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2700 | https://www.youtube.com/watch?v=WrEcQsa-1ME | Monotonicity Theorem
Khan Academy
9090000 subscribers
1155 likes
Description
229816 views
Posted: 12 Jun 2008
Using the monotonicity theorem to determine when a function is increasing or decreasing.
65 comments
Transcript:
0 seconds Welcome back. Well, I've been requested
to do several problems by our friend [? Akosh. ?] So I thought I would
keep doing them. I'm skipping around
a little bit. Because I think if I did
all of them, it would just generate too many videos. But I encourage all of you all
to let me know if you feel that there's something that
I might have missed. But anyway, I'm just going
through some of the problems he gave. So this one-- and I actually
almost find it funny, because they use such formal language
for something that's actually a pretty intuitive concept. Well they say, use the
monotonicity theorem to find where the given function is
increasing, and when it is decreasing. Monotonicity theorem. It make sound very serious. Well all the monotonicity
theorem-- at least if I'm remembering things properly--
all it says is when the derivative is positive, when f
prime of x is greater than 0, your function is increasing. And when the derivative is
negative, your function is decreasing. And why am I kind of
disparaging of this? Well what's the derivative? The derivative is the slope. And you learned in Algebra
1, that if the slope is positive, the graph of
the line is increasing. And if the slope is
negative, the graph of the line is decreasing. And the only thing that's
different now is that this function isn't necessarily
a line, it's a function. So it could be a curve. But having a positive slope
means that the value of the function is increasing. For every change it x, there's
a positive change in y. And similarly, a negative slope
says that for every change in x, no matter how small, there's
a negative change in y. So there's nothing fancy here. But let's see, they
gave some examples. So they want us to use this
theorem to find where the given function is increasing,
and when it is decreasing. So the first function
they give is f of x is equal to 3x plus 3. What's the derivative of this?
f prime of x is equal to 3. For any value of x
this is a positive. Right? F prime of x is positive
for all values of x. So this thing, using the
monotonicity theorem is positive. Sorry, it's increasing
for all values of x. And you could have done
that in the 9th grade. How? Well you would have said, this
is the slope y-intercept. The slope is 3. So this is an
increasing function. The function will look
something like this. The y-intercept is 3. And the slope is 3. So this function looks
something like that. And so it's increasing over
from minus infinite to positive infinite over that interval,
if you wanted to be fancy. But this is really just a lot
of fancy terminology to talk about, I think, something
fairly straightforward. Anyway, the next one looks a
little bit more interesting. 3 minutes H of z is equal to z to the
fourth over 4, minus-- and I hope I'm reading this right--
4 z to the third over 6. So let's see if we can figure
out when this is increasing, and when this is decreasing,
when the function is increasing or decreasing. So what's the
derivative of this? H prime of z is equal to 4. This will cancel out. So it's z to the third,
4z to the third over 4. So it's z to the third minus 3
times 4 is 12z squared over 6. So it's 2 2z squared. So we just have to figure out
when is this function greater than 0, and when is this
function less than 0. And to figure this out, we
really just have to break out our algebra toolkit to figure
out well one, when does this function equal 0? So let's set z-cubed minus
2z squared is equal to 0. We could factor
out a z-squared. And then we have z
minus 2 is equal to 0. So we know that either
z-squared is equal to 0, or z minus 2 is equal to 0. So the points at which the
derivative is 0, so H prime of 0 is equal to 0,
and H prime of 2. And that's just our Algebra 2. So H prime of 2 is equal to 0. And so we just have to figure
out what happens in the interval maybe when we're
below 0, between 0 and 2, and then above 2. So what is the derivative? So when we're below 0,
let's just take a point. Let's say H prime of negative
1 is equal to what? That is equal to negative 1 to
the third is negative 1, minus 2 times negative 1 squared. Well negative 1 squared
is just positive 1. So it's minus 2. So it equals minus 3. OK, let's take a point
in between these two. Well 1 is in between them. So let's just take H prime
of 1 is equal to 1 minus 2. 5 minutes, 42 seconds Lets make sure I'm
doing that right. Because it's 1 squared. So it equals negative
1, H prime of 1. And then what happens when we
go of z values greater than 2. So let's try 3. H prime of 3 is equal to what? H prime of 3 is equal to 27--
what's minus 2 times 9. It's equal to 9,
right, 18, right. So it's positive. So, what do we know? We know that in all fairness,
this is kind of an interesting problem. Because the derivative
is negative. So if we were to draw the
derivative of this, what are the interesting points? It's 0. So that's the x-axis. 6 minutes, 41 seconds That's 0. And then we had 2. And I'm going to draw
the derivative. I'm not drawing the
actual function. Because the monotonicity
theorem, we care about whether the derivative
is positive or negative. So what this tells us is
that the derivative, it was a negative when we're
to left of 0, right? Because we just
took a test point. Because we know it's 0 here. But then what happened? It's not like it
got positive then. It went back negative
again when we tried to derivative it once. It probably does
something like this. 7 minutes, 17 seconds Now what would it be? Then it goes up at 2 where
it's greater than 0. Right. I was confusing
myself for a second. Right. So the derivative, it starts. But it never goes
above right there. So the derivative probably
has a bit of a maximum point there or something. This is the
derivative, remember. That's why I was
confusing myself. And then it probably
flips around, and then goes above there. But from the monotonicity
theorem, what do we care about? We care about the intervals
where the derivative is positive, and the
derivative is negative. So if the derivative is
positive for all of these values, for z
is greater than 2. So we could say, using the
monotonicity theorem, this function is increasing
when z is greater than 2. So let me write that. 8 minutes, 24 seconds And then we can say that the
function is decreasing. We could say flat or
decreasing, right? Because the slope of the
function is 0 right here. And remember this is the
graph of the derivative. So we could say flat
or decreasing when z is less than 2. 8 minutes, 47 seconds You normally do consider flat,
monotonic in one direction. So you could say monotonic
increasing would still include something that
kind of flattens out. Well anyway, I'm
pushing ten minutes. And I don't think I have time
for the next problem, which I might do in the next video. But hopefully you found
that a little bit helpful, and not too confusing. I will see you in
the next video. 9 minutes, 13 seconds |
2701 | https://en.wikipedia.org/wiki/Medulloblastoma | Jump to content
Medulloblastoma
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From Wikipedia, the free encyclopedia
Most common type of primary brain cancer in children
Medical condition
| Medulloblastoma | |
--- |
| | |
| CT scan, showing a tumorous mass in the posterior fossa, giving rise to obstructive hydrocephalus, in a six-year-old girl | |
| Pronunciation | /məˌdʌloʊblæˈstoʊmə/ |
| Specialty | Neuro-oncology, neurosurgery |
| Symptoms | Headaches Nausea Vomiting Tiredness Clumsiness Dizziness Change in vision Handwriting problems |
| Usual onset | Between 5 and 9 years old |
| Prognosis | Five-year survival rate: 72.1% |
| Frequency | About 500 children diagnosed annually in the United States |
Medulloblastoma is a common type of primary brain cancer in children. It originates in the part of the brain that is towards the back and the bottom, on the floor of the skull, in the cerebellum, or posterior fossa.
The brain is divided into two main parts, the larger cerebrum on top and the smaller cerebellum below towards the back. They are separated by a membrane called the tentorium. Tumors that originate in the cerebellum or the surrounding region below the tentorium are, therefore, called infratentorial.
Historically, medulloblastomas have been classified as a primitive neuroectodermal tumor (PNET), but it is now known that medulloblastoma is distinct from supratentorial PNETs and they are no longer considered similar entities.
Medulloblastomas are invasive, rapidly growing tumors that, unlike most brain tumors, spread through the cerebrospinal fluid and frequently metastasize to different locations along the surface of the brain and spinal cord. Metastasis all the way down to the cauda equina at the base of the spinal cord is termed "drop metastasis".
The cumulative relative survival rate for all age groups and histology follow-up was 60%, 52%, and 47% at 5 years, 10 years, and 20 years, respectively, with children doing better than adults.
Signs and symptoms
[edit]
Signs and symptoms are mainly due to secondary increased intracranial pressure due to blockage of the fourth ventricle and tumors are usually present for 1 to 5 months before diagnosis is made. The child typically becomes listless, with repeated episodes of vomiting, and a morning headache, which may lead to a misdiagnosis of gastrointestinal disease or migraine. Soon after, the child will develop a stumbling gait, truncal ataxia, frequent falls, diplopia, papilledema, and sixth cranial nerve palsy. Positional vertigo and nystagmus are also frequent, and facial sensory loss or motor weakness may be present. Decerebrate attacks appear late in the disease.
Extraneural metastasis to the rest of the body is rare, and when it occurs, it is in the setting of relapse, more commonly in the era prior to routine chemotherapy.
Pathogenesis
[edit]
Medulloblastomas are usually found in the vicinity of the fourth ventricle, between the brainstem and the cerebellum. Tumors with similar appearance and characteristics originate in other parts of the brain, but they are not identical to medulloblastoma.
Although medulloblastomas are thought to originate from immature or embryonal cells at their earliest stage of development, the cell of origin depends on the subgroup of medulloblastoma. WNT tumors originate from the lower rhombic lip of the brainstem, while SHH tumors originate from the external granular layer.
Currently, medulloblastomas are thought to arise from cerebellar stem cells that have been prevented from dividing and differentiating into their normal cell types. This accounts for the histologic variants seen on biopsy. Both perivascular pseudorosette and Homer Wright pseudorosette formations are highly characteristic of medulloblastomas and are seen in up to half of cases. The classic rosette with tumor cells around a central lumen can be seen.
In the past, medulloblastoma was classified using histology, but integrated genomic studies have revealed that medulloblastoma is composed of four distinct molecular and clinical variants termed WNT/β-catenin, Sonic Hedgehog, Group 3, and Group 4. Of these subgroups, WNT patients have an excellent prognosis and group 3 patients have a poor prognosis. Also, a subgroup-specific alternative splicing further confirms the existence of distinct subgroups and highlights the transcriptional heterogeneity between subgroups. Amplification of the Sonic Hedgehog pathway is the best characterized subgroup, with 25% of human tumors having mutations in Patched, Sufu (Suppressor of Fused Homolog), Smoothened, or other genes in this pathway. Medulloblastomas are also seen in Gorlin syndrome as well as Turcot syndrome. Recurrent mutations in the genes CTNNB1, PTCH1, MLL2, SMARCA4, DDX3X, CTDNEP1, KDM6A, and TBR1 were identified in individuals with medulloblastoma. Additional pathways disrupted in some medulloblastomas include MYC, Notch, BMP, and TGF-β signaling pathways.[excessive citations]
Diagnosis
[edit]
The tumor is distinctive on T1- and T2-weighted MRI with heterogeneous enhancement and a typical location adjacent to and extension into the fourth ventricle. Histologically, the tumor is solid, pink-gray in color, and is well circumscribed. The tumor is very cellular, with high mitotic activity, little cytoplasm, and a tendency to form clusters and rosettes.
The Chang staging system can be used in making the diagnosis.
DNA methylation profiling of medulloblastoma allows robust sub-classification and improved outcome prediction using formalin-fixed biopsies.
Correct diagnosis of medulloblastoma may require ruling out atypical teratoid rhabdoid tumor.
Cerebellar medulloblastoma in an adult
Cerebellar medulloblastoma in an adult
Treatment
[edit]
Treatment begins with maximal surgical removal of the tumor. The addition of radiation to the entire neuraxis and chemotherapy may increase the disease-free survival. This combination may permit a 5-year survival in more than 80% of cases. Some evidence indicates that proton beam irradiation reduces the impact of radiation on the cochlear and cardiovascular areas and reduces the cognitive late effects of cranial irradiation.
The presence of desmoplastic features such as connective tissue formation offers a better prognosis. Prognosis is worse if the child is less than 3 years old, degree of resection is inadequate, or if any CSF, spinal, supratentorial, or systemic spread occurs. Dementia after radiotherapy and chemotherapy is a common outcome appearing two to four years following treatment. Side effects from radiation treatment can include cognitive impairment, psychiatric illness, bone growth retardation, hearing loss, and endocrine disruption. Increased intracranial pressure may be controlled with corticosteroids or a ventriculoperitoneal shunt. An approach to monitor tumor development and treatment response by liquid biopsy is promising, but remains challenging.
Chemotherapy
[edit]
Chemotherapy is often used as part of treatment. Evidence of benefit, however, is not clear as of 2013. A few different chemotherapeutic regimens for medulloblastoma are used; most involve a combination of lomustine, cisplatin, carboplatin, vincristine, or cyclophosphamide. In younger patients (less than 3–4 years of age), chemotherapy can delay, or in some cases possibly even eliminate, the need for radiotherapy. However, both chemotherapy and radiotherapy often have long-term toxicity effects, including delays in physical and cognitive development, higher risk of second cancers, and increased cardiac disease risks.
Outcomes
[edit]
Array-based karyotyping of 260 medulloblastomas resulted in the following clinical subgroups based on cytogenetic profiles:
Poor prognosis: gain of 6q or amplification of MYC or MYCN
Intermediate: gain of 17q or an i(17q) without gain of 6q or amplification of MYC or MYCN
Excellent prognosis: 6q and 17q balanced or 6q deletion
Transcriptional profiling shows the existence of four main subgroups (Wnt, Shh, Group 3, and Group 4).
Very good prognosis: WNT group, CTNNB1 mutation
Infants good prognosis, others intermediate: SHH group, PTCH1/SMO/SUFU mutation, GLI2 amplification, or MYCN amplification
Poor prognosis: Group 3, MYC amplification, photoreceptor/GABAergic gene expression
Intermediate prognosis: Group 4, gene expression of neuronal/glutamatergic, CDK6 amplification, MYCN amplification
Survival
[edit]
The historical cumulative relative survival rate for all age groups and histology follow-up was 60%, 52%, and 47% at 5 years, 10 years, and 20 years, respectively. Patients diagnosed with a medulloblastoma or PNET are 50 times more likely to die than a matched member of the general population.
A population-based (SEER) 5-year relative survival rates indicated 69% overall: 72% in children (1–9 years) and 67% in adults (20+ years). The 20-year survival rate is 51% in children. Children and adults have different survival profiles, with adults faring worse than children only after the fourth year after diagnosis (after controlling for increased background mortality). Before the fourth year, survival probabilities are nearly identical. Long-term sequelae of standard treatment include hypothalamic-pituitary and thyroid dysfunction and intellectual impairment. The hormonal and intellectual deficits created by these therapies causes significant impairment of the survivors.[self-published source?]
In current clinical studies, the patients are divided into low-, standard- and high-risk groups:
Depending on the study, healing rates of up to 100% are achieved in the low-risk group (usually WNT-activated). The current efforts are therefore moving in the direction of reducing the intensity of the therapy and thus the negative long-term consequences while confirming the high healing rates.
In the HIT-SIOP PNET 4 study, in which 340 children and adolescents of the standard-risk group between the ages of four and 21 from several European countries participated, the 5-year survival rate was between 85% and 87% depending on the randomization. Around 78% of the patients remained without relapse for 5 years and are therefore considered to be cured. After a relapse, the prognosis was very poor. Despite intensive treatment, only four of 66 patients were still alive 5 years after a relapse.
A US study involved 161 patients between the ages of three and 21 with a high-risk profile. Depending on the randomization, half of the patients additionally received carboplatin daily during the radiation. The 5-year survival rate of patients with carboplatin was 82%, those without 68%. The European SIOP PNET 5 study is currently taking place and will run until April 2024, in which an attempt is made to confirm the promising results with carboplatin during irradiation in the standard risk group.
Epidemiology
[edit]
Medulloblastomas affect just under two people per million per year, and affect children 10 times more than adults. Medulloblastoma is the second-most frequent brain tumor in children after pilocytic astrocytoma and the most common malignant brain tumor in children, comprising 14.5% of newly diagnosed brain tumors. In adults, medulloblastoma is rare, comprising fewer than 2% of CNS malignancies.
The rate of new cases of childhood medulloblastoma is higher in males (62%) than females (38%), a feature that is not seen in adults. Medulloblastoma and other PNETs are more prevalent in younger children than older children. About 40% of medulloblastoma patients are diagnosed before the age of five, 31% are between the ages of 5 and 9, 18.3% are between the ages of 10 and 14, and 12.7% are between the ages of 15 and 19.
Research models
[edit]
Using gene transfer of SV40 large T-antigen in neuronal precursor cells of rats, a brain tumor model was established. The PNETs were histologically indistinguishable from the human counterparts and have been used to identify new genes involved in human brain tumor carcinogenesis. The model was used to confirm p53 as one of the genes involved in human medulloblastomas, but since only about 10% of the human tumors showed mutations in that gene, the model can be used to identify the other binding partners of SV40 Large T- antigen, other than p53.
In a mouse model, high medulloblastoma frequency appears to be caused by the down regulation of Cxcl3, with Cxcl3 being induced by Tis21. Consistently, the treatment with Cxcl3 completely prevents the growth of medulloblastoma lesions in a Shh-type mouse model of medulloblastoma. Thus, CXCL3 is a target for medulloblastoma therapy.
References
[edit]
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^ "Selected Childhood Primary Brain and Central Nervous System Tumor Incidence Rates by Major Histology Groupings, Histology and Gender" (PDF). Central Brain Tumor Registry of the United States, 1998–2002. Archived from the original (PDF) on 2007-09-27. Retrieved 2007-03-14.
^ "Selected Childhood Primary Brain and Central Nervous System Tumor Age-Specific Incidence Rates" (PDF). Central Brain Tumor Registry of the United States, 1998–2002. Archived from the original (PDF) on 2007-09-27. Retrieved 2007-03-14.
^ Eibl RH, Kleihues P, Jat PS, Wiestler OD (March 1994). "A model for primitive neuroectodermal tumors in transgenic neural transplants harboring the SV40 large T antigen". The American Journal of Pathology. 144 (3): 556–64. PMC 1887088. PMID 8129041.
^ Ohgaki H, Eibl RH, Wiestler OD, Yasargil MG, Newcomb EW, Kleihues P (November 1991). "p53 mutations in nonastrocytic human brain tumors". Cancer Research. 51 (22): 6202–5. PMID 1933879.
^ Eibl RH, Schneemann M (February 2023). "From TP53 Mutations to Molecular Classification and Liquid Biopsy". Biology. 12 (2): 267. doi:10.3390/biology12020267. PMC 9952923. PMID 36829544.
^ Farioli-Vecchioli S, Cinà I, Ceccarelli M, Micheli L, Leonardi L, Ciotti MT, et al. (October 2012). "Tis21 knock-out enhances the frequency of medulloblastoma in Patched1 heterozygous mice by inhibiting the Cxcl3-dependent migration of cerebellar neurons". The Journal of Neuroscience. 32 (44): 15547–64. doi:10.1523/JNEUROSCI.0412-12.2012. PMC 6621585. PMID 23115191.
^ Ceccarelli M, Micheli L, Tirone F (2016). "Suppression of Medulloblastoma Lesions by Forced Migration of Preneoplastic Precursor Cells with Intracerebellar Administration of the Chemokine Cxcl3". Frontiers in Pharmacology. 7: 484. doi:10.3389/fphar.2016.00484. PMC 5159413. PMID 28018222.
External links
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Brain and Spinal Tumors: Hope Through Research (National Institute of Neurological Disorders and Stroke)
Medulloblastoma Images Archived 2009-01-30 at the Wayback Machine MedPix Medical Image Database
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| Classification | D ICD-O: M9470/3 OMIM: 155255 MeSH: D008527 DiseasesDB: 31105 SNOMED CT: 443333004 |
| External resources | eMedicine: neuro/624 ped/1396 radio/434 Orphanet: 616 |
| v t e Tumours of the nervous system | |
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| Endocrine | | | | --- | | Sellar: | Craniopharyngioma Pituicytoma | | Other: | | |
| CNS | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- | | Neuroepithelial (brain tumors, spinal tumors) | | | | | | | | | | | | | | --- --- --- --- --- --- | | Glioma | | | | --- | | Astrocyte | Astrocytoma + Pilocytic astrocytoma + Pleomorphic xanthoastrocytoma + Subependymal giant cell astrocytoma + Fibrillary astrocytoma + Anaplastic astrocytoma + Glioblastoma | | Oligodendrocyte | Oligodendroglioma Anaplastic oligodendroglioma | | Ependyma | Ependymoma Subependymoma | | Choroid plexus | Choroid plexus tumor + Choroid plexus papilloma + Choroid plexus carcinoma | | Multiple/unknown | Oligoastrocytoma Gliomatosis cerebri Gliosarcoma | | | Mature neuron | Ganglioneuroma: Ganglioglioma Retinoblastoma Neurocytoma Dysembryoplastic neuroepithelial tumour Lhermitte–Duclos disease | | CNS embryonal tumors | Medulloblastoma Atypical teratoid rhabdoid tumor Embryonal tumour with multilayered rosettes | | | Meninges | Meningioma Hemangiopericytoma | | Hematopoietic | Primary central nervous system lymphoma | |
| PNS: | Neuroblastoma + Esthesioneuroblastoma + Ganglioneuroblastoma Nerve sheath tumor Cranial and paraspinal nerves + Neurofibroma + Neurofibromatosis Neurilemmoma/Schwannoma + Acoustic neuroma Malignant peripheral nerve sheath tumor |
| Other | WHO classification of the tumors of the central nervous system |
| Note: Not all brain tumors are of nervous tissue, and not all nervous tissue tumors are in the brain (see brain metastasis). | |
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2702 | https://www.torontomu.ca/content/dam/canadian-conference-computational-geometry-2022/papers/CCCG2022_paper_57.pdf | CCCG 2022, Toronto, ON, Canada, August 25–27, 2022 Diverse Non Crossing Matchings Neeldhara Misra† Harshil Mittal‡ Saraswati Girish Nanoti§ Abstract A perfect matching M on a set P of n points is a col-lection of line segments with endpoints from P such that every point belongs to exactly one segment.
A matching is non-crossing if the line segments do not cross. Two matchings M and N are said to be com-patible if there are no crossings among any pair of line segments in M ∪N. We introduce a notion of diverse non-crossing matchings: a pair of perfect matchings M and N are k-diverse if, for every p ∈P, the distance be-tween the matched partners of p in M and N is at least k. In this contribution, we describe a polynomial time algorithm to determine if a set of points in convex posi-tion admits two compatible and perfect NCMs that are k-diverse. For points in convex position, we also show that if a perfect matching M is given as input, then we can determine, in polynomial time, if another perfect matching N exists that is compatible with M and is such that M and N are k-diverse. Finally, we also es-tablish that every point set in general position admits a pair of compatible and perfect NCMs. The first two re-sults also hold for bichromatic points, and we also give a characterization for when a bichromatic point set in convex position admits a pair of perfect and disjoint NCMs.
1 Introduction Matching problems involve partitioning a set of objects into pairs subject to some constraints. For example, in the context of graphs, we are given a binary relation over the set of objects and require the pairs to be re-lated. In a geometric setting, the set typically consists of geometric objects (Aloupis et al., 2013), and such problems have received a lot of attention because of their practical relevance.
Our focus is on the setting of matching points using line segments. In particular, given a set P of n points in the plane R2, we are interested in matching them The authors are grateful for support from the SERB-MATRICS grant MTR/2017/001033, the SERB-ECR grant ECR/2018/002967, and CSIR. We also thank the anonymous referees for their detailed comments.
†IIT Gandhinagar , neeldhara.m@iitgn.ac.in ‡IIT Gandhinagar , mittal harshil@iitgn.ac.in §IIT Gandhinagar , nanoti saraswati@iitgn.ac.in with straight line segments. We focus on perfect non-crossing matchings (NCMs), i.e, matchings where every point is matched and no two line segments cross. Un-less otherwise mentioned, we assume that all match-ings are perfect.
It turns out that any collection of points admits a NCM and that this can be found in O(n log n) time (Hersh-berger and Suri, 1990; Lo et al., 1994). Many studies on NCMs focus on optimizing some structural property of the matching, such as the maximum, minimum, or av-erage edge length. Two NCMs M and N are said to be disjoint if every point has a different matched part-ner in both matchings, and compatible if the segments in M ∪N do not cross.
For optimization problems, the decision or search ver-sion of the question seeks to find some optimal solu-tion, while the counting version asks to enumerate all optimal solutions. In many application scenarios, the former is not sufficient, while the latter is too demand-ing in terms of computational expense. This motivates the notion of demanding not all but a select collection of solutions. In most applications, the requirement is not just for a multitude of solutions, but for an “inter-esting” collection of solutions: for example, informally speaking, solutions that are minor variations of one an-other and are very similar may not be very useful in most settings.
The existence of a diverse collection of solutions has been explored in several settings recently.
Studies on diverse solutions have focused on a wide array of problems including, but are not limited to, vertex cover (Baste et al., 2022), matchings (Fomin et al., 2020), stable matchings (Ganesh et al., 2021), matroids (Fomin et al., 2021), satisfiability (Nadel, 2011), Kemeny rank aggregation (Arrighi et al., 2021), etc.
To propose that we find “diverse” solutions, we need a notion of distance between solutions. In the setting of matchings between points in R2, a natural notion of “distance between matchings M and N” would be an aggregation of the distance between the matched part-ners of all the points in the two matchings. The aggre-gation function that we work with picks out the small-est such distance. In particular, using M(·) to denote the matched partner of a point p in a matching M, we define the distance between two matchings M and N over a point set P as minp∈P d(M(p), N(p)). Note that 34th Canadian Conference on Computational Geometry, 2022 the we have used the term “distance” informally and this function does not satisfy the triangle inequality.
We say that a collection of matchings M is k-diverse for some positive number k if the distance between ev-ery pair of matchings in M has a distance of at least k between them. Throughout our discussions, we focus on the problem of finding two matchings.
Our Contributions.
We propose the following natural computational questions: DIVERSE NCMS (DIVERSE COMPATIBLE NCMS) Input: A set P of 2n points and a positive ratio-nal number k.
Question: Does P admit two perfect matchings that are k-diverse and compatible, i.e., two DC-NCM’s?
ANOTHER DIVERSE NCM (ANOTHER DIVERSE COMPATIBLE NCM) Input: A set P of 2n points, a perfect matching M over P, and a positive rational number k.
Question: Is there a perfect NCM N over P such that M and N are k-diverse (and compatible)?
We first show that any monochromatic point set P in general position with an even number of points such that |P| ⩾4 admits two compatible NCMs. Note that this is easy to see for points in convex position: a set of alternating edges on the convex hull and the remain-ing edges of the convex hull form a pair of compatible matchings. For points in general position, we gener-alize this idea by considering the layer decomposition and peeling off convex layers with an even number of points, and carefully matching across layers when we encounter layers with an odd number of points. We also characterize bichromatic point sets in convex posi-tion that admit two disjoint non-crossing matchings1.
Theorem 1 (Disjoint Matchings) Any point set P in general position admits two compatible perfect NCMs. A bichromatic point set P in convex position admits two dis-joint and perfect NCMs if and only if the orbit of each point contains at least two points of the opposite colour.
We next propose the computational problem of find-ing a matching that is diverse with respect to and, op-tionally, compatible with a given matching. We show that when points are in convex position, we can find 1We refer the reader to Section 2 for the formal definitions of the terminology used here.
such a matching in polynomial time.
We use a dy-namic programming approach here, considering sub-problems corresponding to contiguous subintervals of the convex hull.
Theorem 2 (Another Diverse Matching) For both monochromatic and bichromatic points in convex position, the problems ANOTHER DIVERSE NCM and ANOTHER DIVERSE COMPATIBLE NCM admit polynomial time algorithms.
Finally, we consider the problem of finding a pair of diverse and compatible matchings. We demonstrate a polynomial time algorithm for points in convex posi-tion. For this algorithm, we note that any solution can be viewed equivalently as a collection of disjoint non-overlapping polygons. We prove a structural lemma which shows that there always exists an optimal so-lution consisting of polygons with a constant number of sides. We can then leverage this to come up with a dynamic programming algorithm that considers, as be-fore, subproblems corresponding to contiguous subin-tervals of the convex hull, and makes progress by guessing all possible choices for the polygon that the first point on the subinterval belongs to.
Theorem 3 (Diverse Compatible Matching) For both monochromatic and bichromatic points in convex posi-tion, DIVERSE COMPATIBLE NCMS admits a polynomial time algorithm.
Related Work.
The task of finding a matching that minimizes the length of the longest edge is called the bottleneck NCM problem and is known to be NP-complete in general and tractable for points in convex position and other special cases, and has been well-studied for monochromatic and bichromatic points (Abu-Affash et al., 2014; Carlsson et al., 2015; Savi´ c and Stojakovi´ c, 2017, 2022; Biniaz et al., 2014).
Other variants of the problem such as those which involve minimizing the length of the shortest edge or maximizing the length of the longest edge are tractable (Mantas et al., 2021). Finally, to the best of our knowledge, the complexity of finding a matching that maximizes the length of the shortest edge is open.
In the context of the setting where we have a point set and a matching, it was conjectured by Aichholzer et al.
(2009) that for every perfect matching M of a point set P such that |P| is a multiple of four, there is another perfect matching, N of P such that M and N are com-patible. This was subsequently proved by Ishaque et al.
(2012) using a constructive argument that also leads to an efficient method for constructing the matching N. It is also known that the conjecture does not hold when |P| is not a multiple of four Aichholzer et al. (2009).
CCCG 2022, Toronto, ON, Canada, August 25–27, 2022 Organization of the paper.
Due to lack of space, we defer the proofs of Theorem 1 and Theorem 2 to the full version of the paper. We provide most of the de-tails towards showing Theorem 3 in Section 3, only de-ferring the argument of correctness and remarks about the bichromatic case to the full version.
2 Preliminaries In the setting of monochromatic points, we use P typ-ically to denote a set of 2n points in R2 with n > 1.
When we work with points in general position, we will use P to denote the convex hull of P.
In case of convex point sets, we label the points of P by p0, p1, . . . , p2n−1 in positive (counterclockwise) direc-tion around the convex hull. To simplify the notation, we will generally use only indices when referring to points. We write {i, . . . , j} to represent the sequence i, i + 1, i + 2, . . . , j −1, j. All operations are calculated modulo 2n. Note that i is not necessarily less than j, and that {i, . . . , j} is not the same as {j, . . . , i}.
A bichromatic set of points is a point set P equipped with a coloring function c : P →{0, 1} that classifies each point as either “red” (points for which c(p) = 0) or “blue” (points for which c(p) = 1). We usually de-note these sets by R and B respectively, with P = R ∪B and |R| = |B| = n, and again, we assume n > 1.
We say that two line segments s and t in the plane cross if there is a point on the plane which is not an endpoint of either s and t that belongs to both s and t. In partic-ular, note that if s = t, then s and t cross each other.
The convex hull of a point set is the smallest convex polygon that contains all the points of it. The convex layers or the onion decomposition of a set of points are a sequence of nested convex polygons having the points as their vertices. The outermost one is the convex hull of the points and the rest are formed in the same way recursively. The innermost layer may be degenerate, consisting only of one or two points. The number of polygons in onion decomposition of a point set is called its layer depth.
A perfect matching on the set P is a set of n straight line segments whose endpoints are points in P such that each point is the endpoint of exactly one line seg-ment. For bichromatic points sets, we further require that each line segment has one red and one blue end-point. If the line segments do not cross, we refer to such a matching as a (bichromatic) non-crossing matching.
All matchings are both perfect and non-crossing unless mentioned otherwise.
We usually use the notation M or N to refer to match-ings. With a slight abuse of notation, given a matching M over P and a point p ∈P, we use M(p) to denote the matched partner of p in M, that is, the point q such that the segment connecting p and q belongs to M. Two matchings M and N are called disjoint if the matched partners of all points in p are different in both, i.e, for all p ∈P, we have that M(p) ̸= N(p), and compati-ble if the segments in the multiset M ∪N do not cross.
Note that all compatible matchings are disjoint, while the converse may not be true.
We define the distance between two matchings M and N over a point set P, denoted DP(M, N), as minp∈P(dist(M(p), N(p))), where dist(·, ·) denotes the Euclidean distance between two points. We also refer to this as the diversity of the set {M, N} or the diversity between M and N. Further, we say that a pair of match-ings M and N over P are k-diverse if DP(M, N) ≥k.
Note that if M and N are not disjoint, then they are 0-diverse. If the point set P is clear from the context, we may drop the subscript P from the notation for dis-tances and diversity.
We now introduce some terminology that is relevant to bichromatic point sets.
Definition 1 (Balanced, Blue-heavy, Red-heavy) A set of points is balanced if it contains the same number of red and blue points. If the set has more red (blue) points than blue (red), we say that it is red-heavy (blue-heavy).
Lemma 1 (Savi´ c and Stojakovi´ c (2022)) Every balanced set of points can be matched.
Definition 2 (Feasible pair.) We say that (i, j) is a feasi-ble pair if there exists a matching containing (i, j). We refer to i as a feasible neighbour of j and vice versa.
Lemma 2 (Savi´ c and Stojakovi´ c (2022)) A pair (i, j) is feasible if and only if i and j have different colors and {i, . . . , j} is balanced.
Definition 3 (Functions o+ and o−.) [Savi´ c and Sto-jakovi´ c (2022)] By o+(i) we denote the first point starting from i in the positive direction such that (i, o+(i)) is feasible. By o−(i) we denote the first point starting from i in the negative direction such that (o−(i), i) is feasible.
As we assume that the given point set is bal-anced, Lemma 2 guarantees that both o+ and o−are well-defined. It also turns out that o−is the inverse function of o+ as mentioned in Savi´ c and Stojakovi´ c (2022). We denote the composition of o+ function k times on a point p as ok(p) and also use the notation o(p) to mean o+(p).
Definition 4 (Orbit) [Savi´ c and Stojakovi´ c (2022)] An orbit of i, denoted by O(i), is defined by O(i) := n ok(i) : k ∈Z o . By O(P) we denote the set of all orbits of a convex point set P, that is O(P) := {O(i) : i ∈P}.
34th Canadian Conference on Computational Geometry, 2022 3 DC-NCM for points in Convex Position Suppose that the points of P are in convex position. Let F be a collection of even-length simple convex poly-gons, each of length ≥4. We say that F is a feasible collection of polygons on P if the following hold true: • Every p ∈P is a vertex of exactly one polygon in F, and every polygon in F has all its vertices in P.
• No edge of a polygon in F crosses an edge of an-other polygon in F.
For any even-length simple polygon T of length ≥4, • let partners(T) denote the set of all unordered pairs {u, v} of vertices of T such that exactly one vertex of T appears between u and v, when one traverses from u to v in counter-clockwise direc-tion along the boundary of T.
• let quality(T) denote the minimum of dist(u, v) over all pairs {u, v} in partners(T).
For any feasible collection F of polygons on P, let quality(F) denote the minimum of quality(T) over all polygons T in F.
Note that for any k > 0, the following are equivalent: • There exists a pair of compatible perfect NCMs M and N on P such that DP(M, N) ≥k.
• There exists a feasible collection F of polygons on P such that quality(F) ≥k.
This claim follows from the fact that the union of the line segments in any pair of compatible NCMs over P is a collection of even-length simple convex polygons whose vertices partition P and do not cross, i.e, a feasi-ble collection of polygons on P.
Thus, our goal is to find a collection F of feasible poly-gons on P for which quality(F) is maximized.
Let A and B be non-empty sets of real numbers. We say that A dominates B if for every x ∈A, we have x ≥y for some y ∈B. Note that • If A ⊆B, then A dominates B.
• A dominates B if and only if min(A) ≥min(B).
Our dynamic programming algorithm relies on the fol-lowing structural lemma, which says the following: if F is a feasible collection of polygons on P with quality s, then we can find a (potentially different) F ′ which is a feasible collection of polygons on P whose quality is no worse than s, and further, every polygon in F ′ has four or six vertices. This allows us to devise a poly-nomial time algorithm based on “guessing” the nature of the polygons that the points belong to in some final solution.
Figure 1: Breaking up a polygon on 4q points.
Our proof for the structural lemma considers two sce-narios. First, when the number of vertices of a poly-gon T in F is a multiple of four, we simply “break” it into four-length polygons. In this situation, we intro-duce no new pairs into the set of matched partners (i.e, partners(T′) ⊆partners(T) for any T′ generated by the breaking procedure) , and so the quality of the solution is not affected. The other situation is that the number of vertices of a polygon T in F is of the form 4q + 2. In this case, we find a six-length polygon T′ and organize the remaining 4q −4 points into (q −1) polygons as in the previous case. The choice of T′ is made carefully so as to ensure that the overall quality of the solution thus obtained is no worse than the original.
Lemma 3 Let F be a feasible collection of polygons on P.
Then, there exists a feasible collection F ′ of polygons on P such that • Every polygon in F ′ has length either 4 or 6.
• quality(F ′) ≥quality(F) Proof.
Let F ′ be a family of polygons on P obtained from F as follows: For each polygon T in F of length > 6, Case 1: T has length 4q, for some integer q ≥2 Let 0, 1, 2 . . . , 4q −1 denote the vertices of T, appear-ing in that order as one traverses in counter-clockwise direction along its boundary. We replace T with q sim-ple convex polygons T0, T1, . . . , Tq−1, each of length 4, where CCCG 2022, Toronto, ON, Canada, August 25–27, 2022 Figure 2: The even and odd polygons Teven and Todd.
• T0 has vertices 0, 1, 2, 3.
• T1 has vertices 4, 5, 6, 7.
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• Tq−1 has vertices 4q −4, 4q −3, 4q −2, 4q −1.
Let 0 ≤j ≤q −1. Note that partners(Tj) = n {4j, 4j + 2}, {4j + 1, 4j + 3} o ⊆partners(T) So, we have min {u,v}∈partners(Tj) dist(u, v) ≥ min {u,v}∈partners(T) dist(u, v) That is, quality(Tj) ≥quality(T).
Thus, min 0≤j≤q−1 quality(Tj) ≥quality(T) Case 2: T has length 4q + 2, for some integer q ≥2 Let 0, 1, 2, . . . , 4q + 1 denote the vertices of T, appear-ing in that order as one traverses in counter-clockwise direction along its boundary.
Let Teven and Todd denote the simple convex poly-gons, each of length 2q + 1, on the vertices 0, 2, 4, . . . , 4q and 1, 3, 5, . . . , 4q + 1 respectively. Let e0, e2, e4, . . . , e4q denote the interior angles of Teven, at the vertices 0, 2, 4, . . . , 4q respectively.
Let o1, o3, o5, . . . , o4q+1 de-note the interior angles of Todd, at the vertices 1, 3, 5, . . . , 4q + 1 respectively.
For any simple convex polygon, since its exterior an-gles sum up to 2π, at most two of them are > 2π 3 . So, at most two of its interior angles are < π 3 .
Figure 3: Breaking up a polygon on 4q + 2 points.
Thus, each of Teven and Todd has at most two in-terior angles that are < π 3 .
That is, at most two of e0, e2, . . . , e4q are < π 3 , and at most two of o1, o3, . . . , o4q+1 are < π 3 . So, among the ≥5 pairs of angles (e0, o1), (e2, o3), (e4, o5), . . . , (e4q, o4q+1), there is at least one pair, say (e2ℓ, o2ℓ+1), such that each of e2ℓ and o2ℓ+1 is ≥π 3 .
We replace T with a simple convex polygon T0 of length 6, and q −1 simple convex polygons T1, . . . , Tq−1, each of length 4 (c.f. Figure 3), where: • T0 has vertices 2ℓ−2, 2ℓ−1, 2ℓ, 2ℓ+ 1, 2ℓ+ 2, 2ℓ+ 3.
• T1 has vertices 2ℓ+ 4, 2ℓ+ 5, 2ℓ+ 6, 2ℓ+ 7.
• T2 has vertices 2ℓ+ 8, 2ℓ+ 9, 2ℓ+ 10, 2ℓ+ 11.
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• Tq−1 has vertices 2ℓ+ 4q −4, 2ℓ+ 4q −3, 2ℓ+ 4q −2, 2ℓ+ 4q −1.
Here, the additions are modulo 4q + 2.
34th Canadian Conference on Computational Geometry, 2022 Let 1 ≤j ≤q −1. Note that partners(Tj) = n {2ℓ+ 4j, 2ℓ+ 4j + 2}, {2ℓ+ 4j + 1, 2ℓ+ 4j + 3} o ⊆partners(T) So, we have min {u,v}∈partners(Tj) dist(u, v) ≥ min {u,v}∈partners(T) dist(u, v) That is, quality(Tj) ≥quality(T).
Next, we show that quality(T0) ≥quality(T).
Let A := n dist(u, v) | {u, v} ∈partners(T0) o B := n dist(u, v) | {u, v} ∈partners(T) o Note that partners(T0) \ partners(T) = n {2ℓ+ 3, 2ℓ−1}, {2ℓ+ 2, 2ℓ−2} o Consider the triangle formed by the points 2ℓ−1, 2ℓ+ 1, 2ℓ+ 3. Here, as o2ℓ+1 ≥π 3 , dist(2ℓ+ 3, 2ℓ−1) ≥min dist(2ℓ−1, 2ℓ+ 1), dist(2ℓ+ 1, 2ℓ+ 3) .
Consider the triangle formed by the points 2ℓ−2, 2ℓ, 2ℓ+ 2. Here, as e2ℓ≥π 3 , dist(2ℓ+ 2, 2ℓ−2) ≥min dist(2ℓ−2, 2ℓ), dist(2ℓ, 2ℓ+ 2) .
Also, note that partner(T) contains the pairs {2ℓ−1, 2ℓ+ 1}, {2ℓ+ 1, 2ℓ+ 3}, {2ℓ−2, 2ℓ}, {2ℓ, 2ℓ+ 2}.
Therefore, A dominates B and so, min(A) ≥min(B).
That is, min {u,v}∈partners(T0) dist(u, v) ≥ min {u,v}∈partners(T) dist(u, v) Thus, quality(T0) ≥quality(T).
Hence, we have min 0≤j≤q−1 quality(Tj) ≥quality(T), and this concludes the proof.
□ Based on the lemma, we have the following dynamic programming approach: Let 0, 1, . . . , 2n −1 denote the points of P in counter-clockwise order. For every 0 ≤ i, j ≤2n −1 such that (j −i) is odd, let Qi,j denote the set of points {i, i + 1, . . . , j} and let: T(i, j) = max DQij(M, N) if j −i ≥3, −∞ if j −i = 1, +∞ if j −i < 0; where the max is taken over all pairs of disjoint com-patible perfect NCMs M and N over the point set Qi,j.
Note that T(0, 2n −1) is the value of the optimal solu-tion. We compute and store T(i, j)′s using the follow-ing recurrence: T(i, j) = max(α(i, j), β(i, j)) if j −i ≥5, α(i, j) if j −i = 3, −∞ if j −i = 1, +∞ if j −i < 0, where α(i, j) is given by: max i < p1 < p2 < p3 ⩽j : p1 −i is odd p2 −p1 is odd p3 −p2 is odd min dist (i, p2) , dist (p1, p3) , T (i + 1, p1 −1) , T (p1 + 1, p2 −1) , T (p2 + 1, p3 −1) , T (p3 + 1, j) and β(i, j) is given by: CCCG 2022, Toronto, ON, Canada, August 25–27, 2022 max i<q1<q2<q3<q4<q5⩽j: q1 −i is odd q2 −q1 is odd q3 −q2 is odd q4 −q3 is odd q5 −q4 is odd min dist (i, q2) , dist (q1, q3) , dist (q2, q4) , dist (q3, q5) , dist (q4, i) , dist (q5, q1) , T (i + 1, q1 −1) , T (q1 + 1, q2 −1) , T (q2 + 1, q3 −1) , T (q3 + 1, q4 −1) , T (q4 + 1, q5 −1) , T (q5 + 1, j) .
We remark that the recurrences are well-defined. The overall intuition for the recurrences above is the fol-lowing: fix an arbitrary solution that has the property guaranteed by Lemma 3. We attempt to “guess” the type and vertices of the polygon that the first point be-longs to in this solution. For each fixed guess, we have a natural partition of the remaining points into smaller sub-instances (see Figures 4 and 5). It is easy to identify invalid guesses, by which we mean a polygon which is such that there is no solution that contains it.
For any valid guess, the recurrence gives us the best possible extension, i.e, the best possible diversity achievable among solutions that contain the guessed polygon. All that remains is to pick the best choice among all choices of polygons that contain the first point.
The overall running time is polynomially bounded because we only have to worry about poly-gons with a constant number of vertices. We make this argument more explicit in the Appendix. We also note that the running time of our algorithm is O(n7) since the DP table has O(n2) indices and the computation at each index is O(n5).
We now sketch the correctness of the dynamic pro-gramming approach proposed in the context of The-orem 3. Consider the subproblem given by the points i, i + 1, . . . , j. Consider the space of all solutions S that have the property guaranteed by Lemma 3 and parti-tion it into two parts: S4 ⊆S consists of all solutions where the point i belongs to a polygon with four sides; and S6 ⊆S consists of all solutions where the point i belongs to a polygon with six sides.
Let A⋆and B⋆denote arbitrary optimal solutions among all the solutions in S4 and S6, respectively. Fur-ther, let a⋆and b⋆denote the corresponding costs. Note that the cost of the optimal solution for this subproblem is max(a⋆, b⋆).
We now argue that α(i, j) correctly computes the value of a⋆. Once again, for every choice of points i < p1 < p2 < p3 ≤j in (Qi+1,j 3 ), let S4 denote the set Figure 4: An example of how a base polygon divides the subproblem on Qi,j further into four smaller in-stances.
of all solutions in S4 where the polygon containing the point i also contains the points p1, p2, p3. Note that if it is not the case that p1 −i is odd and p2 −p1 is odd and p3 −p2 is odd , then S4 = ∅, since for any such combination of points, there is no valid so-lution containing the polygon formed by the points {i, p1, p2, p3}.
For any valid combination, we know that the best solution in S4 is captured by taking the union of the best solutions for the fol-lowing subinstances: (i + 1, p1 −1), (p1 + 1, p2 −1), (p2 + 1, p3 −1), and (p3 + 1, j) corresponding to the four “chunks” of points “carved out” by the polygon (see Figure 4); along with the polygon formed by the points {i, p1, p2, p3}. Note that there are no points in-side the polygon whose vertices are {i, p1, p2, p3} since the original point set is in convex position. Further, note that it is reasonable to consider these subinstances independently since no solution that contains the poly-gon formed by {i, p1, p2, p3} will contain a polygon with points from two distinct segments among the seg-ments listed above. The proof can now be completed using a standard strong induction argument, and we defer the details to the full version.
4 Concluding Remarks We introduced the notion of diverse non-crossing matchings. While we show that DIVERSE COMPATIBLE NCMS can be solved in polynomial time for points in convex position, the complexity of the closely related problem DIVERSE NCMS (where we drop the demand for compatibility from the solution matchings) remains open even for convex point sets. The complexity of all problems considered for more general inputs remains open. We also believe that exploring other notions of diversity, based on either different aggregation func-34th Canadian Conference on Computational Geometry, 2022 Figure 5: An example of how a base polygon divides the subproblem on Qi,j further into six smaller in-stances.
tions (e.g, sum instead of minimum), or other notions of distance (different from Euclidean), would also pose interesting directions for future research. We also pro-pose to study the problems proposed here for more than two matchings.
References Abu-Affash, A. K., Carmi, P., Katz, M. J., and Tra-belsi, Y. (2014). Bottleneck non-crossing matching in the plane. Computational Geometry, 47(3):447–457. (↑2) Aichholzer, O., Bereg, S., Dumitrescu, A., Garc´ ıa, A., Huemer, C., Hurtado, F., Kano, M., M´ arquez, A., Rap-paport, D., Smorodinsky, S., Souvaine, D., Urrutia, J., and Wood, D. R. (2009). Compatible geometric match-ings. Computational Geometry, 42(6):617–626. (↑2) Aloupis, G., Cardinal, J., Collette, S., Demaine, E. D., Demaine, M. L., Dulieu, M., Fabila-Monroy, R., Hart, V., Hurtado, F., Langerman, S., Saumell, M., Seara, C., and Taslakian, P. (2013).
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Di-versity in Kemeny rank aggregation: A parameterized approach.
In Proceedings of the Thirtieth International Joint Conference on Artificial Intelligence, IJCAI, pages 10–16. ijcai.org. (↑1) Baste, J., Fellows, M. R., Jaffke, L., Masar´ ık, T., de Oliveira Oliveira, M., Philip, G., and Rosamond, F. A. (2022).
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Bottleneck bichromatic plane matching of points.
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In Proceedings of the 31st International Symposium on Algo-rithms and Computation, ISAAC, pages 26:1–26:12. (↑1) Fomin, F. V., Golovach, P. A., Panolan, F., Philip, G., and Saurabh, S. (2021). Diverse collections in matroids and graphs. In Proceedings of the 38th International Sympo-sium on Theoretical Aspects of Computer Science, (STACS) 2021, pages 31:1–31:14. (↑1) Ganesh, A., HV, V. P., Nimbhorkar, P., and Philip, G.
(2021).
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In Kowalik, L., Pilipczuk, M., and Rzazewski, P., editors, Proceedings of the 47th International Workshop on Graph-Theoretic Concepts in Computer Science (WG), pages 94– 105. Springer. (↑1) Hershberger, J. and Suri, S. (1990). Applications of a semi-dynamic convex hull algorithm.
In Proceedings of the 2nd Scandinavian Workshop on Algorithm Theory (SWAT), pages 380–392. Springer. (↑1) Ishaque, M., Souvaine, D. L., and T´ oth, C. D. (2012).
Disjoint compatible geometric matchings. Discrete and Computational Geometry, 49(1):89–131. (↑2) Lo, C., Matousek, J., and Steiger, W. L. (1994). Algo-rithms for ham-sandwich cuts. Discrete and Computa-tional Geometry, 11:433–452. (↑1) Mantas, I., Savic, M., and Schrezenmaier, H. (2021).
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In Proceedings of the 7th International Conference on Al-gorithms and Discrete Applied Mathematics, CALDAM, pages 151–164. Springer. (↑2) Nadel, A. (2011). Generating diverse solutions in SAT.
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Open AccessReview
Pathologic and Therapeutic Schwann Cells
by
Michael R. Shurin
Michael R. Shurin
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Sarah E. Wheeler
Sarah E. Wheeler
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Hua Zhong
Hua Zhong
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Yan Zhou
Yan Zhou
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Division of Clinical Immunopathology, Department of Pathology, University of Pittsburgh and University of Pittsburgh Medical Center, Pittsburgh, PA 15213, USA
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Department of Respiratory and Critical Care Medicine, Shanghai Chest Hospital, Shanghai Jiao Tong University School of Medicine, Shanghai 200025, China
Author to whom correspondence should be addressed.
Cells 2025, 14(17), 1336;
Submission received: 22 July 2025 / Revised: 20 August 2025 / Accepted: 25 August 2025 / Published: 28 August 2025
(This article belongs to the Special Issue Emerging Roles of Glial Cells in Human Health and Disease)
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Abstract
Schwann cells (SCs) are the primary glial cells of the Peripheral Nervous System (PNS), which insulate and provide protection and nutrients to the axons. Technological and experimental advances in neuroscience, focusing on the biology of SCs, their interactions with other cells, and their role in the pathogenesis of various diseases, have paved the way for exploring new treatment strategies that aim to harness the direct protective or causative properties of SCs in neurological disorders. SCs express cytokines, chemokines, neurotrophic growth factors, matrix metalloproteinases, extracellular matrix proteins, and extracellular vesicles, which promote the inherent potential of the injured neurons to survive and accelerate axonal elongation. The ability of SCs to support the development and functioning of neurons is lost in certain hereditary, autoimmune, metabolic, traumatic, and toxic conditions, suggesting their role in specific neurological diseases. Thus, targeting, modifying, and replacing SC strategies, as well as utilizing SC-derived factors and exosomes, have been considered novel therapeutic opportunities for neuropathological conditions. Preclinical and clinical data have demonstrated that SCs and SC-derived factors can serve as viable cell therapy for reconstructing the local tissue microenvironment and promoting nerve anatomical and functional recovery in both peripheral and central nerve injury repair, as well as in peripheral neuropathies. However, despite the promising successes of genetic engineering of SCs, which are now in preclinical and clinical trials, improving tactics to obtain ‘repair’ SCs and their products from different sources is the key goal for future clinical success. Finally, further development of innovative therapeutic approaches to target and modify SC survival and function in vivo is also urgently needed.
Keywords:
Schwann cells; neuropathy; cellular therapy; neurotrauma; nerve repair; neuropathic pain; neuroinflammation
Graphical Abstract
1. Introduction
Schwann cells (SCs), the primary glial cells in the peripheral nervous system (PNS), exhibit two key phenotypes: myelinating and non-myelinating. They originate from embryonic Schwann cell precursors (SCPs) derived from the neural crest, which travel and proliferate along the PNS axonal tracts. SCPs differentiate into immature SCs, which then become pro-myelinating SCs that are eventually converted into either myelinating or non-myelinating SCs . During embryogenesis, SCs safeguard neuronal longevity and participate in axonal pathfinding. Simultaneously, they orchestrate the architectural setup of the developing nerves, including the blood vessels and all layers of the peripheral nerves . Later, SCs enable the transmission of neural impulses, deliver nutrients and shelter for neurons, guide axons in nerve repair, and control inflammatory and immune responses in different tissues and organs by enwrapping both myelinated and non-myelinated axons and cross-talking with cells in the local microenvironment . While most nerve fibers are unmyelinated in the PNS, all nerve fibers of the PNS are enveloped by SCs: axons are either bound by a single-layered SC cell membrane to make non-myelinated nerve fibers or enfolded by several coatings of SC membranes to form myelinated nerve fibers [4,5]. Although non-myelinating SCs do not make myelin, they participate in essential functions of the PNS, including the maintenance of axonal metabolism and control of neuropathic pain. Non-myelinating SCs also possess the potential for myelination, as they can transit to repair SCs and initiate the repair process during nerve injury. Non-myelin-forming SCs may likely act as the “first responders” to traumatic insults or illnesses in their local environment .
SCs participate in the pathogenesis of various dysmetabolic, traumatic, and hereditary conditions and diseases of the PNS (Figure 1). SCs may initiate pain and modulate pain sensitivity in both physiological and pathological conditions [7,8]. The role of specific receptors, ion channels, and bioactive molecules in mediating neuropathic pain by SCs has been well described . Notably, a specialized mesh-like network of cutaneous SCs that conveys noxious thermal and mechanical sensitivity has been recently reported . These sensory SCs could transmit nociceptive information to the closely associated unmyelinated nociceptive nerves. In addition, SCs, specifically perisynaptic non-myelinating SCs, participate in the formation and function of the neuromuscular junction, are involved in polyneuronal innervation and synapse elimination, and can alter the synaptic transmission .
SCs are also known to function as immunomodulating cells because they can process and present antigens, produce cytokines and chemokines, and express pattern recognition receptors, such as Toll-like receptors (TLRs) and the nucleotide-binding and oligomerization domain (NOD)-like receptors (NLRs) [12,13]. SCs were identified in the thymus and spleen, where they directly interact with lymphocytes and dendritic cells under both normal and pathological conditions [14,15]. For instance, the interaction between SCs and macrophages is crucial for orchestrating tissue regeneration after peripheral nerve injury . Human repair SCs are phagocytic; express MHC-II, CD40, CD80, CD86, B7H3, and PD-L1; release chemoattractants, matrix remodeling proteins, and pro- and anti-inflammatory cytokines; and may serve as regulators of T-cell immunity in nerve regeneration . Interestingly, in the bone marrow, non-myelinating SCs may be an essential component of the bone marrow niche responsible, at least in part, for maintaining hematopoietic stem cell (HSC) hibernation by regulating activation of latent transforming growth factor beta (TGF-β), as autonomic nerve denervation has been reported to reduce the number of TGF-β-producing SCs, leading to rapid loss of HSCs from the bone marrow .
SCs play a key role in PNS nerve regeneration after trauma or injury by controlling nerve fiber regeneration, myelination, and axonal guidance . They also participate in the formation of a “nerve bridge” between the proximal and distal stumps of a damaged nerve, seen as a 3D structure of fibroblasts, perineurial and inflammatory cells, and a matrix . The metabolic switch in SCs and metabolic coupling between SCs and axons deliver pyruvate and lactate to damaged neurons to prevent axonal degeneration . These complex steps of Wallerian degeneration, or programmed axon degeneration, and axon regeneration are orchestrated by several SC-targeting factors, which polarize SCs into a reparative phenotype and control SC interaction with neurons, immune cells, fibroblasts, and endothelial cells [20,22]. Dedifferentiation, activation, proliferation, and migration of SCs upon injury, recognized as SC plasticity , result in the recruitment of immune cells, breakdown of myelin, elimination of debris, removal of dead cells, promotion of axonal recovery, and subsequent myelination of regenerated nerve fibers [23,24]. However, the neural repair ability of SCs may be reduced with age .
Furthermore, the incredible plasticity of repair SCs is not limited to their role in PNS nerve regeneration. Contemporary data have revealed that SCs possess a broad capacity to promote the restoration and regeneration of various adult tissues, in addition to PNS axons . For example, the use of a mouse model of excisional cutaneous wounding helps reveal that SCs may also induce adult skin wound healing: genetic excision of SCs resulted in suspended wound contraction and closure, inhibited myofibroblast formation, and weakened skin re-epithelialization after damage . These data may also highlight the importance of better understanding the connections between melanocytes and SCs in development and transformation . Furthermore, new data have demonstrated the dedifferentiation of SCs after neighboring alveolar bone injury and their contribution to bone regeneration . At the molecular level, the injury-induced activation of SCs is associated with the expression of various factors that regulate epithelial–mesenchymal transitions (EMT), stemness, and differentiation states linked to and contributing to injury-induced tissue remodeling .
Understanding the role of SCs in the functioning of the PNS and their interactions with various cell types in different tissues and organs, as well as their contribution to the pathogenic mechanisms of neurodegenerative diseases, peripheral neuropathy, neuroinflammation, certain infections, pain, and cancer, is not surprising. In the PNS, for instance, various insults, such as mutations, lesions, autoimmune reactivity, and infections, can damage the myelination process, alter the functions of SCs, and ultimately lead to neurodegeneration (Figure 1). Distraction of SCs is seen in different peripheral demyelinating disorders, such as Charcot–Marie–Tooth (CMT) disease, where myelinating SC dysfunction is attributed to the mutations of genes encoding the myelin proteins and leads to segmental demyelination (the loss of myelin sheaths after their proper development), dysmyelination (improper development of myelin sheaths), muscle dystrophy, and sensory loss . Another example is hyperglycemia-induced SC damage in diabetic neuropathy that results in the loss of trophic support from SCs and damage to both sensory and motor nerves . Interestingly, even though demyelinating diseases have different etiologies, such as inherited, autoimmune, inflammatory, or toxic flaws in myelin development, the concept of the “demyelinating SCs” as a specific cell phenotype required for myelin sheath clearance has been recently announced [33,34].
Thus, a thorough understanding of SC behavior in nerve regeneration, wound healing, neuropathies, and associated pathophysiological conditions may provide a practical tool to accelerate and direct the regenerative process and create the grounds for the development of new glia-centric treatment approaches to counteract axonal loss, pain, neuroinflammation, and potentially tumorigenesis . Glial cells are approximately 10 times more numerous than neurons . Remarkably, several reports suggest that SCs and SC-like cells are among the most promising cell subsets for cell-based therapies [30,37,38,39,40].
2. Schwann Cells in CNS Pathology
The nervous system’s hierarchical organization enables the control of peripheral functions and facilitates a wide range of interactions between the central nervous system (CNS) and the PNS. The confirmed presence of SCs in the CNS in various pathophysiological conditions, along with their reported beneficial effects on central remyelination, opens exciting opportunities for developing rational therapeutic approaches that utilize SCs in CNS repair.
2.1. Schwann Cells for Treating Central Neurotrauma
Traumatic brain injury and spinal cord injury are devastating conditions that lead to profound disability and many related complications. Due to the limited regenerative capacity of the CNS, there is currently no effective cure for these disorders. Developed treatments, such as cell transplantation, could be an encouraging opportunity for treating affected patients, based on the proven role of resident SCs in the remyelination of the spinal cord after injury. In 1981, Duncan et al. demonstrated in a focal demyelination mouse model that SCs can myelinate spinal cord axons . In 1988, Berry et al. reported that SCs participated in axon regeneration within peripheral nerve grafts transplanted into the CNS . Even though SCs do not occur endogenously in the CNS, many new data suggest that non-myelinating and myelinating SCs can be engaged from the PNS to help myelin repair after CNS injury [15,43,44,45]. Using the model of demyelination and paraplegia in rats, a spontaneous entry of SCs into the spinal cord, along with axonal remyelination, concomitant functional recovery from paraplegia, and SC myelin replacement by oligodendrocyte myelin, has been demonstrated . Furthermore, endogenous SCs can be encouraged to migrate by minimal insults within the spinal cord and integrate with astrocytes under certain conditions . Interestingly, new data demonstrate that SC migration to demyelinated CNS lesions may be guided by blood vessels . Oligodendrocyte precursor cells (OPCs) in the CNS may undertake fate-switching to differentiate into remyelinating SCs after demyelination and nerve damage [49,50,51]. These and other data revealed that resident SCs could reverse a critical neurological deficit caused by CNS demyelination, strongly supporting the capacity of SCs to contribute to central nerve injury repair . Intriguingly, it has been postulated that SC-remyelinated axons may be more resistant to demyelinating diseases of the CNS .
The benefits of transplanting cultured SCs for spinal cord injury therapy have been thoroughly investigated in experimental animals since the early 1990s. Numerous preclinical and clinical studies provide evidence that the transplantation of SCs is a feasible therapeutic approach for spinal cord repair (Figure 1 and Figure 2). The utility of SC transplantation over the last few decades has primarily focused on paradigms related to spinal cord injury and, to some extent, traumatic injury of the optic nerves and peripheral nerves [52,53,54]. For instance, it has been demonstrated that SC transplantation promotes axon regeneration from retinal ganglion cells and peripheral-type myelination in the injured optic nerve or spinal cord in a rat model [55,56]. Many studies have successfully utilized SCs for spinal cord repair and advancing functional rehabilitation in animal models of spinal cord injury [57,58,59,60,61].
It was estimated that more than 70% of engrafted SCs were myelinating in the chronically injured spinal cord. In addition to myelinating axons, grafted SCs engage resident SCs to myelinate the injured axons . Furthermore, grafted SC participation in remyelination and promotion of nerve restoration has been demonstrated in a rat cerebral hemorrhage model . The administration of SCs into the injured spinal cord has been reported to reduce tissue loss, support the survival of damaged neurons, encourage axonal regeneration, and accelerate myelination of axons, thereby improving sensorimotor function . For instance, one report revealed a two-fold rise in the number of protected neurons within the lesion after SC transplantation . The ability of SCs to restrain the pro-inflammatory microglia and macrophage phenotypes after intraspinal transplantation has also been reported [30,64], which may contribute significantly to the neuroprotective effects of SCs within the subacute spinal cord injury setting. Administration of SCs may also reduce the activation of inflammasome complexes . Finally, in a xenotransplant model, human SCs have been shown to persist within the contused nude rat spinal cord for six months after transplantation, exhibiting no evidence of tumorigenicity, showing a low proliferation rate, and displaying a controlled biodistribution to the lesion .
Results of several clinical trials suggest that autologous SC transplantation appears safe. No adverse effects were observed two years after SC injection [67,68,69,70,71,72,73]. Thus far, the practicability of autologous SCs for spinal cord injury has progressed through Phase 1 clinical trials [69,74]. For instance, adding SCs into the bioengineered PNS conduits increased regeneration by bridging nerve grafts. Santamaría et al. reported optimized safety and efficiency results of intralesional cell delivery for subacute human spinal cord injury, which allowed the development of new procedures for cell delivery into patients with chronic spinal cord injury. Key parameters of the transfer technique included accurate localization of the injury site, stereotaxic strategies to control needle trajectory, the process of entrance into the spinal cord, spinal cord motion reduction, the volume and density of the SC suspension, frequency of delivery, and regulation of shear stresses on cells . Recent clinical trials have confirmed earlier reports on the feasibility and safety of autologous human SC transplantation in participants with chronic spinal cord injury .
It is crucial to notice that the efficacy of remyelination of injured CNS axons by grafted SCs has been reported to be limited because of different aspects of the hostile injury-induced microenvironment and specific SC properties, like the weak migration of SCs into the CNS: SCs do not migrate extensively, display poor long-term survival, and remyelinate only some axons in the vicinity of the site of transplantation [78,79]. Astrocytes are known to limit the mobility of SCs in the CNS [80,81]. Another SC motility inhibitory mechanism may be the potential of myelin-associated glycoprotein (MAG) to suppress SC migration and induce SC death . However, the regenerative ability of the nerve grafts could be enhanced by using genetically modified SCs expressing growth factors and other known neuroprotective and neuroregenerative molecules (Figure 2), like glial cell-derived neurotrophic factor (GDNF), neurotrophin-3 (NT-3), polysialyltransferase (PST), POU domain class 6 Homeobox 1 (POU6F1), etc. [83,84,85]. For instance, CNS migration of retrovirally transfected SCs expressing the polysialylated form of the neural cell adhesion molecule (NCAM) was significantly increased without altering their differentiation program or impairing their myelinating ability . Furthermore, the grafting of these SCs significantly enhanced functional recovery after spinal cord injury . Similar results were obtained with SCs retrovirally transduced to overexpress the cell adhesion molecule L1, which is known to support neurite outgrowth and is implicated in the myelination process. Transplantation of these SCs enhances early events in spinal cord repair after injury, as evidenced by faster locomotor recovery in mice .
Interestingly, experimental data suggest a higher efficacy of a combination strategy with SC transplantation, like, for instance, the paired elevation of cyclic AMP (cAMP) levels or paired administration of neuroprotective agents or growth factors, which improves the effect of SCs after spinal cord injury [89,90,91,92]. Similarly, amplified neurotrophin secretion by the implanted SCs, virally transfected to produce a bifunctional neurotrophin molecule D15A, which mimics both NT-3 and brain-derived neurotrophic factor (BDNF), alone or together with chondroitinase ABC resulted in a significantly augmented number of axons in the contusion site of the injured spinal cord, promoting axonal regeneration and locomotion [93,94].
Furthermore, the suspension of SCs in bioactive matrices can improve the longevity of transplanted SCs and upregulate their ability to assist axonal regeneration in the injured spinal cord [95,96]. In the model of complete spinal cord transections, the administration of SCs and Matrigel, or SCs and a fibrous piezoelectric polyvinylidene fluoride trifluoroethylene conduit, resulted in the detected regeneration of brainstem axons [95,97]. Co-transplantation of SCs and other cell types, such as bone marrow mesenchymal stromal cells or neural stem cells (NSCs), is also considered a more effective therapy than transplantation of SCs alone following spinal cord injury [38,40]. In addition, SC transplantation as a treatment for spinal cord injury can be supported by combination with electroacupuncture. Tan et al. reported that electroacupuncture improves the survival and proliferation of transplanted SCs, inhibits SC apoptosis, and increases the expression of neuregulin 1 (NRG1) type III in SCs . These effects were associated with promoting axonal regeneration, altering astrogliosis by inducing the extension of astrocytic processes at the SC graft interface, and facilitating remyelination after spinal cord injury [96,97].
Although our understanding of the molecular mechanisms by which transplanted SCs employ a reparative effect on the injured spinal cord is quite limited, the proven clinical benefits include improvement of axon regeneration, remyelination of newborn or sparing axons, control of the inflammatory response, and preservation of the survival of damaged tissue [40,73]. Therefore, multiple barriers still must be overcome, including standardization of human SC manufacturing and understanding of their longevity in the injured tissue after transplantation, to enable feasible and successful treatments for spinal cord injury patients. Finally, despite the established clinical and safety improvements in human SC transplantation for spinal cord injury resulting from recent advances in understanding SC biology and improvements in SC cultures from human nerves, SC transplantation as a treatment strategy for peripheral polyneuropathies has not been systematically followed.
2.2. Schwann Cells for Treating Non-Traumatic CNS Diseases
Various human neurological disorders, such as multiple sclerosis, Parkinson’s disease, Huntington’s disease, or Alzheimer’s disease, result from damage or loss of neurons and glial cells in the brain or spinal cord. Demyelinating CNS diseases can arise due to traumatic insults, genetic mutations, metabolic defects, autoimmune reactions, hypoxic conditions, infectious diseases, or exposure to inflammatory mediators or toxins . These conditions disrupt axon–glial interactions and often lead to secondary demyelination. However, in multiple sclerosis, for instance, demyelination is a primary event, and axonal myelin-preservation signals are still reported. Furthermore, hereditary myelinating defects originate from mutations in structural myelin genes or associated transcription factors.
Various demyelinating disorders are accompanied by continuous and disabling myelin loss . CNS demyelination often results in severe functional impairment, which explains a great interest in designing treatments that promote repair in the CNS and patient recovery. Remyelination is a main issue for avoiding neurodegeneration and irreversible failures of function . Like spinal cord injury models, the involvement of SCs in CNS remyelination during CNS inflammation has also been reported. In a model of inflammatory degenerative CNS neuropathy in dogs, dedifferentiated or injury-responsive SCs were detected in cerebral and cerebellar white and gray matter lesions, as well as in the brainstem, which was associated with a local reduction in axonal damage . A similar response of SCs to axonal damage was described in the canine distemper virus model of multiple sclerosis . Fascinatingly, examination of SC remyelination in the rodent models of ethidium bromide-induced areas of demyelination in the spinal cord following endogenous and transplant-mediated remyelination revealed the CNS, rather than PNS, origin for most SCs that remyelinate damaged CNS axons .
This evidence of gliogenesis in the lesioned CNS is of special clinical importance, as it highlights the role of regeneration-promoting cells that can be engaged for therapeutic purposes. Cell replacement therapy is a promising approach for myelin repair. While various experimental studies demonstrated the convincing repair potential of grafted myelin-forming cells, such as OPC, olfactory ensheathing cells (specialized glial cells of the CNS and PNS showing the characteristics of astrocytes and SCs), and neural and embryonic stem cells, SCs are among the most promising contenders for autologous transplantation . Such conclusions are further supported by recent data indicating that glial cells may play a crucial role in non-cell autonomous neurodegeneration in neurological disorders, contributing to glial crosstalk and its impact on neurons. Particular emphasis has been given to glial cells in neurologic autoimmune disorders because autoantibodies, in addition to targeting neurons, may also target glial cells, causing a disparity in neural networks and synaptic homeostasis, as well as stimulation of neuroinflammation [104,105].
There are more than 30 identified autoimmune diseases of the nervous system, mediated, in general, by CNS- or PNS-targeting antibodies . Autoimmune disorders of the CNS include demyelinating disorders, such as multiple sclerosis and neuromyelitis optica, as well as paraneoplastic and autoimmune encephalomyelitis. In addition to genetic predisposition, the breakdown of immune tolerance and the progression of autoimmune responses may be triggered by infections. For instance, the results obtained in some multiple sclerosis models suggest that prolonged Epstein–Barr virus (EBV) or cytomegalovirus (CMV) infections may induce autoreactive memory T cells that can be reactivated after CNS injury . New data have added strong evidence that EBV infection may cause MS [108,109]. One possible mechanism is molecular mimicry, where EBV protein sequences resemble myelin and other CNS proteins, triggering an autoimmune response against myelin and CNS antigens. Other possibilities may include a dysregulated immune response toward EBV, leading to enhanced bystander activation and an inadvertent autoreactive response, or infection-induced brain tissue damage, resulting in an immune response to immune-privileged tissue antigens.
Cell replacement therapy with normal or genetically engineered SCs delivered to the diseased or injured brain may provide the foundation for developing efficient new treatment approaches for various human neurological diseases (Figure 2).
2.2.1. Multiple Sclerosis and Schwann Cells
Multiple sclerosis is a chronic, demyelinating disease of the CNS characterized by inflammation, demyelination, and axonal damage resulting from the infiltration of immune cells into the CNS and the autoimmune targeting of oligodendrocytes and myelin. The pathological characteristic of multiple sclerosis is the existence of focal demyelinated lesions detected in the CNS . Although spontaneous remyelination might happen in multiple sclerosis, as was demonstrated by the analysis of postmortem tissues from multiple sclerosis patients [111,112], its efficiency, or remyelination index, varies substantially among patients and correlates with functional outcome . Even though accessible therapies can control multiple sclerosis inflammation, they have little effect on progressive neuropathology. In addition, none of the pharmacological agents tested for improving intrinsic myelination pathways delivered reliable termination of demyelination or myelin repair [114,115]. At the same time, the results of transplanting myelin-forming cells have presented convincing evidence for the concept that prompt remyelination protects axons from degeneration and fosters morphological and functional recovery in multiple sclerosis [100,115].
Earlier data demonstrating that spinal cord multiple sclerosis lesions often contain regenerating myelin sheaths formed by SCs [80,116,117] have been confirmed later , supporting the potential of SCs to contribute to CNS repair in multiple sclerosis. Interestingly, SCs have also been observed in the CNS of neuromyelitis optica patients . Together, this highlights a critical need for feasible pro-remyelinating strategies to advance tissue repair and reduce long-term axonal damage. SC transplantation has been discussed as a promising approach that may aid clinical recovery in individuals with multiple sclerosis .
Preclinical studies using experimental autoimmune encephalomyelitis (EAE) rodents showed that transplantation of olfactory unsheathing cells and SCs into the demyelinating area markedly augmented the repair and regeneration of myelin [37,121,122]. For instance, allografted SCs not only survived and migrated within the parenchyma but also intermingled with astrocytes in demyelinated lesions, aligned with axons, and formed P0 protein internodes, i.e., remyelinated lesions in the inflamed CNS. Noticeably, this was associated with a marked decrease in animal mortality . These experimental results align with early data indicating SC involvement in CNS remyelination in EAE animals [49,124]. Notably, accelerating remyelination in the EAE model is sufficient to maintain axonal integrity and neuronal function .
New data indicate that triggering receptor expressed on myeloid cells-2 (TREM2), detected on microglia, regulates the clearance of myelin debris and remyelination . TREM2 agonistic antibodies promoted myelin uptake and degradation in a model of multiple sclerosis, supporting remyelination and axonal integrity . Interestingly, SCs also express TREM2, which plays a critical role in the energy metabolism of SCs and in assisting nerve regeneration . Thus, TREM2 agonists may be potential therapeutic tools for targeting SCs or improving SC-based therapies in different demyelinating neuropathies.
Much less is known about the therapeutic potential of SCs for other CNS autoimmune diseases, such as neuromyelitis optica spectrum disorders or myelin-oligodendrocyte glycoprotein antibody-related diseases, which manifest as optic neuromyelitis, acute demyelinating encephalomyelitis, or cortical encephalitis. This is despite the involvement of CNS glial cells in their pathogenesis having been proven, suggesting that they may be potential clinical targets for innovative treatment approaches [129,130]. Regardless, inhibiting demyelination and promoting myelin sheath regeneration are crucial therapeutic strategies for treating autoimmune CNS diseases.
Thus, while the existence of SCs in the CNS in specific pathophysiological settings has been repeatedly confirmed , the signals that hinder or permit SCs to take control of central axon remyelination are not yet wholly understood . Identification of the CNS factors that regulate SC longevity and motility in the central parenchyma, such as the astrocyte-derived molecules or myelin components, for instance, is crucial to design reasonable treatment approaches for the utilization of SCs in CNS repair to complement oligodendrocyte remyelination and to reduce axonal loss.
2.2.2. Parkinson’s Disease and Schwann Cells
Parkinson’s disease is a common neurodegenerative disorder characterized pathologically by intracytoplasmic inclusions or Lewy bodies composed of α-synuclein aggregates in neurons.
The oxidant–antioxidant theory, as one of several hypotheses proposed to explain the etiology of Parkinson’s disease, asserts that oxidative damage triggered by reactive oxygen species (ROS) can promote dopaminergic neuron degeneration . Accumulating results revealed that oxidized lipids, proteins, and nucleic acids accumulate in the brain tissues of patients with Parkinson’s disease, suggesting that free radicals may affect both α-synuclein aggregation and disease progression . Although a few pathogenic pathways have been implicated in Parkinson’s disease, including mitochondrial dysfunction and oxidative stress, altered iron metabolism and ferroptosis, failure of the ubiquitin–proteasome system, and impaired protein clearance, the processes of apoptosis and/or autophagy, and neuroinflammation, their interaction, and cellular involvement are not yet entirely understood . Neuroinflammation is a notable pathological indication of Parkinson’s disease, and reactive microglia could scavenge abnormal α-synuclein produced by neurons.
The loss of peripheral nerve fibers and activation of SCs with the increased expression of inflammatory cytokines IL-1β, IL-6, and TNF-α were also observed in Parkinson’s disease patients, demonstrating the potential involvement of SCs in the pathophysiological process of the disease [135,136]. Remarkably, α-synuclein immunoreactivity was detected in SCs near spinal nerve roots , and phosphorylated α-synuclein deposits were seen in SCs . New results confirmed that phosphorylated α-synuclein presence in vagus nerve SCs caused activation of the TLR2/MyD88/NF-κB signaling pathway, leading to neuroinflammatory responses and Parkinson’s disease autonomic dysfunction [139,140,141]. These data suggest that abnormal SC α-synuclein accumulation may be an additional pathogenic biomarker for Parkinson’s disease and a new target for autonomic dysfunction in Parkinson’s disease by blocking the SC phosphorylated α-synuclein/TLR2 signaling pathway. Furthermore, in animal studies of Parkinson’s disease, it has been reported that glucagon-like peptide-1 (GLP-1) receptor agonists, such as exendin-4, provide neuroprotective effects in the brain—defense of dopaminergic neurons in the substantia nigra, block of dopamine loss in the basal ganglia, and preservation of motor control . Remarkably, exendin-4 has been reported to support the survival, proliferation, and migration of SCs, as well as myelination, in the DRG neuron/SC co-culture system [143,144]. One clinical trial in patients with Parkinson’s disease revealed that exendin-4 had beneficial effects on cognition and memory .
Importantly, the therapeutic capability of SCs for Parkinson’s disease has also been examined in experimental animal models, based on their proven neuroregenerative properties in the PNS . For instance, in the 1-methyl-4-phenyl-1,2,3,6-tetrahydropyridine- and 6-hydroxydopamine (6-OHDA)-induced models of Parkinson’s disease, it was reported that PNS grafts could survive in the CNS, increase dopamine concentrations in the brain, and improve behavioral tests in rodents, which was associated with the secretion of basement membrane components by SCs [147,148]. Next, the transplantation of GDNF- or FGF-expressing SCs and nigra grafts in 6-OHDA-lesioned rats markedly improved the survival of dopaminergic neurons [149,150]. Moreover, SCs grafted into the brains of 6-OHDA-induced Parkinson’s disease in mice diminished the damage to dopaminergic neurons . The results of SC co-drafting studies were also confirmed in a primate model of Parkinson’s disease . Notably, one human clinical trial investigates whether autologous SCs can be used to restore damaged dopaminergic neurons in the CNS . Thus, intracerebral cell-based therapy by administration of autologous PNS-derived SCs is an encouraging therapeutic approach that may prevent or inhibit Parkinson’s disease progression. However, further clinical studies are needed to confirm these initial findings.
2.2.3. Alzheimer’s Disease and Schwann Cells
Alzheimer’s disease is a neurodegenerative disorder manifested by extracellular amyloid beta plaques in the grey matter, intraneuronal hyperphosphorylated tau filaments, neuronal death, synapse elimination, and brain atrophy, and is associated with free radical and oxidative stress, metabolic dysregulation, and upregulated pro-inflammatory cytokines . For instance, upregulated generation of reactive oxygen and reactive nitrogen species and defective mitochondrial dynamic balance can lead to the misfolding of amyloid-beta and hyperphosphorylation of tau proteins, affecting tau protein kinase activation and phosphatase inhibition, thus causing the formation of neurofibrillary tangles, a hallmark of Alzheimer’s disease . Dysfunctional mitochondria may further release ROS, increase the production of pro-apoptotic proteins, and cause apoptosis of neurons .
Recent clinical data emphasize myelin destruction as a key and particularly aggravating factor in disease pathogenesis . Oligodendrocyte death and demyelination are thought to happen secondary to neurodegeneration in Alzheimer’s disease . Although the results of many studies support the concept that the transplantation of stem cells, which may transdifferentiate into neuronal lineage, could improve synaptic plasticity, cognitive performance, and neurogenesis circuitry in Alzheimer’s disease models [158,159], SC-based treatment approaches have not been widely evaluated.
Yan et al. investigated the effect of transplanting SCs into the brains of rats with Alzheimer’s disease in a model . First, they demonstrated that, in the in vitro co-culture system, SCs significantly promoted NSC proliferation, differentiation, and synaptic formation compared to nerve growth factor (NGF) alone. In vivo experiments revealed that the transplantation of either NSCs alone or (NSCs + SCs) reduced glial fibrillary acidic protein (GFAP) and S100β but increased choline acetyltransferase mRNA expression and markedly endorsed learning and memory in affected rats. However, the effect of (NSCs + SCs) brain co-grafting was significantly more robust. The authors concluded that co-transplantation of NSCs and SCs extensively lowered astrocyte counts and enhanced cholinergic neuron counts, while accelerating the recovery of learning and memory function in Alzheimer’s disease rats compared to grafting NSCs alone . Notably, these results were partially confirmed when it was reported that co-culturing SCs and NSCs supported the differentiation of NSCs into neurons and the secretion of higher concentrations of neurotrophic factors, such as BDNF and glial cell line-derived neurotrophic factor (GDNF) .
Furthermore, in preclinical animal studies of Alzheimer’s disease, it has been reported that GLP-1 receptor agonists, like exendin-4, liraglutide, and lixisenatide, cross the blood–brain barrier and provide neuroprotective effects in the CNS—reduction of β-amyloid plaques, oxidative stress, and the chronic inflammatory response, and averting damage of synapses and memory . Interestingly, exendin-4, a GLP-1R agonist, is known to support the survival, proliferation, and migration of SCs, as well as myelination, in the DRG neuron/SC co-culture system [143,144]. The recent development of micromolecular synthetic neurotrophin mimetics, selectively acting on neurotrophin receptors, offers a unique opportunity for novel therapeutic approaches to neurodegenerative diseases . For instance, BDNF mimetics, known to mediate pro-myelinating effects in the nervous system, may have beneficial therapeutic potential for Alzheimer’s disease . Finally, TREM2 is known to be involved in the amyloid pathology of Alzheimer’s disease through the regulation of glial cell proliferation, homing, and phagocytosis of amyloid plaques . TREM2 agonists have been shown to enhance microglial cell survival and function, offering a potential avenue for developing TREM2-targeting therapies for Alzheimer’s disease . At the same time, TREM2 is highly expressed in SCs, regulating SC metabolism and myelinating function . This may provide an additional indication of further evaluation of modified SC-based treatments in Alzheimer’s disease.
However, many more experimental studies are needed to prove the feasibility of using SCs for mono- or combinational treatment of Alzheimer’s disease.
2.2.4. Huntington’s Disease
Huntington’s disease, the common autosomal-dominant neurodegenerative disease, presents with severe motor, cognitive, and psychiatric symptoms. A mutation of the HTT gene, which is associated with the synthesis of a toxic, misfolded huntingtin protein, leads to specific pathological features such as neuronal dysfunction and subsequent cell death . Stem cell therapy, as a new treatment for Huntington’s disease, utilizes different types and sources of stem cells, such as mesenchymal stem cells (MSCs), NSCs, embryonic stem cells (ESCs), and reprogrammed stem cells or induced pluripotent stem cells (iPSCs) . Stem cells can potentially replace lost neurons, improve neuron regeneration, and deliver pro-survival factors. Another advantage of autologous stem cell transplantation is its origin from the patient, which alleviates the risk of immunological rejection and averts the need for immunosuppression therapy . Interestingly, new animal studies have demonstrated that Huntington’s disease mutation present in iPSCs can be corrected before the transplantation of these cells, without limiting their successful differentiation into neurons .
Several ongoing clinical trials investigate the safety and efficacy of autologous stromal cells in patients with Huntington’s disease . Although no data have revealed the effect of SC transplantation in patients with Huntington’s disease so far, many new encouraging results can predict the approach of a new era of Huntington’s disease therapeutics.
2.2.5. Leukodystrophies
Leukodystrophies are heterogeneous neurodegenerative diseases that primarily affect myelination in the CNS. However, some forms of leukodystrophy are characterized by progressive demyelination of both the CNS and PNS, causing severe neurological symptoms . For instance, axonal degeneration in both the CNS and PNS has been described in Krabbe disease (globoid cell leukodystrophy) and metachromatic leukodystrophy. At present, more than 30 forms of leukodystrophies have been categorized as caused by mutations in myelin- or oligodendrocyte-specific genes or in other white matter components, including astrocytes, neuronal axons, microglia, and brain vasculature .
While the results from animal models and human preclinical studies suggest that glial cell therapy may serve as a cell-based remedy for glial-driven leukodystrophies in the near future, no data have demonstrated the feasibility of using SCs for these diseases . However, new studies may accelerate the development of this approach.
In metachromatic leukodystrophy, a mutation in the ARSA gene, encoding the lysosomal enzyme arylsulfatase A, results in the accumulation of sulfatides in multiple cell types, including SCs, leading to cytotoxicity and demyelination of most of the nerve fibers of the CNS and PNS . Metachromatic material, accumulating in peripheral nerves in patients with metachromatic leukodystrophy, consists of SCs and endoneural macrophages loaded with typical lysosomal inclusions of sulfatides or inclusion bodies. SCs and macrophages die, and demyelination of myelin in the PNS follows . These are also morphological alterations of the endoplasmic reticulum and mitochondria in SCs. Several types of inclusion bodies were seen in SCs on sural nerve biopsy. In addition, the endoplasmic reticulum, mitochondria, and lysosomes in SCs demonstrated significant ultrastructural alterations, suggesting that a subcellular metabolic insufficiency of SCs in metachromatic leukodystrophy may be responsible for the PNS pathology . Similarly, “abnormally dense mitochondria with thickened cristae” in SCs were reported in another ultrastructural analysis of nerve biopsies from a patient with metachromatic leukodystrophy .
Elevation of sulfatide synthesis in oligodendrocytes and SCs of ARSA−/− mice has been reported to significantly upregulate the accumulation of sulfatides and the development of myelin pathology in the CNS and PNS . The sulfatides are structurally normal but disrupt myelin metabolism.
New data on modeling mitochondrial metabolism in metachromatic leukodystrophy showed that human SCs transfected to overexpress ROS in response to sulfatides and displaying bioenergetic and mitochondrial defects could be treated with metformin, which repaired their metabolic activity and diminished ROS production . Interestingly, analysis of myelination in mouse axons by SCs transplanted from human sural nerves of healthy donors and metachromatic leukodystrophy patients revealed the formation of metachromatic granules within SCs in the leukodystrophic grafts . This demonstrated that the grafted SCs remained ARSA deficient during nerve regeneration and were unable to utilize the enzyme produced by the surrounding cells. At the same time, fibroblasts isolated from a metachromatic leukodystrophy patient and genetically modified to overexpress ARSA were able to competently transfer ARSA to oligodendrocytes and SCs through the inner membrane in vitro , suggesting a few therapeutic directions.
In the second classic genetic leukodystrophy—globoid cell leukodystrophy, or Krabbe disease, there are genetic defects in a lysosomal hydrolase, galactosylceramidase, which catabolizes the myelin sphingolipid galactosylceramide. This causes reactive astrocytic gliosis, the rapid and nearly complete disappearance of myelin and myelin-forming oligodendrocytes and SCs, and infiltration of macrophages (“globoid cells”) . A progressive dysfunction of SCs was detected in the animal model of Krabbe disease [181,182].
While bone marrow or hematopoietic stem cell transplantation and mesenchymal stem cell therapy have proven the feasibility and potential effectiveness in leukodystrophies, SC-based approaches have not yet been developed [171,183,184]. However, in one study, the clinical effect of mesenchymal stem cell infusion in metachromatic leukodystrophy patients has been explained by the mesenchymal stem cell differentiation into SCs in vivo .
2.2.6. Schwann Cells in Amyotrophic Lateral Sclerosis
Amyotrophic lateral sclerosis (ALS), or Lou Gehrig’s disease in the USA, is a terminal neurodegenerative disorder and the most common form of motor neuron disease. Although approximately 50 potentially pathogenic and several causative genes have been identified in patients with ALS, the etiology of ALS appears heterogeneous and is not well understood [186,187]. The neuropathological signature of ALS is characterized by the loss of neuromuscular connections, axonal retraction, and the progressive degeneration of both upper motor neurons and lower motor neurons in the CNS, accompanied by astrogliosis and microgliosis . Emerging evidence has demonstrated common extra-motor involvement in ALS, implicating the PNS as a converging point. An increasing amount of data reveal the presence of sensory and autonomic dysfunction in ALS, which has been reported in up to 30% of cases . For instance, 80% of ALS patients displayed a drop in intraepidermal nerve fiber density on skin biopsy . Similar signs of axonal degeneration and peripheral motor and sensory dysfunction have been reported in ALS animal models [191,192]. Thus, it is not surprising to read “ALS is a well-known peripheral neurodegenerative disease” .
However, the pathogenic significance of SCs in distal motor axonopathy is poorly understood. In ALS patients, motor axon loss in the peripheral nerves causes a Wallerian-like degeneration characterized by denervated SCs and immune cell infiltration. Clinical and preclinical results suggest that peripheral axons are lost before the death of cell bodies in the CNS [194,195]. However, peripheral nerve pathology in ALS demonstrates a chronic and progressive degenerative and inflammatory phenotype. It is possible that chronically activated SCs also trigger or orchestrate an inflammatory mechanism in ALS.
Immunohistochemical analysis of SCs on sections of the sciatic nerves from ALS subjects demonstrated a substantial enhancement of GFAP and S100b staining compared with sections from control nerves, suggesting a significant SC reactivity . Differential GFAP and S100β staining in SC subsets resembled denervated or repair phenotypes, such as those observed in Wallerian degeneration. New data confirmed that cultured SCs prepared from sural nerve biopsy material obtained from an ALS patient appeared senescent; however, when treated with exosomes isolated from a cadaver donor SC culture, their growth potential in vitro was significantly improved . A recently identified subset of SCs expressing Adamtsl1, Cldn14, and Pmp2 that preferentially myelinate large-caliber motor axons has been reported to be reduced in both ALS model mice and ALS patient nerves [198,199].
Sciatic nerves from superoxide dismutase (SOD1)G93A rats, resembling ALS, also showed that S100b+ and GFAP+ SCs correspond to two different—myelinating and denervated—cell populations . Cytoplasmic and mitochondrial SOD1 convert superoxide radicals into molecular oxygen and hydrogen peroxide, and mutations in SOD1, associated with alterations in SOD1 functionality and/or aberrant SOD1 aggregation, contribute to ALS pathogenesis. Rodent transgenic ALS models, developed by expressing a human SOD1 transgene with ALS-associated mutation G93A, reproduce the major phenotypic features of human ALS . Importantly, the development of motor neuron disease is not due to a loss of SOD1 function, as many mutant forms of SOD1 retain nearly normal or even elevated SOD1 activity [201,202]. Thus, while neither upregulated expression of wild-type SOD1 nor obliteration of endogenous SOD1 caused motor neuron disease, it is evident that the disease is induced by an acquired toxicity of mutant SOD1 independent of its dismutase activity .
Increased SOD1 activity can ameliorate the production of ROS in the mitochondrial intermembrane space, contributing to mitochondrial damage in SOD1G93A rats . Therefore, mechanisms critical for ALS progression may include mitochondrial dysfunctions, excitotoxicity, oxidative stress, and changed Ca2+ metabolism. The effects of SOD1 G93A mutation in non-neuronal cells, such as glial cells, in ALS may be associated with the altered redox balance and perturbed expression of Ca2+ transporters that may be responsible for altered mitochondrial Ca2+ fluxes .
Interestingly, sciatic nerves from ALS rats also showed upregulated endoneurial expression of colony-stimulating factor 1 (CSF1) and IL-34: CSF1 was expressed by a subset of phagocytic S100+ p75NTR+ SCs engulfing myelin debris, while IL-34 was expressed by a subset of denervated GFAP+ SCs with the morphology of repair SCs, which also expressed c-Kit (stem cell factor, SCF, receptor). In the sciatic nerves of ALS donors, CSF1 and IL-34 were detected in elongated SC-like cells . Most importantly, CSF1-R and c-Kit receptor inhibitor masitinib prevented SC reactivity in degenerating nerves and ameliorated sciatic nerve pathology in ALS rats.
Furthermore, ALS-derived mouse SCs did not affect 3D axonal migration in vitro, whereas it was strongly enhanced by normal SCs . Analysis of ALS mice demonstrated that disease onset is associated with a significant initiation of GFAP expression in SCs, suggesting their distress . iNOS immunoreactivity was also increased in SCs on peripheral nerves in SOD1 mutant mice . SCs expressing a dismutase active mutant SOD1 were shown to reduce disease progression in ALS mice, as the removal of mutant SOD1 reduced survival. This suggests a link between slow disease progression in ALS mice and a protective impact of dismutase active mutant SOD1 in SCs . Importantly, SCs transfected with mutant SOD1 expressed low levels of peroxiredoxin 1, and the expression of peroxiredoxin 1 mRNA was significantly decreased in the lumbar spinal cord of SOD1G93A mice . Peroxiredoxin 1, an antioxidant molecule derived from SCs, protected motor neurons from hydrogen peroxide-induced cell death, suggesting that the reduction of peroxiredoxin secreted from SCs contributes to increased ROS and accompanies motor neuronal death in ALS . Similarly, peroxiredoxin 6 secreted by SCs has been shown to significantly inhibit neuron apoptosis and improve neurological recovery in different models .
However, while myelinating SCs may attempt to remyelinate wounded axons, the ongoing motor neuron degeneration in ALS overcomes this regenerative competence, leading to axonal nerve dysfunction . In addition, the loss of myelinating SCs, which are prone to apoptosis, as observed in the SOD1-G93A/TDP-43 model , also decreases the overall support available to motor neuron axons . Dysfunctional SCs in the SOD1-G93A model display modified gene expression, low axonal metabolic support, and the failure to remove debris from degenerating axons .
Non-myelinating SCs may also participate in muscle weakness and atrophy associated with the peripheral nerve dysfunction seen in ALS patients. Evidence shows that Remak SCs are demoted in the SOD1-G93A model of ALS, suggesting that Remak SC dysfunction may precede or accompany motor neuron degeneration . It has also been reported that the denervation status correlates with the S100+ terminal SC loss from the neuromuscular junction in the Thy1-hTDP-43WT mouse model of ALS . Terminal SCs are non-myelinating SC types that are crucial for synapse development and maintenance . Early and permanent functional alterations of perisynaptic SCs in the neuromuscular junction were also detected in the SOD1-G37R mouse model of ALS, recapitulating many aspects of the human disease [216,217]. Significantly, similarly to axonal SCs, perisynaptic SCs contribute to neuromuscular junction reinnervation following denervation by adopting a repair phenotype , which may be regulated by muscarinic acetylcholine receptors (mAChRs) . Comparative transcriptomic analysis of perisynaptic and other SCs concluded that synaptic SCs may be responsible for age-associated degeneration of the neuromuscular junction .
An analysis of terminal SCs in mice expressing the G93A-SOD1 mutation allowed the conclusion that these SCs may play a role in motor terminal degeneration and denervation in motor neuron pathology . Disease progression is associated with a loss of complete S100 and P75NTR immunoreactivity in denervated neuromuscular junctions, suggesting a loss of perisynaptic SCs. This was confirmed in different animal models and human biopsies [220,221,222]. Interestingly, it has been suggested that perisynaptic SCs may play a pathogenic role in ALS . This might be mediated by a compromised synthesis of the neuromuscular junction regulator agrin, or secretion of matrix metalloproteinase (MMP)-degrading agrin , or generation of CXCL12α (stromal cell-derived factor, SDF-1) in perisynaptic SCs , or SC-induced Connective Tissue Growth Factor-related pathways .
Together, these results support the notion that SCs are promising targets for therapeutic intervention in ALS. Thus, despite attracting limited consideration, SC abnormalities in ALS hold potential for innovative therapies targeting pro-myelinating pathways. As mentioned above, c-Kit signaling may drive some SC abnormalities in ALS . Masitinib, an inhibitor of the protein tyrosine kinase c-kit and other receptors, lowered SC reactivity and immune cell infiltration in nerves in SOD1-G93A rats . Phase 2b/3 and Phase 3 clinical trials involving ALS patients treated with masitinib demonstrated a slowing of disease progression rates [187,226]. In line with this is a first case report of an 81-year-old patient with rapidly progressive ALS who was treated with weekly intravenous infusions of allogeneic SC-derived exosomes to potentially restore impaired SC and motor neuron function . The results revealed a trend for clinical stabilization during the infusion period, suggesting an interesting approach to address damaged SCs and probably neuropathy in patients with ALS.
3. Schwann Cells in PNS Pathology
Peripheral nerve disorders can be genetic or acquired as a result of traumatic, toxic, metabolic, infectious, or autoinflammatory conditions. As non-myelinating and myelinating SCs ensheathe the axons of peripheral nerves, playing a critical role in PNS development, physiology, and regeneration, the loss or dysfunction of SCs is an etiologic or pathogenic feature of many diseases and conditions that compromise the PNS (Figure 1 and Figure 2). Understanding the interplay between SC plasticity/polarization and peripheral neuropathies could reveal novel strategies for treating and managing sensory, motor, or mixed PNS pathology.
3.1. Schwann Cells for Treating PNS Injury
PNS injuries, because of traction, contusion, compression, ischemia, total/partial section, or complications of infection, disorders, or toxicity, often promote sensory and motor disability and neuropathic pain, representing many cases of chronic disability in younger and otherwise healthy individuals . Biological therapy, utilizing various cellular, molecular, and bioengineering approaches, may be the solution for enhancing the outcome of peripheral nerve repair after surgery or rehabilitation [228,229].
Given their role in the pathophysiological repair response and remyelination after peripheral nerve injury, the application or targeting of SCs for improved outcomes has been considered for many years. For example, Guenard et al. in 1992 transplanted SCs into a sciatic nerve defect in rats and revealed improved axonal regeneration and myelin formation . Later, the clinical application of SCs from different sources and SC-like cells for accelerating nerve regeneration was repeatedly tested and characterized [19,229,231,232]. The first human experience with autologous SCs in 2016–2017, to supplement nerve repair after complete transection of sciatic nerves by a boat propeller injury and after partial damage of the sciatic nerve by a gun wound of the leg, revealed that patients regained complete or partial sensory and motor function recovery [233,234]. New studies now focus on the use of allogeneic stem cell-derived SCs and SC-like cells to promote axonal regeneration and repair peripheral nerve injury-induced neuropathy .
These data demonstrate that cell replacement-based therapeutic strategies utilizing exogenous SCs can be curative . There are two common cell replacement approaches for peripheral nerve repair: the isolation of repair SCs for cell transplants and the controlled differentiation of pluripotent stem cells or the lineage conversion of available somatic cells into induced SCs . However, experimental and clinical data demonstrate that direct SC transplantation frequently results in relatively low therapeutic efficacy, particularly for long-distance peripheral nerve lesions. This is because cell administration may induce cell damage and instant loss in the circulatory system. The interaction between SCs and the extracellular matrix also plays a crucial role in peripheral nerve repair . In addition, while SCs are susceptible to local milieu signals due to their plasticity, they cannot maintain their repair phenotype for a prolonged period because the signals gradually alter . Given the limited repair ability of SCs for extended peripheral nerve defects, it is essential to complement SCs through gene engineering, co-transplantation with other cells, or by combining them with bioactive or structural scaffolds . Thus, specific biomaterials have been engineered as functional carriers to mimic the compositional and topological signs of the extracellular matrix (ECM) and maintain the biological properties of SCs for effective cell transplantation .
Ultimately, a comprehensive understanding of myelin clearance pathways is crucial for identifying cellular mechanisms that can be targeted to promote myelin clearance and enhance remyelination and recovery after PNS injury. Using the SC conditional knockout of the calcineurin B model, Reed et al. showed that calcineurin may regulate autophagy in SCs and play an important role in myelin clearance after injury . Likewise, NGF could initiate autophagy in dedifferentiated SCs, upregulate clearance of myelin debris, and promote axon and myelin regeneration at an early stage of peripheral nerve injury .
These data provide novel insights into the mechanisms of pharmacological therapy for activating SC autophagy after peripheral nerve injury, offering a potential avenue for its preclinical testing and verification.
3.1.1. Combinational Schwann Cells Transplantation for Treating PNS Injury
Different biomaterials that mimic the natural regeneration environment may protect SCs and help cell integration into the injured CNS or PNS during SC transplantation. Such biomaterials should be biocompatible, biodegradable, and permeable, allowing cell migration, adhesion, and encapsulation. Various natural biomaterials like fibrin, chitosan, collagen, laminin, and fibronectin, as well as synthetic biomaterials like poly (glycolic acid), polylactic acid, and poly (lactic acid–glycolic acid) copolymerization (PLGA), offer feasible cell-to-cell interaction and outstanding physical properties [243,244]. SCs can be resuspended in a pre-gel solution for in situ gelation or pre-encapsulated in a hydrogel to prevent serious cell damage during administration. For example, collagen hydrogel can provide crucial extracellular matrix molecules and simulate the nanofibrous structure of native nerves to support SC functions. It has been shown that a tethered type-1 collagen gel helps maintain the expression of growth and guiding factors from SCs and promotes nerve regeneration in a rat sciatic nerve defect model . Chitosan conduits combined with collagen, laminin-, or fibronectin-based matrices significantly improved the therapeutic effects of SCs in bridging long nerve gaps in different animal models [246,247].
The role of plasma fibrin in peripheral nerve regeneration has also been demonstrated that the blood–nerve barrier can be disrupted after nerve injury, inducing fibrinogen leakage into the wounded area and cleavage by thrombin to form fibrin cables. These fibrin cables may stimulate SC polarization to a regenerative phenotype and provide physical guidance for SC migration [248,249]. Schuh et al. used a fibrin hydrogel as a transplantation platform for SCs and reported that they mixed fibrin with collagen to create an engineered neural tissue. They proved that adding fibrin aided marked improvement of SC survival and endorsed nerve growth both in vitro and in vivo when compared to collagen alone . Polysaccharides, like hyaluronic acid- and alginate-based hydrogel that can host exogenous SCs, have been used as bridging materials for peripheral nerve injury repair . An alternative to natural biomaterials is synthetic hydrogels, which utilize polyethylene glycol, poly(2-hydroxyethyl methacrylate), and poly[N-(2-hydroxypropyl)-methacrylamide]. However, they are not frequently used alone as SC carriers .
Curcumin, the main active ingredient in turmeric from the root of Curcuma longa, may accelerate the repair of sciatic nerve injury in rats by lowering SCs apoptosis and increasing their proliferation and myelination [251,252]. Interestingly, the addition of curcumin-encapsulated chitosan nanoparticles to the neural guidance conduit prepared by poly-L-lactic acid (PLLA) and surface-modified multi-wall carbon nanotubes (MWCNTs) and filled with SCs significantly improved the regeneration and functional recovery of injured peripheral nerves . Resveratrol, a natural plant compound, can promote sciatic nerve crush injury recovery by promoting the autophagy of SCs . Interestingly, to address local hypoxia in the early stages of peripheral nerve injury, Ma et al. used perfluorotributylamine-based oxygen-carrying fibrin hydrogel in combination with SCs and concluded that enhanced survival of SCs accelerated axonal regeneration, remyelination, and recovery of the motor and sensory function of the regenerating nerves .
Recent experimental animal and preclinical human studies revealed that the transplantation of SCs in combination with axon guidance channel technology, like nerve scaffolds, e.g., decellularized nerve conduits, promotes the repair of injured peripheral nerves and thus represents an applicable strategy for the management of peripheral nerve injury [256,257]. Many studies have reported the development of different types of conduits to increase the therapeutic efficacy of SC transplantation to repair peripheral nerve injuries [240,258]. For instance, the rate and distance of axonal regeneration across a 10 mm nerve defect were increased when SCs were added to a synthetic matrix peptide hydrogel . Increased efficacy of peripheral nerve regeneration, using an electrically conductive biodegradable porous neural guidance conduit, produced from PLLA, MWCNT, and gelatin nanofibrils coated with erythropoietin-loaded chitosan nanoparticles, for transplantation of allogeneic SCs, has also been reported . Similarly, using collagen tubes confirmed that adding SCs to a guidance channel markedly increased the gap space that can be repaired after extended segmental defects of peripheral nerves . Huang et al. have developed a composite nerve conduit with a PLGA hollow tube as the outer layer and gelatin methacryloyl encapsulated with vascular endothelial growth factor (VEGF)-A transfected SCs as the inner layer . This approach provided a comprehensive solution for improved angiogenesis and nerve regeneration.
A rat model of laryngeal nerve injury was used to determine the efficacy of SCs and neural stem cells co-transplantation with a laminin–chitosan–PLGA nerve conduit to repair 5 mm long recurrent nerve injuries . The results demonstrated that the combination approach promoted nerve regeneration markedly better than the autograft method, the use of SC/nerve conduit only, or the neural stem cells/nerve conduit only approaches . Remarkably, SC-aligned scaffolds demonstrated better regeneration success in a 15 mm nerve defect in the rat model than MSC-aligned constructs . A novel PLGA copolymer foam conduit of longitudinally aligned 60–550 µm channels and luminal surfaces promoting SC adherence, when mixed with SCs and implanted across a 7 mm gap in the rat sciatic nerve, demonstrated a strong regenerative capability and may be used to promote precisely guided neural regeneration . Similar results were reported after testing SC-encapsulated chitosan–collagen hydrogel nerve conduits . Treatment of PLGA nerve guidance conduits seeded with SCs and implanted into a 10 mm sciatic nerve gap with low-intensity ultrasound resulted in larger areas of axonal myelination and improved histological recovery . Remarkably, low-intensity ultrasound is known to upregulate the proliferation of SCs and the expression of myelinating genes in SCs .
Altogether, experimental and clinical data suggest that SC engrafting and targeting for peripheral nerve injury is a promising regenerative strategy that may restore neurological function (Figure 1). However, special attention should be given to evaluating SC myelination and axonal growth outcomes of scaffold-based therapies for peripheral nerve injury in experimental animal models . It is also essential to notice that SCs play a fundamental role in the selectivity of nerve regeneration: mature SCs exhibit modality-specific phenotypes and mechanisms, allowing them to differentially control the selective regeneration of motor and sensory axons .
3.1.2. Modified Schwann Cells for Treating PNS Injury
Transplantation of modified or engineered SCs has emerged as a promising cell-based therapy to facilitate peripheral nerve recovery and remodel the microenvironment, thereby maintaining nerve homeostasis. One of these promising approaches is the genetic engineering of SCs, which will be reviewed later. Several examples of other therapeutic SC modifications are presented here.
Magnetic actuation induced by applying superparamagnetic iron oxide nanoparticles (SPIONs) to SCs can also encourage and preserve repair-like phenotypes of SCs, which is associated with the promotion of nerve regeneration and functional recovery in rat sciatic nerve injury models . In the same model, short-term low-frequency electrical stimulation upregulated BDNF expression in the injured sciatic nerves, inducing earlier SC myelination. It elevated the number of myelinated fibers and the myelin sheath thickness . Brief low-frequency electrical stimulation may enhance regeneration even after delayed nerve repair . It is known that SCs respond to electrical stimulation by increasing the proliferation and expression of NGF, GDNF, and BDNF, which can be associated with the augmented outgrowth of injured axons in vivo . A magnetic force-based mechanical stimulation of SCs demonstrated a compelling ability to enhance neurite outgrowth in vitro and nerve regeneration in vivo .
Furthermore, new data revealed that the three-dimensional aggregation of SCs upregulated their acquisition of a repair phenotype and that SC spheroid-derived secretome promoted neurite outgrowth in dorsal root ganglion (DRG) neurons . Transplantation of preassembled SC spheroids in rats with a sciatic nerve transection improved the restoration of injured nerves and motor functional recovery.
Notably, recent reports have demonstrated that inhibiting pyroptosis in SCs, induced by PNS injury, promotes the healing of the sciatic nerve and the restoration of motor function in rats . Pyroptosis is a peculiar form of programmed cell death that could hinder peripheral nerve regeneration. It was reported that silencing Rab32, a small GTPase protein family member, using an adeno-associated virus diminished nerve injury-induced SC pyroptosis and stimulated peripheral nerve regeneration.
Another agent affecting SCs and supporting PNS regeneration is melatonin. It could promote the expression of Parkin, an E3 ubiquitin ligase that mediates mitophagy—an important cytoprotective mechanism of the removal of impaired mitochondria. Melatonin has been shown to suppress SC apoptosis, reduce the production of mitochondrial ROS, and promote myelin regeneration and peripheral nerve repair . Suppression of ROS production by Parkin-mediated mitophagy can reduce SC death associated with peripheral nerve injury . Similarly, it was reported earlier that melatonin improved the proliferation and migration of SCs during peripheral nerve injury, thus advancing nerve regeneration [278,279]. SCs, expressing melatonin membrane receptors MT1 and MT2, respond to melatonin via a complex signaling cascade, stimulation by upregulated expression of GDNF, dedifferentiation, migration, and reduced glial scar formation . Selective activation of cannabinoid receptor 2 (CB2) may also be considered as a therapeutic approach to promoting the remyelination process following peripheral nerve injury. Pharmacological activation of CB2 by specific agonists upregulates the expression of pro-myelination-related genes, including Sox10, Egr2, and Tprv4, in SCs, and alters the expression of Myelin protein zero (MPZ) and myelin basic protein (MBP). In the sciatic nerve injury model, activation of CB2 increases SC proliferation and pro-myelination, improves myelin thickness, and accelerates the remyelination of injured peripheral nerves. At the same time, genetic ablation of SC-specific CB2 intensified pain and motor dysfunction in this model . The authors concluded that SC-expressed CB2 is essential for peripheral nerve regeneration.
Shen et al. have recently constructed zero-dimensional Black phosphorus quantum dots modified with the antioxidant β-carotene and evaluated their potential for peripheral nerve repair . These quantum dots activated the PI3K/Akt and Ras/ERK1/2 signaling pathways in SCs, thereby affecting their pro-inflammatory potential and promoting the functional recovery of neurons by enhancing axon remyelination and regeneration, as well as facilitating intraneural angiogenesis in rat and dog models of peripheral nerve injury.
These and other experimental data hold great potential for enhancing the effectiveness of cell delivery and the subsequent therapeutic outcomes of SC-based transplantation approaches in stimulating peripheral nerve regeneration.
3.2. Schwann Cells as Targets and Tools for Non-Traumatic Peripheral Neuropathy
Peripheral neuropathy of different etiologies, including trauma and side effects of diseases and treatments, is a frequently encountered neurological problem that leads to sensory and motor disorders (Figure 1). The treatment of these conditions primarily focuses on specific clinical symptoms and surgical interventions. Despite the latest advancements in special medications and surgical techniques, functional improvement remains disappointing. Thus, developing, designing, and evaluating novel treatment approaches for peripheral neuropathies is well justified.
Schwann Cells in Diabetic Neuropathy
Diabetic neuropathy, one of the most common complications of diabetic patients, is a unique neurodegenerative disease of the PNS that differently targets sensory axons, autonomic axons, and later motor axons. The number of diabetic neuropathy cases has tripled globally since 1990, suggesting that it is the fastest growing of all neurologic conditions . Although it is not initially thought of as demyelinating neuropathy, SCs are affected by chronic hyperglycemia, and more severe cases exhibit features of demyelination [284,285]. Advanced glycated end products, inflammation, and oxidative stress can damage SCs, decrease the production of NGF, and thus provoke axonal degeneration [286,287,288,289]. For instance, SCs, damaged by oxidative stress-induced mitochondrial dysfunction, demonstrated significantly decreased viability and increased apoptotic death associated with Bcl2, NF-κB, mTOR, and Wnt signaling [290,291]. A specific form of metabolic remodeling, involving aberrant ketogenesis within SCs, has been recently reported in streptozotocin-induced type I diabetes mellitus . This maladaptive peripheral ketogenesis depends on cannabinoid type-1 receptor signaling, as its silencing or pharmacological inhibition rebalances SC metabolism, reduces histopathological abnormalities, and improves neuropathic symptoms.
In sural nerve biopsy samples from patients with diabetes and progressive worsening of neuropathy, the full range of ultrastructural abnormalities, as well as reactive, degenerative, and proliferative changes in SCs, have been repeatedly reported . Segmental demyelination, abnormal myelin sheaths, and remyelination have been demonstrated in various animal models of diabetic neuropathy [294,295,296,297]. High glucose significantly upregulated the expression of peroxide and superoxide, as well as the pro-inflammatory cytokines IL-1β, IL-6, and tumor necrosis factor alpha (TNF-α), in cultured SCs through the TLR4/NF-κB pathway . Significantly, targeted activation of the liver X receptor signaling pathway or specific inhibition of the ROS-producing enzyme NADPH oxidase-4 in vivo and in vitro reduces diabetes-induced ROS production in SCs, thereby reversing the functional alterations of peripheral nerves and restoring the homeostatic profiles of MPZ and peripheral myelin protein 22 (PMP22) . Targeting the peroxisomal biogenesis pathway in SCs with, for instance, palmitic acid may also be an effective strategy for preventing and treating diabetic neuropathy .
Although the pathogenesis of diabetic peripheral neuropathy is multifactorial, numerous data suggest an important role of SCs in the pathogenesis of diabetic neuropathy [288,289]. Evaluating the transcriptional profiles and intercellular communication of SCs in the nerve microenvironment of a mouse prediabetes and neuropathy model, Eid et al. demonstrated that neuropathic SCs develop a pro-inflammatory and insulin-resistant phenotype under prediabetic conditions . The results are not limited by SC dysfunction, such as impaired paranodal barrier function, injured myelin, decreased antioxidative capacity, and declined neurotrophic support for axons . For instance, thioredoxin-interacting protein, which is associated with oxidative stress and inflammation, may mediate SC dysfunction in diabetic peripheral neuropathy via PI3K/Akt pathway-mediated autophagy and apoptosis . Collecting evidence proves that SC death due to high glucose is a part of the pathological process of diabetic peripheral neuropathy [287,304,305]. Bax and cleaved caspase-3 levels were significantly higher, while Bcl-2 levels were reduced considerably in SCs cultured in a high-glucose medium . Interestingly, the overexpression of nuclear factor erythroid 2-related factor 2 (NRF2) in SCs inhibited hyperglycemia-induced SC apoptosis by modulating the TLR4/NF-κB signaling pathway . High glucose-induced ferroptosis of SCs has also been recently demonstrated by inhibiting the NRF2 signaling pathway . As oxidative and endoplasmic reticulum stress; nitrification; autophagy; and NRF2, PI3K/AKT, ERK, and Wnt/β-catenin signaling are involved in the pathological pathways of high glucose-induced SC apoptosis, verification of the therapeutic targeting of these mechanisms in SCs should help identify practical approaches to treating diabetic peripheral neuropathy.
Alkaloids and polysaccharides from Aconite (Fuzi), a well-known traditional Chinese medicinal herb used for the treatment of diabetes and paralysis, were effective in accelerating the nerve conduction velocity in diabetic rats and protected SCs against high-glucose injury by decreasing superoxide anion and peroxide levels and regulating Bax, Bcl-2, cytochrome C, caspase-3, and caspase-9 pathways [307,308]. Jatrorrhizine, a protoberberine plant alkaloid with anti-inflammatory activity, may enhance SC myelination in diabetic mice by controlling the NRG1–ErbB2–PI3K–AKT pathway .
Next, as shown above, NRF2 expression may reduce cell death in SCs under hyperglycemic conditions . In the streptozotocin-induced diabetic mouse model, trehalose, a naturally occurring disaccharide, prevented nerve conduction velocity deficits and sciatic nerve damage . It also blocked high glucose-induced oxidative damage and apoptosis in SCs, thereby reducing the high glucose-induced expression of pro-apoptotic mitochondrial proteins, including Bax, Puma, Bak, and Bim. In a similar model in rats, exendin-4, a synthetic glucagon-like peptide-1 receptor agonist, inhibited peripheral nerve degeneration in association with GLP-1 receptor activation, anti-apoptotic effects, and restoration of cAMP levels in SCs .
Some data showed that a 6-week oral treatment with minocycline, a known inhibitor of SC activation, significantly improved peripheral and autonomic neuropathy in type 2 diabetic patients . Interestingly, a new study found that bupropion, an aminoketone antidepressant, reduced glucotoxicity in SCs and blocked SC apoptosis and myelin damage in treated hyperglycemic mice .
Based on data showing that neurotrophins, erythropoietin, neurosteroids, and ascorbic acid may prevent and reduce behavioral, electrophysiological, and pathological changes in polyneuropathy animal models, it has been suggested that SCs could be considered therapeutic targets for hormones and vitamins . Erythropoietin decreased intracellular ROS levels in SCs in high-glucose culture conditions, increased cell viability, and reduced the apoptotic rate of SCs [314,315]. NGF has been reported to block sciatic nerve degeneration and demyelination in diabetic rats and significantly reduce high glucose-induced endoplasmic reticulum stress and subsequent apoptosis in SCs via the PI3K/Akt/GSK3β and ERK1/2 signaling pathways . Bone morphogenetic protein 5 (BMP5) may also enhance diabetic peripheral neuropathy by enhancing mitochondrial function and reducing apoptosis in SCs . Interestingly, taurine, a natural semi-essential amino acid, can improve the morphology of the damaged myelin sheath and restrain SC apoptosis in the sciatic nerve of diabetic rats, which presumably was mediated by upregulated expression of NGF and phosphorylation levels of Akt and GSK3β in SCs . Additionally, histone deacetylase inhibitors can increase BDNF expression in SCs, thereby ameliorating diabetic peripheral neuropathy
Furthermore, the transplantation of neural crest-like cells, differentiated into SC-like cells, into diabetic mice effectively improved the impaired vascular and neuronal functions . Similarly, stem cells from various sources, such as bone marrow-derived cells, endothelial progenitor cells, and mononuclear cells, are capable of successfully reversing the symptoms of diabetic polyneuropathy due to their neuroprotective effects and neovascularization . For example, the administration of SC-like cells, prepared from human tonsil-derived mesenchymal stem cells, in BKS-db/db mice (a type 2 diabetes model) may alleviate diabetic neuropathy through the remyelination and recovery of sensory neurons . Notably, the impact of different cell therapies on SC function in diabetic neuropathy has been repeatedly reported, including decreased SC apoptosis, promotion of SC proliferation and viability, and increased myelin formation and myelin-related protein expression, as well as restoration of myelination . An improved understanding of the role of SC plasticity in diabetes should lead to new strategies for treating and managing diabetic peripheral neuropathy [323,324]. For instance, immortalized SC lines have been suggested as valuable tools for therapeutic approaches to diabetic peripheral neuropathy .
3.3. Schwann Cells in Immune-Mediated Neuropathies
Inflammatory or immune-mediated neuropathies refer to heterogeneous disorders with pathogenic immune cell infiltration of peripheral nerves, accompanied by demyelination and axonal degeneration (Figure 1) . By networking with immune cells, SCs contribute to the creation of immune responses that may lead to inflammatory neuropathies, such as autoimmune demyelinating diseases. The clinical picture ranges from Guillain–Barré syndrome (GBS) to chronic inflammatory demyelinating polyneuropathy (CIDP). GBS and CIDP are two of the most common inflammatory diseases of the PNS . Acute inflammatory demyelinating polyneuropathy (AIDP), as the GBS subtype, is highly prevalent in the USA and Europe. Although the origin of the autoimmunity in many autoimmune demyelinating diseases has not been precisely elucidated, clinical and experimental data provide clear evidence that the immune response, facilitated partly by SCs, often has a destructive role and promotes disease development both in inflammatory neuropathies and in many hereditary neuropathies [327,328]. SC expression of molecules associated with the antigen processing and presenting machinery, as well as TNF-α and IL-1, is upregulated in inflammatory conditions . On the other hand, SCs can become targets of an autoimmune response in certain inflammatory neuropathies, like the subtypes of GBS (Figure 1).
Recent data suggest that during the segmental demyelination with live myelin-forming SCs, macrophage infiltration is responsible for the myelin removal by its phagocytosis. Importantly, macrophage-attacked SCs transdifferentiated into inflammatory demyelinating SCs with a distinctive demyelination pathology—myelin uncompaction [34,329]. This pathway may represent a novel autophagy-mediated myelin clearance mechanism by SCs in inflammatory demyelinating neuropathies . These findings provide an important basis for a better understanding of the pathobiology of demyelinating peripheral neuropathies and the development of novel diagnostic and therapeutic modalities.
It is important to mention that peripheral inflammatory neuropathy might develop in a small cohort of cancer patients treated with immune checkpoint inhibitors targeting CTLA-4 or PD-1/PD-L1 signaling in T cells. The reported neuropathy cases commonly resemble acute or chronic inflammatory demyelinating or vasculitic neuropathies [331,332,333,334]. Acute neurotoxicity is also documented in up to 30% of cancer patients during CAR-T-cell therapy, which is associated with “immune effector cell-associated neurotoxicity syndrome” or cytokine release syndrome [335,336]. A case of chronic inflammatory demyelinating polyneuropathy developed in a patient with metastatic malignant melanoma after treatment with interferon-α has also been reported . Additionally, paraneoplastic neuropathies are also immune-mediated due to the immune reaction to malignant cells expressing onconeural antigens and mimicking molecules expressed by neurons.
3.3.1. Schwann Cells in Guillain–Barré Syndrome
GBS and its AIDP variant are well-known post-infectious immune-mediated polyneuropathies and the representatives of autoimmune neuroinflammatory diseases of the PNS. In many cases, axonal and demyelinating GBS variants arise after infection and implicate molecular mimicry mechanisms relying upon the structural similarity between pathogens and heterogeneous host autoantigens [338,339]. An important pathophysiological pathway, thus, is complement-mediated nerve damage caused by autoantibodies targeting the peripheral nerves, with anti-ganglioside and anti-galactocerebroside antibodies being the most common in GBS [129,340]. Approximately 60% of patients with GBS demonstrate anti-ganglioside antibodies in their sera . Notably, anti-GM1, -GM1b, -GD1a, and -GalNAc-GD1a antibodies are often detected in GBS patients after Campylobacter jejuni infection; anti-GM2 antibodies after CMV infection; and anti-galactocerebroside antibody after Mycoplasma pnuemoniae . In Miller Fisher syndrome, a GBS variant, 80–90% of patients display antibodies to GQ1b that target cranial motor nerves and may be associated with molecular mimicry after Campylobacter jejuni infection [343,344]. The elevated levels of pro-inflammatory cytokine IFN-γ and TNF-α expression correlated with extreme clinical severity, whereas the anti-inflammatory IL-4, IL-10, and TGF-β correlated with recovery . New results demonstrated that blood and CSF autoreactive CD4+ and CD8+ T cells recognize peripheral nerve myelin antigens P0, P2, and PMP22 in patients with GBS .
Up to 25% of GBS patients have circulating autoantibodies recognizing SCs and binding to the distal tips (leading lamella) of SCs . However, another study reported that 13% of GBS patients showed strong IgG reactivity against SCs . In addition, the deposition of activated complement components has been reported on the surface of SCs in AIDN patients . These and other reports advocate that SCs can serve as targets of immune-mediated demyelination . In experimental autoimmune neuritis, an animal model of GBS, an activated complement complex can be detected on SCs before myelin degradation, suggesting the role of this pathway in demyelination . In GBS, autoantibodies to MPZ, P2, gliomedin, and ganglioside GM1, expressed by SCs, may elicit the complement cascade and SC lysis. Notably, the demyelinating variants of GBS are the consequence of SC and myelin sheath injuries. Analysis of axonopathy pathways in transgenic mice expressing GM1 either solely in neurons or glia confirmed that the roles of both primary and secondary axonal injury are mediated by complement-fixing anti-GM1 ganglioside antibodies that target peripheral nerve axonal and SC membranes, respectively .
On the other hand, complement components and complement regulatory molecules can be produced and expressed by SCs [12,350,353,354], suggesting an important immune regulatory function of SCs in immune-mediated neuropathies. The immunomodulatory activity of activated SCs also plays a substantial role in GBS pathogenesis, as well as other infection-associated PNS inflammation. This may be the result of either indirect pathogen activation of cellular constituents in peripheral nerves, e.g., SCs, or direct SC infection . Crucial to such a response are SCs, as they are the providers of pivotal components of the adaptive and innate immune reactions against invading pathogens. For instance, infection of human SCs with alphaviruses, known to be associated with neurological complications, caused a significant alteration of the expression of genes encoding different immunoregulatory factors in SCs, signifying their potential role in the pathogenesis of GBS after alphavirus infection . Furthermore, SCs in sural nerve biopsies from GBS patients demonstrated upregulated expression of molecules associated with the antigen processing and presenting machinery . Finally, animal models of GBS have demonstrated that perisynaptic SCs, rather than macrophages and neutrophils, are the main cellular population responsible for clearing axonal debris following distal nerve damage .
Although the role of SCs in the neuroinflammatory pathogenesis of GBS is well established, the full scope of potential targeting of SCs or the feasibility of their therapeutic application in GBS remains to be fully deciphered. New data may accelerate these developments.
3.3.2. Schwann Cells in Chronic Inflammatory Neuropathies
The most common forms of chronic inflammatory neuropathies are chronic inflammatory demyelinating polyradiculoneuropathy, multifocal motor neuropathy, and polyneuropathy associated with monoclonal gammopathy of unknown significance (MGUS). CIDP, as a demyelinating autoimmune disorder in the PNS, is characterized by symmetric, progressive limb weakness and sensory loss. CIDP is closely related to AIDP and may be considered its chronic counterpart. Related autoantibodies commonly damage the myelin sheath in the PNS, causing segmental demyelination and remyelination: more than 40% of patients with CIDP display antibodies targeting components of myelinated nerves . Circulating IgG autoantibodies recognizing proliferating, non-myelinating SCs were detected in more than 25% of CIDP patients . Antibodies against integral peripheral myelin components, such as MPZ, PMP2, and PMP22, are also relatively common . Another report confirmed that a high percentage of CIDP patients presented a strong IgG or IgM reactivity against the myelin components, although the patterns of IgG or IgM fixation on myelinating SCs were different, suggesting that discrete myelin antigens may be recognized by autoantibodies . Myelin in peripheral nerves expresses LM1. Gangliosides GM1 and GD1b are expressed in higher quantities in motor and sensory nerves, respectively. GM1 is also widely expressed on SC microvilli . Autoantibodies binding to LM1, GM1, GD1b, and ganglioside complexes containing LM1 have been recognized in patients with CIDP . Circulating IgM anti-GM2 antibodies were detected in patients with multifocal motor neuropathy with an earlier disease onset . They also specifically target SCs and activate complement, similarly to IgM anti-GM1 on motor neurons.
New evidence suggests that immune recognition of additional antigens expressed in the noncompact myelin may also be involved in the pathogenesis of CIDP . Specifically, identifying IgG4 antibodies against the node of Ranvier structure in patients with CIDP represented a significant advance in the field . Importantly, these findings suggest that IgG4 antibodies with paranodal or nodal reactivity are pathogenic, with minimal contribution from inflammatory reactions, immune cells, or complement.
SCs may play a role in antigen presentation in inflammatory neuropathies. Analysis of the expression of co-stimulatory molecules on SCs in different types of immune-mediated neuropathies revealed the constitutive expression of the co-stimulatory molecule BB-1 on unmyelinating SCs in all studied nerves. However, prominent upregulation of BB-1 expression on the myelinating SCs was detected only in CIDP . Of note, SCs expressing BB-1 also expressed HLA-DR. Furthermore, in CIDP patients’ biopsies, but not those from healthy controls, SCs expressed the adhesion/T-cell stimulatory molecule CD58 (lymphocyte function-associated antigen 3, LFA-3) .
Finally, SCs in CIPD may lose their plasticity and neuroregenerative potential. Human and animal SCs treated with serum from CIDP patients display imbalanced expression of BDNF, GDNF, and NGF . The fact that SC abnormalities were preventable by granulocyte-macrophage colony-stimulating factor (GM-CSF), which is typically lower in CIDP sera, suggests its role in facilitating the loss of SC growth support in CIDP and may open up novel therapeutic interventions.
3.3.3. Monoclonal Gammopathy-Associated Peripheral Neuropathies
Plasma cell dyscrasia and other hematological disorders may be associated with a variety of neuropathies of both axonal and demyelinating types. This includes anti-MAG neuropathy, POEMS (polyneuropathy, organomegaly, endocrinopathy, M protein, skin changes) syndrome, and immunoglobulin light-chain amyloidosis, all of which exhibit characteristics of paraneoplastic disorders. Anti-MAG and light-chain amyloidosis can cause painful sensory neuropathy, often accompanied by autonomic features, whereas POEMS syndrome can result in axonal, demyelinating, or mixed neuropathy [367,368].
Neuropathies were also observed in lymphoma, MGUS, Waldenstrom macroglobulinemia, and multiple myeloma. Pathogenesis is likely a direct effect of monoclonal immunoglobulins on the peripheral nerve, leading to a demyelinating process. Multiple myeloma may induce various paraneoplastic neuropathies, including sensory–motor, axonal, and demyelinating neuropathies. In paraproteinemia, monoclonal IgG or IgA often triggers axonal lesions, while IgM induces demyelinating problems. Monoclonal IgM deposits are noticed in the widened lamellae of myelin fibers, and myelin debris is included in SCs . Waldenstrom macroglobulinemia can result in peripheral neuropathy in up to 47% of patients, who mostly complain of sensory loss . Predominant motor neuropathies are less common and associated with a high titer of IgM antibodies that recognize ganglioside GM1 and MAG . Amyloidosis is associated with up to 40% of neuropathies in myeloma, specifically, with free λ light chains .
Neurological complications—sensory, axonal, peripheral neuropathies or sensory and motor multiple mononeuropathies—have been described in 20 percent of cryoglobulinemia cases . Cryoglobulinemic neuropathy is also referred to as paraneoplastic, with typical monoclonal IgM precipitation . However, the role of SCs in the pathogenesis of hematological disease-associated neuropathies has not yet been established. A comprehensive understanding of SC involvement in immune-mediated nerve damage is critical and would be constructive in developing potential targeted therapies.
3.4. Schwann Cells in Infection-Induced Demyelination and Neuropathy of the PNS
Neurotropic viruses that can infect the CNS and cause encephalitis, inflammatory immune disorders, or meningitis are well characterized and include herpes simplex virus (HSV)-1/2, CMV, varicella-zoster virus (VZV), Ebola virus, and rabies virus, and several Arbo-, entero-, and Picornaviruses [373,374]. EBV, HSV-1, and herpesvirus have been linked to demyelinating diseases of the CNS [375,376]. Demyelination is among the most noticeable neurological consequences of SARS-CoV-2 infection in both the CNS and PNS . As discussed above, GBS and CIDP are major PNS demyelinating diseases induced by viral infection or abnormal immune system function. The pathogenesis of the infection-induced demyelination may range from direct infection and lysis of glial cells with degeneration of myelin to immune damage of myelin or supporting cells by cell- or antibody-mediated immune cytotoxicity due to viral antigen presentation on infected cells, myelin epitope spreading into the circulation, or molecular mimicry between virus and myelin antigens.
3.4.1. Schwann Cells in Infection-Associated Neuropathies of the PNS
Various viral infections have been associated with peripheral demyelinating disease, including infections with cytomegalovirus, EBV, human immunodeficiency virus (HIV), and VZV [378,379,380,381]. HSV and VZV have been demonstrated to infect SCs in vitro; however, the clinical significance of this pathway for direct SC infection remains to be proven . SCs are known to be involved in the pathogenesis of certain infectious peripheral inflammatory neuropathies . However, whether the immune reaction is mediated by intracellular SC infection or SC activation via pattern recognition receptors or pro-inflammatory cytokines and chemokines remains unclear. For instance, infection of human SCs with alphaviruses, known to be associated with neurologic complications, caused a significant alteration of expression of genes encoding different immunoregulatory factors in SCs, signifying their potential role in the pathogenesis of GBS after alphavirus infection .
A common source of distal peripheral neuropathy is HIV . The involvement of SCs in HIV-associated distal sensory polyneuropathy has also been reported. The HIV-1 envelope glycoprotein gp120 can activate SCs to release pro-inflammatory cytokines and chemokines, thereby mediating neurotoxicity . Deterioration of small myelinated and unmyelinated nerve fibers during HIV infection might be linked to activated SCs and attracted infected macrophages .
Persistent infection of immature SCs with lymphocytic choriomeningitis mammarenavirus blocked the ability of differentiated cells to form compact myelin sheaths without causing SC apoptosis or cytopathic defects .
Chronic infection with Trypanosoma cruzi (Chaga’s disease), an intracellular protozoan parasite able to invade different cell types, including SCs , is associated with degeneration of parasympathetic ganglia and partial functional loss of autonomic nervous system innervation. T. cruzi expresses a trans-sialidase/parasite-derived neurotrophic factor to mimic mammalian neurotrophic factors and bind to tyrosine kinase receptors to invade SCs [355,386].
Corynebacterium diphtheriae, also known as the Klebs–Löffler bacillus, is a pathogenic bacterium that causes diphtheria. The diphtheria toxin alters protein function in the host by inactivating the elongation factor EF-2. The diphtheria toxin is known to damage SCs both in vitro and in vivo [387,388]. Diphtheric neuropathy is characterized by vacuolation and fragmentation of myelin sheaths in the PNS. Secreted protein exotoxin could gain access to endoneurial fluid, bind to the SC membrane receptor, and catalyze ADP-ribosylation and inactivation of an elongation factor required for SC protein synthesis. Elongation factor 2 (EF-2), a member of the GTP-binding translation elongation factor family, permits the transfer of the peptidyl tRNA/mRNA complex from the ribosome to the peptidyl site during protein synthesis. Inhibition of myelin synthesis can lead to diphtheritic polyneuropathy [389,390].
Thus, since peripheral neuropathy caused by pathogens is a significant source of clinical disability, new studies are required to determine whether therapeutic inhibition of SC infection and SC pro-inflammatory activation is effective in treating chronic infection-associated neuropathy .
3.4.2. Leprosy and Schwann Cells
Leprosy, or Hansen’s disease, is a rare infection caused by obligate intracellular pathogens Mycobacterium leprae or Mycobacterium lepromatosis . M. leprae has been shown to have a unique affinity for infecting and persisting within both non-myelinating and myelinating SCs, leading to PNS nerve injury, including demyelination and a loss of nerve function [392,393]. Neuropathy, including sensory loss or paresthesia, and related disabilities are the main medical concerns of leprosy. Many aspects of the pathogenesis of lepromatous neuropathy remain unclear, despite significant advancements in knowledge about the bacteriological, histopathological, immunological, and clinical features of M. leprae infection .
Intracellular M. leprae infection of SCs results in the subversion of glucose/lactate and lipid/cholesterol metabolism, mitochondrial dysfunction, expression of neurotrophic factors, dysbalanced production and secretion of cytokines, and upregulation of the expression of MMP and ECM proteins [393,395,396,397,398,399,400]. Interestingly, elevated expression of both pro-inflammatory TNF-α, IL-1, IL-6, IL-8 and anti-inflammatory TGF-β, IL-10 cytokines has been reported in SCs challenged with bacilli [393,401,402,403]. M. leprae also causes significant alterations in gene expression, leading to myelin dismantling, elevated SC plasticity, dedifferentiation, and proliferation [403,404,405].
Cytokines, chemokines, and MMP released from infected SCs and activated immune effector cells, as well as SC apoptosis induced by cytotoxic T cells recognizing M. leprae antigens expressed on SCs, trigger epithelioid granuloma formation and edema, ultimately prompting axonal atrophy and demyelination [400,406,407,408]. In addition to the standard therapeutic regimens for treating leprosy infection, new therapeutics for leprosy neuropathy are needed, and candidate vaccines have shown clear advantages in preventing nerve damage [394,409].
3.5. Schwann Cells in Inherited Peripheral Neuropathies
Inherited peripheral neuropathies comprise over 100 genetically defined disorders, and the majority of affected patients suffer from the demyelinating form of the disease, with a primary pathophysiology in SCs . Other inherited peripheral neuropathy subgroups incorporate hereditary motor neuropathies and hereditary sensory and autonomic neuropathies, where motor and sensory neurons are affected. When SCs degenerate, such as in Charcot–Marie–Tooth disease type 1 (CMT1), peripheral nerves are demyelinated, and axons are injured (Figure 1). Clinically, this is characterized by progressive distal muscle weakness and atrophy, sensory deficits, loss of deep tendon reflexes, and reduced nerve conduction velocities.
3.5.1. Schwann Cells in Charcot–Marie–Tooth Disease
Charcot–Marie–Tooth disease is a clinically diverse and genetically heterogeneous group of inherited peripheral nerve disorders with different primary cells involved in its pathogenesis and more than 1000 mutations in over 100 distinct CMT disease-associated genes . Some CMT forms are frequently caused by mutations in SC-expressed genes [412,413]. CMT1 is triggered by myelinating SC dysfunction, and CMT2 results from axonal deficits. The most common form of CMT disease, CMT1A, is caused by a duplication of the gene encoding the peripheral myelin protein 22 (PMP22). Pathological overexpression of PMP22 mRNA and protein in SCs may be the principal mechanism of atypical myelin formation and axonal damage, as well as the abnormal trophic support required for the maintenance of axons [414,415].
CMT1B is an outcome of point mutations in the P0 gene, encoding MPZ . The periaxin gene encodes L- and S-periaxin, which are responsible for maintaining the stability of myelin in myelinating SCs and are mutated in CMT disease type 4F and Dejerine–Sottas syndrome, also known as CMT3 . Duplication of specific segments containing the PMP22 gene, a nearby gene RAI1, and sometimes additional genes, underlies many significant features of Yuan–Harel–Lupski syndrome, which is characterized by multiple neurological problems.
Common PMP22 mutations affect PMP22 protein trafficking in SCs , which may be seen as cytoplasmic protein accumulation in SCs of the sciatic nerve in CMT patients and modeling mice [418,419]. These PMP22 ‘aggresomes’ may be toxic, as they can cause endoplasmic reticulum (ER) stress, SC apoptosis, and demyelination . Interestingly, co-culturing PMP22 transgenic SCs with neurons induced abnormal SC differentiation, trafficking, and motility and impaired myelin formation . New data have identified the first SC-specific protein, TMPRSS5 (spinesin), which is elevated in the blood of CMT1A patients and may provide a promising disease marker . Transmembrane Serine Protease 5 demonstrated preferential expression in SCs, probably because of its regulation by the Sox10 transcription factor .
Utilizing CMT1A mice overexpressing human PMP22, Prior et al. detected that PMP22 decreased the expression of genes associated with lipid and cholesterol metabolism dose-dependently . Similar lipidomic profiles and reduced expression of genes involved in lipid metabolism were confirmed in human CMT1A patients’ SCs prepared from iPSCs. This suggests that PMP22 regulates the lipid composition of the plasma membrane and lipid storage homeostasis in SCs in CMT disease. Thus, targeting the PMP22 gene expression or lipid metabolism may hold some promise as a prospective therapeutic approach for CMT1A patients.
Progesterone, a neuroactive steroid, plays a crucial role in SC physiology through the classic intracellular progesterone receptors, membrane progesterone receptors, and gamma-aminobutyric acid (GABA) type A receptors. Steroids have been shown to affect SC morphology, proliferation, differentiation, motility, and myelination [425,426]. Progesterone may regulate myelin gene expression . A progesterone antagonist has been shown to lower the overexpression of PMP22 and to improve clinical signs in PMP22 transgenic rats . Oral L-serine supplementation was tested in mice and humans. It decreased neurotoxic 1-deoxy-sphingolipid levels, which were associated with improved sensory symptoms and strengthening of the upper and lower extremities in patients with hereditary sensory and autonomic neuropathy [429,430]. Similarly, curcumin could improve clinical and neuropathological signs in the CMT mouse model by relieving ER stress and promoting SC differentiation [431,432].
In a proof-of-concept study, Passage et al. demonstrated a beneficial effect of high-dose ascorbic acid in an animal model of CMT1A . Ascorbic acid-treated mice that overexpress the PMP22 gene showed reduced pathology in sciatic nerves, improved behavioral measures, and an increased life span compared to untreated mice. Based on these encouraging findings, clinical trials testing ascorbic acid in CMT1A have been carried out, which, however, did not demonstrate promising results [313,434,435].
Neurotrophin 3, given in either a peptide form or via adeno-associated virus (AAV)-based gene transfer, demonstrated therapeutic efficacy in CMT [414,436,437]. NT-3 is a valuable autocrine agent supporting SC survival and differentiation in the absence of axons, which plays an essential role in the early stages of myelination associated with regeneration in the adult peripheral nerves .
The usefulness of stem cell therapy for CMT has also been demonstrated by treating C22 or Tr-J mice, a model for CMT1A, with human tonsil-derived mesenchymal stem cells differentiated into SC-like cells, which the authors refer to as neuronal regeneration-promoting cells [439,440]. The significant improvement in sciatic nerve regeneration and motor function and the increased number of myelinated axons following the transplantation of SC-like cells suggest the feasibility of this approach for clinical research on CMT1A.
Viral vectors have been employed in preclinical models for treating different CMT forms by expression of promising trophic factors, like NT-3, or by targeting the specific genes in neurons or SCs [441,442]. SC-targeting gene therapy is discussed below.
3.5.2. Schwann Cells and Hereditary Neuropathy with Liability to Pressure Palsies (HNPP)
Deletion of one copy of the PMP22 gene is the most frequent genetic cause of HNPP, a recurrent, episodic demyelinating neuropathy characterized by tingling, numbness, or loss of muscle function . As a result of this deletion, the amount of PMP22 protein produced may be decreased by up to 50%. HNPP can also be caused by the PMP22 gene mutations, leading to an unusually small and unstable protein production. The deficiency of the PMP22 protein affects the structure of myelin, impairing nerve transmission and leading to disruptions in nerve signaling . New findings suggest that PMP22 dosage affects not only myelinating SCs but also non-myelinating SCs. Sural nerve biopsy of HNPP patients demonstrated increased unmyelinated axons in a single axon-containing non-myelinating SC subunit . Focal thickening of the myelin sheath, seen as sausage-like swellings (tomacula) on nerve biopsy specimens, is a characteristic feature of HNPP .
Because numerous studies have revealed that the dosage of PMP22 may determine the type and severity of the related neuropathy, such as CMT or HNPP, treatment strategies should target the direct pathways affected by the PMP22 gene dosage. As discussed above about CMT treatment, progesterone and ascorbic acid are known regulators of PMP22 expression in SCs. Gene replacement therapies are under evaluation for HNPP. High-throughput screening of cell lines expressing PMP22 reporters should facilitate the selection of novel candidates for pharmacological modulation of PMP22 expression. Another therapeutic strategy focuses on preserving SC-axonal interactions and their intimate connection by providing trophic factor support to degenerating axons in neurodegenerative diseases, including CMT and HNPP. This approach also involves manipulating SC-axonal signal transduction pathways (Figure 2 and Figure 3).
3.6. Schwann Cells and Other Types of Peripheral Neuropathy
Peripheral neuropathies may occasionally be associated with gastrointestinal diseases . For instance, more than two-thirds of patients with inflammatory bowel disease endured axonal neuropathy with sensory predominance . In ulcerative colitis, the most common neuropathology is acute inflammatory demyelinating polyradiculoneuropathy, whereas patients with Crohn’s disease often present with axonal motor and sensory neuropathy [448,449]. Neuropathy was detected in up to 23% of people with celiac disease, and these patients are at a three-fold increased risk of chronic inflammatory demyelinating neuropathy [446,450]. On the other hand, SC functions in digestive system diseases are well described . Glial cells are crucial in maintaining the intestine’s physiological functions, including nutrient absorption, barrier integrity, and immune modulation . It was also suggested that in the enteric microenvironment, the injury- and inflammation-associated upregulation of neurotrophic factors and cytokines may cause the transdifferentiation of mature enteric glia and recruitment and reprogramming of SCs, which then would regulate immune responses and perform neurogenesis [451,452]. Thus, targeting neurogenic mechanisms offers a promising approach to developing innovative strategies for acquired damage to the enteric nervous system. Notably, as an independent nervous system within the gastrointestinal tract, the enteric nervous system is a highly sophisticated neural network comprising over 100 million neurons and more than 400 million enteric glial cells .
Inflammatory neuropathies of the enteric nervous system are characterized by a dense infiltrate of immune cells associated with the neural microenvironment, leading to neuronal dysfunction and degeneration, sometimes including a complete loss of enteric neurons . A recent study has reported a good example of therapeutic transplantation of SCs in gastrointestinal neuropathy. Pan et al. utilized an animal model of Hirschsprung disease—a congenital disorder characterized by the failure of the enteric nervous system to develop in the distal part of the intestine due to the failure of neural crest-derived precursors to colonize the developing intestine . SCs were isolated, expanded in cultures, and transplanted to syngeneic animals with colonic aganglionosis. Results demonstrating the ability of SCs to engraft and restore contractile function in the aganglionic recipient smooth muscle suggest that extrinsic nerve-derived neuronal precursors could serve as an autologous source of neurons having the capability to regain innervation in the aganglionic bowel . GDNF treatment of mice in a similar model stimulated gliogenesis and neurogenesis, which, at least in part, was due to SCs that functioned as stem cells .
These studies suggest that SCs associated with extrinsic gut innervation are a potential target for therapies aimed at restoring the enteric nervous system in the gut.
4. Peripheral Neuropathic Pain and Schwann Cells
Peripheral neuropathic pain is a pathophysiological condition caused by a primary lesion or dysfunction in the PNS. Neuropathic pain may be a consequence of trauma, viral infections, treatment, endocrine dysfunctions, cancer, or neurologic disorders, among others. Interestingly, pain is a serious and common problem in patients with multiple sclerosis, although the primary mechanism is still uncertain [457,458]. While multiple sclerosis is generally considered a CNS disease, some data indicate that PNS may also be involved [459,460].
Growing evidence suggests that non-neuronal cells, particularly glial cells, are also involved in the development and resolution of pain or pain syndrome. Specifically, SCs, by providing the immediate response to nerve trauma or injury, act as a key player in the induction and maintenance of neuropathic pain, and various receptors, ligands, ion channels, and factors expressed by SCs are involved in the regulation of different pain conditions . For instance, the neurological manifestations of Fabry disease include both PNS and CNS involvement, with glycolipid deposits in SCs, DRG, and CNS neurons. Patients with Fabry disease experience chronic debilitating pain and peripheral sensory neuropathy . New data have demonstrated that changes in signaling between SCs and sensory neurons may trigger peripheral sensory nerve dysfunction . The authors found that Fabry SCs released an increased amount of S100 calcium-binding protein A10, also known as p11, which caused sensory neuron hyperexcitability.
PNS injury- or pathology-activated SCs have been shown to contribute to the increased neuronal activity that may result in chronic pain . Under these conditions, SC-derived factors may enhance (inducible nitric oxide synthase (iNOS), nitric oxide (NO), iNOS/NO, TNF-α) or reduce (low-density lipoprotein receptor-related protein 1 (LRP1, CD91), erythropoietin, basal lamina components) pain sensitivity. Importantly, these pain-controlling effects of SCs can occur independently of demyelination or axonal degeneration . For instance, activated SCs destroy the myelin sheath and produce factors that provoke hypersensitivity to pain, leading to a reduced pain threshold that triggers trigeminal neuralgia—a severe type of paroxysmal neuralgia with tender symptoms and sensations that are among the most painful . Interestingly, the results of recent studies suggest that targeting TRPA1 channels in SCs may provide an innovative therapeutic approach to fibromyalgia-linked headaches [466,467].
Several pathways may be involved in SC-mediated neuropathic pain . Following nerve injury, the activation of TNF-α receptors on SCs upregulates the expression of the P2X4 receptor, an ATP-gated cation channel that mediates Ca2+ influx and increases BDNF release. BDNF activates the tropomyosin receptor kinase B (TrkB), leading to the downregulation of the K+-Cl− cotransporter KCC2 and the upregulation of the N-methyl-D-aspartate (NMDA) receptor, both of which contribute to the occurrence of pain. Increased Ca2+ influx enhances the release of the excitatory neurotransmitter glutamate, which activates NMDA and α-amino-3-hydroxy-5-methyl-4-isoxazolepropionic acid (AMPA) receptors, thereby mitigating hyperalgesia and hypersensitivity. Furthermore, NGF from injured SCs activates the TrkA receptors, leading to the upregulation of substance P (SP), calcitonin gene-related peptide (CGRP), and the transient receptor potential cation channel subfamily V member 1 (TRPV1) receptors, also known as capsaicin receptors. SP triggers the neurokinin 1 (NK1) receptor, and CGRP reduces the degradation of SP, leading to pain sensation. NGF can also increase the expression of BDNF, thereby affecting pain perception.
Recent results have demonstrated that LRP1 in SCs may control RAG expression by sensory neurons, which may be linked to chronic pain [468,469]. This supports the notion that SCs may be an essential target for blocking pain following peripheral nerve damage.
