardana K, Godse K, Patel K, Rengasamy M, Rudramurthy S, Dogra S. Expert Consensus on The Management of Dermatophytosis in India (ECTODERM India). BMC Dermatol. 2018 Jul 24;18(1):6. [PMC free article: PMC6057051] [PubMed: 30041646]
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Ilkit M, Durdu M. Tinea pedis: the etiology and global epidemiology of a common fungal infection. Crit Rev Microbiol. 2015;41(3):374-88. [PubMed: 24495093]
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Lipner SR, Scher RK. Onychomycosis: Clinical overview and diagnosis. J Am Acad Dermatol. 2019 Apr;80(4):835-851. [PubMed: 29959961]
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Wang R, Song Y, Du M, Yang E, Yu J, Wan Z, Li R. Skin microbiome changes in patients with interdigital tinea pedis. Br J Dermatol. 2018 Oct;179(4):965-968. [PubMed: 29704463]
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Shemer A, Gupta AK, Amichai B, Baum S, Barzilai A, Farhi R, Kaplan Y, MacLeod MA. Increased Risk of Tinea Pedis and Onychomycosis Among Swimming Pool Employees in Netanya Area, Israel. Mycopathologia. 2016 Dec;181(11-12):851-856. [PubMed: 27435974]
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Mazza M, Refojo N, Davel G, Lima N, Dias N, Passos da Silva CMF, Canteros CE., Mycology Network of the Province of Buenos Aires (MNPBA). Epidemiology of dermatophytoses in 31 municipalities of the province of Buenos Aires, Argentina: A 6-year study. Rev Iberoam Micol. 2018 Apr-Jun;35(2):97-102. [PubMed: 29606407]
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Tsunemi Y, Takehara K, Miura Y, Nakagami G, Sanada H, Kawashima M. Specimens processed with an extraction solution of the Dermatophyte Test Strip can be used for direct microscopy. Br J Dermatol. 2017 Sep;177(3):e50-e51. [PubMed: 27943256]
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Vlahovic TC. Onychomycosis: Evaluation, Treatment Options, Managing Recurrence, and Patient Outcomes. Clin Podiatr Med Surg. 2016 Jul;33(3):305-18. [PubMed: 27215153]
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Drago L, Micali G, Papini M, Piraccini BM, Veraldi S. Management of mycoses in daily practice. G Ital Dermatol Venereol. 2017 Dec;152(6):642-650. [PubMed: 29050446]
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Sahoo AK, Mahajan R. Management of tinea corporis, tinea cruris, and tinea pedis: A comprehensive review. Indian Dermatol Online J. 2016 Mar-Apr;7(2):77-86. [PMC free article: PMC4804599] [PubMed: 27057486]
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Taheri A, Davis SA, Huang KE, Feldman SR. Onychomycosis treatment in the United States. Cutis. 2015 May;95(5):E15-21. [PubMed: 26057514]
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Gupta AK, Paquet M, Simpson FC. Therapies for the treatment of onychomycosis. Clin Dermatol. 2013 Sep-Oct;31(5):544-54. [PubMed: 24079583]
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Roana J, Mandras N, Scalas D, Campagna P, Tullio V. Antifungal Activity of Melaleuca alternifolia Essential Oil (TTO) and Its Synergy with Itraconazole or Ketoconazole against Trichophyton rubrum. Molecules. 2021 Jan 17;26(2) [PMC free article: PMC7830555] [PubMed: 33477259]
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Kably B, Launay M, Derobertmasure A, Lefeuvre S, Dannaoui E, Billaud EM. Antifungal Drugs TDM: Trends and Update. Ther Drug Monit. 2022 Feb 01;44(1):166-197. [PubMed: 34923544]
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Nigam PK, Syed HA, Saleh D. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Oct 29, 2023. Tinea Pedis. [PubMed: 29262247]
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Al Balushi KA, Alzaabi MA, Alghafri F. Prescribing Pattern of Antifungal Medications at a Tertiary Care Hospital in Oman. J Clin Diagn Res. 2016 Dec;10(12):FC27-FC30. [PMC free article: PMC5296449] [PubMed: 28208876]
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Del Rosso JQ. Onychomycosis of Toenails and Post-hoc Analyses with Efinaconazole 10% Solution Once-daily Treatment: Impact of Disease Severity and Other Concomitant Associated Factors on Selection of Therapy and Therapeutic Outcomes. J Clin Aesthet Dermatol. 2016 Feb;9(2):42-7. [PMC free article: PMC4771389] [PubMed: 27047631]
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Sasagawa Y. Internal environment of footwear is a risk factor for tinea pedis. J Dermatol. 2019 Nov;46(11):940-946. [PMC free article: PMC6900014] [PubMed: 31436337]
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Bowman J. Investigate all the options in skin care. Prof Nurse. 2004 Jul;19(11):43. [PubMed: 15317339]
Disclosure:Pramod Nigam declares no relevant financial relationships with ineligible companies.
Disclosure:Hasnain Syed declares no relevant financial relationships with ineligible companies.
Disclosure:Dahlia Saleh declares no relevant financial relationships with ineligible companies.
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Feingold KR, Ahmed SF, Anawalt B, et al., editors. Endotext [Internet]. South Dartmouth (MA): MDText.com, Inc.; 2000-.
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Pharmacological Causes of Hyperprolactinemia
Tea Shehu Kolnikaj, MD, Madalina Musat, MD, Rezvan Salehidoost, MD, and Márta Korbonits, MD, PhD, DSc, FRCP.
Author Information and Affiliations
Tea Shehu Kolnikaj, MD
Faculty of Technical Medical Sciences, University of Medicine, Tirana, Albania
Email: moc.oohay@jakinlokt
Madalina Musat, MD
National Institute of Endocrinology, Carol Davila University of Medicine and Pharmacy, Bucharest, Romania
Email: moc.oohay@tasumdm
Rezvan Salehidoost, MD
Centre for Endocrinology, William Harvey Research Institute, Barts and the London School of Medicine and Dentistry, Queen Mary University of London, UK,
Email: moc.liamg@tsoodihelas.navzer
Márta Korbonits, MD, PhD, DSc, FRCP
Centre for Endocrinology, William Harvey Research Institute, Barts and the London School of Medicine and Dentistry, Queen Mary University of London, UK,
Email: ku.ca.lumq@stinobrok.m
Corresponding author.
Last Update: January 4, 2024.
ABSTRACT
Hyperprolactinemia represents a multifaceted endocrine disorder with both physiological and pathological causes. The increased use of anti-psychotic and anti-depressant medications has increased the role pharmaceutical agents play in inducing hyperprolactinemia, being the most frequent cause of hyperprolactinemia in clinical practice. This has particularly impacted females, who demonstrate a higher susceptibility to drug-induced hyperprolactinemia. Of these medications, anti-psychotics, neuroleptic-like medications, anti-depressants, and histamine receptor type 2 antagonists, emerge as the most prominent culprits. Furthermore, opioids, some anti-hypertensive agents, proton pump inhibitors, estrogens, and other less potent hyperprolactinemia-inducing medications are recognized as potential contributors to drug-induced hyperprolactinemia. Many herbal medicines are reported as lactogenic, but their ability to cause hyperprolactinemia remains unclear. This review endeavors to elucidate the intricate mechanisms underlying the induction of hyperprolactinemia by pharmacological agents. We have included available data on the prevalence and extent of drug-induced changes in prolactin levels. We have also included data on herbal agents. We have highlighted where controversial data are identified. Although a detailed exploration of how these medications impact prolactin regulation is beyond the scope of this chapter, this review aims to deepen our understanding of the interplay between pharmacological agents and their effects on prolactin levels, contributing to valuable insights, refined therapeutic approaches, and better patient care. For complete coverage of all related areas of Endocrinology, please visit our on-line FREE web-text, WWW.ENDOTEXT.ORG.
INTRODUCTION
The most common cause of consistently high prolactin levels is drug-induced hyperprolactinemia. The overall incidence is higher in women compared to men. Drug-induced causes tripled during the 20-year follow-up period, reflecting the increased prevalence of psychoactive drug use (1). Several drugs have been reported to induce hyperprolactinemia, either by inhibiting dopamine receptors or their actions or by directly stimulating prolactin secretion (2).
High levels of prolactin can be attributed to various physiological factors, such as pregnancy and breast-feeding, while minor increases in prolactin levels may also occur during ovulation, after sexual intercourse, during periods of stress, exercise, after food intake, or in association with irritation of the chest wall and breast stimulation. Pathological causes can be related to hypothalamo-pituitary disorders or non-hypothalamo-pituitary disorders. Hypothalamo-pituitary disorders include prolactin-secreting pituitary tumors (including lactotroph tumors, mammo-somatotroph tumors, mature pluri-hormonal PIT1-lineage tumors, immature PIT1-lineage tumors, acidophil stem cell tumors, multi-hormonal pituitary tumors, mixed somatotroph, and lactotroph tumors, and pluri-hormonal tumors) (3); hypothalamic and pituitary stalk compression or damage (non-prolactin-secreting pituitary adenomas, craniopharyngiomas, meningiomas, germinomas, granulomas, metastasis, Rathke cleft cysts, hypophysitis, radiation, surgery, and trauma); infiltrative pituitary disorders; pituitary hyperplasia (McCune-Albright, Carney complex, X-LAG). Other causes include primary hypothyroidism; adrenal insufficiency; systemic diseases such as chronic renal failure and liver cirrhosis; polycystic ovary syndrome; neurogenic causes (chest wall trauma or surgery, herpes zoster); seizures; untreated severe phenylketonuria; pseudocyesis (false pregnancy); autoimmune diseases (lupus, rheumatoid arthritis, multiple sclerosis, systemic sclerosis, Behcet’s disease, polymyositis); cancers (breast, ovarian, colon, hepatocellular) (4). During the diagnostic process of hyperprolactinemia, it is crucial to consider the possibility of macroprolactinoma and the ‘hook’ effect, although the latter is usually not relevant in drug-induced cases (5).
Drug-induced hyperprolactinemia is often characterized by prolactin levels ranging from 25 to 100 ng/mL (530-2130 mIU/L). However, certain medications including metoclopramide, risperidone, amisulpride, and phenothiazines can lead to prolactin levels surpassing 200 ng/mL (4255 mIU/L) (6). On similar doses of prolactin-raising anti-psychotics, women with chronic use are more likely to develop hyperprolactinemia than men, reaching significantly higher prolactin levels, with mean levels of 50 ng/mL (1065 mIU/L) (7,8). Younger age was associated with higher prolactin levels in women, but not in men (8,9). Route of drug administration is important, with prolactin levels returning to normal after cessation of the drug: within 2-3 weeks after stopping oral treatment, but no sooner than 6 months after discontinuation of intramuscular depot administration (10).
This chapter will encompass a comprehensive discussion of all pharmacological causes as well as some alternative factors contributing to changes in prolactin levels especially hyperprolactinemia.
EPIDEMIOLOGY
Drug-induced hyperprolactinemia is the most common cause of consistently high prolactin levels. A retrospective follow-up study conducted in Scotland, involving 32,289 hyperprolactinemic individuals from 1993 to 2013, concluded that within the non-pregnancy-related group, the most prominent cause was drug-induced hyperprolactinemia (45.9%), followed by pituitary disorders (25.6%), macroprolactinoma (7.5%), and hypothyroidism (6.1%). Nevertheless, 15% of cases were deemed idiopathic. The overall incidence was higher in women aged 25-44 years old compared to men (1). Female predominance is reported in other studies with a female: male ratio of 5.9:1 and the mean age at diagnosis of hyperprolactinemia is 40 (range 14–85) years (2,11).The position of hyperprolactinemia as a side effect of medications has been assessed in a French Pharmacovigilance Database from 1985 to 2000, which reported 159 cases of hyperprolactinemia out of 182,836 adverse drug reactions (11). The rates of hyperprolactinemia related to therapeutic drug classes were recorded as 31% associated with anti-psychotics, 28% with neuroleptic-like drugs (medications with a similar mechanism of actions as neuroleptics, but used for different purposes, for example movement disorders or anti-emetics), 26% with anti-depressants, 5% with histamine receptor type 2 (H2-receptor) antagonists, and 10% with other drugs.
PROLACTIN CONTROL MECHANISMS
Prolactin is a polypeptide primarily produced in the anterior pituitary gland, with secondary production occurring in other tissues such as the gonads, mammary gland, endometrium, prostate, lymphocytes, hematopoietic cells, skin, brain, retina, inner ear cochlea, decidua, pancreas, liver, endothelium, and adipose tissue (12–14). In breast and prostate cancer, prolactin has even been proposed as a tumor marker (15,16). Prolactin acts through prolactin receptors (PRLR), which belong to the family of cytokine receptors associated with the non-receptor tyrosine Janus kinase 2. PRLR can activate the JAK-STAT (Janus kinase-signal transducer and activator of transcription) pathway, MAPK (mitogen-activated protein kinase), PI3 (phosphoinositide 3-kinase), Src kinase, as well as the Nek3 / Vav2 / Rac1 serine / threonine kinase pathway (17). There are different isoforms of this receptor: a long isoform, intermediate isoform, 2 short isoforms S1a and S1b which are formed by alternative splicing and partial deletion of exons 10 and 11, and soluble PRLR (18). These different isoforms are expressed in different tissues, mostly studied in rats. The long isoform is mainly expressed in the adrenal glands, kidneys, mammary glands, small intestine, bile ducts, choroid plexus, and pancreas whereas the short isoform is in the liver and ovaries (19). Prolactin possesses nearly 300 functions apart from lactation including neuroprotection and neurogenesis, offspring recognition by both parents, adipose and weight homeostasis, islet functions, immune regulation, angiogenesis, osmoregulation, and mitogenesis (20).
The secretion of prolactin produced by lactotroph cells in the anterior pituitary gland has a circadian rhythm with higher levels during sleep and lower levels during wakefulness (21). Even though pulsatility frequency does not significantly change over 24 hours, the amplitude of pulses is higher during night and day sleep, while wakefulness is associated with an immediate offset of active secretion. Prolactin is lower during the rapid eye movement stage of sleep (22).
The synthesis and secretion of prolactin is under the complex control of peptides, steroid hormones, and neurotransmitters, which can act as inhibitory or stimulatory factors, either by a direct effect on lactotroph cells or by indirect pathways through inhibition of dopaminergic tracts, and are widely studied in mammals (2). Dopamine plays a crucial role in inhibiting prolactin secretion. Dopamine can bind the five types of dopamine receptors (G-protein coupled receptors): DRD1, DRD2, DRD3, DRD4 and DRD5, while lactotroph cells express mainly D2 receptors. Dopamine can reach the pituitary via three pathways (Figure 1): through the tuberoinfundibular dopaminergic (TIDA) system, the tuberohypophysial tract (THDA), and the periventricular hypophyseal (PHDA) dopaminergic neurons (23). TIDA neurons originate from the rostral arcuate nucleus of the hypothalamus and release dopamine into the perivascular spaces of the medial eminence and through long portal vessels dopamine reaches the anterior pituitary gland. The THDA neurons originate in the rostral arcuate nucleus and project into the medial and posterior pituitary lobes and release dopamine at these sites. From THDA tract dopamine then reaches lactotroph cells through the short portal vessels (24). PHDA neurons originate in the periventricular nucleus and axons terminate in the intermediate lobe and dopamine release follows the same direction as from the THDA neurons. Prolactin-inhibiting neurons are considered to be a functional unit working synchronously (23). The binding of dopamine to D2 receptors on the plasma membrane of lactotroph cells inhibits prolactin protein, PRL gene transcription, as well as lactotroph proliferation (24). The release of prolactin through exocytosis of prolactin secretory granules is influenced by dopamine through various pathways. Specifically, D2 receptors are coupled with pertussis toxin-sensitive G proteins, which subsequently inhibit adenylate cyclase activity, resulting in decreased levels of cyclic adenosine monophosphate (cAMP) (25).
Figure 1.
Prolactin – Central Nervous System Regulation.
Additionally, the activation of potassium (K+) channels occurs, leading to a reduction in voltage-gated calcium (Ca2+) currents and inhibition of inositol phosphate production. Collectively, these intracellular signaling events culminate in a decrease in the concentration of free calcium ions (Ca2+) resulting in membrane hyperpolarization, ultimately inhibiting the exocytosis of prolactin from its granules (26). The inhibition of PRL gene transcription occurs when D2 receptors are activated, leading to the inhibition of MAPK or protein kinase C pathways. This activation results in a reduction of phosphorylation events on Ets family transcription factors. These transcription factors play a crucial role in the stimulatory responses of thyrotropin-releasing hormone (TRH), insulin, and epidermal growth factor (EGF) on prolactin expression. Moreover, the Ets family transcription factors interact with the PIT1 protein, which is essential for cAMP-mediated PRL gene expression (27). Dopamine exerts anti-mitogenic effects by activating D2 receptors through multiple pathways. These include the inhibition of MAPK (mitogen-activated protein kinase) signaling, protein kinase A signaling, and stimulation of phospholipase D activity. Additionally, dopamine engages a pertussis toxin-insensitive pathway, activates the extracellular signal-regulated kinases 1 and 2 (ERK1/2) pathway, and inhibits the AKT/protein kinase B pathway (28–31).
In addition to the dopaminergic inhibitory system, the γ-aminobutyric acid (GABA)-ergic tuberoinfundibular system, culminating in the median eminence, exhibits inhibitory properties, albeit of lesser potency compared to the dopaminergic system, while also having a role in prolactin modulation. GABA-B receptors are discernible both within the anterior pituitary gland, contributing to the maintenance of low prolactin levels, and in TIDA neurons which can be powerfully inhibited by GABA via hyperpolarization, consequently contributing to an elevation in prolactin levels (32,33).
Prolactin itself has two negative feedback effects: through short-loop feedback regulation it enhances the activity of TIDA neurons, where both long and short forms of the PRLR are expressed, with the long isoform being predominant in the arcuate and periventricular nuclei, regulating tyrosine hydroxylase (a rate-limiting enzyme in dopamine synthesis) leading to increase of dopamine release, which inhibits prolactin, as well as autocrine inhibition (34,35). Several other local factors influence prolactin release within pituitary gland as shown in Figure 1.
Other prolactin inhibitory factors include somatostatin, acetylcholine, endothelins, gastrin, and growth hormone, while stimulatory factors include thyrotropin-releasing hormone (TRH) (as seen in primary hypothyroidism), angiotensin II, vasopressin, oxytocin, VIP, galanin, and estrogen.
Experiments conducted on rats to elucidate the relationship between the adrenergic system and the regulation of prolactin secretion have focused on stimulating or inhibiting α and β adrenergic receptors. Functional hyperprolactinemia is a complex hormonal interplay of stress-induced neuroendocrine changes involving the dopamine, serotonin and adrenergic systems (36,37). Evidence suggests that the mediobasal hypothalamus and preoptic-anterior hypothalamus harbor the primary adrenoreceptors (38,39). Injecting the α2 agonist clonidine into the mediobasal hypothalamus resulted in a dose-dependent increase in prolactin secretion. This effect was counteracted by the blockade of idazoxan (α2 antagonist). Similarly, the stimulation of prolactin release was induced by isoprenaline (β agonist) and notably attenuated by the β antagonist propranolol. The β2 agonist salbutamol also exhibited efficacy in stimulating prolactin secretion. Conversely, adrenergic agonists such as noradrenaline (mixed α and β), phenylephrine (α1), and tyramine (sympathomimetic) in the mediobasal hypothalamus, failed to elicit an effect on prolactin secretion.
Within the preoptic anterior hypothalamus, noradrenaline and adrenaline were found to stimulate prolactin secretion (40). However, the administration of the α1 agonist phenylephrine failed to stimulate prolactin, indirectly suggesting that the stimulatory effect of noradrenaline in the preoptic anterior hypothalamus is likely due to its action at α2 sites. α2 agonism has been shown to reduce the function of tuberoinfundibular dopaminergic neurons leading to increase prolactin production and secretion (41). Consequently, it was inferred that the activation of α2 and β adrenoceptors in the mediobasal hypothalamus and α2 adrenoceptors in the preoptic-anterior hypothalamus, proximal to prolactin-regulating neurons, leads to heightened prolactin secretion, while the action of α1 in the mediobasal hypothalamus may be inhibitory (42).
Cholinergic activation may have opposite roles in rodents and humans. Cholinergic agonists suppress prolactin release induced by morphine in rats, suggesting that the central cholinergic system has an inhibitory effect on the prolactin release triggered by morphine or β-endorphine, but this cholinergic inhibition does not occur through catecholaminergic neurons (43). Conversely, in humans, cholinomimetic drugs can increase prolactin levels associated with raised plasma β-endorphin, suggesting a stimulatory interplay of cholinergic factors and endogenous opioids on prolactin levels (44), although circulating opioids may not directly relate to central levels.
TIDA neurons express estradiol and progesterone receptors. Estradiol action leads to reduced secretion of dopamine into the portal blood system and mediates a prolactin surge. Progesterone, in addition, suppresses dopamine release being responsible for the plateau phase of the surge (23). Estrogen specifically affects prolactin synthesis by influencing lactotroph cell sensitivity, expression of pituitary dopamine receptor downregulation, and the expression of the prolactin receptor gene (2,34). Ghrelin, a hormone involved in metabolic balance, directly stimulates prolactin secretion at the pituitary level (45).
Tachykinins (substance P, neurokinins A, and B, neuropeptide K, neuropeptide ϒ) can act directly on the lactotroph cell and indirectly within the hypothalamus or posterior pituitary. They have a multifaceted impact on prolactin secretion, with both stimulatory and inhibitory effects. They can stimulate prolactin secretion by stimulating and potentiating the release of oxytocin, vasopressin, TRH, VIP, serotonin and glutamate, and by inhibiting GABA. Tachykinins through paracrine actions can directly increase prolactin within the anterior pituitary. They can also increase dopamine but the overall effect is prolactin elevation. Under specific circumstances, the stimulation of dopamine release can be prominent leading to a decrease in prolactin (46). Endogenous opioids are involved in regulating prolactin secretion, particularly during stressful situations, by reducing the activity of tuberoinfundibular dopaminergic neurons mediated by μ-, κ‑, and δ- opioid receptors, resulting in increased prolactin release (47). Prolonged nicotine exposure has been associated with desensitization of dopamine receptors, diminished dopamine turnover, and a decrease in their abundance within the nigrostriatal pathways (48). These alterations have been suggested to contribute to a diminished prolactin response to opiate blockade observed in individuals who smoke. Similar to opioids, histamine has been shown to induce prolactin production predominantly through inhibiting dopaminergic and stimulating serotoninergic and vasopressinergic neurons (49).
Serotoninergic pathways originating from the dorsal raphe nucleus play a physiological role in mediating nocturnal surges and suckling-induced prolactin rises through a serotonin interaction via serotonin type 1 and 2 receptors (5-hydroxytryptamine receptors, 5HT1, and 5HT2). 5-HT could either release a PRL-releasing factor or inhibit dopamine release. The paraventricular nucleus, where serotoninergic pathways terminate, contains postsynaptic serotonin 5-HT1A, 5-HT2, and 5-HT2C receptor subtypes, and possibly 5-HT3 receptors (50). It was shown that the prolactin-releasing effect of serotonin probably occurs mostly via 5-HT1C / 2 receptors because ritanserin (an elective 5-HT1C / 2 receptor antagonist) opposed this effect (51). Serotonin stimulation of prolactin–releasing factor (PRF) neurons in the paraventricular nucleus leads to PRF release (like VIP and oxytocin) mediating hyperprolactinemia. Moreover, serotoninergic stimulation of GABAergic neurons in the tuberoinfundibular-GABA system has been shown to inhibit TIDA cells which contain 5-HT1A receptors, therefore inhibiting dopamine synthesis/release resulting in increased prolactin secretion (52).
Oxytocin, through the posterior pituitary and vasoactive intestinal peptide (VIP) in the anterior pituitary, play significant roles in enhancing PRL gene transcription and modulating dopamine inhibition. Animal studies suggest a potential mediation of VIP by oxytocin to stimulate prolactin secretion (53).
The extensive hormonal regulation of prolactin renders it susceptible to various disturbances caused by different classes of medications.
CLINICAL CHARACTERISTICS
Persistent hyperprolactinemia is associated with disturbances of the gonadal axis leading to interruptions of gonadotrophin-releasing hormone pulsatility and inhibition of luteinizing hormone and follicle-stimulating hormone release (54). Clinical manifestations attributed to hyperprolactinemia predominantly stem from the suppression of the gonadal axis. In premenopausal women, a spectrum of menstrual cycle dysfunctions is observed, spanning from luteal phase shortening to complete amenorrhea, often correlating with elevated prolactin levels. Secondary amenorrhea can be due to hyperprolactinemia in up 30% of patients, and up to 75% of patients with amenorrhea and galactorrhea (55). Beyond these effects, an array of hypoestrogenic indicators may manifest, including symptoms like vaginal dryness, diminished libido, and decreased energy levels. Galactorrhea can be present in up to 80% of females (55,56). In men, the impact of hyperprolactinemia is manifested through a decrease in libido, ranging from diminished sexual desire to oligospermia or even azoospermia attributed to hypogonadotropic hypogonadism. Notably, erectile dysfunction may arise, primarily attributed to the direct inhibitory influence of dopamine, and can be potentially reversed through the administration of dopamine agonists (57). Gynecomastia, on the contrary, is a manifestation of secondary hypogonadism rather than elevated prolactin levels, whereas galactorrhea is rare in men (21).
In both genders infertility can be observed, with diminished bone mineral density. In females, bone mineral density is significantly decreased in women with amenorrhea and increases during treatment and menstrual cycle restoration (58). Additionally, in cases where hyperprolactinemia is attributed to a mass, accompanying clinical indications may encompass headaches, visual field disturbances, cranial nerve palsies, and hypopituitarism. Notably, these manifestations may be the only clinical features in post-menopausal women (21,59).
PSYCHOTROPIC MEDICATIONS
Anti-psychotics and neuroleptic-like drugs are psychotropic medications which primarily exert their anti-psychotic effects through the blockade of DRD2 and D4 receptors in the mesolimbic area. Newer classes of anti-psychotics block 5HT2 and sometimes noradrenergic α1 or α2 receptors (4,35,60). Blockade of D2 receptors in the hypothalamic tuberoinfundibular system and lactotroph cells results in disinhibition of prolactin secretion leading to hyperprolactinemia, being the most common drugs known to induce hyperprolactinemia (61) (Figure 2). On the contrary, strong binding to D2 receptors can extend the half-life of dopamine by approximately 50%. This effect is achieved through two primary mechanisms: direct blockade of the dopamine transporter (DAT) and antagonism of D2 autoreceptors. These processes collectively result in reduced reuptake of dopamine, prolonging its presence in the synaptic area and further stimulating an upregulation of receptors. However, it is important to note that chronic use of anti-psychotics can lead to the reversal of upregulation of DAT (mRNA and protein), potentially contributing to treatment resistance and potentially lower prolactin elevations in the long term. Nonetheless, it is worth mentioning that anti-psychotics typically exhibit a lower affinity for dopamine transporter blockade compared to selective DAT blockers such as nomifensine (62).
Figure 2.
Mechanisms of drug-induced hyperprolactinemia with selected examples (adopted from La Torre et al. (2). In addition to opiates, cholinomimetics, PPIs and smoking indirectly also stimulate the opioid receptors. PPIs, Protein pump inhibitors; TCAs, tricyclic (more...)
