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12800 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7098569/ | Maternal gatekeepers: How maternal antibody Fc characteristics influence passive transfer and infant protection - PMC
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PLoS Pathog
. 2020 Mar 26;16(3):e1008303. doi: 10.1371/journal.ppat.1008303
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Maternal gatekeepers: How maternal antibody Fc characteristics influence passive transfer and infant protection
Stephanie N Langel
Stephanie N Langel
1 Duke Human Vaccine Institute, Duke University Medical Center, Durham, North Carolina, United States of America
2 Department of Pediatrics, Duke University Medical Center, Durham, North Carolina, United States of America
Find articles by Stephanie N Langel
1,2,, Claire E Otero
Claire E Otero
1 Duke Human Vaccine Institute, Duke University Medical Center, Durham, North Carolina, United States of America
2 Department of Pediatrics, Duke University Medical Center, Durham, North Carolina, United States of America
Find articles by Claire E Otero
1,2, David R Martinez
David R Martinez
3 Department of Epidemiology, University of North Carolina at Chapel Hill School of Public Health, Chapel Hill, North Carolina, United States of America
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3, Sallie R Permar
Sallie R Permar
1 Duke Human Vaccine Institute, Duke University Medical Center, Durham, North Carolina, United States of America
2 Department of Pediatrics, Duke University Medical Center, Durham, North Carolina, United States of America
Find articles by Sallie R Permar
1,2,
Editor: Matthew J Evans 4
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1 Duke Human Vaccine Institute, Duke University Medical Center, Durham, North Carolina, United States of America
2 Department of Pediatrics, Duke University Medical Center, Durham, North Carolina, United States of America
3 Department of Epidemiology, University of North Carolina at Chapel Hill School of Public Health, Chapel Hill, North Carolina, United States of America
4 Mount Sinai School of Medicine, UNITED STATES
I have read the journal's policy and have the following conflicts: SRP serves as a consultant for Pfizer, Sanofi, Moderna, and Merck vaccines and has a sponsored program on preclinical cytomegalovirus vaccine development with Merck and Moderna. All other authors declare no competing interests.
✉
E-mail: stephanie.langel@duke.edu (SNL); sallie.permar@duke.edu (SRR)
Roles
Matthew J Evans: Editor
Collection date 2020 Mar.
© 2020 Langel et al
This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
PMC Copyright notice
PMCID: PMC7098569 PMID: 32214394
Introduction
Maternal antibodies (MatAbs) passively transferred across the placenta and into breast milk are critical for protection against infectious disease and immune development during the first year of life . Passive transfer in the placenta and mammary gland (MG) is dependent on MatAbs binding to crystallizable fragment (Fc) receptors (FcRs) on polarized epithelial cells. For example, immunoglobulin G (IgG) transfers through the placenta by Fc domain binding to the Fc receptor neonatal (FcRn) on syncytiotrophoblasts , providing the fetus with a systemic source of protective IgG antibodies . Additionally, maternal dimeric immunoglobulin A (dIgA) antibodies transfer into breast milk by binding to the polymeric immunoglobulin receptor (pIgR) on MG epithelial cells through the antibody joining chain (J-chain) and provide immune protection in the gut while shaping microbiota colonization [4,5]. Yet MatAbs can interfere with the neonatal immune response, particularly after vaccination . This Pearl explores the role of monomeric IgG, the only antibody isotype to cross the placenta, and polymeric IgA, the major antibody species in breast milk, and their Fc domain characteristics on passive transfer to and functional activity in the newborn.
The IgG Fc domain and its effector functions in the context of MatAb passive transfer
Antibodies contain 2 domains that exert a wide range of effector functions. The antigen-binding fragment (Fab) domain binds foreign antigens and drives antibody diversity , whereas the Fc is responsible for initiating innate immune cell activation and passive antibody transfer . The classical FcRn-driven IgG transport mechanism is responsible for shuttling IgG within acidified endosomes across the syncytiotrophoblast cell barrier from maternal to fetal circulation (Fig 1A) . Once in the neonate, the IgG Fc domain can engage classical type I Fc gamma (Fcγ) receptors (activating [FcγRI, FcγRIIa, FcγRIIc, FcγRIIIa, FcγRIIIb]; inhibitory [FcγRIIb]) or complement to mediate nonneutralizing functions like antibody-dependent cell-mediated cytotoxicity (ADCC) and antibody-dependent cellular phagocytosis (ADCP), or complement-dependent cytotoxicity (CDC), respectively (Fig 1A) . Nonclassical type II FcRs are C-type lectin receptors, including CD209 (DC-SIGN) and CD23, which bind IgG to facilitate immune complex formation . Considering each family of Fc receptors initiates distinct effector functions, the diversity of the Fc domain allows tailoring of nonneutralizing Fc-mediated activity to protect against viruses like HIV, influenza, and cytomegalovirus [10–12]. Alternatively, pathogens such as dengue virus utilize complement and FcR pathways for antibody-dependent enhancement of disease .
Fig 1. Maternal antibody passive transfer and functional activity in the neonate.
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(A) IgG passive transfer in the placenta influences FcγR-mediated cell cytotoxicity, phagocytosis, and complement activation in the developing fetus/newborn. (B) IgA passive transfer in the mammary gland results in FcαR- and IgA-mediated cell activation and microbiota regulation, respectively. Fab, antigen-binding fragment; Fc, crystallizable fragment; FcαR, Fc alpha receptor; FcRn, Fc receptor neonatal; FcγR, Fc gamma receptor; IgA, immunoglobulin A; IgG, immunoglobulin G; J-chain, joining chain; pIgR, polymeric immunoglobulin receptor.
The IgG Fc domain mediates considerable heterogeneity of its effector functions depending on the subclass and glycan profile. For example, each IgG subclass (IgG1-4) has one N-glycosylation site in each CH2 domain, an important binding site for FcγRs (Fig 2). Interestingly, there are up to 36 possible antibody glycan profiles that could theoretically be present on each CH2 domain. This allows for combinatorial diversity of the Fc region with 144 different potential functional states for the 4 IgG subclasses . This is relevant in the context of maternal–fetal immunity, as FcRn has different binding affinities to each IgG subclass, which may reflect their placental transfer efficiency . Additionally, recent data suggest that Fc glycan profiles create antibody transfer hierarchies in the placenta of both healthy and HIV-infected pregnant women. For example, in healthy pregnant women, there is a shift toward IgG galactosylated antibodies, which have higher FcRn-binding affinity, are more efficiently transferred across the placenta, and enhance natural killer (NK) cell degranulation and chemokine secretion . Additionally, binding of tetanus toxoid–specific IgG to placental FcγRIIa H131, FcγRIIa R131, and FcγRIIIa F158 (but not canonical FcRn) was positively associated with placental IgG transfer efficiency in HIV-infected women, suggesting that noncanonical placental FcRs may also play a role in IgG placental transfer [17,18]. Fc-mediated differential selection of IgG antibodies in the placenta is likely an adaptive evolutionary mechanism to passively transfer the most effective antibodies to the infant, which can be altered by disease status.
Fig 2. Schematic representation of IgA and IgG glycosylation.
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N-linked glycosylation is depicted as yellow circles, whereas O-linked glycosylation is depicted as green stars. IgA, immunoglobulin A; IgG, immunoglobulin G; sIgA2, secretory IgA.
Do IgA Fc region characteristics influence IgA passive transfer or effector function in breast milk?
IgA antibodies bind their own unique Fc receptors that facilitate epithelial cell transcytosis and innate immune cell activation. dIgA antibodies are composed of 2 monomers, linked by a 15-kDa J-chain. Transport of dIgA into breast milk is dependent on C-terminal binding of the J-chain to a portion of pIgR, known as the secretory component, on the basolateral surface of MG epithelial cells . Without the J-chain, IgA antibodies are secreted as monomers and are not actively transported across the mucosal epithelium . After transport of the J-chain/pIgR complex to the apical portion of the cell, pIgR is cleaved, releasing secretory IgA (sIgA) into breast milk and other mucosal fluids (Fig 1B) . IgA also binds to Fc alpha receptor (FcαR) on the surface of myeloid cells. Monomeric serum IgA induces inhibitory signals, whereas IgA immune complexes have increased avidity to and cross-link FcαRI, resulting in proinflammatory responses . The dominant immunoglobulin class in breast milk sIgA has decreased affinity for FcαRI likely due to steric hindrance from the attached secretory component . Although the opsonic activity of sIgA is poor compared with monomeric and dIgA , sIgA can initiate macrophage phagocytosis and neutrophil respiratory burst [26,27]. Further defining the anti- and proinflammatory effects of IgA subclass–FcR interactions would allow fine tuning of breast milk immunity and may represent an attractive therapeutic strategy.
Considering breast milk sIgA protects from pathogenic insult and facilitates maturation of the microbiota in early life , understanding breast milk antibody Fc–mediated effector functions is integral to neonatal intestinal health (Fig 1B). This is further supported by the fact that bacteria have evolved mechanisms to block IgA and FcαR interactions . Additionally, it is likely that the more complex and extensive glycosylation pattern of IgA antibodies (Fig 2) impacts effector function in milk. Indeed, mucosal secretions, including breast milk, contain mostly IgA2 , which has 2 [IgA2m(1)] or 3 [IgA2m(2)] additional conserved N-glycans compared with IgA1 , which dominates in serum . Recent evidence demonstrates that IgA glycan–bacteria interactions regulate gut microbiota composition and metabolism as well as retrograde transport of sIgA immune complexes back to the lamina propria independent of antibody–epitope interactions [33–35]. Additionally, the sialic acid in IgA antibody’s C-terminal tail competes with receptor binding to some viruses, providing an innate line of defense against infection . This is relevant to breast milk IgA passive transfer, considering that the leading causes of severe pediatric gastroenteritis worldwide (rotavirus and norovirus) both utilize sialic acid receptors for intestinal infection [37,38]. Studies are needed to define the mechanisms of Fc-mediated IgA effector functions in breast milk, including their interactions with the developing infant microbiome and protection against intestinal viral infections.
MatAb interference is influenced by MatAb Fc domain–receptor interactions
Despite the well-documented benefits of MatAbs on early life immunity , a mounting body of evidence indicates that MatAbs can inhibit immune responses to certain infant vaccinations [6,39]. A recent meta-analysis demonstrated that MatAbs acquired transplacentally inhibited antibody responses to priming vaccinations and these effects were not overcome by administration of a booster dose . This highlights the “window of susceptibility” that exists for infants when MatAbs are not high enough for seroprotection yet still interfere with infant vaccine responses.
Multiple mechanisms have been proposed to describe both Fab- and Fc-mediated MatAb interference. These include live virus vaccine neutralization, inhibition of B-cell responses by epitope masking , and IgG Fc binding to FcγRIIB . Kim and colleagues demonstrated that B-cell responses to a live attenuated measles vaccine were inhibited by passively transferred measles-specific IgG antibodies in a FcγRIIB-dependent manner, suggesting that IgG Fc region characteristics contribute to suppression of the immune response . Additionally, removing the glycans from IgG2b abolished its immunosuppressive activity both in vitro and in vivo [42,43]. Considering that the mechanisms of MatAb interference likely differ depending on vaccine type (live attenuated, inactivated, subunit), route of delivery (oral, subcutaneous [SQ], intramuscular [IM]), and adjuvant formulation, defining glycan-dependent passive transfer of maternal IgG antibody subclasses in the placenta is crucial for developing effective maternal immunization strategies. Although less studied, IgA antibodies in breast milk may also contribute to interference of immune responses to oral infant vaccines such as rotavirus and poliovirus [44,45]. Indeed, the 2 oral rotavirus vaccines Rotarix and Rotateq demonstrate lower efficacy and immunogenicity in infants from some low- and middle-income countries (LMICs) [46,47] where women tend to have higher titers of antirotavirus IgA antibodies and neutralizing activity in milk [48,49]. The high rotavirus neutralizing activity in breast milk of women from LMICs may partially explain the decrease in rotavirus vaccine efficacy; however, IgA MatAb interference is not well defined .
Although infant CD4+ T-cell responses are mostly unaffected by MatAb interference [6,39], recent work demonstrated that MatAbs dampen mucosal T-cell responses against commensal bacteria and limit the expansion of T follicular helper (T FH) cells in the germinal center (GC) . The premature decline in GC T FH cells resulted in the reduction or prevention of plasma cell and memory B-cell generation in a MatAb- and antigen dose–dependent manner . Interestingly, at low or intermediate titers, MatAbs did not prevent the induction of memory B cells, suggesting a gradient effect of MatAbs on infant immune responses . Defining the functional consequences of MatAb gradients will be essential for infant vaccine design and immunization timing. For example, and in addition to previously discussed mechanisms, there is evidence that preexisting antibodies can promote higher affinity antibody responses due to competitive binding in the GC and increased uptake and antigen presentation through immune complexes (ICs) in a Fc glycosylation–dependent manner [53,54]. However, more research is needed to determine the effects of these mechanisms in the setting of passively transferred MatAbs and infant GC responses.
Harnessing MatAb Fc region characteristics and receptor interactions to fine-tune maternal immunizations that maximize infant protection
Passive transfer of MatAbs is central to pathogen protection and immune system development in early life. However, MatAb-mediated interference dampens antibody responses to vaccinations, leaving children more susceptible to infections while increasing transmission rates to unvaccinated cohorts. Recent work demonstrated that maternal IgG antibodies are differentially transferred across the placenta in an Fc glycan–dependent manner [16,17]. Considering vaccination strategies could direct antigen-specific antibody glycosylation , defining MatAb glycan profiles represents an adaptable and powerful mechanism to fine-tune maternal immunizations that maximize infant protection while limiting MatAb interference. Future studies are needed to determine (1) how maternal vaccination and their distinct adjuvant mixtures alter IgG Fc domain glycosylation and whether certain glycan profiles are associated with IgG Fc-mediated MatAb interference and (2) whether or not the IgA Fc domain regulates passive transfer in the MG or effector function in breast milk. Defining the molecular mechanisms of Fc-mediated functional activity at the maternal–fetal/neonatal interface is critical for developing next-generation maternal vaccines and antibody-based therapeutics to improve the health of the mother–neonatal dyad.
Funding Statement
SNL is supported by an NIH National Institute of Allergy and Infectious Diseases (NIAID: Ruth L. Kirschstein National Research Service Award T32 AI007392. CEO is supported by an NIH NIAID Ruth L. Kirschstein National Research Service Award T32 CA009111. DRM is supported by an NIH NIAID Ruth L. Kirschstein National Research Service Award T32 AI007151 and a Burroughs Wellcome Postdoctoral Enrichment Award (www.bwfund.org). SRP is supported by NIH NIAID R01 AI122909, P01 AI129859, and P01 AI117915. The funders had no role in study design, data collection, analysis, decision to publish, or preparation of the manuscript. The content is solely the view of the authors and does not necessarily represent the official views of the National Institutes of Health.
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12801 | https://www.med.tu.ac.th/department/surgery/wp-content/uploads/sites/3/2022/04/Medulloblastoma-case-NeuroSx.pdf | Medulloblastoma: A case study and literature review Authors Tipok Vivattanasarn 5929640208 Peemmawat Phanichwong 5929640240 Lapol Herabat 5929640190 Nichapa Lerthirunvibul 5929640273 Pimpakarn Trarungreang 5929640158 6th year medical students Chulabhorn International College of Medicine (CICM) Department of Neurosurgery, Thammasat University Hospital CASE INFORMATION Case: 16 year old Thai female from Lopburi Chief complaint: Ataxia 3 weeks Present illness: Three weeks prior to the admission, the patient missed a step while going down the stairs leading to a laceration wound on her right hand. She attributed the cause to uncoordinated body movements as she expresses that her leg movements were slower than usual. She did not experience any headache or dizziness, loss of consciousness, memory loss, weakness, or seizures.
Two weeks prior to the admission, the patient started developing generalized bilateral headaches described as dull-aching that lasted all day long. The headache was aggravated by eye straining and reached a pain score of 7-8 out of 10, which was not alleviated by paracetamol and NSAIDs. Associated symptoms included postprandial nausea and vomiting along with decreased appetite, new onset of persistent bilateral blurry vision of distant objects, and seeing bouncing images also known as oscillopsias. She denied previous problems with her eyesight and never needed corrective lenses. In addition, as a naturally fast speaker, she complained of dysarthria and a significantly slower speech. She also developed ataxia and unsteady gait, but was able to ambulate with a walker. She noticed that her gait became slower and her step length was shorter.
Lastly, she also had difficulties writing and carrying things with her hands as it was constantly shaking on movement; her left hand was more affected (non-dominant hand). She complained about appetite loss but she did not weigh herself recently. There were no symptoms of vertigo, tinnitus, facial asymmetry, hearing abnormalities, sensory impairment, muscle weakness nor fever. She also reported recently to have urinary incontinence with the urge to void but failed to do so before being able to reach the toilet.
She had no known underlying disease and no current medications or herbal drug use. She had a history of combined oral contraceptive use for 2 months, but stopped using 10 months ago.
She denied a history of trauma or surgery, and smoking or alcohol drinking. There was no family history of brain tumor. She is currently a student in her last year of high school with good academic performance. She lives with her mother, aunt and younger sister who do not have similar symptoms. There is no family history of cancer. The family owns a woodworking supply store.
Physical examination: -V/S : BP 129/90 mmHg, PR 96 bpm, RR 18/min, BT 36.7, SpO2 98% -Body weight: 44 kg, Height: 155 cm -GA : Alert, good consciousness, well-cooperative -HEENT : No pale conjunctiva, anicteric sclera, no palpable cervical lymph nodes -CVS: Normal S1S2, no murmur, capillary refill < 2 seconds, full and regular pulse -RS: Clear and equal breath sound both lungs -Abdomen : soft, no guarding, not tender -Skin: no mass, no rash -Neurological examination: -Consciousness: -Alert, oriented to time/place/person, GCS E4V5M6 -Cranial nerves: -CN II, III : Pupil 4 mm RTL right eye, sluggishly RTL left eye, RAPD negative, VF intact, VA 20/200 both eyes (pinhole was not available), fundoscopy: cannot see optic disc and cup due to voluntary eye movements, horizontal diplopia -CN III, IV, VI : Full EOM, no ptosis, saccadic pursuit -CN V : equal sensation of V1, V2, V3, intact masseter muscle both sides -CN VII : Symmetrical facial movements, orbicularis oculi and orbicularis oris muscle is intact -CN VIII : Equal hearing both ears by finger rubs, Weber and Rinne tests were not tested -CN IX, X : Decreased gag reflex, uvula at midline -CN XI: Intact trapezius muscle both sides -CN XII : No tongue deviation -Motor: grade V all extremities, decreased muscle tone -Sensory: intact all extremities -Proprioception: normal sense of distal joints -Reflexes: -DTR 2+ all extremities -Babinski’s sign: plantarflexion both sides -Cerebellar signs -Finger to nose test: positive left side > right side -Heel to shin test: positive left side > right side -Nystagmus : 3rd degree jerking (Horizontal) nystagmus of both eye -Truncal ataxia to the left -Ataxic gait to the left -Dysdiadochokinesia positive left upper extremity -Dysmetria of left upper extremity -No wide-based gait, tandem gait not performed -Overshoot test: positive left > right -Rebound phenomenon: positive left > right -Slow and slurred speech Pertinent findings: ● Ataxia 3 weeks ● Headache 2 weeks ● Ataxic Dysarthria 2 weeks ● FTN, Heel to shin, dysdiadochokinesia positive ● 3rd degree jerking horizontal nystagmus Problem list: ● Ataxia with cranial nerve involvement and cerebellar signs Localisation The patient initially presented with truncal ataxia (unsteadiness on standing walking and sitting with broad-based gait with both eyes open and closed as this can be attributed to the lesion initially originating at the cerebellar vermis (Geraint et al., 2010). Later, the patient started noticing her limb dysmetria (handwriting change, unsteady hands on carrying or grabbing object) 1 week after her truncal ataxia symptoms. According to physical examination, her left limb was significantly affected compared to her right limb suggesting that the lesion was on the left of the cerebellum hemisphere (Geraint et al., 2010). This is true for both dysmetria which is overshooting, rebound phenomenon, abnormal finger to nose and heel to shin test, and dysdiadochokinesia in which the left hand is unable to perform alternating movement when compared to the right hand of the left extremities when compared to the right. The progressive nature of the symptoms from truncal ataxia to limb dysmetria suggests that the lesion originated from the cerebellar vermis and began to grow outward, which is consistent with a growing mass lesion. Nonetheless, early onset of autoimmune diseases such as multiple sclerosis or infection such as acute viral cerebellitis could not be excluded.
Other possible locations that can cause ataxia include sensory and vestibular ataxia, which can be difficult to distinguish. The differences between proprioception and cerebellar ataxia can be seen from physical examination such as gait and Romberg’s sign. With the impairment of proprioception, patients tend to walk with a stepping gait by raising their legs high to compensate for their proprioception with visual input (Zhang Q, et al., 2021). On the contrary, in cerebellar ataxia, the gait will be more clumsy, and is unable to walk straight (Zhang Q, et al., 2021). Moreover, the Romberg test may help differentiate between cerebellar ataxia and sensory ataxia. By standing with eyes open there is visual compensation to maintain postural stability in sensory ataxia cases(Zhang Q, et al., 2021). This is not the case for cerebellar ataxia as the patient can not stand even with eyes open. By closing the eye it eliminates the visual input leaving only proprioception to maintain balance, which instability can be seen in both sensory ataxia and cerebellar ataxia cases (Zhang Q, et al., 2021). The presence of nystagmus is also suggestive that the lesion does not involve sensory ataxia. On the other hand, vestibular ataxia can also present with ataxia and nystagmus, which are within the differential diagnosis for the location. Vestibular ataxia can be separated into two types including central and peripheral vestibular ataxia, which can be distinguished by performing the head impulse, the characteristic of nystagmus, and the test of skewed. In addition, the negative head impulse test, intact vestibulo-ocular reflex, rules out the possibility of peripheral vestibular ataxia. However, in this case, the vestibulo-ocular reflex is intact, which can be seen in both central vestibular and cerebellar ataxia. In addition, the 3rd-degree horizontal nystagmus, in this case, can be seen in both central vestibular ataxia and cerebellar ataxia. Whereas peripheral nystagmus is often fatigable, unidirectional nystagmus, suppressed by fixation and is not constant when compared to central (Strupp M, et al., 2011). The absence of tinnitus, hearing loss, and other cranial nerve involvement may also support that the lesion does not involve the vestibular nucleus. Moreover, vestibular ataxia does not have limb dysmetria and ataxic dysarthria.
Her speech patterns are monotonous with abnormal separation, some variability between the pitch and intensity, which is prominent for ataxic dysarthria. The cerebellar vermis and paravermis, especially the superior cerebellar hemisphere bilaterally, are responsible for ataxic dysarthria speech (Spencer K and Slocomb D, 2017).
Nystagmus is also helpful in localizing the lesion, a third-degree bilateral nystagmus is suggestive that the lesion is located within the central nervous system. The inability to inhibit visual fixation is also supportive evidence that the lesion is central. The normal response for the head impulse test also ruled out the peripheral cause of nystagmus. Since there is also saccadic pursuit along with gaze-evoked nystagmus and impaired visual fixation suppression of nystagmus the lesion can be localized at the flocculus to paraflocculus region of the cerebellar (Strupp M, et al., 2011). However, the absence of the pathognomonic down-beating nystagmus does not rule out the lesion at the flocculus area as it can be found in 80% of the cases that involve the flocculus region (Strupp M, et al., 2011). The absence of isolated horizontal saccadic paralysis excluded the lesion at the dorsal pons as it does not involve the cranial nerve, such as facial and abducens or the PPRF region.
Two weeks before her admission, her constant progressive headache along with nausea and vomiting can be attributed to the progressiveness of the cerebellar lesion that obstructs the 4th ventricle causing an increase in intracranial pressure. Furthermore, the recent onset of urinary incontinence is also suggestive of hydrocephalus as it stretches the medial frontal cortex causing hyperactive detrusor muscle from absent or reducing central inhibition from the brain (Rendtorff R, et al., 2012). However, if the incontinence is from the pontine micturition center in the dorsal aspect of the pons, then the patient should not feel the urge to void causing symptoms of neurogenic bladder (Rahman M., et al., 2021).
Differential diagnoses Differential diagnose`s Supportive evidences Opposing evidences Cerebellar mass lesion: progressive cerebellar ataxia with sign of increased ICP Medulloblastoma (Geraint et al., 2010) -Progressive outward growth from cerebellar vermis to cerebellar hemisphere -usually extend into 4th ventricle -developing over a few weeks -Age group commonly at 3-4 years old then 8-9 years old Cerebellar Pilocytic astrocytoma (Knight J. 2021) (Geraint et al., 2010) -Age group commonly before 20 years old -Progressive outward growth from cerebellar vermis to cerebellar hemisphere -usually extend into 4th ventricle -Usually develop gradually over many months Hemangioblastoma (Geraint et al., 2010) (Ibrahim D and Gaillard F., 2008) -Progressive outward growth from cerebellar vermis to cerebellar hemisphere -Can develop hydrocephalus from obstruction -Age group is middle aged, 30 -60 years old -commonly originate in the cerebellar hemisphere Infection: progressive cerebellar ataxia with headache (sign of increased ICP in abscess is possible) Cerebellar abscess (Shaw MD and Russell JA., 1975) -Progressive outward growth from cerebellar vermis to cerebellar hemisphere -usually extend into 4th ventricle -developing over a few weeks -Absent risk factors such as cyanotic heart disease, chronic sinusitis or otitis media, dental caries, etc.
-fever and stiff neck negative but can not rule out as it is not sensitive Acute Viral/ post viral cerebellitis (Yildirim M, et al., 2020) (Van Samkar A, et al., 2017) -Compatible age group -Symptoms is compatible (ataxia, headache, -vomiting, nystagmus, dysmetria) -Fever are rarely seen present in (47%) -In history of recent viral infection -Usually patient would recover within weeks (7-15 days) - headache should present at the same time as ataxia.
Autoimmune: progressive cerebellar ataxia with compatible age Multiple sclerosis (Clinically isolated syndrome) (Efendi H. 2016) -Age is compatible -Can present with cerebellar sign of ataxia, dysmetria, multidirectional nystagmus -attack usually persist 2-3 weeks - can not rule out in first attack of the disease -Should not have sign of increase ICP (ex. headache) Congenital cause: progressive cerebellar ataxia Congenital Cerebellar cyst (Bosemani T, et al., 2015) -Progressive outward growth from cerebellar vermis -can potentially cause hydrocephalus -Usually found in patient that are younger than the age of 1 years old Autosomal Recessive ataxia (Class III: Adolescent-Onset Ataxia) (Fogel BL. 2012) -Age group is compatible before age of 20 -Present with gait ataxia -should also have cutaneous manifestation - limb dysmetria should be spared -does not have sign of increase ICP -rare Vascular cause: subacute cerebellar ataxia due to acute on top chronic due to low flow. (very unlikely) Posterior circulation stroke (Barinagarrementeria F., 1997) -Subacute onset can be from low flow.
- clinical of dysmetria and ataxia with nystagmus -Should present with other cranial nerve involvement such as facial nerve or dysphagia with hoarseness.
-Numbness of the face should be ipsilateral to the ataxia with contralateral hemiparesthesia.
-no atherosclerosis risk factor, illicit drug use and hypercoagulability risk factor -no history of trauma to cause dissection -age is too young Table 1: List of differential diagnoses of adolescent with cerebellar ataxia Management: Investigations included CT brain non-contrast which revealed an isodense lesion with dimensions of 4.4 x 3.7 x 4.1 cm at cerebellar vermis involving both cerebellar hemispheres, more predominantly on the left, with compression of the fourth ventricle and dilatation of the third ventricle and both lateral ventricles, as shown in Fig. 1. Later, an MRI scan of the brain was completed showing a 3.4 x 3.6 x 4.0 cm hyperintense lesion at cerebellum involving the roof of fourth ventricle with obstructive hydrocephalus and compression of medulla oblongata, as shown in Fig. 2. An MRI of the whole spine was also included and showed no spinal lesions indicating no spinal metastasis. These findings suggest that the patient has medulloblastoma that causes mass effect and obstructive hydrocephalus, which explain all the symptoms that the patient has.
Other differential diagnoses include astrocytoma and ependymoma. Neurosurgeon was therefore consulted for proper management.
Figure 1. CT brain without contrast in axial and sagittal view.
Figure 2. MRI T2-weighted image in axial and sagittal view.
● Definite treatment ○ Suboccipital craniotomy with tumor removal ○ Radiation and chemotherapy post operation ● Supportive treatment ○ Seizure prophylaxis: Phenytoin 100 mg IV q 8 hours ○ Brain edema control: Dexamethasone 4 mg IV q 6 hours ○ Pain control: Morphine 3 mg IV prn q 4 hours ● The patient underwent operation on 25/11/64 ● Resected tissue was sent for pathological analysis; pending result.
● Postoperative: ○ Patient is alert (E4VtM6) but still intubated due to respiratory tract infection ○ Cerebellar sign cannot yet be investigated due to poor cooperation ○ CT Brain Contrast was done on postoperative day 1 which revealed a 1.6 x 1.5 x 1.5 cm enhancing lesion at cerebellar vermis, likely residual tumor with intraventricular hemorrhage and hydrocephalus, as shown in Fig. 3.
○ MRI pending Figure 3. CT brain without contrast in axial and sagittal view on postoperative day 1.
MEDULLOBLASTOMA INFORMATION Introduction: Medulloblastomas are the most common malignant pediatric brain tumor and occur primarily in the cerebellum (McNeil et al., 2002). Medulloblastomas are embryonal tumors and are classically described as the most common type of primitive neuroectodermal tumor (PNET).
This is due to the fact that they are histologically identical to tumors (pineoblastoma, neuroblastomas, and retinoblastomas) located in other locations that are believed to have derived from progenitor subependymal neuroepithelial cells undergoing cancerous transformation.
Patients have the manifestation of increased ICP symptoms and medulloblastoma is most commonly present in the cerebellum.
Epidemiology: Among the primary tumors of the central nervous system in young people under 19 years old, medulloblastoma has the highest prevalence, contributing to about 10 percent. Most occur in the first decade of life; peak incidence is between five and nine years of age, but there is a second peak around age 30. The diagnosis of medulloblastoma is made in nearly 70 percent of patients aged less than 20 years old. Due to its embryonic origin, medulloblastoma is atypical after the fourth decade of life (Roberts et al., 1991).
Risk Factors: Chromosomal abnormalities are consistent with almost half of medulloblastomas, notably the deletion of 17p chromosome that contains the tumor suppressor gene TP53 (Raffel et al., 1993). Mutations in these genes lead to an incline in the evolution of medulloblastoma via defects in pathways essential in the pathogenesis of both sporadic and inherited tumors.
Certained genetic conditions are corresponding with a greater risk of developing medulloblastoma which include the followings: Nevoid basal cell carcinoma syndrome (NBCCS), Turcot syndrome (a subtype of familial adenomatous polyposis (FAP) or Lynch syndrome), and BRCA1 gene mutations (Waszak et al., 2018).
Clinical Presentation: Medulloblastoma patients usually have the components of intracranial pressure and cerebellar abnormalities developing over a duration of weeks to a few months. Magnetic resonance imaging (MRI) typically demonstrates an enhancing midline or paramedian cerebellar mass are typically demonstrated in Magnetic resonance imaging (MRI). Also, there is evidence of tumor dissemination through the subarachnoid space either by imaging or cerebrospinal fluid (CSF) examination in approximately one-third of the patients (Smoll & Drummond, 2012).
Symptoms — Patients usually present to the hospital with the chief complaint of increased intracranial pressure symptoms, including nocturnal or morning headaches, nausea, vomiting, and alteration of mental status. The midline tumors may result in gait ataxia or truncal instability. On the other hand, limb clumsiness or incoordination result from tumors located in the lateral cerebellar hemispheres. Dizziness and double vision are common symptoms that can be caused by cerebellar, brainstem, or cranial nerve involvement.
Physical examination —The position of the mass in the posterior fossa contributes to various neurologic exam findings. Truncal or gait ataxia such as a broad-based gait or trouble with heel-to-toe walking, head titubation and nystagmus are usually the evidence of midline tumors. A demonstration of dysmetria on finger-to-nose testing, intention tremor, and problems with heel-to-shin testing in patients with lateral cerebellar tumors.
Cranial nerve deficits may occur in combination with the aforementioned signs, either from direct involvement of specific nerves or from cranial nerve dysfunction due to increased intracranial pressure. For instance, elevated intracranial pressure can induce dysfunction of the abducens nerves, leading to diplopia from lateral gaze palsy. Prolonged elevation of intracranial pressure can contribute to papilledema and complete or partial loss of vision.
Definitive diagnosis: Histopathologic confirmation obtained from surgical resection is required for a diagnosis of medulloblastoma. Due to the fact that maximal safe resection is an integral part of the management of medulloblastoma as well as other posterior fossa tumors, biopsies are not routinely performed in suspected cases by imaging (Thompson et al., 2018).
Investigation Other than history taking and physical examinations, further investigations are helpful in identifying the lesion, localising the mass, performing staging, and planning for further treatment. Magnetic Resonance Imaging (MRI) of the brain with gadolinium (gd) contrast media is the investigation of choice in medulloblastoma. Other investigations that should be done would be MRI spine and cerebrospinal fluid (CSF) analysis for cytology. Investigations that can be done in adjunct would be collecting blood for laboratory work up to find supportive evidence of secondary causes of the patient’s clinical presentation but are not significant in the diagnosis and staging of medulloblastoma. As for CT scan with contrast, it is another imaging that could be done since it will show a hyperdense lesion if a mass is identified. However, it could not be used to identify leptomeningeal seeding.
Investigation Findings Result interpretation MRI brain T1: Hypointense lesion T1 with gd: Hyperintense lesion T2: Similar intensity to grey matter Identify mass presence and location Exclude subarachnoid metastasis MRI spine T1 with gd: Hyperintense lesion along spinal cord Helps include/exclude leptomeningeal seeding along the spine CSF cytology Presence of tumor cells in CSF analysis suggests at least M1 stage in Chang staging system Table 2: Summary of investigations done when medulloblastoma is suspected Figure 4. MRI spine T1 fat suppressed in the sagittal view showing drop metastasis of medulloblastoma along the spinal cord at the L1 level (image courtesy of Radiopedia) Differential diagnosis of posterior fossa tumor Once an MRI was done, and a mass can be seen at the posterior fossa, a handful of differential diagnoses could be made depending on the age and setting of the patient. Some example of such tumor are listed below: Age group Differential diagnoses Comments Children Medulloblastoma Most common Astrocytoma ● Pilocytic astrocytoma ● Brainstem glioma Teratoma Ependymoma Rhabdoid tumor More common in infants Adult Metastatic tumor Hemangioblastoma Astrocytoma Medulloblastoma Ependymoma Lymphoma Lipoma Schwannoma Most common Most common primary Young adults Young adults Young adults Table 3: Differential diagnoses of posterior fossa tumor based on age group Subgroups Medulloblastoma has 4 subgroups that it could be divided into, namely Wnt activating, SHH (Sonic Hedgehog) activating, Group 3, and Group 4. These subgroups were brought about by a consensus conference in Boston in the year 2010 . An agreement was reached and 4 main subgroups were identified based on factors such as their clinical features, genetic buildup, and their gene expression. It is believed that each subgroup could be further divided into subsets but they are not well characterized yet.
Subtype Abnormal expression Immunomarker Metastasis % Wnt WNT pathway activation c-MYC DKK1 0% SHH SHH pathway activation c-MYC SFRP1 7% Group 3 Photoreceptor pathway Neuronal differentiation c-MYC NPR3 75% Group 4 Neuronal differentiation KCNA1 31% Table 4: Molecular characteristics of each subtypes of medulloblastoma (Monje et. al., 2011) Subtype Age Location Clinical manifestation Male: Female Wnt Evenly distributed Cerebellar Peduncle Limb dysmetria, Truncal ataxia, defective proprioception 1 : 1.7 SHH < 3 y.o. and > 16 y.o.
Cerebellar Hemisphere Cerebellar Vermis Ipsilateral limb dysmetria Truncal ataxia 1: 0.5 Group 3 3 - 10 y.o.
Cerebellar Vermis Truncal ataxia 1: 0.7 Group 4 Evenly distributed Cerebellar Vermis Truncal ataxia 1: 0.5 Table 5: Clinical characteristics of each subtypes of medulloblastoma (Dofour et.al., 2011) Once the diagnosis is made, past beliefs suggest that staging should be done in order to predict the prognosis of the disease and to help plan for the treatment of choice for each patient.
A widely used staging criteria is the Change Staging for Medulloblastoma. The two main criteria that the Chang staging used are the tumor size and its extent of metastasis. This is summarized in the table below. However, modern approach to selection of treatment is based on risk stratification rather than Chang staging, hence this staging system is not as useful in planning the patient’s treatment, but rather help to understand the progression of the tumor instead.
Chang staging for medulloblastoma T1 Tumor diameter < 3 cm T2 Tumor diameter ≥ 3 cm T3a Tumor diameter > 3 cm with extension T3b Tumor diameter > 3 cm with unequivocal extension into brainstem T4 Tumor diameter > 3 cm with extension past aqueduct of Sylvius and/or past foramen magnum M0 No metastasis M1 CSF analysis shows tumor cells M2 Tumor extends beyond primary site M3 Gross nodular seeding in spinal subarachnoid space M4 Metastasis beyond the cerebrospinal axis Table 6: Chang Staging for Medulloblastoma In the pediatric population with medulloblastoma, risk categorization is done in order to help predict the prognosis of the disease and to determine the treatment regimen for each patient.
The risk is separated into 2 groups; average risk and high risk, as summarized in table below.
Risk Category Average (must fulfill all of the following) > 3 years of age No metastasis detected Less than 1.5cm2of residual mass High (at least 1 of the following) < 3 years old Overt metastasis based on CSF cytology Seeding identified on MRI More than 1.5cm2 of residual mass Table 7: Risk category of childhood medulloblastoma Treatment Treatment of medulloblastoma mainly depends on the patient’s risk group and age. For the best possible outcome, gross total resection (GTR) of the tumor has shown to lead to better prognosis compared to other methods. However in actual practice, GTR could not be regularly done due to the risk of extensive damage done to nearby normal brain parenchyma, hence it was suggested that the surgical resection done should be maximum safe tumor resection instead.
Since there are chances of residual tumor, newer techniques to treat medulloblastoma is to perform trimodality treatment, meaning that surgical resection, radiotherapy and chemotherapy should all be done. However, certain side effects of treatment were observed throughout the years and research has been done to minimize side effects while keeping the effectiveness up to standard (i.e. no statistical significance in changes of event-free 5 years survival rate). (Packer RJ, et. al., 1999).
Side effects of radiotherapy Hematotoxicity and delayed growth The radiation may reach the vertebral column, hence slowing down one’s growth and hematologic component production.
Neurotoxicity Cognitive impairment is regularly reported to have increased in those who underwent full brain radiotherapy.
Ototoxicity Cochlea and hypothalamus are of close proximity to medulloblastoma lesions, hence they are often damaged when extensive radiotherapy was done.
Decrease quality of life Burden from the effects of radiotherapy could cause the patient’s quality of life to decrease at least during the treatment phase. Extent depends on the individual's opinions.
Table 8: Side effects of radiotherapy for medulloblastoma (Merchant et. al., 2013) Standard risk patient The most preferred treatment regimen for patients with standard risk is to perform resection of the tumor first. After maximal safe tumor resection, reduced-dose radiotherapy with concurrent chemotherapy course is suggested, and after completing the course should adjuvant chemotherapy be done. This method was developed by a non-randomized pilot study by Packer RJ et.al., in 1999. The suggested regimen is as the following: 1) Maximum safe tumor resection 2) Radiotherapy: CSI 23.4Gy + Vincristine (weekly) 3) Radiotherapy: Posterior fossa boost up to 55.8Gy + Vincristine (weekly) 4) Chemotherapy: Vincristine + CCNU + Cisplatin With this revised regimen, the patient significantly showed a decrease in side effects (such as intelligence quotient (IQ), attention capabilities, and reading abilities), while the effectiveness is as good as the regimen where 36 Gy of radiation was used.
High risk patient, age > 3 y.o.
Similarly, these patients are suggested to undergo a process of tumor resection, followed by radiotherapy along with cisplatin. However, high risk patients have a high rate of recurrent tumor, hence CSI radiation should be at full dose of 36 Gy.
1) Maximal safe resection 2) Radiotherapy: CSI 36 Gy + Vincristine (weekly) 3) Radiotherapy: Posterior fossa boost up to 55.8Gy + Vincristine (weekly) 4) Chemotherapy: Cisplatin-based regimen High risk, age < 3 y.o.
The difference between high risk patients who are older than 3 y.o. And those who are younger than 3 y.o. Is that the risk imposed by radiotherapy outweighs the benefit, hence it is suggested for the radiotherapy course to be delayed until one of the following criteria is met.
1. Disease continue to progress while on chemotherapy 2. Already completed 2 years of chemotherapy 3. Age reached 3 years old Risk group Treatment Note Standard risk and adult patients 1) Maximal safe resection 2) CSI 23.4 Gy + Vincristine 3) Post. fossa boost + Vincristine Reduced-dose radiotherapy is suggested as it reduces side effects while treatment effectiveness remains 4) Cisplatin-based chemotherapy the same High risk, > 3 yo 1) Maximal safe resection 2) CSI 36 Gy + Vincristine 3) Cisplatin-based chemotherapy Full-dose radiotherapy is preferred since the benefit outweighs the side effects High risk, < 3 yo 1) Maximal safe resection 2) Cisplatin-based chemotherapy 3) Delayed Radiotherapy Children under 3 y.o. are advised against radiotherapy Table 9: Summary of treatment of medulloblastoma according to risk and age group Conclusion: Medulloblastoma is a tumor which primarily affects the cerebellum, hence , apart from symptoms of increased intracranial pressure due to mass effect, central and peripheral ataxia can be observed. The imaging modality of choice would be MRI brain with gadolinium , MRI spine with gadolinium, and spinal fluid for CSF cytology. As for staging, in the past, Chang staging would be done to help assess the extent of the tumor’s nature. Nowadays, risk stratification is recommended to be done instead since it affects the treatment. The main treatment modality would be to surgically remove the mass via maximum safe tumor resection approach, followed by radiotherapy and chemotherapy.
Reference 1. Barinagarrementeria F, Amaya LE, Cantú Carlos. Causes and mechanisms of cerebellar infarction in young patients. Stroke. 1997;28(12):2400–4.
2. Bosemani T, Orman G, Boltshauser E, Tekes A, Huisman TA, Poretti A. Congenital abnormalities of the posterior fossa. RadioGraphics. 2015;35(1):200–20.
3. Dufour C, Beaugrand A, Pizer B, Micheli J, Aubelle M-S, Fourcade A, et al. Metastatic medulloblastoma in childhood: Chang's classification revisited [Internet]. International journal of surgical oncology. Hindawi Publishing Corporation; 2012 4. Efendi H. Clinically isolated syndromes: Clinical characteristics, differential diagnosis, and management. Noro Psikiyatri Arsivi. 2016;52(Suppl.1):1–11.
5. Fogel BL. Childhood cerebellar ataxia. Journal of Child Neurology. 2012;27(9):1138–45.
6. Geraint. LKWBIF, Bone I, Lindsay KW. Neurology and neurosurgery illustrated (5th edition). Elsevier Health Sciences; 2010.
7. Ibrahim D, Gaillard F. Haemangioblastoma (Central Nervous System). Radiopaediaorg.
2008; 8. Knight J. Pilocytic astrocytoma [Internet]. StatPearls [Internet]. U.S. National Library of Medicine; 2021 [cited 2021Nov28]. Available from: 9. McNeil DE, Coté TR, Clegg L, Rorke LB. Incidence and trends in pediatric malignancies medulloblastoma/primitive neuroectodermal tumor: A seer update. Medical and Pediatric Oncology. 2002;39(3):190–4.
10. Medulloblastoma in children - chulacancer [Internet]. [cited 2021Nov28]. Available from: 11. Pediatric oncologyeducation materials [Internet]. Pediatric Oncology Education Materials. [cited 2021Nov28].
12. Raffel C, Thomas GA, Tishler DM, Lassoff S, Allen JC. Absence of p53 mutations in childhood central nervous system primitive neuroectodermal tumors [Internet]. OUP Academic. Oxford University Press; 1993 [cited 2021Nov28].
13. Rahman M. Neuroanatomy, pontine micturition center [Internet]. StatPearls [Internet].
U.S. National Library of Medicine; 2021 [cited 2021Nov27]. Available from: 14. Rendtorff R, Novak A, Tunn R. Normal pressure hydrocephalus as cause of urinary incontinence – a shunt for incontinence. Geburtshilfe und Frauenheilkunde.
2012;72(12):1130–1.
15. Roberts RO, Lynch CF, Jones MP, Hart MN. Medulloblastoma: A population-based study of 532 cases. Journal of Neuropathology and Experimental Neurology.
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16. Shaw MD, Russell JA. Cerebellar abscess. A review of 47 cases. Journal of Neurology, Neurosurgery & Psychiatry. 1975;38(5):429–35.
17. Smoll NR, Drummond KJ. The incidence of medulloblastomas and primitive neurectodermal tumours in adults and children. Journal of Clinical Neuroscience.
2012;19(11):1541–4.
18. Spencer K, Slocomb D. The neural basis of ataxic dysarthria. The Cerebellum.
2007;6(1):58–65.
19. Strupp M, Hüfner K, Sandmann R, Zwergal A, Dieterich M, Jahn K, et al. Central oculomotor disturbances and nystagmus. Deutsches Aerzteblatt Online. 2011; 20. Taylor MD, Northcott PA, Korshunov A, Remke M, Cho Y-J, Clifford SC, et al.
Molecular subgroups of Medulloblastoma: The current consensus [Internet]. Acta neuropathologica. Springer-Verlag; 2012 21. Thompson EM, Bramall A, Herndon JE, Taylor MD, Ramaswamy V. The clinical importance of medulloblastoma extent of resection: A systematic review. Journal of Neuro-Oncology. 2018;139(3):523–39.
22. Van Samkar A, Poulsen MN, Bienfait HP, Van Leeuwen RB. Acute Cerebellitis in adults: A case report and review of the literature. BMC Research Notes. 2017;10(1).
23. Waszak SM, Northcott PA, Buchhalter I, Robinson GW, Sutter C, Groebner S, et al.
Spectrum and prevalence of genetic predisposition in medulloblastoma: A retrospective genetic study and prospective validation in a clinical trial cohort. The Lancet Oncology.
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24. Yamauchi M, Okada T, Okada T, Yamamoto A, Fushimi Y, Arakawa Y, et al. Differential diagnosis of posterior fossa brain tumors: Multiple discriminant analysis of TL-SPECT and FDG-Pet [Internet]. Medicine. Wolters Kluwer Health; 2017 25. Yildirim M, Gocmen R, Konuskan B, Parlak S, Yalnizoglu D, Anlar B. Acute cerebellitis or postinfectious cerebellar ataxia? clinical and imaging features in Acute Cerebellitis.
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26. Zhang Q, Zhou X, Li Y, Yang X, Abbasi QH. Clinical recognition of sensory ataxia and cerebellar ataxia. Frontiers in Human Neuroscience. 2021;15. |
12802 | https://phys.libretexts.org/Bookshelves/Nuclear_and_Particle_Physics/Nuclear_and_Particle_Physics_(Walet)/09%3A_Relativistic_Kinematics/9.01%3A_Lorentz_Transformations_of_Energy_and_Momentum | Skip to main content
9.1: Lorentz Transformations of Energy and Momentum
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: Aug 9, 2024
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9: Relativistic Kinematics
9.2: Invariant Mass
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: 15052
Niels Walet
University of Manchester
( \newcommand{\kernel}{\mathrm{null}\,})
As you may know, like we can combine position and time in one four-vector x=(x⃗ ,ct), we can also combine energy and momentum in a single four-vector, p=(p⃗ ,E/c). From the Lorentz transformation property of time and position, for a change of velocity along the x-axis from a coordinate system at rest to one that is moving with velocity v⃗ =(vx,0,0) we have
x′t′=γ(v)(x−v/ct),=γ(t−xvx/c2),(9.1.1)(9.1.2)
we can derive that energy and momentum behave in the same way,
p′xE′=γ(v)(px−Ev/c2)=muxγ(|u|),=γ(v)(E−vpx)=γ(|u|)m0c2.(9.1.3)(9.1.4)(9.1.5)
To understand the context of these equations remember the definition of γ
γ(v)=11−β2−−−−−√
and
β=vc.
In Equation 9.1.5, we have also re-expressed the momentum energy in terms of a velocity u⃗ . This is measured relative to the rest system of a particle, the system where the three-momentum p⃗ =0.
Now all these exercises would be interesting mathematics but rather futile if there was no further information. We know however that the full four-momentum is conserved, i.e., if we have two particles coming into a collision and two coming out, the sum of four-momenta before and after is equal,
Ein1+Ein2p⃗ in1+p⃗ in2=Eout1+Eout2,=p⃗ out1+p⃗ out2.
9: Relativistic Kinematics
9.2: Invariant Mass |
12803 | https://artofproblemsolving.com/wiki/index.php/Extrema?srsltid=AfmBOoqSv-_W9h55emWA36BaZagd8Ta-opQ2bm6zUJB0h5IZZNCCCZdI | Art of Problem Solving
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Extrema
The upper and lower bounds of a real valued function are of interest in several situations in pure as well as applied Mathematics
Absolute Extrema
Let be a set
Let
Let the set be bounded
Then is called the Absolute or Global maximum of
and is called the Absolute or Global minimum of
Local Extrema
Let
Let
is said to be a Local maximum iff such that
is said to be a Local minimum iff such that
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Function
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Derivative
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12804 | https://journals.physiology.org/doi/abs/10.1152/physrev.00025.2008 | PHYSIOLOGY.ORG
Models and Mechanisms of Hyperalgesia and Allodynia
Jürgen Sandkühler
Jürgen Sandkühler
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Published Online:01 Apr 2009
Abstract
Hyperalgesia and allodynia are frequent symptoms of disease and may be useful adaptations to protect vulnerable tissues. Both may, however, also emerge as diseases in their own right. Considerable progress has been made in developing clinically relevant animal models for identifying the most significant underlying mechanisms. This review deals with experimental models that are currently used to measure (sect. ii) or to induce (sect. iii) hyperalgesia and allodynia in animals. Induction and expression of hyperalgesia and allodynia are context sensitive. This is discussed in section iv. Neuronal and nonneuronal cell populations have been identified that are indispensable for the induction and/or the expression of hyperalgesia and allodynia as summarized in section v. This review focuses on highly topical spinal mechanisms of hyperalgesia and allodynia including intrinsic and synaptic plasticity, the modulation of inhibitory control (sect. vi), and neuroimmune interactions (sect. vii). The scientific use of language improves also in the field of pain research. Refined definitions of some technical terms including the new definitions of hyperalgesia and allodynia by the International Association for the Study of Pain are illustrated and annotated in section i.
I. ABOUT THIS REVIEW
Hyperalgesia and to some degree allodynia are frequent symptoms of disease and may be useful adaptations for better protection of vulnerable tissues. Enhanced sensitivity for pain may, however, persist long after the initial cause for pain has disappeared, then pain is no longer a symptom but rather a disease in its own right. Changes of signal processing in the nervous system may contribute to or may become the sole cause for hyperalgesia and allodynia. It appears that sensitization of nociceptive Aδ- and C-fiber nerve endings rarely outlast the primary cause for pain and is restricted to the area of injury and thus may be considered adaptive. In contrast, central changes in the processing of nociceptive information may potentially outlast their trigger events for days, months, and perhaps years and may spread to sites somatotopically remote from the primary cause of pain. Thus central mechanisms constitute one of the causes for pain chronicity and pain amplification in pain patients. In this review I address animal models that are currently used to measure (sect. ii) or to induce (sect. iii) hyperalgesia and allodynia in animals. Context-sensitive expression of hyperalgesia is discussed in section iv. Cellular elements that are indispensable for the induction and/or the expression of hyperalgesia and allodynia are summarized in section v. Peripheral mechanisms contributing to hyperalgesia and allodynia have been reviewed extensively (194, 587). Reorganization of sensory processing in cortical areas may also be long-lasting (138, 518, 519, 532). The peripheral, spinal, and supraspinal elements that are essential for hyperalgesia and allodynia are listed in section v. This review focuses on spinal mechanisms of hyperalgesia and allodynia (sect. vi) and relevant mutual neuron-immune interactions (sect. vii). The new International Association for the Study of Pain (IASP) definitions of technical terms from 2008 are explained and used.
A. Definitions
In an early definition hyperalgesia was considered “a state of increased intensity of pain sensation induced by either noxious or ordinarily non-noxious stimulation of peripheral tissue” (169). Thus no distinction was made between hyperalgesia and allodynia. Later, the IASP took over the task to improve the use of language in the pain field by implementing a task force on taxonomy (recommendations from 1994, revised 2008). For pain elicited by normally nonpainfully stimuli, the made-up word allodynia was coined by Professor Paul Potter of the Department of the History of Medicine and Science at The University of Western Ontario (see definitions on the IASP homepage: ). In the year 2008, the IASP modified many of the definitions from 1994 substantially. With respect to the definition of “hyperalgesia,” the original definition experienced a revival, and the term allodynia is now reserved to those forms of pain only that are clearly caused by excitation of low-threshold sensory nerve fibers. Some of the current definitions of the IASP task force are reproduced here in quotes and modified only if useful for the purpose of the present review.
1) Allodynia: “Pain in response to a nonnociceptive stimulus.” The IASP task force comments on this term: “This term should only be used, when it is known that the test stimulus is not capable of activating nociceptors. At present, dynamic tactile allodynia to tangential stroking stimuli, e.g., brushing the skin is the only established example. Future research may present evidence for other types of allodynia. Whenever it is unclear, whether the test stimulus may or may not activate nociceptors, hyperalgesia is the preferred term.” Thus allodynia refers largely to pain evoked by Aβ-fibers (see sect. viF and Fig. 1) or low-threshold Aδ- and C-fibers.
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2) Analgesia: “Absence of pain in response to stimulation which would normally be painful.”
3) Central sensitization: “Increased responsiveness of nociceptive neurons in the central nervous system to their normal or subthreshold afferent input.” Central sensitization is a popular phrase in pain literature, but unfortunately, it is used in many different and sometimes inconsistent ways. At present, a generally accepted definition does not exist. The IASP task force for taxonomy suggests the above quoted definition. This proposal clearly defines a phenomenon but not its functional meaning. Nociceptive neurons comprise a heterogeneous cell group with putatively many different and sometimes opposing functions, including a large group of inhibitory interneurons. Thus enhanced responsiveness of some of these neurons could contribute to hyperalgesia. On the other hand, enhanced responsiveness of inhibitory nociceptive neurons may well lead to stronger feedback inhibition and analgesia, while still other neurons may not contribute to the experiences of pain but rather to altered motor or vegetative responses to a noxious stimulus. Another often used definition implies that “central sensitization” would be an enhanced responsiveness of neurons in the central nervous system leading to hyperalgesia (76). In this definition “central sensitization” necessarily leads to pain amplification due to enhanced neuronal responsiveness. A causal relationship between firing rates of any type of neurons in the central nervous system and the perceived intensity of pain can, however, presently not be ensured. At best, a tight correlation may be shown. Thus none of the presently proposed central mechanisms of hyperalgesia would strictly fulfill the latter definition of “central sensitization.” In the literature “central sensitization” is often not clearly defined, and sometimes two mutually exclusive definitions are used within the same publication.
4) Hyperalgesia: “Increased pain sensitivity.” IASP task force comment (2008): “Hyperalgesia may include both a decrease in threshold and an increase in suprathreshold response. In many cases it may be difficult to know whether or not the test stimulus is capable of activating nociceptors. Therefore, it is useful to have an umbrella term (hyperalgesia) for all types of increased pain sensitivity.” See also Figure 1 for this new distinction between hyperalgesia and allodynia.
5) Hyperalgesia, primary: Hyperalgesia at the site of injury. It is often believed that primary hyperalgesia is mainly due to sensitization of nociceptive nerve endings. Recent evidence suggests, however, that altered processing in the central nervous system is equally important.
6) Hyperalgesia, secondary: Hyperalgesia in an area adjacent to or remote of the site of injury. This form of hyperalgesia is not caused by sensitization of nociceptive nerve endings but solely due to changes in the processing of sensory information in the central nervous system. While the induction of secondary hyperalgesia requires activity in nociceptive nerve fibers, its maintenance is independent of an afferent barrage as local aesthetic block of the injured site preempts but does not reverse secondary hyperalgesia.
7) Hyperalgesia, referred: Hyperalgesia may not only exist within an area of tissue damage but also in the skin (head zone) remote from the inner organ or muscle which is affected.
8) Long-term potentiation: Long-term potentiation of synaptic strength (LTP) is an intensively studied model of neuronal plasticity. It is defined as an increase in synaptic efficacy that outlasts the duration of the conditioning stimulus for at least 30 min (early LTP), a few hours, or days to months (late LTP). Synaptic strength can be quantitatively assessed by measuring changes of the postsynaptic membrane potential or postsynaptic currents in response to a presynaptic stimulus. Normally synaptic strength is measured as the amplitude or area under the curve of postsynaptic excitatory or inhibitory potentials or currents. Alternatively, extracellular recordings of field potentials are being used. Action potential firing and polysynaptic responses not only depend on the strength of synaptic transmission but also on intrinsic membrane properties (e.g., action potential thresholds) and network properties and can thus not be used to quantify synaptic strength and changes thereof.
9) Nociceptive stimulus: “An actually or potentially tissue damaging event transduced and encoded by nociceptors.”
10) Nociceptor: “A sensory receptor that is capable of transducing and encoding noxious stimuli.”
11) Noxious stimulus: “An actually or potentially tissue-damaging event.”
12) Pain: “An unpleasant sensory and emotional experience associated with actual or potential tissue damage, or described in terms of such damage”.
13) Pain versus nociception: The above definitions of pain and its derivates require a conscious subject that is able to experience pain. The molecular, cellular, and systemic mechanisms which deal with the processing of pain-related information, its amplification, or depression are called nociceptive, pro-nociceptive, and anti-nociceptive, respectively. Pain is just one of many possible end points of nociception. Others include but are not limited to withdrawal reflexes, vegetative and hormonal responses, and vocalization, all of which normally accompany pain experience but may under experimental and some pathological conditions be observed in the absence of pain experience, e.g., in the intact but deeply anesthetized subject or in lesioned animals.
The experience of pain is not a phenomenological entity but rather a multidimensional process that may include to varying degrees sensory-discriminative aspects and emotional-aversive components, all of which involve activation of different brain areas and neuronal ensembles. Thus, when reporting “hyperalgesia,” this always implies a contextual definition which is, in a strict sense, only valid for the experimental context in which it was assessed. In the literature, it often does not reflect a measure of pain but one of its surrogates, i.e., signs of amplified nociception, e.g., exaggerated withdrawal reflexes.
Hyperalgesia and allodynia are classified according to the type of stimulus which elicits the sensation of pain. Thermal (heat or cold) stimuli or mechanical brush, pinch, or pressure stimuli are most often used. In addition, moving (dynamic) or static mechanical touch stimuli are being used. Thereby, mechanical and thermal (heat or cold) hyperalgesia and mechanical dynamic allodynia can be differentiated (see also Fig. 1).
The mechanisms underlying the various forms of hyperalgesia and allodynia are not alike (see Fig. 3) but may differ with respect to molecular genetic, physiological, and pharmacological profiles (271, 359).
It is now generally accepted that excitation of thin high-threshold (i.e., nociceptive) primary afferent nerve fibers which are weakly myelinated (Aδ-fibers) or unmyelinated (C-fibers) triggers nociceptive pain. But not all Aδ- and C-fibers are nociceptive as some respond to low-level natural stimuli. On the other hand, almost all thick and heavily myelinated Aβ-fibers are low threshold, and only few Aβ-fibers may be nociceptive. Thus selective excitation of Aβ-fibers, e.g., by electrical nerve stimulation, normally does not evoke pain. The roles of the many different types of spinal dorsal neurons for a pain sensation are much less clear. Spinal dorsal horn neurons have been classified by some of their properties. A popular scheme is based on the type of excitatory mono- and/or polysynaptic afferent input.
14) High-threshold spinal neurons (nociceptive specific neuron, nociceptor specific neuron, class 3 neuron): These neurons selectively respond to stimulation of primary afferent nerve fibers with high thresholds, i.e., to nociceptive nerve fibers. Thus nociceptor specific neurons are exclusively driven by nociceptors.
15) Wide-dynamic range spinal neurons (multireceptive neuron; class 2 neuron): These neurons nonselectively respond to both primary afferents with high and with low thresholds, i.e., to nociceptive nerve fibers and to touch fibers for example.
16) Low-threshold neuron spinal neurons (class 1 neurons): These neurons respond to primary afferent nerve fibers with low thresholds. They have no excitatory input from nociceptors, thus increasing stimulation intensity into the noxious range does not lead to substantial increases in excitation.
This popular classification scheme rests on the response properties of spinal dorsal horn neurons to natural stimulation within the neurons’ receptive field or electrical stimulation of afferent nerve fibers. These response profiles are, however, not static but context sensitive and may change with the level of the membrane potential so that, e.g., lowering membrane potential of “nociceptor specific” neurons may result in their transformation into “multireceptive” neurons (572). Furthermore, when comparing the incidence of recordings made from either low-threshold or wide-dynamic range neurons in awake, drug-free cat, wide-dynamic range neurons are much less often encountered as expected from acute preparations in anesthetized animals (82). When barbiturates are applied, the likelihood of recording from wide-dynamic range neurons increases (82, 83). Another technique to group neurons by their electrophysiological properties including sensory input is the cluster analysis (280).
17) Spinal neuron: Projection neuron: Another way of grouping neurons is by their supraspinal projection. In vivo this can be achieved by electrical deep brain stimulation of the ascending axon and recording the antidromic action potential discharge. Depending on the location of the stimulation electrodes, neurons are then classified according to the ascending tract they project to, e.g., the spinothalamic tract or dorsal column pathway or by their presumed projection territory, e.g., the ventrolateral thalamus (581). One should, however, keep several caveats in mind: 1) The area from which antidromic spikes can be elicited is not necessarily the area of termination. It is equally well possible that fibers of passage are excited. 2) An identified projection area must not necessarily be the only or even the main projection area of the neuron under study, as some spinal dorsal horn neurons project to more than one supraspinal site. 3) Neurons for which no supraspinal projection area could be identified still could be neurons with a projection that just was missed.
For in vitro recordings spinal projection neurons can be identified by retrograde transport of a fluorescent marker such as DiI. This marker can then easily be detected in spinal cord slices prepared 3–4 days after dye injection using a fluorescence microscope (202, 204).
18) Peripheral sensitization: “Increased responsiveness and reduced threshold of nociceptors to stimulation of their receptive fields.”
19) Wind-up: Wind-up is an electrophysiological phenomenon seen in some nociceptive neurons in response to repetitive stimulation of primary afferent C-fibers. When C-fibers are stimulated at frequencies between 0.5 and 5 Hz, some postsynaptic neurons respond with an increasing discharge rate to the first 10–30 stimuli (i.e., in the first few seconds of an ongoing noxious stimulation). Thereafter, the response reaches a plateau or may decline. Wind-up is seen under normal experimental conditions, i.e., in the absence of any intentional inflammation, trauma, or nerve injury and thus constitutes a normal coding property of some nociceptive spinal dorsal horn neurons. Consequently, wind-up per se is not a mechanism of hyperalgesia. However, lowering the wind-up threshold frequency or enhancing the wind-up response may indicate that some form of signal amplification has been induced, for example, LTP of synaptic strength between primary afferent C-fibers and spinal dorsal horn neurons. Thus enhanced wind-up may be a useful marker of increased responsiveness of some spinal dorsal horn neurons to C-fiber stimulation.
B. Hyperalgesia and Allodynia as Symptoms
The proper function of the nociceptive system enables and enforces protective behavioral responses such as withdrawal or avoidance to acutely painful stimuli. In case of an injury, the vulnerability of the affected tissue increases. The nociceptive system adapts to this enhanced vulnerability by locally lowering the nociceptive thresholds and by facilitation of nocifensive responses, thereby adequate tissue protection is ensured. The behavioral correlates of these adaptations are allodynia and hyperalgesia. Thus neither hyperalgesia nor allodynia is per se pathological or a sign of an inadequate response but may rather be an appropriate shift in pain threshold to prevent further tissue damage. Painful syndromes are typical for a large number of diseases and pain intensity if often used by the patients and their health professionals to evaluate the progression of the disease or the success of the therapy.
C. Hyperalgesia and Allodynia as Diseases
The intensity, the duration, and/or the location of pain may not always adequately reflect any known underlying cause. For example, hyperalgesia and allodynia may persist long after the initial cause for pain, e.g., an injury or an inflammation has healed completely. Furthermore, hyperalgesia and allodynia may occur due to dysfunction of parts of the peripheral or central nervous system. Thus, when the location, the duration, or the magnitude of pain, hyperalgesia, and/or allodynia has become maladaptive rather than protective, then pain is no longer a meaningful homeostatic factor or symptom of a disease but rather a disease on its own right.
II. METHODS TO ASSESS HYPERALGESIA OR ALLODYNIA
Because pain cannot be measured directly in animals, it is essential to use quantifiable, sensitive, and specific surrogates of pain sensation. A number of different surrogates have been suggested to fulfill these criteria. One should, however, keep in mind that any reaction to a painful stimulus is not necessarily evidence for a concomitant sensation of pain. Thus none of these tests measures hyperalgesia or allodynia directly, but rather enhanced nociception. This distinction is, however, rarely made in the literature or in this review.
Signs of evoked pain in animals include withdrawal of a paw or the tail from the stimulus source, vocalization upon sensory stimulation, reduced locomotion, or agitation. Motor reflexes may not only be elicited by noxious stimulation but also by innocuous stimuli (470), i.e., may not be specific for nociception. Furthermore, any form of behavior may be modulated by the motor system which constitutes a potential confounding effect (430).
Suggested signs of spontaneous pain include audible and ultrasonic vocalization, conditional place avoidance, analgesic self-administration, excessive grooming, and self-mutilation of a limb, to name a few. These parameters are rarely used in animal studies (360). Autotomy of an affected limb has also been considered a sign of spontaneous pain after nerve injury (75). A systematic methodological review of animal models of pain is provided by Le Bars et al. (275). The impact of strain differences in mice for nociceptive tests is discussed by Mogil et al. (361). Table 1 summarizes presently used methods to assess hyperalgesia and allodynia.
TABLE 1. Methods to assess hyperalgesia or allodynia
| Modality | Test Name (Most Common) | Test Method | Testing Site | Outcome Parameter | Reference Nos. |
| Mechanical | von Frey | Application of nonnoxious calibrated static hairs on skin | Hindpaw, face | Force threshold to elicit paw withdrawal (static mechanical hyperalgesia) | 66, 109 |
| | Randal Sellito | Application of linearly increasing mechanical force in noxious range on skin | Hindpaw | Force threshold to elicit paw withdrawal from noxious stimulus (mechanical hyperalgesia) | 18, 424 |
| | Unnamed | Innocuous brushing, stroking of skin | Hindpaw | Time latency to elicit paw withdrawal or nociceptive behaviors (dynamic mechanical allodynia) | 131, 599 |
| | Unnamed | Noxious mechanical stimulation to viscera | Visceral organs (colon, bladder) | Thresholds or number or strength of muscle contractions, autonomic responses (hyperalgesia) | 381 |
| Heat | Tail flick | Application of radiant heat on tail or immersion of tail in hot water | Tail | Time latency to elicit tail withdrawal (heat hyperalgesia) | 122, 183 |
| | Plantar Hargreave's | Application of radiant heat on skin | Hindpaw | Time latency to elicit paw withdrawal (heat hyperalgesia) | 171, 608 |
| | Hot plate | Animal placed on heated metal plate | Hindpaw (forepaws) | Time latency to elicit nociceptive or escape behavior (heat hyperalgesia) | 275, 341 |
| Cold | Acetone | Application of acetone on skin | Hindpaw | Duration/intensity of nociceptive behaviors (cold hyperalgesia) | 71, 548 |
| | Cold plate | Animal placed on cooled metal plate | Hindpaw | Time latency to elicit nociceptive or escape behaviors (cold hyperalgesia) | 11, 209 |
| | Cold water | Animal placed in shallow cold water bath | Hindpaw | Time latency to elicit nociceptive or escape behaviors, duration and intensity of nociceptive behaviors (cold hyperalgesia) | 20, 340 |
| Electrical | Unnamed | Electrical current application | Various: tail, paw, viscera, dental pulp | Withdrawal thresholds, vocalization, escape latency (allodynia) | 43, 45, 275, 507 |
According to the new definition by the IASP: in the previous literature, this was labeled “mechanical allodynia” (see also section i and Fig 1).
Enlarge table
TABLE 1. Methods to assess hyperalgesia or allodynia
Enlarge table
III. METHODS TO INDUCE HYPERALGESIA OR ALLODYNIA IN ANIMALS
A. Animal Welfare Issues: Replace, Reduce, Refine
Most countries have issued animal welfare acts (see, for example, those of Sweden, The Netherlands, Switzerland, or Germany), and most scientific journals enforce full compliance with local and institutional regulations for animal welfare before considering any manuscript for publication. The IASP has issued ethical guidelines for the investigation of experimental pain in conscious animals (see These guidelines are, however, from 1982 and outdated by now. A revision incorporating contemporary research is desirable. Furthermore, several publications have also addressed this important topic (see, e.g., Refs. 31, 137, 182, 489, 495). General information on animal welfare issues can be obtained from a number of sources including See Table 2 for a summary of contemporary models to study hyperalgesia and allodynia in animals.
TABLE 2. Animal models of pain
| Human Relevance/Disease | Model Primary Mechanism | Induction Method | Model Name (Most Commonly Used) | Nociception Produced --- | | | | | | | | | Reference Nos. |
| | | | | Mechanical allodynia | Mechanical hyperalgesia | Heat hyperalgesia | Cold hyperalgesia | Other behaviors | |
| Peripheral inflammation and peripheral neurogenic inflammation | Chemical stimulation of primary afferents | Injection of inflammatory agent in hindpaw | Complete Freund's adjuvant (CFA) | | • | • | | | 279 |
| | | | Carageenan | | • | • | | | 244, 332 |
| | | | Mustard oil | | • | • | | | 74, 426 |
| | | | Bee venom | • | • | • | | Hindpaw flinching, licking, lifting | 67, 68, 270 |
| | | | Formalin | | | | | Hindpaw flinching, licking in two phases | 2, 439 |
| | | | Capsaicin | | • | • | | Hindpaw flinching | 152, 591 |
| Arthritis | Inflammation | Inflammatory mediator injection into joint or into tail | Adjuvant-induced mono-arthritis | ○ | • | • | | Altered gait, stance, spontaneous pain | 78, 87, 175 |
| | | | Adjuvant-induced poly-arthritis | | • | | | Other systemic changes | 84, 487 |
| Postoperative pain | Surgical mechanical trauma | Surgery | Incision model | • | • | | | | 47, 132 |
| | | | Ovariohysterectomy | • | • | ○ | | Abdominal postures | 156, 273 |
| Sunburn, burn injury | Cell damage by irradiation or thermal injury | UV-B (290–320 nm) dermal irradiation or prolonged noxious heat application on skin | UV model | | • | • | • | | 98, 405 |
| | | | Burn or thermal injury model | | • | • | | | 302, 386 |
| Ischemia-reperfusion injury, complex regional pain syndrome (CRPS), compartment syndrome, peripheral ischemic disease | Ischemia | Temporary hindpaw ischemia and reperfusion or vascular occlusion | Chronic postischemia pain | | • | ○ | • | | 79, 474 |
| Neuropathic pain, CRPS, nerve entrapment | Traumatic nerve injury | Various methods (constriction, ligation, transection) to injure various peripheral nerves (spinal, sciatic, saphenous) or facial nerves (trigeminal, mental) | Chronic constriction injury (CCI; Bennett), | • | • | • | • | Hindpaw guarding, altered weight bearing | 35, 509, 599 |
| | | | Spinal nerve ligation (SNL; Chung) | • | • | • | • | Hindpaw licking, lifting | 71, 131, 240 |
| | | | Partial sciatic nerve ligation (PSNL; Seltzer) | | • | • | | Hindpaw guarding and licking | 315, 473 |
| | | | Spared nerve injury (SNI) | | • | • | • | Hindpaw guarding and altered weight bearing | 101, 477 |
| | | | Sciatic nerve crush | | • | | | | 100, 106 |
| | | | Cryoneurolysis | | • | | | Autotomy, hyperesthesia | 103 |
| | | | Phototoxicity | | • | • | • | | 257 |
| | | | Distal nerve injury | | • | | | | 486 |
| | | | Complete nerve transection | | | | | Autotomy | 554 |
| Trigeminal neuralgia | Traumatic nerve injury | Various methods to injure infraorbital nerve (CCI, ischemic injury) or trigeminal ganglia | Infraorbital nerve injury | • | • | • | | | 15, 205, 550 |
| | | | Trigeminal ganglion compression | • | | | | | 6 |
| Temporomandibular joint inflammation or orofacial pain | Orofacial inflammation | Acute injection of inflammatory agent (complete Freund's adjuvant, carageenan) in temporomandibular joint or face | Rat or mouse temporomandibular joint pain or orofacial pain | | • | | | Face grooming, scratching | 206, 607 |
| Diabetic neuropathy, Postherpetic neuralgia, acute inflammatory demyelinating polyradiculo-neuropathy | Secondary (disease) neuropathy | Administration of streptozotocin to induce diabetes, inoculation with herpex simplex virus type I, immunization with peripheral myelin P2 peptide | Streptozotocin-induced diabetes | • | • | • | • | | 90, 133, 314 |
| | | | Herpes simplex virus inoculation | • | • | • | | | 97, 460, 508 |
| | | | Experimental autoimmune neuritis | | • | • | | | 358 |
| Cancer pain | Compression and inflammation of tissue by cancer cells | Intramedullary, intraplantar, or intragingival injection of cancer cells | Experimental osteolytic sarcoma | • | • | | | Muscle hyperalgesia | 142, 551 |
| | | | Experimental squamous cell carcinoma | | • | • | | | 373, 479 |
| | | | Experimental melanoma | | • | • | | | 617, 461 |
| Inflammatory bowel syndrome | Visceral pain | Injection of irritants into hollow organs or visceral mechanical distention | Injection of acetic acid, capsaicin, mustard oil, turpentine, zymosan into hollow organs | | • | | | Abdominal contraction (writhing) or other visceral nociceptive behaviors, referred hyperalgesia | 264, 613 |
| Muscle pain | Peripheral acidosis | Injection of acid in gastrocnemius muscle | Muscle pain | | • | ○ | | | 146, 490 |
| Fever, central nervous system inflammatory diseases | Generalized immune system activation | Systemic, intrathecal, or central lipopolysaccharide/inflammatory mediator administration | Sickness syndrome | | • | • | | Nonspecific manifestations of inflammation and infection (fever, drowsiness), see sect. viiA | 326, 561 |
| Spinal cord injury neuropathic pain | Spinal cord lesions | Ischemic or traumatic contusion, compression, or transection of spinal cord | Spinal cord injury | • | • | • | • | | 124, 167, 609 |
| Neuritis, neuropathic pain | Neuritis | Acute injection or perineuronal administration of inflammatory agents directly on nerves | Sciatic inflammatory neuritis | | • | ○ | ○ | | 65, 556 |
| Migraine | Various | Various manipulations: neurovascular, electrical, genetic | Various | • | • | • | | Various measures of spontaneous pain | 38, 123 |
| Drug-induced neuropathic pain | | Systemic administration of clinically used therapeutic agent | Chemotherapeutic agents (vincristine, paclitaxel) | • | • | • | • | See sect. iiiB | 415, 515 |
| | | | Antiretroviral agents | | • | | ○ | | 222, 555 |
Solid circles indicate nociception produced, open circles indicate nociception tested but not produced, and empty spaces indicate nociception not tested or controversial effects.
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TABLE 2. Animal models of pain
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B. Drug-Induced Hyperalgesia and Allodynia
Drug-induced pain amplification or pain generation is relevant both in preclinical studies where they serve as tools for animal models of pain and in the clinical situation where they may be unwanted effects of therapeutics including chemotherapeutics and opioids. It is an intriguing yet unproven hypothesis that many of the substances that are sufficient to induce hyperalgesia and/or allodynia may do so by increasing the free cytosolic Ca2+ concentration in neuronal and nonneuronal cells, e.g., in spinal dorsal horn. For example, intrathecal injection of a calcium ionophore (A23187) or of a calcium channel agonist (BAY K 8644) may facilitate the second, but not the first phase of the Formalin test (77). Furthermore, nociceptive behavior that can be induced by intrathecal injection of a neurokinin 1 receptor agonist is blocked by intrathecal injection of dantrolene, which reduces the release of calcium from intracellular stores, or by intrathecal injection of thapsigargin, which inhibits the reticular Ca2+-ATPase thereby blocking intracellular calcium storage (16). Likewise, in diabetic mice, intrathecal application of ryanodine, which blocks Ca2+ release from Ca2+/caffeine-sensitive microsomal pools, increases tail-flick latencies (391). See Table 3 for a summary of substances that induce hyperalgesia and/or dynamic mechanical allodynia when injected into the intrathecal space.
TABLE 3. Intrathecal chemical induction of hyperalgesia or allodynia
| Class of Substance | Mode of Action | Compound | Behavioral Tests Shown to Produce Nociception --- | | | | | | | | | Reference Nos. |
| | | | Mechanical allodynia | Mechanical hyperalgesia | Heat hyperalgesia | Cold hyperalgesia | Other behaviors | |
| Opioids | μ-Opioid receptor activation | Morphine | | • | • | | | 111, 321, 537, 585 |
| | | DAMGO | | • | • | | | 538 |
| Plant alkaloid-based chemotherapeutics | Mitotic inhibitors | Paclitaxel | | • | • | • | Normal motor | 22, 415 |
| | | Vincristine | | • | ○ | | | 10, 60 |
| Platinum-based chemotherapeutics | Inhibition of DNA synthesis | Cisplatin | | • | | | | 21 |
| | | Oxaliplatin | | • | | • | | 294 |
| Glutamate receptor agonists | Activation of NMDA receptors | NMDA | | • | • | | Scratching and biting | 112, 503, 598 |
| | | AMPA | | | • | | | 598 |
| | | Quisqualate | | • | • | | Scratching and biting | 503, 610 |
| | | Kainate | | | ○ | | Scratching and biting | 73, 154 |
| | Activation of class I metabotropic glutamate receptor | DHPG | | • | • | | Scratching and biting | 113, 134, 135 |
| Substances acting on nitric oxide metabolism | Nitric oxide donors | Sodium nitroprusside | | | • | | | 245 |
| | | Hydroxylamine | | | • | | | 245 |
| | Substrate of nitric oxide synthase | l-Arginine | • | • | • | | | 333, 353 |
| | Target of nitric oxide | Cell-permeable analogs of cGMP: 8-bromo-cGMP, Db-cGMP | | | • | | | 148 |
| ATP | Agonist at P2X receptor | ATP, α,β-methylene-ATP | | • | | | | 375 |
| Cytokines | Agonist at specific cell-surface receptors, CX3C receptor 1 | Spinal fractalkine (CXC3CL1) | | • | • | | | 348 |
| | Binding to TNF-R1, TNF-R2, IL1R1 (CD121a), IL1R2 (CD121b), IL6R (CD126), glycoprotein 130 (gp130, IL6ST, IL6β or CD130) | TNF-α | | | • | | | 428 |
| | | IL-1β | | • | • | | | 428, 504 |
| | | IL-6 | | • | • | ○ | | 549, 557 |
| | Agonist at interferon (IFN)-γ receptor | IFN-γ | • | | | | | |
| Lipopolysaccharides | Binds the CD14/TLR4/MD2 receptor complex, which promotes secretion of pro-inflammatory cytokines | LPS | | • | • | | | 52, 334 |
| Lipids | | Platelet activating factor (PAF) | • | • | • | | | 366, 367 |
| Prostanoids | Binding to PGE2, EP1, EP3, EP4 receptors | PGE1 | | • | • | | | 451 |
| | | PGE2, PGD2, | | • | • | | Writhing | 357, 527 |
| | | PGF2α | | • | ○ | | Agitation | 356, 527 |
| Neurotrophic factors | Binding to TrkA, TrkB, p75 receptors, GFRα1 | NGF | | ○ | • | ○ | | 53, 313 |
| | | BDNF | | | • | | | 158, 408 |
| | | NT-3 | | | ○ | | | 313 |
| | | GDNF | | ○ | ○ | | | 46 |
| Endogenous peptides | Agonist at NK-1, NK-2, NK-3 receptors | Substance P | | • | • | | Vocalization | 108, 328, 604–606 |
| | | Neurokinin A, B | | | • | | | 605 |
| | Binding to CGRP1 and CGRP2 receptors | Calcitonin gene-related peptide (CGRP) | | • | | | | 393, 501 |
| | Unknown | Galanin (porcine) | | • | ○ | | | 258 |
| | Unknown | Cocaine- and amphetamine-regulated transcript (CART) peptide | | | • | | | 390 |
| | Agonist action at κ-opioid subtype and bradykinin receptors | Dynorphin A | • | • | • | • | | 263, 274, 539 |
| | Binding to ORL-1 receptor | Nociceptin (Orphanin FQ) | • | • | • | | | 168, 394, 429 |
| | Binding at the kinin B1 and B2 receptors | Bradykinin | | | • | | | 130 |
| Other proteins | Complexing to D1D2 CD4 | Glycoprotein: HIV type I: Gp120 | | • | • | | | 350 |
| | Activation of protease-activated receptors (PARs) | Coagulation protein: thrombin | | ○ | • | | | 251 |
| | ADP-ribosylation and thereby inactivation of Gi proteins | Soluble protein exotoxin: pertussis toxin (PTX) | | • | • | | | 584 |
| | | Fibronectin | | • | | | | 523 |
| | Agonists of two G protein-coupled receptors: prokineticin receptor 1 (PKR1) and PKR2 | Secretory protein Bv8 | | • | • | | | 380 |
| Inhibitory neurotransmitter inhibitors | Blocking glycine or GABAA receptors | Strychnine, | • | | | | | 303 |
| | | Bicucculine | • | | | | | 303 |
| Protein kinase activators | Activation of protein kinase C and direct actions on ion channels | Phorbol ester | | • | • | | | 396 |
| Receptor tyrosine kinases | Eph receptor tyrosine kinases | Ephrins (ephrinB1-Fc, ephrinB2-Fc) | | ○ | • | | | 493 |
| Alkyl-phospholipids | Binding at PAF receptor, phospholipids activation | PAF | • | • | • | | | 367 |
Solid circles indicate nociception produced, open circles indicate nociception tested, and empty spaces indicate nociception not tested or controversial effects.
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TABLE 3. Intrathecal chemical induction of hyperalgesia or allodynia
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C. Diet-Induced Hyperalgesia or Allodynia
An early report showed that rats fed a tryptophan-poor corn diet have reduced levels of brain serotonin and display enhanced responsiveness to electric shock. This diet-induced hyperalgesia can be reversed by feeding the animals diets with adequate amounts of tryptophan, or by systemic injections of this amino acid (306).
In rats fed an Mg2+-deficient diet for 10 days, Mg2+ levels in plasma and cerebrospinal fluid fall after a few days and recover within 1 day after feeding a normal diet. Mechanical hyperalgesia (Randall-Sellito pressure test) is significant at day 10, the first day tested, and outlasts the period of Mg2+-deficient diet for at least 10 days (13). Hyperalgesia induced by Mg2+ deficiency is partially reversed by NMDA receptor blockade (119).
Chronic administration of a diet in which all choline is replaced by N-aminodeanol, an unnatural choline analog, results in mechanical hyperalgesia in rats along with other classical hypocholinergic symptoms, i.e., progressive loss of learning and memory capacities, hyperkinesis, and hyperactivity (210).
D. Anxiety Level Modulates Pain Sensitivity
In contrast to acute fear, which may lead to stress-induced analgesia (see, e.g., Ref. 285), anxiety may enhance pain sensitivity (410, 433). In some groups of human pain patients, pain sensitivity may positively correlate with basal anxiety levels. Similarly, different strains of rats may also display different levels of baseline anxiety when assessed by the acoustic startle response and open-arm exploration in the elevated plus-maze assay. In these tests, Wistar-Kyoto rats reveal higher anxiety levels than Sprague-Dawley or Fisher-344 rats. When innocuous pressure stimuli are applied to the normal or to the sensitized colon, these strains of rats differ in their responsiveness. Both under normal conditions and after sensitization, high-anxiety Wistar-Kyoto rats respond with significantly more abdominal contraction to colon distension, suggesting that genetically determined anxiety levels are associated with higher visceral sensitivity (160).
E. Chronic Stress Induces Hyperalgesia and Allodynia
Repeated exposure to a cold environment, i.e., to a nonnoxious stressful situation, causes mechanical hyperalgesia (pressure test) that outlasts the stressful period for 3 days (462). Similarly, chronic, but not acute, restraint stress leads to thermal hyperalgesia in the tail-flick assay (145).
IV. GENERAL CONDITIONS THAT INFLUENCE INDUCTION OF HYPERALGESIA OR ALLODYNIA
Both normal nociceptive behavior and susceptibility for the development of hyperalgesia and allodynia may vary between species, strains, sex, and the diet fed, indicating substantial genetic and dietetic impacts (483).
A. Gender
Carrageenan injections into a hindpaw at the day of birth affects nociception at adulthood, and this may be different in male and female rats. When a persistent inflammation is induced in adult rats with an intraplantar injection of complete Freund's adjuvant, neonatally injured females display stronger inflammatory hyperalgesia compared with neonatally injured males and controls (269). There are also significant gender differences with respect to the susceptibility to develop neuropathic symptoms. Female Sprague-Dawley and Long-Evans rats display increased hypersensitivity following nerve root injury compared with males. No sex differences were observed, however, in Holtzman rats (261). It is generally believed that endogenous sex steroids play a key role in mediating these sex differences in nociception (260).
B. Genotype
In eight different strains or lines of Sprague-Dawley rats, baseline nociceptive responses to heat and mechanical stimulation as well as heat hyperalgesia, mechanical hyperalgesia, and autotomy following partial sciatic nerve ligation vary greatly. Rats tested included “genetically epilepsy-prone rats,” “high autotomy selection line,” “low autotomy selection line,” “flinders sensitive line,” “Lewis rats” (an inbred line), “Fisher 344” (an inbred line), and “Sabra rats,” an outbred line. Baseline nociceptive thresholds are highest in “genetically epilepsy-prone rats” and lowest in “Fischer 344” rats. Autotomy scores are lowest in “Lewis rats” and highest in “high autotomy rats” (484). Likewise, baseline nociceptive responses and tactile hyperalgesia after an ischemic lesion of the sciatic nerve are different in four strains or lines of rats: “Sprague-Dawley,” “Wistar-Kyoto,” “spontaneously hypertensive,” and “Dark-Agouti rats” and two substrains of “Sprague-Dawley” rats supplied from two different vendors (Sprague-Dawley-BK and Sprague-Dawley-DK). Nerve lesions lead to cold hyperalgesia in “Wistar-Kyoto” and “Sprague-Dawley-BK” rats only. “Sprague-Dawley-DK” rats develop more severe mechanical hyperalgesia than “Sprague-Dawley-BK” rats (597). Complete Freund's adjuvant-induced thermal hyperalgesia is stronger in “Fisher 344” rats than in “Sprague-Dawley” or “Lewis” rats (619).
Similar differences in nociceptive behavior can be demonstrated in different strains of mice. Baseline paw withdrawal thresholds vary from 0.3 to 1.5 g when 15 different strains of mice are tested. These strains of mice also differ with respect to opioid-induced mechanical hyperalgesia. The degree of hyperalgesia ranges from 30 to 85% reduction in mechanical nociceptive thresholds (290). Likewise, from 10 different mouse strains, one strain displayed an especially robust mechanical hyperalgesia following paclitaxel treatment, while another strain was fully resistant to this treatment (491). Some of the species and strain differences in response to manipulations of the sciatic nerve may be due to differences in sciatic nerve anatomy (434).
Mutations in single genes may also have profound effects on nociception. For example, the reeler gene is an autosomal recessive mutation that may naturally occur in humans. When a similar mutation is induced in mice, the protein product Reelin, which is a large secreted extracellular matrix type protein, is missing. This protein is involved in proper neuronal positioning during development. The phenotype of mutant mice includes thermal hyperalgesia in the Hargreaves test but reduced sensitivity to noxious mechanical stimuli (von Frey hairs) (8, 547).
C. Age
A large body of evidence suggests that in neonatal and young rats nociception is quantitatively and qualitatively different compared with the adults (see reviews by Fitzgerald and colleagues, Refs. 136, 378). For example, secondary hyperalgesia may be induced only at later developmental stages in contrast to primary hyperalgesia. Mustard oil or capsaicin induce primary hyperalgesia at all postnatal days tested as assessed by electromyography flexion reflex recordings in response to mechanical stimuli at a hindpaw. In contrast, secondary hyperalgesia cannot be demonstrated at postnatal day 3 but is evident at postnatal days 10 and 21 (552). Likewise, neuropathic pain behavior is not observed in neonatal rats. In the spared nerve injury model, tactile hyperalgesia does not develop if surgery is performed before the fourth postnatal week. This delayed susceptibility to painful neuropathies may be caused by an immature response profile of spinal microglia (369). On the other hand, intraplantar injections of endothelin-1 produce a longer lasting mechanical hyperalgesia in young (postnatal day 7) rats than in adults (330).
D. Diet
Composition of the diet may have profound effects on hyperalgesia and allodynia induced by nerve lesions. In one study, “Wistar” rats were fed with a casein-based, fat-free diet for 1 wk preceding partial sciatic nerve ligation and in addition either hemp oil (20% omega-3 polyunsaturated fatty acids) or corn oil (0.7% omega-3 level). An omega-3-rich diet was associated with a stronger heat hyperalgesia, but tactile hyperalgesia was not different between dietary groups (404). When rats are fed since weaning a diet containing 85% soy protein, partial sciatic nerve ligation evokes less severe mechanical hyperalgesia (von Frey hairs) and thermal hyperalgesia (Hargreaves test) compared with animals with a normal diet. When a soy-rich diet is terminated 15 h before nerve lesion, rats develop full hyperalgesia (480). In rats fed with a synthetic polyamine-deficient diet, unilateral injection of carrageenan into a hindpaw induces a bilateral mechanical hyperalgesia (Randall-Sellito pressure test) that is less pronounced than in control animals (125).
Male mice with a prolonged restriction of caloric intake (60% of ad libitum) show fewer licking or biting responses in the Formalin test. Also, they show longer response latencies in the hot-plate test. In ad libitum control mice but not in caloric-restricted mice, partial tail amputation induces thermal hyperalgesia. Injections of collagen subcutaneously lead to thermal hyperalgesia (Hot plate test) in some strains. This collagen-induced arthritic hyperalgesia can be blocked reversibly during 9–15 wk of caloric restriction (170).
In rats treated with streptozotocin to induce diabetic polyneuropathy, thermal hyperalgesia (Hargreaves test) and mechanical hyperalgesia (von Frey thresholds) are reduced in those animals that received a 2% taurine-supplemented diet for 6–12 wk. Imaging of Ca2+ gradients in sensory neurons suggests that impaired Ca2+ homeostasis in diabetic rats is partially reversed by taurine supplemented diet (287).
V. CELLULAR SYSTEMS THAT ARE INDISPENSABLE FOR HYPERALGESIA AND ALLODYNIA
A classical proof for the involvement of a particular element for a given function is by selective inactivation or destruction of that element. In pain research, much information has been gained from targeted inactivation or destruction of brain regions or fiber tracts. The selective destruction of a well-defined group of neuronal or nonneuronal cells by the cell toxin saporin is a novel, powerful tool to assess their role for hyperalgesia and allodynia in the behaving animals (12, 249, 318, 379, 383).
Figure 2 summarizes cellular elements that are required for the full expression of hyperalgesia and/or allodynia in some animal models of inflammatory or neuropathic pain. Obviously not all these elements have been tested in all models and most likely are also not required for all forms of enhanced pain sensitivity.
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Sensory nerve fibers consist of the following: 1) capsaicin-sensitive C-fibers (61, 99, 234, 237, 336, 354, 355, 395, 476, 481), 2) IB4-sensitive C-fibers (513, 513), and 3) vagal afferents (310, 567).
Spinal cord cells consist of the following: 4) spinal dorsal horn neurons that express the neurokinin I receptor (318, 383, 542, 545, 575, 611), 5) microglia (192, 276, 375, 423, 502, 564, 618), and 6) astrocytes (213, 387, 624).
Spinal cord fiber tracts consist of the following: 7) dorsal columns (185, 239, 397, 399, 447), 8) anterior lateral quadrant (139, 153, 544), and 9) lateral funiculus [397; see monograph by Willis for a review (580)].
Brain nuclei consist of the following: 10) rostral ventromedial medulla (411, 458, 533, 534, 540); 11) nuclei reticularis gigantocellularis (151, 347, 541, 569); 12) thalamic nuclei, ventrobasal complex (446, 622); 13) anterior cingulate cortex (29, 114, 262, 431); and 14) ventrolateral orbital area (29).
Efferent nerve fibers consist of the following: 15) sympathetic postganglionic neurons (4, 17, 236, 241, 284, 342, 432, 435, 472, 482).
VI. SPINAL MECHANISMS OF HYPERALGESIA AND ALLODYNIA
The work summarized in the previous sections demonstrates that the pathogenesis of hyperalgesia and allodynia may have important spinal components. Section viA summarizes some of the global changes that have been observed in association with the development of hyperalgesia or allodynia. The subsequent sections (sect. vi, B–G) describe spinal mechanisms that are likely involved in the generation or maintenance of hyperalgesia or allodynia.
A. General Changes in Spinal Cord After Induction of Hyperalgesia and Allodynia
A large number of conditions may cause hyperalgesia and some also dynamic mechanical allodynia as outlined above. It is possible that each condition may trigger a characteristic set of changes within the central nervous system. The functional consequences of these changes may vary from being necessary or sufficient for induction of hyperalgesia or allodynia; others may facilitate, inhibit, or prevent changes in pain sensitivity; and still others may be unrelated epiphenomena. An early review on some of these activity-dependent neuroplastic changes in spinal dorsal horn is provided by Dubner and Ruda (118). Here, I focus on three conditions that trigger hyperalgesia and/or allodynia: 1) supramaximal electrical stimulation of sensory nerve fibers. This conditioning stimulus can be used in humans and in behaving experimental animals, in acute preparations and in vitro. 2) Activation of transient receptor potential vanilloid 1 channels on a subset of C-fibers by capsaicin is a commonly used model for afferent-induced secondary hyperalgesia in humans, behaving animals, and acute preparations. 3) The chronic constriction injury of the sciatic nerve is a widely used model for peripheral neuropathic pain.
1. Changes induced in spinal dorsal horn by electrical nerve stimulation
Electrical stimulation of sensory nerves at C-fiber intensity causes spinal release of amino acids including aspartate, glutamate, asparagine, serine, glycine, threonine, alanine, and taurine (400). Furthermore, a number of neuropeptides are released including substance P (266, 282, 465), galanin (85), calcitonin gene-related peptide (464), endomorphins (102), nociceptin (576), and dynorphin A (199). Neurotrophic factors such as brain-derived neurotrophic factor may be released in the spinal cord upon electrical stimulation of sensory nerves in a frequency-dependent manner. Release of brain-derived neurotrophic factor requires high-frequency stimulation at C-fiber strength (100 Hz) (282), whereas low-frequency stimulation is ineffective (1 or 2 Hz) (282, 553).
Electrical nerve stimulation at C-fiber but not at Aβ-fiber intensity also leads to posttranslational modification of proteins in spinal neurons including phosphorylation of extracellular signal-regulated kinase (212). Stimulation of dorsal roots at C-fiber intensity with low frequencies (0.05–10 Hz) (143) or higher frequencies [50 Hz (212) or 100 Hz (283, 595)] induces phosphorylation of extracellular receptor-activated kinase in superficial but not in deep spinal dorsal horn [See also Ji and Suter (214) for a review]. Activation of C-fibers by electrical stimulation (or by capsaicin) leads to an 8- to 10-fold increase in extracellular signal-regulated kinase phosphorylation in superficial spinal dorsal horn in vitro (231).
After the initial study by Hunt et al. (198), a large number of reports confirmed the transsynaptic induction of protein products of immediate-early genes such as c-fos in spinal neurons following sensory stimulation (582; for review, see Coggeshall, Ref. 80). Electrical nerve stimulation at C-fiber strength (1 Hz for 6–8 h) causes spinal upregulation of c-Fos protein (291) but no observable changes in gene expression for calcium/calmodulin-dependent protein kinase IIα, or glutamate decarboxylase (291). The pattern and the intensity of c-Fos labeling in spinal dorsal horn depends on the duration of electrical nerve stimulation; brief stimuli (seconds) cause transient labeling in superficial laminae only while longer lasting stimulation (hours) also leads to labeling in deeper layers (50). Electrical stimulation of sciatic nerve at Aδ/C but not at A-fiber intensity leads to expression of transcription factors c-Jun, Jun B, Fos B, and Krox-24 mainly in superficial layers of spinal dorsal horn and of c-Fos and Jun D throughout spinal dorsal horn (179). Expression of c-Fos in dorsal horn neurons is used as an activity marker but provides probably not sufficient evidence for neuronal plasticity and long-term changes in nociception (456). The functional role of enhanced expression of immediate-early genes in neurons of the spinal cord is still largely unknown.
2. Changes induced by capsaicin
Activation of transient receptor potential vanilloid 1 receptor channels on fine primary afferent nerve fibers by subcutaneous injections of capsaicin causes a large number of global changes within spinal dorsal horn. This includes the release of neurotransmitters and modulators such as glutamate (529), substance P (49, 296, 600), calcitonin gene-related peptide (107, 377), somatostatin (258), and nitric oxide (593).
Capsaicin injections trigger posttranslational changes in spinal dorsal horn cells such as phosphorylation of AMPA receptor subunit GluR1 (374), as well as phosphorylation of NMDA receptor 1 through protein kinase C and protein kinase A (630) in spinal dorsal horn neurons, including those with a projection to the thalamus (631).
Furthermore, capsaicin injections significantly increase the phosphorylation levels of enzymes and transcription factors such as cAMP response element-binding protein (594) and calcium/calmodulin-dependent protein kinase II (126) in the ipsilateral side of the spinal cord. Phosphorylation of extracellular signal-regulated kinase in the superficial spinal dorsal horn in vitro increases 8- to 10-fold following capsaicin injections. The extracellular signal-regulated kinase induction is reduced by blockade of NMDA, AMPA/kainate, group I metabotropic glutamate receptor, neurokinin-1, and tyrosine receptor kinase receptors and by inhibitors of protein kinase A or protein kinase C (231).
c-Fos protein is detected in neurons of ipsilateral spinal dorsal horn after subcutaneous injections of capsaicin (590), including spinothalamic tract neurons and postsynaptic dorsal column neurons (398).
Perineuronal injections of capsaicin near the tibial nerve reduce the number of cells in spinal dorsal horn with GABA immunoreactivity (571).
Intracolonic installation of capsaicin causes referred hyperalgesia in mice and recruitment of GluR1 (but not GluR2/3) AMPA receptor subunits to the plasma membrane fraction of spinal cells within 10 min. At 180 min, the increase is 3.7-fold (144).
3. Changes induced by chronic constriction injury of sciatic nerve
In the chronic constriction injury model of rats, glutamate and aspartate contents are increased on the ipsilateral side of the dorsal horn to nerve ligation on days 4, 7, and 14 after nerve injury (229). Likewise, in spinal dorsal horn of hyperalgesic rats with a sciatic nerve ligation (473), extracellular levels of glutamate and aspartate are more than doubled as revealed by microdialysis (94). Furthermore, chronic constriction injury leads to enhanced levels of 5-hydroxytryptamine (serotonin) and norepinephrine bilaterally in spinal cord (463), as well as enhanced content of neuropeptides such as neuropeptide Y (86) and galanin (85). In contrast, substance P immunoreactivity is decreased in ipsilateral spinal dorsal horn 60 days after chronic constriction injury. In the contralateral dorsal horn, calcitonin gene-related peptide and substance P immunoreactivities also decrease 60 days after chronic constriction injury (225). Neuropeptide changes may persist in spinal cord despite resolving mechanical hyperalgesia 100–120 days after chronic constriction injury. Substance P and galanin immunoreactivities are still decreased by ∼30% ipsilaterally in laminae I and II of the dorsal horn compared with sham-operated animals, while calcitonin gene-related peptide and neuropeptide Y contents in laminae I and II are no longer different from controls by this time (371). A number of proteins are either up- or downregulated after chronic constriction injury or other models of neuropathic pain. A systematic review on the proteomics in neuropathic pain research is provided by Niederberger and Geisslinger (384).
As soon as 3 days after a chronic constriction injury of the sciatic nerve, the number of GABA- and glutamate decarboxylase-immunoreactive cells decrease bilaterally to the nerve injury. At 1 wk after chronic constriction injury, the number of GABA-immunoreactive cells continues to decline bilaterally, returning to near normal numbers on the side contralateral to the nerve injury by 8 wk after the nerve injury. The number of glutamate decarboxylase immunoreactive cells begins to increase bilaterally to the nerve injury at 1 wk after chronic constriction injury and continues to increase significantly in numbers over normal values by 8 wk after the nerve injury (121). A quantitative stereological analysis of the proportions of neurons in laminae I, II, and III of the rat dorsal horn that show GABA and/or glycine immunoreactivity 2 wk after chronic constriction injury does, however, not reveal any loss of inhibitory interneurons (413), suggesting that GABA synthesis is downregulated under these conditions and that not a loss of GABAergic neurons accounts for reduced GABA immunoreactivity (see also sect. viD1aI).
Nerve injury may also alter expression or binding properties of cell surface receptors. μ-Opioid receptor binding is increased 2–5 days postinjury, bilaterally to the injury in laminae V and X but only ipsilaterally in laminae I-II. Binding returns to control levels within 10 days. δ-Opioid receptor binding declines gradually over 2–10 days postinjury. κ-Opioid receptor binding displays an increase in ipsilateral laminae I–II and in contralateral lamina X but no change on either side in lamina V, followed by a rapid decrease in κ-opioid receptor binding in all three areas on both sides of the spinal cord by day 5 postinjury (499).
Substance P binding significantly increases ipsilateral to the chronic constriction injury in laminae I/II at 5–20 days after injury and in lamina V 5 days after injury (1), while calcitonin gene-related peptide binding remains unchanged (149).
Fractalkine receptor CX3CR1, which is expressed by microglia in the basal state, is upregulated in a regionally specific manner 10 days after chronic constriction injury, while immunoreactivity and mRNA levels of its ligand remain unchanged in dorsal horn (543).
Synaptic proteins may also be altered in spinal dorsal horn of rats with a chronic constriction injury of sciatic nerve. The period of mechanical and thermal hyperalgesia parallels the duration of enhanced expression of scaffolding proteins Homer and Shank in the postsynaptic density in ipsilateral spinal dorsal horn (346). The distribution and content of synaptophysin are altered following chronic constriction injury as evaluated by immunohistochemistry, Western blotting, and densitometry. Synaptophysin is increased in the ipsilateral dorsal horn with a peak level on day 14 and then returns to baseline on day 21 post-chronic constriction injury. Interestingly, synaptophysin levels correlate temporally with thermal but not with mechanical hyperalgesia (72).
Levels and activation of enzymes in spinal cells are also altered in the course of a chronic constriction injury of sciatic nerve. Three to 14 days after chronic constriction injury of sciatic nerve, total calcium/calmodulin-dependent protein kinase II immunoreactivity is enhanced in spinal cord, and this is preceded by an increase in phosphorylated calcium/calmodulin-dependent protein kinase II immunoreactivity beginning on day 1 (96). Three and 10 days after chronic constriction injury, membrane-bound protein kinase C is increased bilaterally in the lumbar spinal cord (L1–L5) laminae I–IV and V–VI (319). Eight days after chronic constriction injury, protein kinase C-γ immunoreactivity is increased bilaterally in the spinal cord dorsal horn (322).
The number of phosphorylated p38-immunoreactive microglia increases in the laminae I–IV and IX of the spinal cord ipsilateral to a chronic constriction injury (242). Tactile hyperalgesia and activation of microglia may, however, not be closely time locked after chronic constriction injury. When scoring glial responses subjectively by changes in cell morphology, cell density, and intensity of immunoreactivity with specific activation markers (OX-42 and anti-glial fibrillary acidic protein for microglia and astrocytes, respectively), microglial responses are not pronounced in the chronic constriction injury lesioned rats. Spinal astrocytic rather than microglial responses appear to correlate more closely with pain behaviors in rats with a chronic constriction injury (81).
Synaptosomal contents of glutamate and aspartate are enhanced by 45% bilaterally in spinal cord 12 days after unilateral chronic constriction injury to sciatic nerve (492).
The number of apoptotic cells marked by the TUNEL technique plus Hoechst double labeling increases in the ipsilateral dorsal horn of the spinal cord 8 and 14 days following chronic constriction injury compared with the contralateral side and to naive and sham-treated animals (573). Following chronic constriction injury, morphological changes in the ipsilateral L4–L5 lamina II include cell loss and increased TUNEL-positive profiles and reactive gliosis. However, the total number of neurons is apparently unchanged 2 wk after chronic constriction injury when using the quantitative stereological optical dissector method and NeuN immunostaining (412).
Markers for cell activity also change. The amplitude and frequency of spontaneous and miniature excitatory postsynaptic currents increase in superficial dorsal horn neurons of rats with a chronic constriction injury of sciatic nerve for 13–25 days (28). Sciatic nerve ligation produces a bilateral increase in spinal cord 2-[14C]deoxyglucose metabolic activity in four sampling regions (laminae I–IV, V–VI, VII, and VIII–IX) of lumbar segments compared with sham-operated rats (320). The expression of c-Fos protein in spinal cord is also upregulated after chronic constriction injury (224) and may have a biphasic time course (see, however, Ref. 62). The highest number of c-Fos-positive neurons occurs during the first week after chronic constriction injury, followed by a decline at 7 and 14 days and reappearance at day 28 following injury. This biphasic time course does, however, not resemble the monophasic time course of tactile hyperalgesia in the chronic constriction injury model (211). In another study, Fos-positive cells were found bilaterally throughout laminae III–X at all time points examined up to 55 days after surgery in both chronic constriction injury and sham-operated animals (372). The number of c-Fos-positive cells in the ipsilateral spinal cord was positively correlated with the degree of hyperalgesia in one study (195).
Potential spinal mechanisms causing enhanced neuronal responsiveness are shown in Figure 3 and include direct facilitation along the chain of excitation (Fig. 3, A–C) or alteration in physiological modulation of spinal nociception, i.e., less than normal inhibition (Fig. 3, D, F, and J), conversion from inhibition to excitation (Fig. 3, E and G), or stronger than normal excitation (Fig. 3, H–J). Generation of epileptiform activity, burstlike discharges, and synchronous discharges could also amplify nociception (Fig. 3K). If pain should be caused by a unique pattern of discharges of individual neurons (pattern theory) and/or by a characteristic pattern of active versus silent neurons (population coding), any of the above cellular mechanisms could contribute to altered pain perception.
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B. Synaptic LTP
1. Definitions
LTP is a much studied cellular model of synaptic plasticity. It is generally defined as the long-lasting but not necessarily irreversible increase in synaptic strength (42). At least two different stages of LTP can be distinguished depending upon its duration and the signal transduction pathways involved. Early-phase LTP is independent of de novo protein synthesis and lasts for up to 3 h. Late-phase LTP involves protein synthesis and lasts longer than 3 h, up to the life span of an animal. Short-term potentiation of synaptic strength lasts less than half an hour. Synaptic strength is the magnitude of the postsynaptic response (i.e., the postsynaptic potential or the postsynaptic current, but not action potential firing, see below) in response to a presynaptic action potential. LTP can be expressed pre- and/or postsynaptically, i.e., synaptic strength can increase if the release of neurotransmitter(s) is enhanced and/or if the postsynaptic effects of the neurotransmitter(s) become stronger (295). LTP at synapses in hippocampus is the prime model for learning and memory formation (42). Recent studies have shown that LTP can also be induced in pain pathways and may contribute to hyperalgesia caused by inflammation, trauma, or neuropathy (453). This section deals with the latter form of LTP.
2. How to measure LTP properly
LTP is measured as an increase in monosynaptically evoked postsynaptic currents or potentials in response to a single presynaptic action potential. LTP is often studied in in vitro preparations which allow reliable recordings of synaptic strength. Whole cell patch-clamp recording is now the most often used technique. It enables some control over the composition of the intracellular fluid of the postsynaptic neurons, which may be advantageous to study postsynaptic mechanisms of LTP. If, however, a diffusible mediator is involved and dialysis of the postsynaptic neuron has to be avoided, perforated patch-clamp recordings or intracellular recordings with sharp electrodes can be used. To evaluate LTP at the first synapses in nociceptive pathways, transverse slices with long dorsal roots attached can be prepared from lumbar spinal cord of rats or mice to study monosynaptic, Aδ-fiber or C-fiber evoked excitatory postsynaptic potentials or currents in identified dorsal horn neurons (202, 425).
Some aspects of LTP can only be studied in the entire animal with primary afferent nerve fibers and descending pathways from the brain intact. In vivo C-fiber-evoked field potentials can be measured in superficial spinal dorsal horn, e.g., in response to high-intensity electrical stimulation of the sciatic nerve for up to 24 h (300). These extracellularly recorded field potentials reflect summation of postsynaptic, mainly monosynaptically evoked currents but not action potential firing (300, 469).
Monitoring presynaptic activity at synapses of primary afferent nerve fibers is technically quite demanding. In an attempt to monitor presynaptic activity in primary afferents, optical recording techniques have been utilized. Some voltage-sensitive dyes can be anterogradely transported in primary afferents to the central terminals mainly in lamina I (203) and may serve as an indicator for presynaptic electrical activity but not for transmitter release.
LTP cannot be directly investigated by recording action potential discharges of postsynaptic neurons, as action potential firing not only depends on synaptic strength but also on membrane excitability and the balance between excitatory and inhibitory input to the neuron. For the same reasons, polysynaptically evoked responses can generally not be used to study synaptic strength and changes thereof.
3. Stimuli that induce LTP in pain pathways
A) HIGH-FREQUENCY ELECTRICAL NERVE STIMULATION.
The most frequently used form of conditioning stimulation to induce LTP at synapses in the brain consists of high-frequency electrical stimulation (∼100 Hz) of an input pathway. Likewise, LTP can be induced at spinal synapses of small-diameter primary afferents by conditioning high-intensity, high-frequency burstlike stimulation (typically 100 Hz bursts given several times for 1 s at C-fiber strength) both in vitro and in vivo. In spinal cord slice preparations, both Aδ-fiber (425) and C-fiber (202, 204) evoked responses are potentiated by high-frequency stimulation when postsynaptic neurons are mildly depolarized to −70 to −50 mV. The same high-frequency stimulation induces, however, long-term depression (LTD) of Aδ-fiber-evoked responses if cells are hyperpolarized to −85 mV, suggesting that the polarity of synaptic plasticity is voltage dependent (425).
Neurons in spinal cord lamina I which express the neurokinin 1 receptor play a pivotal role for hyperalgesia in behaving animals (318, 383). Most of these neurons send a projection to supraspinal areas. Interestingly, high-frequency stimulation induces LTP selectively at C-fiber synapses with lamina I neurons that express the neurokinin 1 receptor and send a projection to the parabrachial area (202), and vice versa, high-frequency stimulation fails to induce LTP at synapses with neurons which express the neurokinin 1 receptor and send a projection to the periaqueductal gray or at synapses with neurons that do not express the neurokinin 1 receptor and which have no identified supraspinal projection (202, 204).
High-frequency stimulation at C-fiber intensity of sciatic nerve fiber afferents induces LTP of C-fiber, but not Aβ-fiber evoked field potentials in superficial spinal dorsal horn of adult, deeply anesthetized rats (299, 300, 307). In contrast, conditioning high-frequency stimulation at A-fiber intensity fails to induce LTP of either A- or C-fiber evoked field potentials in intact animals. In spinalized animals, conditioning high-frequency stimulation at Aδ-fiber intensity induces, however, LTP of C-fiber evoked field potentials (298). Likewise, in rats with a spinal nerve ligation but not in control animals, high-frequency stimulation at a low intensity (10 V, 0.5-ms pulses) induces LTP of C-fiber evoked field potentials, whereas high-intensity high-frequency stimulation (30 V, 0.5-ms pulses) is effective in both control and in neuropathic animals (596). This suggests that the threshold for inducing LTP is lowered under various neuropathic conditions.
B) LOW-FREQUENCY ELECTRICAL NERVE STIMULATION.
For most of the C-fiber afferents it is not typical to discharge at rates as high as 100 imp/s. Some C-fibers may, however, discharge at these high rates but only for short periods of time, e.g., at the beginning of a noxious mechanical stimulus (165). Many C-fibers discharge at considerably lower rates, ∼1–10 imp/s, e.g., in response to an inflammation or an injury (422). Conditioning stimulation within this lower frequency band is successfully used to induce LTP at C-fiber synapses. In a spinal cord-dorsal root slice preparation, conditioning electrical low-frequency stimulation (2 Hz for 2–3 min, C-fiber strength) of dorsal root afferents induces LTP selectively at C-fiber synapses with lamina I neurons that express the neurokinin 1 receptor and project to the periaqueductal gray (204). C-fiber synapses with lamina I neurons which express the neurokinin 1 receptor and project to the parabrachial area or with no identified supraspinal projection are, in contrast, not potentiated by low-frequency stimulation (204). Thus the pattern and the frequency of discharges in C-fibers determine which synapses at the origin of different ascending pain pathways are potentiated.
In spinal cord slices from neonatal rats, field potentials evoked by electrical stimulation in the tract of Lissauer are potentiated by repetitive burstlike stimulation at 10 Hz (514). Some authors could induce LTP at synapses in deep spinal dorsal horn in slices from young (3–6 day old) but not older (9–16 day old) rats (147) in contrast to a recent study where a robust LTP was induced in superficial dorsal horn by low-frequency stimulation at C-fiber synapses in more mature animals (21- to 28-day-old rats) (204).
In deeply anesthetized adult rats with their spinal cords left intact, low-frequency stimulation (at 2 Hz for 2–3 min) of sciatic nerve fibers at C-fiber intensity but not at Aδ-fiber intensity also triggers LTP of C-fiber evoked potentials (204).
Thus high-frequency stimulation and low-frequency stimulation may have fundamentally different effects on LTP induction at different C-fiber synapses. This finding is in line with previous reports also illustrating that the frequency of afferent barrage in C-fibers may have qualitatively different effects in spinal cord. For example, brain-derived neurotrophic factor is released from primary afferents in spinal cord slices in an activity-dependent manner by high-frequency stimulation at 100 Hz but not by 1-Hz low-frequency stimulation of primary afferent nerve fibers, while substance P is also released by low-frequency stimulation (282).
C) NATURAL NOXIOUS STIMULATION INDUCES LTP.
At synapses in the brain, LTP induction requires synchronous, high-frequency presynaptic activity or pairing of low-level presynaptic activity with strong postsynaptic depolarization. At least some of the C-fiber synapses are apparently unique in that LTP can be induced by low-frequency stimulation and by natural, low- or high-frequency, asynchronous and irregular discharge patterns in sensory nerve fibers. In animals with spinal cord and descending pathways intact, intraplantar, subcutaneous injections of capsaicin (100 μl, 1%) or Formalin (100 μl, 5%) induce slowly rising LTP (204).
Some forms of low-level afferent input can induce LTP only if descending, presumably inhibitory pathways are interrupted or weakened. Noxious radiant heating of the skin at a hindpaw induces LTP in spinalized animals but not in animals with spinal cord intact (455). Likewise, repetitive, noxious squeezing of the skin or the sciatic nerve induces LTP of C-fiber evoked field potentials only in spinalized rats (455). These findings indicate that endogenous antinociceptive systems not only raise thresholds for nociception but also those for the induction of LTP.
D) PHARMACOLOGICAL INDUCTION OF LTP.
At C-fiber synapses LTP can also be induced in the absence of any presynaptic activity. Spinal application of a dopamine 1/dopamine 5 receptor agonist (SKF 38393) in vivo induces a slowly developing LTP of C-fiber-evoked field potentials which lasts for at least 10 h and which is blocked by a dopamine 1/dopamine 5 antagonist (SCH 23390) (602). In spinalized, deeply anesthetized, adult rats, superfusions of spinal cord segments with NMDA, substance P, or neurokinin A are all sufficient to induce LTP of C-fiber evoked field potentials (297). With spinal cord and descending (inhibitory) pathways intact, spinal applications of NMDA, substance P, or neurokinin A fail, however, to induce LTP of C-fiber evoked field potentials (297). When applied spinally to rats, tumor necrosis factor-α may also potentiate synaptic strength in C-fibers (301). In vitro, bath application of serotonin (10–50 μM) may initially depress and after wash-out potentiate responses for more than 30 min of some neurons in laminae I-III of spinal dorsal horn slices evoked by stimulating near the dorsolateral margin of the spinal cord (184).
E) LTP OF A-FIBER EVOKED RESPONSES.
A-fiber evoked spinal field potentials are depressed by conditioning 50-Hz stimulation of sciatic nerve fibers. After systemic application of the GABAA receptor antagonist bicuculline, the same conditioning stimulus now produces LTP rather than LTD (345). Similarly, 50-Hz conditioning stimulation produces short-lasting potentiation followed by LTD in control animals but LTP in animals with a chronic constriction injury of sciatic nerve (344). Topical application of muscimol, a GABAA receptor agonist, to spinal cord prevents tetanus-induced LTP of A-fiber evoked field potentials in animals with a chronic constriction injury (343). This again suggests that the polarity of synaptic plasticity is context sensitive and not solely dominated by the type of afferent input.
4. Signal transduction pathways of LTP at C-fiber synapses
In principle, LTP could be induced and/or expressed by presynaptic or by postsynaptic mechanisms or by any combination thereof (Fig. 3A). At present, there is clear evidence for a postsynaptic, Ca2+-dependent form of LTP induction in spinal cord lamina I neurons (202, 204). Indirect evidence suggests that in addition excitability of presynaptic terminal of primary afferents may be enhanced after LTP-inducing stimuli (203) and that synaptosomal level of aspartate and glutamate, but not that of glycine or GABA, is elevated in rats with a chronic constriction injury of the sciatic nerve (492). These findings are compatible with a presynaptic contribution to synaptic plasticity in spinal dorsal horn.
Induction of LTP at C-fiber synapses requires coactivation of neurokinin 1 and neurokinin 2 receptors (300), opening of ionotropic glutamate receptors of the NMDA type (202, 204, 299), opening of T-type voltage-gated calcium channels (202, 204), and activation of group I but not group II or III metabotropic glutamate receptors (23). Activation of neurokinin 1 receptors by substance P may directly enhance single NMDA channel opening (292) and NMDA receptor-mediated currents in lamina I neurons (202), and all this may lead to a substantial rise in postsynaptic [Ca2+]i. It is presently unknown if Ca2+ influx through Ca2+-permeable AMPA receptors is required for LTP induction in pain pathways. Some indirect evidence suggests, however, that this might be the case (172, 612).
In any case, a rise in postsynaptic [Ca2+]i is essential for LTP induction, and the magnitude in [Ca2+]i rise is linearly correlated with the magnitude of LTP in vitro (202). Recent data demonstrate that LTP-inducing stimuli cause substantial rise in [Ca2+]i in lamina I neurons not only in slice preparations, but also in intact animals (204). Not surprisingly therefore, signal transduction involves Ca2+-dependent pathways including activation of protein kinase C, calcium/calmodulin-dependent protein kinase II, protein kinase A, phospholipase C, inositol trisphosphate receptors, mitogen-activated protein kinase, nitric oxide synthase, and ephrin-EphB2 receptor tyrosine kinase signaling (202, 204, 493, 595, 601, 621).
When assessed with voltage-sensitive dyes, the presynaptic facilitation of electrical activity in primary afferents after LTP-inducing stimuli is partially sensitive to inducible nitric oxide synthase inhibitor (AMT), a blocker of glial cell metabolisms (monofluoroacetic acid, MFA), and a metabotropic glutamate receptor group I antagonist (LY367385) (203).
Inhibition of protein synthesis in spinal cord by either cycloheximide or anisomycin selectively inhibits the maintenance of the late phase of spinal LTP but does not affect either LTP induction or baseline responses of C-fiber evoked field potentials (190).
Potential targets of these signaling pathways are synaptic proteins, including glutamate receptors of the AMPA subtype. And, indeed, already 5 min after capsaicin injections that induce LTP (204), the AMPA receptor subunit GluR1 (at Ser-831 and Ser-845) in spinal dorsal horn becomes phosphorylated for at least 60 min (128) via activation of protein kinases A and C (127) and via calcium/calmodulin-dependent protein kinase II (126). Capsaicin injections also trigger the translocation of GluR1-containing AMPA receptors to the postsynaptic membrane of nonpeptidergic nociceptive primary afferent synapses (272). Phosphorylation of the GluR1 subunit is an essential step of LTP at glutamatergic synapses (44, 277).
Importantly, the very same signal transduction pathways are required for full expression of hyperalgesia in animal models of inflammatory and neuropathic pain (335, 407, 452, 578).
5. Prevention of LTP in pain pathways
LTP induction can be prevented by blockade of any of the above-mentioned essential elements of signal transduction for LTP. In mature rats, deep (surgical) level of anesthesia with either urethane, isoflurane, or sevoflurane is, however, insufficient to preempt LTP induction of C-fiber evoked field potentials (36). In contrast, the noble gas xenon, which has NMDA receptor blocking and anesthetic properties, also prevents induction of LTP at C-fiber synapses in intact rats (37). LTP can also be prevented by low-dose intravenous infusion of μ-opioid receptor agonist fentanyl (36). Similarly, LTP of spinal field potentials elicited by stimulation in the tract of Lissauer in spinal cord slices is blocked by [d-Ala2,N-MePhe4,Gly-ol]-enkephalin (DAMGO), a more specific agonist at these receptors (514). Activation of spinal α2-adrenoreceptors by clonidine (150) or spinal application of the benzodiazepine diazepam (191) also prevents LTP induction in vivo.
Functional blockade of glial cells by intrathecal administration of fluorocitrate changes the polarity of high-frequency stimulation induced synaptic plasticity. When high-frequency stimulation is given 1 h but not 3 h after fluorocitrate, LTD but no LTP of C-fiber evoked field potentials is induced (307).
6. Reversal of LTP in pain pathways
LTP of C-fiber evoked field potentials can be reversed by brief, high-frequency conditioning electrical stimulation of sciatic nerve fibers at Aδ-fiber intensity (298). Reversal of LTP by Aδ-fiber stimulation is time dependent and effective only when applied 15 or 60 min but not 3 h after LTP induction (616).
Spinal application of either neurokinin 1 or neurokinin 2 receptor antagonists 1–3 h after high-frequency stimulation, i.e., after LTP is established, does not affect maintenance of LTP (300), suggesting that activation of these receptors, which are required for the induction of LTP, are not essential for its maintenance.
7. Functional role of LTP in pain pathways
Modulation of synaptic strength is a powerful mechanism to control signal flow in selected pathways. A typical consequence of LTP at excitatory synapses would be an increase in action potential firing of the same and perhaps also of downstream neurons in response to a given stimulus. And indeed, LTP-inducing conditioning stimuli have been found to facilitate action potential firing of multireceptive neurons in deep dorsal horn (3, 174, 403, 445, 546). This is likely due to LTP at the first synapse in the nociceptive pathway, but other mechanisms of facilitation should not be excluded. Action potential firing would also be enhanced if membrane excitability is increased, i.e., the thresholds for action potential firing are lowered, and this has been shown for nociceptive neurons in deep spinal dorsal horn. Furthermore, discharges increase also if inhibition is less effective or if inhibition is even reversed and becomes excitatory, e.g., due to a reversal of the anion gradient in the postsynaptic neuron (88, 89).
High-frequency stimulation of sciatic nerve fibers which induces LTP at synapses of C-fibers in spinal cord has behavioral consequences in rats and causes thermal hyperalgesia at the ipsilateral hindpaw for 6 days (621). This suggests that LTP at C-fiber synapses has an impact on nociceptive behavior of laboratory animals and humans (see next paragraph).
8. Perceptual correlates of LTP in pain pathways in human subjects
An indispensable proof for any proposed mechanism of hyperalgesia is an appropriate correlate in humans. And, indeed, conditioning high-frequency stimulation of cutaneous peptidergic afferents in humans causes increased pain perception in response to electrical test stimuli applied through the same stimulation electrode (246). Noxious stimulation with punctate mechanical probes in skin adjacent to the high-frequency stimulation conditioning skin site uncovers a marked (2- to 3-fold) increase in pain sensitivity, i.e., secondary hyperalgesia (246). Touching the skin around the conditioning stimulation electrode with a soft cotton wisp evokes pain only after high-frequency stimulation. Thus high-frequency stimulation also induces secondary mechanical dynamic allodynia, possibly involving heterosynaptic mechanisms in humans (248). Hyperalgesia at the conditioned site but not secondary hyperalgesia at adjacent skin areas is prevented by pretreatment with ketamine (247), a clinically used substance which, among other effects, also blocks NMDA receptors.
Interestingly, all thermal modalities comprising cold and warm detection thresholds, cold and heat pain thresholds, as well as pain summation (perceptual wind-up) remain unaltered after conditioning high-frequency stimulation of peptidergic skin nerve fibers (268).
When verbal pain descriptors are used to evaluate pain in addition to its perceived intensity after high-frequency stimulation, a significant long-term increase in scores for sensory but not for affective descriptors of pain is detected (166). Within the sensory descriptors, those describing superficial pain, those for heat pain, and those for sharp mechanical pain are all potentiated. The authors conclude that brief painful stimuli rarely have a strong affective component and that perceived pain after high-frequency stimulation exhibits predominantly a potentiation of the C-fiber-mediated percept hot and burning (166).
In human subjects, conditioning low-frequency stimulation causes also an increased pain sensitivity in the area around the low-frequency stimulation conditioned skin site but a depression of pain evoked by stimulation through the same electrode (246).
LTP at synapses between primary afferent C-fibers and a group of nociceptive neurons in spinal cord lamina I which express the neurokinin 1 receptor for substance P is a potential mechanism underlying some forms of pain amplification in behaving animals and perhaps human subjects. Both LTP and hyperalgesia involve the same essential elements, i.e., primary afferent peptidergic C-fibers and lamina I neurons which express the neurokinin 1 receptor. Indirect evidence suggests that ongoing activity in primary afferent C-fibers is essential not only for evoked, but also for spontaneous neuropathic and inflammatory pain (110). Furthermore, induction protocols, pharmacological profile, and signal transduction pathways are virtually identical (453).
C. Intrinsic Plasticity
Active and passive membrane properties determine the input-output relationship of all neurons. Thus changes of electrical membrane properties constitute another powerful means to modulate signal transmission in neuronal networks. A single neuron may integrate the information from 104 synapses, and the resultant output is conveyed by action potentials that are typically generated at the axon hillock and nearby somatic membrane. Changes in membrane excitability of neurons will globally or locally modulate the throughput from synapses impinging on the dendrites and the soma of the postsynaptic neuron. In analogy of synaptic plasticity, long-lasting changes in membrane excitability are called “intrinsic plasticity,” which adds to the computational power of neurons. Intrinsic plasticity may include but is not limited to postsynaptic changes in thresholds for action potential firing (Fig. 3B), changes in discharge patterns and accommodation of firing (Fig. 3C), and presynaptic changes in action potential shape (width and height). Intrinsic plasticity may be global or local, e.g., restricted to some dendrites, axonal branches, or presynaptic terminals. In striking contrast to the comprehensive literature on the modulation of membrane properties of primary afferent nerve fibers, little is known about intrinsic plasticity of nociceptive neurons in the central nervous system and their role for hyperalgesia and allodynia.
Spinal dorsal horn neurons may have quite distinct membrane and discharge properties when grouped by morphology (159, 289, 418), supraspinal projection (193, 441), lamina location of their cell bodies (442, 506), type of afferent input in vivo (572) and in vitro (304), transmitter content (116, 176, 217, 466), or developmental stage (181, 450). The ionic basis for some of the discharge and membrane properties of spinal dorsal horn neurons has been explored (337, 338, 441, 448, 449, 583).
1. Voltage-dependent sodium channels
Dissociated lumbar spinal dorsal horn neurons show the characteristic fast-activating and fast-inactivating sodium currents. Spinal cord contusion injury leads to a shift of the steady-state activation and inactivation of the sodium current towards more depolarized potentials. The increased persistent sodium current and ramp current is consistent with an upregulation of voltage-gated sodium channels of the Nav1.3 subtype within dorsal horn neurons that has been observed after spinal cord contusion injury (Fig. 3C) (163, 265). Likewise, several days after a chronic constriction injury of the sciatic nerve, i.e., at a time point when hyperalgesia is fully expressed, Nav1.3 channels are upregulated in dorsal horn nociceptive neurons. Extracellular recordings reveal enhanced responsiveness of spinal dorsal horn neurons to natural sensory stimuli (164). In another study, chronic constriction injury did not affect resting membrane potential, rheobase, or input resistance of neurons recorded in superficial spinal dorsal horn in slices (28). Neurons in spinal cord laminae III–VI, i.e., in deep dorsal horn, express, however, intrinsic plasticity. In these neurons, associative spike pairing stimulation induces a long-lasting increase in membrane excitability as assessed by lowering the threshold for action potential firing and an increase in the number of action potential firing in response to current injection or synaptic stimulation. Enhanced excitability depends on activation of NMDA receptors and a rise in postsynaptic [Ca2+]i (238).
2. Voltage-dependent potassium currents
Activation of either protein kinase A or protein kinase C reduces transient outward (A-type) potassium currents (188) and strongly enhances membrane excitability of dorsal horn neurons in cultured neurons from superficial spinal dorsal horn of mice (Fig. 3C), possibly via activation of extracellular signal-regulated kinase (187). Membrane excitability of spinal dorsal horn neurons is dampened by activation of Kv4.2 channels. The activation of the extracellular signal-regulated kinase pathways leads to hyperexcitability of spinal dorsal horn neurons in normal mice but not in Kv4.2 knock-out mice. These knock-out mice also show reduced hyperalgesia in the second phase of the Formalin test and after carrageenan injections into a paw (186).
3. Plateau potentials
Plateau potentials are intrinsic mechanisms for input-output amplification. The resulting intense firing and prolonged afterdischarges in response to nociceptive stimulation of neurons in layer V in a spinal cord slice preparation depend on nonlinear intrinsic membrane properties (365). Plateau potentials are rarely found under control conditions in spinal dorsal horn neurons in vitro (<10% of lamina V neurons) (105) or in vivo [4/33 neurons (215)]. Pharmacological activation of group I metabotropic glutamate receptors (by 1S,3R-ACPD) converts tonic firing neurons into plateau firing neurons (46% of lamina V neurons) (Fig. 3C). In contrast, activation of GABAB receptors inhibits plateau responses (444) and converts plateau firing back to tonic firing (105). The antagonistic actions of group I metabotropic glutamate receptors versus GABAB receptors is mediated by inwardly rectifying potassium channels (Kir3) (105). Simultaneous activation of metabotropic glutamate receptors and blockade of GABAB receptors induces rhythmic firing. Switching the discharge mode from tonic to plateau potentials amplifies and improves faithful transmission, whereas rhythmic bursting results in poor transmission capabilities (105). In addition to a role of GABAB receptors, GABAA receptors also inhibit plateau potentials, as bicuculline (50 μM) may facilitate plateau responses (444). Inhibition of plateau properties is also observed in the presence of tetrodotoxin, suggesting a direct action on the neuron under study.
Induction of a unilateral peripheral inflammation with complete Freund's adjuvant leads to hyperalgesia, but the principal passive and active membrane properties and the firing patterns of ipsilateral spinal lamina I neurons are not different in transverse spinal cord slices taken from control rats or rats with an inflammation at a hindpaw (370).
Chronic constriction injury of one sciatic nerve leads to tactile and thermal hyperalgesia in transgenic mice which express the enhanced green fluorescent protein (EGFP) under the promoter of glutamate decarboxylase 67 to label GABAergic neurons. In transverse slices from lumbar spinal cord, membrane excitability of lamina II GABAergic neurons from neuropathic or sham-treated animals is indistinguishable, suggesting that intrinsic plasticity of these neurons is not an essential mechanism of neuropathic pain (467).
D. Changes of Inhibitory Control
Spinal nociceptive neurons are under permanent and powerful inhibitory control, which is indispensable for orderly processing of sensory information in spinal dorsal horn and for a normal perception of pain. Inhibitory systems in spinal dorsal horn serve four principle functions to maintain proper nociception [see Fig. 4 and a recent review for details (454)]: 1) attenuation of the responses of nociceptive neurons to maintain the proper response levels during nociception; 2) muting nociceptive neurons in the absence of noxious stimuli, thereby preventing spontaneous pain; 3) separating labeled lines for nociceptive and nonnociceptive information to prevent cross-talk between sensory modalities; and 4) limiting the spread of excitation to somatotopically adequate areas of the central nervous system.
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The major fast inhibitory neurotransmitters in spinal dorsal horn are GABA and glycine acting on ionotropic, Cl−-permeable GABAA or glycine receptors or metabotropic (G protein-coupled) GABAB receptors. GABA and glycine are coreleased at some inhibitory synapses in spinal cord (221) including laminae I–II, at least in immature rats (235). Junctional codetection by glycine and GABAA receptors ceases, however, by adulthood, leaving pure glycinergic postsynaptic responses in lamina I and either glycinergic or GABAergic responses at equal proportions in lamina II (235). Another cotransmitter at GABAergic synapses is ATP. ATP is coreleased in a subset of GABAergic but not glutamatergic neurons and evokes excitatory synaptic currents in cultured spinal cord lamina I–III neurons (218). Ongoing release of ATP and hydrolysis to adenosine depresses GABA-mediated inhibitory postsynaptic currents through action on adenosine receptors (218).
1. GABAergic systems
A) MODULATION OF THE SPINAL GABAERGIC SYSTEMS.
The spinal GABAergic systems can be modulated by neuropathies, inflammation, pharmacological means, and hormones. GABA-mediated neurotransmission may be altered by changes in release probability, number of release sites, and diffusion. The speed by which GABA is removed from the synaptic cleft may also change. Furthermore, the number, the location, and the subunit composition of synaptic GABA receptors may be modulated, e.g., by phosphorylation as reviewed (Fig. 3D) (69). Changes in the anion gradient of postsynaptic neurons may convert GABA-induced hyperpolarization into depolarization (see sect. viD1cIB).
I) Modulation by neuropathies. A) Peripheral and spinal nerve injuries. Chronic constriction injury of the sciatic nerve induces complex changes in the GABAergic system, but apparently neither the number of GABAergic neurons in spinal dorsal horn nor their electrophysiological properties change. In fact, in rats which develop thermal hyperalgesia following chronic constriction injury of sciatic nerve, the number of neurons in laminae I–III with GABA or glycine immunoreactivity is not different from controls, as evaluated with unbiased stereological methods 14 days after nerve injury (413). Likewise, in rats with a spared nerve injury, the levels of GABA, the vesicular GABA transporter, or the β3-subunit of the GABAA receptor at synapses in the medial part of the superficial dorsal horn are not different from controls (414). In mice that express EGFP under the glutamate decarboxylase 67 promoter, the active and passive membrane properties of identified spinal GABAergic neurons can be assessed quantitatively. In mice with a chronic constriction injury of the sciatic nerve and severe mechanical and thermal hyperalgesia action potential thresholds and widths, membrane resting potential and membrane input resistance as well as firing patterns are all unchanged compared with sham-treated animals. This suggests that changes in membrane excitability or discharge patterns of GABAergic neurons in spinal cord lamina II are unlikely causes for pain in the chronic constriction injury model (467).
Other important features of the GABAergic system do, however, change under the conditions of neuropathy. GABA-like immunoreactivity of neuronal profiles is severely reduced mainly in ipsilateral laminae I–II but also on the contralateral side already 3 days after chronic constriction injury of the sciatic nerve (201). At 3 wk following chronic constriction injury, GABA immunoreactivity is almost absent bilaterally. Some recovery begins at 5 wk and is almost complete on the contralateral but not ipsilateral side at 7 wk (201). The number of GABA immunoreactive neurons is reduced 7 days after a partial injury of the tibial nerve, not only in the termination area of the tibial but also of the peroneal nerve ipsi- and contralateral to the lesion site (278). The reduced immunoreactivity is likely due to diminished GABA synthesis, as the number of GABAergic neurons remains stable and GABAergic neurons do not express caspase-3, an indicator of apoptotic cell death (278). Indeed, glutamate decarboxylase 65 but not glutamate decarboxylase 67 protein levels decrease 6 days up to 4 wk after chronic constriction injury and for even longer in the spared nerve injury model (362). After unilateral transection of a sciatic nerve, the number of neurons in spinal dorsal horn with a detectable immunoreactivity for GABA and the GABA content in spinal homogenates decreases 2–4 wk after neurectomy (59). In contrast, spinal content of GABA is enhanced bilaterally 1–30 days after a unilateral chronic constriction injury of sciatic nerve in the rat (463). The reasons for these discrepant results are presently unknown. Daily pretreatment with intrathecal MK-801 to block spinal NMDA receptors abolishes increases in GABA and glycine levels in spinal cord ipsilateral to the chronic constriction injury of sciatic nerve and prevents development of hyperalgesia (463).
In animals with a unilateral chronic constriction injury of the sciatic nerve, spinal levels of GABA transporter GAT-1 are reduced bilaterally to ∼40% 7 days after the ligature compared with controls (343, 478). This should lead to a reduction of GABA in the terminals in the spinal dorsal horn. This is in line with the observation that in spinal cord slices taken from rats with spinal nerve ligation potassium-induced release of GABA is reduced compared with sham-operated controls (281). In contrast, GAT-1 downregulation does not lead to detectable changes in synaptosomal contents of GABA which are unchanged in spinal cord 12 days after unilateral chronic constriction injury of the sciatic nerve (492). A recent study found on the other hand an upregulation of GAT-1 in spinal dorsal horn of rats with a chronic constriction injury of the sciatic nerve (95). In this study, pharmacological blockage of GAT-1 reduced tactile and thermal hyperalgesia.
In any way, postsynaptic GABAergic inhibition seems to be impaired in spinal dorsal horn of neuropathic rats. But the animal model used may be of importance. The proportion of neurons in superficial spinal dorsal horn in vitro that express primary afferent-evoked inhibitory postsynaptic currents is diminished in animals with chronic constriction injury and spared nerve injury but not in animals with a sciatic nerve transection (362). Likewise, amplitudes and durations of inhibitory postsynaptic currents are reduced after chronic constriction injury and spared nerve injury but not after sciatic nerve transection (362). After spared nerve injury but not after sciatic nerve transection inhibitory postsynaptic current kinetics are changed, then resembling mostly glycinergic but not GABAergic currents, suggesting a preferential loss of GABAergic inhibition (362). Similarly frequency but not amplitude of GABAergic but not glycinergic miniature inhibitory postsynaptic currents is reduced after chronic constriction injury or spared nerve injury, which is also consistent with diminished GABAergic release (Fig. 3F) (362).
Unilateral chronic constriction injury of the sciatic nerve also has effects at the receptor level on primary afferent nerve fibers. The number of GABAA-receptor γ2 subunit mRNA-positive medium to large size neurons in ipsilateral L4/L5 dorsal root ganglion neurons is reduced after chronic constriction injury (389). This suggests that GABAA receptors may be downregulated at the central terminals of primary afferent nerve fibers. If so, the sensitivity of these terminals to GABA should be diminished. And indeed, the mean depolarization elicited by GABA on normal dorsal roots is significantly reduced following sciatic axotomy, dorsal root axotomy, or crush injury. In contrast, chronic sciatic crush injury has no effect on the GABA sensitivity of dorsal root terminals (243).
Two to four weeks after a unilateral neurectomy of the sciatic nerve, GABAB receptor binding in lamina II of the spinal cord is downregulated. In contrast, GABAA binding is enhanced following nerve transection (58).
There is, however, also evidence suggesting that spinal GABAergic inhibition may be enhanced under some conditions of neuropathy. Potency of GABAA receptor blocker bicuculline to enhance Aδ- and C-fiber evoked responses of spinal dorsal horn neurons is higher in rats with a spinal nerve ligation (254). Furthermore, in rats with a chronic constriction injury of the sciatic nerve, activation of GABAA receptors may lead to a depolarization of postsynaptic neurons rather than to an inhibition as discussed in section viD1cIB.
B) Spinal cord trauma and ischemia. The number of GABA immunoreactive cells in lumbar spinal dorsal horn of rats with a transient spinal cord ischemia and mechanical hyperalgesia is reduced bilaterally at 2–3 days but not at 14 days after injury (615). This suggests a reversible reduction of GABA content rather than a loss of GABAergic neurons. And, indeed, after a spinal cord hemisection at the lower thoracic level, the GABA-synthesizing enzyme GAD67 is reduced bilaterally in laminae I and II of the lumbar spinal dorsal horn (162). This possibly translates into reduced GABAergic function as responses of multireceptive neurons 1–2 segments caudal to the lesion are less strongly inhibited by GABA (117).
Seven days following contusion injury of the thoracic spinal cord and development of mechanical hyperalgesia (von Frey thresholds), impaired GABAergic inhibition may affect various sensory modalities differentially. Iontophoretic application of bicuculline in normal animals results in reversible increases in mechanoreceptive field sizes, spontaneous firing rates, and responses to brushing and pinching the skin. In allodynic rats, bicuculline application also enlarges receptive field sizes but has little or no effect on responses to brushing or pinching the skin (117). This suggests that tonic GABAergic inhibition of dynamic mechanical and noxious mechanical input may be reduced in these neurons.
Finally, after spinal cord injury, the network effects of GABAA receptor activation may switch from inhibition to facilitation. In rats with a thoracic spinal cord injury but not in normal rats, blockade of GABAA receptors by iontophoretic application of bicuculline reduces rather than enhances afterdischarges of deep dorsal horn multireceptive neurons to noxious skin pinching (but not to brushing) (117). This suggests that tonic activation of GABAA receptors directly or indirectly facilitates afterdischarges in hyperalgesic rats. In section viD1c synaptic mechanisms are described by which GABAergic inhibition may turn into excitation.
II) Modulation by inflammation. The spinal GABAergic system may also be modulated by peripheral inflammation. For example, GABAB receptor subtypes 1 and 2 (GABAB1/2) mRNA levels are increased bilaterally in the dorsal horn of the spinal cord 24 h after Formalin injection into a hindpaw (329). This upregulation does, however, not translate into an increased GABAB receptor function, at least when determined by its activation of G proteins (457). GABAB1 but not GABAB2 receptors also increase in dorsal root ganglion ipsilaterally but not contralaterally to the injection site (329).
Inflammation may further result in rapid regulation of GABA transporters, as GABA uptake is increased in synaptosomes from mouse spinal cord as soon as 20 and 120 min after subcutaneous injection of Formalin into a hindpaw (189).
Interestingly, the overall modulatory effect of spinal GABAA receptors on behavioral nociceptive thresholds may be reversed during an inflammation induced by complete Freund's adjuvant in rats. In normal rats, intrathecal application of GABAA receptor agonist muscimol increases and GABAA receptor antagonist gabazine lowers nociceptive thresholds. In rats with an inflammation, the effects are inverted (19).
III) Pharmacological modulation of the spinal GABAergic system. The activity of spinal GABAergic neurons, the synthesis, and the release of GABA and the properties and functions of GABA receptors can all be modulated pharmacologically.
Single dose but not repeated systemic administration of morphine at analgesic doses enhances GABA content and glutamate decarboxylase activity in rat spinal dorsal horn (259). However, an acute application of selective μ-opioid receptor agonist DAMGO selectively depresses GABAergic and glycinergic inhibitory postsynaptic currents in lamina II neurons in vitro, probably via a presynaptic mechanism (363). Sustained opioid exposure may lead to apoptotic death of neurons in spinal cord, many of which express glutamic acid decarboxylase for the synthesis of GABA (323).
Indirect evidence suggests that norepinephrine and phenylephrine may excite GABAergic neurons as they enhance the frequency of action potential-dependent spontaneous GABAergic inhibitory postsynaptic currents in dorsal horn neurons (25). There is also direct evidence of an excitatory action of norepinephrine acting on α1-adrenoreceptors on GABAergic neurons. In perforated whole cell patch-clamp recordings from lamina II neurons, bath application of norepinephrine directly depolarizes GABAergic neurons identified by the expression of EGFP in transgenic mice. This action of norepinephrine is partially selective for GABAergic neurons as 42% of those but only 5% of the unidentified neurons are depolarized (M. Gassner and J. Sandkühler, unpublished observations).
Release of GABA is enhanced by activation of spinal muscarinic receptors, as determined by enhanced frequencies of spontaneous GABAergic inhibitory postsynaptic currents recorded from lamina II neurons in spinal cord slices (26). Similarly, frequency of GABAA receptor-mediated miniature inhibitory postsynaptic currents in lamina II neurons increases after acetylcholine application. This effect is blocked by atropine (286). Norepinephrine (and α1-adrenoreceptor agonist phenylephrine) enhances frequency of GABAergic miniature inhibitory postsynaptic currents with a twofold greater efficacy than glycinergic miniature inhibitory postsynaptic currents. Postsynaptic responses to GABA or glycine are not affected, nor are frequencies of miniature excitatory postsynaptic currents changed by norepinephrine (27). α2-Adrenoreceptor agonist clonidine and β-adrenoreceptor agonist isoproterenol are without effect (27). Potassium-stimulated release of GABA is facilitated by brain-derived neurotrophic factor in an adult rat isolated dorsal horn preparation (408) by a yet unknown mechanism. In adult pentobarbital anesthetized rats, spinal release of GABA as detected by microdialysis is enhanced by up to 80% by intrathecal application of selective 5-HT3 receptor agonist 1-phenylbiguanide (230).
The release of GABA can be blocked by various substances. Nocistatin selectively blocks neurotransmitter release equally from inhibitory GABAergic or glycinergic spinal dorsal horn interneurons by 50% via a pertussis toxin-sensitive mechanism (614). Glutamatergic transmission is, in contrast, not affected. Bath application of adenosine reduces amplitudes of evoked GABAergic and glycinergic inhibitory postsynaptic currents of rat spinal lamina II neurons and diminishes frequency but not amplitudes of spontaneous inhibitory postsynaptic currents in vitro (603), suggesting presynaptic suppression of inhibitory transmission. The A1 receptor antagonist 8-cyclopentyl-1,3-dimethylxanthine (CPT) reverses this inhibition.
Once GABA is released, its effects can still be modulated at the receptor level. For example, protein kinase C phosphorylation of the β1-, γ2S-, and γ2L-subunits of the GABAA receptor attenuates GABA-induced currents (256). This protein kinase C-dependent modulation may play a role for afferent-induced facilitation of spinal nociception. In the monkey, iontophoretic application of GABA or muscimol near the recording site of multireceptive lumbar spinal dorsal horn neurons reduces responses to pinching the skin. Pharmacological activation of protein kinase C reduces this inhibition by GABA or muscimol. The inhibition by GABA is almost absent when these agonists are applied 15 min after intradermal injection of capsaicin (which activates protein kinase C in spinal neurons). Inhibition returns to normal ∼1.5 h after capsaicin injection (293). The inhibition by muscimol is not consistently affected (293). Proinflammatory cytokines may be involved as bath application of either interleukin-1 or interleukin-2 inhibits the frequency of spontaneous inhibitory postsynaptic currents in lamina II neurons (232). The subunit composition of the GABAA receptor can be modulated within days, e.g., in oxytocin neurons during pregnancy and lactation. Predominance of the α1-subunit reveals fast channel gating kinetics while predominance of α2-subunits slows kinetics (48).
In rats with a L5 spinal nerve ligation, mechanical and thermal hyperalgesia is reduced by intrathecal brain-derived neurotrophic factor (281). Bath application of brain-derived neurotrophic factor restores impaired GABA release in spinal cord slices of these rats (281).
B) MODULATION OF HYPERALGESIA AND ALLODYNIA BY THE SPINAL GABAERGIC SYSTEM.
I) Facilitation of hyperalgesia and allodynia by GABA receptor blocker. Blockade of spinal GABAA receptors by intrathecal application of bicuculline at doses that do not produce hyperalgesia also do not affect phase 1 of Formalin response. In contrast, the number of flinches and scored pain behavior is enhanced in the interphase period and in phase 2 when bicuculline is given either before or 7 min after Formalin injections (226). Thus expression of the second phase of the Formalin test may be tonically attenuated by spinal GABAergic inhibition. Pretreatment with intrathecal GABAA receptor agonists isoguvacine or muscimol decreases flinches in both phases of the Formalin test (226).
II) Reversal of hyperalgesia and allodynia by GABA receptor agonists. A number of independent studies show that spinal GABAergic inhibition is impaired in neuropathic animals and that spinal application of GABAA or GABAB receptor agonists may reverse neuropathic symptoms. Hyperalgesia induced by spared nerve injury is reversed by subcutaneous injections of GABAA receptor agonists gaboxadol or muscimol but not isoguvacine (436). In rats, subcutaneous or intrathecal applications of GABAB receptor agonists (l-baclofen or CGP35024) reverse mechanical hyperalgesia induced by partial sciatic nerve ligation but not by intraplantar injection of complete Freund's adjuvant (402), suggesting that neuropathic but not inflammatory mechanical hyperalgesia is sensitive to GABAB receptor blockade.
In rats, ligation of spinal nerves L5 and L6 results in tactile hyperalgesia and reduced withdrawal latencies to noxious skin heating. Intrathecal GABAA receptor agonist isoguvacine reverses tactile hyperalgesia up to 75% of the maximal possible effect. This action of isoguvacine is completely blocked by intrathecal bicuculline or phaclofen (312). Thermal hyperalgesia is also reversed by intrathecal isoguvacine, while intrathecal GABAB receptor agonist baclofen disturbs motor behavior (312).
Intrathecal application of a single dose of either GABAA receptor agonist muscimol or GABAB receptor agonist baclofen reverses tactile hyperalgesia in rats with spinal nerve ligation for 2–5 h. Intrathecal injections of antagonists at GABAA receptors (bicuculline) or GABAB receptors (CGP 35348) at doses that fully block the actions of the respective agonists do not change tactile hyperalgesia (200). This indicates absence of tonic GABAergic inhibition in hyperalgesic rats. When neuronal cells bioengineered to synthesize GABA are transplanted in the lumbar subarachnoid space of rats with a chronic constriction injury of the sciatic nerve, both tactile and thermal hyperalgesia are reversed when transplants are placed either 1 or 2 wk after partial nerve injury. Later graft placements are ineffective (500). One study suggests that even a single dose of GABA may have profound and lasting effects on neuropathic pain. In the rat, chronic constriction injury model tactile and thermal hyperalgesia are permanently reversed by a single dose of GABA given intrathecally 1 or 2 wk but not 3–4 wk after nerve ligation. (120). Specifically targeting spinal GABAA receptors containing the α2- and/or α3-subunits reveals antinociception with minor motor effects (250). Likely, the beneficial effects of GABA receptor agonists in animal models of neuropathic pain translate into the clinic. In five human patients with neuropathic, including phantom limb pain, continuous intrathecal baclofen improved pain scores throughout the observation periods of 6 to 20 mo (632).
III) Paradoxical excitation of nociceptive neurons by GABA. Activation of GABAA receptors opens a Cl− ion channel. The direction of Cl− flux is generally determined by level of the Cl− equilibrium potential (ECl) with respect to the resting membrane potential (Vrest) of the cell. In most neurons of mature animals, ECl is more negative than the Vrest. In neurons, potassium-chloride cotransporters and sodium-potassium-chloride cotransporters are the two classes of cation-chloride transporters that regulate Cl− transport. Normally the potassium-chloride cotransporter reduces the concentration of K+ and Cl− while the sodium-potassium-chloride cotransporters increase intracellular Na+, K+, and Cl− within neurons (see Ref. 421 for a review). The continuous removal of Cl− from the cells via a potassium-chloride cotransporter keeps ECl more negative than the Vrest. Thus increasing the Cl− conductance by activation of GABAA receptors will lead to a Cl− influx and hyperpolarization. GABAergic depolarization and eventually excitation can be seen under conditions where ECl is less negative than the resting membrane potential. This may occur when the potassium-chloride cotransporter becomes insufficient. This results in a Cl− efflux and membrane depolarization rather than an influx into the cell upon activation of GABAA receptors. During development and minutes to weeks after trauma of cultured hypothalamic or cortical neurons, GABA may have a depolarizing effect (536). Thus the level of the chloride concentration gradient across the GABAA receptor expressing postsynaptic cell membrane determines if GABA is hyper- or depolarizing (89).
These general biophysical principles, of course, also apply to the membrane of primary afferent nerve terminals. Here, ECl is, however, normally less negative than the resting membrane potential, also in mature animals. This is due to the activity of sodium-potassium-chloride cotransporters and regularly results in a Cl− efflux and membrane depolarization. Thus, under normal conditions, activation of GABAA receptors leads to a depolarization of the terminals of primary afferent nerve fibers. This primary afferent depolarization is not strong enough to cause action potential firing (i.e., an excitation) under normal conditions. Primary afferent depolarization rather inactivates voltage-gated ion channels that are required for the release of neurotransmitter(s) from the terminals. Therefore, moderate depolarization of terminals by GABA may cause presynaptic inhibition.
A) GABAergic excitation of primary afferent nerve terminals. Cervero and co-workers (63, 64) propose a mechanism of dynamic mechanical allodynia that is triggered by low-threshold mechanosensitive Aβ-fiber afferents. If the GABAergic presynaptic depolarization is much enhanced under the conditions of an inflammation or a neuropathy, then the threshold for activation of voltage-gated sodium channels might be passed and action potential discharges will be elicited in these terminals (Fig. 3G). These action potentials may trigger the release of excitatory neurotransmitter. Some of the GABAergic interneurons that impinge on nociceptive nerve terminals can be excited by Aβ-fibers. Therefore, nociceptive specific dorsal horn neurons could be indirectly excited by activity in Aβ-fibers (63, 64, 577). Following an inflammation, the upregulation of the sodium-potassium-chloride (Na+, K+, 2Cl− type I) cotransporter in primary afferents leads to an excessive depolarization of primary afferent terminals by GABA and cross-excitation between low- and high-threshold primary afferents (421). This finding is in line with the above hypothesis (see, however, Ref. 559).
B) GABAergic excitation of spinal lamina I dorsal horn neurons. In rats with a chronic constriction injury of the sciatic nerve and dynamic mechanical allodynia, expression of a potassium-chloride exporter (K+-Cl− cotransporter-2) in spinal dorsal horn is reduced to about half of the control levels (89). In streptozotocin-induced diabetic rats, neuropathy is also accompanied by a reduced immunoreactivity for K+-Cl− cotransporter-2 in laminae I and II (364). This may cause a shift of ECl from normally −75 to −50 mV. Under these conditions, GABAA-receptor activation results in a Cl− efflux and membrane depolarization rather than a Cl− influx into the cell (Fig. 3E). In some neurons, the resulting depolarization may be sufficient to evoke action potential firing. Furthermore, spinal cord lesions may lead to downregulation of Na+-K+-Cl− cotransporter-1 and K+-Cl− cotransporter-2 in the lesion epicenter (93), and in rats, thoracic spinal cord injuries lead to downregulation of the K+-Cl− cotransporter-2 in the lumbar spinal dorsal horn and to reduced GABAergic inhibition (305). Downregulation of K+-Cl− cotransporter-2 in spinal cord and attenuation of GABAergic inhibition or its conversion into excitation may contribute to dynamic mechanical allodynia and to mechanical and thermal hyperalgesia in rats with a diabetic neuropathy (220). Pharmacological blockade of the potassium-chloride exporter in spinal cord slices of naive rats also converts inhibitory action of GABA into an excitation in ∼30% of the lamina I neurons, suggesting that a shift in anion reversal potential can be caused by reduced activity of the potassium-chloride exporter. In intact rats, this leads to mechanical and thermal hyperalgesia (89). Taken together, these results suggest a novel mechanism of GABA receptor-mediated hyperalgesia in neuropathic animals through inversion in polarity of GABAA receptor-mediated action on nociceptive spinal dorsal horn lamina I neurons from inhibition to excitation (89, 421).
Spinal microglia appear to be involved in this process. Stimulation of microglia with ATP causes release of brain-derived neurotrophic factor from activated microglia. Brain-derived neurotrophic factor binding to its TrkB receptor on lamina I neurons is essential for changing the anion gradient and conversion of GABAergic inhibition into excitation (88). A similar brain-derived neurotrophic factor-dependent downregulation of the K+-Cl− cotransporter-2 was observed in spinal dorsal horn of rats with a peripheral inflammation (complete Freund's adjuvant) (620).
2. Glycinergic systems
In addition to spinal GABAergic inhibition, spinal glycinergic interneurons also modulate neuronal activity in spinal dorsal horn. Some changes in the glycinergic system have been observed under conditions of experimental hyperalgesia (see Fig. 3, D and F). Glycine may bind to the Cl−-permeable glycine receptor, a member of the nicotinic acetylcholine receptor family of ligand-gated ion channels. Taurine is another agonist at this receptor with perhaps even higher efficacy than glycine in neurons of spinal cord lamina II (592). At glycinergic synapses in superficial spinal dorsal horn, release of glycine may be inhibited by presynaptic GABAB receptors (70).
A) MODULATION OF SPINAL GLYCINERGIC SYSTEMS.
The spinal content of glycine, like that of GABA, is enhanced bilaterally 1–30 days after a unilateral chronic constriction injury of sciatic nerve in rats (463). The number of neurons in lamina I, lamina II, or lamina III with glycine immunoreactivity is, however, not different from controls, as evaluated with unbiased stereological methods 14 days after chronic constriction injury of sciatic nerve (413). It is not known at present if the electrophysiological properties of glycinergic neurons change under conditions of neuropathy or inflammation. The number and function of glycine receptors do, however, change. For example, a unilateral sciatic nerve constriction leads to a bilateral reduction in the number of glycine receptors in rat spinal dorsal horn (488). Furthermore, protein kinase C phosphorylation of the α- and β-subunits of the glycine receptor attenuates glycine-induced currents (535). Iontophoretic application of glycine near the recording site of multireceptive lumbar spinal dorsal horn neurons in monkeys reduces responses to pinching the skin. Activation of protein kinase C (by phorbol 12-myristate 13-acetate) reduces this inhibition by glycine. The inhibition by glycine (or GABA) is almost absent when the agonist is applied 15 min after intradermal injection of capsaicin (which activates protein kinase C in spinal neurons). Inhibition returns to normal ∼1.5 h after capsaicin injection (293).
The Cl− conductance of glycine receptor channels strongly increases via activation of Gs but not Go or Gi proteins when cAMP or protein kinase A is included into the pipette solution (494). cAMP enhances channel open probability but not mean channel open times or channel conductance, nor binding affinity of glycine to its receptor (494). The authors suggest that the monoamines 5-HT acting on 5-HT1 receptors and norepinephrine acting on α2-adrenergic receptors could change cAMP levels in target cells and thereby the cellular responses to glycine through protein phosphorylation. In a spinal cord slice preparation from young rats, glycine receptor-mediated currents are enhanced by 5-HT in superficial spinal dorsal horn neurons (288) via activation of 5-HT2 receptors. Inhibition of protein kinase C but not inhibition of cAMP-dependent protein kinase A blocks this 5-HT-mediated potentiation of glycinergic currents. A membrane-permeable diacylglycerol analog, like 5-HT, enhances glycine receptor-mediated currents. Thus 5-HT likely activates protein kinase C and potentiates glycinergic currents via a diacylglycerol-dependent pathway (288).
Prostaglandin E2 is released in spinal dorsal horn during peripheral inflammation and may depress spinal glycinergic inhibition via the α2 subtype (173). The inhibitory (strychnine-sensitive) glycine receptor is a specific target of prostaglandin E2. In fact, prostaglandin E2, but not prostaglandin F2α, prostaglandin D2, or prostaglandin I2, reduces inhibitory glycinergic synaptic transmission in spinal dorsal horn in low nanomolar concentrations, whereas GABAA, AMPA, and NMDA receptor-mediated transmissions remain unaffected (5).
ATP acting on P2X receptors enhances the frequency of glycinergic miniature inhibitory postsynaptic currents in dissociated trigeminal neurons. Substance P alone is without effect. The combination of ATP and substance P does, however, reduce ATP-induced facilitation by a presynaptic interaction (558), suggesting that substance P may indirectly diminish glycinergic inhibition. Consistently, in lamina I neurons recorded in rats with an inflamed hindpaw (complete Freund's adjuvant, which releases substance P in superficial spinal dorsal horn), the number of glycinergic miniature inhibitory postsynaptic currents is strongly reduced (370).
B) MODULATION OF HYPERALGESIA AND ALLODYNIA BY THE SPINAL GLYCINERGIC SYSTEM.
After a unilateral chronic constriction injury of the sciatic nerve, the potency of glycine receptor antagonist strychnine increases. Intrathecal doses of strychnine that are subthreshold in control animals do, however, lower thermal threshold for withdrawal reflexes ipsi- but not contralateral to the nerve injury (463). Daily pretreatment with intrathecal MK-801 to block spinal NMDA receptors abolishes increases in glycine potency and prevents development of hyperalgesia (463). Furthermore, tactile hyperalgesia in the partial sciatic nerve ligation model is attenuated when the reuptake of glycine either by neuronal glycine transporter 2 or glial glycine transporter 1 is impaired. This was shown by intrathecal injections of inhibitors or knockdown of spinal glycine transporters by siRNA that reduce mechanical hyperalgesia in mice (368).
E. Changes in Descending Modulation
In recent years, considerable evidence has accumulated showing that spinal nociception may be facilitated by descending pathways (347, 417, 505, 532, 561). Inflammation not only causes hyperalgesia in the area immediately surrounding the primary injury (i.e., secondary hyperalgesia) but may also cause more generalized hyperalgesia at areas well apart from the lesion site. For example, inflammation at a hindpaw facilitates nociceptive withdrawal reflexes at the tail (54). Similarly, Formalin injected into the tail enhances responses of lumbar spinal dorsal horn neurons to noxious heating of a hindpaw (41). Both secondary hyperalgesia and the remote sensitization require a spino-bulbo-spinal loop with a descending facilitatory arm (Fig. 3H).
A number of behavioral studies show that neurons in the rostroventral medulla are required for full expression of hyperalgesia in different animal models of inflammation and neuropathy. Secondary hyperalgesia caused in rats by mustard oil involves activation of glutamate receptors of the NMDA type and subsequent activation of nitric oxide synthase in the rostroventral medulla (530). Secondary thermal hyperalgesia induced either by intra-articular carrageenan/kaolin injection into the knee or by topical mustard oil application to the hindleg is completely blocked by bilateral rostral medial medulla lesions produced by the soma-selective neurotoxin ibotenic acid (534). Bilateral destructions of cells in the nucleus reticularis gigantocellularis with ibotenic acid lead to an attenuation of hyperalgesia and a reduction of inflammation-induced spinal c-Fos expression (569). Likewise, mechanical hyperalgesia induced by spinal nerve ligation in rats is reversed by local anesthetic block in the rostroventral medulla (406). Spinal nerve ligation induces tactile and thermal hyperalgesia; both are blocked by bilateral injections of lidocaine (51) or a cholecystokinin type B receptor antagonist (L 365,260) into the rostroventral medulla (255).
Descending facilitation and inhibition of behavioral and dorsal horn neuronal responses to noxious stimulation can be triggered from the same sites in rostroventral medulla. Electrical stimulation at low intensities (5–25 μA) is faciliatory while higher intensities (≥50 μA) are inhibitory (626, 628). Likewise, injections of small doses (0.03 pmol) of neurotensin in the rostroventral medulla trigger descending facilitation of multireceptive and nociceptor specific neurons in rat lumbar spinal dorsal horn (531). High doses (≥300 pmol) induce descending inhibition. Microinjection of cholecystokinin into the rostroventral medulla of naive rats also produces a robust mechanical and a more modest thermal hyperalgesia (255). The same studies also identified sites in the rostroventral medulla from which only facilitation or inhibition could be elicited.
A neuronal group exists in the rostroventral medulla which increases its firing rates just before the onset of the nociceptive tail-flick reflex. These neurons were termed “ON-cells” and probably mediate descending facilitation. In contrast, “OFF-cells” cease firing shortly before the tail-flick reflex occurs and may be involved in descending inhibition. The roles of ON- and OFF-cells have been reviewed (177, 327, 417). When μ-opioid receptor expressing cells in the rostroventral medulla are selectively destroyed, then spinal nerve ligation no longer induces mechanical and thermal hyperalgesia (416). This is compatible with an involvement of ON-cell in the rostroventral medulla as firing of these neurons is depressed by a μ-opioid receptor agonist (DAMGO) (178). Sensory responses of ON- and OFF-cells are altered in rats with a spinal nerve ligation. Both neuron types exhibit novel responses to innocuous mechanical stimulation and enhanced responses to noxious mechanical stimulation. This neuronal hypersensitivity correlates with mechanical and thermal hyperalgesia in these rats (57).
Interactions between glial cells and neurons are involved by the activation of descending facilitation from the rostral ventromedial medulla following peripheral nerve injury. Chronic constriction injury of the rat infraorbital nerve leads to an early and transient reaction of microglia and a prolonged reaction of astrocytes in that brain region. Microinjections of microglial and astrocytic inhibitors that prevent glial cell activation also attenuate mechanical hyperalgesia at 3 and 14 days after nerve injury (570).
An early study reported that electrical stimulation in dorsolateral funiculus of decerebrated rats produces largely excitatory effects on projection neurons in contralateral spinal lamina I (331). Bilateral lesions of the dorsolateral funiculus abolish descending inhibition by electrical stimulation or neurotensin microinjection without, however, affecting descending facilitation (531, 628). This suggests that descending facilitation and inhibition can be induced from the same brain stem sites but employ separate descending pathways. Others have found that lesions in ventrolateral funiculus (629) attenuate descending facilitation.
Descending facilitation of behavioral and spinal neuronal responses to noxious stimuli cannot only be induced from the rostroventral medulla (227, 532, 626, 628) but also from more rostral sites in the brain including the anterior cingulate cortex (55), pretectal (427) or dorsal reticular nuclei (14), and periaqueductal gray (406). It has been suggested that the final common descending facilitatory pathway originates from the rostroventral medulla (417) and contributes to enhanced pain sensitivity (532).
ON-cells may be at the origin of the descending facilitatory arm of a spino-bulbo-spinal positive-feedback loop. The ascending arm may arise from a small but well-defined group of spinal lamina I projection neurons. These neurons express the neurokinin 1 receptor for substance P (505) and activity-dependent long-term potentiation at synapses with primary afferent C-fibers (see sect. viB and Refs. 202, 204). Selective ablation of these lamina I neurons reduces mechanical and thermal hyperalgesia by inflammation or nerve injury (318, 383) and the descending facilitation of spinal wide-dynamic range neurons (505).
Descending facilitation involves activation of spinal receptors for serotonin (627). The 5-HT3 receptor subtype appears to mediate descending facilitation originating from spinal neurokinin 1 receptor expressing cells (505). Spinal microglia and astrocytes also play a role. Microglia may be activated by neurotransmitter(s) such as excitatory amino acids or substance P either released from primary afferents and/or from fibers descending from the rostroventral medulla to spinal dorsal horn (561).
Both descending inhibition and facilitation may serve to temporarily adapt the general pain responsiveness to the individual needs. Descending facilitation contributes to generalized hyperalgesia and allodynia as components of the “sickness response” to infection and inflammation (see sect. viiA) (561). This may promote healing. If descending facilitation is inadequate with respect to strength or duration, it may become a cause for chronic pain. The mechanisms of generalized facilitation of nociception may have relevance for some human pain patients as it has been suggested that in fibromyalgia patients endogenous pain modulatory systems are impaired (223).
In addition to enhanced descending facilitation as a promoter of hyperalgesia and allodynia, peripheral inflammation may also enhance descending inhibition that counteracts the development of hyperalgesia. For example, 4 h after a unilateral carrageenan injection into a hindpaw of rats, thermal hyperalgesia is enhanced when the locus coeruleus/subcoeruleus from which descending noradrenergic fibers originate is lesioned bilaterally but not unilaterally (309).
F. Aβ-Fiber-Induced Pain (Mechanical Allodynia)
Touch-evoked pain is a hallmark of neuropathic pain. There is now clear evidence that impulses in large myelinated Aβ-fibers may contribute to mechanical allodynia in animal models and in pain patients (56, 157).
1. Phenotypic switch in Aβ-fibers
Under normal conditions, stimulation of primary afferent Aβ-fibers fails to facilitate spinal nociception and does not induce hyperalgesia or allodynia. In the course of an inflammation, some large myelinated Aβ-fibers may switch their phenotype and begin to synthesize substance P. Upon activation, Aβ-fibers may then release substance P into the spinal dorsal horn, and this may, e.g., via extrasynaptic spread of substance P, contribute to facilitation of spinal nociception and enhanced responsiveness of spinal nociceptive neurons (382). Before considering a “phenotypic switch,” it is important to ensure that the markers used for identifying an afferent fiber type do not change also under the experimental conditions. For example, neurofilament (NF) 200 kDa remains a good marker for A-fiber neurons, and isolectin B4 and substance P remain good markers for C-fiber neurons after chronic constriction injury (440).
After sciatic nerve transection, substance P immunoreactivity is induced in medium- and large-sized dorsal root ganglia cells and reduced in small-sized cells (385). The expression of preprotachykinin mRNA encoding substance P and related peptides is strongly upregulated in large A-type neurons of the rat dorsal root ganglion following unilateral chronic constriction injury of the sciatic nerve (325). After intraplantar injection of complete Freund's adjuvant, mechanical stimulation of the inflamed skin or electrical stimulation at Aβ-fiber intensity of sensory nerve fibers innervating the inflamed tissue leads to a slowly progressing facilitation of flexor motor responses (308). Stimulation of a peripheral nerve at A-fiber intensity normally does not cause afterdischarges in spinal multireceptive neurons in spinalized rats. After a chronic constriction injury of sciatic nerve, however, afterdischarges do occur and can be blocked by a neurokinin 1 receptor antagonist (CP-99,994) (409). This newly acquired capacity of Aβ-fibers to enhance spinal nociception may be due to a novel expression of substance P in these primary afferents, thereby switching their phenotype to one resembling nociceptive C-fibers (382; see also Fig. 3L). In three nerve injury models (sciatic nerve transection, spinal nerve ligation, and chronic constriction injury), substance P is, however, not upregulated to any detectable degree, and stimulation of Aβ-fibers does not cause neurokinin 1 receptor internalization in spinal dorsal horn, challenging the view that substance P would be released from Aβ-fibers under these conditions (196).
In addition to substance P, a large number of other molecules are also either up- or downregulated in dorsal root ganglion neurons in various animal models of pain, as reviewed by Ueda (528), Hucho and Levine (194), and Woolf and Ma (587).
2. Sprouting of Aβ-fibers
After nerve injury but not under normal conditions, impulses in Aβ-fibers may elicit pain sensation. Thus information in large myelinated primary afferent must gain access to the nociceptive system. And, indeed, in neuropathic animals, c-Fos expression, an indication for neuronal activity, is increased in lamina II dorsal horn neurons following repeated touch stimuli. This suggests that low-threshold mechanosensitive fibers may now directly or indirectly activate nociceptor specific lamina II neurons (40). Likewise, in animal models of neuropathic pain (sciatic nerve transection or chronic constriction injury), Aβ-fiber-mediated input to the nociceptive superficial dorsal horn increases substantially (252, 253, 392). An attractive hypothesis was suggested by Woolf et al. (588) who reported that application of the neural tracer horseradish peroxidase to a peripheral nerve results in transganglionic transport to the central terminals of the labeled axons. When the B unit of cholera toxin is conjugated to horseradish peroxidase, normally only myelinated afferents are labeled. When this conjugate is applied to an intact nerve of rats, the marker is consequently found selectively in laminae I, III, and deeper dorsal horn, which matches the known termination of myelinated primary afferents (588). In animals with a transection of a peripheral nerve, labeling was also found in lamina II, which is normally devoid of A-fiber terminals. This was interpreted as sprouting of A-fibers into an area normally occupied by C-fibers only. This interpretation was substantiated by intracellular labeling of low-threshold primary afferents (588). This labeling method was used in a number of subsequent studies from the same (115, 316, 317) and other laboratories (34, 376, 401, 485) and revealed similar results. Later studies found, however, that cholera toxin B subunit may not be a reliable marker for myelinated fibers following peripheral nerve injury but rather taken up indistinctively by small and large size dorsal root ganglion neurons (475). Thus, after nerve transection, the marker is taken up also by fine primary afferents (516), including unmyelinated C-fibers (459) and cholera toxin B subunit, then no longer selectively labels A-fibers. These authors conclude that after peripheral nerve injury, the label found in lamina II is largely due to the uptake of the marker by injured C-fibers but not due to sprouting of A-fibers (459). This conclusion is in line with recent studies which found only very limited sprouting of single, identified Aβ-fiber afferents after nerve injury (30, 197). Taken together, these studies challenge the hypothesis that sprouting of Aβ-fibers into the superficial laminae after nerve section is substantial (30, 197, 459, 516).
3. Opening of polysynaptic excitatory synaptic pathways
An alternative explanation for novel Aβ-fiber input to superficial spinal dorsal horn neurons is the opening of preexisting polysynaptic pathways between Aβ-fiber afferents that terminate in deeper dorsal horn and nociceptive neurons in superficial spinal dorsal horn (Fig. 3J). Recent studies suggest that indeed some forms of neuropathy or inflammation may facilitate polysynaptic low-threshold input to neurons in laminae I and II of spinal dorsal horn (24, 468). In transversal lumbar spinal dorsal horn slices, electrical stimulation or microinjection of glutamate [which does not excite (sprouted) fibers of passage] into the deep dorsal horn or stimulation dorsal roots at Aβ-fiber intensity excites only very few neurons in the superficial dorsal horn of control animals. In contrast, numerous neurons in the superficial dorsal horn are excited in slices taken from animals with a spared nerve injury (468). Taken together, these results suggest that Aβ-fiber afferents excite interneurons in lamina III, which via polysynaptic pathways trigger excitation of superficial dorsal horn neurons in neuropathic but not in control animals leading to touch-evoked pain (468).
G. Other Potential Mechanisms of Hyperalgesia and Allodynia
1. Sprouting of fine primary afferents
During development, spinal termination patterns of primary afferents including nociceptive C-fibers are finely tuned. Transgene overexpression of nerve growth factor in spinal dorsal horn results in sprouting of a subpopulation of nociceptive primary afferents that express substance P and calcitonin gene-related peptide in spinal dorsal horn (Fig. 3I). This C-fiber sprouting is accompanied by mechanical and thermal hyperalgesia (438). Sprouting of C-fiber afferents has been investigated in some detail by using calcitonin gene-related peptide immunoreactivity for peptidergic and isolectin B4 immunoreactivity as marker for small nonpeptidergic fibers. Following rhizotomy of L4–S1 dorsal roots and injury of the saphenous nerve in rats, L2, L3 dorsal root afferents may regenerate differentially. Isolectin B4 labeling is not much different in lamina II of denervated spinal cord segments. In contrast, labeling for calcitonin gene-related peptide is much enhanced in segments denervated by rhizotomy in a nonsomatotopic manner (33). The authors conclude that peptidergic (calcitonin gene-related peptide-positive) C-fibers sprout vigorously while nonpeptidergic (isolectin B4 positive) C-fibers remain stable after peripheral nerve injury (33). Collateral sprouting, i.e., sprouting of uninjured axons into the denervated territory, not only requires nerve growth factor. In addition, an intact intermediate filament network within nerve fibers is also essential for collateral sprouting of small-diameter primary afferent nerve fibers. Disruption of intermediate filament network in transgenic mice significantly impairs the ability of uninjured small-sized dorsal root ganglion neurons to sprout collateral axons into adjacent denervated skin (32).
Possibly repulsive guidance cues such as semaphoring 3A play a role in limiting sprouting of a subgroup of C-fiber afferents. This repellent is thought to restrict termination of nerve growth factor-responsive nociceptive afferents to superficial laminae. Reduced sprouting of calcitonin gene-related peptide and substance P-containing axons leads to decreased mechanical hyperalgesia tested with von Frey filaments (510). Thermal hyperalgesia is, in contrast, not significantly affected by semaphorin3A (510).
2. “Wind-up” of action potential firing
Some of the neurons in spinal dorsal horn with excitatory input from primary afferent C-fibers display the phenomenon of “wind-up,” i.e., the increase in the number of action potential discharges in response to repetitive C-fiber stimulation. When C-fiber afferents are stimulated at low frequencies (0.3–5 Hz), postsynaptic responses increase with almost each stimulus until a saturation level is reached (339). This is the case after ∼10–30 C-fiber stimuli, i.e., after 5–60 s. Thus wind-up is a short-lasting phenomenon that enhances action potential firing of some spinal dorsal horn neurons during the first few seconds of an ongoing noxious stimulus. Thereafter, responses no longer increase but may rather decrease. Wind-up is a form of temporal summation of action potential discharges, due to the summation of excitatory postsynaptic potentials. Temporal summation occurs when the duration of excitatory postsynaptic potentials is longer than the interspike intervals of the presynaptic C-fiber discharges. Since NMDA receptor-mediated postsynaptic currents typically prolong excitatory postsynaptic potentials, it is not surprising that wind-up is sensitive to NMDA receptor blockage (589). Wind-up can be observed in normal animals, i.e., in the absence of any pathological changes of spinal nociception. Thus wind-up is a feature of the normal coding properties of some spinal dorsal horn neurons and per se not a sign of “sensitization” in spinal dorsal horn. In other words, the absence or presence of wind-up cannot be used as an indicator for any form of abnormal pain amplification, and consequently, wind-up is not a cellular mechanism of hyperalgesia or chronic pain. A physiological function of wind-up could be to enforce a nocifensive response during a sustained noxious stimulus that triggers discharges in C-fibers at low rates. If the nocifensive response is not triggered within the first few seconds, wind-up will increase the discharge frequencies of some spinal dorsal horn neurons possibly beyond threshold for a response, e.g., a withdrawal reflex. This interpretation is in line with the observation that wind-up can be seen in motoneurons (589) and in motor reflexes (141).
Importantly, a number of changes that may lead to pain amplification may also lead to changes in the properties of wind-up. For example, LTP at synapses between C-fibers afferents and second-order neurons will lead to larger and longer-lasting excitatory postsynaptic potentials and thus may result in stronger wind-up and in lowering of the wind-up threshold frequency of a given neuron. Likewise, increased membrane excitability, i.e., lowering the threshold for action potential firing and/or less negative membrane potentials or changes in discharge patterns from single spiking to burst discharges, all would result in stronger wind-up in response to repetitive C-fiber stimulation. Thus, while wind-up by itself cannot be used as a proof of alterations in spinal nociception changes in the incidence of neurons that display wind-up, increase in wind-up strength or decrease in wind-up threshold frequencies may all indicate that some forms of facilitation occurred in spinal nociceptive pathways.
Wind-up of action potential discharges likely increases activity-dependent Ca2+ influx into the respective neurons and thereby Ca2+-dependent signal transduction pathways. One of the many consequences of which could be activity-dependent changes in synaptic strength, membrane excitability, and discharge patterns. However, wind-up is not necessary for the induction of long-term changes in excitability in most spinal dorsal horn neurons (for review, see also Refs. 180, 586).
Perceptual correlates of action potential wind-up can be studied in normal human subjects (437, 496), in human subjects with an experimental hyperalgesia (471, 560), and in pain patients (267, 497) when repetitive noxious stimuli are given at a frequency that is compatible with the window of wind-up frequencies, i.e., if the interval between C-fiber stimuli is no longer than 3 s. A number of studies suggest that NMDA-receptor-dependent wind-up of C-fiber-evoked second pain is stronger in patients with fibromyalgia compared with normal controls (419, 498).
3. Epileptiform activity in nociceptive pathways
Paroxysmal forms of neuropathic pain share some key features with epileptic seizures. Both can be triggered by harmless sensory stimuli and once started they have a rather stereotyped progression. Another common feature is the refractory period, i.e., during the immediate time after an attack no new attack can be evoked. If not adequately treated, both may end up in a status, i.e., a series of attacks without complete recovery between the attacks. Last but not least, both can be treated successfully by anticonvulsant drugs (443). Epileptiform activity, i.e., highly synchronized, rhythmic discharges of populations of neurons, has been observed in the nociceptive system of the spinal dorsal horn (Fig. 3K). Patch-clamp recordings from individual neurons and Ca2+ imaging of multiple single neurons in a slice preparation of the rat lumbar spinal cord revealed epileptic activity in response to bath application of the potassium channel blocker 4-aminopyridine (4-AP) (443). 4-AP is often used to induce epileptic activity in the cerebral cortex. In the spinal cord, epileptiform activity was also observed in lamina I neurons with a direct projection to the parabrachial area, i.e., in neurons that are directly involved in neuropathic pain behavior in animals (318). Bath application of μ-opiate receptor agonist DAMGO or α2-adrenoreceptor agonist clonidine does not or only weakly attenuates epileptiform activity. In contrast, antiepileptic drugs such as phenytoin, carbamazepine, and valproate are strongly effective (443).
4. Enriched responsiveness of spinal nociceptive neurons
A large number of studies have shown that nociceptive spinal dorsal horn neurons become more excitable by peripheral inflammation or nerve injuries. This includes enhanced responsiveness to normally innocuous natural or electrical nerve stimuli, expansion of low-threshold cutaneous receptive fields, enhanced responses to noxious stimuli, and development of spontaneous action potential discharges. There is still no agreement on the differential roles of the various classes of spinal dorsal horn neurons for acute, inflammatory, and/or neuropathic pain. For example, in intact rats, surgical incision of the hairy skin and subsequent suturing causes mechanical and thermal hyperalgesia from 30 min post incision to 3–5 days (228). In decerebrated, spinalized rats, the same injury triggers enhanced responses of wide-dynamic range, low-threshold and high-threshold spinal dorsal horn neurons during the injury, elevated background activity in wide-dynamic range but not in low-threshold or high-threshold neurons for 30 min and expansion of cutaneous low- and high-threshold mechanoreceptive fields in wide-dynamic range but not low-threshold or high-threshold neurons (228). High-threshold neurons develop responsiveness to low-threshold input only after spinal bicuculline but not after incision (228). The authors conclude that enhanced excitability of wide-dynamic range but not high-threshold or low-threshold neurons mediate mechanical and thermal hyperalgesia after injury of hairy skin (228). Neither of these neuronal cell types investigated is, however, a functionally homogeneous group but comprise excitatory and inhibitory neurons, interneurons, and projection neurons.
Some studies were performed on dorsal horn neurons with a verified projection to brain areas such as spino-thalamic tract neurons. Results from these studies suggest that enhanced neuronal activity in dorsal horn may likely affect nociceptive processing in the brain. These studies have been reviewed extensively before (39, 92, 140, 420, 579, 580, 625).
VII. IMMUNE-CENTRAL NERVOUS SYSTEM INTERACTIONS
A. The Sickness Syndrome
Nonspecific manifestations of inflammation and infection may include fever, drowsiness, and often an increased sensitivity to painful stimuli. A peripheral immune challenge leads to the production of proinflammatory cytokines such as tumor necrosis factor, interleukin-1, and interleukin-6. These peripheral mediators trigger the de novo synthesis of proinflammatory cytokines by cells within the central nervous system including the spinal cord, mainly microglia, and astrocytes (324, 562, 565). These processes subsequently cause the sickness syndrome. The immune-to-brain communication involves blood-borne signaling and neural pathways via sensory vagus nerve fibers (562) and relays in nucleus tractus solitarius and in the ventromedial medulla and descending pathways in the dorsolateral funiculus of the spinal cord (567). The sickness syndrome can be induced experimentally in animals by intravenous injections of pyrogens such as lipopolysaccharides. Systemic (intraperitoneal) injection of lipopolysaccharides produces thermal hyperalgesia, as revealed by decreased tail-flick latencies (326). Hyperalgesia is accompanied by enhanced spinal cord levels of interleukin-1, a product of glial cell activation (see also review by Watkins and Maier, Ref. 561). However, intrathecal administration of interleukin-1 fails to induce hyperalgesia, while intraperitoneal or intracerebroventricular injections are effective (311, 568). A systemically injected single dose of lipopolysaccharides further induces within 6 h muscle hyperalgesia as measured by the grip force assay in mice (233).
A potential mechanism for immune-to-brain communication (563) arising from the abdomen is discussed by Goehler and colleagues (155). An ascending-descending loop may involve the nucleus tractus solitarius-nucleus raphe magnus-spinal cord dorsolateral funiculus circuit (567). Spinal microglia may be activated by neurotransmitter(s) released from nucleus raphe magnus-spinal cord pathways such as excitatory amino acids or substance P. The respective receptors are all expressed by spinal microglia and astrocytes, and ligand binding activates these glial cells in vitro (561).
Subcutaneous injection of Formalin into the dorsum of one hindpaw induces thermal hyperalgesia also at distant sites, e.g., at the tail as measured by the tail-flick reflex (564, 574). This remote hyperalgesia is not mediated solely by circuitry intrinsic to the spinal cord, but rather involves activation of centrifugal pathways originating within the brain and descending to the spinal cord via pathway(s) outside of the dorsolateral funiculus. At the level of the spinal cord, this hyperalgesic state is mediated by an NMDA-nitric oxide cascade, since hyperalgesia can be abolished by administration of either an NMDA antagonist (d-2-amino-5-phosphonovalerate) or a nitric oxide synthesis inhibitor (l-NAME) (574). The decrease in tail-flick latency is further prevented by intrathecal fluorocitrate, which is a glial metabolic inhibitor. A human recombinant interleukin-1 receptor antagonist or an antibody directed against nerve growth factor, i.e., inhibition of products of glial cell activation, are also effective (564). Thus sickness and inflammation-induced hyperalgesia involve overlapping central nervous system circuits and signal transduction pathways (561).
B. Role of Spinal Glia for Allodynia and Hyperalgesia
Work of the recent years has shown that abnormal pain sensitivity involves altered function of neuronal network in spinal dorsal horn and that activated spinal glial cells act as an intermediary between the initial insult and long-term neuronal plasticity leading to pain amplification. Glial cells, i.e., microglia, astrocytes, and oligodendrocytes, constitute ∼70% of the cell population in brain and spinal cord. The physiology of microglial cells has been reviewed recently (129). Spinal microglia and astrocytes are both immunoeffector cells of the central nervous system that are activated following nerve injury or inflammation (104, 520, 566). Furthermore, the number of microglial cells (9) and the number of astrocytes (278) rise in spinal dorsal horn ipsilateral to a peripheral nerve injury. Selective pharmacological blockade of glial cell functions prevents and reverses abnormal pain sensitivity.
1. Activation of spinal glial cells
Microglia can be activated rapidly by neuronal activity (129). One candidate for neuron-glia interaction is the glial excitatory chemokine fractalkine, which is expressed on the extracellular surface of spinal neurons and spinal sensory afferents. After it is released upon strong neuronal excitation, e.g., in response to an insult, it binds to CX3C receptors mainly expressed by microglia. Intrathecal fractalkine causes mechanical and thermal hyperalgesia, while intrathecal fractalkine receptor antagonist delays onset of mechanical and thermal hyperalgesia following chronic constriction injury or inflammatory neuropathy of sciatic nerve (352).
Microglia may also be activated by neurotransmitters such as excitatory amino acids or substance P either released from primary afferents, spinal dorsal horn cells or supraspinal descending fibers, and by ATP, nitric oxide, prostaglandins, and heat shock protein. The respective receptors are expressed not only by spinal microglia but also on astrocytes [520, 561; see also review by Watkins and Maier (562)].
Microglia activation is not a stereotype process but involves various combinations of proliferation and morphological changes, upregulation of surface antigens such as major histocompatibility complex classes I and II antigens, cellular adhesion molecules, cluster determinants 4 and 45 and integrin alpha M, P2X4 receptors (512), and elevated expression of complement receptor 3. From central nervous system injury models it has been suggested that microglia activation releases substances that then activate astrocytes (512). Activation of astrocytes involves hypertrophy and upregulation of the expression of glial fibrillary acidic protein. Astrocytes (but not microglia) closely appose synapses from which they receive signals and which function they modify. For example, activated astroglia may take up less than normal glutamate near excitatory synapses, thereby enhancing its effects on excitatory neurotransmission and neurotoxicity, which in spinal dorsal horn might contribute to hyperalgesia (566).
Already 4 h after L5 spinal nerve transection, the earliest time point investigated, microglial activation markers toll-like receptor 4 and cluster determinant 14 are all upregulated at the mRNA level as assessed by real-time reverse transcription polymerase chain reaction (512). Microglia thus constitute the first noticeable immune responses in spinal cord to several types of peripheral stimuli. Markers remain elevated for 14 days and decline by 28 days. Immunohistochemistry reveals an increase in the number of activated microglial cells as determined by OX42 and glial fibrillary acidic protein in astrocytes of ipsilateral dorsal and ventral horns as early as 2 days after partial sciatic nerve transection lasting for 84 days which parallels the time course of mechanical hyperalgesia (91).
Activation of P2X4 receptor, an ATP-gated ion channel, selectively expressed in microglia of the spinal cord is upregulated after L5 spinal nerve ligation. Likewise, subcutaneous injections of diluted Formalin cause an increase in P2X4 receptor expression on activated microglia in ipsilateral dorsal horn which peaks at day 7 after the injection (161). This suggests that not only nerve injury but also inflammation may trigger expression of this ATP-gated ion channel. In addition, G protein-coupled purinoreceptors also play a role. Intrathecal application of a glial P2Y12 receptor blocker (AR-C69931MX) prevents development of and reverses established tactile hyperalgesia in rats with a tight ligation of a L5 spinal nerve (517). In mice lacking the P2Y12 receptor, tactile hyperalgesia but not normal responses to mechanical stimuli is impaired (517). Toll-like receptors are also expressed on microglia and appear to be essential for their activation by peripheral nerve injury. Antisense knockdown of toll-like receptor 3 in spinal cord attenuates activation of spinal microglia and development of tactile hyperalgesia in rats with a L5 spinal nerve lesion (388).
The functional and phenotypic pattern of activation strongly depends on the type of peripheral stimulus. For example, major histocompatibility complex class II and CC chemokine receptor 2 are all upregulated in spinal cord microglia following spinal nerve ligation but not following peripheral inflammation [see review by Tsuda et al. (520)]. In streptozotocin-induced diabetic rats, tactile hyperalgesia develops that is accompanied by several characteristic changes of activated microglia in the dorsal horn, including increases in Iba1 and OX-42 labeling, hypertrophic morphology. Extracellular signal-regulated protein kinase and Src-family kinase are both activated exclusively in microglia (524). The astrocyte marker glial fibrillary acidic protein is upregulated starting on postoperative day 4 through day 28 (512). Taken together, it has been suggested that microglia is the initial immunoeffector cell sensor that, if inhibited prior to the onset of astrocytic activation, may prevent mechanical hyperalgesia in various models of neuropathy (512).
2. Substances release by activated microglia
Upon activation microglia produce and release cytokines, prostaglandins, leukotrienes, nitric oxide, reactive oxygen intermediates, proteolytic enzymes, and excitatory amino acids such as glutamate (104). Many of the substances produced and released by microglia and astrocytes may mediate hyperalgesia, including nitric oxide, prostaglandins, interleukin-1, brain-derived neurotrophic factor, nerve growth factor, arachidonic acid, and excitatory amino acids such as glutamate (561). Other molecules that are expressed in spinal microglia such as chemotactic cytokine receptor 2, cannabinoid receptor subtype 2, and major histocompatibility complex class II protein may also modulate neuropathic pain.
Platelet activating factor is another substance released from stimulated microglia cells (208) and by neurons in culture upon stimulation with glutamic acid. It is a potent chemotactic factor for microglia which express receptors for platelet activating factor (7).
3. Pain-related behavior modulated by activated microglia
An early report has shown that peripheral inflammation of a hindpaw by intraplantar zymosan injections leads to mechanical and thermal hyperalgesia which involves spinal glia: selective inhibition of glial metabolism by intrathecal administration of fluorocitrate results in a marked, but reversible, attenuation of the persistent thermal and mechanical hyperalgesia (334, 351). Furthermore, intrathecal injection of fractalkine, which selectively activates spinal microglia, is sufficient to induce tactile and thermal hyperalgesia. Blockade of CX3C receptors to which fractalkine binds reverses hyperalgesia when applied 5–7 days after a chronic constriction injury of the sciatic nerve (352).
Proinflammatory cytokines can stimulate the production of multiple components of the complement cascade. Interruption of complement cascade by intrathecal injection of soluble human complement receptor type 1 reverses mechanical hyperalgesia by sciatic inflammatory neuritis and by chronic constriction injury of sciatic nerve and by intrathecal injection of the human immunodeficiency virus-gp120 (526) without affecting normal responses to touch.
Tight ligation of L4/5 spinal nerves leads to activation of spinal p38 mitogen-activated protein kinase (216, 521), which in spinal cord is selectively expressed in activated microglia. Depression of spinal p38 mitogen-activated protein kinase by intrathecal injection of a blocker (SB203580) has no effect on basal nociceptive responses but reverses established mechanical hyperalgesia after spinal nerve ligation (216, 521). Likewise, mechanical hyperalgesia is attenuated by intrathecal inhibition of a p38 mitogen-activated protein kinase or P2X4 receptor blocker or antisense oligonucleotide pretreatment (207). p38 activation may regulate the expression of inducible nitric oxide synthase, cyclooxgygenase-2, and cytokines in microglia through transcriptional and translational effects. Two weeks after an injury of L5, spinal nerve activated microglia are detected and dually phosphorylated active form of p38 mitogen-activated protein kinase and P2X4 ATP receptors are upregulated in microglia in the ipsilateral spinal dorsal horn (207).
Intrathecal injection of minocycline, an inhibitor of microglial cell activation, inhibits mRNA expression of interleukin-1β, tumor necrosis factor-α, interleukin-1β-converting enzyme, tumor necrosis factor-α-converting enzyme, interleukin-1 receptor antagonist, and interleukin-10 as well as mechanical hyperalgesia induced by either sciatic inflammatory neuritis of by intrathecal injection of human immunodeficiency virus-1 gp120 (276).
Pathogens are detected by specific receptors such as the toll-like receptor 4, which is selectively expressed by microglia. Intrathecal antisense oligonucleotides directed against the expression of toll-like receptor 4 or knock out of toll-like receptor 4 gene reduces mechanical and thermal hyperalgesia following L5 nerve transection in mice (511).
Following L5 spinal nerve ligation, extracellular signal-regulated kinase, a mitogen activated protein kinase, is transiently (for <6 h) activated in neurons of spinal dorsal horn and at days 1–10 in spinal microglia and later in astrocytes as well (623). Mechanical hyperalgesia is reduced when a high dose of an extracellular signal-regulated kinase inhibitor (PD98059) is injected intrathecally on days 2, 10, or 21 (623), suggesting its involvement in maintenance of neuropathic pain.
When microglia grown in culture and activated by ATP are injected intrathecally in rats, mechanical hyperalgesia similar to that after spinal nerve ligation is induced, while inactive microglia have no effect (522). Antisense oligonucleotide targeting the ATP receptor P2X4 diminishes tactile hyperalgesia after spinal nerve ligation. Blockade of spinal P2X1–4 receptors by intrathecal injections of 2′,3′-O-(2,4,6-trinitrophenyl)adenosine 5-triphosphate (TNP-ATP) temporarily reverses tactile hyperalgesia 7 days after spinal nerve ligation (522). Interestingly, blockade of spinal P2X1,2,3,5,7 receptors (with PPADS) inhibits pain-related behavior in the first and second phase of the Formalin test and the responses to capsaicin (525) but fails to affect tactile hyperalgesia after spinal nerve ligation (522).
Glia but not neurons express a receptor for interleukin-10. Its activation suppresses release of proinflammatory cytokines. After intrathecal injection, interleukin-10 has a short half-life of ∼2 h. Gene therapy with an adenoviral vector encoded human interleukin-10 prevents and reverses mechanical and thermal hyperalgesia by chronic constriction injury of sciatic nerve and mechanical hyperalgesia by either sciatic inflammatory neuropathy or by intrathecal injection of gp120, an envelope glycoprotein of human immunodeficiency virus-1, all of which activates spinal glia (349). These effects last for more than a week but are absent at 3 wk. Normal responses to heat or touch are not affected by this form of gene therapy (349).
Intrathecal morphine application for 5 days but not single intrathecal morphine injection leads to elevated levels of interleukin-1 mRNA and protein in spinal dorsal horn 2 h but not 24 h after discontinuation of morphine (219). Coadministration of morphine and an interleukin-1 receptor antagonist enhances morphine analgesia and reduces development of tolerance to morphine and morphine-induced mechanical and thermal hyperalgesia (219). The intrathecal injection of neutralizing antibody against the fractalkine receptor has similar effects (219). Thus activation of spinal glial cells is an important intermediate step in the pathogenesis of chronic pain of various origins.
4. Concluding remarks
Considerable progress has been made in developing clinically relevant animal models of hyperalgesia and allodynia. Available models cover inflammatory, traumatic, and neuropathic forms of acute or chronic pain. Standardization of animal models across laboratories has much improved the reproducibility of published work. Many efforts are presently being made to unfold the diverse pain mechanisms at the network, cellular, synaptic, and molecular levels. It is highly unlikely that a unifying model will be developed that may explain all forms of hyperalgesia and allodynia. It is more likely that a number of mechanisms are active in parallel and/or in sequence and that a characteristic pattern of mechanisms will be identified for a given pain syndrome. Thus a single “magic pain killer” will hardly be the future treatment of choice; rather, mechanism-based multimodal treatments that match the particular phase of pain development will be successful. Much of the future progress in the field of experimental pain research will rest on the work that is summarized in this review.
GRANTS
My work was supported by grants from the Austrian Science Fund (FWF), the Vienna Science and Technology Fund (WWTF), the Oesterreichische Nationalbank (OeNB), and the Medizinisch-Wissenschaftlicher Fonds des Bürgermeisters der Bundeshauptstadt Wien.
I sincerely thank Lila Czarnecki for excellent maintenance of our literature database, Drs. Dimitris Xanthos and Tino Jäger for help with the tables, and all members of the Department of Neurophysiology for valuable discussions and comments on earlier versions of the manuscript.
Address for reprint requests and other correspondence: J. Sandkühler, Dept. of Neurophysiology, Center for Brain Research, Medical University of Vienna, Spitalgasse 4, A-1090 Vienna, Austria (
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12805 | https://stl.boost.org/random_shuffle.html | random_shuffle
random_shuffle
Category: algorithmsComponent type: function
Prototype
Random_shuffle is an overloaded name; there are actually two random_shuffle functions. template RandomAccessIterator>
void random_shuffle(RandomAccessIterator first, RandomAccessIterator last);
template RandomAccessIterator, class RandomNumberGenerator>
void random_shuffle(RandomAccessIterator first, RandomAccessIterator last,
RandomNumberGenerator& rand)
Description
Random_shuffle randomly rearranges the elements in the range [first, last): that is, it randomly picks one of the N! possible orderings, where N is last - first. There are two different versions of random_shuffle. The first version uses an internal random number generator, and the second uses a Random Number Generator, a special kind of function object, that is explicitly passed as an argument.
Definition
Defined in the standard header algorithm, and in the nonstandard backward-compatibility header algo.h.
Requirements on types
For the first version:
RandomAccessIterator is a model of Random Access Iterator
For the second version:
RandomAccessIterator is a model of Random Access Iterator
RandomNumberGenerator is a model of Random Number Generator
RandomAccessIterator's distance type is convertible to RandomNumberGenerator's argument type.
Preconditions
[first, last) is a valid range.
last - first is less than rand's maximum value.
Complexity
Linear in last - first. If last != first, exactly (last - first) - 1 swaps are performed.
Example
const int N = 8;
int A[] = {1, 2, 3, 4, 5, 6, 7, 8};
random_shuffle(A, A + N);
copy(A, A + N, ostream_iterator(cout, " "));
// The printed result might be 7 1 6 3 2 5 4 8,
// or any of 40,319 other possibilities.
Notes
This algorithm is described in section 3.4.2 of Knuth (D. E. Knuth, The Art of Computer Programming. Volume 2: Seminumerical Algorithms, second edition. Addison-Wesley, 1981). Knuth credits Moses and Oakford (1963) and Durstenfeld (1964). Note that there are N! ways of arranging a sequence of N elements. Random_shuffle yields uniformly distributed results; that is, the probability of any particular ordering is 1/N!. The reason this comment is important is that there are a number of algorithms that seem at first sight to implement random shuffling of a sequence, but that do not in fact produce a uniform distribution over the N! possible orderings. That is, it's easy to get random shuffle wrong.
See also
random_sample, random_sample_n, next_permutation, prev_permutation, Random Number Generator
Copyright © 1999 Silicon Graphics, Inc. All Rights Reserved.TrademarkInformation |
12806 | https://math.stackexchange.com/questions/4203173/expected-number-of-successes-when-sampling-n-times-from-a-binomial-distribution | combinatorics - Expected number of successes when sampling N times from a binomial distribution, stopping after M successes? - Mathematics Stack Exchange
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Expected number of successes when sampling N times from a binomial distribution, stopping after M successes?
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I'm trying to figure out the equation for the expected number of successes E(X)E(X) when sampling from a binomial distribution with success probability p p. In this sampling scheme, we either sample N N times, or until we observe M M successes, whichever comes first.
For example, if N=10 N=10 and M=3 M=3, the results might look like
| # | Run 1 | Run 2 | Run 3 |
--- --- |
| 1 | T (1 success) | F | F |
| 2 | F | F | T (1 success) |
| 3 | T (2 successes) | T (1 success) | T (2 successes) |
| 4 | F | F | F |
| 5 | F | F | T (3 successes) |
| 6 | F | F | (stop) |
| 7 | T (3 successes) | T (2 successes) | |
| 8 | (stop) | F | |
| 9 | | F | |
| 10 | | F | |
| Total Successes | 3 | 2 | 3 |
I know that if the sampling scheme were only sampling N N times, then E(X)=N p E(X)=N p. Similarly, if the sampling scheme was sampling until M M successes, naturally E(X)=M E(X)=M. However I'm having difficulty figuring out what the combination of the two stopping criteria would make E(X)E(X).
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edited Jul 20, 2021 at 14:50
shoteyes
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asked Jul 20, 2021 at 13:39
Michael YangMichael Yang
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You calculate the probability of seeing 0,1,2,3,…M 0,1,2,3,…M successes and use that to compute the expected value.Ross Millikan –Ross Millikan 2021-07-20 14:09:18 +00:00 Commented Jul 20, 2021 at 14:09
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If M≥N M≥N, then the experiment will never reach M M successes before there are N N trials, so the expected value would be N p N p. The case where M<N M<N is less trivial.
Assuming M<N M<N, you can separate the probabilities in the sum for the expected value
E(X)=∑k=0 M k P(there are k successes)E(X)=∑k=0 M k P(there are k successes)
based on whether the experiment reaches N N trials or not.
If the experiment reaches N N trials, then the probability of k k successes is the usual binomial distribution calculation of
(N k)p k(1−p)N−k.(N k)p k(1−p)N−k.
If the experiment does not reach N N trials, then there must have been M M successes, and the M M-th success appears before the N N-th trial. Summing these probabilities based on where the M M-th success appears, we get
∑k=M N−1(k M)p M(1−p)k−M.∑k=M N−1(k M)p M(1−p)k−M.
We can put these two observations together to express the expected value as
E(X)=∑k=0 M k(N k)p k(1−p)N−k+M∑k=M N−1(k M)p M(1−p)k−M.E(X)=∑k=0 M k(N k)p k(1−p)N−k+M∑k=M N−1(k M)p M(1−p)k−M.
The first sum can be rewritten as
N p∑k=0 M−1(N−1 k)p k(1−p)N−k−1 N p∑k=0 M−1(N−1 k)p k(1−p)N−k−1
using the same technique one does to prove that the expected value of the binomial distribution is N p N p; but other than that, there probably isn’t a nicer way to express these partial sums of a binomial expansion.
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answered Jul 20, 2021 at 14:59
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12807 | https://marshareb.github.io/files/essay_2.pdf | Pell’s Equation History, Algorithms, and Open Problems JAMES MARSHALL REBER December 1, 2017 Abstract In this talk, we discuss finding integral solutions to the Pell equation, which is a Diophantine equation of the form x2 −Dy2 = 1 for integral solutions x, y ∈Z and D > 0 a square-free integer. In particular, we focus on finding fundamental solutions for the Pell equation, which in turn will give us all solutions for the Pell equation.
History Diophantine equations were of great interest to the Greeks.
In particular, Archimedes (287 - 212 BC) had a fascination with finding solutions to a partic-ular Diophantine equation, which we now know is a particular Pell equation.
The problem has to do with calculating the number of cattle of particular colors in a herd which belonged to the sun God, the poem itself which outlines this problem is 40 lines. Letting W, B, D, Y denote white, black, dappled, and yel-low cows, and w, b, d, y denote their respective colors but for cows. Then for example, one of these restriction was W = 1 2 + 1 3 b + y.
This was not the hard part; one could solve this system of restrictions in order to find a particular number of cows and cattle using linear algebra. The real kicker of this problem was what followed, and in particular Archimedes said that if one could solve this, then one could consider themselves "... most skilled in numbers." Archimedes then said that the number of white bulls added to the number of black pulls must be a square number, and the nummber of dappled 1 cows added to the number of yellow bulls must be a triangular number. This could eventually be reformatted into u2 −(606)(7766)2v2 = 1 where we have D + y = v(v + 1)/2 and u = 2v + 1. Many tried to solve this problem, using a variety of different techniques, but what was really required in order to solve this was an efficient algorithm to find a Fundamental solution to the Pell equation; that is, the smallest integer tuple (u, v) so that they satisfy the equation. This was not an easy task, and it wasn’t until Amthor in 1880 discovered the way to find it that the problem was finally satisfied (Amthor es-tablished that the number would be extremely large and thus nearly impossible to calculate by hand; it wasn’t until the advent of computers that we actually had the exact answer).
The key issues with the Pell equation is two fold; do solutions exist to the equation, and if so how do we find these solutions? The existence issue wouldn’t be realized for long after Archimedes, but finding a solutions was of much interest to Indian mathematicians in 628 B.C.. Brahmagupta (598 - 670 AD) discovered the identity of composition of forms (referred to at the time as samsa). Using this, Brahmagupta found an ad hoc way to discover fundamental solutions to the Pell equation. Later, Bhaskara II found a more cyclic algorithm to find solutions to solve the Pell equation, which resembles the modern PQA algorithm. The history of the Pell equation fell silent until Fermat posed the question, "Given any number whatever that is not a square, there are also given an infinite number of squares such that, if the square is multiplied into the given number and unity is added to the product, the result is a square." Brounecker (1620-1684) found an algorithmic solution method to this prob-lem, but Euler would later mistakenly attribute this to Pell (hence the name).
Thus, we see that Brounecker, Euler, Brahmagupta, and Brahska II all found similar methodologies to finding these fundamental solutions, but none of them were concerned about whether or not this would always succeed in finding a solution for all D > 0 which are not square. This was later rectified by Lagrange, although Weil claims that Fermat had an unpublished solution. Using Algebraic Number Theory, the algorithms built by Euler, Lagrange, et al. were improved upon and made more efficient, and it evolved into what we now refer to as the PQA algorithm (or sometimes as the continued fraction method). We now explore the specifics of what the Pell equation entails.
Existence and Finding a Fundamental Solution.
Lagrange was able to notice two main facts about solutions to the Pell equation that allowed him to generate a good algorithm for finding all solutions. First, he 2 noticed that the set of solutions is a cyclic group generated by something called a fundamental unit, and second he noticed that this fundamental unit is related to the continued fraction expansion of the quadratic surd; or, in other words, if the equation is of the form x2 −dy2 = 1, then the fundamental solution is related to to the continued fraction expansion of √ d.
To start, we define the norm mapping, which we’ll be using throughout.
Definition 1. If K is an extension of a field F, and K/F is separable, then for any α ∈K we have NK/F(α) = n ∏ i=1 σi(α), where the σi are all of the isomorphisms of K into the algebraic closure ¯ F of F fixing the elements of F.
Notice that if d > 0 is a square-free integer, we have that Q( √ d)/Q is a Galois extension. Moreover, Gal(Q( √ d)/Q) = {Id, σ} where σ : √ d 7→− √ d; i.e. the conjugate mapping. We can then simplify our definition of the norm mapping to our more specific case.
Definition 2. If d > 0 is a square-free integer, K = Q( √ d), and F = Q, then we define the norm mapping to be NK/F(α) = α¯ α.
There is a special feature of the norm that allows us to determine the units within our field. We need a special feature of the norm mapping in order to establish this, though.
Remark. Throughout, we’ll be using N to denote the norm mapping instead of NK/F.
Lemma 1. The norm mapping is multiplicative; that is, for a, b ∈K defined as prior, we have N(ab) = N(a)N(b).
Proof. This readily follows by the multiplicativity of conjugates. Notice N(ab) = ab ¯ ab = ab¯ a¯ b = a¯ ab¯ b = N(a)N(b).
■ Using this, we can determine the units using the norm mapping.
Corollary 1. We have that a ∈K is a unit if and only if N(a) = ±1.
Now, we go back to the Pell equation. Recall that it is an equation of the form x2 −dy2 = 1. We can factor the left hand side in the order Z[ √ d] to be (x − √ dy)(x + √ dy) = 1. Let α = x + √ dy. Then, using the norm, we can rewrite this as N(α) = 1. Thus, we can see that the norm is related to the units of the order Z[ √ d]. Define G = {α ∈K : N(α) = 1}. We’ll observe that this is an abelian group.
Lemma 2. G as defined prior is an Abelian group.
3 Proof. We need to go through the five requirements for an Abelian group. It’s closed; that is, if a, b ∈G we have ab ∈G, since N(a) = N(b) = 1 and so N(a)N(b) = N(ab) = 1. Next, we notice that there is an identity; we clearly have N(1) = 1. Inverses follows since a¯ a = N(a) = 1, so dividing both sides gives us ¯ a = 1/a. We then see that ¯ a ∈Z[ √ d] and N(¯ a) = 1. Associativity fol-lows, since N(a)(N(b)N(c)) = (N(a)N(b))N(c) clearly. Finally, commutativity follows since N(ab) = N(a)N(b) = N(b)N(a) = N(ba).
■ So we see that the set of solutions for the Pell equation forms a group. How do we know that this group is nontrivial though? Moreover, can we show that this group is cyclic and therefore generated by a cyclic element? If it is a non-trivial group, what is the size of the group (or, informally, how many equations are there)? Finally, if the group is cyclic, how do we find the generator? These questions (suggestive as they may be) did not follow in this order throughout history; as discussed, the second and third question were answered before Lagrange was able to really answer the first. We will go through them as stated.
Lemma 3. The equation x2 −dy2 = 1, where d > 1 and square-free, has at least one solution where y ̸= 0.
Proof. To prove this, we will need a lemma that we’ll state without proof.
Lemma 4. There are infinitely many solutions of the equation x2 −dy2 = k, d, k ∈Z and d square-free, where x, y ∈Z>0 for some k with |k| < 1 + 2 √ d.
We utilize the proof by LeVeque . Using this lemma, there is an integer for which one of the two equations N(a) = ±k has infinitely many solutions a in Z[ √ d]. Since there are only finitely many residue classes mod k in Z[ √ d], some residue classes must contain at least three of these solutions. Let N(a1) = N(a2) = ±k and a1 ≡a2 (mod k), but that a1 ̸= ±a2. Then a1 ¯ a2 = a2 ¯ a2 ≡0 (mod k), so that b = a1 ¯ a2/k is an element of Z[ √ d]; that is, it has integral components. Since N(b) = b¯ b = a1 ¯ a2 · ¯ a1a2 k2 = N(a1)N(a2) k2 = 1, b yields a solution of the equation. If the second component of b were 0, then N(b) = 1 would imply that b = ±1, whence a1 ¯ a2 = ±k = ±a1 ¯ a1 ¯ a2 = ±¯ a1 a2 = ±a1, contrary to hypothesis.
■ 4 With the existence of at least one solution, we are led to the next question – are these solutions generated by a single element? It turns out yes, and it’s generated by what we call the fundamental solution; that is, the smallest such solution.
Lemma 5. Let d > 0 be a square-free integer, and assume that there is a solution to x2 −dy2 = 1. Then there exists a unique f ∈G so that f > 1 and for all a ∈G, we have a = ± f n for some n ∈Z.
Proof. This follows the proof stated in LeVeque . Let a ∈G. We notice that a, 1/a, −a and −1/a give four solutions to Pell’s equation, differing only by signs of x and y. So we need to show that every a > 1 with N(a) = 1 is of the form a = f n where f is the fundamental solution; i.e. the smallest such solution.
Since a > 1 and we chose f to be minimal, we have a ≥f. Hence, there is a positive integer n such that f n ≤a < f n+1. Now notice that a/ f n = a ¯ f n is in Z[ √ d], and N(a/ f n) = 1. Thus, a/ f n = b gives an integral solution of Pell’s equation. From the definition of n, we have 1 ≤b < f and by how we defined f we cannot have 1 < b < f. Hence, b = 1 and a = f n, as we required.
■ Remark. This result should seem very similar to Dirichlet’s Unit theorem. This is because it is, in fact, a special case of it. However, this was proven before Dirichlet’s Unit theorem, and so we provide the proof without using this result.
So we have that the group has two generators, −1 and f where f is the fundamental solution, and f is of infinite order, that is, there are infinitely many solutions. We now have the final question; how do we find such a solution?
This is where the method of continued fractions comes into play. In order to discuss the method of continued fractions, we must make some preliminary definitions.
Definition 3. Given any irrational number x, define the sequence of irrational numbers recursively by x0 = x, xk+1 = 1 xk −⌊xk⌋for k = 0, 1, 2, . . .
in terms of the floor function. The continued fraction of x is the infinite nested fraction a0 + 1 a1 + 1 1 a2+··· where ak = ⌊xk⌋is an integer.
We really only want to deal with finitely many terms, so this leads us to another definition.
Definition 4. Denote the nth convergent as the quantity which is obtained by the first n terms of the continued fraction {a0; a1, a2, . . . , an−1} = a0 + 1 a1 + 1 a2+ 1 ···+ 1 an−1 .
5 Lagrange was able to use these continued fractions to establish his theorem.
Theorem 1. (Lagrange) Fix d > 0 square-free.
(a) The continued fraction of √ d is in the form √ d = {a0; a1, a2, . . . , ah−1, 2a0} where the bar means that the sequence repeats indefinitely.
(b) Write the hth convergent of the continued fraction above as the rational number {a0; a1, a2, . . . , ah−1} = uh vh .
Then uh and vh are positive integers which satisfy the relation u2 h −dv2 h = (−1)h.
Moreover, using this we get how to find the fundamental solution.
Corollary 2. Let h denote the smallest period of the continued fraction sequence. Then we have that the fundamental solution f is f = ( uh + vh √ d, if h ≡0 (mod 2) u2h + v2h √ d = (uh + vh √ d)2, if h ≡1 (mod 2).
Instead of proving these results, we’ll show an example of it in action.
Example 1. Let d = 7. Then we have the sequence {2; 1, 1, 1, 4, 1, 1, 1, 4, . . .}.
Moreover, we see h = 4 (cutting it off before 4), and so we calculate 2 + 1 1+ 1 1+1 = 8 3; or, in other words, uh = 8, vh = 3, and our fundamental solution is 8 + 3 √ 7.
Final Thoughts and Open Problems From prior, we saw that in the case x2 −dy2 = 1, we have infinitely many solutions and we have a "good" algorithm in order to find all of these solutions.
What about the equation x2 −dy2 = k for arbitrary k ∈Z? It turns out that Lagrange’s method gives us a good algorithm for also finding out if there are solutions in the case k = −1, and if there are solutions Lagrange’s theorem also gives us a way to find all of the solutions. This case is dubbed the negative Pell equation. In fact, Lagrange’s continued fraction method can be used to generate all solutions to the Pell equation for arbitrary k, although it is not always guaranteed to have a solution (such as in the case where k = −1).
Thus, Pell’s equation shows us that we have a "good" algorithm for finding a generator for the units of the order O = Z[ √ d] ⊆K = Q( √ d). One then wonders is there always a good algorithm for finding the generator for the units of any arbitrary order? It turns out that currently there is no such algorithm. It is, in fact, the "holy grail" of algebraic number theory to find such an algorithm.
6 There is still yet more to the Pell equation. This short presentation has really only skimmed the topic as a whole. For more details on Pell’s equation, I recommend reading "Fundamentals of Number Theory" and "Solving the Pell equation" . LeVeque gives a good introduction to the topic, while Jacobson and Williams really go into depth on the topic as a whole.
References 1. Hendrick W. Lenstra, Jr. "Solving the Pell equation" MSRI Publications, 44, 2008.
2. Jarrod A. Cunningham, Nancy Ho, Karen Lostritto, Jon A. Middleton, Nikia T. Thomas. "On Large Rational Solutions of Cubic Thue Equations: What Thue Did to Pell" Rose-Hulman Undergraduate Mathematics Journal: Vol. 7: Iss. 2, Article 6, 2006.
3. Keith Conrad. "Dirichlet’s Unit Theorem" Publisher: Author.
4. Mansfield Merriman. "The Cattle Problem of Archimedes" Popular Science Monthly, 67: 660-665, 1905.
5. Michael J. Williams and Hugh C. Williams. "Solving the Pell Equation" Springer, 2009.
6. Seung Hyun Yang. "Continued Fractions and Pell’s Equation" University of Chicago REU Papers, 2008.
7. William J. LeVeque. Fundamentals of Number Theory. Addison-Wesley Publish-ing Co., Reading, Mass-London-Amsterdam, 1977.