SCs can also relieve neuropathic pain by remyelinating injured nerves . Therefore, therapeutic targeting or transplantation of SCs is of great interest and importance in designing novel therapeutic approaches to neuropathic pain. Transplantation of microencapsulated SCs next to the injured sciatic nerve in the rat model of chronic constriction injury significantly improved the healing of myelin sheaths in the injury areas compared to those without SCs . Notably, SC administration decreased the expression of ATP receptors P2X2 and P2X3 that transmit algesia and nociception signals by sensory neurons.
Inhibition of SC pannexin 1 attenuates neuropathic pain by suppressing inflammatory responses . Conversely, genetically targeting Grin1, which encodes the essential GluN1 NMDA receptor subunit in SCs, prompts hypersensitivity in pain processing in the absence of nerve injury .
Another vital aspect of SC regulation of pain sensation is associated with cancer pain, often resulting from the damage to the surrounding organs and erosion of neural tissues during tumor growth, invasion, metastasis, or from cancer treatment. Almost 50% of patients with cancer, especially at advanced stages, suffer from chronic pain, which is often persistent and intense . Accumulating evidence suggests the role of SCs in cancer pain via, for instance, increased secretion of pro-nociceptive mediators such as TNF-α, NGF, glutamate, BDNF, TLR ligands, and IL-6 [474,475,476]. Although the use of SCs in cancer pain therapy is in its early stages of development, numerous studies have reported the therapeutic effects of SCs on neuropathic pain . For instance, transient receptor potential cation channel subfamily V member 4 (TRPV4) activation in SCs has been shown to mediate mechanically induced pain in both in vitro and in vivo mouse cancer models. Conversely, TRPV4 inhibition decreased mechanical nociception in tumor-bearing mice in a dose-dependent manner . Moreover, selective in vivo SC transduction and functional block of Piezo1 channel activity in SCs using an AAV vector have been shown to alleviate mechanical hypersensitivity following nerve injury in rats . Therefore, SC-based approaches might initiate new directions in cancer pain relief.
5. Schwann Cells as a Therapeutic Target for Peripheral Neuropathies
Many PNS diseases affect the SC function and differentiation phenotype (Figure 1). With recent progress in understanding the mechanisms of crosstalk between SCs and peripheral neurons, it is clear that SCs play a central role in the pathogenesis of various inherited, metabolic, and inflammatory neuropathies [9,288,313]. The best example is the well-characterized role of SCs in the pathogenesis of CMT disease, diabetic neuropathy, and neuropathic pain. The progression of neurologic autoimmune diseases, such as GBS, which damages the myelin sheath of the PNS, ultimately affects the function of SCs, complicating the primary pathogenic mechanisms .
A growing body of evidence suggests that targeting abnormal SCs in various types of demyelinating neuropathies may be clinically beneficial (Figure 2). Historically, neurotransmitters, neurosteroids, and neurohormones have presented valuable opportunities for investigating how pharmacological interventions targeting SCs can protect or stimulate the recovery of regenerative functions in SCs in peripheral nerve disorders [313,479]. For example, inhibition of neuronal nitric oxide synthase or activation of heme oxygenase, which regulates oxidative stress in SCs, may protect against peripheral nerve degeneration [480,481]. Minocycline, a broad-spectrum antimicrobial tetracycline, has been shown to protect SCs from ischemia-like injury . Interestingly, lithium represents a fascinating pharmacological agent, unique in its ability to block the onset of myelination without promoting myelin degradation and SC dedifferentiation, while maintaining the integrity of pre-existing myelinated fibers . Roflumilast, a selective phosphodiesterase-4 inhibitor, which is commonly used to treat chronic obstructive pulmonary disease, plaque psoriasis, and atopic dermatitis, has been reported to promote the differentiation of SC to a myelinating phenotype associated with increased expression of myelin proteins, such as MBP and MAG . Treated SCs improved axonal outgrowth, which was accompanied by faster myelination.
Furthermore, recent studies have demonstrated that antineoplastic drugs, such as Epothilone B, a chemotherapeutic microtubule inhibitor, inhibit the PI3K/Akt signaling pathway, thereby stimulating SC motility without affecting SC survival and thus promoting axonal regeneration after peripheral nerve injury . Deferoxamine, an iron and aluminum chelator used for the treatment of various diseases, including acute iron poisoning and hemochromatosis, could stimulate SC viability, proliferation, and migration and upregulate the expression of myelin-related and nerve growth-promoting genes in SCs, therefore promoting peripheral nerve regeneration .
Multiple data also suggest that new approaches should be developed to expand the repair and supportive functions of SCs and avoid their deterioration. Focusing on signaling pathways, transcription factors, epigenetic mechanisms, and microRNA networks in SCs under various pathophysiological conditions is highly justified to elucidate the mechanisms of selectively regulating SC injury response, plasticity, and recovery potential for therapeutic purposes [30,487]. For instance, new data highlight the importance of epigenetic control of gene expression in normal and repair SCs, which regulate cell differentiation and myelinating function [488,489]. The acetylation of histones, mediated by histone acetylases and histone deacetylases (HDACs), is a key epigenetic pathway, and HDAC inhibitors can improve nerve regeneration outcomes . Furthermore, epigenetic regulation by HDACs in SCs after injury accelerates the de- and redifferentiation pathways in SCs, supporting several stages of nerve repair, at least partially, by controlling the expression of inflammatory cytokines . The HDAC6 inhibitor, CKD-504, regulates chaperone heat shock proteins HSP90 and HSP70, which are involved in the folding/refolding of PMP22. CKD-504 treatment reestablished myelination in both MSC-derived SCs from CMT1A patients and in the sciatic nerves of C22 mice, thereby amending the axonal integrity of the sciatic nerve and resulting in behavioral, electrophysiological, and histological progress in C22 mice . Thus, targeting epigenetic regulation of resident repair SCs by HDAC inhibitors after peripheral nerve injury may represent an interesting and promising approach. In fact, new data show that the ablation of HDAC8 in SCs accelerates the regrowth of sensory axons and the recovery of sensory function, through the control of the E3 ubiquitin ligase TRAF7, destabilization of hypoxia-inducible factor 1-alpha (HIF1α), and phosphorylation of c-Jun in SCs ensheathing sensory axons .
Multiple signaling pathways in SCs may serve as potential targets. Low-density lipoprotein receptor-related protein 4 (LRP4) is expressed in SCs and may control peripheral nerve regeneration, as conditional knockout of LRP4 accelerated demyelination and enhanced the proliferation of SCs in injured nerves, probably by downregulating the Krox-20 and MPZ pathways . This indicates that LRP4 in SCs may be a potential therapeutic target for peripheral nerve recovery. Carvacrol, a primary component of the herb Origanum vulgare, can successfully protect against experimental peripheral nerve degeneration by inhibiting the upregulation of transient receptor potential melastatin M7 (TRPM7) in SCs, suggesting its potential pharmacological prophylactic application . Many additional examples are presented in other sections discussing SCs in specific pathological conditions.
Targeting the immunomodulatory function of SCs [327,328,496] may serve as another interesting therapeutic approach. For instance, myeloid-derived suppressor cells (MDSCs), a mixed population of regulatory immune cells, may help resolve inflammatory responses and enhance functional recovery in models of injury and inflammation-mediated degeneration . Given recent data demonstrating that activated repair-like SCs can attract and upregulate the immunosuppressive activity of MDSCs and control the immunosuppressive and exhausting phenotype of T cells , targeting SCs in inflammatory neuropathies may serve as an alternative therapeutic approach.
6. Gene Therapy to Target and Modify Schwann Cells
Due to their ability to support and control axonal regeneration and remyelination following nerve injury and certain types of demyelinating neuropathies, SCs are the essential clinical target in related diseases and medical conditions (Figure 2 and Figure 3). Identifying causative genes or elucidating the molecular mechanisms of SC dysfunction in certain neuropathies has provided the basis for designing gene-modifying therapies. Novel developments in gene therapy methodologies have permitted the transfer of a functional replica of a disease-causing gene and the correction, replacement, or silencing of disease-causing genes in gene-targeting strategies .
6.1. Gene Therapeutic Targeting of Schwann Cells in CMT Disease
Gene therapy approaches, such as gene silencing or gene replacement strategies, targeting SCs have been relatively well evaluated in CMT disease and CMT animal models, demonstrating promise. This is because many genes involved in demyelinating forms of CMT disease are expressed primarily or exclusively by myelinating SCs. Additionally, many knockout mouse models of different CMT forms recapitulate the main features of CMT disease and provide an appropriate model to evaluate targeted genetic therapies to SCs .
Sargiannidou et al. and Kagiava et al. were probably the first to report that a single into-sciatic nerve or intrathecal injection of a lentiviral vector with a myelin-specific promoter resulted in targeted expression of connexin 32 in adult myelinating SCs throughout the PNS in Gjb1−/− mice, a genetically authentic model of CMT1X. They improved the nerve pathology [502,503]. Many studies confirmed the therapeutic potential of this gene addition. Gjb1−/− mice treated with lentiviral vectors encoding the GJB1 gene after the onset of peripheral neuropathy showed improved motor performance at 10 months . Intrathecal injection of lentiviral vectors encoding the GJB1 gene in transgenic Cx32 knockout mice harboring interfering T55I, R75W, or N175D CMT1X mutations to reflect different mutations in human CMT disease showed different levels of improvement in motor performance and fiber myelination . In another study, Sh3tc2−/− mice, representing CMT disease type 4C, were treated with an intrathecal injection of a lentiviral vector to drive the expression of the human SH3TC2 cDNA under the control of the Mpz promoter specifically in myelinating SCs . Gene replacement therapy significantly improved motor performance and nodal molecular architecture and reduced blood neurofilament light chain levels, a marker of neuroaxonal injury. In the same animal model of CMT4C, the utilization of an AAV9 vector and the MPZ promoter to achieve SC-targeted expression of SH3TC2 also demonstrated treatment improvements in motor performance tests, along with a decreased ratio of demyelinated fibers and elevated myelin thickness . Lee et al. used a lentivirus encoding miR-381 to reduce the expression of PMP22 in C22 mice, a CMT1A mouse model injected intraneurally distal to the sciatic notch . Treated mice demonstrated increased motor nerve conduction velocity and improved structural abnormalities in the myelination of the sciatic nerves.
Thus far, parallel studies have determined the feasibility of using AAV vectors to target SCs in CMT disease . A single lumbar intrathecal injection of the AAV-based gene construct in knockout CMT mice for targeted expression of connexin32 in SCs resulted in extensive gap junction protein biodistribution in the PNS, improved myelination and motor performance, and decreased inflammation in peripheral nerves . Similar data on SC-targeted delivery of viral vectors encoding specific proteins confirmed the previous proof of principle for a clinically adaptable gene therapy approach to treat CMT, both before and after the onset of neuropathy [511,512].
Silencing mutated or overexpressed genes is another encouraging approach to amend the phenotype of certain forms of CMT . Indeed, small interfering RNA (siRNA), AAV-delivered small hairpin inhibitory RNA (shRNA), and AAV-based microRNA (miRNA) have been tested to downregulate PMP22 mRNA expression in SCs in various CMT1A model systems . Intra-nerve administration of an AAV vector expressing shRNAs targeting PMP22 in the sciatic nerve resulted in transgene expression in resident myelinating SCs in mice, rats, and non-human primates . This gene therapy upregulated myelination and prevented motor and sensory losses for over 12 months in a rat model of CMT1A. AAV vector-based delivery of miR871 by lumbar intrathecal injection into C61-het mice, another CMT1A model, effectively transduced SCs in peripheral nerves and reduced PMP22 mRNA and protein expression . This was associated with markedly improved functional outcomes, nerve conduction velocities, and ameliorated myelin pathology in the lumbar roots and femoral motor nerves.
Interestingly, new data showed that non-viral delivery of siRNA that specifically and selectively reduced the expression level of the PMP22 mutant allele alleviated the demyelinating neuropathic phenotypes of CMT in Tr-J mice, which was associated with the reversion of the low viability of SCs, improved motor function and muscle volume, and amplified expression of myelinating proteins MBP and MPZ . Similarly, the administration of siRNA PMP22 conjugated to squalene nanoparticles caused normalization of PMP22 protein levels, regeneration of myelinated axons and myelin compaction, and restored locomotor activity in two transgenic CMT1A mouse models . Furthermore, antisense oligonucleotides that bound to the 3′-untranslated region of PMP22 could restore myelination and motor nerve function in two murine CMT1A models .
6.2. Genetic Engineering of Schwann Cells for Nerve Injury Treatment
Genetic modification of SCs can control the expression of specific factors required for peripheral nerve regeneration (Figure 2). For instance, the knockdown of histone deacetylases in SCs abrogated the expression of the transcriptional regulators of myelination SOX10 and Krox20 . Transfection of SCs to overexpress NGF, GDNF, or the transcription factor c-Jun promoted not only the upregulated expression of neurotrophic factors, which in turn led to neurite outgrowth through both autocrine and paracrine mechanisms, but also improved regenerative cellular processes . GDNF-expressing SCs enhanced the formation of myelin sheath and nerve conductance . Transplantation with SCs overexpressing nuclear factor NRF2 increased vascularity and nerve density in crushed sciatic nerves of rats, which was associated with improved survival of SCs . SCs transduced to overexpress fibroblast growth factor (FGF) demonstrated an enhanced ability to promote neuron regeneration after sciatic nerve injury . In a similar model, implanted SCs overexpressing NT-3 prevented neuron death, increased axon regeneration, and improved the damaged nerve function .
Transfection of SCs with a c-Jun-encoding lentiviral vector resulted in the upregulated expression of various growth factors, including GDNF, NGF, BDNF, and artemin . It significantly increased the ability of SCs administered with poly (ε-caprolactone) nerve conduits to improve the survival and elongation of injured sensory neurons and the sciatic function index, and alleviate target muscle atrophy and muscle action potential .
SC targeting via intrasciatic injection of adeno-associated vectors is another promising gene therapy strategy for peripheral nerve regeneration. AAV2/8 preferentially transduced SCs in vivo, and injection of AAV2/8 encoding the ciliary neurotrophic factor (CNTF) resulted in upregulated expression of myelin proteins P0 and PMP22 after transduction of injured sciatic nerves. This was associated with significantly improved axonal regeneration and the compound muscle action potential . New findings demonstrated that mechanosensitive ion channels Piezo1 and Piezo2 are highly expressed in SCs [525,526]. Genetic ablation experiments revealed that Piezo1 is an inhibitor of radial and longitudinal myelination in SCs, while Piezo2 is required for myelin formation . Selective in vivo SC transduction via intrasciatic delivery of AAV encoding shRNA against Piezo1 diminished the development of mechanical hypersensitivity after nerve injury .
Interestingly, it was reported that elevated long non-coding RNA (lncRNA) metastasis-associated lung adenocarcinoma transcript 1 (MALAT1) expression in an injured sciatic nerve promoted the proliferation and migration of SCs by sponging miR-129-5p and increasing BDNF expression and secretion . Sciatic nerve injury may also result in the upregulation of long non-coding RNA (lncRNA) axon regeneration-associated transcript (lncARAT) in SCs and SC exosomes, contributing to axonal regeneration. LncARAT absorbs miRNA-329-5p in macrophages, causing STAT-1/6-dependent elevated expression of cytokine-inducible regulator, suppressor of cytokine signaling (SOCS-2). This, in turn, induces pro-regenerative activity in macrophages . Thus, lncARAT in SCs may be a promising therapeutic tool for peripheral nerve repair. In contrast, miR-328a-3p, which is also upregulated after nerve injury, inhibits SC proliferation, migration, and viability . Lowering miR-328a-3p levels, for instance, by specific lncRNA, might support SC function after nerve damage and stimulate nerve regeneration. Remarkably, it has recently been reported that reducing miR-146a-5p in SCs and subsequently in SC exosomes leads to type 1 macrophage polarization via the TRAF6/NF-κB pathway, affecting peripheral nerve injury .
These and similar findings open new opportunities for utilizing genetic engineering of SCs to treat peripheral nerve damage.
6.3. Gene Therapeutic Targeting of Tumorigenic Schwann Cells
The additional potential object for targeted gene therapy is the tumorigenic SCs that trigger neurofibromatosis and schwannomatosis, as discussed below. For instance, high efficacy in transducing primary human SCs isolated from a plexiform neurofibroma using selected recombinant AAV vector variants has been recently described . Furthermore, numerous approaches for gene-targeted therapies have been introduced and tested in models related to neurofibromatosis and schwannomatosis, opening the path to clinical trials .
Schwannomas are also tempting targets for gene therapy due to their slow growth rate and legibility for localization. For neurofibromatosis type 1 and schwannomatosis, gene-targeted therapies are commonly intended to upregulate the levels of functional proteins in cells that express either one or no functional copies of a tumor suppressor gene . Gene therapy tactics for NF2-related schwannomas comprise a suicide gene, gene replacement, and combination gene knockdown and replacement approaches . The treatment of nude mice bearing human and mouse schwannomas with an oncolytic replication-conditional herpes simplex virus vector induced tumor shrinkage or complete tumor regression . In 2010, HSV-1 amplicon vector-based intratumoral delivery of caspase-1 under the SC promoter was reported to cause regression of schwannoma in a xenograft mouse model . Further published results showed that utilizing the P0 promoter, selectively expressed in the SC lineage, effectively avoids transgene expression in non-Schwann-lineage cells, such as neurons, and prevents any neural toxicity . Injection of an AAV vector, encoding the pro-apoptotic gene (caspase-1) under the P0 promoter in nude mice with the sciatic nerve implanted immortalized human schwannoma cells, induced tumor regression. Even more, the same gene therapy caused tumor regression and prevented tumor-associated pain in another NF2 xenograft model . An AAV1-based vector encoding gasdermin, a substrate of caspases and an effector molecule for pyroptosis, under the P0 promoter also demonstrated high antitumor efficacy in the same tumor model . Bai et al. also confirmed the feasibility of gene replacement therapy for NF1-related tumors by achieving beneficial transduction efficacies and functionality of AAV-based transduction of SCs to restore NF1-affected signaling pathways .
Ahmed et al. studied the apoptosis-associated speck-like protein containing a caspase recruitment domain (ASC) as a putative schwannoma tumor suppressor and reported that ASC expression was downregulated in 80% of the tested human schwannoma cells . ASC transfection of tumor cells induced schwannoma cell death associated with triggered caspase activation. AAV-mediated intratumor gene therapy in a human xenograft schwannoma model suppressed tumor growth by increasing tumor cell apoptosis and fixed tumor-associated pain without noticeable toxicity. The effectiveness of this schwannoma gene therapy approach was proven in a murine schwannoma model, suggesting that it may be translated to human clinical trials . A gene replacement strategy via intratumoral administration of an AAV vector encoding merlin in human NF2-null SC-derived tumors in the sciatic nerve of nude mice also demonstrated high efficacy, inducing tumor regression associated with tumor cell apoptosis .
Knockdown and overexpression experiments with SC lines, including NF1-associated malignant peripheral nerve sheath tumor cells, NF1-heterozygous deficient SCs, and NF1-deficient plexiform neurofibroma cells, to target Forkhead box M1 (FOXM1), a classical proliferation-associated transcription factor, showed that FOXM1 could mediate human malignant peripheral nerve sheath tumor development and could be a promising therapeutic target for tumor treatment .
Together, these data successfully prove the concept of SC targeting for gene silencing and gene replacement and suggest that these tools could benefit different neuropathological conditions and disorders involving SC functionality [531,532,541].
7. Schwann Cell-Derived Factors for Therapeutic Use
SC-secreted molecules maintain axonal integrity and promote regeneration, while also controlling various cells in the neuronal microenvironment by regulating physiological processes such as cell differentiation, proliferation, homing, and gene expression. Protein expression profiling of primary human SCs identified almost 20,000 unique peptides corresponding to more than 1550 individual proteins . SC-secreted signals, collectively referred to as the secretome, include, but are not limited to, growth and neurotrophic factors, cytokines, exosomes, and ECM proteins and factors. These are currently being tested from an application viewpoint as SC-derived biomaterials for nerve injury repair, axonal remyelination, and neuroprotection.
The SC secretome plays a crucial role in orchestrating Wallerian degeneration and promoting axonal regeneration . Indeed, the diminished ability of SCs to sustain the secretion of neurotrophic factors over a prolonged period may explain the collapse of regeneration in continually denervated nerves. Furthermore, SCs are known to produce neurosteroids that can, directly and indirectly, control myelin formation via binding to the classical steroid receptor in SCs and neurons [544,545,546].
7.1. Schwann Cell Extracellular Vesicles and microRNA for Therapeutic Applications
Extracellular vesicles (EVs) are a common term for several types of small-diameter vesicles produced and secreted by different cells that contain numerous molecules such as proteins, lipids, and nucleic acids (DNA, microRNA, circRNA, lncRNA) . EV therapy offers advantages over cell-based therapy, such as ease of engineering and a lower risk of tumorigenesis . Consequently, as a cell replacement therapy, isolated EVs and related engineering derivatives have been developed and applied to the diagnosis and treatment of different pathological conditions, including neurological diseases [548,549,550,551].
Exosomes are EVs with diameters ranging from 30 to 150 nm, derived from the endoplasmic reticulum. They play a fundamental role in cellular communication and are promising vehicles for cell-specific molecule and drug delivery . SC exosomes are important for axonal protection and regeneration following nerve injury. The exosomes produced by repaired SCs, unlike exosomes from differentiated SCs, upregulate the expression of miRNA-21, c-Jun, and Sox2 in SCs, which stimulate differentiation and migration of SCs and support axonal regeneration . EVs secreted by SCs may also encourage axonal elongation and regeneration of neurons via the Akt/mTOR/p70S6K pathway . SC EVs have been reported to promote axonal regeneration in vitro and advance regeneration after sciatic nerve damage in vivo [552,555]. The potential therapeutic value of SC-derived exosomal vesicles has been highlighted by promising engineering strategies to customize SC exosomes for specific activities and demonstrations of how SC exosomes provide exceptional clinical advantages over SC transplantation for repairing the wounded spinal cord [556,557].
SC exosomes have been reported to improve functional recovery after spinal cord injury. For instance, Pan et al. found that SC exosomes promote axonal protection and functional recovery after spinal cord injury by upregulating autophagy, reducing apoptosis via the EGFR/Akt/mTOR signaling pathway, augmenting Toll-like receptor 2 expression in astrocytes, and decreasing the deposition of chondroitin sulfate proteoglycan [558,559]. Newer data from that team revealed that milk fat globule-epidermal growth factor-factor 8 (MFG-E8) within SC exosomes is responsible for suppressing M1 polarization and stimulating M2 polarization of macrophages via the SOCS3/STAT3 signaling pathway, thereby improving the inflammatory microenvironment and inhibiting neuronal apoptosis . In vitro and in vivo experiments have also demonstrated that SC exosomes can be taken up by vascular endothelial cells and promote functional recovery after spinal cord injury by enhancing angiogenesis through the delivery of integrin-β1 . SC-derived EVs have been shown to significantly improve motor and sensory function recovery in rats with long-distance sciatic nerve defects treated with chitosan nerve conduits . The ability of SC EVs to improve regeneration of the damaged sciatic nerve in vivo and promote neurite outgrowth of sensory and motor neurons in vitro has been demonstrated by multiple studies [554,563,564,565,566].
Recently, Nishimura et al. reported that the systemic administration of human SC exosomes significantly decreased the overall contusion volume, microglial activation, secondary inflammatory injury, and histopathological damage in rats with traumatic brain injury . A new study has confirmed multilayered therapeutic advantages of human SC-derived exosomes, targeting both acute and chronic neuroinflammatory pathways to foster functional recovery.
Jugular infusion of SC EVs after traumatic brain injury reduced the acute pro-inflammatory reaction in the ipsilateral cortex and hippocampus, as assessed by a decrease in inflammasomes and inhibition of the STAT3/pSTAT3/SOCS3 pathway. Reduction of cerebral edema and contusion volumes confirmed the diminishing of subacute histopathological changes and a decrease in microglial activation. This was associated with the preservation of both sensorimotor function and cognitive performance. Moreover, a decline in circulating neutrophils indicated a reduction in chronic systemic inflammation .
SC-derived exosomes have also been reported to improve peripheral neuropathies. Normal exogenous SC-derived exosomes have been shown to ameliorate peripheral neuropathy in mice induced by abnormal microRNA biogenesis in resident SCs . MicroRNAs are non-coding RNAs that regulate post-transcriptional protein expression and affect nearly 60% of mammalian genes . Wang et al. investigated the therapeutic potential of exosomes derived from healthy SCs in treating diabetic peripheral neuropathy. They reported that the intravenous injection of SC EVs in Type 2 diabetic db/db mice with peripheral neuropathy significantly improved the sciatic nerve conduction velocity and thermal and mechanical sensitivity . The corrected neuropathy was associated with the extension of epidermal nerve fibers, remyelination of sciatic nerves, and reversion of diabetes-reduced mature form of miRNAs 21, 27a, and 146a in sciatic nerve tissues. In vitro data revealed that SC exosomes stimulated neurite outgrowth in diabetic DRG neurons and altered the migration of SCs in response to high glucose . New results showed functional impairments in SCs cultured from the sural nerve biopsy of an ALS patient. But their treatment with exosomes isolated from cadaver donor SCs markedly recovered their growth potential in vitro .
Interestingly, mechanical stimulation of SCs may alter the microRNA composition of their nanoscale vesicles by upregulating miR-23b-3p, which reduces neuronal neuropilin 1 expression responsible for neurite outgrowth in vitro and nerve regeneration in vivo . Additionally, neuron-to-glia communication may regulate the content of SC-derived pro-regenerative exosomes and their transport to axons, thereby modifying axonal elongation. ATP release from activated sensory neurons has been shown to stimulate P2Y receptors in repair SCs, thereby increasing the level of miRNA-21 present in SC EVs .
Likewise, SC-derived exosomes may restore paclitaxel-induced peripheral neuropathy. The results of in vitro experiments demonstrated that SC EVs protected the DRG neurons from paclitaxel damage, while in vivo data revealed that SC exosomes ameliorated paclitaxel-induced plantar intraepidermal nerve fiber loss, DRG injury, and mechanical nociceptive sensitization in rats . The mechanistic experiments uncovered that SC exosomes facilitated axonal regeneration and protected injured neurons by increasing miR-21, which suppresses the phosphatase and tensin homolog (PTEN) signaling pathway, recovering paclitaxel-induced peripheral neuropathy.
Thus, given its versatile contribution to angiogenesis, neurogenesis, axon remodeling, and anti-inflammatory response, SC exosomes may be a promising next-generation therapeutic modality for improving neurorestorative treatments. Notably, a recent study has established a feasible framework for the isolation and thorough omics characterization of SC EVs, which, importantly, is consistent with their therapeutic properties in neurological applications .
7.2. Schwann Cell-Derived Extracellular Matrix Proteins and Factors
During the development and regeneration processes, SCs secrete extracellular matrix factors containing a diversity of macromolecules, such as collagens, laminins, proteoglycans, elastin, fibronectin, entactin, etc. Remarkably, in the PNS, fibronectin is primarily expressed by SCs and located mainly within the perineurium . ECM components produced and secreted by SCs are deposited beside SCs to support the basal lamina sheets surrounding individual axon-SC units. SC ECM thus regulates axonal growth during development and regeneration . In addition, both the secretion and assembly of ECM components by SCs control SC polarization and are required to develop the key SC functions—ensheathment and myelination . Surprisingly, SCs can regulate their ECM microenvironment to safeguard directional neuron migration. It has been shown that sensory neurons and motoneurons require the SC-derived Sox2-dependent fibronectin matrix to migrate along the oriented SCs . Even more importantly, Sox2 and fibronectin are co-expressed in pro-regenerative SCs in vivo during sciatic nerve regeneration in a time-dependent manner.
As SC transplantation is considered the gold standard in cell-based treatment for peripheral nerve injuries, ECM molecules from SCs are one of the most appropriate biomaterials for nerve repair [236,238]. Such cell-free transplantation methods, along with their combinations, offer excellent promise for clinical translation.
Gu et al. developed a chitosan/silk fibroin-based, SC-derived, ECM-modified scaffold suitable for peripheral nerve tissue engineering applications . Its regenerative properties were verified when it was used to bridge a 10 mm gap in the rat sciatic nerve model. Additionally, blood and histopathological parameters confirmed the safety of scaffold modification by SC-derived ECM. Other studies have confirmed these promising results [579,580]. Interestingly, ascorbic acid can stimulate the secretion of ECM by SCs, and the myelinating ability of SCs expressing collagen I, collagen IV, fibronectin, and laminin has also been demonstrated .
ECM substantially impacts the behavior and activity of resident cells, including SCs, maintaining the physiological functions of tissues and organs and their healing potential. Diverse components extracted from the ECM material of peripheral nerves, which are usually bioactive, biocompatible, and well tolerated in vivo, can induce SC proliferation and alignment, thus serving as an alternative to SCs as cell-free scaffolds for nerve regeneration . Importantly, ECM proteins modulate SC morphology, activity, and protein expression and may accelerate SC conversion to a regenerative phenotype . Several studies have reported that compositional decellularized nerve ECM initiates SC alignment and nerve fiber remyelination in vitro and in vivo [583,584,585,586]. Different ECM scaffolds extracted from PNS materials provide suitable platforms for SC transplantation, offering a feasible alternative to cell-free nerve grafts .
Thus, further studies addressing the interaction between ECM cues and SCs, including SC-derived ECM constituents, may offer novel cell-based and cell-free approaches to augment PNS regeneration and improve neuromodulation treatments.
8. Schwann Cells in Wound Healing
Wound healing is a complex physiological process that requires synchronized and orchestrated communication between numerous cell types with distinct functions to restore the original homeostatic conditions. Substantial experimental and clinical data suggest that innervation plays a key role in wound healing, tissue regeneration, and organ repair pathways . For instance, the absence of innervation is known to impair skin healing, and individuals with peripheral neuropathy often exhibit a failure to heal [588,589]. The skin is a tissue densely innervated by sympathetic autonomic and primary afferent sensory nerve fibers providing pain, pruritus, thermal, and tactile sensations . The skin nerve fibers are accompanied by myelinating and non-myelinating SCs, which, in addition to their well-known function, may also provide nociceptive and touch sensation [7,10,591].
Neuromediators, including neuropeptides, neurohormones, cytokines, and growth factors, are released by immune cells, keratinocytes, and nerve terminals in response to skin lesions and peripheral nerve damage, and are also involved in various stages of wound healing . While the dissemination of SCs from the disrupted PNS nerves into the granulation tissue within the wounded skin, their proliferated and dedifferentiated, have been well studied (see above), recent data demonstrated that genetic ablation of SCs significantly deferred wound contraction and closure, reduced development of myofibroblasts, and diminished re-epithelization in the repairing skin . Similarly, tempered injury responses and delayed SC dedifferentiation have been reported to impair skin wound healing . It has been established that early activation of SCs after injury creates a permissive microenvironment, while the later redifferentiation of SCs ensures the formation of a myelin sheath. This modulation of the SC phenotype is controlled by c-Jun and Sox2—regulators of SC reprogramming, and Krox-20 and Sox10—regulators of SC myelination .
Recent analysis of SC activity using a cell atlas of the wound healing process, generated by integrating single-cell RNA sequencing, revealed that the Wnt signaling pathway regulates SC dedifferentiation at an early stage of wound repair. Wnt inhibition blocked SC dedifferentiation, resulting in defective repair . Furthermore, SC-secreted TGF-β endorsed the motility of keratinocytes and fibroblasts, and TGF-β improved the healing of chronic wounds with impaired SCs . Injury-activated SCs regulate the formation of myofibroblasts via paracrine modulation of TGF-β signaling . Similarly, analysis of skin repair pathways in other models has demonstrated that suspended wound healing in diabetes may be due to a weakened SC repair reaction and diminished effects on myofibroblast differentiation . In addition, a subset of specialized profibrotic SCs with reprogrammed genetic activity may promote the formation of abnormal scars, such as hypertrophic scars or keloids, during the nerve repair process in the injured skin . These SCs were not associated with axons, displayed a nonclassical repair-like phenotype, and supported the overproduction of the extracellular matrix. Therefore, the authors hypothesized the existence of two types of repair/repair-like SCs: Type 1, which appears temporarily after acute injury, and Type 2 with profibrotic properties, which may persist for a prolonged time after tissue regeneration . New data demonstrate that SCs in keloid tissue of the rabbit ear scar model significantly upregulate the expression of the insulin-like growth factor binding protein 5 (IGFBP5) gene, which is positively associated with keloid fibroblast proliferation, migration, invasion, angiogenesis, and cell cycle progression . It is vital to note that identifying the specific set of genes in so-called keloidal SCs should help understand the contribution of SCs to the pathogenesis of other fibrotic disorders, which have devastating consequences for various organs, including the heart, lungs, liver, kidneys, and skin.
These data discovered a novel role for cutaneous SCs. They demonstrated that SCs could be used as a therapeutic target or treatment tool to improve dermal wound healing and reduce scar formation . Collectively, numerous findings from other experimental wound healing and regeneration models demonstrate the pivotal role of nerve-associated SCs in blastema-mediated regeneration, digit regeneration, wound healing, and organ regeneration, as SCs provide trophic factor support, a source of progenitor cells, nerve repair coordination, and assist skin wound closure [587,600,601]. SCs can also promote vascular regeneration owing to the expression of various pro-angiogenic factors.
SCs are also known to contribute to bone regeneration. In the alveolar bone injury model with tooth extraction, dedifferentiated repair SCs either formed new axons or transdifferentiated into osteoblasts during the healing process . Scaffold vascularization and osteogenesis in a femur defect model induced by bone grafts and endothelial cells was significantly accelerated after the addition of SCs . Similar effects on angiogenesis and osteogenesis can be achieved by utilizing SC-derived exosomes . SC exosomes have been reported to upregulate the proliferation, osteogenic, and neurogenic differentiation of human periodontal ligament cells, as well as the expression of angiogenic factors in vascular endothelial cells . Moreover, in a rat model of periodontal bone defects, SC exosomes accelerated the attraction of endogenous cells, controlled neural and vascular neogenesis, and facilitated periodontal bone regeneration . Importantly, accelerated formation of bones by SC-derived exosomes in vivo was dose-dependent . The elimination of SCs by denervation significantly delays the healing process in various models [605,606]. Interestingly, new research has revealed that dental pulp stem cells can reduce pyroptosis and mitochondrial ROS production in SCs through mitochondrial transfer, thereby enhancing nerve regeneration . A comprehensive understanding of SC involvement in regenerative processing across different tissues should provide a better perspective for integrating these pathways into a therapeutic framework to promote tissue and organ repair.
Finally, one unusual aspect of SCs’ reparative function is their potential association with a significant clinical phenomenon: the formation of surgery-associated metastasis. The fact that the surgical insult might trigger or accelerate locoregional and distant tumor recurrence has been acknowledged for a long time, and more recent evidence demonstrated that the surgical operation or even tissue biopsy may generate a permissive environment for tumor growth and the appearance of metastases [608,609,610,611,612,613]. Although the hypothesis that primary tumors can inhibit the growth of metastases—“concomitant resistance” and “cancer hormesis”—is still under discussion [614,615,616,617,618], numerous experimental studies provide more mechanistical results. The mechanisms of surgery-induced metastatic processing involve the promotion of wound healing and functional recovery connected with the stimulation of immunosuppressive, inflammatory, and pro-angiogenic pathways, as well as the promotion of single circulating malignant cells and the disbalance of dormant niches and escape from dormancy [619,620,621,622].
Furthermore, it is well established that nerve axons and SCs are present in human and animal tumors and are capable of accelerating tumor growth and progression (see below). As discussed above, a traumatic injury to neuronal axons within the resected tumor and the surrounding tissue initiates Wallerian degeneration and nerve injury-induced “repair” SCs. In addition, tumor-associated SCs exhibit a repair-like or injury response-like phenotype, as described below. It was also speculated that glial cell failure might happen if adaptive strategies established by tumor-activated SCs collapsed or if the program of SC-driven tissue repair terminated in a maladaptive response. This can then promote or support the metastatic disease . Tumor-associated “repair-like” SCs encourage metastasis in surgery- or biopsy-related wound healing even more strongly because surgical stress and wound healing signaling contribute to the higher degree of SC activation, proliferation, motility, and dedifferentiation in the regenerating tumor microenvironment . However, the role of SCs in metastasis development as an outcome of wound healing has never been experimentally proven. Further studies are needed, as the results should provide a new target for safe and expedited wound healing and tissue regeneration following cancer-associated surgical procedures.
9. Schwann Cells in Cancer
The clinical aspects of tumor innervation and the role of the PNS in the development, growth, and progression of solid tumors have been investigated for decades . While the direct and indirect modulation of tumor growth and metastasis by the autonomic and sensory neurotransmitters and neuropeptides has been proven in numerous preclinical and clinical studies [624,625,626,627], the involvement of SCs in the formation of the tumor neuroenvironment and maintenance of the neuro–immune axis in cancer has been documented only recently [498,628,629,630]. These studies have revealed the functional role of SCs in promoting the proliferation and motility of neoplastic cells, modulating antitumor immune responses, and conveying pain sensations in cancer. Altogether, this suggests that SCs are legitimate targets for anticancer therapy and could also be used as a therapeutic tool for treating at least certain types of malignancies.
9.1. Schwann Cells in Tumorigenesis
The regenerative, immunomodulating, and pro-angiogenic properties of SCs, as well as their functional and phenotypic plasticity, make SCs an exceptional target for malignant cells to exploit and convert into supporters of carcinogenesis and metastasis . All published data support this notion, and several pathways have been identified that are involved in SCs’ ability to modulate the tumor microenvironment and promote tumor growth, spread, and progression to metastatic disease.
First, SCs stimulate the proliferation, motility, and invasiveness of tumor cells. In addition, SCs support both epithelial–mesenchymal and mesenchymal–epithelial transitions of malignant cells [632,633]. For instance, in vitro and in vivo, SCs supported the proliferation and migration of colon cancer cells by secreting NGF and activating the TrkA/ERK/ELK1/ZEB1 signaling pathway in tumor cells or via the activated NF-κB/IL-8 axis . SCs could stimulate pancreatic cancer growth and invasion by releasing MMP-2, tissue inhibitor of metalloproteinases-2, galectin-3 binding protein, cathepsin D, proteoglycan biglycan, plasminogen activator inhibitor-1, galectin-1, or IL-6 [636,637]. SCs could also support the invasion and migration of prostate and pancreatic malignant cells .
Similar data have been reported for the lung cancer models. Exosomal miRNA-21-5p from human SCs (i) increased the proliferation, motility, and invasiveness of several human non-small cell lung cancer (NSCLC) cell lines in vitro by targeting metalloprotease inhibitor RECK in tumor cells and (ii) augmented human lung cancer cell progression and lymph node metastasis in vivo . Human SCs have also been reported to enhance the proliferation and dissemination of various human small cell lung cancer (SCLC) cell lines in vitro and in vivo . Mouse model experiments revealed that SCs promoted lung cancer cell proliferation and motility via upregulating M2 macrophage polarization or through secretory CXCL5/CXCR2-mediated activation of the PI3K/AKT/GSK-3beta/Snail-Twist signaling pathway in malignant cells [633,641]. Notably, SCs also stimulate the epithelial–mesenchymal transition (EMT) of malignant cells, supporting their invasive potential and metastasis formation, as has been reported for lung, pancreatic, salivary adenoid cystic carcinoma, and cholangiocarcinoma [633,637,642,643,644]. Therefore, understanding the metastasis pathways in demyelinating diseases, commonly comorbid with primary cancers, should offer new approaches to recognize how damaged SCs may influence and slow metastasis formation .
Importantly, SCs can chemoattract cancerous cells and stimulate perineural invasion or perineural spread—malignant cell dissemination in and along nerve bundles [629,646,647,648]. Tumor-activated SCs may facilitate the spread of cancerous cells along nerves by forming tracks for migration within the nerve and reorganizing malignant cells into chains [649,650]. For instance, the neuropeptide neuromedin B, produced by cervical cancer, can reprogram SCs to initiate perineural invasion by secreting CCL2 and directing axon regeneration . Cervical cancer cells also upregulated the dedifferentiation of SCs, which in turn promoted perineural invasion by producing the chemokine FGF-17 and by degrading ECM with cathepsin S and MMP-12 . The perineural invasion in prostate cancer was also associated with SCs . SC-produced NGF and TGF-β may be related to a more aggressive phenotype of cholangiocarcinoma [644,654]. The presence of SCs has been shown to increase the aggressiveness of oral cancer cells by stimulating their proliferation, extracellular matrix breakdown, and altering cellular metabolism .
Second, SCs contribute to proper vascular remodeling. New data revealed that GFAP+ cells, derived from nerve-associated tissue-resident SCs, surrounded blood vessels in the microenvironments of melanoma, prostate, and breast cancer, and that genetic depletion of resident SCs attenuated tumor development by regulating angiogenesis and immunosurveillance . The pro-angiogenic effects of SCs may include the promotion of endothelial cell proliferation, migration, and tube formation [602,657]. They may be mediated by SC-derived exosomes containing integrin-β1 [561,658]. In addition, multi-omics and bioinformatics analysis of non-myelinating SCs in gastrointestinal tumors revealed their association with “Activating invasion and metastasis” and “Inducing angiogenesis” .
Third, SCs can regulate the activity, polarization, and function of effector and regulatory immune cells within the tumor microenvironment, thereby manipulating the immunosuppressive potential of the tumor milieu. Recent data suggest that intratumoral SCs can attract immature myeloid cells, conventional dendritic cells, macrophages, and T cells and polarize them into immunosuppressive MDSC, regulatory DC, M2 tumor-associated macrophages (TAMs) and regulatory and exhausted T cells [496,498,499,641,659]. While the interaction between tumor-associated SCs and NK cells has not yet been documented, SC-derived TGF-β, PgE2, IL-6, and other molecules are known to control the activation and function of NK cells, suggesting a likely crosstalk between SCs and NK cells in the tumor environment . New bioinformatic data confirm the association of non-myelinating SCs with MDSC, monocytes, neutrophils, regulatory T cells, and NK cells in gastrointestinal tumors . SCs can also interact with mast cells, contributing to the formation of the tumor immunoenvironment . These data prove that SCs actively participate in the recruitment and alteration of immune cells at the tumor site, forming, shaping, and maintaining the immunosuppressive and tolerogenic tumor immunoenvironment that facilitates immune escape, protection, proliferation, and spreading of malignant cells [629,660].
Fourth, SCs promote the conversion of cancer-associated fibroblasts (CAFs) into inflammatory CAFs, characterized by a more malignant phenotype, which accelerates tumor progression . SC-derived pleiotrophin could upregulate CAF proliferation and collagen synthesis . Ephrin, neuregulin-1 (NRG1), and tenascin-C signaling may play a role in SC and fibroblast communication [664,665,666].
Thus, various recent studies have revealed an active and essential role of SCs in promoting carcinogenesis and tumor growth [667,668,669]. Accumulating experimental data also suggest that the tumor-associated SCs in the tumor microenvironment play foundational roles in the metastatic transformation and metastatic dissemination of cancerous cells . This implies that manipulation of SCs could supply a helpful approach to improve outcomes for patients with cancer .
9.2. Schwann Cell Alterations and Reprogramming in Cancer
SCs have been identified and characterized within the tumor microenvironment in multiple animal and human samples. These tumor-associated or tumor-reactive SCs demonstrated signs of a dynamic process of activation, proliferation, denervation, morphological flexibility, adaptive cellular reprogramming, increased motility, EMT, and stemness. Reprogramming and adopting a dedifferentiated phenotype make the functional state of tumor-activated SCs similar to that of repair SCs seen during nerve trauma and wound repair processes. Therefore, they have been termed ‘repair-like’ SCs [628,640,659,667,670,671]. However, the ‘injury response phenotype’ may be a more appropriate term (Figure 1).
The retrospective in silico analysis of tumor biopsies revealed that high expression of genes associated with SCs correlated with worse prognosis in melanoma patients . The clinicopathological analysis similarly demonstrated that SCs were enriched in human colon cancer tissues and were associated with tumor metastasis and poor prognosis [36,634]. The evaluation of histological sections of pancreatic ductal adenocarcinoma also showed enrichment of SCs compared to non-tumor tissues, which was associated with a worse prognosis and could serve as an independent prognostic factor for overall survival [637,662]. A similar correlation between SC density and patient survival has been reported for lung cancer and cholangiocarcinoma . Importantly, the abundance of tumor-associated SCs in pancreatic ductal adenocarcinoma correlated with tumor-promoting immune cells’ infiltration and unfavorable patient outcomes .
A multi-omics evaluation of tumor-associated adaptively reprogrammed SCs confirmed the expression of various transformation markers associated with activation, dedifferentiation, proliferation, and denervation. Altered expression or secretion of cytokines, chemokines, growth factors, and other molecules by “repair-like” or “injury-responding” SCs has also been repeatedly demonstrated. Otani et al. found that tumor-activated SCs could express catecholamine-synthesizing enzymes, produce adrenaline, and establish an adrenergic microenvironment, which potentiates the chemoresistance of lung cancer cells . Upregulation of chemoresistance in pancreatic adenocarcinoma cells by SC-conditioned medium has also been reported . Melanoma-reprogrammed SCs exhibited dysregulated lipid oxidation, characterized by increased production of prostaglandins and lipoxins, which suppressed the activation of antitumor T cells . Lung cancer-reprogrammed SCs can be characterized by the differentially expressed genes and microRNA . Colorectal carcinoma factors upregulate the expression of IL-8 in SCs, which in turn upregulates tumor cell motility . Pancreatic cancer cell-derived exosomes can upregulate the motility of SCs and thus facilitate perineural invasion . Furthermore, the crosstalk between SCs and TAM is associated with increased perineural invasion and poor survival rates in these patients. TAM could activate SCs via the FGF/PI3K/Akt/c-myc/GFAP pathway, while SC-derived IL-33 could drive macrophages into the perineural milieu and assist in their M2 polarization . Analysis of injury-induced neuroma confirms the presence of SCs in the repair status and significant infiltration by several subsets of macrophages .
Recent studies have also revealed the regulatory effects of glycometabolism reprogramming, which may control various intracellular signaling pathways at both the transcriptional and post-transcriptional levels, of the inflammatory responses within glial cells in the tumor milieu . For instance, increased dietary intake of palmitic acid supports a pro-metastatic state in malignant cells, associated with SC activation and elevated tumor innervation . Interestingly, one study suggested that SCs in premalignant lesions undergo adaptive reprogramming during early pancreatic cancer development, potentially encouraging a protective antitumor neuronal response .
Overall, a better understanding and streamlining of SC reprogramming pathways in the tumor milieu provide new targets and tools for destabilizing and destroying the tumor microenvironment as an additional means of a combinational cancer therapy approach.
9.3. Schwann Cells in Peripheral Neuropathies in Cancer Patients
Peripheral neuropathies are prevalent in patients with cancer and are a leading cause of cancer-related pain. Twenty percent of cancer pain is likely purely neuropathic in origin . Cancer-associated neuropathies can be a direct or indirect complication of cancer or cancer treatment. The direct complications are due to nerve compression or infiltration by cancerous cells. The indirect cancer consequences may be the result of nutritional and metabolic deficits, coagulopathy, or infections, or, most often, they are due to neurotoxic chemotherapeutic agents—chemotherapy-induced peripheral neuropathies. Iatrogenic nerve injuries include radiotherapy-induced nerve damage or post-surgical complications. Less frequently, neoplastic neuropathies appear as a paraneoplastic neurological syndrome with immune-mediated manifestation. Paraneoplastic neuropathies may occur during the tumor’s growth and progression but are more likely seen as the tumor’s first manifestation associated with its identification .
9.3.1. Schwann Cells in Chemotherapy-Induced Peripheral Neuropathy
Chemotherapy-induced peripheral neuropathy is a common dose-limiting side effect seen in cancer patients treated with neurotoxic chemotherapy . It is often sensory-predominant with pain syndrome. For instance, PNS lesions are a common complication of breast cancer treatment, with up to 80% of survivors experiencing signs of PNS damage [685,686]. Less frequently, cancer patients develop paraneoplastic, immune-mediated, or neoplastic neuropathies. Chemotherapeutic drugs, such as anti-microtubule agents (taxanes), vinca alkaloids, platinum compounds, proteasome inhibitors (bortezomib), and thalidomide, are neurotoxic and have a high likelihood of inducing sensory or sensorimotor neuropathy or peripheral demyelination. Interestingly, patients with Charcot–Marie–Tooth disease can have increased susceptibility and severity of chemotherapy-induced peripheral neuropathy [687,688].
Experimental clinical data demonstrate that chemotherapy-induced peripheral neuropathy, associated with sensory and motor nerve abnormalities, ref. may be at least partly mediated by the impairment of SCs. Notably, axonal degeneration (i.e., Wallerian degeneration) plays a major role in the pathogenesis of many neurodegenerative disorders, including chemotherapy-induced peripheral neuropathy . However, there is a lack of evidence to show the role of SC transplantation in chemotherapy-induced neurotoxicity. Related studies have focused on the damage caused by chemotherapeutic agents to SCs and on stem cell therapy for chemotherapy-induced peripheral neuropathy .
Bortezomib-induced peripheral neuropathy is associated with remarkable SC demyelination in sciatic nerves resulting from impaired lysosomal function and induction of autophagy in SCs . In vitro, bortezomib can augment the dedifferentiation of cultured SCs . Mechanistical data showed that bortezomib-induced endoplasmic reticulum (ER) damage to SCs was accompanied by the downregulation of myelin gene expression and the expression of macrophage chemoattractants, which together suggest that adaptive responses of SCs to bortezomib-induced ER stress may participate in the development of bortezomib-induced peripheral neuropathy . Paclitaxel-induced peripheral neuropathy in vivo stimulated the expression of activating transcription factor 3 (ATF3; a marker of cellular injury) in myelinating SCs in the sciatic nerve . Cisplatin and oxaliplatin induce disruption of myelin formation and mitochondrial dysfunction in SCs, while paclitaxel induces SC dedifferentiation characterized by increased expression of p75 and galectin-3, and reduced expression of MBP . Similarly, cisplatin and carboplatin in vitro triggered mitochondrial dysfunction and dedifferentiation in SCs . Epirubicin-docetaxel therapy may also induce SC dedifferentiation and macrophage infiltration in sciatic nerves . These and similar results suggest that the cytotoxic effects of chemotherapy on SCs may be important in the development of neurotoxic neuropathy in conjunction with their immediate damage to peripheral neurons. Thus, agents that inhibit drug-induced SC damage are expected to have therapeutic potential for this common pathology .
Although many strategies have been established, no specific intervention is presently endorsed for the prevention or relief of chemotherapy-induced peripheral neuropathy. Koyanagi et al. have recently demonstrated for the first time that SC-derived galectin-3, the β-galactoside binding protein serving as an immune cell chemoattractant, may play an important pro-nociceptive role in the development of taxanes (paclitaxel and docetaxel)-induced peripheral neuropathy via macrophage infiltration into the peripheral nerves . The absence or pharmacologic inhibition of galectin-3 in experimental animals prevented paclitaxel-induced macrophage infiltration and mechanical hypersensitivity. Thus, therapies targeting SC-derived galectin-3 might represent a unique tactic to suppress taxane-related chemotherapy-induced peripheral neuropathy.
Remarkably, vitamin C was able to rescue SCs treated with bortezomib by inhibiting the cytotoxicity in SCs . Vitamin C is critical for the SC myelination process, as it has been shown to upregulate 10 pro-myelinating genes in SCs. Dietary vitamin C deficiency may trigger peripheral nerve hypomyelination . Recent screening for SC-targeting agents as candidates for chemotherapy-induced neuropathy treatment revealed that a selective phosphodiesterase PDE3 inhibitor, cilostazol, blocked paclitaxel-induced dedifferentiation of cultured SCs via cAMP/EPAC signaling and reduced sciatic nerve SC dedifferentiation in a mouse model of paclitaxel-induced peripheral neuropathy . Moreover, a new study reveals that dihydromyricetin mitigates peripheral neuropathy and the toxic effects on SCs induced by Bortezomib, a proteasome inhibitor used in the treatment of multiple myeloma. Dihydromyricetin, a flavanonol extracted from the Japanese raisin tree, restores lysosome–autophagy activity in Bortezomib-affected SCs through the ERK/TFEB pathway, thereby significantly improving mechanical allodynia, sciatic nerve conduction, and demyelination associated with the adverse reaction of Bortezomib in the peripheral nervous system .
Although no data support the notion that SC transplantation might be a potential strategy for treating chemotherapy-induced peripheral neuropathy, new data suggest that mesenchymal stem/stromal cell therapy has a beneficial effect on improving symptoms [235,703]. It would be essential to determine whether MSCs can differentiate into SC-like cells after chemotherapy in vivo.
9.3.2. Paraneoplastic Neuropathies
Paraneoplastic neuropathies are autoimmune and reflect the development of immune reactions to ectopically expressed antigens on neoplastic cells that mimic onconeural (intracellular) and neuronal surface (synaptic) molecules expressed by neurons and glial cells . Among peripheral paraneoplastic neurological syndrome cases, sensory neuronopathy is the most frequent pathology. In contrast, other types of neuropathies, such as sensory and sensorimotor polyneuropathy, demyelinating neuropathies, and autonomic neuropathies, are less common . Many onconeural antibodies have been well characterized, with anti-Hu and anti-Yo antibodies being the most common. While most identified onconeural antibodies are associated with CNS diseases, anti-Hu antibodies are linked with pure sensory neuronopathy . Notably, almost 30% of patients with presumed paraneoplastic neuropathies do not demonstrate autoantibodies, as their antibodies have not yet been identified . For example, analysis of autoantibodies in a cohort of patients with paraneoplastic neuropathies revealed that a significantly higher proportion of patients with sensory–motor neuropathy demonstrated serum IgM antibodies against SCs compared to healthy controls . In a case report of a patient with paraneoplastic cerebellar degeneration associated with para-ovarian adenocarcinoma, circulating antibodies recognizing SCs have also been demonstrated .
9.4. Targeting Schwann Cells in Cancer
Tumor-induced activation of SCs may serve as a new target to weaken the tumor-protecting microenvironment. While surgical or pharmacological denervation of the local or systemic tumor milieu cannot be considered a feasible treatment approach to target tumor-associated neurons and SCs, at least today, focusing on activated SCs in cancer might provide a new reasonable direction. Although no clinical data directly verify that SC-based therapy is a realistic cancer treatment, we predict a thriving advancement of this approach. Recent results have shown that neuroblastoma cells respond to the primary repair SC secretome by enhancing neuronal differentiation and reducing proliferation . Although cancer-specific, this proof-of-principle study supports the concept of selectively targeting intratumoral SCs as a promising approach for therapeutic intervention. This concept capitalizes on the unique liabilities of tumor-associated SCs, aiming to interrupt their supportive role in tumor progression and possibly improve the efficacy of standard cancer treatments.
Promising therapeutic strategies include the hampering of SC reprogramming, depletion of functionally blocked SCs, and disturbance of SC crosstalk with tumor-infiltrating immune cells . As an example, noncytolytic lymphocytic choriomeningitis virus has been shown to target SCs preferentially, but not neurons, via alpha-dystroglycan expressed on SCs, rendering them defective or incapable of forming compact myelin sheaths . This infection did not induce SC apoptosis or cytotoxicity but disrupted the assembly of SC-axon units. While it is unknown whether the localized infection of tumor-associated SCs may block their protumorigenic activity, these results open an opportunity to control SC function in the cancer milieu.
Depleting tumor-activated SCs with a small-molecule compound may affect the immune resistance of malignant cells, dramatically altering the immunosuppressive tumor microenvironment. In the in vivo pancreatic ductal adenocarcinoma model, SC reduction markedly improved the efficacy of immune checkpoint inhibitors and blocked tumor growth . Genetic ablation of SCs in the melanoma model downregulated tumor growth and angiogenesis, which were associated with increased tumor infiltration by cytotoxic T cells, decreased infiltration by immunosuppressive regulatory cells, and thus upregulated immune surveillance .
Targeting tumor-associated SCs to prevent or suppress SC-composed perineural invasion is another promising clinical implication. The preclinical data demonstrated that a single fraction of low-dose radiation therapy may induce SC death and impair perineural invasion of non-irradiated malignant cells in both in vitro and in vivo nerve perineural invasion models . Inhibition of perineural invasion in pancreatic cancer in vivo can also be achieved by controlling c-Jun activity with a c-Jun inhibitor in tumor-activated SCs . Furthermore, knowing that tumor-derived NGF may activate SC autophagy, which in turn promoted perineural invasion in pancreatic cancer, Zhang et al. reported that targeting NGF and inhibiting autophagy blocked SC autophagic flux, stimulated SC apoptosis, and decreased the invasion of malignant cells to nerve fibers, showing marked therapeutic effects in the perineural invasion in a pancreatic cancer model .
Collectively, these studies support the notion that situating intratumoral SCs as promising therapeutic targets and manipulating SCs, as well as revealing the key mechanisms of the SC–cancer relationship, may provide a practical approach to improving cancer management and outcomes for cancer patients .
10. Tumorigenic Schwann Cells
Different studies revealed that overexpression of growth factor ligands or receptors, including EGFR, PDGF, NRG, SCF, HGF, and TGF-β1, in Schwann lineage cells, especially in combination with other gene mutations, like NF1, could significantly upregulate the formation of SC tumors [712,713]. For example, transgenic mice that lack both the PTEN and neurofibromatosis 1 (NF1) genes in SCs and SC precursor cells have demonstrated an augmented development of neurofibroma and high-grade peripheral nerve sheath tumors . Transgenic mice overexpressing the growth factor NRG1 in SCs developed malignant peripheral nerve sheath tumors . Transgenic mice heterozygous for a Trp53-null allele and overexpressing EGFR in SCs showed a substantial rise in neurofibroma and high-grade peripheral nerve sheath tumor formation. Similarly, modulation of EGFR and TP53 expression in immortalized human SCs markedly enhanced proliferation and anchorage-independent growth in vitro .
There are three main types of nerve sheath tumors: schwannoma, neurofibroma, and perineurioma. Schwannomas are exclusively composed of SCs, while neurofibromas comprise neoplastic SCs or Schwann-like perineural cells and other cells, including fibroblasts, neurons, and immune and endothelial cells . Therefore, all schwannomas are densely positive for the S-100 protein, while neurofibromas may have more inconsistent S-100 expression.
Perineuriomas, a rare entity of peripheral nerve sheath tumors, in contrast to schwannomas or neurofibromas, are stained positive for epithelial membrane antigen (MUC1/CD227), but not for S100 protein; positive staining for neuron filament protein demonstrates axon fibers surrounded by layers of perineural tumor cells . Perineurioma is a benign, slow-growing peripheral nerve sheath tumor that typically occurs in adolescence or young adulthood . Although mutations in the TRAF7 gene and the NF2 gene have been associated with some perineurioma cases, a better understanding of the genetic signature of these relatively rare lesions should lead to targeted therapy and improved patient care .
Of note, a recent finding that chromaffin cells of the adrenal medulla are formed from SC precursors may improve our understanding of the origin of neuroblastoma and pheochromocytoma, which most commonly arise from the adrenal gland region. Furthermore, finding that neuroblastoma cells, derived from aggressive tumors, respond to repair SC secreted factors with upregulated neuronal differentiation and decreased proliferation might help identify promising candidates for the therapy of aggressive neuroblastoma.
10.1. Neurofibromatosis
Neurofibromatosis type 1, also known as von Recklinghausen syndrome, is a tumor predisposition genetic syndrome associated with a mutated copy of the NF1 tumor suppressor gene on chromosome 17q11.2, which encodes neurofibromin, controlling cell growth and proliferation via the RAS signaling pathway . Neurofibromas, especially cutaneous neurofibromas, are the most common tumors in patients with neurofibromatosis type 1, although they can also develop malignant peripheral nerve sheath tumors and gliomas . The hallmark lesion is the neurofibroma, a benign peripheral nerve sheath tumor, and malignant tumors of the CNS and PNS [723,724]. Mutations in NF1 in SCs are the initial tumor-forming event that leads to either complete loss or substantial reduction of neurofibromin function, resulting in hyperactivation of RAS and multiple downstream signaling pathways [712,725]. The key role of SCs in the development of neurofibromas was also demonstrated by showing that homozygous loss of NF1 in SCs is sufficient to induce tumors in different mouse models [726,727]. Ex vivo experiments also indicated that neurofibromin-deficient SCs had a significant growth advantage . Furthermore, NF1−/− SCs exhibit superior pro-inflammatory transcriptional programs, producing various inflammatory cytokines and paracrine factors that mediate immune cell recruitment to the neurofibroma site . Interestingly, analysis of Nf1-deficient SCs demonstrated that although neurofibromas could originate from adult SCs, the nerve environment could switch from tumor-suppressive to tumor-promoting at a site of injury .
Lesions of SC origin may pose a risk for malignant transformation into malignant peripheral nerve sheath tumors, which are aggressive and invasive soft tissue sarcomas with a poor prognosis derived from neurofibromatous SCs . TAZ/YAP hyperactivation caused by Lats1/2-deficiency may be an oncogenic signaling hub that reprograms SCs to a cancerous, progenitor-like phenotype and promotes hyperproliferation, leading to malignant peripheral nerve sheath tumors . Proteomic analysis of pathogenic SCs from malignant peripheral nerve sheath tumors identified over 17,000 distinctive peptides corresponding to more than 1500 individual proteins . DAVID Genetic Association Database analysis, which explores the associations between the identified proteins and diseases, revealed that many SC-derived proteins were associated with cancer, neurological disorders, and immunological and infectious diseases.
Although treatment options, in addition to surgery, for neurofibromas and peripheral nerve sheath tumors are limited, various promising candidate therapeutic targets of signaling pathways for neurofibromatosis type 1-related malignant peripheral nerve sheath tumors have emerged in recent years [734,735]. However, there is an important concept of using cautiously engineered isogenic human Schwann lineage cell lines harboring neurofibromatosis type 1 applicable mutations to search for mutation-definite medication sensitivities and artificial lethal genetic interactions to determine novel therapeutic approaches [736,737].
The intricate interactions between tumor-origin SCs and neurofibroma-associated fibroblasts characterize neurofibromatosis type 1-associated plexiform neurofibroma. A recent study found that mutant SC-derived pleiotrophin, also known as heparin-binding brain mitogen, may be responsible for excessive collagen deposition by fibroblasts, a pathological hallmark of plexiform neurofibroma . Inhibition of nucleolin, a pleiotrophin receptor, signaling reversed collagen synthesis in neurofibroma-associated fibroblasts, suggesting that targeting this pathway offers a new therapeutic strategy for plexiform neurofibroma. Interestingly, SCs deficient in neurofibromin, encoded by the NF1 gene, are biomechanically incompatible, as they lack sensitivity to the biomechanical microenvironment, which likely plays a role in tumor initiation and progression . NF1 is also required to conserve mitochondrial respiratory function in SCs through the stabilization of NADH-associated oxidative phosphorylation and electron transfer , confirming that it may serve as a promising drug target.
Neurofibromatosis 2 is less common than neurofibromatosis type 1 and is triggered by mutations of the NF2 gene in chromosome 22, which encodes the tumor suppressor, merlin, and closely related cytoskeletal ERM proteins (ezrin, radixin, moesin) . Mice expressing a mutated NF2 protein showed a high prevalence of SC-derived tumors and SC hyperplasia . Vestibular schwannomas are the hallmark of neurofibromatosis 2, though non-vestibular schwannomas of the cranial, spinal, and peripheral nerves are also common, as are meningiomas.
New important data demonstrated that NF2 inactivation drove the activation of p21-activated kinases (PAKs) signaling, which initiated NF1-mutant SC tumor dedifferentiation and resistance to therapy . A comprehensive analysis of clinical human and experimental animal data suggest that NF1 or NF2 loss in SCs is necessary but not sufficient for transformation because other genetic harms and non-cellular independent factors play a crucial role in tumorigenesis . A better understanding of the mechanisms triggering transformation and metastasis in SC lineages will eventually improve outcomes for patients who develop SC tumors.
10.2. Schwannomas
Schwannomas are the usual sporadic nervous system tumors and hallmarks of a tumor predisposition syndrome—neurofibromatosis type 2. Schwannomas (or neurilemmomas) are generally benign SC tumors that grow without the entrapment of axons, in contrast to other SC-derived tumors, like neurofibromas, which involve multiple fascicles of a nerve . Less than 1% of schwannomas may become malignant, degenerating into neurofibrosarcoma. Schwannomas are caused by mutations in the loss-of-function NF2 tumor suppressor gene or a loss of chromosome 22, where the NF2 gene is located. The NF2 gene encodes merlin (schwannomin). Merlin is involved in various signaling pathways, including PI3K/Akt/mTORC1, Ras/Raf/MEK/ERK, receptor tyrosine kinase, and Hippo signaling, to regulate cell growth, proliferation, and survival. Mutations in the NF2 gene result in the production of a nonfunctional merlin protein, leading to uncontrolled cell proliferation and tumorigenesis .
Furthermore, in schwannomas, the longitudinal cues that naturally coordinate intrinsically polarized signaling in SCs are absent, and cells exhibit extremely varied polarized surface content. Recent data show that NF2-deficient SCs with unstable polarity can adopt different phenotypic statuses exhibiting enhanced multipolarity and exaggerated cytoskeletal responses to Nrg1 availability . Differential signaling triggered by the two polarity states could also explain the heterogeneous therapeutic responsiveness observed for mTORC1- and EGFR-inhibiting drugs in schwannoma patients .
Schwannomas can develop in the cranial nerves or myelinated peripheral nerves, with nearly 25% of these tumors originating in the head and neck nerve structures. Among the cranial nerves, the vestibular, trigeminal, and hypoglossal nerves are most often affected by schwannomas .
10.2.1. Schwannomas of Cranial Nerves and Spinal Cord
Schwannomas originating from SCs of cranial or spinal nerves are benign tumors . The most common cranial schwannomas developing on the eighth cranial nerve are vestibular schwannomas. Schwannomas of other cranial nerves, such as the trigeminal, facial, and lower cranial nerves, are significantly less common.
A vestibular schwannoma, or acoustic neuroma, according to Rudolf Virchow, is a benign tumor that arises from SCs lining the vestibulocochlear nerve. These schwannomas may occur anywhere along the axons of the nerve from the glial–Schwann sheath junction . Histologically, abnormal SCs exhibit several characteristic microscopic features, including biphasic architecture, nuclear palisading, a fibrous capsule, and degenerative signs . Although the tumoral expansion of SCs in vestibular schwannoma may be one of the causes of affected hearing, demyelination of SCs, as the main morphologic damage of the auditory nerve, and the regulation of long-term survival and function of the auditory nerve, may be directly involved in hearing loss .
ScRNA-seq and exome sequencing data identified repair-like and MHC-II antigen-presenting SCs associated with myeloid cell infiltrate in the tumor microenvironment, implicating a nerve injury-like process. Injury-like SCs support tumor growth by attracting myeloid cells through CSF1 signaling pathways, which may be therapeutically targeted . Of note, data showing that CSF1-R and c-Kit receptor inhibitor masitinib could prevent SC reactivity in degenerating nerves and ameliorate sciatic nerve pathology in ALS rats provide an opportunity for pharmacological treatment of vestibular schwannomas. Furthermore, vestibular schwannomas that express markers of immature and denervated SCs also express NRG1 and activated ErbB2. Blocking anti-NRG1 and anti-ErbB2 inhibitory antibodies could inhibit SC proliferation, suggesting an autocrine pathway that stimulates tumor growth . Thus, the inhibition of constitutive NRG/ErbB signaling provides therapeutic potential for patients with vestibular schwannomas. Similar pathological mechanisms may also be targeted in neurofibromatosis type 1 SC neoplasms known as malignant peripheral nerve sheath tumors [754,755].
Interestingly, recent data have demonstrated that losartan, an FDA-approved antihypertensive drug that blocks fibrotic and inflammatory signaling, can improve hearing loss in patients with neurofibromatosis type 2 by reducing fibrosis . Although these experiments utilized mouse models, analysis of patient samples and data confirmed that losartan treatment normalized the schwannoma microenvironment, suggesting that these findings provide a rationale for prospective clinical trials .
Spinal schwannomas can be located intradurally, extradurally, or rarely intramedullary. Schwannomas can grow extramedullary as spinal cord tumors, representing one of the most common primary intradural extramedullary neoplasms, with malignant peripheral nerve sheath tumors being a less common entity . Spinal schwannomas are among the most common intradural spinal cord tumors in adults, accounting for about 30–50% of such lesions . They typically arise from spinal nerve roots and are associated with neurofibromatosis . Extradural schwannomas can be distinguished from other nerve sheath tumors growing within the spinal canal by their clinicopathological features and their unlikely origin from the nerve root .
10.2.2. Peripheral Schwannomas
Non-cranial nerve schwannomas are among the most common peripheral nerve sheath tumors in adults and can occur at a variety of anatomical locations. Up to 50% of such tumors appear in the head and neck. They are less common than schwannomas in cranial nerves and may present with unique symptoms. Peripheral schwannomas are benign, isolated, non-invasive, and encapsulated tumors that originate from differentiated SCs. Immunohistochemically, all peripheral schwannomas express S100 protein . Almost 90% of peripheral schwannomas arise sporadically and are associated with the dysregulation of the tumor suppressor protein Merlin. However, familial tumor syndromes, like neurofibromatosis type 2 and schwannomatosis, can also cause peripheral schwannomas in 10% of cases .
Malignant transformation of benign schwannoma is rare; however, malignant alternates of schwannomas occur and account for approximately 5–10% of all soft tissue sarcomas . Malignant peripheral schwannomas are more common than their cranial equivalents: they have been described in all somatic nerves of the PNS, including both peripheral and spinal nerves .
10.3. Schwannomatosis
Schwannomatosis is a rare inherited disorder characterized by the predisposition for developing multiple intracranial, spinal, or peripheral nerve schwannomas, with the most common symptom being chronic debilitating pain and neurological dysfunction. Affected individuals inevitably develop schwannomas, typically affecting vestibular nerves . Schwannomatosis is commonly regarded as the least common form of neurofibromatosis . Although historically, schwannomatosis is usually considered a form of neurofibromatosis type 2 , recently developed molecular and clinical diagnostic criteria suggest that these two conditions are clinically and genetically distinct entities, although they share many common features . Neurofibromatosis type 2 and schwannomatosis may share the same trigger—loss-of-function mutations in the NF2 gene. However, schwannomatosis is often genetically heterogeneous, and the NF2 gene mutation is not always involved. Mutations of other tumor suppressor genes, such as SMARCB1 or LZTR1, on chromosome 22 have been identified in familial schwannomatosis cases. Nevertheless, schwannomatosis patients might still be misdiagnosed with neurofibromatosis type 2 or other nerve tumor diseases .
Unfortunately, the results of chromosome analyses of SCs in schwannomatosis or schwannomatosis predisposition syndrome are very restricted, which limits the development of potential cell and gene therapy approaches for treating schwannomatosis. However, there are xenograft mouse models with human NF2-related schwannomatosis tumors in the sciatic nerve of nude mice. Direct intratumoral injection of viral vectors supplying a functional copy of the mutated or inactivated NF2 gene to augment functional Merlin protein re-expression in NF2-deficient tumor cells under the control of the SC-specific promoter P0 resulted in both the death of tumor cells and a significant regression of well-established tumors . However, models focusing on SMARCB1 or LZTR1 mutations and their preclinical evaluation have not yet been developed. Nevertheless, the recent development of immortalized SC lines from human schwannomatosis tumors that retain their essential genotype, phenotype, and cell growth patterns should stimulate future studies on the molecular aspects, signaling, and cellular pathways, as well as therapeutic interventions, in different variants of schwannomatosis .
10.4. Schwannosis
The proliferation of ectopic SCs within the CNS parenchyma, or schwannosis, is not associated with a neoplastic process; however, in early life, it may be coupled with human neurofibromatosis type 2. In addition, transplanted SCs and resident SCs after injury may form schwannosis, which is different from SC tumors such as schwannoma since they normally form myelin sheaths and axons. A few animal cases confirm this pathological proliferation of SCs, described as proliferating cells with solid intracytoplasmic immunoreactivity for SC cell markers, such as MPZ and periaxin, suggesting the formation of PNS myelin within the spinal cord .
Interestingly, schwannosis, as the aberrant proliferation of SCs and nerve fibers, was also described in both human and experimental spinal cord injuries [771,772]. Based on the expression of GFAP and chondroitin sulfate proteoglycan (CSPG), Bruce et al. reported evidence of schwannosis in almost 50% of patients with spinal cord injury who survived 24 h to 24 years after injury . A case of schwannosis in the brainstem has also been reported .
A benign proliferation of SCs and partial myelination of CNS axons may follow different chronic stimuli, including traumatic, degenerative, malignant, and compression lesions. Schwannosis may also result from a developmental abnormality, such as ectopia, during ontogenesis [774,775,776,777].
10.5. Transmissible Schwann Cell Cancers
Devil facial tumor disease (DFTD), an aggressive clonally transmissible non-viral malignancy, is a contagious cancer representing a very rare pathogenic phenomenon of cancerous cells spreading between genetically distinct hosts. DFTD affects Tasmanian devils (Sarcophilus harrisii), carnivorous marsupials endemic to Tasmania, during social biting behaviors and persists in the population in two distinct lineages of cancer: Devil Facial Tumor 1 and 2 (DFT1, DFT2) . The inherent plasticity of SCs helps explain how a common progenitor cell can give rise to apparent contagious cancers. Proteomic and gene expression profiles confirmed that DFT1 and DFT2 are derived from SCs at different stages of differentiation and emerge in different individuals [779,780]. DFT2 reveals a molecular signature of less-differentiated SCs that express immunological markers associated with nerve repair . Though DFT1 cells primarily display a myelinating phenotype, and DFT2 cells exhibit deactivation of myelination pathways, both malignant SC populations show upregulation of pathways associated with SC injury responses .
DFT1 tumors express many proteins associated with SC differentiation and myelination, such as S-100, MBP, MPZ, nestin, PMP22, and the nerve growth factor receptor (NGFR) . DFT2 cells express S-100, SOX10, nesting, and NGFR . Analysis of Tasmanian devil normal PNS samples revealed the expression of a spectrum of myelin genes, suggesting that these functions are intact and SCs are conserved .
Cancer cells are transmitted as malignant allografts among Tasmanian devils, with no detectable induction of immune-mediated allograft rejection. The loss of MHC class 1 molecule expression on the surface of DFT1 cells, resulting from epigenetic downregulation of several components of the MHC class I antigen processing machinery, cannot explain the phenomenon of immune escape, as DFT2 cells express MHC class I molecules. New biochemical data revealed that MHC molecules on DFT2 cells are similar to those of the infected host devils, explaining how malignant SCs may overcome antitumor and allogeneic immunological barriers . Further experiments demonstrated that cancerous DFT2 SCs might slowly evolve to lose their MHC molecules, becoming as contagious as DFT1 cells. New insights into regulating MHC-I and MHC-II antigen processing and presentation in malignant SCs that overcome allogeneic barriers should open up opportunities for treating DFTD, which has a nearly 100% fatality rate . The development of several vaccination platforms is in progress now [785,786]
10.6. Other Schwann Cell Tumors
Mucosal SC hamartomas, introduced in 2009 and considered gastrointestinal schwannomas , are benign, intramucosal tumors of mesenchymal origin commonly located in the distal colon . The term was used to distinguish them from true “neuromas” and “neurofibromas” . SC hamartomas were also found in the gallbladder, the esophagogastric union, and the antrum, and they are generally asymptomatic. Mucosal Schwann cell hamartomas are primarily observed as small polyps, and diagnosis is made after excluding other lesions that resemble spindle cell proliferation and other neuronal tumors . The identification relies on the distinctive clinicopathologic, histologic, and immunohistochemical features. These SC tumors are characteristically positive for S-100 protein, but CD117/c-Kitneg, CD227/MUC1neg, CD34neg, neurofilament proteinlow/neg, and negative for smooth muscle actin staining, which together distinguish mucosal SC hamartomas from other gastrointestinal neoplasms .
Because SC hamartomas are rare and benign, no preclinical models or clinical studies have focused on the pharmacological targeting of tumorigenic SCs. There are no data on isolation, culture, or phenotypic and functional characterization of SCs from these tumors.
11. Conclusions and Future Directions
SCs demonstrate incredible plasticity and a regulable and reversible adaptive cellular reprogramming response to differential stimuli. They are proven to play an etiological and/or pathogenic role in a spectrum of peripheral nerve pathological conditions and diseases. In inherited demyelinating neuropathy, traumatic peripheral neuropathies, immune-mediated neuropathies, and infection-induced demyelination of the PNS, as well as peripheral neuropathic pain, wound healing, and cancer development and progression, pathologically dedifferentiated, dysmyelinated, or repolarized SCs constitute the main nerve pathology. A significant number of evidence demonstrates the beneficial therapeutic roles of SCs and SC-like cells in peripheral neuropathy. Targeting pathological SCs, implicated in specific diseases, is a promising approach to treating peripheral neuropathies (Figure 1 and Figure 2). SC-based therapy also represents an attractive and efficient tactic for treating peripheral neuropathy (Figure 3). Although a growing body of evidence suggests that SCs are a promising target and encouraging tool for advanced experimental and clinical studies, several critical caveats and unresolved issues require attention.
Given their proven role in tissue repair, transplanted SCs are widely evaluated as a new tool for nerve repair in preclinical studies and clinical trials. However, SC transplantation is limited by the necessity of a nerve biopsy to isolate SCs and the lengthy cell culture time required to prepare a sufficient number of cells . Additionally, autologous SC generation protocols are hindered by the limited proliferative activity of adult nerve-associated SCs in vitro. This might result in SC senescence ex vivo, seriously reducing the regenerative capacity [39,73,792]. Improved strategies are also needed to effectively obtain and/or produce functional repair SCs in a shorter timeframe. Understanding these requirements, nerve-derived SCs cannot be considered the optimal source for clinical purposes . Hence, alternative sources of SCs are expected to facilitate the faster development of efficient glial support cell therapies (Figure 3) and are thus being actively explored .
One alternative approach is the in vitro differentiation of SCs from other cell types. Cell sources, differentiation techniques, functional cell characterization, and protocol efficiency have been described and compared . Stem cells represent a valuable resource for SC generation . Mesenchymal stem/stromal cells that can be instantly harvested from different tissues, like skin, bone marrow, umbilical cord, and adipose tissue, are an excellent source of SC-like cells . For example, SCs prepared from bone marrow stromal cells were similar to sciatic nerve-derived SCs in their ability to support axonal regeneration in several animal models [235,794]. Similarly, SC-like cells prepared from subcutaneous adipose-derived stem cells demonstrated higher engraftment and migration capacity than naïve SCs after transplantation into the injured sciatic nerve . SC precursors can be obtained from human pluripotent stem cells cultured and differentiated with TGF-β and GSK-3 inhibitors and neuregulin-1 NRG1 .
Easily accessible human skin fibroblasts may represent another unusual source of SCs, as recently reported using the direct conversion-reprogramming approach to prepare SC precursors proficient in differentiating into functional SCs . Injection of these SCs enabled the repair of a crushed sciatic nerve in a mouse model. Direct conversion technology, which bypasses the pluripotent stage and irreversibly alters cell types, has demonstrated its utility for SC generation by modulating specific transcription factors involved in SC development [237,799].
The therapeutic utilization of SCs also requires an improved understanding of the mechanisms of SC-induced tissue and nerve regeneration and remyelination. The increasing evidence on the capacity of SCs to support wound healing and tissue regeneration is a thrilling field of future investigation. Available results suggest that some SC-mediated effects on regeneration are indirect, resulting from the release of factors and paracrine signaling mechanisms. It is crucial to determine whether SCs are the primary providers of these communications and whether these results can be succeeded through other direct means. Moreover, combining SCs or SC-derived factors with peripheral nerve repair biomaterials appears to be a promising addition to efficiently guide the regeneration of axons and facilitate the myelination of nerve fibers. The utilization of Artificial Intelligence technology, equipped with Machine Learning and Deep Learning, may help develop neural and engineered regenerative biomaterials, thereby increasing the potential to enhance the field of SC-based treatments and their clinical applicability.
Another essential consideration is identifying the SC subtypes that mediate the reported or expected therapeutic effects. The targeted or transplanted SCs can vary, depending on the type of tissues, injury, and repair pathway. The associated vector of future investigation is the detection of biomarkers to prospectively harvest SCs with a repair phenotype . Our understanding of SC alterations with aging and how these changes contribute to their dysfunction, disease pathogenesis, and therapeutic use remains limited [800,801].
Understanding the specific signal transduction and metabolic pathways within normal and pathogenic SCs that are regulated by the mRNA–microRNA–lncRNA competitive endogenous RNA network is a promising clinical research direction. These studies should identify novel intracellular targets that enable the stabilization of functionally abnormal SCs and the generation of therapeutic SCs with a specific pattern of functions. Anticipated results may also help prepare therapeutic cell-free SC-derived exosomes with the required content and activity. By leveraging newly developed technologies, a quickly expanding toolbox for in vitro and ex vivo modeling, modern bioengineering-based approaches, and readily available bioinformatic and database resources, it is possible to create high-throughput, adaptive systems for characterizing and modulating SCs from individual patients. Altogether, this will lead to significant improvements in our understanding of neuropathy pathogenesis and treatments, as well as in personalized medicine, SC-targeted therapeutics, and healthcare overall.
Author Contributions
M.R.S., S.E.W., H.Z. and Y.Z. contributed substantially to the discussion of the content, writing the article, and reviewing and editing the manuscript before submission. All authors have read and agreed to the published version of the manuscript.
Funding
This work was supported by the Department of Pathology, University of Pittsburgh Medical Center, Shanghai Chest Hospital, Shanghai Municipal Health Commission (No. 2022YQ039, to Y.Z.), and Shanghai “Rising Stars of Medical Talents” Youth Development Program (to Y.Z.).
Conflicts of Interest
The authors declare no competing interests. The research was conducted in the absence of any commercial or financial relationships that could be construed as potential conflicts of interest.
Abbreviations
| | |
--- |
| 6-OHDA | 6-hydroxydopamine |
| AAV | adeno-associated virus/viral (vector) |
| AIDP | acute inflammatory demyelinating polyneuropathy |
| ALS | amyotrophic lateral sclerosis |
| AMPA | α-amino-3-hydroxy-5-methyl-4-isoxazolepropionic acid |
| ARAT | axon regeneration-associated transcript |
| BDNF | brain-derived neurotrophic factor |
| BMP5 | bone morphogenetic protein 5 |
| CAF(s) | cancer-associated fibroblast(s) |
| cAMP | cyclic adenosine 3′,5′-monophosphate |
| CB2 | cannabinoid receptor 2 |
| CGRP | calcitonin gene-related peptide |
| CIDP | chronic inflammatory demyelinating polyneuropathy |
| CMT | Charcot–Marie–Tooth Disease |
| CMV | cytomegalovirus |
| CNS | central nervous system |
| CNTF | ciliary neurotrophic factor |
| CSF | cerebrospinal fluid |
| CSF1 | colony-stimulating factor 1 |
| CSPG | chondroitin sulfate proteoglycan |
| DFT1 | Devil Facial Tumor 1 |
| DFT2 | Devil Facial Tumor 2 |
| DFTD | Devil facial tumor disease |
| DRG | dorsal root ganglion |
| EAE | experimental autoimmune encephalomyelitis |
| EBV | Epstein–Barr virus |
| ECM | extracellular matrix |
| EMT | epithelial–mesenchymal transition |
| ER | endoplasmic reticulum |
| ESC(s) | embryonic stem cell(s) |
| EV | extracellular vesicles |
| FGF | fibroblast growth factor |
| FOXM1 | Forkhead box M1 |
| GABA | gamma-aminobutyric acid |
| GBS | Guillain–Barré syndrome |
| GDNF | glial cell-derived neurotrophic factor |
| GFAP | glial fibrillary acidic protein |
| GLP-1 | glucagon-like peptide-1 |
| GM-CSF | granulocyte-macrophage colony-stimulating factor |
| HDAC | histone deacetylases |
| HIF1α | hypoxia-inducible factor 1-alpha |
| HIV | human immunodeficiency virus |
| HNPP | hereditary neuropathy with liability to pressure palsies |
| HSC(s) | hematopoietic stem cell(s) |
| HSV | herpes simplex virus |
| IGFBP(s) | insulin-like growth factor binding protein(s) |
| iNOS | inducible nitric oxide synthase |
| iPSC(s) | induced pluripotent stem cell(s) |
| LFA-3 | lymphocyte function-associated antigen 3 |
| lncARAT | lncRNA axon regeneration-associated transcript |
| lncRNA | long non-coding RNA |
| LRP1 | low-density lipoprotein receptor-related protein 1 |
| LRP4 | low-density lipoprotein receptor-related protein 4 |
| mAChR | muscarinic acetylcholine receptor |
| MAG | myelin-associated glycoprotein |
| MALAT1 | metastasis-associated lung adenocarcinoma transcript 1 |
| MBP | myelin basic protein |
| MFG-E8 | milk fat globule-epidermal growth factor 8 |
| MGUS | monoclonal gammopathy of unknown significance |
| miRNA | microRNA |
| MMP | matrix metalloproteinase |
| MPZ | myelin protein zero |
| MSC(s) | mesenchymal stem cell(s) |
| MWCNT | multi-walled carbon nanotube |
| NCAM | neural cell adhesion molecule |
| NF1 | neurofibromatosis type 1 |
| NGF | nerve growth factor |
| NGFR | nerve growth factor receptor |
| NK1 | neurokinin 1 |
| NLR(s) | NOD-like receptor(s) |
| NMDA | N-methyl-D-aspartic acid |
| NO | nitric oxide |
| NOD | nucleotide-binding and oligomerization domain |
| NOS | nitric oxide synthase |
| NRG1 | neuregulin-1 |
| NRF2 | nuclear factor erythroid 2-related factor 2 |
| NSC(s) | neural stem cell(s) |
| NSCLC | non-small cell lung cancer |
| NT | neurotrophin |
| OPC(s) | oligodendrocyte progenitor cell(s) |
| P0 | myelin protein 0 |
| P2 | myelin protein 2 |
| PAK | p21-activated kinase |
| PLGA | high-molecular-weight copolymer of lactic and glycolic acid |
| PLLA | poly-L-lactic acid |
| PMP22 | peripheral myelin protein 22 |
| PNS | peripheral nervous system |
| POU6F1 | POU domain class 6 Homeobox 1 |
| PST | polysialyltransferase |
| PTEN | phosphatase and tensin homolog (pathway) |
| ROS | reactive oxygen species |
| SC(s) | Schwann cell(s) |
| SCF | stem cell factor |
| SCLC | small cell lung cancer |
| SCP(s) | Schwann cell precursor(s) |
| SDF1 | stromal cell-derived factor 1 |
| shRNA | small hairpin inhibitory RNA |
| SOCS-2 | suppressor of cytokine signaling |
| SOD1 | superoxide dismutase |
| SP | substance P |
| TAM | tumor-associated macrophage |
| TGF-β | transforming growth factor-β |
| TLR(s) | Toll-like receptor(s) |
| TNF-α | tumor necrosis factor-alpha |
| TREM2 | triggering receptor expressed on myeloid cells 2 |
| TrkB | tropomyosin receptor kinase B |
| TRPM7 | transient receptor potential melastatin M7 |
| TRPV1 | transient receptor potential cation channel subfamily V member 1 |
| TRPV4 | transient receptor potential cation channel subfamily V member 4 |
| VEGF | vascular endothelial growth factor |
| VZV | varicella zoster virus |
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Figure 1. Alteration of Schwann cell longevity, phenotype, and function in different diseases. SCs are implicated in a wide range of diseases and pathological conditions due to their inherent potential for phenotypic and functional plasticity, as well as their ability to interact with various subsets of surrounding cells. Many inherited, toxic, autoimmune, metabolic, or pharmacological stimuli can damage SC motility, myelin production, and survival. In contrast, other stimuli, such as trauma, hormones, infection, or inflammation, induce SC injury response. Associated adaptive cellular reprogramming of SCs is seen as their denervation, dedifferentiation, proliferation, and phenotypic flexibility. Both SC damage and adaptive reprogramming are associated with significant and often specific alterations in cellular pathways, including neurotoxic, neurogenic, inflammatory, matrix remodeling, wound healing, and cell death pathways. Together with abundant experimental results, these data suggest that SCs can serve as therapeutic targets or cell transfer therapy for various types of neuropathies and associated pathophysiological conditions.
Figure 1. Alteration of Schwann cell longevity, phenotype, and function in different diseases. SCs are implicated in a wide range of diseases and pathological conditions due to their inherent potential for phenotypic and functional plasticity, as well as their ability to interact with various subsets of surrounding cells. Many inherited, toxic, autoimmune, metabolic, or pharmacological stimuli can damage SC motility, myelin production, and survival. In contrast, other stimuli, such as trauma, hormones, infection, or inflammation, induce SC injury response. Associated adaptive cellular reprogramming of SCs is seen as their denervation, dedifferentiation, proliferation, and phenotypic flexibility. Both SC damage and adaptive reprogramming are associated with significant and often specific alterations in cellular pathways, including neurotoxic, neurogenic, inflammatory, matrix remodeling, wound healing, and cell death pathways. Together with abundant experimental results, these data suggest that SCs can serve as therapeutic targets or cell transfer therapy for various types of neuropathies and associated pathophysiological conditions.
Figure 2. Employment of Schwann cell targeting and Schwann cell transplantation alone or in combination with known neuroprotective and neuroregenerative strategies for the treatment of CNS and PNS injury and neuropathy. Many pathological conditions, such as PNS and CNS trauma, injury, and demyelination, as well as neuropathies of different origins, chronic pain syndrome, and cancer, can benefit from targeting abnormal SCs or transplanting naïve or bioengineered SCs, as has been shown in numerous preclinical and clinical studies. The transfer of SCs alone or in combination with different scaffolds or neurotrophic factors significantly accelerates and improves the regeneration and myelination of injured spinal cord and peripheral nerves. Natural and pharmacological agents block high glucose-induced oxidative damage and apoptosis in SCs, enhance SC myelination, and thereby inhibit peripheral nerve degeneration, accelerating nerve conduction velocity. Many pharmacological agents, inhibitors of signal transduction pathways, and natural plant molecules can stimulate SC viability, proliferation, and migration and upregulate the expression of myelin-related and nerve growth-promoting genes in SCs, therefore promoting peripheral nerve regeneration. Gene therapy approaches, such as gene silencing or gene replacement strategies, targeting SCs demonstrate promise in treating inherited peripheral neuropathies, like Charcot–Marie–Tooth disease. SC-derived exosomal vesicles may have therapeutic value in repairing the wounded spinal cord and chemotherapy-induced peripheral neuropathy. SCs can also relieve neuropathic pain by remyelinating injured nerves. Although the role of SCs in the pathogenesis of various neuropathies is well established, and the applicability of SCs in treating these diseases has been proven, the full scope of potential SC targeting or the feasibility of their therapeutic application in CNS and PNS neuropathic conditions remains to be fully deciphered. BDNF, brain-derived neurotrophic factor; CNTF, ciliary neurotrophic factor; GDNF, glial cell line-derived neurotrophic factor; L1 (L1CAM), transmembrane protein member of the L1 protein family; MSC, mesenchymal stromal cell; NCAM, neural cell adhesion molecule; NT, neurotrophin; NSC, neural stem cell; POU6F1, POU domain class 6, transcription factor 1; PST, polysialyltransferase.
Figure 2. Employment of Schwann cell targeting and Schwann cell transplantation alone or in combination with known neuroprotective and neuroregenerative strategies for the treatment of CNS and PNS injury and neuropathy. Many pathological conditions, such as PNS and CNS trauma, injury, and demyelination, as well as neuropathies of different origins, chronic pain syndrome, and cancer, can benefit from targeting abnormal SCs or transplanting naïve or bioengineered SCs, as has been shown in numerous preclinical and clinical studies. The transfer of SCs alone or in combination with different scaffolds or neurotrophic factors significantly accelerates and improves the regeneration and myelination of injured spinal cord and peripheral nerves. Natural and pharmacological agents block high glucose-induced oxidative damage and apoptosis in SCs, enhance SC myelination, and thereby inhibit peripheral nerve degeneration, accelerating nerve conduction velocity. Many pharmacological agents, inhibitors of signal transduction pathways, and natural plant molecules can stimulate SC viability, proliferation, and migration and upregulate the expression of myelin-related and nerve growth-promoting genes in SCs, therefore promoting peripheral nerve regeneration. Gene therapy approaches, such as gene silencing or gene replacement strategies, targeting SCs demonstrate promise in treating inherited peripheral neuropathies, like Charcot–Marie–Tooth disease. SC-derived exosomal vesicles may have therapeutic value in repairing the wounded spinal cord and chemotherapy-induced peripheral neuropathy. SCs can also relieve neuropathic pain by remyelinating injured nerves. Although the role of SCs in the pathogenesis of various neuropathies is well established, and the applicability of SCs in treating these diseases has been proven, the full scope of potential SC targeting or the feasibility of their therapeutic application in CNS and PNS neuropathic conditions remains to be fully deciphered. BDNF, brain-derived neurotrophic factor; CNTF, ciliary neurotrophic factor; GDNF, glial cell line-derived neurotrophic factor; L1 (L1CAM), transmembrane protein member of the L1 protein family; MSC, mesenchymal stromal cell; NCAM, neural cell adhesion molecule; NT, neurotrophin; NSC, neural stem cell; POU6F1, POU domain class 6, transcription factor 1; PST, polysialyltransferase.
Figure 3. Basic strategies and clinical benefits of therapeutic Schwann cells. Numerous clinical and experimental studies have demonstrated the benefits of targeting pathogenic SCs and utilizing autologous or allogeneic naïve and engineered SCs, as well as their products, as a promising novel therapeutic approach. Proven clinical responses have been documented for various neuropathological conditions associated with neurotrauma, nerve injury, neurotoxic and neuroinflammatory disorders, neuropathic pain, demyelinating states, as well as wound healing and cancer-related abnormalities. The regenerative effect conferred by SCs is not limited to strong capabilities in accelerating axon repair, regeneration, and remyelination. Transplantation of reprogrammed or engineered SCs, application of SC exosomes, and pharmacological targeting or redifferentiation of resident SCs and their precursors in preclinical and clinical studies can improve the control of pain sensation, inflammatory response, EMT, and tissue regeneration, thereby preserving damaged tissue and supporting injured neurons. The utilization of Artificial Intelligence technology should accelerate the development of practical regenerative biomaterials, thereby increasing the potential of SC-based treatments and their clinical applicability. It remains to be established what the optimal conditions are for utilizing SC-based approaches and crucial signal transduction pathways in pathological and therapeutic SCs, which should be targeted for treating specific neurological diseases and abnormalities. AI, Artificial Intelligence; EMT, epithelial–mesenchymal transition; OPC, oligodendrocyte progenitor cells.
Figure 3. Basic strategies and clinical benefits of therapeutic Schwann cells. Numerous clinical and experimental studies have demonstrated the benefits of targeting pathogenic SCs and utilizing autologous or allogeneic naïve and engineered SCs, as well as their products, as a promising novel therapeutic approach. Proven clinical responses have been documented for various neuropathological conditions associated with neurotrauma, nerve injury, neurotoxic and neuroinflammatory disorders, neuropathic pain, demyelinating states, as well as wound healing and cancer-related abnormalities. The regenerative effect conferred by SCs is not limited to strong capabilities in accelerating axon repair, regeneration, and remyelination. Transplantation of reprogrammed or engineered SCs, application of SC exosomes, and pharmacological targeting or redifferentiation of resident SCs and their precursors in preclinical and clinical studies can improve the control of pain sensation, inflammatory response, EMT, and tissue regeneration, thereby preserving damaged tissue and supporting injured neurons. The utilization of Artificial Intelligence technology should accelerate the development of practical regenerative biomaterials, thereby increasing the potential of SC-based treatments and their clinical applicability. It remains to be established what the optimal conditions are for utilizing SC-based approaches and crucial signal transduction pathways in pathological and therapeutic SCs, which should be targeted for treating specific neurological diseases and abnormalities. AI, Artificial Intelligence; EMT, epithelial–mesenchymal transition; OPC, oligodendrocyte progenitor cells.
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Summary of First Week
Summary of Implantation
Summary of Second Week
Birth Defects Resulting from Abnormal Neurulation
Summary of Third Week
Case 4-5
Summary of Fourth to Eighth Weeks
Case 5-1
Procedures for Assessing Fetal Status
Cell Cultures and Chromosomal Analysis
Summary of Fetal Period
Summary of Placenta and Fetal Membranes
Case 7-2
Case 7-4
Case 8-4
Case 8-3
Summary of Pharyngeal Apparatus, Face, and Neck
Case 9-2
Case 9-4
Summary of Respiratory System
Summary of Alimentary System
Summary of Urogenital System
Summary of Cardiovascular System
Summary of Skeletal System
Case 14-1
Case 14-2
Summary of Muscular System
Case 15-4
Summary of Limb Development
Case 16-2
Case 16-3
Case 16-4
Summary of Nervous System
Case 17-3
Case 17-4
Development of Eyes and Related Structures
Summary of Eye Development
Case 18-1
Summary of Ear Development
Teratology: Study of Abnormal Development
Summary of Birth Defects
The Developing Human-Clinically Oriented Embryology by Keith L. Moore
Sakeena Batool
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Abstract
Keith L. Moore
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This essay describes the approach and early results of the collaborative Embryo Project and its on-line encyclopedia ( The project is based on a relational database that allows federated searches and inclusion of multiple types of objects targeted for multiple user groups. The emphasis is on the history and varied contexts of developmental biology, focusing on people, places, institutions, techniques, literature, images, and other aspects of study of embryos. This essay introduces the ways of working as well as the long-term goals of the project. We invite others to join the effort, both in this particular project and in joining together in digital collection, archiving, and knowledge generation at the borders of biology and history.
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Embryology
Fer Sánchez
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The Human Embryo - A Brief Biological and Philosophical Biography
Amrita Bhagia, Henry Travers
South Dakota Medicine, 2023
The human embryo is dimensionally complex. As the details of its developmental biology and pathobiology became more established, widely divergent concepts about the embryo emerged in culture, religion, morals, ethics, and law1 and today underlie worldwide controversies about the very meaning of human life. Our investigation began with our belief that historical research into the evolution of our biological and philosophical understanding of the embryo could provide a basis for approaching those controversies. We hypothesized that scientific understandings of conception and fetal development historically influenced the social, cultural, philosophical and legal status of the embryo. We explored the conceptual divergence between embryology and philosophical domains that began in the Renaissance. We confined ourselves to embryology within western civilizations and philosophical and theological doctrine from a predominantly Christian perspective.
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Full Text (PDF) - The International Journal of Developmental
Manuela Monti
2011
The unresolved debate about frozen embryos has left open the discussion on "what to do with them". There are only three ways to deal with frozen embryos: 1) to leave them frozen indefinitely; 2) to defrost and discard them and 3) to use them for research. In this paper, we suggest that the application of current scientific knowledge, instead of inappropriately referring to ethical principles or to the concept of person, could help with the decision about what to do with hundreds of thousands of frozen embryos, thus bringing the sensitive debate on bioethical issues to shared practical solutions. We face a new individual only when a new functional copy of his genome is formed. In both natural and artificial animal and plant reproduction, this principle applies. This status occurs in humans at the 4-8 cell stage. Acknowledgement of this factual datum would allow advocates of all religious and ideological beliefs to defend their principles and to realign their positions to a setting within the boundaries of current scientific knowledge.
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Atlas of Human Embryology: from Oocytes to Preimplantation Embryos
Sidney Solis Pineda
atlas de células germinales a blastocisto
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New insights into human early embryo development: a particular theoretical study
Hassen Chaabani
International Journal of Modern Anthropology, 2018
The experimental research on the subject of human early embryo development has been remained insufficient because of ethical norms and legal constraints. In this context, the carrying out of theoretical studies could speed up the slow knowledge progression on this subject. Thus, I present here a particular model of theoretical study based on a synthesis of selected published experimental results combined with what has been provided from my interpretation of scientific signs masked in some Qur‟an verses. In the obtained detailed scenario, I consider that the third cleavage, resulting in 8 blastomeres, is coupled to a particular rearrangement in which 2 daughter cells seem move down and 2 others move up; while the 4 remaining cells seem stay at their initial position (I called them “4 HI cells”). Nevertheless, I consider that the fourth cleavage, resulting in 16 blastomeres, is coupled to a rearrangement that gives the impression of a harmonious descent: in fact, one of each two daugh...
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Objective: To describe the times associated with the morphological changes that occur in the embryo during preimplantation development based on the largest sample size described with time lapse. Design: Cohort study. Setting: University-affiliated private center. Patient(s): A total of 9,530 embryos from 1,806 intracytoplasmic sperm injection (ICSI) cycles. Intervention(s): None. Main Outcome Measure(s): Using a time-lapse system, embryo images were acquired for at least 68 hours, in some cases reaching 120-130 hours. Embryo cleavage time points up to 8-cell-stage (t2-t8) as well as morulae (tM) and blastocyst formation (tB) were registered in hours after ICSI. Additionally, duration of the cell cycle (cc) and synchrony (s) of the second and third cell cycles were defined. Finally, four subgroups of embryos were considered: the ''regular divisions'' group excluded embryos with a direct cleavage from 1 to 3 or 2 to 5 cells, and the ''viable 8-cell,'' the ''viable blastocyst,'' and ''implanted embryos'' groups included only embryos viable to the 8-cell stage, blastocyst stage, or transferred and successfully implanted, respectively. Result(s): Averages of times in the general population were: t2 ¼ 27.9 hours, t3 ¼ 38.2 hours, t4 ¼ 40.7 hours, t5 ¼ 51.0 hours, t6 ¼ 54.1 hours, t7 ¼ 56.7 hours, t8 ¼ 59.1 hours, tM ¼ 86.6 hours, tB ¼ 104.1 hours, cc2 ¼ 10.3 hours, cc3 ¼ 12.8 hours, s2 ¼ 2.7 hours, and s3 ¼ 9.9 hours. Comparison between groups showed significant differences between regular divisions and viable 8 cells for t2, t3, t5, cc2, cc3, s2, and s3; between 8 cells and blastocyst for t5, t8, tM, cc3, and s2; and between blastocyst and implanted embryos for t8, tM, tB, and s2. Differences in timing related to morphology of cleavage-and blastocyst-stage embryos were detected. Conclusion(s): A time-lapse monitoring system applied to embryology allows accuracy and objectivity when defining the basis of embryo development within a clinic. The sample size is the largest ever described that provides consistent information about the normal distribution of embryo developmental timings. (Fertil Steril Ò 2013;100:1602-9. Ó2013 by American Society for Reproductive Medicine.)
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Hans J. ten Donkelaar
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2705 | https://www.quora.com/Why-are-the-formulas-of-sphere-and-hemisphere-different | Why are the formulas of sphere and hemisphere different? - Quora
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Geometry
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Math Geometry
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Why are the formulas of sphere and hemisphere different?
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Rohan Awasthi
Data Science Enthusiast, School Level Maths Gold Medalist · Author has 945 answers and 800.1K answer views
·4y
They are not different, they are similar because when the sphere is cut into half, it becomes a hemisphere.
This is a sphere, and the below one is the hemisphere.
Volumes:
Sphere -> 4 / 3 πr³
Hemisphere -> 2 / 3 π r³
Surface Areas:
Sphere -> 4πr²
Hemisphere( Curved) -> 2πr²
Hemisphere( Total) -> 3πr²
So, as you can see, the hemisphere is having half of all the formulas of the sphere, except the total surface area, because, in the total surface area, you get an extra circular base of area πr², which you need to add to the curved surface area, except it, all the formulas are halved.
I hope it helps.
Continue Reading
They are not different, they are similar because when the sphere is cut into half, it becomes a hemisphere.
This is a sphere, and the below one is the hemisphere.
Volumes:
Sphere -> 4 / 3 πr³
Hemisphere -> 2 / 3 π r³
Surface Areas:
Sphere -> 4πr²
Hemisphere( Curved) -> 2πr²
Hemisphere( Total) -> 3πr²
So, as you can see, the hemisphere is having half of all the formulas of the sphere, except the total surface area, because, in the total surface area, you get an extra circular base of area πr², which you need to add to the curved surface area, except it, all the formulas are halved.
I hope it helps.
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More answers below
What is the difference between a hemisphere and a half sphere?
What is the volume formula for a hemisphere?
What's the formula to find the scale factor of two different spheres?
What are the different hemispheres of the earth?
How is a sphere and a hemisphere defined?
Justin Rising
PhD in statistics · Author has 12.1K answers and 26.5M answer views
·Updated 7y
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Why is half of a circle called a semi-circle, but half of a sphere called a hemisphere?
"semi" and "hemi" are respectively Latin and Greek prefixes meaning "half". The word "circle" is of Latin origin, so it gets the Latin prefix, but the word "sphere" is of Greek origin, so it gets the Greek prefix.
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Chris Richardson
Studied Physics · Author has 7.2K answers and 4M answer views
·9y
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How is a sphere and a hemisphere defined?
You need to define a fixed point and a length (let’s call it r for radius).
The surface of the sphere is every point that is the distance r from the fixed point. Imagine you pinch a piece of string at both ends and holding one hand stationary you wave the other hand all over the place keeping the string taught. Every position of the moving hand is a point on the sphere with r being the length of the string.
How you communicate that mathematically depends on your chosen system of coordinates.
If you want to work in cartesian coordinates (with three axes all at right angles to each other i.e. x, y,
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You need to define a fixed point and a length (let’s call it r for radius).
The surface of the sphere is every point that is the distance r from the fixed point. Imagine you pinch a piece of string at both ends and holding one hand stationary you wave the other hand all over the place keeping the string taught. Every position of the moving hand is a point on the sphere with r being the length of the string.
How you communicate that mathematically depends on your chosen system of coordinates.
If you want to work in cartesian coordinates (with three axes all at right angles to each other i.e. x, y, z) then, by the Pythagoras Theorem, r = sqrt (x^2 + y^2 +z^2). Practically that means that if sqrt (x^2 + y^2 +z^2) equals your given distance, r, then it is a point on the surface of the sphere and all points that satisfy the equation define the surface.
If you want to work in spherical coordinates (with a distance, r, and two different angles both perpendicular to a chosen direction) then the sphere is defined by the distance, r, from your fixed point for all values of both angles. Spherical coordinate system
A hemisphere is simply defined by cutting the volume of the sphere in to two equal volumes using a flat plane. This can be done in an infinite number of ways depending on the plane you choose.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·1y
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Can you explain the differences between a sphere, hemisphere, cone, cylinder, cube, and cuboid?
To explain (in words) the differences between these things would be an utterly futile exercise. The meaningful way to answer this question is to show pictures of what these shapes actually are like!
Actually, I quite enjoyed drawing these!
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To explain (in words) the differences between these things would be an utterly futile exercise. The meaningful way to answer this question is to show pictures of what these shapes actually are like!
Actually, I quite enjoyed drawing these!
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What is the center of the sphere formula?
How can you make a hemisphere from a sphere?
How do hemisphere and sphere differ?
What is the difference between a sphere, hemisphere, and half-sphere?
What is the similarity between a sphere and a hemisphere?
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Surface area and volume formulas for a sphere and a hemisphere differ because a hemisphere is not just half of a sphere in every geometric sense: some measures split linearly while others require adding or removing the flat circular face that the cut creates. Explain in three short steps.
Volume splits exactly in half
Volume of a sphere: V_sphere = (4/3)πr³.
Cutting a sphere by a plane through its center divides its solid into two congruent halves, so each hemisphere has exactly half the sphere’s volume:
V_hemisphere = (1/2)·(4/3)πr³ = (2/3)πr³.
Reason: volume is an additive measure; two c
Continue Reading
Surface area and volume formulas for a sphere and a hemisphere differ because a hemisphere is not just half of a sphere in every geometric sense: some measures split linearly while others require adding or removing the flat circular face that the cut creates. Explain in three short steps.
Volume splits exactly in half
Volume of a sphere: V_sphere = (4/3)πr³.
Cutting a sphere by a plane through its center divides its solid into two congruent halves, so each hemisphere has exactly half the sphere’s volume:
V_hemisphere = (1/2)·(4/3)πr³ = (2/3)πr³.
Reason: volume is an additive measure; two congruent solids sum to the whole.
Curved surface area also splits in half
Total curved surface area of a sphere: A_sphere = 4πr².
The curved (spherical) surface of a hemisphere is exactly half of the sphere’s curved surface:
curved A_hemisphere = (1/2)·4πr² = 2πr².
Reason: the spherical surface is partitioned into two congruent curved patches by the great circle.
The flat face changes the total surface area
A hemisphere includes an extra flat circular face of area πr² (the cross-section where the cut was made).
Total surface area of a hemisphere (including the base): A_total_hemisphere = curved area + base area = 2πr² + πr² = 3πr².
Thus the hemisphere’s total surface area is not half of the sphere’s total surface area; it’s half the curved part plus the new flat face.
Summary formulas
Sphere: volume (4/3)πr³; surface area 4πr².
Hemisphere: volume (2/3)πr³; curved surface area 2πr²; total surface area (including base) 3πr².
Intuition: “Half” applies to quantities that are additive and split by the cut (volume, curved area), but the cut creates a new boundary (the flat face) that must be counted in surface area, so the hemisphere’s total surface is not a simple half of the sphere’s total surface.
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Vivek Shaw
I love Proofs
·8y
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How do I derive the formula for the surface area of the sphere?
Consider the diagram below of a hemisphere of a sphere
Let us first try to find out the surface area of this hemisphere whose Radius is R.
Let the center of the sphere be O as shown above in the diagram
So we have OA = R, (radius of the sphere) .
Let us consider 2 circular strips BE and CD as shown in the diagram above in blue color. Let the height of this circular strip i.e CB or DE be infinitesimally small.
Let us first find the surface area of this circular strip.
Suppose we bring out this circular strip. it will be in the form of a cylinder with no top and bottom.Since CB is very small we ca
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Consider the diagram below of a hemisphere of a sphere
Let us first try to find out the surface area of this hemisphere whose Radius is R.
Let the center of the sphere be O as shown above in the diagram
So we have OA = R, (radius of the sphere) .
Let us consider 2 circular strips BE and CD as shown in the diagram above in blue color. Let the height of this circular strip i.e CB or DE be infinitesimally small.
Let us first find the surface area of this circular strip.
Suppose we bring out this circular strip. it will be in the form of a cylinder with no top and bottom.Since CB is very small we can consider CB to be perpendicular to BF as shown in the diagram above.
Let us now find the radius of this cylinder or circular strip BCDE.
From the right angle triangle OBF, we have angle OBF = Θ
So BF = RCosΘ.
We know that the surface area of a cylinder is 2∗p i∗r∗h 2∗p i∗r∗h Here r = BF = RCosΘ
Let us find h = CB.
Consider the triangle OBC.
Let the angle BOC = dΘ.
Since CB is very small we can consider CB as perpendicular to OB.
So OBC is a right angled triangle.
Also OC = R.
So BC = RSin(dΘ)
So surface area of the cylinder or circular strip BCDE is 2∗p i∗r∗h=2∗p i∗R C o s Θ∗R S i n(d Θ)2∗p i∗r∗h=2∗p i∗R C o s Θ∗R S i n(d Θ)
We know that when angle(dΘ) is very small , we have Sin(dΘ) = dΘ
So this
small A r e a=d A A r e a=d A = 2∗p i∗R C o s Θ∗R S i n(d Θ)=2∗p i∗R C o s Θ∗R d Θ=2∗p i∗(R 2)∗C o s Θ∗d Θ.2∗p i∗R C o s Θ∗R S i n(d Θ)=2∗p i∗R C o s Θ∗R d Θ=2∗p i∗(R 2)∗C o s Θ∗d Θ.
Now to cover the full surface area of the half of the sphere we have to vary Θ from 0 to pi/2.
So,
Now for both halves of the sphere, the area is
Edit 1: Sorry for my poor drawing.
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Lokesh Kumar
BE in Computer Engineering, Netaji Subhas Institute of Technology (Graduated 2021)
·7y
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How is the height of a hemisphere equal to its radius?
Just scroll through images and you will find the answer.
ok,what do you think,what its height should be..
D→diameter,R→ radius
height=D=2R
2.Now cut it into half
So what its height should be now.
of course ,2R/2=R
Conclusion: height of hemisphere=its radius
image source: google images
Continue Reading
Just scroll through images and you will find the answer.
ok,what do you think,what its height should be..
D→diameter,R→ radius
height=D=2R
2.Now cut it into half
So what its height should be now.
of course ,2R/2=R
Conclusion: height of hemisphere=its radius
image source: google images
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Geoff Mangum
Founder, PuttingZone in 26 countries worldwide. (2000–present) · Author has 3.8K answers and 3.1M answer views
·3y
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What is the difference between a hemisphere and a half sphere?
If you mean volume, a “half sphere” has half the volume of another sphere. Ordinarily, hemisphere means the geometrical object that remains after a sphere is bisected by a plane that includes the center of a sphere and this defines two equal top and bottom hemispheres. Each hemisphere also has 1/2 the volume of the sphere so divided.
So the “half sphere” is a sphere of 1/2 volume whereas a hemisphere is half of a sphere with a flat plane for its bottom.
The half sphere so defined has 1/2 volume and
V = 4/3•pi•r^3 (sphere)
r = CUBE ROOT(V • 3/4 / pi) (sphere)
V = 1/2•4/3 • pi•r^3 (half sphere)
r = CU
Continue Reading
If you mean volume, a “half sphere” has half the volume of another sphere. Ordinarily, hemisphere means the geometrical object that remains after a sphere is bisected by a plane that includes the center of a sphere and this defines two equal top and bottom hemispheres. Each hemisphere also has 1/2 the volume of the sphere so divided.
So the “half sphere” is a sphere of 1/2 volume whereas a hemisphere is half of a sphere with a flat plane for its bottom.
The half sphere so defined has 1/2 volume and
V = 4/3•pi•r^3 (sphere)
r = CUBE ROOT(V • 3/4 / pi) (sphere)
V = 1/2•4/3 • pi•r^3 (half sphere)
r = CUBE ROOT(V • 6/4 / pi) (half sphere)
r(sphere) / r(half sphere) = CUBE ROOT(3) / CUBE ROOT(6) = 0.793
The half sphere radius is 0.793 times the sphere radius when the half sphere volume is 1/2 the volume of a sphere.
.
0.793^3 = 1/2
V = 4/3•pi•r^3
r = 1 (sphere) so r^3 = 1 (sphere) and 1/2V = 4/3•pi•1/2 so r’^3 = 1/2 (half sphere r’).
r’^3 = 1/2
r’ = CUBE ROOT(1/2)
r’ = 0.793
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Rick Cooper
Studied Electrical Engineering (BSEE) (Graduated 1972) · Author has 434 answers and 289.1K answer views
·10mo
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Is it possible to calculate the volume of a hemisphere with only the surface area and length known? If so, what is the formula?
I say yes. And here’s how I approach it:
You want to deduce the volume of half a sphere (“hemi-” right?). And you’re given the surface area of this half-sphere.
These are the two useful relationships regarding s pheres:
Surface area = 4 π r^2 (square units)
Volume = (4/3) · π · r^3 (cubical units)
So you can use the relationships this way: You have a surface area given, which you can use to find the radius of the sphere, (which is the same as that of the hemisphere).
Knowing the radius will allow you to compute the volume.
Remember that you’re working with half a sphere, so
set the given, that is, half
Continue Reading
I say yes. And here’s how I approach it:
You want to deduce the volume of half a sphere (“hemi-” right?). And you’re given the surface area of this half-sphere.
These are the two useful relationships regarding s pheres:
Surface area = 4 π r^2 (square units)
Volume = (4/3) · π · r^3 (cubical units)
So you can use the relationships this way: You have a surface area given, which you can use to find the radius of the sphere, (which is the same as that of the hemisphere).
Knowing the radius will allow you to compute the volume.
Remember that you’re working with half a sphere, so
set the given, that is, half the total surface area, equal to 4 π r^2 square units;
divide both sides by 4π and
take the (positive) square root, and you now know the radius of the hemisphere! [The number of square units given equals (1/2)4 π r^2, 1/2 of the total surface area.]
Equation 2 yields the total volume of the sphere, once you know r, the radius. The volume of your hemi /half-sphere is therefore (1/2) (4/3) · π · r^3 cubical units
Q.E.D.
I can’t think of what “length” could be involved here. Surface area is in 2 dimensions, and volume is in 3D. A Length is on dimensional. Hmm. ? Distance of the sphere’s center from some given coordinate in three-space?
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Neil Kelcey
Former Public School Math Teacher (retired) at Toronto District School Board (1971–2002) · Author has 86 answers and 243.2K answer views
·6y
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The formula of a surface area hollow hemisphere?
Q posed: The formula of a surface area hollow hemisphere?
I will assume that my diagram below is what you had in mind? And that a “bowl” is a common name for this shape?
Recall that the
• SA of a sphere is: SA = 4πr²
• Area of a circle is: A = πr²
Let “R” represent the radius of the “LARGE” sphere/circle
Let “r” represent the radius of the “SMALL” sphere/circle
Then the total SA of the bowl will be the outer larger sphere surface, plus the smaller inner sphere surface, and plus the “ring” edge on the top.
Hope this helps! Cheers! :)
Continue Reading
Q posed: The formula of a surface area hollow hemisphere?
I will assume that my diagram below is what you had in mind? And that a “bowl” is a common name for this shape?
Recall that the
• SA of a sphere is: SA = 4πr²
• Area of a circle is: A = πr²
Let “R” represent the radius of the “LARGE” sphere/circle
Let “r” represent the radius of the “SMALL” sphere/circle
Then the total SA of the bowl will be the outer larger sphere surface, plus the smaller inner sphere surface, and plus the “ring” edge on the top.
Hope this helps! Cheers! :)
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Anshul Sharma
Medical student (MEDICO) (2023–present)
·6y
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What is the formula for a hollow hemisphere?
Area Of Hemisphere
A Hemisphere
Definition :
A hemisphere is the half part of a sphere. We can find many of the real-life examples of the hemispheres such as our planet Earth can be divided into two segments the southern & northern hemispheres and much more. In the case of the hemispheres, we primarily find out the curved surface area and the total surface area. The area of the outer portion is constituted as the curved surface area while the total surface area is obtained by addition of curved surface area and the area of the upper circle of the hemisphere.
The Hemispheres can either be of a Soli
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Area Of Hemisphere
A Hemisphere
Definition :
A hemisphere is the half part of a sphere. We can find many of the real-life examples of the hemispheres such as our planet Earth can be divided into two segments the southern & northern hemispheres and much more. In the case of the hemispheres, we primarily find out the curved surface area and the total surface area. The area of the outer portion is constituted as the curved surface area while the total surface area is obtained by addition of curved surface area and the area of the upper circle of the hemisphere.
The Hemispheres can either be of a Solid type or can be Hollow.
Area of Hemisphere Formulae:
Curved surface area of hemisphere = 2πr2
As the Hemisphere is the half part of a sphere, therefore, the curved surface area is also half that of the sphere.
Total surface area of the hemisphere– While calculating the total surface area of a hemisphere, we need to consider the base of the hemisphere which is circular in shape. Thus the total surface area of a hemisphere is equal to
total surface area= curved surface area + area of the base circle
= 2πr2+πr2
= 3πr2
Area of Hollow Hemisphere = Area of External Hemisphere – Area of Internal Hemisphere
Real Life Examples of Hemispheres:
Cups
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Stephen Bell
semi-retired writer on technology topics. Interested in language and maths (particularly solid geometry) and historical belief systems particularly in the occult/esoteric area. · Author has 3K answers and 7.7M answer views
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Why is half of a circle called a semi-circle, but half of a sphere called a hemisphere?
Then there’s the French, who use “demi-”. The demi prefix is Latin again, but a more recent arrival in the English language through Norman French; eg demigod, demi-tasse.
What the Americans call a 64th-note in music is in British parlanc...
Continue Reading
Then there’s the French, who use “demi-”. The demi prefix is Latin again, but a more recent arrival in the English language through Norman French; eg demigod, demi-tasse.
What the Americans call a 64th-note in music is in British parlanc...
Upvote ·
9 4
Varun Pai
Science enthusiasit, Maths lover, Polyglot & a casual quoran · Author has 263 answers and 631.5K answer views
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How do you calculate the formula for the hemisphere's surface area?
For a hemisphere of radius r,
Curved surface area CSA =
Total surface area TSA = CSA + (area of the base)
For the derivation of this formula, visit this website Derivation of Formula for Total Surface Area of the Sphere by Integration. The derivation is for sphere, you can apply same logic for a hemisphere.
Continue Reading
For a hemisphere of radius r,
Curved surface area CSA =
Total surface area TSA = CSA + (area of the base)
For the derivation of this formula, visit this website Derivation of Formula for Total Surface Area of the Sphere by Integration. The derivation is for sphere, you can apply same logic for a hemisphere.
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9 2
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verb (1)
as in do well
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thrives
verb as in do well
Synonyms Antonyms
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bloom
blossom
boom
develop
flourish
grow
mushroom
prosper
shine
succeed
Strong matches
advance
arrive
batten
burgeon
increase
progress
radiate
rise
score
wax
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bear fruit
get ahead
get fat
get on
get places
get there
grow rich
make a go
shoot up
turn out well
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2708 | https://onlinelibrary.wiley.com/doi/10.1111/j.1462-5822.2007.00981.x | Macrophage complement receptors and pathogen clearance - Van Lookeren Campagne - 2007 - Cellular Microbiology - Wiley Online Library
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Cellular Microbiology
Volume 9, Issue 9 pp. 2095-2102
Free Access
Macrophage complement receptors and pathogen clearance
Menno Van Lookeren Campagne,
Corresponding Author
Menno Van Lookeren Campagne
Departments of Immunology,
E-mail menno@gene.com; Tel. (+1) 650 225 1755; Fax (+1) 650 225 8221.Search for more papers by this author
Christian Wiesmann,
Christian Wiesmann
Protein Engineering and
Search for more papers by this author
Eric J. Brown,
Eric J. Brown
Microbial Pathogenesis, Genentech, 1 DNA Way, South San Francisco, CA 94080, USA.
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Menno Van Lookeren Campagne,
Corresponding Author
Menno Van Lookeren Campagne
Departments of Immunology,
E-mail menno@gene.com; Tel. (+1) 650 225 1755; Fax (+1) 650 225 8221.Search for more papers by this author
Christian Wiesmann,
Christian Wiesmann
Protein Engineering and
Search for more papers by this author
Eric J. Brown,
Eric J. Brown
Microbial Pathogenesis, Genentech, 1 DNA Way, South San Francisco, CA 94080, USA.
Search for more papers by this author
First published: 30 July 2007
Citations: 253
About
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Sections
Summary
Introduction
Complement C3 opsonization
Structure function of complement C3 receptors on phagocytic cells
Kupffer cells, complement and pathogen recognition
The role of CR3 and CRIg in complement-mediated phagocytosis
Conclusion and future perspective
Acknowledgements
References
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Summary
Phagocytosis, an important mechanism of the host-defence system and a primary function of macrophages, is facilitated by opsonization, a process by which serum components tag pathogens for recognition by neutrophils and macrophages. Complement component C3 is central to opsonization. Its first cleavage product, C3b, forms the multisubunit enzyme, C3bBb, which proteolytically cleaves additional C3 molecules on the pathogen surface. C3b is further degraded to iC3b, C3c and C3dg, products that serve as ligands for selective complement receptors on leukocytes. This receptor–ligand interaction subsequently modulates immune responses or directly targets the pathogen for clearance by phagocytosis. Although a central role for C3 in phagocytosis of certain pathogens is well accepted, the receptors orchestrating the phagocytic response have not been well characterized. The recent structures of C3 and its breakdown products have increased our insights into the molecular basis of complement activation and recognition by their receptors. Here we review the biology of macrophage receptors for C3 fragments and discuss their role in the host response to pathogens.
Introduction
Macrophages have an innate capacity to recognize pathogen-expressed surface structures, so-called molecular patterns (reviewed by Taylor et al., 2005). While direct pathogen recognition is a fundamental aspect of innate immunity, opsonization allows diversification of the phagocyte recognition repertoire (Stuart and Ezekowitz, 2005). The process of recognition and phagocytosis involves multiple ligand–receptor interactions, and it is now clear that various opsonins, including immunoglobulins, collectins and complement components, guide the cellular activities required for pathogen internalization (reviewed by Aderem and Underhill, 1999; Underhill and Ozinsky, 2002). A good example is the cross-talk between Fc- and complement-receptors in mediating efficient phagocytosis of immune complexes (Zhou and Brown, 1994; Caron and Hall, 1998; Skoberne et al., 2006).
Complement-mediated opsonization depends on triggering of the classical, lectin and alternative pathways of complement activation (reviewed by Walport, 2001). All three share the common step of activating the central component C3, but they differ in initial biochemical steps leading to C3 activation. The classical pathway is activated by antibodies bound to the pathogen surface, which in turn bind C1q, the first component of the classical pathway. This sets off a serine protease cascade that ultimately cleaves C3 to its active form, C3b. The lectin pathway is activated after recognition of molecular patterns by lectin proteins. To date, three members of this pathway have been identified: mannose-binding lectins (MBLs), ficolins and the C-type lectin, SIGN-R1. Following pattern recognition, these lectins associate with serine proteases, which act like C1 in the classical pathway, activating component C4 leading to the central C3 step. In addition to recognizing bacteria surface structures, SIGN-R1 can also activate the pathway by binding to C1q (Kang et al., 2006). Finally, the alternative pathway contrasts with both the classical and lectin pathways in that it is activated by direct reaction of the internal C3 ester with the pathogen surface. Initial C3 binding to an activating surface leads to rapid amplification of C3b deposition through the action of the alternative pathway proteases Factor B and Factor D. Importantly, C3b deposited by either the classical or the lectin pathway also can lead to further amplification of C3 proteolytic cleavage through the actions of Factors B and D.
Complement C3 opsonization
In all three pathways of complement activation, the pivotal step in opsonization is conversion of the component C3 to C3b. C3 itself is a complex and flexible protein with a molecular weight of 185 kDa. It contains two chains that originate from a single chain precursor with a total of 13 distinct domains (Janssen et al., 2005). The core of the molecule is formed by eight so-called macroglobulin (MG) domains, which constitute the lion's share of the α and β chains of C3 (Fig. 1A). The small helical domain C3a or anaphylatoxin, is connected to the N-terminus of the alpha-chain via the α'NT linker. Inserted into the core of the alpha-chain are the CUB (C 1r/C1s, U egf and B one mophogenetic protein-1), and TED domains, the latter containing the thioester bond that allows covalent association of C3b with pathogen surfaces. The C345C domain at the C-terminus of the C3 α-chain was identified by homology to the C345C domain in C5b that is required for complex formation with the membrane attack complement components C6 and C7 (Thai and Ogata, 2003).
Figure 1
Open in figure viewerPowerPoint
Structural basis of C3 opsonization and receptor recognition.
A. Conformational changes upon activation and degradation of C3 and recognition by various human complement receptors. Native C3 consists of eight macroglobulin domains (ovals in far left drawing), which constitute the core of the α- (purple) and β- (green) chains of C3. The thioester domain (TED), CUB-domain (orange), C3a domain (blue), α'NT linker (orange thick line) and the C345C domain (dark blue) form the remaining modules of C3. Cleavage of C3 by the enzyme ‘C3 convertase’ releases C3a and induces a major shift in the overall conformation of the remaining C3b molecule. This conformational change facilitates interaction of the ester (red triangle) in the thioester domain (TED) with pathogen surfaces. These structural rearrangements allow for further cleavage by factor I and cofactors and exposes binding sites for CR1-4 and CRIg, initiating phagocytosis or adaptive immune responses.
B. Left: Ribbon diagrams of C3b in complex with CRIg. The colour scheme of A is used except for the C345C domain that is depicted here in pink; the surface of CRIg is shown in yellow. Right: Structural changes in the C3 β-chain upon activation of C3 exposes a binding site for CRIg on C3b. The superposition of C3 and C3b is based on the Cα atoms of their MG6 domains. The MG3 and MG6 domains of C3 are shown in white, the same domains of C3b in green. The surface of CRIg is shown in yellow, with atoms that contact the MG3 and MG6 domains of C3b coloured blue and pink respectively. Cleavage of C3 into C3a and C3b triggers numerous conformational changes. Among these conformational changes is the rotation of the MG3 domain in respect to MG6 indicated by the red arrow. This movement allows CRIg to form numerous contacts with MG3 conferring specificity of CRIg for C3b. Adapted from Wiesmann et al. (2006).
Cleavage of C3 into the small protein C3a and the large C3b complex by enzymes of the complement cascades results in dramatic conformational changes exposing a thioester bond that allows covalent association of C3b with pathogen surfaces (Fig. 1A). This is the initial step in complement opsonization. Subsequent proteolysis of the bound C3b produces iC3b, C3c and C3dg, fragments that are recognized by different receptors (Ross and Medof, 1985). This cleavage abolishes the ability of C3b to further amplify C3b deposition and activate the late components of the complement cascade, including the membrane attack complex, which is capable of direct membrane damage. However, macrophage phagocytic receptors recognize C3b and its fragments preferentially. Due to the versatility of the ester-bond formation, C3-mediated opsonization is central to pathogen recognition and receptors for the various C3 degradation products therefore play an important role in the host immune response (Holers et al., 1992). Comparison of C3 and C3b structures demonstrates not only the large conformational rearrangements upon activation that expose the TED, but identify numerous additional new surfaces of the molecule C3b that can interact with cellular receptors (Abdul Ajees et al., 2006; Janssen et al., 2006; Wiesmann et al., 2006).
Structure function of complement C3 receptors on phagocytic cells
There are now three recognized gene superfamilies of complement receptors: The short consensus repeat (SCR) modules that code for CR1 and CR2, the β 2 integrin family members CR3 and CR4, and the immunoglobulin Ig-superfamily member CRIg.
CR1 (CD35) is a ∼200 kDa glycoprotein consisting of 30 Short Consensus Repeats (SCRs) and plays a major role in immune complex clearance. SCRs are modular structures of about 60 amino acids, each with two pairs of disulphide bonds that provide structural rigidity. High affinity binding to both C3b and C4b occurs through two distinct sites, each composed of three SCRs (reviewed by Krych-Goldberg and Atkinson, 2001). The structure of the C3b binding site, contained within SCR 15–17 of CR1 (site 2), has been determined by NMR, revealing that the three modules are in an extended head-to-tail arrangement with flexibility at the 16–17 junction (Smith et al., 2002). Structure-guided mutagenesis identified a positively charged surface region on module 15 that is critical for C4b binding. This patch, together with basic side chains of module 16 exposed on the same face of CR1, is required for C3b binding. The main function of CR1, first described as an immune adherence receptor (Rothman et al., 1975), is to capture ICs on erythrocytes for transport and clearance by the liver (Taylor et al., 1997). CR1 plays a role in phagocytosis by neutrophils, but not by tissue macrophages (Sengelov et al., 1994). In addition to its role in clearance of immune complexes, CR1 is a potent inhibitor of both classical and alternative pathway activation through its interaction with the respective convertases (Weisman et al., 1990; Krych-Goldberg et al., 1999; Krych-Goldberg and Atkinson, 2001).
CR2 (CD21) binds iC3b and C3dg and is the principal complement receptor that enhances B cell immunity (Weis et al., 1984). Uptake of C3d-coated antigen by B cells results in enhanced signalling via the B cell antigen receptor, lowering the threshold of B cell activation and providing an important survival signal (Matsumoto et al., 1991). The CR2 binding site on iC3b has been mapped partly to the interface between the TED and the MG1 domains (Clemenza and Isenman, 2000). In the mouse, CR1 and CR2 are two products of the same gene formed by alternative splicing. They are primarily associated with B-lymphocytes and follicular dendritic cells and regulate B-cell responses (Molina et al., 1996). The mouse functional equivalent of CR1, Crry, inactivates the classical and alternative pathway enzymes and acts as an intrinsic regulator of complement activation rather than as a phagocytic receptor (Molina et al., 1992; Ahearn et al., 1996).
CR3 and CR4 are transmembrane heterodimers composed of an alpha subunit (CD11b or α M and CD11c or α X respectively) and a common beta chain (CD18 or β 2), and are involved in adhesion to the extracellular matrix and to other cells as well as in recognition of iC3b. They belong to the integrin family and perform functions not only in phagocytosis, but also in leukocyte trafficking and migration, synapse formation and costimulation (reviewed by Ross, 2000). Integrin adhesiveness is regulated through a process called inside-out signalling, transforming the integrins from a low- to a high-affinity binding state (Liddington and Ginsberg, 2002). In addition, ligand binding transduces signals from the extracellular domain to the cytoplasm. Integrin structure and function has been reviewed recently (Luo et al., 2007). The multiple ligands for CR3, namely iC3b, β-glucan and ICAM-1, seem to bind to partially overlapping sites contained within the I domain of CD11b/CD18 (Diamond et al., 1990; Zhang and Plow, 1996; Balsam et al., 1998). Its specific recognition of the proteolytically inactivated form of C3b, iC3b, is predicted from structural studies that locate the CR3 binding sites to residues that become exposed upon unfolding of the CUB domain in C3b which occurs upon α′ chain proteolytic cleavage (Nishida et al., 2006).
CRIg [also referred to as Z39Ig (Langnaese et al., 2000) and VSIG4 (Vogt et al., 2006)] is a recently identified complement receptor of the immunoglobulin superfamily. CRIg expression is restricted to a subset of tissue resident macrophages. These include liver Kupffer cells, interstitial macrophages in the heart, synovial lining macrophages in the joint and foam cells in atherosclerotic plaques (Helmy et al., 2006; Lee et al., 2006; Vogt et al., 2006). Human CRIg is in addition highly expressed on Hofbauer cells in the placenta, on adrenal gland macrophages and on alveolar macrophages (Helmy et al., 2006). The highly restricted expression of CRIg on tissue resident macrophages and its absence from circulating cells and a number of macrophage cell lines may explain why this receptor has gone unnoticed for such a long time. In addition to its role as a complement receptor, CRIg has been reported to regulate T-cell proliferation, but the mechanism of this T-cell modulatory effect remains unknown (Vogt et al., 2006; Zang and Allison, 2006). Unlike most other C3b binding partners (Taniguchi-Sidle and Isenman, 1994; Lambris et al., 1996; Oran and Isenman, 1999; Janssen and Gros, 2007), CRIg binds primarily to the β-chain and not to the α-chain of C3b. The selectivity of CRIg for C3 fragments, and not native C3, can be explained based on the recently solved structure of CRIg bound to C3b (Wiesmann et al., 2006). MG3 and MG6 domains of the C3b β-chain contribute 30% and 40% of the total buried interface (Fig. 1B). The MG3 domain has rotated by 15° when comparing C3 to C3c (Janssen et al., 2005) and C3b (Fig. 1B). The C3b : CRIg structure shows that this reorientation of MG3, along with the movement of a helical section in the linker (LNK) region, is necessary to complete the CRIg binding site, thus explaining why CRIg cannot bind to C3.
Kupffer cells, complement and pathogen recognition
The Kupffer cells, residing within the lumen of the liver sinusoids, form the largest population of macrophages in the reticuloendothelial system. Although Kupffer cells have markers in common with other tissue resident macrophages, they perform specialized functions geared towards efficient clearance of gut-derived bacteria, microbial debris, bacterial endotoxins, immune complexes and dead cells present in the circulation (Bilzer et al., 2006). Efficient binding of pathogens to the KC surface is a crucial step in the first-line immune defence (Benacerraf et al., 1959). Finally, a central role for Kupffer cells in pathogen clearance is illustrated by significantly increased mortality in Kupffer cell-depleted mice challenged with bacteria (Hirakata et al., 1991).
The only complement C3 receptors identified on mouse Kupffer cells so far are CRIg and CR3 (Helmy et al., 2006), while human Kupffer cells show additional expression of CR1 and CR4 (Hinglais et al., 1989). Both CRIg and CR3 on Kupffer cells contribute to binding to iC3b opsonized particles in vitro (Helmy et al., 2006). In vivo, CR3 does not participate in immediate recognition of iC3b-coated pathogens, but rather contributes to clearance indirectly by recruitment of neutrophils through interaction with neutrophil-expressed ICAM-1 (Rogers and Unanue, 1993; Conlan and North, 1994; Gregory et al., 2002). CRIg contributes to clearance of pathogens at an early stage following infection by capturing pathogens in transit through the liver sinusoidal lumen (Helmy et al., 2006). This difference in the biology is partly reflected by a difference in binding characteristics. The efficient capture of C3b/iC3b-coated particles in the sinusoidal lumen is likely the result of high avidity multivalent interactions between CRIg molecules concentrated at the tip of macrophage membrane extensions and multimers of C3b and iC3b on the pathogen surface. Moreover, while CR3 only binds iC3b-coated particles, CRIg binds to C3b, the first C3 cleavage product formed on serum-opsonized pathogens (Gordon et al., 1988; Croize et al., 1993). Binding to C3b in addition to iC3b ensures immediate pathogen recognition through CRIg. Thus, while both CRIg and CR3 are expressed on Kupffer cells, they show distinct binding properties and ligand specificity, resulting in distinct modes of pathogen recognition.
The role of CR3 and CRIg in complement-mediated phagocytosis
Traditionally, the engagement of complement receptors on macrophages was thought to trigger a distinct form of phagocytosis, characterized by ‘sinking’ of the particle into the cell without eliciting an inflammatory response (Aderem et al., 1985; Brown, 1992; Kaplan, 1977). Recent reports have implicated several signalling intermediates, including the tyrosine kinase Syk and Rho GTPases, in phagocytosis of complement-opsonized particles through CR3 (Caron and Hall, 1998; Shi et al., 2006). CRIg harbours a relatively large cytoplasmic domain containing putative internalization motifs, but it is unknown if CRIg can signal upon encounter of C3-opsonized particles. CRIg localization on endosomes and its redistribution upon contact with larger C3-opsonized particles suggests involvement of CRIg in some of the steps in the phagocytic process (Fig. 2). Phagocytosis of numerous or large particles requires rapid replenishment of plasma membrane. Endosomes have been shown to constitute the source of this newly added membrane (Bajno et al., 2000; Cox et al., 2000). Once complement-opsonized particles contact the macrophage surface (Fig. 2A, 1), CRIg-expressing endosomes are recruited to sites of particle contact (2) where they aid in delivering membrane to the forming phagosome (3). Thus, CRIg delivered to the ‘phagocyte synapse’ (Stuart and Ezekowitz, 2005) ensures improved binding of a particle to the macrophage cell surface. The role of CRIg may extend beyond recognition of complement-opsonized particles, as CRIg is also localized to the membrane of the maturing phagosomes, translocating with their cargo from the cell periphery to the lysosomal compartment (4). CRIg does not colocalize with lysosome-associated membrane protein 1 (5) suggesting that CRIg eventually separates from the maturing phagosome and re-enters the endosomal compartment prior to or during phagosome-lysosome fusion (6). The directed transport of CRIg-positive endosomes to the sites of particle contact and the subsequent association of CRIg with the phagosome membrane is suggestive of an active role of CRIg in complement-mediated phagocytosis. The interaction of CRIg intracellular domain with cytosolic signalling intermediates may direct the trafficking of endosomes and improve adhesion to complement-opsonized particles. While Kupffer cells can internalize pathogens (Helmy et al., 2006), we have no evidence that this is critical for clearance. Whether binding of the pathogen alone is sufficient for subsequent pathogen elimination by accessory cells are aspects that warrant further investigation.
Figure 2
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A role for CRIg in binding and phagocytosis of C3-opsonized pathogens.
A. Model of CRIg involvement in phagocytosis of complement C3-opsonized particles. (1) Exposure of the particle (pink sphere) to serum and opsonization with C3b (yellow ovals), (2) binding of the opsonized particle to the macrophage surface, (3) recruitment of CRIg-expressing, transferrin-positive endosomes (green with blue content) to the site of particle contact, (4) localization of CRIg to the phagosome membrane, (5) separation of CRIg from the phagosome upon phagosome-lysosome (red; Lamp-1+) fusion and (6) re-entry of CRIg to the recycling endosome pool.
B. CRIg (green) and transferrin (blue) colocalize (cyan) in peripheral recycling endosomes (arrows) in a human monocyte-derived macrophage in the absence of C3-opsonized particles. Macrophages were incubated with anti-CRIg antibodies (green channel) and transferrin (blue channel) for 10 min at 37°C, fixed, permeabilized and incubated with anti-Lamp-1 antibody (red channel).
C. Upon encounter with C3 opsonized sheep red blood cells, CRIg (green) is recruited to the phagosome membrane (arrows) but does not colocalize with the Lamp-1+ lysosomes (red). Macrophages were incubated with IgM- and C3- opsonized sheep erythrocytes for 2 h at 37°C in the presence of transferrin (blue channel). Following fixation and permeabilization, cells were stained with polyclonal antibodies to CRIg (green channel) and Lamp-1 (red channel). Thin arrows show CRIg-positive phagosomes, arrowheads show erythrocytes in mature phagosomes colocalizing with Lamp-1, but not with CRIg. Scale bar = 20 μm. Adapted from Helmy et al. (2006) with permission from Cell Press.
The constitutive recycling of CRIg and its endocytosis of ligand in resting macrophages fits with a role during the initial phase of a bacterial infection. In addition, CRIg may be involved in the constitutive removal of C3-opsonized apoptotic cells and cell debris, thus preventing local inflammation. In contrast, CR3 internalizes ligand through a macropinocytotic process only after cross-linking of the receptor (Carpentier et al., 1991). Moreover, CR3 is directly implicated in leukocyte trafficking and induces pro-inflammatory signalling upon engagement of the receptor by ligand (Xia et al., 1999). CR3 is expressed on neutrophils, monocytes, macrophages and subsets of lymphocytes and is upregulated upon stimulation with cytokines and PMA (Ross, 2000). Thus, CRIg and CR3 differ in subcellular and cellular localization and regulation of cell-surface expression.
It is intriguing to speculate that, as a major complement receptor on Kupffer cells, which are challenged with bacteria and bacterial products essentially continuously, the role for CRIg is to mediate efficient pathogen clearance without inducing an inflammatory response. When CR3 becomes engaged, a systemic inflammatory response, with accompanying toxicity for the host as well as the infecting organism, ensues. We theorize that each of the different C3 receptors on macrophages has a distinct role. The main role for CR1 is to recognize C3b without leading to internalization. This allows removal of immune complexes from the circulation, preventing pathologic deposition. The role for CRIg may be to mediate removal of potentially pathologic agents through internalization. Thus, CR1 and CRIg may have important roles in maintenance of homeostasis in metazoans living in an environment of potential pathogens. Engagement of the integrin complement receptors CR3 and CR4 is likely a signal that the homeostatic mechanisms have been overwhelmed, and a systemic pathologic response is required. As understanding of the signalling by each of these receptors deepens, insight into their basic functions will arise.
Conclusion and future perspective
Complement research has entered a new and exciting era highlighted by structural biology (Janssen and Gros, 2007) and the identification of additional complement-interacting proteins (Roozendaal and Carroll, 2006). This has provided us with new insights into how complement is activated and subsequently recognized by specific transmembrane receptors. In this review we have focused on macrophage receptors for complement C3 fragments with special attention to the ligand–receptor interactions that trigger phagocytosis. This review touches upon a fraction of the plethora of biological functions of complement and their breakdown products that are now emerging. From a therapeutic point of view, increasing our knowledge of the structure and function of complement components will further guide the rational design of selective inhibitors for therapeutic use.
Acknowledgements
We would like to thank Nico Ghilardi and Suzie Scales for critically reading the manuscript.
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Publication History
Issue Online: 30 July 2007
Version of Record online: 04 June 2007
Editorial history: Received 4 April, 2007; revised 11 May, 2007; accepted 14 May, 2007.
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2709 | https://byjus.com/jee/ie-irodov-solutions-part-1-laws-of-conservation-of-energy-momentum-and-angular-momentum/ | Students can find the solutions to Problems In General Physics I.E. Irodov - Laws of Conservation of Energy, Momentum, and Angular Momentum on this page. When Physical Fundamentals of Mechanics is considered, Laws of Conservation of Energy, Momentum, and Angular Momentum is an important topic. Mechanics deals with motion and forces producing motion. The important topics include work and power of the force, increment of the kinetic energy of a particle, the relationship between the force of a field and the potential energy of a particle in the field, Law of momentum variation of a system, equation of motion of the system's centre of inertia, etc.
Revising these solutions will help students to easily score better ranks in the JEE exam and other competitive exams.
Students can find the solutions to Problems In General Physics I.E. Irodov - Laws of Conservation of Energy, Momentum, and Angular Momentum on this page. When Physical Fundamentals of Mechanics is considered, Laws of Conservation of Energy, Momentum, and Angular Momentum is an important topic. Mechanics deals with motion and forces producing motion. The important topics include work and power of the force, increment of the kinetic energy of a particle, the relationship between the force of a field and the potential energy of a particle in the field, Law of momentum variation of a system, equation of motion of the system's centre of inertia, etc.
Revising these solutions will help students to easily score better ranks in the JEE exam and other competitive exams.
I.E. Irodov Solutions on Laws of Conservation of Energy, Momentum, and Angular Momentum
1. A particle has shifted along some trajectory in the plane xy from point 1 whose radius vector r1 = i + 2j to point 2 with the radius vector r2 = 2i - 3j. During that time the particle experienced the action of certain forces, one of which being F = 3i + 4j. Find the work performed by the force F. (Here r1, r2 and F are given in SI units).
1. A particle has shifted along some trajectory in the plane xy from point 1 whose radius vector r1 = i + 2j to point 2 with the radius vector r2 = 2i - 3j. During that time the particle experienced the action of certain forces, one of which being F = 3i + 4j. Find the work performed by the force F. (Here r1, r2 and F are given in SI units).
Solution:
As
(\vec{F})
is constant so the sought work done
(A=\vec{F}.\Delta \vec{r}= \vec{F}.(\vec{r}_{2}-\vec{r}_{1}))
Or A =
((3\vec{i}+4\vec{j}).((2\vec{i}-3\vec{j})-(\vec{i}+2\vec{j})))
= 17 J
2. A locomotive of mass m starts moving so that its velocity varies according to the law v = a√s where a is a constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the locomotive during the first t seconds after the beginning of motion.
2. A locomotive of mass m starts moving so that its velocity varies according to the law v = a√s where a is a constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the locomotive during the first t seconds after the beginning of motion.
Solution:
Differentiating v(s) with respect to time
dv/dt = a/2√s)ds/dt = (a/2√s)a√s
= a2/2
= w
(As locomotive is in unidirectional motion)
Hence force acting on the locomotive F = mw = ma2/2
Let, at v = 0 at t = 0 then the distance covered during the first t seconds
s = ½ wt2 = ½ (a2/2)t2 = a2t2/4
Hence the sought work, A = Fs = (ma2/2)(a2t2)/4 = ma4t2/8
3. The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T = as2, where a is a constant. Find the force acting on the particle as a function of S.
3. The kinetic energy of a particle moving along a circle of radius R depends on the distance covered s as T = as2, where a is a constant. Find the force acting on the particle as a function of S.
Solution:
We have T = ½ mv2 = as2
Or v2 = 2as2/m (1)
Differentiating Eq.(1) with respect to time
2vwt = 4asv/m or wt = 2as/m (2)
Hence net acceleration of the particle
w = √(wt2 + wn2)
= √[(2as/m)2 + (2as2/mR)2]
= (2as/m)√(1 + (s/R))2
Hence the sought force, F = mw = 2as√(1 + (s/R)2)
4. A body of mass m was slowly hauled up the hill (Fig. 1.29) by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and the coefficient of friction k.
4. A body of mass m was slowly hauled up the hill (Fig. 1.29) by a force F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and the coefficient of friction k.
Solution:
Let vector F makes an angle θ with the horizontal at any instant of time (Fig.). Newton’s second law in projection form along the direction of the force, gives : F = kmg cos θ + mg sin θ (because there is no acceleration of the body.)
As
(\vec{F})
↑ ↑
(d\vec{r})
the differential work done by the force
(\vec{F})
,
dA =
(\vec{F}.d\vec{r}= F\: ds)
= kmg ds cos θ + mg ds sin θ
= kmg dx + mg dy
Hence
(A = kmg\int_{0}^{l}dx + mg\int_{0}^{h}dy)
= kmgl + mgh
= mg(kl + h)
5. A disc of mass m = 50 g slides with the zero initial velocity down an inclined plane set at an angle α = 300 to the horizontal; having traversed the distance l = 50 cm along the horizontal plane, the disc stops. Find the work performed by the friction forces over the whole distance, assuming the friction coefficient k = 0.15 for both inclined and horizontal planes.
5. A disc of mass m = 50 g slides with the zero initial velocity down an inclined plane set at an angle α = 300 to the horizontal; having traversed the distance l = 50 cm along the horizontal plane, the disc stops. Find the work performed by the friction forces over the whole distance, assuming the friction coefficient k = 0.15 for both inclined and horizontal planes.
