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15200	https://pmc.ncbi.nlm.nih.gov/articles/PMC1233987/	Fatal Family Outbreak of Bacillus cereus-Associated Food Poisoning - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice J Clin Microbiol . 2005 Aug;43(8):4277–4279. doi: 10.1128/JCM.43.8.4277-4279.2005 Search in PMC Search in PubMed View in NLM Catalog Add to search Fatal Family Outbreak of Bacillus cereus-Associated Food Poisoning Katelijne Dierick Katelijne Dierick Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Katelijne Dierick 1,, Els Van Coillie Els Van Coillie Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Els Van Coillie 2, Izabela Swiecicka Izabela Swiecicka Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Izabela Swiecicka 3, Geert Meyfroidt Geert Meyfroidt Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Geert Meyfroidt 4, Hugo Devlieger Hugo Devlieger Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Hugo Devlieger 5, Agnes Meulemans Agnes Meulemans Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Agnes Meulemans 6, Guy Hoedemaekers Guy Hoedemaekers Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Guy Hoedemaekers 7, Ludo Fourie Ludo Fourie Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Ludo Fourie 8, Marc Heyndrickx Marc Heyndrickx Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Marc Heyndrickx 2, Jacques Mahillon Jacques Mahillon Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Find articles by Jacques Mahillon 9 Author information Article notes Copyright and License information Institute of Public Health—Food Section, 1050 Brussels, Belgium,1 Ministry of the Flemish Community, Center for Agricultural Research—Department Animal Product Quality, 9090 Melle, Belgium,2 Department of Microbiology, Institute of Biology, University of Bialystok, 15-950 Bialystok, Poland,3 Department of Intensive Care Medicine,4 Department of Paediatrics,5 Department of Emergency Medicine, University Hospital Gasthuisberg, 3000 Leuven, Belgium,6 Department of Paediatrics, Ziekenhuis Maas en Kempen, Maaseik, Belgium,7 Federal Agency for the Safety of the Food Chain, Brussels, Belgium,8 Laboratory of Food and Environmental Microbiology, Université catholique de Louvain, B-1348 Louvain-la-Neuve, Belgium 9 Corresponding author. Mailing address: Institute of Public Health, 14 Juliette Wijtsman Str., B-1050 Brussels, Belgium. Phone: 00 32 2 642 51 53. Fax: 00 32 2 642 53 27. E-mail: Katelijne.Dierick@iph.fgov.be. Received 2004 Nov 23; Revised 2004 Dec 6; Accepted 2005 Apr 9. Copyright © 2005, American Society for Microbiology PMC Copyright notice PMCID: PMC1233987 PMID: 16082000 Abstract Bacillus cereus is a well-known cause of food-borne illness, but infection with this organism is not commonly reported because of its usually mild symptoms. A fatal case due to liver failure after the consumption of pasta salad is described and demonstrates the possible severity of the emetic syndrome. CASE REPORT In August 2003, five children of a family became sick after eating pasta salad. The pasta salad was prepared on a Friday and taken to a picnic on the following Saturday; the remainders had been stored in the fridge until the following Monday evening, when they were served for supper to the children. Because the pasta salad had an unusual smell, three children (B14, G10, and G9) ate only a small quantity. At 6 h after the meal the youngest girl (G7), 7 years old, started vomiting. She complained of respiratory distress and was taken to the emergency department of a local hospital. Upon arrival, her brothers and sisters started vomiting as well. Because the clinical condition of two children (G7 and B9) deteriorated rapidly, they were intubated and mechanically ventilated. All children were transported to the University Hospital in Leuven. During transfer, G7 had severe pulmonary hemorrhage and needed continuous resuscitation. Upon arrival she was moribund with coma, diffuse bleeding, and severe muscle cramps. She died within 20 min, at 13 h after the meal. On autopsy Bacillus cereus was detected in her gut content but also in the spleen, probably by postmortem translocation of the bacterium. A postmortem liver biopsy showed microvascular and extensive coagulation necrosis. Her initial laboratory values showed severe metabolic acidosis and liver failure. All four other children were affected, although to different degrees (Table 1). The 9-year-old boy (B9) was transferred to the pediatric intensive care unit, where mechanical ventilation and invasive hemodynamic monitoring were continued. After fluid resuscitation, his blood lactate levels gradually went down. Basic treatment for liver failure consisted of vitamin K supplementation, oral and rectal lactulose, oral neomycin, and high-dose acetylcysteine. At 24 h after the start of the treatment his aspartate transaminase and alanine transaminase levels peaked at 12,254 U/liter and 8,656 U/liter, respectively; his prothrombin time went down to 21.5%. Thereafter, the hepatic function recovered. He gradually recovered consciousness, and he was successfully extubated. Two sisters (G9 and G10) were treated with fluid resuscitation and bicarbonate substitution. Both gradually recovered. The 14-year-old brother (B14) was kept under observation. Subsequent blood samples showed no deterioration of hepatic function. The surviving children could leave the hospital within 8 days. TABLE 1. Laboratory values for the five children on admission \| Parameterb \| Value for childa: \| :---: \| \| G7 \| B9 \| G9 \| G10 \| B14 \| \| Arterial pH \| \| 7.163 \| 7.369 \| 7.375 (7.35-7.45) \| \| \| Bicarbonate (mmol/liter) \| 9.1 \| 10.2 \| 13.8 \| 10.8 (22.0-29.0) \| 17.7 (22.0-29.0) \| \| Lactate (mmol/liter) \| \| 9.5 \| 3.4 \| 8.8 (0.5-1.6) \| 5.95 (0.4-2.0) \| \| Ammonia (μmol/liter) \| >2,000 \| 107 \| 30 \| 112 (11-32) \| \| \| AP (U/liter) \| 163 \| 268 \| 547 \| 611 (<720) \| 822 (<936) \| \| AST (U/liter) \| 3,123 \| 1,858 \| 239 \| 5,327 (<32) \| 42 (<38) \| \| ALT (U/liter) \| 2,636 \| 5,181 \| 447 \| 4,101 (<31) \| \| \| γGT (U/liter) \| 16 \| 24 \| 19 \| 44 (<35) \| \| \| Total bilirubin \| 0.28 \| 1.23 \| 0.52 \| 1.94 (<1.00) \| \| \| Conjugated bilirubin \| \| 0.66 \| \| 1.49 (0.0-0.5) \| \| \| LDH (U/liter) \| 5,110 \| 9,101 \| 1,481 \| 5,684 (240-480) \| \| \| Prothrombin time (%) \| <10 \| 23.1 \| 83.7 \| 30.6 (70-100) \| >100 (70-100) \| Open in a new tab a Age specific normal values for our laboratory are in parentheses. Normal values for G7, B9, and G9 are identical to the values for G10. b AP, alkaline phosphatase; AST, aspartate aminotransferase; ALT, alanine aminotransferase; γGT, γ-glutamyltransferase; LDH, lactate dehydrogenase. In six food samples and the vomit of the deceased girl B. cereus was detected. The samples were not heat treated before analysis. After 24 h of incubation, the colonies were pink (mannitol negative) with a precipitation zone (lecithinase positive) on MYP agar (Oxoid Ltd., Basingstoke, England), beta-hemolytic on Columbia agar (Oxoid), and positive in motility agar. Gram staining on Columbia agar-grown colonies showed gram-positive rods with nondeforming subterminal spores. The highest B. cereus count (10 7 to 10 8 CFU/g) was found in the pasta salad, the lowest count in the vomit (2.0 × 10 2 CFU/g). From each positive sample, three (or four) isolates were phenotypically confirmed as B. cereus by use of the API 50 gallery. Further characterization of the 22 isolates obtained consisted of repetitive sequence-based PCR (rep-PCR) (6), pulsed-field gel electrophoresis (PFGE) of genomic DNA (5), and PCR analysis of a marker for emetic toxin (1). Rep-PCR has been shown before to be useful for outbreak investigation (8). The typing of the isolates was focused on the food and vomit isolates, because the postmortem spleen isolate was not available at that moment. With rep-PCR these isolates could be divided into four groups representing rep types 1 to 4 (Fig. 1); all isolates obtained from the vomit, together with one isolate obtained from the pasta salad and one from the pasta bowl, were clustered within rep type 1. Rep type 4 consisted of 11 isolates, obtained from the pasta salad but also from another type of boiled pasta, chocolate milk, and semiskimmed milk, which indicates cross-contamination during handling of the food. PFGE revealed four distinct restriction patterns (PFGE types A to D) (Fig. 1 and 2). The majority of the isolates (n = 16) were classified as PFGE type C, which included isolates from the boiled pasta, the pasta salad, chocolate milk, vomit, semiskimmed milk, and the bowl. All these PFGE type C isolates corresponded to rep type 1 or 4 (Fig. 1), indicating a high discriminatory potential of rep-PCR. The above-described data indicate that more than one strain was present in this intoxication case, and they illustrate the importance of obtaining multiple isolates from even one food sample (Fig. 1). Detection of B. cereus emetic and enterotoxin production was performed with cytotoxicity assays (2, 4). Both tested isolates of rep type 1 (vomit and pasta salad) and the three tested isolates from rep type 4 produced the emetic toxin and were also positive in the emetic toxin-specific PCR assay (Fig. 1). FIG. 1. Open in a new tab Cluster analysis of the (GTG)5 fingerprints and toxin production of B. cereus isolates. As listed in the Isolate a column, the following isolates have been deposited in the BCCM/LMG Bacteria Collection (Ghent University, Ghent, Belgium): 5975a (LMG 22728); 5965c (LMG 22729); 5972a (LMG 22730); 5969a (LMG 22731); 5958c (LMG 22732); 5964a (LMG 22733). Toxin titer b column: emetic and enterotoxin titers detected on the basis of cytotoxicity assays. cer c column: emetic toxin-specific marker detected in the PCR assay. The reference strain for emetic toxin production is identified with a superscript “d” suffix in the Isolate a column. The reference strain for enterotoxin production is identified with a superscript “e” suffix. ND, not determined. FIG. 2. Open in a new tab Representative PFGE patterns (A to D) of AscI (A)- and NotI (B)-digested genomic DNA of B. cereus isolates. M, yeast chromosome PFGE marker; m, lambda 48.5-kb size marker. The values on the left and right are molecular mass markers in kilobases. Although Bacillus cereus is a well-known cause of food-borne illness it is not commonly reported because of its usually mild symptoms. It can cause two types of food poisoning known as the emetic and the diarrheal types. The emetic type is caused by a heat-stable toxin, named cereulide, preformed in the food. Only one fatal case has been reported up to now (9). The present results provided evidence for B. cereus food poisoning of five children of one Belgian family. The clinical data and the rapid onset of symptoms, together with the microbiological and molecular study, pointed to B. cereus as the causative agent. It has been demonstrated that B. cereus from the pasta salad, the vomit of the deceased girl, and the pasta bowl produced identical patterns on the basis of both analyses (rep type 1 and PFGE type C). Although the presence of cereulide in the pasta salad was not directly demonstrated, its production at a high level was indirectly proven in the cytotoxicity test of the isolates. These results were confirmed by PCR (1), which amplifies a DNA fragment whose presence is specific for cereulide-producing strains. All the isolates classified as PFGE type C and rep type 1 reacted positively with these primers, indicating the presence of cereulide-related genes. Also, the rest of the isolates of PFGE type C but pertaining to rep type 4 harbored the cereulide genetic determinants. Thus, although no isolate of rep type 4 was detected in the vomit, these isolates could also have produced the toxin in the pasta salad. The present case illustrates the possible severity of the emetic syndrome and the importance of adequate refrigeration of prepared food. Because the emetic toxin is preformed in the food and not inactivated by heat treatment (7) it is important to prevent growth and the production of cereulide during storage. Some B. cereus strains are known to be psychrotrophic and to have the highest emetic toxin production between 12- and 15°C (3). In this case, the temperature of the fridge where the pasta salad was stored was 14°C. This allowed B. cereus to grow to a count of more than 10 8 CFU/g in 3 days with a probably very high toxin production that may explain the fatal outcome. REFERENCES 1.Ehling-Schulz, M., M. Fricker, and S. Scherer. 2004. Identification of emetic toxin producing Bacillus cereus strains by a novel molecular assay. FEMS Microbiol. Lett. 232:189-195. [DOI] [PubMed] [Google Scholar] 2.Finlay, W. J., N. A. Logan, and A. D. Sutherland. 1999. Semiautomated metabolic staining assay for Bacillus cereus emetic toxin. Appl. Environ. Microbiol. 65:1811-1812. 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15201	https://www.wordsclarity.com/dictionary/castigate	castigate \| Meaning, Synonyms, Keydifference & Examples \|... ☰ Home About Contact Words Have Power, Let’s Unlock It! Every word tells a story, sparks an idea, and shapes the way we communicate. Dive into a world where meanings come alive, synonyms open new possibilities, and subtle distinctions make all the difference! Supercharge your vocabulary, sharpen your writing, and fall in love with the art of words. Ready to explore? Let’s begin! Word of the Day Fight A violent confrontation or struggle between people or groups, which can be physical or verbal, and ranges from small-scale personal disputes to large-scale conflicts like wars. Know its synonyms with key difference Find the Right Word with Clarity Search🎤 Voice Search castigate🔊 Meaning: To reprimand or criticize someone severely. Key Difference: Castigate implies a harsh, often public rebuke aimed at correcting behavior, whereas its synonyms may vary in intensity, context, or tone. Examples: The teacher castigated the student for plagiarizing the essay, emphasizing the importance of originality. The media castigated the politician for his unethical practices, demanding accountability. Synonyms of castigate: Similar words with related meanings. reprimand🔊 Meaning: To express formal disapproval of someone's behavior. Key Difference: Reprimand is more formal and often official, while castigate carries a stronger emotional tone. Examples: The manager reprimanded the employee for repeatedly arriving late to work. The judge reprimanded the lawyer for disrespecting the court. berate🔊 Meaning: To scold or criticize angrily. Key Difference: Berate focuses on the anger behind the criticism, whereas castigate implies a more systematic or public rebuke. Examples: She berated her teammate for missing the crucial shot in the game. The coach berated the players for their lack of discipline. chastise🔊 Meaning: To punish or criticize severely, often physically or verbally. Key Difference: Chastise can imply physical punishment, while castigate is strictly verbal or written. Examples: In the past, teachers would chastise students with a ruler for misbehaving. The editorial chastised the government for its slow response to the crisis. rebuke🔊 Meaning: To express sharp disapproval or criticism. Key Difference: Rebuke is less severe and more immediate than castigate, which is often prolonged and public. Examples: The principal rebuked the students for vandalizing school property. She rebuked her friend for spreading rumors. censure🔊 Meaning: To formally criticize or condemn. Key Difference: Censure is often used in official contexts, while castigate can be both formal and informal. Examples: The Senate voted to censure the senator for misconduct. The committee censured the organization for violating ethical guidelines. upbraid🔊 Meaning: To scold or criticize severely. Key Difference: Upbraid is more archaic and less commonly used than castigate, which is more contemporary. Examples: The king upbraided his advisor for failing to foresee the rebellion. She upbraided her brother for his reckless behavior. lambaste🔊 Meaning: To criticize or reprimand harshly. Key Difference: Lambaste is more colloquial and intense, while castigate is more structured and formal. Examples: The critic lambasted the film for its poor screenplay and acting. He lambasted his opponents during the debate. excoriate🔊 Meaning: To criticize severely and bitterly. Key Difference: Excoriate implies a more brutal and relentless criticism than castigate. Examples: The journalist excoriated the corporation for exploiting workers. The review excoriated the novel for its lack of originality. admonish🔊 Meaning: To warn or reprimand firmly. Key Difference: Admonish is gentler and more cautionary, whereas castigate is punitive and severe. Examples: The mother admonished her child for running into the street. The officer admonished the driver for speeding. Conclusion: ✅ Castigate is best used when a strong, often public reprimand is needed to correct behavior or highlight wrongdoing. ✅ Reprimand can be used in formal settings where structured criticism is required. ✅ Berate is suitable when anger drives the criticism, such as in personal disputes. ✅ Chastise fits historical or disciplinary contexts where severe punishment is implied. ✅ Rebuke works for immediate, sharp disapproval in everyday situations. ✅ Censure is ideal for official condemnations, such as in political or organizational settings. ✅ Upbraid is a more literary choice for dramatic or historical narratives. ✅ Lambaste is useful in informal yet intense criticism, like in reviews or debates. ✅ Excoriate should be reserved for extreme cases of bitter denunciation. ✅ Admonish is best for gentle warnings rather than harsh reprimands. 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15202	https://pmc.ncbi.nlm.nih.gov/articles/PMC10136633/	Incidence and Risk Factors for Extremity Osteoradionecrosis after Limb-Sparing Surgery and Adjuvant Radiotherapy - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Cancers (Basel) . 2023 Apr 17;15(8):2339. doi: 10.3390/cancers15082339 Search in PMC Search in PubMed View in NLM Catalog Add to search Incidence and Risk Factors for Extremity Osteoradionecrosis after Limb-Sparing Surgery and Adjuvant Radiotherapy Yun-Jui Lu Yun-Jui Lu 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Software, Validation, Formal analysis, Investigation, Data curation, Writing – original draft, Writing – review & editing, Visualization Find articles by Yun-Jui Lu 1, Chun-Chieh Chen Chun-Chieh Chen 2 Department of Orthopedic Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Conceptualization, Resources Find articles by Chun-Chieh Chen 2, Shih-Heng Chen Shih-Heng Chen 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Find articles by Shih-Heng Chen 1, Cheng-Hung Lin Cheng-Hung Lin 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Validation, Project administration Find articles by Cheng-Hung Lin 1, Yu-Te Lin Yu-Te Lin 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Validation, Supervision Find articles by Yu-Te Lin 1, Chih-Hung Lin Chih-Hung Lin 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Supervision Find articles by Chih-Hung Lin 1, Chung-Chen Hsu Chung-Chen Hsu 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Conceptualization, Validation, Resources, Writing – review & editing Find articles by Chung-Chen Hsu 1, Editor: Kazutaka Kikuta Author information Article notes Copyright and License information 1 Department of Plastic and Reconstructive Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan 2 Department of Orthopedic Surgery, Chang Gung Memorial Hospital, College of Medicine, Chang Gung University, Taoyuan 333, Taiwan Correspondence: hsu.chungchen@gmail.com Roles Yun-Jui Lu: Software, Validation, Formal analysis, Investigation, Data curation, Writing – original draft, Writing – review & editing, Visualization Chun-Chieh Chen: Conceptualization, Resources Cheng-Hung Lin: Validation, Project administration Yu-Te Lin: Validation, Supervision Chih-Hung Lin: Supervision Chung-Chen Hsu: Conceptualization, Validation, Resources, Writing – review & editing Kazutaka Kikuta: Academic Editor Received 2023 Mar 5; Revised 2023 Apr 5; Accepted 2023 Apr 11; Collection date 2023 Apr. © 2023 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: PMC10136633 PMID: 37190268 Abstract Simple Summary This study investigated osteoradionecrosis (ORN), a complication after radiotherapy, in 198 patients with extremity soft tissue sarcoma (STS). The incidence rate of extremity ORN was 3.5%, mostly located in lower extremities. Clinical presentations included chronic ulcers, soft tissue necrosis, sinus discharge, bone nonunion, and pathological fractures. The ORN group had a significantly higher total radiation dose and greater use of intraoperative periosteal stripping. Repeat surgeries and subsequent soft tissue reconstruction or limb amputation were performed as treatments. Careful surveillance should be taken to manage the risk of ORN in extremity STS patients. Abstract Osteoradionecrosis (ORN) is a major complication after radiotherapy. Most studies on ORN have focused on patients with mandibular lesions, with few studies including patients with extremity soft tissue sarcoma (STS). We included 198 patients with extremity STS who underwent limb-sparing surgery and adjuvant radiotherapy between 2004 and 2017. The incidence rate of extremity ORN was 3.5% (7/198), with most lesions (6/7) located in the lower extremities. The mean follow-up time was 62 months. Clinical presentations included chronic ulcers, soft tissue necrosis, sinus discharge, bone nonunion, and pathological fractures. Compared with the non-ORN group, the ORN group had a significantly higher total radiation dose (68 Gy vs. 64 Gy, p = 0.048) and greater use of intraoperative periosteal stripping (p = 0.008). Repeat surgeries and subsequent soft tissue reconstruction or limb amputation were performed as treatments. The risk and management of ORN in patients with extremity STS was ignored previously. Because the disease is complex and affects both clinicians and patients, careful surveillance should be undertaken. Keywords: osteoradionecrosis, extremity sarcoma, soft tissue sarcoma, radiation therapy 1. Introduction Soft tissue sarcoma (STS) is rare, accounting for approximately 1% of all adult malignancies. Nearly half of all STS occur in the extremities . More than 100 STS subtypes have been identified, and pathological diagnosis is performed in accordance with the World Health Organization (WHO) classification guidelines . Although different types of treatments are available, surgical resection remains the standard treatment for STS. Limb-sparing therapy is attempted first to achieve satisfactory functional outcomes and quality of life. Radiation therapy (RT) is an adjunctive treatment administered preoperatively or postoperatively to achieve local control and reduce the local recurrence rate of STS, particularly for tumors with a positive margin and high histological grade. New techniques, such as intensity-modulated RT (IMRT) and proton beam therapy, have advantages over older techniques in reducing peripheral tissue damage . Nevertheless, early and late toxicities of RT may occur, with common ones being edema, dermatitis, compromised wound healing, fibrosis, pathological fracture, and secondary malignancy. Osteoradionecrosis (ORN) is a complication that occurs most frequently in patients with head and neck cancer after radiation exposure. ORN was first described by Ewing in 1926 and is characterized by an irradiated bone that does not heal over 3–6 months, without any evidence of persisting or recurrent tumor . The disease is pernicious, and treatment involves multidisciplinary decisions. Published articles have focused mainly on patients with head and neck cancer, and only a few of them have discussed other regions. To better understand the bone response to radiation doses in extremity STS, we evaluated extremity ORN, as well as its incidence, presentation, treatment, and risk factors. 2. Materials and Methods We conducted a retrospective study between 1 January 2004 and 31 December 2017 in a single medical center. This study was approved by the institutional review board of Chang Gung Medical Foundation. A total of 391 patients with extremity STS were classified using the International Classification of Diseases, Tenth Edition, Clinical Modification codes C47.1, C47.2, C49.1, and C49.2. First, we excluded patients with missing data or who had not undergone surgery. Subsequently, patients who did not receive limb-sparing surgery and adjuvant RT with curative intent were excluded. Finally, we included and analyzed 198 patients. Figure 1 displays the study diagram. Figure 1. Open in a new tab Flowchart of the selection and grouping of patients with extremity STS. Data on the demographics, tumor characteristics, treatment details, and clinical outcomes of the patients were collected in encrypted form. Tumor staging and grading were coded in accordance with the American Joint Committee on Cancer Staging Manual, Eighth Edition , and the French Federation of Cancer Centers Sarcoma Group histological grading system guidelines, respectively. The pathological report was reviewed, and the tumors were classified in accordance with the WHO classification guidelines . The surgical margin was categorized as negative, close (<1 cm), or positive. The dose, fraction, modality, and timespan of RT were recorded. 2.1. ORN Identification The medical records of each patient with extremity STS were reviewed in detail because the initial clinical presentation may be subtle, and diagnostic codes were unreliable. Clinical presentations ranged from pain, chronic ulceration, and fistula formation with purulent discharge to bone exposure and pathological fracture. Several diagnostic images, primarily those obtained using magnetic resonance imaging (MRI), were examined by a radiologist and two clinicians. MRI images were generally obtained using sequences of T1-weighted image (T1 WI), T2-weighted image (T2 WI), short tau inversion recovery (STIR), and T1-weighted image with gadolinium enhancement. The bone marrow may have low signal intensity on T1 WI, heterogeneous intensity on T2 WI, and hyperintensity on STIR images in ORN patients [8,9]. The increased signal intensity of the surrounding tissue in contrast-enhanced T1 WI indicated the inflammatory process or superimposed infection. In advanced disease, cortical defect, bony sequestrum, or pathological fracture would be presented . The type of operative procedure, such as debridement or sequestrectomy, raised suspicion for ORN. The histopathology of ORN is characterized by diffuse bony necrosis with a hypovascular marrow space, nonviable periosteum, and nonexistent osteoclasts [11,12]. Figure 2 demonstrates the pathological findings of one of the ORN patients. Figure 2. Open in a new tab There is extensive necrosis of bony trabeculae and bone marrow tissue. No prominent inflammation is seen. 2.2. Statistical Analysis The dataset was first tested for normality using the Kolmogorov–Smirnov or Shapiro–Wilk test. The difference between the groups (ORN and non-ORN) was evaluated using Fisher’s exact test and the Mann–Whitney U test for categorical and continuous variables, respectively. Two-tailed tests were used, and the significance level was set to 0.05. All analyses were performed using SPSS 26 (SPSS Inc., Chicago, IL, USA). 