7 |
12808 | https://www.methanol.org/wp-content/uploads/2017/02/Training-Module-MI-LIAM-20150915-vfinal-EN.pdf | Page | 1 Methanol Poisoning Community & Medical Education Programs Model Pilot Program Information Developed by The Methanol Institute (MI) & The L.I.A.M Charitable Fund Created: 28 July 2015 Last Reviewed/Updated: 15 September 2015 Page | 2 Contents Background ............................................................................................................................................. 4 Legal Status ............................................................................................................................................. 5 Mission .................................................................................................................................................... 6 Values ...................................................................................................................................................... 6 1. New Country ....................................................................................................................................... 7 1.2. Who can assist you to build your starting networks? ...................................................................... 8 1.3 Build Allegiances ............................................................................................................................... 9 1.4 Build on what has gone before ......................................................................................................... 9 1.5 Know your audience ......................................................................................................................... 9 1.6 Work with people who want to change.......................................................................................... 10 1.7 Listen to their advice and be willing to move slowly ...................................................................... 10 1.8 Provide concrete tools to assist ...................................................................................................... 10 1.9 Know when to push and when to hold back, revisit, revise ........................................................... 11 1.10 Do not ignore the “What’s in it for me?” factor ........................................................................... 11 1.11 Be prepared for setbacks and unexpected outcomes .................................................................. 11 1.12 Acknowledge that the process/change may happen in unexpected ways................................... 12 Community Education Program (CEP) ................................................................................................. 13 Specific Points and Aids ....................................................................................................................... 13 2 Community Education Aims: .............................................................................................................. 13 2.1 Where to start? ............................................................................................................................... 13 2.3 Needs Analysis ................................................................................................................................ 14 2.4 Starting in the chosen region ........................................................................................................ 155 2.5 Community Education Plan ............................................................................................................. 16 Medical Education Program (MEP): ..................................................................................................... 16 Specific Points and Aids ....................................................................................................................... 16 3 Medical Education Aims: .................................................................................................................... 16 Page | 3 3.1 Where to start? ............................................................................................................................... 16 3.3 Needs Analysis ................................................................................................................................ 18 3.4 Starting in the chosen hospital ....................................................................................................... 19 3.5 Medical Education Plan ................................................................................................................... 20 4. Training Equipment & Resources ..................................................................................................... 21 4.1 Media coverage of Global Methanol Poisoning Cases .................................................................. 21 4.2 Medical Reference Materials ......................................................................................................... 22 4.2.1 First Aid for Suspected Methanol Poisoning .............................................................................. 23 4.4 Examples of Information Flyers ..................................................................................................... 29 4.6 Community Education Presentation (CEP) Examples ................................................................... 30 4.7 Medical Education Presentation (MEP) Examples ....................................................................... 30 5. Timelines ........................................................................................................................................... 31 Page | 4 Background The Methanol Institute (MI) is the global trade association which serves one of the world's most vibrant and innovative industries. Founded in 1989 to further the advancement of the methanol industry, MI now serves its members in every corner of the globe from our Singapore headquarters and regional offices in Washington DC, Brussels, and Beijing. The global methanol industry impacts our daily lives in countless ways. Methanol is an essential chemical building block for numerous products that touch our daily lives, and in an era focused on alternative energy, it is an emerging energy resource. Each day, roughly 70,000 metric tonnes of methanol (23 million gallons or 87 million liters) is shipped from one continent to another, enough product to fill 777 rail cars. Today, our membership includes the world’s leading methanol producers, technology companies, distributors, terminal operators and shippers. The methanol industry’s leading companies have joined to support MI as the most effective way to interact with governments, NGOs and potential new customers throughout the global marketplace. MI collaborates with each of our member companies to create solutions for the global industry, and to ensure that already vibrant markets for methanol and its numerous derivatives continue to thrive. An important part of MI’s work and that of our members is providing resources to the communities in which we work about safe handling and use of methanol. Since November 2013, MI and The L.I.A.M Charitable Fund have worked closely to develop community and medical education programs (CEP/MEP) in Indonesia aimed at protecting the public from risks of consuming adulterated alcoholic beverages. The L.I.A.M Charitable Foundation (Lifesaving Initiatives About Methanol) is an Australian organisation that was established after the death of 19 year old Liam Davies in January 2013. Liam was poisoned with methanol that had been added to a drink served from a labelled bottle of vodka that he bought from a bar in Gili Trawangan, Indonesia. He was initially treated in Lombok, but his case was misdiagnosed and the correct treatment was not given. Liam was air-lifted back to Perth, was diagnosed with methanol poisoning within 10 minutes of being at the hospital, all necessary medical treatment was given but too much time had passed and Liam died 4 days later Liam’s life could have been saved with correct, timely medical treatment in Indonesia. Liam’s parents did not want any other family to experience what had happened to them so they developed the LIAM Charitable Foundation. The primary objective of the Foundation was to prevent further accidental deaths from methanol poisoning. This has been done through a campaign of community education and development, as well as medical training for hospital and clinical staff. The aim is to implement systemic change, such that the positive effects of the program are sustainable and long lasting. The LIAM Foundation first began work in 2013, and has rapidly built a reputation in Bali and Lombok for high quality and well-targeted educational programs. LIAM has MOUs with both RSUP Sanglah Hospital and the School of Public Health at Udayana University. LIAM works collaboratively with Page | 5 selected staff from both Sanglah Hospital and Udayana University to provide education and training. There are two parts to the education program: the first a program for community education (for the general community and primary health centre staff) and the second an education program for clinical staff within hospital settings. Contacts: The Methanol Institute (MI) 10 Anson Road #32-10 International Plaza Singapore 079903 Contact: Phone: +65 6325 6300 e-mail: sg@methanol.org Tim & Lhani Davies The L.I.A.M Charitable Fund Legal Status: Not for Profit Charitable Foundation Australian Registered Business Name: L.I.A.M Charity 2 Hasper Place, Marmion Perth, Western Australia 6020 Australia Phone: +61 8 92038145 e-mail: monty.rock@bigpond.com Page | 6 Mission The highest rate of deaths from methanol poisoning are in Indonesia are among males aged 18-30. In third and second world countries, such deaths, often of the primary income earners, bring both immediate and long term pain to families. The loss of income and support can have devastating effects. Mass poisonings in small villages have seen poverty increase as families’ ability to support themselves is taken away. In many traditional communities of developing countries, widowhood is the beginning of ‘social death’ for women and their children. It is not enough that they have lost their husbands (the main income provider), but widowhood also robs them of their status and leaves them to a life of extreme discrimination and stigma. To save one life is to save many. We will never rid the world of methanol poisoning, but we can train medical staff to save lives, and educate about the dangers of incorrect distillation. We can warn people to make the correct drinking choices. Values The LIAM Charity’s strong, grounded values are unnegotiable. This has seen our program succeed in the past and will continue to aid growth and enable us to enact change in the world moving forward. The Liam Charity believes in doing what is right, not what is easy. We strive for: o Empowerment – giving people power through knowledge o Consistency – all programs presented with the same message o Teamwork – pushing forward from all levels o Flexibility – maintain the ability to evolve and change to fit the need o Professionalism o Commitment to Success o Respect Page | 7 1. New Country So you are wanting to tackle the enormous issue of methanol poisoning in a certain country? BEFORE you step foot in that country you must do the most important work to make your training a success. What do you know about? o Local political environment and their views on alcohol and drinking o Local religions and their views on alcohol and drinking o Cultural beliefs and sensitivities o Major do’s and don’ts to keep yourself and all your staff safe o Recent high risk issues for travelers o Laws on foreign aid workers Do you know someone that has worked in this country before? If so, talk to them extensively about issues they came up against, what excited them and what frustrated them when dealing with local governments, law enforcement officials, and other people in positions of power. Head to your own country’s respective embassy or consulate and also ask the above questions. How can you facilitate and fund ways for locals to educate locals? Engage, plan, meet, discuss, brainstorm and network! Making long term and systemic change is never easy. Be clear with your vision, set your 12-month plan, focusing on realistic outcomes for that time but be prepared for idea changes. As your start moving forward, what seemed like a great idea may not fit your location. That is OK, but how do you adjust and tailor your ideas to fit your unique local situation, and what might be other options to help achieve your goals? Be flexible and open minded, in order to mould your training program to the environment in which you are working. If something works well, then expand on that, and see whether you can transfer that into your other training modules. Think about how you would feel if someone who had never lived in your home country suddenly came into your country or workplace and started dictating that they knew best. A successful program requires knowledge sharing, training and funding to bring sustainable changes on many levels. This can only be achieved if we empower and work alongside local stakeholders to understand well their needs, to assess where we can help to achieve their needed outcomes, and to create a framework through which they can achieve sustainable, long-term results among local communities. Page | 8 1.2 Who can assist you to build your starting networks? Sadly, methanol poisoning is predominately an issue in second/third-world countries. Lack of access to legitimate alcoholic spirits, high taxes on legitimate alcohol, and/or cultural bans on the production and consumption of alcohol often lead to black market production of moonshine to satisfy demand by many consumers. Incorrect distillation practices or intentional blending of chemicals – including, but not limited to –methanol – is seen worldwide. Global hotspots where adulterated alcohol health problems and fatalities are occurring include: Indonesia, Vietnam, India, Kenya, Nigeria, and Pakistan, among others. Unfortunately, the need for methanol awareness and poisoning prevention training is only continuing to grow as incidences of methanol-laced products in high-risk markets like these continue to grow. Depending on the needs of each market and the level of support required for community education versus medical education and training, your approach may be different in how you plan to establish a Community Education Program (CEP) versus a Medical Education Program (MEP). You may decide to start with one program in the first 12 months, and introduce the second program in the second year as your network and contacts expand. It’s most likely that medical and community education assistance are already being provided by international bodies in your affected region in regard to other issues. Ask your foreign affairs departments to put you in contact with a trustworthy source already well entrenched, including those who are already doing substantial work in the public healthcare sector. Engaging with and using the services of a major training hospital is your aim for your MEP. Doing the same with a major university that specialises in Community Health is your aim for your CEP. Engage with federal and local government health departments, alcohol regulation agencies, educational ministries and local departments, police, etc. to ensure that they authorise, support, and promote the methanol poisoning awareness programs. Be sure to take time to sell the program to people at all levels – it is of no help if the CEO or other senior leaders think the program is great but the workers who will deliver the program are not on board. The same can be said if the decision makers are not sold on the program. You need to ensure before any training takes place that you have the permission of all necessary governing bodies as well as the necessary workplace agreements and insurance coverages. Page | 9 (Please note: If you are able to work under a Memorandum of Understanding (MOU) with existing bodies, your staff may have all necessary protection and payment through their primary employer. Do not make assumptions – check all details thoroughly and get written acknowledgement.) 1.3 Build Allegiances It takes time to get to know how an organisation works, who the key people are, (i.e. the informal power structure). Don’t hurry to get a lot done straight away. Take your time looking at what’s going on, talk to people and listen to what they have to say. Even if the standards and practices are very different and not as high as you are used to, make sure that you are respectful of the work environment and local cultures, and do not openly criticise what you see is happening around you until you understand things better. If you are invited out socially by the staff you are working with, go if you are able to do so. Changes in practice are built on good relationships, so take time to build your relationships before trying to make major changes. 1.4 Build on what has gone before Many developing countries have had a lot of people trying to assist in the past. Do your best to find out what has been done, and what the local counterparts felt about how it went. Again, this may take time, as they may initially be afraid to tell you the truth if things have not gone so well. Try and explore what went well and why they felt this was the case, and assess what are the change priorities of the local counterparts. Try and find something where there is local buy-in and ownership – this will help you to be successful. 1.5 Know your audience First of all, you will need to really think about who your audience is – i.e. who are you working with, reporting to, and who has a vested interest in the success or failure of your endeavours? You may need to use different strategies to uncover what each of the various stakeholders/interest groups really want and need. o Try and find one or two people in your immediate work area who are happy to give you information about the culture and where the various players sit within the organization. o Be prepared to ask many questions to uncover what outcomes people expect/want. o Try and assess similarly expectations of the organization’s senior leaders. o Also try and find out what the priorities are of the people with whom you are working most closely. You may often find these are very different. Part of a development role may be bridging the gap between the two sets of expectations. Be prepared to be wrong and to listen and learn. Page | 10 1.6 Work with people who want to change As with anywhere, there will be staff who are keen to change and staff who are happy with the status quo. As far as possible, identify people who are keen and willing to change, and work with them. They are insiders and know the culture and the challenges – together you can be a good team. In many cases, the local staff will need you (as an outsider) to give credibility to their ideas. While this is sometimes uncomfortable for someone from a Western culture, it may be the only way that staff who are relatively junior can initiate changes in their practices. As you are working, you will find it relatively easy to identify the keen staff, so get to know them and find out what they would like to do to improve organizational practices. 1.7 Listen to their advice and be willing to move slowly While identifying the keen staff is relatively easy, actually getting change to happen is a lot harder. As a Westerner you will probably feel that once the decision to change has been made, the plan of action will soon follow. However, once you start to try and drive the change forward, you are likely to find that there are many unforeseen obstacles thrown up in the way. Westerners may tend to believe that things happening more or less in a linear fashion – i.e., once we decide to make a change and the idea has been agreed by whatever authority is necessary, then we can just go ahead and do it! This is unlikely to be so in a developing country. In many cases, people are not used to moving ideas from discussion to practice. They may lack the authority to agree to any of the changes that are mooted and they may be very nervous about doing anything that is different in case it is ‘wrong.’ The culture of blame can be very strong. Listen to what your counterparts say about the obstacles and work with them to develop strategies to overcome them. This will take time. 1.8 Provide concrete tools to assist Combined with the lack of training in project/change implementation, you may find that your counterparts have not had experience in developing tools that will assist in implementation, evaluation, and accountability. Be prepared to develop drafts of these types of documents on your own and then to work with local staff to see how they can be used to assist the change process. Page | 11 1.9 Know when to push and when to hold back, revisit, revise Remember that ideas and practices that may be familiar to you may be completely foreign to your counterparts. Sometimes you will feel you’ve got agreement about change only to find that your ideas have been put on hold or overtaken by something else that is seen by everyone else as more important. While you may feel really frustrated and at times undervalued, try to remember that you are assisting others to move towards a goal and that they will do it in their own way and at their own pace. Generally it is better to sit quietly to one side and wait until the time seems right to start again. Occasionally gently raise the idea and see whether or not anyone is interested in progressing it. If there is no interest, then wait. You may need to explain and re-explain what you mean, as local staff may say that they understand when actually they do not. They may not wish to offend you, or perhaps “lose face” by admitting that they do not understand something. If something you suggest keeps getting blocked, then leave it and move on – you can come back to it later if it still seems necessary. 1.10 Do not ignore the “What’s in it for me?” factor It is human nature to resist change – although one way to make it effective is to think about how it will improve the life of those involved, or how it will contribute to them being recognised as innovators and leaders. Sell this to the staff likely to be affected, and persuade them about the value of the identified change. This is slightly different in a developing country, where local staff may not want to ‘put their heads up’ and be recognised or stand out. In a country where there are no consequences for poor practice and where doing the same things each day causes no difficulties in terms of performing satisfactorily at work, addressing this factor can be challenging. The proposed change needs to benefit all participants in some way – e.g., by making their work life easier, by giving them better knowledge and professional recognition, or by creating a culture of accountability. It is important that you try and identify the benefits for all stakeholders before attempting any changes to practice, as these can be powerful persuaders. 1.11 Be prepared for setbacks and unexpected outcomes Even if you are culturally sensitive, build good relationships, work hard and so forth, you still may find that nothing happens. What can you do? Firstly, recognise that this is not an unusual occurrence. Secondly, try and find out what else is going on in the work place, what new government directives have happened or what new priorities have been identified. Other activities will often get in the way of your work. Try to remember that the staff you are working with have full-time positions and they are subject to all sorts of pressures that we may find difficult to understand. Their day-to-day work is often more important than making changes to their practice. You have a full-time job designed to change practice – they are often involved in change, in addition to everything else they are expected to do. Therefore, it is not surprising that our ideas might often be put to one side in favour of other things. The trick is to be ready to assist when (for example) an outside accreditation body mandates a Page | 12 change which is similar to something you’ve been trying to achieve for the past six months. Provide whatever tools and advice you have and the change may well happen in a completely different way to that which you originally anticipated. Remember that – without you there sowing the seeds – there would be nowhere for your counterparts to start the change process. Don’t despair if it feels as if your ideas have been taken and used without recognition or credit being given. The point of development work is for your counterparts to develop the knowledge and skills to make effective changes to practice. If you see improvements, applaud them (and then look for ways to make the changes even better). 1.12 Acknowledge that the process/change may happen in unexpected ways Finally, change can occur in a number of ways. You may find that you have made a proposal for change a number of months previously which, despite getting endorsement did not happen, so you have not pursued it further. Suddenly, one day you see that staff seem to be doing what you suggested all those months ago, and you wonder, “What happened?” It could be that a local staff member has been away on a course and has come back with this ‘new’ idea that has now been adopted with great enthusiasm. So you think to yourself, “Why am I here?” Don’t be downhearted – it could be that the change which is occurring now has only happened because you planted the initial idea months ago. People may have since had time to get used to it, and when they see it has been endorsed by others elsewhere, then they are willing to go along with the revised practice. Page | 13 Community Education Program (CEP) Specific Points and Aids 2. Community Education Aims: o Improving people’s knowledge and awareness about the dangers of drinking incorrectly distilled alcohol or mixing chemicals with alcohol o Improving producers awareness about the importance of safe distillation practices o Improving health worker’s knowledge of methanol poisoning symptoms and how to handle victims of methanol poisoning to avoid deadly outcomes. 2.1 Where to start? You need to undertake an MOU (Memorandum of Understanding) with a community-focused educator. This could be a foreign or local agency who works specifically with community issues and is based in your region. LIAM undertook in Indonesia– to work with a university faculty member who specialises in community health. He/she will guide you on your starting points of who you need to meet, and what approvals you require before you start. Some ideas are: o Government Department of Health o Government Department of Food and Beverage o Government Department of Education o Government Department of Tourism o Government Department of Law Enforcement You may need to have meetings at both the local and regional levels to let people in power know who you are, why you are educating about methanol poisoning and how you need their support and assistance. Find out which official ratifications you will require. It is also important to consider engaging with federal government ministries simultaneously. There may be a need to effectuate national policies for things like the use of ethanol in methanol poisoning treatment, or to enable medical schools around the country to have medical poisoning included in their curricula, etc. An awareness workshop followed by lunch could be a great way to engage and educate about: o What is methanol poisoning? o How are locals being poisoned? o Who is your organization, and why you are seeking to develop these projects? o How the program works (train the trainer, locals helping locals) o How methanol poisoning awareness and education will save lives o Question and Answer time (i.e., what information do they have already and where do they see the greatest need for your involvement?) Page | 14 This is the start of you building up your own network, so take the time to talk with as many people as possible. Ask who they think you should be contacting and what poisonings have they heard about. Ask, who are the key stakeholders who have been involved in helping to treat those affected by adulterated alcohol and methanol poisonings? Ask if you can meet with them individually to follow up from the workshop. Build that relationship. 2.2 Train your trainers Your educators need to have a thorough understanding of all information required for them to be effective within the communities. Dispelling myths around methanol poisoning deaths can be tricky, especially in countries where alcohol consumption is considered culturally or religiously unacceptable (some religions believe deaths from methanol poisoning are a punishment for a wrong doing). Explain the medical facts and then slowly, softly reinforce that the medical facts are accepted all around the world, citing specific examples of countries in which these treatment practices are taking place. Your educators will need a thorough knowledge of: o Who are your funding donors/sponsors, and what are their business and CSR activities and goals? o What is methanol and what should it be used for? o The basics about methanol, what it is, how it is used for industrial applications, and why it is improperly added into alcoholic drinks, as well as its effects on the human body (i.e., toxicology). It is important also to understand the distillation process for making alcohol, and why methanol can inadvertently be produced from making alcohol incorrectly. How does the human body process methanol, and what are the symptoms of methanol poisoning? o Frontline intervention techniques for health centers and community members o Ethanol as a blocker for methanol poisoning o Dialysis treatment, why it is required, and how it works, as well as other treatment options (e.g., Fomepizole) o The long-term effects of methanol poisoning on the body o The impact of methanol poisoning on communities o Speaking tools and styles, and how to engage an audience with different learning styles o Do what is right, not what is easy 2.3 Needs Analysis Media reports, information from local health departments, hospitals involved in methanol poisoning treatment, and members of the public have steered your attention to a specific region within that country. Do not discredit the information that can be sourced from a taxi driver, waitress, expatriate, business owner, or hotel staff member, as well as your own new staff on the ground Page | 15 within your selected country. Be open-minded and receptive to all information from these and other sources, then assess and filter what will be most appropriate for your particular situation. Putting this together with facts and statistics will be needed to build a clear picture of where to start, making sure there is a definite need for the program as well as a willingness to support and take part. Some communities can become ‘over-assisted’ with many international aid agencies pushing into the same high-risk regions. While the needs analysis is being carried out, focus on: o Hospital departments required to treat methanol poisoning, i.e. ambulance, triage, Accident and Emergency, dialysis unit o How easy was it to initiate and maintain contact with department managers and staff? o How forthcoming are management and staff with assistance and information? Sell the program so they see the need as well as the benefits the hospital will receive. o Is there a person in each department who could be your focal contact point moving forward? o Be clear on your goals, objectives and scope of the program – engage people so they understand the need, objectives and positive outcomes. Alleviate all suspicion of what the program could be. o Choose the right tools to engage the audience depending on education, learning methods – keeping in mind time, manpower, and budget constraints. o Flexibility o Professionalism We have found it advantageous to run a mini training program on our first visit to a region. This gives factual information to officials and community leaders about methanol poisoning, what the program will offer, and how taking part in our program will make their region a better and safer place to be part of. You may need to spend a few days in these communities to build up trust. Hopefully this will make it easier to get the necessary information from which you can develop an appropriate Needs Analysis. 2.4 Starting in the chosen region Make sure you have built trust, communication lines, and realistic expectations before launching any programs in your selected regions. This may take multiple calls and visits to the region. Start slowly with a realistic training timeline. Have you considered? o Ensuring all appropriate local authorities have given you authorization to run the programs o Your educators are confident in the message they are delivering. o The village/region has confirmed dates and locations for the trainings. o Specific learning methods will have the greatest impact on regions, and these might be vastly different than those undertaken in the West (i.e., a mix of interactive question and answer sessions, educational videos, traditional dancers or cultural performers, written materials, social and online media, web sites, etc.). o The safety of staff and attendees Page | 16 2.5 Community Education Plan The Community Education Program (CEP) is made up of three training modules: Module 1 - Health Worker Training (HWT) – educating local health workers about methanol, its uses, methanol poisoning symptoms, diagnosis tools and treatment focusing on front-line intervention, and ethanol for the treatment of methanol poisoning. (The depth of medical information to be provided will vary according to participants’ needs). Allow for and encourage extensive Q&A time. Module 2a – Community & Village Education (CVE) Module 2b – Community Student Education (CSE) Improve socialization using education videos, traditional dancers, interactive discussions, and written material, all providing information about methanol poisoning, symptoms, and how to help a friend. Module 3 - Producer Training – Teaching alcohol producers safe distillation practices, how to monitor the temperature during distillation to reduce the production of methanol, as well as general safe practices and the risks to communities from drinking adulterated alcohol. In many countries, locally produced alcohol (i.e., moonshine, arak, ogogoro, etc.) is used for cultural and religious ceremonies and is often legal and also often the only means of income for local communities. Programs seeking to address methanol poisonings from improperly distilled alcohol should not seek to eradicate this from communities, but rather to help local producers develop safe alcohol based on sound scientific principles and measured preparations that will mitigate the risk of improperly developed spirits entering the community Medical Education Program (MEP): Specific Points and Aids 3. Medical Education Aims: o Improving people’s knowledge and awareness of the dangers of methanol poisoning o Develop skills for the recognition and treatment of methanol poisoning in the emergency departments of hospitals. o Improve the skills of key clinical staff in treating those who have been poisoned and who need hemodialysis and intensive care. o Build a long-term strategy to address prevention and early intervention of unnecessary deaths of local and international people from methanol poisoning. 3.1 Where to start? You should undertake an MOU (Memorandum of Understanding) with a major training hospital. This could be a private or public one. Public hospital staff may be better connected to the community health department of universities, which will aid getting into regional hospitals. Does the hospital have a nursing director who could be the face of your program? Motivated, connected medical staff need to be engaged and chosen to host and run your medical programs. You are going Page | 17 to need doctors and nurses from differing specialties to provide your program with a diverse range of medical information. They could be: o Paramedic o Triage Nurse o Accident and Emergency (A&E) Doctor o Nephrologist o Toxicologist o Nursing Director o Medical Educator They will guide you on your starting points of who you need to meet and what approvals you require before you start. Some ideas are: o Government Department of Health o Government Department of Food and Beverage o Government Department of Education o Government Department of Tourism o Government Department of Law Enforcement You are becoming another training tool for the hospital – you will be sharing their staff, so be mindful of the commitment the program will require away from their normal job. Engage their bosses so that everyone is fully aware of the medical benefits –not only the hospital (great for accreditation ratings) – but for the individual team member gaining further medical training. Your first needs analysis will be on your hosting hospital. Be aware that you may need to provide some basic triage systems, patient assessment on arrival models, etc. Be clear on what you are bringing to the hospital. Meet with the hospital bosses and let them know who you are, find out official ratifications you will require, tell them why you are educating about methanol poisoning and how you need their support and assistance. This should also include understanding their facilities and if they are lacking in certain equipment or items which are critical toward the treatment of methanol poisoning victims. An awareness workshop followed by lunch could be a great way to engage and educate about: o What is methanol poisoning? o How are locals being poisoned? o Who you are o How the program works (train the trainer, locals helping locals) o How educating about methanol poisoning will save lives o Q&A time (what information do they have?) This is the start of you building up your own network, so take the time to talk with as many people as possible. Ask who they think you should be contacting, and what poisonings they have heard of. Ask if you can meet with them individually to follow up from the workshop. Build that relationship. Page | 18 3.2 Train your trainers Your educators need to have a thorough understanding of all information required for them to be effective within the hospitals. Dispelling myths around methanol poisoning deaths can be tricky. Explain the medical facts and then – slowly and softly, reinforce that the medical facts are accepted all around the world and share some best practices happening in other countries facing methanol poisoning challenges. (Some religions believe deaths from methanol poisoning are a punishment for a wrong-doing). Your educators will need thorough knowledge of: o Who are your funding donors/sponsors, and what are their business and CSR activities and goals? o What is methanol and what should it be used for? o The basics about methanol, what it is, how it is used for industrial applications, and why it is improperly added into alcoholic drinks, as well as its effects on the human body (i.e., toxicology). It is important also to understand the distillation process for making alcohol, and why methanol can inadvertently be produced from making alcohol incorrectly. How does the human body process methanol, and what are the symptoms of methanol poisoning? o Frontline intervention techniques for health centers and community members o Diagnostic tools o Ethanol as a blocker for methanol poisoning o Dialysis treatment and other options (e.g., Fomepizole) o The long-term effects of methanol poisoning on the body o Speaking tools and styles, and how to engage an audience with different learning styles o Do what is right, not what is easy 3.3 Needs Analysis Hospitals in developing countries differ widely in terms of resources available and the training and experience of doctors, nurses, and other medical professionals. You will need to look at a wide range of issues, beyond simply asking about methanol poisoning cases being treated. Ask yourself whether you have received politically correct responses or valuable, honest answers. Remember not to criticise or react to processes being different to what you are expecting. You are assessing the baseline from where to begin your work, and again this baseline can vary widely from hospital to hospital and among various local community clinics. Media reports, information from local health departments, hospitals involved in methanol poisoning treatment, and members of the public have steered your attention to a specific region within that country. Do not discredit the information that can be sourced from a taxi driver, waitress, expatriate, business owner, or hotel staff member, as well as your own new staff on the ground Page | 19 within your selected country. Be open-minded and receptive to all information from these and other sources, then assess and filter what will be most appropriate for your particular situation. Putting this together with facts and statistics will be needed to build a clear picture of where to start, what level of treated cases the hospital has seen, and what level the hospital is at across all necessary departments. While the needs analysis is being carried out, focus on: o How easy was it to initiate and maintain contact with village elders, regional government representatives, etc.? o How forthcoming were locals with assistance and information? Sell the program so that they see the need for the methanol education program. o Logistics – is it relatively easy to get in and out of the region? Consider time/cost factors for travel, visas, etc. o Is there a village person or community elder who could be your focal contact point moving forward? o It is important to involve local religious leaders (e.g., imams) who wield considerable influence in local decision-making circles. o Be clear on your goals, objectives and scope of the program – engage people so they understand the need, objectives and positive outcomes. Alleviate all suspicion of what the program could be. o Choose the right tools to engage the audience depending on education and learning methods – keeping in mind time, manpower, and budget constraints. o Flexibility o Professionalism We found it advantageous to run two toxicology workshops first. This enabled us to discuss methanol poisoning within the realms of other poisonings, while enabling us to engage with the doctors and nurses using a broader range of topics. If you do not already have a hospital insider, this process will be slow. Being patient and methodical will ensure that you lay the correct foundation for your program. 3.4 Starting in the chosen hospital Make sure you have built trust, communication lines, and realistic expectations. This may take multiple calls and visits to the hospital. Start slowly with a realistic training timeline. Have you considered? o Have all the necessary authorities given you authorization to run the programs? o Are your educators confident on the message they are delivering? o Has the hospital confirmed dates and assisted with attendees? o Have learning methods been catered for with a mix of interactive Q&A sessions, educational videos, written materials, online & social media tools, and information resources that hospital staff can access anytime as they undertake their work? o Safety of staff and attendees? Page | 20 3.5 Medical Education Plan The Medical Education Program (MEP) is made up of three training modules: Module 1 – Regional Hospital/Clinic Methanol Training (HCT) – What methanol is, what it is used for, methanol poisoning symptoms, diagnosis tools and treatment focusing on front line intervention, ethanol for the treatment of methanol poisoning, dialysis treatment and why it is necessary to compliment ethanol treatment. (Depth of medical information will vary according to participants’ needs). This is based on regional hospitals transferring patients to major hospital for dialysis. Module 2 – Major Hospital training (MHT) – What methanol is, what it is used for, methanol poisoning symptoms, diagnosis tools and treatment focusing on Ambulance, A&E, and Dialysis. Ethanol for the treatment of methanol poisoning, dialysis treatment, and why both dialysis treatment and ethanol or Fomepizole treatment must be used together. (Depth of medical information will vary according to participants’ needs). Module 3 – Doctors working outside of hospitals – Front-line intervention training for General Practitioners, doctors in-residence at hotels, and Community Doctors. Your training may have differing focuses along the way (e.g., workshops for paramedics, A&E training, and dialysis training). Start within your host hospital and expand to neighboring hospitals as time permits. Don’t forget social media as a way to ‘advertise’ the program. Get articles in the paper and on TV, and build close relationships with appropriate media to showcase not only the methanol poisoning issue, but also the training you are providing and how you are working with local teams in a long-term commitment to save lives. Get people talking about methanol poisoning, and identify ways to bring the topic out into the open, especially in markets where it might be culturally or religiously sensitive or prohibited. Page | 21 4. Training Equipment Below you will find various information and examples to assist you in putting together and running your training workshops. Some is specific to either CEP or MEP, some is generic. Knowing your audience prior to your presentation will dictate what you require. Remember, make your training workshops dynamic and engaging. 4.1 Media Coverage of Global Methanol Poisoning Cases Indonesia: Bali (July 2015) 1. 2. India (June 2015) 1. 2. 3. Nigeria (2015) Kenya (2015) Pakistan (2014) 1. Indonesia (2014) 1. 2. Indonesia (2013) 1. Page | 22 2. 3. 4. Czech Republic (2012) Indonesia (July 2011) 1. Canada (2000) 1. Vietnam 1. 2. 3. 4. These are just a few samples of news articles available that have covered the methanol poisoning issue globally. Page | 23 4.2 Medical Reference Materials 4.2.1: First Aid for Suspected Methanol Poisoning Prevention is the most effective measure to counter this significant public health issue. Non-commercially distilled ethanol (without regulated quality and safety controls) poses a potentially lethal risk for anyone who ingests it. The only definitive treatment for methanol poisoning is haemodialysis. Haemodialysis helps to maintain the body’s chemical balance – including substances like potassium, sodium and chloride – and helps to keep a patient’s blood pressure under control. For poisoned patients without immediate access to haemodialysis, or for those awaiting haemodialysis, the following is a stepwise guide to management. 1. When to suspect methanol poisoning? Headache, blurred vision, rapid or deep breathing, drowsiness and/or confusion after drinking illicit alcohol are all warning signs of poisoning. These symptoms can occur 12-24 hours after exposure, and all of these patients require tertiary hospital care. 2. Arrange transfer to a major hospital with dialysis facilities 3. Give the patient ethanol to drink – This blocks toxicity from methanol and can reduce further poisoning. For adults, give 1.8ml of spirits per 1 kg of body weight ( for a 70kg adult administer three 40mL shots) of spirit such as Vodka, Gin, or Whisky, with a maintenance dose of 0.40ml/kg (for a 70kg adult one 40ml shot) per hour. This will stop the poisoning from getting worse while transferring for haemodialysis. In U.S. customary units, this would equate to giving an adult patient an initial 0.06 ounces of spirits for each 2.2 lbs of body weight (i.e., for an adult who weighs 154 lbs, administer three shots of spirits at 1.35 oz. per shot) with a maintenance dose per hour of 0.1 oz./lb (i.e., for a 154 lb adult, administer one 1.35 oz. shot per hour). If the patient is drowsy or unconscious, airway protection with intubation should be performed where possible. If not possible, the patient should be administered oral ethanol in the safest way possible, which would include sitting the patient upright and administering ethanol via a nasogastric tube. This is a time-critical situation for a life threatening emergency, and this management guide should be commenced as soon as possible with any person who has suspected methanol poisoning. Endorsed by: Dr Malcolm Johnston-Leek MBBS FACEM Director St. John Ambulance Service, NT Director Pre hospital National Critical Care and Trauma centre, Darwin, NT, Australia Page | 24 Adapted from guidelines by: Dr Mark Monaghan MBBS FACEM Clinical Toxicologist Fiona Stanley Hospital, Perth, Western Australia 4.2.2: Management of Suspected Methanol Poisoning Methanol Poisoning Treatment Guidelines Summarized from the literature Methanol poisonings occur frequently globally, and in recent years outbreaks have occurred in Cambodia, Czech Republic, Ecuador, Estonia, India, Indonesia, Iran, Kenya, Libya, Nicaragua, Norway, Pakistan, Turkey and Uganda. The size of these outbreaks has ranged from 20 to over 800 victims, with case fatality rates of over 30%.1 Since 1949, the treatment of choice for methanol poisoning has been ethanol,2 which acts to block the metabolism of methanol into formaldehyde and formic acid. Because of concerns about the safety of administering ethanol, a new drug, Fomepizole, was developed.1 However, due to the high cost of use (estimated to be in the order of $4000USD per person),3 ethanol remains the drug of choice in many countries.4,5 Key facts1 Methanol is a widely available chemical with a range of uses including as a solvent, in chemical synthesis and as a fuel. Methanol is not poisonous when not ingested, however, once ingested, it is metabolised to highly toxic compounds, which can cause blindness, coma and metabolic disturbances that can be life-threatening.
6 Victims often only seek medical care after a significant delay, mainly because there is a latent period between ingestion and toxic effects. Late medical care contributes to the high level of morbidity and mortality seen in many methanol poisoning outbreaks.7 Outbreaks of methanol poisoning occur when methanol is added to illicitly- or informally-produced alcoholic drinks. Because patients with methanol poisoning often need intensive medical care, outbreaks of methanol poisoning can rapidly overwhelm medical facilities.
8 Outbreaks have occurred in all regions in recent years. Methanol (also known as methyl alcohol, wood alcohol, wood spirits and carbinol), is a widely available chemical. Methanol has many industrial applications and is also found in a number of household products including varnishes, antifreeze, windscreen wash, and fuel for model aircraft. Globally, approximately 225 million litres of methanol is used each day.9 Outbreaks of methanol poisoning arise from the consumption of adulterated, counterfeit or unregulated production of spirit drinks. Signs and Symptoms The major toxic effects do not manifest until methanol has been metabolized to formic acid and this has accumulated to toxic levels. There is, therefore, a latent period between the consumption Page | 25 of methanol and the onset of symptoms and signs. Co-ingestion of ethanol will delay metabolism and further delay the onset of toxicity for many hours.
1 In the first few hours the patient may become drowsy, unsteady and disinhibited; however, since poisoning often occurs in the context of drinking alcohol this may not be noticed. After a variable period of time, victims start to develop headache, vomiting, abdominal pain and vertigo. They may start to hyperventilate and feel breathless. Vision is often affected, with blindness in severe cases. Coma, convulsions, and death from respiratory arrest may ensue.
5,7 Patients who survive may suffer permanent visual impairment.
10, 11,12 Diagnosis Methanol as an alcohol is rapidly absorbed through gastro‐intestinal tract, so the average absorption half-life is 5 minutes and reaches maximum serum concentration within 30 – 60 minutes and will dissolve well in body water.5 Methanol is not toxic by itself, but its metabolites are toxic.9 Methanol is metabolised in different phases mainly in the liver. The initial enzyme that it is metabolised to is alcohol dehydrogenase (ADH). Clinical Manifestations Clinical manifestations of poisoning with methanol start within 0.5 – 4 hours of ingestion and include nausea, vomiting, abdominal pain, confusion, drowsiness and central nervous system suppression. Patients usually do not seek help at this stage.5 After a latent period of 6 – 24 hours (depending on the dose absorbed), decompensated metabolic acidosis occur which induces blurred vision, photophobia, changes in visual field, accommodation disorder, diplopia, blindness and less commonly nystagmus.5,13,14 Blurred vision with unaltered consciousness is a strongly suspicious of methanol poisoning.5,14 Co‐ingestion of ethanol, can delay symptoms of methanol poisoning for more than 24 hours and sometimes up to 72 hours. Severe metabolic acidosis with anion gap and increased osmolality strongly suggest methanol and or ethylene glycol poisoning. Severity of clinical manifestations and mortality are strongly associated with the severity of central nervous system depression and metabolic acidosis, but not with serum methanol concentration.5 As poisoning progresses the serum concentration of methanol decreases while that of its metabolites, including formate, increases. Metabolic acidosis with a high anion gap (a measure of the difference between positively charged ions and negatively charged ions in plasma) is typical of methanol poisoning. The measurement of formate is a simpler analysis than that of methanol, 13,14 however, this test may not be readily available in developing countries, where diagnosis will need to be made on the patient’s history, presenting symptoms and the presence of severe metabolic acidosis with anion gap. Treatment The main principles of treatment are to prevent further metabolism of methanol, correct metabolic abnormalities and provide other supportive care. Metabolism of methanol can be blocked by the administration of ethanol or fomepizole.1,4, 5,6, 14 FIRST AID15 Give the patient ethanol to drink14,15 Page | 26 This blocks toxicity from methanol and can prevent further poisoning. To adults give 1.8mL/kg (or a 70kg adult three 40mL shots) of spirit such as Vodka, Gin, Whisky, with a maintenance dose of 0.40ml/kg (for a 70kg adult one 40mL shot) per hour. This will stop the poisoning from getting worse while transferring for haemodialysis. If the patient is drowsy or unconscious, airway protection with intubation and hyperventilation should be performed where possible. If not possible, the patient should be administered oral ethanol in the safest way possible, which would include sitting the patient upright and administering ethanol via a nasogastric tube.,514,15 Arrange transfer to a major hospital with dialysis facilities All of these patients require tertiary hospital care. Poisoning can be confirmed with a simple formate assay OR through assessment of clinical history and assessment of the patient’s arterial blood gases to assess the level of metabolic acidosis,5,15 see details below.14 A. Asymptomatic patients, normal blood gas: Observe. B. pH>7.2, HCO3>20: Give bicarbonate. Observe minimum 24 hours C. pH 7.0-7.2, HCO3 10-20: Give bicarbonate, ethanol (or Fomepizole), consider Haemodialysis D. pH<7.2, HCO3<10: Give bicarbonate, Fomepizole (or ethanol), Haemodialysis, folinic acid Folinic acid If available this should be administered orally at a dose of 2mg/kg.14 In Conclusion Currently there are two antidotes used to block ADH metabolism: ethanol, a competitive ADH substrate, and Fomepizole, an ADH inhibitor. Supportive measures may include the correction of acidosis with sodium bicarbonate, intubation and mechanical ventilation and the use of extracorporeal elimination such as haemodialysis.
4,5,14,15 Despite concerns about the safety and use of oral ethanol, very few adverse effects have been reported in the literature4,5,16. When compared to Fomepizole, the number of adverse events is the same for both treatments.4,16 Historically, ethanol has been used as an antidote and is still standard therapy in some centres, due to its low cost and physician familiarity.4 However, for ethanol to be an effective antidote, most experts feel the serum level must be carefully titrated and maintained between 22 and 33 umol/L.4 Fomepizole, on the other hand, is administered as a weight-based fixed dose at regular intervals, without the need for monitoring of serum levels4, however, the cost of the drug may make its use impractical.3 The consequences of not treating people who may have been poisoned through the ingestion of methanol are so serious, that the risk of not treating is greater than the risk of treating patients with oral ethanol. The mortality from poisoning reported in the literature is over 30% with many others suffering severe and permanent brain damage and blindness. While follow-up data is scarce, a study which followed up a group of patients from Estonia16 found that, damage to patients seen at the time of initial diagnosis was still present six years later. Further, apparently new neurological and visual complications were also identified in Page | 27 approximately one third of the patients. Some 35% of the patients initially discharged with sequelae and 29% discharged without sequelae were dead six years later. Because of the severe consequences of not treating suspected methanol poisoning it seems that treatment with a readily available and relatively cheap antidote such as ethanol is a socially responsible, evidence-based and practical treatment. Many of the victims of methanol poisoning are young men who could otherwise have led productive and useful lives. Prepared by Prof Di Brown, AO, RN, BAppSci, GDipHEd, PhD, MACN Professorial Fellow, Charles Darwin University References 1. World Health Organisation, Methanol Poisoning Outbreaks, Information Note, July 2014. Geneva: WHO. 2. Agner, K. Hook, O, Von Porat B. The treatment of methanol poisoning with ethanol with report of two cases. QJ Stud Alcohol, 1949 Mar; 9(4): 515-22. 3. Deon P. Druteika, B.S., Peter J. Zed, Pharm.D., Mary H. H. Ensom Role of Fomepizole in the Management of Ethylene Glycol Toxicity, Pharmacotherapy. 2002;22(3) 4. Beatty, L. Green, R. Magee,K, Ze P. Review Article A Systematic Review of Ethanol and Fomepizole Use in Toxic Alcohol Ingestions. Emergency Medicine International Volume 2013, Article ID 638057, 14 pages. Hindawi Publishing Corporation 5. Ministry of Health and Medical Education, Office Bureau of Psychosocial Health and Addiction Deputy for Health, Clinical Guideline for Treatment of Methanol Poisoning, Substance Abuse Prevention and Treatment, Iran: Teheran, 2009 %20%D8%A7%D8%B9%D8%AA%DB%8C%D8%A7%D8%AF/Executive%20Summary% 20of%20Methanol%20Poisoning_03.pdf 6. Barceloux, DG et al . Methanol Guidelines - AACT/EAPCCT American Academy of Clinical Toxicology Practice Guidelines on the Treatment of Methanol Poisoning. Journal of Toxicology - Clinical Toxicology. 2002; 40 (4): 415-446. 7. Hovda, KE et al. Methanol outbreak in Norway 2002-2004: epidemiology, clinical features and prognostic signs. Journal of Internal Medicine. 2005; 258:181-190. 8. Paasma, R et al. Methanol mass poisoning in Estonia: Outbreak in 154 patients. Clinical Toxicology. 2007; 45: 152-157. 9. Zakharov, S. Pelclova,D. Urban, P et al. Czech mass methanol outbreak 2012: Epidemiology, challenges and clinical features Clinical Toxicology, Informa Healthcare USA, Inc. 2014: 1556-3650 10. Methanol Institute. Methanol Basics. accessed 27 June 2015 11. Giovanetti, F. Methanol poisoning among travellers to Indonesia. Travel Medicine and Infectious Disease. 2013; 11(3):190-193. 12. Gee P, Martin E. Toxic cocktail: methanol poisoning in a tourist to Indonesia. Emergency Medicine Australasia. 2012;24: 451-453. accessed 27 June 2015 13. Hovda, KE et al. Anion and osmolal gaps in the diagnosis of methanol poisoning: clinical study in 28 patients. Intensive Care Medicine. 2004; 30:1842-46. Page | 28 14. Hovda, KE et al. Methanol Poisoning at a Glance. Guidelines Published for Medicins San Frontiers (nd). 15. Monaghan, M. Director of Emergency and Clinical Toxicologist. Management of Suspected Methanol Poisoning, Supporting health workers and non-tertiary medical clinics. Guidelines for first aid treatment, May 2015. Perth: Emergency Department, Fiona Stanley Hospital. 16. Paasma, R et al. Methanol poisoning and long term sequelae - a six years follow-up after a large methanol outbreak. BMC Clinical Pharmacology. 2009; 9: 5. S. Fomepizole was added to the WHO Essential Medicines List in 2013. Page | 29 4.4: Examples of Information Flyers o Banners o Note pads o Public education flyers o Public education postcards o Public education warnings (business cards) Go to www.methanol.org and www.Liamcharity.com to download these examples, and/or contact The LIAM Charitable Fund monty.rock@bigpond.com) or Methanol Institute (MI – sg@methanol.org) for a copy of these resources. 4.5: Training Videos Training Video #1 – Community Village education video example Training Video #2 – Medical Education Video example Go to www.methanol.org and www.Liamcharity.com to download these examples, and/or contact The LIAM Charitable Fund monty.rock@bigpond.com) or Methanol Institute (MI – sg@methanol.org) for a copy of these resources. Below is a link to a television documentary that was aired in Australia the month after Liam died. This documentary may give you an insight into the struggles families dealing with methanol poisoning may encounter within themselves, as well as with in-country authorities. “Poison in Paradise:” Page | 30 4.6 Community Education Presentation (CEP) Examples Module 1 Health worker training (PowerPoint example) Module 2a Community and Village education (PowerPoint example) Module 2b Community Student Education (PowerPoint example) Module 3 Producer training (PowerPoint example) Go to www.methanol.org and www.Liamcharity.com to download these examples, and/or contact The LIAM Charitable Fund monty.rock@bigpond.com) or Methanol Institute (MI – sg@methanol.org) for a copy of these resources. 4.7 Medical Education Presentation (MEP) Examples Module 1 Regional hospital/clinic (PowerPoint example) Module 2 Major general hospitals (PowerPoint example) Module 3 Doctors working outside of hospitals (PowerPoint example) Go to www.methanol.org and www.Liamcharity.com to download these examples, and/or contact The LIAM Charitable Fund monty.rock@bigpond.com) or Methanol Institute (MI – sg@methanol.org) for a copy of these resources. Page | 31 5. Timeline We have outlined below a list of targets you will be wanting to reach in five years. Keeping in mind you will have times where things come together well for you but you will also come up against resistance to change, uncertainty in regard to your intentions, as well as authorities not wanting to be the ‘one’ who makes the final decision. This is intended as a guideline only. If you are able to move ahead more quickly, then please do! It is very important from day one that you pay extra attention to building honest and strong working relationships with authorities, and set the culture and ethos that your team will follow, reminding them regularly that what you do today paves the way for future plans. Five-year timeline New Country/Region pre-planning -Understand legal requirements. Lay your Foundations -Obtain work permits, visas. 0-3 months -Study political, religious and regional climates. -Reference section 1-1.12 training module. MOU with major training hospital 3-6 months (MEP) -Use existing contacts with foreign aid groups to sell the need for the training programs. Host general poisoning treatment workshops focusing on methanol poisoning to build relationships and contacts. -Reference section 3-3.3. Build hospital competencies 6-12 months (ongoing) -Develop a Needs Analysis. -Conduct specialty-specific training e.g., A&E processes, medical staff. Provide training including ICU and Dialysis. -Reference section 3.2-3.5. Page | 32 Five-year timeline (continued) Other Hospital, clinic, Specialist training 8 months (ongoing) -Expand the training to other hospitals, clinic nurses, doctors, specialists. Provide front line medical training to the Police, increasing both awareness of methanol poisoning cases and front line medical knowledge. Invite medical staff from the region to workshops, not just your hospital. Workshops can work for 50-120 people. -Arrange medical staff exchanges. -Reference 3.2-3.5. Engage University 10 months (CEP) -Use relationships built from hospital programs to engage community health faculties. -Reference 1-1.12. -Reference 2-2.2. MOU with University 12-16 months -Develop a Needs Analysis. -Reference 2.1-2.5. Community Education Training 16 months (ongoing) -Once you start your first community training, word will spread. Take note of other areas having methanol poisoning issues. -Conduct health clinic training first (use team members from the Hospital program to assist with medical front line education. -Begin community education Training. -Begin producer training. Page | 33 Five-year timeline (continued) MEP 24-36 months -Expand your program to another region, taking into consideration logistics expenses, cultural differences, etc. You may wish to train a whole new team or use your medical education team from your original hospital. CEP 28-52 months -Expand your program to more regions. Concentrate on areas with high methanol poisonings recorded, as well as known producer areas. MEP 36-48 months -You should be seeing the benefit of your train the trainer model. Provide refresher courses inviting medical staff from your past clinics and hospitals. -Assess if your sustainability focus will be effective. If not, alter training to ensure sustainability once you finish. -Arrange staff exchanges to your two hospitals from regional hospitals. MEP 40-60 months -The last 16 months will see your program evolve to a more pro-active focus while winding back your medical training. You are wanting to ensure that future medical staff have correct methanol poisoning information while they are students. -Guest speaking at medical conferences gives you wider audience. -Is methanol poisoning medical treatment part of curriculum at medical universities? If not, engage the education department to make changes. -Employ a lecturer to provide methanol poisoning treatment presentations to doctors and nurses. -Maximise your social media outlets to keep people thinking about methanol poisoning risks. Page | 34 Five-year timeline (continued) CEP 52-60 months -The last year will see you shift your focus to ensure sustainability. Refresher courses are needed for community clinics and producers to reinforce the methanol poisoning information. -Ensure that methanol poisoning treatment and knowledge is part of the education department curriculum for community health faculties as well as education faculties. -Engage a lecturer to provide methanol poisoning education presentations to community health students, hospitality students, tourism students, and teacher students. -Maximise your social media outlets. Push messages on mobile phones prior to festive events. As stated, this timeline is to be used as a guide only. Model your training with sustainability as the foundation. You may have enough budget and manpower to engage more hospitals and regions, so do not limit yourself. Ensure you get media coverage of as many training workshops as possible, this doubles as education and advertising. If you hear of methanol poisoning cases ensure you maximise the media coverage to raise awareness. Cultural sensitivity, clear vision and pushing for what is right are what you carry forward! |
12809 | https://fiveable.me/key-terms/intermediate-algebra/ordinate | Ordinate - (Intermediate Algebra) - Vocab, Definition, Explanations | Fiveable | Fiveable
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key term - Ordinate
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Definition
The ordinate is the vertical coordinate of a point on a two-dimensional coordinate system. It represents the position of a point along the y-axis, which is the vertical axis perpendicular to the horizontal x-axis.
5 Must Know Facts For Your Next Test
The ordinate is the vertical coordinate of a point on a graph, representing its position along the $y$-axis.
The ordinate, along with the abscissa (horizontal coordinate), uniquely identifies the location of a point in a Cartesian coordinate system.
In the slope-intercept form of a linear equation, $y = mx + b$, the ordinate of the $y$-intercept is represented by the value of $b$.
The ordinate is essential for graphing linear equations, as it determines the vertical position of the points that make up the line.
Analyzing the ordinates of points on a graph can provide information about the behavior and characteristics of the linear equation, such as its rate of change (slope) and $y$-intercept.
Review Questions
Explain the role of the ordinate in the Cartesian coordinate system.
The ordinate is the vertical coordinate of a point in a Cartesian coordinate system. It represents the position of a point along the $y$-axis, which is perpendicular to the horizontal $x$-axis. Together, the ordinate and abscissa (horizontal coordinate) uniquely identify the location of a point in the two-dimensional plane. The ordinate is essential for graphing linear equations, as it determines the vertical position of the points that make up the line.
Describe how the ordinate is used in the slope-intercept form of a linear equation.
In the slope-intercept form of a linear equation, $y = mx + b$, the ordinate of the $y$-intercept is represented by the value of $b$. The $y$-intercept is the point where the line crosses the $y$-axis, and its ordinate is the $b$ value in the equation. Analyzing the ordinate of the $y$-intercept can provide information about the behavior and characteristics of the linear equation, such as its rate of change (slope) and the vertical position where the line intersects the $y$-axis.
Evaluate the importance of the ordinate in graphing and analyzing linear equations.
The ordinate is crucial for graphing and analyzing linear equations. The ordinate, along with the abscissa, determines the location of points on the graph, which is essential for visualizing the behavior of the linear equation. By analyzing the ordinates of points on the graph, you can gain insights into the characteristics of the linear equation, such as its rate of change (slope) and the vertical position of the $y$-intercept. Understanding the role of the ordinate is vital for interpreting the properties of linear equations and their graphical representations.
Related terms
Abscissa:
The abscissa is the horizontal coordinate of a point on a two-dimensional coordinate system. It represents the position of a point along the x-axis.
Cartesian Coordinate System:A Cartesian coordinate system is a two-dimensional plane defined by perpendicular x and y axes, used to locate points in space by their coordinates.
Slope-Intercept Form:
The slope-intercept form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept, representing the ordinate of the point where the line crosses the $y$-axis.
"Ordinate" also found in:
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Elementary Algebra
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12810 | https://www.emra.org/emresident/article/methemoglobinemia | Methemoglobinemia: A Case of the Blues EMRA
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Methemoglobinemia: A Case of the Blues
8/12/2024Phillip A. Barnett, DO, Devan K. Barnett, Douglas R. Bethea
Methemoglobinemia is a rare but serious — yet treatable — hematological condition that can lead to death if not identified in a timely manner.
In methemoglobinemia, the iron in hemoglobin becomes oxidized. In a normal erythrocyte, the ferrous (Fe 2+) state allows oxygen to easily offload to oxygenate tissues. However, in methemoglobinemia, the iron on the hemoglobin is oxidized and becomes the ferric (Fe 3+) state.
When this happens, the oxygen does not get picked up by the hemoglobin and, therefore, does not oxygenate the tissue. Furthermore, if only a portion of the red blood cell contains the ferric state and the other contains the ferrous state, the ferrous state is still unable to pick up the oxygen as it causes a left shift in the hemoglobin curve (Figure 1). This causes decreased oxygen delivery to the tissues.1
Figure 1: Example of left shift resulting in hemoglobin not being able to pick up oxygen in methemoglobinemia.
Methemoglobinemia can be acquired or hereditary.2 Topical benzocaine is a known cause of acquired methemoglobinemia and is applied for transesophageal echocardiograms, among other procedures and uses. Cyanosis, a low pulse oximetry reading, and a high partial pressure of arterial oxygen on arterial blood gas (ABG) should aid in diagnosis.
Methemoglobinemia is a serious condition and can lead to death. Although benzocaine is known to cause methemoglobinemia, it is still rare.3
In a large case series study, methemoglobinemia occurred in 1 out of 1,499 transesophageal echocardiograms (TEE). In another study — a random sample of 190 patients with the same sedation, sex, body mass index, and left ventricular systolic function — patients who developed methemoglobinemia were found to be anemic, hospitalized, or with acute systemic infection.3
An ABG is crucial in diagnosing methemoglobinemia and should be the first test. Most ABGs can detect methemoglobinemia by its absorbance spectrum of 631 nanometers (expressed as a percentage).
Partial pressure of arterial oxygen (pO 2) is an indicator of how much oxygen is being delivered to the tissue and how much dissolved oxygen is in the blood. Methemoglobinemia causes a falsely elevated pO 2.4
Pulse oximetry cannot determine how hypoxic a patient truly is in methemoglobinemia. The wavelength of the pulse oximetry can absorb the light of methemoglobinemia and is unable to determine if red blood cells (RBC) are saturated or not.5 An oxygen saturation of 85% is caused by a high percentage of methemoglobinemia.6
A retrospective series, performed in 2 teaching hospitals involving 138 patients with acquired methemoglobinemia, revealed the majority were caused by dapsone. However, 5 of the participants were, in fact, associated with 20% benzocaine spray with a mean methemoglobinemia level of 43.8%. Furthermore, it was shown that 94% of all the subjects in this study were anemic.7
Case ReportThe patient in our case was a 76-year-old Caucasian woman with multiple comorbidities, including a history of type 1 diabetes and a lumbar laminectomy 2 months prior. She presented to the ED with recurrent surgical site pain and a fever of 101 degrees Fahrenheit. Her previous laminectomy was complicated by methicillin-resistant staphylococcus aureus (MRSA), necessitating incision and drainage and treatment with intravenous daptomycin for 6 weeks. Blood cultures were obtained, and the patient was discharged due to negative workup in the ED. Two days later, the blood cultures were found to be positive for MRSA, and she was called to return to the hospital.
The patient was admitted that day for bacteremia and diabetic ketoacidosis (DKA). DKA was effectively treated with intravenous (IV) insulin and IV fluids, and it resolved 2 days later. Because of the positive MRSA blood cultures, the patient was started on daptomycin twice a day. Infectious disease and orthopedic specialties were consulted to assist in management. An MRI was performed upon the orthopedist’s recommendation. The MRI revealed a deep lumbar fluid collection.
The patient was taken to the operating room 2 days after admission for excisional debridement of lumbar wound. Infectious disease recommended 6 additional weeks of IV daptomycin, followed by an indefinite course for suppression if cultures were sensitive to daptomycin.
A transthoracic echocardiogram was ordered 1 day before the orthopedic surgery and exhibited findings concerning for a small vegetation that was mobile on the posterior leaflet of the mitral valve. As a result, cardiology was consulted. Cardiology recommended a TEE. The patient was taken to the catheterization lab 4 days later for a TEE. There, she received benzocaine spray before the procedure.
After the TEE, she became very cyanotic and was having difficulty breathing, with an O 2 saturation of 85-88%. She was placed on bilevel positive airway pressure (BIPAP) ventilation with FiO 2 of 100%. At that time, an ABG was drawn, and the blood appeared dark. The ABG showed pH of 7.48, pCO 2 of 44.9 mmHg, pO 2 of 574 mmHg, and HCO 3 of 33.2 nmol/L.
A computed tomography pulmonary angiogram was performed to rule out pulmonary embolism; it was negative. Despite BIPAP of FiO 2 of 100%, the patient remained diffusely cyanotic. She was transferred to the ICU, and another ABG — testing specifically for methemoglobin levels — was completed. Her methemoglobin was found to be 42.3% (Figure 2). She received 1 dose, 100 mg (1.5 mg/kg), of methylene blue.
Figure 2: Example of patient’s arterial blood gas.
Another ABG was done 1 hour later. Methemoglobin level was 0.5%, showing improvement. She was taken off BIPAP after the methylene blue infusion, as her hypoxia resolved. She was tested for glucose 6 phosphate dehydrogenase (G6PD) deficiency, which resulted as 262, the normal value being 127-427 U/10E12 red blood cells. TEE showed no endocarditis. The MRSA was found to be sensitive to daptomycin, and this regimen was continued.
The patient remained clinically stable and was moved out of the ICU the following day. Three days later, a peripherally inserted central catheter (PICC) line was placed, and she was discharged home where she would continue to receive daily antibiotic infusions.
DiscussionDetermining the cause of methemoglobinemia is important because the cause determines treatment. In fact, first-line therapy is contraindicated in some types of inherited forms.8
In acquired methemoglobinemia, it is important to diagnose promptly and begin treatment immediately to resolve hypoxia. Medications, whether used appropriately or in settings of overdoses, are the main contributors to acquired methemoglobinemia.9
In our case, topical benzocaine was the culprit. However, how benzocaine causes methemoglobinemia isn’t well-known.2
The studies described earlier were similar to our presented case. Oxygen saturation of 85-88%, abnormally high pO 2 of 574 mmHg, BIPAP with FiO 2 of 100% that did not resolve hypoxia, and dark-appearing blood were all characteristic signs of methemoglobinemia, as seen in the aforementioned studies. Our patient’s methemoglobinemia was 42.3%, similar to the mean values in previous studies, specifically acquired methemoglobinemia after TEE.7 Furthermore, she had bacteremia, which further increased her likelihood of acquiring methemoglobinemia.3 Also, 1 dose of methylene blue caused her to revert to her previous state, further confirming an acquired etiology.
If methemoglobin level is 20-45%, dizziness, dyspnea, tachycardia, coughing, weakness, and headache can occur. If the level is greater than 45%, decreased levels of consciousness and fatigue can occur. When above 55%, cardiac arrhythmias or failure, acidosis, seizure, and coma can occur. If 70% is reached, death can occur.
First-line treatment (if levels are above 30%) is 1-2 mg/kg of IV methylene blue.10 Methylene blue reduces the oxidized state of hemoglobin back to the ferrous state (Fe 2+), allowing the hemoglobin to readily pick up oxygen again, resulting in offloading to tissues.
Vitamin C works in the same manner and can be used to treat if methylene blue is contraindicated. Some contraindications for methylene blue are G6PD deficiency because it can cause hemolytic anemia, and pregnancy (category X) because it can lead to intestinal atresia and fetal death, especially if given in the second trimester.11 Furthermore, G6PD deficiency should be considered if the patient does not improve after the first dose of methylene blue. An exchange transfusion can be considered if methylene blue does not resolve methemoglobinemia.10
Because acquired methemoglobinemia is a serious condition caused by an array of medications,9 emergency physicians as well as other care providers should know how methemoglobinemia presents, and all should be familiar with the treatment of choice.
ConclusionMethemoglobinemia — a serious but treatable condition that presents with generalized cyanosis, an oxygen saturation percentage of mid- to high 80s, high pO 2, dark-colored blood, and hypoxia resistant to 100% oxygen — can result from topical benzocaine. It is important to consider methemoglobinemia and perform an ABG if a patient presents similarly. If such signs are present along with a methemoglobin level of 30% or greater, and methemoglobinemia is determined to be acquired, then methylene blue is the treatment of choice.
Take-Home Points
Consider methemoglobinemia if oxygen saturation level is mid- to high 80s, cyanosis is not improving with 100% FiO 2, and dark-colored blood is present.
Treat with methylene blue when methemoglobin is 30% or above and methemoglobinemia is determined to be acquired.
Methylene blue is contraindicated in pregnancy and G6PD deficiency. Vitamin C and exchange transfusions are alternative treatments.
References
Falkenhahn M, Kannan S, O'Kane M. Unexplained acute severe methaemoglobinaemia in a young adult. Br J Anaesth. 2001; 86(2):278-280. doi: 10.1093/bja/86.2.278.
Cortazzo JA, Lichtman AD. Methemoglobinemia: A review and recommendations for management. J Cardiothoracic Vascular Anesthesia. 2014; doi: 10.1053/j.jvca.2013.02.005.
Kane GC, Hoehn SM, Behrenbeck TR, et al. Benzocaine-induced methemoglobinemia based on the Mayo Clinic experience from 28,478 transesophageal echocardiograms: Incidence, outcomes, and predisposing factors. Arch Intern Med. 2007 Oct 8;167(18):1977-82. doi: 10.1001/archinte.167.18.1977.
Cefalu JN, Joshi TV, Spalitta MJ, et al. Methemoglobinemia in the Operating Room and Intensive Care Unit: Early recognition, pathophysiology, and management. advanced therapy. 2020;37(5):1714-1723. doi: 10.1007/s12325-020-01282-5.
Mack E. Focus on diagnosis: co-oximetry. Pediatric Review. 2007;28(2):73-74. doi: 10.1542/pir.28-2-73.
Barker SJ, Tremper KK, Hyatt J. Effects of methemoglobinemia on pulse oximetry and mixed venous oximetry. Anesthesiology. 1989;70(1):112-7. doi: 10.1097/00000542-198901000-00021.
Ash-Bernal R, Wise R, Wright SM. Acquired methemoglobinemia: A retrospective series of 138 cases at 2 teaching hospitals. Medicine (Baltimore). 2004;83(5):265-273. doi: 10.1097/01.md.0000141096.00377.3f.
Rianprakaisang T, Blumenberg A, Hendrickson RG. Methemoglobinemia associated with massive acetaminophen ingestion: A case series. Clin Toxicol (Phila). 2020; 58(6):495-497. doi: 10.1080/15563650.2019.1657883.
Hartshorn EA, Rodriguez LF, Smolik LM, et al. Benzocaine-Induced Methemoglobinemia: Report of a Severe Reaction and Review of the Literature. Annals of Pharmacotherapy. 1994;28(5):643-649. doi:10.1177/106002809402800515
David RS, Sawal NS, Hamzah MN, et al. The Blood Blues: A review on methemoglobinemia.Journal of Pharmacology and Pharmacotherapeutics. 2018;9(1):1-5. doi: 4103/jpp.JPP_79_17
Bistas E, Sanghavi DK. Methylene Blue. In:StatPearls. Treasure Island (FL): StatPearls Publishing; June 26, 2023. Available from:
Kumar, R. Oxyhemoglobin Dissociation Curve . MD Published April 16, 2017. Accessed May 2024.
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12811 | https://mathoverflow.net/questions/152166/varieties-invariant-under-affine-transformations | Skip to main content
Varieties invariant under affine transformations
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This question stems from this question on Mathematics stack exchange. The top answer there provides a geometric solution; I'm curious though about an algebro-geometric solutions.
Let me rephrase that question a bit: Let V be a variety in affine n-space Ank, over field k, given by an ideal I. How one would derive from I the subgroup G of affine transformations that leave V invariant?
In particular, if V is a hypersurface and I is a principal ideal (f) what properties of f determine G?
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Probably, the easiest method is this (at least in characteristic zero, which I will assume henceforth): Suppose that V⊂Ank is the set of zeros of a polynomial ideal I⊂k[x1,…,xn], say, generated by some finite set {f1,…,fm}⊂k[x1,…,xn]. Let a be the (finite dimensional) Lie algebra of affine vector fields on Ank (i.e., the set of derivations of k[x1,…,xn] that preserve the subspace of polynomials of degree at most 1). Consider the linear map Φ:a→km⊗k[x1,…,xn]/I defined as
Φ(X)=([X(f1)]I,…,[X(fm)]I).
The kernel of Φ, say gI⊂a, is the space of affine symmetry vector fields of the ideal I, and it is the Lie algebra of the connected (in the appropriate sense) subgroup G0I of the set of affine transformations of Ank that preserve I, i.e., it is the irreducible component containing the identity of the algebraic group GI that consists of all affine symmetries of I. In particular, G0I is an algebraic subgroup of the group of affine transformations of Ank, and gI is its (Zariski) tangent space at the identity. (Computing the full symmetry group GI is not a linear problem and is much harder in general, though one does know that G0I is a normal subgroup of GI and that the quotient group GI/G0I is finite.)
In practice, if I is complicated, computing [X(p)]I in an efficient way requires using Gröbner bases or some such tool. In all cases, though, the computation of gI reduces to the computation of the kernel of a linear map between finite-dimensional vector spaces.
When I is generated by a single polynomial f, one is just asking whether the 'remainder' of X(f) 'divided by' f is zero. (Here, I am intending the use of a so-called 'multivariate division algorithm' à la Buchberger's algorithm using a Gröbner basis, though a more naïve degree approach will work, too.) This is fairly easy to compute explicitly using a Gröbner basis, one, say, that uses some total order on the monomials in k[x1,…,xn], so the 'continuous' symmetries of a polynomial are not hard to compute using just about any symbolic algebra program.
For example, given f of degree d>0, you can always choose coordinates y1,…,yn so that
f=(yn)d+c1(y1,…,yn−1)(yn)d−1+⋯+cd(y1,…,yn−1)
where each ci is a polynomial of degree at most i. Clearly, there is a unique linear map q:a→k such that
X(f)=q(X)f+R(X,f)
where R(X,f) has degree at most d in the yi and at most degree d−1 in yn.
Now, X(f) is a multiple of f if and only if R(X,f)=0, and, for a given f, this is a finite number of linear equations for X.
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edited Dec 18, 2013 at 17:16
answered Dec 18, 2013 at 11:28
Robert BryantRobert Bryant
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What do you mean at the end by saying "k[x1,…,xn] is a Euclidean ring"? The ring of polynomials in more than one variable over a field is not a Euclidean domain (i.e., no division algorithm) since it is not a PID. Maybe you just had in mind that the coefficient ring k is a field (a trivial type of Euclidean domain).
– KConrad
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@KConrad: You are right; I'm not using 'Euclidean' in its standard sense, which is not a good idea. I just meant that there is an effective (multivariate) division algorithm for this ring, using, say, a total monomial ordering, à la Buchberger's algorithm using Gröbner bases. I'll edit my answer to reflect this.
– Robert Bryant
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12812 | https://math.libretexts.org/Bookshelves/Calculus/CLP-4_Vector_Calculus_(Feldman_Rechnitzer_and_Yeager)/03%3A_Surface_Integrals/3.03%3A_Surface_Integrals | Skip to main content
3.3: Surface Integrals
Last updated
: Mar 2, 2022
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3.2: Tangent Planes
3.4: Interpretation of Flux Integrals
Page ID
: 91906
Joel Feldman, Andrew Rechnitzer and Elyse Yeager
University of British Columbia
( \newcommand{\kernel}{\mathrm{null}\,})
We are now going to define two types of integrals over surfaces.
Integrals that look like are used to compute the area and, when is, for example, a mass density, the mass of the
Integrals that look like with being a
unit vector
that is perpendicular to at are called
flux
integrals. We shall see in §3.4, that when is the velocity field of a moving fluid and is the density of the fluid, then is the rate at which fluid is crossing the
surface
Parametrized Surfaces
Suppose that we wish to integrate over part, of a
surface
that is parametrized by We start by cutting up into small pieces by drawing a bunch of curves of constant (the blue curves in the figure below) and a bunch of curves of constant (the red curves in the figure below).
Concentrate on any one the small pieces. Here is a greatly magnified sketch.
For example, the lower red curve was constructed by holding fixed at the value varying and sketching and the upper red curve was constructed by holding fixed at the slightly larger value varying and sketching So the four intersection points in the figure are
Now if
(where and are any small constants) then, by Taylor expansion,
Applying this three times, once with once with and once with
We have dropped all Taylor expansion terms that are of
degree
two or higher in The reason is that, in defining the integral, we take the
limit
Because of that
limit
, all of the dropped terms contribute exactly to the integral. We shall not prove this. But we shall show, in the optional §3.3.5, why this is the case.
The small piece of our
surface
with corners is approximately a parallelogram with sides
Denote by the angle between the vectors and The
base
of the parallelogram, has length and the height of the parallelogram is So the area of the parallelogram is1
Furthermore, and are
tangent
vectors to the curves and respectively. Both of these curves lie in So and are
tangent
vectors to at and the
cross product
is perpendicular to at We have found both and where is a unit
normal vector
to the
surface
.
Equation 3.3.1
For the parametrized
surface
The sign in 3.3.1 is there because there are two unit normal vectors at each point of a
surface
, one on each side of the
surface
. Typically, the application itself tells you which of the two normal vectors should be used. We shall see many examples shortly.
Graphs
The
surface
which is the graph can be parametrized by
As
we have
So, 3.3.1 gives the following.
Equation 3.3.2
For the
surface
Similarly, for the
surface
and for the
surface
Again, in any given application, some care must be taken in choosing the sign in 3.3.2, so as to get the appropriate
normal vector
.
The formulae like in 3.3.2 have geometric interpretations. The red parallelogram in the sketch
represents a little piece of our
surface
. It has area The blue parallelogram in the same sketch represents the projection of the red parallelogram onto the -plane. It has area The
vector
in the sketch is a unit normal for the red parallelogram. We have seen that it is parallel to
so that the angle between and obeys
The geometric interpretation of is that the area of a little piece of
surface
is the area of its projection on the -plane times the factor where is the angle between (which is perpendicular to the
surface
) and (which is perpendicular to the -plane). Notice that
when is close to zero, which corresponds the being almost constant and our
surface
being almost parallel to the -plane, reduces to almost
On the other hand, in the
limit
which corresponds to and/or becoming infinite and our
surface
becoming perpendicular to the -plane, becomes “infinity times” In this case, we should represent our
surface
either in the form or in the form rather than in the form
Surfaces Given by Implicit Equations
Finally suppose that the
surface
is given by the equation with a constant. Suppose further that at some point on the
surface
Then near that point we may solve2 the equation for as a
function
of and That is, the
surface
also obeys for a
function
that satisfies
near the point. Differentiating this with respect to and gives, by the
chain rule
,
which implies
and
So, by (3.3.1),
Equation 3.3.3
For the
surface
when
Similarly, for the
surface
when
and for the
surface
when
If, for some point we have we also have problem! Often this is a sign that our
surface
is not
smooth
at and in fact does not have a
normal vector
there. For an example of this, see Example 3.2.2.
Examples of
We'll start by computing, in several different ways, the
surface area
of the hemisphere
(with ). You probably know, from high school, that the answer is But you have probably not seen a derivation of this answer. Note that, since on the hemisphere, the set of 's for which there is a with on the hemisphere is exactly
Example 3.3.4. Area of a hemisphere — using cylindrical coordinates
Let's parametrize the hemisphere using as parameters the polar coordinates of the cylindrical coordinates3
and then apply 3.3.1. In cylindrical coordinates the equation becomes and the condition is
So the hemisphere can be parametrized by
Note that we selected the positive solution of in order to satisfy the condition that Since
3.3.1 yields
So the area of the hemisphere is
as it should be.
Example 3.3.5. Area of a hemisphere — using an implicit equation
This time we'll compute the area of the hemisphere by using that, if is on the hemisphere, then with Since
3.3.3 yields
So the area is To evaluate this integral, we switch to polar coordinates, substituting This gives
We already showed, in Example 3.3.4, that the value of this integral is
Example 3.3.6. Area of a hemisphere — using spherical coordinates
Of course “integrating over a
sphere
” cries out for spherical coordinates. So this time we parametrize the hemisphere using as parameters the angular coordinates of the spherical coordinates
and then apply 3.3.1. In spherical coordinates the equation becomes just and the condition is So the hemisphere can be parametrized4 by
Since
3.3.1 yields
and
So the area of the hemisphere is
There is an easier way to do this, using a little geometry.
Example 3.3.7. Area of a hemisphere — using spherical coordinates again
We are now going to again compute the
surface area
of the hemisphere using spherical coordinates. But this time instead of determining using the canned formula 3.3.1, we are going to read it off of a sketch.
Sketch the part of the hemisphere that is in the first octant, Slice it up into small pieces by drawing in curves of constant (the blue lines in the figure below) and curves of constant (the red lines in the figure below).
Each piece is approximately a little rectangle. Concentrate on one of them, like the piece with the thick sides in the figure above. The area, of that piece is (essentially) the product of its height and its width. Each of the two sides of the piece is
a segment of a circle of radius (a fat blue line in both the figure above and in the figure on the left below)
that subtends an angle
and hence is the fraction of a full circle of radius and hence is of length
The top of the piece is
a segment of a circle of radius (a fat red line in both the figure above and in the figure on the right below)
that subtends an angle
and hence is the fraction of a full circle of radius and hence is of length
These are drawn in the figure below.
So the area of our piece is
This is exactly the same formula that we found for in Example 3.3.6 so that we will, yet again, get that the area of a hemisphere of radius is (Phew!)
But wait! We can do it again, by yet another method!
Example 3.3.8. Area of a hemisphere — using
We'll compute the area of the hemisphere one last time5. This time we'll use that the equation of the hemisphere is
So 3.3.2 yields
So the area is We already found, in Example 3.3.5, that the value of this integral in
Let's do some more substantial examples, where the
integrand
is not 1.
Example 3.3.9
Evaluate where is the part of the cone with
Solution 1
: We can express as
Now since
3.3.2 gives6
Our integral is then
Since we are integrating over a circular
domain
, let's convert to polar coordinates.
Remembering7 that the integral of or over a full period is we end up with
Solution 2
: We may parametrize8 the cone by
Then because
3.3.1 yields9
So our integral becomes
We evaluated this integral in Solution 1. So again
Let's do something more celestial.
Example 3.3.10
Consider a spherical shell of radius with mass density per unit area. Think of it as a hollow planet10. We are going to determine the gravitational force that it exerts on a particle of mass a distance away from its centre. This particle can be either outside the shell ( ) or inside the shell ( ). We can choose the coordinate system so that the centre of the shell is at the origin and the particle is at By Newton's law of gravitation, the force exerted on the particle by a tiny piece of the shell of
surface area
located at is
Here is the gravitational constant, is the mass of the tiny piece of shell, is the mass of the particle
and is the
vector
from the particle to the piece of shell. If we
work
in spherical coordinates, as we did in Example 3.3.6,
and
The total force is then
Note for future reference that the square root in is the positive square root because is the length of which is positive.
This integral is a little different than other integrals that we have encountered so far in that the
integrand
is a
vector
. By definition11,
so we just have to compute the three components separately.
In our case, the and components
are both zero12 because so that
To evaluate this integral we substitute
When and when so
Recalling that is the positive square root,
If so that
If so that
The moral13 is
if the particle is inside the shell, it feels no gravitational force at all, and
if the particle is outside the shell, it feels the same gravitational force as it would if the entire mass of the shell () were concentrated at the centre of the shell.
Example 3.3.11. Optional — Gravity Train
The “Gravity Train”14 refers to the following curious, though admittedly not very practical, thought experiment.
Pretend that the Earth is a perfect
sphere
of radius and that it has a constant mass density
Pick any two distinct points on the
surface
of the Earth. Call them and
Bore a tunnel straight through the Earth from to
Place a train in the tunnel at Assume that the only forces acting on the train are gravity, and a normal force, that the tunnel imposes on the train to keep it in the tunnel. In particular, there are no frictional forces, like air resistance, and the train does not have an engine. Release the train and assume that it does not melt as it passes through the centre of the Earth.
What happens?
We'll simplify our analysis of the motion of the train by picking a convenient coordinate system.
First translate our coordinate system so that the centre of the Earth, call it is at the origin,
Then rotate our coordinate system about the origin so that the origin, and all lie in the -plane.
Then rotate our coordinate system about the -axis so that and have the same -coordinate So the coordinates of and are Let's suppose that is at and is at It really doesn't matter which is which, but we can always arrange that it is at by rotating around the -axis by if necessary.
The - and -coordinates of the train are always fixed at and respectively. So let's call the -coordinate at time and look at the -
component
of Newton's law of motion.
It is
because the normal force has no
component
. Recall that Newton's law of gravity says that
where is the gravitational constant, is the
vector
from to the train, and is the mass of the train. In this case, because of our computation in Example 3.3.10, the train only feels gravity from shells of the Earth that are inside the train, so that is the mass of the
part of the Earth whose distance to the centre of the Earth is no more than So
and
so that
This is exactly the
differential equation
of
simple harmonic motion
. We have seen it before in Example 2.2.7. Except for the constant it is identical to the equation solved in Example A.9.4 of the Appendix A.9, entitled “Review of
Linear
Ordinary
Differential
Equations”. The general solution is
with and being arbitrary constants. If we release the train, from rest, at then and so that and
The train reaches when That is, when So the transit time, from to obeys
Notice that this transit time depends only on the gravitational constant and the density of the Earth In particular it is completely independent of
where and are and, in particular,
how close together and are, and also of
the radius of the Earth.
In the case of the Earth, the transit time is about 42 minutes.
Optional — Dropping Higher Order Terms in
In the course of deriving 3.3.1, that is, and formulae for
we approximated, for example, the vectors
where is bounded15 by a constant times and is bounded by a constant times That is, we assumed that we could just drop and
So we approximated
where the length of the
vector
is bounded by a constant times We'll now see why dropping terms like does not change the value of the integral at all16.
Suppose that our
domain
of integration consists of all 's in a rectangle of width and height as in the figure below.
Subdivide the rectangle into a grid of small subrectangles by drawing lines of constant (the red lines in the figure) and lines of constant (the blue lines in the figure). Each subrectangle has width and height Now suppose that in setting up the integral we make, for each subrectangle, an error that is bounded by some constant times
Because there are a total of subrectangles, the total error that we have introduced, for all of these subrectangles, is no larger than a constant times
When we define our integral by taking the
limit
of the Riemann sums, this error converges to exactly
Exercises
Stage 1
1
Let and Denote by the part of the
surface
with
Find the
surface area
of without using any calculus.
2. Find the
surface area
of by using (3.3.2).
2
Let Denote by the triangle with vertices and
Find the
surface area
of in three different ways, each using (3.3.2).
2. Denote by the projection of onto the -plane. (It is the triangle with vertices and ) Similarly use to denote the projection of onto the -plane and to denote the projection of onto the -plane. Show that
3
Let Denote by the part of the
cylinder
with and
Find the
surface area
of without using any calculus.
2. Parametrize byFind the
surface area
of by using (3.3.1).
Stage 2
4
Let be the part of the
surface
lying inside the
cylinder
Find the
moment
of inertia of about the -axis, that is,
5 ✳
Find the
surface area
of the part of the paraboloid which lies above the --plane.
6 ✳
Find the area of the portion of the cone lying between the planes and
7 ✳
Determine the
surface area
of the
surface
given by over the square
8 ✳
To find the
surface area
of the
surface
above the
region
we integrate What is
Consider a “Death Star”, a ball of radius centred at the origin with another ball of radius centred at cut out of it. The diagram below shows the slice where
The Rebels want to paint part of the
surface
of Death Star hot pink; specifically, the concave part (indicated with a thick line in the diagram). To help them determine how much paint is needed, carefully fill in the missing parts of this integral:
2. What is the total
surface area
of the Death Star?
9 ✳
Find the area of the cone between and
10 ✳
Find the
surface area
of that part of the hemisphere which lies within the
cylinder
11
The
cylinder
cuts out a portion of the upper half of the cone Compute
12
Find the
surface area
of the torus obtained by rotating the circle (the circle is contained in the -plane) about the -axis.
13
A spherical shell of radius is centred at the origin. Find the
centroid
(i.e. the centre of mass with constant density) of the part of the
sphere
that lies in the first octant.
14
Find the area of that part of the
cylinder
lying outside
15 ✳
Let and be positive constants, and let be the part of the conical
surface
where Consider the
surface integral
Express as a
double integral
over a disk in the -plane.
2. Use the parametrization etc., to express as a
double integral
over a suitable
region
in the -plane.
3. Evaluate using the method of your choice.
16
Evaluate, for each of the following, the
flux
where is the outward normal to the
surface
and the
surface
is the
sphere
2. and is the
surface
of the rectangular box
3. and is the
surface
of the solid cone
17 ✳
Let be the part of the
surface
that lies above the square
Find
Find the
flux
of upward through
18 ✳
Let be the part of the
surface
that lies above the square in the -plane.
Find
Find the
flux
of upward through
19 ✳
Find the area of the part of the
surface
that lies above
20 ✳
Let be spherical cap which consists of the part of the
sphere
which lies under the plane Let Calculate
21 ✳
Find a parametrization of the
surface
of the cone whose
vertex
is at the point and whose
base
is the circle in the -plane. Only the cone
surface
belongs to not the
base
. Be careful to include the
domain
for the parameters.
2. Find the -coordinate of the centre of mass of the
surface
from (a).
22 ✳
Let be the
surface
of a cone of height and
base
radius The
surface
does not include the
base
of the cone or the interiour of the cone. Find the centre of mass of
Locate the cone in a coordinate system so that its
base
is in the -plane, and its
vertex
on the -axis. So the
vertex
will be the point The
base
is a circle of radius in the -plane with centre at the origin. The cone
surface
is characterized by the fact that for every point of the distance from the -axis and the distance from the -plane add up to
23 ✳
Let be the portion of the elliptical
cylinder
lying between the planes and and let denote the outward normal to Let Calculate the
flux integral
directly, using an appropriate parameterization of
24 ✳
Evaluate the
flux integral
where and is the part of the paraboloid that lies above the triangle is oriented so that its unit normal has a negative -
component
.
25 ✳
Evaluate the
surface integral
where is the part of the
sphere
for which
26 ✳
Let be the
surface
given by the equation
lying between the planes and Evaluate the integral
27 ✳
Let be the part of the paraboloid lying above the -plane. At has density
Find the centre of mass of
28 ✳
Let be the part of the plane
that lies in the first octant oriented so that has a positive
component
. Let
Evaluate the
flux integral
29 ✳
Find the net
flux
of the
vector field
upwards (with respect to the -axis) through the
surface
parametrized for
30 ✳
Let be the
surface
obtained by revolving the curve , around the -axis, with the of having pointing toward the -axis.
Draw a picture of and find a parameterization of
Compute the integral
Compute the
flux integral
where
31 ✳
Compute the net outward
flux
of the
vector field
across the boundary of the
region
between the spheres of radius and radius centred at the origin.
32 ✳
Evaluate the
surface integral
where is the part of the cone where and
33 ✳
Compute the
flux integral
where
and is the part of the paraboloid lying inside the
cylinder
with
orientation
pointing downwards.
34 ✳
Let the thin shell consist of the part of the
surface
with and Find the mass of if it has
surface
density given by kg per unit area.
35 ✳
Let be the portion of the paraboloid that satisfies Its unit
normal vector
is so chosen that Find the
flux
of out of
36 ✳
Let denote the portion of the paraboloid for which Orient so that its unit normal has a positive
component
. Let
Evaluate the
surface integral
37
Let be the boundary of the apple core bounded by the
sphere
and the hyperboloid Find the
flux integral
where and is the outward normal to the
surface
Stage 3
38 ✳
Consider the
surface
given by the equationFind an equation for the
tangent plane
to at the point
2. Compute the integralwhere is the part of the
surface
from (a) lying between the planes and
39 ✳
Let be a
function
on such that all its first order partial derivatives are continuous. Let be the
surface
for some Assume that on Let be the
gradient field
Let be a
piecewise smooth curve
contained in (not necessarily closed). Must it be true that Explain why.
2. Prove that for any
vector field
40 ✳
Give parametric descriptions of the form for the following surfaces. Be sure to state the domains of your parametrizations.
The part of the plane in the first octant
The cap of the
sphere
for
The hyperboloid for
Use your parametrization from part (a) to compute the
surface area
of the cap of the
sphere
for
41 ✳
Let be the part of the
sphere
where oriented away from the origin.
Compute
Compute
42 ✳
Let be the part of the
surface
which lies in the first octant. Find the
flux
of downwards through where
As we mentioned above, the approximation below becomes exact when the
limit
is taken in the definition of the integral. See the optional §3.3.5.
This is called the implicit
function
theorem. We will not prove it. But it is not so hard to understand why it is true, if one thinks in terms of the Taylor expansion of about the point. For simplicity, let's suppose that the point is and happens to be exactly equal to its first order Taylor expansion about That is, for some constants Since is on the
surface
, As we can easily solve for as a
function
of and Namely The general proof is based on the fact that, under reasonable hypotheses, the first order Taylor expansion is a good approximation to near
3. The symbols are the standard mathematics symbols for the cylindrical coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering.
4. As we have noted before, the
spherical coordinate system
really breaks down at because gives the same point, namely the north for all values of We should really treat our integral like an
improper integral
, first integrating over and then taking the
limit
However the breakdown of the
spherical coordinate system
at just like the breakdown of polar coordinates at rarely causes problem and it is routine to skip the “
improper integral
” step.
5. We promise!
6. This answer for is a very clean. Think about why. Hint: review the discussion following 3.3.2.
7. If you have forgotten why, sketch the graph.
8. We did so previously, with different variable names, in Example 3.2.2.
9. Again the formula for is very neat. Think about why.
10. A favourite of science fiction and fantasy writers. Plug “subterranean fiction” into your favourite search engine. While you're at it, also try “gravity train”. We'll look at it in the optional Example 3.3.11.
11. Under this definition we still have
12. Think about why the and components should both be zero. Think symmetry.
13. These two results appeared in Isaac Newton's Principia Mathematica (1687). They are known as Newton's “superb theorems”.
14. The British physicist and architect (he was Surveyor to the City of London and chief assistant to Christopher Wren) Robert Hooke (1635--1703) wrote about the gravity train idea in a letter to Isaac Newton. A gravity train was used in the 2012 movie Total Recall.
15. Remember the error in the Taylor polynomial approximations.
16. See the optional §1.1.6 of the CLP-2 text for an analogous argument concerning Riemann sums.
3.2: Tangent Planes
3.4: Interpretation of Flux Integrals |
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An ant starts at one vertex of a tetrahedron. Each minute, it walks along a random edge to an adjacent vertex.
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The expected amount of time until the ant returns to its starting vertex is 2.5 minutes.
What is vertex?
Vertex in math is the point at which two lines, curves, or edges meet. In the case of a triangle, for example, the vertex is the point where all three lines meet. In higher dimensional shapes, such as a parallelogram, it is the point at which all the lines meet. Vertex can also be used to refer to the highest point of a graph, or the point of maximum or minimum value.
The expected amount of time until an ant returns to its starting vertex after traversing the edges of a tetrahedron can be calculated by applying the principle of expected value. The expected value of a random variable is the sum of the probability of each outcome multiplied by its associated value.
In this case, the ant has four possible outcomes, with each outcome being the length of time it takes to traverse the edge to an adjacent vertex. Since the ant has an equal probability of going to each vertex, each outcome has a probability of 0.25. Thus, the expected value can be calculated as:
Expected Value = (1 minute x 0.25) + (2 minutes x 0.25) + (3 minutes x 0.25) + (4 minutes x 0.25)
Expected Value = 2.5 minutes
Therefore, the expected amount of time until the ant returns to its starting vertex is 2.5 minutes.
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The expected time for the ant to return to its starting vertex on a tetrahedron is 2.5 minutes. This is calculated using a model of random walks and the properties of symmetry present in the tetrahedron's structure. The proportionality of movement to adjacent vertices establishes a consistent expected time frame.
Explanation
To determine the expected amount of time until an ant returns to its starting vertex on a tetrahedron, we can model the ant's movement as a random walk. A tetrahedron has 4 vertices and each vertex is connected to 3 other vertices via edges. Therefore, if the ant is at one vertex, it has an equal probability of moving to any of the 3 adjacent vertices in the next minute.
Modeling the Situation:
Let the expected time to return to the starting vertex from that vertex be denoted as E.
Considering the Moves:
In the first minute, the ant moves to one of the 3 adjacent vertices. The expected time to get back to the original vertex from any adjacent vertex is now E + 1 minute (since it took one minute to move to the adjacent vertex).
Setting Up the Equation:
The equation can be structured as follows:
E=1+3 1(E+E+E)
The three terms inside the parentheses each represent the time from one of the three adjacent vertices back to the starting vertex.
Simplifying the Equation:
This simplifies to:
E=1+E
By solving this equation, we rearrange it to:
3 E=1+3 E 0=1+2 E =>2 E=1
Thus:
E=2 5=2.5 minutes
Conclusion: The expected amount of time until the ant returns to its starting vertex is 2.5 minutes. This calculation highlights how properties of symmetry and equal probabilities contribute to expected values in random walks.
Examples & Evidence
For example, if the ant chooses to walk along edges randomly, it may move from vertex A to vertex B, then to vertex C, and finally back to A, which is a possible sequence contributing to its total expected return time. Further simulations or experiments could explore different starting vertices and edge choices to demonstrate the consistency of the expected value across trials.
This conclusion can be verified by applying the principles of probability and expected value calculations found in probability theory and random processes, which are established mathematical concepts.
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12814 | https://www.taylorfrancis.com/chapters/mono/10.1201/b12815-11/gas-absorption-nayef-ghasem | Gas Absorption | 11 | Computer Methods in Chemical Engineering | Nay
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- Gas Absorption
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Gas Absorption
ByNayef Ghasem
BookComputer Methods in Chemical Engineering
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First Published 2011
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Pages 62
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ABSTRACT
Absorption is a process that refers to the transfer of a gaseous pollutant from the gas phase to the liquid phase. Absorbers are used extensively in the chemical industry for separation and puri”cation of gas streams, as product recovery and as pollution control devices. The absorption process can be categorized as physical or chemical. Physical absorption occurs when the absorbed compounds dissolves in the solvent. Chemical absorption occurs when the absorbed compounds and the solvent react. Examples are separations of acid gases such as CO2 and H2S from natural gas using amine as solvents. Chemical engineers need to be able to design gas absorbers that produce a treated gas of a desired purity with an optimal size and liquid “ow. This can be based on existing correlations and, when required, laboratory and/or pilot plant data. For gas absorption, the two most frequently used devices are the packed tower and the plate tower. Both these devices, if designed and operated properly, can achieve high collection ef”- ciencies for a wide variety of gases. The primary outcomes of the design procedures are to determine the diameter of the column and the tower height [1-3].
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12815 | https://rpg.stackexchange.com/questions/6266/is-it-possible-to-turn-tensers-floating-disk-into-a-chariot | spells - Is it possible to turn Tenser's Floating Disk into a chariot? - Role-playing Games Stack Exchange
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Is it possible to turn Tenser's Floating Disk into a chariot?
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Is it possible to turn Tenser's Floating Disk into a chariot in AD&D 1st edition?
There are several problems that I'm not sure can be solved:
How would the harness be attached to the disk?
Can a mount pull the disk at a higher movement rate than the disk can move by default, without causing the spell to break?
How difficult is it to pull the disk – how much weight does it count as?
It seems that a naive interpretation has the disk weight at nearly nothing, especially since it can move rather quickly, and I can think of some rope/knot patterns that would hold even the most slippery of disks – meaning a mount could move at maximum speed while pulling the disk.
spells
adnd-1e
vehicles
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edited May 3, 2022 at 12:37
Someone_Evil♦
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asked Feb 16, 2011 at 1:37
blueberryfieldsblueberryfields
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From a 4e PoV, I've considered quite a few ways to effectively enbiggen the platform. I don't suppose you could ask similar questions for other editions? :)Brian Ballsun-Stanton –Brian Ballsun-Stanton 2011-02-16 07:38:02 +00:00 Commented Feb 16, 2011 at 7:38
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No speed limit is given except when it moves by direction of the caster (6" rate). It would seem unnecessary to secure it since it remains 6' from the caster... but you could certainly envelop it with a net or similar rope arrangement. Move the caster at high speed, and the magic will keep the disc nearby.
It doesn't seem to be possible to cut holes in the magical disc, so attaching a handle or wagon-tongue is an unexplained situation. Think of it as unbreakable as a wall of force, I'd guess.
Note that if the caster is on horseback, the caster should position the disc to one side (6' away of course), lest it impact the horse's rear hooves; it maintains a height of 3' above ground. (Watch out for potholes and other terrain issues; it will bob around, matching the contour, and aprupt changes may affect its load.)
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answered Feb 16, 2011 at 1:52
ExTSRExTSR
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It may be usable as a component of a chariot.
Looking at @Seven's answer, a null-grav plate would be incredibly helpful if included as part of a chariot's construction. From a simple point of view, you could replace one or more wheels or axles with floating disks, providing a levitation effect for transit over rough terrain. If the disk is linked to the caster via a specific distance, have the caster sit in the "cart" so created. While this may not be the fastest on-road travel, there are quite a few "hovercraft" opportunities that a construction like this would support. The rough principle is a braced construction that expands the effective area of the disk(s) by using it/them for support.
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answered Feb 16, 2011 at 7:41
Brian Ballsun-StantonBrian Ballsun-Stanton
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But how do you tether the "plate" to the chariot? I think the word disk is misleading, it seems the correct way to think about it more like an electro-magnetic field which creates a disk like area rather than as a physical object.Jon Hopkins –Jon Hopkins 2011-02-21 13:16:14 +00:00 Commented Feb 21, 2011 at 13:16
You don't need to tether at all. If it's a zone of "we'll ignore physics" and is providing the quite literal impulse to the cart, the cart moves with the zone. To propel the zone by moving the cart, and assuming the zone moves with the caster, then the caster must merely be seated in the cart.Brian Ballsun-Stanton –Brian Ballsun-Stanton 2011-02-22 01:11:22 +00:00 Commented Feb 22, 2011 at 1:11
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It can't be all that slippery, or things would slide off it all the time, and one would think there would have to be some kind of reflection of that in the rules for it. So tension-based attachment mechanisms like clamps or vises should work reasonably well.
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answered Feb 16, 2011 at 2:27
chaoschaos
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I think if the player comes up with a creative way to effectively do it...and since the caster I feel should be able to make slight changes to the spell as they are casting it, I feel the caster should be allowed to do it.
Course I play more freeform styled games where slight bending of the rules is expected!
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answered Apr 10, 2011 at 13:25
m.s. jacksonm.s. jackson
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This is what AD&D spellcasting is all about.migo –migo 2011-04-10 19:44:39 +00:00 Commented Apr 10, 2011 at 19:44
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No, it can't be used as a chariot.
The Disk is described as a "circular plane of null-gravity" (PHB, p. 68). It operates not by being a physical barrier upon which things are heaped and held up by a physical surface, but rather it prevents things that are in contact with its plane from falling. Think of it as a movable circular area that grants levitation to things touching it from above, and less as a physical disk. Being an "bare" effect that lacks an accompanying physical object, there's nothing there that you could hope to manipulate with ropes, and so therefore it cannot be pulled.
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answered Feb 16, 2011 at 2:40
SevenSidedDieSevenSidedDie
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I personally believe nothing can be attached to a disk because it is indeed some sort of space with different physical properties. That is exactly why it's perfect for a 'floating' chariot.
All you need is somebody who can cut wood. Just make a round platform slightly larger then the disk. Make sure that the edges are thicker then the rest of the platform. Cast disk, put platform over the disk, attach harnas/ropes/whatever to the wooden platform, cast phantom steed and enjoy a custom chariot.
You can make this a whole lot better and stronger if you're willing to put some time and effort in it. As long as you're basically making some sort of platform/lid to put over a disk it works fine. It really is not that complicated.
For the people out there that want something awesome at lvl 5.
Get 4 players to each play a small creature with at least 2 who can cast disk.
10ft long 6ft wide At the front end make a box large enough for 2 mastiffs per player (mastiffs are legal mounts for small creatures). So with 4 players that means 8 mastiffs have to be able to fit in the box. You don't need any joints or whatever. Just 1 large box with the platform attached at the back end. Make sure that the platform at the end is high enough so that you can look over your 'engine' box. Put a repeater bolt thrower on it. Cast enough disks to reduce the effective weight of the construction+passengers to almost 0. For extra defense cast all defensive spells on it that you have.
You've just made a 20ish feet long medieval tank. Because it weighs nothing turning is a breeze. The engine is 'safe' in a box. Your whole party is standing at least 4 feet in the air on a platform with a repeater bolt thrower and you can run as fast as you can throw sausages in the engine.
When you hit lvl 6+ you can make some seriously awesome tanks this way. And for those wondering yes this was the beginning of a beautiful story :D
edit: tip: disk floats at 3 feet above the ground. Disk that have to much weight on them dispell. Make sure you to put your tank on some supports that are just a fraction above where the disks are hanging. When all disks are in place, throw some food to your mastiffs, they will pull the thing from their supports and make it land on the disks. You can go crazy here and make retractable supports or use simple blocks of stone. It's just something to think about when you're building a tank.
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edited Mar 19, 2017 at 17:57
answered Mar 19, 2017 at 17:39
EdwinEdwin
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12816 | http://blog.zacharyabel.com/tag/9-point-circle/ | # Three-Cornered Things
Zachary Abel's Math Blog
Tag Archives: 9-point circle
Three-Cornered Deltoids
Geometry9-point circle, animations, deltoids, Morley's theorem, Steiner deltoid, triangle geometryZachary Abel
[This is the third and final post in this series on triangle geometry. See the previous posts on Morley’s theorem and the 9-point circle.]
For our final exploration in this series, let’s again begin with our triangle ABC and a point P on the circumcircle of the triangle, i.e., the circle through the three vertices. If we drop P directly onto the three lines of the triangle at right angles, then by coincidence, these three points lie on a single line, called the Simson line of point P.
Just for fun, let’s draw point Q diametrically opposite from P on the circumcircle, and let’s also look at Q‘s Simson line. How do these two Simson lines interact? Somewhat surprisingly, these lines intersect at right angles. The phenomenal part is that this point of intersection lies on the 9-point circle!
Furthermore, as P and Q move around the circumcircle, the intersection of their Simson lines moves around the 9-point circle at twice the speed and in the opposite direction:
This story gets even more unbelievable when we look at how the Simson lines move in this animation. As it turns out, the Simson lines trace a curve in the shape of a deltoid, which is like an equilateral triangle with curved sides. The deltoid traced here is called Steiner’s deltoid.
And finally, here’s an incredible fact that ties everything together: If we draw the equilateral triangle around this deltoid, then the edges are parallel to the edges of the equilateral Morley triangle(s).
Holy Morley!
Notes
Note that we may have to extend the lines beyond the triangle. [↩]
Specifically, a deltoid is the shape that results when you roll a circle inside a circle three-times larger and trace the path of a single point. See Wikipedia:Deltoid_curve for more information. [↩]
Several Sneaky Circles
Geometry9-point circle, Feuerbach, orthocenter, triangle geometryZachary Abel
In last week’s exploration we uncovered many equilateral triangles hiding in a general triangle ABC. This week we’ll find circles.
Draw any triangle ABC. The line AHA perpendicular to BC is called the altitude though A, with HA at the foot of the altitude. (See the image below.) It can be shown that the three altitudes AHA, BHB, and CHC meet at a single point, H, called the orthocenter of triangle ABC.
Now, let’s put dots at a few important points on our triangle: First, dot the three altitude feet, HA, HB, and HC. Next, dot the midpoints of the sides of the triangle, which we’ll call MA, MB, and MC. Finally, dot the Euler points EA, EB, and EC, which are halfway between H (the orthocenter) and the vertices of our triangle. Today’s first miracle is that all nine of these dots lie on a single circle, aptly named the 9-point circle:
Recall that it only takes three points to determine a circle, so nine on a single circle is very far from coincidence!
Another way to coax a circle out of triangle ABC is to “inscribe” one in the triangle by drawing a circle tangent to the three sides. This circle is called the incircle. Our second miracle is that the incircle and the 9-point circle, which were constructed in very different ways, nevertheless play nicely together: they are tangent! The point where they touch is known as the Feuerbach point of triangle ABC, named after Karl Feuerbach who discovered and proved this difficult fact in 1822.
But there are three other circles that, like the incircle, are tangent to all three lines of the triangle. These “exscribed” circles are called the excircles of triangle ABC, and as luck (or extraversions) would have it, these are also tangent to the 9-point circle:
But wait, there’s more! It turns out that triangles ABC and HBC have the same 9-point circle, and so if we look at the incircle and excircles of triangle HBC, these will also be tangent to our 9-point circle. The same is true for triangles HCA and HAB, so we have found 16 circles all tangent to the nine-point circle!
Notes
Recall that we used extraversions last week to go from one Morley triangle to another. Here, extraversions interchange the incircle and excircles. [↩] |
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Yield to Maturity (YTM)
Yield to Maturity (YTM)
Comprehensive Overview of Yield of Maturity and How to Calculate It
Updated May. 14, 2025
7m Read Time
What is Yield to Maturity?
How to Calculate Yield to Maturity (YTM)?
Yield to Maturity Formula (YTM)
Accurately Estimating YTM with AI
What Is the Difference Between APY and YTM?
Yield to Maturity vs. Coupon Rate vs. Current Yield
What is a Good Yield to Maturity (YTM)?
Yield to Maturity Calculator (YTM)
Mastering Yield to Maturity
What is Yield to Maturity?
The Yield to Maturity (YTM) represents the expected annual rate of return earned on a bond under the assumption that the debt security is held until maturity.
From the perspective of a bond investor, the YTM is the anticipated total return received if the bond is held to its maturity date and all coupon payments are made on time and are then reinvested at the same interest rate.
Generating
The yield to maturity is the expected annual rate of return earned on a bond, assuming the debt security is held until maturity.
The yield to maturity is calculated by the following formula: [Annual Coupon + (FV – PV) ÷ Number of Compounding Periods] ÷ [(FV + PV) ÷ 2].
The YTM metric offers bondholders with the option to estimate the return on a bond instrument, as well as measure the impact on the portfolio return.
The yield on bonds is inversely related to the market interest rate, meaning that the higher the YTM, the less sensitive the bond prices are to interest rate fluctuations.
How to Calculate Yield to Maturity (YTM)?
Yield to maturity is one of the most frequently used returns metrics for evaluating potential bond and fixed-income investments by investors.
The YTM is the estimated annual rate of return that a bond is expected to earn until reaching maturity, with three notable assumptions:
Assumption 1 → The return assumes the bond investor held onto the debt instrument until the maturity date.
Assumption 2 → All the required interest payments and principal repayment were made on schedule.
Assumption 3 → The coupon payments were reinvested at the same rate as the yield-to-maturity (YTM).
The yield to maturity on a bond is its internal rate of return (IRR) – i.e. the discount rate which makes the present value (PV) of all the bond’s future cash flows equal to its current market price.
The YTM metric facilitates comparisons among different bonds and their expected returns, which helps investors make more informed decisions on how to manage their bond portfolios.
Even for bonds consisting of different maturities and coupon rates, the YTM enables comparisons to be made since the YTM is expressed as an annualized rate regardless of the bond’s years to maturity.
Yield to Maturity Formula (YTM)
The formula for calculating the yield to maturity (YTM) is as follows.
Yield to Maturity (YTM) = [Annual Coupon + (FV – PV) ÷ Number of Compounding Periods)] ÷ [(FV + PV) ÷ 2]
The components of the yield to maturity equation consist of the following inputs:
Coupon Payment (C) → Determined by the coupon rate of the bond, or “interest rate”, the annual coupon payment is the periodic payment distributed by the bond issuer to the bondholders. In general, the higher the coupon rate attached to the bond, the higher the yield, all else being equal.
Face Value (FV) → The face value of a bond (i.e. the par value) is the amount to be repaid to a bondholder on the date of maturity.
Present Value (PV) → The present value (PV) of the bond refers to the current market price and how much investors are willing to pay for the bond in the open market as of the present date, which may be higher (or lower) than the bond’s FV based on the market conditions and supply/demand.
Maturity Date → The pre-specified date on which the issuer is contractually obligated to repay the principal – from this date, the number of years to maturity can be derived.
Number of Compounding Periods (n) → The number of compounding periods refers to the number of payments made in one year multiplied by the number of years to maturity (e.g. five years until maturity and semi-annual coupon payments would mean n = 10 periods).
Accurately Estimating YTM with AI
YTM represents the expected annual return on a bond held to maturity, incorporating both coupon payments and principal repayment. As fixed-income portfolios grow more complex, artificial intelligence is playing a pivotal role in modeling bond performance, forecasting interest rate shifts, and optimizing portfolio decisions. Wall Street Prep created the AI for Business & Finance Certificate Program in partnership with Columbia Business School Executive Education to equip professionals with the tools to integrate AI into bond valuation and fixed-income analysis. Discover how AI-driven interest rate forecasting and automated bond analytics are reshaping how investors evaluate YTM and credit risk.
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What Is the Difference Between APY and YTM?
Annual Percentage Yield (APY) is used to express the actual return on deposit accounts like savings accounts or certificates of deposit (CDs), including the effect of compounding interest. YTM represents the estimated return on a bond if held until maturity, considering both interest payments and any changes in the bond’s price.
Yield to Maturity vs. Coupon Rate vs. Current Yield
The yield to maturity, as mentioned earlier, is the annualized return on a debt instrument based on the total payments received from the date of initial purchase until the maturation date.
In comparison, the current yield on a bond is the annual coupon income divided by the current price of the bond security.
An important distinction between a bond’s YTM and its coupon rate is the YTM fluctuates over time based on the prevailing interest rate environment, whereas the coupon rate is fixed.
The relationship between the yield to maturity and coupon rate (and current yield) are as follows.
Yield to Maturity (YTM) < Coupon Rate and Current Yield → The bond is being sold at a “premium” to its par value.
Yield to Maturity (YTM) > Coupon Rate and Current Yield → The bond is being sold at a “discount” to its par value.
Yield to Maturity (YTM) = Coupon Rate and Current Yield → The bond is said to be “trading at par”.
What is a Good Yield to Maturity (YTM)?
By understanding the YTM formula, investors can better predict how changing market conditions could impact their portfolio holdings based on their portfolio strategy and existing investments.
Considering yields rise when prices drop (and vice versa), investors can project yield-to-maturity (YTM) on portfolio investments to guide better decision-making.
The YTM can also enable debt investors to assess their degree of exposure to interest rate risk, which is defined as the potential downside caused by sudden changes in interest rates.
The relationship between the current YTM and interest rate risk is inversely proportional, which means the higher the YTM, the less sensitive the bond prices are to interest rate changes.
The most noteworthy drawback to the yield-to-maturity measure is that YTM does NOT account for a bond’s reinvestment risk. The bond’s coupon payments are assumed to be reinvested at the same rate as the YTM, which may not be an option in the future given uncertainties regarding the markets.
In effect, if coupons were to be reinvested at lower rates than the YTM, the calculated YTM is going to turn out to have been inaccurate, as the return on the bond would have been overstated.
The standard YTM formula is also meant to be an approximation as opposed to a precise figure – for instance, the YTM is prone to error due to the potential for unexpected events such as if the bondholder decides not to reinvest all coupon payments or if the bond is called early (i.e. repaid prior to maturity).
However, the benefits related to comparability tend to outweigh the drawbacks, which explains the widespread usage of YTM across the debt markets and fixed-income investors.
Yet, the YTM’s assumptions that all coupon payments are made as scheduled, and that interest is reinvested at the same rate are nonetheless risky, simplified assumptions.
Yield to Maturity Calculator (YTM)
In this section, we’ll walk you through modeling exercises to help you better understand how to calculate Yield to Maturity using the formula. By applying the YTM formula to real-life bond scenarios, you will gain hands-on experience in estimating bond returns. The downloadable Excel template provided will allow you to practice these calculations, giving you a tool to explore different bond scenarios and deepen your understanding of how YTM works in a practical, interactive way.
You can access the modeling exercises by filling out the form below.
1. Bond Pricing Assumptions
Suppose we’re tasked with calculating the YTM on a corporate bond issuance using the following set of assumptions.
Face Value of Bond (FV) = $1,000
Annual Coupon Rate (%) = 6.0%
Number of Years to Maturity = 10 Years
Price of Bond (PV) = $1,050
We’ll also assume that the bond issues semi-annual coupon payments.
2. Coupon Rate and Interest Payment Calculation Example
Given those inputs, the next step is to calculate the semi-annual coupon rate, which we can calculate by dividing the annual coupon rate by two.
Semi-Annual Coupon Rate (%) = 6.0% ÷ 2 = 3.0%
Then, we must calculate the number of compounding periods by multiplying the number of years to maturity by the number of payments made per year.
Number of Compounding Periods (n) = 10 × 2 = 20
As for our last input, we multiply the semi-annual coupon rate by the face value of the bond (FV) to arrive at the semi-annual coupon of the bond, i.e. the semi-annual interest payment.
Semi-Annual Coupon (C) = 3.0% × $1,000 = $30
3. Yield to Maturity Calculation Example (YTM)
With all required inputs complete, we can calculate the semi-annual yield to maturity (YTM).
Semi-Annual Yield to Maturity (YTM) = [$30 + ($1,000 – $1,050) ÷ 20] ÷ [($1,000 + $1,050) ÷ 2]
Semi-Annual YTM = 2.7%
In the final part of our bond rate of return analysis exercise in Excel, the only remaining step is to convert our semi-annual YTM to an annual percentage rate, i.e. the annualized yield to maturity (YTM).
Annual Yield to Maturity (YTM) = 2.7% × 2 = 5.4%
In conclusion, the implied yield to maturity (YTM) in our hypothetical bond issuance, expressed on an annual basis, comes out to 5.4%.
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Mastering Yield to Maturity
Yield to Maturity is a vital concept in bond investing and valuation, providing a comprehensive measure of a bond’s potential return if held until maturity. Understanding how to calculate YTM, whether manually or using tools like the YTM calculator and downloadable Excel template, is essential for making informed investment decisions.
Unlike other yield metrics, YTM accounts for both the bond’s price and its future cash flows, offering a more accurate assessment of its total return. With the practical tools provided, investors can deepen their understanding and confidently analyze bond investments.
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A Problem Solver's Paradise ===========================
Poles and Polars - Another Useful Tool!
by Potla, Mar 22, 2012, 2:45 PM
As the title suggests, this post is going to deal with the different aspects of poles and polars, which is a really great tool in case of proving problems using cyclic quadrilaterals and a lot of circular reasoning can be exploited by this. It is easier to deal with lines than with circles, so it is better to work with inversion, poles and polars etc - that's why projective geometry is actually better in many aspects compared to plain euclidean geometry!
Anyways, for the time being let us know the definitions of poles and polars. Before continuing further, you have to note that a pole is a point, and a polar is a line (which is the opposite of Kedlaya, but yeah, notations don't matter much. ) In any case, thanks to Rijul Saini for giving me a hand on this post, which we delayed for more than 8 months. At last, this is getting published on the blog, so Cheers, everyone!
Pole of a line
Take the circle with respect to which you're applying the polar map. Now, drop a perpendicular to the given line from the centre of to the given line. Name that point . Now, invert with respect to the circle. We get a point which lies on the same line as Then, is the Pole of the given line.
Polar of a point
Take the circle with respect to which you're applying the polar map. Let the centre of the circle be Now, invert the given point with respect to the circle and name the image as
Now, Draw the line perpendicular to which passes through the point Then, this line is the Polar of the point
Now let us look into some theorems and lemmas that we will be using while solving problems.
Theorem 1:(La Hire's theorem)
Every point is the pole of its polar, and every line is the polar of its pole.
If lies on the polar of then lies on the polar of
Proof:
(i) Again, direct from the definition.
(ii) is a point on the polar of First, extend to the polar of and name that point If then we are through. Otherwise, let the inverse of the point wrt the circle be Now, Now, by similar triangles, we have and so we are through.
Theorem 2:
Three points are collinear if and only if their polars are concurrent.
Proof:
I shall prove only the forward direction. The reverse direction is entirely analogous.
Let the three points be and their polars be Now, let the line through be and let the pole of be . Since, lie on the polar of therefore, must lie on the polar of i.e. lines Thus, the lines are concurrent at
Now let us get to know the poles and polars of some points and lines that are considered to be important.
Polar of a point outside a circle with respect to itself.
Take an arbitrary point outside a circle. Since the inverse of the point actually lies inside the circle, the polar will be a secant of the circle. Now, let us assume that the inverse point of w.r.t. is The line perpendicular to and passing through will intersect the circle in two points which will be symmetric w.r.t So let one of these two intersections be and since we note that so this leads to ie:
Theorem 3:
The polar of a point lying outside the circle is actually the chord of contact of the point with respect to
Polar of a point lying inside the circle does not have any special property that can be exploited alike to the previous one. However, we can take any arbitrary two chords passing through , find their poles and join the two corresponding points to obtain the polar of that point
Now, let us move on to Theorem 4. The first name was coined by me, and the second one by Rijul Saini. Don't assume that they are the original names of the theorem.
Theorem 4
(The fundamental theorem of poles and polars, or The theorem of two pascals)
Take a quadrilateral which is cyclic. If then the polar of with respect to is the line joining and
Proof 1.
The second name almost gives it away. This problem, however, was directly quoted and asked to prove in the Turkish TST 1993.
Anyways, using Pascal's theorem in we obtain are collinear. Again using the same in we obtain that are collinear. Therefore it is noted that the points are collinear. Denote this line as
Since passes through and which is the polar of we are done.
Proof 2.
This uses harmonic divisions.
Let us take If the tangent to at touches it at respectively such that $M\in\overhat{DC}$, denote and
It can be shown easily that and are harmonic. Then using we note that concur; and are collinear. Similarly we obtain that and are collinear. Since therefore are collinear.
So lies on the polar of w.r.t By symmetry lies on the polar of w.r.t So the intersection of these two polars will be , and the polar of this pole will be We are done!
There are some useful results that are correlated to harmonic divisions and poles and polars, which I explain in the following few theorems.
Theorem 5 (a).
If is the pole of a line w.r.t then any line through is cut harmonically by and
Proof.
Let be the centre of and let
Let be the circle passing through Then its centre is the midpoint of If then is the pole of So $OP\cdpt OQ=OC^2.$ Then and are orthogonal circles, which means that is tangent to Therefore we obtain and since is the midpoint of from Theorem 1 - Harmonic Divisions, we are done.
Theorem 5 (b).
If two lines are self-conjugate lines and if then are harmonic with the two tangents drawn to the circle from
Proof.
Let the poles of be if the two tangents from meet the circle in
Since the polar of passes through from La Hire's theorem we see that the polar of passes through and similarly, Then are collinear. Now, from Theorem we are done.
Theorem 6.
Finally, if four points form a harmonic range then their polars will create a harmonic pencil.
Proof.
Let be the pole of and let
If we draw four perpendicular lines perpendicular to respectively, then we have an interesting result. The polar of passes through so from La Hire's, the polar of passes through , and therefore is the polar of with respect to our aforementioned circle. It is obvious that the pencils and are equiangular pencils, and since are harmonic, therefore are also harmonic.
Wow, that was enough of theory for a day already! Let us quickly look into some applications of our Dual Pascal and try to find boring solutions to some problems!
Let be the diameter of a circle with two points on its circumference such that is closer to than If and intersect at outside the circle, and if the tangents at to the circle meet at then show that
Solution
Let We already know that(from Theorem 4), lies on the polar of The polar of is we see that lies on the polar of So is the polar of , and therefore we are done.
Let be a tangential quadrilateral with incentre Let the opposite sides of meet each other at ie let Let its incircle touch the sides at respectively. If then show that
Solution
Note that are tangents to the incircle of so that is the polar of Similarly the polar of is Therefore the pole of the line is and we are done.
(China 2006 Western MO) is the diameter of a circle with centre is a point on extended. A line through cuts the circle with centre at is a diameter of which has centre Let Prove that lie on a circle.
Solution
Let So the polar of with respect to the circle passes through and So and it is obvious that are collinear.
So by considering the power of we have and so are concyclic points.
(IMO 1985) A circle with center passes through the vertices and of triangle and intersects segments and again at distinct points and respectively. The circumcircles of triangles and intersects at exactly two distinct points and Prove that
Solution
Note that the spiral similarity centered at which sends to and to sends the midpoint of (say, ) to that of (say, ). So is the center of unique spiral similarity that sends to
and to , and thus it follows that are concyclic. Again since so are concyclic, and is the diameter of the common circle. So, we are done.
Circles and meet at points and Circle centered at meets circles and in four distinct points and such that is a convex quadrilateral. Lines and meet at Lines and meet at Prove that
Solution
Actually this is equivalent to our last problem. Note that if then from simple angle chasing we can show that Also it is obvious that passes through All of this implies that as desired.
Let be a triangle with incentre . Reflections of in are Prove that are concurrent.
Note: This is a direct consequence of the isogonic theorem. For details, see here.
Solution
I will use a lemma.
Lemma.
If is the intouch triangle of and if is the incentre of and if is the triangle directly homothetic to such that ; then is perspective to
Now let us come back to our original problem.
Let be, as usual, the intouch triangle of Then let be the incircle of Since we see that the polar of with respect to is the perpendicular bisector of Similarly, the polar of w.r.t is the perpendicular bisector of and the polar of w.r.t is the perpendicular bisector of
Let these perpendicular bisectors intersect each other to form a triangle
Denote by the circles Then is the intersection of and Since is the radical axis of and and since is the radical axis of and therefore is the radical centre of
Since where is the inradius of we note that has the same power with respect to and So all lie on the radical axis of these two circles, and therefore, collinear.
From here it is also obvious that and so and are directly similar and perspective around So they are homothetic.
Applying the aforementioned lemma, we note that the intersections are collinear.
Since and are tangents to we again note that is the pole of w.r.t Also since the polar of w.r.t is it follows from the La Hire's theorem that the pole of w.r.t is
Due to symmetry we note that are respectively the poles of We already have seen that are collinear. Therefore it is obvious that meet at the pole of w.r.t We are done.
(Polish MO Second Round 2012) Let be a triangle with and ,-incentre, -circumcentre. Prove that perpendicular bisector of , line and line have a common point.
Solution
Assume that the incircle touches at and that Since therefore and are equilateral. Also therefore and the incircle of bisects
Now, we claim that is the perpendicular bisector of
Proof of claim.
Let be the reflection of in Then note that we have therefore lies on Now we already have which give leading to the fact that bisects This readily implies that bisects
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Coming back to our main proof, note that we have as the perpendicular bisector of so passes through the pole of wrt Since pole of is and because is the perpendicular bisector of we are done.
(2012 Indonesia Round 2 TST 4: P2) Let be a circle with center , and let be a line not intersecting . is a point on such that is perpendicular with . Let be an arbitrary point on different from . Let and be distinct points on the circle such that and are tangents to . Let and be the foot of perpendiculars from to and respectively. Let be the intersection of and . As moves, determine the locus of .
Solution(hatchguy)
Let be the inverse of wrt I claim is the midpoint of . Clearly, since is in the polar of then is in the polar of , and therefore are collinear. It is easy to see that lie on a circle with diameter . From this, it's very natural to think of droping a perpendicular from to . Let be the foot of this perpendicular. By Simson's line theorem, we have are collinear. Also, using that is cyclic we easily get The last because . Hence and we get that is midpoint of since is a right triangle. We are done.
Let incircle of be tangent to in . A line from that is parallel to meets in and a line from that is parallel to meets in . If be midpoint of , show that is on
Solution
From simple angle chasing, we see that This leads to the fact that lie on a circle with as diameter. Thus, leading to being collinear. Let be the orthocentre of Then note that polar of passes through the pole of wrt which is Also since it follows that Similarly, Therefore, the pole of wrt is Thus, and leading to
Let then note that we have is isosceles, and because therefore is the midpoint of Thence Similarly Using all these, we see that lie on a line parallel to
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Result:
This problem teaches us that the polar of the orthocentre of wrt bisects and This is a very useful fact. Don't ask me why, because even I don't know where it may come in handy.
(Romania MOM 2012)Let be a triangle and let and denote its incentre and circumcentre respectively. Let be the circle through and which is tangent to the incircle of the triangle ; the circles and are defined similarly. The circles and meet at a point distinct from ; the points and are defined similarly. Prove that the lines and are concurrent at a point on the line .
Solution(r1234)
Firstly, we'll cite a lemma:
Lemma.
Let be a line and be circle.Suppose is the pole of wrt .Now let be inscribed in .Let be the circumcevian triangle of wrt .Then is the perspective axis of and .
Proof of lemma.
Let .
Now applying Pascal's theorem on the hexagon we get are collinear.
Again and are perspective.So are collinear.Hence we conclude that are collinear.But this line is nothing but the polar of i.e .Hence we conclude that
Coming back to the main proof,
Clearly is the radical axis of , is the radical axis of and is the radical axis of .So by radical axis theorem are concurrent.
Let be the incircle of and is its intouch triangle. are the touch points of with respectively.Now let tangents at meets at respectively.Then its easy to show that is the radical axis of and .So is the pole of wrt and similar for others.So concur at the pole of wrt .Now consider the circles .Then by radical axis theorem the lines are concurrent ,say at .Then is the pole of wrt .Since are collinear, their polars i.e are concurrent.So and the triangle formed by the polars of are perspective wrt the perspective axis of .But according to our lemma this perspective axis is the polar of the perspective point of , i.e the radical axis of .So we conclude that concur at the pole of the radical axis of wrt .Since we conclude that the pole of the radical axis of lies on .Hence concur on .
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[/asy]](
As for exercises, I won't give any for this post. I will just write down some references from where you can get plenty of problems. ^_^
References.
Cyclic Quadrilaterals - The Big Picture by Yufei Zhao.
Power of a Point by Yufei Zhao.
Circles by Yufei Zhao.
Poles and Polars by Kin Y. Li.
Mathscope Topic on Poles and Polars by Hoàng Quốc Khánh.
Introduction to the geometry of triangle by Paul Yiu.
This post has been edited 2 times. Last edited by levans, Jan 15, 2016, 4:35 AM
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Thank you very much Potla!! I enjoyed it very much!!!! Anyway, Th1(ii) is maybe typo...
by Bigwood, Mar 23, 2012, 10:46 AM
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by AdiletR, Apr 22, 2014, 5:02 AM
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My blog has moved to Wordpress, so please check here:
by Potla, Jun 26, 2015, 8:18 AM
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Very good post !!
by MathGan, Jul 7, 2016, 7:00 PM
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it's very usefull blog thank you very much
by MELSSATIMOV40, Aug 21, 2024, 2:24 PM
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Compiled problems
Potla
Archives
March 2012
Moving the Blog to Wordpress
Poles and Polars - Another Useful Tool!
December 2011
A digression on Calculus... From introductory to advanced?
March 2011
A nice Functional Equation
February 2011
Harmonic Divisions - A Powerful & Rarely Used Tool!
August 2010
Different maddening numbers
February 2010
Colourings
January 2010
My own problems - all in one place
November 2009
Topic Of Nhocnhoc, Maths.vn
Garden
October 2009
Chebyshev's Inequality
Rearrangement Inequality
August 2009
Iran 1996
Geometry - a powerful technique
June 2009
The Floor function / Box function
May 2009
Huygens Inequality
April 2009
Holder's Inequality
Based on an identity
March 2009
Properties of triangles
Four inequalities of my own!
Titu's Lemma
February 2009
Interesting thing
Arithmetic
A Ridiculous Inequality
Geometric Inequality
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first shouts in 2025!
by NicoN9, Apr 25, 2025, 12:56 PM
dang 2 year bump (can the person who next visits PM me? I want to see when the next person visits)
by roribaki, Jan 6, 2024, 9:18 PM
wholsome
by samrocksnature, Aug 24, 2021, 4:16 PM
wholesome
by centslordm, Jul 2, 2021, 2:08 PM
cool blog
by lneis1, Feb 20, 2021, 6:11 AM
Hello.Very nice blog!
by Functional_equation, Aug 29, 2020, 2:20 PM
Great blog!
by freeman66, Jun 3, 2020, 1:25 AM
Hello. Nice blog!
by mathboy282, Apr 10, 2020, 1:39 AM
Masterpiece
by Kamran011, Mar 18, 2020, 6:35 PM
Hello! Why am I looking through random blogs?
by bingo2016, Nov 21, 2019, 2:40 AM
I am late to find this wonderful gem
by RAMUGAUSS, Nov 13, 2019, 7:32 AM
I WILL NECROPOST
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by Davis1234567890, Sep 15, 2019, 2:21 AM
This is a nice blog and it is very useful!
by arandomperson123, Jun 7, 2016, 2:32 AM
bump,,,,
by ythomashu, Jan 15, 2016, 12:51 AM
Incredible
by DanielL2000, Jul 24, 2014, 9:34 PM
Are you still in school?
by harekrishna, Feb 8, 2010, 6:58 PM
ANNOUNCEMENT
My Regional Tests are tomorrow (29 th Nov.), and after that I have my school exams, so it seems as if this blog will not be updated for around a week. Sorry for the inconvenience.
Sincerely,
Potla
(creator and owner of this blog )
by Potla, Nov 27, 2009, 7:23 PM
Nice. Keep working
by mathson, Oct 6, 2009, 11:29 AM
Not at all.
by Potla, Sep 4, 2009, 10:35 AM
Wow! You are very smart.
by Maybach, Sep 1, 2009, 12:56 AM
good going
by turikachi, May 16, 2009, 3:40 AM
Because of my exams..... I must do some work now.
by Potla, Apr 9, 2009, 9:51 AM
well, why no activity here?
by Ramchandran, Apr 6, 2009, 1:46 PM
Thanks for all of yours' kind( )coordination, but I cannot update this bulky( : ) blog till my exams are over, so I want the contrib to continue their job (I am making peine a contrib too).......
by Potla, Mar 3, 2009, 12:56 PM
Really nice solutions.......
However Iliked ochas and murgis problem more because of its simplicity.
however mathias DK sol is just awe strucking
by UKO, Feb 14, 2009, 4:47 AM
@ mathias DK really your proof is beautifull
I didnot wanted to be rude .never mind.
by murgi, Feb 14, 2009, 4:12 AM
@murgi: I just wanted to provide a more beatiful proof.
by Mathias_DK, Feb 13, 2009, 11:08 PM
I couldnot understand why such acomplicated proof is necessary?
when there exits such a simple proof.
by murgi, Feb 13, 2009, 5:11 PM
From now on, this is the place of problem solvers.
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Advanced Inorganic Chemistry: A Comprehensive Text [5 ed.] 0471849979, 9780471849971
Advanced Inorganic Chemistry: A Comprehensive Text [5 ed.] 0471849979, 9780471849971
This text incorporates many new chemical developments, particularly the more recent theoretical advances in the interpre
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###### Advanced Experimental Inorganic Chemistry 9781032789910 This book is divided into four parts: - Part I deals with Qualitative Inorganic Analysis. Systematic procedures of anion 1,025 127 3MB Read more
###### Advanced Inorganic Chemistry [6 ed.] For more than a quarter century, Cotton and Wilkinson's Advanced Inorganic Chemistry has been the source that stude 4,787 483 54MB Read more
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Author / Uploaded
F. Albert Cotton
Geoffrey Wilkinson
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ri>,yu ADVANCED INORGANIC CHEMISTRY A Comprehensive Text
FIFTH EDITION
RADCLIFFE SCIENCE LIBRARY Parks Road Oxford 0X1 3QP
(2)72851 (renewals)
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ADVANCED INORGANIC CHEMISTRY Fifth Edition F. ALBERT COTTON W. T. DOHERTY-WELCH FOUNDATION DISTINGUISHED PROFESSOR OF CHEMISTRY TEXAS A AND M UNIVERSITY COLLEGE STATION, TEXAS, USA
GEOFFREY WILKINSON SIR EDWARD FRANKLAND PROFESSOR OF INORGANIC CHEMISTRY IMPERIAL COLLEGE OF SCIENCE AND TECHNOLOGY UNIVERSITY OF LONDON, ENGLAND
A WILEY-INTERSCIENCE PUBLICATION
JOHN WILEY & SONS New York • Chichester • Brisbane • Toronto • Singapore
Chemistry, the most cosmopolitan of sciences, the most secret of arts. D’Arcy Wentworth Thompson La Chimie cree son objet. Marcelin Berthelot
Copyright © 1988 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada. Reproduction or translation of any part of this work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Library of Congress Cataloging in Publication Data
Cotton, F. Albert (Frank Albert), 1930Advanced inorganic chemistry: a comprehensive text/F. Albert Cotton and Geoffrey Wilkinson.—5th ed., completely rev. from the original literature, p. cm. “A Wiley-Interscience publication.” Includes bibliographies and index. ISBN 0-471-84997-9 1. Chemistry, Inorganic. I. Wilkinson, Geoffrey Sir 1921II. Title. QD151.2.C68 1988 546—dcl9 Printed in the United States of America 10
987654321
87-20728 CIP
Preface to the Fifth Edition
The main purpose of this book is the same as that of previous editions: to provide the student with the background necessary to comprehend current research literature in inorganic and certain aspects of organometallic chem¬ istry. One of the most evident differences between this and earlier editions of the book is the absence of much theoretical material previously included. There are three reasons for this. From a practical point of view, the continuing rapid growth of the chemistry as such required the addition of important new facts to all of the descriptive material. Since we set great store by keeping this book in the format of one volume, something had to go. We found very little that could be omitted in the older descriptive material, and condensation has its limits if readability is to be maintained. Elimination of theoretical material, or at least severe reduction in the space devoted to it, seemed unavoidable. Second, over the years, much of the theoretical background has found its way into textbooks used at a lower (American undergraduate) level, for example, our own Basic Inorganic Chemistry, 2nd edition (1987). This is true of ligand field theory, spectra and magnetism of transition metal complexes, and simple ideas about chemical bonds. Thus, removal of such material from this book can be regarded as the elimination of unnecessary repetition. Third, in our own minds we have, over the years, become less persuaded of the value of certain types of theorizing. Theories come and go. Twentyfive years ago, when we began writing these books, there were widely accepted ideas about bonding that have by now fallen into disuse, and even disrepute. Thus we felt obliged to make space for facts at the expense of theoretical material. We again thank those who have been kind enough to make suggestions for improvements and welcome any comments on this edition. We remind readers of the guidelines under which references are supplied. These are largely from primary journals from mid-1979 to mid-1987. Many statements are documented by earlier references cited in earlier editions. Where there are two authors, both names are given; when more, the first starred author or in absence of a star, the first is given. We have generally not quoted references to the appropriate sections of the authoritative handbooks namely, Gmelin Handbook of Inorganic Chemistry, v
vi
PREFACE TO THE FIFTH EDITION
Springer Verlag, Berlin, and the Pergamon Press Comprehensive Treatises on Inorganic Chemistry, Coordination Chemistry, and Organometallic Chemistry, which should be familiar to the dedicated student. F. Albert Cotton College Station, Texas
Geoffrey Wilkinson
London, England
Preface to the First Edition
It is now a truism that, in recent years, inorganic chemistry has experienced an impressive renaissance. Academic and industrial research in inorganic chemistry is flourishing, and the output of research papers and reviews is growing exponentially. In spite of this interest, however, there has been no comprehensive text¬ book on inorganic chemistry at an advanced level incorporating the many new chemical developments, particularly the more recent theoretical advances in the interpretation of bonding and reactivity in inorganic compounds. It is the aim of this book, which is based on courses given by the authors over the past five to ten years, to fill this need. It is our hope that it will provide a sound basis in contemporary inorganic chemistry for the new generation of students and will stimulate their interest in a field in which trained personnel are still exceedingly scarce in both academic and industrial laboratories. The content of this book, which encompasses the chemistry of all of the chemical elements and their compounds, including interpretative discussion in the light of the latest advances in structural chemistry, general valence theory, and, particularly, ligand field theory, provides a reasonable achieve¬ ment for students at the B.Sc. honors level in British universities and at the senior year or first year graduate level in American universities. Our expe¬ rience is that a course of about eighty lectures is desirable as a guide to the study of this material. We are indebted to several of our colleagues, who have read sections of the manuscript, for their suggestions and criticism. It is, of course, the authors alone who are responsible for any errors or omissions in the final draft. We also thank the various authors and editors who have so kindly given us per¬ mission to reproduce diagrams from their papers: specific acknowledgements are made in the text. We sincerely appreciate the secretarial assistance of Miss C. M. Ross and Mrs. A. B. Blake in the preparation of the manuscript. F. A. Cotton Cambridge, Massachusetts
G. Wilkinson London, England
Vll
Contents
PART ONE
Survey of Principles
1 2
Concepts in Stereochemistry and Bonding Introduction to Ligands and Complexes
3 35
PART TWO
The Chemistry of the Main Group Elements
3 4 5 6 7 8 9 10 11 12
13 14 15 16
Hydrogen The Group IA(1) Elements: Li, Na, K, Rb, Cs, Fr The Group IIA(2) Elements: Be, Mg, Ca, Sr, Ba, Ra Boron: Group IIIA(13) The Group IIIA(13) Elements: Al, Ga, In, T1 Carbon: Group IYA(14) The Group IVA(14) Elements: Si, Ge, Sn, Pb Nitrogen: Group VA(15) The Group VA(15) Elements: P, As, Sb, Bi Oxygen: Group VIA(16) The Group VIA(16) Elements: S, Se, Te, Po The Group VIIA(17) Elements: F, Cl, Br, I, At The Group VIIIA(18) Elements: He, Ne, Ar, Kr, Xe The Group IIB(12) Elements, Zn, Cd, Hg
87 123 143 162 208 234 265 305 382 444 491 544 587 597
PART THREE
The Chemistry of the Transition Elements
17 18
Survey of the Transition Elements The Elements of the First Transition Series A. Titanium: Group IVB(4), 652
625 649
ix
X
CONTENTS B. C. D. E. F. G. H.
19
The Elements of the Second and Third Transition Series A. B. C. D. E. F. G. H. I.
20 21
Vanadium: Group IVB(5), 665 Chromium: Group VIB(6), 679 Manganese: Group VIIB(7), 697 Iron: Group VIII(8), 711 Cobalt: Group VIII(9), 724 Nickel: Group VIII(IO), 741 Copper: Group IB(ll), 755
776
Zirconium and Hafnium: Group IVB(4), 777 Niobium and Tantalum: Group VB(5), 787 Molybdenum and Tungsten: Group VIB(6), 804 Technetium and Rhenium: Group VIIB(7), 847 The Platinum Group Metals, 868 Ruthenium and Osmium: Group VIII(8), 878 Rhodium and Iridium: Group VIII(9), 900 Palladium and Platinum: Group VIII(10), 917 Silver and Gold: Group IB(ll), 937
The Lanthanides; Also Scandium (IIIB,3) and Yttrium (IIIB,3) Actinium, Thorium, Protactinium, and the Actinide Elements
955 980
PART FOUR
Survey of Selected Areas: Metal Carbonyls, Metal-Metal Bonds and Clusters, Hydrides, Organometallic Compounds, Oxidative-Addition and -Insertion Reactions, Homogeneous Catalysis, Reaction Mechanisms, and Bioinorganic Chemistry 22 23 24 25
Transition Metal Carbon Monoxide Compounds Metal-to-Metal Bonds and Metal-Atom Clusters Transition Metal Compounds with Bonds to Hydrogen Compounds with Transition Metal Single, Double, and Triple Bonds to Carbon
26
Compounds of Transition Metals with Alkenes, Alkynes, and Delocalized Hydrocarbon Systems Oxidative-Addition and Migration (Insertion) Reactions Homogeneous Catalytic Synthesis of Organic Chemicals by Transition Metal Complexes Reaction Mechanisms of Transition Metal Complexes Bioinorganic Chemistry
27 28 29 30
1021 1052 1097 1122 1152 1186 1224 1283 1335
Appendices
1 2
Units and Fundamental Constants Ionization Enthalpies of the Atoms
1379 1381
CONTENTS
3 4 5
Enthalpies of Electron Attachment (Electron Affinities) of Atoms Ionic Radii Basic Concepts of Molecular Symmetry; Character Tables
Index
Xl
1383 1385 1389
1411
1
Abbreviations in Common Use
Chemicals, Ligands, Radicals, etc. Ac acac acacH AIBN am Ar aq ATP
acetyl, CH3CO acetylacetonate anion acetylacetone azoisobutyronitrile ammonia (or occasionally an amine) aryl or arene (ArH) aquated, H20 adenosine triphosphate
9-BBN bipy Bu Bz
9-borabicyclo[3,3 ,l]nonane 2,2'-dipyridine, or bipyridine butyl (n-, i-, s-, or t-\ normal, iso, secondary, or tertiary butyl)
C,lmn COD or cod COT or cot Cp cy
benzyl cryptate ligand with /-, m- and n- membered bridges (cf. p. 52) cycloocta-1,5-diene cyclooctatetraene cyclopentadienyl, C5H5 cyclohexyl
diphos
1,4-diaza-l,3-butadienes, RN=CH—CH=NR 1.2- bis(diethylphosphino)ethane 1.2- bis(diethylphosphino)methane o-phenylenebisdimethylarsine, o-C6H4(AsMe2)2 diethylenetriamine, H2N(CH2CH2NH)2H diethylene glycol dimethyl ether, CH30(CH2CH20)2CH3 {[2,2-dimethyl-l,3-dioxolan-4,5-diyl)bis (methylene)]bis(diphenylphosphine)} any chelating diphosphine, but usually l,2-bis(diphenyl-
DME DMF or dmf
phosphino)ethane, dppe dimethoxyethane (also glyme) A,A'-dimethylformamide, HCONMe2
dab or dad depe depm diars dien diglyme diop
dmg dmgH2 dmpe dmpm
dimethylglyoximate anion dimethylglyoxime 1.2- bis(dimethylphosphino)ethane 1.2- bis(dimethylphosphino)methane xiii
XIV
DMSO or dmso dppe DPPH dppm E EDTAH4 EDTAH£„
en Et Fc Fp
ABBREVIATIONS IN COMMON USE
dimethyl sulfoxide, Me2SO l,2-bis(diphenylphosphino)ethane diphenylpicrylhydrazyl bis(diphenylphosphino)methane electrophile or element ethylenediaminetetraacetic acid anions of EDTAH4 ethylenediamine, H2NCH2CH2NH2 ethyl ferrocenyl (Fc' for substituted Fc) Fe(CO)2Cp
glyme (= DME) hfa HMPA
hexafluoroacetylacetonate anion hexamethylphosphoric triamide, OP(NMe2)3
L
ligand
M
central (usually metal) atom in compound macrocyclic polyether with ra-membered ring and n oxygen atoms (cf. p. 51) methyl mesityl tris-(2-dimethylaminoethyl)amine, N(CH2CH2NMe2)3
m-C-n
Me Mes Me6tren NBD or nbd NBS
ethylene glycol dimethyl ether, CH3OCH2CH2OCH3
NTAH3
norbornadiene N-bromosuccinimide bis-(2-diphenylphosphinoethyl)amine, HN(CH2CH2PPh2)2 tris-(2-diphenylphosphinoethyl)amine, N(CH2CH2PPh2)3 nitrilotriacetic acid, N(CH2COOH)3
OAc oep ox
acetate anion octaethylporphyrin oxalate ion, Q>0^
Pc Ph phen pn PNP (= np2) porph
phthalocyanine phenyl, C6H5 1,10-phenanthroline propylenediamine (1,2-diaminopropane) bis-(2-diphenylphosphinoethyl)amine, HN(CH2CH2PPh2)2 porphyrin (or any porphyrin) tris-(2-diphenylphosphinoethyl)phosphine, P(CH2CH2PPh,), [(Ph3P)2N] + propyl (n- or i-) pyridine pyrazolyl
np2 np3
pp2
PPN + Pr py pz
ABBREVIATIONS IN COMMON USE
XV
QAS
tris-(2-diphenylarsinophenyl)arsine, As(o-C6H4AsPh2)3
QP
tris-(2-diphenylphosphinophenyl)phosphine, P(o-C6H4PPh2)3
R RF
alkyl (preferably) or aryl group perfluoro alkyl group
S sal sal2en or salen
solvent salicylaldehyde bis-salicylaldehydeethylenediimine
TAN TAP TAS
tris-(2-diphenylarsinoethyl)amine, N(CH2CH2AsPh2)3 tris-(3-dimethylarsinopropyl)phosphine, P(CH2CH2CH2AsMe: bis-(3-dimethylarsinopropyl)methylarsine, MeAs(CH2CH2CH2AsMe2)2 tetracyanoethylene 7,7,8,8-tetracyanoquinodimethane A, A, A', A'-tetraethylethylenediamine
TCNE TCNQ teen terpy TFA THF or thf THT or tht tmen tn tol TPN (= np3) TPP tren trien triflate (ion) TSN TSP TSeP TTA X
terpyridine trifluoroacetic acid tetrahydrofuran tetrahydrothiophene A, A,A',A'-tetramethylethylenediamine (alsoTMED, tmed) l,3-diaminopropane(trimethylenediamine) tolyl (CH3C6H4) tris-(2-diphenylphosphinoethyl)amine, N(CH2CFI2PPh2)3 meso-tetraphenylporphyrin tris-(2-aminoethyl)amine, N(CH2CF12NH2)3 triethylenetetraamine, (-CH2NHCH2CH2NH2)2 cf3so3tris-(2-methylthiomethyl)amine, N(CH2CH2SMe)3 tris-(2-methylthiophenyl)phosphine, P(o-C6Fl4SMe)3 tris-(2-methylselenophenyl)phosphine, P(o-C6H4SeMe)3 2-thenoyltrifluoroacetone, C4H3SCOCH2COCF3 halogen or pseudohalogen
Miscellaneous
A asym
angstrom unit, 10”10 m asymmetric or antisymmetric
bcc BM bp
body centered cubic Bohr magneton boiling point
ccp CFSE CFT CIDNP
cubic close packed crystal field stabilization energy crystal field theory chemically induced dynamic nuclear polarization
4
xvi
ABBREVIATIONS IN COMMON USE
cm-1 CT
wave number charge transfer
dec d-
decomposes dextorotatory
ESCA
electron spectroscopy for chemical analysis (= XPE, X-ray photoelectron spectroscopy) electron spin (or paramagnetic) resonance
esr or epr eV
electron volt
FT
Fourier transform (for nmr or ir)
8 (g) h hep HOMO Hz
g-values gaseous state
ICCC ir IUPAC
International Coordination Chemistry Conference infrared International Union of Pure and Applied Chemistry
l(1) LCAO LFSE LFT LUMO
levorotatory liquid state linear combination of atomic orbitals ligand field stabilization energy ligand field theory lowest unoccupied molecular orbital
MAS nmr MO mp
magic angle spinning nmr molecular orbital melting point
NOE nmr
nuclear Overhauser effect nuclear magnetic resonance
PE
photoelectron (spectroscopy)
R
gas constant
(s) SCE SCF SCF-Xa-SW sp or spy str sub sym
solid state saturated calomel electrode self-consistent field self-consistent field, Xa, scattered wave (form of MO theory) square pyramid(al) vibrational stretching mode sublimes symmetrical
tbp
trigonal bipyramid(al)
U uv
lattice energy ultraviolet
VB
valence bond
Planck’s constant hexagonal close packed highest occupied molecular orbital hertz, s'1
ABBREVIATIONS IN COMMON USE
atomic number molar extinction coefficient frequency (cm-1 or Hz) magnetic moment in Bohr magnetons magnetic susceptibility Weiss constant
'
ADVANCED INORGANIC CHEMISTRY A Comprehensive Text
FIFTH EDITION
1
_
_
SURVEY OF PRINCIPLES
4
Chapter One
Concepts in Stereochemistry and Bonding 1-1.
Scope and Purpose of This Chapter
The inorganic chemist is called upon to correlate or rationalize an enormous range of factual data concerning the structures of molecules and extended solid arrays. A principal basis for such correlations is an understanding of chemical bonds. In a general sense, the more precisely and profoundly we can describe the electronic structures of compounds, the more fully and re¬ liably we can predict their structures and properties. All chemists should therefore learn as much about molecular quantum mechanics and its appli¬ cations as their time and talents permit. In practice, however, experimental chemists cannot approach the questions that arise every day in their experi¬ mental work from the point of view of rigorous theory. Rougher and readier approaches must be used. In this chapter we shall very briefly survey some of the main, qualitative, generally accepted concepts that are useful in understanding and assimilating the wealth of factual material to follow in this book. This survey is not intended to provide an introduction of these topics for the uninitiated. 1-2.
Structures of Solids
The great majority of solids of uniform purity and composition fall into one of the following categories: 1. Molecular Solids. These consist of discrete molecules making van der Waals contacts (perhaps also some hydrogen bonds) with one another. While the packing forces in the solid (or liquid) state undoubtedly cause minor perturbations to the structures of the molecules, it is usually satisfactory to ignore them and regard the molecule as an isolated entity. Thus, the solid state structure problem is not effectively different from the molecular structure problem. 2. Metals. These consist of close-packed arrays of atoms that have one or more relatively loosely bound electrons which become delocalized into energy bands. It is this type of delocalized electronic structure that accounts for the
3
4
SURVEY OF PRINCIPLES
characteristic physical properties of metals: good electrical and thermal con¬ ductivities, luster, strength, malleability (in many cases), and so on. From a chemical point of view, we shall note only one facet of the metallic state, namely, the existence in the close-packed structure of interstices, mainly tetrahedral and octahedral, into which small nonmetal atoms (H, B, C, N, etc.) can fit. When a given set of interstices is so occupied we have an interstitial phase or interstitial compound, such as Fe3C, which is essential to the hard¬ ening of steel. These interstitial phases are usually less conducting but harder and more refractory than the parent metals. 3. Covalent Infinite Arrays. Substances of this kind are found among both elements and compounds. Elements that form extended covalent (as opposed to metallic) arrays are boron, all the Group IV (14) elements except lead, also phosphorus, arsenic, selenium, and tellurium. All other elements form either only metallic phases or only molecular ones. Some of these elements, of course, have allotropes of the metallic or molecular type in addition to the phase or phases that are extended covalent arrays. For example, tin has a metallic allotrope (white tin) in addition to that with the diamond structure (gray tin), and selenium forms two molecular allotropes containing Se8 rings, isostructural with the rhombic form and the monoclinic form of sulfur. For tellurium we have a situation on the borderline of metallic behavior. The structures of the principal allotropic forms of all the elements are discussed in detail as the chemistry of each element is treated. For illustrative purposes, we shall mention here only one such structure, the diamond struc¬ ture, since this is adopted by several other elements and is a point of reference for various other structures. It is shown from two points of view in Fig. 1-1. The structure has a cubic unit cell with the full symmetry of the group Td. However, for some purposes it can be viewed as a stacking of puckered infinite
FIG. 1-1. The diamond structure seen from two points of view, (a) The conventional cubic unit cell, (b) A view showing how layers are stacked; these layers run perpendicular to the body diagonals of the cube.
CONCEPTS IN STEREOCHEMISTRY AND BONDING
5
layers. It will be noted that the zinc blende structure (Fig. 1-2) can be regarded as a diamond structure in which one half the sites are occupied by Zn2+ (or other cation) and the other half are occupied by S2_ (or other anion) in an ordered way. In the diamond structure itself all atoms are equivalent, each being surrounded by a perfect tetrahedron of four others. The electronic structure can be simply and fairly accurately described by saying that each atom forms a localized two-electron bond to each of its neighbors.
anions.
6
SURVEY OF PRINCIPLES
As soon as one changes from elements, where the adjacent atoms are identical and the bonds are necessarily nonpolar, to compounds, there enters the vexatious question of when to describe a substance as ionic and when to describe it as covalent. No attempt is made here to deal with this question in detail for the practical reason that, very largely, there is no need to have the answer—even granting, for the sake of argument only, that any such thing as “the answer” exists. Suffice it to say that bonds between unlike atoms all have some degree of polarity and (1) when the polarity is relatively small it is practical to describe the bonds as polar covalent ones, and (2) when the polarity is very high it makes more sense to consider that the substance consists of an array of ions. 4. Ionic Structures. No substance is perfectly ionic; there is always some sharing of electron density between adjacent positive and negative ions (i.e., some covalence). However, many substances can be understood in a practical way by employing the approximation that they consist essentially of arrays of oppositely charged ions. In discussing structures from this point of view it is necessary to have numerical values for the radii of the ions. Such radii can be estimated only by empirical and at least partially equivocal methods. Ap¬ pendix 4 contains a set of radii that are widely used together with a few remarks on how they were derived. Figure 1-2 shows six of the most important structures found among essen¬ tially ionic substances. In an ionic structure each ion is surrounded by a certain number of ions of the opposite sign; this number is called the coordination number of the ion. In the first three structures shown, namely, the NaCl, CsCl, and CaF2 types, the cations have the coordination numbers 6, 8 and 8, respectively. We now ask why a particular compound crystallizes with one or another of these structures. To answer this, we first recognize that ignoring the pos¬ sibility of metastability, which seldom arises, the compound will adopt the arrangement providing the greatest stability, that is, the lowest energy. The factors that contribute to the energy are the attractive force between oppo¬ sitely charged ions, which will increase with increasing coordination number, and the forces of repulsion, which will increase very rapidly if ions of the same charge are “squeezed” together. Thus the optimum arrangement in any crystal should be the one allowing the greatest number of oppositely charged ions to “touch” without requiring any squeezing together of ions with the same charge. The ability of a given structure to meet these requirements will depend on the relative sizes of the ions.) Let us analyze the situation for the CsCl structure. We place eight negative ions of radius r around a positive ion with the radius r+ so that the M+ to stance is r+ + r and the adjacent ions are just touching. Then the A to X distance, a, is given by
CONCEPTS IN STEREOCHEMISTRY AND BONDING
7
or
Now, if the ratio r~/r+ is > 1.37, the only way we can have all eight X- ions touching the M+ ion is to squeeze the X" ions together. Alternatively, if r~/r+ is > 1.37, and we do not squeeze the X“ ions, they cannot touch the M+ ion and a certain amount of electrostatic stabilization energy will be unattainable. Thus, when r~lr+ becomes equal to 1.37, the competition be¬ tween attractive and repulsive forces is balanced, and any increase in the ratio may make the CsCl structure unfavorable relative to a structure with a lower coordination number, such as the NaCl structure. In the NaCl structure, in order to have all ions just touching but not squeezed, with radius r~ for X- and r+ for M+ we have
2r
-
V2(r+ + r )
which gives for the critical radius ratio — = 2.44 If the ratio r~/r+ exceeds 2.44, the NaCl structure becomes disfavored, and a structure with cation coordination number 4, for which the critical value of r"/r+ is 4.44, may be more favorable. To summarize, in this simple ap¬ proximation, packing considerations lead us to expect the various structures to have the following ranges of stability in terms of the r~/r+ ratio: CsCl and CaF2 structures NaCl and rutile structures ZnS structures
1 < r lr+ < 1.37 1.37 < r~/r+ < 2.44 2.44 < r !r+ < 4.44
Obviously, similar reasoning may be applied to other structures and other types of ionic compounds. The idea that crystal structures should be consistent with ionic size ratios, according to geometrical relationships such as those just examined, is a useful one, but it has only semiquantitative validity.1 The radius ratio rules appear to hold about two-thirds of the time. Reasons for frequent failure are the flexibility of ions (i.e., the fact that even for a specified coordination number, the radius is not precisely defined), the role of lattice energies, as well as non-Coulombic factors such as ion polarizabilities and covalence. The more common ionic crystal structures shown in Fig. 1-2 are mentioned repeatedly throughout the text. The rutile structure, named after one mineralogical form of Ti02, is very common among oxides and fluorides of the MF2 and M02 types (e.g., FeF2, NiF2, Zr02, and Ru02), where the radius 'L. C. Nathan, J. Chem. Educ., 1985, 62, 215.
i
8
SURVEY OF PRINCIPLES
ratio favors coordination number 6 for the cation. Similarly, the zinc blende and wurtzite structures, named after two forms of zinc sulfide, are widely encountered when the radius ratio favors four coordination, and the fluorite structure is common when eight coordination of the cation is favored. When a compound has stoichiometry and ion distribution opposite to that in one of the structures just mentioned, it may be said to have an anti structure. Thus compounds such as Li20, Na2S, and K2S, have the antifluorite structure in which the anions occupy the Ca2+ positions and the cations the F“ positions of the CaF2 structure. The antirutile structure is sometimes encountered also. Structures with Close Packing of Anions. Many structures of halides and oxides can be regarded as close-packed arrays of anions with cations in the octahedral and/or tetrahedral interstices. Even the NaCl structure can be thought of in this way (ccp array of Cl“ ions with all octahedral interstices filled), although this is not ordinarily useful. CdCl2 also has ccp Cl~ ions with every other octahedral hole occupied by Cd2+, and Cdl2 has hep l~ ions with Cd2+ ions in half the octahedral holes. It is noteworthy that the CdCl2 and Cdl2 structures (the latter appears in Fig. 1-3) are layer structures. The par¬ ticular pattern in which cations occupy half the octahedral holes, is such as to leave alternate layers of direct anion-anion contact. Corundum, the a form of A1203, has an hep array of oxide ions with twothirds of the octahedral interstices occupied by cations and is adopted by many other oxides (e.g., Ti203, V203, Cr203, Fe203, Ga203, and Rh203). The Bil3 structure has an hep array of anions with two-thirds of the octahedral
FIG. 1-3. A portion of the Cdl2 structure. Small spheres represent metal cations.
CONCEPTS IN STEREOCHEMISTRY AND BONDING
9
holes in each alternate pair of layers occupied by cations, and it is adopted by FeCl3, CrBr3, TiCl3, VC13, and many other AB3 compounds. As indicated, all the structures just mentioned are adopted by numerous substances. The structures are usually named in reference to one of these substances. Thus we speak of the NaCl, CdCl2, Cdl2, BI3, and corundum (or a-Al203) struc¬ tures. Some Mixed Oxide Structures. There is a vast number of oxides (and also some stoichiometrically related halides) having two or more different kinds of cation. Most of them occur in one of a few basic structural types, the names of which are derived from the first or principal compound shown to have that type of structure. Three of the most important such structures are now described. 1. The Spinel Structure. The compound MgAl204, which occurs in Nature as the mineral spinel, has a structure based on a ccp array of oxide ions. Oneeighth of the tetrahedral holes (of which there are two per anion) are occupied by Mg2+ ions and one-half of the octahedral holes (of which there is one per anion) are occupied by Al3+ ions. This structure, or a modification to be discussed later, is adopted by many other mixed metal oxides of the type MIIM2II04 (e.g., FeCr204, ZnAl204, and CoI1Co^II04), by some of the type MivM^04 (e.g., TiZn204 and SnCo204),and by some of the type M^MVI04 (e.g., Na2Mo04 and Ag2Mo04). This structure is often symbolized as A[B2]04, where brackets enclose the ions in the octahedral interstices. An important variant is the inverse spinel structure, B[AB]04, in which half the B ions are in tetrahedral interstices and the A ions are in octahedral ones along with the other half of the B ions. This often happens when the A ions have a stronger preference for octahedral coordination than do the B ions. As far as is known, all MIVM^04 spinels are inverse (e.g., Zn[ZnTi]04), as are many of the MnM^04 ones (e.g., FeIII[CoIIFeIII]04, FeIII[FeIIFeI!I]04, and Fe [NiFe]04), as well. There are also many compounds with disordered spinel structures in which only a fraction of the A ions are in tetrahedral sites (and a corresponding fraction in octahedral ones). This occurs when the preferences of both A and B ions for octahedral and tetrahedral sites do not differ markedly. 2. The llmenite Structure. This is the structure of the mineral ilmenite (FenTilv03). It is closely related to the corundum structure except that the cations are of two different types. It is adopted by AB03 oxides when the two cations, A and B, are of about the same size, but they need not be of the same charge so long as their total charge is +6. Thus in ilmenite itself and in MgTi03 and CoTi03 the cations have charges +2 and +4, whereas in a-NaSb03 the cations have charges of +1 and +5. 3. The Perovskite Structure. The mineral perovskite (CaTi03) has a struc¬ ture in which the oxide ions and the large cation (Ca2+) form a ccp array with the smaller cation (Ti4+) occupying those octahedral holes formed exclusively by oxide ions, as shown in Fig. 1-4. This structure is often slightly distortedin CaTi03 itself, for example. It is adopted by a great many AB03 oxides in
4
10
SURVEY OF PRINCIPLES
INTRODUCTORY TOPICS
•
Small cation
© Large cotion
O
Oxide or halide ion
FIG. 1-4. The perovskite structure.
which one cation is comparable in size to 02~ and the other much smaller, with the cation charges variable so long as their sum is +6. It is found in SrnTiIV03, BanTiIV03, LamGam03, and NaINbv03, and KINbv03, and also in some mixed fluorides (e.g., KZnF3 and KNiF3). 1-3.
Coordination Numbers
The term coordination number normally refers to the number of anions (or ligands) that are nearest to a cation in an ionic solid, or a complex. For each coordination number, there are two or more geometrical arrangements. In this section we survey the important coordination numbers and geometries. We shall consider in detail coordination numbers 2 to 9, discussing under each the principal ligand arrangements. Higher coordination numbers will be discussed only briefly as they occur much less frequently. Coordination Number 2. There are two geometric possibilities, linear and bent. If the two ligands are identical, the general types and their symmetries are linear, L M L, bent, L M—L, C2v- This coordination number is, of course, found in numerous molecular compounds of divalent elements, but is relatively uncommon otherwise. In many cases where stoichiometry might imply its occurrence, a higher coordination number actually occurs because some ligands form “bridges” between two central atoms. In terms of the more conventional types of coordination compound—those with a rather metallic element at the center—it is restricted mainly to some com¬ plexes of Cu1, Ag1, Au1, and Hg11. Such complexes have linear arrangements of the metal ion and the two ligand atoms, and typical ones are [ClCuCll~, [H3NAgNH3] + , and NCHgCN. The gold(I) halides present good examples of linear two coordination; they consist of zigzag chains of the type (1-1) The metal atoms in cations such as [U02]2+, [U02] + , and [Pu02]2+, which are linear, may also be said to have coordination number 2, but these oxo cations interact fairly strongly with additional ligands and their actual coor¬ dination numbers are much higher; it is true, however, that the central atoms
11
CONCEPTS IN STEREOCHEMISTRY AND BONDING
Au
Au
Au
X = Cl, Br
(1-D have a specially strong affinity for the two oxygen atoms. Linear coordination also occurs in the several trihalide ions, such as I3~ and ClBrCL. Coordination Number 3. The two most symmetrical arrangements are planar (1-IIa) and pyramidal (1-IIb), with D3h and C3v symmetry, respectively. Both these arrangements are found often among molecules formed by trivalent central elements. Among complexes of the metallic elements this is a rare coordination number; nearly all compounds or complexes of metal cations with stoichiometry MX3 have structures in which sharing of ligands leads to a coordination number for M that exceeds three. There are, however, a few exceptions, such as the planar Hgl3“ ion that occurs in [(CH3)3S+][HgI-f], the MN3 groups that occur in Cr(NR2)3 and Fe(NR2)3, where R = (CH3)3Si, and various gold(I) complexes, some of which are shown in Fig. 1-5, which are intermediate between two and three coordinate. In a few cases (e.g., C1F3 and BrF3), a T-shaped form (l-III) of three coordination (symmetry C2v) is found.
B—A—B B (l-III)
(1-IIb)
(1-IIa)
This is a highly important coordination number, occurring in hundreds of thousands of compounds, including, inter alia, most of those formed by the element carbon, essentially all those formed by silicon, germanium, and tin, and many compounds and complexes of other elements. There are three principal geometries. By far the most prevalent is tetrahedral geometry (1-IV), which has symmetry Td when ideal. Tetrahedral complexes or molecules are almost the only kind of four-coordinate ones formed by nonCoordination Number 4.2
B B (1-IV)
B
B
B
B
(1-V)
2M. C. Favas and D. L. Kepert, Prog. Inorg. Chem., 1980, 27, 325.
B I
B (1-VI)
B
12
SURVEY OF PRINCIPLES
(d)
(e)
FIG. 1-5. Examples of irregular three coordination of Au1. (a) (Ph3P)Au(S2CNEt2), best regarded as two coordinate; (b) (Ph3PAu)2(C4S4) in which there are two identical Au environments per molecule, (c) [(Ph3P)Au(bipy)] + ; (d) (Me2PhP)2AuSnCl3; (e) [CH2(PPh2)2AuCl]2. [Reproduced by permission from P. G. Jones, Gold Bull., 1981 4, 102 ]
uTT 6 ements; whenever the central atom has no electrons in its valence shell orbitals except the four pairs forming the a bonds to ligands, these bonds are disposed in a tetrahedral fashion. With many transition metal complexes square geometry (1-V) occurs because of the presence of additional valence shell electrons and orbitals (i.e., partially filled d orbitals), although there are also many tetrahedral complexes formed by the transition metals In some cases (e.g., with Ni , Co11, and Cu11 in particular), there may be only a small difference in stability between the tetrahedral and the square arrangement and rapid interconversions may occur. 6
Square complexes are also found with nontransitional central atoms when here are two electron pairs present beyond the four used in bonding; these
ieQab°r,and
and fiPM b6l0W thC Plane °f the molecule- Examples are XeF4 and (IC13)2. Similarly, when there is one “extra” electron pair, as in SF4 the irregular arrangement of symmetry C2o(l-VI) is adopted. Coordination Number 5. Though less common than numbers 4 and 6,
CONCEPTS IN STEREOCHEMISTRY AND BONDING
13
coordination number 5 is still very important.2,3 There are two principal ge¬ ometries, and these may be conveniently designated by stating the polyhedra that are defined by the set of ligand atoms. In one case the ligand atoms lie at the vertices of a trigonal bipyramid (tbp) (1-VII), and in the other at the vertices of a square pyramid (sp) (1-VIII). The tbp belongs to the symmetry group D3h; the sp belongs to the group C4„. It is interesting and highly im¬ portant that these two structures are similar enough to be interconverted
(1-VII)
(1-VIII)
without great difficulty. Moreover, a large fraction of the known five-coor¬ dinate complexes have structures that are intermediate between these two prototype structures. This ready deformability and interconvertibility gives rise to one of the most important types of stereochemical nonrigidity (cf. Section 29-12). On the whole the tbp seems to be somewhat more common than the sp, but there is no general predictive rule. For example the [MC15]3_ ions (M = Cu, Cd, Hg) are tbp, but [InCl5]2- and [T1C15]2~ are sp, and there is one compound that contains both tbp and sp [Ni(CN)5]3~ ions in the same crystal. Pentagonal planar coordination, as in [Te(S2COEt)3]~ where two ligands are bidentate and one monodentate, is very unusual. It seems to be due to the presence of two stereochemically active lone pairs (see Section 1-5 for explanation). Coordination Number 6. This is perhaps the most common coordination number, and the six ligands almost invariably lie at the vertices of an oc¬ tahedron or a distorted octahedron. The very high symmetry, group Oh, of the regular octahedron, is discussed in detail in Appendix 5. There are three principal forms of distortion of the octahedron. One is tetragonal, elongation or contraction along a single C4 axis; the resultant symmetry is only Ddh. Another is rhombic, changes in the lengths of two of the C4 axes so that no two are equal; the symmetry is then only Dlh. The third is a trigonal distortion, elongation or contraction along one of the C3 axes so that the symmetry is reduced to D3d. These three distortions are illustrated in Fig. 1-6. The tetragonal distortion most commonly involves an elongation of one C4 axis and, in the limit, two trans ligands are lost completely, leaving a
3R. R. Holmes, Prog. Inorg. Chem., 1984, 32, 119.
14
SURVEY OF PRINCIPLES
FIG. 1-6. The three principal types of distortion found in real octahedral complexes.
square, four-coordinate complex. The trigonal distortion transforms the oc tahedron into a trigonal antiprism. Another type of six-coordinate geometry, much rarer but nonetheless im¬ portant, is that in which the ligands lie at the vertices of a trigonal prism; the ideal symmetry is D3h. This arrangement has never been observed in a discrete ML6 complex but only in complexes with chelating ligands and in a few metal sulfides, namely, MoS2 and WS2, where it was first seen many years ago, and more recently in MM^S6 (M = Mn, Fe, Co, Ni; M' = Nb Ta) The chelate complexes that best exemplify this type of coordination contain RC(Se)—C(hSe)Re
^
1,2'dlSelenolene
tyPe
ligands,
RC(S)—C(S)R,
Structures lying between the extremes of trigonal prismatic and antiprismatic are sometimes found. As shown in Fig. 1-7, we may define the range o structures according to a twist angle , which is 0° for the prism and 60° for the antiprism. Ligands such as the dithiolenes (1-IX) and the tropolonato anion (1-X), which are somewhat inflexible and have too short a “bite” (the distance between the two atoms bonded to the metal atom) to reach across
CONCEPTS IN STEREOCHEMISTRY AND BONDING
15
FIG. 1-7. A six-coordinate structure intermediate be¬ tween the trigonal prism and antiprism projected down the threefold axis. The twist angle 4> is measured in the plane of projection.
the distance between vertices of an octahedron, sometimes dictate a 4> angle of < 60°.
There are three important geometric arrange¬ ments, as shown in Fig. 1-8. Both experimental data and theory indicate that, except where a bias might arise from the requirements of a particular polydentate ligand, these three structures are of similar stability. Moreover, in¬ terconversions are not likely to be seriously hindered, so that seven-coordinate complexes should be prone to stereochemical nonrigidity, as is often observed. Coordination Number 8.5 There are three especially important idealized structures: the cube (Oh), the square antiprism (DJ), and the triangulated Coordination Number 7.4
FIG. 1-8. The three important geometries for seven coordination, (a) Pentagonal bipyramid (Dih); (b) Capped octahedron (C3„); (c) Capped trigonal prism (
T
2
4D L. Kepert, Prog. Inorg. Chem., 1979, 25, 41. 5D. L. Kepert, Prog. Inorg. Chem., 1978, 24, 179; J. K. Burdett et ai, Inorg. Chem., 1978,17
,
2553.
16
SURVEY OF PRINCIPLES
dodecahedron (D2d)• All three are depicted in Fig. 1-9, which also shows how each of the latter two can be obtained by distortions of the cube. The cube occurs very infrequently in discrete complexes, although it is found in various solid arrays (e.g., the CsCl structure). Since each of the other two structures, which can be so easily obtained from it, allow the same close metal-ligand contacts while alleviating the ligand-ligand repulsions, their energetic supe¬ riority over the cube is understandable. The dodecahedron can be viewed as a pair of interpenetrating tetrahedra: a flattened one defined by the B vertices and an elongated one defined by the A vertices. There are also several nonequivalent sets of edges, such as those marked m in Fig. 1-9. The m edges are generally those spanned when there are four bidentate ligands with a short bite. Detailed analysis of the energetics of M X and X—X interactions suggests that there will in general be little difference between the energies of the square antiprism and the dodecahedral arrangement, unless other factors, such as the existence of chelate rings, energies of partially filled inner shells,' exceptional opportunities for orbital hydridization, or the like, come into play.’ Both arrangements occur quite commonly, and in some cases [e.g., the M(CN)g“ (M = Mo or W;n = 3 or 4) ions] the geometry varies from one kind to the other with changes in the counterion in crystalline salts, on changing from crystalline to solution phases, and on changing the oxidation state of the metal atom [e.g., TaCl4(dmpe)2 is square antiprismatic while TaCl4(dmpe)2+ is dodecahedral].63
FIG. 1-9. The two most important ways of distorting the cube: (a) to produce a square antiprism; (b) to produce a dodecahedron.
6aF. A. Cotton et al., Inorg. Chem., 1983, 22, 770.
CONCEPTS IN STEREOCHEMISTRY AND BONDING
17
A form of eight coordination, which is a variant of the dodecahedral ar¬ rangement, is found in several compounds containing bidentate ligands in which the two coordinated atoms are very close together (ligands said to have a small “bite”), such as N03~ and 02_. In these, the close pairs of ligand atoms lie on the m edges of the dodecahedron (see Fig. 1-9h); these edges are then very short. Examples of this are the Cr(02)4~ and Co(N03)4~ ions and the Ti(N03)4 molecule. Three other forms of octacoordination, which occur much less often and are essentially restricted to actinide and lanthanide compounds, are the hex¬ agonal bipyramid (D6h) (1-XI), the bicapped trigonal prism (D3h) (1-XII) and the bicapped trigonal antiprism (D3d) (1-XIII). The hexagonal bipyramid is restricted almost entirely to the oxo ions, where an OMO group defines the axis of the bipyramid, though it is occasionally found elsewhere. B
(l-Xl)
(l-xil)
Although feasible, there is only one case of isolated geometric isomers. These are cis- and trans-SmI2[0(CH2CH20Me]2]2 where the cis has distorted dodecahedral and the trans, bicapped trigonal antiprismatic geometry.6b Higher Coordination Numbers.7 Of these, only nine displays appreciable regularity of form. The tricapped trigonal prism, shown in Fig. 1-10, is rather common, being found, for example, in the [M(H20)9]3+ ions of the lanthanides and [ReH9]2-. Another idealized structure, which is rarer, is that of a square antiprism capped on one rectangular face. Even higher coordination num¬ bers, 10 to 12, are sometimes found for the largest metal ions. In general, these do not conform to any regular geometry, although for 10 coordina¬ tion a bicapped square antiprism is sometimes found, for example, in K4[Th(02CC02)4(H20)2] • 2H20. A distorted icosahedral arrangement for 12 coordination is found in [Ce(N03)6]2- and [Pr(naph)6]3+, where naph is 1,8naphthyridine. The small bite of these bidentate ligands makes possible the high coordination number. 6bA. Sen et al., Inorg. Chem., 1987, 26, 1821. 7M. C. Favas and D. L. Kepert, Prog. Inorg. Chem., 1981, 28, 309.
18
SURVEY OF PRINCIPLES
FIG. 1-10. The structure of many nine-coordinate complexes.
1-4.
Cage and Cluster Structures
The formation of polyhedral cages and clusters is now recognized as an im¬ portant and widespread phenomenon, and examples may be found in nearly all parts of the Periodic Table. This section mentions each of the principal polyhedra and gives illustrations. Further details may be found under the chemistry of the particular elements and especially in Chapter 23. A cage or cluster is in a certain sense the antithesis of a complex; yet there are many similarities due to common symmetry properties. In each type of structure a set of atoms defines the vertices of a polyhedron, but in the one case the complex—these atoms are each bound to one central atom and not to each other, whereas in the other—the cage or cluster—there is no central atom and the essential feature is a system of bonds connecting each atom directly to its neighbors in the polyhedron. There are, however, some examples of clusters that also have a central atom, sometimes C or N and sometimes a metal atom. Examples of the former are discussed in Section 23-5, while examples of the latter are presented in Sections 23-8 and 19-1-8. F To a considerable extent the polyhedra found in cages and clusters are the same as those adopted by coordination compounds (e.g., the tetrahedron trigonal bipyramid, and octahedron), but there are also others (see especially the polyhedra with six vertices), and cages with more than six vertices are far more common than coordination numbers > 6. It should be noted that
19
CONCEPTS IN STEREOCHEMISTRY AND BONDING
triangular clusters, as in [Re Cl12]3“ or Os (CO)12, though not literally polyhedra, are not essentially different from polyhedral species such as Mo C18+ or Ir (CO)12, respectively. Just as all ligand atoms in a set need not be identical, so the atoms making up a cage or cluster may be different; indeed, to exclude species made up of more than one type of atom would be to exclude the majority of cages and clusters, including some of the most interesting and important ones. Four Vertices. Tetrahedral cages or clusters have long been known for the P4, As4, and Sb molecules and in more recent years have been found in polynuclear metal carbonyls such as Co (CO)12, Ir (CO)12, [ti -C H Fe(CO)]4, RSiCo (CO)9, Fe (CO)i3“, Re (CO) H4, and a number of others; B C1 is another well-known example and doubtless many more will be encountered. Five Vertices. Polyhedra with five vertices are the trigonal bipyramid (tbp) and the square pyramid (sp). Both are found among the boranes and carboranes (e.g., the tbp in B C H and the sp in B H9), as well as among the transition elements. Examples of the latter are Os (CO (tbp) and Fe S (CO (sp). Six Vertices. Octahedral cages and clusters are numerous, especially among the transition metals. Examples are Rh (CO and [Co (CO)14]4-, as well as the metal halide type clusters, M X and M X12, for M = Nb, Ta, Mo, W, and Re. The B H62- and B^He species are also octahedral, although B H is a pentagonal pyramid. Less regular geometries are also known such as the bicapped tetrahedron in Os (CO)i and capped square pyramid in 3
6
3
4
4
4
3
4
4
3
4
5
2
5
12
2
4
5
4
5
5
3
5
)16
)9
6
6
8
)16
6
6
6
6
10
6
8
H2Os6(CO)18.
Such polyhedra are relatively rare. The isoelectronic B H7~ and B QH species have pentagonal bipyramidal (DSh) structures. The Os (CO)2i molecule has a capped octahedron of Os atoms with three CO groups on each Os. Eight Vertices. Eight-atom polyhedral structures are very numerous. By far the most common polyhedron is the cube; this is in direct contrast to the situation with eightfold coordination, where a cubic arrangement of ligands is extremely rare because it is disfavored relative to the square antiprism and the triangulated dodecahedron in which ligand-ligand contacts are reduced. In the case of a cage compound, of course, it is the structure in which contacts between atoms are maximized that will tend to be favored (provided good bond angles can be maintained), since bonding rather than repulsive inter¬ actions exist between neighboring atoms. The only known cases with eight like atoms in a cubic array are the hy¬ drocarbon cubane (C H8) and a few metal atom cluster species such as Cu (z-MNT)r ion [i-MNT = S CC(CN)n. The other cubic systems all involve two different species of atom that alternate as shown in (1-XIV). In all cases either the A atoms or the B atoms or both have appended atoms or groups. The following list collects some of the many cube species, the elements at the alternate vertices of the cube being given in bold type. Seven Vertices.
7
5
7
7
8
8
2
20
SURVEY OF PRINCIPLES
A / A-
-B
B(1-XIV) A (and appended groups) Mn(CO)3 Os(CO)3 PtMe3 or PtEt3 CHjZn T1 T,5-C5H5Fe Me3AsCu PhAI Co(CO)3 FeSR Mo(H2Q)3
B (and appended groups) SEt O Cl, Br, I, OH och3 och3
s I NPh Sb S
s
Although the polyhedron in cubane, or in a similar molecule, may have the full Oh symmetry of a cube, the A4B4-type structures can have at best tetrahedral, Td, symmetry since they consist of two interpenetrating tetrahedra. It must also be noted that only when the two interpenetrating tetrahedra happen to be exactly the same size will all the ABA and BAB angles be equal to 90°. Since the A and the B atoms differ, it is not in general to be expected that this will occur. In fact, there is, in principle, a whole range of bonding possibilities. At one extreme, represented by [('n5-C5H5)Fe(CO)]4, the mem¬ bers of one set of atoms (the Fe atoms) are so close together that they must be considered to be directly bonded, whereas the other set (the C atoms of the CO groups) are not at all bonded among themselves but only to those in the first set. In this extreme, it seems best to classify the system as having a tetrahedral cluster (of Fe atoms) supplemented by bridging CO groups. At the other extreme are the A4B4 systems in which all A—A and B—B distances are too long to admit of significant A—A or B—B bonding; thus the system can be regarded as genuinely cubic (even if the angles differ somewhat from 90°). This is true of most of the systems listed previously The atoms, however, in the smaller of the two tetrahedra tend to have some amount of direct interaction with one another, thus blurring the line of de¬ marcation between the cluster and cage types. A relatively few species are known in which the polyhedron is at least soSesT’ ^‘™"«ulat«cld°decah‘;dron (Fig. l-9b). These are the boron species B8H8 , BeQHg, and B8C18. nertJ1!eS; With nine vertices rare. Representative ones are 9 (in Bi24C126), B9H9 , and B7C2H9, all of which have the tricapped trigonal
CONCEPTS IN STEREOCHEMISTRY AND BONDING
21
prism structure (Fig. 1-10), and Sn ~, which is a square antiprism capped on one square face. Ten Vertices. Species with 10 vertices are well known. In B10Hio and BgQHio the polyhedron (1-XV) is a square antiprism capped on the square faces (symmetry Did). But there is a far commoner structure for 10-atom cages that is commonly called the adamantane structure after the hydrocarbon adamantane (C10H16), which has this structure; it is depicted in 1-XVI and consists of two subsets of atoms: a set of four (A) that lie at the vertices of a tetrahedron and a set of six (B) that lie at the vertices of an octahedron. The entire assemblage has the Td symmetry of the tetrahedron. From other points of view it may be regarded as a tetrahedron with a bridging atom over each edge or as an octahedron with a triply bridging atom over an alternating set of four of the eight triangular faces. 9
(1-XVI)
(l-XV)
The adamantane structure is found in dozens of A4B6-type cage compounds formed mainly by the main group elements. The oldest recognized examples of this structure are probably the phosphorus(III) and phosphorus(V) oxides, in which we have P406 and (0P)406, respectively. Other representative ex¬ amples include P4(NCH3)6, (OP)4(NCH3)6, As4(NCH3)6, and (MeSi)4S6. Eleven Vertices. Perhaps the only known eleven-atom cages are BuH?f and BcAHn. Twelve-atom cages are not widespread but play a dom¬ inant role in boron chemistry. The most highly symmetrical arrangement is the icosahedron (1-XVII), which has 12 equivalent vertices and Ih symmetry. Icosahedra of boron atoms occur in all forms of elemental boron, in and in the numerous carboranes of the B10C2H12 type. A related polyhedron, the cuboctahedron (1-XVIII) is found in several borides of stoiTwelve Vertices.
chiometry MB12.
(l-XVIl)
(1-XVIII)
i
22
SURVEY OF PRINCIPLES
1-5.
The Valence Shell Electron Pair Repulsion (VSEPR) Model8
In the VSEPR model the arrangement of bonds around a central atom is considered to depend on how many valence shell electron pairs, each occu¬ pying a localized one- or two-center orbital, are present, and on the relative sizes and shapes of these orbitals. The first rule is as follows: Rule l. The pairs of electrons in a valence shell adopt that arrangement which maximizes their distance apart; that is, the electron pairs behave as if thev repel each other.9
Each electron pair is assumed to occupy a reasonably well-defined region of space, and other electrons are effectively excluded from this space. For electron pairs in the same valence shell, the arrangements that maximize their distance apart are listed in Table 1-1. To apply rule 1 for the qualitative prediction of molecular shapes where only single bonds and unshared pairs are concerned, we compute the total number of electron pairs, bonding and nonbonding, select the appropriate arrangement in Table 1-1, and assign the electron pairs to it. In the cases of two, three, or four pairs, the results are immediately obvious as shown in Fig. 1-11. For example, an AB4 molecule must be tetrahedral, an AB,E molecule (E represents an unshared pair) must be pyramidal, and an AB-Emolecule must be bent. There are no known exceptions to these predictions. Table 1-2 lists a few molecules and ions of the types AB:E: and AB-.E. sivins the angles. Molecules with five electron pairs usually have trigonal bipyramidal struc¬ tures, as predicted, but when some of these electron pairs are lone pairs (LP s) rather than bonding pairs (BP's), unambiguous prediction requires another rule.
TABLE 1-1 Predicted Arrangements of Electron Pairs in One Valence Shell Number of pairs 2 3 4
Linear
5 6 7
Trigonal bipvramid Octahedron
8 9
I"
Polyhedron defined
Equilateral triangle Tetrahedron
Monocapped octahedron Square antiprism Tricapped trigonal prism
, 1984. 106, 7700.
^Q^C^eDna’ 1. CW £., 1984. 61 771; H. O. DesSevo „ al..
L. S. Bartell and Y. Z. Barshad, J. Am. Chem. Soc.,
23
CONCEPTS IN STEREOCHEMISTRY AND BONDING
8
E
FIG. 1-11. Prediction of the shapes of molecules when the valence shell of the central atom contains two, three, or four electron pairs.
Rule 2. A nonbonding pair occupies more space on the “surface” of the central atom than a bonding pair.
This may also be expressed symbolically by writing LP-LP > LP-BP > BP-BP, meaning that the pair-pair repulsive forces decrease in that order. This is an understandable rule in the sense that the bonding pairs are drawn away from the surface of A by the other atom, B, and thus occupy a smaller solid angle. With this rule we can deal with the AB4E, AB3E2, and AB2E3 cases. Typical examples are shown in Fig. 1-12. The larger size of the LP s means that they rather than the BP’s occupy the equatorial positions, because they are less crowded there. A secondary effect seen in these cases, as well as in the data of Table 1-2, is that B—A—B angles are always less than the ideal (i.e., 109.5, 120, 90°) ones because the LP-BP angles are prone to be larger. For species with six pairs of valence shell electrons on the central atom TABLE 1-2 Angles (deg) in Some AB2E2 and AB3E Molecules
2^2
Molecule
Angle
Molecule
Angle
Molecule
Angle
h2o
104.5 92.2
nh3 ph3 AsH3 SbH3
107.3 93.3
nf3
102.1 97.8 96.2
h2s H2Se H2Te of2 OCl2
91 89.5 103.2 111
91.8 91.3
pf3 AsF3 SbF3
88
Molecule
Angle
PC13 AsC13 SbCl3
100.3 98.7 99.5
4
24
SURVEY OF PRINCIPLES
F F
87.
F
FIG. 1-12. Three molecules representing the AB4E, AB3E2, and AB2E3 types.
the structures of AB5E and AB4E2 type molecules can also be explained by invoking rule 2. In the AB5E type the four equatorial B atoms make angles of _ J >Mr5 / >Md
-Fe
Fer-—
O^O / 1/ 1/ —-Cu-Cu—
/l (2-IV)
(2-V)
(2-VI)
0'
I
''O
I
Mo=Mo
/| (2-VII)
(2-VIII)
The standard notation for bridging ligands employs the descriptors p (eta) and p, (mu). The prefix pn indicates that a ligand is using n of its atoms to form bonds to metal atoms (p is used also for nonbridging ligands, see Sections 2-10 to 2-12.), while the prefix p implies that a ligand bridges two metal atoms. If a ligand bridges three, four, or more metal atoms, p,3, or p,4, and so on, are used. For example, in (2-IV), the chlorine atoms are p,-Cl, while in (2-V) the oxygen atom is p3-0. In (2-VII), the carboxylato ligand is jx,p2 since it bridges two metal atoms (hence p) employing two different oxygen atoms (hence p2). Some unidentate ligands have two or more different donor sites so that the possibility of linkage isomerism arises. Some important ligands of this type, which are called ambidentate ligands, are M—no2 M—ONO m—SCN M—NCS
Nitro Nitrito 5-Thiocyanato V-Thiocyanato
(M—CN (M—NC R2S — M, R2SO - - M
O
Cyano Isocyano S and O-Bonded Dialkyl sulfoxide
The next sections discuss primarily the chemistry of simple donor ligands. Much of this chemistry applies equally well to main group metal ions (e.g. Na+ Ca2+, Ga3+, or Cd2+) and to transition metal ions. It is a chemistry largely of aqua ions, nitrogen donor ligands, such as ammonia or ethylenediamine, and halide ions, and it is chemistry of metal ions in positive oxidation states, usually 2+ and 3 + . Later sections consider complexes that have ir-bonding ligands and also compounds that are called tt complexes, which are those formed by unsatu¬ rated organic molecules. This is a chemistry largely of transition metals, often in formally low oxidation states such as -1,0, and +1. The borderline between rr-bonding and non-u-bonding ligands is bv no means clearly defined. Also the terms “nonclassical” versus “classical” are
INTRODUCTION TO LIGANDS AND COMPLEXES
39
of limited validity, since Werner and his contemporaries studied complexes of cyanide ion and of tertiary phosphines; even pyridine, which they used extensively, is not solely a simple donor. STABILITY OF COMPLEX IONS IN AQUEOUS SOLUTION 2-2.
Aqua Ions
In a fundamental sense metal ions simply dissolved in water are already complexed—they have formed aqua ions. The process of forming in aqueous solution what we more conventionally call complexes is really one of displacing one set of ligands, which happen to be water molecules, by another set. Thus the logical place to begin a discussion of the formation and stability of complex ions in aqueous solution is with the aqua ions themselves. From thermodynamic cycles the enthalpies of plunging gaseous metal ions into water can be estimated and the results, 2 x 102 to 4 x 103 kJ mol 1 (see Table 2-1), show that these interactions are very strong indeed. It is of im¬ portance in understanding the behavior of metal ions in aqueous solution to know how many water molecules each of these ions binds by direct metaloxygen bonds. To put it another way, if we regard the ion as being an aqua complex [M(H20)Jn + , which is then further and more loosely solvated, we wish to know the coordination number x and also the manner in which the x water molecules are arranged around the metal ion. Classical measurements of various types—for example, ion mobilities, apparent hydrated radii, en¬ tropies of hydration—fail to give such detailed information because they cannot make any explicit distinction between those water molecules directly bonded to the metal—the x water molecules in the inner coordination sphere and additional molecules that are held less strongly by hydrogen bonds to the water molecules of the inner coordination sphere. There are, however, ways of answering the question in many instances, ways depending, for the most TABLE 2-1 Enthalpies of Hydration0 of Some Ions (kJ mol ') H+ Li + Na+ K+ Rb + Cs +
-1091 -519 -406 -322 -293 -264
TP Be2+
-473 -326 -2494
Mg2+
-1921
Ag +
Ca2+ Sr2+ Ba2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+
-1577 -1443 -1305 -1904 -1841 -1946 -1996 -2105 -2100 -2046
Cd2+ Hg2+ Sn2+ Pb2+ Al3+ Fe3+ F" CL Br" L
-1807 -1824 -1552 -1481 -4665 -4430 -515 -381 -347 -305
“Absolute values are based on the assignment of -1091 ± 10 kJ moL1 to H+ (cf.H. FHallfwdl and S. C. Nyburg, Trans. Faraday So,, 1963, 59, 1126). Each value probably has an uncertainty of at least lOn kJ mol"1, where n is the charge of the ion.
40
SURVEY OF PRINCIPLES
part, on modern physical and theoretical developments. A few illustrative examples will be considered here. For the transition metal ions, the spectral and, to a lesser degree, magnetic properties depend on the constitution and symmetry of their surroundings. For example, the Co11 ion forms both octahedral and tetrahedral complexes. Thus we might suppose that the aqua ion could be either [Co(H20)6]2+ with octahedral symmetry, or [Co(H20)4]2+ with tetrahedral symmetry. It is found that the spectrum and the magnetism pf Co11 in pink aqueous solutions of its salts with noncoordinating anions such as C104~ or N03“ are very similar to the corresponding properties of octahedrally coordinated Co11 in general, and virtually identical with those of Co11 in such hydrated salts as Co(C104)2 • 6H20 or CoS04-7H20 where from X-ray studies octahedral [Co(H20)6]2+ ions definitely exist. Complementing this, the spectral and magnetic properties of the many known tetrahedral Co11 complexes, such as [CoC14]2“, [CoBr4]2~, [Co(NCS)4]2~, and [py2CoCl2], which are intensely green, blue, or purple, are completely different from those of Co11 in aqueous solution. Thus there is little doubt that aqueous solutions of otherwise uncomplexed Co11 contain predominantly well-defined, octahedral [Co(H20)6]2+ ions, further hy¬ drated, of course. Evidence of similar character can be adduced for many of the other transitional metal ions. For all the di- and tripositive ions of the first transition series, the aqua ions are octahedral [M(H20)6]2(°r 3)+ species although in those of Cr11, Mnm, and Cu11 there are definite distortions of the octahedra because of the Jahn-Teller effect, f Information on aqua ions of the second and third transition series, of which there are only a few, however, is not so certain. It is probable that the coordination is octahedral in many but higher coordination numbers may occur. For the lanthanide ions, M3+(aq) ’ it is certain that the coordination number is higher. For ions that do not have partly filled d shells, evidence of the kind men¬ tioned is lacking, since such ions do not have spectral or magnetic properties related in a straightforward way to the nature of their coordination spheres. We are therefore not sure about the state of aquation of many such ions, although nmr and other relaxation techniques have now supplied some such information. It should be noted that, even when the existence of a welle med aqua ion is certain, there are vast differences in the average length of time that a water molecule spends in the coordination sphere, the so-called mean residence time. For Cr111 and Rh”< this time is so long that when a solution of [Cr(H20)6]J+ in ordinary water is mixed with water enriched in O, many hours are required for complete equilibration of the enriched solvent water with the coordinated water. From a measurement of how many molecules of H20 in the Cr'" and Rh'» solutions fail immediately to exchange w,th the enriched water added, the coordination numbers of these ions by water were shown to be 6. These cases are exceptional, however. Most other STwwlTm a‘S0
qUaWi,il!S °'te,rahedrai CofUOW”; see T. J. Swift, /„„„.
m°'eCUleS “ adOPt
lha‘ « — '» . degeneracy in valence
INTRODUCTION TO LIGANDS AND COMPLEXES
41
aqua ions are far more labile, and a similar equilibration would occur too rapidly to permit the same type of measurement. This particular rate problem is only one of several that are discussed more fully in Section 29-3. Aqua ions are all more or less acidic; that is, they dissociate in a manner represented by the equation [M(H20)J"+ - IM(H20)x_1(0H)t-1(QH)] A
[M(H20)]
The acidities vary widely, as the following KA values show:
M in [M(H20)6]" +
ka
Alni Crm FeIn
1.12 x lO"5 1.26 x 10"4 6.3 x lO'3
Coordinated water molecules in other complexes also dissociate in the same way, for example, [Co(NH3)5(H20)]3+ = [Co(NH3)5(OH)]2+ + H+
K «
[Pt(NH3)4(H20)2]4+ = [Pt(NH3)4(H20)(0H)]3+ + H+
K~ IQ-2
2-3.
10-57
Formation Constants of Complexes
The thermodynamic stability of a complex can be indicated by an equilibrium constant relating its concentration to the concentration of other species when the system has reached equilibrium. The kinetic stability of a species refers to the speed with which transformations leading to the attainment of equilibrium will occur. This section considers problems of thermodynamic stability, that is, the nature of equilibria once they are established. If in a solution containing aquated metal ions M and unidentate ligands L, only soluble mononuclearcomplexes are formed, the system at equilibrium may be described by the following equations and equilibrium constants: M + L - ML
Kx -
[ML] [ML2]
ML + L - ML2
K2 - [ML][L]
ml2 + l - ml3
k3 - [MU][L]
[ML3]
[MLjy] MLjv-i + L — MLjy
[MLw_3][L]
SURVEY OF PRINCIPLES
42
There will be N such equilibria, where N represents the maximum coordi¬ nation number of the metal ion M for the ligand L, and N may vary from one ligand to another. For instance, Al3+ forms A1C14“ and AIF^- and Co2+ forms CoCl^- and Co(NH3)|+, as the highest complexes with the ligands indi¬ cated. Another way of expressing the equilibrium relations follows: [ML]
M + L = ML
[M][L]
M + 2L = ML,
[ML,] ' [M][L]2 [ML3]
M + 3L = ML3
M + NL = ML
[m][l]3 [ML]
' ML]"
Since there can be only N independent equilibria in such a system, it is clear that the K- s and the (3,’s must be related. The relationship is indeed rather obvious. Consider, for example, the expression for (33. Let us multiply both numerator and denominator by [ML][ML,] and then rearrange slightly: = P3
[ML3]
[ML] [ML,]
[M][L]3
[ML] [ML,]
[ML]
[ML,]
[ML3]
[M][L] ‘ [ML][L] ' [ML,][L] It is not difficult to see that this kind of relationship is perfectly general, namely, i=k
P/: — ^1K K ••• Kk — 2
3
n i=
1
u
The Ki s are called the stepwise formation constants (or stepwise stability constants), and the (3,’s are called the overall formation constants (or overall stability constants); each type has its special convenience in certain cases. In all of the previous equilibria we have written the metal ion without specifying charge or degree of solvation. The former omission is obviously of no importance, for the equilibria may be expressed as above whatever the c arges. Omission of the water molecules is a convention that is usually convenient and harmless. It must be remembered when necessary. See for example, the discussion of the chelate effect, in the next section. ith only a few exceptions, there is generally a slowly descending pro¬ gression in the values of the K?s in any particular system. This is illustrated
INTRODUCTION TO LIGANDS AND COMPLEXES
43
by the data for the Cdn-NH3 system where the ligands are uncharged and by the Cdn-CN~ system where the ligands are charged. Cd2+ + nh3
[Cd(NH3)2]2+
K = 10210
[Cd(NH3)2]2+ + nh3
[Cd(NH3)3]2+
K - 10144
[Cd(NH3)3]2+ + nh3
[Cd(NH3)4]2+
K = 10093
Cd2+ + CN"
[Cd(CN)] +
K =
[Cd(CN)] + + CN"
[Cd(CN)2]
K = 10512
[Cd(CN)3]-
K - io4-63
[Cd(CN)4]2-
K =
[Cd(CN)3]- + CN-
o
[Cd(CN)2] + CN"
(3. = 10712)
oo
[Cd(NH3)]2+ + nh3
t-H
K = 102-65
uo
[Cd(NH3)]2+
1Q3.65
(P. = 10188)
Thus, typically, as ligand is added to the solution of metal ion, ML is first formed more rapidly than any other complex in the series. As addition of ligand is continued, the ML2 concentration rises rapidly, while the ML con¬ centration drops, then ML3 becomes dominant, ML and ML2 becoming un¬ important, and so forth, until the highest complex ML, is formed, to the nearly complete exclusion of all others at very high ligand concentrations. These relationships are conveniently displayed in diagrams such as those shown in Fig. 2-1. A steady decrease in Kt values with increasing i is to be expected, provided there are only slight changes in the metal-ligand bond energies as a function of i, which is usually the case. For example, in the Ni2+-NH3 system to be discussed next, the enthalpies of the successive reactions Ni(NH3)I_1 + NH3 = Ni(NH3), are all within the range 16.7 to 18.0 kJ mol-1.
FIG. 2-1. Plots of the proportions of the various complexes Cd(CN)c+ as a function of the ligand concentration. 4
ac = [Cd(CN)c]/total Cd
2ac = 2 [Cd(CN)J/total Cd c=0
[Reproduced by permission from F. J. C. Rossetti, in Modem Coordination Chemistry, J. Lewis and R. G. Wilkins, Eds., Interscience, 1960, p. 10.
SURVEY OF PRINCIPLES
44
There are several reasons for a steady decrease in Kt values as the number of ligands increases: (1) statistical factors, (2) increased steric hindrance as the number of ligands increases if they are bulkier than the H20 molecules they replace, (3) Coulombic factors, mainly in complexes with charged lig¬ ands. The statistical factors may be treated in the following way.2 Suppose, as is almost certainly the case for Ni2+, that the coordination number remains the same throughout the series [M(H20)/v] ••• [M(H20)A,„L„] ••• [ML,]. The [M(H20)Af„L„] species has n sites from which to lose a ligand, whereas the species [M(H20)^„ + 1L„_i] has (TV — n + 1) sites at which to gain a ligand. Thus the relative probability of passing from [M(H20)A,„ + 1L„1] to [M(H20)n„L„] is proportional to (TV — n + 1 )/n. Similarly, the relative probability of passing from [M(H20)A,_nLn] to [M(H20)Af_n_1L„+1] is pro¬ portional to (TV — n)/ (« + 1). Hence on the basis of these statistical con¬ siderations alone, we expect „
_ TV — ft 0+1
N
" ~ ft + 1 "
—
n
+
1
ft
n(N — ft)
" (ft + 1 )(TV - ft + 1)
In the Ni2+-NH3 system (TV = 6), we find the comparison between ex¬ perimental ratios of successive constants and those calculated from the pre¬ vious formula to be as shown in Table 2-2. The experimental ratios are con¬ sistently smaller than the statistically expected ones, which is typical and shows that other factors are also of importance. There are cases where the experimental ratios of the constants do not remain constant or change monotonically; instead, one of them is singularly large or small. There are several reasons for this: (1) an abrupt change in coordination number and hybridization at some stage of the sequence of complexes, (2) special steric effects that become operative only at a certain stage of coordination, and (3) an abrupt change in electronic structure of the metal ion at a certain stage of complexation. Each of these is now illustrated. Values of K^/K2 are anomalously low for the halogeno complexes of mer¬ cury (II); HgX2 species are linear, whereas [HgX4]2+ species are tetrahedral. Presumably the change from sp to sp3 hybridization occurs on going from
TABLE 2-2 Comparison of Experimental and Statistical Formation Constants of Ni2+-NH3 Complexes
k2ik, k3/k2 kjk3 kjka KJK,,
Tor a more elaborate discussion
Experimental
Statistical
0.28 0.31
0.417 0.533 0.562 0.533
"
0.29 0.36
0.2
0.417
see R. Pizer, Inorg. Chem., 1984, 23, 3027.
INTRODUCTION TO LIGANDS AND COMPLEXES
45
HgX2 to [HgX3]~. K3/K2 is anomalously small for the ethylenediamine com¬ plexes of Zn11, and this is believed to be due to the change from tetrahedral to octahedral coordination if it is assumed that [Zn en2]2+ is tetrahedral. For the Ag+-NH3 system K2 > Ku indicating that the linear structure is probably attained with [Ag(NH3)2]+ but not with [Ag(NH3)(H20)3(or 5)] + . With 6,6'-dimethyl-2,2'-bipyridine (2-IX), many metal ions that form tris2,2'-bipyridine complexes form only bis or mono complexes, or, in some cases, no isolable complexes at all, because of the steric hindrance between the methyl groups and other ligands attached to the ion.
H3C
ch3 (2-IX)
In the series of complexes of Fen with 1,10-phenanthroline (and also with 2,2'-bipyridine), X3 is greater than K2. This is because the tris complex is diamagnetic (i.e., the Fe2+ ion has the low-spin state t2g) whereas in the mono and bis complexes, as in the aqua ion, there are four unpaired electrons. This change from the t\ge] to the t\g causes the enthalpy change for addition of the third ligand to be anomalously large because the eg electrons are antibonding. 2-4.
The Chelate and Macrocyclic Effects3
The term “chelate effect” refers to the enhanced stability of a complex system containing chelate rings as compared to the stability of a system that is as similar as possible but contains none or fewer rings. As an example, consider the following equilibrium constants: Ni2+(aq) + 6NH3(aq) = [Ni(NH3)6]2+(aq)
log p = 8.61
Ni2+(aq) + 3en(aq) = [Ni en3]2+(aq)
log P = 18.28
The system [Ni en3]2+ in which three chelate rings are formed is nearly 1010 times as stable as that in which no such ring is formed. Although the effect is not always so pronounced, such a chelate effect is a very general one. To understand this effect, we must invoke the thermodynamic relation¬ ships: AG° = -RT In p A G° = AH° - TAS° Thus P increases as AG° becomes more negative. A more negative AG° can result from making AH° more negative or from making AS° more positive. 3J. J. R. Frausto da Silva, J. Chem. Educ., 1983, 60, 390.
SURVEY OF PRINCIPLES
46
As a very simple case, consider the reactions, and the pertinent thermo¬ dynamic data for them, given in Table 2-3. In this case the enthalpy difference is well within experimental error; the chelate effect can thus be traced entirely to the entropy difference. In the example first cited, the enthalpies make a slight favorable contri¬ bution, but the main source of the chelate effect is still to be found in the entropies. We may look at this case in terms of the following metathesis: [Ni(NH3)6]2+(aq) + 3 en(aq) = [Ni en3]2+(aq) + 6NH3(aq)
log 3 = 9.67
for which the enthalpy change is —12.1 kJ mol-1, whereas — T&S° = —55.1 kJ mol-1. The enthalpy change corresponds very closely to that expected from the increased crystal field stabilization energy of [Ni en3]2+, which is estimated from spectral data to be —11.5 kJ mol-1 and can presumably be so explained. As a final example, which illustrates the existence of a chelate effect despite an unfavorable enthalpy term, we may use the reaction [Ni en2(H20)2]2+(aq) + tren(aq) = [Ni tren(H20)2]2+(aq) + 2 en(aq) log 3 = 1.88
[tren = N(CH2CH2NH2)3]
For this reaction we have AH° = +13.0, -TAS° = -23.7, and AG° = -10.7 (all in kJ mol-1). The positive enthalpy change can be attributed both to greater steric strain resulting from the presence of three fused chelate rings in Ni tren, and to the inherently weaker M—N bond when N is a tertiary rather than a primary nitrogen atom. Nevertheless, the greater number of chelate rings (3 vs. 2) leads to greater stability, owing to an entropy effect that is only partially canceled by the unfavorable enthalpy change. Probably the main cause of the large entropy increase in each of the three cases we have been considering is the net increase in the number of unbound molecules ligands per se or water molecules. Thus although 6 NH3 displace 6 H20, making no net change in the number of independent molecules, it takes only 3 en molecules to displace 6 HzO. Another more pictorial way to look at the problem is to visualize a chelate ligand with one end attached to the metal ion. The other end cannot then get very far away, and the probTABLE 2-3 Two Reactions Illustrating a Purely Entropy-Based Chelate Effect Cd2+(aq) + 4CH3NH2(aq) = [Cd(NH2CH3)4]2+(aq)
log p = 6.52
Cd2+(aq) + 2H2NCH2CH2NH2(aq) = [Cd(en)2l2+(aq)
log p = 10.6
Ligands
AH0 (kJ moL1)
AS° (J moL1 deg’1)
TA5° (kJ moLb
4CH3NH2 2 en
57-3 -56.5
—67.3 +14.1
20.1 42
4D. R. Rosseinsky, J. Chem. Soc. Dalton Trans., 1979, 732.
AG° (kJ mol'1) -37.2 -60.7
INTRODUCTION TO LIGANDS AND COMPLEXES
47
ability of it, too, becoming attached to the metal atom is greater than if this other end were instead another independent molecule, which would have access to a much larger volume of the solution. The latter view provides an explanation for the decreasing magnitude of the chelate effect with increasing ring size, as illustrated by data such as those shown below for copper complexes of H2N(CH2)2NH2 and H2N(CH2)3NH2(tn): [Cu en2]2+(aq) + 2tn(aq) = [Cu tn2]2+(aq) + 2 en(aq)
log 3 = -2.86
Of course, when the ring that must be formed becomes sufficiently large (seven membered or more), it becomes more probable that the other end of the chelate molecule will contact another metal ion than that it will come around to the first one and complete the ring. Table 2-4 summarizes the factors influencing the stabilities of complexes. The Macrocyclic Effect. This term refers to the greater thermodynamic stability of a complex with a cyclic polydentate ligand when compared to the complex formed by a comparable noncyclic ligand. A representative com¬ parison would be between the following pair:
L,
L2
The formation of zinc (II) complexes by these two ligands, that is, the reactions (2-1) have been shown5 to have the following thermodynamic Zn2+(aq) + L,(aq)- ZnL?+(aq)
(2-1)
parameters:
Li log K -AH° (kJ moL1) AS° (J deg'1 mol'1)
11.25 44.4 66.5
U 15.34 61.9 85.8
The gross macrocyclic effect is evident in the increase of 4.09 units in log K. It can be seen that the overall effect is of both enthalpic and entropic origin. The relative importance of these two contributions varies from case 5M. Micheloni and P. Paoletti, Inorg. Chim, Acta, 1980, 43, 109.
SURVEY OF PRINCIPLES
48
TABLE 2-4 Factors Influencing Solution Stability of Complexes'1 Enthalpy effects
Entropy effects Number of chelate rings
Variation of bond strength with electronegativities of metal ions and
Size of chelate ring
ligand donor atoms Ligand field effects Steric and electrostatic repulsion between ligands in complex Enthalpy effects related to conformation of uncoordinated ligand Other Coulombic forces involving chelate ring formation Enthalpy of solution of ligands Change in bond strength when ligand is charged (same donor and acceptor atom)
Changes of solvation on complex formation Arrangement of chelate rings Entropy variations in uncoordinated ligands Effects resulting from differences in configurational entropies of the ligand in complex compound Entropy of solution of ligands Entropy of solution of coordinated metal ions
“From R. T. Meyers, Inorg. Chem., 1978, 17, 952.
to case (see ref. 5 for a selection of comparisons), with enthalpy more often providing the dominant contribution. Cryptate ligands afford a similar enhancement of stability in some cases, but the availability of many ligand conformations makes this a much more complicated situation.6 Some cryptate selectivity effects for alkali and alkaline earth metal ions are mentioned in Chapters 4 and 5.
TYPES AND CLASSIFICATION OF LIGANDS 2-5.
Multi- or Polydentate Ligands
Regardless of whether Tr-bonding is involved, ligands can have various denticities, and we now illustrate some of the more important types. Bidentate Ligands. These are very common and can be classified ac¬ cording to the size of the chelate ring formed as in the following examples:
Three-membered
M
X? X)
6H.-J. Buschmann, Inorg. Chim. Acta, 1985, 98, 43; 1985, 102, 95
INTRODUCTION TO LIGANDS AND COMPLEXES
Four-membered
^'CMe
H2
Five-membered
M
M'
R\
/,N^ CH2
/
M Vch2 h2
Six-membered
R
TCNR2
c=cI
s
cC/R
R H €=N
R
^CH R
Tridentate Ligands.
49
O—M
Some are obligate planar such as
Acylhydrazones of salicylaldehyde
and many similar ones where maintenance of the it conjugation markedly favors planarity. Such ligands must form complexes of the types (2-X) or (2XI).
There are also many flexible tridentate ligands such as .CH2CH2NH2 HN^ XH2CH2NH2 Diethylenetriamine (dien)
50
SURVEY OF PRINCIPLES
(CH2)3-As(CH3)2 CH3As^
(CH2)3-As(CH3)2 Bis(3-dimethylarsinylpropyl)methylarsine (triars)
which are about equally capable of meridional (2-X and 2-XI) and facial (2XII) coordination.
(2-XII)
Quadridentate Ligands.
There are four main types:
Open-Chain, Unbranched
H2N(CH2)2NH(CH2)2NH(CH2)2NH2 Triethylenetetramine (trien)
Schiff bases derived from acetylacetone, for example, H3C^
yCH2)2
C—N HCf
c-=o~
CH3 ^N-C S)nCH
//
"O-C
\
CH:,
Tripod Ligands. These are of the type X(—Y)3, where X is nitrogen phosphorus, or arsenic, the Y groups are R2N, R2P, R2As, RS, or RSe, and the connecting chains ( ) are (CH2)2, (CH2)3, or o-phenylene. Some common ones are N(CH2CH2NH2)3
tren
N[CH2CH2N(CH3)2]3
Meetren np3
N[CH2CH2P(C6H5)2]3 P[o-C6H4P(C6H5)2]3 N(CH2CH2SCH3)3 As[o-C6H4As(C6H5)2]3
qp xsn QAS
The tripod ligands are used particularly to favor formation of trigonal-bipvrami a comp exes of divalent metal ions, as shown schematically in (2-XIII) ut they do not invariably give this result. For instance, whereas Ni(np3)I2 is tngonal-bipyramidal, Co(np3)I2 is square pyramidal (2-XIV).
INTRODUCTION TO LIGANDS AND COMPLEXES
(2-XIII)
51
(2-XIV)
Macrocyclic. These may be (a) planar with unsaturated rings as in por¬ phyrin (2-XV) and its derivatives, although, as discussed later in more detail (Section 10-12) the metal atom may be out of the plane of the nitrogen donor atoms, or (b) puckered with saturated rings as in the macrocycles (2-XVI) and (2-XVII).
u
(2-XVII)
(2-XVI)
Pentadentate and Higher-Dentate Ligands. Perhaps the best known hexadentate ligand is ethylenediaminetetraacetate (EDTA4“), which can also be pentadentate as EDTAH3". CH,COO“
"OOCCH,
NCH,CH,N^
^NCH,CH,N
cH,axr
“OOCCH, (EDTA)4-
CH,COOH
"OOCCH,
CH2COO~
"OOCCH, (EDTAH)-
Other important multidentate ligands are the crown ethers and cryptates. These are cyclic and polycyclic ligands that form their most important com¬ plexes with alkali and alkaline earth ions. The macrocyclic polyethers, com¬ monly called crown ethers, are typified by (2-XVIII) and (2-XIX). Since the systematic names for such ligands are very unwieldy, a special nomenclature is used, in which (2-XVIII) and (2-XIX) are called, respectively, 15-crown-5 or 15-C-5 and dibenzo-18-crown-6. These examples should serve to show the rules for the simple nomenclature. Crown ethers with as many as 10 oxygens are known and several [e.g., (2-XIX)] are commercially available.
(2-XVIII)
(2-XIX)
52
SURVEY OF PRINCIPLES
The cryptates are bicyclic species, most of which have the general formula (2-XX). Again a simplified code for naming them is a practical necessity. They are called “cryptate-mmn, ” where m and n are as defined in (2-XX). One of the commonest is cryptate-222.
n^^o tt(NO) inter¬ actions is thus made up of three electrons from M and one from NO. In effect, NO contributes three electrons to the total bonding configuration under circumstances where CO contributes only two. Thus for purposes of formal electron “bookkeeping,” the ligand NO can be regarded as a three-electron donor in the same sense as the ligand CO is considered a two-electron donor. This leads to the following very useful general rules concerning stoichiometry, which may be applied without specifically allocating the difference in the number of electrons to any particular (i.e., ct or tt) orbitals: 1 Compounds isoelectronic with one containing an M(CO)„ grouping are those containing M'(CO)„_,(NO), M"(CO)„_2(NO)2, and so on, where M', M" and so on, have atomic numbers that are 1, 2, ... , and so on, F4 ligand suggest that the carbon atoms bound to the metal approach tetrahedral hybridization. Indeed, it is possible to formulate the bonding as involving two normal 2c-2e metal-carbon bonds in a metallacycle (2-XXXIa) with approximately sp3 hybridized carbon. A number of other molecules that have multiple bonds and can be bound to metals in the tp fashion can be regarded as forming metallacycles (2-XXXIb) through (2XXXIe).
(2-XXXIa)
(2-XXXIb)
(2-XXXIc)
(2-XXXId)
(2-XXXIe)
31S. Miya and K. Saito, Inorg. Chem., 1981, 20, 287. 32N. Kasai et al., J. Am. Chem. Soc., 1983,105, 2482;
6, 1211.
M. F. Rettig etal., Organometallics, 1987,
SURVEY OF PRINCIPLES
74
Actually, the metallacycle view and the -n-donor view are neither incom¬ patible nor mutually exclusive but are complementary, with a smooth gra¬ dation of one description into the other. The one to be preferred in any given case depends on the extent to which the double bond of the ligand has been reduced to a single bond. From a formal point of view, however, the metal¬ lacycle view entails a problem with oxidation state. For example, a compound such as Ni(C2F4)(CO)3 could be regarded as a nickel(II) rather than a nickel(O) complex. Clearly, in a compound such as Pt(C2H4)3, it would be absurd to propose Pt^. It is best to regard molecules bound sideways as .neutral ligands that do not alter the formal oxidation state. Conjugated Alkenes. When two or more conjugated double bonds are engaged in bonding to a metal atom the interactions become more complex, though qualitatively the two types of basic, synergic components are involved. The case of the 1,3-butadiene unit is an important one and shows why it would be a drastic oversimplification to treat such cases as simply collections of separate monoolefin-metal interactions. Two extreme formal representations of the bonding of 1,3-butadiene to a metal atom are possible (Fig. 2-11). The structure (b) would imply that bonds 1-2 and 3-4 should be longer than bond 2-3. In C4H6Fe(CO)3 the bond lengths are approximately the same and 13C—H coupling constants in the nmr spectra indicate that the hybridization at carbon still approximates to sp2. However, in some other compounds of conjugated cyclic alkenes, the pattern is of the long-short-long type, indicating some contribution from this extreme structure.
(b) FIG. 2-11. Two extreme formal representations of the bonding of a 1,3-butadiene group to a metal atom, (a) implies that there are two more or less independent monoolefin metal inter¬ actions; (b) depicts J- Chem. Soc. Chem. Commun., 1987, 606; M. Kaneko etal, ibid.,
An,.
Ch‘m'
^- 1987' “• 2411 A' L Fr“k
HYDROGEN
89
These systems are discussed in detail in Chapters 24, 27, and 28 and their use in hydrogenation of organic substances described. Some transition metal com¬ plexes of dihydrogen are also known (Section 24-7). In many chemical reactions, H2 is evolved. Gas phase Raman spectroscopy provides an easy method for detection of H2, HD, and D2 in submicromolar quantities.2 3-2.
The Bonding of Hydrogen
The chemistry of hydrogen depends mainly on three electronic processes: 1. Loss of the Valence Electron. The Is valence electron may be lost to give the hydrogen ion H + , which is merely the proton. Its small size (,r ~ 1.5 x 10“13 cm) relative to atomic sizes (r ~ 10~8 cm) and its small charge result in a unique ability to distort the electron cloud surrounding other atoms; the proton accordingly never exists as such, except in gaseous ion beams; in condensed phases it is invariably associated with other atoms or molecules. 2. Acquisition of an Electron. The hydrogen atom can acquire an electron, attaining the Is2 structure of He, to form the hydride ion H~. This ion exists as such essentially only in the saline hydrides formed by electropositive metals (Section 3-13). 3. Formation of an Electron-Pair Bond. The majority of hydrogen com¬ pounds contain an electron-pair bond. The number of carbon compounds of hydrogen is legion, and most of the less metallic elements form numerous hydrogen derivatives. The chemistry of many of these compounds is highly dependent on the nature of the element (or the element plus its other ligands) to which hydrogen is bound. Particularly dependent is the degree to which compounds undergo dissociation in polar solvents and act as acids: HX
H+ + X
Also important for chemical behavior is the electronic structure and coor¬ dination number of the molecule as a whole. This is readily appreciated by considering the covalent hydrides BH3, CH4, NH3, OH2, and FH. The first not only dimerizes (to be discussed) but is a Lewis acid, methane is chemically unreactive and neutral, ammonia has a lone pair and is a base, water can act as a base or as a very weak acid, and FH is appreciably acidic in water. Except in H2 itself, where the bond is homopolar, all other H X bonds possess polar character to some extent. The dipole may be oriented either way, and important chemical differences arise accordingly. Although the term “hydride” might be considered appropriate only for compounds with H neg¬ ative, many compounds that act as acids in polar solvents are properly termed
2
R. J. H. Clark et al.,
J.
Am. Chem. Soc., 1985,107, 2626.
90
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
covalent hydrides. Thus although HC1 and HCo(CO)4 behave as strong acids in aqueous solution, they are gases at room temperature and are undissociated in nonpolar solvents. 4. Unique Bonding Features. The nature of the proton and the complete absence of any shielding of the nuclear charge by electron shells allow other forms of chemical activity that are either unique to hydrogen or particularly characteristic of it. Some of these are the following, which are discussed in some detail subsequently: (a) The formation of numerous compounds, often nonStoichiometric, with metallic elements. They are generally called hydrides but cannot be regarded as simple saline hydrides. (b) Formation of hydrogen bridge bonds in electron-deficient com¬ pounds such as in (3-1) or transition metal complexes as in (3-II). 0
o
C .c°
' v/\v '■ H
H" N/ (3-1)
c
.0
OC—Cr'—H—Or—CO
0°
f C 0
r I CF c
0
(3-H)
The bridge bonds in boranes and related compounds are discussed in Chap¬ ter 6; the transition metal compounds are described in detail in Chapter 24. (c) The Hydrogen Bond. This bond is important not only because it is essential to an understanding of much other hydrogen chemistry but also because it is one of the most intensively studied examples of intermolecular attraction. Hydrogen bonds dominate the chemistry of water, aqueous solutions, hydroxylic solvents, and OH-containing species generally, and they are of crucial importance in biological systems, being responsible inter alia for the linking of polypeptide chains in proteins and base pairs in nucleic acids.
THE HYDROGEN BOND, HYDRATES, HYDROGEN IONS, AND ACIDS 3-3.
The Hydrogen Bond3
The concept of the hydrogen bond is due to Huggins, Latimer, and Rodebush in Berkeley in 1920. It is the term given to the relatively weak secondary
.M,
° J°^en’ J\ Chem- Educ > 1982, 59, 362 and references quoted; Hydrogen Bonds
i 3S9e(gas phased"mer„rW ^
C U C—C > secondary CH §> primary CH. Even molecular H2 may be protonated, since H2-D2 exchange is observed in the superacids, probably through a planar H3+ transition state. The reactions possibly involve pentacoordinate carbonium ions, for example, where the two hydrogen atoms are bound to carbon by closed three-center bonds. .H R.CH + H
r,c:
CR3+ + H'
‘
H
The carbonium ions can undergo complex reactions. Thus methane can give carbonium ions with C—C bonds by condensation reactions of the type: CH 4
H+ ^
H, + CH,
CH.
C,H 7
H, + C,Hs+
and so on
CH3+ ions have been detected by trapping with CO and subsequent hydrolysis of the acylium ion with water to give acetic acid, for example, ch3+ + CO — ch3co+
ch3co2h + H+
The carbonylation of aromatics is also catalyzed by super acids.22
21J. Derynck
et al., Chem. Rev., 1982, 82, 591; M. Bassir et al, New J. Chem., 1987, 11, 437.
CH3F—SbF5 in S02F2 will methylate N20 to give CH30—N=N + which is stable below - 30°C G. A. Olah et al., J. Am. Chem. Soc., 1986, 108, 2054. 22G. Olah etal., J. Org. Chem., 1985, 50, 1436.
111
HYDROGEN
The halogen-acid-Al2X6 systems are similar to HF—SbF5.23 Either can isomerize alkanes and catalyze hydrocarbon acetylation or alkylation. 3-12.
Properties of Some Common Strong Acids
A collection of some properties of the more common and useful pure strong acids is found in Table 3-6. Hydrogen Fluoride. The acid is made by the action of concentrated H2S04 on CaF2 and is the principal source of fluoride compounds (Chapter 14). It is available in steel cylinders, with purity approximately 99.5%; it can be purified further by distillation. Although liquid HF attacks glass rapidly, it can be handled conveniently in apparatus constructed either of copper or Monel metal or of materials such as polytetrafluoroethylene (Teflon or PTFE) and Kel-F (a chlorofluoro polymer). The high dielectric constant is characteristic of hydrogen-bonded liquids. Since HF forms only a two-dimensional polymer, it is less viscous than water. In the vapor, HF is monomeric above 80°C, but at lower temperatures the physical properties are best accounted for by an equilibrium between HF and a hexamer, (HF)6, which has a puckered ring structure. Crystalline (HF)„ has zigzag chains (Fig. 3-2). After water, liquid HF is one of the most generally useful solvents. Indeed in some respects it surpasses water as a solvent for both inorganic and organic compounds, which often give conducting solutions as noted previously; it can also be used for cryoscopic measurements. The self-ionization equilibria in liquid HF are 2HF F" + HF
H2F+ + F“ HF2"
HF
K ~ 1010
H2F3"
and so on
TABLE 3-6 Properties of Some Strong Acids in the Pure State Acid HF HC1 HBr HI hno3 hcio4 HS03F h2so4
mp (°C) -83.36 -114.25 -86.92 -50.85 -41.59
-112 -88.98 10.3771
bp (°C) 19.74 -85.09 -66.78 -35.41 82.6 (109°C extrap.) 162.7 -270 d“
k“
(temperature, °C)
1.6
X
10-6 (0)
3.5 1.4 8.5 3.77
X
10-9 (-85) 10-10 (-84) 10-10 (-45) 10“2 (25)
1.085 1.044
X
X X X
X
10-4 (25) 10~2 (25)
tb (temperature, °C)
84 (0) 14.3 (-114) 7.33 (-86) 3.57 (-45)
-120 (25) 110 (20)
“Specific conductance (ohm-1 cnr1)- Values are often very sensitive to impurities. ^Dielectric constant divided by that of a vacuum. “Constant-boiling mixture (338°C) contains 98.33% of H2SQ4; d = with decomposition.
MD. FSrca§iu, Acc. Chem. Res., 1982, 15, 46.
112
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
The formation of the stable hydrogen-bonded anions accounts in part for the extreme acidity. In the liquid acid the fluoride ion is the conjugate base, and ionic fluorides behave as bases. Fluorides of M+ and M2+ are often appre¬ ciably soluble in HF, and some such as T1F are very soluble. The only substances that function as “acids” in liquid HF are those such as SbFs previously noted, which increase the concentration of H2F+. The latter ion appears to have an abnormally high mobility in such solutions. Reactions in liquid HF are known that also illustrate amphoteric behavior, solvolysis, or complex formation. Although HF is waterlike, it is not easy, because of the reactivity, to establish an emf series, but a partial one is known. In addition to its utility as a solvent system, HF as either liquid or gas is a useful fluorinating agent, converting many oxides and other halides into fluorides. Hydrogen Chloride, Bromide, and Iodide. These acids are quite similar but differ from HF; they are normally pungent gases; in the solid state they have hydrogen-bonded zigzag chains and there is probably some hydrogen bonding in the liquid. Hydrogen chloride is made by the action of concentrated H2S04 on concentrated aqueous HC1 or NaCl; HBr and HI may be made by catalytic reaction of H2 + X2 over platinized silica gel or, for HI, by inter¬ action of iodine and boiling tetrahydronaphthalene. The gases are soluble in a variety of solvents, especially polar ones. The solubility in water is not exceptional; in moles of HX per mole of solvent at 0°C and 1 atm the solu¬ bilities in water, 1-octanol, and benzene, respectively, are HC1, 0.409, 0.48, 0.39; HBr, 1.00, 1.30, 1.39; HI, 0.065, 0.173, 0.42. The self-ionization is very small: 3HX
H2X+ + HX2“
Liquid HC1 has been fairly extensively studied as a solvent, and many organic and some inorganic compounds dissolve giving conducting solutions: B + 2HC1 ^=± BH+ + HC12The low temperatures required and the short liquid range are limitations, but conductimetric titrations are readily made. Salts of the ion H2C1+ have not been isolated, but salts of the HCl^ and HBr2- ions, which have X—H—X distances of 3.14 and 3.35 A, respectively, are not uncommon. These distances, like that in HF2 (2.26 A), are ~0.5 A shorter than the sum of van der Waals radii and suggest that there are strong hydrogen bonds.