Solution:
Let s be the distance covered by the disc along the incline, from the Eq. of increment of M.E. of the disc in the field of gravity :
M.E of the disc in the field of gravity: ∆T + ∆U = Afr
0 + (-mgs sin α) = -kmg cos α s - kmgl
Or s = kl/sin α - k cos α (1)
Hence the sought work
Afr = -kmg (s cos α + l)
Afr = -klmg/(1 - k cot α) (using the eqn (1))
On putting the values Afr = -0.05J
6. Two bars of masses m1 and m2 connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to k. What minimum constant force has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar?
Solution:
Let x be the compression in the spring when the bar m2 is about to shift. Therefore at this moment spring force on m2 is equal to the limiting friction between the bar m2 and horizontal floor. Hence
kx = k m2g [where k is the spring constant (say)] (1)
For the block m1 from work-energy theorem: A = ∆T = 0 for minimum force. (A here includes the work done in stretching the spring.)
So Fx - ½ kx2 - kmgx = 0
Or kx/2 = F - km1g (2)
From (1) and (2)
F = kg(m1 + m2/2)
7. A chain of mass m = 0.80 kg and length l = 1.5 m rests on a rough-surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals η = 1/3 of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table?
Solution:
From the initial condition of the problem the limiting friction between the chain lying on the horizontal table equals the weight of the over hanging part of the chain, i.e
ληlg = kλ(1-η)lg (where λ is the linear mass density of the chain)
So k = η/(1-η) (1)
Let (at an arbitrary moment of time) the length of the chain on the table is x. So the net friction force between the chain and the table, at this moment :
fr = kN = kλxg (2)
The differential work done by the friction forces:
dA =
(\vec{f}_{r}.d\vec{r})
= -fr ds
= -kλxg(-dx)
= λg(η/(1-η))x dx (3)
Note that here we have written ds = -dx, because ds is essentially a positive term and as the length of the chain decreases with time, dx is negative)
Hence the sought work done
(A = \int_{(1-\eta )l}^{0}\lambda g\frac{\eta }{1-\eta }x\: dx)
= -(1-η)η mgl/2
= -1.3J
8. A body of mass m is thrown at an angle α to the horizontal with the initial velocity v0. Find the mean power developed by gravity over the whole time of motion of the body, and the instantaneous power of gravity as a function of time.
Solution:
The velocity of the body, t seconds after the beginning of the motion becomes
(\vec{v}=\vec{v}_{0}+g\vec{t})
The power developed by the gravity
(m\vec{g})
at that moment, is
(P= m\vec{g}.\vec{v})
= mg(gt - v0 sin α) (1)
As
(m\vec{g})
is a constant force, so the average power.
P = A/T =
(\frac{m \vec{g}.\Delta \vec{r}}{T})
Where
(\Delta \vec{r})
is the net displacement of the body during time of flight.
As,
(m\vec{g})
perpendicular
(\Delta \vec{r})
So P = 0
9. A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as wn = at2, where a is a constant. Find the time dependence of the power developed by all the forces acting on the particle, and the mean value of this power averaged over the first t seconds after the beginning of motion.
Solution:
We have wn = v2/R = at2 or v = √(aR)t
t is defined to start from the beginning of motion from rest.
Instantaneous power,
(P=\vec{F}.\vec{v})
=
(m(w_{t}\hat{u_{t}}+w_{n}\hat{u_{t}}).(\sqrt{aR}t\hat{u_{t}}))
Where
(\hat{u_{t}})
and
(\hat{u_{t}})
are unit vectors along the direction of tangent (velocity) and normal respectively)
So P = mwt√(aR)t = maRt
Hence the sought average power
(P=\frac{\int_{0}^{t}P\: dt}{\int_{0}^{t}dt}=\frac{\int_{0}^{t}maRt\; dt}{t})
P = maRt2/2t = maRt/2
10. A small body of mass m is located on a horizontal plane at point O. The body acquires a horizontal velocity v0. Find:
10. A small body of mass m is located on a horizontal plane at point O. The body acquires a horizontal velocity v0. Find:
(a) the mean power developed by the friction force during the whole time of motion, if the friction coefficient k = 0.27, m = 1.0 kg, and v0 = 1.5 m/s;
(b) the maximum instantaneous power developed by the friction force, if the friction coefficient varies as k = αx, where α is a constant, and x is the distance from the point O.
Solution:
Let the body m acquire the horizontal velocity v0 along positive x-axis at the point O.
(a) Velocity of the body t seconds after the beginning of the motion,
(\vec{v}=\vec{v}_{0}+\vec{w}t)
=
((v_{0}-kgt)\vec{i})
Instantaneous power P =
(\vec{F}.\vec{v})
= -kmg(v0 - kgt)
From Eq. (1), the time of motion T = v0/kg
Hence sought average power during the time of motion
(P=\frac{\int_{0}^{T}-kmg(v_{0}-kgt)dt}{T})
= -kmgv0/2
= -2W (on substitution)
From Fx = mwx
-kmg = mwx = mvxdvx/dx
Or vxdvx = -kgdx
= -αgx dx
To find v(x), let us integrate the above equation
(\int_{v_{0}}^{v}v_{x}dv_{x}=-\alpha g\int_{0}^{x}x\: dx)
Or v2 = v02 - αgx2 ...(1)
(\vec{P}=\vec{F}.\vec{v})
= -mαxg√(v02 - αgx2) ..(2)
For maximum power, (d/dt)(√(v02x2 - λgx4) = 0 which yields x = v0/√(2αg)
Putting this value of x, in Eq. (2) we get,
Pmax = -½ mv02√(αg)
11. A small body of mass m = 0.10 kg moves in the reference frame rotating about a stationary axis with a constant angular velocity ω = 5.0 rad/s. What work does the centrifugal force of inertia perform during the transfer of this body along an arbitrary path from point 1 to point 2 which are located at the distances r1 = 30 cm and r2 = 50 cm from the rotation axis?
Solution:
Centrifugal force of inertia is directed outward along radial line, thus the sought work.
A =
(\int_{r_{1}}^{r_{2}}m\omega ^{2}r\: dr)
= ½ mω2(r22 - r12)
= 0.20 T (on substitution)
12. A system consists of two springs connected in series and having the stiffness coefficients k1 and k2. Find the minimum work to be performed in order to stretch this system by ∆l.
Solution:
Since the springs are connected in series, the combination may be treated as a single spring of spring constant.
K = K1K2/(K1 + K2)
From the equation of increment of M.E, ∆T + ∆U = Aext
0 + ½ k∆l2 = A
Or A = ½ (K1K2/(K1 + K2))∆l2
13. A body of mass m is hauled from the Earth's surface by applying a force F varying with the height of ascent y as F = 2 (ay - 1) mg, where a is a positive constant. Find the work performed by this force and the increment of the body's potential energy in the gravitational field of the Earth over the first half of the ascent.
Solution:
First, let us find the total height of ascent. At the beginning and the end of the path of the velocity of the body is equal to zero, and therefore the increment of the kinetic energy of the body is also equal to zero. On the other hand, according to the work-energy theorem, ∆T is equal to the algebraic sum of the works A performed by all the forces, i.e. by the force F and gravity, over this path. However, since ∆T = 0 then A = 0. Taking into account that the upward direction is assumed to coincide with the positive direction of the y-axis, we can write
(A=\int_{0}^{h}(\vec{F}+m\vec{g}).d\vec{r}=\int_{0}^{h}(F_{y}-mg)dy)
=
(mg\int_{0}^{h}(1-2ay)dy)
= mgh(1-ah)
= 0 where h = 1/a.
The work performed by the force F over the first half of the ascent is
Af =
(\int_{0}^{\frac{h}{2}}F_{y}dy= 2mg\int_{0}^{\frac{h}{2}}(1-ay)dy)
= 3mg/4a
The corresponding increment of the potential energy is
∆U = mgh/2 = mg/2a
14. The potential energy of a particle in a certain field has the form U = a/r2- b/r, where a and b are positive constants, r is the distance from the centre of the field. Find:
(a) the value of r0 corresponding to the equilibrium position of the particle; examine whether this position is steady;
(b) the maximum magnitude of the attraction force; draw the plots U (r) and Fr (r) (the projections of the force on the radius vector r).
Solution:
From the equation Fr = -dU/dr we get Fr = [-2a/r3 + b/r2]
(a) we have at r = r0, the particle is in equilibrium position. i.e.Fr = 0 so, r0 = 2a/b.
To check, whether the position is steady (the position of stable equilibrium), we have to satisfy
d2U/dr2 >0
We have d2U/dr2 = [6a/r4 - 2b/r3]
Putting the value of r = r0 = 2a/b, we get
d2U/dr2 == b4/8a3, (as a and b are positive constant)
So d2U/dr2 = b2/8a3 > 0
which indicates that the potential energy of the system is minimum, hence this position is steady.
(b) we have Fr = -dU/dr = [-2a/r3 + b/r2]
For F, to be maximum, dFr/dr = 0
So r = 3a/b and then Fr(max) = -b3/27a2
As Fr is negative, the force is attractive.
15. In a certain two-dimensional field of force the potential energy of a particle has the form U = αx2 + βy2 , where α and β are positive constants whose magnitudes are different. Find out:
15. In a certain two-dimensional field of force the potential energy of a particle has the form U = αx2 + βy2 , where α and β are positive constants whose magnitudes are different. Find out:
(a) whether this field is central;
(b) what is the shape of the equipotential surfaces and also of the surfaces for which the magnitude of the vector of force F = const.
Solution:
(a) We have Fx = ∂U/dx = -2αx and Fy = -∂U/dy = -2βy
So
(\vec{F}= 2\alpha x\vec{i}-2\beta y\vec{i})
And Fy = 2√(α2x2 + β2y2)
For a central force
(\vec{r}\times \vec{F}=0)
Here
(\vec{r}\times \vec{F}=(x\vec{i}+y\vec{j})\times (-2\alpha x\vec{i}-2\beta y\vec{j}))
=
(-2\beta xy\vec{k}-2\alpha xy\vec{k})
≠ 0
Hence the force is not a central force.
(b) As U = αx2 + βy2
So, Fx = ∂U/∂x = -2αx and Fy = -∂U/∂y = -2βy
So F = √(Fx2 + Fy2)
= √(4α2x2 + 4β2y2)
According to the problem
F = 2 √(α2x2 + β2y2) = C (constant)
Or (α2x2 + β2y2) = C2/2
Or x2/β2 + y2/α2 = C2/2α2β2 = k (say)
Therefore the surfaces for which F is constant is an ellipse.
For an equipotential surface U is constant.
So αx2 + βy2 = C0 (constant)
Or x2/√β2 + y2/√α2 = C0/αβ = K0 (constant)
Hence the equipotential surface is also an ellipse.
16. There are two stationary fields of force F = ayi and F = axi + byj, where i and j are the unit vectors of the x and y axes, and a and b are constants. Find out whether these fields are potential.
Solution:
Let us calculate the work performed by the forces of each field over the path from a certain point 1 (x1, y1) to another certain point 2 (x2 , y2).
(i) dA =
(\vec{F}.d\vec{r}= ay\vec{i}.d\vec{r}=ay \: dx)
Or A =
(a\int_{x_{1}}^{x_{2}}y\: dx)
(ii) dA =
(\vec{F}.d\vec{r}= (ax\vec{i}+by\vec{j}).d\vec{r}=ax\: dx+by\: dy)
Hence A =
(\int_{x_{1}}^{x_{2}}ax\: dx+\int_{y_{1}}^{y_{2}}by\: dy)
In the first case, the integral depends on the function of type y (x), i.e. on the shape of the path. Consequently, the first field of force is not potential. In the second case, both the integrals do not depend on the shape of the path. They are defined only by the coordinate of the initial and final points of the path, therefore the second field of force is potential.
17. A body of mass in is pushed with the initial velocity v0 up an inclined plane set at an angle α to the horizontal. The friction coefficient is equal to k. What distance will the body cover before it stops and what work do the friction forces perform over this distance?
Solution:
Let s be the sought distance, then from the equation of increment of M.E.
∆T + ∆U = Afr
(0 - ½ mv02) + (mg s sin α) = -kmg cos α s
Or s = (v02/2g)/(sin α + k cos α)
Hence Afr = -kmg cos α s
= -kmv02/2(k + tan α)
18. A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion (Fig. 1.30). What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to?
18. A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion (Fig. 1.30). What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to?
Solution:
Velocity of the body at height h, vh = √(2g(H - h), horizontally (from the figure given in the problem). Time taken in falling through the distance h.
t = √(2h/g) (as initial vertical component of the velocity is zero.)
Now s = vht = √(2g(H + h)) ×√(2h/g)
= √(4(Hh - h2))
For smax, (d/ds)(Hh - h2) = 0, which yields h = H/2
Putting this value of h in the expression obtained for s, we get
smax = H
19. A small body A starts sliding from the height h down an inclined groove passing into a half-circle of radius h/2 (Fig. 1.31). Assuming the friction to be negligible, find the velocity of the body at the highest point of its trajectory (after breaking off the groove).
Solution:
To complete a smooth vertical track of radius R, the minimum height at which a particle starts, must be equal to 5/2 R (one can prove it from energy conservation). Thus in our problem body could not reach the upper most point of the vertical track of radius R/2. Let the particle A leave the track at some point O with speed v (Fig.). Now from energy conservation for the body A in the field of gravity :
mg(h - h(1 + sin θ)/2) = ½ mv2
Or v2 = gh(1 - sin θ) ..(1)
From Newton's second law for the particle at the point O; Fn = mwn,
N + mg sin θ = mv2/(h/2)
But, at the point O the normal reaction N = 0
So v2 = (gh/2) sin θ (2)
From (3) and (4), sin θ = ⅔ and v = √(gh/3)
After leaving the track at O, the particle A comes in air and further goes up and at maximum height of its trajectory in air, it’s velocity (say v') becomes horizontal (Fig.). Hence, the sought velocity of A at this point.
v’ = v cos (90 - θ) = v sin θ = ⅔ √(gh/3)
20. A ball of mass m is suspended by a thread of length l. With what minimum velocity has the point of suspension to be shifted in the horizontal direction for the ball to move along the circle about that point? What will be the tension of the thread at the moment it will be passing the horizontal position?
Solution:
Let the point of suspension be shifted with velocity vA in the horizontal direction towards left then in the rest frame of point of suspension the ball starts with same velocity horizontally towards right Let us work in this frame. From Newton’s second law in projection form towards the point of suspension at the upper most point (say B) :
mg + T = mvB2/l
Or T = mvB2/l - mg ..(1)
Condition required to complete the vertical circle is that T ≥ 0. But (2)
½ mvA2 = mg(2l) + ½ mvB2
So vB2 = vA2 - 4gl ..(3)
From (1), (2) and (3)
T = m(vA2 - 4gl)/l - mg ≥ 0
Or vA ≥ √(5gl)
From the equation Fn = mwn at point C
T’ = mvc2/l ..(4)
Again from energy conservation
½ mvA2 = ½ mvc2 + mgl ..(5)
From (4) and (5)
T = 3mg |
2710 | https://web.ma.utexas.edu/users/m408n/AS/LM2-6-3.html | | | | |
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2712 | https://www.statology.org/binomial-vs-geometric/ | Two commonly used distributions in statistics are the binomial distribution and the geometric distribution.
This tutorial provides a brief explanation of each distribution along with the similarities and differences between the two.
The Binomial Distribution
The binomial distribution describes the probability of obtaining k successes in n binomial experiments.
If a random variable X follows a binomial distribution, then the probability that X = k successes can be found by the following formula:
P(X=k) = nCk pk (1-p)n-k
where:
n: number of trials
k:number of successes
p: probability of success on a given trial
nCk:the number of ways to obtain k successes in n trials
For example, suppose we flip a coin 3 times. We can use the formula above to determine the probability of obtaining 0 heads during these 3 flips:
P(X=0)= 3C0 .50 (1-.5)3-0 = 1 1 (.5)3 = 0.125
The Geometric Distribution
The geometric distribution describes the probability of experiencing a certain amount of failures before experiencing the first success in a series of binomial experiments.
If a random variable X follows a geometric distribution, then the probability of experiencing k failures before experiencing the first success can be found by the following formula:
P(X=k) = (1-p)kp
where:
k:number of failures before first success
p:probability of success on each trial
For example, suppose we want to know how many times we’ll have to flip a fair coin until it lands on heads. We can use the formula above to determine the probability of experiencing 3 “failures” before the coin finally lands on heads:
P(X=3)= (1-.5)3(.5) = 0.0625
Similarities & Differences
The binomial and geometric distribution share the following similarities:
The outcome of the experiments in both distributions can be classified as “success” or “failure.”
The probability of success is the same for each trial.
Each trial is independent.
The distributions share the following key difference:
In a binomial distribution, there is a fixed number of trials (i.e. flip a coin 3 times)
In a geometric distribution, we’re interested in the number of trials required until we obtain a success (i.e. how many flips will we need to make before we see Tails?)
Practice Problems: When to Use Each Distribution
In each of the following practice problems, determine whether the random variable follows a binomial distribution or geometric distribution.
Problem 1: Rolling Dice
Jessica plays a game of luck in which she keeps rolling a dice until it lands on the number 4. Let X be the number of rolls until a 4 appears. What type of distribution does the random variable X follow?
Answer: X follows a geometric distribution because we’re interested in estimating the number of rolls required until we finally get a 4. This is not a binomial distribution because there is not a fixed number of trials.
Problem 2: Shooting Free-Throws
Tyler makes 80% of all free-throws he attempts. Suppose he shoots 10 free-throws. Let X be the number of times Tyler makes a basket during the 10 attempts. What type of distribution does the random variable X follow?
Answer: X follows a binomial distribution because there is a fixed number of trials (10 attempts), the probability of “success” on each trial is the same, and each trial is independent.
Additional Resources
Binomial Distribution Calculator
Geometric Distribution Calculator
Zach Bobbitt
Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike. My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.
One Reply to “Binomial vs. Geometric Distribution: Similarities & Differences”
For the formule of geometric distribution
Do you miss “-1” in the power,
i mean (X=k) = p(1-p)^(k-1)
Reply
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Streptococcus Pyogenes
Sidrah Kanwal; Pradeep Vaitla.
Author Information and Affiliations
Authors
Sidrah Kanwal1; Pradeep Vaitla2.
Affiliations
1 Allama Iqbal Medical College, Lahore, Pakistan
2 University of Mississippi Medical Center
Last Update: July 31, 2023.
Continuing Education Activity
Streptococcus pyogenes is a major human-specific bacterial pathogen that causes a wide array of manifestations ranging from mild localized infections to life-threatening invasive infections. Ineffective treatment of S. pyogenes infections can result in the postinfectious sequela acute rheumatic fever and post-streptococcal glomerulonephritis. Moreover, it causes invasive infections like necrotizing fasciitis and toxic shock syndrome that is associated with and high morbidity and mortality. This activity describes the bacterium Streptococcus pyogenes (S. pyogenes) and reviewed the epidemiology, clinical manifestations of S. pyogenes infections, evaluation, and management of clinical infections caused by S. pyogenes.
Objectives:
Identify the etiology of S. pyogenes.
Outline the appropriate history, physical, and evaluation of Streptococcal pharyngitis.
Review the management options available for Streptococcal pharyngitis.
Describe the interprofessional team strategies for improving care coordination and communication to improve the outcome of severe invasive infections caused by S. pyogenes.
Access free multiple choice questions on this topic.
Introduction
Streptococcus pyogenes is a major human-specific bacterial pathogen that causes a wide array of manifestations ranging from mild localized infections to life-threatening invasive infections. Ineffective treatment of S. pyogenes infections can result in the postinfectious sequela acute rheumatic fever and post-streptococcal glomerulonephritis. Moreover, it causes invasive infections like necrotizing fasciitis and toxic shock syndrome that is associated with and high morbidity and mortality.
Streptococci are gram-positive, catalase-negative, coagulase-negative cocci that occur in pairs or chains. They are divided into three groups by the type of hemolysis on blood agar: beta-hemolytic (complete lysis of red cells), a hemolytic (green hemolysis), and gamma-hemolytic (no hemolysis). Beta-hemolytic streptococci are characterized as group A streptococci (Streptococcus pyogenes) and group B streptococci (Streptococcus agalactiae).
Etiology
S. pyogenes is a gram-positive, catalase-negative, oxidase negative, β-hemolytic streptococci. It is a facultative anaerobe, grows best in 5 to 10% carbon dioxide, and forms pinpoint colonies on blood agar plates. Lancefield serological grouping system is used to differentiate group A streptococci (GAS) from other streptococci. The type A antigen of S. pyogenes is a polysaccharide which comprised of N-acetylglucosamine attached to a rhamnose polymer backbone. M protein is the major surface protein presents on S. pyogenes cell wall, and GAS strains are further divided into different serotypes based on the antigenic characteristics of the M protein. More than 80 different serotypes have been identified on the basis of M protein.. Depending upon the postinfectious sequelae, S. pyogenes are categorized into two classes: Class I and Class II. Class I strains cause rheumatic fever while class II strains result in acute glomerulonephritis.
S. pyogenes usually colonizes, pharynx, anus, and genital mucosa. Infections caused by S. pyogenes are highly contagious. Transmission can occur through airborne droplets, hand contact with nasal discharge or with objects or surfaces contaminated with bacteria, skin contact with contaminated lesions, or contaminated food sources. GAS Strains may acquire access to the skin via abrasions and skin lesions and may lead to erysipelas or cellulitis. GAS can cause infection in muscle and fascia resulting in myositis and necrotizing fascitis usually following a mild trauma and can result in toxic shock syndrome. S. pyogenes can cause the infection of the vaginal mucosa and uterus causing septicemia. Skin lesions have been identified as the most common predisposing factor for severe S. pyogenes infections. Crowded settings like military camps, nursing houses, and schools cause ease of transmission of the organism and result in epidemics of group A streptococci infection.
Epidemiology
The frequency of the infections caused by S. pyogenes varies in different parts of the world, depending upon clinical manifestations of the infections. GAS has always been associated with serious diseases that result in high morbidity and mortality. GAS infections declined by the mid of the 20th century. However, by the end of 1980, severe group A streptococcus (GAS) infections remerged. Over the last two decades, non-suppurative and suppurative complications of the S. pyogenes infection have increased. This increase in disease burden can be attributed to the multitude of factors including, change of virulence and resistance to antibiotics. GAS infections and its complications vary between poorly developed and well-developed countries. In under-developed countries, the prevalence of rheumatic heart disease (RHD) and the incidence of deaths associated with RHD are high. Conversely, in well-developed countries, the incidence of deaths due to invasive GAS infection is high.
The global burden of severe S. pyogenes infections is 18.1 million cases, with 1.78 million new cases per year. The worldwide prevalence of RHD is at least 15.6 million cases, with the incidence of 282,000 new cases per year. Approximately 233,000 deaths per year are attributed to RHD. Each year approximately 663,000 new cases of invasive GAS disease have been reported with 163,000 deaths per year. 616 million cases of sore throat infection worldwide per year can be attributed to S. pyogenes; also, 111 million cases of skin infection in children of developing countries. . In the United States, 15% to 30% of cases of pharyngitis in children and 5% to 20% pharyngitis cases per year in adults are due to S. pyogenes. Moreover, the resurgence of cases of ARF in the children of middle class families of the United States has been observed.
Throat infection is more common in temperate areas, and its incidence increases in late winter and early spring. Impetigo is more prevalent in children of humid climates. It has also been observed that the severity of invasive skin infections caused by GAS escalated from January to April and correlates with the host susceptibility to acquiring a serious infection.
Toxicokinetics
Multiple virulence factors are responsible for clinical manifestations of the S. pyogenes. Bacterial capsule consisted of hyaluronic acid provides protection against phagocytosis. M protein, lipoteichoic acid, and protein F are responsible for the attachment of the bacteria to host cells. M protein is also responsible for inhibiting opsonization by binding to complement regulators and to fibrinogen. M protein is the most important virulence factor for S. pyogenes since experiments have shown that M mutants cannot survive in phagocyte-containing human blood. S. pyogenes also produce exotoxins, like a pyrogenic (erythrogenic) toxin, which is responsible for the rash of scarlet fever and toxic shock syndrome. Other virulent factors include streptokinase, streptodornase, hyaluronidase, and streptolysins, which help in the invasion of tissues.
History and Physical
History and physical findings will vary depending upon the type of infection acquired; nonetheless, an accurate history and proper clinical evaluation are required to reach an accurate clinical diagnosis of S. pyogenes infection.
Sore throat is usually a major complaint in the case of streptococcus pharyngitis. The most common clinical findings for Streptococcal pharyngitis include sudden onset of fever, malaise, pharyngeal exudate, tender cervical lymphadenopathy and, enlarged tonsils.
In children, impetigo is one of the most common skin infections. S. pyogenes cause non-bullous impetigo. Typically the itchy reddish rash appears around mouth or nose that becomes fluid-filled blister later on. Blisters rupture easily and are covered with honey-colored crust. The lesions are usually well-localized and affect exposed areas of the body: face and lower extremities. There are typically no systemic manifestations of the impetigo.
Patient with scarlet fever usually presents with high-grade fever, sore throat, strawberry-like tongue, and a blanchable, papular, non-confluent rash. The rash typically lasts for 7 to 10 weeks, follows by desquamation. Desquamation can only be observed on the palms and soles.
Soft tissue invasive infections due to S. pyogenes mostly present with shock and multiorgan failure.
Necrotizing fasciitis due to group A streptococcus (S. pyogenes) is a deep-seated infection of the subcutaneous tissue that causes rapid destruction of fascia and fat. Systemically or locally immunocompromised individuals are at increased risk for developing necrotizing fasciitis. Other risk factors include surgical procedures, burns, blunt trauma, minor laceration, and childbirth. Localized pain, necrosis of the infected skin lesion, swelling, redness, edema, increased heart rate, and fever are the typical manifestations of necrotizing fasciitis. In the advanced stage of the disease, a picture of septic shock can be present.
Evaluation
Streptococcus pyogenes, also known as group A streptococcus (GAS) is a leading cause of pharyngitis in children and adolescents. Clinicians should use clinical and epidemiological findings to determine the likelihood of GAS pharyngitis. Infectious Disease Society of American (IDSA) recommends that a rapid antigen detection test (RADT) should be used as the first-line measure to assist the clinicians in the diagnosis of GAS pharyngitis. It is recommended to obtain throat cultures in children with negative RADT results to prevent the development of complications. A combined approach using validated clinical criteria such as modified Centor score or the FeverPAIN score along with RADT is an effective strategy to reduce the cost of unnecessary testing and unwarranted antibiotics.
The gold standard test for the detection of GAS is throat culture; however, it is not cost-effective and can delay the treatment. The pyrrolidinyl arylamidase activity (PYR) test is used to distinguish S. pyogenes from other beta-hemolytic streptococci and tests for the presence of the enzyme pyrrolidinyl aminopeptidase.
Anti-streptolysin O (ASO) and anti-DNase B (ADB) titers indicate the previous streptococcal infection and can be used for the diagnosis of post-streptococcal complications.
Treatment / Management
The drug of choice for treatment of bacterial pharyngitis is oral penicillin for 10 days or IM benzathine penicillin. This treatment is cost-effective and has a narrow spectrum of activity.
In patients with penicillin allergy, macrolides and first-generation cephalosporins can be used. However, some strains of S. pyogenes have developed resistance to macrolides and macrolides are used as third-line of treatment for Streptococcal throat infection.
Severe invasive S. pyogenes infections can be treated with vancomycin or clindamycin. Intravenous antibiotic therapy and surgery for the removal of necrotic tissue are recommended in the case of soft tissue skin infection by S. pyogenes.
Differential Diagnosis
Streptococcal pharyngitis should be differentiated from throat infections due to parainfluenza virus, rhinovirus, coxsackievirus, adenovirus, etc), Mycoplasma species, Corynebacterium diphtheria, and Epstein-Barr virus.
Scarlet fever can be confused with measles and rubella. However, the absence of symptoms of upper respiratory tract infection and confluent rash in measles can help to differentiate these diseases.
Impetigo by S. pyogenes needs to be differentiated from impetigo infection by staphylococcus aureus. S. pyogenes cause non-bullous impetigo while S. aureus leads to bullous impetigo.
Prognosis
According to a report of the World Health Organization (WHO) GAS is the ninth leading infectious etiology of human mortality. The majority of deaths are due to invasive infections and RHD, especially in third world countries. Severe GAS infections are high in affluent countries ranging from 14% to 19%. Streptococcal pharyngitis typically resolves in 7 to 10 days. However, poor compliance with treatment can result in post-infection complications.
Complications
Complications of S. pyogenes infections can be divided into suppurative and non-suppurative complications.
Suppurative complications include peritonsillar abscess, peritonsillar cellulitis, retropharyngeal abscess, otitis media and sinusitis, uvulitis, cervical lymphadenitis, meningitis, and brain abscess, arthritis, endocarditis, osteomyelitis, and liver abscess.
Non-suppurative complications are rheumatic fever, post-streptococcal glomerulonephritis, PANDAS (pediatric autoimmune neuropsychiatric disorders associated with streptococcal infections), Sydenham chorea, and other autoimmune movement disorders.
Deterrence and Patient Education
Patients must be counseled about the importance of completing the course of antibiotics. They should be advised to follow up with a health care provider and to maintain personal hygiene. The act of handwashing can severely decrease the chance of acquiring skin infections due to S. pyogenes.
Pearls and Other Issues
Ineffective treatment of skin or pharyngeal infection with group A streptococcus leads to acute rheumatic fever. Prevention of recurrent episodes of ARF and the development of potential serious outcomes of rheumatic heart disease require secondarily prophylaxis of penicillin. Secondary prophylaxis is the most effective method of reducing the incidence of ARF and RHD. Nevertheless, primary prophylaxis is gaining popularity throughout the globe. Primary prevention comprises of prompt treatment of streptococcal throat or skin infection via a single injection of benzathine penicillin or oral penicillin for 10 days after confirmation of S. pyogenes infection. Not only does prompt antibiotic treatment reduce the burden of ARF, but it also reduces the incidence of suppurative complications like otitis media.
Enhancing Healthcare Team Outcomes
While the general health care provider is almost always involved in the care of patients with streptococcal infection, caring for patients with the severe streptococcal disease, it is important to consult with an interprofessional team of specialists that include a general surgeon, plastic surgeon, radiologist, infectious disease, and speciality trained nurses.
Necrotizing fasciitis is a life-threatening condition, and due to the complexity of this disease, a team approach is required. Since necrotizing fasciitis is a surgical emergency, it is required that the patient should be admitted to a surgical intensive care unit, where the medical staff is skilled in performing debridement and, if needed, reconstructive surgery. Emergent surgical consultation is required to explore and debride the necrotic tissue. In addition, surgical exploration identifies the aetiology and extent of necrosis. CT and MRI may be helpful to locate the primary foci of infection. Hemodynamic stability should be maintained.
In patients with toxic shock syndrome, monitoring of cardiac output is important. Due to the high incidence of acute respiratory distress syndrome (55%) in patients with severe Streptoccoal soft tissue infection, intubation, and ventilatory support are needed. Greater than 50% of patients with severe soft tissue infections can develop acute renal failure; hence, dialysis may be needed.
Review Questions
Access free multiple choice questions on this topic.
Click here for a simplified version.
Comment on this article.
Figure
Streptococcus pyogenes Bacteria. This illustration depicts a photomicrograph of a specimen highlighting chain-linked S pyogenes bacteria. Public Health Image Library, Public Domain, via Centers for Disease Control and Prevention (more...)
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: Disclosure: Sidrah Kanwal declares no relevant financial relationships with ineligible companies.
: Disclosure: Pradeep Vaitla declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
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Kanwal S, Vaitla P. Streptococcus Pyogenes. [Updated 2023 Jul 31]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Continuing Education Activity
Introduction
Etiology
Epidemiology
Toxicokinetics
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Prognosis
Complications
Deterrence and Patient Education
Pearls and Other Issues
Enhancing Healthcare Team Outcomes
Review Questions
References
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2714 | https://retainup.co/discounting-impacts-profits/ | Skip to content
How Discounting Impacts Your Profits?
Written by Rahi Jain
Sep 04, 2024
Pricing & Discounting, Profitable Growth
By discounting 10% off, you might be giving half of your profits away.
Or you need to sell twice as much to make up for lost profit.
For most brands, a 10% discount can lower profits by 33% to 70%.
In this example, for $100 in revenue,
Marketing (30%) – $30COD (COGs + Shipping + Payment) – $30OpEx (20%) – $20
Net Profit – $20
Now with a 10% discount
Keeping everything the same, the net profit comes down to $10.
To generate profits of $10,000, we need to sell 1,000 units compared to 500 units.
As discounting directly eats up the revenue, keeping all the costs the same, it can seriously impact profits.
The only positive impact is the reduction of marketing costs, which is also nullified by an increase in OpEx.
Also, in this case, marketing costs need to go down by 33% to justify a 10% revenue loss, which doesn’t happen.
I am not against discounting.
Brands can really use discounting to increase their average order value and conversion rate.
I have written about it over here.
The problem is not realizing the impact of it on the bottom line.
The solution is to run smart discounting.
Targeted offers for each customer segment
Discount on minimum purchase to increase AOV
Discount on selected inventory
Running short sales to create urgency
In short, to get all the benefits of discounting without getting burnt.
Most importantly, accounting for discounts and see their impact on the bottom line.
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Energy units, kinetic energy, unit conversion, joule, kilogram, meter, second
Explanation
Kilojoule per kilogram (kJ/kg) is a unit of specific energy, which means energy per unit mass. The unit m²/s² (square meters per square second) is a unit of specific energy as well, because kinetic energy per unit mass can be expressed as velocity squared (m²/s²).
Since 1 joule (J) = 1 kg·m²/s², dividing joules by kilograms (J/kg) results in m²/s².
Step-By-Step Solution
Step 1
Recall the definition of joule:
1 J=1 kg⋅m2/s2
Step 2
Given the unit kJ/kg, convert kilojoule to joule:
1 kJ=1000 J
Step 3
Express kJ/kg in terms of J/kg:
1 kJ/kg=1 kg1000 J=1000 J/kg
Step 4
Since 1 J = 1 kg·m²/s², dividing by kg gives:
1 J/kg=1 kg1 kg⋅m2/s2=1 m2/s2
Step 5
Therefore,
1 kJ/kg=1000 m2/s2
Final Answer
1 kJ/kg = 1000 m²/s²
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| Question Text | Conversation of kj/kg into m2/s2 |
| Updated On | Jun 26, 2025 |
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| Subject | Smart Solutions |
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2716 | https://math.stackexchange.com/questions/4281140/visually-explaining-this-probability-union-rule | Stack Exchange Network
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Visually explaining this probability union rule
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$\begingroup$
I'm trying to visually wrap my head around the following equivalence of the union probability rule: $$ P(A)+P(B)-P(A)P(B)=P(A)+(1-P(A))P(B) $$
I understand that $(1-P(A))=P(A')$ and the whole equivalence makes sense to me algabraically.
However, I have sketched these Venn diagrams to try and visualise the equivalence, and what I am struggling to understand is that if I was to combine the Venn diagrams for $P(A)$ and $P(A')P(B)$ then I would have $P(A)+P(B)$, not $P(A)+P(B)-P(A)P(B)$.
Where am I tripping up in my reasoning?
probability
discrete-mathematics
inclusion-exclusion
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edited Oct 19, 2021 at 17:46
RobPratt
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asked Oct 19, 2021 at 14:11
laurencejplaurencejp
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$\endgroup$
2
1
$\begingroup$ Are $A$ and $B$ necessarily independent? If not, your statements and diagrams do not work for intersections (or unions) $\endgroup$
Henry
– Henry
2021-10-20 01:12:09 +00:00
Commented Oct 20, 2021 at 1:12
1
$\begingroup$ @Henry, to be clear: if $A$ and $B$ are mutually exclusive, then (1) they must be dependent, and (2) $P(A\cup B)\equiv P(A)+P(B).$ $\endgroup$
ryang
– ryang
2021-10-20 05:29:18 +00:00
Commented Oct 20, 2021 at 5:29
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3 Answers 3
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$\begingroup$
$P(A)$ is the probability of $A$ occurring, and $P(B)$ is the probability of $B$ occurring.
The probability of either $A$ or $B$ occurring is $P(A \cup B)$.
But $P(A \cup B)$ does not in general equal $P(A) + P(B)$.
If you look in your diagrams you see that $P(A) + P(B)$ counts $P(A \cap B)$ twice.
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answered Oct 19, 2021 at 14:22
Prime MoverPrime Mover
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correction: $P\Big(A\cup B\Big)\equiv P(A)+P(B)-P(A\cap B)$
correction: $P\Big(A\cap B\Big)\not\equiv P(A)\,P(B)$
correction: $P\Big((A\cup B)\setminus(A\cap B)\Big)\equiv P(A)+P(B)-2\, P(A\cap B)$
$P\Big(A'\cap B\Big)\equiv P\Big(B\setminus A\Big)\equiv P(B)-P(A\cap B)\not\equiv P(A')\,P(B)$
Note:
$\large {P\setminus Q}$ (read $P$ minus $Q$) means to subtract $P\cap Q$ from $P.$
$\large {P\setminus Q}\:$ is also written as $\:\large{P-Q};$ the former is a better idea, to discourage conflating set operations and arithmetic operations (how you were conflating $\cup$ and $+$ and then $\cap$ and $\times$).
2. $P(A\cap B)=P(A)\,P(B)\,$ if and only if events $A$ and $B$ are independent.
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edited Oct 19, 2021 at 22:00
answered Oct 19, 2021 at 18:04
ryangryang
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$\begingroup$
You can't use visual diagrams to show these kinds of numerical identities the way you do. To be precise, with your visual method of looking at certain regions as unions, intersections, or complements of more basic regions, and translating that directly into numerical operations of addition, multiplication, and subtraction from one, you would get that $P(A \cup B) = P(A) + P(B)$ and $P(A \cap B) = P(A)P(B)$, but that's not true in general.
As others have pointed out: $P(A\cup B)$ equals $P(A)+P(B)$ only when $A$ and $B$ are mutually exclusive events. If not, you end up double-counting. As the most basic example example, your method gets the result that $P(A)=P(A\cup A) =P(A) + P(A)$ which is of course not true.
Now, as also pointed out, you could still use a visual method and handle such double-counting by using different shades of color in the same region.
However, this does not fix the even more problematic implicit assumption in your approach that $P(A \cap B) =P(A)P(B)$, which is only true if $A$ and $B$ are independent. Again, as a trivial case where they are not, your method would say that $P(A) =P(A\cap A)=P(A)P(A)$ which again is clearly not true.
And, for this problem, there is no visual bandaid: you really can't use a qualitative relationship between events to depict their quantitative dependency. Or, more to the point: you can't infer the qualitative formula from the visual depiction.
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edited Oct 20, 2021 at 20:17
answered Oct 19, 2021 at 14:42
Bram28Bram28
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$\endgroup$
4
$\begingroup$ If you color as translucent, it becomes clear there is some double counting occuring. $\endgroup$
qwr
– qwr
2021-10-20 03:01:35 +00:00
Commented Oct 20, 2021 at 3:01
1
$\begingroup$ @qwr Sure, but none of that solves the main problem that you cannot use these diagrams to add or multiply numerical probabilities $\endgroup$
Bram28
– Bram28
2021-10-20 03:11:56 +00:00
Commented Oct 20, 2021 at 3:11
$\begingroup$ You can add probabilities for disjoint events $\endgroup$
qwr
– qwr
2021-10-20 03:15:30 +00:00
Commented Oct 20, 2021 at 3:15
$\begingroup$ @qwr Again, that's true. I incorporated your comments into my Answer. Thanks! $\endgroup$
Bram28
– Bram28
2021-10-20 15:32:27 +00:00
Commented Oct 20, 2021 at 15:32
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2717 | https://electionlab.mit.edu/articles/modeling-voting-service-times-machine-logs | Modeling Voting Service Times with Machine Logs | MIT Election Lab
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Modeling Voting Service Times with Machine Logs
By Jacob Jaffe, Charles Stewart III, and Jacob Coblentz07.27.2018
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Challenges to estimating how long it takes to mark a ballot
A novel approach to estimating ballot-marking time
What influences ballot-marking times? Ballot design, voter behavior, and demographics
TheMIT Election Data and Science Labhelps highlight new research and interesting ideas in election science, and is a proud co-sponsor of theElection Sciences, Reform, & Administration Conference(ESRA).
Jacob Jaffe, Charles Stewart III, and Jacob Coblentz recently presented a paper at the 2018 ESRA conference entitled, “Modeling Voting Service Times with Machine Logs.” Here, they summarize their analysis from that paper.
After the election in 2012, the long lines at polling places on Election Day were an issue highlighted by many, including then-President Barack Obama.
This focus led academics and election administrators on a search for ways to measure how long it takes voters to complete tasks associated with voting. Measuring these tasks, such as marking ballots and checking in, is the first step to improving our understanding of how polling places can operate more efficiently.
Challenges to estimating how long it takes to mark a ballot
There are many challenges to measuring how long it takes voters to cast a ballot; in general, these challenges stem from two factors. First, the time that individual voters take to mark a ballot varies considerably. If we want reliable measurements about how long it takes theaveragevoter to mark a ballot, we need to time a lot of voters as they cast a ballot. Second, ballots vary from precinct-to-precinct. Therefore, to get an accurate assessment of ballot-marking times across different types of ballots, we need to observe voters voting on a wide variety of ballots.
Both of these factors raise the cost of getting accurate measurements of ballot-marking times.
A novel approach to estimating ballot-marking time
Inthis paper, we develop a new approach to estimating ballot-marking times that overcomes these challenges. We then use these estimates to study the ways in which ballot design, voter behavior, and voter demographics influence how long it takes voters to mark ballots.
South Carolina uses an electronic voting machine (termed a direct-recording electronic machine, or DRE) manufactured by Election Systems & Software (ES&S) called the iVotronic. Our method relies on the machine-log files that monitor and record the “events” that occur on the iVotronics on Election Day. An event can be when the machine is turned on, a memory card is inserted, or a vote is cast. Whenever such an event occurs, the machine’s log file records the exact time and date of the event.
Unfortunately, the iVotronic machine-log file only records the time when a ballot is cast, and not the time when the voter begins to mark the ballot. Before we began any analysis of the log files, then, we had to decide on a method that would allow us to measure the time each voter took to mark a ballot. We did this by looking specifically at log files from times when a polling place was most likely to be at full capacity.
If a voting machine is started up and begins to be used when a polling place is at full capacity — that is, if there is a line of voters waiting for machines, and all the machines are being used — then the time between successive ballot-casting time stamps on a machine will approximate how long it takes to mark a ballot. An important feature of our method helps figure out when machines are most likely being used to capacity, so that ballot-casting time stamps from those periods can be used to estimate ballot-marking times.
Prior research, confirmed by our own work, has suggested that voting machines are most likely to be used at full capacity early in the day. Therefore, one option we explore is measuring ballot-marking times between 7:15am and 7:30am, when polls opened. Another option we apply is identifying and measuring ballot-marking time on machines that were used a minimum number of times (four, in this case) in specific fifteen-minute periods. Both of these options diverge from previous analyses of high-capacity times, which have only used timing data from the period of time after the polls have closed, but voters are still standing in line to vote.
To help us understand how the design of a ballot influences marking times, we need to know what, exactly, is on every ballot. We can ascertain this using another file produced by the iVotronics: the “ballot image” or “cast-vote record” file (CVR). The CVR file records the candidate choices made by each voter — they are the ballots that are counted on election night. These ballots are randomly entered onto the CVRs, so that ballot secrecy is preserved. CVR files give us rich information about which candidates and offices appear on every ballot, and the behaviors of voters that might influence how long it takes to mark a ballot, such as whether they use the straight party option or “under-vote” particular races.
One final source of data is the precinct-level demographic information made available by the state of South Carolina. This includes the total number of voters, the number of voters over 65, and the number of non-white voters.
What influences ballot-marking times? Ballot design, voter behavior, and demographics
Overall, we were able to estimate the ballot-marking times from over 1,000 precincts in South Carolina in the 2016 election. Depending on the precise rules we use to select the data we use to estimate marking times, the average marking time statewide ranged from 2.8 to 3.5 minutes
We perform a number of multiple regressions to quantify the degree of association between ballot features, voter behavior, and demographics, on the one hand, and ballot-marking times, on the other. The length of the ballot certainly influences how long it takes voters to mark a ballot. We estimate that each additional office on a ballot increases the ballot-marking time between 1.3 and 1.7 seconds, and that each additional candidate increases the ballot-marking time between 0.2 and 1.2 seconds. Marking the straight-party choice reduces ballot-marking times by up to half a minute. Ballot-marking times are longer in precincts with more elderly and non-white voters, and shorter in precincts that have many voters per voting machine.
Even after controlling for ballot design, voter behavior, and demographics, there remains considerable variation across counties in how long it takes to mark ballots. This suggests the need for further research to understand how county-specific practices, such as poll worker training and precinct layout, influences ballot-marking times.
Measures such as the one we develop in this paper are important, because they offer election officials a way to measure service times in polling places unobtrusively. If measurements are taken automatically as voters count their ballots, officials will have a richer source of data for the management of the polls. Academics benefit from such measures, as well. With richer data about the behavior of voters as they cast ballots, it is possible for scholars to better understand how election administration, voter behavior, and demographics come together to influence polling place dynamics.
Jacob Jaffe is a graduate researcher at the MIT Election Lab, and a Ph.D. candidate in Political Science at MIT.
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Charles Stewart III is the founding director of the MIT Election Data + Science Lab. He is the Kenan Sahin Distinguished Professor of Political Science at MIT, co-director of the Stanford-MIT Project on a Healthy Election, and a co-director of the Caltech/MIT Voting Technology Project.
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Jacob Coblentz was a Senior Research Support Associate at the MIT Election Data and Science Lab until 2019. His research interests include political methodology and using modern statistical methods to analyze how elections are conducted. He graduated from George Washington University in 2017 with degrees in mathematics and political science.
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2719 | https://www.quora.com/How-can-you-prove-that-tan-A-tan-B-tan-C-tan-A-tan-B-tan-C | Something went wrong. Wait a moment and try again.
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5
How can you prove that
tan
A
+
tan
B
+
tan
C
=
tan
A
tan
B
tan
C
?
Carlos Eŭ Th
Triple IMO bronze medalist
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Author has 5.9K answers and 4.4M answer views
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6y
If A,B,C are angles of a triangle, then C=π−(A+B).
Now: tan(π−θ)=−tanθ so tanC=−tan(A+B).
tan(A+B)=sin(A+B)cos(A+B)=sinAcosB+cosAsinBcosAcosB−sinAsinB.
If we divide each term the right by cosAcosB we will have:
tan(A+B)=tanA+tanB1−tanAtanB.
If are angles of a triangle, then .
Now: so .
Where
If we divide each term the right by we will have:
.
So:
Note that if are not angles of a triangle, then this is not true, generally.
Related questions
What is the formula of tan (A + B + C)?
How do you prove that
?
How can I prove that
How do you prove that
?
How can I prove that
?
Satya Parkash Sud
Former Professor at Himachal Pradesh University Shimla (1986–2002)
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2y
The equation,
tan A + tan B + tan C = tan A tan B tan C,
only holds when A, B, C are angles of a triangle ie A + B + C = 180°
Therefore
A + B = 180° - C
Taking tangents of both sides we get
tan ( A + B) = tan (180° - C),
=> (tan A + tan B)/(1 - tan A tan B) = - tan C [because tan(180 - C) - tan C]
=> ...
Kevin Simonson
M.S. in Digital Logic Design & Theoretical Computer Science, Paul G. Allen School of Computer Science & Engineering (Graduated 1994)
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David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.2K answers and 520.3K answer views
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6y
Where A, B, and C, are any three values? It doesn’t work for A = 1.0, B = 1.1, and C = 1.2, where angles are expressed in radians. With these values tan(A) = 1.55740772465490223052, tan(B) = 1.96475965724865195095, and tan(C) = 2.5721516221263189354. The sum of the three tangents is 6.09431900402987311687 and the product of the three tangents is 7.87060871604436557251. Those two values are NOT equal.
Dean Rubine
Been doing high school math since high school, circa 1975
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6y
Presumably you mean given triangle ABC, or
Note that so we focus on the numerator on the right being zero.
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Related questions
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[math] \displaystyle \frac{3\tan x - \tan^3 x}{1-3\tan ^2 x} = \tan 3x [/math]
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[math]\tan 6 \tan 42 \tan 66 \tan 78 =1 [/math]
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[math]\tan x + \tan (60°+x) + \tan (120°+x) =3\tan (3x) [/math]
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Assistant
Bot
·
1y
To prove the identity
[math]\tan A + \tan B + \tan C = \tan A \tan B \tan C[/math]
we can use the fact that [math]A[/math], [math]B[/math], and [math]C[/math] are the angles of a triangle. This means that:
[math]A + B + C = \pi[/math]
We can use the tangent addition formula and the properties of tangent for angles summing to [math]\pi[/math].
Step 1: Use the tangent of a sum
From the tangent addition formula, we know:
[math]\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}[/math]
Step 2: Find [math]\tan C[/math]
Since [math]C = \pi - (A + B)[/math], we have:
[math]\tan C = \tan(\pi - (A + B)) = -\tan(A + B)[/math]
Using the tangent addition formula:
[math]\tan C = -\frac{\tan A + \tan B}{1 - \tan A \tan B}[/math]
Step 3: Substitute [math]\tan C[/math]
To prove the identity
[math]\tan A + \tan B + \tan C = \tan A \tan B \tan C[/math]
we can use the fact that [math]A[/math], [math]B[/math], and [math]C[/math] are the angles of a triangle. This means that:
[math]A + B + C = \pi[/math]
We can use the tangent addition formula and the properties of tangent for angles summing to [math]\pi[/math].
Step 1: Use the tangent of a sum
From the tangent addition formula, we know:
[math]\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}[/math]
Step 2: Find [math]\tan C[/math]
Since [math]C = \pi - (A + B)[/math], we have:
[math]\tan C = \tan(\pi - (A + B)) = -\tan(A + B)[/math]
Using the tangent addition formula:
[math]\tan C = -\frac{\tan A + \tan B}{1 - \tan A \tan B}[/math]
Step 3: Substitute [math]\tan C[/math] into the original identity
Now, we substitute [math]\tan C[/math] into the left-hand side of the identity:
[math]\tan A + \tan B + \tan C = \tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B}[/math]
Step 4: Combine the terms
Let [math]x = \tan A + \tan B[/math], then we have:
[math]x - \frac{x}{1 - \tan A \tan B}[/math]
To combine these, we find a common denominator:
[math]= \frac{x(1 - \tan A \tan B) - x}{1 - \tan A \tan B} = \frac{x - x + x \tan A \tan B}{1 - \tan A \tan B} = \frac{x \tan A \tan B}{1 - \tan A \tan B}[/math]
Step 5: Substitute back for [math]x[/math]
Now substitute back [math]x = \tan A + \tan B[/math]:
[math]= \frac{(\tan A + \tan B) \tan A \tan B}{1 - \tan A \tan B}[/math]
Step 6: Compare with the right-hand side
Now we need to show that
[math]\frac{(\tan A + \tan B) \tan A \tan B}{1 - \tan A \tan B} = \tan A \tan B \tan C[/math]
Using our earlier expression for [math]\tan C[/math]:
[math]\tan A \tan B \tan C = \tan A \tan B \left(-\frac{\tan A + \tan B}{1 - \tan A \tan B}\right) = -\frac{(\tan A + \tan B) \tan A \tan B}{1 - \tan A \tan B}[/math]
Final Step: Equate both sides
From our calculations, we see that:
[math]\tan A + \tan B + \tan C = -\frac{(\tan A + \tan B) \tan A \tan B}{1 - \tan A \tan B}[/math]
This confirms that:
[math]\tan A + \tan B + \tan C = \tan A \tan B \tan C[/math]
Thus, the identity is proven.
[math]\tan A + \tan B + \tan C = \tan A \tan B \tan C[/math]
when [math]A + B + C = \pi[/math].
Zeke Gum
B.S in Bachelor of Science in Chemistry, University of the Philippines (Graduated 2019)
·
6y
Provided angles A,B and C adds up to 180 degrees: A+B+C= 180
A+B=180-C
Tan(A+B)=Tan(180-C)
[tanA+tanB]/[1-tanAtanB]=-tanC
tanA+tanB=-tanC+tanAtanBtanC
tanA+tanB+tanC=tanAtanBtanC
or
C = 180 - (A+B)
tan A + tan B =
= (sin A/ cos A) + (sin B/ cos B)
= ( sin A cos B + cos A sin B)/ (cos A cos B)
= sin (A+B) / (cos A cos B)
= sin C / (cos A cos B)
tan A + tan B + tan C =
= ( sin C / (cos A cos B)) + (sin C / cos C)
= sin C ( cos C + cos A cos B )/ (cos A cos B cos C )
cos C = cos ( 180 - (A+B))
= - cos (A+B)
= - cos A cos B + sin A sin B
cos C + cos A cos B = sin A sin B
hence,
= sin C (s
Provided angles A,B and C adds up to 180 degrees: A+B+C= 180
A+B=180-C
Tan(A+B)=Tan(180-C)
[tanA+tanB]/[1-tanAtanB]=-tanC
tanA+tanB=-tanC+tanAtanBtanC
tanA+tanB+tanC=tanAtanBtanC
or
C = 180 - (A+B)
tan A + tan B =
= (sin A/ cos A) + (sin B/ cos B)
= ( sin A cos B + cos A sin B)/ (cos A cos B)
= sin (A+B) / (cos A cos B)
= sin C / (cos A cos B)
tan A + tan B + tan C =
= ( sin C / (cos A cos B)) + (sin C / cos C)
= sin C ( cos C + cos A cos B )/ (cos A cos B cos C )
cos C = cos ( 180 - (A+B))
= - cos (A+B)
= - cos A cos B + sin A sin B
cos C + cos A cos B = sin A sin B
hence,
= sin C (sin A sin B ) / (cos A cos B cos C )
= tan A tan B tan C
Enrico Gregorio
Associate professor in Algebra
·
Author has 18.4K answers and 16M answer views
·
6y
This holds if and only if [math]A+B+C=\pi[/math].
Suppose [math]C=\pi-(A+B)[/math]. Then, recalling that [math]\tan(\pi-x)=-\tan x[/math] and the addition formula for the tangent,
[math]\tan C=-\tan(A+B)=-\dfrac{\tan A+\tan B}{1-\tan A\tan B}\tag{}[/math]
so that
[math]\begin{align}\tan A+\tan B+\tan C &=\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A\tan B} \ &= (\tan A+\tan B)\left(1-\frac{1}{1-\tan A\tan B}\right) \ &= (\tan A+\tan B)\frac{-\tan A\tan B}{1-\tan A\tan B} \ &= \tan A\tan B\left(-\frac{\tan A+\tan B}{1-\tan A\tan B}\right) \ &= \tan A\tan B\tan C\end{align}\tag{}[/math]
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Doug Dillon
Ph.D. Mathematics
·
Author has 12.4K answers and 11.4M answer views
·
6y
Originally Answered: How can you prove that tanA+tanB+tanC = tanAtanBtanC?
·
Not always true. A, B, C must be angles of a triangle.
Agneebha Ghosh
Love Physics and exhumes innovative meanings from problems
·
7y
Related
If A+B+C = 180° , then (tan A + tan B + tan C) / (tanA.tanB.tanC) equals how much?
Actually, this question is an application of the 'Trigonometric Ratios of Compound Angles' where we have to apply the sides A,B,C as the sides of a triangle…
The detailed method to solve the above question is attached as follows
Cheers!!!
Hope it helps!!
P.S:- Please upvote!!!! It's damn encouraging……as I am a beginner..!!!!
Actually, this question is an application of the 'Trigonometric Ratios of Compound Angles' where we have to apply the sides A,B,C as the sides of a triangle…
The detailed method to solve the above question is attached as follows
Cheers!!!
Hope it helps!!
P.S:- Please upvote!!!! It's damn encouraging……as I am a beginner..!!!!
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Trevor
B.A. with MMath in Mathematics, University of Cambridge (Graduated 2023)
·
Upvoted by
Diganta Bhaskar
, M.Sc Theoretical Physics & Mathematics, Ramakrishna Mission Vivekananda Educational and Research Institute,be… · Author has 1.3K answers and 5.5M answer views
·
Updated 7y
Related
How do I prove that
[math] \frac{sin(A+B)}{sin(A-B)} = \frac{tan(A+B)}{tan(A-B)}[/math]
?
Original question:
How do I prove that [math]\displaystyle \frac {\sin (A-B)}{\sin (A+B)} = \frac {\tan A - \tan B}{\tan A + \tan B}[/math]?
Geometrically:
Let [math]\angle BAE = a[/math] and [math]\angle EAC = \angle DAE = b[/math]. The triangle is constructed such that [math]AE \perp BC[/math].
Consider the area ratio of [math]\triangle ABD[/math] to [math]\triangle ABC[/math].
[math]\displaystyle \frac {[\triangle ABD]}{[\triangle ABC]} = \frac {\frac 12 (AB)(AD) \sin (a-b)}{\frac 12 (AB)(AC) \sin (a+b)} = \frac {\sin (a-b)}{\sin (a+b)} \tag{1}[/math]
However apart from considering the areas in that way, we can also do it more elementarily:
[math]\displaystyle \frac {[\triangle ABD]}{[\triang[/math]
Original question:
How do I prove that [math]\displaystyle \frac {\sin (A-B)}{\sin (A+B)} = \frac {\tan A - \tan B}{\tan A + \tan B}[/math]?
Geometrically:
Let [math]\angle BAE = a[/math] and [math]\angle EAC = \angle DAE = b[/math]. The triangle is constructed such that [math]AE \perp BC[/math].
Consider the area ratio of [math]\triangle ABD[/math] to [math]\triangle ABC[/math].
[math]\displaystyle \frac {[\triangle ABD]}{[\triangle ABC]} = \frac {\frac 12 (AB)(AD) \sin (a-b)}{\frac 12 (AB)(AC) \sin (a+b)} = \frac {\sin (a-b)}{\sin (a+b)} \tag{1}[/math]
However apart from considering the areas in that way, we can also do it more elementarily:
[math]\displaystyle \frac {[\triangle ABD]}{[\triangle ABC]} = \frac {BD}{BC} \tag{2}[/math]
because they have the same height [math]AE[/math] .
Tweaking (2) for a little bit:
[math]\displaystyle \frac {BD}{BC} = \frac {\frac {BE-DE}{AE}}{\frac {BE+EC}{AE}} = \frac {\tan a-\tan b}{\tan a+\tan b} \tag{3}[/math]
Combining (1) and (3), we have the desired equality. Of course we have assumed both [math]a[/math] and [math]b[/math] to be acute, but you can generalize it to any kind of angle.
Algebraically:
[math]\displaystyle \begin {align} \frac {\sin (a-b)}{\sin (a+b)}&= \frac {\sin a\cos b-\cos a\sin b}{\sin a\cos b+\cos a\sin b} \ \;& = \frac {\sin a\cos b-\cos a\sin b}{\sin a\cos b+\cos a\sin b} \div \frac {\cos a \cos b}{\cos a \cos b} \ \;& = \frac {\tan a - \tan b}{\tan a + \tan b} \end {align} \tag{}[/math]
Shreshth Mehrotra
Author has 137 answers and 418.6K answer views
·
7y
Related
How can I prove that
[math]\frac{\sin(a+b)}{\sin(a-b)} = \frac{\tan a + \tan b}{\tan a -\tan b}[/math]
?
Abhinav Sankar
B.Tech from National Institute of Technology, Kurukshetra (Graduated 2021)
·
9y
Related
If A+B+C = 180° , then (tan A + tan B + tan C) / (tanA.tanB.tanC) equals how much?
This question is an application of compound angle formula of tan and sign of trigonometric ratios. Solution is attached. There is a shortcut to do it as well, however i insist that while practicing at home, please dont use such techniques. You may use them in exams as and when applicable.
As A+B+C=180; it must in general satisfy a triangle. You can just check for an equilateral triangle by putting A=B=C=60.
Method is attached:
This question is an application of compound angle formula of tan and sign of trigonometric ratios. Solution is attached. There is a shortcut to do it as well, however i insist that while practicing at home, please dont use such techniques. You may use them in exams as and when applicable.
As A+B+C=180; it must in general satisfy a triangle. You can just check for an equilateral triangle by putting A=B=C=60.
Method is attached:
Cheers!
Enrico Gregorio
Associate professor in Algebra
·
Author has 18.4K answers and 16M answer views
·
1y
Related
How do you prove that tan A+tan B+tan C =tanAtanBtanC, in a triangle ABC?
In a nonright triangle, you want to say.
Let [math]a=\tan A[/math] and [math]b=\tan B.[/math] Now
[math]\tan C=\tan(\pi-A-B)=-\tan(A+B)=\dfrac{a+b}{ab-1}[/math]
Therefore
[math]\tan A+\tan B+\tan C=a+b+\dfrac{a+b}{ab-1}=(a+b)\dfrac{ab-1+1}{ab-1}=ab\dfrac{a+b}{ab-1}[/math]
But the last term is again [math]\tan C,[/math] so we proved
[math]\tan A+\tan B+\tan C=\tan A\tan B\tan C[/math]
What about the cotangents?
Let [math]x=\cot A,y=\cot B,z=\cot C,[/math] so we have
and therefore
that is
[math]\cot A\cot B+\cot B\cot C+\cot C\cot A=1[/math]
which holds for any triangle. Indeed, if [math]A=\pi/2,[/math] it becomes [math]\cot B\cot C=1[/math] which is true, because [math]B[/math] and [math]C[/math]
In a nonright triangle, you want to say.
Let [math]a=\tan A[/math] and [math]b=\tan B.[/math] Now
[math]\tan C=\tan(\pi-A-B)=-\tan(A+B)=\dfrac{a+b}{ab-1}[/math]
Therefore
[math]\tan A+\tan B+\tan C=a+b+\dfrac{a+b}{ab-1}=(a+b)\dfrac{ab-1+1}{ab-1}=ab\dfrac{a+b}{ab-1}[/math]
But the last term is again [math]\tan C,[/math] so we proved
[math]\tan A+\tan B+\tan C=\tan A\tan B\tan C[/math]
What about the cotangents?
Let [math]x=\cot A,y=\cot B,z=\cot C,[/math] so we have
[math]\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{xyz}[/math]
and therefore
[math]xy+yz+zx=1[/math]
that is
[math]\cot A\cot B+\cot B\cot C+\cot C\cot A=1[/math]
which holds for any triangle. Indeed, if [math]A=\pi/2,[/math] it becomes [math]\cot B\cot C=1[/math] which is true, because [math]B[/math] and [math]C[/math] are complementary.
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2720 | https://stackoverflow.com/questions/27970185/find-line-that-is-tangent-to-2-given-circles | Stack Overflow
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Find line that is tangent to 2 given circles
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I've got a situation in which I have 2 circles (C1 and C2) and i need to find the line equation for the line that is tangent to both of these circles.
So as far as i'm aware, given a single point (P1) and C2's point and radius it is possible to quite easily get 2 possible points of tangency for C2 and P1 to make 2 line equations. But as i don't have P1, only the knowledge that the point will be one of a possible 2 points on C1, i'm not sure how to calculate this.
I assume it will be something along the lines of getting the 2 tangent line equations of C1 that are equal to the same of C2.
Both circles can have any radius, they could be the same or they could be hugely different. They will also never overlap (they can still touch though). And I'm looking for the 2 possible internal tangents.
Oh, and also, visuals would be very helpful haha :)
math
line
geometry
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edited Jan 16, 2015 at 1:02
Cimbali
11.5k11 gold badge4343 silver badges7676 bronze badges
asked Jan 15, 2015 at 18:12
Luke BourneLuke Bourne
30533 silver badges1313 bronze badges
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more suited for stackexchange
advocateofnone
– advocateofnone
2015-01-15 18:19:42 +00:00
Commented Jan 15, 2015 at 18:19
There are 4 possible lines tangent to both circles. Which one do you want ? Also, are both circles of the same size ? I use 2 of them (the outer tangents) in this post, otherwise the wikipedia article Tangent lines to circles is a good read.
Cimbali
– Cimbali
2015-01-15 19:20:35 +00:00
Commented Jan 15, 2015 at 19:20
1
There is also the degenerate case of C1 = C2, in which there are infinitely many solutions.
Zéychin
– Zéychin
2015-01-15 21:06:04 +00:00
Commented Jan 15, 2015 at 21:06
both circles can have any radius, they could be the same or they could be hugely different. They will also never overlap (they can still touch though). And I'm looking for the 2 possible internal tangents. Probably should have included that, my bad!
Luke Bourne
– Luke Bourne
2015-01-15 21:47:47 +00:00
Commented Jan 15, 2015 at 21:47
Add a comment |
2 Answers 2
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Let O be the intersection point between the line through the centers and the tangent.
Let d be the distance between the centers and h1, h2 be the distances between O and the centers. By similarity, these are proportional to the radii.
Hence,
h1 / h2 = r1 / r2 = m, h1 + h2 = d,
giving
h1 = m d / (1 + m), h2 = d / (1 + m).
Then the coordinates of O are found by interpolating between the centers
xo = (h2.x1 + h1.x2) / d yo = (h2.y1 + h1.y2) / d
and the angle of the tangent is that of the line through the centers plus or minus the angle between this line and the tangent,
a = arctan((y2 - y1)/(x2 - x1)) +/- arcsin(r1 / h1).
You can write the implicit equation of the tangent as
cos(a).y - sin(a).x = cos(a).yo - sin(a).xo.
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edited Jan 16, 2015 at 11:41
answered Jan 16, 2015 at 10:11
user1196549user1196549
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1
(source: imag.fr)
So we are going to use a homothetic transformation. If the circles C and C' have respectively centres O and O', and radius r and r', then we know there exists a unique homothetic transformation with centre J and ratio a, such that :
a = |JO|/|JO'| = r/r'
Noting AB is the vector from A to B, and |z| the norm of a vector z.
Hence you get J, knowing that it is between O and O' which we both already know.
Then with u the projection of JR on JO', and v the decomposition on its orthogonal, and considering the sine s and cosine c of the angle formed by O'JR, we have
|u| = |JR| c |v| = |JR| s c^2 + s^2 = 1
And finally because the triangle JRO' is right-angled in R :
s = r' / |JO|'
Putting all of this together, we get :
J = O + OO' / |OO'| a / (a+1) if |OJ| == r and |O'J| == r' then return the orthogonal line to (OO') passing through J |JR| = ( |JO'|^ - r'^2 ) s = r' / |JO'| c = ( 1 - s^2 ) u = c |JR| OO' / |OO'| w = (-u.y, u.x) % any orthogonal vector to u v = s |JR| w / |w| return lines corresponding to parametric equations J+t(u+v) and J+t(u-v)
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edited Sep 25, 2019 at 3:15
Glorfindel
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answered Jan 16, 2015 at 1:47
CimbaliCimbali
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2721 | https://courses.lumenlearning.com/suny-osbiology2e/chapter/cancer-and-the-cell-cycle-2/ | 10. Cell Reproduction
Cancer and the Cell Cycle
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By the end of this section, you will be able to do the following:
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Explain how mutant tumor suppressors cause cancer
Cancer comprises many different diseases caused by a common mechanism: uncontrolled cell growth. Despite the redundancy and overlapping levels of cell-cycle control, errors do occur. One of the critical processes monitored by the cell-cycle checkpoint surveillance mechanism is the proper replication of DNA during the S phase. Even when all of the cell-cycle controls are fully functional, a small percentage of replication errors (mutations) will be passed on to the daughter cells. If changes to the DNA nucleotide sequence occur within a coding portion of a gene and are not corrected, a gene mutation results. All cancers start when a gene mutation gives rise to a faulty protein that plays a key role in cell reproduction.
The change in the cell that results from the malformed protein may be minor: perhaps a slight delay in the binding of Cdk to cyclin or an Rb protein that detaches from its target DNA while still phosphorylated. Even minor mistakes, however, may allow subsequent mistakes to occur more readily. Over and over, small uncorrected errors are passed from the parent cell to the daughter cells and amplified as each generation produces more non-functional proteins from uncorrected DNA damage. Eventually, the pace of the cell cycle speeds up as the effectiveness of the control and repair mechanisms decreases. Uncontrolled growth of the mutated cells outpaces the growth of normal cells in the area, and a tumor (“-oma”) can result.
Proto-oncogenes
The genes that code for the positive cell-cycle regulators are called proto-oncogenes. Proto-oncogenes are normal genes that, when mutated in certain ways, become oncogenes—genes that cause a cell to become cancerous. Consider what might happen to the cell cycle in a cell with a recently acquired oncogene. In most instances, the alteration of the DNA sequence will result in a less functional (or non-functional) protein. The result is detrimental to the cell and will likely prevent the cell from completing the cell cycle; however, the organism is not harmed because the mutation will not be carried forward. If a cell cannot reproduce, the mutation is not propagated and the damage is minimal. Occasionally, however, a gene mutation causes a change that increases the activity of a positive regulator. For example, a mutation that allows Cdk to be activated without being partnered with cyclin could push the cell cycle past a checkpoint before all of the required conditions are met. If the resulting daughter cells are too damaged to undergo further cell divisions, the mutation would not be propagated and no harm would come to the organism. However, if the atypical daughter cells are able to undergo further cell divisions, subsequent generations of cells may accumulate even more mutations, some possibly in additional genes that regulate the cell cycle.
The Cdk gene in the above example is only one of many genes that are considered proto-oncogenes. In addition to the cell-cycle regulatory proteins, any protein that influences the cycle can be altered in such a way as to override cell-cycle checkpoints. An oncogene is any gene that, when altered, leads to an increase in the rate of cell-cycle progression.
Tumor Suppressor Genes
Like proto-oncogenes, many of the negative cell-cycle regulatory proteins were discovered in cells that had become cancerous. Tumor suppressor genes are segments of DNA that code for negative regulator proteins, the type of regulators that, when activated, can prevent the cell from undergoing uncontrolled division. The collective function of the best-understood tumor suppressor gene proteins, Rb, p53, and p21, is to put up a roadblock to cell-cycle progression until certain events are completed. A cell that carries a mutated form of a negative regulator might not be able to halt the cell cycle if there is a problem. Tumor suppressors are similar to brakes in a vehicle: Malfunctioning brakes can contribute to a car crash!
Mutated p53 genes have been identified in more than 50 percent of all human tumor cells. This discovery is not surprising in light of the multiple roles that the p53 protein plays at the G1 checkpoint. A cell with a faulty p53 may fail to detect errors present in the genomic DNA ((Figure)). Even if a partially functional p53 does identify the mutations, it may no longer be able to signal the necessary DNA repair enzymes. Either way, damaged DNA will remain uncorrected. At this point, a functional p53 will deem the cell unsalvageable and trigger programmed cell death (apoptosis). The damaged version of p53 found in cancer cells, however, cannot trigger apoptosis.
Art Connection
Figure 1. The role of normal p53 is to monitor DNA and the supply of oxygen (hypoxia is a condition of reduced oxygen supply). If damage is detected, p53 triggers repair mechanisms. If repairs are unsuccessful, p53 signals apoptosis. A cell with an abnormal p53 protein cannot repair damaged DNA and thus cannot signal apoptosis. Cells with abnormal p53 can become cancerous. (credit: modification of work by Thierry Soussi)
Human papillomavirus can cause cervical cancer. The virus encodes E6, a protein that binds p53. Based on this fact and what you know about p53, what effect do you think E6 binding has on p53 activity?
E6 activates p53
E6 inactivates p53
E6 mutates p53
E6 binding marks p53 for degradation
The loss of p53 function has other repercussions for the cell cycle. Mutated p53 might lose its ability to trigger p21 production. Without adequate levels of p21, there is no effective block on Cdk activation. Essentially, without a fully functional p53, the G1 checkpoint is severely compromised and the cell proceeds directly from G1 to S regardless of internal and external conditions. At the completion of this shortened cell cycle, two daughter cells are produced that have inherited the mutated p53 gene. Given the non-optimal conditions under which the parent cell reproduced, it is likely that the daughter cells will have acquired other mutations in addition to the faulty tumor-suppressor gene. Cells such as these daughter cells quickly accumulate both oncogenes and non-functional tumor-suppressor genes. Again, the result is tumor growth.
Link to Learning
Watch an animation of how cancer results from errors in the cell cycle.
Section Summary
Cancer is the result of unchecked cell division caused by a breakdown of the mechanisms that regulate the cell cycle. The loss of control begins with a change in the DNA sequence of a gene that codes for one of the regulatory molecules. Faulty instructions lead to a protein that does not function as it should. Any disruption of the monitoring system can allow other mistakes to be passed on to the daughter cells. Each successive cell division will give rise to daughter cells with even more accumulated damage. Eventually, all checkpoints become nonfunctional, and rapidly reproducing cells crowd out normal cells, resulting in a tumor or leukemia (blood cancer).
Art Connections
(Figure) Human papillomavirus can cause cervical cancer. The virus encodes E6, a protein that binds p53. Based on this fact and what you know about p53, what effect do you think E6 binding has on p53 activity?
E6 activates p53
E6 inactivates p53
E6 mutates p53
E6 binding marks p53 for degradation
Show Solution
(Figure) D. E6 binding marks p53 for degradation.
Review Questions
___________ are changes to the order of nucleotides in a segment of DNA that codes for a protein.
Proto-oncogenes
Tumor suppressor genes
Gene mutations
Negative regulators
Show Solution
C
A gene that codes for a positive cell-cycle regulator is called a(n) _____.
kinase inhibitor.
tumor suppressor gene.
proto-oncogene.
oncogene.
Show Solution
C
A mutated gene that codes for an altered version of Cdk that is active in the absence of cyclin is a(n) _____.
kinase inhibitor.
tumor suppressor gene.
proto-oncogene.
oncogene.
Show Solution
D
Which molecule is a Cdk inhibitor that is controlled by p53?
cyclin
anti-kinase
Rb
p21
Show Solution
D
Free Response
Outline the steps that lead to a cell becoming cancerous.
Show Solution
If one of the genes that produces regulator proteins becomes mutated, it produces a malformed, possibly non-functional, cell-cycle regulator, increasing the chance that more mutations will be left unrepaired in the cell. Each subsequent generation of cells sustains more damage. The cell cycle can speed up as a result of the loss of functional checkpoint proteins. The cells can lose the ability to self-destruct and eventually become “immortalized.”
Explain the difference between a proto-oncogene and a tumor-suppressor gene.
Show Solution
A proto-oncogene is a segment of DNA that codes for one of the positive cell cycle regulators. If that gene becomes mutated so that it produces a hyperactivated protein product, it is considered an oncogene. A tumor suppressor gene is a segment of DNA that codes for one of the negative cell cycle regulators. If that gene becomes mutated so that the protein product becomes less active, the cell cycle will run unchecked. A single oncogene can initiate abnormal cell divisions; however, tumor suppressors lose their effectiveness only when both copies of the gene are damaged.
List the regulatory mechanisms that might be lost in a cell producing faulty p53.
Show Solution
Regulatory mechanisms that might be lost include monitoring of the quality of the genomic DNA, recruiting of repair enzymes, and the triggering of apoptosis.
p53 can trigger apoptosis if certain cell-cycle events fail. How does this regulatory outcome benefit a multicellular organism?
Show Solution
If a cell has damaged DNA, the likelihood of producing faulty proteins is higher. The daughter cells of such a damaged parent cell would also produce faulty proteins that might eventually become cancerous. If p53 recognizes this damage and triggers the cell to self-destruct, the damaged DNA is degraded and recycled. No further harm comes to the organism. Another healthy cell is triggered to divide instead.
Glossary
oncogene
: mutated version of a normal gene involved in the positive regulation of the cell cycle
proto-oncogene
: normal gene that when mutated becomes an oncogene
tumor suppressor gene
: segment of DNA that codes for regulator proteins that prevent the cell from undergoing uncontrolled division
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2722 | https://www.youtube.com/watch?v=1goHxBPRhSI | Simplest Radical Form
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let's talk about simplest radical form a radical expression is in its simplest form when three conditions are met first no radicans have perfect square factors other than one second no radicand appears in a fraction and third no radicals appear in the denominator of a fraction here we have the fraction Square Ro TK of 8 / 2 this can be further simplified to < TK of 2 4 / 2 we can pull out the four so now we have 2 < TK of 2 / by two and finally we can cancel out the twos to Simply get square < TK of two now our radical expression has been simplified to simplest radical form |
2723 | https://planetmath.org/alternatingsum | alternating sum
An alternating sum is a sequence of arithmetic operations in which each addition is followed by a subtraction, and viceversa, applied to a sequence of numerical entities. For example,
| |
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An alternating sum is also called an alternating series.
Alternating sums are often expressed in summation notation with the iterated expression involving multiplication by negative one raised to the iterator. Since a negative number raised to an odd number gives a negative number while raised to an even number gives a positive number (see: factors with minus sign), essentially has the effect of turning the odd-indexed terms of the sequence negative but keeping their absolute values the same. Our previous example would thus be restated
| |
| |
If the operands in an alternating sum decrease in value as the iterator increases, and approach zero, then the alternating sum converges to a specific value. This fact is used in many of the best-known expression for or fractions thereof, such as the Gregory series:
| |
| |
Other constants also find expression as alternating sums, such as Cahen’s constant.
An alternating sum need not necessarily involve an infinity of operands. For example, the alternating factorial of is computed by an alternating sum stopping at .
References
1 Tobias Dantzig, Number: The Language of Science, ed. Joseph Mazur. New York: Pi Press (2005): 166
| | |
--- |
| Title | alternating sum |
| Canonical name | AlternatingSum |
| Date of creation | 2013-03-22 17:35:30 |
| Last modified on | 2013-03-22 17:35:30 |
| Owner | PrimeFan (13766) |
| Last modified by | PrimeFan (13766) |
| Numerical id | 7 |
| Author | PrimeFan (13766) |
| Entry type | Definition |
| Classification | msc 11B25 | |
2724 | https://www.reddit.com/r/AskPhysics/comments/1etaew5/conservative_vs_nonconservative_forces/ | Conservative vs nonconservative forces : r/AskPhysics
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•1 yr. ago
Puzzleheaded-Cod4073
Conservative vs nonconservative forces
I am confused as to how the formula W = Fcosxd applies to conservative and non conservative forces. Would "d" be the displacement when calculating work done by a conservative force, and would it be distance for a nonconservative force(instead of displacement, since the path matters)? If the latter is true, why do people use the formula Wnc = change in KE + change in PE if you can just use the first formula just with d as distance? This is given that the force is constant in both magnitude and direction.
Thanks for your time.
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2725 | https://www.grammarflip.com/curriculum/apostrophes-individual-vs-joint/ | Skip to content
Apostrophes: Individual Ownership vs. Joint Ownership
How Do You Use Apostrophes to Show Individual Ownership vs. Joint Ownership?
Apostrophes are punctuation marks that can be used to show possession. When a compound subject shows possession, the two or more subjects can have either joint or individual ownership of the object. One apostrophe on the last subject indicates joint ownership, which means that all members of the compound subject own the object. When each member of the compound subject has its own apostrophe, they have individual ownership, meaning that each member of the compound subject has its own object.
Some examples of apostrophes with individual ownership vs. joint ownership would be the following:
Joint Ownership
Foster and Lydia’s first class of the day is math.
Eunice and Tony’sparents are both soccer coaches.
Marcel and Lacy’sneighborhood is having an egg hunt on Saturday.
Because there is joint ownership, only one apostrophe and -s is needed at end of the last owner.
Individual Ownership
Kyree’s and Everett’sfathers went to high school together.
Christina’s and Amir’s bikes were both locked up in front of the library.
Judy’s, Marie’s, and Hector’s quizzes all had perfect scores.
Because there is individual ownership, an apostrophe and -s is needed at end of both of the owners.
Long-Form Videos: Apostrophes with Individual vs. Joint Ownership
Long-form instructional video lessons allow students to engage with grammar concepts in more depth and detail.
This format provides students with a stronger foundation and a more comprehensive understanding of apostrophes with individual and joint ownership.
Short-Form Videos: Apostrophes with Individual vs. Joint Ownership
Short-form videos are an excellent way to review grammar concepts. Our two-minute instructional videos help students review the concept of apostrophes with individual ownership vs. joint ownership to further solidify their understanding.
Memorable Images: Apostrophes with Individual vs. Joint Ownership
The use of images to connect visual cues with concepts makes it simpler for students to grasp and remember key ideas. GrammarFlip’s memorable images create visual associations that make apostrohpes with individual ownership vs. joint ownership more engaging and easier to retain.
Definition Cards: Apostrophes with Individual vs. Joint Ownership
Definition cards reinforce grammar concepts by providing clear and concise explanations that students can easily reference for quick review and better retention. GrammarFlip’s definintion cards help students review the concept of apostrophes with individual ownership vs. joint ownership to further solidify their understanding.
Music: Basic Apostrophe Rules
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Download a Free Worksheet on Apostrophes: Individual vs. Joint Ownership!
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2726 | https://dictionary.cambridge.org/us/pronunciation/english/liaison | How to pronounce LIAISON in English
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English Pronunciation
{{word}} {{#beta}} Beta{{/beta}}
English Grammar English–Spanish Spanish–English
English pronunciation of liaison
liaison
How to pronounce liaison
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UK/liˈeɪ.zɒn/
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US/liˈeɪ.zɑːn/
More about phonetic symbols
Sound-by-sound pronunciation
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UK/liˈeɪ.zɒn/liaison
Your browser doesn't support HTML5 audio /l/ as in Your browser doesn't support HTML5 audio l ook
Your browser doesn't support HTML5 audio /i/ as in Your browser doesn't support HTML5 audio happ y
Your browser doesn't support HTML5 audio /eɪ/ as in Your browser doesn't support HTML5 audio d ay
Your browser doesn't support HTML5 audio /z/ as in Your browser doesn't support HTML5 audio z oo
Your browser doesn't support HTML5 audio /ɒ/ as in Your browser doesn't support HTML5 audio s o ck
Your browser doesn't support HTML5 audio /n/ as in Your browser doesn't support HTML5 audio n ame
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US/liˈeɪ.zɑːn/liaison
Your browser doesn't support HTML5 audio /l/ as in Your browser doesn't support HTML5 audio l ook
Your browser doesn't support HTML5 audio /i/ as in Your browser doesn't support HTML5 audio happ y
Your browser doesn't support HTML5 audio /eɪ/ as in Your browser doesn't support HTML5 audio d ay
Your browser doesn't support HTML5 audio /z/ as in Your browser doesn't support HTML5 audio z oo
Your browser doesn't support HTML5 audio /ɑː/ as in Your browser doesn't support HTML5 audio f a ther
Your browser doesn't support HTML5 audio /n/ as in Your browser doesn't support HTML5 audio n ame
(English pronunciations of liaison from the Cambridge Advanced Learner's Dictionary & Thesaurus and from the Cambridge Academic Content Dictionary, both sources © Cambridge University Press)
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/ˌbəʊ.bə ˈtiː/
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/ˌboʊ.bə ˈtiː/
a drink, originally from East Asia, consisting of tea mixed with milk or fruit flavours, to which small balls of tapioca (= a plant substance) or jelly are added. The drink is then shaken to create bubbles.
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2727 | https://www.nature.com/scitable/knowledge/library/eutrophication-causes-consequences-and-controls-in-aquatic-102364466/ | Published Time: Tue, 09 Jan 2024 11:31:21 GMT
Eutrophication: Causes, Consequences, and Controls in Aquatic Ecosystems | Learn Science at Scitable
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Eutrophication: Causes, Consequences, and Controls in Aquatic Ecosystems
By:Michael F. Chislock(Department of Fisheries and Allied Aquacultures, Auburn University),Enrique Doster(Department of Animal Sciences, Auburn University),Rachel A. Zitomer(Department of Biological Sciences, Humboldt University)&Alan E. Wilson(Department of Fisheries and Allied Aquacultures, Auburn University)©2013 Nature Education
)
Citation:Chislock,M.F.,Doster,E.,Zitomer,R.A.&Wilson,A.E.(2013)Eutrophication: Causes, Consequences, and Controls in Aquatic Ecosystems.Nature Education Knowledge 4(4):10
Eutrophication is a leading cause of impairment of many freshwater and coastal marine ecosystems in the world. Why should we worry about eutrophication and how is this problem managed?
Aa)Aa)Aa)
Introduction
Eutrophication is characterized by excessive plant and algal growth due to the increased availability of one or more limiting growth factors needed for photosynthesis (Schindler 2006), such as sunlight, carbon dioxide, and nutrient fertilizers. Eutrophication occurs naturally over centuries as lakes age and are filled in with sediments (Carpenter 1981). However, human activities have accelerated the rate and extent of eutrophication through both point-source discharges and non-point loadings of limiting nutrients, such as nitrogen and phosphorus, into aquatic ecosystems (i.e., cultural eutrophication), with dramatic consequences for drinking water sources, fisheries, and recreational water bodies (Carpenter et al. 1998). For example, aquaculture scientists and pond managers often intentionally eutrophy water bodies by adding fertilizers to enhance primary productivity and increase the density and biomass of recreationally and economically important fishes (Figure 1) via bottom-up effects on higher trophic levels (Boyd & Tucker 1998). However, during the 1960s and 1970s, scientists linked algal blooms to nutrient enrichment resulting from anthropogenic activities such as agriculture, industry, and sewage disposal (Schindler 1974). The known consequences of cultural eutrophication include blooms of blue-green algae (i.e., cyanobacteria, Figure 2), tainted drinking water supplies, degradation of recreational opportunities, and hypoxia. The estimated cost of damage mediated by eutrophication in the U.S. alone is approximately $2.2 billion annually (Dodds et al. 2009).
Figure 1
Aquaculture ponds, such as these channel catfish production ponds, typically contain high concentrations of nutrients, such as nitrogen and phosphorus, due to regular feeding of the fish. Consequently, these ponds are also plagued by recurring cyanobacterial blooms and hypoxia.
© 2013 Nature Education Photos by Mark Peterman and Michael Chislock. All rights reserved. ;)
Figure 2
Sampling Lago de Pátzcuaro during a cyanobacterial bloom with Ilyana Berry, Dr. John Berry, and Dr. Fernando Bernal-Brooks.
© 2013 Nature Education Photo by Alan Wilson. All rights reserved. ;)
Consequences
The most conspicuous effect of cultural eutrophication is the creation of dense blooms of noxious, foul-smelling phytoplankton that reduce water clarity and harm water quality (Figure 2). Algal blooms limit light penetration, reducing growth and causing die-offs of plants in littoral zones while also lowering the success of predators that need light to pursue and catch prey (Lehtiniemi et al. 2005). Furthermore, high rates of photosynthesis associated with eutrophication can deplete dissolved inorganic carbon and raise pH to extreme levels during the day. Elevated pH can in turn ‘blind' organisms that rely on perception of dissolved chemical cues for their survival by impairing their chemosensory abilities (Figure 3) (Turner & Chislock 2010). When these dense algal blooms eventually die, microbial decomposition severely depletes dissolved oxygen, creating a hypoxic or anoxic ‘dead zone' lacking sufficient oxygen to support most organisms. Dead zones are found in many freshwater lakes including the Laurentian Great Lakes (e.g., central basin of Lake Erie; Arend et a l. 2011) during the summer. Furthermore, such hypoxic events are particularly common in marine coastal environments surrounding large, nutrient-rich rivers (e.g., Mississippi River and the Gulf of Mexico; Susquehanna River and the Chesapeake Bay) and have been shown to affect more than 245,000 square kilometers in over 400 near-shore systems (Diaz & Rosenberg 2008). Hypoxia and anoxia as a result of eutrophication continue to threaten lucrative commercial and recreational fisheries worldwide.
Figure 3
Helisoma trivolvis (left) and Physa acuta (right) are two of the most common freshwater snails in North America. Both species use chemical cues to detect predators such as molluscivorous fish and typically respond by hiding under rocks and logs or in shallow water.
© 2013 Nature Education Photo by Andy Turner. All rights reserved. ;)
Some algal blooms pose an additional threat because they produce noxious toxins (e.g., microcystin and anatoxin-a; Chorus and Bartram 1999). Over the past century, harmful algal blooms (HABs) have been linked with (1) degradation of water quality (Francis 1878), (2) destruction of economically important fisheries (Burkholder et al. 1992), and (3) public health risks (Morris 1999). Within freshwater ecosystems, cyanobacteria are the most important phytoplankton associated with HABs (Paerl 1988). Toxigenic cyanobacteria, including Anabaena, Cylindrospermopsis, Microcystis, and Oscillatoria (Planktothrix), tend to dominate nutrient-rich, freshwater systems due to their superior competitive abilities under high nutrient concentrations, low nitrogen-to-phosphorus ratios, low light levels, reduced mixing, and high temperatures (Downing et al. 2001; Paerl & Huisman 2009; Paerl and Paul 2012). Poisonings of domestic animals, wildlife (Figure 4), and even humans by blooms of toxic cyanobacteria have been documented throughout the world and date back to Francis' (1878) first observation of dead livestock associated with a bloom of cyanobacteria . Furthermore, cyanobacteria are responsible for several off-flavor compounds (e.g., methylisoborneal and geosmin) found in municipal drinking water systems as well as in aquaculture-rased fishes, resulting in large financial losses for state and regional economies (Crews & Chappell 2007). In addition to posing significant public health risks, cyanobacteria have been shown to be poor quality food for most zooplankton grazers in laboratory studies (Wilson et al. 2006; Tillmanns et al. 2008), thus reducing the efficiency of energy transfer in aquatic food webs and potentially preventing zooplankton from controlling algal blooms.
Figure 4
A dead African buffalo (Syncerus caffer) found in a reservoir with a dense bloom of the toxic cyanobacterium Microcystis at the Loskop Dam Nature Reserve in South Africa.
© 2013 Nature Education Photo by Jannie Coetzee. All rights reserved. ;)
Eutrophication is also associated with major changes in aquatic community structure. During cyanobacterial blooms, small-bodied zooplankton tend to dominate plankton communities, and past observational studies have attributed this pattern to anti-herbivore traits of cyanobacteria (e.g., toxicity, morphology, and poor food quality) (Porter 1977). However, the biomass of planktivorous fish is often positively related to nutrient levels and ecosystem productivity. Piscivorous fishes (e.g., bass, pike) tend to dominate the fish community of nutrient-poor, oligotrophic lakes, while planktivorous fishes (e.g., shad, bream) become increasingly dominant with nutrient enrichment (Jeppesen et al. 1997). Thus, an alternative explanation for the lack of zooplankton control of cyanobacterial blooms could include consumption of zooplankton by planktivores.
Controls
Given the widespread extent of water quality degradation associated with nutrient enrichment, eutrophication has and continues to pose a serious threat to potable drinking water sources, fisheries, and recreational water bodies. Although many municipalities have passed legislation to regulate point-source loading of nutrients, eutrophication and cyanobacterial blooms are still prevalent in surface waters around the world (Smith & Schindler 2009). Predicted climate change and human population growth has the potential to further degrade water quality and quantity, and there is an immediate need by water resource managers to understand how to minimize the intensity and frequency of algal and cyanobacterial blooms (Paerl & Paul 2012).
Water resource managers routinely employ a variety of strategies to minimize the effects of cultural eutrophication, including (1) diversion of excess nutrients (Edmondson 1970), (2) altering nutrient ratios (Downing et al. 2001), (3) physical mixing (Huisman et al. 2004), (4) shading water bodies with opaque liners or water-based stains, and (5) application of potent algaecides and herbicides (Boyd & Tucker 1998). In general, these strategies have proven to be ineffective, costly, and/or impractical, especially for large, complex ecosystems (but see Edmondson 1970). Water quality can often be improved by reducing nitrogen and/or phosphorus inputs into aquatic systems, and there are several well-known examples where bottom-up control of nutrients has greatly improved water clarity. However, nutrient reduction can be difficult (and expensive) to control, especially in agricultural areas where the algal nutrients come from nonpoint sources. Furthermore, in lakes where external loading of nutrients has been reduced, internal loading of nutrients from sediments may prevent improvements in water quality (Søndergaard et al. 2003). The use of algaecides, such as copper sulfate, is also effective at reducing HABs temporally (Boyd & Tucker 1998). However, algaecides are expensive to apply, do not control the primary cause of the problem (i.e., abundant resources for primary producers) and pose risks to humans, livestock, and wildlife, in addition to harming a variety of non-target aquatic organisms.
Another alternative for improving water quality in nutrient-rich lakes has been biomanipulation - the alteration of a food web to restore ecosystem health (Shapiro et al. 1975). The basic premise is that secondary consumers (planktivorous fishes) are removed either through the addition of tertiary consumers (piscivorous fishes) or harvesting, which allows for the dominance of large-bodied, generalist grazers (e.g., Daphnia) to control phytoplankton (Figure 5). When planktivorous fishes are abundant and there is no predation refuge (e.g., oxygenated hypolimnion) for large-bodied zooplankton, less efficient small-bodied zooplankton grazers (e.g., rotifers and herbivorous copepods) typically dominate zooplankton communities, thus allowing for the overgrowth of phytoplankton (i.e., algal bloom). Many past studies have shown strong correlations between the size structure of zooplankton communities and phytoplankton abundance. These data support the notion that predatory top-down forces can have important implications for aquatic communities and ecosystems. With that said, fish-centric biomanipulation effects on water quality are typically short-lived (i.e., weeks to months), most obvious in small, easily-managed systems (i.e., ponds), and impacted by resource availability, namely phosphorus and nitrogen (Benndorf 1990; Carpenter et al. 1995).
Figure 5
Daphnia are generalist herbivores that play a powerful role in the structure and function of lake ecosystems and can have large effects on phytoplankton abundance, water clarity, and water quality when abundant.
© 2013 Nature Education Photo by Alan Wilson. All rights reserved. ;)
Conclusions
Despite dramatic improvements in water quality as a result of large-scale efforts to reduce nutrient enrichment (e.g., Clean Water and Safe Drinking Water Acts in the 1970s), cultural eutrophication and concomitant HABs continue to be the leading cause of water pollution for many freshwater and coastal marine ecosystems and are a rapidly growing problem in the developing world (Smith & Schindler 2009). Given that the demand for freshwater resources is expected to increase dramatically, protecting diminishing water resources has become one of the most pressing environmental issues and will likely become more complicated as climate change, species invasions, and pollution further degrade water quality and quantity (Schindler 2006). Control and management of cultural eutrophication is a complex issue and will require the collective efforts of scientists, policy makers, and citizens to reduce nutrient inputs, to develop effective, long-term biomanipulation techniques, and to eventually restore aquatic communities.
Glossary
anoxia: Lack of dissolved oxygen in water
biomanipulation: The alteration of a food web to restore ecosystem health Eutrophication - elevated primary production
HAB: Harmful algal bloom; abundant phytoplankton
hypoxia: Reduced dissolved oxygen concentration in water that stresses an organism
internal loading: Release of nutrients, such as phosphorus and nitrogen, from sediments during low oxygen concentration conditions
photosynthesis: Conversion of inorganic carbon (carbon dioxide) to organic carbon (glucose) by a primary producer
References and Recommended Reading
Arend, K. K. et al. Seasonal and interannual effects of hypoxia on fish habitat quality in central Lake Erie. Freshwater Biology 56, 366-383 (2011).
Benndorf, J. Conditions for effective biomanipulation - conclusions derived from whole-lake experiments in Europe. Hydrobiologia200, 187-203 (1990).
Boyd, C. E. & Tucker, C. S. Pond aquaculture water quality management. Norwell MA: Kluwer (1998).
Burkholder J. M. et al. New 'phantom' dinoflagellate is the causative agent of major estuarine fish kills. Nature358, 407-410 (1992).
Carpenter, S. R. Submersed vegetation: an internal factor in lake ecosystem succession. The American Naturalist118, 372-383 (1981).
Carpenter, S. R. et al. Biological control of eutrophication in lakes. Environmental Science & Technology29, 784-786 (1995).
Carpenter, S. R. et al. Nonpoint pollution of surface waters with phosphorus and nitrogen. Ecological Applications8, 559-568 (1998).
Chorus, I. & Bartram, J. (Eds). Toxic cyanobacteria in water: a guide to their public health consequences, monitoring, and management. London UK: E & FN Spon (1999).
Crews, J. R. & Chappell, J. A. Agriculture and natural resources U.S. catfish industry outlook. Auburn AL: Auburn University (2007).
Diaz, R. J. & Rosenberg, R. Spreading dead zones and consequences for marine ecosystems. Science321, 926-929 (2008).
Dodds, W. K. et al. Eutrophication of U.S. freshwaters: analysis of potential economic damages. Environmental Science and Technology43, 12-19 (2009).
Downing, J. A. et al. Predicting cyanobacteria dominance in lakes. Canadian Journal of Fisheries and Aquatic Sciences58, 1905-1908 (2001).
Edmondson, W. T. Phosphorus, nitrogen, and algae in Lake Washington after diversion of sewage. Science169, 690-691 (1970).
Francis, G. Poisonous Australian lake. Nature18, 11-12 (1878).
Huisman J. et al. Changes in turbulent mixing shift competition for light between phytoplankton species. Ecology85, 2960-2970 (2004).
Jeppesen, E. et al. Top-down control in freshwater lakes: the role of nutrient state, submerged macrophytes and water depth. Hydrobiologia342/343, 151-164 (1997).
Lehtiniemi, M. et al. Turbidity decreases anti-predator behaviour in pike larvae, Esox Lucius. Environmental Biology of Fishes73, 1-8 (2005).
Morris, J. G. Harmful algal blooms: an emerging public health problem with possible links to human stress on the environment. Annual Review of Energy and the Environment24, 367-390 (1999).
Paerl, H. W. Nuisance phytoplankton blooms in coastal, estuarine, and inland waters. Limnology and Oceanography33, 823-847 (1988).
Paerl, H. W. & Huisman, J. Climate change: a catalyst for global expansion of harmful cyanobacterial blooms. Environmental Microbiology Reports1, 27-37 (2009).
Paerl, H. W. & Paul, V. J. Climate change: links to global expansion of harmful cyanobacteria. Water Research46, 1349-63 (2012).
Porter, K. G. The plant-animal interface in freshwater ecosystems. American Scientist65, 159-170 (1977).
Schindler, D. W. Eutrophication and recovery in experimental lakes: implications for lake management. Science174, 897-899 (1974).
Schindler, D. W. Recent advances in the understanding and management of eutrophication. Limnology and Oceanography51,356-363 (2006).
Shapiro, J. et al. Biomanipulation: An ecosystem approach to lake restoration. In Water quality management through biological control (pp. 85-96). Eds. Brezonik, P. L. & Fox, J. L. Gainesville, FL: University of Florida (1975).
Smith, V. H. & Schindler, D. W. Eutrophication science: where do we go from here? Trends in Ecology and Evolution24,201-207 (2009).
Søndergaard, M. et al. Role of sediment and internal loading of phosphorus in shallow lakes. Hydrobiologia506-509, 135-145 (2003).
Tillmanns, A. R. et al. Meta-analysis of cyanobacterial effects on zooplankton population growth rate: species-specific responses. Fundamental and Applied Limnology171, 285-295 (2008).
Turner, A. M. & Chislock, M. F. Blinded by the stink: nutrient enrichment impairs the perception of predation risk by freshwater snails. Ecological Applications20, 2089-2095 (2010).
Wilson, A. E. et al. Effects of cyanobacterial toxicity and morphology on the population growth of freshwater zooplankton: meta-analyses of laboratory experiments. Limnology and Oceanography51, 1915-1924 (2006).
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2728 | https://joshuasiktar.medium.com/triangular-numbers-and-complete-graphs-4e42f6ba00ec | Triangular Numbers and Complete Graphs | by Joshua Siktar | Medium
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Triangular Numbers and Complete Graphs
Joshua Siktar
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Dec 11, 2021
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I had some extra time during my mathematical reasoning class this semester, so I gave my classes a brief lecture on the relationship between triangular numbers and properties of complete graphs. Students had seen both of these concepts, and some related ones, throughout the semester but not in this more integrated fashion. I also felt it would be helpful review for the final exam since I could touch on a bunch of things we had covered throughout the semester. With that in mind, I decided to turn the exposition into a blogpost.
What are triangular numbers?
Triangular numbers are a sequence of numbers that begin as follows:
As with any sequence of numbers this one goes on forever, in a never ending list, and hence the collection of triangular numbers are of size aleph naught.But how do we know what the next triangular numbers are? We need a formula for this task, and we can look for one of two types of formulas. A recursive formula is one that generates a term in a sequence by relying directly on one or more previous terms in the sequence. An example of a recursive formula for the triangular numbers is
On the other hand, a non-recursive formulais a formula where previous terms in the sequence do not come into play in calculating a term you want to know about. An example of a non-recursive formula for the triangular numbers is
The defining difference between these two formulas is that the non-recursive formula is a function only of n. You plug…
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Vet Ophthalmol
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. 2003 Dec;6(4):329-32.
doi: 10.1111/j.1463-5224.2003.00315.x.
The effect of topical administration of atropine sulfate on the normal equine pupil: influence of age, breed and gender
J L Davis1,T Stewart,E Brazik,B C Gilger
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1 Department of Clinical Sciences, College of Veterinary Medicine, North Carolina State University, Raleigh, NC, USA.
PMID: 14641831
DOI: 10.1111/j.1463-5224.2003.00315.x
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The effect of topical administration of atropine sulfate on the normal equine pupil: influence of age, breed and gender
J L Davis et al. Vet Ophthalmol.2003 Dec.
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. 2003 Dec;6(4):329-32.
doi: 10.1111/j.1463-5224.2003.00315.x.
Authors
J L Davis1,T Stewart,E Brazik,B C Gilger
Affiliation
1 Department of Clinical Sciences, College of Veterinary Medicine, North Carolina State University, Raleigh, NC, USA.
PMID: 14641831
DOI: 10.1111/j.1463-5224.2003.00315.x
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Objectives: The purpose of this study was to determine the influence of age, breed and gender on vertical pupil diameter (VPD) following a single dose of 1% atropine sulfate ophthalmic solution in the normal equine eye.
Animals studied: Thirty-two horses of various ages, breeds and genders were included. The horses had no history or clinical signs of ophthalmic disease. All horses studied had darkly pigmented irides.
Procedures: Two milligrams of 1% atropine sulfate ophthalmic solution was topically administered as a single dose in the right eye of each horse on Day 0. The VPD (mm) was measured in both eyes using digital calipers prior to treatment and every 24 h after administration for 2 weeks (Days 1-14). Duration of effect on VPD was then calculated for treated and untreated eyes. Data were also analyzed for effect of age, breed and gender on mean VPD, maximum VPD and time to maximum VPD.
Results: The VPD in the treated eye was significantly elevated compared to baseline measurements and compared to the untreated eye at all time points. Arabians had a greater mean VPD at Day 0 and on several days following treatment. Females had greater mean VPD compared to males on 5 out of 15 days.
Conclusions: Duration of mydriasis after administration of 1% atropine sulfate ophthalmic solution in the normal equine eye is greater than 14 days. Horses of the Arabian breed and female horses may be more sensitive to effects of cholinergic blockade in the eye.
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2731 | https://cims.nyu.edu/gcl/papers/tosun2007sou.pdf | Eurographics Symposium on Geometry Processing (2007) Alexander Belyaev, Michael Garland (Editors) Shape Optimization Using Reflection Lines E. Tosun, Y. I. Gingold, J. Reisman, D. Zorin New York University, USA Abstract Many common objects have highly reflective metallic or painted finishes. Their appearance is primarily defined by the distortion the curved shape of the surface introduces in the reflections of surrounding objects.
Reflection lines are commonly used for surface interrogation, as they capture many essential aspects of reflection distortion directly, and clearly show surface imperfections that may be hard to see with conventional lighting.
In this paper, we propose the use of functionals based on reflection lines for mesh optimization and editing. We describe a simple and efficient discretization of such functionals based on screen-space surface parameterization, and we demonstrate how such discrete functionals can be used for several types of surface editing operations.
1. Introduction Many human-made surfaces have highly reflective finishes: cars, kitchen appliances, lamps and jewelry are common ex-amples. The appearance of such objects is primarily defined by the reflections of other objects. Reflections are quite sen-sitive to surface shape and depend on local surface quantities (normals and curvatures) as well as viewer location.
Reflection lines are a widely used interrogation tool for sur-faces. Conceptually, these are obtained by computing reflec-tions of a set of long linear parallel light sources, aligned with a fixed direction. Visualizing such reflections make many types of surface irregularities apparent. Reflection lines can be thought of as a special type of reflected envi-ronment, capturing the distortion introduced by the curved shape of the surface for a particular direction of features in the environment. Reflection lines are widely used in the automotive industry, and were also found to be useful in biomedical engineering as a tool for cornea shape recon-struction [HBKM96]. Advances in graphics hardware made interactive reflection line rendering widely accessible and easy to implement.
Figure 1: An example of reflection line optimization.
The process of evaluating surface quality, for which reflec-tion lines are most commonly used, is complimentary to shape design. In most cases, the designer defines the sur-face by manipulating spline or subdivision control points, or other types of shape controls, then evaluates the quality using interrogation tools and repeats the process until the desired quality is achieved. The controls of the shape have an indirect effect on the quality measure and in the case of reflective surfaces, it may be hard to guess how the shape should be modified to achieve a desired effect. A common alternative is to formulate the surface editing problem as an optimization problem minimizing a quality functional mea-suring deviation from desired behavior.
Reflection lines provide a convenient framework for build-ing such functionals for reflective surfaces. For reasons we discuss in Section 2, arbitrary manipulation of reflections is not possible. Furthermore, recovering the surface from an arbitrarily chosen distortion of a reflected image is not al-ways possible either. However, in most cases, it is possible to find a surface producing a given pattern of reflection lines.
By choosing a reflection line direction, the user chooses what feature direction in the environment can be considered most important. For example, horizontal and vertical lines are most common in urban and indoor environments, and it makes sense to use these directions for surface optimization.
In this paper we present a system for interactive surface modeling based on reflection line manipulation. We show how to discretize reflection lines on arbitrary meshes, and demonstrate that a relatively simple discretization is suffi-cient for shape modeling purposes, provided that a suitable normal estimation algorithm is used. We propose a numer-ical technique for solving the problem at rates adequate for c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines interactive surface manipulation. Our approach is based on two insights. First, an arbitrary mesh can be locally param-eterized over the image plane away from silhouette edges; this makes it possible to reduce the number of degrees of freedom used in optimization, and greatly simplifies ex-pressions. Second, we observe that a simple triangle-based discretization of second-order quantities using only vertex degrees of freedom can be used to compute second-order derivatives of the surface parameterization, which leads to a fast and efficient matrix assembly. We demonstrate how re-flection line manipulation can be used to smooth and warp reflection lines, change reflection line density, and create sur-faces with a desired reflection line pattern.
Related work. Different types of reflection lines were extensively studied in the geometric modeling literature.
Some of the earliest work is described in [Kla80], where a differential-geometric description of reflection lines and an analytic expression for the variation of these lines is de-rived; [KK88] describes a technique for adjusting families of spline curves defining a surface based on reflection line changes. Reflection lines were also used as a surface inter-rogation tool in [Poe84, HHS∗92, GLW96, GOZ95, TF97].
Theisel has shown that isophotes and reflection lines can be viewed as subclasses of a more general class [The01].
Functionals similar to reflection-line functionals are com-mon in shape-from-shading vision literature (e.g. [Hor86]); however, the goal there is to reconstruct an unknown surface entirely from possibly noisy image data, rather than modify an existing mesh.
Surface fairing functionals based on characteristic line pat-terns are explored in [HBKM96] and [LGS99] (reflection lines) and [YBP97, CF98] (highlight lines). A technique for manipulating reflection line shape directly is described in [DB04]. In all cases, these techniques were applied to NURBS surfaces. The formulation we use is closest to [LGS99], which applies reflection line optimization to B-spline height fields; we apply a similar formulation to gen-eral meshes and present algorithms that allow us to achieve interactive performance.
Various types of fairing functionals for surfaces and their discretizations for spline surfaces, subdivision surfaces, and meshes are considered in [CG91,WW94,Kob00,SK01] and many other papers. Fairing functionals are used increasingly in interactive surface deformation settings, in particular for general meshes (e.g. [BK04]; see [BS07] for an excellent survey). An efficient, robust and accurate discretization of the Hessian of the function defined on an arbitrary mesh is central to our work. This problem is closely related to the problem of defining shape operators on meshes; discrete ge-ometry approaches (e.g. [PP93, MDSB03, CSM03]) play an important role in making variational techniques for meshes sufficiently fast for interactive applications. [GGRZ06] con-tains a detailed survey of different types of approaches; our approach builds on [GSH∗04].
Using user-defined reflection fields for surface optimization is similar to gradient and Laplacian deformation techniques ( [SLCO∗04], [YZX∗04]) in that the optimization functional depends on the initial mesh geometry (in our case, through the reflection function).
2. Reflection functionals In this section we present the relevant basic mathematics of reflections and functionals based on reflection lines. The formulations we use are similar to the ones that were used in [LGS99] for optimization of reflection lines of tensor-product B-spline height fields.
2.1. Reflection line function We consider a somewhat simplified formulation of reflec-tion lines, with both the viewer and the light sources located at infinity. The reflection line pattern in our model is created by long line-shaped light sources aligned with a unit length vector a (Figure 2). Each light source can be identified by a direction in the plane perpendicular to a. If we fix a zero di-rection, each direction corresponds to an angle θ in the range −π ...π. For a point p of a surface, let n be the normal, and let v be the view direction, which we assume to be nonparal-lel to a. In this notation, the reflection direction at p is given by r = (2/|n|2)((n·v)n−v). (We do not assume the normal to be unit length).
Figure 2: Vectors used in the definition of the reflection line function θ.
We define the reflection line function to be a scalar function on the surface which assigns to each point the angle θ be-tween a zero direction and the direction d to the linear light source corresponding to reflected direction r. This direction is obtained by projecting r to the plane P perpendicular to a: d = r −(r · a)a. Let va be the projection of the viewing direction v to the plane P, and let a⊥be perpendicular to va in the plane P, va = [v−(v·a)a]norm, a⊥= a×va where [·]norm denotes normalization. We use va as the zero direction; in this case the reflection line function is given by θ = arctan((r·a⊥),(r·va)) (1) where arctan(y,x) produces values in the range −π ...π. A reflection line is defined by a constant θ value. As reflection lines are view-dependent, it makes more sense to consider them as functions on the image plane, rather than the sur-face itself. The function is defined everywhere except at the points where r is parallel to the light direction a. The gradi-ent of θ is of primary importance: the direction of the reflec-tion lines is perpendicular to ∇θ. (Note that |∇θ| measures the local density of reflection lines.) c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines Coordinate formulation. One of the properties distin-guishing the reflection line optimization from most fairing problems is that there are fixed spatial directions a and v which are a part of the problem formulation. Projections to these directions are natural choices of variables. We observe that for silhouette points, for which the normal is perpen-dicular to the view direction, perturbing the surface does not affect the reflection line function corresponding to these points; i.e. one cannot optimize the lines near the silhouettes without moving the silhouettes which is best done by tech-niques of the type described in [NSACO05]. This suggests that projection to the image plane leads to the natural param-eterization for the problem, as in this case silhouette points will form boundaries for optimization regions.
We choose the coordinate system aligned with the image plane (a⊥, av, v), where av is the normalized projection of a to the plane perpendicular to v (Figure 2). The coordinates along the three axes are x,y,z: we use the standard conven-tion for y to be perpendicular to the image, x to be horizon-tal, and z to be the view direction. For a vector t, we denote (t·a⊥) = tx, (t·av) = ty, and (t·v) = tz.
Using this notation we reduce the components of our ex-pression to r1 = (r · a⊥) = 2nznx and r2 = (r · va) = −2nzny sinα +(nz2 −nx2 −ny2)cosα, where α is the angle between v and va.
If we regard the surface as locally parametrized over the im-age plane, i.e. given by z = f(x,y), we take the non-unit length normal to be n = (fx, fy,1), where fx and fy are derivatives of f in two directions, and the expression for θ = arctan(r1,r2) further simplifies to θ(x,y) = arctan 2 fx,−2 fy sinα +(1−f 2 x −f 2 y )cosα .
(2) 2.2. Optimization problems Given a user-defined reflection function our goal is to deter-mine a surface which approximates this field as closely as possible. Mathematically, we can formulate the problem in several ways: exact match of the reflection function, mini-mization of the difference between the desired and actual re-flection function, and minimization of the difference in line directions and density captured by the gradient of the reflec-tion function (cf. [LGS99]). We briefly review these options here, with emphasis on the allowable boundary conditions.
One can observe that if θ(x,y) is given, Equation 2 is a first-order PDE which can be solved using the characteristic ODE system ˙ x = Fpx, ˙ y = Fpy, ˙ px = −Fx, ˙ py = −Fy, where F(x,y, px, py) = (1 −f 2 x −f 2 y )tanθ∗(x,y) −2 fx. With suit-able assumptions on the left-hand side and boundary con-ditions, the solution exists. However, as the system is first-order, only initial value problems generally have solutions.
In particular one cannot expect the problem to have solu-tions if the values are prescribed on the boundary of a patch.
We also note that if instead of specifying θ we had speci-fied the reflection vector r, the resulting system of two PDEs would not necessarily have a common solution even with no boundary conditions.
If the boundary of a region is fixed, the best we can do is to minimize the difference in reflection functions. Instead of fitting the angle values θ(x,y), we avoid the problems with the discontinuity of θ values by using the functional Z S(cosθ −cosθ∗)2 +(sinθ −sinθ∗)2dxdy, (3) where the integral is over the image plane projection of the region of interest, with projection assumed to be one-to-one.
The Euler-Lagrange equation for this problem is second-order; therefore one hopes to be able to solve the problem with Dirichlet data on the boundary, but not with Neumann data. This implies that one cannot expect the solution to blend smoothly with the rest of the surface if the optimiza-tion is performed only on a small area.
Finally, instead of fitting the function values one can fit the gradient of the reflection function to the gradient of the de-sired function: Minimize Z S(∇θ −∇θ∗)2dxdy, θ|∂S = θ0, ∂ ∂nθ|∂S = ϕ0, (4) where ∂/∂n is the derivative along the boundary normal.
In this case, the corresponding Euler-Lagrange equation is fourth-order, similar to the PDE for the thin-plate energy, and one can prescribe both Dirichlet and Neumann bound-ary conditions ensuring smooth transition between the op-timized patch and the surface. In this energy, we need an expression for ∇θ. As r1 = 2 fx and r2 = −2 fy sinα + (1 − f 2 x −f 2 y )cosα, ∇θ = r2∇r1 −r1∇r2 r2 1 +r2 2 All problems that we have considered assume that a reflec-tion line function θ(x,y) is prescribed.
3. Discretization and numerical methods We aim to design a discretization of problem 4 which bal-ances accuracy, robustness and efficiency required by inter-active applications.
A C1 finite-element or arbitrary mesh C1 spline discretiza-tion would be most straightforward but is relatively ex-pensive. We use a more efficient and easier to implement alternative which combines a discrete geometric approach with finite differences. We use triangle-centered discretiza-tion stencils for both first and second-order derivatives which leads to simple discretization of Equation 4. While there is no rigorous convergence guarantee by construction, we show excellent behavior for most mesh types.
c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines Reduction to parametric case. As it was discussed in the previous section, our optimization problems can be solved in a functional setting by using surface parameterization over the image plane. However, while obtaining such a parame-terization is computationally expensive for high-order sur-faces (e.g. subdivision surfaces and splines); for meshes, the parametrization is easily obtained by a simple linear trans-formation: we rotate the coordinate system so that the image plane coincides with the (x,y) plane and the projection of the light direction a to the image plane is aligned with the y axis, i.e. use the coordinate system (v,av,a⊥) (Figure 2). As one cannot reliably control reflections near silhouette points, we fix all vertices close to silhouettes, i.e. fix vertices of all tri-angles with normals n, for which |n·v| < ε (we use ε = 0.02 in all cases).
For surfaces with no silhouettes (e.g. nearly flat patches) additional boundary conditions are necessary: typically, we want the modified surface patch to join smoothly with the rest of the surface. After preprocessing, the mesh is decom-posed into disjoint pieces, each of which is a piecewise linear height field over the image plane.
Our functionals depend on the components of the gradi-ent and Hessian of f, which we discretize next. We use triangle-centered discretization, i.e. a single value of the gra-dient or Hessian is assigned to each face rather than vertex.
This leads to simple formulas for the gradients and Hes-sians, and makes it possible to consider a minimal num-ber of special cases. Each discretization associates a 2 × 6 and 3×6 matrix of coefficients G and H with each triangle.
if fT = (f(p1), f(p2), f(p3), f(q1), f(q2), f(q3)) (Figure 3) then G fT and H fT yield the gradient and Hessian respec-tively.
Discretizing gradients. To discretize the gradient ∇f = (fx, fy) over the image plane, we use standard piecewise-linear continuous finite elements.
Figure 3: Vectors used in the gradient and Hessian defini-tions; all points are in the image plane. The vectors ti j are perpendicular to corresponding triangle sides and have the same length as these sides.
We observe that the gradient can be found in the form ∑i citii where ci are determined by ∑i ci(tii · v j) = f j −fk, and (i, j,k) is a cyclic permutation of (1,2,3). This yields ∇discr fT = 1 2A ∑ i=1,2,3 f(pi)tii (5) where f(pi) denotes the value of f at vertex pi. The coef-ficients tii/2A do not change and need to be computed only when the surface is rotated.
Discretizing Hessians. Discretizing Hessians is consider-ably more difficult: while for gradients a piecewise linear approximation depending only on function values at triangle vertices is adequate, for second derivatives one needs to use more vertices, or introduce additional degrees of freedom.
As the total number of derivatives of order ≤2 is six, one needs at least six degrees of freedom per stencil to capture local behavior correctly.
Most discretizations of second-order quantities (typically, curvature) used in geometric modeling are vertex-centered, which is inconvenient for our purposes. To be compatible with the gradient discretization, we use a triangle-centered stencil shown in Figure 3. † For triangles without vertices of valence three, it has six de-grees of freedom, exactly the number needed for discretizing the Hessian.
On this stencil, an approximation to the Hessian can be con-structed in a number of different ways. We use a combination of two approaches.
Triangle-averaged discretization. Cohen-Steiner and Morvan [CSM03] describes a general approach to comput-ing shape operators for meshes by averaging elementary shape operators corresponding to edges. While convergence of this technique was only established in the integral sense, and for a restricted class of meshes, simplified versions of this approach were shown to work well in practice.
The most common example is the well-known cotangent formula [PP93], which (for small deformations) is equiv-alent to expressions of [CSM03] summed over a single ring of edges around a vertex [HPW05]. Similarly, the triangle-averaged discretization on the stencil of Figure 3 introduced in [GSH∗04] uses averaging over three edges of a triangle. By linearizing this formula, we obtain the following expression for the Hessian: 1 A ∑ i,j,j̸=i 1 A j f(qj)tii ⊗ti j +∑ i 1 Ai f(pi)tii ⊗tii !
(6) † Another possible option is to use a single triangle and add edge-based degrees of freedom as it was done in [GGRZ06]. We have experimented with a linearized version of this discretization. In con-trast to the general curvature discretization (3 coordinates per ver-tex), addition of edge degrees of freedom in our setting (one degree of freedom per vertex) adds a significant computational cost. Fur-thermore, stability of the nonlinear solve is decreased which further decreases performance.
c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines where tij are side perpendiculars to the triangles of the mesh projected to the image plane, shown in Figure 3. A distinc-tive feature of this discretization is its robustness and sim-plicity: only for triangles with very small area may the coef-ficients in the formula be large.
This is a consistent (converging to the correct values) dis-cretization of the Hessian for special types of meshes.
Specifically, the Hessian is consistent for meshes in which vertices qi are reflections of pi with respect to the centers of opposite edges i = 1,2,3. This includes regular meshes and any affine transformations of regular meshes.
For general meshes, the discretization introduces mesh-dependent error in Hessian approximation, as shown in Fig-ure 5. For different types of meshes, the results are mesh-dependent, no matter how fine the mesh is. Importantly for our application, the errors are low-frequency while high-frequency errors have the most effect on the visual quality of results.
regular 4-8 polar distort. half 3-12 irregular Figure 4: Mesh types used in convergence experiments.
Figure 5: Convergence experiments: A spherical surface patch was recovered from an analytically computed reflec-tion function gradient for different mesh connectivities and resolutions. Three discretization types are shown: triangle-averaged, quadratic fit and hybrid. Quadratic interpolation cannot be applied to meshes with vertices of valence 3; op-timization also fails on higher resolution irrregular meshes because it contains stencils with all vertices close to a conic.
The error is measured relative to the size of the object along the view direction.
Quadratic interpolation discretization. An alternative is to use a finite-difference approach to discretization: we compute a quadratic function Q satisfying Q(pi) = f(pi), Q(qi) = f(qi), i = 1...3, and use its quadratic term coef-ficients to estimate the Hessian. The advantage of this ap-proach is that by construction it is consistent whenever the quadratic function is defined. This is not sufficient for con-vergence of the discrete problem solutions to the continu-ous solution, (see [GGRZ06]) but improves independence of the result from mesh connectivity. Unfortunately, the ap-proach is significantly less robust and the following proposi-tion holds: For six points wi ∈R2, i = 1...6, there is a unique quadratic function satisfying Q(wi) = zi, for arbitrary choice of zi, if and only if these six points are not on the same conic (see Appendix.) Whenever six points of the stencil are close to a com-mon conic, the coefficients of the quadratic interpolant become large and Hessian estimation becomes highly unreliable.
Hybrid discretization. As it can be seen from the plots above, quadratic interpolation yields good estimates in most cases, but it is not robust. In practice, we observe that we have several triangles per mesh for which the six-point sten-cil is close to a conic, and quadratic interpolation produces low-quality results. To solve this problem, we combine two techniques: the triangle-averaged scheme is used when the quadratic interpolation is unstable, i.e. if the stencil contains five points or less, or vertices are close to a conic. We eval-uate stability for a specific six-point stencil by comparing the magnitude of the discrete Hessian coefficients to 1/l2 max, where lmax is the maximal edge length in the stencil. If any coefficient exceeds C/l2 max (we use C = 5), we use triangle-averaged discretization instead of the quadratic interpola-tion. As it can be seen from the convergence plots, the re-sulting scheme retains the accuracy of the quadratic fit and yet does not suffer from its robustness problems, although it produces large errors for meshes with many degenerate cases. Such meshes appear to be unusual. Our stability crite-rion is motivated by the observation that if the function value is 0 at all but one vertex of a stencil, and is of the order l2 max (i.e. squared distance to other vertices) at that one vertex, one can expect to get second derivative magnitudes on the order of 1 (for a mesh close to regular). Coefficients much larger than 1/l2 max lead to instability.
While numerically the discretization is more accurate, we note that we have observed few differences in visual quality when using the triangle-averaged discretization alone. Fi-nally, while we found this discretization adequate for the functionals considered in this paper, its performance for thin-plate or Willmore energy is not as good ( [JR07]).
Discretizing normals. An essential part of an interactive system supporting reflection manipulation is rendering of reflection lines and environment maps. Hardware environ-ment maps use vertex normals to compute reflected direc-tions and look up values in the environment texture. As Fig-ure 6 shows, standard vertex normal computation techniques c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines produce low-quality results for complex meshes. Instead, we use a local fit to obtain better normals. For every vertex p, we collect a ring N1 of triangles around it, and all triangles edge-adjacent to N1. The minimal number of vertices in such a configuration is six, unless the whole mesh has a smaller number of vertices. We compute an initial normal ninit, and project vertices to the plane perpendicular to ninit.
Let w be the vector of projected point positions of length K, and let f(w) be the vector of function values at these points. For each vertex, we precompute a 2 × K matrix of coefficients Cnorm mapping the vector f(w) to the linear co-efficients of a best fit quadratic function in the coordinate system with origin at vertex p, with z axis aligned with the initial normal (when it is not defined uniquely, we choose the quadratic function with minimal norm of coefficients; the co-efficients are computed using LAPACK function GESDD).
These coefficients define a plane passing through the ver-tex, and we use the normal to this plane as our final normal.
As long as we do not change the plane we have used for local normal estimation, the normal approximation can be recomputed rapidly as the surface is modified. If the motion is large, the quality deteriorates, and ninit needs to be recom-puted.
analytic averaged face quadratic fit Figure 6: Comparison of the vertex normal quality; the surface is obtained by sampling points from a cylinder.
While face averages do not perform well for this mesh, our quadratic fit procedure yields results visually indistinguish-able from using analytic normals.
3.1. Numerical implementation The reflection-based functionals are more complex than most expressions commonly used for surface optimization, therefore computing Hessians and gradients are also more expensive. The energy cannot be replaced by a linearized functional, because then it would not capture the shape of reflection lines in most cases. Either a full non-linear New-ton solve for the energy minimum or a gradient-only method would be prohibitively expensive, the former due to Hessian computation, and the latter because of the large number of iterations required.
To improve performance we use an inexact Newton method with line search. Instead of a full Hessian computation, we compute the Hessian once for the linearized problem, and use it instead of the full Hessian optimization at all iterations.
If we assume small values of fx and fy a simple calculation shows that the equation for the reflection function reduces to θlin(x,y) = 2 fx cosα.
The gradient of θlin remains simple, and is, up to a constant, [fxx, fxy]. For the quadratic energy based on θlin, the Hessian is easy to compute and does not depend on function values.
As the gradient problem is fourth-order and the condi-tion number of the Hessian matrices grows as N4, iterative solvers are not efficient; instead, we use a direct solver (PAR-DISO). The direct solver performs a sparse LU factorization of the matrix, and solves the system using backsubstitution.
As the Hessian matrix does not change, the matrix factor-ization needs to be performed only once. For each nonlinear iteration, only gradients need to be recomputed. As shown in Figure 7, while using an approximate Hessian increases the number of nonlinear iterations required, each iteration becomes much faster, and there is a considerable net win in performance. This performance improvement depends on the complexity of the target reflection function gradient ∇θ∗ and varies in the range 2× to 10×. When the target function is smooth, which is the most common case (forward opti-mization in Figure 7), the speedups are higher, while for less smooth targets speedups are lower.
Figure 7: Speedups from approximate Hessian computa-tion, dependence on the mesh size. Two model problems: creation and elimination of a bump on a cylinder. Times are given for a Pentium D 3GHz processor.
4. Reflection line manipulation experiments Different types of reflection manipulation using discrete re-flection line functionals follow the same general pattern.
First, the user selects an area to modify and specifies the boundary conditions. Any boundary segment may be free, fixed, or clamped, the latter means that two layers of ver-tices are fixed at the boundary to ensure a smooth match of c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines the surface with the rest of the mesh. Then the target reflec-tion function gradient is defined using user input and the sur-face minimizing the reflection functional is computed. The last two steps may be repeated in the interaction loop.
Density and direction change. The simplest type of reflec-tion line manipulation is requesting a fixed line direction and density in an area, i.e. specifying a fixed target ∇discrθ∗ everywhere. The energy minimization in this case attempts to modify reflection lines as requested, while maintaining a smooth or continuous join of the selected patch with the rest of the surface. The user can also adjust a fall-off curve c(t) ≥1 which determines how gradual the transition be-tween the modified area and the rest of the mesh should be.
The energy of each triangle is scaled by c(d) where d is the distance to the center of influence.
Examples of this type of manipulation are shown in Figure 9, and Figure 11. In Figure 9, one can see how adjusting reflec-tion line density makes it possible to control the appearance of reflections, in particular the sizes of reflected objects. Ro-tating the desired reflection line direction at a point is shown in Figure 9 on the right. In this example, a surface imperfec-tion creates a “twist” in the reflection line field which can be removed by local rotation. Figure 11, demonstrates the dif-ference between conventional smoothing and reflection line optimization: typically, smoothing algorithms flatten the sur-face which does not necessarily improve the reflection line shape.
Smoothing. Isotropic or directional smoothing can be used to transform initial ∇θ values to the target values. This op-eration is similar to reflection line smoothing described in [LGS99]. Directional smoothing is particularly useful, as it straightens reflection lines without changing the behavior in the orthogonal direction. An example is shown in Fig-ure 12, left. In this example, the initial reflection function was smoothed using Laplacian smoothing in the image do-main, and then used as the target reflection function.
initial surface warp target gradientsoptimized surface Figure 8: Warping stages: initial reflection lines, warped re-flection function θ, target ∇θ∗computed per triangle, reflec-tion lines on the optimized surface.
Warping. In the case of warping, the goal is to apply an ar-bitrary user-specified transformation to the reflection lines.
In our current implementation, the transformation of reflec-tion lines is specified by a two-dimensional spline, however any other image warping technique can be used. The general formula for the transformed reflection function is θwarp(s,t) = θinit(w−1(s,t)) where w(s,t) : R2 →R2 is the warping function.
Observe that the inverse of the warping function needs to be computed, which can be relatively expensive for smooth de-formations. To avoid explicit inversion of w(s,t), we imple-ment the warp using texture mapping and image-domain op-erations. Rather than attempting to transform ∇discrθ values needed by the functional directly, we transform reflection function values, and then compute the gradient. First, the re-flection function θ values are computed, interpolated to ver-tices, and used as color values to render the mesh to a texture θinit. A two-dimensional spline w(s,t) : R2 →R2 is created.
Initially, the control points are equispaced so w is an identity.
As the user moves control points, the spline is rendered to a new texture θwarp, using θinit as a texture map. For image-space mesh vertex positions we sample ∆θ = θwarp −θinit, and compute its gradient for each triangle using the gradi-ent formula 5, with appropriate corrections applied to values to eliminate the jump between −π and π. We add resulting ∇discr∆θ values to the initial ∇discrθ values. Note that by construction ∆θ = 0 if w is the identity function. An exam-ple of warping is shown in Figure 10; where the goal is to improve the shape of reflections at a part of a car hood.
Finally, one can use our technique to create surfaces approx-imating an arbitrary reflection line pattern, as shown in Fig-ure 12, right. Any grayscale image can be used to specify ∇θ∗, as long as it is sufficiently smooth (otherwise, the ap-proximation is likely to be poor).
5. Conclusions We have described a simple and efficient approach to dis-cretizing reflection line based functionals on meshes and demonstrated how these functionals can be used in an in-teractive system to optimize the shape of reflective surfaces.
One limitation of the proposed approach (which is also re-sponsible for its comparatively high efficiency) is that the vertices of the mesh move only in the direction perpendicular to the image plane. This means that small scale surface de-tails which make the projection to the image plane not one-to-one cannot be removed, and creates a disturbance in the surface during optimization. Although it can be applied to large perturbations, the technique is best suited for smaller adjustments of surfaces that are already relatively smooth.
While we tried to eliminate obvious inefficiencies in our im-plementation, our code is far from optimal.
Our discretization can be easily combined with other surface optimization techniques, to be applied simultaneously or as a postprocess.
Acknowledgements. The authors would like to thank Robb Bifano, Jeff Han, and Harper Langston for their help with the paper and video production. This work was partially supported by the award NSF CCR-0093390, IBM Faculty Parternship award and a Rudin Foundation fellowship.
Appendix: Quadratic interpolation on a six-point stencil The quadratic interpolation problem for a six-point sten-cil requires solving a linear system, which may be singu-c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines decrease density increase density before after Figure 9: Left: Changing reflection line density. A fixed reflection line direction is specified with density decreased at the top and increased at the bottom. Right: reflection line untwisting. The reflection function gradient is rotated to get desired appearance.
before after Figure 10: Reflection line warping on a car hood. An intermediate warp is shown in the middle. Note that the change in the shape is barely perceptible but the change in the reflection is substantial.
c ⃝The Eurographics Association 2007.
E. Tosun, Y. Gingold, J. Reisman, D. Zorin / Shape Optimization Using Reflection Lines Figure 11: Prescribing fixed reflection line direction on a car, compared to Laplacian smoothing. Note that Laplacian smooth-ing retains reflection line wiggles.
before after Figure 12: Left: Reflection line smoothing on a faucet; the initial reflection line function is smoothed and used as the target reflection line function. Right: Reconstructing a surface with predefined reflection line pattern based on an blurred image.
lar. A simple geometric condition for testing singularities of point configurations can be derived from geometric consid-erations, along with an explicit expression for the quadratic interpolant, which simplifies implementation.
We denote the six points of the stencil in the image plane by wi, i = 0...5. We need to find a quadratic function Q, such that Q(wi) = zi, i = 0...5. Assume that all points are given in homogeneous coordinates. Then we can define Q as a 3×3 matrix: Q(wi) = wT i Qwi. We observe that by linear-ity of the problem it is sufficient to solve it for zi = δi j, for j = 0...5, to obtain six basis matrices Q j. Assume j = 0, i.e. zi = 0 for i ≥1. It well known that any five points in the plane are on a conic. Note that the problem of finding Q0 is equivalent to the problem of finding such conic: Q0 vanishes at all points wi, i ≥1, i.e. defines a conic passing through these points, and conversely given a nontrivial conic with matrix M passing through wi, i ≥1, we obtain Q0 as M/(wT 0 Mw0). If wT 0 Mw0 = 0, either there are multiple con-ics passing through the points, or the system has no solution; in either case, the system for Qi is singular.
The conic matrix M can be computed using a matrix form of the Braikenridge-Maclaurin construction, [CG67]. Specifi-cally, define li,i+1 = wi × wi+1 (3d cross product applied to the homogeneous point representation). Let R(a) be the skew symmetric matrix satisfying Rx = a × x. Then M0 is given by R(w1)R(l34)R(r)R(l23)R(w5), where r = l12 × l45 and Qi are obtained by cyclically permuting wi and applying the same formula.
c ⃝The Eurographics Association 2007.
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2732 | https://arxiv.org/pdf/2401.16768 | arXiv:2401.16768v2 [math.CO] 3 Jul 2025
The transversal game
ˇZarko Randelovi´ c (July 2025)
Abstract
Two players take turns claiming empty cells from an n × n grid. The first player (if any) to occupy a transversal (a set of n cells having no two cells in the same row or column) is the winner. What is the outcome of the game given optimal play? Our aim in this paper is to show that for n ≥ 4 the first player has a winning strategy. This answers a question of Erickson.
1 Introduction
Suppose that player 1 and player 2 play a game on an n × n grid. They play alternating moves. Player 1 goes first. Each player when it is their move writes their letter (say
X for player 1 and O for player 2) in one of the unoccupied cells of the grid. The first player to occupy a set of n cells having no two cells in the same row or column (meaning that those n cells only have the letter of that particular player) is the winner. If neither player does that and the grid is filled the game is a draw. Such positional games have been studied widely (some examples include , , ). Our aim is to prove that player 1 wins if n ≥ 4 (it is easy to check that it is a draw for n = 2 and n = 3). This question was posed by Erickson . We call the above described game the transversal n-game
where n × n is the size of the grid.
Theorem 1. For n = 1 and n ≥ 4 player 1 wins the transversal n-game.
We say that a transversal is a set of n cells on an n × n grid having no two cells in the same row or column. We will also use [ k] to denote the set {1, 2, .., k } for any k.When considering an n × n grid we will label the cells of the grid with ordered pairs (a, b ) with 1 ≤ a, b ≤ n. The cell ( a, b ) will be the cell in the a-th row from top to bottom and the b-th column from left to right (similar to the labeling in matrices). We will give an example of a game for n = 4. We will have player X (i.e. player 1) make his first move in the cell (1 , 1). Then player O (i.e. player 2) will make his first move in (2 , 2). Then player X plays in (2 , 3) trying to make a threat on the next move. Then player O plays in (4 , 4) which blocks a potential threat and gives O a possible chance to make his own threat on the next move depending on what X plays. Now X
Mathematical Institute of the Serbian Academy of Sciences and Arts, Kneza Mihaila 36, Belgrade 11000, Serbia. Email: zarko.randjelovic@turing.mi.sanu.ac.rs
1plays in (4 , 2) making a threat on (3 , 4). Now, O defends by playing in (3 , 4). Now X
plays in (3 .1) making a threat on (1 , 4). Since O has no threat, O must play in (1 , 4). Now X plays in (2 , 4) and makes a threat on both (1 , 3) and (3 , 3) (labeled W below). Since O still has no threats and cannot defend both of X’s threats, X will win on the next move. Subscripts below indicate which move it is for that player.
X1 W O4
O1 X2 X5
X4 W O3
X3 O2
We give a brief outline of the strategy. In the proof of Theorem 1 we will have player
X making a threat as early as possible. From then on the key point is for player X to make moves that help to achieve his goal but without allowing player O to make threats. Player X will start making threats as soon as he has occupied n − 1 cells of the grid and will continue to make new threats forcing O to block those threats. We mention that in the weaker Maker-Breaker version of this game (where X wins if he occupies the cells of a transversal and O wins as long as X does not make a transver-sal) X wins if n ≥ 4. Indeed, this may be proved by induction. If X can win for n ≥ 4while making the first move in the top-left cell then if we have an ( n + 1) × (n + 1) grid suppose X plays on (1 , 1). If O plays in row 1 or column 1 then X can win by induction after removing the first row and column. Otherwise if O plays in ( a, b ) with a, b > 1 then we can have X playing in ( a, c ) for c̸ = 1 , b . Now after removing row a and column c X
can win by induction since he has already occupied the top-left cell. One can check that
X can win with starting move (1 , 1) for n = 4 by a case analysis.
2 Proof of Theorem 1
We will say that the value of a position is ‘player 1’ if the game is a player 1 win, ‘player 2’ if a player 2 win, and ‘draw’ if it is a draw (all with perfect play of course). Also we say that the value of a game is the value of its starting position. Permuting rows or columns also permutes the transversals. This means that permuting rows or columns does not change the value of a position. Notice that reflection with respect to the main diagonal of the grid (i.e. cells ( i, i )) also does not change the value of a position. We define good
transformations of the grid to be any combination of reflecting with respect to the main diagonal and permuting rows or columns. Let us say that regardless of whose move it is a player has a threat on cell ( a, b ) if (a, b ) is empty and they would win on their next turn (if they have one) by playing on (a, b ). If a player can not win on their next move we say that that player has no threats. For completeness, and to give the reader a feel for the game, we show that for n = 3 the value is a draw.
Proposition 2. The value of the transversal 3-game is a draw.
2Proof. We will show that both X and O can guarantee at least a draw. It is easy to see that X can guarantee at least a draw by a strategy stealing argument, similar to Lemma 4 in . Now we show that O can guarantee a draw. By permuting rows and columns every opening move for X is equivalent so we may assume that X plays on (1 , 1). Then O plays on (2 , 2). Now by symmetry with respect to the main diagonal we may assume that X
plays on either (1 , 2) , (1 , 3) , (2 , 3) or (3 , 3). We deal with each case individually. If X plays on (1 , 2) then O plays on (2 , 3). Now if X does not play on (2 , 1) then O
plays there next which blocks off a row, but if X does play on (2 , 1) then O plays on (3 , 3). Now again X has no threats and if X does not play on (1 , 3) then O can block off a column but if X does play on (1 , 3) then O plays on (3 , 2) which blocks all transversals.
X1 X2 X4
X3 O1 O2
O4 O3
If X plays on (1 , 3) then O plays on (2 , 3). Now if X does not play on (2 , 1) then O plays there next which blocks off a row but if X does play on (2 , 1) then O plays on (3 , 2). Now again if X does not play on (1 , 2) then O can block off a column but if X does play on (1 , 2) then O plays on (3 , 3) which blocks all transversals.
X1 X4 X2
X3 O1 O2
O3 O4
If X plays on (2 , 3) then O plays on (3 , 2). Now if X does not play on (1 , 2) then O plays there next which blocks off a column but if X does play on (1 , 2) then O plays on (3 , 1). Now again if X does not play on (3 , 3) then O can block off a row but if X does play on (3 , 3) then O plays on (2 , 1) which blocks all transversals.
X1 X3
O4 O1 X2
O3 O2 X4
Finally if X plays on (3 , 3) then O plays on (3 , 2). Now if X does not play on (1 , 2) then
O plays there next which blocks off a column but if X does play on (1 , 2) then O plays on (2 , 1). Now again if X does not play on (2 , 3) then O can block off a row but if X
does play on (2 , 3) then O plays on (3 , 1) which blocks all transversals.
X1 X3
O3 O1 X4
O4 O2 X2
So player O can always guarantee at least a draw. □
Now we move onto Theorem 1 where n ≥ 4.
Proof of Theorem 1. The case n = 1 is trivial so suppose that n ≥ 4. The general 3strategy of the proof will be as follows. After the first 2 n − 2 moves in total (up to a permutation of rows and columns) X will occupy the cells of the main diagonal except for the bottom right cell and O will occupy at least two cells from the last column, one of which will be forced to be ( n, n ). Then X will make threats and will only need up to four more moves to win depending on where O has played. Player X starts by playing on (1 , 1). Now we will show by induction that player X can guarantee that the following holds after k+1 X-moves and k O -moves for all 1 ≤ k ≤ n−3: Up to some good transformations we have that:
• There are Xs in cells ( a, a ) for 1 ≤ a ≤ k + 1,
• There are no Os in cells ( a, b ) with a, b > k + 1, ()
• There is at least one cell ( a, b ) with an O with b = k + 2 We first show this for k = 1. Since X played in (1 , 1) we may assume by applying good transformations that O played either in (2 , 2) or (1 , 2) In either case X will play in (2 , 3) and we will swap columns 2 and 3. Now we have that (1 , 1) , (2 , 2) have Xs and either (1 , 3) or (2 , 3) has an O. This shows the case k = 1. Now suppose that we can do this for 1 ≤ k ≤ n − 4 So after k + 1 X-moves and kO-moves () is satisfied. Suppose now that O plays on cell ( a, b ). If either a ≤ k + 1 or b ≤ k + 1 then X will play in ( k + 2 , k + 3) and we will swap columns k + 2 and k + 3. This will ensure () is satisfied after k + 2 X-moves and k + 1
O-moves. Suppose now that a, b > k + 1. Now if b > k + 2 then X will play in ( a, k + 2) and we will swap rows a and k + 2 and then if b̸ = k + 3 swap columns k + 3 and b. If b = k + 2 then we will swap columns k + 2 and k + 3. Now we do the same as in the case with
b > k +2. In either case we ensure that () is satisfied which completes the induction step.
X
. . . OX
By above player X can guarantee that () is satisfied for k = n − 3. After n − 2 X-moves and n − 3 O-moves cells ( n − 1, n ), (n, n ) are empty. Now if O plays on ( n, n − 1) X will play on ( n, n ), otherwise X will play on ( n − 1, n ). Either way X will have a threat on (n − 1, n − 1) or ( n, n − 1). Since O has no threats as there are only n − 2 Os so far O
must defend by blocking X’s threat. After a total of 2 n − 2 moves by swapping columns
n − 1 and n and if needed swapping rows n − 1 and n by () we can make sure that X
occupies cells ( i, i ) for 1 ≤ i ≤ n − 1 while O occupies cell ( n, n ), another cell ( a, n ) for
a < n and some other cells. At this point after a total of 2 n − 2 moves let TO, T X be the sets of cells with an O, X respectively. Now let
SO = {i ∈ [n]|(i, n )̸ ∈ TO ∧ (n, i )̸ ∈ TO}
4Since |TO| = n − 1 we know that |SO| ≥ 1. But also since ( n, n ), (a, n ) ∈ TO we know that
|SO| ≤ n − 2. Now notice that since there are at least two Os in the final column, there are at least 2 columns without Os. This means that O currently has no threat. By using the same permutation on the first n−1 rows and the first n−1 columns we can guarantee that SO = [ s] for some s ≤ n − 2. For a set A ⊂ [n] we denote TA = {(i, i )|i ∈ [n]\A}.Now we split into three cases depending on s and n.Case 1: s = 1 , n ≥ 5. Let r = |{ i < n |(i, n ) ∈ TO}| . By above 1 ≤ r ≤ n − 2. Since each pair of cells {(i, n ), (n, i )} for 2 ≤ i ≤ n − 1 has an O in at least one cell and (n.n ) has an O but |TO| = n − 1, we must have that every such pair has exactly one O.In fact these together with ( n, n ) must be all the Os in the grid. Now similar to above by using the same permutation on rows and columns we may assume that without loss of generality
TO = {(n, i )|2 ≤ i ≤ n − r − 1} ∪ { (i, n )|n − r ≤ i ≤ n}.
Now X plays on (1 , n ) and since (1 , n ), (n, 1) , T {1,n } form a transversal X makes a threat on ( n, 1). So O must defend the threat on ( n, 1). Since all Os are in row n and column nO has no threats. Now X plays on ( n, n −1) and since ( n, n −1) , (1 , n ), (n−1, 1) , T {1,n −1,n }
form a transversal O must defend on ( n − 1, 1). Since all Os are in columns 1 , n and row
n O has no threats. This is because to have a threat O must have at least n − 1 ≥ 4 cells no two of which share a row or column. Now we split this into two subcases. Case 1a: r = n − 2. Now X plays on (2 , 1). We have that cells (2 , 1) , (1 , n ), (n, 2) and T{1,2,n } form a transversal, but also ( n − 1, 2) , (2 , 1) , (1 , n ), (n, n − 1) together with
T{1,2,n −1,n } form a transversal. Thus X has a threat on both ( n − 1, 2) and ( n, 2). Since
O cannot defend both threats X will win on the next move.
X Xn
Xn+2 X OX O
. . . ...
On+1 W X OOn W Xn+1 On−1
Case 1b: 1 ≤ r ≤ n−3. Now X plays on ( n−1, 2). Since ( n, n −1) , (n−1, 2) , (2 , n ), T {2,n −1,n }
form a transversal and (2 , 1) , (1 , n ), (n, n − 1) , (n − 1, 2) , T {1,2,n −1,n } form a transversal we have that X has threats on both (2 , 1) and (2 , n ). Since O has no threats X will win on the next move.
X Xn
W X W
. . .
XX O
. . . ...
On+1 Xn+2 X OOn O . . . O Xn+1 On−1
5Case 2: 2 ≤ s ≤ n − 2. Since there is at least one O in the cells ( i, n ), (n, i ) for all
s + 1 ≤ i ≤ n the number of Os in the first s columns is at most ( n − 1) − (n − s) = s − 1. Since s ≥ 2 the number of ordered pairs ( i, j ) such that 1 ≤ i, j ≤ s and i̸ = j is
s(s − 1) > s − 1. Thus there are b, c ∈ [s] such that b̸ = c and ( b, c ) is an unoccupied cell. Now X plays in ( n.b ). Since ( n, b ), (b, n ), T {b,n } form a transversal O must play in (b, n ). Since there were already Os in the final column and there were at least 2 columns without Os we still have 2 columns without Os and hence O has no threats. Now X plays on ( c, n ). Since ( c, n ), (n, c ), T {c,n } form a transversal and ( n, b ), (b, c ), (c, n ), T {b,c,n } form a transversal X has threats on both ( n, c ) and ( b, c ). This means that X will win on the next move.
X
. . .
X W On
. . .
X Xn+1
. . .
XXn W On−1
Case 3: s = 1 , n = 4. Since column 4 has at least two Os by using the same permu-tation of rows and columns we can assume that there is an O in (3 , 4). Now X plays on (1 , 4) which forces O to defend on (4 , 1). Now O has no threats since all Os are in row 4 and column 4 and to have a threat O would need 3 cells no two of which share a row or column. Since s = 1 and (3 , 4) has an O we know that exactly one of (2 , 4) and (4 , 2) have an O. Thus X can play on (4 , 3). Since (3 , 1) , (1 , 4) , (4 , 3) , (2 , 2) is a transver-sal X has a threat on (3 , 1) and hence O must defend on (3 , 1). We split into two subcases: Case 3a: If (2 , 4) has an O. Notice that O still has no threat and X will play on (2 , 1). Now since (2 , 1) , (1 , 4) , (4 , 3) , (3 , 2) is a transversal and also (2 , 1) , (1 , 4) , (4 , 2) , (3 , 3) is a transversal X has threats on both (3 , 2) and (4 , 2) and will hence win on the next move.
X X4
X6 X OO5 W X OO4 W X5 O3
Case 3b: If (4 , 2) has an O we still have that O has no threats so X can play on (3 , 2). Since (3 , 2) , (2 , 4) , (4 , 3) , (1 , 1) is a transversal and (3 , 2) , (2 , 1) , (1 , 4) , (4 , 3) is also a transversal
X has threats on both (2 , 1) and (2 , 4) and will hence win on the next move.
X X4
W X WO5 X6 X OO4 O X5 O3
This covers all cases and thus we have shown that for all n ≥ 4 player X has a winning strategy. This concludes the proof. 63 Related questions
We now know that player X wins for n ≥ 4 so a natural question to ask is how many transversals are needed for player X to win. Suppose that we have a family F of transver-sals on an n×n grid and suppose that the goal of the game for both players was to occupy
n cells of some transversal in F. We will call this game the transversal F-game. It is unclear if our strategy is optimal in terms of how many transversals are needed to guarantee that player X wins. So one might ask what is the minimal possible f (n)such that there is a family F with |F| = f (n) where player X has a winning strategy in the transversal F-game? Another natural question to ask is what is the minimal number g(n) such that for every family F of at least g(n) transversals player X wins? It is clear that f (n) ≤ g(n) ≤ n!. We will give an argument why f (n) < p(n)n!2n for some polynomial p and also why g(n) < n ! − (n!) 1/5−O(1 /n ). We will be using a very similar strategy as in the proof of Theorem 1 but being a bit more careful when choosing the first n − 1 moves.