The potency of anti-psychotics and neuroleptic-like medications to induce a rise in prolactin levels varies (Table 1). The level of prolactin increase depends on the anti-psychotic drug (different affinity and selectivity for dopamine receptors; blood-brain barrier penetrating capability, degree of serotoninergic inhibition), the dose administered, and the patient's age and sex (34,35). Lastly, polymorphisms in genes related to dopamine receptors (such as DRD1, DRD2, DRD3) (63), dopamine transporters (SLC6A3), and dopamine-metabolizing enzymes (such as monoamine oxidase and catechol-O-methyltransferase) have been associated with individual variations in response to anti-psychotic treatment and the development of side effects, including hyperprolactinemia (64).
Although direct evidence establishing the involvement of adrenergic receptors in hyperprolactinemia caused by antipsychotic and antidepressant medications remains unproven, indirect indications, as elucidated in Figure 1, suggest the potential implication of these receptors. It is plausible that adrenergic receptors might play a partial role in the hyperprolactinemia induced by these medications.
Table 1.
Medications and Their Ability to Cause Hyperprolactinemia
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| Cluster Name | Subclass mechanism of action | Medications | Prolactin increment | Frequency of prolactin increment (61,65) |
--- ---
| Anti-Psychotics |
| First generation anti-psychotics | Antagonize/block dopamine receptors, especially D2 receptors. Can block α1 adrenergic receptors. | Butaperazine | UP to 2-3-fold normal range with dozes 60 mg/daily. Higher in women (66) | High |
| Chlorpromazine | Up to 3-fold with initiation of treatment, up to 2-fold in long-term treatment) (67) | Moderate/ High |
| Flupenthixol | Up to 2-3-fold during the first month, and normalization in the next few months (68) | High |
| Fluphenazine | Up to 3-fold with initiation of treatment, up to 2-fold in long-term treatment (67). Up to 40-foldof the upper end of the normal range (69) | High |
| Haloperidol | Up to 9-fold at the beginning of treatment (3-fold in long-term treatment) (70) | High |
| Loxapine | Up to 3-foldof the upper end of the normal range in women (66) | Moderate |
| Perphenazine | Up to 40-fold of the upper end of the normal range (69) | Moderate |
| Pimozide | ? | Moderate |
| Prochlorperazine | ? | ? |
| Promazine | Up to 4-fold of the upper end of the normal range (69) |
| Thiordiazine | Up to 3-fold with initiation of treatment, up to 2-fold in long-term treatment) (67) | High |
| Thiothixene | Up to 3-fold with initiation of treatment, up to 2-fold in long-term treatment (71) | Moderate/ High |
| Trifluoperazine | ? | Moderate |
| Veralipride | Up to 10 time increment, transient (72) | High |
| Zuclopenthixole | ? | ? |
| Second generation anti-psychotics | Dopamine receptors blockade especially D2 receptors, serotonin (5-HT) receptor blockade, glutamate modulation, can antagonize α1 or α2 adrenergic receptors and histamine receptors. | Amisulpiride | Up to 10-fold at the beginning of treatment and remained elevated during treatment but lower levels (68) | Case reports |
| Aripiprazole | Reduce prolactin levels (73) | Case reports / No effect/ Reduced prolactin |
| Asenapine | Up to 2-fold increment and rarely with higher doses up to 4-fold (35) | Low, Moderate |
| Brexpiprazole | Mild increment (74) | Low |
| Clozapine | Mild (up to 2-fold) and transient (75) | Case reports or No effect |
| Iloperidone | Mild increment, transient (76) | Case reports or No effect |
| Levosulpiride | Up to 15-fold normal range (77) | Case reports / Moderate for galactorrhea (78) |
| Lurasidone | Up to 10-fold normal range (79)/ no effect (80) | Case reports or No effect |
| Molindone | ? | Moderate |
| Olanzapine | Mild (up to 2-fold) and transient (75) | Low |
| Paliperidone | 2-10-fold for depot formulations (81) | High |
| Perospirone | None (82) | None/ Case reports |
| Quetiapine | Mild and transient (75) | Low |
| Risperidone | 2-10-fold | High (83) |
| Sertindole | Mild and transient (75) | ? |
| Sulpiride | Up to 6-7-fold from baseline, dose dependent effect (84) | High |
| Thiethylperazine | ? | ? |
| Ziprasidone | Up to 4-fold from baseline and transient (35,75) | Low |
| Neuroleptic-Like Medications |
| Block D2 receptors | Domperidone | Up to 10-fold (85,86) | High |
| Droperidol | Significant increment after 10 minutes of administration, with peak at 20 minutes (87) | ? |
| Metoclopramide | Up to 15-fold (2) | High |
| Anti-Depressants |
| TCAs | Block the reuptake of both serotonin and noradrenaline. | Amitriptyline | 2-fold increment on dosage 200/300mg (88) | Low |
| Amoxapine | 3,5-fold to baseline (89) | High |
| Clomipramine | Up to 3-fold increment from baseline (90) | High |
| Desipramine | Just above the normal limit with 100 mg oral administration (91) | Low, Controversial |
| Imipramine | Up to 4-fold normal range (69) | Controversial |
| Nortriptyline | 2-fold in the first 2 weeks in one patient (88) | None or Low |
| SSRI | Block the reuptake of serotonin. | Citalopram/Escitalopram | Up to 3-fold increment (52) | None or Low (rare reports), Controversial data |
| Fluoxetine |
| Fluvoxamine |
| Paroxetine |
| Sertraline |
| SNRI | Block the reuptake of both serotonin and noradrenaline. | Duloxetine | Up to 2-fold normal range (92) | Case reports |
| Milnacipran | Not increased risk of hyperprolactinemia (93) | None |
| Venlafaxine | Up to 2-fold normal range, dose related (94) | Case reports |
| MAO inhibitors | Inhibit the enzyme. Monoamine oxidase, which breaks down serotonin, noradrenaline, and dopamine, though increasing their levels. | Clorgyline | Up to 2-fold from baseline (95) | Low |
| Pargyline | Up to 3-fold from baseline (95) | Low |
| Phenelzine | Unclear elevation, galactorrhea (96) | Low/ Case reports |
| Atypical anti-depressants | Inhibit noradrenaline and dopamine reuptake. | Bupropion | No significant change (80) | Case reports |
| Increases the release of both serotonin and noradrenaline. | Mirtazapine | No significant change (80) | Case reports |
| Serotonin modulators | Modulate serotonin receptors in the brain to enhance serotonin transmission. | Indoramine | (97) | Case report |
| Nefazodone | Mild increment from baseline only at acute administration (98) | None/ Case reports |
| Trazodone | Up to1.5-fold from baseline (99) | None, Low |
| Vortioxetine | Up to 2-fold elevation (100) | Case reports |
| Selective noradrenaline reuptake inhibitor | Inhibit reuptake of norephinephrine. | Reboxetine | Up to 2-fold from baseline (101) | Case reports |
| NMDA receptor antagonist | Block NMDA receptors though influencing glutamate neurotransmission. | Esketamine | ? | None |
| Gastric Acid Reducers |
| H2 receptor antagonists | H2 receptor antagonists. | Cimetidine | Up to 3-fold after 400 mg IV infusion (102) | Low |
| Ranitidine | Mild increment only in high IV doses (103) | Low |
| Protein pump inhibitors (PPIs) | Inhibit the activity of the proton pump (H+/K+ ATPase) in the stomach's parietal cells. | Esomeprazole | ? | Case reports or No effect |
| Lansoprazole | 4-fold increment from baseline (104) |
| Omeprazole | No significant change (105) |
| Pantoprazole | No significant change (106) |
| Rabeprazole | No significant change (107) |
| Opioids |
| They activate opioid receptors. Main types of opioid receptors: mu (μ), delta (δ), and kappa (κ). | Apomorphine | By acting as dopamine agonist it lowers prolactin (108) | None |
| Heroin | Elevated in addiction (within normal range) compared to healthy control or during abstinence (109) | Moderate in addicted patients that have values over 25 ng/mL |
| Methadone | Mild increment, transient increases for several hours following the administration (110) | ? |
| Morphine | Up to 2-fold increment from baseline (111) | High |
| Antihypertensives |
| It decreases the release of noradrenaline. | Methyldopa | 3-4-fold (65) up to 40-fold normal range (69) | Moderate |
| Inhibit the storage of neurotransmitters like noradrenaline and serotonin in nerve cells, though decreasing their release. | Reserpine | 2.5-fold increment from baseline (112) Up to 40-fold normal range (69) | High |
| Block calcium channels in cardiac and smooth muscle cells. | Verapamil | 2-fold (113) | Low |
| Estrogens |
| By using as contraceptives they suppress sexual axis. | Estradiol infusion | 3-4-fold, dose-dependent, way of administration is important (oral and IV) (114) | Low |
| Estradiol withdrawal | ? |
| Gonadotropins and GNRH Agonists |
| Same as endogenous components, used for fertility induction. | hCG | Up to 4-fold increment, transient (115) | High |
| hMG | Up to 2.7-fold increment from baseline, transient (116) | High, Transient |
| GnRH agonist. | Leuprolide acetate | 1.5-fold higher prolactin in compared to hMG alone, transient (117) | High |
| Other Drugs |
| Benzodiazepines | Enhances the effects of GABA in the brain. | Diazepam | Mild, dose-dependent (118) | Controversial |
| Anxiolytics | Serotonin receptor agonist. | Buspirone | 2-fold (119) | Case report or No effect |
| α-2 adrenergic agonist. | Clonidine | ? | Case reports |
| Anticonvulsant | Block sodium channels in nerve cells. | Carbamazepine | Less than 2-fold in sleep entrained (120,121) | ? |
| Phenytoin | Controversial, it can also lower prolactin levels (122) | ? |
| Enhances the effects of GABA in the brain. | Phenobarbital | Controversial (123) |
| Valproic Acid | Controversial, it can also lower prolactin (124) | Case reports |
| Mood stabilizer | Decrease dopamine release and glutamate, increase GABA inhibition. | Lithium Carbonate | Controversial, no effect (183) | None |
| Antimigraine medication | Calcium channel blocker. | Flunarizine | Mild increment, up to 1.5-fold from baseline (125) | Case reports |
| Weight loss medications | Increase the release of serotonin and inhibit its reuptake. | Fenfluramine | Mild increment within normal range in previews non-hyperprolactinemic patients (126) | High |
| Inhibit the reuptake of serotonin, noradrenaline, and dopamine. | Sibutramine | 4-fold (2,127) | Case report |
| Anticholinesterase inhibitors | Reversible acetylcholinesterase inhibitor. | Physostigmine salicylate | Less than 100 ng/mL (44). | Low |
| Prokinetic medication | Stimulate serotonin receptors in the gut. | Cisapride | High increment (up to 200 ng/dL) but in co-administration of other drug inducing hyperprolactinemia (128) | Case reports |
| Antihistaminic with sedative and antiemetic properties | Block histamine receptors. | Promethazine | ? | ? |
| Central Nervous System Stimulants | Increase the release and reduce the reuptake of noradrenaline and dopamine in the brain. | Amphetamine | Mild, only during withdrawal (129) | ? |
| Methylphenidate | No effect (130) | Case reports/ No effect |
| ADHD medication | α -2 adrenergic agonist. | Guanfascine | Controversial, it can also lower prolactin (131) | Case reports |
| Decongestant | Sympathomimetic amine, predominantly α-1 agonist | Pseudoephedrine | Lower prolactin levels (132) | Case reports |
| Rheumatoid arthritis medications | Reduce inflammation, modify immune response. | Bucillamine | Mild increment within normal range (133) | Case report |
| Penicillamine | ? | Case reports |
| Osteoporosis medication | Monoclonal antibody that inhibit the receptor activator of nuclear factor kappa-B ligand (RANKL). | Denosumab | ? | Case reports |
| Substance of abuse | Blocks the reuptake of noradrenaline, dopamine, and serotonin in the brain. | Cocaine | Decrease prolactin levels (134)Mild increment only during withdrawal (129) | Case reports |
| Increases the release and inhibits the reuptake of serotonin and to some extent, dopamine and noradrenaline. | Ecstasy | Mild or no effect (135) | ? |
| Stimulates nicotinic acetylcholine receptors, leading to the release of neurotransmitters like dopamine and noradrenaline. | Smoking | Mild increment, transient (136) | Moderate |
| Anti-HIV medications | Protease inhibitors that prevent the cleavage of viral proteins and thereby inhibiting viral replication. | Ritonavir / Saquinavir | Mild (137) | Case reports |
| Radiotherapy | Use of high-energy radiation to damage the DNA within the targeted cells. | Intracranial radiotherapy | ? | Moderate (138) |
: Frequency of increase to abnormal prolactin levels with chronic use: high: >50%; moderate: 25 to 50%; low: <25%; none or low: case reports. The effect may be dose-dependent. Drugs marked with blue have controversial data or decrease prolactin levels as explained in the table. Where we could not identify reliable data for the parameters in the table we added a question mark. First-generation anti-psychotics, non-selective dopamine receptors antagonists. Second-generation anti-psychotics.
Anti-Psychotics
Anti-psychotics are traditionally classified as first- and second-generation, but more recently a new classification taxonomy has been developed by McCutcheon et al. to express different receptor affinity of different anti-psychotics. Due to the impossibility to include in this new classification all drugs that cause hyperprolactinemia, we have used the old classification (Table 1) (139,140).
The first-generation anti-psychotics are typically associated with more severe hyperprolactinemia (2-3-fold increment), whereas second-generation drugs have lower D2 affinity and stronger blockade of 5HT2A receptors leading to milder prolactin elevations (1-2-fold), except risperidone, paliperidone, and amisulpiride. Amisulpiride has the greatest potential to cause hyperprolactinemia of all anti-psychotics (4).
The first-generation anti-psychotics, such as fluphenazine and haloperidol, act as non-selective dopamine receptors antagonists (2,10). The therapeutic effects on psychotic symptoms occur through D2 and D4 receptor binding in the mesolimbic area, while side effects are mediated by D2 blockade in the striatal area (linked to extrapyramidal effects) and in the hypothalamic infundibular system (linked to hyperprolactinemia) in more than 50% of patients. A clinical trial involving 69 patients examined the effects of various anti-psychotic medications on prolactin levels, including chlorpromazine, depot haloperidol, fluphenazine, zuclopenthixol, sulpiride, pimozide, droperidol, and flupenthixol. The study found a significant elevation in prolactin levels only in females, with a mean level of 1106 mIU/L (52 ng/mL) compared to the normal range of <480 mIU/L (22.6 ng/mL). In males, the mean prolactin levels were within the normal range, which may be attributed to the significantly lower total daily dose of chlorpromazine used in males (199.0-220.1 mg/day) compared to females (384.4-302.48 mg/day, P<0.05) (7).
Second-generation anti-psychotics with lower D2 affinity led to milder prolactin elevations (1-2-fold), except for paliperidone, risperidone, and amisulpiride whose effect on prolactin is similar to the first-generation neuroleptics. Chlorpromazine, loxapine, olanzapine and quetiapine have variable effects on prolactin secretion, while aripiprazole, clozapine, iloperidone, lurasidone have little or no effect on prolactin secretion (35).
An important factor contributing to variations in the induction of hyperprolactinemia by different anti-psychotic medications is the blood-brain barrier. Permeability glycoprotein transporter (P-gp), coded by the ABCB1 gene, is expressed in various tissues including in the cells of the blood-brain barrier. P-gp plays a role in actively transporting hydrophobic drugs with a molecular weight greater than 400 Da out of the brain, thus protecting the brain from these medications; therefore, this protein can change drug bioavailability (141,142).
The affinity of risperidone, paliperidone, and amisulpiride (prolactin rises up to 10-fold with these drugs) for P-gp is approximately twice that of olanzapine and chlorpromazine (prolactin rise is up to 3-fold with these drugs), and four times greater than haloperidol and clozapine (prolactin rise can be high initially but usually reduces with time) (143). The higher affinity of risperidone, paliperidone, and amisulpiride to P-gp could, among other mechanisms, partly explain the greater induction of hyperprolactinemia by these drugs, as P-gp does not allow them to enter the brain via the blood-brain barrier. Therefore, the portal circulation of the anterior pituitary delivers a somewhat higher concentration of these drugs to the lactotrophs, which are located outside the blood-brain barrier, to inhibit the D2 receptors (144).
Aripiprazole can act as a partial agonist at D2 receptors and display partial agonist activity at 5HT1A receptors, while also acting as an antagonist at 5HT2A receptors. Antagonism at these receptors can help to normalize prolactin levels since 5HT2A receptor activation has been associated with increased prolactin release. That is why it is considered a prolactin secretion modulator (145). Its role in prolactin levels has been investigated in a study involving both retrospective and prospective components (146). The retrospective part of the study included 30 patients undergoing risperidone treatment, when it was observed that after 6 months of treatment, prolactin levels remained high although somewhat lower than at the start of observation. In the prospective part of the study, 30 other patients were divided into two groups: one group receiving risperidone alone at a daily dosage of 2-4 mg and the other group receiving a combination of risperidone and aripiprazole at a daily dosage of 5-10 mg. The group receiving adjunctive aripiprazole exhibited significantly lower serum prolactin levels compared to the risperidone-only group at weeks1 (914±743 vs 1567±1009 mU/L), 2 (750±705 vs 1317±836 mUI/L) and 6 (658±590 vs 1557±882 mUI/L). Notably, during aripiprazole treatment, prolactin levels at weeks 1, 2, and 6 were significantly lower than at baseline (P< 0.05) (at baseline patients were treated with risperidone as monotherapy), suggesting that aripiprazole may effectively alleviate risperidone-induced hyperprolactinemia. Similar findings supporting the role of aripiprazole in reducing prolactin levels have been reported in other studies (147). Combination therapy presents a promising therapeutic approach for adjunctive treatment or for transitioning from risperidone to mitigate hyperprolactinemia (146).
More recently, a new medication SEP-363856, a trace amine-associated receptor 1 (TAAR1) and 5HT1A agonist, has been developed to treat schizophrenia. Its mechanism of action is not based on D2 antagonism, and has a favorable effectiveness and tolerability profile, without causing hyperprolactinemia (148). This category of medication serves as compelling evidence for the significant involvement of dopamine receptors in drug-induced hyperprolactinemia, and it is a future viable therapeutic choice for patients experiencing adverse effects associated with hyperprolactinemia.
DRUG-INDUCED HYPERPROLACTINEMIA IN PEDIATRIC PATIENTS
Anti-psychotic medications have been found to induce hyperprolactinemia in the pediatric population as well as in adults. In a trial involving 35 children and adolescents with early-onset psychosis, primarily diagnosed with childhood-onset schizophrenia or psychotic disorder not otherwise specified, prolactin levels were measured after a 3-week washout period, as well as after 6 weeks of treatment with haloperidol, olanzapine, and clozapine (149). Following the 6-week treatment period, haloperidol (9 of 10 patients – mean age 13.4 years) and olanzapine (7 of 10 patients– mean age 15.9 years) resulted in prolactin levels above the upper limit of normal. The mean increase was 5.2-fold for haloperidol and 2.4-fold for olanzapine. The prolactin response did not show statistically significant differences between females and males treated with haloperidol and olanzapine. Clozapine (22 patients, mean age 14.7) caused a small but significant rise in females (1.2-fold) but levels remained in the normal range for all patients. There was no rise in males. Why this difference between females and males is only on clozapine remains unclear and difficult to explain. However, in a study involving 36 girls aged 8-17 years, mean prolactin levels were higher in girls compared to boys, with the most significant increase occurring around the age of 13y, correlating with menarche. A highly significant correlation was found between increases in plasma prolactin and estradiol levels between the ages of 11 and 13 years. Girls with long menstrual cycles (>28 days) between the ages of 14 and 16 years had higher prolactin levels (p<0.05) (150). Even though the mean age of patients on clozapine was above 13 years, we do not possess information on the duration of the menstrual cycle of those girls, as if it is longer, the physiologic estrogenization because of the longer menstrual cycle can impact the range of prolactin elevation. In any case, the population sample size was relatively small to draw definitive conclusions and to provide answers why this happens only in clozapine patients (149).The authors concluded that the prolactin response in male children and adolescents treated with haloperidol or olanzapine was significantly higher than that observed in adult males. However, the prolactin response in female children and adolescents after haloperidol treatment did not differ significantly from that of adult females in similar studies, possibly due to the adult similarity of estrogen status seen in female adolescents (149). Aripiprazole in another study showed a lesser prolactin increase than olanzapine, quetiapine, and risperidone, similar to the adult population (151).
In another clinical trial involving 396 children and adolescents (aged 14.0 ± 3.1 years), the impact of anti-psychotic medications on prolactin levels was studied. The medications involved risperidone, olanzapine, quetiapine, and aripiprazole. Risperidone caused the highest incidence of hyperprolactinemia (93.5%) and had the highest peak prolactin levels (median = 56.1 ng/mL) followed in order by olanzapine, quetiapine, and aripiprazole. Menstrual disturbances were the most prevalent side effect (28.0%), particularly with risperidone (35.4%). Notably, severe hyperprolactinemia was associated with decreased libido, erectile dysfunction, and galactorrhea (152).
In conclusion, a comprehensive meta-analysis comprising 32 randomized controlled trials with a total of 4643 participants, with an average age of 13 years, has demonstrated that risperidone, paliperidone, and olanzapine are associated with a significant increase in prolactin levels among children and adolescents. Conversely, aripiprazole is linked to a notable decrease in prolactin levels in this age group. It is worth noting that haloperidol was not included in these studies, resulting in an absence of evidence regarding its prolactin-related effects in this population (153).
These findings underscore that haloperidol, risperidone, paliperidone and olanzapine are potent inducers of hyperprolactinemia in children and adolescents, mirroring observations in the adult population. A comprehensive listing of medications associated with hyperprolactinemia in children can be found in Table2.
Table 2.
Drugs Reported to Induce Hyperprolactinemia in Children and Adolescents
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| Medication class | High>50 percent of patients | Moderate25-50 percent of patients | Low<25 percent of patients | Case reports |
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| Anti-psychotics, first-generation 'typical' | Fluphenazine (154)Haloperidol (149,155) | Chlorpromazine (156)Loxapine (157)Pimozide (158,159) |
| Anti-psychotics, second-generation 'atypical' | Paliperidone (160,161)Risperidone (152,155,162–164) | Asenapine (165)Molindone (166)Olanzapine (149,152)Lurasidone (167,168) | Ziprasidone (169)Quetiapine (152,162) | Clozapine (149)Aripiprazole (152,170)Amisulpride (171)Brexpiprazole (172) |
| Anti-depressants | Clomipramine (173) | Desipramine (174) | Bupropion (175)Citalopram (176)Escitalopram (177)Fluoxetine (178)Sertraline (179)Duloxetine (177)Paroxetine (180)Venlafaxine (181) |
| Anti-emetics and gastrointestinal medications | Metoclopramide (182–184)Domperidone (185,186) | Omeprazole (187)Lansoprazole (187)Cisapride |
| Others | Fenfluramine (188) | Estrogens (189)Triptorelin (190) | Clonidine (191)Methylphenidate (181)Guanfacine (181)Valproic acid (181)Penicillamine (181) |
: Aripiprazole is a partial agonist at the type 2 dopamine receptor and display partial agonist activity at the type 1A serotonin receptor (5HT1A) and antagonist at 5HT2A receptor. It can be used in combination with other psychotropic medications to reduce prolactin levels. Aripiprazole itself can sometimes cause mild hyperprolactinemia (192).
CLINICAL MANAGEMENT OF ANTI-PSYCHOTIC-INDUCED HYPERPROLACTINEMIA
Drug-induced hyperprolactinemia should be considered in the differential diagnosis of elevations of prolactin levels, sometimes greater than 200 µg/L (4260 mIU/L). Particularly when serum prolactin levels exceed 80-100 µg/L (1700-2130 mIU/L), pituitary magnetic resonance imaging (MRI) should be performed to rule out the presence of any underlying pituitary or hypothalamic masses that may contribute to hyperprolactinemia (193). According to the guidelines from the Endocrine Society, in symptomatic patients suspected of having drug-induced hyperprolactinemia, it is recommended the first test to diagnose drug-induced hyperprolactinemia is the discontinuation of the medication for 3 days or switch to an alternative drug (e.g. a prolactin-sparing anti-psychotic (e.g. aripiprazole), or an anti-psychotic with lower dopamine antagonist potency (Table 1) followed by retesting of serum prolactin levels.
However, any discontinuation or substitution of anti-psychotic agents should be done in consultation with the patient's psychiatric physician. If discontinuation is not possible or if the onset of hyperprolactinemia does not coincide with therapy initiation, obtaining a pituitary MRI is recommended (despite prolactin levels) to differentiate between medication-induced hyperprolactinemia and hyperprolactinemia caused by a pituitary or hypothalamic mass (194).
Before initiating treatment with an anti-psychotic medication, it is advised that clinicians inquire about the patient's previous treatment experience, sexual dysfunction, menstrual history (including irregularities and menopausal status), as well as any history of galactorrhea. Additionally, obtaining a baseline prolactin level is recommended before starting treatment (195). This pre-treatment screening for hyperprolactinemia can help determine whether subsequently elevated prolactin levels are due to medication-induced factors (196) and make the diagnosis easier without the need to perform further imaging.
While treatment of hyperprolactinemia in patients receiving anti-psychotics may not always be necessary, in cases where clinical hypogonadism is evident, several options are available (193). The initial step is to discontinue the drug if clinically feasible. If discontinuation is not possible, switching to a similar anti-psychotic that does not cause hyperprolactinemia is suggested. If neither of these options is feasible, cautious administration of a dopamine agonist may be considered in consultation with the patient's physician (194). It is worth noting that these interventions only result in the normalization of prolactin levels in approximately half of the patients receiving anti-psychotics, and careful psychiatric monitoring is required due to the possibility of psychosis exacerbation with dopamine agonists (193).
For patients with long-term hypogonadism demonstrated by hypogonadal symptoms or low bone mass, the use of estrogen or testosterone replacement is recommended. In rare instances where patients receiving anti-psychotics also present with a pituitary tumor, treatment options primarily revolve around tumor-specific interventions, considering especially optic chiasm compression. When a non-functioning tumor is suspected, the above options to eliminate the drug-induced component of hyperprolactinemia should be considered, with supervised short-term cessation of the medication to clarify drug-induced or tumor-derived hyperprolactinemia (193).
In addition to the interventions mentioned earlier, addressing fertility concerns in patients with drug-induced hyperprolactinemia may require further measures. Normalization of prolactin levels through medication adjustment or dopamine agonist therapy can often lead to the restoration of fertility. In situations where fertility is not spontaneously regained, fertility treatment with gonadotrophins or assisted reproductive procedures may be necessary.
NEUROLEPTIC-LIKE MEDICATIONS
Metoclopramide and domperidone are anti-emetic and gastrointestinal motility agents known also as neuroleptic-like medications which can increase pituitary prolactin secretion and breast milk production by a dopamine antagonistic action (Figure 1).
Metoclopramide is a central and peripheral D2 receptor antagonist (197). Its administration is followed by an acute increase in prolactin levels up to 15-fold above the baseline that persists in chronic administration of the drug (2). Even though the majority of patients treated with metoclopramide develop hyperprolactinemia (Table 1), related symptoms such as amenorrhea, galactorrhea, gynecomastia, and impotence remain unclear (2).
Domperidone is a peripheral D2 antagonist (it does not cross the blood-brain barrier) used for treating intestinal motility disorders, especially for the prevention of gastrointestinal discomfort with dopaminergic treatment in Parkinson's disease (198), as it antagonizes the D2 receptors in the upper gastrointestinal tract. It also reaches the D2 receptors on lactotroph cells inducing hyperprolactinemia, and can be used for stimulating lactation, for example in women with preterm infants, or an adoptive parent (199,200), and even in transgender women who wish to breastfeed (201,202). It can cause a 10-fold elevation in prolactin levels with normalization of prolactin levels after three days (85,86). Neuroleptic-like medications are summarized in Tables 1 and 3.