Nitric Acid.24 Nitric acid is made industrially by oxidation of ammonia with air over platinum catalysts. The resulting nitric oxide (Section 10-6) is absorbed in water in the presence of air to form N02, which is then hydrated. The normal concentrated aqueous acid (—70% by weight) is colorless but 24C. C. Addison,
Chem. Rev., 1980, 80, 21.
HYDROGEN
113
often becomes yellow as a result of photochemical decomposition, which gives N02: 2HN03
2N02 + H20 + |02
The so-called fuming nitric acid contains dissolved NOz in excess of the amount that can be hydrated to HN03 + NO. Pure nitric acid can be obtained by treating KN03 with 100% H2S04 at 0°C and removing the HN03 by vacuum distillation. The pure acid is a col¬ orless liquid or white crystalline solid; the latter decomposes above its melting point according to the previous equation for the photochemical decomposi¬ tion, hence must be stored below 0°C. The acid has the highest self-ionization of the pure liquid acids. The initial protolysis 2HN03
H2N03+ + NOf
is followed by rapid loss of water: h2no3+
h2o + no2+
so that the overall self-dissociation is 2HN03
N02+ + NOf + HzO
Pure nitric acid is a good ionizing solvent for electrolytes, but unless they produce the N02+ or NOf ions salts are sparingly soluble. In dilute aqueous solution, nitric acid is approximately 93% dissociated at 0.1 M concentration. Nitric acid of concentration below 2 M has little oxidizing power. The concentrated acid is a powerful oxidizing agent and, of the metals, only Au, Pt, Ir, and Re are unattacked, although a few others such as Al, Fe, Cu are rendered passive, probably owing to formation of an oxide film; magnesium alone can liberate hydrogen and then only initially from dilute acid. The attack on metals generally involves reduction of nitrate. Aqua regia (~3 vol. of cone. HC1 + 1 vol. of cone. HN03) contains free chlorine and C1NO, and it attacks gold and platinum metals, its action being more effective than that of HN03 mainly because of the complexing function of chloride ion. Red fuming nitric acid contains N204.25 Some metals, notably tantalum, are quite resistant to HN03 but dissolve with extreme vigor if HF is added, to give TaFf or similar ions. Nonmetals are usually oxidized by HN03 to oxo acids or oxides. The ability of HN03, especially in the presence of con¬ centrated H2S04, to nitrate many organic compounds, is attributable to the formation of the nitronium ion, NOf (Section 10-6). Gaseous F1N03 has a planar structure (Fig. 3-7), although hindered rotation of OH relative to N02 probably occurs.
25M. F. A. Dove
et al., J. Chem. Soc. Dalton Trans., 1985, 2551.
114
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
1.206 A
FIG. 3-7. The structure of nitric acid in the vapor.
Perchloric Acid. This is available in concentrations 70 to 72% by weight. The water azeotrope with 72.5% of HC104 boils at 203°C, and although some Cl2 is produced, which can be swept out by air, there is no hazard involved. The anhydrous acid is best prepared by vacuum distillation of the concentrated acid in the presence of the dehydrating agent Mg(C104)2; it reacts explosively with organic material. The pure acid is stable at room temperature for only 3 to 4 days, decomposing to give HC104H20 (84.6% acid) and C1207. The most important applications of aqueous HC104 involve its use as an oxidant. However, at concentrations below 50% and temperatures not ex¬ ceeding 50 to 60°C, there is no release of 02. The hot concentrated acid oxidizes organic materials vigorously or even explosively; it is a useful reagent for the destruction of organic matter, especially after pretreatment with, or in the presence of, H2S04 or HN03. The addition of concentrated HC104 to organic solvents such as ethanol should be avoided where possible, even if the solutions are chilled. Perchlorate salts of organic or organometallic cations or complexes with organic ligands must be handled with extreme caution; substitution of C104~ by CF3SO3" is always advisable. Sulfuric Acid. This is prepared on an enormous scale by the lead chamber and contact processes. In the former, S02 oxidation is catalyzed by oxides of nitrogen (by intermediate formation of nitrosylsulfuric acid, H0S020N0); in the latter, heterogeneous catalysts such as Pt are used for the oxidation! Pure H2S04 is a colorless liquid that is obtained from the commercial 98% acid by addition first of S03 or oleum and then titration with water until the correct specific conductance or melting point is achieved. The phase diagram of the H2S04-H20 system is complicated, and eutectic hydrates such as H2S04H20 (mp 8.5°C) and H2S04-2H20 (mp -38°C) oc¬ cur. In pure crystalline H2S04 there are S04 tetrahedra with S—O distances 1.42,1.43,1.52, and 1.55 A, linked by strong H-bonds. There is also extensive H-bonding in the concentrated acid. In the gas phase the structure is 02S(0H)2 which is essentially that in the liquid although the latter is extensively H-bonded.26 Pure H2S04 shows extensive self-ionization resulting in high conductivity. The equilibrium 2H2S04 ^ H3S04+ + HS04-
Kwc = 1.7 x lO"4 mol2 kg~2
R. D. Suenram et al., J. Am. Chem. Soc., 1981, 103, 2561.
26
115
HYDROGEN
is only one factor, since there are additional equilibria due to dehydration: 2H2S04 ;= h2o + h2so4 h2s2o7 + h2so4
h3o+ + hs2o7
^10°C = 3.5 x 10-5 mol2 kg-2
h3o+ + hso4-
^10°C — 1 mol kg-1
h3so4+ + HS2Of
^M0°C = 7 x 10“2 mol kg-1
Estimates of the concentrations in 100% H2S04 of the other species present, namely, H30 + , HS04~, H3S04+, HS2Of, and H2S207 can be made; for ex¬ ample, at 25°C, HS04- is 0.023M. Pure H2S04 and dilute oleums have been greatly studied as solvent systems, but interpretation of the cryoscopic and other data is often complicated. Sulfuric acid is not a very strong oxidizing agent, although the 98% acid has some oxidizing ability when hot. The concentrated acid reacts with many organic materials, removing the elements of water and sometimes causing charring, for example, of carbohydrates. Many substances dissolve in the 100% acid, often undergoing protonation. Alkali metal sulfates and water also act as bases. Organic compounds may also undergo further dehydration reactions, for example: c2h5oh ^ c2h5oh2+ + hso4-
c2h5hso4 + h3o+ + hso4-
Because of the strength of H2S04, salts of other acids may undergo solvolysis, for example: nh4cio4 + h2so4
nh4+ + hso4 + hcio4
There are also examples of acid behavior. Thus H3B03, which behaves ini¬ tially as a base, gives quite a strong acid: H3B03 + 6H2S04 b(hso4)3 + hso4-
B(HS04)3 + 3H30+ + 3HSCU b(hso4)4-
The addition of S03 to H2S04 gives what is known as oleum or fuming sulfuric acid (S03)„-H20; the constitution of concentrated oleums is contro¬ versial, but with equimolar ratios the major constituent is pyrosulfuric (disulfuric) acid (H2S207). At higher concentrations of S03, Raman spectra indicate the formation of H2S3O10 and H2S4013. Pyrosulfuric acid has higher acidity than H2S04 and ionizes thus: 2H2S207 ;==i H2S3O10 + H2S04
H3S04+ + HS3O10
The acid protonates many materials; HC104 behaves as a weak base, and CF3C02H is a nonelectrolyte in oleum. Fluorosulfuric Acid. Fluorosulfuric acid is made by the reaction: S03 + HF = FS03H or by treating KHF2 or CaF2 with oleum at ~250°C. When freed from HF by sweeping with an inert gas, it can be distilled in glass apparatus. Unlike
116
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
CISO3H, which is explosively hydrolyzed by water, FSO3H is relatively slowly hydrolyzed. Fluorosulfuric is one of the strongest of pure liquid acids. It is commonly used in the presence of SbF5 as a protonating system, as noted (Section 3-11). An advantage over other acids is its ease of removal by distillation in vacuum. The self-ionization 2FS03H
FS03H2+ + FSO3-
is much lower than for H2S04 and consequently interpretation of cryoscopic and conductometric measurements is fairly straightforward. In addition to its solvent properties, FSO3H is a convenient laboratory fluorinating agent. It reacts readily with oxides and salts of oxo acids at room temperature. For example, K2Cr04 and KC104 give Cr02F2 and C103F, re¬ spectively. Trifluoromethanesulfonic Acid. This very strong (~H0 = 15.1), useful acid (CF3SO3H, bp 162°C) is often given the trivial name “triflic” acid and its salts called “triflates.” It is very hygroscopic and forms the monohydrate CF3S03HH20, mp 34°C. Its salts are similar to perchlorates but nonexplo¬ sive; the CF3SO3-- ion is less likely than C104“ to be found disordered in crystal structures. The ion is also a useful leaving group in many organic and inorganic syntheses.27 Related and most useful protonating acids are (CF3S02)2CH2, (CF3S02)2CHPh, and (CF3S02)2NH that also give noncoordinating anions like CF3SO3-.28 Other Acids. Although “tetrafluoroboric” and “hexafluorophosphoric” acids do not exist as such but only in the form of oxonium salts, for example, H502+BF4~, nevertheless in nonaqueous solvents such as carboxylic anhy¬ drides or ethers there can be strong acid behavior. Tetrafluoroboric acid in diethyl ether is a useful strong acid, [Et2OH]+BF4-. Finally it may be noted that there are some solids that behave as superacids with H0 -16.0. An example is the solid obtained by treating Zr(OH)4 with concentrated H2S04 and calcining the product at >500°C.29a They have uses in catalysis.296
BINARY METALLIC HYDRIDES Figure 3-8 represents a rough attempt to classify the various hydrides. 3-13.
The Hydride Ion H"; Saline Hydrides
The formation of the unipositive ion H+ (or H30 + , etc.) suggests that hy¬ drogen should be classed with the alkali metals in the Periodic Table. On the 27See, for example, A. M. Sargeson et al., Inorg. Chem., 1984, 23, 2940. MA. R. Siedle et al., J. Am. Chem. Soc., 1984, 106, 1510. 29aM. Hino and K. Arata, J. Chem. Soc. Chem. Commun., 1985, 112 29bG. M. Kramer and G. B. McVicker, Acc. Chem. Res., 1986, 19, 78.
117
HYDROGEN
H Li Na K
i
Be Mg
Ti V Cr Mn Fe Co Ni ' Zr Nb Mo Tc Ru Rh Pd ! Ba La-Lu Hf Ta W Re Os Ir Pt } i Ra Ac U ,Pu Transition metal hydrides Saline i i hydrides i Rb Cs Fr
Ca Sr
Sc Y
Cu Ag Au
B | Al ZniGa Cd Tn l Hg
C Si Ge Sn
Tl S Pb
'He F !Ne S Cl [ Ar As Se BrlKr Sb Te 1 'iXe Bi Po At |Rn N P
0
l Borderline iCovalent hydrides l hydridesj I
I
FIG. 3-8. A classification of hydrides. The starred elements are transition elements for which complex molecules containing M—M bonds are known.
other hand, the formation of the hydride ion might suggest an analogy with the halogens. Such attempts at classification of hydrogen with other elements can be misleading. The tendency of the hydrogen atom to form the negative ion is much lower than for the more electronegative halogen elements. This may be seen by comparing the energetics of the formation reactions: iH2(g)
— H(g)
A H = 218 kJ mol-1
2Br2(g)
— Br(g)
A H = 113 kJ mol-1
H(g) + e
— H-(g)
A H = -67 kJ mol-1
Br(g) + e
— Br (g)
A H = -345 kJ mol-1
aH2(g) + e
— H-(g)
AH = + 151 kJ mol-1
|Br2(g) + e
—> Br"(g)
A H = -232 kJ mol"1
Thus, owing to the endothermic character of the H ion, only the most electropositive metals—the alkalis and the alkaline earths form saline or saltlike hydrides, such as NaH and CaH2. The ionic nature of the compounds is shown by their high conductivities just below or at the melting point and by the fact that on electrolysis of solutions in molten alkali halides hydrogen is liberated at the anode. X-ray and neutron diffraction studies show that in these hydrides the H ion has a crystallographic radius between those of F- and Cl". Thus the electrostatic lattice energies of the hydride and the fluoride and chloride of a given metal will be similar. These facts and a consideration of the Born— Haber cycles lead us to conclude that only the most electropositive metals can form ionic hydrides, since in these cases relatively little energy is required to form the metal ion. . The hydrides and some of their physical properties are given in Table 3-7. The heats of formation compared with those of the alkali halides, which are about 420 kJ mol"1, reflect the inherently small stability of the hydride ion. TT ... ... For the relatively simple two-electron system in the H ion, it is possible to calculate an effective radius for the free ion, 2.08 A. It is of interest to
118
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
TABLE 3-7 The Saltlike Hydrides and Some of Their Properties M—H Heat of formation
distance
Structure
— A//, (298 K)(kJ moL1)
(A)
NaCl type NaCl type NaCl type NaCl type NaCl type Slightly distorted hep Slightly distorted hep Slightly distorted hep Rutile type
91.0 56.6 57.9 47.4 49.9 174.5 177.5 171.5 74.5
2.04 2.44
Salt LiH NaH KH RbH CsH CaH2 SrH2 BaH2 MgH2
2.85 3.02 3.19 2.33 2.50 2.67
Apparent radius of h-(A)° 1.36 1.47 1.52 1.54 1.52 1.35 1.36 1.34 1.30
“See text. ^Although half the H" ions are surrounded by four Ca2+ and half by three Ca2+, the Ca—H distances are the same.
compare this with some other values, specifically, 0.93 A for the He atom, —0.5 A for the H atom, 1.81 A for the crystallographic radius of Cl~, and 0.30 A for the covalent radius of hydrogen, as well as with the values of the “apparent” crystallographic radius of H“ given in Table 3-7. The values in the table are obtained by subtracting the Goldschmidt radii of the metal ions from the experimental M—H distances. The value 2.08 A for the radius of free H' is at first sight surprisingly large, being more than twice that for He. This results because the H“ nuclear charge is only half that in He and the electrons repel each other and screen each other (-30%) from the pull of the nucleus. Table 3-7 indicates that the apparent radius of H“ in the alkali hydrides never attains the value 2.08 A and also that it decreases markedly with decreasing electropositive character of the metal. The generally small size is probably attributable in part to the easy compressibility of the rather diffuse H ion and partly to a certain degree of covalence in the bonds. Preparation and Chemical Properties. The saline hydrides are prepared by direct interaction at 300 to 700°C. The rates are in the order Li > Cs > K > Na, largely because of the activation energy term in the Arrhenius equation.30 Extremely active hydrides306 of Li, Na, and K can be made by interaction of hydrogen with BuLi + TMEDA in hexane(LiH) or of BuLi + M(Of-Bu) + TMEDA in hexane (NaH and KH). The saline hydrides are crystalline solids, white when pure but usually gray owing to traces of metal. They can be dissolved in molten alkali halides and on electrolysis of such a solution, for example, CaH2 in LiCl + KC1 at 360°C hydrogen is released at the anode. They react instantly and completely with
o' Hl1 and R' J' Pulham> J Chem. Soc. Dalton Trans., 1982, 217. L. Brandsma et al., Angew, Chem. Int. Ed. Engl., 1986, 25, 465.
119
HYDROGEN
even the weakest acids, such as water, according to the reaction MH + H+ - M+ + H2 The standard potential of the H2/H~ couple has been estimated to be — 2.25 V, making H“ one of the most powerful reducing agents known. The hydrides of Rb, Cs, and Ba may ignite spontaneously in moist air. Thermal decomposition at high temperatures gives the metal and hydrogen. Lithium hydride alone can be melted (mp 688°C) and it is unaffected by oxygen below red heat or by chlorine or dry HC1; it is seldom used except for the preparation of the more useful complex hydride LiAlH4 discussed in Section 7-9. Sodium hydride is available as a dispersion in mineral oil; al¬ though the solid reacts violently with water, the reaction of the dispersion is less violent. It is used extensively in organic synthesis and for the preparation of NaBH4. Sodium hydride acts as a strong base for hydrogen abstractions but as a reducing agent it is slow and inefficient though improved by addition of nickel acetate as catalyst or by sodium t-amyloxide.31 It is soluble in aqueous hexamethylphosphoramide giving a blue solution (cf. Section 4-3). Calcium hydride reacts smoothly with water and is a useful source of H2 (1000 cm3/ g) as well as a convenient drying agent for gases and organic solvents. Fi¬ nally it may be noted that the compound 1,8-naphthalenediylbis(dimethylborane) (3-XV) reacts with KH in THF to give a bridged mono¬ hydride.32 The compound also removes H“ from borohydride and other hydridic compounds and has been called a “hydride sponge” by analogy with a “proton sponge,” which is similar but with NMe2 instead of BMe2 groups. It also gives a bridged anion with F~.
(3-XV)
3-14.
Hydrides of a More Covalent Nature
Beryllium and Mg hydrides can be made by thermal decomposition of Be(CMe3)2 and MgEt2, respectively, but MgH2 is best made by interaction of Mg with EtBr in THF containing anthracene followed by hydrogenation using a CrCl3(THF)3 catalyst with H2 under pressure at 60 to 65°C.33 Zinc hydride is similar. >P. Caubere, Angew. Chem. Int. Ed. Engl, 1983, 22, 599 (81 references to NaH in organic
3
synthesis).
32H. E. Katz, J. Am. Chem. Soc., 1985, 107, 1420. 33B. Bogdanovic et. al., Angew. Chem. Int. Ed. Engl.,
1985, 24, 262.
120
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
Beryllium hydride (BeHO is difficult to obtain pure, but it is believed to have a polymeric structure with bridging hydrogen atoms as in the boranes (Chapter b). Magnesium hydride has a rutile-type structure (Table 3-7) like MgF:: it also has a low heat of formation and is less stable thermally than the true saline hydrides. For aluminum there is evidence for A1H, and A1:H0 in the gas phase and A1FT (Section 7-9) forms at least six different solid phases, as well as solid etherates. The a-form. which appears to be the most stable solid form, has A1 atoms octahedraUy surrounded by H atoms in a structure very similar to that of AlFj. Aluminum hydride (A1FC) is a useful reducing agent in organic chemistry, and the products formed are significantly different from those of reduction by LiAlFL. Nitriles can be reduced to amines via complex inter¬ mediates. and the reduction of alkyl halides is slower than with LiAlH, so that carboxyl or ester groups in compounds RCO;H and RCO:R can be reduced preferentially in the presence of R"X. There is no evidence for a comparable gallium hydride, but an unstable viscous oil that shows bands in the infrared spectrum due to Ga—H has been obtained by the reaction:
MeA-GaHj(s) + BF,(g) —i GaH3(l) + Me3NBF3(g) There is also little evidence for hydrides of In and Tl. 3-15.
Transition .Metal Hydrides
The transition metals form a wide variety of compounds that have M—H bonds, including anions such as [ReHQ]:- and [FeHp]4_; these are discussed in detail in Chapter 24. Hydrogen reacts with many transition metals or their alloys on heating, to give compounds commonly called hydrides even though in some cases thev clearly do not contain hydride ions.54 Many of these metal hvdride systems are exceedingly complicated, showing the existence of more than one phase often with wide divergences from stoichiometry! The most extensive studies a\e been made on the most electropositive elements, the lanthanides and actinides, and the titanium and vanadium groups of the d-block elements Satisfactory theoretical understanding of these substances has been slow to de\elop. Simple models emphasizing hvdridic character, or protonic character or. again, covalent character for the hydrogen, have all been discussed Only by elaborate band structure calculations can a true appreciation of their electronic structures and properties be obtained. -.I^“tha"Ife H-Vdri^es- ^ metals such as lanthanum or neodymium react :fa atm ,afd at or stigMy above room temperature, to give black sohds of graphite-like appearance. These products are pyrophoric in air and
m L n CTopks 111 AppUed PhysKS-Vols- 28-29) G Awwd ^ J-
121
HYDROGEN
react vigorously with water. These are the phases MH2 and MH3 which are nonstoichiometric (e.g., LaH2 87). Normally europium and ytterbium give only the dihydride phase, but higher ratios (e.g., YbH2 55) can be obtained at 350°C under pressure. The hydrides appear to be predominantly ionic and to contain M3+ ions even in the MH2 phase where the odd valence electron is probably located in a metallic conduction band as in the so-called dihalides such as Lal2 (Chap¬ ter 20); in YbH2 55 there is some evidence for both Yb2+ and Yb3+. Actinide Hydrides. Thorium and other actinides form complex systems with nonstoichiometric and stoichiometric phases. Uranium hydride35 is of some importance chemically because it is often more suitable for the prep¬ aration of uranium compounds than is the massive metal. Uranium reacts rapidly and exothermically with hydrogen at 250 to 300°C to give a pyrophoric black powder. The reaction is reversible: U + |H2 = UH3
A//? = -129 kJ mol-1
The hydride decomposes at somewhat higher temperatures to give extremely reactive, finely divided metal. A study of the isostructural deuteride by Xray and neutron diffraction shows that the deuterium atoms lie in a distorted tetrahedron equidistant from four uranium atoms; no U—U bonds appear to be present, and the U—D distance is 2.32 A. The stoichiometric hydride UH3 can be obtained, but the stability of the product with a slight deficiency of hydrogen is greater. Some typical useful reactions are the following: UH3
y
Cl, 200°C
H,0 350°C
U02
\
'
UC14
H,S 450°C '
1
US2
HP 400°C '
J- Am- Chem- Soc., 1986, 108, 78; J. Chem. ’soc. Dalton Trans., 1986,
-J. L. Dye etal J. Am. Chem. Soc., 1987,109, 3508; P. P. Edwards et al, J. Chem. Soc. Chem. Commun., 1986, 1444.
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
129
[Na(C,2,2,2)]+ ions with Na- in octahedral holes between the cationic layers. Other examples are [Rb(15-C-5)2]+Na“ and [Cs(18-C-6)2] + e“ where the elec¬ tron is trapped in a nearby spherical cavity radius —2.4 A. Electrides, which contain trapped electrons, may be formed by addition of a second mole of complexing agent, that is, M
L v
ML + esoW
ML
MU + es-olv
but due to their extreme reactivity, isolation is more difficult. However, [Cs (18-C-6)2]+e~ can be obtained as paramagnetic black crystals. For the diamagnetic alkalides, magic angle nmr spectra of 23Na and 133Cs in, for example, [Na(C,2,2,2)]+Na~ and [Cs(18-C-6)]+Cs~ show separate peaks for M+ and M-, while the electride [Cs(18-C-6)2]+e_ shows only a Cs+ band. Nuclear magnetic resonance spectra for 23Na and 37Rb in liquid crown ether (12-C-4) or hexamethylphosphoramide solutions also give sharp signals for M~ suggesting that the anion is not solvated. The ammonia and amine solutions of alkali metals are useful for preparing both organic and inorganic compounds. Thus Li in methylamine shows great selectivity in its reducing properties, but both this reagent and Li in ethylenediamine are quite powerful and can reduce aromatic rings to cyclic mono¬ olefins. Sodium in liquid ammonia is probably the most widely used system for preparative purposes; the solution is moderately stable, but the decom¬ position reaction Na + NH3(1) = NaNH2 + |H2 can occur photochemically and is catalyzed by transition metal salts. Sodium amide can be conveniently prepared by treatment of Na with liquid ammonia in the presence of a trace of FeCl3. Amines react similarly.
COMPOUNDS OF THE GROUP IA(1) ELEMENTS 4-4.
Binary Compounds
The metals react directly with most nonmetals to give one or more binary compounds; they also form alloys and compounds with other metals such as Pb and Sn. Oxides, are obtained by combustion of metal. Although Na normally gives Na202, it will take up further oxygen at elevated pressures and temperatures to form Na02. The per- and superoxides of the heavier alkalis can also be prepared by passing stoichiometric amounts of oxygen into their solutions in liquid ammonia, and ozonides (M03) are also known. The structures of the ions Ol~, 02", and 03' and of their alkali salts are discussed in Sections 125 and 12-6. The increasing stability of the per- and superoxides as the size of
130
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
the alkali ions increases is noteworthy and is a typical example of the stabi¬ lization of larger anions by larger cations through lattice energy effects. ~-/Owing to the highly electropositive character of the metals, the various oxides (and also sulfides and similar compounds) are readily hydrolyzed by water according to the following equations: M20 + H20 = 2M+ + 20HM202 + 2H20 = 2M+ + 20H- + H202 2M02 + 2H20 = 02 + 2M+ + 20H- + H202 The oxide Cs20 has the anti-CdCl2 structure and is the only known oxide with this type of lattice. An abnormally long Cs—Cs distance and a short Cs—O distance imply considerable polarization of the Cs+ ion. Rubidium and Cs form suboxides such as Rb902, Rb12Oz, and Csn03 that are highly colored and often metallic in appearance. Their structures have metal clusters with M—M bonds; for example, in Rb902 there is a confacial bioctahedron of Rb atoms with an O atom in the center of each. Metal clusters M^+ have also been observed in zeolite cages10 exposed to Na or K vapor. The hydroxides (MOH) are white crystalline solids soluble in water and in alcohols. They can be sublimed unchanged at 350 to 400°C, and the vapors consist mainly of dimers (MOH)2. Measurements of the proton affinities of MOH in the gas phase show that the base strength increases from lithium to cesium, but this order need not be observed in aqueous or alcoholic solutions where the base strength of the hydroxide is reduced by solvent effects and hydrogen bonding. In suspension in nonhydroxylic solvents such as 1,2-dimethoxyethane, the hydroxides are exceedingly strong bases and can conveniently be used to deprotonate a wide variety of weak acids such as PH3 (pK ~ 27) or C5H6 (pK ~ 16). The driving force for the reaction is provided by the formation of the stable hydrate: 2KOH(s) + HA = K+A- + K0HH20(s) The alkali metals form a multitude of compounds with the elements of Groups IIIA-VIA (13-16) only a few of which can be thought of in simple ionic terms (i.e., in terms of M+ ions and anions with complete octets, such as Na2S, K3P). The vast majority are far richer in the metalloidal element (e.g., NaP7, SrSi2, and LiGe) and contain complex polynuclear anionic struc¬ tures. An example is Li12Si7, which contains both planar Si5 and Si4 rings.113 These materials are structurally and electronically transitional between ionic compounds and alloys. Most of them can be made either by direct reaction of the elements or by reaction of a liquid ammonia solution of the alkali metals with compounds of the metalloidal components. 10P. P. Edwards et al., J. Chem. Soc. Chem. Commun., 1984, 982; U. Westphal and G. Geismar Z. Anorg. Allg. Chem., 1984, 508, 165. llaK.-F. Tebbe et at, Angew. Chem. Int. Ed. Engl., 1980, 19, 1033.
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
131
Zintl compoundsllb or phases have anions that can often be isolated in crystalline solids by complexation of the alkali metal with cryptands or crowns; some are mentioned in other chapters but typical examples contain Sn^, Sn2Bi2_, or Pbs- ions. The red nitride Li3N can be used as a reducing agent.llc
4-5.
Ionic Salts and M+ Ions in Solution
Salts of the bases MOH are crystalline, ionic solids, colorless except where the anion is colored. For the alkali metal ions the energies required to excite electrons to the lowest available empty orbitals could be supplied only by quanta far out in the vacuum ultraviolet (the transition 5p6 —> 5p56s in Cs+ occurs at —1000 A). However, colored crystals of compounds such as NaCl are sometimes encountered. Color arises from the presence in the lattice of holes and free electrons, called color centers, and such chromophoric distur¬ bances can be produced by irradiation of the crystals with X rays and nuclear radiation. The color results from transitions of the electrons between energy levels in the holes in which they are trapped. These electrons behave in principle similarly to those in solvent cages in the liquid ammonia solutions, but the energy levels are differently spaced and consequently the colors are different and variable. Small excesses of metal atoms produce similar effects, since these atoms form M+ ions and electrons that occupy holes where anions would be in a perfect crystal. The structures and stabilities of the ionic salts are determined in part by the lattice energies and by radius ratio effects, which have been discussed in Chapter 1. Thus the Li+ ion is usually tetrahedrally surrounded by water molecules or negative ions, although Li(H20)6+ has also been found. On the other hand, the large Cs+ ion can accommodate eight near-neighbor Cl“ ions, and its structure is different from that of NaCl, where the smaller cation Na+ can accommodate only six near neighbors. The Na+ ion appears to be sixcoordinate in some nonaqueous solvents. The salts generally have high melting points, electrical conductivity in melts, and ready solubility in water. They are seldom hydrated when the anions are small, as in the halides, because the hydration energies of the ions are insufficient to compensate for the energy required to expand the lattice. Owing to its small size, the Li+ ion has a large hydration energy, and it is often hydrated in its solid salts when the same salts of other alkalis are unhydrated (namely, LiC104-3H20). For salts of strong acids, the lithium salt is usually the most soluble in water of the alkali metal salts, whereas for weak acids the lithium salts are usually less soluble than those of the other alkalis. The large size of the Cs+ and Rb+ ions frequently allows them to form
nbF. U. Axe and D. S. Marynick, Inorg. Chem., 1987, 26, 1658. llcM. Kilner and G. Parkin, J. Organomet. Chem., 1986, 302, 181.
132
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
ionic salts with rather unstable anions, such as various polyhalide anions (Section 14-18) and the superoxides already mentioned. Since few salts are sparingly soluble in water there are few precipitation reactions of the aqueous ions. Generally the larger the M+ ion, the more numerous are its insoluble salts. Thus sodium has very few; the mixed NaZn and Na-Mg uranyl acetates [e.g., NaZn(U02)3(CH3C02)9-6H20], may be precipitated almost quantitatively under carefully controlled conditions from dilute acetic acid solutions. The perchlorates and hexachloroplatinates of K, Rb, and Cs are rather insoluble in water and virtually insoluble in 90% ethanol. These heavier ions may also be precipitated by the ion [Co(N02)6]3_ and various other large anions. Sodium tetraphenylborate NaB(C6H5)4, which is moderately soluble in water, precipitates the tetraphenylborates of K, Rb, and Cs from neutral or faintly acid aqueous solutions. Sodium chloride and some other MX salts can be precipitated from aqueous solution with the ligand (4-1) as, for example, NaL3Cl, in which each Na+ has distorted octahedral coordination by NH2 groups from six different li¬ gands. CH3 ch3
H
H
(4-1)
The primary hydration shell for Li+ in aqueous solution is doubtless tetrahedral. Only tetrahydrated salts are formed except when there is also hydration of the anions. X-ray scattering studies show that the primary hydration number of K+ is 4, and since Na+ forms the stable Na(NH3)4+ ion in liquid ammonia, it too presumably has a first coordination sphere of four water molecules. There is no direct evidence regarding Rb + or Cs+, but a higher number, probably 6, seems likely. In all cases electrostatic forces operate beyond the first shell, and additional water molecules are bound in layers of decreasing firmness. The larger the cation itself, the less it binds outer layers. Thus although the crystallographic radii increase down the group the hydrated radii decrease (Table 4-2). The hydration number of Li+ is very large and Li+ salt solutions generally deviate markedly from ideal solution behavior, showing abnormal colligative prop¬ erties such as very low vapor pressures and freezing points. Also, hydration energies of the gaseous ions decrease. The decrease in size of the hydrated ions is manifested in various ways. The mobility of the ions in electrolytic conduction increases, and so generally does the strength of binding to ionexchange resins. The equilibria The M+ Ions in Solution.
A+(aq) + B+R- = B + (aq) + A+R“
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
133
where R represents the resin and A+ and B+ the cations, have been measured and the order of preference is usually Li+ < Na+ < K+ < Rb+ < Cs+, although irregular behavior does occur in some cases. The usual order may be explained if we assume that the binding force is essentially electrostatic and that under ordinary conditions the ions within the water-logged resin are hydrated approximately as they are outside it. Then the ion with the smallest hydrated radius (which is the one with the largest “naked” radius) will be able to approach most closely to the negative site of attachment and will hence be held most strongly according to Coulomb’s law. The efficiency of separating alkali metal ions on cation exchangers can be increased significantly by adding chelating agents like EDTA to the eluting solution. These agents bind more strongly to the ions that are less strongly held to the resin, thus enhancing the separation factors.
4-6.
Alkali Metal Complexes12
The M+ ions are only very weakly complexed by simple anions, for example, F~ in 1 M fluoride solution, where the order falls Li — Cs.13 Although lithium is an exception, as discussed later, chelation is usually essential to complex formation. Complexes are formed with (3-diketones, nitrophenols, and so on;: some, such as the hexafluoroacetylacetonates, are sublimable at 200°C, though the metal-ligand bonds are doubtless quite polar. The anhydrous (3-diketonates are usually insoluble in organic solvents, indicating an ionic nature, but in the presence of additional coordinating ligands, including water, they may become soluble even in hydrocarbons; for example, sodium benzoylacetylacetonate dihydrate is soluble in toluene and tetramethylethylenediaminelithium hexafluoroacetylacetonate in benzene. This behavior has allowed the development of solvent-extraction proce¬ dures for alkali metal ions. Thus not only can the trioctylphosphine oxide adduct Li(PhCOCHCOPh)[OP(octyl)3]2 be extracted from aqueous solutions into p-xylene, but also this process can be used to separate lithium from other alkali metal ions. Even Cs+ can be extracted from aqueous solutions by 1,1,1trifluoro-3-(2' thenoyl)acetone (TTA) in MeN02-hydrocarbons. Lithium forms a very wide range of complexes with amines, ethers, carboxylates, alkoxides, dialkylamides, and many other ligands whose structures are usually quite different from those of the other M+ ions. Some of these are discussed in Section 4-8. However, we may note that in many of these compounds lithium can have coordination numbers 3-7. Simple examples are LiNCS(16-C-4), which is tetragonal pyramidal while Lil-en3 has Li+ in a distorted octahedron of N atoms. Chloride complex ions
12N. S. Poonia and A. V. Bajaj, Chem. Rev., 1979, 79, 389; B. Dietrich, J. Chem. Educ., 1985, 62, 954 (with macrocycles). 13G. T. Hefter et al., Polyhedron, 1984, 3, 845.
134
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
such as (4-II) have been characterized;
Li(OEt2)2 /C1\
(Et20)2Li^
L
^Li(OEt2)2
Li(OEt2)2
J
(4-II)
LiBF4[OP(NMe2)3] is soluble in benzene and has strong Li—F interactions143 and LiN03(diacetamide) is five-coordinate with Tq1-N03.14b The effectiveness of THF and dimethylethers of ethylene- and diethylene¬ glycols as media for reactions involving sodium metal may be due in part to slight solubility of Na but the driving force is undoubtably the complexation of Na+ by the chelate ethers. The most important ethers are the crowns and cryptates14c (O, N) (Section 12-14), which bind M+ strongly and often with high selectivity. The affinity of such a ligand for an ion is strongly dependent on how well the ion fits into the cavity that the ligand can provide for it. At the same time the strength of complexation and to some extent the selectivity also depend on the solvent. Illustrations of selectivity, provided by cryptate221 and -222 are shown in Fig. 4-1. For cryptate-222, for example, K+ fits the cavity very well, but Li+ and Na+ are too small to make good contacts with the oxygen atoms and Rb + and Cs+ are too large to enter without appreciable steric strain. An example
FIG. 4-1. Stability constants for cryptate-221 and -222 versus alkali ion in MeOH-H20 = 95:5 [Adapted from J. M. Lehn and J. P. Sauvage, J. Am. Chem. Soc., 1975, 97, 6700.] I4aR. Snaith et al., J. Chem. Soc. Chem. Commun., 1986, 127. I4bD. D. Bray and N. F. Bray, Inorg. Chim. Acta, 1986, 111, 639. 14cD. Parker, Adv. Inorg. Chem. Radiochem., 1983, 27, 1.
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
135
FIG. 4-2. The structure of the cation in the salt [RbCi8H36N206]SCNH20. [Reproduced by permission from M. R. Truter, Chem. Br., 1971, 203.]
of an alkali cation (Rb + ) occupying the cavity in cryptate-222, and coordinated by the six oxygen and two nitrogen atoms, is presented in Fig. 4-2. An 18-C-6 crown ether with a side chain terminating in an NH2 group is selective for Na+/K+ in the transport of ions across a liquid membrane, thus mimicing natural ion-selective transport agents such as monensin.15 Many of the natural agents are small polypeptides of which valinomycin (4-III) is another. The structures of two such complexes are shown in Fig. 4-3. /
r-l\
< \
CH(CH3)2
CH(CH3)2
ch3
CH(CH3)2\
0—C—c—N--c—c—o—c—c—-N—c—C— — H H || / H II H || « « H 0 3 0 0 o-
(4-UI) Some ionophores with high preference for Li+ are crown ethers with long aliphatic chains16 and other macrocycles with functional groups may have utility for therapy in brain disorders for which Li+ salts are used. Open chain polyethers can also complex M+ ions.17 Although few complex ions with only N donor ligands are stable for 15W. Pangborn et al., J. Am. Chem. Soc., 1987, 109, 2163. 16T. Shono et al., J. Am. Chem. Soc., 1984, 106, 6978; A. Shanzar et al., J. Am. Chem. Soc.,
1983, 105, 3815. 17D. L. Hughes and J. N. Wingfield, J. Chem. Soc. Dalton Trans., 1984, 1187.
136
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
FIG. 4-3. Structures of the K+ salt of (a) [D-hydroxyisovaleric acid-A-methyl-L-valine]3 or enniatin B and (b) nonactin. [Reproduced by permission from D. A. Fenton, Chem Soc Rev 1977, 6, 325.]
Na+—Cs+, there are many of these for Li+. Thus the effectiveness of Me2N(CH2)2NMe2 (tmen) in increasing the effectiveness of lithium alkyls (Section 4-7) is due to formation of [Li(tmen)2]+ with concomitant increase in nucleophilicity of the anion. Rather unstable ammoniates such as [Na(NH3)4]+ are obtained by the action of NH3(1) on Nal.
Alkali Metal Salts of Complex Anions.
The alkali metal salts of transition metal carbonylate anions such as [Co(CO)4]~ are discussed in detail in Chapter 22. However, it is relevant to note here that in such anions, ir spectra show that there is extensive ion-pair formation and that M+ perturbs the CO or NO ligands on the metal.18 As expected, Li+ causes maximum perturbation ,8M. Y. Darensbourg, Progr. Inorg. Chem., 1985, 33, 221.
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
137
and lowest CO stretching frequencies, via the M+-—O—C—Co or M+— O—N—Fe interaction. No perturbation occurs with large cations of low charge density such as [(Ph3P)2N]+ or [Na crypt]+ . In hydrido anions (Section 26-6) there is also Na+—H—M interaction, which increases CpV(CO)3H~ < HFe(CO)4- < HCr(CO)5- < HW(CO)5- < HMo(CO)4(PR3)-, whereas the Na+—-OCM interaction is in the reverse order. Exceptional compounds soluble in pentane despite being unsolvated by THF, tmen, and so on, are K2[Co(C2H4)2(PMe3)3]4 and Li3[Co(C2H4)(PMe3)3]3 where, in the latter, Li+ is bound both to Co and C2H4.19 Many similar compounds, especially of Fe, Ni and Co are known, for example, Li2Fe(COD)2 and Li(THF)2Ni(C4H6)3.20a The “sandwich” ion CpCo[P(0)(OEt)2]3~ (Section 26-14) acts as a tripod oxygen ligand giving pentane soluble (NaL)3-2H2O.20b The M+ ions can also interact with oxygen atoms of other complexes such as those of Schiff bases like sal2en21:
nT 'p (
Co11^
< + Na — (
N^^O
^Na(THF)4
N^_
2VO sal2en + NaBPh4- [(VOsal2en)2Na] +BPh4“ Alkali metal complexes binding to the oxygen of alkoxides are also known,22 for example (4-1V). Ph
I
(PhO)4W^ /Li(THF)4 O Ph (4-IV)
Finally, some unusual species are the cation23 [Na4(q.-MeOH)4((xH20)2(Me0H)]4+ found with MogO^ and KZr6Ii4 where the potassium is at the center of a Zr6 cluster.24 In [Cs9(18-C-6)14]9+ in a rhodium cluster salt the Cs+ ion is “sandwiched” between the crown ligands.25 19H.-F. Klein et al., J. Chem. Soc. Chem. Commun., 1983, 231. 20aK. Jonas and C. Kruger, Angew. Chem. Int. Ed. Engl., 1980, 19, 520. 20bW. Klaui et al., J. Am. Chem. Soc., 1987, 109, 164. 21C. Floriani et al, J. Chem. Soc. Chem. Commun., 1987, 183, 281; Inorg. Chem., 1986, 25, 4589. 22G. Wilkinson et al., Polyhedron, 1982, 1, 641. »E. M. McCarron, III, and R. L. Harlow, Acta Cryst. (C), 1984, 40, 1140. 24J. D. Smith and J. D. Corbett, J. Am. Chem. Soc., 1984, 106, 4618. “J. L. Vidal et al., Inorg. Chem., 1981, 20, 227.
138 4-7.
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
Organometallic Compounds26
Lithium Alkyls and Aryls. One of the largest uses for lithium metal is the synthesis of organolithium compounds, which are of great utility, generally resembling Grignard reagents, though usually more reactive. They are commonly made by direct interaction of alkyl or aryl chloride in ether, benzene, or petroleum; ethers are slowly attacked, however, by LiR. RC1 + 2Li - RLi + LiCl In ether, MeLi gives complexes such as Li4Me3Br and Li4Me3I; n- or f-butyl lithium in hexane, benzene, or ethers may also be used in metal-hydrogen or metal-halogen exchanges, for example, 2rc-BuLi + ('n5-C5H5)2Fe- (Ti5-C5H4Li)2Fe + 2n-C4H10 n-BuLi + o-Br(N02)C6H4- o-Li(N02)C6H4 + rc-BuBr Organolithiums are very reactive to water and to air, often being sponta¬ neously flammable. They are soluble in hydrocarbons, have high volatility and may be sublimed in vacuum. Lithium alkyls are often considered to be carbanionic in reactions (although radicals are formed in some cases). However, although some solvated lithium compounds, for example, Li(crown)2CHPh2,27 do have isolated planar carbanions, almost all lithium alkyls and aryls are associated in both the solid state and in solutions. There is accordingly a wide variation in reactivities depending on differences in aggregation and ion-pair interactions all of which are solvent dependent. Thus LiCH2Ph, which is monomeric in THF reacts with a given substrate more than 104 times as fast as does (LiMe)4. In noncoordinating solvents f-BuLi is hexameric while in THF there is an equilibrium28: (t-BuLi)4
2(t-BuLi)2
(PhLiEt20)4 is one of the few cases where solid and solution structures are similar.29 The addition of Me2N(CH2)2NMe2 to lithium reagents also increases the nucleophilicity remarkably by complexing the Li+ ion. The structures of the alkyls and related compounds are discussed in Section 4-8. It may be noted that alkyls and aryls of main group and transition metals that are coordinately unsaturated have Lewis acid character (cf. the halides) and in alkylations of halides by LiR, in contrast to the use of MgR2, lithium 26J. L. Wardell, Comprehensive Organometallic Chemistry, Vol. 1, Chapter 2, Pergamon Press, Oxford, 1981; N. Jones, Comprehensive Organic Chemistry, Vol. 3, Chapter 15, Pergamon Press' Oxford, 1979. 27M. M. Olmstead and P. P. Power, J. Am. Chem. Soc., 1985, 107, 2174. 28J. F. McGarrity and C. A. Ugle, J. Am. Chem. Soc., 1985, 107, 1805, 1810. 29P. P. Power et al, Organometallics, 1984, 4, 2117; L. M. Jackman and L. M. Scarmontos /. Am. Chem. Soc., 1984, 106, 4627.
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
139
alkylate anions are often formed, for example, Mo2(02CMe)4 + 8MeLi
[Li(THF)]4[Mo2Me8]
MgPh2 + 2PhLi-> Li[MgPh4] WMe6 + 2MeLi-> Li2[WMeg] The lithium alkylates of copper are of great utility in organic syntheses. Polylithium Compounds. Compounds such as (CH2Li2)„, Li4C3, and Li4C can be made by the action of LiR on MeCN, MeC=CH, and so on, or for (CH2Li2)„ by thermal decomposition of MeLi.30 Organosodium and Potassium Compounds. These are essentially ionic, sparingly soluble if at all in hydrocarbons, and exceedingly air sensitive and reactive. Methyl potassium can be made by the reaction Me2Hg + 2K/Na- 2MeK + Na/Hg It has the NiAs type structure (Fig. 13-4) and isolated methyl groups are probably rotating or disordered. The Na and K compounds have little practical value due to their insolubility but interaction with Mg(OCH2CH2OEt)2 gives benzene soluble species that can be used for arylations.31 They can also be prepared from Na or K dispersed on an inert support, and such solids act as carbanionic catalysts for the cyclization, isomerization, or polymerization of alkenes. The so-called alfin ca¬ talysts for copolymerization of butadiene with styrene or isoprene to give rubbers consist of sodium alkyl (usually allyl) and alkoxide (usually isopropoxide) and NaCl, which are made simultaneously in hydrocarbons. More important are the compounds formed by acidic hydrocarbons such as cyclopentadiene, indene, and acetylenes. These are obtained by reaction with sodium in liquid ammonia or, more conveniently, sodium dispersed in THF, glyme, diglyme, or dimethylformamide (DMF). 3C5H6 + 2Na-> 2C5H5-Na+ + C5H8 RC=CH + Na-> RC=C“Na+ + |H2 Many aromatic hydrocarbons, as well as aromatic ketones, triphenylphosphine oxide, triphenylarsine, azobenzene, and so on, can form highly colored radical anions when treated at low temperatures with sodium or potassium in solvents such as THF. For the formation of such anions it must be possible to delocalize the negative charge over an aromatic system. Species such as the benzenide (C6H6-), naphthalenide, or anthracenide ions can be detected and characterized spectroscopically and by esr. The sodium—naphthalene sys¬ tem Na+[C10H8]- in an ether is widely used as a powerful reducing agent
30R. J. Lagow et al., Inorg. Chem., 1984, 23, 3717; J. Am. Chem. Soc., 1984, 106, 3694. A. Maercker and M. Theis, Angew. Chem. Int. Ed. Engl., 1984, 23, 995. 31C. G. Screttas and M. M. Screttas, Organometallics, 1984, 3, 904.
140
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
(e.g., in nitrogen-fixing systems employing titanium catalysts) and for the production of complexes in low oxidation states. The blue solution of sodium and benzophenone in THF, which contains the ketyl or radical ion, is a useful and rapid reagent for the removal of traces of oxygen from nitrogen. 4-8.
Structures of Organolithiums and Related Lithium Compounds32
As well as the alkyls and aryls, lithium forms dialkylamides (LiNR2); alkoxides (LiOR); dialkylphosphides (LiPR2)33; thiolates (LiSR), and many other com¬ pounds most of which, as contrasted with their Na or K analogues, have exceptional solubility in organic solvents and with few exceptions are oligo¬ mers. Quite generally (a) the Li+ ion is bound to the principal donor atom, for example, C in the alkyls or N in LiNR2 by multicenter bonds and may also interact with other atoms such as H in the organic moiety; (b) lithium may have coordination numbers from 2 to 6 though four is most common; (c) lithium commonly is found in tetrahedra of lithium atoms, that is, Li4 but other polyhedra occur. It is also found in rings such as (LiN)3 that may, in certain cases, have hydrogen bridges between lithium and a metal Li(|x-H)„M. Organolithium Compounds. The structure of the simplest, methyllithium, is shown in Fig. 4-4; that of (EtLi)4 is similar. In both compounds the Li atoms are at the corners of a tetrahedron with the alkyl groups (|x3) centered over the facial planes. Although the CH3 group is symmetrically bound to three Li atoms, in the ethyl, the a carbon of CH2CH3 is closer to one Li atom than the other two.
FIG. 4-4. The structure of (CH3Li)4. (a) Showing the tetrahedral Li4 unit with the CH3 groups located symmetrically above each face of the tetrahedron. [Adapted from E. Weiss and E. A. C. Lucken, J. Organomet. Chem., 1964, 2, 197.] The structure can also be regarded as derived from a cube (b).
32W. N. Setzer and P. von R. Schleyer, Adv. Organomet. Chem., 1985, 24, 353 (an extensive review including compounds with Li—O, Li—N, Li—Si, etc., bonds). 33See, for example, P. P. Power, et al., Inorg. Chem., 1986, 25, 1243; J. Am. Chem. Soc., 1986, 108, 6921.
141
THE GROUP IA(1) ELEMENTS: Li, Na, K, Rb, Cs, Fr
Examples of other structures known are those of Li6(c-C6Hn)6, and (LiPh-OEt2)4 as well as ate complexes such as [Li(THF) ][Li{C(SiMe3) }2] and 4
3
Li4[Mo2Me8]. In all the oligomers, Li-Li bonding is not important although in a dimeric adduct of lithium cyclohexyl, the Li-Li distance, 2.34 A is shorter than that in MeLi (2.57 A) or in the Li2 gas molecule (2.67 A). Here too, as in other compounds where lithium interacts with alkenes, alkynes, and arenes, there is always a high degree of ionic character, despite the often high degree of solubility in alkanes. The precise degree and nature of the oligomeriza¬ tion is probably due to steric factors. The simpler alkyls usually give tetramers, for example, (t-BuLi)4,34a though even in a very bulky alkyl like Li4{MnH(C2H4)[CH2(Me)PCH2CH2PMe2]2}2-2Et2034b there is a central Li4 tet¬ rahedron. In (c-C6H11Li)6-2C6H6 there are Li6 octahedra with cyclohexyl groups bridging on six of the eight faces and benzenes on the other two; [Li(SiMe3)]6 also has an octahedron of lithium atoms. The oligomers are commonly cleaved by basic ligands like Me2N(CH2)2NMe2 or crown ethers. The cLialkylamides and alkoxides show similar features to unsolvated or solvated alkyls and aryls. Some examples of structures are the following. Me3Si
SiMe3
THF
XN
/L^i\
Me3CO^
/OCMe3 Li THF
Li
OEt, Li^ (Me3Si)2N^ /N(SiMe3)2 Li OEt2 (12-C-4)
In the nonsolvated compound, [LiN(CH2Ph)2]3, there is evidence of Li-—HC interaction in the solid state35 (4-V).
(4-V)
34aR. D. Thomas et al., Organometallics, 1987, 6, 565. 34bG. Wilkinson et al., J. Chem. Soc. Dalton Trans., 1985, 921 35R. Snaith et al., J. Chem. Soc. Chem. Commun., 1984, 287.
142
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
Close Li-Me interactions are also found in Li[MnN(SiMe3)2(OCt_Bu3)2].36 It may be noted, however, that low C—H stretching frequencies in the ir spectra of transition metal species are not necessarily indicative of Li--H—C interactions as has sometimes been thought since these are observed not only in (4-V) but also in [(PhCH2)2NLi(OEt2)]2Another type of oligomeric lithium compound is [LiWH5(PMe3)3]4 (Section 24-5) that has a staggered ring of W and Li atoms that are linked by bridges of the type (4-VI).37
H—Li— (4-VI)
Additional References Bach, R. O., Ed., Lithium: Current Applications in Science Medicine and Technology, Wiley, New York, 1985. Brandsma, L. and H. Verkruijsse, Preparative Polar Organometallic Chemistry, Vol. B. M. Trost, Ed., Springer-Verlag, New York, 1986.
1,
Organolithium Compounds. Solvated Electrons, Top. Curr. Chem., 1986, Vol 138. (polylithiated aliphatic hydrocarbons, lithiation reactions, electrochemistry of solvated electrons).
36B. D. Murray and P. P. Power, J. Am. Chem. Soc., 1984, 106, 7011. 37G. Wilkinson et al., J. Chem. Soc. Dalton Trans., 1987, 837.
Chapter Five
The Group IIA(2) Elements: Be, Mg, Ca, Sr, Ba, Ra GENERAL REMARKS 5-1.
General Remarks
Some pertinent data for the elements are given in Table 5-1. Beryllium has unique chemical behavior with a predominantly covalent chemistry, although it forms a cation [Be(H20)4]2+. Magnesium has a chemistry intermediate between that of Be and the heavier elements, but it does not stand in as close relationship with the predominantly ionic heavier members as might have been expected from the similarity of Na, K, Rb, and Cs. It has considerable tendency to covalent bond formation, consistent with the high charge/radius ratio. For instance, like beryllium, its hydroxide can be precipitated from aqueous solutions, whereas hydroxides of the other elements are all mod¬ erately soluble, and it readily forms bonds to carbon. The metal atomic radii are smaller than those of the Group I metals owing to the increased nuclear charge; the number of bonding electrons in the metals is twice as great, so that the metals have higher melting and boiling points and greater densities. All are highly electropositive metals, however, as is shown by their high chemical reactivities, ionization enthalpies, standard electrode potentials and, for the heavier ones, the ionic nature of their compounds. Although the energies required to vaporize and ionize these atoms to the M ions are considerably greater than those required to produce the M+ ions of the Group I elements, the high lattice energies in the solid salts and the high hydration energies of the M2+(aq) ions compensate for this, with the result that the standard potentials are similar to those of the Li-Cs group. The potential E° of beryllium is considerably lower than those of the other elements, indicating a greater divergence in compensation by the hydration energy, the high heat of sublimation, and the ionization enthalpy. As in Group IA(1), the smallest ion crystallographically (i.e., Be2+) has the largest hy¬ drated ionic radius. smaller and considerably less polarizable than the All the M2+ ions are
143
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
144
TABLE 5-1 Some Physical Parameters for the Group IIA(2) Elements Ionization
Element
Electronic configuration
mp (°C)
Be Mg Ca Sr Ba Ra
[He]2s2 [Ne]3.?2 [Ar]4s2 [Kr]5s2 [Xe]6s2 [Rn]7s2
1278 651 843 769 725 700
enthalpies (kJ mol-1)
E° for M2+ (aq) +
1st
2nd
(V)
Ionic radius (A)“
899 737
1757 1450
-1.70 -2.37
0.31 0.78
590 549 503 509
1146 1064
-2.87 -2.89 -2.90 -2.92
1.06 1.27
965 979
2e = M(s)
1.43 1.57
Charge radius 6.5 3.1 2.0 1.8 1.5 1.3
“Ladd radii. '’Estimated.
isoelectronic M+ ions. Thus deviations from complete ionicity in their salts due to polarization of the cations are even less important. However, for Mg2+ and, to an exceptional degree for Be2+, polarization of anions by the cations does produce a degree of covalence for compounds of Mg and makes co¬ valence characteristic for Be. Accordingly only an estimated ionic radius can be given for Be2+; the charge/radius ratio is greater than for any other cation except H+ and B3+, which again do not occur as such in crystals. The closest ratio is that for Al3+ and some similarities between the chemistries of Be and A1 exist. Examples are the resistance of the metal to attack by acids owing to formation of an impervious oxide film on the surface, the amphoteric nature of the oxide and hydroxide, and Lewis acid behavior of the chlorides. How¬ ever, Be shows just as many similarities to zinc, especially in the structures of its binary compounds (see Sections 16-6 to 16-8) and in the chemistry of its organic compounds. Thus BeS is insoluble in water, although A12S3, CaS, and so on, are rapidly hydrolyzed. Calcium, Sr, Ba, and Ra form a closely allied series in which the chemical and physical properties of the elements and their compounds vary system¬ atically with increasing size in much the same manner as in Group IA(1), the ionic and electropositive nature being greatest for radium. Again the larger ions can stabilize certain large anions, e.g., the peroxide and superoxide ions, polyhalide ions, and so on. Some examples of systematic group trends in the series Ca-Ra are (a) hydration tendencies of the crystalline salts increase; (b) solubilities of sulfates, nitrates, chlorides, and so on (fluorides are an exception) decrease; (c) solubilities of halides in ethanol decrease; (d) thermal stabilities of carbonates, nitrates, and peroxides increase; (e) rates of reaction of the metals with hydrogen increase. Other similar trends can be found. All isotopes of radium are radioactive, the longest-lived isotope being 226Ra (a; -1600 years). This isotope is formed in the natural decay series of 238U and was first isolated by Pierre and Marie Curie from pitchblende. Once widely used in radiotherapy, it has largely been supplanted by radioisotopes made in nuclear reactors. The elements zinc, cadmium, and mercury, which have two electrons out-
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
145
side filled penultimate d shells, are classed in Group IIB(12). Although the difference between the calcium and zinc subgroups is marked, zinc, and to a lesser extent cadmium, show some resemblance to beryllium or magnesium in their chemistry. We discuss these elements separately (Chapter 16), but note here that zinc, which has the lowest second ionization enthalpy in the Zn, Cd, Hg group, still has a value (1726 kJ mol”1) similar to that of beryllium (1757 kJ mok1), and its standard potential (-0.76 V) is considerably less negative than that of magnesium. There are a few ions with ionic radii and chemical properties similar to those of Sr2+ or Ba2+, notably those of the +2 lanthanides (Section 20-14) and especially the europous ion (Eu2+) and its more readily oxidized analogues Sm2+ and Yb2+. Because of this fortuitous chemical similarity, europium is frequently found in Nature in Group II minerals, and this is a good example of the geochemical importance of such chemical similarity. Although the differences between the first and second ionization enthalpies especially for beryllium might suggest the possibility of a stable +1 state, there is no evidence to support this. Calculations using Born-Haber cycles show that owing to the much greater lattice energies of MX2 compounds, MX compounds would be unstable and disproportionate: 2MX-» M + MX2 There is some evidence for Be1 in fused chloride melts, for example Be + Be11 - 2Be but no Be1 compound has been isolated. Some studies of the dissolution of Be from anodes suggested Be+ as an intermediate, but subsequent work showed that disintegration of the metal occurs during dissolution, so that the apparent effect is one of the metal going into solution in the +1 state—too much metal is lost for the amount of current passed. The anode sludge, a mixture of Be and Be(OH)2, had been considered to be due to dispropor¬ tionation of Be+, but photomicrography indicates that the beryllium in the sludge is due merely to spallation of the anode. On the other hand, similar studies of anodic dissolution of Mg in pyridine and aqueous salt solutions do provide some evidence for transitory Mg+ ions, which would account for evolution of H2 at or near the anode. Electrically generated Mg+ ions have been used to reduce organic compounds.
BERYLLIUM 5-2.
Covalency and Stereochemistry for Beryllium
As a result of the small size, high ionization enthalpy, and high sublimation energy of beryllium, its lattice and hydration energies are insufficient to pro¬ vide complete charge separation and the formation of simple Be ions, n fact in all compounds whose structures have been determined, even those of the most electronegative elements (i.e., BeO and BeF2), there appears to
146
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
be substantial covalent character in the bonding. On the other hand, to allow the formation of two covalent bonds —Be—, it is clear that unpairing of the two 2s electrons is required. Where free BeX2 molecules occur, the Be atom is promoted to a state in which the two valence electrons occupy two equiv¬ alent sp hybrid orbitals and the X—Be—X system is linear. However, in such a linear molecule the Be atom has a coordination number of only 2 and there is a strong tendency for Be to achieve maximum (fourfold) coordination, or at least threefold coordination. Maximum coordination is achieved in several ways. 1. Polymerization may occur through bridging, as in solid BeCl2 (Fig. 5-1). The coordination of Be is not exactly tetrahedral, since the ClBeCl angles are only 98.2° which means that the BeCl2Be units are somewhat elongated in the direction of the chain axis. In (BeMe2)„ the CBeC angle is 114°. These distortions from the ideal tetrahedral angle are related to the presence (or absence) of the lone pairs on the bridge group or atom that are available for bonding to the metal.1 In the gas phase at high temperatures the chloride (and other halides) is linear, Cl—Be—Cl, but at lower temperatures there are appreciable amounts (—20% at 560°C) of a dimer in which Be is, presumably, three-coordinate. In the compounds of the type M'Be^h (M1 = K, Rb, Tl, NO, NH4) the anion (5-1) resembles a portion of the (BeCl2)n chain.
(5-1)
Alkoxides, [Be(OR)2]„, are usually associated with three- or four-co¬ ordinate Be. For example, [Be(OMe)2]„ is a high polymer, insoluble in hy¬ drocarbon solvents. Even [Be(0-r-Bu)2]3, which is soluble, is trimeric (5-II). The compound {Be[OC(CF3)3]2}2 is a highly volatile dimer. The dialkylamides2 are similar, for example (5-III). Bu
Bu
Bu
Me,
Me
N
N '
Bu
Me2
Me:
(5-II)
(5-III)
(Me3Si)2N (5-IV) lE. Canadell and O. Eisenstein, Inorg. Chem., 1983, 22, 3856. 2H. Noth and D. Schlosser, Inorg. Chem., 1983, 22, 2700.
Be—N(SiMe3); (5-V)
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
7
Cl
Cl
147
7
FIG. 5-1. The structure of polymeric BeCl2 in the crystal. The structure of Be(CH3)2 is similar.
Only when there are bulky groups on O or N are monomers with twocoordinate Be such as (5-IV) and (5-V) obtained; these are isoelectronic with B(OR)2+ and B(NR2)2+. Another example is Be(t-Bu)2. 3. By functioning as Lewis acids, many beryllium compounds attain max¬ imum coordination of the metal atom. Thus the chloride forms etherates [Cl2Be(OR2)2], and complex ions such as [BeF4]2~ and [Be(H20)4]2+ exist. In chelate compounds such as the acetylacetonate [Be(acac)2], four approxi¬ mately tetrahedral bonds are formed, with the C—O and Be—O bond lengths equivalent. In addition to the tetrahedral [BeF4]2_ion, some substances of composition M!Be2F5 contain infinite sheet anions (i.e., not the chains present in the previous chloro analogue) in which there are hexagonal rings of BeF4 tetrahedra sharing corners, structurally analogous to the sheet silicate anions of empirical formula Si202- (cf. Section 9-7). The packing in crystals is almost invariably such as to give beryllium a coordination number of 4, with a tetrahedral configuration. In binary com¬ pounds, the structures are often those of the corresponding zinc compounds. Thus the low-temperature form of BeO has the wurtzite structure (Fig. 1-2); the most stable Be(OH)2 polymorph has the Zn(OH)2 structure, and BeS has the zinc blende structure (Fig. 1-2). Beryllium silicate (Be2Si04) is exceptional among the orthosilicates of the alkaline earths, the rest of which have struc¬ tures giving the metal ion octahedral coordination, in having the Be atoms tetrahedrally surrounded by oxygen atoms. It may be noted that Be with F gives compounds often isomorphous with oxygen compounds of silicon; thus NaBeF3 is isomorphous with CaSi03, and there are five different correspond¬ ing forms of Na2BeF4 and Ca2Si04. In Be phthalocyanine the metal is perforce surrounded by four nitrogen atoms in a plane. This compound constitutes an example of a forced config¬ uration, since the Be atom is held strongly in a rigid environment It is to be noted especially that beryllium compounds are exceedingly poi¬ sonous, particularly if inhaled, and great precautions must be taken in han¬ dling them. 5-3.
Elemental Beryllium
148
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
The gray metal is rather light (1.86 g/cm3) and quite hard and brittle. Since the absorption of electromagnetic radiation depends on the electron density in matter, beryllium has the lowest stopping power per unit mass thickness of all suitable construction materials. It is used for “windows” in X-ray ap¬ paratus and has other special applications in nuclear technology. Like alu¬ minum, metallic beryllium is rather resistant to acids unless finely divided or amalgamated, owing to the formation of an inert and impervious oxide film on the surface. Thus although the standard potential (-1.70 V) would indicate rapid reaction with dilute acids (and even H20), the rate of attack depends greatly on the source and fabrication of the metal. For the very pure metal the relative dissolution rates are HF > H2S04 ~HC1 > HN03. The metal dissolves rapidly in 3M H2S04 and in 5 M NH4F, but very slowly in HN03. Like aluminum, it also dissolves in strong bases, forming what is called the beryllate ion.
5-4.
Binary Compounds
The white crystalline oxide BeO is obtained on ignition of beryllium or its compounds in air. It resembles A1203 in being highly refractory (mp 2570°C) and in having polymorphs; the high-temperature form (>800°C) is exceedingly inert and dissolves readily only in a hot syrup of concentrated H2S04 and (NH4)2S04. The more reactive forms dissolve in hot alkali hydroxide solutions or fused KHS04. Addition of the OH- ion to BeCl2 or other beryllium solutions gives the hydroxide. This is amphoteric, and in alkali solution the beryllate ion, probably [Be(OH)4]2~, is obtained. When these solutions are boiled, the most stable of several polymorphs of the hydroxide can be crystallized. Beryllium halides, all four of which are known, are deliquescent and cannot be obtained from their hydrates by heating, since HX is lost as well as H20. The fluoride is obtained as a glassy hygroscopic mass by heating (NH4)2BeF4. The glassy form has randomly oriented chains of ••• F2BeF2Be ••• similar to those in BeCl2 and BeBr2 but disordered. Two crystalline modifications are known, which appear to be structurally analogous to the quartz and cristobalite modifications of Si02 (Section 9-7). At 555°C BeF2 melts to a viscous liquid that has low electrical conductivity. The polymerization in the liquid may be lowered by addition of LiF, which forms the [BeF4]2~ ion. Beryllium chloride is prepared by passing CC14 over BeO at 800°C. On a small scale the chloride and bromide are best prepared pure by direct inter¬ action in a hot tube. The white crystalline chloride (mp 405°C) dissolves exothermically in water; from HC1 solutions the salt [Be(H20)4]Cl2 can be obtained. Beryllium chloride is readily soluble in oxygenated solvents such as ethers. In melts with alkali halides, chloroberyllate ions [BeCl4]2~ may be formed, but this ion does not exist in aqueous solution. On interaction of Be with NH3 or N2 at 900 to 1000°C the nitride Be3N2 is
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
149
obtained as colorless crystals, readily hydrolyzed by water. The metal reacts with ethylene at 450°C to give BeQ. 5-5.
Complex Chemistry
Oxygen Ligands. In strongly acid solutions the aqua ion [Be(H20)4]2+ occurs, and crystalline salts with various anions can be readily obtained. The water in such salts is more firmly retained than is usual for aquates, indicating strong binding. Thus the sulfate is dehydrated to BeS04 only on strong heat¬ ing, and [Be(H20)4]Cl2 loses no water over P2Os. Solutions of beryllium salts are acidic; this may be ascribed to the acidity of the aqua ion, the initial dissociation being [Be(H20)J12 +
[Be(H20)3(0H)]+ + HH
The addition of soluble carbonates to beryllium salt solutions gives only basic carbonates. Beryllium salt solutions also have the property of dissolving ad¬ ditional amounts of the oxide or hydroxide. This behavior is attributable to the formation of complex species with Be—OH—Be or Be—O Be bridges^ The rapidly established equilibria involved in the hydrolysis of the [Be(H20)4] + ion are very complicated and depend on the anion, the concentration, the temperature, and the pH. The main species, which will achieve four-coor¬ dination by additional water molecules, are considered to be Be2(OH)3+ and Be3(OH)i+ (probably cyclic).3 The [Be3(OH)3]3+ ion predominates at pH 5.5 in perchlorate solution. Various crystalline hydroxo complexes have been isolated. In concentrated alkaline solution the main species is [Be(OH)4]2“. Other complexes of oxygen ligands are mainly adducts of beryllium halides or alkyls with ethers, ketones, and so on [e.g., BeCl2(OEt2)2]. There are also neutral complexes of p-diketones and similar compounds, of which the acetylacetonate is the simplest, and solvated cationic species such as [Be(DMF)4]2+ . The most unusual complexes have the formula Be40(02CR)6 and are formed by refluxing the hydroxide with carboxylic acids. These white crystalline compounds are soluble in organic solvents, even alkanes, but are insoluble in water and lower alcohols. They are inert to water but are hy¬ drolyzed by dilute acids; in solution they are un-ionized and monomeric. They have the structures illustrated in Fig. 5-2. The central oxygen atom is tetrahedrally surrounded by the four beryllium atoms (this being one of the few cases, excepting solid oxides, in which oxygen is four-coordinate), and each beryllium atom is tetrahedrally surrounded by four oxygen atoms. Zinc also forms such complexes, as does the Zr02+ ion, with benzoic acid The zinc complexes are rapidly hydrolyzed by water, in contrast to those of beryllium. The acetate complex has been utilized as a means of purifying beryllium by solvent extraction from an aqueous solution into an organic layer. When BeC 2 3R. N. Sylva et al., J. Chem. Soc. Dalton Trans., 1983, 2001.
150
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
• Be © Central oxygen atom
FIG. 5-2. The structure of the basic carboxylate complexes Be40(02CR)6. Only three RC02 groups are shown.
is dissolved in N204 in ethyl acetate, crystalline Be(N03)2-2N204 is obtained. When heated at 50°C this gives Be(N03)2, which at 125°C decomposes to N204 and volatile Be40(N03)6. The structure of the latter appears to be similar to that of the acetate but with bridging nitrate groups. The basic nitrate is insoluble in nonpolar solvents. The only halogeno complexes are the tetrafluoroberyllates, which are ob¬ tained by dissolving BeO or Be(OH)2 in concentrated solutions or melts of acid fluorides such as NH4HF2. The tetrahedral ion has a crystal chemistry similar to that of S04~, and corresponding salts (e.g., PbBeF4 and PbS04) usually have similar structures and solubility properties. Beryllium (II) fluor¬ ide (BeF2) readily dissolves in water to give mainly BeF2(H20)2 according to 9Be nmr spectra. In 1 M solutions of (NH4)2BeF4 the ion BeF3" occurs from 15 to 20%. The interaction between Cl- and [Be(H20)4]2+ is very small and may be outer sphere in Nature. Other Complexes. The stability of complexes with ligands containing ni¬ trogen or other atoms is lower than those of oxygen ligands. Thus [Be(NH3)4]Cl2 is thermally stable but is rapidly hydrolyzed in water. When BeCl2 is treated with the Li salt of 2,2'-bipyridine, a green paramagnetic complex is formed, which is best regarded as a complex of Be2+ with the bipyridinyl radical anion. Most of the other nitrogen complexes are derived from the hydride (Section 3-13) or organoberylliums, although compounds are known such as [Be(NMe2)2]3, which have a central four-coordinate Be and terminal threecoordinate Be atoms with both bridge and terminal NMe2 groups (5-III). 5-6.
Organoberyllium Compounds4
Although beryllium alkyls can be obtained by the interaction of BeCl2 with lithium alkyls or Grignard reagents, they are best made in a pure state by heating the metal and a mercury dialkyl, for example: HgMe2 + Be
BeMe2 + Hg
4N. A. Bell, Comprehensive Organometallic Chemistry, Vol. 1, Chapter 4, Pergamon Press Oxford, 1981.
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
151
The alkyl can be collected by sublimation or distillation in a vacuum. On the other hand, the aryls are made by reaction of a lithium aryl in a hydrocarbon with BeCl2 in diethyl ether in which the LiCl formed is insoluble, for example: 2LiC6H5 + BeCl2-» 2LiCl | + Be(C6H5)2 The alkyls are liquids or solids of high reactivity, spontaneously flammable in air and violently hydrolyzed by water. The bonding in polymers like (BeMe^ is of the 3c-2e type. The aryls can be made by the reaction: 3BeEt2 + 2B(aryl)3-» 3Be(aryl)2 + 2BEt3 The o- and m-tolyl compounds are dimers, presumed to have structure (5-VI).
The higher alkyls are progressively less highly polymerized; diethyl- and diisopropylberyllium are dimeric in benzene, but the t-butyl is monomeric, the same feature is found in aluminum alkyls. As with several other elements, notably Mg and Al, there are close sim¬ ilarities between the alkyls and hydrides, especially in the complexes with donor ligands. For the polymeric alkyls, especially BeMe2, strong donors such as Et20, Me3N, or Me2S are required to break down the polymeric structure. Mixed hydrido alkyls are known: thus pyrolysis of diisopropylberyllium gives a colorless, nonvolatile polymer: jc(«o-C3H7)2Be- [(«o-C3H7)BeH]x + xC3H6 However, above 100°C the t-butyl analogue gives pure BeH2. With tertiary amines, reactions of the following types may occur. BeMe2 + Me3N- Me3N-BeMe2 2BeH2 + 2R3N-> [R3NBeH2]2 The trimethylamine hydrido complex appears to have structure (5-VII) Hv
S
HU
Bef
Me3N
(5-VII)
^NMe3 Be
VH
152
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
Beryllium alkyls give colored complexes with 2,2'-bipyridine [e.g., bipyBe(C2H5)2, which is bright red]; the colors of these and similar complexes with aromatic amines given by beryllium, zinc, cadmium, aluminum, and gallium alkyls are believed to be due to electron transfer from the M—C bond to the lowest unoccupied orbital of the amine. Beryllium forms cyclopentadienyl compounds, some of which have unor¬ thodox structures because of the small size of the beryllium atom. In (C5H5)BeX molecules, where X is an ordinary univalent atom or group (e.g., Cl, Br, —CH3 or HC^C—), the ring is attached to the metal in a symmetrical, pentahapto fashion giving the entire molecule C5v symmetry. For the very air-sensitive (C5H5)2Be, the structure has been more difficult to ascertain with certainty. A perfect D5d or D5h structure like that found for the transition metal compounds definitely does not occur, and in view of the size of the Be atom, could not be expected. One ring appears to be symmetrically bonded to Be and the other is variously described as cr bonded, “slipped,” or ionically bonded. The beryllium atoms appear to be disordered in the crystals at both 25 and - 120°C, making definitive interpretation impossible.
MAGNESIUM, CALCIUM, STRONTIUM, BARIUM, AND RADIUM 5-7.
Occurrence; The Elements
Except for radium, the elements are widely distributed in minerals and in the sea. They occur in substantial deposits such as dolomite (CaC03MgC03), carnallite (MgCl2-KCl-6H20), and barytes (BaS04). Calcium is the third most abundant metal terrestrially. Magnesium is produced in several ways. An important source is dolomite from which, after calcination, the calcium is removed by ion exchange using seawater. The equilibrium is favorable because the solubility of Mg(OH)2 is lower than that of Ca(OH)2: Ca(OH)2 Mg(OH)2 + Mg2+-> 2Mg(OH)2 + Ca2+ The most important processes for preparation of magnesium are (a) the electrolysis of fused halide mixtures (e.g., MgCl2 + CaCl2 -I- NaCl) from which the least electropositive metal Mg is deposited, and (b) the reduction of MgO or of calcined dolomite (MgO-CaO). The latter is heated with ferrosilicon: CaO-MgO + FeSi = Mg + silicates of Ca and Fe and the magnesium is distilled out. Magnesium oxide can be heated with coke at 2000°C and the metal deposited by rapid quenching of the high-temperature equilibrium, which lies well to the right: MgO + C ^=± Mg + CO
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
153
Magnesium, which currently sells for about twice the price of aluminum, may in the long run replace it in many applications because the supply avail¬ able in seawater is virtually unlimited. Calcium and the other metals are made only on a relatively small scale, by the reaction: 6CaO + 2A1
3Ca + Ca3Al206
or reduction of the halides with sodium. Radium is isolated in the processing of uranium ores; after coprecipitation with barium sulfate, it can be obtained by fractional crystallization of a soluble salt. Magnesium is a grayish-white metal with a surface oxide film that protects it to some extent chemically—thus it is not attacked by water, despite the favorable potential, unless amalgamated. It is readily soluble in dilute acids and is attacked by most alkyl and aryl halides in ether solution to give Grignard reagents. Highly reactive Mg can be made in various ways for use in reductions, for example, by reduction of the halide with molten Na or K or more conveniently by interaction of the metal with anthracene and Mel in THF, which gives an orange “adduct” 5 (cf. synthesis of MgH2, Section 3-14):
Mg
or by decomposition of MgH2 at low pressure and 250 C. Calcium and the other metals are soft and silvery, resembling sodium in their chemical reactivities, although somewhat less reactive. These metals are also soluble, though less readily and to a lesser extent than sodium, in liquid ammonia, giving blue solutions similar to those of the Group IA(1) metals (Section 4-3). These blue solutions are also susceptible to decomposition (with the formation of the amides) and have other chemical reactions similar to those of the Group IA(1) metal solutions. They differ, however, in that moderately stable metal ammines such as Ca(NH3)6 can be isolated on removal of solvent at the boiling point.
5-8.
Binary Compounds
Oxides.
These are white, high melting crystalline solids with NaCl type
lattices. Calcium oxide,6 mp 2570°C is made on a vast scale CaC03 ;=± CaO + C02(g)
AH°298 = 178.1 kJ mol 1
5B. Bogdanovic, Angew. Chem. Int. Ed. Engl., 1985, 24, 262. Chemistry and Technology of 6K. W. Watkins, J. Chem. Educ., 1983, 60, 60; R. S. Boynton, Lime and Limestone, 2nd ed., Wiley, New York, 1980.
154
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
The oxide is relatively inert, especially after ignition at high temperatures, but the other oxides react with water, evolving heat, to form the hydroxides. They also absorb carbon dioxide from the air. Magnesium hydroxide is in¬ soluble in water (~1 x 10'4 g/L at 20°C) and can be precipitated from Mg2+ solutions; it is a much weaker base than the Ca-Ra hydroxides, although it has no acidic properties and unlike Be(OH)2 is insoluble in an excess of hydroxide. The Ca-Ra hydroxides are all soluble in water, increasingly so with increasing atomic number [Ca(OH)2, ~2g/L; Ba(OH)2, —60 g/L at —20°C], and all are strong bases. There is no optical transition in the electronic spectra of the M2+ ions, and they are all colorless. Colors of salts are thus due only to colors of the anions or to lattice defects. The oxides may also be obtained with defects, and BaO crystals with —0.1% excess of metal in the lattice are deep red. Halides. The anhydrous halides can be made by dehydration (Section 148) of the hydrated salts. For rigorous studies, however, magnesium halides are best made by the reaction Mg + HgX2 boiling ethe> MgX2(solv) + Hg Magnesium and calcium halides readily absorb water. The tendency to form hydrates, as well as the solubilities in water, decrease with increasing size, and Sr, Ba, and Ra halides are normally anhydrous. This is attributed to the fact that the hydration energies decrease more rapidly than the lattice energies with the increasing size of M2+. The fluorides vary in solubility in the reverse order (i.e., Mg < Ca < Sr < Ba) because of the small size of the F~ relative to the M2+ ion. The lattice energies decrease unusually rapidly because the large cations make contact with one another without at the same time making contact with the F" ions. The alkaline earth halides are all typically ionic solids, but can be vaporized as molecules, the structures of which are not all linear (cf. Table 1-4). On account of its dispersion and transparency properties, CaF2 is used for prisms in spectrometers and for cell windows (especially for aqueous solutions). It is also used to provide a stabilizing lattice for trapping lanthanide + 2 ions. Carbides. All the metals in the Ca-Ba series or their oxides react directly with carbon in an electric furnace to give the carbides MC2. These are ionic acetylides whose general properties [hydrolysis to M(OH)2 and C2H2, struc¬ tures, etc.] are discussed in Chapter 8. Magnesium at —500°C gives MgQ> but, at 500 to 700°C with an excess of carbon, Mg2C3 is formed, which on hydrolysis gives Mg(OH)2 and propyne and is presumably ionic, that is,
(Mg2+)2(cn.
Other Compounds.
Direct reaction of the metals with other elements can lead to binary compounds such as borides, silicides, arsenides, and sulfides. Many of these are ionic and are rapidly hydrolyzed by water or dilute acids. At —300°C, Mg reacts with N2 to give colorless, crystalline Mg3N2 (resembling Li and Be in this respect). The other metals also react normally to form M3N2, but other stoichiometries are known. An interesting compound is Ca2N, which
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
155
has an anti-CdCl2 type of layer structure, as does Cs20. In Ca2N, however, there is one “excess” electron per formula unit. These excess electrons evi¬ dently occupy delocalized energy bands within metal atom layers, causing a lustrous, graphitic appearance. The hydrides are discussed in Section 3-13; a complex salt KMgH3 has been prepared.
5-9.
Oxo Salts, Ions, and Complexes7
All the elements of Group IIA(2) form oxo salts, those of magnesium and calcium often being hydrated. The carbonates are all rather insoluble in water, and the solubility products decrease with increasing size of M2+. The same applies to the sulfates; magnesium sulfate is readily soluble in water, and calcium sulfate has a hemihydrate 2CaS04H20 (plaster of Paris) that readily absorbs more water to form the very sparingly soluble CaS04-2H20 (gypsum); Sr, Ba, and Ra sulfates are insoluble and anhydrous. The nitrates of Sr, Ba, and Ra are also anhydrous, and the last two can be precipitated from cold aqueous solution by addition of fuming HN03. Magnesium perchlorate is used as a drying agent. For water, acetone, and methanol solutions, nmr studies have shown that the coordination number of Mg2+ is 6, although in liquid ammonia it appears to be 5. The [Mg(H20)6]2+ ion is not acidic and in contrast to [Be(H20)4]2+ can be dehydrated fairly readily; it occurs in a number of crystalline salts. The complexity constants for Mg, Ca, Sr, and Ba vary greatly; there are three main groups.8 1. For small or highly charged anions and certain uni- and bidentate li¬ gands, the constants decrease with increasing crystal radii, Mg > Ca > Sr > Ba. 2. For oxoanions like NOf, SOI", and I04; the order accords with the hydrated radii, Mg < Ca < Sr < Ba. 3. For hydroxycarboxylic, polycarboxylic, and polyaminocarboxyhc acids ligands the order is Mg < Ca > Sr > Ba. The formation constants for cryptates, which show greatly enhanced sta¬ bility and selectivity compared to crown ethers can be ~106 times greater than those of the ethers.9 10 As for Na+ and K+, there are both synthetic and natural ligands with selective affinity for Ca2+, the latter exercising a con¬ trolling effect in Ca2+ metabolism. One example is the antibiotic A23i87, a monocarboxylic add that binds and transports Ca2 across ‘S a tridentate and forms a seven-coordinate complex Ca(N,0,0)2(H2U). Oxygen chelate compounds, among the most important being those of the ethylenediaminetetraacetate (EDTA) type, readily form complexes m alka7N. S. Poonia and A. V. Bajag, Chem. Rev., 1979, 79, 389. 8R. Aruga, Inorg. Chem., 1980, 19, 2895. . •B. G. Cox el at, J. Am Chem. Soc., 1984, 106, 1273; C. K. Schauer and O. P. Anderson, J. Am. Chem. Soc., 1987, 109, 3646. 10G. D. Smith and W. L. Duax, J. Am. Chem. Soc., 1976, 98, 15/8.
156
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
line aqueous solution, for example: Ca2+ + EDTA4- = [Ca(EDTA)]2The complexing of calcium by EDTA and also by polyphosphates is of im¬ portance, not only for removal of calcium ions from water, but also for the volumetric estimation of calcium. In Ca[CaEDTA]-7H20 the coordination in the anion is 8.Ua An L-aspartate complex116 is pharmacologically active for transport of Mg through membranes. Unusual thermally stable and petroleum soluble complexes of Sr and Ba have been madellc by the reaction: Ba(0-t-Bu)2 + [Sn(Q-r-Bu)2]2-» BaSn2(OT-Bu)6 These have a “sandwich” type of structure (5-VIII) with an MOe group and trigonal planar oxygen atoms. Calcium is evidently too small to form such a species.
(5-VIII)
Ammonia and amines (other than cryptates) do not form complexes in solution but the halides give nonstoichiometric adducts, for example, MCl2-nNH3. In liquid ammonia in the presence of anthracene, barium gives the green radical anion,12 Ba(NH3)2(anthracene2)“ . The only nitrogen com¬ plexes of significance are the magnesium tetrapyrroles, the parent compound of which is porphine (5-IX). These conjugated heterocycles provide a rigid planar environment for Mg2+ (and similar) ions. The most important of such derivatives are the chlorophylls and related compounds, which are of tran¬ scendental importance in photosynthesis in plants. The structure of chlorophyll-a, one of the many chlorophylls, is (5-X).
(5-IX) UaB. L. Barnett and V. A Uchtman, Inorg. Chem., 1979, 18, 2674. nbH. Schmidbaur etal., Angew. Chem. Int. Ed. Engl., 1986, 25, 1013. llcM. Veith et al., Angew. Chem. Int. Ed. Engl., 1986, 25, 375. 12G. R. Stevenson and L. E. Schock, J. Am. Chem. Soc., 1983, 105, 3742.
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
157
Phytyl =
(5-X)
In such porphine compounds the Mg atom is formally four-coordinate but further interaction with either water or other solvent molecules is a common, if not universal, occurrence; furthermore, in chlorophyll, interaction with the keto group at position 9 in another molecule is also established. It also appears that five-coordination is preferred over six-coordination as in the structure of magnesium tetraphenylporphyrin hydrate, where the Mg atom is out of the plane of the N atoms and is approximately square pyramidal. Although
FIG. 5-3. Structure illustrating chlorophyll-a-water-chlorophyll-fl interaction. The dimensions of the ring and the phytyl chain are not to scale. [Reproduced by permission from K. Ballschmiter and J. J. Katz, /. Am. Chem. Soc., 1969, 91, 2661].
158
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
Mg and other metalloporphyrins can undergo oxidation by one-electron changes, for Mg it is the macrocycle, not the metal, that is involved. In chlorophylls hydrogen-bonding interactions lead to polymerization (Fig. 5-3); the hydrates may be monomeric or dimeric in benzene, but ordered aggregates of colloidal dimensions are formed in dodecane. Where a polar solvent is not present, association via coordination of the keto group at po¬ sition 9 occurs as in solutions of anhydrous chlorophyll in alkanes. The role of chlorophyll in the photosynthetic reduction of C02 by water in plants is to provide a source of electrons that may continue to be supplied for a time in the dark. Electron spin resonance studies of light-irradiated chlorophyll show that radicals are formed. These are probably of the type (5-XI). The electrons are transmitted through chlorophyll micelles to other intermediates involved in the reduction of C02. % / \
% / N
N
0=0-^
xc=o—tK
/
h,
JC—OH
c, / \
0—Mg
\
^
0—Mg yj—ivif
X=0—-H/
N
Q
\
\ N
/\
\
CH,
/ \ CH,
(5-XI)
In tetraphenylporphyrin magnesium there can be weak binding, for ex¬ ample, by NCS~ in the octahedral.positions.13 Halide Complexes. These are of magnesium and exist mainly in the solid state. Fusion of CsCl and MgCl2 gives CsMgCl3 where MgCl6 is octa¬ hedral with opposite faces shared.14 The halides MgBr2, Mgl2, and CaCl2 are soluble in alcohols and some other organic solvents, as is Mg(C104)2. Ether adducts such as MgBr2(THF)4 and MgBr2(OEt)2)2 exist. The only simple anion is tetrahedral in (Me4N)2MgCl4. However, other complex ions have been obtained.15 Thus interaction of MgCl2(THF)2 and TiCl4(THF)2 gives [(THF)3Mg(|x-Cl)3Mg(THF)3] + [TiCl5THF]~ while Ti, Zr, or Sn tetrachlorides give complexes such as (EtOAc)4Mg(|x-Cl)2TiCl4.
5-10.
Organomagnesium Compounds16
The organo compounds of Ca, Sr, and Ba of the type RMI and ('n5-C5H5)2M can be prepared but are of little general utility compared to those of mag¬ nesium. 13G. A. Rodley el al., Inorg. Chem., 1984, 23, 4242. 14G. L. McPherson and L. A. Martin, J. Am. Chem. Soc., 1984, 106, 6884. 15P. Sobota et al., J. Chem. Soc. Dalton Trans., 1984, 2077; Z. Anorg. Allg. Chem., 1986, 533, 215; D. Gordon and M. G. H. Wallbridge, Inorg. Chim. Acta, 1984, 88, 15; H. J. de Liefde Meijer, J. Organomet. Chem., 1984, 269, 255. 16W. E. Lindsell, Comprehensive Organometallic Chemistry, Vol. 1, Chapter 4, Pergamon Press, Oxford, 1981 (also Ca, Sr, Ba); L. Brandsma and H. Verkruijsse, Preparative Polar Organo¬ metallic Chemistry, Vol. 1, B. M. Trost, Ed. Springer-Verlag, New York, 1986.
THE GROUP IIA(2) ELEMENTS: Be, Mg, Ca, Sr, Ba, Ra
159
Grignard reagents (MgRX) are probably the most widely used of all organometallic compounds in both organic and organometallic synthesis. They are made by interaction of Mg with an organic halide in an ether.17 The reaction is normally fastest with iodides and iodine can be used as an initiator for RC1 or RBr. The dialkyls (R2Mg) can be made by the dry reaction HgR2 + Mg(excess) = Hg + MgR2 or by removal of MgX2 from the Grignard by precipitation with dioxan and then recovery of R2Mg from the solution. They can also be obtained directly from Mg, H2, and alkene (cf. synthesis of A1R3, Section 7-10).518 This involves the catalyzed interaction using anthracene to give MgH2 (Section 3-13) fol¬ lowed by reaction with alkene at 70 to 90°C catalyzed by 1% ZrCl4: MgH2 + 2RCH=CH2 —-> Mg(CH2CH2R)2 The useful sandwich cyclopentadienide, Mg(C5H5)2, can be made by action of cyclopentadiene vapor on hot Mg or by thermal decomposition of C5H5MgBr, made in turn from the action of C5H6 on EtMgBr in ether. The structures of Me2Mg and Et2Mg are linear polymers similar to Me2Be with tetrahedral Mg and 3c-2e bridging alkyl groups. Both R2Mg and RMgX give stable tetrahedral adducts with donors such as A,A,A\A'-tetramethylethylenediamine (TMED). The Nature of Grignard Reagents. There has been prolonged controversy concerning the nature of Grignard reagents. Discordant results have often been obtained because of failure to eliminate impurities, such as traces of water or oxygen, which can aid or inhibit the attainment of equilibrium, and the occurrence of exchange reactions. Although recent work has given a reasonable understanding, the following discussion probably applies only to Grignard reagents prepared under strict conditions, not to those normally prepared without special precautions in the laboratory. First, the structures of several crystalline Grignard reagents have been determined. In PhMgBr(OEt2)2 and EtMgBr(OEt2)2 the Mg atoms are es¬ sentially tetrahedral with structures of the form (5-XII). For less sterically demanding ethers such as THF, five-coordinate species may occur as in MeMgBr(THF)3 (cf. MgBr2(OEt2)2 versus MgBr2(THF)4). Thus it is clear that
,M§V
in crystals the basic Grignard structure is RMgX-n(solvent). R
OEt2
Br
OEt2
(5-XII)
The nature of Grignard reagents in solution is complex and depends crit¬ ically on the alkyl and halide groups and on the solvent, concentration, and
17Y.-H. Lai, Synthesis, 1981, 585. 18B. Bogdanovic, Chem. Br., 1985, 21, 711.
BC13 ~ BH3 ^ BF3 > BMe3. The triiodide is prepared by the action of iodine on NaBH4 or of HI on BC13 at red heat. It is an unstable white solid that polymerizes on standing, is explosively hydrolyzed by water, and acts as a Lewis acid. Tetrachloroborates are obtained by addition of BC13 to alkali chlorides at 7aD. Mootz and M. Steffen, Angew. Chem. Int. Ed. Engl, 1981, 20, 196. ^W. Beck et at., J. Organomet. Chem., 1983, 252, 187.
176
THE CHEMISTRY OF THE MAIN GROUP ELEMENTS
high pressures, by cold milling at room temperatures, or by the reaction [(C2H5)4N] + C1- + BC13-1 [(C2H5)4N]+BCl4The stability of these salts and the corresponding tetrabromoborates and tetraiodoborates is greatest with the largest cations. With a given cation, the stability order is MBC14 > MBBr4 > MBI4, tetraiodoborates occurring only with the largest cations. Mixed ions such as BF3C1_ also exist. 6-6.
Halides with B—B Bonds8
Diboron Tetrahalides (B2X4). These are known for X = F, Cl, Br, and I, but the last has been little studied. The B2F4 molecule has a planar structure in both crystalline and gaseous phases, whereas B2C14, though planar in the crystal has a Du structure in the vapor (as does B2Br4 also). The B—B distances are 1.70 to 1.75 A consistent with these being a single, a, B—B bond. The staggered structure is favored by the X-X nonbonded repulsions, but B—X tt bond conjugation across the B—B unit may help to stabilize the planar structure. The two rotamers are not very different in stability (% -C
T3
TABLE 6-3
The More Important Properties of Boranes0
3 a t-H
C/5
-2 3 p w C/5
•Z 3a! CQ iri
Cu
d-
n
X> -
ON in CN 00
ON i-H
K
P
S
in NO
8 22 o1 O 00
pa
X
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12820 | https://math.stackexchange.com/questions/232309/how-to-interpret-material-conditional-and-explain-it-to-freshmen | Skip to main content
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After studying mathematics for some time, I am still confused.
The material conditional “→” is a logical connective in classical logic. In mathematical texts one often encounters the symbol “⇒”, which is read as “implies” or “if … then ….” It is customary and reasonable to treat “⇒” as the material conditional, i. e. as equivalent to saying “the antecedent is false or the consequence is true (or both)”.
Now there is some controversy about whether the material conditional really captures conditional statements because it doesn't really say anything about a causal connection between the antecedent and the consequence. This is quite often illustrated by the means of statements in natural languages such as “the moon is made out of cheese ⇒ all hamsters are green” – since the moon isn't made out of cheese, is this statement true? This remained problematic to me.
While I came to accept the material conditional as a good way of describing implications and conditionals, I'm having a hard time to explain this usage to freshmen whenever I get asked.
My questions are: How can we best justify the interpretation of “⇒” as a material conditional? Why is it so well-suited for mathematics? How can we interpret or read it to understand it better? Can my confusion about it be led back to some kind of misunderstandig or misinterpretation of something?
I have yet very poor background in mathematical logic (I sometimes browse wikipedia articles about it), but I'd have no problem with a technical answer to this question if it clarifies the situation.
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edited Oct 5, 2016 at 17:35
k.stmk.stm
asked Nov 7, 2012 at 18:29
k.stmk.stm
19.2k44 gold badges3838 silver badges7373 bronze badges
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see this recent post. There, Rick Decker gives a great example to use to convey the "logic" of material implication. There's also a reference to John Corcoran's explication of the "Meanings of Implication", which gives greater justice to how "implies" is used and what those uses means. Also, If you search the site using the tag logic and the word "implies" and/or the words "if then", you'll turn up a LOT of ways to explicate the notion of material implication!
amWhy
–
amWhy
2012-11-07 18:51:42 +00:00
Commented
Nov 7, 2012 at 18:51
See also here and here.
amWhy
–
amWhy
2012-11-07 19:03:42 +00:00
Commented
Nov 7, 2012 at 19:03
1
...and here.
amWhy
–
amWhy
2012-11-07 19:09:38 +00:00
Commented
Nov 7, 2012 at 19:09
@amwhy I stumbled upon some of these postings. And incidentally just now upon the last one you refered to. It really has a potential to clear things up for me. Many thanks! Still, it would be nice to have some other interpretation of the material implication, e. g. way to read it, which captures its truth-functional character better. I guess this is the main part of my question.
k.stm
–
k.stm
2012-11-07 19:22:02 +00:00
Commented
Nov 7, 2012 at 19:22
Are there other kinds of impliations in math? I seen somthing called a "formal implication" but this donst seem well suited. Am I right?
user123124
–
user123124
2015-11-01 14:11:10 +00:00
Commented
Nov 1, 2015 at 14:11
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The original question asked "Why is the material conditional so well-suited for mathematics?" Here's a central consideration which others have not touched on.
One thing mathematicians need to be very clear about is the use of statements of generality and especially statements of multiple generality – you know the kind of thing, e.g. the definition of continuity that starts for any ϵ ... there is a δ ... And the quantifier-variable notation serves mathematicians brilliantly to regiment statements of multiple generality and make them utterly unambiguous and transparent. (It is when we come to arguments involving generality that borrowing notation from logic to use in our mathematical English becomes really helpful.)
Quantifiers matter to mathematicians, then: that should be entirely uncontentious. OK, so now think about restricted quantifiers that talk about only some of a domain (e.g. talk not about all numbers but just about all the even ones). How might we render Goldbach's Conjecture, say? As a first step, we might write
∀n(if n is even and greater than 2, then n is the sum of
two primes)
Note then, we restrict the universal quantifier by using a conditional. So now think about the embedded conditional here.
What if n is odd, so the antecedent of the conditional is false. If we say this instance of the conditional lacks a truth-value, or may be false, then the quantification would have non-true instances and so would not be true! But of course we can't refute Goldbach's Conjecture by looking at odd numbers!! So in these cases, if the quantified conditional is indeed to come out true when Goldbach is right, then we'll have to say that the irrelevant instances of the conditional with a false antecedent come out true by default. Come out "vacuously" true, if you like. In other words, the embedded conditional will have to be treated as a material conditional which is true when the antecedent is false.
So: to put it a bit tendentiously and over-briefly, if mathematicians are to deal nicely with expressions of generality using the quantifier-variable notation they have come to know and love, they will have to get used to using material conditionals too.
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edited Feb 16, 2014 at 14:54
answered Nov 7, 2012 at 22:53
Peter SmithPeter Smith
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Peter, my mistake. What you point out in your comment is correct: you referred to "a" central consideration, not to "the" central consideration i.e. you asserted "∃... vs. (∃!)... central consideration!". I'll delete my comment accordingly and add that I found your post to be "a central" and enriching answer to the post.
amWhy
–
amWhy
2012-11-08 14:42:31 +00:00
Commented
Nov 8, 2012 at 14:42
@amWhy No problem -- so I'll delete my comment too! :-)
Peter Smith
–
Peter Smith
2012-11-08 16:30:35 +00:00
Commented
Nov 8, 2012 at 16:30
I think, this is indeed a key point. Therefore I accept this answer, even though I feel there is more to it. I also find the answer given here, as also mentioned by amWhy, quite enlightening.
k.stm
–
k.stm
2012-11-15 07:47:42 +00:00
Commented
Nov 15, 2012 at 7:47
I agree to answer give by Peter Smith which is accepted as correct answer. I just want to mention that, It took me a while to understand, interpret and comprehend what Peter Smith has to say in his answer. If you really want to get the gist of correct answer then I would recommend to go through following link - gowers.wordpress.com/2011/09/28/basic-logic-connectives-implies If you first read the above given link and then come back and re-read the correct answer by Peter, you will be certainly enlightened :).
zswap
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zswap
2013-09-22 08:00:34 +00:00
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Sep 22, 2013 at 8:00
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Perhaps this will help to capture the truth-functional character of material implication:
The truth-value of an inclusion (subset) relation between sets corresponds to the truth-value of an implication relation, where ⊆ corresponds to the → relation.
E.g., suppose A⊆B. Then if it is true that x∈A, then it must be true that x∈B, since B contains A. However, if x∉A (if it is false that x∈A), it does not mean that then x∉B, since if A⊆B, then B may very well contain elements that A does not contain.
Similarly, suppose we have that p→q. If p is true, then it must be the case that q is true. But if p is false, that does not necessarily mean then that q is necessarily false. (For all we know, perhaps q is true regardless of whether or not p is true.) So q can be true, while p is false.
I don't know if this analogy helps or not. But it was the above analogy (correspondence) that helped me to firmly grasp the logic of material implication.
Here's a more down-to-earth example you may have already stumbled upon:
CLAIM: "If (it rains), then (I'll take an umbrella)":
I'd be lying (my assertion would be false) if (it rains = true), and I do not (take an umbrella).
But perhaps it's cloudy out, and I decide I'll take an umbrella , just in case it rains. In this case:
If it doesn't rain (it rains = false), but I took my umbrella (true), my claim above would not be a lie (it would not be false).
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edited Dec 9, 2012 at 3:47
answered Nov 7, 2012 at 19:56
amWhyamWhy
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An intuitive way to understand the material conditional is as a promise. If you hold up your end of the deal (so the antecedent is true), then I must hold up mine (the conclusion is true). But if you break the promise, then I can do what ever I want without me breaking the promise. And if I am going to hold up my end of the bargain no matter what, it doesn't particularly matter if you hold up yours.
And as far as why it works so well in mathematics, it might be because it is truth functional and mimics some part of what implication means in a non-mathematical context. It doesn't have anything to do with causality, which is often how it is used out of math; it only relies on the truth values of the constituent statements.
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answered Nov 7, 2012 at 23:32
Francis AdamsFrancis Adams
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The issue seems to be with the behaviour of p→q when p is false. If p→q were false when p is false, then you could conclude p from p→q without any extra premises, and therefore guarantee q as well. That betrays the idea of this being a conditional.
If (you stole the cookie) then (you're a horrible person)
Just from this, which most people with a sweet tooth would accept, I could then conclude that you stole the cookie and are in fact a horrible person. I don't need to assert that you stole the cookie separately, I can just conclude it from the truth of this statement. This behaviour loses the important "if" part of the conditional. To capture
if P (happened), then Q (would happen)
we need to allow for vacuous truths. As a positive example:
If (you won the bet) then (I would have paid you)
Is still true whether you actually win the bet or not.
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answered Nov 7, 2012 at 20:12
Robert MastragostinoRobert Mastragostino
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The material conditional P => Q expresses an ordering relationship among two statements such that Q is "not less true" than P. It is only concerned with comparing truth values and not with what P and Q mean nor how they are related.
Its use is intended to prevent us from starting with true assumptions and reaching false conclusions.
How we know that Q is at least as true as P is a different matter. The bare statement P => Q says nothing about how we know it is true; whether it is a bit of useless trivia, a useful working assumption, or a derived conclusion.
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answered Feb 23, 2014 at 5:52
ConfutusConfutus
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The material conditional is often simply defined as follows:
A⟹B ≡ ¬(A∧¬B)
This "definition" turns out to be a theorem of classical propositional logic that can be derived from what might be called "first principles" using a form of natural deduction.
First, we need to prove that (A⟹B)⟹¬(A∧¬B)
A⟹B (Assume)
A ∧ ¬B (Assume, to obtain a contradiction)
A (Elim ∧, 2a)
¬B (Elim ∧, 2b)
B (Elim ⟹, 1, 3)
B∧¬B (Intro ∧, 5, 4)
¬(A ∧ ¬B) (Intro ¬, 2, 6)
(A⟹B)⟹¬(A∧¬B) (Intro ⟹, 1, 7)
Now, we need to prove that ¬(A∧¬B)⟹(A⟹B)
¬(A∧¬B) (Assume)
A (Assume)
¬B (Assume, to obtain a contradiction)
A∧¬B (Intro ∧, 2, 3)
(A∧¬B)∧¬(A∧¬B) (Intro ∧, 4, 1)
¬¬B (Intro ¬, 3, 5)
B (Elim ¬, 6)
A⟹B (Intro ⟹, 2,7)
¬(A∧¬B)⟹(A⟹B) (Intro ⟹, 1, 8)
Combining these two results, we have as required:
A⟹B ≡ ¬(A∧¬B)
Note that the only properties of the implication operator (⟹) that are used here are:
Introduction of the ⟹ operator by means of conditional (or direct) proof resulting in the discharging of an active premise.
Elimination of the ⟹ operator by detachment (modus ponens)
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answered Dec 16, 2022 at 3:54
Dan ChristensenDan Christensen
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Yes. And also thank you for the comment on this you wrote a couple of months ago. I noted and thought about it. This is an extremely valuable answer, adding something new to the discussion. This motivation may be a tad too formal for freshmen, but the fact that the truth table of material implication is already forced in classical logic by the deduction rules we expect of it is in my opinion indeed the most powerful explanation I’ve seen. I mean, it just explains it.
k.stm
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I would like to show that the truth table of the " IF...THEN" operator is not so far from usual natural language conventions.
Remark. Here by " if ...then" I only talk about material implication ( not about logical/strict implication, that is " necessarily (if A then B) " ).
Let us adopt ( if you please) the following principles :
(1) "I cannot have said something wrong about a thing/ subject I did not talk about."
(2) When I talk in a declarative manner , either what I say is wrong/false, or what I say is "ok", "right", "true".
Now suppose I say " IF the winner of the next US Presidential Elections is a democrat, THEN this winner will be a woman."
What did I talk about? I only talked about the case in which the winner is a democrat.
And what did I say about this (possible) situation? I said that this situation will not be accompanied with a situation in which the winner is not a woman.
So the only case in which others are entitled to tell me that I said something wrong is the case in which (1) the winner is a democrat (2) but is not a woman.
In case the winner is a democrat and is a woman: I talked about this situation, and what I said is true.
In case the winner is not a democrat: I did not talk about this case, so, according to our two initial principles, what I said cannot be false relatively to this case, and consequently, my conditional assertion is true.
Remark. - I do not say that the corresponding logical implication is true in cases TT, TF and FF. Though I am right as long as it is not the case that the winner is a democrat and is not a woman, that does not mean that I would be right in saying : if the winner is a democrat, necessarily it will be a woman. This would amount to saying that the case " democrat + not a woman" is impossible. A much stronger claim than the one I have actually made
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edited Apr 20, 2019 at 13:48
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Sure you can have a valid "implication relationship" between two errant pieces of nonsense.
If we have M=>H between moon and hamster populations then the truth table is
M=>H == not(M and not H). In modal logic (many-world) terms this means we allow all worlds to be possible EXCEPT the one that has a cheesy moon and at least one non-green hamster.
This is the main effect of the implication relationship, it acts as a prohibition on one case, indeed this is the essence of an asymmetric dependency. M=>H means M depends on H in the sense that we cannot have M without H. If H is false then M cannot be true. Asymmetric dependency relationships are very common, hence the utility of this connector.
In the other direction it matters not, there are at least two worlds that have a normal moon, and in only one of them the hamsters are all green. You are NOT forced to live in that world!
The Wason selection task is a good illustration of utility of material conditional support for deductive reasoning:
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answered Apr 24, 2013 at 3:50
Keith AllpressKeith Allpress
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I don't understand how such a "test" makes sense. Why does the experimenter want to limit my ability to flip over the cards? The simplest method of solving the problem as stated consists of just flipping over all four cards. I don't have to assume that I know much at all that way. The exercise is NOT an investigation about some system where we only can do so much or some theory. It gets presented as a real-world empirical problem. So, why not use every tool that you can? Flipping over all the cards in this case is by no means unreasonable or confusing. So perhaps...
Doug Spoonwood
–
Doug Spoonwood
2013-07-20 16:59:43 +00:00
Commented
Jul 20, 2013 at 16:59
so many people "fail" this test, because it comes as so easy to spot that the experimenter is not particularly charitable to you. He wants to limit your tools and your ability to solve this problem. There's no background theory or even a hypothesis to possibly explain why this might help you. So why bother? Flip over all the cards and figure it out. The experimenter in this case comes as the person who fails because s/he comes as the person who wants to play arbitrary games with your head instead of actually treating you like a human being.
Doug Spoonwood
–
Doug Spoonwood
2013-07-20 17:02:57 +00:00
Commented
Jul 20, 2013 at 17:02
And the problem " can easily get solved with very little thought and it still takes very little effort to solve it by simply flipping over all the cards. It is not against any law or well-established convention to flip over all 4 cards (and there does not exist any background theory). Only the experimenter imposes a tacit requirement not flip over all 4 cards. Logic preferably will empower people. When you play games like this, I simply do not see how it will do so.
Doug Spoonwood
–
Doug Spoonwood
2013-07-20 17:07:06 +00:00
Commented
Jul 20, 2013 at 17:07
Well after flipping them all, the experimenter could ask "Did you have to flip them all?" and the answer would be just as revealing.
Keith Allpress
–
Keith Allpress
2013-10-20 10:26:52 +00:00
Commented
Oct 20, 2013 at 10:26
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From Wikipedia, the free encyclopedia
Culinary traditions of Shaanxi province, China
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| Shaanxi cuisine |
| Traditional Chinese | 陝西菜 |
| Simplified Chinese | 陕西菜 |
| | Transcriptions | | Standard Mandarin | | Hanyu Pinyin | Shǎnxī cài | |
| Qin cuisine |
| Chinese | 秦菜 |
| | Transcriptions | | Standard Mandarin | | Hanyu Pinyin | Qín cài | |
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| Part of a series on |
| Chinese cuisine |
| Regional cuisines Four Great Traditions Chuan (Sichuan) Lu (Shandong) Yue (Guangdong) Huaiyang (Jiangsu) Eight Great Traditions(+all above) Anhui Fujian Hunan Zhejiang Ten Great Traditions(+all above) Beijing Shanghai Twelve Great Traditions(+all above) Henan Shaanxi Fourteen Great Traditions(+all above) Hubei Liaoning Sixteen Great Traditions(+all above) Tianjin Yunnan New Eight Great Traditions Gansu Hangzhou Jilin Liaoning Ningbo Shaanxi Shanghai Shanxi Beijing and the vicinity Beijing Imperial Aristocrat Tianjin Other regional styles Teochew Guangxi Guizhou Hainan Haipai Hakka Hong Kong Jiangxi Macanese Manchu Northeastern Putian (Henghwa) Qinghai Taiwan Tibetan Xinjiang |
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| Religious cuisines Buddhist Islamic Taoist |
| Ingredients and types of food Main dishes Desserts Noodles |
| Preparation and cooking Stir frying Double steaming Red cooking |
| See also Customs and etiquette List of Chinese desserts List of Chinese dishes List of Chinese restaurants List of restaurants in China |
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Shaanxi cuisine, or Qin cuisine, is derived from the native cooking styles of Shaanxi Province and parts of northwestern China.
Description
[edit]
Shaanxi cuisine makes elaborate use of ordinary ingredients and is characterized by its noodles, lamb/mutton dishes, and heavy use of strong and complex savory flavors such as salt, garlic, onion and vinegar. Sugar is rarely used. The main cooking methods are steaming, frying and stir-frying.
Due to its geographical location between the provinces of Shanxi and Sichuan, the flavors of Shaanxi cuisine include both sour and spicy of Szechuan flavors and the salty flavors of Shanxi. Shaanxi cuisine's primary flavor profile is "fragrant spicy"(香辣).
Shaanxi cuisine uses more noodles than other Chinese cuisines, but Shaanxi noodles are almost always thicker and longer than those of Beijing cuisine, and to a lesser degree, Shanxi cuisine, especially the Biangbiang ones.
The taste of Shaanxi cuisine can be quite spicy. However, this can be diluted by adding soy sauce.[citation needed] Many different types of meat are used in Shaanxi cuisine such as duck, lamb, chicken, and beef. Additionally, there are spiced, vegetarian dishes where no meat is included, resulting in these dishes being considerably spicier.
Regional styles
[edit]
Shaanxi cuisine includes three regional styles:
Northern Shaanxi style is characterized by the wide use of steaming as the method of cooking. The most common meat is pork, although lamb and mutton are also popular. Beans and soups are also enjoyed by many.
Guanzhong style, which uses pork and lamb / mutton equally with heavy meaty flavors and tastes.
Hanzhong style, similar to Szechuan, is marked by spiciness.
Dishes
[edit]
The following is a selected list of dishes in Shaanxi cuisine.
| English | Chinese (Simplified; Traditional) | Pinyin | Image | Notes |
--- ---
| Biangbiang noodles | 面;麵 | biángbiáng miàn | Notable for its name being written with one of the most complex characters used in modern Chinese. |
| Crispy fried noodles | 金线油塔;金線油塔 | jīn xiàn yóu tǎ | |
| Cured beef and lamb | 腊牛羊肉;臘牛羊肉 | là niú yáng ròu | |
| Gourd head | 葫芦头;葫蘆頭 | húlu tóu | A dish made from pig's large intestine. It is called "gourd head" because the pig's large intestine looks like the top of a gourd. It originated as street food during the Northern Song dynasty. Since then, the dish has evolved to include variations with meat fillings inside the pig's large intestine, as well as the inclusion of other ingredients. Pictured with bottle of Ice Peak soda. |
| Guokui | 锅盔;鍋盔 | guōkuī | A kind of pancake made from flour. It is round in shape, about a foot long in diameter, an inch in thickness, and weighs roughly 2.5 kg. It is traditionally presented as a gift by a grandmother to her grandson when he turns one month old. |
| Hanzhong hot rice noodles | 汉中热米皮;漢中熱米皮 | Hànzhōng rè mǐ pí | |
| Lamb paomo | 羊肉泡馍;羊肉泡饃 | yáng ròu pào mó | Bread is shredded into pieces and placed in a boiling beef or mutton broth before serving. |
| Lamb soup | 水盆羊肉 | shuǐ pén yáng ròu | |
| Mingsixi | 明四喜 | míng sì xǐ | As Shaanxi is located far away from the coast, seafood is hardly featured in Shaanxi cuisine. This dish, composed of ingredients such as abalone, sea cucumber, squid, Shaohsing wine and clear chicken broth, is served only at major banquets and festivals in Shaanxi. |
| Qishan saozi noodles | 岐山臊子面;岐山臊子麵 | Qíshān sàozǐ miàn | |
| Zenggao | 甑糕 | zèng gāo[a] | A type of cake (糕; gāo) made from fully-soaked glutinous rice, peeled dates and kidney beans. It is historically made with an ancient Chinese steam cooker known as Zèng (甑), hence the name. |
| Roujiamo | 肉夹馍;肉夾饃 | ròujiāmó; ròugāmó | Often called "Chinese hamburgers", this famous Shaanxi dish can now be found everywhere in China. |
| Shaanxi-style liangpi | 陕西凉皮;陝西涼皮 | Shǎnxī liángpí | Cold wheat noodles with a chili and vinegar sauce. |
| Hulatang | 胡辣汤;胡辣湯 | hú là tāng | Originating from Henan, Hulatang is a rich, peppery soup, widely popular in Xi'an. The Xi'anese variant of this soup additionally includes meatballs, potatoes and carrots among other ingredients. |
| Vermicelli in lamb's blood soup | 粉汤羊血;粉湯羊血 | fěn tāng yáng xiě | A soup composed of ingredients such as congealed lamb's blood, paomo and other condiments. It is eaten with vermicelli and cilantro. |
Notes
[edit]
^ Locally, it is pronounced jìng gāo.
References
[edit]
^ Cosmo, Serena (October 24, 2017). The Ultimate Pasta and Noodle Cookbook: Over 300 Recipes for Classic Italian and International Recipes. Simon and Schuster. ISBN 978-1-60433-733-4.
^ Foo, Susanna (2002). Susanna Foo Chinese Cuisine: The Fabulous Flavors & Innovative Recipes of North America's Finest Chinese Cook. Houghton Mifflin Harcourt. ISBN 978-0-618-25435-4.
^ "Biangbiang Shaanxi street food - Lifestyle - Chinadaily.com.cn". www.chinadaily.com.cn. Retrieved March 21, 2024.
^ 人民音乐出版社编辑部; 陕北民歌博物馆编 (July 1, 2021). 陕北民歌映画. Beijing Book Co. Inc. ISBN 978-7-103-06077-3.
^ Tudoujiang (土豆酱) (June 25, 2015). "Shaanxi on the tip of your tongue (舌尖上的陕西)". Red Meal Net (红餐网) (in Chinese). Retrieved June 26, 2017.
^ a b 秦闻客户端 (November 28, 2024). ""jìng "还是"zèng " "甑糕"的"甑"怎么读" ["jìng" or "zèng" – How the "Zeng" in "Zenggao" is pronounced]. 人民日报 [People's Daily] (in Chinese). Retrieved September 10, 2025.
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| General | History Politics Economy |
| Geography | Cities Guanzhong Shaanbei Shaannan |
| Education | Chang'an University Fourth Military Medical University Northwest A&F University Northwest University Northwestern Polytechnical University Shaanxi Normal University Xi'an Jiaotong University Xidian University |
| Culture | Biángbiáng noodles Cuisine Music Qinqiang opera Three Qins Xintianyou |
| Visitor attractions | Daqin Pagoda Banpo Neolithic village Zhao Mausoleum Mount Hua Mausoleum of the First Qin Emperor and Terracotta Army Museum Bell Tower and Drum Tower of Xi'an Forest of Stone Steles Museum Shaanxi History Museum Qianling Mausoleum |
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12823 | https://www.quora.com/What-is-the-relation-between-angular-velocity-frequency-and-time-period | Something went wrong. Wait a moment and try again.
Angular Frequency
Time Period
Circular Kinematics
Frequency (physics)
Rotational Dynamics
Uniform Circular Motion
Science Physics
5
What is the relation between angular velocity, frequency, and time period?
Rithik T
Studied Physics & Mathematics at S N Central School Kayamkulam
·
2y
FIRST OF ALL, WE CAN UNDERSTAND THE TERMS,
A time period (denoted by the letter 'T') is the amount of time it takes for one complete cycle of vibration to pass through a given point. The time period of a wave decreases as its frequency increases. The time period is measured in 'seconds.'
Frequency and time period have a mathematical relationship that can be expressed as T = 1/f or f = 1/T.The angle made after one revolution is 360° or 2 x π (π =180°) The time period formula is given as:
Frequency is defined as the number of times a repetitive event occurs per unit of time. Ordinary frequen
FIRST OF ALL, WE CAN UNDERSTAND THE TERMS,
A time period (denoted by the letter 'T') is the amount of time it takes for one complete cycle of vibration to pass through a given point. The time period of a wave decreases as its frequency increases. The time period is measured in 'seconds.'
Frequency and time period have a mathematical relationship that can be expressed as T = 1/f or f = 1/T.The angle made after one revolution is 360° or 2 x π (π =180°) The time period formula is given as:
T=2π/ω.
Frequency is defined as the number of times a repetitive event occurs per unit of time. Ordinary frequency, rather than angular frequency, and temporal frequency, rather than spatial frequency, are other names for it. One hertz (Hz) unit of frequency equals one occurrence per second.
Angular Frequency is known as radial or circular frequency which measures the angular displacement per unit time.
HOPE YOU UNDERSTAND.THANK YOU
Related questions
What is the relation between angular speed and time period?
Why do we use the same symbol for angular velocity and angular frequency?
What is difference between angular velocity and angular frequency? And are angular velocity and angular frequency same?
What are the representations for angular velocity and angular frequency?
What are the core differences between angular frequency and angular velocity, both of which are denoted by ω?
DEEPAK SAINI
MSc in Physics, Chaudhary Charan Singh University (Graduated 2019)
·
Author has 113 answers and 76.5K answer views
·
2y
the relation between angular velocity and frequency is
if w is angular velocity and v is frequency then
w = 2×pi ×v
And angular frequency and time period relation is
w = (2×pi) / T
T is time period
v = 1/T
Manasa Murali
2y
We know that
angular velocity(ω)=2π/Time period(T)
We also know
Frequency (f) = 1/T
So
ω=2πf
Javier Garcia-Julve
Lived in Google
·
Author has 997 answers and 2.1M answer views
·
Updated 7y
Related
What is the relation between angular speed and time period?
If I have got your (literal) meaning right, this question is so easy that even I can answer it, and not needing to do any calculations, either.
The angular speed of a moving point is the angle it covers with respect to a center point, per unit of time. If it's moving in a circle, the period is the time it takes it to go through the same point on the circle again. This time will be, obviously, the angle it has covered—that is 360° or 2π radians—divided by the angle it covers in each unit of time, which is the angular speed :
T = 2π/ω
Or, conversely you could say that, in the particular case o
If I have got your (literal) meaning right, this question is so easy that even I can answer it, and not needing to do any calculations, either.
The angular speed of a moving point is the angle it covers with respect to a center point, per unit of time. If it's moving in a circle, the period is the time it takes it to go through the same point on the circle again. This time will be, obviously, the angle it has covered—that is 360° or 2π radians—divided by the angle it covers in each unit of time, which is the angular speed :
T = 2π/ω
Or, conversely you could say that, in the particular case of the period, the angle covered by the moving point is the full circle so, since the angular speed is the ratio of the angle covered— 2π— to the time he has taken—T—, it must be:
ω= 2π/T
Or directly, since the angular speed is the angle covered per unit of time, its product for the time it has been moving—the period in this case—is the angle covered in thet time, the full circle in this case:
2π =ω.T
And that's it. And it's also the last time I do your homework, although I do it gladly. Because I can't think of any other reason to ask this kind of question in Quora.
Related questions
What is the physical interpretation of angular frequency and angular velocity?
What is the difference between angular velocity and angular frequency?
What is angular frequency and angular velocity?
How can the units of angular velocity be shown to be the same as frequency or time period?
Is the angular velocity referred to as the angular frequency? Why?
Kip Ingram
PhD in Electrical Engineering, The University of Texas at Austin Cockrell School of Engineering (Graduated 1992)
·
Author has 20.1K answers and 21.6M answer views
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5y
Related
Is the angular velocity referred to as the angular frequency? Why?
Frequency refers to the number of complete cycles a system makes per second. In circular motion a cycle is one 360-degree trip around the circle. “Angular velocity” is related, but is measured in radians per second (there are 2pi radians in 360 degrees). So they refer to the same quantity, but with different units.
This is a slight misuse of the term “velocity” - you might also think of the angular velocity as the actual velocity of the thing that’s moving in a circle, in which case it’s equal to the radius of the circle multiplied by the radians per second. There’s probably a precise definiti
Frequency refers to the number of complete cycles a system makes per second. In circular motion a cycle is one 360-degree trip around the circle. “Angular velocity” is related, but is measured in radians per second (there are 2pi radians in 360 degrees). So they refer to the same quantity, but with different units.
This is a slight misuse of the term “velocity” - you might also think of the angular velocity as the actual velocity of the thing that’s moving in a circle, in which case it’s equal to the radius of the circle multiplied by the radians per second. There’s probably a precise definition for the term, but I’ve seen it used in both of the ways I just described and I’m not immediately familiar with what the definition is. I just know what’s being talked about when I read a problem; you can tell from context.
Hope this helps!
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Robert Toop
Author has 4.6K answers and 2.9M answer views
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Updated 1y
Related
Why does angular velocity point up?
Quick answer: sometimes it points down, so you have to read this long answer. Do not say TL;DR!
Imagine a disk rotating counterclockwise as you look down on it.
Every little bit of matter in the disk is moving from right to left, then from far to near, then from left to right, then from near to far, repeating the circular path at a frequency f times per second.
A bit’s instantaneous position can be described by the vector R that has constant magnitude |R| equal to the bit’s distance from the disk center. Call that positive scalar “r”. Vector R’s direction is from center to the bit, rotating count
Quick answer: sometimes it points down, so you have to read this long answer. Do not say TL;DR!
Imagine a disk rotating counterclockwise as you look down on it.
Every little bit of matter in the disk is moving from right to left, then from far to near, then from left to right, then from near to far, repeating the circular path at a frequency f times per second.
A bit’s instantaneous position can be described by the vector R that has constant magnitude |R| equal to the bit’s distance from the disk center. Call that positive scalar “r”. Vector R’s direction is from center to the bit, rotating counterclockwise through an angle from 0 to 360 degrees (0 to 2π radians), f times per second. The position vector R always lies in the plane of the disk.
The bit’s instantaneous velocity is the vector V whose magnitude |V| (call it “v”) is the length of our bit’s full circular path (2πr) times frequency f, so v = 2πrf is the scalar distance per second: the speed. Vector V’s direction is perpendicular to the bit’s position vector R (90 degrees = π/2 radians ahead), so V’s angle is π/2 radians greater than R’s angle. Like every bit’s position R, every velocity vector V is in the plane of the disk.
However, each bit in a different 2D place on the disk has a different velocity vector and position vector. We want to describe the rotation of all bits of the disk with one vector whose magnitude is proportional to f, and with one constant direction. Well, the vectors for bit position R and velocity V are all in the disk’s plane, so why not choose our magic vector’s direction to be normal to that plane? There can be just one such vector, and it always points along the disk’s axis of rotation !!
Let’s call this magic vector “angular frequency” or “angular velocity” and give it the lowercase Greek letter “omega”: ω.
Define it to point toward us as we look down on counterclockwise rotation, or point away from us if we look down on clockwise rotation.
Let’s make ω‘s magnitude |ω| = ω = 2πf, the number of radians per second of every bit of the disk.
In general, if given position vector R for a particular bit of the disk, and vector ω, we know the bit’s velocity vector V = ω X R where X is the cross-product of angular frequency and position vectors; the right-hand rule applies. For circular applications like our disk where R and V are perpendicular, a bit at distance r from the center of rotation has tangential speed v = ωr, and centripetal acceleration = ω^2r, using the scalar magnitudes of ω and R. Direction is your problem.
Given bit position and velocity vectors R and V, angular velocity vector ω = (R X V)/(r^2). For circular applications, magnitude ω = v/r, and centripetal acceleration magnitude = v^2/r. As above, direction is your problem.
Neat, is it not?
A D Telang
Former Faculty at Goa Engineering College (1983–2010)
·
Author has 1.1K answers and 1M answer views
·
3y
Related
What are the core differences between angular frequency and angular velocity, both of which are denoted by ω?
We use the term angular velocity in relation to the rotating machine elements like flywheel, shaft, gears etc. The angular velocity in terms of RPM is
ω=2 π n60 rad/sec
where,
n= revolutions per minute
The term angular frequency is used while dealing with the vibratory system. The motion is visualized with the help of rotating phasor (solution of the governing equation leads to it). In one rotation of phasor one cycle is completed. The number of cycles completed in one second, called as frequency in Hertz is
f=ω2 π cycles/sec (Hertz)
where,
ω= angula
We use the term angular velocity in relation to the rotating machine elements like flywheel, shaft, gears etc. The angular velocity in terms of RPM is
ω=2 π n60 rad/sec
where,
n= revolutions per minute
The term angular frequency is used while dealing with the vibratory system. The motion is visualized with the help of rotating phasor (solution of the governing equation leads to it). In one rotation of phasor one cycle is completed. The number of cycles completed in one second, called as frequency in Hertz is
f=ω2 π cycles/sec (Hertz)
where,
ω= angular velocity of the phasor
Frequency of vibration, in Hertz, is related to the angular velocity by a constant 2 π. And therefore the frequency is specified, very often, as angular frequency ‘ω′ in radians/sec
In a way both the angular frequency and angular velocity are same. Both are referring to the rotation. One is used to specify the angular velocity of the phasor (and hence frequency of vibration) and the other for the rotating components mentioned above.
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Jonatan Kelu
B.Sc. in Physics, University of New South Wales (Graduated 2000)
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Author has 62 answers and 252.8K answer views
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6y
Related
Is angular frequency and angular velocity the same?
They’re related but not the same. Angular velocity is a vector quantity, having a magnitude and a direction. The direction of the vector is determined by the right hand rule, where the fingers curl along the path of the rotation and the thumb points in the direction of the angular velocity vector. The angular frequency is a scalar quantity which is equal to the magnitude of the angular velocity vector. The relationship between angular frequency and angular velocity is analogous to the relationship between speed and velocity in linear motion, respectively.
Vishnu T Ganesh
Assistant Executive Engineer at Oil and Natural Gas Corporation Ltd. (ONGC) (2018–present)
·
Author has 57 answers and 380.6K answer views
·
Updated 7y
Related
What is the relationship between angular velocity and rpm?
Both the quantities are used in physics where there is a rotation.
Angular velocity is the rate of change of angle with respect to time. Unit of angular velocity is radians per second (rad/sec).
It's a vector quantity who's direction is given by the right hand thumb rule
dθdt=ω=2πN60
θ is the angle turned in radians
ω is the angular velocity in radians per second (rad/s)
N is the Revolution per Minute (RPM).
Both the quantities are used to express the rotational speed. RPM is used when the rotational speed is very low. (so converting into time for a minute, it
Both the quantities are used in physics where there is a rotation.
Angular velocity is the rate of change of angle with respect to time. Unit of angular velocity is radians per second (rad/sec).
It's a vector quantity who's direction is given by the right hand thumb rule
dθdt=ω=2πN60
θ is the angle turned in radians
ω is the angular velocity in radians per second (rad/s)
N is the Revolution per Minute (RPM).
Both the quantities are used to express the rotational speed. RPM is used when the rotational speed is very low. (so converting into time for a minute, it would give a recognizable quantity). But rad/s is used for reasonably high speed rotation.
1 RPM means, in 1 minute the object completes one revolution and 1 rad/s means in 1 second the object completes in radian. one complete revolution is equals to 2π radians. (6.28.. approximately).
Unit of ω is radians per second (rad/s), dimension of ω is [s−1]
Unit of N is revolutions per minute often called rpm. Dimension of N is also [s−1]
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Victor Mazmanian
Former Associate Prof. Of Physics (Retired) at United States Air Force Academy
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Author has 1.5K answers and 4.7M answer views
·
7y
Related
Why was the concept of angular velocity oiginated?
I am not sure of the exact date, but I could hazard a guess that it was suggested by Greek natural philosophers. Ever since the concept of pi was developed, it became sort of an enigma to observe the rotation of a wheel (and the original screws that were used in water lifting); where the outermost points of a wheel would be turning at a greater speed because of a greater circumference. In essence, the wheel should be broken apart! But it did not. The angular velocity concept was developed to explain why inner points could have the same angular velocity but differing speeds. During Galileo’s er
I am not sure of the exact date, but I could hazard a guess that it was suggested by Greek natural philosophers. Ever since the concept of pi was developed, it became sort of an enigma to observe the rotation of a wheel (and the original screws that were used in water lifting); where the outermost points of a wheel would be turning at a greater speed because of a greater circumference. In essence, the wheel should be broken apart! But it did not. The angular velocity concept was developed to explain why inner points could have the same angular velocity but differing speeds. During Galileo’s era, carousels and merry-go-rounds became popular devices in their exhibiting the fact that on-the-rim horses travel at speeds greater than inner horses but at the same angular velocities!
Les McLean
Ph.D. in Engineering
·
Author has 4.9K answers and 12.3M answer views
·
6y
Related
What does angular velocity mean?
Angular velocity, ω, is the rate of change of angle of an object rotating about an axis.
For an object rotating about an axis, every point on the object has the same angular velocity. The tangential velocity of any point is proportional to its distance from the axis of rotation. Angular velocity has the units rad/s.
This diagram will explain it all.
Angular velocity, ω, is the rate of change of angle of an object rotating about an axis.
For an object rotating about an axis, every point on the object has the same angular velocity. The tangential velocity of any point is proportional to its distance from the axis of rotation. Angular velocity has the units rad/s.
This diagram will explain it all.
Anurag Malakar
Physics is the Usain Bolt of science, it dazzles endlessly and repeatedly
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Author has 57 answers and 838.2K answer views
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10y
Related
What is the relationship between angular velocity and rpm?
Rpm is just a measure of how many times a wheel is revolving per minute. Let's say the wheel is revolving n times per minute.
So, the rpm of the wheel is n
In a single revolution, a point on the wheel rotates an angle=2pi
So, the total angle a point on the wheel is rotating per minute = n2pi
The angular velocity (Omega) is a measure of how fast a wheel is rotating.
Omega=(Angle Rotated)/Time Required
So Omega=2pin/60
Finally we have the relationship,
Angular velocity= (2pirpm)/60
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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2y
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What is the relationship between centripetal force, time period, and velocity?
The relationship between the centripetal force, the time period and velocity is found by using the formula
Fc= (mv^2)/r
Fc= centripetal force
v=angular velocity
the centripetal force is directly proportional to the square of the velocity
The relationship between the centripetal force and the period is found by knowing that
where
v=angular velocity
d= distance around the circular path= the circumference of the circular path.
The circumference of a circle
t=time talent to make a complete turn around the circle which equals the period T
so
replacing v in the original
The relationship between the centripetal force, the time period and velocity is found by using the formula
Fc= (mv^2)/r
where
Fc= centripetal force
m=mass
v=angular velocity
r=radius
so
the centripetal force is directly proportional to the square of the velocity
The relationship between the centripetal force and the period is found by knowing that
v=d/t
where
v=angular velocity
d= distance around the circular path= the circumference of the circular path.
The circumference of a circle
or
c=2πr
t=time talent to make a complete turn around the circle which equals the period T
so
v=2πr/T
replacing v in the original formula with 2πr/T gives
Fc= m{{2πr}/T}}^2/(r)
so the r in the denominator cancels out one of the r in the numerator leaving
Fc= m4π^2(rr)/T^2
From this equation it is seen that the centripetal force is inversely proportional to the square of the period
Related questions
What is the relation between angular speed and time period?
Why do we use the same symbol for angular velocity and angular frequency?
What is difference between angular velocity and angular frequency? And are angular velocity and angular frequency same?
What are the representations for angular velocity and angular frequency?
What are the core differences between angular frequency and angular velocity, both of which are denoted by ω?
What is the physical interpretation of angular frequency and angular velocity?
What is the difference between angular velocity and angular frequency?
What is angular frequency and angular velocity?
How can the units of angular velocity be shown to be the same as frequency or time period?
Is the angular velocity referred to as the angular frequency? Why?
Is angular velocity the frequency?
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12824 | https://ocw.mit.edu/courses/8-223-classical-mechanics-ii-january-iap-2017/resources/mit8_223iap17_lec13/ | (1 3) Dissipation and Damped Oscillators
1 Dissipation
We’ve been trying to ignore it, but in the real world there is friction. Friction means that mechanical energy is converted to thermal energy, and we no longer have a ‘conservative’ system. But we can try.
Rather than just start with a damped oscillation (as in eqn 25.1 in LL), I will motivate a modified Euler-Lagrange equation which includes dissipation, and then use this to arrive at damped oscillations.
Imagine some fraction of kinetic energy is couple to thermal energy per unit time .
L0 (q, q˙) = (T + Tμ) − U = L +
Z
T dt
⇒
d
dt
d
dt
∂L 0
∂q ˙
= d
dt
∂L
∂q ˙
d
dt
Z ∂
∂q ˙ (T ) dt
= d
dt
∂L
∂q ˙
∂
∂q ˙ (T ) = ∂L 0
∂q
given uniform such that ∂L 0
∂q =
∂L
∂q ˙
= ∂L
∂q − ∂
∂q ˙ (T )
∂L
∂q
Generalizing T to any velocity dependent function,
Let D = 1
2
X bjk q˙j q˙k “dissipative function”
j,k
or, for one degree of freedom, simply
1 2
D = bq ˙
2
1modified E-L
d
dt
∂L
∂q ˙
such that ˙p
=
=
∂L
∂q − ∂D
∂q ˙
∂L
∂q − bq ˙
∂L
where ∂q is the conservative force, and bq ˙ is the dissipative force.
Damped systems lose energy with time until they come to rest. The rate of energy loss is given by the dissipation function.
dE d ∂L
dt = dt q˙
∂q ˙ − L
d ∂L ∂L ∂L ∂L ∂L
= q˙ dt
∂q ˙
+
∂q ˙ ¨q −
∂t + ˙q ∂q + ¨q ∂q ˙
= q˙ d
dt
∂L
∂q ˙ − ∂L
∂q normally zero, but...
dE ∂D
dt = −q˙ ∂q ˙ = −2D
Note that the last line is just the rate of work done by friction as force × velocity. All of this is assuming no external driving force (i.e., ∂L = 0 ). ∂t
2 Damped Oscillator
Let’s put this to work on our harmonic oscillator to make a more realistic damped oscillator.
1 12 −for L = mx ˙ kx 2
2 2
1
and D = bx ˙ 2
2
2the equations of motion are
m¨x = −kx − b ˙x, ω2
0
= k
m
or
¨x + 2λ ˙x + ω2
0
x = 0, 2λ = b
m
This differential equation is best solved with complex exponentials, but the solution can be written in real form as
“under damped”
x (t) = ae −λt cos (ω1t + φ) for λ < ω 0
q
where ω1 = ω2 − λ2
0
“over damped”
x (t) = e−λt �a1eβt + a2e−βt for λ < ω0
where β =
q
λ2 − ω2
0
, note β < λ ⇒ decay
3“critically damped”
−λt
(a1 + a2t)x (t) = e for λ = ω0
To complete the picture, we should add a driving force to our damped oscillator. Returning to the equation of motion...
F(t)f
x¨ + 2 λx ˙ + ω0
2
x = = cos ( ωt )m m
⇒ x (t) = a1e−λt cos (ω0t + φ) + a2 cos (ωt + θ)
f
a2 = q 2
m (ω2 − ω2) + 4 λ2ω2
0
2λω
tan θ = ω2 − ω2
0
where a1 and φ come from the initial conditions
Again, the driven solution has 2 parts, one that depends on the initial conditions and another which is the response to the drive. With damping, we see that the first
1
of these decays with time, such that the motion at t is essentially only the λ
driven response.
41
x (t) ' a2 cos (ωt + θ) for t (1)
λ
x (t) a2
→ = (2)
F (t) f
Transfer Functions
5
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8.223 Classical Mechanics II January IAP 2017
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12825 | https://ejrnm.springeropen.com/articles/10.1186/s43055-022-00700-8 | Advertisement
Cranial nerve involvement in mucormycosis in post-COVID patients: a case series
Egyptian Journal of Radiology and Nuclear Medicine
volume 53, Article number: 28 (2022)
Cite this article
2930 Accesses
4 Citations
Metrics details
Abstract
Background
One of the largest outbreaks of rhinosinocerebral mucormycosis (RSCM) occurred in India close to the second wave of the SARS-CoV-2 infection. RSCM is a rare infection caused by several fungal species occurring in immunocompromised subjects. Mucor shows a high propensity to invade the central nervous system. There have been limited studies, mostly isolated case reports, on the neurological manifestations of RSCM. The outbreak of mucormycosis infection was thus the most opportune to study the neurological manifestations and cranial nerve involvement in mucormycosis in greater depths.
Aim of the study
The purpose of the study was to investigate and review the involvement of cranial nerves in a series of cases of rhinosinocerebral mucormycosis associated with the novel coronavirus disease caused by SARS-CoV-2.
Results
It was a retrospective cross-sectional study of seven patients who were undergoing treatment of RSCM with a recent history of coronavirus disease caused by SARS-CoV-2 infection within the last 3 months. Patients with cranial nerve involvement were identified by magnetic resonance imaging (MRI) at a single institution. Demographic details of the patients, clinical presentation, imaging, microbiological and pathological findings were recorded. All subjects had two or more cranial nerves affected by fungal infection. The most commonly involved cranial nerve was found to be the optic nerve followed by the trigeminal nerve and its branches. We document three cases with extensive involvement of the inferior alveolar branch of the mandibular division of the trigeminal nerve (V3), a previously unreported finding. In one case, in addition to the second and fifth cranial nerves, the third, fourth, sixth, seventh, eighth, and twelfth cranial nerves were involved without any sensory or motor long tract involvement, suggestive of Garcin syndrome secondary to intracranial abscesses and skull base osteomyelitis due to invasive fungal infection. This case is of rare occurrence in the literature, and our study provides one such example.
Conclusion
Cranial nerve involvement in patients of mucormycosis tends to have a poor prognosis, both cosmetic and functional. Radical surgeries and aggressive medical management is needed in such cases to improve the outcome.
Background
Since the first reported case of coronavirus infection in December 2019, the disease has spread to almost all countries of the world.
Coronavirus disease produces protean manifestations and complications. Several uncommon phenomena have been reported in association with the disease. These phenomena include several types of co-infections [13
(Epub 2020 May 23)")], new-onset diabetes, strokes in the young, chronic fatigue syndrome and various dermatological conditions [2 Extrapulmonary manifestations of COVID-19. Nat Med 26:1017–1032.
")].
One such rare disease associated with COVID infection is the occurrence of mucormycosis.
Case reports of sporadic RSCM have been reported in the past with the earliest reports by Kurrien [31
")] in 1954 and Smith et al. [4 Cerebral mucormycosis; a report of three cases. AMA Arch Otolaryngol 68(6):715–726.
")] in 1958.
Rhinosinocerebral mucormycosis (RSCM) is a rare infection, caused by several fungal species such as Mucor, Cunninghamella, Rhizopus and Lictheimia, usually associated with immunocompromised patients and diabetic individuals [56
")]. One of the largest outbreaks of RSCM occurred in India close to the second wave of the SARS-CoV-2 infection [6 Rising incidence of mucormycosis in patients with COVID-19: another challenge for India amidst the second wave? Lancet Respir Med 9(8):e77.
")].
RSCM presents with ophthalmic and non-ophthalmic manifestations.
Non-ophthalmic manifestations of RSCM include sinusitis (100%), nasal discharge/ulceration (74%), infranuclear abducens nerve palsy (46%), palatal necrosis (29%), cerebral lobe involvement (20%), and hemiparesis (17%). Orbital involvement was observed in 80% of patients with cavernous sinus thrombosis in 11%, and internal carotid occlusion (ICA) and hydrocephalus in 3% each [70
")].
Involvement of cranial nerves by mucormycosis, signifying intracranial extension, generally carries poor prognosis 89
"), [9 Rhinocerebral Mucormycosis: consideration of prognostic factors and treatment modality. Auris Nasus Larynx 36(3):274–279.
")]. Imaging however allows early identification of cranial nerve involvement and exact delineation of the spread of infection. This allows aggressive medical and surgical management of these complications leading to an overall improvement in the outcome.
Methods
Inclusion criteria for the subjects included subjects consenting for the study, subjects with two or more cranial nerve involvement and those with history of COVID-19 infection preceding mucor infection.
Subjects excluded from the study included those with just a single cranial nerve involvement, pregnant patients, those with mucor infection not preceded by COVID-19 infection and patients less than 18 years of age.
Seven patients of invasive fungal disease with multiple cranial nerve involvement were studied. All seven patients in our study had a prior history of COVID-19 infection. Nasal swabs, in all seven cases, were positive for COVID-19 infection by polymerase chain reaction. Reverse transcriptase-polymerase chain reaction (RT-PCR) from nasopharyngeal swabs for COVID-19 was positive in all subjects. Further, computed tomography (CT) scan of the thorax showed changes typical for COVID-19 pneumonitis in all cases. None of the subjects had received any doses of the COVID vaccine at the time of the disease (Fig. 1).
Axial HRCT thorax image showing ground-glass opacities in subpleural and peripheral regions of bilateral lungs typical for COVID pneumonitis
Evaluation at presentation included a detailed history, otorhinological, ophthalmology, and neurology examination to assess the extent of disease, sinonasal endoscopy with biopsy. Diagnosis was made on KOH mount (Fig. 2) and histopathological examination (Fig. 3).
Potassium hydroxide (KOH) mount showing ribbon-shaped aseptate hyphae (black arrow) showing branching at right angle (blue arrow)
Hematoxylin and eosin stained microsections showing fragments from the sinonasal mucosa with broad aseptate ribbon-like fungal hyphae showing branching at right angles consistent with Mucor species
Magnetic resonance imaging (MRI) was performed using 3 Tesla (T) Siemens (Verio: Siemens Medical System, Erlangen, Germany). MRI of paranasal sinuses (PNS) and brain were obtained to assess the extent of disease. Contrast-enhanced MRI (CE-MRI) protocol of the PNS, brain, and orbits included T2 FLAIR, GRE, DWI, axial T1, T2, STIR coronal, CISS, T1 FS axial post-contrast (3 mm thickness) and T1 FS post-contrast coronal (3 mm thickness) sequences. Patients were started on systemic amphotericin B at the diagnosis of mucormycosis. All the patients underwent surgical debridement for the necrosed tissue. Of these two patients underwent functional endoscopic sinus surgery (FESS) and enucleation of the left orbit and two patients underwent FESS before the MRI was done.
The MRI scans of the subjects were interpreted and reviewed by two radiologists independently, specializing in head, neck and face radiology and neuroradiology and each with an experience of about 5 years.
Case information
Case 1 A 32-year-old man, presented with right-sided facial swelling, lower toothache, painful proptosis of the right globe and loss of vision. MRI showed optic neuritis with trigeminal nerve abscess and extension of inflammation along the inferior alveolar nerve (Fig. 4).
A, B T1 post-contrast sequence in axial and coronal planes showing optic neuritis with changes of orbital cellulitis on the right side. C CISS in coronal plane shows thickened right trigeminal nerve. D T1 post-contrast sequence in coronal plane showing extensive post-contrast enhancement in the pterygoid muscles along the inferior alveolar nerve on the right side
Case 2 A 73-year-old lady, presented with left-sided facial swelling, and left orbital cellulitis, due to mucormycosis. The patient subsequently underwent enucleation and in the postoperative period had purulent discharge from the orbit. MRI confirmed the presence of disease in the optic chiasma along with thrombosis of the cavernous portion of the left ICA (Fig. 5).
A STIR coronal sequence demonstrating hyperintense residual disease (arrow) in the left half of optic chiasma. B Axial diffusion-weighted sequence demonstrating diffusion restriction in the left half of optic chiasma. C T1 post-contrast coronal sequence showing peripheral contrast enhancement (yellow arrow) in the left optic nerve. Also noted is the loss of flow void (red arrow) of the cavernous portion of the left ICA, suggestive of thrombosis
Case 3 A 34-year-old male, presented with sudden onset left-sided facial palsy with painful proptosis of the left eye and complete ophthalmoplegia. A thorough clinical examination revealed that the patient had a hearing deficit in the left ear (which subsequently proved to be of sensorineural type). On asking the patient to protrude his tongue for examination, the tongue was found to deviate toward the left side (involved side), implying left-sided hypoglossal nerve palsy. No evidence of motor or sensory disturbances were found. A clinical diagnosis of Garcin syndrome secondary to fungal infection was made (Fig. 6).
A T1 post-contrast axial sequence and B post-contrast coronal sequence showing post-contrast enhancement along the left hypoglossal nerve (arrows). MRI showing inflammation along the left CP angle with involvement of seventh and eighth cranial nerves as a result. Thickening and extension of inflammation were seen along the internal acoustic meatus, hypoglossal canal and the left inferior alveolar canal with VII, VIII and XII involvement. C CISS in the axial plane demonstrating thickening of the cisternal segments of the seventh-eighth cranial nerve complex on the left side. D T2 axial and E coronal sequences showing hypointense fungal tissue in the left orbit
Case 4 A 60-year-old male presented with headache, painful proptosis and complete ophthalmoplegia of the right eye. MRI brain was suggestive of right orbital cellulitis, with the inflammation seen extending posteriorly till the orbital apex with optic peri-neuritis. The right inferior alveolar nerve was found to be thickened and showing abnormal enhancement (Fig. 7).
T1 post-contrast axial sequence showing thickened enhancing right inferior alveolar nerve (arrow)
Case 5 A 62-year-old male, presented with a left eye proptosis and acute loss of vision. Extensive left orbital cellulitis with diffusion restriction seen within the optic nerve suggests acute ischemic optic neuropathy. Infiltration was noted along the mandibular division trigeminal nerve (V3) through the left foramen ovale and the Meckel’s cave (Fig. 8).
A T2 axial sequence showing changes of orbital cellulitis on the left side with abnormal signal intensity in the intraorbital portion of the left optic nerve. Also noted is the T2 hypointense fungal tissue infiltrating the left orbit and causing severe distortion of the left globe. B Axial diffusion sequence showing diffusion restriction along the intraorbital portion of the left optic nerve
Case 6 A 59-year-old man, presented with left eye proptosis and acute loss of vision. MRI showed extensive left orbital cellulitis, with left optic neuritis with inflammation seen reaching up to the orbital apex. Thickening of the seventh-eighth nerve complex was also seen on the left side.
Case 7 A 38-year-old man, presented with proptosis of the right globe, with redness and chemosis. MRI brain showed orbital cellulitis with optic neuritis on the right side. Involvement of cavernous sinus thrombosis with involvement of trigeminal nerve and subtle thickening of the seventh-eighth nerve complex on the right side. An ill-defined abscess was seen in the right temporal lobe and the right half of the pons (Fig. 9).
A CISS, and B T2 WI sequence showing thickening of the cisternal segment of the seventh-eighth nerve complex on the left side. The underlying right middle cerebellar peduncle shows T2 hyperintense signal. Fluid signal is noted replacing the normal mastoid air cells on the right suggestive of mastoiditis (red arrow). Inflammation is noted along the floor of the left orbit and sphenoid sinusitis. C T1 WI post-contrast axial sequence showing abnormal enhancement along the right cavernous sinus extending into the orbit
Results
The results and observations gathered can be summarized in Table 1.
Out of the seven subjects who were included in the study, six patients were males, while one was female. The ages of the participants spanned from 32 to 73 years. All, but two had a history of steroid use while they were undergoing treatment for COVID infection. The other two patients had a history of diabetes mellitus (> 5 years each), which was largely under control. The glycosylated hemoglobin (HbA1c) of these patients was < 7%.
The mean time from the diagnosis of COVID infection to the occurrence of mucormycosis in the patients was found to be 3.7 weeks. The optic nerve was the most frequently affected. This was followed by the involvement of trigeminal nerves and its branches. Thrombosis of the intracranial portion of the ICA was seen in three patients. The most common imaging feature of fungal neuritis was abnormal thickening and enhancement of the cranial nerve with perineural fat stranding. All cranial nerves in a particular subject were found to be involved only on one side (unilateral affection) and there were no cases wherein bilateral affection by the disease was found.
Discussion
The coronavirus disease 19 (COVID-19) is a highly transmittable and pathogenic viral infection with multiple variants 106
"), [11 COVID-19 infection: origin, transmission, and characteristics of human coronaviruses. J Adv Res 24:91.
")] caused by severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2), which caused global pandemic leading to healthcare crises and severe loss of human life. Fungal infections, including mucormycosis, aspergillosis and invasive candidiasis, have also been reported in patients with severe COVID-19 or those recovering from the disease. India had seen a rise in the rhino-orbital mucormycosis co-infections in COVID-19 patients during the second wave of COVID-19 infection [6 Rising incidence of mucormycosis in patients with COVID-19: another challenge for India amidst the second wave? Lancet Respir Med 9(8):e77.
")]. Mucorales are killed by the mononuclear and polymorphonuclear phagocytes of normal hosts by generation of oxidative metabolites and defensins. This makes patients with neutropenia, those with dysfunctional phagocytes and uncontrolled diabetes particularly susceptible toward the development of invasive mucormycosis [12 Pathogenesis of mucormycosis. Clin Infect Dis 54(Suppl 1):S16-22.
")]. COVID-19 causes profound lymphopenia and in advanced infections viral replication accentuates the inflammatory response and neutrophil and monocyte influx in the bloodstream [13 Pathophysiology, transmission, diagnosis, and treatment of coronavirus disease 2019 (COVID-19): a review. JAMA 324(8):782–793.
")]. This in turn leads to an imbalance between neutrophil and lymphocyte action making the patient more susceptible to systemic fungal infections [14 A case series of invasive mucormycosis in patients with COVID-19 infection. Int J Otorhinolaryngol Head Neck Surg 7(5):867–870.
")]. Further, in a study conducted by Wu et al. [15 SARS-CoV-2 infects human pancreatic β cells and elicits β cell impairment. Cell Metab.
")], it has been postulated that SARS-CoV-2 infection induces apoptosis of β cell of the pancreatic islets thereby attenuating insulin levels and secretion thereby making COVID-19 infection a potential diabetogenic state [16 New-onset diabetes in Covid-19. N Engl J Med 383(8):789–790.
")].
Further, it has been found in several studies that administration of steroids, and biologics such as tocilizumab have been known to lead to fungal infections and candidemia 174
"), [18 Role of tocilizumab for concomitant systemic fungal infection in severe COVID-19 patient: case report. Medicine (Baltimore) 100(12):e25173.
")].
Mucormycosis tends to manifest in one or more of these six forms—Rhinosinocerebral, pulmonary, cutaneous, gastrointestinal and miscellaneous with RSCM being predominant in patients with uncontrolled diabetes and the pulmonary form being noted in patients with hematological malignancies and solid organ transplant recipients [56
")].
RSCM typically starts with nasal involvement and spreads to adjacent sinuses 199
"), [20 Rhinocerebral mucormycosis. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing")]. Around the paranasal air sinuses, there are many skull base foramina and fissures through which these cranial nerves traverse. From the infected paranasal sinuses, the fungal elements can take these avenues to enter the skull base and intracranial compartment. Whilst doing so, within these channels, the fungi affect, infect, invade, infiltrate and may destroy these important structures and result in various cranial nerve palsies. Anosmia, visual loss, ocular palsies, trigeminal neuralgia, and even lower cranial palsies can occur [21 Anti-GQ1b-negative Miller-Fisher syndrome with lower cranial nerve involvement from parasinusoidal aspergilloma. J Infect 52(3):e81–e85.
"),22 Trigeminal neuralgia as unusual isolated symptom of fungal paranasal sinusitis in patients with haematological malignancies. Neurol Sci 33(3):647–652"),23 Contrast-enhanced 3D FIESTA imaging in Tolosa-Hunt syndrome. Headache J Head Face Pain 52(5):822–824.
"),24 Superior orbital fissure syndrome due to sinusitis: report of two cases. Am J Rhinol 19(4):417–420")].
High fatality is because of angioinvasion by mucormycosis with resultant thrombosis of vessels and resultant tissue necrosis . Three patients in the above series also developed cerebral infarcts secondary to ICA thrombosis.
MRI is an indispensable tool that can be used to diagnose mucormycosis infections involving sino-nasal region, orbits, and intracranial extension of the disease. MRI is useful in assessing the intradural extent of the disease, perineural spread, and thrombosis of the cavernous portion of the ICA, caused by fungal infection as well as planning of the surgical margins for resection and post-surgical follow-up [26, 279
")].
Sino-nasal mucormycosis on MRI is seen as mucosal thickening, which appears hypointense on T1WI and variable to hyperintense on T2WI [279
")]. Fungal tissue appears hypointense on T2WI and may show restricted diffusion on DWI. The involved tissues show post-contrast enhancement; however, areas of non-contrast enhancement are noted within the enhancing turbinates and paranasal sinuses known as the black turbinate sign [28 The “Black Turbinate” sign: an early MR imaging finding of nasal mucormycosis. AJNR Am J Neuroradiol 31(4):771–774.
(Epub 2009 Nov 26)")].
MRI is the most sensitive modality to look for cranial nerve involvement due to its superior soft tissue resolution. MRI findings of cranial nerve involvement in mucormycosis may be as subtle as perineural fat stranding in the early stages, or may show thickening and enhancement of the cranial nerves. There may be meningeal enhancement or phlegmonous soft tissue noted along the cranial nerves. Changes of denervation may be seen in the muscles innervated by the cranial nerve, which include T2W hyperintensity and enhancement in the early stages and fatty changes with volume loss in long-standing denervation [295
")]. In some cases, when the nerve involvement is subtle, these denervation changes can provide a clue to neural involvement.
Limitations of the study
This series being a pilot study covers very few patients. Secondly, fungal infection affecting all cranial nerves could not be demonstrated in this study. In case 3, MRI imaging could not convincingly demonstrate affection of the lower cranial nerves, namely, spinal accessory, vagus and the glossopharyngeal nerves.
Conclusion
Mucormycosis is a serious disease and may be fatal if not treated in time. A high index of clinical suspicion, early diagnosis, confirmation with microbiology, staging with cross-sectional imaging and immediate correction of underlying medical disorders and institution of antifungals are important factors for a favorable prognosis. We have highlighted cranial nerve involvement in post-COVID mucormycosis in this case series. Identifying cranial nerve involvement is important in diagnosis and staging of mucormycosis. MRI is the modality of choice for early identification and precise delineation of the extent of disease.
Availability of data and materials
The data and materials supporting the findings of this study are available on request from the corresponding author.
Abbreviations
3 Tesla
Constructive interference in steady-state sequence
Coronavirus disease 2019
Computed tomography
Diffusion-weighted imaging
Functional endoscopic sinus surgery
Fluid-attenuated inversion recovery
Fat saturated
Gradient recalled echo
Glycosylated hemoglobin
Internal cerebral artery
Potassium hydroxide
Magnetic resonance imaging
Paranasal sinuses
Rhinosinocerebral mucormycosis
Reverse transcriptase polymerase chain reactions
Severe acute respiratory syndrome coronavirus 2
Short tau inversion recovery sequence
T1 weighted image
T2 weighted image
Mandibular branch of trigeminal nerve
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Neeti Gupta & Saurabh Dembla
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12826 | https://www.physics-prep.com/index.php/practice-problems-capacitors?view=article&id=1437:practice-problems-capacitance-solutions&catid=43:physics-2-unit-1 | Practice Problems: Capacitance Solutions - physics-prep.com
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Practice Problems: Capacitors Solutions
(easy) Determine the amount of charge stored on either plate of a capacitor (4x10-6 F) when connected across a 12 volt battery.
C = Q/V
4x10-6 = Q/12
Q = 48x10-6 C
(easy) If the plate separation for a capacitor is 2.0x10-3 m, determine the area of the plates if the capacitance is exactly 1 F.
C = ε o A/d
1 = (8.85x10-12)A/(2.0x10-3)
A = 2.3x10 8 m 2
(moderate) Calculate the voltage of a battery connected to a parallel plate capacitor with a plate area of 2.0 cm 2 and a plate separation of 2 mm if the charge stored on the plates is 4.0pC.
Area = 2.0 cm 2(1 m/100cm)2 = 2.0x10-4 m 2
C = ε o A/d
C = (8.85x10-12)(2.0x10-4)/(2.0x10-3)
C = 8.85x10-13
C = Q/V
8.85x10-13 = 4.0x10-12/V
V = 4.5 volts
(easy) A parallel plate capacitor is constructed of metal plates, each with an area of 0.2 m 2. The capacitance is 7.9nF. Determine the plate separation distance.
C = ε o A/d
7.9x10-9 = 8.85x10-12(0.2)/d
d = 2.2x10-4 m = 0.22 mm
(easy) A capacitor (parallel plate) is charged with a battery of constant voltage. Once the capacitor reaches maximum charge, the battery is removed from the circuit. Describe any changes that may take place in the quantities listed here if the plates were pushed closer together.
a. Charge(The charge deposited on the plates doesn't change when the battery is removed and thus the charge and the charge density remains the same as the plates are moved closer together.)
b. Capacitance (Since the capacitance is C= ε o A/d, and the area is not changing, any decrease in plate sepration (d) will cause an increase in capacitance.)
c. Voltage(Because C = Q/V, and the charge doesn't change, an increase in capacitance implies a decrease on voltage.)
d. E-field (Since ΔV = -Ed, the E-field will remain the same as both the voltage and the distance decrease proportionately.)
e. Would any of the answers above change if the battery was not disconnected from the battery? Explain your response.Yes. The voltage would not change if the battery remained connected to the capacitor. The capacitance would still increase because it is based solely on the geometry of the capacitor (C =ε o A/d).The charge would increase because Q = CV and the capacitance increased while the voltage remained the same. Finally, the E-field would increase because E =|ΔV/d| and the distance across the plates decreased while the voltage stayed the same.
(moderate) Random access memory chips are used in computers to store binary information in the form of "ones" and "zeros". One common way to store a "one" is to charge a very small capacitor. Of course, the same capacitor without charge represents a "zero". A memory chip contains millions of such capacitors, each coupled with a transistor (that acts as a switch), to form a "memory cell".A typical capacitor in a memory cell may have a capacitance of 3x10-14 F. If the voltage across the capacitor reading a "one" is 0.5 v, determine the number of electrons that must move on the the capacitor to charge it.
C = Q/V
3x10-14 = Q/(0.5)
Q = 1.5x10-14 C
electrons = Total charge/Charge per electron
electrons = 1.5x10-14/1.6x10-19
electrons = 93750 electrons
(easy) C 1 = 10 F and C 2 = 5 F. Determine the effective capacitance for C 1 and C 2 connected in series and in parallel.
In series:
1/C = 1/C 1 + 1/C 2
1/C = 1/10 + 1/5
C = 3.3 F
In parallel:
C = C 1 + C 2
C = 10 + 5 = 15 F
(moderate) If the two capacitors in question #7 were connected to a 50 volt battery determine the voltage across the capacitors for each connection type.
For the series connection:
The charge on each capacitor is the same as the charge on the effective capacitance.
C = Q/V
3.3 = Q/50
Q = 165 C
For the 10F capacitor:
10 = 165/V
V = 17 volts
For the 5 F capacitor:
5 = 165/V
V = 33 volts
For the parallel connection:
The voltage is the same (50 v) across each capacitor.
(moderate) Evaluate the circuit shown below to determine the effective capacitance and then the charge and voltage across each capacitor.
The equivalent capacitance is 4 μF. The voltage across the equivalent capacitor is 20 volts.
This voltage is also across both of the 2 μF capacitors that were created by the series combinations in each branch.
Find the charge on each 2 μF capacitor:
C = Q/V
2 μF = Q/20
Q = 40 μC
The 4 μF capacitors in each branch have the same charge as the 2 μF capacitors. Use this to find the voltage across each:
C = Q/V
4 μF = 40 μC/V
V = 10 volts
In summary, each of the original 4 μF capacitors have a charge of 40 μC and a voltage of 10 volts.
(moderate) Evaluate the circuit shown below to determine the effective capacitance and then the charge and voltage across each capacitor.
The effective capacitance is 6 μF with a voltage of 100 v.
The voltage across the 4 μF and the 2 μF capacitors is also 100 v
The charge on the 4 μF capacitor:
C = Q/V
4 μF = Q/100
Q = 400 μC
The charge across the 2 μF capacitor:
C = Q/V
2 μF = Q/100
Q = 200 μC
All three 6 μF capacitors also have 200 μC of charge.
Find voltage for the 6 μF capacitors:
C = Q/V
6 μF = 200 μC/V
V = 33.3 v
(moderate) Evaluate the circuit shown below to determine the effective capacitance and then the charge and voltage across each capacitor.
The equivalent capacitance is 6 μF. The voltage across the equivalent capacitance is 40 v as is the voltage across the 3 μF capacitors and is the same as the 1 μF and 2 μF capacitors.
Find the charge on the 1 μF capacitor:
C = Q/V
1 μF = Q/40
Q = 40 μC
Find the charge on the 2 μF capacitor:
C = Q/V
2 μF = Q/40
Q = 80 μC
Find the charge on the 3 μF capacitors:
C = Q/V
3 μF = Q/40
Q = 120 μC
This is the same charge on each of the 6 μF capacitors.
Find the voltage on each of the 6 μF capacitors:
C = Q/V
6 μF = 120 μC/V
V = 20 v
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12827 | https://www.youtube.com/watch?v=Pj1xnR9dlc8 | Subtracting Polynomials
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hi guys it's me teer Goin In Our today's video we will talk about subtracting pols so without further Ado let's do this topic now here are the first two examples on how to subtract pols focus pooms and by the way in subtracting polinomial it will lead you to addition of polinomial so much better about adding pols and description box so without further Ado let's do this topic so what we have here is the first one the quantity of 3x^2 + 5 - x^2 - 1 so what will happen or how are we going to start subtracting these two polinomial this is your first polinomial and this is your second polinomial first thing you need to do is to copy the first set of polinomial so we have here 3 x^2 + 5 then after that after copying the first set of Pol the first polinomial we will change this operation to addition this will become plus and next the second polinomial or all of the signs of this of the terms of this polinomial will be changed so if this is positive it will become negative if this is negative it will become positive so what will happen is this will become the quantity of x^2 from positive x^2 from - 1 it will become + 1 and as you can see as I've told you earlier in subtracting polinomial it will lead you to add addition of pols so now we can do addition of polinomial and we can do it horizontally or vertically but in this case I will do vertical subtraction we have 3 X2 + 5 and then plus X2 + 1 so add them three x^2 + x^2 there is a negative - 1 here so this is 2 x^2 5 + 1 is 6 and this is now the difference of 3x^2 + 5 and x^2 - 1 now let's continue with item number two for number two we have 7 x^2 + 36 - 5 x^2 + x + 9 so what will happen here same thing we need to do copy the first polinomial so we have the quantity of 7 x^2 + 3 6 and then change this operation to addition and lastly change the signs of all the terms of this polinomial from positive it will become 5x^2 this ISX this is -9 or and after doing that let's do vertical multiplication we have 7 X2 + 0 + 36 so why did I add zero here as you can see this one is is a two-term polinomial or binomial terms and we need X for that okay if since X variable we will put plus 0 next plus 5 x^2 and this negative X will be aligned here S zero then minus 9 then do the the addition 7 x^2 - 5 x^2 is definitely 2 x^2 then 0 +x this is just X then thir 36 - 9 this is equal to + 20 7 and this is now the difference of the first of the second polinomial that we have 7 x^2 + 36 - 5x + x + 9 and as easy as that guys now let's continue with the other two examples that we have today so what we have here here are trinomials that we're going to subtract number three 12 a 2 uh 12 a ra to the 5th power - 6 a - 10 a cub so guys we need to copy this one right and we will change this so arrangement we need to um rearrange them from the highest exponent of the variable going down so we will start here 12 a to 5 this one will become instead of -6 a we will have plus 0 because we have your a to the 4th power and [Music] Then followed by -10 a cub we don't have a squ we don't have a squ here so we will proceed with 6 a so what will happen here we will change this to addition this will become negative this will become positive this one is also positive so we will put it here plus -10 oh not no this is -2 it's -2 a the 5th power followed by -14 a 4th power and then we don't have a cub in the second one so this is plus 0 then minus 10 a I have omitted some of the steps space but still we're doing good so first -2 a to the 5th power here so we can add them 12 a 5th power minus 2 a the 5th power this is 10 a to 5 0 + -14 a to the 4th power that is -14 a to the 4th power -10 a a cub + 0 this is -10 a cub and I forgot a here so we have here -6 a + -1 a this is 16 a and this is now the answer so tell me if there's something wrong with our solution and so what we have now is the fourth example let's proceed so I'm going to move this paper first enough space okay see uh we will do this one properly arranged properly arranged so we need to do is to copy first we have 9 R Cub + 5 R 2 + 11 R here change this to addition this one changed to positive this is negative this is negative also okay so plus positive 2 R Cub this is 8 r² this is 9 R add them 9 R Cub + 2 R Cub this is 11 R Cub next 5 r s - 8 r² this is -3 r² 11 R minus a plus 9 R this is positive2 r and this is the answer guys so guys and then you can you can replay this video some so I hope and I will give you example number five as part of our routine I will give you this example for you to solve and I hope comment section what is your answer in this problem so what we have here is quantity of 4 b ^ 2 - 11us b^ 2 + 5 so I hope the so guys if you're new to my channel don't forget to like And subscribe but Bell button for you to be updated latest uploads again it's me teacheron bye-bye |
12828 | https://math.stackexchange.com/questions/3992275/if-a-winning-strategy-does-not-exist-for-player-2-does-it-exist-for-player-1 | set theory - If a winning strategy does not exist for player 2, does it exist for player 1? - Mathematics Stack Exchange
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If a winning strategy does not exist for player 2, does it exist for player 1?
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I have been studying some instances of strategy-stealing. The issue is that I am not sure whether proving the non-existence of a winning strategy for player 2 necessarily proves a winning (or at least tying) strategy for player 1. Wikipedia says
In combinatorial game theory, the strategy-stealing argument is a general argument that shows, for many two-player games, that the second player cannot have a guaranteed winning strategy.
The arguments that I have seen make sense, in terms of proving that player 2 cannot have a winning strategy. For example, the Wikipedia article on Chomp makes the modest claim
The second player therefore cannot have a winning strategy.
However, in Engel's Problem-Solving Strategies, Double Chess is analyzed in problem #18 of the games chapter and the following seemingly grander claim is made:
The rules of chess are changed as follows: Black and White make alternately two legal moves. Show that there exists a strategy for white which guarantees him at least a tie.
The proof is
Suppose B B can win no matter what A A does. On his first move, A A moves one of his knights to any one of the two possible squares and then back to its original position. Now all the pieces are in their original position, but A A has becmoe the second player and must win. Contradiction!
When does the negation of the existence of a winning strategy for player 2 imply the existence of a winning/tying strategy for player 1?
I might be confused because I am actually not sure of how to formally describe a winning strategy, so a rigorous explanation of that might be helpful as a preface. (I considered using alternating existential and universal quantifiers to describe winning strategies, but didn't know how to terminate the sequence.)
P.S. I have included set theory as a tag because, according to the Wikipedia article on determinacy, the subject is a subfield of set theory. However, I am not familiar with this part of set theory and I did not really understand the article.
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asked Jan 20, 2021 at 1:38
FavstFavst
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3 Zermelo's theorem shows that finite games are always determined. So one of the players must have a winning strategy. If it's not the second, then it has to be the first.Asaf Karagila –Asaf Karagila♦ 2021-01-20 01:54:33 +00:00 Commented Jan 20, 2021 at 1:54
@AsafKaragila That's exactly what I was seeking, thanks. Though I am a little surprised by the statement "either the first-player can force a win, or the second-player can force a win, or both players can force a draw." I would have thought that it would end with "... at least one of the players can force a draw."Favst –Favst 2021-01-20 02:17:03 +00:00 Commented Jan 20, 2021 at 2:17
@AsafKaragila I think an explanation of Zermelo's theorem maybe with an extra bit addressing Favst's comment would be a better answer than the two existing ones, if you'd like to write it out.Mark S. –Mark S. 2021-01-20 14:01:21 +00:00 Commented Jan 20, 2021 at 14:01
1 @Mark: I'm not sure that I'm the right person to explain this theorem of Zermelo. :-)Asaf Karagila –Asaf Karagila♦ 2021-01-20 14:02:11 +00:00 Commented Jan 20, 2021 at 14:02
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The Simplest Case
Simple Conditions
For now, let's consider two-player games of perfect information where there is a pre-known bound on the number of moves a game might take, and one of the two players wins in the end, and the first player is fixed (e.g. the player who controls the white pieces must move first).
Strategies
As mentioned in the comments, there is a theorem/collection of related theorems called Zermelo's theorem (in game theory), that is relevant here. Summarized, it says that in a game of this type, either the first player to move or the second player to move must have a "winning strategy" — a function that provides a move for every situation that player could possibly be in (a strategy) which guarantees a win no matter how the opponent plays.
(As an aside, most discussions of "Zermelo's theorem" aren't quite in line with what Zermelo wrote about and how his proof went. See Zermelo and the Early History of Game Theory by Ulrich Schwalbe and Paul Walker for some clarification about the history.)
The Argument
A full formal setup and proof would make this post much longer, and might not add much more conceptual understanding. Briefly, the idea is to use induction on, say, the maximum length of a game. A game with only one turn either has a winning move for the first player, in which case the winning strategy for them is "play one of the winning moves". Or there isn't, in which case the winning strategy for the second player is an empty function since they have no choices to make. For the induction step: if there are any moves to a "non-next player has a winning strategy" situation, the first player has a winning strategy to make one of those moves and then follow the existing strategy. If there are no such moves, then all moves are to a "next player has a winning strategy" position, and the second player can follow that strategy no matter which move the first player makes. Another, less brief, account can be found in Lecture 15 from the Open Yale Course ECON 159: Game Theory, which has accompanying lecture notes and transcript, etc.
Variants
There are many variants on the conditions that go beyond the simplest case, and I won't cover them all here, but I wanted to address the comments about Chess and give a picture of how this stuff can be extended.
Ties
Some games, like "Tic Tac Toe"/"Noughts and Crosses" don't satisfy the condition where someone must win in the end. The game can end in a "tie" (also known colloquially as a "draw", but we make a distinction in combinatorial game theory).
From a game G G like this, we can build a new game G 1 G 1 by breaking ties in favor for first player, and another game G 2 G 2 by breaking ties in favor of the second player. For G 1 G 1, Zermelo's Theorem tells us that either the first player or the second player has a winning strategy in G 1 G 1. This means that the first player can force at least a tie in G G or the second player can force a win in G G. Analogously, Zermelo's Theorem applied to G 2 G 2 tells us that either the second player can force at least a tie in G G or the first player can force a win.
Now it's just a matter of listing the possibilities:
If the first player can force a win, then the second player can't even force "at least a tie".
If the second player can force a win, then the first player can't force "at least a tie".
If neither player can force a win, then by the observations about G 1 G 1 and G 2 G 2 above, it must be that case that both player can force "at least a tie". And since neither can force a win, it must be the case they can simply force a tie.
Repeated positions (like Chess)
Some games like chess have things like repeated board positions, which would seem to throw a wrench in the induction proof of Zermelo's Theorem. However, this is sort of an illusion because of things like the Chess's repetition rules. Chess doesn't allow real repetitions if you include a list of all previous positions with their frequency counts as part of the "position". With this clarified, Chess and similar games are basically just games with ties (though you could complicate the earlier story if you want to distinguish stalemates from repetition draws, etc.)
Repeated positions (for real)
Some other games (much less likely to be actually played) really can repeat positions, and hence can go on forever. For example, maybe Nim-but-you-can-return-objects-too, or "Fair Shares and Varied Pairs" (a description can be found in this mathstrek blog post or Winning Ways). For these so-called "loopy" games, we must consider "drawn" games which last forever, in addition to the ones that end.
It takes some extra work (see IV.4 of Combinatorial Game Theory, for instance), but it turns out there's a tidy description of things with just one more category in addition to first or second player having a winning strategy: those positions where both players can force the game to be drawn. So it's not obvious, but these drawn games basically work out the same as ties.
Partizan Games
If you study more combinatorial game theory, you'll encounter partizan games where players have an extra distinction aside from who moves first. For example, maybe you could consider a position on a chessboard and the "moves for Black" even if it would be White's turn in the original position. For the purposes of things like Zermelo's Theorem (ignoring a lot of good theory), this doesn't really matter. Given a partizan game, you could basically consider G 1 G 1 where Black moves first (and the color whose turn it is is part of the data of the position) and G 2 G 2 where White moves first and then apply a previous theorem to each of those.
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answered Jan 21, 2021 at 2:04
Mark S.Mark S.
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1 Here I give one sketch of the proof more briefly: By this useful Blackboard Note of the above course link, we can just think the game as one decision tree which can be variant with different states for each node based on the context. Then we go from the leaf to the root of the whole tree in the induction process because the leaf state can be easily decided until the state of the root is decided.An5Drama –An5Drama 2024-04-16 12:56:37 +00:00 Commented Apr 16, 2024 at 12:56
The above link is similar to what wikipedia says (See the first question answer) with more details. They both consider 3 cases W(in)/T(ie)/L(ose). This is same as what Ross Millikan says. (Hope the above comments help you.)An5Drama –An5Drama 2024-04-16 13:34:30 +00:00 Commented Apr 16, 2024 at 13:34
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The positions of a perfect information game with no randomness fall into four categories: win for the first player, win for the second player, draw in finite time, infinite play. Not all games will have positions in all categories. Most games, including chess, have something that prevents the game going on forever. In those cases, showing there is no second player win is sufficient to show there is either a first player win or both players can force at least a draw. Chomp has no ties, so once you eliminate a second player win you know there is a first player win.
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answered Jan 20, 2021 at 1:45
Ross MillikanRoss Millikan
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1 My problem is that I am not seeing why "showing there is no second player win is sufficient to show there is either a first player win or both players can force at least a draw" or "once you eliminate a second player win you know there is a first player win."Favst –Favst 2021-01-20 01:47:23 +00:00 Commented Jan 20, 2021 at 1:47
Because we have a disjoint partition of the positions. If you show some categories are empty, all the positions must be in the others.Ross Millikan –Ross Millikan 2021-01-20 01:48:36 +00:00 Commented Jan 20, 2021 at 1:48
1 I'd add to the categories a caveat, that in many cases we reduce this to "novice v. expert" so that the expert has to win, but the novice can force a draw and still feel good about themselves. In other words, we usually treat a draw as a victory of one of the players.Asaf Karagila –Asaf Karagila♦ 2021-01-20 01:53:00 +00:00 Commented Jan 20, 2021 at 1:53
When you mention the four categories, is that in reference to individual instances of a particular game like Chomp, or in reference to different kinds of games among perfect information games?Favst –Favst 2021-01-20 01:53:05 +00:00 Commented Jan 20, 2021 at 1:53
2 This seems to be sweeping Zermelo's theorem under the rug. It's not a hard theorem that, say, the standard P-position/N-position decomposition actually corresponds to first or second player having a winning strategy. But it still is worthy of proof.Mark S. –Mark S. 2021-01-20 13:59:30 +00:00 Commented Jan 20, 2021 at 13:59
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If a strategy stealing argument applies, we can rule out that the second player will win by imagining an (unspecified, hypothetical) strategy that is "optimal" for them, and showing that a sequence of moves from the first player - a sequence of moves which is allowed to incorporate this imagined strategy - is strictly better than the strategy we started out with. Since we assumed that that strategy was optimal to begin with, we show that the first player cannot do worse than the second.
However, we can argue further from this point that, therefore, this stealing is an optimal strategy for the first player. The question is then, in response to this stolen strategy, whether the 2nd player's responses can force a draw or whether they lose. They cannot win, because if they could, the 1st player would have pre-empted them by doing it first.
A game of finite length can only have three outcomes - P1 wins, a draw, or P2 wins. We know P2 cannot win, but we cannot use strategy-stealing alone to show whether P1's stolen strategy can force a win for them or whether P2 can respond in such a way that they draw. But by showing P2 cannot win, we entirely elimate one of the three outcomes and so all possible perfect plays will result in the other two - P1 wins, or draws.
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answered Jan 20, 2021 at 2:02
redroidredroid
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12Converting a Gomoku winning strategy from a small board to a winning strategy on a larger board
4Prove using a strategy stealing argument that player 1 has a winning strategy in the chomp game
0Why does strategy-stealing not work for Go?
12Are there infinitely many α×β α×β Chomp boards where player 2 wins?
0In his winning strategy, the first move of player 1 in an n×n n×n Chomp game must be (2,2)(2,2)
2Finding the winning strategy in this variation of chomp
4Proof that there is a winning strategy for player 1 in Chomp.
1Any tips or idea how to play this combinatorial game?
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12829 | https://noda.ac.cn/datasharing/datasetDetails/64080bf0f64eb66545fa0b7e | 数据发现平台
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Landslide database of loess plateau based on high-resolution image and terrain interpretation
DOI：10.3974/geodb.2020.04.08.V1
CSTR：10441.11.202303.007789
### 2023-03-08 最新更新时间 ### 0 MB 数据量 ### 0 数据访问量 数据评级
数据基本信息
数据时间：
March 2017 - November 2018
空间位置：
The Loess Plateau
关键词：
Loess Plateau Landslide Quaternary Study
学科类别：
170 Earth Sciences - 170.45 geography - 170.4510 physical geography(Including chemical geography、Ecological geography、Geomorphology、Glaciology、Permafrost Science、Desert Science、etc)
主题分类：
Other data - Ground monitoring data - -
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0 条
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数据摘要
From March 2017 to November 2018, the author's team traveled over 7000 km and conducted landslide drone surveys along six typical routes on the Loess Plateau. The route is: (1) Xi'an Baoji Tianshui Qin'an Dingxi Lanzhou Heifangtai Xining; (2) Xi'an Xianyang Pingliang Huining Dingxi; (3) Xi'an Bailuyuan Lantian Jingyang; (4) Xi'an Tongchuan Yan'an Zhidan Yulin Ordos; (5) Xi'an Weinan Heyang; (6) Xi'an Weinan Sanmenxia Yuncheng Lishi Linfen. A total of 108 high-resolution images and topographic data of loess landslides were obtained. At the same time, 199 landslides were interpreted using Google satellite imagery and Japanese AW3D 5m DSM terrain data. Finally, using ArcGIS 10.5 and Global mapper 17.0, information such as geometric and topographic feature parameters of 307 loess landslides was extracted, and landslides were cataloged to obtain a loess plateau landslide database based on high-resolution images and terrain interpretation. The dataset includes: (1) the number, geographic coordinates, maximum length, maximum width, aspect ratio, perimeter, area, elevation difference, average slope, main sliding azimuth, data source, and terrain of 307 landslides in the Loess Plateau. (2) Photographic data of typical landslides. The dataset is stored in. shp,. jpg, and. docx formats, and consists of 17 data files with a data volume of 38.9 MB (compressed into one file, 38.5 MB). The research results based on this dataset have been published in the following publications: (1) The doctoral dissertation database of CNKI, titled "Spatial Pattern of Landslides on the Loess Plateau and Their Impact on Landform Evolution", was published in 2019; (2) Quaternary Research, Volume 38, Issue 2, 2018; (3) Landslides, Volume 15, Issue 3, 2018; (4) Landslides, Volume 16, Issue 3, 2019; (5) Geography, Volume 359, 2020.
数据作者（联系人）信息
姓名：
Hu Sheng
联系人电话：
邮编：
710127
联系人邮箱：
husheng@nwu.edu.cn
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Key Laboratory of Surface System and Environmental Carrying Capacity of Shaanxi Province, Northwest University
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Xi'an, Shaanxi
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Hu Sheng, Qiu Haijun, Wang Ninglian, Cui Yifei, Cao Mingming Loess Plateau Landslide Database Based on High Resolution Image and Terrain Interpretation [J/DB/OL] Electronic Journal of Global Change Data Warehousing, Chinese and English, 2020
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Please refer to "Spatial Pattern of Landslides on the Loess Plateau and Its Impact on Landform Evolution"
数据生成过程描述：
From March 2017 to November 2018, the author's team traveled over 7000 km and conducted landslide drone surveys along six typical routes on the Loess Plateau. The route is: (1) Xi'an Baoji Tianshui Qin'an Dingxi Lanzhou Heifangtai Xining; (2) Xi'an Xianyang Pingliang Huining Dingxi; (3) Xi'an Bailuyuan Lantian Jingyang; (4) Xi'an Tongchuan Yan'an Zhidan Yulin Ordos; (5) Xi'an Weinan Heyang; (6) Xi'an Weinan Sanmenxia Yuncheng Lishi Linfen. A total of 108 high-resolution images and topographic data of loess landslides were obtained. At the same time, 199 landslides were interpreted using Google satellite imagery and Japanese AW3D 5m DSM terrain data. Finally, using ArcGIS 10.5 and Global mapper 17.0, information such as geometric and topographic feature parameters of 307 loess landslides was extracted, and landslides were cataloged to obtain a loess plateau landslide database based on high-resolution images and terrain interpretation.
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Is it possible to evaluate a polynomial for any value x in Casio-fx82ms?
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Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Author has 16.6K answers and 29.4M answer views
·8y
Is it possible to evaluate a polynomial for any value x in Casio-fx82ms?
Polynomials tend to infinity for large x. There will come a point which is beyond the range of the calculator. The higher the degree, the smaller the x for which this occurs. There are two exceptions, a constant, and linear (if y = x) then any x that can be input has a corresponding y.
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John Gilmore
Author has 224 answers and 1.1M answer views
·8y
Remember there are some under used features on nearly every calculator and that includes the fx82ms. You may not have a lot of fancy features but you can still use the ones you have to their best.
If you are evaluating polynomials and want to cut back on the number of keystokes and eliminate some typing errors then an option with the fx82ms is to use its memory capabilities.
The fx82ms has 9 9 memorie
Continue Reading
Remember there are some under used features on nearly every calculator and that includes the fx82ms. You may not have a lot of fancy features but you can still use the ones you have to their best.
If you are evaluating polynomials and want to cut back on the number of keystokes and eliminate some typing errors then an option with the fx82ms is to use its memory capabilities.
The fx82ms has 9 9 memories that allow you to store any variables (numbers) you want to. They are A A to F F, M M, X X and Y Y and are painted in red above various keys on the calculator.
Say we want to store 2.5 2.5 in X X then we to do this we use the keystrokes: SHIFT STO 2.5 2.5 X X.
Now say we want to evaluate f(x)=2 x 3−4 x 2−5 x+7 f(x)=2 x 3−4 x 2−5 x+7 when x=2.5 x=2.5.
Press AC, then type in the polynomial, but when you need to type the X X press the ALPHA button first each time: 2×2× ALPHA [math]X x^{\blacksq...
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Yes — with limitations. The Casio fx-82MS (a basic scientific non-programmable model) cannot store a polynomial symbolically or run a program, but you can evaluate any polynomial at any numeric x using the calculator’s standard arithmetic and memory. Two practical methods:
1) Direct entry (small degree)
Enter the polynomial expression with x replaced by the numeric value, using parentheses for powers and multiplications.
Example: to evaluate 3x^4 − 5x^2 + 2x − 7 at x = 1.3 enter:
(3(1.3)^4)−(5(1.3)^2)+(2(1.3))−7
Pros: simple, no memory needed. Cons: error-prone for high-degree polynomials
Continue Reading
Yes — with limitations. The Casio fx-82MS (a basic scientific non-programmable model) cannot store a polynomial symbolically or run a program, but you can evaluate any polynomial at any numeric x using the calculator’s standard arithmetic and memory. Two practical methods:
1) Direct entry (small degree)
Enter the polynomial expression with x replaced by the numeric value, using parentheses for powers and multiplications.
Example: to evaluate 3x^4 − 5x^2 + 2x − 7 at x = 1.3 enter:
(3(1.3)^4)−(5(1.3)^2)+(2(1.3))−7
Pros: simple, no memory needed. Cons: error-prone for high-degree polynomials and repeated evaluations.
2) Horner’s method (recommended for efficiency and accuracy)
Rewrite polynomial in nested form to minimize operations and rounding error:
For P(x) = a_n x^n + a_{n−1} x^{n−1} + … + a_0, use
P(x) = (...((a_n x + a_{n−1}) x + a_{n−2}) ... ) x + a_0
Procedure on fx-82MS:
Store x in memory (optionally): enter the numeric x, press SHIFT STO → RCL? (use a single memory like M).
Alternatively, just type x value when needed.
Start with accumulator = a_n.
For each next coefficient a_{n−1} ... a_0: multiply accumulator by x, then add the coefficient.
Final accumulator is P(x).
Example: Evaluate 2x^4 − x^3 + 3x^2 − 4x + 5 at x = 2:
Start: accumulator = 2
Step1: 2 2 = 4 → + (−1) = 3
Step2: 3 2 = 6 → + 3 = 9
Step3: 9 2 = 18 → + (−4) = 14
Step4: 14 2 = 28 → + 5 = 33
Pros: fewer keystrokes, numerically stable, suited for high-degree polynomials.
Practical tips
Use parentheses around negative coefficients when typing directly.
Keep x in memory (M) if evaluating the same x multiple times: store with SHIFT STO→M and recall with RCL M.
Watch for overflow/underflow on very large degrees or large x values; results limited by calculator’s 10-digit display and exponent range.
For many repeated evaluations (different x values) consider using Horner’s method with the calculator memory to reduce keystrokes.
Conclusion
The fx-82MS can evaluate any polynomial numerically by direct substitution or, more efficiently, by Horner’s method using its basic arithmetic and memory functions.
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Richard P
MMath in Mathematics, Churchill College, Cambridge (Graduated 2011) · Author has 807 answers and 317.2K answer views
·May 27
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Are there any tools or calculators that can help break down polynomials like f 3(x)=x 8−8 x 6+20 x 4−16 x 2+2 f 3(x)=x 8−8 x 6+20 x 4−16 x 2+2 for easier solution finding?
Interesting!
I will start by referring to the initial 8th-degree polynomial as s(x). I will deduce the reason why the question refers to it as f³(x).
Firstly, a simple graph plot [here and throughout the answer, graphs are drawn using Mathematica] reveals the fact that s(x) = 0 has eight real roots.
Secondly, the s(x) has no odd-degree terms. So we can simplify it by writing u = x^2, leaving a second polynomial [call it g(u)] of degree 4: g(u) = u⁴ - 8u³ + 20u² - 16u + 2.
This plot again shows a symmetrical structure (this time with four real roots), but this time centred on u=2.
So write u = v+2 a
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Interesting!
I will start by referring to the initial 8th-degree polynomial as s(x). I will deduce the reason why the question refers to it as f³(x).
Firstly, a simple graph plot [here and throughout the answer, graphs are drawn using Mathematica] reveals the fact that s(x) = 0 has eight real roots.
Secondly, the s(x) has no odd-degree terms. So we can simplify it by writing u = x^2, leaving a second polynomial [call it g(u)] of degree 4: g(u) = u⁴ - 8u³ + 20u² - 16u + 2.
This plot again shows a symmetrical structure (this time with four real roots), but this time centred on u=2.
So write u = v+2 and we have another 4th-degree polynomial:
h(v) = (v+2)⁴ - 8(v+2)³ + 20 (v+2)² - 16 (v+2) + 2
h(v) = v⁴ - 4v² + 2 [and again there are no odd-degree terms, so write w = v^2]
j(w) = w² - 4w + 2
Write r = w-2
k(r) = r² - 2.
And now we can unwind the entire construction.
k(r) = 0 has roots r= ±√2
r = w-2, so w = r+2
j(w) = 0 has roots w = 2 ± √2
w = v²
h(v) = 0 has roots v = ± √(2 ± √2) [where the ± signs are independent]
u = v+2
g(u) = 0 has roots u = 2 ± √(2 ± √2) [where the ± signs are independent]
u = x²
s(x) = 0 has roots x = ±√(2 ± √(2 ± √2)) [where the ± signs are independent]. Numerically, the roots of s(x) = 0 are close to ±1.1111405, ±1.6629392, ±0.3901806 and ±1.9615706 .
And as for the notation: let f(x) = x² - 2 .
Then f(f(x)) = (x² - 2)² - 2 = x⁴ - 4x² + 2 .
And f(f(f(x))) = (x⁴ - 4x² + 2)² - 2 = x⁸ - 8x⁶ + 20x⁴ - 16x² + 2 = s(x).
So the notation f³(x) simply means f(f(f(x))).
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More answers below
How do evaluate polynomial f(x) =x⁴-7x²+2x-6 at x=1?
What is the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1?
How do I graph a 5th degree polynomial on a calculator?
How do you evaluate √x=−1 x=−1?
Is x²+4x²÷√x a polynomial?
Mes Haftu
BSC in Computer Engineers, Addis Ababa University (Graduated 2023)
·2y
Related
How do I calculate a mod in a scientific calculator (Casio fx-82MS)?
turn ON fx-82MS CASIO calculator
if you need to solve 729 mod 55
write 729
click the button “ab/c”
then 55
the answer will be something like
13_|14_|55
this means quotient = 13, reminder = 14
the answer or the modulus is the middle number
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Gopi Krishna
computer engineer
·9y
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How do I calculate a mod in a scientific calculator (Casio fx-82MS)?
As far as I know, that calculator does not offer mod functions. You can however computer it by hand in a fairly straightforward manner. Ex.
(1)50 mod 3
(2)50/3 = 16.66666667
(3)16.66666667 - 16 = 0.66666667
(4)0.66666667 3 = 2
Therefore 50 mod 3 = 2
Things to Note: On line 3, we got the "minus 16" by looking at the result from line (2) and ignoring everything after the decimal. The 3 in line (4) is the same 3 from line (1).
Hope that Helped.
Edit As a result of some trials you may get x.99991 which you will then round up to the number x+1.
Upvote ·
999 112
9 6
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.3K answers and 1.7M answer views
·3y
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How do you evaluate polynomials for given values?
Let’s assume we have a polynomial of degree 2: p(x) = x^2 - 17x + 34
To evaluate this polynomial for a given value (eg x = 2), you replace every x in the formula with this value
p(2) = (2)^2 - 17(2) + 34 = 4 - 34 + 34 = 4
p(-1) = (-1)^2 - 17(-1) + 34 = 1 + 17 + 34 = 52
If you repeat this process for a lot of different x values, and plot the evaluated values on a y-axis you get the graph of this polynomial:
Continue Reading
Let’s assume we have a polynomial of degree 2: p(x) = x^2 - 17x + 34
To evaluate this polynomial for a given value (eg x = 2), you replace every x in the formula with this value
p(2) = (2)^2 - 17(2) + 34 = 4 - 34 + 34 = 4
p(-1) = (-1)^2 - 17(-1) + 34 = 1 + 17 + 34 = 52
If you repeat this process for a lot of different x values, and plot the evaluated values on a y-axis you get the graph of this polynomial:
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9 6
Darren Lorent
Author has 5.1K answers and 594.9K answer views
·3y
Related
How do I evaluate a polynomial at a specific x-value?
Replace x with the value in question and do the arithmetic.
Example: evaluate the polynomial function at the value x = 2
f(x) = x ³ - 2x ² + 7x - 1
f(2) = (2)³ - 2(2)² + 7(2) - 1 =
8 - 8 + 14 - 1 = 13
Upvote ·
9 4
Robert Smith
3y
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How do polynomial factoring calculators work?
Fill In The Blank Space With What You Want To Factorize.
Type the expression or number in the provided space that you'd want to get factor from factorize calculator.
Compare The Operator Signs (+, -) In Different Terms.
Examine the operator signs used in the expression terms of factoring solver.
Select The 'Factorize It' Option From The Drop-Down Menu In The Factoring Tool.
To acquire the factorization results, click the 'Factorize it' button. Use Now.
Continue Reading
Fill In The Blank Space With What You Want To Factorize.
Type the expression or number in the provided space that you'd want to get factor from factorize calculator.
Compare The Operator Signs (+, -) In Different Terms.
Examine the operator signs used in the expression terms of factoring solver.
Select The 'Factorize It' Option From The Drop-Down Menu In The Factoring Tool.
To acquire the factorization results, click the 'Factorize it' button. Use Now.
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·Updated Sep 4
What are the all the functions of a printer?
Modern printers have evolved far beyond basic document reproduction. Today’s models offer a range of built-in features that streamline both personal and professional tasks. Depending on the device, a printer can serve as a multifunctional hub for printing, scanning, copying and even faxing. Enhanced connectivity allows seamless operation from mobile devices and cloud-based platforms, while efficiency-focused options such as duplex printing and high-yield cartridges help reduce both cost and waste. For users balancing home office duties with general everyday printing, choosing the right combina
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Modern printers have evolved far beyond basic document reproduction. Today’s models offer a range of built-in features that streamline both personal and professional tasks. Depending on the device, a printer can serve as a multifunctional hub for printing, scanning, copying and even faxing. Enhanced connectivity allows seamless operation from mobile devices and cloud-based platforms, while efficiency-focused options such as duplex printing and high-yield cartridges help reduce both cost and waste. For users balancing home office duties with general everyday printing, choosing the right combination of features matters just as much as print quality itself.
For those who need consistent performance and advanced functionality, the HP LaserJet Pro MFP 3302fdw offers robust print, scan and copy features in a compact all-in-one design. It delivers fast output with sharp resolution and supports duplex printing for efficient paper use. Integrated Wi-Fi and mobile compatibility make remote document handling straightforward. The control panel is intuitive and the device is built for moderate-to-heavy workloads, making it a dependable option for home offices with daily print demands.
LaserJet Printers - Black & White or Color Document Printers
If colour reproduction is a priority for photos, marketing materials or vivid documents, the HP Color Laser MFP 179fnw is a versatile alternative. It supports full-colour printing along with scanning, copying and faxing, and offers borderless output to create professional-grade visuals. Though it has a smaller footprint and lower print speeds compared to the 3302fdw, it remains suitable for light-to-medium use in creative or business settings. Its wireless features ensure flexible access from laptops and smartphones, enhancing convenience in dynamic work environments.
Color Laser Printers - HP LaserJet Pro
A printer’s value lies not only in what it prints but in how effectively it adapts to your workflow. HP’s all-in-one models like the HP LaserJet Pro MFP 3302fdw and Color Laser MFP 179fnw deliver a thoughtful mix of performance, reliability and smart features tailored to a variety of personal and professional needs. Selecting the right model comes down to the volume, content and style of your typical print tasks.
I hope this clarifies the different functions of printers for you, and check out the Printer Buying Guide linked below to choose the right model for your printing needs!
Inkjet vs LaserJet vs OfficeJet: HP Printer Buying Guide
By Henry - HP Tech Expert
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99 15
9 2
Ayush Srivastava
PG Diploma in Artificial Intelligence, Indian Institute of Technology, Guwahati (IITG) (Graduated 2022) · Author has 74 answers and 60.9K answer views
·Updated 8y
Related
Mathematics: How do I solve y= cos(x) - x(e^x) using a Casio-991MS?
I tried solving the equation on my CASIO fx-991MS and I could easily solve it. The unit for angle was set to DEGREE (the default) and I used the SOLVE feature. It took some time, but gave an approximate answer, as 1, nonetheless.
Assuming you could input the expression correctly, I'm going to tell you how you could use the SOLVE feature to solve your problem.
Input the expression, and use parentheses if confused about operator precedence.
Now, press the SHIFT button, present on the top left corner. You'll see that an S has appeared on the top left corner of the screen, indicating that the SHIFT b
Continue Reading
I tried solving the equation on my CASIO fx-991MS and I could easily solve it. The unit for angle was set to DEGREE (the default) and I used the SOLVE feature. It took some time, but gave an approximate answer, as 1, nonetheless.
Assuming you could input the expression correctly, I'm going to tell you how you could use the SOLVE feature to solve your problem.
Input the expression, and use parentheses if confused about operator precedence.
Now, press the SHIFT button, present on the top left corner. You'll see that an S has appeared on the top left corner of the screen, indicating that the SHIFT button has been pressed.
Now, press the CALC button, present directly below the SHIFT button.
You'll enter the SOLVE mode, and a prompt will appear asking for a guess value.
Choose an appropriate guess value something like an approximate solution, or at least, not too arbitrary and enter it by pressing the = button.
Now, use the SOLVE feature to get the solution, i.e., press the SHIFT button, followed by the CALC button.
If you have followed above instructions precisely, and chosen an appropriate guess value you'll find that your screen has gone blank, and within moments you'll have the answer.
Upvote ·
9 4
Csaba Tizedes
Experienced calculator user, programmer, collector · Author has 115 answers and 317K answer views
·Updated 6y
Related
How do I use a Casio FX-82MS calculator to complete a table for an equation?
If you want to prepare a table from a with stepsize d of function f(x):
store these values: a STO A, d STO D, then
store A-2D in X: A-2D STO X
type X+D STO X then
press left arrow to edit the above and type to the end the colon (:) and the function f(x)
Press =, you will see the next x value and the display shows Disp anunciator because of colon (:), then press = again, and you will see the f(x) value
Repeat step 5 and you will see the new x values and the f(x) values in order
An example:
You want to prepare a table for f(x)=3x^2–7 from a=-5 by step d=2:
-5 STO A 2 STO D
A-2D STO X
X+D STO X
press left ar
Continue Reading
If you want to prepare a table from a with stepsize d of function f(x):
store these values: a STO A, d STO D, then
store A-2D in X: A-2D STO X
type X+D STO X then
press left arrow to edit the above and type to the end the colon (:) and the function f(x)
Press =, you will see the next x value and the display shows Disp anunciator because of colon (:), then press = again, and you will see the f(x) value
Repeat step 5 and you will see the new x values and the f(x) values in order
An example:
You want to prepare a table for f(x)=3x^2–7 from a=-5 by step d=2:
-5 STO A 2 STO D
A-2D STO X
X+D STO X
press left arrow and type :3X^2–7 (do not forget the colon(:)!!!)
press =, you will see x=-5 and Disp, press = again, you will see 68 (=3×(-5)^2–7)
press =, you will see x=-3 and Disp, press = again, you will see 20 (=3×(-3)^2–7)
press =, you will see x=-1 and Disp, press = again, you will see -1 (=3×(-1)^2–7)
and so on….
Upvote ·
99 19
9 1
Bernard Leak
Firmware Developer (2008–present) · Upvoted by
Kostyantyn Mazur
, PhD Mathematics, New York University (2018) · Author has 5.8K answers and 5M answer views
·6y
Related
Is it possible to algebraically evaluate values of the Lambert W function, or are numerical algorithms (e.g. Newton's method) the only option?
What counts as algebraic evaluation here? The implied contrast between algebraic and numerical methods seems a bit askew. Newton's method is often rather semi-numerical; but the time you plug actual numbers into it, you've formally, rather than numerically, differentiated something. A contrast between algebraic and analytical methods is perhaps more natural, but it hardly falls between some class of elementary functions and Lambert's W-function. Extracting a square root may be algebraic in effect, but most methods for doing it are analytical in detail. Even multiprecision division is likely to
Continue Reading
What counts as algebraic evaluation here? The implied contrast between algebraic and numerical methods seems a bit askew. Newton's method is often rather semi-numerical; but the time you plug actual numbers into it, you've formally, rather than numerically, differentiated something. A contrast between algebraic and analytical methods is perhaps more natural, but it hardly falls between some class of elementary functions and Lambert's W-function. Extracting a square root may be algebraic in effect, but most methods for doing it are analytical in detail. Even multiprecision division is likely to be performed through an analytical approach to evaluating a reciprocal. On the other hand, Lambert's W-function is defined as the inverse of a function which itself is analytically defined. What could “algebraic evaluation” of it mean?
I suspect that hovering behind the question is a distinction between, say, evaluating a convergent power-series, where improving the precision means adjoining additional terms, and hunting through iterations of a predictor/corrector algorithm. One does feel a difference of flavour, but it's quite hard to pin it down to definite rules. What sort of thing are you including as “algebraic evaluation”?
Upvote ·
9 1
9 4
John K WilliamsSon
Calculator nerd loves trying new things on his calculators · Author has 9K answers and 23.4M answer views
·3y
Related
How do I evaluate a function like this “f(x) =x+1 if x=3x-1” in a Casio FX991EX calculator?
If you had asked how to do it on a TI-84, I’d have said:
graph Y1 = x
graph Y2 = 3x – 1
look for the intersection
But the Casio FX991EX does not graph equations on its screen. Instead, as the Casio site tells us:
Need graph and number line representation? The fx-991EX generates a QR code for your expression. Scan the code with any internet-enabled smartphone and immediately see your data plot through ClassPad.net, our free online math platform.
I don’t have this calculator, so I don’t know if you can display two graphs at once.
Here is how I solved x=3x-1 by graphing it.
As you can see, the two lines
Continue Reading
If you had asked how to do it on a TI-84, I’d have said:
graph Y1 = x
graph Y2 = 3x – 1
look for the intersection
But the Casio FX991EX does not graph equations on its screen. Instead, as the Casio site tells us:
Need graph and number line representation? The fx-991EX generates a QR code for your expression. Scan the code with any internet-enabled smartphone and immediately see your data plot through ClassPad.net, our free online math platform.
I don’t have this calculator, so I don’t know if you can display two graphs at once.
Here is how I solved x=3x-1 by graphing it.
As you can see, the two lines intersect at X = ½.
I could then look at f(x) = x + 1, substitute x = ½, and figure out mentally that f(½) = 1½ = 1.5
Yes, I could have graphed that, too, but it was so simple to do it in my head.
Here is how to solve it without a calculator:
Starting equation:
x = 3x – 1
Subtract x from both sides
0 = 2x – 1
Add 1 to both sides
1 = 2x
Divide both sides by 2
½ = x
Solved for x
Other equation:
f(x) = x + 1
Substitute x = ½
f(½) = ½ + 1
f(½) = 1½ = 1.5
Upvote ·
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12832 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6651628/ | Glucagon Receptor Signaling and Glucagon Resistance - PMC
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Int J Mol Sci
. 2019 Jul 5;20(13):3314. doi: 10.3390/ijms20133314
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Glucagon Receptor Signaling and Glucagon Resistance
Lina Janah
Lina Janah
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
Find articles by Lina Janah
1,2,†, Sasha Kjeldsen
Sasha Kjeldsen
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
Find articles by Sasha Kjeldsen
1,2,†, Katrine D Galsgaard
Katrine D Galsgaard
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
Find articles by Katrine D Galsgaard
1,2,†, Marie Winther-Sørensen
Marie Winther-Sørensen
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
Find articles by Marie Winther-Sørensen
1,2, Elena Stojanovska
Elena Stojanovska
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
Find articles by Elena Stojanovska
1,2, Jens Pedersen
Jens Pedersen
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
3 Department of Cardiology, Nephrology and Endocrinology, Nordsjællands Hospital Hillerød, University of Copenhagen, 3400 Hillerød, Denmark
Find articles by Jens Pedersen
1,3, Filip K Knop
Filip K Knop
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
4 Center for Clinical Metabolic Research, Gentofte Hospital, University of Copenhagen, 2900 Hellerup, Denmark
5 Department of Clinical Medicine, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
6 Steno Diabetes Center Copenhagen, 2820 Gentofte, Denmark
Find articles by Filip K Knop
2,4,5,6, Jens J Holst
Jens J Holst
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
Find articles by Jens J Holst
1,2, Nicolai J Wewer Albrechtsen
Nicolai J Wewer Albrechtsen
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
7 Department of Clinical Biochemistry, Rigshospitalet, 2100 Copenhagen, Denmark
8 Novo Nordisk Foundation Center for Protein Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2100 Copenhagen, Denmark
Find articles by Nicolai J Wewer Albrechtsen
1,7,8,
Author information
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Copyright and License information
1 Department of Biomedical Sciences, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
2 Novo Nordisk Foundation Center for Basic Metabolic Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
3 Department of Cardiology, Nephrology and Endocrinology, Nordsjællands Hospital Hillerød, University of Copenhagen, 3400 Hillerød, Denmark
4 Center for Clinical Metabolic Research, Gentofte Hospital, University of Copenhagen, 2900 Hellerup, Denmark
5 Department of Clinical Medicine, Faculty of Health and Medical Sciences, University of Copenhagen, 2200 Copenhagen, Denmark
6 Steno Diabetes Center Copenhagen, 2820 Gentofte, Denmark
7 Department of Clinical Biochemistry, Rigshospitalet, 2100 Copenhagen, Denmark
8 Novo Nordisk Foundation Center for Protein Research, Faculty of Health and Medical Sciences, University of Copenhagen, 2100 Copenhagen, Denmark
Correspondence: nicolai.albrechtsen@sund.ku.dk
†
Contributed equally to the preparation of the manuscript.
Received 2019 May 30; Accepted 2019 Jul 3; Collection date 2019 Jul.
© 2019 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
PMC Copyright notice
PMCID: PMC6651628 PMID: 31284506
Abstract
Hundred years after the discovery of glucagon, its biology remains enigmatic. Accurate measurement of glucagon has been essential for uncovering its pathological hypersecretion that underlies various metabolic diseases including not only diabetes and liver diseases but also cancers (glucagonomas). The suggested key role of glucagon in the development of diabetes has been termed the bihormonal hypothesis. However, studying tissue-specific knockout of the glucagon receptor has revealed that the physiological role of glucagon may extend beyond blood-glucose regulation. Decades ago, animal and human studies reported an important role of glucagon in amino acid metabolism through ureagenesis. Using modern technologies such as metabolomic profiling, knowledge about the effects of glucagon on amino acid metabolism has been expanded and the mechanisms involved further delineated. Glucagon receptor antagonists have indirectly put focus on glucagon’s potential role in lipid metabolism, as individuals treated with these antagonists showed dyslipidemia and increased hepatic fat. One emerging field in glucagon biology now seems to include the concept of hepatic glucagon resistance. Here, we discuss the roles of glucagon in glucose homeostasis, amino acid metabolism, and lipid metabolism and present speculations on the molecular pathways causing and associating with postulated hepatic glucagon resistance.
Keywords: alpha cell, amino acids, diabetes, glucose, hyperaminoacidemia, hyperglucagonemia, liver–alpha cell axis
1. Introduction
In the 1920s, Kimball and Murlin reported the existence of a pancreatic factor other than insulin with effects on glucose homeostasis (GLUCose-AGONist ) . This factor was later identified as glucagon (denoted glucagon 1–29) . Glucagon is a peptide of 29 amino acids with a variety of biological actions including, but not limited to, glucose homeostasis [4,5,6]. A recent PubMed search on glucagon revealed that more than 45,000 articles relating to glucagon have been published, and extensive efforts are currently being made to further study and understand the physiological role(s) of glucagon [4,7].
The biology of glucagon is intriguing not only from a physiological perspective but indeed also from a therapeutic point [1,8,9]. Glucagon receptor antagonism reduces blood glucose in patients with type 2 diabetes, and glucagon co-agonism (with, e.g., an incretin hormone) reduces body weight in overweight type 2 diabetes patients . The complexity of glucagon´ suggested diabetogenic effect is greater than anticipated, as antagonizing its actions affects fasting but not postprandial blood glucose levels . This is in contrast to what one may have suspected, as the ‘hyperglucagonemia’ observed in some type 2 diabetes patients has been reported to be pronounced in the postprandial state (the lack of glucose-induced suppression of alpha cell secretion, which we will come back to later) [12,13,14]. The impact of glucagon therefore differs between fasting and prandial conditions, at least when it comes to glucose homeostasis. Investigating the physiological actions of glucagon on not only glucose homeostasis but, importantly, also on lipid and amino acid metabolism is needed for understanding glucagon biology at fasting and prandial conditions. Specifically, biased (meaning activation of specific intracellular pathways) targeting of glucagon receptor signaling towards, e.g., hepatic glycogenolysis or gluconeogenesis may potentially represent an ‘intelligent pharmacological’ approach exploiting, e.g., the glucose lowering effects of glucagon receptor antagonism without its undesired effect on hepatic lipid metabolism (increased liver fat ).
The purpose of this review is to assess the key aspects of glucagon receptor signaling not only regarding glucose homeostasis but also with respect to amino acid and lipid metabolism. We will also discuss the evidence and potential relevance of the new concept of hepatic ‘glucagon resistance’. We begin this review by introducing the production, secretion, and measurement of glucagon before turning towards glucagon receptor signaling in regards to glucose, amino acid, and lipid metabolism.
2. Processing of Proglucagon
The GCG gene, encoding the glucagon precursor proglucagon, is well conserved across species . Proglucagon has 160 amino acids and is expressed in certain neurons of the brain stem, in intestinal L cells, and in pancreatic alpha cells . Several bioactive peptides, including glucagon-like peptide 1 (GLP-1) and glucagon-like peptide 2 (GLP-2), are cleaved from proglucagon by prohormone convertase(s) in a tissue-specific (or perhaps more accurately enzyme-specific ) manner (Figure 1). The differential processing of proglucagon appears to reflect the enzymatic activities of the two prohormone convertases: prohormone convertase 1/3 (PC1/3) and 2 (PC2) . Proglucagon therefore gives rise to a variety of peptides. Thus, throughout the small and large intestine, proglucagon-producing cells termed L cells are located within the epithelium [20,21] in an ideal position to sense the variety of nutrients and microbial products and convey the information to the rest of the body via the secretion of GLP-1, GLP-2, oxyntomodulin, and glicentin, which contribute to the regulation of appetite, bone resorption, gastrointestinal growth, and glucose homeostasis [1,22,23,24,25]. With co-expression of PC1/3 (e.g., in intestinal L cells), proglucagon is cleaved to form glicentin, oxyntomodulin, GLP-1, and GLP-2; whereas with PC2 expression as in the alpha cells, proglucagon is cleaved to form mainly glucagon and the so-called major proglucagon fragment [26,27,28]. In line with this, mice deficient of PC1/3 are incapable of producing GLP-1, while mice deficient of PC2 cannot produce glucagon [29,30,31,32].
Figure 1.
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Processing and measurement glucagon. Glucagon (proglucagon 33–61) results from prohormone convertase 2 (PC2)-dependent processing of proglucagon (PG 1–160). In the intestine, PG is processed by prohormone convertase 1/3 (PC1/3) activity to form glicentin (1–69), which may be further cleaved into glicentin-related pancreatic polypeptide (GRPP) and oxyntomodulin (33–69). N-terminal directed antibodies will therefore also cross-react with oxyntomodulin whereas C-terminal antibodies react with proglucagon 1–61, and finally antibodies raised against the mid-region of glucagon will potentially bind to all of the aforementioned peptides. Measurement of glucagon may therefore require a sandwich ELISA targeting both termini.
The absolute selectivity of PC1/3 and PC2 remains a matter of discussion. It has been speculated that metabolic stressors such as type 2 diabetes, obesity, and Roux-en-Y gastric bypass surgery may alter the processing profile of proglucagon both in the pancreas and in the gut, but the extent to which this occurs in humans and the clinical relevance of such changes remain unknown .
3. Secretion of Glucagon
Glucagon is secreted in response to a variety of metabolic signals [6,33] such as changes in blood glucose concentrations [2,34], certain amino acids , perhaps free fatty acids , and in response to stress (e.g., activation of the sympathetic nervous system). Here, we shortly discuss some of the currently suggested mechanisms underlying glucose-dependent glucagon secretion. For further insight, please see Ref. [4,38,39,40].
In humans, blood glucose levels are reciprocally correlated to glucagon secretion, and the potential intrinsic glucose-sensing mechanisms of alpha cells have been studied for decades using a variety of techniques [6,41,42,43]. As an example, the physiological roles of sodium and potassium channels have been studied in whole islets and isolated alpha cells using electrophysiological techniques (patch clamping). Paracrine factors also play an important role and some have argued that combining or integrating intrinsic and paracrine factors is needed to uncover the enigmatic mechanism of glucose-induced inhibition of glucagon secretion [44,45].
The mechanisms underlying glucose-induced inhibition of alpha cell secretion are still a matter of debate. One of the proposed intrinsic pathways leading to hypoglycemia-induced glucagon secretion is a decrease in the ATP/ADP ratio, which paradoxically slightly increases K ATP channel activity, leading to voltage-dependent increased activity of P/Q type calcium channels and a subsequent influx of Ca 2+ . In vivo, another important mechanism may be hypoglycemia-induced activity of the pancreatic sympathetic innervation . The potent regulation of glucagon secretion by glucose from isolated perfused pancreas preparations supports a direct effect of hypoglycemia on the alpha cell , unless stimulatory neurons of the intra-pancreatic ganglia are glucose sensitive.
Hyperglycemia-induced attenuation of glucagon secretion may be even more complex due to extra-pancreatic signals such as the glucagonostatic effect of GLP-1 or the glucagonotropic effects of GLP-2 and glucose-dependent insulinotropic polypeptide (GIP), all three of which are secreted in response to the ingestion of carbohydrates. In addition, paracrine signals elicited by glucose, in particular delta cell-secreted somatostatin and beta cell-secreted insulin, may also inhibit glucagon secretion. A recently published study demonstrates that glucokinase, expressed in alpha cells, may play an important role in glucose-regulated glucagon secretion [49,50]. The inhibitory effect of glucose is preserved in whole islets but appears to be lost when alpha cells are isolated and studied selectively . This again points to an important effect of intra-islet factors secreted from the neighboring beta and delta cells. Somatostatin secreted from the pancreatic delta cells and insulin secreted from pancreatic beta cells in response to increased glucose concentrations both inhibit glucagon secretion , although this may depend on the experimental setting . Other intra-islet factors have also been implied in the regulation of glucagon secretion: urocortin 3 , zinc , GABA/L-glutamate , gamma-aminobutyric acid , amylin , and ephrin . Furthermore, several extra-islet signals have been reported to contribute to the regulation of glucagon secretion and include, but are not limited to, GLP-1 , GIP , GLP-2 , ghrelin , and gastrin . Finally, the newer class of glucose-lowering drugs for the treatment of type 2 diabetes, sodium-glucose co-transporter 2 (SGLT-2) inhibitors, have been associated with increased plasma glucagon concentrations. The absolute differences in plasma glucagon concentrations between SGLT-2 inhibitor-treated and placebo-treated patients are small (1–1.5 pmol/L) [65,66], and although some have argued for a direct role of SGLT-2 on the alpha cells [67,68,69,70], current evidence, which includes the results of glucose clamping, seems to point toward an indirect mechanism by which SGLT-2 inhibition increases renal glucosuria, resulting in significant lowering of blood glucose concentrations that causes a slightly greater alpha cell secretion [71,72,73,74,75]. Whether or not increased plasma glucagon after SGLT-2 inhibition has any clinical impact is currently unknown, but some may argue that increasing hepatic glucose production when combined with reduced glucose reuptake in the kidneys may limit the reduction in blood glucose (and hence clinical efficacy).
Clearly then, changes in glucagon secretion induced by glucose, for instance, depend on the experimental conditions and may differ dramatically between experiments in isolated alpha cells, isolated islets, isolated perfused pancreas preparations, and in vivo. Intra- and extra-islet factors implicated in the regulation of glucagon secretion are summarized in Figure 2.
Figure 2.
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Regulation of glucagon secretion. Several factors regulate the secretion of glucagon; most importantly glucose, amino acids, gastrointestinally derived peptides, the autonomic nervous system (extra-islet regulation), and possibly peptides secreted from the alpha, beta, and delta cells (intra-islet regulation), among which at least the inhibitory action of delta cell-derived somatostatin is well established. Black arrows refer to a stimulatory effect on glucagon secretion and red T bars refer to an inhibitory effect.
4. Accurate Measurement of Glucagon
A major limitation to the study of glucagon physiology and pathophysiology has been the hitherto limited accuracy of many immunoassays used to measure glucagon . Some studies have applied additional analytical techniques such as size-exclusion chromatography and mass-spectrometry for characterization of the measured moiety. Falsely increased glucagon immunoreactivity may be due to degradation of circulating glucagon-containing molecules such as glicentin, oxyntomodulin (in vivo as well as post sampling), or proglucagon with resulting formation of (immunoreactive) glucagon. Another explanation may be insufficient specificity, e.g., cross-reactivity of the antibody/antibodies used in the assay with peptides with related sequences to that of glucagon [77,78]. The term hyperglucagonemia may therefore not in all cases relate to increased plasma concentrations of bioactive glucagon but may as well be ascribed to preanalytical and analytical challenges. The analytical and physiological importance of N-terminally elongated glucagon-like moieties is uncertain [78,79,80], but part of the concentration of immunoreactive glucagon measured in plasma with the single antibody C-terminal immunoassays may certainly be due to such molecular forms, in particular proglucagon 1–61 . This may be particularly important in patients with impaired kidney function .
5. Glucagon Receptor Signaling
5.1. Glucagon and Glucose Homeostasis
A fundamental aspect of glucagon signaling is its role as a regulator of glucose homeostasis [43,82]. Increased plasma glucagon levels result in increased hepatic glucose production . Therefore, glucagon is traditionally known as a counter-regulatory hormone to the hypoglycemic effects of insulin . This balance of glucagon and insulin signaling is thus largely responsible for maintaining physiological euglycemia . In hypoglycemic conditions, glucagon secretion increases, which results in increased hepatic glucose production via a number of cellular mechanisms involving suppression of glycogenesis and glycolysis and stimulation of glycogenolysis and gluconeogenesis [86,87] (Figure 3).
Figure 3.
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Glucagon effects on hepatic glucose production. Activation of the glucagon receptor results in adenylate cyclase activation and cAMP formation. The increase in intracellular cAMP levels activates protein kinase A (PKA), which phosphorylates the transcription factor cAMP-response-element-binding (CREB) protein. CREB induces the transcription of glucose 6-phosphatase and phosphoenolpyruvate carboxykinase (PEPCK), two enzymes that contribute to increased gluconeogenesis. PKA phosphorylates (hence activates) the phospho-fructokinase 2 (PFK-2)/fructose 2,6-bisphosphatase (FBPase2) protein. Upon phosphorylation, PFK-2 activity is inhibited while FBPase2 activity is activated. Glucagon thus lowers the level of fructose 2,6-bisphosphate and increases fructose 6-phosphate levels, which suppresses glycolysis and increases gluconeogenesis. Secondly, PKA phosphorylates pyruvate kinase, resulting in increased fructose 1,6 bisphosphate levels and decreased pyruvate levels, which leads to reduced glycolysis. Most importantly, PKA phosphorylates phosphorylase kinase, initiating the glycogenolysis cascade increasing the conversion of glycogen to glucose 1-phosphate. Finally, PKA phosphorylates and inhibits glycogen synthase (glucose-6-phosphatase). The red arrows indicate inhibitory actions of glucagon receptor signaling while black arrows indicate stimulatory actions of glucagon receptor signaling.
Once glucagon binds its seven transmembrane receptor on the plasma membrane of the cell, it leads to conformational changes that activate G αs-coupled proteins . This consequently increases cAMP levels via the activation of adenylate cyclase, which in turn stimulates activation of protein kinase A (PKA) and cAMP response element-binding (CREB) protein . CREB is responsible for inducing transcription of glucose 6-phosphatase and phosphoenolpyruvate carboxykinase (PEPCK), both of which contribute to increased gluconeogenesis. Meanwhile, PKA activation results in a number of intracellular events in addition to phosphorylation of CREB [41,86]. Firstly, it phosphorylates (hence activates) the phospho-fructokinase 2 (PFK-2)/fructose 2,6-bisphosphatase (FBPase2) protein. Upon phosphorylation, the PFK-2 activity is inhibited while the FBPase2 activity is activated, leading to increased fructose 6-phosphate levels. This induces gluconeogenesis, while decreased fructose 2,6 bisphosphate levels result in reduced glycolysis [41,86]. Secondly, PKA activates pyruvate kinase, which increases fructose 1,6 bisphosphate levels and thus lower pyruvate levels, which suppresses glycolysis . Thirdly, PKA activates phosphorylase kinase, which, via activated phosphorylase, increases conversion of glycogen to glucose 1-phosphate in addition to inhibiting glycogen synthase, which overall lowers glycogen levels and results in hepatic glucose release .