Lemma 3. There is a fixed polynomial p such that f (n) < p(n)n!2n for every n ≥ 4.Proof. Let k = ⌊n−22 ⌋. We may assume that n ≥ 8 since we can otherwise just increase
p to make our statement true for all n ≥ 4. We choose arbitrary sets of k rows and k
columns with the only restriction that one of the chosen rows is the 1st row and none of the chosen columns is the 1st column. Let A, B ⊂ [n] be the chosen sets of rows and columns respectively. Consider the family F of all transversals that have at most four cells in A × B. Our goal is to show that player X can win the transversal F-game. We will consider A, B as variable sets so whenever we permute rows or columns or potentially perform a reflection with respect to the main diagonal we also accordingly change A, B
(and of course F). For convenience denote by Ai, B i the sets A, B after exactly 2 i total moves. For clarity we will only perform good transformations after X’s move (before O
makes his next move) just like in the proof of Theorem 1. Since both A, B have size k
the reflection with respect to the main diagonal will just swap those sets and not change their sizes. We first prove the following claim.
Claim. If X can achieve the following after the first 2 n − 2 moves in the transversal
F-game:
• X occupies precisely the cells ( i, i ) for 1 ≤ i ≤ n − 1,
• O occupies ( n, n ) and at least one other cell in the n-th column,
• none of the cells X occupies are in A × B,then X can guarantee a win.
Proof of claim. Suppose that X achieves the required conditions after 2 n − 2 total moves. Now X follows the exact same strategy as in the proof of Theorem 1. Notice that in each case X requires at most four extra moves to win. Since none of the first n − 17cells taken by X are inside A × B we have that all the transversals that X makes threats on prior to winning have at most four cells in A × B and hence belong to F. Thus X
can win the transversal F-game which proves the claim. □
Now we will show that X can indeed achieve the conditions required by the claim. After that we will show that |F| ≤ p(n)n!2n for some fixed polynomial p.For 1 ≤ i ≤ 2k let Ai = {i + 1 , . . . , n } ∩ Ai and Bi = {i + 1 , . . . , n } ∩ Bi. Notice that after a total of 2 i moves X needs to avoid Ai × Bi on his next move. Now we start the game. Player X begins by playing on (1 , 1). Then O plays somewhere. At this point since 1 ∈ A \ B we have that |A1| = k − 1, |B1| = k. We will show by induction that player X can guarantee that the following holds after i X -moves and i O -moves for all 2 ≤ i ≤ 2k:
• There are Xs in cells ( x, x ) for 1 ≤ x ≤ i,
• None of the first i − 1 moves by O are in cells ( x, y ) with x, y > i , ()
• There is at least one cell ( x, y ) with an O with y = i + 1
• |Ai| + |Bi| ≤ 2k − i or |Ai|| Bi| = 0 We first show this for i = 2. Suppose that O plays his first move in some ( a, b ). There are fundamentally two cases to consider depending on whether a = 1 or a̸ = 1. We will show that in either case X can choose some c̸ = 1 , b and play in (max( a, 2) , c ) where we have either max( a, 2) ∈ A1 or c ∈ B1 but not both. Indeed if max( a, 2) ∈ A1 then since
k < n − 2 pick c̸ = 1 , b such that c̸ ∈ B1. If max( a, 2) ̸ ∈ A1 then pick c ∈ B1 with c̸ = 1 , b
which is possible since k > 2. Now we swap columns 2 and c and if a̸ = 1 swap rows 2 and a. If b = 2 then we further swap columns c and 3 and if b̸ = 2 then we swap columns
b and 3. Now we have that (1 , 1) , (2 , 2) have Xs and either (1 , 3) or (2 , 3) has an O. After
O’s next move we also have that |A2| + |B2| ≤ | A1| + |B1| − 1 ≤ 2k − 2 ensuring that () is satisfied in the case i = 2. Now suppose that we can do this for 2 ≤ i ≤ 2k − 1 So after i X -moves and i O -moves () is satisfied. Suppose now that O’s last move was in cell ( a, b ). If |Ai|Bi| = 0 then we do the same step as in Theorem 1 and we have that ( ∗∗ ) is satisfied. Now suppose that |Ai|| Bi|̸ = 0 and either a ≤ i or b ≤ i. Choose any x ∈ A1. We need to choose c ≥ i + 2 such that c ∈ Bi. Since |Bi| ≤ 2k − i and n − i − 1 > 2k − i
there is a c̸ ∈ Bi with c ≥ i + 2. Now X will play in ( x, c ). Then we will swap rows x
and i + 1, columns i + 1 and c and if c̸ = i + 2 we will swap columns i + 2 and c. This will ensure () is satisfied after i + 1 X-moves and i + 1 O-moves. Suppose now that |Ai|| Bi|̸ = 0 and a, b > i . First suppose that a ∈ Ai. Now since
n − i − 1 > 2k − i ≥ Bi there is a c ≥ i + 1 , c ̸ = b such that c̸ ∈ Bi. Now X plays in (a, c ). Swap rows a and i + 1 and swap columns c and i + 1. Next, if b = i + 1 swap columns c and i + 2, but if b > i + 1 swap columns b and i + 2. Either way, the conditions of () are going to be satisfied after 2 i + 2 total moves. Now suppose that a̸ ∈ Ai. If there is a c ∈ Bi \ { b} then we will have X play in ( a, c ). Now we perform the same 8swaps as in the above case (when a ∈ Ai) and achieve () after 2 i + 2 total moves. Since
|Bi|̸ = 0 the only other case is if Bi = {b}. Now since n − i − 1 > 2k − i ≥ | Ai| there is an a′ ≥ i + 1 , a ′̸ ∈ Ai ∪ { a}. Notice that a′̸ = a so we will have X play in ( a′, b ). Now swap columns b and i + 1 and swap rows a′ and i + 1. Further if a = i + 1 swap rows a′
and i + 2, but if a̸ = i + 1 swap rows a and i + 2. Now perform a reflection with respect to the main diagonal. This ensures that () is satisfied after 2 i + 2 total moves which completes the induction step. After 2 k X moves and 2 k O moves we obtain that A2k = ∅ X or B2k = ∅ and hence we can continue as in Theorem 1 to occupy all cells ( i, i ) with 1 ≤ i ≤ n − 1. This ensures that X has avoided A × B throughout all of his first n − 1 moves. Note that this requires at most two more moves since n − 1 − 2k ≤ 2. On O’s ( n − 1)-th move he must play in ( n, n ) since at that point the main diagonal is a transversal in F. This will ensure the conditions of the above claim meaning that X can win the transversal F-game. Now we will estimate |F| . Notice that for any 0 ≤ m ≤ 4 we have that the number of transversals that have exactly m cells in A × B is equal to
Pn,m =
km
k(k − 1) · · · (k − m + 1)( n − k)( n − k − 1) · · · (n − 2k + m + 1) · (n − k)! .
We have that
Pn,m = n!
km
k(k − 1) · · · (k − m + 1) (n − k) · · · (n − 2k + m + 1)
n(n − 1) · · · (n − k + 1) . (1) Since k < n/ 2 we have that km
< k 4 < n4
16
Notice that
k(k−1) ··· (k−m+1)( n−k)( n−k−1) ··· (n−2k+m+1)
k!
n(n−1) ··· (n−k+1)
k!
=
n−kk−m
nk
(2) We want an upper bound on Pn,m so notice that n−kk−m
= n−kn−2k+m
. Since n − 2k + m < 8we have that n−kk−m
< n 8. Note that n
⌊n/ 2⌋
≥ 2n
n+1
since this is the largest ni
binomial coefficient. So nk
=
n
⌊n/ 2⌋
⌊n/ 2⌋
n − k > 12
n
⌊n/ 2⌋
2n
2( n + 1) .
Now from (1) and (2) we get that
Pn,m ≤ n!n4
16
n82( n + 1) 2n = n12 (n + 1) 8
n!2n .
Now the total number of transversals with at most four cells in A × B is Pn, 0 + Pn, 1 +
· · · + Pn, 4 ≤ 5n12 (n+1) 8
n!2n
so taking p(x) = 5x12 (x+1) 8 gives the desired result. We can see that in Lemma 3 we could not get a better bound using the same method by being more careful with the inequalities. We do however think that f (n) is much lower. On the other end Erd˝ os and Selfridge proved that in a game where winning lines have size n there must be at least 2 n−1 winning lines if it is a first player win. Thus
f (n) ≥ 2n−1. However, the example with 2 n−1 lines has exponentially many elements in total and clearly cannot be embedded into a set of transversals. This is far from the only 9obstacle as one may have a game on a set A of size 2 n − 1 where winning lines are all subsets of A with exactly n elements. This is a player 1 win and the number of lines is less than 4 n. This however contains other structures not present in the transversals (such as two winning lines whose intersecting has size n − 1). The following lemma however shows that for the Maker-Breaker version of this game exponentially many lines are in-deed sufficient to ensure a Maker win.
Lemma 4. Let n ≥ 5 be a positive integer. One may choose a family F of at most 10 4 ·3n
transversals in an n × n board so that the Maker-Breaker transversal F-game is a Maker win. Proof. We will first prove that Maker can win even by playing second on all sufficiently large boards and then we will stack many such boards in our n × n board allowing Maker to play several games as the second player. First we show the following claim.
Claim. For any m ≥ 5 Maker can win the transversal m-game by playing second (where Breaker plays first).
Proof of Claim. We first show that if Maker can win on an m × m board then he can also win on an ( m + 1) × (m + 1) board. Indeed this is a simple induction step. By permuting rows or columns we may assume that Breaker plays his first move in cell (1 , 2). Now Maker plays in (1 , 1) and he continues to play on the m × m sub-board consisting of the cells in the set {2, 3, . . . , m + 1 }2. He can make a transversal there giving him a transversal on the full board. To prove the claim we just need to show that Maker can win for m = 5. We may assume that Breaker plays in (1 , 1). Now Maker plays in (1 , 2). If Breaker plays his sec-ond move anywhere in the row 1 or column 2 then Maker can win by forming a transversal in the board obtained by removing row 1 and column 2 where now he plays first. This gives Maker a transversal of the full board. If however Breaker does not play his second move in row 1 or column 2 then there are essentially two cases. By permuting some of the rows 2 , 3, 4, 5 and or columns 3 , 4, 5 we may assume that Breaker’s second move is on (2 , 1) or (2 , 3). We deal with these separately. Case 1: Breaker’s second move is (2 , 1). Now Maker plays in (3 , 1). Now consider the sub-board T with row 3 and column 1 removed. Maker has already played in the top-left of that board while Breaker has not played in those cells. By using the same strategy as in Theorem 1 we know that Maker can win on a 4 × 4 board by playing first so he can guarantee a transversal in T . Together with (3 , 1) he can guarantee a transversal of the full board. Case 2: Breaker’s second move is (2 , 3). Now Maker plays in (2 , 1). Similar to the previous case consider the board T ′ with row 2 and column 1 removed. Maker has played in the top-left of T ′ while Breaker has not played in T ′. Now since Maker can win by playing first on a 4 × 4 board, Maker can form a transversal on T ′. Together with (2 , 1) this gives a transversal of the full board. This completes the proof of the claim. □
Now we move on to the proof of the lemma. Let k, r be integers such that n = 5 k + r,10 where 0 ≤ r ≤ 4. Consider the sets Si = {5i − 4, 5i − 3, 5i − 2, 5i − 1, 5i}2 of cells on the
n×n board for 1 ≤ i ≤ k −1. Let S = {5k −4, 5k −3, . . . , n }2 and V = ( ∪k−1
i=1
Si)∪S. Now let F be the set of all transversals only consisting of cells in V . We will show that Maker can win the transversal F-game. Indeed Maker may play his first move arbitrarily and then play as the second player on each sub-board S1, S 2, . . . S k−1, S separately. Wherever Breaker plays Maker will play on the same sub-board in his next move (if possible) us-ing the winning strategy from the above claim. If Breaker does not play on one of the sub-boards Maker can just pretend that he did and proceed accordingly (extra moves clearly do not hurt Maker). Thus he can form transversals in each of the sub-boards
S, S 1, S 2, . . . S k−1 giving a transversal in F. This means that Maker can guarantee win-ning the transversal F-game. All that remains is to count the number of transversals entirely contained in V . Ev-ery such transversal is composed of one transversal from each Si and S. This yields at most (5!) k · 6 · 7 · 8 · 9 < 10 4 · 3n transversals. This proves the lemma. Even though a Maker win in the Maker-Breaker version does not guarantee a player 1 win in the original ‘Maker-Maker’ version, it seems that it should not be that much harder (possibly with the addition of some extra lines) for player 1 to win. This leads us to the following conjecture.
Conjecture 5. There is a constant c > 1 such that f (n) < c n for large enough n.
Now we move on to g(n). For each positive integer n > 5 define mn as follows. Let
k0 = 2 , k 5 = n and let 2 < k 1 < k 2 < k 3 ≤ n − 1 be integers. Define pi =
ki+1 −1
Q
j=ki
j
for 0 ≤ i ≤ 3. Now let mn be the maximum possible value of min
0≤i≤3
pi over all possible
k1, k 2, k 3. We will show that g(n) ≤ n! − mn − 1. It is not hard to see that
mn − 1 = ( n!) 1/4−O(1 /n ).
Indeed since each term in the product n! = 2 · 3 · . . . · n is at most n one may pick (for
n ≥ 8 ensuring n! > n 4) k1, k 2, k 3, k 4 in that order such that ( n!) 1/4/n ≤ pi ≤ (n!) 1/4 for 0 ≤ i ≤ 2. This will be sufficient since ( n!) 2 = Qni=1 i(n − i) ≥ nn and we also clearly have that mn ≤ (n!) 1/4.
Lemma 6. For all n > 5 we have that g(n) ≤ n! − mn + 1 .Proof. Suppose that F is a family of at least n! − mn + 1 transversals. Let G = Fc be the family of the mn − 1 transversals not included in F. Let 1 < k 1 < k 2 < k 3 < n be such that min
0≤i≤3
pi = mn where pi =
ki+1 −1
Q
j=ki
j. We follow the same strategy as in the proof of Theorem 1 except that we now show: after the first n − 1 X moves when X occupies all but one cell of the diagonal, for each transversal T ∈ G there are at least five of the
n − 1 cells occupied by X but not in T . As before when permuting rows and columns we change G accordingly. Now consider what happens in the inductive proof for ( ∗) in Theorem 1 when X has made k + 1 moves and O has made k moves. If O’s next move is on ( a, b ) where a ≤ k + 1 or b ≤ k + 1 then X can move anywhere in ( k + 2 , c ) where
c ≥ k + 3. Then we can swap columns k + 2 and c and then swap column c and k + 3 to ensure ( ∗). So X had n − k − 2 choices in the same row. Similarly if a, b > k + 1 if 11 say b > k + 2 then X can play in ( a, c ) for any c ≥ k + 2 with c̸ = b. Then we can swap rows a and k + 2 and swap some columns to ensure ( ∗). Likewise if b = k + 2 then we can swap columns k + 2 and k + 3 and do the same as when b > k + 2. Either way there are always at least n − k − 2 choices in the same row for X. With a similar analysis X’s second move has at least n − 2 choices in the same row and certainly his first move has at least n − 1 choices in the same row. Now let Gn = G. We will form Gn−1, Gn−2, . . . Gk3
as follows. Inductively at move k ≤ n − k3 X will move in the cell C which is one of the possible choices described above that is in the least number of transversals in Gn+1 −k.Then we let Gn−k be the set of all transversals in Gn+1 −k containing C. Since there are at least n − k choices in the same row we must have that |G n−k| ≤ |G n+1 −k |
n−k
. Thus
|G k3 | ≤ mn − 1
n−1
Q
j=k3
j< 1so Gk3 is empty and hence the first no transversal in G will contain all cells of the first
n−k3 moves. Similarly we can ensure that for each 1 ≤ i ≤ 3 no transversal of G contains all cells where X made all his moves from move n − ki + 1 to n − ki−1 inclusive. Indeed for each interval we start from G again and reduce it as much as possible at every step. This completes n − 2 X moves and indeed X can make his next move in accordance with Theorem 1. Thus after n − 1 X moves we can ensure that each transversal of G has at least four cells without X. Let S be the set of cells of the first n − 1 X moves. Now
X follows the same strategy as in Theorem 1. Note that all threats that X uses have at most three cells that are not part of the first n − 1 X moves so X can win with the same strategy and guarantee that his winning transversal is in F. This shows that indeed
g(n) ≤ n! − mn + 1. Note that g(n) > (n−2)! since the family consisting of ( n−2)! transversals containing both (1 , 1) and (2 , 2) is clearly not winning for X. Thus g(n) is much bigger than f (n). So how big is g(n)? It seems that we need to remove a large portion of transversals to stop X from winning. We finish with the following conjecture.
Conjecture 7. As n → ∞ we have that g(n) = o(n!) .
References
J´ ozsef Beck. Combinatorial Games: Tic-Tac-Toe Theory . Cambridge University Press, 01 2008. P Erd¨ os and J.L Selfridge. On a combinatorial game. Journal of Combinatorial Theory, Series A , 14(3):298–301, 1973. M. Erickson. Transversal achievement game on a square grid. http: //www.openproblemgarden.org/op/a_transversal_achievement_game_on_a_ square_grid , 2010. Accessed: 2023-11-01. A. Hales and R.I. Jewett. Regularity and positional games. Transactions of the American Mathematical Society , 106:320–327, 06 2010. 12 |
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41.6: Overview of Vertebrate Organ Systems
41: The Animal Body and Principles of Regulation
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Tue, 14 Nov 2023 04:21:53 GMT
41.6.4: Amphibians
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Contents
1. Home
2. Bookshelves
3. Introductory and General Biology
4. Map: Raven Biology 12th Edition
5. 41: The Animal Body and Principles of Regulation
6. 41.6: Overview of Vertebrate Organ Systems
7. 41.6.4: Amphibians
Expand/collapse global location
Map: Raven Biology 12th Edition
Front Matter
1: The Science of Biology
2: The Nature of Molecules and the Properties of Water
3: The Chemical Building Blocks of Life
4: Cell Structure
5: Membranes
6: Energy and Metabolism
7: How Cells Harvest Energy
8: Photosynthesis
9: Cell Communication
10: How Cells Divide
11: Sexual Reproduction and Meiosis
12: Patterns of Inheritance
13: Chromosomes, Mapping, and the Meiosis-Inheritance Connection
14: DNA- The Genetic Material
15: Genes and How They Work
16: Control of Gene Expression
17: Biotechnology
18: Genomics
19: Cellular Mechanisms of Development
20: Genes Within Populations
21: The Evidence for Evolution
22: The Origin of Species
23: Systematics, Phylogeny and Comparative Biology
24: Genome Evolution
25: The Origin and Diversity of Life
26: Viruses
27: Prokaryotes
28: Protists
29: Seedless Plants
30: Seed Plants
31: Fungi
32: Animal Diversity and the Evolution of Body Plans
33: Protostomes
34: Deuterostomes
35: Plant Form
36: Transport in Plants
37: Plant Nutrition and Soils
38: Plant Defense Responses
39: Sensory Systems in Plants
40: Plant Reproduction
41: The Animal Body and Principles of Regulation
42: The Nervous System
43: Sensory Systems
44: The Endocrine System
45: The Musculoskeletal System
46: The Digestive System
47: The Respiratory System
48: The Circulatory System
49: Osmotic Regulation and the Urinary System
50: The Immune System
51: The Reproductive System
52: Animal Development
53: Behavioral Biology
54: Ecology of Individuals and Populations
55: Community Ecology
56: Dynamics of Ecosystems
57: The Biosphere and Human Impacts
58: Conservation Biology
Back Matter
41.6.4: Amphibians
Last updated Nov 14, 2023
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41.6.3: Fishes
41.6.5: Reptiles
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Table of contents
1. Characteristics of Amphibians
2. Evolution of Amphibians
3. Modern Amphibians
1. Urodela: Salamanders
2. Anura: Frogs
3. Apoda: Caecilians
Summary
Footnotes
Glossary
Skills to Develop
Describe the important difference between the life cycle of amphibians and the life cycles of other vertebrates
Distinguish between the characteristics of Urodela, Anura, and Apoda
Describe the evolutionary history of amphibians
Amphibians are vertebrate tetrapods. Amphibia includes frogs, salamanders, and caecilians. The term amphibian loosely translates from the Greek as “dual life,” which is a reference to the metamorphosis that many frogs and salamanders undergo and their mixture of aquatic and terrestrial environments in their life cycle. Amphibians evolved during the Devonian period and were the earliest terrestrial tetrapods.
Link to Learning
Watch this series of Animal Planet videos on tetrapod evolution:
1: The evolution from fish to earliest tetrapod
2: Fish to Earliest Tetrapod
3: The discovery of coelacanth and Acanthostega fossils
4: The number of fingers on “legs”
Characteristics of Amphibians
As tetrapods, most amphibians are characterized by four well-developed limbs. Some species of salamanders and all caecilians are functionally limbless; their limbs are vestigial. An important characteristic of extant amphibians is a moist, permeable skin that is achieved via mucus glands that keep the skin moist; thus, exchange of oxygen and carbon dioxide with the environment can take place through it (cutaneous respiration). Additional characteristics of amphibians include pedicellate teeth—teeth in which the root and crown are calcified, separated by a zone of noncalcified tissue—and a papilla amphibiorum and papilla basilaris, structures of the inner ear that are sensitive to frequencies below and above 10,00 hertz, respectively. Amphibians also have an auricular operculum, which is an extra bone in the ear that transmits sounds to the inner ear. All extant adult amphibians are carnivorous, and some terrestrial amphibians have a sticky tongue that is used to capture prey.
Evolution of Amphibians
The fossil record provides evidence of the first tetrapods: now-extinct amphibian species dating to nearly 400 million years ago. Evolution of tetrapods from fishes represented a significant change in body plan from one suited to organisms that respired and swam in water, to organisms that breathed air and moved onto land; these changes occurred over a span of 50 million years during the Devonian period. One of the earliest known tetrapods is from the genus Acanthostega. Acanthostega was aquatic; fossils show that it had gills similar to fishes. However, it also had four limbs, with the skeletal structure of limbs found in present-day tetrapods, including amphibians. Therefore, it is thought that Acanthostega lived in shallow waters and was an intermediate form between lobe-finned fishes and early, fully terrestrial tetrapods. What preceded Acanthostega?
In 2006, researchers published news of their discovery of a fossil of a “tetrapod-like fish,” Tiktaalik roseae, which seems to be an intermediate form between fishes having fins and tetrapods having limbs (Figure 41.6.4.1). Tiktaalik likely lived in a shallow water environment about 375 million years ago.1
Figure 41.6.4.1: The recent fossil discovery of Tiktaalik roseae suggests evidence for an animal intermediate to finned fish and legged tetrapods. (credit: Zina Deretsky, National Science Foundation)
The early tetrapods that moved onto land had access to new nutrient sources and relatively few predators. This led to the widespread distribution of tetrapods during the early Carboniferous period, a period sometimes called the “age of the amphibians.”
Modern Amphibians
Amphibia comprises an estimated 6,770 extant species that inhabit tropical and temperate regions around the world. Amphibians can be divided into three clades: Urodela (“tailed-ones”), the salamanders; Anura (“tail-less ones”), the frogs; and Apoda (“legless ones”), the caecilians.
Urodela: Salamanders
Salamanders are amphibians that belong to the order Urodela. Living salamanders (Figure 41.6.4.1) include approximately 620 species, some of which are aquatic, other terrestrial, and some that live on land only as adults. Adult salamanders usually have a generalized tetrapod body plan with four limbs and a tail. They move by bending their bodies from side to side, called lateral undulation, in a fish-like manner while “walking” their arms and legs fore and aft. It is thought that their gait is similar to that used by early tetrapods. Respiration differs among different species. The majority of salamanders are lungless, and respiration occurs through the skin or through external gills. Some terrestrial salamanders have primitive lungs; a few species have both gills and lungs.
Unlike frogs, virtually all salamanders rely on internal fertilization of the eggs. The only male amphibians that possess copulatory structures are the caecilians, so fertilization among salamanders typically involves an elaborate and often prolonged courtship. Such a courtship allows the successful transfer of sperm from male to female via a spermatophore. Development in many of the most highly evolved salamanders, which are fully terrestrial, occurs during a prolonged egg stage, with the eggs guarded by the mother. During this time, the gilled larval stage is found only within the egg capsule, with the gills being resorbed, and metamorphosis being completed, before hatching. Hatchlings thus resemble tiny adults.
Figure 41.6.4.2: Most salamanders have legs and a tail, but respiration varies among species. (credit: Valentina Storti)
Link to Learning
View River Monsters: Fish With Arms and Hands? to see a video about an unusually large salamander species.
Anura: Frogs
Frogs are amphibians that belong to the order Anura (Figure 41.6.4.3). Anurans are among the most diverse groups of vertebrates, with approximately 5,965 species that occur on all of the continents except Antarctica. Anurans have a body plan that is more specialized for movement. Adult frogs use their hind limbs to jump on land. Frogs have a number of modifications that allow them to avoid predators, including skin that acts as camouflage. Many species of frogs and salamanders also release defensive chemicals from glands in the skin that are poisonous to predators.
Figure 41.6.4.3: The Australian green tree frog is a nocturnal predator that lives in the canopies of trees near a water source.
Frog eggs are fertilized externally, and like other amphibians, frogs generally lay their eggs in moist environments. A moist environment is required as eggs lack a shell and thus dehydrate quickly in dry environments. Frogs demonstrate a great diversity of parental behaviors, with some species laying many eggs and exhibiting little parental care, to species that carry eggs and tadpoles on their hind legs or backs. The life cycle of frogs, as other amphibians, consists of two distinct stages: the larval stage followed by metamorphosis to an adult stage. The larval stage of a frog, the tadpole, is often a filter-feeding herbivore. Tadpoles usually have gills, a lateral line system, long-finned tails, and lack limbs. At the end of the tadpole stage, frogs undergo metamorphosis into the adult form (Figure 41.6.4.4). During this stage, the gills, tail, and lateral line system disappear, and four limbs develop. The jaws become larger and are suited for carnivorous feeding, and the digestive system transforms into the typical short gut of a predator. An eardrum and air-breathing lungs also develop. These changes during metamorphosis allow the larvae to move onto land in the adult stage.
Figure 41.6.4.4: A juvenile frog metamorphoses into a frog. Here, the frog has started to develop limbs, but its tadpole tail is still evident.
Apoda: Caecilians
An estimated 185 species comprise caecilians, a group of amphibians that belong to the order Apoda. Although they are vertebrates, a complete lack of limbs leads to their resemblance to earthworms in appearance. They are adapted for a soil-burrowing or aquatic lifestyle, and they are nearly blind. These animals are found in the tropics of South America, Africa, and Southern Asia. They have vestigial limbs, evidence that they evolved from a legged ancestor.
Evolution Connection: The Paleozoic Era and the Evolution of Vertebrates
The climate and geography of Earth was vastly different during the Paleozoic Era, when vertebrates arose, as compared to today. The Paleozoic spanned from approximately 542 to 251 million years ago. The landmasses on Earth were very different from those of today. Laurentia and Gondwana were continents located near the equator that subsumed much of the current day landmasses in a different configuration (Figure 41.6.4.5). At this time, sea levels were very high, probably at a level that hasn’t been reached since. As the Paleozoic progressed, glaciations created a cool global climate, but conditions warmed near the end of the first half of the Paleozoic. During the latter half of the Paleozoic, the landmasses began moving together, with the initial formation of a large northern block called Laurasia. This contained parts of what is now North America, along with Greenland, parts of Europe, and Siberia. Eventually, a single supercontinent, called Pangaea, was formed, starting in the latter third of the Paleozoic. Glaciations then began to affect Pangaea’s climate, affecting the distribution of vertebrate life.
Figure 41.6.4.5: During the Paleozoic Era, around 550 million years ago, the continent Gondwana formed. Both Gondwana and the continent Laurentia were located near the equator.
During the early Paleozoic, the amount of carbon dioxide in the atmosphere was much greater than it is today. This may have begun to change later, as land plants became more common. As the roots of land plants began to infiltrate rock and soil began to form, carbon dioxide was drawn out of the atmosphere and became trapped in the rock. This reduced the levels of carbon dioxide and increased the levels of oxygen in the atmosphere, so that by the end of the Paleozoic, atmospheric conditions were similar to those of today.
As plants became more common through the latter half of the Paleozoic, microclimates began to emerge and ecosystems began to change. As plants and ecosystems continued to grow and become more complex, vertebrates moved from the water to land. The presence of shoreline vegetation may have contributed to the movement of vertebrates onto land. One hypothesis suggests that the fins of aquatic vertebrates were used to maneuver through this vegetation, providing a precursor to the movement of fins on land and the development of limbs. The late Paleozoic was a time of diversification of vertebrates, as amniotes emerged and became two different lines that gave rise, on one hand, to mammals, and, on the other hand, to reptiles and birds. Many marine vertebrates became extinct near the end of the Devonian period, which ended about 360 million years ago, and both marine and terrestrial vertebrates were decimated by a mass extinction in the early Permian period about 250 million years ago.
Link to Learning
View Earth’s Paleogeography: Continental Movements Through Time to see changes in Earth as life evolved.
Summary
As tetrapods, most amphibians are characterized by four well-developed limbs, although some species of salamanders and all caecilians are limbless. The most important characteristic of extant amphibians is a moist, permeable skin used for cutaneous respiration. The fossil record provides evidence of amphibian species, now extinct, that arose over 400 million years ago as the first tetrapods. Amphibia can be divided into three clades: salamanders (Urodela), frogs (Anura), and caecilians (Apoda). The life cycle of frogs, like the majority of amphibians, consists of two distinct stages: the larval stage and metamorphosis to an adult stage. Some species in all orders bypass a free-living larval stage.
Footnotes
1 Daeschler, E. B., Shubin, N. H., and Jenkins, F. J. “A Devonian tetrapod-like fish and the evolution of the tetrapod body plan,” Nature 440 (2006): 757–763, doi:10.1038/nature04639,
Glossary
Acanthostega one of the earliest known tetrapods Amphibia frogs, salamanders, and caecilians Anura frogs Apoda caecilians caecilian legless amphibian that belongs to the clade Apoda cutaneous respiration gas exchange through the skin frog tail-less amphibian that belongs to the clade Anura salamander tailed amphibian that belongs to the clade Urodela tadpole larval stage of a frog Urodela salamanders
This page titled 41.6.4: Amphibians is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax.
29.3: Amphibians by OpenStax is licensed CC BY 4.0.
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1996 AIME Problems/Problem 15
Problem
In parallelogram , let be the intersection of diagonals and . Angles and are each twice as large as angle , and angle is times as large as angle . Find .
Contents
Solution
Solution 1 (trigonometry)
Let . Then , , and . Since is a parallelogram, it follows that . By the Law of Sines on ,
Dividing the two equalities yields
Pythagorean and product-to-sum identities yield
and the double and triple angle () formulas further simplify this to
The only value of that fits in this context comes from . The answer is .
Solution 2 (trigonometry)
Define as above. Since , it follows that , and so . The Law of Sines on yields that
Expanding using the sine double and triple angle formulas, we have
By the quadratic formula, we have , so (as the other roots are too large to make sense in context). The answer follows as above.
Solution 3
We will focus on . Let , so . Draw the perpendicular from intersecting at . Without loss of generality, let . Then , since is the circumcenter of . Then .
By the Exterior Angle Theorem, and . That implies that . That makes . Then since by AA ( and reflexive on ), .
Then by the Pythagorean Theorem, . That makes equilateral. Then . The answer follows as above.
See also
| | | |
---
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| Preceded by Problem 14 | Followed by Final Problem |
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2736 | https://pubmed.ncbi.nlm.nih.gov/26731483/ | Placebo-Controlled, Double-Blind, Prospective, Randomized Study on the Effect of Octreotide LAR in the Control of Tumor Growth in Patients with Metastatic Neuroendocrine Midgut Tumors (PROMID): Results of Long-Term Survival - PubMed
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. 2017;104(1):26-32.
doi: 10.1159/000443612. Epub 2016 Jan 6.
Placebo-Controlled, Double-Blind, Prospective, Randomized Study on the Effect of Octreotide LAR in the Control of Tumor Growth in Patients with Metastatic Neuroendocrine Midgut Tumors (PROMID): Results of Long-Term Survival
Anja Rinke1,Michael Wittenberg,Carmen Schade-Brittinger,Behnaz Aminossadati,Erdmuthe Ronicke,Thomas M Gress,Hans-Helge Müller,Rudolf Arnold;PROMID Study Group
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1 Division of Gastroenterology and Endocrinology, University Hospital Marburg (UKGM), Marburg, Germany.
PMID: 26731483
DOI: 10.1159/000443612
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Randomized Controlled Trial
Placebo-Controlled, Double-Blind, Prospective, Randomized Study on the Effect of Octreotide LAR in the Control of Tumor Growth in Patients with Metastatic Neuroendocrine Midgut Tumors (PROMID): Results of Long-Term Survival
Anja Rinke et al. Neuroendocrinology.2017.
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. 2017;104(1):26-32.
doi: 10.1159/000443612. Epub 2016 Jan 6.
Authors
Anja Rinke1,Michael Wittenberg,Carmen Schade-Brittinger,Behnaz Aminossadati,Erdmuthe Ronicke,Thomas M Gress,Hans-Helge Müller,Rudolf Arnold;PROMID Study Group
Affiliation
1 Division of Gastroenterology and Endocrinology, University Hospital Marburg (UKGM), Marburg, Germany.
PMID: 26731483
DOI: 10.1159/000443612
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Background: Somatostatin analogs have been shown to control the growth of well-differentiated metastatic neuroendocrine tumors. Their effect on overall survival is a matter of debate. We analyzed the prognostic significance of early treatment with octreotide LAR and of hepatic tumor load in the PROMID trial cohort.
Patients and methods: Between 2001 and 2008, 85 treatment-naïve patients were randomly assigned to monthly octreotide LAR 30 mg or placebo until tumor progression or death. Post-study treatment was at the discretion of the investigator. Upon disease progression, 38 out of 43 placebo patients (88.4%) received octreotide LAR. For survival, patients were followed until May 2014.
Results: Forty-eight out of 85 patients (56.5%) died. In 38 patients (79.2%), death was tumor related. The median overall survival (84.7 and 83.7 months) was only slightly different in patients assigned to octreotide and placebo [HR = 0.83 (95% CI: 0.47-1.46); p = 0.51]. The median overall survival was 84.7 months for all 85 patients, 107.6 months in the low-tumor-load (n = 64) and 57.5 months in the high-tumor-load (n = 21) subgroups [HR = 2.49 (95% CI: 1.36-4.55); p = 0.002]. There was a trend towards improved overall survival in patients with a low hepatic tumor load receiving octreotide compared to placebo ['median not reached' and 87.2 months; HR = 0.59 (95% CI: 0.29-1.2); p = 0.142].
Conclusion: The extent of tumor burden is a predictor for shorter survival. Overall survival was similar in patients receiving octreotide LAR or placebo treatment at randomization. Crossover of the majority of placebo patients to octreotide LAR may have confounded the data on overall survival.
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2737 | https://www.physics.rutgers.edu/~eandrei/389/xrays/exp.pdf | Crystal Structure Analysis X-ray Diffraction Electron Diffraction Neutron Diffraction Essence of diffraction: Bragg Diffraction Reading: West 5 A/M 5-6 G/S 3 217 Elements of Modern X-ray Physics, 2nd Ed. by Jens Als-Nielsen and Des McMorrow, John Wiley & Sons, Ltd., 2011 (Modern x-ray physics & new developments) X-ray Diffraction, by B.E. Warren, General Publishing Company, 1969, 1990 (Classic X-ray physics book) Elements of X-ray Diffraction, 3rd Ed., by B.D. Cullity, Addison-Wesley, 2001 (Covers most techniques used in traditional materials characterization) High Resolution X-ray Diffractometry and Topography, by D. Keith Bowen and Brian K. Tanner, Taylor & Francis, Ltd., 1998 (Semiconductors and thin film analysis) Modern Aspects of Small-Angle Scattering, by H. Brumberger, Editor, Kluwer Academic Publishers, 1993 (SAXS techniques) Principles of Protein X-ray Crystallography, 3rd Ed. by Jan Drenth, Springer, 2007 (Crystallography) REFERENCES 218 SCATTERING Elastic (E’ = E) X-rays scatter by interaction with the electron density of a material.
Neutrons are scattered by nuclei and by any magnetic moments in a sample.
Electrons are scattered by electric/magnetic fields.
Scattering is the process in which waves or particles are forced to deviate from a straight trajectory because of scattering centers in the propagation medium. p' p q E' E h Momentum transfer: Energy change: Inelastic (E’ ≠ E) q 2 sin 2 p Elastic scattering geometry • Rayleigh (λ >> dobject) • Mie (λ ≈dobject) • Geometric (λ << dobject) • Thompson (X-rays) E pc For X-rays: • Compton (photons + electrons) • Brillouin (photons + quasiparticles) • Raman (photons + molecular vib./rot.) X-RAY SCATTERING • wide-angle diffraction (θ > 5°) • small-angle diffraction (θ close to 0°) • X-ray reflectivity (films) elastic (Thompson, ΔE = 0) inelastic (ΔE ≠ 0) • Compton X-ray scattering • resonant inelastic X-ray scattering (RIXS) • X-ray Raman scattering X-rays: • 100 eV (“soft”) – 100 keV (“hard”) photons • 12,400 eV X-rays have wavelengths of 1 Å, somewhat smaller than interatomic distances in solids Diffraction from crystals!
First X-ray: 1895 Roentgen 1901 Nobel λ (in Å) = 12400/E (in eV) 225 DIFFRACTION Diffraction refers to the apparent bending of waves around small objects and the spreading out of waves past small apertures.
In our context, diffraction is the scattering of a coherent wave by the atoms in a crystal. A diffraction pattern results from interference of the scattered waves.
Refraction is the change in the direction of a wave due to a change in its speed.
W. L. Bragg W. H. Bragg diffraction of plane waves von Laue Crystal diffraction I.
Real space description (Bragg) II. Momentum (k) space description (von Laue) 226 OPTICAL INTERFERENCE δ = nλ, n = 0, 1, 2, … δ = nλ, n = 1/2, 3/2, … δ: phase difference n: order perfectly in phase: perfectly out of phase: BRAGG’S LAW OF DIFFRACTION When a collimated beam of X-rays strikes pair of parallel lattice planes in a crystal, each atom acts as a scattering center and emits a secondary wave. All of the secondary waves interfere with each other to produce the diffracted beam Bragg provided a simple, intuitive approach to diffraction: • Regard crystal as parallel planes of atoms separated by distance d • Assume specular reflection of X-rays from any given plane → Peaks in the intensity of scattered radiation will occur when rays from successive planes interfere constructively 2Θ 228 BRAGG’S LAW OF DIFFRACTION AC sin d ACB 2 sin d ACB n 2 sin n d Bragg’s Law: When Bragg’s Law is satisfied, “reflected” beams are in phase and interfere constructively. Specular “reflections” can occur only at these angles.
No peak is observed unless the condition for constructive interference (δ = nλ, with n an integer) is precisely met: 229 DIFFRACTION ORDERS 1st order: 1 2 sin d 2nd order: 2 2 2 sin d By convention, we set the diffraction order = 1 for XRD. For instance, when n=2 (as above), we just halve the d-spacing to make n=1.
2 2 2 sin d 2 2( / 2)sin d e.g. the 2nd order reflection of d100 occurs at same θ as 1st order reflection of d200 XRD TECHNIQUES AND APPLICATIONS • powder diffraction • single-crystal diffraction • thin film techniques • small-angle diffraction • phase identification • crystal structure determination • radial distribution functions • thin film quality • crystallographic texture • percent crystalline/amorphous • crystal size • residual stress/strain • defect studies • in situ analysis (phase transitions, thermal expansion coefficients, etc) • superlattice structure Uses: POWDER X-RAY DIFFRACTION • uses monochromatic radiation, scans angle • sample is powder → all orientations simultaneously presented to beam • some crystals will always be oriented at the various Bragg angles • this results in cones of diffracted radiation • cones will be spotty in coarse samples (those w/ few crystallites) crystallite no restriction on rotational orientation relative to beam 232 2 sin hkl hkl d 233 Transmission geometry DEBYE-SCHERRER METHOD …or we can use a diffractometer to intercept sections of the cones 234 2 sin hkl hkl d BASIC DIFFRACTOMETER SETUP 235 THETA-2THETA GEOMETRY • X-ray tube stationary • sample moves by angle theta, detector by 2theta 238 POWDER DIFFRACTOGRAMS increasing θ, decreasing d Minimum d?
min / 2 d In powder XRD, a finely powdered sample is probed with monochromatic X-rays of a known wavelength in order to evaluate the d-spacings according to Bragg’s Law.
Cu Kα radiation: λ = 1.54 Å peak positions depend on: • d-spacings of {hkl} • “systematic absences” 240 ACTUAL EXAMPLE: PYRITE THIN FILM FeS2 – cubic (a = 5.43 Å) Random crystal orientations On casual inspection, peaks give us d-spacings, unit cell size, crystal symmetry, preferred orientation, crystal size, and impurity phases (none!) 111 200 210 211 220 311 Cu Kα = 1.54 Å 2 Theta Intensity “powder pattern” 2θ = 28.3° → d = 1.54/[2sin(14.15)] = 3.13 Å = d111 reference pattern from ICDD (384,000+ datasets) d-SPACING FORMULAS 242 POWDER DIFFRACTION Peak positions determined by size and shape of unit cell Peak intensities determined by the atomic number and position of the various atoms within the unit cell Peak widths determined by instrument parameters, temperature, and crystal size, strain, and imperfections 244 we will return to this later… GENERATION OF X-RAYS X-rays beams are usually generated by colliding high-energy electrons with metals.
2p3/2 → 1s Siegbahn notation X-ray emission spectrum + HEAT 246 Generating Bremsstrahlung Generating Characteristic X-rays GENERATION OF X-RAYS Co Kα1 : 1.79 Å Cu Kα1 : 1.54 Å (~8 keV) Mo Kα1 : 0.71 Å / hc h E Side-window Coolidge X-ray tube X-ray energy is determined by anode material, accelerating voltage, and monochromators: 1/2 ( ) C Z Moseley’s Law: 247 SYNCHROTRON LIGHT SOURCES SOLEIL • brightest X-ray sources • high collimation • tunable energy • pulsed operation GeV electron accelerators 249 Bremsstrahlung (“braking radiation”) MONOCHROMATIC X-RAYS Filters (old way) A foil of the next lightest element (Ni in the case of Cu anode) can often be used to absorb the unwanted higher-energy radiation to give a clean Kα beam Crystal Monochromators Use diffraction from a curved crystal (or multilayer) to select X-rays of a specific wavelength 250 DETECTION OF X-RAYS • Point detectors • Strip detectors • Area detectors Detection principles • gas ionization • scintillation • creation of e-h pairs 251 DETECTION OF X-RAYS Gas proportional counters Point detectors 252 Scintillation counters X-RAY DETECTORS Area detectors Charge-coupled devices • film • imaging plate • CCD • multiwire 253 X-RAY DETECTORS Imaging plates 254 photostimulated luminescence from BaFBr0.85I0.15:Eu2+ X-RAY DETECTORS Imaging plates 255 photostimulated luminescence from BaFBr0.85I0.15:Eu2+ tetragonal Matlockite structure 9-coordinate Ba!
The Reciprocal Lattice and the Laue Description of Diffraction 256 Reading: A/M 5-6 G/S 3 PLANE WAVES A wave whose surfaces of constant phase are infinite parallel planes of equal spacing normal to the direction of propagation.
ψ: wave amplitude at point r A: max amplitude of wave k: wave vector r: space vector from arbitrary origin k |k|=2π/λ Amplitude is constant in any plane normal to k because k•r is a constant for such planes: k•r1 = kr1 k•r2 = kr1√2(cos45) = kr1 k r2 wavefront origin k r1 45° k•r is indeed constant on wavefronts THE RECIPROCAL LATTICE The reciprocal lattice of a Bravais lattice is the set of all vectors K such that for all real lattice position vectors R. 1 i e K R R = n1a1 + n2a2 + n3a3 Direct lattice position vectors: Reciprocal lattice vectors: 2 2 3 1 1 2 3 a a b a a a K = hb1 + kb2 + lb3 2 3 1 2 1 2 3 a a b a a a 2 1 2 3 1 2 3 a a b a a a where the primitive vectors of the reciprocal lattice are: and {ni} and {h,k,l} are integers Reciprocal lattice: The set of all wave vectors K that yield plane waves with the periodicity of a given Bravais lattice.
258 is satisfied when K•R = 2πn, with n an integer To verify that the {bi} are primitive vectors of the reciprocal lattice, let’s first show that bi•aj = 2πδij 2 2 2 1 2 3 2 3 1 1 1 1 2 3 1 2 3 a a a a a b a a a a a a a a 2 0 3 1 2 1 1 1 2 3 a a b a a a a a 2 0 1 2 3 1 1 1 2 3 a a b a a a a a Indeed, bi•aj = 2πδij so, K•R = (hb1 + kb2 + lb3)•(n1a1 + n2a2 + n3a3) = 2π(hn1 + kn2 + ln3) = 2π × integer (since cross product of two vectors is perpendicular to both) K is indeed a reciprocal lattice vector WHAT IS A RECIPROCAL LATTICE VECTOR?
The reciprocal lattice is defined at the lattice generated from the set of all vectors K that satisfy for all direct lattice position vectors R. 1 i e K R What is K?
a wave vector of a plane wave that has the periodicity of the direct lattice The direct lattice is periodic (invariant under translation by R) Reciprocal lattice vectors = wave vectors of plane waves that are unity at all direct lattice sites 260 THE RECIPROCAL LATTICE • the reciprocal lattice is defined in terms of a Bravais lattice • the reciprocal lattice is itself one of the 14 Bravais lattices • the reciprocal of the reciprocal lattice is the original direct lattice e.g., simple cubic direct lattice ˆ a 1 a x ˆ a 2 a y ˆ a 3 a z 2 3 2 ˆ ˆ 2 2 a a a 2 3 1 1 2 3 a a b x x a a a 2 ˆ a 2 b y 2 ˆ a 3 b z →simple cubic reciprocal lattice with lattice constant 2π/a →b1 parallel to a1, etc. 261 Crystals with orthogonal axes (cubic, tetragonal, orthorhombic) b1, b2, b3 are parallel to a1, a2, a3, respectively. b3 a3 b1 a1 a2 b2 reciprocal lattice direct lattice 2 ˆ b 2 b y 2 ˆ a 1 b x 2 ˆ c 3 b z ˆ a 1 a x ˆ b 2 a y ˆ c 3 a z 262 RECIPROCAL LATTICE OF FCC IS BCC FCC primitive vectors: 2 3 ˆ ˆ ˆ ( ) 4 1 4 ˆ ˆ ˆ 2 2 ( ) 2 (2) 8 a a a 2 3 1 1 2 3 y z - x a a b y z - x a a a Note: not orthogonal 4 1 ˆ ˆ ˆ ( + ) 2 a 2 b z x - y 4 1 ˆ ˆ ˆ ( + ) 2 a 3 b x y - z →BCC reciprocal lattice with lattice constant 4π/a 263 RECIPROCAL LATTICE OF BCC IS FCC BCC primitive vectors (not orthogonal): 2 3 ˆ ˆ (2 2 ) 4 1 4 ˆ ˆ 2 2 ( ) 2 (4) 8 a a a 2 3 1 1 2 3 y z a a b y z a a a 4 1 ˆ ˆ ( ) 2 a 2 b z + x 4 1 ˆ ˆ ( ) 2 a 3 b x + y →FCC reciprocal lattice with lattice constant 4π/a 264 RECIPROCAL LATTICES • simple orthorhombic →simple orthorhombic • FCC →BCC • BCC →FCC • simple hexagonal →simple hexagonal (rotated) 265 FIRST BRILLOUIN ZONES The Wigner-Seitz cell of the reciprocal lattice is called the first Brillouin zone (FBZ).
Wigner-Seitz cell: primitive cell with lattice point at its center enclosed region is W-S cell for 2D hexagonal lattice d.l. FCC r.l. BCC 1st Brillouin zone: truncated octahedron rhombic dodecahedron d.l. BCC r.l. FCC 1st Brillouin zone: 269 Theorem: For any family of lattice planes separated by distance d, there are reciprocal lattice vectors perpendicular to the planes, the shortest of which has a length of 2π/d.
Conversely, any reciprocal lattice vector K has a family of real-space planes normal to it, separated by d. hk in 2D hkl in 3D here, g = K K and LATTICE PLANES 275 Orientation of a plane is determined by its normal vector It is natural to pick the shortest perpendicular reciprocal lattice vector to represent the normal Miller indices: coordinates of this reciprocal lattice vector i.e., A plane with Miller indices hkl is normal to the reciprocal lattice vector K = hb1 + kb2 + lb3 → Definition #2: directions in k-space (Definition #1 was inverse intercepts in the real lattice) MILLER INDICES OF LATTICE PLANES 276 Proof that K = hb1 + kb2 + lb3 is normal to (hkl) h 1 a AB If K = hb1 + kb2 + lb3 is normal to the plane at left, its dot product with any in-plane vector is zero.
Consider vector AB that lies in the plane.
By vector addition, l 3 a k 2 a h l AB 3 1 a a The dot product, ( ) h k l h l AB K = 3 1 1 2 3 a a b b b 2 2 0 = So the reciprocal vector formed by using the Miller indices of a plane as its components forms a vector in space that is normal to the Miller plane. Furthermore, the length of the shortest vector K is equal to 2π/dhkl.
In the figure above, the spacing between the planes is the projection of : on h K K 1 a 2 2 hkl h d h h K K K K 1 a (hkl) 0 2 hkl K d K → 277 etc. POWDER (DEBYE-SCHERRER) METHOD • single wavelength • fixed powder sample • equivalent to rotating the reciprocal lattice through all possible angles about the origin every point in reciprocal space traces out a shell of radius K Each shell with radius K < 2k intersects the Ewald sphere to form a circle.
All the diffracted beams from a powder lie on the surface of cones 291 Peak intensities depend on (in large part): 1) intensity scattered by individual atoms (form factors) 2) the resultant wave from atoms in unit cell (structure factor) PEAK INTENSITIES In many cases, the intensity from certain planes (hkl) is zero.
• symmetry of crystal causes complete cancellation of beam “systematic absences” • happenstance Possible reasons: Other factors that affect intensity: • scattering angle • multiplicities • temperature factor • absorption factor • preferred orientation 292 MONOATOMIC BASES ( - ) 2 = 1 i ' i n e e k k R up to now we have considered diffraction only from Bravais lattices with single atom bases (i.e., atoms only at the lattice points R).
We found the diffraction condition: =1 i e K R which is the same as: ( ) i F f e K R K R R K The scattering amplitude FK is the sum over the lattice sites: The scattered intensity is proportional to the absolute square of the amplitude: where fR(K) is the “atomic form factor” for a given atom (disc. later).
2 0 I I F K K …this is what is actually measured in an experiment.
Crystals with n atoms in each primitive cell must be further analyzed into a set of scatterers at positions d1, d2 … dn within each primitive cell. ( ) ( ) j i j j F f e K R+d K R K n-ATOM BASES ( ) j j A R R d The positions of the atoms are: making the scattering amplitude: ( ) j i i j j e f e K d K R R K i L e K R R ( ) j i j j f e K d K K “Lattice sum” “Structure factor” of the basis If the structure factor = 0, there is no diffraction peak.
( ) j i j j f e K d K K The structure factor gives the amplitude of a scattered wave arising from the atoms with a single primitive cell. STRUCTURE FACTOR For crystals composed of only one type of atom, it’s common to split the structure factor into two parts: ( ) j f S K K K j i j S e K d K “atomic form factor” “geometric structure factor” S = 0 gives a systematic absence (i.e., absence of expected diff. peak).
295 2 ( ) hkl I S K 1 i e K d n i e K d … 1 j n i j S e K K d The amplitude of the rays scattered at positions d1, …, dn are in the ratios: The net ray scattered by the entire cell is the sum of the individual rays: STRUCTURE FACTORS Geometric structure factor -Adds up scattered waves from unit cell -In particular, no peak when SK = 0 296 For simple cubic: one atom basis (0,0,0) 0 1 i S e K K SIMPLE CUBIC d1 = 0a1 + 0a2 + 0a3 297 Same result as simple monatomic basis For monoatomic BCC: we can think of this as SC with two point basis (0,0,0), (½,½,½) l k h ) 1 ( 1 S = 2, when h + k + l even S = 0, when h + k + l odd (systematic absences) 2 ( ) 0 2 1 ( ) 1 j a i x y z i i j i h k l S e e e e K K K K d MONATOMIC BCC 2 ˆ ˆ ˆ ( ) h k l a K x y z For SC, 298 e.g. consider the powder pattern of BCC molybdenum Powder card shows only even hkl sums b/c Mo is BCC Why?
- Diffraction from other (hkl) results in destructive interference: (100) d100 Beam cancels b/c body center atoms scatter exactly 180° out of phase (200) d200 Strong reflection b/c all atoms lie on 200 planes and scatter in phase S = 4 when h + k, k + l, h + l all even (h, k, l all even or all odd) S = 0 otherwise.
( ) ( ) ( ) 1 i h k i k l i h l S e e e K For monoatomic FCC: SC with four point basis (0,0,0), (½,½,0), (0,½,½), (½,0,½) 4 ( ) ( ) ( ) 0 2 2 2 1 j a a a i x y i y z i x z i i j S e e e e e K K K K K K d MONATOMIC FCC 2 ˆ ˆ ˆ ( ) h k l a K x y z For SC, 300 (hkl) NaCl KCl (100) (110) (111) (200) (210) (211) (220) (221) (300) (310) (311) Once again, there are more systematic absences for isoelectronic ions (e.g., K and Cl) (110) always absent in RS (111) sometimes absent 306 DIAMOND STRUCTURE Diamond: FCC lattice with two-atom basis (0,0,0,), (¼,¼,¼) ( ) 0 4 , , ( /2)( ) , [ ][ ] [1 ][ ] a iK x y z iK diamond FCC i h k l FCC S e e S e S K K K S = 8 h + k + l twice an even number S = 4(1 ± i) h + k + l odd S = 0 h + k + l twice an odd number IFCC : all nonvanishing spots have equal intensity.
Idiamond : spots allowed by FCC have relative intensities of 64, 32, or 0. 308 Only for all even or all odd hkl is S ≠ 0. For these unmixed values, Additional condition: (hkl) Al Si (100) (110) (111) (200) (210) (211) (220) (221) (300) (310) (311) What about zinc blende?
FCC diamond 309 SUMMARY OF SYSTEMATIC ABSENCES crystal structure condition for peak to occur SC any h,k,l BCC h + k + l = even FCC h,k,l all even or all odd NaCl h,k,l all even, or all odd if fA ≠ fB diamond h,k,l all even and twice an even #, or all odd HCP any h,k,l except when h + 2k = 3n and l is odd ( ) j i j j f e K d K K 310 sum =h+k+l=4n and Silicon Observable diffraction peaks for monoatomic crystals 2 2 2 l k h SC: 1,2,3,4,5,6,8,9,10,11,12,… BCC: 2,4,6,8,10,12,...
FCC: 3,4,8,11,12,16,24,… SIMPLE ANALYSIS OF SIMPLE PATTERNS What will we see in XRD patterns of SC, BCC, FCC?
SC FCC BCC We can take ratios of (h2 + k2 + l2) to determine structure.
SIMPLE ANALYSIS OF SIMPLE PATTERNS n d sin 2 2 2 2 l k h a dhkl For cubic crystals: 2 2 1 sin hkl d 2 2 2 2 sin ( ) h k l 2 2 2 2 th peak th peak 2 2 2 2 1st peak 1st peak sin ( ) sin ( ) n n h k l h k l 312 2 2 2 2 2 1 sin sin 33 2 sin sin 22 SIMPLE ANALYSIS OF SIMPLE PATTERNS 110 200 211 α-Fe is cubic. Is it FCC or BCC?
BCC!
What about Al?
2 2 2 2 2 1 sin sin 22.5 1.33 sin sin 19 111 200 220 311 222 400 331 420 FCC!
313 Ex: An element, BCC or FCC, shows diffraction peaks at 2: 40, 58, 73, 86.8,100.4 and 114.7. Determine:(a) Crystal structure?(b) Lattice constant?
(c) What is the element?
2theta theta (hkl) 40 20 0.117 1 (110) 58 29 0.235 2 (200) 73 36.5 0.3538 3 (211) 86.8 43.4 0.4721 4 (220) 100.4 50.2 0.5903 5 (310) 114.7 57.35 0.7090 6 (222) 2 sin 2 2 2 l k h BCC, a =3.18 Å W normalized 314 ELASTIC X-RAY SCATTERING BY ATOMS Atoms scatter X-rays because the oscillating electric field of an X-ray sets each electron in an atom into vibration. Each vibrating electron acts as a secondary point source of coherent X-rays (in elastic scattering).
Thomson relation: The X-ray scattered from an atom is the resultant wave from all its electrons Particle picture: • zero phase difference for forward/backward scattering → scattering factor (form factor, f ) proportional to atomic number, Z • increasingly destructive interference with larger scattering angle (to 90°) • for a given angle, intensity decreases with decreasing X-ray wavelength • max scattering intensity at 2θ = 0 & 180° • gradual decrease to 50% as 2θ approaches 90° 2 1 (1 cos 2 ) 2 I SCATTERING OF X-RAYS BY ATOMS Thomson relation: 2 1 (1 cos 2 ) 2 I scattering angle probabilities for a free electron: Low energy: Thomson High energy: Compton Klein–Nishina formula ATOMIC FORM FACTORS Form factor f = scattering amplitude of a wave by an isolated atom • Z (# electrons) • scattering angle • X-ray wavelength For X-rays, f depends on: consequences: • powder patterns show weak lines at large 2θ. • light atoms scatter weakly and are difficult to see.
0 ( ) ( ) i j j f e d q r q r r 4 sin q with, For θ = 0 (forward scattering), scattering vector q General elastic formula: 0 (0) ( ) j f d # electrons r r = O K+ Cl-Cl θ = 37° 3 3 PEAK WIDTHS Peak shape is a Voigt function (mixture of Gaussian and Lorentzian) Peak width (broadening) is determined by several factors: • natural linewidth of X-ray emission • instrumental effects (polychromatic λ, focusing, detector) • specimen effects 1) crystallite size 2) crystallite strain • Gaussian component arises from natural linewidth and strain • Lorentzian component arises from coherent domain size Pure Lorentzian Pure Gaussian 319 320 FULL WIDTH AT HALF MAXIMUM (FWHM) Instrument and Sample Contributions to the Peak Profile must be Deconvoluted • In order to analyze crystallite size, we must deconvolute: – Instrumental Broadening FW(I) • also referred to as the Instrumental Profile, Instrumental FWHM Curve, Instrumental Peak Profile – Specimen Broadening FW(S) • also referred to as the Sample Profile, Specimen Profile • We must then separate the different contributions to specimen broadening – Crystallite size and microstrain broadening of diffraction peaks 321 SIZE BROADENING Small crystallites (< 200 nm) show broadened diffraction lines Nanocrystal X-ray Diffraction 322 323 Which of these diffraction patterns comes from a nanocrystalline material?
66 67 68 69 70 71 72 73 74 2 (deg.) Intensity (a.u.) These diffraction patterns were produced from the same sample!
• Two different diffractometers, with different optical configurations, were used • The apparent peak broadening is due solely to the instrumentation in this case 324 1 2 3 4 j-1 j j+1 2j-1 2j B 1 2 at Bragg angle, phase lag between two planes = perfectly in phase, constructive B B 1 At some angle Phase lag between two planes: At (j+1)th plane: Phase lag: • Rays from planes 1 and j+1 cancel • Ditto for 2 & j+2, … j & 2j • Net diffraction over 2j planes = 0 2 j The finite size of real crystals results in incomplete destructive interference over some range of angles Crystal with 2j planes Total thickness T T = (2j-1)d The angular range θB to θ1 is the range where diffracted intensity falls from a maximum to zero (half of Bragg peak profile).
Same arguments apply to B 2 So we see diffracted X-rays over all scattering angles between 2θ1 and 2θ2.
– If we assume a triangular shape for the peak, the full width at half maximum of the peak will be B = (2θ1 – 2θ2)/2 = θ1 – θ2 326 If we have more than 2j planes: 1 2 3 4 j-1 j j+1 2j+1 2j+2 B 1 2 If we have fewer than 2j planes: 1 2 3 4 j-1 j j+1 2j-3 2j-2 B 1 2 still zero intensity at θ1 nonzero intensity at θ1 Rays from planes j-1 & j not canceled Rays from new planes are canceled Thinner crystals result in broader peaks!
327 Peak sharpens!
Peak broadens!
Let’s derive the relation between crystal thickness T and peak width B: 2 sin d 1 2 2 sin (2 1) 2 sin (2 1) T j T j 1 2 (sin sin ) T 1 2 1 2 2 (cos( )sin( )) 2 2 T 1 2 2 (cos )( )) .
2 B T cos B T B 1 2 2( ) 2 B Considering the path length differences between X-rays scattered from the front and back planes of a crystal with 2j+1 planes and total thickness T: If we subtract them: Using trig identity: Since and , 1 2 2 B 1 2 1 2 sin( ) 2 2 But, , so 1 2 1 2 1 2 sin sin 2cos sin 2 2 Here, T = 2jd cos B K T B 2 2 2 M R B B B BM: Measured FWHM (in radians) BR: Corresponding FWHM of bulk reference (large grain size, > 200 nm) Readily applied for crystal size of 2-100 nm.
Up to 500 nm if synchrotron is used.
SCHERRER FORMULA A more rigorous treatment includes a unitless shape factor: Scherrer Formula (1918) T = crystallite thickness λ (X-ray wavelength, Å) K (shape factor) ~ 0.9 B, θB in radians Accurate size analysis requires correction for instrument broadening: 329 • The constant of proportionality, K (the Scherrer constant) depends on the how the width is determined, the shape of the crystal, and the size distribution – the most common values for K are: • 0.94 for FWHM of spherical crystals with cubic symmetry • 0.89 for integral breadth of spherical crystals w/ cubic symmetry • 1, because 0.94 and 0.89 both round up to 1 – K actually varies from 0.62 to 2.08 • For an excellent discussion of K, refer to JI Langford and AJC Wilson, “Scherrer after sixty years: A survey and some new results in the determination of crystallite size,” J. Appl. Cryst. 11 (1978) 102-113.
cos B K T B SCHERRER CONSTANT 0.94 cos B T B 330 Suppose =1.5 Å, d=1.0 Å, and =49˚. Then for a crystal 1 mm in diameter, the width B, due to the small crystal effect alone, would be about 2x10-7 radian (10-5 degree), too small to be observable. Such a crystal would contain some 107 parallel lattice planes of the spacing assumed above.
However, if the crystal were only 50 Å thick, it would contain only 51 planes, and the diffraction curve would be very broad, namely about 43x10-2 radian (2.46˚), which is easily measurable.
331 “Incomplete destructive interference at angles slightly off the Bragg angles” DIFFRACTION FROM DISORDERED SOLIDS amorphous solids 367 |
2738 | https://www.stat.purdue.edu/~lfindsen/stat311/Stars%20and%20Bars%20Review.pdf | 1 Stars and Bars – Review This methodology is with replacement and order does not matter. This means that the options can be repeated and we cannot distinguish the options In the book it has the formula for stars and bars as (𝑛+𝑟−1 𝑟 ) I find it easier to use the following formula: (𝑛+𝑟−1 𝑟−1 ) = (𝑛+𝑟−1 𝑛 ). When using this formula, the stars are r and the bars are n. To me, this makes more sense. n is the total number of objects, r is the number we are choosing. I found this information at the following reference: For the example that I was doing in class with 5 letters in groups of 2. Therefore we have 5 letters (bins, r) and 2 positions (stars, n) (𝑛+ 𝑟−1 𝑛 ) = (5 + 2 −1 2 ) = 15 Counting Handout 1 (SB) 1) How many different sets of non-negative numbers x, y and z are solutions for the following equation: x + y + z = 136. stars = n = 136, bins = r = 3 The total number of stars is 136 and we need to put them into 3 bins. (136 + 3 −1 136 ) = (138 136) = 138!
(138 −136)! 136! = 9453 (I was wrong on the handout.) 2) How many ways are there to buy 13 bagels from 17 types if you can repeat the types of bagels? stars (bagels) = n = 13, bins (types) = n = 17 (17 + 13 −1 13 ) = (29 13) = 29!
(29 −13)! 13! = 67,863,915 Counting Handout 2 3) How many ways can a person draw three dice from a bag containing 10 identical dice if after each draw the person puts the drawn dice back into the bag. Oops…This is not stars and bars, if the dice are identical, then how can you tell them apart. So the answer is 1. 4) Suppose that a small pond contains 500 fish, 50 of them tagged. A fisherman catches 10 fish which he cannot tell apart. After each time he catches a fish, he throws the fish back into the pond. Find the number of ways that he can catch 2 tagged fish. stars (fish) = n = 10, bins (tagged or not tagged) = r = 2 (10 + 2 −1 10 ) = (11 10) = 11 (I was wrong on the handout.) 2 5) How many ways can a person place 5 (undistinguishable) 6-side dice into 4 different containers? stars (dice) = n = 5, bins (containers) = r = 4 (4 + 5 −1 5 ) = (8 5) = 56 (I got this one correct!) |
2739 | https://www.youtube.com/watch?v=61bVFHqSYko | TI-Nspire - Solving Quadratic Equations
Lisa McLeod
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this video will show you how to solve a quadratic equation in the ti Inspire calculator from your home screen we are going to choose number one open a new document from here we're going to click number one again I want to add a calculator once I have the calculator open on my page I'm going to click menu and I'm going to try number three algebra and number three again polom tools I'm going to select number one find the roots of a polinomial our example is going to be x^2 + 4x + 3 = 0 the degree of that quadratic is two the highest exponent in that equation is a two so we'll leave this alone I want only my real roots and I can click okay you'll notice A2 is the number the coefficient that is in front of my X squar piece so in the equation x^2 + 4x + 3 = 0 I have a 1 in front of the X squ the A1 is in front of the X that was a four and a0 was my constant that was a three constant has no variable with it at all click okay and then click enter my answers are -3 and -1 so x = -3 and x = -1 |
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Undergraduate Journal of Mathematical Modeling: One + Two
Home > Open Access Journals > UJMM > Vol. 10 > Iss. 1 (2019)
Acid-Base Titration
Authors
David Pierre, University of South Florida
Publication Year
2019
Abstract
Titration is an analytical method used in biomedical sciences and analytical chemistry laboratories to determine the quantity or the concentration of a known or unknown substance. There are many kinds of titrations, but this investigation is fundamentally based on acid-base titration, in which the development of a mathematical titration model is explained and determined. In a titration experiment, the concentration and volume of the base added is proportionally equal to those of the acid. The concentration of the substance is determined by following this principle. Moreover, these calculations are usually approximate. Thus, the construction of a titration model is done to minimize the percent errors, simultaneously describing the effect of titration, and also allowing the determination of the characteristics of the unknown substance used. Furthermore, the values generated by the titration model are compared to the theoretical ones. It allows the verification of the mathematical model constructed. Subsequently, the efficiency of titration model is further tested by using a theoretical problem to exclude the errors from the experiment. This verification displays the accuracy of the accepted titration model. The concentration of the unknown acid is found by using this mathematical model.
Recommended Citation
Pierre, David (2019) "Acid-Base Titration," Undergraduate Journal of Mathematical Modeling: One + Two: Vol. 10: Iss. 1, Article 8. DOI: Available at:
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2741 | https://soarfoundation.net/wp-content/uploads/2020/07/2018%E6%A0%91%E5%8D%8E%E5%B0%8F%E5%A4%A9%E5%9C%B0.pdf | 小天地 2018 知识就是力量---培根 树华北京办公室地址： 北京市海淀区上庄镇白水洼路 东篱下私家庭院，邮编102206 联系电话：13801212049 树华官方网站： www.soarfoundation.net 各位同学： 首先恭贺大家进入高中并加入树华的大家庭。树华基金 会是在美国加州旧金山湾区成立的一个慈善机构，已有23年的历 史。基金会的宗旨是给中国大陆贫困上进的青少年提供奖学金， 使他们有机会完成高中及职业院校教育。基金会的捐款人大多数 为旅美华侨。他们心系祖国，愿意为青少年的教育做出供献。 高中是一个新的人生阶段。在这三年中，你们要学习的不 只是书本上的知识，也要开始注意培养独立思考的能力以及与人合作的习惯。高中 是一个成长的重要过程，也是一个充满乐趣及接受新事物的时段。记得我高中时就 交了几个好友，几十年都在我身旁分享彼此在生活中的欢乐，也是彼此在困难时侯 的精神支柱。希望你们也能够像我一样的幸运，能交到一生的好友。 至于为人，立定目标作一个公正，善良有诚信的人。做事不求一步登天， 不走捷径。事情的成功是靠一步步的努力，好好充实自己，累积经验及学识打下人 生的基础。总之，希望同学们能用一个正确而乐观的心态来面对高中的时光。这三 年将是你们努力学习知识，同时也是一个充满欢乐成长的日子。 会长的话 会长：李静和 新疆 內蒙古 黑龍江 吉林 遼寧 山西 河北 山東 河南 陝西 湖北 江西 湖南 貴州 廣西 廣東 福建 浙江 江蘇 安徽 西藏 青海 雲南 四川 甘肅 寧夏 重慶 台灣 海南 92 天津 北京 39 62 96 132 25 80 47 48 1 1 14 135 77 113 1 72 38 42 52 37 24 38 2017-2018树华获奖学生分布图 高中生：1,036 职大： 52 大学： 177 获奖学生总数：1,265 敦品励学、自强不息、馈乡报国！ 同学们好！ 在2018学年开始之际，首先欢迎新老学生聚集在树华这个充满爱的大家庭 之中。 你们来自全国十几个省市地区，家庭背景各有不同，有着不同的生活经历， 但是你们有一个共同点：都是通过学校老师推荐，经过甄选委员老师选拔的优秀学 生。你们有着同样的目标：以优异的成绩完成高中学业，考入大专院校，最终改变 自身和家庭的命运，实现自己的梦想。树华大家庭的老师们将帮助和陪伴你们一起 圆满地完成高中这一段重要的人生里程。 明年一月至三月期间，你们的甄选委员老师们将联系你们，把《树华奖学 金申请表－2019 （继续获奖学生使用）》邮寄或当面发给你们。 在繁重的高中课 业的同时，同学们务必在四月份之前把表格和附加文件邮寄回或交给各自的甄选委 员老师。 甄选委员老师们各自都有繁忙的工作和生活，希望同学们严格按照申请 表格的要求，按时递交完整的申请材料，如果申请材料不符合要求，你的申请将 被否决。 重申填写表格时的几点注意事项： 1. 必须是自己填写的，字迹工整，这是你给树华奖学金评审老师们的第一 印象。 2. 提供真实的信息。 如果有不明的家庭情况，请注释，但是不要留空白。 比如家庭成员收入一栏，来自农村的同学如果有问题，可以请家长做出大致估计 并注释。 例：“约5000元”， 或 “5000元左右”。 3. 班主任签名一栏， 必须是老师的亲笔签名，否则，申请会被否决。 4. 推荐信必须是推荐老师亲笔签名，自己签名，或假冒签名申请会被否决。 5. 必须有成绩单。 对成绩单的要求是有班级和年级排名的。 如果所在学 校不提供排名成绩单，请在成绩单上注释， 并由班主任老师签名和学校盖章。 申请表格和申请材料如果不符合要求，申请将被否决。 在2017和2018年的申请中， 绝大部分同学递交了一份学习总结或者感谢信。 从中，我们看到了同学们在过去一年中在各方面明显的成熟和进步。可想而知，当 我们把这些文章转交给奖学金捐助人时，他们会多么欣慰！树华的很多捐款人都不 留姓名，他们不需要你为他们做什么，但他们在默默地关注你的成长 。树华作为一 个慈善教育机构，是架起这个美好关爱的桥梁，我们有责任让这些无私的爱心人士 了解他们的奉献没有白费，他们资助的是有品德，有理想，会珍惜，懂感恩，值得帮 助的年轻人。来而不往非礼也，让我们一起对树华的捐款人真诚地说一声感谢！ 祝同学们在新的学年中学业进步，生活充实而且快乐！ 校不提供排名成绩单，请在成绩单上注释， 并由班主任老师签名和学校盖章。 奖学金组负责人陈勤老师的一封信 【编者按】：星星之火 ，可以燎原。张笑枫女士，对中国青少年的教育问题，超 越政治、宗教和地域因素，她无私的关爱，引燃了树华全体同仁，对中国欠发达 地区，青少年教育，长达23年的资助，9000余位青少年完成学业，甚至改变了命运， 而这一数字仍将继续增长，树华也将陪伴他们一同成长⋯⋯ 树华教育基金会，23年來为中国大陆，贫困而优秀的学子，捐了一千多万 美金。当初树华成立的时候，是预想不到的。因为我们并没有大企业或者是大基金 会的背景。 我从台湾来到美国，又是工作，又是家庭非常忙，我们是在海外辛辛苦苦 打拼的“第一代”，那个时候树华的义工，基本上都是台湾人。有人问我为什么要 帮助大陆。是因为八十年代去广州开工厂时，偶然看到的一幕改变了我的人生轨迹。 改革开放后，大家觉得大城市有金坑银坑，南方很多农民进城打工，甚至 还有十几岁的孩子，结果很多人来了，找不到工作，没办法生存，想要回家，其中 一些人，已经花光积蓄，没钱买车票，在车站争着上车。结果广州火车站前一片混 乱，这些孩子们也被卷入其中。这一幕让我感到震撼，究竟是什么情况呢？我开始 思考。 寻寻觅觅，一直到95年我遇见了赵耀渝，她是旧金山金门大学的教授。有 很好的教育背景，工作能力极强，尤其是软件方面。成立基金会，必须要很好很强 的团队，我们两个一拍即合，我说赵教授，你把整个团队做起来，我去找资金，我 们分头行动。 我们分头到大陆，走进深山了解情况，那个时候的交通，很不方便，记得 到云南大果乡，走路要好几个钟头，因为路通不到那里。今天的大果乡，已经不可 同日而语。经过6个月调研，我决定创立树华教育基金会。 回到美国后我到处游说找义工，比如萧又青和她的先生， 后來帮我们带来60万美元的永续基金。斯坦福的一个朋友介绍了 John Macmeeken，他是一个美国人，帮助我们成立基金会，并做 我们多年的法律顾问。那个时候这些人都是中流砥柱，太多太多了 不起的人。我已经没办法讲那么多名字了。我感到，世上好人这么多， 善意的人这么多，诚恳的人这么多，有能力的人这么多，1995年， 水到渠成，树华教育基金会成立了。 不知不觉中，树华已走过23年头，这23年中， 我每天都在感恩，感谢树华来了很多了不起的新人， 现在的树华人才济济，卧虎藏龙，这方面要讲 树华创始人张笑枫女士讲述树华的故事 回到美国后我到处游说找义工，比如萧又青和她的先生， 后來帮我们带来60万美元的永续基金。斯坦福的一个朋友介绍了 John Macmeeken，他是一个美国人，帮助我们成立基金会，并做 我们多年的法律顾问。那个时候这些人都是中流砥柱，太多太多了 不起的人。我已经没办法讲那么多名字了。我感到，世上好人这么多， 善意的人这么多，诚恳的人这么多，有能力的人这么多，1995年， 不知不觉中，树华已走过23年头，这23年中， 谢树华来了很多了不起的新人， 树华义工郭丽文（整理） 几天都讲不完。总之，下一代的缺失就是我们的担忧，我们希望，把人间最美好 的东西带给下一代，这就是我们所有义工最大的愿望。我很感恩，很幸运，我们 帮助了孩子的同时，把我们的关爱传递过去。 我们每年都去探访学生，看到以前帮扶过的孩子，现在变的很靓丽。记 得四川有一个流鼻涕的孩子，离开的时候送给我一个他妈妈亲手做的鞋垫，这是 很大的一个礼物啊，现在还摆在我家里，而现在他已是广西的一位律师了。我们 的孩子很有成就，他们的成就，就是给我们最好的回报，这回报，让我们很有成 就感。我的后半生如此快乐，是因为基金会，那不是简单的给予，而是太多的收获。 现在我们已经七八十岁了，年纪最长的已经九十几岁，很担心基金会工 作人员断层。值得庆幸的是，3年前，树华前董事长高大伟先生把几位年轻人接进 来了，她们是早年从大陆来的留学生，了解国内的情况，能做实事，并且做得非常 好，最可贵的是她们都很有爱心。Lisa，她不眠不夜，带领他的团队，把树华的网 站做的风生水起，可达到商业化程度，我很感恩她所做的一切，这是我望尘莫及 的。Lily，她是我们的副会长，她做事情的时候，做的太多、太投入了，有時候 我会担心她累坏了。每一个义工都非常出色，所以我说，如果人间有一块净土的 话，那就是树华办公室。 在新学年开始之际致同学们 树华辅导组邹立美 同学们好！ 在2018-2019新学年开始之际，我向一千多名得到 奖学金的树华学子表示祝贺，愿你们在新学年成为积极向 上的，快乐的好学生。 我们树华的义工们认真阅读了每一份奖学金申请表， 听到了你们从内心发出的声音。有的同学盼望着政府把公路修到 村门口； 有的同学希望改变环境，让被塑料瓶和废纸污染的林子恢复原有的自然 生态； 有的同学计划着十年以后学成回乡，创造工作机会，让父母不再背井离乡， 外出打工，孩子不再成为留守儿童；有的同学立志当个律师，为弱势群体声张正 义等等。 你们对祖国，对家庭和自己抱有无限的希望，憧憬着美好的未来。 千里之行，始于足下。在你们立下了雄心壮志后，必须以坚韧不拔的精 神，努力去争取实现美好的理想。高中是关键的三年，是艰苦奋斗，但也是快乐 的三年。特别是高一的新同学，你们都是初中学校的佼佼者， 但是现在要面临新 学校，新同学，新教学要求的考验。在任何时候都不要忘记入学前的理想，要做 一个挑战困难，挑战自我的勇敢者。 人的一生会经受无数的考试。每门课的大，小测验都是对自己的一个小小 考验。在美国的学校里，我们常常听到老师的一句话：“Never competing with your 听到了你们从内心发出的声音。有的同学盼望着政府把公路修到 生 命 中 的 美 好 陪 伴 classmate。” 这句话告诉我们，不要和同学攀比。考完马上打听分数，比名次，盲 目攀比只会让自己丧失信心。每个学生有不同的学习方法，理解能力和不同的起点。 你应该做的是同自己的过去比较： 我是不是注重问题的理解了？ 我的学习习惯改进 了没有？ 我是不是处在很好的心态？ 不要急躁，学会释放压力，坚信自己会好起来， 也就是要有自信心。 高中学习中一定要培养良好的学习习惯。除了上课听讲，完成 作业外，一定要阅读教科书和大量的课外读物。只有拥有大量的知识， 你才会容易理解老师讲的内容，你的思维渠道被打开，你的作文才具有 丰富的内涵和精湛的文采。而学习能力提高，会帮助你去思考如何解决问题，积累 经验。在今天这个信息大爆炸的年代，大家都有相同的门槛。你跨进去了，在知识 的海洋里遨游，你就受益一辈子。你止步不前，就孤陋寡闻，落后于时代。有同学 总结物理考不好是公式背得不熟练。固然，你必须记住一些公式。但是学物理这门 自然科学必须要理解物理的基本原理，用概念去分析解释自然现象，然后你方可选 择正确的公式去计算。 目前的教育改革，是要逐步扭转为高考而学习的教育系统的偏差，要废除 以难题选拔人才的现象，从而鼓励学生拥有广泛的知识和解决问题的能力。从目前 改革试点的教材和考题来看，我们已经感觉到了这些正面的效应。 三年的高中校园生活，同学们朝夕相处。很多同学把老师 视为父母，与同学情同手足，建立了一个相互关爱的大家庭。有位 同学在学习总结中这样写道：在心理辅导课上当老师要求学生先闭 上双眼五分钟，思考着长大干什么工作时，他坚定的回答“老师”。 可见教师在学生的眼中的崇高形象。同时我们也发现了有些学 生有点困挠，或者是同学之间有点小小的矛盾，或者是苦恼于 不善言语，不会主动和同学沟通。同学们离开家庭，三年 生活在学校的环境之中。学校就是熔炉，是一个相对简单 的社会。离开社会，任何个人的努力都不可实现的。我们要学会主动融入集体，学 会表达自己。你也可以向老师同学主动要求参加集体的工作(如当个班级课代表，运 动会拉拉队员，宿舍管理员)。 万事开头难。走出了第一步，就会向前进了。不试， 你永远不知道自己还有演讲的能力。同学间要互相包容、相互尊重。在学校，你们 仅仅只有三，四年的年龄差距。但是当你毕业后进入社会，那才是个年龄，学历， 经历都不相同的大集体了。愿你们进入社会再回忆自己的高中生活时，每个人都会 珍惜那一段天真，淳朴，快乐的年华。 最后要想跟大家说的是要注意身体。很多同学在奖学金里拿出500元配眼镜， 或拿钱放在伙食费里。这些都是明智的做法。高中是你们长身体的时候；近视，营 养不良，或有蛔虫都会影响你们的发育，还会严重影响学习效果（头晕，打瞌睡， 记忆力差等）希望引起你们足够的重视（如保证每天吃一个鸡蛋等）。另外要积极 锻炼身体，坚持每天做适合自己的运动。营养加上运动这两方面问题都跟上的话， 你们的学习效率一定会大大提高。 在你们今后的前进道路上，树华会陪伴你们。祝你们在新学年更上一层楼！ 树华之光 我叫李霞，在深圳的一家投资公司做财务总监，是一名新 义工。有幸于2018年初夏参加了树华教育基金会走访广西留守 儿童的考察调研活动。 5月6日在广西桂林和从美国远道而来的树华教育基金会董 事长张笑枫会、会长IRIS、义工等一行七人汇合，开始了此次旅行。10天的行程在 疲惫和感动中结束，这一次特殊的旅途让我久久不能平静，思绪万千。一路走来收 获了满满的感动和力量。这些树华基金的前辈有几位已经年过古稀，依然精神矍铄。 每天和我们一样跋山涉水，每天中午都是3、4元一碗的米粉，为的是节省时间多走 访几间学校。每晚住在条件极差的招待所，除了严重的潮湿霉味不说，还有各种不 知名的小虫子光临我们的住所，肆意爬进鞋子、箱子。要知道这些树华基金的老前 辈都是旅居美国的成功人士，有自己的公司、自己的事业。而他们对饮食起居的要 求简陋随和，从未听他们抱怨过。无论多么艰苦的条件，他们都甘之如饴。8号的 晚上我们赶往雨卜苗寨，天空下着瓢泼大雨，30公里坑坑洼洼的烂泥山路，我们走 了整整三个半小时。我坐在副驾驶位，眼见前方山路到处都有塌方，还不时有石子 飞落。左边是汹涌湍急的雨卜河，右边是陡峻峭壁。我们的车子不是四轮驱动，好 几次在泥泞的山路上失去了方向控制，不住的打滑。司机师傅全神贯注汗如雨下， 车子里安静极了，大家都紧张的注视着望不到头的山路。没人抱怨、没人多言。三 个半小时后当车子终于冲出泥泞开上了平坦的柏油马路时，车里响起了热烈的掌声， 大家都为司机师傅的努力鼓掌，师傅此刻脸上也终于露出了笑容。大家笑言从地狱 回到了天堂。这条地狱之路让我看到了真正有修为的人们的素养，共度难关后他们 首先想到的不是自己，而是给予他人鼓励和感谢。这样温暖的氛围让我看到他们身 上闪耀着光辉的人性美。 这趟地狱之路连平日甚少出错的苹果手机也调皮了起来。我们在车里上下 翻飞，左右腾挪之时苹果手机的计步器居然很认真的给大家计步。大家为这“偷” 来的步数哈哈大笑，刚刚的紧张随着欢快的笑声一扫而空。接下来的行程都是行走 在最艰苦的山路上，试过九曲十八弯、车行水上⋯⋯，天气炎热、蚊虫叮咬都会被 大家乐观积极、无惧艰险的笑声遮去无踪。 树华基金把捐款人的每一分钱都用在了品学兼优但是贫困的孩子身上。此 次探访的所有费用包括往来国际机票、吃住行全部是AA制，都是花自己的钱，不 会动用捐款人的一分钱。这是我见过的最干净的基金会，也是最令人感动折服的。 行程的最后一天大家开了个小会，让每个人对这次的调研提出自己的看法 和建议。我作为新人第一个发了言：“我初次参与树华教育基金会的活动，还没有 资格提出什么看法和建议，但是这次旅行对我个人却是极好的提升的过程。已经开 始懈怠的我在各位前辈身上又汲取了力量，知道自己还需要奔跑，还需要努力。在 各位前辈身上我看到了善良的力量、看到了人性美的光芒、感受到了赠人玫瑰手有 余香的温暖。加入这样的大家庭实在是我的幸运，感恩能与你们同行，感恩能与你 们一起工作，一起帮助他人”。 是的，生命的每一次相遇都是注定的。不惑之年难免懈怠，庆幸在这个时 候遇到了树华。人生的列车停靠过不同的站台，见过形形色色的人，有过不同的旅 伴，希望接下来的旅途都能与树华同行，一同开往理想的彼岸。 树华义工 李霞 各位叔叔阿姨、爷爷奶奶：你们好！ 我是树华学子雷青青。谢谢你们去年来看我们！在这一年里，我过的挺好 的。自从升入高中，所要学得学科变成了9科。说实话，学习压力很大，作业也多， 有时候做也做不完，但还是要做。高一上学期期末考，我考了班级第十名，年级第 三十二名。考的不好的主要是地理、政治，地理没及格，我有些后悔，如果地理再 努力考10分，我就进年级前三十了，就会有奖学金了。发奖学金的那一天我心情很 沉重。我想在这个学校，我一定要考一次前二十名。 在2017年12月份的时候，我参加了全国中学生数理化学科能力展示活动，解 题技能展示的初赛，荣获贵州赛区高一年级物理、数学学科二等奖。我很高兴，但 是老师说这只是初赛，挺简单的。可我还是高兴，我晋级可参加复赛，但由于我们 学校没人去，而且费用是580/人，我就没去了。即使我知道这是个提高自己的机会， 但考虑到我自身准备也不充足，且时间紧迫，我就放弃了。这是我的选择，也是事 实。希望以后不要后悔，现在也要加油。 在2017年12月份左右，我的母亲在工地受伤，大腿骨折和盆骨受伤，住了3 个月的医院。当时我们又将近期末，父亲奔波于医院和家里，我和弟弟周末都去医 院陪我的母亲。弟弟在我期末考前两天发高烧39.9度，和我妈妈一起住院。我在期 末考前一天也发高烧40度，去医院打了一针，顺便看了一下他们，之后又回学校复 习。考试时差点儿睡着，喉咙也痛，说不出话。饭卡也没带，忘家里了。那时的我 只好与朋友共用一张饭卡，把钱充在里面。从那时开始，我那几个星期都没有向父 亲要钱，我取了你们所资助我的那些钱，谢谢你们。 这个学期开始，我都是在用过年所得的500元压岁钱。我想，你们所资助的 这些钱，我要留一部分给三年后的大学费用。三个星期花200元，一个学期将近20 周将近1500元，我应该还能存一点的。希望我的母亲能快点正常走路，希望我的父 亲不要太累，希望我的奶奶长命百岁，希望你们都一世长安。 你们上个学期的时候来过我们学校看我们。你们都很和蔼，我现在却只有 邹立美奶奶的联系方式。虽然不知道你们还记不记得我，但我还是想说：谢谢你们！ 我们的白老师说你们也很辛苦，回去之后大多都生了场病，我们很感动。希望你们 在工作的时候也多注意一下休息，健康很重要。 我想长高，因为我现在只有156.7厘米，而且从初中到开始就没长过了。对 此，我表示有点伤心，我真的想长高。朋友说：“想着吧，只能想着了”。好伤心！ 我还想写好字，我身边的朋友字都写得好看，多多练字。我觉得我身旁的朋友好多 都会才艺表演，会跳舞唱歌，会玩各种乐器。说不羡慕怎么可能，但现在没有条件 去学，也没有时间去学。 说实在的，现在我学习起来很费力。我真的没有前几名的同学聪明，但我 相信勤奋可以弥补天分，天分不可以弥补勤奋。加油！希望这一年一切都变得越来 越好！ 祝你们一世长安！ 贵州黄平县民族中学学生：雷青青 【编者按】：作者的文风非常朴实，如同和家人拉家常一样，讲出了自己的喜忧愿 想，文字不华丽，目标很实际，胜在真诚！愿你长高，祝你成功！ 感谢信 感谢信-杨瑞欣（河北沧州2016届） 【编者按】：在前进的道路上找到短板，采取措施积极克服，这就是成功。成功了， 还记得回头报个平安，说声谢谢，这就是良好的素质！树华的叔叔阿姨们相信你， 祝你能如愿进入心仪的师范学院，将来做一名优秀的人民教师。 尊敬的树华基金会的叔叔、阿姨们，您们好！ 首先祝叔叔、阿姨们身体健康，生活愉快。我是杨瑞欣，感谢叔叔阿姨们 的帮助，是我能够顺利完成高中学业。我向老师们汇报一下我一年的学业成绩。进 入高二后，我们明显加快了课程的进度，在学习中，我也感到了和成绩优秀学生的 差距。为了不使自己的成绩落后，树华的叔叔、阿姨们，我会更加努力，相信你们 的学生、你们的孩子会给你们争气的。我会更加努力，也会对自己提出更高的要求， 每天晚自习比别的同学晚睡一个小时的目的就是补短板，把成绩差的科目追上来。 通过努力，取得了一定的效果，我的物理成绩比以前进步了。我经常请教老师和我 的同学，利用星期天少休息，把一周来所出现的疑难题和错题重新复习一遍，做到 融会贯通，得以巩固，再就是预习一下下一节的课程。通过自己的认真学习，我也 经常得到老师的表扬，也使我更有信心。叔叔、阿姨们，我的最大目标就是能被一 所优秀的师范学校录取，毕业之后当一名优秀教师，为自己的祖国培养合格人才。 叔叔、阿姨们谢谢了! 您的学生：杨瑞欣，2018年3月 树华 李丽 记得有一个小故事曾经深深地打动我。话说巴西有个老爷爷有一天在海边 救助了一个全身覆满石油，而且饿得奄奄一息的企鹅，待到企鹅伤好后放归大海。 此后老爷爷就把这事儿忘了，直到几个月后的一天这只企鹅又出现在老爷爷居住的 岛上，老爷爷欣喜地将企鹅带回家。从此后每年这只企鹅都要游过5000英里的海域， 躲过深海里各种猎手，不惧风暴都要回到这个岛上，和它的救命恩人一起度过快乐 的八个月后，再回到自己的岛屿。读完这则故事后感触很深，动物尚懂得感谢帮助 过自己的恩人，何况人乎？ 人生一世，每个人都依存于社会和群体中，无一例外，从呱呱坠地到垂垂 老矣。我们需要感恩的人很多很多，父母的养育之恩、老师的教育之恩、大自然的 阳光雨露，新朋老友的扶持陪伴...... 要学会感恩每一道照进你生命中的微光，感恩 每一滴送进你干裂嘴唇的甘泉，感恩每一只在你即将沉没时拉你一把的手...... 生活 中的每一段路，没有他们与我们同行，人生的路更难走。 常怀感恩的心不仅是一种良知，更是一种大智慧。感恩的人积极乐观，心 态平和，更能聚集正能量，产生更好的社会心理效应。《诗经》曰：“投我以木桃， 报之以琼瑶”说的就是这种积极美好的社会情绪效应。投桃报李，对现在的你，也 许只是一个发自内心的微笑，只是一声真诚的谢谢！但它体现了你善良的本质，它 传递出了感动和善意。就像小企鹅的故事，老爷爷救助企鹅的时候没有想要它报恩， 但小企鹅感恩的表示却使老爷爷十分“欣喜”！相互的感动，相互的鼓舞，一起努 力向善，我们生活的世界一定会更美好！ 为什么要常怀感恩之心 树华 李丽 所优秀的师范学校录取，毕业之后当一名优秀教师，为自己的祖国培养合格人才。 树华 李丽 【编者按】：李会萍是树华2010级校友，不完整的家庭留给她很多的阴影。她怨恨 过，挣扎过，并试图斩断所有亲情远走高飞，但生活却让她屡屡受挫。反思人生， 学习传统文化，李会萍认识到只有怨恨而没有感恩的心，忘了本，也就断了根！李 会萍希望成为树华的辅导员，用她的经历来帮助学弟学妹，辅导他们，陪伴他们， 帮助他们，使他们毕业之后的人生少一些迷茫，多一些阳光！ 从小我就生活在一个没有亲情的家庭里。或许这样说过于残酷，毕竟父母 给了我生命。母亲在我很小的时候就离开了我们，她是因为父亲的家暴离家出走的， 父亲也并没有因为母亲的离开而担负起一个父亲该有的责任和义务。所以那时的我 是恨父亲的，恨他的无情无义，恨他拆散了这个家，让这个家支离破碎。 那个时候的我很自卑，家里穷，个子又矮，还没钱读书。直到我上五年级 的时候，是我的班主任贺老师和马校长为我联系了树华基金会，才有了我后来每一 年的资助，直到中专毕业。 在我人生低谷的日子里，是树华给了我一份生活的保障 和精神的鼓励。每年收到海外辅导老师的信是我最开心的事。 按理说，因为我比别人多了一份特殊的经历，更应该毕业 后好好工作，回报社会，帮助更多的人，才不会辜负树华 对我的期望。可是，我没有，我忘恩负义。因为那时的我， 没有一颗感恩的心，生活渺茫看不到希望。毕业后的我， 一个人漂泊在外地，不想再回到那个冷冷清清的家，从那时 起便斩断了亲情。一个人，如果忘了家、忘了本、忘了老祖宗， 那她的苦日子也就来临了。我就是这样的，生活不顺，处处碰壁。 回首之前的经历。我终于明白了，一个人如果没有一颗感恩的心，是很可 悲的，也是很可怜的。想想我的父亲，他难道不是一个受害者吗？他从小没有读过 书，不知道做人的道理。所以，没有办法教会我们做人。可是，我读书了，我要学 做人，要做一个对社会有用的人。父亲没有做好的，我要把它做好。很惭愧，我现 在做的也不好。中华传统的美德是孝亲尊师，当一个人开始把父母的事放在心上的 时候，她的生活已经发生变化了。人有善念，天必佑之。当然，我希望我的那些学 弟学妹们，在校期间除了读书之外，一定也要树立起正确的人生观、价值观、世界 观，不要像我，在校期间，除了读书，就看一些言情小说，耗费了青春，虚度了 光阴。 好在已过而立之年的我，知道错了。现在改还来得及。我感恩树华曾经对 我的资助，没有树华，便没有今天的我。也感恩那个苦难的童年，因为它让我懂得 了生活的不易。 李会萍， 2018年6月8日 大树之恩 一个念念不忘旧仇的人，他的伤口将永远难以愈合，尽管那本来是可以痊愈的。 ——〔英〕弗兰西斯·培根 爱是真正促使人复苏的动力。——〔德〕歌德 金句分享 和精神的鼓励。每年收到海外辅导老师的信是我最开心的事。 一个人漂泊在外地，不想再回到那个冷冷清清的家，从那时 起便斩断了亲情。一个人，如果忘了家、忘了本、忘了老祖宗， 那她的苦日子也就来临了。我就是这样的，生活不顺，处处碰壁。 各位叔叔阿姨、爷爷奶奶：你们好！ 距上次分别已有半年之久，很想念你们那慈祥的笑容。每每想起心中都会 有一股暖流涌动，让我有了前进的动力。半年未见，不知你们的身体可好？我很感 谢李丽阿姨会在百忙之中抽出时间给我回信，对于我的问题一一解答，没有觉得我 啰嗦，而且耐心地答复，这让我感到很暖心。还有张阳阿姨，我收到了你的信。谢 谢你喜欢我写的字，其实还有很多不足之处，我会继续加油，下次有机会我会送你 一幅较完美的字。 时间飞逝，转眼间高一生活已经过半。在这半年中，我仿佛经历了人生的 跌宕起伏，落到了谷底。我满怀希望来到了黄中，我以为我会有一个丰富多彩的高 中生活。然而现实却给了我一巴掌，一次次的考试失利，让我伤痕累累。但我没有 失去信心，我相信努力会有收获。我可以放下过去的荣誉，也可以忘记现在的失败， 从头再来。我不能向命运低头，我身上背负着太多人的希望。我有着自己的梦想， 我已没有退路，只能继续向前奔跑，坚持不懈，终有一天会跑到终点。汗水不会白 流，努力也不会白费。对于过去的一年，我成长了不少。懂得了做事情要再三思考， 不能由着性子来。对于所学知识，要做到牢记于心，不能模棱两可。新学期新开始， 我希望自己可以在这个学期取得突破性的进步。毕竟这学期相当重要，也是人生的 一大转折点。面临分科，不得不让我加倍努力，为以后的事业做打算。其实，我到 现在还不清楚自己内心的想法，还在纠结到底学文还是学理？家人们都希望我将来 成为一名医生，自然希望我学理。但我怕学不好让他们失望。我真的不知道该怎么 办了，也许时间会给我一个答案。 去年的那笔奖学金帮了我很大的忙，我用它来交了这学期的学费，减轻了 父母的压力与负担。每天看着爸妈为了家庭生活到处奔波，很是心疼他们。但我也 无能为力，只能努力学习，将来让他们安享晚年。叔叔阿姨、爷爷奶奶，我知道你 们以前的学习条件更差，但你们没有向命运低头，没有被时代的限制条件给压倒。 你们吃苦耐劳才有了今天的生活，我很崇拜你们。我希望将来的自己也能成为像你 们一样的人，能赚很多的钱，去帮助那些贫困的孩子上学，就像你们帮助我一样。 生活还在继续，我还得继续奔跑。我特别喜欢一首歌的歌词“当我必须像 个完美的小孩，满足所有人的期待，你却不讲你的愿望，怕增加我肩上的重量。我 不完美的梦，你陪着我想⋯⋯”，没有人是十全十美的，但我在努力让我变成更好的 自己。我做不到完美，但我可以让我做的每一件事都趋于完美。我没有能力让所有 人都满足，但我会尽力让他们开心。我也没敢承诺我现在做的每一件事都是对的， 但我会让自己不留遗憾，这就是我，一个倔强而又不服输的女孩儿，一个外表乐观 内心却极为脆弱的女孩儿。我会哭，但不是在别人面前；我会累，但我会坚持下去。 不管未来会面对怎样的风险，我都已经做好了准备。我不会让你们失望，更不会让 自己的努力都功亏一篑。我会让未来的自己成为一个优秀的人才，请相信我！ 祝：身体安康，工作顺利！ 【编者按】：一个倔强不服输的女孩。有困惑有挫折，却充满了信心努力前行。赞！ 对高中生来说，选择文理方向是人生一个重大决定，最终还是要自己拿主意。不过， 几个因素供你参考：1. 个人的优势，扬长避短增加杀出重围的机会；2. 是否志在必 得？一击不中，本人的家庭经济条件是否支持复读？3. 个人的爱好，能做你所爱的 事情，幸事也！但应以谋生立足为本考虑能否兼得。祝你好运 我不完美，但我会尽力变成更好的自己 贵州黄平县民族中学学生 李欢 “世界上的重要事情，都是由看似没有希望，还持续努力的人所完成”。 “Most of the important things in the world have been accomplished by people who have kept on trying when there seemed to be no hope at all.” – Dale Carnegie, Motivational Expert 戴尔卡内基 直面生命中的挫折 ——作者：丁本梅（贵州福泉二中高一） 依稀记得你的模样，如今你已不在身旁，我努力奔跑，却和你相隔永远。 也许你曾美丽，而我却从未在意，只剩下那刻骨铭心的病床，你那让我绝望的模样。 那天，我正在学校复习，却接到了班主任的电话： “小梅，你爸爸让你回家一趟， 家里有事。快去吧，孩子！”。 “啊？⋯⋯哦⋯⋯嗯”。 “小兰，我怕”， “没事，别瞎想。一切都会顺利的”。 “⋯⋯”.
背着书包，我和生命一起赛跑，绵绵小雨陪着我，滴进我心中那块乌云。生命在前 面等我，我奋力跑着，步伐快过心跳⋯⋯然而，生命比我先到了终点。我输了，我 哭了。 因为这万箭穿心的痛，因为再也不能拉你温暖的手掌了，因为再也听不到 你的唠叨。生命如此脆弱，如此匆匆，居然等不到我成长的那一天，更等不到你以 我为傲的那一天。 时光路过它自己，速度不缓不急。记忆中的你已不再那么清晰。无数个夜 晚哭着入睡；无数次绝望却又无可奈何；无数次呼喊却无人回应⋯⋯不经意间，我 钻出土壤，长成了一棵小树苗；我破茧而出，成了一只美丽的蝴蝶。我生长着，风 雨赐予我一圈又一圈年轮。我享受着风雨，享受着成长；我飞翔着，风雨让我的翅 膀更有力；我享受着风雨，更享受飞翔。我会勇敢，因为没人替我坚强。 人生如大海，时而风平浪静，时而波涛汹涌；如一片湖，时而鸟语花香， 时而凄凉苍茫。阳光总在风雨后，上天给你多大的挫折，就意味着你要承担多大的 责任。直面生命中的挫折，让我更勇敢；直面生命中的挫折，让我的青春在拼搏中 闪光。 你可以没有“先天下之忧而忧，后天下之乐而乐”的胸襟，但别放弃“长风 破浪会有时，直挂云帆济沧海”的信念。 【编者按】：文章描述了痛失亲人的悲伤和无助，语言质朴，感情真挚感人！对亲人 对朋友，有爱就大声说出来，因为我们可能在“未在意”的时候失去那些对我们最为 宝贵的人，再没有表达的机会。作者也写出了从苦痛中挣扎出来，继续努力积极面对 人生，势要破茧而出，展翅飞翔！积极的态度，强大的内心，令人赞叹！作者原标题 使用了“享受”二字，“享受生命中的挫折”，“享受”二字以为不妥，我们并不 享受这些挫折和灾难，但我们可以面对它、挑战它、战胜它，使自己变得更强大！ 尊敬的树华叔叔阿姨们：你们好！ 天南地北挡不住温情的传递。即使距离遥远，依然 可以传递关爱。感谢这个世界有爱的人，让温暖之花绽放 于漫山遍野。我，感受到了来自树华的温暖，来自远方 的真情。 又是一年春来到，时间是世界上最快而又最慢的东西。当你身临其境时， 仿佛度日如年；可当时过境迁，却又恍如隔世。今年，我18岁了，可以说是个成年 人了。回望去年，17岁的雨季最美，然而在此期间也经历着不少酸甜苦辣。我欢笑 着，我回家了，迎来的是家人温馨的关切与重聚；我伤心着，想到自己的遭遇及现 在家中的情况，我痛恨命运的不公；我感动着，有着朋友的关心与照顾，在生活中 处处是阳光；我心痛着，看着妈妈日渐消瘦的脸颊及布满青丝的白头；我⋯⋯ 现实打败人心，摧残人的躯壳。现实很残酷，可我却没有被命运打败。人 生下来本没有绝对平等，“生于忧患死于安乐”，我虽不能完全领悟到这句话的精 髓，可我却明白苦难磨练人的意志。生活即使很艰难，可条条道路通远方。任何事 总有解决的办法，“船到桥头自然直，车到山前必有路”，我的生活，痛并快乐着。 已经步入高二下学期，仿佛高考这场战役即将开始。我恐惧，可同时我又期待着。 与时间赛跑固然很累，过程也很痛苦，我希望结果会让自己满意。可如今的我已经 越来越力不从心，成绩似乎一直在倒退。我越是努力，结果却适得其反。我不知道 自己该怎样做，是我努力不够吗，还是老师方法不对？我在积极地寻找其中的原因， 我也清楚现在不拼命更待何时？妈妈不识字，有些事我就是和她说，可能她也不懂。 我总是学会把委屈憋着，不让眼泪流下来，因为我知道学会面对才是最好的解决办法。 我很感谢自己与树华的偶遇，也很感激素不相识的捐助人对我的帮助。我 们从未见过面，但却在冥冥之中传递着一份真情。我收到的不仅仅是一份助学金， 更是一份沉甸甸的温暖，流淌于心间。每当这个时候，我都会打电话告诉爸爸妈妈， 让他们明白原来这个世界上，有爱的人真的很多。于是，接下来的几个月中，我将 不需要他们寄生活费，为家里节省了开支。高三即将来临，收到的奖学金我也会用 在购买更多的学习资料及用具上。此外，每月节省下来的，回家我也会给爸爸妈妈 买些东西，感谢他们在背后对我默默的支持。 人总是会长大。过去一年，我学会很多，遭遇了很多。我懂得了世态变化， 明白了奋发向上的重要性。我领悟了穷且益坚的道理，继续着不坠青云之志的骨气。 我是鹰，云中有志；我有汗，汗中有盐。千言万语说不尽我心中的感激之情，唯有 此感谢信一篇。希望捐助人可以真正了解到我的生活。同时，感谢您的友爱，谢谢！ 天南地北挡不住温情的传递。即使距离遥远，依然 我是鹰，云中有志 云南昭通盐津二中高二学生 黎世丹 【编者按】：该生常怀感恩之心，树立了正确的人生目标。成长从来不是一帆风顺 的，有乐也有痛，但只要有青云之志并坚持奋进，你必有展翅蓝天，搏击长空之时。 成长的烦恼： 人生中只有一条路不能拒绝，那就是成长！ 成长的过程中，很多人有不善言谈，不善与人交往的烦恼，融入困难。要 克服交往焦虑的害羞表现，找到与人交往的正确打开方式，建议你读读戴尔·卡内基 的书。戴尔·卡内基（1888-1955）是美国励志作家及演说家，创立了一系列的行销、 演讲、企业训练、人际关系、及励志成长等训练课程。他的畅销书《卡内基沟通与 人际关系-如何赢取友谊与影响他人》是害羞的我打开与人交往之门的第一把钥匙。 他的书浅显易读，容易实践。祝你尽快掌握与人交流、交往的正确打开方式！ 人生中只有一条路不能拒绝，那就是成长！ 成长的过程中，很多人有不善言谈，不善与人交往的烦恼，融入困难。要 好 书 推 荐 【编者按】：贝多芬名言：“卓越的人一大优点是，在不利与艰难的遭遇里百折不 挠”。你过去的一年虽然遇到人生挫折，但是，通过自身的努力以及树华的帮助， 最终走出了阴霾。在迷茫的时候懂得寻求帮助是明智的。张佐，继续加油，祝你如 愿考上你梦寐以求的大学-清华大学。 尊敬的树华捐款人： 我是高中生张佐，家住中国云南省昭通市，现就读于云南昭通第一中学。 过去的一年中，我经历了人生最大的挫折，但幸亏有树华老师的帮助与鼓励，我才 没有放弃学业。 去年，我因为过度紧张，没有调节好心态，导致了精神方面的疾病“焦虑 症”、“强迫症”，造成身体素质下降，又得了肺结核，所以我不得不休学。当我 感觉一切都放弃了我时，只有“树华”没有忘记我，“树华”的老师亲自打电话劝 导我、鼓励我，使我重新有了面对生活与未来的勇气。于是我去了昆明打零工挣钱 买药，接受精神科医生的治疗。经过多种尝试性的治疗之后，终于有了很大好转。 现在我更加注重充足睡眠，锻炼身体，保持心态积极乐观，感觉生活与学习都有 很大的乐趣。 我非常喜欢自然科学，但学校没有这方面的书，外面借也十分麻烦，我得 到奖学金后就买了很多自然科学方面的书与课内辅导书，每天再忙都会抽空阅读， 觉得生活开始过的充实了。因为自己比较喜欢运动，又买了两双质量较好的运动鞋。 长时间用眼使我眼睛近视了，又花了奖学金去配了一副五百元的眼镜。上面这些如 果没有奖学金的话，是不可能去买的。“树华”的资助人给了我这个机会，而且也 解决了我几个月的生活费，父母的负担更是减轻了。 我想给捐助人说说我现在的生活学习情况。我并不只把目光放在课内知识， 课外会疯狂的阅读世界名著、自然科普书，也培养了许多对生活有益的兴趣爱好， 比如说打篮球、骑自行车和音乐与唱歌。在班上积极帮助同学，为班级做贡献，我 主动放弃了做班长的机会，做了劳动委员，改变了我们班卫生差的状况。我的成绩 也快速上升。我的梦想是考上清华大学，我会加倍努力一点一点靠近目标。 假如没有“树华”的倾情相助，我现在恐怕就没有读书机会了。假如没有 “树华”的资助人，我现在也不会有那么多学习资源与无忧的生活。 诚挚的感谢您们！ 云南高一学生张佐写给树华的感谢信 诚挚的感谢您们！ 张佐， 2018年4月10日 学习英语有无诀窍?
LEARN ENGLISH 树华的义工大都是早年来美的留学生，就是说母语都不是 英语。在“外国”生存，大家都经历了背单词、记语法，练听力，张开嘴的过程。 学习英语到底有无诀窍？热议之下，认为学习英语没有捷径，或者说学习语言都 是没有捷径的，要下功夫！ 虽然没有捷径可走，但如何下功夫使巧劲的确有经验之谈，归纳如下： I. 背单词：必须背单词。单词就像建筑材料，没有足够的砖头，没法儿盖房子。 怎么背记住得快且不容易忘？有一些方法： 1. 合成词记忆法（football，足+球=足球；fireball, 火+球=火球，snowball， 雪+球=雪球，还有像bedroom，classroom，cleanroom 卧室，教室，洁净室等等）； 2. 单词中的规律，前缀／后缀：单词中的前缀，dis-否定 （dislike 不喜欢, disobey不遵守, disappear不见了）；mis-否定（mistake 错拿 -延伸为错误， misspell 错写，misaligned 错位）。单词中的后缀，self-自身（myself, yourself 我自己你自己） less-否定（selfless 无我，homeless 无家，fearless 无畏）。 这些规律并不是适用于所有的英语单词，要“记住”如何正确使用。 3. 词根记忆法，比如 act, action, active, actively, activity, activate, deactivate， 行动，行为，活跃的，活动，激活，锁住，都是源自“act”这个词根。找到词根 帮助你记忆相关的单词。 4. 反复书写，重复记忆。把单词默写下来帮助记忆，如果能写在纸条上、 卡片上时常拿出来温习，循环记忆更有助于记牢。 5. 发声记忆法，很多单词读音是有规律的，会读也就会写了，要习惯大声 读出来！比如后缀tion读“审”nation, action, notion, attention 等等单词要记住的只 有前面部分，容易多了。 II. 实景学习法：包括看视频，警示牌等等生活中的场景帮助记忆联想，如果视 频有字幕，可以尝试打开字幕看一遍，再关掉字幕听一遍，让耳朵熟悉生词的读法。 III. 多听新闻广播：播音员音色标准，吐字清晰，速度适中，较少使用俚语，容 易听懂。如果能够使用暂停和重播，记下一些关键词汇，更有助于听懂全部内容。 IV. 大声地读出来：语言语言，是需要说出来的，要像小孩子一样敢于模仿，重 复，通过发声，让眼睛、嘴巴、耳朵和手配合记忆，会事半功倍。 V. 应用学习法：参加英语角，英语读书小组，英语学习伙伴。有人陪伴进步更快。 Ms. Snow 是树华的英语辅导老师。Ms. Snow非常有语言天分，精通日语和英语， 她的发音标准，无论在日本还是美国，都被当作本地人！以下是她的经验之谈： 根据我自己学习语言的经验，一定要有学语言的环境，进步才会快。平时 如果有英语的广播节目或电视节目可以经常听经常看，因为好的听力对学语言至关 重要。我自己每天做家事或开车都听国家广播电台的新闻或访谈节目，我以后也会 留意录下一些合适的语音供同学们练听力，多听就会增加语感，这对学语言非常重 要。如果看到外国人，你们也要鼓起勇气跟他们讲话，只要经常说，你们的英语会 越来越好的。祝同学们学习快乐！ “Hello，how are you? My name is......
SPEAK 为了帮助同学们学习英语，树华辅导组开始编制 网上英语课堂。为测试课堂内容是否符合学生的 水平，是否能提高学生学英语的兴趣。我们邀请 了部分愿意朗读，愿意与大家分享学习心得的树 华学生参加到树华英语课堂义工群。以下是部分同学两周以来的体会： 杨梅英：我很喜欢“树华课堂义工群”。喜欢和各位同学一起学习，喜欢 老师布置的和学校的不一样的作业。我喜欢在里面可以积极发言，弄懂每一个知识 点。其实每次在上课的时候，我的内心都有点激动。因为在学校，老师绝对不会 跟我们讲世界杯。在“树华课堂义工群”里，我见到的很不一样，课堂内容很新颖。 杨梅英： 我喜欢把自己不会的显露出来，让别人来指出。这是取得进步的 比较快捷的一种方式。但在学校，同学们都挺忙的，有的是不好意思说，大概只 有少数喜欢这样做，所以我很感谢树华给我送来这么一大拨可爱的，喜欢指出彼此 错误的同学，和这个可以相互交流的平台。 雷青青：这次的学习感觉良好，每天学一点英语使我感觉一天变得充实， 并激发出我想学好的热情。还没开始这些天学英语的时候，我很懒，每天睡到早 上十点左右，一整天都不知道自己要做什么。但我参加了这次活动后，我开始六点 钟起床，觉得自己今天有事情做，并想提前去把一些要做的事做了。 聂闽敏：我比群里的同学都晚来报到，入群的第一个作业，便碰上了极难 的英语听力，虽然只有短短十几秒，过多的生词和地道的英语发音让我几度想要 放弃，但在老师的循循善诱下，我们进行了一场探索，那是一场充满汗水与喜悦 探索。我想我永远也不会忘记，老师一个单词，一个单词的提醒我们，一步一步 引导我们的情景。古人有云“有志者事竟成”，我想的确是这样的，但愿在以后 的学习生活中，我能牢记此警句，时刻以正确的心态对待学习。 谢兴丽：I have to say that I have gained a lot from joining the English volunteer group. I can learn English with friends from all over the world every day. Although we don't know each other well, we can learn from each other. I feel very grateful these days. I have realized that my listening skill is limited through various exercises teacher provided to us. Therefore, I will practice my listening and oral skills often from now on. 树华辅导组的日常工作职责有： 对于树华学生发来的的电子邮件，尽快给予回复；回复由 北京办事处转来的信件；辅导或解答学习中遇到的具体问题，介绍， 交流同学的学习经验；协同奖学金组理清学生申请表中的一些有出入 的信息等等。总之，同学们需要和树华或者树华捐款人沟通，都可以联络辅导组， 我们会将你的疑问转发给相关部门，也会将你的问候转发给你的捐款人。树华辅 导组的邮箱地址是：mentor@soaronline.org. 令人欣慰的是，树华已经毕业的很多同学饮水思源，他们感恩树华，愿为 树华的助学事业出一点力。如青海的冯永珍同学即将在青海大学硕士研究生毕业， 他愿意做树华的辅导员。又如宁夏的李会萍同学，非常关心树华的学弟学妹，愿 为树华捐款出力。辅导组把李会萍校友写给树华的“大树之恩”刊登在了本刊第 10页。希望她的人生感悟能对同学们有所启发。 为了帮助同学们学习英语，树华辅导组开始编制 网上英语课堂。为测试课堂内容是否符合学生的 水平，是否能提高学生学英语的兴趣。我们邀请 网上 课堂 组 A proud and confdent genius makes a bet with an idiot. The genius says, "Hey idiot, every question I ask you that you don't know the answer, you have to give me $5. And if you ask me a question and I can't answer yours I will give you $5,000." The idiot says, "Okay." The genius then asks, "How many continents are there in the world?" The idiot doesn't know and hands over the $5. The idiot says, "Now my turn: what animal stands with two legs but sleeps with three?" The genius tries and searches very hard for the answer but gives up and hands over the $5000. The genius says, "Dang it, I lost. By the way, what was the answer to your question?" The idiot hands over $5. A boy asks his father, "Dad, are bugs good to eat?" "That's disgusting. Don't talk about things like that over dinner," the dad replies. After dinner the father asks, "Now, son, what did you want to ask me?" "Oh, nothing," the boy says. "There was a bug in your soup, but now it’s gone." “Laughing is one of the best exercises, it’s like runing inside your mind. You can do it almost anywhere and it’s even better with a friend.” “Dang it” means "darn it" or "oh blast" and it's a very polite little slang term you use when you're frustrated and would rather not swear, curse, or be profane. "My computer crashed again, dang it!" 它意味着“糟了”，这是一个非常有礼貌的小俚语，当你感到沮丧而 不愿发誓，诅咒或亵渎时，你会使用它。 “我的计算机又宕机了，糟糕透顶！” 笑 话 两 则 致高三的同学们： 今年是你们最后一次获得树华高中奖学金的学年。明年此时，你们已 经高中毕业，将奔赴生命中的下一站。无论是开始大学生活，还是进入其他 工作岗位，都请你们写一封信给树华捐款人，告知你们的去向并说声感谢。 树华捐款人陪伴你们成长，更关心你们的未来。我们希望树华学子都是有礼 貌，懂珍惜的年轻人。祝愿你们的人生道路越走越宽！ 树华大学奖学金 树华教育基金会大学奖学金项目，资助的对象是在高中阶段接受过树华高 中奖学金并已考取大学或者高级职业大学的学生。 学生就读的院校必须是教育部、 中央部级、或者省教育厅认可的公立大学或者公立高等职业大学。 每份奖学金的金 额为人民币四千六百元一年，资助期为一至两年，需要每年申请。 树华教育基金会鼓励同学们申请，但是由于大学奖学金项目不在基金会预算 之内，所以不能保证每一个申请者都能获奖。奖学金的授予将优先考虑一年级和二 年级的学生。 大学奖学金申请截止日期为每年的十月十五日。申请材料包括申请表格、 自传、师长的推荐信、所在大专院校的在学证明、高考分数或者前一学期成绩单。 家庭的贫困证明对申请此项奖学金也有帮助。 有意申请大学奖学金的同学们可用微信联系树华北京办事处的商荣梅老师， 她的微信号码是：13801212049。请将申请材料直接递交到树华北京办事处。 树华奖学金组 树华辅导组 I.
II. 感谢有你 我是来自山西太原的义工郭丽文，退休前是一名高级工程师。今年6月我 接受了李静和会长的邀请，主持《树华小天地》期刊的工作。与我来说，这份工作 带给我更多的是责任和义务。我会全心全意努力工作。除了通过《树华小天地》 这个窗口把树华的理念传递给同学们，也希望利用这个平台更多地与你们分享、交 流，沟通、互动，把《树华小天地》办成大家喜爱的期刊。如果你们有什么好的 想法和建议，欢迎与《树华小天地》联系。树华基金会辅导组的邮箱是： mentor@soaronline.org。 现在讲讲我与树华的缘分。这要追溯到三年前了。2015年11月，我退休后 来美国探亲。看到妹妹丽春每周五开车单程近一个小时到树华总部办公室做电脑 和网络的维护工作，解答义工同仁的各种技术问题。每周还会花至少两天时间做 网站和数据库的开发和维护，审核学生申请资料，解决现有问题，并与树华同仁 探讨、规划树华未来的发展，出钱出力。目睹了她对树华无私的倾情投入，我非 常感动。她经常说，自己是知识改变命运的受益者，让贫困地区的孩子通过知识 改变命运，是她及树华同仁们最大的心愿。 我被树华的大爱深深打动，萌发了在大陆这边为树华做一点力所能及的事 情的打算。我把想法告诉妹妹后，她非常高兴，把我介绍给了树华同仁，经过树 华前辈的培训，加深了我对树华的了解。正式认识树华、加入树华后，我参加了 2016西安甄选委员交流会，2016年度树华总部感恩午餐会，2017、2018年学生申请 评审工作，2017年贵州和湖南学生探访活动、2018年广西留守儿童探访，以及树华 每月一次的办公会议等。 随着了解的深入，知道树华的捐款人和义工都是像你我这样的普通人，并 没有富甲一方，但他们为树华忘我工作的态度，对帮助中国大陆贫困学生的愿望 及他们投入的精力和财力，在没有了解树华之前是难以想象的。23年来，树华帮 助9000多名家境贫寒的优秀学子顺利完成了基础教育。树华义工们投入大量的时 间和精力，帮助有志的优秀学子。希望你们也能通过努力和树华的帮助改变自己的 命运及家庭的状况。在树华这个平台上有幸和你们结识，我非常珍惜这个机会， 希望你们也和我一样珍惜这个机会。 为了更好的与你们沟通、交流，树华设有辅导组。如果你在学习或生活中 遇到任何困难，不妨与辅导组的夏老师及他的团队联系，他们将很高兴为大家排 忧解难。辅导组邮箱 mentor@soaronline.org。 最后愿你们像一棵树，树华的树：春天，吐一山淡淡的清香; 夏天，洒一 抹如泉的阴凉; 秋天，举一树甜甜的青果; 冬天，积蓄能量蓄势待发⋯⋯ 郭丽文 2018.7.13. 树华辅导组有奖征文活动 参赛规则如下： 一. 参赛资格： 所有树华在校高中生（高一到高三）均可参加。 二. 三道征文题目任选一题写一篇文章，文体不限，字数约800。 要求原创，抄袭作品自动丧失资格。 三. 征文题目： 1. 对我影响最深的人； 2. 我的求学之路； 3. 我能为家乡做什么？ 四. 征文投递： 文稿署名，并请注明你的年级及树华ID，甄选委员姓名， 上传到辅导组的QQ邮箱：3374904246@qq.com.
五. 征文截止日期（收到日期）： 2019年2月20日以前。 六. 奖项设置： 1. 一等奖3名，每人奖励 ¥800元； 2. 二等奖8名，每人奖励 ¥500元； 3. 三等奖12名，每人奖励 ¥300元。 七. 评审： 树华辅导组在征文截止后将对收到的文章进行统一评审， 并将在3月15日以前在树华官方网站公布获奖名单。请关注 树华网站：www.soarfoundation.net 八. 奖金的发放： 获奖名单除了将在树华官网宣布以外，也将由树华北京办 公室通知获奖学生本人及甄选委员。奖金将于3月底以前转 入获奖学生登记的账号。 九. 作品使用权： 所有参赛文章，获奖与否，树华均有权用作宣传树华助学 相关的所有活动。如你对此条款有异议，请用QQ邮箱与 辅导组联系。 十. 如对上述规则有任何疑问，可写信给辅导组询问。 树 华 辅导组 邮 箱 3374904246@qq.com mentor@soaronline.org 湖南中方县一中学生、甄选委员吴玉芝老师（二排左一）、刘子松校长（后排右一） 树华义工2017年10月贵州湖南学生探访合影 贵州黄平中学学生、甄选委员吴绍祥（后排左一）白文秀老师（前排右一） 甄选委员王荣乐老师（左图后右五）；贵州福泉中学学生、甄选委员赵齐兵老师（右图中） |
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2013 AMC 12B Problems/Problem 17
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (no Cauchy-Schwarz)
5 Solution 4 (geometrical approach)
6 Solution 5
7 Solution 6
8 Solution 7
9 See also
Problem
Let and be real numbers such that
What is the difference between the maximum and minimum possible values of ?
Solution 1
Note that . Now, by Cauchy-Schwarz, we have that . Therefore, we have that . We then find the roots of that satisfy equality and find the difference of the roots. This gives the answer, .
Note: We arrived at as we consider the two sets of numbers and . From Cauchy-Schwarz, we have that . which is how we got the second part of the inequality.
~minor edit by Yiyj1
Solution 2
This is similar to the first solution but is far more intuitive. From the given, we have This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have Substitution of the above results and some algebra yields This quadratic inequality is easily solved, and it is seen that equality holds for and .
The difference between these two values is .
Solution 3 (no Cauchy-Schwarz)
From the first equation, we know that . We substitute this into the second equation to find that This simplifies to , which we can write as the quadratic . We wish to find real values for and that satisfy this equation. Therefore, the discriminant is nonnegative. Hence, or . This factors as . Therefore, , and by symmetry this must be true for and as well.
Now and satisfy both equations, so we see that must be the minimum possible value of . Also, and satisfy both equations, so we see that is the maximum possible value of . The difference between these is , or .
Solution 4 (geometrical approach)
From the given, we have and . The first equation is a line with x and y intercepts of and the second equation is a circle centered at the origin with radius . Intuitively, if we want to find the minimum / maximum such that there still exist real solutions, the two graphs of the equations should be tangent.
Thus, we have that , which simplifies to . Solving the quadratic, we get that the values of for which the two graphs are tangent are and . Thus, our answer is .
Solution 5
Draw the sphere and the plane represented by the two equations in Cartesian space, with the -axis representing . The intersection between the sphere and plane is a circle. We wish to find the point on the circle where is minimized and the point where is maximized. Looking at the graph, it is clear by symmetry that when is maximized or minimized. Thus, we can set . This gives us the following system of equations:
Solving gives , which are the maximum and minimum values of respectively. The answer follows from here.
Solution 6
We can consider , , and to be solutions to a cubic equation. Then, given our information, we have and , so our cubic equation looks like this: where can be any real number.
Since this cubic has real solutions, it must have both a maximum and a minimum (noticed by looking at the graph). Then the greatest solution is maximized when the other two solutions are the same and is minimized when the other two solutions are the same.
Thus, equating and , we have andSolving this, we get the quadratic implying the answer.
Solution 7
Subtracting from the first equation yields . Subtracting from the second equation yields . Thus we have the equations Squaring the first equation yields Subtracting the second equation from this one yields Thus we have the system of equations We can reverse Vieta's Formulas, to get that and are roots of the quadratic equation(in ) Because we have , we have that the discriminant of this quadratic equation must be nonnegative. The discriminant is and it can be factored as . Since we have , we must have . If , then we have that . If , then . If , then . If , then . If , then . Thus our inequality holds if and only if , and the maxmimum value is , whilst the minimum value is . Thus the difference between the maximum and minimum values is
-vsamc
See also
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2744 | https://chemistry.stackexchange.com/questions/37280/whats-the-difference-between-%CE%94g-and-%CE%94g | Skip to main content
What's the difference between ΔG° and ΔG°'?
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In biochemistry contexts, one often sees ΔG∘′ ("delta G naught prime"), rather than the normal standard free energy change ΔG∘ ("delta G naught").
What's the difference between the two quantities?
Is there a formal definition of the two terms?
thermodynamics
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edited Jun 27, 2019 at 13:52
orthocresol
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R.M.R.M.
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Can you give us a quote in context, perhaps from one of the books in which you've seen it? That would help in getting an answer, I think.
Todd Minehardt
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I don't think it is a useful question for the simple reason similar notation nuances appears in all kinds of texts with wide variety of meaning.
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The prime usually denotes a standard free energy that corresponds to an apparent equilibrium constant where the concentration (or activity) of one or more constituents is held constant.
For example, for HA↽−−⇀AX−+HX+ the equilibrium constant is K=[AX−][HX+][HA] and the corresponding standard free energy change is ΔG∘=−RTln(K). If you know the value of K and the start concentration of HA then you can compute the equilibrium concentrations of HA, AX−, and HX+.
However, if the pH is held constant then [HX+] is no longer a free variable and the apparent equilibrium constant is K′=[AX−][HA] and the corresponding standard free energy change is ΔG′∘=−RTln(K′). So ΔG′∘=−RTln(K/[HX+])=ΔG∘+RTln[HX+]
In biochemistry there is often several important constituents in addition to HX+ that are held constant such as MgX++, phosphate, etc.
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answered Sep 15, 2015 at 10:33
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If you could list the constituents in biochemistry which are normally considered to have constant (unit) activity, that would improve the answer.
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Is there a formal definition of the two terms?
One form of the fundamental equation of thermodynamics is:
dU=TdS−PdV+∑iμidNi
In this equation, the total internal energy has canonical variables V, S, and dNi, where S is the total entropy (in units of JK), V is the total volume, and Ni is the number of moles of each molecular species present. T is temperature; P is pressure, and μi is the chemical potential of species i. The equation implies that if we were to know an equation that gave U as a function of S, V, and all the Ni we would know everything about the system. However, this is inconvenient for two reasons. First, S and V are extensive variables. Make the system bigger without changing its composition, and S and V increase. Second and more importantly, it is often difficult to hold S constant when doing experiments. The same is true of V. (We live in a constant pressure atmosphere.)
Taking the Legendre transform of U with respect to variables S and V gives a new fundamental equation:
dG=−SdT+VdP+∑iμidNi
This equation means that if we knew a function that gave the Gibbs free energy as a function of T, P, and all the Ni, we could easily compute all thermodynamic properties of the system.
Say we're interested in the thermodynamics of ATP hydrolysis:
ATP+HX2O↽−−⇀ADP+Pi
This equation is really better written as
A−PX3OX10HX3+HX2O⟶A−PX2OX7HX2+HX3POX4
But of course in a buffer at pH 7, the conditions where many biochemical reactions occur, there really won't be HX3POX4 etc., there will be dissociation of protons HX+ and formation of anions like HX2POX4X− etc. So now all those reactions will have to be tracked too. The number of protons released by ATP is not the same as released by inorganic phosphate, and this is generally true. During a reaction, it is difficult to hold the number NHX+ of protons constant, but through judicious choice of buffers etc. it is possible to hold the chemical potential of protons constant (i.e. do experiments at constant pH). Under such conditions, it makes sense to continue the Legendre transformations one step further:
dG′=−SdT+VdP−NHX+dμHX++∑i≠HX+μidNi
ΔG∘′ is a Legendre transform of ΔG∘ with respect to the number of protons in the system.
Robert Alberty's paper from 1994 is a good place for further reading.
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answered Sep 16, 2015 at 13:13
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If I could accept two answers, I'd also accept yours for the Alberty reference, in particular as he apparently was the head of the IUPAC-IUBMB (International Union of Pure and Applied Chemistry-International Union of Biochemistry and Molecular Biology) panel on Biochemical Thermodynamics nomenclature (official panel recommendations), which also has a discussion of the issue.
R.M.
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@Curt F. I really like your answer since it takes into account of the Legendre transformation. Could you please elaborate on the intuition or motivation behind using such a transformation? Thanks for your answer.
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The biochemistry convention does not assume all solutions are 1 M. If you did this then you would have [H+] = 1 M. This simply never happens in biochemical reactions. Instead, we assume pH = 7. We also assume that water has an activity of 1 even though its concentration is 55 M.
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The use of ΔG∘ does not assume that all solutions are 1 M. It assumes that the standard state is 1 M ideal solution. Also, activity is always defined as γ[X]/[X]∘ which is always 1 for pure substances.
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Delta G naught means that the reaction is under standard conditions (25 celsius, 1 M concentraion of all reactants, and 1 atm pressure). Delta G naught prime means that the pH is 7 (physiologic conditions) everything else is the same. The concentration of [H+] now isn't 1 molar because 1 molar concentration would be an extremely low pH (0). Delta G naught prime is just like Delta G naught but for biology.
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answered Jun 26, 2019 at 23:32
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The prime has nothing to do with whether a concentration is held constant. The prime is used to indicate that some species, typically protons or ions such as Mg, are being set to a value other than the standard 1M concentration for use in a reference free energy. Although the prime has been adopted by some in the biochemical community and was endorsed in 1994 by IUPAC, the prime is not used in other communities for good reason - it is much better to explicitly state what concentrations are being used for the reference free energy. When using a reference free energy other than the standard free energy, concentrations can vary or be held fixed. In most biochemical situations, it is useful to keep the pH fixed since buffers are used. But this is a separate issue.
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The key difference between ΔG° and ΔG°' is the concentration of protons, or [H+], in the solution. ΔG° is the standard Gibbs free energy change, which is measured under standard conditions, which include a temperature of 298 K, a pressure of 1 atm, and all reactants and products at a concentration of 1 M. ΔG°', on the other hand, is the standard Gibbs free energy change under physiological conditions, which means that the pH of the solution is 7.
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Perchloric Acid
Perchloric acid is a corrosive superacid. When heated or highly concentrated, it has powerful oxidizing properties and forms potentially explosive mixtures with organic compounds. Available in various reagent grades, it is used to prepare perchlorate salts.
Its oxidizing power increases with concentration and temperature. Heated above 150°C it reacts violently with organic material and can cause explosions. The monohydrate of perchloric acid (85%) is a good room temperature oxidizer.
Perchloric acid is volatile and corrosive to metals and tissue, and closed containers holding the acid can rupture violently with prolonged exposure to heat. In its anhydrous form, perchloric acid can explode spontaneously, and concentrated forms are unstable. Other dangers include:
The acid is very hygroscopic and produces heat when combined with water
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Inhaling its vapors or mist causes a burning sensation in the nose and throat along with lung irritation and coughing
Prolonged or excessive exposure can cause severe coughing and vomiting
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Although perchloric acid is not explosive in solution, it is still corrosive and harmful to your breathing. It should always be handled inside a chemical fume hood. At elevated temperatures, vapors from perchloric acid can condense on surfaces in chemical hood ductwork and form perchlorate salts that can be shock-sensitive and pose a serious explosion hazard.
Perchlorate crystals are explosive, and detonation can occur by heat, flame, friction, percussion, or another chemical reaction. Something as small as the vibration of the blower motor can cause a violent reaction. Fortunately, perchlorate salts dissolve in water and water neutralizes perchloric acid. Therefore, special hoods that offer complete washdown are required for work with perchlorate crystals.
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Perchloric Acid, 70% (Certified ACS), Fisher Chemical™
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CAS: 7732-18-5 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
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| PubChem CID | 24247 |
| CAS | 7732-18-5 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
2
Perchloric Acid (TraceMetal™ Grade), Fisher Chemical™
CAS: 7732-18-5,7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
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Specifications
| PubChem CID | 24247 |
| CAS | 7732-18-5,7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
3
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Perchloric Acid Solution, N/10 (0.1N) (Certified), Fisher Chemical™
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
4
Perchloric Acid (Optima™), Fisher Chemical™
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CAS: 7732-18-5 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7732-18-5 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
5
Perchloric Acid, 60% (Certified ACS), Fisher Chemical™
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CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
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Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
6
Perchloric Acid, 8% (w/v) Aqueous, Ricca Chemical
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CAS: 7732-18-5,7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
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Specifications
| PubChem CID | 24247 |
| CAS | 7732-18-5,7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
7
Perchloric acid, 70%, 99.9985% (metals basis)
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
8
Perchloric Acid, 70%, ACS, 69.0-72.0%, Spectrum™ Chemical
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CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
9
Perchloric Acid, 0.100 N (N/10) in Glacial Acetic Acid, Ricca Chemical
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CAS: 64-19-7 Molecular Formula: C2H4O2 Molecular Weight (g/mol): 60.05 MDL Number: MFCD00036152 InChI Key: QTBSBXVTEAMEQO-UHFFFAOYSA-N IUPAC Name: acetic acid SMILES: CC(O)=O
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Specifications
| CAS | 64-19-7 |
| Molecular Weight (g/mol) | 60.05 |
| MDL Number | MFCD00036152 |
| SMILES | CC(O)=O |
| IUPAC Name | acetic acid |
| InChI Key | QTBSBXVTEAMEQO-UHFFFAOYSA-N |
| Molecular Formula | C2H4O2 |
10
Perchloric acid, for analysis, ca. 70% solution in water
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
11
Perchloric Acid, TraceSELECT™, 67 to 72%, Honeywell Fluka™
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
12
Perchloric Acid, ACS, 60-62%, Thermo Scientific Chemicals
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
13
Perchloric acid, pure, 0.1M solution in acetic acid
CAS: 7601-90-3 | ClHO4 | 100.45 g/mol
Pricing & Availability
Specifications
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| UN Number | 2920 |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Density | 1.0600g/mL |
| PubChem CID | 24247 |
| Name Note | 0.1 M Solution in Acetic Acid |
| Fieser | 01,796; 02,309; 03,220; 06,453; 09,279; 11,402 |
| Formula Weight | 100.46 |
| Color | Colorless |
| Physical Form | Liquid |
| Chemical Name or Material | Perchloric acid |
| Grade | Pure |
| SMILES | OCl(=O)=O |
| Merck Index | 15, 7267 |
| CAS | 64-19-7 |
| Health Hazard 3 | GHS P Statement: Wear protective gloves/protective clothing/eye protection/face protection. IF SWALLOWED: rinse mouth. Do NOT induce vomiting. Wear eye protection/face protection. IF IN EYES: Rinse cautiously with water for several minutes. Remove contact lenses, if present and easy to do. Continue rinsing. Immediately call a POISON CENTER or doctor/physician. IF ON SKIN: Wash with plenty of soap and water. IF INHALED: Remove victim to fresh air and keep at rest in a position comfortable for breathing. Keep away from heat/sparks/open flames/hot surfaces. - No smoking. WARNING: The information provided on this web site was developed in compliance with European Union (EU) regulations and is correct to the best of our knowledge, information and belief at the date of its publication. The information given is designed only as a guide for safe handling and use. It is not to be considered as either a warranty or quality specification. |
| MDL Number | MFCD00011325 |
| Health Hazard 2 | GHS H Statement: Causes severe skin burns and eye damage. Flammable liquid and vapor. |
| Solubility Information | Solubility in water: insoluble. |
| Flash Point | 40°C |
| Packaging | Plastic Bottle |
| Health Hazard 1 | Danger |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| Molecular Formula | ClHO4 |
| Specific Gravity | 1.06 |
14
Perchloric acid, ACS reagent, ca. 70% solution in water
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
15
Perchloric acid, ACS, 69.0-72.0%, redistilled
CAS: 7601-90-3 Molecular Formula: ClHO4 Molecular Weight (g/mol): 100.45 MDL Number: MFCD00011325 InChI Key: VLTRZXGMWDSKGL-UHFFFAOYSA-N Synonym: unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 PubChem CID: 24247 ChEBI: CHEBI:29221 IUPAC Name: perchloric acid SMILES: OCl(=O)=O
Pricing & Availability
Specifications
| PubChem CID | 24247 |
| CAS | 7601-90-3 |
| Molecular Weight (g/mol) | 100.45 |
| ChEBI | CHEBI:29221 |
| MDL Number | MFCD00011325 |
| SMILES | OCl(=O)=O |
| Synonym | unii-v561v90bg2,perchloric acid,perchloric acid acid by mass forbidden,perchloric acid, acs reagent,perchlorsaeure,perchloric acid,ueberchlorsaeure,hyperchloric acid,hclo4 |
| IUPAC Name | perchloric acid |
| InChI Key | VLTRZXGMWDSKGL-UHFFFAOYSA-N |
| Molecular Formula | ClHO4 |
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2746 | https://www.mathway.com/popular-problems/Trigonometry/300434 | Enter a problem...
Trigonometry Examples
Popular Problems
Find Amplitude, Period, and Phase Shift y=sin(x)
Step 1
Use the form to find the variables used to find the amplitude, period, phase shift, and vertical shift.
Step 2
Find the amplitude .
Amplitude:
Step 3
Find the period of .
Tap for more steps...
Step 3.1
The period of the function can be calculated using .
Step 3.2
Replace with in the formula for period.
Step 3.3
The absolute value is the distance between a number and zero. The distance between and is .
Step 3.4
Divide by .
Step 4
Find the phase shift using the formula .
Tap for more steps...
Step 4.1
The phase shift of the function can be calculated from .
Phase Shift:
Step 4.2
Replace the values of and in the equation for phase shift.
Phase Shift:
Step 4.3
Divide by .
Phase Shift:
Phase Shift:
Step 5
List the properties of the trigonometric function.
Amplitude:
Phase Shift: None
Vertical Shift: None
Step 6
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2747 | https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/03%3A_Composition_of_Substances_and_Solutions/3.03%3A_Molarity | Skip to main content
3.3: Molarity
Last updated
: Aug 26, 2023
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3.2: Determining Empirical and Molecular Formulas
3.4: Other Units for Solution Concentrations
Page ID
: 414614
OpenStax
OpenStax
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Learning Objectives
By the end of this section, you will be able to:
Describe the fundamental properties of solutions
Calculate solution concentrations using molarity
Perform dilution calculations using the dilution equation
Preceding sections of this chapter focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see Figure ). This section will describe one of the most common ways in which the relative compositions of mixtures may be quantified.
Solutions
Solutions have previously been defined as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. A more thorough treatment of solution properties is provided in the chapter on solutions and colloids, but provided here is an introduction to some of the basic properties of solutions.
The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution.
A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).
Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:
Example : Calculating Molar Concentrations
A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?
Solution
Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:
Exercise
A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?
Answer
: 0.05 M
Example : Deriving Moles and Volumes from Molar Concentrations
How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example ?
Solution
Rearrange the definition of molarity to isolate the quantity sought, moles of sugar, then substitute the value for molarity derived in Example , 0.375 M:
Exercise
What volume (mL) of the sweetened tea described in Example contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?
Answer
: 80 mL
Example : Calculating Molar Concentrations from the Mass of Solute
Distilled white vinegar (Figure ) is a solution of acetic acid, CH3CO2H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?
Solution
As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. Since the mass of solute is provided instead of its molar amount, use the solute’s molar mass to obtain the amount of solute in moles:
Exercise
Calculate the molarity of 6.52 g of CoCl2 (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.
Answer
: 0.674 M
Example : Determining the Mass of Solute in a Given Volume of Solution
How many grams of NaCl are contained in 0.250 L of a 5.30-M solution?
Solution
The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example :
Finally, this molar amount is used to derive the mass of NaCl:
Exercise
How many grams of CaCl2 (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?
Answer
: 5.55 g CaCl2
When performing calculations stepwise, as in Example , it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example , the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If the guard digit had not been retained, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.
In addition to retaining a guard digit for intermediate calculations, rounding errors may also be avoided by performing computations in a single step (see Example ). This eliminates intermediate steps so that only the final result is rounded.
Example : Determining the Volume of Solution Containing a Given Mass of Solute
In Example , the concentration of acetic acid in white vinegar was determined to be 0.839 M. What volume of vinegar contains 75.6 g of acetic acid?
Solution
First, use the molar mass to calculate moles of acetic acid from the given mass:
Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:
Combining these two steps into one yields:
Exercise
What volume of a 1.50-M KBr solution contains 66.0 g ?
Answer
: 0.370 L
Dilution of Solutions
Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure ).
Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, a solution of lesser concentration may be prepared. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents.
A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the number of moles of solute in a solution (n) is equal to the product of the solution’s molarity (M) and its volume in liters (L):
Expressions like these may be written for a solution before and after it is diluted:
where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, n1 = n2. Thus, these two equations may be set equal to one another:
This relation is commonly referred to as the dilution equation. Although this equation uses molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used as long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form:
where (C) and (V) are concentration and volume, respectively.
Link to Learning
Use the simulation to explore the relations between solute amount, solution volume, and concentration and to confirm the dilution equation.
Example : Determining the Concentration of a Diluted Solution
If 0.850 L of a 5.00-M solution of copper nitrate, Cu(NO3)2, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution?
Solution
The stock concentration, C1, and volume, V1, are provided as well as the volume of the diluted solution, V2. Rearrange the dilution equation to isolate the unknown property, the concentration of the diluted solution, C2:
Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), the diluted solution’s concentration is expected to be less than one-half 5 M. This ballpark estimate will be compared to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields:
This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M).
Exercise
What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-M solution of CH3OH to 500.0 mL?
Answer
: 0.102 M CH3OH
Example : Volume of a Diluted Solution
What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr?
Solution
Provided are the volume and concentration of a stock solution, V1 and C1, and the concentration of the resultant diluted solution, C2. Find the volume of the diluted solution, V2 by rearranging the dilution equation to isolate V2:
Since the diluted concentration (0.12 M) is slightly more than one-fourth the original concentration (0.45 M), the volume of the diluted solution is expected to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields:
The volume of the 0.12-M solution is 0.041 L (41 mL). The result is reasonable and compares well with the rough estimate.
Exercise
A laboratory experiment calls for 0.125 M HNO3. What volume of 0.125 M HNO3 can be prepared from 0.250 L of 1.88 M HNO3?
Answer
: 3.76 L
Example : Volume of a Concentrated Solution Needed for Dilution
What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH?
Solution
Given are the concentration of a stock solution, C1, and the volume and concentration of the resultant diluted solution, V2 and C2. Find the volume of the stock solution, V1 by rearranging the dilution equation to isolate V1:
Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), the volume of the stock solution is expected to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields:
Thus, 0.314 L of the 1.59-M solution is needed to prepare the desired solution. This result is consistent with the rough estimate.
Exercise
What volume of a 0.575-M solution of glucose () can be prepared from 50.00 mL of a 3.00-M glucose solution?
Answer
: 0.261 L
3.2: Determining Empirical and Molecular Formulas
3.4: Other Units for Solution Concentrations |
2748 | https://medistudygo.com/blogs/dental/indications-contraindications-tooth-extraction?srsltid=AfmBOooYh6WQ_TXkHSJ9bJ5ZAVopIaJXvcuhmaASt7pUmGiw90nJJ_Ym | Complete Guide to Indications and Contraindications for Tooth Extracti – Medi Study Go
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Complete Guide to Indications and Contraindications for Tooth Extraction
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Related Resources
Comprehensive Guide to Exodontia: Principles, Procedures, and Best Practices
Dental Extraction Instruments & Techniques: A Complete Guide
Surgical Extraction Procedures: A Comprehensive Guide
Managing Complications in Tooth Extraction
Post-Extraction Care and Healing
Introduction
The decision to extract a tooth should never be taken lightly, as preservation of natural dentition should always be the primary goal in dentistry. However, there are numerous clinical scenarios where tooth extraction becomes the most appropriate treatment option. Understanding the clear indications and contraindications for exodontia is essential for proper treatment planning and optimal patient care.
This guide will help you make evidence-based decisions about when extraction is necessary and when alternative treatments should be considered. This knowledge is particularly valuable for NEET preparation and clinical practice.
Indications for Tooth Extraction
1. Teeth with Severe Pathology
Severe Caries (49% of Extraction Cases)
Teeth with extensive crown destruction beyond restoration
Deep subgingival caries that make restoration impossible
Failed endodontic treatment with persistent pathology
Severe Periodontal Disease (41% of Extraction Cases)
Excessive mobility (Grade III)
Severe bone loss (>75% of supporting bone)
Teeth serving as source of periodontal infection
Strategic extractions to improve overall periodontal prognosis
Severe Attrition, Abrasion, and Erosion
When tooth structure is compromised beyond restoration
When vertical dimension is severely affected
2. Pulpal and Periapical Pathology
Acute/chronic pulpitis or pulpal necrosis when endodontic treatment is not feasible
Teeth with large periapical lesions without possibility of successful endodontic treatment
Failed root canal treatment with persistent symptoms
3. Dental Trauma Cases
Teeth with severe crown-root fractures
Teeth with vertical root fractures
Luxated teeth with poor prognosis
Teeth in the line of jaw fractures (may complicate fracture healing)
4. Impacted and Malposed Teeth
Impacted teeth causing pathology or at risk of pathology:
Pericoronitis (inflammation around partially erupted teeth)
Cyst or tumor formation
Resorption of adjacent teeth
Unexplained pain
Over-retained deciduous teeth impeding eruption of permanent successors
Supernumerary teeth causing crowding or preventing normal eruption
5. Orthodontic Considerations
Strategic extraction for relief of crowding
Extractions to facilitate orthodontic movement
Teeth with poor prognosis that may complicate orthodontic treatment
Serial extractions in mixed dentition cases
6. Prosthodontic Reasons
Malposed teeth interfering with prosthetic design
Teeth compromising stability of removable prostheses
Teeth with poor prognosis that may jeopardize comprehensive restorative treatment
7. Pre-Radiation Therapy
Teeth with poor prognosis in the radiation field
Preventive extractions to reduce risk of osteoradionecrosis
8. Pre-Cardiac Surgery
Teeth with infections or poor prognosis before cardiac surgery
Prevention of infective endocarditis
Teeth at risk of becoming source of infection for patients with prosthetic heart valves
9. Financial Considerations
While not an ideal indication, financial limitations may sometimes necessitate extraction over more costly restorative options. This decision should be made carefully, with full informed consent and consideration of long-term consequences.
Contraindications for Tooth Extraction
Contraindications to extraction can be classified as absolute (where extraction should not be performed) or relative (where extraction should be postponed or modified).