Table 3.
H2 Receptor Antagonists, Opioids, Anti-Hypertensives, PPIs, Estrogens, and Other Drugs and Their Ability to Cause Hyperprolactinemia.
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| Medication class | High>50% of patients | Moderate25-50% of patients | Low<25% of patients | Case reports |
--- ---
| Anti-emetic and gastrointestinal | DomperidoneMetoclopramide | Prochlorperazine | EsomeprazoleOmeprazoleLansoprazoleCisapride |
| H2-receptor antagonists | CimetidineRanitidine |
| Anti-hypertensives | Methyldopa | Verapamil |
| Others | FenfluramineOpioids | Estrogens | Protease inhibitorsCocaine BucillamineClonidineMethylphenidateGuanfascineValproic AcidPenicillamine |
ANTI-DEPRESSANTS
Anti-depressants can be classified based on their structure and mechanism of action into tricyclic anti-depressants (TCAs), selective serotonin reuptake inhibitors (SSRIs), serotonin-noradrenaline reuptake inhibitors (SNRIs), monoamine oxidase (MAO) inhibitors, atypical anti-depressants, serotonin modulators, selective noradrenaline reuptake inhibitor, and NMDA (N-Methyl-D-Aspartate) receptor antagonists (203). Data on their ability to cause hyperprolactinemia are controversial, as described below. The two main mechanisms, both related to elevated serotoninergic tonus, explain how anti-depressants can induce hyperprolactinemia by indirect modulation of prolactin release by serotonin and serotonin stimulation of GABAergic neurons (50) Adrenergic receptors involvement in their ability to cause hyperprolactinemia remains unclear. (Figure 1).
Data on the incidence of hyperprolactinemia with anti-depressant medications, especially SSRIs, MAO inhibitors, and some TCAs, suggest that they can cause modest and generally asymptomatic hyperprolactinemia (4,193). Their ability to cause hyperprolactinemia is summarized in Table 1.
TCAs
TCAs, such as amitriptyline, desipramine, clomipramine, and amoxapine, can induce sustained mild hyperprolactinemia (2) (Table 1). They manifest their mechanism of action by blocking the reuptake of noradrenaline and serotonin, through increasing serotoninergic tonus, leading to hyperprolactinemia. Amitriptyline's ability to cause hyperprolactinemia seems to be dose-dependent. A dosage of 150-250 mg/day to 5 patients showed no effect on prolactin levels after 3-7 weeks (204), whereas a dosage of 200-300 mg/day caused a 2-fold increment of prolactin levels in two of nine chronic treated patients (88). In 13 patients with depression taking amitriptyline or desipramine, prolactin levels were studied after intravenous injection of tryptophan (serotonin precursor). It was observed that tryptophan-induced prolactin elevation was significantly increased compared with a preceding placebo period (205). A similar tryptophan test was performed with clomipramine 20 mg vs placebo in 6 normal subjects. Levels of prolactin increased after tryptophan infusion in pretreated patients with clomipramine (206),suggesting serotonin involvement in this hyperprolactinemia.
The effect of the traditional TCA imipramine on serum prolactin levels is controversial. A five-week double-blind study on patients with depression taking imipramine did not show any significant change in serum prolactin (207). However, another study in young healthy men showed that imipramine’s effect on prolactin is dose-dependent (usually therapeutic dose is 50-150 mg): oral administration of 100 mg, but not of 40 mg, led to a consistent rise in prolactin levels after 3 hours of administration, but the rise was mild (maximum 25.85 ng/mL (550 mUI/L) (91).
Data regarding nortriptyline's ability to induce sustained hyperprolactinemia are lacking. Anyway, over a 4-6 week treatment of 8 patients with nortriptyline up to 150 mg/daily, no difference was found between placebo and the medication group. In only one patient was observed a transitional 2-fold elevation in the first 2 weeks (88).
Amoxapine, which is an anti-depressant with neuroleptic properties as well, is found to increase prolactin levels in 10 patients approximately 3.5-fold compared to baseline and more than desipramine in 12 patients, where almost no difference with baseline was observed (89). The proposed mechanism of hyperprolactinemia involves the blockade of the D2 receptor in tuberoinfundibular neurons or the anterior pituitary gland (2).
SSRIs
SSRIs enhance serotonin activity via inhibition of neuronal serotonin reuptake. This could be the most prominent mechanism leading to a prolactin elevation. A review of 13 case reports showed prolactin levels between 28 and 60 ng/mL (595 and 1276 mUI/L) (52). In a French study, 27 of 159 cases (17%) had SSRI-induced hyperprolactinemia with sertraline being the most prominent, followed by fluoxetine, paroxetine and fluvoxamine. Only citalopram was found not to increase prolactin levels significantly (208). However, in another study fluoxetine, paroxetine, and fluvoxamine were found again to elevate prolactin levels, but they also found citalopram to induce hyperprolactinemia. In this study duloxetine, milnacipran, and sertraline (which was the most prominent in the previous study) were not associated with an increased risk of hyperprolactinemia (93). Fluoxetine-induced hyperprolactinemia was found in patients with major depression (4.5% of men and 22.2% of women) following 12 weeks of fluoxetine treatment (209).Differences in the ability to cause hyperprolactinemia can be attributed to the variations in the affinity of the SSRIs for dopamine, histamine, and GABA receptors.
The Nurses’ Health Study and its follow-up study assessed anti-depressant use and circulating prolactin levels in 610 women (including 267 anti-depressant users) with two measurements of prolactin an average of 11 years apart (210). In this study, mean prolactin levels were similar among SSRI users (13.2 µg/L, (280 mUI/L), 95% CI 12.2-14.4), users of other classes of anti-depressants (12.7 µg/L (270 mUI/L), 95% CI 11.0-14.6), and non-users (13.1µg/L, (278 mUI/L), 95% CI 12.8-13.4) (210). However, the duration and dosage of anti-depressant use at the time of prolactin sampling had not been assessed, as the participants had only responded as current anti-depressant users/non-users on a questionnaire.
MAO Inhibitors
Monoamine oxidase is an enzyme responsible for breaking down neurotransmitters such as serotonin, noradrenaline, and dopamine in the brain. Inhibition of this enzyme is expected to increase the levels of all those neurotransmitters. Even though increased dopamine is suspected to be related to lower prolactin levels, probably the serotonin increment prevails and that is why this class of medications is related to hyperprolactinemia; or dopamine and serotonin increment neutralize each other and no difference in prolactin levels is seen.
MAO inhibitors with serotoninergic activity (pargyline and cordyline) can cause modest and generally asymptomatic hyperprolactinemia (2,61). Phenelzine was observed to persistently increase prolactin levels in 4 of11 patients, which returned to normal during a placebo week and rose again in all 4 patients after treatment restart (88).
Atypical Anti-Depressants
Mirtazapine's mechanism of action is different from other anti-depressants, as it does not inhibit the reuptake of serotonin or noradrenaline, but it increase the release of serotonin and noradrenaline (211). In any case, prolactin levels did not show any difference in 8 healthy male subjects pre- and post-mirtazapine 15 mg oral administration (212).
Serotonin Modulators
Trazodone acts through dual inhibition of serotonin reuptake and serotonin type 2 receptors, coupled with antagonism of histamine and α-1-adrenergic receptors (213). In 12 patients with depression, 150 mg of trazodone for 3 weeks caused significantly higher prolactin levels after 12 hours, 1 week, and 2 weeks of treatment compared to baseline. Higher levels were after the first week 15,3+/-8,5 ng/mL (325 +/- 180 mUI/L) compared to baseline 9,1 +/- 5,6 ng/mL (194+/-119 mUI/L) (99).
Selective Noradrenaline Reuptake Inhibitors
Reboxetine is a selective noradrenaline reuptake inhibitor that was shown to increase prolactin in healthy men after acute administration, but this effect can be reversed if reboxetine is simultaneously administered with α2-blocker mirtazapine, suggesting a role of α2-receptors in the enhancement of prolactin release after reboxetine (214). In any case, conflicting data are present even with this drug, with one other study showing no difference between pre-treatment and after-treatment prolactin levels in patients taking up to 8 mg reboxetine for 4 weeks in 17 patients (215). This discrepancy can be due to the limited number of patients or other neuroendocrine mechanisms still less explored.
SNRI medications are described to cause only mild and rare elevations on prolactin levels, whereas (94), esketamine (NMDA receptor antagonist)is not described to cause hyperprolactinemia.
Summary
In summary, controversial data are available on anti-depressant-induced hyperprolactinemia. Routine monitoring of prolactin levels in patients taking anti-depressants is not recommended unless symptoms related to prolactin increase (in premenopausal women: menstrual cycle dysfunction leading to amenorrhea, oligomenorrhoea, anovulatory cycles, low libido, and energy; in men: erectile dysfunction, decreased energy and libido, decreased muscle mass, decreased body hair; and for both of them osteopenia, galactorrhea and infertility) occur (50,193). If the prolactin serum level is elevated (> 25 µg/L (531 mUI/L), in these patients a differential diagnosis is needed. The proposed approach is to withdraw the anti-depressant drug slowly over 2 weeks and replace it with another anti-depressant less likely to cause hyperprolactinemia, then reassess symptoms and prolactin levels after 2-4 weeks (193). If the serum prolactin remains elevated, other causes of hyperprolactinemia should be addressed by an endocrinologist. Another approach is to perform a pituitary MRI if the replacement of the anti-depressant is difficult to manage.
GASTRIC ACID REDUCERS
Histamine-Receptor Inhibitors
Histamine, a CNS neurotransmitter, binds to both H1 and H2 receptors. It can stimulate prolactin secretion via H1 receptors by inhibiting the dopaminergic system. On the contrary, histamine can also inhibit prolactin secretion via H2 receptors using a non-dopaminergic mechanism involving β-endorphin, vasoactive intestinal peptide, vasopressin, or TRH (216), all of which act as prolactin-releasing factors (49) (Figure 1).
In the French Pharmacovigilance Study, H2 receptor antagonists were found to contribute to 5% of drug-induced hyperprolactinemia, primarily with ranitidine (odds ratio = 4.43; 95% CI: 1.82–10.8) (11). Other studies have shown that H2-receptor antagonists such as cimetidine and ranitidine can elevate prolactin levels (217,218). Specifically, cimetidine caused a three-fold increase in prolactin levels after a 400 mg IV infusion, although this effect was not observed with oral administration of 800 mg cimetidine in healthy individuals (102) (Table 1,3).
Interestingly, it has been observed that systemic administration of the H2 agonist impromidine does not prevent cimetidine-induced hyperprolactinemia. In contrast, pre-administration of benzodiazepines or GABA lowered the prolactin response. This suggests that cimetidine-induced hyperprolactinemia may be mediated through neurotransmitters in the GABA-ergic system (219).
Proton-Pump Inhibitors
Proton-pump inhibitors (PPIs) have been recently reviewed regarding the risk of hyperprolactinemia and related sexual disorders observed with long-term use (220). The exact mechanism by which PPIs increase prolactin levels is not fully understood; possible explanations include inhibition of dopamine receptors, interference with other dopamine receptors, involvement of the serotoninergic pathway, modulation of the opioid pathway, and a potential role in decreasing prolactin clearance (220) (Figure 2). In addition, PPIs can increase gastrin levels in chronic use especially in females using high doses (221). As gastrin can act as prolactin-inhibitory factor (108), this antagonistic effect may explain the relatively mild hyperprolactinemia occurring with this class of drugs. Esomeprazole has a mild inhibitory effect on CYP3A4, which leads to decreased metabolism of estrogen, thereby increasing serum estrogen levels which can stimulate the production of prolactin (222). Gynecomastia, impotence, irregular menses, and galactorrhea have been described with PPI use. Hyperprolactinemia occurred less often than sexual disorders, and most cases of hyperprolactinemia were reported with omeprazole, esomeprazole, and lansoprazole use (e.g. 4-fold increment with lansoprazole) (104) (Table 1,3). Pantoprazole and rabeprazole were only sporadically associated with hyperprolactinemia. The authors assert that the occurrence of sexual dysfunction in individuals using PPIs, despite having normal prolactin levels, may be attributed to the development of low vitamin B12 levels, hypomagnesaemia and iron deficiency resulting from PPI usage. These nutritional deficiencies have been implicated in the manifestation of sexual disorders in other studies (220,223,224).
OPIOIDS
Endogenous opioids, morphine, and related drugs, activate ε-, μ-, κ- and δ- opioid receptors in the hypothalamus, modulating pituitary hormone secretion. They do not possess direct effects on pituitary cells (225,226). They inhibit the gonadal axis via ε receptors (GnRH suppression) and stimulate prolactin production by reducing the activity of tuberoinfundibular dopaminergic neurons via μ, κ and δ opioid receptors – mostly μ receptors as the μ receptor opioid antagonist naloxone prior to morphine and methadone use prevents opioid-induced hyperprolactinemia (227). Indirect stimulation of prolactin release can be mediated by stimulating prolactin-releasing factors production (the serotoninergic pathways discussed above) (228) (Figure 2). Hyperprolactinemia can then lead to additional gonadal axis suppression. In addition, opioids can modulate the corticotroph axis via κ and δ opioid receptors and the somatotroph axis via μ, κ and δ opioid receptors (226). Additionally, opioids manifest negative effects on bone health through direct inhibition of osteoblasts by opioids, gonadal axis suppression, altered mental status, and other comorbidities (chronic conditions, smoking, alcohol use) (229).
Acute intravenous or intra-ventricular administration of endogenous opioids leads to a rapid plasma prolactin increase in a dose-dependent manner (44,225). Chronic use of opioids effects on prolactin can vary: oral opioids for chronic pain increase prolactin, but morphine administered intrathecally for chronic non-cancer pain had no effect on prolactin (226). Hyperprolactinemia induced by opioids can be symptomatic: painful gynecomastia, galactorrhea, and hypogonadism have been reported in chronic opioid users. These can be alleviated with discontinuation or reduction of opioid dose, and sometimes dopamine agonists such as bromocriptine (226). Methadone induces a transient increase in prolactin levels, whereas chronic methadone users have normal basal prolactin levels (110), (Table 1, 3).
On the contrary, in a study involving six patients with hyperprolactinemia and amenorrhea, the use of naltrexone, an opioid antagonist, was investigated to determine if blocking endogenous opioids could improve the sexual axis. On the first day of naltrexone administration, significant increases were observed in the mean concentration of luteinizing hormone (LH), LH pulse amplitude, and estradiol levels compared to the control day. This indicated a prompt partial reactivation of the hypothalamic-pituitary-gonadal axis as a result of naltrexone, leading to heightened gonadotrophin levels and subsequent release of estradiol. However, it was found that the effect of opioid antagonism did not result in a sustained increase in estradiol secretion with chronic treatment. Additionally, prolactin levels continued to increase over time (mean prolactin level 255 ± 121 microgram/L), despite the initial improvement in the gonadal axis. This study demonstrated that although prolactin-induced suppression of the gonadal axis can be reversed to some extent by acute opioid antagonism, it is not an effective treatment for revitalizing the gonadal axis in the long term. Possible explanations for this lack of sustained effect include desensitization of the hypothalamic-pituitary unit for the effects of opioid receptor blockade and other disruptors of the axis, which may counteract the positive effects of opioid antagonism (230,231).
For endocrinopathies caused by opioids, including hyperprolactinemia, potential management choices include reducing or discontinuing opioid usage whenever feasible and exploring alternative pain relief therapies for chronic pain situations. Hormonal replacement therapy can be considered for hypogonadism and hypoadrenalism (226).
ANTIHYPERTENSIVES
Some antihypertensive medications including α-methyldopa, verapamil, labetalol, and reserpine, have been associated with hyperprolactinemia. This phenomenon is attributed mainly to the potential inhibition of dopaminergic pathways, highlighting the complex interplay between anti-hypertensive therapy and endocrine function. Other mechanisms are drug-specific and will be explained below.
α -methyldopa is an α-adrenergic inhibitor that leads to the suppression of monoamine synthesis, including noradrenaline, dopamine, and serotonin, which likely contributes to its anti-hypertensive effect. It causes hyperprolactinemia through the inhibition of dopamine synthesis by competitive inhibition of DOPA decarboxylase which transforms L-dopa into dopamine (61) (Figure 2). Long-term treatment resulted in elevated basal prolactin levels (3-4-fold), while a single dose of 750-1000mg reaches a peak of high prolactin level after 4-6 hours of administration (232). Gynecomastia is the most common endocrine side effect.
Calcium channel blockers are other drugs studied for their potential to cause hyperprolactinemia. The dihydropyridine class was found to have no effects on prolactin levels. Whereas, from the non-dihydropyridine class, which mainly blocks L-type of calcium channel receptors in the heart, only verapamil was found to cause 2-fold persistent hyperprolactinemia (and galactorrhea),while drug discontinuation reversed hyperprolactinemia in all patients (113). A clinical trial suggested that verapamil acts by reducing dopamine release in the tuberoinfundibular pathway through calcium influx inhibition (Figure 2), possibly by N-calcium channels which are known to be involved in the regulation of dopamine release and other neurotransmitters (233).
Labetalol is an α- and β-adrenoceptor blocker anti-hypertensive that has been reported to increase prolactin levels when administered intravenously, but not when administered orally (100 or 200 mg) as labetalol cross the blood-brain-barrier only in negligible amounts. The increase in prolactin release caused by intravenous labetalol is not readily explained by its interference with adrenergic receptors. The exact mechanism underlying this effect of the drug is currently unclear, but it is possible that labetalol's ability to block dopamine activity (anti-dopaminergic activity) might be involved in this response (234) (Figure 2). Pre-treatment with levodopa and carbidopa can prevent prolactin response after labetalol (235), suggesting dopamine pathway involvement suppression inside the blood-brain-barrier.
Reserpine is a rauwolfia alkaloid previously used for the treatment of hypertension as well as psychosis, schizophrenia, and tardive dyskinesia; it reduces dopamine by inhibiting their hypothalamic storage in secretory granules (236), and by blockade of vesicular monoamine transporter type 2 in monoamine neurons (237), leading to hyperprolactinemia. Prolactin levels are higher during treatment with reserpine than 6 weeks after discontinuation of the drug. Increased incidence of gynecomastia and breast cancer has also been reported among patients on anti-hypertensive therapy with reserpine (236). The ability of anti-hypertensives to cause hyperprolactinemia is summarized in Tables 1 and 3.
ESTROGENS
Estrogens stimulate prolactin secretion by several mechanisms: They bind to specific intracellular lactotroph cells receptors, though enhancing prolactin gene transcription and synthesis (238). They also inhibit tuberoinfundibular dopamine synthesis, stimulate lactotroph cell hyperplasia, downregulate dopamine receptor expression, and modify lactotroph responsiveness to other regulators (23,239) (Figure 2). Estrogen-induced hyperprolactinemia is dependent on the degree of estrogenization. Higher levels of estrogens in pregnancy and during ovulation increase prolactin levels with the last, contributing to a higher normal range of prolactin in pre-menopausal women.
Studies documenting the incidence of hyperprolactinemia showed that women on oral contraceptives were reported to have higher prolactin levels by 12% to 30% (240,241). Some, but not all, studies suggest that there is a dose-dependent effect (4). No increase in basal prolactin levels is reported during therapy with modern contraceptives with lower amounts of estrogen (242)or estrogen plus cyproterone acetate alone (243).
In transgender patients, estradiol or ethinyl estradiol treatment, the prolactin level rise was dependent on the dose of estrogen, duration of exposure, and alteration of SHBG levels. Estradiol infusion at levels above 10,000 pg/mL for as short as 6-7 hours significantly elevated prolactin levels by 3- to 4-fold, whereas ethinyl estradiol 2 mg/day for 1 month did not consistently elevate prolactin in all patients, which can be due to its ability to increase SHBG binding and maintaining free portion in the normal range (114) (Table 1, 3).
For women on post-menopausal hormone replacement therapies over 2.5 years, serum prolactin measured were within the normal range (244). In another study on 75 women, who were randomly assigned to three groups: control (receiving placebo), transdermal hormonal replacement (biphasic 17β-estradiol and progesterone, natural hormones), and oral ethinyl-estradiol and desogestrel, prolactin levels significantly increased in the oral group, but not in the transdermal group. There was a significant difference in hormone levels: in the oral group, estradiol levels increased five times and estrone levels eleven times. In the transdermal group, estrone and estradiol levels were increased three times (245).
GONADOTROPHINS AND GNRH AGONISTS
In addition to the known prolactin function in lactation, several studies have suggested other benefits of prolactin in oocyte development, formation of corpus luteum and its survival, steroidogenesis and implantation (246). In natural cycles there is a transient increase in late follicular phase of prolactin, but this increment is higher in stimulated cycles (246). In a cohort study were included 79 patients; 60 individuals underwent in vitro fertilization, 14 received clomiphene citrate treatment, and five patients with premature ovarian failure were administered estradiol. During the course of human menopausal gonadotrophin (hMG) treatment, a notable increase in both serum estradiol and prolactin concentrations were observed from early to late follicular days (P < 0.01). Specifically, prolactin levels increased from an initial mean value of 367±38 mIU/L (17.25±1.8 ng/mL) to 991±84 mIU/L (46.6±4 ng/mL) (Table 1). Bromocriptine effectively mitigated the increase in prolactin levels but was associated with a significant elevation in estradiol levels (P < 0.05) because prolactin itself works as a controller of estradiol increment. Clomiphene treatment led to a significant increase in serum estradiol levels (P < 0.01) but a significant decrease in serum prolactin concentrations during the late follicular phase (P < 0.01), indicating disruption of the estradiol-prolactin feedback mechanism. Among patients with premature ovarian insufficiency, serum prolactin concentrations increased concomitantly with rising serum of estradiol concentrations (after estradiol administration). Additionally, it was observed that the presence of prolactin significantly reduced estradiol production by granulosa cells (P < 0.05) (116).
An increment of prolactin levels is found even after hCG administration with a maximum prolactin level of 93.2 ng/mL; 1983 mIU/L (115).Notably, knowing that prolactin is a stress hormone, during assisted procedures it is increased, but this is a transitory increment without consequences in fertility outcome (247).
Not only gonadotrophins but also GnRH agonists are widely used during invitro fertilization to maintain a controlled and synchronized ovarian stimulation. Use of leuprolide acetate (GnRH agonist) concomitantly with hMG, resulted in higher prolactin and estradiol levels in comparison with patients receiving only hMG (prolactin 24.2 vs 16.8 ng/mL; 515 vs 358 mIU/L) (117). In another randomized study, along protocol with 0.1 mg subcutaneous triptorelin starting from day 10 of the preceding stimulation cycle and short protocol, where 0.1 mg subcutaneous triptorelin is given in the stimulating cycle, were compared. Prolactin levels were measured at 9 am in the first day of hCG administration. The long protocol correlated with higher prolactin levels (31.3 ± 16.9 vs 23.7 ± 11 ng/mL; 666 ± 359 vs 504 ± 234 mIU/L) (248).
In children, GnRH agonists are used in precocious puberty (CPP)as well as growth hormone deficiency (GHD)who do not properly respond to exogenous growth hormone treatment. In a study involving 119 children with CPP and 93 with GHD, treated with triptorelin or leuprolide, prolactin levels were measured before and every six months for 6 years for CPP group and for 2 years for GHD group. Moreover, prolactin levels were checked after 6 and 12 months of treatment withdrawal. In this study was concluded that even though prolactin levels were higher in triptorelin treated patients (only 3.8% developed hyperprolactinemia in triptorelin group which was solved after withdrawal – baseline 12.5 ± 3.7 ng/mL (266 ± 79 mIU/L) to max 45.6 ± 4.5 ng/mL; 970 ± 96 mIL/L), no significant difference was found in prolactin in basal condition and during GNRH agonist treatment in CPP and GHD (190) (Table 2).
OTHER DRUGS
A lot of other drugs have been reported to cause mild (less than 2-fold increment) increases in prolactin levels. A synthesized visualization of these mechanisms is shown in Figure 2.
The acute administration of buspirone, an anxiolytic medication, was investigated in a study involving 8 healthy volunteers. The findings revealed an increase in plasma prolactin levels across all participants compared to the baseline levels observed in 8 control subjects. During the study, blood samples were collected at 30-minute intervals over a duration of 2 hours. The zenith of prolactin levels was observed between minutes 90 and 120 for all individuals, with the maximum elevation reaching 37 ng/mL (787 mIU/L) (249). It is noteworthy that the augmentation of prolactin is believed to exhibit a dose-dependent relationship. Furthermore, it was observed that chronic usage of buspirone did not lead to significant alterations in prolactin levels, indicating a potential adaptation to the acute changes induced by the medication. The underlying mechanism responsible for this phenomenon is posited to involve both serotoninergic and dopaminergic implications (119).
Carbamazepine, a widely used anticonvulsant, was examined in a cohort comprising 4 patients with complex partial seizures undergoing chronic carbamazepine treatment (200 mg administered three times daily). Blood samples were collected at intervals of 2 hours. Additionally, a group of 5 patients with untreated epileptic seizures participated, wherein a thyrotrophin-releasing hormone (TRH) stimulation test was performed both prior to and 35-50 days post the administration of 200 mg carbamazepine three times daily. Blood samples were obtained 10, 30, and 60 minutes following intravenous injection of 200µg TRH. Furthermore, 4 normal volunteer subjects were included in the study. On the first day, a placebo was administered, followed by the administration of 400 mg carbamazepine at 8 AM on the second day. Blood samples were collected at baseline on both days and subsequently at hourly intervals until 4 PM. After a span of two weeks, a nocturnal study was conducted, spanning from 6 PM to 6 AM. The investigation revealed that there were no discernible alterations in spontaneous prolactin release or TRH-stimulated prolactin levels. However, a slight increase in sleep-entrained prolactin values was observed, while retaining the secretory circadian rhythm. Given that the release of prolactin during sleep is largely attributed to serotoninergic activity, it is plausible that the modest increment (less than 2-fold) may implicate serotoninergic modulation (working as a serotonin-releasing factor and reuptake inhibitor) facilitated by carbamazepine (120,121).
Sympathomimetic amines fenfluramine and sibutramine, formerly used for appetite suppression due to their stimulatory effect on the synaptic concentration of serotonin, have been shown to induce hyperprolactinemia as a result of increased serotoninergic activity and postsynaptic stimulation of 5HT2Areceptors. In a case report, after starting sibutramine, a 38-year-old female patient developed hyperprolactinemia (prolactin levels 46 and 89.6 ng/mL (978 and 1906 mUI/L) with amenorrhea and galactorrhea. Discontinuation of sibutramine, confirmed by a sella MRI, led to rapid normalization of prolactin levels within 15 days, and symptoms resolved during a 90-day follow-up (2,127).
Cholinomimetic drugs have been reported controversially in the literature regarding their ability to cause hyperprolactinemia. However, in collaborative studies from the National Institute of Mental Health and the University of California, San Diego, three separate experiments were conducted involving volunteers of different genders and ages. In the first experiment, nine volunteers received physostigmine salicylate at 33 µg/kg, while in the second experiment, eleven male volunteers were given 22 µg/kg of physostigmine salicylate. The third experiment involved six volunteers receiving 3 mg of arecoline hydrobromide. Placebo saline was administered in all experiments as well. It was shown that intravenous injection of physostigmine or arecoline can elevate prolactin correlating with raised β-endorphin levels in the blood. Prolactin elevation was less than 100 ng/mL (2127 mUI/L). Cholinergic activation in the hypothalamus, particularly focusing on β-endorphin, might help in explaining how peptides modify primary neurochemical effects on hormone regulation in the hypothalamus and pituitary (44).