3. Results A total of 198 patients with extremity STS who were treated using limb-sparing surgery and postoperative RT were included. Most patients were middle-aged, with a median age of 51.5 years, and had no considerable chronic illness (Charlson comorbidity index ≤ 3). Of these patients, 113 were men (57.1%), 85 were women (42.9%), and 41 were smokers (20.7%). The primary tumor was more commonly located in the lower extremity than in the upper extremity (155 patients vs. 43 patients), with the thigh being the most commonly affected anatomical region. A total of 174 patients had primary tumor occurrence, and 24 presented with recurrent disease. All patients with tumor relapse had no history of RT, and all patients underwent wide excision of the tumor with limb preservation. Lymphovascular invasion occurred in 31 patients (15.7%) due to locally advanced growth or adjacency, and periosteal stripping was performed in 14 patients (7.1%) during surgery. Soft tissue reconstruction with local flap transfer (4.0%) or microsurgical reconstruction (4.0%) was performed in patients with unfeasible primary closure. The clinical and tumor characteristics are presented in Table 1. The most common type of tumor histopathology was undifferentiated pleomorphic sarcoma (45 patients, 22.7%) and myxoid liposarcoma (26 patients, 13.1%), followed by synovial sarcoma (25 patients, 12.6%), myxofibrosarcoma (24 patients, 12.1%), leiomyosarcoma (10 patients, 5.1%), and malignant peripheral nerve sheath tumors (10 patients, 5.1%). The R0 tumor resection was achieved in more than half of the surgical group; however, 42.9% of patients with positive surgical margins remained. Adjuvant RT was initiated immediately after the patient recovered and the surgical wounds healed. The total dose and field size of the RT were determined after being reviewed by the tumor board in our institute. Regular surveillance was arranged, and the mean follow-up period was 62 months. Of the 198 patients, 7 (3.5%) had ORN, and the clinical characteristics of each patient with ORN are listed in Table 2. No significant differences in age, sex, body mass index, comorbidity, smoking status, tumor location, tumor dimension, histopathology, and surgical margin were observed between the ORN and non-ORN groups. However, periosteal stripping was significantly higher in the ORN group than in the non-ORN group (p = 0.008). Furthermore, the median of the total radiation dose was higher in the ORN group than in the non-ORN group (68.0 Gy vs. 64.0 Gy, p = 0.048). Table 1. Patient characteristics. \| \| All \| Without ORN \| With ORN \| p \| :---: :---: \| \| 198 \| 191 \| 7 \| \| \| Sex \| \| \| \| 0.654 \| \| Male \| 113 (57.1%) \| 109 (57.1%) \| 4 (57.1%) \| \| \| Female \| 85 (42.9%) \| 82 (42.9%) \| 3 (42.9%) \| \| \| Age, median (IQR) \| 51.5 (28) \| 52.0 (28) \| 51.0 (33) \| 0.971 \| \| BMI, median (IQR) \| 24.3(5.4) \| 24.2 (5.3) \| 27.6 (5.3) \| 0.121 \| \| Smoker \| 41 (20.7%) \| 38 (19.9%) \| 3 (42.8%) \| 0.157 \| \| Alcohol \| 19 (9.6%) \| 18 (9.4%) \| 1 (14.3%) \| 0.512 \| \| Betel nuts \| 5 (2.5%) \| 4 (2.1%) \| 1 (14.3%) \| 0.166 \| \| Hypertension \| 64 (32.3%) \| 61 (31.9%) \| 3 (42.9%) \| 0.683 \| \| CCI \| \| \| \| 0.301 \| \| ≤3 \| 167 (84.3%) \| 162 (84.8%) \| 5 (71.4%) \| \| \| >3 \| 31 (15.7%) \| 29 (15.2%) \| 2 (28.6%) \| \| \| Tumor \| \| \| \| 0.203 \| \| Primary \| 174 (87.9%) \| 169 (88.5%) \| 5 (71.4%) \| \| \| Recurrent \| 24 (12.1%) \| 22 (11.5%) \| 2 (28.6%) \| \| \| Location \| \| \| \| 0.238 \| \| Upper limb, proximal \| 24 (12.1%) \| 22 (11.5%) \| 2 (28.6%) \| \| \| Upper limb, distal \| 19 (9.6%) \| 19 (9.9%) \| 0 \| \| \| Lower limb, proximal \| 107 (54.0%) \| 105 (54.9%) \| 2 (28.6%) \| \| \| Lower limb, distal \| 48 (24.2%) \| 45 (23.6%) \| 3 (42.9%) \| \| \| Depth \| \| \| \| 0.452 \| \| Superficial \| 47 (23.7%) \| 46 (24.1%) \| 1 (14.3%) \| \| \| Deep \| 144 (72.7%) \| 138 (72.3%) \| 6 (85.7%) \| \| \| Re-resection \| 30 (15.2%) \| 29 (15.2) \| 1 (14.3%) \| 0.713 \| \| Periosteal stripping \| 14 (7.1%) \| 11 (5.8%) \| 3 (42.8%) \| 0.008 \| \| Lymphovascular invasion \| 31 (15.7%) \| 29 (15.2%) \| 2 (28.6%) \| 0.301 \| \| Tumor classification \| \| \| \| 0.644 \| \| T1 \| 56 (28.3%) \| 54 (28.3%) \| 2 (28.6%) \| \| \| T2 \| 77 (38.9%) \| 74 (38.7%) \| 3 (42.8%) \| \| \| T3 \| 30 (15.2%) \| 30 (15.7%) \| 0 \| \| \| T4 \| 34 (17.2%) \| 32 (16.8%) \| 2 (28.6%) \| \| \| Histologic grade \| \| \| \| 0.63 \| \| G1 \| 48 (24.2%) \| 47 (24.6%) \| 1 (14.3%) \| \| \| G2 \| 53 (26.8%) \| 50 (26.2%) \| 3 (42.8%) \| \| \| G3 \| 86 (43.4%) \| 83 (43.5%) \| 3 (42.8%) \| \| \| Margin \| \| \| \| 0.242 \| \| Negative \| 33 (16.7%) \| 33 (17.3%) \| 0 \| \| \| Close \| 80 (40.4%) \| 78 (40.8%) \| 2 (28.6%) \| \| \| Positive \| 85 (42.9%) \| 80 (41.9%) \| 5 (71.4%) \| \| \| Pathology \| \| \| \| 0.630 \| \| Undifferentiated pleomorphic sarcoma \| 45 (22.7%) \| 42 (21.9%) \| 3 (42.8%) \| \| \| Myxoid liposarcoma \| 26 (13.1%) \| 25 (13.1%) \| 1 (14.3%) \| \| \| Synovial sarcoma \| 25 (12.6%) \| 24 (12.6%) \| 1 (14.3%) \| \| \| Myxofibrosarcoma \| 24 (12.1%) \| 24 (12.6%) \| 0 \| \| \| Leiomyosarcoma \| 10 (5.1%) \| 9 (4.7%) \| 1 (14.3%) \| \| \| MPNST \| 10 (5.1%) \| 10 (5.2%) \| 0 \| \| \| Other \| 58 (29.3%) \| 57 (29.8%) \| 1 (14.3%) \| \| \| Dosage (Gy), median (IQR) \| 64.0 (6.0) \| 64.0 (6) \| 68.0 (6) \| 0.048 \| \| Adjuvant chemotherapy \| 17 (8.6%) \| 16 (8.4%) \| 1 (14.3%) \| 0.472 \| \| Repeated RT \| 16 (8.1%) \| 13 (6.8%) \| 3 (42.8%) \| 0.057 \| Open in a new tab Statistically significant; BMI, body mass index; CCI, Charlson comorbidity index; IQR, interquartile range; MPNST, malignant peripheral nerve sheath tumor; ORN, osteoradionecrosis. Table 2. Characteristics of patients with extremity ORN. \| Case \| Sex \| Age \| Location \| Diagnosis \| Margin \| Radiation Dose (Gy) \| Lag Time (m) \| Presentation \| Management \| Outcome \| :---: :---: :---: :---: :---: \| 1 \| M \| 76 \| Left lower leg \| Undifferentiated pleomorphic sarcoma \| Positive \| 66 \| 3 \| Chronic ulcers, pathological fracture \| Debridement, sequestrectomy, bone graft, ORIF, flap reconstruction \| Nonunion bone \| \| 2 \| F \| 69 \| Left ankle \| Leiomyosarcoma \| Positive \| 60 \| 13 \| Chronic ulcers, bone exposure \| Debridement, flap reconstruction \| Healed wound \| \| 3 \| F \| 26 \| Left hip \| Undifferentiated pleomorphic sarcoma \| Positive \| 74 \| 67 \| Periprosthetic fracture, persistent sinus discharge \| Debridement, hip joint arthroplasty, flap reconstruction \| No infection sign \| \| 4 \| M \| 36 \| Left upper arm \| Synovial sarcoma \| Positive \| 70 \| 14 \| Persistent sinus discharge, wound poor healing \| Debridement, synovectomy \| Shoulder disarticulation \| \| 5 \| M \| 45 \| Left thigh \| Myxoid liposarcoma \| Close \| 70 \| 23 \| Persistent sinus discharge, wound poor healing, pathological fracture \| Debridement, HBO therapy \| AK amputation \| \| 6 \| F \| 57 \| Left lower leg \| Dedifferentiated liposarcoma \| Positive \| 70 \| 3 \| Nonunion bone \| Debridement, bone graft, HBO therapy, ORIF, flap reconstruction \| BK amputation \| \| 7 \| M \| 48 \| Right thigh \| Undifferentiated pleomorphic sarcoma \| Close \| 60 \| 36 \| Persistent sinus discharge, soft tissue necrosis, nonunion bone \| Debridement, sequestrectomy, ORIF, flap reconstruction, pentoxifylline, tocopherol \| Bone union, wound healed \| Open in a new tab AK, above the knee; BK, below the knee; ORIF, open reduction and internal fixation; HBO, hyperbaric oxygen. The ORN group consisted of 4 male and 3 female patients, with a mean age of 51 years. Of the 7 necrotic lesions, 6 were in the lower extremity, and 1 was in the left shoulder joint. Tumor differentiation was mainly high grade, and R0 resection was not achieved. The average radiation dose in the ORN group was 67.4 Gy, and the time to ORN onset ranged from 3 to 67 months. Initial manifestations were chronic ulcer in 2 patients, persistent sinus discharge in 3 patients, and periprosthetic fracture and nonunion fracture in 2 patients (Table 2). ORN Case Presentations A 45-year-old man with left lateral thigh myxoid liposarcoma (pT4N0M0, stage IIIB) underwent an operation and postoperative RT with a total radiation dose of 70 Gy. The patient presented with chronic non-healing wounds on the left lateral thigh with persistent sinus and abscess discharge approximately 2 years after the last radiation course. He received wide excision of the necrotic tissue and hyperbaric oxygen (HBO) therapy. However, due to the pathological fracture with persistent sinus discharge and chronic pain, he underwent high above-knee amputation 4 years after treatment (Figure 3). Figure 3. Open in a new tab (a) Contrast-enhanced T1-weighted image over a 21-month follow-up after RT revealing soft tissue inflammation, bone marrow edema, and skin breakdown with sinus drainage. (b) MRI 3.5 years after RT, revealing lesion expansion. (c) MRI 4 years after RT, revealing severe tissue inflammation on the medial and lateral thigh, bone marrow edema, and one cortical defect in the left lateral femoral shaft. (d) MRI 4.5 years after RT revealing a pathological fracture of the left femoral shaft. Figure 4 shows the case of a 48-year-old man with a recurrent right posterior thigh mass and pulmonary metastases. The initial tumor was resected approximately 10 years ago, and the patient received radical excision for the recurrent tumor, which was diagnosed as undifferentiated pleomorphic sarcoma with a close margin. IMRT was performed with a total radiation dose of 60 Gy, and adjuvant chemotherapy was administered concurrently for systemic control. However, 36 months after RT, the surgical site began to break down with progressive soft tissue necrosis and turbid discharge. The patient underwent repeated surgical debridement, sequestrectomy, local flap transfer (four times), and interlocking nail fixation for a pathologically fractured femoral bone in the following years. After the Masquelet-induced membrane technique was performed by an orthopedic surgeon, the bone achieved fair union. Figure 4. Open in a new tab (a) Right posterior thigh intramuscular tumor, 4 × 5 × 6.9 cm. (b) Right femur osteonecrosis, fluid collection, and marrow replacement. (5 years after RT); (c) Full-thickness soft tissue necrosis with secondary infection and purulent discharge (3.5 years after RT); (d) pathological fracture of the right femoral shaft (5.5 years after RT). Another patient, a 57-year-old woman with dedifferentiated liposarcoma of the left lower leg (pT4N0M0, stage IIIB), presented with increasing leg size. Initial surgery removed the tumor along with most of the fibula bone. RapidArc RT was administered with a total dose of 70 Gy. However, the patient encountered a traffic accident 3 months after the treatment, resulting in a fracture of the left distal tibial bone that was fixed using an interlocking nail. Despite these efforts, neither the bone nor the wound healed. The patient developed soft tissue infection, sinus discharge, and infected nonunion, and underwent a series of debridement, sequestrectomy, and free tissue transfer to cover wound defects, as well as tibiotalocalcaneal fusion for the nonunion site. The treatment spanned over 3 years and involved 11 surgeries. Unfortunately, the patient ultimately required a below-knee amputation due to relapsed infection episodes. In this study, extremity ORN stabilized after a series of surgical treatments in four out of seven patients. The cure rate for patients with extremity ORN was approximately 57% (4/7). However, three patients required limb amputation due to uncontrollable disease. 4. Discussion ORN was first described as radiation osteitis by Ewing in 1926 . In ORN, the exposed bone does not heal over 3 months after irradiation, and persistent or recurrent tumors are not evidenced . In 1983, Marx proposed the widely adopted “three H” theory to describe ORN pathophysiology . According to this theory, hypoxia, hypocellularity, and hypovascularization occur in the irradiated tissue, resulting in tissue breakdown and a non-healing wound. In recent decades, clinical investigations on ORNs have mainly focused on patients with head and neck cancers. In one study, the prevalence rate of mandibular ORN was approximately 5–15%; however, the rate varies significantly in different studies . ORN of other body parts, such as the temporal bone, chest wall, and gluteal region, has only been mentioned in a few studies [13,14,15]. Studies have described several risk factors associated with mandibular ORN, including high radiation dose, RT techniques, smoking, dental extraction, alcohol use, poor oral hygiene, and the primary tumor location [5,16,17,18,19,20]. A 3.5% (7/198) incidence rate of ORN was observed among patients with extremity STS. Non-parametric tests revealed two factors associated with ORN development, a high radiation dose (68.0 Gy vs. 64.0 Gy) and intraoperative periosteal stripping. The commonly applied dose of adjuvant RT was 50 Gy on average, followed by a boost of 10–16 Gy to the tumor bed . In our institute, we have, thus far, adopted the following protocol: for patients undergoing primary surgery, 60–64 Gy will be prescribed for negative margins, 66–70 Gy will be prescribed for microscopic residual tumors, and 70–74 Gy for gross residual tumors [22,23]. Theoretically, a higher radiation dose may provide better local tumor control in patients with a positive surgical margin and high tumor burden. However, the use of a boost remains a topic of debate considering the selection factors and potential toxicities [24,25]. Holt et al. reported that higher doses of radiation (60 or 66 Gy) are associated with a higher risk of pathological fracture . Most patients with ORN had poor tumor differentiation and a positive surgical margin. No significant difference in tumor stage, grade, or surgical margin status was observed between the ORN and non-ORN groups. A relatively high positive margin rate with a relatively low reresection rate was found. When dealing with extremity soft tissue sarcomas, the surgical goal should be to achieve microscopically negative margins to minimize local recurrence rates, if possible . However, in cases where the tumor is located near critical neurovascular structures, bones, or joints, planned resection with a microscopically positive (R1) margin may be necessary. This approach, combined with postoperative radiation therapy, can still result in outcomes similar to those achieved with negative resection margins , while minimizing morbidity and maximizing postoperative functions . The association between local tissue trauma and ORN has been repeatedly mentioned in studies on the mandibular area . We believe that the effect of periosteal stripping on the appendicular skeleton is similar to that of local tissue trauma. The procedure thins the cortex and damages the periosteal vascularity, which may increase the risk of bone fracture and local ischemia. The resulting ischemia could further deteriorate the hypoxia—hypocellularity—hypovascularization conditions. It has been mentioned as one of the risk factors for pathological fractures after treatment for soft tissue sarcoma . We also found a similar result in our patient group, although the extent of periosteal stripping was not well-documented. The clinical symptoms of ORN vary with the anatomical region, and the general presentation includes pain, paresthesia, chronic ulcers, sinus drainage, tissue necrosis, bone exposure, and pathological fractures. Other specific and serious manifestations, such as trismus and orocutaneous fistula due to ORN of the jaw, otorrhea and hearing loss due to ORN of the temporal bone area, brain abscess, and cerebrospinal fluid leakage due to ORN of the skull base, and thoracic viscera exposure and empyema due to ORN of the chest wall, have been reported [5,15,31]. The presentation of patients with extremity ORN is similar to that of patients with ORN of other anatomical regions. Therefore, a proper diagnosis and identification of the possible cause, especially the possibility of tumor recurrence, are crucial. The ORN onset time is between 4 months and 3 years after RT . However, in this study, the patients had a broad range of presentation times (3–67 months). The presentation of ORN may overlap with other disease entities. For example, postradiation fractures have been reported with an incidence of 1.2–6.4% and with well-established risk factors . These fractures can occur without significant trauma due to changes in the mechanical strength of the bone after irradiation . In our study, eight patients (4.0%) experienced pathological fractures, and five of them were managed as ORN based on the clinical history, imaging, and assessment of soft tissue quality. Patients with ORN tend to have more prominent soft tissue problems. Persistent sinus discharge with secondary infection was also frequently observed in our patient group, which is a hallmark of chronic osteomyelitis . The difference between ORN and chronic osteomyelitis could be challenging in the clinical setting. However, based on the patient’s history of previous radiation therapy and evaluation of pathological tissues , we are more inclined to diagnose ORN, as there is a certain discrepancy in the treatment principles between the two conditions. Unlike the head and neck region, the extremities have a greater amount of soft tissue covering the hard tissue. Although ORN diagnosis is primarily based on clinical manifestations, image analysis plays a critical role in detecting underlying soft tissue change. Different imaging modalities are indicated in the head and neck region, including radiographs, computed tomography, MRI, and positron emission tomography . However, in an extremity evaluation, MRI exhibits superior soft tissue contrast and delineation to other types and, therefore, is the choice of imaging modality . In ORN of the jaw, abnormal marrow signal, cortical destruction, and slight-to-mild irregular enhancement are noted in MRIs . Radiological changes in long bones after RT have been described as muscle atrophy, cortical thinning with associated remodeling, and bone infarcts in a study . In our patients, hypointensity in the bone marrow with mixed signals over the surrounding tissue was frequently observed on T1-weighted images. The hyperintense signal on STIR and T1 WI after contrast enhancement over the periphery reflected persistent inflammation and tissue necrosis. Complete recovery from mandibular ORN with conservative treatment was less than 20% in the 1990s. Therefore, several adjunctive treatments were proposed and trialed. HBO therapy with high oxygen tension stimulates fibroblast function and increases angiogenesis in the hypoxia—hypocellularity—hypovascularization environment . However, in a recent review and clinical trial on mandibular ORN, convincing results were not obtained [41,42]. Two patients underwent HBO therapy in our series, and the disease persisted. The other emerging treatment involved the combination of oral pentoxifylline, tocopherol, and clodronate. Delanian et al., in a review, discussed the effectiveness of this combination in the treatment of radiation-induced fibrosis . The results of this combination treatment for mandibular ORN appeared promising in a recent case report and systemic review [44,45]. However, only one patient received the combination for a short period in our patient group, and the effect was limited. Surgery with pathological tissue excision as well as subsequent soft tissue reconstruction (with a local or free flap) are the most promising management for advanced ORN. In patients with mandibular ORN who underwent free flap reconstruction of the jaw, a high treatment success rate and nourishment of the local vascularity by the transferred soft tissue were observed . Adequate debridement to remove all of the devitalized soft tissue and bone is essential for the initial treatment. We would collaborate with an orthopedic surgeon to either achieve rigid bone fixation or apply the induced membrane technique if a segmental bony defect is present after sequestrectomy . Systemic control with intravenous or oral antibiotics was always required for the secondary infection and was also supplemented with antibiotic-containing beads locally. The decision between locoregional tissue transfer or microsurgical reconstruction would be justified based on the patient’s clinical condition and the defect size. Even after all of the efforts, one patient in the ORN group relapsed and ended up with below-knee amputation due to uncontrollable disease. The limitations of this study include the retrospective approach and relatively small sample size. The progression of the surgical concept and newer radiation modalities may also affect the clinical outcome. Different surgeons may follow different wound management practices, thereby potentially introducing additional bias in this study. 5. Conclusions In this clinical study on extremity ORN, we found that the incidence rate of ORN was 3.5% among patients with extremity STS. Periosteal stripping during the initial operation and a high radiation dose were identified as risk factors for extremity ORN. Compared with mandibular ORN, the presentation of extremity ORN is obscure, making disease management challenging. Treatment options, including HBO, drugs, and surgical excision with soft tissue reconstruction, may be applied in patients with extremity ORN. More clinical studies are warranted for a cautious assessment and early disease prediction. Acknowledgments We thank Wen-Yu Chuang for the assistance with the pathological assessment. Author Contributions Conceptualization, C.-C.C. and C.-C.H.; methodology, S.-H.C.; software, Y.-J.L.; validation, Y.-T.L., C.-H.L. (Cheng-Hung Lin) and C.-C.H.; formal analysis, Y.-J.L.; investigation, Y.-J.L.; resources, C.-C.C. and C.-C.H.; data curation, Y.-J.L.; writing—original draft preparation, Y.-J.L.; writing—review and editing, Y.-J.L. and C.-C.H.; visualization, Y.-J.L.; supervision, Y.-T.L. and C.-H.L. (Chih-Hung Lin); project administration, C.-H.L. (Cheng-Hung Lin). All authors have read and agreed to the published version of the manuscript. Institutional Review Board Statement The study was conducted in accordance with the Declaration of Helsinki, and approved by the Institutional Review Board of Chang Gung Medical Foundation (protocol code 202100800B0) approved on 7 June 2021. Informed Consent Statement Informed consent was obtained from all subjects involved in the study. Data Availability Statement The data are available from the corresponding author upon request. 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. Funding Statement This work was supported and funded by Chang Gung Memorial Hospital, grant numbers CORPG3L0381 and CORPG3L0411. Footnotes Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. References 1.Hung G.Y., Yen C.C., Horng J.L., Liu C.Y., Chen W.M., Chen T.H., Liu C.L. Incidences of primary soft tissue sarcoma diagnosed on extremities and trunk wall: A population-based study in Taiwan. Medicine. 2015;94:e1696. doi: 10.1097/MD.0000000000001696. [DOI] [PMC free article] [PubMed] [Google Scholar] 2.Fletcher C., Bridge J., Hogendoorn P., Martens F. Soft Tissue and Bone Tumours: WHO Classification of Tumours. 5th ed. IARC Press; Lyon, France: 2020. [Google Scholar] 3.Alektiar K.M., Brennan M.F., Healey J.H., Singer S. 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Data Availability Statement The data are available from the corresponding author upon request. Articles from Cancers are provided here courtesy of Multidisciplinary Digital Publishing Institute (MDPI) ACTIONS View on publisher site PDF (39.7 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Materials and Methods 3. Results 4. Discussion 5. 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15203	https://www.studocu.com/co/document/universidad-nacional-abierta-y-a-distancia/fisica-general/laboratorio-5/93291917	Laboratorio 5: Análisis del Teorema de Conservación de la Energía en Proyectiles - Studocu Saltar al documento Profesores Universidad Instituto Descubre Iniciar sesión Te damos la bienvenida a Studocu Inicia sesión para acceder a los recursos de estudio Iniciar sesión Registrate Usuario invitado Añade tu universidad o instituto 0 seguidores 0 Subidos 0 upvotes Nuevo Página de inicio Mi Biblioteca AI Notes Ask AI Quiz IA Chats Reciente Todavía no tienes ningún elemento reciente. Mi Biblioteca Asignaturas Todavía no tienes ninguna asignatura. Añade Cursos Libros Todavía no tienes ningún libro. Studylists Todavía no tienes ninguna Studylists. 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Vista previa del texto Teorema de conservación de la energía Andrés Felipe Borda Galvis Juan Camilo Laitón Nuñes Esteban Ricardo Parrado Gomez Miladys Robles Cadena Yolian Alexander Mendoza Rosado 1 Escuela de Ciencias Básicas e Ingeniería, Universidad Nacional Abierta y a Distancia Resumen En esta práctica de péndulo balístico se analiza el comportamiento de proyectiles lanzados a diferentes velocidades utilizando esferas de acero y madera, el propósito es estudiar la conservación del momento y la energía en el sistema, la metodología implica la configuración del péndulo, mediciones de masa y ángulos, así como el cálculo de velocidades teóricas y experimentales, los resultados muestran discrepancias entre las velocidades calculadas y medidas, con porcentajes de error que varían entre 14% y 24%. Las discusiones se centran en las posibles causas de error, como las vibraciones del dispositivo y la resistencia del medio. Se proponen medidas para reducir la desviación, como estabilizar la máquina y mejorar la precisión de las mediciones. Además, se calculan las energías inicial y final del sistema, revelando discrepancias significativas, lo que plantea dudas sobre la conservación de la energía en este contexto. La comparación de los errores experimentales indica una posible mejora en la precisión a lo largo del experimento Palabras clave: incertidumbre de la medida, cinemática. Abstract In this ballistic pendulum practice, the behavior of projectiles launched at different speeds using steel and wooden spheres is analyzed. The purpose is to study the conservation of momentum and energy in the system. The methodology involves the configuration of the pendulum, mass measurements and angles, as well as the calculation of theoretical and experimental velocities, the results show discrepancies between the calculated and measured velocities, with error percentages that vary between 14% and 24%. Discussions focus on possible causes of error, such as device vibrations and medium resistance. Measures are proposed to reduce the deviation, such as stabilizing the machine and improving measurement accuracy. Furthermore, the initial and final energies of the system are calculated, revealing significant discrepancies, raising questions about energy conservation in this context. Comparison of experimental errors indicates possible improvement in precision throughout the experiment Keywords: uncertainty, kinematic. 1. INTRODUCCIÓN La práctica del péndulo balístico es un experimento fundamental en el estudio de la conservación del momento y la energía en sistemas físicos, el propósito de este informe es analizar los resultados obtenidos de una serie de lanzamientos de proyectiles utilizando un péndulo balístico, con el objetivo de investigar la conservación de estas magnitudes durante el movimiento del proyectil, el péndulo balístico consiste en un péndulo simple al cual se le agrega un dispositivo para lanzar proyectiles, cuando el proyectil es lanzado desde el péndulo, este adquiere energía cinética y se eleva a una cierta altura, la altura alcanzada depende de la velocidad de lanzamiento y otros factores físicos del sistema, el marco teórico de este experimento se basa en principios fundamentales de la física, incluyendo la conservación del momento y la energía. 10) cargue de nuevo el lanzador, luego coloque el indicador del ángulo para orientar 2 o 3º menos del alcanzado en el paso anterior (9), esto eliminará la fricción causada por el indicador en el arrastre del péndulo. así, el péndulo moverá sólo el indicador para los últimos grados. 11) dispare el lanzador, y anote el ángulo alcanzado por el péndulo en la tabla 5, al igual que la velocidad de salida de la esfera para tres lanzamientos con la primera velocidad del disparador. 12) calcule la velocidad de disparo teórica a partir de los ángulos obtenidos, usando la ecuación v=( 1 + M m ) √ 2 g Rcm ( 1 −cosθ ) 13) repita los pasos 9 a 11 con las otras dos velocidades del lanzador, hasta completar la tabla 5. 14) calcule el porcentaje de error de la velocidad de disparo experimental respecto a la velocidad calculada con la ecuación (5), usando la ecuación %Error= \|V teo−V exp\| V teo ∗ 100 15) repita el procedimiento con la esfera de madera, y complete la tabla Tabla 5 resultados obtenidos para la esfera de acero Masa M del Péndulo (kg) 0 Masa m del proyectil (kg) 0 Distancia RCM entre el pivote y el CM (m) 0 m Resultado para la primera velocidad de lanzamiento Angulo Máximo (º) Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje de error(% 23º 3 m/s 2 m/s 17% 23º 3 m/s 2 m/s 18% 23º 3 m/s 2 m/s 14% Resultado para la segunda velocidad de lanzamiento Angulo Máximo (º) Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje de error(% 32º 4 m/s 3 m/s 16% 32º 4 m/s 3 m/s 16% 36º 4 m/s 3 m/s 24% Resultado para la tercera velocidad de lanzamiento Angulo Máximo (º) Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje de error(% 44º 5 m/s 4 m/s 20% 48º 6 m/s 4 m/s 24% 44º 5 m/s 4 m/s 22% Masa M del Péndulo (kg) 0 Masa m del proyectil (kg) 0 Distancia RCM entre el pivote y el CM (m) Resultado para la primera velocidad de lanzamiento Angulo Máximo (º) Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje 11º 3 m/s 2 m/s 12º 3 m/s 2 m/s 12º 3 m/s 2 m/s Resultado para la segunda velocidad de lanzamiento Angulo Máximo (º) Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje 16º 4 m/s 3 m/s 17º 4 m/s 4 m/s 17º 4 m/s 3 m/s Resultado para la tercera velocidad de lanzamiento Angulo Máximo (º) Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje 21º 5 m/s 4 m/s 22º 6 m/s 5 m/s 22º 6 m/s 5 m/s Tabla 5 resultados obtenidos para la esfera de madera 18) Analice críticamente los porcentajes de error obtenidos en las tablas 5 y 5. ¿Cuáles son las posibles causas de error que ocasionan estas desviaciones? Durante la ejecución del experimento utilizando la máquina destinada para tal propósito, se observó detenidamente que, tras activar el dispositivo para realizar un disparo y registrar los datos pertinentes, la fuerza con la que el proyectil era expulsado resultaba considerablemente elevada. Esta fuerza generaba vibraciones en el mecanismo de la máquina, lo que provocaba que la aguja encargada de indicar el valor del ángulo utilizado en el experimento retrocediera notablemente. Este fenómeno se repetía en cada instancia de toma de datos para cada una de las esferas utilizadas en el experimento. ¿Cómo podría reducirse la desviación? Para reducir la desviación causada por las vibraciones en el mecanismo de la máquina durante la toma de datos, es crucial implementar medidas que estabilicen el sistema, esto podría lograrse colocando la máquina sobre una base amortiguadora para absorber las vibraciones, además, se puede reforzar o estabilizar el propio mecanismo de medición para hacerlo menos susceptible a las oscilaciones. ¿Cómo modificaría usted el experimento para obtener mejores resultados? Para mejorar los resultados del experimento y minimizar el impacto de las vibraciones en los datos obtenidos, consideraría varias modificaciones, primero, ajustaría el diseño de la máquina para que sea más robusto y tenga una estructura que reduzca las vibraciones, además, utilizaría un sistema de sujeción más estable para el péndulo y el proyectil, también invertiría en un sistema de medición más avanzado y preciso, menos susceptible a las vibraciones, implementaría técnicas de análisis de datos avanzadas para corregir los efectos de las vibraciones en los resultados. 19) Usando los resultados de las tablas 5 y 5, calcule el valor promedio de los ángulos máximos y la velocidad de disparo experimental, con el fin de calcular las energías mecánicas inicial y final, y registre sus resultados en la tabla 5. Calcule la energía inicial del sistema como la energía cinética del proyectil, 𝐸𝑖=12𝑚𝑣2, y la energía final como la energía potencial del sistema péndulo-proyectil, 𝐸𝑓=(𝑀+𝑚)𝑔(1+𝑐𝑜𝑠θ). Use en ambos casos los valores experimentales. Esfera de acero Velocidad de lanzamiento Angulo máximo promedio (°) Velocidad de disparo experimental promedio (m/s) Energía mecánica inicial (J) Energía mecánica final (J) Po err Primera 23° 2 m/s 1 3 76 Segunda 33° 3 m/s 2 2 18 Tercera 45° 4 m/s 3 2 14 Esfera de madera Velocidad de lanzamiento Angulo máximo promedio (°) Velocidad de disparo experimental promedio (m/s) Energía mecánica inicial (J) Energía mecánica final (J) Po err Primero 11° 2 m/s 0 2 72 Segundo 16° 3 m/s 1 2 59 Tercero 21° 5 m/s 1 2 47 Tabla 5 valores promedios del ángulo máximo y velocidad de disparo para analizar la conservación de la energía 20) De acuerdo con los resultados de la tabla 5, ¿cuál es su conclusión sobre el teorema de conservación de la energía? A partir de los resultados presentados en la tabla 5, surge una inquietud respecto a la aplicación del teorema de conservación de la energía en el sistema estudiado, la discrepancia notable entre la energía mecánica inicial y final en varios casos sugiere que, en este contexto específico, la conservación de la energía podría no cumplirse de manera estricta, esto plantea la necesidad de un análisis más detallado para comprender las causas subyacentes de esta discrepancia y evaluar la validez del teorema en este contexto particular. ¿Qué aspectos pueden ocasionar la desviación entre la energía mecánica inicial y la energía mecánica final? 2. .RESULTADOS Y DISCUSIÓN En la sección de Resultados y Discusión se presentan los datos recopilados durante la práctica del péndulo balístico, incluyendo ángulos máximos, velocidades de disparo teóricas y experimentales, así como porcentajes de error. Estos resultados se organizan en tablas y gráficos para una mejor comprensión. Se lleva a cabo una discusión detallada de los datos, observando posibles tendencias o discrepancias con el modelo teórico, como la resistencia del aire o errores experimentales. Se comparan los resultados con las expectativas teóricas y se evalúa la validez del modelo utilizado. Esta sección proporciona una visión completa de los resultados y su significado en el contexto de la física de conservación del momento y la energía. Los datos presentados son fundamentales para sacar conclusiones significativas sobre el experimento realizado. 