During periods of short-term fasting (less than ~12 h), glucose levels in the bloodstream rely mainly on glycogenolysis, but over time as glycogen stores become depleted, the maintenance of euglycemia increasingly depends on gluconeogenesis . Because the effect of glucagon on gluconeogenesis relies on gene transcription, the physiological effects do not result until hours after glucagon secretion, whereas the stimulatory effect of glucagon on glycogenolysis has observable effects within minutes due to the rapid phosphorylation cascade . Given that gluconeogenesis is a substrate-dependent mechanism and glucagon does not feed gluconeogenesis via provision of, e.g., muscle-derived amino acids (to date no glucagon receptor has been identified on human skeletal cells), other mechanisms, including catecholamines and cortisol [91,92,93], must act in order to maintain blood glucose during long-term fasting. However, glucagon stimulates hepatic amino acid metabolism , resulting in an increased flux of amino acids into the hepatocytes and is thereby also able to provide substrates in the form of gluconeogenic amino acids. The relative contribution of glucagon to glucose production therefore varies depending on glycogen content and the availability of gluconeogenic substrates in the circulation. This is in concordance with studies showing that during prolonged fasting (when glycogen stores are depleted, and the hepatic glucose production depends on gluconeogenesis), the effect of glucagon on hepatic glucose production is weak [83,91,92]. Glucagon may therefore be ineffective in regards to hepatic glucose production when glycogen stores are exhausted, e.g., during starvation and prolonged exercise [84,88].
This is exemplified by the findings of attenuated glucose production in response to exogenous glucagon after one week of low carbohydrate diet (LCD) compared to the response after a week of high carbohydrate diet (HCD) in patients with insulin pump-treated type 1 diabetes [95,96]. The lowered glucose production in response to glucagon is likely to be due to the reduced storage of hepatic glycogen in the LCD group [97,98]. However, acute glucagon administration after a period of LCD caused a significantly higher increase in free fatty acid (FFA) and ketone body plasma concentrations compared to glucagon administration after the HCD. This indicates that the LCD may have promoted the use of fat over carbohydrates, resulting in increased fat oxidation and ketogenesis , supporting the conclusion that glucagon regulates glucose homeostasis mainly through glycogenolysis rather than gluconeogenesis. These results suggest that diet carbohydrate content must be accounted for when treating hypoglycemia with low-dose glucagon, as the amount of glycogen is crucial for the effect of glucagon on blood glucose concentrations .
The above-mentioned studies are of clinical interest as glucagon is used to treat hypoglycemia—in most cases due to exogenous hyperinsulinemia (iatrogenic). Furthermore, low dose glucagon injection is currently being investigated in connection with the development of dual-pumps (glucagon + insulin infusion) [101,102,103,104,105]. Recent studies suggest that the amount of glucose produced (assumable by hepatic glucose production) after glucagon injection may not be influenced by body weight—in the same study, insulin levels across patients and testing time points remained constant. Therefore, the capacity for glucagon to appropriately regulate plasma glucose concentrations may, as suggested, not be affected by body weight and hence the existence of glucagon resistance (see the Section regarding glucagon resistance). In contrast, studies in animal models of hepatic steatosis have reported an attenuated effect of glucagon on glucose production thereby possibly linking pathophysiological traits of the liver to glucagon resistance [106,107]. Preliminary data from our group did, however, not support the notion of glucagon resistance with respect to its ability to stimulate hepatic glucose production in humans, but in contrast revealed an impaired effect on amino acid metabolism .
As alluded to above, it is poorly understood how glucagon secretion is suppressed during hyperglycemia [109,110,111,112]. In studies examining the effect of carbohydrates on glucagon secretion, glucagon secretion decreases in response to these stimuli, thereby deeming glucagon redundant for glucose homeostasis during these conditions [113,114]. (Orally administered glucose has, however, been reported to increase plasma concentrations of glucagon in healthy subjects and in patients with type 2 diabetes ). This is, however, not the case when high-fat or protein-rich meals are orally administered, as these stimuli increase glucagon secretion [113,116]. In a recent study of the importance of glucagon as a postprandial hormone in healthy individuals, oral administration of whey protein increased both endogenous glucose production and glucose disposal to the same degree . Despite protein ingestion resulting in a concomitant increase in insulin and glucagon concentrations in the blood, euglycemia was preserved throughout the trial . These data suggest that glucagon may be important for glucose flux during the postprandial state after protein ingestion.
A known characteristic of patients with type 2 diabetes is dysregulated glucagon secretion and many have consequently focused on glucagon antagonism [117,118]. Genetic and pharmacologic inhibition of glucagon signaling leads to lowering of fasting plasma glucose concentrations by approximately 1–2 mmol/L in mice and in humans [11,119,120,121,122]. These observations are in line with the early suggestion by Unger and colleagues that increased glucagon signaling is diabetogenic [123,124]. However, administration of a glucagon receptor antagonist (GRA) does not seem to affect postprandial glucose excursions in patients with type 2 diabetes , indicating that glucagon is important for glucose homeostasis in the fasting state rather than postprandially. GRAs are currently being pursued as a possible therapeutic target for the treatment of diabetes, though clinical data regarding different antagonists also comprise side effects including increases in liver fat, blood pressure, and low-density lipoprotein (LDL) cholesterol . On the other hand, glucagon agonism has also been suggested as a potential therapeutic strategy for diabetes treatment due to its potential effect on appetite and insulin secretion . Furthermore, co-infusion of glucagon and GLP-1 in overweight humans did not induce hyperglycemia or glucose intolerance but led to a greater decrease in food intake compared to that seen with GLP-1 alone . Additionally, acute glucagon agonism in rodents shows improved glucose tolerance and improved glucose-stimulated insulin secretion after initial glucagon-induced hyperglycemia . These data point to a more complicated interplay between peptide hormones in the regulation of glucose homeostasis than hitherto thought, and several therapeutic strategies involving glucagon signaling are currently being investigated for diabetes treatment. A recent study from our group focused on the pharmacological effects of GRA on postprandial blood glucose in combinatory effect with an insulin receptor antagonist (IRA) . During administration of both GRA and IRA, mice displayed blood glucose responses to a glucose challenge comparable to the vehicle group whereas (and as expected) the blood glucose response to the same challenge was higher after IRA only and lower after GRA only. This further supports the notion that for appropriate glucose homeostasis, both hormones are important .
There is no doubt that glucagon is important for the maintenance of normal glucose levels in humans . However, studies carried out during the last decades suggest that glucagon is also a key regulator of hepatic amino acid metabolism, which, as discussed in the following Section, may be linked with hepatic steatosis/fibrosis and potentially dispose towards the development of type 2 diabetes .
5.2. Glucagon and Amino Acid Metabolism
Glucagon is a powerful stimulus for hepatic amino acid turnover [128,129] by induction of increased activity of enzymes in the urea cycle . Through cAMP-PKA-CREB protein-mediated effects, glucagon regulates several enzymes in the urea cycle at a transcriptional level [129,131]. This is important considering that the capacity of ureagenesis is determined by enzyme activity , and long-term regulation of ureagenesis depends on the synthesis rate of the five necessary enzymes (shown in Figure 4). Glucagon also activates the transcription of system A amino acid transporters present in the hepatocyte membrane, thus allowing increased amino acid uptake and substrate availability for ureagenesis . Thus, because ureagenesis is a substrate-regulated pathway, glucagon further induces ureagenesis by providing substrates via increased amino acid uptake. For short-term regulation of ureagenesis, acute activation of at least one of the five enzymes is necessary. N-acetyl glutamate (NAG) is the obligatory allosteric activator of carbamoyl phosphate synthetase-1 (CPS-1), and NAG formation, by N-acetyl glutamate synthetase (NAGS), is therefore an important factor for short-term regulation of the urea cycle. Steady-state levels of NAG are determined by glutamate and acetyl-CoA concentrations (the substrates for NAG), and activators of NAGS . Glucagon regulates the transcription of NAGS , but the rapid increase in NAG concentration induced by glucagon signaling allows rapid activation of CPS-1 and may thus regulate the urea cycle within minutes. Thus, glucagon may acutely regulate hepatic amino acid metabolism via increased ureagenesis [5,128,135,136,137,138,139], and upon pharmacological blockade of glucagon receptor signaling, using a GRA, plasma concentrations of amino acids increase and ureagenesis decrease [135,137,140]. In accordance with this, inhibition of glucagon signaling also reduces the expression of genes involved in hepatic amino acid uptake and metabolism [121,141,142], resulting in hyperaminoacidemia [100,135,140,143]. The rapid changes in urea production upon glucagon signaling cannot be explained by transcriptional changes, and suggest that glucagon also allosterically activates the urea cycle; however, further studies are needed to clarify such a mechanism that may include acetylation , phosphorylation , sirtuins , and/or increased activity of amino acid transporters [121,147]. A detailed map covering the molecular mechanisms of how glucagon receptor signaling may lead to increased hepatic amino acid metabolism remains to be generated.
Figure 4.
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Glucagon’s effect on hepatic amino acid metabolism and ureagenesis. In the hepatocytes, transaminases catalyze the cleavage of the amino group in the form of ammonia, leaving behind the alpha-keto-acid of the respective amino acid. The amino group is transferred to α-ketoglutarate, yielding glutamate. In addition, glutamine is transported via the circulation to the hepatocytes where glutaminase converts glutamine to glutamate and ammonium, which is deposited in the hepatocyte mitochondria. Alanine aminotransferase transfers the amino group from alanine to α-ketoglutarate forming glutamate and leaving pyruvate behind. Glutamate enters the mitochondria of the hepatocyte where glutamate dehydrogenase catalyzes the cleavage of the amino group yielding ammonia and α-ketoglutarate. The ammonium enters the urea cycle by conversion to carbamoyl phosphate, catalyzed by carbamoyl phosphate synthase-1 (CPS-1). Aspartate aminotransferase catalyzes the formation of aspartate via transamination of oxaloacetate and glutamate. Aspartate thus functions as the second nitrogen donor in the urea cycle. The urea cycle starts with N-acetyl-L-glutamate (NAG) formation, which is catalyzed by N-acetyl glutamate synthase (NAGS). NAG activates CPS-1, which catalyzes the formation of ammonium to carbamoylphosphate. To form citrulline, the carbamoyl group is transferred from carbamoyl phosphate to ornithine, which is catalyzed by ornithine transcarbamoylase. The next step requires argininosuccinate synthase and argininosuccinase, which convert citrulline and aspartate to arginosuccinate and subsequently to arginine and fumarate. Urea is then produced when arginase cleaves arginine. Glucagon receptor signaling increases urea cycle activity by activating the cAMP-protein kinase A (PKA)-cAMP response element-binding (CREB) protein-pathway, resulting in transcription of urea cycle enzymes and amino acid transporters, the latter serving to increase substrate availability. Glucagon may also allosterically activate the urea cycle by additional PKA mediated phosphorylations (see text).
Secretion of glucagon is powerfully and rapidly stimulated by protein-containing meals [94,148]. Thus, glucagon may serve to regulate postprandial amino acid metabolism in accordance with its acute effect on amino acid metabolism. Postprandial insulin secretion during intake of protein-rich meals also needs to be taken into consideration. Conventionally, the secretion of insulin and glucagon is viewed as being inversely regulated, which is consistent with the opposite effects of the hormones on glucose metabolism. This is, however, not the case upon protein ingestion, as the secretion of both insulin and glucagon increase in response to amino acids [114,116]. To exemplify this further, consider two situations: If only carbohydrates are ingested, plasma concentrations of glucagon will drop in healthy individuals, reaching values close to zero (or to the detection limits of the glucagon assay applied), whereas protein-rich meals are associated with marked increases in glucagon secretion. In the latter situation, the glucagon response is thought to compensate for the potential hypoglycemia induced by an amino acid-driven insulin secretion. However, both mechanisms may also serve to limit amino acid excursions; insulin, by increasing tissue uptake and deposition of amino acids in newly synthesized proteins, and glucagon, by enhancing amino acid metabolism [116,149] (whereas the glucagon receptor is not expressed on human skeletal muscle cells, precluding direct actions of glucagon). In the aforementioned study using both GRA and IRA administration in mice during intravenous administration of amino acids, only glucagon antagonism increased plasma amino acid concentrations compared to vehicle-treated controls, whereas insulin antagonism did not . These findings indicate that clearance of amino acids during the postprandial state depends relatively more on glucagon receptor signaling than on insulin. Whether this also holds true in humans, we do not yet know.
Studies involving genetic or pharmacological ablation of glucagon receptor signaling in rodents are consistently associated with hyperglucagonemia, hyperaminoacidemia, and alpha cell hyperplasia [121,135,150]. In fact, both global and liver-specific elimination of glucagon receptor signaling in mice induces alpha cell hyperplasia. It is especially interesting that hepatic elimination of the glucagon receptor is sufficient to induce an increased alpha cell mass , pointing to a direct or indirect signaling loop between the liver and the pancreas. As previously stated, glucagon regulates amino acid metabolism via increased ureagenesis, and when glucagon signaling is inhibited, hyperaminoacidemia occurs. Hyperaminoacidemia is therefore suggested as the factor leading to alpha cell hyperplasia [121,135,141]. This endocrine feedback loop, in which glucagon induces hepatic amino acid metabolism and amino acids in turn stimulate glucagon secretion, has been named the liver–alpha cell axis [7,18]. The glucagonotropic amino acids, alanine and glutamine, may be of particular importance in this feedback loop in mediating alpha cell secretion and growth, respectively . Glucagon receptor blockade also increases the expression of several amino acid transporters in the plasma membrane of alpha cells . Upon pharmacological inhibition of glucagon receptor signaling the pancreatic expression of the amino acid transporter, Slc38a5, increased, and in mice deficient of this amino acid transporter, the development of alpha cell hyperplasia was prevented . These findings did, however, not translate into human islets implanted into mice despite an observed alpha cell proliferation following pharmacological glucagon receptor inhibition . Rather, glucagon receptor inhibition in mice with implanted human islets elicited an increased pancreatic expression of the amino acid transporter Slc38a4 . When investigating molecular mechanisms responsible for increasing alpha cell mass in both rodents and implanted human islets after glucagon receptor blockade, the mechanistic target of the rapamycin (mTOR) pathway has been suggested to be involved [121,151]. However, this is a relatively broad signaling cascade involved in multiple cellular functions including mitosis, and therefore it is of interest to search for potential, more specific downstream targets of amino acid induced alpha cell proliferation.
The aforementioned observations of hyperaminoacidemia, alpha cell hyperplasia, and hyperglucagonemia mimic the manifestations of some metabolic diseases. Both alpha and beta cell proliferation may be increased in islets from patients with recent-onset type 1 diabetes . Furthermore, some, but not all, subjects with type 2 diabetes show elevated glucagon concentrations in the fasting state and sometimes fail to suppress postprandial glucagon secretion [124,153,154]. The idea that not all type 2 diabetes patients display hyperglucagonemia was further nourished when type 2 diabetic patients were compared to patients with non-alcoholic fatty liver disease (NAFLD) , showing that NAFLD rather than diabetes was associated with hyperglucagonemia and hyperaminoacidemia .
That glucagon is a powerful regulator of amino acid metabolism further becomes evident in cases of extreme glucagon excess and deficiency . Extreme hyperglucagonemia, as observed in patients with glucagon-producing tumors (glucagonomas), is associated with accelerated hepatic amino acid turnover and ureagenesis [90,157,158], resulting in severe hypoaminoacidemia [41,159,160,161] and increased plasma concentrations of urea [143,162]. In studies of the plasma metabolome and proteome , the major changes associated with glucagonomas are low levels of plasma amino acids. Glucagonomas are not, as a rule, linked to disturbed glucose metabolism; instead, patients often develop necrolytic migratory erythema, a skin disease associated with severe hypoaminoacidemia , which resolves upon intravenous infusions of amino acid [160,166]. Patients with inactivating glucagon receptor mutations (pseudohypoglucagonemia) show hyperaminoacidemia and hyperglucagonemia [167,168] and do not develop severe hypoglycemia. Likewise, pancreatectomized humans show hyperaminoacidemia, which can be normalized by glucagon administration . Collectively, this suggests that one dominant effect of glucagon signaling is both acute (normally postprandial) and chronic regulation of amino acid metabolism; the two may be differentially regulated, and the involved mechanisms, as discussed above, need further exploration.
5.3. Glucagon and Lipid Metabolism
The glucagon receptor has, as discussed above, been considered a target for glucose-lowering therapy in type 2 diabetic patients. However, in a phase 2 clinical trial of GRAs (e.g., LY2409021), clinical concerns were raised. Despite the ability of a GRA to effectively lower blood glucose, blocking the glucagon receptor resulted in negative effects on lipid-related processes—the GRA-treated patients had higher total cholesterol, increased hepatic fat fraction (measured by magnetic resonance imaging), and weight gain as compared to controls . Therefore, questions have been raised regarding the relationship between glucagon signaling and lipid metabolism (Figure 5).
Figure 5.
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Glucagon effect on lipid metabolism. Activation of the glucagon receptor results in adenylate cyclase-mediated cAMP formation. cAMP accumulation activates cAMP-responsible-binding-protein (CREB), inducing transcription of carnitine acyl transferase-1 (CPT-1) and other genes required for beta-oxidation. CPT-1 promotes conversion of fatty acids to acylcarnitines, which are transported into the mitochondria and broken down to acetate. Acetate and CoA react to form acetyl-CoA, which enters the citric acid cycle. cAMP accumulation activates protein kinase A (PKA), which leads to inactivation of acetyl-CoA carboxylase and thus suppression of malonyl-CoA formation and a disinhibition of beta-oxidation. Thus, glucagon promotes increased beta-oxidation and a decreased fatty acid synthesis and, in turn, very-low-density lipoprotein (VLDL) release.
Glucagon is known to act mainly on the hepatocytes, which is in line with the fact that the expression levels of glucagon receptors are highest in the liver. Adipocyte expression of the glucagon receptor is possible, but the expression seems to be restricted to rats rather than humans. Therefore, glucagon may primarily regulate lipid metabolism via hepatic signaling. This may involve the following mechanisms. Firstly, when glucagon binds to its receptor on the hepatocyte, it activates cAMP, which in turn accumulates and activates CREB protein. As a consequence, transcription of carnitine acyl transferase (CPT-1) is increased, which enables conversion of fatty acids to acylcarnitines whereby beta-oxidation is activated , resulting in the formation of acetate [170,171]. Acetate and CoA react to yield acetyl-coA, which in turn interacts with oxaloacetate to form citrate (thereby inhibiting glycolysis), which enters the citric acid cycle . As a result, hepatic glucagon signaling increases fatty acid catabolism, inhibits glycolysis, and stimulates the citric acid cycle. Secondly, when glucagon binds to its receptor on the hepatocyte, it induces PKA-dependent phosphorylation, which leads to inactivation of acetyl-CoA carboxylase, which functions to catalyze the formation of malonyl-CoA. Because malonyl-CoA is responsible for inhibiting CPT-1, and thus beta-oxidation, lowering malonyl-CoA levels results in a diversion of FFAs to beta-oxidation rather than re-esterification to triglycerols (TGs). Typically, when FFAs are re-esterified, they are stored as TGs and released from hepatocytes into circulation as very-low-density lipoprotein (VLDL). Taken together, glucagon decreases de novo fatty acid synthesis and, consequentially, VLDL release. Lastly, it is possible that glucagon signaling increases the AMP/ATP ratio necessary to activate AMP-activated kinase . This leads to transcriptional activation of peroxisome proliferator-activated receptor-α , which induces transcription of beta-oxidation related genes such as CPT-1 and acetyl-CoA oxidase [174,175].
There is additionally in vivo support of the role of glucagon in hepatic lipid metabolism . When mice were administered 30 μg/kg of glucagon acutely, this resulted in decreased FFA and TG plasma concentrations as well as reduced hepatic TG content and secretion. Furthermore, an injection of 10 μg of glucagon every eighth hour for 21 days resulted in a 70% and 38% decrease in plasma concentrations of TG and phospholipids, respectively . In further support of this, synthesis and release of TG were inhibited by glucagon in cultured hepatocytes and in isolated hepatocytes [178,179]. Glucagon also decreased hepatic VLDL synthesis in rats . In humans, hyperglucagonemia (during a pancreatic clamp) resulted in reduced hepatic lipoprotein particle turnover , and there has also been indications of a glucagon-mediated increase in hepatic beta-oxidation .
In addition to the clinical trials, which demonstrated negative effects on lipid metabolism by GRA administration in humans, similar results have been observed in rodent in vivo studies. In both rats and diabetic (db/db) mice , treatment with a glucagon antisense oligonucleotide (impairing glucagon signaling) resulted in increased amounts of hepatic fat. In the livers of global glucagon receptor knockout (Gcgr −/−) mice, decreased gluconeogenesis and citric acid cycle activity, as well as increased glycolysis, was observed, potentially leading to increased hepatic lipogenesis due to decreased acetyl-CoA oxidation and accumulation. This is supported by reports of upregulation of genes involved in lipogenesis [93,185] and downregulation of enzymes involved in beta-oxidation (e.g., CPT-1) . With regards to the development of steatosis in Gcgr −/− mice, there are conflicting results [93,186]. When administered a glucagon/GLP-1 receptor co-agonist, type 2 diabetic and obese rodents displayed reduced hepatic steatosis, increased hormone-sensitive lipase (HSL) activity in adipocytes, and improved dyslipidemia [187,188,189,190,191,192,193]. The inhibitory effect on hepatic lipogenesis and the stimulatory effect on beta-oxidation would seem to be mediated particularly by glucagon receptor signaling rather than GLP-1-mediated effects [187,188]. As a result, a number of clinical studies are evaluating the potential for glucagon/GLP-1 receptor co-agonists for obesity and type 2 diabetes treatment .
One final aspect regarding the relationship between glucagon signaling and lipid metabolism is the effect in adipocytes. In adipocytes, lipolysis depends on PKA-dependent phosphorylation of HSL [194,195,196] and perilipins, which are present on the surface of lipid droplets . Upon phosphorylation of perilipins, the protein CGI-58 dissociates, which in turn activates adipose triglycerol lipase, which is responsible for converting TGs to diaglycerols. Additionally, phosphorylated perilipins bind HSL and allow it to access the area of the lipid droplet where diacylglycerols are converted to monoglycerols. Monoglycerol lipase hydrolyzes the monoglycerols, resulting in FFAs [198,199,200,201,202] and glycerol, which are released into the circulation. Circulating FFAs and glycerol levels are therefore sensitive indicators of the rate of lipolysis . In rat adipocytes, glucagon receptor mRNA has been detected [204,205], supporting an effect of glucagon on HSL [206,207] and subsequently lipolysis in rat adipocytes [208,209,210,211,212,213,214]. Despite this evidence in rats, there is no clear evidence of expression of a glucagon receptor on human adipocytes, and a lipolytic effect of glucagon at physiological plasma concentrations has been difficult to observe in humans [181,215,216,217,218,219,220,221]. Lipolytic responses to supra-physiological glucagon concentrations have been reported in humans (some resulting in concentrations upwards of >1000 pM) [222,223]. However, this is possibly a result of glucagon-stimulated secretion of catecholamines . The lipolytic effect of supra-physiological glucagon concentration was abolished by insulin in some studies [222,223,225,226,227], and this is consistent with the powerful anti-lipolytic effects of insulin, also found in humans [208,209,225,228,229]. Therefore, if any such lipolytic effect of glucagon on human adipocytes exists, it is only of physiological interest during low insulin secretion [220,228,230].
6. Glucagon Resistance and Potential Biomarkers
Does glucagon resistance exist and is it measurable? We hypothesize that glucagon resistance is a molecular phenomenon characterizing impaired physiological effects of glucagon on glucose, amino acid and/or lipid metabolism, which for each of the three listed systems may occur independently. It is important to stress that the hypothesis is new and rests on relatively few observations with limited number of participants. That said, we believe it is important to discuss glucagon resistance as it may aid our understanding of glucagon biology and its contribution to metabolic diseases and potentially allow us to understand the current challenges with glucagon agonism/antagonism by designing targeted therapies like, for example., glucagon-induced hepatic glucose production without tampering with its effect on amino acid and lipid metabolism.
Obesity (and malnutrition) has been suggested as a potential inducer of glucagon resistance in animal models [106,107,231,232], but to what extent this is recapitulated in humans is not yet clear. One study in humans with biopsy-verified hepatic steatosis was presented in 2018 at the American Diabetes Association Scientific Session, reporting that fat infiltration may cause glucagon resistance regarding amino acid metabolism whereas the acute effect of glucagon on hepatic glucose production was preserved . This observation would be in agreement with findings of an attenuated capacity for ureagenesis in individuals with hepatic steatosis [233,234]. One explanation may be that impaired liver function/structure directly affects the transcription of enzymes related to ureagenesis such as CPS-1 or that the effect of glucagon receptor signaling on the same components is impaired. In support of the latter, glucagon receptor knockout (be it liver specific or not) alters the hepatic transcriptome including lowered levels of ureagenesis-related transcripts [7,142,235]. A key regulator of gluconeogenesis is substrate availability, as discussed earlier, and not primarily hormonal systems (insulin and glucagon), whereas ureagenesis may be more dependent on hormonal regulation (but obviously still dependent on a certain flux of substrates) [5,128,138,140]. It will therefore be important to dissect the molecular signals that separate gluconeogenesis from ureagenesis and how these are affected by glucagon receptor signaling. Protein kinase c has been suggested as a potential mediator of glucagon receptor desensitization (measured as impaired production of cAMP, the second-messenger of glucagon receptor activation) . Glucagon receptor internalization [237,238,239] may be a potential mechanism driving glucagon resistance, although others have found glucagon receptor internalization to be minimal compared to, e.g., insulin receptor internalization. It has also been suggested that lipids may interfere with glucagon receptor signaling [81,240] (a case for glucagon resistance was made using adipocytes, relevant for rats ). We believe that an important step is to differentiate between chronic (changing protein levels of, e.g., CSP-1) and acute effects of glucagon (e.g., phosphorylation of NAG)—both requires studies examining unbiased changes in the proteome and post-translational modifications, respectively. Further evidence is clearly required to pinpoint impaired glucagon receptor signaling.
How can glucagon resistance be measured and thereby further investigated? In analogy with hepatic insulin resistance, it would be logical to look at effector (insulin/C-peptide) and outcome (glucose) concentrations in plasma. Here, the corresponding parameters would be plasma concentrations of glucagon and amino acids. As discussed above, insulin concentrations might also be considered.
Both measures are sensitive to the accuracy of the measurement methodologies applied (e.g., cross-reactivity to other proglucagon-derived peptides may create false-positive hyperglucagonemia). In addition glucagon levels may be dependent on intact renal function (hyperglucagonemia in patients with renal disease may be composed of not only 1–29 glucagon but also considerable amount of 1–61 ). The glucagon/insulin ratio may, on the other hand, be affected by insulin resistance (be it hepatic or peripheral), resulting in increased beta cell secretion (that eventually will fall in line with beta cell failure) and may be invalid in the cases exogenous insulin (late stage type 2 diabetes or type 1 diabetes). That said, to evaluate glucagon’s impact on, for example, glucose production in individuals with and without hepatic steatosis, the manipulation of the glucagon/insulin ratio by somatostatin-induced pancreas clamps and continuous infusion of insulin and glucagon may be a tool to investigate glucagon resistance . Therefore, we speculate that plasma concentrations of glucagon, preferably measured by accurate sandwich ELISA techniques , are the golden standard to assess existence of hyperglucagonemia and hence potentially also glucagon resistance. Two recent studies from our group have found support of a glucagon-alanine index that captures the lack of glucagon-induced ureagenesis [155,156], which seems to be specifically impaired in patients with hepatic steatosis and fibrosis [233,234]. Of note, alanine is also involved in the alanine-glucose shuttle between the skeletal muscle and the liver, enabling substrate flow to hepatic gluconeogenesis, and given that glucagon receptor signaling is lacking in skeletal muscle cells (as they do not express the glucagon receptor), the proposed index may therefore reflect the liver–alpha cell axis rather than a muscle–liver–alpha cell axis. However, this will need independent validation by other research groups and larger cohorts (although the index was formed on the basis of results from >1400 individuals).
7. Outlook and Conclusions
In this review, we have discussed the essential aspects of glucagon biology with an emphasis on glucagon resistance as a new, potential physiological concept. We provide an overview of the three major biological areas of glucagon receptor signaling: glucose homeostasis, amino acid metabolism, and lipid metabolism. The two latter areas are not as well characterized as the former, and future mechanistic studies involving glucagon agonism/antagonism may be helpful to delineate the physiological importance of glucagon in these areas. Key physiological questions remain unanswered: Are all amino acids affected by glucagon receptor signaling or is it only those with glucagonotropic effects? Does glucagon have a physiological effect on lipid metabolism in humans? The concept of glucagon resistance must be further investigated before conclusions about its relevance in metabolic diseases can be reached. Nevertheless, further studies of the concept will be useful and important for our understanding of glucagon pathobiology.
Research during the coming years may provide answers to whether or not glucagon agonism/antagonism will reach clinical application for treatment of metabolic diseases such as obesity and type 2 diabetes . Interestingly, glucagon agonism (low-dose glucagon) is likely to be applied together with insulin in the dual pump systems development for type 1 diabetes therapy . Furthermore, as glucagon receptor antagonism may result in dyslipidemia, hyperaminoacidemia and hepatic lipid deposition, it has been suggested that glucagon agonism may be of therapeutic relevance for treatment of type 2 diabetes . Future studies will tell whether this holds true, but it seems evident that targeting glucagon receptor signaling alone may be inappropriate because of possible worsening of glycemic control), but perhaps either biased glucagon receptor signaling or glucagon agonism in combination with another therapeutic agent (e.g., GLP-1) may solve this problem.
Thus, glucagon biology remains a challenging but an exciting research object .
Acknowledgments
We appreciate the discussions on this topic with Liselotte Gluud and Kirstine Bojsen-Møller (Hvidovre Hospital, Hvidovre, Denmark).
Abbreviations
CPS-1 carbamyl phosphate synthetase-1
CPT-1 carnitine acyl transferase-1
CREB cAMP response element-binding
FBPase2 Fructose 2,6-bisphosphatase
FFA Free fatty acid
Gcgr −/− Glucagon receptor knockout
GLP-1 Glucagon-like peptide-1
GLP-2 Glucagon-like peptide-2
GIP Glucose-dependent insulinotropic polypeptide
GRA Glucagon receptor antagonist
HCD High carbohydrate diet
HSL Hormone sensitive lipase
IRA Insulin receptor antagonist
LCD Low carbohydrate diet
LDL Low density lipoprotein
NAFLD Non alcoholic fatty liver disease
NAG N-acetyl glutamate
NAGS N-acetyl glutamate synthetase
PEPCK Phosphoenolpyruvatecarboxykinase
PFK-2 Phospho-fructokinase 2
PKA Protein kinase A
TG Triglyceride
VLDL Very-low density lipoprotein
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Funding
NNF Tandem Programme (31526), NNF Project support in Endocrinology and Metabolism—Nordic Region (34250), and Excellence Emerging Investigator Grant—Endocrinology and Metabolism (NNF19OC0055001).
Conflicts of Interest
The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript; or in the decision to publish the results.
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Abstract
1. Introduction
2. Processing of Proglucagon
3. Secretion of Glucagon
4. Accurate Measurement of Glucagon
5. Glucagon Receptor Signaling
6. Glucagon Resistance and Potential Biomarkers
7. Outlook and Conclusions
Acknowledgments
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Ovarian Cancer Stages
After a woman is diagnosed with ovarian cancer, doctors will try to figure out if it has spread, and if so, how far. This process is called staging. The stage of a cancer describes how much cancer is in the body. It helps determine how serious the cancer is and how best to treat it. Doctors also use a cancer's stage when talking about survival statistics.
On this page
[show] [hide]
How is the stage determined?
Clinical vs. pathologic stage for ovarian cancer
Stages of ovarian cancer
How is the stage determined?
One of the goals of surgery for ovarian cancer is to take tissue samples for diagnosis and staging. To stage the cancer, samples of tissues are taken from different parts of the pelvis and abdomen and examined in the lab.
The 2 systems used for staging ovarian cancer, the FIGO (International Federation of Gynecology and Obstetrics) system and the AJCC (American Joint Committee on Cancer) TNM staging system, are basically the same.
They both use 3 factors to stage (classify) this cancer:
The extent (size) of the tumor (T): Has the cancer spread outside the ovary or fallopian tube? Has the cancer reached nearby pelvic organs like the uterus or bladder?
The spread to nearby lymph nodes (N): Has the cancer spread to the lymph nodes in the pelvis or the para-aortic lymph nodes (the lymph nodes around the aorta, the main artery that runs from the heart down along the back of the abdomen and pelvis)?
The spread (metastasis) to distant sites (M): Has the cancer spread to fluid around the lungs (malignant pleural effusion) or to distant organs such as the liver or bones?
Numbers or letters after T, N, and M provide more details about each of these factors. Higher numbers mean the cancer is more advanced. Once a person’s T, N, and M categories have been determined, this information is combined in a process called stage grouping to assign an overall stage.
Clinical vs. pathologic stage for ovarian cancer
The stage groupings described below use the pathologic stage (also called the surgical stage). It is determined by examining tissue removed during surgery.
Sometimes, if surgery is not possible right away, the cancer will be given a clinical stage instead. This is based on the results of a physical exam, biopsy, and imaging tests done before surgery. For more information see Cancer Staging.
Stages of ovarian cancer
Ovarian cancer stages range from stage I (1) through IV (4). As a rule, the lower the number, the less the cancer has spread. A higher number, such as stage IV, means cancer has spread more. Although each person’s cancer experience is unique, cancers with similar stages tend to have a similar outlook and are often treated in much the same way.
The system described below is the most recent AJCC system effective January 2018. It is the staging system for ovarian cancer, fallopian tube cancer, and primary peritoneal cancer.
Cancer staging can be complex, so ask your doctor to explain it to you in a way you understand.
Stage I (stage grouping: T1, N0, M0)
The cancer is only in the ovary (or ovaries) or fallopian tube(s) (T1). It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage IA (T1a, N0 M0)
The cancer is in one ovary, and the tumor is confined to the inside of the ovary; or the cancer is in one fallopian tube and is only inside the fallopian tube. There is no cancer on the outer surfaces of the ovary or fallopian tube. No cancer cells are found in the fluid (ascites) or washings from the abdomen and pelvis (T1a). It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage IB (T1b, N0, M0)
The cancer is in both ovaries or fallopian tubes but not on their outer surfaces. No cancer cells are found in the fluid (ascites) or washings from the abdomen and pelvis (T1b). It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage IC (T1c, N0, M0)
The cancer is in one or both ovaries or fallopian tubes and any of the following are present:
The tissue (capsule) surrounding the tumor broke during surgery, which could allow cancer cells to leak into the abdomen and pelvis (called surgical spill). This is stage IC1.
Cancer is on the outer surface of at least one of the ovaries or fallopian tubes or the capsule (tissue surrounding the tumor) has ruptured (burst) before surgery (which could allow cancer cells to spill into the abdomen and pelvis). This is stage IC2.
Cancer cells are found in the fluid (ascites) or washings from the abdomen and pelvis. This is stage IC3.
It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage II (stage grouping: T2, N0, M0)
The cancer is in one or both ovaries or fallopian tubes and has spread to other organs (such as the uterus, bladder, the sigmoid colon, or the rectum) within the pelvis or there is primary peritoneal cancer (T2). It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage IIA (T2a, N0, M0)
The cancer has spread to or has invaded (grown into) the uterus or the fallopian tubes, or the ovaries. (T2a). It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage IIB (T2b, N0, M0)
The cancer is on the outer surface of or has grown into other nearby pelvic organs such as the bladder, the sigmoid colon, or the rectum (T2b). It has not spread to nearby lymph nodes (N0) or to distant sites (M0).
Stage IIIA1 (stage grouping: T1 or t2, N1, M0)
The cancer is in one or both ovaries or fallopian tubes, or there is primary peritoneal cancer (T1) and it may have spread or grown into nearby organs in the pelvis (T2). It has spread to the retroperitoneal (pelvic and/or para-aortic) lymph nodes only. It has not spread to distant sites (M0).
Stage IIIB (T3b, N0 or N1, M0)
There is cancer in one or both ovaries or fallopian tubes, or there is primary peritoneal cancer and it has spread or grown into organs outside the pelvis. The deposits of cancer are large enough for the surgeon to see but are no bigger than 2 cm (about 3/4 inch) across. (T3b).
It may or may not have spread to the retroperitoneal lymph nodes (N0 or N1), but it has not spread to the inside of the liver or spleen or to distant sites (M0).
Stage IIIC (T3c, N0 or N1, M0)
The cancer is in one or both ovaries or fallopian tubes, or there is primary peritoneal cancer and it has spread or grown into organs outside the pelvis. The deposits of cancer are larger than 2 cm (about 3/4 inch) across and may be on the outside (the capsule) of the liver or spleen (T3c).
It may or may not have spread to the retroperitoneal lymph nodes (N0 or N1), but it has not spread to the inside of the liver or spleen or to distant sites (M0).
Stage IVA (stage grouping: any T, any N, M1a)
Cancer cells are found in the fluid around the lungs (called a malignant pleural effusion) with no other areas of cancer spread such as the liver, spleen, intestine, or lymph nodes outside the abdomen (M1a).
Stage IVB (any T, any N, M1b)
The cancer has spread to the inside of the spleen or liver, to lymph nodes other than the retroperitoneal lymph nodes, and/or to other organs or tissues outside the peritoneal cavity such as the lungs and bones (M1b).
Other staging categories you may see
TX: Main tumor cannot be assessed due to lack of information.
T0: No evidence of a primary tumor.
NX: Regional lymph nodes cannot be assessed due to lack of information.
Written by
References
Developed by the American Cancer Society medical and editorial content team with medical review and contribution by the American Society of Clinical Oncology (ASCO).
American Joint Committee on Cancer. Ovary, Fallopian Tube, and Primary Peritoneal carcinoma. In: AJCC Cancer Staging Manual. 8th ed. New York, NY: Springer; 2017:681-690.
Prat J; FIGO Committee on Gynecologic Oncology. Staging classification for cancer of the ovary, fallopian tube, and peritoneum. Int J Gynecol Obstet. 2014;124(1):1-5.
Last Revised: August 8, 2025
American Cancer Society medical information is copyrighted material. For reprint requests, please see our Content Usage Policy.
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12834 | https://artofproblemsolving.com/wiki/index.php/Trigonometric_identities?srsltid=AfmBOorWjaFptrp-kFuXi7Ys9fhZL9asRkKgIWKQ0AOPWXcwOyy3uTBD | Art of Problem Solving
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Trigonometric identities
In trigonometry, trigonometric identities are equations involving trigonometric functions that are true for all input values. Trigonometric functions have an abundance of identities, of which only the most widely used are included in this article.
Contents
[hide]
1 Pythagorean identities
2 Angle addition identities
3 Double-angle identities
3.1 Cosine double-angle identity
4 Half-angle identities
5 Product-to-sum identities
6 Sum-to-product identities
7 Other identities
7.1 Triple-angle identities
7.2 Even-odd identities
7.3 Conversion identities
7.4 Euler's identity
7.5 Miscellaneous
8 Resources
9 See also
Pythagorean identities
The Pythagorean identities state that
Using the unit circle definition of trigonometry, because the point is defined to be on the unit circle, it is a distance one away from the origin. Then by the distance formula, . To derive the other two Pythagorean identities, divide by either or and substitute the respective trigonometry in place of the ratios to obtain the desired result.
Angle addition identities
The trigonometric angle addition identities state the following identities:
There are many proofs of these identities. For the sake of brevity, we list only one here.
Euler's identity states that . We have that By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas.
To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by , and simplify. as desired.
Double-angle identities
The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting . Doing so yields:
Cosine double-angle identity
Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle identity given above.
In addition, the following identities are useful in integration and in deriving the half-angle identities. They are a simple rearrangement of the two above.
Half-angle identities
The trigonometric half-angle identities state the following equalities:
The plus or minus does not mean that there are two answers, but that the sign of the expression depends on the quadrant in which the angle resides.
Consider the two expressions listed in the cosine double-angle section for and , and substitute instead of . Taking the square root then yields the desired half-angle identities for sine and cosine. As for the tangent identity, divide the sine and cosine half-angle identities.
Product-to-sum identities
The product-to-sum identities are as follows:
They can be derived by expanding out and or and , then combining them to isolate each term.
Sum-to-product identities
Substituting and into the product-to-sum identities yields the sum-to-product identities.
Other identities
Here are some identities that are less significant than those above, but still useful.
Triple-angle identities
All of these expansions can be proved using trick and perform the angle addition identities. Same for and for .
Even-odd identities
The functions , , , and are odd, while and are even. In other words, the six trigonometric functions satisfy the following equalities:
These are derived by the unit circle definitions of trigonometry. Making any angle negative is the same as reflecting it across the x-axis. This keeps its x-coordinate the same, but makes its y-coordinate negative. Thus, and .
Conversion identities
The following identities are useful when converting trigonometric functions.
All of these can be proven via the angle addition identities.
Euler's identity
Euler's identity is a formula in complex analysis that connects complex exponentiation with trigonometry. It states that for any real number , where is Euler's constant and is the imaginary unit. Euler's identity is fundamental to the study of complex numbers and is widely considered among the most beautiful formulas in math.
Similar to the derivation of the product-to-sum identities, we can isolate sine and cosine by comparing and , which yields the following identities:
They can also be derived by computing and . These expressions are occasionally used to define the trigonometric functions.
Miscellaneous
These are the identities that are not substantial enough to warrant a section of their own.
Resources
Table of trigonometric identities
List of Trigonometric Identities
See also
Trigonometry
Trigonometric substitution
Proofs of trig identities
Retrieved from "
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Trigonometry
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Solution - Cumulative probability in the standard normal distribution
Other Ways to Solve
Step-by-step explanation
1. Find the cumulative probability of the z-scores values up to 1.64
Use the positive z-table to find the value corresponding to 1.64. This value is the cumulative probability of the area to the left of 1.64.
| | | | | | | | | | | |
--- --- --- --- ---
| Z | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| 0.0 | 0.5 | 0.50399 | 0.50798 | 0.51197 | 0.51595 | 0.51994 | 0.52392 | 0.5279 | 0.53188 | 0.53586 |
| 0.1 | 0.53983 | 0.5438 | 0.54776 | 0.55172 | 0.55567 | 0.55962 | 0.56356 | 0.56749 | 0.57142 | 0.57535 |
| 0.2 | 0.57926 | 0.58317 | 0.58706 | 0.59095 | 0.59483 | 0.59871 | 0.60257 | 0.60642 | 0.61026 | 0.61409 |
| 0.3 | 0.61791 | 0.62172 | 0.62552 | 0.6293 | 0.63307 | 0.63683 | 0.64058 | 0.64431 | 0.64803 | 0.65173 |
| 0.4 | 0.65542 | 0.6591 | 0.66276 | 0.6664 | 0.67003 | 0.67364 | 0.67724 | 0.68082 | 0.68439 | 0.68793 |
| 0.5 | 0.69146 | 0.69497 | 0.69847 | 0.70194 | 0.7054 | 0.70884 | 0.71226 | 0.71566 | 0.71904 | 0.7224 |
| 0.6 | 0.72575 | 0.72907 | 0.73237 | 0.73565 | 0.73891 | 0.74215 | 0.74537 | 0.74857 | 0.75175 | 0.7549 |
| 0.7 | 0.75804 | 0.76115 | 0.76424 | 0.7673 | 0.77035 | 0.77337 | 0.77637 | 0.77935 | 0.7823 | 0.78524 |
| 0.8 | 0.78814 | 0.79103 | 0.79389 | 0.79673 | 0.79955 | 0.80234 | 0.80511 | 0.80785 | 0.81057 | 0.81327 |
| 0.9 | 0.81594 | 0.81859 | 0.82121 | 0.82381 | 0.82639 | 0.82894 | 0.83147 | 0.83398 | 0.83646 | 0.83891 |
| 1.0 | 0.84134 | 0.84375 | 0.84614 | 0.84849 | 0.85083 | 0.85314 | 0.85543 | 0.85769 | 0.85993 | 0.86214 |
| 1.1 | 0.86433 | 0.8665 | 0.86864 | 0.87076 | 0.87286 | 0.87493 | 0.87698 | 0.879 | 0.881 | 0.88298 |
| 1.2 | 0.88493 | 0.88686 | 0.88877 | 0.89065 | 0.89251 | 0.89435 | 0.89617 | 0.89796 | 0.89973 | 0.90147 |
| 1.3 | 0.9032 | 0.9049 | 0.90658 | 0.90824 | 0.90988 | 0.91149 | 0.91308 | 0.91466 | 0.91621 | 0.91774 |
| 1.4 | 0.91924 | 0.92073 | 0.9222 | 0.92364 | 0.92507 | 0.92647 | 0.92785 | 0.92922 | 0.93056 | 0.93189 |
| 1.5 | 0.93319 | 0.93448 | 0.93574 | 0.93699 | 0.93822 | 0.93943 | 0.94062 | 0.94179 | 0.94295 | 0.94408 |
| 1.6 | 0.9452 | 0.9463 | 0.94738 | 0.94845 | 0.9495 | 0.95053 | 0.95154 | 0.95254 | 0.95352 | 0.95449 |
| 1.7 | 0.95543 | 0.95637 | 0.95728 | 0.95818 | 0.95907 | 0.95994 | 0.9608 | 0.96164 | 0.96246 | 0.96327 |
| 1.8 | 0.96407 | 0.96485 | 0.96562 | 0.96638 | 0.96712 | 0.96784 | 0.96856 | 0.96926 | 0.96995 | 0.97062 |
| 1.9 | 0.97128 | 0.97193 | 0.97257 | 0.9732 | 0.97381 | 0.97441 | 0.975 | 0.97558 | 0.97615 | 0.9767 |
| 2.0 | 0.97725 | 0.97778 | 0.97831 | 0.97882 | 0.97932 | 0.97982 | 0.9803 | 0.98077 | 0.98124 | 0.98169 |
| 2.1 | 0.98214 | 0.98257 | 0.983 | 0.98341 | 0.98382 | 0.98422 | 0.98461 | 0.985 | 0.98537 | 0.98574 |
| 2.2 | 0.9861 | 0.98645 | 0.98679 | 0.98713 | 0.98745 | 0.98778 | 0.98809 | 0.9884 | 0.9887 | 0.98899 |
| 2.3 | 0.98928 | 0.98956 | 0.98983 | 0.9901 | 0.99036 | 0.99061 | 0.99086 | 0.99111 | 0.99134 | 0.99158 |
| 2.4 | 0.9918 | 0.99202 | 0.99224 | 0.99245 | 0.99266 | 0.99286 | 0.99305 | 0.99324 | 0.99343 | 0.99361 |
| 2.5 | 0.99379 | 0.99396 | 0.99413 | 0.9943 | 0.99446 | 0.99461 | 0.99477 | 0.99492 | 0.99506 | 0.9952 |
| 2.6 | 0.99534 | 0.99547 | 0.9956 | 0.99573 | 0.99585 | 0.99598 | 0.99609 | 0.99621 | 0.99632 | 0.99643 |
| 2.7 | 0.99653 | 0.99664 | 0.99674 | 0.99683 | 0.99693 | 0.99702 | 0.99711 | 0.9972 | 0.99728 | 0.99736 |
| 2.8 | 0.99744 | 0.99752 | 0.9976 | 0.99767 | 0.99774 | 0.99781 | 0.99788 | 0.99795 | 0.99801 | 0.99807 |
| 2.9 | 0.99813 | 0.99819 | 0.99825 | 0.99831 | 0.99836 | 0.99841 | 0.99846 | 0.99851 | 0.99856 | 0.99861 |
| 3.0 | 0.99865 | 0.99869 | 0.99874 | 0.99878 | 0.99882 | 0.99886 | 0.99889 | 0.99893 | 0.99896 | 0.999 |
| 3.1 | 0.99903 | 0.99906 | 0.9991 | 0.99913 | 0.99916 | 0.99918 | 0.99921 | 0.99924 | 0.99926 | 0.99929 |
| 3.2 | 0.99931 | 0.99934 | 0.99936 | 0.99938 | 0.9994 | 0.99942 | 0.99944 | 0.99946 | 0.99948 | 0.9995 |
| 3.3 | 0.99952 | 0.99953 | 0.99955 | 0.99957 | 0.99958 | 0.9996 | 0.99961 | 0.99962 | 0.99964 | 0.99965 |
| 3.4 | 0.99966 | 0.99968 | 0.99969 | 0.9997 | 0.99971 | 0.99972 | 0.99973 | 0.99974 | 0.99975 | 0.99976 |
| 3.5 | 0.99977 | 0.99978 | 0.99978 | 0.99979 | 0.9998 | 0.99981 | 0.99981 | 0.99982 | 0.99983 | 0.99983 |
| 3.6 | 0.99984 | 0.99985 | 0.99985 | 0.99986 | 0.99986 | 0.99987 | 0.99987 | 0.99988 | 0.99988 | 0.99989 |
| 3.7 | 0.99989 | 0.9999 | 0.9999 | 0.9999 | 0.99991 | 0.99991 | 0.99992 | 0.99992 | 0.99992 | 0.99992 |
| 3.8 | 0.99993 | 0.99993 | 0.99993 | 0.99994 | 0.99994 | 0.99994 | 0.99994 | 0.99995 | 0.99995 | 0.99995 |
| 3.9 | 0.99995 | 0.99995 | 0.99996 | 0.99996 | 0.99996 | 0.99996 | 0.99996 | 0.99996 | 0.99997 | 0.99997 |
A z-score of 1.64 corresponds to an area of 0.9495
p(z<1.64)=0.9495
The cumulative probability that z<1.64 is 94.95%
2. Find the cumulative probability for the z-scores values greater than 1.64
To find the cumulative probability of the values greater than 1.64, we need to subtract the cumulative probability of the values less than 1.64 from the total probability under the curve, which is equal to 1:
1−0.9495=0.0505
p(z>1.64)=0.0505
The cumulative probability of z>1.64 is 5.05%
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The normal distribution is easy to work with. Many statistical tests rely on it. Moreover, these tests work well even when the distribution is only approximately normal. For example, if a set's mean and standard deviation are known, and the set follows the normal distribution, we can easily convert between percentiles and raw scores.
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A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but no one had ever heard of the question before, and no one knew the answer.
I hope this is an original question, but seeing how natural it is, I doubt this is the first time someone has asked it.
First, some motivation. Take PP any nonzero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative P′P′ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write P=a⋅∏ri=1(X−αi)miP=a⋅∏ri=1(X−αi)mi where the αi (i=1,…,r)αi (i=1,…,r) are the different roots of PP, and mimi the corresponding multiplicities, and evaluate P′P=∑imiX−αiP′P=∑imiX−αi on a root ββ of P′P′ which is not also a root of PP. You'll end up with an expression of ββ as a convex combination of α1,…,αrα1,…,αr. It is worth mentioning that all the convex coefficients are >0>0, so the new root cannot lie on the edge of the convex hull of PP's roots.
Now fix PP a certain nonzero complex polynomial, and consider ΠΠ, its primitive (antiderivative) that vanishes at 0: Π(0)=00: Π(0)=0 and Π′=PΠ′=P. For each complex ωω, write Πω=Π−ωΠω=Π−ω, so that you get all the primitives of PP. Also, define for any polynomial QQ, Conv(Q)Conv(Q), the convex hull of QQ's roots.
MAIN QUESTION: describe
Hull(P)=⋂ω∈CConv(Πω)Hull(P)=⋂ω∈CConv(Πω).
By the property cited above, Hull(P)Hull(P) is a convex compact subset of the complex plane that contains Conv(P)Conv(P), but I strongly suspect that it is in general larger.
Here are some easy observations:
replacing PP (resp. ΠΠ) by λPλP (resp. λΠλΠ) will not change the result, and considering P(aX+b)P(aX+b) will change Hull(P)Hull(P) accordingly. Hence we can suppose both PP and ΠΠ to be monic. The fact that ΠΠ is no longer a primitive of PP is of no consequence.
the intersection defining Hull(P)Hull(P) can be taken for ωω ranging in a compact subset of CC: as |ω|→∞|ω|→∞, the roots of ΠωΠω will tend to become close to the (deg(P)+1)(deg(P)+1)-th roots of ωω, so for large enough ωω, their convex hull will always contain, say, Conv(Π)Conv(Π).
Hull(P)Hull(P) can be explicitly calculated in the following cases:
P=XnP=Xn, PP of degree 11 or 22. There are only 2 kinds of degree 22 polynomials: two simple roots or a double root. Using z→az+bz→az+b, one only has to consider P=X2P=X2 and P=X(X−1)P=X(X−1). The first one yields {00}, which equals Conv(X2)Conv(X2), the second one gives [0,1]=Conv(X(X−1))[0,1]=Conv(X(X−1)).
Also, if ΠΠ is a real polynomial of odd degree n+1n+1 that has all its roots real and simple, say λ1<μ1<λ2<⋯<μn<λn+1λ1<μ1<λ2<⋯<μn<λn+1, where I have also placed PP's roots μ1,…,μnμ1,…,μn, and if you further assume that Π(μ2j)≤Π(μn)≤Π(μ1)≤Π(μ2j+1)Π(μ2j)≤Π(μn)≤Π(μ1)≤Π(μ2j+1) for all suitable jj (a condition that is best understood with a picture), then Hull(P)=Conv(P)=[μ1,μn]Hull(P)=Conv(P)=[μ1,μn]: just vary ωω between [Π(μn),Π(μ1)][Π(μn),Π(μ1)]; the resulting polynomial ΠωΠω is always split over the real numbers and you get
[μ1,μn]=Conv(P)⊂Hull(P)⊂Conv(ΠΠ(μ1))∩Conv(ΠΠ(μn))==[μ1,…]∩[…,μn]=[μ1,μn]
[μ1,μn]=Conv(P)⊂Hull(P)⊂Conv(ΠΠ(μ1))∩Conv(ΠΠ(μn))==[μ1,…]∩[…,μn]=[μ1,μn]
The equation Πω(z)=Π(z)−ω=0Πω(z)=Π(z)−ω=0 defines a Riemann surface, but I
don't see how that could be of any use.
Computing Hull(P)Hull(P) for the next most simple
polynomial P=X3−1P=X3−1 has proven a challenge, and I can only conjecture what
it might be.
Computing Hull(X3−1)Hull(X3−1) requires factorizing degree 4
polynomials, so one naturally tries to look for good values of ωω,
the ωω that allow for easy factorization of Πω=X4−4X−ωΠω=X4−4X−ω---for instance, the ωω that produces a double root. All that remains to be done
afterwards is to factor a quadratic polynomial. The problem is symmetric,
and you can focus on the case where 1 is the double root (i.e., ω=−3ω=−3).
Plugging in the result in the intersection, and rotating twice, you obtain the following superset of
Hull(X3−1)Hull(X3−1): a hexagon that is the intersection of three similar isoceles
triangles with their main vertex located on the three third roots of unity 1,j,j21,j,j2
QUESTION: is this hexagon equal to Hull(X3−1)Hull(X3−1)?
Here's why I think this might be.
Consider the question of how the convex hulls of the roots of ΠωΠω vary as ωω varies.
When ω0ω0 is such that all roots of Πω0Πω0 are simple, then the inverse function theorem shows that the roots of ΠωΠω
with ωω in a small neighborhood of ω0ω0 vary holomorphically ∼∼
linearly in ω−ω0ω−ω0: z(ω)−z(ω0)∼ω−ω0z(ω)−z(ω0)∼ω−ω0. If however ω0ω0 is such that Πω0Πω0
has a multiple root z0z0 of multiplicity m>1m>1, then a small variation of ωω
about ω0ω0 will split the multiple root z0z0 into
mm distinct roots of ΠωΠω that will spread out roughly as
z0+c(ω−ω0)1mz0+c(ω−ω0)1m, where cc is some nonzero coefficient. This
means that for small variations, these roots will move at much higher velocities
than the simple roots, and they will constitute the major contribution to the variation of
Conv(Πω)Conv(Πω); also, they spread out evenly, and (at least if the
multiplicity is greater or equal to 33) they will tend to increase the convex hull
around z0z0. Thus it seems not too unreasonable to conjecture that the convex hull
Conv(Πω)Conv(Πω) has what one can only describe as
critical points at the ω0ω0 that produce roots with
multiplicities. I'm fairly certain there is a sort of calculus on convex sets that
would allow one to make this statement precise, but I don't know see what it could be.
Back to X3−1X3−1: explicit calculations suggest that up to second order, the double root 11 of X4−4X+3−hX4−4X+3−h for |h|<<1|h|<<1 splits in half nicely (here ω=−3+hω=−3+h), and the convex hull will continue to contain the aforementioned hexagon.
QUESTION (Conjecture): is it true that
Hull(P)=⋂ω∈MRConv(Πω)Hull(P)=⋂ω∈MRConv(Πω), where
MRMR is the set of all ω0ω0 such that
Πω0Πω0 has a multiple root, i.e., the set of all Π(αi)Π(αi) where the
αiαi are the roots of PP?
All previous examples of calculations agree with this, and I have tried as best I can to justify this guess heuristically.
Are you aware of a solution? Is this a classical problem? Is anybody brave enough to
make a computer program that would compute some intersections of convex hulls
obtained from the roots to see if my conjecture is valid?
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edited Jun 3, 2018 at 11:23
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13 revs, 4 users 79%Olivier Bégassat
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In other words: If PP is a polynomial, what points lie in the convex hull of the roots of every antiderivative of PP, besides the roots of PP and their convex hull? It's a good question, and your partial results are interesting too, but your write-up is somehow long and hard to read.
– Greg Kuperberg
Commented
Mar 18, 2011 at 9:23
5
Another way to word the question: what is the intersection of the convex hull of level sets {z|Q(z)=ω}{z|Q(z)=ω} for a polynomial Q? Here P is the derivative of Q. By chance, I've discussed this question a bit with Tan Lei; she made some nice movies of how the convex hulls of level sets vary with ω. (Also, it's fun to look at their diagrams interactively manipulated in Mathematica). If I get my thoughts organized I'll post an answer.
– Bill Thurston
Commented
Mar 18, 2011 at 12:13
3
It's a nice question. A suggestion for readability : replace Π (which looks like a product) with a roman letter.
– François Brunault
Commented
Mar 18, 2011 at 12:17
1
This reminds me Hildebrandt's Theorem (it does not answer the question, however): If P is the characteristic polynomial of a matrix A, the intersection of the numerical ranges W(B) where B runs over the matrices conjugated to A equals the convex hull of the roots of P.
– Denis Serre
Commented
Mar 18, 2011 at 15:17
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First, a counterexample to your conjecture. Let Π=x4+x3+4x2+4x=x(x+1)(x2+4), so P=4x3+3x2+8x+4. The critical values of ω are 1.06638,3.89455+2.87687i,3.89455−2.87687i, and by inspection (using Mathematica) we see that for each of these values of ω, Conv(Πω) contains a neighborhood of 0.
Now for a calculus on convex sets. Every convex set is the intersection of a set of halfplanes. Call a halfplane in this collection essential if removing all of the halfplanes in an open set of halfplanes (in the halfplane topology) containing it from our set of halfplanes makes the intersection of the halfplanes in our set bigger. We wish to find a characterization of the essential halfplanes of Hull(P).
First of all, I claim that any essential halfplane of Hull(P) occurs as an essential halfplane of Conv(Πω) for some ω. This follows from continuity - for any open set around our essential halfplane there is some ω, take the limit of a subsequence of these ωs...
Now, suppose that the halfplane Re(x)≤0 occurs as an essential halfplane of some Conv(Πω), i.e. there are at least two roots of Πω with real part 0, and the rest of the roots have negative real part. If the number of roots on the line Re(x)=0 (counted with multiplicity) is two, then by holomorphicity we can always find a direction to move ω so that either both roots move to the left, or both roots stay on the line Re(x)=0 and move towards eachother. If we can ever make both roots move to the left, then clearly the halfplane Re(x)≤0 is not an essential halfplane of Hull(P), otherwise we keep pushing the roots towards eachother until either they run into eachother or until a third root hits the line Re(x)=0. In any case, we see that if a halfplane is essential for Hull(P), then there is some ω such that the halfplane is essential for Conv(Πω) and such that at least three roots (counted with multiplicity) of Πω are on the boundary of the halfplane, or two of the roots are equal and Πω has no other roots.
So if we let L be the set of ωs such that three of the roots of Πω lie on a line, we get that Hull(P)=∩ω∈LConv(Πω) if degP≥2.
Edit: Actually, I think there is a problem with this. It's conceivable that two roots are on the line Re(x)=0 and have derivatives (with respect to ω) pointed in opposite directions, such that we can't simply push them towards eachother. For instance, the map from one root to the other root could, locally, look like the fractional linear transform sending the left halfplane to a circle contained in the right halfplane and tangent to the line Re(x)=0 at the other root. So, we may need to enlarge the set L to contain also those ωs for which the ratio of the derivatives of two of the roots (with respect to ω) is a negative real number.
Edit 2: It turns out that this isn't an issue. Call the two roots on the line Re(x)=0 r1 and r2. Suppose that locally, r1(ϵ)=ϵ, r2(ϵ)=i−mϵ+aϵk+O(ϵk+1), a≠0, m>0. Note that if we had r2(ϵ)=i−mϵ, then the intersection of the halfplanes corresponding to r1(ϵ),r2(ϵ) and r1(−ϵ),r2(−ϵ) would be contained in the halfplane Re(x)≤0, and the intersection of their boundaries would be located at i/(m+1). Now if k is even, then the correction term shifts the intersection of the boundaries by aϵk/(m+1)+O(ϵk+1), so if we choose ϵ small such that aϵk is real and negative, then we see that the halfplane Re(x)≤0 is not essential. If k is odd, then if we choose ϵ small such that Re(ϵ)<0 and aϵk is a positive real times i, then r2(ϵ) is shifted up and r2(−ϵ) is shifted down, so the intersection of the boundaries will be shifted to the left (draw a picture), so again the halfplane Re(x)≤0 is not essential.
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Hi Zeb, thank you for your answer. I should use mathematica to convince myself, but I see now that problems may arise when 3 roots are aligned as is the case for −2i,0,2i: they force Hull(P) to lie inside the closed halp plane ℜ(z)≤0, yet all the convex envelopes I considered may very well contain 0 in their inside. I understand that you can say any essential halfplane H of Hull(P) arises as one of the essential half planes of Conv(πω): there must half planes Hω close to H, otherwise you'd have an angular point
– Olivier Bégassat
Commented
Mar 18, 2011 at 18:02
where there is no essential half plane, or else all essential half planes of the Πω would contour a sort of disk with positive radius. And therefore there are always essential half planes for certain ω that are close to H
– Olivier Bégassat
Commented
Mar 18, 2011 at 18:08
if the derivatives with respect to ω of the two simple roots you counsider are such that their ratio is in mathbbC∖R−, then you can indeed push the half plane to the left for a suitable choice of perturbation ω+h, if they are opposites you should be able to push them towards eachother. I don't really understand the problem you mention in your edit.
– Olivier Bégassat
Commented
Mar 18, 2011 at 18:33
The problem with pushing the roots towards eachother is that the higher order derivatives might screw things up in such a way that we can neither move both roots to the left nor push them directly towards eachother... I found a workaround, though (my second edit).
– zeb
Commented
Mar 18, 2011 at 19:39
I still wonder if you can at least show that only finitely many such root aligning values of ω need to be considered. Take the example of X3−1 and it's renormalized primitive X4−4X. Take ω that gives a multiple root (such as ω=−3). Take r<<1 and let θ vary, the two roots for −3+r.eiθ (that split out of the double root 1) will turn around 1, performing a semicircle as the exponential completes a complete circle. Continuity and the near staticness of the 2 other roots show that there will be many values of ω that produce alignment.
– Olivier Bégassat
Commented
Mar 19, 2011 at 0:25
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I. First I want to share some computer experiments of H.H. Rugh. The following image supports the positive answer of the
QUESTION: is this hexagon equal to Hull(X3−1)?
triangle (as a new user I was not allowed to use image tags).
A scilab program testing this problem can be found at
roots-dancing. In particular an example z6−3z3+z similar to the starting
example of Zeb, showing that Hull(P) can be strictly smaller then ⋂vConv(fv)
for v ranging the critical values of f (shown in blue polygons).
See smaller.
II. Here is a proof (communication of Rugh) of the same statement of Zeb (with f=Π):
If we let L be the set of v's such that three of the roots of fv lie on a line, we get Hull(P)=⋂v∈LConv(fv) if degf≥3.
The underlying idea is very similar to that of Zeb.
Statements: Let f be a polynomial of degree at least 3. Assume that a0 and b0 are two distinct simple roots of f(z)−v0. Then for v in a small neighborhood of v0, there are two simple roots
a(v), b(v) of f(z)−v with a(v0)=a0 and b(v0)=b0. In this case,
If for some fixed complex number t≠0,1,
the holomorphic function v↦ta(v)+(1−t)b(v) is a constant c, then c is a critical point of f and f has a rotational symmetry about c.
If the segment [a0,b0] is a boundary edge of the polygon Conv(fv0) and no other point in the line through a0,b0 is
mapped to v0, then
2.1 for v sufficiently close to v0, the segment [a(v),b(v)] is a boundary edge of the polygon Conv(fv),
and for any t∈]0,1[, the
map v↦ta(v)+(1−t)b(v) is an open mapping.
2.2. The line through a0,b0 is outside ⋂vConv(fv).
Proof.
1. Replacing f(z) by f(z−c) if necessary, we may assume c=0.
Note that a↦b(f(a)) is defined and holomorphic in a neighborhood of a0, satisfying that b(f(a0))=b0 and f(b(f(a)))=f(a).
It follows that ta+(1−t)b(f(a))≡0 so b(f(a))=tat−1.
Therefore f(a)=f(tat−1) in a neighborhood of a, thus in the entire complex plane.
Comparing the coefficients we conclude that tt−1 is a root of unity and f′(0)=tt−1f′(0).
Using tt−1≠1 we get f′(0)=0. An example is z6+3z4−5z2.
2.1. The condition means that all the other points in f−1(v0) are contained in one of the open half planes delimited by the line through a0,b0. This
is clearly an open condition.
Now if ta(v)+(1−t)b(v)≡c, by Point 1 c must be a critical point and a center of symmetry and
Conv(fv0) would have been symmetric with respect to c. This is not possible by our assumption that all the other
points of f−1(v0) (and there is at least one due to the assumption
on the degree of f and on the simplicity of a0,b0 as roots of fv0) are on one side of the line through a0,b0.
2.2. We may look at the open set W=⋃vOuter(fv) where Outer(fv) is the complement of Conv(fv).
We may assume a0,b0 are on the imaginary axis and all the other points of f−1(v0) are
on the left half plane. We know already iR−[a0,b0]⊂Outer(fv0)⊂W.
Fix some t∈[0,1] and let z0=ta0+(1−t)b0. We may assume z0=0. Now v↦z(v)=ta(v)+(1−t)b(v) is
open. Choose a path v(s) such that z(v(s)) is negative real. Then
the segment [a(v(s)),b(v(s))] passes through z(v(s)) and remains almost vertical for sufficiently small s, so has z0 on its right side. By 2.1 we may conclude
z0∈Outer(fv(s))⊂W. qed.
III. Finally I want to share some numerical experiments (with the help of Jos Leys)
illustrating a refinement (communication by Thurston) of Gauss-Lucas property.
Consider a polynomial f as a branched covering of the complex plane. Denote by C the convex hull of the critical points of f. It is called {\em the critical convex} of f. The following statements are equivalent:
(1) For any v∈C, we have Conv(fv)⊃C.
(2) The map f is surjective on any closed half plane H intersecting C.
(1)⟹(2). Assume f(H)∌v. Then f−1(v) is contained in C−H which is an open half plane, in particular convex.
Then Conv(f|v) is also contained in C−H. So Conv(f|v)⊅C, contradicting (1).
(2)⟹(1). Assume that Conv(f|v) does not contain the entire set C. Then there is a
closed half plane H intersecting C but disjoint from Conv(f|v). Then f(H)∌v, that is, f is not surjective on H, contradicting (2).
Now the refinement (I'll leave the proof to Thurston if somebody requires) is that if one takes a supporting line L of C
there is a region on the outer half space of L on which f is a bijection onto C (injective in the interior and bijective on the union of the interior with
half of the boundary arc).
In fact this region is bounded by two geodesic rays of the conformal metric |f′(z)|⋅|dz|
tangent to L at a critical point (if the critical point is simple, otherwise
the region is even smaller). This means that we use the euclidean length in the range to measure tangent vectors in the domain. Geodesics in this metric are the pullbacks by f of straight lines.
The movie bijectivity is made by Jos Leys to illustrate this result.
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This problem has been considered before:
The notes to chapter 4 of
Rahman/Schmeißer: Analytic Theory of Polynomials, Oxford University Press, 2002
mention this problem, state that conv(P) is a proper subset of hull(P) in general, and give two references:
1) J. L. Walsh: The location of Critical Points of Analytic and Harmonic Functions,
AMS Colloquium Publications Volume 34, 1950, p. 72
2) E. Chamberlin and J. Wolfe: Note on a converse of Lucas's theorem, Proceedings of the American Mathematical Society 5, 1954, pp. 203 - 205
I had a look at both of them, the paper of Chamberlin and Wolfe can be obtained online via the AMS for free.
The relevant paragraph in Walsh's book is 3.5.1 (starts on page 71). I state the 4 theorems given there for convenience:
1) conv(P) = hull(P) if P is of degree 1 or 2
2) conv(P) = hull(P) if P is of degree 3 and its zeros are collinear
3) There exists a P of degree 3, such that conv(P) is a proper subset of hull(P); example P(z)=z3+1.
4) There exists a P of degree 4 with real zeros, such that conv(P) is a proper subset of hull(P); example P(z)=(z2−1)2.
Chamberlin and Wolfe prove, that
1) Any point on the boundary of hull(P) is on the boundary of conv(Πω) for some ω.
2) If a side of any conv(Πω) contains a point of hull(P) and only two zeros of Πω , counting multiplicities, then P is of degree 1. (Here a side is equal to conv(Πω), if the zeros of Πω are collinear.)
3) Vertices of conv(P) need not lie on the boundary of hull(P), example
Π(z)=z2(z+1)(z2−2az+1+a2), where a is positive and sufficiently small.
4) Even if P is cubic, hull(P) need not be determined by its primitives with multiple zeros, example P(z) = 4z3+9/2z2+2z+3/2.
While I have corrected what I considered a minor typo in the example Π(z)=z2(z+1)(z2−2az+1+a2), I did no thorough proofreading.
Walsh gives no special references to further work in section 3.5.1 and the only reference provided by Chamberlin and Wolfe is to Walsh's book cited above.
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The link to the AMS paper is link . Morris Marden gives result 2) of Chamberlin and Wolfe as exercise 8 to paragraph 6 of the 1966 edition of his "Geometry of polynomials" on page 24 with a reference to Chamberlin and Wolfe.
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We can characterize those P for which Hull(P)=Conv(P).
First suppose that the roots of P do not lie on a line. We prove that Hull(P)=Conv(P) if and only if there is an antiderivative Q of P for which Q=A2B, and the roots of B lie in the convex hull of A. The "if" is immediate and left to the reader.
Note that a root β of P can lie on the boundary of Conv(Πω) only if β is a root of Πω, or all the roots of Πω lie on a line. This latter case of course is impossible if the roots of P do not lie on a line.
Let β be a root of P. We claim that for every neighborhood N of Π(β) there is a neighborhood M of β such that Conv(Πy)⊃M for every y∈C∖N. This certainly holds for any given M when y lies outside a large compact subset of C, so we can think of y ranging over a compact set. For each y≠Π(β), there is a ball of positive radius around β in Conv(Πy), and the size of the maximal such ball varies continuously, so the claim follows.
Now, suppose that β and γ are adjacent vertices (extreme points) of Conv(P). Suppose γ is not a root of ΠΠ(β). Then γ lies in the interior of Conv(ΠΠ(β)), and we can find γ′ in the interior of Conv(ΠΠ(β)) so that ¯γ′β∩Conv(P)=β. Then ¯γ′β⊂Conv(Πy) for all y in a suitable neighborhood N of Π(β). On the other hand, M⊂Conv(Πy) for a suitable neighborhood M of β and all y∉N. Therefore ¯γ′β∩M⊂Hull(P),
and hence Conv(P)⊊Hull(P).
So if Conv(P)=Hull(P), then Π(β)=Π(γ) for all adjacent vertices β,γ of Conv(P). Letting Q be Π(β) for any extreme point β of Conv(P), we see that every extreme point of Conv(P) is a root of Q, and hence a double root of Q. Moveover, if Conv(P)=Hull(P), then Conv(P)⊇Conv(Q), so Q=A2B, where the roots of B lie in the convex hull of the roots of A, the extreme points of Conv(P).
Now suppose that the roots of P lie on a line. We can assume that P is real (with positive leading term) and all of the roots of P are real as well. Let β and γ be the least and greatest root of P, respectively. Then, by a similar analysis, Hull(P)=Conv(P) if and only if
Πy has all real roots for some value of y,
The roots of ΠΠ(β) have real part at least β, and
The roots of ΠΠ(γ) have real part at most γ.
You can easily verify that 1, 2, and 3 hold when P has degree 3. I would be tempted to conjecture that 2 and 3 always hold (when the roots of P are real).
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What is the time constant in an RC circuit? What value should it be close to or equal to, and why?
André Giguère
Electrical engineer from Polytecnic Montréal
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Author has 480 answers and 288K answer views
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2y
What is the time constant in an RC circuit? What value should it be close to or equal to, and why? (13 Feb. 2023)
Short answer
The RC time constant is the time it takes to charge a capacitor to 63.2% of its maximum value for a given stable voltage.
Long answer
The formula for charging a capacitor is
Q(t) the cumulative charge which is a function of time
C the value of the capacitor
ε₀ the dielectric constant
R the value of the resistance
e = 2.7182 the Euler constant
When choosing the time equal to RC we say that we have the value of one time constant and if we apply this value in the formula o
What is the time constant in an RC circuit? What value should it be close to or equal to, and why? (13 Feb. 2023)
Short answer
The RC time constant is the time it takes to charge a capacitor to 63.2% of its maximum value for a given stable voltage.
Long answer
The formula for charging a capacitor is
Q(t) the cumulative charge which is a function of time
C the value of the capacitor
ε₀ the dielectric constant
R the value of the resistance
e = 2.7182 the Euler constant
When choosing the time equal to RC we say that we have the value of one time constant and if we apply this value in the formula of Q(t).
The precision of the time constant is as precise as the value of the components we use, but the value of "e" being an irrational number, 0.632 is also an irrational number and its precision depends on the number of decimals you use to represent "e".
Finally, if we put 0.632 in %, we get 63.2 %.
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What are the applications of time constant of an RC circuit?
Muhammad Usman Sarwar
IT and Network Expert (2010–present)
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Originally Answered: What is the time constant for a RC circuit?
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In simple answer the it is the time required to charge the capacitor to 63.5 % of capacity. Now this is little confusing, let me make it simple. The capacitor’s job is to supply voltage to the circuit in case of the absence of the battery. In other words it has to store the voltage before it has to supply it. Now how much voltage it can store in how much time that is basically determined through the time constant.
“The time constant defines the amount of time it takes for a capacitor to charge to 63% of the supply voltage which is charging it “
For example :-
Lets suppose we have below mentioned
In simple answer the it is the time required to charge the capacitor to 63.5 % of capacity. Now this is little confusing, let me make it simple. The capacitor’s job is to supply voltage to the circuit in case of the absence of the battery. In other words it has to store the voltage before it has to supply it. Now how much voltage it can store in how much time that is basically determined through the time constant.
“The time constant defines the amount of time it takes for a capacitor to charge to 63% of the supply voltage which is charging it “
For example :-
Lets suppose we have below mentioned details in our circuit.
Battery = 9 Volts
Resistance = 3 KΩ
Capacitor = 1000µF
One time constant, τ=RC=(3KΩ)(1000µF)=3 seconds. So after 3 seconds, the capacitor is charged to 63% of the 9 volts that the battery is supplying it, which would be approximately 5.67 volts.
The content above is taken from the below link.
RC Time Constant of a Capacitor
Tom Quetchenbach
B.S. in Electrical Engineering, California Institute of Technology (Caltech) (Graduated 2007)
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Originally Answered: What is the RC time constant in electronics?
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The RC time constant is, as its name implies, the product of resistance and capacitance, which has units of time.
When a resistor is connected across a fully charged capacitor, the RC time constant is the time it will take to discharge to approximately 1e≈36.8% of its original charge. When a fully discharged capacitor is charged by an ideal voltage source through a resistor, the RC time constant is the time it will take to charge up to 1−1e≈62.3% of its final charge. (Since charge is proportional to voltage, you could make these statements about voltage inste
The RC time constant is, as its name implies, the product of resistance and capacitance, which has units of time.
When a resistor is connected across a fully charged capacitor, the RC time constant is the time it will take to discharge to approximately 1e≈36.8% of its original charge. When a fully discharged capacitor is charged by an ideal voltage source through a resistor, the RC time constant is the time it will take to charge up to 1−1e≈62.3% of its final charge. (Since charge is proportional to voltage, you could make these statements about voltage instead of charge.)
Very approximately, the time constant is the time it will take for a capacitor to get 2/3 charged or discharged (because 62.3% is close to 2/3).
For example, suppose a 100 microfarad capacitor has a voltage of 10.00 volts at open circuit (nothing connected across the terminals). Then, at time t=0, a resistor of 1000 ohms is connected across the terminals. How long until the voltage across the capacitor has decreased to 3.68 volts? The answer is the RC time constant, R⋅C=(100×10−6 F)⋅(1000 Ω)=0.1 s.
Why do you need to know when a capacitor is 62.3% charged? Usually you don't, but the time constant gives you an idea of the timescales involved in a circuit, and it's easy to calculate the time required for other percentages, by multiplying by a constant.
The time constant is also related to the cutoff frequency:
fc=12πRC
which is the frequency at which a filter attenuates the input signal by 50%.
Gurpreet Singh
Author has 67 answers and 178.8K answer views
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10y
Originally Answered: What is the RC time constant in electronics?
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If a series RC circuit is fed by a DC source, then just in the beginning of the circuit closing, there will be transient phenomenon taking place.
solution for the above mentioned equations is as below-
graph for such a solution typically looks like as below-
So, that means the current flow is not constant with time. High in the start and decays exponentially; asymptotically approaches zero. The time when t=RC the current will be (V/R) x (1/e).V/R is the maximum current value.
1/e is approximately 0.37.
So RC in a series RC circuit is the time in which current will decay down to approx 37% of it
If a series RC circuit is fed by a DC source, then just in the beginning of the circuit closing, there will be transient phenomenon taking place.
solution for the above mentioned equations is as below-
graph for such a solution typically looks like as below-
So, that means the current flow is not constant with time. High in the start and decays exponentially; asymptotically approaches zero. The time when t=RC the current will be (V/R) x (1/e).V/R is the maximum current value.
1/e is approximately 0.37.
So RC in a series RC circuit is the time in which current will decay down to approx 37% of its maximum value.
As per requirements of electronic circuit, the suitable value of R & C is selected to get desired current vs time characteristic etc etc.
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Related questions
How do you find the time constant for RL and RC circuits?
What is the effect of a large value of R in an RC circuit?
What is the difference between a time constant and an RC circuit?
What is the time delay in an RC circuit?
How do you calculate the rise time for an RC circuit?
Lee Felsenstein
Analog and digital design engineer for 50 years, working in discrete components.
·
Author has 2.3K answers and 7.1M answer views
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8y
Originally Answered: What is the time constant for a RC circuit?
·
No. It’s more like the time to charge to 63.2% of full charge. Here’s why:
Current through the resistor is the voltage difference across the resistor divided by the resistance (Ohm’s law). Current through the capacitor is given by the equation I = C (dV/dt). In the series connection this combines to:
(Vo - V) / R = C (dV/dt)
The solution to this differential equation is:
(1 - (V/Vo)) = exp (-t/ (RC)
(I won’t detail this solution here)
As V (the voltage on the capacitor) increases, the quantity (Vo - V) decreases (the voltage across the resistor). This means that as the capacitor voltage increas
No. It’s more like the time to charge to 63.2% of full charge. Here’s why:
Current through the resistor is the voltage difference across the resistor divided by the resistance (Ohm’s law). Current through the capacitor is given by the equation I = C (dV/dt). In the series connection this combines to:
(Vo - V) / R = C (dV/dt)
The solution to this differential equation is:
(1 - (V/Vo)) = exp (-t/ (RC)
(I won’t detail this solution here)
As V (the voltage on the capacitor) increases, the quantity (Vo - V) decreases (the voltage across the resistor). This means that as the capacitor voltage increases the current decreases.
Meaning - the capacitor will never become “fully charged” (V/Vo = 1), or 0 = exp (-t/(RC)). According to the exponential equation that would require t to be infinity!
The “time constant” (R C) in the equation is the time at which the exponent is -1. At that time (1-V/Vo) = 1/e = 0.368 . V/Vo then is (1–0.368) = 0.632 .
This may seem like a strange number, but let’s look at the situation where V/Vo = 0.5 , or 0.5 = exp(-t/(RC)). Taking the natural log (ln) of both sides yields:
-0.69 = (-t/(RC)), or 0.69 = t / (RC) . The verbal expression for this is “at the half-charge point the time is 0.69 time constants”.
We resolve the problem of the charge never coming to an end by defining an acceptable threshold at which we declare the capacitor to be “fully charged”. This threshold will be 95% for t/RC = 3 and 99.2% for t/RC = 5.
Welcome to the wonderful world of exponentials!
Gregory Diana
Worked at University of KwaZulu-Natal
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Author has 2.4K answers and 5.4M answer views
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Updated 10y
Originally Answered: What is the RC time constant in electronics?
·
There is no difference in the RC time constant whether it be electrical or electronics or even mechanical, chemical or otherwise. There are many analogies but the the RC or FIRST ORDER time constant represents both the rate and ratio of the energy stored in the electric field to that dissipated as heat in doing so. No matter what combination of R and C one chooses for the rate the ratio remains constant. The RC time constant is the time or rate for the capacitor voltage to change by ~63% from the previous or between each such time interval. This is a simple way to avoid elegant but somewhat u
There is no difference in the RC time constant whether it be electrical or electronics or even mechanical, chemical or otherwise. There are many analogies but the the RC or FIRST ORDER time constant represents both the rate and ratio of the energy stored in the electric field to that dissipated as heat in doing so. No matter what combination of R and C one chooses for the rate the ratio remains constant. The RC time constant is the time or rate for the capacitor voltage to change by ~63% from the previous or between each such time interval. This is a simple way to avoid elegant but somewhat unnecessary math and a quick way to check in exams and test. Back of the matchbox stuff :)
To see this consider an RC circuit with
the capacitor initially discharged and a voltage of 100V applied.
After time RC Vc=63%100= 63V after time 2RC Vc =63+63%(100-63) = 63+63%37 = 63+23.31 = 86.31 V and so on.
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Jonathan Bernal
A curious someone looking for ways to learn more and better.
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10y
Originally Answered: What is the RC time constant in electronics?
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The answers that other have provided are indeed very informative. I would like to complement what others have said by saying that the RC time constant does not only represent the time it takes for a capacitor to discharge but also the time it takes for it to become fully charged.
Oftentimes, at least in textbooks, the RC time constant is denoted by the Greek letter tau. To observe the behavior of RC circuits, it is more convenient to calculate the value of tau for any given circuit and then observe the amount of energy stored in the capacitor after some multiple of RC. Capacitors usually become
The answers that other have provided are indeed very informative. I would like to complement what others have said by saying that the RC time constant does not only represent the time it takes for a capacitor to discharge but also the time it takes for it to become fully charged.
Oftentimes, at least in textbooks, the RC time constant is denoted by the Greek letter tau. To observe the behavior of RC circuits, it is more convenient to calculate the value of tau for any given circuit and then observe the amount of energy stored in the capacitor after some multiple of RC. Capacitors usually become fully charged or discharged after 5tau.
Manjunath Pai H
Former 30+ yrs in Electronics Engg and loved it.
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Author has 2.7K answers and 6.7M answer views
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9y
Originally Answered: What is the time constant for a RC circuit?
·
No . It is just like a bank interest rate, a constant. To see what is your amount in the bank after some time, say after a few months or years, you need to know the starting amount, the bank interest rate and the time. Same here. You should know the starting voltage, the time and the rate of growth, determined by the time constant RC. Actually in our case, the growth rate is 1/RC.
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With the various HTML editors available today there are several considerations to keep in mind when deciding on which is the right fit for your company including ease of use, SEO controls, high performance hosting, flexible content management tools and scalability. Webflow offers all those features and allows you to experience the power of code – without writing it.
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Loring Chien
Electrical Engineer for 45 years & IEEE Sr. Life member
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Author has 67.9K answers and 250M answer views
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10y
Originally Answered: What is the RC time constant in electronics?
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In simple terms its about R x C and is expressed in seconds (or milliseconds or microseconds). where R is in ohms and C in Farads.
Its indicative of delays and filter constants.
In more detail it expresses the time for a capacitor fed by a voltage source and a series resistor to reach a certain percentage (~63%) of the voltage source value. Or similarly to discharge to ~37% from 100%.
This is chosen (rather than computing the time it takes to charge up to 100% of the voltage source) because it theoretically takes forever to ooze all the way up to the voltage source value.
Amarish Pandey
Technical Consultant (2016–present)
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Author has 108 answers and 204.9K answer views
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10y
Originally Answered: What is the RC time constant in electronics?
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www.electronics-tutorials.ws/rc/rc_1.html
RC as name suggest is combination of Resitance and capacitence of a circuit. It is basically used to measure charging and discharging time of any electrical or electronic circuit.
Kev Go
Training Electricians (2012–present)
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Author has 11.4K answers and 19.7M answer views
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5y
What is the time constant of an RC circuit? Show that the product RC has the dimension of time.
tc = R x C
The time constant of any series conncted R/C circuit is shown by the graphs below.
At 1tc the voltagev is 63.2% of maximum and current is 36.8 % of maximum.
The same curves apply to the discharge of a capacitor in the RC circuit.
tc = R x C
The time constant of any series conncted R/C circuit is shown by the graphs below.
At 1tc the voltagev is 63.2% of maximum and current is 36.8 % of maximum.
The same curves apply to the discharge of a capacitor in the RC circuit.
Ankit Goyal
GATE Guru (AIR-1), AVP Unacademy, An avid learner of Science
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Author has 352 answers and 4.9M answer views
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2y
Related
What is the physical significance of the time constant of an RC circuit?
In an RC circuit, the RC time constant means the time required to charge a capacitor to about 63 percent of the maximum voltage in an RC circuit when the circuit is excited by a step voltage.
( source-Google)
For more queries and discussions, you can inbox me on Quora.
From the pen of,
Ankit Goyal, the GATE guru & AIR-1
In an RC circuit, the RC time constant means the time required to charge a capacitor to about 63 percent of the maximum voltage in an RC circuit when the circuit is excited by a step voltage.
( source-Google)
For more queries and discussions, you can inbox me on Quora.
From the pen of,
Ankit Goyal, the GATE guru & AIR-1
LetsCrackIt
Raymond Woodward
Extra class licensee
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Author has 161 answers and 26.6K answer views
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Aug 17
Related
What is the time constant in a circuit, and why is it important for understanding how capacitors charge and discharge?
First, keep in mind that it takes some amount of time, however small, for things to happen in an electric circuit. In the beginning, when voltage is first applied across a capacitor, the capacitor is essentially a short circuit. Maximum current is flowing through it. The amount of current depends upon the resistance in the circuit, which includes whatever resistance might be in series with the capacitor, the resistance of the capacitor itself, and the resistance of the voltage source because no voltage source is capable of supplying infinite current.
Once current starts flowing “through” the ca
First, keep in mind that it takes some amount of time, however small, for things to happen in an electric circuit. In the beginning, when voltage is first applied across a capacitor, the capacitor is essentially a short circuit. Maximum current is flowing through it. The amount of current depends upon the resistance in the circuit, which includes whatever resistance might be in series with the capacitor, the resistance of the capacitor itself, and the resistance of the voltage source because no voltage source is capable of supplying infinite current.
Once current starts flowing “through” the capacitor, the voltage measured across it starts to rise. As it rises, the current starts to diminish, and this process continues until the voltage across the capacitor essentially equals the supply voltage and the current essentially drops to zero. With little resistance in the circuit, this process happens rather quickly. Increasing the resistance means that the current flow is reduced, and the time it takes to fully charge the capacitor is longer.
The time constant is the combination of the capacitor and the resistances involved, and we say that it takes five time constants to consider the capacitor as fully charged.
Related questions
How does the time constant of an RC or RL circuit end up with a unit of seconds?
What is the time constant of an RC circuit? Show that the product RC has the dimension of time.
What is the value of C in the circuit if the resistance in an RC circuit of time constant 2 seconds is 1000 ohms?
What is the time constant for a RC circuit?
What are the applications of time constant of an RC circuit?
How do you find the time constant for RL and RC circuits?
What is the effect of a large value of R in an RC circuit?
What is the difference between a time constant and an RC circuit?
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How do you calculate the rise time for an RC circuit?
Why does the current in a series RC circuit remain constant over time?
Why is it that in an RL circuit the time constant is L/R and in an RC circuit the time constant is equal to 1/RC?
What is the value of a capacitor for an RC time constant?
What is meant by RC time constant?
Why does the time constant of an RC circuit lie between 63% to 37%?
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12839 | https://projecteuclid.org/journals/real-analysis-exchange/volume-29/issue-1/On-t-convex-functions/rae/1149860187.pdf | Real Analysis Exchange Vol. 29(1), 2003/2004, pp. 219228 Kazimierz Nikodem, Department of Mathematics, University of Bielsko-Biaªa, ul. Wilowa 2, 43-309 Bielsko-Biaªa, Poland.
email: knik@ath.bielsko.pl Zsolt Páles, Institute of Mathematics, University of Debrecen, H-4010 Debrecen, Pf. 12, Hungary. email: pales@math.klte.hu ON t-CONVEX FUNCTIONS∗ Abstract The main results of the paper, answering an open problem raised in , show that t-convexity can also be characterized in terms of a lower second-order generalized derivative. As a consequence, we obtain that t-convexity is also a localizable convexity property.
1 Introduction A real-valued function f : I →R de ned on an interval I ⊆R is called t-convex if f(tx + (1 −t)y) ≤tf(x) + (1 −t)f(y) for x, y ∈I, (1) where t is a xed element of the open unit interval ]0, 1[. If (1) holds with t = 1/2, then f is said to be Jensen-convex or midpoint convex (cf. ).
Obviously, any convex function is t-convex, however there are nonconvex but t-convex functions. By a result of Kuhn , t-convexity always implies Jensen-convexity (cf. also for a more elementary proof) but, for every irrational t, there exists a Jensen-convex but not t-convex function.
A related functional inequality is f((1 −t)x + ty) + f(tx + (1 −t)y) ≤f(x) + f(y) for x, y ∈I.
Functions satisfying the above inequality are called t-Wright-convex (see for the origin of this notion). It is obvious that t-convex functions are also Key Words: t-convexity, t-Wright-convexity, localizable convexity properties, general-ized 2nd-order derivative Mathematical Reviews subject classi cation: Primary 26A51, 39B62 Received by the editors December 13, 2002 Communicated by: B. S. Thomson ∗This research has been supported by the Hungarian Scienti c Research Fund (OTKA) Grant T-043080 and by the Higher Education, Research and Development Fund (FKFP) Grant 0215/2001.
219 220 Kazimierz Nikodem and Zsolt Páles t-Wright-convex, however, depending on the algebraic character of t, t-Wright convexity can be equivalent and also non-equivalent to t-convexity. (See the paper of Maksa, Nikodem, and Páles for further details.) In a recent paper , Gilányi and Páles proved that t-Wright-convexity can be characterized in terms of a properly chosen generalized second-order derivative. Due to this characterization, it turns out that t-Wright-convexity is also localizable; i.e., a function is t-Wright-convex on I if and only if each point of I possesses a neighborhood such that the function restricted to this neighborhood is t-Wright-convex.
The main results of the paper, answering an open problems raised in , show that t-convexity can also be characterized in terms of a lower second-order generalized derivative. As a consequence, we obtain that t-convexity is also a localizable convexity property.
2 Second-Order Divided Dierences For an arbitrary function f : I →R de ne the second-order divided dierence of f at three pairwise distinct points x, y, z of I by f[x, y, z] := f(x) (y −x)(z −x) + f(y) (x −y)(z −y) + f(z) (x −z)(y −z).
Obviously, the above expression is symmetric in x, y, z.
The Mean Value Theorem of divided dierences is recalled in the following Lemma (cf. , ).
Lemma 1. Let f : I →R be a twice dierentiable function on I. Then, for all distinct elements x, y, z of I, there exists a point ξ ∈co{x, y, z} such that f[x, y, z] = f ′′(ξ) 2 .
It is an immediate consequence of the above Lemma that if f is a second-degree polynomial of the form f(x) = a + bx + cx2, then f[x, y, z] = c for all pairwise distinct x, y, z in I.
The next result oers an identity called the chain formula for chains of divided dierences of second-order. A generalization of this result for higher-order divided dierences can be found in [8, Lemma XV.2.2, pp. 376377.].
Lemma 2. (Chain Formula) Let x0 < x1 < · · · < xn (n ≥2) be arbitrary points in I. Then, for each xed 0 < j < n, there exist positive constants λ1, . . . , λn−1 with λ1 + · · · + λn−1 = 1 such that n−1 X i=1 λif[xi−1, xi, xi+1] = f[x0, xj, xn] (2) On t-Convex Functions 221 holds for all functions f : I →R. Moreover, λi = (xi+1 −xi−1)(xi −x0) (xn −x0)(xj −x0) if 1 ≤i < j, xi+1 −xi−1 xn −x0 if i = j, (xi+1 −xi−1)(xn −xi) (xn −x0)(xn −xj) if j < i ≤n −1.
(3) For the sake of completeness, we provide a simple proof for the above lemma which uses a completely dierent argument than that of [8, Lemma XV.2.2].
Proof. First of all observe that if (2) holds for some function f and f ∗: I →R is a function satisfying f(xi) = f ∗(xi) for i = 0, 1, . . . , n, then (2) is also satis ed by f ∗instead of f. To utilize this observation, we show that, for every function f : I →R, there exist constants a, c0, c1, . . . , cn−1 such that the function f ∗: I →R de ned by f ∗(x) := a + n−1 X i=0 ci(x −xi)+ (4) satis es f(xi) = f ∗(xi) (i = 0, 1, . . . , n), (5) where the positive part t+ of a real number t is de ned by t+ := max(0, t).
Indeed, we can easily see that (5) is equivalent to the following system of linear equations f(x0) = a, f(x1) = a + c0(x1 −x0), f(x2) = a + c0(x2 −x0) + c1(x2 −x1), .
.
.
f(xn) = a + c0(x2 −x0) + c1(x2 −x1) + · · · + cn−1(xn −xn−1), which can be solved recursively for a, c0, c1, . . . , cn−1. Thus, due to the above observation, (2) is satis ed for all functions f if and only if it is valid for all functions f ∗of the form (4). Since the identity (2) is linear in f, it is su cient to show that (2) is valid for the functions f ∗ −1(x) := 1, f ∗ 0 (x) := (x −x0)+, . . . , f ∗ n−1(x) := (x −xn−1)+.
222 Kazimierz Nikodem and Zsolt Páles Clearly, (2) holds with f = f ∗ −1 and f = f ∗ 0 identically (because these func-tions are polynomials of degree at most one on the interval [x0, xn] and hence their second-order divided dierences equal 0). Observe also that f ∗ k is an at most rst degree polynomial on the interval [xi−1, xi+1] if k is dierent from i. Therefore, substituting f = f ∗ k into (2), we obtain λkf ∗ k[xk−1, xk, xk+1] = f ∗ k[x0, xj, xn] for (k = 1, . . . , n−1). Hence, with the choice λk := f ∗ k [x0,xj,xn] f ∗ k [xk−1,xk,xk+1] for (k = 1, . . . , n −1), (2) holds for f = f ∗ k (k = 1, . . . , n −1).
Thus it holds also for all functions of the form (4). Now a simple computation yields that λ1, . . . , λk are of the form (3) and then the inequalities λk > 0 can be checked directly.
Finally, substituting f(x) = x2 into (2), it follows that λ1 + · · · + λn−1 = 1 also holds.
An obvious consequence of the previous lemma is the following result which we call the chain inequality in the sequel.
Corollary 1. (Chain Inequality) Let f : I →R and x0 < x1 < · · · < xn (n ≥2) be arbitrary points in I. Then, for all xed 0 < j < n, min 1≤i≤n−1 f[xi−1, xi, xi+1] ≤f[x0, xj, xn] ≤ max 1≤i≤n−1 f[xi−1, xi, xi+1].
3 Convexity Triplets It is easy to check that a function f : I →R is convex if and only if f[x, y, z] ≥ 0 for (x, y, z) ∈I3 with x < y < z. Motivated by this characterization of convexity, a triplet (x, y, z) in I3 with x < y < z is called a convexity triplet for the function f : I →R if f[x, y, z] ≥0 and the set of all convexity triplets of f is denoted by C(f). Using this terminology, f is t-convex if and only if x, tx + (1 −t)y, y , x, (1 −t)x + ty, y ∈C(f) for x, y ∈I with x < y.
Applying the chain inequality established in Corollary 1, we can deduce the following chain rule for convexity triplets.
Corollary 2. (Chain Rule) Let f : I →R and x0 < x1 < · · · < xn (n ≥2) be arbitrary points in I such that (xi−1, xi, xi+1) is in C(f) for all i = 1, . . . , n−1.
Then (x0, xj, xn) ∈C(f) (6) for all 0 < j < n.
Proof. We have that f[xi−1, xi, xi+1] ≥0 for all i = 1, . . . , n −1. Therefore, by the chain inequality, f[x0, xj, xn] ≥0; i.e., (6) holds for all j = 1, . . . , n − 1.
On t-Convex Functions 223 As applications of the above Corollary, we derive two well known results on the connection of t- and Jensen-convexity. The second statement of the next Corollary was proved by Kuhn by using Rodé's theorem. An elementary proof for this fact was rst found by Daróczy and Páles .
Corollary 3. Let f : I →R. If f is Jensen-convex, then it is t-convex for all rational t in ]0, 1[. Conversely, if f is t-convex for some t ∈]0, 1[, then it is also Jensen-convex.
Proof. For the rst statement, assume f is Jensen-convex and let t = j/n where 0 < j < n are integers.
Let x, y be xed and assume that y < x.
(The case x < y can be treated similarly.) De ne xi by xi := i nx + n−i n y for (i = 0, . . . , n). Then x0 = y < x1 < · · · < xn = x, furthermore, xi−1+xi+1 2 = xi for all i = 1, . . . , n −1. Therefore, by the Jensen-convexity of f, we have that (xi−1, xi, xi+1) belongs to C(f). Hence, by the chain rule for convexity triplets, we get that (6) holds. Thus f[x0, xj, xn] ≥0; i.e., f[y, tx + (1 −t)y, x] ≥0, which shows that f is t-convex, indeed.
To prove the converse, assume that f is t-convex for some t ∈]0, 1[. To prove the Jensen-convexity of f, let x, y ∈I with x < y be arbitrary. De ne the points x0, x1, x2, x3, x4 by x0 := x, x1 := tx+(1−t)x + y 2 , x2 := x + y 2 , x3 := tx + y 2 +(1−t)y, x4 := y.
Then, for i = 1 and for i = 3, we obviously have xi = txi−1 + (1 −t)xi+1.
Furthermore x2 = (1 −t)x1 + tx3.
Hence, due to the t-convexity of f, (xi−1, xi, xi+1) ∈C(f) for i = 1, 2, 3. Thus, by the chain rule, we get that (x0, x2, x4) ∈C(f); i.e., f is Jensen-convex.
4 Main Results Our main results oer mean value theorems in terms of the lower 2nd-order generalized derivatives de ned by δ2f(ξ) := lim inf (x,y)→(ξ,ξ) ξ,u∈co{x,y} 2f[x, u, y] for ξ ∈I, (7) δ2 tf(ξ) := lim inf (x,y)→(ξ,ξ) ξ∈co{x,y} 2f[x, tx + (1 −t)y, y] for ξ ∈I, (8) where, in the second de nition, t ∈]0, 1[ is a xed parameter. Clearly, δ2 tf(ξ) ≥δ2f(ξ) (9) 224 Kazimierz Nikodem and Zsolt Páles for all ξ ∈I and t ∈]0, 1[. One can also easily show that if f is twice continu-ously dierentiable at ξ, then δ2 tf(ξ) = δ2f(ξ) = f ′′(ξ).
Theorem 1. (Mean Value Inequality for t-convexity) Let I ⊆R be an interval, f : I →R, t ∈]0, 1[, and let x, y ∈I with x ̸= y. Then there exists a point ξ ∈co{x, y} such that tf(x) + (1 −t)f(y) −f(tx + (1 −t)y) t(1 −t)(x −y)2 = f[x, tx+(1−t)y, y] ≥δ2 tf(ξ) 2 . (10) Proof. In the sequel, a triplet (x, u, z) ∈I3 will be called a t-triplet if either u = tx + (1 −t)y or u = (1 −t)x + ty. Let x and y be distinct elements of I. Without loss of generality, we may assume that x < y. In what follows, we intend to construct a sequence of t-triplets (xn, un, yn) such that x = x0 ≤x1 ≤x2 ≤. . . , y = y0 ≥y1 ≥y2 ≥. . . , xn < un < yn (n ∈N), (11) |yn −xn| ≤ max(t, 1 −t) n|y −x| (n ∈N), (12) and f[x, u, y] = f[x0, u0, y0] ≥f[x1, u1, y1] ≥f[x2, u2, y2] ≥. . . .
(13) De ne (x0, u0, y0) = (x, tx + (1 −t)y, y) and assume that we have constructed (xn, un, yn). Now set zn,0 := xn, zn,1 := (1−t)xn+tun, zn,2 := un, zn,3 := tun+(1−t)yn, zn,4 := yn.
Then, clearly, (zn,0, zn,1, zn,2) and (zn,2, zn,3, zn,4) are t-triplets. On the other hand, we have that un = snxn + (1 −sn)yn, where either sn = t or sn = 1 −t.
Thus, snzn,1 + (1 −sn)zn,3 = sn (1 −t)xn + t(snxn + (1 −sn)yn) + (1 −sn) t(snxn + (1 −sn)yn) + (1 −t)yn = un; that is, (zn,1, zn,2, zn,3) is also a t-triplet.
Using the Chain Inequality, we get that there exists an index i ∈{1, 2, 3} such that f[xn, un, yn] ≥f[zn,i−1, zn,i, zn,i+1]. Finally, let (xn+1, un+1, yn+1) := (zn,i−1, zn,i, zn,i+1).
The sequence so constructed clearly satis es (11) and (13). We prove (12) by induction. It is obvious for n = 0. Assume that it holds for n. Then |yn+1 −xn+1| ≤max 1≤i≤3 |zn,i+1 −zn,i−1| = max(1 −sn, 1 −t, sn)|yn −xn| = max(t, 1 −t)|yn −xn| ≤ max(t, 1 −t) n+1|y −x|.
On t-Convex Functions 225 Thus (12) is also veri ed.
Due to the monotonicity properties of the sequences (xn), (yn) and also (12), there exists a unique element ξ ∈[x, y] such that T∞ i=0[xn, yn] = {ξ}.
Then, by (13), we get that f[x, u, y] ≥lim inf n→∞f[xn, un, yn] ≥ lim inf (v,w)→(ξ,ξ) ξ∈co{v,w} f[v, tv + (1 −t)w, w] = δ2 tf(ξ) 2 , which completes the proof of the theorem.
Corollary 4. (Mean Value Inequality for convexity) Let I ⊆R be an interval, f : I →R, and let x, u, y ∈I with x < u < y. Then there exists a point ξ ∈[x, y] such that f[x, u, y] ≥δ2f(ξ) 2 .
Proof. Choose t ∈]0, 1[ so that u = tx+(1−t)y. Then, by Theorem 1, there exists ξ ∈[x, y] such that (10) holds. Therefore, by (9), f[x, u, y] = f[x, tx + (1 −t)y, y] ≥δ2 tf(ξ) 2 ≥δ2f(ξ) 2 .
If one replaces f by −f in the above results, then mean value inequality for the upper 2nd-order generalized derivatives can be deduced that are de ned via (7) and (8) with limsup instead of liminf.
As an immediate consequence of the above theorem, we get the following characterization of convexity and t-convexity.
Corollary 5. Let t ∈]0, 1[. A function f : I →R is t-convex (resp. convex) on I if and only if δ2 tf(ξ) ≥0 (resp. δ2f(ξ) ≥0) for ξ ∈I.
Proof. If f is t-convex, then, clearly δ2 tf ≥0. Conversely, if δ2 tf is nonneg-ative on I, then, by the previous Theorem f[x, tx + (1 −t)y, y] ≥0 for all x, y ∈I; i.e., f is t-convex. A similar argument shows that the convexity of f is characterized by the nonnegativity of δ2f.
Another obvious but interesting consequence of Corollary 5 is that the t-convexity property (and also convexity) is localizable in the following sense.
Corollary 6. Let t ∈]0, 1[. A function f : I →R is t-convex (resp. convex) on I if and only if, for each point ξ ∈I, there exists a neighborhood U of ξ such that f is t-convex (resp. convex) on I ∩U.
The localizability of convexity for upper semicontinuous functions was also proved in .
In the literature, there are some other de nitions for local 226 Kazimierz Nikodem and Zsolt Páles convexity and local Jensen-convexity. For instance, following Cardinali and Papalini , we call a function f : I →R J∗-convex at a point p ∈I if there is a neighborhood U of p such that f x+p 2 ≤f(x)+f(p) 2 for x ∈U. Another de nition is motivated by Kostyrko . We say that a function f : I →R is locally Jensen-convex at a point p ∈I if there exists a positive number δ such that f(p) ≤f(p −h) + f(p + h) 2 for 0 < h < δ.
Note, however, that neither J∗-convex functions, nor locally Jensen-convex functions in the sense of Kostyrko need not be Jensen-convex. For instance, the function g : R →R de ned by g(x) := |x| for −1 ≤x < 1 and then extended periodically to R is J∗-convex but not Jensen-convex. The function h : R →R de ned as h(x) := x −[x] for noninteger x and h(x) := 1/2 for integer x is locally Jensen-convex in the sense of Kostyrko but it is not Jensen-convex (cf.
).
The following corollary derives t-convexity from a formally weaker prop-erty, namely from the local γ-th order approximate t-convexity. A function f : I →R is called approximately t-convex of order γ on I (where γ ≥0) if there exists a nonnegative constant c such that f(tx + (1 −t)y) ≤tf(x) + (1 −t)f(y) + ct(1 −t)|x −y|γ (14) for all x, y ∈I. If the above inequality holds for all t ∈[0, 1] with a constant c independent of t, then we say that f is an approximately convex function of order γ. If each point of I has a neighborhood such that f restricted to this neighborhood is approximately t-convex (resp. convex) of order γ, then we say that f is locally approximately t-convex (resp. convex) of order γ.
Approximately convex functions of rst-order were introduced by Páles in . First-order approximately Jensen-convex functions were investigated by Házy and Páles .
The next result shows that local approximate t-convexity (resp. local ap-proximate convexity) of order higher than 2 is equivalent to t-convexity (resp.
convexity). It is also related to a result of Rolewicz , stating that if a function f is γ-paraconvex; that is, f(tx + (1 −t)y) ≤tf(x) + (1 −t)f(y) + c|x −y|γ for x, y ∈I, t ∈[0, 1] and γ > 2, then it is convex.
Corollary 7. Let t ∈]0, 1[. Assume that, for some γ > 2, f : I →R is a locally approximately t-convex (resp. convex) function of order γ. Then f is t-convex (resp. convex).
On t-Convex Functions 227 Proof. We prove the statement concerning t-convexity. Let ξ ∈I be arbi-trary. By the assumption, there exists a neighborhood U of ξ and c ≥0 such that (14) holds for all x, y ∈U ∩I. Then f[x, tx + (1 −t)y, y] = tf(x) + (1 −t)f(y) −f(tx + (1 −t)y) t(1 −t)(x −y)2 ≥−c|x −y|γ−2 for x, y ∈I with x ̸= y. Thus, upon taking the liminf as (x, y) →(ξ, ξ), we get that δ2 tf(ξ) = lim inf (x,y)→(ξ,ξ) ξ∈co{x,y} 2f[x, tx + (1 −t)y, y] ≥ lim inf (x,y)→(ξ,ξ) ξ∈co{x,y} −2c|x −y|γ−2 = 0.
Therefore, by Corollary 5, f is t-convex.
We note that, in Corollary 7, the lower bound 2 for γ cannot be improved, because the function f(x) = −x2 is obviously approximately convex of order 2 and not t-convex for any t ∈]0, 1[.
References T. Cardinali and F. Papalini, Una estensione del concetto di midpoint convessitá per multifunzioni, Riv. Math. Univ. Parma, (4) 15 (1989), 119131.
Z. Daróczy and Zs. Páles, Convexity with given in nite weight sequences, Stochastica, 11 (1987), no. 1, 512.
A. Gilányi and Zs. Páles, On Dinghas-type derivatives and convex func-tions of higher order, Real Anal. Exchange, 27 (2001/2002), no. 2, 485 493.
A. Házy and Zs. Páles, On approximately midconvex functions, Bull. Lon-don Math. Soc., accepted.
G. Hämmerlin and K. H. Homan, Numerische Methematik, Springer, BerlinHeidelberg, 1994.
F. B. Hildebrand, Introduction to Numerical Analysis, McGrawHill, New York, 1956.
P. Kostyrko, On a local form of Jensen's functional equation, Aequationes Math., 30 (1986), no. 1, 6569.
228 Kazimierz Nikodem and Zsolt Páles M. Kuczma, An Introduction to the Theory of Functional Equations and Inequalities, Pa«stwowe Wydawnictwo Naukowe Uniwersytet l¡ski, WarszawaKrakówKatowice, 1985.
N. Kuhn, A note on t-convex functions, General Inequalities, 4 (Ober-wolfach, 1983) (W. Walter, ed.), International Series of Numerical Math-ematics, vol. 71, Birkhäuser, BaselBostonStuttgart, 1984, 269276.
Gy. Maksa, K. Nikodem, and Zs. Páles, Results on t-Wright convexity, C.
R. Math. Rep. Acad. Sci. Canada, 13 (1991), no. 6, 274278.
Zs. Páles, Nonconvex functions and separation by power means, Math.
Ineq. Appl., 3 (2000), no. 2, 169176.
Zs. Páles, On approximately convex functions, Proc. Amer. Math. Soc., 131 (2003), no. 1, 243252.
T. Popoviciu, Les fonctions convexes, Herman et Cie, Paris, 1944.
A. W. Roberts and D. E. Varberg, Convex Functions, Academic Press, New YorkLondon, 1973.
S. Rolewicz, On γ-paraconvex multifunctions, Math. Japonica, 24 (1979), no. 3, 293300.
E. M. Wright, An inequality for convex functions, Amer. Math. Monthly, 61 (1954), 620622. |
12840 | https://math.stackexchange.com/questions/3425541/finding-the-maximum-value-of-elements-to-be-selected-in-a-grid-zio-2009-p1 | Skip to main content
Finding the maximum value of elements to be selected in a grid - ZIO 2009, P1
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Hello Community! The above problem you see is a problem I got wrong. :( This is ZIO 2009, P1.
I tried the problem and miserably found the wrong answer as 20. Here is how my approach goes - part (a): Notice that the largest element in the whole grid is 16 which appears two times. I may be a good decision to start there to maximize the score but unfortunately, it is covered by only negative numbers. Although if we try to start, with the upper 16, we get value: 16−9+13=20. Similarly, starting with other big numbers we observe that the value gets even lesser so the answer must be 20. However, like most trial and error attempts is optimization problems, this is wrong as the answer is 29.
Now the main question I have for this problem is: How do we ensure maximum value? Is there some sort of an algorithm or something that we can follow and can be assured to have found the maximum value? Note that this problem is from a pen and paper exam where you are given 10 minutes to solve one sub-case (that is 30 minutes for this whole problem), so complete trial and error is of no use at all.
I asked a similar problem on MSE only: link but haven't got any answers till now... Any help there would also be appreciated.
The answers are 29,9,20.
I would be grateful if anyone could help.. Thanks!
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Starting in the upper left corner, replace each number x with x plus the larger of the number above it and the number to the left of it. In (a) this results in
−210−61−8−1−67319−46−719100−61423−42−2147
You must exit at the bottom row or the rightmost column, and you want to exit at the biggest exit number, which is the 23 in the bottom row. Now trace your way back to the left and up from that 23, always choosing the larger of the two possible numbers. This takes you left to 10, then left to 19 (or up to 19, it doesn't matter), then up to 3, up to 7, left (or up) to −6, up to 10, up to −2. The smallest number on the way was the −6, so that path will give you 23−(−6)=29, which is the maximum.
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answered Nov 8, 2019 at 6:51
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Let a(i,j) be the given values, and let b(i,j) be the maximum value attainable starting from cell (i,j). Then
b(i,j)=a(i,j)+⎧⎩⎨⎪⎪⎪⎪⎪⎪0max(0,b(i,j+1))max(b(i+1,j),0)max(b(i+1,j),b(i,j+1))if i=j=nif i=n and j<nif i<n and j=nif i<n and j<n
The resulting values of b(i,j) for part (a) are:
```
23 18 17 20 16
25 13 16 1 12
13 29 6 13 6
23 16 20 -2 10
13 20 4 13 -16
```
The largest value is b(3,2)=29.
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12841 | https://www.sciencealert.com/forget-what-you-know-science-says-you-actually-can-get-pregnant-while-pregnant | Forget What You Know, You Actually Can Get Pregnant While Pregnant
In a disturbing twist that goes against everything we learned in sex education, a recent New York TimesQ&A confirmed that it's actually possible for a woman to get pregnant… while she's already pregnant.
As far-fetched as that sounds, the science on this one actually lines up. In a creepy quirk of the human body, under very exceptional circumstances, a woman can continue to ovulate while pregnant and can conceive another child - something known as 'superfetation'.
That means a woman can have two foetuses developing inside her at the same time, both at different stages of development. And we sort of wish we could immediately scrub that knowledge from our brains.
Before you panic, this is highly unlikely to happen to you or anyone you know. In fact, according to a paper published in 2008 in the European Journal of Obstetrics and Gynaecologyit's only happened 10 times in the scientific literature… ever.
Many claims of double pregnancies over the years that have gone on to be debunked, but, according to the researchers, at least 10 of those seem to be the real, evidence-backed deal.
And since that paper came out, there have been a few other cases reported.
In 2009, a couple from Arkansas became pregnant after already conceiving two and a half weeks earlier. Both babies were delivered healthily by c-section on 2 December, with one baby measurably a fortnight more premature than the other.
And in 2015, an Australian couple gave birth to two girls that were 10 days apart in age at birth.
So, how is this even possible? Superfetation is actually quite common in mammals outside of humans, and has been seen in species including rodents, rabbits, horse, sheep, and kangaroos.
Sometimes these mammals have two uteri to facilitate the double pregnancy, or sometimes their menstrual cycle simply continues during pregnancy. It's even considered a handy reproductive strategy in some species.
But, in humans, superfetation appears to be a very rare accident - that's because, as soon as a woman becomes pregnant, her body actively blocks a second pregnancy from happening.
"Ordinarily, the release of eggs ceases once a woman is pregnant, and the hormonal and physical changes of pregnancy work together to prevent another conception," C. Clairborne Ray explained in a New York Times' science Q&A this week.
But for some reason, in superfetation, a pregnant women still manages to ovulate. A male's sperm then manages to fertilise that egg, somehow bypassing the the mucus plug that blocks up a woman's cervix once she's conceived.
Finally, implantation has to occur - which is an incredibly delicate process even in ordinary pregnancies. And when a woman is already pregnant, her hormones should make the uterus an unfavourable environment for another fertilised egg to implant (not to mention that there wouldn't be much room).
"In order for superfetation to occur in humans … it would appear that three seemingly impossible things need to happen," Khalil A. Cassimally reported for Scientific Americanback in 2011.
"Ovulation must take place during an ongoing pregnancy, semen must somehow find its way through the blocked cervix to the oviduct, via the occupied uterus and finally, the conceptus has to successfully implant itself in an unsuspecting already-occupied uterus."
Unlike twins - which occur either when a fertilised egg splits into two, or when two eggs are fertilised by two sperm at the same time - superfetation leads to a woman being pregnant with an additional foetus that's younger than the existing pregnancy.
So far, no cases have been reported of the age gap being greater than a few weeks.
According to the 2008 paper in the European Journal of Obstetrics and Gynaecology,whenever superfetation was confirmed to occur, the two foetuses had a separate amniotic sac. And they differed in size throughout the pregnancy and after birth.
Given the small sample size scientists have to work with, it's not yet clear why superfetation sometimes occurs, and whether whether there are any risk factors that can increase the odds of the phenomenon.
But Casimally notes that reports of superfetation are more common in women who've undergone fertility treatments, which could explain how one or two of those checkpoints get passed.
Despite how rare superfetation is, however, surprisingly, most babies conceived through the strange accident end up surviving.
The main risk is the fact that the babies are still born at the same time, despite their age difference - so one of the babies has the additional challenge of being born premature.
If your brain can handle it, you can find out more about superfetation here and here. The more you know…
Get the latest science first |
12842 | https://en.wikipedia.org/wiki/Formal_charge | Jump to content
Formal charge
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From Wikipedia, the free encyclopedia
Hypothetical charge assigned to an atom in a molecule based on its valence shell
In chemistry, a formal charge (F.C. or q), in the covalent view of chemical bonding, is the hypothetical charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. In simple terms, formal charge is the difference between the number of valence electrons of an atom in a neutral free state and the number assigned to that atom in a Lewis structure. When determining the best Lewis structure (or predominant resonance structure) for a molecule, the structure is chosen such that the formal charge on each of the atoms is as close to zero as possible.
The formal charge of any atom in a molecule can be calculated by the following equation:
where V is the number of valence electrons of the neutral atom in isolation (in its ground state); L is the number of non-bonding valence electrons assigned to this atom in the Lewis structure of the molecule; and B is the total number of electrons shared in bonds with other atoms in the molecule. It can also be found visually as shown below.
Formal charge and oxidation state both assign a number to each individual atom within a compound; they are compared and contrasted in a section below.
Examples
[edit]
Example: CO2 is a neutral molecule with 16 total valence electrons. There are different ways to draw the Lewis structure
Carbon single bonded to both oxygen atoms (carbon = +2, oxygens = −1 each, total formal charge = 0)
Carbon single bonded to one oxygen and double bonded to another (carbon = +1, oxygendouble = 0, oxygensingle = −1, total formal charge = 0)
Carbon double bonded to both oxygen atoms (carbon = 0, oxygens = 0, total formal charge = 0)
Even though all three structures gave us a total charge of zero, the final structure is the superior one because there are no charges in the molecule at all.
Pictorial method
[edit]
The following is equivalent:
Draw a circle around the atom for which the formal charge is requested (as with carbon dioxide, below)
Count up the number of electrons in the atom's "circle." Since the circle cuts the covalent bond "in half," each covalent bond counts as one electron instead of two.
Subtract the number of electrons in the circle from the number of valence electrons of the neutral atom in isolation (in its ground state) to determine the formal charge.
The formal charges computed for the remaining atoms in this Lewis structure of carbon dioxide are shown below.
It is important to keep in mind that formal charges are just that – formal, in the sense that this system is a formalism. The formal charge system is just a method to keep track of all of the valence electrons that each atom brings with it when the molecule is formed.
Usage conventions
[edit]
In organic chemistry convention, formal charges are an essential feature of a correctly rendered Lewis–Kekulé structure, and a structure omitting nonzero formal charges is considered incorrect, or at least, incomplete. Formal charges are drawn in close proximity to the atom bearing the charge. They may or may not be enclosed in a circle for clarity.
In contrast, this convention is not followed in inorganic chemistry. Many workers in organometallic and a majority of workers in coordination chemistry will omit formal charges, unless they are needed for emphasis, or they are needed to make a particular point. Instead a top-right corner ⌝ will be drawn following the covalently-bound, charged entity, in turn followed immediately by the overall charge.
The top-right corner ⌝ is sometimes replaced by square brackets enclosing the entire charged species, again with the total charge written in the upper right corner, just outside the brackets.
This difference in practice stems from the relatively straightforward assignment of bond order, valence electron count, and hence, formal charge for compounds only containing main-group elements (though oligomeric compounds like organolithium reagents and enolates tend to be depicted in an oversimplified and idealized manner), but transition metals have an unclear number of valence electrons so there is no unambiguous way to assign formal charges.
Formal charge compared to oxidation state
[edit]
The formal charge is a tool for estimating the distribution of electric charge within a molecule. The concept of oxidation states constitutes a competing method to assess the distribution of electrons in molecules. If the formal charges and oxidation states of the atoms in carbon dioxide are compared, the following values are arrived at:
The reason for the difference between these values is that formal charges and oxidation states represent fundamentally different ways of looking at the distribution of electrons amongst the atoms in the molecule. With the formal charge, the electrons in each covalent bond are assumed to be split exactly evenly between the two atoms in the bond (hence the dividing by two in the method described above). The formal charge view of the CO2 molecule is essentially shown below:
The covalent (sharing) aspect of the bonding is overemphasized in the use of formal charges since in reality there is a higher electron density around the oxygen atoms due to their higher electronegativity compared to the carbon atom. This can be most effectively visualized in an electrostatic potential map.
With the oxidation state formalism, the electrons in the bonds are "awarded" to the atom with the greater electronegativity. The oxidation state view of the CO2 molecule is shown below:
Oxidation states overemphasize the ionic nature of the bonding; the difference in electronegativity between carbon and oxygen is insufficient to regard the bonds as being ionic in nature.
In reality, the distribution of electrons in the molecule lies somewhere between these two extremes. The inadequacy of the simple Lewis structure view of molecules led to the development of the more generally applicable and accurate valence bond theory of Slater, Pauling, et al., and henceforth the molecular orbital theory developed by Mulliken and Hund.
See also
[edit]
Oxidation state
Valence (chemistry)
Coordination number
References
[edit]
^ Jump up to: a b Hardinger, Steve. "Formal Charges" (PDF). University of California, Los Angeles. Archived from the original (PDF) on 12 March 2016. Retrieved 11 March 2016.
^ Jump up to: a b c d "Formal Charge". Royal Society of Chemistry. Retrieved 10 December 2021.
^ "Chapter 48, Organometallic Chemistry". Organic chemistry. Clayden, Jonathan. Oxford: Oxford University Press. 2001. pp. 1311-1314. ISBN 0198503474. OCLC 43338068.{{cite book}}: CS1 maint: others (link)
Retrieved from "
Categories:
Chemical bonding
Electric charge
Hidden categories:
CS1 maint: others
Articles with short description
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12843 | https://www.wyzant.com/resources/answers/924171/i-m-not-sure-about-this-how-do-i-prove-the-midline-theorem-what-is-the-diag | WYZANT TUTORING
Eric J.
I'm not sure about this. How do I PROVE the midline theorem? What is the diagonal supposed to indicate? A clear explanation could really help. Thank you!
536.If M and N are the midpoints of the non-parallel sides of a trapezoid, it makes sense to call the segment MN the mid- line of the trapezoid. Why? (It actually should be called the midsegment, of course. Strange to say, some textbooks call it the median). Suppose that the parallel sides of a trapezoid have lengths 7 and 15. What is the length of the midline of the trape- zoid? Notice that the midline is divided into two pieces by a diagonal of the trapezoid. What are the lengths of these pieces? Does it matter which diagonal is drawn?
Mark M.
03/26/23
1 Expert Answer
Paul M. answered • 03/26/23
BS Mathematics, MD
First, it is easy to show that MN is parallel to the parallel sides of the trapezoid.
Then if you drop perpendiculars from the ends of the of the small base to the larger base, you can use similar triangles to show that the length of MN is 11.
Each diagonal divides MN into 2 segments and the diagonal is bisected by MN
Without the height of the trapezoid I think you do not have enough information to calculate the length of the diagonal.
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12845 | https://irrationalityblog.wordpress.com/getting-centered/the-four-centers-of-a-triangle/the-orthocenter/ | The Orthocenter
The orthocenter of a triangle is defined as the intersection of its altitudes. The altitude of a triangle is a segment passing through a vertex that is perpendicular to the line forming the opposite side.
Note: The orthocenter of a triangle isn’t always located on the inside of the triangle. It can either be inside the triangle, on a vertex of the triangle, or outside the triangle. The reason that it isn’t necessarily located inside the triangle is because the altitudes of a triangle can be located outside of the triangle, which means that the intersection of the altitudes can also lie outside the triangle.
I’ve broken down a list of specific triangles to take a look at those specific orthocenters:
Isosceles
Triangle ABC is an isosceles triangle, such that segments AB and AC are congruent. Point G is the orthocenter.
The altitude through point A is an axis of symmetry.
The altitude through A bisects angle A.
ΔABF ≅ ΔACF
The altitude through point A bisects segment BC.
Equilateral
Triangle ABC is an equilateral triangle (i.e. all sides and angles are congruent). Point G is the orthocenter.
Each altitude also bisects the side it intersects.
Each altitude is an axis of symmetry.
ΔABD ≅ ΔBCF ≅ ΔCAE
All triangles formed by the altitudes (triangles CGD, CGE, BGE, BGF, AGF, AGF) are congruent.
Acute
Triangle ABC is an acute triangle (i.e. all angles are less than 90°). Point G is the orthocenter.
Point G will always be located inside of the triangle.
Right
Triangle ABC is a right triangle, where A is the right angle.
The orthocenter of any right triangle is the same point as the vertex of the right angle of the triangle. In this figure, the orthocenter is point A.
Obtuse
Triangle ABC is an obtuse triangle, where B is greater than 90°. Point G is the orthocenter.
Point G will always be located outside of the triangle, “behind” point B.
Other
There are no special properties of the orthocenter of a scalene triangle.
Click here to go back to the Four Different Types of Centers page.
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12846 | https://www.math.columbia.edu/~woit/eulerformula.pdf | Euler’s Formula and Trigonometry Peter Woit Department of Mathematics, Columbia University September 10, 2019 These are some notes first prepared for my Fall 2015 Calculus II class, to give a quick explanation of how to think about trigonometry using Euler’s for-mula. This is then applied to calculate certain integrals involving trigonometric functions.
1 The sine and cosine as coordinates of the unit circle The subject of trigonometry is often motivated by facts about triangles, but it is best understood in terms of another geometrical construction, the unit circle.
One can define Definition (Cosine and sine). Given a point on the unit circle, at a counter-clockwise angle θ from the positive x-axis, • cos θ is the x-coordinate of the point.
• sin θ is the y-coordinate of the point.
The picture of the unit circle and these coordinates looks like this: 1 Some trigonometric identities follow immediately from this definition, in particular, since the unit circle is all the points in plane with x and y coordinates satisfying x2 + y2 = 1, we have cos2 θ + sin2 θ = 1 Other trignometric identities reflect a much less obvious property of the cosine and sine functions, their behavior under addition of angles. This is given by the following two formulas, which are not at all obvious cos(θ1 + θ2) = cos θ1 cos θ2 −sin θ1 sin θ2 sin(θ1 + θ2) = sin θ1 cos θ2 + cos θ1 sin θ2 (1) One goal of these notes is to explain a method of calculation which makes these identities obvious and easily understood, by relating them to properties of exponentials.
2 The complex plane A complex number c is given as a sum c = a + ib where a, b are real numbers, a is called the “real part” of c, b is called the “imaginary part” of c, and i is a symbol with the property that i2 = −1. For any complex number c, one defines its “conjugate” by changing the sign of the imaginary part c = a −ib The length-squared of a complex number is given by cc = (a + ib)(a −ib) = a2 + b2 2 which is a real number.
Some of the basic tricks for manipulating complex numbers are the following: • To extract the real and imaginary parts of a given complex number one can compute Re(c) =1 2(c + c) Im(c) = 1 2i(c −c) (2) • To divide by a complex number c, one can instead multiply by c cc in which form the only division is by a real number, the length-squared of c.
Instead of parametrizing points on the plane by pairs (x, y) of real numbers, one can use a single complex number z = x + iy in which case one often refers to the plane parametrized in this way as the “com-plex plane”. Points on the unit circle are now given by the complex numbers cos θ + i sin θ These go around the circle once starting at θ = 0 and ending up back at the same point when θ = 2π. Now the picture is 3 A remarkable property of complex numbers is that, since multiplying two of them gives a third, they provide something new and not at all obvious: a consistent way of multiplying points on the plane.
We will see in the next section that multiplication by a point on the unit circle of angle θ will have an interesting geometric interpretation, as counter-clockwise rotation by an angle θ.
3 Euler’s formula The central mathematical fact that we are interested in here is generally called “Euler’s formula”, and written eiθ = cos θ + i sin θ Using equations 2 the real and imaginary parts of this formula are cos θ = 1 2(eiθ + e−iθ) sin θ = 1 2i(eiθ −e−iθ) (which, if you are familiar with hyperbolic functions, explains the name of the hyperbolic cosine and sine).
In the next section we will see that this is a very useful identity (and those of a practical bent may want to skip ahead to this), but first we should address the question of what exactly the left-hand side means. The notation used implies that it is “the number e raised to the power iθ” and a striking example of this is the special case of θ = π, which says eiπ = −1 which relates three fundamental constants of mathematics (e, i, π) although these seem to have nothing to do with each other.
The problem though is that the idea of multiplying something by itself an imaginary number of times does not seem to make any sense.
To understand the meaning of the left-hand side of Euler’s formula, it is best to recall that for real numbers x, one can instead write ex = exp(x) and think of this as a function of x, the exponential function, with name “exp”.
The true signficance of Euler’s formula is as a claim that the definition of the exponential function can be extended from the real to the complex numbers, preserving the usual properties of the exponential. For any complex number c = a + ib one can apply the exponential function to get exp(a + ib) = exp(a) exp(ib) = exp(a)(cos b + i sin b) 4 The trigonmetric addition formulas (equation 1) are equivalent to the usual property of the exponential, now extended to any complex numbers c1 = a1+ib1 and c2 = a2 + ib2, giving ec1+c2 =ea1+a2ei(b1+b2) =ea1+a2(cos(b1 + b2) + i sin(b1 + b2)) =ea1+a2((cos b1 cos b2 −sin b1 sin b2) + i(sin b1 cos b2 + cos b1 sin b2)) =ea1(cos b1 + i sin b1)ea2(cos b2 + i sin b2) =ec1ec2 It is possible to show that eiθ = cos θ + i sin θ has the correct exponential property purely geometrically, without invoking the trigonometric addition for-mulas. One can do this by showing that multiplication of a point z = x + iy in the complex plane by eiθ rotates the point about the origin by a counter-clockwise angle θ. It then follows that multiplication by the product of eiθ1 and eiθ2 will be counterclockwise rotation by an angle θ1 + θ2, implying the correct exponential property eiθ1eiθ2 = ei(θ1+θ2) To show that multiplication by eiθ will give a rotation by θ, one can argue as follows. One can easily see that multiplication by eiθ rotates the point z = 1 along the unit circle by an angle θ, taking (in terms of real coordinates) (1, 0) →(cos θ, sin θ) This is also true for the point z = i, which gets taken to i(cos θ + i sin θ) = −sin θ + i cos θ. In terms of real coordinates on the plane, this is (0, 1) →(−sin θ, cos θ) and the rotation looks like this: 5 An arbitrary point on the plane is a linear combination of the points (1, 0) and (0, 1), and one can see that multiplication by eiθ will act as rotation by θ on any such linear combination, knowing that it does so for the cases of (1, 0) and (0, 1).
Two other ways to motivate an extension of the exponential function to complex numbers, and to show that Euler’s formula will be satisfied for such an extension are given in the next two sections.
3.1 eiθ as a solution of a differential equation The exponential functions f(x) = exp(cx) for c a real number has the property d dxf = cf One can ask what function of x satisfies this equation for c = i. Using the derivatives of the cosine and sine one finds d dx(cos x + i sin x) = −sin x + i cos x = i(cos x + i sin x) so cos x + i sin x has the correct derivative to be the desired extension of the exponential function to the case c = i.
3.2 eiθ and power series expansions By the end of this course, we will see that the exponential function can be represented as a “power series”, i.e. a polynomial with an infinite number of terms, given by exp(x) = 1 + x + x2 2! + x3 3! + x4 4! + · · · There are similar power series expansions for the sine and cosine, given by cos θ = 1 −θ2 2! + θ4 4! + · · · and sin θ = θ −θ3 3! + θ5 5! + · · · Euler’s formula then comes about by extending the power series for the expo-nential function to the case of x = iθ to get exp(iθ) = 1 + iθ −θ2 2! −iθ3 3! + θ4 4! + · · · and seeing that this is identical to the power series for cos θ + i sin θ.
6 4 Applications of Euler’s formula 4.1 Trigonometric identities Euler’s formula allows one to derive the non-trivial trigonometric identities quite simply from the properties of the exponential. For example, the addition for-mulas can be found as follows: cos(θ1 + θ2) =Re(ei(θ1+θ2)) =Re(eiθ1eiθ2) =Re((cos θ1 + i sin θ1)(cos θ2 + i sin θ2)) = cos θ1 cos θ2 −sin θ1 sin θ2 and sin(θ1 + θ2) =Im(ei(θ1+θ2)) =Im(eiθ1eiθ2) =Im((cos θ1 + i sin θ1)(cos θ2 + i sin θ2)) = cos θ1 sin θ2 + sin θ1 cos θ2 Multiple angle formulas for the cosine and sine can be found by taking real and imaginary parts of the following identity (which is known as de Moivre’s formula): cos(nθ) + i sin(nθ) =einθ =(eiθ)n =(cos θ + i sin θ)n For example, taking n = 2 we get the double angle formulas cos(2θ) =Re((cos θ + i sin θ)2) =Re((cos θ + i sin θ)(cos θ + i sin θ)) = cos2 θ −sin2 θ and sin(2θ) =Im((cos θ + i sin θ)2) =Im((cos θ + i sin θ)(cos θ + i sin θ)) =2 sin θ cos θ 7 4.2 Derivatives of trigonometric functions Writing the cosine and sine as the real and imaginary parts of eiθ, one can easily compute their derivatives from the derivative of the exponential. One has d dθ cos θ = d dθRe(eiθ) = d dθ(1 2(eiθ + e−iθ)) = i 2(eiθ −e−iθ) = −sin θ and d dθ sin θ = d dθIm(eiθ) = d dθ( 1 2i(eiθ −e−iθ)) =1 2(eiθ + e−iθ) = cos θ 4.3 Integrals of exponential and trigonometric functions Three different types of integrals involving trigonmetric functions that can be straightforwardly evaluated using Euler’s formula and the properties of expo-nentials are: • Integrals of the form Z eax cos(bx)dx or Z eax sin(bx)dx are typically done in calculus textbooks using a trick involving two inte-grations by parts. They can be more straightforwardly evaluated by using Euler’s formula to rewrite them as integrals of complex exponentials, for 8 instance Z eax cos(bx)dx =Re( Z eaxeibxdx) =Re( Z e(a+ib)xdx) =Re( 1 a + ibe(a+ib)x) + C =Re( a −ib a2 + b2 eaxeibx) + C =Re( a −ib a2 + b2 eax(cos(bx) + i sin(bx))) + C = 1 a2 + b2 eax(a cos(bx) + b sin(bx)) + C • Integrals of the form Z cos(ax) cos(bx)dx, Z cos(ax) sin(bx)dx or Z sin(ax) sin(bx)dx are usually done by using the addition formulas for the cosine and sine functions.
They could equally well be be done using exponentials, for instance (assuming a ̸= b) Z cos(ax) cos(bx)dx = Z 1 2(eiax + e−iax)1 2(eibx + e−ibx)dx =1 4 Z (ei(a+b)x + ei(a−b)x + e−i(a−b)x + e−i(a+b)x)dx =1 2 Z (cos((a + b)x) + cos((a −b)x))dx =1 2( 1 a + b sin((a + b)x) + 1 a −b sin((a −b)x)) + C • When a and b are integers m, n, and one integrates over an interval of size 2π (for instance [−π, π]), the above integrals give very simple results.
This is due to the fact that Z +π −π eimxeinxdx = Z +π −π ei(m+n)xdx = ( 0 if m ̸= n 2π if m = −n One can show this by integrating the exponential, or more simply by noticing that the real and imaginary parts of the answer will, for m ̸= −n, be given by integrating a cosine and sine over m + n periods, This gives zero since the area under the curves is the same above and below the x-axis. For m = −n, the integrand is just 1, so the integral is the length of the interval of integration.
9 • Integrals of the form Z cosm x dx, Z cosm x sinn x dx or Z sinm x dx are performed in calculus textbooks by a combination of use of the sub-stitution u = sin(x) or u = cos(x), of the identity cos2 x + sin2 x = 1 to turn even powers of the cosine into even powers of the sine (and vice-versa), as well as the double angle formulas for the cosine and sine. Such methods are often the simplest ones, but one can also do such integrals by expressing them in terms of exponentials. For example Z cos3 x dx = Z (1 2(eix + e−ix))3dx =1 8 Z (e3ix + 3eix + 3e−ix + e−i3x)dx =1 4 Z (cos(3x) + 3 cos(x))dx = 1 12 sin(3x) + 3 4 sin(x) + C Note that this technique will typically give answers in a different form than the technique used in the book, giving not powers of the cosine or the sine, but something equivalent related to these by multiple-angle formulas.
4.4 Polar coordinates Instead of Cartesian coordinates x and y, one can parametrize points in the plane by polar coordinates r (the distance from the origin) and θ (the angle with the positive x axis. A point with polar coordinates (r, θ) has (x, y) coordinates x = r cos θ, y = r sin θ 10 The point with polar coordinates (r, θ) has a simple complex coordinate, which, using Euler’s formula, will be given by z = x + iy = r cos θ + ir sin θ = reiθ 5 Problems 1. Compute Z eax sin(bx)dx for arbitrary real constants a and b.
2. Compute Z cos(ax) sin(bx)dx for arbitrary real constants a and b.
3. Compute R cos3 x dx using Euler’s formula, and show that the result is the same as what one gets by the textbook method (using a substitution u = sin x).
11 |
12847 | https://www.lessonplanet.com/teachers/ck-12-foundation-algebra-multi-step-equations-with-like-terms-general-math-7th-10th | Ck 12: Algebra: Multi Step Equations With Like Terms Unit Plan for 9th - 10th Grade | Lesson Planet
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Ck 12: Algebra: Multi Step Equations With Like Terms
Curated by ACT
[Free Registration/Login may be required to access all resource tools.] In this lesson students learn how to solve multi-step equations by combining like terms. Students examine guided notes, review guided practice, watch instructional videos and attempt practice problems.
3 Views 0 Downloads CCSS:Adaptable
Concepts
equations, like terms, solving equations
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Additional Tags
algebra, combine like terms, multistep equations, ck 12, ck-12, ck-12 foundation, ck12
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Classroom Considerations
Knovation Readability Score: 2 (1 low difficulty, 5 high difficulty)
Common Core
8.EE.C.7.b
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12848 | https://brilliant.org/wiki/wilsons-theorem/ | Wilson's Theorem
Jubayer Nirjhor,
Eeshan Banerjee,
Arron Kau,
and
Yash Jain
Ankit Chabarwal
A Former Brilliant Member
Rodrigo Paniza
Abdulrahman El Shafei
charisse mae tumapon
Calvin Lin
Jimin Khim
Eli Ross
contributed
Wilson's theorem states that
a positive integer n>1 is a prime if and only if (n−1)!≡−1(modn).
In other words, (n−1)! is 1 less than a multiple of n. This is useful in evaluating computations of (n−1)!, especially in Olympiad number theory problems. For example, since we know that 101 is a prime, we can conclude immediately that 100!=101k−1 for some integer k.
Contents
Explanation of Wilson's Theorem
Proof of Wilson's Theorem
Applications of Wilson's Theorem
Additional Problems with Wilson's Theorem
Explanation of Wilson's Theorem
This statement means two things, which are as follows:
(1) For a positive integer n (>1), if (n−1)!≡−1(modn), then n is a prime.
(2) If p is a prime number, then (p−1)!≡−1(modp) holds.
Notice how (1) provides us with a way to check if a number is prime. But this is really inefficient as factorials grow really fast. So it's hard to compute the modulus for sufficiently large n even on a computer. Fortunately, we've got better primality tests to save the world! But (2) is helpful in easing out computations and cracking several Olympiad number theory problems.
Let's verify Wilson's theorem for small values:
n2345678910(n−1)!12624120720504040320362880(n−1)!(modn)122406000 Is prime?yesyesnoyesnoyesnonono
Evaluate 30!(mod899).
First, factorize 899=29×31.
We have 30!=30×29×28!≡0(mod29).
From Wilson's theorem, 30!≡−1(mod31).
Hence, by the Chinese remainder theorem, we get that 30!≡464(mod899). □
Explain why 17!≡1(mod19).
From Wilson's theorem, we know that 18!≡−1(mod19). Hence, dividing both sides by 18≡−1(mod19), we conclude that 17!≡1(mod19). □
In fact, the above example generalizes to the following:
If p is a prime, then (p−2)!≡1(modp). □
If p is a prime, then by Wilson's theorem, we know that (p−1)!≡−1≡p−1(modp). Dividing by p−1=0, we get that
(p−2)!≡1(modp).
Hence, (p−2)!≡1(modp) for all primes p. □
Evaluate 97!(mod101).
Since 101 is prime, we know that 100!≡−1(mod101). Hence, 97!≡(−1)(−2)(−3)−1≡61(mod101).
We need to find the multiplicative inverse of 6, which is equal to 17. Thus
97!≡17(mod101). □
Proof of Wilson's Theorem
A positive integer n (>1) is a prime if and only if (n−1)!≡−1(modn). □
At first glance it seems that proving (1) is a really difficult job, but proving (2) shouldn't be that hard. Surprisingly, the situation is exactly opposite. Proofs of (1) and (2) are included separately below.
(1): Assuming n a composite number, we show a contradiction. If n is a composite number then it has at least one divisor d less than n, that is d≤n−1. But since (n−1)! is the product of all positive integers from 1 to n−1, the product must contain d and thus be divisible by d. So we have (n−1)!≡0(modd). Also (n−1)!≡0≡−1(modd) since d∣n, contradicting the hypothesis. So n can't be composite, hence prime. □
(2): Consider the field Zp. This is just the set of integers modulo p, i.e. contains all integers from 0 to p−1. All the operations are done in modulo p. For example, in this field 5+(p−2)=3 since 5+(p−2)=p+3≡3(modp). Now consider the polynomial f(x)=xp−1−1, which clearly has p−1 roots by the fundamental theorem of algebra. Also xp−1−1≡0(modp) for all 1≤x≤p−1 by Fermat's little theorem (See worked example 3 here) since p is a prime. So in Zp, these must be the p−1 roots of f. Hence we can write
xp−1−1=k=1∏p−1(x−k)=(−1)p−1k=1∏p−1(k−x)=k=1∏p−1(k−x)
because for odd primes, p−1 is even, implying (−1)p−1=1, and for even prime 2 we have (−1)p−1=−1≡1(mod2). Now simply plugging in x=0 in the last equation we get
−1=k=1∏p−1k=(p−1)!
in Zp. That means (p−1)!≡−1(modp). □
Applications of Wilson's Theorem
Prove that 437∣(18!+1).
First notice that 437=19×23 and so it suffices to prove that
18!≡−1(mod19),(mod23).
Now since 19 is a prime, 18!≡−1(mod19) immediately follows by Wilson's theorem. Also noting that 23 is a prime, we have
24×18!=≡=18!×(−1)×(−2)×(−3)×(−4)18!×19×20×21×2222!≡Wilson’s−1≡−24(mod23),
and thus 18!≡−1(mod23) as well.
Therefore, 437∣(18!+1). □
(ARML 2002)
Let a∈N such that
1+21+31+⋯+231=23!a.
Find a(mod13).
Rewrite the equation as
a=23!+223!+323!+⋯+2323!.
Clearly all the terms in the right side are integers. Also except 1323!, all the quotients contain the factor 13 and thus are divisible by 13. Therefore, we get
a=≡≡=23!+223!+323!+⋯+2323!1323!=1×2×⋯×12×14×⋯×23(1×2×⋯×12)×(1×2×⋯×10)12!×10!≡Wilson’s12×10!(mod13).
So we know that a≡12×10!(mod13). Now we use 11!≡1≡66(mod13), which follows by Wilson's. Hence 10!≡6(mod13). Finally we have
a≡12×10!≡12×6≡7(mod13),
which is the answer. □
Let p be an odd prime. Let A={a1,a2,…,ap} and B={b1,b2,…,bp} be complete sets of residue classes modulo p. Show that the set {a1b1,a2b2,…,apbp} is not a complete set of residue classes.
We will prove by contradiction. Suppose there exist sets A,B which give us a complete set of residue classes.
First, if there exists i=j such that ai≡bj≡0, then aibi≡ajbj≡0, which would not give us a complete set of residue classes. Thus, we may assume that ai≡bi≡0(modp). WLOG, i=p.
By Wilson's theorem, we get that i=1∏p−1ai≡−1(modp) and i=1∏p−1bi≡−1(modp).
If {aibi}i=1p−1 form a complete set of non-zero residue class, then we must have
−1≡i=1∏p−1aibi≡i=1∏p−1aii=1∏p−1bi≡(−1)×(−1)≡1(modp).
Since p is an odd prime, p>2 and we have −1≡1(modp), which is a contradiction. Hence, {aibi} is not a complete set of residue classes. □
Find the remainder when 2016!−2015! is divided by 2017.
You may use the fact that 2017 is prime.
The correct answer is: 2015
Additional Problems with Wilson's Theorem
What is the remainder when 18! is divided by 19?
Notation: ! denotes the factorial notation. For example, 8!=1×2×3×⋯×8.
The correct answer is: 18
Find the GCD of (19!+19,20!+19).
Note: GCD stands for the greatest common divisor.
1) Prove that if n (>4) is composite, then (n−1)!≡0(modn).
2) Let p=4k+1 be a prime. Show that there exists an integer n such that n2≡−1(modp).
3) Determine all positive integers a and n such that
(a−1)!+1=an.
4) (ISL 2000) Find all the integer solutions of (am+1)∣(a+1)n.
5) (IMO 1999/4 ) Solve for prime p, with x an integer such that x≤2p and
xp−1((p−1)x+1).
Cite as:
Wilson's Theorem.
Brilliant.org.
Retrieved
20:42, August 12, 2025,
from |
12849 | https://edia.app/worksheets/algebra_2/arithmetic_and_geometric_sequences/recursive_formulas_arithmetic | Algebra II > Arithmetic & geometric sequences
Recursive formulas for arithmetic sequences
This Algebra II arithmetic & geometric sequences worksheet generates free practice problems on recursive formulas for arithmetic sequences.
Recursive formulas for arithmetic sequences Worksheet
Name:
Date:Period:
1.
Write the recursive formula for the sequence below.
8
,
−
1
,
−
10
,
−
19
,
...
2.
Write the recursive formula for the sequence below.
−
10
,
−
8
,
−
6
,
−
4
,
...
3.
Write the recursive formula for the sequence below.
6
,
10
,
14
,
18
,
...
4.
Write the recursive formula for the sequence below.
12
,
14
,
16
,
18
,
...
5.
Given the recursive formula, find
b
4
.
{
b
1
=
−
2
b
n
=
b
n
−
1
+
5
6.
Given the recursive formula, find
f
6
.
{
f
1
=
−
6
f
n
=
f
n
−
1
−
4
7.
Given the recursive formula, find
b
3
.
{
b
1
=
8
b
n
=
b
n
−
1
+
5
8.
Given the recursive formula, write the common difference.
{
f
1
=
−
11
f
n
=
f
n
−
1
+
1
Free resources at edia.app
© 2025 Edia Learning, Inc. All Rights Reserved.
9.
Given the recursive formula, write the common difference.
{
d
1
=
−
1
d
n
=
d
n
−
1
−
8
10.
Given the recursive formula, write the common difference.
{
b
1
=
−
9
b
n
=
b
n
−
1
+
6
Free resources at edia.app
© 2025 Edia Learning, Inc. All Rights Reserved.
Recursive formulas for arithmetic sequences Answer key
1.
Write the recursive formula for the sequence below.
8
,
−
1
,
−
10
,
−
19
,
...
e
1
=
8
e
n
=
e
n
−
1
−
9
2.
Write the recursive formula for the sequence below.
−
10
,
−
8
,
−
6
,
−
4
,
...
e
1
=
−
10
e
n
=
e
n
−
1
+
2
3.
Write the recursive formula for the sequence below.
6
,
10
,
14
,
18
,
...
f
1
=
6
f
n
=
f
n
−
1
+
4
4.
Write the recursive formula for the sequence below.
12
,
14
,
16
,
18
,
...
e
1
=
12
e
n
=
e
n
−
1
+
2
5.
Given the recursive formula, find
b
4
.
{
b
1
=
−
2
b
n
=
b
n
−
1
+
5
b
4
=
13
6.
Given the recursive formula, find
f
6
.
{
f
1
=
−
6
f
n
=
f
n
−
1
−
4
f
6
=
−
26
7.
Given the recursive formula, find
b
3
.
{
b
1
=
8
b
n
=
b
n
−
1
+
5
b
3
=
18
8.
Given the recursive formula, write the common difference.
{
f
1
=
−
11
f
n
=
f
n
−
1
+
1
1
9.
Given the recursive formula, write the common difference.
{
d
1
=
−
1
d
n
=
d
n
−
1
−
8
−
8
10.
Given the recursive formula, write the common difference.
{
b
1
=
−
9
b
n
=
b
n
−
1
+
6
6
Free resources at edia.app
© 2025 Edia Learning, Inc. All Rights Reserved.
1.
Write the recursive formula for the sequence below.
8
,
−
1
,
−
10
,
−
19
,
...
2.
Write the recursive formula for the sequence below.
−
10
,
−
8
,
−
6
,
−
4
,
...
3.
Write the recursive formula for the sequence below.
6
,
10
,
14
,
18
,
...
4.
Write the recursive formula for the sequence below.
12
,
14
,
16
,
18
,
...
5.
Given the recursive formula, find
b
4
.
{
b
1
=
−
2
b
n
=
b
n
−
1
+
5
6.
Given the recursive formula, find
f
6
.
{
f
1
=
−
6
f
n
=
f
n
−
1
−
4
7.
Given the recursive formula, find
b
3
.
{
b
1
=
8
b
n
=
b
n
−
1
+
5
8.
Given the recursive formula, write the common difference.
{
f
1
=
−
11
f
n
=
f
n
−
1
+
1
9.
Given the recursive formula, write the common difference.
{
d
1
=
−
1
d
n
=
d
n
−
1
−
8
10.
Given the recursive formula, write the common difference.
{
b
1
=
−
9
b
n
=
b
n
−
1
+
6
1.
Write the recursive formula for the sequence below.
8
,
−
1
,
−
10
,
−
19
,
...
e
1
=
8
e
n
=
e
n
−
1
−
9
2.
Write the recursive formula for the sequence below.
−
10
,
−
8
,
−
6
,
−
4
,
...
e
1
=
−
10
e
n
=
e
n
−
1
+
2
3.
Write the recursive formula for the sequence below.
6
,
10
,
14
,
18
,
...
f
1
=
6
f
n
=
f
n
−
1
+
4
4.
Write the recursive formula for the sequence below.
12
,
14
,
16
,
18
,
...
e
1
=
12
e
n
=
e
n
−
1
+
2
5.
Given the recursive formula, find
b
4
.
{
b
1
=
−
2
b
n
=
b
n
−
1
+
5
b
4
=
13
6.
Given the recursive formula, find
f
6
.
{
f
1
=
−
6
f
n
=
f
n
−
1
−
4
f
6
=
−
26
7.
Given the recursive formula, find
b
3
.
{
b
1
=
8
b
n
=
b
n
−
1
+
5
b
3
=
18
8.
Given the recursive formula, write the common difference.
{
f
1
=
−
11
f
n
=
f
n
−
1
+
1
1
9.
Given the recursive formula, write the common difference.
{
d
1
=
−
1
d
n
=
d
n
−
1
−
8
−
8
10.
Given the recursive formula, write the common difference.
{
b
1
=
−
9
b
n
=
b
n
−
1
+
6
6
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Decimal To Percent Chart:
Get going through the chart below that is packed with most common decimal to percent ratios:
DecimalPercent (%)
0.011 %
0.022 %
0.033 %
0.044 %
0.055 %
0.066 %
0.077 %
0.088 %
0.099 %
0.1010 %
0.2020 %
0.3030 %
0.4040 %
0.5050 %
0.6060 %
0.7070 %
0.8080 %
0.9090 %
1.00100 %
How To Turn a Decimal Into a Percent?
Let’s pay heed to the following examples to learn better how to alter decimals into percents. Stay focused!
Example # 01:
How to convert a decimal to a percent given below:
0.323
Solution:
0.323 = 0.323 100
0.323 = 32.3 %
Example # 02:
Change decimal to percent using shortcut method given as follows:
0.2333
Solution:
As the given decimal number is:
0.2333
Moving decimal 2 times right:
23.33 %
Which is the required answer and can also be checked using this best decimal to percent calculator.
Example # 03:
How to convert decimals to percents given below:
0.23 and 1.345
Solution:
Here we have:
0.23 = 0.23 100
0.23 = 23%
Now;
1.345 = 1.345 100
1.345 = 134.5%
How Decimal To Percent Calculator Works?
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Corresponding percentage form of the given decimal number
Detailed calculations involved in the phenomenon
FAQ’s:
What is 0.007 as a percent?
The percent equivalent of the decimal 0.007 is 0.7%
How do you turn 0.4 to percent?
0.4 = 0.4 100
0.4 = 4%
What is the importance of fraction decimals in your daily life?
Fraction decimals are a very important concept in mathematics as they allow you to do many calculations on a daily basis. These include distributing pizza slices among friends, telling clock time, getting to know what fraction of baking soda you are required with to make a cake, and other like that. Moreover, decimals allow you to attain precision when whole numbers get insufficient to provide.
Why do we use decimal number system?
There is only one pet reason behind using this number system which is that human beings have only 10 fingers.
Conclusion:
Converting decimals to percents allows you to understand numbers properly and compare various fractions percentages as well. This is why this best decimal to fraction calculator is designed to make you feel comfortable while doing decimal to percentage calculations.
References:
From the source of Wikipedia: Percentage, Calculations, Decimal, Decimal notation
From the source of Khan Academy:decimals to percents, percents to decimals & fractions
From the source of Lumen Learning: Decimals and Fractions as Percents
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12852 | https://math.stackexchange.com/questions/2476928/proving-a2-b2-%E2%87%94-a-b | inequality - Proving: $a^2 < b^2 ⇔ |a| < |b|$ - Mathematics Stack Exchange
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Proving: a 2<b 2⇔|a|<|b|a 2<b 2⇔|a|<|b|
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Asked 7 years, 11 months ago
Modified7 years, 11 months ago
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This question shows research effort; it is useful and clear
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I started studying mechanical engineering and it works perfectly fine for me but i stumbled across this problem:
a 2<b 2⇔|a|<|b|
a 2<b 2⇔|a|<|b|
I found a solution but that took me a full piece of paper and I am sure it can be solved quicker. I would be very happy if someone could help me proving the (obvious equation).
Greetings, Finn
inequality
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edited Oct 17, 2017 at 17:17
Guy Fsone
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asked Oct 17, 2017 at 15:17
Finn EggersFinn Eggers
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Because f(x)=√x f(x)=x−−√ increases and from here we obtain: a 2<b 2⇔√a 2<√b 2⇔|a|<|b|.
a 2<b 2⇔a 2−−√<b 2−−√⇔|a|<|b|.
Your second inequality: We need to prove that |a+b|+|a−b|≥|a|+|b|
|a+b|+|a−b|≥|a|+|b|
or (a+b)2+2|a 2−b 2|+(a−b)2≥a 2+2|a b|+b 2
(a+b)2+2|a 2−b 2|+(a−b)2≥a 2+2|a b|+b 2
or a 2−2|a b|+b 2+2|a 2−b 2|≥0
a 2−2|a b|+b 2+2|a 2−b 2|≥0
or (|a|−|b|)2+2|a 2−b 2|≥0,
(|a|−|b|)2+2|a 2−b 2|≥0,
which is obvious.
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edited Oct 17, 2017 at 16:41
answered Oct 17, 2017 at 15:18
Michael RozenbergMichael Rozenberg
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That's it? I feel really stupid right now Finn Eggers –Finn Eggers 2017-10-17 15:20:14 +00:00 Commented Oct 17, 2017 at 15:20
@Finn Eggers It's all. I used also that √x 2=|x|x 2−−√=|x|.Michael Rozenberg –Michael Rozenberg 2017-10-17 15:22:48 +00:00 Commented Oct 17, 2017 at 15:22
I am currently stuck with an other inequality. It would be great if you could also help me with this one: |a + b| + |a−b| ≥ |a| + |b|Finn Eggers –Finn Eggers 2017-10-17 15:57:24 +00:00 Commented Oct 17, 2017 at 15:57
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12853 | https://www.nagwa.com/en/videos/687138210451/ | Question Video: Finding the Union of Two Sets of Numbers Involving Interval Notation | Nagwa
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Question Video: Finding the Union of Two Sets of Numbers Involving Interval Notation Mathematics • Second Year of Preparatory School
Given that 𝑋 = [−7, −6] and 𝑌 = [3, ∞), find 𝑋 ∪ 𝑌.
03:53
Video Transcript
Given that 𝑋 is equal to the closed interval from negative seven to negative six and 𝑌 is equal to the left-closed right-open interval from three to ∞, find 𝑋 union 𝑌.
In this question, we are given two sets, 𝑋 and 𝑌, in interval notation. And we need to find the union of these two sets. To do this, we can start by recalling that a union of sets means that we take all of the elements in both sets. So, we say that 𝑎 is a member of the union of 𝑋 and 𝑌 if 𝑎 is a member of 𝑋 or if 𝑎 is a member of 𝑌.
To find the union of these two intervals, we should start by determining exactly which numbers are members of each set. First, we note that 𝑋 is a closed interval. So, the endpoints of the interval are included. This means we want to include all real numbers between negative seven and negative six including the endpoints. We can follow a similar process for 𝑌. We note that 𝑌 is closed at three and unbounded above. So, the set 𝑌 includes all real values greater than or equal to three.
At this point, we could find the union of 𝑋 and 𝑌 using set builder notation. However, when working with intervals, it is often easier to visualize the sets using a number line first. First, 𝑋 is the set of real numbers between negative seven and negative six, including the endpoints. We can represent this on a number line using solid dots at negative seven and negative six to show that they are included in the set.
We can apply a similar process to 𝑌. We want to include all of the values greater than or equal to three in the set. So, we sketch a solid dot at three and a line to the right of three. We include an arrow on the right to show that the set is not bounded above.
There are a few different ways of finding an expression for the union of these sets. One way is to find a set that includes both 𝑋 and 𝑌. For instance, we can see that the set of all real values greater than or equal to negative seven encompasses both 𝑋 and 𝑌.
However, we can see that this set includes all of the real values between negative six and three, which are not elements of 𝑋 or 𝑌. So, we need to remove these values from our set to find the union of 𝑋 and 𝑌. This means that we want to remove the open interval from negative six to three from our set to find the union of 𝑋 and 𝑌.
Finally, we can recall that we can remove the elements of a set from another set by using the set minus operation. This means that we can write the union of 𝑋 and 𝑌 as the left-closed, right-open interval from negative seven to ∞ minus the open interval from negative six to three. It is also worth noting that we can use parentheses or reversed brackets to show that an interval does not include the endpoints. Both notations are common, and it is personal preference which one to use.
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Differentiability and continuity of f(x)=sin|k x|f(x)=sin|k x|
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QUESTION(MCQ) :Let f(x)=sin|k x|,x f(x)=sin|k x|,x is real. then
1.f is continuous nowhere.
2.f is continuous and differentiable everywhere except at integral values of k
3.f is continuous and differentiable everywhere.
4.f is continuous but nowhere differentiable.
MY VIEW :Easily discarding option 1,3and 4.But for option 2, I was trying to solve through drawing the graph. First checking for k=1,first I draw the graph on positive x axis and then making the graph symmetrical about y axis, I draw the graph on negative x axis. Sharpness arises only at x=0. Also for k=2,we first draw the graph on positive x axis by shrinking the graph 2 times along x axis and then making it symmetrical about y axis, I draw the graph. Sharpness arises only at x=0 .
HOW IS OPTION 2 CORRECT?PLEASE GIVE ARGUMENTS.
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edited Nov 7, 2018 at 12:08
amWhy
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Uttam SahuUttam Sahu
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The question is not properly stated. Is f f a function of x x or k k?. Does 2) sat that f f is differentiable at all points x x if k k is not an integer? If that is the interpretation then all options are false .(If k=0 k=0 then 3) is true.Kavi Rama Murthy –Kavi Rama Murthy 2018-10-31 09:46:39 +00:00 Commented Oct 31, 2018 at 9:46
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If k=0 k=0 the correct answer is (3). Let now k≠0 k≠0.
Since |k x|=|k|⋅|x||k x|=|k|⋅|x|, without loss of generality we can suppose k>0 k>0. Then, the function f(x)f(x) is just a rescaled version of g(x)=sin|x|g(x)=sin|x|. This function is continuous everywhere, and differentiable infinitely many times everywhere except at x=0 x=0, where there is a cusp. So, none of the options is correct.
However, if the function was not f(x)=sin|k x|f(x)=sin|k x|, but rather f(x)=|sin(k x)|f(x)=|sin(k x)|, it would have cusps at k x=n π k x=n π with n∈Z n∈Z, i.e. at integral values of π/k π/k. This function would be everywhere continuous and infinitely many times differentiable everywhere except at the cusps x n=n π/k x n=n π/k. This is something similar to (but not quite the same as) answer (2).
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edited Nov 7, 2018 at 12:03
answered Oct 31, 2018 at 10:27
Khalid Wenchao YjiboKhalid Wenchao Yjibo
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Yes, your argument should be correct . The book has given answer as option( 2.). Actually no option is correct for the stated question.Uttam Sahu –Uttam Sahu 2018-10-31 11:14:07 +00:00 Commented Oct 31, 2018 at 11:14
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For k=0 k=0 your function is identically zero and it is continuous and differentiable everywhere.
For k≠0 k≠0 your function
f(x)=|sin(k x)|f(x)=|sin(k x)|
is continuous everywhere because it is the composite function of two continuous functions.
Your function is differentiable everywhere except where it takes the value of zero.
That is where
sin(k x)=0⟺k x=n π⟺x=n π/k sin(k x)=0⟺k x=n π⟺x=n π/k
for integer values of n n
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answered Nov 7, 2018 at 12:18
Mohammad Riazi-KermaniMohammad Riazi-Kermani
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12857 | https://theteacherscafe.com/parentheses-brackets-and-braces-in-math-expressions-easier-version/ | Parentheses, Brackets and Braces in math Expressions - Easier Version
Evaluate Expressions with Parentheses, Brackets & Braces - This worksheet is the easier version and will work for fifth grade and up.
Evaluate Expressions with Parentheses, Brackets & Braces Answers - Just to save you time!
CCSS.Math.Content.5.OA.A.1 - Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols. |
12858 | https://fluentwords.net/en/collocations/meander/en | Typical usage patterns for 'meander' | FluentWords
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Top 10 collocations for meander
| # | Collocations for meander | Example Sentence |
---
| 1 | river meander | The river is known as the plain , meandering and extremely slow . |
| 2 | stream meander | The beautiful streammeandered across the valley like a silver ribbon , bordered by rushes and other aquatic plants . |
| 3 | it meander | It is sometimes referred to as the rectangular survey system , although non rectangular methods such as meandering can also be used . |
| 4 | creek meander | The creek has been channelized in some places , and meanders , islands and natural channel formations have been destroyed . |
| 5 | path meander | A single path then meanders through a shaded avenue to the main Park Gates . |
| 6 | to meander through valley | The river passes through the city and meanders through the valley , moving onward and deepening in the Wular Lake . |
| 7 | meander of the river seine | Geography A farming village situated in a meander of the river Seine , some west of Rouen , on the D65 road . |
| 8 | swastika meander | With relieved swastikameanders however , some examples of affinity with comparable mosaics from sites possessing Orpheus pavements is apparent . |
| 9 | to meander north | Next , in the 1740s and 1750s , French settlers meandered north from the Pointe Coupee Post in south Louisiana and named many of north Louisiana 's bayous and prairies . |
| 10 | brook meander | Springs of sweet water gushed from the hill-sides , and a beautiful brook , overshadowed by the lofty forest , meandered at its base . |
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river meander stream meander creek meander to meander through valley swastika meander to meander north
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to meander + Adverb
e.g. to meander north
north west
Subject + meander +(e)s
e.g. river meander +(e)s
river stream it creek path swastika brook course road trail archaeoprepona streamlet channel the trail depression the road
to meander + Prep + Noun
e.g. to meander through valley
through valley through forest through park through plain through town through it through meadow
Verb + Prep + meander
e.g. to situate in meander
to situate in
Noun + Prep + meander
e.g. antioch on the meander
antioch on the
meander + Prep + Noun
e.g. meander of the river seine
of the river seine
Try these words
north, west, river, stream, creek, path, swastika, brook, course, road, trail, archaeoprepona, streamlet, channel, depression
Some more examples with collocations
Adjective + smoker:
heavy smoker
inveterate smoker
lifelong smoker
habitual smoker
to smoke + Object:
to smoke pipe
to smoke cigarette
to smoke marijuana
to smoke cigar
hot + Noun:
hot water
hot day
hot iron
hot coffee
Noun + shop:
gift shop
coffee shop
repair shop
barber shop
memory + Verb + [s]:
memory fades
memory fails
memory serves
memory remains
[to] Verb [a/the] light:
to throw the light
to see the light
to shed the light
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12859 | https://www.cs.cmu.edu/~odonnell/boolean-analysis/lecture21.pdf | Analysis of Boolean Functions (CMU 18-859S, Spring 2007) Lecture 21: Berry-Esseen Theorem April. 3, 2007 Lecturer: Ryan O’Donnell Scribe: Suresh Purini 1 Berry-Esseen Theorem In this class we study a simplified version of the Berry-Esseen theorem, which quantifies how “close” are the distributions of the two random variables Q(x1, · · · , xn) and Q(g1, · · · , gn), where x1, · · · , xn are uniform random bits ±1 and g1, · · · , gn are Gaussian random variables with mean 0 and variance 1, and Q is a multivariate polynomial of degree 1 or in other words it is an affine function. In the next class, we study a version of Berry-Esseen theorem where Q is a multi-linear polynomial. This theorem is used to prove that “Majority is Stablest”. The following is the main theorem of this class.
Theorem 1.1 Let Q(u1, · · · , un) = n P i=1 αiui, αi ∈R, be a linear polynomial over formal variables u1, · · · , un. Also assume that n P i=1 α2 i = 1 and α2 i ≤τ, ∀i ∈[n]. Let x1, · · · , xn be i.i.d uniform random ±1 bits. Let the random variable g be normally distributed with mean 0 and variance 1, i.e., g ∼N(0, 1). Then the random variables Q(x1, · · · , xn) and g are “close” in distribution. In particular 1. ∀to ∈R, |Pr[Q(x1, · · · , xn) < t0] −Pr[g < t0]| ≤O(τ).
2. |E[|Q(x1, · · · , xn)|] −E[|g|]| ≤O(τ).
We first understand some of the ideas used in the proof of theorem 1.1.
Idea 1: The first idea is to replace the Gaussian random variable g by another Gaussian random variable with same mean and variance but looks very similar to the random variable Q(x1, · · · , xn). Let g1, · · · , gn be i.i.d Gaussian random variables with mean 0 and variance 1. Consider the random variable Q(g1, · · · , gn) ∼ n P i=1 αigi. It turns out that Q(g1, · · · , gn) ∼ N(0, Pn i=1 α2 i ) ∼N(0, 1)1 is also Gaussianly distributed. Now we can compare the distributions Q(x1, · · · , xn) and Q(g1, · · · , gn) which at least appear to look alike.
1If X ∼N(0, 1), then αX ∼N(0, α2).
If X ∼N(µ, σ2) and Y ∼N(ν, τ2), then X + Y ∼ N(µ + ν, σ2 + τ 2). In other words, sum of Gaussian random variables is Gaussian whose mean and variance are respectively equal to sum of the individual means and variances of Gaussian random variables used in summation.
Refer of normally distributed random variables 1 Idea 2: We shall try to see a generic way to say that two random variables X and Y are close in distribution. For all “nice” test functions ψ : R →R, |E[ψ(X)] −E[ψ(Y )]| ≤“small”.
First note that if the functions ψt0(t) = 1 if t < t0 0 otherwise and ψ2(t) = |t| fit in the definition of “nice”, then by letting X = Q(x1, · · · , xn) and Y = Q(g1, · · · , gn) we shall get statements 1 and 2 of theorem 1.1 respectively. We shall later see that the above two functions ψ1 and ψ2 do not fit in our definition of “nice” but they can be approximated by “nice” functions.
Let us look at some examples of “nice” functions with respect to the random variables X = Q(x1, · · · , xn) and Y = Q(g1, · · · , gn). Note that if g ∼N(0, 1), then E[g2k+1] = 0 and E[g2k] = (2k −1) · (2k −3) · · ·5 · 3 · 1. In particular E[g4] = 3. These facts are repeatedly used in the following examples.
1. Let ψ(t) = a + bt. Then E[ψ(X)] −E[ψ(Y )] = E[a + bX] −E[a + bY ] = b(E[X] −E[Y ]) = b(E[ n P i=1 αixi] −E[ n P i=1 αigi]) = 0 2. Let ψ(t) = a + bt + ct2. Then E[ψ(X)] −E[ψ(Y )] = E[a + bX + cX2] −E[a + bY + cY 2] = c(E[( n P i=1 αixi)2] −E[( n P i=1 αigi)2]) = 0 3. Let ψ(t) = t3. Then E[ψ(X)] −E[ψ(Y )] = E[( n P i=1 αixi)3] −E[( n P i=1 αigi)3 = P i,j,k αiαjαkE[xixjxk] −P i,j,k αiαjαkE[gigjgk] = 0 At this point one might conjecture that all polynomials are nice functions. But it turns out it is not “quite” true as we see in the next example.
2 4. Let ψ(t) = t4. Then E[ψ(X)] −E[ψ(Y )] = E[( n P i=1 αixi)4] −E[( n P i=1 αigi)4] = P i,j,k,l αiαjαkαlE[xixjxkxl] −P i,j,k,l αiαjαkαlE[gigjgkgl] = n P i=1 α4 i − n P i=1 α4 i E[g4 i ] Since the fourth moment of N(0, 1) is 3, we have that |E[ψ(X)] −E[ψ(Y )]| ≤ 2 n P i=1 α4 i ≤ 2 (m i ax α2 i )( n P i=1 α2 i ) ≤τ.
We now formally define “nice” functions.
Definition 1.2 A function ψ : R →R is B-nice, B ∈R+, if ψ is smooth and |ψ′′′′(t)| ≤B, ∀t.
Remark 1.3 By bounding the fourth derivative of a function at all points, we can see that we have an upper bound on the remainder term in the Taylor’s theorem. So if the bound on the fourth derivative of a function is small at all points, then we get a very good approximation for the function at any point by using the first four terms in the Taylor series expansion of the function.
Idea 3: The final idea required to prove our main theorem is “hybridization” of random vari-ables. Let Zi = α1g1 + · · · + αigi + αi+1x1 + · · · + αnxn. So Z0 = X (= Pn i=1 αixi). And Zn = Y (= Pn i=1 αigi) We shall show that for any B-nice function ψ |E[ψ(Zi−1] −E[ψ(Zi)]| ≤O(Bα4 i ) ∀i = 1 · ·n Then by telescoping together with triangle inequality we get |E[ψ(X) −E[ψ(Y )]| ≤ n P i=1 O(Bα4 i ) ≤ O((m i ax α2 i )( n P i=1 α2 i )) ≤ O(Bτ) We shall now recall Taylor’s theorem which we shall use in proving the next theorem.
3 Theorem 1.4 (Taylor’s theorem) For all smooth functions f and for any r ∈N, there exists y ∈[x, x + ϵ], such that f(x + ϵ) = f(x) + ϵf ′(x) + ϵ2 2!f ′′(x) + · · · + ϵr−1 (r −1)!f (r−1)(x) + ϵr r!f (r)(y) Theorem 1.5 For all B-nice functions ψ |E[ψ(Zi−1)] −E[ψ(Zi)]| ≤O(Bτ).
Proof: Write R = α1g1 + · · · + αi−1gi−1 + αi+1xi+1 + · · · + αnxn. Then Zi−1 = R + αixi and Zi = R + αigi Note that xi, gi, R are mutually independent. We want to bound |E[ψ(R + αixi)] −E[ψ(R + αigi)]|. Since ψ is a B-nice function we have ψ′′′′(t) ≤B, ∀t. This gives us the following ∀t, ϵ > 0, ψ(t + ϵ) = ψ(t) + ψ′(t)ϵ + ψ′′(t)ϵ2 2 + ψ′′′(t) 6 ϵ3 + {≤B 24ϵ4} Hence |E[ψ(X)] −E[ψ[Y ]]| = |E[ψ(R + αixi)] −E[ψ(R + αigi)]| = |E[ψ(R) + ψ′(R)(αixi) + ψ′′(R)(αixi)2 2 + ψ′′′(R) 6 (αixi)3 + {≤B 24(αixi)4}] − E[ψ(R) + ψ′(R)(αigi) + ψ′′(R)(αigi)2 2 + ψ′′′(R) 6 (αigi)3 + {≤B 24(αigi)4}]| ≤ E[{ B 24(αixi)4} + { B 24(αigi)4}] ≤ B 24α4 i + B 243α4 i ≤ O(Bα4 i ) ≤ O(Bτ) 2 Recall that we mentioned before that if the threshold function ψt0(t) = 1 if t < t0 0 otherwise and the absolute value function ψ2(t) = |t| fit in the definition of B-nice functions we have our Berry-Esseen theorem proved. We can see that they are not B-nice functions. However, they can be approximated by B-nice functions. We use this fact prove the Berry-Esseen theorem.
Claim 1.6 ∀to ∈R and ∀λ, 0 < λ < 1 2, there exists a O( 1 λ4)-nice function ψt0,λ : R →R which approximates ψt0 in the following sense: ψt0,λ = 1 for t < t0 −λ; ψt0,λ(t) = 0 for t > t0 + λ and 0 ≤ψt0,λ(t) ≤1 for |t −t0| ≤λ.
4 We give a proof sketch for the first part of the Berry-Esseen theorem, ∀t0 ∈R, Pr[X < t0] −Pr[g < t0]| ≤O(τ), where X = n P i=1 αixi, with xi’s as i.i.d uniform random ±1 bits and g ∼N(0, 1). Also Pn i=1 α2 i = 1 and α2 i ≤τ, ∀i ∈[n].
Proof Sketch: E[ψt0−λ,λ(X)] ≤Pr[X < t0] ≤E[ψt0+λ,λ(X)] By Berry-Essen theorem we have E[ψt0−λ,λ(X)] = E[ψt0−λ,λ(g)] ± O( τ λ4) and E[ψt0+λ,λ(X)] = E[ψt0+λ,λ(g)] ± O( τ λ4). But E[ψt0+λ,λ(g)] = Pr[g < t0 + λ], which is within O(λ) of Pr[g < t0].
Therefore we have |Pr[X < t0] −Pr[g < t0]| ≤O( τ λ4) + O(λ) By taking λ = τ 1 5, we have |Pr[X < t0] −Pr[g < t0]| ≤O(τ 1 5).
5 |
12860 | https://www.pasco.com/resources/lab-experiments/137?srsltid=AfmBOorJIbSxSyR-HYZegdS_G6l7IxPTGoozxfz_E2ehhYYmosfrkcIB | Free Webinar SeriesView Schedule — Register Today »
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12861 | https://www.youtube.com/watch?v=J0ShQzX1Ukk | How to Write Formulas for Monatomic Ions: Practice Problems
Dean Covalt
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Transcript:
hello everyone and welcome back my name is Mr Koval and in this video I'm going to show you how to write formulas for monoatomic ions let's get into this so in my previous video I showed you how to go from the formula or to the name for a monoatomic ion but now here we're going to go in the reverse Direction here we're given the name and we want to write the formula we want to write the ion itself with the charge and so that's what we're going to do so here um they give us the name so we're going to translate the name into the formula in this case it uh seems to be probably a little bit easier to do that so here um starting with phosphide ion remember if it has an IDE ending and we know it's a monoatomic ion then it means that that's going to be a negative ion so that means we're going to have to write the name of the element so from fosi the element's name is going to be phosphorus phosphorus right so so phosphorus uh is written so phosphorus so that's the name of the element from which the name phosphide came from and so uh now we need to know the charge so so the charge is going to be uh well we put symbol first so let's put the symbol here so P for phosphorus and so you need to be able to identify the charge from your periodic table so there's a couple ways to do this you can memorize uh what group it's in and memorize what the uh what what the charge is for elements that are typical in that group right so you can remember oh it's uh phosphorus is in group three or I'm sorry group five so therefore it's going to have a neg -3 charge or if it's in group six it's going to have a negative -2 charge if it's in group seven it'll have a negative one charge remember the rule for identifying the charges in groups uh five through eight is the group number minus 8 that's going to give you the negative charge or you can remember that um all elements want to be like the noble gases so if you are in group five then you want to have a full outer shell and the easiest way to [Music] oops sorry about that so the easiest way to be like a noble gas is to either gain or lose electrons um to to get the noble gas configuration like the noble gas that is closest to your element so for phosphorus the closest element is going to be Neon right or not neon I'm sorry let me get my is it argon sorry so argon so in order to be like argon argon has a full outer shell with eight this has a veence electrons of five five veence electrons so to get the eight it would gain three so or it could lose five but it'll be a lot harder to lose five electrons so it's easier to gain three so that's why if it gains three more electrons it would have a three minus charge so three minus so that is the formula for phosphide ion uh for B we have the lithium ion so uh in this case we don't have the IDE ending so that indicates that we have a positive ion a metal and so if we look up lithium on the periodic table we find that it's in group one so again if you know the rules for positive ions in groups one two and three then you would know that oh the uh the charge on the ion is is re uh equal to the group number right so the positive one charge would correspond to group number one if you're in group number two that would correspond to a two plus charge group number three that would correspond to a three plus charge or once again the better way is to understand that these charges are coming from Gaining or losing electrons in this case it's a metal so it's going to lose electrons to be like the noble gas so um for lithium it's in group one so it can lose one electron to be like a helium right so helium is the closest noble gas to lithium and so to be like helium it just needs a lose that one electron um another possibility it could gain seven more electrons that but that's highly unlikely um and harder to do so lithium is going to be more like that's closest noble gas so lithium would lose one electron and therefore it has a plus one charge so the formula would be the lithium symbol with a plus charge on it so that would be the form formula for the lithium ion next we have Cobalt 2 ion so Cobalt notice there's no ID ending so this is a positive ion it's a metal uh this is one of the transition elements so we have a Roman numeral so we're going to write the symbol for Cobalt and in the Roman numeral remember the Roman numeral tells us to charge on the positive on the metal ion so that would mean that this has a 2+ charge next we have the bromide ion notice we have the ID ending and so that indicates that we have a negative ion and so we're going to have to write the uh formula or the uh symbol for this element which is bromine so bromine is the element we're going to write the symbol BR and since we uh know where it is on the periodic table we can figure out what charge is going to be it's a main group element or a representative element which means it's going to only have that one charge and so we look on the periodic table and we see bromine is in group seven so either you know the rule which is the group number minus 8 so 7 - 8 is is negative one so it has a negative one charge or you can remember that oh it's in group seven it wants to gain one more electron to be like the closest normal gas so it's going to have a negative one charge so we put a negative and so there's our formula for the bromide ion next we have aluminum ion or for for our friends across the lake it would be alumin aluminium I can't even and say it it's it's hard to pronounce aluminium ion uh so here again we have aluminum aluminum is the metal so it's not there's no ID ending here so that means it's a positive ion it's a metal so we're just going to write down the formula or symbol for the element that's going to be Al and then finally uh what charge is the aluminum I ion going to have well again you can look at the periodic table and say oh it's in group three so the rule there is if you're in groups one two and three then the group number is equal to the charge so the charge is equal to the group number uh aluminum is in group three so therefore it's a thre plus charge or the other way you can think about it is it can lose three electrons to be like the closest noble gas losing three negative electrons means that you're going to have three protons extra compared to the electrons so you get a three plus charge okay finally we have gold one ion the gold one ion again gold is a transition element it's a transition metal the the one in the parentheses the Roman numeral tells us to charge on it uh this is again a transition element and so we would expect a Roman numeral remember there are those exceptions silver zinc cavan um so we're going to write the symbol for gold and the Roman numeral tells us to charge which is a one so we're just going to put a plus and that ladies and gentlemen is how you come up with the right formula for your ion's name this is for monatomic ions and again if you like this video then please like share and 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12862 | https://ocw.mit.edu/courses/6-061-introduction-to-electric-power-systems-spring-2011/c26f6204e05bd4c47875a96452982b6f_MIT6_061S11_ch8.pdf | X Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.061 Introduction to Power Systems Class Notes Chapter 8 Electromagnetic Forces and Loss Mechanisms ∗ J.L. Kirtley Jr. 1 Introduction This section of notes discusses some of the fundamental processes involved in electric machin ery. In the section on energy conversion processes we examine the two major ways of estimating electromagnetic forces: those involving thermodynamic arguments (conservation of energy) and field methods (Maxwell’s Stress Tensor). In between these two explications is a bit of description of electric machinery, primarily there to motivate the description of field based force calculating methods. The subsection of the notes dealing with losses is really about eddy currents in both linear and nonlinear materials and about semi-empirical ways of handling iron losses and exciting currents in machines. 2 Energy Conversion Process: In a motor the energy conversion process can be thought of in simple terms. In “steady state”, electric power input to the machine is just the sum of electric power inputs to the different phase terminals: Pe = viii i Mechanical power is torque times speed: Pm = T Ω And the sum of the losses is the difference: Pd = Pe −Pm ∗ c 2003 James L. Kirtley Jr. 1 Mechanical Electro-Converter Mechanical Power Out Electric Power In Losses: Heat, Noise, Windage,... Figure 1: Energy Conversion Process It will sometimes be convenient to employ the fact that, in most machines, dissipation is small enough to approximate mechanical power with electrical power. In fact, there are many situations in which the loss mechanism is known well enough that it can be idealized away. The “thermodynamic” arguments for force density take advantage of this and employ a “conservative” or lossless energy conversion system. 2.1 Energy Approach to Electromagnetic Forces: Magnetic Field System + v -f x Figure 2: Conservative Magnetic Field System To start, consider some electromechanical system which has two sets of “terminals”, electrical and mechanical, as shown in Figure 2. If the system stores energy in magnetic fields, the energy stored depends on the state of the system, defined by (in this case) two of the identifiable variables: flux (λ), current (i) and mechanical position (x). In fact, with only a little reflection, you should be able to convince yourself that this state is a single-valued function of two variables and that the energy stored is independent of how the system was brought to this state. Now, all electromechanical converters have loss mechanisms and so are not themselves conser vative. However, the magnetic field system that produces force is, in principle, conservative in the sense that its state and stored energy can be described by only two variables. The “history” of the system is not important. It is possible to chose the variables in such a way that electrical power into this conservative 2 Z X Z system is: dλ P e = vi = i dt Similarly, mechanical power out of the system is: dx P m fe = dt The difference between these two is the rate of change of energy stored in the system: dWm = P e −P m dt It is then possible to compute the change in energy required to take the system from one state to another by: a Wm(a) −Wm(b) = idλ −fedx b where the two states of the system are described by a = (λa, xa) and b = (λb, xb) If the energy stored in the system is described by two state variables, λ and x, the total differential of stored energy is: ∂Wm ∂Wm dWm = dλ + dx ∂λ ∂x and it is also: dWm = idλ −fedx So that we can make a direct equivalence between the derivatives and: ∂Wm fe = − ∂x This generalizes in the case of multiple electrical terminals and/or multiple mechanical termi nals. For example, a situation with multiple electrical terminals will have: dWm = ikdλk −fedx k And the case of rotary, as opposed to linear, motion has in place of force fe and displacement x, torque T e and angular displacement θ. In many cases we might consider a system which is electricaly linear, in which case inductance is a function only of the mechanical position x. λ(x) = L(x)i In this case, assuming that the energy integral is carried out from λ = 0 (so that the part of the integral carried out over x is zero), λ 1 1 λ2 Wm = λdλ = 0 L(x) 2 L(x) This makes 1 ∂ 1 fe = −2 λ2 ∂x L(x) 3 0 X 0 X 0 0 Note that this is numerically equivalent to 1 ∂ fe = −2 i2 ∂x L(x) This is true only in the case of a linear system. Note that substituting L(x)i = λ too early in the derivation produces erroneous results: in the case of a linear system it is a sign error, but in the case of a nonlinear system it is just wrong. 2.1.1 Coenergy We often will describe systems in terms of inductance rather than its reciprocal, so that current, rather than flux, appears to be the relevant variable. It is convenient to derive a new energy variable, which we will call co-energy, by: W = λiii −Wm m i and in this case it is quite easy to show that the energy differential is (for a single mechanical variable) simply: dW = m k λkdik + fedx so that force produced is: ∂Wm fe = ∂x Consider a simple electric machine example in which there is a single winding on a rotor (call it the field winding and a polyphase armature. Suppose the rotor is round so that we can describe the flux linkages as: λa = Laia + Labib + Labic + M cos(pθ)if 2π λb = Labia + Laib + Labic + M cos(pθ − )if 3 2π λc = Labia + Labib + Laic + M cos(pθ + )if 3 2π 2π λf = M cos(pθ)ia + M cos(pθ − )ib + M cos(pθ + ) + Lfif 3 3 Now, this system can be simply described in terms of coenergy. With multiple excitation it is important to exercise some care in taking the coenergy integral (to ensure that it is taken over a valid path in the multi-dimensional space). In our case there are actually five dimensions, but only four are important since we can position the rotor with all currents at zero so there is no contribution to coenergy from setting rotor position. Suppose the rotor is at some angle θ and that the four currents have values ia0, ib0, ic0 and if0. One of many correct path integrals to take would be: Z ia0 W = + Laiadia 0 Z ib0 0 (Labia0 + Laib) dib 4 m The result is: 0 1 2 W m = La i2 a0 + i2 b0 + ico + Lab (iaoib0 + iaoic0 + icoib0) 2 2π 2π 1 + i i cos(pθ) + i 2 M f0 a0 b0 cos(pθ − ) + ic0 cos(pθ + ) + Lfi 3 3 2 f0 ZZ Z ! Z ic0 + (Labia0 + Labib0 + Laic) dic 0 Z if0 2π 2π + M cos(pθ)ia0 + M cos(pθ − )ib0 + M cos(pθ + )ic0 + Lfif dif 0 3 3 If there is no variation of the stator inductances with rotor position θ, (which would be the case if the rotor were perfectly round), the terms that involve La and L(ab) contribute zero so that torque is given by: ∂W 0 2π 2π Te = m = −pMif0 ia0 sin(pθ) + ib0 sin(pθ − ) + ico sin(pθ + ) ∂θ 3 3 We will return to this type of machine in subsequent chapters. 2.2 Continuum Energy Flow At this point, it is instructive to think of electromagnetic energy flow as described by Poynting’s Theorem: S ~ = E ~ H ~ × Energy flow S ~, called Poynting’s Vector, describes electromagnetic power in terms of electric and magnetic fields. It is power density: power per unit area, with units in the SI system of units of watts per square meter. To calculate electromagnetic power into some volume of space, we can integrate Poyting’s Vector over the surface of that volume, and then using the divergence theorem: nda = Sdv P = − S ~ · ~ − vol r · ~ Now, the divergence of the Poynting Vector is, using a vector identity: r · S ~ = r · E ~ × H ~ = H ~ · r × E ~ −E ~ · r × H ~ ∂B ~ = −H ~ · ∂t −E ~ · J ~ The power crossing into a region of space is then: P = Z E ~ J ~ + H ~ ∂B ~ dv vol · · ∂t Now, in the absence of material motion, interpretation of the two terms in this equation is fairly simple. The first term describes dissipation: E ~ J ~ = E ~ 2σ = J ~ 2ρ · | | | | 5 The second term is interpreted as rate of change of magnetic stored energy. In the absence of hysteresis it is: ∂Wm ~ ∂B ~ = H ∂t · ∂t Note that in the case of free space, ∂B ~ ∂H ~ ∂ 1 H ~ · ∂t = µ0H ~ · ∂t = ∂t 2µ0|H ~ | 2 which is straightforwardedly interpreted as rate of change of magnetic stored energy density: 1 2 Wm = 2µ0|H| Some materials exhibit hysteretic behavior, in which stored energy is not a single valued function of either B ~ or H ~ , and we will consider that case anon. 2.3 Material Motion In the presence of material motion ~ v, electric field E ~ 0 in a “moving” frame is related to electric field E ~ in a “stationary” frame and to magnetic field B ~ by: E ~ 0 = E ~ + ~ v B ~ × This is an experimental result obtained by observing charged particles moving in combined electric and magnetic fields. It is a relatavistic expression, so that the qualifiers “moving” and “stationary” are themselves relative. The electric fields are what would be observed in either frame. In MQS systems, the magnetic flux density B ~ is the same in both frames. The term relating to current density becomes: E ~ J ~ = E ~ 0 v B ~ J ~ · −~ × · We can interpret E ~ 0 J ~ as dissipation, but the second term bears a little examination. Note · that it is in the form of a vector triple (scalar) product: −~ v × B ~ · J ~ = −~ v · B ~ × J ~ = −~ v · J ~ × B ~ This is in the form of velocity times force density and represents power conversion from electro magnetic to mechanical form. This is consistent with the Lorentz force law (also experimentally observed): F ~ = J ~ B ~ × This last expression is yet another way of describing energy conversion processes in electric machinery, as the component of apparent electric field produced by material motion through a magnetic field, when reacted against by a current, produces energy conversion to mechanical form rather than dissipation. 6 0 Z Z Z r × I 0 Z Z I 0 Z 2.4 Additional Issues in Energy Methods There are two more important and interesting issues to consider as we study the development of forces of electromagnetic origin and their calculation using energy methods. These concern situations which are not simply representable by lumped parameters and situations that involve permanent magnets. 2.4.1 Coenergy in Continuous Media Consider a system with not just a multiplicity of circuits but a continuum of current-carrying paths. In that case we could identify the co-energy as: λ(~ a)dJ ~ · d~ a W = m area where that area is chosen to cut all of the current carrying conductors. This area can be picked to be perpedicular to each of the current filaments since the divergence of current is zero. The flux λ is calculated over a path that coincides with each current filament (such paths exist since current has zero divergence). Then the flux is: λ(~ a) = B ~ d~ n · Now, if we use the vector potential A ~ for which the magnetic flux density is: B ~ = A ~ the flux linked by any one of the current filaments is: λ(~ a) = A ~ d~ · where d~ is the path around the current filament. This implies directly that the coenergy is: A ~ dd ~ J ~ W d~ a = · · m area J Now: it is possible to make d~ coincide with d~ a and be parallel to the current filaments, so that: d ~ A ~ Jdv W = · m vol 2.4.2 Permanent Magnets Permanent magnets are becoming an even more important element in electric machine systems. Often systems with permanent magnets are approached in a relatively ad-hoc way, made equivalent to a current that produces the same MMF as the magnet itself. The constitutive relationship for a permanent magnet relates the magnetic flux density B ~ to magnetic field H ~ and the property of the magnet itself, the magnetization M ~ . B ~ = µ0 H ~ + M ~ 7 r × Z r × ZZ Z Now, the effect of the magnetization is to act as if there were a current (called an amperian current) with density: J ~∗ = M ~ Note that this amperian current “acts” just like ordinary current in making magnetic flux density. Magnetic co-energy is: W 0 m = vol Mdv A ~ · r × d ~ Next, note the vector identity r · C ~ × D ~ = D ~ · r × C ~ −C ~ · r × D ~ So that: Z Z d ~ W 0 = vol −r · A ~ × dM ~ dv + r × A ~ · Mdv m vol Then, noting that B ~ = A ~: W 0 m = − A Md~ s + B Mdv ~ × d ~ vol ~ · d ~ m The first of these integrals (closed surface) vanishes if it is taken over a surface just outside the magnet, where M ~ is zero. Thus the magnetic co-energy in a system with only a permanent magnet source is Z W 0 = B ~ Mdv d ~ vol · Adding current carrying coils to such a system is done in the obvious way. 2.5 Electric Machine Description: Actually, this description shows a conventional induction motor. This is a very common type of electric machine and will serve as a reference point. Most other electric machines operate in a fashion which is the same as the induction machine or which differ in ways which are easy to reference to the induction machine. Consider the simplified machine drawing shown in Figure 3. Most (but not all!) machines we will be studying have essentially this morphology. The rotor of the machine is mounted on a shaft which is supported on some sort of bearing(s). Usually, but not always, the rotor is inside. I have drawn a rotor which is round, but this does not need to be the case. I have also indicated rotor conductors, but sometimes the rotor has permanent magnets either fastened to it or inside, and sometimes (as in Variable Reluctance Machines) it is just an oddly shaped piece of steel. The stator is, in this drawing, on the outside and has windings. With most of the machines we will be dealing with, the stator winding is the armature, or electrical power input element. (In DC and Universal motors this is reversed, with the armature contained on the rotor: we will deal with these later). In most electrical machines the rotor and the stator are made of highly magnetically permeable materials: steel or magnetic iron. In many common machines such as induction motors the rotor and stator are both made up of thin sheets of silicon steel. Punched into those sheets are slots which contain the rotor and stator conductors. 8 Stator Stator Conductors Rotor Rotor Conductors Bearings Shaft End Windings Air Gap Figure 3: Form of Electric Machine Figure 4 is a picture of part of an induction machine distorted so that the air-gap is straightened out (as if the machine had infinite radius). This is actually a convenient way of drawing the machine and, we will find, leads to useful methods of analysis. What is important to note for now is that the machine has an air gap g which is relatively small (that is, the gap dimension is much less than the machine radius r). The machine also has a physical length l. The electric machine works by producing a shear stress in the air-gap (with of course side effects such as production of “back voltage”). It is possible to define the average air-gap shear stress, which we will refer to as τ. Total developed torque is force over the surface area times moment (which is rotor radius): T = 2πr2< τ > Power transferred by this device is just torque times speed, which is the same as force times surface velocity, since surface velocity is u = rΩ: Pm = ΩT = 2πr < τ > u If we note that active rotor volume is πr2, the ratio of torque to volume is just: T = 2 < τ > Vr Now, determining what can be done in a volume of machine involves two things. First, it is clear that the volume we have calculated here is not the whole machine volume, since it does not include the stator. The actual estimate of total machine volume from the rotor volume is actually quite complex and detailed and we will leave that one for later. Second, we need to estimate the value of the useful average shear stress. Suppose both the radial flux density Br and the stator surface current density Kz are sinusoidal flux waves of the form: Br = √ 2B0 cos (pθ −ωt) 9 Stator Core Stator Conductors In Slots Rotor Conductors In Slots Air Gap Figure 4: Windings in Slots Kz = √ 2K0 cos (pθ −ωt) Note that this assumes these two quantities are exactly in phase, or oriented to ideally produce torque, so we are going to get an “optimistic” bound here. Then the average value of surface traction is: 1 Z 2π < τ >= BrKzdθ = B0K0 2π 0 This actually makes some sense in view of the empirically derived Lorentz Force Law: Given a (vector) current density and a (vector) flux density. In the absence of magnetic materials (those with permeability different from that of free space), the observed force on a conductor is: F ~ = J ~ B ~ × Where J ~ is the vector describing current density (A/m2) and B ~ is the magnetic flux density (T). This is actually enough to describe the forces we see in many machines, but since electric machines have permeable magnetic material and since magnetic fields produce forces on permeable material even in the absence of macroscopic currents it is necessary to observe how force appears on such material. A suitable empirical expression for force density is: F ~ = J ~ × B ~ − 1 2 H ~ · H ~ rµ where H ~ is the magnetic field intensity and µ is the permeability. Now, note that current density is the curl of magnetic field intensity, so that: F ~ = r × H ~ × µH ~ −2 1 H ~ · H ~ rµ 1 = µ r × H ~ × H ~ −2 H ~ · H ~ rµ And, since: H ~ H ~ r × H ~ × H ~ = H ~ · r H ~ −2 1 r · 10 Fk = µHiHk − δik X H2 ∂x n i 2 n the Kroneker delta δ = 1 if i = k, 0 otherwise. ! ! Z Z ZZ P I I X force density is: 1 1 F ~ = µ H ~ · r H ~ −2µr H ~ · H ~ −2 H ~ · H ~ rµ 1 = µ H ~ · r H ~ −r µ H ~ · H ~ 2 This expression can be written by components: the component of force in the i’th dimension is: X ∂ ∂ 1 X Fi = µ Hk ∂xk Hi −∂xi 2µ Hk 2 k k Now, see that we can write the divergence of magnetic flux density as: B ~ = X ∂ µHk = 0 r · k ∂xk and X ∂ X ∂ X ∂ µ Hk ∂xk Hi = ∂xk µHkHi −Hi ∂xk µHk k k k but since the last term in that is zero, we can write force density as: ∂ µ where we have used ik Note that this force density is in the form of the divergence of a tensor: ∂ Fk = Tik ∂xi or F ~ = T r · In this case, force on some object that can be surrounded by a closed surface can be found by using the divergence theorem: f ~ = ~ Tdv = T ~ nda · vol Fdv = vol r · or, if we note surface traction to be τi = k Tiknk , where n is the surface normal vector, then the total force in direction i is just: f ~ = τida = Tiknkda s k The interpretation of all of this is less difficult than the notation suggests. This field description of forces gives us a simple picture of surface traction, the force per unit area on a surface. If we just integrate this traction over the area of some body we get the whole force on the body. Note that 11 P n o n o n o n o 3 this works if we integrate the traction over a surface that is itself in free space but which surrounds the body (because we can impose no force on free space). Note one more thing about this notation. Sometimes when subscripts are repeated as they are here the summation symbol is omitted. Thus we would write τi = k Tiknk = Tiknk. Now, if we go back to the case of a circular cylinder and are interested in torque, it is pretty clear that we can compute the circumferential force by noting that the normal vector to the cylinder is just the radial unit vector, and then the circumferential traction must simply be: τθ = µ0HrHθ Simply integrating this over the surface gives azimuthal force, and then multiplying by radius (moment arm) gives torque. The last step is to note that, if the rotor is made of highly permeable material, the azimuthal magnetic field is equal to surface current density. Tying the MST and Poynting Approaches Together y x Field Region Contour Figure 5: Illustrative Region of Space Now that the stage is set, consider energy flow and force transfer in a narrow region of space as illustrated by Figure 5. The upper and lower surfaces may support currents. Assume that all of the fields, electric and magnetic, are of the form of a traveling wave in the x- direction: Re ej(ωt−kx) . If we assume that form for the fields and also assume that there is no variation in the z- direction (equivalently, the problem is infinitely long in the z- direction), there can be no x- directed currents because the divergence of current is zero: r J ~ = 0. In a magnetostatic system this is true of · electric field E ~ too. Thus we will assume that current is confined to the z- direction and to the two surfaces illustrated in Figure 5, and thus the only important fields are: E ~ = ~ izRe Ezej(ωt−kx) H ~ = ~ ixRe Hxej(ωt−kx) + ~ iyRe Hyej(ωt−kx) We may use Faraday’s Law (r × E ~ = −∂B ~ ) to establish the relationship between the electric ∂t and magnetic field: the y- component of Faraday’s Law is: jkE = −jωµ0H z y or ω E = z −k µ0Hy 12 n o n o n o 4 The phase velocity uph = ω k is a most important quantity. Note that, if one of the surfaces is moving (as it would be in, say, an induction machine), the frequency and hence the apparent phase velocity, will be shifted by the motion. We will use this fact shortly. Energy flow through the surface denoted by the dotted line in Figure 5 is the component of Poynting’s Vector in the negative y- direction. The relevant component is: Sy = E ~ × H ~ y = EzHx = − ω k µ0HyHx Note that this expression contains the xy component of the Maxwell Stress Tensor Txy = µ0HxHy so that power flow downward through the surface is: ω S = −Sy = µ0HxHy = uphTxy k The average power flow is the same, in this case, for time and for space, and is: 1 µ0 < S >= Re {E H∗ x} = uph Re HyHx ∗ 2 z 2 We may choose to define a surface impedance: E Z = z s −Hx which becomes: H Zs = −µ0uph y = −µ0uphR Hx where now we have defined the parameter R to be the ratio between y- and x- directed complex field amplitudes. Energy flow through that surface is now: 1 1 S = −s Re {EzHx ∗ } = 2Re |H |2Z x s Simple Description of a Linear Induction Motor g j(ω t - k x) K e µ zs y x µ σ u s Figure 6: Simple Description of Linear Induction Motor The stage is now set for an almost trivial description of a linear induction motor. Consider the geometry described in Figure 6. Shown here is only the relative motion gap region. This is bounded by two regions of highly permeable material (e.g. iron), comprising the stator and shuttle. On the surface of the stator (the upper region) is a surface current: K ~ s = ~ izRe Kzsej(ωt−kx) 13 n o The shuttle is, in this case, moving in the positive x- direction at some velocity u. It may also be described as an infinitely permeable region with the capability of supporting a surface current with surface conductivity σs, so that Kzr = σsEz. Note that Ampere’s Law gives us a boundary condition on magnetic field just below the upper surface of this problem: Hx = Kzs, so that, if we can establish the ratio between y- and x- directed fields at that location, < Txy >= µ0 yHx ∗ µ 2 0 |Kzs| 2Re {R} Re H = 2 Note that the ratio of fields Hy/Hx = R is independent of reference frame (it doesn’t matter if we are looking at the fields from the shuttle or the stator), so that the shear stress described by Txy is also frame independent. Now, if the shuttle (lower surface) is moving relative to the upper surface, the velocity of the traveling wave relative to the shuttle is: ω us = uph −u = s k where we have now defined the dimensionless slip s to be the ratio between frequency seen by the shuttle to frequency seen by the stator. We may use this to describe energy flow as described by Poynting’s Theorem. Energy flow in the stator frame is: Supper = uphTxy In the frame of the shuttle, however, it is Slower = usTxy = sSupper Now, the interpretation of this is that energy flow out of the upper surface (Supper) consists of energy converted (mechanical power) plus energy dissipated in the shuttle (which is Slower here. The difference between these two power flows, calculated using Poynting’s Theorem, is power converted from electrical to mechanical form: Sconverted = Supper(1 −s) Now, to finish the problem, note that surface current in the shuttle is: K = E0 σs = −usµ0σsH zr z y where the electric field E0 is measured in the frame of the shuttle. z We assume here that the magnetic gap g is small enough that we may assume kg 1. Ampere’s Law, taken around a contour that crosses the air-gap and has a normal in the z- direction, yields: ∂Hx g = Kzs + Kzr ∂x In complex amplitudes, this is: −jkgHy = Kzs + Kzr = Kzs −µ0usσsHy 14 or, solving for Hy. jKzs 1 Hy = kg 1 + jµ usσs 0 kg Average shear stress is | |2 ( ) | | µ 2 0usσs µ0 µ0 < T >= Re n H H o K = zs j µ0 K k xy y x Re = zs g 2 2 kg 1 + j µ0usσs 2 kg 2 µ0usσ kg 1 + s kg n o 5 Surface Impedance of Uniform Conductors The objective of this section is to describe the calculation of the surface impedance presented by a layer of conductive material. Two problems are considered here. The first considers a layer of linear material backed up by an infinitely permeable surface. This is approximately the situation presented by, for example, surface mounted permanent magnets and is probably a decent approximation to the conduction mechanism that would be responsible for loss due to asynchronous harmonics in these machines. It is also appropriate for use in estimating losses in solid rotor induction machines and in the poles of turbogenerators. The second problem, which we do not work here but simply present the previously worked solution, concerns saturating ferromagnetic material. 5.1 Linear Case The situation and coordinate system are shown in Figure 7. The conductive layer is of thicknes T and has conductivity σ and permeability µ0. To keep the mathematical expressions within bounds, we assume rectilinear geometry. This assumption will present errors which are small to the extent that curvature of the problem is small compared with the wavenumbers encountered. We presume that the situation is excited, as it would be in an electric machine, by a current sheet of the form Kz = Re Kej(ωt−kx) H x Permeable Surface Conductive Slab y x Figure 7: Axial View of Magnetic Field Problem In the conducting material, we must satisfy the diffusion equation: = r2H µ0σ∂ ∂t H 15 where the skin depth is: s 2 δ = ωµ0σ To obtain surface impedance, we use Faraday’s law: ∂B r × E = −∂t which gives: ω Ez = −µ0 H k y Now: the “surface current” is just Ks = −Hx so that the equivalent surface impedance is: E ω Z = z = jµ0 coth αT −Hx α In view of the boundary condition at the back surface of the material, taking that point to be y = 0, a general solution for the magnetic field in the material is: n o Hx = Re A sinh αyej(ωt−kx) k Hy = Re j A cosh αyej(ωt−kx) α where the coefficient α satisfies: α2 = jωµ0σ + k2 and note that the coefficients above are chosen so that H has no divergence. Note that if k is small (that is, if the wavelength of the excitation is large), this spatial coefficient α becomes 1 + j α = δ A pair of limits are interesting here. Assuming that the wavelength is long so that k is negligible, then if αT is small (i.e. thin material), ω 1 Z jµ0 = → α2T σT On the other hand as αT →∞, 1 + j Z → σδ Next it is necessary to transfer this surface impedance across the air-gap of a machine. So, with reference to Figure 8, assume a new coordinate system in which the surface of impedance Zs is located at y = 0, and we wish to determine the impedance Z = −Ez/Hx at y = g. In the gap there is no current, so magnetic field can be expressed as the gradient of a scalar potential which obeys Laplace’s equation: H = −rψ 16 and r 2ψ = 0 Ignoring a common factor of ej(ωt−kx), we can express H in the gap as: H = −ky x jk ψ eky + ψ + − e H = −k ψ eky − −ky y ψ + − e At the surface of the rotor, Ez = −HxZs or −ωµ0 ψ + −ψ − = jkZs ψ + ψ + − and then, at the surface of the stator , ψ ekg Ez ω ψ e−kg − + Z = = jµ − − 0 H k ψ ekg + ψ e−kg x + − ( ) Kz y x g Surface Impedance Z s Figure 8: Impedance across the air-gap A bit of manipulation is required to obtain: ω ekg (ωµ0 −jkZs) −e−kg (ωµ0 + jkZs) Z = jµ0 k ekg (ωµ0 −jkZ ) + e−kg (ωµ0 + jkZ ) s s It is useful to note that, in the limit of Zs →∞, this expression approaches the gap impedance ωµ0 Zg = j k2g and, if the gap is small enough that kg 0, → Z →Zg||Zs 17 6 Iron Electric machines employ ferromagnetic materials to carry magnetic flux from and to appropriate places within the machine. Such materials have properties which are interesting, useful and prob lematical, and the designers of electric machines must deal with this stuff. The purpose of this note is to introduce the most salient properties of the kinds of magnetic materials used in electric machines. We will be concerned here with materials which exhibit magnetization: flux density is something other than B ~ = µ0H ~ . Generally, we will speak of hard and soft magnetic materials. Hard materials are those in which the magnetization tends to be permanent, while soft materials are used in magnetic circuits of electric machines and transformers. Since they are related we will find ourselves talking about them either at the same time or in close proximity, even though their uses are widely disparite. 6.1 Magnetization: It is possible to relate, in all materials, magnetic flux density to magnetic field intensity with a consitutive relationship of the form: B ~ = µ0 H ~ + M ~ where magnetic field intensity H and magnetization M are the two important properties. Now, in linear magnetic material magnetization is a simple linear function of magnetic field: M ~ = χmH ~ so that the flux density is also a linear function: B ~ = µ0 (1 + χm) H ~ Note that in the most general case the magnetic susceptibility cm might be a tensor, leading to flux density being non-colinear with magnetic field intensity. But such a relationship would still be linear. Generally this sort of complexity does not have a major effect on electric machines. 6.2 Saturation and Hysteresis In useful magnetic materials this nice relationship is not correct and we need to take a more general view. We will not deal with the microscopic picture here, except to note that the magnetization is due to the alignment of groups of magnetic dipoles, the groups often called domaines. There are only so many magnetic dipoles available in any given material, so that once the flux density is high enough the material is said to saturate, and the relationship between magnetic flux density and magnetic field intensity is nonlinear. Shown in Figure 9, for example, is a “saturation curve” for a magnetic sheet steel that is sometimes used in electric machinery. Note the magnetic field intensity is on a logarithmic scale. If this were plotted on linear coordinates the saturation would appear to be quite abrupt. At this point it is appropriate to note that the units used in magnetic field analysis are not always the same nor even consistent. In almost all systems the unit of flux is the weber (W), which 18 Figure 9: Saturation Curve: Commercial M-19 Silicon Iron is the same as a volt-second. In SI the unit of flux density is the tesla (T), but many people refer to the gauss (G), which has its origin in CGS. 10,000 G = 1 T. Now it gets worse, because there is an English system measure of flux density generally called kilo-lines per square inch. This is because in the English system the unit of flux is the line. 108 lines is equal to a weber. Thus a Tesla is 64.5 kilolines per square inch. The SI and CGS units of flux density are easy to reconcile, but the units of magnetic field are a bit harder. In SI we generally measure H in amperes/meter (or ampere-turns per meter). Often, however, you will see magnetic field represented as Oersteds (Oe). One Oe is the same as the magnetic field required to produce one gauss in free space. So 79.577 A/m is one Oe. In most useful magnetic materials the magnetic domaines tend to be somewhat “sticky”, and a more-than-incremental magnetic field is required to get them to move. This leads to the property called “hysteresis”, both useful and problematical in many magnetic systems. Hysteresis loops take many forms; a generalized picture of one is shown in Figure 10. Salient features of the hysteresis curve are the remanent magnetization Br and the coercive field Hc. Note that the actual loop that will be traced out is a function of field amplitude and history. Thus there are many other “minor loops” that might be traced out by the B-H characteristic of a piece of material, depending on just what the fields and fluxes have done and are doing. Now, hysteresis is important for two reasons. First, it represents the mechanism for “trapping” magnetic flux in a piece of material to form a permanent magnet. We will have more to say about that anon. Second, hysteresis is a loss mechanism. To show this, consider some arbitrary chunk of 19 Courtesy of United States Steel Corporation. (U.S. Steel). U.S. Steel accepts no liability for reliance on any information contained in the graphs shown above.