Bucillamine, an analogue of D-penicillamine used as an antirheumatic drug in Japan, has been reported to induce hyperprolactinemia (109 ng/mL (2319 mUI/L)) after 30 months of treatment start, associated with gynecomastia and galactorrhea in one case report. The mechanism remains unclear (133).
‘Ecstasy’ (MDMA) was shown to increase prolactin secretion in rhesus monkeys by stimulating serotonin release and by direct-acting as a 5HT2A agonist (250); In nine studies, five of them observed an increase in prolactin levels due to the intervention. However, in the remaining studies, there was no significant change in prolactin levels, and these unresponsive results tended to occur when a lower dose of the intervention was used on average. This suggests a potential relationship between the dosage of the intervention and its effect on prolactin levels (135).
Smoking, particularly the consumption of high-nicotine cigarettes, has been associated with a significant acute elevation in prolactin levels, ranging from 50% to 78% above the baseline, within 6 minutes after smoking. These elevated levels persist for approximately 42 minutes and return to baseline within 120 minutes of initiating smoking (136). The underlying mechanism probably involves the stimulation of rapid prolactin release through the augmentation of endogenous opioids, which subsequently inhibits dopamine release (251). However, prolonged nicotine exposure leads to desensitization of dopamine receptors, and lowers dopamine turnover (48) probably contributing to hyperprolactinemia. It has been hypothesized that the increased incidence of osteopenia and osteoporosis could be at least partly related to this effect (252).
Recently, an association has been reported between HIV-1 protease inhibitors and the adverse effect of galactorrhea and hyperprolactinemia in four HIV-1 infected women treated with indinavir, nelfinavir, ritonavir, or saquinavir. The cause of this unexpected toxicity could be attributed to several possible mechanisms: 1) Protease inhibitors may enhance the stimulatory effects of prolactin due to their inhibition of the cytochrome P450 system, leading to longer half-life of prolactin; 2) opportunistic infections in AIDS patients may induce cytokine-driven prolactin production by pituitary or immune cells; 3) protease inhibitors might exert direct endocrine effects on the pituitary or hypothalamus (253,254). To explore mechanisms of hyperprolactinemia induced by protease inhibitors, experiments were conducted using rat pituitary cells and hypothalamic neuronal endings. The results showed that both ritonavir and saquinavir could directly stimulate prolactin secretion, while not affecting dopamine release. This suggests that these protease inhibitors might interact with specific mammalian proteins in the anterior pituitary involved in prolactin secretion, leading to the observed galactorrhea and hyperprolactinemic effect (137).
Regarding chemotherapy and immunosuppression, there are some controversial data on the effect of chemotherapy and immunosuppression on prolactin levels, as significant prolactin increases are not frequent and usually mild (2). Prolactin and growth hormone have been involved as part of a cytokine system in the recovery of the immune response after chemotherapy and bone marrow transplantation (255). In a study of 20 breast cancer patients undergoing high-dose chemotherapy and autologous stem-cell transplantation, plasma prolactin levels increased within and 30 days after transplant, yet still remaining within the normal range. The use of antiemetic drugs further raised prolactin levels. Patients in continuous complete remission after transplantation exhibited higher prolactin levels, while elevated prolactin did not impact disease-free survival, suggesting potential for further research into post-transplant immune response (256).
Radiotherapy for intracranial germ cell tumors was shown to induce hyperprolactinemia with a prevalence of 35.3% (138).
Other drug inducing hyperprolactinemia are described in Table 1.
DRUGS REPORTED TO DECREASE PROLACTIN LEVELS OR HAVE AN EQUIVOCAL EFFECT
Several medications, beyond the established treatments like cabergoline, bromocriptine, carbidopa and levodopa, have reported effects on reducing prolactin levels. For instance, pseudoephedrine, an α-adrenergic stimulant primarily affecting α1 receptors, shares structural similarities with amphetamine and moderately stimulating dopamine release in the brain by acting on D2 receptors in the pituitary, consequently lowering prolactin levels. Studies have indicated pseudoephedrine's potential to decrease milk production, at least partly attributed to its effect on prolactin levels through dopaminergic actions in the pituitary (132). Moreover, indirect evidence suggests that α-1 receptors stimulation leads to decreased prolactin levels (41).
Amphetamine was seen to produce a poor prolactin suppressant effect in either normal- or hyperprolactinemic subjects. The proposed mechanism of prolactin lowering potential is due to their ability to stimulate the release of dopamine (257). However, during the withdrawal period of cocaine use, hyperprolactinemia has been observed, probably due to a decrease in dopamine levels, leading to dysregulation in the dopamine system and increased prolactin. Moreover, during withdrawal, prolactin can be secreted as a stress hormone (129).
Guanafascine, an α2 adrenergic agonist, used to treat ADHD, has been shown to decrease prolactin levels. In a longitudinal study spanning three years involving 15 patients diagnosed with hyperprolactinemia, the noteworthy suppressive impact of guanfacine on prolactin levels suggests potential involvement of hypothalamic or extrahypothalamic adrenergic pathways in the intricate regulation of prolactin secretion (131). Even though α2 stimulation has been shown to increase prolactin levels in rats, this is not fully understood in humans making the explanation in this case confusing (42).
The impact of benzodiazepines (BDZ) on prolactin secretion is a subject of debate. Research findings have yielded conflicting results. Some studies conducted on both non-epileptic patients and healthy volunteers have not detected significant modifications in prolactin levels following BDZ treatment (258). A study on 30 adolescent patients with schizophrenia with gradually increasing doses of diazepam to a maximum of 100-400 mg/day, with 4 weeks of treatment, showed that only doses higher than 250 mg/day give a significant but mild increase in prolactin levels. Proposed mechanisms are inhibition of TIDA neurons by activation of the GABA system, or activation of the endorphin-ergic system leading to hyperprolactinemia (118). On the contrary, diazepam was found to suppress the secretion of prolactin in vitro through one of two mechanisms: it either strengthens the direct inhibitory action of GABA on prolactin release, or it hinders a benzodiazepine-sensitive Ca2+-calmodulin dependent protein kinase at micromolar concentrations leading to a reduction of prolactin secretion (259).
Moreover, phenytoin, an anticonvulsant impeding sodium channels in nerve cells, have generated conflicting data regarding their impact on prolactin levels. In animal studies, phenytoin showcased a rapid decline in both prolactin release and mRNA concentrations, functioning as a partial T3 agonist by binding to T3 nuclear receptors (260). However, clinical observations revealed elevated resting levels of prolactin in phenytoin-treated patients compared to untreated counterparts. Remarkably, responses to metoclopramide and bromocriptine remained unaltered, indicating a limited effect of phenytoin on the D2 receptors present on lactotrophs (261). Even the conclusions drawn from these findings remain contentious. Evidence suggests that phenytoin treatment may enhance the growth hormone response to levodopa, implying a phenytoin-induced dopaminergic activity at the hypothalamic-pituitary level (122). More specifically, it is postulated that phenytoin might enhance dopamine receptor sensitivity by inhibiting the Ca2+ calmodulin complex. This effect could contribute to reduced prolactin secretion (122). On the contrary, other studies have not demonstrated any notable alterations in prolactin levels due to phenytoin administration (262). The discordant outcomes surrounding phenytoin's impact on prolactin levels underscore the complexity of its effects and necessitate further investigation for conclusive insights into its mechanisms of action.
In another comprehensive study involving 126 subjects, both males and females, with generalized or partial epilepsy receiving phenobarbital as monotherapy or in combination with phenytoin or benzodiazepines, a distinct pattern emerged. Specifically, the administration of phenobarbital, either alone or in combination, resulted in elevated prolactin levels, but this elevation was found to be statistically significant only in the male participants. Notably, knowing that an epileptic attack itself can cause hyperprolactinemia, those data remain confusing. The proposed mechanism in this study is phenobarbital interaction with GABA receptors, leading to increased prolactin levels (263). However other studies do not show any change in prolactin levels (123).
Valproic acid, an anticonvulsant working as a central stimulant of GABAergic neurons, has demonstrated the ability to reduce prolactin basal levels as well as TRH-stimulated prolactin levels. This is indirect proof of the synergically acting of GABA neurons with dopaminergic tracts (124). However, in another study, no effect of valproic acid on prolactin levels was noticed during the night (264).
Lithium carbonate, a pharmaceutical agent employed as a mood stabilizer, has undergone investigation in different studies, as it decreases dopamine release and glutamate, and increases inhibitory GABA (265). One of them encompassed a longitudinal examination involving 9 patients diagnosed with bipolar disorder. The focus of this study was the assessment of plasma prolactin levels before and 12 hours after the evening administration of lithium. Evaluations were conducted on days 1, 6, 8, 13, 30, 60, and 90. Notably, this investigation yielded no discernible correlation between lithium concentration and prolactin levels, and no statistically significant alterations in prolactin levels were observed. The second part of this study adopted a cross-sectional design, involving 26 patients with an established history of long-term lithium treatment spanning durations of 3 months to 20 years. A comparative analysis revealed that prolactin levels, measured at 9 AM following a one-hour period of rest, did not demonstrate elevation in comparison to 16 controls. It is noteworthy that in both studies, lithium concentrations ranged from 0.4 to 1.4 mmol/L (normal range 0.5-1.2 mmol/L) (266). Additionally, the administration of lithium did not exert an impact on the plasma prolactin response to thyrotrophin-releasing hormone (TRH) stimulation compared to pre-treatment levels (267). The combined findings from these investigations provide compelling evidence that lithium does not contribute to hyperprolactinemia, thereby distinguishing it from medications with such an effect.
Cocaine has been shown to decrease prolactin levels beginning at 30-min following cocaine administration reaching statistical significance at the 90- and 120-minute time points (134).
Those medications are mentioned in Table 1. Their mechanism of altering prolactin levels is summarized in Figure 2.
HERBAL MEDICINES AFFECTING PROLACTIN LEVELS
In Table 4 is list of herbal medicines has been used traditionally to stimulate lactation (268). However, firm scientific evidence that they actually induce hyperprolactinemia is scarce.
Table 4.
Lactogenic Herbs (268)
View in own window
| Family name | Species name | Common name |
---
| Amaryllidaceae | Allium sativum | Garlic |
| Annonaceae | Xylopia aethiopica | African Pepper or Ethiopian Pepper |
| Asclepiadaceae | Secamoneafzelii
| Costaceae | Costusafer | African Ginger |
| Euphorbiaceae | Euphorbia hirta | Asthma Plant or Tawa-Tawa |
| Euphorbia thymifolia | Petty Spurge |
| Hymenocardiaacida | African Almond or Honeytree |
| Plagiostylesafricana
| Ricinus communis | Castor Bean Plant |
| Leguminosae | Tamarindus indica | Tamarind |
| Acacia nicolita
| Desmodiumadscendens
| Malvaceae | Hibiscus sabdariffa | Roselle or Red Sorrel |
| Gossypium herbaceum | Cotton Plant |
| Moraceae | Milicia excelsa | African Teak or Iroko |
| Ficus species | Ficus or Fig trees |
| Musaceae | Musa paradisiaca | Plantain |
| Ranunculaceae | Nigella sativa | Black Cumin or Black Seed |
| Actaea (Cimiciguga) racemose | Black Cohosh |
| Solanaceae | Solanum torvum | Turkey Berry or Devil's Fig |
| Verbanaceae | Lippia multiflora | Bush Tea or False Green Tea |
| Zingiberaceae | Aframomummelegueta | Grains of Paradise or Alligator Pepper |
| Fabaceae | Trifolium pratense | Red Clover |
| Trigonella foenum-graecum | Fenugreek |
| Apiaceae | Foeniculum vulgare | Fennel |
Some herbs are known to decrease prolactin levels. For example, chaste tree (Vitex agnus-castus) decrease prolactin levels by activating to D2-receptors and suppressing prolactin release, as shown in in vitro experiments on lactotroph cell cultures and in in vivo animal experiments (269).
Another herb, Mucuna pruriens, which is a natural source of l-dihydroxyphenylalanine (a dopamine precursor) is found to decrease prolactin levels in humans (270). Vitamin B6 (pyridoxine), by acting as a coenzyme in dopamine synthesis and aspartame, a sweetener metabolized in phenylalanine (dopamine precursor), have been shown to interfere with milk production by reducing prolactin levels (271). Ashgawanda (Withania somnifera) is found to decrease prolactin levels up to 12% (272).
Moreover, oral zinc is found to decrease prolactin levels below the normal range in all 17 subjects with normal prolactin levels, in scenario of increased zinc levels in the blood (273).
While none of the mentioned herbs are currently established within clinical guidelines for specifically lactogenic or prolactin-reducing purposes, ongoing research and anecdotal evidence suggest potential roles for these botanicals as adjunctive therapies.
CONCLUSION
In summary, this review underscores the significant role of drug-induced hyperprolactinemia in causing higher prolactin levels and provides detailed insights into how pharmaceutical agents contribute to this effect. However, understanding the complex mechanisms behind drug-induced hyperprolactinemia is still a work in progress. More research is needed to delve deeper into these mechanisms and gain better insights. These efforts will contribute to refining treatment strategies and improving patient care.
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Kolnikaj TS, Musat M, Salehidoost R, et al. Pharmacological Causes of Hyperprolactinemia. [Updated 2024 Jan 4]. In: Feingold KR, Ahmed SF, Anawalt B, et al., editors. Endotext [Internet]. South Dartmouth (MA): MDText.com, Inc.; 2000-.
In this Page
ABSTRACT
INTRODUCTION
EPIDEMIOLOGY
PROLACTIN CONTROL MECHANISMS
CLINICAL CHARACTERISTICS
PSYCHOTROPIC MEDICATIONS
NEUROLEPTIC-LIKE MEDICATIONS
ANTI-DEPRESSANTS
GASTRIC ACID REDUCERS
OPIOIDS
ANTIHYPERTENSIVES
ESTROGENS
GONADOTROPHINS AND GNRH AGONISTS
OTHER DRUGS
DRUGS REPORTED TO DECREASE PROLACTIN LEVELS OR HAVE AN EQUIVOCAL EFFECT
HERBAL MEDICINES AFFECTING PROLACTIN LEVELS
CONCLUSION
REFERENCES
LINKS TO WWW.ENDOTEXT.ORG
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Contents
1 Definition
2 Properties
3 Factorization of decimal repunits
4 Repunit primes
4.1 Decimal repunit primes
4.2 Algebra factorization of generalized repunit numbers
4.3 The generalized repunit conjecture
5 History
6 Demlo numbers
7 See also
8 Footnotes
8.1 Notes
8.2 References
9 References
10 External links
Repunit
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From Wikipedia, the free encyclopedia
Numbers that contain only the digit 1
Repunit prime
| No. of known terms | 11 |
| Conjectured no. of terms | Infinite |
| First terms | 11, 1111111111111111111, 11111111111111111111111 |
| Largest known term | (108177207−1)/9 |
| OEIS index | A004022 Primes of the form (10^n − 1)/9 |
In recreational mathematics, a repunit is a number like 11, 111, or 1111 that contains only the digit 1 — a more specific type of repdigit. The term stands for "repeated unit" and was coined in 1966 by Albert H. Beiler in his book Recreations in the Theory of Numbers.[note 1]
A repunit prime is a repunit that is also a prime number. Primes that are repunits in base-2 are Mersenne primes. As of October 2024, the largest known prime number 2136,279,841 − 1, the largest probable prime R8177207 and the largest elliptic curve primality-proven prime R86453 are all repunits in various bases.
Definition
The base-b repunits are defined as (this b can be either positive or negative)
Thus, the number Rn(b) consists of n copies of the digit 1 in base-b representation. The first two repunits base-b for n = 1 and n = 2 are
In particular, the decimal (base-10) repunits that are often referred to as simply repunits are defined as
Thus, the number Rn = Rn(10) consists of n copies of the digit 1 in base 10 representation. The sequence of repunits base-10 starts with
: 1, 11, 111, 1111, 11111, 111111, ... (sequence A002275 in the OEIS).
Similarly, the repunits base-2 are defined as
Thus, the number Rn(2) consists of n copies of the digit 1 in base-2 representation. In fact, the base-2 repunits are the well-known Mersenne numbers Mn = 2n − 1, they start with
: 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, ... (sequence A000225 in the OEIS).
Properties
Any repunit in any base having a composite number of digits is necessarily composite. For example,
: R35(b) = 11111111111111111111111111111111111 = 11111 × 1000010000100001000010000100001 = 1111111 × 10000001000000100000010000001,
: since 35 = 7 × 5 = 5 × 7. This repunit factorization does not depend on the base-b in which the repunit is expressed.
: Only repunits (in any base) having a prime number of digits can be prime. This is a necessary but not sufficient condition. For example,
: R11(2) = 211 − 1 = 2047 = 23 × 89.
If p is an odd prime, then every prime q that divides Rp(b) must be either 1 plus a multiple of 2p, or a factor of b − 1. For example, a prime factor of R29 is 62003 = 1 + 2·29·1069. The reason is that the prime p is the smallest exponent greater than 1 such that q divides bp − 1, because p is prime. Therefore, unless q divides b − 1, p divides the Carmichael function of q, which is even and equal to q − 1.
Any positive multiple of the repunit Rn(b) contains at least n nonzero digits in base-b.
Any number x is a two-digit repunit in base x − 1.
The only known numbers that are repunits with at least 3 digits in more than one base simultaneously are 31 (111 in base-5, 11111 in base-2) and 8191 (111 in base-90, 1111111111111 in base-2). The Goormaghtigh conjecture says there are only these two cases.
Using the pigeon-hole principle it can be easily shown that for relatively prime natural numbers n and b, there exists a repunit in base-b that is a multiple of n. To see this consider repunits R1(b),...,Rn(b). Because there are n repunits but only n−1 non-zero residues modulo n there exist two repunits Ri(b) and Rj(b) with 1 ≤ i < j ≤ n such that Ri(b) and Rj(b) have the same residue modulo n. It follows that Rj(b) − Ri(b) has residue 0 modulo n, i.e. is divisible by n. Since Rj(b) − Ri(b) consists of j − i ones followed by i zeroes, Rj(b) − Ri(b) = Rj−i(b) × bi. Now n divides the left-hand side of this equation, so it also divides the right-hand side, but since n and b are relatively prime, n must divide Rj−i(b).
The Feit–Thompson conjecture is that Rq(p) never divides Rp(q) for two distinct primes p and q.
Using the Euclidean Algorithm for repunits definition: R1(b) = 1; Rn(b) = Rn−1(b) × b + 1, any consecutive repunits Rn−1(b) and Rn(b) are relatively prime in any base-b for any n.
If m and n have a common divisor d, Rm(b) and Rn(b) have the common divisor Rd(b) in any base-b for any m and n. That is, the repunits of a fixed base form a strong divisibility sequence. As a consequence, If m and n are relatively prime, Rm(b) and Rn(b) are relatively prime. The Euclidean Algorithm is based on gcd(m, n) = gcd(m − n, n) for m > n. Similarly, using Rm(b) − Rn(b) × bm−n = Rm−n(b), it can be easily shown that gcd(Rm(b), Rn(b)) = gcd(Rm−n(b), Rn(b)) for m > n. Therefore, if gcd(m, n) = d, then gcd(Rm(b), Rn(b)) = Rd(b).
Factorization of decimal repunits
(Prime factors (or prime powers) parenthesized and colored (red) are "new factors", i. e. the prime factor (or power) divides Rn but does not divide Rk for all k < n) (sequence A102380 in the OEIS)
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | --- | | R1 = | 1 | | R2 = | (11) | | R3 = | (3) · (37) | | R4 = | 11 · (101) | | R5 = | (41) · (271) | | R6 = | 3 · (7) · 11 · (13) · 37 | | R7 = | (239) · (4649) | | R8 = | 11 · (73) · 101 · (137) | | R9 = | 3(2) · 37 · (333667) | | R10 = | 11 · 41 · 271 · (9091) | | | | | --- | | R11 = | (21649) · (513239) | | R12 = | 3 · 7 · 11 · 13 · 37 · 101 · (9901) | | R13 = | (53) · (79) · (265371653) | | R14 = | 11 · 239 · 4649 · (909091) | | R15 = | 3 · (31) · 37 · 41 · 271 · (2906161) | | R16 = | 11 · (17) · 73 · 101 · 137 · (5882353) | | R17 = | (2071723) · (5363222357) | | R18 = | 32 · 7 · 11 · 13 · (19) · 37 · (52579) · 333667 | | R19 = | (1111111111111111111) | | R20 = | 11 · 41 · 101 · 271 · (3541) · 9091 · (27961) | | | | | --- | | R21 = | 3 · 37 · (43) · 239 · (1933) · 4649 · (10838689) | | R22 = | 11(2) · (23) · (4093) · (8779) · 21649 · 513239 | | R23 = | (11111111111111111111111) | | R24 = | 3 · 7 · 11 · 13 · 37 · 73 · 101 · 137 · 9901 · (99990001) | | R25 = | 41 · 271 · (21401) · (25601) · (182521213001) | | R26 = | 11 · 53 · 79 · (859) · 265371653 · (1058313049) | | R27 = | 3(3) · 37 · (757) · 333667 · (440334654777631) | | R28 = | 11 · (29) · 101 · 239 · (281) · 4649 · 909091 · (121499449) | | R29 = | (3191) · (16763) · (43037) · (62003) · (77843839397) | | R30 = | 3 · 7 · 11 · 13 · 31 · 37 · 41 · (211) · (241) · 271 · (2161) · 9091 · 2906161 | |
Smallest prime factor of Rn for n > 1 are
: 11, 3, 11, 41, 3, 239, 11, 3, 11, 21649, 3, 53, 11, 3, 11, 2071723, 3, 1111111111111111111, 11, 3, 11, 11111111111111111111111, 3, 41, 11, 3, 11, 3191, 3, 2791, 11, 3, 11, 41, 3, 2028119, 11, 3, 11, 83, 3, 173, 11, 3, 11, 35121409, 3, 239, 11, ... (sequence A067063 in the OEIS)
Repunit primes
For a more comprehensive list, see List of repunit primes.
The definition of repunits was motivated by recreational mathematicians looking for prime factors of such numbers.
It is easy to show that if n is divisible by a, then Rn(b) is divisible by Ra(b):
where is the cyclotomic polynomial and d ranges over the divisors of n. For p prime,
which has the expected form of a repunit when x is substituted with b.
For example, 9 is divisible by 3, and thus R9 is divisible by R3—in fact, 111111111 = 111 · 1001001. The corresponding cyclotomic polynomials and are and , respectively. Thus, for Rn to be prime, n must necessarily be prime, but it is not sufficient for n to be prime. For example, R3 = 111 = 3 · 37 is not prime. Except for this case of R3, p can only divide Rn for prime n if p = 2kn + 1 for some k.
Decimal repunit primes
Rn is prime for n = 2, 19, 23, 317, 1031, 49081, 86453, 109297 ... (sequence A004023 in OEIS). On July 15, 2007, Maksym Voznyy announced R270343 to be probably prime. Serge Batalov and Ryan Propper found R5794777 and R8177207 to be probable primes on April 20 and May 8, 2021, respectively. As of their discovery, each was the largest known probable prime. On March 22, 2022, probable prime R49081 was eventually proven to be a prime. On May 15, 2023, probable prime R86453 was eventually proven to be a prime. On May 26, 2025, probable prime R109297 was eventually proven to be a prime.
It has been conjectured that there are infinitely many repunit primes and they seem to occur roughly as often as the prime number theorem would predict: the exponent of the Nth repunit prime is generally around a fixed multiple of the exponent of the (N−1)th.
The prime repunits are a trivial subset of the permutable primes, i.e., primes that remain prime after any permutation of their digits.
Particular properties are
The remainder of Rn modulo 3 is equal to the remainder of n modulo 3. Using 10a ≡ 1 (mod 3) for any a ≥ 0,n ≡ 0 (mod 3) ⇔ Rn ≡ 0 (mod 3) ⇔ Rn ≡ 0 (mod R3),n ≡ 1 (mod 3) ⇔ Rn ≡ 1 (mod 3) ⇔ Rn ≡ R1 ≡ 1 (mod R3),n ≡ 2 (mod 3) ⇔ Rn ≡ 2 (mod 3) ⇔ Rn ≡ R2 ≡ 11 (mod R3).Therefore, 3 | n ⇔ 3 | Rn ⇔ R3 | Rn.
The remainder of Rn modulo 9 is equal to the remainder of n modulo 9. Using 10a ≡ 1 (mod 9) for any a ≥ 0,n ≡ r (mod 9) ⇔ Rn ≡ r (mod 9) ⇔ Rn ≡ Rr (mod R9),for 0 ≤ r < 9.Therefore, 9 | n ⇔ 9 | Rn ⇔ R9 | Rn.
Algebra factorization of generalized repunit numbers
If b is a perfect power (can be written as mn, with m, n integers, n > 1) differs from 1, then there is at most one repunit in base-b. If n is a prime power (can be written as pr, with p prime, r integer, p, r >0), then all repunit in base-b are not prime aside from Rp and R2. Rp can be either prime or composite, the former examples, b = −216, −128, 4, 8, 16, 27, 36, 100, 128, 256, etc., the latter examples, b = −243, −125, −64, −32, −27, −8, 9, 25, 32, 49, 81, 121, 125, 144, 169, 196, 216, 225, 243, 289, etc., and R2 can be prime (when p differs from 2) only if b is negative, a power of −2, for example, b = −8, −32, −128, −8192, etc., in fact, the R2 can also be composite, for example, b = −512, −2048, −32768, etc. If n is not a prime power, then no base-b repunit prime exists, for example, b = 64, 729 (with n = 6), b = 1024 (with n = 10), and b = −1 or 0 (with n any natural number). Another special situation is b = −4k4, with k positive integer, which has the aurifeuillean factorization, for example, b = −4 (with k = 1, then R2 and R3 are primes), and b = −64, −324, −1024, −2500, −5184, ... (with k = 2, 3, 4, 5, 6, ...), then no base-b repunit prime exists. It is also conjectured that when b is neither a perfect power nor −4k4 with k positive integer, then there are infinity many base-b repunit primes.
The generalized repunit conjecture
A conjecture related to the generalized repunit primes: (the conjecture predicts where is the next generalized Mersenne prime, if the conjecture is true, then there are infinitely many repunit primes for all bases )
For any integer , which satisfies the conditions:
.
is not a perfect power. (since when is a perfect th power, it can be shown that there is at most one value such that is prime, and this value is itself or a root of )
is not in the form . (if so, then the number has aurifeuillean factorization)
has generalized repunit primes of the form
for prime , the prime numbers will be distributed near the best fit line
where limit ,
and there are about
base-b repunit primes less than N.
is the base of natural logarithm.
is Euler–Mascheroni constant.
is the logarithm in base
is the th generalized repunit prime in baseb (with prime p)
is a data fit constant which varies with .
if , if .
is the largest natural number such that is a th power.
We also have the following 3 properties:
The number of prime numbers of the form (with prime ) less than or equal to is about .
The expected number of prime numbers of the form with prime between and is about .
The probability that number of the form is prime (for prime ) is about .
History
Although they were not then known by that name, repunits in base-10 were studied by many mathematicians during the nineteenth century in an effort to work out and predict the cyclic patterns of repeating decimals.