3. CONCLUSIONES la práctica del péndulo balístico proporciona una comprensión profunda de la conservación del momento y la energía. sin embargo, se observan desviaciones significativas entre los valores teóricos y experimentales, lo que cuestiona la aplicabilidad del teorema de conservación de la energía en este caso específico. se destacan la importancia de mejorar la precisión experimental y la necesidad de una evaluación más detallada de las fuentes de error. estos hallazgos subrayan la complejidad de los sistemas físicos y la importancia de la precisión y la repetibilidad en la experimentación científica REFERENCIAS “Physics for Scientists and Engineers With Modern Physics, 9th Edition - Detalles de registro - EBSCO Discovery Service.” “Questionable Authenticity of Some Problems in ‘Fundamentals of Physics’ by Halliday, Resnick and Walker: An Initial Study of Students’ Critical Considerations - Detalles de registro - EBSCO Discovery Service.” “Mechanics: An Introduction to Different Coupling Strategies and Acceleration Techniques - Detalles de registro - EBSCO Discovery Service.”. “Classical Mechanics - Detalles de registro - EBSCO Discovery Service.” Taylor Laboratorio 5: Análisis del Teorema de Conservación de la Energía en Proyectiles Descarga Descarga Herramientas de IA Ask AI Opción múltiple Flashcards Video quiz Lección de audio 2 0 Guardar Laboratorio 5: Análisis del Teorema de Conservación de la Energía en Proyectiles Asignatura: Física General 999+documentos Universidad: Universidad Nacional Abierta y a Distancia Información Más información Descarga Descarga Herramientas de IA Ask AI Opción múltiple Flashcards Video quiz Lección de audio 2 0 Guardar INFORME DE LABORATORIO COMPONENTE PRÁCTICO Teorema de conservación de l a energía Andrés Felipe Borda Galvis Juan Camilo Laitón Nuñes Esteban Ricardo Parrado Gomez Miladys Robles Cadena Yolian Alexander Mendoza Rosado 1 Escuela de Ciencias Básicas e Inge niería, Universidad Nacional Abierta y a Distancia Resumen En esta práctica de péndulo balístico se analiza el comportamiento de proyectiles lanzados a diferentes velocidades utilizando esferas de acer o y madera, el propósito es estudiar la conservación del momento y la energía en el sistema, la metodología implica la configuración del péndulo, mediciones de masa y ángulos, así como el cálculo de velocidades teóricas y experimentales,los resultados muestran discrepancias entre las velocidades calculadas y medidas,con porcentajes de error que varían entre 14%y 24%.Las discusiones se centran en las posibles causas de error, como las vibraciones del dispositivo y la resistencia del medio.Se proponen medidas para reducir la desviación,como estabilizar la máquina y mejorar la precisión de las mediciones.Además,se calcula n las energías inicial y final del sistema,revelando discrepancias significativas,lo que plantea dudas sobre la conservación de la energía en este contexto.La comparación de los errores experimentales indica una posible mejora en la precisión a lo largo del experimento Palabras clave: incertidumbre de la medida, cinemática. Abstract In this ballistic pendulum practice, the behavior of projectiles launched at different speeds using steel and wooden spheres is analyzed.The purpose is to study the conserva tion of momentum and energy in the system.The methodology invol ves the configuration of the pendulum,mass measurements and angles,as well as the calculation of theoretical and experimental velocities,the results show discrepancies between the calculated and m easure d velocities,with error percentage s that vary between 14%and 24%. Discussions focus on possible causes of error, such as device vibrations and medium resistanc e.Measures are proposed to reduce the deviation,such as stabilizing t he machine and improving measurement accuracy.Furthermore,the initial and final energies of the system are calculated,revealing significant discrepancies,raising questions about energy conserva tion in this context.Comparison of experimental errors indicates possible improvement in pre cision throughout the experiment Keywords: uncertainty, kinematic. INFORME DE LABORATORIO COMPONENTE PRÁCTICO 1.INTRODU CCIÓN La práctica del péndulo balístic o es un experimento fundamental en el estudio de la conservación del momento y la energía en sistemas físicos, el propósito de este i nforme es analizar los resultados obtenidos de una serie de lanzamient os de proyectiles utilizando un péndulo balístico, con el objetivo de investigar la conservación de estas magnitudes durante el movimie nto del proyectil, el péndulo balístico consiste en un péndulo simple al cual se le agrega un disposit ivo para lanzar proyectiles, cuando el proyectil es lanzado desde el péndulo, este adquiere energía ciné tica y se eleva a una cierta altura, la altura alcanzada depende de la velocida d de lanzamiento y otros factores físicos del sistema, el marco teórico de este experimento se basa en principios fundamenta les de la física, incluyendo la conservación del momento y la energía. INFORME DE LABORATORIO COMPONENTE PRÁCTICO 1.MONTAJE Y PROCEDIMIENTO EXP ERIMENTAL 1)coloque el lanzador de proyectiles en el montaje del péndulo balí stico al nivel del capturador de la bola. asegúrese de que el péndulo cuelgue vertic almente con respecto al lanzador. 2)sujete la base del péndulo a la mesa. 3)ubique el péndulo a 90° con respecto a la horizontal de la parte superi or del dispositivo, luego cargue el lanzador de proyectiles con el balín. permit a al péndulo colgar libremente, y mueva el indicador del ángulo hasta que la marca de 0° coinc ida con la línea del péndulo. 4)quite el péndulo de la base desatornilla ndo y quitando el eje del pivote. mida la masa del péndulo (m) y la masa del proyectil de acero (m) y regístrelos en la tabla 5.1. 5)encuentre el centro de masa del sistema pivote-varilla-péndulo cuando el proyectil se encuentra dentro, que reemplazará la longitud del péndulo 𝐿 en la ecuación (5.3). para ell o, utilice una cuerda, cuelgue el péndulo de la cuerda hasta que se equil ibre horizontalmente y marque el punto de equilibrio sobre el péndulo. este punto corresponde al cent ro de masa del sistema pivote-varilla-péndulo (ver figura 5.2). 6)mida la distancia desde el pivote hasta el cent ro de masa (encontrado en el numeral 5), y regístrelo como rcm en la tabla 5.2 (ver figura 5.2). 7)reensamble el péndulo, y asegúrese que quede bien armado (soli cite ayuda al tutor de la práctica). asegúrese de que el indicador de ángulo esté a la derecha del péndulo.col oque el lanzador de proyectiles en el montaje del péndulo balístic o al nivel del capturador de la bola. asegúrese de que el péndulo cuelgue vertical mente con respecto al lanzador. 8)cargue el lanzador con la esfera de acero y realice un primer disparo, usando la primera velocidad de lanzamiento que ofrece el disparador. observe y registre el ángul o alcanzado. 9)cargue de nuevo el lanzador con la misma esfera y a la primera veloci dad de lanzamiento. coloque el indicador del ángulo para orientar 2 o 3º menos del alcanz ado en el paso anterior (8), lo que eliminará la fricción causada por el indic ador en el arrastre del péndulo. así, el péndulo moverá sólo el indicador para los últim os grados. luego, dispare la esfera de acero, y registre el ángulo máximo alcanza do por el péndulo en la tabla 5.1, al igual que la velocidad de salida de la esfera, registrada por el medidor digital del dispositi vo “péndulo balístico (figura 5.1)”. ¿Demasiado largo para leerlo en el teléfono? Guárdalo para leerlo más tarde en el ordenador Guardar en una Studylist INFORME DE LABORATORIO COMPONENTE PRÁCTICO 10)cargue de nuevo el lanzador, luego coloque el indicador del ángulo para orientar 2 o 3º menos del alcanzado en el paso anterior (9), esto elim inará la fricción causada por el indicador en el arrastre del péndulo. así, el péndulo moverá sólo el indicador para los últ imos grados. 11)dispare el lanzador, y anote el ángulo alcanza do por el péndulo en la tabla 5.1, al igual que la velocidad de salida de la esfera para tres lanza mientos con la primera velocidad del disparador. 12)calcule la velocidad de disparo teórica a partir de los ángulos obteni dos, usando la ecuación v= ( 1+M m ) √ 2 g R cm ( 1−cosθ ) 13)repita los pasos 9 a 11 con las otras dos velocida des del lanzador, hasta completar la tabla 5.1. 14)calcule el porcentaje de error de la velocidad de disparo experim ental respecto a la velocidad calculada con la ecuación (5.1), usando la ecuaci ón %Err or= \| V teo−V exp \| V teo ∗100 15)repita el procedimiento con la esfera de madera, y comple te la tabla Tabla 5.1 resultados obtenidos para la esfera de acero Masa M del Péndulo (kg) 0.135kg Masa m del proyectil (kg) 0.028kg Distancia R CM entre el pivote y el CM (m) 0.187 m Resultado para la primera velocidad de lanzamiento Angulo Máximo (º)Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje de error(%) 23º 3.14 m/s 2.58 m/s 17.8% 23º 3.14 m/s 2.56 m/s 18.4% 23.5º 3.21 m/s 2.60 m/s 14% Resultado para la segunda velocidad de lanzamiento Angulo Máximo (º)Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje de error(%) 32.5º 4.41 m/s 3.70 m/s 16.04% 32.5º 4.41 m/s 3.70 m/s 16.04% 36º 4.87 m/s 3.69 m/s 24.2% Resultado para la tercera velocidad de lanzamiento Angulo Máximo (º)Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje de error(%) 44º 5.90 m/s 4.71 m/s 20% 48º 6.41 m/s 4.82 m/s 24.8% 44.5º 5.97 m/s 4.65 m/s 22% INFORME DE LABORATORIO COMPONENTE PRÁCTICO Masa M del Péndulo (kg) 0.135kg Masa m del proyectil (kg) 0.012kg Distancia R CM entre el pivote y el CM (m) Resultado para la primera velocidad de lanzamiento Angulo Máximo (º)Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje d 11.5º 3.21 m/s 2.76 m/s 12º 3.35 m/s 2.82 m/s 12º 3.35 m/s 2.73 m/s Resultado para la segunda velocidad de lanzamiento Angulo Máximo (º)Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje d 16.5º 4.60 m/s 3.86 m/s 17º 4.74 m/s 4.03 m/s 17º 4.74 m/s 3.95 m/s Resultado para la tercera velocidad de lanzamiento Angulo Máximo (º)Velocidad de disparo teórica(m/s) Velocidad de disparo experimental (m/s) Porcentaje d 21º 5.84 m/s 4.99 m/s 22º 6.12 m/s 5.12 m/s 22º 6.12 m/s 5.09 m/s Tabla 5.2 resultados obtenidos para la esfera de madera INFORME DE LABORATORIO COMPONENTE PRÁCTICO 16)A partir de los datos registrados en las tablas 5.1 y 5.2, determi ne las velocidades de lanzamiento promedio para cada caso, tanto teóri cas como experimentales, para analizar la conservación de momento y energía en el experime nto. Calcule la velocidad del conjunto péndulo proyectil usando la ecuación (5.1), el momento lineal inicial como 𝑃𝑖=𝑚𝑣 y el momento lineal final como 𝑃𝑓=(𝑀+𝑚)𝑉. Complete las tablas 5.3 y 5.4 para las esfera s de acero y de madera, respectivamente. 17)A partir de los datos registrados en las tablas 5.1, 5.2, 5.3 y 5.4, determi ne las energías inicial y final del sistema para cada caso, con el objet ivo de analizar la conservación de la energía. Velocidad de lanzamiento Velocida d de disparo promedio (teórica) (m/s) Velocidad de disparo promedio (experime ntal) (m/s) Velocidad del conjunto péndulo proyectil (teórica) (m/s) Velocidad del conjunto péndulo proyectil (experimental ) (m/s) Momento lineal inicial (kg m/s) Momento lineal final (k m/s) Primera 3.16 m/s 2.58 m/s 0.54 m/s 0.44 m/s 0.088(kg m/s)0.088(kg m/ Segunda 4.56 m/s 3.69 m/s 0.78 m/s 0.63 m/s 0.127(kg m/s)0.127(kg /m Tercera 6.09 m/s 4.72 m/s 1.04 m/s 0.81 m/s 0.170(kg m/s)0.169(kg m/ Tabla 5.3 Análisis de la conservación del momento de line al para la esfera de acero Velocidad de lanzamiento Velocida d de disparo promedio (teórica) (m/s) Velocidad de disparo promedio (experimen tal) (m/s) Velocidad del conjunto péndulo proyectil (teórica) (m/s) Velocidad del conjunto péndulo proyectil (experiment al) (m/s) Momento lineal inicial (kg m/s) Moment o lineal fin m/s) Primera 3.30 m/s 2.77 m/s 0.26 m/s 0.22 m/s 0.039(kg m/s)0.038(kg m Segunda 4.69 m/s 3.94 m/s 0.38 m/s 0.32 m/s 0.056(kg m/s)0.055(kg/ Tercera 6.02 m/s 5.06 m/s 0.49 m/s 0.41 m/s 0.072(kg m/s)0.072(kg m Tabla 5.4 Análisis de la conservación del momento de line al para la esfera de madera Document gaat hieronder verder Descubre más de: Física GeneralUniversidad Nacional Abierta y a Distancia 999+documentos Ir al curso 24 Anexo 1 - Tarea 1: Ejercicios de Cinemática en Física Física General Ejercicios obligatorios 100% (10) 16 Ejercicios de Tarea 3 - Física General 100413 - Edwin Angarita Física General Trabajo tutorial 95% (21) 22 Anexo 1 - Tarea 3: Ejercicios de Estabilidad Dinámica y Movimiento Física General Preparación de examen 100% (9) 12 Tarea 2: Principios de Mecánica y Termodinámica - Física Aplicada Física General Apuntes 100% (9) 9 Tarea 2: Dinámica y Energía de la Máquina de Atwood - UNAD Física General Práctica 100% (9) 1 Fenómenos físicos y la aplicación en seguridad y salud en el trabajo Física General Apuntes 100% (8) Descubre más de: Física GeneralUniversidad Nacional Abierta y a Distancia999+documentos Ir al curso 24 Anexo 1 - Tarea 1: Ejercicios de Cinemática en Física Física General 100% (10) 16 Ejercicios de Tarea 3 - Física General 100413 - Edwin Angarita Física General 95% (21) 22 Anexo 1 - Tarea 3: Ejercicios de Estabilidad Dinámica y Movimiento Física General 100% (9) 12 Tarea 2: Principios de Mecánica y Termodinámica - Física Aplicada Física General 100% (9) 9 Tarea 2: Dinámica y Energía de la Máquina de Atwood - UNAD Física General 100% (9) 1 Fenómenos físicos y la aplicación en seguridad y salud en el trabajo Física General 100% (8) INFORME DE LABORATORIO COMPONENTE PRÁCTICO 18)Analice críticamente los porcentajes de error obtenidos en las tablas 5.1 y 5.2. ¿Cuáles son las posibles causas de error que ocasiona n estas desviaciones? Durante la ejecución del experiment o utilizando la máquina destinada para tal propósito, se observó detenidamente que, tras activar el disposit ivo para realizar un disparo y registrar los datos pertinentes, la fuerza con la que el proyectil era expulsado resultaba considerabl emente elevada. Esta fuerza generaba vibraciones en el mecani smo de la máquina, lo que provocaba que la aguja encargada de indicar el valor del ángulo utilizado en el experim ento retrocediera notablemente. Este fenómeno se repetía en cada instancia de toma de datos para cada una de las esferas uti lizadas en el experimento. ¿Cómo podría reducirse la desviaci ón? Para reducir la desviación causada por las vibracione s en el mecanismo de la máquina durante la toma de datos, es crucial implement ar medidas que estabilicen el sistema, esto podría lograrse colocando la máquina sobre una base amortiguadora para absorber las vibraciones, además, se puede reforzar o estabilizar el propio mecanismo de medic ión para hacerlo menos susceptible a las oscilaciones. ¿Cómo modificaría usted el experimento para obtener mejores re sultados? Para mejorar los resultados del experime nto y minimizar el impacto de las vibraciones en los datos obtenidos, consideraría varias modifi caciones, primero, ajustaría el diseño de la máquina para que sea más robusto y tenga una estructura que reduzca las vibracione s, además, utilizaría un sistema de sujeción más estable para el péndulo y el proyecti l, también invertiría en un sistema de medición más avanzado y preciso, menos susceptible a las vibracione s, implementaría técnicas de análisis de datos avanzadas para corregir los efectos de las vibraciones en los resultados. 19)Usando los resultados de las tablas 5.1 y 5.2, calcul e el valor promedio de los ángulos máximos y la velocidad de disparo experimental, con el fin de calcul ar las energías mecánicas inicial y final, y registre sus resulta dos en la tabla 5.5. Calcule la energía inicial del sistema como la energía cinét ica del proyectil, 𝐸𝑖=12 𝑚𝑣 2, y la energía final como la energía potencial del sistema péndulo-proyectil, 𝐸𝑓=(𝑀+𝑚)𝑔(1+𝑐𝑜𝑠 θ). Use en a mbos casos los valores experimentale s. Esfera de acero 1 de 10 Compartir Descarga Descarga Más de:Física General Más de: Física GeneralUniversidad Nacional Abierta y a Distancia 999+documentos Ir al curso 24 Anexo 1 - Tarea 1: Ejercicios de Cinemática en Física Física General Ejercicios obligatorios 100% (10) 16 Ejercicios de Tarea 3 - Física General 100413 - Edwin Angarita Física General Trabajo tutorial 95% (21) 22 Anexo 1 - Tarea 3: Ejercicios de Estabilidad Dinámica y Movimiento Física General Preparación de examen 100% (9) 12 Tarea 2: Principios de Mecánica y Termodinámica - Física Aplicada Física General Apuntes 100% (9) Más de: Física GeneralUniversidad Nacional Abierta y a Distancia999+documentos Ir al curso 24 Anexo 1 - Tarea 1: Ejercicios de Cinemática en Física Física General 100% (10) 16 Ejercicios de Tarea 3 - Física General 100413 - Edwin Angarita Física General 95% (21) 22 Anexo 1 - Tarea 3: Ejercicios de Estabilidad Dinámica y Movimiento Física General 100% (9) 12 Tarea 2: Principios de Mecánica y Termodinámica - Física Aplicada Física General 100% (9) 9 Tarea 2: Dinámica y Energía de la Máquina de Atwood - UNAD Física General 100% (9) 1 Fenómenos físicos y la aplicación en seguridad y salud en el trabajo Física General 100% (8) Recomendado para ti 16 Ejercicios de Tarea 3 - Física General 100413 - Edwin Angarita Física General Trabajo tutorial 95% (21) 16 Unidad 2 - Tarea 2 - Dinámica y energía Física General Trabajo tutorial 100% (8) 40 Informe practico fisica general Física General Trabajo tutorial 100% (6) 13 2359 ejercicios - resueltos Física General Trabajo tutorial 100% (4) 16 Ejercicios de Tarea 3 - Física General 100413 - Edwin Angarita Física General 95% (21) 16 Unidad 2 - Tarea 2 - Dinámica y energía Física General 100% (8) 40 Informe practico fisica general Física General 100% (6) 13 2359 ejercicios - resueltos Física General 100% (4) 15 Tarea 2 - Física General y Geometría Vectorial - Daniela Ramirez Física General 100% (3) 24 Anexo 1 - Tarea 1: Ejercicios de Cinemática en Física Física General 100% (10) Otros estudiantes también vieron Tarea 1 1026 100413 A 761 106023 - Unidad 1 Fase 1: Generalidades de Proyectos de Investigación Tarea 1: Análisis de Cinemática en Física General 100413 Anexo 1 - Tarea 1: Ejercicios de Cinemática en Física FIS101 - Conversión de Unidades y Magnitudes Físicas en Física I Anexo 1 - Tarea 3: Estabilidad Dinámica y Flujo de Movimiento Otros documentos relacionados Fisica Electronica taera 3 Fisica Linea DE Tiempo Historia DE LA Física timeline Timetoast timelines Tarea 3 fisica electronica II2022 Guía de Aprendizaje - Tarea 4 - Energía en Transformación - Física General Resolución 2643 de 03 Jun 2025: Sanción a Carbones Las Indias S.A.S. 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15204	https://www.nagwa.com/en/videos/276120537381/	Question Video: Recognising Graphs That Show Inverse Proportion Mathematics • Third Year of Preparatory School Which of the graphs shown represents inverse variation? [A] Graph A [B] Graph B [C] Graph C [D] Graph D Video Transcript Which of the graphs shown represents inverse variation? And then we have four graphs (A), (B), (C), and (D) plotted in the first quadrant of the coordinate plane. In order to answer this question, we’ll remind ourselves what we mean by the term “inverse variation.” Suppose we have the variables 𝑦 and 𝑥. If these are inversely proportional to one another, or if they represent inverse variation, we say that 𝑦 is proportional to one over 𝑥. The corresponding equation we use to describe this relationship is 𝑦 equals 𝑘 over 𝑥. Now, of course, 𝑦 could also be inversely proportional to 𝑥 squared. In this case, we say that 𝑦 is proportional to one over 𝑥 squared. And the corresponding equation is 𝑦 equals 𝑘 over 𝑥 squared. This holds for 𝑦 being inversely proportional to the square root of 𝑥, the cube of 𝑥, and so on. So essentially, we’re looking to find the graph of an equation of the form 𝑦 equals 𝑘 over 𝑥 to the 𝑛th power, where 𝑛 must be a positive number. And so there are some graphs that we can disregard straight away. Graph (C) is a straight line. That means its general form is 𝑦 equals 𝑚𝑥 plus 𝑏. But it passes through the origin, the point zero, zero, so, in fact, since its 𝑦-intercept is zero, we can write it as 𝑦 equals 𝑚𝑥 or 𝑦 equals 𝑘𝑥. This, in fact, is an example of two variables that are in direct proportion to one another. And so we disregard option (C). Similarly, option (D) looks like it could be of the form 𝑦 equals 𝑘 times the square root of 𝑥. And looking at a couple of coordinates here, for instance, four, two and one, one, we can certainly verify that these coordinates satisfy this equation. This is an example of 𝑘 being directly proportional to the square root of 𝑥. And so this does not represent inverse variation, and we disregard (D). Similarly, equation (B) could be of the form 𝑦 equals 𝑘𝑥 squared. It certainly looks like a quadratic. And it’s a reflection of our previous graph in the line 𝑦 equals 𝑥. Once again, this is a graph that represents direct variation. It’s 𝑦 varies directly with 𝑥 squared, and so we disregard option (B). And so this must leave option (A) only. And this makes a lot of sense when we think about what we know about the graph of, say, 𝑦 equals one over 𝑥. It looks like this, and it has asymptotes given by the 𝑦- and 𝑥-axes. Similarly, the graph of 𝑦 equals one over 𝑥 squared looks like this. In fact, if we think about the shape of the curve purely in the first quadrant, the graph of 𝑦 equals 𝑘 over 𝑥 to the 𝑛th power for positive values of 𝑛 will always have this shape. And so the answer is (A). Graph (A) then represents inverse variation: 𝑦 is inversely proportional to some power of 𝑥. Lesson Menu Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company Content Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
15205	https://math.mcmaster.ca/~boden/students/Karimi-PhD.pdf	Alternating Virtual Knots Alternating Virtual Knots By HOMAYUN KARIMI, M.SC. A Thesis Submitted to the school of Graduate Studies in Partial Fulﬁlment of the Requirements for the Degree of Doctor of Philosophy McMaster University c ⃝Copyright by Homayun Karimi, September 2018 DOCTOR OF PHILOSOPHY (2018) McMaster University (Mathematics) Hamilton, Ontario TITLE: Alternating Virtual Knots AUTHOR: Homayun Karimi, B.Sc. (Amirkabir University), M.Sc. (McMaster University) SUPERVISOR: Dr. Hans U. Boden NUMBER OF PAGES: 1, 102 ii Abstract In this thesis, we study alternating virtual knots. We show the Alexander polynomial of an almost classical alternating knot is alternating. We give a characterization theorem for alternating knots in terms of Goeritz matrices. We prove any reduced alternating diagram has minimal genus, and use this to prove the ﬁrst Tait Conjecture for virtual knots, namely any reduced diagram of an alternating virtual knot has minimal crossing number. iii Acknowledgements I would like to express my gratitude to my supervisor Prof. Hans U. Boden. He walked me through many diﬀerent areas of Knot Theory and Low Dimensional Topology. The amount of support and encouragement which I have received from him is beyond words. I would like to thank my parents for their support, Dr. Ian Hambleton and Dr. Andrew Nicas for the wonderful courses and being members of my Supervisory Committee, David Duncan for helping me in learning the basics of gauge theory, Lindsay White for providing T EXtemplate and computations of Alexander and Jones polynomials, and Amanda Tost for her encouragement during the ﬁnal year. iv Contents 1 Introduction 1 2 Classical Knot Theory 7 2.1 Knots, Links and their Invariants . . . . . . . . . . . . . . . . . 7 2.2 Alternating Knots . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3 The Tait Conjectures . . . . . . . . . . . . . . . . . . . . . . . . 19 3 Virtual Knot Theory 25 3.1 Four Equivalent Deﬁnitions . . . . . . . . . . . . . . . . . . . . 25 3.2 Invariants of Virtual Knots . . . . . . . . . . . . . . . . . . . . . 31 3.3 Checkerboard Knots . . . . . . . . . . . . . . . . . . . . . . . . 33 3.4 Parity Projection . . . . . . . . . . . . . . . . . . . . . . . . . . 38 4 Signatures for Checkerboard Knots 41 4.1 Goeritz Matrices and Signatures . . . . . . . . . . . . . . . . . . 41 4.2 Signatures and Connected Sum . . . . . . . . . . . . . . . . . . 43 4.3 Virtual Unknotting Operations . . . . . . . . . . . . . . . . . . 45 4.4 Signatures and Mirror Images . . . . . . . . . . . . . . . . . . . 48 5 Alternating Virtual Knots 52 5.1 Alternating Virtual Knots . . . . . . . . . . . . . . . . . . . . . 52 5.2 Characterization of Alternating Virtual Links . . . . . . . . . . 58 5.3 The Virtual Tait Conjecture . . . . . . . . . . . . . . . . . . . . 63 6 Khovanov Homology 69 6.1 Khovanov Homology for Classical Knots . . . . . . . . . . . . . 69 6.2 Khovanov Homology for Virtual Knots . . . . . . . . . . . . . . 73 6.3 Khovanov Homology and Alternating Virtual Links . . . . . . . 78 v 7 Problem List and Further Studies 85 Appendices 87 A Alternating Patterns 88 B Alternating Virtual Knots and Their Invariants 90 C Table of Jones Polynomials 94 vi List of Figures 2.1 The classical Reidemeister moves. . . . . . . . . . . . . . . . . . 8 2.2 The 0- and 1-smoothing of a crossing. . . . . . . . . . . . . . . . 9 2.3 A positive and a negative crossing. . . . . . . . . . . . . . . . . 9 2.4 Arc labels at a crossing. . . . . . . . . . . . . . . . . . . . . . . 11 2.5 A band move. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.6 A type A and a type B crossing (left) and a type I and a type II crossing (right). . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.7 A removable crossing. . . . . . . . . . . . . . . . . . . . . . . . . 17 2.8 The ﬂype move. In this picture T refers to a tangle diagram. . . 19 3.1 The virtual Reidemeister moves. . . . . . . . . . . . . . . . . . . 26 3.2 Reidemeister moves for Gauss diagrams. . . . . . . . . . . . . . 27 3.3 A Gauss diagram and virtual knot diagram for the almost clas-sical knot K = 4.105. . . . . . . . . . . . . . . . . . . . . . . . . 27 3.4 Modifying N ′(D) at a virtual crossing. . . . . . . . . . . . . . . 29 3.5 The virtual knots 2.1 (left) and 3.5 (right), notice that both have virtual genus 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.6 A checkerboard coloring of the knot 3.5 on a torus. . . . . . . . 34 3.7 The projection of an alternating diagram. . . . . . . . . . . . . 40 4.1 The unknotting operations. . . . . . . . . . . . . . . . . . . . . 46 4.2 A coloring and the sc operation. . . . . . . . . . . . . . . . . . . 46 4.3 A coloring and the or operation. . . . . . . . . . . . . . . . . . . 47 4.4 A coloring and the cc operation. . . . . . . . . . . . . . . . . . . 47 4.5 A colored crossing in D and −D. . . . . . . . . . . . . . . . . . 48 5.1 The source-sink orientation. . . . . . . . . . . . . . . . . . . . . 55 5.2 The Gauss diagram for 5.37. . . . . . . . . . . . . . . . . . . . . 65 6.1 A single cycle smoothing, with induced map zero. . . . . . . . . 73 vii 6.2 An x-marker for a crossing and the corresponding smoothings. . 74 6.3 The knot K = 3.7. . . . . . . . . . . . . . . . . . . . . . . . . . 76 6.4 The cube of resolutions for K = 3.7. . . . . . . . . . . . . . . . 77 6.5 A Gauss diagram and virtual knot diagram for 5.2426. . . . . . 82 6.6 A Gauss diagram and virtual knot diagram for 5.2427. . . . . . 82 A.1 Alternating patterns. . . . . . . . . . . . . . . . . . . . . . . . . 89 viii List of Tables 2.1 Table of sign, incidence number and type for a crossing c. . . . . 16 4.1 The eﬀect of applying sc, or, cc to the crossing c. . . . . . . . . 47 5.1 Almost classical knots with non-alternating Alexander polynomial. 56 5.2 Alternating almost classical knots. . . . . . . . . . . . . . . . . . 62 5.3 Signatures and genus of some almost classical knots. . . . . . . . 63 6.1 String decoration and corresponding maps. . . . . . . . . . . . . 75 6.2 The basis elements for the chain complex. . . . . . . . . . . . . 77 6.3 The image of the basis elements. . . . . . . . . . . . . . . . . . . 78 A.1 Alexander determinant for each alternating pattern. . . . . . . . 88 B.1 Alternating knots and their invariants. . . . . . . . . . . . . . . 93 C.1 Alternating virtual knots and their Jones polynomial. . . . . . . 