Z Z ZZ Z Z Z Z ZZ Z ZZZ ZZZ I Magnetic Field Coercive Field Hc Remanent Flux Density Br Saturation Field H s Saturation Flux Density Bs Flux Density Figure 10: Hysteresis Curve Nomenclature material for which we can characterize an MMF and a flux: ~ d ~ F = NI = H ell · V Φ = dt = B ~ dA ~ N Area · Energy input to the chunk of material over some period of time is ~ d~ d ~ d ~ w = V Idt = FdΦ = H B A dt t · · Now, imagine carrying out the second (double) integral over a continuous set of surfaces which are perpendicular to the magnetic field H. (This IS possible!). The energy becomes: ~ d ~ w = H Bdvol dt t · and, done over a complete cycle of some input waveform, that is: w = Wmdvol vol Wm = H ~ dB ~ t · That last expression simply expresses the area of the hysteresis loop for the particular cycle. Generally, for most electric machine applications we will use magnetic material characterized as “soft”, having as narrow a hysteresis loop (and therefore as low a hysteretic loss) as possible. At the other end of the spectrum are “hard” magnetic materials which are used to make permanent magnets. The terminology comes from steel, in which soft, annealed steel material tends to have narrow loops and hardened steel tends to have wider loops. However permanent magnet technology has advanced to the point where the coercive forces possible in even cheap ceramic magnets far exceed those of the hardest steels. 20 ~ n o 6.3 Conduction, Eddy Currents and Laminations: Steel, being a metal, is an electrical conductor. Thus when time varying magnetic fields pass through it they cause eddy currents to flow, and of course those produce dissipation. In fact, for almost all applications involving “soft” iron, eddy currents are the dominant source of loss. To reduce the eddy current loss, magnetic circuits of transformers and electric machines are almost invariably laminated, or made up of relatively thin sheets of steel. To further reduce losses the steel is alloyed with elements (often silicon) which poison the electrical conductivity. There are several approaches to estimating the loss due to eddy currents in steel sheets and in the surface of solid iron, and it is worthwhile to look at a few of them. It should be noted that this is a “hard” problem, since the behavior of the material itself is difficult to characterize. 6.4 Complete Penetration Case t y x z Figure 11: Lamination Section for Loss Calculation Consider the problem of a stack of laminations. In particular, consider one sheet in the stack represented in Figure 11. It has thickness t and conductivity σ. Assume that the “skin depth” is much greater than the sheet thickness so that magnetic field penetrates the sheet completely. Further, assume that the applied magnetic flux density is parallel to the surface of the sheets: jωt B = ~ izRe √ 2B0e Now we can use Faraday’s law to determine the electric field and therefore current density in the sheet. If the problem is uniform in the x- and z- directions, ∂Ex ∂y = −jω0B0 Note also that, unless there is some net transport current in the x- direction, E must be anti symmetric about the center of the sheet. Thus if we take the origin of y to be in the center, electric field and current are: E = −jωB0y x J = −jωB0σy x Local power dissipated is P (y) = ω2B0 2σy2 = |J|2 σ 21 To find average power dissipated we integrate over the thickness of the lamination: Z t Z t 2 2 2 2 1 ω2B0 2t2σ < P >= P (y)dy = ω2B0 2σ y 2dy = t 0 t 0 12 Pay attention to the orders of the various terms here: power is proportional to the square of flux density and to the square of frequency. It is also proportional to the square of the lamination thickness (this is average volume power dissipation). As an aside, consider a simple magnetic circuit made of this material, with some length and area A, so that volume of material isA. Flux lined by a coil of N turns would be: Λ = NΦ = NAB0 and voltage is of course just V = jwL. Total power dissipated in this core would be: 1 V 2 ω2B0 2t2σ = Pc = A12 Rc where the equivalent core resistance is now A 12N 2 Rc = σt2 6.5 Eddy Currents in Saturating Iron The same geometry holds for this pattern, although we consider only the one-dimensional problem (k 0). The problem was worked by McLean and his graduate student Agarwal . They → assumed that the magnetic field at the surface of the flat slab of material was sinusoidal in time and of high enough amplitude to saturate the material. This is true if the material has high permeability and the magnetic field is strong. What happens is that the impressed magnetic field saturates a region of material near the surface, leading to a magnetic flux density parallel to the surface. The depth of the region affected changes with time, and there is a separating surface (in the flat problem this is a plane) that moves away from the top surface in response to the change in the magnetic field. An electric field is developed to move the surface, and that magnetic field drives eddy currents in the material. H B B 0 Figure 12: Idealized Saturating Characteristic 22 Z Assume that the material has a perfectly rectangular magnetization curve as shown in Figure 12, so that flux density in the x- direction is: Bx = B0sign(Hx) The flux per unit width (in the z- direction) is: −∞ Φ = Bxdy 0 and Faraday’s law becomes: ∂Φ Ez = ∂t while Ampere’s law in conjunction with Ohm’s law is: ∂Hx = σEz ∂y Now, McLean suggested a solution to this set in which there is a “separating surface” at depth ζ below the surface, as shown in Figure 13 . At any given time: Hx Jz = = Hs(t) 1 + y ζ σEz = Hs ζ y B B s s x Separating Surface Penetration Depth Figure 13: Separating Surface and Penetration Depth That is, in the region between the separating surface and the top of the material, electric field Ez is uniform and magnetic field Hx is a linear function of depth, falling from its impressed value at the surface to zero at the separating surface. Now: electric field is produced by the rate of change of flux which is: ∂Φ ∂ζ Ez = = 2Bx ∂t ∂t Eliminating E, we have: ∂ζ Hs 2ζ = ∂t σBx 23 s and then, if the impressed magnetic field is sinusoidal, this becomes: dζ2 H0 dt = sin ωt σB0 | | This is easy to solve, assuming that ζ = 0 at t = 0, s 2H0 ωt ζ sin = ωσB0 2 Now: the surface always moves in the downward direction (as we have drawn it), so at each half cycle a new surface is created: the old one just stops moving at a maximum position, or penetration depth: 2H0 δ = ωσB0 This penetration depth is analogous to the “skin depth” of the linear theory. However, it is an absolute penetration depth. The resulting electric field is: 2H0 ωt Ez = cos 0 < ωt < π σδ 2 This may be Fourier analyzed: noting that if the impressed magnetic field is sinusoidal, only the time fundamental component of electric field is important, leading to: 8 H0 Ez = (cos ωt + 2 sin ωt + . . .) 3π σδ Complex surface impedance is the ratio between the complex amplitude of electric and magnetic field, which becomes: E 8 1 Z = z = (2 + j) s H 3π σδ x Thus, in practical applications, we can handle this surface much as we handle linear conductive surfaces, by establishing a skin depth and assuming that current flows within that skin depth of the surface. The resistance is modified by the factor of 3 16 π and the “power factor” of this surface is about 89 % (as opposed to a linear surface where the “power factor” is about 71 %. Agarwal suggests using a value for B0 of about 75 % of the saturation flux density of the steel. Semi-Empirical Method of Handling Iron Loss Neither of the models described so far are fully satisfactory in describing the behavior of laminated iron, because losses are a combination of eddy current and hysteresis losses. The rather simple model employed for eddy currents is precise because of its assumption of abrupt saturation. The hysteresis model, while precise, would require an empirical determination of the size of the hysteresis loops anyway. So we must often resort to empirical loss data. Manufacturers of lamination steel sheets will publish data, usually in the form of curves, for many of their products. Here are a few ways of looking at the data. 24 7 0 A low frequency flux density vs. magnetic field (“saturation”) curve was shown in Figure 9. Included with that was a measure of the incremental permeability dB µ = dH In some machine applications either the “total” inductance (ratio of flux to MMF) or “incremental” inductance (slope of the flux to MMF curve) is required. In the limit of low frequency these numbers may be useful. For designing electric machines, however, a second way of looking at steel may be more useful. This is to measure the real and reactive power as a function of magnetic flux density and (sometimes) frequency. In principal, this data is immediately useful. In any well-designed electric machine the flux density in the core is distributed fairly uniformly and is not strongly affected by eddy currents, etc. in the core. Under such circumstances one can determine the flux density in each part of the core. With that information one can go to the published empirical data for real and reactive power and determine core loss and reactive power requirements. Figure 14: Real and Apparent Loss: M19, Fully Processed, 29 Ga Figure 14 shows core loss and “apparent” power per unit mass as a function of (RMS) induction for 29 gage, fully processed M-19 steel. The two left-hand curves are the ones we will find most useful. “P ” denotes real power while “Pa ” denotes “apparent power”. The use of this data is quite straightforward. If the flux density in a machine is estimated for each part of the machine and the mass of steel calculated, then with the help of this chart a total core loss and apparent power can 25 Courtesy of United States Steel Corporation. (U.S. Steel). U.S. Steel accepts no liability for reliance on any information contained in the graphs shown above.
q Table 1: Exponential Fit Parameters for Two Steel Sheets 29 Ga, Fully Processed M-19 M-36 Base Flux Density B0 1 T 1 T Base Frequency f0 60 Hz 60 Hz Base Power (w/lb) P0 0.59 0.67 Flux Exponent B 1.88 1.86 Frequency Exponent F 1.53 1.48 Base Apparent Power 1 V A0 1.08 1.33 Base Apparent Power 2 V A1 .0144 .0119 Flux Exponent 0 1.70 2.01 Flux Exponent 1 16.1 17.2 be estimated. Then the effect of the core may be approximated with a pair of elements in parallel with the terminals, with: Rc = q|V |2 P Xc = q|V |2 Q Q = Pa 2 −P 2 Where q is the number of machine phases and V is phase voltage. Note that this picture is, strictly speaking, only valid for the voltage and frequency for which the flux density was calculated. But it will be approximately true for small excursions in either voltage or frequency and therefore useful for estimating voltage drop due to exciting current and such matters. In design program applications these parameters can be re-calculated repeatedly if necessary. “Looking up” this data is a it awkward for design studies, so it is often convenient to do a “curve fit” to the published data. There are a large number of possible ways of doing this. One method that has bee found to work reasonably well for silicon iron is an “exponential fit”: B F B f P ≈P0 B0 f0 This fit is appropriate if the data appears on a log-log plot to lie in approximately straight lines. Figure 15 shows such a fit for the same steel sheet as the other figures. For “apparent power” the same sort of method can be used. It appears, however, that the simple exponential fit which works well for real power is inadequate, at least if relatively high inductions are to be used. This is because, as the steel saturates, the reactive component of exciting current rises rapidly. I have had some success with a “double exponential” fit: 0 1 B B VA ≈VA0 + VA1 B0 B0 To first order the reactive component of exciting current will be linear in frequency. 26 M-19, 29 Ga, Fully Processed 0.01 0.1 1 10 100 10 100 1000 10000 Loss, W/Lb 0.1 T 0.3 T 0.5 T 0.7 T 1.0 T Flux Density Frequency, Hz Figure 15: Steel Sheet Core Loss Fit vs. Flux Density and Frequency In the disk that is to be distributed with these notes there are a number of data files representing properties of different types of nonoriented sheet steel. The format of each of the files is the same: two columns of numbers, the first is flux density in Tesla, RMS, 60 Hz. The second column is watts per pound or volt-amperes per pound. The materials are denoted by the file names, which are generally of the format: “M-Mtype-Proc-Data-Gage.prn”. The coding is relatively dense because of the short file name limit of MSDOS. Mtype is the number designator (as in M-19). Proc is “f” for fully processed and “s” for semiprocessed. Data is “p” for power, “pa” for apparent power. Gage is 29 (.014” thick), 26 (.0185” thick) or 24 (.025” thick). Example: m19fp29.prn designates loss in M-19 material, fully processed, 29 gage. Also on the disk are three curve fitting routines that appear to work with this data. (Not all of the routines work with all of the data!). They are: 1. efit.m implements the single exponential fit of loss against flux density. Use: in MATLAB type efit . The program prompts fit what (name.prn) ==> Enter the file name for the material designator without the .prn extension. The program will think about the problem for a few seconds and put up a plot of its fit with points noting the actual data. Enter a and a summary of the fit turns up, including the 27 fit parameters and an error indication. These programs use MATLAB’s fmins routine to minimize a mean-squared error as calculated by the auxiliary function fiterr.m. 2. e2fit.m implements the double exponential fit of apparent power against flux density. Use is just like efit. It uses the auxiliary function fit2err.m. 3. pfit.m uses the MATLAB function polyfit to fit a polynomial (in B) to the data. Most of the machine design scripts enclosed with the material for this special summer subject employ the exponential fits for core iron developed here. References W. MacLean, “Theory of Strong Electromagnetic Waves in Massive Iron”, Journal of Applied Physics, V.25, No 10, October, 1954 P.D. Agarwal, “Eddy-Current Losses in Solid and Laminated Iron”, Trans. AIEE, V. 78, pp 169-171, 1959 28 LAMINATION STEELS THIRD EDITION Excerpts specially prepared for the Massachusetts Institute of T echnology by the Electric Motor Education and Research Foundation Non-Oriented Silicon Steels Summary Graphs Magnetization – B vs. H T otal Core Loss – Pc vs. B AK Steel 7500 3500 25000 Di-Max M-19 20000 1000 15000 1.5 Tesla 1000 Fully Processed 100 .014 inch 10000 1 Tesla 100 5000 10 (.36 mm, 29 gauge) 10 1 1 Summary Graphs 1000 0.1 0.1 Magnetization Curves Data Core Loss Curves Magnetic Flux Density in Gausses – B 500 0.01 Core Loss in Watts per Kilogram – Pc Core Loss in Watts per Pound – Pc 0.01 0.001 0.001 0.0001 0.0001 0.00001 1 Tesla 1.5 Tesla 0 1000 5000 10000 15000 20000 0.00001 100 40 Data 0.001 0.01 0.1 1 10 100 1000 0.1 1 10 100 1000 10000 100000 Exciting Power Data Spreadsheet Other Thicknesses .0185 inch .025 inch AK Steel Product Info AK Steel Non-Oriented Silicon Steel Menu Non-Oriented Silicon Steels Menu Lamination Steels Main Menu Magnetic Flux Density in Gausses – B Magnetizing Force in Oersteds – H Magnetizing Force in Amperes per Meter – H Magnetization curves for this material, DC through 2000 hertz T otal core loss curves for this material, 50 through 2000 hertz All non-oriented silicon steels All non-oriented silicon steels All other materials All other materials Summary magnetization and total core loss curves for as-sheared .014 inch (.36 mm, 29 gauge) Di-Max M-19 fully processed cold-rolled non-oriented silicon steel showing their relation to these properties for other materials found in Lamination Steels Third Edition. See the following pages for detailed graphs and data values. Producer: AK Steel, Middletown, Ohio, USA, www.aksteel.com. Primary standard: ASTM A677 36F155. Information on this page is not guaranteed or endorsed by The Electric Motor Education and Research Foundation. Confirm material properties with material producer prior to use. © 2007 The Electric Motor Education and Research Foundation. MIT OCW excerpts prepared October 2008. This and the following five pages are excerpted from the Laminations Steels Third Edition CD-ROM published by the Electric Motor Education and Research Foundation and are intended for use in the Massachusetts Institute of T echnology OpenCourseWare program. Unauthorized duplication and distribution of this document in violation of the OpenCourseWare license is prohibited. Incorporation of this information in other publications or software, in whole or in part, in violation of the OpenCourseWare license or without specific authorization from Electric Motor Education and Research Foundation, is prohibited. Courtesy of the Electric Motor Education and Research Foundation. Used with permission. Please use the following citation when referring to these pages: Sprague, Steve, editor. 2007. Lamination Steels Third Edition, A Compendium of Lamination Steel Alloys Commonly Used in Electric Motors. South Dartmouth, Massachusetts: The Electric Motor Education and Research Foundation. CD-ROM. Non-Oriented Silicon Steels: AK Steel Di-Max M-19, Fully Processed, .014 inch (.36 mm, 29 gauge), MIT OCW Excerpts. Lamination Steels Third Edition is © 2007 by the Electric Motor Education and Research Foundation; ISBN 0971439125. Information about the complete CD-ROM can obtained from: The Electric Motor Education and Research Foundation, Post Office Box P182, South Dartmouth, Massachusetts 02748 USA tel: 508.979.5935 fax: 508.979.5845 email: info@smma.org www.smma.org E M E R F L S T ' E ' ' Non-Oriented Silicon Steels E E M E R F L S T ' ' ' AK Steel Di-Max M-19 Fully Processed .014 inch (.36 mm, 29 gauge) Magnetization Curves Summary Graphs Magnetization Data Core Loss Curves Data Exciting Power Data Spreadsheet Other Thicknesses .0185 inch .025 inch AK Steel Product Info AK Steel Non-Oriented Silicon Steel Menu Non-Oriented Silicon Steels Menu Lamination Steels Main Menu Magnetic Flux Density in Gausses – B Magnetization – B vs. H – by Frequency 22000 20000 15000 10000 5000 2000 1000 Magnetizing Force in Oersteds – H Magnetizing Force in Amperes per Meter – H LAMINATION STEELS THIRD EDITION Excerpts specially prepared for the Massachusetts Institute of T echnology by the Electric Motor Education and Research Foundation 10 100 1000 10000 100000 0.1 1 10 100 1000 0.4 0.5 0.6 0.7 0.8 0.9 1 1.25 1.5 2 2.5 3 3.5 4 5 6 7 8 9 10 O e 30 35 40 50 60 70 80 90 100 125 150 200 250 300 350 400 500 600 700 800 1000 A/m 10 15 20 25 30 35 40 50 60 70 80 90 100 125 150 200 250 300 350 400 500 600 700 800 1000 Oe 800 1000 1500 2000 3000 4000 5000 6000 8000 10000 15000 20000 30000 40000 50000 60000 80000 A/m 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 11000 12000 13000 14000 15000 G 15000 16000 17000 17000 18000 19000 20000 21000 G 1 Tesla 1.5 Tesla 50 60 100 150 200 300 400 600 1000 1500 2000 Frequency in Hertz 50 60 100 150 200 300 400 600 1000 1500 2000 Hertz DC DC DC DC T ypical DC and derived AC magnetizing force of as-sheared .014 inch (.36 mm, 29 gauge) Di-Max M-19 fully processed cold-rolled non-oriented silicon steel. See magnetization data page for data values. DC curve developed from published and AC curves from previously unpublished data for Di-Max M-19 provided by AK Steel, 2000. AC magnetization data derived from exciting power data; see exciting power data page for source data and magnetization data page for conversion information. Chart prepared by EMERF , 2004. Information on this page is not guaranteed or endorsed by The Electric Motor Education and Research Foundation. Confirm material properties with material producer prior to use. © 2007 The Electric Motor Education and Research Foundation. MIT OCW excerpts prepared October 2008. This page is excerpted from the Laminations Steels Third Edition CD-ROM published by the Electric Motor Education and Research Foundation and is intended for use in the Massachusetts Institute of T echnology OpenCourseWare program. Unau thorized duplication and distribution of this document in violation of the OpenCourseWare license is prohibited. Please refer to the Summary Graphs page, reached by the link at left, for additional information concerning this document. Non-Oriented Silicon Steels E E M E R F L S T ' ' ' AK Steel Di-Max M-19 Fully Processed .014 inch (.36 mm, 29 gauge) Magnetization Data Summary Graphs Magnetization Curves Core Loss Curves Data Exciting Power Data Spreadsheet Other Thicknesses .0185 inch .025 inch AK Steel Product Info AK Steel Non-Oriented Silicon Steel Menu Non-Oriented Silicon Steels Menu Lamination Steels Main Menu Magnetic Flux Density in Gausses – B LAMINATION STEELS THIRD EDITION Excerpts specially prepared for the Massachusetts Institute of T echnology by the Electric Motor Education and Research Foundation Magnetization – B vs. H DC and Derived AC Magnetizing Force in Oersteds and Amperes per Meter at Various Frequencies – H Oe A/m DC 50 Hz 60 Hz 100 Hz 150 Hz 200 Hz 300 Hz 400 Hz 600 Hz 1000 Hz 1500 Hz 2000 Hz 0.333 26.5 0.334 26.6 0.341 27.1 0.349 27.8 0.356 28.3 0.372 29.6 0.385 30.6 0.412 32.8 0.485 38.6 0.564 44.9 0.642 51.1 1000 0.401 31.9 0.475 37.8 0.480 38.2 0.495 39.4 0.513 40.8 0.533 42.4 0.567 45.1 0.599 47.7 0.661 52.6 0.808 64.3 0.955 76.0 1.09 86.9 2000 0.564 44.9 0.659 52.4 0.669 53.2 0.700 55.7 0.739 58.8 0.777 61.8 0.846 67.3 0.911 72.5 1.04 82.8 1.30 103 1.56 124 1.80 143 4000 0.845 67.3 0.904 71.9 0.916 72.9 0.968 77.0 1.03 82.0 1.09 87.1 1.21 96.4 1.33 105 1.55 124 2.00 159 2.48 198 2.95 235 7000 1.34 106 1.25 99.3 1.26 101 1.32 105 1.40 112 1.48 118 1.65 131 1.82 145 2.17 173 2.87 228 3.70 294 4.53 361 10000 2.06 164 1.71 136 1.72 137 1.78 141 1.86 148 1.94 155 2.13 169 2.33 185 2.74 218 3.66 291 4.77 380 5.89 469 12000 2.95 235 2.21 176 2.22 177 2.27 181 2.34 186 2.42 193 2.61 208 2.82 224 3.24 258 4.27 340 5.50 438 13000 5.47 435 3.51 279 3.51 279 3.57 284 3.63 289 3.69 294 3.86 307 4.13 329 14000 13.9 1109 8.28 659 8.31 662 8.37 666 8.37 666 8.48 675 8.65 689 9.74 775 15000 22.8 1813 13.6 1084 13.6 1081 13.8 1095 13.7 1092 13.8 1096 14.1 1122 16.5 1313 15500 35.2 2802 21.6 1718 21.7 1728 21.8 1735 21.8 1738 21.9 1742 16000 50.9 4054 32.4 2577 32.5 2587 32.6 2597 32.5 2590 32.6 2594 16500 70.3 5592 46.1 3670 46.2 3680 46.4 3692 46.6 3712 46.6 3711 17000 122 9711 18000 202 16044 19000 394 31319 20000 1112 88491 21000 T ypical DC and derived AC magnetizing force of as-sheared .014 inch (.36 mm, 29 gauge) Di-Max M-19 fully processed cold-rolled non-oriented silicon steel. DC values in Oersteds from published AK Steel documents. AC values in Oersteds developed from previously unpublished exciting power information provided by AK Steel, 2000. AC values have been derived from RMS Exciting Power using the following formulas: 88.19 × Density (g/cc) × RMS Exciting Power (VA/lb) Magnetizing Force in Oersteds = Magnetic Flux Density (kG) × Frequency (Hz) Density of M-19 = 7.65 g/cc Values in Amperes per meter = Oersteds × 79.58 See exciting power data page for AC exciting power source data. Magnetizing force formula developed by AK Steel; use only for deriving magnetizing force of AK Steel non-oriented silicon steel. Data table preparation, including conversion of data values, by EMERF , 2004. Information on this page is not guaranteed or endorsed by The Electric Motor Education and Research Foundation. Confirm material properties with material producer prior to use. © 2007 The Electric Motor Education and Research Foundation. MIT OCW excerpts prepared October 2008. This page is excerpted from the Laminations Steels Third Edition CD-ROM published by the Electric Motor Education and Research Founda tion and is intended for use in the Massachusetts Institute of T echnology OpenCourseWare program. Unauthorized duplication and distribu tion of this document in violation of the OpenCourseWare license is prohibited. Please refer to the Summary Graphs page, reached by the link at left, for additional information concerning this document. Non-Oriented Silicon Steels E E M E R F L S T ' ' ' AK Steel Di-Max M-19 Fully Processed .014 inch (.36 mm, 29 gauge) Core Loss Curves Summary Graphs Magnetization Curves Data Core Loss Data Exciting Power Data Spreadsheet Other Thicknesses .0185 inch .025 inch AK Steel Product Info AK Steel Non-Oriented Silicon Steel Menu Non-Oriented Silicon Steels Menu Lamination Steels Main Menu Core Loss in Watts per Kilogram – Pc Core Loss in Watts per Pound – Pc T otal Core Loss – Pc vs. B – by Frequency 200 400 100 200 100 10 10 1 1 0.1 0.1 0.01 0.02 Magnetic Flux Density in Gausses – B LAMINATION STEELS THIRD EDITION Excerpts specially prepared for the Massachusetts Institute of T echnology by the Electric Motor Education and Research Foundation 1000 2000 5000 10000 15000 19000 1 Tesla 1.5 Tesla 3000 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000 15000 16000 17000 18000 G 0.04 0.06 0.08 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10 12 14 16 18 20 25 30 35 40 W/lb 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10 12 14 16 18 20 25 30 35 40 50 60 70 80 90 100 W/kg 0.2 0.4 0.6 0.8 1 2 4 6 8 10 12 14 16 18 20 25 30 35 40 50 60 70 80 90 100 125 150 175 W/lb 0.4 0.6 0.8 1 2 4 6 8 10 12 14 16 18 20 25 30 35 40 50 60 70 80 90 100 125 150 175 200 225 250 275 300 350 400 W/kg Hertz 50 60 100 150 200 300 400 600 1000 1500 2000 Frequency in Hertz 50 60 100 150 200 300 400 600 1000 1500 2000 T ypical total AC core loss of as-sheared .014 inch (.36 mm, 29 gauge) Di-Max M-19 fully processed cold-rolled non-oriented silicon steel. See core loss data page for data values. Curves developed from previously unpublished information provided by AK Steel, 2000. Chart prepared by EMERF , 2004. Information on this page is not guaranteed or endorsed by The Electric Motor Education and Research Foundation. Confirm material properties with material producer prior to use. © 2007 The Electric Motor Education and Research Foundation. MIT OCW excerpts prepared October 2008. This page is excerpted from the Laminations Steels Third Edition CD-ROM published by the Electric Motor Education and Research Foundation and is intended for use in the Massachusetts Institute of T echnology OpenCourseWare program. Unau thorized duplication and distribution of this document in violation of the OpenCourseWare license is prohibited. Please refer to the Summary Graphs page, reached by the link at left, for additional information concerning this document. Magnetic Flux Density in Gausses – B Non-Oriented Silicon Steels E E M E R F L S T ' ' ' AK Steel Di-Max M-19 Fully Processed .014 inch (.36 mm, 29 gauge) Core Loss Data Summary Graphs Magnetization Curves Data Core Loss Curves Exciting Power Data Spreadsheet Other Thicknesses .0185 inch .025 inch AK Steel Product Info AK Steel Non-Oriented Silicon Steel Menu Non-Oriented Silicon Steels Menu Lamination Steels Main Menu LAMINATION STEELS THIRD EDITION Excerpts specially prepared for the Massachusetts Institute of T echnology by the Electric Motor Education and Research Foundation T otal Core Loss – Pc vs. B Core Loss in Watts per Pound and Watts per Kilogram at Various Frequencies – Pc W/lb W/kg 50 Hz 60 Hz 100 Hz 150 Hz 200 Hz 300 Hz 400 Hz 600 Hz 1000 Hz 1500 Hz 2000 Hz 0.008 0.0176 0.009 0.0198 0.017 0.0375 0.029 0.0639 0.042 0.0926 0.074 0.163 0.112 0.247 0.205 0.452 0.465 1.02 0.9 1.98 1.45 3.20 1000 0.031 0.0683 0.039 0.0860 0.072 0.159 0.119 0.262 0.173 0.381 0.300 0.661 0.451 0.994 0.812 1.79 1.79 3.94 3.37 7.43 5.32 11.7 2000 0.109 0.240 0.134 0.295 0.252 0.555 0.424 0.934 0.621 1.37 1.09 2.39 1.64 3.60 2.96 6.52 6.34 14.0 11.8 26.1 18.5 40.8 4000 0.273 0.602 0.340 0.749 0.647 1.43 1.11 2.44 1.64 3.61 2.92 6.44 4.45 9.81 8.18 18.0 17.8 39.1 33.7 74.3 54.0 119 7000 0.494 1.09 0.617 1.36 1.18 2.61 2.04 4.50 3.06 6.74 5.53 12.2 8.59 18.9 16.2 35.7 36.3 80.0 71.5 158 117 257 10000 0.687 1.51 0.858 1.89 1.65 3.63 2.86 6.30 4.29 9.46 7.83 17.3 12.2 26.9 23.5 51.8 54.3 120 109 240 179 395 12000 0.812 1.79 1.01 2.23 1.94 4.28 3.36 7.41 5.06 11.2 9.23 20.3 14.4 31.8 27.8 61.3 65.1 143 132 291 13000 0.969 2.14 1.21 2.66 2.31 5.09 4.00 8.82 6.00 13.2 10.9 24.1 17.0 37.5 14000 1.16 2.56 1.45 3.19 2.77 6.11 4.76 10.5 7.15 15.8 13.0 28.7 20.1 44.4 15000 1.26 2.77 1.56 3.44 2.99 6.59 5.15 11.4 7.71 17.0 13.9 30.7 21.6 47.6 15500 1.34 2.96 1.67 3.67 3.18 7.01 5.47 12.0 8.19 18.0 16000 1.42 3.13 1.76 3.89 3.38 7.44 5.79 12.8 8.67 19.1 16500 1.49 3.29 1.85 4.08 3.54 7.80 6.09 13.4 9.13 20.1 17000 2.00 4.40 18000 T ypical total AC core loss of as-sheared .014 inch (.36 mm, 29 gauge) Di-Max M-19 fully processed cold-rolled non-oriented silicon steel. Watts per pound values from previously unpublished information provided by AK Steel, 2000. Data table preparation, including conversion of data values, by EMERF , 2004. Watts per kilogram values developed using this formula: Watts per Kilogram = Watts per Pound × 2.204 . Information on this page is not guaranteed or endorsed by The Electric Motor Education and Research Foundation. Confirm material properties with material producer prior to use. © 2007 The Electric Motor Education and Research Foundation. MIT OCW excerpts prepared October 2008. This page is excerpted from the Laminations Steels Third Edition CD-ROM published by the Electric Motor Education and Research Foundation and is intended for use in the Massachusetts Institute of Technology OpenCourseWare program. Unau thorized duplication and distribution of this document in violation of the OpenCourseWare license is prohibited. Please refer to the Summary Graphs page, reached by the link at left, for additional information concerning this document. Magnetic Flux Density in Gausses – B Non-Oriented Silicon Steels E E M E R F L S T ' ' ' AK Steel Di-Max M-19 Fully Processed .014 inch (.36 mm, 29 gauge) Exciting Power Data Summary Graphs Magnetization Curves Data Core Loss Curves Data Spreadsheet Other Thicknesses .0185 inch .025 inch AK Steel Product Info AK Steel Non-Oriented Silicon Steel Menu Non-Oriented Silicon Steels Menu Lamination Steels Main Menu LAMINATION STEELS THIRD EDITION Excerpts specially prepared for the Massachusetts Institute of T echnology by the Electric Motor Education and Research Foundation Exciting Power Exciting Power in Volt-amps per Pound and Volt-amps per Kilogram at Various Frequencies V-A/lb V-A/kg 50 Hz 60 Hz 100 Hz 150 Hz 200 Hz 300 Hz 400 Hz 600 Hz 1000 Hz 1500 Hz 2000 Hz 0.025 0.055 0.030 0.066 0.051 0.112 0.078 0.172 0.106 0.234 0.165 0.364 0.228 0.503 0.366 0.807 0.719 1.58 1.25 2.76 1.90 4.20 1000 0.07 0.154 0.085 0.187 0.147 0.324 0.228 0.503 0.316 0.696 0.504 1.11 0.710 1.56 1.18 2.59 2.40 5.28 4.25 9.36 6.48 14.3 2000 0.195 0.430 0.238 0.525 0.415 0.915 0.657 1.45 0.921 2.03 1.51 3.32 2.16 4.76 3.70 8.15 7.70 17.0 13.9 30.5 21.4 47.1 0.469 1.03 0.57 1.26 1.00 2.21 1.60 3.53 2.27 5.00 3.77 8.31 5.50 12.1 9.67 21.3 20.8 45.7 38.7 85.2 61.3 135 4000 7000 0.925 2.04 1.12 2.48 1.96 4.32 3.12 6.88 4.39 9.68 7.33 16.2 10.8 23.8 19.3 42.5 42.5 93.7 82.2 181 134 296 10000 1.52 3.34 1.83 4.04 3.16 6.96 4.96 10.9 6.91 15.2 11.4 25.0 16.6 36.5 29.2 64.4 65.1 143 127 280 210 462 12000 2.13 4.69 2.57 5.66 4.38 9.65 6.77 14.9 9.34 20.6 15.1 33.2 21.7 47.8 37.5 82.7 82.3 181 159 350 13000 3.64 8.02 4.37 9.63 7.41 16.3 11.3 24.9 15.3 33.8 24.0 52.9 34.3 75.6 9.20 20.3 11.1 24.4 18.6 41.0 27.9 61.5 37.7 83.1 57.7 127 86.6 191 14000 15000 15.6 34.5 18.7 41.3 31.6 69.6 47.3 104 63.3 140 97.2 214 152 334 15500 25.6 56.4 30.9 68.1 51.7 114 77.7 171 104 229 16000 39.6 87.3 47.7 105 79.8 176 119 263 159 351 16500 58.1 128 69.9 154 117 258 176 389 235 518 17000 T ypical RMS Exciting Power of as-sheared .014 inch (.36 mm, 29 gauge) Di-Max M-19 fully processed cold-rolled non-oriented silicon steel. Volt-amps per pound values from previously unpublished information provided by AK Steel, 2000. Data table preparation, including conversion of data values, by EMERF , 2004. Volt-amps per kilogram developed using this formula: Volt-amps per kilogram = Volt-amps per pound × 2.204 . Information on this page is not guaranteed or endorsed by The Electric Motor Education and Research Foundation. Confirm material properties with material producer prior to use. © 2007 The Electric Motor Education and Research Foundation. MIT OCW excerpts prepared October 2008. This page is excerpted from the Laminations Steels Third Edition CD-ROM published by the Electric Motor Education and Research Foundation and is intended for use in the Massachusetts Institute of T echnology OpenCourseWare program. Unau thorized duplication and distribution of this document in violation of the OpenCourseWare license is prohibited. Please refer to the Summary Graphs page, reached by the link at left, for additional information concerning this document. MIT OpenCourseWare 6.061 / 6.690 Introduction to Electric Power Systems Spring 2011 For information about citing these materials or our Terms of Use, visit: |
12863 | https://mathmonks.com/monomial/dividing-monomials | Dividing Monomials - Rules and Examples
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Solved Examples
Last modified on January 10th, 2024
chapter outline
Solved Examples
Home » Algebra » Monomial » Dividing Monomials
Dividing Monomials
Dividing monomials is a way of simplifying that involves the division of their coefficients followed by that of their variables.
Here, we use the exponent rule a m a n=a m−n, where a ≠ 0, while dividing the variables with the same base and different exponents.
Let us divide the monomials, 15p 7 and 3p 2.
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Step 1: Writing In Quotient Form
15 p 7 3 p 2
Step 2: Dividing the Coefficients
= (15 3)(p 7 p 2)
= 5(p 7 p 2)
Step 3: Dividing the Variables
= 5p 7-2
Thus, the quotient is 5p 5.
Let us consider another example 18 m n 3 3 m 2
It is written as (18 3)(m n 3 m 2)
On dividing the coefficients, 6(m n 3 m 2)
Now, applying the exponent rules, we get 6m 1-2 n 3 = 6m-1 n 3
Thus, the quotient is 6 n 3 m.
Solved Examples
Divide the monomials 15pq by -5p.
Solution:
On dividing 15pq by -5p, we get 15 p q−5 p
= −(15 5)(p q p)
Applying the exponents rule, we get -3p 1-1 q
Thus, the quotient is -3q
Divide 25p 15 q 13 by 5p 24 q 4.
Solution:
On dividing the given monomials, we get 25 p 15 q 13 5 p 24 q 4
= (25 5)(p 15 q 13 p 24 q 4)
Applying th exponents rule, we get 5 p 15−24 q 13−4
= 5 p−9 q 9
= 5 q 9 p 9
Thus, the quotient is 5(q p)9
Simplify 64 a 9 b 11 c 5 8 a 15 c 2
Solution:
Here, 64 a 9 b 11 c 5 8 a 15 c 2
= (64 8)(a 9 b 11 c 5 a 15 c 2)
= 8(a 9−15 b 11 c 5−2)
= 8(a−6 b 11 c 3)
= 8 b 11 c 3 a 6
Thus, 64 a 9 b 11 c 5 8 a 15 c 2 is simplified to 8 b 11 c 3 a 6
What will be the coefficient of the quotient of 16m 4 y 2 and 24my 2?
Solution:
On dividing the given monomials, the quotient is written as 16 m 4 y 2 24 m y 2
Now, by dividing the coefficients and the variables individually, we get
(16 24)(m 4 y 2 m y 2)
Applying the exponents rule, we get
(2 3)(m 4−1 y 2−2)
= (2 3)m 3, which is the required quotient.
Thus, the coefficient of the quotient is 2 3
Last modified on January 10th, 2024
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12864 | https://www.bigideasmath.com/protected/content/ipe/grade%207/additional_topics/msfl7wbad_14.pdf | 98 Chapter 14 Linear Inequalities Lesson 14.1 An inequality is a mathematical sentence that compares expressions. It contains the symbols <, >, ≤ , or ≥ . To write an inequality, look for the following phrases to determine where to place the inequality symbol.
Key Vocabulary inequality solution of an inequality solution set graph of an inequality EXAMPLE Writing an Inequality 1 A number w minus 3.5 is less than or equal to −2. Write this sentence as an inequality.
A number w minus 3.5 is less than or equal to −2. w − 3.5 ≤ −2 An inequality is w − 3.5 ≤ −2.
Write the word sentence as an inequality. 1. A number b is fewer than 30.4. 2. Twice a number k is at least − 7 — 10 .
A solution of an inequality is a value that makes the inequality true. An inequality can have more than one solution. The set of all solutions of an inequality is called the solution set.
Exercises 6 – 9 Reading The symbol ≥ / means “is not greater than or equal to.” Value of x x + 5 ≥ −2 Is the inequality true?
−6 −6 + 5 ≥ ? −2 −1 ≥ −2 ✓ yes −7 −7 + 5 ≥ ? −2 −2 ≥ −2 ✓ yes −8 −8 + 5 ≥ ? −2 −3 ≥ / −2 ✗ no Inequality Symbols Symbol < > ≤ ≥ Key Phrases ● is less than ● is fewer than ● is greater than ● is more than ● is less than or equal to ● is at most ● is no more than ● is greater than or equal to ● is at least ● is no less than Section 14.1 Writing and Graphing Inequalities 99 EXAMPLE Checking Solutions 2 Tell whether −4 is a solution of the inequality. a. x + 8 < −3 b. −4.5x > −21 x + 8 < −3 Write the inequality. −4.5x > −21 −4 + 8 < ? −3 Substitute −4 for x. −4.5(−4) > ? −21 4 < / −3 ✗ Simplify. 18 > −21 ✓ 4 is not less than −3. 18 is greater than −21. So, −4 is not a solution So, −4 is a solution of the inequality. of the inequality.
Tell whether −6 is a solution of the inequality. 3. c + 4 < −1 4. 5 − m ≤ 10 5. 21 ÷ x ≥ −3.5 The graph of an inequality shows all of the solutions of the inequality on a number line. An open circle is used when a number is not a solution. A closed circle is used when a number is a solution. An arrow to the left or right shows that the graph continues in that direction.
Exercises 11–16 EXAMPLE Graphing an Inequality 3 Graph y ≤ −3.
Graph the inequality on a number line. 6. b > −8 7. g ≤ 1.4 8. r < − 1 — 2 9. v ≥ √— 0.09 Exercises 17–20 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 Test a number to the left of −3.
y = −4 is a solution.
Test a number to the right of −3.
y = 0 is not a solution.
Use a closed circle because −3 is a solution.
−7 −6 −5 −4 −3 −2 −1 0 1 2 3 Shade the number line on the side where you found the solution.
Exercises 14.1 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= 100 Chapter 14 Linear Inequalities 1. VOCABULARY Would an open circle or a closed circle be used in the graph of the inequality k < 250? Explain. 2. DIFFERENT WORDS, SAME QUESTION Which is different? Write “both” inequalities. w is greater than or equal to −7. w is no less than −7. w is no more than −7. w is at least −7. 3. REASONING Do x ≥ −9 and −9 ≥ x represent the same inequality? Explain.
Write an inequality for the graph. Then, in words, describe all the values of x that make the inequality true. 4. −3 0 3 6 9 12 15 18 5. −7 −6 −5 −4 −3 −2 −1 Write the word sentence as an inequality.
6. A number x is no less than −4. 7. A number y added to 5.2 is less than 23. 8. A number b multiplied by −5 is at most − 3 — 4 . 9. A number k minus 8.3 is greater than 48. 10. ERROR ANALYSIS Describe and correct the error in writing the word sentence as an inequality.
Tell whether the given value is a solution of the inequality. 11. s + 6 ≤ 12; s = 4 12. 15n > −3; n = −2 13. a − 2.5 ≤ 1.6; a = 4.1 14. −3.3q > −13; q = 4.6 15. 4 — 5 h ≥ −4; h = −15 16. 1 — 12 − p < 1 — 3 ; p = 1 — 6 Graph the inequality on a number line. 17. g ≥ −6 18. q > 1.25 19. z < 11 1 — 4 20. w ≤ − √— 289 21. DRIVING When you are driving with a learner’s license, a licensed driver who is 21 years of age or older must be with you. Write an inequality that represents this situation. Twice a number c is at least − 4 — 9 . 2c ≤ − 4 — 9 ✗ 1 2 3 Section 14.1 Writing and Graphing Inequalities 101 Solve the equation. Check your solution. 28. r − 12 = 3 29. 4.2 + p = 2.5 30. n − 3π = 7π 31. MULTIPLE CHOICE Which linear function relates y to x ? ○ A y = −0.5x − 3 ○ B y = 2x + 3 ○ C y = 0.5x − 3 ○ D y = 2x − 3 Tell whether the given value is a solution of the inequality.
22. 3p > 5 + p; p = 4 23. y — 2 ≥ y − 11; y = 18 24. VIDEO GAME RATINGS Each rating is matched with the inequality that represents the recommended ages of players. Your friend is old enough to play “E 10+” games. Is your friend old enough to play “T” games? Explain.
25. SCUBA DIVING Three requirements for a scuba diving training course are shown.
a. Write and graph three inequalities that represent the requirements.
b. You can swim 10 lengths of a 25-yard pool. Do you satisfy the swimming requirement of the course? Explain.
26. LUGGAGE On an airplane, the maximum sum of the length, width, and height of a carry-on bag is 45 inches. Find three different sets of dimensions that are reasonable for a carry-on bag.
27. A number m is less than another number n. The number n is less than or equal to a third number p. a. Write two inequalities representing these relationships. b. Describe the relationship between m and p. c. Can m be equal to p ? Explain.
x −1 0 1 2 y −5 −3 −1 1 x q 3 x q 6 x q 10 x q 13 x q 17 6 3 13 17 10 h w The ESRB rating icons are registered trademarks of the Entertainment Software Association.
102 Chapter 14 Linear Inequalities Lesson 14.2 Addition Property of Inequality Words If you add the same number to each side of an inequality, the inequality remains true.
Numbers −3 < 2 Algebra x − 3 > −10 + 4 + 4 + 3 + 3 1 < 6 x > −7 Subtraction Property of Inequality Words If you subtract the same number from each side of an inequality, the inequality remains true.
Numbers −3 < 1 Algebra x + 7 > −20 − 5 − 5 − 7 − 7 −8 < −4 x > −27 These properties are also true for ≤ and ≥ .
Study Tip You can solve inequalities the same way you solve equations. Use inverse operations to get the variable by itself. Study Tip To check a solution, you check some numbers that are solutions and some that are not. EXAMPLE Solving an Inequality Using Addition 1 Solve x − 6 ≥ −10. Graph the solution. x − 6 ≥ −10 Write the inequality. + 6 + 6 Add 6 to each side. x ≥ −4 Simplify. The solution is x ≥ −4. −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 x ≥ −4 Check: x = −5 is not a solution.
Check: x = 0 is a solution.
Solve the inequality. Graph the solution. 1. b − 2 > −9 2. m − 3.8 ≤ 5 3. 1 — 4 > y − 1 — 4 Undo the subtraction. Section 14.2 Solving Inequalities Using Addition or Subtraction 103 EXAMPLE Solving an Inequality Using Subtraction 2 Solve −8 > 1.4 + x. Graph the solution. −8 > 1.4 + x Write the inequality. − 1.4 − 1.4 Subtract 1.4 from each side. −9.4 > x Simplify. The solution is x < −9.4. −10.0−9.9 −9.8 −9.7 −9.6 −9.5 −9.4 −9.3 −9.2 −9.1 −9.0 x < −9.4 Solve the inequality. Graph the solution. 4. k + 5 ≤ −3 5. 5 — 6 ≤ z + 2 — 3 6. p + 0.7 > −2.3 Reading The inequality −9.4 > x is the same as x < −9.4.
Exercises 6 –17 EXAMPLE Real-Life Application 3 On a train, carry-on bags can weigh no more than 50 pounds. Your bag weighs 24.8 pounds. Write and solve an inequality that represents the amount of weight you can add to your bag.
Words Weight of your bag plus amount of weight you can add is no more than the weight limit.
Variable Let w be the possible weight you can add.
Inequality 24.8 + w ≤ 50 24.8 + w ≤ 50 Write the inequality. − 24.8 − 24.8 Subtract 24.8 from each side. w ≤ 25.2 Simplify. You can add no more than 25.2 pounds to your bag. 7. WHAT IF? Your carry-on bag weighs 32.5 pounds. Write and solve an inequality that represents the possible weight you can add to your bag.
Undo the addition.
Exercises 14.2 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= 104 Chapter 14 Linear Inequalities 1. REASONING Is the inequality r − 5 ≤ 8 the same as 8 ≤ r − 5? Explain. 2. WHICH ONE DOESN’T BELONG? Which inequality does not belong with the other three? Explain your reasoning. c + 7 — 2 ≤ 3 — 2 c + 7 — 2 ≥ 3 — 2 3 — 2 ≥ c + 7 — 2 c − 3 — 2 ≤ − 7 — 2 Graph the inequality. 3. x ≥ − 6 4. p < 2 5. 4 ≤ w Solve the inequality. Graph the solution. 6. y − 3 ≥ 7 7. t − 8 > −4 8. n + 11 ≤ 20 9. a + 7 > −1 10. 5 < v − 1 — 2 11. 1 — 5 > d + 4 — 5 12. − 2 — 3 ≤ g − 1 — 3 13. m + 7 — 4 ≤ 11 — 4 14. 11.2 ≤ k + 9.8 15. h − 1.7 < −3.2 16. 0 > s + π 17. 5 ≥ u − 4.5 18. ERROR ANALYSIS Describe and correct the error in graphing the solution of the inequality. 19. PELICAN The maximum volume of a great white pelican’s bill is about 700 cubic inches. a. A pelican scoops up 100 cubic inches of water. Write and solve an inequality that represents the additional volume the bill can contain. b. A pelican’s stomach can contain about one-third the maximum amount that its bill can contain. Write an inequality that represents the volume of the pelican’s stomach.
1 2 5 ≥ x − 5 10 ≥ x ✗ 8 9 10 11 12 13 Section 14.2 Solving Inequalities Using Addition or Subtraction 105 Solve the equation. 28. 6 = 3x 29. r — 5 = 2 30. 4c = 15 31. 8 = 2 — 3 b 32. MULTIPLE CHOICE Which fraction is equivalent to 3.8? ○ A 5 — 19 ○ B 19 — 5 ○ C 12 — 15 ○ D 12 — 5 Write and solve an inequality that represents the value of x.
20. The perimeter is less 21. The base is greater 22. The perimeter is less than 16 feet. than the height. than or equal to 5 feet.
4 ft 4 ft x x + 2 10 m x 12 in.
12 in.
10 in.
10 in.
23. REASONING The solution of w + c ≤ 8 is w ≤ 3. What is the value of c ? 24. FENCE The hole for a fence post is 2 feet deep. The top of the fence post needs to be at least 4 feet above the ground. Write and solve an inequality that represents the required length of the fence post. 25. VIDEO GAME You need at least 12,000 points to advance to the next level of a video game. a. Write and solve an inequality that represents the number of points you need to advance. b. You fi nd a treasure chest that increases your score by 60%. How does this change the inequality? 26. POWER A circuit overloads at 1800 watts of electricity. A microwave that uses 1100 watts of electricity is plugged into the circuit. a. Write and solve an inequality that represents the additional number of watts you can plug in without overloading the circuit. b. In addition to the microwave, what two appliances in the table can you plug in without overloading the circuit? 27. The maximum surface area of the solid is 15π square millimeters. Write and solve an inequality that represents the height of the cylinder.
Appliance Watts Clock radio 50 Blender 300 Hot plate 1200 Toaster 800 TIME LEFT: 1 min.
CURRENT SCORE: 4500 h 2 mm 106 Chapter 14 Linear Inequalities 14 Study Help Make a four square to help you study these topics. 1. inequality ( > ) 2. inequality ( ≤ ) 3. inequality ( ≥ ) 4. solving an inequality using addition 5. solving an inequality using subtraction After you complete this chapter, make four squares for the following topics. 6. solving an inequality using multiplication 7. solving an inequality using division You can use a four square to organize information about a topic. Each of the four squares can be a category, such as defi nition, vocabulary, example, non-example, words, algebra, table, numbers, visual, graph, or equation. Here is an example of a four square for an inequality.
“Sorry, but I have limited space in my four square. I needed pet names with only three letters.” INEQUALITY (<) Graph: To graph x < 2, draw an open circle at x = 2. Then draw an arrow pointing to the left.
Definition: A mathematical sentence that uses an inequality symbol (<).
Example: Words: A number x is less than 2.
Symbols: x < 2 Vocabulary: Key phrases that can represent an inequality: • i s less than • i s fewer than −3 −2 −1 0 1 2 3 Sections 14.1– 14.2 Quiz 107 Quiz 14.1– 14.2 Write the word sentence as an inequality. 1. A number x plus 1 is less than −13. 2. A number t minus 1.6 is at most 9.
Tell whether the given value is a solution of the inequality. 3. 12n < −2; n = −1 4. y + 4 < −3; y = −7 Graph the inequality on a number line. 5. x > −10 6. y ≤ 3 — 5 7. w < 6.8 Solve the inequality. 8. x − 2 < 4 9. g + 14 ≥ 30 10. h − 1 ≤ −9 11. s + 3 > −7 12. v − 3 — 4 < 0 13. 3 — 2 < p + 1 — 2 14. WATERCRAFT You must be at least 14 years old to operate a personal watercraft in Florida. Write an inequality that represents this situation. 15. REASONING The solution of x − a > 4 is x > 11. What is the value of a ? 16. MP3 PLAYER Your MP3 player can store up to 8 gigabytes of media. You transfer 3.5 gigabytes of media to the MP3 player. Write and solve an inequality that represents the amount of memory available on the MP3 player. 17. LIFEGUARD Three requirements for a lifeguard training course are shown. a. Write and graph three inequalities that represent the requirements. b. You can swim 350 feet. Do you satisfy the swimming requirement of the course? Explain.
Lifeguard Training Requirements Swim at least 100 yards Tread water for at least 5 minutes Swim 10 yards or more underwater without taking a breath LIFEGUARDS NEEDED LIFEGUARDS NEEDED Take Our Training Course NOW!!!
108 Chapter 14 Linear Inequalities Lesson 14.3 Remember Multiplication and division are inverse operations.
EXAMPLE Solving an Inequality Using Multiplication 1 Solve x — 8 > −5. Graph the solution. x — 8 > −5 Write the inequality. 8 ⋅ x — 8 > 8 ⋅ (−5) Multiply each side by 8. x > −40 Simplify. The solution is x > −40. Solve the inequality. Graph the solution. 1. a ÷ 2 < 4 2. n — 7 ≥ −1 3. −6.4 ≥ w — 5 Undo the division.
−90 −80 −70 −60 −50 −40 −30 −20 −10 0 10 x > −40 Check: x = 0 is a solution.
Check: x = −80 is not a solution.
Multiplication and Division Properties of Inequality (Case 1) Words If you multiply or divide each side of an inequality by the same positive number, the inequality remains true.
Numbers −6 < 8 6 > −8 2 ⋅ (−6) < 2 ⋅ 8 6 — 2 > −8 — 2 −12 < 16 3 > −4 Algebra x — 2 < −9 4x > −12 2 ⋅ x — 2 < 2 ⋅ (−9) 4x — 4 > −12 — 4 x < −18 x > −3 These properties are also true for ≤ and ≥ . Section 14.3 Solving Inequalities Using Multiplication or Division 109 EXAMPLE Solving an Inequality Using Division 2 Solve 3x ≤ −24. Graph the solution. 3x ≤ −24 Write the inequality. 3x — 3 ≤ −24 — 3 Divide each side by 3. x ≤ −8 Simplify. The solution is x ≤ −8. Solve the inequality. Graph the solution. 4. 4b ≥ 36 5. 2k > −10 6. −18 > 1.5q Undo the multiplication.
Exercises 10 –18 A negative sign in an inequality does not necessarily mean you must reverse the inequality symbol. Only reverse the inequality symbol when you multiply or divide both sides by a negative number.
Common Error −16 −14 −12 −10 −8 −6 −4 −2 0 2 4 x ≤ −8 Check: x = 0 is not a solution.
Check: x = −10 is a solution.
Multiplication and Division Properties of Inequality (Case 2) Words If you multiply or divide each side of an inequality by the same negative number, the direction of the inequality symbol must be reversed for the inequality to remain true.
Numbers −6 < 8 6 > −8 (−2) ⋅ (−6) > (−2) ⋅ 8 6 — −2 < −8 — −2 12 > −16 −3 < 4 Algebra x — −6 < 3 −5x > 30 −6 ⋅ x — −6 > −6 ⋅ 3 −5x — −5 < 30 — −5 x > −18 x < −6 These properties are also true for ≤ and ≥ .
110 Chapter 14 Linear Inequalities EXAMPLE Solving an Inequality Using Multiplication 3 Solve y — −3 > 2. Graph the solution. y — −3 > 2 Write the inequality. −3⋅ y — −3 < −3 ⋅ 2 Multiply each side by −3. Reverse the inequality symbol. y < −6 Simplify. The solution is y < −6. Undo the division.
EXAMPLE Solving an Inequality Using Division 4 Solve −7y ≤ −35. Graph the solution. −7y ≤ −35 Write the inequality. −7y — −7 ≥ −35 — −7 Divide each side by −7. Reverse the inequality symbol. y ≥ 5 Simplify. The solution is y ≥ 5. −1 0 1 2 3 4 5 6 7 8 9 y ≥ 5 Check: y = 6 is a solution.
Check: y = 0 is not a solution.
Solve the inequality. Graph the solution. 7. p — −4 < 7 8. x — −5 ≤ −5 9. 1 ≥ − 1 — 10 z 10. −9m > 63 11. −2r ≥ −22 12. −0.4y ≥ −12 Undo the multiplication.
Exercises 27–35 −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 y < −6 Check: y = −9 is a solution.
Check: y = 0 is not a solution. Section 14.3 Solving Inequalities Using Multiplication or Division 111 Exercises 14.3 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= 1. VOCABULARY Explain how to solve x — 6 < −5. 2. WRITING Explain how solving 2x < −8 is different from solving −2x < 8. 3. OPEN-ENDED Write an inequality that is solved using the Division Property of Inequality where the inequality symbol needs to be reversed.
Use a table to solve the inequality. 4. 4x < 4 5. −2x ≤ 2 6. −5x > 15 7. x — −3 ≥ 1 8. x — −2 > 5 — 2 9. x — 4 ≤ 3 — 8 Solve the inequality. Graph the solution. 10. 3n > 18 11. c — 4 ≤ −9 12. 1.2m < 12 13. −14 > x ÷ 2 14. w — 5 ≥ −2.6 15. 5 < 2.5k 16. 4x ≤ − 3 — 2 17. 2.6y ≤ −10.4 18. 10.2 > b — 3.4 19. ERROR ANALYSIS Describe and correct the error in solving the inequality. Write the word sentence as an inequality. Then solve the inequality. 20. The quotient of a number and 3 is at most 4. 21. A number divided by 8 is less than −2. 22. Four times a number is at least −12. 23. The product of 5 and a number is greater than 20. 24. CAMERA You earn $9.50 per hour at your summer job. Write and solve an inequality that represents the number of hours you need to work in order to buy a digital camera that costs $247. x — 2 < −5 2 ⋅ x — 2 > 2 ⋅ (−5) x > −10 ✗ 1 2 24.
112 Chapter 14 Linear Inequalities 25. COPIES You have $3.65 to make copies. Write and solve an inequality that represents the number of copies you can make. 26. SPEED LIMIT The maximum speed limit for a school bus is 55 miles per hour. Write and solve an inequality that represents the number of hours it takes to travel 165 miles in a school bus.
Solve the inequality. Graph the solution.
27. −2n ≤ 10 28. −5w > 30 29. h — −6 ≥ 7 30. −8 < − 1 — 3 x 31. −2y < −11 32. −7d ≥ 56 33. 2.4 > − m — 5 34. k — −0.5 ≤ 18 35. −2.5 > b — −1.6 36. ERROR ANALYSIS Describe and correct the error in solving the inequality. 37. CRITICAL THINKING Are all numbers greater than zero solutions of −x > 0? Explain. 38. TRUCKING By law, the maximum height (including freight) of a vehicle in Florida is 13.5 feet. a. Write and solve an inequality that represents the number of crates that can be stacked vertically on the bed of the truck. b. Five crates are stacked vertically on the bed of the truck. Is this legal? Explain.
Write and solve an inequality that represents the value of x. 39. Area ≥ 102 cm2 40. Area < 30 ft2 12 cm x 10 ft x −4m ≥ 16 −4m — −4 ≥ 16 — −4 m ≥ −4 ✗ 4 3 h 3.5 ft 28 in.
Not drawn to scale Section 14.3 Solving Inequalities Using Multiplication or Division 113 Solve the equation.
48. −4w + 5 = −11 49. 4(x − 3) = 21 50. v — 6 − 7 = 4 51. m + 300 — 4 = 96 52. MULTIPLE CHOICE Which measure can have more than one value for a given data set? ○ A mean ○ B median ○ C mode ○ D range 41. TRIP You and three friends are planning a trip. You want to keep the cost below $80 per person. Write and solve an inequality that represents the total cost of the trip. 42. REASONING Explain why the direction of the inequality symbol must be reversed when multiplying or dividing by the same negative number.
43. PROJECT Choose two musical artists to research. a. Use the Internet or a magazine to complete the table.
b. Find the average number of copies sold per month for each CD.
c. Use the release date to write and solve an inequality that represents the minimum average number of copies sold per month for each CD.
d. In how many months do you expect the number of copies of the second top selling CD to surpass the current number of copies of the top selling CD? Artist Name of CD Release Date Current Number of Copies Sold 1.
2. Describe all numbers that satisfy both inequalities. Include a graph with your description.
44. 3m > −12 and 2m < 12 45. n — 2 ≥ −3 and n — −4 ≥ 1 46. 2x ≥ −4 and 2x ≥ 4 47. m — −4 > −5 and m — 4 < 10 g h.
the table.
month s 114 Chapter 14 Linear Inequalities Lesson 14.4 You can solve multi-step inequalities the same way you solve multi-step equations.
EXAMPLE Solving Two-Step Inequalities 1 a. Solve 5x − 4 ≥ 11. Graph the solution. 5x − 4 ≥ 11 Write the inequality. + 4 + 4 Add 4 to each side. 5x ≥ 15 Simplify. 5x — 5 ≥ 15 — 5 Divide each side by 5. x ≥ 3 Simplify. The solution is x ≥ 3. −3 −2 −1 0 1 2 3 4 5 6 7 x ≥ 3 Check: x = 0 is not a solution.
Check: x = 4 is a solution. b. Solve y — −6 + 7 < 9. Graph the solution. y — −6 + 7 < 9 Write the inequality. −7 −7 Subtract 7 from each side. y — −6 < 2 Simplify. −6 ⋅ y — −6 > −6 ⋅ 2 Multiply each side by −6. Reverse the inequality symbol. y > −12 Simplify. The solution is y > −12.
−2 −4 −6 −8 −10 −12 −14 −16 −18 0 2 y > −12 Solve the inequality. Graph the solution. 1. 4b − 1 < 7 2. 8 + 9c ≥ −28 3. n — −2 + 11 > 12 Step 1: Undo the subtraction.
Step 2: Undo the multiplication.
Exercises 5–10 Section 14.4 Solving Multi-Step Inequalities 115 EXAMPLE Standardized Test Practice 2 Which graph represents the solution of −7(x + 3) ≤ 28? −7(x + 3) ≤ 28 Write the inequality. −7x − 21 ≤ 28 Use Distributive Property. + 21 + 21 Add 21 to each side. −7x ≤ 49 Simplify. −7x — −7 ≥ 49 — −7 Divide each side by −7. Reverse the inequality symbol. x ≥ −7 Simplify. The correct answer is ○ B .
EXAMPLE Real-Life Application 3 You need a mean score of at least 90 to advance to the next round of the trivia game. What score do you need on the fi fth game to advance?
Use the defi nition of mean to write and solve an inequality. Let x be the score on the fi fth game. 95 + 91 + 77 + 89 + x —— 5 ≥ 90 352 + x — 5 ≥ 90 Simplify. 5 ⋅ 352 + x — 5 ≥ 5 ⋅ 90 Multiply each side by 5. 352 + x ≥ 450 Simplify. − 352 − 352 Subtract 352 from each side. x ≥ 98 Simplify. You need at least 98 points to advance to the next round.
Solve the inequality. Graph the solution. 4. 2(k − 5) < 6 5. −4(n − 10) < 32 6. −3 ≤ 0.5(8 + y) 7. WHAT IF? In Example 3, you need a mean score of at least 88 to advance to the next round of the trivia game. What score do you need on the fi fth game to advance?
Remember The mean in Example 3 is equal to the sum of the game scores divided by the number of games.
Exercises 12–17 The phrase “at least” means greater than or equal to.
○ A −10 −9 −8 −7 −6 −5 −4 ○ B −10 −9 −8 −7 −6 −5 −4 ○ C 4 5 6 7 8 9 10 ○ D 4 5 6 7 8 9 10 Exercises 14.4 9+(-6)=3 3+(-3)= 4+(-9)= 9+(-1)= 116 Chapter 14 Linear Inequalities 1. WRITING Compare and contrast solving multi-step inequalities and solving multi-step equations. 2. OPEN-ENDED Describe how to solve the inequality 3(a + 5) < 9. 3. For what values of k will the 4. For what values of h will the perimeter of the octagon be surface area of the solid be less than or equal to 64 units? greater than 46 square units? k k k 1 2 k 1 2 4 4 4 4 5 3 h Solve the inequality. Graph the solution.
5. 7b + 4 ≥ 11 6. 2v − 4 < 8 7. 1 − m — 3 ≤ 6 8. 4 — 5 < 3w − 11 — 5 9. 1.8 < 0.5 − 1.3p 10. −2.4r + 9.6 ≥ 4.8 11. ERROR ANALYSIS Describe and correct the error in solving the inequality.
Solve the inequality. Graph the solution. 12. 6( g + 2) ≤ 18 13. 2( y − 5) ≤ 16 14. −10 ≥ 5 — 3 (h − 3) 15. − 1 — 3 (u + 2) > 5 16. 2.7 > 0.9(n − 1.7) 17. 10 > −2.5(z − 3.1) 18. ATM Write and solve an inequality that represents the number of $20 bills you can withdraw from the account without going below the minimum balance.
1 2 x — 4 + 6 ≥ 3 x + 6 ≥ 12 x ≥ 6 ✗ 74 ft 8 ft Section 14.4 Solving Multi-Step Inequalities 117 Find the area of the circle. 26. 10 mm 27. 25 in. 28. 66 m 29. MULTIPLE CHOICE What is the volume of the cube? ○ A 8 ft3 ○ B 16 ft3 ○ C 24 ft3 ○ D 32 ft3 Solve the inequality. Graph the solution.
19. 5x − 2x + 7 ≤ 15 + 10 20. 7b − 12b + 1.4 > 8.4 − 22 21. TYPING One line of text on a page uses about 3 — 16 of an inch. There are 1-inch margins at the top and bottom of a page. Write and solve an inequality to fi nd the number of lines that can be typed on a page that is 11 inches long.
22. WOODWORKING A woodworker builds a cabinet in 20 hours. The cabinet is sold at a store for $500. Write and solve an inequality that represents the hourly wage the store can pay the woodworker and still make a profi t of at least $100.
23. FIRE TRUCK The height of one story of a building is about 10 feet. The bottom of the ladder on the fi re truck must be at least 24 feet away from the building. Write and solve an inequality to fi nd the number of stories the ladder can reach.
24. DRIVE-IN A drive-in movie theater charges $3.50 per car. The drive-in has already admitted 100 cars. Write and solve an inequality to fi nd the number of cars the drive-in needs to admit to make at least $500. 25. For what values of r will the area of the shaded region be greater than or equal to 9(π − 2)?
r 2 ft Quiz 14.3 – 14.4 Solve the inequality. Graph the solution. 1. x ÷ 4 > 12 2. n — −6 ≥ −2 3. −4y ≥ 60 4. −2.3 ≥ p — 5 Write the word sentence as an inequality. Then solve the inequality. 5. The quotient of a number and 6 is more than 9. 6. Five times a number is at most −10.
Solve the inequality. Graph the solution. 7. 2m + 1 ≥ 7 8. n — 6 − 8 ≤ 2 9. 2 − j — 5 > 7 10. 5 — 4 > −3w − 7 — 4 11. FLOWERS A soccer team needs to raise $200 for new uniforms. The team earns $0.50 for each fl ower sold. Write and solve an inequality to fi nd the number of fl owers it must sell to meet or exceed its fundraising goal. 12. PARTY You buy lunch for guests at a party. You can spend no more than $100. You will spend $20 on beverages and $10 per guest on sandwiches. Write and solve an inequality to fi nd the number of guests you can invite to the party. 13. BOOKS You have a gift card worth $50. You want to buy several paperback books that cost $6 each. Write and solve an inequality to fi nd the number of books you can buy and still have at least $20 on the gift card. 14. GARDEN The area of the triangular garden must be less than 35 square feet. Write and solve an inequality that represents the value of b.
118 Chapter 14 Linear Inequalities 10 ft b ms.
an t or eral Chapter Test 119 Chapter Test 14 Write the word sentence as an inequality. 1. A number j plus 20.5 is greater than or equal to 50. 2. A number r multiplied by 1 — 7 is less than −14.
Tell whether the given value is a solution of the inequality. 3. v − 2 ≤ 7; v = 9 4. 3 — 10 p < 0; p = 10 5. −3n ≥ 6; n = −3 Solve the inequality. Graph the solution. 6. n − 3 > −3 7. x − 7 — 8 ≤ 9 — 8 8. −6b ≥ −30 9. y — −4 ≥ 13 10. 3v − 7 ≥ −13.3 11. −5(t + 11) < −60 12. VOTING U.S. citizens must be at least 18 years of age on Election Day to vote. Write an inequality that represents this situation. 13. GARAGE The vertical clearance for a hotel parking garage is 10 feet. Write and solve an inequality that represents the height (in feet) of the vehicle. 14. LUNCH BILL A lunch bill, including tax, is divided equally among you and fi ve friends. Everyone pays less than $8.75. Write and solve an inequality that describes the total amount of the bill. 15. TRADING CARDS You have $25 to buy trading cards online. Each pack of cards costs $4.50. Shipping costs $2.95. Write and solve an inequality to fi nd the number of packs of trading cards you can buy. 16. SCIENCE QUIZZES The table shows your scores on four science quizzes. What score do you need on the fi fth quiz to have a mean score of at least 80? Test 1 2 3 4 5 Score (%) 76 87 73 72 ?
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How to calculate exchange rates
How to calculate exchange rates
Here you will learn about how to calculate exchange rates, including how to convert between American dollars and foreign currencies.
Students will first learn how to calculate exchange rates as part of ratios and proportions in 6 th grade and 7 th grade.
Every week, we teach lessons on how to calculate exchange rates to students in schools and districts across the US as part of our online one-on-one math tutoring programs. On this page we’ve broken down everything we’ve learnt about teaching this topic effectively.
What is an exchange rate?
An exchange rate is the rate at which the money of one country can be changed for the money of another country. It can also be referred to as a foreign exchange rate and be seen as the price of one currency expressed in terms of another currency.
In the United States, our currency is the American dollar. The exchange rate tells us the value of \$1 in terms of a foreign country’s currency. Different countries have different currencies.
For example,
Some major currencies are…
[FREE] How To Calculate Exchange Rates Worksheet (Grade 6 to 7)
Use this worksheet to check your grade 6 to 7 students’ understanding of calculating exchange rates. 15 questions with answers to identify areas of strength and support!
[FREE] How To Calculate Exchange Rates Worksheet (Grade 6 to 7)
Use this worksheet to check your grade 6 to 7 students’ understanding of calculating exchange rates. 15 questions with answers to identify areas of strength and support!
You can use a known equivalent value between two currencies to calculate the exchange rate.
For example,
If £10 is equal to \$12.70 \; USD, what is the exchange rate from £ \; (GBP) to \$ \; (USD)?
When calculating the currency exchange rate from £ \; (GBP) to \$ \; (US dollars ), you want to know how many \$ are equal to £1.
This is the ratio of \$ to £, so set up the rate as \cfrac{\$ 12.70}{£ 10}.
In this case, the base currency is the £ \; (GBP), so divide both parts by 10, rounding the \$ \; (USD) to the nearest cent:
\cfrac{\$ 12.70 \div 10}{£ 10 \div 10}=\cfrac{\$ 1.27}{£ 1}
The exchange rate from £ \; (GBP) to \$ \; (USD) is 1.27.
The exchange rates (currency conversions) are constantly changing depending on the country’s economic status. It is the role of the central bank to monitor these fluctuations and adjust the country’s monetary policy accordingly, which in turn impacts the currency’s value in the forex trading market (foreign exchange market).
Using a currency’s exchange rate, you can convert between currencies.
To do this, you need to either multiply or divide the quantity you are trying to convert with the exchange rate.
To convert from US dollars (USD) to Mexican pesos (MXN) , you must multiply by the exchange rate.
For example,
So \$15 \; USD would be \$251.40 \; MXN because \$ 15 \; U S D \times 16.76=\$ 251.4 \; M X N.
For example,
The exchange rate from \$ \; (USD) to £ \; (GBP) is 1.28.
To convert a British pound (GBP) into US dollars (USD) , you must divide by the exchange rate.
So, £40 would be \$31.25 because £ 40 \; G B P \div 1.28=\$ 31.25 \; U S D
What is an exchange rate?
Common Core State Standards
How does this relate to 6 th grade math and 7 th grade math?
How to calculate exchange rates
In order to calculate exchange rates:
How to calculate exchange rates examples
Example 1: converting from EUR / USD
€9 is equal to \$9.87 \; USD. What is the exchange rate from € \; (EUR) to \$ \; (USD)?
When calculating the currency exchange rate from € \; (EUR) to \$ \; (USD or US dollars ), you want to know how many \$ are equal to €1. This is the ratio of \$ to €, so set up the rate as \cfrac{\$ 9.87}{€ 9}.
2Divide both parts by the base currency.
In this case, the base currency is the euro \, € \; (EUR), so divide both parts by 9, rounding the \$ \; (USD) to the nearest cent:
\cfrac{\$ 9.87 \div 9}{€ 9 \div 9}=\cfrac{\$ 1.10}{€ 1}
3State the final exchange rate with the correct currency symbols.
The exchange rate from € \; (EUR) to \$ \; (USD) is 1.10.
Example 2: converting from AUD / USD
\$9.09 \; AUD is equal to \$6 \; USD. What is the exchange rate from \$ \; (USD) to \$ \; (AUD)?
Use the information given to set up a rate.
When calculating the currency exchange rate from \$ \; (USD) to \$ \; (AUD), you want to know how many \$ \; (AUD) are equal to \$1 \; (USD).
This is the ratio of \$ \; (AUD) to \$ \; (USD) so set up the rate as \cfrac{\$ 9.09 \text { AUD }}{\$ 6 \text{ USD }}.
Divide both parts by the base currency.
In this case, the base currency is \$ \; (USD), so divide both parts by 6, rounding the \$ \; (AUD) to the nearest cent:
\cfrac{\$ 9.09 \, A U D \div 6}{\$ 6 \, U S D \div 6}=\cfrac{\$ 1.52 \, A U D}{\$ 1 \, U S D}
State the final exchange rate with the correct currency symbols.
The exchange rate from \$ \; (USD) to \$ \; (AUD) is 1.52.
Example 3: converting from JPY / USD
¥26 \; JPY is equal to \$0.18 \; USD. What is the exchange rate from \$ \; (USD) to ¥ \; (JPY)?
Use the information given to set up a rate.
When calculating the currency exchange rate from \$ \; (USD) to ¥ \; (JPY), you want to know how many ¥ \; (JPY) are equal to \$1 \; (USD).
This is the ratio of ¥ to \$, so set up the rate as \cfrac{¥ 26}{\$ 0.18}.
Divide both parts by the base currency.
In this case, the base currency is \$ \; (USD), so divide both parts by 0.18, rounding the ¥ \; JPY is to the nearest whole:
\cfrac{¥ 26 \div 0.18}{\$ 0.18 \div 0.18}=\cfrac{¥ 144}{\$ 1}
State the final exchange rate with the correct currency symbols.
The exchange rate from \$ \; (USD) to ¥ \; (JPY) is 144.
How to calculate with exchange rates
In order to calculate with exchange rates:
Example 4: converting from USD to GBP
If \$1 \; USD = £0.78 \; GBP, convert \$506 to £ \; (GBP).
Use the information given to set up an equation.
\$ 506=£___
Multiply or divide by the given exchange rate.
Since each dollar is equal to £0.78, multiply the USD to find the number of £ \; (GBP).
State the final amount with the correct currency symbol.
\$ 506=£394.68
Example 5: word problem
After a trip to Costa Rica, Joel is exchanging ₡15,000 \; CRC to \$ \; (USD). The exchange rate is \$1 \; USD = ₡545 \; CRC.
Use the information given to set up an equation.
₡15,000 = \$___
Multiply or divide by the given exchange rate.
Since ₡545 \; CRC is equal to \$1 \; USD, divide the CRC by 545 to find the number of \$ \; (USD).
State the final amount with the correct currency symbol.
₡15,000 = \$27.52 Rounded to the nearest cent
Example 6: comparing different currencies
In the USA a computer costs \$790. The same computer costs £575 in the UK. Given that the current exchange rate is £1 = 1.35 \; USD, where is the computer cheaper?
Use the information given to set up an equation.
You are asked to compare £575 and \$790. To compare them, you must convert them to the same currency. In this case, you can choose to either convert £575 into US dollars (USD) or \$790 into Great British pounds (GBP).
Let’s convert £575 to American dollars.
£575 = \$ ___
Multiply or divide by the given exchange rate.
Since £1 \; GBP is equal to \$1.35 \; USD, multiply the £ \; (GBP) by 1.35 to find the number of \$ \; (USD).
State the final amount with the correct currency symbol.
£575 = \$776.25
The computer costs \$776.25 in the UK and \$790 in the United States.
Therefore, it is cheaper to buy the computer in the UK.
[FREE] How To Calculate Exchange Rates Worksheet (Grade 6 to 7)
Use this worksheet to check your grade 6 to 7 students’ understanding of calculating exchange rates. 15 questions with answers to identify areas of strength and support!
[FREE] How To Calculate Exchange Rates Worksheet (Grade 6 to 7)
Use this worksheet to check your grade 6 to 7 students’ understanding of calculating exchange rates. 15 questions with answers to identify areas of strength and support!
Teaching tips for how to calculate exchange rates
Easy mistakes to make
Related ratio lessons
Practice how to calculate exchange rates questions
1) \$18 \; USD is equal to \$23.92 \; CAD. What is the exchange rate from \$ \; (USD) to \$ \; (CAD).
When calculating the currency exchange rate from \$ \; (USD) to \$ \; (CAD), you want to know how many \$ \; (CAD) are equal to \$1 \; USD, so set up the rate as \cfrac{\$ 23.92 \; C A D}{\$ 18 \; U S D}.
In this case, the base currency is \$ \; (USD), so divide both parts by 18, rounding the \$ \; (CAD) to the nearest cent:
\cfrac{\$ 23.92 \; C A D \div 18}{\$ 18 \; U S D \div 18}=\cfrac{\$ 1.33 \; C A D}{\$ 1 \; U S D}
The exchange rate from \$ \; (USD) to \$ \; (CAD) is 1.33.
2) ¥20 \; CNY is equal to \$2.79 \; USD. What is the exchange rate from ¥ \; (CNY) to \$ \; (USD).
When calculating the currency exchange rate from ¥ \; (CNY) to \$ \; (USD), you want to know how many \$ \; (USD) are equal to ¥1 \; CNY, so set up the rate as \cfrac{\$ 2.79}{¥ 20}.
In this case, the base currency is ¥ \; (CNY), so divide both parts by 20, rounding the \$ \; (USD) to the nearest cent:
\cfrac{\$ 2.79 \div 20}{¥ 20 \div 20}=\cfrac{\$ 0.14}{¥ 1}
The exchange rate from ¥ \; (CNY) to \$ \; (USD) is 0.14.
3) ฿275 \; THB is equal to \$8 \; USD. What is the exchange rate from ฿ \; (THB) to \$ \; (USD).
When calculating the currency exchange rate from ฿ \; (THB) to \$ \; (USD), you want to know how many \$ \; (USD) are equal to ฿1 THB, so set up the rate as \cfrac{\$ 8}{฿ 275}.
In this case, the base currency is ฿ \; (THB) so divide both parts by 275 , rounding the \$ \; (USD) to the nearest cent:
\cfrac{\$ 8 \div 275}{\$ 275 \div 275}=\cfrac{\$ 0.03}{฿ 1}
The exchange rate from ฿ \; (THB) to \$ \; (USD) is 0.03.
4) Given the exchange rate between US dollars (USD) and New Zealand dollars (NZD) is \$1 \; USD = \$1.63 \; NZD, convert \$45 \; USD to New Zealand dollars (NZD). Round to the nearest cent.
\$45 \; USD = \$ \rule{0.8cm}{0.25mm} \; NZD
Since each US dollar is equal to \$1.63 \; NZD, multiply the USD by 1.63 to find the number of \$ \; (NZD).
\$45 \; USD = \$73.35 \; NZD
5) In China, an iPad costs ¥2,500 (Chinese Yuan). The same iPad costs \$300 in the US. Given that the current exchange rate is \$1 = ¥7.18, where is the iPad cheaper? And by how much?
The US by ¥346
China by ¥155
The US by ¥42
China by ¥34
\$300 \; USD = ¥ \rule{0.8cm}{0.25mm} \; CNY
Since each US dollar is equal to ¥7.18 CNY, multiply the USD by 7.18 to find the number of ¥ \; CNY).
\$300 \; USD = ¥2,154 \; CNY
This US price is less than the price in China. To find out how much, subtract: ¥2,500 \; – \; ¥2,154 = ¥346.
6) After a trip to Brazil, Ju is exchanging \$750 \; BRL to \$ \; (USD). The exchange rate is \$1 \; USD = \$4.79 \; BRL. How many US dollars will Ju have after exchanging?
\$750 \; BRL = \$ \rule{0.8cm}{0.25mm} \; USD
Since each US dollar is equal to \$4.79 \; BRL, divide the USD by 4.79 to find the number of \$ \; (USD). Round to the nearest cent.
\$750 \; BRL = \$156.58 \; USD
How to calculate exchange rates FAQs
When comparing a currency in terms of another, the two currencies together form a currency pair.
For many people and companies, all money transactions are done in their countries’ currency. However, there are times when they need to make foreign transactions.For example, when people complete money transfers through a financial institution to relatives who live outside the US. Or whenever a company exports their product to another country, who is responsible for paying for the export.
Services that exchange money make a profit by charging at a rate slightly higher than the market exchange rate.
With a debit card, a customer spends money that they already have in an account. With a credit card, a customer spends money on credit. Because a customer is using credit and not their money, the card has interest rates, and the user is charged interest at the end of every month if the card is not fully paid off.
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12866 | https://www.statology.org/binomial-distribution-assumptions/ | The Three Assumptions of the Binomial Distribution
by Zach Bobbitt Posted on
The binomial distribution is a probability distribution that is used to model the probability that a certain number of “successes” occur during a fixed number of trials.
The binomial distribution is appropriate to use if the following three assumptions are met:
Assumption 1: Each trial only has two possible outcomes.
We assume that each trial only has possible two outcomes. For example, if we flip a coin 100 times, each time there can only be two possible outcomes – heads or tails.
Assumption 2: The probability of success is the same for each trial.
We assume that the probability of achieving a “success” is the same for each trial. For example, the probability of a coin landing on heads is 0.5 for any given flip. This probability does not change from one coin flip to the next.
Assumption 3: Each trial is independent.
We assume that each trial is independent of every other trial. For example, the outcome of one coin flip does not affect the outcome of another coin flip. The flips are independent.
The following examples show various scenarios that meet the assumptions of the binomial distribution.
Example 1: Number of Free Throws Made
Suppose a basketball player is known to make 70% of his free throws attempts. If he makes 20 attempts, this scenario can be modeled using the binomial distribution.
This scenario meets each of the assumptions of the binomial distribution:
Assumption 1: Each trial only has two possible outcomes.
For each free throw attempt, there are only two possible outcomes – a make or a miss.
Assumption 2: The probability of success is the same for each trial.
The probability that the player makes a free throw on each attempt is the same – 70%. This does not change from one attempt to the next.
Assumption 3: Each trial is independent.
Each free throw attempt is independent of every other attempt. Whether or not a player makes one attempt does not affect whether he makes another attempt.
Example 2: Number of Side Effects
Suppose it’s known that 5% of adults that take a certain medication experience negative side effects. Suppose a medical profession then gives this medication to 100 adults in a given month.
This scenario meets each of the assumptions of the binomial distribution:
Assumption 1: Each trial only has two possible outcomes.
For each adult that receives the medication, there is only two possible outcomes – they experience negative side effects or they do not.
Assumption 2: The probability of success is the same for each trial.
The probability that each adult experiences a negative side effect is the same – 5%.
Assumption 3: Each trial is independent.
The outcome for each adult is independent. Whether or not one adult experiences negative side effects does not affect whether or not another adult does as well.
Example 3: Number of Shopping Returns
Suppose it’s known that 10% of all customers who walk into a shop are there to make a return. Suppose 200 people enter a store in a given day and the manager records the number who are there to make a return.
This scenario meets each of the assumptions of the binomial distribution:
Assumption 1: Each trial only has two possible outcomes.
Each time a customer enters the shop, there are only two reasons they may be there – to make a return or not.
Assumption 2: The probability of success is the same for each trial.
The probability that a given customer is there to make a return is the same – 10%.
Assumption 3: Each trial is independent.
The outcome for each customer is independent. Whether or not one customer is there to make a return does not affect whether or not another customer is there to make a return.
Additional Resources
The following tutorials offer additional information on the binomial distribution:
An Introduction to the Binomial Distribution Binomial Distribution Calculator 5 Real-Life Examples of the Binomial Distribution
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.
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12867 | https://ocw.mit.edu/courses/2-60j-fundamentals-of-advanced-energy-conversion-spring-2020/acf2690c40525bd6ac26b3f1e43eb542_MIT2_60s20_lec8.pdf | Lecture # 8 Electrochemical Thermodynamics 1 Ahmed F. Ghoniem February 26, 2020 • Electrochemical reactions • Electrodes and electrolytes • Fuel cell components • Work generated by a fuel cell • Voltage and Ideal Efficiency © by Ahmed F. Ghoniem 1 2 ηcar = 1−n TF T / TF T −1 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 70% for TF / T = 8 Heat engine (heat to work) efficiency Stream at stream at Environment at HEAT ENGINE w Environment at Reservoir at HEAT ENGINE w ηcar = 1−TL TH = 87.5% forTH / TL = 8 Ideal thermo-mechanical efficiency using practically achievable/manageable temperatures is 70-85%!
© by Ahmed F. Ghoniem QL Environment at (T, p) Products at (T, p) Products at TF HEAT ENGINE fuel at (T, p) air at (T, p) Adiabatic combustor W Ideal Generator a wmax.adi b.comb.eng. = wmax.chem.eng −T ΔSad.comb ηbest adiabatic.comb ( ) ≈wmax.c G omb.eng Δ R ≈wmax.chem.eng ΔGR −T ΔSad.comb ΔGR = 75% Engines running on adiabatic combustion Keeping the reaction isothermal and in equilibrium with the environment produces maximum work Q − W = Hout − Hin = ΔH R Q T = Sout − Sin = ΔSR Q = TΔSR heat added −W = (H − TS)out − (H − TS)in = ΔH R − TΔSR W = −ΔGR work produced.
The isothermal reaction produces work = Gibbs free energy of reaction, and rejects heat = T . entropy of reaction TΔSR = ΔH R −ΔGR = Q for typical exothermic reactions, both ΔH R and ΔGR < 0 and heat is mostly rejected (but can also be added) depending on T © by Ahmed F. Ghoniem 3 Q, Heat transfer in Products stream @ T, p W, work out Oxidizer stream @ T, p Fuel stream @ T, p Chemical reactions @ T Environment @ T How can we perform such an isothermal reaction with work transfer? 1 The overall Reaction: H2 + O2 ⇒ H2O 2 can be performed in a Redox Pair (reduction-oxidation), or two electrochemical "half reactions", across an electronically non-conducting material, leading to the formation of charged species; _ HydrogenOxidation: H2 ⇒ 2H+ + 2e− Hydrogen loses e Hydrogen Reduction: 1 O2 + 2e- + 2H+ ⇒ H2O Reactants gain e 2 _ H+ ions diffuse through the electrolyte (acidic, +ve ion (proton) transport medium (PEMFC) e- moves through an external resistance A useful note: the general definition of oxidation and reductions: Oxidation is Loss of electron (loss of -ve charge or becoming positive) Reduction is Gain of electron (gain of -ve charge of becoming negative) © by Ahmed F. Ghoniem 4 _ HydrogenOxidation: H2 ⇒ 2H + + 2e − Hydrogen loses e Hydrogen Reduction: 1 O2 + 2e - + 2H + ⇒ H2O Reactants gain e 2 _ Electric Electrolyte, current only ions are through a Hydrogen in resistance + -in anode cathode allowed to migrate across Oxygen Galvanic or voltaic cell. At finite current, ΔV < Δε.
In galvanic cells the anode is the -ve electrode and the cathode is the +ve electrode. Water out in acidic cells © by Ahmed F. Ghoniem 5 + - anode e ΔV cathode e cation anion Δε Negative electrode Positive electrode Another redox pair uses an alkaline electrolyte (transports -ve ions): 1 Overall Reaction: H2 + O2 ⇒ H2O 2 1 →O2 Oxygen Reduction, Cathode: O2+2e 2 Hydrogen Oxidation, Anode: H2 + 1 O2- → H2O + 2e− 2 O2 ions move through the electrlyte from the cathode to the anode CO Electrochemical Oxidation 1 Overall Reaction: CO + O2 ⇒ CO2 2 The Redox Pair (two half reaction) Oxidation, Anode: CO + O2- → CO2 + 2e− 1 →O2-Reduction, Cathode: O2+2e 2 Can combine H2 and CO in a single cell © by Ahmed F. Ghoniem 6 Hydrogen or carbon monoxide Electric current through a resistance + - Oxygen in anode cathode only O2- ions can pass through this electrolyte Water or carbon dioxide Open circuit Work: ' " (−ΔGR (T , p )) = ∑νi (h ˆ i (T ) − T s ˆi (T , p)) −∑νi (h ˆ (T ) − T s ˆ (T , p)) i i react prod ⎛∏ Xi νi " ⎞ ⎛ p ⎞ ⎜ prod ⎟ = (−ΔGR o (T )) −σ ℜT ℓn ⎠ ⎟ −ℜT ℓn ⎝ ⎜ po ⎜ ⎟ ⎜∏ Xi νi ' ⎟ ⎝ react ⎠ o ' o " o ΔGR (T ) = ∑νi (h ˆ i (T ) − Ts ˆi (T )) −∑νi (h ˆ i (T ) − Ts ˆi (T )) react prod Important remarks: 1. Reactants are introduced separately. 2. Products mix with one of the reactant stream 3. Or products leave separately through the electrolyte. Equilibrium or Open-circuit Efficiency: W ΔGR,H2O max = = ηOC ΔH R,H2O ΔH R,H2O 7 © by Ahmed F. Ghoniem For separate streams for hydrogen, oxygen and water: All at 1 atm o ⎡ 1 ⎤ ΔGR (T ) =(h ˆ (T ) − Ts ˆo (T )) − (h ˆ (T ) − Ts ˆo (T )) + (h ˆ (T ) − Ts ˆo (T )) H2O ⎣ ⎢ H2 2 O2 ⎦ ⎥ s ˆo (T ) = s ˆ(T , p = 1 atm) ΔGR,H2O ηOC = HHVH2O @ T = 300 K: water leaving as liquid: ηOC = 237/286)=83% (water leaving as vapor: ηOC = 228/242=94%) @ T = 500 K: water leaving as vapor: ηOC = 219 / 242 = 76.5% @ T = 1000 K: water leaving as vapor: ηOC = 193 / 245 = 67.3% © Source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see • Reversible work produced by H2/O2 cell and a simple Carnot engine. • Based on the HHV (284 kmol H2). Cross over point is 950 K. • Comparison is not necessarily meaningful. © by Ahmed F. Ghoniem 8 Open Circuit Cell Potential: w ˆ = Δg ˆR = Δε ς (work of moving charge ς across a potential difference Δε) max ς = neς Na = neℑa e− ne: number of electrons produced when oxidizing one fuel molecule Na = 6.023 ×1023 mole−1 (Avogadro's number) ς e− = −1.602 ×10−19 Coulombs/electron ℑa = ς e− Na = 9.6485 ×104 Coulombs/mole (Faraday's number) for the hydrogen-oxygen, ne = 2, and ς =2ς e− Na = 2ℑa , @ 300K, Δε = 1.18 volts with water leaving in vapor form Δε = 1.23 volts with water leaving in liquid form. For the methanol-oxygen reaction, Δεo = 1.21 V. © by Ahmed F. Ghoniem 9 The Nernst Equation: effect of pressure and fuel concentration ⎛∏( pi ) νi " ⎞ ⎛∏ Xi νi " ⎞ ℜT ⎜ prod ⎟ σ ℜT ⎛ p ⎞ ℜT ⎜ prod ⎟ Δε( p,T ) = Δεo(T )− ℑ ℓn⎜ νi " ⎟ = Δεo(T )− ℓn ⎝ ⎜ ⎠ ⎟− ℓn⎜ νi ' ⎟ ne ⎜∏( pi ) ⎟ neℑa po neℑa ⎜∏ Xi ⎟ ⎝ ⎠ ⎝ react ⎠ react " ' = Δεo(T )+ Δε ( p,T )+ Δε (Xi ,T ) where σ = ∑νi −∑νi p conc prod react ℜT ⎛ 1 ⎞ for a hydrogen-oxygen cell: Δεconc = ℓn( XH2 ) + ℓn( XO2 ) − ℓn( XH2O ) 2ℑa ⎝ ⎜ fuel 2 oxy sep/ fuel/oxy ⎠ ⎟ Equation applied at one point (under equilibrium, things are subtle). Lower reactants concentrations decrease the OC potential, especially at higher T. Using air instead of oxygen also penalizes the potential, Δεconc,O2 = 2.15 ×10−5 T ℓn(0.21) = −0.012@350K Using products of methane reforming as fuel: CH4 + 2H2O → CO2 + 4H2 XH2 = 0.8, Δεconc = 2.15 ×10−5 T ℓn(0.8) this reduces the OC by Δε = 0.00168V (@350K ) conc,H2 © by Ahmed F. Ghoniem 10 Impact of fuel, concentration, temperature and pressure P=1, thick lines P=10 thin line Colors for different fuel concentrations © PECS. All rights reserved. This content is excluded from our Creative Commons license. For more information, see Hanna, Lee, Shi, and Ghoniem, PECS, 40 (2014) 74-111 11 © by Ahmed F. Ghoniem Impact of fuel, concentration, temperature and pressure © Source unknown. All rights reserved. This content is excluded from our Creative Commons license. For more information, see © PECS. All rights reserved. This content is excluded from our Creative Commons license. For more information, see Hanna, Lee, Shi, and Ghoniem, PECS, 40 (2014) 74-111 © by Ahmed F. Ghoniem 12 Fuel Cell Components Also known as membrane-electrode-assembly (MEA), and made of one physical plate with anode and electrode material deposited on both side. The membrane is a polymer (nafion) for low T cells and a ceramic plate for high T cells. Images courtesy of DOE. DOE, Fuel Cell Handbook,2004 © by Ahmed F. Ghoniem 13 Electrochemical Reactions and Types of Fuel Cells 1 Overall Reaction: H2 + O2 ⇒ H2O 2 (A) H2 ⇒ 2H+ + 2e− and (C) 1 O2 + 2H+ + 2e- ⇒ H2O, acidic electrolyte (PEM cell) 2 (A) H2 + O2−⇒ H2O + 2e− and (C) 1 O2 + 2e- ⇒ O2− , alkaline electrolyte (SOFC cell) 2 (A) H2 + 2OH−⇒ 2H2O + 2e− and (C) 1 O2 + H2O+2e- ⇒ 2OH− , alkaline electrolyte (Alkaline cell with humidified air) 2 1 Overall Reaction: CO + O2 ⇒ CO2 2 (A) CO + O2−⇒ CO2 + 2e− and (C) 1 O2 + 2e- ⇒ O2− , alkaline electrolyte (SOFC cell) 2 Overall Reaction: CH4 + 2O2 ⇒ CO2 + 2H2O (A) CH4 + 4O2−⇒ CO2 + 2H2O + 8e− and (C) 2O2 + 8e- ⇒ 4O2− , alkaline electrolyte (SOFC cell) in all, two electrons are produced per oxygen atom. © by Ahmed F. Ghoniem 14 Fuel Cell Types Fuel cell Proton Exchange Alkaline Phosphoric Acid Molten Carbonate Solid Oxide Electrolyte Polymer ion exchange membrane potassium hydroxide in asbestos liquid phosphoric acid in SiC liquid molten carbonate in LiAlO2 Perovskites Electrode Carbon Transition metals Carbon Nickels and nickel oxides perovskites/ metal cermet Catalyst Platinum Platinum Platinum Electrode material Electrode material Interconnect Carbon or metal Metal Graphite Stainless steel of nickel Nickel, ceramics Temperature 40 - 80 °C 65 - 220 °C 205 °C 650 °C 600 -1000 °C Charge Carrier H+ OH H+ CO3= O= fuel Hydrogen Hydrogen Hydrogen Hydrocarbon hydrocarbon 15 Materials for Solid Oxide Fuel Cells In the left bubble, oxygen is reduced at the cathode and oxygen ions are conducted through the electrolyte. Oxygen ions move into the anode (right bubble), where they are used to oxide the fuel at the three-phase boundary TPB). Electrons released in the charge-transfer reactions are conducted through the anode (metal), to the external circuit. © PECS. All rights reserved. This content is excluded from our Creative Commons license. For more information, see LaMnO3 (lanthanum manganese oxide) is used to catalyze the oxygen reduction reaction on the cathode side YZS (Yittria stabilized zirconia) is use as an ion transport membrane Ni (nickel) is used to catalyze the fuel oxidation reaction on the anode side. See: Hanna, Lee, Shi, and Ghoniem, PECS, 40 (2014) 74-111 16 Solid Oxide Fuel Cells High T cells use regular metals as catalysts Electrolyte: YSZ, Anode: Ni-YSZ, Cathode: Sr-doped LaMnO3 YSZ = yttria – stabilized zirconia 20µm Porous YSZ 5kV x350 Dense YSZ Impregnation LSM Metals Metal Oxides Courtesy of Raymond J. Gorte, University of Pennsylvania. Used with permission. 17 Fuel Utilization and its impact on the Open Circuit Potential and Cell Efficiency ' ' ν fl χ fl +νox χox →ν " p χ p , ⎛ νo ⎞ ( ) −σ ℜT ⎛ p ⎞ ℜT X fl Xox ' Δε = Δε o ℓn ℓn " ⎝ ⎜ ⎠ ⎟ + ⎜ ⎜ ν p ⎟ ⎟ neℑa po neℑa ⎝ Xp ⎠ For a SOFC where products form concentrations of fuel and oxidizer decrease between inlet and outlet in the fuel channel as both are consumed accordignto their stoichiometric ratio n fl1 − n fl 2 Partial Fuel utilization: ϕ = n fl1 @ Inlet: X fl1 = 1 , Xox1 = 0.21, 1+ nd1 ' 1−ϕ (1−ϕ )νox @ exit: X fl 2 = , Xox2 = ' ' 1−ϕ + nd1 +ν " pϕ 3.76νox +(1−ϕ )νox using values at exit gives lower Δε 18 © by Ahmed F. Ghoniem Open Circuit potential for different fuel utilization and for 50% oxygen utilization, cell is fueled by hydrogen produced by SMR (Fuel Cell Explained, Laramie etal.) If products of methane-water reforming are used: CH4 + 2H2O → CO2 + 4H2 The fuel mixture has XH 2 = 0.8 Oxidizer is air, XO2−inlet = 0.21 Much air is flown to ensure that oxidation is not limited by oxygen Assume that @ exit, 50% utilized of oxygen, XO2−exit = 0.105 There advantages to running the cell at low T, but chemistry is slow and we need a precious metal catalyst, which make it expensive and sensitive to fuel impurities. © by Ahmed F. Ghoniem 19 • In this example, products are mixed with fuel in the fuel channel, reducing the fuel concentration towards the exit. • Methane can also be ”naturally” reformed internally, hydrogen and CO are more electro-chemically active CH4 + Steam/Oxygen CO+H2+O 2- → CO2+H2O+4e Solid electrolyte (Oxygen ion conductor) Anode (catalyst) Cathode O 2-Electrons, e -O2+4e - → 2O 2 Oxygen (Air), O2 Courtesy Elsevier, Inc., Used with permission. Kee et al, J. Power Sources. © by Ahmed F. Ghoniem 20 Fuel Cell Performance at Finite Current (Power) Conditions Faraday's Law: I = neℑan ! f or i = neℑa j f 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 0 0.5 1 1.5 2 Current Density (A/cm^2) Cell Potential (V) Equilibrium Voltage 1.17 V Activation Overpotential Ohmic Overpotential Mass Transfer Overpotential Mass transfer overpotential significant at high current Relative contributions depend on design and operating conditions: • Catalysis, type and density. • Thickness of electrodes and membrane. • Water management (in PEM cells). © by Ahmed F. Ghoniem 21 MIT OpenCourseWare 2.60J Fundamentals of Advanced Energy Conversion Spring 2020 For information about citing these materials or our Terms of Use, visit: |
12868 | https://math.mit.edu/research/highschool/primes/circle/documents/2022/Sophia%20&%20Jaeyi.pdf | Group Theory Jaeyi Song and Sophia Hou Abstract In the MIT PRIMES Circle (Spring 2022) program, we studied group theory, often following Contemporary Abstract Algebra by Joseph Gallian. In this paper, we start by introducing basic ideas relating to group theory such as the definition of a group, cyclic groups, subgroups, and quotient groups.
We then introduced the notions of homomorphisms, as well as generators and relations. Finally, we delved into two fun and interesting problems that address generators and relations.
1 Groups Definition 1.1. A group (G, ∗) is a set G with a binary operation ∗that has three requirements satisfied: 1. Associativity: a ∗(b ∗c) = (a ∗b) ∗c for all elements a, b, c ∈G.
2. Identity: there is an element e ∈G in which a ∗e = e ∗a = a for all elements of G. The identity for groups under multiplication is 1, under addition it is 0.
3. Inverse: For every element a ∈G, there is the inverse of a (let’s say b) that satisfies a ∗b = b ∗a = e.
Remark. Usually the group operation ∗will be multiplication, so we will often just omit writing ∗.
Example 1.2. The group (Z/nZ, +), which is the set {0, 1, 2, . . . , n −1} under addition taken modulo n, is a group under addition because it satisfies all 3 require-ments. First, it is associative because addition is associative. The identity is 0 and the inverse of x is n −x.
Example 1.3. The group {1, 3, 7, 9} (mod 10) is a group under multiplication be-cause it fulfills all the three requirements above. For multiplication the identity is 1, which is included in the set. Associativity is fulfilled since multiplication as an operation itself is associative, and the inverse requirement a ∗b = b ∗a = e is also true for all elements.
Example 1.4. The group {1, 2, 4, 7, 8, 11, 13, 14} (mod 15) is a group under multi-plication as well, for the same reasons as above.
Example 1.5. The real numbers under addition, denoted by (R, +) is a group. In this case, the identity would be 0 since the identity for addition is always 0. Under addition, the inverse of an element x is just −x.
1 Example 1.6. The rational numbers under addition, or (Q, +) is also a group for similar reasons.
Example 1.7. Integers under addition (Z, +) is a group because it conforms to all group requirements.
Example 1.8. The set Mat2(R) is a group under addition because the identity is the zero matrix 0 0 0 0 and the inverse of an element a b c d is −a −b −c −d .
Example 1.9. However, GL(2, R) is a group under multiplication because it fulfills all the requirements for groups listed above. Matrix multiplication is associative.
The multiplicative matrix identity is 1 0 0 1 , which is in GL(2, R). The inverse of a 2 by 2 matrix a b c d is 1 ad−bc d −b −c a .
Example 1.10. The free group on two elements ⟨a, b⟩consists of all words formed by a, b, a−1, b−1. It is associative because it is essentially concatenation of words.
The identity is the empty word, usually denoted e. The inverse of every word can be formed by reversing the order and then taking the inverse of each letter.
Remark. The free group with two elements is not commutative.
Remark. A similar process can be applied to a free group on three elements ⟨a, b, c⟩.
Non-example 1.11. However, the natural numbers under multiplication (N, ×) is not a group because it is not closed for inverses.
Non-example 1.12. The set Mat2(R) is not a group under multiplication because not every matrix has an inverse. For example, 0 0 0 0 does not have a multiplicative inverse because the determinant is 0.
Definition 1.13. We define the order of G to be the number of elements in G, and write it as |G|.
Definition 1.14. The order of an element g ∈G is defined to be the smallest positive integer n such that gn = e, the identity in G. We write this as ordG(g).
Proposition 1.15. For any group element a, ak = e if and only if ordG(a)|k.
Proposition 1.16. If a and b are elements of a finite group G and ab = ba, ordG(ab) divides ordG(a) · ordG(b).
Example 1.17. The order of the identity in any group is always 1, because e1 = e already.
Example 1.18. The order of 1 in Z/nZ under addition is n because n is the smallest positive integer k, such that adding k 1’s gives you 0.
Example 1.19. From Example 1.4, the order of 2 in the group is 4 because 4 is the smallest positive integer k such that 2k ≡1 (mod 15).
Example 1.20. The order of any element in ⟨a, b⟩which is not identity (see Exam-ple 1.10) is infinity.
2 1.1 Cyclic groups Definition 1.21. Cyclic groups are a special type of group in which every element can be written as iterated copies of a single element a, called a generator of G. For example, if the operation is multiplication, then every element is a power of a. A cyclic group G generated by a is written as G = ⟨a⟩.
Proposition 1.22. Subgroups of cyclic groups are cyclic as well.
Proposition 1.23. For any group element a ∈G, ordG(a) = |⟨a⟩|.
Proposition 1.24. In a finite cyclic group, the order of an element divides the order of a group.
Remark. Cyclic groups can be finite or infinite, however every cyclic group follows the shape of Z/nZ, which is infinite if and only if n = 0 (so then it looks like Z).
Example 1.25. The group Z/6Z = {0, 1, 2, 3, 4, 5} (mod 6) is a cyclic group, and cyclic subgroups generated by the following elements are listed below: • ⟨1⟩= {1, 2, 3, 4, 5, 0} = Z/6Z.
• ⟨2⟩= {2, 4, 0}.
• ⟨3⟩= {3, 0}.
• ⟨4⟩= {4, 2, 0}.
• ⟨5⟩= {5, 4, 3, 2, 1, 0} = Z/6Z.
• ⟨0⟩= {0}, only has one element.
Remark. Notice that the cyclic subgroups 1 and 5 generate the entire group which means that they are the generators of this group.
Example 1.26. The group Z/7Z = {0, 1, 2, 3, 4, 5, 6} (mod 7) is a cyclic group for similar reasons. In Z/7Z every nonzero element generates the group and thus can be considered a generator.
To generalize the previous two examples, we have the following.
Proposition 1.27. The group Z/nZ is cyclic under addition. The generators of this group are all integers x such that x is relatively prime to n.
2 Subgroups and Quotient Groups 2.1 Subgroups Definition 2.1. A subgroup is a subset H of a group G that is closed under the operation of G, inverses, and contains the identity. It then becomes a group in its own right. Note that associativity is inherited from the parent group and the other two axioms are verified by definition.
3 Example 2.2. In Z/10Z, the subset {2, 4, 6, 8, 0} is a subgroup under addition because the identity exists and is 0 and the inverse of 2 is 8 and the inverse of 4 is 6. It is also associative because addition is associative.
Example 2.3. The subset 0, 2, 4, 6 ⊂Z/8Z is a subgroup (under addition) since it has identity, inverse, and associativity.
Alternatively, we may use the Finite Subgroup Test, see below.
Example 2.4. The subset 1, 4 ⊂(Z/5Z)× is a subgroup as it fulfills all the require-ments.
Example 2.5. Another example similar to the previous one is the subset 1, 5, 7, 11 ⊂ (Z/12Z)×.
Example 2.6. The subset Q>0, which is the multiplicative group of positive rational numbers, is a subgroup of (R>0, ×), the multiplicative group of positive real numbers.
This is because the identity of Q>0 is 1 and the inverse of x ∈Q>0 is 1/x, which is still a positive rational number. Multiplication is also associative.
Non-example 2.7. The positive integers are not a subgroup of Z, which is the additive group of integers. This is because the inverse of 2 is −2, which is not in the positive integers and thus, every element does not have a inverse.
Non-example 2.8. The set a b c d where a, b, c, and d are all positive real numbers is not a subgroup of Mat2(R) because the identity, which is the zero matrix 0 0 0 0 is not in the set.
It turns out that it’s easy to check if finite subsets are subgroups.
Proposition 2.9. The Finite Subgroup Test shows that if H is a nonempty finite subset of a group G and if H is closed under the operation of G, then H is a subgroup of G.
The idea is that for any element h ∈H, we may take all of its powers. Since they are all in H but H is finite, they must start repeating, so ha = hb for a ̸= b and we have that h has finite order and hence an inverse in H.
Proposition 2.10. Let G be a group and let a be any element of G. Then, ⟨a⟩is a subgroup of G.
Definition 2.11. The center, Z(G) of a group G is the subset of elements in G that commute with every element in G: Z(G) = {a ∈G|ax = xa for all x ∈G}.
Proposition 2.12. The center of a group G is a subgroup of G.
Proof. The identity e is in Z(G). In addition, if a, b ∈Z(G), then (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab) for all x in G. Thus, ab ∈Z(G). Lastly, if a ∈Z(G), 4 then xa−1 = exa−1, = a−1axa−1, = a−1(ax)a−1, = a−1(xa)a−1, = a−1x(aa−1), = a−1xe, = a−1x.
Thus, a−1 ∈Z(G) for all a ∈Z(G).
Somewhat weaker than the condition that an element must commute with every element of G, is the condition that it only must commute with some specified element of G.
Definition 2.13. Let a be a fixed element of a group G. The centralizer of a in G, C(a), is the set of all elements in G that commute with a. In other terms, C(a) = {g ∈G | ga = ag}.
Proposition 2.14. For all a in a group G, the centralizer of a is a subgroup of G.
2.2 Cosets Definition 2.15. Let G be a group and H be a nonempty subset of G. For any a ∈G, the set {ah|h ∈H} is denoted by aH and is called the left coset; the right coset Ha is defined similarly.
Proposition 2.16. The coset aH = H if and only if a ∈H.
Proof. Suppose aH = H. Then, a = ae ∈aH = H. Next, we assume that a ∈H.
aH ⊆H and H ⊇aH are both true.
The former is true because H is closed.
The latter is through the following proof. Let h ∈H, then a−1h ∈H and because h = eh = (aa−1)h = a(a−1h) ∈aH. Thus, aH = H.
Proposition 2.17. We have that aH = bH if and only if a ∈bH.
Proof. If aH = bH, then a = ae ∈aH = bH. If a ∈bH, then a = bh for some h ∈H, and therefore, aH = (bH)H = b(hH) = bH.
Proposition 2.18. The cosets are either disjoint or coincide completely: aH = bH or aH ∩bH = ∅.
Proof. If there is an element c in aH ∩bH, then cH = aH and cH = bH.
Proposition 2.19. A coset aH is a subgroup of G if and only if a ∈H; i.e., the only coset which is a subgroup is the identity coset H.
Proof. If aH is a subgroup, then it contains e. Thus, aH ∩eH ̸= ∅and as a result aH = eH = H. This means that a ∈H. If a ∈H, then aH = H.
Example 2.20. The cosets of H = {0, 3, 6} in Z/9Z are: 5 • 0 + H = 3 + H = 6 + H = {0, 3, 6}; • 1 + H = 4 + H = 7 + H = {1, 4, 7}; • 2 + H = 5 + H = 8 + H = {2, 5, 8}.
Example 2.21. The cosets of H = {. . . , −4, 0, 4, 8, . . . } = 4Z in (Z, +) are • 0 + H = {. . . , −4, 0, 4, 8, . . . }; • 1 + H = {. . . , −3, 1, 5, 9, . . . }; • 2 + H = {. . . , −2, 2, 6, 10, . . . }; • 3 + H = {. . . , −1, 3, 7, 11, . . . }.
Notice that these cosets act like the group Z/4Z.
2.3 Lagrange Theorem One very crucial result in basic group theory is Lagrange’s theorem.
Theorem 2.22 (Lagrange). If G is a finite group and H is a subgroup of G, then: 1. |H| divides |G| 2. The number of distinct left (also, right) cosets of H in G is |G|/|H| 2.4 Normal Subgroups Definition 2.23. A subgroup H of a group G is called a normal subgroup of G if aH = Ha for all a in G. Note that every subgroup in an abelian group is normal!
Proposition 2.24. The following conditions are equivalent: 1. H is a normal subgroup of G.
2. gHg−1 ⊆H for all g ∈G.
3. The normalizer of H in G (the set of elements whose conjugation action pre-serves H) is G, i.e. NG(H) = G.
4. There exists a homomorphism φ from G to another group such that H = ker(φ).
Example 2.25. Let H = 1. Then by (2) the trivial subgroup is always a normal subgroup of any group G. This is because gHg−1 will be {g1g−1} = {1} = H, which is a subset of H.
Example 2.26. Let H = G. Then by (2) the whole group is always a normal subgroup of any group G. This is shown by gHg−1 = gGg−1 = G = H.
Example 2.27. The group SL(2, R) of 2×2 matrices with determinant 1 is a normal subgroup of GL(2, R) (the group of 2 × 2 matrices with nonzero determinants).
If x ∈GL(2, R) and h ∈SL(2, R), then det(xhx−1) = (det x)(det h)(det x)−1 = (det x)(det x)−1 = 1. Thus, xhx−1 ∈H and therefore, xHx−1 ⊆H.
6 Example 2.28. The center Z(G) of a group is a normal subgroup because for every a ∈G and h ∈Z(G), ah = ha (by definition).
Example 2.29. The alternating group An of even permutations is a normal sub-group of Sn.
2.5 Quotient Subgroups Definition 2.30. Let G be a group and let H be a normal subgroup of G. The set G/H = {aH|a ∈G} is a group under the operation (aH)(bH) = abH.
Remark. Note that this is note true if H is not normal!
Example 2.31. Let 4Z = {0, ±4, ±8, ...} ⊂Z, as in Example 2.21. The quotient group consists of the cosets of 4Z in Z, which in turn behave like the elements 0, 1, 2, 3 modulo 4. The quotient group is Z/4Z, which matches our usual description of this group.
Example 2.32. Let nZ = {0, ±n, ±2n, ...} ⊂Z. Then the quotient group Z/nZ is {0 + nZ, 1 + nZ, 2 + nZ, 3 + nZ, ..., n −1 + nZ} = {0, 1, 2, . . . , n −1} taken modulo n.
Example 2.33. Let Sn be the permutation group and An ⊆Sn be the alternating group (of even permutations). Then Sn/An ∼ = {±1} ∼ = Z/2Z with the identification given by the sign of the permutation.
3 Group Homomorphisms Definition 3.1. A homomorphism ϕ from group G to a group G′ is a function that preserves the group’s operation. The following requirements must be satisfied: 1. ϕ(ab) = ϕ(a)ϕ(b) and for all a, b ∈G.
2. The identity maps to identity, i.e. ϕ(eG) = eG′.
Definition 3.2. The kernel of a homomorphism ϕ from group G to a group G′ (with identity e) is the set {g ∈G | ϕ(g) = e}.
Remark. For subgroups as well, many of its original features are preserved under the image of a homomorphism. For instance, if H is abelian, then ϕ(H) is also abelian. If H is normal in G, then ϕ(H) is also normal inside ϕ(G).
Proposition 3.3. If ϕ is a group homomorphism from G to G′ then ker ϕ is a normal subgroup for G. Conversely, every normal subgroup is the kernel of some group homomorphism from G (to varying targets).
Proof. We want to show that axa−1 ∈ker ϕ, i.e. that ϕ(axa−1) = e, for x ∈ker ϕ and any a ∈G. But we have ϕ(axa−1) = ϕ(a)ϕ(x)ϕ(a−1) = ϕ(a)eϕ(a)−1 = e, so ker ϕ is a normal subgroup.
Conversely, a normal subgroup N of G satisfies the condition that the cosets G/N form a group. Therefore in the canonical map G →G/N, the kernel is N, so N is indeed the kernel of some homomorphism.
7 We now list some properties of groups under homomorphisms.
Proposition 3.4. Let ϕ be a homomorphism from a group G to a group G′ and let g be an element of G. Then the following statements are true.
1. ϕ carries the identity of G to the identity of G′.
2. ϕ(gn) = ϕ(g))n for all n in Z.
3. If |g| is finite, then |ϕ(g)| divides |g|. If |G| is finite, then |ϕ(g)| divides |g| and |ϕ(G)|.
4. ker ϕ is a normal subgroup of G 5. ϕ(a) = ϕ(b) if and only if ab−1 ∈ker ϕ.
6. If ϕ(g) = g′, then ϕ−1(g′) = {x ∈G | ϕ(x) = g′} = g ker ϕ.
We have some additional properties of subgroups under homomorphisms.
Proposition 3.5. Let ϕ be a homomorphism from a group G to a group G′ and let H be a subgroup of G. Then the following statements are true.
1. ϕ(H) = {ϕ(h) | h ∈H} is a subgroup of G′.
2. If H is cyclic, then ϕ(H) is cyclic.
3. If H is abelian, then ϕ(H) is abelian.
4. If H is normal in G, then ϕ(H) is normal in ϕ(G) (but not necessarily G′!).
5. If | ker ϕ| = n, then ϕ is an n to 1 mapping from G onto ϕ(G).
6. If H is finite, then |ϕ(H)| divides |H|.
7. ϕ(Z(G)) is a subgroup of Z(ϕ(G)).
8. If K′ is a subgroup of G′ then ϕ−1(K′) = {k ∈G | ϕ(k) ∈K′} is a subgroup of G.
9. If K′ is a normal subgroup of G′, then ϕ−1(K′) = {k ∈G | ϕ(k) ∈K′} is a normal subgroup of G .
10. If ϕ is onto and ker ϕ = {e}, then ϕ is an isomorphism from G to G′.
Example 3.6. The function f : G →H defined by f(g) = 1 for all g ∈G is a homomorphism. This is also called the “trivial homomorphism,” and it shows that G is a normal subgroup of G.
Example 3.7. An example of a homomorphism is the mod 3 map Z →Z/3Z.
This can be seen since x + y (mod 3) = x (mod 3) + y (mod 3); for instance, 5 + 2 (mod 3) = 1 and 5 (mod 3) + 2 (mod 3) = 1 as well. Clearly, identity maps to identity since 0 maps to 0 (mod 3) = 0.
8 Example 3.8. The determinant map det : GL(2, R) →R×, where matrix A 7→ det A is a homomorphism because the identity matrix 1 0 0 1 maps to 1 and det(A) det(B) = det(AB).
Example 3.9. The nth power map f : Q× →Q× defined by f(x) = xn is a group homomorphism because 1 7→1 and f(xy) = (xy)n = xnyn = f(x)f(y).
Example 3.10. The absolute value function f : C∗→R>0 is a homomorphism for similar reasons.
Non-example 3.11. The function f : Z →Z defined by f(x) = x + 1 is not a group homomorphism since f(x + y) = x + y + 1 ̸= f(x) + f(y) = x + y + 2.
Non-example 3.12. The function f : Q× →Q× is defined by f(x) = 3x is not a group homomorphism because f(xy) = 3xy ̸= f(x) · f(y) = 9xy.
Non-example 3.13. The function f : Z →Z is defined by f(x) = x2 is not a group homomorphism because f(x+y) = (x+y)2 = x2 +2xy+y2 ̸= x2 +y2 = f(x)+f(y).
Non-example 3.14. The function f : GL(2, R) →R× sending a b c d 7→b is not a group homomorphism because the identity matrix, 1 0 0 1 7→0, but the identity of R× is 1, so the identity does not map to the identity. Furthermore, 0 does not even exist in the group R×.
3.1 Group Isomorphisms Definition 3.15. An isomorphism is a group homomorphism that is bijective.
For such a homomorphism G →G′, we say that G and G′ are isomorphic.
Generally, a group G can be proven to be isomorphic to group G′ through the following three steps: (1) Mapping: Determine a candidate for the isomorphism; define a homomorphism ϕ from group G to group G′.
(2) Injective: Prove ϕ is injective, so that no two elements map to the same element in G′.
(3) Surjective: Prove that ϕ is surjective, so that for any element g′ in G, we can find an element g in G such that ϕ(g) = g′.
Theorem 3.16 (First Isomorphism Theorem). Let ϕ be a group homomorphism from G to G′. Then the mapping from G/ ker ϕ to ϕ(G), given by g ker ϕ 7→ϕ(g) is an isomorphism.
Corollary 3.17. If ϕ is a homomorphism from a finite group G to G′, then |G|/| ker ϕ| = |ϕ(G)|.
Corollary 3.18. If ϕ is a homomorphism from a finite group G to G′, then |ϕ(G)| divides |G| and |G′|.
9 Proposition 3.19. Let ϕ : G →G′ be an isomorphism. Then the following state-ments are true.
1. ϕ carries the identity of G to the identity of G′.
2. For every integer n and for every group element a in G, ϕ(an) = [ϕ(a)]n, or in the additive form, ϕ(na) = nϕ(a).
3. For any elements a and b in G, a and b commute if and only if ϕ(a) and ϕ(b) commute as well.
4. G = ⟨a⟩if and only if G′ = ⟨ϕ(a)⟩.
5. Isomorphisms preserve orders, as ordG(a) = ordG′(ϕ(a)) for all a in G.
6. For a fixed integer k and a fixed group element b in G, xk = b has the same number of solutions in G as xk = ϕ(b) has in G′.
7. If G is finite, then G and G′ have exactly the same number of elements of every order.
8. ϕ−1 is an isomorphism from G′ to G.
9. G is abelian if and only if G′ is abelian.
10. G is cyclic if and only if G′ is cyclic.
11. If K is a subgroup of G then ϕ(K) = {ϕ(k) | k ∈K} is a subgroup of G′.
12. If K′ is a subgroup of G′ then ϕ−1(K′) = {g ∈G | ϕ(g) ∈K′} is a subgroup of G.
13. ϕ(Z(G)) = Z(G′).
In other words, if two groups G and G′ are isomorphic, we can really think about them as the same group, with the identification given by the isomorphism.
Definition 3.20. An automorphism of G is an isomorphism from a group G to itself.
Definition 3.21. Let G be a group and let a ∈G. Then, the function ϕa defined by ϕa(x) = axa−1 for all x in G is called the inner automorphism of G induced by a, and is an automorphism of G.
Proposition 3.22. The set of automorphisms of a group Aut(G) forms a group (under the operation of function composition), and the set of inner automorphisms of a group Inn(G) forms a subgroup of this group.
Remark. The group Aut(G) was first studied by O. Holder in 1893 and also inde-pendently by E.H. Moore in 1894.
Proposition 3.23. For every positive integer n, Aut(Z/nZ) is isomorphic to (Z/nZ)×.
Theorem 3.24 (Cayley 1854). Every finite group is isomorphic to a subgroup of a permutation group.
10 Example 3.25. Let G be real numbers under addition (R, +) and let G′ be the positive real numbers under multiplication, (R>0, ×). Then G and G′ are isomorphic under the mapping f(x) = ex. It is one-to-one since ex = ey →log ex = log ey, thus x = y. To prove onto, we must find a value such that f(x) = y will be fulfilled.
This value is log y. Finally, it is operation preserving since f(x + y) = e(x + y) = ex · ey = f(x)f(y). Thus, it is an isomorphism.
Non-example 3.26. The mapping from R under addition to itself mapped by f(x) = x3 is not an isomorphism since it is not operation preserving.
In other words, (x + y)3 ̸= x3 + y3 for certain values of x and y.
4 Generators and Relations Definition 4.1. Let S be some (finite, for now) set of symbols. The words formed by S are just finite-length concatenations of s and s−1 for all s ∈S.
Example 4.2. Suppose S = {x}. Then some examples of words are xx−1x, x, and x−4.
Example 4.3. Suppose S = {a, b}. Then some examples of words are abaa−1b, aba−1ab−1, and a3b−5a−2.
There’s one minor problem with words: xx−1 should not be any different that the empty word. We’ll remedy that by putting the following equivalence relation of words.
Definition 4.4. For any words u, v of S, we say that u ∼v if v can be obtained from u by a finite sequence of insertions or deletions of words of the form xx−1 or x−1x where x ∈S.
Proposition 4.5 (Equivalence classes form a group). Let S be a set of distinct symbols. For any word u, let Wu denote the set of all words on S equivalent to u.
Then the set of all equivalence classes of elements is a group under the operation u′ ∗v′ = uv′. This is called the free group on S.
To state it again: Definition 4.6. The free group on elements {x1, . . . , xn}, denoted by ⟨x1, x2, ..., xn⟩, consists of all finite-length words formed by x1, x2, ..., xn, x−1 1 , x−1 2 , ..., x−1 n under the equivalence class described above.
Proposition 4.7. The free group on one element is Z.
Example 4.8. For an example of a free group with two elements, see example 1.10.
Example 4.9. A free group with three elements would be ⟨a, b, c⟩.
Definition 4.10. Consider the free group on n elements, x1, x2, ..., xn. Let r1, r2, ..., rm be elements in this group (these are just words). The group ⟨x1, x2, ..., xn | r1, r2, ..., rm⟩ is the quotient we get by setting each ri equal to identity. Simply put, the presen-tation of a group, G, is an expression of G in terms of generators and relations.
11 Remark. We can also describe the prior group by considering the smallest normal subgroup N containing r1, . . . , rm. Then ⟨x1, x2, ..., xn | r1, r2, ..., rm⟩∼ = ⟨x1, . . . , xn⟩/N.
Theorem 4.11. Every group is a homomorphic image of a free group. In other words, every group has a presentation in terms of generators and relations.
Proof. Every group has presentation ⟨{xg | g ∈G} | xgxg′ = xgg′∀g, g′ ∈G⟩. (Note that the number of generators and relations may be infinite). We are essentially taking a generator for every element of G, as we don’t know which elements generate G. In addition, we impose a relation among the generators for every relation between elements of G.
Remark. The construction above results in a large amount of relations to check by hands. Generators and relations are important because they allow us to determine how many homomorphisms exist between two groups without checking all necessary relations between all elements.
Proposition 4.12 (Dyck). Let G = ⟨a1, a2, ..., an|w1 = w2 = ... = wt = e⟩and let G′ = ⟨a1, a2, ..., an|w1 = w2 = ... = wt = wt+1 = ... = wt+k = e⟩. Then G′ is a homomorphic image of G.
In other words, we can really think about generators and relations as taking elements which generate a group, subject to conditions/relations that the generators must satisfy. Dyck’s theorem tells us that by imposing more relations, we get a quotient group! (So if we want more elements to be zero, we just have to quotient them out.) Example 4.13. The group Z/3Z has a presentation ⟨x | x3 = e⟩. The x represents the element 1, so x3 = e just means that 1 + 1 + 1 = 0 (mod 3).
Example 4.14. The group Z2 has a presentation ⟨x, y | xy = yx⟩. The x and y represent elements (1, 0) and (0, 1), and the relation xy = yx just means that x and y commute, i.e. that (1, 0) + (0, 1) = (0, 1) + (1, 0).
Example 4.15. The symmetric group S4 has presentation ⟨x1, x2, x3 | x2 1 = x2 2 = x2 3 = (x1x2)3 = (x2x3)3 = (x1x3)2 = e⟩.
The xi represents the transpositions (i, i + 1).
Notice how these presentations are much simpler than in the constructive proof of Theorem 4.11!
5 Interesting Problems Lastly, we’ll consider two interesting problems.
12 5.1 Homophones Let’s consider the free group generated by 26 generators, say a, b, c, d, ..., x, y, z.
Now impose the relations of homophones: that is, for every pair of words which are homophones, set them equal (i.e. read and red, so read = red, where the generators are being multiplied). What is this group?
Answer: There are many ways to arrive at the same answer. Here is one plausible solution.
(1) by = bye = ⇒e = 1 (2) see = sea = ⇒a = 1 (3) buy = by = ⇒u = 1 (4) fir = fur = ⇒i = 1 (5) whole = hole = ⇒w = 1 (6) hour = our = ⇒h = 1 (7) in = inn = ⇒n = 1 (8) knot = not = ⇒k = 1 (9) die = dye = ⇒y = 1 (10) ad = add = ⇒d = 1 (11) all = awl = ⇒l = 1 (12) arc = ark = ⇒c = 1 (13) ate = eight = ⇒g = 1 (14) base = bass = ⇒s = 1 (15) berry = bury = ⇒r = 1 (16) boos = booze = ⇒s = 1 (17) bat = batt = ⇒t = 1 (18) check = cheque = ⇒q = 1 (19) idle = idol = ⇒o = 1 (20) lam = lamb = ⇒b = 1 (21) coo = coup = ⇒p = 1 (22) faze = phase = ⇒f = 1 (23) genes = jeans = ⇒j = 1 (24) flex = flecks = ⇒x = 1 13 (25) gamma = gama = ⇒m = 1 All letters except v are identity. According to Merriam Webster, there are also no relations in v, so it turns out the quotient group is just ⟨v⟩∼ = Z.
Remark. According to Prof. Etingof, there is a paper which claims that there is a homophone which also uses v, which then implies that the group is actually trivial.
You can ask similar questions about alphabets in other languages as well.
5.2 Unraveling a string Suppose we have two (infinitely tall) telephone poles. Pavel is a man of chaos and takes a very strong metal chain and loops it around these telephone poles, then fuses the two ends together. He needs a configuration so that you cannot simply pull the chains away from the poles and drag them away; as is, they are knotted around the poles.
This can be done with the following configuration: However, he is nice and allows that if they remove a single pole - any pole! - the chain will fall away and can simply be removed from the remaining pole with no problems. Can you find such a configuration? What about 3 poles? What about n poles?
For two poles: A loop (beginning and ending at this base point) going counterclockwise around the left pole is denoted as a and a loop going counterclockwise around the right pole is denoted as b.
Loops (beginning and ending at the base point, up to homotopy) form a group by concatenation, with inverse being the reverse direction of the loop. This is called the fundamental group: in this case, the group is ⟨a, b⟩. The inverse is given by 14 reversing the direction of the loop; for example, a−1 is a clockwise motion around the right pole and b−1 is a clockwise motion around the left pole.
Let us reformulate the question in terms of group theory.
• A loop is an element of this group ⟨a, b⟩.
• A loop that is entangled around the poles and cannot be removed is an element that is not the identity.
• Removing the left pole is the same as setting a to be the identity element.
Similarly, removing the right pole is the same as setting b to be the identity element.
• We must find an element x ∈⟨a, b⟩that is not identity, but when either a or b is set to identity, x becomes the identity.
One element that satisfies these conditions is aba−1b−1, shown below.
The expression aba−1b−1 represents a configuration that follows the requirments of the problem. In the current state with two poles, the chain cannot be taken out.
However, when the pole on the right is removed, the expression becomes bb−1, which can be simplified to 1. The same can be said about the other pole.
For three poles: The concept is very similar, but we now have motion c, which is around the third pole counterclockwise. The expression (aba−1b−1)c(aba−1b−1)−1c−1 is the configuration for 3 poles. By taking out any one of the three poles, we see that the expression reduced to just 1.
For n poles: Let a1, a2, . . . , an be the generators of the fundamental group, where ai is the counterclockwise loop around the ith pole.
Let xn−1 be the solution representing n −1 poles. The element xn−1anx−1 n−1a−1 n represents the solution for n poles.
Why? When either one of the poles from 1 to n −1 are removed, xn−1 becomes the identity and the element becomes ana−1 n , which is identity. If the nth pole is removed, the element becomes xn−1x−1 n−1, which is also identity.
Try drawing this out for n > 3: you will find it very hard to have discovered by hand! Maybe group theory is quite useful after all.
15 Acknowledgements We would like to thank the MIT math department, the PRIMES program at MIT, Pavel Etingof and Slava Gerovitch, who are the organizers and founders of PRIMES, Marisa Gaetz and Mary Stelow who helped organize and manage the PRIMES Circle program and for their feedback on this paper, and our mentor Merrick Cai for his continuous help and support.
16 |
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Algebra I: High School
Coefficient | Definition, Types & Examples
Contributors: Riley Kench, DaQuita Hester
Author
Author:
Riley Kench
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Instructor
Instructor:
DaQuita Hester
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Learn what a coefficient is through explanations and examples. Also, explore different types of coefficients in this lesson. Finally, learn about common numbers that are coefficients and those that are not coefficients. Updated: 11/21/2023
Table of Contents
What is a Coefficient in Mathematics?
Explanation of Coefficients in Algebraic Expressions
Exploration of Different Types of Coefficients in Mathematics
Coefficient of a Term
Coefficient and Variable
Coefficient vs. Constant
Understanding the Role of Coefficients in Algebraic Expressions
Lesson Summary
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FAQ
What's an example of a coefficient?
Let's take the expression 3x + 5 as an example. Here, the number 3 is getting multiplied by the letter x. So, the special number 3 is called the coefficient of the term 3x.
What is a simple definition of coefficient?
A coefficient in mathematics is a numerical factor that multiplies a variable. The variable represents an unknown, and the number associated with it is termed the coefficient. When a variable appears alone without a visible coefficient, it is considered to have a coefficient of 1.
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What is a Coefficient in Mathematics? -------------------------------------
What is a coefficient? A coefficient is a number that is multiplied by a variable. That is, a number being multiplied by a variable, such as x, y, or z, is called a coefficient. Moreover, any variable without a written number is said to have a coefficient of one. Coefficients appear everywhere in mathematics, especially where equations are being used. Here is an example:
In y=6 x+2, the coefficient of 6 x is 6.
Another term that requires explaining is the leading coefficient. A leading coefficient is the coefficient of the term containing the highest power of x. In a polynomial equation, the highest power of x usually comes first which is why its coefficient is called leading.
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Explanation of Coefficients in Algebraic Expressions ----------------------------------------------------
The value of a coefficient can take on any real or imaginary value. A coefficient can be an imaginary number, a real number, a positive number, a negative number, an integer, a fraction (fractional coefficient), a whole number, or a decimal.
Guidelines on How to Find the Coefficient in an Algebraic Expression
Here are the steps to find the coefficient in an algebraic expression:
Identify the term: Locate the specific term in the algebraic expression for which you want to find the coefficient.
Look for the Numerical Factor: The coefficient is the numerical factor in front of the variable(s) in the term. Identify and note down this numerical value.
Consider Implicit Coefficients: If there is no explicit numerical factor, assume an implicit coefficient of 1. For example, in the term "xy," the coefficient is 1.
Examples of Coefficients in Mathematics
Coefficients can be any of these values because they are being multiplied by a variable and any value can be multiplied by a variable. Below are examples of these possible coefficients:
f(z)=2 i z+3 has an imaginary coefficient of 2 i.
y=π x has a real coefficient of π. This is an irrational number as well.
y=5 x+9 has a positive coefficient of 5.
y=−3 x−7 has a negative coefficient of −3.
y=9 x−9 has an integer coefficient of 9.
y=2 3 x−1 has a fraction as a coefficient: 2 3.
y=x has one as a coefficient. This is a whole number as well.
y=0.666 x−1 has a decimal as a coefficient.
These examples demonstrate that a coefficient can take any real or imaginary value. Moreover, the examples presented above are all linear equations even though coefficients arise in many different equations. Any number being multiplied by any variable, no matter its power, is considered a coefficient.
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Exploration of Different Types of Coefficients in Mathematics -------------------------------------------------------------
Two principal types of coefficients appear in mathematical terms: numerical coefficients and literal coefficients. A numerical coefficient and a literal coefficient:
The numerical coefficient is the numerical value being multiplied by the variable(s).
The literal coefficient is the coefficient containing any literal numbers, or symbols representing numbers.
Here is an example:
Numerical vs. Literal Coefficient
In the image above, the numerical coefficient is 5 3 while the literal coefficient is x 2 y 3. The numerical part is the numerical coefficient while the remaining literal parts form the literal coefficient.
Numerical Coefficient
The numerical coefficient is the numerical value being multiplied by the variable. It will always be a number and can only be a numerical value. This is also the typical coefficient defined earlier in this lesson. Here is an example of a term with a numerical coefficient:
Numerical Coefficient
The numerical coefficient in the above image is 14 because it is the numerical value being multiplied by the variables.
Literal Coefficient
Before defining a literal coefficient, the literal number must be looked at. A literal number is any symbol that is being used to represent a numerical value. Now, a literal coefficient is any coefficient that contains a literal number and no numerical values. Any non-numerical value can serve as the literal coefficient. That is, any symbol or group of symbols can be a literal coefficient. Also, the literal coefficient is the coefficient to the numerical value in the term in the same way the numerical coefficient is the coefficient to the variables in the term. Here is an example:
Literal Coefficient
In the image above, the numerical coefficient is known. The literal coefficient is the remaining variable portions of the term: m 2 n.
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Coefficient of a Term ---------------------
The coefficient of a variable was already defined, but what about other numbers that appear in an equation? Another mathematical designation that arises is the term. A term in an equation is a number, variable, product or quotient of numbers and variables. Here is the makeup of terms:
Coefficient: A coefficient attached to a variable is part of a term.
Variable: Any variable, or symbol representing an unknown value, is also part of a term.
Constant: Any numerical value by itself is considered a term. That is, any value that is not a coefficient nor a variable is called a constantand is considered a term.
In the above list, the variable is defined as any non-numerical symbol that stands for an unknown value. Usually, a variable is a letter. On the other hand, a constant is any numerical value that is not being multiplied by any variable. It remains unchanged in the equation and is therefore called constant.
Here is an example:
Example of Terms
In the above image, the variables are x and y. The coefficient of x is 3 and the constant is 2.
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Coefficient and Variable ------------------------
In general, a coefficient always appears with a variable. Otherwise, the number would just be a constant. It takes a variable to make the number a coefficient. Now, what would the coefficient be if the variable has no number attached to it? There is still one variable there so the coefficient of said variable will be one. That is:
A coefficient always comes with a variable.
A number without a variable is called a constant.
A variable without a coefficient automatically has a coefficient of one (any variable with no apparent numerical multiplication has a coefficient of one).
Here is an example:
Example of Variable
In the above image, the variables are x and y, and the coefficient of x is 5.
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Coefficient vs. Constant ------------------------
The coefficient was defined as the numerical value being multiplied by a variable; or, as any number being multiplied by a variable. If a value is not being multiplied by a variable, it is called a constant. This is because the value of this constant remains unchanged. Here is an example:
Example of Constant
In this image, the variables are x and y. The coefficient of x is 7 because it is being multiplied to the x variable while the constant term is 7 because it is not being multiplied by any variable.
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Understanding the Role of Coefficients in Algebraic Expressions ---------------------------------------------------------------
Coefficients play a vital role in algebra where they serve in solving equations, determining the number of solutions, etc.
Scaling factors: Coefficients in algebraic expressions act as scaling factors, determining the magnitude of each term. Larger coefficients result in a greater impact on the overall expression.
Equation Solving: Coefficients are important while solving equations. They indicate the operations needed to isolate variables and find solutions.
Balancing Equations: Coefficients play a crucial role in preserving the equality of equations. Operations on both sides of an equation involve manipulating coefficients to balance the equation.
System of equations: While solving the system of equations, the relationship between the coefficients of variables is used to determine the number of solutions of the system.
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Lesson Summary --------------
In this lesson, the definition of coefficientwas examined in depth. In short, a coefficient is any numerical value or number, that is being multiplied by a variable or variables. Moreover, the coefficient can take on many different values. For example, it can be real, imaginary, positive, negative, integer, fractional, whole, decimal, rational, or irrational. Many numbers can appear in an equation but any groupings of numbers, variables, or both are considered a term. Any number by itself is called a constant because it remains unchanging and is not being multiplied by a variable. The variable was defined as the symbol that stands for an unknown number.
Finally, the difference between numerical and literal coefficients was explored. In general, the numerical coefficient is the number being multiplied by the variable or variables. It is always a number. The literal coefficient is the coefficient that contains at least one literal number. It is generally the remaining variable product in the term. Together, these two coefficients form a total term in an algebraic equation or expression. Here are the highlights:
A coefficient is any number or numerical value being multiplied to a variable.
A coefficient can take many different values that are both real and imaginary.
A term is any number, variable, or product of numbers and variables that appear in an equation or expression.
A constant term is a single number with no variable. It is constant because, without a variable, it will always be unchanged.
A variable is any symbol that is used to represent a numerical value or number.
A numerical coefficient is the number that is multiplied to a variable or variables. It will always be a number.
A literal coefficient is the coefficient that contains at least one literal number; furthermore, a literal number is any symbol that represents a number.
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Video Transcript
Coefficients Defined
Coefficients are the multiplicative numbers located directly in front of a variable, such as x or y. If a number in an equation is not connected with a variable, that number is not considered a coefficient. Instead, it is called a constant. Coefficients can be positive or negative and real or imaginary, as well as decimals, fractions, or whole numbers. Let's take a look at some examples:
In the term 5 x, the coefficient is the number 5. This coefficient is a whole number that is positive and real.
In the term -3 y, the coefficient is the number -3. This coefficient is a whole number that is negative and real.
For an imaginary coefficient, let's examine the term 2 ix. The coefficient for this term is 2 i, which is an imaginary whole number.
Now, let's look at 3.5 z. For this term, 3.5 is our decimal coefficient.
It is important to note that all variables will have a coefficient. If a variable is written without a coefficient, then it is assumed to have a coefficient of 1; frequently, when the coefficient is 1, it is not written. An example of this would be in the equation x + 6. There is no coefficient presented with the variable, so the term x has a coefficient of 1.
Leading Coefficients
A leading coefficient is the coefficient of a term with the highest degree, or largest exponent, within an equation. For example, the degree of the equation 6 y^4 + 5 y^2 - 2 y + 1 is 4, because that is the largest exponent. Therefore, the leading coefficient is 6.
The leading coefficient can also give you information about the graph. Specifically, it tells you how wide or narrow a quadratic equation will be and how steep a linear equation will be. Additionally, the leading coefficient will tell you which direction the graph will face. Let's take a look at a few examples:
To begin, let's compare the graph of y = x^2 to the graph of y = 2 x^2.
y = x^2
y = 2x^2
You can see that the graph of y = 2 x^2 is skinnier than the graph of y = x^2, which has a coefficient of 1. From this, we can determine that the larger the coefficient is, the skinnier or steeper the graph will be.
Now, let's compare the graph of y = 3 x to the graph of y = -3 x.
y = 3x
y = -3x
We can see that the graph of y = 3 x increases from left to right, while the graph of y = -3 x decreases from left to right. From this observation, we can conclude that having a negative coefficient will change the direction of the graph.
Lesson Summary
Coefficients are located directly in front of the variable in a term. The leading coefficient, which is located on the term with the largest exponent, tells you how steep a linear equation will be or how wide or narrow a quadratic equation will be, as well as which direction the graph will face.
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Algebra I: High School
19 chapters
163 lessons
1 flashcard sets
Chapter 1
High School Algebra: Solving Math Word Problems
Solving Word Problems: Steps & Examples 5:54 min
Solving Word Problems with Multiple Steps 5:30 min
Restating Word Problems Using Words or Images 4:32 min
Personalizing a Word Problem to Increase Understanding 4:50 min
Chapter 2
High School Algebra: Percent Notation
Percentage | Definition & Calculation 4:20 min
Converting Percent to Decimal | Overview & Examples 2:37 min
Converting Percent Notation to Fraction Notation 3:16 min
Changing Between Decimals and Percents 4:53 min
Decimal to Fraction | Conversion & Examples 7:32 min
Converting Fractions to Percents 3:43 min
Chapter 3
High School Algebra: Calculations, Ratios, Percent & Proportions
Ratios & Rates | Differences & Examples 6:37 min
How to Solve Problems with Money 8:29 min
Proportion | Definition, Formula & Types 6:05 min
Calculations with Ratios and Proportions 5:35 min
Percents: Definition, Application & Examples 6:20 min
How to Solve Word Problems That Use Percents 6:30 min
Combination in Mathematics | Definition, Formula & Examples 7:14 min
Permutation Definition, Formula & Examples 6:58 min
How to Solve Problems with Time 6:18 min
Distance Equations | Formula, Calculation & Examples 6:31 min
Chapter 4
High School Algebra: Real Numbers
Types of Numbers & Its Classifications 6:56 min
Graphing Rational Numbers on a Number Line | Chart & Examples 5:02 min
Notation for Rational Numbers, Fractions & Decimals 6:16 min
The Order of Real Numbers: Inequalities 4:36 min
Finding the Absolute Value of a Real Number 3:11 min
Chapter 5
High School Algebra: Exponents and Exponential Expressions
How to Use Exponential Notation 2:44 min
Scientific Notation | Definition, Conversion & Examples 6:49 min
Simplifying and Solving Exponential Expressions 7:27 min
Exponential Expressions & The Order of Operations 4:36 min
Multiplying Exponents | Overview, Methods & Rules 4:07 min
Dividing Exponential Expressions 4:43 min
The Power of Zero: Simplifying Exponential Expressions 5:11 min
Negative Exponents: Writing Powers of Fractions and Decimals 3:55 min
Power of Powers: Simplifying Exponential Expressions 3:33 min
Chapter 6
High School Algebra: Properties of Exponents
Properties of Exponents | Formula & Examples 5:26 min
How to Define a Zero and Negative Exponent 3:13 min
Simplifying Expressions with Exponents | Overview & Examples 4:52 min
Rational Exponents | Definition, Calculation & Examples 3:22 min
Simplifying Algebraic Expressions with Rational Exponent 7:41 min
Chapter 7
High School Algebra: Radical Expressions
Square Root | Definition, Formula & Examples 7:05 min
Estimating Square Roots | Overview & Examples 5:10 min
Simplifying Square Roots When not a Perfect Square 4:45 min
Simplifying Square Root Expressions | Steps & Examples 7:03 min
Division and Reciprocals of Radical Expressions 5:53 min
Radicands and Radical Expressions 4:29 min
Evaluating Square Roots of Perfect Squares 5:12 min
Factoring Radical Expressions 4:45 min
Simplifying Square Roots of Powers in Radical Expressions 3:51 min
Multiplying then Simplifying Radical Expressions 3:57 min
How to Divide Radicals, Square Roots & Rational Expressions 7:07 min
Simplifying Square Roots | Overview & Examples 4:49 min
Rationalizing the Denominator | Overview & Examples 7:01 min
Addition and Subtraction Using Radical Notation 3:08 min
How to Multiply Radical Expressions 6:35 min
Solving Radical Equations | Overview & Examples 6:48 min
Solving Radical Equations with Two Radical Terms 6:00 min
Chapter 8
High School Algebra: Algebraic Expressions and Equations
What is the Correct Setup to Solve Math Problems?: Writing Arithmetic Expressions 5:50 min
Common Math Formulas | Overview, Uses & Importance 7:08 min
Algebraic Expression | Definition, Operations & Examples 5:12 min
Evaluating Algebraic Expressions | Rules & Examples 7:27 min
Combining Like Terms in Algebraic Expressions 7:04 min
Practice Simplifying Algebraic Expressions 8:27 min
Negative Signs and Simplifying Algebraic Expressions 9:38 min
Writing Equations with Inequalities: Open Sentences and True/False Statements 4:22 min
Algebra Equations | Formula, Types & Examples 7:28 min
Defining, Translating, & Solving One-Step Equations 6:15 min
Solving Equations Using the Addition Principle 5:20 min
Multiplication Principle | Definition, Equations & Examples 4:03 min
Solving Equations Using Both Addition and Multiplication Principles 6:21 min
Collecting Like Terms On One Side of an Equation 6:28 min
Solving Equations Containing Parentheses 6:50 min
Solutions to Systems of Equations | Overview & Examples 4:45 min
Translating Words into Algebraic Expressions | Phrases & Examples 6:31 min
How to Solve One-Step Algebra Equations in Word Problems 5:05 min
Solving Multiple Step Equations | Explanation, Steps & Examples 5:44 min
Solving Algebra Word Problems | Multi-Step Equations & Examples 6:16 min
Algebra Terms Flashcards
Chapter 9
High School Algebra: Algebraic Distribution
Distribution in Math | Definition, Process & Examples 5:39 min
Distributing First vs. Adding First: Differences & Examples 6:44 min
Distributing Positive and Negative Signs 5:56 min
Distributing Algebraic Expressions with Numbers and Variables 7:57 min
Changing Negative Exponents to Fractions 6:24 min
Fractional Exponents | Definition, Rules & Examples 6:38 min
Distribution of More Than One Term in Algebra 6:12 min
Chapter 10
High School Algebra: Properties of Functions
Function in Math | Definition & Examples 7:57 min
Transformations: How to Shift Graphs on a Plane 7:12 min
How to Add, Subtract, Multiply and Divide Functions 6:43 min
Domain & Range of a Function | Definition, Equation & Examples 8:32 min
How to Compose Functions 6:52 min
Inverse Functions | Definition, Methods & Calculation 6:05 min
Applying Function Operations Practice Problems 5:17 min
Chapter 11
High School Algebra: Working With Inequalities
Inequality Signs in Math | Symbols, Examples & Variation 7:09 min
Graphing Inequalities | Definition, Rules & Examples 7:59 min
Inequality Notation | Overview & Examples 8:16 min
Graphing Inequalities | Overview & Examples 12:06 min
Solve & Graph an Absolute Value Inequality | Formula & Examples 8:02 min
Absolute Value Inequalities | Definition, Calculation & Examples 9:06 min
Translating Math Sentences to Inequalities 5:36 min
Chapter 12
High School Algebra: Linear Equations
Linear Equations | Definition, Formula & Solution 7:28 min
Applying the Distributive Property to Linear Equations 4:18 min
Forms of a Linear Equation | Overview, Graphs & Conversion 6:38 min
Abstract Algebraic Examples and Going from a Graph to a Rule 10:37 min
Undefined & Zero Slope Graph | Definition & Examples 4:23 min
Parallel vs Perpendicular vs Transverse Lines Overview & Examples 6:06 min
Parallel & Perpendicular Lines | Equation, Graph & Examples 6:07 min
Linear Equation | Parts, Writing & Examples 8:58 min
System of Equations in Algebra | Overview, Methods & Examples 8:39 min
How Do I Use a System of Equations? 9:47 min
Nonlinear Function | Definition, Examples & Graphs 6:03 min
Chapter 13
High School Algebra: Factoring
Factoring in Algebra | Definition, Equations & Examples 5:32 min
Finding the Prime Factorization of a Number | Meaning & Examples 5:36 min
Using Prime Factorizations to Find the Least Common Multiples 7:28 min
Equivalent Expressions and Fraction Notation 5:46 min
Using Fraction Notation: Addition, Subtraction, Multiplication & Division 6:12 min
Factoring Out Variables: Instructions & Examples 6:46 min
Combining Numbers and Variables When Factoring 6:35 min
Transforming Factoring Into A Division Problem 5:11 min
Factoring by Grouping | Definition, Steps & Examples 7:46 min
Chapter 14
High School Algebra: Quadratic Equations
Quadratic Equation | Definition, Formula & Examples 5:13 min
Solving Quadratics: Assigning the Greatest Common Factor and Multiplication Property of Zero 5:24 min
Quadratic Function | Formula, Equations & Examples 9:20 min
How to Solve Quadratics That Are Not in Standard Form 6:14 min
Solving Quadratic Inequalities Using Two Binomials 5:36 min
Chapter 15
High School Algebra: Graphing and Factoring Quadratic Equations
Tables & Graphs in the Real World | Uses & Examples 5:50 min
Scatter Plot Graph | Overview, Uses & Examples 7:17 min
Parabola | Definition & Parabolic Shape Equation 4:36 min
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Applications of Algebraic Expressions in Real Life
Mathematics is all around us, and Algebraic Expressions are one of its most powerful tools. From calculating expenses to designing video games, algebra plays a crucial role in solving real-world problems. At 98thPercentile, we believe that understanding Algebraic Expressions in Real Life can make learning math more engaging and meaningful. Let’s explore how algebra is used in everyday situations!
What Are Algebraic Expressions?
An Algebraic Expression is a combination of numbers, variables (like x or y), and operations (such as +, −, ×, ÷). For example, 3x + 5 is an algebraic expression where x is a variable. These expressions help us represent real-life situations mathematically, making problem-solving easier.
How Are Algebraic Expressions Used in Daily Life?
1. Budgeting and Shopping
Imagine you’re at a store buying chocolates. Each chocolate costs $2, and you want to buy x number of them. The total cost can be represented as:
Total Cost = 2x
If you also buy a drink for $3, the expression becomes:
Total Cost = 2x + 3
This simple Algebraic Expression helps you calculate expenses and manage your pocket money efficiently.
2. Travel and Distance Calculations
Planning a road trip? Algebra can help! If a car travels at a speed of 60 miles per hour, the distance covered in t hours is:
Distance = 60t
This formula helps estimate travel time and fuel costs, making trips smoother.
3. Sports and Fitness
Athletes and coaches use algebra to track performance. For example, if a runner improves their speed by 2 seconds every week, their progress can be modeled as:
New Time = Current Time — 2x (where x is the number of weeks)
This helps in setting realistic fitness goals.
4. Cooking and Recipes
Ever adjusted a recipe for more or fewer people? If a cake recipe requires 2 cups of flour for 4 servings, the amount needed for n servings is:
Flour Needed = (2/4) × n = 0.5n
Algebra ensures you get the measurements right every time!
5. Technology and Video Games
Video game designers use Algebraic Expressions to create movement, scores, and levels. For example, a character’s jump height (h) might depend on the button press time (t):
h = 10t — 5t²
This makes gameplay dynamic and exciting!
6. Construction and Architecture
Builders use algebra to calculate materials needed. If a wall requires 50 bricks per row and has r rows, the total bricks are:
Total Bricks = 50r
This prevents shortages and saves costs.
7. Business and Profit Calculations
Entrepreneurs use algebra to predict profits. If selling lemonade at 1percupwithacostof1percupwithacostof0.50 per cup, the profit (P) for x cups sold is:
P = 1x — 0.5x = 0.5x
This helps in making smart business decisions.
8. Science and Engineering
Scientists use algebra to model experiments. For example, the growth of bacteria (B) over time (t) might follow:
B = 100 × 2^t
This helps in medical and environmental research.
Why Learning Algebraic Expressions Matters
Understanding Algebraic Expressions in Real Life helps in:
At 98thPercentile, we make learning algebra fun with interactive lessons and real-world examples. Our expert tutors help students master Algebraic Expressions with ease, preparing them for academic and career success.
Conclusion
From shopping to gaming, Algebraic Expressions are everywhere! Learning how to use them not only improves math skills but also equips students with tools for everyday life. Whether calculating expenses, planning trips, or designing games, algebra makes life easier and more efficient.
Ready to explore the power of algebra? Join 98thPercentile today and unlock the magic of math in real life!
By understanding Algebraic Expressions in Real Life, students can see math as a helpful friend rather than a challenging subject. With practical applications in finance, sports, technology, and more, algebra is truly a skill for life. Let 98thPercentile guide you on this exciting mathematical journey!
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Why does 1 plus 1 equal 2?
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Sachin Parekh
Contributing to Entropy Since 1996 ! · Upvoted by
Michael Harrison
, M.S. Mathematics, Louisiana State University (2007)
·11y
Originally Answered: Why does 1+1=2? ·
Yesterday I came across this Calvin And Hobbes comic strip and I think this may help you.
Here's a quote on Mathematics which makes sense to me.
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.
Albert Einstein
Continue Reading
Yesterday I came across this Calvin And Hobbes comic strip and I think this may help you.
Here's a quote on Mathematics which makes sense to me.
As far as the laws of mathematics refer to reality, they are not certain, and as far as they are certain, they do not refer to reality.
Albert Einstein
Upvote ·
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More answers below
Why is 2 plus 2 the same as 2 times 2?
What does 1 1/2" + 1 1/2" equal to?
If 1 plus 1 equals 2, why does 2 plus 2 equal 4?
What is the answer to 1 plus 1\2?
What is 1/4 plus 1/4? 1/4 plus 1/4 is equal to 1/2.
Randy C. Hamilton
Lives in Central Kansas USA " The Real Wild West" (1950–present) · Author has 16.2K answers and 1.2M answer views
·Jul 9
Originally Answered: Why does 1+1=2 and 1×1=1? ·
Get yourself some beans or coins and do a visual. Or go back to first grade and stay awake this time.
Upvote ·
9 2
Nick Picard
JD, University of Victoria · Author has 2.2K answers and 4.5M answer views
·6y
Originally Answered: Does 1+1 really equal 2 or can we prove that 1+1=something else? ·
Those little squiggles on your screen that you call “numbers”? You want to know what they do, how they’re related?
Numbers do exactly what we agree that they do. If you want numbers to do something different, you would use different symbols.
So let’s say you wanted to explore a mathematical parallel universe where 1+1=3. You could do that, and I bet you could make it work, because your curious and motivated! The problem is? Everyone would find it way too confusing. The first person you showed it to would say, “I don’t think you should use the + symbol because you’re using it to mean something to
Continue Reading
Those little squiggles on your screen that you call “numbers”? You want to know what they do, how they’re related?
Numbers do exactly what we agree that they do. If you want numbers to do something different, you would use different symbols.
So let’s say you wanted to explore a mathematical parallel universe where 1+1=3. You could do that, and I bet you could make it work, because your curious and motivated! The problem is? Everyone would find it way too confusing. The first person you showed it to would say, “I don’t think you should use the + symbol because you’re using it to mean something totally different. Use another symbol, like € or something.”
So that’s what mathematicians do. They have invented worlds where there are things a lot like addition or numbers a lot like 1 but they call them by different names. Because calling them the same way you and I do would be ridiculously confusing.
So 1+1=2. But don’t let that get you down. Explore abstract algebra. That’s the field of math where you can break all the rules. Have fun!
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Timothy T. Jones
Studied Mathematics at Johns Hopkins University · Author has 3K answers and 570.1K answer views
·7y
Originally Answered: What is 1 plus 2? ·
this is an one of the great unsolved questions of classical mathematics and although many of the worlds most high-quality intellects have spent lifetimes working on this there’s been real no forward progress in several hundred years —there’s really very little hope in my opinion that we will ever know the answer to this
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What is 1 plus 1, other than the answer 2?
Is it possible to prove that 1+1 is 2?
How do you solve this problem? 15-(6-4) (2)
What is 1/10 plus 1/4 plus 1/5 plus 2/15?
What was the answer to 1/2+1/2+1/2+1/2 ?
John Timmers
"Philosopher", artsy type, and probably some other stuff. Mostly harmless. · Author has 81 answers and 380.5K answer views
·Updated 8y
Originally Answered: Why is one plus two equal to three? ·
Well, there are a number of views about this in the philosophy of mathematics. I'll discuss two. One is called Platonism, and it is the view that abstract objects have some sort of being or existence. So 1 exists, and so do functions, and all that sort of stuff--whatever they are. Then there is non-Platonism. Usually, non-Platonists deny that abstract objects exist and account for numbers in terms of how language works.
There's a lot to the philosophy of mathematics, and I suggest checking out The Stanford Encyclopedia of Philosophy about all that. It's a great resource. Here's the link: Philos
Continue Reading
Well, there are a number of views about this in the philosophy of mathematics. I'll discuss two. One is called Platonism, and it is the view that abstract objects have some sort of being or existence. So 1 exists, and so do functions, and all that sort of stuff--whatever they are. Then there is non-Platonism. Usually, non-Platonists deny that abstract objects exist and account for numbers in terms of how language works.
There's a lot to the philosophy of mathematics, and I suggest checking out The Stanford Encyclopedia of Philosophy about all that. It's a great resource. Here's the link: Philosophy of Mathematics
Aside from all that one can analyze 1 + 2 = 3 further. There are two ways to do so I'm familiar with: 1) in terms of sets and 2) in terms of the successor function.
So, you can analyze the statement "1 + 2 = 3" further in these ways:
1) Take any set containing one object, and combine it with any other set that contains two objects, and you have yourself a set that contains three objects. Yippee!
2) Let's introduce some notation: when I add this guy, ' , next to a number, it means the successor of that number (the number that comes after it).
Here we go:
i) 1 = 0' (1 is the successor of 0, or 1 comes after 0)
ii) 2 = 1' (2 is the successor of 1, or 2 comes after 1)
iii) 3 = 2' (3 is the successor of 2, or 3 comes after 2)
iv) 2 = 0'' (2 is the successor of the successor of 0, or 2 comes after the number that comes after 0)
v) 3 = 0''' (3 is the successor of the successor of the successor of 0, or 3 comes after the number that comes after the number that comes after 0)
Therefore, by substituting equals for equals:
1 + 2 = 3 is equivalent to 0' + 0'' = 0'''
Here are two renditions of that for you:
a.) The successor of 0 added to the successor of the successor of 0 equals the successor of the successor of the successor of 0.
b.) The number that comes after 0 added to the number that comes after the number that comes after 0 equals the number that comes after the number that comes after the number that comes after 0.
Hope that was fun.
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Mark Gardner
Copyeditor, Proofreader, Fact-checker at Self-Employment (2016–present) · Author has 418 answers and 102.8K answer views
·1y
So, why DOES 1 plus equal 2? Basically, our personal self-convincing of this comes from the ancient notion of aggregating collections of things, long before the notion of ‘plus’ or ‘add’, maybe even before the notion of ‘number’ took firm hold in infant vocabularies. You have a stone in each hand, you put them both together on the corner table-rock in the back of the cave, and the notion of ‘two’ is born.
However, man is man and is a curious creature — as you have shown with your question. Eventually the concepts grow (especially when they are useful, so commerce was certainly a driver for arit
Continue Reading
So, why DOES 1 plus equal 2? Basically, our personal self-convincing of this comes from the ancient notion of aggregating collections of things, long before the notion of ‘plus’ or ‘add’, maybe even before the notion of ‘number’ took firm hold in infant vocabularies. You have a stone in each hand, you put them both together on the corner table-rock in the back of the cave, and the notion of ‘two’ is born.
However, man is man and is a curious creature — as you have shown with your question. Eventually the concepts grow (especially when they are useful, so commerce was certainly a driver for arithmetic, just as the use of area and volume was a driver for geometry, the other primal ‘mathematics’). The growth leads to more advanced manipulations and methods (like column addition, long division, sum-of-partial-products multiplication, etc., etc.). The growth also leads to questions just like yours, about the whole process and collection of processes.
As to your particular question, it’s a bit like Jang’s “What if one plus one didn’t equal two?” in the movie ‘Lucy’, or like Ogre’s “What if … uhh … C-A-T really spelled dog?” in the movie ‘Revenge of the Nerds’. But it is an important foundational question in mathematics, and Russell and Whitehead spent 397 pages answering your question, obviously a bit much to put here, in the first volume of their Principia Mathematica. I recommend you watch this video: https:/ /
Peano also took a stab at it with his axioms; you might enjoy watching https:/ /
However, I DON’T recommend you read their book — it’s way too obscure for most of us, me included.
Good luck.
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Andrew Chamberlain
I'm good at math, I wish our education system was too. · Author has 144 answers and 313.8K answer views
·9y
Originally Answered: Why does one plus one always equal two? ·
The fact that you used the words "one" and "two" instead of the Arabic numerals means that it is because of natural number definitions (i.e. the definition of 2 is 1+1).
However, if you had asked "Why does 1 + 1 always equal 2?", there would be a different answer.
"1" always means a value of one, unless it is in base zero, but base zero is very strange and I don't really even understand it. In base two, "1" means one just like in most bases, but "2" does not exist in base two.
There are a couple concrete rules for bases (and if you apply them to base zero you will see why it is so strange) and t
Continue Reading
The fact that you used the words "one" and "two" instead of the Arabic numerals means that it is because of natural number definitions (i.e. the definition of 2 is 1+1).
However, if you had asked "Why does 1 + 1 always equal 2?", there would be a different answer.
"1" always means a value of one, unless it is in base zero, but base zero is very strange and I don't really even understand it. In base two, "1" means one just like in most bases, but "2" does not exist in base two.
There are a couple concrete rules for bases (and if you apply them to base zero you will see why it is so strange) and they are:
(the "base number" of base ten is ten, of base nine it is nine, etc.)
I: "Base Number" = 10
II: "Base Number" minus "one" = highest number that can be represented with a single digit in said base
III: "Base Number" = the number of unique digits used in said base
Just as clarification, the "10" in rule I does not necessarily mean "ten". This directly leads to the point that in base two, "10" means "two". See, the base number (two) equals 10, as in "two". 1 + 1 = 10
So, "1 + 1" doesn't always equal "2", but "one plus one" always equals "two".
Upvote ·
9 9
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Rob Weir
Former Senior Technical Staff Member at IBM (company) (1991–2018) · Author has 19.3K answers and 36.1M answer views
·10y
Originally Answered: Why is 1 + 1 = 2? ·
This is page 86 of Volume II of Russell and Whitehead's Principia Mathematica. It proves that 1+1=2.
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This is page 86 of Volume II of Russell and Whitehead's Principia Mathematica. It proves that 1+1=2.
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Roger Pickering
Spent 6 years at 2 universities doing maths · Author has 14.9K answers and 5.9M answer views
·1y
It has to equal something and a long time ago someone decided to call it "two".
Evidently that idea has caught on extremely well.
Upvote ·
9 1
Martyn Hathaway
BSc in Mathematics, University of Southampton (Graduated 1986) · Author has 4.7K answers and 6.7M answer views
·7y
Originally Answered: What is the answer to 1 plus 1\2? ·
What is the answer to 1 plus 1\2?
As I’ve no idea what the ‘\’ sign means, I can’t answer this!
Now, if only you had used the ‘/’ sign!
Tell you what, I’m assume you mistyped and meant to use ‘/’.
Case 1:
Assuming you are using the ‘/’ to represent division, then 1/2 means one divided by two, which is known as a half. Thus 1 plus 1/2 is equivalent to saying one more than a half, which is one and a half. This number can be expressed in several ways. For example:
1 1 2 1 1 2
3 2 3 2
1.5 1.5
Case 2:
Prior to the decimalisation of British currency in 1971, the pound was divided into 20 shillings each of
Continue Reading
What is the answer to 1 plus 1\2?
As I’ve no idea what the ‘\’ sign means, I can’t answer this!
Now, if only you had used the ‘/’ sign!
Tell you what, I’m assume you mistyped and meant to use ‘/’.
Case 1:
Assuming you are using the ‘/’ to represent division, then 1/2 means one divided by two, which is known as a half. Thus 1 plus 1/2 is equivalent to saying one more than a half, which is one and a half. This number can be expressed in several ways. For example:
1 1 2 1 1 2
3 2 3 2
1.5 1.5
Case 2:
Prior to the decimalisation of British currency in 1971, the pound was divided into 20 shillings each of 12 pence. Went writing amounts less than a pound, the convention was to write the number of shillings, then a ‘/’ followed by the number of pennies - if there were no pennies, ‘-’ was used, not ‘0’.
Thus 1/2 means 1 shilling and two pennies (which was said as ‘one shilling and tuppence’).
Now, we are seemingly adding this to 1; but 1 what?
If it was 1 shilling, it would have been written as 1/-
So, it must be 1 penny.
So, adding 1 penny to one shilling and tuppence, we get one shilling and thruppence (apparently, some people may have said ‘threppence’), which would be written as:
1/3
Sure, case 1 is by far the more likely scenario, and probably would have been more likely even in the pre-decimal UK; but, if there is any ambiguity in your question, don’t be surprised if someone assumes you mean something you didn’t intend!
Upvote ·
9 1
9 1
Abhishek Sharma
mechanical engineer, farmer, reader, coder(rookie), DIY projects maker
·9y
Originally Answered: Why does 1+1 is equal to 2? ·
Please read carefully.
If N and N' are any real numbers then N+N' should be equal to N"((hypothetically).
Because we know that if we walk N km and then another N' km then summation of these two distances must be greater than N and N'.
N">=N+N'
Similarly put N equals to any number and write that number.
If that number is 1 (assume) and another number N' is also equals to 1
Then N+N'=1+1
Please see that in starting we assume that N">N'>N then by addition rule we can say that
N+N'=N"
1+1=N"
Now again check that how much N"is equals to?
If it is 1 write 1, if it is not then go for higher no and then wri
Continue Reading
Please read carefully.
If N and N' are any real numbers then N+N' should be equal to N"((hypothetically).
Because we know that if we walk N km and then another N' km then summation of these two distances must be greater than N and N'.
N">=N+N'
Similarly put N equals to any number and write that number.
If that number is 1 (assume) and another number N' is also equals to 1
Then N+N'=1+1
Please see that in starting we assume that N">N'>N then by addition rule we can say that
N+N'=N"
1+1=N"
Now again check that how much N"is equals to?
If it is 1 write 1, if it is not then go for higher no and then write 2.
Because we know from assumptions
9>8>7>6>5>4>3>2>1>0
So 1+1 is always equals to 2..
Upvote ·
9 2
9 1
Sean Luo
A 13-year-old that's quite smart.
·6y
Originally Answered: What does 1+1 equal to, 2 or 3? ·
Place one eraser on your paper. Count it and you get one eraser. Add another eraser to the paper. Count it. There are now two erasers on the paper. 1+1 equals to 2, not 3 because adding just one eraser to a paper that already had an eraser on it will not magically make a third eraser pop up on the paper.
Upvote ·
9 1
David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by
Michael Jørgensen
, PhD in mathematics and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 9.9K answers and 68.4M answer views
·11y
Originally Answered: Why does 1+1=2? ·
2 is the name of 1+1. No matter how you answer this question, it comes down to that. There's nothing deep or profound about it. 2 is what we call 1+1.
Upvote ·
999 424
9 9
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12872 | https://www.nextgurukul.in/wiki/concept/kerala/class-8/maths/part-2-ratio/ratios-and-proportions/3957646 | Typesetting math: 100%
Notes On Ratios and Proportions - Kerala board Class 8 Maths
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Ratio
Ratios are used to compare quantities. To compare two quantities, the units of the quantities must be the same. Ratios help us to compare quantities and determine the relation between them. We write ratios in the form of fractions and then compare them by converting them to like fractions. If these like fractions are equal, then the ratios are said to be equivalent.
e.g. Cost of 6 pens is Rs 90. What would be the cost of 10 such pens?
Solution: Cost of 6 pens = Rs 90
Cost of 1 pen = 90 ÷ 6 = Rs 15
Hence, cost of 10 pens = 10 × 15 = Rs 150.
Proportion
When two ratios are equivalent, the four quantities are said to be in proportion.
Ratio and proportion problems can be solved by using two methods, the unitary method and equating the ratios to make proportions, and then solving the equation.
Unitary method
Unitary method is the method of finding the value of one unit (unit rate) at first and then the value of required number of units.
Percentages
Percentage is another method used to compare quantities. Percent is derived from the Latin word ‘per centum’, which means per hundred. Percentage is the numerator, of a fraction, whose denominator is hundred. Percent is represented by the symbol - %.
e.g. 21 100 21 100 or 21%
Summary
Ratio
Ratios are used to compare quantities. To compare two quantities, the units of the quantities must be the same. Ratios help us to compare quantities and determine the relation between them. We write ratios in the form of fractions and then compare them by converting them to like fractions. If these like fractions are equal, then the ratios are said to be equivalent.
e.g. Cost of 6 pens is Rs 90. What would be the cost of 10 such pens?
Solution: Cost of 6 pens = Rs 90
Cost of 1 pen = 90 ÷ 6 = Rs 15
Hence, cost of 10 pens = 10 × 15 = Rs 150.
Proportion
When two ratios are equivalent, the four quantities are said to be in proportion.
Ratio and proportion problems can be solved by using two methods, the unitary method and equating the ratios to make proportions, and then solving the equation.
Unitary method
Unitary method is the method of finding the value of one unit (unit rate) at first and then the value of required number of units.
Percentages
Percentage is another method used to compare quantities. Percent is derived from the Latin word ‘per centum’, which means per hundred. Percentage is the numerator, of a fraction, whose denominator is hundred. Percent is represented by the symbol - %.
e.g. 21 100 21 100 or 21%
Videos
References
LearnNext - Ratios and Proportions open_in_new
Emathzone - Concept of Ratio open_in_new
Basic-mathematics - Ratios open_in_new
Freemathhelp - Proportions and Ratios open_in_new
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12873 | https://www.physicsforums.com/threads/difference-between-zero-and-identically-zero.924360/ | General Math
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
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1
PrathameshR
: 35
: 3
What's the difference between something (eg a function or a matrix) becoming zero and it becoming identically zero?
Illustrations will be helpful. Thanks
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Hi PrathameshR ;)
There is no real mathematical distinction.
Identical zero is merely an emphasis to indicate it's 'more' zero than might otherwise be thought.
When we say that a function is identical to zero, we want to emphasize that we really mean the zero-function, which is zero everywhere in its domain.
Saying that a function is zero should mean the same thing (that it's the zero-function), but some authors are a bit sloppy, and they might mean that the function just becomes zero for a certain value in its domain.
Same thing for a matrix.
A matrix that is zero means the same thing as a matrix that is identical to zero - it's the matrix with only zeroes.
That is as opposed to a matrix that has a zero.
3
PrathameshR
: 35
: 3
I like Serena said:
Hi PrathameshR ;)
There is no real mathematical distinction.
Identical zero is merely an emphasis to indicate it's 'more' zero than might otherwise be thought.
When we say that a function is identical to zero, we want to emphasize that we really mean the zero-function, which is zero everywhere in its domain.
Saying that a function is zero should mean the same thing (that it's the zero-function), but some authors are a bit sloppy, and they might mean that the function just becomes zero for a certain value in its domain.
Same thing for a matrix.
A matrix that is zero means the same thing as a matrix that is identical to zero - it's the matrix with only zeroes.
That is as opposed to a matrix that has a zero.
This really helped. Thanks
4
Mark44
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I like Serena said:
There is no real mathematical distinction.
Identical zero is merely an emphasis to indicate it's 'more' zero than might otherwise be thought.
I'm going to disagree slightly with this.
For example the equation is a true statement only when x = 1. OTOH, the equation is true for all real values of x. The expression is identically zero.
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I like Serena
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Mark44 said:
I'm going to disagree slightly with this.
For example the equation ##x^2 - 2x + 1 = 0## is a true statement only when x = 1. OTOH, the equation ##x^2 - 2x + 1 - (x - 1)^2 = 0## is true for all real values of x. The expression ##x^2 - 2x + 1 - (x - 1)^2## is identically zero.
This is indeed exactly where the ambiguity occurs.
When we write , it's somewhat ambiguous if we're talking about a specific function value, or about the function in general.
Literally speaking, the expression is not a function - it's a specific function value.
It's just that it's not uncommon that an author intends the corresponding function.
So saying is identical to zero is intended to mean that is zero.
Or alternatively that the function given by is zero.
The word identical here is used to disambiguate, although it's not really what the word identical means (mathematically it just means the same thing as equal to).
Note that 'zero' here is ambiguous as well, since it's not clear whether it's a function value that is zero, or the function itself that is the zero-function.
6
Mark44
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I like Serena said:
This is indeed exactly where the ambiguity occurs.
When we write ##x^2 - 2x + 1 - (x - 1)^2##, it's somewhat ambiguous if we're talking about a specific function value, or about the function in general.
I didn't say anything about functions, and explicitly wrote "the expression .
I like Serena said:
Literally speaking, the expression ##x^2 - 2x + 1 - (x - 1)^2## is not a function - it's a specific function value.
What I wrote was not in the context of functions, but if you treat it as such, with , then this is a function that is identically zero. I.e., . That was not my intent, though. Instead, I was distinguishing between an expression that is zero for a particular value of the variable (conditional equality) versus another one that was identically zero.
I like Serena said:
It's just that it's not uncommon that an author intends the corresponding function.
So saying ##x^2 - 2x + 1 - (x - 1)^2## is identical to zero is intended to mean that ##x\mapsto x^2 - 2x + 1 - (x - 1)^2## is zero.
Or alternatively that the function given by ##f(x)=x^2 - 2x + 1 - (x - 1)^2## is zero.
The word identical here is used to disambiguate, although it's not really what the word identical means (mathematically it just means the same thing as equal to).
Note that 'zero' here is ambiguous as well, since it's not clear whether it's a function value that is zero, or the function itself that is the zero-function.
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Mark44 said:
I didn't say anything about functions, and explicitly wrote "the expression ##x^2 - 2x + 1 - (x - 1)^2##.
What I wrote was not in the context of functions, but if you treat it as such, with ##f(x) = x^2 - 2x + 1 - (x - 1)^2##, then this is a function that is identically zero. I.e., ##f(x) \equiv 0##. That was not my intent, though. Instead, I was distinguishing between an expression that is zero for a particular value of the variable (conditional equality) versus another one that was identically zero.
Whether we use the word 'function' or not, when we talk about an expression to be identical to zero, we mean the function that assigns a value according to the given expression to each value that is in its domain instead of the actual expression. (And without specification of the domain, it is assumed to be minus any points for which the expression is not defined.)
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Mark44
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I like Serena said:
Whether we use the word 'function' or not, when we talk about an expression to be identical to zero, we mean the function that assigns a value according to the given expression to each value that is in its domain instead of the actual expression.
I don't agree that the concept of functions necessarily needs to be part of such a discussion. One can write without either implicitly or explicitly stating that the left side is a function. I'm not saying it's wrong to do so, just that it's not necessary.
I like Serena said:
(And without specification of the domain, it is assumed to be ##\mathbb R## minus any points for which the expression is not defined.)
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Mark44 said:
I don't agree that the concept of functions necessarily needs to be part of such a discussion. One can write ##\forall x \in \mathbb R, \sin^2(x) + \cos^2(x) - 1 = 0## without either implicitly or explicitly stating that the left side is a function. I'm not saying it's wrong to do so, just that it's not necessary.
When we use the '=' operator, we need to have an equivalence relation that corresponds to it.
And when we add and/or multiply values, we need something like a field (typically ) in which those are defined.
And when we have an expression with a variable, we need a context to compare it to anything.
That context can either be the field with its associated equality operator (for the 'expression' in your example).
Or it's a function space with its associated equality operator (the 'function' that I refer to).
TL;DR, we leave out that it's a function, but it's a function nonetheless to algebraically treat it as we do.
10
FactChecker
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"identically zero" means that for all legitimate values of variables, the result is zero. That is quite different from saying that something equals zero for some particular values of variables.
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StoneTemplePython
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One more example for OP's original question:
I won't insist on the exclusivity of usage here, just that the common way of using "identically zero" is with respect to polynomials. The idea is that a degree polynomial is completely characterized by unique data points. (You can prove this using Vandermonde Matrices -- I assume some underlying field with characteristic zero here, probably or for convenience.)
At most of those unique data points may be zeros (aka roots), except in the degenerate case where you're dealing with the zero polynomial (i.e. every thing is identically zero).
i.e. a 'regular' degree polynomial may be written as
if you somehow find that this polynomial has (at least) unique zeros, that means you in fact have the zero polynomial, i.e.
I'm not totally comfortable calling the zero matrix "identically zero". I suppose you could argue that the underlying linear transformation has this property. However, note that for an x matrix filled entirely with zeros, its characteristic polynomial (like that of all nilpotent matrices) is still
which is not identically zero.
12
WWGD
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Maybe a way of making the distinction is whether we have an equation or an equality/identity with the same formula. By the Fundamental Thm. of Algebra, above equation ( in one variable) may only have at most 2 zeros.
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Mark44
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FactChecker said:
"identically zero" means that for all legitimate values of variables, the result is zero. That is quite different from saying that something equals zero for some particular values of variables.
I agree completely. There is no need to bring in functions, fields, equivalence operators, or other concepts from advanced algebra.
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12874 | https://www.cs.cmu.edu/~venkatg/teaching/codingtheory/notes/notes4.pdf | Introduction to Coding Theory CMU: Spring 2010 Notes 4: Elementary bounds on codes January 2010 Lecturer: Venkatesan Guruswami Scribe: Venkatesan Guruswami We now turn to limitations of codes, in the form upper bounds on the rate of codes as a function of their relative distance. We will typically give concrete bounds on the size of codes, and then infer as corollaries the asymptotic statement for code families relating rate and relative distance. All the bounds apply for general codes and they do not take advantage of linearity. However, for the most sophisticated of our bounds, the linear programming bound, which we discuss in the next set of notes, we will present the proof only for linear codes as it is simpler in this case.
We recall the two bounds we have already seen. The Gilbert-Varshamov bound asserted the ex-istence of (linear) q-ary codes of block length n, distance at least d, and size at least qn Volq(n,d−1).
Or in asymptotic form, the existence of codes of rate approaching 1 −hq(δ) and relative distance δ. The Hamming or sphere-packing bound gave an upper bound on the size (or rate) of codes, which is our focus in these notes. The Hamming bound says that a q-ary code of block length n and distance d can have at most qn Volq(n,⌊(d−1)/2⌋) codewords. Or in asymptotic form, a q-ary code of relative distance δ can have rate at most 1 −hq(δ/2) + o(1).
As remarked in our discussion on Shannon’s theorem for the binary symmetric channel, if the Hamming bound could be attained, that would imply that we can correct all (as opposed to most) error patterns of weight pn with rate approaching 1 −h(p). Recall that there are perfect codes (such as the Hamming codes) that meet the Hamming bound. However, these codes have very small distance (3 in the case of Hamming codes). A generalization of Hamming codes called binary BCH codes (the acronym stands for the code’s independent inventors Hocquenghem (1959) and Bose and Ray-Chaudhuri (1960)) show that when d is a fixed constant and the block length is allowed to grow, the Hamming bound is again tight up to lesser order terms. However, we will improve upon the Hamming bound and show that its asymptotic form (for any relative distance bounded away from zero) cannot be attained for any fixed alphabet. The proof method has some connections to list decoding, which will be an important focus topic later in the course.
1 Singleton bound We begin with the simplest of the bounds: Theorem 1 Let C be a code of block length n and minimum distance d over an alphabet of size q.
Then |C| ≤qn−d+1.
Proof: Suppose not, and |C| > qn−d+1. By the pigeonhole principle there must be two codewords c1, c2 ∈C, c1 ̸= c2 that agree on the first n −d + 1 locations. But then ∆(c1, c2) ≤d −1 < d, contradicting the hypothesis that C has minimum distance d. □ 1 This gives an alphabet-independent asymptotic upper bound on the rate as a function of relative distance.
Corollary 2 The rate R and relative distance δ of a code satisfy R ≤1 −δ + o(1).
Though really simple, the Singleton bound is tight in general — we will later see an algebraic family of codes called Reed-Solomon codes which achieve the Singleton bound and have dimension n −d + 1 and minimum distance d. The family of codes which meet the Singleton bound are called maximum distance separable (MDS) codes.
However, Reed-Solomon and other MDS codes will be (necessarily) defined over an alphabet that grows with the block length. For code families over a fixed alphabet such as binary codes, substantial improvements to the Singleton bound are possible. We turn to such bounds next.
2 The Plotkin bound The Gilbert-Varshamov bound asserts the existence of positive rate binary codes only for relative distance δ < 1/2. The Hamming bound on the other hand does not rule out positive rate binary codes even for δ > 1/2, in fact not even for any δ < 1. Thus there is a qualitative gap between these bounds in terms of identifying the largest possible distance for asymptotically good binary codes. We now prove an upper bound which shows that the relative distance has to be at most 1/2 (and thus the Hamming bound is quite weak for large δ) unless the code has very few codewords (and in particular has vanishing rate).
While the same underlying ideas are involved, the proofs are simpler to present for the binary case, so we will focus on binary codes. We will state the bound for the q-ary case and leave the details as an exercise. Our proofs reduce the task of bounding the size of the code to bounding the number of pairwise far-apart unit vectors in Euclidean space, and then use a geometric argument for the latter task.
We first state the geometric lemma we need.
Lemma 3 Let v1, v2, . . . , vm be m unit vectors in Rn.
1. Suppose ⟨vi, vj⟩≤−ϵ for all 1 ≤i < j ≤m. Then m ≤1 + 1 ϵ.
2. Suppose ⟨vi, vj⟩≤0 for all 1 ≤i < j ≤m. Then m ≤2n.
Proof: We only prove the first part, and leave the second as an (interesting) exercise. Note that bound of 2n is best possible, as we can take n orthonormal vectors and their negations. For the first part, we have 0 ≤⟨ m X i=1 vi, m X i=1 vi⟩= m X i=1 ∥vi∥2 + 2 X 1≤i n/2, then |C| ≤ 2d 2d−n.
2. If d ≥n/2, then |C| ≤2n.
Proof: Let m = |C| and let c1, c2, . . . , cm ∈{0, 1}n be the codewords of C.
By hypothesis ∆(ci, cj) ≥d for 1 ≤i < j ≤m. We will map the codewords into unit vectors vi ∈Rn, i = 1, 2, . . . , m, such that the angle between every pair of vectors is at least 90 degrees (i.e., their dot product ⟨vi, vj⟩< 0). These vectors are defined as follows: vi = 1 √n (−1)ci, (−1)ci, · · · , (−1)ci[n] , where ci[ℓ] is the ℓ’th bit of the codeword ci. It is easy to check that ⟨vi, vj⟩= 1 n(n −2∆(ci, cj)) ≤n −2d n .
When d ≥n/2, these dot products are non-positive, and by the second part of Lemma 3, we can bound m ≤2n.
For the first part, if 2d > n, then ⟨vi, vj⟩≤−2d−n n < 0, and therefore by the first part of Lemma 3, we can bound m ≤1 + n 2d −n = 2d 2d −n .
□ The above shows that a code family of relative distance δ ≥1/2 + γ can have at most O(1/γ) codewords. Thus a code family cannot have relative distance strictly bounded away from 1/2 with a number of codewords that is growing with the block length. In particular, such code families have zero rate. We now prove that this is also the case if the relative distance is 1/2.
We now use a “puncturing” argument, which implies that the complement of the feasible rate vs relative distance region is convex, to derive an upper bound of rate for relative distances δ < 1/2.
Theorem 5 If a binary code C has block length n and distance d < n/2, then |C| ≤d · 2n−2d+2.
Proof: Let ℓ= n −2d + 1 and S = {1, 2, . . . , ℓ}. For each a ∈{0, 1}ℓ, define the subcode Ca to be the subcode of C consisting of all codewords which have a in the first ℓpositions, projected on Sc = {1, 2, . . . , n} \ S. Formally, Ca = {c|Sc | ci = ai for 1 ≤i ≤ℓ}. Each Ca is a binary code of block length n −ℓ= 2d −1. Note that since C has distance at least d, so does Ca. By Theorem 4, |Ca| ≤2d. Of course, |C| = P a∈{0,1}ℓ|Ca|. So we conclude |C| ≤2d · 2ℓ= d · 2n−2d+2. □ We record the asymptotic implication of the above upper bound as: 3 Corollary 6 The rate R of a binary code of relative distance δ must satisfy R ≤1 −2δ + o(1).
The above arguments can be extended to the q-ary case. The idea is to map q symbols to q unit vectors whose pairwise dot product is exactly − 1 q−1.
Exercise 1 Let C be a q-ary code of block length n and minimum distance at least d.
1. If d > (1 −1/q)n, then |C| ≤ qd qd−(q−1)n.
2. When d < (1 −1/q)n, |C| ≤q3d q−1qn−qd/(q−1).
Deduce that the R of a q-ary code of relative distance δ must satisfy R ≤1 − q q−1δ + o(1).
Here is a plot of the Gilbert-Varshamov lower bound on rate, and the Hamming and Plotkin upper bounds on rate, for binary codes. On the horizontal axis is the relative distance δ ∈[0, 1] and the vertical axis is the rate R ∈[0, 1]. Any (R, δ) point under the leftmost curve (the Gilbert-Varshamov bound) is achievable, and any (R, δ) point above either the Plotkin bound (the straight line) or the Hamming bound is not achievable. Notice that the Hamming bound is stronger than the Plotkin bound for low distances (or high rates). We now proceed to a bound that improves both the Hamming and Plotkin bounds.
4 3 Johnson and Elias-Bassalygo bounds Recall that Aq(n, d) denotes the size of the largest q-ary code of block length n and distance d. We denote by Aq(n, d, w) the size of a largest constant weight code of block length n and distance d all of whose codewords have Hamming weight w. We also denote by A′ q(n, d, w) the largest size of a code of block length n and distance d all of whose codewords have Hamming weight at most w. For the case q = 2, we just denote these quantities by A(n, d), A(n, d, w), and A′(n, d, w) respectively.
On your problem set, you are asked to prove the following (just for the binary case).
Exercise 2 Prove that A(n, d) ≤ 2n (n w)A(n, d, w). More generally, prove that Aq(n, d) ≤ qn (n w)(q−1)w Aq(n, d, w) The above gives a method to upper bound the size Aq(n, d) of unrestricted codes via upper bounds on the size Aq(n, d, w) of constant weight codes. Note that when w < d/2, Aq(n, d, w) = 1, so the Hamming like upper bound Aq(n, d) ≤ qn (n w)(q−1)w is a special case of this. In general the larger the w as a function of d for which we can prove a good upper bound on A(n, d, w) (as either a constant or a polynomial function of n), the better the upper bound on Aq(n, d).
We will now prove such an upper bound, in fact for the more general quantity A′ q(n, d, w). Such a bound, called the Johnson bound, is intimately connected to list decoding. Proving that A′ q(n, d, w) is small, say at most L which is either a constant or bounded by nO(1), implies that for every q-ary code C of block length n and distance d, every Hamming ball of radius w contains at most L codewords of C. In other words, if a codeword is transmitted and at most w errors corrupt it, then one can list decode a small list of at most L candidate codewords one of which must equal the original codeword. Of course, for w < d/2, we have L = 1, and the key here is that one can have w ≫d/2 and still ensure a small worst-case list size L.
Once we prove the Johnson bound, we will deduce our desired bound on Aq(n, d), called the Elias-Bassalygo bound after their inventors, by appealing to Exercise 2. Our proofs of the Johnson bound will be geometric in nature, relying on Lemma 3. We prove the bounds for binary codes, and leave the extension to larger alphabets as exercises.
3.1 Binary codes Theorem 7 (Binary Johnson bound) For integers 1 ≤w ≤d ≤n/2, if w ≤1 2(n− p n(n −2d)), then A′(n, d, w) ≤2n. (We will often refer to the quantity 1 2(n − p n(n −2d)) as J2(n, d), the (bi-nary) Johnson radius.) Proof: Let C = {c1, . . . , cm} ⊆{0, 1}n be a code such that ∆(ci, cj) ≥d for i ̸= j, and wt(ci) ≤w for each i = 1, 2, . . . , m. We will map the codewords ci into vectors vi ∈Rn similarly to the proof of the Plotkin bound (Theorem 4), except we won’t normalize them to unit vectors here: vi = (−1)ci, (−1)ci, · · · , (−1)ci[n] , where ci[ℓ] is the ℓ’th bit of the codeword ci. Likewise the all 0’s vector is mapped to the vector r ∈Rn 5 r = (1, 1, · · · , 1) .
Let α > 0 be a parameter to be picked later. The parameter α will be picked so that all the pairwise dot products ⟨vi −αr, vj −αr⟩are nonpositive for i ̸= j. Now ⟨vi −αr, vj −αr⟩ = n −2∆(ci, cj) + α2n + α(2wt(ci) −n + 2wt(cj) −n) ≤ n −2d + α2n + 2α(2w −n) .
The latter quantity is at most 0 provided 4w ≤2n − αn + n −2d α .
The choice α = p n(n −2d) maximizes the right hand side, and leads to the requirement w ≤n 2 − p n(n −2d) 2 which is met by hypothesis about w. Therefore, for this choice of α, ⟨vi −αr, vj −αr⟩≤0 for 1 ≤i < j ≤m. are nonpositive for i ̸= j. Appealing to (the second part of) Lemma 3, we can conclude m ≤2n, and thus A′(n, d, w) ≤2n. □ Remark 8 It is possible to improve the upper bound on A′(n, d, w) for w < J2(n, d) from 2n to n by noting that for the choice of parameters ⟨vi −αr, r⟩> 0 for each i. This together with the nonpositive pairwise dot products can be used to improve the geometric upper bound on the number of vectors from 2n to n.
For w slightly less than the Johnson radius J2(n, d), one can sharpen the upper bound on A′(n, d, w) to a constant independent of n.
Exercise 3 Prove that when w ≤n 2 − √ n(n−2d+2d/L) 2 , A′(n, d, w) ≤L.
(Hint: Follow the above proof, and pick parameters so that the first part of Lemma 3 can be used.) Together with Exercise 2, we thus have the following upper bound, called the Elias-Bassalygo bound, on the size (rate) of codes of certain distance (relative distance).
Theorem 9 For integers 1 ≤d ≤n/2, A(n, d) ≤n2n+1 n J2(n,d) where J2(n, d) = (n − p n(n −2d))/2.
Thus a binary code family of relative distance δ has rate at most 6 1 −h 1 − √ 1 −2δ 2 + o(1) .
3.2 Statement for larger alphabets As with the Plotkin bound, we leave the extension of the Johnson and Elias-Bassalygo bounds to larger alphabets exercises. The hint is to map q symbols into appropriate vectors in Rq so that large distance between codewords translates into small dot product between their associated vectors.
Exercise 4 For all integers q ≥2 and 1 ≤d ≤(1 −1/q)n, A′ q(n, d, w) ≤n(q −1) provided w < Jq(n, d) = n 1 −1 q 1 − s 1 − qd (q −1)n !
.
Further, if w ≤n 1 −1 q 1 − s 1 −qd(1 −ϵ) (q −1)n !
then A′ q(n, d, w) ≤1/ϵ.
Together with Exercise 2, this gives the Elias-Bassalygo upper bound for q-ary codes: Theorem 10 A q-ary code family of relative distance δ has rate at most 1 −hq (1 −1/q) 1 − s 1 − qδ q −1 !
+ o(1) .
3.3 Alphabet oblivious bound for Johnson radius When discussing list decoding of codes such as Reed-Solomon codes which are defined over an alphabet size that grows with the block length, it will be useful to have the following ”alphabet oblivious” version of the Johnson bound. (This version is also a simpler and often good enough approximation to the q-ary Johnson radius when q is somewhat large.) Theorem 11 Let C ⊆Σn be a code of block length n and distance d. Then the following hold: 1. Every Hamming ball of radius at most J(n, d) = n − p n(n −d) in Σn has at most O(n|Σ|) codewords of C.
7 2. Every Hamming ball of radius at most n − p n(n −d + dϵ) in Σn has at most 1/ϵ codewords of C.
The above statement follows from Exercise 4 by verifying that for every q ≥2 and 0 ≤x ≤1−1/q, 1 − √ 1 −x ≤(1 −1/q) r 1 − qx q −1 .
We conclude with a plot that adds the Elias-Bassalygo upper bound to the earlier plot.
Note that this bound improves on both the Hamming and Plotkin bounds, but for small distances the difference between the Hamming the Elias-Bassalygo bounds is small.
8 |
12875 | https://naomi.princeton.edu/wp-content/uploads/sites/744/2021/03/Pais_thesis_main-min.pdf | Emergent Collective Behavior in Multi-Agent Systems: An Evolutionary Perspective Darren Pais A Dissertation Presented to the Faculty of Princeton University in Candidacy for the Degree of Doctor of Philosophy Recommended for Acceptance by the Department of Mechanical and Aerospace Engineering Adviser: Professor Naomi Ehrich Leonard November 2012 c ⃝Copyright by Darren Pais, 2012.
All rights reserved.
Abstract The study of collective behavior involves the analysis of interactions among a set of agents that yield collective outcomes at the level of the group. The behavior is said to be emergent when it cannot be understood simply as the sum of its constituent parts. Further, group-level outcomes can in turn influence individual interactions.
The complexity of this interplay makes the study of emergence challenging and excit-ing. This dissertation is focused on the study of emergent collective behavior from the perspective of evolution. Evolution is a simple yet powerful algorithm, which when acting on interacting entities in a dynamic environment, yields an array of fascinating behavior as manifest in the natural world. Natural collectives display a wide variety of cooperative behavior and have evolved to efficiently manage the inherent tradeoff between robust behavior and adaptability to dynamic environments. These properties have motivated the design of bio-inspired algorithms for sensing and decision-making in robotic collectives. In this work, we study the evolutionary mechanisms for co-operation and tradeoffmanagement in biological collectives, with a focus on four related topics: replicator-mutator dynamics, collective migration, collective pursuit and evasion, and decision-making dynamics in swarms.
The replicator-mutator dynamics define a canonical model from evolutionary the-ory and have recently been used to study the evolution of language and the behavioral dynamics of social networks. While the analysis of stable equilibria of these dynamics has been a focus in the literature, we prove that certain conditions suffice for the equa-tions to exhibit stable limit cycles. These cycles correspond to oscillations of grammar dominance in language evolution and to oscillations in behavioral preferences in so-cial networks. For the collective migration problem, it is well-established that a small group of leaders can guide a large swarm of followers. It is less clear how presumably self-interested individuals have evolved to take on such divergent roles. We design a network-based evolutionary model to understand the evolution of leadership in migra-tion, with a focus on the role of network topology on the emergent dynamics. Pursuit and evasive behaviors are ubiquitous in biology and are key drivers for collective motion. We use computational simulations and analytical calculations to study a co-evolving pursuit and evasive system, and incorporate the evolved strategies in a cyclic pursuit-evasion collective motion model. The ‘stop-signaling’ inhibitory mechanism has been recently shown to be critical to the decentralized decision-making dynamics in honeybee swarms. We investigate bifurcations in a model of swarm decision-making as a function of the stop-signal and the values of different alternatives, and present a comprehensive analysis of the dynamics of the model.
iii Acknowledgements First, I would like to acknowledge my advisor Naomi Leonard. Naomi is a brilliant scientist, an immensely creative researcher, and above all, a wonderful and caring person. Naomi’s gentle persuasion to always think deeply and clearly about a prob-lem, her discerning taste for exciting research questions, and her extraordinary ability to make connections across diverse areas, have had an immense impact on my de-velopment and growth as a graduate student. I am extremely grateful to Naomi for allowing me the freedom to pursue my interests and for helping me build the confi-dence to tackle challenging new problems. I will fondly remember the countless hours spent discussing ideas in Naomi’s office and will always be grateful for her consistent and detailed edits of numerous paper drafts and thesis chapters. Naomi’s constant support and encouragement have made for a truly rewarding graduate school experi-ence.
My committee members Phil Holmes and Simon Levin have been important sources of guidance and support over the years. Much of this thesis rests on fun-damental tools and ideas that I first encountered in classes that I took with Phil and Simon, early on in graduate school. I am grateful for the interest that they have taken in my research and my career. I am also tremendously grateful for their time and effort spent in reading my thesis and providing meticulous feedback. The process of editing this thesis with help from Naomi, Phil and Simon has been an invaluable learning experience.
I am grateful to my thesis defense examiners Rob Stengel and Howard Stone for taking an interest in my work and for being excellent role models. I appreciate their willingness and enthusiasm to discuss a range of topics including teaching, careers, jobs, and research styles; these discussions have left a lasting impression on me.
I immensely enjoyed the opportunity to assist Rob in teaching his ‘robotics and intelligent systems’ course. Playing regular ‘noon hoops’ basketball with Howard at Dillon Gym has been a unique privilege; Howard has the same impressive creativity and energy on the basketball court that he does as a professor.
The Mechanical and Aerospace engineering (MAE) department at Princeton is a vibrant and stimulating environment, and has been the prefect place to learn the fundamentals of dynamics and control (D&C), and consider broad applications.
The quality of teaching and breath of research in D&C at Princeton is fantastic. I would like to acknowledge the D&C faculty (Naomi, Phil, Rob, Jeremy Kasdin, Mike Littman and Clancy Rowley) and my D&C classmates for helping me build a strong iv foundation and for introducing me to their diverse and exciting research interests including planet finding, fluid flow control and neuronal network dynamics.
The work on honeybee swarms in Chapter 7 has come out of a recent collaboration with James Marshall and Patrick Hogan at the University of Sheffield, which began in 2011 at the Mathematical Biosciences Institute in Ohio.
Working with James and Patrick has been an extraordinary privilege. I have learnt a tremendous amount from them, not only about the biology of honeybee swarms, but also about successful approaches to interdisciplinary research and collaboration. I am grateful for the many stimulating discussions we have had over the past two years; these discussions have impacted several areas of this thesis. The work in Chapter 5 is inspired in large part by the array of captivating experimental and computational work by Iain Couzin and his collaborators, particularly, Vishu Guttal and Colin Torney. Iain is an extremely gifted speaker and scientist. Interacting with him, Vishu, Colin, and the rest of the Couzin group over the years has been tremendously rewarding.
I would like to acknowledge two former postdoctoral research scholars in our group, Carlos Caicedo-N´ u˜ nez and Ming Cao. Carlos and I collaborated closely on the work presented in Chapter 4; I appreciate his algebraic tenacity and enthusiasm for new ideas. Ming was a collaborator on my first graduate school research project on tensegrity-based formation control, and was an important mentor during my ini-tial years as a graduate student. I am grateful to both Carlos and Ming for their friendship over the years. I am also grateful for the friendship and support from all the members of the Leonard group, both past and present. Ben Nabet, Dan Swain, Andy Stewart, Kendra Cofield, Stephanie Goldfarb, George Young, Tian Shen and Paul Reverdy have all been good friends and always available to discuss new ideas.
The Princeton campus and the MAE department have been extremely supportive and nurturing environments in which to think, interact and grow. I am grateful for the many close friends that I have made here, beginning with my experience as a first-year resident at the Graduate College. I hesitate to name all these individuals, lest I mistakenly leave someone out. Suffice to say that each of them has impacted me in a unique way and I hope we continue to remain in touch for many years to come. I want to thank Anand Ashok, Katie Fitch, Paul Reverdy, David Turnbull and George Young for helping me proof-read the final draft of this thesis. Of course, any missed typos you may now find are their fault. I am grateful to the MAE graduate administrators Jessica O’Leary and Jill Ray for making the paperwork seamless, and for always having something encouraging to say, even on the toughest days.
v My work as a graduate fellow at the McGraw Center for Teaching and Learning has been extremely rewarding. I would like to acknowledge the McGraw center staff (Carol Porter, Nic Voge, JeffHimpele, Sandra Moskovitz) for their valuable guidance.
Much of this thesis was written in the comfort of the beautiful Princeton home of Daisy Fitch and Professor Val Fitch. I thank them for the extraordinary opportunity to spend my final Princeton summer in such a calm and peaceful environment; this thesis would perhaps have taken significantly longer to complete without it.
Finally, I would like to thank my family for their love, support, and encouragement over the years. I thank my sister for always helping me keep things in perspective and for helping me stay balanced. She has been a constant cheerleader of my successes and I value our relationship greatly. I thank my parents for giving us an open-minded multicultural childhood that was fun and dynamic, with never a dull moment. Most importantly, I thank them for helping us prioritize the right things in life and teaching us the immense joy and value of learning. My parents always encouraged us to dream big and I will forever be grateful to them for giving us the resources and ability to freely follow our dreams, wherever that may lead.
My time at Princeton was supported by several generous fellowships, for which I am deeply grateful: the Gordon Y. S. Wu first-year fellowship, the Martin Sum-merfield memorial graduate fellowship, the Britt and Eli Harari fellowship for an international student, and the Harold W. Dodds honorific fellowship. My research was also supported in part by grants from the Office of Naval Research, the Air Force Office of Scientific Research, the Army Research Office, and the National Science Foundation.
This dissertation is labeled T-3247 in the records of the Department of Mechanical and Aerospace Engineering.
vi To my parents.
vii Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iii Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xi 1 Introduction 1 1.1 Overview of Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 1.1.1 Replicator-Mutator Dynamics . . . . . . . . . . . . . . . . . .
4 1.1.2 Collective Migration . . . . . . . . . . . . . . . . . . . . . . .
5 1.1.3 Pursuit and Evasion . . . . . . . . . . . . . . . . . . . . . . .
7 1.1.4 Swarm Decision-Making . . . . . . . . . . . . . . . . . . . . .
7 1.2 Contributions and Thesis Outline . . . . . . . . . . . . . . . . . . . .
8 2 Background 10 2.1 Evolutionary Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . .
10 2.2 Dynamical Systems Tools . . . . . . . . . . . . . . . . . . . . . . . .
16 2.3 Graph Theory Tools . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 2.4 Stochastic Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 2.4.1 Random Points on a Simplex . . . . . . . . . . . . . . . . . .
25 3 Replicator-Mutator Dynamics in the Plane 26 3.1 Model Description . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 3.2 Motivation for Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 3.3 Bifurcations with N = 2 Strategies . . . . . . . . . . . . . . . . . . .
32 3.4 Planar Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33 4 Replicator-Mutator Dynamics in Higher Dimensions 41 4.1 Hopf Bifurcation Calculation . . . . . . . . . . . . . . . . . . . . . .
42 4.2 Criticality of Hopf Bifurcation . . . . . . . . . . . . . . . . . . . . .
45 4.3 Illustration of Bifurcations . . . . . . . . . . . . . . . . . . . . . . .
47 viii 4.4 One-Parameter Multi-Cycles . . . . . . . . . . . . . . . . . . . . . .
49 4.4.1 Case 2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .
50 4.5 Extensions and Generalizations . . . . . . . . . . . . . . . . . . . . .
54 4.6 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58 5 Evolutionary Dynamics of Collective Migration 59 5.1 Model Description . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61 5.2 Evolutionary Dynamics in the All-to-all Limit . . . . . . . . . . . . .
66 5.2.1 Adaptive Dynamics Calculations . . . . . . . . . . . . . . . .
67 5.2.2 Adaptive Dynamics Results . . . . . . . . . . . . . . . . . . .
68 5.2.3 Evolutionary Simulations . . . . . . . . . . . . . . . . . . . . .
70 5.3 Evolutionary Dynamics with Limited Social Interactions . . . . . . .
72 5.3.1 Fast Timescale Results . . . . . . . . . . . . . . . . . . . . . .
72 5.3.2 Slow Timescale Evolutionary Dynamics . . . . . . . . . . . .
77 5.4 Dynamic Nodes and Bifurcations . . . . . . . . . . . . . . . . . . . .
80 5.5 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84 6 Coevolutionary Dynamics of Pursuit and Evasion 87 6.1 Dynamics of Pursuit and Evasion . . . . . . . . . . . . . . . . . . . .
89 6.2 Evolutionary Dynamics . . . . . . . . . . . . . . . . . . . . . . . . .
93 6.2.1 Monte-Carlo Simulations . . . . . . . . . . . . . . . . . . . . .
94 6.2.2 Theoretical Analysis . . . . . . . . . . . . . . . . . . . . . . .
96 6.3 Collective Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99 6.4 Final Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101 7 Decision-Making Dynamics in Honeybee Swarms 103 7.1 Model Description . . . . . . . . . . . . . . . . . . . . . . . . . . . .
104 7.2 Symmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
106 7.3 Asymmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
108 7.4 Separation of Timescales . . . . . . . . . . . . . . . . . . . . . . . .
112 7.5 Stochastic Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . .
114 7.6 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117 8 Final Remarks 120 8.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
120 8.2 Common Themes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
122 8.3 Looking Ahead . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123 ix A Calculation of Lyapunov coefficient 127 B Supporting material for Chapter 3 128 C Supporting material for Chapter 4 130 C.1 Proof of Lemma 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
130 C.2 Proof of Lemma 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
131 C.3 Proof of Lemma 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
133 C.4 Criticality analysis for Corollaries 4.1 and 4.2 . . . . . . . . . . . . .
135 D Supporting material for Chapter 5 136 E Supporting material for Chapter 6 138 E.1 Proof of Lemma 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
138 E.2 Lemma E.1 used in Theorem 6.1 . . . . . . . . . . . . . . . . . . . . .
139 F Supporting material for Chapter 7 140 G Edge Detection 146 Bibliography 148 x List of Figures 1.1 Robustness vs. adaptability in collective systems . . . . . . . . . . . .
2 2.1 Constant vs. dynamic fitness landscapes ⋆ . . . . . . . . . . . . . .
12 2.2 Simplex phase space . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 2.3 Pairwise invasibility plots . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.4 Bifurcations in one-dimensional systems . . . . . . . . . . . . . . . .
17 2.5 Canonical Hopf bifurcation ⋆ . . . . . . . . . . . . . . . . . . . . . .
18 2.6 Illustration of cusp catastrophe . . . . . . . . . . . . . . . . . . . . .
19 2.7 Illustration of graph adjacency and Laplacian . . . . . . . . . . . . .
20 2.8 Circulant graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 2.9 Spatially embedded graphs . . . . . . . . . . . . . . . . . . . . . . . .
22 2.10 Illustration of one-dimensional SDEs . . . . . . . . . . . . . . . . . .
23 2.11 Uniformly distributed points on ∆2 . . . . . . . . . . . . . . . . . . .
25 3.1 Simulation of RM dynamics with random payoff. . . . . . . . . . . .
31 3.2 Simulation of RM dynamics with random payoffand LCs . . . . . . .
31 3.3 Limit cycle for language dynamics . . . . . . . . . . . . . . . . . . . .
32 3.4 Bifurcations for N = 2 ⋆ . . . . . . . . . . . . . . . . . . . . . . . .
33 3.5 One parameter graphs and bifurcations with mutation (Q1) ⋆. . . .
34 3.6 One parameter graphs and bifurcations with mutation (Q2) ⋆. . . .
35 3.7 Phase portraits for all-to-all payoffand N = 3 ⋆ . . . . . . . . . . .
36 3.8 Phase portraits for directed cycle payoffand N = 3 ⋆ . . . . . . . .
37 3.9 ℓ1(α, β) in three dimensions for N = 3 . . . . . . . . . . . . . . . . .
39 3.10 Limit cycle bifurcations for N = 3 ⋆ . . . . . . . . . . . . . . . . . .
39 3.11 Heteroclinic cycle ⋆ . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 4.1 Two-parameter circulant graph . . . . . . . . . . . . . . . . . . . . .
42 4.2 Criticality of bifurcation ⋆ . . . . . . . . . . . . . . . . . . . . . . .
46 4.3 Effect of parameters α and β on bifurcations . . . . . . . . . . . . . .
47 4.4 Illustration of an infeasible Hopf bifurcation point . . . . . . . . . . .
48 4.5 One-parameter circulant graphs . . . . . . . . . . . . . . . . . . . . .
49 xi 4.6 Multi-cycles for N = 6 ⋆ . . . . . . . . . . . . . . . . . . . . . . . .
51 4.7 N = 3 limit cycles for non-circulant payoffmatrices . . . . . . . . . .
55 4.8 N > 3 limit cycles for non-circulant payoffmatrices ⋆ . . . . . . . .
56 4.9 Limit cycles for the dynamics (3.2) with random payoffgraphs ⋆ . .
57 5.1 Migratory performance as a function of investment kD . . . . . . . .
63 5.2 Steady-state migration speed . . . . . . . . . . . . . . . . . . . . . . .
66 5.3 Evolutionary singular strategies k∗ . . . . . . . . . . . . . . . . . . .
69 5.4 Simulations of the evolutionary dynamics in the all-to-all limit ⋆ . .
71 5.5 Minimal root sets of social graph . . . . . . . . . . . . . . . . . . . .
74 5.6 Effect of social graph topology on evolutionary outcomes . . . . . . .
78 5.7 Evolutionary equilibria for lattice and random graphs . . . . . . . . .
80 5.8 Bifurcations of the adaptive node dynamics with N = 2 nodes ⋆ . .
83 5.9 Bifurcations of the adaptive node dynamics with N = 10 nodes . . . .
84 5.10 Nodal dynamics for the undirected star ⋆ . . . . . . . . . . . . . . .
84 5.11 Role of topology in determining locations of leaders ⋆ . . . . . . . .
85 6.1 Cartoon trajectories of a pursuer and an evader . . . . . . . . . . . .
90 6.2 Simulations for the 9 pairs of competing pursuit & evasive strategies .
92 6.3 Monte-Carlo simulations . . . . . . . . . . . . . . . . . . . . . . . . .
95 6.4 Simulation of smooth dynamics (6.5) . . . . . . . . . . . . . . . . . .
99 6.5 Sensing topology for cyclic pursuit and evasion . . . . . . . . . . . . .
100 6.6 Simulations of pursuit-evasion collective dynamics . . . . . . . . . . .
101 6.7 Pursuit-evasion chains for large N . . . . . . . . . . . . . . . . . . . .
102 7.1 From microscopic to mean-field equations ⋆ . . . . . . . . . . . . . .
105 7.2 Pitchfork bifurcation for the symmetric case . . . . . . . . . . . . . .
108 7.3 Bifurcation set for the asymmetric case ⋆ . . . . . . . . . . . . . . .
109 7.4 Projected equilibria illustrating bifurcations ⋆ . . . . . . . . . . . .
110 7.5 Full three-parameter bifurcation set . . . . . . . . . . . . . . . . . .
111 7.6 Timescale separation in the stop-signalling dynamics ⋆ . . . . . . .
114 7.7 Stochastic dynamics comparisons ⋆ . . . . . . . . . . . . . . . . . .
116 7.8 Stochastic dynamics simulations ⋆ . . . . . . . . . . . . . . . . . . .
118 8.1 Three instances of hysteresis . . . . . . . . . . . . . . . . . . . . . . .
123 8.2 Exploration, cyclical domination, exploitation . . . . . . . . . . . . .
125 ⋆denotes figures in color xii Chapter 1 Introduction Seek simplicity, and distrust it. -Alfred North Whitehead (1861-1947) Emergent collective behavior involves interactions between individual agents that yield distinct patterns at the level of the group. Emergent systems have group level out-comes that cannot be understood simply as the superposition of their constituent elements, instead emergent group behavior is nonlinearly related to individual inter-actions. Moreover, just as individual actions affect group outcomes, group outcomes feed back to affect individual actions. This coupling between the microscopic indi-vidual level and the macroscopic group level makes the study of emergent behavior vibrant, exciting, and challenging. The past few decades have seen significant re-search activity in applying computational and analytical tools to studying emergent phenomena in a wide variety of applications. These include ecology [72, 49], cellular biology [127, 21], animal behavior , disease dynamics [63, 20], climate , eco-nomics and more recently, robotic swarms [61, 65] and social networks [146, 4] (the few examples cited here are a small sample from a vast literature).
For problems in biology, the evolutionary approach involves studying the fascinat-ing array of observed behavior in natural collectives (flocks, schools, herds, etc.) from the perspective of evolution by natural selection. This approach provides important insights into the mechanisms that drive group behavior in natural collectives. The development of mathematical models to explain evolutionary puzzles such as coop-eration and altruism [62, 105, 84] in swarms, flocks, and schools, continues to be an active area of research (see [88, 89, 70] for a recent debate on the topic).
In this thesis, we use a set of relatively simple and tractable evolutionary mod-els to study emergent behavior in selected natural collective systems. We focus our study on four areas described in §1.1, and utilize three related evolutionary models in our analysis: replicator-mutator dynamics for a single population with discrete 1 strategies, adaptive dynamics for a single population with continuous strategies, and coevolutionary replicator dynamics for a two-population system with discrete strate-gies. Each of these models is described in detail in §2.1.
Robustness Adaptability Engineered Formations Emergent Social Phenomena Biological Collectives e.g. tensegrity based formation control e.g. collective migration (Ch 5), collective pursuit and evasion (Ch 6) and honeybee swarms (Ch 7) e.g. replicator-mutator dynamics (Ch 3 and Ch 4) Figure 1.1: A classification of collective systems based on robustness and adaptability metrics.
Collective systems can be classified based on a large variety of metrics (system size, heterogeneity, network architecture, dynamics, etc.), but looking broadly, ro-bustness and adaptability are two key metrics that enable a rough classification as illustrated in Figure 1.1. Robustness refers to the ability of collective systems to re-ject disturbances and perform specific tasks accurately, predictably, and repeatedly, in stochastic environments. Adaptability refers to the ability of collective systems to continually modify their behavior in reaction to a dynamic environment and to solve a range of problems such as foraging for resources, migrating to new locations, and avoiding threats or prey. Highly adaptive systems have the ability to learn from past experiences and to find innovative solutions to novel problems.
Robustness and adaptability constitute a fundamental tradeoffin engineered col-lective systems in the sense that systems designed to be significantly robust and pre-dictable for specific tasks (industrial robots for example) are inherently unadaptive, and vice-versa. In Figure 1.1, the three rectangles at the corners of the robustness-adaptability space represent three classes of collective systems and allow us to connect the four focus areas of this thesis: 2 • Bottom-right; engineered formations: These systems are designed to be highly robust so that they can perform specific tasks reliably (often with provable guar-antees on performance). However, they have limited adaptability to problems outside their specific domain of design. Examples include the wide variety of industrial robotic systems and formation control in mobile autonomous robotic collectives. One example of recent work in our group in this area is the use of tensegrity-based control laws (involving simultaneous attraction and repulsion between agents) to stabilize the shapes of formations of autonomous vehicles [78, 94].
• Top-left; emergent social phenomena: These collective systems are highly adap-tive in dynamic environments and can display macroscopic behavior that rapidly cascades between extremes. This macroscopic behavior is often unpredictable, and in certain cases can be inherently chaotic, resulting in a limited robust-ness. Examples include the popular area of social network dynamics, cascades in financial markets, and the dynamics of behavioral preferences and fashion trends. In Chapters 3 and 4 we study a model of behavioral preferences in social networks (known as replicator-mutator dynamics) that fits the paradigm of this quadrant.
• Top-right; biological collectives: One of the main reasons that bio-inspired robotic collective behavior has gained tremendous traction over the last two decades is because many biological systems possess the unique ability to act both robustly and adaptively. The seemingly complex emergent collective be-havior observed in these biological systems has frequently been shown to be the consequence of simple individual rules at the microscopic level. These rules, and the emergent behavior, have been shaped by evolutionary dynamics on gen-erational timescales. In Chapters 5 and 6 we study two ubiquitous biological collective phenomena, namely, collective migration, and pursuit and evasion, respectively. In Chapter 7 we focus on the decision-making dynamics in swarms of honeybees. A honeybee swarm is a particularly good example of a system in this quadrant, where robust and accurate decision-making in dynamic environ-ments has been shaped by the evolutionary forces of reproduction and colony survival. Individuals in a honeybee swarm can be radically adaptive. For ex-ample, scout bees that are involved in locating nesting sites for the swarm are, in fact, forager bees that have switched behavior from seeking bright blossoms to searching for dark nesting crevices .
3 This thesis is the product of interdisciplinary research drawing from ideas in evo-lutionary biology, animal behavior, engineering and applied mathematics. The back-ground material in Chapter 2 introduces key tools from these areas that are used throughout the thesis. Along with the exciting quest of understanding the complexity of biological collectives, research in our group also strives to draw ideas and principles from biology that can be applied to the design of robotic collectives. From the per-spective of Figure 1.1, this effort involves utilizing ideas from the top-right quadrant to push engineered systems (bottom right quadrant) up the adaptability axis, while still managing the adaptability/robustness tradeoff. This is no easy challenge and will remain an area of research emphasis going forward as autonomous collective systems are tasked with solving increasingly complex problems. As a result, together with a focus on understanding collective dynamics in the four areas described below, we also consider how this understanding inspires engineering design in each case.
1.1 Overview of Topics 1.1.1 Replicator-Mutator Dynamics∗ The replicator-mutator dynamics define a canonical model from evolutionary theory and represent the evolution of a discrete number of strategies in a single large popu-lation. The dynamics have received significant attention recently as a model for the evolution of language. They also provide a simple model for the analysis of behavior dominance in social networks where replication is akin to imitation of individuals subscribed to successful behaviors in a population, and mutation is akin to random error in behavior selection. Much of the analysis of the dynamics has focused on stable equilibria and their bifurcations. In this thesis we focus on the existence of structurally stable limit cycles of the dynamics and prove that Hopf bifurcations oc-cur, yielding these cycles. Stable limit cycles correspond to sustained oscillations in strategy dominance across some or all of the population. The form of the dynamics considered, and the interpretation of the oscillations, depends on the applications of interest; the following are three motivating applications.
a) The replicator-mutator dynamics have been used in the development of a mathe-matical framework for the evolution of language . For a large population, the strategies represent different grammars in the population and mutations reflect er-rors in grammar transmission or learning from one generation to the next. A key ∗The discussion in this subsection appears verbatim in .
4 result is the bifurcation of the equilibria from a state where several grammars co-exist in a population to a state of high grammatical coherence as mutations in the population decrease (or equivalently, the fidelity of learning increases) [57, 75, 74].
Limit cycles of the replicator-mutator dynamics correspond to oscillations in the dominance of the different grammars in the population. As noted in , oscilla-tions appear to be more realistic than stable equilibria for the language dynamics with timescales on the order of several centuries.
b) The replicator-mutator dynamics were recently proposed [90, 44] as a model for behavior adoption in social networks, with a focus on the emergence of dominance of particular behaviors in these networks. Simulations of the evolutionary social network model show a transition from the dominance of a single strategy (behav-ior), to the coexistence of several strategies, to the eventual collapse of dominance, as the extent of mutation in the network increases. Limit cycles of the replicator-mutator dynamics correspond to oscillations of behavior preference in this context, for example cycles in trends or fashions.
c) The replicator-mutator dynamics can also be used to model decision-making dy-namics in networked multi-agent systems. It has been shown that simple mod-els with pairwise interactions between agents and noisy imitation of successful strategies reduce (under certain conditions) to the replicator-mutator dynamics [40, 7, 135, 134]. Recent papers have employed the replicator-mutator equations to model wireless multi-agent networks [145, 130]. Hopf bifurcations of replicator-mutator dynamics in this context address the exploration versus exploitation tradeoff: few mutations favor fast convergence to a decision (exploitation) whereas extensive mutations favor exploration of the decision space. In an intermediate range, mutations can lead to limit cycles, which enable dynamic examination of alternative choices.
1.1.2 Collective Migration Collective migration is a natural phenomenon common in a number of species in-cluding birds, fish, invertebrates and mammals [119, 42, 22, 6]. Animals migrate by leveraging a variety of environmental cues such as nutrient and thermal gradients, magnetic fields, odor cues, or visual markers [132, 30, 140, 133]. Measuring these stochastic environmental signals is complicated and requires the investment of time and energy, and the development of necessary physiological and sensory machinery.
5 Animals migrating collectively also have the ability to leverage social information from neighbors in the group . One way of doing so is by imitating invested neighbors (via consensus processes such as attraction and alignment of heading) and thereby effectively achieving good migratory performance, without paying the measurement and processing cost.
Using agent-based models of collective migration, Couzin et al. have shown that a small group of designated leaders is capable of guiding a larger of group of naive, socially interacting followers. This situation is similar, for example, to the way in which a small number of informed scout bees directs a large swarm of uninformed conspecifics to a new nest site [116, 120]. The ability of followers in such migratory swarms to leverage the investments made by leaders in the group, and to gain the benefits of successful migration without paying the associated costs, is perplexing from an evolutionary perspective (this question is related to the broader puzzle of the evolution of cooperation).
Our study of the evolutionary dynamics of collective migration is motivated in large part by a recent paper by Guttal and Couzin that addresses this evolutionary question. Simulations in show that the coexistence of leaders and followers in migratory populations is a stable emergent outcome of the evolutionary dynamics for a large region of the parameter space studied. The authors of also examine the role of anthropogenic influences on evolved population migration patterns by studying the impact of increasing habitat fragmentation on the collective dynamics.
High levels of habitat fragmentation make it increasingly difficult for individuals to measure external cues; migration is gradually lost because of the higher costs of reaching more distant destinations . Simulations in illustrate a hysteretic effect in restoring lost migration ability in the population - once migration ability is lost for a threshold level of fragmentation, much greater habitat recovery is necessary for the population to recover the ability to migrate.
We are also motivated by the paper by Torney et al. that analytically vali-dates the evolutionary branching simulated in by studying a mean-field model of migration dynamics. The mean-field model in effectively prescribes an all-to-all interconnection between the agents in the migratory system and serves as the starting point for our work, which focuses explicitly on the role of the structure of a limited interaction network on evolved outcomes. Indeed, the structure of the interaction network between agents in a collective has been shown to be critical to the perfor-mance of the collective, and to the emergent outcomes observed as a consequence of the local interactions.
6 1.1.3 Pursuit and Evasion† Pursuit and evasion behaviors play a critical role in predator foraging, prey survival, mating and territorial battles in several species. Species such as bats and dragonflies have evolved sophisticated dynamical strategies such as motion camouflage to disguise themselves as stationary during aerial pursuit [77, 27]. Studies on migratory canni-balistic locusts have revealed that pursuit and evasive behavior among conspecifics is integral to the formation of mass-moving migratory bands in dense swarms [35, 5].
Recent experimental work on the dynamics of coordinated predator pursuit and prey evasion among schooling fish has shown that collective behavior, among both preda-tors and prey, plays a vital role in predator hunting and prey evasion under condi-tions of considerable informational constraints (such as dynamic ocean environments) [37, 45].
The pervasiveness of pursuit and evasion in nature motivates the examination of winning strategies from an evolutionary perspective. Recently, Wei et al. used the evolutionary approach to study pursuit games, with dynamics derived in . The authors of [147, 50] use Monte-Carlo simulations and analytical calculations to study three pursuit strategies competing against a field of deterministic or random nonre-active evasive strategies (an evader with a nonreactive strategy has dynamics that are uncoupled from those of the pursuer). The three chosen pursuit strategies (classi-cal, constant bearing and motion camouflage) are biologically inspired. The authors show convergence of the evolutionary game dynamics between the three strategies to pure motion camouflage and motivate this result by empirical observations of motion camouflage in hoverflies, dragonflies and bats . We build on the work in by studying the coevolution of the three strategies of pursuit from playing against three distinct evasive strategies, two of which are reactive strategies (an evader with a reactive strategy has dynamics that are coupled to those of the pursuer).
1.1.4 Swarm Decision-Making Honeybee colonies reproduce by casting out swarms, each of which comprises a queen accompanied by several thousand worker bees. A small fraction of the worker bees are known as scout bees and perform the task of locating suitable nest sites for the swarm by engaging in a decentralized democratic decision-making process of choosing among several competing options. This process involves the famous waggle dance in which scout bees advertise the location and quality of a suitable nest site by †The discussion in this subsection is adapted from .
7 performing a distinctive dance on the surface of the swarm. The book by Seeley provides an engaging description of the waggle dance, as well as a detailed discussion of the organization and behavior of honeybee swarms.
In a recent paper, Seeley et al. have shown that scouts send inhibitory stop-signals to other scouts advertising alternative nest sites, thereby causing these scouts to cease dancing. This cross-inhibitory process has been shown to be critical to the ability of swarms to make decisions effectively, particularly when choosing between competing options of near-equal value. In this thesis we study bifurcations in a model of honeybee swarm decision-making and illustrate the critical role played by stop-signal inhibition in enabling swarms to manage the speed-accuracy tradeoff inherent to most decision-making problems. We show that an intermediate evolved level of stop-signaling is necessary for swarms to effectively make decisions when presented with both equal and unequal alternatives. Our analysis also shows that cross-inhibition is a potentially valuable mechanism for enabling effective collective decision-making in decentralized artificial swarms.
1.2 Contributions and Thesis Outline The chapters of this thesis are organized according to the four topics described in §1.1. Chapter 2 comprises background material and establishes the mathematical notation that will be used throughout the thesis. Background material is presented in four main areas: Evolutionary Dynamics, Dynamical Systems, Graph Theory and Stochastic Processes.
Chapters 3 and 4 focus on limit cycles and Hopf bifurcations of the replicator-mutator dynamics. The analysis in Chapter 3 is restricted to N = 3 strategies and the corresponding planar phase space. The restriction to N = 3 is convenient for visualization and allows us to motivate the general results for N ≥3 strategies to follow in Chapter 4. In Chapter 4 we prove conditions for the existence of stable limit cycles arising from multiple distinct Hopf bifurcations of the dynamics in the case of circulant fitness matrices. In the noncirculant case we illustrate how stable limit cycles of the dynamics are coupled to embedded directed cycles in the payoff graph. We study special conditions where multiple cycles in the payoffgraph yield multiple stable limit cycle attractors. The stability of the limit cycles is determined by an analytical calculation of the first Lyapunov coefficient of the dynamics.
In Chapter 5 we study the role of the social interconnection network on the evo-lutionary dynamics of collective migration. We design a networked migration model 8 and study evolution and adaptation as a function of network topology. Our model has two timescales: the fast timescale corresponds to fitness/utility calculations and the slow timescale corresponds to the evolution/adaptation of the network. We present a comprehensive analysis of the all-to-all limit of the model and prove conditions for population branching into leaders and followers. For networks with limited connec-tivity, we derive analytical tools for computing fitness on the fast timescale and show a minimum connectivity threshold necessary for branching. We also study a simple model of selfish local adaptation of nodes on a graph, and illustrate bifurcations in the dynamics as a function of increasing cost. We show the prominent role played by network topology in determining the location of leaders in the adaptive network.
Chapter 6 focuses on the coevolutionary dynamics of pursuit and evasion. We consider an evolutionary game between three strategies of pursuit (classical, con-stant bearing, motion camouflage) and three strategies of evasion (classical, random, optical-flow based). Pursuer and evader agents are modeled as self-propelled steered particles with constant speed and strategy-dependent heading control. We use Monte-Carlo simulations and theoretical analysis to show convergence of the evolutionary dynamics to a pure strategy Nash equilibrium of classical pursuit versus classical eva-sion. We extend our work to consider a novel pursuit and evasion based collective motion scheme, motivated by collective pursuit and evasion in bands of migrating cannibalistic locusts.
In Chapter 7 we study the collective decision-making dynamics of honeybee swarms.
The cross-inhibitory stop-signalling mechanism has been shown to be critical to the decision-making dynamics in swarms of house-hunting honeybees.
We study a model of stop-signal based collective decision-making and present a comprehensive picture of the dynamics and bifurcations of this model. We prove a separation of timescales in the decision-making process and show how swarms must evolve to an intermediate level of stop-signalling to address a fundamental speed-accuracy tradeoff.
We also present several stochastic simulations to help elucidate the decision-making process.
Chapter 8 presents our conclusions and topics for future work. We discuss how some of the analysis and conclusions of this thesis inspire algorithms for control and decision-making in decentralized collective artificial systems.
9 Chapter 2 Background In this chapter we describe some of the main mathematical tools that will be used in the rest of the thesis and we establish notation. Each of the areas discussed in the sections to follow is a significant domain of research by itself. Consequently, this chapter is not intended to be a comprehensive presentation of these areas, but rather an introduction to selected tools from these areas that will be useful going forward.
The cited references provide more detail.
In §2.1, we discuss models of evolutionary dynamics and make connections with game theory, genetic algorithms, and optimization. §2.2 focuses on bifurcations in continuous dynamical systems, including the Hopf bifurcation theorem, which features prominently in Chapters 3 and 4. §2.3 introduces some notation and results from graph theory and §2.4 introduces some basic results from stochastic dynamics; these are employed in Chapters 5 and 7.
Basic Notation: Matrices are denoted in capital letters and vectors are denoted in boldface lowercase letters. mij denotes the (i, j) element of matrix M (M = [mij] ∈ RM×N) and xi denotes the ith element of vector x x = [ x1 · · · xN ]T ∈RN . 1 and 0 denote the vectors of ones and zeros respectively. D = [dij] = diag(x) denotes a diagonal matrix with elements of vector x on the main diagonal, i.e. dii = xi and dij = 0 for all i ̸= j. The N × N identity matrix is given by IN = diag(1N).
2.1 Evolutionary Dynamics Evolutionary dynamics [142, 87, 41, 141] are, broadly speaking, an effort to cast the basic tenets of Darwinian natural selection (replication, competition, strategy dependent fitness, mutation) in a mathematical framework that can be simulated, 10 interpreted, and often rigorously analyzed. Our understanding of the evolutionary process has its roots in Darwin’s three simple postulates [142, 16]: 1. Like tends to beget like, and there is heritable variation in traits associated with each type of organism. (replication and mutation) 2. Among organisms there is a struggle for existence. (competition) 3. Heritable traits influence the struggle for existence. (strategy dependent fitness) These postulates inherently define a game-theoretic interaction between individuals in the population as a function of their strategies (expressed as phenotypes or traits), and their interactions with the environment and other individuals. These strategies and interactions map to payoffs, which in turn translate to reproductive fitness. The game theoretic mechanism implies that evolutionary solutions are not necessarily optimal in terms of maximizing the fitness of the population as a whole.
Nonetheless, the ability of evolutionary dynamics to shape natural systems to-wards a fascinating array of effective solutions has inspired powerful tools for opti-mization in engineering design, e.g. genetic algorithms . In genetic optimization algorithms, agents imitate the evolutionary process to search for local optima on a constant landscape as shown in Figure 2.1.
This is in contrast with biological (game-theoretic) evolutionary dynamics where the strategies interact and influence the landscape on which they are evolving (see Figure 2.1). In this thesis, we focus on studying the outcomes of models of game-theoretic evolutionary dynamics, while also making connections with how these solutions inspire the design of engineered collective systems.
The replicator dynamics are the simplest model of evolutionary dynamics for a large population comprised of N sub-populations, each subscribed to a different competing strategy. These differential equations model the game theoretic interac-tions among the sub-populations and determine how each sub-population changes in size as a consequence of these interactions. Let xi (i = 1, · · · , N) denote the fraction of individuals in the population subscribed to strategy i N P i=1 xi = 1 . The replicator dynamics are given by ˙ xi = xi(fi(x) −φ); φ = f Tx, (2.1) where fi(x) denotes the fitness function of individuals with strategy i and φ is the population average fitness. The average fitness φ is obtained by summing over the individual fitness terms fi(x) weighted by their appropriate fractions xi. For pairwise 11 Figure 2.1: (A) Optimization picture: evolutionary optimization dynamics on a constant fitness landscape; population evolves on the landscape towards the fitness peak. (B) Game-theoretic picture: evolutionary dynamics on an adaptive landscape; the population co-evolves with the dynamic landscape. (adapted from Fig. 1 in ) encounters in a large population, the fitness functions are linear [87, 135] and f = Ax, where A ∈RN×N is known as the payoffmatrix.