It was found very early on that for any prime p greater than 5, the period of the decimal expansion of 1/p is equal to the length of the smallest repunit number that is divisible by p. Tables of the period of reciprocal of primes up to 60,000 had been published by 1860 and permitted the factorization by such mathematicians as Reuschle of all repunits up to R16 and many larger ones. By 1880, even R17 to R36 had been factored and it is curious that, though Édouard Lucas showed no prime below three million had period nineteen, there was no attempt to test any repunit for primality until early in the twentieth century. The American mathematician Oscar Hoppe proved R19 to be prime in 1916, and Lehmer and Kraitchik independently found R23 to be prime in 1929.
Further advances in the study of repunits did not occur until the 1960s, when computers allowed many new factors of repunits to be found and the gaps in earlier tables of prime periods corrected. R317 was found to be a probable prime circa 1966 and was proved prime eleven years later, when R1031 was shown to be the only further possible prime repunit with fewer than ten thousand digits. It was proven prime in 1986, but searches for further prime repunits in the following decade consistently failed. However, there was a major side-development in the field of generalized repunits, which produced a large number of new primes and probable primes.
Since 1999, four further probably prime repunits have been found, but it is unlikely that any of them will be proven prime in the foreseeable future because of their huge size.
The Cunningham project endeavours to document the integer factorizations of (among other numbers) the repunits to base 2, 3, 5, 6, 7, 10, 11, and 12.
Demlo numbers
D. R. Kaprekar has defined Demlo numbers as concatenation of a left, middle and right part, where the left and right part must be of the same length (up to a possible leading zero to the left) and must add up to a repdigit number, and the middle part may contain any additional number of this repeated digit. They are named after Demlo railway station (now called Dombivili) 30 miles from Bombay on the then G.I.P. Railway, where Kaprekar started investigating them. He calls Wonderful Demlo numbers those of the form 1, 121, 12321, 1234321, ..., 12345678987654321. The fact that these are the squares of the repunits has led some authors to call Demlo numbers the infinite sequence of these, 1, 121, 12321, ..., 12345678987654321, 1234567900987654321, 123456790120987654321, ..., (sequence A002477 in the OEIS), although one can check these are not Demlo numbers for p = 10, 19, 28, ...
See also
All one polynomial — Another generalization
Goormaghtigh conjecture
Repeating decimal
Repdigit
Wagstaff prime — can be thought of as repunit primes with negative base
Footnotes
Notes
^ Albert H. Beiler coined the term "repunit number" as follows:
A number which consists of a repeated of a single digit is sometimes called a monodigit number, and for convenience the author has used the term "repunit number" (repeated unit) to represent monodigit numbers consisting solely of the digit 1.
References
^ Beiler 2013, pp. 83
^ For more information, see Factorization of repunit numbers.
^ Maksym Voznyy, New PRP Repunit R(270343)
^ Sloane, N. J. A. (ed.). "Sequence A004023 (Indices of prime repunits: numbers n such that 11...111 (with n 1's) = (10^n - 1)/9 is prime.)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
^ "PrimePage Primes: R(49081)". PrimePage Primes. 2022-03-21. Retrieved 2022-03-31.
^ "PrimePage Primes: R(86453)". PrimePage Primes. 2023-05-16. Retrieved 2023-05-16.
^ "PrimePage Primes: R(109297)". PrimePage Primes. 2025-05-27. Retrieved 2025-05-27.
^ Chris Caldwell. "repunit". The Prime Glossary. Prime Pages.
^ Deriving the Wagstaff Mersenne Conjecture
^ Generalized Repunit Conjecture
^ a b Dickson & Cresse 1999, pp. 164–167
^ Francis 1988, pp. 240–246
^ Kaprekar 1938a, 1938b, Gunjikar & Kaprekar 1939
^ Weisstein, Eric W. "Demlo Number". MathWorld.
References
Beiler, Albert H. (2013) , Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, Dover Recreational Math (2nd Revised ed.), New York: Dover Publications, ISBN 978-0-486-21096-4
Dickson, Leonard Eugene; Cresse, G.H. (1999), History of the Theory of Numbers, Volume I: Divisibility and primality (2nd Reprinted ed.), Providence, RI: AMS Chelsea Publishing, ISBN 978-0-8218-1934-0
Francis, Richard L. (1988), "Mathematical Haystacks: Another Look at Repunit Numbers", The College Mathematics Journal, 19 (3): 240–246, doi:10.1080/07468342.1988.11973120
Gunjikar, K. R.; Kaprekar, D. R. (1939), "Theory of Demlo numbers" (PDF), Journal of the University of Bombay, VIII (3): 3–9
Kaprekar, D. R. (1938a), "On Wonderful Demlo numbers", The Mathematics Student, 6: 68
Kaprekar, D. R. (1938b), "Demlo numbers", J. Phys. Sci. Univ. Bombay, VII (3)
Kaprekar, D. R. (1948), Demlo numbers, Devlali, India: Khareswada
Ribenboim, Paulo (1996-02-02), The New Book of Prime Number Records, Computers and Medicine (3rd ed.), New York: Springer, ISBN 978-0-387-94457-9
Yates, Samuel (1982), Repunits and repetends, FL: Delray Beach, ISBN 978-0-9608652-0-8
External links
Weisstein, Eric W. "Repunit". MathWorld.
The main tables of the Cunningham project.
Repunit at The Prime Pages by Chris Caldwell.
Repunits and their prime factors at World!Of Numbers.
Prime generalized repunits of at least 1000 decimal digits by Andy Steward
Repunit Primes Project Giovanni Di Maria's repunit primes page.
Smallest odd prime p such that (b^p-1)/(b-1) and (b^p+1)/(b+1) is prime for bases 2<=b<=1024
Factorization of repunit numbers
Generalized repunit primes in base -50 to 50
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9390 | https://www.khanacademy.org/math/ap-statistics/gathering-data-ap/sampling-methods/v/techniques-for-generating-a-simple-random-sample | Use of cookies
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9391 | https://www.youtube.com/watch?v=_2T7upeV32A | Q4 Ratios - Mixtures - Rule of Alligation | EA and GMAT Quant Practice Question
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The given question is a medium difficulty EA / GMAT problem solving question in ratios and mixtures. At the core, the concept tested is weighted average. You have to find the weightages used to compute the weighted average.
Any weighted average question in which the weightage has to be computed can be quickly solved using a process called Rule of Alligation - which is explained in this question.
Question
In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?
A. 1 : 2
B: 2 : 1
C. 1: 3
D. 3: 1
E: 2 : 3
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Transcript:
[Music] hey everyone this is a medium difficulty ratio mixtures question it's a medium difficulty question but if you understand how to solve it in one of the two methods that we're going to be looking at today this question should not take you anything more than 15 seconds right that's I'm not joking it's 15 seconds is what you will take to crack this question with the second method that we'll be learning as the part of solving this question before we get started one quick thing have you subscribed to the channel so thank you so much if you have not subscribed please subscribe to the channel right away and turn on notifications right let's get started let's read this question in what ratio should a 20% methal alcohol solution be mixed with a 50% methal alcohol solution lesser methal alcohol in the first one more in the second one so you're going to get a final solution which is going to have something between 20 and 50 that's expected how much does it have it's got 40% methyl alcohol got 20% methyl alcohol with 50% methy alcohol you put them together in some ratio that what you get net net has got 40% methy alcohol right this is the ratio in which we should basically get one of these is the answers right quickly run through using the first method which is a typical weighted average method I'm going to assign a few variables solution one we'll take it to be X lit in volume y lit in volume so essentially what we are trying to find out is if this is 11 lit and this is 19 L we're mixing them in the ratio of 11 is to 19 so you're trying to find out what is this x is to Y right that is what our final answer is all going to be right now the strength of the first solution is 20% So 20% of X lit which if I write it in terms of decimal will be2 20 upon 100 which is 0.2x is the amount of is the volume of methal alcohol in the first mixture second one while letter is the volume strength is 50% so half of it which is 0.5 of what volume of y l is a amount of is a quantum of quantity of methyl alcohol that we have so how much methyl alcohol do we totally have from the two Solutions 0.2x Plus 0.5 y from the first one from the second one net net what is the total volume we have X letter of the first solution y of the second one so x + y is the total volume we have and what do they tell us they're saying that once you mixed it in some ratio this x to Y ratio the final solution has got 40% methal alcohol in it so it's 40% of what 40 percentage of x + y so this final solution is going to have 40 upon 100 which is 0.4 X+ y this is the quantity of methal alcohol when I look at it as the final concentration and final volume but this methal alcohol that I'm talking about where did it come from 2x came from the first solution 0 5 y came from the second one so the 2x +5 Y is the amount of methyl alcohol when I look at it as how I mixed it this is the amount of meth alcohol when I look at it as to how much I have it finally these two should obviously be equal to each other so this is equal to 0.2x + 0.5 y solve this we have one equation in two variables we're not trying to find out what is X what is y we're actually trying to find out X is to Y the ratio we're actually trying to find out X upon y let's see how to go about it expand this 0.4x + 0.4 Y is equal to 0.2x + 0.5 y take this y term to the right hand side and the X term to the left hand side 0.4x minus 0.2x will leave us with a 0.2x is let's see what do you have equal to sign yes we do have an equal to sign equal to what on the right hand side on the right hand side you have 0.5 y minus 0.4 Y which is equal to a 0.1 Y what did we say we trying to find out we trying to find out X is to Y which is s X upon y bring the Y to the denominator we'll be having X upon y here take the 0.2 to the right hand side's denominator so that will make it as 0.1 upon 0.2 so X upon y is1 upon point2 which is nothing but 1 by 2 so in what ratio should we mix this x is to Y which is to be mixed in the ratio of 1 is 2 right this is the answer we look at what the correct answer is in a subsequent slide now let's see if we can solve it using method 2 this is called Rule of alligation we have done it extensively in our online course but I'll quickly run you through this to get you familiar the low concentration is 20% so let's write it as 20 the high concentration is 50 so what is the mean or the weighted average concentration that we have we have a concentration of 40 so low high and let's call it as the mean that we have what ratio should I mix it high minus mean is to mean minus low right but this pictorial representation actually makes it easy for you to get to the answer in a JY 50 - 40 is a 10 this is mean minus low which is 40 - 20 which is equal to 20 so in what ratio are we mixing it 10 is to 20 or 1 is 2 is this the same answer that we got here yes 1 is2 look at it this all the time it takes you low concentration High concentration mean High minus mean is to mean minus low is the ratio in which you're going to mix the low and high so it takes you exactly 15 seconds 10 20 50 40 50 - 40 is 10 40 - 20 is 20 10 is to 20 or 1 is 2 is the answer quickly look at the answer options 1 is 2 is Choice a this says Choice B Choice a is the correct answer to this question now let's look at a bonus question in what ratio should a mixture which comprises 1/3 methal alcohol be mixed right what ratio should have 1/3 methal alcohol solution be mixed with a 34 methal alcohol solution so that the resultant solution has 50 % methal alcohol in it try and solve this bonus question even to start with using the rule of allegation method right so set the rule of allegation low concentration High concentration and what is the mean 50% which is a mean right it as 1 by two right you have 1/3 3/4 1 by two figure out how this basically pans out see with which of these five answer options is right and post your answers to the comment section of this video best wishes for a GMAT preparation [Music] |
9392 | https://math.umd.edu/~immortal/MATH406/lecturenotes/ch9-1.pdf | Math 406 Section 9.1: The Order of an Integer and Primitive Roots 1. Introduction (to Chapter 9): Consider in basic algebra: 3x = 7 ⇔ x = log3 7 How might this function, if at all, in modular arithmetic, say mod 10?
3x ≡7 mod 10 ⇔ Hmmm...
In this example we can find a solution x = 3 by trial-and-error. But a different example fails to have a solution: 9x ≡7 mod 10 ⇔ Hmmm...no such x...
This notion, of determining when we can find powers in modular arithmetic and what those powers are, is important in mathematics and computer science and is known as the discrete logarithm problem. It is extremely difficult when the modulus is large. For example 35x ≡ 14536 mod 34571 has solution x = 458 but that’s not obvious at all.
The approach to understanding these problems is to go back to Euler’s Theorem to see, for a given base, what sorts of results we can achieve by raising that base to different powers.
2. Nonpositive Powers It’s worth pausing to note that if gcd (a, m) = 1 then we know that a has a multiplicative inverse mod m and so we can write the notation a−1 to refer to that inverse. In other words: aa−1 ≡1 mod m From here we can use all sorts of negative powers as long as we understand we mean inverses.
So for example a−3 can be thought of either as (a−1)3 (the cube of the inverse of a) or (a3)−1 (the inverse of the cube of a). These are the same thing.
Everything is as expected with this notation.
In addition if we say that a0 ≡1 mod m then we can make sense of all exponents when gcd (a, m) = 1.
3. The Order of an Integer: Given a modulus m and an integer a with gcd (a, m) = 1 Euler’s Theorem tells us that aφ(m) ≡1 mod m. It does not however tell us that φ(m) is the lowest power which yields 1.
This leads to the following: (a) Definition: Given a modulus m and an integer a with gcd (a, m) = 1 we define the order of a mod m, denoted ordma to be the smallest positive integer n such that an ≡1 mod m.
Note: The order of a depends not just on a but also on the modulus m. Sometimes we say simply “The order of a” when the modulus is clear but it’s always relevant.
Example: Consider a = 3 and m = 11. Euler’s Theorem (and Fermat’s Little Theorem) tell us that 310 ≡1 mod 11 but observe that 10 is not the first power for which we get 1.
To find the order of 3 mod 11 we observe: 31 ≡1 mod 11 32 ≡9 mod 11 33 ≡5 mod 11 34 ≡4 mod 11 35 ≡1 mod 11 Thus ord113 = 5.
The order of an integer underlies the pattern under which powers of the integer repeat.
For example since ord113 = 5 this means that 3x repeats when x repeats mod 5, for example 34 ≡39 mod 11 because 4 ≡9 mod 5.
This idea leads to the following theorems. For all of the following assume gcd (a, m) = 1.
(b) Theorem 1: For x ∈Z+ and gcd (a, m) = 1 we have: ax ≡1 mod m ⇐ ⇒x ≡0 mod ordma ⇐ ⇒ordma | x.
Example: We have 3x ≡1 mod 11 iffx ≡0 mod 5 iff5 | x, so x = ..., −15, −10, 5, 0, 5, 10, 15, ....
Proof: The second ⇐ ⇒is just the defintion. For the first...
⇒Assume ax ≡1 mod m use the Division Algorithm to write x = q(ordma)+r and then we have: 1 ≡ax = aordmaq ar ≡ar mod m and since 0 ≤r < ordma we have r = 0.
⇐If ordma | x then x = α · ordma for some α ∈Z and then ax = aordmaα ≡1 mod m.
QED (c) Corollary: We have ordma | φ(m).
Proof: Since aφ(m) ≡1 mod m this follows from the previous theorem.
QED Note: This can be used to help find orders more quickly. For example if want to know ordma we need only check the divisors of φ(m).
Example: To find ord112 we note φ(11) = 10 so we only need to check 21, 22 and 25 since if none of those work then it must be 210.
(d) Theorem 2: We have ax ≡ay mod m iffx ≡y mod ordma.
Proof: ⇒If ax ≡ay mod m then WLOG assume x > y and then cancel ay (coprimality guar-antees we can) to get ax−y ≡1 mod m and so then ordma | (x −y).
⇐If ordma | (x −y) then WLOG assume x > y and then x = y + α · ordma for some α ∈Z+ and then: ax = ay aordmaα ≡ay mod m QED Understanding: This tells us that although the base works mod m, the exponent works mod ordma.
Example: For example when m = 20 noting that ord203 = 4 and ord209 = 2 we can write: 63102 · 10983 ≡3102 · 983 ≡32 · 91 mod 20 Here the bases 63 and 109 reduce mod 20, the exponent 102 reduces mod ord203 = 4 and the exponent 83 reduces mod ord209 = 2.
4. Primitive Roots: We know that given a modulus m and an integer a with gcd (a, m) = 1 we have ordma ≤φ(m) (in fact it divides φ(m)) but we are especially lucky when we have ordma = φ(m). The reason why this is lucky will be explained soon.
(a) Definition: Given a modulus m and an integer a with gcd (a, m) = 1 we say that a is a primitive root mod m if ordma = φ(m).
Example: If m = 11 then a = 6 is a primitive root mod 11 because ord116 = 10 = φ(11) which can be verified by noting that 61 ≡6, 62 ≡3 and 65 ≡10. Remember why we only need to check these, it’s because we know the order divides φ(11) = 10 and so since it’s not 1,2 or 5 it must be 10.
(b) Importance: Think of a primitive root as a “best possible” base in that powers of a primitive root will achieve all values coprime to m.
Example: We saw that 6 is a primitive root mod 11 and observe that: {61, 62, 63, 64, 65, 66, 67, 68, 69, 610} ≡{6, 3, 7, 9, 10, 5, 8, 4, 2, 1} | {z } Got all the coprimes!
mod 11 This is clarified in the following theorem: (c) Theorem: If r is a primitive rood mod m then the set r, r2, r3, ..., rφ(m) is a reduced residue set mod m.
Note: Recall this means that this set contains φ(m) integers all of which are coprime to m and none of which are equivalent to each other mod m.
Proof: They are all coprime to m since if gcd (m, rk) ̸= 1 for some k then if some prime p divided both then it would divide rk and hence it would divide r, contradicting gcd (r, m) = 1.
If we had ri ≡rj mod m then i ≡j mod ordmr = φ(m) so that i = j because each is nonstrictly between 1 and φ(m).
QED (d) Existence of Primitive Roots: Interestingly if we start with a modulus m there may or may not be any primitive roots mod m. For example m = 8 has no primitive roots since it can be easily checked that φ(8) = 4 but ord81 = 1, ord83 = 2, ord85 = 2 and ord87 = 1 and so we never get ord8a = φ(8).
However if there is a primitive root then usually there are several. We’ll show how many in steps: (e) Theorem: Given a modulus m and an integer a with gcd (a, m) = 1 We have: ordm(ak) = ordma gcd (ordma, k) Obscure Note: For those in MATH403 this is the same as the result from cyclic groups which states that |ak| = |a| gcd (|a|,k).
Proof: First, note that: akordma/gcd (ordma,k) = aordmak/gcd (ordma,k) ≡1k/gcd (ordma,k) mod m ≡1 mod m This tells us that: ordm(ak) ≤ ordma gcd (ordma, k) Second, note that by definition of the order of ak we have: ak ordm(ak) = akordm(ak) ≡1 mod m and so: ordma k ordm ak From whence it follows that: ordma gcd (ordma, k) k gcd (ordma, k)ordm ak Since the gcd of the two fractions is 1 we then know that: ordma gcd (ordma, k) ordm(ak) and so ordma gcd (ordma, k) ≤ordm(ak) The two results together give us...
QED (f) Corollary: Suppose r is a primitive root mod m, then rk is a primitive root mod m iffgcd (k, φ(m)) = 1.
Example: We saw that r = 6 is a primitive root mod 11. Thus we know that since φ(11) = 10 that all primitive roots can be found using 6k with gcd (k, φ(11)) = gcd (k, 10) = 1.
This yields k = 1, 3, 7, 9 and thus the set of primitive roots mod 11 are {61, 63, 67, 69} ≡ {6, 7, 8, 2} mod 11.
Proof: Well rk is a primitive root iffordm rk = φ(m) = ordmr and by the theorem this is iff gcd (ordmr, k) = 1 which is iffgcd (φ(m), k) = 1.
QED (g) Corollary: If there is a primitive root mod m then there are φ(φ(m)) of them.
Example: There are φ(φ(11)) = φ(10) = 4 primitive roots mod 11.
Proof: Let r be one primitive root. Since powers of r form a reduced residue set mod m we know that all other integers coprime to m may be written as rk for some k then by the previous corollary we know that rk is also a primitive root iffgcd (k, φ(m)) = 1 and there are φ(φ(m)) such k.
QED |
9393 | http://blog.felixbreuer.net/2012/06/11/hol.html | Formal proof - first steps with HOL Light - Felix Breuer's Blog
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Formal proof - first steps with HOL Light
Jun 11 th, 2012
Recently, I have philosophized about how the mathematical community needs to move beyond (new) theorems as their currency of research. One different form of currency that I personally find particularly interesting are formal proofs. So, in the last few weeks I have spent some time getting my feet wet with one of the formal proof systems out there: HOL Light.
To learn HOL light, I decided to pick a simple theorem (from the theory of formal power series) and try to formalize it, both in a procedural and in a declarative style. In this blog post, I document what I have learned during this exercise.
My point of view is that of an everyday mathematician, who would like to use formal proof systems to produce formal proofs to go along with his everyday research papers. I realize of course, that this is not practical yet, given the current state of formal proof systems. But my goal is to get an idea of just how far away this dream is from reality.
During my experiments the hol-info mailing list was a huge help. In particular, I would like to thank Freek Wiedijk, Petros Papapanagiotou and John Harrison.
I will begin by giving a short summary of the informal proof and its two formal versions. This should give you a general idea of what all of this is about. If you want more details, you can then refer to the other parts of this post. Here is an outline:
Summary: One Theorem and Three Proofs
Meeting HOL Light
Formal Definitions
A Procedural Proof
Mizar in HOL: miz3
A Declarative Proof
Making Declarative Proofs more Self-Contained
So, without further ado, let's start with a summary.
Summary: One Theorem and Three Proofs
The theorem that I want to prove formally is simply that every formal power series with non-zero constant term has a reciprocal - and this reciprocal has a straightforward recursive definition. In everyday, informal mathematical language, this theorem and its proof might be stated like this.
Theorem. Let f=\sum_{n\geq 0} a_n x^n be a formal power series with a_0\not = 0. Then, the series g=\sum_{n\geq 0} b_n x^n with \begin{eqnarray} b_0 & = & \frac{1}{a_0} \cr b_n & = & -\frac{1}{a_0}\sum_{k = 1}^n a_k b_{n-k} \; \forall n\geq 1 \end{eqnarray}
is a reciprocal of f, i.e., f\cdot g = 1.
Proof. To see that g is a reciprocal of f, we have to show that f\cdot g = \sum_{n \geq 0} \sum_{k=0}^n a_k b_{n-k} = 1
which means that for all n\in\mathbb{Z}{\geq 0} we have \sum{k=0}^n a_k b_{n-k} = \left{ \begin{array}{ll} 1 & \text{if } n=0 \cr 0 & \text{if } n\geq 1 \end{array} \right. To show this, we proceed by induction. If n=0, then we have \sum_{k=0}^n a_k b_{n-k} = a_0 b_0 = a_0 \frac{1}{a_0} = 1 as desired. For the induction step, let n\in\mathbb{Z}{\geq 0} and assume that the identity holds for n. We are now going to show that it also holds for n+1. To this end we calculate \begin{eqnarray} \sum_{k=0}^{n+1} a_k b_{n+1-k} & = & a_0 b_{n+1} + \sum_{k=1}^{n+1} a_k b_{n+1-k} \ &=& a_0 \left( -\frac{1}{a_0}\sum_{k = 1}^{n+1} a_k b_{n+1-k} \right) + \sum_{k=1}^{n+1} a_k b_{n+1-k} \ &=& 0. \end{eqnarray}_End of Proof.
This proof is very simple and straightforward. Therefore, many authors would try to be more terse. For example, in Wilf's classic text book with the wonderful title generatingfunctionology this theorem is Proposition 2.1 and its proof does not even state the above calculation explicitly. Personally, I think the above proof is just about the right length.
You may have noticed that in the above proof, we did not use the induction hypothesis in the induction step. Indeed, the above proof can be done just using case analysis, without induction at all. I will nonetheless stick with the induction version of the proof, because it is more idiomatic when formalized in HOL Light.
Now, let's see how the above proof looks when rendered in a formal proof language. In fact, we are going to do this proof in two formal proof languages: a procedural and a declarative one.
We first need to make some definitions for working with generating functions (aka formal power series). I chose to model a generating function f simply as a function f:\mathbb{N}\rightarrow\mathbb{R}. So, if f corresponds to \sum_{n\geq 0} a_n x^n in the power series notation, then f(n)=a_n. With this approach, addition, multiplication and various other things can be defined as follows. See the section on Formal Definitions below for details.
let gf_plus = new_definition `gf_plus = \(f:num->real) (g:num->real) (n:num). (f n) + (g n) :real`;; let gf_times = new_definition `gf_times = \(f:num->real) (g:num->real) (n:num). (sum (0 .. n) (\k. (f k)(g (n-k)))) :real`;; let gf_one = new_definition `gf_one = \(n:num). if n = 0 then &1 else &0`;; let gf_reciprocal = new_definition `gf_reciprocal = \(f:num->real) (g:num->real). gf_times f g = gf_one`;; let recip = define `recip f (0:num) = ((&1:real)/(f 0) :real) /\ recip f (n+1) = -- (&1/(f 0)) sum (1..(n+1)) (\k. (f k) recip f ((n + 1) - k))`;;
With these definitions, the theorem we are trying to prove is this:
`!(f:num->real). ~(f (0:num) = (&0:real)) ==> gf_reciprocal f (recip f)`
Note how the recursive formula for the reciprocal of f has been wrapped in the definition of recip. The relation gf_reciprocal is used abbreviate the statement that recip f is a reciprocal of f.
Now we are ready to give the two formal versions of our above proof of this theorem. We start with the procedural version. It looks like this:
let GF_RECIP_EXPLICIT_FORM = prove (`!(f:num->real). ~(f (0:num) = (&0:real)) ==> gf_reciprocal f (recip f)`, GEN_TAC THEN DISCH_TAC THEN REWRITE_TAC[gf_reciprocal;gf_times;gf_one] THEN REWRITE_TAC[FUN_EQ_THM] THEN INDUCT_TAC THEN ASM_SIMP_TAC[NUMSEG_SING;SUM_SING] THEN CONV_TAC NUM_REDUCE_CONV THEN REWRITE_TAC[recip] THEN SIMP_TAC[REAL_ARITH `x &1 / x = x / x`] THEN ASM_SIMP_TAC[REAL_DIV_REFL] THEN REWRITE_TAC[MATCH_MP (SPEC `(\k. f k recip f (x - k))` SUM_CLAUSES_LEFT) (SPEC_ALL LE_0);SUB_0;ADD_0;ADD_AC;ADD1;recip] THEN (FIRST_ASSUM (ASSUME_TAC o (MATCH_MP (REAL_FIELD `!x. ~(x = &0) ==> !a. x -- (&1 / x) a = -- a`)))) THEN ASM_REWRITE_TAC[] THEN SIMP_TAC[REAL_FIELD `--a + a = &0`] THEN ASM_SIMP_TAC[ARITH_RULE `~(x + 1 = 0)`]);;
I will explain this proof in detail in the section A Procedural Proof below. For now, the important thing to note here is simply this: Presented in this way, the procedural proof is completely unintelligible. To make sense of procedural proofs, they have to be run interactively. I will walk you through such an interactive run of this proof below and try to explain what is going on in the process. The gist of it is this: Procedural proofs like this one consist of a sequence of proof tactics. Tactics are essentially functions like REWRITE_TAC that are written in OCaml and that produce a part of the proof as their output. They may take pre-proved theorems like FUN_EQ_THM as their arguments.
This way of writing proofs is opposite to the way humans write informal proofs. Take for example an equational proof, i.e., a proof done by rewriting a term step by step to show an equality. Humans would write down the intermediate expressions obtained during the process. In the procedural style, you write down the rule you apply to move from one expression to the next.
While in some ways, this procedural style of carrying out a proof may in fact be closer to how humans think about proofs, this procedural notation has huge drawbacks when it comes to communicating proofs. To me, the most important drawback is that the meaning of the proof is hidden almost entirely within the library of tactics and theorems that it is built upon. Without constantly referring to a dictionary of all these ALL_CAPS_SYMBOLS it impossible to read or write these proofs at all. This is problematic insofar as it adds an additional layer of complexity to the problem of reading or writing a proof: You not only have to know the subject matter, you also have to bind yourself to a particular "API". This not only makes proofs less human readable, it also makes procedural proofs stored in this fashion "bit-rot at an alarming rate".