Absolute Contraindications
1. Systemic Conditions
Uncontrolled diabetes mellitus
Severe uncontrolled hypertension
Recent myocardial infarction (within 6 months)
Unstable angina pectoris
Uncontrolled congestive cardiac failure
End-stage renal disease with severe uremia
Advanced or uncompensated liver disease
Uncontrolled bleeding disorders (hemophilia, severe thrombocytopenia)
Acute leukemia with severe neutropenia
Uncontrolled adrenal insufficiency
Recent cerebrovascular accident (stroke)
2. Local Conditions
Teeth in the area of active malignant lesions (risk of metastasis)
Teeth in previously irradiated areas (risk of osteoradionecrosis)
Pregnancy (first trimester and late third trimester)
Relative Contraindications
1. Cardiovascular Conditions
Controlled hypertension
History of myocardial infarction (>6 months ago)
Stable angina
History of rheumatic heart disease
Prosthetic heart valves (requires prophylaxis)
2. Blood Disorders
Controlled bleeding disorders
Patients on anticoagulant therapy (may require modification)
Controlled anemia
3. Metabolic Disorders
Controlled diabetes mellitus
Controlled renal disease
Controlled thyroid disorders
4. Local Factors
Acute infections (pericoronitis, dentoalveolar abscess)
Acute ulcerative gingivitis
Radiation therapy (within 6 months to 1 year)
5. Special Physiologic States
Pregnancy (second trimester is safest for emergency procedures)
Menstruation (increased estrogen can cause excessive bleeding)
Extreme old age (with compromised physical and mental conditions)
Management of Tooth Extraction in Medically Compromised Patients
1. Cardiovascular Disorders
Hypertension
Measure blood pressure before extraction
For BP <180/110 mmHg: Proceed with normal protocol
For BP >180/110 mmHg: Postpone elective extractions
Short appointments in morning
Effective pain control
Limit epinephrine in anesthetics
Ischemic Heart Disease
Consult with cardiologist
Schedule short morning appointments
Effective stress and anxiety reduction
Nitrous oxide-oxygen sedation if necessary
Limit epinephrine (0.04 mg maximum)
Prophylactic nitroglycerin before procedure for angina patients
Post-Myocardial Infarction
Defer elective procedures for 6 months
Mandatory medical consultation
Oxygen supplementation during procedure
Continuous vital sign monitoring
Prosthetic Valves
Antibiotic prophylaxis per current guidelines
International Normalized Ratio (INR) check for patients on anticoagulants
Target INR: 2.0-3.0 for most cases
2. Bleeding Disorders
Patients on Anticoagulants
For warfarin patients: Check INR (ideal range 2.0-3.5)
For newer direct oral anticoagulants: Consult prescribing physician
Local hemostatic measures (gelfoam, tranexamic acid)
Careful surgical technique with minimal trauma
Post-extraction compression for at least 30 minutes
Avoid NSAIDs for pain management
Hemophilia and von Willebrand's Disease
Mandatory hematologist consultation
Factor replacement therapy as recommended
Antifibrinolytic agents (tranexamic acid)
Local hemostatic measures
Extended post-operative monitoring
3. Endocrine Disorders
Diabetes Mellitus
Morning appointments
Ensure patient has taken medication and eaten
Check blood glucose before procedure (ideal: 70-180 mg/dL)
Prophylactic antibiotics for poorly controlled cases
Close post-operative follow-up
Adrenal Insufficiency
Stress dose steroids before procedure
Morning appointments
Prophylactic antibiotics in some cases
Careful post-operative monitoring
Thyroid Disorders
Postpone elective procedures in uncontrolled cases
For hyperthyroidism: Risk of thyroid storm
For hypothyroidism: Risk of prolonged drug action
4. Respiratory Disorders
Asthma
Ensure patient brings inhaler
Avoid triggers (anxiety, certain medications)
Consider premedication with bronchodilators
Oxygen availability during procedure
Chronic Obstructive Pulmonary Disease (COPD)
Short appointments
Semi-upright chair position
Careful use of sedatives (risk of respiratory depression)
Avoid excessive water spray
5. Liver Disease
Cirrhosis
Coagulation studies (PT, INR, PTT)
Reduced doses of many medications
Risk of impaired drug metabolism
Prophylactic antibiotics for severe cases
6. Pregnancy
First Trimester
Avoid elective procedures
Emergency care only with minimal radiographs
No teratogenic medications
Second Trimester (Ideal time for treatment)
Short appointments
Left lateral position to prevent supine hypotension
Careful medication selection
Third Trimester
Increasing risk of supine hypotensive syndrome
Stress reduction protocols
Left lateral positioning
Consultation with obstetrician
7. Neurological Disorders
Epilepsy
Ensure medication compliance
Stress reduction protocols
Morning appointments when seizure threshold is highest
Avoid bright lights if photosensitive
Post-Stroke
Defer elective extractions for 6 months after CVA
Consultation with neurologist
Consider discontinued anticoagulants/antiplatelets
Careful BP monitoring
Special Considerations for Medically Complex Patients
Pre-Extraction Assessment
Thorough medical history review
Consultation with specialists when indicated
Medication review and potential modifications
Laboratory tests as indicated by medical condition
Risk-benefit analysis and informed consent
Modifications to Standard Extraction Technique
Shorter appointments
Appropriate timing (usually morning)
Stress reduction protocols
Modified anesthetic approaches
Modified hemostasis techniques
Prophylactic medications as indicated
Post-Extraction Considerations
More frequent follow-up
Clear emergency instructions
Medication adjustments
Recognition of potential complications specific to medical condition
Conclusion
The decision to extract a tooth requires careful consideration of both oral and systemic factors. A thorough understanding of indications and contraindications helps clinicians make appropriate treatment decisions that balance immediate needs with long-term consequences.
For medically compromised patients, interdisciplinary collaboration with medical specialists is essential to ensure safe extraction procedures with minimal complications. Modifications to standard protocols based on specific medical conditions can significantly reduce risks and improve outcomes.
This comprehensive approach to treatment planning for extractions is essential knowledge for success in the NEET MDS examination and for providing optimal patient care in clinical practice.
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2749 | https://pubmed.ncbi.nlm.nih.gov/30903611/ | Impact of calcifications on diagnostic yield of endoscopic ultrasound-guided fine-needle aspiration for pancreatic ductal adenocarcinoma - PubMed
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Indian J Gastroenterol
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. 2019 Apr;38(2):128-133.
doi: 10.1007/s12664-019-00941-y. Epub 2019 Mar 22.
Impact of calcifications on diagnostic yield of endoscopic ultrasound-guided fine-needle aspiration for pancreatic ductal adenocarcinoma
Anoop K Koshy1,Rao B Harshavardhan1,Ismail Siyad1,Rama P Venu2
Affiliations Expand
Affiliations
1 Department of Gastroenterology, Amrita Institute of Medical Sciences, AIMS PO, Ponekkara, Kochi, 682 041, India.
2 Department of Gastroenterology, Amrita Institute of Medical Sciences, AIMS PO, Ponekkara, Kochi, 682 041, India. harshavardhanrao1985@gmail.com.
PMID: 30903611
DOI: 10.1007/s12664-019-00941-y
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Impact of calcifications on diagnostic yield of endoscopic ultrasound-guided fine-needle aspiration for pancreatic ductal adenocarcinoma
Anoop K Koshy et al. Indian J Gastroenterol.2019 Apr.
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. 2019 Apr;38(2):128-133.
doi: 10.1007/s12664-019-00941-y. Epub 2019 Mar 22.
Authors
Anoop K Koshy1,Rao B Harshavardhan1,Ismail Siyad1,Rama P Venu2
Affiliations
1 Department of Gastroenterology, Amrita Institute of Medical Sciences, AIMS PO, Ponekkara, Kochi, 682 041, India.
2 Department of Gastroenterology, Amrita Institute of Medical Sciences, AIMS PO, Ponekkara, Kochi, 682 041, India. harshavardhanrao1985@gmail.com.
PMID: 30903611
DOI: 10.1007/s12664-019-00941-y
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Introduction: Chronic calcific pancreatitis (CCP) is a major risk factor for pancreatic ductal adenocarcinoma (PDAC) and is common in southern India. Endoscopic ultrasound fine-needle aspiration (EUS-FNA) is useful for tissue acquisition in patients with solid pancreatic lesions. Multiple factors may affect the diagnostic yield of FNA samples. The present study was performed to assess the impact of pancreatic calcifications on the diagnostic yield of EUS-FNA in PDAC.
Methods: All patients with confirmed PDAC from January 2013 to December 2017 were included. CCP was diagnosed based on typical imaging characteristics with or without evidence of pancreatic insufficiency along with surgical histopathology reports showing features of chronic pancreatitis. The diagnostic yield and adequacy of cellularity were assessed by a pathologist who was blinded and were compared between the two groups: group 1: PDAC patients with no evidence of CCP and, group 2: PDAC patients with CCP.
Results: A total of 122 patients were included in the study. The diagnostic yield was lower in patients in group 2 (n = 42, 25 [59.52%]) as compared to those in group 1 (n = 80, 63 [78.75%]) (p-value = 0.01). On multivariate analysis, only the presence of calcifications was found to have an independent association with diagnostic yield (odds ratio 3.83 [95% confidence interval 1.22-11.9]).
Conclusions: CCP had a significant impact on the diagnostic yield of EUS-FNA for pancreatic adenocarcinoma. Novel techniques and newer technology that may mitigate the negative effect of calcification on diagnostic yield of EUS-FNA in patients with CCP.
Keywords: Chronic calcific pancreatitis; Diagnosis; Endoscopic ultrasound; Pancreatic adenocarcinoma.
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References
Gastrointest Endosc. 2005 Jun;61(7):854-61 - PubMed
Gastroenterology. 2001 Feb;120(3):682-707 - PubMed
Pancreas. 2013 Jan;42(1):20-6 - PubMed
CA Cancer J Clin. 2012 Jan-Feb;62(1):10-29 - PubMed
Trials. 2018 Jan 17;19(1):45 - PubMed
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2750 | https://lucianopirri.altervista.org/Varie/Ingegneria%20Meccanica/Auto%20in%20una%20curva%20sopraelevata.html | Auto in una curva sopraelevata - La pista di Nardò
Teoria
Quando un veicolo affronta una curva una forza centripeta è necessaria per tenerlo in carreggiata. Se la strada è orizzontale la forza centripeta è fornita dall'aderenza (attrito statico) dei pneumatici con la strada.
Molto spesso le curve sono sopraelevate in modo che il peso del veicolo contribuisca alla creazione della forza centripeta necessaria alla tenuta di strada. Questo consente anche di ridurre il consumo dei pneumatici delle auto.
La fisica del fenomeno, oltre alla accelerazione di gravità , coinvolge i seguenti parametri:
$v$ - velocità dell'automobile
$mu$ - coefficiente di aderenza o di attrito statico
$theta$ - angolo di inclinazione del profilo stradale
$r$ - raggio di curvatura
Nel sistema di riferimento non inerziale solidale al guidatore, la macchina è soggetta a 4 forze: la forza peso $P$ diretta lungo la verticale, la reazione normale $R_n$ lungo la normale $n$, la forza centrifuga $F_c$ lungo la radiale, la forza di attrito statico $A$ lungo lasse tangenziale $t$.
Nel sistema solidale al guidatore lauto è ferma e le 4 forze si equilibrano
$vec P + vec F_c + vec R_n + vec A = 0$
In condizioni di equilibrio le forze formano un poligono chiuso.
I vettori sono a due a due perpendicolari tra loro.
In questo schema l'inclinazione è poca e l'auto tende sbandare verso l'esterno della curva quindi la forza d'attrito è diretta verso l'interno.
| |
| |
In questo schema l'inclinazione è troppa e l'auto tende sbandare verso l'interno della curva quindi la forza d'attrito è diretta verso l'esterno.
| |
| |
Se l'angolo diventa di 90° si è realizzato il dispositivo noto come rotore diffuso in nei parchi divertimento. Una persona è appoggiata ad una parete circolare che inizia a ruotare, quando la velocità di rotazione è adeguata viene tolto il pavimento sotto i suoi piedi e la persona non cade ma rimane attaccata alla parete perché trattenuta dalla forza d'attrito.
| |
| |
Il coefficiente di aderenza trasversale richiesto è:
$mu=|vecA|/|vec R_n|$
Proiettando lequazione vettoriale lungo gli assi $n$,$t$ otteniamo:
| |
| |
Adesso deriviamo le espressioni esplicite per le varie grandezze.
Componente della forza centrifuga parallela alla strada, diretta verso l'esterno della curva, che tende a mandare fuori strada e che quindi è da equilibrare:
$F_(cp) := (mv^2)/rcos(theta) $
Componente della forza centrifuga normale alla strada che contribuisce a schiacciare l'auto sulla carreggiata:
$F_(cn) := (mv^2)/rsin(theta) $
Componente della forza peso parallela alla strada, è diretta verso l'interno della curva e quindi contribuisce ad equilibrare la forza centrifuga:
$P_p= mgsin(theta)$
Componente della forza peso normale alla strada che contribuisce a schiacciare l'auto sulla carreggiata:
$P_n= mgcos(theta)$
La forza d'attrito massima $A$ è diretta in verso opposto al possibile movimento, quindi è diretta come $P_p$ ed è proporzionale alla forza normale alla carreggiata
| |
| $A = muR_n = mu(P_n + F_(cn))$ |
| $ A = mu(mgcos(theta)+(mv^2)/rsin(theta)) $ |
All'equilibrio si ha che la componente parallela della forza peso aiuta la forza d'attrito ad equilibrare la forza centrifuga:
| | |
--- |
| $F_(cp)$ | $= P_p+ A $ |
| $(mv^2)/rcos(theta)$ | $ = mgsin(theta) +$ |
| $+ mu(mgcos(theta)+(mv^2)/rsin(theta)) $ |
Che può essere riscritta come:
| | |
--- |
| $ v^2/(gr) = tan(alpha) + mu(1+v^2/(gR)tan(alpha)) $ | (1) |
Dalla precedente ricaviamo le espressioni cercate.
| | |
--- |
| $ r = -v^2/g(mutan(theta)-1)/(mu+tan(theta)) $ | (2) |
| $mu=( v^2 - rgtan(theta) )/(v^2tan(theta)+rg)$ | (3) |
| $ v=sqrt( rg(mu+tan(theta))/(1- mutan(theta) ) ) $ | (4) |
| $theta=atan((v^2-mugr)/(v^2mu+gr))$ | (5) |
Occorre notare che per l'inclinazione $theta$ deve essere $theta < atan(mu)$ altrimenti l'auto da ferma (cioè senza forza centrifuga) scivolerebbe verso il basso per effetto del solo peso.
Esempi
Esempio Un'automobile affronta una curva alla velocità $v=##v=72## uukph = #v=vuukph# uum//uus $, la pendenza della sezione stradale è $theta = ##theta=10##° = #theta=thetapi/180# uurad$. il coefficiente di aderenza è $mu=#mu=0.22#$ ed infine $g=#g# uuN//uukg$. Calcolare il raggio di curvatura minimo della strada, al disotto del quale si finisce fuori strada.
Soluzione. Dall'equazione (2) otteniamo:
| |
| $ r = -v^2/g(mutan(theta)-1)/(mu+tan(theta)) $ $ r = -##v##^2/##g##(##mu##tan(#theta#)-1)/(##mu##+tan(#theta#)) $ $ r = # r = -v^2/g(mutan(theta)-1)/(mu+tan(theta)) # uum$ |
Il raggio è positivo per $mutan(theta)-1 < 0$ ovvero per $theta < atan(1/mu) = #tMax=atan(1/mu)# uurad = #tMax180/pi#°$
| |
| |
La zona dei valori ammissibili è sotto la curva dei valori limite.
Esempio Un auto viaggia alla velocità $v=##v=72## uukph = #v=vuukph# uum//uus $ ed affronta una curva di raggio di curvatura $r=##r=100## uum$ ed inclinata rispetto allorizzontale di un angolo $theta = ##theta=10##° = #theta=thetapi/180# uurad$. Calcolare quale deve essere il valore minimo del coefficiente di attrito statico fra piano stradale e pneumatici affinché lauto non sbandi.
Soluzione. Dall'equazione (3) otteniamo il valore minimo del coefficiente di aderenza:
| |
| $mu=( v^2 - rgtan(theta) )/(v^2tan(theta)+rg)$ $mu=( ##v##^2 - ##r###g#tan(#theta#) )/(##v##^2tan(#theta#)+##r###g#)$ $ mu=#mu=( vv - rgtan(theta) )/(vvtan(theta)+rg)# $ |
Esempio Una curva ha un raggio di curvatura $r=##r=100## uum$, una sopraelevazione di $theta=##theta=10##° = #theta=thetapi/180# uurad$. Calcolare la velocità massima con cui si può percorrere la curva nei due casi: 1) assenza di attrito 2) con attrito $mu=0.22$
Soluzione
Prima di tutto verifichiamo se l'auto può essere parcheggiata, cioè star ferma, sulla curva sopraelevata. L'angolo d'attrito corrispondente coefficiente di attrito $mu$ vale
$phi = arctan(mu) = # atan(mu) # uurad $
quindi dato che la pendenza della strada $theta < phi$ c'è da attendersi che la macchina anche se rallentasse fino a fermarsi rimarrebbe in carreggiaia.
1) Nel primo caso poniamo nell'equazione (4) $mu=##mu=0.0##$ ed otteniamo
| |
| $ v=sqrt( rg(mu+tan(theta))/(1- mutan(theta) ) ) $ $ v=sqrt( rgtan(theta) ) $ $ v=sqrt( ##r###g#tan(#theta#) ) $ $ v = #v=sqrt( rgtan(theta) ) # uum//uus = #v3.6# uukm//uuh $ |
La velocità è illimitata per £\theta=\pi/2£ cioè $90°$.
2) Nel secondo caso $mu=#mu=0.22#$ otteniamo
| |
| $ v=sqrt( rg(mu+tan(theta))/(1- mutan(theta) ) ) $ $ v=sqrt( ##r###g#(#mu#+tan(#theta#))/(1- #mu#tan(#theta#) ) ) $ $v= #v=sqrt( rg(mu+tan(theta))/( 1-mutan(theta)) ) # uum//uus = #v3.6# uukm//uuh $ |
La velocità è illimitata per $theta = atan(1/mu) = #tMax=atan(1/mu)# uurad = #tMax180/pi#°$
| |
| |
Le due zone dei valori ammissibili sono sotto le due curve dei valori limite.
Soluzione. Dall'equazione (5) otteniamo:
| |
| $theta=atan((v^2-mugr)/(v^2mu+gr))$ $theta=atan((#v#^2-#mu##g###r##)/(#v#^2#mu#+#g###r##)) $ $theta=#theta=atan((vv-mugr)/(vvmu+gr))# uurad = #theta180/pi#°$ |
La normativa italiana
Riprendiamo l'equazione (1). Nelle strade ordinarie la pendenza è di qualche punto percentuale quindi possiamo trascurare nel termine entro parentesi il termine $tan(alpha)$ rispetto all'unita. L'equazione di equilibrio dell'auto al limite dello sbandamento, cioè quando tutta l'aderenza è impegnata, si può riscrivere:
$v^2 = r g(tan(alpha)+mu)$
Esprimendo la velocità $V$ in $uukm//uuh$ si ha che $3.6^29.81 approx 127$ quindi la precedente può essere riscritta :
$ V^2 = 127r(tan(alpha)+mu)$
ed ancora, ponendo la pendenza $q=tan(alpha)$
$ V^2 = 127r(q+mu)$
Sostituendo nella precedente il valore massimo della pendenza e i valori massimi dell'aderenza si ricava il raggio minimo:
$ r_"min" = V^2/(127(q_"max"+mu_"max")) $
I valori di aderenza trasversale proposti dalla normativa CNR del 1980 sono:
| | | | | | | |
--- --- ---
| V [km/h] | 40 | 60 | 80 | 100 | 120 | 140 |
| μ | 0.24 | 0.17 | 0.13 | 0.11 | 0.10 | 0.09 |
Per quanto riguarda la pendenza la normativa CNR del 1980 propone $0.025 < q < 0.07 $ e per il raggio massimo $r_"max"=5r_"mim"$ infine nel seguente abaco è riportato il raggio minimo in funzione della velocità e della pendenza.
Ad oggi (2021) la normativa del 1980 è soggetta ad una importante revisione perché i valori sono ritenuti eccessivamente cautelativi. ©Ing. Luciano Pirri - giovedì 16 dicembre 2021 - Agg. |
2751 | https://openstax.org/books/chemistry/pages/17-1-balancing-oxidation-reduction-reactions | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Chemistry
17.1 Balancing Oxidation-Reduction Reactions
Chemistry17.1 Balancing Oxidation-Reduction Reactions
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Learning Objectives
By the end of this section, you will be able to:
Define electrochemistry and a number of important associated terms
Split oxidation-reduction reactions into their oxidation half-reactions and reduction half-reactions
Produce balanced oxidation-reduction equations for reactions in acidic or basic solution
Identify oxidizing agents and reducing agents
Electricity refers to a number of phenomena associated with the presence and flow of electric charge. Electricity includes such diverse things as lightning, static electricity, the current generated by a battery as it discharges, and many other influences on our daily lives. The flow or movement of charge is an electric current (Figure 17.2). Electrons or ions may carry the charge. The elementary unit of charge is the charge of a proton, which is equal in magnitude to the charge of an electron. The SI unit of charge is the coulomb (C) and the charge of a proton is 1.602 10−19 C. The presence of an electric charge generates an electric field. Electric current is the rate of flow of charge. The SI unit for electrical current is the SI base unit called the ampere (A), which is a flow rate of 1 coulomb of charge per second (1 A = 1 C/s). An electric current flows in a path, called an electric circuit. In most chemical systems, it is necessary to maintain a closed path for current to flow. The flow of charge is generated by an electrical potential difference, or potential, between two points in the circuit. Electrical potential is the ability of the electric field to do work on the charge. The SI unit of electrical potential is the volt (V). When 1 coulomb of charge moves through a potential difference of 1 volt, it gains or loses 1 joule (J) of energy. Table 17.1 summarizes some of this information about electricity.
| Common Electrical Terms |
| Quantity | Definition | Measure or Unit |
| Electric charge | Charge on a proton | 1.602 10−19 C |
| Electric current | The movement of charge | ampere = A = 1 C/s |
| Electric potential | The force trying to move the charge | volt = V = J/C |
| Electric field | The force acting upon other charges in the vicinity | |
Table 17.1
Figure 17.2 Electricity-related phenomena include lightning, accumulation of static electricity, and current produced by a battery. (credit left: modification of work by Thomas Bresson; credit middle: modification of work by Chris Darling; credit right: modification of work by Windell Oskay)
Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced chemical equation were relatively simple. In this section, we will concentrate on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important partly for balancing more complicated reactions and partly because many aspects of electrochemistry are easier to discuss in terms of half-reactions. There are alternate methods of balancing these reactions; however, there are no good alternatives to half-reactions for discussing what is occurring in many systems. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier.
Electrochemical reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. It helps to see this in an actual problem. Consider the following unbalanced oxidation-reduction reaction in acidic solution:
We can start by collecting the species we have so far into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction. Each of these half-reactions contain the same element in two different oxidation states. The Fe2+ has lost an electron to become Fe3+; therefore, the iron underwent oxidation. The reduction is not as obvious; however, the manganese gained five electrons to change from Mn7+ to Mn2+.
In acidic solution, there are hydrogen ions present, which are often useful in balancing half-reactions. It may be necessary to use the hydrogen ions directly or as a reactant that may react with oxygen to generate water. Hydrogen ions are very important in acidic solutions where the reactants or products contain hydrogen and/or oxygen. In this example, the oxidation half-reaction involves neither hydrogen nor oxygen, so hydrogen ions are not necessary to the balancing. However, the reduction half-reaction does involve oxygen. It is necessary to use hydrogen ions to convert this oxygen to water.
17.3
The situation is different in basic solution because the hydrogen ion concentration is lower and the hydroxide ion concentration is higher. After finishing this example, we will examine how basic solutions differ from acidic solutions. A neutral solution may be treated as acidic or basic, though treating it as acidic is usually easier.
The iron atoms in the oxidation half-reaction are balanced (mass balance); however, the charge is unbalanced, since the charges on the ions are not equal. It is necessary to use electrons to balance the charge. The way to balance the charge is by adding electrons to one side of the equation. Adding a single electron on the right side gives a balanced oxidation half-reaction:
17.4
You should check the half-reaction for the number of each atom type and the total charge on each side of the equation. The charges include the actual charges of the ions times the number of ions and the charge on an electron times the number of electrons.
17.5
If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent.
Now return to the reduction half-reaction equation:
17.6
The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) (1) + (8) (+1)], or +7, while the total charge on the right side is [(1) (+2) + (4) (0)], or +2. The difference between +7 and +2 is five; therefore, it is necessary to add five electrons to the left side to achieve charge balance.
17.7
You should check this half-reaction for each atom type and for the charge, as well:
17.8
Now that this half-reaction is balanced, it is easy to see it involves reduction because electrons were gained when was reduced to Mn2+. In all reduction half-reactions, electrons appear as reactants (on the left side). As discussed in the earlier chapter, the species that was reduced, in this case, is also called the oxidizing agent. We now have two balanced half-reactions.
17.9
It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go on to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives
17.10
The electrons do not appear in the final answer because the oxidation electrons are the same electrons as the reduction electrons and they “cancel.” Carefully check each side of the overall equation to verify everything was combined correctly:
17.11
Everything checks, so this is the overall equation in acidic solution. If something does not check, the most common error occurs during the multiplication of the individual half-reactions.
Now suppose we wanted the solution to be basic. Recall that basic solutions have excess hydroxide ions. Some of these hydroxide ions will react with hydrogen ions to produce water. The simplest way to generate the balanced overall equation in basic solution is to start with the balanced equation in acidic solution, then “convert” it to the equation for basic solution. However, it is necessary to exercise caution when doing this, as many reactants behave differently under basic conditions and many metal ions will precipitate as the metal hydroxide. We just produced the following reaction, which we want to change to a basic reaction:
17.12
However, under basic conditions, normally reduces to MnO2 and iron will be present as either Fe(OH)2 or Fe(OH)3. For these reasons, under basic conditions, this reaction will be
17.13
(Under very basic conditions will reduce to instead of MnO2.)
It is still possible to balance any oxidation-reduction reaction as an acidic reaction and then, when necessary, convert the equation to a basic reaction. There are very few examples in which the acidic and basic reactions will involve the same reactants and products. However, balancing a basic reaction as acidic and then converting to basic will work. To convert to a basic reaction, it is necessary to add the same number of hydroxide ions to each side of the equation so that all the hydrogen ions (H+) are removed and mass balance is maintained. Hydrogen ion combines with hydroxide ion (OH−) to produce water.
Let us now try a basic equation. We will start with the following basic reaction:
17.14
Balancing this as acid gives
17.15
In this case, it is necessary to add two hydroxide ions to each side of the equation to convert the two hydrogen ions on the left into water:
17.16
17.17
Note that both sides of the equation show water. Simplifying should be done when possible. In this case, it is necessary to remove one H2O from each side of the reaction arrow.
17.18
Again, check each side of the overall equation to make sure there are no errors:
17.19
Everything checks, so this is the overall equation in basic solution.
Example 17.1
Balancing Acidic Oxidation-Reduction Reactions
Balance the following reaction equation in acidic solution:
17.20
Solution
This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction.
17.21
Starting with the oxidation half-reaction, we can balance the chromium
17.22
In acidic solution, we can use or generate hydrogen ions (H+). Adding seven water molecules to the left side provides the necessary oxygen; the “left over” hydrogen appears as 14 H+ on the right:
17.23
The left side of the equation has a total charge of [2 (+3) = +6], and the right side a total charge of [−2 + 14 (+1) = +12]. The difference is six; adding six electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):
17.24
Checking the half-reaction:
17.25
Now work on the reduction. It is necessary to convert the four oxygen atoms in the permanganate into four water molecules. To do this, add eight H+ to convert the oxygen into four water molecules:
17.26
Then add five electrons to the left side to balance the charge:
17.27
Make sure to check the half-reaction:
17.28
Collecting what we have so far:
17.29
The least common multiple for the electrons is 30, so multiply the oxidation half-reaction by five, the reduction half-reaction by six, combine, and simplify:
17.30
17.31
Checking each side of the equation:
17.32
This is the balanced equation in acidic solution.
Check your learning
Balance the following equation in acidic solution:
17.33
Answer:
Example 17.2
Balancing Basic Oxidation-Reduction Reactions
Balance the following reaction equation in basic solution:
17.34
Solution
This is an oxidation-reduction reaction, so start by collecting the species given into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction
17.35
Starting with the oxidation half-reaction, we can balance the chromium
17.36
In acidic solution, we can use or generate hydrogen ions (H+). Adding one water molecule to the left side provides the necessary oxygen; the “left over” hydrogen appears as five H+ on the right side:
17.37
The left side of the equation has a total charge of , and the right side a total charge of [−2 + 5 (+1) = +3]. The difference is three, adding three electrons to the right side produces a mass- and charge-balanced oxidation half-reaction (in acidic solution):
17.38
Checking the half-reaction:
17.39
Now work on the reduction. It is necessary to convert the four O atoms in the MnO4− minus the two O atoms in MnO2 into two water molecules. To do this, add four H+ to convert the oxygen into two water molecules:
17.40
Then add three electrons to the left side to balance the charge:
17.41
Make sure to check the half-reaction:
17.42
Collecting what we have so far:
17.43
In this case, both half reactions involve the same number of electrons; therefore, simply add the two half-reactions together.
17.44
17.45
Checking each side of the equation:
17.46
This is the balanced equation in acidic solution. For a basic solution, add one hydroxide ion to each side and simplify:
17.47
Checking each side of the equation:
This is the balanced equation in basic solution.
Check Your Learning
Balance the following in the type of solution indicated.
(a)
(b)
(c)
(d) Identify the oxidizing agents in reactions (a), (b), and (c).
(e) Identify the reducing agents in reactions (a), (b), and (c).
Answer:
(a) (b) (c) (d) oxidizing agent = species reduced: Cu2+, Cu(OH)2, Ag+ (e) reducing agent = species oxidized: H2, H2, Fe.
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Chebyshev polynomials
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The Chebyshev polynomials, named after Pafnuty Chebyshev, are sequences of polynomials (of orthogonal polynomials) which are related to de Moivre's formula and which are easily defined recursively, like Fibonacci or Lucas numbers. One usually distinguishes between Chebyshev polynomials of the first kind, which are denoted , and Chebyshev polynomials of the second kind, which are denoted . The letter T is used because of the alternative transliterations of the name Chebyshev as Tchebychev (French) or Tschebyschow (German.)
Contents
[hide]
1 Chebyshev polynomials of the first kind
1.1 Triangle of coefficients of Chebyshev polynomials of the first kind
1.1.1 Rows of coefficients of Chebyshev polynomials of the first kind
1.1.1.1 Row sums of coefficients of Chebyshev polynomials of the first kind
1.1.2 Columns of coefficients of Chebyshev polynomials of the first kind
1.1.2.1 Columns of absolute values of coefficients of Chebyshev polynomials of the first kind
1.1.3 Rising diagonals of coefficients of Chebyshev polynomials of the first kind
1.1.3.1 Rising diagonal sums of coefficients of Chebyshev polynomials of the first kind
1.1.4 Falling diagonals of coefficients of Chebyshev polynomials of the first kind
2 Chebyshev polynomials of the second kind
2.1 Triangle of coefficients of Chebyshev polynomials of the second kind
2.1.1 Rows of coefficients of Chebyshev polynomials of the second kind
2.1.1.1 Row sums of coefficients of Chebyshev polynomials of the second kind
2.1.2 Columns of coefficients of Chebyshev polynomials of the second kind
2.1.2.1 Columns of absolute values of coefficients of Chebyshev polynomials of the second kind
2.1.3 Rising diagonals of coefficients of Chebyshev polynomials of the second kind
2.1.3.1 Rising diagonal sums of coefficients of Chebyshev polynomials of the second kind
2.1.4 Falling diagonals of coefficients of Chebyshev polynomials of the second kind
3 See also
4 Notes
Chebyshev polynomials of the first kind
The Chebyshev polynomials of the first kind are defined by the recurrence relation
The first few Chebyshev polynomials of the first kind are
A closed-form formula (would be a Binet formula of the first type, but for the factor ) (Cf. Lucas numbers#Binet's closed-form formula) giving the Chebyshev polynomials of the first kind is
where and are the roots of the quadratic polynomial in terms of
The ordinary generating function for is
The exponential generating function for is
Triangle of coefficients of Chebyshev polynomials of the first kind
Chebyshev polynomials of the first kind
| | | | | | | | | | | | | | | |
| 0 | | | | | | | | | | | | | | |
| 1 | | | | | | | | | | | | | | |
| 2 | | | | | | | | | | | | | | |
| 3 | | | | | | | | | | | | | |
| 4 | | | | | | | | | | | | | | |
| 5 | | | | | | | | | | | | | | |
| 6 | | | | | | | | | | | | | | |
| 7 | | | | | | | | | | | | | | |
| 8 | | | | | | | | | | | | | | |
| 9 | | | | | | | | | | | | | | |
| 10 | | | | | | | | | | | | | | |
| 11 | | | | | | | | | | | | | | |
| 12 | | | | | | | | | | | | | | |
The triangle of coefficients of Chebyshev polynomials of the first kind gives the infinite sequence of finite sequences
: {{1}, {1}, {-1, 2}, {-3, 4}, {1, -8, 8}, {5, -20, 16}, {-1, 18, -48, 32}, {-7, 56, -112, 64}, {1, -32, 160, -256, 128}, {9, -120, 432, -576, 256}, {-1, 50, -400, 1120, -1280, 512}, {-11, 220, -1232, 2816, -2816, 1024}, {1, -72, 840, -3584, 6912, -6144, 2048}, {13, -364, 2912, -9984, 16640, -13312, 4096}, ...}
Triangle of coefficients of Chebyshev polynomials of the first kind . (Cf. A008310)
: {1, 1, -1, 2, -3, 4, 1, -8, 8, 5, -20, 16, -1, 18, -48, 32, -7, 56, -112, 64, 1, -32, 160, -256, 128, 9, -120, 432, -576, 256, -1, 50, -400, 1120, -1280, 512, -11, 220, -1232, 2816, -2816, 1024, 1, -72, 840, -3584, 6912, -6144, 2048, 13, -364, 2912, -9984, 16640, -13312, 4096, ...}
Rows of coefficients of Chebyshev polynomials of the first kind
Row sums of coefficients of Chebyshev polynomials of the first kind
Columns of coefficients of Chebyshev polynomials of the first kind
Columns of absolute values of coefficients of Chebyshev polynomials of the first kind
Compare with the (2,1)-Pascal triangle columns.
Rising diagonals of coefficients of Chebyshev polynomials of the first kind
(...)
Rising diagonal sums of coefficients of Chebyshev polynomials of the first kind
Except for the 0th rising diagonal, which sums to 1, the rising diagonal sums are all 0. Thus the j th rising diagonal sums to .
Falling diagonals of coefficients of Chebyshev polynomials of the first kind
(...)
Chebyshev polynomials of the second kind
The Chebyshev polynomials of the second kind are defined by the recurrence relation
The first few Chebyshev polynomials of the second kind are
A closed-form formula (would be a Binet formula of the second type, except that the exponents are instead of ) (Cf. Fibonacci numbers#Binet's closed-form formula) giving the Chebyshev polynomials of the second kind is
where and are the roots of the quadratic polynomial in terms of
The ordinary generating function for is
The exponential generating function for is
Triangle of coefficients of Chebyshev polynomials of the second kind
Chebyshev polynomials of the second kind
| | | | | | | | | | | | | | | |
| 0 | | | | | | | | | | | | | | |
| 1 | | | | | | | | | | | | | | |
| 2 | | | | | | | | | | | | | | |
| 3 | | | | | | | | | | | | | |
| 4 | | | | | | | | | | | | | | |
| 5 | | | | | | | | | | | | | | |
| 6 | | | | | | | | | | | | | | |
| 7 | | | | | | | | | | | | | | |
| 8 | | | | | | | | | | | | | | |
| 9 | | | | | | | | | | | | | | |
| 10 | | | | | | | | | | | | | | |
| 11 | | | | | | | | | | | | | | |
| 12 | | | | | | | | | | | | | | |
The triangle of coefficients of Chebyshev polynomials gives the infinite sequence of finite sequences
: {{1}, {2}, {-1, 4}, {-4, 8}, {1, -12, 16}, {6, -32, 32}, {-1, 24, -80, 64}, {-8, 80, -192, 128}, {1, -40, 240, -448, 256}, {10, -160, 672, -1024, 512}, {-1, 60, -560, 1792, -2304, 1024}, {-12, 280, -1792, 4608, -5120, 2048}, {1, -84, 1120, -5376, 11520, -11264, 4096}, ...}
Triangle of coefficients of Chebyshev polynomials . (Cf. A008312)
: {1, 2, -1, 4, -4, 8, 1, -12, 16, 6, -32, 32, -1, 24, -80, 64, -8, 80, -192, 128, 1, -40, 240, -448, 256, 10, -160, 672, -1024, 512, -1, 60, -560, 1792, -2304, 1024, -12, 280, -1792, 4608, -5120, 2048, 1, -84, 1120, -5376, 11520, -11264, 4096, ...}
Rows of coefficients of Chebyshev polynomials of the second kind
Row sums of coefficients of Chebyshev polynomials of the second kind
Columns of coefficients of Chebyshev polynomials of the second kind
Columns of absolute values of coefficients of Chebyshev polynomials of the second kind
Compare with the (1,1)-Pascal triangle columns.
Rising diagonals of coefficients of Chebyshev polynomials of the second kind
(...)
Rising diagonal sums of coefficients of Chebyshev polynomials of the second kind
The rising diagonal sums are alternatively 1 and 0.
Falling diagonals of coefficients of Chebyshev polynomials of the second kind
(...)
See also
Special cases of the Chebyshev polynomials
Notes
Jump up ↑ Chebyshev polynomials were first presented in: P. L. Chebyshev (1854) "Théorie des mécanismes connus sous le nom de parallélogrammes," Mémoires des Savants étrangers présentés à l’Académie de Saint-Pétersbourg, vol. 7, pages 539–586.
Jump up ↑ Weisstein, Eric W., Chebyshev Polynomial of the First Kind, from MathWorld—A Wolfram Web Resource.
Jump up ↑ Weisstein, Eric W., Chebyshev Polynomial of the Second Kind, from MathWorld—A Wolfram Web Resource.
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2753 | https://math.stackexchange.com/questions/4395081/function-satisfying-sort-of-differential-inequality-fx-x-le-cfx | real analysis - Function satisfying (sort of) differential inequality: $|f'(x) x| \le Cf(x)$ - Mathematics Stack Exchange
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Function satisfying (sort of) differential inequality: |f′(x)x|≤C f(x)|f′(x)x|≤C f(x)
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During my research I stumbled upon some inequality which I tried to tackle using cutoff functions. Ideally, I would need a C 1 C 1 (at least) function f:[δ,2 δ]→[0,1]f:[δ,2 δ]→[0,1], with δ>0 δ>0, such that f(δ)=1,f(2 δ)=0 f(δ)=1,f(2 δ)=0 and, most importantly,
|f′(x)x|≤C f(x)∀x∈(δ,2 δ),C>0.|f′(x)x|≤C f(x)∀x∈(δ,2 δ),C>0.
It would also be nice if the function was decreasing (since we could use f′(x)≤0 f′(x)≤0 and simplify the inequality) , but this is not required. Nevertheless, I somehow struggle to find an explicit function satisfying those conditions, mostly due to the fact that |f(x)|→0|f(x)|→0 as x→2 δ x→2 δ, implying f′(x)→0 f′(x)→0 as well for x→2 δ x→2 δ. All examples I had in mind seem to fail because (for them):
lim x→2 δ|f′(x)x||f(x)|=∞.lim x→2 δ|f′(x)x||f(x)|=∞.
This led me to believe that maybe such functions cannot exist. Then again, I can't see why that would be true. The question is then to either
Provide an example of such function, or a hint towards such example;
Prove that such function cannot exist (or, once again, hint towards a proof).
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asked Mar 3, 2022 at 11:43
Wham Bang Shang-a-LangWham Bang Shang-a-Lang
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Such a function does not exist, and it suffices to assume that f f is differentiable, the continuity of the derivative is not needed.
If f f satisfies |f′(x)x|≤C f(x)|f′(x)x|≤C f(x) on some interval [a,b]⊂(0,∞)[a,b]⊂(0,∞) then
(x C f(x))′=C x C−1 f(x)+x C f′(x)≥C x C−1 f(x)−x C C f(x)x=0,(x C f(x))′=C x C−1 f(x)+x C f′(x)≥C x C−1 f(x)−x C C f(x)x=0,
so that x↦x C f(x)x↦x C f(x) is non-decreasing.
It follows that if f(a)>0 f(a)>0 then f(x)>0 f(x)>0 on the entire interval [a,b][a,b].
In particular, f(δ)=1 f(δ)=1 and f(2 δ)=0 f(2 δ)=0 is not possible.
How did I c0me up with this? If we first assume that f(x)>0 f(x)>0 on some interval [δ,b][δ,b] then the given inequality can be written as
−C x≤f′(x)f(x)≤C x−C x≤f′(x)f(x)≤C x
and integration from δ δ to x x gives
−C(log x−log δ)≤log(f(x))≤C(log x−log δ)−C(logx−logδ)≤log(f(x))≤C(logx−logδ)
or
(δ x)C≤f(x)≤(x δ)C.(δ x)C≤f(x)≤(x δ)C.
The left inequality shows that f f is bounded below by some strictly positive function. Writing the lower bound as
x C f(x)≥δ C x C f(x)≥δ C
suggest to consider x C f(x)x C f(x) in the first place, and that makes the initial assumption f(x)>0 f(x)>0 obsolete.
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edited Mar 3, 2022 at 12:30
answered Mar 3, 2022 at 12:08
Martin RMartin R
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Thanks, I came up with the same thing for C=1 C=1, shame I didn't think of general case. Could it be rescued if we swap C f(x)C f(x) with C|f(x)|C|f(x)| (i.e. we no longer assume the range of f f is [0,1][0,1] and allow it to have negative values). The first inequality of your answer wouldn't necessarily produce zero at the right hand side.Wham Bang Shang-a-Lang –Wham Bang Shang-a-Lang 2022-03-03 12:47:42 +00:00 Commented Mar 3, 2022 at 12:47
1 @WhamBangShang-a-Lang: If f f can have any sign then the second approach gives (δ x)C≤|f(x)|≤(x δ)C(δ x)C≤|f(x)|≤(x δ)C on any interval on which f f is not zero. This again implies that, on any interval, f f is either identically zero or nowhere zero.Martin R –Martin R 2022-03-03 13:04:54 +00:00 Commented Mar 3, 2022 at 13:04
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Physiology, Pulmonary Vasoconstriction
Maqsood Khan; Stephen J. Bordes; Ian V. Murray; Sandeep Sharma.
Author Information and Affiliations
Authors
Maqsood Khan1; Stephen J. Bordes2; Ian V. Murray3; Sandeep Sharma4.
Affiliations
1 Kentucky College of Osteopathic Medicine
2 Department of Surgery, Louisiana State University Health Sciences Center
3 Texas A&M University
4 Mery Fitzgerald Hospital
Last Update: April 17, 2023.
Introduction
Pulmonary vasoconstriction is a physiological phenomenon and mechanism in response to alveolar hypoxia or low oxygen partial pressures in the pulmonary arterioles and, to some extent, the pulmonary venules. Pulmonary vasoconstriction redirects blood flow within the vasculature away from poorly ventilated parts of the lungs towards better-ventilated portions. Ventilation and perfusion (V/Q) matching is a physiological process that influences gas exchange in the lung, as the lung attempts to efficiently pair oxygenated (ventilated) regions with areas of sufficient blood supply (perfusion). In low-oxygen states, pulmonary vessels constrict in an attempt to shunt blood to better-ventilated regions of the lung. Poor oxygen availability has profound and overarching systemic ramifications manifesting in a plethora of pathologies starting within the lungs itself. Maintaining correct and appropriate oxygen homeostasis is a critical component for systemic stability and functioning, and the process begins within the pulmonary vasculature. While many details pertaining to pulmonary vasoconstriction are not fully understood, the mechanism involves the activity of ion channels as well as several molecular and chemical agents.
Issues of Concern
Pulmonary vasoconstriction is a topic of concern as an inappropriate reflex can result in pulmonary hypertension. Chronic pulmonary hypertension may lead to sequelae of hypoxemia, right heart failure, and venous congestion.
Organ Systems Involved
Hypoxic pulmonary vasoconstriction is a pulmonary vasculature mechanism. However, inappropriate and chronic vasoconstriction can lead to pulmonary hypertension, which has systemic ramifications. Pulmonary hypertension increases the work of the heart. The right ventricle, which propels poorly oxygenated blood into the pulmonary artery, works harder to overcome the increasing pressures in the pulmonary circuit, leading to right ventricular hypertrophy in most cases. The grim sequelae of pulmonary hypertension may lead to the classic 'cor pulmonale' presentation in which right heart failure leads to venous congestion.
Mechanism
Hypoxic pulmonary vasoconstriction relies on the appropriate functioning and response of the pulmonary vasculature in the presence of diminished oxygen availability. Approximately 250 million alveoli are present within each lung. Each one is a functional unit that serves to deliver inhaled oxygen from the atmosphere to the blood and expulsion of carbon dioxide from the blood to the atmosphere. The alveoli interact with the pulmonary capillaries, allowing for gas exchange.
Theories of the Vasoconstriction Reflex
The vasoconstriction reflex is triggered in states of hypoxia. There are contrasting views as to the vascular structures that first detect hypoxia. The classical understanding suggests that decreased oxygen levels are initially detected within the pulmonary artery, whereas a new concept postulates that low oxygen levels are detected in the alveoli. This latter concept further suggests that gap junctions throughout the pulmonary endothelium transmit signals to the pulmonary arterioles, causing them to constrict.
Original Understanding of the Vasoconstriction Reflex
The original mechanism is thought to involve voltage-gated potassium and calcium channels. These channels are located in the smooth muscle cells of the pulmonary arteries and are very sensitive to low oxygen states. In addition to the critical roles of potassium and calcium in hypoxic pulmonary vasoconstriction, there are indications that there could be other ion channels contributing to the mechanism. These ion channels are transient receptor potential vanilloid 4 (TRPV4) and transient receptor potential canonical 6 (TRPC).
Related Testing
Echocardiography
Echocardiography uses ultrasound to visualize the four chambers of the heart and provides insight regarding inappropriate internal pressures or hypertrophy. In addition to the size of the right atrium and ventricle, the echocardiogram evaluates right ventricular ejection fraction through the tricuspid annular plane systolic excursion (TAPSE), another important consideration when diagnosing hypertension in the pulmonary vasculature.
Right Heart Catheterization
If preliminary findings warrant further investigation, a cardiologist will conduct a right heart catheterization to confirm the diagnosis. A right heart catheterization provides information regarding the mean pulmonary artery pressure, right atrial pressure, and pulmonary artery occlusion pressure. It also can provide empirical information that would be used to calculate a transpulmonary gradient and pulmonary vascular resistance.
Different Variables and Calculations
In a healthy individual at rest, the mean pulmonary artery pressure (mPAP) is 14 +/- 3 mmHg while a resting (mPAP) of greater than 25 mmHg would meet the criteria for a diagnosis of pulmonary hypertension. Pulmonary artery occlusion pressure (PAOP) and mean pulmonary artery pressure (mPAP) are variables in the calculations for determining transpulmonary gradient and pulmonary vascular resistance.
The equation to calculate pulmonary vascular resistance: (mPAP-PAOP)/(CO).
The equation to calculate the transpulmonary gradient: mPAP-PAOP.
Pulmonary artery occlusion pressure is one of the most sensitive findings of right heart catheterization.
Additional Testing
Additional testing needs to be completed to rule out other disease processes that could contribute to pulmonary hypertension. Potential etiologies of pulmonary hypertension include connective tissue diseases, obstructive sleep apnea, embolism, and left heart failure, to name a few. These include spiral CT or ventilation-perfusion (VQ) scans to rule out the presence of thromboembolic activity within the lung's vasculature. Pulmonary function testing identifies and categorizes any intrinsic lung disease that may be contributing. High-resolution CT imaging can be helpful when ruling out pulmonary parenchymal disease.
Pathophysiology
Classification of Pulmonary Hypertension
The complete pathophysiology of pulmonary hypertension (PH) is not entirely understood. The current etiologies of pulmonary hypertension are stratified into five groups.
The first group is idiopathic PH, likely related to inheritable and intrinsic pulmonary arterial pathology.
The second group is PH due to left heart disease, causing a build-up of volume and pressure in the pulmonary vasculature.
The third group is PH due to pathology within or related to the lungs. This group includes several different diseases such as COPD, sleep apnea, inflammatory scarring resulting from interstitial lung disease, chronic exposure to a high altitude environment, and certain developmental abnormalities.
The fourth group is associated with frequent occurrence of embolic deposition in the lung's vessels or thrombosis known as chronic thromboembolic pulmonary hypertension (CTEPH).
The fifth group is designated for any remaining unclear or unestablished etiologies in addition to multifactorial causes. Examples of multifactorial processes that could contribute to or cause PH are metabolic disorders pertaining to the thyroid, hematological disorders, and systemic pathologies that involve the respiratory system, such as sarcoidosis.
Idiopathic Pulmonary Hypertension
Of these groups, idiopathic PH is the least understood. There are two prevailing concepts regarding the progression of idiopathic PH. The most prevalent understanding is related to a derangement in the production of local vasoconstrictors and dilators, denoted as the vasoconstriction theory. The vasodilators include prostacyclin and nitric oxide, and the vasoconstrictors include thromboxane and endothelin.
Roles of Chemical and Molecular Agents in Idiopathic Pulmonary Hypertension
In the vasoconstriction theory, nitric oxide (NO) plays a critical role in maintaining appropriate vascular tone. It is produced and regulated by two enzymes, nitric oxide synthase II and III. Patients with idiopathic pulmonary hypertension are found to have diminished levels of NO. Prostacyclin is a product of arachidonic acid metabolism. It is an effective vasodilator and exerts its effects on platelets as well. In patients experiencing pulmonary remodeling, there are decreased levels of prostacyclin and the enzyme prostacyclin synthase. On the contrary, endothelin 1, a systemic vasoconstrictor, is found in abnormally high quantities in idiopathic PH, which contradicts this theory.
Recent Understanding of Idiopathic Pulmonary Hypertension
A more recent understanding is centered on the concept of endothelial and smooth muscle cell activity in the context of arterial remodeling. Bone morphogenetic receptor type 2 (BMR2) is involved with cell growth, proliferation, and osteogenesis. It is believed that a mutation in BMR2 is directly involved in idiopathic PH. Another contributing factor may involve an inappropriately high sensitivity to depolarization in vascular smooth muscle cells. This alteration in the resting potential setpoint may be due to potassium channel irregularities that cause excess availability of calcium.
Sleep Apnea and Pulmonary Hypertension
A distinct relationship exists between obstructive sleep apnea (OSA) and the presence of pulmonary hypertension (PH). OSA is not accepted as a direct singular cause of PH; however, an association does exist. It is believed that OSA induces hypoxia during sleep, which creates a cycle of oxygen desaturation that leads to decreased nitric oxide synthase, endothelin derangements, and an increase in sympathetic tone.
Clinical Significance
Diagnosis
Hypoxic pulmonary vasoconstriction clinically manifests as pulmonary hypertension. To properly diagnose pulmonary hypertension, clinical criteria must be met in the absence of a primary cause, and the degree of pulmonary hypertension must be determined. Early detection is paramount for long-term morbidity and mortality. The initial process typically begins with an ECG and echocardiogram, followed by pulmonary function tests and cardiac catheterization. Right heart pathology such as hypertrophy should be expected in chronic cases.
Nonsurgical Treatment
Therapy is catered to the patient's individual profile. Pharmaceutical regimens are not always indicated for patients; however, anticoagulation is recommended for all diagnoses of idiopathic pulmonary hypertension if there are no contraindications. If pharmacological therapy is indicated, agents such as phosphodiesterase-5 inhibitors, calcium channel blockers, prostacyclin analogs, endothelin receptor antagonists, and cyclic GMP agonists can be utilized. If pulmonary vasoconstriction is refractory to pharmacological therapy or if the severity of the disease is extreme, surgery may be a treatment option. Patients with obstructive sleep apnea should be screened for pulmonary hypertension, and likewise, pulmonary hypertensive patients should be evaluated for sleep apnea due to the association of both diseases. Continuous positive airway pressure (CPAP) may treat OSA and concurrent pulmonary hypertension to a lesser degree.
Surgical Treatment
Surgical options include atrial septostomy, which relieves right heart pressures via a right to left atrial shunt. Lung transplantation may be explored in severe cases.
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References
1.
: Delong C, Sharma S. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): May 1, 2023. Physiology, Peripheral Vascular Resistance. [PubMed: 30855892]
2.
: Di Mise A, Wang YX, Zheng YM. Role of Transcription Factors in Pulmonary Artery Smooth Muscle Cells: An Important Link to Hypoxic Pulmonary Hypertension. Adv Exp Med Biol. 2017;967:13-32. [PubMed: 29047078]
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: Pahal P, Sharma S. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Apr 10, 2023. Idiopathic Pulmonary Artery Hypertension. [PubMed: 29489262]
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: Guazzi M. Pulmonary Hypertension and Heart Failure: A Dangerous Liaison. Heart Fail Clin. 2018 Jul;14(3):297-309. [PubMed: 29966628]
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: Olschewski A, Berghausen EM, Eichstaedt CA, Fleischmann BK, Grünig E, Grünig G, Hansmann G, Harbaum L, Hennigs JK, Jonigk D, Kuebler WM, Kwapiszewska G, Pullamsetti SS, Stacher E, Weissmann N, Wenzel D, Schermuly RT. Pathobiology, pathology and genetics of pulmonary hypertension: Update from the Cologne Consensus Conference 2018. Int J Cardiol. 2018 Dec 01;272S:4-10. [PubMed: 30314839]
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: Turzo M, Vaith J, Lasitschka F, Weigand MA, Busch CJ. Role of ATP-sensitive potassium channels on hypoxic pulmonary vasoconstriction in endotoxemia. Respir Res. 2018 Feb 13;19(1):29. [PMC free article: PMC5810061] [PubMed: 29433570]
7.
: Lyle MA, Davis JP, Brozovich FV. Regulation of Pulmonary Vascular Smooth Muscle Contractility in Pulmonary Arterial Hypertension: Implications for Therapy. Front Physiol. 2017;8:614. [PMC free article: PMC5572347] [PubMed: 28878690]
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: He X, Song S, Ayon RJ, Balisterieri A, Black SM, Makino A, Wier WG, Zang WJ, Yuan JX. Hypoxia selectively upregulates cation channels and increases cytosolic [Ca2+] in pulmonary, but not coronary, arterial smooth muscle cells. Am J Physiol Cell Physiol. 2018 Apr 01;314(4):C504-C517. [PMC free article: PMC5966788] [PubMed: 29351410]
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: Grimmer B, Kuebler WM. The endothelium in hypoxic pulmonary vasoconstriction. J Appl Physiol (1985). 2017 Dec 01;123(6):1635-1646. [PubMed: 28860164]
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: Thenappan T, Ormiston ML, Ryan JJ, Archer SL. Pulmonary arterial hypertension: pathogenesis and clinical management. BMJ. 2018 Mar 14;360:j5492. [PMC free article: PMC6889979] [PubMed: 29540357]
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: Brinkman JE, Reddy V, Sharma S. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Apr 3, 2023. Physiology of Sleep. [PubMed: 29494118]
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: Schmid E, Hilberath JN, Blumenstock G, Shekar PS, Kling S, Shernan SK, Rosenberger P, Nowak-Machen M. Tricuspid annular plane systolic excursion (TAPSE) predicts poor outcome in patients undergoing acute pulmonary embolectomy. Heart Lung Vessel. 2015;7(2):151-158. [PMC free article: PMC4476769] [PubMed: 26157741]
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: Hashmi MF, Modi P, Basit H, Sharma S. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Feb 19, 2023. Dyspnea. [PubMed: 29763140]
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: Sharma S, Hashmi MF. StatPearls [Internet]. StatPearls Publishing; Treasure Island (FL): Dec 22, 2022. Partial Pressure Of Oxygen. [PubMed: 29630271]
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: Mushaben EM, Hershey GK, Pauciulo MW, Nichols WC, Le Cras TD. Chronic allergic inflammation causes vascular remodeling and pulmonary hypertension in BMPR2 hypomorph and wild-type mice. PLoS One. 2012;7(3):e32468. [PMC free article: PMC3302893] [PubMed: 22427841]
: Disclosure: Maqsood Khan declares no relevant financial relationships with ineligible companies.
: Disclosure: Stephen Bordes declares no relevant financial relationships with ineligible companies.
: Disclosure: Ian Murray declares no relevant financial relationships with ineligible companies.
: Disclosure: Sandeep Sharma declares no relevant financial relationships with ineligible companies.
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2755 | https://byjus.com/maths/decimal-to-binary-questions/ | Decimal to binary questions with solutionsis given here to practice decimal to binary conversions. After knowing how to convert a base-10 number into a base-2 number, these questions will help you practice number system conversions.
There are different types of numbers in the number system:
Decimal Numbers – base 10
Binary numbers – base 2
Octal numbers – base 8
Hexadecimal numbers – base 16.
Steps to convert decimal to binary numbers:
| |
| Step I: Divide the given number by 2. Note the remainders obtained. Step II: For an even number, the remainder is 0, and for an odd number, the remainder is 1. Step III: Successively divide the obtained result by 2 and note the remainders till the last dividend is 1. Step IV: The binary equivalent of the decimal number is obtained by writing the remainders in reverse order – MSB (Most Significant Bit) to LSB (Least Significant Bit). |
Decimal to Binary Conversions Questions with Solution
Solve the following questions to practice decimal to binary conversions.
Question 1: Convert 27810 into a binary number.
| | | | |
--- --- |
| Division by 2 | | | Binary Bit |
| 278 ÷ 2 | | | 0 (LSB) |
| 139 ÷ 2 | | | 1 |
| 69 ÷ 2 | | 1 | 1 |
| 34 ÷ 2 | | 0 | 0 |
| 17 ÷ 2 | 8 | 1 | 1 |
| 8 ÷ 2 | 4 | 0 | 0 |
| 4 ÷ 2 | 2 | 0 | 0 |
| 2 ÷ 2 | 1 | 0 | 0 |
| 1 ÷ 2 | 0 | 1 | 1 (MSB) |
∴ (278)10 = (100010110)2
Also check with decimal to binary calulator.
Question 2: Convert 18010 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 180 ÷ 2 | 90 | 0 | 0(LSB) |
| 90 ÷ 2 | 45 | 0 | 0 |
| 45 ÷ 2 | 22 | 1 | 1 |
| 22 ÷ 2 | 11 | 0 | 0 |
| 11 ÷ 2 | 5 | 1 | 1 |
| 5 ÷ 2 | 2 | 1 | 1 |
| 2 ÷ 2 | 1 | 0 | 0 |
| 1 ÷ 2 | 0 | 1 | 1(MSB) |
∴ (180)10 = (10110100)2
Question 3: `Convert 5610 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 56 ÷ 2 | 28 | 0 | 0(LSB) |
| 28 ÷ 2 | 14 | 0 | 0 |
| 14 ÷ 2 | 7 | 0 | 0 |
| 7 ÷ 2 | 3 | 1 | 1 |
| 3 ÷ 2 | 1 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 | 1 |
∴ (56)10 = (111000)2
Question 4: Convert 107310 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 1073 ÷ 2 | 536 | 1 | 1 (LSB) |
| 536 ÷ 2 | 268 | 0 | 0 |
| 268 ÷ 2 | 134 | 0 | 0 |
| 134 ÷ 2 | 67 | 0 | 0 |
| 67 ÷ 2 | 33 | 1 | 1 |
| 33 ÷ 2 | 16 | 1 | 1 |
| 16 ÷ 2 | 8 | 0 | 0 |
| 8 ÷ 2 | 4 | 0 | 0 |
| 4 ÷ 2 | 2 | 0 | 0 |
| 2 ÷ 2 | 1 | 0 | 0 |
| 1 ÷ 2 | 0 | 1 | 1 |
∴ (1073)10 = (10000110001)2
Question 5: Convert 8110 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 81 ÷ 2 | 40 | 1 | 1(LSB) |
| 40 ÷ 2 | 20 | 0 | 0 |
| 20 ÷ 2 | 10 | 0 | 0 |
| 10 ÷ 2 | 5 | 0 | 0 |
| 5 ÷ 2 | 2 | 1 | 1 |
| 2 ÷ 2 | 1 | 0 | 0 |
| 1 ÷ 2 | 0 | 1 | 1(MSB) |
∴ (81)10 = (1010001)2
Also Read:
Binary to Decimal Conversion
Octal to Binary Conversion
Hexadecimal to Binary Conversion
Hexadecimal to Decimal Conversion
Question 6: Convert 40310 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 403 ÷ 2 | 201 | 1 | 1(LSB) |
| 201 ÷ 2 | 100 | 1 | 1 |
| 100 ÷ 2 | 50 | 0 | 0 |
| 50 ÷ 2 | 25 | 0 | 0 |
| 25 ÷ 2 | 12 | 1 | 1 |
| 12 ÷ 2 | 6 | 0 | 0 |
| 6 ÷ 2 | 3 | 0 | 0 |
| 3 ÷ 2 | 1 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 | 1(MSB) |
∴ (403)10 = (110010011)2
Question 7: Convert 50810 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 508 ÷ 2 | 254 | 0 | 0(LSB) |
| 254 ÷ 2 | 127 | 0 | 0 |
| 127 ÷ 2 | 63 | 1 | 1 |
| 63 ÷ 2 | 31 | 1 | 1 |
| 31 ÷ 2 | 15 | 1 | 1 |
| 15 ÷ 2 | 7 | 1 | 1 |
| 7 ÷ 2 | 3 | 1 | 1 |
| 3 ÷ 2 | 1 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 | 1(MSB) |
∴ (508)10 = (111111100)2
Question 8: Convert 11110 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 111 ÷ 2 | 55 | 1 | 1(LSB) |
| 55 ÷ 2 | 27 | 1 | 1 |
| 27 ÷ 2 | 13 | 1 | 1 |
| 13 ÷ 2 | 6 | 1 | 1 |
| 6 ÷ 2 | 3 | 0 | 0 |
| 3 ÷ 2 | 1 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 | 1(MSB) |
∴ (111)10 = (1101111)2
Question 9: Convert 127810 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 1278 ÷ 2 | 639 | 0 | 0(LSB) |
| 639 ÷ 2 | 319 | 1 | 1 |
| 319 ÷ 2 | 159 | 1 | 1 |
| 159 ÷ 2 | 79 | 1 | 1 |
| 79 ÷ 2 | 39 | 1 | 1 |
| 39 ÷ 2 | 19 | 1 | 1 |
| 19 ÷ 2 | 9 | 1 | 1 |
| 9 ÷ 2 | 4 | 1 | 1 |
| 4 ÷ 2 | 2 | 0 | 0 |
| 2 ÷ 2 | 1 | 0 | 0 |
| 1 ÷ 2 | 0 | 1 | 1(MSB) |
∴ (1278)10 = (10011111110)2
Question 10: Convert 14510 into a binary number.
Solution:
| | | | |
--- --- |
| Division by 2 | Quotient | Remainder | Binary Bit |
| 145 ÷ 2 | 72 | 1 | 1 (LSB) |
| 72 ÷ 2 | 36 | 0 | 0 |
| 36 ÷ 2 | 18 | 0 | 0 |
| 18 ÷ 2 | 9 | 0 | 0 |
| 9 ÷ 2 | 4 | 1 | 1 |
| 4 ÷ 2 | 2 | 0 | 0 |
| 2 ÷ 2 | 1 | 0 | 0 |
| 1 ÷ 2 | 0 | 1 | 1 (MSB) |
∴ (145)10 = (10010001)2
| Related Articles: | |
--- |
| Permutation and Combination Questions | Square Root Questions |
| Logarithm Questions | Profit and Loss Questions |
Practice Questions on Decimal to Binary Conversions
Convert 15510 into a binary number.
Convert 37510 into a binary number.
Convert 100010 into a binary number.
Convert 1001110 into a binary number.
Convert 7410 into a binary number.
Keep visiting BYJU’S to get more such Maths lessons in a simple, concise and easy to understand way. Also, register at BYJU’S – The Learning App to get complete assistance for Maths preparation with video lessons, notes, tips and other study materials.
Leave a Comment Cancel reply |
2756 | https://www.youtube.com/watch?v=5psyoEDqVhA | Calculator Tutorial 13: Trigonometry on a scientific calculator
Science Made Simple
7200 subscribers
2859 likes
Description
574973 views
Posted: 20 Oct 2016
This video shows you how to do sin, cos and tan calculations on a scientific calculator.
192 comments
Transcript:
let's look now at how to do trigonometric functions for the trigonometry lesson so on the calculator we have a number of trig functions and if I zoom in you'll be able to see better so here we've got our s our cos or cosine and our tan or [Music] tangent when doing Trigon trigonometry we're wanting to use those buttons again as with everything in the same order as we would see the question written down on the page so if I wanted to find the answer to sign 30 for example I would go to the sign button here press that and we can now see on the screen it's come up with sign and an Open Bracket type in my 30 close the bracket press equals and sin 30 is [Music] 0.5 clear that cos let's do cos 30 again so exactly the same I press C it's come up on the screen with an open bracket put in 30 close the bracket equals and there we've got cos 30 and the same with tan if you get numbers that you're not expecting with trigonometric functions chances are your calculator will be set to radians radians is just another way of measuring angles but it's not one we use in this course so before doing any trigonometry on your calculator check that you've got a little D there [Music] if you go to setup I'll do that again more slowly so setup here in yellow so it's in yellow so I need my shift and then click setup if you go to setup you can see there's a variety of different um formats to have your calculator in number three here de that means degrees number four is radians if your calculator is in radians if I just show you you'll be able to see at the top a little r but we want it in degrees so if I go back to setup go shift set up and then I want number three for degrees you can now see we're back in degrees |
2757 | https://www.tiger-algebra.com/en/solution/quadratic-equations-by-factoring/3x%5E2-3%3D0/ | Copyright Ⓒ 2013-2025
tiger-algebra.com
This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here.
Solution - Solving quadratic equations by factoring
Other Ways to Solve
Step-by-step explanation
1. Factor out the greatest common factor to get perfect squares
3x2−3=0
Factor out of the terms on the left side:
3·x2-3·1=0
3·(x2-1)=0
2. Find the factors
3·(x2-1)=0
Because both x2 and -1 are perfect squares, rewrite the equation using the difference of perfect squares formula:
a2-b2=(a-b)(a+b) :
3(x+1)(x-1)=0
The factors of 3x2−3=0 are 3 , (x+1) and (x−1).
3. Find the roots of the quadratic equation
Find the roots of:
3x2−3=0
using its factored form:
3(x+1)(x−1)=0
If
3(x+1)(x−1)=0
Then
(x+1)=0
and/or
(x−1)=0
Solve each factor for x:
Factor 1:
(x+1)=0
Subtract from both sides:
(x+1)-1=0-1
Simplify the arithmetic:
x=0−1
Simplify the arithmetic:
x=−1
Factor 2:
(x-1)=0
Add to both sides:
(x-1)+1=0+1
Simplify the arithmetic:
x=0+1
Simplify the arithmetic:
x=1
4. Graph
How did we do?
Why learn this
In their most basic function, quadratic equations define shapes like circles, ellipses and parabolas. These shapes can, in turn, be used to predict the curve of an object in motion, such as a ball kicked by a football player or a shot fired out of a cannon.
When it comes to an object’s movement through space, what better place to start than space itself, with the revolution of planets around the sun in our solar system? The quadratic equation was used to establish that planets’ orbits are elliptical, not circular. Determining the path and speed an object travels through space is possible even after it has come to a stop: the quadratic equation can calculate how fast a vehicle was moving when it crashed. With information like this, the automotive industry can design brakes to prevent collisions in the future. Many industries use the quadratic equation to predict and thus improve their products’ lifespan and safety.
Terms and topics
Related links
Latest Related Drills Solved
Copyright Ⓒ 2013-2025
tiger-algebra.com |
2758 | https://chemistry.stackexchange.com/questions/15789/distillation-simple-vs-fractional | organic chemistry - Distillation: Simple vs Fractional - Chemistry Stack Exchange
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Distillation: Simple vs Fractional
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Here is the excerpt of my lab manual:
"Simple distillation is more useful for purifying a liquid that contains either a non-volatile impurity or small amounts of higher or lower boiling impurities. Fractional distillation allows for several condensation-condensation cycles in a single operation. It can be used to separate liquids with comparable volatilities and to purify liquids that contain relatively large amounts of volatile impurities."
So, why does simple distillation put emphasis on small amounts and fractional distillation, on large volumes?
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asked Sep 1, 2014 at 16:00
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So, why does simple distillation put emphasis on small amounts and fractional distillation, on large volumes?
It's not small and large volumes of material being distilled, but rather small and large volumes of impurities present in the material to be distilled. Both simple and fractional distillations can be used to distill large quantities of material.
The efficiency of a distillation column is measured in "height-equivalent theoretical plates" (HETP), the more plates the more efficient the column, the more efficient the column the more likely that it will be able to separate two closely boiling materials. If you have a small amount of a lower boiling impurity in your solution and you want to separate it from a large amount of the desired material, then you can probably get away with using a simple distillation column and discarding some of the forerun (the first drops of liquid that distill over). On the other hand, if there is a large amount of impurity and it boils close to the desired product, then you will need a column with a higher efficiency (fractional distillation column) to effect a good separation between the two fractions (or "cuts" as they're sometimes called) and get pure product.
To sum up, perhaps the original question would be better written as
So, why does simple distillation put emphasis on small amounts and fractional distillation, on large amounts of impurities?
and the answer would be that even an inefficient (fewer HETP) simple distillation column has enough power (enough theoretical plates) to separate a small amount of impurity from a large amount of product, especially if you are willing to discard the initial fore-run from the distillation. If there is a larger amount of impurity, then we may need a more powerful (higher efficiency column) to produce the desired separation.
The higher the efficiency of a distillation column, the greater its ability to separate two things. If we have to separate two materials that are both present in large quantity, or distill at boiling points close together, then we need a more efficient column than if they boiled further apart or one was present in only a small amount and we were willing to throw away the forerun (e.g. throw away some of the desired product in that early fraction that contained the impurity). When we call a column a fractionation column, we're really saying that it has greater separating efficiency than a simple distillation column. The more efficient the column, the better the job it will do on difficult separations.
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edited Sep 1, 2014 at 19:37
answered Sep 1, 2014 at 16:19
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MIT Calculus Revisited: Calculus of Complex Variables
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2760 | https://www.scienzaeconoscenza.it/blog/scienza_e_fisica_quantistica/che-cosa-sono-i-fotoni | Che cosa sono i fotoni?
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Che cosa sono i fotoni?
Scienza e Fisica Quantistica
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"Condividi questo articolo")79 condivisioni
Cosa sono i fotoni? Quali le caratteristiche particolari di queste interessanti particelle? Come si studiano?
Antonella Ravizza - 15/06/2019
La luce da sempre ha attratto la curiosità dell’uomo: di che cosa è fatta? Perché è così brillante? Il segreto sta nel fotone, una piccolissima particella di luce! Cerchiamo di scoprire insieme le principali caratteristiche di questa interessante particella elementare.
Quando aveva 16 anni, Albert Einstein sognò davanti ad uno specchio di cavalcare un raggio di luce. Il giovane sognatore intuì che non sarebbe riuscito a vedersi riflesso nello specchio, perché, stando sopra alla luce, si sarebbe mosso esattamente alla sua velocità; per potersi specchiare avrebbe dovuto superare la velocità della luce stessa. Qualche tempo dopo lo stesso Einstein, studente al Politecnico di Zurigo, si rese conto che la velocità della luce è una costante.
Cosa sono i fotoni?
Il termine fotone deriva dal greco e fu introdotto per la prima volta da Gilbert Lewis nel 1926. Il fotone si indica con la lettera greca γ ed è associato ad ogni radiazione elettromagnetica. Pur essendo un fenomeno ondulatorio, la radiazione elettromagnetica ha anche una natura quantizzata che le consente di essere descritta come un flusso di fotoni.
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Il fotone è una particella che ha vita infinita: può essere creato e distrutto dall’interazione con altre particelle, ma non può decadere spontaneamente. Pur non avendo massa, è influenzato dalla gravità e possiede energia; nel vuoto si muove alla velocità della luce (c=300000 km/s circa), mentre nella materia si comporta in modo diverso e la sua velocità può scendere al di sotto di c. In effetti, quando interagisce con altre particelle acquisisce massa e non si muove più alla velocità della luce. Bohr ipotizzò che un atomo può emettere un’ onda elettromagnetica (o radiazione) solo quando un elettrone si trasferisce da un’orbita con energia maggiore (Ei) a un’ orbita con energia minore (Ef).
L’energia dell’onda elettromagnetica emessa è: E= Ei-Ef . Dal momento che sia Ei sia Ef possono assumere solo valori ben definiti, l’energia della radiazione elettromagnetica emessa dall’atomo non può avere qualsiasi valore, ma solo quantità discrete, dette quanti di energia: i fotoni. Quindi la materia è in grado di emettere o assorbire energia raggiante solo sotto forma di pacchetti energetici. Einstein calcolò l’energia associata ad ogni fotone e vide che era proporzionale alla frequenza dell’onda elettromagnetica.
Onda o particella? La doppia natura del fotone
Prima delle scoperte della prima metà del XX secolo, onde e particelle sembravano concetti opposti: un'onda riempie una regione di spazio, mentre un elettrone o uno ione hanno una locazione ben definita. Su scala atomica, in effetti, la distinzione diventa confusa: le onde hanno alcune proprietà delle particelle e viceversa.
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Scienza e Conoscenza - N. 59 - Ebook SCOPRI
Effettivamente il fotone mostra una duplice natura, sia corpuscolare, sia ondulatoria: a seconda della strumentazione usata per rilevarlo, si comporta come una particella, o si comporta come un’onda. L’esperimento dell’effetto fotoelettrico (quel fenomeno per cui si ha emissione di elettroni da parte di un corpo colpito da onde elettromagnetiche) suggerisce la natura corpuscolare della luce, mentre i fenomeni di diffrazione e di interferenza suggeriscono una natura ondulatoria.
Per valutare come la luce passi attraverso un telescopio, si calcola il suo moto come se la luce fosse un’onda. Però, quando la stessa onda cede la sua energia a un singolo atomo, risulta che essa si comporta come una particella. Indipendentemente dal fatto che un raggio di luce sia più brillante o debole, la sua energia viene trasmessa in quantità delle dimensioni di un atomo (il fotone) la cui energia dipende soltanto dalla lunghezza d'onda. Le osservazioni hanno mostrato che tale “dualità” onda-particella esiste anche in direzione opposta.
Un elettrone dovrebbe avere, in ogni istante, posizione e velocità ben definite; ma la fisica quantistica ci dice che una precisione in osservazioni di questo tipo non può essere ottenuta, e ci suggerisce che il moto può essere descritto come un'onda. Il dualismo onda-particella era considerato paradosso fino all’introduzione completa della meccanica quantistica, che descrisse in maniera unificata i due aspetti. La radiazione si comporta come un’onda quando si propaga nello spazio, mentre si comporta come particella quando interagisce con la materia.
Si introducono quindi nuove quantità e notazioni: un'onda elettromagnetica di lunghezza d'onda λ percorre una distanza di c metri ogni secondo. La sua frequenza ν, cioè il numero di oscillazioni in su e giù ogni secondo, si può ottenere dividendo c per la lunghezza d'onda: ν = c/ λ . Una legge fondamentale della fisica quantistica dice che l'energia E in joule di un fotone di frequenza ν è: E = hν, dove h = 6,624 10-34joule-sec è la "costante di Planck".
Luce solida e computer quantistici
Oggi si sa molto di più: i ricercatori dell’Università di Princeton sono riusciti a rallentare i fotoni e a creare una stranissima e nuova forma di luce: la luce solida! Hanno cioè creato un cristallo fatto non di atomi ma di fotoni, cioè di particelle che costituiscono la luce (fotoni congelati).
Hanno ottenuto un agglomerato di 100 miliardi di atomi di materiale superconduttore come fosse un atomo artificiale; nelle sue vicinanze hanno fatto passare un filo superconduttore contenente fotoni. La luce ha potuto, così, “solidificarsi”, cambiando natura con un processo che è stato paragonato a una transizione di fase, cioè simile a quando un gas si condensa per diventare liquido o solido.
Lo scopo finale dei ricercatori è la realizzazione di un computer quantistico capace di effettuare calcoli molto più complessi di quelli che risolvono i computer tradizionali. Chissà, magari tra qualche anno, un altro giovane sognatore potrà effettivamente cavalcare un cristallo di luce solida e rendere realtà il sogno del piccolo “grande” Einstein!
Antonella Ravizza
Antonella Ravizza si è laureata in fisica nucleare all’Università degli Studi di Pavia, con la quale ha mantenuto rapporti di collaborazione.... Leggi la biografia
Antonella Ravizza si è laureata in fisica nucleare all’Università degli Studi di Pavia, con la quale ha mantenuto rapporti di collaborazione. È docente di Fisica presso l’Istituto “A. Cesaris” di Casalpusterlengo (LO).Nominata tutor del Presidio Scientifico della provincia di Lodi per l’insegnamento delle scienze sperimentali, si... Leggi la biografia
"Condividi questo articolo")79 condivisioni
Vuoi lasciare un commento?
VnT
postato da Danny il 21/12/2020
Qui dentro c'è la risposta al viaggio nel tempo
La massa del fotone
postato da Ernesto Bertagna il 15/04/2020
Come ha esposto nell'articolo sopra, il fotone mostra una duplice natura, sia corpuscolare, sia ondulatoria. Einstein definisce fotone un pacchetto di energia o quanto nell'esperimento dell'Effetto Fotoelettrico. Ma dalla sua nota equazione abbiamo che materia è equivalente ad energia e viceversa. Ma allora mi chiedo perchè il fotone è considerato di massa nulla? Grazie.
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2761 | https://ijmttjournal.org/public/assets/volume-64/number-2/IJMTT-V64P512.pdf | International Journal of Mathematics Trends and Technology (IJMTT) – Volume 64 Number 2 – December 2018 ISSN: 2231 – 5373 Page 74 Early Proofs of Fermat‟s Little Theorem and Applications R.P.Sah#1, U.N.Roy#2, A.K.Sah#3, S.K.Sourabh#4 #1Dept of Mathematics, R Lal College, Lakhisarai, Munger, Bihar, India #2Dept of Mathematics, BRM College, Munger, Bihar, India #3Head, Department of Mathematics, Marwari College, Bhagalpur, Bihar, India #4 UDCA, T.M. Bhagalpur University, Bhagalpur, Bihar, India Abstract In this paper we study some of the early proofs of Fermat’s Little Theorem. Many of the original sources to the proofs of these theorems are obscure. The sequence of the proofs appears chronologically, in order to display how the proofs involved throughout the 17th -21th centuries. Keywords –Fermat, Little,terminology, emphasize, sketchy. I. INTRODUCTION Fermat‟s Little Theorem is one of the jewels of Number Theory and to mark the 400th anniversary of Fermat‟s birth. This little theorem, which was relatively easy to prove, but it has a vast range of mathematical consequences and one major practical application.Bernhard Frenicle de Bessy dated October 18, 1640; Pierre de Fermat first revealed his result. He did not, however, provide a proof, stating to Frenicle. The first proof of Fermat‟s Little Theorem was given nearly 100 years later by Euler. Burton also tells us that Leibnitz did not receive this recognition, „for he left an identical argument in an unpublished manuscript sometime before 1683.” From Beginning Number Theory by Neville Robbins [13, p.102], it is a proof found in many number theory textbooks, and we see later that it is essentially equivalent to Euler‟s first two proofs.In Dickson‟s History of the Theory of Numbers [7, Chapter 3], the original works to prove Fermat‟s Little Theorem are given. Among these proofs, worked by Leibniz, Euler, Lambert, Ivory, and Thue. Enter Leibniz in 1894 attention was called to a collection of unpublished manuscripts located in the Hanover Library attributed to Gottfried Wilhelm von Leibniz (1646-1716), most famous as one of the creators of calculus as well as for his philosophical theory of monads. We usually do not think of Leibniz as a Pioneer of number theory. However, among his works found in the Hanover Library are results believed to have been attained prior to 1683 which include proofs of Both Fermat‟s Little Theorem and Wilson‟s Theorem. Euler‟s first published proof of Fermat‟s little theorem before 60 years. A. The use of ‘little’ in the theorem When theorem is start, to be called Fermat‟s little theorem. The question is arise, who first use the word „Little‟ in the Fermat‟s little theorem. Actually not everyone calls it so. In Vol 1 of Dickson‟s monumental three volume [History of Mathematics] Theory of Numbers, there is an entire chapter devoted to „Fermat‟s and Wilson‟s Theorems. Hardy and Wright, Davenport, Nagell, simply use „Fermat‟s Theorem‟. And Sierpinski calls it „Simple theorem of Fermat‟ in 1964, a selection of problems in the theory of numbers. Of course everyone knows that Wilson‟s theorem is only one such theorem but „Fermat‟s theorem. There are several claimants to find the beautiful result –to name but one- that every prime p, with p = 1 (mod 4), is represent table by p = a2 + b2 for some integers a andb, could well claim to be Fermat‟s theorem. B. Fermat’s ‘little’ Theorem with Examples If p be a prime and a be any integer then the number ap– a is an integer multiple of p. It is expressed in modular arithmetic as ap ≡a(mod p) For example: If a= 2, p = 7, 27 = 128 and 128 – 2 = 7× 18 is an integer multiple of 7. In non-congruence language: Let p be any prime, and a be any integer not divisible by p, then a(p−1) leaves remainder 1 on division by p. Symbolically, ap−1 ≡1(mod p) Illustration: Let p =7 and a=2, then 𝑎(𝑝−1) = 26 = 64 = 7 × 9 + 1. 1) Theorem:Let k be the order of a in the group Zp x Then by Lagrange, k divides the order of the group (p – 1) . International Journal of Mathematics Trends and Technology (IJMTT) – Volume 64 Number 2 – December 2018 ISSN: 2231 – 5373 Page 75 So, p – 1= km, for some m. A more general result is Xα(N) ≡1(mod N), for X and N coprime, where α(N)Euler‟s function is, the number of integers strictly smaller than N coprime to N. Elegant group-theoretic proof too. ap−1 ≡akm ≡ ak m ≡1m ≡1 (mod p) The following proof is given Beginning Number Theory by Neville Robbins [13, p.102]. It is a proof found in many number theory textbooks, and we see later that it is essentially equivalent to Euler‟s first two proofs. Proof: Let 𝑆= {𝑎|𝑎𝑝≡𝑎 𝑚𝑜𝑑 p } for p prime and a ∈ N. Then 0∈ S because Op = O for all p. So Op ≡ O (mod p ). Now assume k ∈S and 𝑘𝑝≡𝑘 (modp). Then we want to show that for k + 1 ∈S , (𝑘+ 1 )𝑝≡(k+1)(modp). By Binomial Theorem, 𝑘+ 1 𝑝= 𝑘 𝑝+ 1𝑝+ 𝑝 𝑗 𝑘 𝑝−𝑗 𝑝−1 𝑗 =1 ≡ (k+1) (mod p) If gcd (a,p) = 1, then by cancellation 𝑎𝑝≡𝑎 (𝑚𝑜𝑑 𝑝), implies 𝑎𝑝−1 ≡1 𝑚𝑜𝑑 𝑝 . If a is negative, then 𝑎 ≡𝑟 (𝑚𝑜𝑑 𝑝) for some r, where 0 ≤ r ≤ p – 1. Thus 𝑎𝑝≡𝑟𝑝≡𝑟≡𝑎 𝑚𝑜𝑑 𝑝 . 2) Theorem:Let p be a prime and let 𝑥= 𝑎1 + 𝑎2 + ⋯+ 𝑎𝑚.and a number𝑥𝑝− 𝑎i 𝑝 𝑚 i−1 , show that 𝑝| 𝑥𝑝− 𝑎i 𝑝 𝑚 𝑖 1 . Proof:We know by the Multinomial Theorem, 𝑥𝑝= (𝑎1 + 𝑎2 + ⋯+ 𝑎𝑚)𝑝= 𝑝 𝑘1, 𝑘2, … , 𝑘𝑚 𝑎1 𝑘1.
𝑘1+⋯+𝑘𝑚=𝑝 𝑎2 𝑘2 … 𝑎𝑚 𝑘𝑚. where 𝑝 𝑘1,𝑘2,…,𝑘𝑚 = 𝑝!
𝑘1!…𝑘𝑚! , when 𝑘𝑖 ≠𝑝 for any i, then 𝑘i < 𝑝 for all i. Then there is no factor of p in the denominator of any coefficient, but there is a factor of p in the numerator. Thus for 𝑘𝑖≠p for all 𝑖, 𝑝!
𝑘1!…𝑘𝑚! ≡ 0 (𝑚𝑜𝑑 𝑝).Thus x𝑝− 𝑎𝑖 𝑝≡ (𝑎1 𝑝 𝑚 𝑖=1 + 𝑎2 𝑝… + 𝑎𝑚 𝑝) − 𝑎1 𝑝+ 𝑎2 𝑝… + 𝑎𝑚 𝑝 ≡0(𝑚𝑜𝑑 𝑝) Thus 𝑝|x𝑝− 𝑎𝑖 𝑝 𝑚 𝑖=1 Taking, 𝑎1 = 𝑎2 = ⋯= 𝑎𝑚= 1. Then since 𝑥= 𝑎1 + 𝑎2 + ⋯+ 𝑎𝑚, it follows that 𝑝|(𝑥𝑝−𝑥), for any integer x (depending on the value of m). Thus 𝑥𝑝−𝑥≡0 𝑚𝑜𝑑 𝑝 , So𝑥𝑝≡𝑥 𝑚𝑜𝑑 𝑝 . If gcd (x,p)= 1, then by cancellation 𝑥 𝑝−1 ≡1 𝑚𝑜𝑑 𝑝 . C. Euler’s Proof with help of Fermat’s Little theorem Fermat‟s little theorem is an important property of integers to a prime modulus. For prime p an any a∈ Z such that 𝑎≢0( 𝑚𝑜𝑑 𝑝), 𝑎𝑝−1 ≡1( 𝑚𝑜𝑑 𝑝). If we want to extend Fermat‟s little theorem to a composite modulus, a false generalization would be: if a ≢ 0 (mod m) then 𝑎𝑚−1≡ 1 (mod m). For a counter example, take m = 15 and a=2: 214≡ 4 ≢1(mod 15). A correct extension of Fermat„s little theorem to non-prime module requires a new way of thinking about the hypothesis in Fermat‟s little theorem. For prime p,a≢ 0 (mod p), gcd( a, p ) = 1. But these two conditions are not equivalent when p is replaced with a composite number. It is the relative primarily point of view on the right that lets Fermat‟s little theorem be extended to a general modulus, as Euler discovered. 1) Theorem:For m≥2 in 𝑍+and any a∈Zsuch that (a ,m) =1, 𝑎φ(m) ≡1 (mod m) Where φ(m) is the number of invertible integers modulo m. When m = p is prime, all non-zero integers modulo p are invertible, International Journal of Mathematics Trends and Technology (IJMTT) – Volume 64 Number 2 – December 2018 ISSN: 2231 – 5373 Page 76 So φ(p ) = p - 1 and Euler‟s theorem becomes Fermat‟s little theorem. Now we compute φ(m). Consider m = 12. To count the number of invertible integers modulo 12, write down a set of representatives for integers modulo 12, such as 1,2,3,4,5,6,7,8,9,10,11,12. The numbers here which are invertible modulo 12 are 1,5,7,11, So φ(12) = 4. From Euler‟s theorem for m = 12 says 𝑎4 ≡1 (mod 12) when (a, 12) = 1. Being invertible modulo m is the same as being relatively prime to m, i.e. we can solve ax ≡ 1 (mod m) for x, exactly when (a, m) = 1, so we can describe φ(m) concretely as follows: φ(m ) = { a:1 ≤ a≤ m,(a,m) = 1 }. Here is a small table of values derived from this formula. where φ(m) is even for m> 2. We can explain this by observing that when a mod m is invertible, so is (-a) mod m. Therefore, using the standard representatives modulo m, invertible numbers modulo m come in pairs as {a, m – a}. Suppose a = m ─ a, then a = m/2, so m is even. But (m/2, m) = m/2, and this is bigger than 1 when m >2, so m/2 is not invertible modulo m. Thus, when m >2, the invertible numbers modulo m fall into pairs, and that shows φ(m) is even. 2) Theorem:If a primep and a ≢0 mod p. To show 𝑎𝑝−1 ≡1 mod p, consider non-zero integers modulo p in the standard range: S= {1,2,3,….,p-1} We will compare S with the set obtained by multiplying the elements of S by a: aS= { a1 , a2 ,a.3,…., a(p-1) }. The elements of S represent the nonzero numbers modulo p. The key point is that the elements of aSalso represent the nonzero numbers modulo p. For any b≢ 0 mod p, we can solve the equation ax ≡ b mod p since a mod p is inventible and a solution x is nonzero modulo p too (since b mod p is nonzero). Choosing z to lie between 1 and p-1,ax∈aS, so b mod p is represented by an element of aS. There arep -1 elements in aS, so aS is a set of representatives for the nonzero numbers modulo p. Since S and aS, when reduced modulo p, becomes the same thing, the product of the numbers in each set must be the same modulo p: 1.2.3……..(p - 1)≡a(a.2)(a.3)…(a(p - 1)) mod p. Pulling the (p – 1) copies of a to the front of the product on the right, we get 1.2.3 … … (𝑝 − 1) ≡𝑎𝑝−1(1.2.3 … … 𝑝−1 𝑚𝑜𝑑 𝑝 Now we cancel each of 1,2,3,…p –1 on both sides (since they are all invertible modulo p) and we are left with1 ≡𝑎𝑝−1 (mod p). 3) Theorem:Replace the condition “non-zero modulo p” with “relatively prime to m.” For positive integer m and a that (a,m) =1. We consider the units modulo m in the standard range: S = u1. u2.u3 … … uφ m . Where 1 ≤𝑢𝑖≤𝑚−1, 𝑢𝑖, 𝑚 = 1, and the uj‟s are distinct. (If m=p is prime we can use ui = 𝑖 for all i, but in general there isn‟t a simple formula for the ith unit modulo m.) We will compare S with the set obtained by multiplying the elements of S by a: 𝑎𝑆= 𝑎𝑢1. 𝑎𝑢2. 𝑎𝑢3 … . . 𝑎𝑢φ m . Since (a,m)=1a mod m is a unit and therefore a S consists of units modulo m. We will show that aS represents all the units modulo m. Given any unit b mod m, the congruence ax ≡ b(modm) is solvable since a mod m is invertible. The solution x is a unit modulo m and placing x between 1 and m – 1 makes ax a member of aS. Thus b mod m is represented by an element of aS. Since aS has size φ(m), it is a set of representatives for the units modulo m. Since the members of S and aS agree modulo m, the product of the numbers in each set must be the same modulo m: 𝑢1𝑢2𝑢3𝑢φ(m) ≡ 𝑎𝑢1 𝑎𝑢2 𝑎𝑢3 𝑎𝑢φ m 𝑚𝑜𝑑 𝑚. Pulling the φ(m) copies of a to the product on the right, we get 𝑢1𝑢2𝑢3 … . 𝑢φ(m) ≡𝑎φ m 𝑢1𝑢2𝑢3 … . 𝑢φ(m)𝑚𝑜𝑑 𝑚. m 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 φ( m) 1 2 2 4 2 6 4 6 4 10 4 12 6 8 8 16 International Journal of Mathematics Trends and Technology (IJMTT) – Volume 64 Number 2 – December 2018 ISSN: 2231 – 5373 Page 77 Now we cancel each Ui on both sides (since they are all invertible modulo m) and we are left with 1 ≡𝑎φ(m) mod m. Passing from Fermat‟s little theorem to Euler‟s theorem amounted to replacing non-zero numbers modulo a prime p with invertible numbers (not the non-zero numbers) for a general modulus m. There is a common notation for these numbers in elementary number theory courses. 𝑈𝑚= {𝑎 𝑚𝑜𝑑 𝑚∶(𝑎, 𝑚) = 1} The notation Um comes the fact that invertible numbers mod m are called units mod m. Example: We have U5 ={1, 2, 3, 4} and U18 = {1, 5, 7, 11, 13, 17 }. When p isUp = {1,2,3, … , 𝑝−1}. The functionφ(m) does not vary in a simple manner from one integer to the next. This is typical of functions that arise in number theory which are based on divisibility. The right way to think about φ(m) is by thinking about positive integers not in terms of the usual m → m+1 paradigm, but in terms of the progression Primes →prime powers→ general case. This progression is how the integers are best arranged from the viewpoint of divisibility Primes are the building we can get formulas for φ(m)directly from its definition in the first two cases of the above progression: φ(p)=p – 1 since there are p – 1 integers from 1 to p which are relatively prime to p. φ(𝑝k)=𝑝k −𝑝k−1 = 𝑝k−1(𝑝−1) for prime p and k≥1 since among the integers from 1 to 𝑝𝑘, those which are not relatively prime to 𝑝k are the multiples of p: p, p2, p3,…𝑝𝑘 There are 𝑝𝑘−1 such numbers, so we subtract this from𝑝𝑘 to compute 𝜑(𝑝𝑘) = 𝑝𝑘−𝑝𝑘−1 . What about φ(m) when m is composite? We will treat one case here, which is important in elementary cryptography: m= pq is a product of two different primes. If1 ≤ a ≤ pqand (a, pq) =1, then a is neither a multiple of p nor a multiple of q. The multiples of p in this range are p, 2p,…,qp and the multiples of q in this range are q, 2q,….pq. There are q numbers in the first case and p numbers in the second case. The two lists only overlap at pq(indeed, a positive integer divisible by p and q is divisible by pq, so can‟t be less than pq). Therefore, to compute φ(pq), we take away from pq the number of terms in both lists without double-counting the common term: φ(pq) =pq – p – q + 1 = p(q - 1) – 1(q - 1) = (p - 1)(q –1) This is interesting: φ(pq) = φ(p)φ(q) for different primes p and q. This formula is false when 𝑝= 𝑞, φ(𝑝2) = 𝑝2 −𝑝= 𝑝(𝑝−1) rather than (𝑝−1)2 With these formulas, we can make Euler‟s theorem more explicit for certain modulo. Example: When p is prime, (𝑎, 𝑝2) = 1 ⟹ 𝑎𝑝 𝑝−1 ≡1 mod 𝑝2 Example: When p and q are different primes, (𝑎, 𝑝𝑞) = 1 ⟹ 𝑎𝑝 𝑝−1 (𝑞−1) ≡1 mod 𝑝𝑞 Example will be crucial for the RSA cryptosystem. 4) Theorem:Any reduced form fraction a/b with (10, b) = 1 can be written as a fraction with denominator 10d- 1for some≥1. Moreover, the period length of the decimal expansion for a/b is the smallest 𝑑≥1 such that 10d ≡1(mod b). In particular, 𝑑≤𝜑 𝑏 and the period length is independent of the numerator a. Proof: Let the fraction be a/b, where (10, b) = 1. By Euler‟s theorem, 10φ (b) ≡ 1 (mod b). That means 10 φ (b)–1 is a multiple of b, so we can rewrite a/b asa fraction with denominator 10φ (b) – 1. Let 𝑑≥1 be minimal such that 10d ≡1(mod b), so 𝑑≤𝜑 𝑏 . Write 10d-1 = bn, so 𝑎 𝑏 = 𝑎𝑛 𝑏𝑛 = 𝑎𝑛 10𝑑−1. Since a/b <1, an 1, subtracting its integer part leaves a fraction between 0 and 1 which still has b as its reduced form denominator, so the period of the decimal expansion of a/b is still completely determined by b. There are further interesting questions worth asking about decimal expansions: Which numbers have finite decimal expansions (such as 5/16=.3125). Which numbers have periodic decimal expansions with an initial non repeating block (such as 7/15=.466666…). If we compare all the reduced proper fractions with the same denominator b, they may all have the same expansion as 1/b except for a shift, e.g. 1/7 =. 142857, 2/7 =. 285714, 3/7 =. 428571, 4/7 =. 571428, 5/7 =. 714285, 6/7 =. 857142, We can explain when all the reduced fractions with denominator b have this feature.For some denominators, more than one digit sequence occurs. e.g., there are two possibilities when the denominator is 13: 1 13 =. 076923, 2/13 =. 153846, 3/13 =. 230769, 4/13 =. 307692, 5/13 =. 384615, 6/13 =. 461538, 7/13 =. 538461, 8/13 =. 615384, 9/13 =. 692307, 10/13 =. 769230, 11/13 =. 846153, 12/13 =. 923076, Every decimal expansion here is a shift of the expansion for 1/13 or 2/13. If we collect numerators of fractions above whose decimal expansions have the same digit sequence, the 12 numerators fall into two sets of size 6: {1,3,4,9,10,12} and {2,5,6,7,8,11}. II. CONCLUSION In this paper we have discussed some of the early proofs of Fermat‟s Little Theorem. We have also demonstrated the Fermat‟s Little Theorem through examples. REFERENCES Hardy, G. H. and Wright, E. M. An Introduction to the Theory of Numbers, 5th ed. Oxford, England: Clarendon Press, 1979. International Journal of Mathematics Trends and Technology (IJMTT) – Volume 64 Number 2 – December 2018 ISSN: 2231 – 5373 Page 79 Nagell, T. "Fermat's Theorem and Its Generalization by Euler." Introduction to Number Theory. New York: Wiley, pp. 71-73, 1951. Springer Science+BusineHazewinkel, Michiel, ed. (2001) , "Fermat's little theorem", Encyclopedia of MathematicssSéroul, R. "The Theorems of Fermat and Euler." Programming for Mathematicians. Berlin: Edwards, Harold M., Fermat‟s Last Theorem (A Genetic Introduction to Algebraic Number Theory). Springer-Verlag (1977). Media B.V. / Springer-Verlag, p. 15, 2000. Dickson, Leonard E. D., History of the Theory of Numbers, Volume 1 (First published in 1919). Chelsea Publishing Company (1971). Kluwer Academic Publishers, ISBN 978-1-55608-010-4 Albert, A. Adrian (2015) , Modern higher algebra, Cambridge UniversityPress, ISBN 978-1-107-54462-8. Burton, David M. (2011), The History of Mathematics / An Introduction (7th ed.), McGraw-Hill, ISBN 978-0-07-338315-6 |
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G I A N C O L I
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V o l u m e n 1
F
B
Desplazamiento
Este lanzador de béisbol está a punto de acelerar
la bola a una gran velocidad al ejercer una fuerza
sobre ella.Él realizará un trabajo sobre la bola conforme
ejerza la fuerza durante un desplazamiento de tal vez varios
metros,que van desde atrás de su cabeza hasta el punto en
que suelte la bola con el brazo estirado hacia el frente.El
trabajo total realizado sobre la bola será igual a la energía
cinética adquirida por la bola,un resultado
conocido como el principio trabajo-energía.
A 1
2 mv 2 B
136
CAPÍTULO 6
T r a b a j o y e n e r g í a
H asta ahora se ha estudiado el movimiento de traslación de un objeto en
términos de las tres leyes del movimiento de Newton.En este análisis,la
fuerza ha jugado un papel central como la cantidad que determina el mo-
vimien to.En este cap ítulo y el siguiente,se re alizará un análisis a lternativo del movi-
miento de traslación de los objetos,en términos de las cantidades energía y cantidad
de movimiento.Lo importante de la energía y la cantidad de movimiento es que se
conservan.Esto es,en circunstancias bastante generales,permanecen constantes.El
hecho de que existan cantidades que se conservan no sólo nos brinda una compren-
sión más profunda de la naturaleza del mundo,sino que también nos ofrece otra
forma de plantear la resolución de problemas prácticos.
Las leyes de conservación de la energía y la cantidad de movimiento son espe-
cialmente valiosas al tratar con sistemas de muchos objetos,en los que una conside-
ración detallada de las fuerzas implicadas sería difícil,si no es que imposible.Estas
leyes son aplicables a un amplio rango de fenómenos,incluido el mundo atómico y
subatómico,donde las leyes de Newton no se aplican.
SECCIÓN 6–1 T rabajo realizado por una fuerza constante 137
T rabajo
Este capítulo está dedicado a los conceptos fundamentales de trabajo y energía.
Estas dos cantidades son escalares,pues no tienen una dirección asociada,lo que a
menudo hace más fácil trabajar con ellas en comparación con las cantidades vecto-
riales.
T rabajo r ealizado por una fuerza constante
La palabra trabajo tiene varias acepciones en el lenguaje cotidiano.Pero en física,al
trabajo se le ha dado un significado muy específico para describir lo que se logra
cuando una fuerza actúa sobre un objeto,y éste se mueve a lo largo de una distan-
cia.Específicamente,el trabajo realizado sobre un objeto por una fuerza constante
(tanto en magnitud como en dirección) se define como el producto de la magnitud
del desplazamiento por el componente de la fuerza paralelo al desplazamiento.En
forma de ecuación,esto se expresa como
donde F∑∑es el componente de la fuerza constante paralelo al desplazamiento
T ambién se escribe
(6–1)
donde F es la magnitud de la fuerza constante,d es la magnitud del desplazamiento
del obje to y u e s el ángul o ent re la s dire ccion es de la f uerza y el despl azami ento (figu-
ra 6-1).En la ecuació n 6-1 apar ece el fac tor cos u por que es el compo-
nente de qu e es pa rale lo a El tra baj o es un a ca nti dad esc alar:só lo tien e m agn itud,
que puede ser positiva o negativa.
Primero consideremos el caso en el que el movimiento y la fuerza están en la
misma dirección,de modo que u⫽0 y cos u⫽1;en este caso,W⫽Fd.P or ejemplo,
si alguien empuja un carrito del supermercado repleto con mercancías a lo largo de
50 m,ejerciendo una fuerza horizontal de 30 N sobre él,realiza 30 N ⫻50 m ⫽1500
N⭈m de trabajo sobre el carrito.
Como muestra este ejemplo,en unidades SI,el trabajo se mide en newton-me-
tros (N⭈m).Esta unidad tiene un nombre especial:j o u l e(J) ⫽1 J ⫽1 N⭈m.En el
sistema cgs,la unidad de trabajo se llama erg,y se define como 1 erg ⫽1 dina⭈cm.
En unidades inglesas,el trabajo se mide en pies-libras.Es fácil demostrar que 1 J ⫽
10 7 erg ⫽0.7376 ft⭈lb.
d
B
.
F
B
F co s u
A=F∑∑B
W=Fd cos u,
d
B
.F
B
W=F
∑∑d,
6–1
Definición de trabajo
(para fuerza constante).
F
B
F cos u =F
u
d
B
FIGURA 6–1 Una persona que jala una caja a lo largo del suelo.El trabajo realizado por la fuerza es
donde es el desplazamiento.
d
B
W=Fd cos u,F
B
Unidad para trabajo:el joule.
m
P
F
B
g
B
d
B
138 CAPÍTULO 6 T rabajo y energía
Es posible ejercer una fuerza sobre un objeto y aún así no efectuar trabajo.P or
ejemplo,si sostiene en reposo sobre las manos una gran bolsa llena de víveres,no
efectú a trabajo sobre ella.Ejerce una fuerza sobre la b olsa,pero el desplazam iento
de la bo lsa es cero,de m odo q ue el trab ajo qu e rea liza sobre la b olsa e s W⫽0.Pa ra
efectuar trabajo se necesita tanto una fuerza como un desplazamiento.T ampoco
realiza trabajo sobre la bolsa de provisiones si la lleva conforme camina horizonta l-
men te a lo lar go de l pis o con velo cidad const ante,como se m uestra en l a fig ura 6-2.
No se requiere fuerza horizontal para mover la bolsa con velocidad constante.La
persona que aparece representada en la figura 6-2 ejerce sobre la bolsa una fuerza
hacia arriba igual a su peso.Pero esta fuerza hacia arriba es perpendicular al des-
plazam iento hori zontal de l a bolsa y,por e nde,no t iene que ve r con ese movimiento.
En consecuencia,la fuerza hacia arriba no realiza trabajo.Esta conclusión proviene
de la definición de trabajo (ecuación 6-1):W⫽0,porque u⫽90°y cos 90°⫽0.De
es ta f orm a,cua ndo una fue rza par tic ular es per pen dicu lar al des plaz ami ento,tal fuerza
no realiza ningún trabajo.(Cuando comienza a caminar o deja de hacerlo,exist e una
aceleración horizontal y brevemente ejerce una fuerza horizontal,por lo que reali-
za trabajo sobre la bolsa).
Cuando se manejan problemas relacionados con trabajo,al igual que con la
fuerza,es necesario especificar si se habla acerca del trabajo efectuado por un obje-
to específico o si éste se realiza sobre un objeto específico.T ambién es importante
señalar si el trabajo efectuado se debe a una fuerza particular (y cuál es),o si el tra-
bajo total (neto) es efectuado por la fuerza neta sobre el objeto.
EJEMPLO 6–1 T rabaj o efectua do sobre u na caja.Una pe rsona jala por 40 m
una caja de 50 kg a lo largo de un suelo horizontal mediante una fuerza constante
F P⫽100 N,que actúa en un ángulo de 37°,como se indica en la figura 6-3.El sue-
lo es rugoso y ejerce una fuerza de fricción F fr⫽50 N.Determine a) el trabajo
efectuado por cada fuerza que actúa sobre la caja y b) el trabajo neto efectuado
sobre la caja.
PL AN TE AM IEN TO Se elige el sistema coordenado de modo que sea el vector
que representa el desplazamiento de 40 m (esto es,a lo largo del eje x).Como se
aprecia en la figura 6-3,sobre la caja actúan cuatro fuerzas:la que ejerce la perso-
na,la fuerza de fricción debida al suelo;el peso de la caja,;y la fuerza
normal,,ejercida hacia arriba por el suelo.La fuerza neta sobre la caja es la su-
ma vectorial de estas cuatro fuerzas.
SOLUCIÓN a) El trabajo realizado por las fuerzas gravitacional y normal es cero,
puesto que son perpendiculares al desplazamiento (u⫽90°en la ecuación 6-1):
El trabajo realizado por es
El trabajo realizado por la fuerza de fricción es
El ángulo entre el desplazamiento y la fuerza es de 180°porque apuntan en
direcciones opuestas.Como la fuerza de fricción es opuesta al movimiento (y cos
180°⫽⫺1),el trabajo realizado por la fricción sobre la caja es negativo.
F
B
fr
x
B
=(50 N)(40 m)(–1)=–2000 J.
W
fr=F
fr x cos 180°
W
P=F
P x cos u=(100 N)(40 m)cos 37°=3200 J.
F
B
P
W
N=F
N x cos 90°=0.
W
G=mgx cos 90°=0
x
B
F
B
N
m g
B
F
B
fr
F
B
P;
x
B
F
B
P
FIGURA 6–2 La persona no realiza
trabajo sobre la bolsa de alimentos
pues es perpendicular al
desplazamiento .
d
B
F
B
P
P R E C A U C I Ó N
Fuerza sin trabajo
P R E C A U C I Ó N
Hay que indicar si el trabajo
es realizado sobre o por un
objeto.
SECCIÓN 6–1 T rabajo realizado por una fuerza constante 139
y
x
37°(40 m)
fr
S S
P
S S
m
N
S S
θ
=
F
B
F
B
F
B
g
B
x
B
FIGURA 6–3 Ejemplo 6-1.Una
caja de 50 kg jalada a lo largo del
suelo.
W n e t o es el trabajo efectuado por
todas las fuerzas que actúan sobre el
objeto.
(b) El trabajo neto se puede calcular de dos formas equivalentes:
1.El trabajo neto efectuado sobre un objeto es la suma algebraica del trabajo efec-
tuado por cada fuerza,puesto que el trabajo es un escalar:
2.El trabajo neto también se puede calcular determinando primero la fuerza neta
sobre el objeto y luego tomando su componente a lo largo del desplazamiento:
Entonces el trabajo neto es
En la dirección vertical (y) no existe desplazamiento ni trabajo realizado.
En el ejemplo 6-1,la fricción realizó un trabajo negativo.En general,el trabajo
efe ctuad o por una f uerza es n egati vo sie mpre que l a fue rza (o el c ompon ente de és ta,
F∑∑) actúa en la dirección opuesta a la dirección del movimiento.Además,se puede
ver que,cuando el trabajo efectuado por una fuerza sobre un objeto es negativo,tal
fuerza intenta frenar al objeto (y lo frenaría si ésa fuese la única fuerza actuante).
Cuando el trabajo es positivo,la fuerza en cuestión intenta aumentar la rapidez del
objeto.
EJERCICIO A Una fuerza ,que forma un ángulo u con la horizontal como en la fi-
gur a 6-1 o en la 6-3,arr astra una c aja a travé s del suelo.Si la mag nitud de se m antie ne
constante,pero el ángulo u aumenta,el trabajo efectuado por a) permanece igual,
b) aumenta,c) disminuye,d) primero aumenta,pero luego disminuye.
F
B
P
F
B
P
F
B
P
=1200 J.
=(100 N cos 37°-50 N)(40 m)
W n eto=A F n eta B x x=A F P co s u-F fr B x
A F n et o B x=F P c os u-F f r.
=1200 J.
=0+0+3200 J-2000 J
W
neto=W
G+W
N+W
P+W
fr
1.Dibuje un diagrama de cuerpo libre que muestre to-
das las fuerzas que actúan sobre el objeto que elige es-
tudiar.
2.Elija un sistema coordenado xy.Si el objeto está en
movimiento,es conveniente elegir una de las direccio-
nes coordenadas como la dirección de una de las fuer-
zas,o como la dirección del movimiento.[Así,para un
objeto sobre un plano inclinado,puede elegirse un eje
coordenado que sea paralelo al plano.]
3.Aplique las ley es de Newton para determinar cual-
quier fuerza desconocida.
4.Enc uentr e el trabaj o efe ctuado por un a fue rza e specí fi-
ca so bre el objeto,media nte l a ecua ción
para una fuerza constante.Note que el trabajo efec-
tua do es negat ivo c uando una fuerz a tien de a opone rse
al desplazamiento.
5.Para encontrar el trabajo neto efectuado sobre el ob-
jeto,hay dos posibilidades:a) encontrar el trabajo
ef ectu ado por ca da f uer za y su mar los res ult ado s al ge-
br aica men te;o b) e ncon tra r la fu erza ne ta s obr e el ob je-
to,F neta,y luego usarla para encontrar el trabajo neto
efectuado,que,para una fuerza neta constante,es:
W ne to=F ne ta d c o s u.
W=Fd cos u
RESOLUCIÓN DE PROBLEMAS T rabajo
PRECAUCIÓN
T rabajo negativo
140 CAPÍTULO 6 T rabajo y energía
EJEMPLO 6–2 T rabajo sobre una mochila.a) Determine el trabajo que un
excursionista debe efectuar sobre una mochila de 15.0 kg para llevarla a la cima de
una colina de altura h⫽10.0 m,como se observa en la figura 6-4 a.Determine
también b) el trabajo realizado por la gravedad sobre la mochila y c) el trabajo ne-
to ef ectua do sob re es ta úl tima.P or s impli cidad,vamos a sup oner que e l mov imient o
es suave y a velocidad constante (es decir,la aceleración es despreciable).
PLANTEAMIENTO Siga explícitamente paso a paso el recuadro de resolución de
problemas.
SOLUCIÓN
1.Dibuje un diagrama de cuerpo libre.En la figura 6-4 b se indican las fuerzas so-
bre la mochila:la fuerza de gravedad,,que actúa hacia abajo;y E,la fuerza
que el excursionista debe ejercer hacia arriba para sostener la mochila.Como se
supone que la aceleración es despreciable,las fuerzas horizontales sobre la mo-
chila también son despreciables.
2.Elija un sistema coordenado.Nos interesa el movimiento vertical de la mochil a,
así que se elija la coordenada y como positiva verticalmente hacia arriba.
3.Aplique las ley es de Newton.Al aplicar la segunda ley de Newton a la dirección
vertical de la mochila,se obtiene
P or tanto,
4.Encuen tre el t rabajo efect uado por un a fuerza e specífica.a) Pa ra calcula r el tra-
bajo r ealizado p or el excu rsionista s obre la mo chila,se escribe la ecuación 6-1
como
y se advierte,a partir de la figura 6-4 a,que d cos u⫽h.De modo que el traba-
jo efectuado por el excursionista es
Note que el trabajo efectuado sólo depende del cambio en la elevación y no del
ángulo de la colina,u.El excursionista realizaría el mismo trabajo al levantar la
mochila verticalmente la misma altura h.
b) El trabajo efectuado por la gravedad sobre la mochila es (a partir de la ecua-
ción 6-1 y la figura 6-4 c)
Como se obtiene
NOT A El trabajo efectuado por la gravedad (que aquí es negativo) no depende
del ángulo de la pendiente,sólo de la altura vertical h de la colina.Esto se debe a
que la gravedad actúa verticalmente,así que sólo el componente vertical del des-
plazamiento aporta al trabajo efectuado.
5.Encuentre el trabajo neto efectuado.a) El trabajo neto realizado sobre la mo-
chila es W neto⫽0,pues la fuerza neta sobre la mochila es cero (se supone que
no acelera significativamente).T ambién es posible determinar el trabajo neto
efectuado sumando el trabajo realizado por cada fuerza:
NOT A A un cuando el trabajo neto efectuado por todas las fuerzas sobre la mochi-
la sea cero,el excursionista sí realiza trabajo sobre ésta,igual a 1470 J.
W
neto=W
G+W
E=–1470 J+1470 J=0.
=–(15.0 kg)A 9.80 m 兾 s 2 B(10.0 m)=–1470 J.
=–mgh
W
G=F
G d(–cos u)=mg(–d cos u)
cos
(180°-u)=–cos u,
W
G=F
G d cos
(180°-u).
=(147 N)(10.0 m)=1470 J.
W
E=F
E(d cos u)=F
E h=mgh
W
E=F
E(d cos u),
F
E=mg=(15.0 kg)
A 9.80 m 兾 s 2 B=147 N.
F
E-mg=0.
©F
y=ma y
F
B
m g
B
➥RESOLUCIÓN DE PROBLEMAS
El trabajo efectuado por la gravedad
depende de la altura de la colina y no
del ángulo de la pendiente.
y
x
θ
a)
b)
h
180°
θ
θ
c
)
θ
180°−
m
m
H
H
F
B
F
B
g
B
g
B
d
B
d
B
FIGURA 6–4 Ejemplo 6-2.
Luna
T ierra
G
F
B
v
B
FIGURA 6–5 Ejemplo 6–3.
SECCIÓN 6–3 Energía cinética y el principio trabajo-energía 141
EJEMPLO CONCEPTUAL 6–3¿La T ierra realiza trabajo sobre la Luna?La
Luna gira alrededor de la T ierra en una órbita casi circular,y se mantiene en ella
gracias a la fuerza gravitacional ejercida por la T ierra.¿La gravedad realiza a) un
trabajo positivo,b) un trabajo negativo o c) ningún trabajo sobre la Luna?
RES PUEST A La fuerza gravitacional ejercida por la T ierra sobre la Luna (figura
6-5) actúa hacia la T ierra y proporciona su aceleración centrípeta,hacia dentro a lo
largo del radio de la órbita de la Luna.El desplazamiento de la Luna en cualquier
momento es tangente al círculo,en la dirección de su velocidad,perpendicular al
radio y perpendi cular a la fuerza de gravedad.Po r tanto,el ángul o u entre la f uerza
y el desplazamiento instantáneo de la Luna es de 90°,y el trabajo efectuado
por la g raveda d de la T ierra sobre la L una m ientr as es tá en órbit a es,por t anto,cero
(cos 90°⫽0).Es por esto por lo que la Luna,al igual que los satélites artificiales,
permanece en órbita sin gasto de combustible:no se necesita efectuar trabajo neto
contra la fuerza de gravedad.
T rabajo r ealizado por una fuerza variable
Si la fuerza que actúa sobre un objeto es constante,el trabajo efectuado por esa
fuerza se calcula mediante la ecuación 6-1.Pero,en muchos casos,la fuerza varía en
magnitud o dirección durante el proceso.P or ejemplo,mientras un cohete se aleja
de la T ierra,se realiza cierto trabajo para superar la fuerza de gravedad,la que va-
ría como el cuadrado inverso de la distancia desde el centro de la T ierra.Otros
ejemplos son la fuerza ejercida por un resorte,que aumenta con la cantidad de elon-
gación,o el trabajo realizado por una fuerza variable al jalar una caja o un carrito a
la parte superior de una colina dispareja.
El trabajo realizado por una fuerza variable se puede determinar gráficamente.
El procedimiento es igual al que se utiliza para determinar el desplazamiento cuan-
do se conoce la velocidad como función del tiempo (sección 2-8).Para determinar el
trabajo realizado por una fuerza variable,se grafica F∑∑(⫽F cos u,el componente de
paralelo a la dirección del movimiento en cualquier punto) como función de la
distancia d,como en la figura 6-6 a.Se divide la distancia en pequeños segmentos
¢d.Para cada segmento,se indica el promedio de F∑∑mediante una línea horizontal
punteada.Entonces el trabajo realizado por cada segmento es que es
el área de un rectángulo ¢d de ancho y F∑∑de alto.El trabajo total efectuado para
mover el objeto una distancia total d⫽d B⫺d A es la suma de las áreas de los rec-
tángulos (cinco en el caso que se muestra en la figura 6-6 a).Generalmente,debe es-
timarse el valor promedio de F∑∑para cada segmento,y entonces se hace una
aproximación razonable del trabajo realizado.Si la distancia se subdivide en muchos
segmen tos más,¢d se p uede hacer cada vez má s pequeño y la estima ción del tr abajo
efectuado será más precisa.En el límite cuando ¢d tiende a cero,el área total de los
muchos rectángulos estrechos se aproxima al área bajo la curva (figura 6-6 b).Esto
es,el trabajo rea lizado por una fuerza variable p ara mov er un objet o entre dos puntos
es igual al área bajo la curva de F∑∑contra d entre dichos puntos.
Energía cinética y el principio trabajo-energía
Energía es uno de los conceptos más importantes en ciencia.A unque no se puede
dar una definición general simple de energía en unas cuantas palabras,sí es posible
definir cada tipo específico de energía de manera bastante sencilla.En este capítulo
se define la energía cinética de traslación y algunos tipos de energía potencial.En
capítulos posteriores se examinarán otros tipos de energía,como la relacionada con
el calor (capítulos 14 y 15).El aspecto crucial de los diferentes tipos de energía es
que la suma de todos ellos,la energía total,es la misma antes y después de cualquier
proceso;es decir,la “energía”es una cantidad que se conserva.
Para los propósitos de este capítulo,la energía se define en la forma tradicional
como “la capacidad de realizar trabajo”.Esta definición simple no es muy precisa ni
tampoco es válida para todos los tipos de energía.†Sin embargo,es válida para la
6–3
¢W=F
∑∑¢d,
F
B
6–2
F
B
G
200
100
0
200
100
a)
d A d B
Distancia,d
0 d A d B
Distancia,d
b)
∆d 4
F (N)F (N)
(F )4
FIGURA 6–6 El trabajo realizado
por una fuerza F se calcula
considerando a) la suma de las áreas
de los rectángulos;b) el área bajo la
curva de F∑∑contra d.
†Con frecuencia,la energía asociada con el calor no está disponible para realizar trabajo,como se
verá en el capítulo 15.
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Vo l u m e n 1
F B Desplazamiento Este lanzador de béisbol está a punto de acelerar la bola a una gran velocidad al ejercer una fuerza sobre ella. Él realizará un trabajo sobre la bola conforme ejerza la fuerza durante un desplazamiento de tal vez varios metros, que van desde atrás de su cabeza hasta el punto en que suelte la bola con el brazo estirado hacia el frente. El trabajo total realizado sobre la bola será igual a la energía cinética adquirida por la bola, un resultado conocido como el principio trabajo-energía. A 12 mv 2 B 136
CAPÍTULO 6
Trabajo y energía
H
asta ahora se ha estudiado el movimiento de traslación de un objeto en términos de las tres leyes del movimiento de Newton. En este análisis, la fuerza ha jugado un papel central como la cantidad que determina el mo- vimiento. En este capítulo y el siguiente, se realizará un análisis alternativo del movi- miento de traslación de los objetos, en términos de las cantidades energía y cantidad de movimiento. Lo importante de la energía y la cantidad de movimiento es que se conservan. Esto es, en circunstancias bastante generales, permanecen constantes. El hecho de que existan cantidades que se conservan no sólo nos brinda una compren- sión más profunda de la naturaleza del mundo, sino que también nos ofrece otra forma de plantear la resolución de problemas prácticos. Las leyes de conservación de la energía y la cantidad de movimiento son espe- cialmente valiosas al tratar con sistemas de muchos objetos, en los que una conside- ración detallada de las fuerzas implicadas sería difícil, si no es que imposible. Estas leyes son aplicables a un amplio rango de fenómenos, incluido el mundo atómico y subatómico, donde las leyes de Newton no se aplican.
m F P B gB d B 138 CAPÍTULO 6 Trabajo y energía Es posible ejercer una fuerza sobre un objeto y aún así no efectuar trabajo. Por ejemplo, si sostiene en reposo sobre las manos una gran bolsa llena de víveres, no efectúa trabajo sobre ella. Ejerce una fuerza sobre la bolsa, pero el desplazamiento de la bolsa es cero, de modo que el trabajo que realiza sobre la bolsa es W 0. Para efectuar trabajo se necesita tanto una fuerza como un desplazamiento. Tampoco realiza trabajo sobre la bolsa de provisiones si la lleva conforme camina horizontal- mente a lo largo del piso con velocidad constante, como se muestra en la figura 6-2. No se requiere fuerza horizontal para mover la bolsa con velocidad constante. La persona que aparece representada en la figura 6-2 ejerce sobre la bolsa una fuerza hacia arriba igual a su peso. Pero esta fuerza hacia arriba es perpendicular al des- plazamiento horizontal de la bolsa y, por ende, no tiene que ver con ese movimiento. En consecuencia, la fuerza hacia arriba no realiza trabajo. Esta conclusión proviene de la definición de trabajo (ecuación 6-1): W 0, porque u 90° y cos 90° 0. De esta forma, cuando una fuerza particular es perpendicular al desplazamiento, tal fuerza no realiza ningún trabajo. (Cuando comienza a caminar o deja de hacerlo, existe una aceleración horizontal y brevemente ejerce una fuerza horizontal, por lo que reali- za trabajo sobre la bolsa). Cuando se manejan problemas relacionados con trabajo, al igual que con la fuerza, es necesario especificar si se habla acerca del trabajo efectuado por un obje- to específico o si éste se realiza sobre un objeto específico. También es importante señalar si el trabajo efectuado se debe a una fuerza particular (y cuál es), o si el tra- bajo total (neto) es efectuado por la fuerza neta sobre el objeto. EJEMPLO 6–1 Trabajo efectuado sobre una caja. Una persona jala por 40 m una caja de 50 kg a lo largo de un suelo horizontal mediante una fuerza constante FP 100 N, que actúa en un ángulo de 37°, como se indica en la figura 6-3. El sue- lo es rugoso y ejerce una fuerza de fricción Ffr 50 N. Determine a) el trabajo efectuado por cada fuerza que actúa sobre la caja y b) el trabajo neto efectuado sobre la caja. PLANTEAMIENTO Se elige el sistema coordenado de modo que sea el vector que representa el desplazamiento de 40 m (esto es, a lo largo del eje x). Como se aprecia en la figura 6-3, sobre la caja actúan cuatro fuerzas: la que ejerce la perso- na, la fuerza de fricción debida al suelo; el peso de la caja, ; y la fuerza normal, , ejercida hacia arriba por el suelo. La fuerza neta sobre la caja es la su- ma vectorial de estas cuatro fuerzas. SOLUCIÓN a) El trabajo realizado por las fuerzas gravitacional y normal es cero, puesto que son perpendiculares al desplazamiento (u 90° en la ecuación 6-1): El trabajo realizado por es El trabajo realizado por la fuerza de fricción es El ángulo entre el desplazamiento y la fuerza es de 180° porque apuntan en direcciones opuestas. Como la fuerza de fricción es opuesta al movimiento (y cos 180° 1), el trabajo realizado por la fricción sobre la caja es negativo. F B x fr B Wfr = Ffr x cos 180° = ( 50 N)( 40 m)(– 1 ) = – 2000 J. WP = FP x cos u = ( 100 N)( 40 m) cos 37° = 3200 J. F B P WN = FN x cos 90° = 0. WG = mgx cos 90° = 0 x B F B N F mgB B F fr B P ; xB F B P FIGURA 6–2 La persona no realiza trabajo sobre la bolsa de alimentos pues es perpendicular al desplazamiento d. F B B P P R E C A U C I Ó N Fuerza sin trabajo P R E C A U C I Ó N Hay que indicar si el trabajo es realizado sobre o por un objeto.
SECCIÓN 6–1 Trabajo realizado por una fuerza constante 139 y x 37 ° (40 m) fr SS P SS m N SS θ = F B F B F B gB xB FIGURA 6–3 Ejemplo 6-1. Una caja de 50 kg jalada a lo largo del suelo. Wneto es el trabajo efectuado por todas las fuerzas que actúan sobre el objeto. (b) El trabajo neto se puede calcular de dos formas equivalentes:
El trabajo neto efectuado sobre un objeto es la suma algebraica del trabajo efec- tuado por cada fuerza, puesto que el trabajo es un escalar:
El trabajo neto también se puede calcular determinando primero la fuerza neta sobre el objeto y luego tomando su componente a lo largo del desplazamiento: Entonces el trabajo neto es En la dirección vertical (y) no existe desplazamiento ni trabajo realizado. En el ejemplo 6-1, la fricción realizó un trabajo negativo. En general, el trabajo efectuado por una fuerza es negativo siempre que la fuerza (o el componente de ésta, F∑∑) actúa en la dirección opuesta a la dirección del movimiento. Además, se puede ver que, cuando el trabajo efectuado por una fuerza sobre un objeto es negativo, tal fuerza intenta frenar al objeto (y lo frenaría si ésa fuese la única fuerza actuante). Cuando el trabajo es positivo, la fuerza en cuestión intenta aumentar la rapidez del objeto. EJERCICIO A Una fuerza , que forma un ángulo u con la horizontal como en la fi- gura 6-1 o en la 6-3, arrastra una caja a través del suelo. Si la magnitud de se mantiene constante, pero el ángulo u aumenta, el trabajo efectuado por a) permanece igual, b) aumenta, c) disminuye, d) primero aumenta, pero luego disminuye. F B P F B P F B P = 1200 J. = ( 100 N cos 37° - 50 N)( 40 m) Wneto = AFneta Bx x = AFP cos u - FfrBx AFneto Bx = FP cos u - Ffr. = 1200 J. = 0 + 0 + 3200 J - 2000 J Wneto = WG + WN + WP + Wfr
Dibuje un diagrama de cuerpo libre que muestre to- das las fuerzas que actúan sobre el objeto que elige es- tudiar.
Elija un sistema coordenado xy. Si el objeto está en movimiento, es conveniente elegir una de las direccio- nes coordenadas como la dirección de una de las fuer- zas, o como la dirección del movimiento. [Así, para un objeto sobre un plano inclinado, puede elegirse un eje coordenado que sea paralelo al plano.]
Aplique las leyes de Newton para determinar cual- quier fuerza desconocida.
Encuentre el trabajo efectuado por una fuerza específi- ca sobre el objeto, mediante la ecuación para una fuerza constante. Note que el trabajo efec- tuado es negativo cuando una fuerza tiende a oponerse al desplazamiento.
Para encontrar el trabajo neto efectuado sobre el ob- jeto, hay dos posibilidades: a) encontrar el trabajo efectuado por cada fuerza y sumar los resultados alge- braicamente; o b) encontrar la fuerza neta sobre el obje- to, Fneta, y luego usarla para encontrar el trabajo neto efectuado, que, para una fuerza neta constante, es: Wneto = Fneta d cos u. W = Fd cos u
# RESOLUCIÓN DE PROBLEMAS Trabajo
P R E C A U C I Ó N Trabajo negativo
Luna Tierra F G B vB FIGURA 6–5 Ejemplo 6–3. SECCIÓN 6–3 Energía cinética y el principio trabajo-energía 141 EJEMPLO CONCEPTUAL 6–3 ¿La Tierra realiza trabajo sobre la Luna? La Luna gira alrededor de la Tierra en una órbita casi circular, y se mantiene en ella gracias a la fuerza gravitacional ejercida por la Tierra. ¿La gravedad realiza a) un trabajo positivo, b) un trabajo negativo o c) ningún trabajo sobre la Luna? RESPUESTA La fuerza gravitacional ejercida por la Tierra sobre la Luna (figura 6-5) actúa hacia la Tierra y proporciona su aceleración centrípeta, hacia dentro a lo largo del radio de la órbita de la Luna. El desplazamiento de la Luna en cualquier momento es tangente al círculo, en la dirección de su velocidad, perpendicular al radio y perpendicular a la fuerza de gravedad. Por tanto, el ángulo u entre la fuerza y el desplazamiento instantáneo de la Luna es de 90°, y el trabajo efectuado por la gravedad de la Tierra sobre la Luna mientras está en órbita es, por tanto, cero (cos 90° 0). Es por esto por lo que la Luna, al igual que los satélites artificiales, permanece en órbita sin gasto de combustible: no se necesita efectuar trabajo neto contra la fuerza de gravedad.
Trabajo realizado por una fuerza variable
Si la fuerza que actúa sobre un objeto es constante, el trabajo efectuado por esa fuerza se calcula mediante la ecuación 6-1. Pero, en muchos casos, la fuerza varía en magnitud o dirección durante el proceso. Por ejemplo, mientras un cohete se aleja de la Tierra, se realiza cierto trabajo para superar la fuerza de gravedad, la que va- ría como el cuadrado inverso de la distancia desde el centro de la Tierra. Otros ejemplos son la fuerza ejercida por un resorte, que aumenta con la cantidad de elon- gación, o el trabajo realizado por una fuerza variable al jalar una caja o un carrito a la parte superior de una colina dispareja. El trabajo realizado por una fuerza variable se puede determinar gráficamente. El procedimiento es igual al que se utiliza para determinar el desplazamiento cuan- do se conoce la velocidad como función del tiempo (sección 2-8). Para determinar el trabajo realizado por una fuerza variable, se grafica F∑∑ ( F cos u, el componente de paralelo a la dirección del movimiento en cualquier punto) como función de la distancia d, como en la figura 6-6a. Se divide la distancia en pequeños segmentos ¢d. Para cada segmento, se indica el promedio de F∑∑ mediante una línea horizontal punteada. Entonces el trabajo realizado por cada segmento es que es el área de un rectángulo ¢d de ancho y F∑∑ de alto. El trabajo total efectuado para mover el objeto una distancia total d dB dA es la suma de las áreas de los rec- tángulos (cinco en el caso que se muestra en la figura 6-6a). Generalmente, debe es- timarse el valor promedio de F∑∑ para cada segmento, y entonces se hace una aproximación razonable del trabajo realizado. Si la distancia se subdivide en muchos segmentos más, ¢d se puede hacer cada vez más pequeño y la estimación del trabajo efectuado será más precisa. En el límite cuando ¢d tiende a cero, el área total de los muchos rectángulos estrechos se aproxima al área bajo la curva (figura 6-6b). Esto es, el trabajo realizado por una fuerza variable para mover un objeto entre dos puntos es igual al área bajo la curva de F∑∑ contra d entre dichos puntos.
Energía cinética y el principio trabajo-energía
Energía es uno de los conceptos más importantes en ciencia. Aunque no se puede dar una definición general simple de energía en unas cuantas palabras, sí es posible definir cada tipo específico de energía de manera bastante sencilla. En este capítulo se define la energía cinética de traslación y algunos tipos de energía potencial. En capítulos posteriores se examinarán otros tipos de energía, como la relacionada con el calor (capítulos 14 y 15). El aspecto crucial de los diferentes tipos de energía es que la suma de todos ellos, la energía total, es la misma antes y después de cualquier proceso; es decir, la “energía” es una cantidad que se conserva. Para los propósitos de este capítulo, la energía se define en la forma tradicional como “la capacidad de realizar trabajo”. Esta definición simple no es muy precisa ni tampoco es válida para todos los tipos de energía.† Sin embargo, es válida para la
6–
¢W = F∑∑ ¢d, F B
6–
F B G 200 100 0 200 100 a) dA dB Distancia, d 0 dA dB Distancia, d b) ∆d 4 F (N) F (N) (F ) 4 FIGURA 6–6 El trabajo realizado por una fuerza F se calcula considerando a) la suma de las áreas de los rectángulos; b) el área bajo la curva de F∑∑ contra d. †Con frecuencia, la energía asociada con el calor no está disponible para realizar trabajo, como se verá en el capítulo 15.
142 CAPÍTULO 6 Trabajo y energía Definición de energía cinética energía mecánica que se estudia en este capítulo y sirve para entender la conexión fundamental entre trabajo y energía. A continuación se definirá y analizará uno de los tipos básicos de energía: la cinética. Un objeto en movimiento puede efectuar trabajo sobre otro al golpearlo. Una bala de cañón en vuelo realiza trabajo sobre una pared de ladrillos a la que derriba; un martillo en movimiento efectúa trabajo sobre un clavo que introduce en la ma- dera. En cualquier caso, un objeto en movimiento ejerce una fuerza sobre un segundo objeto que experimenta un desplazamiento. Un objeto en movimiento tiene la capa- cidad de efectuar trabajo y, por lo mismo, puede decirse que tiene energía. La energía del movimiento se llama energía cinética. El término cinético proviene de la palabra griega kinetikos, que significa “movimiento”. Para obtener una definición cuantitativa de la energía cinética, consideremos un objeto rígido de masa m que se mueve en una línea recta con una rapidez inicial v 1. Para acelerarlo uniformemente a una rapidez v 2 , sobre él se ejerce una fuerza neta constante Fneta paralela a su movimiento sobre un desplazamiento d (figura 6-7). Entonces el trabajo neto efectuado sobre el objeto es Wneto Fnetad. Se aplica la se- gunda ley de Newton, Fneta ma, y se emplea la ecuación 2-11c, que ahora se escri- be como con v 1 como la rapidez inicial y v 2 como la rapidez final. Se resuelve para a en la ecuación 2-11c, luego, se sustituye esto en Fneta ma y se determina el trabajo efectuado: o (6–2) La cantidad se define como la energía cinética (EC) de traslación del objeto: (6–3) (A esta energía cinética se le llama “de traslación” para distinguirla de la energía ci- nética de rotación, que se estudiará en el capítulo 8). La ecuación 6-2, expresada aquí para el movimiento unidimensional con fuerza constante, es válida en general para el movimiento de traslación de un objeto en tres dimensiones e incluso si la fuerza varía. La ecuación 6-2 se reescribe como: Wneto EC 2 – EC 1 o Wneto ¢EC. (6–4) La ecuación 6-4 (o la ecuación 6-2) es un resultado importante conocido como el principio trabajo-energía. Con palabras, se enuncia como: El trabajo neto realizado sobre un objeto es igual al cambio en la energía ciné- tica del mismo. Hay que hacer notar que se utilizó la segunda ley de Newton, Fneta ma, donde Fneta es la fuerza neta, es decir, la suma de todas las fuerzas que actúan sobre el objeto. Así entonces, el principio trabajo-energía es válido sólo si W es el trabajo neto efec- tuado sobre el objeto; esto es, el trabajo realizado por todas las fuerzas que actúan sobre el objeto. El principio trabajo-energía es una reformulación muy útil de las leyes de Newton. Nos dice que, si sobre un objeto se realiza un trabajo neto W (positivo), la energía cinética del objeto aumenta por una cantidad W. El principio también es cierto para la situación inversa: si el trabajo neto W realizado sobre un objeto es negativo, la ec = 12 mv 2. 1 2 mv 2 Wneto = 12 mv 22 - 12 mv 12. Wneto = Fneta d = mad = m a v 22 - v 12 2 d b d = m a v 22 - v 12 2 b a = v 22 - v 12 2 d , v 22 = v 12 + 2 ad, net 1 F net B F B v vB B d B FIGURA 6–7 Una fuerza neta constante Fneta acelera un autobús desde la rapidez v 1 hasta la rapidez v 2 a lo largo de un desplazamiento d. El trabajo neto realizado es Wneto Fnetad. PRINCIPIO TRABAJO-ENERGÍA PRINCIPIO TRABAJO-ENERGÍA PRECAUCIÓN Trabajo-energía es válido sólo para trabajo neto.
144 CAPÍTULO 6 Trabajo y energía (d = 20 m) b) (d = ?) a) v 1 = 60 km/h v1= 120 km/h v 2 = 0 v 2 = 0 F B F B d B d B FIGURA 6–10 Ejemplo 6-6. Energía potencial F Í S I C A A P L I C A D A Distancia de frenado de un automóvil r rapidez inicial al cuadrado EP gravitacional EJEMPLO CONCEPTUAL 6–6 Trabajo para detener un automóvil. Un au- tomóvil que viaja a 60 kmh puede frenar hasta detenerse dentro de una distancia d de 20 m (figura 6-10a). Si el automóvil viaja el doble de rápido, es decir, a 120 kmh, ¿cuál es su distancia de frenado (figura 6-10b)? Se supone que la fuerza de frena- do máxima es aproximadamente independiente de la rapidez. RESPUESTA Como la fuerza de frenado F es aproximadamente constante, el tra- bajo necesario para detener el automóvil, Fd, es proporcional a la distancia reco- rrida. Se aplica el principio trabajo-energía, pero considerando que y están en direcciones opuestas y que la rapidez final del automóvil es cero: Entonces Por tanto, como la fuerza y la masa son constantes, se sabe que la distancia de fre- nado, d, aumenta con el cuadrado de la rapidez: Si la rapidez inicial del automóvil se duplica, la distancia de frenado es (2) 2 4 ve- ces mayor, esto es, 80 m. EJERCICIO B ¿La energía cinética puede ser negativa alguna vez?
Energía potencial
Acabamos de analizar cómo es que un objeto tiene energía en virtud de su movi- miento, a lo que se le llama energía cinética. Pero también es posible tener energía potencial, que es la energía asociada con las fuerzas que dependen de la posición o configuración de un objeto (u objetos) en relación con su entorno. Existen varios ti- pos de energía potencial (EP), y cada uno está asociado con una fuerza particular. El resorte de un juguete de cuerda es un ejemplo de un objeto con energía po- tencial. El resorte adquirió su energía potencial gracias al trabajo que sobre él reali- zó la persona que dio cuerda al juguete. Conforme el resorte se desenrolla, ejerce una fuerza y efectúa trabajo para hacer que el juguete se mueva. Quizá el ejemplo más común de energía potencial sea la energía potencial gravi- tacional. Un pesado ladrillo que se mantiene elevado en el aire tiene energía potencial en virtud de su posición en relación con la Tierra. El ladrillo elevado tiene la capaci- dad de realizar trabajo, por lo que, si es liberado, caerá hacia el suelo por la fuerza gravitacional, y puede realizar trabajo sobre, por ejemplo, una estaca a la que clava- rá en el suelo. Busquemos ahora la forma para la energía potencial gravitacional de un objeto cerca de la superficie de la Tierra. Para que un objeto de masa m se eleve verticalmente, sobre él debe aplicarse una fuerza ascendente (por ejemplo, por la mano de una persona) y que sea por lo menos igual a su peso, mg. Para elevarlo sin aceleración un desplazamiento vertical de altura h, desde la posición y 1 hasta la y 2
6–
d r v 2. = 0 - 12 mv 12. –Fd = ¢ec = 12 mv 22 - 12 mv 12 Wneto = Fd cos 180° = –Fd. d B F B
(ejercida por la mano) y 1 y 2 h
m ext FG m B F B d B gB FIGURA 6–11 Una persona ejerce una fuerza ascendente Fext mg para elevar un ladrillo desde y 1 hasta y 2. SECCIÓN 6–4 Energía potencial 145 en la figura 6-11 (la dirección hacia arriba es positiva), una persona debe realizar trabajo igual al producto de la fuerza externa necesaria, Fext mg hacia arriba, y el desplazamiento vertical h. Esto es, (6–5a) La gravedad también actúa sobre el objeto conforme se mueve desde y 1 hasta y 2 , y realiza trabajo sobre él igual a donde u 180° porque y apuntan en direcciones opuestas. De este modo (6–5b) Si ahora se permite que el objeto parta del reposo y se le deja en caída libre bajo la ac- ción de la gravedad, adquiere una velocidad dada por v 2 2 gh (ecuación 2-11c) des- pués de caer una altura h. Entonces tiene energía cinética y, si golpea una estaca, puede realizar trabajo sobre ella igual a mgh (principio trabajo-energía). Así, para elevar un objeto de masa m a una altura h se requiere una cantidad de trabajo igual a mgh (ecuación 6-5a). Y, una vez en la altura h, el ob- jeto tiene la capacidad de realizar una cantidad de trabajo igual a mgh. Por tanto, la energía potencial gravitacional de un objeto, debida a la gravedad de la Tierra, se define como el producto del peso mg del objeto y su altura y sobre cierto nivel de referencia (como el suelo): (6–6) Cuanto más alto esté con respecto al piso, el objeto tendrá más energía potencial gravitacional. Ahora se combina la ecuación 6-5a con la 6-6: (6–7a) Esto es, el trabajo efectuado por una fuerza externa para mover al objeto de masa m desde el punto 1 hasta el punto 2 (sin aceleración) es igual al cambio en la ener- gía potencial entre las posiciones 1 y 2. De manera alternativa, se puede expresar el cambio en la energía potencial, ¢EP, en términos del trabajo realizado por la gravedad misma. Comenzando por la ecuación 6-5b, se obtiene (6–7b) Esto es, el trabajo realizado por la gravedad conforme el objeto de masa m se mue- ve desde el punto 1 hasta el punto 2 es igual al negativo de la diferencia en la ener- gía potencial entre las posiciones 1 y 2. La energía potencial pertenece a un sistema y no a un solo objeto en particular. La energía potencial está asociada con una fuerza, y una fuerza sobre un objeto siempre la ejerce algún otro objeto. Por eso, la energía potencial es una propiedad del sistema como un todo. Para un objeto elevado a una altura y sobre la superficie de la Tierra, el cambio en la energía potencial gravitacional es mgy. Aquí el sistema es el objeto más la Tierra, y las propiedades de ambos están implicadas: objeto (m) y Tierra (g). La energía potencial gravitacional depende de la altura vertical del objeto sobre algún nivel de referencia (ecuación 6-6). En algunas situaciones, es posible que se pregunte desde qué punto hay que medir la altura y. La energía potencial gravita- cional de un libro que se mantiene elevado sobre una mesa, por ejemplo, depende de si se mide y desde lo alto de la mesa, desde el suelo o desde algún otro punto de re- ferencia. Lo que es físicamente importante en cualquier situación es el cambio en la WG = – Aep 2 - ep 1 B = – ¢ep. WG = –mg Ay 2 - y 1 B Wext = ep 2 - ep 1 = ¢ep. Wext = mg Ay 2 - y 1 B epgrav = mgy. 1 2 mv 2 = 1 2 m( 2 gh) = mgh, = –mg Ay 2 - y 1 B. WG = –mgh d B F B G WG = FG d cos u = mgh cos 180°, = mg Ay 2 - y 1 B. Wext = Fext d cos 0° = mgh EP gravitacional P R E C A U C I Ó N La energía potencial pertenece a un sistema, no a un objeto particular.
SECCIÓN 6–4 Energía potencial 147 Consideremos ahora otro tipo de energía potencial, la asociada con los materia- les elásticos. Esto incluye una gran variedad de aplicaciones prácticas. Piense, por ejemplo, en el simple resorte de alambre que se ilustra en la figura 6-13. El resorte tiene energía potencial cuando se le comprime (o alarga) de modo que, cuando se le libera, puede realizar trabajo sobre una pelota, como se observa. Para mantener un resorte alargado o comprimido una cantidad x desde su longitud natural o de equi- librio (sin deformar), se requiere que la mano ejerza una fuerza sobre el resorte, FP, que es directamente proporcional a x. Esto es, donde k es una constante, conocida como constante de rigidez de resorte y es una medida particular de la rigidez del resorte. El resorte alargado o comprimido ejerce una fuerza FR en la dirección opuesta sobre la mano, como se observa en la figura 6-14: (6–8) A esta fuerza a veces se le llama “fuerza restauradora” porque el resorte ejerce su fuerza en la dirección opuesta al desplazamiento (de ahí el signo menos), y actúa para regresarlo a su longitud natural. La ecuación 6-8 se conoce como la ecuación del resorte o como ley de Hooke, y es precisa para resortes en tanto x no sea dema- siado grande. Para calcular la energía potencial de un resorte estirado, vamos a determinar el trabajo que se requiere para estirarlo (figura 6-14b). Tal vez quiera emplear la ecuación 6-1 para el trabajo que se realiza sobre él, W Fx, donde x es la cantidad que se estira desde su longitud natural. Pero esto sería incorrecto dado que la fuer- za FP ( kx) no es constante, sino que varía con la distancia, y se vuelve mayor con- forme el resorte se estira cada vez más, como se muestra gráficamente en la figura 6-15. De modo que hay que usar la fuerza promedio Puesto que FP varía lineal- mente (desde cero en la posición natural, hasta kx cuando se estira hasta x), la fuer- za promedio es donde x aquí es la cantidad final estirada (que se designa como xf en la figura 6-15 para mayor claridad). Entonces, el trabajo realizado es Entonces, la energía potencial elástica es proporcional al cuadrado de la cantidad estirada: (6–9) Si un resorte se comprime una distancia x desde su longitud natural, la fuerza pro- medio una vez más es y de nuevo la energía potencial está dada por la ecuación 6-9. En consecuencia, x puede ser la cantidad comprimida o la cantidad es- tirada desde la longitud natural del resorte.† Note que, en el caso de un resorte, el punto de referencia para EP cero se elige en la posición natural del resorte. f = 12 kx, ep elástica = 12 kx 2. W = fx = A 12 kxB(x) = 12 kx 2. f = 12 [0 + kx] = 12 kx, f. FS = –kx. FP = kx, EP de un resorte elástico a) b) c) FIGURA 6–13 a) Un resorte puede almacenar energía (EP elástica) cuando se le comprime como en b) y puede realizar trabajo cuando se le libera como en c). † También se puede obtener la ecuación 6-9 a partir de la sección 6-2. El trabajo realizado, y por tanto ¢EP, es igual al área bajo la gráfica de F contra x en la figura 6-15. Esta área es un triángulo (som- breado en la figura 6-15) de altura kx y base x, por lo que su área (como para todo triángulo) es igual a 12 (kx)(x) = 12 kx 2. a) b) c) x x = 0 x P R R P F B F B F B F B xf 1 F 2 —= kx f FP
kx F 0 x FIGURA 6–15 Conforme un resorte se estira (o comprime), la fuerza necesaria aumenta linealmente a medida que x aumenta: gráfica de F kx contra x desde x 0 hasta x xf. EP elástica FIGURA 6–14 a) Resorte en posición natural (sin alargar). b) Una persona alarga el resorte al ejercer una fuerza hacia la derecha (dirección positiva). El resorte jala de vuelta con una fuerza R, donde FR kx. c) La persona comprime el resorte (x 0) al ejercer una fuerza hacia la izquierda; el resorte empuja de vuelta con una fuerza FR kx, donde FR 0 porque x 0. F B P F B F B P
148 CAPÍTULO 6 Trabajo y energía 1 2 fr fr fr F B F B F B FIGURA 6–16 Una caja se empuja por el suelo desde la posición 1 hasta la posición 2 a través de dos trayectorias, una recta y otra curva. La fuerza de fricción siempre está exactamente en la dirección opuesta a la del movimiento. Por tanto, para una fuerza de fricción con magnitud constante, Wfr - Ffrd, de modo que si d es mayor (como para la trayectoria curva), entonces W es mayor. El trabajo realizado no depende sólo de los puntos 1 y 2. TABLA 6–1 Fuerzas conservativas y no conservativas Fuerzas Fuerzas conservativas no conservativas Gravitacional Fricción Elástica Resistencia del aire Eléctrica Tensión sobre una cuerda Motor o propulsión de un cohete Empujón o jalón de una persona La EP sólo está definida para una fuerza conservativa. En cada uno de los ejemplos anteriores de energía potencial (desde un ladrillo sostenido a una altura y hasta un resorte estirado o comprimido), un objeto tiene la capacidad o potencial de efectuar trabajo aun cuando en realidad no lo esté reali- zando. Estos ejemplos muestran que la energía se puede almacenar, para uso poste- rior, en la forma de energía potencial (figura 6-13, por ejemplo, para un resorte). Es importante hacer notar que existe una única fórmula universal para la energía cinética de traslación de un objeto, pero no existe una fórmula única para energía potencial. En vez de ello, la forma matemática de la energía potencial de- pende de la fuerza implicada.