95 ix Chapter 1 Introduction In knot theory, the objects are knots, (links, and other knotted objects such as braids, graphs, doodles, virtual knots, etc.), and the goal is to develop a precise and useful mathematical understanding of these objects. The key problems are, how can we best represent a knot (or link, or braid, etc.) mathematically? When are two representatives equivalent as knots (or links, or braids, etc.), and how to tell when they are diﬀerent? This is the classiﬁcation problem, and here invariants play the central role. The early pioneers such as P.G. Tait and C.N. Little studied the classiﬁcation problem for knots without the aid of knot invariants, and in retrospect it is amazing that they were able to produce a tabulation of knots up to 10 crossings which eﬀectively solved the classiﬁcation problem for low-crossing knots without access to such technology. Of course these early results were empirical, being based on the many examples and the mathematical intuition they had gleaned. It was only through the later work of M. Dehn, H. Seifert, J. Alexander, and K. Reidemeister that their results were placed on a sound mathematical footing. For instance, Alexander’s polynomial is remarkably strong for low-crossing knots, and it distinguishes a great many knots in these early tabulations. Nevertheless, despite this early success and the many spectacular break-throughs in knot theory that have occurred since then, the classiﬁcation prob-lem for knots remains an open and active area of research. In the late 1980s, the advent of the Jones polynomial enabled powerful and elegant solutions to the classiﬁcation problem for alternating knots. Here, a knot is called alternating if it admits a diagram whose crossings alternate between over and under crossing as one travels around the knot. Up to seven crossings, all knots are alternat-ing, and the ﬁrst non-alternating knot in the table is 819. As a result, there are 1 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics relatively few non-alternating knots of low-crossing number. Not surprisingly, in producing his ﬁrst tabulations of knots, Tait was largely concerned with the classiﬁcation problem for alternating knots. This would appear to be an easier problem, since it concerns a subset of all knots. However, it is not immediately apparent whether a given knot is a member of this subset. For instance, it is obvious that the standard diagram for 819 is not alternating, but it is at all not obvious that 819 cannot be represented by any alternating diagram. Knot invariants help to address this question, but as mentioned, Tait did not have the beneﬁt of invariants. Instead, he formu-lated three far-reaching conjectures which, when proved 100 years later, greatly facilitated the classiﬁcation of alternating knots. The ﬁrst two conjectures were settled through deep results of Kauﬀman, Murasugi, and Thistlethwaite, each giving an independent argument involving the newly discovered Jones polyno-mial [Kau87, Mur87a, Thi87]. Tait’s third conjecture was solved a few years later in a joint paper by Menasco and Thistlethwaite [MT93]. Details on the Tait conjectures for classical knots will be given in Section 2.3, for now we note that collectively, the results of Kauﬀman, Menasco, Murasugi, and Thistlethwaite in [Men84, Kau87, Mur87a, Thi87, MT93] provide a strategy for classifying al-ternating knots and links. Indeed, this problem is much more tractable than the harder problem of classiﬁcation of knots in general. For instance, the latter problem has been achieved for prime knots by computational methods up to 16 crossings in [HTW98], whereas the former has been solved up to 23 crossings, see the srticles [RFS04a, RFS04b] and the online program [RF06]. The Tait conjectures further suggest that alternating knots can be under-stood in terms of one suitably chosen representative diagram. For instance, for prime alternating knots, any minimal crossing diagram is necessarily alternat-ing, and any two such diagrams representing the same knot are related by a sequence of ﬂypes. Furthermore, a reduced alternating diagram represents a composite knot if and only if the diagram is visibly composite [Men84]. Conse-quently, for alternating knots, the crossing number is additive under connected sum, i.e., it satisﬁes c(K#J) = c(K) + c(J). This formula is conjectured to hold for all knots, but the general case is a diﬃcult open problem that serves to illustrate how much more we know about alternating knots. The resolution of the Tait conjectures, especially the third conjecture [MT91b, MT93], provide an algorithm for distinguishing alternating knots and links. In-deed, given a collection of alternating knot and/or link diagrams, one can use the results to determine which of the diagrams are prime, which are reduced, 2 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics and ultimately which among the prime and reduced ones are equivalent. What these results do not provide is a means for generating a complete col-lection of all alternating knots diagrams. We call that the problem of “genera-tion”. This is a combinatorial problem, and in fact it can be rephrased entirely as a problem about graphs using the following well-known correspondence. Every classical knot diagram is checkerboard colorable, and a coloring de-termines a graph, called the Tait graph. The Tait graph has one vertex for each black region, and edges between vertices whenever the corresponding black re-gions meet at a crossing. Obviously, performing a crossing change has no eﬀect on the Tait graph, thus the map from knot diagrams to Tait graphs is not faithful. However, it becomes faithful if one restricts attention to alternating knots, (see [Mur96] for more details). The same holds for alternating links, and in this way we see that the generation problem can be recast as a problem about graphs. This problem has been well-studied, and in fact an inductive scheme for generating all alternating knots (and links!) is contained in the papers [RFS04a, RFS04b, RF04]. Virtual knots were introduced by Kauﬀman in [Kau99], and they grew out of the study of quantum topology and ﬁnite type invariants of knots. They represent a natural generalization of classical knots to knots in thickened sur-faces. In fact it is well known that any two classical knots that are equivalent as virtual knots are in fact equivalent as classical knots, thus classical knot theory embeds faithfully into virtual knot theory. This was ﬁrst proved by Goussarov, Polyak, and Viro [GPV00], but it is also an immediate consequence of Kuper-berg’s theorem [Kup03]. Not surprisingly, many of the standard invariants of classical knots extend in a natural way to virtual knots, but in many cases, the extensions are unsatisfactory. For instance, for classical knots, the knot group GK and its peripheral structure give a complete invariant [Wal68], whereas there are many nontrivial virtual knots whose knot group and peripheral structure are “trivial,” meaning that GK ∼ = Z. In a similar way, the Jones polynomial VK(t) extends to virtual knots in a completely straightforward way, but again we ﬁnd the resulting invariant is not nearly as powerful in the virtual setting as it was for classical knots. For instance, it is simple to construct nontrivial virtual knots with trivial Jones polynomial, whereas for classical knots it is an open problem whether the only knot with trivial Jones polynomial is the un-knot. (Among alternating classical knots, the Jones polynomial is an unknot detector!) We can ask about the generation problem for alternating virtual knots. Here 3 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics we describe a way to generate all alternating virtual knot diagrams (as Gauss diagrams): Take 2n points and place them on a core circle, and then number them 1, 2, 3, · · · , 2n as we go counterclockwise around the circle. Now we draw arrows, starting at the odd points and ending at the even points. This produces an “alternating pattern,” and any choice of signs for the chords gives us an alternating virtual knot. The pattern is determined by a special kind of partition of the set {1, 2, · · · , 2n}, namely one with exactly n subsets, each of which contains 2 elements, one even and the other odd. Of course, not all the diagrams are unique, and note that we need to mod out by rotational symmetry. In this thesis, we study properties of alternating virtual knots, motivated in part by problems 15 and 16 in [FIKM14]. View a virtual knot as a knot in a thickened surface, taken up to diﬀeomorphism and stablization of the sur-face. A knot diagram on a surface is called alternating if its crossings alternate between over and under as one travels around the knot, and a virtual knot is called alternating if it can be represented by an alternating knot diagram on a thickened surface. Clearly with this deﬁnition, any knot that is alternating as a classical knot remains alternating when viewed as a virtual knot, but it is not immediately clear whether the converse is true. Question 1.1. If K is non-alternating as a classical knot, does it remain non-alternating as a virtual knot? Put another way, does there exist a classical knot which is not alternating but which admits an alternating virtual knot diagram? Apart from the Tait Conjectures, there are numerous results about invariants of classical knots that are useful in recognizing whether or not a given knot admits an alternating diagram. That is because many classical invariants take a very special form on alternating knots. For example, Murasugi proved that the Alexander polynomial of an alternating knot has non-zero coeﬃcients which are alternating in sign. Thus, any knot whose Alexander polynomial does not have this form cannot admit an alternating diagram. The following theorem gives an analogue of Murasugi’s result for alternating virtual knots which are almost classical (see Deﬁnition 3.10). Theorem 1.2. If L is an almost classical alternating link with a connected alternating diagram D, then its Alexander polynomial ∆L(t) is alternating. The Khovanov homology is a very powerful invariant for classical knots, 4 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics and in [Lee05], Lee provides a structure theorem for the Khovanov homology of alternating links. Given an alternating link L with signature σ, she proves that its Khovanov homology is supported in the two lines j = 2i −σ ± 1, where i and j are homological and quantum degrees, respectively. In this thesis, we establish the following generalization of her result for alternating virtual links: Proposition 1.3. If D is a connected alternating virtual link diagram with genus g, and signatures σξ, σξ∗, then its Khovanov homology is supported in the g + 2 lines: j = 2i −σξ∗+ 1, j = 2i −σξ∗−1, . . . , j = 2i −σξ −1. One central problem in knot theory is to characterize alternating knots in a useful or eﬀective way. The solution is to give necessary and suﬃcient conditions for a given knot to admit an alternating diagram. In [Gre17], Greene provides an answer to this question in terms of spanning surfaces, and thus he character-izes alternating classical knots as those admitting positive and negative deﬁnite spanning surfaces. We prove an analogue of Greene’s theorem, summarized in the next theorem and giving necessary and suﬃcient conditions for a virtual knot to admit an alternating virtual knot diagram. Theorem 1.4. Suppose K is a connected checkerboard colorable virtual link with virtual genus gv(K). Then D is an alternating diagram for K if and only if D admits a checkerboard coloring ξ with dual coloring ξ∗such that 1) sg(D) = gv(K), 2) σξ(D) −σξ∗(D) = 2gv(K), 3) Gξ(D) is negative deﬁnite (or empty) and Gξ∗(D) is positive deﬁnite (or empty), where Gξ and Gξ∗refer to the Goeritz matrices associated to the checkerboard colorings ξ and ξ∗, respectively (cf. Section 4.1). Sometimes in virtual knot theory, one can deﬁne invariants which have no analogue in the classical case or are trivial on all classical knots. One exam-ple is the virtual genus of a knot, which is zero for all classical knots. Using homological parity projection, it follows that for alternating virtual knots, this invariant can be computed from any reduced alternating diagram. Theorem 1.5. Suppose D is a reduced alternating virtual knot diagram for a virtual knot K. Then D is a minimal genus diagram. As a consequence of this, in Corollary 5.29 we address Question 1.1 by show-ing that any classical knot which admits an alternating virtual knot diagram is 5 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics necessarily alternating as a classical knot. Finally we have the Tait Conjectures for virtual knots. Unfortunately, the Jones polynomial is not strong enough to give a virtual analogue of the ﬁrst Tait Conjecture. Nevertheless using the previous theorem and a result of Adams et al. ([AFLT02, Theorem 1.1]), we obtain a positive solution to the ﬁrst Tait Conjecture for virtual knots, as follows: Theorem 1.6. Let D be a reduced alternating knot diagram for a virtual knot K. Then D has minimal crossing number. The outline of this thesis is as follows: In Chapter 2 we brieﬂy introduce clas-sical knot theory and some of classical knot invariants including the Jones and Alexander polynomials, knot signature and Gordon and Litherland’s method for computing the signature. Then we discuss alternating classical knots, state the Tait Conjectures, and outline a proof of the ﬁrst and second Tait Conjectures. In Chapter 3, we introduce virtual knot theory, various ways of describing them and some of their invariants. Then we discuss the checkerboard colorable knots, diﬀerent ways of describing this notion and why all these descriptions are equivalent to each other. At the end we introduce parity projection, which we will use later in the proof of Theorem 1.5. Although the notion of signature has been deﬁned for checkerboard knots, its behaviour under operations on knot diagrams has not been fully explored. In Chapter 4, we describe the signatures for virtual knots and study the eﬀect of taking mirror images and other operations on knot diagrams. In Chapter 5, ﬁrst we use the classical theorem of Bott and Mayberry to prove Theorem 1.2, and the weak form of the ﬁrst Tait Conjecture for alternating virtual links. Then we state and prove Theorems 1.4, 1.5 and 1.6. We will also apply Theorem 1.4 to decide which almost classical knots up to 6 crossings are alternating. We devote Chapter 6 in its entirety to Khovanov homology. In the ﬁrst sec-tion, we discuss Khovanov homology and Rasmussen’s invariant. Then we de-scribe Tubbenhauer’s approach to deﬁne Khovanov homology for virtual knots. We give one example of Khovanov homology of a virtual knot computed by this method. Then we prove Theorem 1.3. We list some open problems related to alternating virtual knots in Chapter 7, and at the end in the appendices we list the computations for some of the invariants of alternating virtual knots up to six crossings. 6 Chapter 2 Classical Knot Theory In this chapter, we present a brief introduction to the mathematical theory of knots and links. We recall deﬁnitions of classical invariants, including the knot group, Alexander module, Jones polynomial, and Seifert invariants. We also present the Tait conjectures for alternating knots and give a survey of the celebrated results of Kauﬀman, Murasugi, and Thistlethwaite. 2.1 Knots, Links and their Invariants In this section, we review the basic notions for classical knots and links and introduce their invariants. We begin with a review of the Kauﬀman bracket and Jones polynomial. We then recall the knot group and Alexander module, as well as invariants derived from Seifert surfaces, including the determinant, signature, and nullity. We also review Gordon and Litherland’s method of computing signatures of knots and links in terms of spanning surfaces and Goeritz matrices. Classical knot theory studies the embeddings of S1 in S3. Deﬁnition 2.1. A link L of m components is a subset of S3 consisting of m disjoint, piecewise linear, simple closed curves. A link of one component is called a knot. Deﬁnition 2.2. Two links L1 and L2 in S3 are equivalent if there is an ori-entation preserving, piecewise linear homeomorphism h : S3 →S3 such that h(L1) = L2. We can project a link onto a plane. For each link component, we have a closed curve in the projection. In general, these curves have intersection points, 7 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics but we can arrange the intersection points to be transverse double points. This is called a regular projection. To each intersection point, we record the extra information of which arc is above and which is below. This is called a link diagram. For such diagrams, there are three moves, called Reidemeister moves. Two links in S3 are equivalent if and only if their associated diagrams are related by a sequence of Reidemeister moves and planar isotopy. Figure 2.1: The classical Reidemeister moves. Deﬁnition 2.3. A link L ⊂S3 with at least two components is called split if there is a 2-sphere in S3∖L separating S3 into two balls, each of which contains a component of L. A link diagram D in the plane is a split diagram if there is a simple closed curve in R2 ∖D separating R2 into two regions, each containing part of D, i.e. the diagram is disconnected. A link is split if and only if it admits a split diagram. An invariant for links is a function which assigns an algebraic object (a number, a group, etc) to every link. This object should be “the same” for two equivalent links (equal numbers, isomorphic groups, etc). An example of an invariant for knots is the crossing number, deﬁned as follows. Deﬁnition 2.4. The minimal number of crossings taken over all diagrams of a given knot or link K is called the crossing number of K. Using invariants, we can distinguish diﬀerent links. In fact the only way to show two links are diﬀerent is to ﬁnd an invariant which assigns diﬀerent objects to them. There are many diﬀerent invariants for links. Some of these invariants are very coarse. For example the Arf invariant, only assigns 0 or 1 to a knot. Some of them can be deﬁned, only for knots. Some give us more information about knots and links. One way to test the strength of an invariant is to see if it detects the unknot. This means, if the invariant for a knot is the same as it is for the unknot, then that knot is equivalent to the unknot. 8 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics For example, a deep result of Kronheimer and Mrowka shows that Khovanov homology is an unknot detector [KM11]. It is an open problem whether there exists a nontrivial knot K whose Jones polynomial is trivial, i.e. whether the Jones polynomial is an unknot detector. The Kauﬀman bracket and Jones polynomial Let D be a diagram for a link L. For each crossing, there are two ways to resolve that crossing. One is called the 0-smoothing, and the other is the 1-smoothing, according to the following picture. c 0 1 Figure 2.2: The 0- and 1-smoothing of a crossing. Resolving all the crossings of D in both ways, we obtain 2n states, and each state is a link diagram with no crossings. If s is a state, we denote by \|s\|, the number of cycles in this state. For a state s, let i(s) be the number of 0-smoothings minus the number of 1-smoothings. Then Kauﬀman bracket is deﬁned as follows: ⟨D⟩= X s Ai(s)(−A−2 −A2)\|s\|−1 , with the sum taken over all states. One can show that the Kauﬀman bracket is invariant under Reidemeister two and three moves. ε(c) = +1 c ε(c) = −1 c Figure 2.3: A positive and a negative crossing. 9 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Deﬁnition 2.5. For a link diagram D, the writhe of D, denoted w(D) is deﬁned as n+(D) −n−(D), where n+(D) and n−(D) are the number of positive and negative crossings in D, respectively. The writhe w(D) depends on the diagram and does not give a well-deﬁned invariant of the underlying knot or link. Deﬁnition 2.6. For a link L with diagram D, the Jones polynomial of L is given by VL(t) = (−A)−3w(D) ⟨D⟩ t1/2=A−2 . This normalization is chosen so that the right hand side of the above equation is invariant under all three Reidemeister moves. It follows that the polynomial VL(t) is independent of the diagram used. Remark 2.7. If the link L has an odd number of components, VL(t) is a Laurent polynomial over the integers. If the number of components is even, VL(t) is t1/2 times a Laurent polynomial (see [Jon85, Theorem 2]). Deﬁnition 2.8. Given a Laurent polynomial P in one variable, let M(P) denote its maximum degree and m(P) its minimum degree. The span of P is deﬁned to be the diﬀerence between the maximum and minimum degrees, i.e. span P(t) = M(P) −m(P). Notice that if L has an even number of components, then by Remark 2.7, t−1/2VL(t) is a Laurent polynomial and span VL(t) means the span of t−1/2VL(t). The Knot Group and Alexander Module Another important invariant of knots and links is the knot group, which is deﬁned as the fundamental group of the complement of the knot or link. For a knot or a link K, this group is denoted GK. Thus, GK = π1(XK) where XK is the result of removing an open tubular neighborhood of K from S3. Here we describe the Wirtinger presentation of GK. Let K be a knot or a link in S3, and let D be a regular projection of it with n crossings. Enumerate the arcs of D by x1, . . . , xm and the crossings by c1, . . . , cn. For each crossing, labelled as in Figure 2.4, we have the relation ri = xλ(i)x−1 ν(i)x−1 ρ(i)xν(i). The Wirtinger presentation for GK is as follows. GK = ⟨x1, . . . , xm \| r1, . . . , rn⟩. 10 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics xν(i) xρ(i) xλ(i) Figure 2.4: Arc labels at a crossing. Here we are using the right-handed meridian convention. Notice that start-ing with a regular projection of a link K, we can ﬁnd a link which is equivalent to K. Consider a small neighborhood of each crossing in the projection plane, and let v be a normal vector to the plane. Push the over-crossing arc into S3 in the direction of v. We ﬁx a base-point above the projection plane, i.e. in the v direction. Choose m loops, denoted x1, . . . , xm, and suppose xi goes under the i-th arc in a direction which forms a positive crossing. If we use the other direction, the convention is called left-handed meridian, and it leads to another presentation for GK, which is isomorphic to the former. For this convention we have ri = xλ(i)xν(i)x−1 ρ(i)x−1 ν(i). In order to deﬁne the Alexander module, we brieﬂy recall Fox diﬀerentiation. Let Fm be the free group on m generators, so elements of Fm are words in x1, . . . , xm. For j = 1, . . . , m, the Fox derivative ∂/∂xj is an endomorphism of Z[Fm], the group ring, deﬁned so that ∂/∂xj(1) = 0 and ∂ ∂xj (xi) = ( 1 if i = j, 0 otherwise. Further, given words w, z ∈Fm, the Fox derivative satisﬁes the Leibnitz rule: ∂ ∂xj (wz) = ∂ ∂xj (w) + w ∂ ∂xj (z). These relations completely determine ∂/∂xj on every word w ∈Fm, and it is extended linearly to the group ring Z[Fm]. We use this to describe the construction of the Alexander module associated to a knot or link K. Let G′ K = [GK, GK] and G′′ K = [G′ K, G′ K] be the ﬁrst and second commutator subgroups, then the Alexander module is the quotient G′ K/G′′ K. It is a ﬁnitely generated module over Z[t, t−1], the ring of Laurent polynomials, and it is determined by the Fox Jacobian matrix A as follows. Here, 11 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics A is the n × m matrix with ij entry equal to ∂ri ∂xj x1,...,xm=t. In particular, the Fox Jacobian is obtained by Fox diﬀerentiating the relations ri with respect to the generators xj and applying the abelianization map xj 7→t for j = 1, . . . , m. We deﬁne the k-th elementary ideal Ek as the ideal of Z[t, t−1] generated by all (n −k) × (n −k) minors of A. The matrix A depends on the choice of a presentation for GK, but the associated sequence of elementary ideals {0} = E0 ⊂E1 ⊂. . . ⊂En = Z[t, t−1] does not. For any knot or link K, the ﬁrst elementary ideal E1 is a principal ideal, and the Alexander polynomial ∆K(t) is deﬁned as the generator of E1. The Alexander polynomial is well-deﬁned up to multiplication by ±tk for k ∈Z. It is obtained by taking the determinant of the Alexander matrix, which is the (n −1) × (n −1) matrix obtained by removing a row and column from A. Seifert Invariants Many useful invariants of knots and links can be deﬁned in terms of Seifert surfaces. A Seifert surface for a link K is an orientable, compact, connected surface with boundary K. This allows us to deﬁne an invariant called the Seifert genus of K, denoted g(K), to be the minimum genus over all the compact orientable surfaces in S3 which cobound K. One way to construct a Seifert surface is to apply Seifert’s algorithm to a diagram for K. We take an oriented projection of the link, cut each crossing open in the manner that preserves orientation, attach disks to each resulting circle (the so-called Seifert circuits), and connect the disks with half-twisted bands at each crossing. An orientable surface has trivial normal bundle (and trivial [−1, 1]-bundle) in S3. Since the ﬁrst Stiefel-Whitney class of the surface is zero, there is no obstruction to trivialize the normal bundle. Using the bicollaring S × [−1, 1] for S, and we can deﬁne the positive and negative push-oﬀs for curves in S. We identify S with S × {0}. For a curve in the surface, the positive push-oﬀ is a parallel copy in S × {1}, and the negative push-oﬀis the parallel copy in S × {−1}, deﬁned similarly. The Seifert pairing is deﬁned on H1(S)×H1(S), and takes (a, b) to lk(a, b+), 12 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics where b+ is the positive push-oﬀof b, and lk denotes the linking number. Choose a basis for H1(S), and let V be the matrix of the pairing with respect to this basis. We call V a Seifert matrix. Deﬁnition 2.9. Let L be a link with a connected Seifert surface S and Seifert matrix V according to a choice of basis for H1(S). (i) The determinant det(V −tV τ) is equal to the Alexander polynomial deﬁned earlier, thus setting ∆L(t) = det(V −tV τ) gives an alternative deﬁnition of the Alexander polynomial. (ii) The absolute value of the Alexander polynomial at −1 is called the deter-minant of L: det(L) = \|∆L(−1)\| = \| det(V + V τ)\|. Since the matrix V + V τ is symmetric, it has real eigenvalues, and we deﬁne its signature, sig(V + V τ), to be the number of positive eigenvalues minus the number of negative eigenvalues. We also deﬁne its nullity to be the dimension of its kernel. Deﬁnition 2.10. (i) Given an oriented link L, the signature of L is deﬁned to be the signature of the matrix V + V τ and is denoted σ(L). (ii) The nullity of L is deﬁned to be nullity(V + V τ) and is denoted N(L). Note that every classical knot K has det(K) an odd integer, thus the nullity N(K) = 0. In that case, the signature σ(K) is an invariant of knot concordance, which is introduced next. Knot Concordance Here we introduce the notion of smooth concordance for knots. Recall that an annulus is a 2-manifold A homormorphic to S1 × [0, 1]. Deﬁnition 2.11. Two knots K0 and K1 are called smoothly concordant if there is a smoothly embedded annulus A →S3×[0, 1] whose boundary is −K0×{0}⊔ K1 × {1}. A knot is smoothly slice if it is concordant to the unknot. Notice that the Euler characteristic of the annulus S1 ×[0, 1] is zero (a genus zero surface with two boundary components). We can allow the surface to have a higher genus, and in that case, K0 and K1 are called cobordant. Any two knots are cobordant. To see this, take a Seifert surface for K0 in S3 × {0}, and push the interior of it into the interior of S3 ×[0, 1], similarly for K1 in S3 ×{1}. The connected sum of the two surfaces provides a cobordism from K0 to K1. 13 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics When a knot K is slice, it bounds a smoothly embedded disk in D4. We deﬁne the slice genus of K, denoted g4(K), to be the minimum genus of all the compact orientable surfaces in D4 which bound K. It is clear that the slice genus g4(K) is less than or equal to the Seifert genus g(K). Similarly we say two oriented links L0 and L1 with m components are smoothly concordant if there is a smooth embedding of annuli (⊔m i=1Ai → S3 × [0, 1] whose boundary is −L0 × {0} ⊔L1 × {1}. Any connected cobordism between two knots K0 and K1 can be described in terms of elementary cobordisms. In fact any connected cobordism S can be decomposed into union of a ﬁnite sequence of births, deaths and saddles. A birth is a cobordism from the unknot to the empty set, a death is the cobordism from the empty set to the unknot. One can perform a saddle to arcs in a nontrivial knot or link. There are two types of saddles: (i) joining or fusion type, and (ii) separating or ﬁssion type. We can visualize a cobordism by a movie. We denote S3 × {t} by S3 t , then for a cobordism S, we can assume the height function h : S →[0, 1] is Morse. For each t, h−1(t) = S3 t ∩S. At a regular value t, h−1(t) is a knot or link in S3 t . Suppose S is a saddle between L0 and L1 with diagrams D0 and D1 respectively. To describe the saddle in terms of the movie, we choose two parallel arcs in D0 and connect them by attaching a band as in Figure 2.5, this is called a band move. saddle Figure 2.5: A band move. The Euler characteristic of a birth and a death is 1, and for the saddle, the Euler characteristic is −1. If S decomposes into b births, d deaths and s saddles, then χ(S) = b + d −s. A cobordism S is a concordance if and only if b + d = s. Spanning Surfaces and the Goeritz Matrix In general, a spanning surface for L is an unoriented (and possibly non-orientable) compact, connected surface which cobounds L. Similar to Seifert’s algorithm, we describe a way to construct spanning surfaces for a given link L. 14 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Project the link L onto the surface of S2, to obtain a projection D. Without specifying the over and under crossing information, D is just a 4-valent graph, and it divides S2 into regions. A coloring of D is an assignment of two colors (black and white) to these regions, in a checkerboard manner, i.e. if two regions share an edge, they should have diﬀerent colors. It is a well-known fact that any classical link diagram admits a checkerboard coloring, and in Section 3.3, we explain this from the point of view of virtual knot theory. Now connect the black regions by half twisted bands at each crossing, the result is a spanning surface for L, which we call it the black surface. Similarly, we can deﬁne the white surface. Deﬁnition 2.12. A diagram D for a knot K is called a special diagram if it admits a checkerboard coloring such that the black surface is oriented. Every knot admits a special diagram. See Proposition 13.15 in [BZH14]. There is a convenient way to use spanning surfaces to compute the signature of a link introduced by Gordon and Litherland in [GL78]. For a crossing c, we have the following pictures: c η(c) = 1 c η(c) = −1 c type I c type II Figure 2.6: A type A and a type B crossing (left) and a type I and a type II crossing (right). They are a type A crossing with η(c) = +1 and a type B crossing with η(c) = −1 crossing in a colored link diagram, and a type I and a type II crossing in an oriented, colored link diagram, respectively. We call η(c) the incidence number of the crossing c. Thus, to each crossing, we have three binary quantities: the sign ε(c), incidence number η(c), and type. The next lemma implies that any two of these quantities determine the third. Lemma 2.13. For a crossing c, let ε(c) be the sign of the crossing, and η(c) be the incidence number. If ε(c)η(c) = +1, then the crossing is of type II, otherwise it is of type I. 15 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Proof. This Lemma follows from the Table 2.1. ε(c) η(c) type + +1 II + −1 I − +1 I − −1 II Table 2.1: Table of sign, incidence number and type for a crossing c. We enumerate the white regions of S2 ∖\|D\| by X0, X1, . . . , Xm. Let C(D) denote the set of all crossings of D. For each pair i, j ∈{1, 2, . . . , m}, let Cij(D) = {c ∈C(D) \| c is adjacent to both Xi and Xj} and deﬁne gij =        − P c∈Cij(D) η(c), for i ̸= j, − m P k=0;k̸=i gik, for i = j. The pre-Goeritz matrix of D is deﬁned to be the symmetric integral matrix G′(D) = (gij)0≤i,j≤m, and the Goeritz matrix of D is denoted G(D) and deﬁned to be the principal minor (gij)1≤i,j≤m obtained by removing the ﬁrst row and column from G′(D). In [GL78], Gordon and Litherland deﬁned the correction term µ(D) = X c is type II η(c) and established a formula for the signature of the link in terms of the signature of the Goeritz matrix and the correction term: σ(L) = sig(G(D)) −µ(D). 2.2 Alternating Knots In this section, we review classical results of Crowell, Murasugi, and Thistleth-waite on the Alexander and Jones polynomials of alternating knots and links. 