John Maynard Smith’s pioneering work made formal connections between classical game theory and evolutionary dynamics. Particularly important was May-nard Smith’s definition of evolutionarily stable strategies (ESSes), which are equi-libria of an evolutionary dynamical system that are uninvadable by other competing strategies in the environment, and hence stable in an evolutionary sense. From a game theoretic perspective, ESSes are a subset of the Nash equilibria of a game: they satisfy both the Nash best reply condition and evolutionary uninvadability. Lyapunov stable equilibria of the replicator dynamics (2.1) are Nash equilibria of the corresponding game . Further, all ESSes of the replicator dynamics are asymptotically stable .
Although the replicator dynamics have proved to be a powerful tool in analyzing a variety of classical games from an evolutionary perspective, they do not model mutation, a key ingredient of selection theory. Mutation can be included by adding the possibility that individuals spontaneously change from one strategy to another.
12 This yields the replicator-mutator dynamics [9, 92] given by ˙ xi = N X j=1 xjfj(x)qji −xiφ; φ = f Tx, (2.2) where qij denotes the probability of mutation from strategy i to strategy j N P j=1 qij = 1 !
.
The replicator-mutator dynamics have played a prominent role in evolutionary theory and contain as limiting cases many other important equations in biology ; these include models of language evolution , autocatalytic reaction networks , and population genetics . The dynamics have also recently been employed to model social and multi-agent network interactions . The replicator-mutator equations have been shown in to be equivalent to the generalized Price equation from evolutionary genetics [105, 106]. The standard replicator dynamics (2.1) can be obtained from the replicator-mutator dynamics (2.2) in the limit of zero mutation (qii = 1 for all i and qij = 0 for all i ̸= j).
Biological evolutionary models typically consider small mutation probabilities while larger mutation probabilities are more common in models of language dynamics and social interactions. Both sets of dynamics (2.1) and (2.2) evolve on the (N −1)-dimensional simplex phase space (see Figure 2.2) defined as ∆N−1 = x ∈RN | xi ≥0, xT1 = 1 .
In chapters 3 and 4 we focus on stable limit cycle attractors for the replicator-mutator dynamics (2.2).
Figure 2.2: Illustration of the simplices ∆1 and ∆2.
The replicator and replicator-mutator equations model the evolutionary dynamics for a discrete number of strategies. However, certain problems require the consider-ation of a continuous strategy space, and can be studied using tools from adaptive 13 dynamics [25, 26, 18]. Adaptive dynamics are particularly well-suited for studying the evolution of a one-dimensional trait in a population undergoing small mutations in strategy. Consider a continuous strategy space parameterized by k ∈[a, b] ⊂R.
The fitness of a resident population with strategy kR is given by FR(kR). The adap-tive dynamics approach considers the consequences of a small group of mutants with strategy kM invading the resident population. Let FM(kR, kM) denote the fitness of the mutants in the environment of the residents. The relative fitness of the mutants with respect to the residents is known as the differential fitness and is given by S(kR, kM) = FM(kR, kM) −FR(kR).
(2.3) The two-parameter function S allows us to predict which mutant strategies can invade a particular resident population. For example, for a given resident strategy kR, the values of kM that result in S > 0 correspond to the mutant strategies that when rare, can invade the established resident population. Further, a study of the selection landscape S can help us predict when we expect to see an evolutionarily stable monomorphic population (all individuals having the same strategy) and when we expect to see opportunities for branching (speciation into sets of individuals having different strategies) in evolutionary simulations.
The evolutionary dynamics of the resident strategy kR are given by dkR dt = α ∂S ∂kM kM=kR =: α g(kR), (2.4) where g(kR) is known as the selection gradient and α > 0 is related to the extent of mutation in the population. As the strategy kR of the population evolves according to (2.4), the small mutations kM about the resident strategy kR also evolve accord-ingly. The sign of the selection gradient g(kR) is indicative of the direction in which the strategy of the population will evolve. In particular, the strategy of the popula-tion will increase for g(kR) > 0 and decrease for g(kR) < 0. Evolutionary singular strategies k∗correspond to a vanishing selection gradient g(k∗) = 0. Singular strate-gies are evolutionary attractors for the population (also known as Convergent Stable Strategies (CSS)) if they satisfy the condition ∂g ∂kR kR=k∗ < 0.
(2.5) 14 CSS stable attractive strategies k∗can be either monomorphic evolutionary stable strategies for the population, or can be branching opportunities where the population speciates. The condition for a CSS singular strategy to be a branching point for the population is given by ∂2S ∂k2 M kM=kR=k∗ > 0.
(2.6) Figure 2.3: Pairwise invasibility plots. Dark regions correspond to differential fitness (2.3) S > 0 and white regions to S < 0. The resident population on the main diagonal moves in the direction of the white arrows shown, corresponding to the dynamics (2.4). The red vertical lines mark singular strategies. Singular strategies with arrows pointing towards them are convergence stable (CSS). Singular strategies with vertical lines passing through regions that are exclusively white are evolutionarily stable (ESS).
Figure 2.3 shown pairwise invasibility plots (PIP) corresponding to the CSS and branching cases described above. PIPs provide a geometric method for analyzing evolutionary dynamics. The population strategy at any instant is distributed about a point on the main diagonal of each plot, with resident strategy kR and small mu-tations kM about the resident strategy. The population distribution moves in the direction of the arrows shown on the plot, eventually arriving at a CSS attractor, or at the boundaries of the strategy space. At the CSS attractor, the population can remain monomorphic when attractor is at a fitness maximum, or can branch when the attractor is at a local fitness minimum. We employ PIPs to study collective migration in Chapter 5.
The replicator (2.1), replicator-mutator (2.2), and adaptive dynamics (2.3), (2.5) model the evolution of strategies in a single large population. However certain co-evolutionary processes, such as interactions between predators and prey for example, involve the coupled evolutionary dynamics between two distinct populations. The single-population models described here can be extended to study such coevolution-15 ary processes; we will consider an extension of the replicator dynamics to model the evolutionary dynamics of pursuit and evasion in Chapter 6.
2.2 Dynamical Systems Tools A dynamical system is a mathematical model describing the time evolution of a set of variables which collectively define the state of the system at any given point in time. There is a rich set of tools in the dynamical system literature (for example see [32, 59, 125]) for computing and understanding the solutions to these systems, including the study of bifurcations. A bifurcation is a qualitative change in the solu-tions of a dynamical system as a result of the smooth change in a system parameter (known as the bifurcation parameter). Bifurcations that are detected by studying small neighborhoods of equilibria and limit cycles of dynamical systems are known as local bifurcations. In Chapters 3, 4, 5 and 7, we study local bifurcations of specific continuous-time dynamical systems, and in this section we introduce the kinds of bifurcations that appear in these chapters.
Consider the continuous-time dynamical system with state x ∈RN and bifurcation parameter µ ∈R given by ˙ x = f(x, µ), (2.7) with the vector field f : RN × R 7→RN. The range of behavior for one-dimensional systems (x ∈R) is limited; solutions either approach stable equilibria or diverge to infinity. Nonetheless, the analysis of certain bifurcations in higher dimensional systems can be reduced to studying one-dimensional normal forms (nonlinear analog of matrix diagonalization or matrix Jordan normal form [32, 59]) by using the center manifold theorem (see for details). The four main bifurcations for one-dimensional systems are given below, and illustrated in Figure 2.4.
(a) Saddle-Node: Normal form ˙ x = µ + x2. Equilibria exist for µ < 0. As µ →0− an unstable saddle and a stable sink collide and disappear.
(b) Transcritical: Normal form ˙ x = µx−x2. A stable branch and an unstable branch of equilibria exist for all µ ̸= 0. The branches exchange stability at µ = 0.
(c) Supercritical Pitchfork: Normal form ˙ x = µx−x3. The x = 0 equilibrium is stable for µ < 0 and unstable for µ > 0. There are two additional stable equilibria for µ > 0, which appear at the pitchfork bifurcation point µ = 0.
16 (d) Subcritical Pitchfork: Normal form ˙ x = µx + x3. The x = 0 equilibrium is stable for µ < 0 and unstable for µ > 0. There are two additional unstable equilibria for µ < 0.
Figure 2.4: Illustration of the four canonical bifurcations in one-dimensional systems.
Solid curves are stable equilibria and dashed curves are unstable. (a) Saddle-Node (b) Tran-scritical (c) Supercritical pitchfork (d) Subcritical pitchfork.
In addition to stable and unstable equilibria, N(≥2)-dimensional systems (x ∈ RN) can also have periodic orbits, limit cycles, homo/hetero-clinic connections and chaotic attractors (chaotic attractors may exist only for N ≥3, see the Poincar´ e-Bendixson theorem in and Peixoto’s theorem in [32, 100]). Bendixson’s criterion provides sufficient conditions for the nonexistence of periodic orbits for planar systems (x ∈R2) .
Theorem 2.1. (Bendixson’s Criterion) An autonomous planar vector field (2.7) de-fined on a simply connected region D ⊂R2 has no periodic orbits lying entirely in D if ∂f1 ∂x1 + ∂f2 ∂x2 is not identically zero and does not change sign in D.
One-parameter bifurcations in dynamical systems with a one- or two-dimensional state space can be conveniently plotted in two and three dimensions respectively.
Constructing bifurcation plots is more challenging for higher dimensional (N ≥3) systems, for which we can either plot projections in two or three dimensions (as in Chapters 4 and 5), or construct reduced dimensional approximations of the dynamics (as in Chapter 7).
For high dimensional systems, our focus in Chapter 4 will be on the existence of stable limit cycles, arising out of Hopf bifurcations of the dynamics. The normal form of the Hopf bifurcation is given by ˙ x1 = µx1 −x2 + σx1(x2 1 + x2 2) ˙ x2 = x1 + µx2 + σx2(x2 1 + x2 2), (2.8) where σ = ±1 (σ = sign(ℓ1(0)) from Theorem 2.2 below). The Hopf bifurcation point is µ = 0. The origin is asymptotically stable for µ < 0 and unstable for µ > 0. The 17 bifurcation is supercritical for σ = −1 and subcritical for σ = +1. The supercritical bifurcation results in stable limit cycles of radius √µ for µ > 0 as shown in Figure 2.5. For general N ≥3 dimensional systems, the Hopf bifurcation theorem [32, 59] µ x1 x2 x1 x2 µ = −2 µ = 3 µ = −2 µ = 3 Figure 2.5: Bifurcation plot (left) for the supercritical Hopf bifurcation normal form (2.8).
Blue curves are stable equilibria, red curves are unstable equilibria and magenta curves are stable limit cycles. The two right plots are phase portraits for µ = −2 and µ = 3, corresponding to the grey slices on the bifurcation plot.
provides sufficient conditions for the existence of stable limit cycles arising out of supercritical Hopf bifurcations.
Theorem 2.2. (Hopf bifurcation) Suppose that the system ˙ x = f(x, µ), x ∈RN, µ ∈R, has an equilibrium (x0, µ0) and the following properties are satisfied: • (H1) The Jacobian Dxf|(x0,µ0) has a simple pair of pure imaginary eigenvalues λ(µ0) and λ(µ0) and no other eigenvalues with zero real parts, • (H2) d dµ(Re λ(µ)) (µ=µ0) ̸= 0.
Then the dynamics undergo a Hopf bifurcation at (x0, µ0) resulting in a one-parameter family of periodic solutions. The system is locally topologically equivalent to the nor-mal form (2.8) near the origin. The stability of the periodic solutions is given by the sign of the first Lyapunov coefficient of the dynamics σ = sign ℓ1|(x0,µ0) . If ℓ1 < 0 then these solutions are stable limit cycles and the Hopf bifurcation is supercritical, while if ℓ1 > 0 the periodic solutions are repelling.
18 A key challenge is determining the right sets of coordinate transformations to con-vert the center manifold dynamics to the normal form (2.8); details of the calculation of the Lyapunov coefficient ℓ1 are provided in Appendix A.
µ1 µ2 xeq Bistable Figure 2.6: Cusp catastrophe bifurcation plot. The surface corresponds to the equilibria of the cusp catastrophe normal form (2.9) plotted as a function of the bifurcation parameters µ1 and µ2. The planar projection on the µ1 −µ2 surface shows the bifurcation set of the system, including the region with bistability. Outside this region, the system has one stable equilibrium.
Our discussion on bifurcations so far has focused on bifurcations as a single pa-rameter µ is varied. There exist several classes of multi-parameter bifurcations, one of which is the two-parameter cusp catastrophe , the normal form of which is given by ˙ x = x3 −µ1x −µ2.
(2.9) Figure 2.6 shows the equilibria and bifurcation set of this system as a function of bi-furcation parameters µ1 and µ2. Saddle node bifurcations occur along the bifurcation set defined by µ2 = −2 3 √ 3µ3/2 1 except at the point µ1 = µ2 = 0, which is a pitchfork bifurcation point (pitchfork bifurcation moving along the line µ2 = 0). This results in a region of bistability (two stable equilibria), as shown in the plot. We will encounter systems with a cusp catastrophe in Chapter 3 (Figure 3.4) and in Chapter 7 (Figure 7.3).
19 2.3 Graph Theory Tools Graphs are a convenient tool for representing interactions between agents in a multi-agent network. For a graph with N nodes, the structure of the graph is encoded in an N × N non-negative adjacency matrix A = [aij]. Each aij > 0 (i ̸= j) is the weight of a directed edge from node i to node j. Each aii > 0 corresponds to the weight of a self cycle at node i. An edge exists from node i to node j if and only if aij > 0.
Let Ni denote the set of neighbors of node i, i.e., j ∈Ni if and only if aij > 0. For unweighted graphs, aij ∈{0, 1}. A graph is said to be undirected if its adjacency matrix is symmetric, i.e., aij = aji for all i, j. For a graph with adjacency matrix A, the Laplacian matrix of the graph is given by L = diag(A1)−A. Let G(L) denote the directed graph with Laplacian matrix given by L. Figure 2.7 illustrates the adjacency and Laplacian matrices for a particular graph.
Figure 2.7: Adjacency and Laplacian matrices for a specific directed graph with N = 6 nodes. Graph edge weights are chosen such that the Laplacian matrix is normalized to zeros and ones on the main diagonal.
In biological systems, individual agents such as birds in a flock, or fish in a school, are represented as nodes on a graph and interactions between agents are represented as directed edges with a prescribed (possibly state dependent) weight. For example, an edge from node i to node j with weight aij could represent that agent i senses agent j with strength aij. Similarly, in robotic networks, nodes can represent individual robots with edges corresponding to communication or sensing links between them.
The Laplacian matrix features prominently in the multi-agent systems literature to model the continuous-time consensus or agreement process [111, 91, 47]. Consider a state vector x ∈RN with dynamics given by ˙ x = −Lx.
(2.10) 20 Here xi corresponds to the state of each node of the network and the dynamics (2.10) correspond to agents on the graph updating their state to reach agreement with the mean of their neighbors (as defined at the beginning of this section). The dynamics (2.10) converge to the agreement subspace α1 for some scalar α if and only if the directed graph corresponding to L (G(L)) is connected. Connectivity of G(L) requires that there exists at least one node, labeled the root, such that a directed path exists from every other node of the network to the root [111, 64, 110]. For example, the graph in Figure 2.7 is connected with root node 3. In Chapter 5 we discuss graph connectivity and consensus in more detail in the context of collective migration.
Figure 2.8: Illustration of circulant graphs for N = 6 nodes where each node has one outgoing edge.
In Chapters 3 and 4 we focus on a specific class of graph topologies with circulant adjacency matrices. A circulant matrix is fully specified by its first row; the subse-quent rows are cyclic permutations of the first row to the right with offset given by the row index. Let circulant matrix C be given by C = Circulant(c1, c2, · · · , cN) = c1 c2 · · · · · · cN cN c1 c2 ...
.
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.
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...
...
...
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· · · cN c1 c2 c2 · · · · · · cN c1 .
(2.11) In Figure 2.8 we illustrate three circulant graph topologies. The symmetry of circulant graphs, and corresponding well-known properties such as an analytical characteriza-tion of eigenvalues and eigenvector of circulant adjacency and Laplacian matrices, make these topologies particularly well-suited for analysis .
Graphs can also be constructed by considering the spatial embedding of nodes and using spatial metrics to define the existence and weights of edges. This is particularly 21 useful when considering models of collective motion where agents interact with others within a specific distance from them (distance metric) [14, 15], or with a specific number of nearest neighbors (topological metric) [3, 10]. Graphs with the distance metric are undirected by definition if the specific distance is the same for every agent; see Figure 2.9 for an illustration.
Figure 2.9: Spatially embedded graphs with N = 20 nodes. The blue circles indicate node positions, both graphs have an identical set of nodes. For the graph on the left, each node is connected to its three nearest neighbors. For the graph on the right, each node is connected to all nodes within the distance indicated by the dashed line segment.
2.4 Stochastic Dynamics It is often the case that in modeling physical systems one chooses to ignore or ne-glect the role of noise and disturbances that drive these systems, either by a rigorous calculation, or informally. Indeed a large part of the field of control theory is de-voted to designing controllers for systems that minimize the impact of external noise and disturbances, thereby enabling such systems to track and converge efficiently.
Nonetheless, the lack of predictability is inherent to a large variety of natural phe-nomena. In this section, we will be concerned with noisy (stochastic) systems that have limited predictability: the mean motion of these stochastic systems is often easy to determine, but the fluctuations about this mean motion are not predictable. As we will see in Chapters 5 and 7, quantifying the magnitude of these fluctuations is important for understanding the dynamics of specific collective systems.
22 The simplest one-dimensional stochastic system can be represented using an Ito stochastic differential equation (SDE) dx = α(x, t)dt + β(x, t)dW.
(2.12) Here x denotes the stochastic state variable, α(x, t) and β(x, t) are the drift and noise intensity functions respectively, and dW is the standard Wiener increment.
The Wiener process dx = dW ⇐ ⇒x(t) = W(t) has mean x(0) and variance that grows linearly with time. There is a one-to-one correspondence between the SDE repre-sentation of the stochastic process (2.12) and the Fokker-Planck (FP) representation of the evolution of the probability density of the state x with time. Specifically, let p(x, t) represent the probability density function of the state x at time t with initial condition x(t = 0) = x0. Then p(x, t) evolves according to ∂p ∂t = −∂ ∂x [α(x, t)p(x, t)] + 1 2 ∂2 ∂x2 β(x, t)2p(x, t) , (2.13) with initial distribution p(x, 0) = δ(x −x0). There is a complete equivalence between the SDE (2.12) and the FP equation (2.13) with drift coefficient α(x, t) and diffusion coefficient β(x, t)2 .
(a) WP (b) OU (c) DDM t t t x x x Figure 2.10: Simulations of the stochastic dynamics (2.12) dx = (a + bx)dt + σdW with initial condition x0 = 2; each figure shows 50 sample trajectories with the ensemble mean plotted in the darker dashed curve. (a) Wiener Process (WP) with parameters a = b = 0, σ = 1. (b) Ornstein Uhlenbeck process (OU) with parameters a = −5, b = 25 and σ = 1.
(c) Drift Diffusion process (DDM) with parameters a = 0, b = 1, σ = 1. Note that only the stable OU process (b) has finite steady-state variance. We use the first order Euler-Mayurama method for all simulations of SDEs .
Let N(µ, σ2) denote a Gaussian distribution with mean µ and variance σ2. If α = 0 and β = σ, the system (2.12) reduces to the standard Wiener process (x(t) = σW(t) or equivalently p(x, t) = N(x0, σ2t)) illustrated in Figure 2.10(a). For a linear drift term 23 α(x, t) = ax+b and constant noise term β(x, t) = σ the equation (2.12) corresponds to the one-dimensional Ornstein-Uhlenbeck (OU) process. For a < 0 the OU process is stable with a stationary steady state solution with mean E[x]ss = −b/a and variance σ2 ss = σ2/(2a). Equivalently lim t→∞p(x, t) = N(−b a , σ2 2a). The stable OU is illustrated in Figure 2.10(b). The SDE with α = b and β = σ is the continuum limit of the random walk model and is also known as the drift diffusion model (DDM), widely studied as a model for optimal decision making in neural systems . This model does not converge to a steady state solution, but rather has solution mean and variance that grow linearly with time, i.e., p(x, t) = N(x0 + bt, σ2t). The DDM is illustrated in Figure 2.10(c).
For a multi-dimensional stochastic state vector x ∈RN, the multivariate stochas-tic dynamics are given by dx = α(x, t)dt + β(x, t)dW , (2.14) where α : RN × R 7→RN, β : RN × R 7→RN, and dW ∈RN is the multi-dimensional Wiener increment. Similar to (2.13), the mutivariate FP equation equivalent to (2.14) is given by ∂p ∂t = − X i ∂ ∂xi [αi(x, t)p(x, t)] + 1 2 X i X j ∂2 ∂xi∂xj n β(x, t)β(x, t)T ij p(x, t) o , (2.15) where p(x, t) is the multivariate probability density function for the state x(t) with initial condition x(t = 0) = x0.
Let N (µ, Σ) denote the multivariate Gaussian distribution with mean µ and covariance matrix Σ (symmetric, positive semi-definite). The multivariate Wiener process is given by α = 0 and β = B where B is a constant matrix; this process has solution p(x, t) = N (x0, BBT). α = Ax + b and β = B results in the multivariate OU process dx = (Ax + b)dt + BdW , (2.16) where A ∈RN×N and B ∈RN×N are constant square matrices and b ∈RN is a constant vector. If A is Hurwitz (all eigenvalues in the open left half of the complex plane), then the process (2.16) has a steady-state mean given by E[x]ss = −A−1b and state state covariance matrix Σss given by the solution to the system of linear equations (Lyapunov equation) AΣss + ΣssAT = −BBT.
(2.17) 24 We employ the system of equations (2.17) to study the evolutionary dynamics of a networked migration model in Chapter 5.
Setting A = −L, b = b1 and B = σI in (2.16) results in a coupled version of the DDM studied recently in . Here L is the positive semi-definite Laplacian ma-trix from (2.10) that encodes the networked coupling between different drift-diffusing decision-making units.
2.4.1 Random Points on a Simplex UNIFORM EXPONENTIAL (a) (b) Figure 2.11: Random distributions of points on ∆2.
The red points in R3 are drawn randomly from a uniform distribution in (a) and an exponential distribution in (b). The intersection of the line joining each red point and the origin, with the simplex ∆2, is marked in blue (this is the geometric illustration of the division by sum normalization).
For several of the simulations in the chapters to follow, we pick initial conditions that are uniformly randomly distributed on the simplex ∆N−1.
A seemingly rea-sonable method to do this would be to choose N independent uniformly randomly distributed values on the interval [0, 1] and divide each by the total sum of the values to get a simplex vector. This method, however, results in points that are clustered in the middle of the simplex as illustrated in Figure 2.11. The correct method to do this is to draw N values independently from an exponential distribution and then normalize. The resultant vectors are then uniformly distributed on the simplex, as illustrated geometrically in Figure 2.11. We refer the reader to the Dirichlet distri-bution described in for more details.
25 Chapter 3 Replicator-Mutator Dynamics in the Plane As discussed in Chapters 1 and 2, the replicator-mutator dynamics define a canonical model from evolutionary theory and have been recently applied to model the evolution of language and the decision-making dynamics of social networks. In this chapter, we study a form of the replicator-mutator dynamics and prove necessary and sufficient conditions for the existence of stable limit cycles for N = 3 competing strategies; we generalize these results to N ≥3 strategies in Chapter 4. Stable limit cycles correspond to sustained oscillations in strategy dominance across some or all of the population.
The form of the dynamics considered, and the interpretation of the oscillations, depends on the applications of interest; three motivating applications are discussed in §1.1. The results presented in this chapter have been published in .
The analysis of the replicator-mutator dynamics in the literature has focused primarily on stable limiting equilibrium behavior where the fitness terms are assumed to have a lot of symmetry (e.g. [75, 85, 90]). However, the recent N = 2 analysis in (discussed in §3.3) shows that the symmetric case is structurally unstable and that breaking symmetry in fitness yields qualitatively different bifurcations of the dynamics. Further, in the authors illustrate that the replicator-mutator dynamics exhibit limit cycles and chaos for specific model parameter values. In this chapter, we show that the limiting behaviors of the replicator-mutator dynamics are tied to the structure of the fitness model, and we prove how breaking symmetry yields some of the richer outcomes simulated in . It is known that the replicator dynamics for N ≥4 can generate limit cycles and chaos for particular choices of fitness [76, 124, 86].
Here we investigate the role that both fitness and mutation play in generating limit cycles for the replicator-mutator dynamics.
26 With mutation strength as the bifurcation parameter, we prove that Hopf bifurca-tions can occur for the replicator-mutator dynamics with N = 3 and characterize the existence of stable limit cycles using an analytical derivation of the Hopf bifurcation points and the corresponding first Lyapunov coefficients [32, 59]. It is important to emphasize that the limit cycles discussed in this chapter are fundamentally different from the rock-paper-scissors game dynamics cycles of the replicator equations (e.g.
[121, 53, 41]). This is because the limit cycles of the replicator-mutator dynamics here are driven in part by mutation, and exist as a consequence of bifurcations that occur as mutation strength changes. Indeed, in the absence of mutation, the corresponding replicator game dynamics would yield stable equilibria at the simplex boundaries for the models studied here.
3.1 Model Description∗ Consider a large population of agents and N distinct strategies Si, i = 1, 2, · · · , N.
Let xi ∈[0, 1] be the fraction of individuals in the population with strategy Si such that N X i=1 xi = 1. Let the population distribution vector x = [x1 , · · · , xN]T. The fitness fi of agents with strategy Si is given by fi = N X k=1 bikxk.
Let f = [f1 , · · · , fN]T. Then f = Bx, where B = [bij] ∈RN×N, and the average population fitness is φ = f Tx = xTBx. B is known as the payoffmatrix where bij ≥0 represents the payoffto an agent with strategy Si on interacting with an agent with strategy Sj.
We assume that payoffs are all non-negative and that agents get a maximal payoff (normalized to 1) on interacting with others subscribed to the same strategy. Hence B satisfies bii = 1 and bij ∈[0, 1) for i ̸= j.
(3.1) As noted in , the payoffmatrix B can be interpreted from a graph theoretic perspective as the adjacency matrix of a directed graph. The nodes of the graph correspond to the strategies Si. The diagonal elements of B (bii = 1) correspond to self-cycles at each node. Each of the non-zero off-diagonal elements bij corresponds to ∗This sections is presented verbatim as in .
27 the weight of a directed edge from node Si to node Sj. Symmetric payoffmatrices B correspond to undirected graphs. This graph theoretic viewpoint is important in our work, particularly as a tool to visualize the structure of the payoffmatrix. Consider the following condition: Condition 3.1. Every row and column of B has at least one non-zero offdiagonal element.
From the graph theoretic perspective, Condition 1 requires that every node of the graph has at least one outgoing and one incoming link; this ensures that there are no isolated disconnected nodes of the payoffgraph. We will restrict to examining graphs that satisfy Condition 1.
Next we define qij to be the probability that agents with strategy Si mutate (spontaneously change) to strategy Sj.
Note that since X j qij = 1, the mutation matrix Q = [qij] is row stochastic. The elements of the mutation matrix Q are defined in terms of a mutation parameter µ ∈[0, 1]. The mutation parameter represents the probability of error in replication. For example, µ = 0 denotes perfect replication and no mutation whereas µ = 1 denotes pure mutation.
In this chapter, we use two specific models for the mutation matrix Q. The first model defines the mutation probabilities qij as a function of the payoffs bij and the mutation strength µ as follows: qii = (1 −µ), qij = µbij P i̸=j bij for i ̸= j.
(Q1) The form of qij in (Q1) is motivated by the graph theoretic perspective on the replicator-mutator dynamics and is a generalization of the structured mutational models in [60, 56]. Intuitively, this model implies that spontaneous mutation to al-ternative strategies is weighted in favor of strategies that yield higher payoff. The mutation models in [90, 60, 75, 85] are special cases of (Q1) in which the payoffmatrix B is symmetric. We call (Q1) the dependent mutation model since (Q1) is dependent on B.
The second mutation model we consider corresponds to a uniform random prob-ability of mutating to alternative strategies as follows: qii = (1 −µ), qij = µ N −1 for i ̸= j.
(Q2) We call (Q2) the independent mutation model since (Q2) is independent of B.
28 There are several alternative possibilities for the mutation matrix Q. Our choice of (Q1) and (Q2) enables a comparison between independent and dependent mutation models, and represents two generic models that are popular in the literature and meaningful in the context of our motivating applications.
The strategies Si, payoffs bij and mutation probabilities qij can be interpreted in each of our motivating contexts from §1.1: a) For the evolution of language, each Si is a specific grammar in the population and bij is the probability that a sentence spoken at random by individuals with grammar Sj can be parsed by individuals with grammar Si. Higher values of the diagonal terms qii = 1 −µ of the mutation matrix Q correspond to more effective language transmission or learning, and the off-diagonal terms qij correspond to mutation probabilities to alternative grammars.
b) In social networks, each Si represents a particular behavior in a population and bij represents the degree to which agents with behavior Si are attracted to behavior Sj. Higher values of the mutation probabilities qij correspond to a greater tendency for individuals to explore and adopt alternative behaviors in the population.
c) In multi-agent decision-making, each Si represents an alternative choice for the group and the bij represent the perceived relative advantage of choice Sj for agents currently subscribed to choice Si. The mutation terms qij model errors in the decision-making process, or agent random exploratory behavior.
The replicator-mutator dynamics describe the dynamics of the population distri-bution x as a result of replication driven by fitness f and mutation driven by Q: ˙ xi = N X j=1 xjfj(x)qji −xiφ =: gi(x); φ = f Tx.
(3.2) The replicator-mutator dynamics (3.2) can be derived as the limit of a simple stochastic error-prone imitation process, where agents imitate successful strategies proportional to relative payoffs (fi/φ) and mutate to alternative strategies with prob-abilities qij; see [40, 7, 135] for details. As illustrated in [135, 7], there exist several possible microscopic imitation mechanisms that yield alternatives to the replicator-mutator dynamics in the limit. For this chapter, we focus on the replicator-mutator dynamics as these are popular in the literature and hence allow for comparisons with past work [75, 90, 60, 76].
29 The dynamics (3.2) evolve on the (N −1)-dimensional simplex phase space as follows. Define the n-simplex as ∆n = x ∈Rn+1 | xi ≥0, xT1 = 1 , where 1 is a column vector of ones of appropriate dimension. Let g(x) : RN →RN be g(x) = [g1(x), · · · , gN(x)]T where gi(x) is defined in (3.2). One can compute directly from (3.2) that xT1 = 1 = ⇒1Tg(x) = 0. Hence xT1 = 1 is an invariant hyperplane for the dynamics. Further, the non-negative orthant of RN is a trapping region for the dynamics; this follows from the fact that ˙ xi|xi=0 ≥0. The intersection of the invariant hyperplane and the non-negative orthant of RN is the simplex ∆N−1.
Hence ∆N−1 is a trapping region for the replicator-mutator dynamics (3.2) (see Figure 2.2 for an illustration of ∆1 and ∆2).
Given the restriction to the simplex ∆N−1, the N-dimensional dynamics (3.2) can be reduced to an (N −1)-dimensional system of equations: ˙ xi = hi(˜ x), i ∈{1, 2, · · · , N −1}, hi(˜ x) := gi x1, x2, · · · , xN−1, 1 − N−1 X j=1 xj !
, (3.3) where ˜ x = [x1, · · · , xN−1]T and h : RN−1 →RN−1.
For µ = 0, the replicator-mutator dynamics reduce to the replicator dynamics (2.1) with payoffmatrix B (3.1). Since B is diagonally dominant, the corresponding replicator dynamics are fairly constrained .
In particular, the vertices of the simplex (pure strategies xi = 1, xj̸=i = 0) are asymptotically stable, and are also the only evolutionarily stable states.
3.2 Motivation for Cycles† Our motivation to prove the existence of limit cycles in replicator-mutator dynamics comes in part from simulations of the dynamics (3.2) for random payoffmatrices B (bij chosen from the uniform distribution on the interval [0, 1) for i ̸= j), which frequently exhibit oscillations. Figure 3.1 shows one simulation of the dynamics that is typical for mutation matrix (Q1) or (Q2). The dynamics in this simulation illustrate the transition from dominance of a single strategy (Figure 3.1(a)), to the coexistence †This section is presented verbatim as in , except for the last paragraph and Figure 3.3.
30 of several strategies (Figure 3.1(b)), to eventually the collapse of dominance (Figure 3.1(c)), as the extent of mutation (parameterized by µ) increases. The study of this shift in dominance, as a consequence of bifurcations in the dynamics, has received significant attention in the literature (e.g. [75, 85, 57, 60]).
(a) µ = 0.01 (b) µ = 0.15 (c) µ = 0.4 Figure 3.1: A first typical simulation of the dynamics (3.2) for N = 20 nodes and bij ∈ [0, 1) chosen randomly from the uniform distribution. The dynamics transition from (a) a highly coherent state for small µ, to (b) coexistence for intermediate µ, and eventually to (c) a mixed collapse of dominance for large µ.
Figure 3.2 shows another simulation of the replicator-mutator dynamics that is also typical for mutation matrix (Q1) or (Q2).
The dynamics in this simulation also transition from dominance of a single strategy (Figure 3.2(a)) to collapse of dominance (Figure 3.2(c)). However, unlike the first simulation, the dynamics ex-hibit sustained oscillations in strategy dominance at intermediate values of mutation strength µ (Figure 3.2(b)).
(a) µ = 0.01 (b) µ = 0.15 (c) µ = 0.4 Figure 3.2: A second typical simulation of the dynamics (3.2) for N = 20 nodes and bij ∈[0, 1) chosen randomly from the uniform distribution. The dynamics transition from (a) a highly coherent state for small µ, to (b) oscillating dominance for intermediate µ, and eventually to (c) a mixed collapse of dominance for large µ.
We are also motivated in part by simulations of the replicator-mutator dynamics in that illustrate stable limit cycles for specific asymmetric payoffmatrices (Figure 31 3.3). The authors in study language dynamics and are particularly motivated by the observation that oscillations appear more realistic than stable equilibria for the language dynamics with timescales on the order of several centuries.
Figure 3.3: Illustration of a limit cycle (left simplex) for the dynamics (3.2) with payoff and mutation matrix as shown (from Figure 1 in ).
3.3 Bifurcations with N = 2 Strategies‡ Before we move on to study the N = 3 (planar) case, we summarize the results from for dynamics (3.2) with N = 2 strategies. To simplify notation, define b12 := b1 and b21 := b2. With this notation and following the reduction (3.3), we have the one-dimensional system ˙ x1 = h1(x1) = x1f1q11 + x2f2q21 −x2 1f1 −x1x2f2 = x1 [b1 + x1(1 −b1)] (1 −µ −x1) + (1 −x1) [1 + x1(b2 −1)] (µ −x1).
(3.4) ‡This sections is presented verbatim as in .
32 Figure 3.4 shows the equilibria of the dynamics (3.4), and their stability, as a func-tion of the bifurcation parameter µ. The bifurcation plot shows a transition from bistability to a mixed equilibrium via a pitchfork bifurcation in the case b1 = b2. The pitchfork bifurcation is structurally unstable; for b1 ̸= b2 a saddle-node bifurcation occurs at a critical value µc as shown in Figure 3.4. Three branches of equilibria exist.
One of the branches remains stable for all µ and approaches x1 = 0.5 as µ approaches 1. The other two branches exist for µ < µc and collide in a saddle-node bifurcation at µc. Note that b1 ̸= b2 corresponds to a directed payoffgraph between the two nodes.
Figure 3.4: Three bifurcation plots for N = 2 nodes with parameters b1 = 0.2 and b2 = 0.2, 0.25 and 0.5. Blue curves are the stable equilibria and the red curves are the unstable equilibria. Similar to Figure 14 of .
A key observation from Figure 3.4 is that directed payoffgraphs yield qualitatively different bifurcations when compared to undirected (symmetric) payoffs; this feature persists in higher dimensions. The equilibria and bifurcations plotted in Figure 3.4 correspond to vertical slices through a cusp catastrophe, the normal form of which is given in §2.2 and plotted in Figure 2.6.
3.4 Planar Analysis§ To build intuition for our general results in Chapter 4, we focus here on the bifurca-tions of the dynamics (3.2) as a function of the bifurcation parameter µ for N = 3 §This sections is presented verbatim as in except for the discussion of the topology cases, which is taken from , and Figures 3.7 and 3.9, which are new.
33 strategies. Because the simplex is two-dimensional for N = 3 (3.3), it is easier than in higher dimensions to prove necessary and sufficient conditions for limit cycles and to visualize codimension-one bifurcations in three dimensions. We will show a transition from multiple stable dominant equilibria to a unique stable mixed equilibrium for in-creasing µ, and prove conditions for stable limit cycles to exist over an intermediate range of µ.
Consider the dynamics (3.2) with N = 3 and with the payoffparameters bij in (3.1) set to be either 0 or equal to a constant value b ∈(0, 1). There are five non-isomorphic graph topologies with three nodes that satisfy the connectivity specified by Condition 1 and have edges of identical weight b; these are shown in Figure 3.5.
Figure 3.5 also shows the analytically computed¶ bifurcation plots for each of the topologies as a function of the mutation strength µ for mutation matrix (Q1) and payoffmatrix (3.1). The corresponding plots for mutation matrix (Q2) are shown in Figure 3.6.
Figure 3.5: Bifurcation plots for the N = 3 case of dynamics (3.2), constant edge weights b = 0.2 and mutation matrix (Q1). The x-axis in each plot is the mutation strength µ, blue and red curves are stable and unstable equilibria, respectively, and the magenta curves are stable limit cycles. The three-node graphs in each subplot have adjacency matrix B with self-cycles (not shown) at each node. Panels (a) and (e) with circulant payoffmatrices each have an unstable central equilibrium x1 = x2 = x3 = 1 3 which stabilizes for large enough µ (see Lemma 4.1).
¶Equilibria and nullclines are solved using the MATLAB symbolic toolbox.
34 Figure 3.6: Bifurcation plots for the N = 3 case of dynamics (3.2), constant edge weights b = 0.2 and mutation matrix (Q2). The x-axis in each plot is the mutation strength µ, blue and red curves are stable and unstable equilibria, respectively, and the magenta curves are stable limit cycles. The three-node graphs in each subplot have adjacency matrix B with self-cycles (not shown) at each node.
Note that for µ = 0 the only stable equilibria for the replicator-mutator dynamics with payoffs (3.1) are the three pure strategy dominant equilibria at the corners of the triangle simplex. In all the subplots in Figures 3.5 and 3.6, bifurcations yield a unique mixed strategy interior equilibrium for increasing µ. The transition from the dominant equilibria to the mixed equilibrium for increasing µ depends strongly on the topology of the payoffgraph B under consideration; there are three distinct cases: 1. All-to-all Interconnection: The replicator-mutator dynamics with all-to-all in-terconnection and identical weights is studied in detail in . For N = 3 the payoffand mutation matrices are B = 1 b b b 1 b b b 1 and Q1 = Q2 = 1 −µ µ 2 µ 2 µ 2 1 −µ µ 2 µ 2 µ 2 1 −µ .
The bifurcation plot Figure 3.5(a) (or identically 3.6(a)) has two bifurcation points µCA1 = 2(1 −b) 3(2 + b) and µCA2 = 6 + 2b 1 −b − s6 + 2b 1 −b 2 −4.
(3.5) At µ = µCA1 the equilibrium xmix,3 = 1 313 changes stability via an S3-symmetric transcritical bifurcation . At µ = µCA2 six equilibria disappear via three symmetric saddle-node bifurcations.
Thus for µ > µCA2 the only remaining equilibrium is the stable xmix,3. Figure 3.7 shows phase portraits of the dynamics with the all-to-all payofftopology, for various choices of µ.
35 (a) µ = 0 (b) µ = 0.1 < µC1 (c) µ = 0.25 ∈(µC1, µC2) (d) µ = 0.3 > µC2 Figure 3.7: Phase portraits for dynamics (3.2) and with all-to-all payoffas in Figure 3.5(a) or Figure 3.6(a). The figure on the left of each of the four sub-figures shows nullclines (red ˙ x1 = 0, green ˙ x2 = 0 and magenta ˙ x3 = 0), vector field (grey arrows) and fixed points (filled circles are stable, unfilled circles are unstable). The figure on the right of each of the four sub-figures shows sample trajectories for randomly chosen initial conditions. The color scale indicates the magnitude of the flow (vector field) with hot colors corresponding to fast flow. b = 0.2 for this set of plots which gives µCA1 = 0.2424 and µCA2 = 0.2540 from (3.5).
2. Limited Interconnections: The bifurcation plots for graphs in Figures 3.5(b)-(d) and Figures 3.6(b)-(d) each have a stable branch of equilibria for all µ. They also have two other stable and four unstable equilibria at µ = 0 which disappear in saddle-node bifurcations as µ increases. Perturbations of the structurally un-stable symmetric all-to-all case yield bifurcations that are qualitatively similar to the limited interconnection cases, much like the N = 2 bifurcations in Figure 3.4.
3. Directed Cycle Interconnection: The bifurcation plots in Figures 3.5(e) and 3.6(e) correspond to a directed cycle interconnection between nodes with payoff and mutation matrices B = 1 b 0 0 1 b b 0 1 , Q1 = 1 −µ µ 0 0 1 −µ µ µ 0 1 −µ and Q2 = 1 −µ µ 2 µ 2 µ 2 1 −µ µ 2 µ 2 µ 2 1 −µ .
36 The equilibrium xmix,3 exists for all values of µ ∈[0, 1].
Three symmetric saddle-node bifurcations occur at µ = µC1 as shown in Figure 3.5(e); because of symmetry, the unstable manifold of each saddle-node lies in the stable manifold of the next, forming a degenerate heteroclinic cycle with three non-hyperbolic saddle-nodes. This gives rise to stable limit cycles as µ increases further. A Hopf bifurcation at µ = µC2 where xmix,3 changes stability from an unstable to a stable focus and is surrounded by stable limit cycles for µ < µC2. Figure 3.8 shows phase portraits of the dynamics with the directed cycle payoff, for various choices of µ.
(a) µ = 0 (b) µ = 0.1 < µC1 (c) µ = 0.25 ∈(µC1, µC2) (d) µ = 0.35 > µC2 Figure 3.8: Phase portraits for dynamics (3.2) and with directed cycle topology as in Figure 3.5(e) and mutation matrix (Q1). The figure on the left of each of the four sub-figures shows nullclines (red ˙ x1 = 0, green ˙ x2 = 0 and magenta ˙ x3 = 0), vector field (grey arrows) and equilibria (filled circles are stable, unfilled circles are unstable).
The figure on the right of each of the four sub-figures shows sample trajectories for randomly chosen initial conditions. The color scale indicates the magnitude of the flow (vector field) with hot colors corresponding to fast flow. b = 0.2 for this set of plots which gives µC1 = 0.2136 and µC2 = 0.3.
Stable limit cycles of the dynamics exist in a wide region of parameter space for circulant payoffmatrices B, for which the directed cycle topology in Case 3 above is a special case. Here we state necessary and sufficient conditions for the existence of 37 stable limit cycles for (3.2) with circulant payoffmatrix given by B := BC,3 = Circulant(1, α, β) = 1 α β β 1 α α β 1 , {α, β} ∈[0, 1) and α + β > 0.
Our choice of circulant payoffmatrix BC,3 is in part for analytical tractability; we discuss limit cycles for non-circulant payoffs in Chapter 4. Lemma 3.1 provides nec-essary conditions for the existence of limit cycles for (3.2) with N = 3 and payoff matrix BC,3. Corollary 3.1 provides sufficient conditions for Hopf bifurcations and stable limit cycles and is a special case of the more general result in Theorem 4.1 to follow in Chapter 4.
Lemma 3.1. The dynamics (3.2) with payoffmatrix BC,3 have no closed orbits in the simplex ∆2 for µ > (2 −α −β)(α + β) 6(α + β + αβ) =: µ01 for mutation (Q1), and µ > 2(2 −α −β) 3(4 + α + β) =: µ02 for mutation (Q2).
Proof. The simplex ∆2 (a simply connected subset of R2) is a trapping region for the dynamics (3.2) (see §3.1). The divergence of the vector field on ∆2 is negative semi-definite for µ > µ0i, i ∈{1, 2} (see Lemmas B.1 and B.2 in Appendix B). Therefore Bendixson’s Criterion (Theorem 2.1) implies that no closed orbits can lie in ∆2 for µ > µ0i.
Corollary 3.1. Equilibrium xmix,3 of the dynamics (3.2) with N = 3 strategies, payoffmatrix BC,3, mutation matrix (Qi) (i = 1, 2) and bifurcation parameter µ, undergoes a supercritical Hopf bifurcation at µ = µ0i leading to stable limit cycles for µ < µ0i if α ̸= β and additionally if 2α +2β +5αβ +α2 +β2 ̸= 2 for mutation matrix (Q1).
Proof. The proof relies on satisfying the conditions of the Hopf Bifurcation Theorem 2.2.
This is shown for N ≥3 in Theorem 4.1.
For N = 3, the first Lyapunov coefficient is given by ℓ1(α, β) = 3(α+β−2) ω0i , where, ω0i = |˜ ωi|, ˜ ωi = ( (α−β)(α2+β2+2α+2β+5αβ−2) 6 √ 3(α+β+αβ) i = 1 (α−β)(1+α+β) √ 3(4+α+β) i = 2.
38 This follows from the calculation of ℓ1 in Lemma 4.4. Supercriticality follows from ω0i ̸= 0 = ⇒ℓ1 < 0. See Figure 3.9 for a plot of ℓ1(α, β).
α α β β ℓ1 ℓ1 0 0 0 0 0 0 1 1 0.5 0.5 0.5 0.5 1 1 −50 −100 −50 −100 (Q1) (Q2) Figure 3.9: Plot of ℓ1(α, β) from Corollary 3.1.
Figures 3.5(e) and 3.6(e) show limit cycles for the specific case of BC,3 with α = b and β = 0.
Figure 3.10 shows three more limit cycle bifurcation plots for non-zero α and β and mutation matrix (Q1).
Interestingly, for the parameter values selected in Figure 3.10(b) stable limit cycles coexist with multiple stable equilibria.
This coexistence of stable equilibria and stable limit cycles implies that different initial conditions can yield qualitatively distinct limiting behavior even with fixed parameters for the dynamics (i.e., without bifurcations).
Figure 3.10: Bifurcation plots for the dynamics (3.2), payoffmatrix BC,3 and parameters α and β as shown. The existence of Hopf bifurcations and stable limit cycles for the set of parameter choices follows from Corollary 3.1. Note the coexistence of stable equilibria with stable limit cycles in panel (b).
While the focus of this chapter and the one to follow is on (local) Hopf bifurca-tions of the dynamics, we note that the dynamics also have global bifurcations. For example, for µ increasing from zero in the bifurcation plot in Figure 3.10(b), a hetero-clinic cycle containing three hyperbolic saddle points exists at µ = 0.1. A heteroclinic 39 bifurcation yields stable limit cycles as µ increases further. Phase portraits illustrat-ing the heteroclinic cycle and one of the stable limit cycles are shown in Figure 3.11.
In contrast, for the bifurcation plot in Figure 3.5(e), the directed cycle at µ = µC1 occurs because the unstable manifold of each of the non-hyperbolic saddle-nodes lies in the stable manifold of the next.
Figure 3.11: Phase portraits for dynamics (3.2) and parameters α = 0.95 and β = 0.05 as in Figure 3.10(b) and mutation matrix (Q1). The top row of plots shows sample trajectories for randomly chosen initial conditions. The color scale indicates the magnitude of the flow (vector field) with hot colors corresponding to fast flow. The bottom row of plots shows nullclines (red ˙ x1 = 0, green ˙ x2 = 0 and magenta ˙ x3 = 0), vector field (grey arrows) and equilibria (filled circles are stable, unfilled circles are unstable). For µ = 0.1 (center pair), a heteroclinic connection exists between the three hyperbolic saddles. For µ = 0.14 ∈(0.1, 0.19) (right pair), stable limit cycles coexist with three stable sinks.
In the next chapter we will consider bifurcations of the replicator-mutator dynam-ics in the general N ≥3 case and prove conditions for the existence of stable limit cycles. We will also consider perturbations of the circulant fitness cases studied in this chapter, along with more general payoffgraph topologies.
40 Chapter 4 Replicator-Mutator Dynamics in Higher Dimensions ∗ In Chapter 3 we looked at Hopf bifurcations for the replicator-mutator dynamics (3.2) with N = 3 strategies and circulant payoffmatrix BC,3. While the focus on three strategies was convenient for visualization, the simulations in Figure 3.2 indicate that the dynamics have stable limit cycles in higher dimensions as well (N ≥4). In this chapter, we show that this is indeed the case by proving two main results. Theorem 4.1 shows that the dynamics undergo multiple Hopf bifurcations at distinct bifurcation points for N ≥3, and Lemma 4.4 provides analytical conditions for the stability of the limit cycles arising from these Hopf bifurcations.
We focus on the dynamics with circulant payoffmatrix BC,N ∈RN×N, N ≥3 given by BC,N := Circulant(1, α, 0, · · · , 0, β), {α, β} ∈[0, 1) and α + β > 0, (4.1) and mutation matrices (Q1) and (Q2).
The directed graph corresponding to the payoffmatrix BC,N is illustrated in Figure 4.1.
We have chosen to study the circulant two-parameter payoffstructure BC,N both for purposes of tractability and to gain important insights regarding Hopf bifurcations of the dynamics in N dimensions. In particular, we show the existence of multiple Hopf bifurcations as several distinct pairs of eigenvalues cross the imaginary axis with increasing mutation parameter µ in §4.1. Further, the criticality of the bifurcations, and correspondingly the existence of stable limit cycles, depends on the choice of parameters α and β, unlike the N = 3 case where all existing bifurcations are super-∗Sections 4.1–4.5 are presented verbatim as in .
41 1 2 3 4 N N−1 BC,N = 1 α 0 · · · 0 β β 1 α 0 0 β 1 α .
.
.
.
.
.
...
...
...
0 0 β 1 α α 0 · · · 0 β 1 α β . . . Figure 4.1: Graph topology corresponding to the payoffmatrix BC,N from (4.1).
critical; this is shown in §4.2. The illustrations in §4.2 and §4.3 show that the choice of the mutation matrix (Q1) or (Q2) plays an important role in determining the exis-tence and criticality of the Hopf bifurcations. In §4.4 we extend the results from the previous sections to study multi-cycle dynamics for one-paramter graphs and show the existence of multiple stable limit cycle attractors. §4.5 focuses on generalizations to non-circulant and random payoffgraphs and shows the tight connection between the existence of embedded cycles in these graphs and corresponding limit cycles of the dynamics.
4.1 Hopf Bifurcation Calculation For payoffmatrix BC,N, the equilibrium xmix,N = 1 N 1N = h 1 N · · · 1 N iT ∈RN under-goes Hopf bifurcations. Lemma 4.1 shows that xmix,N is always an equilibrium of (3.2) for circulant B.
Lemma 4.1. If the payoffmatrix B is circulant, then xmix,N is an equilibrium of the replicator-mutator dynamics (3.2) with mutation matrix (Q1) or (Q2).
Proof. Suppose B is circulant.
Then 1 is an eigenvector of B with eigenvalue rB = N X j=1 b1j, i.e., B1 = rB1.
Matrix Q is also circulant by construction from (Q1) and (Q2).
This means that N X j=1 qji = N X j=1 qij = 1.
Let x = xmix,N.
Then f = Bxmix,N = 1 N B1 = rB N 1. From (3.2), ˙ xi xmix,N = 1 N N X j=1 fj qji −1 N = rB N 2 N X j=1 qji −rB N 2 = 0, and xmix,N is an equilibrium.
42 To proceed with the analysis, we start by calculating Dxg|xmix,N , the Jacobian matrix of the dynamics evaluated at the equilibrium point xmix,N. We then prove conditions for the existence of ⌊N−1 2 ⌋pairs of complex conjugate eigenvalues of the Jacobian.
We prove that each of these pairs of complex eigenvalues has distinct real part and hence each pair crosses the imaginary axis at different values of the bifurcation parameter µ. We show that each such crossing satisfies the conditions of the Hopf Bifurcation Theorem 2.2.
The (i, j) entry of the Jacobian Dxg|xmix,N , denoted h Dxg|xmix,N i ij, is given by 1 N (2 + α + β) qji + αqj−1,i + βqj+1,i − 2 N + δij (1 + α + β) , (4.2) where δij is the Kronecker delta and the indices i, j are denoted modulo N, i.e.
1 ≡N + 1, 0 ≡N, etc. For circulant fitness B, the Jacobian Dxg|xmix,N is also circulant.
Let ωN = cos 2π N + i sin 2π N be a complex, primitive N th root of unity. Let ωN,k = ωk N = cos 2π N k + i sin 2π N k for any integer k. For a circulant matrix M = [mij] ∈RN×N, let λk(M) = N X j=1 m1j ω j−1 N,k .
(4.3) Then, the N eigenvalues of M are {λk(M), λk+1(M), · · · , λN+k−1(M)} for any k .
Lemma 4.2 provides necessary and sufficient conditions for the existence of complex eigenvalues for the Jacobian Dxg|xmix,N .
Lemma 4.2. The Jacobian Dxg|xmix,N has N−1 2 pairs of complex conjugate eigen-values if and only if α ̸= β and µ ̸= α + β 2 (1 + α + β) for mutation (Q1), or, µ ̸= N −1 N for mutation (Q2).
Proof. The proof relies on the cyclic properties of complex roots of unity. Details are in Appendix C.1.
Note that if the conditions in Lemma 4.2 are not satisfied (either for the mutation (Q1) or for the mutation (Q2)), then the eigenvalues of the Jacobian are strictly real.
When the conditions are satisfied, ωN,k is complex if and only if λk Dxg|xmix,N is complex. There are N−1 2 complex conjugate pairs among the ωN,k for k = 1, · · · , N.
For N = 3, the one complex pair is associated with the unique Hopf bifurcation point as seen in Figures 3.5(e), 3.6(e) and 3.10.
43 To prove the existence of Hopf bifurcations we need to show that conditions (H1) and (H2) of Theorem 2.2 are satisfied. We begin by calculating critical values of the bifurcation parameter µ corresponding to pairs of eigenvalues crossing the imaginary axis. Since mutation matrices (Q1) and (Q2) have entries that are linear in µ, the entries of the Jacobian Dxg|xmix,N are also all linear in µ. In order to simplify the notation, we set h Dxg|xmix,N i 1j = γj + µηj, j = 1, · · · , N, (4.4) where both γj and ηj are independent of µ. Using this notation, we compute the bifurcation points for the dynamics in Lemma 4.3.
Lemma 4.3. The pair of complex conjugate eigenvalues λr, λN−r of the Jacobian Dxg|xmix,N , for each r = 1, · · · , ⌊N−1 2 ⌋, is purely imaginary if and only if µ = − " N X j=1 γj cos 2π N (j −1)r # " N X j=1 ηj cos 2π N (j −1)r #−1 =: µ0,r, (4.5) and µ0,r satisfies the conditions of Lemma 4.2. Further, the bifurcation points µ0,r are distinct, i.e. µ0,k ̸= µ0,l when k ̸= l.
Proof. The proof is in Appendix C.2.
From Lemma 4.3, a unique pair of eigenvalues of Dxg|(xmix,N ,µ0,r) is purely imaginary at each µ0,r; this implies condition (H1). Lemma 4.3 also implies that d dµRe (λr) = PN j=1 ηj cos 2π N (j −1)r ̸= 0, which is condition (H2).
We can now collect these results and state our main theorem.
Theorem 4.1. The equilibrium point xmix,N with payoffmatrix BC,N undergoes ⌊N−1 2 ⌋Hopf bifurcations, with the rth r = 1, · · · , ⌊N−1 2 ⌋ such bifurcation located at µ0,r = − " N X j=1 γj cos 2π N (j −1)r # " N X j=1 ηj cos 2π N (j −1)r #−1 when α ̸= β and µ0,r ̸= α + β 2 (1 + α + β) for mutation (Q1), or, µ0,r ̸= N −1 N for mutation (Q2).
Remark 4.1. Equation (4.5) gives an analytic expression for µ0,r corresponding to a unique pair of purely imaginary eigenvalues of the Jacobian. However, not all values 44 of µ0,r are feasible. That is, there might be pairs (α, β) ∈[0, 1) × [0, 1) that yield bifurcation points µ0,r outside the feasible parameter range 0 ≤µ0,r ≤1 of our model.
This is illustrated in §4.3.
4.2 Criticality of Hopf Bifurcation In Theorem 4.1 we proved conditions for the existence of Hopf bifurcations for the replicator-mutator dynamics with payoffBC,N. In this section we study the criticality of the bifurcations (and correspondingly the existence of stable limit cycles for the dynamics) by computing an analytical expression for the first Lyapunov coefficient ℓ1|(xmix,N ,µ0,r) in Lemma 4.4.
Lemma 4.4. Let A0 = Dxg|(xmix,N ,µ0,r). Then A0 has a pair of purely imaginary eigenvalues λr (A0) = i ˆ ω and λN−r (A0) = −i ˆ ω, where ˆ ω ∈R is calculated from (4.3).
Define t = r sign (ˆ ω) and ω0 = |ˆ ω|. The first Lyapunov coefficient of the dynamics (3.2) with payoffBC,N evaluated at the fixed point xmix,N and bifurcation point µ0,r is given by ℓ1|(xmix,N ,µ0,r) = 1 2ω0 Re (T1 + T2) , where, T1 = −2N 2 + (α + β)(ωt N + ω−t N ) and T2 = 2λt(QT) λ2t(QT) 2 i ω0 −λ2t(A0) 1 + αωt N + βω−t N 2 + 1 + ω3t N βω−2t N + αω−t N .
A0 is the Jacobian Dxg|(xmix,N ,µ0,r) with purely imaginary eigenvalue λr (A0) = i ˆ ω calculated from (4.3).
Proof. The function λk(M) is defined for a general integer k and square matrix M in Equation (4.3). We exploit the circulant structure of the dynamics to obtain this analytical result. The details of the calculation are in Appendix C.3.
Lemma 4.4 allows us to study the criticality of the Hopf bifurcations at each of the bifurcation points µ0,r as a function of the parameters (α, β) ∈[0, 1) × [0, 1).
In Figure 4.2 we plot regions of positive and negative ℓ1 as a function of α and β. For each of the subplots in Figure 4.2, black denotes negative l1 (supercritical Hopf bifurcation, stable limit cycles) and white denotes positive l1 (subcritical Hopf bifurcation, repelling periodic solutions). Gray denotes the unfeasibility region for µ0,r (either µ0,r < 0 or µ0,r > 1, see Remark 4.1). The red curves correspond to critical points µ0,r that do not satisfy the conditions of Lemma 4.2.
45 Figure 4.2: Criticality of the Hopf bifurcations as a function of parameters α and β for N = 3, 4, 5, 6 and 8.
46 Figure 4.2 illustrates the effect of the number of strategies N and payoffparameters α and β on the existence and criticality of Hopf bifurcations for the dynamics. Several different cases exist. For example, there are cases corresponding to a supercritical bifurcation throughout (α, β) ∈[0, 1) × [0, 1) (as when N = 3), and cases for which the bifurcation is subcritical on a subset of the parameter space (as when N = 4 and with mutation matrix (Q2)).
The regions corresponding to infeasible critical points (µ0,r outside the range [0, 1]) can be connected as when N = 8, r = 3, or disconnected, as when N = 6, r = 1 and with mutation matrix (Q1). Some cases are illustrated in §4.3.
4.3 Illustration of Bifurcations Figure 4.3: Effect of parameters α and β on bifurcations. Subplots labeled (a)–(e) are bifurcation plots for the dynamics with N = 3 strategies, mutation matrix (Q1), α = 0.1 and β as shown. The top left subplot shows the criticality and existence of the Hopf bifur-cations (taken from Figure 4.2: N = 3, (Q1)) with parameters corresponding to subplots (a)–(e) marked. Subplot (b) corresponds to the all-to-all payoffmatrix; the corresponding bifurcations are discussed in Section 3.4 with bifurcation points given by (3.5).
The existence and criticality of Hopf bifurcations computed in Lemma 4.3 and Theorem 4.1 vary as a function of parameters α and β in ways that may not be 47 Infeasible Figure 4.4: Bifurcation plots for the dynamics with N = 6 strategies, parameters α = 0.8, β = 0.05, and mutation matrix as indicated on each subplot. In the case of the left subplot with mutation matrix (Q1), the Hopf bifurcation point µ0,1 = 1.24 lies outside the feasible range µ ∈[0, 1].
immediately obvious. In this section, we explore the parameter dependence of the Hopf bifurcations using a set of selected simulations to help illustrate this variation.
Figure 4.3 shows bifurcation plots for the dynamics with N = 3 and mutation matrix (Q1). Parameter α is set to 0.1 and β is varied between 0 and 1. Looking at the corresponding criticality plot in Figure 4.3 (reproduced from Figure 4.2), we expect that the bifurcation is supercritical for all β except at two points labeled (b) and (d). These are precisely the points that violate the conditions of Lemma 4.2 and Theorem 4.1. i.e., at (b), α = β = 0.1 and at (d), µ0,1 = α+β 2(1+α+β) for α = 0.1 and β = 0.58. As a result, the bifurcation plots show the existence of stable limit cycles for all values of β along the line α = 0.1, except at the points (b) and (d). In Figure 4.3 stable limit cycles are apparent in Figures 4.3(a), 4.3(c) and 4.3(e), but not in 4.3(b) and 4.3(d). The payofftopology corresponding to the parameters in 4.3(b) is fully symmetric, with a bifurcation plot analogous to Figure 3.5(a).
Figure 4.4 shows the bifurcation plots for the dynamics with N = 6, α = 0.8, β = 0.05, and mutation matrices (Q1) and (Q2). The corresponding criticality plots in Figure 4.2 show that the supercritical Hopf bifurcation point µ0,1 lies outside the feasible range µ ∈[0, 1] for (Q1) and inside the feasible range for (Q2), for the chosen parameters. This is illustrated in Figure 4.4; the left plot shows a Hopf bifurcation at µ = 1.249 while the right plot shows a Hopf bifurcation at µ = 0.363. The left plot in Figure 4.4 also illustrates that infeasible supercritical bifurcation points can yield stable limit cycles within the range of feasible µ.
48 4.4 One-Parameter Multi-Cycles In §4.1 and §4.2 we focused on a particular two-parameter circulant payoffstructure given by (4.1) and illustrated in Figure 4.1. In this section we leverage the results from the previous sections to study the dynamics corresponding to a class of circulant payoffstructures with each node having a single outgoing edge. For simplicity of presentation, we consider only mutation matrix (Q1) in this section. We show that for a particular set of topologies in this class, the dynamics exhibit multiple simultaneous Hopf bifurcations about distinct fixed points. The analysis in this section points to the fact that the dynamics with payoffgraphs having multiple embedded cycles can have multiple distinct stable limit cycle attractors; we explore these multi-cycle dynamics more generally in §4.5.
Consider the dynamics (3.2) where the payoffmatrix B is given by B := BN,k = Circulant (1, a1, · · · , aN−1) , with ak = α and as = 0 for s ̸= k.
(4.6) Let gcd(a, b) denote the greatest common divisor of a and b. Two graphs with payoff Figure 4.5: Graph topologies corresponding to circulant payoffmatrix BN,k from (4.6) for N = 5, 6, 15. Three cases are shown, Case 1 corresponds to simple cycles, Case 2 to multiple cycles, and Case 3 to connected pairs of vertices (only exists for N even). Note that multiple values of k can yield the same graph topology modulo a vertex relabeling; non-isomorphic topologies have distinct d.
49 matrices BN,k1 and BN,k2 are isomorphic if and only if gcd(N, k1)=gcd(N, k2). Hence, among the payoffmatrices BN,k, the set of matrices BN,d where d belongs to the set of proper divisors of N, corresponds to a set of non-isomorphic graph topologies. We split the set of graph topologies with payoffBN,d (d a proper divisor of N) into three distinct cases as described below: (1) d = 1, the graph is a directed cycle containing all vertices (2) 1 < d < N/2, the graph consists of d disjoint cycles, each of length N/d (3) d = N/2, the graph consists of N/2 disjoint pairs of connected vertices. This case exists only for N even.
Figure 4.5 illustrates the three cases of graph topologies for N = 5, 6, and 15.
In addition to xmix,N, the dynamics with payoffmatrix BN,d and mutation matrix (Q1) have d equilibria denoted xj,d,N and given by xj,d,N = h 0T j−1 d N 0T d−j · · · 0T j−1 d N 0T d−j iT ∈RN, j = 1, · · · , d.
(4.7) In Case 1, d = 1, and correspondingly j = 1, and x1,1,N = xmix,N. For a given N and d ̸= 1, the d equilibria xj,d,N are cyclically and symmetrically spaced around xmix,N. Case 1 is studied in detail in §4.1 and obtained by setting β = 0 (a simple one-parameter cycle). For the topology with pairs of connected nodes in Case 3, the Jacobian of (3.2) evaluated at the equilibrium xmix,N, or also at any of the equilibria xj,N/2,N, is real and symmetric and therefore has only real eigenvalues. Thus the system cannot have Hopf bifurcations for these equilibria in this case. We henceforth focus on Case 2 and study the dynamics with payofftopologies comprising multiple cycles.
4.4.1 Case 2 Analysis We begin the analysis of the dynamics with multi-cycle graph topologies by first look-ing at the case N = 6, d = 2 (payoffB6,2), before generalizing to higher dimensions.
The two disjoint cycles in the graph corresponding to B6,2 suggest that the behavior of the system might be similar to that observed for the N = 3 cycles in Chapter 3. Indeed, simulations of the dynamics shown in Figure 4.6 suggest the existence of two stable limit cycle attractors, each dominated exclusively by three strategies, corresponding to the connected nodes of the graph. Further, simulations of the phase 50 (a) Trajectories of the dynamics (3.2) for N = 6, d = 2, and B = B6,2 = Circulant (1, 0, α, 0, 0, 0), and for two different random initial conditions. Red corresponds to components x1, x3, x5 while blue corresponds to components x2.x4, x6.
(b) Phase portrait for dynamics with payoffB6,2 and 50 different initial conditions showing two stable limit cycle attractors.
Figure 4.6: Simulation of the dynamics (3.2) for N = 6, payoffB6,2 and µ = 0.25. Panel (a) shows two typical trajectories for the system while panel (b) illustrates the limit cycle attractors in a decoupled phase space. The red trajectories correspond to the components x1, x3 and x5, while the blue trajectories correspond to x2, x4 and x6.
space for 50 different randomly selected initial conditions, as in Figure 4.6(b), indi-cate that the two limit cycles are the only stable attractors for the dynamics for an appropriate range of bifurcation parameter µ.
The linearization of the system at equilibrium xmix,6 violates condition (H1) of Theorem 2.2 (i.e., complex eigenvalues of the Jacobian have algebraic multiplicity greater than one). However the dynamics with payoffB6,2 have two other equilibria 51 (as in (4.7)) given by x1,2,6 = h 1/3 0 1/3 0 1/3 0 iT , x2,2,6 = h 0 1/3 0 1/3 0 1/3 iT .
The simulations in Figure 4.6 suggest that the dynamics undergo Hopf bifurcations at each of these equilibria. In Corollary 4.1, which follows from Corollary 3.1, we prove that this is indeed the case by showing that x1,2,6 and x2,2,6 undergo two simultaneous Hopf bifurcations at the critical point µ = 2−α 6 .
Corollary 4.1. The system (3.2) with payoffmatrix B6,2 and mutation matrix (Q1) has equilibria x1,2,6 and x2,2,6 that undergo supercritical Hopf bifurcations at the critical point µ = 2−α 6 with α ̸= √ 3 −1.
Proof. Here we analyze the equilibrium x1,2,6. The analysis for x2,2,6 is similar. The Jacobian Dxg|x1,2,6 is given by Dxg|x1,2,6 = 1−2α−6µ 9 0 −2+α−3αµ 9 0 3αµ+6µ−2−2α 9 0 0 −α−1 3 0 0 0 0 3αµ+6µ−2−2α 9 0 1−2α−6µ 9 0 −2+α−3αµ 9 0 0 0 0 −α−1 3 0 0 −2+α−3αµ 9 0 3αµ+6µ−2−2α 9 0 1−2α−6µ 9 0 0 0 0 0 0 −α−1 3 .
Permuting rows and columns (i.e. reindexing the nodes), this matrix can be rewritten as the block matrix M6,2 given by M6,2 = 1−2α−6µ 9 −2+α−3αµ 9 3αµ+6µ−2−2α 9 3αµ+6µ−2−2α 9 1−2α−6µ 9 −2+α−3αµ 9 −2+α−3αµ 9 3αµ+6µ−2−2α 9 1−2α−6µ 9 03×3 03×3 −1+α 3 I3×3 , (4.8) which has the same eigenvalues as Dxg|x1,2,6. The upper diagonal block of M6,2 is the same as the Jacobian of the system (3.2) for N = 3, payoffBC,3 with β = 0, mutation (Q1) and evaluated at equilibrium h 1/3 1/3 1/3 iT . The lower diagonal block is a Hurwitz matrix. Also the two blocks are decoupled, hence the eigenvalues of M6,2 are given by the union of the sets of eigenvalues of each block. A pair of the 52 eigenvalues crosses the imaginary axis resulting in a Hopf bifurcation for precisely the conditions given in Corollary 3.1 with β = 0 and mutation matrix (Q1) (i.e., a critical point µ = 2−α 6 and complex eigenvalue condition α ̸= √ 3 −1). In Appendix C.4, we leverage the reindexing and decoupling (as in M6,2) to compute the first Lyapunov coefficient for the dynamics and show that the Hopf bifurcations in this case are supercritical.
Following the intuition developed from the analysis of the two-cycle dynamics for N = 6 above, and the illustrations in Figure 4.6, we now extend the analysis to general N. Just as in the N = 6 case, a decoupling in the Jacobian allows us to prove the existence of multiple Hopf bifurcations about the d fixed points xj,d,N.
The (m, n) entry for the Jacobian of the system (3.2) with payoffBN,d and mutation matrix (Q1) is given by [Dxg]mn = (2xn + α xn+d) qnm + αxn−dqn−d,m −xm [2xn + α (xn−d + xn+d)] − N X k=1 x2 k + α N X k=1 xkxk+d !
δmn.
(4.9) Evaluating the Jacobian from (4.9) at the equilibrium xj,d,N, and rearranging its rows and columns analogous to (4.8), we obtain the matrix MN,d = AN1×N1 0N2×N1 0N1×N2 −1+α N1 IN2×N2 , where N1 = N d , N2 = N −N1, and A is the Jacobian of the system (3.2) with payoffBN1,1 (simple cycle), mutation (Q1) and evaluated at equilibrium x1,1,N1 = 1 N11N1 = xmix,N1.
MN,d has a block diagonal structure with N −N1 eigenvalues equal to −1+α N1 . Its remaining N1 eigenvalues are given by the eigenvalues of the circulant matrix A. The Jacobian (4.9) evaluated at each of the d equilibria xj,d,N is similar to MN,d, hence making the Hopf bifurcation analysis of all of these equilibria equivalent. The matrix A is precisely the Jacobian studied in §4.1 for the case N = N1 and β = 0; hence the existence of Hopf bifurcations follows from Theorem 4.1. The criticality of each of these d simultaneous Hopf bifurcations of equilibria xj,d,N is analogous to the criticality calculations in §4.2 and is computed in Appendix C.4. Corollary 4.2 below summarizes the bifurcation result described above and is the multi-cycle extension to Theorem 4.1. To simplify notation, analogous to (4.4) we set [A]1n = γn + µηn.
53 Corollary 4.2. The system (3.2) with payoffmatrix BN,d (with N ≥6, d = 1, · · · , N/2, d a proper divisor of N) and mutation matrix (Q1) has d equilibria xj,d,N (j = 1, · · · , d) that concurrently undergo N−d 2d Hopf bifurcations, with the rth of such bifurcations located at µ0,r = − N/d X n=1 γn cos 2π N (n −1)rd N/d X n=1 ηn cos 2π N (n −1)rd −1 for r = 1, · · · , ⌊N−d 2d ⌋, if α > 0 and µ0,r ̸= α 2(1+α).
In this section we have shown the coexistence of multiple stable limit cycles for the dynamics, when the underlying circulant payoffgraph topologies have multiple distinct cycles. In the following section we investigate the connections between cycles in the payoffgraph topology and limit cycles in the dynamics for more general payoff structures.
4.5 Extensions and Generalizations Studying the fully general model (3.2), even with N = 3 strategies, is highly complex.
This complexity motivated our restriction to circulant payoffmatrices of the form BC,N (4.1) and BN,d (4.6) for the analysis in §4.1 and §4.4. These results might lead one to conclude that the circulant structure of payoffmatrix B is a necessary condition for Hopf bifurcations of the dynamics. In this section we illustrate that this is not the case. We show examples of limit cycles for selected noncirculant payoffmatrices, first for N = 3 strategies, and then for N ≥4. The simulations in this section illustrate a tight connection between the topology of the payoffgraphs and the existence of stable limit cycles for the dynamics. In particular, embedded cycles in the payoff graph appear to be necessary for (and often lead to) the existence of limit cycles, and amplitude and frequency of limit cycles appear to be related to symmetries in the graph.
Consider 3 × 3 payoffmatrices B satisfying (3.1) and Condition 3.1 that have directed links of two kinds: strong links with weights b and weak links with weights ϵb where b ∈(0, 1) and 0 < ϵ ≪1. There are 73 corresponding non-isomorphic graph topologies in the set . Figure 4.7 shows stable limit cycles for four topologies in this set corresponding to noncirculant payoffmatrices. Each of these topologies has an embedded directed cycle.
54 Figure 4.7: Limit cycles for noncirculant payoffmatrices B, N = 3, and mutation matrix (Q1). The solid arrows in the graphs are strong links with weight b and the dashed arrows are weak links with weight ϵb. Parameters for all plots are b = 0.2 and ϵ = 0.1.
We next look at selected noncirculant payofftopologies with N ≥4 nodes in Figure 4.8. The bifurcation plots (middle panel in Figure 4.8) are obtained by simulating the dynamics for a range of different values of mutation parameter µ and random initial conditions. Stable equilibria are marked blue and limit cycles are marked magenta.
Also shown in the right panel of the figure are limit cycle trajectories of the dynamics for specific chosen values of µ in the magenta range. The colors of the nodes in the topologies (left panel) match the colors of the corresponding trajectories.
Several interesting features can be observed in the plots in Figure 4.8. In topology (a), the connection between the amplitude of oscillation of a given strategy in the trajectory plot and the location of the corresponding node on the graph is apparent.
In particular, the cyan node, which is part of two directed cycles, has a significantly higher oscillation amplitude than the green node, which is part of only one directed cycle. The symmetry of topology (b) about the red node leads to the existence of two stable limit cycle attractors, much like the illustrations in §4.4. On the other hand, topology (c) has two embedded cycles but only one limit cycle attractor for both (Q1) and (Q2) mutation matrices.
In Figure 4.9, we go a little further and consider random payoffgraph topologies having strong links with weight b and weak links with weight ϵb, much like in Figure 4.7. We simulate the dynamics for a set of different random graphs and a range of 55 Figure 4.8: Limit cycles for the dynamics (3.2) with noncirculant payoffgraph topology.
The left panel shows the payoffgraph; all edges have equal weight b = 0.7. The center panel shows the bifurcation plot for each topology and mutation matrix (Q1) or (Q2) as indicated.
The bifurcation plot is obtained by simulating the dynamics for 120 values of the mutation parameter µ in the range shown on the x-axis of each plot. For each µ, the dynamics are simulated for 12 randomly chosen initial conditions and the limiting set (stable equilibria or limit cycles) is obtained. Stable equilibria are plotted in blue and limit cycles are plotted in magenta. The right panel shows trajectory plots of the dynamics for a value of µ chosen in the magenta (limit cycle) range of the bifurcation plot (µ = 0.2 in (a), (b) and (c) (Q2); µ = 0.4 in (c) (Q1)). The colors of each trajectory match the colors of the nodes on the corresponding payoffgraph. For topology (b), the two trajectory plots correspond to different initial conditions.
56 Figure 4.9: Limit cycles for the dynamics (3.2) with random payoffgraphs. The left panel shows the payoffgraph with two types of edges: strong edges (solid lines) with a weight of b and weak edges (dashed lines) with a weight of ϵb; here b = 0.7 and ϵ = 0.1. The center panel shows the resulting trajectories with mutation (Q2) and suitable µ (0.2 in (a), 0.25 in (b) and 0.27 in (c)). The right panel highlights the interconnection between nodes corresponding to the dominant components of the limit cycle trajectories. The color of each of the nodes on the payoffgraph matches the color of the corresponding trajectory in the center panel. In each case, it is observed that there is a directed cycle between the dominant component nodes.
values for the mutation parameter µ, and focus specifically on graphs and parameters that induce a limit cycle oscillation for the dynamics. We show three such examples in Figure 4.9. For each set of simulated limit cycle trajectories, the dominant com-ponents are obtained; dominant components are defined as those having a relatively high oscillation amplitude, or correspondingly a trajectory with standard deviation above a set threshold. The main observation we make here is that the existence of limit cycles for the dynamics is tied to the existence of a directed cycle between nodes of the payoffgraph. In all our simulations of random graphs, for both mutation ma-trices (Q1) and (Q2), we consistently find that the dominant components of stable limit cycles correspond to the existence of at least one directed cycle between the 57 corresponding nodes of the payoffgraph, as illustrated in the right panel of Figure 4.9.
The purpose of the simulations in this section is to illustrate that one can break symmetry in the payoffgraph significantly (as compared to the modest symmetry breaking that yields the circulant structures analyzed in this chapter) and still obtain stable limit cycles for the dynamics, even in the case of random graph topologies, as long as the topologies have at least one embedded cycle. In addition to limit cycles, the dynamics can also have chaotic attractors as found in [74, 76]. We have not seen chaotic attractors for the payofftopologies and mutation matrices (Q1) and (Q2) studied in this chapter; we conjecture that alternative mutation models such as the one used in will exhibit chaotic dynamics.
4.6 Final Remarks Much of the existing analysis of the replicator-mutator dynamics has been focused on stable equilibria.
The analysis in the literature has also primarily considered payoffand mutation matrices that are symmetric, which correspond to undirected payoffgraph topologies. Recent work on a graph theoretic model of language dynamics has shown that the graph connectivity plays a critical role in determining the location of bifurcation points in the dynamics, but the restriction to undirected graphs confines the range of limiting behavior to stable equilibria. In , it is shown that considering asymmetric payoffand mutation matrices (corresponding to directed graphs) can yield limit cycle behavior and even chaos for replicator-mutator dynamics.
Here we have proved conditions guaranteeing stable limit cycles in the replicator-mutator dynamics. These arise as a consequence of Hopf bifurcations for N ≥3 strategies and for circulant payoffmatrices. From a graph perspective, we showed how breaking symmetry by considering directed graphs allows for oscillatory limiting behavior. We emphasize that the limit cycles are not restricted to circulant payoffs, but can exist for noncirculant payoffs as shown in §4.5.
The simulations in §4.5 illustrate the connection between embedded directed cycles in the payoffgraph and the existence of stable limit cycles for the dynamics. A Hopf bifurcation analysis of these more general cases is an intended future direction. We will also explore the effects of the structure of the mutation matrix, beyond (Q1) and (Q2) considered here, as a step towards understanding the transition to chaos illustrated in . Further directions of future research, including applications to the design of decision-making protocols in multi-agent robotic systems, are discussed in Chapter 8.
58 Chapter 5 Evolutionary Dynamics of Collective Migration Collective migration is a ubiquitous natural phenomenon common in a number of species including birds, fish, invertebrates and mammals [119, 42, 22, 6]. The mi-gratory process is often an adaptive response to conditions such as competition for resources in a dynamic environment, seasonal variability, and selection of new habi-tats for breeding, for example [132, 119]. Animals perform such migratory tasks by leveraging a variety of environmental cues such as nutrient and thermal gradients, magnetic fields, odor cues, or visual markers [132, 30, 140, 133]. Measuring these stochastic environmental signals is complicated and requires the investment of time and energy, as well as the development of necessary physiological and sensory machin-ery such as vision in insects and vertebrates and chemical signaling in bacteria . Along with environmental cues, animals migrating collectively also have the ability to leverage social information from neighbors in the group . One way of doing so is by imitating invested neighbors (via consensus processes such as attraction and alignment of heading) and thereby effectively achieving good migratory perfor-mance, without paying the measurement and processing cost. The interplay between costly information acquisition from the environment and relatively less expensive so-cial information from the group raises two pertinent questions regarding leadership and social interactions in migratory populations.
The first question relates to the migratory performance of large groups in the presence of a limited number of leaders, i.e., can a subset of informed leaders effectively lead a large group? The authors in address this question using individual-based simulations (involving attraction, repulsion and alignment between individuals) and demonstrate that in a group of socially interacting individuals, a small fraction of 59 informed leaders can effectively determine the direction of travel of a large group of uninformed followers. Further, they show that for a fixed fraction of leaders, the accuracy of group direction improves with increasing group size.
The second question relates to the evolution of leadership in collective migration, i.e., under what conditions is the coexistence of invested leaders and social follow-ers stable in an evolutionarily sense? This question is especially pertinent when the cost of investing in signal acquisition is sufficiently high; followers can leverage the investments made by leaders via social interactions, without paying the investment costs. In a recent paper (also see related commentary ), the authors address this question using an individual-based model (similar to ) and evolutionary sim-ulations, and show that under certain conditions, the specialization of groups into coexisting leaders and followers (also known as branching or speciation) is a stable evolutionary outcome.
Individual-based bio-inspired models such as those used in [14, 33] provide im-portant insights about the relevant biological processes being modeled, but are chal-lenging to analyze. In particular, it can be difficult to ascertain what aspects of the low-level interactions (social network topology, cost functions, movement dynamics, etc.) are most critical to the observed high-level behavior of interest (evolution of leadership for example). This analysis can be done more effectively by constructing appropriate simpler (lower-dimensional) models that capture the relevant high-level behavior, while also being tractable for analysis. In , the authors construct one such lower-dimensional mean-field approximation to the evolutionary model studied in , and using tools from evolutionary adaptive dynamics [25, 26] prove conditions for the branching of a migrating population into leader and follower groups.
While simpler models have the advantage of analytical tractability, this comes at the cost of abstracting some potentially important features of the corresponding detailed model. In the case of the migration model in , the mean-field approach effectively prescribes an all-to-all social interaction topology between the individuals in order to reduce dimensionality; this ignores the potentially important role of limited social interactions. Indeed, it has been shown [98, 151, 3, 10] that network topology plays a critical role in determining outcomes in biological as well as robotic collectives.
In this chapter, we design a model to study the evolution of collective migration that is explicitly dependent on the social interaction graph topology. We present a comprehensive analysis of the all-to-all limit of the model, recovering the speciation results of the mean-field analysis in , and demonstrating the hysteretic effect associated with recovering lost migration ability described in [33, 34]. We then go 60 on to study the effect of topology on the evolutionary model, and show a minimum social connectivity threshold necessary for the evolution of leadership in migration.
Our evolutionary model has two timescales. The fast timescale corresponds to the stochastic migration dynamics and individual fitness computations as a consequence of migratory performance. The slow timescale corresponds to the evolutionary dy-namics of the population and changes in population strategy distribution as a conse-quence of replication and mutation. A key advantage of our model is that it allows for analytically computing fitness on the fast timescale by solving a set of coupled linear equations. For the detailed model in , the fitness computation requires extensive Monte-Carlo simulations using agent-based models. On the other hand, the detailed model has the advantage of incorporating a time-varying social interconnection topol-ogy between individuals, which eventually results in fission-fusion spatial dynamics (where groups constantly merge and split) described in ; the model studied in this chapter assumes a time-invariant social interaction graph for fitness computation. In ongoing work on generalizing the model presented here to time-varying graphs, we intend to investigate the observed fission-fusion result of the detailed model.
There is significant recent interest in understanding the mechanisms of interaction, signal processing and information transfer in evolved natural collectives in order to design better algorithms for collective motion and decision-making in robotic groups [128, 98, 66, 142]. Robotic collectives often have resource constraints (analogous to the measurement and processing cost for migration tracking), operate in dynamic and stochastic environments, and interact with neighbors on some networked graph topology.
In Section 5.4, we consider the migration model from this multi-agent robotic perspective and study a simple model of adaptive dynamic nodes on a network.
We illustrate the critical role that the structure of the interaction graph plays in determining the location of leaders (highly invested nodes) in the adaptive network and in bifurcations in the nodal dynamics as a function of increasing cost.
The outline of this chapter is as follows. In Section 5.1 we present our graph-dependent evolutionary migration model and study its all-to-all limit in Section 5.2.
We study evolutionary dynamics with limited interconnection in Section 5.3 and focus on adaptive dynamic nodes in small networks in Section 5.4.
5.1 Model Description Our model is derived from the mean-field migration model in with two key modifications; we explicitly account for a limited social interaction graph topology 61 in the dynamics and we introduce a slightly modified social noise model to allow for analytical fitness computations as a function of graph topology and individual investments, as described below.
Consider a set of N agents indexed by i ∈{1, · · · , N} that each intend to control their direction of migration represented by a stochastic scalar variable xi ∈R, thereby tracking some desired ‘true’ direction µ ∈R with high fidelity. Accurate tracking of the desired direction µ over time may correspond to benefits such as improvement in environmental conditions for foraging, predator evasion, early access to breeding grounds, etc. Following , the stochastic dynamics of each agent are given by dxi = kidxDi + (1 −ki)dxSi, (5.1) where dxDi and dxSi are the driven tracking and social consensus stochastic processes respectively. ki ∈[0, 1] is an adaptive parameter that tunes the level of investment made by agent i in the driven and social processes. ki = 1, for example, corresponds to an individual i fully invested in the tracking process and ignoring social information from neighbors, while ki = 0 corresponds to the individual exclusively leveraging so-cial information without tracking the environmental signal. The ki’s are the adaptive evolutionary parameters in the simulations and analysis in the sections to follow.
The driven process dxDi is modeled as an Ornstein-Uhlenbeck stochastic process [137, 24] of the form dxDi = −kDi(xi −µ)dt + σDdWDi.
(5.2) Here, the parameter kDi ≥0 corresponds to the gain associated with tracking, σ2 D > 0 is the noise intensity associated with measuring the environmental signal µ, and dWDi represents the standard Wiener increment.
For kDi > 0, the process (5.2) has a steady-state mean and variance given by E [xi] = µ, E (xi −µ)2 = σ2 D 2kDi .
(5.3) Higher values of tracking gain kDi result in lower steady-state variance in migration direction xi, which corresponds to improved tracking. Figure 5.1 shows the spatial dy-namics of migrating agents with driven process (5.2) and varying levels of investment kDi, illustrating the effect of increasing investment on improved tracking performance.
We use basic tools from graph theory [111, 91, 47] (see also §2.3) to model the social consensus process. Individuals are modeled as nodes on a directed social inter-62 Figure 5.1: Migratory performance as a function of investment kD. Agents are modeled as steered particles with constant speed, headings xi given by (5.2), and starting at the origin with random orientations. Each plot shows 500 trajectories with agents each having the same value of investment kDi = kD as indicated and noise parameter σD = 1. The blue arrow shows the desired direction of migration µ. Increasing investment kD results in improved tracking performance as agent trajectories track the blue arrow µ more effectively.
connection graph with adjacency matrix A = [aij] ∈RN×N. A directed edge in the graph from individual i to individual j is read as ‘i can sense j’. Let Ni denote the set of neighbors of individual i (i.e., the set of agents that individual i can sense), and ∥Ni∥denote the cardinality of this set (number of neighbors). For the social model in this work, we assume that agents weight their neighbors equally by distributing a total weight normalized to 1. This corresponds to the adjacency matrix aij = 0 if i = j ∥Ni∥+ if i ̸= j, (5.4) where ∥Ni∥+ is the pseudoinverse of ∥Ni∥(∥Ni∥+ = 0 when ∥Ni∥= 0, ∥Ni∥+ = 1/∥Ni∥otherwise). The Laplacian matrix of the graph corresponding to the adjacency matrix A is given by L = diag(A1) −A, where 1 is a vector of ones of appropriate dimension. In subsequent sections we use the terms individual or node/agent, and population or network, interchangeably.
The social consensus process dxSi is dependent on the social interaction graph Laplacian L, the gain associated with the social process kSi ≥0, and the noise associated with measuring the social signal σSi > 0 as follows: dxSi = −kSiLixdt + σSidWSi.
(5.5) 63 In the social process (5.5), dWSi is the standard Wiener increment, and Li denotes the ith row of the Laplacian matrix L. Let x be the vector of individual directions xi. Then Lix = ∥Ni∥+ P j∈Ni(xi −xj).
Following the setup in , we make a simplification to reduce the parameter space to one dimension by assuming that the gains are proportional to the relative investments in each process, i.e., kDi = ki and kSi = 1 −ki.
Substituting this simplification in (5.2), (5.5) and (5.1) we have dxi = kidxDi + (1 −ki)dxSi = −k2 i (xi −µ)dt −(1 −ki)2Lixdt + q k2 i σ2 D + (1 −ki)2σ2 Si dWi.
(5.6) In order to simplify the notation we make a coordinate transformation and define the normalized direction ˜ xi as ˜ xi = xi −µ σD , and correspondingly ˜ x = x −µ1 σD .
(5.7) Substituting the transformation (5.7) in (5.6) and observing that Li1 = 0 we have the normalized dynamics d˜ xi = −k2 i ˜ xidt −(1 −ki)2Li˜ xdt + s k2 i + (1 −ki)2σ2 Si σ2 D dWi.
(5.8) The social noise term σSi reflects the difficulty that agents have in extracting social information gained from interactions with neighbors. In , it is assumed that this difficulty (magnitude of σSi) decreases as the ordering or coherence of the population increases. Here we take a slightly different local (and graph dependent) approach and relate the social noise term for an individual agent to the average investment of the neighbors of that agent (i.e. average magnitude of parameter kj, j ∈Ni). Specifically, agents that interact socially with neighbors having a high level of investment, have a correspondingly lower social noise term, and are hence better able to extract social information from their neighbors. The specific relationship can take many forms, but for simplicity we use a linear relationship between the ratio σ2 Si/σ2 D and the average neighborhood investment, σ2 Si σ2 D = β2(1 −knbhd,i), (5.9) where knbhd,i is the average value of the investment made by the neighbors of agent i and β2 is a social noise scaling parameter. In vector form, knbhd = Ak.
64 The stochastic system (5.8) can be written compactly in matrix form as d˜ x = −(K1 + K2L)˜ x dt + SdW , (5.10) where the diagonal matrices K1, K2 and S are given by K1 = diag(k2 i ), K2 = diag ((1 −ki)2) and S = diag p k2 i + β2(1 −k2 i )(1 −knbhd,i) .
As discussed in [33, 132], the long-term migratory performance of an individual with dynamics (5.8) can be computed by projecting the steady-state distribution of directions xi in the desired direction of migration µ. This quantity is given by exp −σ2 ss,i 2 , where σ2 ss,i is the steady-state variance of xi, and corresponds to the expected migration speed of an individual i in the desired direction µ. The fitness or utility of an agent as a function of the steady state migratory performance and the level of investment ki is defined in [33, 132] as Fi = exp −σ2 ss,i 2 exp −ck2 i , (5.11) where the second term of the fitness function models the cost associated with in-vestment in tracking with a scaling cost parameter c > 0. The choice of cost func-tion (5.11) is not unique; the multiplicative exponential form is chosen for analytical tractability. Simulations in [132, 33] show that reasonable variations of fitness func-tion yield qualitatively comparable results. The saturating form of the performance function exp −σ2 ss,i 2 as a function of investment can be interpreted as modeling the diminishing returns of increasing investment. Further, the quadratic form of the cost ck2 i implies that higher investments in the driven process are increasingly costly. In the absence of any social interactions, the optimal strategy for solitary migrating individuals can be found by maximizing (5.11) with respect to kDi resulting in kD,opt = 3 r σ2 D 8c .
(5.12) While disconnected individuals adopt the optimal strategy from (5.12), the pres-ence of social interactions between individuals dramatically alters this picture; along with having the ability to invest in measuring the environmental signal, migrating individuals can leverage relatively less expensive social information available from neighbors by effectively imitating or flocking with their neighbors. We explore this dynamic in the sections to follow.
65 5.2 Evolutionary Dynamics in the All-to-all Limit As a first step in analyzing the social migration model (5.10), we consider the limit of all-to-all interconnection (aij = 1 N−1 for all i ̸= j in (5.4)) in a large population (labeled the resident population with subscript R), with all individuals having a com-mon level of investment kR > 0. This limit corresponds to the mean-field dynamics analyzed in . By the law of large numbers, the average direction of population migration in the limit of large N is the same as the desired migration direction µ after the decay of transients (i.e., in steady-state, lim N→∞Lx = x −µ1). Substituting L˜ x = ˜ x in (5.10), the dynamics of an individual in the population are given by d˜ xR = − k2 R + (1 −kR)2 ˜ xR dt + q k2 R + β2(1 −kR)3 dW.
(5.13) The corresponding steady-state variance of an individual’s direction is given by σ2 ss,R = k2 R + β2(1 −kR)3 2 [k2 R + (1 −kR)2], (5.14) with steady-state migration speed (performance) given by exp(−σ2 ss,R/2). In Figure 5.2 we plot this steady-state migration speed as a function of investment kR for varying social noise term β. As defined in Equation (5.9), the parameter β reflects Figure 5.2: Steady-state migration speed as a function of resident population investment parameter kR and noise parameter β for a large population with all-to-all interconnection and common investment kR.
66 the strength of the noise from social interactions relative to the noise associated with the tracking process. In Figure 5.2 we see that the migration performance saturates at high levels of investment kR, and remains low over greater kR ranges, for large β.
We use β > 2 in this work to model noisier social interactions relative to tracking (consistent with ); this provides an incentive for individuals to invest in the tracking process.
Now consider the evolution of strategies for such an all-to-all connected popu-lation. A key part of any evolutionary algorithm is the computation of fitness of individuals in the population as a function of strategy distribution, model parame-ters, environmental conditions, and other such features. In certain cases (such as the all-to-all limit here), fitness can be analytically computed, which allows for an explicit calculation of the outcomes of the evolutionary process using tools from adaptive dy-namics [25, 26, 18]. Adaptive dynamics are well-suited for studying the evolution of a continuous one-dimensional trait in a population undergoing small mutations (see §2.1 for a detailed discussion).
5.2.1 Adaptive Dynamics Calculations Using (5.14) and (5.11), the fitness of an individual in the resident population with dynamics (5.13) is given by FR(kR) = exp −k2 R + β2(1 −kR)3 4(2k2 R −2kR + 1) −ck2 R .
(5.15) Consider a small population of mutants with strategy kM interacting with each other and with all the residents. The mutants (owing to their small numbers) will experience the same social noise as the residents so that they have dynamics given by d˜ xM = − k2 M + (1 −kM)2 ˜ xM dt + q k2 M + β2(1 −kR)(1 −kM)2 dW.
(5.16) Correspondingly, the fitness of individuals in the mutant population is given by FM(kR, kM) = exp −k2 M + β2(1 −kR)(1 −kM)2 4(2k2 M −2kM + 1) −ck2 M .
(5.17) The relative fitness of the mutant strategy in the environment of the resident is known as the differential fitness and is given by (2.3), S(kR, kM) = FM(kR, kM) −FR(kR).
(5.18) 67 The two-dimensional function S allows us to predict which mutant strategies can invade a particular resident population. For example, for a given resident strategy kR, the values of kM that result in S > 0 correspond to the mutant strategies that when rare can invade the established resident population. Further, a study of the selection landscape S can help us predict when we can expect to see an evolutionarily stable monomorphic population (all individuals having the same strategy) and when we can expect to see opportunities for branching in evolutionary simulations (resulting in leader (ki ≈1) and follower (ki ≈0) populations) as we vary the cost c associated with strategy investment.
The evolutionary dynamics of the resident strategy kR are given by (2.4), where g(kR) is the selection gradient (note that the timescale t associated with (2.4) corre-sponds to slow evolutionary time and is different from the fast timescale associated with the stochastic migrations dynamics (5.10) and (5.13)). For the differential fitness S defined in (5.18), (5.15) and (5.17), singular strategies k∗(defined as g(k∗) = 0) are given by the solutions to the polynomial equation (details in Appendix D) k∗ β2(1 −k∗) −1 (k∗−1) + 4ck∗(2k2 ∗−2k∗+ 1)2 = 0.
(5.19) A singular strategy k∗is known as a Convergent Stable Strategy (CSS) if it is locally asymptotically stable for the dynamics (2.4), which is equivalent to satisfying the con-dition (2.5). As discussed in §2.1, a CSS strategy k∗can be either locally evolutionary stable (local ESS) for the population, or can be a branching opportunity where the population speciates. For S defined in (5.18), the branching condition (2.6) evaluates to (details in Appendix D) 0 < k∗< 5 − √ 7 6 ≈0.3923.
(5.20) The branching condition in (5.20) is exactly the same as that obtained in (Equa-tion (18)). Parameters β and c that yield CSS singular strategies k∗∈ 0, 5− √ 7 6 result in speciated populations via evolutionary branching.
5.2.2 Adaptive Dynamics Results The bifurcation diagram in Figure 5.3 summarizes the singular strategy condition (5.19), CSS condition (2.5) and branching condition (5.20) to obtain four distinct sets of evolutionary outcomes (in ranges A, B, C and D) for our model for increasing cost parameter c (calculations in Appendix D): 68 Figure 5.3: Evolutionary singular strategies k∗as a function of cost parameter c. The two sets of singular strategies defined by (5.19) are plotted in blue. One set corresponds to k∗= 0 and the other corresponds to the curve given by the equation c = (1−k∗)[β2(1−k∗)−1] 4(2k2 ∗−2k∗+1)2 .
Solid curves are CSS strategies, and dashed curves are unstable singular strategies. The regions marked (A)-(D) correspond to the descriptions in the text. Analytical derivations for the cost parameters β2−1 4 , c1 and c2 that divide the regions are given in Appendix D.
(A) Monomorphic: For 0 < c < β2−1 4 there exists only one CSS strategy.
This strategy is also evolutionarily stable (evolutionary stability since k∗> 5− √ 7 6 ), resulting in a monomorphic population with strategy k∗. The k∗= 0 singular strategy is not convergent stable.
(B) Two local CSS’s that are each evolutionarily stable: For β2−1 4 < c < c1 there exist two convergent stable strategies, one of which is the fully social strategy k∗= 0. Both singular strategies are locally evolutionarily stable. c1 is defined implicitly as the solution to the equation g(5− √ 7 6 ) c=c1 = 0.
(C) Branching: For c1 < c < c2 the convergent stable interior (k∗∈(0, 1)) singular strategy satisfies the branching condition (5.20) resulting in the speciation of the population.
(D) Collapse of Migration: c > c2 represents the high cost scenario in which the only singular strategy that exists is the convergent stable fully social strategy k∗= 0 and the population does not develop any migration ability.
69 These four cases constitute a comprehensive picture of the evolutionary dynamics of the migration model (5.10) in the all-to-all limit, and encompass key features of the branching calculations in and evolutionary simulations in . The existence of two locally evolutionary stable attractors in case (B) above implies that the evolu-tionary dynamics can potentially yield speciated outcomes in the population without evolutionary branching (case (C)). We illustrate this using simulations in §5.2.3.
A hysteretic effect associated with restoring population migration ability once destroyed is apparent in Figure 5.3. In particular, once migration ability in the pop-ulation is lost for high cost parameter c > c2, the cost parameter needs to be reduced below the level β2−1 4 (< c1 < c2) for migration ability to be regained. This compares to the simulations in [33, 34] where agent-based models are used to study the effect of habitat fragmentation on the evolution of migration. In these simulations, the authors study the impacts of progressively more fragmented habitats on migratory outcomes, and show that once migration ability is lost for a threshold level of frag-mentation, much greater habitat recovery is necessary to restore lost migration ability (a hysteretic effect). Higher levels of habitat fragmentation are comparable to higher cost parameter c.
5.2.3 Evolutionary Simulations∗ In Figure 5.4 we show the pairwise invasability plots (PIPs) [25, 26, 18] (see Figure 2.3 in §2.1 for a description) of the differential fitness S(kR, kM) for increasing cost c to illustrate the four sets of outcomes described above. These plots show the sign of S as a function of the resident and mutant population strategies. Dark regions correspond to differential fitness S > 0 and allow mutant invasions; white regions correspond to S < 0 and prohibit mutant invasions. Comparing the PIPs from Figure 5.4 to the canonical classes of PIPs in Figure 2.3, we see conditions for an initially monomorphic population for low values of cost parameter c, a branching speciated solution for intermediate values of c, and for high values of c we see conditions that prevent individuals from developing any significant investment, i.e., the population is unable to develop any significant migration ability.
We confirm our predictions from the adaptive dynamics analysis by running evo-lutionary simulations in the case of a monomorphic initial condition and also a uni-formly randomly distributed initial condition as shown in Figure 5.4 (bottom two rows). These simulations comprise the roulette-wheel selection and small mutation ∗Results from this subsection were presented at the SIAM Conference on Applications of Dy-namical Systems, Snowbird, UT, 2011.
70 Figure 5.4: Evolutionary dynamics for the migration model with all-to-all interconnection, noise parameter β = 3, and with increasing cost parameter c as indicated in each column.
The top row shows the pairwise invasability plots. Black regions correspond to differential fitness S(kR, kM) > 0 (mutants can invade) and white regions to S < 0. The red vertical lines pass through convergent stable interior strategies kR = k∗. The middle row of plots are evolutionary simulations starting with a monomorphic population with strategy k = 0.5; hot colors correspond to high population density. The bottom row of plots are also evolutionary simulations, but having an initial population with a uniformly randomly distributed strategy k ∈[0, 1]. We use N = 2000 individuals for these simulations; corresponding to β = 3, boundary cost parameters from Figure 5.3 are c1 = 2.48 and c2 = 2.77.
operations on each generation of a population of N = 2000 individuals with all-to-all social graph, dynamics (5.10), and fitness (5.11).
The columns in Figure 5.4 from left to right correspond to the cases (A)-(D) respectively. In case (A), both initial conditions result in a monomorphic evolutionary outcome. In case (B), the polymorphic solution for the evolutionary simulation with random initial conditions (Column 2, last Row) is a consequence of the stability of the k∗= 0 singular strategy, and not as a consequence of branching, as in case (C).
Case (D) corresponds to the collapse of migration with all individuals having an insignificant level of investment.
71 The analysis in this section shows the range of evolutionary outcomes for the migration model with all-to-all social interconnection. We are particularly motivated by conditions that result in the speciation or branching of the population into invested leaders and social followers. In the following section we study the role that limited social interconnection topology plays in the emergence of this evolutionary branching.
5.3 Evolutionary Dynamics with Limited Social Interactions† In this section we study the role that graph topology plays in the evolutionary dy-namics of the migration model (5.10). While the all-to-all topology assumption of §5.2 allows for a detailed analysis of the evolutionary dynamics in a large population, it is unrealistic for most biological and decentralized artificial systems. In §5.3.1, we relax the all-to-all assumption and study the migration model (5.10) with limited interconnection. We prove two main results. Theorem 5.1 provides necessary and suf-ficient conditions for a population with limited interconnections to develop the ability to migrate. Theorem 5.2 presents an analytical technique for computing migratory performance (solution to fast timescale dynamics) as a function of interconnection topology and individual node investments.
In §5.3.2 we utilize the analytical fast timescale calculation from Theorem 5.2 to simulate evolutionary dynamics for three classes of limited interconnection topologies: undirected ring lattices, undirected random graphs, and directed random graphs.
In each case, the simulations show a minimum connectivity threshold necessary for speciated evolutionary outcomes in the population.
5.3.1 Fast Timescale Results As we will see in Theorem 5.2, the steady-state probability distribution of the state ˜ x for the migration dynamics (5.10) exists if and only if the zero equilibrium of the noise-free form of the dynamics (5.10) are asymptotically stable. The noise-free form of the dynamics (5.10) are given by ˙ ˜ x = M ˜ x, where M = −(K1 + K2L).
(5.21) †Results from this section were presented at the SIAM Conference on Applications of Dynamical Systems, Snowbird, UT, 2011.
72 Intuitively, the asymptotic stability of the zero equilibrium of the noise-free dy-namics (5.21) corresponds to the population developing the ability to collectively migrate since ˜ x →0 = ⇒ x →µ1. The lack of asymptotic stability for (5.21) implies that there exist individuals in the population that do not have an ability to migrate because their direction states xi diverge away from the desired direction µ, and correspondingly, the steady-state variance of xi given by σ2 ss,i does not exist. The linear dynamics (5.21) are asymptotically stable if and only if matrix M is Hurwitz , i.e., all eigenvalues of M have real part strictly less than 0. Lemmas 5.1 and 5.2 provide necessary and sufficient conditions, respectively, for matrix M to be Hurwitz.
Lemma 5.1. If matrix M is Hurwitz, then there exists some node i such that ki > 0.
Proof. If the condition is not satisfied, i.e., ki=0 for all i ∈{1, · · · , N}, then M = −L.
The Laplacian matrix L has a zero eigenvalue associated with the eigenvector 1 and hence M is not Hurwitz.
Lemma 5.2. If ki > 0 for all i ∈{1, · · · , N}, then M is Hurwitz.
Proof. This can be checked by a straightforward application of the Gershgorin circle theorem . The Laplacian matrix L is given by L = [lij]. The eigenvalues of M lie in the union of N disks, each centered about −k2 i −(1−ki)2lii with radius (1−ki)2lii.
If ki > 0 for all i, then the disks all lie strictly in the left half complex plane.
Lemmas 5.1 and 5.2 are extreme cases of investment that can be interpreted in the context of migration as follows. Lemma 5.1 says that regardless of social interconnec-tion topology, the population does not develop migration ability unless at least one individual is invested in acquiring the external signal µ. Lemma 5.2 says that if all individuals in the population are invested, then the population develops the ability to migrate in the direction µ, for any social graph. In Theorem 5.1 we derive necessary and sufficient conditions for the population to develop migration ability (matrix M Hurwitz) that depend explicitly on the topology of the social interconnection graph.
The proof of Theorem 5.1 requires the following lemma from (see also [111, 64]).
Lemma 5.3. For a general Laplacian matrix ˜ L = [˜ lij] ∈RN×N given by ˜ lij ≤0 for i ̸= j and N X j=1 ˜ lij = 0 for all i ∈{1, · · · , N}, the following conditions are equivalent: (i) ˜ L has a simple zero eigenvalue and all of the other eigenvalues have positive real parts.
73 (ii) The directed graph G(˜ L) of ˜ L (graph with adjacency matrix ˜ A = [˜ aij], ˜ aii = 0 for all i and ˜ aij = −˜ lij for all i ̸= j) has a directed spanning tree (see Definition 5.1 below).
(iii) For z ∈RN, the dynamics ˙ z = −˜ Lz converge asymptotically to α1 for some scalar α.
Proof. See Lemma 3.1 in , Theorem 2 in , and Lemma 2 in .
Definition 5.1. Spanning Tree: A directed graph has a directed spanning tree (con-dition (ii) in Lemma 5.3) if there exists at least one node k on the graph such that a directed path exists from every other node on the graph to node k. Node k is known as a root node of the graph.
For undirected connected graphs, every node is a root node. For general (con-nected or disconnected) directed graphs, one can define a root set that is accessible from every other node in the network, i.e. there is a directed path from every node to at least one node in the root set. Let R(˜ L) denote a minimal root set (set with smallest cardinality) of the graph with Laplacian ˜ L (denoted G(˜ L)). Note that for a given graph, the set R(˜ L) is not necessarily unique. For example, for an undirected connected graph, R(˜ L) = {i} for any node i (see Figure 5.5(a) for an illustration).
Figure 5.5: Illustrations of the root set R(L) and the conditions of Theorem 5.1. In each graph, the set of nodes labeled 1, · · · , N and solid arrows correspond to the social graph G(L). The complete set of nodes labeled 0, 1, · · · , N and all the arrows correspond to the augmented graph G(ˆ L). Node 0 represents the external signal. All three graphs shown have a spanning tree rooted at node 0, and hence satisfy the conditions of Theorem 5.1. (a) Social graph is undirected and connected, hence R(L) = {1} , {2} , {3} , {4} , or {5}. (b) Directed social graph with R(L) = {3} . (c) Directed social graph with R(L) = {2, 3, 6} or {2, 3, 7}.
74 Theorem 5.1. Matrix M from (5.21) is Hurwitz if and only if there exists a minimal root set R(L) such that kj > 0 for all nodes j ∈R(L), where L is the Laplacian matrix of the social graph with adjacency matrix (5.4).
Proof. Define the normalized external signal state (xi = µ in (5.7)) ˜ x0 = 0 and the augmented state vector z = [˜ x0 ˜ x]T. Consider the dynamics ˙ z = " ˙ ˜ x0 ˙ ˜ x # = − " 0 0N×1 −K11 −M # " ˜ x0 ˜ x # = −ˆ Lz.
(5.22) Then ˆ L satisfies the properties of general Laplacian matrices given in Lemma 5.3.
The graph corresponding to Laplacian ˆ L is the same as the graph corresponding to Laplacian K2L with an additional node (labeled 0, with state ˜ x0) having incoming links with weights k2 j from all nodes j = 1, · · · , N (see Figure 5.5 for an illustration, links to node 0 are shown as dashed arrows).
Since node 0 of G(ˆ L) has no outgoing links, from Definition 5.1 we have the following condition, G(ˆ L) has a spanning tree ⇐ ⇒R(ˆ L) = {0} .
(5.23) We claim the following G(ˆ L) has a spanning tree = ⇒∃R(L) s.t. kj > 0 for all j ∈R(L).
(5.24) We prove the statement above by contradiction. Assume that G(ˆ L) has a spanning tree and for each root set R(L), there exists a node j such that kj = 0. Since G(ˆ L) has a spanning tree, R(ˆ L) = {0}, which means that there is a directed path from every node to node 0. Now consider any root set R(L). Since kj = 0, node j in R(L) can only reach node 0 by a path to a node m / ∈R(L), for which km > 0. However, if such a path exists, then the set R(L) where node j is replaced with node m is another root set. By assumption we must have km = 0. Thus there exists no directed path from node j to node 0. Hence G(ˆ L) does not have a spanning tree and we have proved the claim.
Consider any root set R(L) and assume that kj > 0 for all j ∈R(L). Then all nodes j ∈R(L) are connected to node 0 of G(ˆ L). For all nodes m / ∈R(L), either km > 0 and m has a direct link to node 0, or km = 0 in which case m has a link to at least one other node on a directed path to the root node 0, via an element of R(L).
75 Hence ∃R(L) s.t. kj > 0 for all j ∈R(L) = ⇒G(ˆ L) has a spanning tree .
(5.25) Combining (5.23), (5.24) and (5.25), we have the condition, ∃R(L) s.t. kj > 0 ∀j ∈R(L) ⇐ ⇒G(ˆ L) has a spanning tree ⇐ ⇒R(ˆ L) = {0} .
(5.26) From Lemma 5.3, G(ˆ L) has a spanning tree if and only if the dynamics (5.22) converge asymptotically to α1 for some scalar α. However in (5.22), the state ˜ x0 = 0 is invariant and hence α = 0. The linear dynamics (5.21) converge asymptotically to 0 if and only if matrix M is Hurwitz . This gives the condition G(ˆ L) has a spanning tree ⇐ ⇒M is Hurwitz.
(5.27) Combining (5.26) and (5.27) we have the desired result.
Sets of root nodes R(L) are illustrated for a set of simple social graphs in Figure 5.5. The directed connected graph in Figure 5.5(b) has a single root node, whereas the graph in Figure 5.5(c) has two possible minimal sets of root nodes.
We now return to the noisy migration model given by the system of stochastic equations (5.10). In order to compute the fitness of an individual in a migratory collective using (5.11), a computation of the steady-state variance of the individuals dynamics σ2 ss,i is necessary. This quantity in turn depends on the level of investment of the individuals in the network (represented by the vector k) and the underlying topology of the social interconnection graph G(L), and can be computed by solving the matrix Lyapunov equation, as shown in Theorem 5.2 (see also [151, 103] for related results on the consensus and drift-diffusion models respectively).
Theorem 5.2. For the stochastic dynamics (5.10) with social graph G(L), suppose there exists an R(L) such that kj > 0 for all j ∈R(L). Then the system of stochastic differential equations (5.10) has steady-state mean lim t→∞E [˜ x(t)] = 0 and steady-state covariance matrix Σ = lim t→∞E ˜ x(t)T ˜ x(t) given by the solution to the Lyapunov equa-tion (K1 + K2L)Σ + Σ(K1 + K2L)T = SST.
Proof. The system (5.10) is a multivariate Ornstein-Uhlenbeck process with mean given by E [˜ x(t)] = exp (Mt) E [˜ x(0)] .
76 By Theorem 5.1, M is Hurwitz and hence lim t→∞E [˜ x(t)] = 0.
The covariance matrix of ˜ x is given by E (˜ x(t) −E [˜ x(t)])(˜ x(t) −E [˜ x(t)])T = exp(Mt)E ˜ x(0)˜ xT(0) exp(Mt) + t Z 0 exp(M(t −s))SST exp(M(t −s))ds.
Since M is Hurwitz, the steady-state covariance matrix is given by Σ = lim t→∞E ˜ x(t)T ˜ x(t) = lim t→∞ t Z 0 exp(M(t −s))SST exp(M(t −s))ds, which as is shown in is the solution to the Lyapunov equation MΣ + ΣM T + SST = 0.
(5.28) 5.3.2 Slow Timescale Evolutionary Dynamics The Lyapunov equation (5.28) in Theorem 5.2 allows us to compute the migratory performance (and correspondingly fitness (5.11)) of individuals since the diagonal terms of the steady-state covariance matrix Σ are the individual steady-state vari-ances σ2 ss,i. In this section, we use evolutionary simulations based on fast timescale fitness calculations from (5.28) to study the role that graph connectivity plays in the evolution of branching.
We focus on three classes of social graph topologies of which one is ordered (ring lattice) and two are random (undirected and directed). In each class, a single param-eter controls the level of connectivity of the graph. The three classes of graphs used are listed below: • Undirected Ring Lattice: A graph with N nodes, each connected to K nearest neighbors, K/2 on each side, for K even. An undirected edge exists between nodes i and j if and only if 0 < min {|i −j|, N −|i −j|} ≤K/2. The graph is connected for K ≥2.
• Random Undirected (Erd˝ os-R´ enyi) [19, 82]: Undirected graph with N nodes.
Every edge in the graph exists randomly with a uniform probability p. The 77 expected number of neighbors of a node is E[K] = Np for large N. The graph is almost surely connected if p > ln(N)/N or equivalently E[K] > ln(N).
• Random Directed [83, 2]: Directed graph with N nodes.
Each node has a probability p of having a directed link to every other node in the network. The expected number of neighbors of a node is E[K] = Np for large N.
Figure 5.6: Effect of social graph topology on the evolutionary outcomes of the migration model. The left plot shows the equilibrium strategy distribution as a function of number of nearest neighbors for the ring lattice graph model with N = 400 nodes and parameters β = 3 and c = 2.6; bright colors correspond to higher population density. The two plots on the right labeled (a) and (b) are evolutionary simulations, the steady state conditions in these plots correspond to the red dashed slices in the left plot. Notice that the speciated two-strategy equilibrium exists only once the graph connectivity exceeds a threshold number of neighbors (≈18 for parameters chosen here).
For each class of topologies, the parameters K and p allow us to explore a range of connectivities; for K = p = 0, the social graphs are fully disconnected and individuals must resort to solitary migration with a monomorphic optimal strategy (5.12). For K →N −1 (for the ring lattice) and p →1 (for the random graphs), the social graph is fully connected, resulting in the leader-follower speciated evolutionary equilibrium for certain parameter choices as discussed in Section 5.2. Between these two connec-tivity extremes, intuition suggests that an intermediate level of limited connectivity can provide adequate information flow in the network for followers to leverage the investments made by leaders, thereby resulting in a speciated population. In Figures 5.6 and 5.7 we confirm this intuition by showing that the transition from a monomor-78 phic, to a branched evolutionary solution as a function of topology (parameterized by K and p), occurs at an intermediate threshold level of connectivity.
For the simulation in Figure 5.6 we use the ring lattice topology for the social graph with N = 400 individuals and choose a set of parameters for which the all-to-all social graph is known to have a two-strategy solution, β = 3 and c = 2.6 (see Figure 5.4, column 3). For a range of values of the number of neighbors K = 2, 4, · · · , 100, we compute the fitness of individuals using (5.28) and (5.11) and evolve the strategies of the populations to an evolutionary steady state. This steady-state strategy distribu-tion is plotted as a function of number of neighbors K in the figure. We see that there exists an intermediate threshold for social connectivity (given by K ≈18) that allows for adequate information flow in the network to result in evolutionary branching. In particular, social graphs that are much sparser (fewer edges) than the all-to-all case analyzed in Section 5.2 can yield speciated evolutionary outcomes.
In Figure 5.7 we present simulations similar to those described for Figure 5.6 above, for all three classes of social graph topologies (ring lattice, undirected ran-dom, directed random) and for N = 200, 400, 600. In each case, we see a threshold connectivity for evolutionary branching that is higher than the minimum thresholds for the social graph to be connected. We also see that the number of individuals in the population N does not have a significant affect on this threshold in the range considered. Further, the location of the threshold is dependent on the class of graph being considered; the two classes of random graphs have lower thresholds than the ordered ring lattice.
The simulations in this section illustrate the significant effect of limited social graph connectivity on the evolution of branching in migration. In particular, we show that social connectivity above threshold levels can yield speciated outcomes in the evolutionary dynamics. The thresholds depend on the classes of social graph topology being considered.
Determining the analytical minimum connectivity threshold as function of parameters β, c and class of graph, is a topic of ongoing work.
We also intend to look at other classes of graph topologies such as spatially embedded graphs with a topological metric on connectivity (such as in ), and classes of small-world graphs parameterized by a single rewiring parameter [146, 82] (the ring lattice and undirected random graphs considered here are two extreme limits of the Watts-Strogatz small world model).
79 Ring Lattice Random Undirected Random Directed N=200 N=400 N=600 N=200 N=400 N=600 N=200 N=400 N=600 Figure 5.7: Evolutionary equilibria as a function of social graph topology for ring lattice, random undirected and random directed graph models with parameters β = 3 and c = 2.6; number of nodes N are shown on each plot. The minimum number of neighbors (or mean number for the random graphs) is independent of the number of nodes N in these plots, ≈18 for the ring lattice, ≈9 for the random undirected graphs and ≈8 for the random directed graphs.
5.4 Dynamic Nodes and Bifurcations Our analysis of the collective migration model (5.10) thus far has focused on the evolutionary perspective, typically considering the dynamics of networks with large numbers of nodes. In this section we shift focus and consider the model from an adaptive perspective on much smaller dynamic networks. This analysis is motivated in part by questions about leadership, task assignment and robust adaptive behavior in multi-agent robotic systems [128, 98, 107]. We consider a simple model of greedy local optimization by nodes on a graph, which yields individual adaptation of investments ki. For this model we show bifurcations as a function of cost that yield leader-follower 80 emergent behavior as equilibria of the adaptive process. We also illustrate the critical role played by graph topology in determining the location of leaders in the network of adaptive nodes using several examples.
Consider a system of interconnected agents with fast-timescale tracking dynamics given by (5.10). Further, suppose that each agent seeks to maximize its local utility function by adapting its investment parameter ki. We assume that the utility function for each agent Ui is given by the fitness function (5.11), Ui = exp −σ2 ss,i 2 exp −ck2 i .
(5.29) In this setting, the utility function for a focal agent i depends on the agent’s invest-ment ki, as well as the investments of other nodes in the network; we assume that the agent can measure its own utility, but not the investments of other nodes. Given the measured local utility Ui, agents modify their investments ki on a slow timescale by climbing the gradient of their local utility to reach its local maximum, dki dt = ∂Ui ∂ki .
(5.30) Our goal is to study the outcomes of this simple adaptive process by computing equilibria of the dynamics defined by (5.30), (5.29) and (5.10), and studying their bifurcations. Let ˜ Ui = ln Ui and consider the non-homogeneous time scaling τ = t R 0 Ui(σ)dσ. Then the dynamics (5.30) transform to ˙ ki = dki dτ = ∂˜ Ui ∂ki = ∂ ∂ki −σ2 ss,i 2 −ck2 i , i = {1, · · · , N} .
(5.31) The dynamics (5.31) are identical to (5.30), modulo the time-scaling, and are nota-tionally simpler. In particular, the equilibria of (5.31) are identical to those of (5.30).
We focus on these dynamics for the remainder of the section.
We first look at the simplest case of the dynamics with N = 2 nodes, with an all-to-all graph. The steady-state covariance matrix Σ (from (5.28)) in this case can be computed analytically as Σ = (Det M)SST + [−M + (Tr M)] SST [−M + (Tr M)]T −2(Det M)(Tr M) , (5.32) 81 where matrices M and S are defined in (5.10) and (5.21). For each pair {i, j} = {1, 2}, {2, 1}, the diagonal elements of Σ from (5.32) are given by σ2 ss,i = fij (ki −1) 4 + fji (2 (kj −1) kj + 1) 2 + fji (kj (3kj −2) + 1) k2 i −2k2 jki + k2 j 4 ((ki −1) ki + (kj −1) kj + 1) (kj (3kj −2) + 1) k2 i −2k2 jki + k2 j , (5.33) where fij = k2 j −β2 (ki −1) (kj −1) 2 .
Substituting (5.33) in (5.31), we compute the equilibria keq of the dynamics (5.31) and (5.10) and their stability as a function of increasing cost parameter c. Analyt-ical expressions of the equilibria are complicated, but the equilibria are illustrated for β = 3 in Figure 5.8.
For low cost, both individuals make a significant equal investment corresponding to the symmetric equilibrium keq,1 = keq,2 ≫0. As cost increases, the level of this equilibrium investment decreases and eventually a pair of stable leader-follower equilibria appear via two saddle-node bifurcations. As the cost increases further, the symmetric stable equilibrium loses stability in a subcriti-cal pitchfork bifurcation, leaving the leader-follower pair of stable equilibria and an unstable symmetric saddle equilibrium.
For larger all-to-all networks with the dynamics (5.31) and (5.10), bifurcations in cost c yield generalizations of the leader-follower equilibria for the N = 2 case: the fraction of nodes in the leader populations decreases with increasing cost parameter c as a consequence of several bifurcations in the dynamics. This is illustrated for N = 10 nodes in Figure 5.9.
The topology of the social graph and the cost parameter c together play an im-portant role in determining the location of leader and follower nodes on the graph at equilibrium. This is illustrated for an undirected star graph in Figure 5.10. At low values of cost c, the fringe nodes of the star invest strongly in the external signal and the central node leverages these neighbors as a follower with small investment.
At intermediate cost c all individuals make similar investments, while at high cost c the central node adapts to become the leader with all the fringe nodes leveraging this investment as followers.
We show equilibrium outcomes for three more graph topologies in Figure 5.11. For the undirected ring lattice, alternate nodes adapt to become leaders and followers, while for more complicated nearest-neighbor type topologies, the precise connection between topology and location of leaders is more challenging to interpret. This inter-pretation requires the development of a graph- and investment-dependent metric to 82 1 2 c k1 k1 k1 k2 k2 k2 c = 15 c = 17.5 c = 20 c1 c2 keq,i Figure 5.8: Bifurcations for the adaptive node dynamics (5.30) with N = 2 nodes, an all-to-all social graph, and noise parameter β = 3. The top plot shows the two components of keq (equilibria of the dynamics (5.30) and (5.10)) as a function of the cost parameter c. Stable sinks are marked blue and unstable saddles are marked red. The inset shows a zoomed in view of the region with 15 ≤c ≤20 marked in the dotted square. The dashed lines in the inset c1 ≈16.7 and c2 ≈18.2 denote the saddle-node and pitchfork bifurcation points respectively. The row of bottom plots are phase portraits for the dynamics with parameter c as indicated; the circles are stable sinks and the squares are saddles. These plots remains qualitatively the same for different values of β > 2; the bifurcation points c1 and c2 move further to the right for higher β.
rank nodes for their leadership potential, perhaps analogous to the information cen-trality metric used for drift-diffusion stochastic networks in , a topic of ongoing work.
83 keq,i c 10 nodes 8 nodes 2 nodes 4 nodes 6 nodes 10 9 8 7 6 3 4 5 N = 10, β = 3 Figure 5.9: Bifurcations for the adaptive node dynamics (5.30) and (5.10) with N = 10 nodes, an all-to-all social graph, and noise parameter β = 3. The left plot shows stable equilibria of the dynamics as a function of c; trajectories for the points marked with the circle (c = 1), square (c = 3) and triangle (c = 10) are shown on the right. The labels on the left plot indicate the number of leaders in each branch of stable solutions. Notice that bifurcations yield fewer leaders for increasing cost.
c = 3 c = 5 c = 0.1 Figure 5.10: Role of topology and cost parameter for the adaptive node dynamics on the undirected star graph. Parameter c is indicated on each plot. The x-axis of each plot is time and the y-axis is strategy ki. The color-scale corresponds to the magnitude of equilibrium investment keq,i.
5.5 Final Remarks The study of leadership has received significant attention in both biology and multi-agent robotics.
One focus in biology has been on determining conditions for the 84 Figure 5.11: Role of topology in determining locations of leaders on the social graph.
Parameters for all three plots are β = 3 and c = 4.
The colors of nodes on each plot correspond to equilibrium investments ki for the dynamics (5.30) and (5.10) with magnitudes indicated in the color-bar. The left graph is an undirected star with N = 10 nodes. The two right plots show a random spatial embedding of nodes with two different interconnection models. In the middle plot, each node is connected to its three nearest neighbors (topological metric) and in the right-most plot, each node is connected to all neighbors within a fixed radius given by the dashed line drawn (distance metric).
stable evolution of leadership behavior in collective systems, since followers in such systems have good performance at a lower cost. In networked robotic systems, the leader-follower paradigm has been studied in a variety of contexts as a tool to design control protocols that achieve desired performance. In this work we focus on the role that the social graph connectivity plays in a networked model of collective migration.
We use tools from adaptive dynamics to study the all-to-all limit of the evolutionary model and derive bounds for branching of the population into leader and follower groups.
For limited connectivity, we prove necessary and sufficient conditions for convergence of the noise-free migration model, and show that fitness of individuals in the stochastic model can be derived analytically using the Lyapunov equation. For random networks and lattices, we show a minimum connectivity bound that yields evolutionary branching in the population.
In the final section, we study smaller networks inspired by collective robotic systems, and show that the network topology plays a critical role in determining the location of leaders in the adaptive system.
Our goal here was not to design the best system for a given task, as is often the case in the leader-follower work, but rather to understand the role of networked topology in the emergence of leadership. One interesting avenue for future work is to understand where the top-down engineering design approach meets the bottom-up adaptive approach taken in this chapter. This understanding will aid in the design of 85 collective systems that are near optimal, and also robust to failures such as lost nodes or links. For the evolutionary model, further work involves analytically deriving the minimal connectivity threshold for branching, and generalizing to study arbitrary social graphs, beyond the lattice and random networks studied here.
86 Chapter 6 Coevolutionary Dynamics of Pursuit and Evasion Pursuit and evasion behaviors are widely observed in nature and play a critical role in predator foraging, prey survival, mating, and territorial battles in several species.
Species such as bats and dragonflies have evolved sophisticated dynamical strategies such as motion camouflage to disguise themselves as stationary during aerial pursuit [77, 27]. Studies on migratory cannibalistic locusts have revealed that pursuit and evasive behavior among conspecifics is integral to the formation of mass-moving mi-gratory bands in dense swarms [35, 5]. Recent experimental work on the dynamics of coordinated predator pursuit and prey evasion among schooling fish has shown that collective behavior, among both predators and prey, plays a vital role in predator hunting and prey evasion under conditions of considerable informational constraints (such as dynamic ocean environments) [37, 45].
Pursuit-evasion contests have been studied extensively from a game-theoretic per-spective as differential games . Unlike classical (matrix-based) games, the dy-namics of players in differential games are modeled using differential equations, with payoffs corresponding to particular states of the system, or to system trajectories.
In engineering, pursuit and evasion games have received much attention, particularly in the context of missile guidance and avoidance [101, 51] and aircraft pursuit and evasion [108, 81]. The book by Nahin provides a review of the topic along with relevant historical background.
The pervasiveness of pursuit and evasion in nature motivates the examination of winning strategies from an evolutionary perspective. Here one can think of a strategy as a control law that a particular pursuer (evader) employs to capture (escape). Cor-relates of evolutionary fitness, such as time-to-capture, provide natural metrics that 87 connect the dynamics of individual pursuer-evader pairs to evolutionary dynamics of populations comprising individuals playing different strategies. ESSes of the pursuit-evasion game point to strategies or behaviors one would expect to observe in nature.
Recently, Wei et al. used the evolutionary approach to study pursuit games with dynamics derived in . The authors of [147, 50] use Monte-Carlo simulations and analytical calculations to study three pursuit strategies competing against a field of deterministic or random nonreactive evasive strategies (an evader with a nonreactive strategy has dynamics that are uncoupled from those of the pursuer). The three chosen pursuit strategies (classical, constant bearing and motion camouflage) are bi-ologically inspired. The authors show convergence of the evolutionary game dynamics between the three strategies to pure motion camouflage and motivate this result by empirical observations of motion camouflage in hoverflies, dragonflies and bats .
In this chapter, we build on the work in by studying the coevolution of the three strategies of pursuit from playing against three distinct evasive strategies, two of which are reactive strategies (an evader with a reactive strategy has dynamics that are coupled to those of the pursuer). In contrast to the analysis in Chapters 3, 4 and 5, which involve the evolutionary dynamics of a single population, the analysis in this chapter involves the interaction between two distinct populations (pursuers and evaders). This will require us to look at an extension of the standard replicator equations to a double simplex phase space.
We use Monte-Carlo simulations and theoretical analysis to show convergence to an equilibrium of classical pursuit versus classical evasion. We point out that extend-ing the ‘games against nature’ approach (evolution of pursuers with respect to nonreactive evaders) to competitions between two sets of strategies does not result in a motion camouflage as the winning pursuit strategy, as in .
Indeed, the environment of evasive strategies that a pursuer population encounters is critical to determining the winning pursuit strategy. It is anticipated that analysis of strat-egy spaces different again from those studied in the present chapter will yield other interesting evolutionary outcomes.
We explore the winning strategies (classical pursuit and classical evasion) in a collective motion model with agents pursuing and evading designated neighbors on a cyclical interaction topology. This exploration is motivated by collective motion in cannibalistic locusts and has strong parallels to prior work in cyclic pursuit [23, 67, 69, 68]. Simulation results suggest a rich set of solutions for this collective motion model.
88 The outline of this chapter is as follows. §6.1 describes the planar dynamics for pursuit and evasive agents and the different pursuit and evasion strategies under consideration. In §6.2 we study the coevolutionary dynamics of the two populations, and in §6.3 we focus on the collective motion model with the winning strategies.
NOTATION: For notational convenience, the Euclidean plane R2 is identified with the complex plane C. Thus (x, y) ∈R2 ≡x + iy ∈C. For two complex numbers c1, c2 ∈C, the complex inner product is defined as ⟨c1, c2⟩:= Re(c1c∗ 2), the real part of c1c∗ 2, where c∗ 2 is the complex conjugate of c2. |c1| is the complex modulus of c1.
M # denotes the element-wise inverse of matrix M = [mij], i.e. m# ij = 1/mij.
6.1 Dynamics of Pursuit and Evasion∗ We study a two-agent planar pursuit and evasion problem where each agent is mod-eled as a self-propelled steered particle with constant speed and with angular velocity determined by the interaction between the particles. We consider three pursuit behav-iors: classical, constant bearing and motion camouflage, and three evasive behaviors: classical, random motion, and optical-flow based . The choice of the three pursuit behaviors is motivated by work in and , where it is proved that if the speed of the pursuer is greater than that of the evader, the pursuer captures the evader in finite time. Here ‘capture’ means that the Euclidean distance between the pursuer and evader reaches a designated minimum.
Consider a pursuer and an evader moving on the complex plane with positions rP = xP + iyP ∈C and rE = xE + iyE ∈C, and headings θP ∈S1 and θE ∈S1, respectively. The dynamics of the two-agent system are given by ˙ rP = eiθP , ˙ θP = uP ˙ rE = νeiθE, ˙ θE = uE.
(6.1) Here, the speed of the pursuer is normalized to be 1 and the evader has a constant positive speed ν < 1. We define the baseline vector r as the relative position of pursuer with respect to evader, i.e., r = rP −rE.
∗Sections 6.1 – 6.3 are presented verbatim as in with some minor modifications.
89 Figure 6.1: Cartoon trajectories of a pursuer and an evader. Pursuer position rP and evader position rE at time t0 are shown as circles. The corresponding velocities eiθP and νeiθE (and the vectors ieiθP and νieiθE normal to these) are shown as dotted arrows. Also shown is the baseline vector r. Solid trajectories correspond to t < t0 and dashed trajectories to t > t0.
Figure 6.1 shows the positions and velocity vectors for each particle, and the baseline vector. Note that ˙ r = eiθP −νeiθE. We define the three pursuit control laws following (with some change of notation): • Classical pursuit†: uP = −η r |r|, ieiθP , (P1) where η is a constant gain.
• Constant bearing pursuit†: uP = −η r |r|, ieiφeiθP , (P2) where φ ∈(−π/2, π/2) is a chosen constant bearing angle.
†Pursuer strategies (P1) and (P2) are modified slightly for clarity from those used previously in . We have dropped the second terms in each control law; these terms were only relevant in the fast initial transient dynamics and do not affect the results described here.
90 • Motion camouflage pursuit: uP = −µ r |r|, i ˙ r , (P3) where µ is a constant gain.
We define the three evasion control laws as follows: • Classical evasion: uE = −η r |r|, ieiθE , (E1) where η is a constant gain.
• Random motion evasion: Piecewise linear paths with turns every α time units, and turning rate uE selected uniformly randomly from [−κ, κ] at every turn.
(E2) • Optical-flow based evasion: uE = −η tan−1 ˙ θ , (E3) where θ is the complex argument of r and ˙ θ = −1 |r|2 ⟨r, i ˙ r⟩.
Intuitively, classical pursuit (evasion) involves the pursuer (evader) aligning its velocity vector with the baseline. In constant bearing pursuit, the pursuer maintains a constant bearing angle φ between its velocity vector and the baseline, whereas in motion camouflage, the pursuer contracts the magnitude of the baseline, while leaving the argument of the baseline unchanged. In [147, 50], the authors use elegant geometric ideas to show that the pursuit control laws (P1)-(P3) provably correspond to the desired pursuit strategies described above. This is done by defining pursuit manifolds for each strategy (‘states of the interacting system that satisfy particular relative position and velocity criteria’) and proving convergence to these manifolds for sufficiently high gains η and µ. In optical-flow based evasion, the evader reacts to the changes in the argument of the baseline vector; these changes are intended to mimic optical flow generated by the pursuer on the retina of the evader .
Figure 6.2 shows a simulation of the three pursuit and three evasion strategies pitted against one another. Note that the baseline vectors remain roughly parallel in each case of motion camouflage pursuit (bottom row of panels in Figure 6.2). The evader 91 trajectory in optical-flow based evasion is a straight line when competing against a motion camouflage pursuit strategy.
P P P P P P P P P E E E E E E E E E (P1) (P2) (P3) (E3) (E2) (E1) Figure 6.2: Simulated trajectories of each of the nine pairs of competing pursuit and eva-sive strategies. The rows correspond to pursuit control laws (P1), (P2) and (P3) respectively and the columns correspond to evasive control laws (E1), (E2) and (E3) respectively. For example, the plot in the middle corresponds to constant bearing pursuit versus random mo-tion evasion. The starting positions are indicated with ‘P’ and ‘E’. In all plots the evader (E) starts at the origin with θE(0) = 0. The pursuers (P) in columns 1 and 3 start at rP (0) = 5 + 3i with a heading θP (0) = π. In the second column the pursuers start at ran-dom positions and with random headings. The straight lines in each plot are snapshots of the baseline vector at specific points in time.
For all control laws uP and uE defined above, Lemma 6.1 ensures that capture is always possible in finite time.
Lemma 6.1. Consider dynamics (6.1) and control laws (P1)-(P3) and (E1)-(E3).
For every capture radius ϵ > 0 and every initial condition rP(0), rE(0) such that |r(0)| = |rP(0) −rE(0)| > ϵ, there exists a finite capture time T such that |r(T)| = ϵ.
92 Proof. Refer to [147, 50] for proof. Note that the evasive controls (E1)-(E3) satisfy the continuity and boundedness assumptions of Proposition 3.3 in .
The central question we ask in this chapter is which strategies win out in a co-evolutionary contest between the three proposed pursuit strategies and the three proposed evasive strategies? In the authors studied the evolutionary dynamics of the three pursuit strategies (P1)-(P3) playing against an environment of nonreac-tive deterministic or random evasive strategies such as (E2). In the following section we consider an evolutionary scenario in which the pursuit strategies coevolve with reactive evasive strategies (E1) and (E3) as well.
6.2 Evolutionary Dynamics To model the evolutionary dynamics of pursuer-evader interactions, fitnesses are de-termined by the cumulative effect of several one-on-one contests between pursuers and evaders such that a long time-to-capture for a particular contest corresponds to a high evader fitness and a low pursuer fitness. Consider a pursuer population represented by the population vector p = h p1 p2 p3 iT . Here pi, i ∈{1, 2, 3}, corresponds to the fraction of individuals in the population playing strategy (Pi). Hence the vector p is restricted to the simplex ∆2. Similarly, the evader population is represented by the population vector q = h q1 q2 q3 iT which is also restricted to ∆2. The fitness vectors for the pursuer and evader populations are denoted by fP ∈R3 + and fE ∈R3 + respectively, where fPi is the fitness of pursuit strategy (Pi) and fEj is the fitness of evasive strategy (Ej) . Define population mean fitness by c fP = pTfP and c fE = qTfE.
We can now write down a discrete update equation for each population that depends on the relative fitnesses of the different strategies in that population. For transition from generation g to generation g+1, we have for i = 1, 2, 3 and j = 1, 2, 3: pi(g + 1) = pi(g)fPi c fP qj(g + 1) = qj(g)fEj c fE .
(6.2) One can verify that the equations (6.2) ensure that each population vector remains in the simplex ∆2. Intuitively, strategies with fitness greater than the population mean fitness are favored. Given an expression for the fitness vectors, we can study the outcomes of the dynamics (6.2). As in we use time-to-capture as a measure 93 of fitness. The fitness of the set of pursuer strategies fP depends on the distribution of evaders in the population q and fE depends on the distribution of pursuers in the population p. This implies that the equations (6.2) are coupled. We simulate the evolutionary dynamics (6.2) using Monte-Carlo calculations that determine fitnesses.
We then perform a theoretical analysis.
6.2.1 Monte-Carlo Simulations We follow the setup in for our Monte-Carlo experiments. The important step is the construction of the capture time matrix T ∈R3×3 such that tij > 0 represents the time-to-capture for pursuit strategy (Pi) competing against evasive strategy (Ej).
Lemma 6.1 gives us that all elements of T are positive and finite. To construct T, we perform nine simulations, one for each element of T, such that each simulation has a pursuer and an evader starting from the same initial conditions. In each simulation the evader starts at the origin with a heading of zero. The pursuer’s initial position is chosen from a uniform distribution on the square [−10, 10] × [−10, 10], and its initial heading from a uniform distribution on S1. The other parameters are the same for each simulation: η = µ = 10, ν = .6, ϵ = 0.05, φ = 0.3, α = 0.2, and κ = 2.
The results presented here remain qualitatively consistent for reasonable variations of these parameters. A detailed study of the effect of each parameter on capture times is a direction of future work.
For each generation, we compute ten time matrices T k = tk ij , k ∈{1, . . . , 10}, such that each matrix corresponds to a different choice of pursuer initial conditions and evader random trajectory for column 2 (note that T k denotes the kth time matrix, not to be confused with matrix multiplication). The average matrix T = tij is defined by tij = 1 10 10 X k=1 tk ij. Let T(g) denote the average time matrix computed at generation g. For matrices M(g) = T #(g) and N(g) = T T(g), the fitness vectors are defined by fP(g) = M(g)q(g) fE(g) = N(g)p(g).
(6.3) The inverse and direct relationships between the time matrix and fitness for pursuers and evaders, respectively, ensure that high capture times have asymmetric fitness consequences for pursuers and evaders.
Fitnesses (6.3) also encode the frequency dependence and coupling of the evolutionary dynamics (6.2) since the fitness of a 94 pursuer (evader) strategy depends on the population distribution of evader (pursuer) strategies. Another way of interpreting equations (6.3) is from the perspective of a focal pursuer (evader) employing a specific strategy in a given generation. The expected fitness of this individual depends on expected interactions with each evasive (pursuit) strategy, which in turn depends on the population distribution of evaders (pursuers).
Equations (6.2) and (6.3) give us the necessary tools to simulate the pursuit-evasion dynamics defined on the direct product of two simplexes. This is done for a set of 50 randomly chosen pairs of initial distributions p(0) and q(0). Each set of initial conditions is propagated using equations (6.2) and (6.3) for 100 generations with new fitness matrices M(g) and N(g) calculated at each generation. The results of the simulation are plotted in Figure 6.3. Note that the trajectories eventually converge to the point corresponding to pure classical pursuit and pure classical evasion, i.e.
peq = qeq = h 1 0 0 iT . Further, the Monte-Carlo simulations point to a structure in the matrix T (and correspondingly in M = T # and N = T T) which we state below in Conjecture 6.1 without proof.
Motion Camouflage Classical Pursuit Constant Bearing Random Motion Classical Evasion Relative Heading Change Optical Flow Figure 6.3: Monte-Carlo simulation of equations (6.2) and (6.3). The simplex on the left corresponds to pursuit strategies and the simplex on the right to evasive strategies. The curves on each simplex evolve in pairs (p, q) ∈∆2 × ∆2; there are 50 pairs in all corre-sponding to 50 different initial conditions (p(0), q(0)). Each pair of trajectories comprises 100 generations (iterations of equation (6.2)) and hence 100 × 10 = 1000 evaluations of matrix T. The trajectories all eventually converge to pure classical pursuit (left figure) and pure classical evasion (right figure), indicating that classical pursuit and classical evasion is the evolutionarily stable equilibrium of the dynamics.
95 Conjecture 6.1. The first column of matrix T is dominant, i.e. ti1 > ti2 and ti1 > ti3 for all i. Further, in the first column t11 < t21 and t11 < t31.
The claim in Conjecture 6.1 is made as a consequence of Monte-Carlo computa-tions of matrix T and consistent observations of the proposed matrix structure. A rigorous proof of Conjecture 6.1 requires careful analytical computations of capture times for the different strategies, a topic for future work. The structure of Conjec-ture 6.1 provides a useful tool for analyzing the evolutionary dynamics on the direct product of two simplexes and for proving convergence properties. We investigate this in the following subsection.
6.2.2 Theoretical Analysis For some small time step ∆t > 0, we can rewrite the equations (6.2) as follows: 1 ∆t (pi(g + 1) −pi(g)) = pi(g)fPi −c fP ∆t c fP 1 ∆t (qj(g + 1) −qj(g)) = qj(g)fEj −c fE ∆t c fE .
In the limit ∆t →0 and with a change of timescale, we arrive at the set of differential equations ˙ pi = pi c fP fPi −c fP ˙ qj = qj c fE fEj −c fE .
(6.4) Consider a constant matrix T that obeys the structure of Conjecture 6.1. Further let M = T # and N = T T. Defining fitness vectors fP = Mq and fE = Np analogous to equation (6.3), and substituting into (6.4) we get ˙ pi = pi pTMq (Mq)i −pTMq ˙ qj = qj qTNp (Np)j −qTNp .
(6.5) Equations (6.5) are a form of the replicator dynamics (see Chapter 2) for two interact-ing populations with fitnesses defined by linear functions of the population distribu-tions. Critical to arriving at equations (6.5) is the assumption that T is constant and thus M and N are constant, which is justified by a ‘law of large numbers’ argument 96 . Further, the assumption makes the analysis of equations (6.5) tractable, and hence allows us to formally investigate the convergence shown in the Monte-Carlo experiments.
The system of equations (6.5) is a four-dimensional system evolving on ∆2 × ∆2.
There are several possible solutions of the dynamics on these simplexes. For instance, all vertex pairs (pairs of pure strategies) are fixed points. Further, a strategy that is initially absent does not emerge, i.e., pi(0) = 0 = ⇒pi(t) = 0, ∀t, and the same holds for qj (replicator dynamics are said to be non-innovative as they lack mutation). In order to investigate the coupled replicator dynamical system (6.5), we first study the simpler single population replicator dynamics given by ˙ qi = qi qTf fi −qTf , for i = 1, 2, 3.
(6.6) The fitness functions for the system of equations (6.6) are assumed to satisfy the following properties: • Property 1: fi ≡fi(t), i = 1, 2, 3, are each distinct functions of time, i.e., fi ̸= fj pointwise. If this were not the case then populations i and j would be indistinguishable from the perspective of evolutionary dynamics.
• Property 2: The functions fi(t) are each globally Lipschitz, bounded and posi-tive for all t ≥0.
• Property 3: The functions fi(t) have a single dominant fitness; without loss of generality, f3(t) > f2(t) and f3(t) > f1(t) for all t ≥0.
Lemma 6.2. Assume initial conditions are restricted to the domain D = {q ∈∆2|q3 > 0}.
The dynamics (6.6), satisfying Properties 1-3, have a unique asymptotically stable equilibrium point qeq = h 0 0 1 iT attracting all initial conditions in D.
Proof. The proof is in Appendix E.1.
The case of Lemma 6.2 with each fi constant is considered in . We now return to the coupled set of equations (6.5) and state the main theorem of this section. Here we employ the dominant structure of matrix T, assuming that Conjecture 6.1 holds, to prove convergence.
Theorem 6.1. Assume initial conditions are restricted to the domain D2 = {(p, q) ∈∆2 × ∆2|p1 > 0, q1 > 0}. Let matrix T satisfy Conjecture 6.1 with M = T # 97 and N = T T. Then, the coupled replicator dynamics (6.5) have a unique asymp-totically stable equilibrium point peq = qeq = h 1 0 0 iT attracting all initial conditions in D2.
Proof. The invariance of the domain D2 with respect to the dynamics (6.5) follows from the invariance of the domain D in the proof of Lemma 6.2. The column dominant structure of matrix T implies that the first element of the fitness vector fE = Np = T Tp is dominant. That is, regardless of the population distribution p at any time instant, fE1 > fE2 and fE1 > fE3. Hence we can use Lemma 6.2 to conclude that regardless of the evolution of the population vector p, the population vector q will asymptotically converge to qeq = h 1 0 0 iT .
Asymptotic stability implies that for every ϵ > 0 there exists a time t1 ≥0 such that t > t1 = ⇒∥q −qeq∥< ϵ. From the calculations in Lemma E.1 in Appendix E, ∥q −qeq∥< ϵ = ⇒fP1 > fP2 and fP1 > fP3 for ϵ = min 2(m11 −m21) (m11 −m21) + ∥M∥1 , 2(m11 −m31) (m11 −m31) + ∥M∥1 .
Hence we can apply Lemma 6.2 again to the dynamics of p to conclude that after time t1, the population vector p will also asymptotically converge to peq = h 1 0 0 iT .
In Figure 6.4 we simulate the dynamics in equations (6.5), for a particular cal-culation of matrix T, showing smooth convergence to the unique stable equilibrium.
This can be compared to the Monte-Carlo simulations of Figure 6.3. From a game-theoretic perspective, the stable equilibrium point corresponds to the single unique pure Nash equilibrium of the bi-matrix game, for payoffmatrices M and N.
In this section we have performed a computational and theoretical analysis of the evolutionary dynamics corresponding to the competition between a population of pursuers and a population of evaders, each having three strategies. We have shown convergence to a unique stable pure-strategy equilibrium corresponding to classical pursuit and classical evasion. This differs from the solution in , where the au-thors study the three pursuit strategies as a ‘game against nature’, each competing independently against nonreactive evasive strategies. There the evolutionary dynam-ics converge to the motion camouflage pursuit strategy. The competition of pursuers with a reactive set of evaders introduces a rich set of possibilities for evolutionary outcomes. Indeed, by comparison to the results of , we see that making avail-able alternative evader strategies affects the evolved strategies for both pursuers and 98 Motion Camouflage Classical Pursuit Constant Bearing Random Motion Classical Evasion Relative Heading Change Optical Flow Figure 6.4: Simulation of smooth dynamics (6.5). The capture time matrix T is chosen to be the mean of 50 Monte-Carlo computations of capture time matrices T k; M = T # and N = T T . The plot on the left shows the population vector p and the one on the right shows q. 50 pairs of trajectories are plotted with initial conditions chosen from a uniform random distribution on the space of two simplexes ∆2 × ∆2. Comparisons to the Monte-Carlo simulations in Figure 3 show strong similarities.
evaders. We note that one need not restrict to the three chosen pursuit or evasive strategies; other choices of strategies may be appropriate depending on the context.
We now shift focus to employing pursuit and evasive behaviors for collective mo-tion. In the following section we examine classical pursuit and classical evasion as this pair constitutes the evolutionary equilibrium of our strategy space.
6.3 Collective Motion The multi-agent dynamics described in this section are motivated in part by the in-triguing dynamics of social forging in groups of cannibalistic migratory locusts [5, 112]: individuals pursue other conspecifics in front of them and evade individuals approach-ing from behind. Inspired by these dynamics of cannibalistic locusts, and also by prior work on cyclic pursuit [67, 69, 68], we study the collective dynamics of steered parti-cles with each exhibiting classical pursuit and evasive behaviors simultaneously.
Consider a system of N agents indexed by j = 1, . . . , N, each having position rj = xj + iyj and heading θj. The agents are steered particles with constant speed v > 0 and steering control uj. Similar to equations (6.1), the kinematics of the agents 99 are given by ˙ rj = veiθj, ˙ θj = uj.
(6.7) For each agent we define baseline vectors bj+ = rj+1 −rj and bj−= rj−1 −rj. Note that j + 1 and j −1 are defined mod N; i.e., bN+ = r1 −rN and b1−= rN −r1. The control law uj is given by uj = ujP + ujE = K ieiθj, bj+ ∥bj+∥ −βK ieiθj, bj− ∥bj−∥ , (6.8) for scalar gain K > 0 and scaling parameter β > 0. The first term in the control law (6.8) is a classical pursuit term, with agent j pursuing agent j + 1 by attempting to align its heading with the baseline between j and j + 1 (i.e., bj+). The second term is an evasive term with agent j evading agent j −1 by attempting to align its heading anti-parallel to the baseline between j and j −1. The dynamics correspond to a cyclical interaction (sensing) topology between agents as illustrated in Figure 6.5.
Figure 6.5: Sensing topology for cyclic pursuit and evasion. An arrow from agent j to agent k should be read as ‘j senses k’. Agents pursue the agent immediately ahead and evade the agent immediately behind.
Simulations of the collective dynamics show several interesting outcomes: • For β < 1, stable circular motions exist with agents traveling equally spaced around a circle of radius v K(1−β) sin(π/N). This is illustrated for a formation of N = 8 agents in Figure 6.6(a).
• At β = 1 we observe a bifurcation. Specifically, the steady circular motions disappear and the agents diverge into an incoherent state for β > 1 as shown in Figure 6.6(b).
• For β < 1, the circular motions are not the only stable steady motions. We also observe convergence to regular figure eight weaving patterns as illustrated 100 in Figure 6.6(c); the initial conditions in Figure 6.6(c) are the only change from the simulation in Figure 6.6(a). Observation of such weaving patterns was also reported for pure cyclic pursuit .
(a) (b) (c) Figure 6.6: (a) Convergence to a circle for a group of N = 8 agents with β < 1. Param-eters are K = 5, v = 1 and β = 0.5. (b) Divergence into an incoherent state for β ≥1, i.e., evasion is stronger than pursuit. Parameters are K = 5, v = 1 and β = 1.2. (c) Weaving figure eight solution for N = 8 agents and β < 1. Parameters are K = 5, v = 1 and β = 0.5.
These results suggest that for stable circular motions to emerge from cyclic pur-suit and evasion, the pursuit action must be stronger than the evasive action. This is consistent with in which stable vortices for pursuit agents are observed. The au-thors note that ‘pursuit facilitates the formation of clusters’ (cohesion) whereas ‘escape (evasion) leads to a homogenization of density’ (dispersion).
In the case of a large number of agents following the pursuit-evasion dynamics (6.7), (6.8), the agents settle quickly into a cyclical chain-like formation. The chain starts offwith an arbitrary shape (dependent on initial conditions) and continually deforms (changes shape) on a slow time scale as illustrated in Figure 6.7. Simulations run for extended time do not show convergence of the formation to a regular motion; instead they show the continued slow deformation of the chain. An analysis of the chain-like formations as shown in Figure 6.7, perhaps using continuum methods and knot theory, is another topic of future work.
6.4 Final Remarks In this chapter we have considered an evolutionary game of three strategies of pursuit against three strategies of evasion, two of which are reactive. Monte-Carlo simulations 101 Figure 6.7: Chains for large N, here N = 50. The plots from left to right show snapshots at indicated times. The plot on the right is a composite showing trajectories of all agents over time. Parameters are K = 5, v = 1 and β = 0.5.
of the evolutionary dynamics, involving fitness computations for the interactions be-tween the different pursuer and evader pairs, show convergence to a stable equilibrium of classical pursuit and classical evasion. Using the structure of the fitness matrices observed in these simulations, we analytically prove the convergence of replicator dynamics to the same classical pursuit and classical evasion equilibrium as in our Monte-Carlo simulations. This effort builds on prior work where it was shown that replicator dynamics converge to motion camouflage pursuit when competing against nonreactive evasion strategies. Our result provides an interesting contrast to the earlier result and further illustrates that the consequences of evolutionary dynamics depend significantly on the space of strategies considered.
Motivated by the outcome of the evolutionary game, and by the behavior of cannabilistic locusts, we have investigated a novel control scheme involving agents performing simultaneous pursuit and evasion on cyclical interaction topologies. In the case that the pursuit gain is larger than the evasion gain, simulations indicate local convergence to circular motion formations of specified radius, as well as local convergence to more complex weaving patterns. Exploring the use of different pursuit and evasive behaviors and their corresponding collective outcomes is a topic of future work.
102 Chapter 7 Decision-Making Dynamics in Honeybee Swarms A honeybee colony∗is a fascinating natural example of an emergent system where an aggregation of thousands of individuals functions together as a composite integrated unit. Other examples include the gigantic colonies of leafcutter ants , army ants , or fungus growing termites [1, 116]. The colony functions as a truly decentralized system where local interactions among the bees yield group-level structure and func-tion, critical to survival. A typical bee colony is composed mostly of (predominantly) sterile female worker bees, all of whom are daughters of the one fertile queen bee that lives in their midst . A small set of male drone bees in the colony performs the fundamental reproductive task of mating with young queen bees from other nearby colonies. The worker bees perform most of the colony survival tasks including for-aging, rearing the queen for reproduction, hive construction, and scouting for new potential nest sites during swarming. The scout bees in the swarm are among the most experienced, and correspondingly oldest, forager workers . The focus of this chapter is on the decision-making dynamics of nest site selection among these scout bees during the colony swarming process. The book by Seeley is an excellent reference for an in-depth look at all of the other captivating elements of honeybee colonies, in addition to the swarm decision-making discussed here.
Honeybee colonies reproduce by casting swarms, each of which comprises a queen accompanied by several thousand worker bees. A small fraction of the worker bees are known as scout bees and perform the task of locating suitable nest sites for the ∗While we use the term honeybee quite generally, the experimental evidence for the features of honeybees and swarms described in this chapter comes from Apis mellifera, the most commonly domesticated species .
103 swarm by engaging in a democratic decision-making process of choosing among several competing options. This process involves the famous waggle dance in which scouts advertise the location and quality of suitable nest sites by performing a distinctive dance on the surface of the swarm (the waggle dance is also used in foraging to signal flower locations). The collective nest site selection process among the scouts is known to have several of the typical elements of decentralized decision-making systems: commitment of individuals to options after sampling, recruitment of uncommitted individuals by those already subscribed to an option, and decay of preference affiliation over time (akin to the ‘leaking’ in leaky-accumulator models of neuronal decision-making ). Seeley et al. in a recent paper have shown that scouts also send inhibitory stop signals to other scouts advertising alternative nest sites, thereby causing these scouts to cease dancing. This cross-inhibitory process has been shown to be critical to the ability of swarms to make decisions effectively, particularly when choosing between competing options of near-equal value.
In this chapter we study bifurcations in an analytical model of honeybee swarm decision-making from and illustrate the critical role played by stop-signal inhibi-tion in enabling swarms to manage the speed-accuracy tradeoffin the decision-making problem. The model from is described in §7.1. §7.2 summarizes the analysis of the symmetric case of options with equal value from and illustrates a pitch-fork bifurcation in the dynamics.
In §7.3 we present a general description of the bifurcations of the model for the asymmetric case of options with unequal value, and illustrate how the pitchfork bifurcation in the symmetric case is one slice through a more general cusp catastrophe bifurcation set corresponding to the dynamics. In §7.4 we show that the model has two timescales and derive an analytical expression for the slow manifold of the dynamics. We study a stochastic version of the model in §7.5 and summarize our findings in §7.6.
7.1 Model Description We study a mean-field model of the swarm decision-making dynamics derived in .
We restrict to looking at the decision-making process of choosing between two options A and B for analytical tractability. Also, while scouts might advertise several options (on the order of 10 ) early on in the selection activity, only two or three options usually remain in the more advanced stages. The model has states yA, yB and yU corresponding to the fraction of scouts subscribed to (i.e., dancing for) option A, the fraction subscribed to option B, and the fraction uncommitted (U) to either option, 104 respectively. The vector y = [yA yB yU]T is a simplex vector satisfying yi ≥0 for all i ∈{A, B, U} and yA + yB + yU = 1. The dynamics have four main processes: • Commitment: Uncommitted scout bees commit to an option upon indepen-dently encountering one in the scouting process at rate γi for option i ∈{A, B}.
• Recruitment: Scout bees committed to a particular option recruit uncommitted scouts to subscribe to that option (using the waggle dance) at rate ρi for option i ∈{A, B}.
• Decay: The number of waggle dance circuits declines over time due to the fading interest of a committed scout at rate αi for option i ∈{A, B}. Decay has been shown to be essential to consensus forming in the decision-making dynamics .
• Stop-signal: By the cross-inhibitory mechanism, scouts subscribed to a partic-ular option inhibit those subscribed to alternative options at rate σi for option i ∈{A, B}. The stop-signal is known to reduce waggle dancing and is typically delivered as a vibrational signal by the sender butting her head against the dancer .
COMMITMENT RECRUITMENT DECAY STOP-SIGNAL U γA →A U γB →B A + U ρA →2A B + U ρB →2B A αA →U B αB →U A + B σA →A + U B + A σB →B + U ˙ yA = γAyU + ρAyAyU −αAyA −σByAyB ˙ yB = γByU + ρByByU −αByB −σAyByA Figure 7.1: Connections between the microscopic (top) and macroscopic (bottom) descrip-tions of the swarm decision-making dynamics. The set of four categories on the top show the four processes that are part of the decision-making dynamics (commitment, recruitment, decay, stop-signal), and the microscopic rate equations corresponding to each (for example U γA →A denotes the transition from U to A at rate γA). The bottom pair of macroscopic mean-field equations (7.1) is color-coded to match each of the four processes.
105 The microscopic rate equations for the four processes described above are shown in Figure 7.1 (top). The corresponding mean-field equations are given by dyA dt = γAyU + ρAyAyU −αAyA −σByAyB dyB dt = γByU + ρByByU −αByB −σAyByA.
(7.1) These equations have been derived rigorously in as a system-size expansion [139, 24] of the microscopic master equations . Note that ˙ yU = −˙ yA −˙ yB.
The nest sites of swarms are typically tree cavities that provide good protection from predators and from harsh environmental conditions. Scout bees evaluate poten-tial nest sites on a variety of metrics including cavity volume, cavity entrance size, height of the entrance from the ground, and the presence of earlier combs . Fur-ther, various studies [31, 29, 116] have shown that scout bees have an innate scale of absolute nest site goodness or value that encapsulates these metrics. We use vi > 0 to denote this absolute value of nest site i ∈{A, B}. Following the parameterization used previously in , we assume that all of the rates γi, ρi, and αi may depend on the value vi of the potential nest site with which they are associated. We set the commitment and recruitment rates γi = vi and ρi = vi, and the decay rates αi = 1 vi.
The parameterization in also assumes that the stop-signal rate σi is independent of site choice and hence σA = σB = σ. Substituting for these rates in (7.1) we obtain a pair of quadratic differential equations in three parameters vA, vB, and σ: dyA dt = −1 vA yA + vAyU(1 + yA) −σyAyB dyB dt = −1 vB yB + vByU(1 + yB) −σyAyB.
(7.2) Our analysis of bifurcations and timescale separation in the sections to follow focuses on this pair of coupled equations.
7.2 Symmetric Case The symmetric case of equal options vA = vB = v was studied in ; we summarize the results here. Substituting vA = vB = v in (7.2), the symmetric dynamics are 106 given by dyA dt = −yA v + vyU(1 + yA) −σyAyB dyB dt = −yB v + vyU(1 + yB) −σyAyB.
(7.3) The dynamics (7.3) have three sets of equilibria (the fourth equilibrium set is always outside the simplex ∆2) denoted (yAeq, yBeq) and given by p1 = 2v2 1 + v2 + √ 1 + 2v2 + 9v4 + 4v3σ, 2v2 1 + v2 + √ 1 + 2v2 + 9v4 + 4v3σ p2 = 1 2 −1 2v2 + 1 2v2 r (v2 −1)2 −4v3 σ , 1 2 −1 2v2 −1 2v2 r (v2 −1)2 −4v3 σ !
(7.4) p3 = 1 2 −1 2v2 −1 2v2 r (v2 −1)2 −4v3 σ , 1 2 −1 2v2 + 1 2v2 r (v2 −1)2 −4v3 σ !
.
These equilibria correspond to a pitchfork bifurcation in the dynamics with a bifur-cation set (parameterized by v): σ∗= 4v3 (v2 −1)2.
(7.5) For σ < σ∗, the equilibria p2 and p3 from (7.4) are imaginary and p1 is stable. A pitchfork bifurcation occurs at σ = σ∗resulting in two stable equilibria p2 and p3 and one unstable equilibrium p1 for σ > σ∗, as illustrated in Figure 7.2 (see also Case 1 in Figure 7.4).
The pitchfork bifurcation describes an important feature of the stop-signal based cross-inhibition on the decision dynamics. For values of stop-signal below the thresh-old σ∗, the only stable equilibrium of the dynamics is the symmetric (yA = yB) equilibrium resulting in a deadlock in the decision-making process with equal frac-tions of the population subscribed to each option. This situation is suboptimal from a speed-accuracy perspective since the time spent in deadlock between equal options does not yield any greater accuracy in decision-making. Ideally, we would like for the system to be able to spontaneously ‘flip a coin’ between the equal alternatives in order to make a fast decision. This is exactly what is enabled by the cross-inhibitory mechanism. Post bifurcation (σ > σ∗), the two stable equilibria correspond to a choice of one of the two alternatives, and the system dynamics are attracted to one or the other based on initial conditions, thereby breaking deadlock (see inset phase portraits in Figure 7.3).
107 Stop Signal σ 1 fixed point 3 fixed points A B U A B U Quality v v = 2, σ = 1 v = 3, σ = 5 σ∗= 4v3 (v2 −1)2 Figure 7.2: Pitchfork bifurcation set for the stop-signalling dynamics with equal alterna-tives (7.3). The dark region below the curve σ∗= 4v3 (v2−1)2 has a single stable equilibrium; the light region above the curve has three equilibria, two of which are stable. The pitchfork bifurcation for adequately high stop-signal σ breaks the deadlock in decision-making between equal alternatives. Insets show typical phase portraits in each region.
7.3 Asymmetric Case Looking back at the bifurcations for the N = 2 case of the replicator-mutator dynam-ics shown in Figure 3.4, we see that as symmetry is broken, the (structurally unstable) pitchfork bifurcation disintegrates into a saddle-node bifurcation and a stable branch of equilibria. These bifurcation plots correspond to slices through the cusp catastro-phe, plotted in Figure 2.6. In a qualitatively similar sense, the symmetric pitchfork bifurcation for the stop-signaling dynamics from §7.2 disintegrates into saddle nodes for vA ̸= vB, with a corresponding cusp catastrophe bifurcation set. Before we plot this bifurcation set, the following change of variables helps simplify notation: let mean quality v = vA+vB 2 and quality difference ∆v = vA −vB. Then vA = v + ∆v 2 and vB = v −∆v 2 .
(7.6) 108 Substituting (7.6) in (7.2) gives dyA dt = −2yA 2v + ∆v + v + ∆v 2 yU(1 + yA) −σyAyB dyB dt = −2yB 2v −∆v + v −∆v 2 yU(1 + yB) −σyAyB.
(7.7) The bifurcation set for the dynamics (7.7) is plotted for a chosen value of v in Figure A dominates B dominates bistability ∆v σ v = 5 Pitchfork bifur pt Saddle-Node bifur set Case 1: ∆v = 0 Case 2: ∆v = 0.25 Case 3: σ = 5 (i) (ii) (iii) (iv) (v) (v) (iv) (iii) (ii) (i) A B U ∆v Figure 7.3: Bifurcation diagram for the stop-signaling dynamics (7.7). The dashed red curves are the saddle-node bifurcation set for ∆v ̸= 0, the blue circle is the pitchfork bi-furcation point for ∆v = 0. The horizontal and vertical lines mark three bifurcation cases as indicated in the legend (see also Figure 7.4). Case 1 corresponds to a pitchfork bifurca-tion, Case 2 corresponds to a saddle-node bifurcation, and Case 3 corresponds to hysteresis (these can be seen clearly in Figure 7.4). The phase portraits marked (i)-(v) correspond to the squares marked on the ∆v −σ space. The bifurcation set is qualitatively identical to that of the cusp catastrophe shown in Figure 2.6. This plot corresponds to v = 5; plots for other values of v are qualitatively similar as shown in Figure 7.5.
109 7.3†.
The corresponding sets for other values of v are qualitatively identical (see also Figure 7.5). The set divides the equilibria of the dynamics into three regions: bistability with two stable equilibria close to the A and B corners of the simplex and an unstable central equilibrium; A dominates with a single equilibrium such that yAeq > yBeq; B dominates with a single equilibrium such that yBeq > yAeq. Figure 7.3 also shows typical phase portraits of the dynamics at various points in the ∆v −σ parameter space, illustrating the pitchfork and saddle-node bifurcations.
σ σ ∆v Projected Equilibria Projected Equilibria Projected Equilibria v = 5 ∆v = 0 v = 5 ∆v = 0.25 v = 5 σ = 5 Case 1 Case 2 Case 3 Figure 7.4: Bifurcations of the three cases from Figure 7.3. The vertical axis on each plot is given by 1 2 + yAeq−yBeq 2 , and corresponds to equilibria of the dynamics projected orthogonally onto the yU = 0 boundary of the simplex. Blue curves correspond to stable sinks and red curves to unstable saddles.
Figure 7.4 shows bifurcation plots of the dynamics corresponding to the three cases (slices of the cusp catastrophe bifurcation set) from Figure 7.3. The two-dimensional equilibria of the dynamics denoted (yAeq, yBeq) are plotted in one dimension (vertical axis ‘Projected Equilibria’) in Figure 7.4 by projecting each equilibrium orthogonally onto the yU = 0 boundary of the simplex. This projection is given by 1 2 + yAeq−yBeq 2 .
Figure 7.4 clearly shows the pitchfork bifurcation (Case 1), saddle-node bifurcation (Case 2), and hysteresis (Case 3) of the cusp catastrophe.
The full bifurcation set of the dynamics (7.2) in three parameters vA, vB and σ is plotted in Figure 7.5. The bifurcation set in Figure 7.3 is a slice through this three-parameter set corresponding to v = vA+vB 2 = constant (i.e., orthogonal to ∆v = 0).
The bottom two plots in Figure 7.5 are level curves of the three parameter bifurcation set corresponding to the σ −∆v (bottom left) and v −∆v (bottom right) planes.
The vertical axis in the level curve plots titled |∆v| corresponds to the minimum †Analytical expressions for the equilibria of the dynamics in the asymmetric case are cumbersome to write down, but solvable using symbolic manipulation software. The bifurcation set is computed using analytical calculations of the equilibria of the dynamics and an iterative algorithm to determine the boundary of the set. This algorithm is described in Appendix G.
110 absolute difference in alternatives necessary for a unique equilibrium (A dominates or B dominates). Hence the region of parameter space below each level curve has bistability, and the region above has a unique stable equilibrium close to the correct alternative (the one with higher value).
Bistable σ∗= 4v3 (v2 −1)2 σ vB vA vA vB Stop Signal σ Minimum |∆v| v = 3 v = 4 v = 5 v = 6 Minimum |∆v| σ = 1 σ = 2 σ = 4 σ = 8 Average Value v σ v = 6 σ = 4 Bistable Figure 7.5: Bifurcation set for the dynamics (7.2) in three parameters vA, vB and σ. The top panels show two views of the three-dimensional bifurcation set. A vertical cut through this set orthogonal to the line vA = vB ⇐ ⇒∆v = 0 gives the two-dimensional set shown in Figure 7.3. Hot colors correspond to higher values of σ. The bottom left plot shows level curves corresponding to a series of vertical slices of the the three-dimensional set orthogonal to ∆v = 0. The bottom right plot shows level curves corresponding to a series of horizontal slices of the three-dimensional set parallel to the vA −vB plane.
There are four key decision-making features that are apparent in the bifurcation plots in Figures 7.3-7.5. First, while bistability is favorable for ‘fast’ decision-making when alternatives have near-equal value (plot (iii) in Figure 7.3), it is not favor-able when alternatives are adequately different (|∆v| adequately large) since the ‘dis-tracting’ attractor close to the incorrect low value alternative can lead to errors in decision-making (large σ Case 2 in Figure 7.4).
111 Second, for a given mean value v, the minimum |∆v| necessary to precipitate a bifurcation in the dynamics resulting in a single attractor increases, with increasing stop-signal σ (Figure 7.5 bottom left); i.e. high-values of stop-signal make the col-lective decision dynamics less sensitive to differences between options. Another way of interpreting this is that for a high value of stop-signal, a high magnitude of differ-ence in options is necessary (low sensitivity) to prevent the swarm from picking the incorrect low-value alternative by falling into the distracting attractor.
Third, for a given value of stop-signal σ, the minimum |∆v| for a single attractor grows asymptotically linearly with increasing v (Figure 7.5 bottom right). This is sim-ilar to Weber’s law of just-noticable differences from psychology which states that ‘minimum difference in stimulus intensity required to discriminate between sources varies linearly with average intensity’ .
Fourth, the level of stop-signal σ introduces a tradeoffin the collective dynamics between speed and sensitivity. High values of stop-signal enable fast dynamics when alternatives are of nearly equal value, but make the collective insensitive to differences between alternatives. Lower values of stop-signal promote sensitivity, at the cost of potentially expensive deadlock when alternatives are nearly equal. This tradeoffcan potentially lead to intermediate evolved levels of stop-signal cross-inhibition in the decision-making dynamics.
7.4 Separation of Timescales The phase portraits in Figure 7.3 show a clear separation of timescales in the dynamics 7.7. In phase portrait (iii) for example, trajectories converge quickly onto the slow manifold corresponding to the heteroclinic connections between the central saddle and the boundary stable equilibria. The dynamics then evolve slowly on (near) this slow manifold. In Appendix F, we use singular perturbation theory to compute a rigorous analytical approximation to the slow manifold of the dynamics; these results are summarized below.
Singular perturbation theory requires the identification of a small parameter in the model and a coordinate transformation that converts the dynamics into the singular perturbation ‘standard form’. In Appendix F, we use ϵ := 1/v as the small parameter (thereby assuming large v) and convert the dynamics to the standard form using the 112 coordinate transformation (yA, yB) 7→(x, z) given by z = (1 + yA)(1 + yB) x = 1 + yA 1 + yB .
We then go on to show that the slow manifold is given analytically by the implicit expression σ 2vyAyB = yU(1 + yA)(1 + yB) 3 −yU (7.8) in the (yA, yB) coordinates, and by the explicit expression z = dˆ x2 + 6 + √ D 2(2 + d)ˆ x !2 =: h(x), (7.9) where d = σ v , ˆ x = √x + 1 √x and D = (dˆ x2 + 6)2 −4dˆ x2(2 + d), in the (x, z) coordinates.
In Figure 7.6, we plot a comparison of the analytically derived slow manifold (7.8) (or equivalently (7.9)) and trajectories of the dynamics (7.7) for various combinations of parameters σ and ∆v, for large v (taken to be v = 10 for this plot). The match between the approximation and the trajectories is excellent, except for the case when both ∆v and σ are large. This case violates the limiting conditions of the singular perturbation calculation, as discussed in Appendix F.
In Appendix F we also compute the slow-timescale dynamics on the slow manifold and show that the equilibria of these dynamics match those of the two-dimensional system (7.7) in certain limits (symmetric vA = vB, σ →0 and σ = v). A more general analysis of the slow dynamics and their equilibria is a topic of future work.
One of the main reasons to pursue the timescale separation results discussed in this section is to attempt to reduce the two-dimensional system (7.7) to a one-dimensional description. This reduction is particularly relevant when comparing the swarm dy-namics described here with classical models of binary decision-making which are often in the form of one-dimensional stochastic differential equations (see the description of the drift diffusion model in §2.4 for example). We leverage the slow manifold calcu-lation made here to explore this connection in more detail in a paper in preparation .
113 ∆v = 0 ∆v = 0 ∆v = O(1) ∆v = O(1) ∆v = O(v) ∆v = O(v) σ = O(1) σ = O(1) σ = O(1) σ = O(v) σ = O(v) σ = O(v) Figure 7.6: Comparison between the analytically computed slow manifold h(x) plotted in magenta and simulations of the stop-signaling dynamics (7.7). The match between the analytical slow manifold and the simulations is excellent, except for the case ∆v = O(v), σ = O(v). For this set of plots, v = 10, O(v) ≡10 and O(1) ≡1.
7.5 Stochastic Dynamics Our analysis thus far has focused on bifurcations and timescale separation of the deterministic mean-field decision-making model (7.1). It is important to recognize, however, that the evaluations made by individual scout bees of nest site values vi are inherently noisy. Experiments have shown that there is much overlap in distributions of dance strength (and correspondingly judgements of value) between scouts advertis-ing a medium-quality and a high-quality nest site during the decision-making process . Despite this noisy individual reporting of quality, there are clear differences in mean qualities reported, resulting in a sharp distinction at the swarm level.
In this section, we move beyond the deterministic model (7.1), and build a stochas-tic model to capture this inherent variability. The dynamics of this model can then be compared to other standard stochastic models of decision-making, such as the drift diffusion model (Figure 2.10(c)) which has been shown to optimally address the speed-accuracy tradeoffin two-alternative forced choice tasks . We assume that the rates γi, ρi and αi in (7.1) that depend on the stochastic measurement vi are each subject to variability. Similar to the setup in , we add independent white noise terms of variance k2 to each of these rates (for example αi 7→αi + kηαi where ηαi 114 is a Gaussian random variable with mean 0 and variance 1) to obtain the stochastic decision-making dynamics dyA = (γAyU + ρAyAyU −αAyA −σByAyB) dt + k q y2 U + y2 A + y2 Uy2 A dW dyB = (γByU + ρByByU −αByB −σAyByA) dt + k q y2 U + y2 B + y2 Uy2 B dW.
(7.10) The parameter k in (7.10) sets the level of stochasticity or noisiness in the decision-making process; higher values of k correspond to noisier evaluations.
In Figure 7.7 we simulate the stochastic dynamics (7.10) using the parameteriza-tion used previously, γi = vi, ρi = vi, and αi = 1 vi, for a finite time interval t ∈[0, 30].
Parameters v, ∆v and σ are chosen to correspond to the three cases shown in Figures 7.3 and 7.4. Each simplex in Figure 7.7 has 50 trajectories initialized at the origin.
The dark dashed lines plotted are quorum decision thresholds (yi = 0.7, i = A, B); a decision for a particular alternative A or B is assumed to be made when a trajectory first crosses the corresponding threshold. The simulations illustrate some key features of the stochastic decision-making dynamics: • Deadlock and speed (Case 1): This case corresponds to equal alternatives ∆v = 0. At low values of stop-signal σ, the population is in a state of deadlock with a majority of trajectories (75%) unable to reach a decision boundary (top row left). At intermediate levels of stop-signal, the bifurcation yields a much more successful decision-making outcome with most (82%) of the trajectories successfully breaking deadlock (top row middle). Speed increases (i.e. deci-sion time shortens) at higher levels of stop-signal with all trajectories breaking deadlock (top row right).
• Intermediate stop-signal for accuracy (Case 2): This case corresponds to un-equal alternatives ∆v = 0.25.
At low values of stop-signal, the population is in a state of near-deadlock with a single attractor slightly closer to corner A than B; 60% of the trajectories do not reach a threshold (center row left).
At intermediate levels of stop-signal, the bifurcation yields excellent decision-making performance with a vast majority (98%) of the trajectories reaching the correct threshold (center row middle). At higher levels of stop-signal, the increased speed of decision-making comes at the expense of poor accuracy; only 70% of trajectories reach the correct threshold (center row right). This perfor-mance degrades with increasing σ, illustrating a classical speed-accuracy trade-115 Case 1 Case 2 Case 3 Figure 7.7: Simulations of the stochastic dynamics (7.10) with parameterization γi = vi, ρi = vi, αi = 1 vi , k = 0.1, and for time t ∈[0, 30]. Each simplex shows 50 trajectories initialized at the origin yA = yB = 0 (correspondingly yU = 1).
Circles are equilibria (filled stable, hollow unstable) of the deterministic dynamics (7.7), and the dashed lines are quorum thresholds yA = 0.7 and yB = 0.7. The inset bar plots show the fraction of trajectories reaching each threshold; ‘none’ indicates trajectories reaching neither threshold in the given time (plotted in cyan). For ∆v = 0, trajectories reaching either threshold are plotted in blue. For ∆v ̸= 0 trajectories reaching the correct threshold (one with higher vi) are plotted in green, those reaching the incorrect threshold are in red. Parameter values appear below each plot.
off. However, unlike the speed-accuracy tradeofffor the linear DDM model (also §2.4), here the tradeoffcomes from nonlinear dynamics.
• Hysteresis (Case 3): The decision swings from B, to bistability, to A, as ∆v varies from −2 to 2. Such continuous variations in quality difference ∆v are unrealistic in a swarm decision dynamics, but may be more reasonable in other 116 decision systems (some examples are discussed in ). Experiments and anal-ysis of neuronal decision-making models, for example, have shown that strong stimuli (similar to large ∆v in this case) result in both faster and more accurate choices than weaker stimuli [38, 109, 99].
A detailed analysis of the speed-accuracy tradeoffdescribed in Case 2 above is a topic on ongoing work.
We are also in the process of investigating the role of parameter variations during the decision-making process for systems with two or more alternatives. Two such examples are shown in Figure 7.8: • Breaking symmetric deadlock (Figure 7.8 left): The decision-making process is between two equal alternatives with value vA = vB = 3. The stop-signal σ ramps up linearly from an initial value of σ = 0 to a final value of σ = 3.
The increasing stop-signal enables the deadlock to be broken when σ > σ∗≈1.7 with one option randomly winning out. This simulation is motivated in part by the fact that elapsed time might influence the computations that underlie decision processes since prolonged deliberation can be expensive .
In models of neuronal decision making [38, 11, 12], time-dependent ‘urgency’ signals (potentially analogous to the time-dependent stop-signal simulated here) have been suggested as a way to impose a soft deadline on deliberation.
• Deadlock enables better final outcome (Figure 7.8 right): For a system tuned to intermediate levels of stop-signal (σ = 1 here), options of low quality result in deadlock. This enables the system to wait for other potentially better options to arrive into the mix and precipitate a decision. In this simulation, a third high-quality option (vC = 4) enters at t = 30 and dominates.
7.6 Discussion The decision-making process in honeybee swarms is rooted in the requirement that the swarm comes to a unanimous consensus decision about a suitable future home.
The unanimity of the decision is critical because each swarm has only one queen, and because fragmentation of the swarm in the migration process to the new nest can be extremely costly, and potentially fatal. Experiments have shown that swarms are not only able to come to this consensus decision, but nearly always pick the best possible option of those available. This is remarkable given that the swarm decision-making process is completely decentralized and inherently stochastic.
117 Time Population Frequency yi σ v = 3 ∆v = 0 σ∗≈1.7 Time Time Population Frequency yi Figure 7.8: Simulations of the stochastic dynamics. (left) A deadlocked population is able to converge to a decision by picking one of two equal alternatives by slowly ramping up the stop-signal; the critical value of stop-signal for the pitchfork bifurcation is marked on the bottom plot. (right) For time t < 30 the population is deadlocked while choosing between two low-value (v = 1) equal alternatives; a third high-value alternative (vC = 4) enters the picture at t = 30, precipitating a decision in favor of this alternative.
As discussed in , there are two hypotheses for building consensus in a system.
The first is known as the compare and convert hypothesis and corresponds to agents comparing their current state to neighbors and switching states to match those of neighbors with better outcomes.
This is reminiscent of the imitation mechanism of the replicator-mutator dynamics in Chapters 3 and 4. The second is known as the retire and rest hypothesis in which agents have a decaying (leaking) interest in an option, thereby making each participant highly flexible in the decision-making process.
Experiments have shown that scouts in a honeybee swarm do not directly compare the options available, but instead rely on their decentralized interactions (recruitment, commitment, etc.) to produce a consensus outcome. It has also been shown that scouts decrease the number of dance circuits performed over time resulting in a de-cayed interest in the corresponding option. In addition to this decay, in this chapter we have shown the critical role played by the stop-signal cross-inhibitory mecha-nism in enabling effective swarm decision-making. Stop-signaling enables breaking deadlocks between equal alternatives and tunes the sensitivity of the dynamics to differences between alternatives. We prove a separation of timescales in the decision-118 making model, thereby enabling a reduction of the dynamics to a one-dimensional system. Leveraging this reduction to make detailed comparisons between the stochas-tic swarm dynamics derived here, and other classical models of neuronal collective decision-making, is an important avenue of future work.
119 Chapter 8 Final Remarks This thesis studies emergent collective behavior in selected biological systems from the perspective of evolution by natural selection. Our main aim has been to understand the mechanisms that endow biological collectives with extraordinary robustness and adaptability, and to leverage this understanding to inspire effective decision-making and collective motion protocols in artificial decentralized systems. We study four related topics: replicator-mutator dynamics, collective migration, pursuit and evasion, and swarm decision-making.
The main conclusions from each of these topics are presented in §8.1. §8.2 comprises some of the common themes between the topics that have emerged through the analysis. We look ahead to future directions of investigation in §8.3.
8.1 Conclusions Replicator-mutator dynamics: Much of the existing analysis of the replicator-mutator dynamics has focused on stable equilibria. The analysis in the literature has also primarily considered payoffand mutation matrices that are symmetric, which cor-respond to undirected payoffgraph topologies. Here we prove conditions such that stable limit cycles in the replicator-mutator dynamics arise as a consequence of Hopf bifurcations for N ≥3 strategies and circulant payoffmatrices. From a graph the-oretic perspective, we show how breaking symmetry by considering directed graphs allows for oscillatory limiting behavior. We emphasize that the limit cycles are not restricted to circulant payoffs, but can exist for more general noncirculant cases as well. The simulations in Chapter 4 illustrate the structural stability in the dynamics to perturbations of circulant payoff, and show the tight connections between embed-120 ded directed cycles in the payoffgraph and the existence of stable limit cycles of the dynamics.
Collective migration: The study of leadership has received significant attention in both biology and multi-agent robotics. Agent-based simulations and experiments with fish schools have shown that a small group of leaders is capable of guiding the motion of a large group of followers.
The ability of followers to leverage the investments made by leaders and essentially operate as free-riders in the system is an evolutionary paradox. In this work we focus on the role that the social graph connectivity plays in the evolutionary dynamics of a networked model of collective migration. We use tools from adaptive dynamics to study the all-to-all limit of the evolutionary model and derive bounds for branching of the population into leader and follower groups. For limited connectivity, we prove necessary and sufficient conditions for convergence of the noise-free migration model, and show that fitness of individuals in the stochastic model can be derived analytically using the Lyapunov equation. For random networks and lattices, we show a minimum connectivity bound that yields evolutionary population branching into leaders and followers. We also study a simple model of greedy adaptive nodes on small networks, inspired by collective robotic systems. We show that the network topology plays a critical role in determining the location of leaders in the adaptive system; we study bifurcations in node leadership as a function of investment costs.
Pursuit and evasion: Building on previous work that focuses on the evolutionary dynamics of three pursuit strategies playing against an environment of nonreactive evaders, we study an evolutionary game of three strategies of pursuit against three strategies of evasion, two of which are reactive. Monte-Carlo simulations of the evo-lutionary dynamics show convergence to a stable equilibrium of classical pursuit and classical evasion. Using the structure of the fitness matrices observed in these sim-ulations, we analytically prove the convergence of replicator dynamics to the same classical pursuit and classical evasion equilibrium as in our Monte-Carlo simulations.
We then go on to incorporate the winning pursuit and evasive strategies in a novel collective motion scheme and show conditions for the convergence of the dynamics to circular formations and more complex weaving patterns.
Swarm decision-making: We study bifurcations in a mean-field model of honeybee swarm decision-making and illustrate the critical role played by stop-signal cross-inhibition in enabling swarms to break deadlocks between equal alternatives and manage sensitivity to differences between unequal options. We carefully illustrate the cusp catastrophe bifurcation set of the dynamics and prove a timescale separation that 121 reduces the model to a one-dimensional system. We derive a stochastic version of the model and study its dynamics using simulations. These simulations illustrate the role of stop-signalling in managing the speed-accuracy tradeoffin the swarm decision dynamics.
8.2 Common Themes In the introduction to the thesis, we made an effort to connect the various topics studied by looking at the systems from a robustness vs. adaptability perspective, as summarized in Figure 1.1. In this section, we identify some of the common themes that have emerged in our investigations of each topic. These themes are bifurcations, timescale separation, consensus and hysteresis.
Bifurcations: Bifurcations feature prominently in this thesis and include Hopf bifurcations studied in Chapters 3 and 4, cusp catastrophes studied in Chapter 7, bifurcations of collective dynamics in Chapter 6, and more complicated bifurcations of numerous equilibria as a function of graph topology in Chapter 5. These bifurca-tions correspond to macroscopic models of collective dynamics and represent signifi-cant changes in macroscopic emergent behavior as a consequences of relatively small changes in microscopic parameters governing inter-agent interactions. The nonlinear relationship between the microscopic and macroscopic descriptions that yields such bifurcations is an important hallmark of emergent systems.
Timescale Separation: Evolutionary models such as those studied in Chapters 3-6 inherently have two timescales. The fast timescale in these models corresponds to fitness computations as a function of ecological interactions. The slow timescale corresponds to the evolutionary process that modifies traits in the population as a result of fitness dependent replication and mutation. For example, for the migration problem in Chapter 5, the fast timescale corresponds to the stochastic migration dynamics and the slow timescale to the evolutionary change of agent strategies or investments. A very different kind of timescale separation is shown in Chapter 7. Here the fast timescale corresponds to individuals being attracted to an invariant decision manifold which corresponds roughly to the recruitment of uncommitted individuals to the decision making process. The slow timescale represents the dynamics along this manifold towards a collective decision.
Consensus: Consensus is critical to the functioning of most multi-agent coopera-tive systems and has inspired a vast literature in engineering, physics, mathematics and other fields.
In fact, most systems (even those with complicated inter-agent 122 dynamics) that have emergent coordinated behavior can be approximated to first order by some form of simple linear consensus dynamics. In this thesis, consensus appears in the zero-mutation limit of the replicator-mutator model in Chapters 3 and 4, the stable OU process of the migration dynamics in Chapter 5, and the successful decisions of the swarm dynamics in Chapter 7.
Hysteresis: Hysteresis is a remarkable feature that has appeared several times in this thesis. In a basic sense, hysteresis corresponds to a delay associated with the restoration of a macroscopic state of a system with respect to variations in a bifur-cation parameter. Specific parameters of the replicator-mutator Hopf bifurcations yield hysteretic curves (Figure 3.10(b) in Chapter 3 for example). Hysteresis was also observed for the adaptive evolutionary dynamics of migration in Chapter 5 and as part of the cusp catastrophe for the swarm dynamics in Chapter 7. In the context of animal behavior, hysteresis has been associated with collective memory. This refers to the phenomenon in which previous history of group structure influences collective be-havior as individual interactions change, even though individuals have no knowledge of what that history is . Figure 8.1 is a composite plot of the various hysteretic curves shown throughout the thesis.
µ xi ∆v Projected Equilibria 0 1 c k∗ Figure 8.1: Three examples of hysteresis in the thesis.
(left) from Figure 3.10(b) in Chapter 3, (center) from Figure 5.3 in Chapter 5, (right) from Figure 7.4 in Chapter 7.
The shaded rectangles in each plot mark regions of bi(multi)-stability.
8.3 Looking Ahead In a broad sense, the focus of robotic design and engineering over the past few decades has been on creating robust platforms that are configured to solve specific problems reliably, often with provable guarantees on performance.
The rapid expansion of communication, distributed sensing, networking, learning algorithms, and advanced 123 software, however, have enabled a new generation of intelligent and interconnected multi-agent systems. These systems are often comprised of individual platforms or agents that are simple and robust, but can have collective emergent group behavior that is highly complex.
Two areas that are currently undergoing this transition from the focus on the individual agent, to the focus on both the individual as well as the collective, are autonomous driving for cars on the highway, and collective swarm robotics. In au-tonomous driving [138, 52], one of the main challenges involves the design of control protocols that leverage inter-agent interactions (sensing and communication between vehicles) to produce optimal group-level outcomes (efficient traffic flow and safe high-way maneuvers). The alignment of individual interests (“I want to get there fast”) and group outcomes (overall traffic flow) poses an important challenge in such systems, and relates back to the evolutionary paradox of cooperation in natural swarms.
The developing area of swarm robotics focuses on using hundreds of relatively cheap and expendable robotic platforms that have limited sensing and communica-tion capabilities, to perform collective tasks with a high degree of parallelism [79, 114].
These systems are designed to have significant flexibility and adaptability for appli-cations such as foraging for information in dynamic and hazardous environments, distributed sensing, and distributed task allocation (the Robobees project at Harvard is a nice example [150, 79]).
A key challenge in this area is the design of inter-agent control laws that yield provable collective solutions while still maintaining the adaptability and scalability of the system.
The impressive robust and adaptive behavior of biological collectives serves as an important source of inspiration in the design of bio-inspired algorithms for artifi-cial multi-agent systems, including the two examples above. This is because natural collectives possess several attributes that are highly desirable for these artificial sys-tems.
The honeybee swarms studied in Chapter 7 are an excellent example: the decision-making in these swarms is completely decentralized, extensively accurate, highly adaptive, and robust to noise and disturbances. The analysis in Chapter 7 shows that cross-inhibition is a critical component of effective decision-making dynam-ics in honeybee swarms. We propose that cross-inhibition is a potentially important ingredient (along with recruitment, commitment and decay) for effective decision-making in robotic swarms as well. Hence, the study of cross-inhibition in the context of robotic swarms is a topic for future investigation.
Another important feature of natural collectives is the ability to rapidly transi-tion between different regimes of behavior (collective foraging to prey evasion, for 124 example) without any centralized control. In macroscopic models, this feature mani-fests itself as bifurcations in the dynamics of the system as a function of usually one or two parameters. For the replicator-mutator model studied in Chapters 3 and 4, for example, bifurcations as a function of mutation strength µ produce limit cycles.
These bifurcations can be interpreted in the context of artifical multi-agent system as a transition from exploitation of a single option, to the cyclical examination of options, to the exploration of all options, as discussed in the Chapters 1 and 3, and illustrated in Figure 8.2. This ability to shape the macroscopic behavior of a collective system (comprising large numbers of agents) by tuning a single parameter is useful for applications. A detailed examination of the bifurcations studied in this thesis, applied to artificial collective systems, is an important future direction.
µ µC1 µC2 Exploit Explore Cyclic Domination Figure 8.2: Exploration vs. exploitation. As bifurcation parameter µ increases from µ = 0, the dynamics transition from exploitation (only one dominant option µ < µC1), to cyclical domination of options µ ∈(µC1, µC2), to exploration (all options have equal fractions of agents µ > µC2). Adapted from Figure 3.5.
At the end of Chapter 5 we studied a simple model of greedy adaptation of nodes on a network and showed the emergence of leadership through bifurcations in the adaptive dynamics as a function of cost.
We also showed the critical role played by interconnection topology in determining the locations of leaders in the network.
We did not address, however, the connection between the equilibria of the adaptive process, and optimal group solutions. This is an area worth investigating to help elu-125 cidate the connections between a local bottom-up process such as greedy adaptation and a global top-down optimal design approach. The fundamental tradeoffhere is that local approaches, while being highly adaptive, reconfigurable to compensate for loss of agents, and computationally inexpensive, can come at the cost of suboptimal group performance. Understanding this tradeoffcarefully, especially as a function of graph topology, is another potential future direction.
The rich variety of collective dynamics in swarms, flocks, schools and herds has in-spired a generation of scientists and engineers. Recent advances in experimental and computational technology have enabled a careful examination of the mechanisms that produce the observed rich emergent behavior, including through the lens of evolution.
The ongoing quest to better understand the interaction patterns in biological collec-tives, coupled with innovative bio-inspired ideas for artificial multi-agent systems, will continue to be an exciting area of research going forward.
126 Appendix A Calculation of Lyapunov coefficient As stated in Theorem 2.2, the sign of the first Lyapunov coefficient ℓ1|(x0,µ0) evaluated at the fixed point x0 and bifurcation point µ0 determines the criticality of the Hopf bifurcation. What follows are the expressions for calculating ℓ1|(x0,µ0) as presented in .
Consider the N-dimensional dynamical system ˙ x = f(x, µ) where x ∈RN and µ ∈R. Let A0 = Dxf|(x0,µ0), where x0 ∈RN, µ0 ∈R. A0 has two purely imaginary complex conjugate eigenvalues, given by ±i ω0, where ω0 > 0. Define T1, T2 and T3 as T1 = ⟨p, C (q, q, q)⟩ T2 = p, B q, (2i ω0 −A0)−1 B (q, q) T3 = −2 p, B q, A−1 0 B (q, q) .
Here ⟨r, s⟩= r · s is the complex inner product between two complex vectors, and q and p are respectively the normalized eigenvector and adjoint-eigenvector of A0 satisfying A0q = i ω0q, AT 0 p = −i ω0p, and normalization ⟨p, q⟩= 1. B and C are high dimensional tensors given by B (r, s) = B1 (r, s) B2 (r, s) .
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BN (r, s) , Bi (r, s) = X k,l ∂2fi ∂xk∂xl x=x0 rksl C (r, s, t) = C1 (r, s, t) C2 (r, s, t) .
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CN (r, s, t) , Ci (r, s, t) = X k,l,m ∂3fi ∂xk∂xl∂xm x=x0 rksltm 127 The first Lyapunov coefficient ℓ1|(x0,µ0) is given by ℓ1|(x0,µ0) = 1 2ω0 Re(T1 + T2 + T3).
Appendix B Supporting material for Chapter 3 Lemma B.1. The divergence of the vector field g(x) restricted to the simplex ∆N−1 is given by ∇· g(x) x∈∆N−1 = ∇· h(˜ x) = 1T (1 −µ)B + ST x −xT NB + BT x, where S = Q ◦B, the element-wise product of Q and B.
Proof. The divergence is given by ∇· h(˜ x) = N−1 X i=1 ∂hi ∂xi = N X i=1 ∂gi ∂xi − N X i=1 ∂gi ∂xN .
(B.1) We substitute for gi(x) from (3.2) in the first term of the difference in (B.1) and using (3.1) and (3.3) we have N X i=1 ∂gi ∂xi = X i ∂ ∂xi " xi(fiqii −φ) + X j̸=i xjfjqji # = X i " fiqii + xiqii ∂fi ∂xi −φ −xi ∂φ ∂xi + X j̸=i xjqjibji # = (1 −µ)1TBx + (1 −µ) −Nφ −xT ∂φ ∂x + X i X j̸=i xjqjibji 128 = (1 −µ)1TBx + (1 −µ) −Nφ −xT(B + BT)x + X i X j̸=i xjsji = 1T (1 −µ)B + ST x −xT (N + 1)B + BT x.
(B.2) where the last equality follows by X i X j̸=i xjsji = 1TSTx −(1 −µ). Computing the second term in the difference in (B.1) and using N X i=1 qji = 1 we have N X i=1 ∂gi ∂xN = ∂ ∂xN " N X i=1 N X j=1 xjfjqji −xiφ # = ∂ ∂xN " (1 − N X i=1 xi)φ # = −φ = −xTBx.
(B.3) Substituting (B.2) and (B.3) in (B.1) we get the desired result.
Lemma B.2. Let B be circulant and invertible and define the row sums of B and B◦B as rB := N X j=1 bij and rB◦B := N X j=1 b2 ij respectively, for any row i. Then the divergence ∇· h(˜ x) ≤0 on the simplex ∆N−1 if µ ≥(N −rB)(rB −1) N (r2 B −rB◦B) for mutation (Q1) , µ ≥ (N −1)(N −rB) N (N + NrB −2rB) for mutation (Q2).
Proof. From Lemma B.1, the divergence ∇· h(˜ x) is negative semi-definite on the simplex if max x∈∆N− 1 1T (1−µ)B+ST x ≤min x∈∆N− 1 xT NB+BT x.
(B.4) The term on the left hand side (LHS) of (B.4) is the maximum of a convex combination of non-negative scalars and hence evaluates to LHS = max i N X j=1 (1 −µ)bji + sij = (1 −µ) max i N X j=1 bji + (1 −µ) + µ max i X j̸=i bijqij 129 = (1 −µ)(1 + rB) + µ rB◦B−1 rB−1 for mutation (Q1) (1 −µ)(1 + rB) + µ(rB−1) N−1 for mutation (Q2) (B.5) The term on the right hand side (RHS) of (B.4) is the minimum of a quadratic form that is positive on the simplex. Given that B is circulant and invertible (an N × N circulant matrix B of the form (3.1) is always invertible for N prime ), this quadratic form has an isolated minimum at xmix = 1 N 1. Thus, RHS = min x∈∆N−1 xT NB + BT x = N + 1 N 2 1TB1 = N + 1 N rB.
(B.6) Substituting (B.6) and (B.5) in (B.4) and some rearranging gives the desired result.
Appendix C Supporting material for Chapter 4 C.1 Proof of Lemma 4.2 Proof. To simplify notation in this proof, we denote the Jacobian Dxg|xmix,N as matrix A. Since A is circulant, its eigenvalues are given by (4.3). From (4.3) we see that the eigenvalue λk(A) cannot be complex when ωk N is real. To guarantee complex ωk N, let k be given by k ∈ {1, · · · , N −1} N odd {1, · · · , N −1} \ { N 2 } N even.
One can then verify that λN−k(A) = λk(A). Hence, as long as the λk(A) are complex, A has N−1 2 complex conjugate pairs of eigenvalues. Now we compute Im (λk(A)) 130 and hence obtain conditions for the existence of complex λk(A). The calculations differ slightly between mutation matrices (Q1) and (Q2) as shown below. We obtain a simplified expression for the imaginary component of the eigenvalues by grouping identical terms of A and using the identity N X j=1 ωjk N = 0.
• Mutation (Q1): Im (λk(A)) = (a12 −a1N) sin 2π N k N = 3, 4, 5 (a12 −a1N) sin 2π N k + (a13 −a1,N−1) sin 2π N 2k N ≥6 • Mutation (Q2): Im (λk(A)) = (a12 −a1N) sin 2π N k N ≥3 Substituting for the aij terms from (4.2), Im (λk(A)) = 0 ⇐ ⇒ (α −β) 1 −µ −µ (2+α+β) α+β = 0 mutation (Q1) (α −β) 1 −µ − µ N−1 = 0 mutation (Q2).
(C.1) The conditions of the Lemma follow from the expressions in (C.1).
C.2 Proof of Lemma 4.3 Proof. To simplify notation in this proof, we denote the Jacobian Dxg|xmix,N as matrix A. From Lemma 4.2, for r = 1, · · · , N−1 2 , λr(A) is complex. Using the notation a1j = γj + µηj we obtain Re (λr(A)) = N X j=1 γj cos 2π N (j −1)r + µ N X j=1 ηj cos 2π N (j −1)r which is zero if and only if µ = − " N X j=1 γj cos 2π N (j −1)r # " N X j=1 ηj cos 2π N (j −1)r #−1 =: µ0,r.
(C.2) Before we proceed, we need to establish that µ0,r is indeed well defined; that is, the denominator in (C.2) is non-zero. Let dr denote the denominator of µ0,r. For 131 N = 3, dr = 6(α + β + αβ) ̸= 0 for mutation (Q1) and dr = 3(4 + α + β) ̸= 0 for mutation (Q2); for N = 4, dr = 2(α + β + 2αβ) ̸= 0 for mutation (Q1) and dr = 4(2 + α + β) ̸= 0 for mutation (Q2). For N ≥5, by grouping identical terms, using the identity N X j=1 ωjr N = 0 and replacing the expressions for ηj in terms of α and β we have, dr ̸= 0 ⇐ ⇒ 2 (α + β) cos 2π N r −1 + 2αβ cos 2π N 2r −1 ̸= 0 mutation (Q1) cos 2π N r ̸= −2+α+β α+β mutation (Q2).
The conditions above can be verified to always hold given that the cosine function is bounded between −1 and 1 and α and β satisfy the conditions in (4.1).
Finally, we establish that if r, s = 1, · · · N−1 2 , r ̸= s then µ0,r ̸= µ0,s, i.e. the bifurcation points are distinct. If N = 3, 4, Lemma 4.2 establishes that there is only one bifurcation point. For N = 5, the two critical points can be shown to be distinct by a direct calculation. Here we show the distinctness of the critical points in the cases N ≥6. Using (C.2), µ0,r ̸= µ0,s ⇐ ⇒ PN j=1 γj cos 2π N (j −1)r PN j=1 ηj cos 2π N (j −1)r ̸= PN j=1 γj cos 2π N (j −1)s PN j=1 ηj cos 2π N (j −1)s .
(C.3) By grouping identical terms, using the identity N X j=1 ωjr N = 0, and replacing the expressions for γj and ηj in terms of α and β, we find that (C.3) ⇐ ⇒ (1 + α + β + 2αβ) (α + β) +2αβ cos 2π N r + cos 2π N s +2 (α + β) αβ cos 2π N r cos 2π N s ̸= 0 mutation (Q1) cos 2π N r ̸= cos 2π N s mutation (Q2).
(C.4) For mutation (Q1), the left hand side of the inequality in (C.4) can be bounded below by (α −β)2 + (α + β) > 0.
For mutation (Q2), the condition in (C.4) is equivalent to the initial hypothesis of r ̸= s. The distinctness result now follows.
132 C.3 Proof of Lemma 4.4 Proof. Here we compute the terms T1, T2 and T3 from Appendix A to obtain a simplified analytical expression for the first Lyapunov coefficient ℓ1|(xmix,N ,µ0,r). For a circulant matrix M ∈RN×N, let {(λk, vk)} be an eigenvalue–right eigenvector pair (Mvk = λkvk), where vk = h 1 ωk N ω2k N · · · ω(N−1)k N iT and λk (M) = N X j=1 m1j ω(j−1)k N .
(C.5) We compute ℓ1|(xmix,N ,µ0,r) as a function of the parameters α and β. From Appendix A, the Jacobian Dxg|(xmix,N ,µ0,r) is denoted by A0 with eigenvalue λr(A0) = i ˆ ω. Let ω0 = |ˆ ω|, t = r sign (ˆ ω), and q = vt. Note that A0q = i ω0q.
Computing T1 Direct calculation and simplification gives Ci (q, q, q) = −2N 2 + (α + β) ωt N + (α + β) ω−t N ω(i−1)t N .
Hence C (q, q, q) = −2N 2 + (α + β) ωt N + (α + β) ω−t N q, which leads to T1 = ⟨p, C (q, q, q)⟩= −2N 2 + (α + β) ωt N + (α + β) ω−t N ⟨p, q⟩ = −2N 2 + (α + β) ωt N + ω−t N .
(C.6) Computing T2 We compute B (q, q) = 2 1 + αωt N + βω−t N QTv2t.
Since Q is circulant, so is QT and v2t is a right eigenvector. Then B (q, q) = 2 1 + αωt N + βω−t N λ2t QT v2t.
For each eigenvector-eigenvalue pair v, λ of A0, direct calculation shows that 1/ (2i ω0 −λ) is the corresponding eigenvalue for the eigenvector v of (2i ω0I −A0)−1, 133 where I denotes the identity matrix. Then, (2i ω0I −A0)−1 B (q, q) = 2 1 + αωt N + βω−t N λ2t QT 2i ω0 −λ2t (A0) v2t.
Since B (x, κy) = κB (x, y) for any κ ∈C, then B q, (2i ω0 −A0)−1 B (q, q) = 2 1 + αωt N + βω−t N λ2t QT 2i ω0 −λ2t (A0) B (q, v2t) .
A calculation similar to that for B (q, q) gives Bi (q, v2t) = βω−2t N + αω−t N + 2 + βωt N + αω2t N N X j=1 qji ω(j−1)t N .
Hence B (q, v2t) = βω−2t N + αω−t N + 2 + βωt N + αω2t N λt QT q.
This implies that B q, (2i ω0 −A0)−1 B (q, q) = 2λt QT λ2t QT 2i ω0 −λ2t (A0) 1 + αωt N + βω−t N βω−2t N + αω−t N + 2 + βωt N + αω2t N q, and T2 = p, B q, (2i ω0 −A0)−1 B (q, q) = 2λt QT λ2t QT 2i ω0 −λ2t (A0) 1 + αωt N + βω−t N βω−2t N + αω−t N + 2 + βωt N + αω2t N .
(C.7) Computing T3 We show that B (q, q) = 0.
Bi (q, q) = −2 −(α + β) ωt N + ω−t N + 2 + (α + β) ωt N + ω−t N N X j=1 qji = 2 + (α + β) ωt N + ω−t N " −1 + N X j=1 qji # = 0, (C.8) 134 where the last equality comes from the fact that Q is a doubly-stochastic matrix.
This implies T3 = −2 p, B q, A−1 0 B (q, q) = −2 ⟨p, B (q, 0)⟩= 0.
Combining the previous expressions for T1, T2 and T3, the result follows.
C.4 Criticality analysis for Corollaries 4.1 and 4.2 In this section we establish that the Lyapunov coefficient at each of the d concur-rent Hopf Bifurcations of the equilibria xj,d,N is identical to that at the equilibrium xmix,N/d for the simple cycle payoffBC,N/d with β = 0 (i.e. BN/d,1). The mutation matrix used is (Q1). Let N/d = N1 and N2 = N −N1.
In order to simplify the calculations, consider the payoffmatrix ˆ BN,d given by ˆ BN,d = BN1,1 0N1×N1 · · · 0N1×N1 0N1×N1 BN1,1 · · · 0N1×N1 .
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0N1×N1 0N1×N1 · · · BN1,1 .
ˆ BN,d is obtained by relabeling the graph nodes corresponding to BN,d such that index labels for connected nodes are consecutive. The payoffgraph topology induced by ˆ BN,d is isomorphic to that of BN,d, see Figure 4.5. The dynamics (3.2), with payoff ˆ BN,d and mutation (Q1) have equilibria ˆ xj = h 0T N1(j−1) 1 N11T N1 0T N1(d−j) iT , which correspond to the equilibria xj,d,N.
The Jacobian of the system above evaluated at the equilibrium ˆ x1 is precisely MN,d in (4.8). Using this definition of payoff, we compute the first Lyapunov coefficient as described in Appendix A. We focus on equilibrium ˆ x1; the analysis for the other ˆ xj is equivalent.
The eigenvalues λk and eigenvectors vk of a circulant matrix are defined in (C.5).
Let the Jacobian MN,d evaluated at the critical point µ0,r (defined in Corollary 4.2) be denoted as ˆ A0 = A0 0N2×N1 0N1×N2 −1+α N1 IN2×N2 , 135 with eigenvalue λr (A0) = i ˆ ω.
Let ω0 = |ˆ ω|, t = r sign(ˆ ω), q = vt (A0), ˆ q = h qT 0T d(N1−1) iT and ⟨ˆ p, ˆ q⟩= 1. Note that A0q = i ω0q and hence ˆ A0 ˆ q = i ω0 ˆ q. Let Q be the (Q1) mutation matrix corresponding to BN,d and ˆ Q be the (Q1) mutation matrix corresponding to ˆ BN,d.
With these definitions, we follow the calculations in Appendix A and compute each of the terms T1, T2 and T3 given below. A comparison of each of these terms, to the corresponding terms in Appendix C.3 shows that the Lyapunov coefficient is identical (when N 7→N1, β 7→0 in (C.6), (C.7) and (C.8)).
T1 = ˆ p, C ˆ q, ˆ q, ˆ q = −2N1 2 + αωt N1 + αω−t N1 , T2 = ˆ p, B ˆ q, 2i ω0 −ˆ A0 −1 B (ˆ q, ˆ q) = 2λt QT λ2t QT 2i ω0 −λ2t (A0) 1 + αωt N1 αω−t N1 + αω2t N1 + 2 , T3 = −2 D ˆ p, B ˆ q, ˆ A−1 0 B ˆ q, ˆ q E = 0.
Appendix D Supporting material for Chapter 5 In this appendix we discuss the details of the adaptive dynamics analysis from Section 5.2. Define the function G(kR, kM) = k2 M + (1 −kM)2β2(1 −kR) 4(2k2 M −2kM + 1) + ck2 M.
Then the differential fitness from (2.3) is given by S = exp[−G(kR, kM)] −exp[−G(kR, kR)].
136 The selection gradient g(kR) is given by g(kR) = ∂S ∂kM kM=kR = −exp[−G(kR, kR)] kM(1 −kM)[1 + β2(kR −1)] 2(2k2 M −2kM + 1)2 + 2ckM .
Solving for the singular strategy condition g(k∗) = 0 gives the expression (5.19) k∗(1 −k∗)[1 + β2(k∗−1)] + 4ck∗(2k2 ∗−2k∗+ 1)2 = 0.
This expression has two sets of solutions that are plotted in Figure 5.3. One set corresponds to k∗= 0 and the other is defined implicitly by the equation c = (k∗−1)[1 + β2(k∗−1)] 4(2k2 ∗−2k∗+ 1)2 =: ˜ c(k∗).
(D.1) To determine conditions for evolutionary branching we compute ∂2S ∂k2 M kM=kR=k∗ = −exp [−G(k∗, k∗)] (1 −β2 + β2k∗) (3k∗−10k2 ∗+ 6k3 ∗) 2 (1 −2k∗+ 2k2 ∗)3 .
Hence the branching condition ∂2S ∂k2 M kM=kR=k∗> 0 corresponds to (β2(1 −k∗) −1) k∗(3 −10k∗+ 6k2 ∗) 2 (1 −2k∗+ 2k2 ∗)3 > 0.
(D.2) The zeros of function above in the range k∗∈[0, 1] are 0, 5− √ 7 6 , and 1 − 1 β2.
A derivative test shows that the condition (D.2) is satisfied for k∗∈ 0, 5− √ 7 6 .
The critical cost parameter c1 in Figure 5.3, corresponds to the maximum singular value k∗for branching and is given from (D.1) by c1 = ˜ c 5− √ 7 6 . The parameter c2 is determined by calculating the local maximum of the function ˜ c(k∗) as seen in Figure 5.3. We use the notation ˜ c(kcrit) = c2. kcrit can be calculated analytically and is given by a particular root of a cubic equation. The analytical expression for kcrit (and correspondingly c2) is cumbersome and hence is left out of this text. Nonetheless, the sketch in Figure 5.3 clearly conveys the main ideas.
Finally we discuss the convergence stability of the singular strategies. The con-vergence stability condition is given by (2.5). For the k∗= 0 singular strategy, ∂g ∂kR kR=0 = exp −β2 4 β2 −1 −4c 2 , hence ∂g ∂kR kR=0 < 0 ⇐ ⇒c > β2 −1 4 .
137 For the second singular strategy curve defined implicitly by (D.1), the derivative term in the convergence stability condition evaluates to ∂g ∂kR kR=k∗ = −k∗[3 −10k∗+ 6k2 ∗+ 2β2 (−1 + 5k∗−6k2 ∗+ 2k3 ∗)] exp [−G(k∗, k∗)] 2 (1 −2k∗+ 2k2 ∗)3 The interior root of ∂g ∂kR kR=k∗= 0 is precisely the value kcrit that maximizes ˜ c(k∗) (since c = ˜ c(k∗) ≡g(k∗) = 0); this root corresponds to c2 (see above). Hence one can verify that the singular strategies corresponding to the curve (D.1) are stable for k∗> kcrit and unstable for k∗< kcrit as shown in Figure 5.3.
Appendix E Supporting material for Chapter 6 E.1 Proof of Lemma 6.2 Proof. D is invariant with respect to the dynamics (6.6) since all boundaries of D are invariant (qi ∈{0, 1} = ⇒˙ qi = 0), and further q3(0) > 0 = ⇒q3(t) > 0 for all t > 0 since ˙ q3 = q3f3 qTf −q3 ≥−q3 = ⇒q3(t) ≥q3(0)e−t > 0.
Using the constraint qT13 = 1, we restate the dynamics (6.6) as a two-dimensional system, ˙ q1 = q1 b f (f31(t)(q1 −1) + f32(t)q2) ˙ q2 = q2 b f (f32(t)(q2 −1) + f31(t)q1) , (E.1) where f32(t) = f3(t) −f2(t) > 0, f31(t) = f3(t) −f1(t) > 0, and b f = qTf = f3 −q1f31 −q2f32. One can check that the only three equilibria of the system (E.1) in 138 ∆2 (given properties (1)-(3)) are the vertices (qeq1, qeq2) = (0, 0), (0, 1), (1, 0), of which (0, 0) is the only equilibrium in D.
Linearization about the (0, 0) equilibrium gives the dynamics, " ˙ q1 ˙ q2 # = " −f31/f3 0 0 −f32/f3 # " q1 q2 # .
This non-autonomous linear system in diagonal form can be solved easily as qi(t) = qi(0)exp − Z t 0 f3i(t) f3(t) dt , i = 1, 2.
For i = 1, 2, lim t→∞qi(t) = 0 and hence the (0, 0) equilibrium point of the non-autonomous system (E.1) is locally asymptotically stable by Theorem 4.13 of .
To prove that the invariant domain D is the region of attraction for the asymptoti-cally stable (0, 0) equilibrium point of (E.1), we use a Lyapunov function V = q1 +q2.
V is positive definite on D with a unique minimum: V = 0 ⇐ ⇒q1 = q2 = 0. We compute ˙ V = ˙ q1 + ˙ q2 = 1 b f (q1f31 + q2f32)(q1 + q2 −1) < 0.
Since ˙ V is negative definite on the domain D, by Theorem 4.9 of we have that D is the region of attraction for the equilibrium point qeq1 = 0, qeq2 = 0 (and qeq3 = 1).
Hence the equilibrium point qeq = h 0 0 1 iT is the asymptotically stable limit for all q(0) ∈D.
E.2 Lemma E.1 used in Theorem 6.1 Lemma E.1. Let q ∈∆2, qeq = h 1 0 0 iT and fP = Mq, where M = T # and T satisfies Conjecture 6.1. Then ∥q −qeq∥< ϵ = ⇒fP1 > fP2 and fP1 > fP3, where ϵ ≤min n 2(m11−m21) (m11−m21)+∥M∥1, 2(m11−m31) (m11−m31)+∥M∥1 o .
Proof. ∥q −qeq∥1 < ϵ = ⇒q1 > 1 −ϵ 2, q2 < ϵ 2 and q3 < ϵ 2. Suppose that ϵ ≤ 2(m11 −m21) (m11 −m21) + ∥M∥1 = ⇒ϵ < 2(m11 −m21) (m11 −m21) + (m22 + m23) = ⇒(m11 −m21) 1 −ϵ 2 > (m22 + m23)ϵ 2 139 = ⇒(m11 −m21)q1 > m22q2 + m23q3 = ⇒(m11 −m21)q1 > m22q2 + m23q3 −m12q2 −m13q3 = ⇒ h m11 m12 m13 iT q > h m21 m22 m23 iT q, or fP1 > fP2.
(E.2) Similarly one can show that ϵ ≤ 2(m11 −m31) (m11 −m31) + ∥M∥1 = ⇒fP1 > fP3.
(E.3) Combining (E.2) and (E.3) we get the desired result.
Appendix F Supporting material for Chapter 7 This appendix comprises the details of the timescale separation calculation for the stop-signalling dynamics (7.7), dyA dt = −2yA 2v + ∆v + v + ∆v 2 yU(1 + yA) −σyAyB dyB dt = −2yB 2v −∆v + v −∆v 2 yU(1 + yB) −σyAyB.
We perform a nonlinear coordinate transformation of the dynamics (7.7) and apply singular perturbation theory to separate timescales and derive an analytic expression for the slow manifold (heteroclinic connections). Assuming large v, define ϵ := 1 v as the small parameter for the timescale separation calculations. We follow the notation used in Chapter 11 of and apply Tikhonov’s Theorem (Theorem 11.1 in ) to prove the timescale separation.
140 Standard Singular Perturbation Model Consider the coordinate transformation (yA, yB) 7→(x, z) given by z = (1 + yA)(1 + yB) x = 1 + yA 1 + yB .
(F.1) The transformation is well-defined on the domain (yA, yB) ∈∆2 since the Jacobian of the linearization has non-zero determinant −2(1+yA) 1+yB on ∆2. The inverse transformation is given by yA = √zx −1 yB = rz x −1.
(F.2) Note that (yA, yB) ∈∆2 implies that z ∈ 1, 9 4 and x ∈ 1 2, 2 . Level curves in x, z coordinates on the simplex are illustrated in the figure below.
z = 2 z = 1.75 z = 1.5 z = 1.25 x = 1 x = 1.5 x = .75 Level curves in x,z coordinates.
Define the functions α(x, z) and β(x, z) as α(x, z) = √zx + rz x = 2 + yA + yB (F.3) β(x, z) = √zx − rz x = yA −yB.
(F.4) 141 Then, yAyB = z + 1 −α, yA + yB = α −2, and yU = 3 −α.
(F.5) Computing the time derivative of z and substituting from (7.7), (F.3), (F.4) and (F.5) we get ˙ z = ˙ yB(1 + yA) + ˙ yA(1 + yB) = 2 4v2 −∆v2 [∆v β(x, z) −2v(2z −α(x, z))] + 2vz(3 −α(x, z)) −σα(x, z)(z + 1 −α(x, z)).
(F.6) Computing the time derivative of x and substituting from (7.7), (F.3), (F.4) and (F.5) we get ˙ x = ˙ yA(1 + yB) −˙ yB(1 + yA) (1 + yB)2 = x∆v(4z −2α(x, z)) −4xvβ(x, z) 4v2z −z(∆v)2 + x∆v(3 −α(x, z)) + σxβ(x, z) z (z + 1 −α(x, z)).
(F.7) With small parameter ϵ = 1/v and ratio d = σ/v, the transformed dynamics (F.6), (F.7) can be written as a singular perturbation problem in standard form, ϵdz dt = 2z(3 −α) −dα(z + 1 −α) + 2ϵ3β∆v 4 −ϵ2(∆v)2 −4(2z −α)ϵ2 4 −ϵ2(∆v)2 =: g(x, z, ϵ) (F.8) dx dt = x(3 −α)∆v + σβx z (z + 1 −α) − 4βxϵ 4z −zϵ2(∆v)2 + x(4z −2α)ϵ2∆v 4z −zϵ2(∆v)2 =: f(x, z, ϵ).
(F.9) To ensure that the ϵ →0 limit is well-defined we assume lim v→∞ ∆v v = lim ϵ→0 ϵ∆v = 0.
(F.10) Slow Manifold Calculation The slow manifold is given by the root of g(x, z, 0) = 0.
g(x, z, 0) = 0 = ⇒2z(3 −α(x, z)) −dα(x, z)(z + 1 −α(x, z)) = 0 142 = ⇒2zyU −dα(x, z)(z + 1 −α) = 0 = ⇒2(1 + yA)(1 + yB)yU = d(3 −yU)(yAyB) = ⇒d 2yAyB = yU(1 + yA)(1 + yB) 3 −yU .