Fuerzas conservativas y no conservativas
El trabajo realizado contra la gravedad al mover un objeto desde un punto hasta otro no depende de la trayectoria que se siga. Por ejemplo, toma el mismo trabajo ( mgy) elevar un objeto de masa m verticalmente una cierta altura que llevarlo ha- cia arriba por un plano inclinado sin frición de la misma altura vertical, como en la figura 6-4 (ejemplo 6-2). Las fuerzas como la gravedad, para las que el trabajo efec- tuado no depende de la trayectoria que se siga, sino sólo de las posiciones inicial y final, se llaman fuerzas conservativas. La fuerza elástica de un resorte (u otro mate- rial elástico) en el que F kx, también es una fuerza conservativa. Un objeto que parte en un punto determinado y regresa a ese mismo punto bajo la acción de una fuerza conservativa no tiene trabajo neto aplicado sobre él porque la energía poten- cial es la misma al principio y al final del trayecto. La fricción, por otra parte, es una fuerza no conservativa porque el trabajo que realiza depende de la trayectoria. Por ejemplo, cuando una caja se mueve a través de un piso desde un punto a otro, el trabajo realizado depende de si la trayectoria seguida es recta, curva o en zigzag. Como se observa en la figura 6-16, si una caja se empuja desde el punto 1 hasta el punto 2 a lo largo de una trayectoria semicircular (más larga) en lugar de seguir la trayectoria recta, se realiza más trabajo para vencer la fricción. Esto se debe a la mayor distancia y, a diferencia de la fuerza gravitacional, a que la fuerza de fricción siempre está en dirección opuesta a la del movimiento. (El término cos u en la ecuación 6-1 siempre es, para la fuerza de fricción, cos 180° 1 en todos los puntos de la trayectoria). Así que el trabajo realizado por la fricción en la figura 6- 16 no depende sólo de los puntos 1 y 2. Otras fuerzas que son no conservativas inclu- yen la fuerza ejercida por una persona y la tensión sobre una cuerda (tabla 6-1).
6–
1 2 mv 2 , Como la energía potencial es energía asociada con la posición o configuración de los objetos, sólo tiene sentido si puede establecerse de manera única para un punto dado. Esto no se puede hacer con las fuerzas no conservativas porque el trabajo reali- zado depende de la trayectoria seguida (como en la figura 6-16). En consecuencia, la energía potencial sólo se puede definir para una fuerza conservativa. Por eso, aunque la energía potencial siempre está asociada con una fuerza, no todas las fuerzas tienen una energía potencial. Por ejemplo, no existe energía potencial para la fricción. EJERCICIO C Un objeto sobre el que actúa una fuerza constante F se mueve desde el punto 1 hasta el punto 2 y luego de regreso. El trabajo realizado por la fuerza F en el tra- yecto completo es de 60 J. ¿A partir de esta información puede determinar si F es una fuerza conservativa o una no conservativa? Ahora es posible extender el principio trabajo-energía (que se explicó en la sec- ción 6-3) para incluir la energía potencial. Supongamos que varias fuerzas actúan so- bre un objeto que puede experimentar movimiento de traslación. Y supongamos No existe energía potencial para la fricción.
h mitad EC, mitad EP y 1 = h y 2 = 0 y toda EP toda EC 150 CAPÍTULO 6 Trabajo y energía Conservación de la energía mecánica cuando sólo actúa la gravedad En la siguiente sección se podrá constatar la gran utilidad del principio de con- servación de la energía mecánica en varias situaciones, y cómo con frecuencia es más fácil de usar que las ecuaciones cinemáticas o las leyes de Newton. Después de ello se analizará cómo se pueden incluir otras formas de energía en la ley general de conservación de la energía, como la asociada con fuerzas no conservativas.
Resolución de problemas a partir
de la conservación de la energía mecánica
Un ejemplo simple de la conservación de la energía mecánica (considerando des- preciable la resistencia del aire) es una roca a la que se le deja caer bajo la acción de la gravedad desde una altura h sobre el suelo, como se ilustra en la figura 6-17. Si la roca parte desde el reposo, toda la energía inicial es energía potencial. Conforme la roca va en descenso, la energía potencial disminuye (puesto que y disminuye), pe- ro la energía cinética de la roca aumenta para compensar, de modo que la suma de las dos permanece constante. En cualquier punto a lo largo de la trayectoria, la energía mecánica total está dada por donde y es la altura de la roca sobre el suelo en un instante dado y v es su rapidez en ese punto. Si se deja que el subíndice 1 represente la roca en un punto a lo largo de su trayectoria (por ejemplo, el punto inicial) y que el subíndice 2 represente al- gún otro punto, entonces se puede escribir energía mecánica total en el punto 1 energía mecánica total en el punto 2 o (véase también la ecuación 6-12a) [sólo EP gravitacional] (6–13) Justo antes de que la roca golpee el suelo, donde y 0, toda la energía potencial ini- cial se habrá transformado en energía cinética. EJEMPLO 6–8 Caída libre de una roca. Si la altura original de la roca en la figura 6-17 es y 1 h 3 m, calcule la rapidez de la roca cuando va a 1 m sobre el suelo. PLANTEAMIENTO Un planteamiento es usar las ecuaciones cinemáticas del capí- tulo 2. En vez de ello, se aplicará el principio de conservación de energía mecánica (ecuación 6-13) suponiendo que sobre la roca sólo actúa la gravedad. Se elige el suelo como el nivel de referencia (y 0). SOLUCIÓN En el momento de liberación (punto 1) la roca se encuentra en la po- sición y 1 3 m y está en reposo: v 1 0. Se desea encontrar v 2 cuando la roca es- té en la posición y 2 1 m. La ecuación 6-13 da Las m se cancelan; si se establece que v 1 0 y se resuelve para , se encuentra y La rapidez de la roca cuando está 1 m sobre el suelo es 6 ms hacia abajo. NOTA La velocidad en el punto 2 es independiente de la masa de la roca. EJERCICIO D Resuelva el ejemplo 6-8 mediante el principio trabajo-energía aplicado a la roca, sin el concepto de energía potencial. Enuncie todas las ecuaciones que utilice, comenzando con la ecuación 6-4. Una forma simple de visualizar la conservación de la energía es con la ayuda de una “cubeta de energía”, como la que se ilustra en la figura 6-18. Por ejemplo, en cada punto de la caída de la roca, la cantidad de energía cinética y energía potencial se indica como si hubiese dos materiales de diferentes colores en la cubeta. La cantidad total de material en la cubeta ( energía mecánica total) permanece constante. v 2 = 2 39 ms = 6 ms. = 2 A9 ms 2 B C(3 m) - (1 m)D = 39 m 2 s 2 , v 22 = 2 g Ay 1 - y 2 B v 22 1 2 mv 1 2 + mgy 1 = 1 2 mv 2 2 + mgy 2. 1 2 mv 1 2 + mgy 1 = 1 2 mv 2 2 + mgy 2. E = ec + ep = 12 mv 2 + mgy
6–
y = 3 m y = 1 m y = 0 y v = 0 v = 7 m/s v = 6 m/s EC EC EP EP FIGURA 6–18 Cubetas de energía (para el ejemplo 6-8). La energía cinética se indica con gris y la energía potencial con azul. El total (EC EP) es el mismo para los tres puntos que se representan. La rapidez en y 0, justo antes de que la roca golpee el suelo, es 32 A9 ms 2 B A3 mB = 7 ms. FIGURA 6–17 Mientras cae una roca, su energía potencial cambia a energía cinética.
C
uando una tetera con agua fría se coloca sobre el quemador caliente de una estufa, la temperatura del agua aumenta. Se dice que el calor “fluye” del quemador caliente hacia el agua fría. Cuando dos objetos a diferentes tem- peraturas se ponen en contacto, el calor fluye espontáneamente del más caliente al más frío. El flujo espontáneo de calor es en la dirección que tiende a igualar la tem- peratura. Si los dos objetos se mantienen en contacto el tiempo suficiente para que sus temperaturas se igualen, se dice que los objetos están en equilibrio térmico y a partir de entonces ya no existirá más flujo de calor entre ellos. Por ejemplo, cuando se coloca por primera vez un termómetro en la boca de una persona, el calor fluye de la boca hacia el termómetro. Cuando este último alcanza la misma temperatura que el interior de la boca, entonces el termómetro y la boca están en equilibrio, y ya no fluye más calor. Con frecuencia, calor y temperatura se confunden. Son conceptos muy diferen- tes, y en este capítulo se hará una clara distinción entre ellos. Se comenzará por de- finir y usar el concepto de calor. También se explicará cómo se usa el calor en calorimetría y cómo participa en los cambios de estado de la materia y los procesos de transferencia de calor: conducción, convección y radiación. 384
CAPÍTULO 14
Calor
Cuando hace frío, la ropa abrigadora actúa como aislante para reducir la pérdida de calor del cuerpo hacia el exterior mediante conducción y convección. La radiación del calor proveniente de una fogata calienta tanto al cuerpo como a la ropa. El fuego también transfiere energía di- rectamente mediante conducción de calor hacia los alimentos que se co- cinan. El calor, al igual que el tra- bajo, representa una transferencia de energía. Por eso se define como una transferencia de energía causada por una diferencia de temperatura. Otro concepto útil es el de energía interna U, que es la suma de todas las energías de las moléculas de un sistema.
386 CAPÍTULO 14 Calor F Í S I C A A P L I C A D A Quema de calorías Energía interna El resultado de Joule fue crucial porque extendió el principio trabajo-energía para incluir procesos que implican calor. Sus estudios también condujeron al esta- blecimiento de la ley de la conservación de energía, que se estudiará con detalle en el siguiente capítulo. EJEMPLO 14–1 ESTIMACIÓN Quema de calorías adicionales. Una perso- na come demasiado helado y pastel en el orden de 500 Calorías. Para contrarrestar este exceso, quiere hacer una cantidad equivalente de ejercicio subiendo escaleras o una montaña. ¿Qué altura debe escalar? Para este cálculo, considere que la masa de la persona es de 60 kg. PLANTEAMIENTO El trabajo W que se necesita hacer para subir escaleras es igual al cambio en energía potencial gravitacional: donde h es la altura vertical escalada. SOLUCIÓN 500 Calorías es igual a 500 kcal, lo que en joules equivale a El trabajo realizado al escalar una altura vertical h es W = mgh. Se resuelve para h: Éste es un gran cambio en altura (arriba de 11,000 ft). NOTA El cuerpo humano no transforma energía con el 100% de eficiencia; más bien, lo hace con un 20% de eficiencia. Como se explicará en el capítulo 15, algu- na energía siempre se “desperdicia”, así que en realidad, la persona en cuestión tendría que ascender sólo aproximadamente (0)(3600 m) L 700 m, que todavía es mucho (alrededor de 2300 ft de ganancia de altura).
Energía interna
La suma de la energía de todas las moléculas de un objeto constituye su energía in- terna. (En ocasiones se usa el término energía térmica para significar lo mismo). Ahora se introduce el concepto de energía interna, puesto que ayudará a clarificar varias ideas en torno al calor.
# Distinción entre temperatura, calor y energía interna
La teoría cinética permite hacer una clara distinción entre temperatura, calor y energía interna. La temperatura (en kelvins) es una medida de la energía cinética promedio de moléculas individuales. La energía interna se refiere a la energía total de todas las moléculas en el objeto. (En consecuencia, dos lingotes de hierro de igual masa pueden tener la misma temperatura, pero dos de ellos tienen el doble de energía térmica de la que tiene uno solo). El calor, finalmente, se refiere a una transferencia de energía de un objeto a otro como resultado de una diferencia en temperatura. La dirección del flujo de calor entre dos objetos depende de sus temperaturas, no de cuánta energía interna tenga cada uno. De este modo, si 50 g de agua a 30°C se ponen en contacto (o se mezclan) con 200 g de agua a 25°C, el calor fluye des- de el agua a 30°C hacia el agua a 25°C, aun cuando la energía interna del agua de 25°C sea mucho mayor puesto que hay mayor cantidad de ella.
# Energía interna de un gas ideal
Ahora se calculará la energía interna de n moles de un gas monoatómico (un átomo por molécula) ideal. La energía interna, U, es la suma de las energías cinéticas de traslación de todos los átomos. Esta suma es justo igual a la energía cinética prome-
14–
h = W mg
2 106 J ( 60 kg)A9 ms 2 B = 3600 m. ( 500 kcal)A4 103 JkcalB = 2 106 J. W = ¢PE = mgh, P R E C A U C I Ó N Distinción entre calor, energía interna y temperatura P R E C A U C I Ó N La dirección del flujo de calor depende de la temperatura (no de la cantidad de energía interna) elsolucionario
SECCIÓN 14–3 Calor específico 387 a) b) FIGURA 14–2 Además de energía cinética de traslación, las moléculas pueden tener a) energía cinética de rotación y b) energía vibratoria (tanto cinética como potencial). Energía interna de gas monoatómico ideal Relación entre transferencia de calor y cambio de temperatura Calor específico dio por molécula por el número total de moléculas, N: Usando la ecuación 13-8, se puede escribir como o (recuerde la sección 13-9) [gas monoatómico ideal] (14–1) donde n es el número de moles. Por tanto, la energía interna de un gas ideal sólo de- pende de la temperatura y del número de moles de gas. Si las moléculas del gas contienen más de un átomo, entonces también deben considerarse las energías de rotación y vibratoria de las moléculas (figura 14-2). La energía interna será más grande a una temperatura dada que para un gas monoató- mico, pero todavía será sólo una función de la temperatura para un gas ideal. La energía interna de los gases reales también depende principalmente de la temperatura, pero cuando los gases reales se desvían del comportamiento de gas ideal, su energía interna depende también un poco de la presión y del volumen (en virtud de la energía potencial atómica). La energía interna de los líquidos y sólidos es bastante complicada, pues inclu- ye energía potencial eléctrica asociada con las fuerzas (o enlaces “químicos”) entre los átomos y moléculas.
Calor específico
Si el calor fluye a un objeto, la temperatura de éste aumenta (si se supone que no hay cambio de fase). Pero, ¿cómo sube la temperatura? Eso depende de varios fac- tores. Ya desde el siglo XVIII, los experimentadores habían reconocido que la canti- dad de calor Q requerida para cambiar la temperatura de un material dado es proporcional a la masa m del material presente y al cambio de temperatura T. Es- ta notable simplicidad en la naturaleza se expresa en la ecuación (14–2) donde c es una cantidad característica del material llamada calor específico. Puesto que c = Qm T, el calor específico está indicado en unidades de JkgC° (la unidad SI adecuada) o kcalkgC°. Para el agua a 15°C y una presión constante de 1 atm, c = 4 103 JkgC° o 1 kcalkgC°, pues, por la definición de caloría y de joule, toma 1 kcal de calor elevar la temperatura de 1 kg de agua en 1 C°. La tabla 14- proporciona los valores de calor específico para otras sustancias a 20°C. Los valores de c dependen en cierta medida de la temperatura (así como ligeramente de la pre- sión), pero, para cambios de temperatura que no son demasiado grandes, c general- mente se puede considerar constante. EJEMPLO 14–2 Cómo el calor transferido depende del calor específico. a) ¿Cuánta entrada de calor se necesita para elevar la temperatura de una barrica vacía de 20 kg, hecha de hierro, desde 10°C hasta 90°C? b) ¿Y cuánta si la barri- ca está llena con 20 kg de agua? PLANTEAMIENTO Se aplica la ecuación 14-2 a los diferentes materiales que par- ticipan en este caso. SOLUCIÓN a) El sistema es la barrica de hierro sola. A partir de la tabla 14-1, se sabe que el calor específico del hierro es 450 JkgC°. El cambio en la tempera- tura es (90°C - 10°C) = 80°C. Por tanto, b) El sistema es la barrica más el agua. El agua sola requeriría o casi 10 veces lo que requiere una masa igual de hierro. El total, para la barrica más el agua, es 720 kJ + 6700 kJ = 7400 kJ. NOTA En b), la barrica de hierro y el agua experimentan el mismo cambio de temperatura, T = 80 C°, pero sus calores específicos son diferentes. Q = mc ¢T = ( 20 kg)( 4186 Jkg C°)( 80 C°) = 6 106 J = 6700 kJ, Q = mc ¢T = ( 20 kg)( 450 Jkg C°)( 80 C°) = 7 105 J = 720 kJ. Q = mc ¢T,
14–
U = 32 nRT, U = 32 NkT G = 12 mO = 32 kT, U = N A 12 mOB. TABLA 14–1 Calores específicos (a 1 atm de presión constante y 20°C, a menos que se indique de otra manera) Calor específico, c Sustancia Aluminio 0 900 Alcohol (etílico) 0 2400 Cobre 0 390 Vidrio 0 840 Hierro o acero 0 450 Plomo 0 130 Mármol 0 860 Mercurio 0 140 Plata 0 230 Madera 0 1700 Agua Hielo 0 2100 Líquido 1 4186 Vapor 0 2010 Cuerpo humano (promedio) 0 3470 Proteína 0 1700 (110°C) (15°C) (–5°C)
( calg C°) Jkg C°
kcalkg C°
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2763 | https://aleph0.substack.com/p/an-exciting-result-in-the-world-of | An exciting result in the world of elliptic curves!
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An exciting result in the world of elliptic curves!
A recent finding of an elliptic curve with rank at least 29!
Aleph 0
Aug 31, 2024
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A quick question for you!
Are you struggling with career stagnation and job insecurity due to insufficient advanced math skills?
If this is you, I’d love to ask you few questions about your experience. This is in preparation for a program I’m building to help working professionals in math-heavy careers, and it’s purely for my own research.
If you're open to chatting, please fill out this form here. Thanks in advance!
A recent breakthrough about elliptic curves
Now let’s get into the math.
Mathematicians Elkies and Klagsbrun have recently announced the following result:
There exists an elliptic curve over the rationals with rank at least 29.
This breaks a previous record which was established by Elkies in 2006 of an elliptic curve with rank at least 28.
In this issue, I’d like to explain what this result is.
What are elliptic curves?
An elliptic curve is a curves of the following form:
y 2=x 3+A x+B
where A and Bare constants. The fundamental question that mathematicians want to understand about elliptic curves is:
How many rational solutions does this curve have?
For example, below I’ve highlighted the rational solutions to this elliptic curve as red dots. This curve below has infinitely many rational points.
The group law on an elliptic curve
In the 1900s, Poincare figured out a neat trick to generate new rational points on an elliptic curve. Here’s how it goes:
Suppose you have two rational points P and Q on the elliptic curve. Connect them with a line.
That line will intersect the elliptic curve at a third point. Call that third point R.
Now reflect R across the x-axis. Call that point P+Q.
If P and Q are rational points, then the sum P+Q (defined as above) will also be a rational point. This 3-step process is called the chord-and-tangent process. Given two rational points P and Q, this gives a way to combine them to produce a third point, which we denote P+Q.
The rank of an elliptic curve
Given an elliptic curve E, we let E(Q) denote the set of all rational points on E. Here are three elliptic curves, along with the set of rational points:
Example 1:
E:y 2=x 3−2 E(Q)={n⋅(3,5):n∈Z}
Example 2:
E:y 2=x 3−3024 x+46224 E(Q)={n⋅(−24,324)+m⋅(12,108):m,n∈Z}
Example 3:
E:y 2=x 3−x E(Q)is finite!
The rank of an elliptic curve is the “number of copies of the integers” in E(Q). So:
The first curve has rank 1 since there is one copy of the integers in E(Q).
The second curve has rank 2 since there are two copies of the integers in E(Q).
The last curve has rank 0 since E(Q) is finite.
How big can the rank get?
Here is a recent chronology of “rank records” of elliptic curves over the rationals.
In 1938, Billing produced an elliptic curve of rank at least 3.
In 1982, Mestre produced an elliptic curve of rank at least 12.
In 1993, Nagao produced an elliptic curve of rank at least 20.
In 2006, Elkies produced an elliptic curve of rank at least 28.
(This is just a sample of the rank results. If you’re curious, you can see the full table here.)
And now — 18 years later — Elkies and Klagsbrun have broken this record after 18-years, producing an elliptic curve of rank at least 29. They’ve explicitly written down the equation for the curve and produced 29 independent points of infinite order.
It is a wide open question whether there are elliptic curves over the rationals with arbitrarily large rank.
Problem of the Week
Here is this week’s challenge problem:
Prove that if you add up the reciprocals of a sequence of consecutive positive integers, the numerator of the sum (in lowest terms) will always be odd. For example,
1 7+1 8+1 9=191 504.
If you have a solution to this problem,submit it herefor a chance to be featured in the next issue of this newsletter.
Thanks for reading and happy learning! Until next time,
Adithya
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Business Without Borders
Aug 31, 2024
Very nice newsletter I really enjoyed this one. I need to stay more up yo date on math news and breakthroughs
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VK
Sep 2, 2024
I think there is an error in the problem of the week. The fractions should be added, not multiplied:
1/7+1/8+1/9 = 191/504
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2764 | https://math.stackexchange.com/questions/2334235/prove-inequality-1-1-tk-le-exp-k-t | exponential function - Prove Inequality $(1 - 1/t)^k \le \exp(-k/t)$ - Mathematics Stack Exchange
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Prove Inequality (1−1/t)k≤exp(−k/t)(1−1/t)k≤exp(−k/t)
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I have seen the inequality bellow in an online lecture video. I cannot see how it holds for any k≥0 k≥0.
Inequality:
(1−1 t)k≤exp(−k t)(1−1 t)k≤exp(−k t)
where t t is a positive constant and k>0 k>0.
Is this related to some well-known property?
inequality
exponential-function
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edited Jun 23, 2017 at 23:49
Sahiba Arora
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asked Jun 23, 2017 at 23:45
André GomesAndré Gomes
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For k k even and t<1 2 t<1 2 it does not hold.user228113 –user228113 2017-06-23 23:51:17 +00:00 Commented Jun 23, 2017 at 23:51
Hint: Look at the Taylor expansion of exp(−1 t)exp(−1 t).user392395 –user392395 2017-06-24 00:51:19 +00:00 Commented Jun 24, 2017 at 0:51
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This has to do with the definition of the number e e.
e=lim n→∞(1+1 n)n e=lim n→∞(1+1 n)n
More generally,
e x=lim n→∞(1+x n)n e x=lim n→∞(1+x n)n
for any real number x x. Moreover, the expression (1+x/n)n(1+x/n)n is increasing in n n. Hence,
1−1 t≤(1+−1/t n)n≤e−1/t 1−1 t≤(1+−1/t n)n≤e−1/t
Raising both sides to the k k th power yields the result.
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answered Jun 24, 2017 at 0:02
Jonathan DavidsonJonathan Davidson
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One has 1−1/t≤e−1/t 1−1/t≤e−1/t because e−x e−x is convex, so it lies above its tangent line. Raise both sides to the power k k.
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answered Jun 23, 2017 at 23:47
ChappersChappers
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But where is the point that the curve kiss the line ?Tuong Nguyen Minh –Tuong Nguyen Minh 2024-05-24 10:27:40 +00:00 Commented May 24, 2024 at 10:27
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2765 | https://www.ahajournals.org/doi/pdf/10.1161/CIRCEP.108.789081 | Basic Science for the Clinical Electrophysiologist Cardiac Ion Channels
Augustus O. Grant, MB, ChB, PhD T he analysis of the molecular basis of the inherited cardiac arrhythmias has been the driving force behind the iden-tification of the ion channels that generate the action poten-tial. The genes encoding all the major ion channels have cloned and sequenced. The studies have revealed greater complexity than heretofore imagined. Many ion channels function as part of macromolecular complexes in which many components are assembled at specific sites within the mem-brane. This review describes the generation of the normal cardiac action potential. The properties of the major ionic currents are the examined in detail. Special emphasis is placed on the functional consequences of arrhythmia-associated ion channel mutations. The review concludes with a glimpse of the directions in which this new electrophysiol-ogy may lead.
The Cardiac Action Potential
The normal sequence and synchronous contraction of the atria and ventricles require the rapid activation of groups of cardiac cells. An activation mechanism must enable rapid changes in heart rate and also respond to the changes in autonomic tone. The propagating cardiac action potential fulfils these roles. Figure 1 illustrates the 5 phases of the normal action potential: 1. Phase 4, or the resting potential, is stable at 90 mV in normal working myocardial cells. 2. Phase 0 is the phase of rapid depolarization. The membrane potential shifts into positive voltage range. This phase is central to rapid propagation of the cardiac impulse (conduction velocity,
1 m/s). 3. Phase 1 is a phase of rapid repolarization. This phase sets the potential for the next phase of the action potential. 4. Phase 2, a plateau phase, is the longest phase. It is unique among excitable cells and marks the phase of calcium entry into the cell. 5. Phase 3 is the phase of rapid repolarization that restores the membrane potential to its resting value. 1The action potentials of pacemaker cells in the sinoatrial (SA) and atrioventricular (AV) nodes are significantly differ-ent from those in working myocardium. The membrane potential at the onset of phase 4 is more depolarized ( 50 to 65 mV), undergoes slow diastolic depolarization, and gradually merges into phase 0. The rate of depolarization in phase 0 is much slower than that in the working myocardial cells and results in slow propagation of the cardiac impulse in the nodal regions (
0.1 to 0.2 m/s). Cells in the His-Purkinje system may also show phase 4 depolarization under special circumstances. The characteristics of the action po-tential change across the myocardial wall from endocardium, midmyocardium, to epicardium. Epicardial cells have aprominent phase 1 and the shortest action potential. The action potential duration is longest in the midmyocardial region. 2 The average duration of the ventricular action poten-tial duration is reflected in the QT interval on the ECG. Factors that prolong the action potential duration (eg, a decrease in outward K currents or an increase in inward late Na current) prolong the action potential duration and the QT interval on the ECG. The QT interval of males and females is equal during early childhood. However, at puberty the inter-val of males shortens. 3 Studies have focused on the longer QT interval of females and the possible reduction in K
channel function. However, a definitive conclusion has not been made.
General Properties of Ion Channels
The generation of the action potential and the regional differences that are observed throughout the heart are the result of the selective permeability of ion channels distributed on the cell membrane. The ion channels reduce the activation energy required for ion movement across the lipophilic cell membrane. During the action potential, the permeability of ion channels changes and each ion, eg, X, moves passively down its electro-chemical gradients ( V[V mVx,] where Vm is the membrane potential and V x the reversal potential of ion X) to change the membrane potential of the cell. The electrochemical gradient determines whether an ion moves into the cell (depolarizing current for cations) or out of the cell (repolarizing current for cations). Homeostasis of the intracellular ion concentrations is maintained by active and coupled transport processes that are linked directly or indi-rectly to ATP hydrolysis. Ion channels have 2 fundamental properties, ion perme-ation and gating. 4 Ion permeation describes the movement through the open channel. The selective permeability of ion channels to specific ions is a basis of classification of ion channels (eg, Na , K , and Ca 2 channels). Size, valency, and hydration energy are important determinants of selectiv-ity. The selectivity ratio of the biologically important alkali cations is high. For example, the Na :K selectivity of From the Cardiovascular Division, Department of Medicine, Duke University Medical Center, Durham, NC. Correspondence to Augustus O. Grant, Box 3504, Cardiovascular Division, Department of Medicine, Duke University Medical Center, Durham, NC 27710. E-mail grant007@mc.duke.edu
(Circ Arrhythmia Electrophysiol . 2009;2:185-194.)
© 2009 American Heart Association, Inc.
Circ Arrhythmia Electrophysiol is available at DOI: 10.1161/CIRCEP.108.789081
185 Downloaded from by on September 29, 2025 sodium channels is 10:1. Ion channels do not function as simple fluid-filled pores, but provide multiple binding sites for ions as they traverse the membrane. Ions become dehy-drated as they cross the membrane as ion-binding site interaction is favored over ion–water interaction. Like an enzyme–substrate interaction, the binding of the permeating ion is saturable. Most ion channels are singly occupied during permeation; certain K channels may be multiply occupied. The equivalent circuit model of an ion channel is that of a resistor. The electrochemical potential, V is the driving force for ion movement across the cell membrane. Simple resistors have a linear relationship between V and current I (Ohm’s Law, I V/ R Vg, where g is the channel conduc-tance). Most ion channels have a nonlinear current-voltage relationship. For the same absolute value of V, the magni-tude of the current depends on the direction of ion movement into or out of the cells. This property is termed rectification and is an important property of K channels; they pass little outward current at positive (depolarized) potentials. The molecular mechanism of rectification varies with ion channel type. Block by internal Mg and polyvalent cations is the mechanism of the strong inward rectification demonstrated by many K channels. 5Gating is the mechanism of opening and closing of ion channels and is their second major property. Ion channels are also subclassified by their mechanism of gating: voltage-dependent, ligand-dependent, and mechano-sensitive gating. Voltage-gated ion channels change their conductance in response to variations in membrane potential. Voltage-dependent gating is the commonest mechanism of gating observed in ion channels. A majority of ion channels open in response to depolarization. The pacemaker current channel ( Ifchannel) opens in response to membrane hyperpolarization. The steepness of the voltage dependence of opening or activation varies between channels. Sodium channels increase their activation by e-fold (2.73) for 4 mV of depolarization; in contrast, the K channel activation increase e-fold for 5 mV of depolarization. 4Ion channels have 2 mechanism of closure. Certain chan-nels like the Na and Ca 2 channels enters a closed inacti-vated state during maintained depolarization. To regain their ability to open, the channel must undergo a recovery process at hyperpolarized potentials. The inactivated state may also be accessed from the closed state. Inactivation is the basis for refractoriness in cardiac muscle and is fundamental for the prevention of premature re-excitation. The multiple mecha-nisms of inactivation are discussed below. If the membrane potential is abruptly returned to its hyperpolarized (resting) value while the channel is open, it closes by deactivation, a reversal of the normal activation process. These transitions may be summarized by the following state diagram (as proposed for the Na channel 6 ):
The C 3 I transition may occur from multiple closed states. However, because these states are nonconducting, the kinet-ics of transition between them are difficult to resolve with certainty. Ligand-dependent gating is the second major gating mech-anism of cardiac ion channels. The most thoroughly studied channel of this class is the acetylcholine (Ach)-activated K
channel. Acetylcholine binds to the M-2 muscarinic receptor and activates a G protein–signaling pathway, culminating in the release of the subunits G
i and G
. The G
subunit activates an inward-rectifying K channel, IKAch that abbre-viates the action potential and decreases the slope of diastolic depolarization in pacemaker cells. IKAch channels are most abundant in the atria and the SA and atrioventricular nodes.
IKAch activation is a part of the mechanism of the vagal control of the heart. The ATP-sensitive K channel, also termed the ADP-activated K channel, is a ligand-gated channel distributed abundantly in all regions of the heart. The open probability of this channel is proportional to the [ADP]/ [ATP] ratio. This channel couples the shape of the action potential to the metabolic state of the cell. Energy depletion
Figure 1. Membrane currents that generate the a normal action potential. Resting (4), upstroke (0), early repolarization (1), pla-teau (2), and final repolarization are the 5 phases of the action potential. A decline of potential at the end of phase 3 in pace-maker cells, such as the sinus node, is shown as a broken line. The inward currents, INa ,ICa , and If, are shown in yellow boxes; the sodium-calcium exchanger (NCX) is also shown in yellow. It is electrogenic and may generate inward or outward current.
IKAch ,IK1 ,Ito ,IKur ,IKr , and IKs are shown in gray boxes. The action potential duration (APD) is approximately 200 ms. Repro-duced with permission from Stanley and Carlsson. 44
186 Circ Arrhythmia Electrophysiol April 2009 Downloaded from by on September 29, 2025 during ischemia increases the [ADP]/[ATP] ratio, activates
IK ATP , and abbreviates the action potential. The abbreviated action potential results in less force generation and may be cardioprotective. This channel also plays a central role in ischemic preconditioning. The mechanosensitive or stretch-activated channels are the least studied. They belong to a class of ion channels that can transduce a physical input such as stretch into an electric signal through a change in channel conductance. Acute cardiac dilatation is a well-recognized cause of cardiac arrhythmias. Stretch-activated channel are central to the mechanism of these arrhythmias. Blunt chest wall impact at appropriately timed portions of the cardiac cycle may also result in PVCs or ventricular fibrillation (the VF of commotio cordis). The channels that transduce the impact into an electric event are unknown. The major ion channels that shape the action potential have been cloned and sequenced. Table 1 lists the clones of the primary
-subunits of the major ion channels. Over the past 2 decades, the focus of research has been the relationship between channel structure and function, including the molec-ular underpinnings of the permeation and gating processes. Recent studies have focused on molecular suprastructures of which ion channels are a part. 7,8 The channels are not randomly distributed in the membrane, but tend to cluster at the intercalated disc in association with modulatory subunits. The sodium channel has a binding site for the structural protein ankryin and mutations that affects its binding site result in LQTS or Brugada syndrome. 9
Sodium Channels
Sodium channels are the arch-type of voltage-gated ion channels. 10 The human cardiac sodium channel hNa V 1.5 is a member of the family of voltage-gated sodium channels (hNa V 1 to 9). The channel consists of a primary
and multiple secondary
-subunits. When studied in a mamma-lian expression system, the
-subunit of hNa V 1.5 is sufficient to generate sodium current with features characteristic of the current in native cells. The
1-subunit increases the level of expression and alters the gating of the neuronal sodium channel. An analogous role of the
1 subunit for the cardiac sodium channel has not been established. The sodium channel consists of 4 homologous domains, DI – DIV 11,12 arranged in a 4-fold circular symmetry to form the channel (Figure 2). Each domain consists of 6 membrane-spanning segments, S1 through S6. The membrane-spanning segments are joined by alternating intra- and extracellular loops. The loops between S5 and S6 of each domain termed the P loops curve back into the membrane to form the pore. Each S4 segment has a positively charged amino acid at every third or fourth position and acts as the sensor of the transmem-brane voltage. The movement of these charges across the membrane during channel gating generates small currents that can be recorded at high resolution. Transmission of the voltage sensor transition to S-5 has been suggested as the critical element of channel gating. The highly potent neurotoxin tetrodotoxin (TTX) and the systematic mutation of residues in the loop have enabled the tentative identification of the amino acid residues that are critical for ion permeation; these residues include aspartate, glutamate, lysine, and alanine (D, E, K, and A) contributed by D1 through D4, respectively. The lysine (K) in domain III is critical for Na:K selectivity. Mutation of multiple residues in D4 renders the channel noncation selective. Each sodium channel opens very briefly ( 1 ms) during more than 99% of depolarizations. 13,14 The channel occasion-
Table 1. Membrane Currents That Generate the Action Potential
Current Description AP Phase Activation Mechanism Clone Gene
-subunit of action potential inward current channels
INa Sodium current Phase 0 Voltage, depolarization Na v1.5 SCN5A
ICa,L Calcium current, L-type Phase 2 Voltage, depolarization Ca v1.2 CACNA1C
ICa,T Calcium current, T-type Phase 2 Voltage, depolarization Ca v3.1/3.2 CACNA1G
-subunit of action potential outward (K ) current channels Ito,f Transient outward current, fast Phase 1 Voltage, depolarization KV 4.2/4.3 KCND2/3 Ito,s Transient outward current, slow Phase 1 Voltage, depolarization KV 1.4/1.7/3.4 KCNA4 KCNA7 KCNC4
IKur Delayed rectifier, ultrarapid Phase 1 Voltage, depolarization KV 1.5/3.1 KCNA5 KCNC1
IKr Delayed rectifier, fast Phase 3 Voltage, depolarization HERG KCNH2
IKs Delayed rectifier, slow Phase 3 Voltage, depolarization KVLQT1 KCNQ1
IK1 Inward rectifier Phase 3 & 4 Voltage, depolarization Kir 2.1/2.2 KCNJ2/12
IKATP ADP activated K current Phase 1 & 2 ADP /ATP 1Kir 6.2 (SURA) KCNJ11
IKAch Muscarinic-gated K current Phase 4 Acetylcholine Kir 3.1/3.4 KCNJ3/5
IKP Background current All Phases Metabolism, stretch TWK-1/2 KCNK1/6 TASK-1 KCNK3 TRAAK KCNK4
IFPacemaker current Phase 4 Voltage, hyperpolarization HCN2/4 HCN2/4
Grant Cardiac Ion Channels 187 Downloaded from by on September 29, 2025 ally shows alternative gating modes consisting of isolated brief openings occurring after variable and prolonged laten-cies and bursts of openings during which the channel opens repetitively for hundreds of milliseconds. The isolated brief openings are the result of the occasional return from the inactivated state. The bursts of openings are the result of occasional failure of inactivation. 13,15 Sodium channel muta-tions that favor these slow gating modes are the basis of a subgroup of the long QT syndromes (LQT3). 16 Sodium channel inactivation is a multifaceted process that may occur in the time frame of milliseconds, seconds, or tens of seconds, depending on the duration of the antecedent depolarization. 17 In response to depolarization lasting tens of milliseconds, the process is fast. Intermediate and slow inactivation develops over hundreds of milliseconds, for example during the course of the normal action potential and in response to trains of action potentials. The fraction of channels available for opening (1-the inactivated fraction), denoted by h , varies from 1 at 90 mV to zero at 40 mV. The structural basis of fast sodium channel inactivation resides in the interdomain linker between DIII and DIV (ID111/IV). The primary amino acid sequence of this region is highly conserved between species and sodium channel subtypes. The tertiary structure of the region has been resolved by NMR spectroscopy. 18 The putative form is that of a tilting disk that folds into the membrane to occlude the pore. The amino acid triplet isoleucine, phenylalanine, methionine (IFM) is crucial for inactivation; the mutation IFM 3 QQQ abolishes inactivation. 19 The receptor site to which the triplet binds has not been identified. The carboxyl terminus also plays an important role in sodium channel inactivation. The cardiac sodium channel has consensus sites for phos-phorylation by protein kinase (PKA), protein kinase C (PKC), and Ca-calmodulin kinase. Data on the effects of PKA on the
INa are controversial, with some studies reporting a decrease in current whereas others report an increase. 20 –22 Phosphor-ylation of the channel by PKC results in a decrease in INa .Modulation the Na channel by glycerol-3phosphate dehy-drogenase like1 kinase was recently established by the identification of a kindred with Brugada syndrome and a mutation in the enzyme. 23 In vitro expression showed that enzyme action is associated with a decrease in INa .Mutations in cardiac sodium channel gene SCN5A have been associated with LQTS, Brugada syndrome, primary cardiac conduction system disease (PCCP), and dilated car-diomyopathy (Table 4). 24 The long QT syndrome is the result of defects in inactivation that enhance the late component of sodium current. The late component of current is more sensitive to block by class 1 antiarrhythmic drugs than the peak current. Mexiletine and flecainide decrease the late component of sodium current and restore the QT interval toward normal. 25,26 They have been used to treat patients with LQT3, particularly in the neonatal period and in children when ICD implantation may prove technically challenging. Sodium channel mutations have been described in 20% of patients with Brugada syndrome. 27 The mutations reduce the Na current as a result of synthesis of nonfunctional proteins, failure of the protein to be targeted to the cell membrane or accelerated inactivation of the channel. As a subgroup, the patients with Na channel mutations that produce Brugada syndrome have H-V interval prolongation at electrophysiol-ogy study. The mechanism of ST segment elevation and T wave inversion in the syndrome is controversial. One group view the syndrome as primarily a repolarization abnormality 28 ;others view the Na channel variant as a conduction defect. 29 Slow conduction from endocardium to epicardium results in delayed epicardial activation. The sequence of transmural repolarization is reversed, resulting in the ST-T wave changes. The mutations associated with primary cardiac conduction disease also reduce the Na current. 30 The clinical syndromes include sinus node dysfunction, atrial standstill, AV block, and fascicular (infra-Hisian) block. Overlap syn-dromes of LQT3, Brugada syndrome, and PCCD may occur in the same kindred or individual. 31 The mechanisms by
Figure 2. Putative transmembrane organization of the sodium channel. The channel consists of 4 homologous domains, D1 through D1V. The amino and carboxyl termini are intracellular. The positive charges ( ) on the fourth transmembrane segment are evident, as are the extended extracellular loops between S5 and S6 of each domain. Examples of loci at which functionally characterized muta-tions cause Brugada syndrome, LQT3, the overlap syndrome (Brugada syndrome/LQT3), and isolated conduction system disease are also shown. Reproduced with permission from Herbert and Chahine. 46
188 Circ Arrhythmia Electrophysiol April 2009 Downloaded from by on September 29, 2025 which Na defects result in dilated cardiomyopathy are not well understood. 32 Long standing conduction delay and asyn-chronous contraction may be contributory. The cardiac sodium channel is the substrate for the action of class 1 antiarrhythmic drugs (Table 3). Open and inacti-vated channels are more susceptible to block than resting channels. The differential block may be the result of a difference in binding affinity or state-dependent access to the binding site. 33,34 Binding of antiarrhythmic drug occurs pri-marily during the action potential. This block is dissipated in the interval between action potentials. Because a fast heart rate is associated with abbreviation of the diastolic period and insufficient time for recovery, block accumulates (ie, it is use-dependent). Class 1 antiarrhythmic drugs may be classi-fied according to the kinetics of unbinding, with various drugs showing fast, intermediate, or slow unbinding kinetics. 35
Calcium Channels
Calcium ions are the principal intracellular signaling ions. They regulate excitation– contraction coupling, secretion, and the activity of many enzymes and ion channels. [Ca 2]i is highly regulated despite its marked fluctuation between systole and diastole. Calcium channels are the principal portal of entry of calcium into the cells; a system of intracellular storage sites, and transporters such as the sodium-calcium exchanger (NCX), also play important roles in [Ca 2]iregulation. In cardiac muscle, 2 types of Ca 2 channels, the L- (low threshold type) and T-type (transient-type), transport Ca 2 into the cells. The L-type channel is found in all cardiac cell types. The T-type channel is found principally in pace-maker, atrial, and Purkinje cells. The unqualified descriptor Ca 2 channel refers to the L-type channel. Table 2 contrasts the properties of the two types of channels. A combination of as many as 5 subunits,
1 ,
2,
,
, and , unite to form the channel in its native state. The
1c subunit, Ca v1.2, is the cardiac-specific subunit. The
subunit increases channel expression 10-fold and accelerates the activation and inactivation kinetics. Ca 2 channels have a similar structure to the sodium channel: 4 homologous domains each consisting of 6 membrane-spanning segments. The P-loop of each domain contributes a glutamate residue (E) to the pore structure. These residues (EEEE) are critical for calcium selectivity; the channel can be converted from a Ca 2-sensitive channel to one with high monovalent cation sensitivity by mutating a glutamate residue. 36 Several molec-ular mechanisms contribute to a complex system of inactiva-tion. Membrane depolarization decreases the fraction (d ) of channels available for opening; d varies from 1 at 45 mV to 0 at zero mV. The carboxyl terminus has multiple Ca 2 binding sites and Ca-calmodulin– dependent kinase activity. Ca 2 in the immediate vicinity of the channel and phosphorylation also play roles in the inactivation of the channel. Reuptake of Ca 2 by the sarcoplasmic reticulum during prolonged depolarization can result in the recovery from Ca 2-dependent inactivation and enable secondary de-polarization. This may be the basis for the early afterdepo-larizations, EADs that trigger polymorphic VT in LQTS. The overall kinetics of the Ca channel is important in controlling contractility in response to various patterns of stimulation. At low (depolarized) membrane potentials, recovery of ICa from inactivation between action potentials is slow; ICa declines in response to repetitive stimulation and a negative staircase of contractility is observed. At normal resting potentials, recov-ery of ICa from inactivation is fast, and ICa may increase progressively during repetitive stimulation. This positive staircase or rate-dependent potentiation of contractility is Ca 2-dependent. It is the result of enhanced loading of the sarcoplasmic reticulum and may be facilitated by calmodulin kinase II– dependent phosphorylation. Timothy syndrome is a multi-system disease with LQTS, cognitive abnormalities, immune deficiency, hypoglycemia, and syndactyly that is the result of mutations of Ca V 1.2. 37 The mutation of glycine to arginine converts a neighboring serine to a consensus site for phosphorylation by calmodulin kinase. The phosphorylation of this site promotes a slow gating mode of the calcium channel, increasing Ca 2 entry and resulting in cytotoxicity. 38 A sudden death syndrome that combines the features of Brugada syndrome, including the characteristic ECG pattern, and a short QT interval has been described recently. 39 The syndrome results from a loss of function of the
1- or
2b subunit of the L-type Ca 2 channel. The Ca 2 channel is the target for the interaction with class IV antiarrhythmic drugs. The principal class IV drugs are the phenylalkylamine, verapamil and the benzothiazepine, dilti-azem. Both drugs block open and inactivated Ca 2 channels;
Table 2. A Comparison of the L-Type and T-Type Ca 2 Channels
L-Type T-Type Activation Range Low Em ( 30 mV) High Em ( 60 mV) Inactivation Range Low Em ( 40 mV) Hyperpolarized Voltage dependence Slow Fast
Ca 2idependent Yes No Pharmacologic sensitivity Dihydropyridnes Yes No Cd High Low NiLow High Isoproternol Yes No
Table 3. Classification of Antiarrhythmic Drug Actions
Class Unbinding Kinetics Effect on APD Example 1A Intermediate APD 1Procainamide 1B Fast APD 3Lidocaine 1C Slow APD 2,3Flecainide II (
-blocker) APD Metoprolol III (K channel blocker) APD 1Dofetelide, Ibutilide IV (Ca channel blocker) Slow APD 1,3Verapamil
Grant Cardiac Ion Channels 189 Downloaded from by on September 29, 2025 they cause use-dependent block of conduction in cells with Ca 2-dependent action potentials such those in the SA- and AV nodes and slow the sinus node rate. However, the hypotensive effects of verapamil may cause an increase in sympathetic tone and increase the heart rate. A third class of Ca 2 channel blockers, the dihydropyridines, block open Ca 2 channels. However, the kinetics of recovery from block is sufficiently fast that they produce no significant cardiac effect but effectively block the smooth muscle Ca 2 channel because of its low resting potential.
Potassium Channels
Cardiac K channels fall into 3 broad categories: Voltage-gated ( Ito , IKur , IKr , and IKs ), inward rectifier channels ( IK1 ,
IKAch , and IKATP ), and the background K currents (TASK-1, TWIK-1/2). It is the variation in the level of expression of these channels that account for regional differences of the action potential configuration in the atria, ventricles, and across the myocardial wall (endocardium, midmyocardium, and epicardium). K channels are also highly regulated and are the basis for the change in action potential configuration in response to variation in heart rate. Voltage-gated K channels consist of principal
-subunits and multiple
-subunits. The channel functional units also include the complementary proteins K V -channel associated protein, KChAP, and the K V channel interacting protein, KChIP. The major subfamilies of
-subunits include K V N.x (n 1 to 4), the HERG channel (gene KCNH2 ), and KvLQT1 (gene KCNQ1 ). They are important in generating outward current in the heart. Members of the K V N.x subfamily may coassemble to form hetero-multimers through conserved amino-terminal domains. In contrast, members of the HERG and K v LQT1 subfamilies assemble as homotetramers. The
-subunits that coassemble to form the various types of K
channels and their role in the generation of the action potential are summarized in Table 1. Most
-subunits have been cloned and sequenced. They have oxio-reductase activ-ity. The
-subunits can generate voltage dependent K
current when expressed in heterogonous systems. However, the accessory subunits are required to recapitulate the K
currents seen in native cells. KChAP ( KCHAP ) and KChIP (KCNIP2 ) may increase channel activity independent of transcription and alter channel kinetics. The structure of voltage-gated K channels is similar to 1 of the 4 domains of voltage-gated Na and Ca 2 . The amino acid sequence glycine-tyrosine-glycine GYG is the sequence requirement for K selectivity. The transient outward current is composed of a K current
Ito1 and a Ca 2-activated chloride current, Ito2 . The former has fast and slow components, Ito ,f and Itos . I tof is the principal subtype expressed in human atrium; Ito ,f and Itos are expressed in the ventricle. Myocardial regions with relatively short action potentials such as the epicardium, right ventricle, and the septum have higher levels of Ito expression. Compared to other voltage-gated K channels, activation of Ito is fast (activation time constant 10 ms). The rate of inactivation is variable and highly voltage-dependent.
-adrenergic stimula-tion reduces Ito in human myocytes through PKA-dependent phosphorylation. Chronic
-adrenergic stimulation and an-giotensin II also reduces channel expression. The influence of a reduction of Ito on the action potential duration varies with species; in rodents, a reduction in Ito prolongs the action potential duration. In large mammals, a reduction in Ito shifts the plateau to more positive potentials increasing the activa-tion of the delayed rectifier and promoting faster repolariza-tion. In a rodent model of hypothyroidism the action potential prolongation is associated with a reduction of Ito . The current is also reduced in human heart failure but is associated with a prolongation of the action potential duration. Because the level of the plateau is set by Ito , modulators that decrease Ito shift the plateau into the positive range of potentials. This decreases the electro-chemical driving force for Ca 2 and hence ICa .The delayed rectifier K currents IKur , IKr , and IKs are slowly activating outward currents that play major roles in the control of repolarization. The deactivation of these channels is sufficiently slow that they contribute outward current throughout phase 3 repolarization. IKur is highly expressed in atrial myocytes and is a basis for the much shorter duration of the action potential in the atrium. IKr is differentially ex-pressed, with high levels in the left atrium and ventricular endocardium. IKs is expressed in all cell types, but is reduced in midmyocardial myocytes. These cells have the longest action potential duration across the myocardial wall. The
-subunits that make up the delayed rectifier currents are summarized in Table 1.
-subunits are associated with IKr and
IKs . MinK-Related Peptide-1 (MiRP-1) and MinK are the most thoroughly studied. MiRP-1 and MinK are single-membrane spanning peptides with extracellular amino ter-mini. The
-subunits are nonconducting but regulate
-subunit function, including gating, response to sympathetic stimulation, and drugs.
-adrenergic stimulation regulates IKr through activation of protein kinase A and elevation of c-AMP. The former effect is inhibitory; the latter is stimula-tory through binding to the cyclic nucleotide binding domain of the channel.
-adrenergic stimulation is inhibitory.
-adrenergic stimulation increases IKs through PKA-dependent phosphorylation. This action involves a complex of PKA, protein phosphatase1, and the adaptor protein yotaio. 7 Ion channel mutations that disrupt the function of the complex result in the action potential prolongation of LQT1.
-adrenergic blockers indirectly regulate this complex and are important therapeutic options in LQT1. The inward rectifier channel current Ik1 sets the resting membrane potential in atrial and ventricular cells. Channel expression is much higher in the ventricle and protects the ventricular cell from pacemaker activity. The strong inward rectification of the Ik1 limits the outward current during phases 0, 1, and 2 of the action potential. This limits the outward current during the positive phase of the action potential and confers energetic efficiency in the generation of the action potential. Because block of the outward current by intracellular Mg and the polyamines is relieved during repolarization, Ik1 makes a significant contribution to phase 3 repolarization. The acetylcholine-activated K channel is a member of the G protein– coupled inward rectifying potassium channels. The channel is highly expressed in the SA and AV nodes, and
190 Circ Arrhythmia Electrophysiol April 2009 Downloaded from by on September 29, 2025 atria, but low in ventricle. Activation of IKAch hyperpolarizes the membrane potential and abbreviates the action potential. Phase 4 depolarization of pacemaker cells is slowed. The channel structure is similar to that of IK1 . The binding of acetylcholine to the M2 muscarinic receptor activates the G protein G i and the release of the subunits G i and G . The dissociated G
subunit binds to the channel and activities it. The binding of adenosine to the P1 receptor also results in the release of G
and activation of the channel. Methylxanthines such as theophylline block the P1 receptor and antagonized the effects of adenosine. Coexpression of the inward rectifier K channel Kir6.x and sulfonylurea receptor yield channels with properties similar to the native IKATP .Mutations of the genes encoding cardiac K channels are the principal causes of arrhythmias that result from abnormal repolarization (Table 4). 40 Mutations of KCNQ1, the gene encoding K vLQT1, and KCNH2, the gene encoding HERG, account for more than 80% of autosomal dominant LQTS (Romano-Ward syndrome). Bilateral neurosensory deafness is a part of the autosomal recessive form (Jervell and Lange-Nielsen syndrome). Mutations of the
-subunit KCNJ2 and the
-subunits of KVLQT1 and HERG are minor causes of LQTS. A majority of these mutations have a dominant negative effect, coassembling with normal sub-units, but impairing their function. Polymorphisms of the genes encoding the K channels may increase susceptibility to drug-induced LQTS. Gain of function mutations of IKr , IKs ,
IK1 cause marked acceleration of repolarization and the short QT syndrome. Mutations of these subunits have also been associated with familial atrial fibrillation. Cardiac K channels are the targets for the action of class III antiarrhythmic drugs. The HERG channel is very suscep-tible to block by a broad range of drugs that are not primarily used to treat cardiac arrhythmias, including antipsychotics, and the macrolide antibiotics. The potent K channel block-ing action of quinidine, procainamide, and disopyramide account for their QT- prolonging action and occasionally torsade de pointes. HERG blockers produce greater blockade
Table 4. The Genetic Basis of Inherited Arrhythmias
Type Gene Protein Functional Alteration LQTS LQT1 KCNQ 1 IKs potassium channel
-subunit Loss of function LQT2 KCNH 2 IKr potassium channel
-subunit Loss of function LQT3 SCN 5 A INa cardiac sodium channel
-subunit Gain of function LQT4 ANK 2 Ankyrin-B Loss of function LQT5 KCNE 1 IKs potassium channel
-subunit Loss of function LQT6 KCNE 2 IKr potassium channel
-subunit Loss of function LQT7 KCNJ 2 IK1 potassium channel subunit Loss of function LQT8 CACNA 1 C Calcium channel
-subunit Gain of function Short QT syndrome S4T1 KCNH 2 IKr potassium channel
-subunit Gain of function S4T2 KCNQ 1 IKs potassium channel
-subunit Gain of function S4T3 KCNJ 2 IK1 potassium channel subunit Gain of function Brugada syndrome BrS1 SCN 5 A INa sodium channel
-subunit Loss of function BrS2 GPD1L G-3PD 1 Altered function BrS3 CACNA1C ICa calcium channel
-subunit Loss of function BrS4 CACNB2b ICa calcium channel
-subunit Loss of function BrS other ANKRYN-B Ankryin-B Altered function Catecholaminergic polymorphic VT RYR 2 Cardiac ryanodine receptor Gain of function CASQ 2 Calsequestrin Gain of function Idiopathic sick sinus syndrome HCN 4 If pacemaker channel subunit Loss of function SCN 5 A INa cardiac sodium channel
-subunit Loss of function Cardiac conduction disease SCN 5 A INa cardiac sodium channel
-subunit Loss of function Familial atrial fibrillation KCNQ 1 IKs potassium channel
-subunit Gain of function KCNE 2 IKr potassium channel
-subunit Gain of function KCNJ 2 IK1 potassium channel subunit Gain of function KCNH 2 IKr potassium channel
-subunit Gain of function
Grant Cardiac Ion Channels 191 Downloaded from by on September 29, 2025 at slow heart rates; block tends to dissipate during the rapid heart rates of a tachycardia, so called reverse-use dependence. Amiodarone is exceptional in that it produces K channel blockade that shows little use dependence. Although antiar-rhythmic drugs have fallen out of favor for the management for ventricular tachycardia, they retain an important role in the prevention of recurrences of atrial fibrillation. The dis-covery of the atrial-specific distribution of IKur has made this channel a target for novel therapies for atrial fibrillation. The drug vernakalant is a IKur /Na channel blocker and is under-going review by the FDA for the acute termination of atrial fibrillation.
Hyperpolarization-Activated Cyclic Nucleotide Gated Channel
Autorhythmicity is one of the most characteristic features of cardiac cells and resides in the pacemaker cells of the specialized conducting system, including the SA and AV nodes, and His-Purkinje system. Pacemaker activity initiates and sustains electric activity of the heart independent of the underlying innervation. Phase 4 diastolic depolarization is characteristic of pacemaker cells. Many ion channels contrib-ute to phase 4 depolarization: the K channel current acti-vated during the preceding action potential, a background Na current, the sodium-calcium exchange, the If channel, and the L- and T-type Ca 2 channels. However, the If current channel is unique to this process. Unlike other voltage-gated channels, If is activated by hyperpolarization negative to 40 mV. The channel is not very selective for Na over K and has a reversal potential (Er) of 10 to 20 mV. Therefore, it carries inward current throughout the range of pacemaker potential. The phase 4 depolarization reduces the membrane to the threshold for the regenerative activation of ICa,T and ICaL .The genes encoding If channels have been cloned and sequenced in the past decade. If channels (hyperpolarization-activated cyclic nucleotide gated [HCN]I–HCN4) are mem-bers of a family of cyclic nucleotide activated voltage-gated channels. Although members of this family of channels are expressed in heart and brain, HCN2 and HCN4 are expressed in the heart. Expression is both developmentally and region-ally regulated. Neonatal cardiac ventricular myocytes that show pacemaker activity predominantly expressed HCN2. Expression of HCN2 declines in adulthood. HCN4 is the subtype primarily expressed in the sinus node, AV node, and ventricular conducting system. Knockout of the HCN4 gene is embryonic lethal. Channels are formed by the assembly of 4
-subunits, each with a structure analogous to that of the voltage gated K channels. Binding of cAMP to this domain shift the voltage dependence of If activation to more depo-larized potentials and increase the rate of pacemaker dis-charge. Protons shift the activation of If to more hyperpolar-ized potentials and slow pacemaker activity. The If channel is the target for a new class of bradycardic agents, eg, ivabra-dine. They have the advantage over
blockers in that they slow the heart rate without the disadvantage of negative inotropy or hypotension. They have proved effective in the management of patients with chronic stable angina. Isolated reports of mutations in the HCN4 gene have appeared recently. One kindred had idiopathic sinus brady-cardia and chronotrophic incompetence. 41 Severe bradycar-dia, QT prolongation, and torsade de pointes have been described in another family. If expression is upregulated in cardiac hypertrophy and congestive heart failure. This re-sponse may contribute to the arrhythmias observed in these disease states.
Gap Junction Channels
That cardiac tissues are made up of discrete cells was a seminal observation in cell biology. The rapid conduction of the cardiac impulse required the presence of low resistance connections between cardiac cells. 42 The propagation veloc-ity in a uniform cable is inversely related to the intercellular (internal) resistance; low internal resistance favors rapid conduction. Gap junction channels form the low resistance connections between cardiac cells. 43 In the young, gap junc-tions are widely distributed over the surface of cells. Cells become elongated and arranged in parallel bundles in the adult heart, and gap junctions become localized principally at the ends of cells. The density of gap junctions is lower at the lateral margins of cells, particular conduction system myo-cytes. This nonuniform distribution of gap junctions changes the pattern and safety of conduction. Propagation occurs rapidly through the cytoplasm of cells and slows at the intercellular junctions, ie, conduction is discontinuous. Con-duction is faster in the longitudinal direction, with velocity ratios of 3 to 8 for longitudinal direction compared with the transverse direction. The higher density of gap junction in the longitudinal direction results in unloading of the excitatory current during propagation. Longitudinal conduction is more likely to fail. Conduction is more sustained in the transverse direction and can occur at very slow rates. Each of a pair of neighboring cells contributes hemi-channels or connexons to the junction. The connexons are made up of 6 connexins. These are the fundamental building blocks of the junction. Three types of connexins are ex-pressed in heart and are defined on the basis of their molecular weight: connexin 40, connexin 43, and connexin 45 (molecular weights 40, 43, and 45 kDa). Connexin 43 is the principal connexin expressed in the heart. Regional differences in the type and distribution of connexins are important determinants of the passive spread of excitation over boundaries such as those of the SA and AV nodes. The connexins may assemble as homomultimeric or heteromulti-meric channels. The conductance of gap junctional channels is regulated in health and disease states. Protein kinase A phosphorylation and low pH decrease junctional conductance. The latter may be an important contributing factor to slow conduction during acute ischemia. With aging, the density of gap junctions declines and cells become separated by connective tissue septa. This favors the occurrence of slow conduction, frac-tionated extracellular electrograms, and block. Gap junctions are the targets for a new class of antiarrhyth-mic drugs. 44 An antiarrhythmic peptide (AAP) inhibits ische-mia-induced conduction slowing. An analog rotigaptide pre-vents ischemia-induced ventricular tachycardia.
192 Circ Arrhythmia Electrophysiol April 2009 Downloaded from by on September 29, 2025 Future Directions
The cloning and sequencing of the ion channel genes that regulate the action potential hold the promise that these genes could be manipulated to treat arrhythmias. Proof of principle has been established. The initial problem approached is the control of the ventricular response in atrial fibrillation. -adrenergic blockers are the most widely used drugs used to control the ventricular response in atrial fibrillation. Adren-ergic inhibition decreases intracellular [cAMP] and the Ca 2
current. This would slow conduction over the AV node. Donahue and colleagues developed an indirect strategy to decrease sympathetic activity in AV nodal cells. 45 They inserted the inhibitory G protein G
i
into an adenoviral vector. The adenoviral vector-G
i
construct was infused in the AV nodal artery of pigs with atrial fibrillation. G
i
over expression decreased the heart rate in atrial fibrillation by 20% compared to the drug-free state. Persistence of the effect was limited and the delivery of vector would be challenging in the clinical situation. Sick sinus syndrome is the most common cause for permanent pacemaker implantation. A genetic strategy to treat sinus node failure would be attractive. In an earlier contribution to this series, Rosen and colleagues have pro-vided a state of the art review on the genetic approach to the development of biological pacemakers by manipulating the HCN4 gene. 47 The biological pacemakers have relatively slow rates. The initial effort is focused on biological pace-maker that will complement rather than replace the normal sinus node pacemaker.
Disclosures
Dr Grant has received honoraria from Boston Scientific, Medtronic, St Jude Medical, and Sanofi Aventis.
References
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Antzelevitch C, Brugada P, Borggrefe M, Brugada J, Brugada R, Corrado D, Gussak I, LeMarec H, Nademanee K, Riera ARP, Shimizu W, Schulze-Bahr E, Tan H, Wilde A. Brugada Syndrome: Report of the Second Consensus Conference. Heart Rhythm . 2005;2:429 – 440. 28. Hong K, Bjerregaard P, Gussak I, Brugada R. Short QT syndrome and atrial fibrillation caused by mutation in KCNH2. J Cardiovasc Electro-physiol . 2005;16:394 –396. 29. Zhang, Z-S, Tranquillo J, Neplioueve V, Bursac N, Grant A. Sodium channel Kinetic changes that produce Brugada syndrome or progressive cardiac conduction system disease. Am J Physiol Heart Circ Physiol .2007;292:H399 –H407. 30. Schott, J-J, Alshinawi C, Kyndt F. Cardiac conduction defects associate with mutations in SCN5A. Nat Genet . 1999;23:20 –21. 31. Grant AO, Carboni MP, Neplioueva V, Starmer CF, Memmi M, Napolitano C, Priori S. Long QT syndrome, Brugada syndrome, and conduction system disease are linked to a single sodium channel mutation. 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Circ Arrhythmia Electrophysiol. 2008;1:54 – 61. KEY WORDS : action potentials electrocardiography electrophysiology
genetics ion channels
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2766 | https://courses.lumenlearning.com/dcccd-collegealgebracorequisite/chapter/exponential-equations-with-like-bases/ | Exponential Equations | Math 1314
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Math 1314
Unit 4
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Exponential Equations
Learning Outcomes
Solve an exponential equation with a common base.
Rewrite an exponential equation so all terms have a common base then solve.
Recognize when an exponential equation does not have a solution.
Use logarithms to solve exponential equations.
Exponential Equations
The first technique we will introduce for solving exponential equations involves two functions with like bases. The one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where b>0,b≠1 b>0,b≠1, b S=b T b S=b T if and only if S= T.
In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
For example, consider the equation 3 4 x−7=3 2 x 3 3 4 x−7=3 2 x 3. To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x:
3 4 x−7=3 2 x 3 3 4 x−7=3 2 x 3 1 Rewrite 3 as 3 1.3 4 x−7=3 2 x−1 Use the division property of exponents.4 x−7=2 x−1 Apply the one-to-one property of exponents.2 x=6 Subtract 2 x and add 7 to both sides.x=3 Divide by 3.3 4 x−7=3 2 x 3 3 4 x−7=3 2 x 3 1 Rewrite 3 as 3 1.3 4 x−7=3 2 x−1 Use the division property of exponents.4 x−7=2 x−1 Apply the one-to-one property of exponents.2 x=6 Subtract 2 x and add 7 to both sides.x=3 Divide by 3.
tip for success
Note the use of the exponent rules and the one-to-one property of exponents in the example above. Return to the review section for this module or to Algebra Essentials anytime if you need a refresher.
A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations
For any algebraic expressions S and T, and any positive real number b≠1 b≠1,
b S=b T if and only if S=T b S=b T if and only if S=T
How To: Given an exponential equation Of the form b S=b T b S=b T, where S and T are algebraic expressions with an unknown, solve for the unknown
Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form b S=b T b S=b T.
Use the one-to-one property to set the exponents equal to each other.
Solve the resulting equation, S= T, for the unknown.
Example: Solving an Exponential Equation with a Common Base
Solve 2 x−1=2 2 x−4 2 x−1=2 2 x−4.
Show Solution
2 x−1=2 2 x−4 The common base is 2.x−1=2 x−4 By the one-to-one property the exponents must be equal.x=3 Solve for x.2 x−1=2 2 x−4 The common base is 2.x−1=2 x−4 By the one-to-one property the exponents must be equal.x=3 Solve for x.
Try It
Solve 5 2 x=5 3 x+2 5 2 x=5 3 x+2.
Show Solution
x=–2 x=–2
Rewriting Equations So All Powers Have the Same Base
Sometimes the common base for an exponential equation is not explicitly shown. In these cases we simply rewrite the terms in the equation as powers with a common base and solve using the one-to-one property.
For example, consider the equation 256=4 x−5 256=4 x−5. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents along with the one-to-one property to solve for x:
256=4 x−5 2 8=(2 2)x−5 Rewrite each side as a power with base 2.2 8=2 2 x−10 To take a power of a power, multiply the exponents.8=2 x−10 Apply the one-to-one property of exponents.18=2 x Add 10 to both sides.x=9 Divide by 2.256=4 x−5 2 8=(2 2)x−5 Rewrite each side as a power with base 2.2 8=2 2 x−10 To take a power of a power, multiply the exponents.8=2 x−10 Apply the one-to-one property of exponents.18=2 x Add 10 to both sides.x=9 Divide by 2.
How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it
Rewrite each side in the equation as a power with a common base.
Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form b S=b T b S=b T.
Use the one-to-one property to set the exponents equal to each other.
Solve the resulting equation, S= T, for the unknown.
Example: Solving Equations by Rewriting Them to Have a Common Base
Solve 8 x+2=16 x+1 8 x+2=16 x+1.
Show Solution
8 x+2=16 x+1(2 3)x+2=(2 4)x+1 Write 8 and 16 as powers of 2.2 3 x+6=2 4 x+4 To take a power of a power, multiply the exponents.3 x+6=4 x+4 Use the one-to-one property to set the exponents equal to each other.x=2 Solve for x.8 x+2=16 x+1(2 3)x+2=(2 4)x+1 Write 8 and 16 as powers of 2.2 3 x+6=2 4 x+4 To take a power of a power, multiply the exponents.3 x+6=4 x+4 Use the one-to-one property to set the exponents equal to each other.x=2 Solve for x.
Try It
Solve 5 2 x=25 3 x+2 5 2 x=25 3 x+2.
Show Solution
x=–1 x=–1
Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base
Solve 2 5 x=√2 2 5 x=2.
Show Solution
2 5 x=2 1 2 Write the square root of 2 as a power of 2.5 x=1 2 Use the one-to-one property.x=1 10 Solve for x.2 5 x=2 1 2 Write the square root of 2 as a power of 2.5 x=1 2 Use the one-to-one property.x=1 10 Solve for x.
Try It
Solve 5 x=√5 5 x=5.
Show Solution
x=1 2 x=1 2
Q & A
Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?
No. Recall that the range of an exponential function is always positive. While solving the equation we may obtain an expression that is undefined.
Example: Determine When an Equation has No Solution
Solve 3 x+1=−2 3 x+1=−2.
Show Solution
This equation has no solution. There is no real value of x that will make the equation a true statement because any power of a positive number is positive.
Analysis of the Solution
The figure below shows that the two graphs do not cross so the left side of the equation is never equal to the right side of the equation. Thus the equation has no solution.
Try It
Solve 2 x=−100 2 x=−100.
Show Solution
The equation has no solution.
Using Logarithms to Solve Exponential Equations
Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall that since l o g(a)=l o g(b)l o g(a)=l o g(b) is equivalent to a= b,we may apply logarithms with the same base to both sides of an exponential equation.
applying the one-to-one property of logarithms
Note in the paragraph above the reiteration of the one-to-one property of logarithms. When applying logarithms with the same base to both sides of an exponential equation, we often use the common logarithm, log log or the natural logarithm, ln ln. The choice is yours which to use in most situations, but if either base in a given exponential equation is 10 10, use log log or if the base is e e, use ln ln to take advantage of the identity property of logarithms.
How To: Given an exponential equation Where a common base cannot be found, solve for the unknown
Apply the logarithm to both sides of the equation.
If one of the terms in the equation has base 10, use the common logarithm.
If none of the terms in the equation has base 10, use the natural logarithm.
Use the rules of logarithms to solve for the unknown.
Example: Solving an Equation Containing Powers of Different Bases
Solve 5 x+2=4 x 5 x+2=4 x.
Show Solution
5 x+2=4 x There is no easy way to get the powers to have the same base.l n 5 x+2=l n 4 x Take ln of both sides.(x+2)l n 5=x l n 4 Use the power rule for logs.x l n 5+2 l n 5=x l n 4 Use the distributive property.x l n 5−x l n 4=−2 l n 5 Get terms containing x on one side, terms without x on the other.x(l n 5−l n 4)=−2 l n 5 On the left hand side, factor out x.x l n(5 4)=l n(1 25)Use the properties of logs.x=l n(1 25)l n(5 4)Divide by the coefficient of x.5 x+2=4 x There is no easy way to get the powers to have the same base.l n 5 x+2=l n 4 x Take ln of both sides.(x+2)l n 5=x l n 4 Use the power rule for logs.x l n 5+2 l n 5=x l n 4 Use the distributive property.x l n 5−x l n 4=−2 l n 5 Get terms containing x on one side, terms without x on the other.x(l n 5−l n 4)=−2 l n 5 On the left hand side, factor out x.x l n(5 4)=l n(1 25)Use the properties of logs.x=l n(1 25)l n(5 4)Divide by the coefficient of x.
Try It
Solve 2 x=3 x+1 2 x=3 x+1.
Show Solution
x=l n 3 l n(2 3)x=l n 3 l n(2 3)
Q & A
Is there any way to solve 2 x=3 x 2 x=3 x?
Yes. The solution is x = 0.
Equations Containing e e
One common type of exponential equations are those with base e. This constant occurs again and again in nature, mathematics, science, engineering, and finance. When we have an equation with a base e on either side, we can use the natural logarithm to solve it.
How To: Given an equation of the form y=A e k t y=A e k t, solve for t t
Divide both sides of the equation by A.
Apply the natural logarithm to both sides of the equation.
Divide both sides of the equation by k.
Example: Solve an Equation of the Form y=A e k t y=A e k t
Solve 100=20 e 2 t 100=20 e 2 t.
Show Solution
100=20 e 2 t 5=e 2 t Divide by the coefficient 20.l n 5=l n e 2 t Take ln of both sides.l n 5=2 t Use the fact that l n(x)and e x are inverse functions.t=l n 5 2 Divide by the coefficient of t.100=20 e 2 t 5=e 2 t Divide by the coefficient 20.l n 5=l n e 2 t Take ln of both sides.l n 5=2 t Use the fact that l n(x)and e x are inverse functions.t=l n 5 2 Divide by the coefficient of t.
Analysis of the Solution
Using laws of logs, we can also write this answer in the form t=l n√5 t=l n 5. If we want a decimal approximation of the answer, then we use a calculator.
tip for success
Just as you have done when solving various types of equations, isolate the term containing the variable for which you are solving before applying any properties of equality or inverse operations. That’s why, in the example above, you must divide away the A A first.
Remember that the functions y=e x y=e x and y=l n(x)y=l n(x) are inverse functions.
Therefore, l n(e x)=x l n(e x)=x for all x x, and e l n(x)=x e l n(x)=x for x>0 x>0.
Try It
Solve 3 e 0.5 t=11 3 e 0.5 t=11.
Show Solution
t=2 l n(11 3)t=2 l n(11 3) or l n(11 3)2 l n(11 3)2
Q & A
Does every equation of the form y=A e k t y=A e k t have a solution?
No. There is a solution when k≠0 k≠0, and when y y and [latex] A [/latex] are either both 0 or neither 0 and they have the same sign. An example of an equation with this form that has no solution is 2=−3 e t 2=−3 e t.
Example: Solving an Equation That Can Be Simplified to the Form y=A e k t y=A e k t
Solve 4 e 2 x+5=12 4 e 2 x+5=12.
Show Solution
4 e 2 x+5=12 4 e 2 x=7 Subtract 5 from both sides.e 2 x=7 4 Divide both sides by 4.2 x=l n(7 4)Take ln of both sides.x=1 2 l n(7 4)Solve for x.4 e 2 x+5=12 4 e 2 x=7 Subtract 5 from both sides.e 2 x=7 4 Divide both sides by 4.2 x=l n(7 4)Take ln of both sides.x=1 2 l n(7 4)Solve for x.
Try It
Solve 3+e 2 t=7 e 2 t 3+e 2 t=7 e 2 t.
Show Solution
t=l n(1√2)=−1 2 l n(2)t=l n(1 2)=−1 2 l n(2)
Extraneous Solutions
Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when taking the logarithm of both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.
tip for success
Always check solutions to logarithmic equations to see if they satisfy the domain restriction on the logarithm.
Example: Solving Exponential Functions in Quadratic Form
Solve e 2 x−e x=56 e 2 x−e x=56.
Show Solution
e 2 x−e x=56 e 2 x−e x−56=0 Get one side of the equation equal to zero.(e x+7)(e x−8)=0 Factor by the FOIL method.e x+7=0 or e x−8=0 If a product is zero, then one factor must be zero.e x=−7 or e x=8 Isolate the exponentials.e x=8 Reject the equation in which the power equals a negative number.x=l n 8 Solve the equation in which the power equals a positive number.e 2 x−e x=56 e 2 x−e x−56=0 Get one side of the equation equal to zero.(e x+7)(e x−8)=0 Factor by the FOIL method.e x+7=0 or e x−8=0 If a product is zero, then one factor must be zero.e x=−7 or e x=8 Isolate the exponentials.e x=8 Reject the equation in which the power equals a negative number.x=l n 8 Solve the equation in which the power equals a positive number.
Analysis of the Solution
When we plan to use factoring to solve a problem, we always get zero on one side of the equation because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation e x=−7 e x=−7 because a positive number never equals a negative number. The solution x=l n(−7)x=l n(−7) is not a real number and in the real number system, this solution is rejected as an extraneous solution.
Try It
Solve e 2 x=e x+2 e 2 x=e x+2.
Show Solution
x=l n 2 x=l n 2
Q & A
Does every logarithmic equation have a solution?
No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.
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2767 | https://www.soinc.org/sites/default/files/StrawTowerChallenge.pdf | For Grades 5-12 OCTOBER 2007 © 2007 Science Olympiad, Inc.
CHICAGO SCIENCE IN THE CITY STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Children First Office of High Schools and High School Programs Department of High School Teaching & Learning · 125 South Clark Street, 12th Floor · Chicago, Illinois 60603 Telephone 773/553-3540 · FAX 773/553-2148 CHICAGO PUBLIC SCHOOLS Arne Duncan Chief Executive Officer David G. Gilligan Chief High School Officer Careda Taylor Chief of Staff Lydia C. Nantwi Gerod Sherley Administrators Michael C. Lach Officer September 4, 2007 Dear Teachers and Principals, The City of Chicago, in collaboration with a dynamic group of businesses, civic, cultural and community organizations is pleased to announce Chicago’s second annual Chicago Science in the City beginning the week of October 2, 2007. This two-week compilation of events is designed to encourage Chicagoans to explore and experience the various fields of science and to raise public awareness of the importance of science in our past, present and future. Last year’s expedition was very successful and brought together 1,500 participants and 5,000 spectators to experience the kick-off event alone. As one way of supporting Chicago Science in the City: 2007, we have partnered with Science Olympiad to create the Straw Tower Challenge. For this Challenge, we are asking that teams of students in grades 5-12 build the tallest tower possible out of a limited supply of drinking straws and tape. These towers not only have to be tall but also they must support the weight of a tennis ball. Your school can publish its results, and track the progress of other schools, on the Straw Tower Challenge website. By participating in the Challenge your school is eligible to win various prizes as well as be invited to bring their towers to one of the Chicago Science in the City: Neighborhood Science Carnivals for a more public display of your work. To help you complete this Challenge, your school has been provided with two sets of tower building materials, this is enough material to involve approximately 25 to 30 students in the Challenge, and this booklet of lessons focusing on tower building and the engineering of structures. Science Olympiad designed these lessons to be used throughout the two weeks of Chicago Science in the City: 2007 either as part of the regularly scheduled classroom instruction or during after-school or at lunch “club” time. In addition, the College of Engineering at the University of Illinois Urbana-Champaign will be providing undergraduate engineering majors to help students as they design and build their towers. These engineering mentors will be “email pals” of the students and offer help and guidance as the students strive to create the tallest tower. If you have any questions, need additional materials, or need assistance implementing this activity, elementary schools should contact Ms. Chandra James, Office of Math & Science at 773-553-6192 and high schools should contact Dr. John F. Loehr, Department of High School Teaching and Learning at 773-535-8080 ext. 155. Thank you in advance for your cooperation to make this effort a success. The hard work and dedication you display are always appreciated. Sincerely, Michael C. Lach PAGE - 1 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE DESCRIPTION: To build the tallest tower out of 50 straws that will hold a designated load for 3 seconds. NUMBER OF PARTICIPANTS: 5-6 APPROXIMATE TIME: 120 minutes total over two weeks MATERIALS PROVIDED: 50 drinking straws, masking tape, tennis ball, measuring tape Optional Materials: Scissors and a pencil; additional load items Event P arameters: 1. The tower must be constructed using only the drinking straws and masking tape. There will be enough materials to build 3-5 towers.
2. The straws can be interconnected directly by sliding one end into another straw. Straws may be bent, cut or slit. 3. Students may not completely cover their structure with masking tape as reinforcement; tape is used at joints only. Tape may not be used to secure the tower to a table, base, ceiling or surface. You may not make a leg, brace, cone or column out of tape.
4. The tower must accommodate a standard tennis ball. The student must be able to balance the ball on the tower at the loading point for the tower to qualify.
Measuring Your Achievement: 1. Remember to measure your tower height before loading.
2. The height of the tower will be measured again once the tennis ball is placed on the structure. Use a flat surface to test the structure.
3. The tower that supports the load at the highest point is ranked first. If the tower does not support the load, it will be ranked BELOW the tower that did support the load based on tower height.
4. Once you have collected the data on your schools’ towers please visit the Chicago Science in the City website and enter the data from the top 3 towers. In addition, pictures can be submitted by following the instructions posted at the site.
The Science Behind Structures: 1. Instructions for Teachers - Lists National and Illinois Standards alignment and tips for using this lesson plan.
2. Extension Activities - Optional extensions for students.
3. Implementation Schedule - A two-week calendar with suggested daily activities.
4. Tower Log - Record observations and plan, design and evaluate your structure.
5. Tower Sketch - A grid to plan your team’s design.
6. Basic Tower Design - Covers legs, columns and bracing.
7. Simple Structures - How shapes like triangles and squares form the parts of a tower.
8. Analysis of Structures and Joints - Tips and engineering terms for joining your straws.
Resources: Science in the City Homepage University of Illinois at Urbana-Champaign College of Engineering Chicago Architecture Foundation Science Olympiad Home P age STRAW TOWERS You may copy all of these pages for school use.
PAGE - 2 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE With all the excitement surrounding the construction of several new skyscrapers in Chicago (the 92-story Trump International Hotel & Tower, Santiago Calatrava’s 2000-foot-tall Chicago Spire), the Science in the City “Straw Tower Challenge” reflects the rich architectural and engineering history of our great hometown. Your challenge as a teacher will be to turn a handful of straws and a roll of tape into a meaningful science activity. You might find skepticism among your students who will ask, “Can we really build a tall tower with straws?” But once presented with the materials, skyscrapers will arise from desks, floors and tabletops, reaching unexpected heights of creativity.
We at Science Olympiad have provided you with a special set of instructions, containing a detailed daily calendar, extension activities for students, a tower log, a sketch sheet and four pages of engineering tips that will help you form the building blocks of a successful tower design. (Please feel free to copy these pages for your internal school use.) Science Olympiad takes pride in aligning all its events to National Science Standards as well as State Standards, and the Straw Tower Challenge meets both: STRAW TOWER BUILDING – The objective of this event is to design and build a straw tower with the highest structural efficiency capable of supporting a standard load. Teams should maintain and submit a log at check-in containing data to help them improve future designs.
M.E.1.b-d E. Science and Technology – An understanding of science and technology establishes connections between the natural and designed world linking science and technology.
M.E.1 Abilities of technological design b. Design a solution or product. c. Implement a proposed design. d. Evaluate completed technological designs or products.
Illinois Learning Standard 11.B concerning knowledge and application of the concepts, principles and processes of technological design.
Below are a few sample ways to utilize the following engineering tip pages. Be sure to have students record all observations in the Tower Log: Basic Tower Design - Ask students about towers they have seen that look like the Basic Tower Design drawing (cell phone towers, electricity towers, the John Hancock Building). Identify the main parts of a tower (legs, columns, braces). Explain the job of each per the handout.
Simple Structures - Take a moment before the first lesson to assemble a few of the shapes with straws that will form the building blocks of a sample tower (square, triangle, pyramid and cube). Let the students touch and examine the shapes. Push, pull and bend the shapes to see how they react and analyze the strength of each shape. Discuss why triangles are stronger and show how they are used as reinforcing cross braces on towers. (How about making a column by bundling cables?) Analysis of Structures - Now illustrate the concepts of what will happen when you apply stress, pressure or a load to a tower. Use your shapes as a base and test with your hand or an object. What can students do in their design to reduce strain, stress, bending or shearing? Let each student experiment with a straw. Joints - Demonstrate and practice the proper way to wrap tape around the straws. Build a butt joint with adjoining straws for the longest section. Will it bend at the joint? Overlap the tips of two straws then tape. Is it stronger than the butt joint? Slip one straw into another and wrap the slip joint with tape. The section is shorter, but is it stronger? This exercise is vital to tower assembly and maximizing tower height.
INSTRUCTIONS FOR TEACHERS PAGE - 3 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Optional Extension Activities On all of these challenges, remember to take pictures and upload them to the Straw Towers Photo Gallery located on the Science in the City website at www.chicagoscienceinthecity.org. The website also has a Leader Board which will continuously update your progress against other schools or classrooms within your school. See how your tower stacks up against the rest!
University of Illinois College of Engineering “Ask A Student” E-Mail Program for Straw Tower Challenge - Chicago Public Schools students participating in the Straw Tower Challenge will be able to exchange emails with civil engineering students from the University of Illinois at Urbana-Champaign to ask questions and seek advice on their team projects. Simply access the Straw Tower Challenge portion of the Science In The City website and click on “Ask A Student.” You will be directed to an email address of a UI administrator who will forward your email to an assigned mentor. CPS students will then directly communicate with the civil engineering students, who will also be attending the Science Carnivals on Saturday October 6 and 13 in the Science Olympiad booth.
Build a Chicago Landmark out of straws - Test your skill at modeling by choosing a famous Chicago building and recreate it with straws! Start with an easy geometric shape like the Sears Tower or the Hancock Tower and move on to tougher assignments like the Water Tower, the Prudential Building or even the Frank Gehry-designed band shell at Millennium Park.
Use other materials to test - Hit it out of the park! Instead of a tennis ball, test your structure with a baseball, mini basketball, football or volleyball. As you increase the load, think about the changes you will need to make to your structure to accommodate the additional weight. You can also place a cup on the top of your tower and add marbles, paper clips or pennies one by one to determine the quantity your structure will hold. Design your own Tower Log for The Cup Challenge.
Use other materials to build - Don’t limit your imagination to drinking straws to build a tower. Use other materials you will commonly find around school or home, like cardboard, strips of wood, Tinker Toys, K’nex or LEGOS, to build your own creation.
EXTENSION ACTIVITIES Average Weights of Common Straw Towers Load Items: Penny 2.5g Ping Pong Ball 2.7g Quarter 5.7g Golf Ball 45g Tennis Ball 58g Apple 130g Baseball 145g Hockey Puck 162g Orange 288g Basketball 600g PAGE - 4 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE IMPLEMENTATION SCHEDULE - OCTOBER 2007 Monday Week 1 1 2 3 4 5 6 8 9 10 11 12 13 Week 2 Tuesday 10am - 2pm Science in the City Kickoff in Daley Plaza Wednesday Thursday Friday Saturday 20 minutes: IDENTIFY THE PROBLEM - Introduce activity and materials, set goals. Assemble student teams and assign jobs. Log on to the SITC web site. Discuss Basic Tower Design (p.7).
20 minutes: ANALYZE & GENERATE SOLUTIONS - Go over Simple Structures, Analysis of Structures and Joints (p. 8-10) and begin sketching out designs (p. 6). Work as a team to brainstorm shapes to hold the tennis ball atop your Straw Tower - grid, teepee, raft, shelf, basket, rooftop etc. Incorporate all ideas! Make sure you consider a way to stop the ball from rolling off. Use 10-15 straws to begin assembling 2-D and 3-D shapes: squares, triangles, cubes, pyramids (p.8).
20 minutes: SELECT & PLAN SOLUTIONS - Settle on a design idea based on trial and error. Enter data into Tower Log (p.5) and begin testing weights on short structures. Build levels with remaining straws. Upload photos onto SITC web site.
20 minutes: IMPLEMENT SOLUTION - Assemble multi-tiered or multi-level tower. Test load. Evaluate joints and discuss ways to use tape to make the strongest joints. Enter data into Tower Log.
20 minutes: EVALUATE SOLUTION - Make adjustments and record engineering and construction successes and failures in Tower Log. Finalize design and test. Take photos!
20 minutes: Utilize Student and Teacher resources page. Compare your tower to others within your school to declare which is the tallest tower that holds the tennis ball. Finalize your Tower Log and turn in to teacher. Upload final photo. Congratulations on building a Straw Tower!
Extension: Try to build a Chicago landmark out of straws.
Extension: Use other materials to build your tower.
Extension: Use other materials to test your tower.
Science in the City Science Carnival #1 Science in the City Science Carnival #3 Science in the City Science Carnival #2 PAGE - 5 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Part One - Information Collected During Construction Date: School Name: Team Members Straw Boss: Architect: Engineer: Information Manager: Construction Workers: Number of Straws Used: Number of Joints on Straw Tower: Part Two - Information Collected During Testing Height of Tower: Height at top of tennis ball: Did the Straw Tower hold the tennis ball? Yes No Part Three - Information Collected After Testing How did the Straw Tower fail and what could you do to improve your design?
Where did the Straw Tower fail? (list location of failure and ID on drawing above) What one thing would you change about the design next time?
Jobs - Each member of the team should choose a job from the following list: Straw Boss - Serve as the Project Manager and supervise the project, making sure the job is getting built on time.
Architect - Complete the drawings and sketch out the design of the tower.
Engineer - Manages building the tower and testing the load.
Information Manager - In charge of all information related to the construction project; complete the Tower Log and take photos of all progress.
Construction Workers - Members of the crew will build all parts of the tower.
STRAW TOWER LOG Sketch your design here and label the parts of your Straw Tower PAGE - 6 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE SKETCH YOUR TOWER PAGE - 7 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Legs/Columns Cross braces Load The object is to support a load at some height above the base.
A tower is made up of: Legs or Columns – Vertical members Extend from the base to the top of the tower Support the load Need to be strong in compression Bend easily because they are long and slender Cross Braces – Horizontal/Diagonal members Connect one leg to another “Break” the legs up into shorter and stronger members Prevent the legs from bending May be under compression or tension Have very little stress as long as the legs remain straight BASIC TOWER DESIGN Note to Teachers: The following four pages provide an overview of the components necessary to build successful Straw Towers. Use these tips as instructional tools to correspond with the activities listed on the Instructions for Teachers on Page 2 and the Implementation Schedule on Page 4.
PAGE - 8 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Square Use 4 drinking straws and tape to form a square. Stand the square on edge and apply a downward force to the top of it. The square deforms easily due to the weakness (in rotation) of the corner joints.
Triangle Make a triangle in the same manner using 3 straws and tape and test it as you did the square. The triangle won’t change shape unless you push hard enough for either a joint or one of the straws to break.
Fortunately a 4-sided figure can be changed into 2 triangles by connecting 2 opposite corners together. Add a rigid cross brace to the square and test it again. Now it is much stronger.
Use squares and triangles as a base to form pyramids and cubes which will be the building blocks of your Straw Tower.
SIMPLE STRUCTURES PAGE - 9 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Definitions Joint – The place where 2 or more members are connected together.
Member – A part of a structure that connects 2 joints.
Strain – The deformation or change of shape that is the result of applied stress.
Stress – The force applied to an object that tries to change its shape. There are 5 different forms of stress: Bending – Stress that tends to deform an object from a straight line.
Compression – Stress that tends to compress an object together.
Shear – Stress that tends to make two parts of an object slide past each other.
Tension – Stress that tends to pull an object apart.
Torsion – Stress that tends to twist an object around an axis.
For our simplified analysis we will only consider bending, compression, and tension.
ANALYSIS OF STRUCTURES PAGE - 10 STRAW STRAW STRAW TOWER TOWER TOWER CHALLENGE CHALLENGE CHALLENGE Slip Joint Butt Joint Lap Joint A joint is where 2 or more members are joined together; Butt Joint – Pieces are joined end to end Lap Joint – Pieces are overlapped Slip Joint – One piece is slipped into the other The strength depends on the type of joint and material used to hold it together.
You can use tape to hold butt and lap joints together.
A lap joint can be held together with string or tape used as glue.
Strength of Taped Joints Butt Joint Bending – Medium Compression – Strong Tension – Weak Lap Joint Bending – Medium Compression – Medium Tension – Medium Slip Joint Bending – Weak Compression – Medium Tension – Weak JOINTS |
2768 | https://www.howtopronounce.com/hackneyed | How to pronounce hackneyed | HowToPronounce.com
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How to Pronounce hackneyed Phonetically (With Audio)?
Learn how to pronounce the English word Hackneyed in english using phonetic spelling and the International Phonetic Alphabet (IPA)
hack-neyed
hak-need
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Meanings for hackneyed
This word is an adjective that describes a word or phrase, that is used a lot over time.
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Synonyms for hackneyed
unoriginal
banal
trivial
corny
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Examples of hackneyed in a sentence
Is the Bharatiya Janata Party back to its hackneyed Hindutva agenda? The question is being raised following the party's decision to entrust the leadership of the party's campaign for the forthcoming state bye-election in Uttar Pradesh to the saffron clad rabble-rouser Mahant Adityanath
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Translations of hackneyed
German : abgedroschen
Spanish : manido
Arabic : مبتذل
French : rebattu
Chinese : 陈腐的
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2769 | https://math.stackexchange.com/questions/1087910/a-result-of-erd%C5%91s-on-increasing-multiplicative-functions | number theory - A result of Erdős on increasing multiplicative functions - Mathematics Stack Exchange
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A result of Erdős on increasing multiplicative functions
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Asked 10 years, 7 months ago
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Erdős proved that if f(n)f(n) is a monotone increasing function from the natural numbers to the positive reals, and f(n)f(n) is completely multiplicative, then there exists some constant C C such that f(n)=n C f(n)=n C for all n n.
I have not been able to find a nice proof of this online and was hoping someone could provide a proof.
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edited Jan 1, 2015 at 22:57
user147263
asked Jan 1, 2015 at 22:22
Prajanan PatelPrajanan Patel
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5 "A result of Erdös" is probably not a good title, given the number of results due to Erdös. "A theorem of Euler", "A lemma of Gauss." =D –Pedro♦ Commented Jan 1, 2015 at 22:28
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Here's a proof I came up with.
Let C=log 2 f(2)C=log 2f(2), note that f(2)≥1 f(2)≥1 and therefore C≥0 C≥0. Proceed by induction: Suppose that f(n)=n C f(n)=n C, and n≥2 n≥2, we will show that f(n+1)=(n+1)C f(n+1)=(n+1)C. Write D=log n+1 f(n+1)D=log n+1f(n+1), then C≤D C≤D, otherwise for sufficiently large k k we will have that f(n k)>f((n+1)k)f(n k)>f((n+1)k), which contradicts the assumption f f is increasing. On the other hand, f(n 2)=f(n)f(n)=n 2 C=(n 2)C f(n 2)=f(n)f(n)=n 2 C=(n 2)C, and since n≥2 n≥2 we have that n 2>n+1 n 2>n+1. So for the same reason as before D≤C D≤C, and therefore D=C D=C.
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edited Jan 1, 2015 at 23:11
answered Jan 1, 2015 at 22:55
Asaf Karagila♦Asaf Karagila
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I'm not quite sure what details are missing. n 2=n⋅n n 2=n⋅n, and f f is multiplicative. –Asaf Karagila♦ Commented Jan 14, 2015 at 18:58
I meant that by a logarithm of f(n+1)f(n+1) and f(n 2)f(n 2) the same reasoning used to conclude that C≤D C≤D can be used again to conclude the opposite inequality. –Asaf Karagila♦ Commented Jan 14, 2015 at 19:20
Simply find a sufficiently large k k such that (n+1)D k>(n 2)C k(n+1)D k>(n 2)C k, which is impossible due to monotony. –Asaf Karagila♦ Commented Jan 14, 2015 at 19:46
Why would we have for large k k, f(n k)>f((n+1)k)f(n k)>f((n+1)k)? Even if D1 n>1), f(n)>1 f(n)>1.
Second, consider g(x)=ln f(a)ln a g(x)=lnf(a)lna. Prove by contradiction that g(x)g(x) is a constant.
Suppose not. Then, we have a,b≠1 a,b≠1 such that g(a)
Since f(b)>1 f(b)>1, thus ln f(b)>0 lnf(b)>0. Also, ln a>0 lna>0, so we can divide and multiply respectively to conclude that
0
As such, there exist positive integers such that ln f(a)ln f(b) The RHS inequality gives us a df(b c)f(a d)>f(b c). Hence, the function is not increasing.
Third, conclude that g(x)g(x) is a constant, and thus f(a)=a C f(a)=a C for some constant C C. Furthermore, C>0 C>0.
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edited Jan 1, 2015 at 23:16
answered Jan 1, 2015 at 22:33
Calvin LinCalvin Lin
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For the record, and then I will leave (and I cleaned all my other comments): your initial (not present as edited during grace period) was something like "Hint: Consider g(x)=ln f(a)ln a g(x)=lnf(a)lna. Prove by contradiction that g(x)g(x) is a constant." Spoilered 'Suppose not.' " This is not a nice proof. –quid Commented Jan 1, 2015 at 23:10
@quid For the record, I was in the midst of writing out the main steps. I wanted to check that spoiler works. It gets tricky with displaying equations in newline mode, which is why nothing else appeared. –Calvin Lin Commented Jan 1, 2015 at 23:11
6 @quid It is entirely appropriate to post a proof outline. This attitude that 1. OP deserves a full proof with all details and 2. An answerer, having provided a proof outline, is to be berated until a full proof is supplied, is frankly ridiculous. –aes Commented Jan 1, 2015 at 23:11
This was unfortunate timing then, as my initial impression was formed based on this version. Sorry. To avoid such problems in the future you might consider using the sandbox on meta. –quid Commented Jan 2, 2015 at 0:43
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2770 | http://www.cse.uaa.alaska.edu/~afkjm/cs411/handouts/mastermethod.pdf | Lecture Notes, CS411 Kenrick Mock Chapter 4: Recurrence Relations : Iterative and The Master Method Iteration Method: Expand the terms into a summation, and solve algebraically Example: T(n)= Theta(1) for n=1 T(n) = 3T(n/4) + n for n>1 T n T n n 4 3 4 4 4 = + We can plug this back into the original recurrence relation: ( ) T n T n n n = + + 3 3 16 4 We can keep on going: ( ) T n T n n n n = + + + 3 3 3 64 16 4 If we stop at this point and do some math: T(n) = 27T(n/64) + 9(n/16) + 3(n/4) + n T n n n n T n ( ) = + + + 3 4 9 16 27 64 There’s a pattern here! If we consider i as the index, where i=1 gives us n+(3/4)n, then we can generalize this as i increases: + + + + + + = i i j j n T n n n n n T 4 3 ...
4 3 ...
16 9 4 3 ) ( How far does i go? Does it increase to infinity? NO at some point we will have to stop. But we already know when we stop – we stop at T(1) because at this point there is no more recursion, we just return a constant number for the amount of work to do. If we stop at T(1), this means we will stop when 1=(n/4i). i n 4 1= n i = 4 log4 n i = So we can now express the recurrence relation as: ) 1 ( 3 ...
4 3 ...
16 9 4 3 ) ( 4 log Θ + + + + + = n i n n n n n T substituting Θ( ) 1 for T(n/4i) since we will only do a constant amount of work on the last iteration. We can summarize this as a single summation. First recall that 3 4 4 3 log log n n = ; this is sublinear since log 43 < 1 ) ( 4 3 ) ( 3 log 1 log 0 4 4 n n n T n i i Θ + = ∑ − = ) ( 4 3 ) ( 3 log 0 4 n n n T i i Θ + ≤ ∑ ∞ = ; up to infinity bigger, so <= applies recall that x x k k = − = ∞ ∑ 1 1 0 ; for x<1 T n n n ( ) ( ) log ≤ − + 1 1 3 4 4 3 Θ T n n n ( ) ( ) log ≤ + 4 4 3 Θ ;T n n o n ( ) ( ) ≤ + 4 ; loose upper bound so use little-o This means that the recurrence is O(n). This method is accurate but can result in a lot of algebra to keep track of; can also get very challenging for more complicated recurrence relations. Second Example: T(n)=1 if n=1 T(n)=4T(n/2)+n if n>1 T(n) =4T(n/2) + n =4(4T(n/4)+n/2)+n =4(4(4T(n/8)+n/4)+n/2)+n =64T(n/8) + 4n +2n +n =n + 2n +4n + 64T(n/8) =n + 2n + 4n + … +2jn + … 4iT(n/2i) ; hard part to figure out What is the last term? When (n/2i)=1 à i=lgn T(n) = n + 2n + 4n + 8n + … 2in + … 4lgn Θ( ) 1 = ) 1 ( 4 2 lg 1 lg 0 Θ + ∑ − = n n i i We know that x x x k m k m = − − + = ∑ 1 0 1 1 Let’s let m=lgn-1. Then: T(n) = ) 1 ( 4 1 2 1 2 lg 1 1 lg Θ + − − + − n n n = n n n n 2 4 1 lg lg ( ) − + Θ = n2 –n + nlg4Θ( ) 1 = 2 n2 Θ( ) 1 –n = Θ( ) n2 Sometimes a recursion tree can help: Recursion Tree: Help to keep track of the iterations Given T(n) = 2T(n/2)+n2 How deep does the tree go? We stop at the leaf, and we know we’re at a leaf when we have a problem of size 1. n 2 T(n/2) T(n/2) n 2 (n/2) (n/2) 2 2 (n/4) 2 (n/4) 2 (n/4) 2 (n/4) 2 n 2 1/2n 2 1/4n 2 1=(n/2i)2 so n2=22i ; n=2i ; i=lgn The amount of work done is then: n i i n 2 0 2 = ∑ lg =Θ( ) n2 ; this is geometrically decreasing in size, so it won’t get any bigger than n 2. One more example: T(n) = T(n/3) + T(2n/3) + n Each level does work of size n; if we just know the height of the tree, i, the total work is ni. The tree stops when the leaf is of size 1. The hard part is to figure out the formula based on the height: n i 2 3 1 = (why pick the 2/3 branch and not 1/3?) n i i = = 1 2 3 3 2 i n = log / 3 2 So the total work is (log ) / 3 2 n n or O(nlog 3/2 n). Master Method: n (n/3) (2n/3) (n/9) (2n/9) (2n/9) (4n/9) n n n If the form of a recurrence is: ( ) ( ) T n aT n b f n a b = + ≥ > , , 1 1 then we can use the Master Method, which is a cookbook-style method for proving the runtime of recurrence relations that fit its parameters. Note that not all recurrence of the above form can be solved through the master method. We won’t prove the master method, but will give an argument as to how it works. In the master method: • a is the number of subproblems that are solved recursively; i.e. the number of recursive calls. • b is the size of each subproblem relative to n; n/b is the size of the input to the recursive call. • f(n) is the cost of dividing and recombining the subproblems. Recursion tree example: T(n)=aT(n/b)+f(n) What is the height of the tree? When f n b f n b n b i n i i i b = → = → = → = ( ) log 1 1 How many leaves are there? a NumberLeaves height = F(n) F(n/b) F(n/b) F(n/b) a F(n/b2) F(n/b2) F(n/b2) a ...
F(n/b2) F(n/b2) F(n/b2) ...
) 1 ( Θ ) 1 ( Θ ...
n b log F(n) aF(n/b) a2F(n/b2) ) ( log a b n Θ ( ) ∑ − = + Θ = 1 log 0 log n j j j a b b b n f a n Total a n b b n a log log = Work at the leaves is : ( ) ( ) Θ Θ 1 n n b b a a log log = Work of dividing and combining is: f n af n b a f n b ( ) ( ) ( ) ...
+ + + 2 2 = a f n b i i i n b ( ) log = − ∑ 0 1 this does not include the cost of the leaves. The total work/runtime T(n) is: ∑ − = + Θ 1 log 0 log ) ( ) ( n i i i a b b b n f a n The time T(n) might be dominated by: 1. The cost of the leaves 2. The cost of the divide/combine or the root 3. Evenly distributed at all the levels The master method tells us what the asymptotic running time will be depending on which cost is the highest (dominates). If the form is: ( ) ( ) T n aT n b f n a b = + ≥ > , , 1 1 Then based on comparing f(n) and n b a log we know the running time given the following three cases: • If f n O n b a ( ) ( ) log = −ε for some constant ε > 0then T n n b a ( ) ( ) log = Θ ; cost of leaves dominates. • If f n n b a ( ) ( ) log = Θ then T n n n b a ( ) ( lg ) log = Θ ; cost is evenly distributed • If f n n b a ( ) ( ) log = + Ω ε for some constant ε > 0 and if af n b cf n ( ) ( ) ≤ for some constant c<1 and all sufficiently large n, then T n f n ( ) ( ( )) = Θ ; divide/conquer or root cost dominates Example: T n T n n ( ) ( ) = + 9 3 So a=9, b=3, f(n)=n Case 1 works for f n O n b a ( ) ( ) log = −ε . We need to prove this relationship by showing that: f n O n b a ( ) ( ) log = −ε n O n O n = = − − ( ) ( ) log3 9 2 ε ε if ε =1 then n=O(n) and case 1 is satisfied. Therefore: T n n n n b a ( ) ( ) ( ) ( ) log log = = = Θ Θ Θ 3 9 2 In this example, the cost of the leaves has dominated the runtime. Example: T n T n n ( ) ( ) = + 2 2 ; Merge Sort So a=2, b=2, f(n)=n Check case 1: Is f n O n b a ( ) ( ) log = −ε ? n O n = − ( ) log2 2 ε n O n = − ( ) 1 ε For any epsilon>0, n is bigger, so case 1 does not work. Check case 2: Is f n n b a ( ) ( ) log = Θ n n n = = Θ Θ ( ) ( ) log2 2 YES therefore: T n n n n n n n b a ( ) ( lg ) ( lg ) ( lg ) log log = = = Θ Θ Θ 2 2 Cost is evenly distributed among leaves and upper part of tree. Example: T n T n ( ) ( ) = + 2 3 1 So a=1, b=3/2, f(n)=1 Case 1 does not work (exercise for the reader) Case 2: Is f n n b a ( ) ( ) log = Θ ? 1 1 3 2 1 0 = = = Θ Θ Θ ( ) ( ) ( ) log / n n YES therefore: T n n n n n n n n b a ( ) ( lg ) ( lg ) ( lg ) (lg ) log log / = = = = Θ Θ Θ Θ 3 2 1 0 Cost is again evenly distributed. Example: T n T n n n ( ) ( ) lg = + 3 4 a=3,b=4,f(n)=nlgn Case 1 and 2 don’t fit (exercise for the reader) Case 3: Is f n n b a ( ) ( ) log = + Ω ε ? n n n n lg ( ) ( ) log .
= = + + Ω Ω 4 3 0 79 ε ε YES, if epsilon =0.21, then n n n lg ( ) = Ω We also need to show the extra condition: Is af n b cf n ( ) ( ) ≤ for c<1? 3 4 3 4 4 3 4 4 3 4 2 f n cf n n n cn n n n cn n n n cn n ( ) ( ) lg lg (lg lg ) lg (lg ) lg ≤ ≤ − ≤ − ≤ YES, if c=¾ then 3 4 2 3 4 n n n n (lg ) lg − ≤ therefore: T n f n n n ( ) ( ( )) ( lg ) = = Θ Θ Example: T n T n n n ( ) ( ) lg = + 4 2 2 So a=4, b=2, f(n)= n n 2 lg Try case 1: Is f n O n b a ( ) ( ) log = −ε ? n n O n n n O n 2 4 2 2 2 lg ( ) lg ( ) log = = − − ε ε NO, for epsilon>0, f(n) is larger. Try case 2: Is f n n b a ( ) ( ) log = Θ ? n n n n 2 4 2 2 lg ( ) ( ) log = = Θ Θ NO, grows smaller than n2. Try case 3: Is f n n b a ( ) ( ) log = + Ω ε ? n n n 2 2 lg ( ) = + Ω ε NO, for epsilon > 0, f(n) is smaller, not bigger. Master method does not work for this recurrence relation! (Solution is Θ( lg lg ) n n 2 by substitution) Selection Problem (Chapter 10): Consider the problem of finding the ith smallest element in a set of n unsorted elements. This is referred to as the selection problem or the ith “order statistic”. If i=1 this is finding the minimum of a set i=n this is finding the maximum of a set i=n/2 this is finding the median or halfway point of a set -- common problem Selection problem defined as: Input: A set of n numbers and a number i, with 1<=i<=n Output: The element x in A that is larger than exactly i-1 other elements in A. How many comparisons are necessary to determine the selection? Say we want to find the minimum: Lower bound of at least n-1 comparisons to see every other element Think as a tournament: Pick contender Contender competes with another (comparison) Winner is the smallest element Every element except the winner must lose one match. This is a simple example to show that we need at least n-1 comparisons, we will use this technique later in more complex examples to show a lower bound. Selecting the ith smallest element: Can do in Θ( lg ) n n time easily by sorting with Merge Sort, and then pick A[i]. But can do better! Consider if the set of n numbers is divided as follows: S1: < p S2: > p p Note that the elements in S1 are not sorted, but all of them are smaller than element p (partition). We know that p is the (|S1| +1)th smallest element of n. We will use this idea later to also sort numbers (known as quicksort). Now consider the following algorithm to find the ith smallest element from Array A: • Select a pivot point, p, out of array A. • Split A into S1 and S2, where all elements in S1 are p • If i=|S1|+1 then p is the ith smallest element. • Else if i<=|S1| then the ith smallest element is somewhere in S1. Repeat the process recursively on S1 looking for the ith smallest element. • Else i is somewhere in S2. Repeat the process recursively looking for the i-|S1|-1 smallest element. Question: How do we select p? Best if p is close to the median. If p is the largest element or the smallest, the problem size is only reduced by 1. • Always pick the same element, n or 1 • Pick a random element • Pick 3 random elements, and pick the median • Other method we will see later How do we partition once we have p? If A contains: [5 12 8 6 2 1 4 3] Can create two subarrays, S1 and S2. For each element x in A, if x=p put it in S2. p=5 S1: [2 1 4 3] S2: [5 12 8 6] This certainly works, but requires additional space to hold the subarrays. We can also do the partitioning in-place, using no additional space: Partition(A,p,r) ; Partitions array A[p..r] x←A[p] ; Choose first element as partition element i←p-1 j←r+1 while true do repeat j←j-1 until A[j]≤ x repeat i←i+1 until A[i]≥x if ij j i Return j. All elements in A[p..j] smaller or equal to x, all elements in A[j+1..r] bigger or equal to x. (Note this is a little different than the initial example, where we split the sets up into < p, p, and > p. In this case the sets are =p. If the pivot point selected happens to be the largest or smallest value, it will also be guaranteed to split off at least one value). This routine makes only one pass through the array A, so it takes time Θ( ) n . No extra space required except to hold index variables. Worst case running time of selection: Pick min or max as partition element, producing region of size n-1. T n T n n ( ) ( ) ( ) = − + 1 Θ subprob time to split Evaluate recurrence by iterative method: T T T ( ) ( ), ( ) ( ) ( ), ( ) ( ) ( ) ( ),...
1 1 2 1 2 3 1 2 3 = = + = + + Θ Θ Θ Θ Θ Θ Θ Θ ( ) i i i n i n = = ∑ ∑ 1 1 =Θ( ) n2 Recursion tree for worst case: Best-case Partitioning: In the best case, we pick the median each time. T n T n n ( ) ( ) ( ) = + 2 Θ Using the master method: a=1, b=2, f(n)=Θ( ) n Case 3: Is f n n b a ( ) ( ) log = + Ω ε ? n 1 n-1 1 n-2 …1 n n n-1 n-2 ) ( 2 n Total Θ = Θ Ω Θ Ω ( ) ( ) ( ) ( ) log n n n n = = + + 21 0 ε ε YES if epsilon between 0 and 1, say 0.5 Also is af n b cf n ( ) ( ) ≤ for c<1? Θ Θ ( ) ( ) n c n 2 ≤ YES if c > ½ So T n f n n ( ) ( ( )) ( ) = = Θ Θ Recursion Tree for Best Case: Average Case: Can think of the average case as alternating between good splits where n is split in half, and bad splits, where a min or max is selected as the split point. Recursion tree for bad/good split, good split: Both are Θ( ) n , with just a larger constant in the event of the bad/good split. So average case still runs in time Θ( ) n . We can solve this problem in worst-case linear time, but it is trickier. In practice, the overhead of this method makes it not useful in practice, compared to the previous method. However, it has interesting theoretical implications. Basic idea: Find a partition element guaranteed to make a good split. We must find this partition element quickly to ensureΘ( ) n time. The idea is to find the median of a sample of medians, and use that as the partition element. n n/2 n/2 n/4 n/4 …1 lgn n n/2 n/4 ) (n Total Θ = 1...
n n/2 n/2 1 (n/2)-1 …1 ~ 2lgn n n/2 n/2 ) (n Total Θ = 1...
((n/2-1)/2 ((n/2)-1)/2 4 / n ≈ New partition selection algorithm: • Arrange the n elements into n/5 groups of 5 elements each, ignoring the at most four extra elements. (Constant time) • Find the median of each group. This gives a list M of n/5 medians. (time Θ( ) n if we use the same median selection algorithm as this one or hard-code it) • Find the median of M. Return this as the partition element. (Call partition selection recursively using M as the input set) See picture of median of medians: Guarantees that at least 30% of n will be larger than pivot point p, and can be eliminated each time! Runtime: T n T n T n O n ( ) ( ) ( ) ( ) = + + 5 7 10 select recurse overhead of split/select pivot subprob The O(n) time will dominate the computation by far resulting in O(n) run time. x Quicksort We can also use the Partition selection algorithm to do sorting, this is called Quicksort. QuickSort(A,p,r) ; Sort A[p..r] if p<r then q← Partition(A,p,r) QuickSort(A,p,q) QuickSort(A,q+1,r) Show tree for sorting example of A=[5 3 2 6 4 1 3 7], use first element as partition: Now do an in-order tree-traversal and we get the list in sorted order. What’s going on if we do this in-place in the array: A = [ 5 3 2 6 4 1 3 7 ] Partition on 5 A = [3 2 4 1 3 ] Partition on 3 A = [6 7 5 ] Partition on 6 A = [1 2] Partition on 1 A = [3 3 4] Partition on 3 A=1 A=2 A = 3 A = [3 4] Partition on 3 A=3 A=4 A=5 A = [6 7] Partition on 6 A=6 A=7 We end up with the sorted array at the end of the recursive steps, following the tree from left-to-right (inorder). All work is done in Partition. Worst case runtime: T n T n n ( ) ( ) ( ) = − + 1 Θ which we know is Θ( ) n2 Best case runtime: T n T n n ( ) ( ) ( ) = + 2 2 Θ which is the same as Merge Sort we know is Θ( lg ) n n Average case: Same argument as before, alternating good and bad splits. Results in same as the best case runtime but with larger constants than the best case, Θ( lg ) n n . Even though Quick Sort has the same average case runtime than Merge Sort (Θ( lg ) n n ), usually Quick Sort has smaller runtime constants than Merge sort, resulting in an overall faster execution time. What if we ran the median of median strategy to find partition point? Still would get Θ( lg ) n n . Random strategy usually best, pick a small # of random elements, and use median of those elements as the partition point. QS(A,1,8): A=5,3,2,6,4,1,3,7 Partition 5: A=3,2,4,1,3,6,5,7 QS(A,1,5) Partition 3: A=2,1,3,3,4,6,5,7 QS(A,1,2) Partition 2: A=1,2,3,3,4,… QS(A,1,1) Terminate QS(A,2,2) Terminate QS(A,3,5) Partition 3: A=1,2,3,3,4,… QS(A,3,3) Terminate QS(A,4,5) Partition 3: A=1,2,3,3,4,… QS(A,4,4) Terminate QS(A,5,5) Terminate QS(A,6,8) Partition 6: A=1,2,3,3,4,6,5,7 QS(A,6,6) Terminate QS(A,7,8) Partition 6: A=1,2,3,3,4,5,6,7 QS(A,7,7) Terminate QS(A,8,8) Terminate |
2771 | https://knowledge.carolina.com/discipline/life-science/introductory-genetics-with-drosophila/ | Published Time: 2024-01-26T04:24:04+00:00
Introductory Genetics with Drosophila - Carolina Knowledge Center
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Introductory Genetics with Drosophila
Publication Date
Categories
January 25, 2024
AP Biology, AP Biology Activities, Biology, Biology Activities, LabSheets, Life Science, Life Science Activities
Carolina LabSheets™
Overview
Drosophila is one of the most popular model organisms for the study of genetics. This LabSheet can be used with a variety of our Drosophila cultures and sets in studies of monohybrid, dihybrid, and sex-linked crosses. The following crosses are especially recommended for introductory genetics activities although many others can be done. See our complete listing of Drosophilaculturesfor additional possibilities.
Download Student Lab Sheet
Monohybrid Crosses
apterous × wild
Needed cultures:172320 apterous (wingless)and172100 wild type. Available asset 171904. Apterous (ap) is recessive to wild type and is located on chromosome 2. The expected F 2 ratio is 3 wild type: 1 apterous.
vestigial × wild
Needed cultures:172460 vestigial (reduced wings)and172100 wild type. Available asset 171905.
Vestigial (ve) is recessive to wild type and is located on chromosome 2. The expected F 2 ratio is 3 wild type: 1 apterous.
sepia × wild
Needed cultures:172575 sepia (dark brown-eyed)and172100 wild type. Sepia (se) is recessive to wild type and is located on chromosome 3. The expected F 2 ratio is 3 wild type: 1 sepia.
Dihybrid Crosses
apterous × sepia
Needed cultures:172320 apterous (wingless)and172575 sepia (dark brown-eyed). Apterous (ap) is recessive to wild type and is located on chromosome 2. Sepia (se) is recessive to wild type and is located on chromosome 3. Apterous flies are wild type for eye color, and the sepia flies are wild type for wings. The expected F 2 ratio is 9 normal wings, red eyes: 3 normal wings, sepia eyes: 3 apterous, red eyes: 1 apterous, sepia eyes.
vestigial × sepia
Needed cultures:172460 vestigial (reduced wings)and172575 sepia (dark brown-eyed). Vestigial (vg) is recessive to wild type and is located on chromosome 2. Sepia (se) is recessive to wild type and is located on chromosome 3. Vestigial flies are wild type for eye color, and the sepia flies are wild type for wings. The expected F 2 ratio is 9 normal wings, red eyes: 3 normal wings, sepia eyes: 3 vestigial, red eyes: 1 vestigial, sepia eyes.
vestigial × ebony
Needed cultures:172460 vestigial (reduced wings)and172500 ebony (dark-bodied). Available asset 171915.
Vestigial (vg) is recessive to wild type and is located on chromosome 2. Ebony (eb) is recessive to wild type and is located on chromosome 3. Vestigial flies are wild type (tan) for body color and the ebony flies are wild type for wings. The expected F 2 ratio is 9 normal wings, normal body color: 3 normal wings, dark body: 3 vestigial, normal body color: 1 vestigial, dark body.
Sex-linkage
white × wild
Needed cultures:172220 white (white eyes)and172100 wild type. Available asset 171910. White (w) is recessive and is located on chromosome 1. Chromosome pair 1 consists of the sex chromosomes. The locus for white is on a position on the X chromosome that has no homolog on the Y; thus, white is inherited only through the X chromosome, making white a sex-linked (or X-linked) gene.
Needed Materials
F1 Drosophila Cultures
empty culture vials and plugs (173076)
vial labels or wax pencils
Formula 4-24® Instant Drosophila Medium (173200)
FlyNap® Kit (173010)
alcohol
Drosophila sorting brushes (173094)
index cards to use as sorting trays
Carolina™ Drosophila Manual (452620)
stereomicroscopes
Optional Materials
Carolina Drosophila Stand (173030)
173050 Drosophila Culture Kit
Drosophila Care Guide
Safety
Ensure that students understand and adhere to safe laboratory practices when performing any activity in the classroom or lab. Demonstrate the protocol for correctly using the instruments and materials necessary to complete the activities, and emphasize the importance of proper usage. Use personal protective equipment such assafety glassesorgoggles,gloves, andapronswhen appropriate. Model properlaboratory safety practicesfor your students and require them to adhere to all laboratory safety rules.
Procedures
Students can work individually or in groups of 2–4.
When your cultures arrive, open the package immediately and inspect the cultures to verify that they have arrived in good condition.
Examine the label on each vial. The label gives the name of the culture (wild, sepia, etc.) and, for mutants, the genetic symbol and chromosome on which the gene is located. For example, sepia se (3). The label is stamped with the date that the parent flies were placed in the vial. Flies should begin emerging 12–14 days after the date on the label. Each culture will produce approximately 100 flies over a 10-day period and will serve 6–8 groups. Either order enough cultures for the class or plan to expand the parent cultures you receive by subculturing.
Observing and Crossing Parent Flies
Students will need virgin females for setting up crosses to obtain their F 1. Clear (remove) all adult flies from the cultures 8 to 12 hours before students are to set up their crosses. This gives time for adults to emerge but not enough time for the females to mate. See the Carolina Drosophila Manual for more details.
vial with medium and plug
vial Label
FlyNap®Kit (for every 6–8 students)
sorting brush
2 index cards
stereomicroscope
Either prepare a vial with medium for each student group or provide materials and instructions to the students so they can prepare their own vials. Materials needed are culture vials with plugs and Formula 4-24®Instant Drosophila Medium. Open the bag of medium and locate the small measuring cup and packet of yeast. Using the small cup, add a level cup of medium to each vial. When all the vials have dry medium, follow with a level cup of cool water. Open the yeast packet and sprinkle 6–8 grains of yeast onto the surface of the medium in each vial. Do not add too much yeast. The yeast produces CO 2, which in large amounts can cause sterility or even death of the flies. If you want students to check for sex-linkage, either provide two vials with medium for each group or have groups do reciprocal crosses (e.g., group 1: brown-eyed females × red-eyed males, group 2: brown-eyed males × red-eyed females, etc.).
Either prepare the fly morgues with alcohol (one morgue is included in each FlyNap®Kit) or provide materials and instructions so the students can do this.
Seven to 10 days after the F 1 vials are set up, the parent flies need to be removed from the cultures. This is done to avoid mating of the F 1 flies with the parents. Materials needed:
FlyNap®Kit (for every 6–8 groups of students)
sorting brush
index card
Observing F 1 Flies and Setting up F 2 Cultures
F 1 flies will begin emerging about 12 to 14 days after the cultures are setup. Since the F 1 flies can only mate among themselves, it is not necessary to collect virgins to set up cultures for the F 2.
vial with medium and plug
vial label
FlyNap®Kit (for every 6–8 groups of students)
sorting brush
index card
stereomicroscope
Seven to 10 days after the F 2 vials are set up, the F 1 flies need to be removed from the F 2 cultures. Materials needed:
FlyNap®Kit (for every 6–8 groups of students)
sorting brush
index card
Scoring F 2 Phenotypes
F 2 flies will begin emerging about 12–14 days after students have set up their vials. Begin scoring phenotypes of the F 2 on the day after they first begin emerging. More females than males emerge on the first day, but the ratio evens out on successive days. Do not return scored flies to the vial; dispose of them in a morgue to prevent their being counted twice. Phenotypes can be scored every other day for up to 10 days. Counts carried beyond 10 days risk including flies of the next generation.
For scoring the phenotypes of the F 2 flies, students will need their F 2 vials and the following materials.
FlyNap®Kit (for every 6–8 groups of students)
sorting brush
index card
stereomicroscope
Optional
To make the activity more of an inquiry, remove the label from the F 1 culture vial and ask the students to analyze the cross without the information provided on the vial.
For a demonstration of the use of FlyNap®, students can watch our Observing Phenotypes with Drosophila melanogaster video.
Chi-square analysis is not included as part of the activity, but it can easily be added. There are several Internet sites that perform the calculation from input data.
Answer Key to Questions Asked on the Student LabSheet
Why is it necessary to remove the parent flies?
To prevent them from breeding with the F 1 flies.
2. Anesthetize the F 2 flies and examine them under a dissecting microscope. Record your observations below.
Observations should include the observed phenotype(s) of the F 1 and a comparison to the phenotypes of the parent flies.
3. What does this tell you about the alleles involved in the cross? Explain your answer.
By comparing the phenotype(s) of the F 1 flies to the parents, students can tell which are due to dominant alleles. They should also be able to tell if they are dealing with a monohybrid or a dihybrid cross. If they are dealing with a sex-linked trait, they should be able to see that the trait is not inherited equally by the male and female flies. Some students may state that they need more information. If so, note what they think is missing as this can reveal misconceptions or gaps in their knowledge.
4. Why is it not necessary to select virgin female flies from the F 1 to make this cross?
All the F 1 flies have the same genotype, and the F 1 flies are supposed to be crossed to produce the F 2.
5. Are you dealing with a monohybrid cross, a dihybrid cross, or a sex-linked cross? Explain your answer. If you need additional information before specifying the nature of the cross you are studying, state the additional information needed and how you will obtain it.
Answers will vary according to the data collected and the student’s background knowledge. Most students should know that if the parent flies differed by two sets of contrasting characteristics that they are dealing with a dihybrid cross. If the phenotypes are not being equally distributed between the male and female flies, they should suspect that they are dealing with a case of sex-linked inheritance.
6. Give here the gene symbols and their corresponding phenotypes that you will use in describing the cross.Note:students can use the accepted gene symbols and names for the phenotypes, or you can make the exercise more of an inquiry by removing the labels from the parent vials and replacing them with a generic label (Parent 1, etc.). Students can then select their own gene symbols and descriptions of the strains. If you allow students to do this, all should agree to use the same symbols. Otherwise, groups will be unable to communicate about their results. Of course, that could be an interesting discovery for them to make. The example shown below is for sex-linked white.
X w+ X chromosome with allele for wild-type eyes
X W X chromosome with allele for white eyes
Y W Y chromosome without locus for eye color
On the basis of your Punnett square, give the expected results for the F 2.
1 white-eyed female: 1 red-eyed female: 1 white-eyed male: to 1 red-eyed male
Sample Data for a Monohybrid Cross
Sample Data for a Dihybrid Cross
Sample Data for a Sex-linked White
Are your results compatible with your expected ratios? Explain.
Answers will vary according to data collected.
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2772 | https://www.cuemath.com/numbers/factors-of-144/ | Factors of 144
Factors of 144 are the list of integers that we can split evenly into 144. There are overall 15 factors of 144, of which 2 and 3 are its prime factors. The Prime Factorization of 144 is 24 × 32.
| What Are the Factors of 144?
| How to Calculate the Factors of 144?
| Factors of 144 in Pairs
| Important Notes
| FAQs on Factors of 144
What Are the Factors of 144?
Factors of a number are the numbers that divide the given number exactly with no remainder. Factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144. So,144 is a composite number as it has factors other than 1 and itself.
How to Calculate the Factors of 144?
We can use different methods like divisibility test, prime factorization, and the upside-down division method to calculate the factors of 144. In prime factorization, we express 144 as a product of its prime factors and in the division method we divide the number by whole numbers or integers. Let us see what all numbers divide 144 exactly without a remainder.
Prime Factorization by Division Method
The upside-down division got its name so because the division symbol is flipped upside down.
Therefore, prime factorization of 144 = 2 × 2 × 2 × 2 × 3 × 3.
2 and 3 are prime factors of 14.
Explore factors using illustrations and interactive examples.
Factors of 144 in Pairs
Factor pairs of 144 are the two numbers which, when multiplied, give the number 144.
Factors of 144 in Pairs | Notation
1 × 144 | (1,144)
2 × 72 | (2,72)
3 × 48 | (3,48)
4 × 36 | (4,36)
6 × 24 | (6,24)
8 × 18 | (8,18)
9 × 16 | (9,16)
12 × 12 | (12,12)
Therefore, factors of 144 in pairs are (1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), and (12, 12).
We can have negative factors also for a given number.
With factors, we are only looking for whole numbers that are equal to or less than the original number.
Important Notes:
Challenging Questions
Factors of 144 Solved Examples
Example 1: The area of the dining room at Thomas Jefferson’s home in Monticello is about 144 square feet. What are the possible dimensions of the room?
Solution:
The area of a rectangle is length × breadth.
Given area = 144 square feet
So, the possible length and breadth are the factor pairs of 144 (as the product of these pairs is 144).
Length of Room (Feet) | Breadth of Room (Feet)
1 | 144
2 | 72
3 | 48
4 | 36
6 | 24
8 | 18
9 | 16
12 | 12
(1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), and (12, 12).
Here, the layout of the rectangular room can be changed by swapping length by breadth. So, there are 16 possible options.
Example 2: Help Ria find the prime factors of 144.
Solution:
The factors of composite number 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, and 144.
The prime factorization of 144 is 1 × 2 × 2 × 2 × 2 × 3 × 3.
The prime factors of 144 are 1, 2, and 3.
Example 3: How many factors are there for 144?
Solution:
The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144. Therefore, 144 has 15 factors.
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FAQs on Factors of 144
What are the Factors of 144?
The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 and its negative factors are -1, -2, -3, -4, -6, -8, -9, -12, -16, -18, -24, -36, -48, -72, -144.
What is the Greatest Common Factor of 144 and 92?
The factors of 144 and 92 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 and 1, 2, 4, 23, 46, 92 respectively.
Common factors of 144 and 92 are [1, 2, 4].
Hence, the GCF of 144 and 92 is 4.
What are the Common Factors of 144 and 101?
Since, the factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 and factors of 101 are 1, 101. Hence, 144 and 101 have only one common factor which is 1. Therefore, 144 and 101 are co-prime.
What is the Sum of all the Factors of 144?
Sum of all factors of 144 = (24 + 1 - 1)/(2 - 1) × (32 + 1 - 1)/(3 - 1) = 403
What numbers are the Pair Factors of 144?
The pair factors of 144 are (1, 144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), (12, 12). |
2773 | https://brilliant.org/wiki/taylor-series-error-bounds/ | Taylor Series - Error Bounds | Brilliant Math & Science Wiki
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July Thomas and Jimin Khim contributed
The Lagrange error bound of a Taylor polynomial gives the worst-case scenario for the difference between the estimated value of the function as provided by the Taylor polynomial and the actual value of the function. This error bound (R n(x))\big(R_n(x)\big)(R n(x)) is the maximum value of the (n+1)th(n+1)^\text{th}(n+1)th term of the Taylor expansion, where M M M is an upper bound of the (n+1)th(n+1)^\text{th}(n+1)th derivative for a<z<x:a<z<x:a<z<x:
R n(x)=M(n+1)!(x−a)n+1.R_n(x)=\frac{M}{(n+1)!}(x-a)^{n+1}.R n(x)=(n+1)!M(x−a)n+1.
Contents
Derivation
Finding M
Calculating Error Bounds
Derivation
The n th n^\text{th}n th degree Taylor polynomial at x=a x=a x=a is
P n(x)=f(a)+f′(a)1!(x−a)+⋯+f n(a)n!(x−a)n.P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n.P n(x)=f(a)+1!f′(a)(x−a)+⋯+n!f n(a)(x−a)n.
Since the Taylor approximation becomes more accurate as more terms are included, the P n+1(x)P_{n+1}(x)P n+1(x) polynomial must be more accurate than P n(x):P_n(x):P n(x):
P n+1(x)=f(a)+f′(a)1!(x−a)+⋯+f n(a)n!(x−a)n+f(n+1)(a)(n+1)!(x−a)n+1=P n(x)+f(n+1)(a)(n+1)!(x−a)n+1.\begin{aligned} P_{n+1}(x) &= f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}\ &= P_n(x) + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. \end{aligned}P n+1(x)=f(a)+1!f′(a)(x−a)+⋯+n!f n(a)(x−a)n+(n+1)!f(n+1)(a)(x−a)n+1=P n(x)+(n+1)!f(n+1)(a)(x−a)n+1.
Since the difference between P n(x)P_n(x)P n(x) and P n+1(x)P_{n+1}(x)P n+1(x) is just that last term, the error of P n(x)P_n(x)P n(x) can be no larger than that term. In other words, the error R n R_n R n is
R n(x)=max(f(n+1)(a)(n+1)!(x−a)n+1).R_n(x) = \max\left( \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} \right).R n(x)=max((n+1)!f(n+1)(a)(x−a)n+1).
Since a a a and n n n are constant in this formula, terms depending only on those constants and x x x are unaffected by the max\max max operator and can be pulled outside:
R n(x)=max(f(n+1)(a))(n+1)!(x−a)n+1.R_n(x) =\frac{\max\big( f^{(n+1)}(a)\big)}{(n+1)!} (x-a)^{n+1}.R n(x)=(n+1)!max(f(n+1)(a))(x−a)n+1.
The largest value obtainable by f n+1 f^{n+1}f n+1 could not possibly exceed the maximum value of that derivative between a a a and x.x.x. Call the x x x value that provides that maximum value z z z and the error becomes
R n(x)=f(n+1)(z)(n+1)!(x−a)n+1.R_n(x)=\frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}.R n(x)=(n+1)!f(n+1)(z)(x−a)n+1.
Finding M
Let M M M be an upper bound on the (n+1)th(n+1)^\text{th}(n+1)th derivative of f(x)f(x)f(x) for the interval between a a a and x x x such that
∣f(n+1)(z)∣≤M\big| f^{(n+1)}(z) \big| \leq Mf(n+1)(z)≤M
for all z∈[a,x].z\in [a, x].z∈[a,x].
The upper bound of the (n+1)th(n+1)^\text{th}(n+1)th derivative on the interval [a,x][a, x][a,x] will usually occur at z=a z=a z=a or z=x.z=x.z=x. If given a defined interval on which to find the error, test the endpoints of the interval.
What is the upper bound of the third derivative of y=sin(x)y = \sin(x)y=sin(x) on the interval [0,2 π]?[0, 2\pi]?[0,2 π]?
The third derivative of y=sin(x)y=\sin(x)y=sin(x) is y(3)=−cos(x),y^{(3)} = -\cos(x),y(3)=−cos(x), which oscillates between -1 and 1. So M=1 M=1 M=1 and
∣f(n+1)(z)∣≤1.□\big| f^{(n+1)}(z) \big| \leq 1.\ _\squaref(n+1)(z)≤1.□
a 4 e a 2 a^4e^{a^2}a 4 e a 2 1 1 1 e a 2 e^{a^2}e a 2 a 4 a^4 a 4
Reveal the answer
What is the supremum of the fourth derivative of f(x)=e a x f(x) = e^{ax}f(x)=e a x on the interval [a,0][a, 0][a,0] for −1<a<0?-1<a<0?−1<a<0?
The correct answer is: a 4 e a 2 a^4e^{a^2}a 4 e a 2
Calculating Error Bounds
In order to compute the error bound, follow these steps:
Step 1: Compute the (n+1)th(n+1)^\text{th}(n+1)th derivative of f(x).f(x).f(x).
Step 2: Find the upper bound on f(n+1)(z)f^{(n+1)}(z)f(n+1)(z) for z∈[a,x].z\in [a, x].z∈[a,x].
Step 3: Compute R n(x).R_n(x).R n(x).
Find the error bound of the Maclaurin polynomialP 3(π 2)P_3\big(\frac{\pi}{2}\big)P 3(2 π) for f(x)=sin(x).f(x) = \sin(x).f(x)=sin(x).
The Maclaurin series is just a Taylor series centered at a=0.a=0.a=0. Follow the prescribed steps.
Step 1:Compute the (n+1)th(n+1)^\text{th}(n+1)th derivative of f(x):f(x):f(x):
Since P 3 P_3 P 3 is being investigated, n=3,n = 3,n=3, so write down the 4 th 4^\text{th}4 th derivative of f(x)=sin(x):f(x) = \sin(x):f(x)=sin(x):f(4)(x)=sin(x).f^{(4)}(x) = \sin(x).f(4)(x)=sin(x).
Step 2:Find the upper bound on f(n+1)(z)f^{(n+1)}(z)f(n+1)(z) for z∈[a,x]:z\in [a, x]:z∈[a,x]:
The Maclaurin series is centered on a=0 a = 0 a=0 and P n(x)=P 3(π 2)P_n(x)=P_3\big(\frac{\pi}{2}\big)P n(x)=P 3(2 π) implies x=π 2:x = \frac{\pi}{2}:x=2 π:f(4)(0)=sin(0)=0 f(4)(π 2)=sin(π 2)=1.\begin{aligned} f^{(4)}(0) &= \sin(0)=0\ f^{(4)}\left(\frac{\pi}{2}\right) &= \sin\left(\frac{\pi}{2}\right)=1. \end{aligned}f(4)(0)f(4)(2 π)=sin(0)=0=sin(2 π)=1. So M=1.M=1.M=1.
Step 3:Compute R n(x):R_n(x):R n(x):
R n(x)=M(n+1)!(x−a)n+1 R 3(π 2)=1(3+1)!(π 2−0)3+1=π 4 384.□\begin{aligned} R_n(x)&=\frac{M}{(n+1)!}(x-a)^{n+1}\ R_3\left(\frac{\pi}{2}\right)&=\frac{1}{(3+1)!}\left(\frac{\pi}{2}-0\right)^{3+1}\ &= \frac{\pi^4}{384}.\ _\square \end{aligned}R n(x)R 3(2 π)=(n+1)!M(x−a)n+1=(3+1)!1(2 π−0)3+1=384 π 4.□
Cite as: Taylor Series - Error Bounds. Brilliant.org. Retrieved 22:10, September 28, 2025, from
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2774 | https://www.govinfo.gov/content/pkg/GOVPUB-C13-84f8d22fa812849b3066386db90cef50/pdf/GOVPUB-C13-84f8d22fa812849b3066386db90cef50.pdf | UNITED STATES DEPARTMENT OF COMMERCE • Charles Sawyer, Secretary NATIONAL BUREAU OF STANDARDS • E U. Condon, Director Charts of Compressibility Factors and Charts Showing Quantities Delivered by Commercial Cylinders for Hydrogen, Nitrogen, and Oxygen by Harold J. Hoge, Cyril H. Meyers, and Robert E. McCoskey Miscellaneous Publication M191 Issued November 17, 1948 For sale by the Superintendent of Documents, U. S. Government Printing Office, Washington 25. D. C.
Price 25 cents Contents Page I. Introduction 1 II. Description of charts 1 III. Examples of use of the charts 3 IV. References 3 ii Charts of Compressibility Factors and Charts Showing Quantities Delivered by Commercial Cylinders, for Hydrogen, Nitrogen, and Oxygen by Harold J. Hoge, Cyril H. Meyers, Robert E. McCoskey Abstract For each of the gases hydrogen, nitrogen, and oxygen, two oharts are given.
The first gives directly the number of standard cubic feet of the gas which a cylinder will deliver.
The second chart gives values of the compressibility factor Z=PV/nRT and of the density p.
Examples of the use of the charts are given, also numerical values of constants used in the preparation of the charts or needed in the solution of problems.
I. Introduction The six charts published herewith present in a form convenient for engineering use some of the results of rather extensive investigations and analyses of the existing data for hydrogen, nitrogen, and oxygen.
The work on hydrogen was performed by Woolley, Scott, and Brickwedde [l].
1 That on nitrogen was done chiefly by R. B. Scott and H. W. Woolley and has not been published.
The work on oxygen was performed by Cyril H. Meyers .
II. Description of Charts The first three charts (1, 2, and 3) give the number of standard cubic feet of gas which a cylinder will deliver when the pressure and temperature of the gas and the volume of the cylinder are known.
The last three charts (4, 5, and 6) give values of the compressibility factor Z=PV/nRT and are useful in solving a wide variety of gas-law problems.
Charts 1, 2, and 3 are for a cylinder volume of 2,640 cubic inches (1.528 cu ft).
This is approxi-mately the minimum volume of the commonest commercial cylinder.
The average volume as given in table 1 is about 1 percent greater.
The quantity which a cylinder will deliver is equal to its contents less the amount remaining in the cylinder when the gage pressure has fallen to zero, that is, when the absolute pressure has fallen to atmospheric.
In preparing charts 1, 2, and 3, gage pressure was taken to be equal to absolute pressure minus 14.696 lb per sq in., and the gas remaining in the cylinder was assumed to be at 70° F.
The accuracy of these charts is greater than the accuracy with which they can be conveniently read.
Hence the only error introduced by the use of these charts is the reading error.
Charts 1, 2, and 3 may be used for cylinders having volumes different from 2,640 cu in. provided the quantities delivered appearing on the charts are mlti plied by F/2640, where V is the volume of the given cylinder in cubic inches.
Table 1 contains data on a number of the more common sizes of com-mercial cylinders normally used for oxygen, hydrogen, nitrogen, helium, neon, argon, and other gases.
This table was compiled from data kindly furnished us by Franklin R. Fetherston of the Compressed Gas Manufacturers' Association.
In general the cylinders listed are manufactured to comply with the Interstate Commerce Commission's 3A or 3AA specifications.
These are seamless drawn cylinders, the A type being steel of less than 0.55 percent carbon content, and the AA type alloy steels which permit the cylinder to be lighter.
In most cases the tolerances of an individual manufacturer on cylinder volume are smaller than the tolerances given in the table, the minimum volume in the table generally coming from one manufacturer and the corresponding maximum volume from another.
The volumes given in the table are only approximate and no estimate of their accuracy can be given.
It is inter-esting to note that about the only volume on which there was general agreement was the 2,640 cu in.
minimum volume of the most common commercial cylinder.
Column 4 of the table contains values of the factor F/2640.
» Figures in brackets indicate the literature references at the end of this paper.
792417—48 1 Table 1.
Data on some common commercial gas cylinders used for oxygen, hydrogen, nitrogen, helium, neon, argon, and other gases Volume Min cu in.
168 278 315 1, 320 2, 340 2, 640 2, 970 3, 955 14, 925 Avg cu in.
173 290 325 1,352 2, 390 2, 666 3, 026 4, 006 15, 000 Max cu in.
180 303 332 1,386 2, 420 2, 720 3, 066 4, 075 15, 075 Multiplying factor » for use with charts 1, 2, 3 0. 0636 . 1053 . 1193 . 5000 . 8864 1 1. 1250 1. 4981 5. 6534 Inside diameter in.
3% 4/ 8 8% 10 Length •> in.
16% 25% 26H 43 51 51 55 56 d 20 ft 6% in.
Service pressure lb per sq in.
2, 015 2, 015 2, 015 f 2, 015 I.
2, 265 2, 000 2, 015 2, 265 2, 400 2, 400 2, 400 Rated capacity for oxygen std cuft 14 23 26 110 125 "220 (206 for N 2) (192 for Hjj) 250 300 400 1, 500 Rated capacity for medical oxygen gal 80 or 95 130 or 165 150 1, 150 or 1, 400 1, 300 • Based in all cases on the minimum volume.
b Over all length, not including valve or protecting cap.
« One standard gallon equals 231/1728 std ft,' but rated capacity for medical oxygen is based on lower pressures than are used for industrial oxygen.
_ V i t 2000 3000 5000 6000 PRESSURE LBS. PER SQ. IN.ABS.
Otwt& O.Tj«»S«.
Own )<«<) tfcewBc««ibilir>(o«twZ-PVflRI.
Pressure NATIONAL BL 1000 1.25 2000 Pressure (lb in" abs) 3000 4000 5000 1.20 A cr :> CL n M 6000 - 1.07 cr ii N Pressure (lb in?
abs)-Okflit 4.
Hydrojfn 901.
CKort S'.tnj lh« ewcuuibjttiy fo<lat Z - PV nRT. |
2775 | https://suttonscreek.com/combination-product-ecosystem/ | Combination Product Ecosystem | Suttons Creek
Combination Product Ecosystem | Suttons Creek
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Creating a supportive combination product ecosystem within your drug world
Often when we are hired by a Pharma or Biotech company to support their drug delivery device development, even in the most sophisticated of companies that have been to the combination product rodeo before, we find that they do not have a culture that fosters combination product development success. There are a few main reasons for this:
Drugs/biologics and devices are developed very differently, and drug companies tend to try to fit a device’s square peg into the drug’s round hole.
Drugs/biologics and devices are regulated similarly, in some cases by the same regulatory body (the FDA), however regulatory schemes are different and the divisions responsible for the review may be different.
Management does not provide sufficient resources and weight to the importance of, and unique processes of, device development.
When embarking on a combination product development program, it’s essential to recognize that each initiative is unique. The complexity and development phase will shape your company’s strategies and understanding certain key factors will play a critical role in building a robust combination product organization. These factors can be broadly categorized into:
Differences in Drug and Device Development Processes:Understanding the distinct timelines, methodologies, and requirements for drug and device development is crucial to aligning these processes effectively.
Differences in Drug and Device Personnel and Cross-Functional Teams:Bridging the gap between professionals with drug-focused expertise and those with device-specific knowledge is vital to fostering effective collaboration.
Differences in Drug and Device Regulatory and Quality Schemes and Strategies:Successfully navigating the regulatory frameworks governing drugs and devices requires a tailored, integrated approach.
Added Complexities of Post-Market Support and Surveillance Activities:Ensuring compliance and addressing post-market requirements for combination products demand careful planning and execution.
The Role of Management in Driving Success:Leadership must prioritize cross-disciplinary integration, resource allocation, and strategic oversight to guide the program toward its objectives.
Below we will discuss these areas, beginning with differences in drug and device development processes, defining key details companies must grasp to better position themselves to manage the inherent challenges of combination product development while ensuring regulatory compliance and market success. Each week, we will add on to this article to complete the blueprint for creating a supportive combination product sub-ecosystem in your drug world.
Differences in Drug and Device Development Processes
The development processes come from two very different approaches.Drug/biologic product formulation centers around discovery; scale up and refinement of the manufacturing process, validating, testing, establishing specs, then tightening the specifications as more data is collected until a safe and effective product is realized.
Meanwhile, device development is fundamentally different.
The design process for medical devices is governed by a subset of regulations, ensuring a comprehensive design control framework. It all starts with identifying the intended use of the device and understanding the needs of the users. These insights are then transformed into Design Input Requirements, which are tailored to the capabilities and conditions of the intended users.
As the development progresses, the device undergoes rigorous testing to ensure that the Design Inputs align with the expected outputs, a process known as Design Verification. Additionally, the device must be validated to confirm that it effectively meets the user needs, referred to as Design Validation. This structured approach not only enhances the safety and efficacy of medical devices but also ensures they are user-friendly and reliable.
The first hurdle companies face is recognizing this difference and not trying to shoehorn device development into the same formulation mentality as drugs and biologics. The second hurdle is making these two different processes work together by identifying critical engagement timings and understanding the differences and similarities in methodology and terminology between drug development (quality by design) and device development (design controls).
Developing a combination product means tackling new research tasks that extend far beyond traditional drug development. These tasks encompass all aspects of the device constituent necessary to bring the product to market. For instance, if your project involves creating a novel device, device development itself becomes a significant undertaking. This process includes engineering, design, and early characterization testing to optimize and ensure the robustness of your device, ensuring functionality and safety in your intended use environment. Even in instances where you select to use an established platform, engineering activities will still be needed to ensure the platform operates with your unique drug product. This can take on various forms of testing and analysis as outlined below.
Usability and Human Factors play a critical role in this development journey. These efforts focus on gathering real-world evidence to confirm that patients can effectively and safely use the device, particularly in home settings.
Design Verification is a crucial step in the product development process where you will generate the data and objective evidence to show your device meets its engineering requirements, often across a range of preconditions and stress conditions (Temperature, physical agitation, humidity, etc.).
Labeling takes on added complexity, as it must cover not only the drug’s information but also comprehensive Instructions for Use (IFU) for the device. Additionally, compliance with laws governing disposal becomes a critical consideration, especially for used or potentially contaminated components like injectors.
Other essential characterization and analysis activities involve compatibility testing, preconditioning or initial stability testing. Preconditioning simulates the aging and transport conditions of the drug-device combination, while stability studies will be critical for establishing the product’s shelf life.
These examples only provide a glimpse into the considerations a company must make. Each of these elements represents a distinct challenge, underscoring the importance of a holistic and well-coordinated approach to combination product development.
Differences in Drug and Device Personnel and Cross-Functional Teams
Device development requires the expertise of engineers.While there are plenty of engineers at Pharma and Biotech companies, they are usually associated with manufacturing operations or research and development functions and have a more scientific or academic mindset. On the device side of things, you typically need engineers who are more heavily mechanically inclined or think in systems and system integration. What may seem like simple development on the surface often turns out to be a highly complex integration of components, systems, processes and decision making that needs to function through your product’s Life Cycle Management activities. Because of this, a development operations mentality and strong combination product experience in the selected device type is required to get the job done right the first time.
The development of a drug delivery system adds new external partners to the mix.Setting up management processes and responsibilities for these new partners, creating the right amount of transparency, and clarifying roles and communication protocols are critical to taming this new beast that just grew some more arms and legs. Below is a list of your new appendages:
Your device constituent will be purchased and developed with an OEM (Original Equipment Manufacturer)
You may contract a CRO (Contract Research Organization) to outsource research services for device clinical or commercial development
Your device constituent inventory will be manufactured and the final product assembled by a CDMO (Contract Development and Manufacturing Organization)
You will most likely outsource Human Factors and Useability Studies to a contract organization
Various testing activities will be outsourced to vendors: sterilization, design verification / stability, biocompatibility
Your IFU (Instructions For Use) will likely be designed by an external expert
Your packaging/IFU will be manufactured by an external vendor
“So, you’re saying I just need to hire a few companies, and the development will take care of itself?” Absolutely not! As the Marketing Authorization Holder, you bear ultimate responsibility for managing and executing the entire process. This includes having the right team on the inside maintaining diligent oversight, fostering effective cross-collaboration, and ensuring timely reviews and summaries at key milestones. Throughout the development journey, it’s crucial to actively identify, mitigate, and control risks specific to your product’s intended use and users. This level of accountability ensures the integrity and success of your combination product.
As if adding new external partners wasn’t challenging enough, internal stakeholders will also have to take on new responsibilities across the business. One example of this will be the Commercial Team. With the introduction of a combination product, this will require the commercial team to have a newfound exposure to device constituent training and education as well as cross-functional training on new Commercial activities and the resulting interactions needed between the Commercial team, Device teams, and suppliers. For example, User Needs and Design Inputs. This is a completely new responsibility for a Combination Product development team. While drugs have a Target Product Profile (TPP) that itemizes all of the drug characteristics needed for the product, User Needs and Design Inputs are unique to medical devices and combination products. User Needs and Design Inputs are typically drafted by the product development or engineering team with advisement and input from the Commercial Team, Clinical Team and other key stakeholders. This input should include providing User and Stakeholder needs to the engineers that relate to the salability and consumer appeal of the device.