16 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics We begin by recalling the deﬁnition of alternating knots and links. Deﬁnition 2.14. A diagram D for a link L is alternating if, when traveling along each component, the over and under crossings alternate. A link L is alternating if it admits an alternating diagram. The main question, due to Ralph Fox, is “What is an alternating knot?” This is a question about how to characterize alternating knots, and in [Gre17], Greene gives a beautiful answer in terms of spanning surfaces. Greene’s result provides a topological characterization of alternating knots, and in Section 5.2, we present a virtual analogue of Greene’s characterization for virtual alternating knots. There is also a topological characterization of alternating knot exteriors by Howie (see [How17]). Another important question is how to determine whether a given knot is alternating? A number of knot invariants take a special form for alternating knots and links, and this can often be used to show that a given knot or link is not alternating. For instance, the Alexander polynomial and the Jones poly-nomial have special properties when computed for alternating knots and links, and each can be used to answer the second question. This is extremely useful as it allows us to use those invariants to determine whether a given knot or link is alternating. Deﬁnition 2.15. Let D be a knot or link diagram. A crossing c in D is called removable if we can ﬁnd a simple closed curve which intersects D only in the double point c. The diagram D is called reduced, if it has no removable crossings. J K Figure 2.7: A removable crossing. We can make a removable crossing disappear by rotating one side of the diagram 180 degrees. In [Ban30], Bankwitz proved a nontriviality result for links admitting re-duced alternating diagrams with at least one crossing. This was reproved by 17 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Crowell in [Cro59] and more recently by Balister et al. in [BBRS01] using graph theoretic methods. Proposition 2.16. Suppose L is a link admitting a reduced alternating diagram with n ≥1 crossings. Then det(L) ≥n. In particular, L is a nontrivial link. Deﬁnition 2.17. Let p(t) = Pn i=m aiti in Z[t, t−1], where m ≤n are integers. If for every i, ai and ai+1 have opposite signs, then p(t) is called an alternating polynomial. If in addition, for every m ≤i ≤n, ai ̸= 0, then p(t) is called a strongly alternating polynomial. Example 2.18. The polynomial p1(t) = t −1 + t−1 + t−3 is alternating, but p2(t) = t −t−1 + t−3 is not. In [Mur58], Murasugi proved the following theorem. Theorem 2.19. The Alexander polynomial ∆K(t) of an alternating knot K is a strongly alternating polynomial with degree 2g, where g denotes the Seifert genus of K. This theorem applies to show several families of knots are non-alternating. For example any torus knot Tp,q with p > q ≥3 is not alternating. Example 2.20. For K = T4,3, we have ∆K(t) = (t12 −1)(t −1) (t3 −1)(t4 −1) = 1 −t + t3 −t5 + t6. Therefore T4,3 is not alternating. In [Cro59] Crowell gave an independent proof of Theorem 2.19 and extended it to links. He also showed that the degree of the Alexander polynomial of a link is equal to twice its Seifert genus. In Section 5.1, we will outline his proof and generalize it to almost classical alternating knots. We conclude this section by stating without proof three useful results due to Murasugi. The ﬁrst theorem was proved in [Mur58], the second in [Mur65], and the third in [Mur89]. Please note that the third result was also proved independently by Thistlethwaite in [Thi87]. Theorem 2.21. If D is an alternating diagram for a knot K, then the Seifert surface obtained by applying Seifert’s algorithm has minimal genus. In other words, the genus of knot is realized by the genus of the given surface. 18 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Theorem 2.22. If L is a special alternating link, i.e. if L admits a special alternating diagram, then \|σ(L)\| is equal to the degree of ∆L(t). Theorem 2.23. The Jones polynomial of a non-split alternating link is alter-nating. 2.3 The Tait Conjectures Peter Guthrie Tait (1831-1901) is arguably the founding father of knot the-ory. He gave the ﬁrst tabulation of knots up to 7 crossings in 1877 and was a close friend of James Clerk Maxwell (of the famous Maxwell’s equations). Tait also corresponded frequently with William Rowan Hamilton, who invented the quaternions. In addition to his work in knot theory and topology, Tait made important contributions to combinatorics. He formulated a conjecture in graph theory (also known as the “Tait conjecture”) which would have implied the four color theorem. (This conjecture was shown to be false by Tutte in [Tut46].) For more about Tait and his life, see [OR]. In this section, we recall three conjectures of Tait and the role played by the Jones polynomial in their resolution. Conjecture 1: Any reduced diagram of a classical alternating link has minimal crossing number. Conjecture 2: An amphicheiral alternating link has zero writhe. Conjecture 3: Given any two reduced alternating diagrams D1 and D2 of an oriented, prime alternating link, D1 can be transformed to D2 by means of a sequence of ﬂype moves (see Figure 2.8). T T Figure 2.8: The ﬂype move. In this picture T refers to a tangle diagram. A Proof of the ﬁrst two Tait Conjectures For alternating links, the Jones polynomial has a special form. This property was used by Kauﬀman [Kau87], Murasugi [Mur87a] and Thistlethwaite [Thi87] to show the ﬁrst Tait conjecture is true. 19 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics The second Tait conjecture was proved by Thistlethwaite [Thi88] and Mura-sugi [Mur87b]. The Tait ﬂyping conjecture was proved by Thistlethwaite and Menasco [MT93]. The Tait ﬂyping conjecture implies that any two reduced diagrams of the same alternating knot have the same writhe, and the second Tait conjec-ture, follows from this. For simpliﬁed and shorter proofs of the ﬁrst and second Tait Conjectures see [Tur87]. We outline the proof of the ﬁrst and the second Tait Conjectures here. For more details, see [Lic97]. We denote by s0, the all 0-smoothing state, and by s1, the all 1-smoothing state. Deﬁnition 2.24. The diagram D is called plus-adequate, if for any state s′ with exactly one 1-smoothing, \|s0\| > \|s′\|. The diagram D is called minus-adequate if, for any state s′ with exactly one 0-smoothing, \|s1\| > \|s′\|. A diagram is called adequate if it is both plus- and minus-adequate. Plus-adequate means, at each crossing, two diﬀerent cycles of s0 meet. Sim-ilarly, minus-adequate means that at each crossing, two diﬀerent cycles of s1 meet. Proposition 2.25. A reduced alternating link diagram is adequate. Proof. Color the diagram, so that each crossing has η(c) = +1. Notice that by Lemma 5.18, all the crossings can have the same incidence number. The white regions corresponds to the cycles of s0. If at a crossing, a white region meets itself, that crossing is removable. This means, the diagram is plus-adequate. The proof of minus-adequate is similar. Lemma 2.26. Let D be a link diagram with n crossings. 1) M(⟨D⟩) ≤n + 2\|s0\| −2, and equality happens, if D is plus-adequate. 2) m(⟨D⟩) ≥−n −2\|s1\| + 2, and equality happens, if D is minus-adequate. Proof. Let ⟨D \| s⟩= Ai(s)(−A−2 −A2)\|s\|−1, then ⟨D⟩= P s ⟨D \| s⟩. Then M(⟨D \| s0⟩) = n+2\|s0\|−2. If a state s′, contains a 0-smoothing, and we change it to a 1-smoothing to obtain s′′, then i(s′′) = i(s′)−2, and \|s′′\| = \|s′\|±1. Thus M(⟨D \| s′′⟩) = M(⟨D \| s′⟩), or M(⟨D \| s′′⟩) = M(⟨D \| s′⟩)−4. This means, for any state s, M(⟨D \| s⟩) ≤M(⟨D \| s0⟩). If D is plus-adequate, and s′ is a state with only one 1-smoothing, then \|s′\| = \|s0\| −1, and M(⟨D \| s′⟩) = M(⟨D \| s0⟩) −4. Therefore M(⟨D⟩) = M(⟨D \| s0⟩) = n+\|s0\|−2. The proof for a minus-adequate diagram is similar. 20 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Corollary 2.27. If D is an adequate diagram, then M(⟨D⟩) −m(⟨D⟩) = (n + 2\|s0\| −2) −(−n −2\|s1\| + 2), = 2n + 2(\|s0\| + \|s1\|) −4. Theorem 2.28. If D is a connected, reduced, alternating, n-crossing link dia-gram for the link L, then span VL = n. Proof. We have 4 span VL = span ⟨D⟩= M(⟨D⟩) −m(⟨D⟩). The diagram D is alternating, so it has the boundary property (Deﬁnition 3.16), and \|s0\| + \|s1\| = n + 2 (Lemma 3.18). Combining this with Corollary 2.27, the result follows. Notice that, for any link L with n crossings, span VL ≤n. Corollary 2.29. The ﬁrst Tait Conjecture is true. Proof. Let D be a connected, reduced, alternating diagram with n crossings for a link L. Then span VL = n. If D′ is another diagram for L with n′ crossings, then n = span VL ≤n′. Example 2.30. For the knot K = 819, VK(t) = t3 + t5 −t8. Then span VK = 8 −3 = 5 < 8. If 819 were alternating, then it would have a diagram with 5 crossings. Using other invariants, 819 is diﬀerent from any other knot with 5 crossings or less. Therefore 819 is not alternating. Also according to [Lic97, Table 6.1], this knot has Alexander polynomial ∆K(t) = 1 −t + t3 −t5 + t6 and det(K) = 1. So Proposition 2.16 applies and shows K is not alternating. (One can also conclude this from Theorem 2.19.) Deﬁnition 2.31. Let D be a link diagram, its r-parallel Dr, is the diagram in which each link component of D is replaced by r copies, all parallel in the plane, and each copy repeating the “over” and “under” behavior of the original link component. Lemma 2.32. If D is plus-adequate, then Dr is also plus-adequate. If D is minus-adequate, then Dr is also minus-adequate. Proof. The all 0-smoothing state s0Dr, is r parallel copies of s0D. Each cycle of s0Dr runs parallel to a cycle of s0D, and there cannot be a crossing in which, a cycle of s0Dr meets itself. 21 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Theorem 2.33. Let D and E be diagrams with nD and nE crossings, respec-tively, for the same link L. Suppose D is plus-adequate, then nD −w(D) ≤ nE −w(E). Proof. Let {Li} be components of L, and Di and Ei be sub-diagrams of D and E, corresponding to Li. Choose non-negative integers µi and vi, such that for any i, w(Di) + µi = w(Ei) + vi. Change D to D∗, by changing each Di to D∗i, by adding to Di, µi positive kinks. Similarly, add vi positive kinks to Ei. Note that D∗is still plus-adequate, and w(D∗i) = w(Di) + µi = w(Ei) + vi = w(E∗i). It follows that w(D∗) = w(E∗), because the writhe of a link diagram, is the sum of writhes of the components, and the sum of the signs of the crossings between diﬀerent components. The second part, is a combination of linking numbers, so remains unchanged. For any r, take Dr ∗and Er ∗. Then w(Dr ∗) = r2w(D∗), because in r-parallel of a diagram, each crossing is replaced by r2 crossings of the same sign. The diagrams Dr ∗and Er ∗, are equivalent, so VDr ∗(t) = VEr ∗(t), also their writhes are equal, therefore their bracket polynomials are equal, ⟨Dr ∗⟩= ⟨Er ∗⟩. For any r, by Lemma 2.26, we have: M(⟨Er ∗⟩) ≤ (nE + X i vi)r2 + 2(\|s0E\| + X i vi)r −2, M(⟨Dr ∗⟩) = (nD + X i µi)r2 + 2(\|s0D\| + X i µi)r −2. If for every positive integer r, ar2 + br −2 ≤cr2 + dr −2 with a, b, c, d ∈Z and b positive, then a ≤c. Suppose to the contrary that a > c, then a ≥c + 1. Choose r > max{b, d}, then: cr2 + dr −2 ≤cr2 + r2 −2 = (c + 1)r2 −2 ≤ar2 −2 < ar2 + br −2, which is a contradiction. Therefore we have: nD + X i µi ≤ nE + X i vi, nD − X i w(Di) ≤ nE − X i w(Ei). 22 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Once again, using the fact that the sum of the signs of crossings of distinct components, is determined by linking numbers of the components of L, the result follows. Corollary 2.34. Let D and E be diagrams with nD and nE crossings, respec-tively, for the same link L. (i) If D is plus-adequate, then the number of negative crossings of D is less than or equal to the number of negative crossings of E. (ii) If D is minus-adequate, then the number of positive crossings of D is less than or equal to the number of positive crossings of E (iii) An adequate diagram has the minimal number of crossings. (iv) Two adequate diagrams of the same link (e.g. reduced alternating dia-grams) have the same writhe. Proof. (i) Let, n+ and n−be the number of positive and negative crossings, respectively. We have nD −w(D) ≤ nE −w(E), n+(D) + n−(D) −(n+(D) −n−(D)) ≤ n+(E) + n−(E) −(n+(E) −n−(E)), n−(D) ≤ n−(E). (ii) Use the negative kinks, in the proof of the theorem. It follows that −nD + X i µi ≥−nE + X i vi = ⇒ nD − X i µi ≤nE − X i vi, nD + w(D) ≤nE + w(E) = ⇒ n+(D) ≤n+(E). (iii) Follows from (i) and (ii). (iv) From (iii), we have nD = nE. It follows from the theorem that nD −w(D) ≤nE −w(E) ⇒w(E) ≤w(D). From (ii), we have nD + w(D) ≤nE + w(E) ⇒w(D) ≤w(E). Therefore w(D) = w(E). 23 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Therefore, this theorem proves the ﬁrst two Tait Conjectures. One application of the ﬁrst Tait Conjecture is to show that the crossing number for alternating knots is additive under connected sum. In general it is not known whether the crossing number is additive under connected sum or not. In [Mur89], Murasugi proves that if D is a reduced alternating link diagram for a non-split link L, then M(VL(t)) + m(VL(t)) = w(D) −σ(L). This means that we can read oﬀthe signature of a non-split alternating link from the Jones polynomial. 24 Chapter 3 Virtual Knot Theory In this chapter, we provide an introduction to virtual knots and links. We begin with four equivalent deﬁnitions of virtual knots as equivalence classes of (i) vir-tual knot diagrams, (ii) Gauss diagrams, (iii) knots in thickened surfaces, and (iv) abstract link diagrams. We explain how to extend many of the invariants from classical to virtual knot theory, including the Jones polynomial and the knot group, and we recall a number of invariants of virtual knots, including the virtual knot group and virtual Alexander polynomial. In contrast to the situation for classical knots, not all virtual knots and links are checkerboard col-orable. We prove a result relating checkerboard colorability to the virtual knot or link satisfying the boundary condition. We also introduce almost classical knots and links and review the general theory of parity and projection, due to Manturov. Throughout this thesis, we will use decimal numbers to refer to virtual knots in Green’s tabulation [Gre04]. 3.1 Four Equivalent Deﬁnitions Kauﬀman introduced the notion of virtual knots and links in [Kau99]. In this section, we review how to deﬁne virtual knots and links via virtual knot dia-grams, Gauss diagrams, abstract link diagrams, and knots and links in thickened surfaces. First description: We consider a collection of immersed closed curves in the plane, with a ﬁnite number of intersection points which all are double points. Record extra information at each intersection point by specifying which one of 25 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics the two strands goes over the other (classical crossing), or we do not specify it by putting a small circle around the intersection point (virtual crossing). This is called a virtual link diagram. A virtual link is an equivalence class of virtual link diagrams modulo the generalized Reidemeister moves and the planar isotopy. The combination of classical and virtual Reidemeister moves are called the generalized Reidemeister moves. See Figures 2.1 and 3.1. Figure 3.1: The virtual Reidemeister moves. Second description: A Gauss diagram is a decorated trivalent graph con-sisting of one or more core circles, oriented counterclockwise, together with a ﬁnite collection of signed, directed chords connecting distinct pairs of points on the circles. Each core circle represents a knotted curve on a surface, and the directed chords, which are also called arrows, connect preimages of the classical crossings of the underlying immersed curve; they point from the over-crossing arc to the under-crossing arc, and their sign indicates the writhe of the classical crossing. A virtual knot or link is then an equivalence class of Gauss diagrams under the equivalence generated by the virtual Reidemeister moves. In [Pol10], Polyak showed that all Reidemeister moves can be generated by the four moves Ω1a, Ω1b, Ω2a and Ω3a (see Figure 3.2). This observation facilitates deﬁning new invariants of classical and virtual knots and links. Starting with a virtual link diagram, we can associate a Gauss diagram to it. Number the classical crossings in an arbitrary order. Choose a point on each component, and a direction along which we start to travel that component. Now for each component draw a circle. If that component has n classical crossings, then choose 2n points on the associated circle. Choose a point on the circle as well, and start traveling counter-clockwise on the circle, while we are traveling on the link diagram component, along the chosen direction. Every time we pass a classical crossing on the diagram, we should pass a point on the circle. If we pass an under-crossing, put an arrow-tail on the associated point on the circle, 26 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics + Ω1a Ω1b + Ω2a + − + − + Ω3a − + + + + − Ω3a + + − Figure 3.2: Reidemeister moves for Gauss diagrams. otherwise we put an arrow-head. We also record the number of the classical crossing which we pass, beside the associated point on the circle. At the end, match the arrow-head and tails with the same number, and record the sign of the classical crossing beside the arrow-head or tail. Example 3.1. Here we have the virtual diagram and the Gauss diagram for the knot 4.105. − − − − Figure 3.3: A Gauss diagram and virtual knot diagram for the almost classical knot K = 4.105. Conversely, given a Gauss diagram D with n chords, enumerate the chords by 1, . . . , n. Draw n disjoint classical crossings in the plane and enumerate them by c1, . . . , cn. For each chord i in D with the sign ε, assign suitable direction to the arcs of ci in order to ci has the sign ε. Start with a point on each 27 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics core circle of D and go around counterclockwise. If you pass a chord’s tail, in the corresponding classical crossing we have to go along the under-crossing arc. Passing a chord’s head, we go along the over-crossing arc. Each core circle of D determines a component of the link diagram. Every time we cross an arc in a point other than the classical crossings, we need to put a circle around the intersection point to indicate it is a virtual crossing. Third description: Deﬁnition 3.2. Let Σ be a closed, oriented surface and I = [0, 1]. The product Σ × I is called a thickened surface. A link L in a thickened surface Σ × I is a ﬁnite collection of disjoint 1-dimensional submanifolds in the interior of Σ × I with each connected component is diﬀeomorphic to a circle. Deﬁnition 3.3. Stable equivalence on links in thickened surfaces is generated by the following operations, which transform a given link L in a thickened surface Σ × I into a new link L′ in a possibly diﬀerent thickened surface Σ′ × I. (i) Let f : Σ × I →Σ′ × I be an orientation-preserving diﬀeomorphism sending the orientation class of Σ to that of Σ′ (this implies that f(Σ × {0}) = Σ′ × {0} and f(Σ×{1}) = Σ′ ×{1}). The link L′ = f(L) in Σ′ ×I is said to be obtained from L in Σ × I by a diﬀeomorphism. (ii) Let h : S0 × D2 →Σ be the attaching region for a 1-handle that is disjoint from the image of L under projection Σ × I →Σ, then 0-surgery on Σ along h is the surface Σ′ = Σ ∖h(S0 × D2) ∪S0×S1 D1 × S1. The link L′ is the image of the link L in Σ′ × I, and we say that it is the link obtained from L by stabilisation. (iii) Destabilisation is the inverse operation, and it involves cutting Σ×I along a vertical annulus A and attaching two copies of D2 × I along the two annuli. In the resulting thickened surface, we keep only the components containing L. Note that in (iii), an annulus A in Σ×I is called vertical if there is an embedded circle γ ⊂Σ such that A = γ × I ⊂Σ × I. An equivalence class under the equivalence relation generated by (i), (ii), and (iii) above is called a virtual link. Starting with a virtual link L in Σ × I, consider the projection p : Σ × I → Σ×{1 2}. By a small isotopy on L we can make sure D = p(L) has only transverse double points. Choosing a point on each component of D and going around that component, we can read oﬀthe Gauss diagram. To obtain a virtual diagram we 28 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics can use the Gauss diagram or we can project D onto a plane (again make sure the intersection points in the image are transverse). Each double point on D determines a classical crossing on the planar diagram. Any other double point is a virtual crossing. Fourth description: Suppose S is a compact oriented surface with boundary. Let D be a link diagram on S with no virtual crossing. We denote by \|D\|, the graph obtained by replacing each classical crossing in D by a four-valent vertex. We say P = (S, D) is an abstract link diagram (ALD), if \|D\| is a deformation retract of S. Let Σ be a closed, connected and oriented surface and f : S →Σ be an orientation preserving embedding. We call (Σ, f(D)) a realization of P. Two abstract link diagrams (S1, D1) and (S2, D2) are said to be abstract R-move equivalent, if there are realizations (Σ, f1(D1)) and (Σ, f2(D2)) on the same surface Σ such that f1(D1) diﬀers from f2(D2) by performing one Reidemeister move on F. Two abstract link diagrams (S1, D1) and (S2, D2) are abstract equivalent, if one can be changed into the other by a ﬁnite sequence of abstract R-moves. A virtual link is an equivalence class of abstract link diagrams modulo ab-stract equivalence. Associated with each virtual link diagram, we can construct an abstract link diagram (see [KK00]). We review that construction here. Let D be a virtual link diagram with n classical crossings and U1, U2, . . . , Un regular neighborhoods of the crossings of D. Put W = cl(R2 −∪n i=1Ui). Thick-ening the arcs and loops of D ∩W, we obtain immersed bands and annuli in W whose cores are D ∩W. Their union together with U1, U2, . . . , Un forms an immersed disk-band surface N ′(D) in the plane. Modifying N ′(D) as shown below at each virtual crossing of D, we obtain a compact oriented surface SD embedded in R3, and a diagram ˜ D on SD corresponding to D. We call the pair P = (SD, ˜ D) the abstract link diagram associated with D. or Figure 3.4: Modifying N ′(D) at a virtual crossing. 29 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics In [KK00], the authors show there is a bijection between the set of virtual link diagrams modulo the generalized Reidemeister moves and the set of abstract link diagrams modulo abstract equivalence. The supporting genus of an ALD P = (Σ, ˜ D) is denoted by sg(P), and is deﬁned to be the minimal genus among the realization surfaces F of P. The supporting genus of a virtual link diagram D is deﬁned to be the supporting genus of the ALD P = (SD, ˜ D) associated with D and denoted by sg(D). The virtual genus of a virtual link L is denoted by gv(L) and deﬁned to be the minimum number among the supporting genus sg(D), where D runs over all virtual link diagrams representing L. Let L be a virtual link. A virtual link diagram D representing L such that sg(D) = gv(L) is called a minimal diagram of L. All these four deﬁnitions are equivalent and we will use them interchange-ably. If we think of S3 as R3 with a point at inﬁnity, then a classical knot K ⊂S3 can be isotoped to be disjoint from the two points {0, ∞}. Thus, we can view it as a knot in the thickened surface S2 × I. The associated virtual knot is independent of the choice of isotopy, and we call such a knot classical. No-tice that a classical knot diagram D (a virtual knot diagram with no virtual crossings) has supporting genus zero, thus any classical knot has virtual genus zero. Therefore a virtual knot is classical if and only if its virtual genus is zero. Kuperberg [Kup03, Theorem 1] proved a strong uniqueness result for minimal genus representatives. Namely, he showed that if K ⊂Σ×I and K′ ⊂Σ′×I are two minimal genus representatives for the same virtual knot, then K′ = f(K) for some diﬀeomorphism f : Σ × I →Σ′ × I as in (i) of Deﬁnition 3.3 above. Deﬁnition 3.4. A virtual link is said to be split, if it admits a virtual link diagram D such that \|D\| is not connected. Otherwise we say the virtual link is connected or non-split. Deﬁnition 3.5. (i) Two given knots K0 ⊂Σ0×I and K1 ⊂Σ1×I in thickened surfaces are virtually concordant if there exists a connected oriented 3-manifold W with ∂W ∼ = −Σ0 ⊔Σ1 and an annulus A ⊂W × I cobounding −K0 ⊔K1. (ii) A knot K ⊂Σ × I is called virtually slice if it is concordant to the unknot. Equivalently, the knot K is virtually slice if there exists a connected 3-manifold W with ∂W ∼ = Σ and a 2-disk ∆⊂W × I cobounding K. We call ∆a slice disk for K. Remark 3.6. Note that a result of Boden and Nagel shows that a classical knot 30 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics K is virtually slice if and only if it is classically slice [BN17]. 3.2 Invariants of Virtual Knots In this section, we review a number of invariants of virtual knots and links. Some of the invariants, like the virtual crossing number and virtual genus, have no analogue in the theory of classical knots. Since the virtual genus was already discussed in the previous section, we recall the deﬁnition of the crossing number. Given a virtual knot or link diagram D, let v(D) denote the number of virtual crossings of D. Then the virtual crossing number of a virtual knot or link K is deﬁned to be v(K) = inf{v(D) \| D is a virtual knot diagram representing K}. As with the virtual genus, we see that a virtual knot or link K is classical if and only if v(K) = 0. Many of the standard invariants of classical knots and links extend in a straightforward way to virtual knots and links. For example, the Kauﬀman bracket ⟨D⟩and Jones polynomial VK(t) can be deﬁned for virtual knots and links in exactly the same way, but the resulting invariants are much less pow-erful than in the classical setting. Indeed, as we shall see, there are nontrivial virtual knots K with trivial Jones polynomial VK(t) = 1, and the knot K can even be taken to be alternating. In fact, in chapter 6 we shall see the same phenomenon occurs for the Khovanov homology for virtual knots, which is to say that Khovanov homology is not an unknot detector for virtual knots. The knot group is another powerful invariant of classical knots which gen-eralizes in a natural way to virtual knots and links by means of Wirtinger presentations. As an invariant of classical knots, the knot group is an unknot detector, indeed the only classical knot K whose knot group is inﬁnite cyclic is the trivial knot. In fact, Waldhausen’s theorem implies that the knot group together with its peripheral structure is a complete invariant of classical knots, which is to say that two classical knots are equivalent if and only if they have isomorphic knot groups with equivalent peripheral structures. Unfortunately, these results fail in the virtual setting; one can construct nontrivial virtual knots with trivial knot group. For a virtual knot K, one can mimic the construction of the Alexander module by regarding the quotient G′ K/G′′ K as a module over Z[t, t−1], This can 31 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics be used to deﬁne elementary ideals and the Alexander polynomial for virtual knots and links. However, in contrast to the case of classical knots, the ﬁrst elementary ideal may not be principal. One way to remedy the situation is to replace the elementary ideals Ek with the smallest principal ideal containing them. For instance, this would suggest a way to deﬁne an Alexander polynomial for a virtual knot K to be a generator of the principal ideal containing E1. However, since the knot group itself is only an invariant of the associated welded knot1, the invariants one obtains in this way will not be very reﬁned. Indeed, the Alexander polynomial for the vast majority of low-crossing virtual knots is trivial. An alternative is to work with the virtual knot group V GK, which was intro-duced in [BDG+15]. In particular, the virtual Alexander polynomial HK(s, t, q) is deﬁned in terms of the elementary ideals of V G′ K/V G′′ K, which is a ﬁnite dimensional module over Z[s±1, t±1, q±1] called the virtual Alexander module. The virtual Alexander polynomial records information about the virtual cross-ing number; for example Theorem 3.4 of [BDG+15] implies that q-width HK(s, t, q) ≤2v(K), thus HK(s, t, q) provides a lower bound of the virtual crossing number of K. Further, the virtual Alexander polynomial is intimately related to the gener-alized Alexander polynomial GK(s, t) of Sawollek and Silver-Williams, Proposi-tions 3.8 and Corollary 4.8 of [BDG+15] show that each one of these polynomials determines the other: GK(s, t) = HK(s, t, 1) and HK(s, t, q) = GK(sq−1, tq). In order to study virtual knot concordance, it would be useful to have a virtual analogue of the knot signature. However, since virtual knots do not generally admit Seifert surfaces, the standard approach does not work to deﬁne signatures for virtual knots. One approach is to focus attention on almost classical knots, which consist exactly of virtual knots that do admit Seifert surfaces. The idea is to deﬁne signatures in the restricted setting of almost classical knot and to use parity projection to extend them to all virtual knots. This approach is carried out in [BCG17a]. 1Two virtual knots or links are said to be welded equivalent if one can be obtained from the other by generalized Reidemeister moves plus the forbidden overpass, which is the move that exchanges two adjacent arrow-tails on a Gauss diagram. 32 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics An alternative approach which was developed by Im, Lee and Lee in [ILL10] is to generalize the method of Gordon and Litherland by deﬁning Goeritz matri-ces and correction terms of checkerboard knots and links. We summarize their construction in Chapter 4, where the reader will ﬁnd a full account of these invariants. 3.3 Checkerboard Knots In this section, we introduce the notion of checkerboard coloring for virtual knots and links. We then recall the notion of the boundary property for virtual link diagrams [Dye16] and relate it to checkerboard colorability. Deﬁnition 3.7. Given P = (F, D), where F is a compact, connected, oriented surface and D is a link diagram on F, a checkerboard coloring ξ is an assignment, to each region of F ∖\|D\|, one of two colors, say black and white, such that any two adjacent regions sharing an edge of \|D\| have diﬀerent colors. A virtual link diagram D is said to be checkerboard colorable if the associated ALD P = (SD, ˜ D) admits a checkerboard coloring. A checkerboard link is a virtual link L which can be represented by a checkerboard colorable virtual link diagram. Given a pair P = (F, D) with checkerboard coloring ξ, deﬁne the dual checkerboard coloring ξ∗to be the one obtained from ξ by interchanging black and white regions. If a virtual link L is checkerboard colorable, then it admits two colorings which are dual to one another. Note that, being checkerboard colorable depends only on the underlying ﬂat virtual link diagram. A ﬂat diagram of a link is the link projection, so we do not specify whether a classical crossing is an over- or under-crossing. Example 3.8. The virtual knot 2.1 is not a checkerboard knot. The virtual knot 3.5 is a checkerboard knot (see Figure 3.5). Deﬁnition 3.9. Suppose c is a chord in a Gauss diagram D, which we draw with c pointing up. We deﬁne the index of c as ind(c) = r+(c) −r−(c) + ℓ−(c) −ℓ+(c), 33 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Figure 3.5: The virtual knots 2.1 (left) and 3.5 (right), notice that both have virtual genus 1. Figure 3.6: A checkerboard coloring of the knot 3.5 on a torus. where r±(c) are the number of ±-chords intersecting c and pointing to the right, and ℓ±(c) are the number of ±-chords pointing to the left. Deﬁnition 3.10. Let D be a Gauss diagram. If ind(c) = 0 for every chord c, then D is called an almost classical diagram. A virtual knot K which admits an almost classical diagram is called an almost classical knot. Deﬁnition 3.11. Let D be a Gauss diagram. If ind(c) = 0 (mod p) for every chord c, then D is called a mod p almost classical diagram. A virtual knot K which admits a mod p almost classical diagram is called a mod p almost classical knot. Observe that a virtual knot diagram D is checkerboard colorable if and only if every chord c of D has ind(c) = 0 (mod 2). Remark 3.12. (i) A virtual knot is checkerboard if and only if it can be rep-resented by a knot in Σ × I which bounds a spanning surface (possibly non-orientable). (ii) A virtual knot is almost classical if and only if it can be represented by a knot in Σ × I which bounds an orientable spanning surface. 