(F.11) (F.11) is an implicit expression for the slow manifold. Define the function ˆ x = √x + 1 √x. Then α(x, z) = √zˆ x. In order to obtain an explicit expression we rewrite (F.11) in the (x, z) coordinates as follows, g(x, z, 0) = 0 = ⇒2z(3 −α(x, z)) −dα(x, z)(z + 1 −α(x, z)) = 0 = ⇒2z(3 −√zˆ x) −d√zˆ x(z + 1 −√zˆ x) = 0 = ⇒6√z −2zˆ x −dˆ xz −dˆ x + dˆ x2√z = 0 = ⇒(2 + d)ˆ xz −(dˆ x2 + 6)√z + dˆ x = 0.
(F.12) (F.12) is quadratic in √z. The solutions to the quadratic are given by √z = dˆ x2 + 6 ± √ D 2(2 + d)ˆ x , (F.13) where the discriminant D = d2ˆ x4 + 36 + 4dˆ x2 −4d2ˆ x2. Hence we have two distinct solutions for the slow manifold given by z = dˆ x2 + 6 + √ D 2(2 + d)ˆ x !2 , dˆ x2 + 6 − √ D 2(2 + d)ˆ x !2 .
(F.14) The second solution in (F.14) lies outside the feasible domain z ∈[1, 9 4] (and corre-spondingly (yA, yB) ∈∆2) and is hence rejected.
To summarize, the slow manifold is given by z = dˆ x2 + 6 + √ D 2(2 + d)ˆ x !2 =: h(x) where ˆ x = √x + 1 √x and D = (dˆ x2 + 6)2 −4dˆ x2(2 + d).
(F.15) 143 Attractively of the Slow Manifold The boundary layer dynamics are given by dy dτ = g(x, y + h(x), 0), (F.16) where x is treated as fixed parameter.
Stability of the boundary layer dynamics requires the exponential stability of its origin, uniformly in the fixed parameter x . To test for exponential stability of the origin, we compute the Jacobian of the dynamics (F.16) evaluated at the origin ∂ ∂yg(x, y + h(x), 0) y=0 = −1 12 r 1296 + (1 + x)2σ (24x + (x −1)2σ) x2 (F.17) and note that ∂ ∂yg(x, y + h(x), 0) y=0 < 0 for all x ∈[1 2, 2].
Reduced Dynamics of the Slow Model The reduced dynamics on the slow manifold defined by (F.15) are given by ˙ x = f (x, h(x), 0) = σx h(x) [h(x) + 1 −α (x, h(x))] β (x, h(x)) + x(3 −α (x, h(x)) ∆v.
(F.18) The general expression for the equilibria of (F.18) is complicated. Nonetheless, ana-lytical solutions can be obtained for two special cases described below.
Special Cases • d = O(ϵ): The slow manifold is given by the d →0 limit of the expression (F.15), lim d→0 h(x) = 9x (x + 1)2, (F.19) which corresponds to yU = 1 −yA −yB = 0. Another way of seeing this is computing g(x, z, 0) from (F.6) in the limit d →0, which gives 2z(3 −α) = 0 ⇐ ⇒3 −α = 0 ⇐ ⇒yU = 0. The reduced dynamics on the slow manifold 144 are given by substituting (F.19) in (F.18) ˙ x = f (x, h(x), 0) = σ(x + 1)2 9 9x (x + 1)2 + 1 −α x, 9x (x + 1)2 β x, 9x (x + 1)2 = σ(x + 1)2 9 9x (x + 1)2 −2 3(x −1) (x + 1) .
(F.20) Equilibria of (F.20) are xeq = −1, 1 2, 1, 2. These equilibria and their stability are summarized in Table F.1. The two stable equilibria, xeq = 1 2 and xeq = 2, correspond to yB = 1 and yA = 1, respectively.
The unstable equilibrium zeq = 1 corresponds to yA = yB = 1 2. The xeq = −1 equilibrium is rejected since the dynamics are only defined for x ∈ 1 2, 2 .
Table F.1: Equilibria and stability for the reduced dynamics (F.20) xeq 1 2 1 2 zeq 2 9 4 2 yAeq = 2xeq−1 xeq+1 0 1 2 1 yBeq = 1 −yAeq 1 1 2 0 ∂f(x,h(x),0) ∂x x=xeq −σ 3 σ 6 −σ 3 Stability Stable Unstable Stable • d = 1, ∆v = 0: A general expression for the equilibria of (F.18), even for the symmetric case ∆v = 0, is challenging to determine. However, for d = 1, analytical expressions for the equilibria and their stability are summarized in Table F.2. In practice, the expressions from Table F.2 hold for a broad range of d as long as v is sufficiently large.
Note that from Equation (7.4) (and assuming σ = O(v)), lim v→∞p1 = lim v→∞ 2 1 v2 + 1 + q 1 v4 + 2 v2 + 9 + 4 , 2 1 v2 + 1 + q 1 v4 + 2 v2 + 9 + 4 = 2 1 + √ 13, 2 1 + √ 13 , 145 Table F.2: Equilibria and stability for the reduced dynamics (F.18) with d = 1 and ∆v = 0 xeq 2 1 1 2 zeq 2 1 18(19 + 5 √ 13) 2 yAeq 1 2 1+ √ 13 0 yBeq 0 2 1+ √ 13 1 ∂f(x,h(x),0) ∂x x=xeq −4σ 15 4− √ 13 3 σ −4σ 15 Stability Stable Unstable Stable which is precisely the unstable equilibrium point corresponding to xeq = 1 in Table F.2. Similarly, for σ = O(v), lim v→∞p2 = (1, 0) and lim v→∞p3 = (0, 1), which are precisely the two stable equilibria corresponding to xeq = 2 and xeq = 1 2 in Table F.2, respectively. Hence, the equilibria of the reduced dynamics (F.18) and their stability match exactly those of the full dynamics computed in (7.4), in the limit as v →∞and σ = O(v).
Appendix G Edge Detection This section describes a simple algorithm to find the boundaries of an object in a specified region R of parameter space.
Though we use the planar case R2 for illustration, the algorithm works generally in RN. The region R is assumed to contain a domain D, the boundary of which we intend to find.
Assumptions: 1. For each point p ∈R ⊂RN, there is a map f : RN 7→{0, 1} such that f(p) = 1 denotes a point inside the domain D and f(p) = 0 denotes a point outside D.
146 2. Initial inside (f(pin) = 1) and outside (f(pout) = 0) points are easy to de-termine, such that the straight line from pout to pin has a single boundary in between.
Algorithm: 1. Initialize a point pin inside D (i.e. f(pin) = 1) and a point pout outside D (i.e.
f(pout) = 0).
2. Compute f(pmid) where pmid = 1 2(pin + pout).
3. If f(pmid) = 1 then set f(pin) = f(pmid). Else set f(pout) = f(pmid).
4. Iterate steps 2 and 3 until ∥f(pin) −f(pout)∥< ϵ for some chosen ϵ.
One iteration of the algorithm above is illustrated in the figure below. A single run of the algorithm to convergence produce one point on the boundary. Successive points may be obtained by running the algorithm on several initializations of pin and pout (potentially randomly). The main computational barrier is the mapping f.
R2 R D pin pmid pout pin pout Illustration of one iteration of the edge detection algorithm.
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Question: Sulfuric acid reacts with potassium hydroxide in an acid-base neutralization reaction: H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2 H2O(l). Calculate the volume, in mL, of 0.100 M sulfuric acid required to neutralize 25.0 mL of 0.1632 M KOH. Provide your answer as a number in decimal (non-scientific) notation below, without units. For full credit,
Sulfuric acid reacts with potassium hydroxide in an acid-base
neutralization reaction:
H2SO4(aq) + 2
KOH(aq) → K2SO4(aq) +
2 H2O(l).
Calculate the volume, in mL, of 0.100 M sulfuric
acid required to neutralize 25.0 mL of 0.1632 M KOH.
Provide your answer as a number in decimal
(non-scientific) notation below, without units. For full
credit, your answer must be within 1% of the correct
value.
Solution: Consider the balanced equation:
HA2SOA4(aq)+2KOH(aq)⟶KA2SOA4(aq)+2HA2O(l)
To determine the volume (in mL) of 0.100 M sulfuric acid (HA2SOA4) required, ...
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Introduction [Music] this is the 15th video in a series that i'm making to support a course in elementary number theory and here we're going to look at solving polynomial congruences so this is quite a bit more technical than solving linear congruences but we still get a nice payoff so we're going to start with something called lagrange's theorem so that involves a polynomial with integer roots so i denote that by f of x is in z adjoin x and it satisfies the condition that the leading coefficient is not divisible by p so by the leading coefficient i mean the coefficient of the highest power of x then the congruence f of x congruent to 0 mod p has at most degree of f of x solutions now we really want to compare this to what's going on over the real numbers so notice that polynomials over the real numbers have at most degree whatever they are solutions over the real numbers so somehow working mod p is fairly similar to working over the real numbers okay so now let's look at the proof of this and this will be by induction on the degree of f of x so let's write that degree of f of x which maybe we'll just set to n just to make it easy okay so let's look at our base case so our base case will be the n equals one case so notice in the n equals one case f of x has the form ax plus b and we know that p does not divide a and so that means we're trying to solve the congruence ax is congruent to minus b mod p but then by a previous video we know exactly how many solutions there are for linear congruence congruences and in this case there is exactly one so i'll just write that down here exactly one solution and notice that one is most definitely equal to the degree of f of x in this case again that's by a previous video and we've done lots of stuff with linear concordances already okay so now let's make an induction hypothesis and the induction hypothesis will go like this we want to suppose our statement is true for all polynomials of degree k which is bigger than or equal to one and then we want to show that it's true for a degree k plus one polynomial so let's consider f of x which is a degree k plus 1 polynomial and i won't write this down but i'll just say it in words where the leading coefficient is not divisible by p so it satisfies the hypothesis over here okay now i want to make the following observation and that is for all integers maybe i'll call them m we have f of x minus f of m has a root at x equals m so we're not working congruent anything at the point where at this point we're just working over polynomials with integer coefficients okay so the fact that it has a root at that point means that we've got some sort of factorization theorem and then again this is like from an algebra class that you've probably taken earlier in your life so that tells us that we can factor f of x minus f of m as x minus m times g of x again that's part of what it means to have a root at x equals m now we want to suppose that we've got some a where f of a is congruent to zero modulo p you might say well what if this congruence doesn't have any solutions then such an a does not exist but that's okay because if this has no solutions well zero is most most definitely smaller than k plus one and so our conclusion over here is satisfied so we're assuming that we have at least one solution to this congruence okay but now this kind of action is possible for any integer m so the fact that it's true for any integer m means it's true for the our integer a so that means we can write f of x as x minus a times g of x mod p notice our f of a term which came from that f of m with the m replaced with a disappears because we're working mod p that was our supposition up here and then again since the degree of f of x is k plus one this is degree one we have the degree of g of x is equal to k so that tells us that g of x has at most k roots modulo p but if g of x has at most k roots modulo p and then a is an additional root for f then that tells us that f of x has at most k plus 1 roots mod p the k that come from g and then the one that comes from this up here okay now before we move on to our next result i want to look at a case when we're working modulo not a prime and see that this is not true notice if we take x squared minus 1 congruent to 0 mod 12 we have four solutions to this and our solutions come from the set 1 5 7 and 11. you can check that all of those square to 1 mod 12 and so that when you subtract 1 you get 0 mod 12. that means there are four roots or i guess i should say at least four roots to this polynomial congruence which is more than the degree of this polynomial which would seem like a problem but 12 is not prime so it's not a problem so in other words this lagrange's theorem really uses the fact that this p is prime okay let's get rid of some of this and then we're going to look at something called hensel's lena Hensels Lemma like i said before now we're going to look at something called hensel's limit which involves solving polynomial congruences modulo a composite number but not just any composite number this must be modulo some power of a prime and the statement and proof are quite technical that being said the proof is really constructive so it gives you a strategy for creating these solutions if you need them okay so let's carefully look at the statement of this lemma let's suppose that f of x is a polynomial with integer coefficients and f evaluated at a is congruent to 0 mod p to the m so in other words we have a root modulo p to the m already furthermore we need to assume that f prime of a is not congruent to 0 mod p notice that's p to the first power just the base power of p okay then you might say well what do we mean by the derivative this is just like the term by term standard derivative of a polynomial then there is a unique t which comes from the set 0 1 2 up to p minus 1 such that f evaluated at a plus t to the p m is congruent to zero mod p to the m plus one so let's look at this this lemma gives us the ability to start with the solution mod p to the m and create a solution mod p to the m plus one so the idea goes like this let's say you want to solve some congruence like f of x is congruent to zero mod p cubed or p to the fourth or p to the fifth you start by solving the congruence f of x is congruent to zero mod p and you do that just like by hand and then hensel's limit or really more appropriately the technique of the proof of hensel's letter gives us a way to boost that solution to a solution of the congruence f of x congruent to 0 mod p squared and then we can repeat that process to boost it to a solution of f of x is congruent to zero mod p cubed and so on and so forth until you get your desired power of p okay good so now that we've gotten this kind of like motivation taken care of let's go to the proof so we want to start by introducing some notation so let's let n equal the degree of f x and then we want to expand f of x as a taylor series at x equals a in other words like centered around x equals a so this is one of the very few times that we use any of the notions from calculus in this class but generally if you're like looking at a number theory class then you've probably taken at least calculus 2. okay so that means we can write f of x as f of a plus f prime of a times x minus a so that's the linear approximation and then we can keep going f double prime of a over 2 factorial x minus a squared all the way up to the nth derivative evaluated at a over n factorial x minus a to the n and for an arbitrary function you may have to worry about some sort of interval or radius of convergence but since f is a polynomial we don't have to worry about that at all and now from here we will evaluate at a special value of x that'll make this simplify quite a bit and that value of x will be equal to t times p to the m plus a or maybe a plus t times p to the m if you want to make it look exactly like this right here okay so let's see what that gives us that will give us f evaluated a plus t times p to the m is equal to f of a so that's obviously not going to change and then f prime of a times t times p to the m notice that a and a cancel when we do that subtraction and then this is going to go all the way up to the nth derivative evaluated at a over n factorial and then t to the n p to the m n so again that's because if we do x minus a this guy right here cancels okay now this might look a little bit problematic because we've got these things in the denominator but this is actually not a problem so maybe let's do that with the following observation and that is the kth derivative evaluated at a over k factorial is always an integer and we can actually prove this by proving something quite a bit simpler and i'll spell that out here so we'll show this is okay for a function which i'll call g of x which is equal to x to the r power you might say well that seems a little bit like a cheat but recall that our function f of x is just a linear combination of powers of x like that so if this holds for x to the r where r is arbitrary then this is going to hold for our function f of x okay so now let's move into the proof of this thing which is impeach with our function g of x so notice we have g the kth derivative of that over k factorial evaluated at a so notice that's going to give us r times r minus 1 times r minus 2 all the way down to r minus k plus 1 over k factorial times a to the r minus k let's talk our way through that so this numerator comes from taking repeated derivatives of this x to the r the first derivative will give us the r the second the r minus one the third the r minus two and then the k will give us that guy way down there then the k factorial came along for the ride then each time we take a derivative our exponent of x lowers by one giving us this a to the r minus k but now let's notice that this stuff that i'm putting in yellow parenthesis is in fact a binomial coefficient that is the binomial coefficient r choose k and then we've got this a to the r minus k going along for free but binomial coefficients are always integers so we're okay in that case okay so no worries all of these things just have integer coefficients so that means this line right here might be something useful to work with let's go over here and look at our theorem or our pencils Theorem limit and notice that our eventual goal is for this object to be congruent to zero mod p to the m plus one so maybe we should take this line right here which i'll star in purple and reduce it modulo p to the m plus one and then somehow argue that we can pick a an appropriate t to make that zero so let's notice that f of a plus t to the t times p to the m will be congruent to f evaluated at a plus f prime evaluated at a times t times p to the m all the other powers of p will disappear because the next smallest is p to the m plus one because the next smallest is p to the two m which is larger than m plus one so that means it's zero mod m plus one so let's just note here that we're working modulo p to the m plus one okay so let's summarize what we've got at the top and then we'll finish this proof off let's recall on the last board we did some trickery with the taylor series and we came up with the following Congruence congruence modulo p to the m plus one we've got f evaluated at a plus t times p to the m is congruent to f of a plus f prime of a t p to the m modulo p to the m plus one and our big goal is for this thing to be congruent to zero mod p to the m plus one so let's see what it would take for that to happen so in particular we need this to be congruent to zero mod p to the m plus one so that will occur if we have f prime of a times t times p to the m is congruent to f of a i should say that is negative f of a mod p to the m plus one okay but that actually might seem a little bit problematic because we're still working modulo p to the m plus one we'd like to work mod something a little bit easier and we can do that with a little bit of a trick so let's notice that f of a is congruent to 0 mod p to the m that was like one of our original assumptions or one of the hypotheses built into the statement of our lemma but what does that tell us that tells us that p to the m divides f of a but that means that f of a divided by p to the m is an integer right so those are all equivalent statements now from there we can take all three parts of this congruence and divide by p to the m so that produces like an equivalent congruence now i think that was something that was proven in a homework exercise earlier so just to see what we're doing here we're going to divide this by p to the m this by p to the m and then this by p to the m so that means this thing right here holds if and only if f prime a times t is congruent to negative f of a over p to the m modulo p now this seems a little bit sketchy because we've got a denominator here and generally we're just working in integers but by that pink discussion over there we're actually okay here now we have to argue that this has a unique solution so how do we know that this has a unique solution i guess i should be more thorough and say i mean a unique solution for this t here well let's notice by assumption f prime of a is not 0 mod p that's one of our assumptions that we haven't done anything with yet so by some previous work involving solving linear congruences we know there is a unique solution for t mod p okay but that means that that solution can be rewritten from 0 1 2 to p minus 1 because those are all the residue classes modulo p so now let's see how we would finish this off we take that unique solution which is a solution to this equation which will be a solution to this equation before we divided by the orange p to the m but if it's the solution to this equation that makes this blue underline zero but the blue underline is congruent to our goal object which means our goal object is congruent to zero mod p to the m plus one that's exactly what we wanted to end up with okay well we're going to do a constructive example using this at the end of the video but before we do that i want to look at one more proposition building off of pencil's limbo we've got a proposition and that is that the polynomial equation x to the p minus 1 congruent to 1 mod p to the m has exactly p minus 1 solutions so notice this is saying that there are p minus one p minus first roots of the number one modulo a power of that prime of p and we can have any power of that prime p okay well how could we prove this with induction and we'll use hensel's lineup that's why we did hansel's lemon okay so let's look at our base case so notice our base case will be the m equals one case but let's recall that for all a from the set one two up to p minus one we know that a to the p minus one is congruent to one modulo p and that's by fermat's little theorem notice that they're exactly p minus 1 numbers that a can come from but there are only p residues modulo p and the one excluded residue is clearly not a solution because zero to the p minus one is congruent to zero it is not congruent to one okay so from moz little theorem takes care of the base case so now let's make an induction hypothesis and the induction hypothesis will go like this so let's suppose for some k bigger than or equal to one we know x to the p minus one congruent to one mod p to the k has exactly p minus one solutions okay cool and now we're going to want to consider the next case and now before we move on to our induction step let's maybe name these solutions so i'm going to name them x1 x2 all the way up to xp minus 1. so that means x1 to the p minus 1 is congruent to 1 mod p to the k and then so on and so forth we've got that for all of those okay and we also want to notice that the other hypotheses of hensel's lemma are also satisfied so let's notice that if we set f of x equal to x to the p minus one minus one then well first of all f evaluated at x i is congruent to 0 mod p to the k that essentially comes for free from our induction hypothesis and then f prime evaluated at x i is equal to p minus 1 times x i to the p minus 2 but let's notice that both of these numbers are relatively prime to p and thus relatively prime to p to the k which means when we take their product it'll still be relatively prime to p to the k and thus it's impossible for it to be congruent to zero mod p to the k so those are the one two hypotheses of hensel's lemma and now we can invoke the conclusion of hensel's luna so hensel's lima will imply that there exists a unique i'll call it ti it's going to come from the set 0 1 2 up to p minus 1 such that we know f evaluated at x i plus t i times p to the k is congruent to 0 mod p to the k plus one so what happened here is we took our solutions mod p to the k and boosted them to solutions mod p to the k plus one and how many solutions do we have so far well we have at least as many solutions as we started with down here we have at least p minus 1 solutions now we want to finish this by showing that we have all of the solutions so let's suppose y is another solution in other words we have f evaluated at y is congruent to zero mod p to the k plus 1. what we want is for this y to really be one of these that we've already constructed okay so how can we get there so this congruence right here tells us that f of y is a multiple of p to the k plus one but if it's a multiple of p to the k plus one that means it's a multiple of p to the k so we know f of y is also congruent to zero mod p to the k but recall we have all of our solutions to that congruence and so that tells us that our y is congruent to x to the i mod p to the k for some i that brings it from this set but now building hensel's lemma back out of this that means there exists a unique i'll call it t from the set 0 1 up to p minus 1 such that y equals x i plus t times p to the k but let's see what we've done we used hensel's lemma to create a ti for the x i and then we used it again to create a t for this y which happened to be congruent to x i but then we know that those t's are unique so this t is the same thing as this t i which tells us this is equal to x i plus t i times p to the k in other words our new solution was actually one of our old solutions and so that means we have all of the solutions for our induction step and that finishes the proof of this proposition okay so we're going to finish this thing off by looking at an example where we construct roots to a polynomial equation over a prime power set up using hensel's limit and the strategy outlined in the proof like i said before we're going to finish this video off by constructing a Example solution to a polynomial congruence using hensel's lemma the polynomial congruence we'll look at is x squared plus 15x plus 31 is congruent to 0 mod 125 so that's clearly equal to 5 cubed and this is laid out in three steps first you solve the congruence mod 5 then you apply hinselma to boost that to a solution mod 25 and then you apply the technique from the proof of tinsel's limb again to boost that to a solution mod 125 in other words we go from 5 to five squared to five cubed and if you needed to keep going you could just reapply hensel's limit over and over again okay so let's first introduce some notation so let's set f of x equal to x squared plus 15x plus 31. and notice that if we reduce this thing mod 5 well 15 is just going to disappear to 0 and 31 is 1. so that means we're really solving x squared plus one is congruent to zero modulo five and there are exactly two solutions to this which you can easily guess and check because 5 is a fairly small number so here we have x equals 2 or x equals 3 and then you also want to check that if you take f prime and you evaluate it at 2 and three you do not get zero that's another one of the hypotheses for hansel's lemma so we could work this off any of these roots and if your goal in the end was to find all of the roots to our polynomial congruence mod 125 you would have to take every branch possible but what we'll do is just go off of this x equals two and then maybe i'll leave as a warm up do the same thing for x equals three and see what final solution you get mod 125 okay great so let's recall that mod 25 we're going to need to solve the following congruence and this was built into the structure of tinsel's limit like i said before so we've got t times f prime of 2 is congruent to minus f of 2 over 5 modulo 5. so again i'll let you look carefully at that proof to see why this is the object we want to look at so we need to calculate these two numbers f prime evaluated at 2 and then f evaluated at 2 divided by 5. okay so let's notice that f prime evaluated at 2 is going to be the same thing as 2x plus 15 evaluated at x equals 2. we just use the power rule there so let's see that's going to give us 4 plus 15 or the number 19. but notice we can reduce that mod 5 because we're working mod 5 so this is congruent to 4 mod 5. so when we get around to it we can replace this number here with just the number 4. okay now next we need to calculate this so minus f of 2 over 5. again we might be worried that this is not an integer but from discretion that we said before this is an integer this is okay okay so let's calculate this this is going to be negative then we have 2 squared which is 4 plus 30 plus 31 and that's going to be all over 5. that's what we get from substituting 2 up here but notice that's going to give us negative 65 over 5 which is negative 13. but again we only need to work modulo 5 so it's useful to notice that that is the same thing as 2 mod 5 it's 2 more than a multiple of five that multiple of five is negative fifteen so just as we did with this k this thing right here we can take this guy right here and exchange it for the number two so that means the congruence that we need to solve is 4t is congruent to 2 mod 5. but it's easy to check that a solution to that is t equals 3 from the theory of linear congruences okay but that means our new solution so i'll just call it new will be equal to our old solution so that was 2 plus 3 times the first power of 5. so in the end that is 17. so in other words at this stage we know f evaluated at 17 is congruent to 0 mod 25 and now we can build this game again starting with 17 and that's what we'll do over here okay so again following the structure of hensel's luma we need to solve the following congruents so we have f prime of 17 times t must be congruent to negative f of 17 over 25 and that has to occur mod 5. so again this is just like some calculation it's not too hard to do but what we get is negative f of 17 over 25 can easily be calculated to be congruent to 2 mod 5 and then f prime evaluated at 17 can easily be calculated to be 4 modulo 5. so that means we need to solve plugging those things in 4t congruent to 2 mod 5. and this is just dumb luck that we end up with the same linear congruence that we did down here don't assume that's always going to happen so that means that our solution for this t is t equals three and that means we can get our final solution so this was like an intermediate solution so our final solution will be our intermediate solution which was 17 [Music] plus our value of t which is 3 times our new power of 5 which was 25 so you can put all that together and see that you get the number 92 and now you can check that f evaluated at 92 is congruent to 0 modulo 125 which is exactly where we want to end up okay so before we finish this thing off maybe i'll give you some warm-up problems based on the stuff that we've talked about okay so our warm-up just really involves the technical aspects of applying hensel's lemma so our goal is to find all solutions to the three polynomial congruences so we've got x squared plus four x plus two is congruent to zero mod seven zero mod forty nine and zero mod 343 okay that's a good place to stop |
12878 | https://www.mathway.com/popular-problems/Precalculus/986269 | Find the Vertex y=x^2-2x-2 | Mathway
Enter a problem...
[x]
Precalculus Examples
Popular Problems
Precalculus
Find the Vertex y=x^2-2x-2
y=x 2−2 x−2 y=x 2-2 x-2
Step 1
Rewrite the equation in vertex form.
Tap for more steps...
Step 1.1
Complete the square for x 2−2 x−2 x 2-2 x-2.
Tap for more steps...
Step 1.1.1
Use the form a x 2+b x+c a x 2+b x+c, to find the values of a a, b b, and c c.
a=1 a=1
b=−2 b=-2
c=−2 c=-2
Step 1.1.2
Consider the vertex form of a parabola.
a(x+d)2+e a(x+d)2+e
Step 1.1.3
Find the value of d d using the formula d=b 2 a d=b 2 a.
Tap for more steps...
Step 1.1.3.1
Substitute the values of a a and b b into the formula d=b 2 a d=b 2 a.
d=−2 2⋅1 d=-2 2⋅1
Step 1.1.3.2
Cancel the common factor of −2-2 and 2 2.
Tap for more steps...
Step 1.1.3.2.1
Factor 2 2 out of −2-2.
d=2⋅−1 2⋅1 d=2⋅-1 2⋅1
Step 1.1.3.2.2
Cancel the common factors.
Tap for more steps...
Step 1.1.3.2.2.1
Factor 2 2 out of 2⋅1 2⋅1.
d=2⋅−1 2(1)d=2⋅-1 2(1)
Step 1.1.3.2.2.2
Cancel the common factor.
d=2⋅−1 2⋅1 d=2⋅-1 2⋅1
Step 1.1.3.2.2.3
Rewrite the expression.
d=−1 1 d=-1 1
Step 1.1.3.2.2.4
Divide−1-1 by 1 1.
d=−1 d=-1
d=−1 d=-1
d=−1 d=-1
d=−1 d=-1
Step 1.1.4
Find the value of e e using the formula e=c−b 2 4 a e=c-b 2 4 a.
Tap for more steps...
Step 1.1.4.1
Substitute the values of c c, b b and a a into the formula e=c−b 2 4 a e=c-b 2 4 a.
e=−2−(−2)2 4⋅1 e=-2-(-2)2 4⋅1
Step 1.1.4.2
Simplify the right side.
Tap for more steps...
Step 1.1.4.2.1
Simplify each term.
Tap for more steps...
Step 1.1.4.2.1.1
Raise −2-2 to the power of 2 2.
e=−2−4 4⋅1 e=-2-4 4⋅1
Step 1.1.4.2.1.2
Multiply 4 4 by 1 1.
e=−2−4 4 e=-2-4 4
Step 1.1.4.2.1.3
Cancel the common factor of 4 4.
Tap for more steps...
Step 1.1.4.2.1.3.1
Cancel the common factor.
e=−2−4 4 e=-2-4 4
Step 1.1.4.2.1.3.2
Rewrite the expression.
e=−2−1⋅1 e=-2-1⋅1
e=−2−1⋅1 e=-2-1⋅1
Step 1.1.4.2.1.4
Multiply−1-1 by 1 1.
e=−2−1 e=-2-1
e=−2−1 e=-2-1
Step 1.1.4.2.2
Subtract 1 1 from −2-2.
e=−3 e=-3
e=−3 e=-3
e=−3 e=-3
Step 1.1.5
Substitute the values of a a, d d, and e e into the vertex form (x−1)2−3(x-1)2-3.
(x−1)2−3(x-1)2-3
(x−1)2−3(x-1)2-3
Step 1.2
Set y y equal to the new right side.
y=(x−1)2−3 y=(x-1)2-3
y=(x−1)2−3 y=(x-1)2-3
Step 2
Use the vertex form, y=a(x−h)2+k y=a(x-h)2+k, to determine the values of a a, h h, and k k.
a=1 a=1
h=1 h=1
k=−3 k=-3
Step 3
Find the vertex(h,k)(h,k).
(1,−3)(1,-3)
Step 4
y=x 2−2 x−2 y=x 2-2 x-2
21 21 24 24 y y
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9
≤
≤
°
°
θ
θ
4
4
5
5
6
6
/
/
^
^
×
×
π
π
1
1
2
2
3
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⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
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12879 | https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-arithmetic-operations/cc-6th-sub-decimals/v/subtracting-decimals-up-to-thousandths-place | Subtracting decimals: 39.1 - 0.794 (video) | Khan Academy
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6th grade math
Course: 6th grade math>Unit 2
Lesson 2: Subtracting decimals
Subtracting decimals: 9.005 - 3.6
Subtracting decimals: 39.1 - 0.794
Subtracting decimals: thousandths
Math>
6th grade math>
Arithmetic with rational numbers>
Subtracting decimals
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Subtracting decimals: 39.1 - 0.794
AZ.Math: 6.NS.B.3
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About About this video Transcript
Learn to subtract decimals by aligning the decimal points and place values. Regroup numbers for easier subtraction, and then perform the subtraction for each place value. Mastering decimal subtraction helps build strong math skills and real-world problem-solving abilities.Created by Sal Khan.
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sebastianbrinezc 5 years ago Posted 5 years ago. Direct link to sebastianbrinezc's post “In the video Sal shows th...” more In the video Sal shows the hundredths place borrows 1 tenth from 11 tenths in the tenths place, then the hundredths place turns out 10 tenths, shouldn't it be 1 tenth? Because we are taken just 1 tenth, the tenths place got 10 tenths after that, because of the hundredths place took just 1 tenth. What is the explanation?
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Elaina163 5 years ago Posted 5 years ago. Direct link to Elaina163's post “Basically you know the te...” more Basically you know the tenths place and hundredths place on a decimal? 0.12, 1 is the tenth place and two is the hundredth. This goes the same for fractions. 1/100 you say it as one hundredth same goes for the hundredth in the example decimal (0.12) except the hundredth is 2 so it'd be equivalent to 2/100. Does that help?
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Nathan C. Fields 10 years ago Posted 10 years ago. Direct link to Nathan C. Fields's post “Is there a good video to ...” more Is there a good video to watch explaining how to borrow from 0s, Sal seemed to do it from the left to the right and I've always learned to do it opposite, but I get these problems wrong because of one or two letters. Thanks!
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Parmi Chanduri 10 years ago Posted 10 years ago. Direct link to Parmi Chanduri's post “I'm not entirely sure if ...” more I'm not entirely sure if there's a separate video just for borrowing from zeros but I wouldn't be surprised if there was! I can give you the basic rundown.
When you borrow from a zero, you aren't actually borrowing from that zero. In fact, that zero borrows from the number next to it. Here's an example:
5.09 - 2.29 = ?
Here, you can see that 0 is needed to 'borrow' from 5 so that 2 can be subtracted from 0. Since 0 borrowed from a number that you need to multiply the place to the left (from the 5's 'point of view') 10 times to get the place you're in. So, now you have
{10}9
29 -
1.80
And you have your answer! Sorry if it was a bit confusing. Hope it helped though!
Keep learning,
Chocomuff
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joannapatidar 5 years ago Posted 5 years ago. Direct link to joannapatidar's post “when we add subtract mult...” more when we add subtract multiply do we line up the deciml points
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jt115821 4 years ago Posted 4 years ago. Direct link to jt115821's post “Subtracting decimals uses...” more Subtracting decimals uses the same setup as adding decimals: line up the decimal points, and then subtract. In cases where you are subtracting two decimals that extend to different place values, it often makes sense to add extra zeros to make the two numbers line up—this makes the subtraction a bit easier to follow.
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Aiden Kim 6 years ago Posted 6 years ago. Direct link to Aiden Kim's post “when were numbers first u...” more when were numbers first used in history?
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David Severin 6 years ago Posted 6 years ago. Direct link to David Severin's post “see www.vedicsciences.net...” more see www.vedicsciences.net/articles/history-of-numbers.html
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Isaiah Walker 7 years ago Posted 7 years ago. Direct link to Isaiah Walker's post “Will you ever have 2 numb...” more Will you ever have 2 numbers that will have 2 decimals. For example: 123.456.778
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Allison Fitzpatrick 3 years ago Posted 3 years ago. Direct link to Allison Fitzpatrick's post “you can't have two decima...” more you can't have two decimals because you can stretch behind the decimal point for an infinite time such as pi which is 3.14159265359... pi is infinite and there is only one whole number and infinite numbers behind the decimal
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Caelan M 6 years ago Posted 6 years ago. Direct link to Caelan M's post “I don't understand how he...” more I don't understand how he's regrouping here. How does taking one off of 39 give you eleven tenths?
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AngelB 3 years ago Posted 3 years ago. Direct link to AngelB's post “This is me from 2023 is a...” more This is me from 2023 is anyone alive and what shows have come out does anyone still use the app?
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IshaanS 2 years ago Posted 2 years ago. Direct link to IshaanS's post “0 rocks!” more 0 rocks!
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kobra 8 months ago Posted 8 months ago. Direct link to kobra's post “im conpuzzled (confused) ...” more im conpuzzled (confused) i did the same thing and got the answer but it said wrong
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1 Marini Carver 3 years ago Posted 3 years ago. Direct link to 1 Marini Carver's post “This helped a lot and was...” more This helped a lot and was very useful. Was it useful for you guys?
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Video transcript
Let's try to calculate 39.1 minus 0.794, and so pause the video and try this on your own. All right, I'm assuming you've given a go at it, so now let's work through it together. So I'm going to rewrite this. It's 39.1 minus-- I'm going to line up the decimals so that I have the right place values below the right place values-- minus-- this 0 is in the ones place, so I'll put it in the ones place-- 0.794. And now we're ready to subtract. Now, how do we subtract 4 from nothingness here, and 9 from nothingness here? Well, the same thing as nothing is a 0. And so now we can start to think about how to subtract. Well, we still have the problem. Well, we're trying to subtract 4 from 0, so we're trying to subtract 9 from 0. So what we could do is take this one tenth and try to regroup it into the hundredths place and the thousandths place. So let's think about this. If we make this-- actually that's not actually going to solve our problem. Well we could do it, but then we're going to have zero tenths, and we're still going to have a problem here. So actually let me go to the ones place. So let me get rid of a ones, so that's eight ones, which is going to be 10 tenths. So that's going to now-- we're going to have 11 tenths. The 10 tenths from here plus 1 is 11 tenths. Now let's take one of those tenths so that we have 10 tenths, and give it to the hundredths. So that's going to be 10 hundredths. And now let's take one of those hundredths-- so now we have nine hundredths-- and give it to the thousandths. So that's going to be 10 thousandths. Now we're ready to subtract. So 10-- let me do this in yellow-- 10 minus 4 is 6. 9 minus 9 is 0. 10 minus 7 is 3. We have our decimal point. 8 minus 0 is 8. And then we have 3 minus nothing is 3. So we're done, 38.306.
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Top Voted |
12880 | https://parwatisingari.com/2015/03/03/mathematical-properties-of-hexagon/ | Parwatisingari's Weblog
Mathematical properties of Hexagon.
Since this article is part of the Hexagon awareness series I am just restating what we all probably learnt at school. This is subjected more research and in depth understanding. Usually when we say hexagon, we mean regular hexagons, so is it here. but hexagonal does not necessarily refer to symmetric hexagons.
A hexagon is a figure with 6 edges and and 6 vertices, a regular hexagon is convex figure with sides of the same leght, and the internal angles or 120 degrees. It has 6 rotational symmetries and 6 reflection symmetries.
Hexagons are the only regular polygons that can be subdivided into another regular polygon.
Hexagons are unique regular polygon such that the distance between the centre and each vertex is equal to the length of each side.
As for number six it is the first perfect number, the second smallest composite number. It is the only number that is both the sum and the product of three consecutive numbers. That is
Hexagons can be tiled or tessellated, in a regular pattern. If like me you wondered what it was, then hexagon can bordered by six other hexagons which can themselves be bordered by six hexagons—including each other and so on ad infinitum, in any direction without leaving empty spaces. This kind of laying can be possible only with squares and triangles. This is probably why it is important in a variety of feilds. Look at our Chinese chekkar boards or the beehives. They are all hexagonal in pattern. Somewhere the hexagonal tessellation is a dual of triangular tessellation these can be converted to one to the other with a little alteration of layout. The hexagonal tessellation, or lattices are they are called, provides an ideal blend of efficiency and strength, and economy of material as compared to the triangle or square.
When it comes to circles, hexagons helps to pack cicles on a flat plane very efficiently. Though we are placing circles, their grid is hexagonal. Since cells tend to most efficient exposure of shapes, they tend to circularity, but lateral compression of the neighbours result in hexagonal tilting. This could be why hexagons are considered as natural shape.
The connect between triangles, and hexagons is a continuation of the relation of the hexagon with circle. Like I shared before, triangles and hexagons are the only two regular polygons. The hexagon can be divided into equilateral triangles while the triangle into subtriangles.
This is quite hectic, to go on, I am just teasing your thoughts as it is the month of Hexagon awareness. Hexagons pop up in unsual places like aromatic chemistry, healing, bio-sciences, mathematics, spirituality whatever. Like I shared before for me the benzene hexagon is the representative of vibrant life.
If you would like to know more – hexagon (Wikipedia.), www.hexnet.org, www.wolframmathworld/hexagon
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12881 | https://zhouquan34.github.io/MathProb/STAT614_Unit4_QuanZhou.pdf | Lecture 4 Instructor: Quan Zhou For more details about the materials covered in this note, see Chapters 5.1 and 5.2 of Resnick and Chapter 1.4 of Durrett .
4.1 Simple functions We first give a formal definition of simple functions. Note that we require them to be real-valued and measurable in the definition.
Definition 4.1. Given a measurable space (Ω, F), a simple function f is a measurable mapping from (Ω, F) to (R, B(R)) with a finite range.
Remark 4.1. Given a simple function f, let {a1, . . . , ak} denote its range, where a1, . . . , ak are distinct real numbers, and Ai = f −1(ai). Then, A1, . . . , Ak are disjoint and measurable, and we can express f by f = Pk i=1 ai1Ai.
Lemma 4.1. Let f, g: (Ω, F) →(R, B(R)) be simple functions. Then, f + g is also a simple function. For any a ∈R, af is also a simple function.
Proof. Try it yourself.
Lemma 4.2. Let f : (Ω, F) →([0, ∞], B([0, ∞])) be a non-negative extended real-valued Borel function.1 There exists a sequence of simple functions {φn} such that 0 ≤φn ↑f pointwisely; i.e.
0 ≤φ1 ≤φ2 ≤· · · ≤f and limn→∞φn(x) = f(x) for every x ∈Ω. Further, if f is bounded, then φn ↑f uniformly.
Proof. Here we only prove the first claim. The idea is that we construct φn by (1) truncating f at height 2n, and (2) rounding f(x) to the greatest integer multiple of 2−n that does not exceed f(x). Explicitly, we define the simple function φn by φn(x) = sup k 2n : k = 0, 1, 2, . . . , k 2n ≤min{f(x), 2n} .
It is not difficult to show that (1) φn ∈[0, ∞), (2) φn is measurable, and (3) φn ≤φn+1 for every n.
To prove φn(x) →f(x), first assume that f(x) < ∞. Then the approximation error |φn(x) −f(x)| vanishes at rate 2−n. If f(x) = ∞, then φn(x) = 2n →∞.
1It can be shown that B(¯ R) = {A ⊂¯ R: A ∩R ∈B(R)}.
Fall 2022 Quan Zhou 4.2 Lebesgue integral Let (Ω, F, µ) be a measure space and f : (Ω, F) →(¯ R, B(¯ R)). Now we show how to construct the Lebesgue integral R Ωfdµ. Other notation often used for denoting Lebesgue integral includes Z Ω f(ω)µ(dω), Z Ω f(ω)dµ(ω), Z fdµ, Z f.
It is better to use the first two for clarity. There are a few ways to construct R Ωfdµ.
Here we present the construction that is commonly used in real analysis books, which is slightly different from the approaches in Resnick’s and Durrett’s books.
Step 1.
Consider a non-negative simple function φ(ω) = Pn i=1 ai1Ai(ω) where a1, . . . , an ∈[0, ∞) and A1, . . . , An are disjoint measurable subsets of Ωthat partition Ω. Define Z Ω φdµ = n X i=1 aiµ(Ai).
(1) Note that the right-hand side makes sense since φ is assumed measurable (and thus Ai ∈F). The definition immediately implies that, for any measurable set A, R Ω1Adµ = µ(A).
Step 2.
For f : (Ω, F) →([0, ∞], B([0, ∞])), define Z Ω fdµ = sup Z Ω φdµ : 0 ≤φ ≤f, and φ is simple .
(2) We need to prove that when f is a non-negative simple function, the two definitions in Step 1 and Step 2 give the same value (see Lemma 4.3).
Equivalently, we can define R Ωfdµ = limn→∞ R Ωφndµ using any sequence of simple functions φn such that 0 ≤φn ↑f. This approach requires us to check that (1) R Ωφndµ does converge, and (2) the limit does not depend on the sequence {φn} we pick [3, §5.2.2]. In the next lecture, we will prove the monotone convergence theorem (MCT) using the definition given in (2). The equivalence of the two definitions then follows from MCT.
Lemma 4.3. For the Lebesgue integral of a non-negative simple function f, the two definitions given in (1) and (2) are the same.
2 Fall 2022 Quan Zhou Proof. Since f is simple, f can be written as f(ω) = Pn i=1 ai1Ai(ω) where a1, . . . , an ∈[0, ∞) and A1, . . . , An are disjoint subsets of Ω. Let I1 = n X i=1 aiµ(Ai).
We need to show that I1 = I2, where I2 = sup Z Ω φdµ : 0 ≤φ ≤f, and φ is simple .
Since f is simple, we immediately have that I1 ≤I2 by choosing φ = f.
Next, for any two non-negative simple functions f, φ such that f ≥φ, one can show that R fdµ ≥ R φdµ using the definition of the Lebesgue integral given in Step 1; see Proposition 4.1(iv). Taking sup on both sides we get I1 ≥I2. Hence, I1 = I2.
Step 3.
For f : (Ω, F) →(¯ R, B(¯ R)), define the positive part and negative part of f as follows: f +(ω) = f(ω) ∨0, f −(ω) = (−f(ω)) ∨0.
Hence, f = f + −f −and |f| = f + + f −. If at least one of R Ωf +dµ and R Ωf −dµ is finite, we say f is quasi-integrable and the integral exists (or is defined). If both are finite, we say f is integrable, which is equivalent to requiring that R Ω|f|dµ < ∞. When the integral exists, we define it by Z Ω fdµ = Z Ω f +dµ − Z Ω f −dµ.
Definition 4.2. When µ is a probability measure, we define E[X] = R ΩXdµ, which is called the expectation of X.
Definition 4.3. For A ∈F, define Z A fdµ = Z Ω 1A fdµ, provided that the right-hand side is defined.
3 Fall 2022 Quan Zhou Example 4.1. Consider the probability space (Ω, F, δx) where δx denotes a unit point mass on x. Then, for any measurable f, R Ωfdδx = f(x). To prove this, first check that it holds true for non-negative simple functions. Next, for f ≥0 and any simple function 0 ≤ϕ ≤f, we have Z Ω ϕ(ω)δx(dω) = ϕ(x) ≤f(x).
Now if we choose ϕ(ω) = f(x)1{x}(ω), the supremum is achieved and thus R fdδx = f(x).
Finally, for a general measurable f, we have R fdδx = f +(x) −f −(x) = f(x).
Example 4.2. Consider the measure space (Ω, P(Ω), #) where Ω= {a1, a2, . . . } and # is the counting measure. Then, for any measurable f, it can be shown that R Ωfd# = P i f(ai).
4.3 Properties of Lebesgue integral Proposition 4.1. Let f, g be Lebesgue integrable functions defined on (Ω, F, µ).
(i) If f ≥0 a.e., then R fdµ ≥0.
(ii) ∀a ∈R, R afdµ = a R fdµ.
(iii) R (f + g)dµ = R fdµ + R gdµ.
(iv) If g ≤f a.e., then R fdµ ≥ R gdµ.
(v) If g = f a.e., then R fdµ = R gdµ.
(vi) | R fdµ| ≤ R |f|dµ.
Remark 4.2. “a.e.” means almost everywhere. When we say some property holds a.e., it means that there exists a set N ∈F with µ(N) = 0 such that the property holds on N c. When µ is a probability measure, we often say almost surely, which is abbreviated as “a.s.”.
Proof of part (iii). We need to prove the claim using the definition of the Lebesgue integral, i.e. the three-step construction. So let’s start from non-negative simple functions and prove the property using the corresponding definition of the Lebesgue integral.
4 Fall 2022 Quan Zhou Step (1). We claim that for two non-negative simple functions f, g, we have R (f + g)dµ = R fdµ + R gdµ. By definition, we can express f, g as f = m X i=1 ai1Ai, g = n X i=1 bi1Bi, where ai, bi are non-negative real numbers and {Ai} (or {Bi}) is a partition of Ω. Hence, we have (f + g)(ω) = m X i=1 n X j=1 (ai + bj)1Ai∩Bj(ω), and clearly f + g is also a non-negative simple function. Further, {Ai ∩Bj : 1 ≤i ≤m, 1 ≤j ≤n} forms a new partition of Ω.
Therefore, by the definition of Lebesgue integral for non-negative simple functions, we have Z (f + g)dµ = m X i=1 n X j=1 (ai + bj)µ(Ai ∩Bj) = m X i=1 ai n X j=1 µ(Ai ∩Bj) + n X j=1 bj m X i=1 µ(Ai ∩Bj) = m X i=1 aiµ(Ai) + n X j=1 bjµ(Bj) = Z fdµ + Z gdµ.
Step (2). We prove that for any two non-negative measurable functions f and g, we also have R (f + g)dµ = R fdµ + R gdµ. We do not use the definition given in (2). Instead, we use the alternative definition discussed in Step 2 in Section 4.2.
(Again, it follows from the monotone convergence theorem, which will be proven in the next lecture.) Choose sequences of non-negative simple functions {fn} and {gn} such that fn ↑f and gn ↑g.
They always exist by 5 Fall 2022 Quan Zhou Lemma 4.2. Since (fn + gn) ↑(f + g), we find that Z (f + g)dµ = lim n→∞ Z (fn + gn)dµ = lim n→∞ Z fndµ + Z gndµ = lim n→∞ Z fndµ + lim n→∞ Z gndµ = Z fdµ + Z gdµ.
Step (3). We first prove that for non-negative integrable functions f1, f2, we have R (f1 −f2)dµ = R f1dµ − R f2dµ. Let f = f1 −f2. Since f = f + −f −, we have f1 + f −= f2 + f +, which gives Z f1dµ + Z f −dµ = Z f2dµ + Z f +dµ.
By definition, R fdµ = R f +dµ − R f −dµ = R f1dµ − R f2dµ.
Now we consider arbitrary integrable functions f, g. Notice that f + g = (f + −f −) + (g+ −g−) = (f + + g+) −(f −+ g−).
Hence, R (f+g)dµ = R (f ++g+)dµ− R (f −+g−)dµ = R fdµ+ R gdµ.
The proof is complete.
Proof of the remaining part(s). Try it yourself.
Proposition 4.2. Assume X, Y are random variables defined on the same probability space such that E[X] and E[Y ] exist (may be equal to infinity).
(i) If E[X+] < ∞and E[Y +] < ∞, then E[X + Y ] = E[X] + E[Y ]. The condition can also be replaced by E[X−] < ∞and E[Y −] < ∞.
(ii) ∀a, b ∈R, E[aX + b] = aE[X] + b.
(iii) If X ≥Y a.e., then E[X] ≥E[Y ].
Proof. Try it yourself.
6 Fall 2022 Quan Zhou Theorem 4.1. Let f : [a, b] →R be Riemann integrable for some −∞< a ≤b < ∞. Then, f is Lebesgue integrable on [a, b] and the two integrals coincide.
Proof. See the textbook.
Remark 4.3. See a real analysis textbook for the construction of Riemann integrals. A function f : [a, b] →R, is Riemann integrable if and only if f is bounded and the set of discontinuity points of f has Lebesgue measure zero (the latter is known as “Lebesgue’s integrability criterion”). Note that an improperly Riemann integrable function f on [a, ∞) or (a, b] may not be Lebesgue integrable.
References Dennis D. Cox. The Theory of Statistics and Its Applications. Unpub-lished.
Rick Durrett. Probability: Theory and Examples, volume 49. Cambridge university press, 2019.
Sidney Resnick. A Probability Path. Springer, 2019.
Halsey Royden and Patrick Michael Fitzpatrick. Real analysis. China Machine Press, 2010.
Terence Tao. Analysis, volume 1. Springer, 2006.
7 |
12882 | https://www.kristakingmath.com/blog/geometric-random-variables | Probability with geometric random variables — Krista King Math | Online math help
Probability with geometric random variables
What are geometric random variables?
Remember that for a binomial random variable X, we’re looking for the number of successes in a finite number of trials.
For a geometric random variable, most of the conditions we put on the binomial random variable still apply:
each trial must be independent,
each trial can be called a “success” or “failure,”
the probability of success on each trial is constant.
The difference is that for a geometric random variable, we’re looking at how many trials we have to use until we get a certain success. For a binomial random variable, we decided ahead of time on a certain number of trials. But for a geometric random variable, we’ll run an infinite number of trials until we get a success.
For example, “flipping a coin until we get heads” could be described by a geometric random variable. It might take just one flip to get heads, but it could take us 5, 10, or (though very, very unlikely) 10,000 flips.
To find the probability that a success S occurs on the nth attempt, when a success has a probability of p, and therefore a failure has a probability of 1−p, we use this formula:
P(S=n)=p(1−p)n−1
If we look closely at this formula, we see that we’re really just multiplying the probability of failure over and over again until the trial right before we have a success, and then multiplying by the probability of a success.
In other words, if we want to find the probability that we get our first success on the 7th trial, then the probability will be
P(success on the 7th trial)=(probability of failure)6(probability of success)1
Notice that the exponents add to 7, since we needed 7 trials to get the first success.
Answering probability questions with geometric random variables
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Probability of winning a prize on the nth play
Example
I’m playing a game where the probability of winning a prize is 0.7. What is the probability that I don’t win a prize until the 4th time I play the game, assuming each game is independent?
We’re looking for the probability that I don’t “succeed” until the 4th “trial,” so we can represent this with a geometric random variable.
Since the probability of success is 0.7, it means the probability of failure is 0.3. Since I fail 3 times, and then succeed once on the 4th game, the probability of this happening is
P(S=4)=(0.3)3(0.7)1
P(S=4)=(0.027)(0.7)
P(S=4)=0.0189
P(S=4)≈2%
There’s an approximately 2% chance that I don’t win a prize until the fourth game.
More than, less than, at most, and at least probability
More than and less than
Less than
Sometimes we can be asked to find the probability that it takes less than a specific number of trials in order to get our first success. For instance, continuing with the example we just worked through, we could be asked to find the probability that it takes us less than 4 games to win a prize.
This is the same as saying that we win a prize on game 1, 2, or 3. If we call a success S, that means we want either S<4 or S≤3, which mean the same thing in the case of a geometric random variable.
P(S<4)=P(S=1)+P(S=2)+P(S=3)
The probability of success is 0.7 and the probability of failure is 0.3. When S=1, that means we have 0 failures before we then have 1 success. When S=2, that means we have 1 failure and then 1 success. When S=3, that means we have 2 failures and then 1 success.
P(S<4)=(0.3)0(0.7)1+(0.3)1(0.7)1+(0.3)2(0.7)1
P(S<4)=(1)(0.7)+(0.3)(0.7)+(0.3)2(0.7)
P(S<4)=0.7+0.21+(0.09)(0.7)
P(S<4)=0.7+0.21+0.063
P(S<4)=0.973
P(S<4)=97.3%
At most
This is slightly different than being asked the probability that it takes us less than 4 games to win a prize. If it takes less than 4 games to win, that means we get a prize in the third game, or earlier. But if it takes us at most 4 games to win, that means we could win a prize in the fourth game. We could write that as S<5 or as S≤4. But either way, we fail no more than 3 times and then succeed in the fourth game, at the latest.
More than
Similarly, we’ll be asked to find the probability that it takes more than a specific number of trials in order to get our first success. For instance, continuing with the same example, we could be asked to find the probability that it takes more than 2 games for us to win a prize.
Remember that all probability distributions add to 1. If we’re looking for the probability that it takes more than 2 trials to win a prize, we can find the probability of winning on the first trial and the probability of winning on the second trial, and then subtract those probabilities from 1, which will give us all the total probability of all outcomes, other than winning on the first or second game.
So the probability that it takes more than 2 games to win is
P(S>2)=1−P(S≤2)
P(S>2)=1−[(0.3)0(0.7)1+(0.3)(0.7)1]
P(S>2)=1−[(1)(0.7)+(0.3)(0.7)]
P(S>2)=1−(0.7+0.21)
P(S>2)=1−0.91
P(S>2)=0.09
P(S>2)=9%
Keep in mind that we also could have written S>2 as S≥3, or S≤2 as S<3.
At least
This is slightly different than being asked the probability that it takes us more than 2 games to win a prize. If it takes more than 2 games to win, that means we don’t get a prize until the third game. But if it takes us at least 2 games to win, that means we could win a prize in the second game. We could write that as S>1 or as S≥2. But either way, we failed once and then succeeded sometimes in the second game or later.
P(S≥2)=1−P(S≤1)
P(S≥2)=1−P(S=1)
P(S≥2)=1−(0.3)0(0.7)1
P(S≥2)=1−(1)(0.7)
P(S≥2)=1−0.7
P(S≥2)=0.3
P(S≥2)≈30%
Mean, variance, and standard deviation
Mean
The mean μX of a geometric random variable, which can also be called the expected value E(X) is given by
μX=E(X)=p1
where the probability of a success on a trial is p, and X is the number of independent trials required to get the first success.
So in our example from this section where we have a 70% chance of winning a prize, the mean is
μX=0.71≈1.43
This means you should expect to win the game if you play about one or two times.
Variance and standard deviation
The variance σX2 of a geometric random variable is given by
σX2=p21−p
and standard deviation is the square root of the variance. Therefore, the variance of the geometric random variable we’ve been working with is
σX2=0.721−0.7
σX2=0.490.3
σX2≈0.61
and the standard deviation is
√σX2≈√0.61
σX≈0.78
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12883 | https://pecheyponderings.wordpress.com/2025/08/22/shantaram-by-gregory-david-roberts/ | Shantaram, by Gregory David Roberts – Book Reviews to Ponder
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Shantaram, by Gregory David Roberts
Posted on August 22, 2025 June 23, 2025 by canadamatt
Nine stars
Returning to read Gregory David Roberts’ epic novel again, I found myself drawn to the complexities and nuances embedded throughout the text. This is not simply a novel, but an epic journey and discovery of a man who seeks to erase his past and rediscover the intricacies he finds within. I could not get enough of this book and hope others will find something within it. While it is long, its richness should also be seen as a benefit for the dedicated and curious reader. A stunning book that has Gregory David Roberts firmly cemented within the inner circle of my favourite authors, at least for the two fiction publications he produced.
As the novel opens, the reader is introduced to Lin, a man who has escaped his Australian jail. Lin arrives in Bombay, hoping to hide in India’s vast population. Early on, Lin is forced to realise that India is a beast unlike any other; culturally, racially, and economically. It is, however, home to many who have the same idea, hiding from their criminal pasts elsewhere. These include Karla Saarinen, a woman who occupies Lin’s mind and dreams from the moment he lays eyes on her.
As Lin befriends others who have recently arrived in-country, seeking to blend into the billions around him with vague and beige backstories, he meets a tour guide, Prabaker (Prabu). Their connection is almost instantaneous, soon becoming an entertaining pair throughout the narrative. Prabu is able to help Lin make numerous connections in and around the city. While they venture out to better explore Bombay and eventually other parts of the country, Lin learns the cultural differences between India and his Australian upbringing.
As Prabu and Lin continue their adventures, the latter finds himself living in the city’s slums and opens a medical clinic to cater to the poorest population, where Lin becomes involved with the shady underworld and black market living. Throughout the book, Lin crosses paths with those whose simple conversations turn philosophical and force him to digest complex analyses to the universe’s most basic concepts. When offered a position working in forged passports by the Bombay Mafia, Lin accepts, if only to explore new pathways to survival. His living in the slums of Bombay prove not only eye opening, but life changing in ways that the reader can only understand by being enveloped in the larger narrative.
Even as Lin is able to build himself up in his new homeland, he is broken by the cruelest and most sadistic Indians, especially when his identity is learned and extradition considered. Roberts offers so much in this narrative that it is hard to summarise or believe that this is the life of a single man on the run. However, where truth ends and fiction commences, the reader is permitted a front seat for everything and the chance to change alongside Lin throughout. A must read by any and all who want to offer up all they feel they know, only to finish the book and question everything.
Set in the early to mid-1980s, the story weaves together a collection of vignettes within Lin’s Indian life, while also telling an overarching story of change and progress. I have read that some criticise Roberts for being too free with his truths and duping the reader, though I must say that fiction is all about embellishment or at least working with clay and forming it into an image of your choosing. Roberts’ writing style is so blunt and yet smooth that the reader cannot help but get lost therein. The daunting size of the book should not deter the interested reader, as the vignettes play out easily and the characters are rich in their backstories and mesh well with the larger tale. Roberts has certainly held back little in this account of his ‘life on the run’, but also offers gaps significant enough to keep scores of questions floating in the minds of the attentive reader. Will these be resolved and if so, how does it all play into the narrative Roberts presents? The second volume of this quasi-memoir should tell more, though the bar has already been set quite high. I am eager to see how the detail will continue and what Roberts has to say with the handful of characters still involved in Lin’s life. This is a brilliant piece of work and I can only imagine what is to come.
I cannot finish this review without commenting on the narrator of the audiobook version of this massive tome. Humphrey Bower brought the story to life, from his melodious Australian accent in the narrative to the countless accents that he brought out to give characters their personality. I adore Bower’s work and his dedication to another favourite author of mine made me wonder, when first I listened, if this was that writer using another name. Powerful and daunting, Bower deserves a shout out for his reading of this piece. I am pleased that I will be able to acquire an audiobook version of the second volume this time around, as I had to physically read it when first I found it years ago!
Kudos, Mr. Roberts for this epic story. With simplistic writing and complex threads, a vast array of readers will surely enjoy this book. Onto the sequel, which one can hope is as exciting and life-altering.
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Tools & Reference>Nephrology
Hypercalcemia Clinical Presentation
Updated: Apr 26, 2023
Author: Mahendra Agraharkar, MD, MBBS, FACP, FASN; Chief Editor: Vecihi Batuman, MD, FASN more...;)
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Sections
Hypercalcemia
Sections Hypercalcemia
Overview
Practice Essentials
Pathophysiology
Etiology
Epidemiology
Mortality/Morbidity
Prognosis
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Presentation
History
Physical Examination
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DDx
Workup
Approach Considerations
Imaging Studies
Electrocardiography
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Treatment
Medical Care
Surgical Care
Consultations
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Guidelines
Medication
Medication Summary
Bisphosphonates
Monoclonal Antibodies, Endocrine
Antidote, hypercalcemia agents
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References;)
Presentation
History
The mnemonic "stones, bones, abdominal moans, and psychic groans" describes the constellation of symptoms and signs of hypercalcemia. These may be due directly to the hypercalcemia, to increased calcium and phosphate excretion, or to skeletal changes.
The presentation in a patient with hypercalcemia varies with how fast and how far the calcium level rises, as well as the sensitivity of the individual to elevated calcium levels. Mild prolonged hypercalcemia may produce mild or no symptoms, or recurring problems such as kidney stones. Sudden-onset and severe hypercalcemia may cause dramatic symptoms, usually including confusion and lethargy, possibly leading quickly to death.
Central nervous system effects include the following:
Lethargy
Weakness
Confusion
Coma
Renal effects include the following:
Polyuria
Nocturia
Dehydration
Renal stones
Renal failure
Gastrointestinal effects include the following:
Constipation
Nausea
Anorexia
Pancreatitis
Gastric ulcer
Cardiac effects include syncope from arrhythmias.
Next:
Physical Examination
Most patients with hypercalcemia do not have any specific findings on physical examination. Those with higher calcium levels may have findings that are more striking. Evidence of the underlying cause may be found, such as a suggestive breast mass in someone with hypercalcemia secondary to malignancy.
Nervous system findings include the following:
Confusion
Hypotonia
Hyporeflexia
Paresis
Coma
Renal findings include the following:
Volume depletion
Signs of renal failure
Gastrointestinal findings include the following:
Fecal impaction (from constipation)
Signs of pancreatitis
Signs of malignancy (eg, enlarged liver or masses)
Cardiovascular findings include the following:
Arrhythmias
Hypotension
Ophthalmic findings may include band keratopathy, which is calcium precipitation in a horizontal band across the cornea in the palpebral aperture.
Previous
Differential Diagnoses
References
Walker MD, Shane E. Hypercalcemia: A Review. JAMA. 2022 Oct 25. 328 (16):1624-1636. [QxMD MEDLINE Link].
Sadiq NM, Naganathan S, Badireddy M. Hypercalcemia. 2022 Jan. [QxMD MEDLINE Link].[Full Text].
Zagzag J, Hu MI, Fisher SB, Perrier ND. Hypercalcemia and cancer: Differential diagnosis and treatment. CA Cancer J Clin. 2018 Sep 21. [QxMD MEDLINE Link].[Full Text].
Hoenderop JG, Nilius B, Bindels RJ. Molecular mechanism of active Ca2+ reabsorption in the distal nephron. Annu Rev Physiol. 2002. 64:529-49. [QxMD MEDLINE Link].
McKay CP, Portale A. Emerging topics in pediatric bone and mineral disorders 2008. Semin Nephrol. 2009 Jul. 29(4):370-8. [QxMD MEDLINE Link].
Guarnieri V, Canaff L, Yun FH, et al. Calcium-Sensing Receptor (CASR) Mutations in Hypercalcemic States: Studies from a Single Endocrine Clinic Over Three Years. J Clin Endocrinol Metab. 2010 Feb 17. [QxMD MEDLINE Link].
Nissen PH, Christensen SE, Wallace A, et al. Multiplex ligation-dependent probe amplification (MLPA) screening for exon copy number variation in the calcium sensing receptor gene: no large rearrangements identified in patients with calcium metabolic disorders. Clin Endocrinol (Oxf). 2009 Nov 11. [QxMD MEDLINE Link].
Luna-Cabrera F, Justicia-Rull EA, Caricol-Pérez MP, et al. Incidence of hypercalcemia, hypercalciuria and related factors in patients treated with recombinant human parathyroid hormone (1-84). Minerva Med. 2012 Apr. 103(2):103-10. [QxMD MEDLINE Link].
Nanmoku K, Shinzato T, Kubo T, Shimizu T, Yagisawa T. Prevalence and predictors of early hypercalcemia after kidney transplantation: a nested case-control study within a cohort of 100 patients. Clin Exp Nephrol. 2018 Aug 18. [QxMD MEDLINE Link].
Vanstone MB, Oberfield SE, Shader L, Ardeshirpour L, Carpenter TO. Hypercalcemia in Children Receiving Pharmacologic Doses of Vitamin D. Pediatrics. 2012 Mar 12. [QxMD MEDLINE Link].
Haridas K, Holick MF, Burmeister LA. Hypercalcemia, nephrolithiasis, and hypervitaminosis D precipitated by supplementation in a susceptible individual. Nutrition. 2020 Jun. 74:110754. [QxMD MEDLINE Link].
Griebeler ML, Kearns AE, Ryu E, Thapa P, Hathcock MA, Melton LJ 3rd, et al. Thiazide-Associated Hypercalcemia: Incidence and Association With Primary Hyperparathyroidism Over Two Decades. J Clin Endocrinol Metab. 2016 Mar. 101 (3):1166-73. [QxMD MEDLINE Link].
Vargas-Poussou R, Mansour-Hendili L, Baron S, et al. Familial Hypocalciuric Hypercalcemia Types 1 and 3 and Primary Hyperparathyroidism: Similarities and Differences. J Clin Endocrinol Metab. 2016 May. 101 (5):2185-95. [QxMD MEDLINE Link].
John SM, Sagar S, Aparna JK, Joy S, Mishra AK. Risk factors for hypercalcemia in patients with tuberculosis. Int J Mycobacteriol. 2020 Jan-Mar. 9 (1):7-11. [QxMD MEDLINE Link].[Full Text].
Gastanaga VM, Schwartzberg LS, Jain RK, Pirolli M, Quach D, Quigley JM, et al. Prevalence of hypercalcemia among cancer patients in the United States. Cancer Med. 2016 Jun 5. [QxMD MEDLINE Link].
Royer AM, Maclellan RA, Daniel Stanley J, Willingham TB, Heath Giles W. Hypercalcemia in the emergency department: a missed opportunity. Am Surg. 2014 Aug. 80(8):732-5. [QxMD MEDLINE Link].
Cafforio P, Savonarola A, Stucci S, De Matteo M, Tucci M, Brunetti AE, et al. PTHrP produced by myeloma plasma cells regulates their survival and pro-osteoclast activity for bone disease progression. J Bone Miner Res. 2014 Jan. 29(1):55-66. [QxMD MEDLINE Link].
Goldner W. Cancer-Related Hypercalcemia. J Oncol Pract. 2016 May. 12 (5):426-32. [QxMD MEDLINE Link].
Vallet N, Ertault M, Delaye JB, Chalopin T, Villate A, Drieu La Rochelle L, et al. Hypercalcemia is associated with a poor prognosis in lymphoma a retrospective monocentric matched-control study and extensive review of published reported cases. Ann Hematol. 2020 Feb. 99 (2):229-239. [QxMD MEDLINE Link].
Alsirafy SA, Sroor MY, Al-Shahri MZ. Hypercalcemia in advanced head and neck squamous cell carcinoma: prevalence and potential impact on palliative care. J Support Oncol. 2009 Sep-Oct. 7(5):154-7. [QxMD MEDLINE Link].
Durant E, Singh A. ST elevation due to hypercalcemia. Am J Emerg Med. 2017 Jul. 35 (7):1033.e3-1033.e6. [QxMD MEDLINE Link].
Sonoda K, Watanabe H, Hisamatsu T, Ashihara T, Ohno S, Hayashi H, et al. High Frequency of Early Repolarization and Brugada-Type Electrocardiograms in Hypercalcemia. Ann Noninvasive Electrocardiol. 2016 Jan. 21 (1):30-40. [QxMD MEDLINE Link].
Schutt RC, Bibawy J, Elnemr M, Lehnert AL, Putney D, Thomas AS, et al. Case report: Severe hypercalcemia mimicking ST-segment elevation myocardial infarction. Methodist Debakey Cardiovasc J. 2014 Jul-Sep. 10 (3):193-7. [QxMD MEDLINE Link].[Full Text].
Makras P, Papapoulos SE. Medical treatment of hypercalcaemia. Hormones (Athens). 2009 Apr-Jun. 8(2):83-95. [QxMD MEDLINE Link].[Full Text].
Turner JJO. Hypercalcaemia - presentation and management. Clin Med (Lond). 2017 Jun. 17 (3):270-273. [QxMD MEDLINE Link].
Brooks M. FDA Approves New Indication for Denosumab (Xgeva). Medscape Medical News. Available at Accessed: December 13, 2014.
Hu MI, Glezerman IG, Leboulleux S, Insogna K, Gucalp R, Misiorowski W, et al. Denosumab for treatment of hypercalcemia of malignancy. J Clin Endocrinol Metab. 2014 Sep. 99(9):3144-52. [QxMD MEDLINE Link].[Full Text].
Bergenfelz AO, Jansson SK, Wallin GK, et al. Impact of modern techniques on short-term outcome after surgery for primary hyperparathyroidism: a multicenter study comprising 2,708 patients. Langenbecks Arch Surg. 2009 Jul 18. [QxMD MEDLINE Link].
Low TH, Clark J, Gao K, et al. Outcome of parathyroidectomy for patients with renal disease and hyperparathyroidism: predictors for recurrent hyperparathyroidism. ANZ J Surg. 2009 May. 79(5):378-82. [QxMD MEDLINE Link].
[Guideline] Dickens LT, Derman B, Alexander JT. Endocrine Society Hypercalcemia of Malignancy Guidelines. JAMA Oncol. 2023 Jan 13. [QxMD MEDLINE Link].[Full Text].
Media Gallery
Investigations flowchart.
Vitamin D metabolism.
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Contributor Information and Disclosures
Author
Mahendra Agraharkar, MD, MBBS, FACP, FASN Clinical Associate Professor of Medicine, Baylor College of Medicine; President and CEO, Space City Associates of NephrologyMahendra Agraharkar, MD, MBBS, FACP, FASN is a member of the following medical societies: American College of Physicians, American Society of Nephrology, National Kidney FoundationDisclosure: Nothing to disclose.
Coauthor(s)
O David Dellinger, III, MD Assistant Professor, Departments of Family Medicine and Internal Medicine, University of Alabama School of Medicine at BirminghamO David Dellinger, III, MD is a member of the following medical societies: American Academy of Hospice and Palliative Medicine, American College of Physicians-American Society of Internal Medicine, American Geriatrics Society, AMDA - The Society for Post-Acute and Long-Term Care Medicine, American Society of Addiction MedicineDisclosure: Nothing to disclose.
Arun Kumar Gangakhedkar, MD, FRACP Consultant in General Pediatrics, Starship Children's Hospital/Waitakere Hospital, New ZealandDisclosure: Nothing to disclose.
Specialty Editor Board
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug ReferenceDisclosure: Received salary from Medscape for employment.
Christie P Thomas, MBBS, FRCP, FASN, FAHA Professor, Department of Internal Medicine, Division of Nephrology, Departments of Pediatrics and Obstetrics and Gynecology, Medical Director, Kidney and Kidney/Pancreas Transplant Program, University of Iowa Hospitals and ClinicsChristie P Thomas, MBBS, FRCP, FASN, FAHA is a member of the following medical societies: American College of Physicians, American Heart Association, American Society of Nephrology, Royal College of PhysiciansDisclosure: Nothing to disclose.
Chief Editor
Vecihi Batuman, MD, FASN Professor of Medicine, Section of Nephrology-Hypertension, Deming Department of Medicine, Tulane University School of MedicineVecihi Batuman, MD, FASN is a member of the following medical societies: American College of Physicians, American Society of Hypertension, American Society of Nephrology, Southern Society for Clinical InvestigationDisclosure: Nothing to disclose.
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12885 | https://madformath.com/calculators/basic-math/base-converters/decimal-to-base-5-converter-with-steps/decimal-to-base-5-converter-with-steps | Decimal to
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Decimal to Base 5 Converter
BASIC MATH CALCULATORS
TOOL: DECIMAL TO BASE 5 CONVERTER (WITH STEPS)
⇩ BASE N TO DECIMAL CONVERTERS ⇩
23456789111213141516
DECIMAL TO BASE 5 CONVERTER (WITH STEPS)
Enter a number.
RESULT
(214)10 = (1324)5
DESCRIPTIONS
WHOLE NUMBER PART
Divide the number repeatedly by 5 until the quotient becomes 0.
When 214 is divided by 5, the quotient is 42 and the remainder is 4.
When 42 is divided by 5, the quotient is 8 and the remainder is 2.
When 8 is divided by 5, the quotient is 1 and the remainder is 3.
When 1 is divided by 5, the quotient is 0 and the remainder is 1.
Write the remainders from bottom to top.
(214)10 = (1324)5
FRACTIONAL PART
OVERALL RESULT
OTHER INFORMATION
Click here to see the binary equivalent of 214
Click here to see the octal equivalent of 214
Click here to see the hexadecimal equivalent of 214
Download Solution
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⇩ BASE N TO DECIMAL CONVERTERS ⇩
23456789111213141516
INFORMATION
DECIMAL TO BASE 5 CONVERSION
WHOLE NUMBERS
We apply the following rules to convert a decimal number to base 5.
We divide the decimal number by 5 repeatedly until the quotient becomes 0.
Starting at the least significant digit, we write the remainders in the same order of divisions.
For example, to convert decimal 42 to base 5, we divide 42 by 5 repeatedly until the quotient becomes 0.
When we divide 42 by 5, the quotient is 8 and the remainder is 2. Thus, 2 is the least significant digit of the base 5 equivalent.
We continue the algorithm with 8.
When we divide 8 by 5, the quotient is 1 and the remainder is 3. Then 3 is the second least significant digit of the base 5 number.
Finally, we divide 1 by
5. When we do this operation, the quotient is 0 and the remainder is 1. Because the quotient is 0, we stop the procedure. Then we write the last
remainder to the most significant digit of the base 5 number. To sum up, the base 5 representation of 42 is 132.
(42)10 = (132)5
DECIMAL NUMBERS
In case, the decimal number is not an integer, we can convert the whole number and fractional parts separately and add the base 5 equivalents up.
To convert the fractional part of a decimal number, we apply the following rules.
We multiply the fractional part by 5 repeatedly until the product becomes an integer or the number of significant digits is sufficient for our calculations.
At each step, we write the integer part of the rightmost digit to the fractional part of the base 5 number. We continue with the fractional part of the product.
For example, to convert 42.304 to base 5, we multiply the fractional part by 5 repeatedly until we find an integer.
0.304 × 5 = 1.52
0.52 × 5 = 2.6
0.6 × 5 = 3
The fractional part of 42.304 is 0.304. When we multiply 0.304 by 5, the result is 1.52. The integer part of
1.52 is 1. Thus we write 1 to the first digit on the RHS of the radix point.
0.1
We continue with the fractional part of 1.52. When we multiply 0.52 by 5, the result is 2.6. We write the integer part of
2.6 to the next digit of the base 5 number.
0.12
The fractional part of 2.6 is 0.6. Therefore, we continue with this number. The product of 0.6 and 5 is equal to 3. Because
3 is an integer we write it to the next digit of the base 5 number and stop multiplications.
(0.304)10 = (0.123)5
Base 5 representation of 42.304 is equal to the sum of base 5 representations of 42 and 0.304. Thus, decimal 42.304 is equal to
132.123 in base 5.
(42.304)10 = (42)10 + (0.304)10
= (132)5 + (0.123)5
= (132.123)5
WHAT IS DECIMAL TO BASE 5 CONVERTER?
Decimal to base 5 converter,
Computes the base 5 equivalent of the entered decimal number and
Describes each step of the conversion for both whole number and fractional parts,
HOW TO USE DECIMAL TO BASE 5 CONVERTER?
You can use decimal to base 5 converter in two ways.
USER INPUTS
You can enter a decimal number to the input box and click on the "CONVERT" button. The result and
explanations appaer below the calculator
#### RANDOM INPUTS
You can click on the DIE ICON next to the input box. If you use this property, a random decimal number is generated and entered to the calculator, automatically.
You can see the result and explanations below the calculator. You can create your own examples and practice using this property.
#### CLEARING THE INPUT BOX
To check the base 5 equivalent of other decimals you can clear the input box by clicking on the CLEAR button under the input box.
#### COPYING & DOWNLOADING THE SOLUTION
You can copy the generated solution by clicking on the "Copy Text" link, appaers under the solution panel.
Even you can download the solution as an image file with .jpg extension if you click on the "Download Solution" link at the bottom of the solution panel. You can share the
downloaded image file.
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12886 | https://ttaomae.github.io/posts/2014/08/solving-the-tetris-cube | Todd Taomae
Solving the Tetris Cube
• Posted in Computer Science
About a year ago someone lent me a Tetris Cube which has been sitting in my office since then. Being a fan of puzzles I thought it would be fun to tinker away at it occasionally during any down time or if I needed to clear my mind. For a few months I would spend a few minutes here and there trying to solve it. After a while I figured that my puzzle solving skills would not be enough and decided to put my programming skills to use. Since then I have hardly touched the puzzle and I haven’t had much time to try to create a solver until recently.
Polyomino Tiling
Rather than jumping straight into solving the Tetris Cube, I decided that a solving a similar, but much easier, problem might be a good place to start. I decided to try solving polyomino tiling puzzles first. Anyone who has ever played or knows anything about Tetris should be familiar with polyominoes. In case you aren’t familiar with the game it involves trying to place various falling polyominoes such that you fill up entire rows with no gaps. Polyominoes are simply shapes made from adjacent squares. Tetris uses tetrominoes which are polyominoes consisting of four squares.
The goal of a polyomino tiling puzzle is similar to Tetris except instead of dealing with random falling shapes, you are given a set of polyominoes which you must place in order to fill up a specified shape. Usually you can modify the polyominoes by rotating them by 90 degrees at a time or by flipping them over.
Backtracking
My initial thought was that this should be easy to solve with a backtracking algorithm. Backtracking is probably better explained in relation to a different problem, such as the eight queens puzzle. In the eight queens puzzle the goal is to place eight queens on a chessboard such that none of the queens can attack any other queen. You might approach this problem by placing a queen on the first column of the first row, then placing a queen on the third column of the second row (the third column is the first one which does not lead to the queens attacking each other), and so on for the following columns. At some point you won’t be able to place any more queens, which is where you begin to backtrack. You remove the last queen that you placed and move it to the next valid column and try to move down the rows again. Each time you’ve tried all possible moves for a single row, you backtrack one more row and then repeat the process again.
In the case of the tiling puzzle, the rows would be analogous to the different polyominoes and the columns would be analogous to the different positions and orientations of the polyominoes. For example, if we are given three pieces A, B, and C. We might place A in the top left corner, then B to the right of that. Then we may find that there is no way to place C so we rotate B and try again. After we’ve tried all rotations of B we can move it and try all rotations again. After we’ve tried all positions and rotations of B we go back to A and rotate it, and so on.
Implementation
I’ve had some experience with backtracking before (to solve Sudoku puzzles) so implementing the algorithm wasn’t difficult. I actually found it more difficult to decide on a good representation of polyominoes and to implement the methods to rotate and flip the polyominoes.
I ended up with a 2-dimensional boolean array for both the puzzle and the polyominoes. The polyominoes additionally had an x and y position to allow them to move.
After everything was implemented I tested my solver on some very simple problems such as fitting a single piece into a grid of the same shape and fitting for square tetrominoes into a 4x4 square.
Once I was confident that my solver was working I tried it on some of the examples shown here and here. I was quite disappointed to find that it took much longer than expected to solve. The 5x8 tetromino puzzle took several minutes and none if the pentomino puzzles finished within the time I let them run (I think I let one run for about an hour before giving up).
My backtracking algorithm at this point was very naive. It would place pieces as I described earlier and only backtrack if there was no space to place the next piece. This would sometimes lead the algorithm down an obviously wrong path. For example, if the first piece is placed in such a way that there is a 1x1 hole in the corner, it is obvious that nothing can fit in there (except a monomino). However the algorithm would continue to fill in other pieces and might be stuck on that path for quite some time. In order to partially alleviate this problem I added a simple heuristic to help prune the search space. Before placing any additional pieces, I would find the size of the smallest connected gap in the puzzle. If the smallest polyomino remaining is larger than the gap, then the algorithm will immediately backtrack. Although this approach was not perfect it did help reduce the run time. However, the pentomino puzzles were still not going to be completed in any reasonable amount of time.
At this point it was pretty clear that this approach probably wouldn’t scale well into three dimensions even if I implemented better heuristics for further pruning the search space. So I decided I would need to find another way.
Algorithm X and Dancing Links
Luckily Wikipedia led me to an answer pretty quickly. I don’t remember exactly how I got there, but it was just a few clicks away from the polyomino page. In particular I was led to the pages on Knuth’s Algorithm X and Dancing Links.
Exact Cover
In order to understand Algorithm X, you must first understand what an exact cover problem is. I will try to explain it without using too much mathematical or technical terms.
Suppose you have a set of things numbered 1 through 10. You are given a set of options that each contain one or more of those things. For example you might have option A = {1, 2, 5, 6}, option B = {2, 3, 7, 10}, and so on. Notice that you can have the same thing in multiple options. The goal of an exact cover problem is to select a set of options such that each thing is chosen exactly once.
If that was too abstract suppose that the 10 things are different kinds of candy. We want to know all the ways that we can pick five pairs of candy such that each one is selected only once. In this case our options would look something like this: {1, 2}, {1, 3}, …{2, 3}, {2, 4}, … {7, 9}, {8, 9}. We want to pick 5 options such that each number appears only once.
We can create a variation of this problem where we have 10 pieces of candy that we want to distribute to 5 children, such that they each get two pieces. Let us further suppose that each child only likes certain types of candy. Since we only want to give each child two pieces of candy, we must be sure that we select each child only once. In order to maintain this restriction we should add 5 more things to our initial set. Let us label them a through e, representing the 5 children.
Suppose child a only likes candies 1, 2, 3, and 4. To represent this we would include the options {a, 1, 2}, {a, 1, 3}, {a, 1, 4}, {a, 2, 3}, {a, 2, 4}, and {a, 3, 4}. We would include similar options for each child depending on what their preferences are.
Believe it or not, a polyomino tiling puzzle can be solved by direct analogy to this problem. Instead of children, we are dealing with the different polyominoes and instead of each child having different possible combinations of candy, each polyomino has different combinations of coordinates on the shape that they are trying to fill. So a square tetromino (often referred to as an ‘O’ tetromino due to its resemblance to the letter) might have options {O, 00, 01, 10, 11}, {O, 10, 11, 20, 21}, and so on, where the first digit of each number represents an x-coordinate and the second digit is a y-coordinate.
Algorithm X
Now that we understand the exact cover problem and how we can use it to solve a polyomino tiling puzzle, we need a way to solve the exact cover problem. It turns out that this problem is actually pretty difficult. In fact, it is NP-complete, which basically means that there is no known way for the problem to be solved in polynomial time. In very simple terms it basically means that as the problem size grows, the runtime of the algorithm will grow very quickly. At this point I simply hoped that my problem was simple enough to be solved in a reasonable amount of time.
Luckily, smarter people than I have worked on solving this problem in a relatively efficient manner. In particular, Donald Knuth developed Algorithm X to solve this problem. Algorithm X relies on a particular representation of the exact cover problem. It turns out that you can represent an exact cover problem using a binary matrix. For example, going back to (a smaller version of) the candy problem, the following matrix might represent a few options.
( \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline & a & b & c & 1 & 2 & 3 & 4 & 5 & 6 \ \hline \bf{A} & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \ \hline \bf{B} & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \ \hline \bf{C} & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \ \hline \end{array} )
In this example, A represents child a receiving candies 1 and 2, B represents child a receiving candies 1 and 3, and C represents child b receiving candies 2 and 5.
Given this representation, we can start to understand Algorithm X. If you prefer pseudo-code, you can simply refer to the Wikipedia page. If you want a complete and in depth discussion of the algorithm, you may want to read Knuth’s paper. In this post I will try to describe it in plain English.
The first step is to choose a column. Any one will do, but Knuth recommends the one with the fewest 1’s. Then we go down that column and stop at every row where there is a 1. For example, if we chose column a, we would stop at rows A and B. Each time we stop at a row, we delete that row as well as any column with a 1 in that row. So, if we stop at row A, we would delete row A and columns a, 1, and 2. The row that we just deleted is part of our partial solution.
We would then repeat this process by selecting a new column (of the ones remaining after the deletions in the previous step). If we reach a point where the matrix is empty, then we’ve found a solution. If we reach a point where there are no columns with any 1’s in it, then we’ve hit a dead end and there is no way to find a solution by continuing to search. Either way, we must backtrack. In this case, that means we undelete everything from the previous step, then move to the next row.
Dancing Links
Luckily for me, Knuth not only developed an algorithm for solving the exact cover problem, he also described an efficient implementation, since a naive implementation would spend a large portion of the time searching for 1’s. It seems he was more interested in the implementation than the algorithm, as evidenced by his complete lack of effort in naming the algorithm.
He refers to the implementation technique as Dancing Links. To understand the name, we must talk a little about the data structure that he uses. In order to avoid having to search for 1’s in what will usually be a sparse matrix, only the 1s will be stored. They are stored as a node of a type of doubly linked list, which is where the “links” part of the name come from. The “dancing” part comes from the fact that in the process of deleting and undeleting rows and columns the links will be undone and redone in a sort of “dance” using the technique outlined in the code below.
// remove node from list node. left. right = node. right; node. right. left = node. left;// node still points to its neighbors// we can make the neighbors point back to node// which adds it back to the list node. left. right = node; node. right. left = node;
I won’t go into full detail about the implementation; again, you can refer to the Wikipedia page or the paper for more details. However, I need to clarify what I meant by a “type of doubly linked list.” You could maybe call it a “double circular doubly linked list”. Again, each 1 in the matrix is represented by a node. Each node, points to the nodes above and below and to the left and right and, since it is circular, of course the bottom node points to the top node and vice versa and similarly for the far left and right nodes. There are some other components, but again, I won’t go into any more detail.
Polyomino Tiling. Again
After completing my implementation of Dancing Links, I once again tried solving some polyomino tiling problems. I would once again need a way to represent the polyominoes as well as a way to convert the tiling problem into an exact cover problem. I decided to use a different approach, both for the practice of creating and implementing different designs and because of the way the problem is now framed. Previously I would check if any polyominoes are overlapping, which involves checking whether or not a polyomino occupies a certain square. Now, constructing each row of the exact cover matrix only requires knowing which squares the polyomino does occupy. I created a simple class to represent the coordinates of a single square of a polyomino. I could have put these into a list, but since I knew each coordinate should only occur once per polyomino, I used a set instead.
The next step was to convert the polyominoes into an exact cover matrix. For rectangular puzzles, I simply used a height and width, but I also provided a way to create irregular shaped puzzles. Of course you also need the puzzle pieces as well as names for each of the pieces so that they can be given a column label on the matrix. For this I used a map of Strings (the labels) to polyominoes (the piece with that label). The remainder of the columns are all the coordinates that are part of the puzzle.
In order to convert a polyomino into a row on the matrix, I check that each coordinate of the polyomino is also a coordinate of the puzzle. If it is not, it is discarded. Otherwise, it is converted into a row by basically setting 1’s in the appropriate columns. Of course I then repeat the process for each transformation of each polyomino, which includes rotations (turning), reflections (flipping), and translations (moving).
I was glad to see that they were solved in seconds rather than hours. I didn’t perform any extensive testing but I did perform some basic sanity checks before scaling up to three dimensions. First I made sure that the solutions actually worked by going through a few manually and making sure that the pieces actually fit in the puzzle properly. I also compared some numbers from my implementation and solutions to those found in Knuth’s Dancing Links paper (where he discusses different exact cover problems, including, but not limited to, polyomino tiling). One particular problem that he focused on was the problem of using one of each pentomino in an 8x8 square with a 2x2 square removed from the middle. He claimed that the exact cover matrix should have 1568 rows, which is exactly how many I had. He reported that there were 65 unique solutions, while I found 520 solutions. However, my solutions are not unique when taking into account rotations and reflections. When taking into account that there are 8 rotations and reflections (including the original) for each solution, my 520 matches his 65.
Unfortunately I have not found a generalized way to eliminate these duplicate solutions. For the particular problem mentioned above, Knuth accomplishes this by restricting certain pieces such that the rotations and reflections are impossible. Basically he restricts a particular piece to a certain quadrant of the puzzle since the rotations and reflections would cause the piece to show up in a different quadrant.
Solving the Soma Cube
It was now almost time to finally solve the Tetris Cube. However, I once again decided that it might be a good idea to take a smaller step towards the final problem instead. This time I tried to solve the Soma Cube. In order to do that I would need to somehow represent the polycubes (which is the name for shapes made up of adjacent cubes, like the ones used in the Tetris Cube) and convert the puzzle into an exact cover matrix. For this I followed pretty much exactly the same process as for the polyomino puzzles.
According to Wikipedia, there are 240 unique solutions to the Soma Cube. However, I came up with 11,520 solutions. When taking into the 24 rotations of each solution (imagine a six-sided die, there are six sides that can face up and you can rotate it by 90 degrees, while keeping the same side up, four times; 6 4 = 24), that brings my number down to 480. I’m not completely sure why it is still off by a factor of 21, but at least it’s an even multiple which probably means that I’ve done everything reasonably correctly.
Parallel Dancing Links
After seeing the Soma Cube solved in a matter of minutes, I was quite disappointed to find that, like my first attempts at solving the polyomino puzzles, it was going to take many hours to complete the Tetris Cube. However, one major difference was that this time it would find all solutions instead of just one. Regardless, I wasn’t quite satisfied with the performance and decided to try to parallelize the algorithm.
My initial idea was fairly simple. During the first pass through the matrix, instead of going down the column one row at a time, you could do several rows in parallel, since the operations are all independent of each other. Sort of. What I mean is that completing one row does not rely on the previous one being completed. However, if you are deleting and undeleting rows and columns in the same Dancing Links structure in parallel, you will quickly run into problems.
The obvious solution would be to make a copy of the Dancing Links. However, before moving onto that solution I did a quick search to see if anyone else has tackled a parallel version of the algorithm and perhaps came up with a better solution. I came across two papers (here and here) which describe similar algorithms which are a slightly more general form of the algorithm I proposed. Instead of simply parallelizing the first pass, you would search up to a certain depth and compute a number of partial solutions. You would then use the partial solutions as a starting point for the parallel search.
Unfortunately neither paper described any way to conveniently manipulate the data structure so I resorted to making a copy for each parallel search. I could not think of an easy way to copy the data structure so I resorted to having it store the exact cover matrix that it was constructed from and simply recreate the Dancing Links from scratch whenever necessary. I would then delete the rows and columns that were covered by the particular partial solution.
Solving the Tetris Cube… Finally
I used my laptop which has a 2.2GHz i7 processor to run the program. It goes up to 2.8GHz with Intel Turbo Boost enabled but I decided to keep it at 2.2GHz so that my laptop didn’t overheat while solving the puzzle with 8 threads (4 cores plus hyper-threading) at full power. After about an hour I found 236136 solutions, which means that, if I’ve done everything correctly, there are probably 9839 unique solutions. This time it divides evenly by 24, but not by 48 (as with the Soma Cube1).
Reflection
I went into this thinking that I would spend a day or two writing a simple backtracking solution and be done with it. I was quite disappointed to find that it would not be so simple. However, it was an interesting journey that I actually enjoyed quite a lot.
While this does show the importance of a good algorithm, I think it is also important to keep in mind that it isn’t necessarily a good idea to jump straight into fastest algorithm. It took me much longer to understand and implement Algorithm X with the Dancing Links structure than it did for me to create the naive backtracking algorithm. If it turned out that the simpler algorithm were good enough for my purposes but I went with the faster but more complicated algorithm first, I might have wasted time researching, understanding, and implementing an unnecessary algorithm. On the other hand, by starting with the simple algorithm I spent a relatively short amount of time on an unnecessary algorithm, but I gained a better understanding of its limitations in the process.
If you are interested, you can find the code on GitHub. It includes the ability to create and solve your own exact cover problems, to create and solve polyomino or polycube puzzles, or to solve the Tetris Cube or Soma Cube.
Since the Tetris Cube matches my expectations about the rotations, it indicates that maybe it is a property of the Soma Cube that gives it extra symmetrical solutions. Looking at the pieces of the Soma Cube, I suspect that it may be the fact that it has two chiral pieces, which means that they are basically mirror images of each other. ↩ ↩2 |
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Convert Cubic Centimeter To Milliliter
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Convert Cubic Centimeter To Milliliter
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Cubic
Centimeter
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1 cubic cm = 1 ml
Convert Cubic Centimeter To Milliliter
Converting cubic centimeters (cm³) to milliliters (mL) is a fundamental process in the realm of volume measurements. Understanding this conversion is crucial in various fields such as science, engineering, medicine, and everyday life where precise volume measurements are essential. This conversion involves a straightforward mathematical relationship between the two units, but exploring the concept in detail can provide valuable insights into the practical applications and significance of this conversion.To begin, let's establish the basic understanding of what cubic centimeters and milliliters represent. A cubic centimeter is a unit of volume that measures the volume of a cube with sides that are one centimeter long. It is commonly abbreviated as cm³. On the other hand, a milliliter is also a unit of volume, but it is based on the metric system and represents one-thousandth of a liter. It is abbreviated as mL. Despite being different units, they are equivalent in value: 1 cm³ is equal to 1 mL.The conversion between cubic centimeters and milliliters is straightforward due to their equivalence. Since 1 cm³ is equal to 1 mL, converting from cubic centimeters to milliliters involves simply using the same numerical value for both. For example, if you have 50 cubic centimeters, it is also equivalent to 50 milliliters. Now, let's delve deeper into the practical significance of converting between these units. In various scientific disciplines such as chemistry, physics, and biology, accurate volume measurements are crucial for conducting experiments, analyzing data, and formulating solutions. Whether it's measuring the volume of a liquid reactant in a chemical reaction or determining the volume of a biological sample in a laboratory assay, precise volume measurements are indispensable. In such contexts, both cubic centimeters and milliliters are commonly used, depending on the preferences of the researcher or the conventions of the field.
About Cubic Centimeter
Cubic centimeters, often abbreviated as cm³, represent a unit of volume in the metric system, widely employed across diverse fields and disciplines due to its simplicity and versatility. One cubic centimeter is defined as the volume of a cube with sides each measuring one centimeter. This unit is particularly significant in scientific research, engineering, medicine, manufacturing, and everyday applications where precise volume measurements are essential. In scientific experiments and research, cubic centimeters are utilized to quantify the volume of liquids, gases, and solids, facilitating accurate data collection and analysis. Engineering disciplines, including mechanical, civil, and aerospace engineering, rely on cubic centimeters to calculate volumes of various components, materials, and fluids, aiding in design, analysis, and optimization processes. In medicine, cubic centimeters are frequently used to measure volumes of medications, bodily fluids, and organs, playing a critical role in dosing, diagnostic imaging, and surgical procedures. Moreover, cubic centimeters find extensive applications in manufacturing industries, where they are employed to determine the volume of raw materials, components, and finished products, ensuring quality control and efficient production processes. Additionally, cubic centimeters are integral to everyday measurements, such as calculating the capacity of containers, fuel tanks, and packaging, as well as determining dimensions in construction and home improvement projects. Understanding and utilizing cubic centimeters facilitate precision, consistency, and accuracy in volume measurements across various sectors, contributing to advancements in science, technology, healthcare, and commerce.
What are the general uses of Cubic centimeters?Cubic centimeters, as a unit of volume in the metric system, find widespread use across numerous industries and applications. In engineering, cubic centimeters are employed to calculate the volume of objects, materials, and fluid systems, aiding in design, analysis, and optimization processes across disciplines such as mechanical, civil, and aerospace engineering. Additionally, in scientific research and experimentation, cubic centimeters are utilized to quantify the volume of liquids, gases, and solids, facilitating accurate data collection and analysis in fields such as chemistry, physics, and biology. In medicine, cubic centimeters play a crucial role in measuring the volume of medications, bodily fluids, and organs, supporting accurate dosing, diagnostic imaging, and surgical procedures. Moreover, in manufacturing and production, cubic centimeters are utilized to determine the volume of raw materials, components, and finished products, ensuring quality control and efficient production processes. Cubic centimeters also have everyday applications, such as calculating container capacities, fuel tank volumes, and packaging dimensions, as well as determining material quantities in construction and home improvement projects. Overall, the general uses of cubic centimeters span across engineering, scientific research, medicine, manufacturing, and everyday measurements, contributing to precision, consistency, and efficiency in volume calculations across diverse fields and industries.
How to Measure Cubic Centimeters To Milliliter?Milliliters = cubic centimeters ÷ 1
ExampleCubic Centimeters To Milliliter ConversionFor example, here's how to convert 5 cubic centimeters to milliliters using the formula above:Milliliters = (5 cm³ ÷ 1) = 5 mLThus, the result is 5 milliliters
About Milliliter
The milliliter (ml), a key unit of volume in the metric system, represents one-thousandth of a liter. Derived from the Latin term "milli" meaning one thousand, the milliliter is a precise measure widely utilized in scientific, medical, and everyday contexts. In scientific experiments, milliliters are commonly used to measure liquid volumes, ensuring accuracy. In the medical field, milliliters express liquid medication volumes, allowing for precise dosages. In daily life, milliliters are used in cooking and baking recipes, providing a standardized way to measure liquids. Product packaging often includes milliliter markings, aiding consumers in understanding liquid quantities. The milliliter's versatility extends across industries, from manufacturing to agriculture, with its standardized representation facilitating clear communication of volumes. In essence, the milliliter's significance lies in its precise yet straightforward measurement, making it a fundamental unit for quantifying liquids across various contexts within the metric system.
How to Measure Milliliter To Cubic Centimeters?cubic centimeters = milliliters ÷ 1
Cubic Centimeter to Milliliter Conversion Table
| Cubic Centimeters | Milliliters |
--- |
| 1 cm³ | 1 mL |
| 2 cm³ | 2 mL |
| 3 cm³ | 3 mL |
| 4 cm³ | 4 mL |
| 5 cm³ | 5 mL |
| 6 cm³ | 6 mL |
| 7 cm³ | 7 mL |
| 8 cm³ | 8 mL |
| 9 cm³ | 9 mL |
| 10 cm³ | 10 mL |
| 11 cm³ | 11 mL |
| 12 cm³ | 12 mL |
| 13 cm³ | 13 mL |
| 14 cm³ | 14 mL |
| 15 cm³ | 15 mL |
| 16 cm³ | 16 mL |
| 17 cm³ | 17 mL |
| 18 cm³ | 18 mL |
| 19 cm³ | 19 mL |
| 20 cm³ | 20 mL |
| 21 cm³ | 21 mL |
| 22 cm³ | 22 mL |
| 23 cm³ | 23 mL |
| 24 cm³ | 24 mL |
| 25 cm³ | 25 mL |
| 26 cm³ | 26 mL |
| 27 cm³ | 27 mL |
| 28 cm³ | 28 mL |
| 29 cm³ | 29 mL |
| 30 cm³ | 30 mL |
| 31 cm³ | 31 mL |
| 32 cm³ | 32 mL |
| 33 cm³ | 33 mL |
| 34 cm³ | 34 mL |
| 35 cm³ | 35 mL |
| 36 cm³ | 36 mL |
| 37 cm³ | 37 mL |
| 38 cm³ | 38 mL |
| 39 cm³ | 39 mL |
| 40 cm³ | 40 mL |
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12888 | https://www.nku.edu/~longa/classes/mat420/days/highlights/highlights2.3.pdf | Section 2.3: Bolzano-Weierstrass Theorem October 19, 2011 Abstract The Bolzano-Weierstrass Theorem says something intuitive: that a set of numbers, of infinite cardinality yet whose elements are bounded in size, is going to have to huddle around at least one point. You can’t pack that many points into a confined space without leaving a clump of them about some point.
We define a couple of new notions, lim sup and lim inf, related to the limits that we’re going to find in such bounded sets with an infinite number of elements.
Definition: limit point: A number x is called a limit point (or cluster point or accumulation point) of a set of real numbers A if, ∀ε > 0, the interval (x −ε, x + ε) contains infinitely many points of A.
Theorem 2-12 (Bolzano-Weierstrass): Every bounded infinite set of real numbers has at least one limit point.
Note: Clearly some bounded infinite sets of real numbers have no more than one limit point (e.g. the set represented by the sequence {2−n}) – here thinking of a sequence as representing a set. This is a potential cause for confusion! Be careful.... We’re transitioning from sequences to general sets.
Proof: (sketch) Since A is bounded, ∃M > 0 / A ⊂[−M, M]. Now cut the interval in half, and choose a half that has an infinite number of elements in it (WLOG A1 ≡[0, M]). Repeat this process over and over, creating a set of nested intervals as in Theorem 2-7, whose widths tend to zero. Thus there is some p such that ∩∞ n=1An = {p}. Using the intervals created, we can create an appropriate sequence.
Theorem 2-13: Consider the sequence {an}. Then L is a subsequential limit of {an} if and only if L satisfies either of the following conditions: (i) There are infinitely many terms of {an} that equal L.
(ii) L is a limit point of the set consisting of the terms of {an}.
1 Note: Here we are connecting a result about sequences and their convergence to a result about sets and cluster points or accumulation points (i.e. limit points) – how suggestive these names are!
Proof: • →: Let’s assume that L is a subsequential limit of {an}. Then by theorem 2-11, ∀ε > 0, ∃infinite number of terms in the interval (L −ε, L + ε).
If there are an infinite number of repeated terms L, then the set {an} may be a singleton (and hence would have no limit point) – but we’ve satisfied the first condition of the right-hand side.
Suppose not, therefore: then there is a subsequence of {an} that converges to L, call it {bn}.
There can be no other value M such that there are an infinite number of terms M in {bn} – otherwise, M would be a subsequential limit point different from L – a contradiction of convergence to L.
Hence we can now construct an infinite set of distinctly different points in any ε−neighborhood of L. Given ε > 0. Let ε0 = ε. Find an element different from L in the interval (L −ε0, L + ε0) (call it e1). Now set ε1 = |e1 −L|. Find an element different from L in the interval (L −ε1, L + ε1) (call it e2). In this way we will create an infinite set of distinctly different points in the interval (L −ε, L + ε), and hence L is a limit point of the terms of {an}.
• ←We assume that L satisfies either of the following conditions: (i) There are infinitely many terms of {an} that equal L, or (ii) L is a limit point of the set consisting of the terms of {an}.
Assume that there are infinitely many terms of {an} that equal L. Then choose those elements of the sequence as the subsequence, which converges to L as a constant sequence.
Hence, assume that there aren’t infinitely many terms equal to L , but that L is a limit point of the set consisting of the terms of {an}. Then, by definition ∀ε > 0, the interval (L −ε, L + ε) contains infinitely many points of A.
Let’s construct a subsequence that converges to L, by induction.
Let εn = 1 2n (this is an arbitrary sequence that approaches 0 – I could just as well have chosen εn = 1 n, for example).
Let s1 ≡an1 be any element in the interval (L −ε1, L + ε1) (of which there are infinitely many). Then to choose the next element, we seek any element in the interval (L −ε2, L + ε2) (of which there are infinitely many); hence we can find the first one past the index n1, for example, and call that index n2: s2 ≡an2.
The inductive proposition is P(k) : sk ≡ank ∈(L −εk, L + εk) ∧nk > nk−1 2 Assuming P(k), we need to show P(k + 1). Consider the subsequence of {an} where n > nk.
Since {an} contains an infinite number of elements in the interval (L −εk+1, L + εk+1, so does the subsequence. Choose one such element, at index nk+1 (> nk), and set sk+1 ≡ank+1.
Then P(k + 1).
Hence, we conclude that there is a subsequence, {ank}k∈I N such that ∀ε > 0 ∃N(ε) such that n > N(ε) →|an −L| < ε. Given any ε , we simply find the element εk = 1 2k < ε, and all the elements of the subsequence beyond the corresponding nk in the sequence we constructed will be with the ε−neighborhood.
Theorem 2-14: Every bounded sequence has a convergent subsequence.
Corollary 2-14: A bounded sequence that does not converge has more than one subse-quential limit point.
Note: We will prove this one as a homework exercise (#17, p. 59).
Theorem 2-15: (i) A sequence that is unbounded above has a subsequence that diverges to ∞.
(ii) A sequence that is unbounded below has a subsequence that diverges to −∞.
Theorem 2-16: A sequence {an} converges if and only if it is bounded and has exactly one subsequential limit point.
Definition: lim sup: Let {an} be a sequence of real numbers. Then lim sup an = lim an is the least upper bound of the set of subsequential limit points of {an}, and lim inf an = lim an is the greatest lower bound of the set of subsequential limit points of {an}.
Note: We show in exercise 16 that the supremum of a set of limit points is a limit point of the sequence, as is the infimum. That means that the lim an and lim an are the max and min of the subsequential limit points.
Theorem 2-17: Let {an} be a bounded sequence of real numbers. Then (i) lim an = L if and only if, ∀ε > 0, there are infinitely many terms of {an} in (L−ε, L+ε) but only finitely many terms of {an} with an > L + ε.
(ii) lim an = K if and only if, ∀ε > 0, there are infinitely many terms of {an} in (K − ε, K + ε) but only finitely many terms of {an} with an < K −ε.
Corollary 2-17: A bounded sequence {an} of numbers converges if and only if lim an = lim an Theorem 2-18: Let {an} and {bn} be bounded sequences. Then 3 (i) lim(an + bn) ≤lim an + lim bn.
(ii) lim an + lim bn ≤lim (an + bn).
Definition: bounded: We say that a function f is bounded if the range of f is a bounded set.
Note: If f is bounded, we denote lub(R(f)) by sup f, and glb(R(f)) by inf f.
Theorem 2-19: Let f and g be bounded functions with the same domain. Then (i) sup(f + g) ≤sup f + sup g (ii) inf f + inf g ≤inf(f + g) Example: Exercise 13, p. 58 (proof of Theorem 2-9, from section 2.1) 4 |
12889 | https://math.stackexchange.com/questions/199430/find-the-necessary-and-sufficient-conditions-on-a-b-so-that-ax2-b-0 | Skip to main content
Find the necessary and sufficient conditions on a, b so that ax2+b=0 has a real solution.
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This question is really confusing me, and I'd love some help but not the answer. :D
Is it asking: What values of a and b result in a real solution for the equation ax2+b=0? a=b=0 would obviously work, but how does x come into play? There'd be infinitely many solutions if x can vary as well (a=1, x=1, b=−1, etc.).
I understand how necessary and sufficient conditions work in general, but how would it apply here? I know it takes the form of "If p then q" but I don't see how I could apply that to the question. Is "If ax2+b=0 has a real solution, then a and b= ..." it?
discrete-mathematics
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edited Sep 20, 2012 at 2:28
Michael Albanese
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asked Sep 20, 2012 at 0:49
Doug SmithDoug Smith
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By the way, if there is a real solution, all (how many?) solutions are real too :)
– Tapu
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I assume the question is "find conditions that are necessary and sufficient to guarantee solutions" rather than "find necessary conditions and also find sufficient sufficient conditions for a solution." If the former is the case, then you're asked for constraints on a and b such that (1) if the conditions are met then ax2+b is zero for some x and (2) if the conditions are not met then ax2+b isn't zero for any x.
So, when will ax2+b have some value x for which the expression is zero? Well, as André suggested, try to solve ax2+b=0 "mechanically". By subtracting we have the equivalent equation ax2=−b and we'd like to divide by a to get x2=−b/a. Unfortunately, we can't do that if a=0, so we need to consider two cases
a=0
a≠0
In the first case, our equation becomes 0⋅x2+b=0, namely b=0 and if b=0 (and, of course a=0) then any x will satisfy this. In other words, there's a solution (actually infinitely many solutions) if a=b=0, as you've already noted.
Now, in case (2) we can safely divide by a to get x2=−b/a. When does this have a solution? We know that x2≥0 no matter what x is, so when will our new equation have a solution? You said you don't want the full answer, so I'll denote the answer you discover by C, which will be of the form "some condition on a and b". Once you've done that, your full answer will be:
ax2+b=0 has a solution if and only if
a=b=0, or
Your condition C.
These are sufficient, since either guarantees a solution, and necessary, since if neither is true, then there won't be a solution (since we exhausted all possibilities for solutions).
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answered Sep 20, 2012 at 2:36
Rick DeckerRick Decker
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I'm a little confused why x^2 = -b/a having a solution is the answer to the problem. And why do we know that x^2 >= 0? Thanks so much for the help, just a little blurp in my knowledge.
– Doug Smith
Commented
Sep 20, 2012 at 3:47
For your first question, remember that you're asked to find conditions on a and b so that ax2+b=0 will have a solution, so in the case that a≠0 that equation will have a solution if and only if x2=−b/a will have a solution. For your second question, note that for any real x, we'll always have x2≥0. (For example, 12=1≥0,32=9≥0,(−2)2=4≥0.) Thus, there will be an x that satisfies x2=−b/a if and only if −b/a≥0.
– Rick Decker
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Hint: Mechanically "solve" the equation. What could go wrong?
As to what the question is asking, suppose that we are given fixed numbers a and b. For example, let a=−3 and b=17. Does the equation have a real solution x? What about if a=−3 and b=−5? (And don't forget about the possibility that one or both of a and b might be 0.) Can we come up with simple conditions on a and b so that just glancing at them tells us whether or not there is a solution?
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edited Sep 20, 2012 at 1:01
answered Sep 20, 2012 at 0:51
André NicolasAndré Nicolas
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A simpler hint might be to note that for real x, we should have x2≥0.
– Tapu
Commented
Sep 20, 2012 at 0:54
@Tapu: That is a very useful hint. It may be better at the "write-up" stage.
– André Nicolas
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Sep 20, 2012 at 0:55
b^2-4ac > 0 or 0 then you have real solutions.
– yiyi
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Sep 20, 2012 at 0:55
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If ax2+b=0, then x2=−ba. The solutions are therefore x=±−b/a−−−−√. For the equation to have a real solution, −b/a must be non-negative. A fraction is non-negative if
the numerator and denominator are positive, or
the numerator and denominator are negative, or
the numerator is 0 and the denominator is not 0.
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answered Sep 20, 2012 at 2:40
Michael HardyMichael Hardy
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Factor de seguridad: Coeficiente de seguridad en el diseño y el uso
Conozca qué es el factor de seguridad, sus ejemplos y su importancia para garantizar la seguridad en el diseño y el uso en la seguridad general de las personas y los lugares de trabajo.
En este artículo
Definición del factor de seguridad
Factores generales de seguridad típicos
Recomendaciones generales
Cálculo: factor de seguridad fórmula
Importancia del factor de seguridad
Selección del
Ejemplos
SafetyCulture (antes iAuditor) para factores de seguridad
Publicación 7 Aug 2025
Artículo de
SafetyCulture Content Team
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6 min de lectura
Definición del factor de seguridad
Cuanto mayor sea el número de FoS, más seguro será el producto o la estructura. Un FoS de 1 indica que una estructura o componente fallará inmediatamente cuando se alcance la carga de diseño y no será capaz de soportar ninguna carga adicional. No se aceptan estructuras o componentes con FoS inferior a uno. Si las consecuencias de un fallo son graves, como la pérdida de vidas o las lesiones físicas, se exigirá una FoS más alta, ya sea por diseño o por ley.
Cuanto mayor sea el número de FoS, más seguro será el producto o la estructura. Un FoS de 1 indica que una estructura o componente fallará inmediatamente cuando se alcance la carga de diseño y no será capaz de soportar ninguna carga adicional. No se aceptan estructuras o componentes con FoS inferior a uno. Si las consecuencias de un fallo son graves, como la pérdida de vidas o las lesiones físicas, se exigirá una FoS más alta, ya sea por diseño o por ley.
En pocas palabras, las estructuras o dispositivos deben ser capaces de soportar más peso del que tendrían en condiciones normales de uso para ser más seguros. FoS indica por qué factor el diseño es seguro.
Factores generales de seguridad típicos
A continuación se muestra una tabla de factor de seguridad típica de los equipos, proporcionada por Engineering ToolBox:
EquipoFactor de seguridad
– FOS –
Componentes de aeronaves 1.5 – 2.5
Calderas 3.5 – 6
Pernos 8.5
Ruedas de hierro fundido 20
Componentes del motor 6 – 8
Eje de alta resistencia 10 – 12
Equipos de elevación – ganchos 8 – 9
Recipientes a presión 3,5 – 6 (especificado en el código de diseño)
Componentes de la turbina – estáticos 6 – 8
Componentes de la turbina – giratorios 2 – 3
Muelle, grande y resistente 4.5
Trabajos de estructura de acero en edificios 4 – 6
Trabajos de acero estructural en puentes 5 – 7
Cables de acero 8 – 9
Recomendaciones generales
Aplicaciones– FOS –
Para su uso con materiales de alta fiabilidad donde las condiciones de carga y ambientales no son severas y donde el peso es una consideración importante 1.3 – 1.5
Para su uso con materiales fiables cuando las condiciones de carga y ambientales no son severas 1.5 – 2
Para su uso con materiales ordinarios cuando las condiciones de carga y ambientales no son severas 2 – 2.5
Para su uso con menos intentos y para materiales frágiles donde la carga y las condiciones ambientales no son severas 2.5 – 3
Para su uso con materiales cuyas propiedades no son fiables y cuyas condiciones de carga y ambientales no son severas, o cuando se utilizan materiales fiables en condiciones difíciles y ambientales 3 – 4
Cálculo: factor de seguridad fórmula
Todos los cálculos miden básicamente lo mismo: cuánta tensión adicional más allá de la carga diseñada puede soportar una estructura. Un cálculo sencillo para el factor es:
Ecuación del factor de seguridad fórmula | SafetyCulture
Hay muchas calculadoras de factor disponibles en línea para calcular los valores de FoS cuando se conocen los valores de tensión máxima y tensión admisible.
Importancia del factor de seguridad
El factor varía según la situación. Los sistemas se diseñan intencionadamente para que sean mucho más fuertes de lo que necesitan para una configuración normal. Esto aumenta la probabilidad de que sigan funcionando incluso en condiciones extremas, como situaciones de emergencia, cargas añadidas, uso excesivo o degradación causada por el desgaste. Además, a continuación se exponen más razones por las que es importante utilizar FoS en el diseño:
Mantiene la funcionalidad de la estructura para el futuro al tiempo que proporciona seguridad adicional para el uso actual
Evita daños a la propiedad, a los trabajadores y a las máquinas
Proporciona protección frente a los riesgos imprevisibles que puedan surgir al utilizar un producto o servicio
Reduce las posibilidades de que un producto falle
Selección del factor de seguridad mecánica de materiales
Según A Textbook of Machine Design, de R.S.Khurmi y J.K.Gupta, la selección del factor de diseño adecuado que debe utilizarse en el diseño de cualquier sistema mecánico se basa en una serie de consideraciones, entre las que se incluyen las siguientes:
Material dúctil o frágil; los materiales dúctiles utilizan el límite elástico; los materiales frágiles utilizan el límite máximo.
El límite elástico determina el FoS hasta el comienzo de la deformación.
Resistencia última: determina la FoS hasta el fallo.
Proceso de fabricación
Tipo de estrés
Condiciones generales de servicio
Forma de las piezas
Ejemplos
A continuación se presentan dos ejemplos de cómo se utiliza el factor de seguridad mecánica de materiales:
Recipientes a presión
Las calderas y los recipientes a presión, así como los sistemas de las centrales nucleares, están sujetos a las directrices de seguridad del Código Internacional de Calderas y Recipientes a Presión de la Sociedad Americana de Ingenieros Mecánicos (ASME ), que controlan el diseño, la fabricación y la inspección de las calderas y los recipientes a presión durante el proceso de construcción. Por su propia naturaleza, los recipientes a presión son potencialmente peligrosos. Es necesario añadir f actores de seguridad y de carga en estructuras de acero para protegerse de la incertidumbre de los fallos en el diseño, los materiales utilizados, la fabricación, la inspección y el funcionamiento.
Sistemas personales de detención de caídas (PFAS)
Los sistemas personales de detención de caídas (PFAS) y otros equipos de protección contra caídas deben construirse con un alto factor de seguridad. El Norma 1915.159 de la Administración de Seguridad y Salud en el Trabajo (OSHA) destaca los criterios para que los conectores y el anclaje sean capaces de sostener una carga de tracción mínima de 3.000 a 5.000 libras (22,24 Kn) por empleado, y el requisito de un sistema personal de detención de caídas completo que mantenga un factor de al menos 2. Si el equipo se va a utilizar en circunstancias difíciles, puede requerir un FoS aún mayor.
SafetyCulture (antes iAuditor) para factores de seguridad
El factor general de seguridad existe como una medida de seguridad destinada a hacer que un producto, un sistema o una estructura sean seguros. SafetyCulturepuede ayudar a los fabricantes, ingenieros e inspectores a realizar inspecciones externas e internas de los equipos y a evaluar su factor para determinar su capacidad de soportar la tensión máxima permitida. Con SafetyCulture, puede:
Realice inspecciones más cómodas utilizando un dispositivo móvil.
Capturar los problemas encontrados durante las inspecciones, como los fallos del sistema. Asigne inmediatamente acciones para problemas urgentes o cualquier fallo de los equipos y reciba actualizaciones del estado en tiempo real.
Comparta los informes de inspección a través de la app con otro personal cualificado para realizar inspecciones de seguridad.
Personalice las plantillas de listas de comprobación de seguridad según los requisitos del sistema o equipo, como en los recipientes a presión, y cumpla las normas especificadas en los códigos industriales.
Proporcionar orientación y formación móvil a los trabajadores que puedan llevar a cualquier lugar como referencia para el uso, la inspección y el almacenamiento adecuados del equipo. Auditoría de fallos de incertidumbre y control de calidad del material durante la fabricación, construcción o montaje.
Related Templates
Listas de comprobación destacadas
Consulte otras listas de comprobación útiles para el factor de seguridad:
Lista de verificación del control de calidad de la fabricación
Listas de comprobación para la inspección de arneses de seguridad
Lista de comprobación del informe de inspección de arneses y eslingas
Listas de comprobación para la inspección de recipientes a presión
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En este artículo
Definición del factor de seguridad
Factores generales de seguridad típicos
Recomendaciones generales
Cálculo: factor de seguridad fórmula
Importancia del factor de seguridad
Selección del
Ejemplos
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12891 | http://math.ucdavis.edu/~romik/data/uploads/teaching/math205a-2018/complex-analysis-2018.pdf | Complex Analysis Lecture Notes Dan Romik Note. I created these notes for the course Math 205A: Complex Analysis I taught at UC Davis in 2016 and 2018. With a few exceptions, the exposition follows the textbook Complex Analysis by E. M. Stein and R. Shakarchi (Prince-ton University Press, 2003). The notes were not heavily vetted for accuracy and may contain minor typos or errors. You can help me continue to improve them by emailing me with any comments or corrections you have.
Acknowledgements. I am grateful to Jianping Pan, Anthony Nguyen, Christo-pher Alexander, Brynn Caddel, Jennifer Brown, and Brad Velasquez for com-ments that helped me improved the notes. Figure 5 on page 23 was created by Jennifer Brown.
Complex Analysis Lecture Notes Document version: April 20, 2018 Copyright c ⃝2018 by Dan Romik Email comments and feedback to romik@math.ucdavis.edu Cover figure: a heat map plot of the entire function z 7→z(z −1)π−z/2Γ(z/2)ζ(z).
Created with Mathematica 10 using code by Simon Woods, available at 2 Contents 1 Introduction 4 2 The fundamental theorem of algebra 5 3 Analyticity, conformality and the Cauchy-Riemann equations 8 4 Power series 12 5 Contour integrals 15 6 Cauchy’s theorem 18 7 Consequences of Cauchy’s theorem 22 8 Zeros, poles, and the residue theorem 29 9 Meromorphic functions and the Riemann sphere 32 10 The argument principle 34 11 Applications of Rouch´ e’s theorem 38 12 Simply-connected regions and Cauchy’s theorem 38 13 The logarithm function 40 14 The Euler gamma function 41 15 The Riemann zeta function 47 16 The prime number theorem 57 17 Introduction to asymptotic analysis 64 Additional reading 74 3 1 Introduction 1. Complex analysis is in my opinion one of the most beautiful areas of mathemat-ics. It has one of the highest ratios of theorems to definitions (i.e., a very low “entropy”), and lots of applications to things that seem unrelated to complex numbers, for example: • Solving cubic equations that have only real roots (historically, this was the motivation for introducing complex numbers by Cardano, who published the famous formula for solving cubic equations in 1543, after learning of the solution found earlier by Scipione del Ferro).
Example. Using Cardano’s formula, it can be found that the solutions to the cubic equation z3 + 6z2 + 9z + 3 = 0 are z1 = 2 cos(2π/9) −2, z2 = 2 cos(8π/9) −2, z3 = 2 sin(π/18) −2.
• Proving Stirling’s formula: n! ∼ √ 2πn(n/e)n.
• Proving the prime number theorem: π(n) ∼ n log n.
• Proving many other asymptotic formulas in number theory and combina-torics, e.g., the Hardy-Ramanujan formula p(n) ∼ 1 4 √ 3n eπ√ 2n/3, where p(n) is the number of integer partitions of n.
• Evaluation of complicated definite integrals, for example Z ∞ 0 sin(t2) dt = 1 2 r π 2 .
• Solving physics problems in hydrodynamics, heat conduction, electrostat-ics and more.
• Analyzing alternating current electrical networks by extending Ohm’s law to electrical impedance.
• Probability and combinatorics, e.g., the Cardy-Smirnov formula in perco-lation theory and the connective constant for self-avoiding walks on the hexagonal lattice.
• It was proved in 2016 that the optimal densities for sphere packing in 8 and 24 dimensions are π4/384 and π12/12!, respectively. The proofs make spectacular use of complex analysis (and more specifically, modular forms).
4 Figure 1: Print Gallery, a lithograph by M.C. Escher which was discovered to be based on a mathematical structure related to a complex function z 7→zα for a certain complex number α, although it was constructed by Escher purely using geometric intuition. See the paper Artful mathematics: the heritage of M.C. Escher, by B. de Smit and H.W. Lenstra Jr. (Notices Amer. Math. Soc.
50 (2003), 446–457).
• Nature uses complex numbers in Schr¨ odinger’s equation and quantum field theory. Why? No one knows.
• Conformal maps, which were used by M.C. Escher (though he had no mathematical training) to create amazing art, and used by others to better understand and even to improve Escher’s work. See Fig. 1.
• Complex dynamics, e.g., the iconic Mandelbrot set. See Fig. 2.
2. In the next section I will begin our journey into the subject by illustrating a few beautiful ideas and along the way begin to review the concepts from undergraduate complex analysis.
2 The fundamental theorem of algebra 3. The Fundamental Theorem of Algebra. Every nonconstant polynomial p(z) over the complex numbers has a root.
I will show three proofs. Let me know if you see any “algebra”. . .
5 Figure 2: The Mandelbrot set. [Source: Wikipedia] 4. Analytic proof. Let p(z) = anzn + an−1zn−1 + . . . + a0 be a polynomial of degree n, and consider where |p(z)| attains its infimum.
First, note that it can’t happen as |z| →∞, since |p(z)| = |z|n · (|an + an−1z−1 + an−2z−2 + . . . + a0z−n|), and in particular lim|z|→∞ |p(z)| |z|n = |an|, so for large |z| it is guaranteed that |p(z)| ≥|p(0)| = |a0|. Fixing some radius R > 0 for which |z| > R implies |p(z)| ≥|a0|, we therefore have that m0 := inf z∈C |p(z)| = inf |z|≤R |p(z)| = min |z|≤R |p(z)| = |p(z0)| where z0 = arg min |z|≤R |p(z)|, and the minimum exists because p(z) is a continuous function on the disc DR(0).
Denote w0 = p(z0), so that m0 = |w0|. We now claim that m0 = 0. Assume by contradiction that it doesn’t, and examine the local behavior of p(z) around z0; more precisely, expanding p(z) in powers of z −z0 we can write p(z) = w0 + n X j=1 cj(z −z0)j = w0 + ck(z −z0)k + . . . + cn(z −z0)n, where k is the minimal positive index for which cj ̸= 0. (Exercise: why can we expand p(z) in this way?) Now imagine starting with z = z0 and traveling away from z0 in some direction eiθ. What happens to p(z)? Well, the expansion gives p(z0 + reiθ) = w0 + ckrkeikθ + ck+1rk+1ei(k+1)θ + . . . + cnrneinθ.
6 When r is very small, the power rk dominates the other terms rj with k < j ≤n, i.e., p(z0 + reiθ) = w0 + rk(ckeikθ + ck+1rei(k+1)θ + . . . + cnrn−keinθ) = w0 + ckrkeikθ(1 + g(r, θ)), where limr→0 |g(r, θ)| = 0. To reach a contradiction, it is now enough to choose θ so that the vector ckrkeikθ “points in the opposite direction” from w0, that is, such that ckrkeikθ w0 ∈(−∞, 0).
Obviously this is possible: take θ = 1 k(arg w0 −arg(ck) + π). It follows that, for r small enough, |w0 + ckrkeikθ| < |w0| and for r small enough (possibly even smaller) |p(z0 + reiθ)| = |w0 + ckrkeikθ(1 + g(r, θ))| < |w0|, a contradiction. This completes the proof.
Exercise. Complete the last details of the proof (for which r are the inequalities valid, and why?) Note that “complex analysis” is part of “analysis” — you need to develop facility with such estimates until they become second nature.
5. Topological proof. Let w0 = p(0). If w0 = 0, we are done. Otherwise consider the image under p of the circle |z| = r. Specifically: (a) For r very small the image is contained in a neighborhood of w0, so it cannot “go around” the origin.
(b) For r very large we have p(reiθ) = anrneinθ 1 + an−1 an r−1e−iθ + . . . + a0 an r−ne−inθ = anrneinθ(1 + h(r, θ)) where limr→∞h(r, θ) = 0 (uniformly in θ). As θ goes from 0 to 2π, this is a closed curve that goes around the origin n times (approximately in a circular path, that becomes closer and closer to a circle as r →∞).
As we gradually increase r from 0 to a very large number, in order to transition from a curve that doesn’t go around the origin to a curve that goes around the origin n times, there has to be a value of r for which the curve crosses 0. That means the circle |z| = r contains a point such that p(z) = 0, which was the claim.
6. Remark. The argument presented in the topological proof is imprecise. It can be made rigorous in a couple of ways — one way we will see a bit later is using Rouch´ e’s theorem and the argument principle. This already gives a hint as to the importance of subtle topological arguments in complex analysis.
7 7. Remark. The topological proof should be compared to the standard calculus proof that any odd-degree polynomial over the reals has a real root.
That argument is also “topological,” although much more trivial.
8. Standard textbook proof using Liouville’s theorem. Recall: Liouville’s theorem. A bounded entire function is constant.
Assuming this result, if p(z) is a polynomial with no root, then 1/p(z) is an entire function. Moreover, it is bounded, since as we noted before lim|z|→∞ |p(z)| |z|n = |an|, so lim|z|→∞1/p(z) = 0. It follows that 1/p(z) is a constant, which then has to be 0, which is a contradiction.
9. Summary. We saw three proofs of FTA. I like the first one best since it is elementary and doesn’t use Cauchy’s theorem or any of its consequences, or subtle topological concepts. Moreover, it is a “local” argument that is based on understanding how a polynomial behaves locally. The other two proofs can be characterized as “global.” It is a general philosophical principle in analysis (that has analogies in other areas, such as number theory) that local arguments are easier than global ones.
3 Analyticity, conformality and the Cauchy-Riemann equations 10. Definition.
A function f(z) of a complex variable is holomorphic (a.k.a.
complex-differentiable, analytic1) at z if f ′(z) := lim h→0 f(z + h) −f(z) h exists.
11. Geometric meaning of holomorphicity in the case f ′(z) ̸= 0: f is locally a rotation and rescaling.
12. Interpretation: analytic functions are conformal mappings where f ′(z) ̸= 0: if γ1 are two differentiable curves such that γ1(0) = γ2(0) = z, f is differentiable at z and f ′(z) ̸= 0, then, denoting v1 = γ′ 1(0), v2 = γ′ 2(0), w1 = (f ◦γ1)′(0), w2 = (f ◦γ2)′(0), we have ⟨v1, v2⟩= Re(v1v2), ⟨w1, w2⟩= ⟨(f ′(γ1(0))γ′ 1(0)), (f ′(γ2(0))γ′ 2(0))⟩ = f ′(z)f ′(z)⟨v1, v2⟩= |f ′(z)|2⟨v1, v2⟩, so, if we denote by θ (resp. ϕ the angle between v1, v2 (resp. w1, w2), we have cos ϕ = ⟨w1, w2⟩ |w1| |w2| = |f ′(z)|2⟨v1, v2⟩ |f ′(z)v1| |f ′(z)v2| = ⟨v1, v2⟩ |v1| |v2| = cos θ.
1Note: some people use “analytic” and “holomorphic” with two a priori different definitions that are then proved to be equivalent; I find this needlessly confusing so I may use these two terms interchangeably.
8 13. Conversely, if f is conformal in a neighborhood of z then (under some additional mild assumptions) it is analytic — we will prove this below after discussing the Cauchy-Riemann equations.
14. Properties of derivatives: under appropriate assumptions (explain them precisely — see Proposition 2.2 on page 10 of [Stein-Shakarchi]), (f + g)′(z) = f ′(z) + g′(z), (fg)(z) = f ′(z)g(z) + f(z)g′(z), 1 f ′ = −f ′(z) f(z)2 , f g ′ = f ′(z)g(z) −f(z)g′(z) g(z)2 , (f ◦g)′(z) = f ′(g(z))g′(z).
15. Denote z = x + iy, f = u + iv. Note that if f is analytic at z then f ′(z) = lim h→0 f(z + h) −f(z) h = lim h→0, h∈R u(x + h + iy) −u(x + iy) h + iv(x + h + iy) −v(x + iy) h = ∂u ∂x + i ∂v ∂x.
On the other hand also f ′(z) = lim h→0 f(z + h) −f(z) h = lim h→0, h∈iR u(x + h + iy) −u(x + iy) h + iv(x + h + iy) −v(x + iy) h = lim h→0, h∈R u(x + iy + ih) −u(x + iy) ih + iv(x + iy + ih) −v(x + iy) ih = −i∂u ∂y −i · i∂v ∂y = ∂v ∂y −i∂u ∂y .
Since these limits are equal, by equating their real and imaginary parts we get the Cauchy-Riemann equations: ∂u ∂x = ∂v ∂y , ∂v ∂x = −∂u ∂y .
16. Conversely, if f = u+iv is continuously differentiable (in the real analysis sense) at z = x + iy and satisfies the C-R equations there, f is analytic at z.
Proof. The assumption implies that f has a differential at z, i.e., in the notation of vector calculus, denoting f = (u, v), z = (x, y)⊤, ∆z = (h1, h2)⊤, we have f(z + ∆z) = u(z) v(z) !
+ ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y ! h1 h2 !
+ E(h1, h2), 9 where E(h1, h2) = o(|∆z|) as |∆z| →0. Now, by the assumption that the C-R equations hold, we also have ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y ! h1 h2 !
= ∂u ∂xh1 + ∂u ∂y h2 −∂u ∂y h1 + ∂u ∂xh2 !
, which is the vector calculus notation for the complex number ∂u ∂x −i∂u ∂y (h1 + ih2) = ∂u ∂x −i∂u ∂y ∆z.
So, we have shown that (again, in complex analysis notation) lim ∆z→0 f(z + ∆z) −f(z) ∆z = lim ∆z→0 ∂u ∂x −i∂u ∂y + E(∆z) ∆z = ∂u ∂x −i∂u ∂y .
This proves that f is holomorphic at z with derivative given by f ′(z) = ∂u ∂x − i ∂u ∂y .
17. Interesting consequence of C-R (1). Theorem: if f = u + iv is conformal at z, continuously differentiable in the real analysis sense, and satisfies det Jf > 0 (i.e., f preserves orientation as a planar map), then f is holomorphic at z.
Proof. In the notation of the proof above, we have as before that f(z + ∆z) = u(z) v(z) !
+ ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y ! h1 h2 !
+ E(h1, h2), where E(h1, h2) = o(|∆z|) as |∆z| →0. The assumption is that the differential map Jf = ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y !
preserves orientation and is conformal; the conclusion is that the Cauchy-Riemann equations are satisfied (which would imply that f is holomorphic at z by the result shown above. So the whole thing reduces to proving the following simple claim about 2 × 2 matrices: Conformality lemma. Assume that A = a b c d !
is a 2 × 2 real matrix. The following are equivalent: (a) A preserves orientation (that is, det A > 0) and is conformal, that is ⟨Aw1, Aw2⟩ |Aw1| |Aw2| = ⟨w1, w2⟩ |w1| |w2| for all w1, w2 ∈R2.
(b) A takes the form A = a b −b a !
for some a, b ∈R with a2 + b2 > 0.
10 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 -4 -2 2 4 (a) (b) (c) Figure 3: The level curves for the (a) real and (b) imaginary parts of z2 = (x2 −y2) + i(2xy). (c) shows the superposition of both families of level curves.
(c) A takes the form A = r cos θ −sin θ sin θ cos θ !
for some r > 0 and θ ∈R. (That is, geometrically A acts by a rotation followed by a scaling.) Proof that (a) = ⇒(b). Note that both columns of A are nonzero vectors by the assumption that det A > 0. Now applying the conformality assumption with w1 = (1, 0)⊤, w2 = (0, 1)⊤yields that (a, c) ⊥(b, d), so that (b, d) = κ(−c, a) for some κ ∈R \ {0}. On the other hand, applying the conformality assumption with w1 = (1, 1)⊤and w2 = (1, −1)⊤yields that (a + b, c + d) ⊥(a −b, c −d), which is easily seen to be equivalent to a2 + c2 = b2 + d2. Together with the previous relation that implies that κ = ±1. So A is of one of the two forms a −c c a !
or a c c −a !
. Finally, the assumption that det A > 0 means it is the first of those two possibilities that must occur.
Exercise. Show also that (b) ⇐ ⇒(c) and that (b) = ⇒(a).
18. Interesting consequence of C-R (2): orthogonality of level curves of u and of v: if f = u + iv is analytic then ∇u · ∇v = (ux, uy) ⊥(vx, vy) = uxvx + uyvy = vyvx −vxvy = 0.
Since ∇u (resp. ∇v) is orthogonal to the level curve {u = c} (resp. the level curve {v = d}, this proves that the level curves {u = c}, {v = d} meet at right angles whenever they intersect.
19. Interesting consequence of C-R (3): Assume that f is analytic at z and twice continuously differentiable there. Then ∂2u ∂x2 + ∂2u ∂y2 = ∂ ∂x ∂u ∂x + ∂ ∂y ∂u ∂y = ∂ ∂x ∂v ∂y −∂ ∂y ∂v ∂x = ∂2v ∂x∂y −∂2v ∂y∂x = 0.
11 -1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0 -2 -1 0 1 2 -2 -1 0 1 2 -1.0 -0.5 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0 (a) (b) (c) Figure 4: The level curves for the real and imaginary parts of z−1 = x x2+y2 − i y x2+y2 .
Similarly (check), v also satisfies ∂2v ∂x2 + ∂2v ∂y2 = 0.
That is, we have shown that u and v are harmonic functions. This is an extremely important connection between complex analysis and the theory of partial differential equations, which also relates to many other areas of real analysis.
20. We will later see that the assumption of twice continuous differentiability is unnecessary.
21. The Jacobian of an analytic function considered as a two-dimensional map: if f = u + iv then Jf = det ux uy vx vy !
= uxvy −uyvx = u2 x + v2 x = |ux + ivx| = |f ′(z)|2.
This can also be understood geometrically (exercise: how?).
4 Power series 22. Power series are functions of a complex variable, defined by f(z) = ∞ X n=0 anzn where (an)∞ n=0 is a sequence of complex numbers, or more generally by g(z) = f(z −z0) = ∞ X n=0 an(z −z0)n.
12 23. Where does this formula make sense? It is not hard to see that it converges absolutely precisely for 0 ≤|z| < R where R = lim sup n→∞|an|1/n −1 .
R is called the radius of convergence of the power series.
Proof. Assume 0 < R < ∞(the edge cases R = 0 and R = ∞are left as an exercise). The defining property of R is that for all ϵ > 0, we have that |an| < 1 R + ϵ n if n is large enough, and R is the minimal number with that property. Let z ∈DR(0). Since |z| < R, we have |z| 1 R + ϵ < 1 for some fixed ϵ > 0 chosen small enough. That implies that for n > N (for some large enough N as a function of ϵ), ∞ X n=N |anzn| < ∞ X n=N 1 R + ϵ |z| n , so the series is dominated by a convergent geometric series, and hence converges.
Conversely, if |z| > R, then, |z| 1 R −ϵ > 1 for some small enough fixed ϵ > 0.
Taking a subsequence (ank)∞ k=1 for which |ank| > 1 R −ϵ nk (guaranteed to exist by the definition of R), we see that ∞ X n=0 |anzn| ≥ ∞ X k=1 |z| 1 R −ϵ nk = ∞, so the power series diverges.
Exercise. Complete the argument in the extreme cases R = 0, ∞.
24. Another important theorem is: power series are holomorphic functions and can be differentiated termwise in the disc of convergence.
Proof. Denote f(z) = ∞ X n=0 anzn = SN(z) + EN(z), SN(z) = N X n=0 anzn, EN(z) = ∞ X n=N+1 anzn, g(z) = ∞ X n=0 nanzn.
The claim is that f is differentiable on the disc of convergence and its derivative is the power series g. Since n1/n →1 as n →∞, it is easy to see that f(z) and g(z) have the same radius of convergence. Fix z0 with |z| < r < R. We wish to 13 show that f(z0+h)−f(z0) h converges to g(z0) as h →0. Observe that f(z0 + h) −f(z0) h −g(z0) = SN(z0 + h) −SN(z0) h −S′ N(z0) + EN(z0 + h) −EN(z0) h + (S′ N(z0) −g(z0)) The first term converges to 0 as h →0 for any fixed N. To bound the second term, fix some ϵ > 0, and note that, if we assume that not only |z0| < r but also |z0 + h| < r (an assumption that’s clearly satisfied for h close enough to 0) then EN(z0 + h) −EN(z0) h ≤ ∞ X n=N+1 |an| (z0 + h)n −zn 0 h = ∞ X n=N+1 |an| h Pn−1 k=0 hk(z0 + h)n−1−k h ≤ ∞ X n=N+1 |an|nrn−1, where we use the algebraic identity an −bn = (a −b)(an−1 + an−2b + . . . + abn−2 + bn−1).
The last expression in this chain of inequalities is the tail of an absolutely con-vergent series, so can be made < ϵ be taking N large enough (before taking the limit as h →0).
Third, when choosing N also make sure it is chosen so that |S′ N(z0)−g(z0)| < ϵ, which of course is possible since S′ N(z0) →g(z0) as N →∞. Finally, having thus chosen N, we get that lim sup h→0 f(z0 + h) −f(z0) h −g(z0) ≤0 + ϵ + ϵ = 2ϵ.
Since ϵ was an arbitrary positive number, this shows that f(z0+h)−f(z0) h →g(z0) as h →0, as claimed.
25. The proof above can be thought of as a special case of the following more conceptual result: if gn is a sequence of holomorphic functions on a region Ω, and gn →g uniformly on closed discs in Ω, g′ n →h uniformly on closed discs on Ω, and h is continuous, then g is holomorphic and g′ = h on Ω. (Exercise: prove this, and explain the connection to the previous result.) 26. Corollary. Analytic functions defined as power series are (complex-)differentiable infinitely many times in the disc of convergence.
27. Corollary. For a power series g(z) = P∞ n=0 an(z −z0)n with a positive radius of convergence, we have an = g(n)(z0) n!
.
In other words g(z) satisfies Taylor’s formula g(z) = ∞ X n=0 g(n)(z0) n!
(z −z0)n.
14 5 Contour integrals 28. Parametrized curves: γ : [a, b] →C. Two curves γ1 : [a, b] →C, γ2 : [c, d] →C are equivalent, denoted γ1 ∼γ2, if γ2(t) = γ1(I(t)) where I : [c, d] →[a, b] is a continuous, one-to-one, onto, increasing function. A “curve” is an equivalence class of parametrized curves.
In practice, we will usually refer to parametrized curves as “curves”, which is the usual abuse of terminology (that one sees in various places in mathematics), in which one blurs the distinction between equivalence classes and their members, remembering that various arguments need to “respect the equivalence” in the sense that they do not depend of the choice of member. (Meta-exercise: think of 2–3 other examples of this phenomenon.) 29. We shall assume all our curves are piecewise continuously differentiable. (More generally, one can assume them to be rectifiable, but we will not bother to develop that theory).
30. Reminder from vector calculus: line integrals of the first and second kind: Z γ u(z)ds = lim max j ∆sj→0 n X j=1 u(zj)∆sj (line integral of the first kind), Z γ F · ds = Z γ P dx + Q dy = lim max j ∆sj→0 n X j=1 P(zj)∆xj + Q(zj)∆yj (F = (P, Q); line integral of the second kind).
31. Standard fact from calculus: the line integrals can be computed as Z γ u(z) ds = Z b a u(γ(t))|γ′(t)| dt, Z γ P dx + Q dy = Z b a F(γ(t)) · γ′(t) dt.
32. Reminder: the fundamental theorem of calculus for line integrals: if F = ∇u then Z γ F · ds = u(γ(b)) −u(γ(a)).
33. Important definition: contour integrals. For a function f = u + iv of a complex variable z and a curve γ, define Z γ f(z) dz = “ Z γ (u + iv)(dx + idy)” = Z γ u dx −v dy + i Z γ v dx + u dy = Z b a f(γ(t))γ′(t) dt (contour integral), Z γ f(z) |dz| = Z γ f(z) ds = Z γ u ds + i Z γ v ds (arc length integral).
15 If γ is a closed curve (the two endpoints are the same, i.e., it satisfies γ(a) = γ(b)), we denote the contour integral as H γ f(z) dz.
34. A special case of an arc length integral is the length of the curve, defined by the integral of the constant function 1: len(γ) = Z γ |dz| = Z b a |γ′(t)| dt.
35. These definitions do not depend on the parametrization of the curve. Indeed, if γ2(t) = γ1(I(t)), then (using the change of variables formula for real integrals) we have that Z γ2 f(z) dz = Z d c f(γ2(t))γ′ 2(t)dt = Z d c f(γ1(I(t)))(γ1 ◦I)′(t) dt = Z d c f(γ1(I(t)))γ′ 1(I(t))I′(t) dt = Z b a f(γ1(τ))γ′ 1(τ) dτ = Z γ1 f(z) dz.
Exercise.
Show that the integral with respect to arc length also does not depend on the parametrization.
36. Proposition (properties of contour integrals). Contour integrals satisfy the following properties: (a) Linearity (as an operator on functions): R γ(αf(z)+βg(z)) dz = α R γ f(z) dz+ β R γ g(z) dz.
(b) Linearity (as an operator on curves): if a contour Γ is a “composition” of two contours γ1 and γ2 (in a sense that is easy to define graphically, but tedious to write down precisely), then Z Γ f(z) dz = Z γ1 f(z) dz + Z γ2 f(z) dz.
Similarly, if γ2 is the “reverse” contour of γ1, then Z γ2 f(z) dz = − Z γ1 f(z) dz.
(c) Triangle inequality: Z γ f(z) dz ≤ Z |f(z)| |dz| ≤len(γ) · sup z∈γ |f(z)|.
Exercise. Prove this claim (part of the exercise is to define precisely the notions of “composition of curves” and “reverse curve”).
16 37. The fundamental theorem of calculus for contour integrals.
If γ is a curve connecting two points w1, w2 in a region Ωon which a function F is holomorphic, then Z γ F ′(z) dz = F(w2) −F(w1).
Equivalently, to compute the contour integral R γ f(z) dz, try to find a primitive to f, that is, a function F such that F ′(z) = f(z) on all of Ω. Then R γ f(z) dz is given by F(w2) −F(w1).
Proof. For smooth curves, Z γ F ′(z) dz = Z b a F ′(γ(t))γ′(t) dt = Z b a (F ◦γ)′(t) dt = (F ◦γ)(t)|t=b t=a = F(γ(b)) −F(γ(a)) = F(w2) −F(w1).
For piecewise smooth curves, this is a trivial extension that is left as an exercise.
38. Corollary. If f = F ′ where F is holomorphic on a region Ω(in that case we say that f has a primitive), γ is a closed curve in Ω, then I γ f(z) dz = 0.
39. A converse to the last claim: if f : Ω→C is a continuous function on a region Ωsuch that I γ f(z) dz = 0 holds for any closed contour in Ω, then f has a primitive.
Proof. Fix some z0 ∈Ω. For any z ∈Ω, there is some path γ(z0, z) connecting z0 and z (since Ωis connected and open, hence pathwise-connected — a standard exercise in topology, see the exercises in Chapter 1 of [Stein-Shakarchi]). Define F(z) = Z γ(z0,z) f(w) dw.
By the assumption, this integral does not depend on which contour γ(z0, z) connecting z0 and z was chosen, so F(z) is well-defined. We now claim that F is holomorphic and its derivative is equal to f. To see this, note that F(z + h) −F(z) h −f(z) = 1 h Z γ(z0,z+h) f(w) dw − Z γ(z0,z) f(w) dw !
−f(z) = 1 h Z γ(z,z+h) f(w) dw −f(z) = 1 h Z γ(z,z+h) (f(w) −f(z)) dw where γ(z, z + h) denotes a contour connecting z and z + h. When |h| is suffi-ciently small so that the disc D(z, h) is contained in Ω, one can take γ(z, z + h) 17 as the straight line segment connecting z and z + h. For such h we get that F(z + h) −F(z) h −f(z) ≤1 h len(γ(z, z + h)) sup w∈D(z,h) |f(w) −f(z)| = sup w∈D(z,h) |f(w) −f(z)| − − − → h→0 0, by continuity of f.
40. Remark. Note that with the last result, if we knew that holomorphic functions are differentiable infinitely many times (the so-called regularity theorem), we could conclude that a function that satisfies the assumption that all its contour integrals on closed contours were 0 is holomorphic. This is in fact true, and is called Morera’s theorem (and is an important fact in complex analysis), but we won’t be able to prove it until we’ve proved Cauchy’s theorem.
41. Example. Compute H |z|=1 zn dz for n ∈Z. What do we learn from the fact that the integral is not zero for n = −1? (Hint: something; but what?) And what do we learn from the fact that it’s 0 when n ̸= −1? (Hint: nothing; but why?) 42. If f is holomorphic on Ωand f ′ ≡0 then f is a constant.
Proof. Fix some z0 ∈Ω. For any z ∈Ω, as we discussed above there is a path γ(z0, z) connecting z0 and z. Then f(z) −f(z0) = Z γ(z0,z) f ′(w) dw = 0, hence f(z) ≡f(z0), so f is constant.
6 Cauchy’s theorem 43. One of the central results in complex analysis is Cauchy’s theorem: Cauchy’s theorem. If f is holomorphic on a simply-connected region Ω, then for any closed curve in Ωwe have I γ f(z) dz = 0.
The challenges are: first, to prove Cauchy’s theorem for curves and regions that are relatively simple (where we do not have to deal with subtle topological considerations); second, to define what simply-connected means; third, which will take a bit longer and we won’t do immediately, to extend the theorem to the most general setting.
44. Two other theorems that are closely related to Cauchy’s theorem are Goursat’s theorem and Morera’s theorem.
18 45. Goursat’s theorem (a relatively easy special case of Cauchy’s theo-rem). If f is holomorphic on a region Ω, and T is a triangle contained in Ω, then H ∂T f(z) dz = 0 (where T refers to the “full” triangle, and ∂T refers to its boundary considered as a curve oriented in the usual positive direction).
46. Morera’s theorem (“the converse of Cauchy’s theorem”). If f : Ω→C is a continuous function on a region Ωsuch that I γ f(z) dz = 0 holds for any closed contour in Ω, then f is holomorphic on Ω.
47. Proof of Goursat’s theorem. The proof idea: “localize the damage”. Namely, try to translate a global statement about the integral around the triangle to a local statement about behavior near a specific point inside the triangle, which would become manageable since we have a good understanding of the local be-havior of a holomorphic function near a point. If something goes wrong with the global integral, something has to go wrong at the local level, and we will show that can’t happen (although technically the proof is not a proof by contradiction, conceptually I find this a helpful way to think about it).
The idea can be made more precise using triangle subdivision. Specifically, let T (0) = T, and define a hierarchy of subdivided triangles order 0 triangle: T (0), order 1 triangles: T (1) j , 1 ≤j ≤4, order 2 triangles: T (2) j,k , 1 ≤j, k ≤4 order 3 triangles: T (3) j,k,ℓ, 1 ≤j, k, ℓ≤4, .
.
.
order n triangles: T (n) j1,...,jn, 1 ≤j1, . . . , jn ≤4.
.
.
.
Here, the triangles T (n) j1,...,jn for jn = 1, 2, 3, 4 are obtained by subdividing the order n−1 triangle T (n−1) j1,...,jn−1 into 4 subtriangles whose vertices are the vertices and/or edge bisectors of T (n−1) j1,...,jn−1 (see Figure 1 on page 35 of [Stein-Shakarchi]).
Now, given the way this subdivision was done, it is clear that we have the equality I ∂T (n−1) j1,...,jn−1 f(z) dz = 4 X jn=1 I ∂T (n) j1,...,jn f(z) dz due to cancellation along the internal edges, and hence I ∂T (0) f(z) dz = 4 X j1,...,jn=1 I ∂T (n) j1,...,jn f(z) dz.
19 That is, the integral along the boundary of the original triangle is equal to the sum of the integrals over all 4n triangles of order n. Now, the crucial observation is that one of these integrals has to have a modulus that is at least as big as the average. That is, we have I ∂T (0) f(z) dz ≤ 4 X j1,...,jn=1 I ∂T (n) j1,...,jn f(z) dz ≤4n I ∂T (n) j(n) f(z) dz where j(n) = (j(n) 1 , . . . , j(n) n ) is some n-tuple chosen such that the second in-equality holds. Moreover, we can choose j(n) inductively in such a way that the triangles T (n) j(n) are nested — that is, T (n) j(n) ⊂T (n−1) j(n−1) for n ≥1, or equivalently j(n) = (j(n−1) 1 , . . . , j(n−1) n−1 , k) for some 1 ≤k ≤4 — to make this happen, choose k to be such that H ∂T (n) (j(n−1),k) f(z) dz is greater than (or equal to) the average 1 4 4 X d=1 I ∂T (n) (j(n−1),d) f(z) dz , which in turn is (by induction) greater than or equal to 1 4 4 X d=1 I ∂T (n) (j(n−1),d) f(z) dz = I ∂T (n−1) j(n−1) f(z) dz ≥4−(n−1) I ∂T f(z) dz.
Now observe that the sequence of nested triangles shrinks to a single point. That is, we have ∞ \ n=0 T (n) j(n) = {z0} for some point z0 ∈T. This is true because the diameter of the triangles goes to 0 as n →∞, so certainly there can’t be two distinct points in the intersection; whereas, on the other hand, the intersection cannot be empty, since the sequence (zn)∞ n=0 of centers (in some obvious sense, e.g., intersection of the angle bisectors) of each of the triangles is easily seen to be a Cauchy sequence (and hence a convergent sequence, by the completeness property of the complex numbers), whose limit must be an element of the intersection.
Having defined z0, write f(z) for z near z0 as f(z) = f(z0) + f ′(z0)(z −z0) + ψ(z)(z −z0), where ψ(z) = f(z) −f(z0) z −z0 −f ′(z0).
The holomorphicity of f at z0 implies that ψ(z) →0 as z →z0. Denote by d(n) the diameter of T (n) j(n) and by p(n) its perimeter. Each subdivision shrinks both the diameter and perimeter by a factor of 2, so we have d(n) = 2−nd(0), p(n) = 2−np(0).
20 It follows that Z ∂T (n) j(n) f(z) dz = Z ∂T (n) j(n) f(z0) + f ′(z0)(z −z0) + ψ(z)(z −z0) dz = Z ∂T (n) j(n) ψ(z)(z −z0) dz ≤p(n)d(n) sup z∈T (n) j(n) |ψ(z)| Finally, combining this with the relationship between H ∂T (0) f(z) dz and | R ∂T (n) j(n) f(z) dz|, we get that Z ∂T (0) f(z) dz ≤p(0)d(0) sup z∈T (n) j(n) |ψ(z)| − − − − → n→∞0, which finishes the proof.
48. Goursat’s theorem for rectangles. The theorem is also true when we replace the word “triangle” with “rectangle”, since a rectangle can be decomposed as the union of two triangles, with the contour integral around the rectangle being the sum of the integrals around the two triangles.
49. Corollary: existence of a primitive for a holomorphic function on a disc.
If f is holomorphic on a disc D, then f = F ′ for some holomorphic function F on D.
Proof. The idea is similar to the proof of the result in 39 above. If we knew that all contour integrals of f around closed contours vanished, that result would give us what we want. As it is, we know this is true but only for triangular contours. How can we make use of that information? [Stein-Shakarchi] gives a clever approach in which the contour γ(z0, z) is comprised of a horizontal line segment followed by a vertical line segment. Then one shows in three steps, each involving a use of Goursat’s theorem (see Figure 4 on page 38 of [Stein-Shakarchi]), that F(z0 +h)−F(z0) is precisely the contour integral over the line segment connecting z0 and z0 + h. From there the theorem proceeds in exactly the same way as before.
50. Corollary: Cauchy’s theorem for a disc. If f is holomorphic on a disc, then H γ f dz = 0 for any closed contour γ in the disc.
Proof. f has a primitive, and we saw that that implies the claimed consequence.
51. Cauchy’s theorem for a region enclosed by a “toy contour”. Repeat the same ideas, going from Goursat’s theorem, to the fact that the function has a primitive, to the fact that its contour integrals along closed curves vanish.
The difficulty as the toy contour gets more complicated is to make sure that the geometry works out when proving the existence of the primitive — see for example the (incomplete) discussion of the case of “keyhole contours” on pages 40–41 of [Stein-Shakarchi].
21 7 Consequences of Cauchy’s theorem 52. Cauchy’s integral formula. If f is holomorphic on a region Ω, and C = ∂D is a circular contour contained in Ω, then 1 2πi I C f(w) w −z dw = f(z) if z ∈D, 0 if z ∈Ω\ D, undefined if z ∈C.
Proof. The case when z / ∈D is covered by Cauchy’s theorem in a disc, since in that case the function w 7→f(w)/(w −z) is holomorphic in an open set containing D. It remains to deal with the case z ∈D. In this case, denote by z0 the center of the circle C. The idea is now to consider instead the integral I Γϵ,δ Fz(w) dw = I Γϵ,δ f(w) w −z dw, where Γϵ,δ is a so-called keyhole contour, namely a contour comprised of a large circular arc around z0 that is a subset of the circle C, and another smaller circular arc of radius ϵ centered at z, with two straight line segments connecting the two circular arcs to form a closed curve, such that the width of the “neck” of the keyhole is δ (think of δ as being much smaller than ϵ); see Fig. 5. Note that the function Fz(w) is holomorphic inside the region enclosed by Γϵ,δ, so Cauchy’s theorem for toy contours gives that I Γϵ,δ Fz(w) dw = 0.
As δ →0, the two parts of the integral along the “neck” of the contour Γϵ,δ cancel out in the limit because Fz is continuous, and hence uniformly continuous, on the compact set D \ D(z, ϵ). So we can conclude that I C Fz(w) dw = I |w−z|=ϵ Fz(w) dw.
The next, and final, step, is to take the limit as ϵ →0 of the right-hand side of this equation, after first decomposing Fz(w) as Fz(w) = f(w) −f(z) w −z + f(z) · 1 w −z , Integrating each term separately, we have for the first term I C f(w) −f(z) w −z dw ≤2πϵ · sup |w−z|=ϵ |f(w) −f(z)| ϵ = 2π sup |w−z|=ϵ |f(w) −f(z)| − − − → ϵ→0 0, 22 C z0 z ϵ δ Figure 5: The keyhole contour used in the proof of Cauchy’s integral formula.
by continuity of f; and for the second term, I |w−z|=ϵ f(z) · 1 w −z dw = f(z) I |w−z|=ϵ 1 w −z dw = 2πif(z) (by a standard calculation, see 41 above). Putting everything together gives H C 1 2πiFz(w) dw = f(z), which was the formula to be proved.
53. Example: in the case when z is the center of the circle C = {w : |w −z| = r}, Cauchy’s formula gives that f(z) = 1 2π I |w−z|=r f(w) dw i(w −z) = 1 2π Z 2π 0 f(z + reit)dt.
In other words, we have proved: Theorem (the mean value property for holomorphic functions). The value of a holomorphic function f at z is equal to the average of its values around a circle |w −z| = r (assuming it is holomorphic on an open set containing the disc |w −z| ≤r).
54. Considering what the mean value property means for the real and imaginary parts of f = u + iv, which are harmonic functions, we see that they in turn also satisfy a similar mean value property: u(x, y) = 1 2π Z 2π 0 u(x + r cos t, y + r sin t) dt.
This is in fact true for all harmonic functions — a fact, known as the mean value property for harmonic functions, that can be proved separately using PDE/real analysis methods, or derived from the above considerations by proving that every harmonic function in a disc is the real part of a holomorphic function.
23 55. Cauchy’s integral formula, extended version. Under the same assump-tions, f is infinitely differentiable, and for z ∈D its derivatives f (n)(z) are given by f (n)(z) = n!
2πi I C f(w) (w −z)n+1 dw.
The fact that holomorphic functions are differentiable infinitely many times is referred to by [Stein-Shakarchi] as the regularity theorem.
Proof. The key observation is that the expression on the right-hand side of Cauchy’s integral formula for f(z) (which is the case n = 0 of the “extended” version) can be differentiated under the integral sign. To make this precise, let n ≥1, and assume inductively that we proved f (n−1)(z) = (n −1)!
2πi I C f(w) (w −z)n dw.
Then f (n−1)(z + h) −f (n−1)(z) h = (n −1)!
2πi I C f(w) · 1 h 1 (w −z −h)n − 1 (w −z)n dw.
It is easily seen that as h →0, the divided difference (w−z−h)−n−(w−z)−n h con-verges to n(w −z)−n−1, uniformly over w ∈C. (The same claim without the uniformity is just the rule for differentiation of a power function; to get the uniformity one needs to “go back to basics” and repeat the elementary algebraic calculation that was originally used to derive this power rule — an illustration of the idea that in mathematics it is important not just to understand results but also the techniques used to derive them.) It follows that we can go to the limit h →0 in the above integral representation, to get f (n)(z) = (n −1)!
2πi I C f(w)n(w −z)−n−1 dz, which is precisely the nth case of Cauchy’s integral formula.
56. Proof of Morera’s theorem. We already proved that if f is a function all of whose contour integrals over closed curves vanish, then f has a primitive F. The regularity property now implies that the derivative F ′ = f is also holomorphic, hence f is holomorphic, which was the claim of Morera’s theorem.
57. As another immediate corollary to Cauchy’s integral formula, we now get an ex-tremely useful family of inequalities that bounds a function f(z) and its deriva-tives at some specific point z ∈C in terms of the values of the function on the boundary of a circle centered at z.
Cauchy inequalities. For f holomorphic in a region Ωthat contains the closed disc DR(z), we have |f (n)(z)| ≤n!R−n sup z∈∂DR(z) |f(z)| (where ∂DR(z) refers to the circle of radius R around z).
24 58. Analyticity of holomorphic functions. If f is holomorphic in a region Ω that contains a closed disc DR(z0), then f has a power series expansion at z0 f(z) = ∞ X n=0 an(z −z0)n, that is convergent for all z ∈DR(z0), where (of course) an = f (n)(z0)/n!.
Proof. The idea is that Cauchy’s integral formula gives us a representation of f(z) as a weighted “sum” (=an integral, which is a limit of sums) of functions of the form z 7→(w −z)−1. Each such function has a power series expansion since it is, more or less, a geometric series, so the sum also has a power series expansion.
To make this precise, write 1 w −z = 1 (w −z0) −(z −z0) = 1 w −z0 · 1 1 − z−z0 w−z0 = 1 w −z0 ∞ X n=0 z −z0 w −z0 n = ∞ X n=0 (w −z0)−n−1(z −z0)n.
This is a power series in z −z0 that, assuming w ∈CR(z0), converges absolutely for all z such that |z −z0| < R (that is, for all z ∈DR(z0)). Moreover the convergence is clearly uniform in w ∈CR(z0). Since infinite summations that are absolutely and uniformly convergent can be interchanged with integration operations, we then get, using the extended version of Cauchy’s integral formula, that f(z) = 1 2πi I CR(z0) f(w) w −z dw = 1 2πi I CR(z0) f(w) ∞ X n=0 (w −z0)−n−1(z −z0)n dw = ∞ X n=0 1 2πi I CR(z0) f(w)(w −z0)n−1 dw !
(z −z0)n = ∞ X n=0 f (n)(z0) n!
(z −z0)n, which is precisely the expansion we were after.
59. Remark.
In the above proof, if we only knew the simple (n = 0) case of Cauchy’s integral formula (and in particular didn’t know the regularity theorem that follows from the extended case of this formula), we would still conclude from the penultimate expression in the above chain of equalities that f(z) has a power series expansion of the form P n an(z−z0)n, with an = (2πi)−1 R CR(z0) f(w)(w− z)−n−1. It would then follow from earlier results we proved that f(z) is differ-entiable infinitely many times, and that an = f (n)(z0)/n!, which would again give the extended version of Cauchy’s integral formula.
25 60. Liouville’s theorem. A bounded entire function is constant.
Proof. An easy application of the (case n = 1 of the) Cauchy inequalities gives upon taking the limit R →∞that f ′(z) = 0 for all z, hence, as we already proved, f must be constant.
61. Theorem. If f is holomorphic on a region Ω, and f = 0 for z in a set containing a limit point in Ω, then f is identically zero on Ω.
Proof. If the limit point is z0 ∈Ω, that means there is a sequence (wk)∞ k=0 of points in Ωsuch that f(wk) = 0 for all n, wk →z, and wk ̸= z0 for all k. We know that in a neighborhood of z0, f has a convergent power series expansion.
If we assume that f is not identically zero in a neighborhood of z0, then we can write the power series expansion as f(z) = X n=0 an(z −z0)k = ∞ X n=m an(z −z0)∞ = am(z −z0)m ∞ X n=0 an+m am (z −z0)n = am(z −z0)m(1 + g(z)), where m is the smallest index such that am ̸= 0, and where g(z) = P∞ n=1 an+m am (z− z0)n is a holomorphic function in the neighborhood of z0 that satisfies g(z0) = 0.
It follows that for all k, am(wk −z0)m(1 + g(wk)) = f(wk) = 0, but for large enough k this is impossible, since wk −z0 ̸= 0 for all k and g(wk) →g(z0) = 0 as k →∞.
The conclusion is that f is identically zero at least in a neighborhood of z0. But now we claim that that also implies that f is identically zero on all of Ω, because Ωis a region (open and connected). More precisely, denote by U the set of points z ∈Ωsuch that f is equal to 0 in a neighborhood of z. It is obvious that U is open, hence also relatively open in Ωsince Ωitself is open; U is also closed, by the argument above; and U is nonempty (it contains z0, again by what we showed above). It follows that U = Ωby the well-known characterization of a connected topological space as a topological space that has no “’clopen” (closed and open) sets other than the empty set and the entire space.
An alternative way to finish the proof is the following. For every point z ∈Ω, let r(z) be the radius of convergence of the power series expansion of f around z. Thus the discs {Dr(z)(z) : z ∈Ω} form an open covering of Ω. Take w ∈Ω (with z0 being as above), and take a path γ : [a, b] →Ωconnecting z0 and w (it exists because Ωis open and connected, hence pathwise-connected). The open covering of Ωby discs is also an open covering of the compact set γ[a, b] (the range of γ). By the Heine-Borel property of compact sets, it has a finite subcovering {Dr(zj)(zj) : j = 0, . . . , m} (where we take w = zm+1. The proof above shows that f is identically zero on Dr(z0)(z0), and also shows that if we know f is zero on Dr(zj)(zj) then we can conclude that it is zero on the next disc 26 Dr(zj+1)(zj+1). It follows that we can get all the way to the last disc Dr(w)(w).
In particular, f(w) = 0, as claimed.
62. Remark. The above result is also sometimes described under the heading zeros of holomorphic functions are isolated, since it can be formulated as the following statement: if f is holomorphic on Ω, is not identically zero on Ω, and f(z0) = 0 for z0 ∈Ω, then for some ϵ > 0, the punctured neighborhood Dϵ(z0) \ {z0} of z0 contains no zeros of f. In other words, the set of zeros of f contains only isolated points.
63. Remark 2. The condition that the limit point z0 be in Ωis needed. Note that it is possible to have a sequence zn →z0 of points in Ωsuch that f(zn) = 0 for all n. For example, consider the function e1/z −1 — it has zeros in every neighborhood of z0 = 0.
64. Corollary. If f, g are holomorphic on a region Ω, and f(z) = g(z) for z in a set with limit point in Ω(e.g., an open disc, or even a sequence of points zn converging to some z ∈Ω), then f ≡g everywhere in Ω.
Proof. Apply the previous result to f −g.
65. The previous result is usually reformulated slightly as the following conceptually important result: Principle of analytic continuation. If f is holomorphic on a region Ω, and f+ is holomorphic on a bigger region Ω+ ⊃Ωand satisfies f(z) = f+(z) for all z ∈Ω, then f+ is the unique such extension, in the sense that if ˜ f+ is another function with the same properties then f+(z) = ˜ f+(z) for all z ∈Ω+.
66. Example. In real analysis, we learn that “formulas” such as 1 −1 + 1 −1 + 1 −1 + . . . = 1 2, 1 + 2 + 4 + 8 + 16 + 32 + . . . = −1 don’t have any meaning. However, in the context of complex analysis one can in fact make perfect sense of such identities, using the principle of analytic continuation! Do you see how? We will also learn later in the course about additional such amusing identities, the most famous of which being 1 + 2 + 3 + 4 + . . . = −1 12, 1 −2 + 3 −4 + . . . = 1 4.
Such supposedly “astounding” formulas have attracted a lot of attention re-cently, being the subject of a popular Numberphile video, a New York Times article, a discussion on the popular math blog by Terry Tao, a Wikipedia article, a discussion on Mathematics StackExchange, and more.
67. A “toy” (but stil very interesting) example of analytic continuation: removable singularities. A point z0 ∈Ωis called a removable singularity 27 of a function f : Ω→C ∪{undefined} if f is holomorphic in a punctured neighborhood of Ω, is not holomorphic at z0, but its value at z0 can be redefined so as to make it holomorphic at z0.
Riemann’s removable singularities theorem.
If f is holomorphic in Ω except at a point z0 ∈Ω(where it may be undefined, or be defined but not known to be holomorphic or even continuous). Assume that f is bounded in a punctured neighborhood Dr(z0) \ {z0} of z0. Then f can be extended to a holomorphic function ˜ f on all of Ωby defining (or redefining) its value at z0 appropriately.
Proof. Fix some disc D = DR(z0) around z0 whose closure is contained in Ω.
The idea is to prove that the Cauchy integral representation formula f(z) = 1 2πi I CR(z0) f(w) w −z dw =: ˜ f(z) is satisfied for all z ∈D \ {z0}. Once we show this, we will set ˜ f(z0) to be defined by the same integral representation, and it will be easy to see that that gives the desired extension.
To prove that the representation above holds, consider a “double keyhole” con-tour Γϵ,δ that surrounds most of circle C = ∂D but makes diversions to avoid the points z0 and z, circling them in the negative direction around most of a circle of radius ϵ. After applying a limiting argument similar to the one used in the proof of Cauchy’s integral formula, we get that 1 2πi I C f(w) w −z = 1 2πi I Cϵ(z) f(w) w −z + 1 2πi I Cϵ(z0) f(w) w −z .
On the right-hand side, the first term is f(z) by Cauchy’s integral formula (since f is known to be holomorphic on an open set containing Dϵ(z)). The second term can be bounded in magnitude using the assumption that f is bounded in a neighborhood of z0; more precisely, we have I Cϵ(z0) f(w) w −z ≤2πϵ sup w∈Cϵ(z0) |f(w)| · 1 |z −z0| −ϵ − − − → ϵ→0 0.
Thus by taking the limit as ϵ →0 we obtain precisely the desired representation for f.
Finally, once we have the integral representation ˜ f (defined only in terms of the values of f(w) for w ∈CR(z0)), the fact that this defines a holomorphic function on all of DR(z0) is easy to see, and is something we implicitly were aware of already. For example, the relevant argument (involving a direct manipulation of the divided differences 1 h( ˜ f(z + h) −˜ f(z))) appeared in the proof of the extended version of Cauchy’s integral formula. Another approach is to show that integrating ˜ f over closed contours gives 0 (which requires interchanging the order of two integration operations, which will not be hard to justify) and then use Morera’s theorem. The details are left as an exercise.
28 68. Definition: Uniform convergence on compact subsets. If f, (fn)∞ n=0 are holomorphic functions on a region Ω, we say that the sequence fn converges to f uniformly on compact subsets if for any compact set K ⊂Ω, fn(z) →fn(z) uniformly on K.
69. Theorem.
If fn →f uniformly on compact subsets in Ωand fn are holo-morphic, then f is holomorphic, and f ′ n →f ′ uniformly on compact subsets in Ω.
Proof. The fact that f is holomorphic is an easy consequence of a combination of Cauchy’s theorem and Morera’s theorem. More precisely, note that for each closed disc D = Dr(z0) ⊂Ωwe have fn(z) →f uniformly on D. In particular, for each curve γ whose image is contained in the open disc D = Dr(z0), Z γ fn(z) dz − − − − → n→∞ Z γ f(z) dz.
By Cauchy’s theorem, the integrals in this sequence are all 0, so R γ f(z) dz is also zero. Since this is true for all such γ, by Morera’s theorem f is holomorphic on D. This was true for any disc in Ω, and holomorphicity is a local property, so in other words f is holomorphic on all of Ω.
Next, let D = Dr(z0) be a disc whose closure D satisfies D ⊂Ω. for z ∈D we have by Cauchy’s integral formula that f ′ n(z) −f ′(z) = 1 2πi I ∂D fn(w) (w −z)2 dw − 1 2πi I ∂D f(w) (w −z)2 dw = 1 2πi I ∂D fn(w) −f(w) (w −z)2 dw.
It is easy to see therefore that f ′ n(z) →f ′(z) as n →∞, uniformly as z ranges on the disc Dr/2(z0), since fn(w) →f(w) uniformly for w ∈∂D ⊂D, and |w −z|−2 ≤(r/2)−2 for z ∈Dr/2(z0), w ∈∂D.
Finally, let K ⊂D be compact. For each z ∈K let r(z) be the radius of a disc Dr(z)(z) around z whose closure is contained in Ω.
The family of discs {Dz = Dr(z)/2(z) : z ∈Ω} is an open covering of K, so by the Heine-Borel property of compact sets it has a finite subcovering Dz1, . . . , Dzn. We showed that f ′ n(z) →f ′(z) uniformly on every Dzj, so we also have uniform convergence on their union, which contains K, so we get that f ′ n →f ′ uniformly on K, as claimed.
8 Zeros, poles, and the residue theorem 70. Definition (zeros). z0 is a zero of a holomorphic function f if f(z0) = 0.
71. Lemma/Definition. If f is a holomorphic function on a region Ωthat is not identically zero and z0 is a zero of f, then f can be represented in the form f(z) = (z −z0)mg(z) 29 in some neighborhood of z0, where m ≥1 and g is a holomorphic function in that neighborhood such that g(z) ̸= 0. The number m is determined uniquely and is called the order of the zero z0, i.e, z0 will be described as “a zero of order m.” Remark 1. In the case when z0 is not a zero of f, the same representation holds with m = 0 (and g = f), so in certain contexts one may occasionally say that z0 is a zero of order 0.
Remark 2. A zero of order 1 is called a simple zero.
Proof. Power series expansions – this is similar to the argument used in the proof that zeros of holomorphic functions are isolated. That is, write the power series expansion (known to converge in a neighborhood of z0) f(z) = ∞ X n=0 an(z −z0)n = ∞ X n=m an(z −z0)n = (z −z0)m ∞ X n=0 am+n(z −z0)n =: (z −z0)mg(z), where m is the smallest index ≥0 such that am ̸= 0. This gives the desired rep-resentation. On the other hand, given a representation of this form, expanding g(z) as a power series shows that m has to be the smallest index of a nonzero coefficient in the power series expansion of f(z), which proves the uniqueness claim.
72. Definition (poles). If f is defined and holomorphic in a punctured neighbor-hood of a point z0, we say that it has a pole of order m at z0 if the function h(z) = 1/f(z) (defined to be 0 at z0) has a zero of order m at z0. A pole of order 1 is called a simple pole.
Remark. As with the case of zeros, one can extend this definition in an obvious way to define a notion of a “pole of order 0”. If f(z) is actually holomorphic and nonzero at z0 (or has a removable singularity at z0 and can be made holomorphic and nonzero by defining its value at z0 appropriately), we define the order of the pole as 0 and consider f to have a pole of order 0 at z0.
73. Lemma. f has a pole of order m at z0 if and only if it can be represented in the form f(z) = (z −z0)−mg(z) in a punctured neighborhood of z0, where g is holomorphic in a neighborhood of z0 and satisfies g(z0) ̸= 0.
Proof. Apply the previous lemma to 1/f(z).
74. Theorem. If f has a pole of order m at z0, then it can be represented uniquely as f(z) = a−m (z −z0)m + a−m+1 (z −z0)m−1 + . . . + a−1 z −z0 + G(z) where G is holomorphic in a neighborhood of z0.
30 Proof. This follows immediately on expressing f(z) as (z −z0)−mg(z) as in the previous lemma and separating the power series expansion of g(z) into the powers (z −z0)k with 0 ≤k ≤m −1 and the powers with k ≥m.
75. Definition.
The expansion a−m (z−z0)m + a−m+1 (z−z0)m−1 + . . . + a−1 z−z0 in the above representation is called the principal part of f at the pole z0. The coefficient a−1 is called the residue of f at z0 and denoted Resz0(f).
76. Exercise. The definitions of the order of a zero and a pole can be consistently unified into a single definition of the (generalized) order of a zero, where if f has a pole of order m at z0 then we say instead that f has a zero of order −m. Denote the order of a zero of f at z0 by ordz0(f). With these definitions, prove that ordz0(f + g) ≥min (ordz0(f), ordz0(g)) (can you give a useful condition when equality holds?), and that ordz0(fg) = ordz0(f) + ordz0(g).
77. The residue theorem (simple version). Assume that f is holomorphic in a region containing a closed disc D, except for a pole at z0 ∈D. Then I ∂D f(z) dz = 2πi Resz0(f).
Proof. By the standard argument involving a keyhole contour, we see that the circle C = ∂D in the integral can be replaced with a circle Cϵ = Cϵ(z0) of a small radius ϵ > 0 around z0, that is , we have I ∂D f(z) dz = I Cϵ f(z) dz.
When ϵ is small enough, inside Cϵ we can use the decomposition f(z) = −1 X k=−m ak(z −z0)k + G(z) for f into its principal part and a remaining holomorphic part.
Integrating termwise gives 0 for the integral of G(z), by Cauchy’s theorem; 0 for the integral powers (z −z0)k with −m ≤k ≤−2, by a standard computation; and 2πia−1 = 2πi Resz0(f) for the integral of r(z −z0)−1, by the same standard computation.
This gives the result.
78. The residue theorem (extended version). Assume that f is holomorphic in a region containing a closed disc D, except for a finite number of poles at z1, . . . , zN ∈D. Then I ∂D f(z) dz = 2πi N X k=1 Reszk(f).
31 Proof.
The idea is the same, except one now uses a contour with multiple keyholes to deduce after a limiting argument that I ∂D f(z) dz = N X k=1 I Cϵ(zk) f(z) dz for a small enough ϵ, and then proceeds as before.
(Note: The above argument seems slightly dishonest to me, since it relies on the assertion that a multiple keyhole contour with arbitrary many keyholes is a “toy contour”; while this is intuitively plausible, it will be undoubtedly quite difficult to think of, and write, a detailed proof of this argument.) 79. The residue theorem (version for general toy contours). Assume that f is holomorphic in a region containing a toy contour γ (oriented in the posi-tive direction) and the region Rγ enclosed by it, except for poles at the points z1, . . . , zN ∈Rγ. Then I γ f(z) dz = 2πi N X k=1 Reszk(f).
Proof.
Again, construct a multiple keyhole version of the same contour γ (assuming that one can believably argue that the resulting contour is still a toy contour), and then use a limiting argument to conclude that I γ f(z) dz = N X k=1 I Cϵ(zk) f(z) dz, for a small enough ϵ. Then proceed as before.
9 Meromorphic functions, holomorphicity at ∞ and the Riemann sphere 80. Definition (meromorphic functions). A meromorphic function on a region Ωis a function f : Ω→C ∪{undefined} such that f is holomorphic except for an isolated set of poles.
81. Definition (holomorphicity at ∞). Let U ⊂C be an open set containing the complement C \ DR(0) of a closed disc around 0. A function f : U →C is holomorphic at ∞if g(z) = f(1/z) (defined on a neighborhood D1/R(0) of 0) has a removable singularity at 0. In that case we define f(∞) = g(0) (the value that makes g holomorphic at 0).
82. Definition (order of a zero/pole at ∞). Let U ⊂C be an open set contain-ing the complement C\DR(0) of a closed disc around 0. We say that a function f : U →C has a zero (resp. pole) of order m at ∞if g(z) = f(1/z) has a zero (resp. pole) at z = 0, after appropriately defining the value of g at 0.
32 83. Conceptually, it is useful to think of meromorphic functions as holomorphic functions with range in the Riemann sphere ˆ C. Let’s define what that means.
84. Definition. The Riemann sphere (a.k.a. the extended complex numbers) is the set ˆ C = C ∪{∞}, equipped with the following additional structure: • Topologically, we think of ˆ C as the 1-point compactification of C; that is, we add to C an additional element ∞(called “the point at infinity”) and say that the neighborhoods of ∞are the complements of compact sets in C. This turns ˆ C into a topological space in a simple way.
• Geometrically, we can identify ˆ C with an actual sphere embedded in R3, namely S2 = n (x, y, z) ∈R3 : x2 + y2 + z −1 2 2 = 1 2 o .
The identification is via stereographic projection, given explicitly by the formula (X, Y, Z) ∈S2 7− → x + iy = X 1−Z + i Y 1−Z if (X, Y, Z) ̸= (0, 0, 1), ∞ if (X, Y, Z) = (0, 0, 1).
See page 88 in [Stein-Shakarchi] for a more detailed explanation. One can check that this geometric identification is a homeomorphism between S2 (equipped with the obvious topology inherited from R3) and ˆ C (with the 1-point compactification topology defined above).
• Analytically, the above definition of what it means for a function on a neighborhood of ∞to be holomorphic at ∞provides a way of giving ˆ C the structure of a Riemann surface (the simplest case of a manifold with a complex-analytic structure). The details can be found in many textbooks and online resources, and we will not discuss them in this course.
85. With the above definitions, the concept of a meromorphic function f : Ω→C ∪ {undefined} can be seen to coincide with the notion of a holomorphic function f : Ω→ˆ C — that is, the underlying concept of the definition is still holomorphicity, but it concerns functions taking values in ˆ C, a different Riemann surface, instead of C.
86. Definition (classification of singularities).
If a function f : Ω→ˆ C ∪ {undefined} is holomorphic in a punctured neighborhood Dr(z0) \ {z0} of z0, we say that f has a singularity at z0 if f is not holomorphic at z0. We classify singularities into three types, two of which we already defined: • Removable singularities: when f can be made holomorphic at z0 by defining or redefining its value at z0.
• Poles.
• Any singularity that is not removable or a pole is called an essential singularity.
33 For a function defined on a neighborhood of ∞that is not holomorphic at ∞, we say that f has a singularity at ∞, and classify the singularity as a removable singularity, a pole, or an essential singularity, according to the type of singularity that z 7→f(1/z) has at z = 0.
87. Theorem (Casorati-Weierstrass theorem on essential singularities). If f is holomorphic in a punctured neighborhood Dr(z0) \ {z0} of z0 and has an essential singularity at z0, the image f(Dr(z0) \ {z0}) of the punctured neigh-borhood under f is dense in C.
Proof.
Assume that the closure f(Dr(z0) \ {z0}) does not contain a point w ∈C. Then g(z) = 1 f(z)−w is a function that’s holomorphic and bounded in Dr(z0) \ {z0}. By Riemann’s removable singularity theorem, its singularity at z0 is removable, so we can assume it is holomorphic at z0 after defining its value there. It then follows that f(z) = w + 1 g(z) has either a pole or a removable singularity at z0, depending on whether g(z0) = 0 or not.
10 The argument principle 88. Definition. The logarithmic derivative of a holomorphic function f(z) is f ′(z)/f(z).
89. Lemma. The logarithmic derivative of a product is the sum of the logarithmic derivatives. That is, Qn k=1 fk ′ Qn k=1 fk = n X k=1 f ′ k(z) fk(z).
Proof. Show this for n = 2 and proceed by induction.
90. Theorem (the argument principle). Assume that f is meromorphic in a region Ωcontaining a closed disc D, such that f has no poles on the circle ∂D.
Denote its zeros and poles inside D by z1, . . . , zn, where zk is a zero of order mk = ordzk(f) (in the sense mentioned above, where mk = m is a positive integer if zk is a zero of order m, and mk = −m is a negative integer if zk is a pole of order m). Then 1 2πi I ∂D f ′(z) f(z) dz = n X k=1 mk = [total number of zeros of f inside D] −[total number of poles of f inside D].
Proof. Define g(z) = n Y k=1 (z −zk)−mkf(z).
34 Then g(z) is meromorphic on Ω, has no singularities zeros on ∂D, and inside D it has no poles or zeros, only removable singularities at z1, . . . , zn (so after redefining its values at these points we can assume it is holomorphic on D). It follows that f(z) = n Y k=1 (z −zk)mkg(z).
Taking the logarithmic derivative of this equation gives that f ′(z) f(z) = n X k=1 mk z −zk + g′(z) g(z) .
The result now follows by integrating this equation and using the residue the-orem (the term g′(z)/g(z) is holomorpic on D so does not contribute anything to the integral).
91. By similar reasoning, the theorem also holds when the circle is replaced by a toy contour γ.
92. Intuitive explanation for the argument principle. Note that the integral in the argument principle can be represented as 1 2πi I γ f ′(z) f(z) dz = 1 2πi Z b a f ′(γ(t))γ′(t) f(γ(t) dt = 1 2πi Z b a (f ◦γ)′(t) (f ◦γ)(t) dt = 1 2πi Z f◦γ 1 w dw, that is, an integral of dw/w over the contour f ◦γ — the image of γ under f.
Now note that the differential form dw/w has a special geometric meaning in complex analysis, namely we have dw w = “d (log w) ” = “d (log |w| + i arg w) ”.
We put these expressions in quotes since the logarithm and argument are not single-valued functions so it needs to be explained what such formulas mean.
However, at least log |w| is well-defined for a curve that does not cross 0, so when integrating over the closed curve f ◦γ, the real part is zero by the fundamental theorem of calculus. The imaginary part (which becomes real after dividing by 2πi) can be interpreted intuitively as the change in the argument over the curve — that is, initially at time t = a one fixes a specific value of arg w = arg γ(a); then as t increases from t = a to t = b, one tracks the increase or decrease in the argument as one travels along the curve γ(t); if this is done correctly (i.e., in a continuous fashion), at the end the argument must have a well-defined value. Since the curve is closed, the total change in the argument must be an integer multiple of 2π, so the division by 2πi turns it into an integer.
Of course, this explanation also explains the name “the argument principle,” which may sound arbitrary and uninformative when you first hear it.
35 93. Connection to winding numbers. What the above reasoning shows is that in general, an integral of the form 1 2πi I γ f(w) w dw over a closed curve γ that does not cross 0 carries the meaning of “the total number of times the curve γ goes around the origin,” with the number being positive if the curve goes in the positive direction around the origin; negative if the curve goes in the negative direction around the origin; or zero if there is no net change in the argument. This number is more properly called the winding number of f around w = 0 (also sometimes referred to as the index of the curve around 0), and denoted Ind0(f) = 1 2πi I γ f(z) z dz.
More generally, one can define the winding number at z = z0 as the number of times a curve γ winds around an arbitrary point z0, which (it is easy to see) will be given by Indz0(f) = 1 2πi I γ f(z) z −z0 dz, assuming that γ does not cross z0.
Note that winding number is a topological concept of planar geometry that can be considered and studied without any reference to complex analysis; indeed, in my opinion that is the correct approach. It is possible, and not especially difficult, to define it in purely topological terms without mentioning contour integrals, and then show that the complex analytic and topological definitions coincide. Try to think what such a definition might look like.
94. Rouch´ e’s theorem. Assume that f, g are holomorphic on a region Ωcontaining a circle γ = C and its interior (or, more generally, a toy contour γ and the region U enclosed by it). If |f(z)| > |g(z)| for all z ∈γ then f and f + g have the same number of zeros inside the region U.
Proof. Define ft(z) = f(z) + tg(z) for t ∈[0, 1], and note that f0 = f and f1 = f + g, and that the condition |f(z)| > |g(z)| on γ implies that ft has no zeros on γ for any t ∈[0, 1]. Denote nt = 1 2πi I γ f ′ t(z) ft(z) dz, which by the argument principle is the number of “generalized zeros” (zeros or poles, counting multiplicities) of ft in U. In particular, the function t 7→nt is integer-valued. If we also knew that it was continuous, then it would have to be constant (by the easy exercise: any integer-valued continuous function on an interval [a, b] is constant), so in particular we would get the desired conclusion that n1 = n0.
36 To prove continuity of nt, note that the function g(t, z) = f ′ t(z)/ft(z) is con-tinuous, hence also uniformly continuous, on the compact set [0, 1] × γ. For s, t ∈[0, 1] satisfying |t −s| < δ, we can write |nt −ns| ≤ 1 2πi I γ |g(t, z) −g(s, z)| · |dz| ≤ 1 2πi len(γ) sup{|g(u, z) −g(v, z)| : z ∈γ, u, v ∈[0, 1], |u −v| < δ}.
Given ϵ > 0, we can choose δ that ensures that this expression is < ϵ if |t−s| < δ, by the uniform continuity. This is precisely what is needed to show that t 7→nt is continuous.
95. Intuitive explanation for Rouch´ e’s theorem: “walking the dog”. The following intuitive explanation for Rouch´ e’s theorem appears in the book Visual Complex Analysis by Tristan Needham. Imagine that you are walking in a large empty park containing at some “origin” point 0 a large pole (in the English sense of a metal post sticking out of the ground, not the complex analysis sense). You start at some point X and go for a walk along some curve, ending back at the same starting point X. Let N denote your winding number around the pole at the origin — that is, the total number of times you went around the pole, with the appropriate sign.
Now imagine that you also have a dog that is walking alongside you in some erratic path that is sometimes close to you, sometimes less close. As you traverse your curve C1, the dog walks along on its own curve C2, which also begins and ends in the same place. Let M denote the dog’s winding number around the pole at the origin. Can we say that N = M? The answer is: yes, we can, provided that we know the dog’s distance to you was always less than your distance to the pole. To see this, imagine that you had the dog on a leash of variable length; if the distance condition was not satisfied, it would be possible for the dog to reach the pole and go in a short tour around it while you were still far away and not turning around the pole, causing an entanglement of the leash with the pole.
Amazingly, the above scenario maps in a precise way to Rouch´ e’s theorem, using the following dictionary: the curve f ◦γ represents your path; the curve (f +g)◦γ represents the dog’s path; g ◦γ represents the vector pointing from you to the dog; the condition |f| > |g| along γ is precisely the correct condition that the dog stays closer to you than your distance to the pole; and the conclusion that the two winding numbers are the same is precisely the theorem’s assertion that f and f + g have the same number of generalized zeros in the region U enclosed by γ (see the discussion above regarding the connection between the integral (2πi)−1 H γ f ′/f dz and the winding number of f ◦γ around 0).
Exercise. Spend a few minutes thinking about the above correspondence and make sure you understand it. You will probably forget the technical details of the proof of Rouch´ e’s theorem in a few weeks or months, but I hope you will remember this intuitive explanation for a long time.
37 96. As another small cryptic remark to think about, the proof of Rouch´ e’s theorem given above can be thought of as an argument about the invariance of a certain integral under the homotopy between two curves. Can you see how?
11 Applications of Rouch´ e’s theorem 97. Topological proof of the fundamental theorem of algebra. At the begin-ning of the course we discussed the topological proof of FTA. We can now make that argument precise using Rouch´ e’s theorem. The details will be assigned as a homework exercise.
98. The open mapping theorem. Holomorphic functions are open mappings, that is, they map open sets to open sets.
Proof. Let f be holomorphic in a region Ω, z0 ∈Ω, and denote w0 = f(z0).
What we need to show is that the image of any neighborhood Dϵ(z0) for ϵ > 0 contains a neighborhood Dδ(z0) of w0 for some δ > 0. Fixing w (visualized as being near w0), denote h(z) = f(z) −w = (f(z) −w0) + (w0 −w) =: F(z) + G(z).
The idea is now to apply Rouch´ e’s theorem to F(z) and G(z). Fix ϵ > 0 small enough so that the disc Dϵ(z0) is contained in Ωand does not contain solutions of the equation f(z) = w0 other than z0 (this is possible, by the property that zeros of holomorphic functions are isolated). Defining δ = inf{|f(z) −w0| : z ∈Dϵ(z0)}, we therefore have that δ > 0 and |f(z) −w0| ≥δ for z on the circle |z −z0| = ϵ.
That means that under the assumption that |w −w0| < δ (i.e., if w is assumed to be close enough to w0), the condition |F(z)| > |G(z)| in Rouch´ e’s theorem will be satisfied for z ∈Dϵ(z0). The conclusion is that the equation h(z) = 0 (or equivalently f(z) = w) has the same number in solutions (in particular, at least one solution) as the equation f(z) = w0 in the disc Dϵ(z0). This was precisely the claim to be proved.
99. Corollary: the maximum modulus principle. If f is a non-constant holo-morphic function on a region Ω, then |f| cannot attain a maximum on Ω.
Proof. Trivial exercise.
12 Simply-connected regions and the general ver-sion of Cauchy’s theorem 100. Definition (homotopy of curves). Given a region Ω⊂C, two parametrized curves γ1, γ2 : [0, 1] →Ω(assumed for simplicity of notation to be defined on [0, 1]) are said to be homotopic (with fixed endpoints) if γ1(0) = γ2(0), γ1(1) = γ2(1), and there exists a function F : [0, 1] × [0, 1] →Ωsuch that 38 (a) F is continuous.
(b) F(0, t) = γ1(t) for all t ∈[0, 1].
(c) F(1, t) = γ2(t) for all t ∈[0, 1].
(d) F(s, 0) = γ1(0) for all s ∈[0, 1].
(e) F(s, 1) = γ1(1) for all s ∈[0, 1].
The map F is called a homotopy between γ1 and γ2. Intuitively, for each s ∈[0, 1] the function t 7→F(s, t) defines a curve connecting the two endpoints γ1(0), γ1(1). As s grows from 0 to 1, this family of curves transitions in a continuous way between the curve γ1 and γ2, with the endpoints being fixed in place.
101. Exercise. Prove that the relation of being homotopic is an equivalence relation.
102. Definition (simply-connected regions). A region Ωis called simply-connected if any two curves γ1, γ2 in Ωwith the same endpoints are homotopic.
103. Remark. A common alternative way to define the notion of homotopy of curves is for closed curves, where the endpoints are not fixed but the homotopy must keep the curves closed as it is deforming them.
The definition of a simply-connected region then becomes a region in which any two closed curves are homotopic. It is not hard to show that those two definitions are equivalent.
104. Theorem. If f is a holomorphic function on a region Ω, and γ1,γ2 are two curves on Ωwith the same endpoints that are homotopic, then Z γ1 f(z) dz = Z γ2 f(z) dz.
105. Proof. See pages 93–95 in [Stein-Shakarchi].
106. Cauchy’s theorem (general version).
If f is holomorphic on a simply-connected region Ω, then for any closed curve in Ωwe have I γ f(z) dz = 0.
107. Proof. Assume for simplicity that γ is parametrized as a curve on [0, 1]. Then it can be thought of as the concatenation of two curves γ1 and −γ2, where γ1 = γ|[0,1/2] and γ2 is the “reverse” of the curve γ|[1/2,1]. Note that γ1 and γ2 have the same endpoints. By the invariance property of contour integrals under homotopy proved above, we have Z γ f(z) dz = Z γ1−γ2 f(z) dz = Z γ1 f(z) dz − Z γ2 f(z) dz = 0.
108. Corollary. Any holomorphic function on a simply-connected region has a prim-itive.
39 13 The logarithm function 109. The logarithm function can be defined as log z = log |z| + i arg z on any region Ωthat does not contain 0 and where one can make a consistent, smoothly varying choice of arg z as z ranges over Ω. It is easy to see that this formula gives an inverse to the exponential function.
For example, if Ω= C \ (−∞, 0] (the “slit complex plane” with the negative real axis removed), we can set Log z = log |z| + i Arg z where Arg z is set to take values in (−π, π). This is called the principal branch of the logarithm. However, sometimes we may want to consider the logarithm function on more strange or complicated regions. When can this be made to work? The answer is: precisely when Ωis simply-connected.
110. Theorem. Assume that Ωis a simply-connected region with 0 / ∈Ω, 1 ∈Ω.
Then there exists a function F(z) = logΩ(z) with the properties: (a) F is holomorphic in Ω.
(b) eF (z) = z for all z ∈Ω.
(c) F(r) = log r (the usual logarithm for real numbers) for all real numbers r ∈Ωsufficiently close to 1.
Proof. We define F as a primitive function of the function z 7→1/z, that is, as F(z) = Z z 1 dw w , where the integral is computed along a curve γ connecting 1 to z.
By the general version of Cauchy’s theorem for simply-connected regions, this integral is independent of the choice of curve. As we have already seen, this function is holomorphic and satisfies F ′(z) = 1/z for all z ∈Ω. It follows that d dz ze−F (z) = e−F (z) −zF ′(z)e−F (z) = e−F (z)(1 −z/z) = 0, so ze−F (z) is a constant function. Since its value at z = 1 is 1, we see that eF (z) = z, as required. Finally, for real r close to 1 we have that F(z) = R r 1 dw w = log r, which can be seen by taking the integral to be along the straight line segment connecting 1 and r.
111. Exercise. Prove that the principal branch of the logarithm has the Taylor series expansion Log z = ∞ X n=1 (−1)n−1 n (z −1)n (|z| < 1).
40 112. Exercise. Modify the proof above to prove the existence of a branch of the logarithm function in any simply-connected region Ωnot containing 0, without the assumption that 1 ∈Ω. In what way is the conclusion weakened in that case?
113. Exercise. Explain in what sense the logarithm functions F(z) = logΩ(z) sat-isfying the properties proved in the theorem above (and its generalization de-scribed in the previous exercise) are unique.
114. Exercise.
Prove the following generalization of the logarithm construction above: if f is a holomorphic function on a simply-connected region Ω, and f ̸= 0 on Ω, then there exists a holomorphic function g on Ω, referred to as a branch of the logarithm of f, satisfying eg(z) = f(z).
115. Definition (power functions and nth roots). On a simply-connected region Ωwe can now define the power function z 7→zα for an arbitrary α ∈C by setting zα = eα log z.
In the special case α = 1/n this has the meaning of the nth root function z 7→z1/n, which satisfies (z1/n)n = e 1 n log zn = en 1 n log z = elog z = z.
Note that if f(z) = z1/n is one choice of an nth root function, then for any 0 ≤k ≤n−1, the function g(z) = e2πik/nf(z) will be another function satisfying g(z)n = z. Conversely, it is easy to see that those are precisely the possible choices for an nth root function.
14 The Euler gamma function 116. The Euler gamma function (often referred to simply as the gamma func-tion) is one of the most important special functions in mathematics.
It has applications to many areas, such as combinatorics, number theory, differential equations, probability, and more, and is probably the most ubiquitous transcen-dental function after the “elementary” transcendental functions (the exponential function, logarithms, trigonometric functions and their inverses) that one learns about in calculus. It is a natural meromorphic function of a complex variable that extends the factorial function to non-integer values. In complex analysis it is particularly important in connection with the theory of the Mellin transform (a version of the Fourier transform associated with the multiplicative group of positive real numbers).
117. Most textbooks define the gamma function in one way and proceed to prove several other equivalent representations of it. However, the truth is that none 41 of the representations of the gamma function is more fundamental or “natural” than the others. So, it seems more logical to start by simply listing the various formulas and properties associated with it, and then proving that the different representations are equivalent and that the claimed properties hold.
Theorem (the Euler gamma function). There exists a unique function Γ(s) of a complex variable s that has the following properties: (a) Γ(s) is a meromorphic function on C.
(b) Connection to factorials: Γ(n + 1) = n! for n = 0, 1, 2, . . ..
(c) Important special value: Γ(1/2) = √π.
(d) Integral representation: Γ(s) = Z ∞ 0 e−xxs−1 dx (Re s > 0).
(e) Hybrid series-integral representation: Γ(s) = ∞ X n=0 (−1)n n!(n + s) + Z ∞ 1 e−xxs−1 dx (s ∈C).
(f) Infinite product representation: Γ(s)−1 = seγs ∞ Y n=1 1 + s n e−s/n (s ∈C), where γ = limn→∞ 1 + 1 2 + 1 3 + . . . + 1 n −log n .
= 0.577215 is the Euler-Mascheroni constant.
(g) Limit of finite products representation: Γ(s) = lim n→∞ n! ns s(s + 1) · · · (s + n) (s ∈C).
(h) Zeros: the gamma function has no zeros (so Γ(s)−1 is an entire function).
(i) Poles: the gamma function has poles precisely at the non-positive integers s = 0, −1, −2, . . ., and is holomorphic everywhere else. The pole at s = −n is a simple pole with residue Ress=−n(Γ) = (−1)n n!
(n = 0, 1, 2, . . .).
(j) Functional equation: Γ(s + 1) = s Γ(s) (s ∈C).
(k) The reflection formula (a surprising connection to trigonometry): Γ(s)Γ(1 −s) = π sin(πs) (s ∈C).
42 118. To begin the proofs, let’s take the formula Γ(s) = Z ∞ 0 e−xxs−1 dx as our working definition of Γ(s). This improper integral is easily seen to con-verge absolutely for Re(s) > 0, since Z ∞ 0 e−xxs−1 dx ≤ Z ∞ 0 e−x|xs−1| dx = Z ∞ 0 e−xxRe(s)−1 dx.
I leave it as an exercise to check (or read the easy explanation in the book) that the function it defines is holomorphic in that region.
119. Next, perform an integration by parts, to get that, again for Re(s) > 0, we have Γ(s + 1) = Z ∞ 0 e−xxs dx = −e−xxs x=∞ x=0 + Z ∞ 0 e−xsxs−1 dx = s Γ(s), which is the functional equation.
120. Combining the trivial evaluation Γ(1) = R ∞ 0 e−x dx = 1 with the functional equation shows by induction that Γ(n + 1) = n!.
121. The special value Γ(1/2) = √π follows immediately by a change of variable x = u2 in the integral and an appeal to the standard Gaussian integral R ∞ −∞e−u2 du = √π: Γ(1/2) = Z ∞ 0 e−xx−1/2 dx = Z ∞ 0 e−u22 du = Z ∞ −∞ e−u2 du = √π.
122. The functional equation can now be used to perform an analytic continuation of Γ(s) to a meromorphic function on C: for example, we can define Γ1(s) = Γ(s + 1) s , which is a function that is holomorphic on Re(s) > −1, s ̸= 0 and coincides with γ(s) for Re(s) > 0. By the principle of analytic continuation this provides a unique extension of Γ(s) to the region Re(s) > −1. Because of the factor 1/s and the fact that Γ(1) = 1 we also see that Γ1(s) has a simple pole at s = 0 with residue 1.
Next, for Re(s) > −2 we define Γ2(s) = Γ1(s + 1) s = Γ(s + 2) s(s + 1) , a function that is holomorphic on Re(s) > −2, s ̸= 0, −1, and coincides with Γ1(s) for Re(s) > −1, s ̸= 0. Again, this provides an analytic continuation of Γ(s) to that region. The factors 1/s(s + 1) show that Γ2(s) has a simple pole at s = −1 with residue −1.
43 Continuing by induction, having defined an analytic continuation Γn−1(s) of Γ(s) to the region Re(s) > −n + 1, s ̸= 0, −1, −2, . . . , −n + 2, we now define Γn(s) = Γn−1(s + 1) s = . . . = Γ(s + n) s(s + 1) · · · (s + n −1).
By inspection we see that this gives a meromorphic function in Re(s) > −n whose poles are precisely at s = −n + 1, . . . , 0 and have the claimed residues.
123. An alternative way to perform the analytic continuation is to separate the inte-gral defining Γ(s) into Γ(s) = Z 1 0 e−xxs−1 dx + Z ∞ 1 e−xxs−1 dx and to note that the integral over [1, ∞) converges (and defines a holomorphic function of s) for all s ∈C, and the integral over [0, 1] can be computed by expanding e−x as a power series in x and integrating term by term. That is, for Re(s) > 0 we have Z 1 0 e−xxs−1 dx = Z 1 0 ∞ X n=0 (−1)n n!
xn+s−1 dx = ∞ X n=0 (−1)n n!
Z 1 0 xn+s−1 dx = ∞ X n=0 (−1)n n!(n + s) The justification for interchanging the summation and integration operations is easy and is left as an exercise. Thus, we have obtained not just an alternative proof for the meromorphic continuation of Γ(s), but a proof of the hybrid series-integral representation of Γ(s), which also clearly shows where the poles of Γ(s) are and that they are simple poles with the correct residues.
124. Lemma. For Re(s) > 0 we have Γ(s) = lim n→∞ Z n 0 1 −x n n xs−1 dx.
Proof. As n →∞, the integrand converges to e−xxs−1 pointwise. Furthermore, the factor 1 −x n n is bounded from above by the function e−x (because of the elementary inequality 1 −t ≤e−t that holds for all real t). The claim therefore follows from the dominated convergence theorem.
125. Lemma. For Re(s) > 0 we have Z n 0 1 −x n n xs−1 dx = n! ns s(s + 1) · · · (s + n).
Proof. For n = 1, the claim is that Z 1 0 (1 −x) xs−1 dx = 1 s(s + 1), 44 which is easy to verify directly. For the general claim, using a linear change of variables and an integration by parts we see that Z n 0 1 −x n n xs−1 dx = ns Z 1 0 (1 −t)nts−1 dt = ns (1 −t)n ts s t=1 t=0 + Z 1 0 n(1 −t)n−1 ts s dt = ns · n s Z 1 0 (1 −t)n−1t(s+1)−1 dt, so the claim follows by induction on n.
126. Corollary. For Re(s) > 0 we have Γ(s) = lim n→∞ n! ns s(s + 1) · · · (s + n).
127. Proof of the infinite product representation for Γ(s). For Re(s) > 0 we have Γ(s)−1 = lim n→∞ s(s + 1) · · · (s + n) n! ns = s lim n→∞e−s log n 1 + s 1 1 + s 2 · · · 1 + s n = s lim n→∞es(Pn k=1 1 k −log n) n Y k=1 1 + s k e−s/k = seγs ∞ Y n=1 1 + s n e−s/n.
128. We now check that the infinite product actually converges absolutely and uni-formly on compact subsets in all of C, so defines an entire function. Let’s start with some preliminary elementary observations on infinite products.
Lemma. For a sequence of complex numbers (an)∞ n=1, we have Q∞ n=1(1+|an|) ∈ (0, ∞) if and only if none of the factors an is equal to −1 and P∞ n=1 |an| < ∞.
Proof. Assume all an’s are not equal to −1, and P n |an| < ∞. In particular all an’s for large enough n satisfy |an| < 1/2, so we can assume without loss of generality that this holds for all n. We therefore have n Y k=1 (1 + ak) = n Y k=1 exp (Log(1 + ak)) = exp n X k=1 Log(1 + ak) !
, where Log(z) denotes the principal branch of the logarithm function, which has the Taylor expansion around z = 1 Log(z) = ∞ X m=1 (−1)m−1 m (z −1)m (|z| < 1).
45 In particular, for z near 0 we have Log(1 + z) = z + O(z2), i.e., for example |z|−10|z|2 ≤| Log(1+z)| ≤|z|+10|z|2 in some sufficiently small neighborhood of z = 0. It follows that if P n |an| < ∞then also P n | Log(1 + an)| < ∞, so the product Qn k=1(1 + ak) converges. Conversely, it is easily seen from the same inequality that if P n | Log(1 + an)| < ∞and an →0 as n →∞ (which would be a consequence of Q n(1+an) converging to a number in (0, ∞)) then P n |an| < ∞ Lemma.
Let (fn)∞ n=1 be a sequence of functions that are holomorphic and nonzero on some region Ω. Then Q∞ n=1(1 + fn) converges absolutely uniformly on compact subsets in Ωto a nonzero holomorphic function if and only if the series P∞ n=1 fn also converges absolutely uniformly on compact subsets in Ω.
Proof. Use the same estimates in the previous proof together with the uni-formity of the convergence on compacts to ensure that the inequalities hold uniformly so the limiting function is holomorphic.
129. Proof that Q∞ n=1 1 + z n e−z/n is an entire function.
∞ X n=1 1 + z n e−z/n −1 = ∞ X n=1 1 + z n 1 −z n + O z2 n2 −1 = ∞ X n=1 O z2 n2 < ∞, The convergence is uniform on compacts on C, but to apply the previous result (which requires the functions to be nonzero) one needs to be a bit more careful and separate out the zeros: for a fixed disc DN+1/2(0) of radius N +1/2 around 0, consider only the product starting at n = N +1 — those functions are nonzero in the disc so the previous result applies to give a function that’s holomorphic and nonzero in DN(0). Then separately the factors (1 + z/n), n = 1, . . . , N contribute simple zeros at z = −1, . . . , −N.
130. Corollary (the reflection formula). Γ(s)Γ(1 −s) = π sin πs.
Proof.
1 Γ(s)Γ(1 −s) = Γ(s)−1(−s)−1Γ(−s)−1 = −1 s · seγs ∞ Y n=1 1 + s n e−s/n · (−s)e−γs ∞ Y n=1 1 −s n es/n = s ∞ Y n=1 1 −s2 n2 = ssin(πs) πs = sin(πs) π , where we used the product representation sin(πz) = πz Q∞ n=1(1−z2/n2) for the sine function derived in a homework problem.
46 131. Alternative derivation of the reflection formula ([Stein-Shakarchi], page 164). By analytic continuation, it is enough to prove the formula for real s in (0, 1). For such s we have Γ(s)Γ(1 −s) = Z ∞ 0 e−tt−sΓ(s) dt = Z ∞ 0 e−tt−s t Z ∞ 0 e−vt(vt)s−1 dv dt = Z ∞ 0 Z ∞ 0 e−t(1+v)vs−1 dv dt = Z ∞ 0 Z ∞ 0 e−t(1+v) dt vs−1 dv = Z ∞ 0 vs−1 1 + v dv = Z ∞ −∞ esx 1 + ex dx (by setting v = ex).
So it is enough to prove that for 0 < s < 1 we have Z ∞ −∞ esx 1 + ex dx = π sin(πs).
This integral can be evaluated using residue calculus; see Example 2 in Section 2.1, Chapter 3, pages 79–81 of [Stein-Shakarchi] for the details.
132. Note that by combining the alternative derivation of the reflection formula given above with the infinite product representation for the gamma function, we get a new proof of the infinite product representation for sin(πz).
15 The Riemann zeta function 133. The Riemann zeta function (often referred to simply as the zeta function when there is no risk of confusion), like the gamma function is considered one of the most important special functions in “higher” mathematics. However, the Riemann zeta function is a lot more mysterious than the gamma function, and remains the subject of many famous open problems, including the most famous of them all: the Riemann hypothesis, considered by many (including myself) as the most important open problem in mathematics.
134. The main reason for the zeta function’s importance is its connection with prime numbers and other concepts and quantities from number theory.
Its study, and in particular the attempts to prove the Riemann hypothesis, have also stimulated an unusually large number of important developments in many areas of mathematics.
135. As with the gamma function, the Riemann zeta function is usually defined on only part of the complex plane and its definition is then extended by analytic continuation. Again, I will formulate this as a theorem asserting the existence of the zeta function and its various properties.
136. Theorem. There exists a unique function, denoted ζ(s), of a complex variable s, having the following properties: 47 (a) ζ(s) is a meromorphic function on C.
(b) For Re(s) > 1, ζ(s) is given by the series ζ(s) = ∞ X n=1 1 ns = 1 + 1 2s + 1 3s + . . . .
(c) Euler product formula: for Re(s) > 1, ζ(s) also has an infinite product representation ζ(s) = Y p 1 1 −p−s , where the product ranges over the prime numbers p = 2, 3, 5, 7, 11, . . ..
(d) ζ(s) has no zeros in the region Re(s) > 1.
(e) ζ(s) has no zeros on the line Re(s) = 1 (this requires a separate proof from the previous claim).
(f) The “trivial” zeros: the zeros of ζ(s) in the region Re(s) ≤0 are pre-cisely at s = −2, −4, −6, . . ..
(g) ζ(s) has a unique pole located at s = 1. It is a simple pole with residue 1.
(h) The “Basel problem” and its generalizations: the values of ζ(s) at even positive integers are given by Euler’s formula ζ(2n) = (−1)n−1(2π)2n 2(2n)!
B2n (n = 1, 2, . . .), where (Bm)∞ m=0 are the Bernoulli numbers, defined as the coefficients in the Taylor expansion z ez −1 = ∞ X m=0 Bm m! zm.
Many of the properties of these amazing numbers were discussed in our homework problem sets.
(i) Values at negative odd integers: we have ζ(−n) = −Bn+1 n + 1.
(j) Functional equation: the zeta function satisfies ζ∗(1 −s) = ζ∗(s), where we denote by ζ∗(s) the symmetrized zeta function ζ∗(s) = π−s/2Γ s 2 ζ(s).
(k) Mellin transform representation: an expression for ζ(s) valid for all s ∈C is π−s/2Γ s 2 ζ(s) = − 1 1 −s −1 s + 1 2 Z ∞ 1 t−s+1 2 + t s−2 2 (ϑ(t) −1) dt, 48 where the function ϑ(t) is one of Jacobi theta series, defined as ϑ(t) = ∞ X n=−∞ e−πn2t = 1 + 2 ∞ X n=1 e−πn2t.
(l) Contour integral representation: another expression for ζ(s) valid for all s ∈C is ζ(s) = Γ(1 −s) 2πi Z C (−x)s ex −1 dx x , where C is a keyhole contour coming from +∞to 0 slightly above the positive x-axis, then circling the origin in a counterclockwise direction around a circle of small radius, then going back to +∞slightly below the positive x-axis.
(m) Connection to prime number enumeration — the “explicit for-mula of number theory”: define Von Mangoldt’s weighted prime count-ing function ψ(x) = X pk≤x log p, where the sum is over all prime powers less than or equal to x. Then for non-integer x > 1, ψ(x) = x − X ρ xρ ρ −log(2π), where the sum ranges over all zeros ρ of the Riemann zeta function. (In most textbooks the sum is separated into two sums, one ranging over the trivial zeros which can be evaluated explicitly, and the other ranging over the much less trivial zeros in the strip 0 < Re(s) < 1.) 137. The explicit formula of number theory illustrates that knowing where the zeros of ζ(s) has important consequences for prime number enumeration. In particular, proving that Re(s) has no zeros in Re(s) ≥1 will enable us to prove one of the most famous theorems in mathematics.
The prime number theorem. Let π(x) denote the number of prime numbers less than or equal to x. Then we have lim x→∞ π(x) x/ log x = 1.
138. The Riemann hypothesis. All the nontrivial zeros of ζ(s) are on the “critical strip” Re(s) = 1/2.
139. Proofs.
To begin the proof, again, let’s take as the definition of ζ(s) the standard representation ζ(s) = ∞ X n=1 1 ns .
49 Since P n |n−s| = P n n−Re(s), we see that the series converges absolutely pre-cisely when Re(s) > 1, and that the convergence is uniform on any half-plane of the form Re(s) > α where α > 1. In particular, it is uniform on compact subsets, so ζ(s) is holomorphic in this region.
140. Similarly, the Euler product Z(s) := Q p(1 −p−s)−1 converges absolutely if and only if the series P p |p−s| = P p p−Re(s) converges, and in particular if Re(s) > 1. It follows that Z(s) is well-defined, holomorphic and nonzero for Re(s) > 1.
141. We now prove that Z(s) = ζ(s). This can be done by manipulating the partial products associated with the infinite product defining Z(s), as follows: ζN(s) := Y p≤N 1 1 −p−s = Y p≤N (1 + p−s + p−2s + p−3s + . . .) = X n=pj1 1 ···pjk k p1,...,pk primes ≤N 1 ns , where the last equality follows from the fundamental theorem of arithmetic, together with the fact that when multiplying two (or a finite number of) infinitely convergent series, the summands can be rearranged and summed in any order we desire. So, we have represented ζN(s) as a series of a similar form as the series defining ζ(s), but involving terms of the form n−s only for those positive integers n whose prime factorization contains only primes ≤N. It follows that |ζ(s) −ζN(s)| ≤ X n>N 1 ns .
Taking the limit as N →∞shows that Z(s) = limN→∞ζN(s) = ζ(s). This proves the validity of the Euler product formula.
142. Corollary: ζ(s) has no zeros in the region Re(s) > 1.
Proof. The Euler product formula gives a convergent product for ζ(s) in this region where each factor (1 −p−s)−1 has no zeros.
143. Theorem (the Poisson summation formula). For a sufficiently well-behaved function f : R →R, we have ∞ X n=−∞ f(n) = ∞ X k=−∞ ˆ f(k), where ˆ f(k) = Z ∞ −∞ f(x)e−2πikx dx.
is the Fourier transform of f.
Proof. Define a function g : [0, 1] →R by g(x) = ∞ X n=−∞ f(x + n), 50 the “periodiciziation” of f. Assume that f(x) is sufficiently well-behaved (i.e., decays fast enough as x →±∞so that g(x) is in turn well-behaved, and has reasonable smoothness properties). In that case, g(x) will have a convergent Fourier expansion of the form g(x) = ∞ X k=−∞ ˆ g(k)e2πikx, where the Fourier coefficients ˆ g(k) can be computed as ˆ g(k) = Z 1 0 g(x)e−2πikx dx.
In particular, setting x = 0 in the formula for g(x) gives the standard result that g(0) = ∞ X k=−∞ ˆ g(k).
However, note that g(0) = P∞ n=−∞f(n), the quantity on the left-hand side of the Poisson summation formula. On the other hand, the Fourier coefficient ˆ g(k) can be expressed in terms of the Fourier coefficients of the original function f(x): ˆ g(k) = Z 1 0 g(x)e−2πikx dx = Z 1 0 ∞ X n=−∞ f(x + n)e−2πikx dx = ∞ X n=−∞ Z 1 0 f(x + n)e−2πikx dx = ∞ X n=−∞ Z n+1 n f(u)e−2πiku du = Z ∞ −∞ f(u)e−2πiku du = ˆ f(k).
Combining these observations gives the result, modulo a few details we’ve glossed over concerning the precise assumptions that need to be made about f(x) (we will only apply the Poisson summation formula for one extremely well-behaved function, so I will not bother discussing those details).
144. Theorem. The Jacobi theta function ϑ(t) satisfies the functional equation ϑ(t) = 1 √ tϑ(1/t) (t > 0).
(Note: equations of this form are studied in the theory of modular forms, an area of mathematics combining number theory, complex analysis and algebra in a very surprising and beautiful way.) Proof. The idea is to apply the Poisson summation formula to the function f(x) = e−πtx2, for which it can be checked that ˆ f(k) = t−1/2e−πk2/t, 51 using a simple change of variables from the standard integral evaluation Z ∞ −∞ e−πx2e−2πixu du = e−πu2 (that is, the fact that the function e−πx2 is its own Fourier transform); this evaluation appears in Example 1, Chapter 2, pages 42–44 in [Stein-Shakarchi].
With the above substitution for f(x) and ˆ f(k), the Poisson summation formula becomes precisely the functional equation for ϑ(t).
145. Exercise. (a) Use the residue theorem to evaluate the contour integral I γN e−πz2t e2πiz −1 dz, where γN is the rectangle with vertices ±(N+1/2)±i (with N a positive integer), then take the limit as N →∞to derive the integral representation ϑ(t) = Z ∞−i −∞−i e−πz2t e2πiz −1 dz − Z ∞+i −∞+i e−πz2t e2πiz −1 dz for the Jacobi theta function.
(b) In this representation, expand the factor (e2πiz −1)−1 as a geometric series in e−2πiz (for the first integral) and as a geometric series in e2πiz (for the second integral). Evaluate the resulting infinite series, rigorously justifying all steps, to obtain an alternative proof of the functional equation for ϑ(t).
146. Lemma. The asymptotic behavior of ϑ(t) near t = 0 and t = +∞is given by ϑ(t) = O 1 √ t (t →0+), ϑ(t) = 1 + O(e−πt) (t →∞).
Proof. The asymptotics as t →∞is immediate from ϑ(t) −1 = 2 ∞ X n=1 e−πn2t ≤2 ∞ X n=1 e−πnt = 2e−πt 1 −e−πt , which is bounded by Ce−πt if t > 10. Using the functional equation now gives that ϑ(t) = t−1/2(1 + O(e−π/t)) = O(t−1/2) as t →0+.
147. Proof of the analytic continuation of ζ(s). Start with the formula Γ s 2 = Z ∞ 0 e−xxs/2−1 dx, valid for Re(s) > 0. A linear change of variable x = πn2t brings this to the form π−s/2Γ s 2 n−s = Z ∞ 0 e−πn2tts/2−1 dt.
52 Summing the left-hand side over n = 1, 2, . . . gives π−s/2Γ s 2 ζ(s) — the func-tion we denoted ζ∗(s) — adding the stronger assumption that Re(s) > 1. For the right-hand side we have that ∞ X n=1 Z ∞ 0 e−πn2tts/2−1 dt. = Z ∞ 0 ∞ X n=1 e−πn2t !
ts/2−1 dt = Z ∞ 0 ϑ(t) −1 2 ts/2−1 dt, where the estimates in the lemma are needed to justify interchanging the order of the summation and integration, and show that the integral converges for Re(s) > 1. Thus we have obtained the representation ζ∗(s) = 1 2 Z ∞ 0 (ϑ(t) −1)ts/2−1 dt = Z ∞ 0 ϕ(t)ts/2−1 dt, where we denote ϕ(t) = 1 2(ϑ(t) −1). Next, the idea is to use the functional equation for ϑ(t) to bring this to a new form that can be seen to be well-defined for all s ∈C except s = 1. Specifically, we note that the functional equation for can be expressed in the form ϕ(t) = t−1/2ϕ(1/t) + 1 2t−1/2 −1 2.
We can therefore write ζ∗(s) = Z 1 0 ϕ(t)ts/2−1 dt + Z ∞ 1 ϕ(t)ts/2−1 dt = Z 1 0 t−1/2ϕ(1/t) + 1 2t−1/2 −1 2 ts/2−1 dt + Z ∞ 1 ϕ(t)ts/2−1 dt = − 1 1 −s −1 s + Z ∞ 1 t−s/2−1/2 + ts/2−1 ϕ(t) dt.
We have derived a formula for ζ∗(s) (one of the formulas claimed in the main theorem above) that is now seen to define a meromorphic function on all of C — the integrand decays rapidly as t →∞so actually defines an entire function, so the only poles are due to the two terms −1/s and 1/(s −1). We have therefore proved that ζ(s) can be analytically continued to a meromorphic function on C.
148. Corollary. The zeta function satisfies the functional equation ζ∗(1 −s) = ζ∗(s).
Equivalently, because of the reflection formula satisfied by the gamma function, it is easy to check that the functional equation can be rewritten in the form ζ(s) = 2sπs−1 sin πs 2 Γ(1 −s)ζ(1 −s).
Proof. The representation we derived for ζ∗(s) is manifestly symmetric with respect to replacing each occurrence of s by 1 −s.
53 149. Corollary. The only pole of ζ(s) is a simple pole at s = 1 with residue 1.