It’s easy for a drug company to dismiss the device as simply “the container for our drug,” but it’s crucial to recognize that the device is what the end user directly interacts with. This interaction plays a significant role in shaping perceptions of the drug’s brand. For teams accustomed to focusing solely on the pharmaceutical aspects, this shift in perspective can feel like learning an entirely new language. However, it’s a critical element of development that cannot be overlooked!
Another group that can be heavily impacted by the transition from drugs to combination products is Strategic Sourcing. Pharmaceutical companies typically rely on Strategic Sourcing or Procurement teams to identify, evaluate, and approve suppliers. For combination products, this process extends beyond traditional raw materials to include highly specialized device manufacturers. Suppliers may range from providers of simple components, such as needles or plungers, to producers of complex finished devices, such as autoinjectors or inhalers. Each new supplier relationship demands rigorous vetting to ensure compliance with regulatory requirements and alignment with the product’s specifications. The Combination Product team often plays a critical role in this process, ensuring that the selected suppliers can meet the unique needs of the combination product.
An essential element in managing device suppliers is the establishment of Quality Agreements, which outline the roles, responsibilities, and quality standards required of each supplier. These agreements are particularly critical for combination products, where the failure of even a single component—such as a needle in an autoinjector—can disrupt the entire supply chain. For example, a supplier’s inability to meet demand or comply with regulatory requirements could delay not only the development and approval process but also the product’s availability in the market. Changes to components, materials or processes additionally must be carefully evaluated and traced back to the programs Design History File (DHF) and Risk Management File (RMF) to ensure it remains current at all times during the products life cycle.
Differences in Drug and Device Regulatory and Quality Schemes and Strategies
There are very specific requirements and regulatory pathways with nuanced approaches needed for each drug/biologic and device combination. There is no cookie cutter regulatory pathway. Many of the niche situations you will find yourself dealing with are not well covered, or not covered at all, in guidance. Even with knowledgeable combination product specialists on the payroll, if they have not worked on that device type, with that particular niche situation before, or in a while, their knowledgebase may be stale. Employee turnover, which tends to be relatively high in the Pharma/Biotech world, can also contribute to a knowledge gap.
To complicate your regulatory strategy further, guidance is continually evolving and differs globally. This makes it necessary to constantly keep up with regulatory intelligence and modify regulatory strategies, having team members who are steadily focused on combination products. For example, below are some of the various permutations and combinations that your combination product could fall within.
Single Entity Products
Drug PMOA — filed as IND (investigational); and NDA (small molecule) or BLA (large molecule)
Device PMOA — filed as IDE (investigational); and PMA or 510(k) depending on classification
Co-Packaged Products
May be one submission (e.g., same as single entity), or
May be a combination of submissions, e.g., 510(k) for device constituent and BLA for drug constituent;
Even with multiple submissions, the PMOA submission needs to include reference to the other licenses
Determinant is typically whether the constituents included in the package are “off- the shelf” and have existing registrations
Cross-Labeled Products
Typically, two separate submissions, connected through labeling
When it comes to quality systems, drug development primarily falls under Good Manufacturing Practices (GMP), emphasizing chemical stability, sterility, and consistent pharmacokinetics. Key processes focus on raw material controls, process validation, and robust batch record documentation to ensure drug safety and efficacy.
In contrast, device development adheres to Quality System Regulations (QSR), particularly 21 CFR Part 820 in the U.S., which prioritize design controls, risk management, and traceability. Device quality systems are heavily driven by iterative testing, human factors engineering, and post-market surveillance to ensure safety and usability.
When developing a combination product, the quality system must account for both GMP and QSR, often requiring a hybrid approach that carefully integrates the two processes in a manner that is compliant yet not overly burdensome. At Suttons Creek, we often encounter clients diving into combination product development without a robust 21 CFR Part 4-compliant quality system. This is obviously not ideal and often needs careful remediation tailored to the company’s specific needs. However, overcorrecting by implementing an overly complex quality system can be just as problematic. We’ve seen companies burdened by cumbersome processes that lead to a chain reaction of non-conformances, CAPAs, and deviations, which may signal to regulators that the system isn’t functioning as intended. The key lies in striking a balance—creating a quality system that ensures compliance while remaining practical and aligned with the organization’s scale and maturity. A right-sized approach, grounded in risk management and streamlined processes, reduces inefficiencies and positions companies to navigate the complexities of combination product development effectively, avoiding delays and ensuring safety and compliance.
Collaboration across teams with expertise in drugs and device quality systems is vital. Establishing a unified quality system that draws from both GMP and QSR ensures that the final product meets all safety, efficacy, functionality and usability requirements.
Added Complexities of Post-Market Support and Surveillance Activities
Upon approval of your product the complex work continues! Below we look at a few specific examples and considerations you and your team need to keep in mind.
Scaling issues with a combination product.
Scaling is different for a combination product from clinical studies through marketing. Clinical studies require a smaller number of devices to be produced than the volume that will be required for commercial launch. The planning for this transition often needs to take places years in advance to ensure a smooth transition to market volumes. Detailed plans need to be made for a weighty supply increase if this product is anticipated to be widely used (think about some big therapy areas, such as ulcerative colitis and Crohn’s disease). These changes might entail not just ordering larger volumes but potentially mold modifications or small design changes as you move from manual or semi-automated processes to fully automated assembly processes. All of this adds both technical and logistic complexity that needs to be managed with a high degree of finesse.
In comparison to a stand-alone drug, a combination product contains drug plus device(es), the associated labeling and packaging, instructional material and even potentially training devices. Careful consideration needs to be paid to all aspects of the process, including drug product supply, device supply capabilities (suppliers), CMO to fill the drug into devices, and packaging and labeling CMOs. All of these finish steps may be done by one manufacturing site or multiple manufacturing sites that need to operate with a high degree of coordination. If your finished combination product requires refrigeration, then transport from one finishing site to another must be controlled. These logistics can add up to real world impact – for example, think of all the press about shortages of weight management drugs like Ozempic, Mounjaro, Zepbound and Wegovy. In many instances with these weight management drugs, the drug itself is available but other elements of the supply chain are causing a bottleneck.
Commercial and Marketing Challenges
Launching a combination product introduces unique challenges that extend beyond traditional pharmaceutical development. These challenges, present during both development and post-approval phases, require strategic planning and cross-functional collaboration. While many decisions are technical in nature, they significantly influence the commercial claims and approaches available to your business. Key considerations include:
Pricing and Reimbursement Strategy:Developing a pricing model for combination products is complex, as it must account for the value of both the drug and device components. Reimbursement pathways can vary significantly by market, requiring robust evidence to support cost-effectiveness, such as real-world data or health economics models.
Shelf Life Considerations:Unlike standalone pharmaceuticals, combination products must address the shelf life of both the drug and the device. Stability studies must evaluate how these components interact over time to ensure safety, efficacy, and usability.
Industrial Design and Branding:The industrial design, including colors, labeling, and ergonomic considerations, is critical for usability and market differentiation. These elements must be defined early in collaboration with engineering teams, as downstream design changes can lead to delays, increased costs, and regulatory hurdles.
Managing Regional Labeling and Language Variations:Global distribution necessitates localized labeling and instructions for use, often in multiple languages. Ensuring compliance with regional regulations and cultural expectations is critical to successful product adoption.
End-User Support and Training:Effective end-user support strategies can drive adoption and adherence while reducing misuse risks. These strategies may include:
Nurse Trainers/Educators: Providing in-home training for patients, particularly for devices with complex functionality.
Web and Phone Support: Offering accessible resources, such as instructional videos or live support, to guide patients through device use.
Integration with Specialty Care Providers: Facilitating connections between patients and physicians or specialty pharmacies for prescriptions and product access.
Note – The FDA closely monitors training materials, websites, and videos to prevent unsupported promotional claims. Training methods may also need to be validated through Summative Human Factors studies to ensure their effectiveness and compliance.
How can management set the combination product team up for success?
It is very easy, based on the proportionate size of the needed drug/biologic development budget and timeline versus the device development budget and timeline, that the device is an ancillary side-project, or even afterthought, of the drug/biologic. Think about how much strategic focus you put into buying a home versus the brainpower you put into selecting a banana at the grocery store. The difference feels like that to Pharma/Biotech leadership. However, drug delivery systems are a very user-centric product, and the combination product’s market success will rely equally on the user’s opinion of the device as it does on the effectiveness of the drug. Inadequate attention to your device constituent can also delay launch when drug/biologic and device processes are not aligned or you run into regulatory approval delays. We have been called in by many a company that ran into problems on the device side that caused their product to not get approved. Changing this mindset is a necessity that many companies realize too late.
To complete our thoughts on setting up a combination product ecosystem within your drug world, below is a listing of the top eight pieces of advice we give our clients:
1.When a drug/biologic company views devices in the above way, it often goes beyond process into a cultural space – making change more difficult and requiring change to start from the top down.
2.The device team cannot be siloed nor tacked on as a sub-team of a drug team. Instead, the device team needs to become an integrated part of an overall product development team, adopting a cross-functional, collaborative approach. This product development team must expand to include all of the additional stakeholders that come with the addition of a device constituent mentioned earlier (OEM, CDMO, supply chain, testing facilities, the Commercial team), ensuring phase-appropriate cooperation and inputs. Below is a listing of internal teams that need to come together for success.
3.The device team needs to have current hands-on experience in the type of device being brought to market. Often Pharma and Biotech companies hire device experts, but they have purely medical device experience and lack the combination product knowledge that will support alignment of development processes and a successful regulatory pathway. This is why it is highly recommended that these companies strategically outsource certain roles and responsibilities to combination product consultants who have worked across many combination products and can bring current best practices to the table.
4.All stakeholders need to understand the basic considerations and critical challenges of drug delivery system development, what their role is within the overall process, and how other functions can affect the work they are doing to bring it to market. When a team is created for a new combination product, trainings should be held to ensure all team members have a basic understanding of the combination product arena and get updated on current best practices and regulatory changes (there are always changes, thanks to the newness of this industry and its constant innovation). This education will support the development of SOPs that create alignment between the various functions within the overall product development team, including external partners, that gives everyone protocols to follow. A solid knowledgebase allows understanding of the work to be done, so these SOPs are written to guide the work without being over prescriptive.
5.Alignment is utterly dependent on cross-functional/stakeholder collaboration and communication systems. For a combination product to get developed successfully, you will be bringing multiple internal departments together, along with external partners. Furthermore, you will be communicating with regulatory bodies that will be guiding deliverables contributed to through collaborative efforts by multiple stakeholders. Since all of these people will not be sitting in the same room together, knowing when to engage is a key element of alignment. Having a communication strategy and agreed upon communication platforms throughout the lifecycle of the product, from strategy creation through postmarket surveillance is critical to getting your combination product on the market…and keeping it there.
6.A full lifecycle approach in which your product team is taking upstream and downstream effects of decision-making into account from strategy through postmarket surveillance is necessary in combination product development. Decisions made about device design can affect how it interacts with the drug, while drug formulation decisions affect the design inputs for device development and combination product regulatory strategy. Market research and marketing decisions made by the Commercial team can affect device design, along with Human Factors and Useability studies and the resulting design change and packaging/IFU decisions made. Device design decisions and Human Factors/Useability studies will affect the user training and support initiatives designed by the Commercial team. And the list goes on…
7.Cultural change management needs to take place to educate and prepare the organization, from the executive suite down to those executing project tasks, to prepare and support the individuals involved for this new mindset.
8.Leadership influence/support of the above. Messaging from the top down is important to ensure the entire product team
Acknowledges that a combination product is more than the sum of its parts from a regulatory and development perspective and that the device is not just some “off the shelf” shell that goes around the drug product at the end (for the record, there is really no such thing as an off the shelf drug delivery device).
Recognizes when the company doesn’t have the needed experience and filling those gaps in strategic minds and execution specialists (knowing there are issues are half the battle).
Buys into a collaborative, cross-functional approach to align stakeholders and drug/device processes.
Supports continual education and cultural change management within the combination product teams (turnover and time away from combination product projects can contribute to a shift away from your desired culture and a constant need for reeducation).
AUTHOR
Bryan Bobo, Principal Consultant, Suttons Creek –With over 9 years of experience in the pharmaceutical industry, Bryan has extensive knowledge of combination product development and what it takes to launch products on the market. He has practical and hands-on experience in various types of drug delivery systems from his years as both a Device Development Engineer as well as project and program management positions. In addition to his direct product development experience, Bryan engages with the global combination product community through his participation on various international standards committees as well as other various industry forums.
Jonathan Amaya-Hodges, Director, Technical Services, Suttons Creek –Jonathan has over 16 years of multidisciplinary experience in regulated medical products (drugs, biologics, medical devices, and combination products) at multiple global companies. He has practical experience in Development/Engineering, Quality Assurance, and Regulatory Affairs for various types of combination products with a focus on drug delivery. Additional background includes digital health (including smart packaging/connected devices and software as a medical device, or SaMD) and in vitro diagnostics, along with clinical development (bridging) and lifecycle management for combination products. Jonathan engages with the global combination product community by speaking at conferences, lecturing in courses, serving key roles within prominent industry organizations, and interfacing with regulators on a variety of topics.
Carolyn Dorgan, Director, Technical Services, Suttons Creek –Carolyn Dorgan has 10+ years of experience in the Medical Device and Combination Product industries, including 6 years at the FDA leading the Infusion Devices team. Carolyn is active in the international regulatory community through speaking engagements and has participated in over a dozen international standards ranging from infusion devices, to needle-based injection systems, on-body delivery systems, and infant incubators. In addition, her unique approach to product development and risk management has been instrumental in working with products for which there are no established performance or regulatory standards which include pediatric medical devices, novel on-body drug delivery systems, and rare disease combination products. Carolyn has a demonstrated ability for working through complex problems at a system level to implement practical solutions using her multi-disciplinary background which includes engineering, software, cybersecurity, human factors, and business.
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Algorithm for generating a cross-sum matrix game
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I'm trying to generate a matrix, like for a cross-sum game, where in a matrix of random numbers, for a given a sum (or a product, depending on a chosen operation) for each row and column, there's exactly 1 way to "deactivate" (meaning, to exclude the number from the final sum or product) correct numbers so that each row and column end up summing the active numbers to the correct sum.
To illustrate this, let's say I have a 3x3 matrix, and chosen sums (numbers next to represent the sum):
```
12 5 3
4 1 2 3 4
9 4 5 6 9
7 7 8 9 7
```
In order to solve this, I would need to deactivate numbers 2, 6, 9 and 8.
One way to generate a matrix with needed sums it to just generate the numbers, and then choose which ones to exclude at random. However, the drawback is that for bigger matrices, like 7x7, 8x8, there's a good possibility that there will be more than 1 solution.
Another solution I'm thinking of is to exclude the numbers that can add up to another for each row / column. For example if a required sum is 5, then 4 2 1 3 would be invalid because (4 + 1 and 3 + 2), but this seems rather complicated and inefficient.
If anyone has any pointers, I'd greatly appreciate it. This seems like it's a solved problem, but I have no idea what to look for.
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asked Jun 12, 2019 at 21:32
StefanStefan
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The fact that 5 can be 1+4 or 3+2 doesn't mean their is differents solutions because the thing is that you have to make a choice taking in account column + line. After 10 minutes testing I am not able to find, in 33, any ambigious situation, could you show an ambigious matrix with differents solutions and not only a line ?PopHip –PopHip 2019-06-13 10:42:00 +00:00 Commented Jun 13, 2019 at 10:42
1 You probably should look to en.wikipedia.org/wiki/Hungarian_algorithmSugar –Sugar 2019-06-13 15:18:05 +00:00 Commented Jun 13, 2019 at 15:18
One possibility is to build a solver for your puzzle. If the solver can't solve the puzzle, then re-shuffle.Jakob Lovern –Jakob Lovern 2019-06-13 23:34:50 +00:00 Commented Jun 13, 2019 at 23:34
1 I feel like you could use Dancing Links on this, but have no idea on how to implement...Jakob Lovern –Jakob Lovern 2019-06-13 23:36:25 +00:00 Commented Jun 13, 2019 at 23:36
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Checking random grids with a solver
For a matrix of limited size, up to 10×10 or so, a simple solver can quickly find the solutions. If there is only one, even the quick'n'dirty solver I wrote in javaScript usually finds in it less than a second.
I used a simple recursive row-by-row solver which works like this:
For each column, iterate over every possible selection of numbers and check whether excluding them gives the column the correct sum. Then check whether any numbers are part of all or none of the valid selections; these will have to be included and avoided in all selections.
For each row, iterate over every possible selection of numbers and check whether excluding them gives the row the correct sum, and whether they contain all of the to-be-included and none of the to-be-avoided numbers identified in the previous step. Store all valid selections per row.
After these preparations, the recursive part of the algorithm is then called:
Receive a matrix with numbers, a list of sums per row, and a list of sums per column.
For the top row, check whether any of the numbers cannot be excluded (because the numbers below it add up to less that the sum for that column).
Iterate over all valid selections of numbers in the top row (as identified in the preparation phase). For each selection, check whether removing it gives the row its correct sum. If it does, recurse with a copy of the matrix with the top row removed, a list of sums per row with the first item removed, and a list of sums per column with the non-excluded numbers in the top row subtracted.
Starting from a pattern like this, where the X's indicate which cell will be excluded:
- - - X - - - X - -
- - - - X - X - - -
X - - - - X - - - -
- X - - - - - - - X
- - X - - - - - X -
- X - - - - - X - -
X - - - - - - - X -
- - - - X - - - - X
- - - X - X - - - -
- - X - - - X - - -
I let the matrix be filled with random numbers from 1 to 9, and then ran the solver on it, and about one in ten attempts results in a grid like this, which has exactly one solution:
```
4 1 3 8 1 3 4 1 1 8 25
9 9 7 8 1 1 3 2 1 7 44
9 8 8 1 5 5 9 2 2 6 41
4 6 8 1 9 2 1 7 1 5 33
9 4 2 4 4 5 8 6 3 8 48
8 5 6 9 6 6 6 4 1 8 50
4 3 2 4 8 7 6 7 9 1 38
6 7 8 1 9 9 9 4 6 7 50
7 7 1 7 9 6 2 7 1 2 36
3 3 8 8 9 2 4 9 6 8 48
50 42 43 36 51 35 45 44 19 48
```
When using only numbers from 1 to 9, grids with only one solution are easy to find for smaller grids (more than half of 8×8 grids have only one solution), but become hard to find for grid sizes over 10×10. Most larger grids have many solutions, like this one which has 16:
```
4 1 5 7 2 2 5 6 5 8 32
5 1 1 6 4 6 5 2 2 9 32
9 2 3 8 7 7 4 8 3 6 41
4 8 1 8 4 3 1 9 7 2 37
4 6 9 8 8 5 8 6 6 5 50
1 5 5 5 1 3 5 7 7 1 28
5 5 1 7 2 9 2 6 3 8 40
9 8 9 2 8 3 1 9 6 8 47
5 1 3 7 1 2 6 1 8 9 34
1 5 1 2 1 1 1 6 4 3 23
33 29 28 46 26 32 32 47 42 49
```
The number of solutions also depends on the number of excluded numbers per row and column. The results shown above are specifically for the pattern with two excluded numbers per row and column. The more excluded numbers, the greater the average number of solutions (I assume with a peak at 50% excluded numbers).
You can of course use a random pattern of cells to be excluded, or choose the numbers by hand, or have random numbers chosen with a certain distribution, or give the matrix any other property that you think will enhance its usefulness as a puzzle. Multiple solutions don't seem to be a big problem for smaller grids, but it is of course best to check for them; I first ran the solver on a grid I had made by hand, and it turned out to have three solutions.
Choosing the excluded values
Because the value of the excluded numbers can be chosen freely, this is the obvious way to improve the chance of a matrix having only one solution. If you choose numbers that don't occur anywhere else in the row and column, or only once, then the percentage of 10×10 grids that have only one solution rises from 10% to 50%.
(This simple method obviously gives a clue about which numbers should be excluded – it's not the numbers that occur several times in a row or column – so it's probably better to use how many times each number occurs in the whole grid, not just in its own row and column.)
You could of course choose excluded values that add up to a number that can't be made with any other combination of values in the row or column, and that would guarantee only one solution. The problem with this is of course that such a grid doesn't really work as a puzzle; there is only ever one way to exclude values and get the correct sum for every row and column. A variant would be to choose excluded values so that the sum of the row or column can be made in exactly two, or three, or ... ways. This would also give you a way to choose the difficulty level of the puzzle.
Sudoku – avoiding duplicate values
The fact that larger grids have a higher chance of having more than one solution is of course linked to using only values for 1 to 9. Grids of 10×10 and greater are guaranteed to have duplicate values in every row and column.
To check whether grids with no duplicate values per row or column are more likely to lead to only one solution, the obvious test data is the Sudoku.
When using random patterns of 1 to 3 cells per row and column to be excluded, around 90% of cross-sum matrix games based on Sudokus have only one solution, compared to around 60% when using random values.
(It could of course be interesting to create puzzles which work both as a Sudoku and as a cross-sum matrix puzzle. For every Sudoku it should be easy to find a visually pleasing pattern of excluded cells that has only one solution.)
Examples
For those who like a challenge (or want to test a solver), here's a cross-sum Sudoku and an 11×11, 12×12 and 13×13 cross-sum matrix puzzle that have just one solution:
```
. 3 . 4 . . . . . 36
. 6 . . 9 . . 4 5 35
4 . . . . . 9 . . 33
. . 3 . . 1 . . . 39
. . . . . 8 2 . 3 29
. 7 . . . 2 6 . 9 40
. 2 . . . . . . . 33
3 . 8 . . . . . . 31
. . 7 . 5 . . 6 4 36
33 34 35 37 27 42 34 32 38
```
```
6 6 5 2 9 4 4 6 7 1 8 44
1 8 1 1 4 7 3 3 3 1 2 25
5 8 7 7 5 5 6 1 7 6 5 43
8 9 6 2 9 1 6 2 9 8 3 59
8 8 2 3 6 3 7 7 5 9 8 53
8 2 7 2 6 2 9 4 7 1 2 47
3 9 2 8 8 4 2 9 3 6 6 50
3 1 8 2 6 4 1 7 9 4 6 42
8 3 6 7 8 5 4 4 2 8 4 46
8 3 8 6 5 7 9 8 6 9 2 59
9 6 8 4 6 2 4 8 5 6 2 49
52 50 47 40 58 34 46 50 54 48 38
```
```
1 5 8 6 6 5 4 9 9 7 7 8 66
5 6 2 5 5 4 8 5 7 7 3 6 54
8 2 8 2 8 6 9 4 9 5 9 9 67
1 2 8 2 3 4 5 8 8 7 6 2 48
8 9 4 8 7 2 8 2 2 3 7 7 57
2 2 1 9 4 1 1 1 5 6 1 5 36
2 1 4 2 9 1 2 8 1 6 9 7 49
3 6 5 7 5 5 7 9 4 7 7 5 59
8 2 3 4 8 2 2 3 3 1 6 1 35
4 2 1 7 7 1 7 9 6 7 9 7 51
7 4 3 2 8 3 6 7 8 3 1 8 54
3 8 9 8 7 6 5 7 1 1 7 3 59
48 45 51 47 62 38 61 59 57 50 60 57
```
```
4 3 9 3 7 6 6 9 7 7 5 9 1 71
2 7 4 7 1 1 9 8 8 3 3 5 4 52
6 9 6 5 6 4 6 7 3 6 6 8 8 68
5 7 8 8 1 5 3 4 5 7 2 9 6 60
5 3 1 3 3 5 4 5 9 1 8 2 7 50
3 8 3 1 8 4 8 2 2 9 7 3 6 58
6 6 9 8 3 5 9 1 4 6 9 8 2 69
8 1 8 2 9 7 1 3 8 5 2 1 5 50
9 9 4 5 4 9 7 1 8 8 1 2 6 60
9 2 4 8 4 5 3 3 7 9 6 1 6 58
5 2 7 6 8 5 6 6 1 3 4 7 2 47
8 3 5 2 7 2 4 5 8 1 2 6 2 49
7 1 7 4 9 2 9 8 9 3 5 2 3 59
66 50 69 50 58 49 64 57 65 66 56 47 54
```
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edited Aug 30, 2019 at 3:23
answered Aug 23, 2019 at 3:10
m69 ''snarky and unwelcoming''m69 ''snarky and unwelcoming''
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2777 | https://ukmt.org.uk/wp-content/uploads/2023/08/SMC_2020_Extended_Solutions.pdf | UKMT UKMT UKMT United Kingdom Mathematics Trust Senior Mathematical Challenge Organised by the United Kingdom Mathematics Trust supported by Solutions and investigations November 2020 These solutions augment the shorter solutions also available online. The shorter solutions sometimes leave out details. The solutions given here are full solutions, as explained below. In some cases alternative solutions are given. There are also many additional problems for further investigation. We welcome comments on these solutions and the additional problems. Please send them to enquiry@ukmt.org.uk.
The Senior Mathematical Challenge (SMC) is a multiple-choice paper. Sometimes you can find the correct answers by working backwards from the given alternatives, or by showing that four of them are not correct. This can be a sensible thing to do in the context of the SMC.
However, this does not provide a full mathematical explanation that would be acceptable if you were just given the questions without any alternative answers. So we aim at including for each question a complete solution with each step explained (or, occasionally, left as an exercise), and not based on the assumption that one of the given alternatives is correct. We hope that these solutions provide a model for the type of written solution that is expected when presenting a complete solution to a mathematical problem (for example, in the British Mathematical Olympiad, the Mathematical Olympiad for Girls and similar competitions).
These solutions may be used freely within your school or college. You may, without further permission, post these solutions on a website that is accessible only to staffand students of the school or college, print out and distribute copies within the school or college, and use them in the classroom. If you wish to use them in any other way, please consult us.
© UKMT November 2020 Enquiries about the Senior Mathematical Challenge should be sent to: SMC, UK Mathematics Trust, School of Mathematics, University of Leeds, Leeds LS2 9JT T 0113 343 2339 enquiry@ukmt.org.uk www.ukmt.org.uk 1 D 2 C 3 A 4 B 5 C 6 E 7 D 8 D 9 B 10 C 11 E 12 B 13 D 14 B 15 A 16 E 17 A 18 E 19 A 20 E 21 C 22 A 23 D 24 B 25 C Senior Mathematical Challenge 2020 Solutions and investigations 1. What is the value of 2020 20 × 20?
A 10.1 B 5.5 C 5.1 D 5.05 E 0.55 Solution D 푁표푡푒: In the absence of a calculator, the best way to tackle this question is to first do some cancelling and then the division. There is more than one way to do this. In our method we twice divide the numerator and denominator by 10.
2020 20 × 20 = 202 2 × 20 = 20.2 2 × 2 = 20.2 4 = 5.05.
For investigation 1.1 What is the value of the following?
(a) 2020 2 × 2, (b) 2020 200 × 200, (c) 2020 2 × 200.
1.2 Find the positive integer 푛such that 2020 푛× 푛= 126.25.
1.3 Find all the positive integers 푛for which 2020 푛× 푛is an integer.
2. What is the remainder when 1234 × 5678 is divided by 5?
A 0 B 1 C 2 D 3 E 4 Solution C The units digit of 1234 × 5678 is the same as the units digit of 4 × 8, and hence is 2. Therefore the remainder when 1234 × 5678 is divided by 5 is 2.
For investigation 2.1 What is the remainder when 1234 × 5678 is divided by 3?
2.2 What is the remainder when 1234 × 5678 is divided by 11?
2.3 The integer 푚has remainder 3 when it is divided by 11. The integer 푛has remainder 4 when divided by 11.
What is the remainder when 푚푛is divided by 11?
© UKMT November 2020 www.ukmt.org.uk 2 Senior Mathematical Challenge 2020 Solutions and investigations 3. A shape is made from five unit cubes, as shown.
What is the surface area of the shape?
A 22 B 24 C 26 D 28 E 30 Solution A The surface of the shape is made up of two faces of the central cube, and five faces of each of the four other cubes.
Therefore the surface is made up of 2 + 4 × 5 = 22 square faces each of size 1 × 1.
Hence the surface area of the shape is 22.
For investigation 3.1 The shape shown is made by adding two unit cubes to the shape of this question. It is made from seven unit cubes.
What is the surface area of the shape?
4. The numbers 푝, 푞, 푟and 푠satisfy the equations 푝= 2, 푝× 푞= 20, 푝× 푞× 푟= 202 and 푝× 푞× 푟× 푠= 2020.
What is the value of 푝+ 푞+ 푟+ 푠?
A 32 B 32.1 C 33 D 33.1 E 34 Solution B We have 푝= 2, 푞= 푝× 푞 푝 = 20 2 = 10, 푟= 푝× 푞× 푟 푝× 푞 = 202 20 = 10.1, and 푠= 푝× 푞× 푟× 푠 푝× 푞× 푟 = 2020 202 = 10.
Therefore 푝+ 푞+ 푟+ 푠= 2 + 10 + 10.1 + 10 = 32.1.
© UKMT November 2020 www.ukmt.org.uk 3 Senior Mathematical Challenge 2020 Solutions and investigations 5. What is √ 123454321?
A 1111111 B 111111 C 11111 D 1111 E 111 Solution C 푁표푡푒: You could not be expected to be able to calculate the value of √ 123454321 without the use of a calculator. So you need to find some other way to select the correct option. We use a method based on the size of the number 123454321.
We have 108 < 123454321 < 1010.
Therefore p 108 < √ 123454321 < p 1010, that is, 104 < √ 123454321 < 105.
The correct option is therefore the only one that is between 104 and 105. Therefore, of the given options, it is 11111 that equals √ 123454321.
For investigation 5.1 The answer given above assumes that one of the given options is correct.
Verify that 11111 = √ 123454321 by checking that 111112 = 123454321.
6. There are fewer than 30 students in the A-level mathematics class. One half of them play the piano, one quarter play hockey and one seventh are in the school play.
How many of the students play hockey?
A 3 B 4 C 5 D 6 E 7 Solution E Because one half of the students play the piano, the number of students is a multiple of 2.
Because one quarter of the students play hockey, the number of students is a multiple of 4.
Because one seventh of the students are in the school play, the number of students is a multiple of 7.
Therefore the number of students is a multiple of 2, 4 and 7. Hence the number of students is a multiple of 28.
Because there are fewer than 30 students in the class, it follows that there are 28 students in the class.
Therefore, because one quarter of the 28 students play hockey, the number of students who play hockey is 7.
© UKMT November 2020 www.ukmt.org.uk 4 Senior Mathematical Challenge 2020 Solutions and investigations 7. Official UK accident statistics showed that there were 225 accidents involving teapots in one year. However, in the following year there were 47 such accidents.
What was the approximate percentage reduction in recorded accidents involving teapots from the first year to the second?
A 50 B 60 C 70 D 80 E 90 Solution D The reduction in the number of teapot accidents in the second year was 225 −47 = 178.
178 as a percentage of 225 is 178 225 × 100 ≈180 225 × 100 = 20 25 × 100 = 20 × 4 = 80.
8. What is the largest prime factor of 1062 −152?
A 3 B 7 C 11 D 13 E 17 Solution D 푁표푡푒: It is not a good idea to attempt to calculate 1062 and 152, then do a subtraction and finally attempt to factorize the resulting answer.
Instead we make use of the standard factorization of the difference of two squares: 푥2 −푦2 = (푥−푦)(푥+ 푦).
We have 1062 −152 = (106 −15)(106 + 15) = 91 × 121 = 7 × 13 × 11 × 11.
Therefore the prime factorization of 1062 −152 is 7 × 112 × 13, from which we see that its largest prime factor is 13.
For investigation 8.1 What is the largest prime factor of 3002 −32?
© UKMT November 2020 www.ukmt.org.uk 5 Senior Mathematical Challenge 2020 Solutions and investigations 9. In 2018, a racing driver was allowed to use the Drag Reduction System provided that the car was within 1 second of the car ahead. Suppose that two cars were 1 second apart, each travelling at 180 km/h (in the same direction!).
How many metres apart were they?
A 100 B 50 C 10 D 5 E 1 Solution B The distance apart of the cars was the distance that a car travelling at 180 km/h travels in 1 second.
There are 1000 metres in one kilometre. Hence there are 180 × 1000 metres in 180 km.
There are 60 × 60 seconds in each hour.
It follows that at 180 km/h a car travels 180 × 1000 metres in 60 × 60 seconds.
Therefore the number of metres that it travels in 1 second is 180 × 1000 60 × 60 = 3 × 1000 60 = 1000 20 = 50.
Therefore when the cars are 1 second apart, they are 50 metres apart.
For investigation 9.1 The Highway Code gives the following table of typical 푠푡표푝푝푖푛푔푑푖푠푡푎푛푐푒푠in metres for motor vehicles travelling at different velocities.
velocity stopping distance 32 km/h 12 m 48 km/h 23 m 64 km/h 36 m 80 km/h 53 m 96 km/h 73 m 112 km/h 96 m For each velocity, how far apart in seconds should two cars travelling in the same direction at that velocity be so that their distance apart is the same as the corresponding stopping distance given in the above table?
9.2 푁표푡푒: The Highway Code adds that “The distances shown are a general guide. The distance will depend on your attention (thinking distance), the road surface, the weather conditions and the condition of your vehicle at the time. ” The Highway Code gives a thinking distance of 12 m for a car travelling at 64 km/h. How much thinking time does that correspond to?
© UKMT November 2020 www.ukmt.org.uk 6 Senior Mathematical Challenge 2020 Solutions and investigations 10. Six friends Pat, Qasim, Roman, Sam, Tara and Uma, stand in a line for a photograph.
There are three people standing between Pat and Qasim, two between Qasim and Roman and one between Roman and Sam. Sam is not at either end of the line.
How many people are standing between Tara and Uma?
A 4 B 3 C 2 D 1 E 0 Solution C We indicate each of the friends by the first letter of their name, and a person whose name we are not yet sure about by an asterisk ().
We can assume, without loss of generality, that, from the point of view of the photographer, Qasim is to the right of Pat. Because there are three people standing between Pat and Qasim, the line is either ∗푃∗∗∗푄 or 푃∗∗∗푄∗.
There are two people between Qasim and Roman. Roman cannot be to the right of Quasim, because there is at most one friend to the right of Quasim. Therefore Roman is to the left of Quasim and the line is either ∗푃푅∗∗푄 or 푃푅∗∗푄∗.
There is one person between Roman and Sam. Therefore either Sam is immediately to the left of Pat, or immediately to the left of Qasim.
If Sam is immediately to the left of Pat, the line would be 푆푃푅∗∗푄.
However this is impossible because Sam is not at either end of the line.
Therefore Sam is immediately to the left of Qasim. Hence the line is either ∗푃푅∗푆푄 or 푃푅∗푆푄∗ with Tara and Uma occupying the two places marked by asterisks.
We see that, however Tara and Uma occupy these two places, the number of people standing between Tara and Uma will be 2.
For investigation 10.1 Specify one additional piece of information that would make it possible to work out the exact order of the six friends from left to right, as seen by the photographer.
© UKMT November 2020 www.ukmt.org.uk 7 Senior Mathematical Challenge 2020 Solutions and investigations 11. Two congruent pentagons are each formed by removing a right-angled isosceles triangle from a square of side-length 1. The two pentagons are then fitted together as shown.
What is the length of the perimeter of the octagon formed?
A 4 B 4 + 2 √ 2 C 5 D 6 −2 √ 2 E 6 Solution E Let 푃, 푄, 푅, 푆, 푇, 푈, 푉and 푊be the vertices of the octagon, as shown in the diagram, and let 푋be the vertex as shown.
The sides 푃푋, 푋푈, 푈푉, 푄푅, 푅푆and 푆푇all have length 1.
Let 푃푊have length 푠. Then, because 푃푊푉is a right-angled isosceles triangle, 푉푊also has length 푠.
Because the pentagons 푃푋푈푉푊and 푄푅푆푇푋are congruent, 푄푋and 푇푋also have length 푠.
It follows that 푃푄and 푇푈each have length 1 −푠.
We can now deduce the the length of the perimeter of the octagon 푃푄푅푆푇푈푉푊is (1 −푠) + 1 + 1 + 1 + (1 −푠) + 1 + 푠+ 푠= 6.
Commentary Note that in order to work out the length of the perimeter we did not need to know the value of 푠. It is, however, not difficult to find the value of 푠. You are asked to do this in Problem 11.1.
For investigation 11.1 Find the length of 푄푋.
11.2 Find the area of the octagon 푃푄푅푆푇푈푉푊.
© UKMT November 2020 www.ukmt.org.uk 8 Senior Mathematical Challenge 2020 Solutions and investigations 12. A three-piece suit consists of a jacket, a pair of trousers and a waistcoat. Two jackets and three pairs of trousers cost £380. A pair of trousers costs the same as two waistcoats.
What is the cost of a three-piece suit?
A £150 B £190 C £200 D £228 E more information needed Solution B We let the cost of a jacket, a pair of trousers and a waistcoat be £퐽, £푇and £푊, respectively.
From the information given in the question we can deduce that 2퐽+ 3푇= 380 (1) and 푇= 2푊 (2) 푁표푡푒: In this problem we have three unknowns 퐽, 푇and 푊, but only two equations. We don’t have enough information to enable us to find the values of 퐽, 푇and 푊. However, we can deduce the value of 퐽+ 푇+ 푊which is what we need to know. We give two methods for doing this.
Method 1 If we subtract equation (2) from equation (1) we obtain 2퐽+ 2푇= 380 −2푊.
Hence 2퐽+ 2푇+ 2푊= 380.
Hence, dividing both sides of this last equation by 2, we obtain 퐽+ 푇+ 푊= 190.
Therefore the cost of a three piece suit is £190.
Method 2 It follows from equation (2) that 푊= 1 2푇. Hence, 퐽+ 푇+ 푊= 퐽+ 푇+ 1 2푇 = 퐽+ 3 2푇 = 1 2 (2퐽+ 3푇) = 1 2 (380), by equation (1), = 190.
Therefore the cost of a three-piece suit is £190.
© UKMT November 2020 www.ukmt.org.uk 9 Senior Mathematical Challenge 2020 Solutions and investigations 13. The number 16! ÷ 2푘is an odd integer. Note that 푛! = 1 × 2 × 3 × · · · × (푛−1) × 푛.
What is the value of 푘?
A 9 B 11 C 13 D 15 E 17 Solution D The number 푘that we seek is such that 16! = 2푘× 푞, where 푞is an odd integer. Thus 푘is the power of 2 that occurs in the prime factorization of 16!.
Method 1 To find the highest power of 2 that is a factor of 16! we need only consider the even numbers that occur in the product 1 × 2 × · · · × 15 × 16 that gives the value of 16!
In the following table we give the powers of 2 that divide each of these factors.
factor 2 4 6 8 10 12 14 16 power of 2 1 2 1 3 1 2 1 4 The power of 2 in the prime factorization of 16! is the sum of the powers in the second row of this table. That is, it is 1 + 2 + 1 + 3 + 1 + 2 + 1 + 4 = 15.
Hence, the required value of 푘is 15.
Method 2 푁표푡푒: In this method we use the formula, usually attributed to the French mathematician Adrian-Marie Legendre (1752-1833), for the highest power of a prime 푝that divides 푛!. This formula makes use of the 푓푙표표푟function which we first explain.
We define the 푓푙표표푟of 푥, written as ⌊푥⌋, to be the largest integer that is not larger than 푥.
For example, ⌊26 7⌋= 2, ⌊4.275⌋= 4, ⌊휋⌋= 3, ⌊7⌋= 7, ⌊0.43⌋= 0 and ⌊−5.23⌋= −6.
According to Legendre’s formula, the highest power of the prime 푝that divides 푛! is given by 푛 푝 + 푛 푝2 + 푛 푝3 + . . . .
At first sight, the formula involves an infinite sum. However, if 푝푘> 푛, we have 0 < 푛 푝푘< 1 and hence 푛 푝푘 = 0. Therefore only a finite number of terms in the above sum are non-zero. The number of non-zero terms in this sum is the largest integer 푘for which 푝푘≤푛.
© UKMT November 2020 www.ukmt.org.uk 10 Senior Mathematical Challenge 2020 Solutions and investigations By Legendre’s formula the highest power of 2 that divides 16! is 푘, where 푘= 16 2 + 16 22 + 16 23 + 16 24 = 16 2 + 16 4 + 16 8 + 16 16 = ⌊8⌋+ ⌊4⌋+ ⌊2⌋+ ⌊1⌋ = 8 + 4 + 2 + 1 = 15.
For investigation 13.1 Why is 15 the only value of 푘for which 16! ÷ 2푘is an odd integer?
13.2 Use Legendre’s formula to find the highest power of 3 that divides 16! .
13.3 What is the highest power of 2 that divides 100! ?
13.4 What is the highest power of 3 that divides 100! ?
13.5 Find the prime factorization of 100!
13.6 Find the least positive integer 푛such that 2100 is a factor of 푛! .
13.7 Prove that Legendre’s formula is correct.
That is, show that the highest power of the prime 푝that divides 푛! is given by the sum 푛 푝 + 푛 푝2 + 푛 푝3 + · · · + 푛 푝푘 , where 푘is the largest integer such that 푝푘≤푛.
13.8 Let 푝be a prime number and let 푛be a positive integer. Find a formula in terms of 푝and 푛for the highest power of 푝that divides (푝푛)!.
© UKMT November 2020 www.ukmt.org.uk 11 Senior Mathematical Challenge 2020 Solutions and investigations 14. Diane has five identical blue disks, two identical red disks and one yellow disk. She wants to place them on the grid opposite so that each cell contains exactly one disk. The two red disks are not to be placed in cells that share a common edge.
How many different-looking completed grids can she produce?
A 96 B 108 C 144 D 180 E 216 Solution B 푁표푡푒: The key to a problem of this kind is deciding the order in which to consider the placing of the differently coloured disks. In this case it is best to consider first the number of different ways the two red disks may be placed, because they are subject to the condition that they should not be put in cells that share an edge.
In the diagram we have labelled the cells so that we can refer to them.
If the first red disk is placed the cell 푃, then the second red disk may be placed in any one of the 5 cells 푅, 푆, 푈, 푉and 푊.
Likewise, if it placed in the any of the cells 푆, 푇and 푄, there are 5 possible cells for the second red disk.
If the first red disk is placed any one of the cells 푄, 푅, 푈and 푉, there are 4 choices for the second red disk.
This gives 5 × 4 + 4 × 4 = 36 ways to place the two red disks, but each possible pair of cells has been counted twice.
Therefore there are 36 ÷ 2 = 18 different ways to place the two red disks.
Once the red disks have been placed, there remain 6 cells in which the yellow disk may be placed.
Once the red and yellow disks have been placed, the 5 blue disks must be placed in the remaining 5 empty cells. This may be done in just 1 way.
This gives 18 × 6 × 1 = 108 different-looking ways of filling the grid.
For investigation 14.1 How many different-looking grids can Diane produce if she has one yellow disk, three identical red disks and four identical blue disks, and there are no restrictions other than that each cell should contain exactly one disk?
14.2 How many different-looking grids can Diane produce if she has one yellow disk, two identical red disks, two identical blue disks and three identical green disks, and there are no restrictions other than that each cell should contain exactly one disk?
14.3 How many different-looking grids can Diane produce if she has one yellow disk, two identical red disks, two identical blue disks and three identical green disks, with each cell containing exactly one disk and the two red disks not in cells that share a common edge?
© UKMT November 2020 www.ukmt.org.uk 12 Senior Mathematical Challenge 2020 Solutions and investigations 15. The shaded area shown in the diagram consists of the interior of a circle of radius 3 together with the area between the circle and two tangents to the circle. The angle between the tangents at the point where they meet is 60°.
What is the shaded area?
A 6휋+ 9 √ 3 B 15 √ 3 C 9휋 D 9휋+ 4 √ 3 E 6휋+ 9 √ 3 4 60◦ Solution A Let 푂be the centre of the circle and let 푃푆and 푃푇be the tangents to the circle, as shown.
In the triangles 푃푆푂and 푃푇푂we have ∠푃푆푂= ∠푃푇푂= 90° because 푃푆and 푃푇are tangents to the circle; 푆푂= 푇푂, because they are radii of the same circle; and the side 푃푂is common.
Therefore, the two triangles are congruent (RHS). Hence ∠푆푃푂= ∠푇푃푂= 30°.
It follows that 푂푆 푃푆= tan 30° = 1 √ 3 .
Hence 푃푆= 푂푆 √ 3 = 3 √ 3.
Using the formula area = 1 2 (base × height), for the area of a triangle, it follows that the area of the triangle 푃푆푂is 1 2 (푂푆× 푃푆) = 1 2(3 × 3 √ 3) = 1 2 (9 √ 3).
This is also the area of the congruent triangle 푃푇푂. Therefore the sum of the areas of these two triangles is 9 √ 3.
The total shaded area is the sum of the areas of these two triangle plus the area of that part of the circle that lies outside the two triangles. Because ∠푆푂푇= 120°, the part of the circle outside the two triangles makes up two-thirds of the circle and hence its area is given by 2 3 (휋× 32) = 6휋.
Hence the shaded area is 6휋+ 9 √ 3.
For investigation 15.1 Explain why ∠푆푂푇= 120°.
15.2 Explain why it follows that the area of the part of the circle outside the two triangles is two-thirds of the total area of the circle.
15.3 The solution above uses the fact that tan 30° = 1 √ 3 .
Explain why this is correct.
15.4 What is the shaded area in the case where ∠푆푃푇= 120°?
© UKMT November 2020 www.ukmt.org.uk 13 Senior Mathematical Challenge 2020 Solutions and investigations 16. Which diagram represents the set of all points (푥, 푦) satisfying 푦2 −2푦= 푥2 + 2푥?
A 푦 푥 B 푦 푥 C 푦 푥 D 푦 푥 E 푦 푥 Solution E We have 푦2 −2푦= 푥2 + 2푥⇔푦2 −푥2 −2푦−2푥= 0 ⇔(푦−푥)(푦+ 푥) −2(푦+ 푥) = 0 ⇔(푦−푥−2)(푦+ 푥) = 0 ⇔푦−푥−2 = 0 or 푦+ 푥= 0 ⇔푦= 푥+ 2 or 푦= −푥.
푁표푡푒: An alternative method is to complete the square on both sides of the equation. See Problem 16.1 below.
We therefore see that the set of points satisfy-ing the equation 푦2 −2푦= 푥2 +2푥is made up of all the points on the line with the equation 푦= 푥+ 2 together with all the points on the line with the equation 푦= −푥.
The line with the equation 푦= 푥+ 2 is the line with slope 1 that goes through the point (0, 2).
The line with the equation 푦= −푥is the line with slope −1 that goes through the point (0, 0).
The diagram that shows these two lines is that in option E.
For investigation 16.1 Show that adding 1 to both sides of the equation 푦2 −2푦= 푥2 + 2푥, provides an alternative way to show that 푦= 푥+ 2 or 푦= −푥.
16.2 What are the coordinates of the point where the lines with equations 푦= 푥+ 2 and 푦= −푥 meet?
16.3 Draw a diagram to represent the set of all points (푥, 푦) which satisfy the equation 푥2 −1 = 푦2 + 2푦.
16.4 Draw a diagram to represent the set of all points (푥, 푦) which satisfy the equation 푥2푦2 + 푥푦= 푥3 + 푦3.
© UKMT November 2020 www.ukmt.org.uk 14 Senior Mathematical Challenge 2020 Solutions and investigations 17. The positive integers 푚, 푛and 푝satisfy the equation 3푚+ 3 푛+ 1 푝 = 17.
What is the value of 푝?
A 2 B 3 C 4 D 6 E 9 Solution A Because 3 푛+ 1 푝 = 17 −3푚, where 푚is an integer, it follows that 3 푛+ 1 푝 is an integer.
Because 푛and 푝are positive integers, 1 < 푛+ 1 푝and therefore 0 < 3 푛+ 1 푝 < 3. Therefore 3 푛+ 1 푝 is either 1 or 2.
If 3 푛+ 1 푝 = 1, then 3푚= 17 − 3 푛+ 1 푝 = 17 −1 = 16. This implies that 푚= 16 3 contradicting the fact that 푚is an integer. Hence 3 푛+ 1 푝 = 2. Therefore 푛+ 1 푝= 3 2 = 1 + 1 2. Hence 푛= 1 and 푝= 2.
Also 3푚= 17 − 3 1 + 1 2 = 17 −2 = 15 and so 푚= 5. We therefore see that 푚= 5, 푛= 1 and 푝= 2.
18. Two circles 퐶1 and 퐶2 have their centres at the point (3,4) and touch a third circle, 퐶3.
The centre of 퐶3 is at the point (0,0) and its radius is 2.
What is the sum of the radii of the two circles 퐶1 and 퐶2?
A 6 B 7 C 8 D 9 E 10 Solution E Let 푃be the centre of the circles 퐶1 and 퐶2, 푂be the centre of the circle 퐶3, and 푅and 푆be the points where 퐶1 and 퐶2, respectively, touch 퐶3. We let the radius of 퐶1 be 푟.
Note that the points 푂, 푃, 푅and 푆lie on a straight line.
By Pythagoras’ Theorem, the distance between the point 푃 with coordinates (3, 4) and the point 푂with coordinates (0, 0) is √ 32 + 42 = √ 25 = 5.
Therefore, 푟+ 2 = 5 and hence 푟= 3.
The radius of the circle 퐶2 is 푟+ 2 + 2 = 3 + 2 + 2 = 7.
Therefore the sum of the radii of 퐶1 and 퐶2 is 3 + 7 = 10.
For investigation 18.1 Prove that the points 푃, 푅, 푂and 푆lie on a straight line.
© UKMT November 2020 www.ukmt.org.uk 15 Senior Mathematical Challenge 2020 Solutions and investigations 19. The letters 푝, 푞, 푟, 푠and 푡represent different positive single-digit numbers such that 푝−푞= 푟and 푟−푠= 푡.
How many different values could 푡have?
A 6 B 5 C 4 D 3 E 2 Solution A Because 푟= 푝−푞, we have 푡= 푟−푠= (푝−푞) −푠= 푝−(푞+ 푠).
It follows that the maximum value of 푡occurs when 푝takes its maximum value, and 푞+ 푠takes its minimum value.
The maximum value of 푝is 9.
Because 푞and 푠have different values, the minimum value of 푞+ 푠is 1 + 2 = 3.
Therefore the maximum possible value of 푡is 9 −3 = 6.
The following table shows that 푡can take all the positive integer values from 1 to 6.
푝 푞 푟 푠 푡 9 1 8 2 6 9 1 8 3 5 9 2 7 3 4 9 2 7 4 3 9 3 6 4 2 9 3 6 5 1 Therefore the number of values that 푡could take is 6.
For investigation 19.1 In how many different ways is it possible to choose different single-digit values for 푝, 푞, 푟 and 푠so that 푡= 1?
19.2 The letters 푝, 푞, 푟and 푠represent different positive single-digit numbers such that 푝−푞= 푟 and 푞−푟= 푠. How many different values could 푠have?
19.3 The letters 푝, 푞, 푟and 푠represent different positive single-digit numbers such that 푝−푞= 푟 and 푝−푟= 푠. How many different values could 푠have?
© UKMT November 2020 www.ukmt.org.uk 16 Senior Mathematical Challenge 2020 Solutions and investigations 20. The real numbers 푥and 푦satisfy the equations 4푦= 1 8( √ 2)푥+2 and 9푥× 3푦= 3 √ 3.
What is the value of 5푥+푦?
A 5 √ 5 B 5 C √ 5 D 1 5 E 1 √ 5 Solution E We have 4푦= (22)푦= 22푦, and 1 8( √ 2)푥+2 = 1 23(2 1 2)푥+2 = 2−(3+ 1 2 (푥+2)).
Therefore from the equation 4푦= 1 8( √ 2)푥+2 we deduce that 2푦= −(3 + 1 2 (푥+ 2)).
Therefore 푦= −1 4푥−2.
(1) Also, 9푥× 3푦= (32)푥× 3푦= 32푥+푦and 3 √ 3 = (31)(3 1 2) = 31+ 1 2 = 3 3 2.
Therefore, from the equation 9푥× 3푦= 3 √ 3 we deduce that 2푥+ 푦= 3 2.
Therefore 푦= −2푥+ 3 2.
(2) From equations (1) and (2) −1 4푥−2 = −2푥+ 3 2.
This last equation may be rearranged to give 7 4푥= 7 2.
Hence 푥= 2.
Therefore, from equation (1), 푦= −5 2.
It follows that 푥+ 푦= 2 + (−5 2) = 2 −5 2 = −1 2.
Therefore 5푥+푦= 5−1 2 = 1 √ 5.
For investigation 20.1 The real numbers 푥and 푦satisfy the equations ( √ 3)푥× 3푦= 5 √ 27 and 8푥× ( √ 2)푦= 3 √ 4.
Find the value of 8푥+푦.
© UKMT November 2020 www.ukmt.org.uk 17 Senior Mathematical Challenge 2020 Solutions and investigations 21. When written out in full, the number (102020 + 2020)2 has 4041 digits.
What is the sum of the digits of this 4041-digit number?
A 9 B 17 C 25 D 2048 E 4041 Solution C Using the expansion (푥+ 푦)2 = 푥2 + 2푥푦+ 푦2, we have (102020 + 2020)2 = (102020)2 + 2 × 102020 × 2020 + 20202 = 104040 + 4040 × 102020 + 4080400.
(1) We now note that when we add the three terms 104040, 4040×102002 and 408040, no two non-zero digits occur in the same column: 100............0000...........0000000 + 4040...........0000000 + 4080400 100............4040...........4080400 It follows that the non-zero digits in the final answer are exactly the non-zero digits in the three terms in (1) above.
Therefore the sum of the digits in the number (102020 + 2020)2 is 1 + 4 + 4 + 4 + 8 + 4 = 25.
Commentary The exact number of 0s in the final answer is not important. What matters is that the non-zero digits in the three terms 104040, 4040 × 102020 and 4080400 are in different columns, and so, when they are added, the non-zero digits in the final answer are exactly the non-zero digits in these terms. However, you are asked to work out the number of 0s in Problem 21.1 below.
For investigation 21.1 Find the positive integers 푚and 푛so that (102020 + 2020)2 = 1 푚 z }| { 000 . . . 000 404 푛 z }| { 000 . . . 000 4080400.
© UKMT November 2020 www.ukmt.org.uk 18 Senior Mathematical Challenge 2020 Solutions and investigations 22. A square with perimeter 4 cm can be cut into two congruent right-angled triangles and two congru-ent trapezia as shown in the first diagram in such a way that the four pieces can be rearranged to form the rectangle shown in the second diagram.
What is the perimeter, in centimetres, of this rectangle?
A 2 √ 5 B 4 √ 2 C 5 D 4 √ 3 E 3 √ 7 Solution A The square has perimeter 4 cm and hence its sides each have length 1 cm.
Let the vertices in the diagram on the left be labelled as shown.
We let the length of 푆푈be 푥cm.
For the pieces to fit together as shown on the right, 푇푉and 푊푅must also have length 푥cm. It follows that the other lengths, in cm, are as shown.
Method 1 For the pieces to fit together the angles 휙and 휃must be equal, as shown on the right of the above diagram. Therefore tan 휙= tan 휃. That is, 1 −2푥 1 −푥= 푥 1.
Hence 1 −2푥= 푥(1 −푥).
This last equation may be rearranged to give 푥2 −3푥+ 1 = 0.
From the standard formula −푏± √ 푏2 −4푎푐 2푎 for the solutions of a quadratic equation, we can deduce that 푥= 3 ± √ 5 2 .
© UKMT November 2020 www.ukmt.org.uk 19 Senior Mathematical Challenge 2020 Solutions and investigations Since 푥< 1 it follows that 푥= 3 − √ 5 2 .
From the diagram on the right above we see that the perimeter of the rectangle is given by (1 −푥) + (2 −푥) + (1 −푥) + (2 −푥) = 6 −4푥= 6 −4 3 + √ 5 2 = 2 √ 5.
Method 2 Since they are made up from the same two triangles and two trapezia, the area of the rectangle equals the area of the square.
Therefore, from the diagram above, we have (1 + (1 −푥)) × (1 −푥) = 1 × 1, that is, (2 −푥)(1 −푥) = 1.
Hence 2 −3푥+ 푥2 = 1 and therefore 푥2 −3푥+ 1 = 0.
Thus, as in Method 1, 푥= 3 − √ 5 2 , and the perimeter of the rectangle is 2 √ 5.
For investigation 22.1 The diagram above shows on the left an 8 × 8 square divided into four pieces which are rearranged to make the 13 × 5 rectangle shown on the right.
However 8 × 8 < 13 × 5.
Where has the additional area come from?
푁표푡푒: This puzzle has been attributed to William Hooper in his book 푅푎푡푖표푛푎푙푅푒푐푟푒푎푡푖표푛푠 of 1774.
© UKMT November 2020 www.ukmt.org.uk 20 Senior Mathematical Challenge 2020 Solutions and investigations 23. A function 푓satisfies 푦3 푓(푥) = 푥3 푓(푦) and 푓(3) ≠0. What is the value of 푓(20) −푓(2) 푓(3) ?
A 6 B 20 C 216 D 296 E 2023 Solution D By putting 푥= 20 and 푦= 3 in the equation 푦3 푓(푥) = 푥3 푓(푦), we have 27 푓(20) = 8000 푓(3).
Hence 푓(20) = 8000 27 푓(3).
By putting 푥= 2 and 푦= 3 in the same equation, we have 27 푓(2) = 8 푓(3). Hence 푓(2) = 8 27 푓(3).
Therefore, 푓(20) −푓(2) = 8000 27 푓(3) −8 27 푓(3) = 8000 27 −8 27 푓(3) = 7992 27 푓(3) = 296 푓(3).
Therefore, because 푓(3) ≠0, it follows that 푓(20) −푓(2) 푓(3) = 296.
For investigation 23.1 What is the value of 푓(46) −푓(23) 푓(23) ?
23.2 Show that there are infinitely many positive integers 푥, 푦and 푧which satisfy the equation 푓(푥) −푓(푦) 푓(푧) = 7.
23.3 Show that for each positive integer 푛the equation 푓(푥) −푓(푦) 푓(푧) = 푛 has either infinitely many solutions in which 푥, 푦and 푧are positive integers, or no such solutions.
23.4 Show that a function 푔satisfies the equation 푦3푔(푥) = 푥3푔(푦) for all real numbers 푥and 푦 if, and only if, there is a constant 푘such that 푔(푥) = 푘푥3, for all real numbers 푥.
© UKMT November 2020 www.ukmt.org.uk 21 Senior Mathematical Challenge 2020 Solutions and investigations 24. In the diagram shown, 푀is the mid-point of 푃푄.
The line 푃푆bisects ∠푅푃푄and intersects 푅푄at 푆. The line 푆푇is parallel to 푃푅and intersects 푃푄at 푇. The length of 푃푄is 12 and the length of 푀푇is 1. The angle 푆푄푇is 120°.
What is the length of 푆푄?
120◦ 푃 푄 푅 푀푇 푆 A 2 B 3 C 3.5 D 4 E 5 Solution B Let 푡be the length of 푆푄.
Let ∠푄푃푆= 푥°.
Because 푃푄has length 12 and 푀is the midpoint of 푃푄, it follows that 푀푃has length 6. We are given that 푀푇has length 1. Therefore 푃푇has length 7, and 푇푄has length 5.
Because 푃푆bisects ∠푅푃푄, we have ∠푅푃푆= ∠푄푃푆= 푥°. Therefore ∠푅푃푄= 2푥°.
Because 푆푇is parallel to 푃푅, we have ∠푆푇푄= ∠푅푃푄= 2푥°.
It follows from the External Angle Theorem [See Problem 24.1], applied to the triangle 푃푇푆, that ∠푃푆푇= 푥°.
Therefore the triangle 푃푆푇is isosceles and hence 푆푇has the same length as 푃푇, namely 7.
We now apply the Cosine Rule to the triangle 푆푇푄. This gives 72 = 52 + 푡2 −2(5푡cos 120°).
Hence, as cos 120° = −1 2, 49 = 25 + 푡2 + 5푡.
Therefore 푡2 + 5푡−24 = 0. Hence (푡−3)(푡+ 8) = 0. Therefore 푡is either 3 or −8. Since 푡 corresponds to a length, 푡> 0. We deduce that 푡= 3.
For investigation 24.1 The 퐸푥푡푒푟푛푎푙퐴푛푔푙푒푇ℎ푒표푟푒푚says that the external angle of a triangle is the sum of the two opposite angles.
In terms of the diagram it says that 훼= 훽+ 훾.
Explain why the External Angle Theorem is true.
24.2 The solution above uses the fact that cos 120° = −1 2. Explain why this is correct.
© UKMT November 2020 www.ukmt.org.uk 22 Senior Mathematical Challenge 2020 Solutions and investigations 25. A regular 푚-gon, a regular 푛-gon and a regular 푝-gon share a vertex and pairwise share edges, as shown in the diagram.
What is the largest possible value of 푝?
A 6 B 20 C 42 D 50 E 100 푚-gon 푛-gon 푝-gon Solution C The sum of the internal angles of a polygon with 푚sides is (푚−2)180° [see Problem 25.1].
Hence each internal angle of the regular 푚-gon is 푚−2 푚 180°, and similarly for the regular 푛-gon and the regular 푝-gon. Therefore, because the sum of the angles at a point is 360°, we have 푚−2 푚 180 + 푛−2 푛 180 + 푝−2 푝 180 = 360.
It follows that 푚−2 푚 + 푛−2 푛 + 푝−2 푝 = 2.
We may rewrite this last equation as 1 −2 푚 + 1 −2 푛 + 1 −2 푝 = 2. Hence 1 푚+ 1 푛+ 1 푝= 1 2.
(1) We note that as 푝is a positive integer, it follows from (1) that 1 푚+ 1 푛< 1 2.
(2) We seek a solution of (1), where 푚, 푛and 푝are positive integers and 푝is as large as possible.
For 푝to be as large as possible, 1 푝needs to be as small as possible. Hence, by (1), 1 푚+ 1 푛needs to be as large as possible, subject to the inequality (2). Thus 푚and 푛need to be as small as possible. Without loss of generality we may assume that 푚≤푛.
By (2), 푚≥3. If 푚= 3, then 1 푛< 1 2 −1 3 = 1 6, and hence 푛> 6. When 푚= 3 and 푛= 7, we have 1 푚+ 1 푛= 1 3 + 1 7 = 10 21. Hence, by (1), 1 푝= 1 2 −10 21 = 1 42 and so 푝= 42. We show that this gives the largest possible value of 푝.
We cannot have 푚= 푛= 4, as this is not compatible with the inequality (2).
If 푛> 푚≥4, we have 1 푚+ 1 푛≤1 4 + 1 5 = 9 20 < 10 21 and so 1 푚+ 1 푛is not as large as possible.
Hence the largest possible value of 푝is given by 푚= 3, 푛= 7 and 푝= 42.
For investigation 25.1 Prove that the sum of the angles of a polygon with 푚sides is (푚−2)180° and hence that each internal angle of a regular 푚-gon is 푚−2 푚 180°.
© UKMT November 2020 www.ukmt.org.uk 23 |
2778 | https://www.ncbi.nlm.nih.gov/books/NBK425560/ | Introduction and Context - Adopting the International System of Units for Radiation Measurements in the United States - NCBI Bookshelf
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National Academies of Sciences, Engineering, and Medicine; Division on Earth and Life Studies; Nuclear and Radiation Studies Board. Adopting the International System of Units for Radiation Measurements in the United States: Proceedings of a Workshop. Washington (DC): National Academies Press (US); 2017 Mar 1.
Adopting the International System of Units for Radiation Measurements in the United States: Proceedings of a Workshop.
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National Academies of Sciences, Engineering, and Medicine; Division on Earth and Life Studies; Nuclear and Radiation Studies Board.
Washington (DC): National Academies Press (US); 2017 Mar 1.
Contents
Hardcopy Version at National Academies Press
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1 Introduction and Context
Dr. Armin Ansari, Centers for Disease Control and Prevention (CDC), and Dr. Matt Heavner, Office of Science and Technology Policy (OSTP), the White House, provided opening remarks on the possible motivation for adopting SI units for radiation measurements in the United States. They provided different perspectives, but both speakers endorsed actions that support scientific and technological advances and improve the nation's ability to respond to large-scale disasters.
Dr. Ansari stated that units of radiation measurements need to be standardized internationally for the purposes of commerce and science, and in his view standardization equates to adoption of SI. He explained that the discussions during this workshop are restricted to adopting SI units for radiation measurements only and not for other measurements such as distance, volume, and temperature. Changing the units for these other measurements would require considerable effort by members of the public to replace a familiar system with a new one.
Dr. Heavner spoke about his office's mandate, which is to advise the President and others within the Executive Office of the President on the effects of science and technology on domestic and international affairs. The three main goals of the office are
1.
Ensure that policy decisions at the White House are informed by the best science.
2.
Understand how policy decisions at the White House affect the scientific enterprise, nationally and internationally.
3.
Lead interagency efforts to develop and implement sound science and technology policies.
Dr. Heavner recognized that these three goals are relevant to adoption of SI units. He also said that communicating information related to radiation can be complex even among experts because of the potential for misuse of radiation quantities, for example, when describing total dose and dose rate. He acknowledged that using two systems of radiation measurements complicates communication.
Go to:
U.S. POLICIES FOR ADOPTING SI UNITS FOR ALL MEASUREMENTS
The United States has been taking steps toward adopting SI units for all measurements for many years. Drs. Ansari and Heavner provided the time line of U.S. policies toward adopting SI. Some milestones not mentioned by the two presenters, likely because of time limitations, were filled in by the rapporteur of this proceedings to provide a more complete timeline to the reader.1
The Metric Act of 1866 legally recognized the metric system in the United States. In 1875, the United States solidified its commitment to the development of the SI by becoming one of the original 17 nations to sign an international agreement, known as the Meter Convention. The Meter Convention established—among other organizations—the General Conference on Weights and Measures (French: Conférence générale des poids et mesures or CGPM), a conference made up of official delegates of member nations. CGPM established the SI, the modern metric system of measurement, in 1960.
In 1968, Congress authorized a 3-year feasibility study of adopting SI. The study was conducted by the U.S. Department of Commerce (De Simone, 1971). It concluded that the United States would eventually join the rest of the world in the use of the metric system and that conversion to the metric system was in the best interests of the country, particularly in view of the importance of foreign trade. This recommendation led to the passage of the Metric Conversion Act of 1975 (PL 94-68; Ford, 1975), which called for a national policy of coordinating and planning for the increased use of the metric system for U.S. trade and commerce.
The adoption of SI was anticipated to become mandatory in the 1970s, but the movement to adopt SI stalled. Congress included new encouragement to adopt SI in the Omnibus Trade and Competitiveness Act of 1988. This legislation aimed to enhance the competitiveness of the U.S. industry in international markets and designated the metric system as the preferred system of weights and measures for U.S. trade and commerce. The legislation states that “the federal government has a responsibility to assist industry, especially small business, as it voluntarily converts to the metric system.”2 To accelerate the use of the metric system within the federal government, President George Bush signed Executive Order 12770, Metric Usage in Federal Government Programs, in 1991. The order required all agencies to develop conversion plans within the period 1993-1996. It also gave the National Institute of Standards and Technology (NIST) the primary responsibility to direct and coordinate the metric transition; the institute's role was advisory only. Several agencies took actions following the 1991 directive toward adopting SI. The Nuclear Regulatory Commission's (NRC's) actions were discussed in some detail at the workshop.
The NRC examined the possibility of exclusively3 using SI units for radiation measurements as early as 1980, when it contracted with Oak Ridge National Laboratory to provide a cost–benefit analysis (CBA) of adoption of SI units in health physics (Eichholz et al., 1980). The agency published its Metrication Policy (57 FR 46202) in 1992 in response to the 1991 Executive Order. The NRC policy “supports and encourages the use of the metric system of measurement by licensed nuclear industry.”4 The NRC publishes new regulations, regulatory guides, and other documents in both SI and conventional units to facilitate use of SI units for radiation measurements. However, the metrication policy requires that event reporting and emergency response communications between the NRC's licensees (including the nuclear power industry), the NRC, and state and local authorities be in conventional units.
The NRC's policy statement also called for the Commission to reexamine whether the metrication policy would need to be modified after 3 years of implementation. In 1995, the Commission decided that the agency did not need to change its policy and considered the agency's conversion to the metric system to be complete. However, NRC staff revisited the topic in 2012 and proposed to the Commission that it increase use of SI units for radiation measurements in the agency's regulations to—among other reasons—increase alignment with international standards and practices by other countries. The proposal was rejected by the Commission by a three-to-two vote. Workshop participant Mr. William Ostendorff, who was an NRC commissioner at the time (he served on the Commission from April 2010 to June 2016), said that the Commission rejected the staff's proposal because there was no evidence that the adoption of SI units for radiation measurements would improve worker and public safety or nuclear plant safety and security. Three of the commissioners stated that they had concerns that adoption of SI units could negatively impact nuclear power plant safety.
As discussed in Chapter 5, many workshop participants suggested that the NRC's refusal to require SI units for radiation measurements is inhibiting federal, state, and local governments from adopting the SI units for radiation measurements.
Go to:
NATIONAL AND INTERNATIONAL RECOMMENDATIONS FOR ADOPTING SI UNITS FOR RADIATION MEASUREMENTS
A number of national and international organizations have endorsed the use of SI units for radiation measurements (see Table 1.1). Two of these national organizations (National Council on Radiation Protection and Measurements [NCRP] and NIST) and one international organization (International Commission on Radiation Units and Measurements [ICRU]) were represented at the workshop. Specifically, Dr. Paul DeLuca, University of Wisconsin School of Medicine, Madison, spoke in his capacity as vice chairman of ICRU; Dr. Steven Simon, National Cancer Institute (NCI), spoke in his capacity as vice president of NCRP's Program Area on Radiation Measurements and Dosimetry; and Dr. Ronaldo Minniti represented NIST.
TABLE 1.1
Endorsement of Use of SI Units by National and International Organizations.
ICRU, which is primarily concerned with the physical aspects of radiation, and the International Commission on Radiological Protection (ICRP), which is primarily concerned with the biological effects of radiation and radiation protection, adopted and recommended the exclusive use of the SI units for radiation measurements in 1975 and 1977, respectively. Dr. Paul DeLuca emphasized that although it is an international organization, ICRU has had several U.S. scientists in leadership roles since the early years of its establishment. Therefore, the opinions and recommendations that this international organization issues also represent the thinking of prominent U.S. scientists, not only scientists from other countries. The main motivation for ICRU to recommend the use of SI units for radiation measurements is to ensure accurate and conformal communications among scientific organizations.
NCRP recommended in 1985 that the United States gradually adopt SI units for radiation measurements, but the recommendation has not been implemented. The NCRP recommended the simultaneous use of SI and conventional units (referred to as present units in that report) for a period of 2 years and the exclusive use of SI and elimination of conventional units over a 5-year transition period, that is, by 1990 (NCRP, 1985). However, in a 2010 guide for decision makers who respond to radiological or nuclear terrorism incidents (NCRP, 2010), NCRP endorsed the use of conventional units in emergency response and recommended that the corresponding SI units be displayed in parentheses. Dr. Simon, NCI, clarified that this newer recommendation was not a reversal of NCRP's 1985 recommendation; it was a necessary adjustment to the previous recommendation for addressing a new threat—radiological and nuclear terrorism incidents. NCRP acknowledged that first responders and decision makers would face difficulties converting from conventional units for surface contamination (disintegrations per minute per unit area) to SI units (Becquerel per square centimeter).
Go to:
SUMMARY
The workshop discussions summarized in this chapter can be organized in the following theme (Theme 1):
The United States has a long history of government policies supporting the adoption of SI units for all measurements. National and international organizations have recommended the exclusive use of SI units for radiation measurements for more than 30 years. However, the United States continues to use conventional units for radiation measurements. As a result, most radiation professionals in the United States must understand both conventional and SI units for radiation measurements and make conversions between the two.
Go to:
Footnotes
1
The rapporteur used primarily this source for information:
2
H.R. 4848 (100th): Omnibus Trade and Competitiveness Act of 1988.
3
Workshop participant Dr. Fred Mettler argued that the term “exclusive” is not correct because there will always be some exceptions. The rapporteur of this Proceedings of a Workshop agrees with this argument and uses the term “exclusively” to denote the predominant use of SI (or conventional) units and that some exceptions may apply.
4
See: NRC's Metrication Policy (57 FR 46202), 1992.
U.S. POLICIES FOR ADOPTING SI UNITS FOR ALL MEASUREMENTS
NATIONAL AND INTERNATIONAL RECOMMENDATIONS FOR ADOPTING SI UNITS FOR RADIATION MEASUREMENTS
SUMMARY
Footnotes
Copyright 2017 by the National Academy of Sciences. All rights reserved.
Bookshelf ID: NBK425560
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National Academies of Sciences, Engineering, and Medicine; Division on Earth and Life Studies; Nuclear and Radiation Studies Board. Adopting the International System of Units for Radiation Measurements in the United States: Proceedings of a Workshop. Washington (DC): National Academies Press (US); 2017 Mar 1. 1, Introduction and Context. Available from:
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Diabetic Gastroparesis
Ganesh S. Aswath; Lisa A. Foris; Ashwini K. Ashwath; Krunal Patel.
Author Information and Affiliations
Authors
Ganesh S. Aswath1; Lisa A. Foris2; Ashwini K. Ashwath; Krunal Patel3.
Affiliations
1 Medical College of Georgia- NW Campus
2 St. George's University
3 Louisiana State University
Last Update: March 27, 2023.
Continuing Education Activity
Gastroparesis is defined by objective delaying of gastric emptying without any evidence of mechanical obstruction. Diabetic gastroparesis is a potential complication that occurs in the setting of poorly controlled diabetes, resulting from dysfunction in the coordination and function of the autonomic nervous system, neurons, and specialized pacemaker cells (interstitial cells of Cajal, ICC) of the stomach and intestine, and the smooth muscle cells of the gastrointestinal tract. This activity reviews the evaluation and management of diabetic gastroparesis and interprofessional team members' role in collaborating to provide well-coordinated care and enhance patient outcomes.
Objectives:
Identify the causes of diabetic gastroparesis.
Describe the pathophysiology of diabetic gastroparesis.
Summarize the treatment of diabetic gastroparesis.
Outline the evaluation and management of diabetic gastroparesis and the role of interprofessional team members in collaborating to provide well-coordinated care and enhance patient outcomes.
Access free multiple choice questions on this topic.
Introduction
Gastroparesis is defined by objective delaying of gastric emptying without any evidence of mechanical obstruction. Diabetic gastroparesis is a potential complication that occurs in the setting of poorly controlled diabetes, resulting from dysfunction in the coordination and function of the autonomic nervous system, neurons, and specialized pacemaker cells (interstitial cells of Cajal, ICC) of the stomach and intestine, and the smooth muscle cells of the gastrointestinal tract.
Etiology
Hyperglycemia (blood glucose greater than 200 mg/dL), commonly seen in poorly controlled diabetes, has been associated with diabetic gastroparesis resulting from neuropathy in the setting of chronic hyperglycemia and does not resolve with improved glycemic control. Acute hyperglycemia, on the other hand, though it can also result in delayed gastric emptying, is often reversible with improved glycemic control.
Gastric emptying requires coordination of fundal tone and antral phasic contraction with simultaneous inhibition of pyloric and duodenal contractions. This coordination also requires interactions between the enteric and autonomic nervous systems, smooth muscle cells, and the stomach's specialized pacemaker cells (myenteric interstitial cells of Cajal; ICCs). The gastric motor dysfunction that is encountered in the setting of diabetes may occur as a result of autonomic neuropathy (both sympathetic and parasympathetic), enteric neuropathy (both excitatory and inhibitory neurons), ICC abnormalities (intrinsic neuropathy), acute blood glucose fluctuations, use of incretin-based medications, or psychosomatic factors. As a result, most patients with diabetes tend to have dysfunction at multiple points in the process of gastric emptying. This includes abnormal postprandial proximal gastric accommodation and contraction, as well as abnormalities in antral motor function.
Epidemiology
Although idiopathic gastroparesis is the most common form of gastroparesis, diabetes is the most common disease associated with the condition. Upper gastrointestinal symptoms are reported in 11% to 18% of patients with diabetes, most of which are associated with delayed gastric emptying. Gastroparesis is seen in approximately 4.8% of individuals with type 1 diabetes, 1% of those with type 2 diabetes, and 0.1% of those without diabetes. Although there is a stronger association between type 1 diabetes and gastroparesis, the incidence of type 2 diabetes is much greater, and therefore, gastroparesis associated with type 2 diabetes is seen more frequently. Additionally, incretin mimetics are used to treat patients with type 2 diabetes, and these medications pose an additional risk factor for developing gastroparesis.
Signs and symptoms of delayed gastric emptying are seen more frequently in individuals with type 1 versus type 2 diabetes and typically in those who have had the disorder for at least five years. It has been observed that gastroparesis typically occurs in patients with a diagnosis of diabetes of at least ten years and is therefore seen more commonly in older individuals with type 2 diabetes.
Pathophysiology
Diabetic gastroparesis occurs as a result of dysfunction in the autonomic and enteric nervous systems. Chronically high levels of blood glucose (or inefficient glucose uptake) lead to neuronal damage resulting in abnormal myenteric neurotransmission (e.g., vagus nerve), impaired inhibitory (nitric oxide) neuronal function, and dysfunctional smooth muscle and pacemaker (interstitial cells of Cajal) cells. Altogether, this dysfunction results in a combination of fewer contractions of the antrum, uncoordinated antro-duodenal contractions, and pyloric spasms, ultimately resulting in delayed gastric emptying (gastroparesis).
Delayed gastric emptying in patients with diabetes, particularly of solids, may also occur in the setting of abnormal small bowel motility, which is thought to occur by a similar mechanism as that which is described in the stomach. Some patients with diabetes may additionally experience changes in gastric compliance, both increased or decreased, which may also contribute to delayed gastric emptying.
In addition to this, serum (postprandial) glucose levels have a direct relationship with gastric emptying. In the setting of diabetic autonomic neuropathy, acute hyperglycemia stimulates gastric electrical activity. In patients with diabetes (without neuropathy) and healthy controls, acute hyperglycemia will instead relax the proximal stomach and suppress gastric electrical activity (e.g., reduced the frequency, propagation, and contraction of the antrum) in both fasting and post-prandial conditions, thereby slowing gastric emptying.
Acute hyperglycemia has also been associated with increased sensitivity in the gastrointestinal tract. This may be responsible for the postprandial dyspepsia (e.g., early satiety, nausea, vomiting, heartburn, bloating, and pain) frequently experienced by patients with diabetic gastroparesis.
Carbohydrate absorption is highly dependent on the speed of gastric emptying through the release of peptides such as glucagon-like peptide-1 and glucose-dependent insulinotropic polypeptide, in which slower gastric emptying results in a higher level of carbohydrate absorption. Therefore, a higher serum glucose level due to delayed gastric emptying can itself lead to the worsening of gastroparesis.
Histopathology
Histopathology does not routinely help in diagnosing a patient with diabetic gastroparesis. However, the findings seen on animal models and studies involving human pathology specimens have helped us understand the disease process better. Full-thickness biopsies, which are needed to diagnose changes in the deeper layers of the stomach wall, are not practical to be used as a diagnostic tool. In special studies done using full-thickness gastric biopsies, the findings included an inflammatory infiltrate, a reduction in the number of nerve cell bodies and ICCs in the myenteric plexus, and fibrosis. Recent studies have pointed at the role of macrophages in the pathogenesis of diabetic gastroparesis.
History and Physical
Nausea is the most common symptom in gastroparesis. Other common symptoms include vomiting, early satiety, postprandial fullness, and bloating. Vomitus often contains undigested chewed food. Both weight loss and weight gain can occur. Wide glycemic fluctuations can also happen in gastroparesis. The timing of the symptoms relative to meals is important. Physical examination is usually nonspecific. However, neuropathy, abdominal distension, and halitosis can present in patients with diabetic gastroparesis.
Factors that may trigger an exacerbation of diabetic gastroparesis include uncontrolled blood glucose levels, medication noncompliance or intolerance, adrenal insufficiency, and infection.
Evaluation
The first step in evaluating patients who present with the symptoms above would be to exclude mechanical obstruction and peptic ulcer disease. All patients should undergo an upper gastrointestinal endoscopy. This should be followed by either a CT scan with oral and intravenous contrast or other imaging such as a small bowel follow-through to exclude obstruction beyond the duodenum. Retained food is seen on endoscopy without mechanical obstruction, and in rare cases, bezoars suggest gastroparesis.
The conventional test that is cost-effective and widely used for gastroparesis diagnosis is measuring gastric emptying by scintigraphy, which is considered the gold standard by many. Solid-phase emptying is usually used to evaluate for gastroparesis. Liquid-phase emptying is not well demonstrated for diagnosing gastroparesis and is more widely used to assess dumping syndrome after gastric surgery. Most centers use a 99 mTc sulfur colloid labeled egg white sandwich as the test meal. Standard imaging is performed at 0,1,2,4 hours postprandially. A four-hour study is more sensitive and accurate in diagnosing gastroparesis, and clinicians should avoid shortened 2- and 3-hour studies. Medications such as opiates and anticholinergics can slow gastric motility, whereas prokinetics and other medications can hasten gastric emptying. Hyperglycemia is known to slow gastric motility, and it is worthwhile to try and achieve euglycemia before performing the test if that is an option.
Breath testing is another validated study for gastroparesis. Most commonly, 13C-labelled octanoate, a medium-chain triglyceride, is bound to a solid meal and ingested. After emptying from the stomach, it is absorbed by the small intestine and metabolized to 13CO2, which is expelled from the lungs during respiration. Currently, it has limited penetrance into the clinical practice and is performed for clinical research and pharmaceutical studies.
Electrogastrography and gastroduodenal manometry are other tests based on myoelectrical activity that are potentially helpful but are not yet available for widespread clinical use.
Treatment / Management
The first step in the treatment of symptomatic gastroparesis is often lifestyle modifications. Optimal glycemic control in patients with diabetes will minimize the effects of hyperglycemia, causing delayed gastric emptying. Smaller, more frequent meals with minimizing carbonated beverages, increasing the liquid content, reducing fats and fiber content are common recommendations. Alcohol and smoking should be discouraged, as they can worsen the symptoms.
Antiemetics are often the first step in gastroparesis patients as they help with the common symptoms of nausea and vomiting. Serotonin (5-HT3) receptor antagonists like ondansetron, phenothiazines (dopamine receptor antagonists) like prochlorperazine and promethazine are commonly used. Side effects include sedation and extrapyramidal effects, and therefore, these are usually used on an as-needed basis.
The cornerstone of pharmacologic therapy in diabetic gastroparesis consists of prokinetic agents.
Metoclopramide, a 5-HT4 receptor activator and dopamine receptor antagonist in the stomach with weak 5-HT3 receptor antagonism in the nervous system, is a commonly used medication that increases the contraction amplitude of esophageal fundic and antral musculature. It is effective for treating gastroparesis for several weeks, but long-term dosing has shown inconsistent results.
Erythromycin binds to the motilin receptors responsible for the initiation of the MMC in the upper gut. Like metoclopramide, erythromycin works well in the short term, but the long-term loss of response is common.
Domperidone is a dopamine (D2) agonist with effects similar to metoclopramide but is much less likely to cause extrapyramidal side effects as it does not cause the blood-brain barrier. Symptomatic patients with diabetes have reported improved domperidone therapy, but again, the effect could be lost in about six weeks. Domperidone was associated with cardiac arrhythmias in the past and is not FDA approved in the United States. Cisapride is a 5-HT4 agonist that might have effects that last up to 1 year, but unfortunately, it was also associated with ventricular arrhythmias, including torsades de pointes, and has been withdrawn from the US market. Both these drugs are available through select investigators as part of clinical trials with strict patient monitoring. In patients who do not respond well to the above therapies alone, combination prokinetic therapies can be tried. Symptomatic control of nausea using antiemetics is another option. Several studies have evaluated pyloric botulinum toxin injection, but these studies were often unblinded and had small numbers of patients.
Gastric electrical stimulation has been studied, but so far, there has not been a significant reduction in symptoms noted consistently across various small studies. It was granted humanitarian approval from the FDA to treat chronic refractory nausea, but it is not available widely and is not routinely reimbursed.
In refractory patients with severe symptoms, a gastrostomy tube for intermittent decompression by venting or suctioning may provide relief. For patients who cannot maintain nutrition with oral intake, placement of a feeding jejunostomy can be effective in providing nutrition, fluids, electrolytes and reduce hospitalizations. A trial of nasojejunal feeding can predict the response to a jejunostomy.
The last resort in refractory gastroparesis is surgery, including a partial gastrectomy with Roux-en-Y gastrojejunostomy and gastric resection. There is not enough data on surgical treatment at this time, and the decision to subject a patient to this should ideally be made by a motility expert.
Differential Diagnosis
Gastric outlet obstruction
Rumination syndrome
Functional dyspepsia
Chronic pancreatitis
Biliary colic
Prognosis
The outcomes of diabetic gastroparesis are not well defined. Prognosis depends on the adequate control of hyperglycemia and compliance with medications. It affects the quality of life and can have significant morbidity if left uncontrolled. Education regarding the need for lifelong management of this condition is essential to improve clinical outcomes.
Complications
Malnutrition is an important complication of diabetic gastroparesis that often needs management with jejunostomy, parenteral nutrition, or surgery. Wide glycemic fluctuations can occur in diabetic gastroparesis, and this can lead to complications such as hypoglycemia and diabetic ketoacidosis or hyperosmolar hyperglycemic state. Nausea and vomiting can lead to aspiration pneumonia.
Deterrence and Patient Education
Patients require education regarding the chronic and often irreversible nature of the disease. Adherence to a "gastroparesis diet," judicious use of medications, and strict control of hyperglycemia are important patient factors that can change the course of the disease and prevent complications. Patient education and consultation with a diabetic educator can greatly improve the clinical outcomes of this disease.
Enhancing Healthcare Team Outcomes
Treatment of diabetic gastroparesis aims to alleviate the associated symptoms and replenish electrolytes, nutrition, and hydration. Frequently used medications include erythromycin (macrolide antibiotic-associated with increased gastrointestinal motility) and metoclopramide (antiemetic and prokinetic), and polyethylene glycol 3350 may additionally be used to provide relief from severe constipation. Patients who continue to experience gastroparesis symptoms despite medical therapy may be candidates for gastric electrical stimulation (GES), wherein an electrical device is implanted into the abdomen. Unfortunately, no treatment works reliably or consistently. Once the condition has been diagnosed, it is progressive and imparts a very poor quality of life.
The management of diabetic gastroparesis is extremely challenging. It is best done with an interprofessional team that includes an internist, endocrinologist, diabetes nurse educator, primary care provider, gastroenterologist, specialty-trained nurses, and pharmacists. Specialty nurses play the crucial role of assisting clinicians with patient education and monitoring at regular intervals to ensure adequate control of symptoms. A diabetes nurse educator can assist the clinical team by educating the patient on adequate glycemic control to prevent the progression of this disease. Clinical pharmacists are essential in monitoring patients for adverse reactions from therapy and to limit potential drug-drug interactions. This interprofessional approach will optimize patient outcomes. [Level 5]
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References
1.
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: Silvers D, Kipnes M, Broadstone V, Patterson D, Quigley EM, McCallum R, Leidy NK, Farup C, Liu Y, Joslyn A. Domperidone in the management of symptoms of diabetic gastroparesis: efficacy, tolerability, and quality-of-life outcomes in a multicenter controlled trial. DOM-USA-5 Study Group. Clin Ther. 1998 May-Jun;20(3):438-53. [PubMed: 9663360]
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: Parkman HP, Yamada G, Van Natta ML, Yates K, Hasler WL, Sarosiek I, Grover M, Schey R, Abell TL, Koch KL, Kuo B, Clarke J, Farrugia G, Nguyen L, Snape WJ, Miriel L, Tonascia J, Hamilton F, Pasricha PJ, McCallum RW. Ethnic, Racial, and Sex Differences in Etiology, Symptoms, Treatment, and Symptom Outcomes of Patients With Gastroparesis. Clin Gastroenterol Hepatol. 2019 Jul;17(8):1489-1499.e8. [PMC free article: PMC6500483] [PubMed: 30404035]
: Disclosure: Ganesh Aswath declares no relevant financial relationships with ineligible companies.
: Disclosure: Lisa Foris declares no relevant financial relationships with ineligible companies.
: Disclosure: Ashwini Ashwath declares no relevant financial relationships with ineligible companies.
: Disclosure: Krunal Patel declares no relevant financial relationships with ineligible companies.
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Aswath GS, Foris LA, Ashwath AK, et al. Diabetic Gastroparesis. [Updated 2023 Mar 27]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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2780 | https://www.youtube.com/watch?v=GYNX5f-JUWY | Algebra - Solving One-Step Equations
yaymath
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Posted: 11 Jun 2020
Before we get into more complex equations, in this video we learn about solving equations in just one step. Mainly, that means using either addition, subtraction, multiplication, or division, to isolate the variable, and solve for it.
Most notably though, I invite my father to join me in this exploration. When I was a child, my dad was instrumental in fostering a lifelong love of math learning. We'd work for hours on many nights, slowly solving problems piece by piece, eventually leading to my deep appreciation for the subject, and his efforts. And what a full-circle moment this is... standing together in the YouTube studios in LA, with ME showing HIM how I teach some of the very concepts he worked on with me, decades ago. The energy in the clip is palpable, and I'm so happy to share it in this way.
Thank you dad, for the gift (and career) of math. I love you.
Learning math can always be fun and connective. Please visit for:
• all videos
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Transcript:
Introduction [Music] all right welcome to yay math dot org and a special edition of yay math in studio where I've invited my twin here also known as my dad to join us for a little math learning lesson on a whim I invited you last night to come here thanks for coming you have to be here yeah this is kind of cool yeah you know that you know why I invited you I was thinking about the first math memories that I have with you do you recall that seventh and eighth grade there's the early game II yeah mr. della ganas that's my invitation to go from the at level class the honors class was in eighth grade and we worked on problems every night remember in the desk in the corner with the little lamp yeah it was like a dark room but the lamp next to the thing yeah I loved it yeah yeah yeah those really those are really cool memories we did tons of word problems like erase it and draw the lines straighter like you introduced me I remember those things like you introduced me to like trying to seek out and do a better way of the problem like do my best like have pride in what I do yeah I remember things like put your name on it have pride in what you put your name at the top and I say those things to the students and and now look where we are could you ever in your wildest dreams Solving OneStep Equations think that we'd be recording a math lesson for maybe thousands of people all over the world just wonderful yeah it's a totally did yeah like I would ever imagine that this is what it would lead to so thanks for that yeah so let's let's do a full circle let's let's teach math maybe the other way around and as you're an engineer retired engineer world class so be fun to see what kind of algebra is stuck you know so no pressure okay all right cool you want to say a math one two three yeah all right rockin cool okay so this is called solving one-step equations basically the ideas you can do all of these in one step officially but what happens a lot with kids you imagine if they're in class they get so caught up with algebra that they forget to use common sense so that's why it's perfect that you're here because week your your of illogical side a problem solving side so I know that maybe looking at some of these you can solve them with common sense and then we'll do algebra as much as possible all right so let's let's see how it goes so this is a number plus nine is 20 can you think of that number eleven right so X is 11 now can you pretend because 11 Plus 9 is 20 can you pretend if you didn't know what X was what application you need to do to both sides of the equation to get X by itself yeah I would think of it as a scale yeah is one side and this is the other side right scale got you yeah so I have a nine oh wait yeah yeah and it's equal 20 so if I take nine-point off of this side yeah I have to take nine pound off of that so right so that would make it balanced balanced it's still balanced right so it doesn't matter what we do as long as we do it to both sides right a reminder that algebra means to restore and rebalance in Arabic Oh to restore and rebalance so that's the literal definition of algebra and so yes pretending we didn't know that X was 11 we take off nine ounces on this side of the scale and nine off this side of the scale the balance is returned so it's like if you take off nine off this side it would go like that right and then you take nine off that and we'll go back to middle and not plus nine minus nine goes to zero leaving X xx minus 9 is 11 nice so what's the opposite function of minus nine mathematically oh you want to give me the answer what's the answer I'm just thinking 29 29 but you know it's interesting that when you have a number yeah on this side when he goes to the other side it changes the sign right that's how you balance it right you have to counteract what over here to cancel it so that's the question I asked students all the time what do we need to do to both sides to get X by itself and the fancy word for getting X by itself is isolate right so put that down I want to isolate the variable and the image that I have in my mind about isolating the variable this is the variable that's the thing we don't know right and think of it like X is at a party and X is kind of being a jerk and everyone wants everyone's like it's like getting away you know like X is just sort of like passive-aggressive so let's go to the kitchen right so what would we need to do to both sides to get X by itself nice good job so let's ramp it up here minus nine plus nine cancels 20 plus nine like you said is 29 all right do you know the answer 20 over nine is correct and what if you does that simplify in any way first of all like can you simplify that fraction good so we'll leave it alone as improper what do we do need to do to both sides that's opposite x 9 divided by 9 that's right don't worry we'll challenge you 9 over 9 is 1 we got a lot to do so 20 over 9 very good can you think of a number such that when you divide it by 9 the result is 20 20 times 9 what 180 180 there you go there you go okay so X is indeed 180 and now what is the opposite function of divided by 9 both side by not exactly so like you said to isolate X we need to do the opposite function right so divided by 9 opposite that is times what is 9 divided by 9 1 there you go so it's like 1 1 X whatever you do to one side we do to the other same with the scale if you multiplied one side it's weighed by nine and multiplied the other size weight by nine balanced good you would that confuse everybody if I say the way we did it or what you do although I like back in Iran right this yeah what you do you put 20 over one a lot of people do that yes 20 over 1 I know you cross multiply nine times honey equal each time one yes you're allowed to cross multiply whenever you have two fractions equal to each other it's a great method yeah and Cross Multiplication it's also beneficial for when you have complex fractions like X minus 1 over 3 equals 2x plus 1 over 6 or something like that this would be great cross multiplication time right because it's like I don't could you do this in your head this is 2 times 4x plus 1 equal X minus 1 so are you doing algebra in your head are you using logic in your right so that's doing algebra right this was in your head you knew that a number plus 9 was 20 that number would have to be 11 so what I'm getting at I don't want you to do it I don't want you do it because I know you that you can do algebra and this is a ratio and then there's another video that covers how to do ratios itself and I might yeah it might not even make sense that's true where is this so but that's a good cross multiplication would be yeah these two multiplied is the same as these two multiplied that's true but this is the idea of like what I'm trying to encourage is to not get carried away with process uh-huh you know and just like logically speaking opposite function is x and then we know what 20 times 9 is it's 2 times 9 which is 18 with a 0 right in your opinion in your opinion based on your schooling and just as a citizen do you think that students should be obligated to solve problems in a certain ways no I should just get the right answer they should get the right answer but you know if you practice a lot a lot a lot yeah you get very good at it right it just comes to natural it's just a practice thing yeah I agree I don't think we should test a method I think we should test results right now and explain yourself that would be good to two-thirds of a number is 12 I saw you looking at it what number multiplied times 3 ah interesting so do you know two thirds of some number what that number will have to be is two thirds of 4 equal to 12 this is great okay so you're doing the math seven let's do the math let's do it so yeah 18 2012 then is a nice so you use the algebra to help you write that was really cool and this is the point thank goodness for algebra thank goodness because if you're like well two-thirds of what you could start to say well I need a number bigger than 12 because two-thirds of it right two-thirds of something like buying bang bang this is one way to do it visually here's thirds like chocolate right and then you have two thirds of it one two right this is 12 so what each of these have to be 6 + 6 + 6 right so the whole number is 18 so this would be using some sort of visual aid for people that like to learn visually but and if that doesn't occur to us totally how do we solve this in one step guess this is awesome this is so great yeah like you know you have a lot of empathy for kids they're like learning and even adults learning this you know if you don't know it we can learn it I'll show you I'll show you how did we get rid of / 9 x 9 cool what do we multiply by here first boy 3 2 3 3 3 halves yeah 3 you could say that - yeah yeah that's doing it in one step could you have done it in two steps as in yeah but we're trying for one just want to play three on both sides and then it was that's gone then / - on the next round right undo undo I can't I can't undo I would love undo right now there undo is my fingers okay there we go so yeah let's do it in one step multiply by three over to this fancy word is called a reciprocal right and you multiply by the fraction or the number that basically cancels everything notice blue over here blue over here yes boom so three and three cancels to one two and two cancels to one this isn't just being formal about it you don't really always have to do that this is over one yeah about this yeah what does 12 become six six two becomes one yeah exactly two and twelve divided into six and one multiplying fractions six times three 18 beautiful okay doing great negative three times what is 12 you got energy your key for - for - for good how would you do it in one step if you didn't know that we can divide both sides by -3 that's right divide both sides by -3 these go negative 3 over itself a positive divided by a negative is always very good yeah 12 over 3 is 4 whoa right mm-hmm some people find this intimidating I had a one of my first mentor teachers a department chair with Kathy Edwards is her name when her students said I hate fractions mmm I'd fetch of the Fed weights she would say just serious should say you need to write an apology letter to fractions ah and it's claimed how sorry you are for hating fractions your whole life fractions is very hurt they will do it 9th grade kids they're like dear fractions I'm sorry for hating you for so long the healing process fractions are just numbers now what would we do to both sides we could subtract the tool with lead over three right to get X by itself good job but then we'll need to deal with this so because it's difficult it's not straightforward to subtract fractions that don't have common denominators so we're gonna make that common common denominator now what would you like the denominators to be with two and three already they're six Bravo all right so can you make this as an improper how many halves in one and a half three three / - that's right so if you had one and a half pizzas that's three halves of pizza one and a half is three halves one two three good job and then how many sixths is that times what I want to make it six like you said three x three top and bottom yes x three top and bottom very good top becomes 3 times 3 9 good job two times three six there's nine over six okay and then let's make this into six as well so multiply top and bottom of this by what number so that the three becomes a six - good very nice multiply across this is negative 4 and then 3 times 2 and 8 is 6 so now we're dealing with this pink stuff 9 6 - 4 6 how many left five five six good job ok so I I'm really not not Double Negative to Subtract happy that you're here so you're happy right this good very good very good so that's called the double negative to subtract a negative becomes what positive positive Bravo so this isn't officially a step this is a simplification like we haven't done anything to the scale in other words do what to both sides - seven-one-seven here you go 7 + 7 good yeah so this is plus 7 here plus 7 here these go and what's negative 12 plus 7 yep you find I get at 5 indeed so we got x equals negative 5 last one on this particular brand I don't know it but do you know 3 and a half of what thing is 6 no I don't know oh he's trying okay you're trying okay so 3 half of something smaller than 6 right because you have like this thing this thing 3 and a half of it so same thing thing half of thing 3 and 1/2 of thing is 6 that's hard that's hard it's hard it's a fraction clearly we and we've apologized to fractions so we're on good terms now very good we're congenial yeah suggestions right now is that convenient for you 6 divided by 3 point 5 if you divide both sides by 3 point 5 right yeah that yeah I mean there's no that's not that's not a crime that's not a crime you know some people are like I need my cat just do it they they're like talking into the phone what 6 divided by 3 point 5 so that technology is there and people can use it that's cool I wonder if there's a way we can craft this using fractions instead of decimals improper 5 improper can you turn that into an improper fraction as in how many halves oh yes can you do that mathematically 3 times what times 2 is 6 plus 1 is 7 did you remember that all the way from school back in the day day or did you do that in your head like like how did you know to do that you just remember we did it forever you did an ad nauseam you would solve hundreds of problems hundreds yeah just like the drill drill yeah drill drill drill drill do you think that's a good learning method that always was pretty good in math I had no fear of it because of that right because I it was drilling me so lots of work volume high volume yeah let's just begin to like it actually I mean you see the problem you can solve it so easily you begin to like it right right yes so you don't let you remove the stigma it's because they get some momentum yeah a lot of people are into that they find beauty in the grind right they call that they grind and they grind the grind and then they find beauty at the other end of that work yeah multiply both sides by what for a two over seven Bravo reciprocal and yeah clearly as we both know it was obvious all along and for some people like to multiply fractions like that what's the result of course that was what I was gonna say that twelve over seven so that's that was the tip of my tongue okay congratulations that was solving equations with one step [Music] |
2781 | https://www.quora.com/You-have-two-different-concentrations-of-the-same-solution-What-would-be-the-final-molarity-and-volume-of-the-solution-if-you-mixed-750mL-of-the-weaker-1-6M-with-0-35L-of-the-stronger-6-0M-version-of-the-solution | Something went wrong. Wait a moment and try again.
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5
You have two different concentrations of the same solution. What would be the final molarity and volume of the solution if you mixed 750mL of the weaker 1.6M with 0.35L of the stronger 6.0M version of the solution?
Ted Krapkat
46 years as a chemist in the cement & aluminum industries.
·
Author has 996 answers and 4.9M answer views
·
Updated 1y
Let the volumes and molar concentrations of the two solutions involved be;-
V1 = volume of the first solution (in L) = 0.75L
M1 = molarity of the first solution (in mol/L) = 1.6M
V2 = volume of the second solution (in L) = ) = 0.35L
M2 = molarity of the second solution (in mol/L) = 6.0M
V3 = volume of the final solution (in L) = (V1 + V2) = (0.75 +0.35) =1.10L
M3 = molarity of the final solution (in mol/L)
Now, using these variables we can write the following equations;-
V1M1 = number of moles of solute in V1 litres of the first solution.
V2M2 = number of moles of solute in V2
Let the volumes and molar concentrations of the two solutions involved be;-
V1 = volume of the first solution (in L) = 0.75L
M1 = molarity of the first solution (in mol/L) = 1.6M
V2 = volume of the second solution (in L) = ) = 0.35L
M2 = molarity of the second solution (in mol/L) = 6.0M
and;-
V3 = volume of the final solution (in L) = (V1 + V2) = (0.75 +0.35) =1.10L
M3 = molarity of the final solution (in mol/L)
Now, using these variables we can write the following equations;-
V1M1 = number of moles of solute in V1 litres of the first solution.
V2M2 = number of moles of solute in V2 litres of the second solution.
V3M3 = number of moles of solute in V3 litres of the final solution.
Since the sum of the numbers of moles of solute in the first and second solutions must equal the total number of moles of solute in the final solution, we can also write this equation as ;-
V1M1 + V2M2 = V3M3 ……. Eqn 1.
Expressing this equation in terms of M3 we have;-
M3 = (V1M1 + V2M2) / V3 ……. Eqn 2.
We can now substitute the values for the variables given above into Eqn 2;-
M3 = (0.751.6 + 0.356.0) / 1.1
M3 = 3.0
So, the molarity of the final mixture is 3.0M and its volume is 1.1L.
Using the algebraic equation;
V1M1 + V2M2 = V3M3
any unknown volume or molarity in this type of problem can be calculated, if the others are known.
Original Question Answered: “You have two different concentrations of the same solution. What would be the final molarity and volume of the solution if you mixed 750mL of the weaker 1.6M with 0.35L of the stronger 6.0M version of the solution?”
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Related questions
What happens to molarity when two solutions containing different concentrations of the same substance are mixed?
What is the molarity of the resulting solution if 500ml of water is added to 300ml of 0.4M H2SO4 solution?
Why does mixing two solutions decrease the concentration of the solutes?
What is the molarity of a solution prepared by diluting 6.20 mL of 1.75 M NaCl solution to a final volume of 12.0 mL?
How would you prepare 0.001M of a solution from a solution whose concentration is 0.1M?
Eric Ressner
Ph.D. in Medicinal Chemistry, University of Kansas (Graduated 1974)
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Author has 6K answers and 7.8M answer views
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4y
You have two different concentrations of the same solution. What would be the final molarity and volume of the solution if you mixed 750mL of the weaker 1.6M with 0.35L of the stronger 6.0M version of the solution?
This and all such similar problems are more algebraic than chemical. What we are being asked about here is a “weighted average”, where the final concentration (the weighted average) depends not only on the concentrations of the inputs but on their volumes as well.
A simple non-chemical analogy would be something like
The population of Biphasicville is composed completely of public scho
You have two different concentrations of the same solution. What would be the final molarity and volume of the solution if you mixed 750mL of the weaker 1.6M with 0.35L of the stronger 6.0M version of the solution?
This and all such similar problems are more algebraic than chemical. What we are being asked about here is a “weighted average”, where the final concentration (the weighted average) depends not only on the concentrations of the inputs but on their volumes as well.
A simple non-chemical analogy would be something like
The population of Biphasicville is composed completely of public school teachers and CEOs. The 120 families of teachers have incomes of $30,000/year each. The 15 families of CEOs have incomes of $1,500,000 each. What is the average income of a Biphasicville family?
See, we can’t just average $30,000 and $1,500,000 because there are more families earning the smaller income. So the calculation of the weighted average would be
(120×$30000)+(15×$1500000)=135x
On the left is the contribution of each of the individual classes, and on the right is the total of all the different classes, 120 + 15 (and there can certainly be more than two if you’re dealt that sort of problem) times the weighted average.
This is exactly like the mixing problem posed above. The final volume is the sum of the input volumes and the final concentration will be the weighted average of all the input solutions, weighted according to their respective volumes. And again here, there can certainly be more than two. But we only have two, so …
(750 mL×1.6 M)+(350 mL×6.0 M)=(1100 mL)x
(and note that we do have to be consistent with our units: mL or L throughout, either one would work, but you can’t mix them).
And finding x is not particularly challenging.
The same kind of calculation also applies to finding the atomic mass of an element when given the masses and abundances of the naturally occurring isotopes. And probably elsewhere as well, but nothing comes immediately to mind.
We are forced to assume that the volumes are additive. This is almost never strictly true, but is probably close enough when our givens — and thus our answer — are limited to two sig figs.
Howard Neilly
Studied at University of the Bahamas (Graduated 1983)
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Author has 3.1K answers and 1.5M answer views
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4y
Originally Answered: If You have two different concentrations of the same solution. What would be the final molarity and volume of the solution if you mixed 750mL of the weaker 1.6M with 0.35L of the stronger 6.0M version of the solution?
·
Let A represent the solution having a molarity of 1.6M and volume of 750mL. Let B represent the solution having a molarity of 6.0M and a volume of 0.35L
First calculate the amount of moles in solution A
Moles in solution A is found by using the equation
MV=moles
Where
M= molarity=1.6M
V=volume= 750mL (volume must be in liters) knowing that 1000ml= 1L, 750mL1L/1000mL= 0.750L
1.60.750=1.2mol
For solution B
molarity=6.0M
volume =0.35L
using the same equation as above gives
6.00.35=2.1mol
Total amount of combined moles= 1.2+2.1= 3.3mol
Total amount of combined volume= 0.750+0.35= 1.1L
Final molarity = moles/v
Let A represent the solution having a molarity of 1.6M and volume of 750mL. Let B represent the solution having a molarity of 6.0M and a volume of 0.35L
First calculate the amount of moles in solution A
Moles in solution A is found by using the equation
MV=moles
Where
M= molarity=1.6M
V=volume= 750mL (volume must be in liters) knowing that 1000ml= 1L, 750mL1L/1000mL= 0.750L
1.60.750=1.2mol
For solution B
molarity=6.0M
volume =0.35L
using the same equation as above gives
6.00.35=2.1mol
Total amount of combined moles= 1.2+2.1= 3.3mol
Total amount of combined volume= 0.750+0.35= 1.1L
Final molarity = moles/volume of solution= 3.3/1.1=3M
Final volume= 1.1L
Final molarity= 3M
Jakob Krämer
Studied Chemistry at Technical University of Berlin
·
Author has 2.6K answers and 1.8M answer views
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4y
Well, we calculate first the number of moles:
c=nV so n=c⋅V
n1=1.6mol/l⋅0.75l=1.2mol
n2=6.0mol/l⋅0.35l=2.1mol
the final number of moles will be 1.2 mol + 2.1 mol= 3.3 mol
the final volume will be 0.75l+0.35l=1.1 l
the final molarity will be c=3.3mol1.1l=3.0mol/l
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Related questions
How do you make a 1/10 dilution of a Molarity solution?
250mL of a 5.0M KCl solution is diluted with water to a final volume of 500mL. What is the final molarity of the solution?
How do I prepare 50 ml 0.1 mM DPPH from a 1 molar D PPH stock solution?
How is the 0.1 molar solution of NaOH prepared?
How will you prepare a 1 molar solution of glucose?
Daniel Iyamuremye
Former Senior Lecturer (Retired) (2000–2018)
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4y
This question cannot be answered for the following reasons:
The two solutions are not the same solutions since you say that one is stronger and the other is weaker???
What kind of solutions are they: acids, bases, salts?
Even if the above two questions are answered, there are two contradictory factors: (I) mixing the two solutions is equivalent to dilution; this may increase the dissolution of the weaker solute! (ii) mixing the two solutions brings about the problem of common ion and decreases the dissolution of the weaker solute!
In conclusion, this Problem cannot be solved
Jose Carlos Gomes
Chemisty, MSc, PhD and retired food science full professor
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Author has 2.3K answers and 2.1M answer views
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4y
Use the equation but make sure that there is no variation on the volume of added solutions. I explain: adding 500 mL of concentrated sulfuric acid to 500 mL of water, the final volume is not 1000 mL, it is lesser. There is a contraction of volume due to the interaction of ions and water molecules. V x M + V1 x M1 = V2 x M2; 750 x 1.6 + 350 x 6.0 = (750+350) x M2; M2=
3
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Armando Viray
B.S Biochemistry from University of Santo Tomas (Graduated 1983)
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Author has 104 answers and 64.3K answer views
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2y
Related
What happens to molarity when two solutions containing different concentrations of the same substance are mixed?
The molar concentration will decrease but not less than the lesser concentration that was added but not more than the one with greater concentration added. In other words the new molarity is in between the two added solutions of different molar concentration. Here is an example:
.5L of 3 M HCl + .3L of 5 M HCl
.5L x 3 M = 1.5 mols
.3L x 5M = 1.5 mols
Total number of mols:
1.5 mols + 1.5 mols = 3 mols HCL
Total Volume: .5L + .3 L = 0.8 L HCl
The new Conc. is:
M = mol/V
M = 3mols/0.8L = 3.75 M HCl
Hence the concentration : 3M < 3.75M < 5M
Trevor Hodgson
Knows English
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Author has 11.8K answers and 12.3M answer views
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Updated 2y
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There is 100mL of solution at 1M concentration. If I take 1mL from the solution, what is the molarity concentration of the 1mL solution?
Have you ever carried out a titration - for example titrating 0.1 M HCl against an unknown molarity NaOH solution ?
Say that you decide to Add the NaOH solution from the burette . You then take a standard 0.1 M solution of HCl ( usually kept in most laboratories in at least 1.0 L bottles) . You use a pipette to transfer 25.0 mL of this standard 0.1 M solution to the Erlenmeyer flask. If you have ever done this then you would say with confidence that the molarity of the 25 mL used is exactly the same as the molarity of the large volume of standard HCl solution .
Simply - the molarity does not dep
Have you ever carried out a titration - for example titrating 0.1 M HCl against an unknown molarity NaOH solution ?
Say that you decide to Add the NaOH solution from the burette . You then take a standard 0.1 M solution of HCl ( usually kept in most laboratories in at least 1.0 L bottles) . You use a pipette to transfer 25.0 mL of this standard 0.1 M solution to the Erlenmeyer flask. If you have ever done this then you would say with confidence that the molarity of the 25 mL used is exactly the same as the molarity of the large volume of standard HCl solution .
Simply - the molarity does not depend on the volume of sample removed from the larger volume .
Now you can answer your own question.
Pick the correct answer : 0.01 M , 0.1 M , 1.0 M or 10M?
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Former Physicist, Retired
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Author has 610 answers and 242.9K answer views
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3y
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What is volume of two solution v1 having molarity 80 M and v2 having molarity 20 M is required that is mixed together to make a solution of 1L solution of 50 molar?
The solution #1 having the volume V1 L and the molarity M1 = 80 mol/L contains V1 M1 mol of solute. The same way solution #2 having the molarity M2 = 20 mol/L contains V2 M2 mol of solute and the mixture V1 + V2 = 1 L with the molarity M = 50 mol/L will contain the total solute V1M1 + V2M2 = (V1 + V2) M. Introducing the numbers we get
V1 + V2 = 1, and
80 V1 + 20 V2 = 50.
The solution of these equations is V1 = 0.5 L and V2 = 0.5 L. This result could also be guessed from the fact that 50 is the middle value between 20 and 80, so the volumes to be mixed would be equal.
Stanton De Riel
B.S. from M I T (Graduated 1990)
·
Author has 717 answers and 246.4K answer views
·
2y
Related
What happens when we mix two solutions of different concentrations?
That depends largely on whether the two solvents of the two solutions are miscible (dissolve in each other). If not, the two solutions stay as two discrete layers. Example: Salt in water, mixed with wax in carbon tetrachloride.
However, if the two solvents are miscible, then they will mix. The end result then depends on whether the two solutes are soluble in the solvent mixture thus produced. If they are, then they, too, mix. The resulting concentration of solute(s) is just the amount(s) of each divided by the final volume (which might not be additive arithmetically!). If either or both of them
That depends largely on whether the two solvents of the two solutions are miscible (dissolve in each other). If not, the two solutions stay as two discrete layers. Example: Salt in water, mixed with wax in carbon tetrachloride.
However, if the two solvents are miscible, then they will mix. The end result then depends on whether the two solutes are soluble in the solvent mixture thus produced. If they are, then they, too, mix. The resulting concentration of solute(s) is just the amount(s) of each divided by the final volume (which might not be additive arithmetically!). If either or both of them are NOT soluble at the levels predicted as mixed, then the partially-insoluble one(s) generally precipitate out (if they are a solid when pure), although they also may sequester a bit of their initial solvent and form a more-concentrated liquid phase. It all depends on what we are mixing!
Lastly, particularly on Quora, one should note that sometimes components of the two solutions may react vigorously when mixed. Then you, too, are likely to end up in the mix, unless you are careful.
Antonio Daniels
Former Learner 2020- Chemist
·
Author has 8K answers and 4.5M answer views
·
1y
Related
I have a 0.012M of 250mL of solution. If I take 25mL out of that solution, does the molarity stays the same?
Yes, it stays the same. What you took out and what you left back will have the same molarity.
VM = mol
left back
0.012M x0. 225 L = 2.7 x 10-3 mol
2.7 x 10-3 mol / 0.225 L = 0.012 M
what you took out
0.012 x 0.025 = 0.003 mol
0.003 mol / 0.025 = 0.012 mol/L
You are only taking out a fraction of the volume of the solution
You are not adding solute either water. Nothing changes.
If the solvent evaporates then the molarity will change
CheeHock Chew
Author has 3.9K answers and 2.6M answer views
·
2y
Related
If 25.0 mL of 0.500 M solution of CuSO₄ is mixed with 55.0 mL of 0.100 M solution of CuSO₄, what is the final concentration? Assume that the volumes are additive.
If 25.0 mL of 0.500 M solution of CuSO₄ is mixed with 55.0 mL of 0.100 M solution of CuSO₄, what is the final concentration? Assume that the volumes are additive.
Let C = final concentration of mixture.
Find the concentration C by using the formula of V1C1 + V2C2 = V3C3, where C1 = 0.5 M, V1 = 25.0 ml, C2 =0.10 M, V2 =55.0 ml, C3 = C M and V3 = (25.0 + 55.0) = 80.0 ml
From the notation, (0.5 x 25) + (0.1 x 55.0) = C x 80.0
That is, 12.5 + 5.5 = 80C. On solving for C = (18/80) = 0.225
Thus, the final concentration of mixture = 0.225 M
Barry Gehm
Former Asst Prof. Of Chemistry/Biochemistry at Lyon College (2003–2024)
·
Author has 13.6K answers and 16.6M answer views
·
8y
Related
We have 49gm of H3PO4 per 500ml of solution. After calculating normality and molarity, which solution is more concentrated than the other and why?
The question is meaningless: there is only one solution, so it is daft to ask which solution is more concentrated.
The number expressing the normality will be greater than the number expressing the molarity, but the question is like asking “A stick is 25.4 cm long, which is equal to 10 inches. Which stick is longer?”
Related questions
What happens to molarity when two solutions containing different concentrations of the same substance are mixed?
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© Quora, Inc. 2025 |
2782 | https://admjournal.luguniv.edu.ua/index.php/adm/article/download/2312/pdf | ©Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 39 (2025). Number 1, pp. 1–35 DOI:10.12958/adm2312
Flip graphs of coloured triangulations of convex polygons
Karin Baur, Diana Bergerova, Jenni Voon, and Lejie Xu
Communicated by V. Mazorchuk
Abstract. A triangulation of a polygon is a subdivision of it into triangles, using diagonals between its vertices. Two different triangulations of a polygon can be related by a sequence of flips: a flip replaces a diagonal by the unique other diagonal in the quadri-lateral it defines. In this paper, we study coloured triangulations and coloured flips. In this more general situation, it is no longer true that any two triangulations can be linked by a sequence of (coloured) flips. In this paper, we study the connected components of the coloured flip graphs of triangulations. The motivation for this is a result of Gravier and Payan proving that the Four-Colour Theorem is equivalent to a property of the flip graph of 2-coloured triangulations: that any two triangulations can be 2-coloured in such a way that they belong to the same connected component of the 2-coloured flip graph.
1. Introduction
A triangulation of a polygon is a subdivision of it into triangles, using diagonals between its vertices. Two different triangulations of a poly-gon can be related by a sequence of flips: a flip replaces a diagonal by the unique other diagonal in the quadrilateral it defines. In this paper,
2020 Mathematics Subject Classification: 05C15, 05A99.
Key words and phrases: triangulations, flips, graph colourings, connected com-ponents.
2 Flip graphs of coloured triangulations
we study m-coloured triangulations and m-coloured flips: we allocate n
colours to the triangles and flip diagonals only if the two triangles in-cident with it have the same colour, say i. The flip assigns the colour
i + 1 (reducing modulo n) to the two introduced triangles. When using colours, it is no longer true that any two triangulations can be linked by a sequence of (coloured) flips. In this paper, we study the connected components of the coloured flip graphs of triangulations. The motivation for this is a result of Gravier and Payan proving that the Four-Colour Theorem is equivalent to the property that for any two triangulations of a convex polygon, one can find a 2-colouring of their triangles in such a way that they belong to the same connected component of the 2-coloured flip graph. This problem is naturally very hard to solve and still wide open. Our contribution sheds light on the size of the connected components under coloured flips and on the shapes of the components. Flip graphs of triangulated surfaces are well known and have been studied a lot, in particular the polygon case is one of the many instances of the Catalan combinatorics. Different generalisations appear in various contexts. Here, we only mention two directions. Edge colouring, their orbits under flipping and the diameters of the flip graphs have been studied in and . Triangulated surfaces and flips also appear in the context of cluster algebras and cluster categories: any (monochro-matic) triangulation of a so-called marked surface gives rise to a cluster in the cluster algebra of the surface and any two clusters are linked by a sequence of flips. A different notion of coloured flips also appears in cluster combinatorics, in the context of higher cluster categories, see and . This article is structured as follows: Section 2 contains the back-ground on triangulated polygons and introduces coloured triangulations. Then it explains the link between coloured triangulations and the Four-Colour theorem. In Section 3, we study the size and structure of the connected components of coloured flip graph. Section 4 contains further observations and a conjecture about the possible occurrence of a triangle in a component of the 2-coloured flip graph: we expect that a triangle cannot appear with two different colours in a single component. In an appendix, we provide a translation of the results we use from , we de-scribe the connected components of the 2-coloured flip graphs for small polygons and compute the size of their components. K. Baur, D. Bergerova, J. Voon, L. Xu 3
2. Background
Here we recall the notions of triangulations of convex polygons. We write
Pn to denote a convex polygon with n vertices.
Definition 2.1 (Triangulation) . A triangulation of Pn is a subdivision of the polygon into triangles, using pairwise non-crossing diagonals. Boundary segments are not considered to be diagonals. Note that any triangulation of Pn decomposes the polygon into n − 2 triangles, using n − 3 diagonals.
Example 2.2. A triangulation given by n − 3 diagonals incident with a common vertex will be called a fan triangulation . An example of a fan triangulation of a hexagon is Figure 1. Figure 1: A fan triangulation of a hexagon. The following result is well-known. We include a proof for convenien-ce. The strategy of the proof is illustrated for n = 8 in Figure 2.
Lemma 2.3. The number of triangulations of a convex (n + 2) -gon is given by the n-th Catalan number Cn = 1
n+1
2nn
.Proof. The proof can be done using an inductive argument. One checks that the claim is true for n = 1. Choose an edge E, and consider the triangle it is a part of. In an ( n+2)-gon, there are n other options for the third vertex of this triangle. All of these reduce the problem to one or two smaller cases, as to the left and right of this triangle, there are smaller polygons of size m − 1 and n + 4 − m respectively, for m = 3 , . . . , n + 2. (For m = 3, there is only a polygon of size n + 1 on the right of the triangle, for m = n + 2, there is only a polygon of size n + 1 on the left of the triangle.) We count the number of triangulations of these two subpolygons and let m run. This gives the total number of triangulations as Cn−1 + C1Cn−2 + ... + Cn−2C1 + Cn−1, which is a well-known recursive formula for the Catalan numbers. 4 Flip graphs of coloured triangulations
Figure 2: Every triangulation of the octagon falls into one of 6 types. Triangulations of a polygon are related by the so-called flip:
Definition 2.4. Let t be a diagonal in the triangulation T of Pn. The union of the two triangles containing t is a quadrilateral. We obtain a new triangulation T ′ by replacing the diagonal t with the other diagonal of that quadrilateral. This local move is called a flip .Any two triangulation of Pn can be linked by a sequence of flips. This is true for more general surfaces and has been proven by several authors independently and for different set-ups. See for example [5, 8, 12] or for triangulations of planar point sets. In the convex polygon case, it can be shown using the procedure of that consists in flipping two triangulations into the same fan triangulation.
Definition 2.5. The flip graph of Pn is the graph whose vertices are triangulations of the polygon, and two vertices T1, T 2 are connected by an edge if and only if there exists a (single) flip linking T1 with T2.
2.1. Coloured triangulations, coloured flips
In this article, we are interested in a generalisation of triangulations: we equip triangulations with a set of colours and define a new flip operation for them. Let m ≥ 1 and let C = {1, . . . , m } be a set of m different colours. If T
is a triangulation of a polygon, we write F (T ) for the set of its triangles (or faces). We write Sm for the symmetric group on m letters.
Definition 2.6 (Colouring) . Let T be a triangulation of a convex poly-gon. By a colouring of T we mean an assignment of colours from 1 , . . . , m
for every triangle of T .
Definition 2.7 (Coloured flip) . Let C = {1, . . . , m } be a set of colours and σ ∈ Sm be a permutation. Let T be a triangulation of a convex polygon with each triangle a colour in C. Let t ∈ T be a diagonal incident with two triangles of the same colour i. Then the σ-flip of T at
t is defined as follows: K. Baur, D. Bergerova, J. Voon, L. Xu 51. Replace t by the flip of t′ in the underlying uncoloured triangula-tion. 2. Change the colours of the two triangles incident with t′ to the colour
σ(i). If the permutation σ is a single cycle of the form (1 , 2, . . . , m ) (i.e. i 7 →
i + 1), we call a σ-flip simply an m-coloured flip .
Definition 2.8. Let P be a convex polygon and let C = {1, . . . , m } be a set of colours, let σ ∈ Sm be a permutation. The coloured flip graph
of P with colours C and permutation σ or the σ-flip graph of P is the graph whose vertices are the coloured triangulations of Pn and whose edges correspond to σ-flips. We will often just call it the flip graph of the polygon. The coloured triangulations are also counted in terms of Catalan numbers. We note that whenever no two adjacent triangles have the same colour, no edge can be flipped and we have an isolated vertex in the flip graph.
Lemma 2.9. Consider a convex n+2 -gon Pn+2 and a set C of m colours. (i) There are Cnmn coloured triangulations of Pn+2 ;(ii) There are Cnm(m−1) n−1 triangulations of Pn+2 where none of the diagonals can be flipped. Proof. (i) Any triangulation of Pn+2 has n triangles, so there are mn
different ways to colour a triangulation. The claim then follows from Lemma 2.3. (ii) We consider the dual graph GT to a given triangulation T of Pn+2 :it has as vertices the triangles in T and an edge between the two vertices of adjacent triangles. This graph is known to be a tree, it has n vertices and at least two leaves. We start by colouring one leaf with one of the m
colours and then proceed to colour adjacent vertices. Since G is a tree, for any new vertex we want to colour, there are m − 1 options. Hence the factor ( m − 1) n−1. The claim then follows with Lemma 2.3.
2.2. Triangulations of polygons and the Four-Colour Theorem
The Four-Colour Theorem is one of the most famous mathematical prob-lems in history. It concerns the question whether four colours are enough 6 Flip graphs of coloured triangulations
to colour any map drawn in the plane. In 1977, Appel, Haken and Koch established that four colours are enough (see and ):
Theorem 2.10 (Four-Colour Theorem) . Any map on R2 can be coloured using four colours such that any two regions sharing an edge are of dif-ferent colours.
This result was proved with the assistance of computers. So far, there is no abstract proof of this theorem. The main motivation of this project is the search for an alternative approach to its proof. In 1999, Elia-hou formulated the following conjecture about signed flips (2-coloured flips) [6, §1]:
Conjecture 1. Let T1, T 2 be two arbitrary triangulations of a convex polygon P . Let C = {1, 2} be two colours. Then there exist colourings of
T1 and of T2 such that there is a sequence of 2-coloured flips between the two coloured triangulations.
Eliahou showed that the signed flip conjecture implies the Four-Colour Theorem. Gravier and Payan later proved that the two are equi-valent , completing the earlier work of Eliahou and of Krychkov (see , a reprint of the original preprint from 1992):
Theorem 2.11 () . Given any two triangulations of a convex polygon, it is possible to find a 2-colouring of each triangulation such that they can be transformed into one another by a sequence of 2-coloured flips if and only if the Four-Colour Theorem holds.
So in order to give an abstract approach to the Four-Colour Theorem, it is enough to give an abstract prove of Conjecture 1. We will recall the proof of Theorem 2.11 in Section 1 of the Appendix.
3. Connected components of coloured flip graphs
We will now show how to reduce the study of coloured flip graphs to permutations which are a single cycle (Lemma 3.1) and the study of its components to 1 or 2 colours (Lemma 3.3).
Lemma 3.1. Every coloured flip graph is the Cartesian product of coloured flip graphs associated to cyclic permutations. Proof. Any permutation can be written as a product of disjoint cycles, say σ = σs . . . σ 2σ1 (where the σi ∈ Sm are single cycle permutations). Let mi be the size of σi. Let T be a triangulation of a polygon and equip it with an m-colouring. K. Baur, D. Bergerova, J. Voon, L. Xu 7For i ∈ { 1, 2, . . . , s } let Di be the union of all triangles of T which are coloured with a colour appearing in σi. Each Di is a possibly dis-connected union of triangles of T . Note that if none of the colours of
σi appear in the m-colouring of T , then Di is empty. See Figure 3 for an illustration with s = 3. Each cycle σi only acts on Di, leaving the rest of the coloured triangulation fixed. Let Gi be the flip graph of the
mi-coloured region Di. Then the m-coloured flip graph of the polygon is a cartesian product of the Gi.763
Figure 3: A 7-coloured triangulated decagon with two regions D1
(shaded) and D3 (white) for σ as in Example 3.2.
Example 3.2. Consider the triangulation of the decagon of Figure 3. Let σ = (1 , 2, 3, 4)(5)(6 , 7) = σ1σ2σ3 be the product of three disjoint cycles. D1 is the union of six triangles, shaded in grey in the figure. D2
is empty and D3 is the union of two triangles (in white in the figure). The coloured flip graph is the product of the flip graph of D1 on the four colours {1, 2, 3, 4}, the trivial flip graph of D2 and of the flip graph of D3
on the two colours {6, 7}.From now on we will concentrate on the case where σ is a single cycle of length m. As a consequence of Lemma 3.1 in order to understand the flip graph of coloured triangulations, it is enough to understand the case from now on we will assume that σ is a single cycle of length m. The next result shows that we can reduce the study of flip graphs to the case with one or two colours.
Lemma 3.3. Let m ≥1 and let σ be a single cycle of length m.(1) If m is odd, then any two triangulations can be m-coloured in such a way that they are linked by a sequence of m-coloured flips. (2) If m is even, two triangulations can be T1 and T2 can be m-coloured in such a way that they are linked by an m-coloured flip sequence if and 8 Flip graphs of coloured triangulations
only if they can be 2-coloured in such a way that they are linked by a
2-coloured flip sequence. Proof. Let T1 and T2 be two triangulations of the same polygon. Let
σ = (1 , 2, . . . , m ) be the single cycle permutation changing colour i to colour i + 1. The idea is to use an uncoloured flip sequence from T1 to
T2 and to replace it with an appropriate coloured sequence. (1) Let m be odd. Let ν = νs · · · ν1 be a sequence of (uncoloured) flips such that T2 = ν(T1). For i = 1 , . . . , s let di be the diagonal of the polygon which will be flipped under νi (the diagonal di is a diagonal of the triangulation νi−1 · · · ν1(T1)). The diagonal di determines a quadrilateral
Di in νi−1 · · · ν1(T1). We write d′
i
to denote the other diagonal of Di.We colour all triangles of T1 with colour 1 and do m coloured flips on D1 (switching between d1 and d′
1
and back while doing so). By doing so, we replace the first flip ν1 by the sequence μd1 . . . μ d′
1
μd1 with m
terms. The resulting triangulation again has colour 1 on all triangles but the diagonal d1 got replaced with d′
1
. Then we iterate, replacing each uncoloured flip νi = νdi by a sequence μdi . . . μ d′
i
μdi of length m of coloured flips. The result of this longer sequence of m-coloured flips is a coloured version of T2.(2) Let m be even. Assume first that there are 2-colourings of T1 and
T2 such that there is a 2-coloured flip sequence ν = νs . . . ν 1 from T1 to
T2, on the two colours {1, 2}. We construct a coloured flip sequence for
σ = (1 , 2, . . . , m ) from ν. As in part (1) of the proof, we denote by di the diagonal of the polygon which is flipped under νi and we let Di be the quadrilateral of the polygon determined by di (it is a quadrilateral in the triangulation νi−1 · · · ν1(T1)). We modify the sequence ν by replacing νi
if needed: If the colour in Di is 1, we keep νi and view it as part of a
σ-flip for the cyclic permutation σ = (1 , 2, . . . , m ). If the colour is 2, we replace νi by the sequence μdi . . . μ d′
i
μdi with m − 1 terms. In both cases, the effect is to replace the diagonal di by d′
i
in Di and the resulting coloured triangulation is the same as νi · · · ν1(T1). We can iterate the procedure and replace (or keep) νi+1 with the same rule. After going through all flips of ν, we obtain a coloured flip sequence from T1 to T2
for m colours, where both T1 and T2 are coloured as for the original 2-coloured sequence. Assume now that there exists a 4-colouring for T1 and T2 and a 4-coloured sequence ν = νs · · · ν1 linking them. We replace every even-numbered colour in T1, T 2 by 2 and any odd-numbered colour by 1. If di
is the diagonal flipped under νi, we define μi to be the same flip, but for K. Baur, D. Bergerova, J. Voon, L. Xu 9
σ = (1 , 2). This provides a 2-coloured flip sequence for the 2-coloured versions of T1 and T2.The strategy to go from a 2-coloured flip sequence to a 4-coloured flip sequence is illustrated in Figure 4. 1 11
ν1ν2
μd1μd1
μd′
1
d2
μ2
Figure 4: A flip sequence ν with two colours, translated into a sequence with four colours (with σ = (1 , 2, 3, 4) = (red, yellow, light blue, green))
Example 3.4. We will again consider an example of coloured hexagon
P6 triangulations, with 2 colours and where σ switches the two colours. In Figure 5, we list all types of connected components of the associated coloured flip graph.
Figure 5: Connected components of the 2-coloured flip graph of P6 up to isometry, excluding the isolated points. 10 Flip graphs of coloured triangulations
From now on, we restrict to 2-coloured flips, i.e. to the case m = 2. We will often choose C = {1, −1} and indicate these colours by + , − in the examples. We often use yellow and red as the two colours. In the statement of the theorem below, we use the notion of a flip sequence: a 2-coloured flip sequence μ = μ2 · · · μ1 (with s > 0) is a sequence of 2-coloured flips, applied successively.
Theorem 3.5. Let G be the coloured flip graph of Pn+2 . Then every connected component of G is either an isolated point or is of size ≥ n.Moreover, if T is a triangulation in a non-trivial connected component and t a diagonal of T , then either t can be 2-coloured flipped or there exists a 2-coloured flip sequence μ = μ2 · · · μ1 (where s ≥ 1) such that
t ∈ μ(T ) and such that t can be colour-flipped in μ(T ).Proof. Let T be a triangulation of Pn+2 , with a colouring. We assume that there is at least one diagonal which can be 2-colour-flipped. We mark the diagonals of T with a blue or a red dot: a diagonal t is marked blue if it can be flipped after some (possibly empty, if it can be flipped immediately) sequence of coloured flips. Diagonals of T which can never be flipped are marked with a red point. By assumption, at least one diagonal is marked with a blue dot. If there exists a diagonal with a red dot, find a triangle with a red and a blue dot on two of its diagonals. Such a triangle always exists (see Remark 3.6). Call these two diagonals
B and R. The two triangles incident with R must be of different colours, since otherwise, we can flip it immediately. We now execute a (coloured) flip sequence in order to flip the edge B. At some point in this sequence (or at the end of the sequence), the colour of one of the triangles incident with edge R has changed colour, and at this point, the edge R can be flipped. This is a contradiction, hence there can be no red dots and every edge can be flipped eventually. Hence there are at least n vertices in this connected component of the coloured flip graph.
Remark 3.6 . Consider a triangulation T where at least one diagonal can be 2-colour flipped. We mark the diagonals of T by a blue dot if the diagonal can eventually be colour-flipped and with a red dot otherwise (as in the proof of Theorem 3.5). Then there exists a triangle with a red and blue dot on two of its sides: We consider the line graph L(T )of T : it has vertices for the diagonals of T and edges between vertices
t, t ′ whenever there is a triangle containing t and t′. We equip this graph K. Baur, D. Bergerova, J. Voon, L. Xu 11 with the two colours. Since the graph L(T ) is connected, there has to be an edge between a red and a blue node. The following result shows that connected components of size n do exist in the coloured flip graph of Pn+2 .
Example 3.7. Consider a fan triangulation T , with alternatingly colou-red triangles apart from at one end. See Figure 6 for an illustration of such a fan triangulation of a decagon. Then the connected component containing T is a single line with n vertices: in every step, only a single 2-coloured flip can be made. Under this, the two triangles of the same colour move from one side of the fan to the other end of the fan.
Figure 6: A coloured fan triangulation of a decagon. We recall the notion of a weighting on the vertices of a triangulated polygon with a 2-colouring from . If we have a coloured triangulation and if △ is a triangle of T , we denote its colour by s(△). Recall that s(△) ∈ {− 1, +1 }.
Definition 3.8. A weighting of the polygon P is given by a choice of a triangulation T and a function p assigning to each vertex of P an element of {− 1, 0, 1} such that there is a 2-colouring of T where for every vertex x
of P , we have p(x) = P
x∈△
s(△) mod 3 (the sum is taken over all triangles incident with the vertex x). Two weightings of coloured triangulated quadrilaterals are shown in Figure 8. The statement that flips do not change the weighting already appears in [7, Text and Figure 3 on page 819]. We include a proof for completeness.
Lemma 3.9. Any two 2-coloured triangulations which are in the same connected component of the flip graph have the same weighting. 12 Flip graphs of coloured triangulations
Proof. Let p be a weighting of a triangulated polygon with 2-colouring. If x is a vertex of the quadrilateral where the flip happens, the flip either changes two triangles with +1 to one triangle with −1 (or vice versa) or two triangles with −1 to one triangle with +1 (or vice versa). In all cases, p(x) remains the same. (See Figure 8). However, there exist 2-colourings which are not flip equivalent but have the same weighting, see Figure 7 for an example.
−−
−
− −
−+
−
−
−
−−
−
− −
−+
−
−
−
Figure 7: Two triangulations which are not flip equivalent but have the same weighting The following statement is mentioned in . We include a proof for completeness.
Theorem 3.10. Given a weighting of a triangulation, there is at most one way to colour it to match the weighting. Proof. Since any triangulation must contain a triangle with 2 of its sides being sides of the polygon, there is a vertex which is only contained in one triangle. At this vertex, if the value is zero then there is no such colouring, and if it is −1 or +1 then this determines the colour of the triangle. Consider removing this triangle, and subtracting off the value it contributes to the neighbouring triangles to give a valuation and triangulation for a ( n−1)-gon, repeat until we either reach a contradiction or have completely coloured the shape. Hence if the colouring exists, it must be unique (see Figure 8).
Theorem 3.11. Any cycle in the coloured flip graph is of even length. Proof. Let T be a 2-coloured triangulation, let X be the number of trian-gles marked + in this triangulation. After every flip, X either increases or decreases by 2, hence every flip changes X by ±2. Therefore, if we K. Baur, D. Bergerova, J. Voon, L. Xu 13
−+
− +++
flip
−+
− +--Figure 8: Example of quadrilateral valuation reach T again after a sequence of coloured flips, this sequence has to have even length, since the number of triangles marked with a + will be equal to X again.
3.1. Structure of the flip graph
In this section, we show properties of the flip graph. In particular, we prove the existence of hypercubes in the flip graph.
Definition 3.12. Let t, t ′ be two diagonals in a triangulation of a convex polygon. If two quadrilaterals have disjoint interiors, we say that the quadrilaterals are disjoint . In this case, we say that the flips of t and of
t′ are independent .A triangulated convex n + 2-gon has n triangles and since two in-dependent flips require two disjoint quadrilaterals, there can be at most
⌊ n
2
⌋ independent flips in any triangulation of Pn+2 .
Example 3.13. Let T be a fan triangulation of Pn+2 where each face is assigned the same colour (see Figure 9 for an example). There are
⌊ n
2
⌋ independent flips in this case: We start by choosing the first two triangles on the left in T . The we continue, choosing quadrilaterals close to the previous one, but independent. We end up with ⌊ n
2
⌋ disjoint quadrilaterals (and possibly one left over triangle). So this coloured triangulation reaches the upper bound of possible independent flips.
Proposition 3.14. Let T be a 2-coloured triangulation of a convex poly-gon. Let G be the connected component of the flip graph containing T .Assume that there are k > 1 diagonals in T which can be 2-coloured flipped and whose quadrilaterals are pairwise disjoint. Then G contains a k-dimensional hypercube. 14 Flip graphs of coloured triangulations
Figure 9: A flip graph of the monochromatic n + 2-gon, n ≥ 8 contains 4 disjointquadrilaterals.
Proof. Denote that k diagonals of T which can be flipped independently by 1 , 2, . . . , k . For any i̸ = j, 1 ≤ i, j ≤ k, the flips μi and μj commute. We consider all the triangulations which can be reached from T by ar-bitrary 2-coloured flips of these k diagonals. In the subgraph of the flip graph they define, each of them has degree k. So they form a subgraph isomorphic to a k-dimensional hypercube as claimed.
Corollary 3.15. For n ≥ 8, the 2-coloured flip graph of Pn+2 contains a connected component which is not planar. Proof. Consider the fan triangulation where every triangle is coloured with the same colour. Let G be the connected component of the coloured flip graph which contains this coloured fan triangulation. Since n ≥ 8, there is a sub-polygon with the same structure as the fan decagon (see figure 8). Hence there are at least four quadrilaterals which can be flipped independently, given by the thick lines. Therefore, G contains a 4-dimensional hypercube by Proposition 3.14. Denote this by Q4. Since
Q4 has the complete bipartite graph K3,3 as a subgraph, and the latter is not planar, G cannot be planar.
Notation. We consider two k-dimensional hypercubes in a connected component of the flip graph to be distinct if they are disjoint or if their intersection is a union of hypercubes of smaller dimension.
Lemma 3.16. Suppose T that is a fan triangulation of a convex (n + 2) -gon where all triangles have the same colour. Let G be the connected component of the coloured flip graph that contains T . Then (i) if n is even, then G contains a n
2
-dimensional hypercube, meeting at least one disjoint 1 ( n
2
− 1) -dimensional hypercube;
1apart from the common vertex with coloured triangulation T.
K. Baur, D. Bergerova, J. Voon, L. Xu 15
(ii) if n is odd, then G contains n+1 2 hypercubes of dimension n−12 ,meeting at the vertex with coloured triangulation T .Proof. We label the vertices of Pn+2 so that the fan is based at vertex 1 and that there are two “boundary triangles” with vertices 1 , 2, 3 and
n + 1 , n + 2 , 1. (i) When we assume n to be even, the maximum number of indepen-dent coloured flips is n
2
. We reach this if we group pairs of adjacent triangles into quadrilaterals starting at the boundary, i.e. with the quadrilateral on the vertices 1 , 2, 3, 4 and continuing with ad-jacent triangles, see Example 3.13. If instead we group the pairs of adjacent triangles of T starting the quadrilateral on the vertices 1, 3, 4, 5, we only get n
2
− 1 disjoint quadrilaterals and find a hy-percube based on n
2
− 1 independent flips. These two hypercubes meet at the monochromatic fan triangulation. (ii) If n is odd, we “ignore” one of the two triangles of T and pair all other triangles to form quadrilaterals, in the same way as in part (i). The triangles we pick for this have vertices 1 , 2ℓ, 2ℓ + 1 where 1 ≤ ℓ ≤ n+1 2 . The remaining triangles form n−12 disjoint quadrilaterals and so we get n−12 independent flips in each case, giving n−12 hypercubes of dimension n−12 , meeting at the coloured triangulation T as their common vertex. Note that a version of Lemma 3.16 can be proved for more gene-ral triangulations: the number of disjoint quadrilaterals in an arbitrary 2-coloured triangulations gives a lower bound on the dimension of ma-ximal dimensional hypercubes it contains. To get the minimal number of independent flips in a monochromatic triangulation, one considers a triangulation with as many “ear triangles” as possible (an ear triangle is a triangle formed by two boundary segments and one diagonal). This way, the number of independent quadrilaterals stays as small as possible. We choose a triangulation built with nested sequence of ear triangles: first draw the diagonals (1 , 3), (3 , 5), etc., ending with ( n − 1, n + 1). Then the diagonals (1 , 5), (5 , 9), etc. There are ⌊ n+2 2 ⌋ ear triangles in such a triangulation. Around half of them will not get matched when forming independent quadrilaterals, depending on the factors of n. A lower bound on the dimension of the largest hypercube in a monochromatic triangulation of this shape is ⌊ n+1 3 ⌋.16 Flip graphs of coloured triangulations
Example 3.17. We illustrate Lemma 3.16 by showing for the monochro-matic fan triangulation of Pn with 6 ≤ n ≤ 9 in Figures 10, 11, 12 and 13. In each figure from left to right, the first graph is the original triangula-tion, and then are the possible hypercubes of different dimensions, and the last one is the combination of all these hypercubes. We number the diagonals in T , and the numbers on edges of the k-dimensional cube represents a flip of that diagonal. The black points represent the fan triangulation. 12 323 3113 3112Figure 10: The connected component with the monochromatic fan triangulation of the hexagon contains two shown 1-dimensional and 2-dimensional cubes. 12 34 4441 1 12 233Figure 11: The connected component with the monochromatic fan trian-gulation of the heptagon with three 2-dimensional cubes.
4. Observations and a conjecture
We conclude this paper by a number of observations and a conjecture. Let Pn+2 be a convex n + 2-gon.
Observation 1. For n ≤ 4 any connected component of the 2-coloured flip graphs of Pn+2 is either a tree, or obtained from adding leaves onto a 4-cycle. See Appendix 2. For n > 4, this is not true anymore. An example is a component for n = 7 in Figure 14. K. Baur, D. Bergerova, J. Voon, L. Xu 17 123 45 2224 4 45 511333355 5511 11333355 55511 111222444Figure 12: The connected component with the monochromatic fan trian-gulation of the octagon has a 3-dimensional cube.
Observation 2. There are connected components of the flip graph which do not have any leaves, see e.g. Figure 14 for n = 7 or Appendix 2 in the case of n = 6. In the examples we considered, no two triangulations in a connected component contained two triangles with the same vertices but with dif-ferent colours. See for example Figure 14 for an illustration. We suspect that this could be true in general:
Conjecture 2. In a connected component of the 2-coloured flip graph, a triangle cannot appear in the same position but with different colours.
APPENDIX
1. Proof of Theorem 2.11
We recall the statement of Theorem 2.11 from the Introduction: the Four-Colour Theorem holds if and only if for any two triangulations of a convex 18 Flip graphs of coloured triangulations
1
2
3 4
5
6
5 5
2
2
5
3
1
2
4
6
5
3
1
2
4
6
Figure 13: The connected component with the monochromatic fan trian-gulation of nonagon contains four 3-dimensional cubes.
Figure 14: A connected component of the flip graph for coloured nonagon triangulations which contains no fans, and has minimum cycle size 20. polygon, one can 2-colour them in a way that there exists a sequence of 2-coloured flips linking the two triangulations. This result by Gravier and Payan motivates the notion of coloured mutation. The work of Gravier and Payan has appeared in French in 2002. For the convenience of the reader, we illustrate their reasoning in this section. We first recall the notions needed. In this section, we will use ‘signed triangulations’ to refer K. Baur, D. Bergerova, J. Voon, L. Xu 19 to 2-coloured triangulations in order to distinguishing from the notion of a colour in the 4-colour theorem.
Figure 15: A connected component for the heptagon, which has three leaves of two different triangulation types.
Definition 1.1. Let P be a convex polygon. We introduce the following definitions:
Let T be a triangulation of P . We write D(T ) for the set of all diagonals of T and F(T ) for its faces (the triangles).
A sub-polygon S ⊂ P is a polygon whose vertices are a subset of those of P , and which respects the cyclic order of the vertices of P .
S − x denotes the sub-polygon induced by all vertices except x.
A signed triangulation of P is a 2-coloured triangulation T of the polygon, i.e. a pair {T, s }, where s : F(T ) → { +1 , −1} is a 2-colouring of the triangles of T . Let s be the signed triangula-tion obtained from s by changing all signs. We write ( T, s ) to denote the class {{ T, s }, {T, s }} .
A signed flip is a 2-coloured flip of a diagonal of a signed triangu-lation.
We recall the definition of a weighting of P (Definition 3.8) and introduce a notation suitable with the other terms of this section: The pair {T, p } where p : V (T ) → {− 1, 0, +1 } is a function on the vertices of T (or of P ) is called a weighting of T is there exists a 2-colouring s of T such that for every vertex x of T we have p(x) = P
t∈F (T)
s(t) mod 3. Similarly as before, if {T, p } is a weighting and 20 Flip graphs of coloured triangulations
s a 2-colouring giving rise to it, we write {T, p } for the weighting associated to s. We use ( T, p ) to denote the valuation p of T , up to exchanging s with s.
A valuation of T is a pair ( T, v ), where v : D(T ) → { 0, 1} assigns 0 or 1 to every diagonal of T .
A colouring of T is a pair ( T, col ), where col is a 4-proper colouring of the vertices of T (i.e., no two vertices adjacent under T share the same colour). We will often use the letters a, b, c, d to indicate the four colours of a colouring. We only consider colourings up to permutation of colours. Let {T, s } be a signed triangulation. The signs determine a weighting of T by definition. There is a natural way to associate a valuation ( T, v )to any signed triangulation {T, s } if a diagonal is incident with two trian-gles of the same sign, its valuation is set to be 0. Otherwise, its valuation is set to be 1. By definition, this procedure associates the same valuation
v to {T, s }. So we can naturally assign a valuation ( T, v ) to ( T, s ).
Example 1.2. See Figure 16 for an example of a signed triangulation
T of heptagon, with associated weighting (on the left), valuation (in the middle) and with a colouring for T (on the right). Notice that “signed triangulations, weighting, valuation and colou-ring” are equivalent notions, up to taking the opposite signs/weights: 1. ( T, s ) ≡ (T, p ). Weightings and 2-colourings are equivalent by defi-nition. 2. ( T, s ) ≡ (T, v ). Any signed triangulation ( T, s ) gives a valuation (T, v ) as we have explained above (for any diagonal xy ∈ T ,
v(xy ) = 0 if and only if the two triangles adjacent to xy have the same sign). Conversely, any valuation ( T, v ) gives rise to two signed triangulations {T, s } and {T, s }.3. ( T, v ) ≡ (T, col ). Given a valuation ( T, v ), we construct a 4-colou-ring col of the vertices of P compatible with T , denoted by col (T, v ): Choose a vertex of degree 2 in T . Such a vertex lies in a triangle which has two boundary edges (every triangulation has at least two such triangles). We colour the three vertices of this triangles in three different colours. We proceed as follows: for any quadrilateral K. Baur, D. Bergerova, J. Voon, L. Xu 21
−
−
++ −
+0++
−
+
−
++
−
+
−
1001
a
b
d
c b
a
d1001Figure 16: A signed triangulation of heptagon with weighting, valuation, colouring. with vertices xyzt , formed by two adjacent triangles sharing the common diagonal yt , we colour x, z in the same colour if and only if the diagonal yt is valued 1 under v. Starting with the above triangle, we thus obtain a colouring of T with (up to) four colours. The colouring col (T, v ) is unique up to permutation of the colours. Reciprocally, starting from ( T, col ), we get a valuation of T by setting a diagonal of any quadrilateral to be 0 if and only if the four vertices of the quadrilateral this diagonal determines are all coloured differently. By the above, it makes sense to write ( T, ε ) where ε is in {s, p, v, col }
as these are all equivalent.