34 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Suppose a Gauss diagram D for a knot K has n chords. Then these chords divide the core circle into 2n arcs. We start on one arc and label the right side of it by w (for white). As we go around the core circle counterclockwise, every time we pass a new arc we change the label from right side to the left and vice versa. Let c be a chord and assume when we pass the tail/head of c, we change the label from right side to the left. When we arrive to the head/tail of c, we should change the label from left to the right. This property should hold for all the chords in order for D to admit a checkerboard coloring. Notice this property holds if and only if the number of arcs at one side (hence at both sides) of c is odd for every chord c. Again this is true if and only if the number of chords intersecting c is even for every chord c. The latter means ind(c) = 0 (mod 2) for every chord c. In [Kam02], Kamada proves that a virtual link diagram is checkerboard colorable if and only if it can be made alternating by changing a ﬁnite number of classical crossings from over-crossing to under-crossing or vice versa. Notice that Kamada’s result implies that every classical link diagram admits a checkerboard coloring, and next we give an alternative, elementary proof of this well-known fact. Suppose D is a classical diagram for a knot K. Pick a crossing c and smooth it in an oriented way, i.e. 0- or 1-smooth c if it is a positive or negative crossing respectively. The result is a two component link diagram. Every double point is either a self-intersection point of one of the components, or is a intersection point of the two components with each other. The self-intersection points correspond to chords on the Gauss diagram which are entirely on one side of the chord c. Now consider an intersection point between the two components. If we ignore one of the link components, then there is a simple closed curve (which is a subset of the second link component) which the ﬁrst link component enters it at the intersection point, hence by the Jordan curve theorem, it has to leave the simple closed curve at some other intersection point. On the Gauss diagram this means if a chord intersects the chord c, there should be another intersecting chord. Therefore the mod 2 index of c is zero. This completes the proof. We introduce the coloring matrix of a knot or link diagram. Given a knot or link diagram D, denote the classical crossings and arcs of D by {c1, · · · , ck} and {a1, · · · , ak} respectively. The k × k coloring matrix M(D) of D can be 35 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics deﬁned as follows: mij(D) =        2, if aj is the over-crossing arc at ci, 1, if aj is an under-crossing arc at ci, 0, otherwise. In particular if aj is the over-crossing and an under-crossing arc at ci, then mij(D) = 2 −1 = 1. Deﬁnition 3.13. Let D be a virtual knot or link diagram with coloring matrix M(D). The absolute value of a principal minor of M(D) is called the Alexander determinant of D. It is also called the knot determinant of D. It follows from [BGH+17] that the Alexander determinant is well-deﬁned for a checkerboard colorable knot or link L. We denote the Alexander determinant of L by det(L). If we remove the sign of all the chords in a Gauss diagram, the remaining object is called a Gauss pattern, deﬁned as follows. Deﬁnition 3.14. A Gauss pattern is a trivalent graph with a core circle con-taining 2n distinct points and n oriented edges which connect the points in pairs (e.g. see Figure A.1 in Appendix A). Proposition 3.15. The Alexander determinant of a link depends only on the underlying Gauss pattern. Proof. Starting with a Gauss pattern, on each core circle enumerate the arcs which are between two consecutive arrow-heads. On the planar diagram for a knot this is the same, as we enumerate the arcs between two under-crossings. Now suppose D1 and D2 are two Gauss diagrams with the same Gauss pattern. Enumerate the arcs and chords of the Gauss pattern. It follows M(D1) and M(D2) are exactly the same matrices. In particular they have equal principal minors. Given a virtual link diagram D, for each classical crossing, we can resolve the crossing into a 0-smoothing or a 1-smoothing (see Figure 2.2). If we resolve all the classical crossings, the resulting diagram is called a state. Then, a state is a virtual link diagram with only virtual crossings, i.e. it is an unknotted diagram of the unlink. For a link diagram with n classical crossings, we have 2n states. In fact, once an ordering of the crossings {c1, . . . , cn} has 36 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics been ﬁxed, the states are in one-to-one correspondence with binary strings of length n. For a given state s, the dual state is denoted ¯ s and it is the one obtained from s by changing all 0-smoothings to 1-smoothings, and vice versa. In other words, if s corresponds to the binary word with i-th entry si ∈{0, 1}, then ¯ s corresponds to the binary word with i-th entry ¯ si = 1 −si. Deﬁnition 3.16. Let D be a virtual link diagram, and (SD, ˜ D) be the abstract link diagram associated with D. Then D has the boundary property if there exists a state s∂such that ∂SD = s∂∪¯ s∂, where ¯ s∂is the dual state of s∂. The following lemma relates the boundary property to checkerboard col-orability. Lemma 3.17. A virtual link diagram D has the boundary property if and only if it is checkerboard colorable. Proof. Suppose D has the boundary property and deﬁne a checkerboard coloring ξ as follows. Let the white regions be those with boundary a component of s∂, and let the black regions be those with boundary a components of ¯ s∂. This gives a checkerboard coloring ξ for D. Conversely, suppose ξ is a checkerboard coloring of the abstract link diagram (SD, ˜ D). Let s∂be the state obtained by performing 0-smoothing to all crossings c with η(c) = +1 and 1-smoothing to all crossings c with η(c) = −1, and let ¯ s∂ be the dual state. Then it can be easily checked that ∂SD = s∂∪¯ s∂, therefore D has the boundary property. Since every alternating virtual link diagram is checkerboard colorable, it follows that every alternating virtual link diagram has the boundary property. Let \|s∂\| and \|¯ s∂\| be the number of components of s∂and ¯ s∂, respectively. Lemma 3.18. Suppose D is a virtual link diagram with n classical crossings. If SD has genus g and D has the boundary property, then \|s∂\| + \|¯ s∂\| = n + 2 −2g. Proof. Attach disks to the boundary components of SD to get a closed surface Σ. There is a cell decomposition on Σ, deﬁned as follows: There is a one-to-one correspondence between the classical crossings of D and 0-cells, bands of SD and 1-cells, and 2-disks that we attached to SD and 2-cells. The Euler characteristic of Σ is 2 −2g. And the number of 0, 1 and 2-cells are n, 2n and \|s∂\| + \|¯ s∂\|, respectively. The lemma now follows. 37 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Remark 3.19. Diﬀerent authors use diﬀerent names for checkerboard colorable knots and links. For instance, in [KNS02], Kamada calls them normal. In [Dye16], Dye refers to them as diagrams satisfying the boundary property, which we have seen is equivalent by Lemma 3.17. In [Rus17], Rushworth calls them even diagrams. Some authors refer to them as mod-2 almost classical diagrams. 3.4 Parity Projection In [Man10], Manturov introduced the notion of parity, and this deep and im-portant idea has led to some of the most striking and far-reaching results in virtual knot theory. In this section, we review the deﬁnition of parity and its associate projection. Note that parity is only deﬁned for knots, not for links. Let D be a diagram category, one whose objects are Gauss diagrams of knots and whose morphisms consist of compositions of Reidemeister moves. Deﬁnition 3.20. A parity is a collection of functions {fD \| D ∈D}, where fD : {chords of D} →{0, 1}, satisfying in the following axioms: Axiom 0: Under any Reidemeister move, the parity of any chord not partici-pating in the move is unchanged. Axiom 1: If c ∈D is a chord which occurs in a Reidemeister 1 move, then fD(c) = 0. Axiom 2: If c1, c2 ∈D are two chords which occur in a Reidemeister 2 move, then fD(c1) = fD(c2). Axiom 3: If c1, c2, c3 ∈D are three chords which occur in a Reidemeister 3 move, and D′ is the new Gauss diagram obtained, after applying the Reidemeis-ter 3 move, then fD(ci) = fD′(c′ i), ∀i. Furthermore, either they are all even, all odd, or exactly two are odd. A chord c is even, if fD(c) = 0, and odd, if fD(c) = 1. Deﬁnition 3.21. The diagram obtained by removing all the odd chords from D is denoted Pf(D) and is called the projection of D with respect to the parity f. We denote by P k f (D), the result of applying k times parity projection to D. Notice for any diagram D there is a positive number k such that P k f (D) = P k+1 f (D) and for this k we denote P k f (D) by P ∞ f (D). Proposition 3.22. If two Gauss diagrams D1 and D2 are equivalent, then Pf(D1) and Pf(D2), are also equivalent. 38 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics For a proof, see [Man10]. Deﬁnition 3.23. Let K be a knot diagram in Σ and c be a crossing of K, and γ be a simple closed curve on Σ. The oriented smoothing of K at c gives rise to two loops in Σ, which we denote by K0 and K1. We say the crossing c is even if either K0 or K1 intersects γ an even number of times, otherwise we say c is odd. Manturov proves in [IMN14], that this deﬁnes a parity for knots in Σ. It is called homological parity. Let P = Pγ denote the associated projection map, then P(K) is the knot with all the odd crossings removed. Let f be the mod 2 Gaussian parity, i.e. f(c) = ind(c) (mod 2) for any chord c ∈D. Then for any Gauss diagram D, the projection Pf(D) is the Gauss diagram obtained by deleting all chords c of D with ind(c) ̸= 0 (mod 2). The following proposition is immediate. Proposition 3.24. For f and Pf as above, Pf(D) = D if and only if D is checkerboard colorable. If f is the mod 2 Gaussian parity, then by the previous proposition, P ∞ f (D) is checkerboard colorable. Proposition 3.25. If K is a checkerboard knot and D is a diagram for K with minimal crossing number, then D is a checkerboard colorable diagram. Proof. Suppose D is not checkerboard colorable, then it contains a chord c with ind(c) = 1 (mod 2). Therefore Pf(D) ̸= D and in fact Pf(D) has fewer crossings. The knot K is checkerboard which means it has a checkerboard colorable diagram D′. The diagrams D and D′ are equivalent and by Proposition 3.22, Pf(D) and Pf(D′) are equivalent. On the other hand, by the previous proposition, Pf(D′) = D′. Hence Pf(D) is a diagram for K with fewer crossings than D, and this is impossible because D has minimal crossing number. The total Gaussian parity, denoted ftot(c), is deﬁned by setting ftot(c) = ( 1 if ind(c) ̸= 0 0 if ind(c) = 0. . It is not diﬃcult to check that ftot satisﬁes the parity axioms. Let Ptot denote the associated projection map on Gauss diagrams. Then an argument analogous to the preceding one shows that Ptot(D) = D if and only if D is an almost classical diagram. We can also prove the following proposition similarly: 39 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Proposition 3.26. If K is an almost classical knot and D is a diagram for K with minimal crossing number, then D is an almost classical diagram. Proposition 3.27. Suppose K is an almost classical knot and D is a diagram for K which has minimal crossing number. If D is a connected sum, then both of the factors are almost classical diagrams. Proof. By Proposition 3.26, D is an almost classical diagram, and it follows that Ptot(D) = D. Let D = D1#D2, and suppose c is a chord in D. Then c is a chord in one of the factors say D1. It is clear from the deﬁnition of the index that ind(c) as a chord in D is the same as ind(c) as a chord in D1. Hence Ptot(D1) = D1 and D1 is almost classical. For D2 the argument is similar. Remark 3.28. If D is an alternating Gauss diagram, Ptot might not be alternat-ing. In Figure 3.7, the diagram on the left is an alternating diagram, but the projection on the right is not. + − + − + − − + − − Figure 3.7: The projection of an alternating diagram. 40 Chapter 4 Signatures for Checkerboard Knots In this chapter, we present virtual analogues of the signature, nullity, and deter-minants, which are invariants deﬁned for checkerboard colorable virtual knots and links. We study the behavior of these invariants under connected sum of checkerboard long knots, showing that the signature is additive and the de-terminant is multiplicative. We relate the signature of virtual knots obtained by taking vertical and horizontal mirror images, and also under performing a crossing change. 4.1 Goeritz Matrices and Signatures In this section, we introduce the signature and determinants of checkerboard knots and links, ﬁrst deﬁned by Im, Lee, and Lee in [ILL10]. Their deﬁnition is similar to the one developed by Gordon and Litherland in [GL78] and described in Section 2.1. Let ξ be a checkerboard coloring for a pair P = (F, D), where F is a closed, oriented and connected surface and D is a link diagram on F. We enumerate the white regions of Fξ ∖\|D\| by X0, X1, . . . , Xm. Let C(D) denote the set of all classical crossings of D on F. For each pair i, j ∈{1, 2, . . . , m}, let Cij(D) = {c ∈C(D) \| c is adjacent to both Xi and Xj} 41 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics and deﬁne gij =        − P c∈Cij(D) η(c), for i ̸= j, − m P k=0;k̸=i gik, for i = j. Just as for classical knots, the pre-Goeritz matrix of D is deﬁned to be the symmetric integral matrix G′ ξ(D) = (gij)0≤i,j≤m, and the Goeritz matrix of D is the principal minor Gξ(D) = (gij)1≤i,j≤m obtained by removing the ﬁrst row and column of G′ ξ(D). For any classical knot diagram, the Goeritz matrices are non-singular, but this does not hold in general for checkerboard colored virtual knot diagrams. The correction term is deﬁned by setting µξ(D) = P c is type II η(c). Deﬁnition 4.1. For a link diagram D and a checkerboard coloring ξ, we deﬁne the signature as follows: σξ(D) = sig(Gξ(D)) −µξ(D). Remark 4.2. If L is a non-split checkerboard link represented by a diagram D of minimal genus and with coloring ξ, then the pair {σξ(D), σξ∗(D)} of signatures is independent of the choice of virtual link diagram and gives a well-deﬁned invariant of the virtual link L (see [ILL10]). Notice the fact that D should be a minimal genus diagram, is essential since using another diagram for L with diﬀerent genus, one might obtain diﬀerent signatures. Example 4.3. We compute the signatures for the knot K = 3.5. The diagram D in Figure 3.5 has supporting genus equal to 1. On the other hand K is not classical, so D is a minimal genus diagram. Let ξ be the coloring in Figure 3.6, and let c1, c2 and c3 be the crossings from left to right. There is only one white region for ξ, therefore Gξ(D) is the empty matrix and its signature is zero. The two crossings c1 and c3 are type II and they have η = −1, thus µξ = −2, and σξ = 2. Now we consider ξ∗by changing the black and white color. There are two white regions for ξ∗and we have: G′ ξ∗= h 2 −2 −2 2 i , it follows Gξ∗= , and sig(Gξ∗) = 1. On the other hand, only c2 is a type II crossing for ξ∗and µξ∗= −1, as a result σξ∗= 2. In a similar way, one can also deﬁne analogues of the determinant for checker-42 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics board links in terms of detξ(D) = \| det Gξ(D)\| and detξ∗(D) = \| det Gξ∗(D)\|. In [ILL10], they prove that the pair (\| det Gξ(D)\|, \| det Gξ∗(D)\|) is independent of the choice of virtual link diagram and gives a well-deﬁned invariant of the checkerboard link L. We call this pair the checkerboard determinants of L. The nullities of L are deﬁned to be (Nξ(L), Nξ∗(L)) = (nullity(Gξ(D)), nullity(Gξ∗(D))). Notice that in general, the checkerboard determinants detξ(K) and detξ∗(K) are not equal to one another, and not equal to det(K) for checkerboard knots and links. Proposition 4.4. For a non-split classical link L, the signatures, determinants and nullities are singletons. Also we have: \|∆L(−1)\| = det(L) = \|detξ(L)\| = \|detξ∗(L)\|. Proof. For the ﬁrst part see [ILL10]. For the second part, suppose V is a Seifert matrix for L, then \|∆L(−1)\| = \| det(V + V τ)\|. Combining [BZH14, Proposition 13.15] and [GL78, Theorem 1], the result follows. 4.2 Signatures and Connected Sum In this section, we study the behavior of the invariants of the previous section under connected sum. Since the operation of connected sum is not well-deﬁned on round virtual knots, we will work with long virtual knots. Our main results are that, under connected sum, the signature is additive and the determinant is multiplicative for long checkerboard knots. Recall that a long virtual knot diagram is a regular immersion of R in R2 which coincides with the x-axis outside of some compact set. Each double point is either a virtual crossing or a classical crossing, and classical crossings have over- and under-crossing arcs indicated as usual. A long virtual knot is deﬁned to be an equivalence class of long virtual knot diagrams modulo the generalized Reidemeister moves. Long virtual knot diagrams are oriented from left to right. If D1 and D2 are two long virtual knot diagrams, the connected sum D1#D2 is deﬁned to be the diagram obtained by concatenating them, with D1 on the left and D2 on the right. This gives a well-deﬁned operation on long virtual knots, and in general, connected sum is not commutative. Suppose D is a long virtual knot diagram which coincides with the x-axis outside of a closed ball B in R2. The closure of D, denoted b D, is the round virtual knot diagram obtained as the union of D ∩B and ∂B ∩{(x, y) \| y ≥0}. 43 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics An elementary argument shows that equivalent long virtual knot diagrams have equivalent closures. Thus, closure gives a well-deﬁned map from long virtual knots to round virtual knots. A long virtual knot diagram is called checkerboard colorable if its closure is checkerboard colorable. One can also view long virtual knots as round virtual knots with a choice of basepoint, which we take to be the point at inﬁnity. For any checkerboard colorable long virtual knot diagram, there is a canonical choice of checkerboard coloring; it is the coloring with white region to the right of the basepoint. We deﬁne the signature of the checkerboard long knot to be the signature of the corresponding round virtual knot with respect to the canonical checkerboard coloring of the long knot. Proposition 4.5. The signature of checkerboard long knots is additive under connected sum. Proof. Let D and D′ be two checkerboard colorable long virtual knot diagrams, and let ξ and ξ′ be the canonical checkerboard colorings of D and D′. Enumerate the white regions for D with X0, . . . , Xm, such that X0 is the region containing the base-point. Similarly we have Y0, . . . , Yℓfor D′, such that Y0 contains the base-point of D′. The pre-Goeritz matrices for D and D′ are the following matrices, respectively: G′ ξ(D) = h−x u uτ A i and G′ ξ′(D′) = h−y v vτ B i , where u and v are row vectors, and x and y are sum of the entries of u and v, respectively, and A = Gξ(D) and B = Gξ′(D′) are the Goeritz matrices of D and D′, respectively. The white regions for D#D′ are Z = X0 ∪Y0, X1, . . . , Xm, Y1, . . . , Yℓ, where Z is the region to the right of the base-point of D#D′. The pre-Goeritz matrix for D#D′ is therefore given by G′ ξ#ξ′(D#D′) = "−x −y u v uτ A 0 vτ 0 B # . Since signatures are additive under block sum, i.e. since sig A 0 0 B = sig(A) + sig(B), it follows that sig (Gξ#ξ′(D#D′)) = sig (Gξ(D)) + sig (Gξ′(D′)) . 44 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Every crossing of D#D′ belongs to either D or D′, and further it is clear that the incidence number and type of each crossing in D#D′ with respect to ξ#ξ′ is the same as it is in D (with respect to ξ) or in D′ (with respect to ξ′). Consequently, it follows that the correction term is additive, i.e. that µξ#ξ′(D#D′) = µξ(D) + µξ′(D′). This completes the proof. Proposition 4.6. The determinant of checkerboard long knots is multiplicative under connected sum. Proof. The result is immediate since det A 0 0 B = det(A) det(B). 4.3 Virtual Unknotting Operations In this section, we introduce the virtual unknotting operations, which consist of four moves which can be used to unknot any virtual knot diagram, and study their eﬀect on the incidence number and type of the crossing. We have four unknotting operations: Chord deletion (cd), crossing change (cc), sign change (sc) and orientation reversal (or) (see [BCG17b]). On a Gauss diagram, we can show them as follows: It is obvious that if we apply cd to a Gauss diagram with n chords, after at most n −1 times applying it, we get the unknot. Using all four operations, we can often unknot a given knot in fewer steps. For example, if on a diagram it happens for two chords that their heads and tails are next to each other, then depending on the signs of the chords and their directions, combining the other three operations would create two chords which can be removed using a Reidemeister 2 move. Starting with a checkerboard diagram with a ﬁxed crossing, if we apply any one of {cc, sc, or} to this crossing, the new diagram is again checkerboard. We will now examine the eﬀect on the incidence number and type of the crossing. Starting with a checkerboard colored diagram, if we apply sc to a chord c, then η(c) remains the same, but its type changes. See Figure 4.2. 45 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics ε cd ε cc −ε ε sc −ε ε or ε Figure 4.1: The unknotting operations. sc Figure 4.2: A coloring and the sc operation. If we instead apply or to a chord c, then η(c) changes sign and the type of the chord also changes, but the sign remains the same. See Figure 4.3. Last but not least, if we apply cc to a chord c, then both the sign ε(c) and the incidence η(c) change signs, but the type remains the same. See Figure 4.4. The table 4.1 summarizes the eﬀect of applying sc, or, cc to the crossing c on its sign ε(c), incidence number η(c), and type. Remark 4.7. Recall a ﬂat knot is the knot diagram without the over- and under-46 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics or Figure 4.3: A coloring and the or operation. cc Figure 4.4: A coloring and the cc operation. sign incidence type sc −ε(c) η(c) −type(c) or ε(c) −η(c) −type(c) cc −ε(c) −η(c) type(c) Table 4.1: The eﬀect of applying sc, or, cc to the crossing c. crossing information. More precisely, ﬂat knots are equivalence classes of virtual knot diagrams up to Reideimeister moves and applications of cc. Flat knots can also be viewed as undecorated Gauss diagrams, but the interpretation of the arrow is diﬀerent from that of a Gauss pattern. In a Gauss pattern, we interpret the arrow as pointing from over to under, whereas in ﬂat knot theory one interprets the arrow as indicating which arc crosses from right to left. Remark 4.8. If we apply sc to any chord in a Gauss diagram, then the Alexander determinant is unchanged. This follows from Proposition 3.15, since if we apply sc to any chord of a Gauss diagram, the underlying Gauss pattern is unchanged. 47 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics 4.4 Signatures and Mirror Images There are three involutions on virtual knots given by orientation reversal and horizontal and vertical mirror symmetry. In this section, we study the eﬀect these involutions have on the signatures of the checkerboard knots. We start with a virtual knot diagram D. In the Gauss diagram, if we apply cc to all chords, the result is called vertical mirror image of D, and we denote it by D∗. If we apply sc to all chords, the result is called horizontal mirror image of D, and we denote it by D†. We can also deﬁne −D, the inverse of D. Finally we can apply or to all chords of D, to obtain D∗†. Lemma 4.9. If D is a minimal genus diagram, then so are −D, D∗and D†. Proof. It is obvious that if D is a minimal genus diagram for K, then −D is a minimal genus diagram for −K. If P = (SD, ˜ D) is the abstract link diagram associated with D, we place it inside {(x, y, z) ∈R3 \| y < 0}, in a way that the projection of ˜ D on the xy-plane, is D. Now reﬂect P with respect to the plane y = 0. The result is the abstract link diagram associated with D†. This shows that if D is a minimal genus diagram, then D† is also minimal genus. Finally, if P = (SD, ˜ D) is the abstract link diagram associated with D, we switch all the over-crossings and under-crossings in ˜ D, to obtain the abstract link diagram for D∗. It follows that if D is a minimal genus diagram, then D∗ is also minimal genus. Suppose ξ is a checkerboard coloring of D and ξ∗is its dual coloring. Notice that a coloring is determined by the underlying ﬂat knot. Therefore we can use the same notation for the colorings of the diagrams of the mirror images and the inverse knot. The following picture, is a colored crossing in D (left), and −D (right). c c Figure 4.5: A colored crossing in D and −D. 48 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Therefore, at each crossing, type and incidence number of that crossing is unchanged. Notice that black and white regions are also unchanged. Thus the two signatures for D and −D are the same. For D∗, based on section 4.3, at each crossing, the type is unchanged but the incidence number changes by a negative sign. As a result, both the Goeritz matrix and the correction term, are multiplied by −1. Thus σξ(D∗) = −σξ(D) and σξ∗(D∗) = −σξ∗(D). For D†, we use the dual coloring ξ∗. Thus at each crossing, the type is unchanged but the incidence number changes by a negative sign. Therefore σξ∗(D†) = −σξ(D), and σξ(D†) = −σξ∗(D). Similarly, for D∗†, we ﬁnd that σξ∗(D∗†) = σξ(D) and σξ(D∗†) = σξ∗(D). We can summarize these observations, in the following proposition. Proposition 4.10. For a virtual knot K with checkerboard coloring ξ, the sig-natures of the mirror images and the inverse knot, are as follows: (σξ(−K), σξ∗(−K)) = (σξ(K), σξ∗(K)), (σξ(K∗), σξ∗(K∗)) = (−σξ(K), −σξ∗(K)), (σξ(K†), σξ∗(K†)) = (−σξ∗(K), −σξ(K)), (σξ(K∗†), σξ∗(K∗†)) = (σξ∗(K), σξ(K)). For virtual knots, since we have diﬀerent mirror images, we can deﬁne dif-ferent notions of a knot being amphichiral. Deﬁnition 4.11. An unoriented virtual knot K is called vertically amphichiral if K∗= K, and is called horizontally amphichiral if K† = K. An oriented virtual knot K is called positive (negative) vertically amphichiral if K∗= K (K∗= −K), and is called positive (negative) horizontally amphichiral if K† = K (K† = −K). We can use Proposition 4.10 to prove the following: Proposition 4.12. For a virtual knot K, if at least one of the signatures σξ(K) or σξ∗(K) is non-zero, then K is not positive vertically amphichiral. If σξ(K)+ σξ∗(K) ̸= 0, then K is not positive horizontally amphichiral. Now if we apply cc to one chord, we want to investigate how the signature changes. Let D+ be a minimal genus diagram with at least one positive crossing. We choose a positive crossing, and apply cc to that. It is not diﬃcult to see that the new diagram is also minimal genus, and we denote it by D−. 49 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics For classical links, both of the signatures are equal to the usual signature (see [GL78]). We can show the following relation holds between the signatures of D+ and D−: σ(D+) ≤σ(D−) ≤σ(D+) + 2. There are several ways to show this. One way has been explained in [Liv04] for τ invariant, and the same argument applies with τ replaced by σ. For checkerboard colorable virtual links, we have the following proposition (cf. [BZH14, Prop. 13.32]), which can be used to calculate the signature. Proposition 4.13. Let Q be a symmetric matrix of rank r over a ﬁeld. There exists a chain of principal minors Mi ,i = 0, 1, . . . , r, such that Mi is a principal minor of Mi+1 and that no two consecutive determinants Mi and Mi+1 vanish (M0 = 1). For any such sequence of minors, σ(Q) = Pr−1 i=0 sign(MiMi+1). Proposition 4.14. Let D+ and D−be as above, and ξ be one of the colorings. Then we have σξ(D+) ≤σξ(D−) ≤σξ(D+) + 2. Proof. We assume the Goeritz matrices are all non-singular. If they are not, we work with a non-singular sub-matrix, and the following proof works in that case as well. We denote by c, the crossing which we change from positive to negative, and consider four cases: Case 1: Two diﬀerent white regions meet at c, we call them X0 and X1, and c in both D+ and D−is a type II crossing. In D+, η(c) = +1, and in D−, η(c) = −1. We have µξ(D+) = µξ(D−) + 2, Gξ(D+) = ha + 1 ∗ ∗ A i , Gξ(D−) = ha −1 ∗ ∗ A i . If det(Gξ(D+)) and det(Gξ(D−)) have the same sign, then sig(Gξ(D+)) = sig(Gξ(D−)), hence σξ(D−) = σξ(D+) + 2. Now suppose the determinants have opposite signs, and assume the Goeritz matrices are r ×r. By Proposition 4.13, 50 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics we have: sig(Gξ(D+)) = r−2 X i=0 sign(MiMi+1) + sign(det(A) det(Gξ(D+))). sig(Gξ(D−)) = r−2 X i=0 sign(MiMi+1) + sign(det(A) det(Gξ(D−))), = r−2 X i=0 sign(MiMi+1) + sign(det(A) det(Gξ(D+)) −(det(A))2). It follows, in this case sig(Gξ(D+)) = sig(Gξ(D−)) + 2, hence σξ(D−) = σξ(D+). Case 2: Same as case 1, except c is a type I crossing. In D+, η(c) = −1, and in D−, η(c) = +1. We have µξ(D+) = µξ(D−), Gξ(D+) = ha −1 ∗ ∗ A i , Gξ(D−) = ha + 1 ∗ ∗ A i . By similar argument, if det(Gξ(D+)) and det(Gξ(D−)) have the same sign, then σξ(D−) = σξ(D+), otherwise σξ(D−) = σξ(D+) + 2. Case 3: Only one white region occurs at c, we call it X0, and c is a type II crossing. In this case, the corresponding Goeritz matrices are equal, but µξ(D+) = µξ(D−) + 2, therefore σξ(D−) = σξ(D+) + 2. Case 4: Same as case 3, except c is a type I crossing. In this case, the Goeritz matrices are equal, and µξ(D+) = µξ(D−), thus σξ(D−) = σξ(D+). 51 Chapter 5 Alternating Virtual Knots In this chapter, we introduce alternating virtual knots and links. Our main result is Theorem 5.19, which gives necessary and suﬃcient conditions for a virtual knot or link to be alternating in terms of its Goeritz matrices being positive and negative deﬁnite. We also prove a virtual analogue of the ﬁrst Tait conjecture by adapting a result of [AFLT02]. about reduced alternating diagrams of knots on surfaces. 5.1 Alternating Virtual Knots In this section, we recall the matrix-tree theorem of Bott and Mayberry [BM54], and use it to adapt Crowell’s proof [Cro59] to show that the Alexander polyno-mial of any almost classical alternating link is alternating. We also outline the proof of [BBRS01, Theorem 2] and generalize it to alternating virtual links. Deﬁnition 5.1. A virtual link diagram D is called alternating if the classical crossings alternate between over-crossing and under-crossing as you go around each component. A virtual link L is called alternating if it admits an alternating diagram. The spectacular results concerning the Jones polynomial of classical alter-nating links are generally not true in the virtual case. For instance, the span of the Jones polynomial is not equal to the crossing number, and in fact there are alternating virtual knots with trivial Jones polynomial. For example, the knot K = 6.90101 is alternating and has Jones polynomial VK(t) = 1. Further, the Jones polynomial is not necessarily alternating for alternating virtual knots. 