Proof. Our representation for ζ∗(s) expresses it as a sum of −1 s, 1 s−1, and an entire function. Thus the poles of ζ∗(s) are simple poles at s = 0, 1 with residues −1 and 1, respectively. It follows that ζ(s) = πs/2Γ(s/2)−1ζ∗(s) has a pole at s = 1 with residue π1/2Γ(1/2)−1 = 1, and a pole (that turns out to be a removable singularity) at s = 0 with residue π0Γ(0)−1 = 0. (That is, the pole of ζ∗(s) at s = 0 is cancelled out by the zero of Γ(s/2).) 150. Corollary. ζ(−n) = −Bn+1/(n + 1) for n = 1, 2, 3, . . ..
Proof. Using the functional equation, we have that ζ(−n) = 2−nπ−n−1 sin(−πn/2)Γ(n + 1)ζ(n + 1) = 2−nπ−n−1 sin(−πn/2)n!ζ(n + 1).
If n = 2k is even, then sin(−πn/2) = sin(−πk) = 0, so we get that ζ(−2k) = 0 (that is, n = 2k is one of the so-called “trivial zeros”).
We also know that B2k+1 = 0 for k = 1, 2, 3, . . ., so the formula ζ(−n) = Bn+1/(n + 1) is satisfied in this case.
If on the other hand n = 2k −1 is odd, then sin(−π(2k −1)/2) = (−1)k, and therefore we get, using the formula expressing ζ(2k) in terms of the Bernoulli numbers (derived in the homework and in the textbook), that ζ(−n) = (−1)k2−2k+1π−2k(2k −1)!ζ(2k) = (−1)k2−2k+1π−2k(2k −1)!(−1)k−1(2π)2k 2(2k)!
B2k = −B2k 2k = −Bn+1 n + 1, so again the formula is satisfied.
151. Corollary. The zeros of ζ(s) in the region Re(s) < 0 are precisely the trivial zeros s = −2, −4, −6, . . ..
152. Proof. We already established the existence of the trivial zeros. The fact that there are no other zeros also follows easily from the functional equation and is left as an exercise.
153. An alternative approach to the analytic continuation of ζ(s). There is a more “down-to-earth” approach to the analytic continuation of ζ(s) based on the standard idea from numerical analysis of approximating an integral by a sum (or in this case going in the other direction, approximating a sum by an integral). The technical name for this procedure, when it is done in a more 54 systematic way, is Euler-Maclaurin summation.
ζ(s) = ∞ X n=1 1 ns = ∞ X n=1 Z n+1 n dx xs + 1 ns − Z n+1 n dx xs = Z ∞ 1 dx xs + ∞ X n=1 Z n+1 n 1 ns −1 xs dx = 1 s −1 − Z ∞ 1 x−s −⌊x⌋−s dx.
This representation is certainly valid for Re(s) > 1. However, note that we have the bound x−s −⌊x⌋−s ≤|s| · ⌊x⌋−Re(s)−1 (x ≥1) by the mean value theorem. Thus, the integral is actually an absolutely conver-gent integral in the larger region Re(s) > 0, and the representation we derived gives an analytic continuation of ζ(s) to a meromorphic function on Re(s) > 0, which has a single pole at s = 1 (a simple pole with residue 1) and is holomorphic everywhere else.
154. An elaboration of this idea using what is known as the Euler-Maclaurin summa-tion formula can be used to perform the analytic continuation of ζ(s) to a mero-morphic function on C by extending it inductively from each region Re(s) > −n to Re(s) > −n −1, as we saw could be done for the gamma function.
An-other approach is to use the analytic continuation for Re(s) > 0 shown above, then prove that the functional equation ζ(1 −s) = ζ∗(s) holds in the region 0 < Re(s) < 1, and then use the functional equation to analytically continue ζ(s) to Re(s) ≤0 (which is the reflection of the region Re(s) ≥1 under the transformation s 7→1 −s).
155. Next, we prove a nontrivial and very important fact about the zeta function that will play a critical role in our proof of the prime number theorem.
Theorem. ζ(s) has no zeros on the line Re(s) = 1.
This theorem can also be thought of as a “toy” version of the Riemann hypoth-esis. If you ever want to try solving this famous open problem, getting a good understanding of its toy version seems like a good idea...
Proof. For this proof, denote s = σ + it, where we assume σ > 1 and t is real and nonzero. The proof is based on investigating simultaneously the behavior of ζ(σ + it), ζ(σ + 2it), and ζ(σ), for fixed t as σ ↘1. Consider the following somewhat mysterious quantity X = log |ζ(σ)3ζ(σ + it)4ζ(σ + 2it)|.
55 We can evaluate “X” as log |ζ(σ)3ζ(σ + it)4ζ(σ + 2it)| = 3 log |ζ(σ)| + 4 log |ζ(σ + it)| + log |ζ(σ + 2it)| = 3 log Y p |1 −p−σ|−1 !
+ 4 log Y p |1 −p−σ−it|−1 !
+ log Y p |1 −p−σ−2it|−1 !
= X p −3 log |1 −p−σ| −4 log |1 −p−σ−it| −log |1 −p−σ−2it| = X p −3 Re Log(1 −p−σ) −4 Re Log(1 −p−σ−it) −Re Log 1 −p−σ−2it , where Log(·) denotes the principal branch of the logarithm function. Now note that for z = a + ib with a > 1 and p prime we have |p−z| = p−a < 1, so −Log(1 −p−z) = ∞ X m=1 p−mz m , and −Re h Log(1 −p−z) i = ∞ X m=1 p−ma m Re h cos(mb log p) + i sin(mb log p) i = ∞ X m=1 p−ma m cos(mb log p).
So we can rewrite X as X = ∞ X n=1 cnn−σ(3 + 4 cos θn + cos(2θn)) where θn = t log n and cn = 1/m if n = pm for some prime p. We can now use a simple trigonometric identity 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2, to rewrite X yet again as X = 2 ∞ X n=1 cnn−σ(1 + cos θn)2.
We have proved a crucial fact, namely that X ≥0, or equivalently that eX = |ζ(σ)3ζ(σ + it)4ζ(σ + 2it)| ≥1.
We now claim that this innocent-looking inequality is incompatible with the existence of a zero of ζ(s) on the line Re(s) = 1. Indeed, assume by contradiction 56 that ζ(1 + it) = 0 for some real t ̸= 0. Then the three quantities ζ(σ), ζ(σ + it) and ζ(σ + 2it) have the following asymptotic behavior as σ ↘1: |ζ(σ)| = 1 σ −1 + O(1) (since ζ(s) has a pole at s = 1), |ζ(σ + it)| = O(σ −1) (since ζ(s) has a zero at s = 1 + it), |ζ(σ + 2it)| = O(1) (since ζ(s) is holomorphic at s = 1 + 2it).
Combining these results we have that eX = |ζ(σ)3ζ(σ + it)4ζ(σ + 2it)| = O((σ −1)−3(σ −1)4) = O(σ −1).
In particular, eX →0 as σ ↘1, in contradiction to the result we proved above that eX ≥1. This proves the claim that ζ(s) cannot have a zero on the line Re(s) = 1.
156. Exercise. The above proof that eX ≥1 (which immediately implied the claim of the theorem) relied on showing that for any prime number p, the corresponding factors in the Euler product formula satisfy the inequality (1 −p−σ)−3|1 −p−σp−it|−4|1 −p−σp−2it|−1 ≥1, and this was proved by taking the logarithm of the left hand-side, expanding in a power series and using the elementary trigonometric identity 3+4 cos θ+cos 2θ = 2(1 + cos θ)2. However, one can imagine a more direct approach that starts as follows: denote x = p−σ and z = p−it = e−it log p. Then the inequality reduces to the claim that (1 −x)3|1 −zx|4|1 −z2x| ≤1 for all x ∈[0, 1] and z satisfying |z| = 1. Since this is an elementary inequality, it seems like it ought to have an elementary proof (i.e., a proof that does not involve logarithms and power series expansions). Can you find such a proof?
16 The prime number theorem 157. The prime number theorem was proved in 1896 by Jacques Hadamard and in-dependently by Charles Jean de la Vall´ ee Poussin, using the groundbreaking ideas from Riemann’s famous 1859 paper in which he introduced the use of the Riemann zeta function as a tool for counting prime numbers. (This was the only number theory paper Riemann wrote in his career!) The history (including all the technical details) of these developments is described extremely well in the classic textbook Riemann’s Zeta Function by H. M. Edwards, which I highly recommend.
The original proofs of the prime number theorem were very complicated and re-lied on the “explicit formula of number theory” (that I mentioned in the previous section) and some of its variants. Throughout the 20th century, mathematicians worked hard to find simpler ways to derive the prime number theorem. This 57 resulted in several important developments (such as the Wiener tauberian the-orem and the Hardy-Littlewood tauberian theorem) that advanced not just the state of analytic number theory but also complex analysis, harmonic analysis and functional analysis.
Despite all the efforts and the discovery of several paths to a proof that were simpler than the original approach, all proofs re-mained quite difficult. . . until 1980, when the mathematician Donald Newman discovered a wonderfully simple way to derive the theorem using a completely elementary use of complex analysis. It is Newman’s proof (as presented in the note Newman’s short proof of the prime number theorem by Don Zagier, Amer.
Math. Monthly 104 (1997), 705–708) that I present here.
158. Define the weighted prime counting functions π(x) = #{p prime : p ≤x} = X p≤x 1, ψ(x) = X pk≤x log p = X p≤x log p log x log p , with the convention that the symbol p in a summation denotes a prime number, and pk denotes a prime power, so that summation over p ≤x denotes summation over all primes ≤x, and the summation over pk denotes summation over all prime powers ≤x. Another customary way to write the function ψ(x) is as ψ(x) = X n≤x Λ(n), where the function Λ(n), called the von Mangoldt function, is defined by Λ(n) = log p if n = pk, p prime, 0 otherwise.
159. Lemma. The prime number theorem π(x) ∼ x log x is equivalent to the statement that ψ(x) ∼x.
Proof. Note the inequality ψ(x) = X p≤x log p log x log p ≤ X p≤x log plog x log p = X p≤x log x = log x · π(x).
In the opposite direction, we have a similar (but slightly less elegant) inequality, namely that for any 0 < ϵ < 1 and x ≥2, ψ(x) ≥ X p≤x log p ≥ X x1−ϵ 1 we have −ζ′(s) ζ(s) = ∞ X n=1 Λ(n)n−s.
Proof. Using the Euler product formula and taking the logarithmic derivative (which is an operation that works as it should when applied to infinite products of holomorphic functions that are uniformly convergent on compact subsets), we have −ζ′(s) ζ(s) = X p d ds(1 −p−s) 1 −p−s = X p log p · p−s 1 −p−s = X p log p (p−s + p−2s + p−3s + . . .) = X p prime ∞ X k=1 log p · p−ks = ∞ X n=1 Λ(n)n−s.
59 161. Lemma. There is a constant C > 0 such that ψ(x) < Cx for all x ≥1.
Proof. The idea of the proof is that the binomial coefficient 2n n is not too large on the one hand, but is divisible by many primes (all primes between n and 2n) on the other hand — hence it follows that there cannot be too many primes, and in particular the weighted prime-counting function ψ(x) can be easily bounded from above using such an argument. Specifically, we have that 22n = (1 + 1)2n = 2n X k=0 2n k !
2n n !
≥ Y n<p≤2n p = exp X n1 log p .
≥exp ψ(2n) −ψ(n) −O(√n log2 n) .
(The estimate O(√n log2 n) for the sum of log p for prime powers higher than 1 is easy and is left as an exercise.) Taking the logarithm of both sides, this gives the bound ψ(2n) −ψ(n) ≤2n log 2 + C1 √n log n ≤C2n, valid for all n ≥1 with some constant C2 > 0. It follows that ψ(2m) = (ψ(2m) −ψ(2m−1)) + (ψ(2m−1) −ψ(2m−2)) + . . . + (ψ(21) −ψ(20)) ≤C2(2m−1 + . . . + 20) ≤C22m, so the inequality ψ(x) ≤C2x is satisfied for x = 2m. It is now easy to see that this implies the result also for general x, since for x = 2m + ℓwith 0 ≤ℓ< 2m we have ψ(x) = ψ(2m + ℓ) ≤ψ(2m+1) ≤C22m+1 ≤2C22m ≤2C2x.
162. Newman’s tauberian theorem. Let f : [0, ∞) →R be a bounded function that is integrable on compact intervals. Define a function g(z) of a complex variable z by g(z) = Z ∞ 0 f(t)e−zt dt (g is known as the Laplace transform of f).
Clearly g(z) is defined and holomorphic in the open half-plane Re(z) > 0. Assume that g(z) has an analytic continuation to an open region Ωcontaining the closed half-plane Re(z) ≥0.
Then R ∞ 0 f(t) dt exists and is equal to g(0) (the value at z = 0 of the analytic continuation of g).
Proof. Define a truncated version of the integral defining g(z), namely gT (z) = Z T 0 f(t)e−zt dt 60 C R δ C+ C-R δ C+ C-' R (a) (b) (c) Figure 6: The contours C, C+, C−and C′ −.
for T > 0, which for any T is an entire function of z.
Our goal is to show that limT →∞gT (0) = g(0). This can be achieved using a clever application of Cauchy’s integral formula. Fix some large R > 0 and a small δ > 0 (which depends on R in a way that will be explained shortly), and consider the contour C consisting of the part of the circle |z| = R that lies in the half-plane Re(z) ≥ −δ, together with the straight line segment along the line Re(z) = −δ connecting the top and bottom intersection points of this circle with the line (see Fig. 6(a)).
Assume that δ is small enough so that g(z) (which extends analytically at least slightly to the right of Re(z) = 0) is holomorphic in an open set containing C and the region enclosed by it. Then by Cauchy’s integral formula we have g(0) −gT (0) = 1 2πi Z C (g(z) −gT (z))eT z 1 + z2 R2 dz z = 1 2πi Z C+ + Z C− !
(g(z) −gT (z))eT z 1 + z2 R2 dz z , where we separate the contour into two parts, a semicircular arc C+ that lies in the half-plane Re(z) > 0, and the remaining part C−in the half-plane Re(z) < 0 (Fig. 6(b)). We now bound the integral separately on C+ and on C−. First, for z lying on C+ we have |g(z) −gT (z)| = Z ∞ T f(t)e−zt dt ≤B Z ∞ T |e−zt| dt = Be−Re(z)T Re(z) , where B = supt≥0 |f(t)|, and eT z 1 + z2 R2 = eRe(z)T 2 Re(z) R (by the trivial identity |1 + eit|2 = |eit(eit + eit)|2 = 2 cos(t), valid for t ∈R).
61 So in combination we have 1 2πi Z C+ (g(z) −gT (z))eT z 1 + z2 R2 dz z ≤(πR) 2B 2πR2 = B R .
Next, for C−, we bound the integral by bounding the contributions from g(z) and gT (z) separately. In the case of gT (z), the function is entire, so we can deform the contour, replacing it with the semicircular arc C′ −= {|z| = R, Re(z) < 0} (Fig. 6(c)). On this contour we have the estimate |gT (z)| = Z T 0 f(t)e−zt dt ≤B Z T −∞ |e−zt| dt = Be−Re(z)T | Re(z)| , which leads using a similar calculation as before to the estimate 1 2πi Z C′ − gT (z)eT z 1 + z2 R2 |dz| |z| ≤B R .
The remaining integral 1 2πi Z C− g(z)eT z 1 + z2 R2 |dz| |z| tends to 0 as T →∞, since the dependence on T is only through the factor eT z, which converges to 0 uniformly on compact sets in Re(z) < 0 as T →∞.
Combining the above estimates, we have shown that lim sup T →∞ |g(0) −gT (0)| ≤2B R .
Since R was an arbitrary positive number, the lim sup must be 0, and the theorem is proved.
163. An application of Newman’s theorem. Take f(t) = ψ(et)e−t −1 (t ≥0), which is bounded by the lemma we proved above, as our function f(t). The associated function g(z) is then g(z) = Z ∞ 0 (ψ(et)e−t −1)e−zt dt = Z ∞ 1 ψ(x) x −1 x−z−1 dx = Z ∞ 1 ψ(x)x−z−2 dx −1 z = Z ∞ 1 X n≤x Λ(n) x−z−2 dx −1 z = ∞ X n=1 Λ(n) Z ∞ n x−z−2 dx −1 z = ∞ X n=1 Λ(n) x−z−1 −z −1 ∞ n −1 z = 1 z + 1 ∞ X n=1 Λ(n)n−z−1 −1 z = − 1 z + 1 · ζ′(z + 1) ζ(z + 1) −1 z (Re(z) > 0).
62 Recall that −ζ′(s)/ζ(s) has a simple pole at s = 1 with residue 1 (because ζ(s) has a simple pole at s = 1; it is useful to remember the more general fact that if a holomorphic function h(z) has a zero of order k at z = z0 then the logarithmic derivative h′(z)/h(z) has a simple pole at z = z0 with residue k). So − 1 z+1 · ζ′(z+1) ζ(z+1) has a simple pole with residue 1 at z = 0, and therefore − 1 z+1 · ζ′(z+1) ζ(z+1) −1 z has a removable singularity at z = 0. Thus, the identity g(z) = − 1 z+1 · ζ′(z+1) ζ(z+1) −1 z shows that g(z) extends analytically to a holomorphic function in the set {z ∈C : ζ(z + 1) ̸= 0}.
By the “toy Riemann Hypothesis” — the theorem we proved according to which ζ(s) has no zeros on the line Re(s) = 1, g(z) in particular extends holomorphi-cally to an open set containing the half-plane Re(z) ≥0. Thus, f(t) satisfies the assumption of Newman’s theorem. We conclude from the theorem that the integral Z ∞ 0 f(t) dt = Z ∞ 0 (ψ(et)e−t −1)dt = Z ∞ 1 ψ(x) x −1 dx x = Z ∞ 1 ψ(x) −x x2 dx converges.
164. Proof of the prime number theorem. We will prove that ψ(x) ∼x, which we already showed is equivalent to the prime number theorem.
Assume by contradiction that lim supx→∞ ψ(x) x > 1 or lim infx→∞ ψ(x) x < 1. In the first case, that means there exists a number λ > 1 such that ψ(x) ≥λx for arbitrarily large x. For such values of x it then follows that Z λx x ψ(t) −t t2 dt ≥ Z λx x λx −t t2 dt = Z λ 1 λ −t t2 dt =: A > 0, but this is inconsistent with the fact that the integral R ∞ 1 (ψ(x) −x)x−2 dx converges.
Similarly, in the event that lim infx→∞ ψ(x) x < 1, that means that there exists a µ < 1 such that ψ(x) ≤µx for arbitrarily large x, in which case we have that Z x λx ψ(t) −t t2 dt ≤ Z x λx λx −t t2 dt = Z 1 λ λ −t t2 dt =: B < 0, again giving a contradiction to the convergence of the integral.
63 17 Introduction to asymptotic analysis 165. In this section we’ll learn how to use complex analysis to prove asymptotic formulas such as n! ∼ √ 2πn n e n (Stirling’s formula), p(n) ∼ 1 4 √ 3n eπ√ 2n/3 (the Hardy-Ramanujan formula), Ai(x) ∼ 1 2√π x−1/4 exp −2 3x3/2 (asymptotics for the Airy function), and more. At the heart of many such results is an important technique known as the saddle point method.
Some related techniques (that are all minor variations on the same theme) are Laplace’s method, the steepest descent method and the stationary phase method.
166. A toy estimate for n!. For x > 0 real, we have xn n! ≤ ∞ X n=0 xn n! = ex, which gives a lower bound n! ≥e−xxn.
for n!. It makes sense to try to get the best lower bound possible by looking for the x where the lower-bounding function is maximal. This happens when 0 = d dx e−xxn = e−x −xn + nxn−1 = e−xxn−1(−x + n), i.e., when x = n. Plugging this value into the inequality gives the bound n! ≥(n/e)n (n ≥1).
This is of course a standard and very easy result. The point of this computation is that, as we shall see below, there is something special about the value x = n that resulted from this maximization operation; when interpreted in the context of complex analysis, it corresponds to a so-called “saddle point,” since it is a local minimum of ex/xn as one moves along the real axis, but it will be a local maximum when one moves in the orthogonal direction parallel to the imaginary axis.
167. First example: Stirling’s formula. Start with the power series expansion ez = ∞ X n=0 zn n! .
As we know very well from our study of Cauchy’s integral formula and the residue theorem, the nth Taylor coefficient can be extracted from the function by contour integration, namely by writing 1 n! = 1 2πi I |z|=r ez zn+1 dz, 64 where the radius r of the circle chosen as the contour of integration is an arbitrary positive number. It turns out that some values of r are better than others when one is trying to do asymptotics. We select r = n (I’ll explain later where that seemingly inspired choice comes from), to get 1 n! = 1 2πi I |z|=n ez zn+1 dz = 1 2πi Z π −π exp neit n−ne−inti dt = 1 2π nn Z π −π exp n(eit −it) dt = en 2π nn Z π −π exp n(eit −1 −it) dt, where we have strategically massaged the integrand (by pulling out the factor en) to cancel out a term in the Taylor expansion of eit, in addition to a term that was already canceled out. For convenience, rewrite this as nn enn! = 1 2π Z π −π exp n(eit −1 −it) dt.
Now noting that n eit −1 −it = −nt2 2 + O(nt3) = (√nt)2 2 + O (√nt)3 √n , for |t| small, we see that a change of variable u = √nt in the integral will enable us to rewrite this as n eiu/√n −1 −iu √n = −u2 2 + O u3 √n .
Performing the change of variable and moving a factor of √n to the left-hand side, the integral then becomes √nnn enn! = 1 2π Z π√n −π√n exp n eiu/√n −1 −iu √n du.
The integrand converges pointwise to e−u2/2 (for u fixed and n →∞), so it’s reasonable to guess that the integral should converge to R ∞ −∞e−u2/2 du = √ 2π, which would lead to the formula √nnn enn! ≈ 1 √ 2π , or n! ≈ √ 2πn n e n , which is precisely Stirling’s formula. However, note that the O(u3/√n) estimate holds whenever t = u/√n is in a neighborhood of 0, and since u actually ranges in [−π√n, π√n], we need to be more careful to get a precise asymptotic result.
65 To proceed, it makes sense to divide the integral into two parts. Denote M = n1/10, and let I = Z π√n −π√n exp n eiu/√n −1 −iu √n du = I1 + I2, I1 = Z M −M exp n eiu/√n −1 −iu √n du, I2 = Z [−π√n,π√n][−M,M] exp n eiu/√n −1 −iu √n du.
We now estimate each of I1 and I2 separately. For I1, we have I1 = Z M −M exp −u2 2 + O u3 √n du = Z M −M e−u2/2 exp O u3 √n du = Z M −M 1 + O u3 √n e−u2/2du = 1 + O(n−1/5) Z M −M e−u2/2du = 1 + O(n−1/5) Z ∞ −∞ −2 Z ∞ M e−u2/2du = 1 + O(n−1/5) √ 2π −O exp −n−1/5 = 1 + O(n−1/5) √ 2π.
For I2, we have |I2| ≤2 Z π√n M exp n eiu/√n −1 −iu √n du = 2 Z π√n M exp n Re eiu/√n −1 du = 2 Z π√n M exp n cos u √n −1 du Now use the elementary fact that cos(t) ≤1 −t2/8 for x ∈[−π, π] (see Fig. 7) to infer further that |I2| ≤2 Z π√n M exp −u2/8 du ≤2π√n exp −n1/5 = O(n−1/5).
Combining the above results, we have proved the following version of Stirling’s formula with a quantitative (though suboptimal) bound: Theorem. As n →∞we have n! = 1 + O(n−1/5) √ 2πn n e n .
168. Second example: the central binomial coefficient. Let an = 2n n = (2n)!
(n!)2 .
A standard way to find the asymptotic behavior for an as n →∞is to use Stirling’s formula. This easily gives that 2n n !
= (1 + o(1)) 4n √πn.
66 -3 -2 -1 1 2 3 -1.0 -0.5 0.5 1.0 Figure 7: Illustration of the inequality cos(t) ≤1 −t2/8.
(Note that this is not too far from the trivial upper bound 2n n ≤(1 + 1)2n = 22n.) It is instructive to rederive this result using the saddle-point method, starting from the expansion (1 + z)2n = 2n X k=0 2n k !
zn, which in particular gives the contour integral representation 2n n !
= 1 2πi I |z|=r (1 + z)2n zn+1 dz.
By the same trivial method for deriving upper bounds that we used in the case of the Taylor coefficients 1/n! of the function ez, we have that for each x > 0, 2n n !
≤(1 + x)2n/xn = exp (log(1 + x) −n log x) .
We optimize over x by differentiating the expression log(1 + x) −n log x inside the exponent and setting the derivative equal to 0. This gives x = 1, the location of the saddle point. For this value of x, we again recover the trivial inequality 2n n ≤22n.
Next, equipped with the knowledge of the saddle point, we set r = 1 in the contour integral formula, to get 2n n !
= 1 2πi I |z|=r (1 + z)2n zn+1 dz = 1 2π Z π −π (1 + eit)2ne−int dt = 1 2π Z π −π exp n 2 log(1 + eit) −t dt.
Now note that the expression in the exponent has the Taylor expansion n(2 log(1 + eit) −t) = 2 log 2 −1 4nt2 + O(nt4) as t →0.
67 Again, we see that a change of variables u = t/√n will bring the integrand to an asymptotically scale-free form. More precisely, we have 2n n !
= 1 2π Z π −π exp n 2 log 2 −1 4nt2 + O(nt4) !
dt = 4n 2π√n Z π −π exp −1 4u2 + O u4 n du.
It is now reasonable to guess that in the limit as n →∞, the pointwise limit of the integrands translates to a limit of the integrals, so that we get the approxi-mation 2n n !
≈ 4n 2π√n Z ∞ −∞ e−u2/4 du = 4n 2π√n2√π = 4n √πn, as required. Indeed, this is correct, but it remains to make this argument precise by breaking up the integral into two parts, a “central part” where the O(u4/n) error term can be shown to be small, and the remaining part that has to be bounded separately.
169. Exercise. Complete this analysis to give a rigorous proof using this method of the asymptotic formula 2n n = (1 + o(1))4n/√πn.
170. Exercise. Repeat this analysis for the sequence (bn)∞ n=1 of central trinomial coefficients, where bn is defined as the coefficient of xn in the expansion of (1 + x + x2)n, a definition that immediately gives rise to the contour integral representation bn = 1 2πi I |z|=r (1 + z + z2)n zn+1 dz.
Like their more famous cousins the central binomial coefficients, these coeffi-cients are important in combinatorics and probability theory. Specifically, an and bn correspond to the numbers of random walks on Z that start and end at 0 and have n steps, where in the case of the central binomial coefficients the allowed steps of the walk are −1 or +1, and in the case of the central trinomial coefficients the allowed steps are −1, 0 or 1; see Fig. 8.
Using a saddle point analysis, show that the asymptotic behavior of bn as n →∞ is given by bn ∼ √ 3 · 3n √πn .
171. A conceptual explanation. In both the examples of Stirling’s formula and the central binomial coefficient we analyzed above, the quantities we were trying to estimate took a particular form, where for some function g(z) we had a(n) = 1 2πi I |z|=r e−ng(z) zn dz z = 1 2πi I |z|=r exp −n(g(z) + log z) dz z = 1 2π Z π −π e−ng(reit)r−ne−int dt = 1 2π Z π −π exp −n(g(reit) + it −log r) dt.
68 10 20 30 40 -5 -4 -3 -2 -1 1 2 10 20 30 40 -2 -1 1 2 3 4 5 (a) (b) Figure 8: An illustration (with n = 40) of the random walks enumerated by (a) the central binomial coefficients and (b) the central trinomial coefficients.
(Sometimes g(z) would actually be gn(z), a sequence of functions that depends on n.) The idea is to choose the contour radius r as the solution to the equation d dz (g(z) + log z) = g′(z) + 1 z = 0.
This causes the first-order term in the Taylor expansion of g(z)+log z around z = r to disappear. One is then left with a constant term, that can be pulled outside of the integral; a second order term, which (in favorable circumstances where this technique actually works) causes the integrand to be well-approximated by a Gaussian density function e−u2/2 near z = r; and lower-order terms which can be shown to be asymptotically negligible.
Geometrically, if one plots the graph of |g(z)+1/z| then one finds the emergence of a saddle point at z = r, and this is the origin of the term “saddle point method.” This phenomenon is illustrated with many beautiful examples and graphical figures in the lecture slides prepared by Sedgewick and Flajolet as an online resource to accompany their excellent textbook Analytic Combinatorics.
The lecture slides can be accessed at
172. Exercise. It is instructive to see an example where the saddle point analysis fails if applied mindlessly without checking that the part of the integral that is usually assumed to make a negligible contribution actually behaves that way. A simple example illustrating what can go wrong is the function f(z) = ez2 = ∞ X n=0 z2n n! = ∞ X n=0 bnzn, where the Taylor coefficients are bn = 1 (n/2)!
n even, 0 otherwise.
69 Clearly any analysis, asymptotic or not, needs to address and take into account the fact that bn behaves differently according to whether n is even or odd. Try to apply the method we developed to derive an asymptotic formula for bn. The method fails, but the failure can easily be turned into a success by noting that there are actually two saddle points, each of which makes a contribution to the integral, in such a way that for odd n the contributions cancel and for even n they reinforce each other. This shows that periodicities are one common pitfall to look out for when doing asymptotic analysis.
173. Exercise. As another amusing example, apply the saddle point method to the function f(z) = 1/(1−z) = P∞ n=0 dnzn, for which the Taylor coefficients dn = 1 are all equal to 1. Can you succeed in deriving an asymptotic formula for the constant function 1?
174. Third example: Stirling’s formula for the gamma function. Our next goal is to prove a stronger version of Stirling’s formula that gives an asymptotic formula for Γ(t), the extension of the factorial function to non-integer arguments.
Specifically, we will prove.
Theorem.
For a real-valued argument t, the gamma function satisfies the asymptotic formula Γ(t) = 1 + O(t−1/5) r 2π t t e t (t →∞).
Proof.
We use a method called Laplace’s method, which is a variant of the saddle-point method adapted to estimating real integrals instead of contour integrals around a circle. Start with the integral formula Γ(t) = Z ∞ 0 e−xxt−1 dx Performing the change of variables x = tu in the integral gives that Γ(t) = tt Z ∞ 0 e−tuut−1 du = tte−t Z ∞ 0 e−tu+tut−1 du = tte−t Z ∞ 0 e−tu+tut−1 du = tte−t Z ∞ 0 e−tΦ(u) du u = t e t I(t), where we define Φ(u) = u −1 −log u, I(t) = Z ∞ 0 e−tΦ(u) du u .
(Again, note that we massaged the integrand to cancel the Taylor expansion of −log u around u = 1 up to the first order.) Our goal is to prove that I(t) = r 2π t + O(t−7/10) as t →∞.
70 1 2 3 4 0.5 1.0 1.5 2.0 2.5 3.0 Figure 9: The function Φ(u) = u −1 −log u.
As before, this will be done by splitting the integral into a main term and error terms. The idea is that for large t, the bulk of the contribution to the integral comes from a region very near the point where Φ(u) takes its minimum. It is easy to check by differentiation that this minimum is obtained at u = 1, and that we have Φ(1) = 0, Φ′(1) = 0, Φ′′(1) = 1, and Φ(u) ≥0 for all u ≥0. See Fig. 9. Denote I1 = Z 1/2 0 e−tΦ(u) du u , I2 = Z 2 1/2 e−tΦ(u) du u , I3 = Z ∞ 2 e−tΦ(u) du u , so that I(t) = I1 +I2 +I3. The main contribution will come from I2, the part of the integral that contains the critical point u = 1, so let us examine that term first. Expanding Φ(u) in a Taylor series around u = 1, we have Φ(u) = (u −1)2 2 + O((u −1)3) for u ∈[1/2, 2] (in fact the explicit bound Φ(u) −(u−1)2 2 ≤(u −1)3 on this interval can be easily checked). As before, noting that t (u −1)2 2 + O((u −1)3) = 1 2( √ t(u −1))2 + O ( √ t(u −1))3 √ t , we see that it is natural to apply a linear change of variables v = √ t(u −1) to 71 bring the integrand to a scale-free, centered form. This results in I2 = 1 √ t Z √ t −1 2 √ t exp −tΦ 1 + v √ t 1 1 + v/ √ t dv = 1 √ t Z √ t −1 2 √ t exp −v2 2 + O v3 √ t 1 + O t √ t dv.
As before, we actually need to split up this integral into two parts to take into account the fact that the O(v3/ √ t) term can blow up when v is large enough.
Let M = t1/10, and denote J1 = 1 √ t Z M −M exp −tΦ 1 + v √ t 1 1 + v/ √ t dv, J2 = 1 √ t Z [−1 2 √ t, √ t][−M,M] exp −tΦ 1 + v √ t 1 1 + v/ √ t dv, so that I2 = J1 + J2. For J1 we have J1 = 1 √ t Z M −M exp −tΦ 1 + v √ t 1 1 + v/ √ t dv = 1 √ t Z M −M e−v2/2 1 + O v3 √ t 1 + O v √ t dv = 1 √ t 1 + O(t−1/5) Z M −M e−v2/2dv = r 2π t 1 + O(t−1/5) , in the last step using a similar estimate as the one we used in our proof of Stirling’s approximation for n!. Next, for J2 we use the elementary inequality (prove it as an exercise) Φ(u) ≥(u −1)2 2 (0 ≤u ≤1), and the more obvious fact that 1/(1 + v/ √ t) ≤2 for v ∈[−1 2 √ t, √ t] to get that J2 ≤2 √ t Z [−1 2 √ t, √ t][−M,M] e−v2/2 dv ≤4 √ t Z ∞ M e−v2/2 dv = O(e−M) = 1 √ tO(t−1/5).
as in our earlier proof. Combining the above results, we have shown that I2 = 1 + O(t−1/5) r 2π t .
Next, we bound I1. Here we use a different method since there is a different source of potential trouble near the left end u = 0 of the integration interval.
Considering first a truncated integral over [ε, 1/2] and performing an integration 72 by parts, we have Z 1/2 ε e−tΦ(u) du u = −1 t Z 1/2 ε d du e−tΦ(u) 1 Φ′(u)u du = −1 t e−tΦ(u) u −1 u=1/2 u=ε −1 t Z 1/2 ε e−tΦ(u) du (u −1)2 .
Taking the limit as ε →0 (and noting that Φ(ε) →+∞in this limit) yields the formula I1 = 2 t e−tΦ(1/2) −1 t Z 1/2 0 e−tΦ(u) du (u −1)2 = O 1 t as t →∞.
Finally, I leave it as an exercise to obtain a similar estimate I3 = O(1/t) for the remaining integral on [2, ∞). Combining the various estimates yields the claimed result that I(t) = I1 + I2 + I3 = 1 + O(t−1/5) r 2π t .
175. The proof above is a simplified version of the analysis in Appendix A of [Stein-Shakarchi]. The more detailed analysis there shows that the asymptotic formula we proved for Γ(t) remains valid for complex t. Specifically, they prove that for complex s in the “Pac-Man shaped” region Sδ = {z ∈C : | arg z| ≥π −δ} (for each fixed 0 < δ < π) the gamma function satisfies Γ(s) = 1 + O(|s|−1/2) √ 2πss−1/2e−s as |s| →∞, s ∈Sδ.
Here, ss−1/2 is defined as exp((s −1/2) Log s), where Log denotes as usual the principal branch of the logarithm function.
73 Additional reading While preparing these notes I consulted the following sources, which contain a large amount of additional interesting material related to the topics we covered.
H. M. Edwards. Riemann’s Zeta Function. Dover Publications, 2001.
P. Flajolet, R. Sedgewick. Analytic Combinatorics. Cambridge University Press, 2009.
P. Flajolet, R. Sedgewick. Analytic Combinatorics Lecture Slides. Online re-source: Accessed March 6, 2016.
B. de Smit, H. W. Lenstra Jr. Artful mathematics: the heritage of M.C. Escher.
Notices Amer. Math. Soc. 50 (2003), 446–457.
E. M. Stein, R. Shakarchi, Complex Analysis. Princeton University Press, 2003.
104 (1997), 705–708.
A. Ivi´ c. The Riemann Zeta-Function: Theory and Applications. Dover Publi-cations, 2003.
D. Zagier. Newman’s short proof of the prime number theorem. Amer. Math.
Monthly 104 (1997), 705–708.
74 |
12892 | https://www.ck12.org/flexi/cbse-math/inverse-trigonometric-functions/how-does-the-graph-of-y-arcsec-x-differ-from-the-secant-function/ | Flexi answers - How does the graph of y = arcsec x differ from the secant function? | CK-12 Foundation
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Inverse Trigonometric Functions
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How does the graph of y = arcsec x differ from the secant function?
Flexi Says:
The secant function, y=sec(x), and the inverse secant function, y=sec−1(x) or y=arcsec(x), are related but have different graphs due to their different properties.
Secant Function
The secant function,y=sec(x), is undefined for x=(2 n+1)π 2, where n is an integer, because these are the values for which the cosine function equals zero. The graph of the secant function consists of a series of U-shaped pieces, each of them occurring between vertical asymptotes.
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Inverse Secant Function
The inverse secant function,y=arcsec(x), is defined only for x≤−1 or x≥1. The graph of the inverse secant function is a reflection of the graph of the secant function over the line y=x, but it's also restricted to two disjoint intervals of the x-axis: (−∞,−1] and [1,∞). This is because the inverse secant function must pass the horizontal line test to ensure that any horizontal line intersects the graph at most once, which is a requirement for the function to be invertible.
i
So, while the secant and inverse secant functions are related, their graphs look quite different due to the properties of each function.
Analogy / Example
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Identities of inverse trigonometric functions with negative angles.Describe the key characteristics of the graph of y = arccos x.Explain the graphical representation of the arccsc function.
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12893 | https://www.dadisp.com/webhelp/mergedProjects/refman2/FncrefAE/BUTTERWORTH.htm | BUTTERWORTH - DADiSP
BUTTERWORTH
Purpose:
Designs a digital IIR Butterworth filter.
Syntax:
BUTTERWORTH(type, order, rate, pb1, pb2, ripple, attn, sb1, sb2, "options")
type-An integer, the filter type. 1:lowpass 2:highpass 3:bandpass 4:bandstop
order-Optional. An integer, the filter length. If not specified or set to -1, the filter order is automatically estimated.
rate-A real, the sample rate of the filter in Hertz.
pb1-A real, the first passband edge in Hertz.
pb2-A real, the second passband edge in Hertz.
ripple-Optional. A real, the passband ripple in dB. Defaults to 3 dB.
attn-Optional. A real, the stopband attenuation in dB. Defaults to 40 dB.
sb1-Optional. A real, the first stopband edge frequency in Hertz. Defaults to pb1 - 0.05 rate
sb2-Optional. A real, the second stopband edge frequency in Hertz. Defaults to pb2 + 0.05 rate
"options"-Optional. A string, zero or more filter options: "analog" :produce analog filter coefficients, rate is ignored "matched_z" :use the matched z method to convert an analog filter prototype to a digital filter
Returns:
A series, the Butterworth filter coefficients in multi-stage cascade format.
Example:
W1: butterworth(1, 1000.0, 100.0)
W2: 20log10(filtmag(W1, 1024));sety(-80, 10)
W1 creates a Butterworth lowpass filter with a sample rate of 1000 Hz, a cutoff frequency of 100 Hz and a stopband attenuation of 40 dB. The stopband frequency defaults to 150 Hz and the passband ripple defaults to 3 dB. W2 displays the frequency response of the filter.
Example:
W1: butterworth(1, 1000.0, 100.0)
W2: gsin(1000, 1/1000, 3) + gsin(1000, 1/1000, 250)
W3: iirfilter(W2, W1)
Creates the same IIR lowpass filter as the previous example. W2 contains a series with two sinusoids and W3 applies the filter to recover the lower frequency sine wave.
Example:
W2: butterworth(1, 1000.0, 100.0, 3.0, 50.0, 130.0)
Creates a similar filter except the stopband attenuation is set to 50 dB and the stopband edge is set to 130 Hz.
Example:
W3: butterworth(3, 18, 1000.0, 200.0, 300.0)
Creates a Butterworth bandpass filter with a sample rate of 1000 Hz, a filter order of 18 and a passband that extends from 200 Hz to 300 Hz.
Example:
W4: butterworth(3, 24, 1000.0, 200.0, 300.0, 2.0, 50.0, 180.0, 320.0)
Creates a similar Butterworth bandpass filter as above except the order is set to 24 (resulting in 121 coefficients), the desired passband ripple is set to 2 dB and the desired stopband attenuation is set to 50 dB. The first stopband edge is 180 Hz and the last stopband edge is set to 320 Hz.
Example:
W1: butterworth(1, 1000.0, 100.0, "analog")
W2: 20log10(filtmag(W1, 1024, "analog"));semilogx;sety(-80, 10)
Same as the first example, except the result is an analog filter. The rate parameter is ignored.
Remarks:
The generic BUTTERWORTH filter specifications are depicted as follows:
Type = 1, Lowpass
Type = 2, Highpass
Type = 3, Bandpass
Type = 4, Bandstop
For filter type 1 and 2 (lowpass and highpass), the band frequencies pb2 and sb2 are omitted.
BUTTERWORTH uses the Bilinear Transform Method to compute the coefficients by converting an analog filter prototype to the digital domain. The filter order refers to the number of resulting poles (2X poles result for type 3 and type 4) and is not equivalent to the number of filter coefficients.
If "matched_z" is specified, the matched z transform is used instead of the BILINEAR transform. The matched z method maps the analog prototype filter poles and zeros to the digital domain with:
z = e sT where T is the sample rate of the digital filter.
The band edges must lie between 0.0 and 0.5 rate (the Nyquist frequency). Overlapping band edges are not permitted.
The filter coefficients are produced in multi-stage bi-quad form suitable for processing by the CASCADE function.
The cascade stages are ordered such that the poles of each stage are closer to the unit circle than the previous stage. The zeros of each stage are chosen to be closest to the poles of the same stage.
If "analog" is specified, the filter coefficients represent an analog filter, with coefficients in cascaded second order stages of analog frequency s. In this case, the rate parameter is ignored.
The transfer function of an analog low pass Butterworth filter with a cut off frequency of ω c is given by the all pole expression:
where n is the filter order and the Butterworth polynomial is defined as:
The gain of the filter in terms of the transfer function is:
The gain at ω c remains fixed at 0.707 or -3 dB for increasing filter order.
A Butterworth filter has a maximally flat frequency response in the passband, though the passband is not necessarily uniformly flat. Butterworth filters generally result in more coefficients than other IIR filter types for a given filter specification.
BUTTERWORTH can be abbreviated BUTTER.
See BANDPASS, BANDSTOP, HIGHPASS and LOWPASS to design linear phase FIR filters using the Remez Exchange method.
BUTTERWORTH requires the DADiSP/Filters Module.
See Also:
BESSEL
BILINEAR
CASCADE
CHEBY1
CHEBY2
DADiSP/Filters
ELLIPTIC
RATE
References:
Oppenheim and Schafer
Discrete Time Signal Processing
Prentice Hall, 1989
Digital Signal Processing Committee
Programs for Digital Signal Processing
I.E.E.E. Press, 1979
Bateman & Yates
D igital Signal Processing Design
Computer Science Press, 1989
DADiSP® Online Reference
Copyright© 1995-2025 DSP Development Corporation All rights reserved. |
12894 | https://www.amazingfishametric.com/tag/weary-pond/ | Weary Pond | The Amazing Fish-a-Metric
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Largemouth bass fishing on Weary Pond, Whitefield, Maine (September 18, 2016)
Posted on September 19, 2016 by Stan
1
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View of Wearey Pond looking north
Weary Pond is a 42-acre body of water located in Whitefield, Maine (see The Maine Atlas and Gazetteer map 13 D2). I try to reach this pond by driving south on Weary Pond Road off Hilton Road in North Whitefield. Weary Pond Road is rough and unimproved. I have to turn around after driving for about half a mile when I hit a stretch that is too bouldery for my little front wheel-drive car. I successfully reach my intended destination by driving north for 0.8 miles on Weary Pond Road off Jewett Lane in Whitefield. Jewett Lane is a solid four-season gravel road, whereas Weary Pond Road from this end is still unmaintained and rough but passable with a normal car. The pond becomes visible on the right through the trees. Park your vehicle as best as possible on the side of the forest trail. A boat launch is not available. Hence, only hand-carried craft can be used and need to be transported for about 300 ft or so through the woods from the road to the pond. But the destination is well worth the effort!!
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12895 | https://help.desmos.com/hc/en-us/articles/4406810279693-Integrals | Integrals – Desmos Help Center
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Use the Desmos Graphing Calculator, Geometry Tool, or 3D Calculator to investigate the beautiful world of integral calculus.
Get started with the video on the right, then dive deeper with the resources and challenges below.
If you'd like to explore the integral graph shown in the video (and open the "visual" folder), click here.
Definite Integrals
Type int (or integral if working in the Geometry tool) in an expression line to bring up an integration template where you can type in a lower bound, upper bound, integrand, and differential (such as dxdx).
Let’s look at an example. If f(x)\=110x2+1f(x)\=110x2+1, and you take the integral from 00 to 33 of f(x)dxf(x)dx, you’ll see this definite integral evaluates to 3.93.9. By clicking on the images, you can open the graphs and explore.
In Desmos 3D you can take double integrals. For example, try setting f(x,y)\=xy2f(x,y)\=xy2 and taking the double integral of f(x,y)dxdyf(x,y)dxdy where the bounds for xx are from 00 to 22 and the bounds for yarefrom\(0yarefrom\(0 to 11.
Indefinite Integrals and Infinite Limits of Integration
It's also possible to graph the output of some indefinite integrals. One way to do so is by including xx in the upper bound, 00 in the lower bound, and integrating with respect to a variable other than xx. Try graphing the integral from 00 to xx of t2dtt2dt.
Desmos will evaluate convergent integrals with infinite limits. Type infinity or infty into either the upper or lower bound! For example, if f(x)\=1x2f(x)\=1x2, then the integral from 11 to infinity of f(x)dxf(x)dx is 11.
Divergent integrals will show undefined. Let’s set f(x)\=1xaf(x)\=1xa with a slider for aa from 11 to 1010. When a\=1a\=1, the integral diverges and shows undefined. Click on the GIF and play the slider to see how the graph of the function and evaluation of the integral changes.
Learn More
Arc Length by Integration Example Graph
Different Integration Bounds Example Graph
Fundamental Theorem of Calculus Example Graph
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12896 | https://whgis-nlsc.moi.gov.tw/Support/FileDownload.ashx?ID=14 | 基本 基本 基本 基本圖 圖圖圖測製說明 測製說明 測製說明 測製說明
壹壹壹壹、 、、、總則 總則 總則 總則
一、基本圖，係包含主要地物、地貌及地理資料之二維基本地形圖。
二、以攝影測量方法測製為原則。
三、比例尺訂為五千分之一。
四、圖幅訂為東西經距 1 分 30 秒，南北緯距 1 分 30 秒。
五、圖號以五萬分之一地形圖之圖幅分為縱橫各 10 幅，共 100 幅，由 1 至 100 依序編號。 例示如下： 9521-Ⅲ-016 （9521-Ⅲ為五萬分之一圖號， 016 為自左向右、由上而下依 序編號） 。
六、量度單位：
（一）長度單位，採用公制。
（二）角度單位，採用一圓周 360 度式。
（三）面積單位，採用公頃（ 10,000 平方公尺） 。
七、坐標系統及高程系統：
（一）坐標系統：使用內政部所定之一九九七坐標系統 (TWD97) 為原則，並採用內政部最 新公告之坐標成果。
（二）高程系統：使用內政部所定之二○○一高程系統 (TWVD2001) 為原則，並採用內政 部最新公告之正高成果。
八、前條所述坐標系統為理論定義，基本圖測製時，應以強制套合至國家法定基準點而實 現該定義。
九、同一圖幅之正射影像應採用同一型式攝影機所拍攝之影像製作。
十、作業區內測繪資料應與外圍已測繪基本圖資料作接邊整合。
十一、數值資料檔包括數值正射影像資料檔、向量資料檔及數值地形模型資料檔等 3 種。
十二、數值資料檔應分別就各該資料內涵、資料精度與品質、使用注意事項及相關資料等， 製作詮釋資料（ Meta Data ），提供使用者參考。前項詮釋資料應依內政部國土資訊系 統詮釋資料相關標準製作。
十三、數值資料檔應提供支援國土資訊系統流通格式為原則，以利資料交換及應用基本圖 測製之地物向量資料。
貳貳貳貳、 、、、作業程序 作業程序 作業程序 作業程序
一、基本圖測製（修測）作業程序，如圖所示。
二、辦理基本圖測製時，因使用之軟、硬體不同等因素，在不降低本規範訂定的品質及內 容標準的條件下，部分工作項目得酌予合併或變更。 擬定測圖計畫
採用既有 合格 影像
正射影像製作
地物測繪 等高線測繪(或內插計算) 數值地形模型測製或修測
空中三角測量
蒐集影像控制區塊 控制測量 底片掃描 （底片式攝影機）
影像下載及處理 (數位式攝影機)
航空攝影
航測控制點佈設 （設置對空標誌）
辦理航空攝影
GPS（及/或IMU） 解算攝影站位置
調繪補測
數值地形圖地理資訊圖層製作
基本圖編纂
詮釋資料製作
成果 繳交
及/或 及/或
影像控制區塊測製 參參參參、 、、、工作項目及方法 工作項目及方法 工作項目及方法 工作項目及方法
一、擬定測圖計畫
（一）依據年度測製（修測）計畫涵蓋地區及工作數量，先期規劃工作進度、人員調派、 儀器設備及材料準備等工作，並蒐集控制點及相關圖籍等資料。
（二）測區範圍規劃時應注意邊緣界線之劃定。凡依控制點布設規定，所應布設地面控制 點之處，應儘量涵蓋有豐富之地物、紋理，以便於選擇天然地物點作為控制點之用， 並有交通路線方便進入布設地面控制點以及實施地面控制測量。
（三）辦理航空攝影前，應擬訂測製地區航空攝影計畫（含航線及控制點對空標誌（簡稱 空標）之規劃） 。航線方向以南北、東西為原則，其內容符合內政部「實施航空測量 攝影及遙感探測管理規則」所規定實施計畫書內容項目。如採用已有合格影像，取 代重新攝影，則應提出相關評估說明。
（四）釐定工作項目及作業細節，並彙整有關資料，擬訂工作計畫書以為作業之依據。
二、航測控制點布設（設置對空標誌）
（一）航測控制點應優先使用測區內現有之已知控制點（一、二、三等衛星控制點、一等 水準點） ，選擇位置符合空中三角測量控制點需求，且透空度良好之點位，並於航空 攝影前在點位上設置對空標誌（以下簡稱空標） ，做為空中三角測量控制之用。
（二）如為基本圖修測，前一版沒有錯誤時，應使用前一版基本圖測製時之控制點作為控 制，並於其上重新設置空標。如該等控制點有移位、毀損或遺失之情況時，應改用 前一版基本圖上可明確辨認之自然點或內政部所測設之影像控制區塊做為控制。
（三）航測控制點之分布，應適合空中三角區域平差之要求，原則如下：
採用衛星定位測量（ Global Positioning System;GPS ）輔助空中三角測量時，測區四 角各布設一組 2 個全控制點，並於測區首尾（航線端處）布設橫貫測區的高程控制 鍊，除測區左右側邊外，鍊上之高程控制點應位於航線重疊區內。高程控制鍊得以 增加橫貫飛行航帶的方式取代，惟此作為高程控制之橫貫航帶內，每片 9 個標準點 位中，至少有 5 個必須與原測圖用航帶連結。除布設控制點外，測區中央尚必須均 勻測設 5 個以上檢核點，供驗證空中三角測量品質之用。
若採 GPS 及慣性測量元件（ Inertial Measurement Unit ；IMU ）直接地理定位 (direct geo-referencing) 時，得省略前項之高程控制部分，僅於測區四角布設全控制點即可。 但是檢核點部分仍然比照前項辦理。另外必須注意控制點高程為正高系統，而直接 定位為幾何高者，必須使用內政部公告之大地起伏模式進行轉換，使空中三角測量 之結果最終為正高系統。
未採用 GPS 輔助空中三角測量時，平面控制點應分布在測區 (空中三角區域平差之 測區 )周圍界線上或界線附近，點與點之間隔約為 2 個航空攝影基線 (基線長以 60 ％之重疊率為準計算 )。高程控制點應成鍊狀分布，鍊之走向與航空攝影之航線大致成 垂直方向，測區兩端應各測高程控制點 1 條，測區內部約每隔 5 個航空攝影基線應 各測高程控制點 1 條。組成高程控制鍊之各點應選在相鄰航帶像片左右重疊範圍內。
（四）空標設置地點之上空，對天頂四周應有 40 度以上之透空度，且需視空標點位之地面 情形，選用耐久及與地面顏色足夠反差之材料，使其在影像上易於辨認量測。
（五）空標之尺寸應使其在影像上可明確辨認為原則。中心標採方形，利用解析測圖儀測 圖時，影像上中心標尺寸應較立體測圖儀之測標直徑略大；採數位影像攝影或測圖 時，中心標在影像上尺寸應不小於 2 個像素（ pixel ）。翼標之寬度與中心標之邊長相 等，長度則為寬度之 2 倍以上。翼標與中心標的間距等於中心標的寬度之 1.5 ～2 倍。 翼標以對稱之四個為原則，相對二翼標軸線交點與中心標之標心之偏差不得大於 5
公分，相鄰二翼標軸線夾角必須為 90 度，最大偏差量不得大於 5 度。如因環境限制， 無法布設四翼標時，可僅布設二翼標，二翼標採成一直線或成互相垂直。成垂直時， 二翼標軸線交點與標心之偏差不得大於 5 公分，二翼標成一直線時，軸線形成之中 點與標心偏差不得大於 5 公分。
（六）空標中心應與控制點位中心一致，最大偏心值不得大於 2 公分。
（七）空標設置完成後，需製作空標紀錄，內容含點號、點名、圖號 (五千分之ㄧ圖幅 )、等級 (已知點 )、點位控制種類、空標形狀、材料、顏色、坐標、建置單位、佈標人 員、日期、位置略圖、交通路線、點位照片等。
（八）實施航空攝影前，應先清點空標設置作業情形，有毀損遺失者，應予恢復。
（九）航空攝影完成後，發現原設置之空標毀損遺失率過高，致影響空中三角測量作業時， 應另覓得明確自然點或影像控制區塊取代空標，以確保空中三角測量精度要求。
（十）為檢核成像品質，測區內至少布設一處幾何解析度及色彩平衡檢定標，背景為純黑， 檢定標顏色與背景成黑白對比，白色及黑色灰度值對比強度應達 2.6 倍以上。解析 度則由 50 公分開始，每種解析度以√ 2 倍數增長設置至 100 公分（ 2 倍地元尺寸， 計 3 種解析度）為止，每種解析度水平方向及垂直方向各設置 3 條長方形標，並分 別近似平行及垂直航線方向，長方形標之ㄧ邊長度與解析度相同，另一邊長度則為 解析度之 3 倍以上，每條長方形標間隔與寬度相同（如下圖） 。
水平方向 垂直方向
三、航空攝影
（一）航空攝影工作應由具航空攝影能力之政府機關或普通航空業辦理。
（二）於二萬五千分之一地形圖上規劃航線，航線方向以南北或東西為原則，可視地形狀 況斟酌決定，航線設計必須確保涵蓋整個測區範圍。
（三）航空攝影機：
應採用精密測圖用之底片式寬角或常角航空攝影機，或精度相當於精密測圖用底片 式攝影機之數位式攝影機。均應附攝影機檢定報告書。
數位式攝影機
(1) 陣 列 式 (area array) 數 位 式 攝 影 機 必 須 分 開 具 有 全 色 域 (panchromatic) 及 多 譜 （multi-spectral ）成像能力，全色域之影像得由多個攝影機模組各自拍攝所得之次 影像 (sub-image) 合成，合成後其側向像幅不得少於 7500 像素 (pixel)( 即側向寬度像 素數目以能涵蓋一基本圖圖幅影像單邊長度以上為原則 )。縱向之像幅應使在組成 立體模型測繪時，具有基高比 (B/H) 不小於 0.3 之能力。多譜影像必須具有至少紅、 綠、藍三波段，其每個像素涵蓋之地元面積，不得大於全色域像素涵蓋地元面積 之 25 倍。
(2) 線列式 (linear array) 掃描數位式攝影機側向像幅不得少於 10000 像素，必須具有同 時前視、後視及垂直向下底視掃描的能力。前視及後視與底視掃描所張之立體視 角不得小於 15 度。至少在天底掃描位置必須具有多譜線列裝置，且其像素尺寸至 多不得大於全色域像素尺寸之 2 倍。
(3) 陣列式攝影機每張影像成像時間必須快到足以拍攝前後重疊達 80 ％的能力。
(4) 數位式攝影機的原始像素尺寸應優於（含） 15 微米，並至少具有 4096 階(12 位元
(bit)) 的原始輻射解析度（輸出得為 8 位元） 。
(5) 合成影像像素之公稱 (nominal) 位置與經校正後實際位置之差不得大於 1/3 像素尺 寸。個別原始攝影機模組影像經透鏡畸變改正後其殘餘透鏡畸變差在光圈為 f/5.6
（含）以下時，不得大於 1/2 像素尺寸。經檢定校正後，原始個別攝影機之率定 焦距、對稱像主點偏移量內方位元素中誤差不得大於 10 微米。
(6) 數位式攝影機必須具備輸出快門曝光時間訊號至外接 GPS 接收器之功能，且時間 訊號必須準確至 1 毫秒。
使用底片式攝影機則需採用航空測量專用像幅 23 公分× 23 公分之彩色底片，且底片 需在有效期限以內。
航空攝影機檢定報告
(1) 航空攝影機必須提出最近 5 年內攝影機檢定報告及合格證明書（經由實驗室或地 面檢定場檢定） ，各項檢定方法均需於檢定報告中詳述。
(2) 數位式攝影機檢定項目至少包含幾何率定 (像主點、透鏡畸變差 )、CCD 幾何位置 精度、輻射率定（像機靈敏度、光圈校正、線性度 (Sensor Linearity) 、雜訊、缺陷 像元 (Defect Pixel Recognition) ）、調制轉換函數（ Modulation Transfer Function ；
MTF ）、像素光譜反應特性曲線等檢定項目。
(3) 底片式攝影機至少包含幾何率定 (像主點、框標率定、透鏡畸變差 )、輻射率定、 離心變形等項目。
若原廠檢定報告內容不全，或對其檢定有疑慮時，得委託本國具有檢定能力之學術 或研究單位予以補充或重作。
（四）像片比例尺：底片式攝影之比例尺不得小於二萬分之一。數位式攝影之比例尺應使 原始像素在地面上之解析度優於（含） 0.5 公尺為原則。
（五）攝影方式：採垂直連續攝影，攝影軸傾斜小於 8 度，航偏角小於 10 度，各航線兩端 各應多拍攝 2 個像對。航線間相鄰影像重疊率 (左右重疊 )為 30% ，航線內相鄰影像 重疊率 (前後重疊 ) 數位式攝影機為 80 ％，底片式攝影機為 60 ％；實際影像重疊率 不得低於以上規定之重疊率 10% 。攝影完成後，應繪製像片涵蓋圖。
（六）攝影時機：儘量選擇晴朗無雲或雲高高於航高，無煙霧濛氣，能見度良好，且太陽 高度大於 30 度以上的時間進行攝影。
（七）底片掃描
採用底片式攝影機執行航空攝影時，應將所拍攝之像片使用精密像片掃描儀掃描數 化，掃描儀幾何精度優於 3 微米以上，輻射解析度在紅、綠、藍三波段均不得少於
256 階(至少為 24 位元之彩色像素 )。
掃描影像時，應依據基本圖成圖影像解析度之規定，選擇適當之掃描解析度。數化 成數位影像後之模糊參數（ blur parameter ）不得大於一個像素尺寸。
（八）數位影像下載及處理：採數位式攝影機，可直接下載並經影像拼接、融合、調色等 步驟處理後得到最終數值影像。
（九）使用底片掃描或數位影像下載處理所產生之數位影像，應符合以下影像品質要求：
1.MTF 在 20 lp/mm( 每公釐 20 線對數）時不得低於 0.4 ；模糊參數（ blur parameter ）不 得大於 1 個像素尺寸。以上各值均應經由檢定標或等同效力之地物檢定之。
影像色調必須均勻及反差足夠 ，全測區內影像中已知最強純白色地物像素的 RGB 值應在 250±5 範圍內，且該像素 RGB 三值之間最大之差不得大於 2；影像中已知最暗 之純黑色地物像素之 RGB 值應在 10±5 範圍內，且該像素 RGB 三值之間最大之差 不得大於 2。相鄰影像中同樣地物的 RGB 值應相同，最大差異不得大於 5（但受日 照方向及不同時期攝影影響的差異不在此限） 。
四、控制測量
（一）本作業方式僅適用於採用航空攝影測量辦理基本圖控制測量，其測設原則如下：
控制測量應依據內政部公告之測量基準與參考系統辦理，其中坐標系統為一九九七 坐標系統（ TWD97 ）2010 年成果，高程系統為二００一高程系統（ TWVD2001 ）。
辦理 GNSS 正高測量時，應使用本中心提供最新之臺灣地區大地起伏模型。
控制點平面中誤差及高程中誤差不得大於 20 公分。
（二）為確保成果品質，控制測量作業所使用之儀器裝備，至少每 3 年送至國家度量衡標 準實驗室或簽署國際實驗室認證聯盟相互承認辦法之認證機構所認證之實驗室校正 一次，並出具校正報告。
（三）辦理控制測量前應檢測已知控制點，已知控制點包含基本控制點及加密控制點，其 辦理原則如下：
清查涵蓋測區範圍及其毗鄰位置之已知控制點，並填載於已知控制點清查結果清 冊，清理後存在並適合進行測量之已知控制點，應可涵蓋測區範圍且至少 5 點以上， 實地查對如發現與原成果表或點之記所載事項不符時，應重新製作已知控制點調查 表陳報本中心。
已知控制點檢測結果符合規範者，即視已知控制點位無變動，可應用於後續控制測 量之依據，檢測結果不合格，造成已知控制點點數不足或無法涵蓋全部測區，應再 另外清查鄰近已知控制點並辦理檢測作業。
表 1 已知控制點檢測規範
控制點 檢測項目 檢 測 標 準
平面
控制點
利用 GNSS 靜態測量或
VBS-RTK 檢測兩相鄰已知 控制點位間之平面距離與 橢球高差，並與公告坐標反 算之水平距離與橢球高差 比較。
高程
控制點
利用 GNSS 正高測量或水準 測量檢測兩相鄰已知水準 點間之正高差，並與公告正 高差比較。
距離不大於 5 公里時，檢 測平面距離較差、橢球高 差、正高差與距離之比值不 大於二萬分之一。
距離大於 5 公里時，檢測 平面距離較差、橢球高差、 正高差不大於 28 公分
+6ppmL ，L 為點位間之公 里數。 （四）新設控制點選點原則如下：
空中攝影前已設置有對空標誌，則使用對空標誌為地面控制點。航空攝影後，對空 標誌毀損或攝影前未設置對空標，則得選擇自然點或影像控制區塊作為空中三角測 量之地面控制點。使用自然點或影像控制區塊作為控制點時，為保障精度及可靠度， 在每一應設置控制點之位置，必須選用 2 個以上自然點或影像控制區塊。另應選擇
5% 圖幅數以上 (不少於 10 個點 )均勻分佈於測區內之檢核點，以供空中三角平差檢 核，若測區中因地形限制無法挑選規定數量之檢核點，經委辦單位同意後得調降檢 核點之數量。
地面控制點選定後，應於實地釘立標誌，並製作點位紀錄表，新設點位編號應以英 文及數字組成 4 至 6 碼為原則。
（五）平面控制測量
平面控制測量可採用衛星定位靜態測量或虛擬基準站即時動態定位測量 （VBS-RTK ），各項作業方法與精度說明如下：
衛星定位靜態測量：全球導航衛星定位系統 (GNSS) 靜態測量，其觀測時間、記錄頻 率、重複觀測及成果精度作業規範如表 2。
表 2、衛星定位靜態測量作業規範
項目 作業規範
觀測時間 連續且同步 ≧60 分鐘 (距離大於 5 公里者應適度延 長觀測時間）
資料記錄速率 5 秒以下
重複觀測 新點重複觀測率 ≧25%
成果精度 基線水平分量 ≦30 毫米 + 6ppm L
基線垂直分量 ≦75 毫米 + 15ppm L
虛擬基準站即時動態定位測量（ VBS-RTK ）：其觀測時間、記錄頻率、重複觀測及 成果精度作業規範如表 3。本作業方式應依本中心「採用虛擬基準站即時動態定位 技術辦理加密控制及圖根測量作業手冊」相關規定辦理，並將所獲得之平面坐標依 該手冊規定之坐標轉換方式轉至 TWD97 坐標系統，至高程坐標仍沿用原虛擬基準 站即時動態定位測量成果高程值。
表 3、VBS-RTK 作業規範
項目 作業規範
資料記錄速率 1 秒
觀測數量 固定 (FIX) 解至少 180 筆以上
重複觀測
至少觀測 2 次，每次至少須間隔 60 分鐘以上， 且兩次坐標較差要符合平面位置較差 ≦30 毫米，高程位置較差 ≦50 毫米。 成果精度 平面中誤差 ≦20 毫米
高程中誤差 ≦50 毫米
（六）高程控制測量
高程控制測量可採用直接水準測量、三角高程測量或 GNSS 正高測量，各項作業方 法與精度說明如下：
直接水準測量：需辦理往返觀測，測段往返閉合差不得大於 20 毫米 （S 為單一測 段長度之公里數，小於 1 公里時閉合差不得大於 20 毫米） 。
三角高程測量：其起點及末端必須附合至已知水準點上，平差改正前每測段閉合差 不得大於 5 公分 （N 為所經邊數） ，測段距離超過 500 公尺時，應作大氣折光及地 球曲率改正。
3.GNSS 正高測量：採用衛星定位靜態測量或虛擬基準站即時動態定位測量之作業方 式(精度要求如表 2、表 3) ，測得高程控制點橢球高，並利用大地起伏模型內插計算 高程控制點之大地起伏值，由橢球高與大地起伏值計算高程控制點正高參考值，另 至少需連測每個高程控制點附近 5 公里內之已知水準點，分析已知水準點之大地起 伏值精度，據以修正高程控制點之正高值。
五、空中三角測量
（一）採用數值立體測圖儀或航測影像工作站量測空中三角連結點及設有空標之平面、高 程控制點。
（二）以光束法量測空中三角結點時，量測中誤差不得大於 10 微米，在坡度達 IV 級以上 之山地或植被覆蓋達 IV 級之林地（坡度及植被覆蓋分級如附錄 1）不得大於 15 微米。
（三）空中三角測量連結點分布
每片的 9 個標準位置上至少量測 2 個點，每一標準位置至少有一量測點與同航帶或 相鄰航帶像片上共軛點相連，不同鄰片允許以不同量測點連結。惟連結相鄰航帶之 連結點必須至少為 4 重點（ 4 光線束） 。當航帶前後重疊大於標準的 60 ％（例如為
80 ％或 90 ％）時，則相鄰航帶間之連結可以不必每片之每一標準位置都與相鄰航帶 相連，而可減至以前後重疊率 60 ％計算之基線距離內，至少有一連結點為原則。
如採影像匹配自動化量測空中三角連結點，得不以上述原則分析連結強度，惟其連 結應符合以下標準（如表 4），且相鄰航帶之間仍應達到以 60 ％重疊率計算基線時， 每一基線距離內至少有一 4 重以上點連結鄰航帶。
表 4 連結點強度標準
前後重疊率
可靠度指標 60 ％ 80 ％ 90 ％
平均多餘觀測數（總多餘 觀測數 /總觀測數） ≧0.55 ≧0.6 ≧0.7
連結點平均光線數（連結 點總光線數 /總連結點數） ≧4 ≧6 ≧7
連結點強度指標（ N 重光 線以上連結點數 /總點數）
(4 重光線以上連結點 點數 )/( 總點數 )≧0.3 (6 重光線以上連結點 點數 )/( 總點數 ) ≧0.3
(8 重光線以上連結點 點數 )/( 總點數 ) ≧0.3 註： 1. 平均多餘觀測數：空中三角測量平差計算時，網系總多餘觀測數除以總觀測值個數後 所得到之一個平均可靠度之指標。
註： 2. 連結點平均光線數：觀測同一連結點的總影像片數，即為該連結點的光線數，亦稱為 連結點重點數。所有連結點的總光線數除以總連結點數，即為連結點平均光線數。
註： 3. 連結點強度指標： N 重光線以上連結點數（ N 指自然數 1.2.3.4 …）除以總點數後所得 到之一強度指標。
（四）空中三角測量平差計算，須分 2 個過程進行。先以最小約制 (或自由網 )平差，以進 行粗差偵測並得到觀測值精度的估值，其觀測值之殘餘誤差均方根值不得大於 10 微米，在坡度達 IV 級以上之山地或植被覆蓋達 IV 級之林地不得大於 15 微米。其次進 行強制附合至控制點上平差，其觀測值之殘餘誤差均方根值不得大於 13 微米，在坡 度達 IV 級以上之山地或植被覆蓋達 IV 級之林地不得大於 20 微米。
六、影像控制區塊測製
（一）以影像方式記錄地面特徵點作為控制之用，亦可視為控制點之影像點之記。影像控 制區塊之影像中心為特徵點的位置，通常一影像控制區塊組應包含多個此特徵點之 共軛影像區塊，由不同拍攝角度所組成之影像控制區塊組有助於提供辨認上的優 勢，亦可提高自動化量測之成功率。
（二）影像控制區塊資料內容包含影像、影像中心點三維坐標及其它屬性，以特徵型態而 言屬於點特徵之控制實體。影像控制區塊來源可分為地面控制點 (包含佈標點及特徵 點)、空中三角測量之連結點或人工選取經由前方交會所得之特徵點。
（三）影像區塊點分佈及密度：將每圖幅平均劃分為 9 個宮格，每宮格至少採 1 點對方式 建置為原則。圖幅涵蓋山區或水域部分，無顯著特徵時則採隨處取樣，並得酌予減 少點對數量。
（四）影像控制區塊選點原則及品質標準如附錄 2。
七、數值地形模型測製
（一）本規範所稱數值地形模型（ Digital Terrain Model ；DTM ）涵蓋了 2 類內容。第一類 是數值高程模型（ Digital Elevation Model ；DEM ），是不含地表植被及人工構造物時 地球表面自然地貌起伏的數值模型。第二類是數值表面模型 (Digital Surface Model ；
DSM) ，是地表最上層覆蓋物 (含人工建物及植被 )表面的模型。
（二）數值地形模型高程點之分布採規則方格網，網格間距以 5 公尺為原則，且應量測地 形特徵點（如山頂、山窪、鞍部等） 、地形特徵線（如山脊線、山谷線）及地形斷線 （地面傾斜角劇烈變化分界線）等資料。
（三）數值地形模型資料以美國資訊交換標準碼（ ASCII ）格式，製作成數值資料檔。數值 資料檔之分幅應與五千分之ㄧ基本圖幅分幅一致，圖檔名稱以取用圖幅號命名為原 則，資料檔格式如附錄 3。
（四）數值地形模型高程中誤差之允許值如附錄 3。
八、等高線測繪
（一）等高線可利用精密解析立體測圖儀、數值航測影像工作站或其他同等精度之航測儀 器直接測繪，或運用數值地形模型資料，以內插計算方式產生。
（二）等高線間隔首曲線為 5 公尺，計曲線為 25 公尺。
（三）等高線為地表面實際高程之連續性表現，遇地物不間斷，測繪時應扣除地面覆蓋物 （如樹木、建築物）之高度，等高線應予製作成數值等高線檔。
（四）以數值地形模型資料內插計算時，應考量地形特徵點、特徵線及地形斷線等資料。
（五）獨立標高點量測
空地及重要交叉路口，必須有標高點。圖上道路距離約以 5 至 10 公分測 1 點。
水田視為等高，原則上每 1 塊田應有 1 獨立標高點，標高點宜註記在田中央。種植 之旱田選擇較平坦處測 1 獨立標高點。
樹林內獨立標高點不可太密，以成圖容易解讀及美觀性為原則，應選擇覆蓋較稀疏 處或地形變化特徵處（如山頭）量測，若是遇有裸露地則必須量測。
量測的獨立高程點與等高線相互關係必須合理。
（六）等高線表示之高程應與由數值地形模型內差所得高程一致，其高程中誤差之允許值 如附錄 4。
九、正射影像製作
（一）利用數值航測影像工作站或同等精度之航測儀器，配合數值地形模型資料作為正射 糾正之高程控制資料，將中心透視投影之影像，逐點糾正成正射影像，並製作數值 正射影像資料檔。
（二）同一圖幅內以採用同一時期之攝影機所拍攝之航空影像製成正射影像為原則。
（三）正射影像製作，其每一像素以使用距離像主點最近之像素為原則產製檔案以基本圖 圖幅為單位，並涵蓋該圖幅範圍與基本圖圖幅相配合，以每幅圖 1 個檔案為原則， 其地元尺寸不得大於 25 公分，正射影像解析度之查核仍以原始影像解析度為準。
（四）測製地區地勢陡峭，於影像較邊緣處投影位移大，加上地勢變化劇烈，正射影像上 植被在糾正時有影像拉扯的現象，必須檢查數值地形模型成果，且儘量選擇合宜拍 攝位置的空照影像來製作正射影像，並進行正射影像鑲嵌。若無合宜的影像可替換 則仍使用原影像，不得在影像拉扯處直接填上重複的紋理影像。
（五）鐵、公路、橋樑等對地圖判讀有重要意義的基礎建設，必須依其實際測量高度進行 正射微分糾正，因而產生之無影像遮蔽區應以鄰影像補足，若無影像可供補足，得 以黑色區塊填補。
（六）正射影像製作使用之數值地形模型資料，其網格間距為 5 公尺以內。
（七）正射影像位於平坦地表面無高差移位的明顯地物點其平面位置中誤差應小於 2.5 公尺。
（八）數值正射影像以彩色影像表示，並需進行無接縫鑲嵌（ mosaic ）及調整全區影像之 色調、亮度一致，整張正射影像的色調應均勻，其明亮度（ intensity, brightness ）的 直方圖分布在 5~250 之範圍（全反射之地物不計入範圍） 。
（九）比照國際照明委員會（ CIE ）定義白色的方式來定義電子檔及出圖色彩的平衡，但僅 做相對平衡的定義即可，亦即測區內已知為白色地物（或無顏色的灰色、黑色地物 均可） ，其在正射影像電子檔中紅、綠、藍三波段的強度值應該相等，在以 24 位元 表示全彩的軟體系統中，紅、綠、藍三波段值間最大的差異不得大於 5。
十、地物測繪
（一）使用儀器及編碼
地物測繪係利用精密解析立體測圖儀、數值航測影像工作站或其他同等精度之航測 儀器以數值立體測圖方式施測。測圖前應先將各地物、地類、地貌以分類編碼，並 依其性質分層施測。
地物、地類、地貌之分層分類請參照附錄 5「基本地形資料分類編碼說明」進行分 類編碼，其圖式依內政部頒佈之「基本地形圖資料庫圖式規格表」規定辦理。
（二）每個立體模型採用像對基高比 （B/H ）不小於 0.3 之立體像對，以保障立體測圖精度。
（三）主要道路、水系、房屋、地類均須測繪，並依分層分類編碼規則製作向量檔。測繪 原則如下：
道路：
(1) 寬度 3 公尺（含）以上皆應測繪，但若為郊區或山區房屋區塊之間的唯一道路， 即使寬度不足 3 公尺亦應測繪。
(2) 道路應以現地之現況及道路實形測繪。
(3) 重要指標性之綠帶應測繪。
水系：
(1) 河、溝、渠等明渠，皆需繪製河岸線，若河流兩岸有明顯堤防或河床有明顯範圍， 以此認定河流寬度；若無明顯河流範圍，則以河川流域面認定河流寬度。若因遭 遇水利構造物或遮蔽等因素導致河流不連貫 ，仍需配合實際狀況使河流合理連貫。
(2) 河流、水道寬度 3 公尺（含）以上皆應測繪，如具有連通性質之水道，雖不足 3
公尺亦應以單線測繪。
(3) 水體面積大於 3 公尺× 3 公尺皆應測繪。若水體與水體之間距小於 5 公尺得合併同 一區塊，大於 5 公尺則須分開測繪。
(4) 河岸線應繪製於河堤或地形變化之崁下、坡下，河岸線之上下游應連貫、完整， 以 95111 繪製河岸線下游範圍內部存在之沙洲、臨時性旱田等地類，請以 95119
類別繪製河川水流線。
房屋：
(1) 單棟房屋大於 5 公尺× 5 公尺皆應測繪。但每 100 公尺× 100 公尺範圍內或每公里道 路沿線所能尋獲之唯一房屋，均應予以繪製。若房屋與房屋間之開放式行人通道 寬度小於 3 公尺得合併同一區塊，大於 3 公尺則須分開測繪。
(2) 房屋區之間所包含之空地面積（如三合院、中庭、停車場、綠地等）小於 100 平方公尺得合併為房屋區之一部分，大於 100 平方公尺則須分開測繪；另房屋邊緣 線小於 5 公尺之折線可省略。
植被覆蓋及農漁養殖（以下簡稱地類）主要分為林地、水田、旱作地、果園、茶園、 養殖池、牧場、鹽田等類別，按地類實際範圍測繪其地類界線，不可僅繪一小段， 區塊大於 25 公尺× 25 公尺須予以繪製，同類範圍之間距若小於 5 公尺者得合併同一 區塊，大於 5 公尺，則須分開測繪，地類判釋以攝影當時情形為依據；空地免予測 繪。
十一、調繪補測
（一）立體測圖所得之向量及編碼資料依地物、地類、地貌等屬性予以分類分層編輯，繪 製稿圖，攜赴實地調繪，以修正立體測圖之錯誤、補充立體測圖時無法辨認、遺漏 或因影像受遮蔽未能於立測時測繪之地物地貌。調繪補測以確認攝影當時情形為原 則，並調查地物、地名、交通系統、水系、人工構造物、地類等名稱，製成調繪稿 圖，以供基本圖編輯使用。 （二）實地調繪前，應先核對現有之航空影像、前一版基本圖、地形圖等相關圖籍資料， 逐一詳實比較，確認調繪內容，並辦理調繪人員講習。
（三）實地調繪範圍及注意事項：
調繪時，均在地物中心位置，或近旁適當易辨識處繪製記號，如有名稱應併予註記。
調繪稿圖應儘量維持圖面清潔、清晰，以利後續編圖使用。
交通系統調繪，包括鐵路、鐵路機車廠、高速鐵路、高速鐵路機車廠、各級公路、 鄉村道、立體交叉道、捷運、捷運機車廠、航站大廈、港管所、燈塔、纜車線和索 道等，及與交通系統有關且長、寬均大於 5 公尺之橋樑、箱涵、隧道口等。
水系調繪，包括河流、溝渠、渡口、水壩、洩洪道、攔河堰、攔沙壩、堤防、瀑布、 碼頭、湖泊、池塘、沼澤、溼地、水庫、蓄水池、島嶼等項，水流系統需加繪水流 方向箭頭。
人工構造物調繪，包括變電所、墓地（不含獨立墓） 、船塢、抽水站（磚石或混凝土 建造之永久性抽水站） 、公用污水處理廠、公用垃圾處理場、公用焚化爐、礦場、儲 油場、天然氣廠、雷達站、衛星資料接收站、無線電台、廣播電台、電視台、回歸 線標等。
地類調繪，包括林地、水田、旱作地、果園、茶園、養殖池、牧場、鹽田等。
地貌調繪，包括山丘、谷地、斷崖等。
地標調繪，包括政府及民意機關、學校、職訓中心、圖書館、博物館、美術館、文 化中心、社教館、研究機構、醫院、衛生所、公立孤兒院、公立養老院、殯儀館、 火葬場、劇院、音樂廳、國家公園、國家森林遊樂區、風景名勝區、公園、遊樂場、 動物園、植物園、旅客服務中心、體育館、體育場、游泳池、海水浴場、紀念 (堂、 館、塔 )等設施、孔廟、古蹟、天文台、氣象站、市場（固定聚集數十個以上攤位之 市場） 、地下街、購物商場（附設大型停車場之百貨公司、大賣場等） 、連鎖便利商 店、郵局、電信局、電力公司服務處、天然氣 (瓦斯 )公司、旅館、金融機構、火車 站、汽車站、捷運車站、高速鐵路車站、交流道、收費站、加油站、公有停車場、 服務區、休息區、機場、港埠、教堂、寺廟、回教寺、加工區、發電廠、造船廠、 自來水廠、自來水公司服務處、外國領事館及駐華辦事處及全國性知名地標。廢棄 或遷移之機關或學校，依現地調繪結果為主，如確實已經廢棄或搬遷，無須註記原 機關或學校之名稱，並於圖面上原位置加註（廢）字。
地名須全部調查註於圖上。有新舊地名者，僅註記新地名。
國防軍事設施不予調繪。
（四）調繪稿圖整理完成，調繪人員須先自行檢查無誤後，在圖幅左下方簽名，註明調繪 完成年月日。送審查人員審查認可，始得移送編繪人員應用。
十二、基本圖編纂
（一）調繪補測完成後，按附錄 5「基本地形資料分類編碼說明」及內政部「基本地形圖 資料庫圖式規格表」規定分幅編纂及圖面整飾（含圖元類別與註記、圖式線號、圖 例、圖廓、方格線、方格線坐標、圖號、比例尺、地名、行政界線、圖幅接合表等） 整理成基本圖向量資料檔。
（二）相鄰圖幅間需予相互接邊，注意圖幅間之線狀物體、界線、等高線、道路到達地、 方格線註記、地標、居住地名稱、河流流向箭頭及其他地物等，必須彼此銜接、吻 合。
（三）行政界線及林班界編繪：行政界線分縣市、鄉鎮市區、村里界線等，可參照內政部 行政區域圖、地方政府行政轄區圖及相關圖籍資料，將行政界線轉繪。林班界線可 參考行政院農業委員會林務局林班圖轉繪。
（四）基本圖測製日期以成果完成審核驗收通過日期為準，表示至年、月。
（五）等高線編繪：計曲線及首曲線應依圖示線號規定編繪，以示區別。相鄰等高線在圖 面上距離小於 0.2 公釐時，應在出圖時截斷等高線，以免線條過於密集。
（六）地名及註記編繪
地名、註記、圖式及圖幅整飾等資料，依「基本地形圖資料庫圖式規格表」之規定 建檔。中文內碼採用 UTF 碼。
註記包括地名、高程、方格線註記、圖廓外說明、點狀地物、線狀地物、區域表面、 山部、控制點及標高點等之名稱及符號等。
註記應置於該地物之中央或附近適當地點，以不遮蓋重要地物為原則。且應儘量避 開地物、方格線等，其他線亦儘量不通過註記及圖式。
圖內地名及各項註記之字體，以採用等線體為原則，高程、方格線註記及其他數字， 採用阿拉伯數字。
圖廓外註記資料，應包括圖名、圖號、版次、圖例、比例尺、測圖說明、圖幅接合 表、行政界線略圖、圖幅位置圖、圖幅經緯度、方格坐標、道路到達地等。
地名及註記字儘量按水平等距排列，由左到右，由上到下排列，若為直列，由右至 左排列；註記字體及大小則依「基本地形圖資料庫圖式規格表」規定繪製，註記位 置及排列選擇以不影響製圖品質及使用者閱讀之便利為原則，並依下列規定辦理：
(1) 註記字列及方向
註記字列分垂直、水平，可依地形物自然形態採用；其排列方向按中文、英 文及阿拉拍數字之不同（如下圖）配合字列註出。
(2) 點狀地物註記
點狀地物註記因無法於範圍內加以註記，則以採用水平字列為宜，但必要時 方可採用垂直字列註出。註記位置之選擇可依下列優先次序選定：
□中文 中文 □ 中文
□
□
中文
(3) 線狀地物註記 線狀地物包括街道、公路、鐵路、雙線與單線河川及類似地物，當寬度足可 容納註記字體時，則註記於二邊線之中央，沿地物之方向等距排列且保持閱讀連 貫性，較長線狀地物應分段重複註記 (以道路為例：以能表達出道路起迄為原則， 應考量各街廓路名表達之完整性，即每隔 1~2 個街廓重複出現路名 )，彎曲線狀地 物註記應儘可能註記在較直線處，避免註記在彎曲部分；當寬度不能容納註記字 體或單線地物時，則於地物上側註記。
(4) 區域表面註記
縣(市)、鄉鎮區 (市)、村里、地類等區域表面註記，以能完全表示該區域之範 疇為原則，註記字應置於該區域之中央，區域之表面需視形狀以水平字列、垂直 字列註記。如為狹長地帶，則循線狀地物註記概略中心方向。
(5) 山部註記
山部註記依實際形態與情況而定，如為山脈、山谷、山脊、峽谷等，可沿一 單純之徐緩弧線或概略中心線，置於略為偏上或偏右之處，以水平字列註記，小 山、山頂、山峰等可註於頂部中央上方（如下圖） ，但不得遮蓋其他突出之細部， 以保持顯示地貌之紋理。
(6) 控制點及標高點註記
控制點、標高點之點號及高程註記應儘量靠近點位，並依實際情形選定優先 註記位置，且不應與其他地物重疊。
（七）基本圖編繪清查完成後，編繪人員需進行自我檢查，並在圖幅檢核表簽名，註明編 繪年月日及所用電腦檔名，送檢查人員檢查。
（八）像片基本圖出圖檔
基本圖編纂完成並經檢查後，將數值正射影像資料檔、基本圖向量資料檔（包括數 值等高線）予以套疊成像片基本圖，並依所選擇之繪圖機出圖格式每幅圖製作出圖 檔，其解析度不得小於 508 dpi(dots per inch) 。
為便於像片基本圖 (紙圖 )資訊讀取，出圖檔套疊圖層顏色及文字註記設定原則如附 錄 6。
水流方向線之處理：基本圖 CAD 成圖內水流方向線可繪於水道雙線之中。製作出圖 檔時為版面美觀須將水流方向線移至水道雙線之外，使水流方向線清楚的呈現於圖 面之上。
十三、數值地形圖地理資訊圖層製作
為利日後各項地理資訊系統應用使用，將數值地形圖向量成果（ CAD 格式） ，進行 圖形物件、屬性資料及位相關係等資料處理，轉置數值地形圖地理資訊圖層，共分為控 制點、行政界、房屋、地標、交通系統、水系、公共事業網路、地貌、國有林界、圖幅 共 10 類主題圖層，各圖層轉置內容以原地形圖向量成果內容為原則，圖層說明如下，其 內容架構如附錄 7。屬性欄位長度可視資料內容酌予調整。
（一）控制點：控制點以點圖元的方式儲存，並以屬性方式紀錄控制點分類及坐標資料。
（二）行政界：行政界線包括縣市界、鄉鎮市區界，應封閉且為面型態，以參考主管機關 現有之行政區域圖資料為原則。
（三）房屋：房屋圖元應封閉且為面型態，需針對位相矛盾進行調整，如房屋不可超過道 路線。 （四）重要地標：以點圖元的方式儲存，並以屬性方式建立地標名稱、分類及坐標資料， 且分成政府及民意機關、文教設施、醫療社福及殯喪設施、公共及紀念場所、生活 機能設施、交通運輸設施、宗教、工廠及其他等類別。
（五）交通：分成鐵路、高鐵、捷運、道路、立體道路、小徑、路網、隧道、橋樑等圖層。
鐵路、高鐵、捷運等圖層幾何型態為線型態，依車站分段，並以屬性方式，將名稱、 類型、來源定義等資料紀錄於資料欄位內，連結至圖元上。
道路圖層為將雙線道路以面圖元表達道路實形，且應連貫及圖元應封閉。
立體道路圖層為高速公路、市區快速道路、高架道路、匝道等不同於一般道路之道 路，以面圖元表達道路實形，且應連貫及圖元應封閉。
小徑圖層為單線道路，幾何型態為線型態。
路網圖層為道路中線，即雙線道路邊緣線等分中心之連線，幾何型態為線型態，路 網建置注意事項如下：
(1) 路段應以鄉鎮市區界、道路等級、道路結構及路段名稱分割，且應考量道路連續 性及完整性，不受道路寬度及遮蔽影響。
(2) 道路中線除國道、快速公路 (含市區高架道路 )及與其平行之平面道路以雙線表示 外，其餘道路以單線表示。
(3) 若遇有高架道路 (或隧道 )與平面道路同時存在時，須同時以雙線繪製高架道路 (或隧道 )與平面道路。此外並以屬性方式，將道路等級、名稱、路寬等資料紀錄於資 料欄位內，連結至圖元上。
(4) 如遇有上下多重疊立體道路時，將上下立體道路中線平行錯開不重疊。
(5) 各平面道路交叉口均需要有節點 (Node) ，即平面交叉路口線圖元需斷線。
(6) 平面道路由高架道路下面穿越，為區分兩者之不同，因此其交叉處不應產生節點。
(7) 遇隧道或車行地下道，無法正確施測道路位置時，於進出口增設節點。
(8) 車道數變更或路寬變更超過 2 公尺處應增加節點。
隧道圖層為面型態，指隧道及車行地下道等道路，並記錄隧道名稱。
橋樑圖層為面型態，並記錄橋樑名稱。
（六）水系：包括河流與水體，分為河流、小河、流域中線、水池湖泊等圖層。
河流圖層為將雙線河流以面圖元表達河流實形，且應連貫及圖元應封閉。
小河圖層為單線河流，幾何型態為線型態。
流域中線圖層為雙線河邊緣線等分中心之連線，為樹狀流域圖。各河流交叉口均需 要有端點，並以屬性方式，將河流類型代碼、河流名稱、等級等資料紀錄於資料欄 位內，連結至圖元上。
水池湖泊圖層為水體圖元，幾何型態為面型態且應封閉。
（七）公共事業網路：圖層為點型態包含電信塔、高壓電塔等。
（八）地貌：以表現地形起伏之高程資料為主，包括等高線（線）及獨立標高點（點）圖 層，等高線必須連續且不可相交。
（九）國有林界：分為國有林事業區界及國有林班界，需封閉為面圖層。
（十）圖幅接合圖層為記錄圖幅編號、名稱及攝影、測製日期等。
十四、詮釋資料製作
依據內政部國土資訊系統之「地理資訊詮釋資料標準」 (TaiWan Spatial Metadata Profile ；TWSMP) 相關規定填寫各項成果之詮釋資料，並利用內政部 「詮釋資料建置系統」 針對詮釋資料資訊、識別資訊、限制資訊、資料品質資訊、資料歷程資訊、空間展示資 訊、供應資訊、範圍資訊、維護資訊、引用資訊、參考系統資訊等類別按規定之項目填 寫（測製日期為全案完成審核驗收日期） 。
十五、成果繳交
基本圖各階段工作完成後，成果繳交項目如下表：
項次 項次 項次 項次 工作項目 工作項目 工作項目 工作項目 成果繳交內容 成果繳交內容 成果繳交內容 成果繳交內容
1 擬定測圖計畫 測圖計畫
2 航測控制點布設 （設置對空標誌）
航線規劃圖
空標紀錄表
航測控制點分佈圖（採農林航空測量所拍攝之影像，無需繳交）
3 航空攝影 1. 航測攝影機檢定報告
航線涵蓋圖
航拍紀錄（攝影日期、天氣資料）
4.GPS 或 GPS/IMU 導航資料（ GPS 輔助空三需檢附）
攝影站坐標（ GPS 輔助空三需檢附）
數位影像資料檔：
(1) 採傳統底片像機拍攝，需附航拍底片、底片掃描檔、掃描儀檢 定報告及內方位量測資料。
(2) 採數位像機拍攝，需附數位原始資料 (raw data) 、轉影像檔所需 資料。
影像檢查紀錄表（採農林航空測量所拍攝之影像需繳交）
4 控制測量 1. 控制測量報告：包含坐標系統、已知點清查及檢測成果、控制點 網絡圖、新設點位統計、測量方式 (觀測時段、參數設定、使用 儀器 )、測量成果。
觀測資料：
(1)GNSS 原始觀測資料需轉換為 RINEX 格式、 GNSS 觀測時段表 （GNSS 靜態測量需附）
(2)VBS-RTK 觀測資料檔（ VBS-RTK 需附）
(3) 水準觀測資料
點位調查表：新設點位及已知控制點位變動者
成果計算報表
(1) 基線成果（含可判斷基線計算品質的指標） 、最小約制網平差成 果、強制附合平差成果、坐標成果（含坐標值及其標準偏差值） （GNSS 靜態測量需附）
(2)VBS-RTK2 測回坐標成果、坐標轉換參數 （VBS-RTK 測量需附）
(3) 水準測量往返閉合差計算報表
(4)GNSS 正高計算報表（採 GNSS 測正高需附）
5 空中三角測量 1. 控制點及連結點展點網系圖
控制點號及像片編號對照表 項次 項次 項次 項次 工作項目 工作項目 工作項目 工作項目 成果繳交內容 成果繳交內容 成果繳交內容 成果繳交內容
像坐標原始量測檔
控制點檔
5.GPS 觀測（ /IMU ）資料（ GPS 輔助空三需檢附）
空中三角平差報表 (含最小約制與強制附合 )7. 空三成果自我檢核紀錄（至少有 5 個檢核點）
影像控制區塊 1. 低解析度索引影像。
影像控制區塊成果檔（含索引檔） 。
數值高程模型 （DEM ）
地形特徵資料檔（ fea 檔）
檔頭資料檔（ hdr 檔）
五千分之一圖幅數值高程模型成果檔（網格檔）
8 數值表面模型 （DSM ）
檔頭資料檔（ hdr 檔）
五千分之一圖幅數值表面模型成果檔（網格檔）
9 正射影像 分幅及全區五千分之ㄧ 24 位元彩色正射影像檔（含 TIFF 、JPEG
及其坐標定位檔等格式）
10 地物測繪 1. 立體測圖原始三維稿圖檔（ DWG 或 DGN 格式）
如為修測可允許引用之舊資料為 2 維資料，但新測之地物仍須保 持為 3 維資料
11 調繪補測 調繪稿圖（需有作業人員簽名及標註日期，可繳交紙圖或 200dpi
掃描檔）
12 基本圖編纂 1. 數值基本地形圖檔（含 DXF 、DWG 及 DGN 格式） ，需有包含完 整圖幅框、去圖幅框、去等高線 (含圖幅框 )及同時去等高線與圖 幅框之數值地形圖檔
像片基本圖出圖檔（含 PS 及 PDF 格式）
13 數值地形圖地理 資訊圖層
數值地形圖地理資訊圖層成果檔（含 ESRI 之 SHP 、MapInfo 之
TAB/MIF/MID 格式） ，需有分幅、分縣市、全區整合資料
14 詮釋資料 1. 基本圖詮釋資料文字檔及 XML 檔
正射影像詮釋資料文字檔及 XML 檔
數值地形圖地理資訊圖層詮釋資料文字檔及 XML 檔
肆肆肆肆、 、、、成果檢查 成果檢查 成果檢查 成果檢查
一、總則
（一）基本圖各階段工作完成後，作業單位應以本章成果檢查項目依相關測製作業規定實 施檢查後，就各檢查項目做成檢查紀錄備查。
（二）檢查作業方法可包括內業檢查與外業檢查，可採人工檢查或程式檢查。檢查數量為 全數辦理檢查或抽樣檢查，抽樣方式除作業中另有訂定外，皆採用 ISO 2859.1-1999
抽樣檢查計畫表（原 MIL －STD －105E 表，以下簡稱抽樣計畫表，如附錄 8），批量 以該批送檢資料實際數量計數，檢查水準為第Ｉ級或第Ⅱ級，採單次或雙次隨機抽 樣辦理。抽樣結果應平均分布於測區，不可集中於測區一隅。不合格數量在允收數
(AC) 以內，則檢查通過；不合格數若達到拒收數 (RE) ，則該項抽樣檢查不通過；採 雙次抽樣時，若第一次檢查不合格數介於允收數 (AC) 與拒收數 (RE) 之間，再進行第 二次抽檢，兩次不合格數量之總和在允收數 (AC) 以內，則檢查通過；若達到拒收數
(RE) ，則該項抽樣檢查不通過。
（三）地形地物成果檢查抽樣以於圖幅內劃設之方格區域為抽樣樣本單元，即將五千分之 ㄧ圖幅之長及寬各等分成 5 段，每圖幅計 25 個方格 (如下圖 )，若方格內測製範圍未 滿一半方格，則該方格不予計數，以該批次送檢成果實際範圍計算方格總數，並以 方格總數進行成果檢查抽樣，所選取方格範圍內之物件全數檢查。每一個物件皆視 為一個檢核點。
01 02 03 04 05
06 07 08 09 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
（四）幾何精度檢查採抽樣方式辦理，幾何精度如採中誤差為標準者，以檢查中誤差為原 則；如採最大誤差限度者，則以檢查相對誤差為原則，並依合格率判定是否合格。
（五）作業單位交付各項作業成果檢查時，該項作業之原始觀測資料應一併交付檢查。
（六）以精度（中誤差）做為幾何精度標準者，雖檢查結果之中誤差通過標準，惟檢查點 位之較差超過中誤差 2 倍時，仍應進行修正；另其餘檢查亦同雖通過標準，惟檢查 所發現之問題，仍必須完全改正，再經復查確定無誤後才可錄存最後成果。
（七）檢查結果如未達合格標準，作業單位應全面重新修正後再辦理復查。
二、成果檢查項目及方法
（一）測圖計畫
檢查測圖計畫內容是否完整描述工作項目及細節？作業方式是否符合相關作業規 定？
（二）航測控制點布設（設置對空標誌）
書面檢查
本項檢查應在航空攝影之前實施，檢查航線規劃圖、航測控制點點位分布是否 符合作業規定？是否設置檢查點？需全數合格，如有不符者作業單位應全面重新修 正後再辦理復查。
實地檢查
(1) 依抽樣計畫表實施抽樣，以空標為樣本單元，檢查水準第Ⅱ級，採單次抽樣，允 收品質水準（ Acceptance Quality Level;AQL ）為 6.5 。
(2) 檢查控制點分布、對空標誌（以下簡稱空標）紀錄格式、空標位置、形狀、尺寸、 材料是否符合規範要求？
(3) 抽樣之空標上述檢查項目有 1 項（含）以上不合格，則該空標為不合格。如不合 格空標數量大於允收數（ AC ），作業單位應全面重新修正後再辦理復查。如果航 空攝影已完成，則作業單位應於未通過檢核之空標附近另覓 2 個自然點補測取代 該空標。
（三）航空攝影
航測攝影機 本項檢查應在航空攝影之前實施，檢查航空攝影機檢定日期及檢定報告書所列 檢定項目是否符合作業規定？需全數合格，如有不符者作業單位應全面重新修正後 再辦理復查。
航空攝影品質
(1) 書面檢查
檢查航線涵蓋圖是否完整涵蓋全測區？攝影日期及天氣是否符合作業規定？ 需全數合格，如有不符者作業單位應全面重新修正後再辦理復查。
(2) 航攝影像檢查
A. 依抽樣計畫表實施抽樣，以單片影像為樣本單元，檢查水準第Ⅱ級，採單次抽 樣，允收品質水準（ Acceptance Quality Level;AQL ）為 1。
B. 檢查像片比例尺、底片掃描 （掃描儀幾何精度及輻射解析度） 、地面像素解析度、 影像重疊率、像片品質（調制轉換函數（ MTF ）、模糊參數、色調）等是否符合 作業規定？影像是否有雲、模糊、陰影過長無法用於測繪？空標是否出現於影 像上，且清晰可辨？
C. 抽樣之航攝影像上述檢查項目有 1 項（含）以上不合格，則該片影像為不合格。 如不合格影像片數大於允收數 （AC ），作業單位應全面重新修正後再辦理復查。 另不合格影像部分，應由作業單位提出合格影像，以取代不合格影像。
（四）平面控制測量
書面檢查
(1) 檢查數量：全數辦理檢查。
(2) 檢查內容：檢查平面控制點點位紀錄、平面控制點展點網系圖、觀測紀錄（含已 知點檢測） 、已知平面控制點檢測成果報表、平面控制測量平差計算成果報表、平 面控制點成果報表等資料是否依作業規定製作繳交？平面控制點密度是否合於作 業規定？
(3) 通過標準：需全數合格，如有不符者建置單位應全面重新修正後再送監審單位復 查。
實地檢查
(1) 檢查數量：抽全數 10% 以上，且不少於 4 點，應於網系相對精度較差的地區優先 檢查。
(2) 檢查內容：
A. 點位設置：實地點位設置情形是否與點位紀錄表記載相符？
B. 成果精度：以 GNSS 靜態定位測量辦理成果精度檢查。
(3) 通過標準：
A. 應 95% 點位實地設置情形是否與點位紀錄表記載相符，視為合格。
B. 抽查點位基線之計算成果與建置單位最小約制網平差計算獲得之坐標（或
VBS_RTK 坐標成果）反算距離比對之較差， 95% （含）以上水平分量、垂直分 量小（含）於 20 公分 +6ppmL （L 為基線長）視為合格。檢查不通過，建置單 位應重新檢查修正後，再送監審單位復查。
（五）高程控制測量
書面檢查 (1) 檢查數量：全數辦理檢查。
(2) 檢查內容：檢查高程控制點點位紀錄、水準路線展點圖、觀測紀錄（含已知點檢 測） 、已知高程控制點檢測成果報表、高程控制測量平差計算成果報表、高程控制 點成果報表等資料是否依作業規定製作繳交？
(3) 通過標準：需全數合格，如有不符者建置單位應全面重新修正後再送監審單位復 查。
實地檢查
(1) 檢查數量：抽全數 10% 以上，且不少於 4 點。
(2) 檢查內容：
A. 點位設置：實地點位設置情形是否與點位紀錄表記載相符？
B. 成果精度：以水準測量辦理成果精度檢查。
(3) 通過標準：
A. 應 95% 點位實地設置情形是否與點位紀錄表記載相符，視為合格。
B. 抽查點位之高程差與建置單位測量之高程差比較，應 95% （含）以上小於 20 公分+6ppmL 視為合格。檢查不通過，建置單位應重新檢查修正後，再送監審單 位復查。
（六）空中三角測量
書面檢查
檢查控制點及連結點展點網系圖、像坐標原始量測資料、空中三角平差報表 (含最小約制與強制附合 )是否符合作業規定？需全數合格，如有不符者作業單位應全面 重新修正後再辦理復查。
上機檢查
(1) 抽樣空中三角測量所使用之影像總片數 2% 。
(2) 檢查空中三角測量重新計算成果 ：利用建置單位所送之影像量測檔及控制點檔 （含 空三 GPS 資料）使用相同之空中三角測量平差軟體重新計算成果與原計算成果比 較是否相符？
(3) 連結點重複量測檢查：
A. 每片影像至少抽查 2 個人工量測連結點，針對同一連結點的所有影像（如 4 重光線連結點需量測該點所在之 4 片影像） ，進行上機重複量測，重複量測值與原 量測值較差之均方根值不大於 10 微米√ 2 倍，在坡度達 IV 級以上或植被覆蓋 達 IV 級以上不大於 15 微米√ 2 倍。
B. 連結點採影像自動匹配量測時，則以人工方式於抽查之影像內 9 個標準點位中 至少重新觀測 2 點，並將觀測所得結果加入原觀測值檔案內，重新平差計算， 以驗證原匹配結果的正確性。重新計算後，最大像坐標改正數增量不得超過上 述連結點量測中誤差的 2 倍。
(4) 檢核點檢查：強制附合平差後，由全數檢核點計算得到之平面及高程坐標均方根 誤差值並依像片比例尺換算至像片坐標上，不得大於上述連結點量測中誤差的 3
倍。
(5) 空中三角測量重新計算成果、連結點重複量測檢查及檢核點檢查皆需全數合格， 如有不符者，作業單位應全面重新修正後再辦理復查。 （七）影像控制區塊
書面檢查
檢查影像控制區塊低解析度索引影像及影像控制區塊成果檔（含索引檔）是否 符合作業規定？需全數合格，如有不符者作業單位應全面重新修正後再辦理復查。
上機檢查
(1) 抽樣以實施共軛點前方交會所計算之影像控制區塊總數 3% 以上。
(2) 針對抽樣的影像控制區塊，進行上機重複量測並計算，所獲得之坐標值與原坐標 值較差之均方根值，不得大於附錄 2 所規定之影像控制區塊精度之√ 2 倍，即平 面位置較差之均方根值不大於 0.5 公尺√ 2 倍，且高程較差之均方根值亦不大於
0.5 公尺√ 2 倍。
(3) 抽樣之影像控制區塊幾何精度不大於上述規定，則檢查通過；若大於上述規定， 則檢查不通過，作業單位應全面重新修正後再辦理復查。
（八）立體測圖品質
每位測圖員完成第一個模型，即送監審單位進行初期檢查，每次檢查取模型內面積 最少 1/4 之方形區域，進行檢查。
於立體模型上檢查是否有缺漏未測繪或屬性錯誤的地物，並進行重複量測地物點平 面位置及高程，地物點重複量測平面位置與原平面位置較差之均方根值不大於 1.25
公尺√ 2 倍；地物點重複量測高程值與原高程值較差之均方根值不大於附錄 4 所定 高程中誤差允許值√ 2 倍。
抽樣模型之幾何精度不符合上述標準或缺漏未測繪或屬性錯誤地物數量超過地物數 量（該模型方形檢查區域內地物數量包含缺漏地物） 10% ，則該模型為不合格。若 抽樣模型不合格，作業單位應對該測圖員加強輔導，並重新測繪該立體模型後再辦 理復查。
（九）數值地形模型 (以下簡稱 DTM) 1. 本項目以內業檢查方式進行，作業單位繳交 DTM 必須同時含有數值高程模型 (以下 簡稱 DEM) 、數值表面模型 (以下簡稱 DSM) 及其檔頭資料檔。
檢查數量：以圖幅數為單位，以該批次繳交資料為檢查母體，按抽樣計畫表之檢查 水準Ⅱ級，雙次抽樣計畫計算檢查數量。
檢查內容：
(1) 涵蓋範圍：與該比例尺圖幅框套疊，檢查 DEM/DSM 是否完全覆蓋該圖幅框，範 圍內如含有不需測繪之區域則不受此檢查限制。
(2) 網格間隔：檔案格式為 ASCII(1) 者須檢查所有記錄之 E、N 坐標間隔是否相同， 並與檔頭檔紀錄一致，且符合本規範參、七、敘述之最小間距；為 ASCII(2) 者僅 檢查檔頭檔之紀錄間隔是否符合前述之最小間距。
(3) 檔案格式：須符合附錄 3「數值地形模型資料檔格式」之要求。
(4)DEM/DSM 模型精度檢查：
A. 各抽驗圖幅至少抽查 15 點，上機重複量測點位高程。
B. 山區重要道路且無植被覆蓋範圍超過 4 個網格之範圍，其植被覆蓋密度（ c）及 植被平均高度（ k）均視為零。
C. 於 DTM 內插計算抽查點位高程值時，如內插之網格地形變化劇烈，應重新以適 當網格間距或以不規則三角網（ TIN ）內插計算高程值。
D. 抽查點位重複量測高程值與原高程值 (網格內插產生 )較差之均方根值不得超過 附錄 4 所定數值高程中誤差允許值√ 2 倍。
通過標準：
(1) 抽驗圖幅於前述四項檢查均須合格，則該圖幅為合格圖幅。
(2) 本項整體判定結果，以圖幅為單位，按抽樣計畫表中檢查水準Ⅱ級，雙次抽樣， 允收品質水準（ AQL ）為 6.5 之標準判定是否通過，檢查不通過時，作業單位應 全面重新修正後再辦理復查。
（十）正射影像
本項目以內業檢查方式進行，作業單位須繳交正射影像及對應並通過檢查之立測草 圖或立體影像模型，檢查內容包含：標準影像檢查、抽樣影像檢查兩部分。
標準影像檢查：
(1) 檢查數量：以該作業區為單位，繳交一幅影像作為標準影像進行檢查。
(2) 檢查內容：
A. 亮度：取則影像中地面範圍 2 公頃之區域，將區域內所有像素紅、綠、藍三波 段值取平均後，統計最大與最小值。
B. 色調品質：以人工方式瀏覽全幅影像，檢查有無鑲嵌線、色調明顯落差與鄰幅 相連色調是否有落差。
C. 色彩平衡：將整圖幅均分九宮格，各抽一點為無色調之地物 (常識中為白色、灰 色、黑色地物 )之像素點，計算紅、綠、藍三波段值之間最大相差量，倘該方格 無此類地物則增加其於方格抽樣數，補足 9 個抽樣點。
(3) 合格標準：
A. 亮度：影像之亮度值範圍須界於 5~250 。
B. 色調品質：以人眼判可否明顯看出為原則，人眼可輕易辨識出則為不合格。
C. 色彩平衡：取樣點之紅、綠、藍三波段值，於 24 位元表示之全彩系統中，彼此 相差之最大值不得超過 5。
D. 前述三項有一項不合格，則標準影像檢查判定不合格。
抽樣影像檢查：
(1) 檢查數量：以圖幅數為單位，以該批次繳交資料為檢查母體，按抽樣計畫表之檢 查水準Ⅱ級，雙次抽樣計畫計算檢查數量。
(2) 檢查內容：
A. 地元尺寸：計算圖幅 (縱坐標差／縱軸像素數 )、(橫坐標差／橫軸像素數 )，再將 之取平均值為該圖幅平均地元尺寸。
B. 連續地物合理性：以人工方式瀏覽全幅影像，重要性依序為：道路、建物、其 他地物、地貌；必須檢查地物完整性，地物、地貌是否扭曲變形 (鐵、公路、橋 樑及對地圖判讀有重要意義的基礎建設必須糾正高差位移 )，影像鑲嵌處是否連 續無縫。
C. 平面位置精度：以正射影像圖套疊線繪地形圖或以立體量測方式量測地物點 （如 道路邊緣交點、田埂交點）平面位置，每幅檢查點數至少 10 點，若圖幅內無足 夠數量之明確點可供量測，則得以擴大至相鄰圖幅內量測。 (3) 合格標準：
A. 地元尺寸：每幅抽檢圖幅之平均地元尺寸不得大於參、九、 （三）規定之大小， 亦不得大於原始掃描像素尺寸乘原像片比例尺。
B. 連續地物合理性：以人眼判可否明顯看出為原則，本項缺失單幅影像不得超過
5 處。
C. 平面位置精度：量測抽樣點於正射影像上之平面位置與線繪地形圖（或以立體 量測方式）平面位置較差之均方根值不得大於 2.5 公尺。
D. 前述三項檢查有一項不合格則該圖幅視為不合格。
E. 以圖幅為單位，按抽樣計畫表中檢查水準Ⅱ級，雙次抽樣，允收品質水準 （AQL ）為 6.5 之標準判定是否合格。
通過標準：
(1) 標準影像檢查與抽樣影像檢查兩部分檢查均須合格，正射影像檢查方為通過，檢 查不通過時，作業單位應全面重新修正後再辦理復查。
(2) 標準影像檢查不需分批檢查，於單一作業區當中僅須檢查一次。
（十一）地形地物
本項目以外業檢查方式進行，作業單位須繳交基本圖成果。
檢查數量：以方格為單位，按肆、一、 （三）之規定劃設，並以該批次送檢成果實際 範圍計算方格總數，並按抽樣計畫表之檢查水準Ｉ級，雙次抽樣計算檢查數量。
檢查內容：
(1) 屬性檢查：檢查方格內地形地物是否缺漏未測繪 （含未調繪補測） 及屬性正確性。
(2) 幾何精度檢查：每個方格中抽選地物總數 10 ％或 5 點以上之明確地物點，重複量 測地物點位坐標或地物點間之相對距離及高程或相對高差。
通過標準：
(1) 屬性檢查：屬性錯誤數量（含缺漏地物）不得超過地物數量（該方格檢查區域內 地物合計數量，包含缺漏地物）之 5% 。
(2) 平面精度：抽查點位重複量測之地物點平面位置與原平面位置較差之均方根值或 地物點間之相對距離與原距離較差之均方根值不大於 1.25 公尺。
(3) 高程精度：抽查點位重複量測高程值與原高程值較差之均方根值或地物點間相對 高差不大於附錄 4 所定高程中誤差允許值。
(4) 單一方格就前三項檢查均須合格，則該方格為合格方格。
(5) 本項整體判定結果，以方格為單位，按抽樣計畫表中檢查水準Ｉ級，雙次抽樣， 允收品質水準（ AQL ）為 6.5 之標準判定是否通過，檢查不通過時，作業單位應 全面重新修正後再辦理復查。
（十二）基本圖編纂
本項目以內業檢查方式進行，作業單位須繳交基本圖成果及其對應之調繪稿圖、出 圖檔與通過檢查之正射影像。
檢查數量：以圖幅數為單位，以該批次繳交資料為檢查母體，按抽樣計畫表之檢查 水準Ⅱ級，雙次抽樣計畫計算檢查數量。
檢查內容：
(1) 調繪物件漏編：將受檢基本圖成果與調繪稿圖比對，以人工方式檢查地形地物是 否遺漏未編輯。
(2) 地形地物接邊：將受檢基本圖與周邊相鄰之圖幅一同展繪，檢查線段是否銜接， 屬性註記是否合理連續。
(3) 圖式及註記設定：將受檢基本圖與正射影像套疊，檢查地形、地物、地貌與正射 影像是否相符，有無漏繪情形，圖式及註記是否符合作業規定。
(4) 圖幅整飾：檢查圖廓外註記資料及圖幅整飾要件是否符合附錄 10 圖幅整飾規格。
(5) 出圖檔設定：檢查出圖檔解析度及套疊圖層顏色、文字註記設定是否符合作業規 定。
通過標準：
(1) 圖面編輯檢查缺失錯誤數量不得超過地物數量 5% （圖幅區域內地物數量，包含缺 漏地物） 。
(2) 圖幅整飾及出圖檔之缺點合計超過 5 處，則該圖幅不合格。
(3) 單一圖幅就前兩項檢查均須合格，則該圖幅為合格圖幅。
(4) 本項整體判定結果，以圖幅為單位，按抽樣計畫表中檢查水準Ⅱ級，雙次抽樣， 允收品質水準（ AQL ）為 6.5 之標準判定是否通過，檢查不通過時，作業單位應 全面重新修正後再辦理復查。
（十三）數值地形圖地理資訊圖層 (以下簡稱 GIS 圖層 )1. 本項目以內業檢查方式進行，作業單位須繳交 GIS 圖層與對應之基本圖 CAD 圖檔， 檢查內容包含：繳交格式檢查、圖層品質檢查兩部分。
繳交格式檢查：
(1) 檢查數量：以該批次繳交資料為檢查對象，以整批 1 式為單位，進行全數檢查。
(2) 檢查內容：
A. 繳交數量： GIS 圖層成果依照資料區劃分為分幅、分縣市 (圖幅為檔案 )、分縣市
(縣市為檔案 )、全區合併等層級；須檢查各層級之繳交數量皆無缺漏。
B. 檔案格式：以該檔案格式之軟體開啟，確認開啟式否正常、非錯誤檔案或空資 料。
C. 涵蓋範圍：僅檢核分幅資料，以圖幅框套疊各圖層，資料範圍需涵括該圖幅框。
D. 命名規則：需符合下列方式命名方式。
分類項目 資料夾路徑命名 檔案命名規則
分幅 分幅\五千圖號 年度_圖幅號_圖層名.shp[99_95171001_Contour.shp]
分縣市(圖幅) 分縣市(圖幅)\縣市名\分幅 年度_圖幅號_圖層名.shp[99_95171001_Contour.shp]
分縣市(整併) 分縣市(整併)\縣市名 縣市碼_圖層名.shp[S_Contour.shp]
全區整合 全區 年度_圖層名.shp[99_Contour.shp]
(3) 合格標準：前述四項檢查項目均須合格，繳交格式檢查方為合格。
圖層品質檢查：
(1) 檢查數量：以圖幅數為單位，以該批次繳交資料為檢查母體，按抽樣計畫表之檢 查水準Ⅱ級，雙次抽樣計畫計算檢查數量。
(2) 檢查內容：
A. 圖層架構：檢查各類別圖層對應正確之英文名稱、圖徵型態 (點、線、面 )、投影 坐標系統、欄位格式 (含名稱、型態、長度 )。B. 圖層內容：
(a) 圖檔轉換完整性：由地形圖圖檔轉換至地理資訊圖層之地物是否有遺漏、形狀 是否有誤。
(b) 圖形破碎：線、面圖元是否有圖形破碎情形。
(c) 圖層接邊：相鄰圖幅間之接續部分，圖形是否有疏漏、錯動及屬性不連續。
(d) 空間位相關係：檢查是否有空圖元、點圖元重疊、線圖元重疊、線自我相交、 相交未斷線、相接未斷線、懸掛節點、虛擬端點、面圖元重疊等位相關係。
(e) 屬性資料格式：檢查字體全 /半形是否正確、有無亂碼、多餘空格、正確代碼 是否正確及其他內容不合理之處。
(3) 合格標準：
A. 圖層架構：應全數合格。
B. 圖層內容：各次項目檢查之缺失量不得超過該次項總數量之 10% （圖幅內所有 圖層之地物合併計算數量包含缺漏地物） 。
C. 圖層架構與圖層內容檢查均須合格，則該圖幅方為合格。
通過標準：
(1) 繳交格式檢查以整批為 1 式計算，須全數合格。
(2) 圖層品質檢查以圖幅為單位，按抽樣計畫表中檢查水準Ⅱ級，雙次抽樣，允收品 質水準（ AQL ）為 6.5 之標準判定是否合格。
(3) 繳交格式與圖層品質兩部分檢查均須合格，本項 GIS 圖層檢查方判定通過，檢查 不通過時，作業單位應全面重新修正後再辦理復查。
（十四）詮釋資料
本項目以內業檢查方式進行，作業單位繳交之詮釋資料須含文字檔 (CSV 格式 )及
XML 格式兩種。
檢查數量：以檔案數為單位，並以該批次繳交資料為檢查對象，每種格式各抽 10%
的檔案進行檢查。
檢查內容：
(1) 繳交數量核對：檢查各資料成果之繳交數量皆無缺漏。
(2) 檔案格式檢查：檢查文字檔 (CSV 格式 )能正確讀入，非錯誤檔案或空資料； XML
須能通過文法驗證與資料結構驗證。
(3) 資料內容檢查：依本中心規定之必填欄位不得缺漏，填寫方式亦須符合本中心之 規定。
通過標準：
(1) 繳交數量核對以整體 1 式計算，須全數合格。
(2) 檔案格式檢查，須全數合格。
(3) 資料內容檢查項目之缺失欄位數不得超過應填欄位數之 10% ，否則該檔案視為不 合格。
(4) 繳交數量核對、檔案格式檢查、資料內容檢查等三項檢查均須合格，本項詮釋資 料檢查方判定通過；檢查不通過時，作業單位應全面重新修正後再辦理復查。 附錄 附錄 附錄 附錄 1：：：：坡度及植被覆蓋密度分級說明 坡度及植被覆蓋密度分級說明 坡度及植被覆蓋密度分級說明 坡度及植被覆蓋密度分級說明
一、坡度分級
I 級 ─ 坡度在 10 ％以下。
II 級 ─ 坡度介於 10 ％至 25 ％之間。
III 級 ─ 坡度介於 25 ％至 50 ％度之間。
IV 級 ─ 坡度介於 50 ％至 100 ％之間。
V 級 ─ 坡度介於 100 ％至 175 ％度之間。
VI 級 ─ 坡度在 175 ％以上。
某網格點上坡度之計算可以由該點及其四鄰共 5 個點密合一平面後，以該平面的最 大坡度為該點的坡度代表值。每一個網格點都有一個坡度值，全體網格點組成一個坡度 模型 (slope model) 。
二、植被覆蓋密度分級
任一點上其受植物覆蓋之厚度超出 DEM 或 DSM 規範中誤差之二倍者則視為受植被 覆蓋。覆蓋密度依由高空向地面觀察時的地面透空率來分級。
I 級 ─ 透空率在 90 ％以上。
II 級 ─ 透空率介於 50 ％至 90 ％之間。
III 級 ─ 透空率介於 20 ％至 50 ％之間。
IV 級 ─ 透空率介於 0％至 20 ％之間。
覆蓋密度分級係以待分析點為中心，半徑在十個網格間距的範圍內計算區內之透空 率。透空率之估計是以人工立體量測該範圍內可看到地面的面積來與範圍所包圍的總面 積比來計算。由於透空率級數不多，且相鄰級數之間的對精度影響的差異僅為平均樹高 的 5％，差異不大，故此估計不必很準確。若估值恰位於相鄰級數之分界點，則以較大 級數計。 附附附附錄 錄錄錄2：：：：影像控制區塊選點原則及品質標準 影像控制區塊選點原則及品質標準 影像控制區塊選點原則及品質標準 影像控制區塊選點原則及品質標準
一一一一、 、、、選擇原則 （一）不易變遷。 （二）坡度較平緩地區。 （三）幾何位置及灰階對比明確。 （四）易於人工或自動方式辨識及量測。 二二二二、 、、、品質標準 品質標準 品質標準 品質標準 影像控制區塊存錄於資料庫中，為方便使用者有效率擷取該影像區塊資料以及充 份使用該影像資料(包括依屬性資料查閱或統計影像特徵點之功能)，除此之外，在不 加重圖資生產複雜性前題下，預留自動化作業(例如影像匹配)所需屬性資料，整體而 言，資料庫應包含以下所列圖資及屬性資料： （一）低解析度索引影像標示影像特徵點位置 此索引影像用途為輔助辨識影像特徵點在大範圍影像中之相對位置，方便執行在待 解算影像上量測控制點的工作。 （二）影像控制區塊(組) 1. 檔案存放路徑: 以五千分之一像片基本圖圖幅為目錄檔名，利用檔名及圖幅坐標 範圍索引檔可方便影像特徵點存放及使用。 2. 檔名。 3. 儲存格式：影像格式，例如JPEG、TIF、RAW等。 4. 成像比例尺：例如兩萬分之一成像比例尺，登錄為1:20000。 5. 共軛點數目： 即多重點個數。為確認影像特徵點量測品質(包括精度與可靠度) ，每個影像控制區塊組必須至少含有三個共軛影像區塊，如此透過前方交會計算 與偵錯程序能確保特徵點影像量測以及特徵點物空間三維坐標品質。 6. 影像尺寸(長x寬像元個數)： 229 x 229像元。 7. 原影像航線方位角： 以徑度量為單位，有效位數至少到小數點後第二位(即至少 展示至”度”單位)。 8. 攝影日期： 依年/月/日格氏，例如當攝影日期為2007年8月3日，則登錄為 2007/08/03。 9. 相機種類(含標示類比式或數位式)及廠牌。 10. 焦距:以公釐為單位。 11. 像元地面解析度：以公尺為單位，有效位數至少到小數點後第二位(即至少展示 至”公分”單位)。 12. 航拍影像空三片號。 13. 影像控制區塊中心點在原影像之像片坐標值(x,y)： 以公釐為單位，有效位數至 少到小數點後第三位(即至少展示至”微米”單位)，搭配焦距能據以反推特徵點 位在原影像中之視角，此資訊有助益於執行影像匹配挑選與待匹配影像相近視角 之控制點。 14. 點位三維坐標系統： 包含標示所選用之平面坐標系統及高程坐標系統。 15. 點位三維坐標值： 依X/Y/Z格式登錄，以公尺為單位，有效位數至少到小數點後 第二位(即至少展示至”公分”單位)。 16. 點位三維坐標中誤差： 呈現點位幾何品質，能據以對所用之控制資料實施精度 分級。依σX/σY/σZ格式登錄，以公尺為單位，有效位數至少到小數點後第三 位(即至少展示至”公釐”單位)。影像控制區塊來源可為地面控制點(包含佈標 點及特徵點)、空中三角測量之連結點或人工選取經由前方交會所得之特徵點。 依不同來源，影像控制區塊精度應符合下列要求： （1）地面控制點：應符合貳、四、控制測量所規定地面控制點之精度要求。 （2）空中三角測量之連結點：應符合貳、五、空中三角測量所規定連結點之精度要 求。 （3）人工選取經由前方交會所得之特徵點：平面位置及高程中誤差皆應優於0.5公尺 。17. 特徵分類碼：區分地上點或物表點，並依幾何再細分為直線交會點、直線端點、 圓心點或其它。 附錄 附錄 附錄 附錄 3：：：：數值 數值 數值 數值地形 地形 地形 地形模型資料檔格式 模型資料檔格式 模型資料檔格式 模型資料檔格式
每一幅 DTM （DEM/DSM ）由檔頭檔案與網格資料檔組成。若加測地形特徵資料則需產 製特徵資料檔。
一、檔頭檔案
以 ASCII 格式紀錄。應包括下列元素：。
圖幅名稱－取與五千分之一基本圖同名 (以文數字表示 )
比例尺等級－ 5000( 比例尺之分母，以整數表示 )
東西向網格間距－ 5( 以公尺為單位 )
南北向網格間距－ 5( 以公尺為單位 )
總網格點數－網格資料檔內的總網格點數 (以整數表示 )。
行數－規則網格在東西方向的總行數 (以整數表示 )
列數－規則網格在南北方向的總列數 (以整數表示 )
圖幅西北隅 E 坐標－以整數表示 (為 5 的倍數 )
圖幅西北隅 N 坐標－以整數表示 (為 5 的倍數 )
圖幅東北隅 E 坐標－以整數表示 (為 5 的倍數 )
圖幅東北隅 N 坐標－以整數表示 (為 5 的倍數 )
圖幅西南隅 E 坐標－以整數表示 (為 5 的倍數 )
圖幅西南隅 N 坐標－以整數表示 (為 5 的倍數 )
圖幅東南隅 E 坐標－以整數表示 (為 5 的倍數 )
圖幅東南隅 N 坐標－以整數表示 (為 5 的倍數 )
生產方式代碼－ 00 航空攝影測量，人工量測。
01 航空攝影測量，自動匹配，經人工編修。
10 光達測量，經人工參照影像輔助資料編修。
11 光達測量，經人工依光達資料本身編修，未參照輔助資料。
12 光達測量，未經人工編修。
20 衛星測量，人工量測。
21 衛星測量，自動匹配，經人工編修。
22 衛星測量，自動匹配，未經人工編修。
生產設備名稱－ AP( 解析測圖儀 )，如 AP-LEICA-SD2000 DPW( 影像工作站 )，如 DPW-VIRTUOZO LIDAR( 光達 )，如 TerraScan
（代碼後之說明必須以短橫線「－」連接起來，以成為連續之一個字） 附錄 附錄 附錄 附錄 4：：：：數值高程 數值高程 數值高程 數值高程中誤差 中誤差 中誤差 中誤差允許值 允許值 允許值 允許值
高程中誤差之允許值，以 22222 kcba ⋅++=
σ 之公式訂定之。其中 a 為常數， b 為地表 坡度分級（如表 1）參數， c 為植被覆蓋密度分級（如表 2）係數， k 為植被平均高度（公尺） 。
訂定 a = 1m ，而 b、c 值，如表 1 及表 2 所列，峭壁、斷崖、峽谷處不列入精度等級。
表 1 b 參數值 (公尺 )
坡度分級 b
I 0.0
II 0.3
III 0.6
IV 1.0
V 3.0
VI 6.0
表 2 c 係數值 (無單位 )
植被覆蓋密度分級 c
I 0.0
II 0.05
III 0.10
IV 0.20 附錄 附錄 附錄 附錄 5：：：：基本地形資料分類編碼說明 基本地形資料分類編碼說明 基本地形資料分類編碼說明 基本地形資料分類編碼說明
一、基本地形資料分類編碼分為以下 10 類：
（一）測量控制點
（二）界線
（三）人工構造物
（四）交通系統
（五）水系
（六）公共事業網路
（七）植被覆蓋及農漁養殖
（八）地貌
（九）地標
（十）圖幅整飾及註記
二、地形資料分類採樹狀階層結構建立，由較上階層的粗略主題，持續分類到較下階層之特 定主題資料分類。每一層級的分類都包括獨特的編碼，藉由組合不同層級的編碼，可唯 一辨別特定主題的資料分類。基本地形圖資料庫為國土資訊系統九大資料庫分組之一， 編定為第 9 大類，故所有地形資料分類之編碼均以 9 為第一層之編碼。地形資料分類之 第二層稱為中類，其內容引用上述之十類地形分類，以一位代碼代表，其中第十類分類 並非以「 10 」記錄，而是記錄為「 0」。中類以下再細分為小類、細類、細目等三個階層。
三、基本地形資料分類編碼表
資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
91000 測量控制點 測量控制點 測量控制點 測量控制點
91100 基準點
91110 大地基準點
91120 絕對重力點
91130 天文點
91140 水準原點
91150 標準基線端點
91160 衛星定位追蹤點
91200 平面控制點
91210 衛星控制點 僅適用國家佈設之一等、二 等、三等及四等衛星控制點
91220 精密導線點
91230 導線點
91240 交會點
91250 自由測站點
91300 高程控制點
91310 一等水準點
91320 水準點 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
91330 驗潮站水準點
91400 重力點
91900 其他
91920 航測控制點
3 91930 衛星定位點
國家佈設之一等、二等、三等 及四等衛星控制點以外，以衛 星定位方式測製之點位，不適 用「五千分之ㄧ」比例尺
92000 界線 界線 界線 界線
92100 國界
92110 國界
92120 未定國界
92200 省、直轄市等界
92300 縣、省轄市、直轄市區等界
鄉、鎮、縣轄市、省轄市區等界
92500 村里界
92600 疑義界線
92700 特種界線
國有林事業區界
92720 林班界
92730 小班界
93000 人工構造物 人工構造物 人工構造物 人工構造物
93100 房屋
93110 永久性房屋(建築區)
93120 建築中房屋 不適用「五千分之ㄧ」 」比例尺
93130 臨時性房屋
93200 牆垣
93210 牆
93211 圍牆
93212 板牆 不適用「五千分之ㄧ」 」比例尺
93213 土牆 不適用「五千分之ㄧ」 」比例尺
93214 施工圍籬 不適用「五千分之ㄧ」 」比例尺
93220 垣 不適用「五千分之ㄧ」 」比例尺
93230 柵欄
93240 網
93250 籬資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
93260 圍 不適用「五千分之ㄧ」 」比例尺
93270 門 不適用「五千分之ㄧ」 」比例尺
93500 生活公共設施及場所
93520 喪葬設施
93523 墓地
93524 獨立墓
93590 其他
93591 塔
93592 亭
93593 水塔
93594 水井 不適用「五千分之ㄧ」 」比例尺
93595 噴泉 不適用「五千分之ㄧ」 」比例尺
93596 消防栓 不適用「五千分之ㄧ」 」比例尺
93597 防空洞 不適用「五千分之ㄧ」 」比例尺
93598 大佛像
93700 工礦設施
93720 工業設施
93721 變電所
93722 船塢
93723 倉庫
93724 油庫
93725 抽水站
93726 堆積場
93730 環保設施
93731 污水處理廠
93732 垃圾處理場
93733 焚化爐
93734 環境品質檢驗站 不適用「五千分之ㄧ」 」比例尺
93740 礦
93741 礦場
93742 溫泉區
93743 冷泉區
93790 其他
93791 輸送管 不適用「五千分之ㄧ」 」比例尺
93792 煙囟 不適用「五千分之ㄧ」 」比例尺 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
93793 油井、瓦斯井
93794 油槽
93795 瓦斯槽
93796 貯存槽
93797 變壓箱座 不適用「五千分之ㄧ」 」比例尺
93798 磚瓦窯 不適用「五千分之ㄧ」 」比例尺
93799 電信箱座 不適用「五千分之ㄧ」 」比例尺
93800 通訊及傳播設施
93810 通訊設施
93811 雷達站
93812 微波中繼站
93813 衛星資料接收站
93814 無線電台
93815 天線
93820 傳播設施
93821 廣播電台
93822 電視台
93823 攝影棚
93890 其他
93891 廣告架 不適用「五千分之ㄧ」 」比例尺
93900 其他
93901 界標 不適用「五千分之ㄧ」 」比例尺
93902 廢墟
93903 靶場
93904 階梯 不適用「五千分之ㄧ」 」比例尺
93905 碉堡 不適用「五千分之ㄧ」 」比例尺
93906 瞭望台(塔)
93907 升旗台 不適用「五千分之ㄧ」 」比例尺
93908 回歸線標
94000 交通系統 交通系統 交通系統 交通系統
94100 鐵路及附屬設施
94110 鐵路
c. 高架鐵路 屬性適用94111、94112及94113 等分類
d. 地下鐵路 屬性適用94111、94112及94113 等分類
e. 建築中鐵路 屬性適用94111、94112及94113 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
等分類
94111 一般鐵路
94112 高速鐵路
94113 輕便鐵路
94120 附屬設施
94122 轉車台
94123 平交道柵欄 不適用「五千分之ㄧ」 」比例尺
94125 鐵路機車廠
94200 道路及附屬設施
94210 道路
a 高架道路 屬性適用94213、94214、 94215、94216及94217等分類
94211 國道
94212 快速公路(含市區高架道路) 含匝道
94213 省道
94214 市區道路(路、街) 含圓環
94215 縣（市）道
94216 鄉（鎮）道路
94217 小徑
94218 建築中道路
94219 產業道路 含農路
a 林道 屬性適用94219分類
94220 附屬設施
94223 立體交叉道
94224 中央分隔島
94225 行人陸橋
94226 行人地下道 不適用「五千分之ㄧ」 」比例尺
94230 人行道 不適用「五千分之ㄧ」 」比例尺
94300 捷運及附屬設施
94310 捷運
a. 高架捷運 屬性適用94310分類
b. 地下捷運 屬性適用94310分類
c. 建築中捷運 屬性適用94310分類
94320 附屬設施
94322 捷運機車廠
94400 路工設施 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
94410 隧道
94411 鐵路隧道
94412 公路隧道
94420 橋樑
a. 鐵橋 屬性適用94421、94422、 94423、94424、94425、94426 及94427等分類
b. 鋼筋混凝土橋 屬性適用94421、94422、 94423、94424、94425、94426 及94427等分類
c. 磚石橋 屬性適用94421、94422、 94423、94424、94425、94426 及94427等分類
d. 木橋 屬性適用94421、94422、 94423、94424、94425、94426 及94427等分類
94421 鐵路橋(吊橋除外)
94422 公路橋(吊橋除外)
94423 車行吊橋
94424 人行吊橋
94425 浮橋
94427 小橋
94430 雜項工程
94431 箱涵
94432 管涵
94433 擋土牆
94434 路堤
94435 路塹
94436 駁坎
94437 橋墩
94500 機場附屬設施
94520 附屬設施
94521 跑道
94522 滑行道
94524 停機棚
94525 修護廠
94526 航站大廈
94527 管制塔台
94600 港灣附屬設施
94620 附屬設施 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
94621 港管所
94622 防波堤
94623 燈塔
94624 港燈
94625 錨地
94626 浮標
94627 沈船浮
94628 消波塊
94700 高速鐵路機車廠
94900 其他
94901 纜車線和索道
94903 國道線號符號
94904 省道線號符號
94905 縣道線號符號
95000 水系 水系 水系 水系
95100 河川及附屬設施
95110 河川
95111 江、河、溪
95112 時令河
95113 乾河
95114 小河
95115 運河
95116 溝、渠
95117 小水溝
95118 暗溝 不適用「五千分之ㄧ」 」比例尺
95119 河川水流線
95120 附屬設施
95121 引水槽
95122 渡口
95124 水壩
95125 洩洪道
95126 水閘
95127 攔河堰、攔沙壩
95129 漁梯
95130 岸邊工程 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
95131 堤防
土堤 屬性適用95131分類
石堤 屬性適用95131分類
混凝土堤 屬性適用95131分類
95132 混凝土塊護岸
95133 蛇籠
95134 土坎
95135 塊石護岸
95140 河岸、河中地形
95141 石磯
95142 沙洲
95143 陡岸
95150 水流性質
95152 瀑布
95153 河川流向
95160 碼頭
95161 渡船碼頭
95162 湖濱碼頭
95163 海濱碼頭
95200 面狀水域
95210 湖泊
95220 池塘
95230 乾池
95240 沼澤
95250 濕地
95260 水庫
95270 蓄水池
95300 海岸
95310 崖岸
95320 海岸線
95400 岸濱及水底地質
95410 岸濱地質
95411 濱
a. 泥濱 屬性適用95411分類
b. 沙濱 屬性適用95411分類 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
c. 礫濱 屬性適用95411分類
d. 岩濱 屬性適用95411分類
95412 珊瑚礁
95420 水底地質
95421 泥
95422 沙
95423 礫
95424 岩床
95900 其他
95901 顯礁
95902 暗礁
95905 島嶼
95906 濱外沙洲
96000 公共事業網路 公共事業網路 公共事業網路 公共事業網路
96100 線路
96110 電力線路
96111 輸送線(高壓線)
96112 配電線(電力線) 不適用「五千分之ㄧ」 」比例尺
96120 電信線路 不適用「五千分之ㄧ」 」比例尺
96200 管路
96210 水管 不適用「五千分之ㄧ」 」比例尺
96220 油管 不適用「五千分之ㄧ」 」比例尺
96230 瓦斯管 不適用「五千分之ㄧ」 」比例尺
96900 其他
96910 塔、桿、燈柱
96911 高壓線塔
96912 電信塔
96913 電線桿 不適用「五千分之ㄧ」 」比例尺
96914 路燈 不適用「五千分之ㄧ」 」比例尺
96920 人孔
96921 電力人孔 不適用「五千分之ㄧ」 」比例尺
96922 電信人孔 不適用「五千分之ㄧ」 」比例尺
96923 自來水人孔 不適用「五千分之ㄧ」 」比例尺
96924 雨污水下水道人孔 不適用「五千分之ㄧ」 」比例尺
97000 植被覆蓋及農漁養殖 植被覆蓋及農漁養殖 植被覆蓋及農漁養殖 植被覆蓋及農漁養殖 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
97100 樹木
97110 點狀
97111 獨立樹 不適用「五千分之ㄧ」 」比例尺
97120 線狀
97121 防風林
97122 行道樹 不適用「五千分之ㄧ」 」比例尺
97130 面狀
97131 針葉林
97132 闊葉林
97133 針、闊葉混合林
97134 灌木林
97135 竹林
97200 草地
97300 農地
97310 水田
97320 旱作地
97330 園、圃
97331 果園
97332 茶園
97336 圃
97400 養殖用地
97410 養殖池
97420 蓄牧
97421 牧場
97422 養雞場
97423 養猪場
97430 鹽田
97900 其他
97910 裸露地
97911 伐跡地
97912 荒地
97913 空地
97920 界線
97921 地類界
97922 田埂 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
98000 地貌 地貌 地貌 地貌
98100 高程起伏
98110 等高線
98111 計曲線
98112 首曲線
98113 間曲線
98114 助曲線
98120 獨立標高點
98130 水深
98131 水深點
98132 等深線
98200 諸地貌
98210 起伏地
98211 凹地
98212 土墩、台地、小丘
98213 斷崖
98220 崩、蝕、風化
98221 雨裂
98222 流土
98223 崩土
98224 惡地
98230 岩床
98231 獨立岩
98232 散岩
98233 露岩
98240 沙丘
98250 洞穴
98260 火山
98900 其他
98910 泥火山
99000 地標 地標 地標 地標
99100 政府及民意機關
99110 行政機關
99111 總統府
99112 中央政府公署 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
99113 省政府
99114 直轄市政府
99115 縣政府
99116 省轄市政府
99117 鄉、鎮、縣轄市、區公所
99118 村里辦公室
99120 民意機關
99121 中央民意機關
99122 省諮議會
99123 直轄市議會
99124 縣議會
99125 省轄市議會
99126 鄉、鎮、縣轄市民代表會
99130 軍事機關
99140 安全機關
99141 警察局隊、派出所、分駐所
99142 監獄、看守所
99143 消防隊
99200 文教設施
99210 學校及訓練機構
99211 大專院校
99212 國民中學、高級中學、高級 職校、完全中學
99213 國民小學
99214 職訓中心
99215 公立幼稚園
99216 特殊學校
99220 陳列及展覽設施
99221 圖書館
99222 博物館
99223 資料及陳列館
99224 文化中心
99225 社教館
99226 美術館
99230 研究機構 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
99300 醫療社福、殯喪設施
99310 醫療及社會福利設施
99311 醫學中心、醫院
99312 衛生所
99313 公立之孤兒院、育幼院
99314 公立之養老院、安養中心
99320 公立之殯儀館
99330 火葬場
99400 公共及紀念場所
99410 休閒設施
a 國家公園 屬性適用99410分類
b 國家森林遊樂區 屬性適用99410分類
c 旅客服務中心 屬性適用99410分類
99411 劇院
99412 音樂廳
99413 活動中心
99414 國家風景區
99415 公園
99416 遊樂場(園)
99417 露天劇場、音樂台
99418 動物園
99419 植物園
99420 健康設施
99421 體育館
99422 體育場
99423 公立游泳池
99424 海水浴場
99430 古蹟及紀念性設施
99431 古蹟
99432 紀念堂（館） 、孔廟
99440 碑、塔、像
99441 紀念碑
99442 紀念塔
99443 紀念像
99444 牌坊 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
99445 牌樓
99450 天文及氣象
99451 天文台
99452 氣象台、測候所
99500 生活機能設施
99510 市場
99511 公有市場
99512 地下街
99513 大賣場、大型百貨公司、 大型
超級市場、大型零售式量販
99514 連鎖便利商店
99520 郵政、電信及電力機構
99521 郵局
99522 電信公司
99523 電力公司服務處
99524 自來水公司服務處
99525 天然氣(瓦斯)公司
99530 金融機構
99540 旅館
99600 交通運輸設施
99610 車站
99611 臺鐵站
99612 長途公共汽車站
99613 捷運站
99614 高鐵站
99620 道路附屬設施
99621 國道及快速公路交流道 （市區
高架道路）
99622 收費站
99623 加油站
99624 公有停車場
a. 地面停車場 屬性適用99624分類
b. 地下停車場 屬性適用99624分類
c. 立體停車場 屬性適用99624分類
99625 國道休息站、服務區 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
99630 機場
99631 陸上機場
99632 水上機場(位)
99640 港灣
99641 商港
99642 漁港
99643 工礦港
99644 軍港
99700 宗教
99710 教堂、寺廟及宗祠
99711 教堂
99712 寺廟
99713 回教寺
99714 宗祠
99800 工廠
99810 工廠
99820 發電廠
水力發電廠 屬性適用99820分類
火力發電廠 屬性適用99820分類
核能發電廠 屬性適用99820分類
風力發電廠 屬性適用99820分類
99830 造船廠
99840 自來水廠
99900 其他
99901 水文站、驗潮站
99902 絕對重力站
99903 衛星定位追蹤站
99904 外國使領館及駐華辦事處
99905 地震測站
99906 監測站
99907 科學園區、工業園區
90000 圖幅整飾及註記 圖幅整飾及註記 圖幅整飾及註記 圖幅整飾及註記
90100 圖廓、方格線及經緯線
90110 圖廓
90111 圖廓 資料庫編碼 中
類
小
類
細
類 地形資料編碼 屬性碼 類別及地形資料名稱 備 註
90112 圖廓註記
90120 方格線
90121 方格線
90122 方格線註記
90130 經緯線
90131 經緯線
90132 經緯線註記
90200 圖廓外整飾
90210 圖名、圖號
90211 圖名
90212 圖號
90220 參考基準
90221 比例尺
90222 高程起算註記
90223 地圖投影坐標系
90224 大地基準
90225 等高線間隔
90230 接合表、偏角圖及其他圖表
90231 圖幅接合表
90232 行政界線略圖
90233 本圖幅位置圖
90234 偏角圖
90240 測製說明
90241 測製時間
90242 主管機關
90243 主辦機關
90244 測製機關
90300 地名及註記
90310 中文地名
90320 英文地名
90330 中文註記
90340 英文註記
90400 圖例
9 90900 其他 附錄 附錄 附錄 附錄 6：：：：出圖檔套疊圖層顏色及文字註記設定原則 出圖檔套疊圖層顏色及文字註記設定原則 出圖檔套疊圖層顏色及文字註記設定原則 出圖檔套疊圖層顏色及文字註記設定原則
以下相關設定建議使用 Autocad 2010 以上版本才具透明度設定功能
一一一一、 、、、圖層 圖層 圖層 圖層顏色 顏色 顏色 顏色設定 設定 設定 設定： [顏色代號為 Autocad 之顏色設定 ]
圖 層 顏色 顏色代號 備 註：
行政界線 黑色 7
林班界 黑色 7
鐵路邊線 黑色 7
高速鐵路邊線 黑色 7
捷運邊線 黑色 7
道路邊線 紅色 1
道路面著色 紅色 1 透明度 (Transparency) ：85
小徑 紅色 1
箱涵、管涵 黑色 7
橋樑 黑色 7
水系邊線 藍色 5
水面著色 淡藍色 4 透明度 (Transparency) ：85
單線水系 藍色 5
水流方向 藍色 5 圖例藍底描白邊
高壓電線 黑色 7
地貌 橘色 30
等高線 灰色 8
方格線 (90121) 白色 255
67 方格線 (90122-67) 紅色 1
二二二二、 、、、線寬 線寬 線寬 線寬設定 設定 設定 設定： [線寬除以下列舉外，餘皆為 0.2 mm]
圖 層 線寬 （mm ） 圖檔內設定
行政界
(僅 92200 、92300 、92400) 0.3 1.5
事業區界 (92710 、92720) 0.2 1.0
小徑 (94216) 0.3 1.5
圖廓線 (90111) 0.3 1.5
計曲線 (98111) 0.2 1.0
首曲線 (98112) 0.1 0.5
方格線 含 67 坐標標註線 0.1 0.5 三三三三、 、、、所有圖例符號 所有圖例符號 所有圖例符號 所有圖例符號（水流方向除外）均以黑色線寬 0.2 mm 之線段進行製作。
四四四四、 、、、文字註記設定 文字註記設定 文字註記設定 文字註記設定： ：：：
所有文字註記字體 - 等線體：華康中黑體。
斜等線體：左斜華康中黑體（圖檔內設定旋轉 340 °）
圖 層 顏色 顏色代
號
圖面字高
（mm ）
圖檔內 圖檔內 圖檔內 圖檔內
設定 設定 設定 設定字高 字高 字高 字高 字體
圖名 黑色 7 10.0 40.0 等線體
直轄市 白色 255 9.0 35.0 等線體
縣、直轄區 白色 255 7.0 27.25 等線體
鄉、鎮、區 白色 255 5.0 19.5 等線體
村、里 白色 255 3.5 13.5 等線體
事業區 白色 255 4.0 15.5 等線體
林班界 白色 255 3.5 13.5 等線體
行政界註記 白色 255 3.0 11.5 等線體
控制點文字 黑色 7 2.0 7.75 等線體
地名、山名 黑色 7 3.0 11.5 等線體
公路 黃色 2 3.0 11.5 等線體
鐵路、捷運 黑色 7 3.0 11.5 等線體
街道 黃色 2 2.5 9.75 等線體
巷道 黃色 2 2.5 9.75 等線體
國道線號 黑色 7 2.0 7.75 等線體
省道線號 黑色 7 2.0 7.75 等線體
縣道線號 黑色 7 2.0 7.75 等線體
海洋 藍色 5 5.0 19.5 斜等線體
湖泊、水庫 藍色 5 4.0 15.5 斜等線體
河川 藍色 5 3.0 11.5 斜等線體
港灣 藍色 5 4.0 15.5 斜等線體
機場 黑色 7 4.0 15.5 等線體
公園、遊樂園 黑色 7 3.0 11.5 等線體
機關、學校 黑色 7 2.5 9.75 等線體
橋樑 黑色 7 2.0 7.75 等線體
植被地類 黃色 2 2.5 7.75 等線體
其他地物 黑色 7 2.0 7.75 等線體
(沙)、 (礫) 藍色 5 2.5 7.75 等線體
(崩)、(露岩)、(雨裂) 橘色 30 2.5 7.75 等線體
等高線 -標高註記 白色 255 1.8 7.5 等線體
獨立標高點 黑色 7 1.8 7.5 等線體
五五五五、 、、、其他注意事項 其他注意事項 其他注意事項 其他注意事項： ：：：
1、建議以配合正射影像 配合正射影像 配合正射影像 配合正射影像之成圖資料進行產製 虛擬道路、新建道路、雲區範圍 …等圖層毋 須顯示。
2、建議崩塌地及雨裂地 [標示 標示 標示 標示 (崩崩崩崩)、（ 、（ 、（ 、（雨裂 雨裂 雨裂 雨裂） ）））]…相關地類、地貌，僅呈現屬性之文字說明， 範圍界毋須顯示。
3、PDF 檔輸出：以 600dpi 、最高列印品質輸出成 A1 尺寸。 附錄 附錄 附錄 附錄 7：：：：數值地形圖地理資訊圖層內容說明 數值地形圖地理資訊圖層內容說明 數值地形圖地理資訊圖層內容說明 數值地形圖地理資訊圖層內容說明
一一一一、 、、、圖層命名規則 圖層命名規則 圖層命名規則 圖層命名規則
為了便於索取識別成果文件，各縣市圖資應該按如下規則命名，各縣市名稱代碼及圖 層名稱代碼請參閱表 1 及表 2：
FILENAME .SHP
表 1 各縣市名稱代碼
縣市名稱 檔名 縣市名稱 檔名
臺北市 A 南投縣 M
臺中市 B 彰化縣 N
基隆市 C 新竹市 O
臺南市 D 雲林縣 P
高雄市 E 嘉義縣 Q
新北市 F 屏東縣 T
宜蘭縣 G 花蓮縣 U
桃園縣 H 臺東縣 V
嘉義市 I 金門縣 W
新竹縣 J 澎湖縣 X
苗栗縣 K 連江縣 Z
表 2 圖層說明
類別 圖層名稱 型態 檔 名
控制點 點 ControlPt
直轄市、縣、省轄市界 面 AdminCity
行政界 鄉、鎮、市、區界 面 AdminTown
房屋 面 Building
地標 點 Landmark
鐵路 線 Railway
高鐵 線 HSR
捷運 線 RTS
道路 (雙線 ) 面 Road
立體道路 面 Hroada
小徑 (單線 ) 線 Path
隧道 面 Tunnel
橋樑 面 Bridge
交 通
路網 線 MidRoad
包括所屬縣市別、年度及圖層名 稱，例如： A_97_ROAD
資料檔格式，例如： SHP 、
TAB 類別 圖層名稱 型態 檔 名
河流 面 River
小河 線 Stream
水池湖泊 面 Lake
水 系
流域中線 線 MidRiver
公共事業網路 點 Tower
等高線 線 Contour
地貌 獨立標高點 點 Spot
國有林事業區界 面 AdminForest
國有林界 國有林班界 面 ForestSub
圖幅 面 FrameIndex
二二二二、 、、、圖層名稱及屬性欄位結構 圖層名稱及屬性欄位結構 圖層名稱及屬性欄位結構 圖層名稱及屬性欄位結構
（一）控制點 ControlPt
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 點序號 數字 10 與控制點空間資料檔之節點序號對應
TerrainID 控制點地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
CNO 控制點點號 文字 20
CName 控制點名稱 文字 20
E_COORD97 TWD97 E 坐標值 數字 8 記錄 TWD97 坐標系統之 E坐標值（公 尺，至整數）
N_COORD97 TWD97 N 坐標值 數字 9 記錄 TWD97 坐標系統之 N坐標值（公 尺，至整數）
OrthoH 高程 H坐標值 數字 5 記錄高程值（正高）（公尺，至整數）
（二）行政界
直轄市、縣、省轄市界 AdminCity
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與縣市界空間資料檔之多邊形序號對 應
CityName 縣市名稱 文字 8
CityCode 縣市代碼 文字 11 主計處縣市代碼
鄉、鎮、市、區界 AdminTown
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與鄉鎮市區界空間資料檔之多邊形序 號對應
CityName 縣市名稱 文字 8
TownName 鄉鎮市區名稱 文字 10 記錄鄉鎮市區名，非全名
TownCode 鄉鎮市區代碼 文字 11 主計處鄉鎮市區代碼
（三）房屋 Building
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與房屋空間資料檔之多邊形序號對應
（四）地標 Landmark
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 點序號 數字 10 與地標點空間資料檔之點序號對應
TerrainID 地標地形編碼 文字 10 依據「基本地形資料分類編碼表」進 欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
行分類編碼
MarkName 地標點名稱 文字 40 填寫地標點全名
（五）交通
鐵路 Railway
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與鐵路空間資料檔之線段序號對應
TerrainID 鐵路地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
RailName 鐵路名稱 文字 30
高鐵 HSR
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與高鐵空間資料檔之線段序號對應
TerrainID 高鐵地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
HSRName 高鐵名稱 文字 30
捷運 RTS
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與捷運空間資料檔之線段序號對應
TerrainID 捷運地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
RTSName 捷運名稱 文字 30
道路 Road
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與道路空間資料檔之多邊形序號對應
立體道路 Hroada
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與立體道路空間資料檔之多邊形序號 對應
小徑 Path
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與小徑空間資料檔之線段序號對應
TerrainID 小徑地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
隧道 Tunnel
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與隧道空間資料檔之多邊形序號對應
TerrainID 隧道地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
TnlName 隧道名稱 文字 30
橋樑 Bridge
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與橋樑空間資料檔之多邊形序號對應
TerrainID 橋樑地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
BrilName 橋樑名稱 文字 30
路網 MidRoad 欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與路網空間資料檔之線段序號對應
TerrainID 路網地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
CityName 縣市名稱 文字 8 該路段所屬的縣市名稱
TownName 鄉鎮名稱 文字 8 該路段所屬的鄉鎮市區名稱
RoadStruct 道路結構碼 文字 2 0：一般道路 1：橋樑 2：隧道
3：匝道 4：高架 5：過水路
RoadNUM 道路編號 文字 8
此欄位儲存路段所屬國道、省道、縣 道、鄉道、市區道路、產業道路等道路 等級與編碼，如： "國1" 、"台3" 、"縣187"
等。
RoadNUM1 道路編號 1 文字 8
若同時有兩種道路等級發生共線時，於 此欄位儲存第二個所屬之省道、縣道、 鄉道、市區道路、產業道路等道路等級 與編碼，如： "台3" 、"縣168" 、"市1" 等。
RoadNUM2 道路編號 2 文字 8
若同時有三種道路等級發生共線，於此 欄位儲存第三個所屬之省道、縣道、鄉 道、市區道路、產業道路等道路等級與 編碼，如： "台3" 、"縣187" 、"市1" 等。
RoadName 道路名稱 文字 20
此欄位儲存路段所屬國道、省道、縣 道、鄉道、市區道路、產業道路等道路 名稱，至於圓環則該圓環名稱。
RoadAlias 道路別名 文字 20
除上述道路名稱外，若道路有
其他一般公認之名稱，皆可存放於此欄 位。
RdNameSect 段名 文字 8 該道路段名，例如：中山路一段，則本 欄為填寫 ”一段 ”。
BriunName 橋樑名、隧道名 文字 20 儲存各座橋樑、隧道之名稱。
RdNameLane 巷 文字 20 此欄位儲存路段所屬巷名，如： "新光 巷"、"19 巷"等。
RdNameNon 弄 文字 16 此欄位儲存路段所屬弄名，如： "1 弄"、
"2 弄"等。
RoadWidth 路寬 數字 4 路段實際寬度（公尺，至整數）
RoadNO 車道數 數字 1 路段車道數
（六）水系
河流 River
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與河流空間資料檔之多邊形序號對應
小河 Stream
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與小河空間資料檔之線段序號對應
TerrainID 小河地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
水池湖泊 Lake
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 多邊形序號 數字 10 與水池湖泊空間資料檔之多邊形序號 對應
TerrainID 水 池 湖 泊 地 形 編 碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
LakeName 水池湖泊名稱 文字 30
Lake_A 水池湖泊面積 數字 10 面積單位平方公尺（至整數） 4. 流域中線 MidRiver
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與流域中線空間資料檔之線段序號對 應
TerrainID 河流地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
RiverLCode 河流等級 文字 8 分中央管河川、縣管河川
RiverLName 河流名稱 文字 30
（七）公共事業網路 Tower
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 點序號 數字 10 與公共事業網路資料檔之點序號對應
TerrainID 線塔地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
（八）地貌
等高線 Contour
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與等高線資料檔之線段序號對應
TerrainID 等高線地形編碼 文字 8 依據「基本地形資料分類編碼表」進行 分類編碼
Height 高程 數字 4 等高線高程值（公尺，至整數）
獨立標高點 Spot
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 點序號 數字 10 與獨立標高點資料檔之點序號對應
Height 高程 數字 4 獨立標高點高程值（公尺，至整數）
（九）國有林界
國有林事業區界 AdminForest
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與國有林事業區界資料檔之線段序號 對應
RegionNAME 事業區名稱 文字 30
國有林班界 ForestSub
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
ID 線段序號 數字 10 與國有林班界資料檔之線段序號對應
RegionNAME 林班界名稱 文字 30
（十）圖幅 FrameIndex
欄位名稱 (英文 ) 欄位名稱 (中文 ) 欄位型態 長度 內容說明
MapID 圖幅編號 文字 8
MapName 圖幅名稱 文字 20
PhotoDate 攝影日期 文字 12
PlotDate 測製日期 文字 12 附錄 附錄 附錄 附錄 8：：：：抽樣檢查計畫表 抽樣檢查計畫表 抽樣檢查計畫表 抽樣檢查計畫表（ （（（摘錄自 摘錄自 摘錄自 摘錄自 ISO 2859.1-1999 表表表表） ）））
表 1 抽樣檢查樣本代字表
一般檢查水準
批量
Ｉ Ⅱ Ⅲ
2-8 9-15 16-25 26-50 51-90 91-150 151-280 281-500 501-1,200 1,201-3,200 3,201-10,000 10,001-35,000 35,001-150,000 150,001-500,000 500,001 以上
Ａ
Ａ
Ｂ
Ｃ
Ｃ
Ｄ
Ｅ
Ｆ
Ｇ
Ｈ
Ｊ
Ｋ
Ｌ
Ｍ
Ｎ
Ａ
Ｂ
Ｃ
Ｄ
Ｅ
Ｆ
Ｇ
Ｈ
Ｊ
Ｋ
Ｌ
Ｍ
Ｎ
Ｐ
Ｑ
Ｂ
Ｃ
Ｄ
Ｅ
Ｆ
Ｇ
Ｈ
Ｊ
Ｋ
Ｌ
Ｍ
Ｎ
Ｐ
Ｑ
Ｒ表 2 單次抽樣計畫表
允收品質水準（ ＡＱＬ）
0.65 1.0 1.5 2.5 4.0 6.5
樣
本
代
字
樣
本
大
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│ 採用箭頭下第一個抽樣計畫 AC= 允收數
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│ 採用箭頭上第一個抽樣計畫 RE= 拒收數 表 3 雙次抽樣計畫表
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0.65 1.0 1.5 2.5 4.0 6.5
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＋ 採用單次抽樣計畫 （或採用下面的雙次抽樣計畫） 附錄 附錄 附錄 附錄 9：：：：英文縮寫名詞定義 英文縮寫名詞定義 英文縮寫名詞定義 英文縮寫名詞定義
地理資訊系統 地理資訊系統 地理資訊系統 地理資訊系統（ （（（Geographic Information Systems ；；；；GIS ）：）：）：）：可 可可可獲取、儲存、分析，並顯示各種 形式地理參考資訊之電腦系統。 1
衛星定位測量或稱全球定位系統 衛星定位測量或稱全球定位系統 衛星定位測量或稱全球定位系統 衛星定位測量或稱全球定位系統 (Global Positioning System ；；；；GPS) ：：：：是一套以衛星訊號為基 礎的導航系統，具有全球性、全天候的精密三維導航與定位能力，由美國國防部為了軍事上 的需求，滿足海上、陸地和空中軍事應用進行高精度定位和導航所建立的系統。全球定位系 統定位的過程，基本上是距離的量測，藉由接收 GPS 衛星所發射的電磁波訊號，量測地面接 收儀與衛星之間的瞬時距離，利用觀測至少 4 顆衛星所得到的瞬時距離，再配合幾何原理求 解地面接收儀之三維坐標。
調制轉換函數 調制轉換函數 調制轉換函數 調制轉換函數（ （（（Modulation Transfer Function ；；；；MTF ）：）：）：）：指攝影成像系統之清晰度或描繪此 系統任何元件（如鏡頭感測器等）特性之參數。在本規範中特指拍攝所得影像的解析力。其 值可由布設地面解析力檢定標而計算得到。
數值地形模型 數值地形模型 數值地形模型 數值地形模型（ （（（Digital Terrain Model ；；；；DTM ）：）：）：）：本規範所稱數值地形模型涵蓋二種內容，第 一種是數值高程模型 (Digital Elevation Model ；DEM) ，第二種是數值表面模型 (Digital Surface Model ，簡稱 DSM) 。
數值高程模型 數值高程模型 數值高程模型 數值高程模型 (Digital Elevation Model ；；；；DEM) ：：：：描述不含植被及人工人工構造物之地表天然 面高程起伏的數值模型，此模型為二維半（ 2.5D ），並以規則網格式離散點所組成。為了便於 資料的管理及後續應用，在此定義離散點是等間距的方格點。
數值 數值 數值 數值表 表表表面模型 面模型 面模型 面模型 (Digital Surface Model ；；；；DSM) ：：：：表示地球表面可見光無法穿透的最上層表面的 二維半（ 2.5D ）模型，它與數值高程模型不同之處乃在表示了人工構造物及長年生植被的最 上層表面。此模型亦是以規則網格式離散點所組成，在此同樣定義離散點是等間距的方格點。
慣性測量元件 慣性測量元件 慣性測量元件 慣性測量元件（ （（（Inertial Measurement Unit ；；；；IMU ）：）：）：）：裝置於載具上的一種電子儀器，用以量 測及提供載具加速度及姿態角資訊。
允收品質水準 允收品質水準 允收品質水準 允收品質水準（ （（（Acceptance Quality Level ；；；；AQL ）：）：）：）：所能接受送驗批量最高不合格率。 附錄 附錄 附錄 附錄 10 ：：：： 圖幅整飾規格 圖幅整飾規格 圖幅整飾規格 圖幅整飾規格
項目 註記內容
圖廓 依經緯度分幅，經差、緯差均為 1’30” 之整數。
依 98 年 11 月 30 日 97 年基本圖第 13 次工作會議決議，略以： 「因九二 一震災後，基本圖的經緯度坐標製作即依取整數秒方式，而非被一分 三十秒整除分幅方式製作。為了災後一致性，本建置案的經緯度坐標 採取整數秒方式製作」 。
圖廓註記 註記於圖廓四隅，緯度註記「××°×× ’×× ”N 」於圖隅之左、右側，經度 註記「××°×× ’×× ”E 」於圖隅之上、下側。字體採等線體，字高 2mm ，距圖廓線 2mm 。
方格線 於圖幅內以每 10 公分（ 500m 之整數）之間隔繪製。
方格線註記 方格線之縱坐標註記於圖廓之左、右側，橫坐標註記於上、下側。字 體採等線體，字高 3mm ，距圖廓線 2mm ，左下角第一條橫線註「×××× ×00mN 」，第一條縱線註「×××× 00mE 」，字體採等線體，字高 1.5mm ，其餘方格線不註記 N、E，僅註記坐標與單位。
圖名 位於圖廓線上方正中，字體採等線體，字高 1cm ，間隔 1cm ，下緣距圖 廓線 1.2cm 。
圖號 位於圖廓線右上方，字體採等線體，字高 5.0mm ，右緣與右圖廓線齊， 下緣距圖廓線 1.7cm 。
比例尺 位於圖廓線下方正中，繪 12cm （現地 600 公尺）長之圖示比例尺，最 下範圍距下圖廓線 3cm 。其上方以 3mm 字高之等線體註記「比例尺： 五千分之一」 ,上緣距圖廓線 1.2cm 。
高程起算註記 註記「高程：標高自臺灣基隆平均海水面為零公尺起算」 。
方格線說明 「方格線：橫麥卡脫投影坐標系統 500 公尺方格」
地圖投影坐標系 註記「投影：橫麥卡脫投影，經差二度分帶，中央子午線東經 121 °E」。
大地基準點及地球 原子
註記 「坐標系統：採用內政部一 0 一年公布之 「一九九七坐標系統 2010
年成果」 (Taiwan Datum 1997, 簡稱 TWD97) 圖廓外斜體紅色 坐標註記係 TWD67 參考坐標」 。
換行註記「地球原子：採用 1980 年國際大地測量學及地球物理學聯合 會(IUGG) 公佈之參考橢球體 (GRS80) 」。
等高線間隔 註記「等高線間隔：首曲線 5 公尺」 。
面積說明 註記 「面積：本圖幅涵蓋地面 (包括水面 )面積為○○○ .○○○○公頃。」
主管機關 說明主管該幅地形圖之機關名稱，註記「主管機關： xxxx 」
主辦機關 說明主辦該幅地形圖製作之機關名稱，註記「主辦機關： xxxx 」
測製機關 說明測製該幅地形圖之機關名稱，註記「測圖機關： xxxx 」
測製時間 該幅地形圖測製之時間，註記「測製時間：民國○○○年○○月」 。
並條列 ○○○年○○月 (航空攝影 )
○○○年○○月 (野外調查 )
○○○年○○月 (測製 ) |
12897 | https://www.shiyue.li/mathcamp/shiyueli-mathcamp-rpec.pdf | Lecture Notes on Rational Points on Elliptic Curves Shiyue Li Mathcamp 2018 Acknowledgment: Over the course of Rational Points on Elliptic Curves class (Week 4) in Canada/USA Mathcamp 2018, these notes are improved and completed via conver-sations with Mira, Aaron, students in the class, and other Mathcamp staff. The notes are based on a very nice treatment of rational points on elliptic curves in [ST15].
Contents 1 Cubic Curves 2 1.1 Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.2 The Weierstrass Normal Form, Singular and Non-Singular Curves . . . . .
3 1.3 The Group Law on Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . .
4 1.4 Exercises for Day 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2 A Hike to Mordell’s Theorem 7 2.1 Mordell’s Theorem for Curves with a Rational Point of Order Two . . . . .
8 2.2 Heights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9 2.3 Descent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 2.4 Exercises for Day 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 2.5 Homomorphisms of Mordell Weil Groups . . . . . . . . . . . . . . . . . . .
13 2.6 Exercises for Day 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 3 Torsions 19 References 23 1 1 Cubic Curves 1.1 Diophantine Equations The theory of Diophantine equations is a branch of number theory that finds integral or rational solution of polynomial equations.
Some natural questions that mathematicians want to ask are: • Are there integral solutions to these equations? If so, how many are there?
• Are there rational solutions to these equations? If so, how many are there?
• If any integral or rational solution exists, can we express it in terms of the coeffi-cients of the equation?
Example 1.1. The vanilla flavor of Diophantine equations are in the form of anxn + an−1xn−1 + · · · + a1x + a0 = 0, where ai’s are integers. By Rational Root Theorem, if p q is a rational solution of the equation above, we have q divides an and p divides a0. This gives us list of candidates to plug in and check.
For Diophantine equations in two variables, things can be more complicated.
Example 1.2. Consider the linear equation in two variables ax + by = c for a, b, c ∈Z.
If gcd(a, b) divides c. We consider the set of all positive numbers of the form S = {ax + by : x, y ∈Z} which by well-ordering principle, has a smallest element d. We can show by Euclidean Algorithm that d is the gcd(a, b) and d divides all s ∈S. Therefore, we have infinitely many integral solutions if gcd(a, b) divides c; otherwise, we have no integral solutions.
There are infinitely many rational solutions to this equation once we see the equation as y = −ax b + c, since every rational number x will give a rational y.
What about Diophantine equations of degree 2?
Example 1.3. Let us find all rational solutions to the equation x2 + y2 = 1.
We need to convince ourselves that if a line ℓpasses through a rational point P on the circle C and has rational slope t, then ℓmust meet the circle C again at another rational point Q.
2 Pick a point P = (1, 0) and let t ∈Q be the slope of the line ℓ: y = tx + 1 that passes through P. Solving the system of equations ( y = tx + 1, x2 + y2 = 1, we get x = −2t 1+t2, and y = 1−t2 1+t2. As t varies over different rational numbers, x, y will take in different rational values.
Definition 1.4. The set of all real solutions to an equation f (x, y) = 0 gives a curve C in R2 called algebraic curve. Algebraic curves of degree 2 are called conics and algebraic curves of degree 3 are called cubics or cubic curves.
Remark 1.5.
• Now algebraic expressions f (x, y) and geometric objects in the plane are in correspondence. Throughout the course, we will use the language – “let C be a curve f (x, y)” or “we have a curve C : f (x, y)” – to indicate that C contains the set of real points in R2 whose coordinates (x, y) satisfies the equation f (x, y) = 0.
• Elliptic curves do not have much to do with ellipses (a family of conic sections that are given by quadratic equations). They are named so because of their first appearance in calculation of arc length of ellipses.
This course will walk you through this beautiful subject of cubic curves, and we will primarily focus on finding the rational points on cubic curves. In the Example above, we see that given a rational point P on a cubic curve, we can draw the tangent line at P, and take the third point of the intersection of the line with the cubic, thereby finding more rational points. This “geometric operation” later on gives an abelian group structure to the set of all rational points on a cubic curve, which is the substance of the celebrated Mordell’s Theorem.
1.2 The Weierstrass Normal Form, Singular and Non-Singular Curves For every cubic curve: C : ax3 + bx2y + cxy2 + dy3 + ex2 + f xy + gy2 + hx + iy + j = 0, for all a, b, c, d, e, f, g, h, i, j in Q. we can use projective geometry to show that there is a simpler-looking curve in Weierstrass normal form: C′ : y2 = f (x) = x3 + a1x2 + a2x + a3 such that C and C′ are birationally equivalent. We won’t elaborate what birational equivalence means, but we know that it is an equivalence relation on the set of all cubic curves. So we only need to study representatives of each equivalence class that look as simple as Weierstrass normal form. Another nice thing about the normal form is that, the curve is symmetric around x-axis.
3 Question 1.6. What do we know about real roots of f (x)?
Analyzing the roots of f (x), we have the following situations: (1) If f (x) has one real root α, C : f (x) = (x −α)(x2 + βx + γ) = 0 for some real number α, β, γ ∈R. C has one component.
(2) If f (x) has three real roots, C = f (x) = (x −α)(x −β)(x −γ) for α, β, γ ∈R.
• If all three of the real roots are distinct, C has two components.
• If two of the real roots are the same, say α = β without loss of generality, then in order to fully understand the geometry of C, we inspect the tangent line of C at each point of C. Let us rewrite y2 −f (x) = F(x, y) and the partial derivatives with respect to x, y are ∂F(x, y) ∂x = −f ′(x) and ∂F(x, y) ∂y = 2y.
Since f (x) = (x −α)2(x −γ) = y2, then y = 0 at x = α and −f ′(α) = −(α −α)2 −2(α −α)(α −γ) = 0, which implies that the partial derivatives are 0. This means that C has an singularity at (α, 0) and we say that C is a singular curve with distinct tangent directions or a self-intersection.
• If all three roots are the same, then by the same reasoning as above, C is a singular curve with a cusp.
Mordell’s Theorem will not hold for these singular curves, and it is not a pity since finding rational points on a singular curve is in fact easy. To find rational points on a singular curve is the same as conics. We can project the curve to a line from the singular point, and a projecting line will only path through one other point on the curve. Hence we have a one-to-one correspondence from rational points on a singular curve to a rational points on a line; see Exercises for Day 1 for examples.
Thus we are only interested in non-singular cubic curves and they have a special name.
Definition 1.7. If a cubic curve C : y2 = f (x) = x3 + ax2 + bx + c has distinct complex roots, then we call this curve an elliptic curve.
1.3 The Group Law on Elliptic Curves In fact, all the points on an elliptic curve form an abelian group. Before we do this, let us admit a few facts from projective geometry that we won’t elaborate here (talk to me at TAU if you are interested).
Assumption 1.8.
(1) The point at infinity O is a rational point.
4 (2) Our non-singular curve C contains all the points on the xy-plane and the point at infinity O.
(3) Every line meets the cubic in three points: • The line at infinity meets the cubic at O three times.
• A vertical line meets the cubic at two points in the xy plane and also at the point O.
• Any non-vertical line meets the cubic in three points in the xy plane. But we have to be careful here.
Now we can define the group structure.
• Addition: for any P, Q ∈C, draw a line through P and Q. The line will intersect with the cubic at a third intersection. Denote the third intersection P ∗Q. Draw another line through P ∗Q and O and the line will intersect with the cubic at the point O ∗(P ∗Q). This is P + Q; or more precisely, we have P + Q = O ∗(P ∗Q).
• From this formula, we see that for any P ∈C, O + P = P + O = O ∗(O ∗P) = O ∗(P ∗O) = P.
So O is the identity of the group.
• What is the negative of a point P on C?
• Associativity of this group law will also hold.
• Addition is abelian.
Therefore, we have obtained a group structure on the points on an elliptic curve.
5 1.4 Exercises for Day 1 The projection of a singular cubic curve onto a line is one-to-one. Hence the rational points on a singular cubic can be put in a one-to-one correspondence with rationals on the line. The following two exercises are two examples.
Exercise 1.9. Find all the rational points of the singular cubic curve C(y2 = x3) by parametrizing all rational x and y using a rational t.
Exercise 1.10. Find all the rational points on the singular cubic curve C(y2 = x2(x + 1)) by letting r = y x for (y, x) ̸= (0, 0).
The following exercises help you understand the group law on elliptic curves!
Exercise 1.11. Show that the group law is associative: for any P, Q, R ∈C, (P + Q) + R = P + (Q + R).
Exercise 1.12. If P and Q are distinct rational points in the xy-plane, prove that the line conncecting them is a rational line.
Exercise 1.13. Let C be a non-singular cubic curve. Let C(Q), C(R), C(C) be the rational points, real points, complex points on C. Show that each of C(Q), C(R), C(C) is an additive group under the addition defined in the group law.
This exercise will give you a formula called “Duplication Formula”, which will be an important tool later.
Exercise 1.14. Compute an explicit expression for the coordinates of 2P in terms of the coordinates for P on C : y2 = x3 + ax2 + bx + c.
Exercise 1.15. Let C : y2 = x3 + ax2 + bx + c be a non-singular cubic curve. Let P = (x, y) be a point on C. Find a polynomial in x whose roots are the x coordinates of the points Q = (x, y) satisfying 3Q = O. 1 Exercise 1.16. Consider the point P = (3, 8) on the cubic curve y2 = x3 −43x + 166.
Compute P, 2P, 3P, 4P and 8P. What do you find?
This exercise shows you an example of birational transformation.
Exercise 1.17. If u, v satisfies the relation u3 + v3 = α, then the quantities x = 12α u+v, y = 36α u−v u+v satisfies y2 = x3 −432α2. This gives a birational transformation from the curve u3 + v3 = α to the curve y2 = x3 −432α2 1 Hint: The relation 3Q = O can be translated into 2Q = −Q.
6 2 A Hike to Mordell’s Theorem Let’s wander where the WiFi is weak Anonymous Throughout, we assume our non-singular cubic curve C is given by a Weierstrass form: y2 = x3 + ax2 + bx + c, where a, b, c are rationals. Recall that the “non-singularity” condition translates into the case that the elliptic curve C : y2 = f (x) crosses x-axis at exactly one or three distinct points; in other words, f (x) has three distinct roots (real or complex). Recall that the points on C have a group structure and O is the identity element of the group.
Yesterday, we have seen that C ∪{O} is an abelian group.
Definition 2.1. An element P of any group element is said to have order n if nP = P + P + · · · + P | {z } n times = O and mP ̸= O for all 1 ≤m < n. If n < ∞, we say that P has finite order; if n = ∞, P has infinite order.
To study furthermore about the structure of points on C, we can ask ourselves the following questions.
• Are there points of finite orders?
• Are there points of order 2?
• Are there points of order 3?
To answer the first question, we want to find P ∈C such that 2P = O and P ̸= O, which is the same as P = −P. These are the points on C that are lying on the x-axis or floating in the vast space of C that we cannot draw on xy-plane. But if we include these complex roots, we get 4 points of order 2: P1 = (α1, 0), P2 = (α2, 0), P3(α3, 0), where α1, α2, α3 are real or complex.
Question 2.2. What is the relation between the three points plus the point O at infinity?
Proposition 2.3. The set of points π2 of order 2 on C : y2 = f (x) can be described as follows.
(i) If we consider π2 ⊆C(C), then π2 ∼ = Z2 × Z2, the Klein Four Group.
7 (ii) If we consider π2 ⊆C(R), then π2 ∼ = Z2 (if f (x) has 1 real root) or π2 ∼ = Z2 × Z2 (if f (x) has 3 real roots).
(iii) If we consider π2 ⊆C(Q), then π2 ∼ = Z2 (if f (x) has 1 real root) or π2 ∼ = Z2 × Z2 (if f (x) has 3 real roots), or π2 = {O} is trivial.
To answer the second question (cf. Exercises for Day 1), we want to find points P such that 3P = O but P ̸= O and 2P ̸= O. If such point P = (x, y) exists, then 2P = (x′, y′) = −P implies that x = x′.
Using the “Duplication Formula” that you found out in Exercises in Day 1 x′ = x4 −2bx2 −8cx + b2 −4bc 4x3 + 4ax2 + 4bx + 4c .
Then x is a root of the polynomial p(x) = 3x4 + 4ax3 + 6bx2 + 12cx + (4ac −b2).
From these calculation you can see that the computation of the Mordell-Weil group could be a total mess.
2.1 Mordell’s Theorem for Curves with a Rational Point of Order Two In this class, we state Mordell’s Theorem for non-singular curves with a rational point of order two and see a bunch of examples that will illustrate the ideas.
Theorem 2.4. Let C be a non-singular cubic curve given by an equation C : y2 = x3 + ax2 + bx, where a and b are integers and C has a rational point of order two. Then the group of rational points, or Mordell-Weil group, C(Q) is a finitely generated abelian group.
Theorem 2.5 (Structure Theorem for Finitely Generatedy Abelian Groups). A finitely generated abelian group G is isomorphic to Z × Z × · · · × Z | {z } r is called rank ×Zp α1 1 × Zpα2 2 × · · · × Zpαn n , where each pi is a prime and each αi is a positive integer.
Mordell’s Theorem gives us some hope in fully describing the rational points on elliptic curves.
Example 2.6. For C : y2 = x3 −5x, C(Q) has rank 1.
Example 2.7. For C : y2 = x3 + x and C : y2 + 4x, C(Q) are finite.
8 Example 2.8. For C : y2 = x3 + px where p ≡7, 11 (mod 16), C(Q) is finite.
In this section, we prove Mordell’s Theorem that the group of rational points on a non-singular cubic is finitely generated for curves that have a rational points of order 2 to start with. To do this, we need to define some useful tools to help us understand rational points on this cubics.
Throughout, we work with non-singular cubic curves C : y2 = x3 + ax2 + bx + c for a, b, c ∈Z.
2.2 Heights Some rational points are complicated, some of them are not.
Definition 2.9. Let x = p q be a rational number written in reduced terms. Then the height H(x) is the maximum of the absolute values of the numerator and the denomi-nator. That is, H(x) = H p q = max{|p|, |q|}.
Question 2.10. Given positive integer m, how many rationals x are there such that H(x) ≤m?
Definition 2.11. For a rational point P = (x, y) on a curve C, we define the height of P to be the height of its x-coordinate. For calculation convenience, we define “small height” h(P) to be log(H(P)).
Note that h(P) is a non-negative real number, so we are really on a hike.
Lemma 2.12. Given a non-negative real number m, the set {P ∈C(Q) : h(P) ≤m} is finite.
Lemma 2.13. Fix P0 on C. For all P ∈C(Q), there exists a constant k0 that only depends on P0, a, b, c such that h(P + P0) ≤2h(P) + k0.
Lemma 2.14. For all P ∈C(Q), there exists a constant k, that only depends on a, b, c such that h(2P) ≥4h(P) −k.
Remark 2.15. The lemmas give us a tool to connect group law and height, which is a number theoretic tool. Lemma 2.13 tells us, if you start at a rational point and add another rational point, the height of the result is under control. Lemma 2.14 tells us, if you keeps doubling a point, the height will blow up eventually.
Lemma 2.16. The index [C(Q) : 2C(Q)] is finite.
9 2.3 Descent In this section, we will outline the proof of Mordell’s Theorem using the four lemmas that we presented.
Lemma 4 told us that the index of 2C(Q) as a subgroup of C(Q) is finite. Take a representative for each of the finitely many cosets of 2C(Q) in C(Q), and call them Q1, . . . , Qn.
For each element P ∈C(Q), P is in the coset of some Qi1. That is, P −Qi1 ∈2Γ ⇐ ⇒P −Qi1 = 2P1.
Iteratively, we have P −Qi1 = 2P1 P1 −Qi2 = 2P2 P2 −Qi3 = 2P3 P3 −Qi4 = 2P4 · · · .
For each Pj in the sequence P1, P2, P3 . . ., we have that h(Pj −Qi) ≤2h(Pj) + ki for all 1 ≤i ≤n. We can let k be the maximum of all ki’s such that h(Pj −Qi) ≤2h(Pj) + ki.
Using Lemma 3, we have 4h(Pj) ≤h(2Pj) + k′ = h(Pj−1 −Qij) + k′ ≤2h(Pj−1) + k + k′.
Thus we know that h(Pj) ≤1 2h(Pj−1) + 1 4(k + k′).
There exists some integer m such that 1 2h(Pm−1) ≤3 4(k + k′), which gives h(Pm) ≤k + k′.
Therefore, P = Qi1 + 2Qi2 + 4Qi3 + · · · + 2m−1Qim + 2mPm. Since h(Pm) is bounded and n is finite, P is generated by {Q1, . . . , Qn} and points R ∈C(Q) such that h(R) ≤k + k′, whose union is a finite set.
This method is called the method of infinite descent, originated from Fermat. He used this to show that x4 + y4 = 1 has no rational solutions with xy ̸= 0. Maybe this was his idea to show that xn + yn = 1 for any n ≥3. But the margin was apparently too small for an infinite descent. Who knows.
10 2.4 Exercises for Day 2 The exercises today will walk you through proving some key steps of the lemmas that we used in class. Throughout, let C : y2 = f (x) = x3 + ax2 + bx + c where a, b, c are integers.
Exercise 2.17. Convince yourself that Lemma 2.12 is true.
Exercise 2.18. To prove Lemma 2.13, we want to study the relationship between P, P0, and P + P0 for P, P0 ∈C(Q). First we notice that the lemma is trivial if P0 = O, and so we let P0 = (x0, y0) ̸= O. Let P = (x, y) ∈C(Q), and we can write x = m M and y = n N such that they are in reduced form with M > 0 and N > 0. We will show that M3 = N2.
(i) Use the fact that P ∈C to deduce that N2 divides M3.
(ii) Show that M divides N2m3 and deduce that M divides N2.
(iii) Show that M2 must divides N2m3, and deduce that M must divide N.
(iv) Show that M3 divides N2m3, so M3 must divide N2.
(v) Now we have N2|M3 and M3|N2 so M3 = N2. Since we showed that M|N, we can let r = N M. Deduce the following equations.
r2 = N2 M2 = M3 M2 = M and r3 = N3 M3 Therefore, we can have x = m r2 and y = n r3.
(vi) Substitute the new expressions for x and y back into C : y2 = f (x) and clear the denominator. Use triangle inequality to show that |n| ≤ q 1 + |a| + |b| + |c|H(P).
That is, we have bounded the numerator of the y-coordinate of P in terms of H(P).
11 (vii) Now we are ready to prove the lemma statement. Let P + P0 = (s, t), Express s in terms of coordinates of P, P0 and a, b, c and when you get some term involving y2 −x3, replace it with ax2 + bx + c. Eventually you get something like s = Ay + Bx2 + Cx + D Ex2 + Fx + G where all the coefficients are integers and are in terms of a, b, c, x0, y0 (that’s good!) (viii) Substituting x = m r2, and y = n r3 into the equation and clearing the denominator, conclude that H(P + P0) = H(s) ≤max(|AK| + |B| + |C| + |D|, |E| + |F| + |G|)H(P)2.
(ix) Taking the log will give us the desired result.
12 2.5 Homomorphisms of Mordell Weil Groups The definition of a good mathematical problem is the mathematics it generates rather than the problem itself.
Andrew Wiles In Exercises for Day 2, we proved Lemma 2.12 and Lemma 2.13. The proofs of Lemma 2.14 use similar ideas as Lemma 2.13 and are rather too technical to be fun.
Today we will outline a proof of Lemma 2.16, which states: The index [C(Q) : 2C(Q)] is finite.
To study 2C(Q), we want to see how any point P ∈C as mapped to 2P. Recall the “Duplication Formula” that you found out in Exercises in Day 1 P = (x, y) 7→2P x 7→x4 −2bx2 −8cx + b2 −4bc 4x3 + 4ax2 + 4bx + 4c .
This map has degree 4. The strategy of the proof is the following.
(i) We construct a degree-2 map φ from C to C and a degree -2 map φ from C to C.
(ii) We show that C and C are isomorphic as cubic curves, hence having the isomor-phic C(Q).
(iii) The composition of φ ◦φ is a degree 4 map and induces a group homomorphism C(Q) →2C(Q).
(iv) Use a lemma to show Lemma 4.
Lemma 2.19. Let A and B be abelian groups, and consider two homomorphisms φ : A →B and φ : B →A. Suppose that for all a ∈A, and b ∈B φ ◦φ(a) = 2a and φ ◦φ(b) = 2b.
Suppose that [B : φ(A)] is finite, [A : φ(B)] is finite, then [A : 2A] ≤[A : φ(B)][B : φ(A)].
Example 2.20. Let A = Z × Z and B = Z × 2Z. Let φ : A →B be (a, b) →(a, 2b) and φ : B →A be (a, 2b) 7→(2a, 2b). Then φ ◦φ((a, b)) = (2a, 2b) and φ ◦φ((a, 2b)) 7→(2a, 4b).
Notice that [Z × 2Z : Z × 2Z] = 1 and [Z × Z : 2Z × 2Z] = 4, and we can check that 4 = [Z × 2Z : 2Z × 2Z] ≤[Z × Z : 2Z × 2Z][Z × Z : 2Z × 2Z] = 4.
13 Proof. Similar to the proof of Mordell’s Theorem, we will take advantage of the fact that [A : φ(B)] and [B : φ(A)] are finite.
Since φ(B) has finite index in A, we can find elements a1, . . . , an representing the finitely many cosets of φ(B) in A. Since φ(A) has finite index in B, we can find elements b1, . . . , bm to represent the finitely many cosets of φ(A) in B.
Recall our goal is to show that the number of cosets of 2A in A is finite. This amounts to showing that for any arbitrary a ∈A, a can be written as the sum of some element that can be chosen from only finitely many elements in A and some element in 2A.
Since φ(B) is finite-index subgroup of A, there exists a coset representative ai such that a −ai ∈φ(B) ⇐ ⇒a −ai = φ(b), for some b ∈B. Since φ(A) has finite index in B, there exists a coset representative bj such that b −bj ∈φ(A) ⇐ ⇒b −bj = φ(a′). Therefore, a = ai + φ(b) = ai + φ(φ(a′) + bj) = ai + φ(bj) + 2a′.
Since the options for the combination of ai and φ(bj) are finite, 2A is indeed a finite index subgroup in A.
To use this Lemma 2.19, we need to construct φ : C(Q) →C(Q) and φ : C(Q) → C(Q) such that (1) φ ◦φ and φ ◦φ are duplication maps.
(2) [C(Q) : φ(C(Q))] is finite.
(3) [C(Q) : φ(C(Q))] is finite.
Now we start our really long journey of proving the last statements. First we recall that our C : y2 = f (x) = x3 + ax2 + bx + c, where a, b, c ∈Z is a non-singular curve with a rational point of order 2. This means that f (x) have a rational root α. Since f (x) is a polynomial with integer coefficients and leading coefficient 1, α must be an integer, by Rational Root Theorem. Making a change of coordinates, we can move (α, 0) to origin, and the curve of interst is now in the form of C : y2 = f (x) = x3 + ax2 + bx where a, b ∈Z.
For Step (1), we define a map between curves first. For C : y2 = x3 + ax2 + bx, define a “conjugate curve” C : y2 = x3 + (−2a)x2 + (a2 −4b)x.
Let φ : C →C be defined by φ(P) = ( y2 x2, y(x2−b) x2 , P ̸= O, (0, 0), O P = O or (0, 0).
14 The kernel of φ is {O, (0, 0)}. Let φ : C →C be defined by φ(P) = ( y2 4x2, y(x2−b) 8x2 , P ̸= O, (0, 0), O P = O or (0, 0).
The kernel of φ is {O, (0, 0)}. One can check that (i) φ, φ are group homomorphisms.
(ii) φ ◦φ ande φ ◦φ are duplication maps.
(iii) φC(Q) and φC(Q) is a homomorphism between C(Q) and C(Q).
To prove Lemma 2.16, we need to show that • [C(Q) : φ(C(Q))] is finite.
• [C(Q) : φ(C(Q))] is finite.
It suffices to show one of these since the other is given by the same map.
We proceed by proving the second one: [C(Q) : φ(C(Q))] < ∞.
Showing this amounts to showing that there is an injective homomorphism from the quotient group C(Q)/φ(C(Q)) into a finite group.
Recall our notation from last time, Q∗= {multiplicative group of Q}, and (Q∗)2 = {x2 : x ∈Q∗}.
Define the map α : C(Q) →Q∗/Q∗2 by α(O) = 1 (mod Q∗2) α(ξ) = b (mod Q∗2) α(x, y) = x (mod Q∗2), x ̸= 0.
It is not obvious that α is a group homomorphism, but you will show this in the homework.
Proposition 2.21. The kernel of α is image of φ(C(Q)).
15 Proof. The kernel of α is the set of elements in C(Q) that get sent to Q∗2. The definition of α is as follows.
α(O) = 1 (mod Q∗2) α(ξ) = b (mod Q∗2) α(x, y) = x (mod Q∗2), x ̸= 0.
We describe points in the image of φ.
(i) O ∈φ(C(Q)) by definition of φ.
(ii) ξ = (0, 0) = ( y2 4x2, y(x2−b) 8x2 ) ∈φ(C(Q)) if and only if the polynomial f (x) has a nonzero rational solution. This implies that in the factorization of f (x) = x(x2 + ax + b), x2 + ax + b has a rational solution, and thus −a ± p a2 −4b 2 is rational. This holds if and only if a2 −4b is a square of a rational number if and only if 4a2 −4(a2 −4b) = 16b ∈Q∗2 ⇐ ⇒b ∈Q∗2.
(iii) P = (x, y) ∈φ(C(Q)) with x ̸= 0 if and only if x is a square square of a rational number.
Proposition 2.22. The image of α is contained in the subgroup Q∗/Q∗2 consisting of the ele-ments {±pβ1 1 pβ2 2 · · · pβt t : βi = 0 or 1, pi divides b}, Proof. Recall in Exercises for Day 1, we know that for every point P = (x, y) ∈C(Q) with x ̸= 0, we can write x = m r2, and y = n r3 in their reduced form where r = N M in reduced form and where M, N are the original denominators for x, y in reduced form. You also remember in Exercises in Day 1 that if you plug in the point (x, y) in the polynomial equation of C and clear the denominator, you get n2 = m3 + am2r2 + bmr4 = m(m2 + amr2 + br4).
Consider the divisors of m, they either divide m2 + amr2 + br4 or don’t divide m2 + amr2 + br4. Those divisors of m that divide m2 + amr2 + br4 must divide b since m and 16 r are coprime. Those divisors of m that do not divide m2 + amr2 + br4 must have even power in the factorization of m. Therefore, α(P) = x = m r2 ≡m ≡±pβ1 1 pβ2 2 · · · pβt t (mod Q∗2) where pi are divisors of b, and βi = 0 or 1 for all i.
If P = (0, 0), then α(T) = b = ±pβ1 1 pβ2 2 · · · pβt t (mod Q∗2) where pi are divisors of b, and βi = 0 or 1 for all i.
Since Lemma 2.16 is true, we have completed the proof of Mordell’s Theorem!!
2.6 Exercises for Day 4 Exercise 2.23.
These exercises will walk you through the proof of α is a group homomorphism from C(Q) to Q∗/Q∗2.
Exercise 2.24. Show that for any point P = (x, y) ∈C(Q), α(−P) = α(P)−1 (mod Q∗2).
Exercise 2.25. Show that if y = rx + s is a line that passes through C in three general points (not including the infinity P), the x-coordinate of the three points x1, x2, x3 satisfy a cubic equation: x3 + Ax2 + Bx + C = 0 . What are A, B, C in terms of r, s, a, b, c?
Exercise 2.26. Using the last exercise, what can you say about x1x2x3? Deduce that if P1 + P2 + P3 = O, then α(P1)α(P2)α(P3) ≡1 (mod Q∗2). This combines with the first exercise shows you that α is indeed a group homomorphism.
Before we compute a concrete example, let us think about an algorithm to compute the rank of Mordell-Weil group.
Exercise 2.27. Recall that the structure theorem for finitely generated abelian group and Mordell’s Theorem combined tell us that C(Q) ∼ = Zr ⊕Zp1α1 ⊕Zp2α2 ⊕· · · ⊕Zpnαn where pi are primes, αi are positive integers. Write out C(Q)/2C(Q). What is [C(Q) : 2C(Q)] (a number), depending on pi for all i?
Exercise 2.28. Consider the set of elements P such that 2P = O in C(Q) (Caution! These are not just elements of order 2). Denote this group as H. How would you describe it using [C(Q) : 2C(Q)] and the rank r of C(Q)?
17 Exercise 2.29. What is the cardinality of H from last exercise, depending on a, b in the polynomial of the curve C?
Exercise 2.30. Assuming the condition for H to have the largest cardinality holds, we can obtain the following formula 2r = |α(C(Q))||a(C(Q))| 4 .
In this formula, α : C(Q) is defined similarly as α but takes input from C(Q). You don’t have to deduce this formula. This is just for you to read for your own sanity and for the next exercises. You are welcome to think about why this formula is true and talk to Shiyue about it.
This exercise will walk you through an example of computing the rank of a Mordell-Weil group C(Q) and C(Q) for C : y2 = x3 −x and C : y2 = x3 + 4x. This exercise will also help you understand the consequence of Mordell’s Theorem.
Exercise 2.31. The above exercises tell us that we need to calculate cardinality of α(C(Q)) and a(C(Q)) to get a hold of the rank. The cardinality of α(C(Q)) and a(C(Q)) depends on factors of b, b, by Proposition 2.22.
• What is a, b, a, b in this case?
• How many ways of factorization that satisfies Proposition 2.22 does b, b have?
• What are the points in C(Q) that get sent to these factorizations?
Exercise 2.32. Do the same thing for α(C(Q)).
Exercise 2.33. Deduce that C(Q) and C(Q) is finite.
18 3 Torsions Given Mordell’s Theorem and the structure theorem of finitely generated abelian groups, we have C(Q) ∼ = Z × Z × · · · × Z | {z } r is called rank ×Zp α1 1 × Zpα2 2 × · · · × Zpαn n .
The rank r of a general C(Q) can be incredibly hard to compute, despite existence of ways to compute for specific curves. This section we study the finite part or the torsion part of C(Q).
Theorem 3.1. Let C : y2 = f (x) = x3 + ax2 + bx + c where a, b, c ∈Z be a non-singular cubic curve. Recall that the discriminant of the cubic polynomial f (x) D = −4a3c + a2b2 + 18abc −4b3 −27c2.
When c = 0 for Mordell’s Theorem for curves with a rational point of order two, D = b2(a2 − 4b). If P = (x, y) is a rational point of finite order, then • x, y are integers; • y = 0 and P has order 2; or y divides D.
Example 3.2. Consider the curve y2 = x3 −x2 + x, and we try to find all of its rational points of finite order. Notice that (0, 0) is a rational point of order two. Since D = b2(a2 −4b) = −3, the possible y values are ±1, ±3.
y = ±1 ⇐ ⇒x = 1 ⇐ ⇒P = (1, ±1), y = ±3 ⇐ ⇒x3 −x2 + x −9 = 0.
Example 3.3. Consider the curve C : y2 = x3 −x = x(x2 −1), which we knew from Exercises for Day 4 that the rank of C(Q) is 0. We try to find all of its rational points of finite order. Notice that (0, 0), (±1, 0) are a rational points of order two. Since D = b2(a2 −4b) = 4, the possible y values are ±1, ±2, ±4.
y = ±1 ⇐ ⇒x3 −x −1 = 0.
y = ±2 ⇐ ⇒x3 −x −4 = 0.
y = ±4 ⇐ ⇒x3 −x −16 = 0.
So C(Q) ∼ = Z4.
Recall that one technique that we have seen to prove that a rational number is in-teger, is to show that the denominator is 1; for example, we used this to show that non-zero rational points of order two for C : y2 = f (x) = x3 + ax2 + bx + c are in-tegers, which allowed us to move one of the integer point on x-axis to the origin. We 19 adopt the same strategy and try to show that if P = (x, y) is a rational point, then the denominators of x, y are 1; in other words, no primes divide the denominators.
To systematically study how primes divide denominators of a rational number, we define the p-order of a rational number as follows.
Definition 3.4. For a prime p, every non-zero rational number q can be written as m npα, where m, n are coprime, and p does not divide n or m. Then the p-order of q is ord (q) = ord m npα = α.
Assume we have a rational point P = (x, y) ∈C(Q) and some p divides the denom-inator of x. Then we have x = m npα and y = u vpβ for some α > 0. Since P ∈C(Q), we can plug in C to get u2 v2p2β = m3 + am2npα + bmn2p2α + cn3p3α n3p3α .
Since p does not divide u2 and v2, and α > 0 and p does not divide m we have p ̸ |(m3 + am2npα + bmn2p2α + cn3p3α).
Therefore, −2β = ord u2 v2p2β = ord m3 + am2npα + bmn2p2α + cn3p3α n3p3α = −3α.
Similarly, if we assume that p divides the denominator of y, then we find the same result. So there exists some ξ > 0 such that if p divides either one of the denominators of x, y, α = 2ξ and β = 3ξ.
Define C(pξ) = {(x, y) ∈C(Q) : ord (x) ≤−2ξ and ord (y) ≤−3ξ}.
The important part of the proof lies in the proof that C(pα) is a subgroup of C(Q).
To show this we perform a change of coordinates and t = x y and s = 1 y.
The curve becomes s = t3 + at2s + bts2 + cs3.
Notice that things have shifed in the following ways: 20 • O becomes the origin.
• The origin becomes O.
• A line ℓ: y = ux + v is ℓ′ : s = −u vt + 1 v.
• Group law almost changed but to get the inverse you connect with the infinity which is the origin now.
Now we can see how this new coordinate system will help us to show things we want. Let p be any prime. Let P = (x, y) be our rational point such that pα divides the denominator of y or x. That is x = m np2(α+i) and y = u vp3(α + i) for i ≥0. After changing coordinates, we have t = x y = mw nu pv+i and s = 1 y = w u p3(v+i).
Now let us give ourselves more language. Let R be the set of all non-zero rational numbers x such that ord (x) ≥0. R has the following properties: • R has unique factorization.
• R has only one prime p.
• The units of R are rational numbers of p-order 0.
• If (t, s) ∈C(pα), then t ∈pαR and s ∈p3αR.
To show that C(pα) is a subgroup of C(Q), we add two points and show that if pα divides the t-coordinate of each numerator of the two rational points, then pα divides the t-coordinate of their sum. Let P1 = (t1, s1) and P2 = (t2, s2). To ease your life, I calculated the followings. If t1 = t2, then P1 = −P2 and P1 + P2 = O ∈C(pα). If t1 ̸= t2, then the line passing s = γt + θ through P1 and P2 will have slope γ = s2 −s1 t2 −t1 = t2 2 + t1t2 + t2 1 + a(t2 + t1)s2 + bs2 2 1 −at2 1 −bt1(s2 + s1) −c(s2 2 + s1s2 + s2 1).
Let P3 = (t3, s3) be −(P1 + P2). Substituting s = γt + θ into the curve, we get that t1 + t2 + t3 = −aθ + 2bαθ + 3cγ2θ 1 + aγ + bγ2 + cγ3.
Another thing we need is θ = s1 −γt1.
We have all the information we need.
21 • The numerator of γ is in p2αR since all t1, s1, t2, s2 ∈p2αR.
• The denominator of γ disregarding 1 is in p2αR.
• θ ∈p3αR since s1 ∈p3αR and α ∈p2αR and t1 ∈pαR.
• t1 + t2 + t3 ∈p3αR.
• t3 ∈pαR since t1, t2 ∈pαR.
Hence C(pα) is a subgroup of C(Q) and t1 + t2 + t3 ∈p3αR.
This means that t3 ≡t1 + t2 (mod p3vR).
The map P = (t, s) 7→t is a map from C(pα) to additive group of rational numbers under mod out the subgrougp p3αR. The kernel of this map is C(p3α). Therefore, we have an injective homomorphism C(pα) C(p3α) →pαR p3αR.
We now want to show that P ∈C(Q) of finite order and P ̸= O does not live in any C(p) for all p.
Proof. Let P have order m and p be any prime. Since P ̸= O, m ̸= 1. Suppose for contradiction that P = (t, s) such that P ∈C(pα), but since P is not infinity, so there exists a µ > 0 such that P ∈C(pµ) but P / ∈C(pµ+1).
If p does not divide m, we have that 0 ≡t(mP) ≡mt(P) (mod p3µR).
Since m is coprime to p, P ∈C(p3α).
If p divides m. We have m = pn for some n. Consider P′ = nP. Then P′ has order p.
Using the same reasoning: Suppose P ∈C(p), then since C(p) is a subgroup of C(Q), P′ ∈C(p). There exists a µ > 0 such that P ∈C(pµ) but P / ∈C(pµ+1).
Then 0 = t(pP′) ≡pt(P′) (mod p3µR).
Therefore, P ∈p3µ−1R, a contradiction.
Therefore, P has integer coordinates.
22 Let P = (x, y) such that P has finite order. Then 2P also has finite order and both P and 2P have integer coordinates. Let 2P = (z, w). The “Duplication Formula”, 2x + z = f ′(x) 2y 2 −a.
Since x, z are integers, f ′(x) 2y are integers, so 2y| f ′(x) and y| f (x) since D = r(x) f (x) + s(x) f ′(x).
Hence y|D.
References [ST15] Joseph Silverman and John Tate. Rational Points on Elliptic Curves. Undergradu-ate Texts in Mathematics. Springer, 2015.
23 |
12898 | https://www.pubs.ext.vt.edu/content/dam/pubs_ext_vt_edu/spes/spes-201/SPES-201.pdf | Conversion Factors Needed for Common Fertilizer Calculations Authored by Mark Reiter, Associate Professor and Extension Soils and Nutrient Management Specialist, Eastern Shore Agricultural Research and Extension Center, Virginia Tech Introduction The world is a big place and farmers, industry, government, and others likely use different units, oxidation states, and measurements when calculating and reporting nutrient use for farming systems. The following table outlines some of the most common conversions needed for nutrient management. For instance, to convert K to K2O, you would multiply your K number by 1.2051. So, a fertilizer being reported as 49.8% K is also commonly reported as 49.8% × 1.2051 = 60% K2O. Therefore, you are equally correct to report muriate of potash (KCl) fertilizer as 49.8% K or 60% K2O, as long as you have the correct unit represented. However, note that fertilizer law generally states that certain oxidation states should be reported for certain nutrients (i.e. K2O must be used on Virginia fertilizer labels). Table 1. Common fertilizer conversions needed for nutrient management calculations. Column 1: Conversion Multiply by Column 2: Multiplication Value Nutrient Sources P to P2O5 Multiply P by 2.2910 P2O5 to P Multiply P2O5 by 0.4365 K to K2O Multiply K by 1.2051 K2O to K Multiply K2O by 0.8301 KCl to K Multiple KCl by 0.5244 KCl to Cl Multiply KCl by 0.4756 K2SO4 to K Multiply K2SO4 by 0.4487 Mg to MgO Multiply Mg by 1.6578 MgO to Mg Multiply MgO by 0.6032 MgCO3 to MgO Multiply MgCO3 by 0.4782 MgO to MgCO3 Multiply MgO by 2.0913 MgSO4 to Mg Multiply MgSO4 by 0.2020 MgCO3 to CaCO3 Multiply MgCO3 by 1.1867 CaO to Ca Multiply CaO by 0.7147 Ca to CaO Multiply Ca by 1.3992 CaCO3 to MgCO3 Multiply CaCO3 by 0.8426 SPES-201NP Virginia Cooperative Extension 2 Column 1: Conversion Multiply by Column 2: Multiplication Value CaCO3 to CaO Multiply CaCO3 by 0.5603 K2SO4 to S Multiply K2SO4 by 0.1840 CaSO4 to Ca Multiply CaSO4 by 0.2938 CaSO4 to S Multiply CaSO4 by 0.2350 SO4 to S Multiply SO4 by 0.3339 S to SO4 Multiply S by 2.9963 NaCl to Cl Multiply NaCl by 0.6066 N to NH3 Multiply N by 1.2158 N to KNO3 Multiply N by 7.2162 NH3 to N Multiply NH3 by 0.8225 N to (NH4)2SO4 Multiply N by 4.7160 (NH4)2SO4 to N Multiply (NH4)2SO4 by 0.2120 (NH4)2SO4 to S Multiply (NH4)2SO4 by 0.2427 N to NH4NO3 Multiply N by 2.8571 NH4NO3 to N Multiply NH4NO3 by 0.3500 Concentration Parts per million (ppm) to pounds per acre (lbs./acre) Multiply ppm by 2.0 Pounds per acre (lbs./acre) to parts per million (ppm) Multiply lbs./acre by 0.5 Percent to gram per kilogram Multiply percent by 10 Gram per kilogram to percent Multiply gram per kilogram by 0.1 Length Mile to kilometer Multiply mile by 1.609 Kilometer to mile Multiply kilometer by 0.621 Foot to meter Multiply foot by 0.304 Meter to foot Multiply meter by 3.28 Area Acre to hectare Multiply acre by 0.405 Hectare to acre Multiply hectare by 2.47 Square foot to square meter Multiply square foot by 0.0929 Square meter to square foot Multiply square meter by 10.76 Virginia Cooperative Extension 3 Column 1: Conversion Multiply by Column 2: Multiplication Value Volume Gallon to liter Multiply gallon by 3.78 Liter to gallon Multiply liter by 0.265 Quart to liter Multiply quart by 0.946 Liter to quart Multiply liter by 1.057 Mass Pound to gram Multiply pound by 454 Gram to pound Multiply gram by 0.00220 Pound to kilogram Multiply pound by 0.454 Kilogram to pound Multiply kilogram by 2.205 U.S. ton to tonne Multiply U.S. ton by 0.907 Tonne to U.S. ton Multiply tonne by 1.102 Yield and Rate Pound per acre to kilogram per hectare Multiply pound per acre by 1.12 Kilogram per hectare to pound per acre Multiply kilogram per hectare to 0.893 Bushel per acre (bu/acre) for 60 lb. bushel to kilogram per hectare Multiply bu/acre by 67.19 Bushel per acre (bu/acre) for 56 lb. bushel to kilogram per hectare Multiply bu/acre by 62.71 Bushel per acre (bu/acre) for 48 lb. bushel to kilogram per hectare Multiply bu/acre by 53.75 Gallon per acre to liter per hectare Multiply gallon per acre by 9.35 Liter per hectare to gallon per acre Multiply liter per hectare by 0.107 Temperature Fahrenheit (°F) to Celsius (°C) Multiply Fahrenheit by 5/9 × (°F - 32) Celsius (°C) to Fahrenheit (°F) Multiply Celsius by (9/5 × °C) + 32 Virginia Cooperative Extension 4 References Alley, M.M. 2000. “Part VIII: Fertilizers.” Agronomy Handbook. Publication 424-100. Virginia Cooperative Extension, Blacksburg. American Society of Agronomy, Crop Science Society of America, and Soil Science Society of America. 2020. “Chapter 7: Units and Measurements.” Publications Handbook & Style Manuel. Madison, WI. Acknowledgements Funding for this work was provided in part by the Virginia Agricultural Experiment Station and the Hatch program of the National Institute of Food and Agriculture, US Department of Agriculture. Visit Virginia Cooperative Extension: ext.vt.edu Virginia Cooperative Extension programs and employment are open to all, regardless of age, color, disability, gender, gender identity, gender expression, national origin, political affiliation, race, religion, sexual orientation, genetic information, veteran status, or any other basis protected by law. An equal opportunity/affirmative action employer. Issued in furtherance of Cooperative Extension work, Virginia Polytechnic Institute and State University, Virginia State University, and the U.S. Department of Agriculture cooperating. Edwin J. Jones, Director, Virginia Cooperative Extension, Virginia Tech, Blacksburg; M. Ray McKinnie, Administrator, 1890 Extension Program, Virginia State University, Petersburg. |
12899 | https://math.mit.edu/research/undergraduate/spur/documents/2013Kamtue.pdf | DOMINO-TILING PROBLEM SPUR FINAL PAPER, SUMMER 2013 SUPANAT KAMTUE MENTOR: BEN YANG Abstract. Given a chessborad with a small number of blocks being removed, we want to figure out in which case it is possible to tile the rest of the board with 2 × 1 dominoes. In this paper, we prove that if the number of removed blocks is small enough, then the tiling is always possible. We give a proof for both an infinite chessboard case and a finite chessboardcase. Moreover, a draft proof for three dimensional case is given as well.
Date: July, 2013.
1 2 SUPANAT KAMTUE MENTOR: BEN YANG 1. Tileability Problem Problem. Given a chessboard with a small number of blocks removed, we want to figure out in which case it is possible to tile the rest of the board with 2×1 dominoes.
The first condition for the board to be tiled with 2 × 1 dominoes is the numbers of black and whites blocks that are left need to be equal. This first condition is a global condition, and certainly is not enough to guarantee the tileability.
Let consider two examples in Figure 1 where no tiling is possible (though it is not obvious in the second example). These two bad examples suggest that the removed blocks should not be too dense in one area. A further question is how dense the removed blocks should be so that we can guarantee the tileabililty.
(a) (b) Figure 1. Bad Examples Let us implement a density-limit condition: in any R × R square, no more than ⌈cR⌉blocks are removed. The constant c is called a density constant.
In this paper we are going to give a proof that we can always tile 2×1 dominoes in a chessboard with some blocks removed in such the way that white and black blocks are remained equally and that the density-limit condition is satisfied.
In section 2, we will discuss about Hall’s matching theorem that we will mainly use in this paper. We will prove the tileability for an infinite chessboard (in section 3) and for a finite n × n chessboard (in section 4). In section 5, we will expand our result from a two-dimensional chess board into a three-dimensional gridboard.
Section 6 is a comments section where we will discuss the ideas beyond this paper.
2. Hall’s Theorem Theorem 1 (Phillip Hall’s Theorem). In a finite bipartite graph with the bipar-titions X and Y , there exists a matching that covers X if and only if every subset A of X is connected to at least |A| vertices in Y .
Hall’s theorem provides a necessary and sufficient condition for the existence of a matching in the graph. It is directly related to our problem here as a chessboard can be viewed as the bipartitions of black blocks and white blocks, and a pair of black and white blocks are connected when they are neighbors (i.e. they share the same edge).
DOMINO-TILING PROBLEM SPUR FINAL PAPER, SUMMER 2013 3 The Hall’s theorem is originally for a finite graph, but is later extended for a infinite graph as followed.
Theorem 2 (Extended Hall’s Theorem). In a bipartite graph with the biparti-tions X and Y such that the degree of every vertex is finite, there exists a matching that covers X if and only if every finite subset A of X is connected to at least |A| vertices in Y .
Before investigating more into our problem, as for the convenience we will clarify the denotations that we will be using in this paper. Denoted by W an arbitrary finite set of white blocks that we want to verify the Hall’s condition. The set of the neighbor black blocks of W is denoted by N(W). The set of blocks that are removed is called E, which is divided into the whites Ew and the blacks Eb.
The Hall’s theorem says that the domino tiling is possible if for any W (suppos-edly W ∩Ew = ∅), |N(W) −Eb| ≥|W| or equivalently, (1) |N(W)| −|W| ≥|N(W) ∩Eb| In the other words, the difference ∆= |N(W)| −|W| needs to be larger than the number of removed black blocks in N(W). Call this number R. Our required Hall’s condition for the tileability is in short ∆≥R.
Moreover, we define an expanded region of W as Exp(W) := W ∪N(W) (the region W and its black neighbors). And we call W to be semi-connected if the expanded Exp(W) is a connected region.
3. Infinite Chessboard What we are going to prove in this section is: for an infinite chessboard, with some blocks removed satisfying a density-limit condition, it is always possible to tile the rest of the board.
We want to verify the Hall’s condition for a set W. Without the loss of generality we can assume the W to be semi-connected. The reason is that if we partition W = W1 ∪W2 where Exp(W1) and Exp(W2) are disjoint, then the inequality ∆≥R can be obtained by verifying W1 and W2 separately.
Consider that a semi-connected set W, so that the connected expanded region Exp(W) occupies row number of rows and col number of columns. Denote m = max(row, col). Then the region Exp(W) can fit in some square m × m. It follows from the density-limit condition that R ≤cm. On the other hand, in each row of the board, the number of black blocks in Exp(W) will be more than the number of white blocks in Exp(W) by at least 1. This gives ∆≥row. Similarly, we will also have ∆≥col, and therefore ∆≥m ≥1 c R.
Choosing c = 1 results in the desired inequality ∆≥R.
4 SUPANAT KAMTUE MENTOR: BEN YANG 4. Finite n × n Chessboard In this section, we are going to give a proof of the main result of this paper.
Problem. For an n×n chessboard, with some blocks are removed so that the same number of black and white blocks are left, and so that the density-limit condition is satisfied, we can guarantee the domino-tiling on the rest of the board.
We will consider only the case that n is even. A proof for n being odd will be in a similar manner.
The case of the finite n × n chessboard is a little more complicated than the infinite one because of the restriction at the boundary of the chessboard.
The inequality ∆≥R is more difficult to verify. We also need to use the fact that the numbers of removed black blocks and white ones are equal (|Ew| = |Eb|).
Consider a set of white blocks W (which does not include any removed white blocks: W ∩EW = ∅). By our previous definition, N(W) is the set of black neighbors of W. Let B = {all black blocks} −N(W) be the set of all black blocks that are not in N(W), and let N(B) be the set of white neighbors of B.
Observe that Exp(W) and Exp(B) are disjoint and the union of both sets con-tains all but a few white blocks of the entire board. Call the set of those white blocks V .
Consider the following inequality: (2) |N(B)| −|B| ≥|N(B) ∩Ew| This inequality is very similar to the Hall’s condition (ineq.(1)): |N(W)| −|W| ≥|N(W) ∩Eb| except that instead of starting with the set W we start with the set B instead.
The inequality (2), in fact, implies our condition (1) because given (2) is true, |N(W)| −|W| = |N(B)| −|B| + |V | ≥|N(B) ∩Ew| + |V | ≥|Ew ∩(N(B) ∪V )| = |Ew| = |Eb| ≥|N(W) ∩Eb| The equation |Ew ∩(N(B) ∪V )| = |Ew| comes from the fact that Ew ∩W = 0 which means Ew ∈{all black blocks} −W = N(B) ∪V .
To verify the Hall’s condition, we can do it from either inequality (1) considering the region Exp(W) directly, or inequality (2) which consider the region Exp(B) instead.
Like in the previous section, we consider the Exp(W) to be connected and oc-cupies row number of rows and col number of columns. Let m = max(row, col), so Exp(W) can be fit in a m × m rectangle. Before continuing on the proof, let us consider the following lemma.
Lemma 1. Define a double-stripe as a rectangle of size 2 × n or n × 2.
In a double-stripe S which contains some part of Exp(W), the number of black blocks from Exp(W) is more than the number of white from Exp(W) blocks, except the case that the entire S is contained in Exp(W).
DOMINO-TILING PROBLEM SPUR FINAL PAPER, SUMMER 2013 5 Or algebraically if Exp(W) ∩S ̸= ∅, then |N(W) ∩S| −|W ∩S| ( = 0, S ⊂Exp(W) ≥1 otherwise Proof of Lemma 1 is straightforward and we will not write it down here.
Back to our main problem, we have 3 separate cases to check as follow: Case 1 There is no double-stripe that is fully occupied by Exp(W). Then in each two consecutive rows which are partially occupied by Exp(W), we will apply the lemma 2. We then have the inequality ∆≥⌊1 2row⌋. In fact, one can check that ∆≥1 2row. Similarly, ∆≥1 2col. Therefore, ∆≥1 2m ≥R The last inequality is from a density-limit condition with a density constant c = 1 2.
Case 2 There is either a 2 × n or n × 2 double-stripe (but not both) that is fully occupied by Exp(W). Without lost of generality, assume that there is such a 2 × n double-stripe, and there is no n × 2 one. Then we will have the inequality ∆≥1 2col = 1 2n ≥R Case 3 There are double-stripes of both types (2 × n and n × 2) which are fully occupied by Exp(W). Then it implies that there is no double-stripe that is fully occupied by Exp(B). Then the inequality (2) can be verified, using a similar argument as in Case 1 (also without loss of generality assume Exp(B) to be connected).
In conclusion, with the condition that the number of blacks and whites are equal and with a density-limit condition (c = 1 2), we verify the Hall’s condition. Then it implies the tileability of the rest of the board. Our problem is proved.
5. Three Dimensional Gridboard The Hall’s theorem can be carried out in the case of a three dimensional grid-board as well. The density-limit condition can be adapted for the three dimensional case: in any R × R × R cube, no more than ⌈cR2⌉blocks are removed. Our ap-proach to prove the inequality ∆≥R is that . One idea that we have is that ∆(the difference between white blocks and their black neighbors) can be approximated by the surface area of Exp(W), and the R (the number of removed black blocks) can be approximated by the volume of the region and the density-limit condition.
Specifically, we will show that ∆= c1δExp(W) ≥inf X C ⌈c2d2⌉≥R where δExp(W) denotes the surface area of the region Exp(W). And the summa-tion is summed over a set C of grid cubes that covers region Exp(W) and d is the side length of each cube.
Next investigate the leftmost inequality. if we slice the region Exp(W) along grid lines that are parallel to z-axis, we will obtain small vertical pieces 1 × 1 × k.
For each of this pieces, the top-end and bottom-end are on the surface of Exp(W).
6 SUPANAT KAMTUE MENTOR: BEN YANG On the other hand, this vertical piece are alternating black and white blocks, and both of the ending block are black (because the region Exp(W) only has black blocks on the boundary. Thus, in this vertical piece, the number of black blocks will be more than the number of white blocks by one. Altogether, we will have 2∆= surface of Exp(W) that is parallel to the xy-plane. A similar equation will hold for the surface that is parallel to the yz-plane(or zx-plane). As the result, 6∆= δExp(W) (a) (b) Figure 2. Vertical Slice And the left equation follows. The rightmost inequality can be obtained immedi-ately from the density-limit condition. What is left to prove is the most important and most difficult part: the inequality δExp(W) ≥inf X C ⌈kd2⌉ We believe that this inequality δS ≥inf P C ⌈kd2⌉is true, and the idea to prove is to find a set of suitable cubes that cover the region. This can be done by staring from a unit grid cube inside S, and we try to enlarge that unit cube into a bigger cube of a suitable size: not too big but contains a good amount of surface area of Exp(W) inside. Then we will delete this cube from our original region Exp(W), and continue investigating the rest region. We do not provide an explicit proof in here.
6. Comments This problem has started with a matching between black and white blocks in a well-mannered graph, like a chessboard.
This can be considered as a partial matching problem: when some blocks are randomly removed and we try to find a matching in the rest of board. One way that this problem can be carried on is to consider a matching in not only a chessboard or a gridboard but a more abstract graph with a similar property. A possible further question is: what is the ”similar property”, in the way that a matching exists in the original graph, but when some points are removed offa matching becomes questioned.
DOMINO-TILING PROBLEM SPUR FINAL PAPER, SUMMER 2013 7 Acknowledgement This paper is made possible through supports from everyone involved, and I would like to especially dedicate my acknowledgement of gratitude toward my men-tor and supervisors as followed.
First and foremost, I would like to thank my mentor, Ben Yang, for his most support. He worked together with me from the very beginning until the end, and came up with many key ideas in this paper.
Secondly, I would like to thank my and Ben’s supervisor, Professor Larry Guth, who passed the original idea of the problem to us, without which this problem would have not been started.
Thirdly, I would like to thank Professor Jacob Fox and Professor Pavel Etingof for adding comments that we can improve on, and for suggesting ways that we can work further.
Finally, I would like to thank my parents for sending the best emotional support oversea. This paper would not be possible if missing any of them.
References Ron Aharoni, Infinite matching theory. Discrete Mathematics 95 (1991), 6-7 |
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