Of course, the procedural style also has its advantages, most importantly the way it lends itself to automation of proof methods. However, as I stated in the beginning, my primary concern is the translation of everyday informal mathematical articles into a formal language. Which brings us to the declarative proof style.
The declarative version of the proof that I came up with is the following:
let GF_RECIP_EXPLICIT_FORM = thm `; !(f:num->real). ~(f (0:num) = (&0:real)) ==> gf_reciprocal f (recip f) proof let f be num->real; assume ~(f 0 = &0) ; !n. sum (0..n) (\k. f k recip f (n - k)) = (if n = 0 then &1 else &0) proof sum (0..0) (\k. f k recip f (0 - k)) = f 0 recip f (0 - 0) by NUMSEG_SING,SUM_SING; .= f 0 recip f 0; .= f 0 &1 / f 0 by recip; .= f 0 / f 0; .= &1 by REAL_DIV_REFL,1; .= (if 0 = 0 then &1 else &0) ; !n. sum (0..n) (\k. f k recip f (n - k)) = (if n = 0 then &1 else &0) ==> sum (0..SUC n) (\k. f k recip f (SUC n - k)) = (if SUC n = 0 then &1 else &0) proof let n be num; assume sum (0..n) (\k. f k recip f (n - k)) = (if n = 0 then &1 else &0); !x. ~(x = &0) ==> !a. x -- (&1 / x) a = -- a ; sum (0..SUC n) (\k. f k recip f (SUC n - k)) = (\k. f k recip f (SUC n - k)) 0 + sum (0 + 1..SUC n) (\k. f k recip f (SUC n - k)) by SUM_CLAUSES_LEFT,LE_0; .= (f 0 recip f (SUC n)) + sum (1..SUC n) (\k. f k recip f (SUC n - k)) by ADD_0,SUB_0,ADD_AC; .= (f 0 recip f (n + 1)) + sum (1..(n + 1)) (\k. f k recip f ((n + 1) - k)); .= (f 0 -- (&1/(f 0)) sum (1..(n + 1)) (\k. (f k) recip f ((n + 1) - k))) + sum (1..(n + 1)) (\k. f k recip f ((n + 1) - k)) by recip; .= -- sum (1..(n + 1)) (\k. (f k) recip f ((n + 1) - k)) + sum (1..(n + 1)) (\k. f k recip f ((n + 1) - k)) by 1,4; .= &0; .= (if SUC n = 0 then &1 else &0); qed; qed by INDUCT_TAC from 2,3; (\n. sum (0..n) (\k. f k recip f (n - k))) = (\n. (if n = 0 then &1 else &0)) by FUN_EQ_THM; qed by gf_reciprocal,gf_times,gf_one,1 from 5; `;;
To my eye, this looks wonderful! This is an entirely formal proof, that
is very close to the informal version,
is not that much longer than the informal version,
is human readable,
uses very few references to an external "dictionary" or "API" (not all steps have to be justified!),
and all external references refer to theorems and not to tactics.
On top of that, I had a much, much easier time writing this proof than writing the procedural version.
Of course there are still many aspects that can be improved. First of all, while this ASCII syntax is quite readable (much more readable than LaTeX anyway), the ability to edit a fully typeset version of the proof would be a pleasant convenience. I will definitely look into creating a custom transformer for Qute for this syntax.
More importantly, some steps in this proof seem unnecessarily small. For example, when calculating sum (0..0) (\k. f k recip f (0 - k)), it would be great to go directly to f 0 recip f 0 without using f 0 recip f (0 - 0) as a stepping stone. In a similar vein, I would like to avoid stating trivial lemmas like !x. ~(x = &0) ==> !a. x -- (&1 / x) a = -- a explicitly, since in the line where it is used, it is easy to infer from context what lemma is needed here. Finally, I think it is very important to reduce API dependencies even further and so I would like to drop explicit references to such "trivial" library theorems as ADD_0 or ADD_AC.
More details about the declarative proof system and the construction of this proof make up the last part of this post.
On the whole, I am extremely impressed with the state of current formal proof systems. While a lot remains to be done until everyday mathematicians not specialized in formal proof can routinely produce formal versions of their papers, this dream appears to be almost within reach.
After this summary, I am now going to get my hands dirty and explain all of this in detail.
Meeting HOL Light
The proof system I used for these experiments is HOL Light. HOL stands for "higher order logic" which basically means classical mathematical logic with the additional feature that quantifiers can range over functions as well and not only over elements of the ground set. (In practice, working with formal higher order logic is not too different from the mix of "colloquial" classical logic and set theory that I, as a mathematician not specialized in logic, am used to working with.) Higher order logic is one of the standard foundations of mathematics used in the formal proof community and several major proof systems such as Isabelle/HOL, HOL4, ProofPower and HOL Light work with it.
In this post, I will not assume familiarity with HOL Light and I hope to keep things reasonably self-contained. Nonetheless, I want to recommend some references as starting points for further reading: Freek Wiedijk has done some very helpful work comparing and contrasting the different provers out there, see for example the list of the Seventeen Provers of the World which he compiled. To get started with HOL Light, I recommend the great HOL Light Tutorial by John Harrison. When doing a proof, it is useful to have the quick reference at hand.
Installing HOL Light is non-trivial. First, you need to have a working installation of OCaml. Second, you will need Camlp5, which has to be compiled with the -strict configuration option. This is often not the case for camlp5 packages that ship with major Linux distributions. Third, you need to fetch HOL Light from its SVN repository and compile it. I will not walk through these steps in detail, but I can recommend a script by Alexander Krauss that does these things for you (and that you can read if you need instructions on how to do this by hand).
After everything is installed, you run OCaml from the HOL Light source directory and then, from the OCaml toplevel, start HOL Light with the command #use "hol.ml";; HOL Light will need about a minute to start and will generate lots of output in the process.
In the rest of this section, I will give a very short summary of the basics of HOL Light. As an introduction to HOL Light, this whirlwind tour is of course woefully inadequate. It is mainly intended as a brief orientation for readers unfamiliar with HOL who do not want to read the tutorial first. Either way, I hope that these brief notes will help in making sense of the exercise that this post is about.
Proofs in HOL Light are typically developed interactively at the OCaml toplevel.
In HOL Light, mathematical expressions in higher order logic are enclosed in backquotes. These expressions have the OCaml type term. For example, if you enter 1 + 1 = 2 at the OCaml toplevel (followed by two semicolons and return) you will see: # `1 + 1 = 2`;; val it : term = `1 + 1 = 2`
The HOL Light syntax for higher order logic is straightforward. For example the principle of induction on the natural numbers \forall P: P(0) \wedge \left( \forall n: P(n) \Rightarrow P(S(n)) \right) \Rightarrow (\forall n. P(n)) would be written as !P. P 0 /\ (!n. P n ==> P (SUC n)) ==> (!n. P n) where SUC is a predefined constant, \forall is written as ! and \wedge is written as /\. Here is the notation for some other useful symbols: \exists is written as ?, \vee is written as \/, \neg is written as ~ and \Leftrightarrow is written as <=>. Lambda expressions are written using \, e.g., \lambda x. x + 1 is written as \x. x + 1.
Higher order logic in HOL Light is typed. As long as HOL Light can figure out the types on its own, you don't need to specify them. If you want to give type information explicitly, you can use the colon. For example, here is the principle of induction on lists: !P:(A)list->bool. P [] /\ (!h t. P t ==> P (CONS h t)) ==> !l. P l
Theorems in HOL Light are objects of type thm. The only way to produce an object of type theorem in HOL Light is to prove that theorem.
Inference rules are special OCaml functions that produce theorems. They encapsulate proof methods. For example ARITH_RULE is a function of type term->thm that uses linear arithmetic methods to prove statements about natural numbers. For example we can produce the theorem \vdash 1 + 1 = 2 by # ARITH_RULE `1 + 1 = 2`;; val it : thm = |- 1 + 1 = 2 or we can produce the theorem x > 0 \vdash x \geq 1 by # UNDISCH (ARITH_RULE `x > 0 ==> x >= 1`);; val it : thm = x > 0 |- x >= 1 where UNDISCH takes a theorem of the form \vdash p \Rightarrow q and produces the theorem p \vdash q.
A goal captures a claim of the form p_1,\ldots,p_n\vdash q that we are currently trying to prove but have not proved yet. In the HOL Light documentation, the distinction is made clear using the ?- notation, e.g., the goal that x > 0 \vdash x \geq 1 might be written as x > 0 ?- x >= 1. On the OCaml toplevel (a stack of goals containing) this goal would be printed as: val it : goalstack = 1 subgoal (1 total) 0 [`x > 0`] `x >= 1` where the expression in square brackets represents the assumption of that goal.
A tactic is a function that (essentially) takes a goal and produces a list of subgoals, such that a proof of all subgoals produces a proof of the original goal. This allows backward proofs. These are the typical way of working with HOL Light.
An example. Suppose we want to prove x\not=0 \Rightarrow 1 \leq x using the lemma 0 < x \Rightarrow 1 \leq x. We store the lemma in an OCaml variable. # let lemma = ARITH_RULE `0 < x ==> 1 <= x`;; val lemma : thm = |- 0 < x ==> 1 <= x Then we set up our goal with the function g. # g `~(x=0) ==> 1 <= x`;; Warning: Free variables in goal: x val it : goalstack = 1 subgoal (1 total) `~(x = 0) ==> 1 <= x` Next we apply the tactic DISCH_TAC using the function e. This reduces the goal \vdash x\not=0 \Rightarrow 1 \leq x to x\not=0 \vdash 0 \Rightarrow 1. # e DISCH_TAC;; val it : goalstack = 1 subgoal (1 total) 0 [`~(x = 0)`] `1 <= x` Now we apply our lemma to "simplify" the conclusion of our goal, using MATCH_MP_TAC which is of type thm->tactic. MATCH_MP_TAC uses the theorem \vdash p \Rightarrow q to reduce the goal \vdash q to \vdash p, making instantiations as necessary. # e(MATCH_MP_TAC(lemma));; val it : goalstack = 1 subgoal (1 total) 0 [`~(x = 0)`] `0 < x` What is now left to prove is the content of the theorem LT_NZ in the library. # LT_NZ;; val it : thm = |- !n. 0 < n <=> ~(n = 0) We make use of this theorem via ASM_REWRITE_TAC of type thm list -> tactic, which rewrites goals by the list of (equational) theorems it is passed, taking assumptions into account. # e(ASM_REWRITE_TAC[LT_NZ]);; val it : goalstack = No subgoals The statement that there are no subgoal left, means that the proof is complete. We can get the theorem we just proved using top_thm. # top_thm();; val it : thm = |- ~(x = 0) ==> 1 <= x (Note that this example is non-sensical insofar as we could have used ARITH_RULE to solve our goal immediately.)
If you make a mistake during one of these goalstack-style proofs, you can use b to back up one step, by calling b();;.
After this minimal introduction to HOL Light, we can now turn to formalizing the above proof.
Formal Definitions
Before we can start with the proof proper, however, we need to make a couple of definitions.
For the purposes of this exercise a formal power series or generating function will be simply a map f:\mathbb{N}\rar\mathbb{R}. Of course you could also view f as a sequence of real numbers or, classically, as the sequence of coefficients of a power series. Addition of formal power series is defined componentwise, i.e., (f+g)(n)=f(n)+g(n), which in HOL Light looks as follows: let gf_plus = new_definition `gf_plus = \(f:num->real) (g:num->real) (n:num). (f n) + (g n) :real`;; This simply means that gf_plus is a function that takes two generating functions and returns the generating function that is the componentwise sum of the arguments.
The product of two generating functions is defined via the Cauchy product rule (f\cdot g)(n) = \sum_{k=0}^n f(k)\cdot g(n-k).
This looks much more natural when phrased in terms of formal series and this is the reason why formal power series are traditionally written in power series notation: \sum_{n\geq 0}a_nx^n \cdot \sum_{n\geq 0} b_n x^n = \sum_{n\geq0} \left(\sum_{k=0}^n a_k b_{n-k}\right) x^n. Personally, I think of this "product" more as a "Minkowski sum", but that will be the subject of another blog post. The HOL Light version of this definition is straightforward: let gf_times = new_definition `gf_times = \(f:num->real) (g:num->real) (n:num). (sum (0 .. n) (\k. (f k)(g (n-k)))) :real`;;
Next, we define a constant gf_one which represents the formal power series 1, i.e., the function f with f(0)=1 and f(n)=0 for all other n. let gf_one = new_definition `gf_one = \(n:num). if n = 0 then &1 else &0`;;
Now, it is straightforward to define a binary relation gf_reciprocal that specifies when a generating function g is the reciprocal of a generating function f, that is, f\cdot g = 1. let gf_reciprocal = new_definition `gf_reciprocal = \(f:num->real) (g:num->real). gf_times f g = gf_one`;;
Finally, the theorem we will prove contains a recursive formula for the reciprocal of a generating function f, namely: \begin{eqnarray} b_0 & = & \frac{1}{a_0} \cr b_n & = & -\frac{1}{a_0}\sum_{k = 1}^n a_k b_{n-k} \; \forall n\geq 1 \end{eqnarray}
We will use the shorthand recip f to denote the function defined by this recursive formula. let recip = define `recip f (0:num) = ((&1:real)/(f 0) :real) /\ recip f (n+1) = -- (&1/(f 0)) sum (1..(n+1)) (\k. (f k) recip f ((n + 1) - k))`;;
Note that these definitions all return theorems. In addition they have the side effect of modifying the OCaml variable the_definitions, see fusion.ml. This has the drawback that you cannot change existing definitions within a running HOL Light session. You will have to restart HOL Light to change definitions. (If you want to learn more about definitions and state in HOL, you may want to look at thesepapers.)
A Procedural Proof
The theorem that we want to prove says that if f(0)\not= 0 then \text{recip}(f), as defined in the previous section, is a reciprocal of f. In HOL this statement reads simply as follows.
!(f:num->real). ~(f (0:num) = (&0:real)) ==> gf_reciprocal f (recip f)
Note that 0 refers to the natural number 0, whereas &0 refers to the real number 0 - which are distinct in HOL. Real number literals always have to be prefixed with & in HOL. The type annotations given above are not essential. We could also write simply !f. ~(f 0 = &0) ==> gf_reciprocal f (recip f).
In this section, we will go through the procedural proof given in the summary at the beginning step by step, and take a look at the goalstack at each point. If you want to expand other proofs in a similar fashion, you may want to take a look at the tool Tactician.
We start out by pushing the theorem we want to prove onto the goalstack.
# g `!(f:num->real). ~(f (0:num) = (&0:real)) ==> gf_reciprocal f (recip f)`;; val it : goalstack = 1 subgoal (1 total) `!f. ~(f 0 = &0) ==> gf_reciprocal f (recip f)`
Next, we eliminate the universal quantifier and the implication, turning f(0) \not = 0 into an assumption, and we expand the definitions.
# e(GEN_TAC);; val it : goalstack = 1 subgoal (1 total) `~(f 0 = &0) ==> gf_reciprocal f (recip f)` # e(DISCH_TAC);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] `gf_reciprocal f (recip f)` # e(REWRITE_TAC[gf_reciprocal;gf_times;gf_one]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] `(\n. sum (0..n) (\k. f k recip f (n - k))) = (\n. if n = 0 then &1 else &0)`
We have to show that two functions are equal. This boils down to showing that their values are equal for all arguments. This is the content of FUN_EQ_THM, which we apply using REWRITE_TAC.
# e(REWRITE_TAC[FUN_EQ_THM]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] `!x. sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`
Now we start working at the heart of the problem. Just as in the informal proof, we are going to use induction, even though a more elementary case analysis would work as well. To use induction we apply INDUCT_TAC which leaves us with two subgoals: The base of the induction and the induction step.
# e(INDUCT_TAC);; val it : goalstack = 2 subgoals (2 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] `sum (0..SUC x) (\k. f k recip f (SUC x - k)) = (if SUC x = 0 then &1 else &0)` 0 [`~(f 0 = &0)`] `sum (0..0) (\k. f k recip f (0 - k)) = (if 0 = 0 then &1 else &0)`
We start working on the base of the induction, i.e., the second subgoal. All we simplify the equality we have to prove, by evaluating the if .. then .. else .. and using the fact that the sum over an interval that contains just 0 can be obtained by evaluating the expression we sum over at 0. This latter fact follows from the lemmas NUMSEG_SING and SUM_SING.
# e(ASM_SIMP_TAC[NUMSEG_SING;SUM_SING]);; val it : goalstack = 1 subgoal (2 total) 0 [`~(f 0 = &0)`] `f 0 recip f (0 - 0) = &1`
The next couple of steps are about calculating with numbers and we use a couple of different methods for doing so. NUM_REDUCE_CONV is a so-called conversion, that can be used to simplify expressions as part of a bigger expression, and conversions can be turned into tactics using CONV_TAC. NUM_REDUCE_CONV evaluates expressions over natural numbers and this allows us to reduce 0-0 to 0, without knowing the name of a lemma in the HOL library that says so. Afterward, we expand recip according to its definition.
# e(CONV_TAC NUM_REDUCE_CONV);; val it : goalstack = 1 subgoal (2 total) 0 [`~(f 0 = &0)`] `f 0 recip f 0 = &1` # e(REWRITE_TAC[recip]);; val it : goalstack = 1 subgoal (2 total) 0 [`~(f 0 = &0)`] `f 0 &1 / f 0 = &1`
Next, we want to go from f(0)\cdot \frac{1}{f(0)} to \frac{f(0)}{f(0)}. Again we don't know the exact name of a lemma in the library that allows us to get there. However, the library does contain a general purpose inference rule REAL_ARTIH that can prove certain identities of real numbers automatically. We use this to prove the lemma we need on the fly. After this, all that is left to show is that \frac{f(0)}{f(0)}=1 under the assumption that f(0)\not=0. The appropriate lemma in the library is REAL_DIV_REFL. To apply it, we use ASM_SIMP_TAC which uses not only the theorem it is passed as argument, but also takes into account the available assumptions. This, then, completes our proof of the base of the induction, and we continue to the next goal on the stack: the induction step.
# e(SIMP_TAC[REAL_ARITH `x &1 / x = x / x`]);; val it : goalstack = 1 subgoal (2 total) 0 [`~(f 0 = &0)`] `f 0 / f 0 = &1` # e(ASM_SIMP_TAC[REAL_DIV_REFL]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] `sum (0..SUC x) (\k. f k recip f (SUC x - k)) = (if SUC x = 0 then &1 else &0)`
The first thing that we want to do here is to split off the first summand of the sum. The theorem in the library that says we can split off the first term in a sum is SUM_CLAUSES_LEFT.
# SUM_CLAUSES_LEFT;; val it : thm = |- !f m n. m <= n ==> sum (m..n) f = f m + sum (m + 1..n) f
We want to use this theorem for f=\lambda k. f(k)\cdot \text{recip}(f,x-k), m=0 and n=n. We can achieve these specializations as follows. Here LE_0 is the theorem \forall n. 0 \leq n and SPEC_ALL LE_0 is simply the theorem 0 \leq n. Note that n refers to a natural number here.
# let sum_lemma = (SPEC `(\k. f k recip f (x - k))` SUM_CLAUSES_LEFT);; val sum_lemma : thm = |- !m n. m <= n ==> sum (m..n) (\k. f k recip f (x - k)) = (\k. f k recip f (x - k)) m + sum (m + 1..n) (\k. f k recip f (x - k)) # let split_sum_lemma = MATCH_MP sum_lemma (SPEC_ALL LE_0);; val split_sum_lemma : thm = |- sum (0..n) (\k. f k recip f (x - k)) = (\k. f k recip f (x - k)) 0 + sum (0 + 1..n) (\k. f k recip f (x - k))
We use our split_sum_lemma to split the sum in question. After some simplifications, the first occurrence of \text{recip} in the expression is in the exact form we need in order to be able to apply the recursive definition of \text{recip}.
# e(REWRITE_TAC[split_sum_lemma]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] `f 0 recip f (SUC x - 0) + sum (0 + 1..SUC x) (\k. f k recip f (SUC x - k)) = (if SUC x = 0 then &1 else &0)` # e(REWRITE_TAC[SUB_0;ADD_0;ADD_AC;ADD1]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] `f 0 recip f (x + 1) + sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) = (if x + 1 = 0 then &1 else &0)` # e(REWRITE_TAC[recip]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] `f 0 --(&1 / f 0) sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) + sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) = (if x + 1 = 0 then &1 else &0)`
Now we want to cancel the two occurrences of f(0) in the numerator and the denominator. Standard facilities for working with expressions of real numbers, such as REAL_RAT_REDUCE_CONV, do not suffice here, as we have to take the assumption that f(0)\not=0 into account. However, the lemma that we need in this situation can be proved automatically by REAL_FIELD.
# let calc_lemma = (REAL_FIELD `!x. ~(x = &0) ==> !a. x -- (&1 / x) a = -- a`);; 2 basis elements and 0 critical pairs 3 basis elements and 1 critical pairs 3 basis elements and 0 critical pairs val calc_lemma : thm = |- !x. ~(x = &0) ==> (!a. x --(&1 / x) a = --a)
We need to specialize this lemma to make use of it. The expression FIRST_ASSUM (ASSUME_TAC o (MATCH_MP calc_lemma)) is idiomatic for this purpose. FIRST_ASSUM tac tries to apply the tactic tac to each assumption in turn, until it finds one on which tac succeeds. In our case ASSUME_TAC o (MATCH_MP calc_lemma) succeeds immediately on the first assumption. MATCH_MP calc_lemma applied to the assumption ~(f 0 = &0) makes the substitution x=f(0) in calc_lemma and produces the theorem !a. f 0 --(&1 / f 0) a = --a. This is then added as an assumption to our goal by ASSUME_TAC. Afterward, ASM_REWRITE_TAC is able to perform the simplification we intended.
# e(FIRST_ASSUM (ASSUME_TAC o (MATCH_MP calc_lemma)));; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] 2 [`!a. f 0 --(&1 / f 0) a = --a`] `f 0 --(&1 / f 0) sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) + sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) = (if x + 1 = 0 then &1 else &0)` # e(ASM_REWRITE_TAC[]);; val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] 2 [`!a. f 0 --(&1 / f 0) a = --a`] `--sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) + sum (1..x + 1) (\k. f k recip f ((x + 1) - k)) = (if x + 1 = 0 then &1 else &0)`
For the last simplification we again create the necessary theorem on the fly, without looking it up in the library.
# e(SIMP_TAC[REAL_FIELD `--a + a = &0`]);; 1 basis elements and 0 critical pairs val it : goalstack = 1 subgoal (1 total) 0 [`~(f 0 = &0)`] 1 [`sum (0..x) (\k. f k recip f (x - k)) = (if x = 0 then &1 else &0)`] 2 [`!a. f 0 --(&1 / f 0) a = --a`] `&0 = (if x + 1 = 0 then &1 else &0)`
The last thing we need to do is to show that the right-hand side of this equality is 0, using the fact that x+1\not=0 for any natural number x. The proof is then complete and we can retrieve the theorem we just proved using top_thm.
# e(ASM_SIMP_TAC[ARITH_RULE `~(x + 1 = 0)`]);; val it : goalstack = No subgoals # top_thm();; val it : thm = |- !f. ~(f 0 = &0) ==> gf_reciprocal f (recip f)
As we can see, running the procedural proof interactively is crucial to understanding what the proof does. However, working through this exercise, I have often found it excruciatingly hard to perform even simple manipulations of the goalstack in the way I intended. My understanding of HOL's library of theorems and tactics has to improve considerably, before I can become effective at this. Fortunately, working with declarative proofs was much easier for me.
Mizar in HOL: miz3
The Mizar system pioneered declarative proof languages, and has a lasting influence on declarative proof languages in other systems, such as the Isar language that is part of Isabelle or the various variants of Mizar implementations for the HOL family of provers. The latest incarnation of Mizar for HOL Light is miz3 by Freek Wiedijk.
As you can see from the declarative version of the proof given in the introduction, Mizar-style proofs are largely self-explanatory. Of course some elements still require explanation, but I will not go through them in detail here. Instead I want to refer you directly to Freek's paper. Instead, I will use this section to give some tips for installing and using miz3. These tips refer to the version of miz3 available at the time of publication of this post (miz3 releases do not appear to be numbered).
You can get miz3 from here. To use miz3, all you need to do is to extract miz3.ml from the archive and place it in your HOL Light source directory.
miz3 comes with a Vim interface. This interface is very lightweight and its use is not required: You can always paste miz3 code into your OCaml toplevel. But these notes are mainly about using miz3 via Vim. To install the Vim interface, you need to
place the two executables miz3 and miz3f from the bin/ subdirectory of the miz3 distribution on your path, and
add source /path/to/exrc to your .vimrc, where the path points to the file exrc from the miz3 distribution.
To use miz3, you need to start the OCaml toplevel using the command ocaml unix.cma. On the toplevel, first run #use "hol.ml";; and then #use "miz3.ml";;. This automatically starts the miz3 server and you can begin using Vim.
In Vim, you can press Ctrl-C RET to send paragraphs of code (delimited by blank lines) to the OCaml toplevel. Error reports are then displayed right in the Vim buffer. However, some errors are not. So you should always keep an eye on your toplevel!
You should keep the paragraphs you send to miz3 minimal: Separate your miz3 proofs and your ( OCaml comments ) with a blank line. Do the same for definitions and proofs. Otherwise miz3 may throw errors that only appear on the toplevel.
GOAL_TAC is another very useful tool to know about in a declarative proof. If you want to know, what exact goal you have to prove at any given step of you declarative proof, you can add by GOAL_TAC to the end of the line. After the declarative proof has been processed by miz3 you can inspect the goal at that point by running p();; at the toplevel. This works no matter if you proof has been successful or not. As GOAL_TAC is a tactic, it is very flexible and can be used in procedural proofs or as part of a more complicated tactic as well.
After these introductory comments, we can now go on to the declarative version of the proof.
A Declarative Proof
Using a vanilla installation of HOL Light and miz3, a declarative version of the proof might look like this.
let GF_RECIP_EXPLICIT_FORM = thm `; !(f:num->real). ~(f (0:num) = (&0:real)) ==> gf_reciprocal f (recip f) proof let f be num->real; assume ~(f 0 = &0) ; !n. sum (0..n) (\k. f k recip f (n - k)) = (if n = 0 then &1 else &0) proof sum (0..0) (\k. f k recip f (0 - k)) = f 0 recip f (0 - 0) by NUMSEG_SING,SUM_SING; .= f 0 recip f 0; .= f 0 &1 / f 0 by recip; .= f 0 / f 0; .= &1 by REAL_DIV_REFL,1; .= (if 0 = 0 then &1 else &0) ; !n. sum (0..n) (\k. f k recip f (n - k)) = (if n = 0 then &1 else &0) ==> sum (0..SUC n) (\k. f k recip f (SUC n - k)) = (if SUC n = 0 then &1 else &0) proof let n be num; assume sum (0..n) (\k. f k recip f (n - k)) = (if n = 0 then &1 else &0); !x. ~(x = &0) ==> !a. x -- (&1 / x) a = -- a by REAL_FIELD; sum (0..SUC n) (\k. f k recip f (SUC n - k)) = (\k. f k recip f (SUC n - k)) 0 + sum (0 + 1..SUC n) (\k. f k recip f (SUC n - k)) by SUM_CLAUSES_LEFT,LE_0; .= (f 0 recip f (SUC n)) + sum (1..SUC n) (\k. f k recip f (SUC n - k)) by ADD_0,SUB_0,ADD_AC; .= (f 0 recip f (n + 1)) + sum (1..(n + 1)) (\k. f k recip f ((n + 1) - k)); .= (f 0 -- (&1/(f 0)) sum (1..(n + 1)) (\k. (f k) recip f ((n + 1) - k))) + sum (1..(n + 1)) (\k. f k recip f ((n + 1) - k)) by recip; .= -- sum (1..(n + 1)) (\k. (f k) recip f ((n + 1) - k)) + sum (1..(n + 1)) (\k. f k recip f ((n + 1) - k)) by 1,4; .= &0; .= (if SUC n = 0 then &1 else &0); qed; qed by INDUCT_TAC from 2,3; (\n. sum (0..n) (\k. f k recip f (n - k))) = (\n. (if n = 0 then &1 else &0)) by REWRITE_TAC,FUN_EQ_THM; qed by REWRITE_TAC,gf_reciprocal,gf_times,gf_one,1 from 5; `;;
Note that this version differs slightly from the declarative proof given at the beginning. The difference is that this version uses additional references to the HOL library, making explicit use of the tactics REAL_FIELD and REWRITE_TAC. The proof given in the introduction is a bit less verbose and it is tactic-free, i.e., it only references theorems and not tactics in the library.