52 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics For example, the knot K = 5.2426 in Figure 6.5 is alternating and has Jones polynomial VK(t) = 1/t2 + 1/t3 −1/t5. Let K be a virtual knot or link. We deﬁne the knot group GK as in Section 2.1. We use Fox derivative to deﬁne the Jacobian matrix A. For virtual knots, the ﬁrst elementary ideal E1 is not necessarily principal. We deﬁne the Alexander polynomial ∆K(t) to be the generator of the smallest principal ideal containing E1. Since Z[t, t−1] is a gcd domain, it is given by taking the gcd of all the (n −1) × (n −1) minors of A . If we remove the i-th row and j-th column of A we denote the corresponding minor by Aij. In [NNST12] and [BNW18], the authors showed for almost classical knots or links, E1 is principal, and the Alexander polynomial ∆K(t) is given by taking the determinant of the (n −1) × (n −1) matrix obtained by removing any row and any column from A. Proposition 5.2. For an almost classical knot or link L, the Alexander deter-minant det(L) is equal to \|∆L(−1)\|. Proof. If D is a diagram for L, the coloring matrix M(D) is exactly the matrix obtained from the Fox Jacobian matrix by replacing t with −1. Remark 5.3. For an almost classical knot K, the knot determinant \|∆K(−1)\| is an odd number (see [BGH+17]). Proposition 5.4. If L is a split checkerboard link then det(L) = 0. Proof. Suppose D = D1 ∪D2 is a split checkerboard diagram for L. In each row of the coloring matrix the non-zero elements are either 2, −1, −1 or 1, −1. It follows the columns add up to zero. We consider a simple closed curve in the plane which separates D into two parts. It follows that the coloring matrix M = M(D) admits a 2 × 2 block decomposition of the form M = hM1 0 0 M2 i , where M1 and M2 are the coloring matrices for D1 and D2, respectively. Since det(M1) = 0 = det(M2), it follows that the matrix obtained by removing a row and column from M also has determinant zero. We state the Bott-Mayberry theorem. For more details see [BZH14] and [BM54]. 53 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Let Γ be a ﬁnite oriented graph with vertices {ci \| 1 ≤i ≤n} and oriented edges {uδ ij}, such that ci is the initial point and cj the terminal point of uδ ij. Notice that δ enumerates the diﬀerent edges from ci to cj. By a rooted tree (with root ci) we mean a subgraph of n −1 edges such that every point ck is terminal point of a path with initial point ci. Let aij denote the number of edges with initial point ci and terminal point cj. Theorem 5.5. Let Γ be a ﬁnite oriented graph without loops (aii = 0). The principal minor Hii of the graph matrix H(Γ) =      (P k̸=1 ak1) −a12 −a13 · · · −a1n −a21 (P k̸=2 ak2) −a23 · · · −a2n . . . . . . . . . . . . −an1 −an2 −an3 · · · (P k̸=n akn)     , is equal to the number of rooted trees with root ci. Corollary 5.6. Let Γ be a ﬁnite oriented loopless graph with a valuation f : {uδ ij} →{−1, 1} on edges. Then the principal minor Hii of the matrix H = [bij], where bij = (P δ f(uδ ij), i ̸= j, −P k̸=i bki, i = j, satisﬁes the following equation: Hii = X f(Tr(i)), where the sum is to be taken over all ci-rooted trees Tr(i), and where f(Tr(i)) = Y uδ kj∈Tr(i) f(uδ kj). For a virtual link diagram, there are (at least) two ways one can associate a 4-valent graph. One way is to consider the diagram D itself. It has vertices for the classical and virtual crossings and edges running from one classical or virtual crossing to the next. This graph is planar. The other way to associate a graph is to consider vertices only for classical crossings. The key diﬀerence is that in general, this graph is not planar. For an alternating diagram D, we describe this graph and an orientation on it as follows: Let D has classical crossings c1, . . . , cn. The vertices of Γ are c1, . . . , cn. At 54 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics each vertex consider two out-going edges corresponding to the over-crossing arc, and two in-coming edges for the under-crossing arcs (see Figure 5.1). This is called the source-sink orientation or the alternate orientation. This orientation is possible because D is alternating, and an out-going edge at the vertex ci, should be an in-coming edge for the adjacent vertex. Remark 5.7. In general, any checkerboard colorable diagram D admits a source-sink orientation. In fact, a diagram is checkerboard colorable if and only if it admits a source-sink orientation (see [KNS02, Proposition 6]). Figure 5.1: The source-sink orientation. Theorem 5.8. Let L be an almost classical alternating link with a connected alternating diagram D. The Alexander polynomial ∆L(t) is alternating. Proof. For the unknot the result is obvious. Assume D has n ≥1 classical crossings. Orient D and enumerate the crossings by c1, . . . , cn. Label the arcs by g1, . . . , gn. At the crossing ci, suppose gν(i) is the over-crossing arc in the upward direction, and gλ(i) and gρ(i) are the left and right under-crossing arcs, respectively. Deﬁne the relation ri = gλ(i)gν(i)g−1 ρ(i)g−1 ν(i). Notice that we are using left-handed meridian convention. Now consider the graph Γ associated with D, with the source-sink orientation on it. Label the edges by uδ ij. Deﬁne the valuation f as follows. At the crossing cj, if uδ ij corresponds to gλ(j), then f(uδ ij) = 1, and if it corresponds to gρ(j), then f(uδ ij) = −t. Deﬁne the matrix H as in the Corollary 5.6. Notice that D is alternating and there is a one-to-one correspondence between the crossings of D and the set of over-crossing arcs. Therefore we can choose to label over-crossing arcs, such that ν(i) = i. The matrix H is the transpose of the Jacobian matrix A. The Alexander polynomial ∆L(t) = Aii = Hii. By the Corollary 5.6, Hii = X Y uδ kj∈Tr(i) f(uδ kj). 55 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Since f(uδ kj) = 1, or −t, the product Q uδ kj∈Tr(i) f(uδ kj) is of the form (−1)ltl and Hii is an alternating polynomial. Therefore ∆L(t) is alternating. Example 5.9. Up to 6 crossings, the following eight almost classical knots do not have alternating Alexander polynomial, therefore by Theorem 5.8 they are not alternating virtual knots. K ∆K(t) 5.2331 t2 −1 + t−1 6.85091 1 + t−1 −t−2 6.85774 t −1 + t−2 6.87548 −t2 + 2t + 1 −t−1 6.87875 t + 1 −2t−1 + t−2 6.89156 2t −1 −t−1 + t−2 6.89812 t2 −2 + 2t−1 6.90099 t −t−1 + t−3 Table 5.1: Almost classical knots with non-alternating Alexander polynomial. The weak form of the ﬁrst Tait Conjecture, namely that every knot having a reduced alternating diagram with at least one crossing is nontrivial, was ﬁrst proved by Bankwitz [Ban30] in 1930; and since then, Menasco and Thistleth-waite [MT91a] and Andersson [And95] published simpler proofs. Here we outline the proof by Balister et al. [BBRS01] and generalize it to alternating virtual links. Notice that for alternating virtual knots, this result has been proved using a diﬀerent method by Cheng in [Che15, Proposition 3.3]. Consider the graph Γ with vertices {c1, . . . , cn} as before. Deﬁnition 5.10. The outdegree of the vertex ci, denoted d+(ci), is the number of edges of Γ with initial point ci. The indegree of the vertex ci, denoted d−(ci), is the number of edges of Γ with terminal point ci. Therefore d+(ci) = n X j=1 aij , d−(ci) = n X j=1 aji. Deﬁnition 5.11. A walk in a graph is an alternating sequence of vertices and edges, starting with a vertex ci and ending with a vertex cj. A walk is called a trail if all the edges in that walk are distinct. A circuit is a trail which starts 56 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics and ends at a vertex ci. An Eulerian circuit is a circuit which contains all the edges of Γ. A graph Γ is called Eulerian if it has an Eulerian circuit. An Eulerian graph is necessarily connected and has d+(ci) = d−(ci) for every vertex. Let ti(Γ) be the number of rooted trees with root ci, then the BEST Theorem is as follows (see [vAEdB51] and [Bol98, Theorem 13]). Theorem 5.12. Let s(Γ) be the number of Eulerian circuits of Γ, then s(Γ) = ti(Γ) n Y j=1 (d+(cj) −1)! In particular, if Γ is a two-in two-out oriented graph, i.e., d+(ci) = d−(ci) = 2 for every i, then by Theorem 5.5 and 5.12, s(Γ) = ti(Γ) = Hii, for every i. A vertex c of a graph Γ is an articulation vertex if Γ is the union of two nontrivial graphs with only the vertex c in common. In particular, a vertex incident with a loop is an articulation vertex. In [BBRS01] Balister et al. proved the following theorem: Theorem 5.13. Let Γ be a connected two-in two-out oriented graph with n ≥2 vertices and with no articulation vertex. Then s(Γ) ≥n. Deﬁnition 5.14. Let Σ be a closed surface, and consider a link L in Σ×I. Let D be the link projection in Σ0 = Σ × {0}. A crossing c of D is called nugatory if there is a simple closed curve on Σ0 that intersects D exactly once at c. Recall that associated with an alternating virtual link diagram D, there is an oriented two-in two-out graph Γ. If D has no nugatory crossings, then Γ has no articulation vertex. Corollary 5.15. Let K be an almost classical alternating knot, and D an al-ternating diagram for K, which has no nugatory crossings with n ≥2 crossings. Then \|∆K(−1)\| ≥n. Proof. By the proof of Theorem 5.8, \|∆K(−1)\| counts ti(Γ) the number of rooted trees with root ci in the oriented graph Γ, associated with the knot diagram D. By Theorem 5.12, ti(Γ) = s(Γ), and the result follows from Theorem 5.13. 57 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Theorem 5.16. Let L be an alternating virtual link and D a connected alter-nating diagram with n ≥2 classical crossings, which has no nugatory crossings. Then for the Alexander determinant of L we have: det(L) ≥n. Proof. Since D is alternating, we can repeat the proof of Theorem 5.8. By Theorem 5.6, the determinant of L counts the number of spanning trees which is equal to s(Γ). The result follows from Theorem 5.13. Corollary 5.17. Suppose a virtual link L admits a connected alternating dia-gram D with no nugatory crossings. If D has n ≥1 classical crossings, then L is not split. Proof. By Theorem 5.16 det(L) ≥n, in particular det(L) ̸= 0. The result follows from 5.4. 5.2 Characterization of Alternating Virtual Links In this section, we prove one of the main results in this thesis, which gives necessary and suﬃcient conditions for a virtual knot to admit an alternating virtual knot diagram. Lemma 5.18. Suppose D is a connected checkerboard colorable link diagram. Then D is alternating if and only if each crossing has the same incidence num-ber, i.e. for any two crossings c1 and c2 of D, we have η(c1) = η(c2). Proof. First assume D is alternating. Consider the Gauss diagram for D. On each core circle, start with the foot of one chord, it is between two arcs, travel counterclockwise and let the right side of the ﬁrst arc and the left side of the second arc be black. Alter the coloring as you pass each arrow-head or tail. The diagram is alternating, thus at all the crossings we see one pattern. Namely, they are all type A crossings. Change the coloring, then all the crossings have type B. Now suppose each crossing c has incidence number η(c) = +1. When we move counterclockwise on a core circle, at the tail of each arrow, ﬁrst we see the black color on the right and then we see it on the left, and at the head of each arrow, ﬁrst we see the black color on the left, then on the right. The diagram D is checkerboard colorable and passing each arrow-head or tail, we have to 58 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics switch the color, thus the diagram should be alternating. If each crossing c has incidence number η(c) = −1, then switch the checkerboard coloring. Thus if D is an alternating diagram, then s∂is the state obtained from D by resolving all the crossings to 0-smoothing and ¯ s∂is all 1-smoothing state. The next result is a generalization of Proposition 4.1. from [Gre17]. Theorem 5.19. Suppose K is a connected checkerboard colorable virtual link with virtual genus gv(K). Then D is an alternating diagram for K if and only if D admits a checkerboard coloring ξ with dual coloring ξ∗such that 1) sg(D) = gv(K), 2) σξ(D) −σξ∗(D) = 2gv(K), 3) Gξ(D) is negative deﬁnite (or empty) and Gξ∗(D) is positive deﬁnite (or empty). Proof. Set g = gv(K) and suppose D has n real crossings. First assume D is alternating. Then D is a minimal genus diagram (see Remark 5.21). Let ξ be the checkerboard coloring in which all the crossings are type B. Suppose Gξ(D) is not empty. Every crossing c of D has incidence number η(c) = −1 with respect to ξ. Therefore all the non-diagonal entries of Gξ(D) are non-negative. In this case negative crossings are type II. So if n−is the number of negative crossings, then µξ(D) = −n−. If Gξ(D) = [gij]1≤i,j≤m, and we have [x1 · · · xm] [gij]1≤i,j≤m   x1 . . . xm  = X 1≤i,j≤m gijxixj. Gξ(D) is symmetric, thus we have X 1≤i,j≤m gijxixj = 2 X 1≤i<j≤m gijxixj + m X i=1 giix2 i , = 2 X 1≤i<j≤m gijxixj − m X i=1 m X k=1,k̸=i gikx2 i − m X i=1 gi0x2 i , = − X 1≤i 0. Throughout the argument, we will use K to denote the associated knot in the thickened surface Σ × I associated to D. By the Theorem 5.28, D is a minimal genus diagram for K. Let D′ be a diagram for K with minimal crossing number, and use K′ to denote the associated knot in the thickened surface Σ′ × I. Using parity, one 65 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics can show that, given any minimal crossing diagram for a virtual knot, its Carter surface has minimal genus (see Theorem 5 in [Man13]). From this, it follows that D′ is also a minimal genus diagram for K. By [Kup03, Theorem 1], there is a diﬀeomorphism f : Σ × I →Σ′ × I such that f(K) = K′. We pick the same model surface for both Σ and Σ′, which we denote by Σ, and we assume that f : Σ × I →Σ × I is the diﬀeomorphism with f(K) = K′. Notice that f(Σ × {0}) = Σ × {0} (see Deﬁnition 3.3). Let π : Σ×I →Σ×{0} be the projection map. We can assume π(K) and D have the same crossing number and π(K) is alternating and reduced. Similarly, we can assume π(K′) and D′ have the same crossing number. Let f0 : Σ →Σ be the induced diﬀeomorphism, i.e. f0(x) = π(f(x, 0)). For (x, t) ∈Σ × I, deﬁne f ′(x, t) = (f0(x), t). This gives a diﬀeomorphism f ′ : Σ × I →Σ × I such that f\|Σ×{0} = f ′\|Σ×{0}. Set h = f ◦(f ′)−1. Notice that h\|Σ×{0} is the identity map. Hatcher proved in [Hat76, Lemma 2], that if Σ is a compact surface other than S2, then PL(Σ×I, rel Σ×{0}) is contractible. Here, PL(Σ×I, rel Σ×{0}) is the space of piecewise linear self-homeomorphisms of Σ × I that are the identity when restricted to Σ×{0}. The corresponding statement in the smooth category, namely that Diﬀ(Σ × I, rel Σ × {0}) is contractible, can be deduced from [Hat76, Lemma 2] and Hatcher’s positive solution to the Smale Conjecture, that is, Diﬀ(D3, ∂D3) is contractible [Hat83]. Since h ∈Diﬀ(Σ×I, rel Σ×{0}), there is a path of diﬀeomorphisms between h and the identity map. It follows that if K ⊂Σ × I is a knot, then K and h(K) are isotopic. Now let K′′ = f ′(K) and D′′ = π(K′′). It is clear from the deﬁnition of f ′ that D′′ = π(K′′) = π(f ′(K)) = f0(π(K)) = f0(D). If we apply a diﬀeomorphism of Σ × {0} to an alternating knot diagram in Σ×{0}, then we obtain another alternating knot diagram with the same crossing number. Thus D′′ is an alternating diagram with the same crossing number as D. On the other hand K′ = f(K) = h(f ′(K)) = h(K′′), hence K′ and K′′ are isotopic as knots in Σ × I. Now by [AFLT02, Theorem 1.1.], c(D′′) ≤c(D′), but D′ has minimal crossing number, so c(D′′) = c(D′). Therefore c(D) = c(D′), and D has minimal crossing number. Deﬁnition 5.31. Suppose D is a virtual link diagram and (SD, ˜ D) the abstract 66 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics link diagram associated with D. Consider the components of ∂SD. A crossing is called proper if four diﬀerent components of ∂SD meet at that crossing. A diagram D is called proper if each crossing is proper. Remark 5.32. Notice that if D is proper, then it has no nugatory crossings. As an aﬃrmative answer for Question 6.5 in [Kam04] in the case of virtual knots, we have: Corollary 5.33. If D is a proper alternating knot diagram for a virtual knot K, then D has minimal crossing number. Proof. The result follows from the Remark 5.32, Theorem 5.28, and [Kam04, Theorem 1.2]. Proposition 5.34. If K is an almost classical knot that is alternating, then it admits an alternating almost classical diagram. Proof. Suppose D is a reduced alternating diagram for K, then by Theorem 5.30, it has minimal crossing number. The result follows from Proposition 3.26. In the next lemma, we will show that the three remaining almost classical knots from Proposition 5.25 are not alternating. In establishing the lemma, we will make use of Theorem 5.30 and Proposition 5.34. Lemma 5.35. The three almost classical knots 6.87188, 6.87310 and 6.87859 are not alternating. Proof. Let K1 = 6.87188, K2 = 6.87310 and K3 = 6.87859. We have det(K2) = det(K3) = 7. Suppose to the contrary that K2 and K3 admit reduced alternating diagrams. Then the diagrams would have minimal crossing number. By Table A.1, the only pattern with 6 or fewer crossings and determinant 7 is Θ5c, but both K2 and K3 have minimal crossing number 6 and they cannot be equivalent to a 5 crossing alternating knot. Therefore they are not alternating. Now det(K1) = 9 and if it were to admit a reduced alternating diagram, it would have one of the patterns Θ5a, Θ3#Θ3 or Θ6g. For a similar reason it cannot have the pattern Θ5a. Suppose D is a reduced alternating diagram for K1, then it has minimal crossing number and by Proposition 3.26, it must be an almost classical diagram. If D has the pattern Θ3#Θ3, then by Proposition 3.27, both factors must be almost classical which means up to mirror image the 67 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics factors are the trefoil knot. It follows that D is a classical diagram, but the virtual genus of K1 is 1, which is a contradiction. It remains to show D does not have the pattern Θ6g. The fact D is almost classical imposes restrictions on the signs of the chords, as we now explain. Referring to the diagram for Θ6g in A.1, list the chords c1, . . . , c6 in counter-clockwise order starting at the top (12 o’clock) and let ε1, . . . , ε6 be their signs in D. Almost classicality implies that ε2 = ε3. For the other chords, there are two possibilities. Either ε1 = ε4 and ε5 = ε6, or ε1 = −ε5 and ε4 = −ε6. Each one of resulting almost classical knots is, up to mirror images, equivalent to one of the knots 6.90214, 6.90217, 6.90219 or 6.90227. However, 6.90227 is classi-cal, and the Alexander polynomials of 6.90214 and 6.90217 are 3t2 −4t + 2 and t3−4t2+3t−1, respectively, which are diﬀerent from ∆K1(t) = 2t3−3t2+3t−1. On the other hand, the Jones polynomial for K1 is t+1/t−1/t2+2/t3−1/t4−1, which is diﬀerent from that of 6.90219, which equals 1. This completes the ar-gument and shows that K1 is not alternating. 68 Chapter 6 Khovanov Homology In this chapter, we introduce Khovanov homology, Lee homology and Ras-mussen’s invariant for classical and virtual knots. We calculate the Khovanov homology for the virtual knot 3.7. We generalize Lee’s theorem about the Kho-vanov homology of alternating classical links to the virtual case. 6.1 Khovanov Homology for Classical Knots In this section, we brieﬂy introduce the Khovanov homology for classical knots and links. For more details see [Tur17] and [BN02]. Khovanov homology is a (1 + 1)-TQFT (topological quantum ﬁeld theory), i.e. it is a functor from the category of compact 1-dimensional manifolds (a collection of circles) with morphisms compact and orientable, 2-dimensional cobordisms (surfaces) between them, into the category of graded vector spaces and graded linear maps. Khovanov introduced the invariant for links in [Kho00]. It is a bigraded homology theory which can be deﬁned and computed in a purely combinatorial way. Khovanov homology is a categoriﬁcation of the Jones polynomial in that its graded Euler characteristic is equal to the unnormalized Jones polynomial. For a link L, we denote its Khovanov homology by Kh∗,∗(L), and we have b χ(Kh∗,∗(D)) = X i,j∈Z (−1)iqjdimKhi,j(D) = b VL(q). Suppose D is a link diagram with n+ positive crossings and n−negative crossings. Let n = n+ + n−and enumerate the crossings by c1, . . . , cn. With 69 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Q as the coeﬃcient ring, we set V = Q1 ⊕QX to be the 2 dimensional vector space with basis {1, X}. Setting the degree of 1 to be +1 and the degree of X to be −1 gives V ⊗n the structure of a graded vector space. This grading will be denoted j and called vertical or quantum grading. If W = L m∈Z Wm is a graded vector space, then a vertical grading shift of W by ℓis deﬁned as W{ℓ} = L m∈Z W ′ m, where W ′ m = Wm−ℓ. We consider the cube of resolutions of D, which is an n-dimensional cube with 2n vertices, one for each state. Here we denote states by α ∈{0, 1}n, which is a binary sequence of length n that indicates how each crossing has been resolved. Let rα and kα be the number of 1’s and cycles in α, respectively. Let Ci,∗(D) be L V ⊗kα{rα + n+ −2n−}, where we take the direct sum over all the states α with rα = i+n−. Here i is called horizontal or homological grading. If C(D) = L i Ci(D), then a horizontal grading shift of C(D) by l is deﬁned as C(D)[l] = L i C′i(D), where C′i(D) = Ci−l(D). We deﬁne the Khovanov complex as CKh(D) = L i,j Ci,j(D). To deﬁne the diﬀerential d, we introduce the product and coproduct maps. Note that henceforth we will suppress the symbol ⊗in writing elements of V ⊗k. ∆: V →V ⊗V, m : V ⊗V →V. 1 7→1X + X1 11 7→1 X 7→XX 1X 7→X X1 7→X XX 7→0 We only deﬁne a map from a state α to a state α′ if α′ obtained from α by changing one 0 to 1. In that case, either two cycles of α merge into one cycle of α′, or one cycle of α splits into two cycles of α′. In the ﬁrst case, we use the product map m, and in the second, we use the coproduct map ∆. For all other cycles of α, we apply the identity. In order to write down all the maps, we ﬁx once and for all an enumeration of the cycles in each state, and these are not changed throughout the calculations. The last step is to assign negative signs to some of the maps. There are many ways to do that, but the homology groups for the diﬀerent choices of signs are all isomorphic. Here we follow the sign convention of [BN02]. Suppose we change 0 to 1 in the m-th spot to obtain α′ from α. In α, we count how many 1’s we have before the m-th spot. If it is an odd number, we assign a negative sign to the associated map. 70 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics For a ﬁxed j the map d2 : Ci,j →Ci+2,j is zero and we obtain a bigraded homology theory denoted by Kh∗,∗(D). In [Lee05], Lee constructs a new complex by modifying the maps ∆and m: ∆′ : V →V ⊗V, m′ : V ⊗V →V. 1 7→1X + X1 11 7→1 X 7→11 + XX 1X 7→X X1 7→X XX 7→1 This results in a new homology theory called Lee homology and denoted Lee(D). Notice the maps no longer preserve the homological degree, thus Lee homology is only graded rather being bigraded. It turns out that Lee(K) ∼ = Q ⊕Q for all knots, nevertheless as we will see the Lee homology contains a nontrivial and powerful invariant s(K) called the Rasmussen invariant. This invariant was introduced by Rasmussen in [Ras10], and we brieﬂy recall its deﬁnition. The quantum degree deﬁnes a decreasing ﬁltration on CKh(K). This induces a ﬁltration on Lee(K), H∗(C) = FnH∗(C) ⊃Fn+1H∗(C) ⊂. . . ⊃FmH∗(C). For x ∈Lee(K), let s(x) be the ﬁltration degree of x, i.e. s(x) = k if x ∈ FkH∗(C) but x does not belong to Fk+1H∗(C). We deﬁne smin(K) = min{s(x) ∈Lee(K) \| x ̸= 0}, smax(K) = max{s(x) ∈Lee(K) \| x ̸= 0}. Rasmussen proves that smax(K) = smin(K) + 2 for all knots, and he deﬁnes s(K) = smin(K) + 1 = smax(K) −1. For a link L, the ﬁltration on CKh(L) induces a spectral sequence with E0 term the Khovanov complex and d0 = dKh. It follows that the E1 term is Kh∗,∗(L). For every m, dm = 0 unless m is a multiple of 4. As a result, for any m ≥0, E4m+1 ∼ = E4m+2 ∼ = E4m+3 ∼ = E4m+4. The E∞page is isomorphic to the Lee homology. For a knot K, it has two copies of Q which are placed on the y-axis. Their location indicates the ﬁltration degree of the generators of the Lee homology. In particular the average of their y-coordinates is equal to s(K). 71 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics In [Lee05], Lee proves that, for any alternating link L, its Khovanov homol-ogy Kh∗,∗(L) is supported in the two lines j = 2i−σ(L)±1. As a result, in the spectral sequence dm = 0 for every m ≥5 and E∞= E5. If K is an alternating knot, then the y-coordinates of the two surviving copies of Q are −σ(K) ± 1. This implies that s(K) = −σ(K). If we have a cobordism S between two links L0 and L1, then S induces a map ϕ′ S : Lee(L0) →Lee(L1) with ﬁltration degree equal to χ(S). We will describe the map ϕ′ S in a moment, but ﬁrst notice that this implies that if K is a knot, then \|s(K)\| ≤2g4(K). The same inequality holds for the knot signature σ(K), and Lee’s theorem tells us that, for alternating knots, the Rasmussen invariant s(K) does not improve the bound on the 4-ball genus that one gets from the knot signature. However, for non-alternating knots, it is no longer true that s(K) = −σ(K), and sometimes the Rasmussen invariant provides a better bound. It should further be noted that Rasmussen’s invariant gives a lower bound on the smooth 4-ball genus, whereas the knot signature gives a bound on the topological 4-ball genus. Example 6.1. For K = 942, s(K) = 0 and σ(K) = 2, thus the signature provides a better bound in this case. On the other hand, for the knot K = 10132, s(K) = −2 and σ(K) = 0, so s(K) gives a better bound for the 4-ball genus. We now describe the map ϕ′ S, and since any cobordism decomposes into a sequence of elementary cobordisms, it suﬃces to deﬁne ϕ′ S for births, deaths, and saddles. In doing that, we will use the maps ι : Q →V (1 7→1) and ε : V →Q (1 7→0 and X 7→1). Note that an elementary cobordism is either a birth, a death, or a saddle. For a birth, we set ϕ′ S = ι. For a death, we set ϕ′ S = ε. As noted previously, a saddle can be either a fusion or joining saddle, or a ﬁssion or splitting saddle. For a fusion saddle, we use the product map m′, and for a ﬁssion saddle we use the coproduct ∆′. In general, the Rasmussen invariant is diﬃcult to compute. However, the calculation simpliﬁes for positive (or negative) knots, as we now explain. Deﬁnition 6.2. A link is called positive if it admits a diagram with only positive crossings. Similarly, a link is negative if it admits a diagram with only negative crossings. . If K is a positive knot with diagram D with n positive crossings, then the Rasmussen invariant is given by s(K) = −k + n + 1, 72 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics where k is the number of cycles in the all 0-smoothing state of D [Ras10]. If K is a negative knot with diagram D with n negative crossings, then the (vertical) mirror image D∗has n positive crossings, and the all 0-smoothing state of D∗is the all 1-smoothing state of D. Since s(K∗) = −k + n + 1, and since the Rasmussen invariant satisﬁes s(K∗) = −s(K) under taking mirror images, it follows that s(K) = k −n −1. 6.2 Khovanov Homology for Virtual Knots In this section, we brieﬂy introduce the Khovanov homology for virtual knots and links. The Khovanov homology for virtual knots and links, ﬁrst was deﬁned by Manturov in [Man04] and only with Z2 coeﬃcients. Later, in [Man07], he deﬁned the Khovanov homology with arbitrary coeﬃcients. In [DKK17] Dye, Kaestner and Kauﬀman reformulated Manturov’s approach and used that to prove a number of results such as a large family of virtual knots with unit Jones polynomial is not classical. Tubbenhauer in [Tub14], used un-oriented TQFT’s to deﬁne a Khovanov homology for virtual knots and links. When one attempts to deﬁne a Khovanov theory for virtual knots, the major problem is the presence of the single cycle smoothing (see Figure 6.1). We need to assign a map to a single cycle smoothing, which we can do by assigning the zero map. In classical Khovanov theory, the signs of maps are chosen in a way to make each face of the cube of resolutions to be anti-commutative. Then this fact enables us to deﬁne a diﬀerential d satisfying d2 = 0. For virtual knots, the existence of single cycle smoothings makes it more diﬃcult to assign signs. Figure 6.1: A single cycle smoothing, with induced map zero. In what follows, we describe Tubbenhauer’s method. We will cover the combinatorial deﬁnitions. To see the discussion about un-oriented TQFT’s and 73 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics for more detail, see [Tub14]. Let Q be the coeﬃcient ring and V = Q1 ⊕QX. Start with a virtual link diagram D with n classical crossings. Resolve all the crossings in both ways to obtain 2n states. The Khovanov chain complex C(D) is deﬁned as before, i.e. we assign V ⊗k to a state with k components. The degree of each element and the grading shifts are deﬁned as before. Whenever two vertices of an edge in the cube of resolutions have the same number of states, then that indicates the presence of a single cycle smoothing. In that case, we assign the zero map to the edge. It remains to deﬁne the joining and splitting maps and the signs. Choose orientations for the cycles of each state. Although we can do this in an arbitrary way, to have less complicated maps at the end, we use a span-ning tree argument. Choose a spanning tree for the cube of the resolution and start with the rightmost vertices and choose orientations for the cycles of cor-responding states. Now remove those vertices and again choose orientations for the rightmost vertices, in a compatible way. That means we compare the two vertices which are joined by an edge of the spanning tree, and orient the cycles of the left vertex as follows. For cycles which are not involved in the join, split or the single cycle smoothing, orient each cycle of the left vertex exactly like the corresponding cycle in the right vertex. For other cycles try to orient them in a way to have the most compatibility. Choose an x-marker for each crossing and the corresponding 0- and 1-smoothings, as in Figure 6.2. We choose either x or x′ and notice that up to rotating the diagram and the corresponding states, there are only these two ways to assign the x-markers. c x x′ 0 x x′ 1 x x′ Figure 6.2: An x-marker for a crossing and the corresponding smoothings. We deﬁne the sign of the non-zero maps as follows. By a spanning tree argument, number the cycles of each state. Suppose we have a joining map from a state s to another state s′, and suppose s has m + 1 cycles. Let the joining map, merges the cycles number a and b in s and the resulting cycle in 74 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics s′ has number c, and let the cycle number a has the x-marker. In an (m+1)-tuple, put a ﬁrst, then b and then place the remaining numbers in an ascending order. Let τ1 be the permutation which takes (1, 2, · · · , m + 1) to this (m + 1)-tuple. Now in an m-tuple, put c ﬁrst, and place the remaining numbers in an ascending order, and let τ2 be the permutation which takes (1, 2, · · · , m) to this m-tuple. Deﬁne the sign of the joining map to be sign(τ1)sign(τ2). The sign of the splitting map is deﬁned, similarly. To deﬁne the maps, we proceed as follows. Maps are deﬁned between the two vertices of an edge of the cube of resolutions. For each edge the smoothing of only one of the crossings is diﬀerent, and we deﬁne a map from the state with 0-smoothing to the state with 1-smoothing. If a cycle of the source state is disjoint from the smoothing change, assign the identity map to it, if its orientation agrees with the orientation of the corresponding cycle in the target state, otherwise assign negative of the identity map. At a small neighborhood of the crossing, there are two parallel strings in each state. Notice that each cycle is oriented. Now if the map looks like ↓↑→⇄, we decorate the four strings in the source and target state with a + sign, and we call this decoration standard. We always rotate the states so the two strings in the source state are vertical, and the two strings in the target state are horizontal. Then we compare the orientation of each string with the orientation of the corresponding one in the standard decoration, if they agree, decorate that string with a + sign, otherwise decorate it with a −sign. We record the diﬀerent cases in Table 6.1. string splitting map string joining map ↓↑→⇄ ∆+ ++ ↓↑→⇄ m++ + ↓↑→⇒ ∆+ −+ ↑↑→⇄ m−+ + ↓↑→⇔ ∆+ +− ↓↓→⇄ m+− + ↓↑→⇆ ∆+ −− ↑↓→⇄ m−− + ↑↓→⇄ ∆− ++ ↓↑→⇆ m++ − ↑↓→⇒ ∆− −+ ↑↑→⇆ m−+ − ↑↓→⇔ ∆− +− ↓↓→⇆ m+− − ↑↓→⇆ ∆− −− ↑↓→⇆ m−− − Table 6.1: String decoration and corresponding maps. Other cases happen only when we have a single cycle smoothing. We describe the map ∆a bc(v) as follows. Multiply v by a, apply ∆, then multiply the ﬁrst 75 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics component of the resulting tensor product by b and the second component by c. Notice that the ﬁrst component of the tensor product, corresponds to the lower string or the string with the x-marker on it. Similarly, we deﬁne the map mbc a . First multiply the ﬁrst component of the tensor product by b and the second component by c, then apply m, at the end multiply the result by a. Remark 6.3. We know that every checkerboard colorable diagram admits a source-sink orientation ([KNS02, Proposition 6]). We can use this orientation to make all the decorations standard. In that case we only need the maps ∆+ ++ and m++ + . Example 6.4. We compute the Khovanov homology for the knot K = 3.7. Figure 6.3 is a diagram for K and Figure 6.4 is the cube of resolutions. Figure 6.3: The knot K = 3.7. The red dots are the x-markers. We enumerate the components of a state in a way that the one which has more x-markers in it be the ﬁrst component. All the m maps are m++ + , and ∆maps are ∆+ ++. A red arrow means the associated map has negative sign. All the maps are a single splitting or joining map except for the state which has 3 components in it. For this state the incoming map is ∆⊗id, and for the outgoing maps, the upper one is ϕ deﬁned as ϕ(a, b, c) = −m(a, c) ⊗b, and the lower one is −id ⊗m. The Khovanov complex is as follows: V ⊗V {−3} →V ⊕V ⊕ V ⊗3 {−2} → V ⊗2 ⊕ V ⊗2 ⊕ V ⊗2 {−1} →V. We record the basis elements of the chain complex in Table 6.2. The image of each basis element is in Table 6.3. 