Remark 1.3 . Let T be a triangulation of a polygon. We comment on the effect of a flip on the notions weighting, valuation and colouring. See Figure 17 for an illustration.
Any flippable diagonal has valuation 0. If we flip it, the new diago-nal also has valuation 0 while the diagonals bounding the corres-ponding quadrilateral change their valuation. 22 Flip graphs of coloured triangulations
The weighting of the vertices remains unchanged under flips.
Any colouring for T is still a colouring for the new triangulation. +
−
0+ +
−
−
− −
+
−
+
flip
+
−
0+ +
−
−
+++
−
+
a
b
d
b a
d
c10 01
flip
a
b
d
b a
d
c01 01Figure 17: The effect of a signed flip on weighting, valuation, colouring. Note that a 3-colour colouring of a triangulation corresponds to the case where each diagonal has value 1, and such signed triangulation is called alternating . Alternating signed triangulations are isolated vertices in the flip exchange graph and so they are not of interest for us.
Definition 1.4. Let ( T, ε ) and ( T ′, ε ′) be two signed triangulations of the same polygon. We write ( T, ε ) ∼ (T ′, ε ′) if there exists a sequence of 2-coloured flips from ( T, ε ) to ( T ′, ε ′). This sequence may be empty (i.e. we allow T = T ′ with ε = ε′). One can check that ∼ is an equivalence relation, we denote the class of ( T, ε ) by [ T, ε ]. Now we are ready to prove Theorem 2.11 which we reformulate as follows:
Theorem 1.5. Let (T, v )̸ = ( T ′, v ′) be signed triangulations of P . Then
(T, v ) ∼ (T ′, v ′) if and only if col (T, v ) = col (T ′, v ′) and it uses 4 colours. Proof of =⇒ of Theorem 1.5. Using Remark 1.3 one can see that a colou-K. Baur, D. Bergerova, J. Voon, L. Xu 23 red flip does not change the colouring of the vertices. Iterating, we get that ( T, v ) ∼ (T ′, v ′) implies col (T, v ) = col (T ′, v ′). Since we as-sumed that the two triangulations are different, the sequence of signed flips needed to go from ( T, v ) to ( T ′, v ′) is not empty, i.e. the flip graph is not a single point and there is at least one diagonal valued with 0. Hence col (T, v ) uses four colours. To prove the converse of the theorem, we first show three lemmas. We have to study the vertices of P and their neighbours. In a triangulated polygon any vertex of P has neighbours on the boundary and potentially neighbours through diagonals of the triangulation. When dealing with the former, we refer to them as neighbours along the boundary (or on the boundary).
Lemma 1.6. Let (T, ε ) be a signed triangulation of a polygon P and x
a vertex of P . Assume that the two neighbours of x along the boundary are the only two neighbours of x with the same colour. Then x has 3 or 4 neighbours, and p(x) = 0 .Proof. Clearly, x cannot have only 2 neighbours in this case as in that case, these would belong to a common triangle with x.Suppose for contradiction that the vertex x has at least five neigh-bours. Then the two neighbours on the polygon are not the only two neighbours of the same colour in T : we can only colour three neigh-bours of x with distinct colours (different from the colour of x). And we would have at least three vertices of the same colour or another pair of neighbours with the same colour. Hence x cannot have more than 4 neighbours. In case x has three neighbours, these four vertices span a quadrilateral (with x) and the diagonal ending at x has value 1 as the other end must be of a different colour. In particular, the two triangles incident with x
have opposite sign and x has weight 0. In case x has four neighbours, the two neighbours which are linked to
x by diagonals must be of two different colours which are also different from the colour of x. In particular, both these diagonals have value 0. Therefore, x is incident with three triangles of the same sign and the weight p(x) is 0 (mod 3).
Lemma 1.7. Let (T, ε ) be a signed triangulation of P and x a vertex of P . If x has no two neighbours of the same colour, then x has 2 or 3 neighbours and the weight p(x) of x is not 0.24 Flip graphs of coloured triangulations
Proof. It is clear that x can only have 2 or 3 neighbours as if there are more, there would be at least two of them with the same colour. In case
x has only two neighours, it is incident with only one triangle and so
p(x) is 1 or 2 (mod 3). So assume that x has three neighbours. In the quadrilateral spanned by x and its three neighbours, T has a diagonal connecting x with the fourth vertex, say y. The vertices x and y have to be of different colour and so all four vertices of this quadrilateral are of different colours. Hence the diagonal xy has value 0. So the two triangles at x are of the same sign and the weight p(x) is in {1, 2} mod 3.
Lemma 1.8. Let (T, ε ) be a signed triangulation of a polygon P . Let x
be a vertex of P . Then p(x) = 0 if and only if its two neighbours on P
have the same colour. Proof. It is enough to consider the full subgraph of the triangulated polygon induced by x (it consists of x, of all vertices connected with x
and of all boundary edges and diagonals connecting them). The idea is to use induction on the degree of the vertex x.(1) If x has no two neighbours of the same colour, then x has degree 2 or 3 and p(x)̸ = 0 by Lemma 1.7. (2) If the two neighbours of x on the polygon are the only neighbours of x with the same colour, then x has degree 3 or 4 and p(x) = 0 by Lemma 1.6. With (1) and (2) we have covered all cases where x has degree 2 or 3 (in degree 3, if there are vertices of the same colour among the neighbours of x, they have to be on the boundary, for a colouring to be valid). So the result holds for vertices x of degree ≤ 3. (3) It remains to check the general situation. Let y1, y 2, . . . , y r be the neighbours of x, with y1 and yr being along the boundary and where
r ≥ 4. See left hand picture of Figure 18. Since r ≥ 4, there are vertices among the yi of the same colour. Pick
yi, yj , i < j − 1 of the same colour such that there are no two vertices of the same colour among yi+1 , . . . , y j−1. Consider the triangulated sub-polygon on the vertices x, y i, y i+1 , . . . , y j . Using the same argument as in Lemma 1.6, we find that either j = i + 2 or j = i + 3 and that the triangles incident with x and that p(x) = 0 in this subpolygon (there are either two triangles of opposite signs or three triangles of the same sign). We then identify yi with yj , getting a new polygon P ′, reducing the degree of x in it, see right hand side of Figure 18. So in the polygon K. Baur, D. Bergerova, J. Voon, L. Xu 25 xy1
y2
yr
yr−1
yj
yixy1
y2
yr
yr−1
yj≡yi
Figure 18: The neighbourhood of x in P and in P ′
P ′, the weight of x is 0 if and only if the two neighbours y1 and yr on the boundary have the same colour. Since the region between yi and yj
contributes by 0 to the weight, the claim holds.
Proof of ⇐= of Theorem 1.5. Assume that there exists a polygon P and two triangulations ( T, v ) and ( T ′v′) of P which provide a counterexample. Let P be minimal with this property. The polygon P has at least 5 vertices (one can check that the theorem is true for 4 vertices). So
col (T, v ) = col (T ′, v ′), this colouring uses all four colours, and there is no sequence of signed flips between these two signed triangulations. Among the vertices of T of degree 2 we choose a vertex x with the property that T − x (the triangulated polygon without the triangle at x) is still coloured with four colours. Such a vertex always exists as P has at least 5 vertices and among them at least two vertices of degree 2. At least one of them satisfies this condition (if one removes a degree 2 vertex y and the remaining colouring only uses 3 colours, one replaces y by another degree 2 vertex in T ). Since all four colours are present in T − x, there exists a diagonal with valuation 0. If there exists a signed triangulation T ′′ in the equivalence class [ T ′, v ′]where x has degree 2, then by minimality of the size of P , we know that for the polygon P − x we have ( T − x, v ) ∼ (T ′′ − x, v ′′ ). But then (T, v ) ∼ (T ′′ , v ′′ ) and the latter is in the equivalence class of ( T ′, v ′), so (T, v ) ∼ (T ′, v ′), a contradiction. So we can assume that x has degree ≥ 3 in every triangulation in [T ′, v ′]. We partition this equivalence class into two sets T1 and T2. We will show that these are both empty, thus proving that no counter-example to the implication ⇐ exists. We define T1 to be the set of all signed triangulations in [ T ′, v ′] having a diagonal of value 0 incident with x. The set T2 are the ones where every 26 Flip graphs of coloured triangulations
diagonal at x has value 1. These are the signed triangulations which are alternating on the subpolygon induced by x and all its neighbours in T ′.(Since the degree of x is at least 3 for any signed triangulation in [ T ′, v ′], there is always at least one diagonal at x). Claim: T1 is empty: From the elements of T1 choose a signed triangulation ( T ′′ , v ′′ ) where
x has minimal degree (this degree is ≥ 3 as we have seen). The two neighbours of x (along the boundary of the polygon) are adjacent in T
(as x has degree 2 in T ) and so have different colour. By Lemma 1.8, this means that p′′ (x)̸ = 0, where p′′ is the weighting of ( T ′′ , v ′′ ). This weighting is the same as that of ( T ′, v ′) and as that of ( T, v ) as their colourings are the same. If there is a diagonal of value 1 incident with x,say xy k (for some k), we flip a diagonal with value 0 next to this diagonal. Then the diagonal xy k has value 0. In this new triangulation, the degree of x has gone down by one and we reach a contradiction. So all diagonals at x must have value 0. We flip the first such diagonal at x (e.g. going clockwise through these diagonals). The result is a triangulation where either x has degree 2 (contradicting that the vertex x has degree > 2for all elements of [ T ′, v ′]) or it has degree 3 and no diagonal of value 0 incident with it, implying that p′′ (x) = 0 (a contradiction to p′′ (x)̸ = 0) or the resulting triangulation is an element of T1 where x has smaller degree. Figure 19 illustrates the first two of these cases. In all three cases, this leads to a contradiction. Therefore, T1 is empty. Claim: T2 is empty: Recall that the signed triangulation of the subpolygon induced by x and its neighbours in T ′ is alternating (all diagonals at x have value 1). For any ( Q, ε ) an element of T2, we write P (Q) the maximal alternating subpolygon (maximal by inclusion) which contains x and its neighbours. Let ( T ′′ , v ′′ ) be an element of T2 which minimizes the size of P (T ′′ ). As
P (T ′′ ) is maximal as alternating signed polygon, the boundary edges of
P (T ′′ ) which are diagonals in the original triangulated polygon have to have value 0. Since col (T ′′ , v ′′ ) = col (T ′, v ′) and all four colours appear, there exists at least one diagonal of value 0 (so such a boundary edge of
P (T ′′ ) has to exist). If we flip this diagonal, we obtain a new triangula-tion S. If this diagonal is incident with two edges (two diagonals or one diagonal and a boundary edge) at x, S belongs to T1 (see Figure 20 for an illustration). However, T1 is empty. Otherwise, S belongs to T2 with P (S) smaller than P (T ′′ ) (see Fi-gure 21 for an illustration), also a contradiction. So T2 is empty. K. Baur, D. Bergerova, J. Voon, L. Xu 27
x
0 011
flip
x
0111
x
1001
flip
x
1100Figure 19: Examples where x becomes a vertex of degree 2 respectively of degree 3.
x
1000
flip
x
0010Figure 20: Example with S ∈ T 1
x
1
1
0
0
flip
x
1
0
0
1
Figure 21: Example with S ∈ T 228 Flip graphs of coloured triangulations
2. Connected components of flip graphs
In this appendix, we describe the connected components of the 2-coloured flip graphs of Pn+2 for n ≤ 6. We omit the isolated vertices.
n = 2 . There is only one type of (non-trivial) connected compo-nents.
◦ ◦
n = 3 . There is only one type of connected components.
◦ ◦ ◦
n = 4 . There are four different shapes of connected components.
◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦◦ ◦ ◦ ◦◦◦◦ ◦ ◦
n = 5 . The seven shapes of the different connected components are:
◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦◦ ◦◦ ◦ ◦◦ ◦◦◦◦ ◦ ◦ ◦◦
n = 6 . The 26 shapes of the different connected components are: K. Baur, D. Bergerova, J. Voon, L. Xu 29
◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦◦ ◦ ◦◦◦ ◦◦◦◦ ◦◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦◦ ◦◦ ◦ ◦◦ ◦◦◦◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦◦◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦◦◦◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦◦◦◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦◦30 Flip graphs of coloured triangulations
◦◦◦ ◦◦ ◦◦ ◦ ◦◦◦ ◦ ◦◦◦ ◦ ◦ ◦ ◦ ◦◦◦◦◦◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦◦ ◦◦ ◦ ◦◦◦ ◦ ◦◦◦ ◦ ◦◦◦◦◦◦ ◦ ◦ ◦ ◦ ◦ ◦◦◦◦◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦◦ ◦ ◦ ◦ ◦◦ ◦◦◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦◦◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦◦K. Baur, D. Bergerova, J. Voon, L. Xu 31
◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦◦◦ ◦◦◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦◦◦◦ ◦ ◦ ◦ ◦◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦32 Flip graphs of coloured triangulations
◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦ ◦◦ ◦ ◦◦ ◦ ◦ ◦
3. Codes for component sizes
The following tables show the number of connected components for the flip graph, for the square, pentagon, hexagon, heptagon, octagon, and nonagon. They were found using a computer search, after generating all triangulations using the same recursive method as Figure 2, and then testing which pairs differ by a flip. Square: n = 2 size 1 2number 4 2Pentagon: n = 3 size 1 3number 10 10 Hexagon: n = 4 size 1 4 5 6number 28 16 12 12 K. Baur, D. Bergerova, J. Voon, L. Xu 33 Heptagon: n = 5 size 1 5 6 9 10 12 number 84 14 28 42 14 42 Octagon: n = 6
size 1 6 7 8 10 12 13 14 15 16 18 number 264 16 16 16 16 64 8 8 16 32 32
Octagon, continued
size 19 20 21 22 23 26 28 29 32 34 36 number 64 40 16 32 32 16 8 16 2 8 4
Nonagon
size 1 7 9 13 15 17 18 21 23 27 28 29 number 858 18 36 36 54 36 36 18 72 126 72 6
Nonagon, continued
size 31 32 33 34 35 36 37 38 41 42 44 number 54 36 18 72 18 108 36 72 36 36 36
Nonagon, continued
size 45 46 53 55 57 59 61 66 70 71 79 number 108 36 54 36 18 54 36 36 36 18 6
Acknowledgments
This write-up is a result of the “Count Me In” summer school, organised by David Jordan, Milena Hering, and Nick Sheridan, funded by ICMS, University of Edinburgh and Glasgow Mathematical Journal Trust. Diana Bergerova, Jenni Voon and Lejie Xu thank Karin Baur for suggesting the topic of the paper and supervising the work as well as for providing diagrams of connected components from work with Mark Parsons. They are also grateful to their tutor and mentor Stefania Lisai for helping the project with her comments and notes. Karin Baur was supported by a Royal Society Wolfson Award,
RSWF \R1 \180004 and by the EPSRC Programme Grant EP/W007509/1.
References
Appel, K., Haken, W.: Every planar map is four colorable. Part I: Discharging. Illinois J. Math. 21 (3), 429–490 (1977). Appel, K., Haken, W., Koch, J.: Every planar map is four colorable. Part II: Reducibility. Illinois J. Math. 21 (3), 491–567 (1977). ijm/1256049012
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Bose, P., Lubiw, A., Pathak, V., Verdonschot, S.: Flipping edge-labelled triangu-lations. Comput. Geom. 68 , 309–326 (2018). 2017.06.005 Aslak Bakke, B., Hugh, Th.: Coloured quiver mutation for higher cluster catego-ries. Adv. Math. 222 (3), 971–995 (2009). 017 Disarlo, V., Parlier, H.: The geometry of flip graphs and mapping class groups. Trans. Amer. Math. Soc. 372 (6), 3809–3844 (2019). tran/7356 Eliahou, Sh.: Signed diagonal flips and the four color theorem. European J. Com-bin. 20 (7), 641–647 (1999). Gravier, S., Payan, Ch.: Flips sign´ es et triangulations d’un polygone. European J. Combin. 23 (7), 817–821 (2002). Hatcher, A.: On triangulations of surfaces. Topology Appl. 40 (2), 189–194 (1991). Kryuchkov, S.I.: The Four Color Theorem and Trees. Preprint at org/abs/1309.6930 (2013) Lawson, Ch.L.: Transforming triangulations. Discrete Math. 3(4), 365–372 (1972). Lubiw, A., Mas´ arov´ a, Z., Wagner, U.: A proof of the orbit conjecture for flipping edge-labelled triangulations. 33rd International Symposium on Computational Geometry 77 . Schloss Dagstuhl. Leibniz-Zent. Inform., Wadern (2017) Mosher, L.: Mapping class groups are automatic. Annals Math. 142 (2), 303–384 (1995). Marsh, B.R., Palu, Y.: Coloured quivers for rigid objects and partial triangu-lations: the unpunctured case. Proc. Lond. Math. Soc. 108 (2), 411–440 (2014). Sleator, D.D., Tarjan, R.E., Thurston, W.P.: Rotation distance, triangulations, and hyperbolic geometry. J. Amer. Math. Soc. 1(3), 647–681 (1988). https:// doi.org/10.2307/1990951
Contact information
K. Baur Ruhr-Universit¨ at Bochum, IB, Universit¨ ats-strasse 150, 44801 Bochum, Germany
E-Mail: ka.baur@me.com
D. Bergerova University of Edinburgh, James Clerk Maxwell Building, Peter Guthrie Tait Rd, Edinburgh EH9 3FD, United Kingdom
E-Mail: D.Bergerova@sms.ed.ac.uk K. Baur, D. Bergerova, J. Voon, L. Xu 35
J. Voon Department of Computer Science, University of Warwick, CV4 7AL, United Kingdom
E-Mail: jennivoon@icloud.com
L. Xu Mathematics Institute, Zeeman Building, University of Warwick, Coventry CV4 7AL, United Kingdom
E-Mail: lejie.xu@warwick.ac.uk
Received by the editors: 10.07.2024 and in final form 01.11.2024. |
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If we have two perpendicular lines,then they always are on same plane,because math has axioms and one of them states if two lines have same point,then on that lines we can always create only one plane and if we have 3 or more lines perpendicular in same point,then at least one line is on different plane,then other two,but this planes always cross each other, because they have same point.
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More answers below
Are two perpendicular lines always, sometimes or never co-planar?
Can you give some examples of perpendicular lines and planes in real life?
What is the relation of a plane and the line perpendicular to it?
Can perpendicular lines be non-coplanar?
What is the definition of a straight line? Does it mean that all points on a straight line lie in one plane? When we draw a perpendicular line, does it also lie in one plane with the given straight line?
Jerry Rufener
Former Flight Test Engineer. · Author has 1.3K answers and 2.4M answer views
·6y
If your question is a 2 dimensional one then - yes; the lines always, without fail lie in a plane.
But I suspect your question is a 3 dimensional one. If you have two lines and the cross product of their unit vectors is 1 then they are perpendicular. If in addition they share one point in common, they lie in the same plane. (If they have more than one point in common you are in a non-euclidean space and I wish you good luck.) Otherwise they are not in the same plane.
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Robert Telarket
Studied Mathematics (college major) (Graduated 1971) · Author has 1.5K answers and 1.1M answer views
·Updated 6y
Originally Answered: Are two perpendicular lines always, sometimes or never co-planar? ·
In the Euclidean plane 2 perpendicular lines therein are always co-planar!
If one line lies in that plane and a second line is perpendicular to it but juts out into space then they're not co-planar in that first plane!
There is also in standard 3-dimensional solid geometry, R×R×R(R the reals), for any sphere therein, a definition for perpendicularity for great circles of that sphere. Since these “perpendicular lines” are contained in the set of points, their sphere, they may or not be co-planar depending on whether there is an isomorphic mapping between the Euclidean plane and the points of the
Continue Reading
In the Euclidean plane 2 perpendicular lines therein are always co-planar!
If one line lies in that plane and a second line is perpendicular to it but juts out into space then they're not co-planar in that first plane!
There is also in standard 3-dimensional solid geometry, R×R×R(R the reals), for any sphere therein, a definition for perpendicularity for great circles of that sphere. Since these “perpendicular lines” are contained in the set of points, their sphere, they may or not be co-planar depending on whether there is an isomorphic mapping between the Euclidean plane and the points of the sphere “preserving” their respective basic spanning properties. If yes then the latter can be said to be co-planar. If not there may still be preservation of co-planarity with perpendicularity.
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9 1
Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994) · Author has 10.6K answers and 23.7M answer views
·5y
Perpendicular lines are lines that meet and form 90∘90∘ angles. Any pair of lines which meet determine a plane containing both lines. It’s the plane containing the meeting point and an arbitrary different point on each line.
Since perpendicular lines meet, the answer to the question is there always exists a plane containing both lines.
It’s possible in three space (and higher dimensions) for two lines to have perpendicular direction vectors but not actually meet; those lines aren’t perpendicular.
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·
1y
Perpendicular lines always lie on the same plane if they intersect. In a three-dimensional space, two lines can be perpendicular and still not be in the same plane if they do not intersect. However, if two lines are perpendicular and intersect, they are guaranteed to be in the same plane.
In summary:
If they intersect, they are always in the same plane.
If they do not intersect, they may or may not be in the same plane (in 3D space).
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Can two perpendicular lines have different slopes?
What does a perpendicular line look like?
What are real life examples of perpendicular lines?
How do you find two points on a plane that always lie on one line?
How can a plane be perpendicular to 2 lines that are of different directions?
Michael Muirhead
Lives in Maple Ridge, BC (2022–present) · Author has 5.1K answers and 6.5M answer views
·5y
Originally Answered: Are two perpendicular lines always, sometimes or never co-planar? ·
Absolutely always by Euclid’s rules.
Being perpendicular means the two lines connect at some point, and the definition of a Euclidian plane can (with ridiculous wordiness) be accurately stated as :
“a surface defined by two lines intersecting, and the lines being revolved about their intersection such that every point on each line finds a corresponding point on the other line”.
Outside of Euclidian defs, the mathematical meaning of “plane” remains the same, but the meaning of “perpendicular” depends on the chosen geometry’s definition of “surface”.
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Robert Nichols
Author has 5K answers and 15.6M answer views
·6y
Originally Answered: Are two perpendicular lines always, sometimes or never co-planar? ·
Are two perpendicular lines always, sometimes or never co-planar?
Always.
Two intersecting lines define a plane. Perpendicular lines always intersect.
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9 1
Devon Johnson
6y
Originally Answered: Are two perpendicular lines always, sometimes or never co-planar? ·
If two perpendicular lines exist, then it is always possible to make a plane that they both fall on by the def of perpendicular lines.
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9 1
Alex Eustis
Ph.D. in Mathematics, University of California, San Diego (Graduated 2013) · Author has 4.6K answers and 23.8M answer views
·4y
Related
Are all perpendicular lines coplanar?
If “perpendicular” means “intersecting at right angles”, then yes, any two perpendicular lines in that sense are coplanar.
If “perpendicular” means “having orthogonal direction vectors”, then no, the lines could be skew. For instance, any two non-coplanar (a.k.a. skew) edges of a cube are perpendicular in this weaker sense. If you need help visualizing this, examine a corner of the room you're standing in, look at the line where one wall meets the floor, and compare it with the line where the other wall meets the ceiling.
As for which meaning is more standard, well, it's kind of a hot mess. Ther
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If “perpendicular” means “intersecting at right angles”, then yes, any two perpendicular lines in that sense are coplanar.
If “perpendicular” means “having orthogonal direction vectors”, then no, the lines could be skew. For instance, any two non-coplanar (a.k.a. skew) edges of a cube are perpendicular in this weaker sense. If you need help visualizing this, examine a corner of the room you're standing in, look at the line where one wall meets the floor, and compare it with the line where the other wall meets the ceiling.
As for which meaning is more standard, well, it's kind of a hot mess. There are approximately 4 billion overloaded meanings of the words “perpendicular”, “orthogonal”, and “normal” in mathematics, some of which have something to do with right angles, some of which have something to do with a generalized version of right angles (i.e. vectors in an inner product space whose product is zero), and some of which have nothing to do with right angles whatsoever. My advice, as always, is to be clear about the exact meaning of the words you use in the context you use them, whenever there is any chance of ambiguity. For instance in the context of 2D Euclidean Geometry there is no possible ambiguity, so feel free to use perpendicular and orthogonal interchangeably and without clarification. However in 3D geometry I would be more careful, for instance either say “intersecting perpendicular lines”, if that's what I mean, or “lines with orthogonal direction vectors” if that's what I mean.
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Michael Paglia
Former Journeyman Wireman IBEW · Author has 33.3K answers and 5.3M answer views
·Feb 10
Related
Do parallel lines exist on the same plane?
They simply must
Just simply
I cant stress it enough
😆
I think 2 lines are that aren't in same plane could be called skew
Hows about that
Thats positively skewy
Continue Reading
They simply must
Just simply
I cant stress it enough
😆
I think 2 lines are that aren't in same plane could be called skew
Hows about that
Thats positively skewy
Upvote ·
9 1
Ryan Saridar
BEng in Electronic Engineering (Course), King's College London (KCL) (Graduated 2021) · Author has 92 answers and 274.4K answer views
·Updated 6y
Related
How do you determine if a line lies on a plane?
Take the vector equation of a line:
→r(λ)=→a+λ→b r→(λ)=a→+λ b→
For a given line to lie on a plane, it must be perpendicular to the normal vector of the plane. If we want to know whether a line lies on the plane or not, we need to look at the part which judges its direction - the vector →b b→ from the equation I quoted above. If our line lies on the plane, then this vector will be parallel to the plane, meaning it will be perpendicular to a normal vector of that plane. Thus, the dot product of →b b→ with the normal vector must be zero:
→b⋅→N=0 b→⋅N→=0
Where →b b→i
Continue Reading
Take the vector equation of a line:
→r(λ)=→a+λ→b r→(λ)=a→+λ b→
For a given line to lie on a plane, it must be perpendicular to the normal vector of the plane. If we want to know whether a line lies on the plane or not, we need to look at the part which judges its direction - the vector →b b→ from the equation I quoted above. If our line lies on the plane, then this vector will be parallel to the plane, meaning it will be perpendicular to a normal vector of that plane. Thus, the dot product of →b b→ with the normal vector must be zero:
→b⋅→N=0 b→⋅N→=0
Where →b b→is the line’s directional vector, and →N N→is a normal vector to the plane.
It’s not enough that the line is parallel to the plane, though - a line can be parallel to the plane, yet still not in it. We must be able to take any point on the line, and any point on the plane, and have the vector between these points be parallel to the plane (and so perpendicular to the normal vector). We can write this in equation form as follows:
(→r(λ)−→P 0)⋅→N=0(r→(λ)−P 0→)⋅N→=0
Where →r(λ)r→(λ)is a vector to any given point on the line, →P 0 P 0→ is a given point on the plane, and →N N→ is a normal vector to the plane.
Substituting in the equation of the line:
(→a+λ→b−→P 0)⋅→N=0(a→+λ b→−P 0→)⋅N→=0
If the line is parallel to the plane, you’ll only have to test the equation for one value of λ λ - and to simplify things, you can choose λ=0 λ=0. This would make the equation into:
(→a−→P 0)⋅→N=0(a→−P 0→)⋅N→=0
To conclude, if →b⋅→N=0 b→⋅N→=0 (the line is parallel to the plane) then (→a−→P 0)⋅→N=0(a→−P 0→)⋅N→=0 must be true for the line to lie in the plane. If →b⋅→N≠0 b→⋅N→≠0, then the line does not lie in the plane.
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David Joyce
edited Euclid's Elements · Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Robby Goetschalckx
, Computer scientist for 11+ years and passionate about math since childhood. · Author has 9.9K answers and 68.4M answer views
·9y
Related
Must two lines actually intersect in order to be perpendicular to one another?
The word “perpendicular” when referring to a pair of lines usually assumes that they lie in the same plane. So the question is: Is there a reasonable definition for noncoplanar lines to be perpendicular, and has that definition ever been used?
Yes, there is such a definition. Consider this labelled cube:
The lines A B A B and C G C G don’t lie in the same plane, so they’re skew lines, yet they seem to be perpendicular in some sense. The sense can be made precise in a way that works in arbitrary dimensions using linear algebra. Let v v be the displacement vector from A A to B B, and let w w be the
Continue Reading
The word “perpendicular” when referring to a pair of lines usually assumes that they lie in the same plane. So the question is: Is there a reasonable definition for noncoplanar lines to be perpendicular, and has that definition ever been used?
Yes, there is such a definition. Consider this labelled cube:
The lines A B A B and C G C G don’t lie in the same plane, so they’re skew lines, yet they seem to be perpendicular in some sense. The sense can be made precise in a way that works in arbitrary dimensions using linear algebra. Let v v be the displacement vector from A A to B B, and let w w be the displacement vector from C C to G G. The angle θ θ between these vectors is 90∘90∘ if and only if their dot product (also called inner product) v⋅w v⋅w is 0 0. That says the directions of the lines A B A B and C G C G are at right angles.
That definition has been used for perpendicular lines on occasion, but it’s not standard. If you do use that definition, be sure to say so because otherwise readers will assume you’re using the standard definition which requires the lines to be coplanar.
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Kumar Pushpesh
MBA from Xavier Institute of Management, Bhubaneswar (Graduated 2019) · Author has 326 answers and 1.2M answer views
·7y
Related
Do two skew lines always lie in parallel planes?
Skew lines are non-intersecting and non-parallel. These are the two requirements that must be checked.
Now, if two planes are not parallel, they’ll intersect along some line. If we draw 2 non parallel lines, one on each plane, then most likely they’ll be skew. Only if both of them cut the intersecting line at the same point, will they be non-skew. Otherwise, they’ll end up being skew.
I’ve tried explaining through an example below. The planes are intersecting, but the lines are skew:
In the figure above, if we shift the lines such that they cut each other exactly on the common line of the planes,
Continue Reading
Skew lines are non-intersecting and non-parallel. These are the two requirements that must be checked.
Now, if two planes are not parallel, they’ll intersect along some line. If we draw 2 non parallel lines, one on each plane, then most likely they’ll be skew. Only if both of them cut the intersecting line at the same point, will they be non-skew. Otherwise, they’ll end up being skew.
I’ve tried explaining through an example below. The planes are intersecting, but the lines are skew:
In the figure above, if we shift the lines such that they cut each other exactly on the common line of the planes, (the case mentioned earlier), then the 2 lines will not be called skew. Here’s the diagram for that:
Just spend some time visualizing the diagrams above. I think you’ll understand what I’m trying to say here.
In case it’s not clear, let me know in the comments.
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9 5
Related questions
Are two perpendicular lines always, sometimes or never co-planar?
Can you give some examples of perpendicular lines and planes in real life?
What is the relation of a plane and the line perpendicular to it?
Can perpendicular lines be non-coplanar?
What is the definition of a straight line? Does it mean that all points on a straight line lie in one plane? When we draw a perpendicular line, does it also lie in one plane with the given straight line?
Can two perpendicular lines have different slopes?
What does a perpendicular line look like?
What are real life examples of perpendicular lines?
How do you find two points on a plane that always lie on one line?
How can a plane be perpendicular to 2 lines that are of different directions?
How many planes are perpendicular to a straight line?
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Related questions
Are two perpendicular lines always, sometimes or never co-planar?
Can you give some examples of perpendicular lines and planes in real life?
What is the relation of a plane and the line perpendicular to it?
Can perpendicular lines be non-coplanar?
What is the definition of a straight line? Does it mean that all points on a straight line lie in one plane? When we draw a perpendicular line, does it also lie in one plane with the given straight line?
Can two perpendicular lines have different slopes?
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Lesson Explainer: Second Derivatives of Parametric Equations Mathematics
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In this explainer, we will learn how to find second derivatives and higher-order derivatives of parametric equations by applying the chain rule.
Parametric equations are a way in which we can express the variables in an equation in terms of another parameter. For example, if we have an equation in terms of the variables and , then we could write parametric equations for these variables in terms of a parameter, , as follows:
Note
Parametric equations can be used in conjunction with any coordinate system, not only Cartesian. For example, if we wanted to parameterize some polar coordinates, we would express and in terms of a parameter.
We can find the derivative of with respect to in terms of the parametric equations using the following definition.
Definition: Derivative of a Parametric Equation
Let and be differentiable functions, such that and are a pair of parametric equations:
Then we can define the derivative of with respect to as when .
The first derivative of an equation can be a very useful tool for finding equations of tangents and normals to the curve or calculating gradients along the curve. The second derivative, or , can also tell us useful information about the concavity of the curve.
You may think that we can find the second derivative by finding the second derivatives of the parametric equations with respect to and dividing by , similar to how we did for the first derivative. However, this does not work as you will see below.
We find the second derivative of with respect to by differentiating the first derivative with respect to :
Finding this second derivative in terms of the parametric equations is not simple, since the equation we have for the first derivative is in terms of our parameter, . In order to perform this differentiation with respect to , we will need to use the chain rule. Recall the definition of the chain rule.
Definition: The Chain Rule
Given a function that is differentiable at and a function that is differentiable at , their composition which is defined by is differentiable at and its derivative is given by
We know how to find in terms of , since we can find the first derivative of a parametric equation. However, we need to differentiate this with respect to to find the second derivative. In order to do this, we will need to use a different form of the chain rule, which is as follows:
Applying this to our equation for the second derivative, we get
Now, we have , which is in terms of ; however, is in terms of . Here, we can use the inverse function theorem, which tells us that, for derivatives which are nonzero,
Substituting this into our equation gives us the following formula for finding the second derivative of parametric equations:
Definition: Second Derivative of a Parametric Equation
Let and be differentiable functions such that and are a pair of parametric equations: Then, we can define the second derivative of with respect to as when .
Let us now look at an example of how we can find the second derivative of a parametric equation.
Example 1: Finding the Second Derivative of Parametric Equations
Given that and , find .
Answer
The first step in finding the second derivative of these parametric equations is to find the first derivative. We can do this by using the formula
First, we can differentiate with respect to . Since is a polynomial in terms of , we use polynomial differentiation. First, multiply each term by its power of , and then reduce the power of by one. This gives us
Similarly, we also need to differentiate with respect to . This gives us
Substituting these back into our formula for the first derivative, we obtain
We are now ready to use the formula for the second derivative, which is as follows:
We have already found , so all we need to do is find the derivative of with respect to . By differentiating, we obtain
We have now found all the parts for our equation for the second derivative; we can substitute them in, which will give us our solution of
We can evaluate the second derivative of a parametric equation at a given point as we can see in the next example.
Example 2: Evaluating the Second Derivative of Parametric Equations at a Given Point
If and , find at .
Answer
Since we are trying to find the second derivative of with respect to , where and are in parametric form, we can use the following equation:
We can start by differentiating and with respect to . Using polynomial differentiation, we get and
Using this, we can find
Next, we need to differentiate with respect to . In doing this, we obtain
We are now ready to find . Substituting what we have found into the formula, we obtain
Now all that we need to do is evaluate our second derivative at the given point, . When we do this, we reach our solution of
The second derivative of an equation can also tell us about the concavity of the function at that point.
Definition: The Concavity of a Function at a Point
For a function, , which is twice differentiable and exists at some ,
if , then is concave upward at ;
if , then is concave downward at ;
if , then may be concave upward or downward, or there may be a point of inflection at .
We will need to do another test to check which is the case.
Let us now look at an example of how we can find the concavity of a parametric curve at a given point.
Example 3: Determining the Concavity of a Parametric Curve at a Given Point
Consider the parametric curve and . Determine whether this curve is concave up, down, or neither at .
Answer
We have been asked about the concavity of a curve, so we will need to evaluate the second derivative of the curve at the given point. Since this is a parametric curve, we can use the following formula to find its second derivative:
Let us start by finding and . We can do this by using trigonometric differentiation. We have that and
So we find that
Now we can differentiate with respect to . Using the differentials of reciprocal trigonometric functions, we have that
Now we are able to substitute these into the formula for the second derivative. In doing this, we obtain
Now that we have found the second derivative of the curve, we need to substitute in the value of the parameter at the point at which we are trying to find the concavity. The given point is , so we substitute this into our equation which gives us
We can see that the second derivative at is negative, so we have that
Therefore, our function must be concave downward at this point.
In our final example, let us see how we can find a function which involves the second derivative of parametric equations.
Example 4: Finding the Second Derivative of a Function Defined by Parametric Equations
If and , find .
Answer
In order to solve this question, we first need to find . We can use the formula
In order to find , we need to find and . We can expand the binomial terms in and to obtain
We differentiate these with respect to to get
Using what we have just found, we can say that
In order to find the second derivative, we need to differentiate with respect to . In order to do this, we will need to use the quotient rule for differentiation. The quotient rule tells us that if we have a function of the form , then we find its derivative using
In our case, and . Therefore, and . Substituting these values into our formula, we obtain
We can simplify the numerator to give us
Now, we are ready to find . Substituting into the formula, we have
In order to find the solution to the problem, all we need to do is multiply by . In doing this, we reach our solution of
We have now seen how to find the second derivative of parametric equations and how we can use it to find the concavity of a parametric curve. Let us recap some key points.
Key Points
We can find the second derivative of parametric equations using the formula where when .
We can use the second derivative to find the concavity of the curve at different points.
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2785 | https://www.youtube.com/watch?v=3Po3nfITsok | Order of operations, more examples | Linear equations | Algebra I | Khan Academy
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Complete a series of order of operations problems, including expressions with variables.
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27 comments
Transcript:
Let's do some order of
operations problems, and for the sake of time I'll do
every other problem. So let's start with 1b. 1b right there. They have 2 plus 7 times 11
minus 12 divided by 3. So just remember, the top
priority is always going to be your parentheses. So you have your parentheses--
Let me write it this way. Your top priority's going to
be your parentheses, after that you're going to have your
exponents, after that you have multiplying and dividing, and
after that you have addition and subtraction. So let's remember that and
tackle these order of operations problems. So priority, there's no
parentheses here, there's no exponents, so the priority's
going to go to multiplication and division. So you could view this as being
equivalent to-- So we're going to do our multiplication
before we do any addition or subtraction, and we're going
to do our division before doing any addition
or subtraction. Problem 1b is exactly equivalent
to this, the parentheses are just-- I'm
reinforcing the notion that I'm going to do my
multiplication and division before I do the addition
and the subtraction. So 7 times 11 is 77, and then
12 divided by 3 is 4. And the rest of the problem was
2 plus this thing, which is 77, minus this thing. And here, since everything is
in addition or subtraction, let's just go left to right. 2 plus 77 is 79 minus 4,
which is equal to 75. So 1b is equal to 75. Let's do 1d. This is a nice hairy problem
right there. So 1d. It says 2 times 3
plus 2 minus 1. Closing two parentheses, all of
that over 4 minus 6 plus 2 minus 3 minus 5. Let's see if we can simplify
this a little bit. As we said, parentheses
take our priority. So let's do the parentheses
first. 2 minus 1. 2 minus 1 is just 1. 3 minus 5. That is minus 2, or negative
2, I should say. 6 plus 2 is 8. Now let's keep looking at the
parentheses to see where we can simplify things. We have this parentheses
right here. So 3 plus this 1 is now going
to be equal to 4. Actually, let me rewrite it. So we're going to have 2 times
this whole expression, 3 plus 1, so it's 2 times 4. That right there is 4. All of that over 4 minus
8, that's negative 4. This right here is negative 4. And then minus this
negative 2. So minus negative 2. 2 times 4 is 8, so this whole
thing simplifies to-- A minus of a negative, that's just
the plus of the plus, the negatives cancel out. So this whole thing simplifies
to 8 divided by negative 4 is negative 2 plus 2. So it equals 0. So this big, hairy thing
simplified to 0. Now let's do 2b. Let me clear some space here. I'll leave the order of
operations stuff there. Let me clear that and
let me clear this. All right, 2b. Evaluate the following expressions involving variables. Fair enough. So they wrote 2y squared, and
they're saying that x is equal to 1, which is irrelevant
because there is no x here, and y is equal to 5. So if y is equal to 5, this
thing becomes the same thing as 2 times 5 squared. And notice, I put parentheses
there. I could have written this as,
this is the same thing as 2 times 5 squared. And if you look at the order of
operations, exponents take priority over multiplication. That's why in my head I just
automatically put those parentheses. We're going to do the
exponent first. So this is 25, so you get 2
times 25 is equal to 50. So that is 2b, this is equal
to-- use a darker color --that is equal to 50. Let's do 2d. They're giving us y squared
minus x, whole thing squared. x is equal to 2 and
y is equal to 1. Well, we just substitute. Where we see a y we put a 1. So this is going to be 1 squared
minus x squared-- Sorry, minus x, not x squared. So we just put a regular
x there. That's where we put a 2. And then all of that squared. Well 1 squared is just
1, so that is just 1. 1 minus 2 is negative 1. And then we're going to want to
square our negative 1, so that will be equal
to positive 1. So that is equal to 1. Negative times a negative
is a positive. All right, let's do 3b. Doing every other problem. I'll do it in yellow. Evaluate the following expressions involving variables. All right. Same idea. So they're giving us
4x over 9x squared. Oh, actually I said I'd
do 3b, I was doing 3a. So here we go. We have z squared over x
plus y plus x squared over x minus y. And they're telling us that x
is equal to 1, y is equal to negative 2, and z
is equal to 4. So let's just do our
substitutions first. So z squared, that's the same
thing as-- I'll do it in a different color --4 squared over
x, 1, plus y, negative 2, plus x squared, that's
1 squared, over x, which is 1, minus y. y is negative 2. So this is going to be equal to
4 squared is 16 over 1 plus negative 2, that's 1 minus 2--
it's just a negative 1 --plus 1 squared, which is 1, over
1 minus negative 2. That's the same thing
as 1 plus 2. So it's 1/3. And so this will be 16 divided
by negative 1. We could write that as that's
equal to negative 16 plus 1/3. Now if we want to actually add
these as fractions we could have a common denominator. Negative 16 is the same thing
as minus 48 over 3, or negative 48 over 3. If you take 48 divided by 3
you'll get 16, and I'm just keeping the negative sign. And then you add
that plus 1/3. We have a common denominator
now, 3. Negative 48 plus 1
is negative 47. So our answer is negative
47 over 3. Problem 3d. Same type of situation. x squared minus z squared over
xz minus 2x times z minus x. x is equal to negative
1, z is equal to 3. Let's do our substitutions. So this is x squared. That's minus 1 squared. Minus z squared, so
minus 3 squared. All of that over x times z. x times z is minus 1 times 3,
minus 2 times x, x is negative 1, times z minus x,
times 3 minus x. x is negative 1 minus x. Wherever we saw an x
we put a minus 1. So this is going to be equal
to-- Remember, you do your exponents first. Well, parentheses first, then exponents. So we have negative 1 squared,
that's just a positive 1. 3 squared, that's just
a positive 9. So our numerator becomes 1 minus
9, that's minus 8 or negative 8. And then our denominator. Negative 1 times 3
is negative 3. And then let's go to our
parentheses here. We have 3 minus negative 1,
that's the same thing as 3 plus plus 1. So that right there becomes 4. So our denominator becomes
negative 3 minus 2 times negative 1 times 4, so
that's negative 8. Minus negative 8. Minus of a negative is the
same thing as a plus. So this whole thing becomes
negative 8 over negative 3 plus 8 is 5. So it's negative 8/5,
minus 8 over 5. All right, let me clear up
some space just so we can reference this problem
properly. Let me clear all of this
out of the way. These are interesting now. Problem 4: insert parentheses in
each expression to make it a true equation. Fascinating. So 4b. You have 12 divided by 4 plus
10 minus 3 times 3 plus 7 is equal to 11. So let's see what happens if we
just do traditional order of operations, and I'll do a
little bit in my head because this is going to take some
experimentation. Oh yeah, this is 4b,
12 divided by 4-- Yep, that's the problem. So if did 12 divided by 4 first,
and we would get 3. So let me just do
this in yellow. So if we did regular order of
operations this would be a 3. This right here would
it be a 9. So you would have 3 plus 10,
which is 13, minus 9, 13 minus 9 is 4 plus 7. Actually, that seems right. Let me make sure I
did that right. 3 plus 10-- Right,
that looks right. So we really just have to do
regular order of operations. So it already looks like
a true equation. So if you do 12 divided by 4
plus 10 minus 3 times 3 plus 7, I think it turns out right. Let me confirm. Make sure I'm not making
a mistake. 12 divided by 4 is 3 plus 10
minus 3 times 3 is 9 plus 7. This is equal to 13 minus 9,
which is equal to-- So all of this is equal to 13 minus 9 is
equal to 4 plus 7 is, indeed, equal to 11. So that one wasn't too bad. You actually wouldn't have to
put any parentheses to make this a true expression. You would just have to follow
the order of operations. But throwing those parentheses
there makes it a little bit easier to read. Let's try 4d. 12 minus 8 minus 4 times
5 is equal to minus 8. So first let's just see what
happens if we did traditional order of operations. If we did traditional order of
operations we would do this 4 times 5 first, which would
give us 20 over there. And then we would have
12 minus 8 is 4. And then we would do 4 minus
20-- No, that's not right. That would give us
negative 16. So that's not going
to be right. So we can't just do traditional
order of operations. Sorry, this is a minus
8 right there. So let's see how we can
experiment with this. Let's try out a couple
of situations. What if we did 12 minus 8 minus
4 and then multiplied that times 5. Let's see what this give us. I'm just experimenting
with parentheses. So if you do 8 minus 4, that
right there would be 8 minus 4 is 4. And then 4 times 5 would be
20, and then 12 minus 20-- yeah, that works. So let me confirm that. So I'm saying I'm going to put
parentheses right there and right there and let's
work it out. You would get 8 minus 4 is 4. So this whole thing was
simplified to 12 minus 4 times 5. And you just do order of
operations, you do multiplication first.
So that is just 20. And if I wanted to make it very
clear I could actually write it like this. I could actually put
another round of parentheses right like that. But order of operations would
tell us to do it anyway. So it becomes 12 minus 20, which
is, indeed, minus 8 or negative 8. |
2786 | https://guides.library.tulsacc.edu/c.php?g=118452&p=771587 | Converting between percents, fractions and decimals - Math Help - LibGuides at Tulsa Community College
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CONVERTING BETWEEN PERCENTS, FRACTIONS AND DECIMALS
In this m odule, you canstudy how to change a fraction into a decimal number or a percent. You can also study how to change from a decimal number to a fraction or a percent, or from a percent to a fraction or a decimal number.
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2787 | https://mathsnoproblem.com/en/approach/bar-modelling | Bar Modelling
Bar modelling is an essential maths mastery strategy. A Singapore-style of maths model, bar modelling allows pupils to draw and visualize mathematical concepts to solve problems.
At a glance
A versatile maths model strategy that can be used across a wide range of concepts and topics
Gives pupils a powerful and adaptable strategy for solving increasingly difficult problems
Allows pupils to understand on a conceptual level what occurs when using complex formulas (for example, algebra)
Draws on the Concrete, Pictorial, Abstract approach
Used extensively in Singapore-style maths mastery textbooks and workbooks
Based on three pedagogical theorists — Jerome Bruner, Zoltan Dienes, and Alan J Bishop
Bar modelling and the CPA approach
The bar model method draws on the Concrete, Pictorial, Abstract (CPA) approach — an essential maths mastery concept. The process begins with pupils exploring problems via concrete objects. Pupils then progress to drawing pictorial diagrams, and then to abstract algorithms and notations (such as the +, -, x and / symbols).
The example below explains how bar modelling moves from concrete maths models to pictorial representations.
As shown, the bar method is primarily pictorial. Pupils will naturally develop from handling concrete objects, to drawing pictorial representations, to creating abstract rectangles to illustrate a problem. With time and practice, pupils will no longer need to draw individual boxes/units. Instead, they will label one long rectangle/bar with a number. At this stage, the bars will be somewhat proportional. So, in the example above, the purple bar representing 12 cookies is longer than the orange bar representing 8 cookies.
The lasting advantages of bar modelling
On one hand, the Singapore maths model method — bar modelling — provides pupils with a powerful tool for solving word problems. However, the lasting power of bar modelling is that once pupils master the approach, they can easily use bar models year after year across many maths topics. For example, bar modelling is an excellent technique (but not the only one!) for tackling ratio problems, volume problems, fractions, and more.
Importantly, bar modelling leads students down the path towards mathematical fluency and number sense. Maths models using concrete or pictorial rectangles allow pupils to understand complex formulas (for example, algebra) on an intuitive, conceptual level. Instead of simply following the steps of any given formula, students will possess a strong understanding of what is actually happening when applying or working with formulas.
The result? A stable, transferable, and solid mathematical framework for approaching abstract concepts. Combined with other [Error: Link is empty → essential maths mastery strategies and concepts ←] ( bar modelling sets students up for long-term maths success.
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2788 | https://www.grc.nasa.gov/www/BGH/realspec.html | Specific Heats - Calorically Imperfect Gas
+ Text Only Site + Non-Flash Version + Contact Glenn Thermodynamics is a branch of physics which deals with the energy and work of a system. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiments. In aerodynamics, we are most interested in thermodynamics for the role it plays in high speed flight and engine design. High temperature real gas effects play a major role in hypersonic aerodynamics. As the flight Mach number increases above Mach = 3, some of the kinetic energy of the moving object is converted into heat. Many fluid dynamics processes, like isentropic flows and shock waves, are altered by real gas effects. On the figure we present some equations that relate the specific heat capacities of air at hypersonic conditions to the specific heat capacities at lower speeds. For air at low speeds, the ratio of the specific heat capacities is a numerical constant equal to 1.4. If the specific heat capacity is a constant value, the gas is said to be calorically perfect and if the specific heat capacity changes with temperature, the gas is said to be calorically imperfect. At subsonic and low supersonic Mach numbers, air is calorically perfect. But under low hypersonic conditions, air is calorically imperfect. The specific heat capacity changes with the temperature of the flow because of excitation of the vibrational modes of the diatomic nitrogen and oxygen of the atmosphere. This computer simulation illustrates molecular vibration: Click on the slider bar, hold the mouse button down and drag to the right to increase the temperature. As the temperature increases, the vibration increases and more energy is associated with the vibration. The equations shown on the figure were developed using the kinetic theory of gases including a simple harmonic vibrator for the diatomic gases. The details of the analysis were given by Eggars in NACA Report 959. A synopsis of the report is included in NACA Report 1135. The equation for the specific heat capacity at constant volume is: cv = (cv)perf (1 + (gamp - 1) [(theta/T)^2 e^(theta/T) /(e^(theta/T) -1)^2]) where cv is the specific heat capacity at constant volume, (cv)perf is the specific heat capacity for a calorically perfect gas, gamp is the ratio of heat capacities for a perfect gas, theta is a thermal constant equal to 5500 degrees Rankine, and T is the static temperature. Similarly, for the specific heat capacity at constant pressure: cp = (cp)perf (1 + ((gamp - 1)/gamp) [(theta/T)^2 e^(theta/T) /(e^(theta/T) -1)^2]) where cp is the specific heat capacity at constant pressure, and (cp)perf is the specific heat capacity for a calorically perfect gas. The ratio of specific heats is designated by gam, which is given by: gam = 1 + (gamp - 1) / ( 1 + (gamp-1) [(theta/T)^2 e^(theta/T) /(e^(theta/T) -1)^2]) The variation of the value of the specific heat capacities with temperature causes the values of many derived functions and processes to also become functions of temperature in hypersonic flows. Activities: Guided Tours Navigation.. Beginner's Guide Home Page + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility CertificationEditor: Tom Benson NASA Official: Tom Benson Last Updated: May 07 2021 + Contact Glenn |
2789 | https://thirdspacelearning.com/us/math-resources/topic-guides/algebra/negative-exponents/ | High Impact Tutoring Built By Math Experts
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In order to access this I need to be confident with:
Exponents Simplifying expressions Order of operations
What are negative exponents?
How to use negative exponents
Common Core State Standards
How to evaluate negative exponents
Negative exponents examples
↓
Example 1: no coefficient in front of the base Example 2: negative exponents Example 3: with a coefficient in front of base Example 4: with a coefficient in front of base Example 5: with fractional exponents Example 6: with fractional exponents
Teaching tips for negative exponents
Easy mistakes to make
Related laws of exponents lessons
Practice negative exponents questions
Negative exponents FAQs
Next lessons
Still stuck?
Math resources Algebra Laws of exponents
Negative exponents
Negative exponents
Here you will learn about negative exponents, including how to simplify and evaluate with negative exponents.
Students will first learn about negative exponents as part of expressions and equations in 8 th grade, and will continue to expand their knowledge through high school.
What are negative exponents?
Negative exponents are powers (also called indices) with a negative sign (minus sign) in front of them.
Examples of negative exponents:
x^{-2}
3^{-4}
2b^{-\frac{1}{2}}
You get negative exponents by dividing two terms with the same base where the first term is raised to a power that is smaller than the power that the second term is raised to. Similarly to how a positive exponent means repeated multiplication, a negative exponent means repeated division.
For example,
x^{3} \div x^{4}=\cfrac{x \times x \times x}{x \times x \times x \times x}
When you cancel the common factors of x,
\cfrac{ \cancel{x} \times \cancel{x} \times \cancel{x}}{ \cancel{x} \times \cancel{x} \times \cancel{x} \times x}
You are left with,
\cfrac{1}{x}
Using the division law of exponents, you know that,
x^{3} \div x^{4}=x^{3-4}=x^{-1}
So,
x^{-1}=\cfrac{1}{x}
[FREE] Negative Exponents Worksheet (Grade 8)
Use this worksheet to check your 8th grade students’ understanding of negative exponents. 15 questions with answers to identify areas of strength and support!
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x
[FREE] Negative Exponents Worksheet (Grade 8)
Use this worksheet to check your 8th grade students’ understanding of negative exponents. 15 questions with answers to identify areas of strength and support!
DOWNLOAD FREE
How to use negative exponents
A negative exponent can be defined as the multiplicative inverse of the base raised to the power, which is of the opposite sign of the given power.
In other words, in order to make the negative exponent positive, put the term over 1 and flip it. It is known as finding the reciprocal of the base (term).
The negative exponent rule states that a number with a negative exponent should be put in the denominator.
For example,
x^{-2}=\cfrac{x^{-2}}{1}=\cfrac{1}{x^{2}} \cfrac{1}{x^{2}}
is the same as
x^{-2}
For example,
2^{-3}=\cfrac{2^{-3}}{1}=\cfrac{1}{2^{3}}=\cfrac{1}{8}
Negative exponents will often be used in conjunction with other exponent laws, including division, parentheses, and multiplication laws.
You can also figure out the value of negative number expressions by identifying patterns. For example, notice how when the power decreases by 1, the answer is half of the answer of the previous expression.
\begin{aligned} & 2^{3}=8 \\ & 2^{2}=4 \\ & 2^{1}=2 \\ & 2^{0}=1 \\ &2^{-1}=\cfrac{1}{2} \end{aligned}
What do you think 2^{-2} will be equal to?
What are negative exponents?
Common Core State Standards
How does this relate to 8 th grade math?
Grade 8: Expressions and Equations (8.EE.A.1)Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 \times 3-5 = 3-3 = \cfrac{1}{33} = \cfrac{1}{27}.
How to evaluate negative exponents
In order to evaluate a negative exponent, you need to:
Put the term over \bf{1}.
Flip the fraction to make the exponent positive.
Simplify, if necessary.
Negative exponents examples
Example 1: no coefficient in front of the base
Simplify and leave your answer in index form.
a^{-4}
Put the term over \bf{1}.
When the exponent is negative, put it over 1,
\cfrac{a^{-4}}{1}
2Flip the fraction to make the exponent positive.
Flip and change the power from -4 to +4.
\cfrac{1}{a^{4}}
Example 2: negative exponents
Simplify and leave your answer in index form.
7^{-9}
Put the term over \bf{1}.
When the exponent is negative, put it over 1,
7^{-9}=\cfrac{7^{-9}}{1}
Flip the fraction to make the exponent positive.
Flip and change the power from -9 to 9.
=\cfrac{1}{7^9}
Simplify, if necessary.
\begin{aligned}& \cfrac{1}{7^9}=\cfrac{1}{7 \times 7 \times 7 \times 7 \times 7 \times 7 \times 7\times 7 \times 7} \\ & =\cfrac{1}{40,353,607}\end{aligned}
Example 3: with a coefficient in front of base
Simplify and leave your answer in index form.
(10 a)^{-3}
Put the term over \bf{1}.
Notice how the exponent affects the entire bracket.
\cfrac{(10 a)^{-3}}{1}
Flip the fraction to make the exponent positive.
Flip and change the power from -3 to +3.
\cfrac{1}{(10 a)^{3}}
Simplify, if necessary.
Simplify the denominator.
=\cfrac{1}{1000 a^{3}}
Example 4: with a coefficient in front of base
Simplify and leave your answer in index form.
3b^{-2}
Put the term over \bf{1}.
Notice how the exponent only affects the variable b.
\begin{aligned}3 b^{-2} &=3 \times b^{-2} \\ &=3 \times \cfrac{b^{-2}}{1}\end{aligned}
Flip the fraction to make the exponent positive.
Flip and change the power -2 to +2.
\begin{aligned}3 \times \cfrac{b^{-2}}{1} &=3 \times \cfrac{1}{b^{2}} \\ &=\cfrac{3}{b^{2}}\end{aligned}
The exponent only applies to the variable b and not the coefficient 3.
Example 5: with fractional exponents
Evaluate
\left(\cfrac{4}{3}\right)^{-2}
Put the term over \bf{1}.
When dealing with fractions, skip to step 2.
Flip the fraction to make the exponent positive.
Flip and change the power from -2 to +2.
\left(\cfrac{4}{3}\right)^{-2}=\left(\cfrac{3}{4}\right)^{2}
Simplify, if necessary.
Simplify the numerator and denominator.
\begin{aligned}&=\cfrac{3^{2}}{4^{2}} \\ &=\cfrac{9}{16}\end{aligned}
Example 6: with fractional exponents
Evaluate
\left(\cfrac{14}{4}\right)^{-2}
Put the term over \bf{1}.
When dealing with fractions, skip to step 2.
Flip the fraction to make the exponent positive.
Flip and change the power from -2 to +2.
\left(\cfrac{14}{4}\right)^{-2}=\left(\cfrac{4}{14}\right)^2
Simplify, if necessary.
Simplify the numerator and denominator.
\begin{aligned}& =\cfrac{4^2}{14^2} \\ & =\cfrac{16}{196} \\ & =\cfrac{4}{49} \end{aligned}
Teaching tips for negative exponents
Introduce the concept using concrete examples that can illustrate it. Using simple numerical examples can make it easy to show how negative exponents relate to taking the reciprocal.
Offer a variety of practice problems to reinforce the concept. This can be done using a worksheet, but find an interactive way to allow students to discuss and problem solve together.
Allow students the opportunity to explore real-life applications of negative exponents in fields such as science, engineering, or finance.
Easy mistakes to make
Confusing integer and fractional powersRaising a term to the power of 2 means you square it.For example, 2^{2}=2 \times 2 Raising a term to the power of \cfrac{1}{2} means we find the square root of it.For example, 2^{\frac{1}{2}}=\pm \sqrt{2}
Thinking indices, powers or exponents are all differentExponents can also be called powers or indices.
Not turning a negative exponent into a positive exponent when flipping the termAfter putting the term over one, make sure to flip the exponent from a negative to a positive exponent.
Making a mistake when writing one over a fractionTypically, when finding the reciprocal, you write the term over one and flip the term. However, when you are finding the reciprocal of a fraction, the steps are a little different. This is why when dealing with fractions it is easier to skip to changing the powers and just flip the fraction.
Related laws of exponents lessons
Exponential notation
Anything to the power of 0
Dividing exponents
Multiplying exponents
Cube root
Square root
Exponent rules
Distributing exponents
Practice negative exponents questions
Simplify. Express your answer in index form.
x^{-5}
x^{5}
\cfrac{1}{x^{-5}}
\cfrac{1}{x^{5}}
\cfrac{1}{5x}
The negative exponent means finding the reciprocal, so
x^{-5}=\cfrac{1}{x^{5}}
Simplify. Express your answer in index form.
(2 b)^{-4}
16b^{-4}
\cfrac{1}{2 b^{4}}
\cfrac{1}{8 b^{4}}
\cfrac{1}{16 b^{4}}
The negative exponent means finding the reciprocal, so
\begin{aligned} & (2 b)^{-4} \\ & =\cfrac{1}{(2b)^{4}} \\ & =\cfrac{1}{2b\times2b\times2b\times2b} \\ & =\cfrac{1}{16 b^{4}} \end{aligned}
Simplify. Express your answer in index form.
3 b^{-3}
\cfrac{1}{b^{3}}
\cfrac{3}{b^{3}}
\cfrac{3}{3b}
\cfrac{1}{3b}
The negative exponent means finding the reciprocal, but this only applies to the variable, so
3 b^{-3}=\cfrac{3}{b^{3}}
Evaluate. Express your answer in index form.
\left(\cfrac{5}{8}\right)^{-1}
-\cfrac{8}{5}
\cfrac{8}{5}
-\cfrac{3}{5}
\cfrac{3}{5}
The negative exponent means finding the reciprocal, which means inverting the fraction, so
\left(\cfrac{5}{8}\right)^{-1}=\cfrac{8}{5}
Evaluate. Express your answer in index form.
9^{-2}
\cfrac{1}{81}
\cfrac{1}{9^2}
\cfrac{1}{18}
-\cfrac{1}{81}
The negative exponent means finding the reciprocal, so
\begin{aligned}& 9^{-2}=\cfrac{1}{9^2} \\ & =\cfrac{1}{9 \times 9} \\ & =\cfrac{1}{81} \end{aligned}
Evaluate. Express your answer in index form.
\left(\cfrac{4}{5}\right)^{-2}
\left(\cfrac{4}{5}\right)^{2}
\cfrac{16}{25}
\cfrac{25}{16}
-\cfrac{25}{16}
The negative exponent means finding the reciprocal, so
\begin{aligned} &\begin{aligned} & \left(\cfrac{4}{5}\right)^{-2}=\left(\frac{5}{4}\right)^2 \\ & =\left(\cfrac{5}{4}\right) \times\left(\cfrac{5}{4}\right) \end{aligned} \\ &\begin{aligned} & =\cfrac{5 \times 5}{4 \times 4} \\ & =\cfrac{5^2}{4^2} \\ & =\cfrac{25}{16} \end{aligned} \end{aligned}
Negative exponents FAQs
Can you have a negative base with a negative exponent?
Yes, the base can be a positive number, negative number, or zero. The negative exponent only affects the power to which the base is raised to.
How are negative exponents related to scientific notation?
Scientific notation involves negative exponents often. They can represent numbers in the form of a \times 10^{-n}, where a is a number between 1 and 10.
What is the difference between a negative power and a positive power?
A positive power results in the multiplication of the base by itself multiple times, while negative powers result in taking the reciprocal of the base raised to the corresponding positive exponent.
Can a base have a zero exponent?
Yes, according to the properties of exponents, the power of zero, any nonzero base raised to the zero power equals 1.
What are the rules of exponents?
Some of the fundamental rules of exponents are:∘ Product rule: \left(a^m \times a^n\right)=a^{m+n} ∘ Quotient rule: \cfrac{a^m}{a^n}=a^{m-n} ∘ Power rule: \left(a^m\right)^n=a^{m \times n}
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Math formulas
Quadratic graphs
Factoring
Radicals
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Introduction
What are negative exponents?
↓
How to use negative exponents
Common Core State Standards
How to evaluate negative exponents
Negative exponents examples
↓
Example 1: no coefficient in front of the base Example 2: negative exponents Example 3: with a coefficient in front of base Example 4: with a coefficient in front of base Example 5: with fractional exponents Example 6: with fractional exponents
Teaching tips for negative exponents
Easy mistakes to make
Related laws of exponents lessons
Practice negative exponents questions
Negative exponents FAQs
Next lessons
Still stuck?
x
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♾️AP Calculus AB/BC Unit 5 Review
5.3 Determining Intervals on Which a Function is Increasing or Decreasing
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AP Calculus AB/BC
Unit 5 – Analytical Applications of Differentiation
Topic: 5.3
♾️AP Calculus AB/BC Unit 5 Review
5.3 Determining Intervals on Which a Function is Increasing or Decreasing
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
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♾️AP Calculus AB/BC
Unit & Topic Study Guides
AP Calculus AB/BC Exams
Unit 1 – Limits and Continuity
Unit 2 – Fundamentals of Differentiation
Unit 3 – Composite, Implicit, and Inverse Functions
Unit 4 – Contextual Applications of Differentiation
Unit 5 – Analytical Applications of Differentiation
Unit 5 Overview: Analytical Applications of Differentiation
5.1 Using the Mean Value Theorem
5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points
5.3 Determining Intervals on Which a Function is Increasing or Decreasing
5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
5.6 Determining Concavity
5.7 Using the Second Derivative Test to Determine Extrema
5.8 Sketching Graphs of Functions and Their Derivatives
5.9 Connecting a Function, Its First Derivative, and its Second Derivative
5.10 Introduction to Optimization Problems
5.11 Solving Optimization Problems
5.12 Exploring Behaviors of Implicit Relations
Unit 6 – Integration and Accumulation of Change
Unit 7 – Differential Equations
Unit 8 – Applications of Integration
Unit 9 – Parametric Equations, Polar Coordinates, and Vector–Valued Functions (BC Only)
Unit 10 – Infinite Sequences and Series (BC Only)
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5.3 Determining Intervals on Which a Function is Increasing or Decreasing
What can the derivative of a function show us about the function itself? Can it tell us when the function increases or decreases? Yes, it can! In this lesson, we’ll delve into how we can use derivatives to determine when a function increases or decreases. 📈
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🕑 When Does a Function Increase or Decrease?
In order to determine the intervals on which a function is increasing or decreasing, we first need to understand the concept of the derivative. The derivative of a function is the rate of change of the function at a given point.
Thus, we know the following:
➕=📈 If the derivative is positive at a certain point (which means the rate of change is positive at that point), the function is increasing at that point.
➖=📉 If the derivative is negative at a certain point, the function is decreasing at that point.
Take a look at this graph to see these trends in action. The gray line represents the function, f f f, and the black line represents its derivative, f′f'f′.
Graph of a function and its derivative to demonstrate the trend. Image Courtesy of Informal Calculus
Now, you may be thinking that finding the interval where a function is increasing or decreasing is as simple as finding the interval where the function’s derivative is positive or negative, respectively. Well, it is! 🪄
Where can a function change from increasing to decreasing and vice versa? It can only change its direction from increasing to decreasing and vice versa at its critical points, points where the function’s derivative equals 0 0 0 or is undefined, and the points where the function itself is undefined.
So, for each of the intervals defined by the points where the function can change behavior, we can determine whether the function is increasing or decreasing on the interval by just plugging a point on that interval into the function’s derivative and seeing if the result is positive or negative. If it’s positive, then the function is increasing on that interval; if it’s negative, then the function is decreasing on that interval. Let’s give it a try!
✏️ Function Behavior: Walkthrough
➡️ Let $h$ be a function defined for all real numbers except $0.$ Also let $h'$ be defined as $h'(x)=\frac{(x+7)}{x^2}$. On which intervals is $h$ increasing?
From what we learned above, we can analyze the intervals where h h h is increasing by looking for the intervals where its derivative h h h is positive.
A function can only change its direction from increasing to decreasing and vice versa at its critical points and the points where the function itself is undefined. Based on the problem statement, we determine that in this case, the only points where h h h can change direction are x=−7 x=-7 x=−7 and x=0 x=0 x=0. We determine this by doing the following and solving:
h′(x)=(x+7)x 2 h'(x)=\frac{(x+7)}{x^2}h′(x)=x 2(x+7)
0=(x+7)x 2 0=\frac{(x+7)}{x^2}0=x 2(x+7)
h′=0 h'=0 h′=0 at x=−7 x=-7 x=−7 so it is a critical point and h h h is undefined at 0 0 0 as stated in the problem.
These two points divide the number line into three intervals:
(−∞,−7),(−7,0),(0,∞)(-\infty,-7),(-7,0),(0,\infty)(−∞,−7),(−7,0),(0,∞)
Let’s evaluate h′h'h′ at each interval to see if it’s positive or negative on that interval and therefore see if h h h is increasing or decreasing on the interval. To evaluate h′h'h′ on an interval, we can choose any x-value within the interval to substitute x x x with and calculate.
In the table below, you’ll notice that we chose x=−8 x=-8 x=−8 for the first interval, x=−1 x=-1 x=−1 for the second interval, and x=1 x=1 x=1 for the third! Plug in, solve, and determine the behavior of the function.
| Interval | x x x | h′(x)h'(x)h′(x) | Verdict |
--- --- |
| (−∞,−7)(-\infty,-7)(−∞,−7) | x=−8 x=-8 x=−8 | h′(−8)=−1 64<0 h'(-8)=-\frac{1}{64} <0 h′(−8)=−64 1<0 | h h h is decreasing |
| (−7,0)(-7,0)(−7,0) | x=−1 x=-1 x=−1 | h′(−1)=6>0 h'(-1)=6 > 0 h′(−1)=6>0 | h h h is increasing |
| (0,∞)(0,\infty)(0,∞) | x=1 x=1 x=1 | h′(1)=8>0 h'(1)=8>0 h′(1)=8>0 | h h h is increasing |
In conclusion, h h h is increasing on the intervals (−7,0)(-7,0)(−7,0) and (0,∞)(0,\infty)(0,∞).
🪜 Steps to Determining Function Behavior
Here are the steps we used if you prefer seeing them in list form:
🎯 Determine the critical points of the function, where f′(x)f'(x)f′(x) equals zero or is undefined.
🔢 Divide the function's domain into intervals based on the critical points determined. This is where you break the number line into pieces to see its behavior!
🔨 For each interval, choose a test point within that interval.
✏️ Evaluate the function's derivative at the chosen test point.
🏁 Interpret the result.
If the result is positive, the function is increasing on that interval.
If the result is negative, the function is decreasing on that interval.
Now you can put these steps into practice!
📝 Function Behavior: Practice Problems
Time to solve some problems! 🔍
❓Function Behavior: Problems
Function Behavior: Question 1
Let f(x)=x 3−27 x f(x)=x^3-27x f(x)=x 3−27 x.
On which interval(s) is f f f decreasing?
Function Behavior: Question 2
Let f(x)=x 4−2 x 2 f(x)=x^{4}-2x^{2}f(x)=x 4−2 x 2.
On which interval(s) is f f f increasing?
✅ Function Behavior: Answers and Solutions
Function Behavior: Question 1
The answer to this problem is (−3,3)(-3,3)(−3,3). Let’s get into why. ⬇️
We can analyze the intervals where f f f is decreasing by looking for the intervals where its derivative f f f is negative.
A function can only change its direction from increasing to decreasing and vice versa at its critical points and the points where the function itself is undefined. So first, take the derivative and find the critical points.
Since the derivative of f f f is f′(x)=3 x 2−27 f'(x)=3x^2-27 f′(x)=3 x 2−27, the only points where f f f can change direction are x=−3 x=-3 x=−3 and x=3 x=3 x=3 as f′(x)=0 f'(x)=0 f′(x)=0 at only these points.
These two points divide the number line into three intervals: (−∞,−3),(−3,3),(3,∞)(-\infty,-3),(-3,3),(3,\infty)(−∞,−3),(−3,3),(3,∞).
Let’s evaluate f′f'f′ at each interval to see if it’s positive or negative on that interval and therefore if f f f is increasing or decreasing on the interval.
| Interval | x x x | f′(x)f'(x)f′(x) | Verdict |
--- --- |
| (−∞,−3)(-\infty,-3)(−∞,−3) | x=−4 x=-4 x=−4 | f(−4)=21>0 f(-4)=21>0 f(−4)=21>0 | f f f is increasing |
| (−3,3)(-3,3)(−3,3) | x=0 x=0 x=0 | f(0)=−27<0 f(0)=-27<0 f(0)=−27<0 | f f f is decreasing |
| (3,∞)(3,\infty)(3,∞) | x=4 x=4 x=4 | f(4)=21>0 f(4)=21>0 f(4)=21>0 | f f f is increasing |
In conclusion, f f f is decreasing on the interval (−3,3)(-3,3)(−3,3).
Function Behavior: Question 2
The answer to this problem is (−1,0)(-1,0)(−1,0) and (1,∞)(1,\infty)(1,∞). Here’s why:
We can analyze the intervals where f f f is increasing by looking for the intervals where its derivative f f f is positive.
A function can only change its direction from increasing to decreasing and vice versa at its critical points and the points where the function itself is undefined.
Since the derivative of f f f is f′(x)=4 x 3−4 x f'(x)=4x^3-4x f′(x)=4 x 3−4 x, which we can rewrite as 4 x(x+1)(x−1)4x(x+1)(x-1)4 x(x+1)(x−1), the only points where f f f can change direction are x=−1 x=-1 x=−1, x=0 x=0 x=0, and x=1 x=1 x=1 as f′(x)=0 f'(x)=0 f′(x)=0 at only these points.
These three points divide the number line into four intervals: (−∞,−1),(−1,0),(0,1),(1,∞)(-\infty,-1),(-1,0),(0,1),(1,\infty)(−∞,−1),(−1,0),(0,1),(1,∞).
Let’s evaluate f′f'f′ at each interval to see if it’s positive or negative on that interval and therefore if f f f is increasing or decreasing on the interval.
| Interval | x x x | f′(x)f'(x)f′(x) | Verdict |
--- --- |
| (−∞,−1)(-\infty,-1)(−∞,−1) | x=−2 x=-2 x=−2 | f(−2)=−24<0 f(-2)=-24<0 f(−2)=−24<0 | f f f is decreasing |
| (−1,0)(-1,0)(−1,0) | x=−0.5 x=-0.5 x=−0.5 | f(−0.5)=1.5>0 f(-0.5)=1.5>0 f(−0.5)=1.5>0 | f f f is increasing |
| (0,1)(0,1)(0,1) | x=0.5 x=0.5 x=0.5 | f(0.5)=−1.5<0 f(0.5)=-1.5<0 f(0.5)=−1.5<0 | f f f is decreasing |
| (1,∞)(1,\infty)(1,∞) | x=2 x=2 x=2 | f(2)=24>0 f(2)=24>0 f(2)=24>0 | f f f is increasing |
In conclusion, f f f is decreasing on the intervals (−1,0)(-1,0)(−1,0) and (1,∞)(1,\infty)(1,∞).
⭐ Closing
Great work! You now have the steps and practice to determine where a function changes from increasing to decreasing.
Frequently Asked Questions
How do I find where a function is increasing or decreasing using derivatives?
You use f′ to tell where f goes up or down. Steps: 1. Find f′(x). Solve f′(x) = 0 and note where f′ is undefined—those are critical points (including endpoints if on a closed interval). (Keywords: first derivative, critical point, stationary point.) 2. Make a sign chart: break the domain into intervals at those critical points. Pick one test point in each interval and evaluate f′ there. If f′(test) > 0, f is increasing on that interval; if f′(test) < 0, f is decreasing. (This is the First Derivative Test idea from the CED: FUN-4.A.1.) 3. Use endpoints and nondifferentiable points as needed (they can affect local/global behavior). If you need to justify local maxima/minima, check sign change of f′ around the critical point. On the AP exam you’ll be asked to justify intervals using f′ (show derivative work, critical points, sign chart or test points). For a handy walk-through and examples, see the Topic 5.3 study guide (Fiveable) here: ( For extra practice, use the AP Calc practice bank (
What's the formula for determining if a function is increasing or decreasing?
The rule is simple: use the first derivative. If f′(x) > 0 on an interval, f is increasing there; if f′(x) < 0, f is decreasing there. Procedure: 1) Compute f′(x). 2) Find critical points by solving f′(x) = 0 or where f′ is undefined. 3) Make a sign chart (test points between critical values and check endpoints if the domain’s closed). 4) Record intervals where f′ is positive (increasing) or negative (decreasing). Use the first-derivative test at each critical point to classify local extrema and remember nondifferentiable points (cusps) can change behavior. This is exactly what the AP CED expects for FUN-4.A (justify using derivative sign and tests). For a quick study guide, see Topic 5.3 ( For more review and practice problems, check the Unit 5 overview ( and thousands of practice questions (
When do I use the first derivative test vs just looking at the graph?
Use the graph when it’s given and clear: if you have f(x) plotted on the exam, you can read where it rises/falls directly (look for left-to-right up = increasing, down = decreasing), and use visible endpoints and cusps. Use the first-derivative test when you only have a formula, need justification, or when the graph is ambiguous (flat spots, cusps, vertical tangents, nondifferentiable points). Procedure: find critical points (where f′ = 0 or undefined), make a sign chart for f′ on intervals between them (or test points), then apply the first derivative test to classify local max/min (f′ changes +→− for a local max, −→+ for a local min). Remember to check endpoints for absolute extrema. This matches CED keywords: critical point, derivative sign, test point, first derivative test, differentiability. For a quick review, see the Topic 5.3 study guide ( and try practice problems ( to get fluent at sign charts and AP-style justifications.
What's the difference between finding critical points and finding increasing/decreasing intervals?
Good question—they’re related but not the same job. - Critical points are the x-values where f′(x) = 0 or f′ is undefined (and where f is in the domain). They’re just candidates—stationary points (f′ = 0) and nondifferentiable points. Finding them is Step 1. (See Topic 5.2 for more on critical points.) - Determining increasing/decreasing intervals uses those critical points plus sign analysis of f′. You make a sign chart (or test points) on each interval between critical points (and include endpoints on closed intervals) to see where f′>0 (f increasing) or f′<0 (f decreasing). That gives the actual intervals of increase/decrease and lets you apply the First Derivative Test for local extrema (Topic 5.4 / FUN-4.A). AP tip: always justify using derivative sign and mention nondifferentiable endpoints if they affect behavior—that’s what the CED expects (FUN-4.A.1). For a quick review, check the Topic 5.3 study guide (Fiveable) here: ( For extra practice, try problems at (
Can someone explain step by step how to find intervals where f(x) is increasing?
Step-by-step: how to find where f(x) is increasing 1. Compute f ′(x). The first derivative gives the function’s rate of change (FUN-4.A). 2. Find the domain of f and f ′. Note points where f′(x)=0 or f′ is undefined—those are critical points. 3. Put the critical points (and any endpoints if you have a closed interval) in order and make intervals between them. 4. Pick one test x-value in each interval and evaluate the sign of f′(test). If f′>0 on an interval, f is increasing there; if f′<0, f is decreasing. (This is the sign-chart / first derivative test idea.) 5. State the final answer as a union of open intervals where f′>0; include endpoints only if the interval is closed and the derivative/sign supports it. 6. Mention nondifferentiable cusps or jumps as special cases (check the function itself). This is exactly what AP expects: use the derivative’s sign to justify increasing/decreasing intervals (see Topic 5.3 study guide for examples) ( For more review across Unit 5, see ( and try practice problems (
I don't understand why f'(x) > 0 means the function is increasing - can someone explain this?
Think of f′(x) as the instantaneous rate of change (slope) of f. If f′(x) > 0 on an interval, every tangent line there has positive slope, so small moves to the right always increase f. More formally, if f is differentiable on (a,b) and f′(x) > 0 for all x in (a,b), the Mean Value Theorem says for any x1 0 and (x2 − x1) > 0, f(x2) − f(x1) > 0, so f(x2) > f(x1). That proves f is increasing on (a,b). In practice on the AP you’ll use a sign chart or test points around critical values (where f′=0 or undefined) and apply the First Derivative Test to identify increasing/decreasing intervals and classify extrema (FUN-4.A in the CED). For a quick refresher and examples, see the Topic 5.3 study guide ( and try practice problems (
How do I solve for where f'(x) > 0 and f'(x) < 0?
Solve f′(x) > 0 or f′(x) < 0 in 4 clear steps: 1. Compute f′(x). Use derivative rules (power, product, chain, etc.). 2. Find critical points: solve f′(x) = 0 and note where f′ is undefined. These split the domain. (Include endpoints of the domain too.) 3. Make a sign chart: pick one test x in each interval between critical points and evaluate f′(test). If f′(test) > 0 that whole interval is increasing; if f′(test) < 0 it’s decreasing. Don’t forget intervals where f′ is undefined—they can change sign. 4. Use the First Derivative Test to classify local extrema: sign change +→− gives a local max, −→+ gives a local min. State interval endpoints separately if the interval is closed. On the AP exam, justify your sign choices (show test points) and include nondifferentiable points as candidates (CED FUN-4.A). For a refresher and examples, see the Topic 5.3 study guide ( and more practice at (
What do I do after I find the critical points to determine increasing/decreasing intervals?
After you find the critical points (where f′ = 0 or f′ is undefined), make a sign chart for f′ to determine increasing/decreasing intervals: 1. List the critical points and any domain/endpoints in order on the number line. 2. Pick one test point in each open interval between them. 3. Evaluate f′ at each test point: - If f′(test) > 0 on an interval, f is increasing there. - If f′(test) < 0, f is decreasing there. 4. Use the First Derivative Test to classify each critical point: - f′ changes + → −: local max. - f′ changes − → +: local min. - No sign change: no local extremum (could be inflection or flat spot). 5. On closed intervals, also check endpoints (candidates for absolute extrema). 6. If f′ is undefined at a point, consider nondifferentiability (cusp, corner) before concluding. This is exactly the FUN-4.A/CED approach—you can review an example and practice problems in the Topic 5.3 study guide ( and try more problems at (
Why do I need to test points in each interval instead of just finding where f'(x) = 0?
Good question—setting f′(x)=0 only finds critical (stationary) points; it doesn’t tell you whether f actually goes up or down on either side. A sign chart (pick one test point in each interval between critical points and non-differentiable points) tells you whether f′>0 (increasing) or f′<0 (decreasing) on that whole interval. That matters because at some critical points f′ changes sign (local max/min) and at others it doesn’t (flat inflection). Also include points where f′ is undefined and endpoints as interval boundaries. On the AP exam you’re expected to justify interval behavior using derivative signs (FUN-4.A, first derivative test), so show the sign work or a quick test-value calculation. For a focused review, see the Topic 5.3 study guide ( and try practice problems (
How do I write interval notation for where a function is increasing or decreasing?
Find where f′(x) > 0 (increasing) and where f′(x) < 0 (decreasing). Steps and how to write them in interval notation: 1. Find domain, then solve f′(x) = 0 and where f′ is undefined → list critical points. 2. Make a sign chart: split the domain at the critical points and pick a test point in each subinterval to evaluate sign of f′. 3. Record intervals where f′>0 (increasing) and where f′<0 (decreasing). Use parentheses for open intervals between critical points: e.g., (−∞, a), (a, b), (b, ∞). If an endpoint is in the domain and the derivative sign on the adjacent open interval shows increasing/decreasing up to that endpoint, include the endpoint with a bracket: [a, b) or (a, b] as appropriate. (You only use a bracket if the endpoint is part of the function’s domain and you want to include it.) 4. State your result: “f is increasing on (−∞, −1) ∪ (2, 5] and decreasing on (−1, 2).” Always justify by citing sign of f′ on each piece (first-derivative test / sign chart). This aligns with FUN-4.A in the CED. For a worked example and practice, see the Topic 5.3 study guide ( and try problems from the Unit 5 page ( or the practice bank (
I'm confused about sign charts - how do I make one for the first derivative?
Make a sign chart for f′ in four simple steps: 1. Find critical numbers: solve f′(x)=0 and note where f′ is undefined. These partition the domain into open intervals (include endpoints if you have a closed interval). (CED keywords: critical point, differentiability, cusp.) 2. List the intervals on a number line and pick one test point in each interval (not the endpoints). (CED keyword: test point.) 3. Evaluate f′ at each test point (or determine sign from a graph/table). Put + if f′>0 (f increasing) or − if f′<0 (f decreasing) on that whole interval. If f′ changes sign at a critical number, use the First Derivative Test to classify local max/min; if it doesn’t change sign, no local extremum. (CED: first derivative test, local maximum, local minimum.) 4. Don’t forget endpoints and places where f isn’t differentiable—they can be candidates for absolute extrema. Also watch for cusps/vertical tangents where f′ is undefined. On the AP exam you may get algebraic, graphical, or tabular f′ information, so be ready to use arithmetic or reasoning. For extra practice and worked examples, see the Topic 5.3 study guide ( and more practice problems (
What happens at critical points - are they included in increasing or decreasing intervals?
Short answer: critical points themselves aren’t counted as points “where a function is increasing” or “decreasing”—you describe increasing/decreasing on intervals, and those intervals are usually open (e.g., (a, c) or (c, b)). A critical point is where f′(x)=0 or f′ is undefined; use it as a boundary when making a sign chart for f′. How to report on the exam (CED/FUN-4.A language): - Find critical points (f′=0 or f′ undefined). - Test sign of f′ on intervals between those critical points (pick test points). If f′>0 on (p,q) then f is increasing on (p,q); if f′<0 then f is decreasing on (p,q). - Don’t include the critical point in the open interval answer. If the problem asks for behavior on a closed domain or wants endpoints, you may state inclusion of endpoints separately (e.g., increasing on [a,c) if a is the domain endpoint). - Watch for non-differentiable features (cusps, corners)—they’re critical points too and must be checked. For practice and a quick refresher, see the Topic 5.3 study guide ( and try problems at (
How do I find increasing and decreasing intervals if the derivative has fractions or radicals?
Short answer: treat fractional or radical derivatives the same way—find where f′ = 0 or undefined, break the domain into intervals, pick test points, and use a sign chart. How-to steps (quick): - First get the algebra under control: simplify f′ as much as possible (factor numerator, combine to a single fraction, and simplify radicals). - Find critical numbers: solve numerator = 0 and note x where denominator = 0 or a radical’s inside becomes invalid (these make f′ undefined). Also keep the original function’s domain in mind. - Make a sign chart: list all critical points and domain endpoints in order. For each open interval, pick an easy test point and evaluate the sign of f′ (you can often just check signs of factors rather than full values). If f′>0 the function’s increasing there; if f′<0 it’s decreasing. - Watch endpoints, points where f′ is undefined (possible cusps/vertical tangents), and use the first-derivative test to classify local extrema (FUN-4.A in the CED). This is exactly what AP expects for Topic 5.3—show the sign reasoning/justification (first-derivative test, critical points). For extra worked examples and practice, see the Topic 5.3 study guide ( and more Unit 5 resources ( For lots of practice problems, try (
Can I use my calculator to find where f'(x) is positive or negative?
Yes—your graphing calculator is a helpful tool, but it’s not a substitute for the math you must justify on the exam. How a calculator helps - Graph f and/or f′ and use the TRACE/ROOT features to find zeros of f′ (critical points) and sample values to see where f′>0 or f′<0. - Use numerical derivative routines or difference quotients to approximate f′(x) at specific x (good for nonsymbolic functions). - Make a sign chart quickly by evaluating f′ (or slopes of f) at test points between critical values. Important AP/CEC notes (CED-aligned) - You still need to justify conclusions about increasing/decreasing using derivative sign, test points, and first-derivative test (FUN-4.A; keywords: derivative sign, critical point, test point). - On the AP exam, calculators are allowed only on certain parts (multiple-choice Part B and some free-response Part A). You must be ready to do analyses without a calculator for non-calculator sections. - Watch for nondifferentiable points (cusps, corners) and endpoints as candidates when deciding intervals. Practice: review Topic 5.3 study guide ( and try problems ( so you can both use the calculator effectively and justify answers without it.
I keep getting the wrong intervals - what are the most common mistakes when finding where functions increase or decrease?
Common mistakes and how to avoid them: - Forgetting the domain/endpoints. Always restrict critical points and sign charts to where f is defined (include endpoints on closed intervals). AP problems expect you to justify behavior on the given domain (CED FUN-4.A). - Missing critical points where f′ is undefined. Points where f′ = 0 or f′ DNE can change sign—include both. - Bad algebra when solving f′(x)=0. Re-check factoring, canceling, and extraneous solutions; plug suspects back into f′. - Skipping the sign chart or using too few test points. Make a sign chart with intervals determined by every critical point and choose one test value per interval to check f′ sign. - Using f instead of f′ (or vice versa) to decide increase/decrease. Remember: f′>0 ⇒ f increasing; f′<0 ⇒ f decreasing (FUN-4.A.1). - Ignoring nondifferentiable cusps/vertical tangents. If f′ DNE, test either side for sign change. - Not reporting intervals cleanly (open vs closed) or not justifying conclusions. On the exam, state your intervals and give brief reasons (first derivative test or sign chart). If you want a step-by-step refresher, check the Topic 5.3 study guide ( and drill problems at
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2791 | https://www.swbh.nhs.uk/wp-content/uploads/2012/07/Anti-thyroid-drugs-ML5742.pdf | Sandwell and West Birmingham Hospitals NHS Trust Anti-thyroid drugs Information and advice for patients Endocrinology Page 1 What are anti-thyroid drugs?
Anti-thyroid drugs, such as carbimazole and propylthiouracil, are medicines used to treat hyperthyroidism. Hyperthyroidism, also called an overactive thyroid, is a condition where the thyroid gland releases too many hormones. The thyroid gland is in the neck and releases hormones which help control growth and the way your body uses energy. When the thyroid gland releases too many hormones it can cause some of the body’s functions to speed up, causing symptoms such as feeling very nervous or anxious, being hyperactive, weight loss and a lump in your neck. What is the benefit of taking anti-thyroid drugs?
The benefit of taking anti-thyroid drugs is that they reduce the amount of hormones your thyroid gland releases, which can improve the symptoms of hyperthyroidism. What are the risks of taking anti-thyroid drugs?
A risk of taking carbimazole is that it can lower the number of white blood cells in your body. These are the cells that fight infection. If you develop any of the following symptoms while taking carbimazole, stop taking it, see a doctor immediately and ask them to do a blood test to check your white cell count: • Sore throat • Temperature of 37.5°C or above • Unexplained bruising • Feeling more tired than usual • Mouth ulcers • Feeling unwell Other possible side effects of carbimazole include stomach irritation and rash. A full list of side effects can be found in the manufacturer’s leaflet that comes with the medicine. A risk of taking propylthiouracil is that it can cause liver problems, although this is rare. If you develop any of the following symptoms while taking propylthiouracil, stop taking it and see a doctor immediately: • Weight loss • Poor appetite • Feeling or being sick • Feeling more tired than usual Sandwell and West Birmingham Hospitals NHS Trust Anti-thyroid drugs Information and advice for patients Endocrinology • Abdominal (tummy) pain • Dark-coloured urine • Itching • A yellow colour to your skin or eyes A full list of possible side effects can be found in the manufacturer’s leaflet that comes with the medicine. What are the risks of not taking anti-thyroid drugs?
If you choose not to take anti-thyroid drugs for your hyperthyroidism then it will become worse and your symptoms will become more severe. You could then experience complications such as an irregular heart rate, heart failure and eventually you could go into a coma which can be life-threatening. Are there any alternatives to anti-thyroid drugs?
Another treatment for hyperthyroidism which may be appropriate for some people is with radioactive iodine which is taken once or twice only, as a tablet or drink. Another treatment for hyperthyroidism is ‘block and replace therapy’. This involves taking a high dose of the anti-thyroid drug carbimazole and the medication thyroxine which is a thyroid hormone. The high dose of carbimazole blocks your thyroid gland from releasing any hormones, and thyroxine then replaces the thyroid hormone that your thyroid gland should be releasing. Surgery to remove part of the thyroid gland is another treatment for hyperthyroidism which may be suitable for some people. Your doctor will discuss these treatments with you if they are suitable for you. Before taking anti-thyroid drugs Before taking carbimazole or propylthiouracil, please let your doctor know if you have any medical conditions or if you are pregnant or breastfeeding. You should also inform them of any other medications you are taking. Please make sure you read the manufacturer’s information leaflet that comes with your medication before taking it. Page 2 Sandwell and West Birmingham Hospitals NHS Trust Anti-thyroid drugs Information and advice for patients Endocrinology How to take the medication Your doctor will tell you which anti-thryoid drug you have been prescribed, how much you need to take of it and how often you need to take it. Anti-thyroid drugs come as a tablet. If you forget to take a dose of your medication, miss out the forgotten dose and then continue taking it as normal when your next dose is due. If you take too much of your medication contact your GP or NHS Direct on 0845 46 47 for advice. When you first start taking anti-thyroid drugs you will be prescribed a high dose. You will then be given an appointment to see the doctor in 6 – 8 weeks after starting it. At the appointment your doctor will ask you about your symptoms and do a blood test to check your thyroid function (how well your thyroid gland is working). If your symptoms and thyroid function are improving then the dose may be reduced, so that you are taking the lowest dose necessary to keep your thyroid function normal. You may need to take anti-throid drugs for 6 – 24 months to reduce your chance of having hyperthyroidism in the future, and will have regular blood tests and check-ups during this time. After this time, if you do not develop any further symptoms and your blood tests remain normal your hospital doctor will discharge you. If you experience any of the symptoms of hyperthyroidism in the future, please see your GP. Precautions Please tell any doctors or nurses treating you that you are taking anti-thyroid drugs. Storing your medication Store your medication: • in the original packaging • in a dry place, at room temperature • out of the sight and reach of children. Further information If you have any questions or concerns about your hyperthyroidism or the treatment please speak to your GP, consultant or specialist nurse.
You can contact your consultant or specialist nurse by calling the hospital switchboard on 0121 554 3801, holding to speak to an advisor, and asking for them.
Page 3 ML5742 Issue Date: May 2018 Review Date: May 2021 Sandwell and West Birmingham Hospitals NHS Trust If you would like to suggest any amendments or improvements to this leaflet please contact the communications department on 0121 507 5303 or email: swb-tr.swbh-gm-patient-information@nhs.net A Teaching Trust of The University of Birmingham Incorporating City, Sandwell and Rowley Regis Hospitals © Sandwell and West Birmingham Hospitals NHS Trust Anti-thyroid drugs Information and advice for patients Endocrinology Page 4 For more information about hyperthyroidism: NHS Choices www.nhs.uk/conditions/Thyroid-over-active Patient UK www.patient.co.uk/health/hyperthyroidism-overactive-thyroid For more information about our hospitals and services please see our website www.swbh.nhs.uk, follow us on Twitter @SWBHnhs and like us on Facebook www.facebook.com/SWBHnhs. Sources used for the information in this leaflet • British National Formulary, section 6.2.2 ‘Anti-thyroid drugs’, accessed online May 2013 • Patient UK Professional Reference, ‘Hyperthyroidism’, December 2012 • Ballinger, ‘Essentials of Kumar and Clark’s Clinical Medicine’, chapter 14, 5th edition, 2012 • Oxford University Press, ‘Oxford handbook of clinical medicine’, chapter 5, 8th edition, 2010 • British Thyroid Association, ‘Hyperthyroidism – clinical features and treatment’, 2007 • Ward and Linden, ‘Physiology at a glance’, chapter 6, 2nd edition, 2005 |
2792 | https://www.patronus.ai/blog/sequential-probability-ratio-test-for-ai-products | Sequential Probability Ratio Test for AI Products
April 25, 2025
Introduction
It’s critical for AI teams to measure the success of new AI product launches, but it can be especially challenging when the product has very few users. Enterprises are actively rolling out new agentic systems to production, but no one we know is considering running A/B tests. This is because traditional A/B testing requires a certain sample size and a fixed test duration, which may be completely impossible with low user volume in the early days. Of course, you also can’t wait for weeks or months till you have enough data. So how do you know how to iterate on your feature set with limited data and under time constraints?
This is where the Sequential Probability Ratio Test (SPRT) can be extremely useful. SPRT is a sequential hypothesis testing method that lets you analyze results dynamically as data arrives, helping you reach conclusions with sacrificing statistical rigor. We will walk through how SPRT is particularly useful for new AI products with limited distribution.
Basic Intuition
At its core, SPRT continually computes a likelihood ratio comparing how likely the observed data is under 2 competing hypotheses:
Null hypothesis H0: the new AI feature has no improvement
Alternative hypothesis H1: the new AI feature has a meaningful improvement
After each new data point (or small chunk of data points), SPRT updates the likelihood ratio (LR) defined as:
[ \Lambda_n = \frac{P(\text{Data}_{1 \dots n} \mid H_1)}{P(\text{Data}_{1 \dots n} \mid H_0)} ]
We can then compare Λn to 2 thresholds – an upper threshold A and a lower threshold B to decide whether to stop:
If Λn >= A: we have enough evidence for H1 (the alternative), so accept H1
If Λn < B: we have enough evidence for H0, so accept H0
If B < Λn < A: the evidence is inconclusive, so continue and collect more data
This process is repeated sequentially until one of the thresholds is crossed, at which point a decision is made. In practice, A and B are chosen based on the desired error rates, with A > 1 and B < 1. Usually:
( A = \frac{1 - \beta}{\alpha} )
( B = \frac{\beta}{1 - \alpha} )
Remember that ɑ is the probability of making a Type I error (usually set to 0.05), and β is the probability of making a Type II error (set to 0.2 if you want 80% power).
The key idea is that SPRT allows you to “stop when you’re confident”, as soon as sufficient evidence has accumulated. You don’t have to wait for a fixed sample size!
SPRT vs. Traditional A/B Testing
How does SPRT differ from traditional A/B testing, and why does it matter?
Sample size and efficiency: in a traditional A/B test, you calculate an upfront sample size (say, via power analysis) and run the experiment until that many observations are collected, regardless of whether the difference was obvious earlier or not. This can lead to collecting more data than necessary. SPRT, by contrast, often stops early when results are clear, avoiding unnecessary samples. On average, SPRT uses fewer samples to reach a conclusion than a fixed test with the same error rates.
Decision timing: A fixed test only makes a decision at the end of the experiment (after N samples). Until then, you typically shouldn’t peek at results, or you risk the “peeking problem” of inflating false positives. SPRT, on the other hand, enables continuous monitoring of results and allows stopping as soon as criteria are met.You can look at your experiment dashboard every day and not worry that you're invalidating the test – if it meets the SPRT stop condition, you can trust the result 😊
Type I and II Errors: Fixed-sample tests guarantee error rates exactly at N samples (under the planned effect size). SPRT guarantees error bounds at whatever sample size it stops. In essence, SPRT does not increase false positives even though you’re checking continuously.
Outcome Interpretation: A fixed-sample test yields a p-value and either reject/fail-to-reject conclusion after N samples. SPRT yields a decision possibly at a random sample count. One additional benefit of SPRT is that it can affirmatively accept H0(with confidence) if evidence for no difference accumulates (when it falls below B). This can be useful information for AI teams (for instance, knowing a new AI feature is genuinely not outperforming the old one, so you might stop investing in that approach).
Why SPRT is Especially Useful for AI Product Experimentation
For AI products or features in their early stagesorlimited rollout, data can be a precious commodity. Here’s why SPRT is particularly useful in this context:
Limited Traffic / Users: New AI features (say a beta version of an agent or AI search feature) can receive limited traffic in the early days – perhaps only a few hundred users try it in a week. A traditional A/B test might tell you that you need, for example, 5k users to detect a moderate improvement with statistical significance. Waiting to accumulate that many users could take a long time. SPRT offers a smarter approach: you start the experiment and let it run sequentially. If the new feature truly has a big impact (positive or negative), SPRT may detect it after 500 or 1000 users, thereby shorteningthe experiment time dramatically 😃 Conversely, if there’s no effect, SPRT might also figure that out sooner and allow you to stop or adjust the feature without wasting time.
Faster Iteration Cycles: Early-stage products undergo rapid iteration. AI teams often want to try a tweak, ship to a small set of users, measure, and iterate. SPRT aligns with iteration speed. As soon as your new AI feature is clearly outperforming the old, you can roll it out wider; if it’s clearly underperforming, you can roll back immediately.
Risk Mitigation (Catch Regressions Early): New AI features can have unexpected behavior like RAG hallucinations (leading to a regression in a KPI). With early users, you want to minimize harm, of course. SPRT will flag a significant drop as soon as enough evidence accumulates. SPRT might detect the significant harm after, say, 200 user interactions and you can roll back immediately, rather than discovering only after reaching 1000 users in a fixed test that you had a serious issue.
Small Effect Detection and Futility: If the improvement from your new AI feature is very small (or non-existent), a fixed sample test might grind out the full sample and still give you a “not statistically significant” result… 😆 With SPRT, you can set a “futility boundary”, effectively the lower threshold B, which will terminate the test early if it’s clear that the results are not trending toward stat sig. For a low-traffic AI product, deciding early that “we’re not seeing a big enough effect” is great because it frees you to pivot to the next idea instead of running a long inconclusive test.
Real World Example: Testing a new feature for a chatbot with limited users
Imagine you’ve developed a new AI chatbot and added a feature where it proactively suggests solutions. You want to test if this new feature improves user satisfaction (measured by a thumbs-up feedback rate) compared to the old version. However, your chatbot is only in beta with ~2000 users total, and only 10% of them might use the new feature in a week – so getting even a few hundred samples could take weeks.
Using a traditional A/B test, you estimate needing ~800 feedback responses in each group to have 80% power to detect, say, a 5% increase in the positive feedback rate at 95% confidence. That might mean running the experiment for 8+ weeks given your traffic. That’s a long time…
Let’s say you use SPRT with ɑ=0.05, β=0.2 (power 0.8) and set H1 as a 5% increase scenario. You might set up the test with boundaries A≈16 and B≈0.21, based on the equations mentioned earlier. The experiment goes live, and you monitor results daily. After each user interaction with the chatbot, the likelihood ratio is updated. A few scenarios could happen:
By week 3, suppose the new feature has gotten 300 users and the data shows a noticeably higher satisfaction rate. The SPRT likelihood ratio crosses above A – meaning the odds of seeing such data if H1 (the improvement) is true vs. if H0 (no improvement) is true are over 16:1. The test stops early and concludes the new feature significantly improves satisfaction. You can confidently roll out the feature to all users a month sooner than planned.
Alternatively, by week 3, maybe the satisfaction rates are about the same or slightly worse with the new feature. The likelihood ratio dips below B, say at 0.15, indicating strong evidence against the new feature delivering the hoped improvement. In fact, this would correspond to statistically detecting a likely regression or no gain. The SPRT would stop and essentially “accept H0,” meaning you decide the feature isn’t worth pursuing (at least not at the effect size you cared about). This saves you from continuing the test for 5 more weeks only to find a null result; you can go back to the drawing board sooner.
If the results are borderline (neither clearly good nor clearly bad), SPRT will continue to gather more data until it reaches a decision or perhaps until a maximum sample or time limit is hit. In the worst case, you might run the full length similar to a fixed test. But in no case will SPRT perform worse than a fixed test in terms of error rates, and in many cases it will have reached a decision much faster.
This example shows you how SPRT is great for the low-data regime of new AI features: it maximizes the info gained from each user interaction and allows you to reach conclusions as soon as the data is there. 🚀
How to use SPRT, as an AI Engineer 🚀
Define hypotheses and minimum effect size. You need to specify H0 and H1. For example, if you expect a 10% lift due to a new AI feature, you need to set H1 to 10%. If you want a range, you could use a mixture approach.
Choose error rates (ɑ, β). As mentioned earlier, usually ɑ is 0.05, and you could choose β = 0.2 if you want 80% power.
Compute SPRT thresholds. Use the formulas mentioned earlier in the blog post.
Track your data sequentially in the Logs feature on Patronus. Implement code that updates the likelihood ratio after each new data point or batch. Check the likelihood ratio against A and B at each step. Continuous monitoring is key!
Automate experiment stopping rules. This could be done with cron jobs or CI/CD. This is so you stop the experiment once A or B is crossed. You could define a maximum sample size and experiment duration too.
Advanced: Mathematical Background of SPRT
Let’s formalize the setup with some notation and math. Suppose we want to test:
H0: Θ = Θ0 (e.g. conversion rate p = p0 for the control or old model),
H1: Θ = Θ1 (e.g. improved conversion rate p = p1 for the new AI-driven feature),
where Θ is some metric or parameter of interest (it could be a click-through rate, conversion probability, mean user rating, etc.). For concreteness, imagine Θ is a conversion probability. Under H0, p=p0; under H1, p=p1. Each user interaction (conversion or no conversion) provides evidence. The likelihood ratio after n interactions (with x successes/conversions out of n) would be:
Λn = P(data | H1) / P(data | H0) = [p1x(1 − p1)n-x] / [p0x(1−p0)n-x],
which is the ratio of the two hypotheses’ likelihoods for the observed outcomes. We initialize ( \Lambda_0 = 1 ) (before any data). After each new observation, we update ( \Lambda_n ) multiplicatively by the factor
[ \frac{P(\text{new data} \mid H_1)}{P(\text{new data} \mid H_0)} ]
(this is equivalent to adding the log-likelihood ratio).
Decision thresholds: We choose thresholds A (upper) and B (lower) based on our tolerances for Type I and Type II errors (false positives and false negatives). Specifically, if we want a significance level (maximal Type I error rate) and power 1-β (so β is the Type II error rate), one convenient choice is:
( A = \frac{1 - \beta}{\alpha} )
( B = \frac{\beta}{1 - \alpha} )
These thresholds guarantee that ( P(\text{false alarm} \mid H_0) \leq \alpha ) and ( P(\text{miss} \mid H_1) \leq \beta ) for the test. For example, if ( \alpha = 0.05 ) and ( \beta = 0.20 ) (i.e. 95% confidence, 80% power), then:
( A = \frac{0.95}{0.05} = 19 \quad\quad B = \frac{0.20}{0.95} \approx 0.21 )
In log terms, we would compare ( \log \Lambda_n ) to ( \log A ) and ( \log B ), which correspond to linear boundaries as data accrues.
How it works: As data comes in, if H1 is true, the likelihood ratio will tend to grow and eventually exceed A; if H0 is true, Λn will tend to shrink below B. The test stops at whichever boundary is hit first. Notably, SPRT can also end by accepting H0 (not just rejecting it); if evidence strongly favors no effect, SPRT explicitly makes that call. This differs from fixed-sample tests where one typically either rejects H0 or “fails to reject” (without a firm conclusion in favor of H0). The thresholds A, B effectively control the Type I/II error rates of these decisions by design.
Optimality: One of the cool things about SPRT is that it is optimal (in the Wald–Wolfowitz sense) for simple hypothesis testing. And actually, Wald proved that no other test with the same error rates uses fewer samples on average than SPRT. So, SPRT minimizes the expected sample size required to reach a decision, among all sequential tests with given ɑ, β. This optimality means that if an effect truly exists, SPRT will detect it as efficiently as possible, and if there is no effect, SPRT will likely recognize that sooner than a fixed-length test. This makes SPRT highly sample-efficient in many scenarios – this is great when data is scarce or costly.
It’s worth noting that the classic SPRT assumes simple hypotheses (exact H0 and H1 parameter values). In practice, if H1 is composite or not a single specified value (e.g. “the new feature could have some positive impact, but we don’t know how much”), you can use variants like mixture SPRT (mSPRT), which average over a range of possible effects.
If you’d like a free SPRT experimentation setup for your AI product, fill out this form below.
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2793 | https://wk.baidu.com/view/0b5243d9ac02de80d4d8d15abe23482fb4da02f0?pcf=2&re=view&bfetype=new | 人教版高中数学(理)高考专题复习辅导讲义(含答案解析)：第一章 合与常用逻辑用语 - 百度文库
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第一章
集合与常用逻辑用语
1.
集合
(1)
集合的含义与表示
①了解集合的含义，体会元素与集合的属于关
系
．
②能用自然语言、
图形语言、
集合语言
(
列举法
或描述法
)
描述不同的具体问题
．
(2)
集合间的基本关系
①理解集合之间包含与相等的含义，能识别给
定集合的子集
．
②在具体情境中，了解全集与空集的含义
．
(3)
集合的基本运算
．
①理解两个集合的并集与交集的含义，会求两
个简单集合的并集与交集
．
②理解在给定集合中一个子集的补集的含义，
会求给定子集的补集
．
③能使用韦恩
(Venn)
图表达集合间的基本关系
及集合的基本运算
．
2
．
常用逻辑用语
(1)
理解命题的概念
．
(2)
了解“若
p
，
则
q
”形式的命题及其逆命题、
否命题与逆否命题，会分析四种命题的相互关系
．
(3)
理解必要条件、充分条件与充要条件的含
义
．
(4)
了解逻辑联结词“或”、
“且”、
“非”的
含义
．
(5)
理解全称量词和存在量词的意义
．
(6)
能正确地对含一个量词的命题进行否定
．
§1.1
集合及其运算
1
．
集合的基本概念
(1)
我们把研究对象统称为
，
把一些元
素组成的总体叫做
．
(2)
集合中元素的三个特性：
，
，
.
(3)
集
合
常
用
的
表
示
方
法
：
和
．
2
．
常用数集的符号
数集
正整
数集
自然
数集
整数
集
有理
数集
实数
集
复数
集
符号
3.
元素与集合、集合与集合之间的关系
(1)
元素与集合之间存在两种关系：
如果
a
是集
合
A
中的元素，就说
a
集合
A
，记作
；如果
a
不是集合
A
中的元素，就说
a__
集合
A
，记作
．
(2)
集合与集合之间的关系：
表示
关系
文字语言
符号语言
相等
集合
A
与集合
B
中的所有元
素都相同
⇔
A
＝
B
子集
A
中任意一个
元素均为
B
中
的元素
或
真子集
A
中任意一个
元素均为
B
中
的元素，且
B
中至少有一个
元素不是
A
中
的元素
或
空集
空集是任何集
合的子集，是
任何
的
真子集
∅
⊆
A
，
∅
B
(
B
≠
∅
)
结论：集合
{
a
1
，
a
2
，…，
a
n
}
的子集有
个，
非空子集有
个，
非空真子集有
个
．
4
．
两个集合
A
与
B
之间的运算
集合的并
集
集合的交
集
集合的补
集
符号
表示
若全集为
U
，
则集合
A
的补集
记为
Venn
图表
示
(
阴影
部分
)
意义
5.
集合运算中常用的结论
(1)①
A
∩
B__A
；
②
A
∩
B_B
；
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2794 | https://en.wikipedia.org/wiki/Lewis_structure | Jump to content
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Contents
(Top)
1 Construction and electron counting
1.1 Miburo method
1.2 Lever method
2 Formal charge
3 Resonance
4 Example
5 Alternative formations
6 Usage and limitations
7 See also
8 References
9 External links
Lewis structure
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From Wikipedia, the free encyclopedia
Diagrams for the bonding between atoms of a molecule and lone pairs of electrons
Lewis structures – also called Lewis dot formulas, Lewis dot structures, electron dot structures, or Lewis electron dot structures (LEDs) – are diagrams that show the bonding between atoms of a molecule, as well as the lone pairs of electrons that may exist in the molecule. Introduced by Gilbert N. Lewis in his 1916 article The Atom and the Molecule, a Lewis structure can be drawn for any covalently bonded molecule, as well as coordination compounds. Lewis structures extend the concept of the electron dot diagram by adding lines between atoms to represent shared pairs in a chemical bond.
Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol. Lines are drawn between atoms that are bonded to one another (pairs of dots can be used instead of lines). Excess electrons that form lone pairs are represented as pairs of dots, and are placed next to the atoms.
Although main group elements of the second period and beyond usually react by gaining, losing, or sharing electrons until they have achieved a valence shell electron configuration with a full octet of (8) electrons, hydrogen instead obeys the duplet rule, forming one bond for a complete valence shell of two electrons.
Construction and electron counting
[edit]
Main article: Electron counting
For a neutral molecule, the total number of electrons represented in a Lewis structure is equal to the sum of the numbers of valence electrons on each individual atom, not the maximum possible. Non-valence electrons are not represented in Lewis structures as they do not bond.
Once the total number of valence electrons has been determined, they are placed into the structure according to these steps:
Initially, one line (representing a single bond) is drawn between each pair of connected atoms.
Each bond consists of a pair of electrons, so if t is the total number of electrons to be placed and n is the number of single bonds just drawn, t − 2n electrons remain to be placed. These are temporarily drawn as dots, one per electron, to a maximum of eight per atom (two in the case of hydrogen), minus two for each bond.
Electrons are distributed first to the outer atoms and then to the others, until there are no more to be placed.
Finally, each atom (other than hydrogen) that is surrounded by fewer than eight electrons (counting each bond as two) is processed as follows: For every two electrons needed, two dots are deleted from a neighboring atom and an additional line is drawn between the two atoms. This represents the conversion of a lone pair of electrons into a bonding pair, which adds two electrons to the former atom's valence shell while leaving the latter's electron count unchanged.
In the preceding steps, if there are not enough electrons to fill the valence shells of all atoms, preference is given to those atoms whose electronegativity is higher.
Lewis structures for polyatomic ions may be drawn by the same method. However when counting electrons, negative ions should have extra electrons placed in their Lewis structures; positive ions should have fewer electrons than an uncharged molecule. When the Lewis structure of an ion is written, the entire structure is placed in brackets, and the charge is written as a superscript on the upper right, outside the brackets.
Miburo method
[edit]
A simpler method has been proposed for constructing Lewis structures, eliminating the need for electron counting: the atoms are drawn showing the valence electrons; bonds are then formed by pairing up valence electrons of the atoms involved in the bond-making process, and anions and cations are formed by adding or removing electrons to/from the appropriate atoms.
A trick is to count up valence electrons, then count up the number of electrons needed to complete the octet rule (or with hydrogen just 2 electrons), then take the difference of these two numbers. The answer is the number of electrons that make up the bonds. The rest of the electrons just go to fill all the other atoms' octets.
Lever method
[edit]
Another simple and general procedure to write Lewis structures and resonance forms has been proposed.[example needed]
This system works in nearly all cases, however there are 3 instances where it will not work[citation needed]. These exceptions are outlined in the table below.
Exceptions
| The exception | How it breaks the system | How to fix the Lewis structure |
| Free radicals (molecules with unpaired valence electrons) | Sum of TVe is an odd number. Bond number is not a whole number. | Round calculated bond number down to the nearest whole number (e.g. 4.5 bonds would round down to 4 bonds). |
| Valence shell deficiency | Does not break the system, must instead memorize when it occurs. | BeX2, BX3, and AlX3. "X" represents hydrogen or halogens. When Be is bonded with 2 other atoms, or when B and Al are bonded with 3 other atoms, they do not form full valence shells. Assume single bonds and use the actual bond number to calculate lone pairs. |
| Expanded octet (only occurs for elements in groups 3–8) | Bond calculation provide too few bonds for the number of atoms in the molecule. | Assume single bonds, use the minimal number of bonds necessary to create the molecule. |
Formal charge
[edit]
Main article: Formal charge
In terms of Lewis structures, formal charge is used in the description, comparison, and assessment of likely topological and resonance structures by determining the apparent electronic charge of each atom within, based upon its electron dot structure, assuming exclusive covalency or non-polar bonding. It has uses in determining possible electron re-configuration when referring to reaction mechanisms, and often results in the same sign as the partial charge of the atom, with exceptions. In general, the formal charge of an atom can be calculated using the following formula, assuming non-standard definitions for the markup used: where
: is the formal charge,
: represents the number of valence electrons in a free atom of the element,
: represents the number of unshared electrons on the atom,
: represents the total number of electrons in bonds the atom has with another.
The formal charge of an atom is computed as the difference between the number of valence electrons that a neutral atom would have and the number of electrons that belong to it in the Lewis structure. Electrons in covalent bonds are split equally between the atoms involved in the bond. The total of the formal charges on an ion should be equal to the charge on the ion, and the total of the formal charges on a neutral molecule should be equal to zero.
Resonance
[edit]
Main article: Resonance structure
For some molecules and ions, it is difficult to determine which lone pairs should be moved to form double or triple bonds, and two or more different resonance structures may be written for the same molecule or ion. In such cases it is usual to write all of them with two-way arrows in between (see § Example below). This is sometimes the case when multiple atoms of the same type surround the central atom, and is especially common for polyatomic ions.
When this situation occurs, the molecule's Lewis structure is said to be a resonance structure, and the molecule exists as a resonance hybrid. Each of the different possibilities is superimposed on the others, and the molecule is considered to have a Lewis structure equivalent to some combination of these states.
The nitrate ion (NO−3), for instance, must form a double bond between nitrogen and one of the oxygens to satisfy the octet rule for nitrogen. However, because the molecule is symmetrical, it does not matter which of the oxygens forms the double bond. In this case, there are three possible resonance structures. Expressing resonance when drawing Lewis structures may be done either by drawing each of the possible resonance forms and placing double-headed arrows between them or by using dashed lines to represent the partial bonds (although the latter is a good representation of the resonance hybrid which is not, formally speaking, a Lewis structure).
When comparing resonance structures for the same molecule, usually those with the fewest formal charges contribute more to the overall resonance hybrid. When formal charges are necessary, resonance structures that have negative charges on the more electronegative elements and positive charges on the less electronegative elements are favored.
Single bonds can also be moved in the same way to create resonance structures for hypervalent molecules such as sulfur hexafluoride, which is the correct description according to quantum chemical calculations instead of the common expanded octet model.
The resonance structure should not be interpreted to indicate that the molecule switches between forms, but that the molecule acts as the average of multiple forms.
Example
[edit]
The formula of the nitrite ion is NO−2.
Nitrogen is the least electronegative atom of the two, so it is the central atom by multiple criteria.
Count valence electrons. Nitrogen has 5 valence electrons; each oxygen has 6, for a total of (6 × 2) + 5 = 17. The ion has a charge of −1, which indicates an extra electron, so the total number of electrons is 18.
Connect the atoms by single bonds. Each oxygen must be bonded to the nitrogen, which uses four electrons—two in each bond.
Place lone pairs. The 14 remaining electrons should initially be placed as 7 lone pairs. Each oxygen may take a maximum of 3 lone pairs, giving each oxygen 8 electrons including the bonding pair. The seventh lone pair must be placed on the nitrogen atom.
Satisfy the octet rule. Both oxygen atoms currently have 8 electrons assigned to them. The nitrogen atom has only 6 electrons assigned to it. One of the lone pairs on an oxygen atom must form a double bond, but either atom will work equally well. Therefore, there is a resonance structure.
Tie up loose ends. Two Lewis structures must be drawn: Each structure has one of the two oxygen atoms double-bonded to the nitrogen atom. The second oxygen atom in each structure will be single-bonded to the nitrogen atom. Place brackets around each structure, and add the charge (−) to the upper right outside the brackets. Draw a double-headed arrow between the two resonance forms.
Alternative formations
[edit]
Two varieties of condensed structural formula, both showing butane
Chemical structures may be written in more compact forms, particularly when showing organic molecules. In condensed structural formulas, many or even all of the covalent bonds may be left out, with subscripts indicating the number of identical groups attached to a particular atom. Another shorthand structural diagram is the skeletal formula (also known as a bond-line formula or carbon skeleton diagram). In a skeletal formula, carbon atoms are not signified by the symbol C but by the vertices of the lines. Hydrogen atoms bonded to carbon are not shown—they can be inferred by counting the number of bonds to a particular carbon atom—each carbon is assumed to have four bonds in total, so any bonds not shown are, by implication, to hydrogen atoms.
Other diagrams may be more complex than Lewis structures, showing bonds in 3D using various forms such as space-filling diagrams.
Usage and limitations
[edit]
Despite their simplicity and development in the early twentieth century, when understanding of chemical bonding was still rudimentary, Lewis structures capture many of the key features of the electronic structure of a range of molecular systems, including those of relevance to chemical reactivity. Thus, they continue to enjoy widespread use by chemists and chemistry educators. This is especially true in the field of organic chemistry, where the traditional valence-bond model of bonding still dominates, and mechanisms are often understood in terms of curve-arrow notation superimposed upon skeletal formulae, which are shorthand versions of Lewis structures. Due to the greater variety of bonding schemes encountered in inorganic and organometallic chemistry, many of the molecules encountered require the use of fully delocalized molecular orbitals to adequately describe their bonding, making Lewis structures comparatively less important (although they are still common).
There are simple and archetypal molecular systems for which a Lewis description, at least in unmodified form, is misleading or inaccurate. Notably, the naive drawing of Lewis structures for molecules known experimentally to contain unpaired electrons (e.g., O2, NO, and ClO2) leads to incorrect inferences of bond orders, bond lengths, and/or magnetic properties. A simple Lewis model also does not account for the phenomenon of aromaticity. For instance, Lewis structures do not offer an explanation for why cyclic C6H6 (benzene) experiences special stabilization beyond normal delocalization effects, while C4H4 (cyclobutadiene) actually experiences a special destabilization. Molecular orbital theory provides the most straightforward explanation for these phenomena.
See also
[edit]
Valence shell electron pair repulsion theory
Molecular geometry
Structural formula
Natural bond orbital
References
[edit]
^ IUPAC definition of Lewis formula
^ Zumdahl, S. (2005) Chemical Principles Houghton-Mifflin (ISBN 0-618-37206-7)
^ G.L. Miessler; D.A. Tarr (2003), Inorganic Chemistry (2nd ed.), Pearson Prentice–Hall, ISBN 0-13-035471-6
^ Lewis, G. N. (1916), "The Atom and the Molecule", J. Am. Chem. Soc., 38 (4): 762–85, Bibcode:1916JAChS..38..762L, doi:10.1021/ja02261a002, S2CID 95865413
^ Miburo, Barnabe B. (1993), "Simplified Lewis Structure Drawing for Non-science Majors", J. Chem. Educ., 75 (3): 317, Bibcode:1998JChEd..75..317M, doi:10.1021/ed075p317
^ Lever, A. B. P. (1972), "Lewis Structures and the Octet Rule", J. Chem. Educ., 49 (12): 819, Bibcode:1972JChEd..49..819L, doi:10.1021/ed049p819
^ Miessler, G. L. and Tarr, D. A., Inorganic Chemistry, 2nd ed., Prentice Hall, 1998, ISBN 0-13-841891-8, pp. 49–53 – Explanation of formal charge usage.
^ Peter Senn (1992). "A Simple Quantum Mechanical Model that Illustrates the Jahn-Teller Effect". J. Chem. Educ. 69 (10): 819. Bibcode:1992JChEd..69..819S. doi:10.1021/ed069p819.
External links
[edit]
Lewis Dot Diagrams of Selected Elements
Lewis structures for all compounds
| v t e Molecular visualization |
| Chemical formulas | | | | --- | | Non-structural formulas | Empirical formula Molecular formula | | Structural formulas | Lewis structure Condensed formula Skeletal formula | |
| Molecular models | Ball-and-stick model Space-filling model CPK coloring |
| Other ways | Molecular graph SMILES InChl |
| v t e Chemical bonding theory |
| Atomic orbital Quantum mechanics + Introduction to quantum mechanics |
| Types of bonds | | | | --- | | By symmetry | Sigma (σ) Pi (π) Delta (δ) Phi (φ) | | By multiplicity | 1 (single) 2 (double) 3 (triple) 4 (quadruple) 5 (quintuple) 6 (sextuple) | | By spin | Triplet Singlet Exchange-coupled | |
| Valence bond theory | | | | --- | | Concepts | Hybrid orbital Resonance Lewis structure | | Constituent units | Covalent bond Lone pair | |
| Molecular orbital theory | | | | --- | | Concepts | Molecular orbital LCAO MO diagram | | Constituent units | Bonding MO Non-bonding MO Antibonding MO | |
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Lewis structure
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2795 | https://www.smartick.com/blog/mathematics/multiplication-and-division/division-basic-concepts/ | Loading [MathJax]/jax/output/HTML-CSS/config.js
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What Is a Division? Basic Concepts
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In today’s post we’re going to learn what a division is, and what we call the numbers that make up a division.
Dividing is the same as sharing in equal parts
For example: Imagine we have 10 cupcakes. We want to put them into 2 boxes so that each box has the same number of cupcakes. How would we divide the cupcakes?
We’ve divided 10 cupcakes between 2 boxesand we’ve put 5 cupcakesin each box.
What makes a division?
A division is made up of different numbers, and each of these numbers has a special name. We want to share 7 pineapples between 2 boxes…
We’ve put 3 pineapples in each box, and we have one pineapple left over.
The total number of pineapples that we have to give out is called the Dividend.
The number of boxes that we want to to put the pineapples into is called the Divisor.
The number of pineapples that we’ve put into each box is called the Quotient.
The number of pineapples that we have left over is called the Remainder.
Types of divisions:
Exact divisions:When the remainder is zero, or in other words, when there’s nothing left over when we’ve finished dividing (sharing).
Integer division:When the remainder is not zero, or, when we have 1 or more than 1 left over when we’ve finished the division (sharing).
And that’s it for today’s post! I hope you’ve enjoyed learning the basic concepts of divisions. If you want to keep practising divisions and much more, you just have to log in to Smartick.
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jonesFeb 12 2023, 9:43 PM
the very area has made division credible and easy to follow
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2796 | https://physics.stackexchange.com/questions/714162/multiplication-and-significant-figures | Stack Exchange Network
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Multiplication and significant figures
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$\begingroup$
What is point of rounding off to significant figures in multiplication. e.g. $3\times123456789=370370367$, if I round off the answer to least sig fig it becomes $400000000$ which has a large error. So what is point of significant figures here?
I really don't understand the importance of sig fig.
conventions
error-analysis
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edited Jun 16, 2022 at 22:45
Qmechanic♦
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asked Jun 16, 2022 at 20:39
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$\begingroup$ No rounding in the example you give: the product of two integers is an integer. $\endgroup$
Ed V
– Ed V
2022-06-16 20:53:11 +00:00
Commented Jun 16, 2022 at 20:53
3
$\begingroup$ Significant figures are a waste of time. It is much easier and cleaner to simply report the absolute uncertainty and use error propagation to report the uncertainty of results of multiplication. $\endgroup$
Jagerber48
– Jagerber48
2022-06-16 22:41:22 +00:00
Commented Jun 16, 2022 at 22:41
3
$\begingroup$ @Jagerber48 Yep! And some physicists are not at all bashful about spending 100 pages or so pounding that home: av8n.com/physics/uncertainty.htm. I like where Denker says People who care about their data dont use sig figs.. $\endgroup$
Ed V
– Ed V
2022-06-16 23:06:18 +00:00
Commented Jun 16, 2022 at 23:06
2
$\begingroup$ @Jagerber48 Soft disagree. The significant figures approach is a useful introduction to the idea of finite-precision data at the high-school level. Compare to the oversimplified model in intro biology courses where brown eyes and blue eyes are used to teach about dominant and recessive genes, even though eye color is a complex phenotype controlled by the interactions of many different genes. For people who havent yet internalized calculus, error propagation is a dark art. $\endgroup$
rob
– rob ♦
2022-06-17 00:01:35 +00:00
Commented Jun 17, 2022 at 0:01
2
$\begingroup$ @Jagerber48 Have you actually done this though? I think you're forgetting what it's like to be a high school student. High school students have trouble remembering significant digits (one simple rule that you can justify to them at their level), let alone memorizing error propagation rules (several equations that seem to come out of nowhere and that they won't have a basis to understand for a few years). It's not going to stick. $\endgroup$
DKNguyen
– DKNguyen
2022-06-17 02:00:08 +00:00
Commented Jun 17, 2022 at 2:00
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I like to teach my students to trust their hunches about significant figures using their emotional response to the rounded number. I call this the anger management method of error estimation.
When I go hiking, I like to scrounge up a walking stick from the deadwood in the forest. The right length for a walking stick is pretty close to my height. I am (to one significant figure) about two meters tall. One meter is too short for a walking stick (thats more like a cane, which serves a different purpose). If you imagine coming across a hiker with a three-meter walking stick, you will probably set yourself to laughing an emotional response.
If you found two candidate walking sticks whose lengths were different by 20 cm, you might stand them up next to yourself and decide that one is a better length than the other. You might even break off a handspan (10 cm) or two to get the length right. But imagine coming across a hiker who has heard somewhere that a walking stick should be as long as he is tall, and he was 181.5 cm tall at his last doctor visit, so he is sitting on the trail with a measuring tape carefully sanding down the knob at the top of his piece of found deadwood. Youd probably have an emotional response to him, too.
So for the length of a walking stick, you really care about the first digit. You probably also care about the second digit. That last centimeter of length, through, is insignificant. The length of a walking stick has about two significant digits.
The trousers I wore yesterday are 110 cm long. (Two meters: not trousers. Half a meter: also not trousers.) If you buy trousers from a fancy shop, the tailor may fuss about the last centimeter of their length, which controls how the cuff folds where it lands at your shoe. If you have a tailor fixing the hem of your trousers, hell mark the correct length with a piece of chalk and then pin the fabric in place with sub-centimeter precision. But if you had a tailor who was fussing over sub-millimeter alterations in the length of your trousers (the pin is between the wrong fibers!), you would have an emotional reaction to him and leave the shop. If your trousers came back from the tailor 10 cm too long or 10 cm too short, you would likewise have an emotional reaction and leave the shop.
The millimeter digit is the first insignificant digit in the length of a pair of trousers: you would possibly be annoyed if the centimeter digit were wrong, but probably not notice if the millimeter digit were wrong. So for my trousers, 110 cm has three significant digits.
Of course, you might be shorter than I am, and your trousers might only be 95 cm long. Do your trousers only have two significant figures? If I decree trousers have three significant figures, have I doomed short people to a world filled with insane submillimeter tailors? No. Trust your intuition, keep an extra digit if you feel squashy about rounding too much, and be aware that there are more robust ways than counting significant figures for people who really need to be sure about their uncertainty analysis.
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answered Jun 17, 2022 at 2:51
rob♦rob
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$\begingroup$ Anger management previously. $\endgroup$
rob
– rob ♦
2022-06-17 02:56:35 +00:00
Commented Jun 17, 2022 at 2:56
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Actually, it becomes $4\times10^8$, not $400000000$. It seems pointless because you are thinking of exact numbers in your example, and if every number in there was an exact number it would be pointless. But in real life at least some of those numbers are going to come from measurements and measurements have uncertainty.
Basically, you do not write down digits that may not be true. Another more explicit method of expressing uncertainty is writing down the error bars like $3\pm0.1$ but even there the fact that $0.1$ was written instead of $0.10$ holds significance.
Significant digits is sort of an implicit way of expressing and interpreting uncertainty when there is nothing explicit to express uncertainty to keep things in check.
If your 3 is a factor of three (exact) then you don't need to round significant digits because it is exactly 3; In other words $3.\bar0$.
But if your 3 is actually a measurement like 3 meters then there is uncertainy in that number. Since you do not write digits you are not sure about and all you wrote down was $3m$ that means the actual number could be anywhere between $3.\bar0$m and $3.\bar9$. If you wrote $3.0$m then that means you were sure of more digits in your measurement and by doing that you are saying the real number is somewhere between $3.\bar0m$ and $3.0\bar9$m
So if we assume that 123456789 is an exact number in your example (or is a measurement accurate to that number of digits) if we multiply them together, then the actual answer could be anywhere between $370 370 367.\bar0$ and $493 827 156.\bar0$
Between 123456789 and 3, the 3 has the most uncertainty in it because it has the least significant digits (the digits you are sure about) therefore it is the weakest link holding back the your calculation. The result of the calculation can't be any more accurate than the 3 is so we use the same number of significant digits in the result.
That is why you cannot say the answer is 400000000, because if you did that would mean you are certain about every zero in that result because that's what you wrote down. Since you are not sure of all those zeroes, you can only say the answer is $4\times10^8$, which as written means the result could be anywhere between $4\times10^8$ and $4.\bar9\times10^8$
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edited Jun 17, 2022 at 4:58
answered Jun 16, 2022 at 22:10
DKNguyenDKNguyen
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$\begingroup$ I'm used to the convention that something like "3m" is treated as the result of rounding so for the actual length x, 2.5m <= x < 3.5m. $\endgroup$
Jasper
– Jasper
2022-06-17 08:24:27 +00:00
Commented Jun 17, 2022 at 8:24
$\begingroup$ I believe that the least count is kept in mind while writing uncertain digit. E.g. 3m =2.9m or 3.1m. so why there is 3m= 3.0 to 3.9 in your answer? $\endgroup$
Level1
– Level1
2022-06-17 09:00:13 +00:00
Commented Jun 17, 2022 at 9:00
$\begingroup$ @Jasper I think there are different schemes because as someone basically said in the comments: significant digits are kind of a sloppy way to handle things. Maybe they should be considered a lie to children to make them more aware of what they are writing down. $\endgroup$
DKNguyen
– DKNguyen
2022-06-17 13:18:34 +00:00
Commented Jun 17, 2022 at 13:18
$\begingroup$ @user359206 That could be true and would make a bit more sense. But it can also depends on the way your measuring instrument works too. There's a lot of different ways to go about things. $\endgroup$
DKNguyen
– DKNguyen
2022-06-17 13:24:09 +00:00
Commented Jun 17, 2022 at 13:24
$\begingroup$ Mathematically $4\times10^8 =400000000$. $\endgroup$
md2perpe
– md2perpe
2022-06-18 06:34:37 +00:00
Commented Jun 18, 2022 at 6:34
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Sometimes the error one might have while conducting measurements is large. Then, you do not need to be performing calculations with accuracy larger than the one allowed by the experimental device. This is the point, if I am not mistaken.
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answered Jun 16, 2022 at 20:45
schris38schris38
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The concept of significant digits is used when working with measured numbers as opposed to exact numbers. It is a relatively simple way to determine how precisely you will know the result of a calculation involving measured numbers.
Say your numbers $3$ and $123456789$ come from measuring lengths by using a method that that has a precision of one unit. Perhaps you're pacing to measure length and you can measure to the nearest metre. The product of $3$ m and $123456789$ m might represent an area in square metres.
The thing is, if you are only measuring to the lengths to the nearest metre then the length you recorded as $3$ m is almost certainly not exactly three metres. The actual length might be anywhere between $2.5$ m and $3.5$ m. Similarly the other length could be up to half a metre more or less, though as percentage uncertainty this is much, much less important. So the actual area could be anywhere from $2.5 \times 123456788.5 = 3.1 \times 10^8$ to $3.5 \times 123456789.5 = 4.3 \times 10^8$. So, if you only know the shorter length to one significant digit, you would not be able to claim to know the area to the nearest square metre i.e. to nine significant digits. Rounding off to one significant digit, i.e. $400000000$, expresses what the calculated product is and also how precisely you know it. It may look like you are introducing a large error in the product, and you would be if you knew the two initial numbers exactly, but if they are measured values then $370370367$ is almost certainly not the correct product.
There are more sophisticated methods of error analysis.
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answered Jun 16, 2022 at 22:05
M. EnnsM. Enns
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$\begingroup$ I'm sure you know this but the way your text is written basically implies that writing $3$m means $3m\pm0.5m$ which is not true. It's an alternate and more explicit approach to expressing the uncertainty than significant digits. Using significant digits for uncertainty $3m$ is more kind of the default way when there is nothing else mentioned so things don't unknowingly go out of control. $\endgroup$
DKNguyen
– DKNguyen
2022-06-16 22:41:23 +00:00
Commented Jun 16, 2022 at 22:41
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$\begingroup$
Play with this Rounding Significant Figures Calculator, and you will understand.
You will have to provide the number of significant figures you want to round to. In your example you rounded to one significant figure.
If you round to four significant figures for example, you will get 370400000 So you will get the first four most significant figures and the rest will be zero. The last significant figure will be rounded: ...37.... will be rounded to ...40....
The number of significant figures you need depends on the application.
If for example, that number is an amount in your bank account, then you want to know it to the last dollar, so you may not want to round at all.
If instead, you are looking at the speed of light trying to get a grasp on its magnitude, then two or three significant figures are enough. Thousands separators will help, as in 299,000,000 m/s. This particular example cries for even more rounding, to 300,000,000 m/s or better 300,000 km/s
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edited Jun 17, 2022 at 2:54
answered Jun 16, 2022 at 21:21
user338333user338333
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Your example can be useful as an exercise in rounding, but is not a good example of how significant figures are used.
A more typical example is a simple electrical circuit with a battery and a resistor. You find the voltage over the resistor be $U = 7.23\ \mathrm{V}$ and the current be $I = 1.37\ \mathrm{mA}$. You use Ohm's law to calculate the resistance and enter the values into your calculator getting $$ R = \frac{U}{I} = \frac{7.23\ \mathrm{V}}{1.37\ \mathrm{mA}} = 5.277372262\ \mathrm{k\Omega} $$ Then you realize that the calculator has rounded or truncated the value and there is in fact an infinite number of decimals. This makes you wonder how many decimals you should write down. So you start thinking...
After a while you realize that also the Volt and Ampere meters have rounded the values. The real voltage is something between $7.225$ and $7.235\ \mathrm{V},$ and the real current is something between $1.365$ and $1.375\ \mathrm{mA}.$ Therefore the true resistance is something between $$ R_\text{min} = \frac{U_\text{min}}{I_\text{max}} = \frac{7.225\ \mathrm{V}}{1.375\ \mathrm{mA}} = 5.254545454\ldots\ \mathrm{k\Omega} $$ and $$ R_\text{max} = \frac{U_\text{max}}{I_\text{min}} = \frac{7.235\ \mathrm{V}}{1.365\ \mathrm{mA}} = 5.300366300\ldots\ \mathrm{k\Omega}. $$ Obviously it's hardly meaningful to write more than one decimal: $5.3\ \mathrm{k\Omega}$ to which both $R_\text{min}$ and $R_\text{max}$ can be rounded. But the value is probably close to the mean value of $R_\text{min}$ and $R_\text{max},$ which is $5.27745587\ldots\ \mathrm{k\Omega}$ and which is quite close to what you calculated first. So perhaps $5.28\ \mathrm{k\Omega}$ is a good answer...? Or as an interval, e.g. a mean value and an error: $5.28\pm 0.023\ \mathrm{k\Omega}$?
It has become a practice to round the result to the minimum number of significant figures among the input values. It's a rule of thumb, not a mathematically exact rule. In real science a more rigorous analysis should be made.
In the given example, both input values have three significant figures so the rule says that we should use that also for the result, i.e. the answer should be given as $5.28\ \mathrm{k\Omega}.$
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answered Jun 20, 2022 at 13:59
md2perpemd2perpe
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$\begingroup$ What if I measure the volume of a cube of length=9cm. How sig fig is going to produce some good answer. I believe the answer would be 729cm³, while using sig fig it would be 700cm³. Are both acceptable? $\endgroup$
Level1
– Level1
2022-06-20 16:38:40 +00:00
Commented Jun 20, 2022 at 16:38
$\begingroup$ @user359206. How big is the error in your measure of the cube side length? Do you think that it's $0.5\ \mathrm{cm}$? I don't think so. Probably it's rather $0.05\ \mathrm{cm}$ (half a millimeter). That means that you have 2 significant figures and should answer with $730\ \mathrm{cm}^3.$ $\endgroup$
md2perpe
– md2perpe
2022-06-21 07:38:01 +00:00
Commented Jun 21, 2022 at 7:38
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2798 | https://ximera.osu.edu/mooculus/calculus3/lagrangeMultipliers/digInLagrangeMultipliers | We give a new method of finding extrema.
Throughout this course, we hope it has become apparent that when given a
problem:
There is more than one way to solve it.
The method of Lagrange multipliers tells us that to maximize a function constrained
to a curve, we need to find where the gradient of the function is perpendicular to the
curve.
Previously, when we were finding extrema of functions when constrained to some
curve, we had to find an explicit formula for the curve. Consider this example from
the previous section:
Let and let the setFind the maximum and minimum values of on .
The first step for solving this problem was to find an explicit formula that
drew the curve . In the case above, we choose:
However, finding a function
that draws the constraining set could be very difficult or even impossible!
If our constraining set had beenour previous method will not work, as
we (at least this author!) cannot find an explicit formula describing the set
above. Nevertheless, there is another way. It is called the method of Lagrange
multipliers. This method is named after the mathematician Joseph-Louis
Lagrange. This method relies on the geometric properties of the gradient vector.
Recall: There are three things you must know about the gradient vector:
.
points in the direction that one must leave in order to see the initial
greatest increase in .
points in the direction that is perpendicular to any level surface of .
It is the last two facts that we will think about now. Below we see level
curves for some function along with a constraining curve that we will call :
Let’s add vectors to our graph that point in the direction of . Since we know that the
gradient vector is perpendicular to level curves, we can do this without computation.
If for some point on the gradient points in the “general” direction of the tangent
vectors of , then cannot give an extremal value of , as moving along will either
increase or decrease the value of . Here’s the upshot:
The only candidates for local extrema occur when the
gradient of is perpendicular to .
How do we find these points? To do this, we will imagine that is a level curve for
some other function , and define as:
now, the candidates for extrema of when
constrained to a curve are found by finding such thatsince that satisfy this
equation are those where the gradient vectors of are perpendicular to the level curve
. This is the essence of the method of Lagrange multipliers.
Lagrange Multipliers Let , , , and let be the constraint, or level set,If has extrema
when constrained to at , thenfor some number .
The first step for solving a constrained optimization problem using the method of
Lagrange multipliers is to write down the equations needed to solve the
problem.
Let and let the setWrite down the three equations one must solve to find the
extrema of when constrained to .
First set . Now is the level curve . We must write
down:Unpacking these formulas we find:Solving these equations for and will give us our candidates for the extrema of on
the set .
The constraint curve used in the example above is called the lemniscate of
Bernoulli.
Working with geometry
Lagrange multipliers tell us that to maximize a function along a curve defined by ,
we need to find where is perpendicular to . In essence we are detecting geometric
behavior using the tools of calculus.
Below we have plotted a curve along with .Find the candidates for the maximum and minimum values for when restricted to .
At the candidates for the extrema, we know that the gradient vector of must beto the curve . Hence we see that the pointsandare our candidates for extrema.
Below we have plotted a curve along with .Find the candidates for the maximum and minimum values for when restricted to .
At the candidates for the extrema, we know that the gradient vector of must beto the curve . Hence we see that the pointsare our candidates for extrema.
Working with algebra
We’ll start with an example we did in our last section.
Let and let the setFind the maximum and minimum values of on .
Start by
setting . The set is now the level curve . Now compute:Write with me:Breaking
this vector equation into components, and adding in the constraint equation, the
method of Lagrange multipliers gives us three equations and three unknowns:To solve this system of equations, first note that if , then . This gives us
two candidates for extrema:Now proceed assuming that . Looking at the
second equationwe can divide both sides by to see that . Now we see that:Solving this quadratic equation for we findNow use the constraint equation to find
-values. From this we gain two more candidate for an extrema:Plugging these points
back into , we find that the minimum value of on is and the maximum value is
.
Lagrange multipliers help out when when constraint set is given by an implicit
function. Let’s see this in an example.
Let and let the setFind the maximum and minimum values of on .
Start by
setting . The set is now the level curve . Now compute:Write with me:Breaking this vector equation into components, and adding in the constraint
equation, the method of Lagrange multipliers gives us three unknowns:To solve this system of equations, first note that if , then . This gives us our first
candidate for an extrema:Now proceed assuming that . Looking at the
second equationwe can divide both sides by to see that . Now we see that:Solving this quadratic equation for we findNow use the constraint equation to find
-values. If is , then is not a real number, so we will discard this solution. If , then .
From this we gain two more candidate for an extrema:Plugging these points back
into , we find that the minimum value of on is and the maximum value is
.
The constraint curve used in the example above is called a Mordell curve. The
Mordell curve is a type of elliptic curve, a central object of study in number theory
and cryptography. For much more information on the Mordell curve, see this paper
by Keith Conrad.
The method of Lagrange multipliers gives a unified method for solving a large class of
constrained optimization problems, and hence is used in many areas of applied
mathematics.
For some interesting extra reading check out:
Unifying a Family of Extrema Problems, W. Barnier and D. Martin,
College Math Journal, November 1997.
An “Extremely” Cautionary Tale, M. Krusemeyer, College Math Journal,
March 2000.
Lagrange Multipliers Can Fail to Determine Extrema, J. Nunemacher,
College Math Journal, January 2003.
On the Genesis of the Lagrange Multipliers, P. Bussotti, Journal of
Optimization Theory and Applications, June 2003.
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2799 | https://unacademy.com/content/jee/study-material/mathematics/right-hand-rule-cross-product/ | An Overview on Right-Hand Rule Cross Product
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JEE Exam » JEE Study Material » Mathematics » Right-Hand Rule Cross Product
Right-Hand Rule Cross Product
In this article we will cover Right-hand Rule Cross Product, The direction of resultant vector, for the cross product of two non-parallel vectors is represented by right hand rule, direction of the resultant vector, resultant vectors is perpendicular to the vectors.
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The right-hand thumb rule for the cross-product of two vectors aids in determining the resultant vector’s direction. The orientation of a vector is the angle it makes with the x-axis, which is its direction. A vector is created by drawing a line with an arrow at one end and a fixed point at the other. The vector’s direction is determined by the direction in which the arrowhead is pointed. Velocity, for example, is a vector. It indicates the magnitude of the object’s movement as well as the direction in which the object is moving.
The right hand rule can be used to determine the direction of the cross product as follows:
The index finger is pointing in the velocity vector v direction. The magnetic field vector B is represented by the middle finger. The thumb is pointing towards the cross product F.
Cross Product
The cross product a b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a magnitude equal to the area of the parallelogram that the vectors span and a direction given by the right-hand rule. The cross product determines how far two vectors differ in their directions.
Right-hand Rule – Cross Product
The right-hand thumb rule for the cross-product of two vectors aids in determining the resultant vector’s direction. Our thumb will point in the direction of the cross product of the two vectors if we point our right hand in the direction of the first arrow and curl our fingers in the direction of the second. The cross product formula for determining the direction of the resultant vector is given by the right-hand thumb rule.
The right-hand rule can be used to determine the direction of the vector that results from the cross product of two vectors. To determine the direction of the cross product of two vectors, we use the following procedure:
Align your index finger to the first vector’s direction (A→)
Align the middle finger in the second vector B direction(B→)
Now the thumb points in the direction of the cross product of two vectors
We must first raise your right hand to see the right hand rule. If it’s our left, it won’t work. Hold index, middle, and thumb perpendicular to each other, as if you were using an x, y, and z coordinate system. Rotate our hand so that your index finger points in the vector a direction and our middle finger points in the vector b direction. The direction of the cross product a x b will be indicated by our thumb.
When calculating a cross product, be cautious about mixing up the vectors. The order matters, though: a x b does not equal b x a.
Which direction of the resultant vector?
The cross product can point in any direction while remaining at right angles to the other two vectors, so we have:
Point your index finger along vector a and your middle finger along vector b with your right hand: the cross product goes in the direction of your thumb.
The resultant’s direction can be determined by measuring the angle the resultant makes with the north-south or east-west vectors. The angle theta (Θ)is marked inside the vector addition triangle in the diagram to the right. Theta is the angle formed by the resultant with respect to west.
The direction of the resultant vector, for the cross product of two non-parallel vectors, is represented by the right hand rule.
Conclusion
In this article we conclude that Cross products are a type of “difference” measure that compares two vectors (in opposition to the dot product which is a measure of the “sameness” between two vectors). The magnitude of a cross product is proportional to the perpendicularity of the two vectors. For the cross product of two non-parallel vectors, the right hand rule is used to represent the direction of the resultant vector. The direction of the resultant vector is denoted by the thumb if we point our index finger along vector A and our middle finger along vector B.
Frequently Asked Questions
Get answers to the most common queries related to the IIT JEE Examination Preparation.
What is the cross-product right-hand rule?
Ans. Right hand rule represents the direction of the resultant vector for the cross product of two non-parallel vect...Read full
How do you figure out which way a cross product will go?
Ans. Point your right hand’s index finger along the first vector. Then, with your middle finger pointing along...Read full
Is the direction of the cross product important?
Ans. This is due to the fact that the cross product operation is non-communicative, which means that order is import...Read full
What is the direction of the cross product resultant vector?
Ans. The right-hand rule can be used to determine the direction of the resultant vector. The vectors are represented...Read full
What is the direction of two vectors' cross product?
Ans. The right hand rule is used to represent the cross product of two non-parallel vectors. The cross product a b i...Read full
Ans. Right hand rule represents the direction of the resultant vector for the cross product of two non-parallel vectors. The thumb represents the direction of the resultant vector if we point our index finger along vector A and our middle finger along vector B.
Ans. Point your right hand’s index finger along the first vector. Then, with your middle finger pointing along the second vector, orient your hand. Make a thumb extension. It indicates the cross product’s direction.
Ans. This is due to the fact that the cross product operation is non-communicative, which means that order is important. Switching the order of the inputs, for example, produces a result that is the polar opposite of your original calculation.
Ans. The right-hand rule can be used to determine the direction of the resultant vector. The vectors are represented by the thumb (u) and index finger (v) held perpendicular to one another, and the direction of the cross vector is indicated by the middle finger held perpendicular to the index and thumb.
Ans. The right hand rule is used to represent the cross product of two non-parallel vectors. The cross product a b is defined as a vector c that is perpendicular (orthogonal) to both a and b, with a magnitude equal to the area of the parallelogram that the vectors span and a direction given by the right-hand rule.
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