Why do I care about a proof being tactic-free? After all, the availability of proof tactics and the corresponding opportunities for automation are a great strength of procedural provers such as HOL Light? Well, on the one hand this exercise is about translating human-written proofs and not about automated reasoning. On the other hand, theorems used in by clauses can be stated in higher order logic and can thus be easily included in a prover-independent library. Tactics, however, depend critically on the prover they were written for and are therefore less portable. I will write another post in the near future on why I think prover independence is so important.
Making Declarative Proofs more Self-Contained
In this section we will see how to tweak miz3 in order to eliminate the tactics REWRITE_TAC and REAL_FIELD from the above declarative proof.
First, we have to examine a bit more closely how miz3 proves each step in a declarative proof:
If the by clause contains only theorems and no tactics (or if it is even empty altogether), then miz3 uses the tactic specified in the variable default_prover to prove this step.
If the by clause contains a tactic, that tactic is used for the proof and the default prover is ignored.
The standard default prover in miz3 is HOL_BY, which in turn was inspired by the prover contained in the original Mizar. HOL_BY resides in the file Examples/holby.ml in the HOL Light source directory (not the miz3 source directory). It is set as the default prover of miz3 at the very top of miz3.ml.
let default_prover = ref ("HOL_BY", CONV_TAC o HOL_BY);;
HOL_BY is of type thm list -> term -> thm whence CONV_TAC o HOL_BY is of type thm list -> tactic, which is the type expected of a default prover.
We now turn to the task of eliminating the explicit reference to REAL_FIELD from the proof. The idea is to improve HOL_BY so that the default prover tries REAL_FIELD automatically. REAL_FIELD is an inference rule. So in order to find a suitable place for integrating REAL_FIELD into HOL_BY, we look for other inference rules in the HOL_BY source. Right at the beginning of the definition of the function HOL_BY itself, we find a bunch of conversions and inference rules. So we just include REAL_FIELD right after REAL_ARITH, like this:
let HOL_BY = let BETASET_CONV = TOP_DEPTH_CONV GEN_BETA_CONV THENC REWRITE_CONV[IN_ELIM_THM] and BUILTIN_CONV tm = try EQT_ELIM(NUM_REDUCE_CONV tm) with Failure _ -> try EQT_ELIM(REAL_RAT_REDUCE_CONV tm) with Failure _ -> try ARITH_RULE tm with Failure _ -> try REAL_ARITH tm with Failure _ -> try REAL_FIELD tm with Failure _ -> failwith "BUILTIN_CONV" in fun ths tm -> ... etc. ...
And indeed, with this addition, we can remove REAL_FIELD from our declarative proof (after reloading miz3.ml).
Eliminating REWRITE_TAC from our proof turns out to be a bit more subtle. If we remove the explicit occurrences of REWRITE_TAC from the proof, we get an inference timeout. Does this imply that REWRITE_TAC can do something that HOL_BY cannot? No. Even if we replace REWRITE_TAC with ALL_TAC, the proof will succeed, even though ALL_TAC is the tactic that does nothing. The answer to this riddle lies in the code that miz3 wraps around the tactics it is given. This code we find in the function tactic_of_by in miz3.ml.
In tactic_of_by we find the block:
(try 0,((FILTER_ASSUMS (fun _,(x,_) -> x <> "=") THEN FILTER_ASSUMS (fun n,(x,_) -> mem x labs or n < hor or (n = 0 & mem "-" labs) or full_asl) THEN MAP_ASSUMS (fun l,th -> l,PURE_REWRITE_RULE sets th) THEN MIZAR_NEXT' (PURE_REWRITE_TAC sets) THEN (fun (asl',w' as g') -> let key = name,(map concl thms,map concl thms'),w' in try if grow then failwith "apply"; let e,th = apply (!just_cache) key in if e = 0 then (ACCEPT_TAC th THEN NO_TAC) g' else if e = 2 then raise Timeout else failwith "cached by" with | Failure "apply" -> try let (_,_,just as gs) = ((fun g'' -> let gs' = TIMED_TAC (!timeout) (tac thms) g'' in if grow then raise (Grown (steps_of_goals g gs')) else gs') THEN REPEAT (fun (asl'',_ as g'') -> if subset asl'' asl' then NO_TAC g'' else FIRST_ASSUM (UNDISCH_TAC o concl) g'') THEN TRY (FIRST (map ACCEPT_TAC thms'')) THEN REWRITE_TAC thms'' THEN NO_TAC) g' in let th = just null_inst [] in just_cache := (key |-> (0,th)) !just_cache; gs with | Grown _ as x -> raise x | x -> if name <> "GOAL_TAC" then just_cache := (key |-> ((if x = Timeout then 2 else 1),TRUTH)) !just_cache; raise x )) g)
And the crucial piece in here is the line:
let gs' = TIMED_TAC (!timeout) (tac thms) g'' in
Here tac refers to either the tactic that was found in the by clause or the default prover. TIMED_TAC tries the tactic it is passed for a fixed amount of time. If the tactic does not succeed in the specified interval, a Timeout exception is raised and the tactics that follow are never even tried. As a consequence a long-running tactic such as HOL_BY may achieve less than a instantaneous tactic such as ALL_TAC, because the exception thrown by the former prevents the last few tactics in tactic_of_by from being called. A workaround, therefore, is to modify the default prover so that HOL_BY is tried for a fixed amount of time strictly less than the timeout. After that HOL_BY is aborted without raising an exception and the subsequent tactics given in tactic_of_by are still executed.
To implement this workaround, we add the following definition to then end of miz3.ml right before the definition of reset_miz3:
let TIMEOUT_HOL_BY = fun ths g -> try TIMED_TAC 1 (CONV_TAC (HOL_BY ths)) g with Timeout -> ALL_TAC g;;
This defines a new tactic TIMEOUT_HOL_BY that tries to run HOL_BY and aborts it silently after 1 second. Next, we set the timeout after which tactics are killed and an exception is raised to 10 seconds (so that this timeout does not interfere with TIMEOUT_HOL_BY) and set the default prover accordingly to TIMEOUT_HOL_BY, by modifying reset_miz3 as follows:
let reset_miz3 h = horizon := h; timeout := 10; default_prover := ("HOL_BY", TIMEOUT_HOL_BY); sketch_mode := false; just_cache := undefined; by_item_cache := undefined; current_goalstack := []; server_up();;
After this, we reload miz3.ml and then run reset_miz3 !horizon;; on the toplevel. Our miz3 can now tackle the tactic-free declarative proof given at the beginning of this post!
As I already wrote at the end of the summary, I am very impressed. With current provers it is possible to write proofs that are at the same time formal, human-readable and reasonably short and self-contained. On top of that, further improvements seem to be within reach.
For me, the biggest pleasant surprise was that a declarative proof in miz3 requires relatively few explicit justifications (by clauses) in each step. I think this is very important, as this reduces the volume of the dictionary that a user has to learn to get started, it decreases the danger of bit-rot and it increases portability. I will write more on these issues in a future blog post.
Posted by Felix Breuer Jun 11 th, 2012formal, math, proof, software
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Introductory Statistics
Homework
Introductory Statistics Homework
Contents Contents
Highlights
Table of contents
Preface
1 Sampling and Data
2 Descriptive Statistics
3 Probability Topics
4 Discrete Random Variables
5 Continuous Random Variables
6 The Normal Distribution
7 The Central Limit Theorem
8 Confidence Intervals
9 Hypothesis Testing with One Sample
10 Hypothesis Testing with Two Samples
11 The Chi-Square Distribution
12 Linear Regression and Correlation
Introduction
12.1 Linear Equations
12.2 Scatter Plots
12.3 The Regression Equation
12.4 Testing the Significance of the Correlation Coefficient
12.5 Prediction
12.6 Outliers
12.7 Regression (Distance from School)
12.8 Regression (Textbook Cost)
12.9 Regression (Fuel Efficiency)
Key Terms
Chapter Review
Formula Review
Practice
Homework
Bringing It Together: Homework
References
Solutions
13 F Distribution and One-Way ANOVA
A | Review Exercises (Ch 3-13)
B | Practice Tests (1-4) and Final Exams
C | Data Sets
D | Group and Partner Projects
E | Solution Sheets
F | Mathematical Phrases, Symbols, and Formulas
G | Notes for the TI-83, 83+, 84, 84+ Calculators
H | Tables
Index
Search for key terms or text.
Close
12.1 Linear Equations ---------------------
57.
For each of the following situations, state the independent variable and the dependent variable.
A study is done to determine if elderly drivers are involved in more motor vehicle fatalities than other drivers. The number of fatalities per 100,000 drivers is compared to the age of drivers.
A study is done to determine if the weekly grocery bill changes based on the number of family members.
Insurance companies base life insurance premiums partially on the age of the applicant.
Utility bills vary according to power consumption.
A study is done to determine if a higher education reduces the crime rate in a population.
Piece-rate systems are widely debated incentive payment plans. In a recent study of loan officer effectiveness, the following piece-rate system was examined:
% of goal reached< 80 80 100 120
Incentive n/a$4,000 with an additional $125 added per percentage point from 81–99%$6,500 with an additional $125 added per percentage point from 101–119%$9,500 with an additional $125 added per percentage point starting at 121%
Table 12.15
If a loan officer makes 95% of his or her goal, write the linear function that applies based on the incentive plan table. In context, explain the y-intercept and slope.
12.2 Scatter Plots ------------------
59.
The Gross Domestic Product Purchasing Power Parity is an indication of a country’s currency value compared to another country. Table 12.16 shows the GDP PPP of Cuba as compared to US dollars. Construct a scatter plot of the data.
| Year | Cuba’s PPP | Year | Cuba’s PPP |
--- --- |
| 1999 | 1,700 | 2006 | 4,000 |
| 2000 | 1,700 | 2007 | 11,000 |
| 2002 | 2,300 | 2008 | 9,500 |
| 2003 | 2,900 | 2009 | 9,700 |
| 2004 | 3,000 | 2010 | 9,900 |
| 2005 | 3,500 | | |
Table 12.16
The following table shows the poverty rates and cell phone usage in the United States. Construct a scatter plot of the data
| Year | Poverty Rate | Cellular Usage per Capita |
---
| 2003 | 12.7 | 54.67 |
| 2005 | 12.6 | 74.19 |
| 2007 | 12 | 84.86 |
| 2009 | 12 | 90.82 |
Table 12.17
61.
Does the higher cost of tuition translate into higher-paying jobs? The table lists the top ten colleges based on mid-career salary and the associated yearly tuition costs. Construct a scatter plot of the data.
| School | Mid-Career Salary (in thousands) | Yearly Tuition |
---
| Princeton | 137 | 28,540 |
| Harvey Mudd | 135 | 40,133 |
| CalTech | 127 | 39,900 |
| US Naval Academy | 122 | 0 |
| West Point | 120 | 0 |
| MIT | 118 | 42,050 |
| Lehigh University | 118 | 43,220 |
| NYU-Poly | 117 | 39,565 |
| Babson College | 117 | 40,400 |
| Stanford | 114 | 54,506 |
Table 12.18
If the level of significance is 0.05 and the p-value is 0.06, what conclusion can you draw?
63.
If there are 15 data points in a set of data, what is the number of degree of freedom?
12.3 The Regression Equation ----------------------------
What is the process through which we can calculate a line that goes through a scatter plot with a linear pattern?
65.
Explain what it means when a correlation has an r 2 of 0.72.
Can a coefficient of determination be negative? Why or why not?
12.5 Prediction ---------------
67.
Recently, the annual number of driver deaths per 100,000 for the selected age groups was as follows:
| Age | Number of Driver Deaths per 100,000 |
--- |
| 16–19 | 38 |
| 20–24 | 36 |
| 25–34 | 24 |
| 35–54 | 20 |
| 55–74 | 18 |
| 75+ | 28 |
Table 12.19
For each age group, pick the midpoint of the interval for the x value. (For the 75+ group, use 80.)
Using “ages” as the independent variable and “Number of driver deaths per 100,000” as the dependent variable, make a scatter plot of the data.
Calculate the least squares (best–fit) line. Put the equation in the form of: ŷ = a + bx
Find the correlation coefficient. Is it significant?
Predict the number of deaths for ages 40 and 60.
Based on the given data, is there a linear relationship between age of a driver and driver fatality rate?
What is the slope of the least squares (best-fit) line? Interpret the slope.
Table 12.20 shows the life expectancy for an individual born in the United States in certain years.
| Year of Birth | Life Expectancy |
--- |
| 1930 | 59.7 |
| 1940 | 62.9 |
| 1950 | 70.2 |
| 1965 | 69.7 |
| 1973 | 71.4 |
| 1982 | 74.5 |
| 1987 | 75 |
| 1992 | 75.7 |
| 2010 | 78.7 |
Table 12.20
Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the ordered pairs.
Calculate the least squares line. Put the equation in the form of: ŷ = a + bx
Find the correlation coefficient. Is it significant?
Find the estimated life expectancy for an individual born in 1950 and for one born in 1982.
Why aren’t the answers to part e the same as the values in Table 12.20 that correspond to those years?
Use the two points in part e to plot the least squares line on your graph from part b.
Based on the data, is there a linear relationship between the year of birth and life expectancy?
Are there any outliers in the data?
Using the least squares line, find the estimated life expectancy for an individual born in 1850. Does the least squares line give an accurate estimate for that year? Explain why or why not.
What is the slope of the least-squares (best-fit) line? Interpret the slope.
69.
The maximum discount value of the Entertainment® card for the “Fine Dining” section, Edition ten, for various pages is given in Table 12.21
| Page number | Maximum value ($) |
--- |
| 4 | 16 |
| 14 | 19 |
| 25 | 15 |
| 32 | 17 |
| 43 | 19 |
| 57 | 15 |
| 72 | 16 |
| 85 | 15 |
| 90 | 17 |
Table 12.21
Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the ordered pairs.
Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
Find the correlation coefficient. Is it significant?
Find the estimated maximum values for the restaurants on page ten and on page 70.
Does it appear that the restaurants giving the maximum value are placed in the beginning of the “Fine Dining” section? How did you arrive at your answer?
Suppose that there were 200 pages of restaurants. What do you estimate to be the maximum value for a restaurant listed on page 200?
Is the least squares line valid for page 200? Why or why not?
What is the slope of the least-squares (best-fit) line? Interpret the slope.
Table 12.22 gives the gold medal times for every other Summer Olympics for the women’s 100-meter freestyle (swimming).
| Year | Time (seconds) |
--- |
| 1912 | 82.2 |
| 1924 | 72.4 |
| 1932 | 66.8 |
| 1952 | 66.8 |
| 1960 | 61.2 |
| 1968 | 60.0 |
| 1976 | 55.65 |
| 1984 | 55.92 |
| 1992 | 54.64 |
| 2000 | 53.8 |
| 2008 | 53.1 |
Table 12.22
Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least squares line. Put the equation in the form of: ŷ = a + bx.
Find the correlation coefficient. Is the decrease in times significant?
Find the estimated gold medal time for 1932. Find the estimated time for 1984.
Why are the answers from part f different from the chart values?
Does it appear that a line is the best way to fit the data? Why or why not?
Use the least-squares line to estimate the gold medal time for the next Summer Olympics. Do you think that your answer is reasonable? Why or why not?
71.
| State | # letters in name | Year entered the Union | Rank for entering the Union | Area (square miles) |
--- ---
| Alabama | 7 | 1819 | 22 | 52,423 |
| Colorado | 8 | 1876 | 38 | 104,100 |
| Hawaii | 6 | 1959 | 50 | 10,932 |
| Iowa | 4 | 1846 | 29 | 56,276 |
| Maryland | 8 | 1788 | 7 | 12,407 |
| Missouri | 8 | 1821 | 24 | 69,709 |
| New Jersey | 9 | 1787 | 3 | 8,722 |
| Ohio | 4 | 1803 | 17 | 44,828 |
| South Carolina | 13 | 1788 | 8 | 32,008 |
| Utah | 4 | 1896 | 45 | 84,904 |
| Wisconsin | 9 | 1848 | 30 | 65,499 |
Table 12.23
We are interested in whether or not the number of letters in a state name depends upon the year the state entered the Union.
Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx.
Find the correlation coefficient. What does it imply about the significance of the relationship?
Find the estimated number of letters (to the nearest integer) a state would have if it entered the Union in 1900. Find the estimated number of letters a state would have if it entered the Union in 1940.
Does it appear that a line is the best way to fit the data? Why or why not?
Use the least-squares line to estimate the number of letters a new state that enters the Union this year would have. Can the least squares line be used to predict it? Why or why not?
12.6 Outliers -------------
The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories of the building (beginning at street level).
| Height (in feet) | Stories |
--- |
| 1,050 | 57 |
| 428 | 28 |
| 362 | 26 |
| 529 | 40 |
| 790 | 60 |
| 401 | 22 |
| 380 | 38 |
| 1,454 | 110 |
| 1,127 | 100 |
| 700 | 46 |
Table 12.24
Using “stories” as the independent variable and “height” as the dependent variable, make a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables?
Calculate the least squares line. Put the equation in the form of: ŷ = a + bx
Find the correlation coefficient. Is it significant?
Find the estimated heights for 32 stories and for 94 stories.
Based on the data in Table 12.24, is there a linear relationship between the number of stories in tall buildings and the height of the buildings?
Are there any outliers in the data? If so, which point(s)?
What is the estimated height of a building with six stories? Does the least squares line give an accurate estimate of height? Explain why or why not.
Based on the least squares line, adding an extra story is predicted to add about how many feet to a building?
What is the slope of the least squares (best-fit) line? Interpret the slope.
73.
Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies to study their population. They gather data for the percent of new sparrow hawks in each colony and the percent of those that have returned from migration.
Percent return:74; 66; 81; 52; 73; 62; 52; 45; 62; 46; 60; 46; 38
Percent new:5; 6; 8; 11; 12; 15; 16; 17; 18; 18; 19; 20; 20
Enter the data into your calculator and make a scatter plot.
Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
Explain in words what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70% of the adults from the previous year have returned. What is the prediction?
The following table shows data on average per capita coffee consumption and heart disease rate in a random sample of 10 countries.
Yearly coffee consumption in liters2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7
Death from heart diseases221 167 131 191 220 297 71 172 211 300
Table 12.25
Enter the data into your calculator and make a scatter plot.
Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
Explain in words what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
Do the data provide convincing evidence that there is a linear relationship between the amount of coffee consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.
75.
The following table consists of one student athlete’s time (in minutes) to swim 2000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days:
| Swim Time | Heart Rate |
--- |
| 34.12 | 144 |
| 35.72 | 152 |
| 34.72 | 124 |
| 34.05 | 140 |
| 34.13 | 152 |
| 35.73 | 146 |
| 36.17 | 128 |
| 35.57 | 136 |
| 35.37 | 144 |
| 35.57 | 148 |
Table 12.26
Enter the data into your calculator and make a scatter plot.
Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
Explain in words what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
A researcher is investigating whether population impacts homicide rate. He uses demographic data from Detroit, MI to compare homicide rates and the number of the population that are White males.
| Population Size | Homicide rate per 100,000 people |
--- |
| 558,724 | 8.6 |
| 538,584 | 8.9 |
| 519,171 | 8.52 |
| 500,457 | 8.89 |
| 482,418 | 13.07 |
| 465,029 | 14.57 |
| 448,267 | 21.36 |
| 432,109 | 28.03 |
| 416,533 | 31.49 |
| 401,518 | 37.39 |
| 387,046 | 46.26 |
| 373,095 | 47.24 |
| 359,647 | 52.33 |
Table 12.27
Use your calculator to construct a scatter plot of the data. What should the independent variable be? Why?
Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot.
Discuss what the following mean in context.
The slope of the regression equation
The y-intercept of the regression equation
The correlation r
The coefficient of determination r2.
Do the data provide convincing evidence that there is a linear relationship between population size and homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.
77.
| School | Mid-Career Salary (in thousands) | Yearly Tuition |
---
| Princeton | 137 | 28,540 |
| Harvey Mudd | 135 | 40,133 |
| CalTech | 127 | 39,900 |
| US Naval Academy | 122 | 0 |
| West Point | 120 | 0 |
| MIT | 118 | 42,050 |
| Lehigh University | 118 | 43,220 |
| NYU-Poly | 117 | 39,565 |
| Babson College | 117 | 40,400 |
| Stanford | 114 | 54,506 |
Table 12.28
Using the data to determine the linear-regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer.
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9395 | https://math.stackexchange.com/questions/179961/intuition-behind-the-concept-of-indicator-random-variables | probability - Intuition behind the concept of indicator random variables. - Mathematics Stack Exchange
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Intuition behind the concept of indicator random variables.
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I am studying Randomized Algorithms chapter in the book "Introduction to Algorithms" by Cormen et al.
In this chapter the book introduces the concept of an indicator random variable and state that the expected value of an indicator random variable as :
I am having difficulty understanding why this is called an indicator random variable, specifically why indicator and random and how this concept is useful in analyzing algorithm timings . It has been some time since I studied probability in school . However , I am aware of the concept behind probability. So you can base your answer on this premise.
As you can see from the diagram all it is saying is that the expected value of an indicator random variable of an event is equal to the probability of that event . We already have the concept of probability , why should we know about this new concept which happens to be the same value as the probability ?
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edited Aug 7, 2012 at 15:35
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As the name implies, an indicator random variable indicates something: the value of I A I A is 1 1 precisely when the event A A occurs, and is 0 0 when A A does not occur (that is, A c A c occurs). Think of I A I A as a Boolean variable that indicates the occurrence of the event A A. This Boolean variable has value 1 1 with probability P(A)P(A) and so its average value is P(A)P(A). In terms of long-term frequencies, I A I A will have value 1 1 on roughly N⋅P(A)N⋅P(A) of N N trials of the experiment, and the long-term average value of I A I A on these N N trials will be approximately P(A)P(A).
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answered Aug 7, 2012 at 15:38
Dilip SarwateDilip Sarwate
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1 but why "random" ?Geek –Geek 2012-08-07 15:41:38 +00:00 Commented Aug 7, 2012 at 15:41
3 Random because you cannot be sure whether the next time you check I A I A, the variable will have value 1 1 or value 0 0, but you can be reasonably sure that over the next 10 6 10 6 observations of I A I A, the observed value of I A I A will be 1 1 roughly 10 6 P(A)10 6 P(A) times.Dilip Sarwate –Dilip Sarwate 2012-08-07 15:46:00 +00:00 Commented Aug 7, 2012 at 15:46
It will have 1 on roughly N⋅P(A)N⋅P(A) in N trials? So you're saying that the probability of having 1 on N outcomes equals N⋅P(A)N⋅P(A) which means N N (this one i get) times the same probability that we want to find, ie. P(A)P(A) (which is P(1)P(1)). This got me confused. You basically are doing a recursive definition for P(1)P(1) on N N trials if i understand correctly.KeyC0de –KeyC0de 2018-07-20 17:31:22 +00:00 Commented Jul 20, 2018 at 17:31
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An indicator function is a function that returns the value of 1 when something is true:
1[A]={1,0,A i s t r u e,A i s f a l s e.1[A]={1,A i s t r u e,0,A i s f a l s e.
Therefore, the expectation is essentially the same thing as computing the expected value of a Bernoulli random variable: the value 1 times the probability that A A is true, plus the value 0 times the probability it is not.
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answered Aug 7, 2012 at 15:38
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9397 | https://www.omnicalculator.com/math/average | Average Calculator
The average calculator will calculate the mean of up to 50 numbers. An interesting aspect of the calculator is that you can see how the mean changes as more values are entered. Before you use the calculator, you should know how to calculate the average, just in case you are without the internet and cannot access this calculator. Note that the mean is the same as average, and we can use these terms interchangeably.
🙋 There are also different methods to estimate the mean value. Our geometric mean calculator will help you understand the concept of the geometric mean and evaluate the result in a second.
How to use the average calculator
Eager to quickly learn how to use our average calculator and make the most of its functionalities? Just follow the steps below:
Start by entering values into the calculator. You can input up to 50 numbers, but you don't need to fill in all the entries if you don't require them.
As you enter your numbers, the calculator will automatically compute the average for you. The mean average is displayed as the sum of all the values you've entered, divided by the total number of values.
The interface is designed to be dynamic. Once you reach the fourth entry, the field for the fifth number will appear automatically, and this will continue as you add more numbers.
There's no need to press a calculate button; the average updates instantly after every entry. So, you can add or remove numbers as needed, and the calculator will adjust the average accordingly.
For instance, if you're looking to calculate the average score of a class test, simply input each student's score into the calculator. If the scores are 56, 75, 88, 45, and 92, the calculator will determine the average to be 71.2.
The calculator can also be used for larger datasets. Suppose you have a set of 50 temperature readings from a science experiment; just keep inputting each reading into the calculator. As you input the 50th reading, the average of all 50 temperatures will be automatically calculated and presented to you.
Explore further to learn more about the mean average concept, its significance in various fields, and how it's mathematically derived.
How to calculate an average
The average of a set of numbers is simply the sum of the numbers divided by their total number. For example, suppose we want an average of 24,55, 17, 87 and 100. Simply find the sum of the numbers: 24 + 55 + 17 + 87 + 100 = 283 and divide by 5to get 56.6. A simple problem like this one can be done by hand without too much trouble, but for more complicated numbers involving many decimal places, it is more convenient to use our calculator. Note that the average rating calculator does a similar math — it calculates an average rating given the number of votes with values from 1 to 5.
Similar concepts involving averages
The weighted average calculator lets you assign weights to each number. A number weighting is an indicator of its importance. A common example of a weighted mean is the grade point average (GPA). Check our dedicated GPA calculator for more details. To do this by hand, follow these steps:
Multiply the value of the letter grade by the number of credits in the class.
Do this for all the classes and take the sum.
Divide the sum by the total number of credits.
Suppose the grades are an A for a 3-credit class, two B's for 4-credit classes, and a C for a 2-credit class. Using the standard value of 4 for an A, 3 for a B, and 2 for a C, the grade point average is GPA = [4×3 + 3×4 + 3×4 + 2×2]/(3 + 4 + 4 + 2) = 40/13 = 3.08
Note that the average calculator will compute the average for all values that are weighted equally, in contrast to the tools linked above. In statistics, we treat the mean as a measure of central tendency.
Behind the scenes of the average calculator
I'm Mateusz, the founder of Omni Calculator, and I brought my extensive expertise to the development of our average calculator. With years of experience managing financial projects, I understand the pivotal role of accurate and efficient analysis in decision-making processes.
The concept of the average calculator was born out of my recognition of the need for a streamlined, intuitive tool that could simplify the calculation of averages for both my team and clients. My goal was to create a calculator that would not only expedite the analysis of data sets but also be accessible to individuals at all levels of statistical knowledge.
Now, I regularly employ the average calculator in my professional toolkit to swiftly compute averages during analysis sessions. This tool has proven invaluable in providing clear, instantaneous insights into complex data sets, enhancing productivity and decision-making accuracy.
In developing the average calculator, we've meticulously ensured the quality and reliability of the content. Each feature is peer-reviewed by experts to guarantee precision and proofread by native English speakers for clarity and accuracy.
FAQs
What are the 4 averages?
The four so-called averages are the mean, median, mode, and range. The mean is what you typically think as the average — found by summing all values and dividing the sum by the number of values. The median is the middle value of the set (or the average of the two middle values if the set has an even number of elements). The mode is the most frequent piece of data, and the range is the difference between the highest and lowest values.
Why do we calculate average?
We calculate averages because they are a useful, handy, and quick way to describe large data. Instead of having to trawl through hundreds or thousands of pieces of data, we have one number that succinctly summarises the whole set. While there are some problems with averages, such as outliers showing an inaccurate average, they are useful to compare data at a glance.
Why are averages misleading?
Averages tend to be distorted by extreme values: even one can change the average dramatically. For instance, suppose that in a group of five people, four make $1,000 per month each while the fifth one earns $16,000. Then their average salary $4,000 exceeds four times the typical salary of $1,000 while being much lower than the highest earning of $16,000. The average salary in this group gives a misleading view on its members' real earnings.
How do I calculate my grade average?
To calculate your grade average:
Multiply each grade by the credits or weight attached to it. If your grades are not weighted, skip this step.
Add all of the weighted grades (or just the grades if there is no weighting) together.
Divide the sum of weighted grades by the sum of the weights. If your grades are not weighted, simply divide the sum of the grades by the number of grades.
The resulting quotient is your final grade average.
How do I calculate a weighted average?
To evaluate a weighted average:
Multiply each number by its weight.
Add all of the weighted numbers together.
Divide the sum of numbers by the sum of weights.
The resulting quotient is the weighted average.
Is average better than mode?
There is no easy answer to whether the average is better than the mode — it depends entirely on your data set. If the data is normally distributed and has no outliers, then you should probably use the average, as it will give you the most representative value. The mode, however, is more robust and will present the most common value, regardless of any outliers. The mode should always be used with categorical data — that is, data with distinct groups — as the groups are not continuous.
What is better, average or median?
Whether you should use the average or the median depends on your data. If the data is normally distributed and has no outliers, then you should probably use the average, although the value will be quite similar to that for the median. If the data is heavily skewed, the median should be used as it is less affected by outliers.
How do you calculate the average percentage in Excel?
Although it is easier to use the Omni Average Calculator, here is a recipe for calculating the average percentage in Excel:
Input your desired data, e.g., from cells A1 to A10.
Highlight all cells, right click, and select Format Cells.
In the Format Cells box, under Number, select Percentages and specify your desired number of decimal places.
In another cell, input =AVERAGE(cell 1, cell 2,…). In our example, this would be =AVERAGE(A1:A10).
Enjoy your average!
Can you average averages?
You can average averages, but it is often very inaccurate and should be done carefully. Let's say you have two countries, one with a population of 10 million and a GDP of $30,000 and another with 10,000 inhabitants and $2,000 GDP. The average GDP per country is $16,000, while the average GDP per person is ~$30,000, both vastly different figures showing vastly different things — so be careful.
Is the average of averages accurate?
The average of averages is not accurate most of the time due to two main factors: lurking variables and weighted averages. Lurking variables are where important information is lost by taking the average of averages. The other issue is not weighting averages when needed. If, say, the number of visitors changes each month, by not weighting against the number of people, information will be lost.
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9398 | https://www.childrenshospital.org/conditions/total-anomalous-pulmonary-venous-return | Current Environment: Production
Home
Total Anomalous Pulmonary Venous Return (TAPVR)
Listen
What is total anomalous pulmonary venous return?
Total anomalous pulmonary venous return (TAPVR), also known as total anomalous pulmonary venous connection (TAPVC), is a rare heart defect in which the blood vessels that drain the lungs (pulmonary veins) are not connected normally to the heart. Instead, the pulmonary veins are redirected abnormally to other chambers of the heart. About 1 in every 20,000 babies is born with TAPVR.
In order to get blood to the body, most babies with TAPVR also have another heart defect, called atrial septal defect, which is a hole from the right atrium to the left atrium.
There are four major types of TAPVR. Each is based on where the pulmonary veins connect to the heart:
Supracardiac, where the pulmonary veins make an abnormal connection above the heart
Cardiac, where the pulmonary veins connect behind the heart
Infracardiac, where the pulmonary veins connect below the heart
Mixed, a combination of any of the connections above
TAPVR can occur with obstruction, meaning that some of the draining blood vessels are obstructed. This can cause high blood pressure in the lungs (pulmonary hypertension) and can be a surgical emergency.
Children with TAPVR will need surgery in infancy to repair the problem.
Total Anomalous Pulmonary Venous Return | Symptoms & Causes
What are the symptoms of total anomalous pulmonary venous return?
Most babies born with total anomalous pulmonary venous return (TAPVR) are very ill soon after birth. Symptoms may include:
A bluish tint to the skin and lips
Trouble breathing
Rapid breathing
Poor feeding or poor growth
If your child has any of these symptoms, your pediatrician may refer you to a pediatric cardiologist for testing.
What are the causes of total anomalous pulmonary venous return?
In many cases, we don’t know what causes TAPVR. It occurs because of abnormal development of the heart’s pulmonary veins during early fetal growth.
Some congenital heart defects may have a genetic link, causing heart problems to occur more often in certain families. Most often, though, this heart defect occurs by chance, with no clear reason for its development.
Total Anomalous Pulmonary Venous Return | Diagnosis & Treatments
How is total anomalous pulmonary venous return diagnosed?
In some cases, total anomalous pulmonary venous return is found before birth on a fetal echocardiogram.
Most babies with total anomalous pulmonary venous return (TAPVR) have symptoms on the day they’re born. Severe TAPVR can usually be diagnosed promptly based on symptoms and tests, including chest X-ray and cardiac ultrasound. These babies are generally admitted immediately to the hospital.
Babies with less severe TAPVR may have symptoms in the first few days of life. If your newborn baby is born with a bluish tint to the skin or is having difficulty breathing, you may be referred to a pediatric cardiologist to determine a diagnosis.
Your baby’s doctor may order one or more additional tests to diagnose or confirm a diagnosis of TAPVR:
Cardiac MRI
Electrocardiogram (ECG or EKG)
Cardiac catheterization
How is total anomalous pulmonary venous return treated?
Newborns with severe TAPVR will need emergency surgery shortly after birth. They often need to be admitted to the cardiac intensive care unit (CICU) and require intensive support with medications and a ventilator (breathing machine).
Some babies with severe TAPVR may need a specialized life support system called ECMO (extracorporeal membrane oxygenation), an advanced technology that functions as a replacement for a critically ill child's heart and lungs.
Babies with less severe TAPVR usually have surgery in the days or weeks after they're diagnosed.
The goal of surgery for TAPVR is to restore normal connections of the pulmonary veins to the heart, alleviate any obstructions or narrowing of the pulmonary veins, to tie up any vessels that have developed and to close the atrial septal defect (ASD).
What is the long-term outlook for children with TAPVR?
Thanks to updates in surgical techniques for repairing TAPVR, the long-term outlook is continually improving.
Children who have had a repair for TAPVR will require lifelong checkups with a cardiologist to make sure their veins remain open. If the veins become narrowed, they may need catheterizations or surgeries to repair these veins.
How we care for total anomalous pulmonary venous return
Our team in the Boston Children’s Department of Cardiac Surgery treat some of the most complex pediatric heart conditions in the world. Our specialized clinicians can often diagnose this condition during fetal echocardiogram.
Our cardiac surgeons have vast experience in repairing this defect, and work with nurses and doctors who are focused on providing expert care after surgery. Adult patients with TAPVR are followed by Boston Children’s cardiologists with special training for adults with congenital heart problems.
Total Anomalous Pulmonary Venous Return | Programs & Services
Programs
Boston Adult Congenital Heart (BACH) and Pulmonary Hypertension
Program
The Boston Adult Congenital Heart (BACH) and Pulmonary Hypertension Program offers a full range of inpatient and outpatient clinical services to adults with congenital heart disease and pulmonary hypertension.
Learn more about Boston Adult Congenital Heart (BACH) and Pulmonary Hypertension
Cardiac Neurodevelopmental Program
Program
The Cardiac Neurodevelopmental Program uses a compassionate, family-centered approach to diagnose and treat neurodevelopmental disorders.
Learn more about Cardiac Neurodevelopmental Program
Cardiovascular 3D Modeling and Simulation Program
Program
The Cardiovascular 3D Modeling and Simulation Program has created and institutionalized a standard of preoperative planning for heart surgeons.
Learn more about Cardiovascular 3D Modeling and Simulation Program
Heterotaxy Program
Program
The Heterotaxy Program serves children with heterotaxy, a rare heart defect.
Learn more about Heterotaxy Program
Outpatient Cardiology
Program
Our clinic provides comprehensive evaluation and coordinated care for infants, children, and adults with various heart, and heart-related illnesses, diseases, and conditions.
Learn more about Outpatient Cardiology
Pulmonary Vein Stenosis Program
Program
Specialists at the Pulmonary Vein Stenosis Program are known for treating the most complex cases of PVS and our expertise in specialized treatments.
Learn more about Pulmonary Vein Stenosis Program
Departments
Cardiac Imaging
Department
The Division of Cardiac Imaging serves children and adults with congenital heart disease.
Learn more about Cardiac Imaging
Cardiac Surgery
Department
The Department of Cardiac Surgery has grown to become the largest pediatric cardiology center in the U.S. and the most specialized in the world.
Learn more about Cardiac Surgery
Cardiology
Department
The Department of Cardiology at the Benderson Family Heart Center is the largest pediatric cardiology center in the United States and one of the most specialized in the world.
Learn more about Cardiology
Centers
Benderson Family Heart Center
The Benderson Family Heart Center treats the full spectrum of heart disorders, including the rarest and most complex congenital heart defects.
Learn more about Benderson Family Heart Center
Total Anomalous Pulmonary Venous Return | Contact Us
Contact the Benderson Family Heart Center
617-355-4278
Email
heart@childrens.harvard.edu
Request an Appointment Request a Second Opinion |
9399 | https://www.academia.edu/83456693/Heat_transfer_a_practical_approach_by_y_a_cengel | (PDF) Heat transfer a practical approach by y a cengel
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Heat transfer a practical approach by y a cengel
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Wael Ayman
2022, Hdudkd
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Abstract
Application Areas of Heat Transfer 3 Historical Background 3
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Heat Transfer Principles and Applications
Manuel Gutierrez Pinto
This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein). Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility.
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Heat transfer--A review of 1993 literature
K. Tamma
1996
This review surveys and characterizes papers comprising various fields of heat transfer that were published in the literature during 1993. It is intended to encompass the English language literature, including English translations of foreign language papers, and also includes many foreign language papers for which English abstracts are available. The literature search was inclusive, however, the great number of publications made selections in some of the review sections necessary. Several conferences during 1993 were devoted to heat transfer or included heat transfer topics in their sessions. They will be briefly discussed in chronological order in this :section. The 6th International Symposium on Transport Phenomena in Thermal Engineering was organized by the Pacific Center of Thermal Fluids engineering in Seoul, Korea on 9-13 May. The program included heat exchangers, electronic equipment cooling and thermal control of equipment among its topics. Papers are included in a bound volume. The 38th ASME International Gas Turbine and Aeroengine Congress and Exposition (ASME Turbo Expo 93) 2427 May at Cincinnati, Ohio, U.S.A. contained in its program sessions on heat transfer, unsteady effects, external, internal and film cooling. Papers are available at the ASME order department. The AIAA 28th Thermophysics Conference, held in Orlando, Florida, U.S.A. presented among others papers on conduction, convection, radiation, phase change and thermal protection of spacecraft. Two conferences 'organized by the Institute for Numerical Methods, University of Swansea, U.K. are of interest to workers in heat transfer: Numerical Methods for Thermal Problems, held 11-16 July and Numerical Methods in Laminar and Turbulent Flow, 18-23 July. Some papers are published in the Journal of the Institute. The Fourth International Conferences on Circulating Fluid Beds in Somerset, Pennsylvania, U.S.A., l-5 August contained a session on heat and mass transfer. Papers are published in the conference proceedings. The conference is sponsored by the American Institute of Chemical Engineers. The 29th National Heat Transfer Conference in Atlanta, Geor-Conduction-convection and$ow The influence of convection and related flow effects in conduction heat transfer included investigation of heat exchanges in heterogeneous flows [173A], conjugate forced convection-conduction analysis [174A], transient heat transfer due to conduction and internal sources in a slab [175A] and conjugate problems dealing with conduction and free convection [176A].
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Essentials of Heat Transfer: Principles, Materials, and Applications, by Massoud Kaviany
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The accuracy of the Content should not be relied upon and should be independently verified with primary sources of information. Taylor and Francis shall not be liable for any losses, actions, claims, proceedings, demands, costs, expenses, damages, and other liabilities whatsoever or howsoever caused arising directly or indirectly in connection with, in relation to or arising out of the use of the Content. This article may be used for research, teaching, and private study purposes. Any substantial or systematic reproduction, redistribution, reselling, loan, sub-licensing, systematic supply, or distribution in any form to anyone is expressly forbidden.
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Heat transfer—a review of 1991 literature
Kim Stelson
International Journal of Heat and Mass Transfer, 1992
INTRODUCTION THIS REVIEW surveys and characterizes papers comprising various fields of heat transfer that were published in the literature during 1991. It is intended to encompass the English language literature, including English translations of foreign language papers, and also includes many foreign language papers for which English abstracts are available. The literature search was inclusive, however, the great number of publications made selections in some of the review sections necessary. Several conferences during 1991 were devoted to heat transfer or included heat transfer topics in their sessions. They will be briefly discussed in chronological order in this section. The 1991 International Seminar on Heat and Mass Transfer in Porous Media was sponsored by the International Center for Heat and Mass Transfer on 20-24 May at Dubrovnik, Yugoslavia. The majority of the papers are available at the Center in a bound volume. The ASME Turbo Expo-Land, Sea, and Air sponsored by the International Gas Turbine Institute and held on 3-6 June at Orlando, Florida included in its program session on ceramic technology, fii cooling, coatings and composites. Printed papers are available through the ASMB order department. The ACHEMA 91, International Meeting on Chemical Engineering and Biotechnology, connecting with a large exhibition, was held at Fra&mt/Main, Germany on !+-15 June. The Second World Conference on Experimental Heat Transfer, Fluid Mechanics and Thermodynamics was organized by its assembly on 23-28 June at Dubrovnik, Yugoslavia in 10 keynote lectures, 7 panel discussions, 25 invited lectures, 209 contributed papers and open forum sessions. Papers are available at Else&r Science Publishing Co. Raymond Viscanta was awarded the first Nusselt-Reynolds Prim. The 26th Thermophysics Conference was organized by the American Institute of Aeronautics and Astronautics on 24-27 June in Honolulu, Hawaii with sessions on spacecraft coatings, solidification and convection, thermal analysis, heat pipes, and hypersonic non equilibrium flow. Conference pmceedings are available and selected papers are published in AIAA journals. 7'7~ 7th International Conference on Numerical Me&oak in Luminar and Turbulent Flow was sponsored by Lockheed Missile and Space Co.
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PART 3 INTRODUCTION TO ENGINEERING HEAT TRANSFER Introduction to Engineering Heat Transfer
Cragon C.
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Part I - Fundamental Principles of Heat Transfer
Whitaker, Stephen
Fundamental Principles of Heat Transfer, 1977
This book is intended to provide a comprehensive treatment of the fundamental aspects of conduction, convection and radiation at an introductory level.
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Heat transfer—a review of 2002 literature
behnam abbasi
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Heat transfer original studies –Theory and applications
Alexandra Raicu
IOP Conference Series: Materials Science and Engineering, 2019
Thermal stresses, thermal fatigue, materials’ constants variation with the temperature, residual stresses and micro-structure of the parts after the solidification process are heat transfer phenomenon related problems. The calculation instrument of a given age of science and technology has a great influence on the mathematical solutions and the according applications in engineering. The paper presents a synthesis, including the authors’ contributions in the heat transfer problems, which may be regarded in a broader context of the distinct branches of engineering that are the fields of expertise of the authors. The first step is to analyse the theoretical aspects to be mastered in order to wisely apply them in concrete engineering problems. In this way, there are presented the basic equations used by the authors in their analytical studies and the background of the numerical methods used in engineering, i.e. the finite difference method and the finite element method. In the following...
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Fundamental Principles of Heat Transfer - Part II
Whitaker, Stephen
This second part of Fundamental Principles of Heat Transfer deals primarily with the applications of convection and radiation. Emphasis is again placed on careful and complete theoretical developments.
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Heat Transfer Research in Sweden: Historical and Current Activities
Bengt Sundén
Heat Transfer Engineering, 2012
This article describes some historical and significant contributions in the field of heat transfer from various Swedish universities and researchers. It also gives an overview of what kind of heat transfer-related research has been going on recently, as well as some current activities. Traditionally, research projects have been associated with the Ph.D. education programs and in these, common theses for the licentiate of engineering and doctor of philosophy appear, besides publications in journals, conference proceedings, and technical reports. The only location with a Ph.D. program in heat transfer and the main research focus on this subject is at Lund University. Nevertheless, heat transfer research is carried out at various universities, but not as the major subject.
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Heat transfer--A review of 2003 literature
K. Tamma
International Journal of …, 2006
The present paper is intended to encompass the English language heat transfer papers published in 2003, including some translations of foreign language papers. This survey, although extensive cannot include every paper; some selection is necessary. Many papers reviewed herein relate to the science of heat transfer, including numerical, analytical and experimental works. Others relate to applications where heat transfer plays a major role not only in man-made devices, but in natural systems as well. The papers are grouped into categories and then into sub-fields within these categories. We restrict ourselves to papers published in reviewed archival journals. Besides reviewing the journal articles in the body of this paper, we also mention important conferences and meetings on heat transfer and related fields, major awards presented in 2003, and books on heat transfer published during the year.
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Heat transfer––a review of 2001 literature
Sean Garrick
International Journal of Heat and Mass Transfer, 2010
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Heat transfer — a review of 1992 literature
K. Tamma
International Journal of Heat and Mass Transfer, 1994
Composite(s) or layered material(s)/anisotropic media Papers addressing heat contact in composite construction, thermal expansion issues, thermal cracks, multi-layered models, influences due to various heat loads and boundary effects, thermal calculations in composite wall(s) and anisotropic media, effective thermal conductivity approximations, multilayered media, graphite fiber/polymer matrix composites, transverse thermal diffusivity evaluations, thermal resistance in multi-layer composites appear in refs. [18A-39A]. Injluence of laser/pulse heat and thermal propagation The effect(s) due to sudden laser impact on materials, pulse heat loading situations and thermal shock(s) are addressed in this subcategory. Of mention are also publications involving thermal wave propagation problems under the influence of a hyperbolic heat conduction mode. Tbe papers in this subcategory are identified in refs. [40A-53A]. Conduction in arbitrary geometries and complex configurations In this subcategory, papers dealing with simplified models for homogeneous cylinders, temperature distribution in journal bearings and spherical ridges and troughs in a plane are addressed. These are identified in refs. [54A-56A]. Models/methods and approaches and numerical studies This subcategory continues to attract a wide range of interest in the development of accurate models, and modeling/analysis approaches including numerical studies for a variety of physical situations involving heat transfer due to conduction. Finite difference, finite element, boundary element methods and the like have been employed for a wide range of research investigations. These appear in refs. [57A-8lA]. Thermo-mechanical problems The influence of temperature effects on materials and components in particular, thermal-stresses and thermally induced stress waves are addressed in this subcategory. Both linear and nonlinear thermal-stress issues are addressed including theoretical/numerical and experimental studies. These papers are identified in refs. [82A-145A]. Inverse heat conduction Inverse heat conduction aspects including development of methods, substitution of multi-dimensional problems, prediction under the influence of heat sources and various types of boundary conditions, regularized solutions, explicitly sometimes, numerical approximations and simulations appear in refs. [146A-148A]. Miscellaneous conduction studies Various types of miscellaneous heat conduction problems have been studied in literature and appear in refs. [149A-178A]. Special applications Specialized applications involving heat conduction via theoretical, numerical/approximate method sand/or experimental investigations are addressed in refs. [179A-209A]. Electronic packaging Various theoretical, experimental and numerical studies dealing with thermal heat transfer characteristics, influence of heat sources, contact issues, prediction of temperature field and the like in microelectronic packaging appear in refs. [210A-232A]. BOUNDARY LAYERS AND EXTERNAL FLOWS The research on boundary layer and external flows during 1992 has been categorized as follows: flows influenced externally, flows with special geometrir Heat transfer-a review of 1992 literature 1287 effects, compressible and high-speed flows, analysis and modeling techniques, unsteady flow effects, films and interfacial effects, flows with special fluid types, and conjugate heat transfer situations. Unsteady eflects External effects Several papers documented the effects of an imposed streamwise pressure gradient [lB, 3B, lOB, IJB-lSB]. Some included the effects of acceleration on the stabiiity of the boundary layer indicating conditions of transition and reiaminarization. One addressed the effect of agitation. Other effects discussed were variations of the thermal boundary condition, raising of the external (free stream) disturbance level, imposition of longitudinal vortex arrays, and application of sonic disturbances [2B, 4B-9B, llB, 12B] Heat transfer-a review of 1992 literature 1289 periodically perturbed flow and heat transfer due to electrohydrodynamical forcing was treated in refrigerant 113. Transient flow was examined in the presence of twisted oval tubes and for time-varying temperature field in a thick-walled pipe [88C-97C]. Multi-phase jlow in ducts Multi-phase flow in ducts was examined in over a wide range of physical situations. Solid-gas two-phase flow was considered in the following studies: submicron particle flow in a cooled laminar tube considering convection, diffusion, and thermophoresis; gas-particle flow of nonisothermal turbulent swirling flow was studied in a cylindrical channel; and mixtures of combustible and non-combustible particles in gas were studied. Gas (typically air) and liquid flow was examined in the presence of wave motion; the interfacial heat transfer was investigated. A new correlation was presented for the air-water flow in horizontal rectangular channels. Air-water flow was also treated in the stratified arrangement found in certain rod bundles. Polydisperse aerosols in cooled laminar flow was studied theoretically and experimentally. Three-phase flow of water (ice-steamliquid) was examined and compared to single-phase flow. A three-phase system of air-water-sand was also studied in the presence of a tube bundle [98C-109C]. Non-Newtonian jlow in ducts Non-Newtonian fluid flow in ducts was a particularly active research area during the year. Power law tluids in concentric annular ducts were examined, where both the heat transfer and pressure drop were considered. Viscous dissipation effects on heat transfer to power law fluids was studied in arbitrary duct cross sections. Non-Newtonian flow was investigated in a variety of geometries including: axisymmetric sudden expansion (with applications to extruston processes and capillary rheometry); flow in a rectangular duct (viscoelastic, inelastic, and polymerizable fluids were considered); Couette flow in an annuli with moving outer cylinder (power law fluids); and viscoelastic fluid flow in a screw-wall channel. A second law analysis of non-Newtonian forced convection was also presented [l lOC-12OC]. Miscellaneous duct jlow A handful of studies did not fit well into the categories highlighted above. These investigations included cryogenic applications (e.g. liquid helium), high-speed gas flows, and fluidized bed channel flow [121C-128C].
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Fundamental Principles of Heat Transfer - Part I
Whitaker, Stephen
This heat transfer text provides a comprehensive treatment of the fundamental aspects of conduction, convection, and radiation. Emphasis is placed on careful and complete theoretical developments, and numerous solved example problems and design problems are included to illustrate practical applications of fundamental principles. The appendices provide properties of materials, tables of mathematical functions, author index, and subject index. (LCL)
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Solutions to Problems in Heat Transfer
Dr. Osama M Elmardi
During my long experience in teaching several engineering subjects I noticed that many students find it difficult to learn from classical textbooks which are written as theoretical literature. They tend to read them as one might read a novel and fail to appreciate what is being set out in each section. The result is that the student ends his reading with a glorious feeling of knowing it all and with, in fact, no understanding of the subject whatsoever. To avoid this undesirable end a modern presentation has been adopted for this book. The subject has been presented in the form of solution of comprehensive examples in a step by step form. The example itself should contain three major parts, the first part is concerned with the definition of terms, the second part deals with a systematic derivation of equations to terminate the problem to its final stage, the third part is pertinent to the ability and skill in solving problems in a logical manner. This book aims to give students of engineering a thorough grounding in the subject of heat transfer. The book is comprehensive in its coverage without sacrificing the necessary theoretical details. The book is designed as a complete course test in heat transfer for degree courses in mechanical and production engineering and combined studies courses in which heat transfer and related topics are an important part of the curriculum. Students on technician diploma and certificate courses in engineering will also find the book suitable although the content is deeper than they might require. The entire book has been thoroughly revised and a large number of solved examples and additional unsolved problems have been added. This book contains comprehensive treatment of the subject matter in simple and direct language. The book comprises eight chapters. All chapters are saturated with much needed text supported and by simple and self-explanatory examples.
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Numerical Heat Transfer, Part A: Applications
A. Liakopoulos
2010
and-conditions-of-access.pdf This article may be used for research, teaching and private study purposes. Any substantial or systematic reproduction, redistribution , reselling , loan or sub-licensing, systematic supply or distribution in any form to anyone is expressly forbidden. The publisher does not give any warranty express or implied or make any representation that the contents will be complete or accurate or up to date. The accuracy of any instructions, formulae and drug doses should be independently verified with primary sources. The publisher shall not be liable for any loss, actions, claims, proceedings, demand or costs or damages whatsoever or howsoever caused arising directly or indirectly in connection with or arising out of the use of this material.
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Heat Transfer Engineering
Aklilu Tesfamichael
Heat Transfer Engineering, 2020
In this study, the cooling and dehumidification performance of a photovoltaic powered thermoelectric air conditioning system is investigated under tropical climate. The system is first investigated under a controlled environment with different flow arrangements of the hot junction cooling air stream. Better temperature reduction was observed when the crossflow arrangement is used. Besides, it delivered the best reduction when 1.7 A current level is supplied to each thermoelectric module (TEM). The system performance was then investigated in a 3.6 m 3 room application under real weather conditions. When 1.7 A current level supplied to each TEM, it delivered 181 W and 0.14 ltr/hour average cooling load and moisture removal rate, respectively. The corresponding indoor room temperature is 4 C less than the unconditioned reference room, 7 to 7.6 C lower than the outdoor in the afternoon, and the indoor relative humidity varies between 58 to 83%. The indoor thermal comfort analysis indicated that the indoor modified predicted mean vote is 0.24 and classified as Class A in the ISO thermal acceptability, coupled with 26.6% of mean dissatisfaction. Hence, the photovoltaic thermoelectric cooling system could be a potential option to substitute the conventional air conditioning system as it is Freon-free and no moving parts.
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Analytical Heat Transfer
Je-chin Han
Analytical Heat Transfer, 2016
This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers.
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