76 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Figure 6.4: The cube of resolutions for K = 3.7. j \ i −2 −1 0 1 (11, 0, 0) 1 (0, 0, 111) (0, 11, 0) 1 (0, 0, 11) (1, 0, 0) (1X, 0, 0) (0, 1, 0) (X1, 0, 0) −1 11 (0, 0, 11X) (0, 1X, 0) X (0, 0, 1X1) (0, X1, 0) (0, 0, X11) (0, 0, 1X) (0, 0, X1) (X, 0, 0) 1X (0, X, 0) (XX, 0, 0) −3 (0, 0, 1XX) (0, XX, 0) X1 (0, 0, X1X) (0, 0, XX) (0, 0, XX1) −5 XX (0, 0, XXX) Table 6.2: The basis elements for the chain complex. It is easy to check d2 = 0. When we take the homology, two copies of Q survive, both in homological degree 0, one in quantum degree 1 and the other in quantum degree −1. Therefore the Khovanov homology of K is isomorphic to the Khovanov homology of the unknot. In [DKK17], Dye, Kaestner and Kauﬀman deﬁne Lee homology and the 77 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics j \ i −2 −1 0 1 1 1 (0, −11, −11) −1 0 1 (1X + X1, 1X + X1, 0) X (−1X −X1, 0, 1X + X1) X −1 (1, 1, 1X1 + X11) (0, −X1, −1X) −X 0 (0, −1X, −1X) −X (0, −X1, −X1) X X (XX, XX, 0) (X, X, 1XX + X1X) (−XX, 0, XX) 0 −3 (0, −XX, 0) 0 (X, X, XX1) (0, 0, −XX) 0 (0, −XX, −XX) −5 (0, 0, XXX) (0, 0, 0) Table 6.3: The image of the basis elements. Rasmussen invariant for virtual knots, and they show that Rasmussens invariant is an invariant of virtual knot concordance. Example 6.5. Table B.1 lists the Rasmussen invariant for the alternating virtual knots up to six crossings. The three virtual knots 6.90115, 6.90150 and 6.90170 all have Rasmussen invariant equal to −2, and as a result we conclude that none of these virtual knots are slice. In [BCG17a], Boden et al. deﬁne slice obstructions in terms of signatures of symmetrized Seifert matrices for almost classical knots, and as an application they show that neither 6.90115 nor 6.90150 are slice. However, the virtual knot 6.90170 is not almost classical, so the Rasmussen invariant not only provides an alternate method to show that 6.90115 and 6.90150 are not slice, it also shows that 6.90170 is not slice, which is a new result. 6.3 Khovanov Homology and Alternating Vir-tual Links Following [Lee05], we seek a relation between Rasmussen’s invariant and signa-tures of alternating virtual knots. If i is the homological degree and j is the quantum degree for Khovanov homology, then H-thinness for classical alternat-ing knots means, j = 2i −σ ± 1, where σ is the signature. This implies that s = −σ, where s is Rasmussen’s invariant. On the other hand, not all virtual 78 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics alternating knots are H-thin. For example the Khovanov polynomial for the knot K = 5.2426 depicted in Figure 6.5 is as follows: 1 q11t3 + 1 q9t3 + 1 q7t2 + 1 q5t2 + 1 q5 + 1 q3, which is supported in three lines j = 2i −1, j = 2i −3 and j = 2i −5. Notice that from Table B.1 (σξ∗, σξ) = (2, 4), and we can write the three lines as: j = 2i −σξ∗+ 1, j = 2i −σξ∗−1, j = 2i −σξ −1. In fact, instead of H-thinness we have: Proposition 6.6. If D is a connected alternating virtual link diagram with genus g, and signatures σξ, σξ∗, then its Khovanov homology is supported in g + 2 lines: j = 2i −σξ∗+ 1, j = 2i −σξ∗−1, . . . , j = 2i −σξ −1. Proof. Following [Lee05], we apply induction on the number of crossings. The base case is trivial. Let D be an alternating virtual link diagram with n crossings. 0 and 1 smooth the last crossing to obtain D(∗0) and D(∗1), re-spectively. We can easily see that they are alternating diagrams. Shift the Khovanov complex n−horizontally, and 2n−−n+ vertically. Denote the re-sulting complex by ¯ C(D) and its homology by ¯ H(D). We denote this shift by ¯ C(D) = C(D)[n−]{2n−−n+}. We have the following short exact sequence: 0 →¯ C(D(∗1))[+1]{+1} →¯ C(D) →¯ C(D(∗0)) →0, which gives a long exact sequence involving ¯ H(D), ¯ H(D(∗0)) and ¯ H(D(∗1))[+1]{+1}, which implies ¯ H(D) is supported inside ¯ H(D(∗0)) and ¯ H(D(∗1)). It suﬃces to show that ¯ H(D) is supported in g + 2 lines with y-intercepts of −\|s∂\| + 2, −\|s∂\|, · · · , −\|s∂\| −2g because after shifting back ¯ H(D), the result follows. The all 0 state of D is the same as the all 0 state of D(∗0). Also the all 1 state of D is the same as the all 1 state of D(∗1). In the all 0 state of D, if we change the resolution of the last crossing from a 0-smoothing to a 1-smoothing, we obtain the all 0 state for D(∗1). Similarly, in the all 1 state of D, if we 79 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics change the resolution of the last crossing from a 1-smoothing to a 0-smoothing, we obtain the all 1 state for D(∗0). These three diagrams, all have the boundary property. D(∗0) and D(∗1), both have n −1 crossings. Thus we have: \|s∂(D)\| + \|¯ s∂(D)\| = n + 2 −2g(D), \|s∂(D(∗0))\| + \|¯ s∂(D(∗0))\| = n + 1 −2g(D(∗0)), \|s∂(D(∗1))\| + \|¯ s∂(D(∗1))\| = n + 1 −2g(D(∗1)). Using the above observations, we can rewrite the last two equations as: \|s∂(D)\| + \|¯ s∂(D(∗0))\| = n + 1 −2g(D(∗0)), \|s∂(D(∗1))\| + \|¯ s∂(D)\| = n + 1 −2g(D(∗1)). Since the genus is an integer, the ﬁrst equation implies that \|¯ s∂(D(∗0))\| cannot be equal to \|¯ s∂(D)\|, so it is either one more, or one less. Similarly, \|s∂(D(∗1))\| is either one more, or one less than \|s∂(D)\|. Thus we have four diﬀerent cases: Case 1: \|¯ s∂(D(∗0))\| = \|¯ s∂(D)\| −1 , \|s∂(D(∗1))\| = \|s∂(D)\| −1 ⇒g(D) = g(D(∗0)) = g(D(∗1)). We use the induction hypothesis. Since \|s∂(D(∗0))\| = \|s∂(D)\| and g(D(∗0)) = g(D), the y-intercepts of the lines for D(∗0), are: −\|s∂(D)\| + 2, −\|s∂(D)\|, · · · , −\|s∂(D)\| −2g(D). The y-intercepts of the lines for D(∗1)[+1]{+1} are the y-intercepts of the lines for D(∗1) minus 1. Since \|s∂(D(∗1))\| = \|s∂(D)\|−1, the y-intercepts of the lines for D(∗1)[+1]{+1} and D(∗0) agree, and they are precisely the numbers that we are looking for. Thus the result follows in this case. Case 2: \|¯ s∂(D(∗0))\| = \|¯ s∂(D)\| + 1 , \|s∂(D(∗1))\| = \|s∂(D)\| −1 ⇒g(D) = g(D(∗0)) + 1 = g(D(∗1)). In this case, there are g(D) + 1 lines for D(∗0), and their y-intercepts are: −\|s∂(D)\| + 2, −\|s∂(D)\|, · · · , −\|s∂(D)\| −2g(D) + 2. On the other hand for D(∗1), the y-intercepts are as before. Hence the union of the supports of D(∗0) and D(∗1)[+1]{+1} is again the desired g(D)+2 lines. Case 3: \|¯ s∂(D(∗0))\| = \|¯ s∂(D)\| −1 , \|s∂(D(∗1))\| = \|s∂(D)\| + 1 ⇒g(D) = 80 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics g(D(∗0)) = g(D(∗1)) + 1. In this case, there are g(D) + 1 lines for D(∗1)[+1]{+1}, and their y-intercepts are: −\|s∂(D)\|, −\|s∂(D)\|, · · · , −\|s∂(D)\| −2g(D). For D(∗0), we have the same g(D) + 2 line, as in case 1. As before, their union is the g(D) + 2 lines with the desired y-intercepts. Case 4: \|¯ s∂(D(∗0))\| = \|¯ s∂(D)\| + 1 , \|s∂(D(∗1))\| = \|s∂(D)\| + 1 ⇒g(D) = g(D(∗0)) + 1 = g(D(∗1)) + 1. Combining case 2 and 3, we see that the result follows. Corollary 6.7. Classical alternating links are H-thin. The results of Kronheimer and Mrowka [KM11] imply that Khovanov ho-mology is an unknot detector for classical knots. For virtual knots, this is not true even for alternating virtual knots. For example 3.7 has the trivial Kho-vanov homology (1 q + q). In fact, many alternating virtual knots have trivial Khovanov homology (which is supported in two lines). Therefore the previous result is very coarse. Let D be a checkerboard virtual link diagram. Apply or to all crossings with η = −1. The result is a diagram in which η = +1 for each chord. Hence, by Lemma 5.18, the new diagram, which we call Dalt, is alternating. However, an application of or does not change the Khovanov homology of the diagram (see [DKK17]), thus it follows that D and Dalt have isomorphic Khovanov homology groups. In particular, starting with any classical diagram, we can change it to an alternating virtual diagram with the same Khovanov homology. We can do the same, starting with any checkerboard colorable diagram. Lemma 6.8. Suppose D is a positive alternating virtual knot. Then s(D) = −σξ∗(D). Proof. We computed σξ∗= β−1−n+, where β is the number of all 0-smoothing state. For any positive knot K, we have s(K) = 1 −β + n+ (see [DKK17]). For a negative knot both Rasmussen’s invariant and the signatures, are neg-ative of the corresponding values for the vertical mirror image (positive knot). It follows that s(K) = −σξ(K). In general it is not true that Rasmussen’s in-variant is the negative of one of the signatures for alternating virtual knots. For 81 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics example, the virtual knot 5.2427 is alternating (see Figure 6.6), has Rasmussen invariant s(K) = −2, and signatures σξ(K) = 4 and σξ∗(K) = 0. − − − − − Figure 6.5: A Gauss diagram and virtual knot diagram for 5.2426. − − − − + Figure 6.6: A Gauss diagram and virtual knot diagram for 5.2427. Deﬁnition 6.9. The underlying Gauss pattern of an alternating virtual knot is called an alternating pattern. Remark 6.10. If K is an alternating virtual knot, then by Proposition 3.15 the Alexander determinant of K depends only on the underlying alternating pattern. Remark 6.11. Notice that some alternating patterns do not contain any classical knot diagrams. For example, the pattern Θ5a in Figure A.1 has no classical diagrams in it. Proposition 6.12. If Θ is an alternating pattern, then it contains at least one almost classical knot. 82 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Proof. Suppose c is a chord in Θ and its arrow-head points upwards. The pattern is alternating which means the number of heads and tails on each side of c are equal. If we exclude the chords which don’t intersect c, then at each side the number of remaining heads and tails are still equal. Therefore, the number of chords intersecting c with their arrow-heads to the left is equal to the number of chords intersecting c with their arrow-heads to the right. If we assign negative sign to each chord, the resulting Gauss diagram is almost classical. By assigning diﬀerent signs to the chords of an alternating pattern with n chords, we obtain 2n alternating knot diagrams (some of them might be equivalent to each other). We arrange them on the vertices of an n-dimensional cube. We can move from one vertex to another one by applying a sequence of sign change operations. If D is an alternating pattern, label the chords by 1, . . . , n. Denote each vertex by Dε1,ε2,...,εn, where εi is the sign of i-th chord. Consider D++···+, and ﬁnd the states of resolution. Denote them by S++···+ i1,i2,...,in, where ik is 0 or 1, according to whether we resolve the k-th chord to a 0-smoothing or a 1-smoothing. Then we have Sε1,ε2,...,εn i1,i2,...,in = S++···+ j1,j2,...,jn, where if εk = +, then jk = ik, and if εk = −, then jk = 1 −ik. In ordinary Khovanov homology, one assigns the zero map to each single cy-cle smoothing. Using diﬀerent maps for single cycle smoothings, one can obtain a more reﬁned version of Khovanov homology for virtual knots. For example, in [Rus17] Rushworth uses this approach to deﬁne a variant theory called doubled Khovanov homology. For virtual links whose cube of resolutions has no single cycle smoothings, then the doubled Khovanov homology is the direct sum of two copies of ordinary Khovanov homology. In that case, the doubled Kho-vanov homology is completely determined by the ordinary Khovanov homology and thus it contains no new information. In [Rus17], Rushworth proves that for all checkerboard colorable diagrams, there is no single cycle smoothing. This fact follows also from [KNS02, Propo-sition 6]. Here we provide a diﬀerent proof of that fact, from which it follows that for checkerboard links, the doubled Khovanov homology is the direct sum of two copies of ordinary Khovanov homology. Lemma 6.13. Let D be an alternating link diagram, and s∂be the all 0-smoothing state, and ¯ s∂the all 1-smoothing state. If we change one 0-smoothing to obtain the state s, the number of components of s∂and s, diﬀers by one. Sim-ilar result holds for ¯ s∂. 83 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Proof. Assume we change the smoothing in the last crossing. We consider D(∗1), which is an alternating diagram and has the boundary property. If D has c crossings, and g and g1 are the genera for D and D(∗1) respectively, we have: \|s∂\| + \|¯ s∂\| = c + 2 −2g, \|s\| + \|¯ s∂\| = c −1 + 2 −2g1, \|s∂\| −\|s\| = 1 + 2(g1 −g). Thus the diﬀerence is an odd number, and the result follows. The proof for the other case is similar. Proposition 6.14. Let D be an alternating link diagram. Then there is no single cycle smoothing in the cube of resolutions for D. Proof. Assume we change a 0-smoothing of the state s to a 1-smoothing at the crossing ci. If for all the other crossings, we have 0-smoothing in s, then this is the previous lemma. Otherwise, we apply sc to the crossings of D, which have been resolved to 1-smoothings in s. Call the new diagram D′. Since the state s is the all 0-smoothing state for D′, the result follows from the previous lemma. Proposition 6.15. Let D be a checkerboard colorable link diagram. Then there is no single cycle smoothing in the cube of resolutions for D. Proof. Assume we change one 0-smoothing of the state s to a 1-smoothing at the crossing ci, and call the resulting state s′. First we consider Dalt. Let C′ = {ci1, . . . , cik} be the set of crossings of D which are changed to obtain Dalt. There are two cases. If ci does not belong to C′, then the edge with vertices s and s′ corresponds exactly to an edge in the cube of resolutions for Dalt, and the result follows. If ci ∈C′, then the same thing happens. The only diﬀerence is the direction of the map in Dalt is reversed, going from s′ to s. The result still holds. Corollary 6.16. If D is a checkerboard colorable link diagram, then the dou-bled Khovanov homology for D is the direct sum of two copies of the ordinary Khovanov homology for D. 84 Chapter 7 Problem List and Further Studies Suppose we have a checkerboard colored diagram. If we apply the unknotting operations on a chord, the question is, how exactly the signatures change. In particular if we start with an alternating diagram, we can study this question. If we apply sc to a chord in an alternating diagram, then the resulting diagram is again alternating. Suppose D is an alternating diagram, with at least one positive crossing. Enumerate the crossings and suppose the last crossing is positive. We have: σξ = 1 −\|¯ s∂(D)\| + n−(D) , σξ∗= \|s∂(D)\| −1 −n+(D). Apply sc to the last crossing to ﬁnd D′. Then s∂(D′) is the same as s∂(D(∗1)), and ¯ s∂(D′) is ¯ s∂(D(∗0)). Similar to the Proposition 6.6, we have four cases. In each case the signatures of D′ is the same as signatures of D, or one or both go up by 2. Problem 7.1. For an alternating diagram, in terms of other invariants, like Alexander polynomials, determine when exactly the signatures change under the sc operation. This problem is motivated by Conway’s result about the signature of the classical knots in [Gil82], which states that if we apply cc to a positive crossing, then the knot signature goes up by 2, if ∆K(−1) changes sign, otherwise the knot signature stays the same. We can also analyze the behavior of the signatures under sc for a checker-board colorable diagram. 85 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Therefore we have the following problem: Problem 7.2. Start with a checkerboard colorable diagram. For each unknotting operation (except cd), analyze the behavior of the signatures. Problem 7.3. What is the relation between the Khovanov homologies of the knot diagrams with the same alternating pattern? Related to the Problem 7.1, we have: Problem 7.4. When we move between diﬀerent vertices of the cube of an alter-nating pattern, determine how Rasmussen’s invariant and the signatures change. Problem 7.5. Do the signatures provide a lower bound on the slice genus of a knot? Is the signature for long virtual knots a concordance invariant? Notice if the signatures were slice obstructions for checkerboard knots, then a direct consequence of Theorem 5.19 would be that non-classical alternating knots are never slice. Problem 7.6. Suppose L is a virtual link and D is an alternating almost classical diagram for it. Apply Seifert’s algorithm to ﬁnd a Seifert surface for D. Does this surface have minimal genus? Problem 7.7. Does Theorem 5.30 admit a converse if one assumes the knot K is prime, i.e. is every minimal crossing diagram of a prime alternating virtual knot is reduced and alternating? There are (at least) two ways to state an analogue of the third Tait Con-jecture for virtual knots. If the tangle diagram T (see Figure 2.8) is purely classical, then we can do a ﬂype move and it will preserve the virtual alternat-ing link type. It is not clear whether this move alone is powerful enough to pass from any minimal crossing diagram to any other. On the other hand, if the tangle diagram T is allowed to contain virtual crossings, then performing this kind of “virtual ﬂype” will not always preserve the link type. For instance, the Kauﬀman ﬂype is used in [Kam17] and leads to K-equivalence of virtual knots and links. The fact that “virtual ﬂyping” can change the alternating virtual knot or link type was also noted by Zinn-Justin and Zuber in [ZJZ04]. Problem 7.8. Is there an analogue of the Tait ﬂyping conjecture for alternating virtual knots? This is Problem 15 in [FIKM14], see also [ZJ06]. 86 Appendices 87 Appendix A Alternating Patterns Up to 6 crossings, there are 15 distinct alternating patterns. Figure A.1 shows their associated Gauss diagrams, and Table A.1 lists the Alexander determinants of the alternating virtual knots according to these patterns. Notice by Remark 6.10, the Alexander determinants depend only on the underlying pattern. Alternating Alexander Pattern Determinant Θ3 3 Θ4 5 Θ5a 9 Θ5b 11 Θ5c 7 Θ5d 5 Θ3#Θ3 9 Θ6a 15 Θ6b 19 Θ6c 17 Θ6d 13 Θ6e 19 Θ6f 11 Θ6g 9 Θ6h 13 Table A.1: Alexander determinant for each alternating pattern. 88 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Θ3 Θ4 Θ5a Θ5b Θ5c Θ5d Θ3#Θ3 Θ6a Θ6b Θ6c Θ6d Θ6e Θ6f Θ6g Θ6h Figure A.1: Alternating patterns. 89 Appendix B Alternating Virtual Knots and Their Invariants The signatures and determinants are computed using a Mathematica program written by Micah Chrisman, and the Khovanov homology (unlisted) is computed using online Mathematica program written by Daniel Tubbenhauer. The Ras-mussen’s invariants are then computed by hand using Lee’s spectral sequence. Virtual knots are grouped according to their alternating patterns. For each pattern, we ﬁx a base-point on the core circle of the Gauss diagram. Starting with the base-point, we travel counterclockwise. For knots with the same alter-nating pattern, instead of writing the entire Gauss code, we only list the signs of the chords. Boldface font is used to indicate that the knot is classical. Virtual Alternating Sign Signatures Determinants Rasmussen Knot Pattern Sequence (σ∗ ξ, σξ) (det∗ ξ, detξ) Invariant 3.6 Θ3 −−− (2, 2) (3, 3) −2 3.7 Θ3 −−+ (0, 2) (1, 2) 0 4.105 Θ4 −−−− (0, 2) (1, 4) −2 4.106 Θ4 −−−+ (0, 2) (3, 2) 0 4.107 Θ4 −−++ (−2, 2) (1, 1) 0 4.108 Θ4 −+ +− (0, 0) (5, 5) 0 5.2426 Θ5a −−−−− (2, 4) (4, 5) −4 5.2427 Θ5a −−−−+ (0, 4) (2, 1) −2 5.2428 Θ5a −−−+ + (0, 2) (7, 2) 0 5.2429 Θ5a −−+ −− (2, 4) (8, 1) −2 5.2430 Θ5a −−+ −+ (0, 2) (5, 4) 0 5.2431 Θ5a −+ −−+ (−2, 2) (1, 3) 0 5.2432 Θ5a −+ −+ − (−2, 2) (1, 3) 0 5.2433 Θ5b −−−−− (0, 4) (1, 5) −4 5.2434 Θ5b −−−−+ (0, 4) (3, 1) −2 90 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Virtual Alternating Sign Signatures Determinants Rasmussen Knot Pattern Sequence (σ∗ ξ, σξ) (det∗ ξ, detξ) Invariant 5.2435 Θ5b −−−+ + (0, 2) (8, 3) 0 5.2436 Θ5b −−+ −+ (−2, 2) (4, 1) 0 5.2437 Θ5c −−−−− (2, 2) (7, 7) −2 5.2438 Θ5c −−−−+ (0, 2) (2, 5) −2 5.2439 Θ5c −−−+ + (0, 2) (4, 3) 0 5.2440 Θ5c −−+ −− (0, 2) (4, 3) 0 5.2441 Θ5c −−+ −+ (−2, 2) (2, 1) 0 5.2442 Θ5c −−+ + − (−2, 2) (2, 1) 0 5.2443 Θ5c −+ + −− (−2, 0) (1, 6) 0 5.2444 Θ5c −+ + + − (−2, 0) (3, 4) 0 5.2445 Θ5d −−−−− (4, 4) (5, 5) −4 5.2446 Θ5d −−−−+ (2, 4) (4, 1) −2 5.2447 Θ5d −−−+ + (0, 2) (3, 2) 0 5.2448 Θ5d −−+ −+ (0, 2) (3, 2) 0 6.89187 Θ3#Θ3 −−−−−− (4, 4) (9, 9) −4 6.89188 Θ3#Θ3 −−−−+− (2, 4) (6, 3) −2 6.89189 Θ3#Θ3 −−−−++ (0, 2) (3, 6) −2 6.89198 Θ3#Θ3 + + + −−− (0, 0) (9, 9) 0 6.90101 Θ3#Θ3 −−+ −−+ (0, 4) (4, 1) 0 6.90102 Θ3#Θ3 −−+ −+− (0, 4) (4, 1) 0 6.90103 Θ3#Θ3 −−+ −++ (−2, 2) (2, 2) 0 6.90104 Θ3#Θ3 −−+ + −− (0, 4) (4, 1) 0 6.90105 Θ3#Θ3 −−+ + −+ (−2, 2) (2, 2) 0 6.90106 Θ3#Θ3 −−+ + +− (−2, 2) (2, 2) 0 6.90107 Θ3#Θ3 −+ −−+− (0, 4) (4, 1) 0 6.90108 Θ3#Θ3 −+ −+ −+ (−2, 2) (2, 2) 0 6.90109 Θ6a −−−−−− (2, 4) (7, 8) −4 6.90110 Θ6a −−−−−+ (2, 4) (13, 2) −2 6.90111 Θ6a −−−−+− (0, 4) (2, 4) −2 6.90112 Θ6a −−−−++ (0, 4) (5, 1) −2 6.90113 Θ6a −−−+ −− (0, 4) (2, 4) −2 6.90114 Θ6a −−−+ −+ (0, 4) (5, 1) −2 6.90115 Θ6a −−−+ +− (0, 2) (4, 11) −2 6.90116 Θ6a −−−+ ++ (0, 2) (10, 5) 0 6.90117 Θ6a −−+ −−− (0, 4) (2, 2) −2 6.90118 Θ6a −−+ −−+ (0, 2) (6, 9) 0 6.90119 Θ6a −−+ −+− (0, 4) (7, 1) 0 6.90120 Θ6a −−+ −++ (−2, 2) (5, 2) 0 6.90121 Θ6a −−+ + −− (0, 4) (7, 1) 0 6.90122 Θ6a −−+ + −+ (−2, 2) (5, 2) 0 6.90123 Θ6a −−+ + +− (0, 2) (12, 3) 0 6.90124 Θ6a −−+ + ++ (−2, 2) (1, 4) 0 6.90125 Θ6a −+ −−−− (0, 4) (4, 2) −2 6.90126 Θ6a −+ −−−+ (0, 2) (8, 7) 0 6.90127 Θ6a −+ −−+− (−2, 4) (1, 1) 0 6.90128 Θ6a −+ −−++ (−2, 2) (4, 3) 0 6.90129 Θ6a −+ −+ −− (−2, 4) (1, 1) 0 6.90130 Θ6a −+ −+ −+ (−2, 2) (4, 3) 0 6.90131 Θ6a −+ −+ +− (−2, 2) (3, 2) 0 6.90132 Θ6a −+ −+ ++ (−2, 2) (1, 6) 0 6.90133 Θ6a −+ + −−− (−2, 2) (6, 1) 0 6.90134 Θ6a −+ + −+− (−2, 2) (3, 4) 0 91 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Virtual Alternating Sign Signatures Determinants Rasmussen Knot Pattern Sequence (σ∗ ξ, σξ) (det∗ ξ, detξ) Invariant 6.90135 Θ6a −+ + −++ (−4, 2) (1, 1) 0 6.90136 Θ6a −+ + + −− (−2, 2) (3, 4) 0 6.90137 Θ6a −+ + + −+ (−4, 2) (1, 1) 0 6.90138 Θ6a −+ + + ++ (−4, 0) (2, 4) 2 6.90139 Θ6b −−−−−− (2, 4) (12, 9) −4 6.90140 Θ6b −−−−−+ (0, 4) (4, 3) −2 6.90141 Θ6b −−−−++ (−2, 4) (1, 1) 0 6.90142 Θ6b −−−+ −+ (−2, 2) (11, 1) 0 6.90143 Θ6b −−−+ ++ (−2, 2) (5, 5) 0 6.90144 Θ6b −−+ −−+ (0, 4) (12, 1) 0 6.90145 Θ6b −−+ −++ (−2, 2) (4, 4) 0 6.90146 Θ6b −+ + −+− (−2, 2) (3, 3) 0 6.90147 Θ6c −−−−−− (0, 4) (1, 8) −4 6.90148 Θ6c −−−−−+ (0, 4) (3, 4) −2 6.90149 Θ6c −−−−+− (0, 4) (3, 4) −2 6.90150 Θ6c −−−−++ (0, 2) (5, 12) −2 6.90151 Θ6c −−−+ −− (0, 4) (3, 2) −2 6.90152 Θ6c −−−+ −+ (0, 4) (8, 1) 0 6.90153 Θ6c −−−+ +− (0, 4) (8, 1) 0 6.90154 Θ6c −−−+ ++ (0, 2) (13, 4) 0 6.90155 Θ6c −−+ −−− (0, 2) (5, 2) −2 6.90156 Θ6c −−+ −−+ (0, 4) (1, 1) 0 6.90157 Θ6c −−+ −+− (−2, 4) (1, 1) 0 6.90158 Θ6c −−+ −++ (−2, 2) (4, 2) 0 6.90159 Θ6c −−+ + −− (−2, 2) (7, 1) 0 6.90160 Θ6c −−+ + −+ (−2, 2) (4, 4) 0 6.90161 Θ6c −+ −+ −− (−2, 2) (9, 1) 0 6.90162 Θ6c −+ −+ −+ (−2, 2) (5, 3) 0 6.90163 Θ6c −+ −+ +− (−2, 2) (5, 3) 0 6.90164 Θ6c −+ −+ ++ (−2, 2) (1, 5) 0 6.90165 Θ6c −+ + −−− (0, 2) (11, 6) 0 6.90166 Θ6c −+ + −−+ (−2, 2) (3, 3) 0 6.90167 Θ6d −−−−−− (2, 4) (4, 9) −4 6.90168 Θ6d −−−−−+ (2, 4) (10, 3) −2 6.90169 Θ6d −−−−+− (0, 4) (2, 3) −2 6.90170 Θ6d −−−−++ (0, 2) (5, 8) −2 6.90171 Θ6d −−−+ +− (0, 2) (5, 4) 0 6.90172 Θ6d −−−+ ++ (0, 0) (13, 13) 0 6.90173 Θ6d −−+ −−+ (0, 4) (6, 1) 0 6.90174 Θ6d −−+ −+− (−2, 4) (1, 1) 0 6.90175 Θ6d −−+ −++ (−2, 2) (3, 3) 0 6.90176 Θ6d −−+ + −− (−2, 4) (1, 1) 0 6.90177 Θ6d −−+ + −+ (−2, 2) (3, 3) 0 6.90178 Θ6d −−+ + +− (−2, 2) (2, 4) 0 6.90179 Θ6d −+ −−−+ (0, 4) (6, 1) 0 6.90180 Θ6d −+ −−+− (−2, 4) (1, 1) 0 6.90181 Θ6d −+ −+ −+ (−2, 2) (3, 3) 0 6.90182 Θ6d −+ −+ +− (−2, 2) (2, 4) 0 6.90183 Θ6d −+ + −++ (−2, 2) (2, 2) 0 6.90184 Θ6d −+ + + +− (−4, 0) (1, 4) 2 6.90185 Θ6e −−−−−− (0, 4) (1, 9) −4 6.90186 Θ6e −−−−−+ (0, 4) (5, 3) −2 92 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Virtual Alternating Sign Signatures Determinants Rasmussen Knot Pattern Sequence (σ∗ ξ, σξ) (det∗ ξ, detξ) Invariant 6.90187 Θ6e −−−−+− (0, 4) (3, 3) −2 6.90188 Θ6e −−−−++ (−2, 4) (1, 1) 0 6.90189 Θ6e −−−+ −+ (−2, 2) (8, 1) 0 6.90190 Θ6e −−−+ ++ (−2, 2) (4, 5) 0 6.90191 Θ6e −−+ −+− (0, 2) (8, 11) 0 6.90192 Θ6e −−+ + −− (0, 4) (11, 1) 0 6.90193 Θ6e −−+ + −+ (−2, 2) (3, 4) 0 6.90194 Θ6e −+ + −+− (0, 2) (16, 3) 0 6.90195 Θ6f −−−−−− (2, 4) (3, 8) −4 6.90196 Θ6f −−−−−+ (0, 4) (2, 2) −2 6.90197 Θ6f −−−−+− (0, 4) (2, 2) −2 6.90198 Θ6f −−−−++ (−2, 2) (4, 1) 0 6.90199 Θ6f −−−+ −+ (−2, 2) (4, 1) 0 6.90200 Θ6f −−−+ ++ (−2, 0) (5, 6) 0 6.90201 Θ6f −−+ −−− (2, 4) (7, 4) −2 6.90202 Θ6f −−+ −−+ (0, 4) (5, 1) 0 6.90203 Θ6f −−+ −+− (0, 4) (5, 1) 0 6.90204 Θ6f −−+ −++ (−2, 2) (2, 3) 0 6.90205 Θ6f −−+ + −− (0, 4) (5, 1) 0 6.90206 Θ6f −−+ + −+ (−2, 2) (2, 3) 0 6.90207 Θ6f −−+ + +− (−2, 2) (2, 3) 0 6.90208 Θ6f −−+ + ++ (−4, 0) (1, 3) 2 6.90209 Θ6f −+ + −−− (2, 2) (11, 11) −2 6.90210 Θ6f −+ + −−+ (0, 2) (8, 3) 0 6.90211 Θ6f −+ + −++ (−2, 0) (5, 6) 0 6.90212 Θ6f −+ + + −+ (−2, 0) (5, 6) 0 6.90213 Θ6f −+ + + ++ (−4, −2) (2, 9) 2 6.90214 Θ6g −−−−−− (0, 2) (1, 8) −2 6.90215 Θ6g −−−−−+ (0, 2) (3, 6) −2 6.90216 Θ6g −−−−+− (0, 2) (3, 6) −2 6.90217 Θ6g −−−−++ (0, 2) (5, 4) 0 6.90218 Θ6g −−−+ −+ (0, 2) (5, 4) 0 6.90219 Θ6g −−−+ +− (0, 2) (5, 4) 0 6.90220 Θ6g −−−+ ++ (0, 2) (7, 2) 0 6.90221 Θ6g −−+ −−− (0, 2) (5, 4) 0 6.90222 Θ6g −−+ −−+ (−2, 2) (3, 1) 0 6.90223 Θ6g −−+ −+− (−2, 2) (3, 1) 0 6.90224 Θ6g −−+ −++ (−2, 2) (2, 2) 0 6.90225 Θ6g −−+ + −+ (−2, 2) (2, 2) 0 6.90226 Θ6g −−+ + +− (−2, 2) (2, 2) 0 6.90227 Θ6g −+ + −−− (0, 0) (9, 9) 0 6.90228 Θ6h −−−−−− (0, 2) (1, 12) −2 6.90229 Θ6h −−−−−+ (0, 2) (5, 8) −2 6.90230 Θ6h −−−−++ (−2, 2) (5, 1) 0 6.90231 Θ6h −−−+ −+ (−2, 2) (5, 1) 0 6.90232 Θ6h −−−+ +− (0, 2) (9, 4) 0 6.90233 Θ6h −−−+ ++ (−2, 2) (2, 2) 0 6.90234 Θ6h −−+ −++ (−2, 2) (3, 3) 0 6.90235 Θ6h −−+ + −+ (−2, 2) (3, 3) 0 Table B.1: Alternating knots and their invariants. 93 Appendix C Table of Jones Polynomials The computations of the Jones polynomial were performed in Matlab with a program written by Lindsay White. Knot Jones Polynomial 3.6 1/t + 1/t3 −1/t4 3.7 1 4.105 1/t + 1/t3 −1/t4 4.106 1 4.107 1 4.108 t2 −t + 1 −1/t + 1/t2 5.2426 1/t2 + 1/t3 −1/t5 5.2427 1/t + 1/t3 −1/t4 5.2428 −t + 2 + 1/t2 −1/t3 5.2429 2/t −1/t2 + 1/t3 −2/t4 + 1/t5 5.2430 1 5.2431 1 5.2432 1 5.2433 1/t2 + 1/t3 −1/t5 5.2434 1/t + 1/t3 −1/t4 5.2435 −t + 2 + 1/t2 −1/t3 5.2436 1 5.2437 1/t −1/t2 + 2/t3 −1/t4 + 1/t5 −1/t6 5.2438 1/t + 1/t3 −1/t4 5.2439 1 5.2440 1 5.2441 1 5.2442 1 5.2443 t2 −t + 1 −1/t + 1/t2 5.2444 1 5.2445 1/t2 + 1/t4 −1/t5 + 1/t6 −1/t7 5.2446 1/t + 1/t3 −1/t4 5.2447 1 5.2448 1 6.89187 1/t2 + 2/t4 −2/t5 + 1/t6 −2/t7 + 1/t8 Knot Jones Polynomial 6.89188 1/t + 1/t3 −1/t4 6.89189 1/t + 1/t3 −1/t4 6.89198 −t3 + t2 −t + 3 −1/t + 1/t2 −1/t3 6.90101 1 6.90102 1 6.90103 1 6.90104 1 6.90105 1 6.90106 1 6.90107 1 6.90108 1 6.90109 1/t2 + 1/t3 −1/t5 6.90110 2/t −2/t2 + 2/t3 −2/t4 + 2/t5 −1/t6 6.90111 1/t + 1/t3 −1/t4 6.90112 1/t + 1/t3 −1/t4 6.90113 1/t + 1/t3 −1/t4 6.90114 1/t + 1/t3 −1/t4 6.90115 t −1 + 1/t −1/t2 + 2/t3 −1/t4 6.90116 −t + 2 + 1/t2 −1/t3 6.90117 2/t −1/t2 + 1/t3 −2/t4 + 1/t5 6.90118 t −1/t2 + 1/t3 6.90119 1 6.90120 1 6.90121 1 6.90122 1 6.90123 t2 −2t + 2 −1/t + 2/t2 −1/t3 6.90124 −t3 + t2 + 2 −1/t 6.90125 1/t + 1/t3 −1/t4 6.90126 1 6.90127 1 94 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics Knot Jones Polynomial 6.90128 1 6.90129 1 6.90130 1 6.90131 t2 −t + 1 −1/t + 1/t2 6.90132 1 6.90133 1 6.90134 1 6.90135 1 6.90136 1 6.90137 1 6.90138 −t4 + t3 + t 6.90139 1/t2 + 2/t3 −1/t4 −1/t5 −1/t6 + 1/t7 6.90140 2/t −1/t2 + 1/t3 −2/t4 + 1/t5 6.90141 1 6.90142 t −1/t2 + 1/t3 6.90143 1 6.90144 2 −1/t −1/t3 + 1/t4 6.90145 t2 −t + 1 −1/t + 1/t2 6.90146 2t2 −2t + 1 −2/t + 2/t2 6.90147 1/t2 + 1/t3 −1/t5 6.90148 1/t + 1/t3 −1/t4 6.90149 1/t + 1/t3 −1/t4 6.90150 t −1 + 1/t −1/t2 + 2/t3 −1/t4 6.90151 2/t −1/t2 + 1/t3 −2/t4 + 1/t5 6.90152 1 6.90153 1 6.90154 t2 −2t + 2 −1/t + 2/t2 −1/t3 6.90155 1/t + 1/t3 −1/t4 6.90156 1 6.90157 1 6.90158 t2 −t + 1 −1/t + 1/t2 6.90159 1 6.90160 1 6.90161 t −1/t2 + 1/t3 6.90162 1 6.90163 1 6.90164 −t3 + t2 + 2 −1/t 6.90165 −1/t −1/t3 + 1/t4 6.90166 t2 −t + 1 −1/t + 1/t2 6.90167 1/t2 + 1/t4 −1/t5 + 1/t6 −1/t7 6.90168 1/t −1/t2 + 2/t3 −1/t4 + 1/t5 −1/t6 6.90169 1/t + 1/t3 −1/t4 6.90170 1/t + 1/t3 −1/t4 6.90171 1 6.90172 −t3 + 2t2 −2t + 3 −2/t + 2/t2 −1/t3 6.90173 1 6.90174 1 6.90175 1 6.90176 1 6.90177 1 6.90178 1 6.90179 1 6.90180 1 6.90181 1 Knot Jones Polynomial 6.90182 1 6.90183 t2 −t + 1 −1/t + 1/t2 6.90184 −t4 + t3 + t 6.90185 1/t2 + 1/t3 −1/t5 6.90186 1/t + 1/t3 −1/t4 6.90187 2/t −1/t2 + 1/t3 −2/t4 + 1/t5 6.90188 1 6.90189 1 6.90190 1 6.90191 t −1/t2 + 1/t3 6.90192 2 −1/t −1/t3 + 1/t4 6.90193 t2 −t + 1 −1/t + 1/t2 6.90194 2t2 −3t + 2 −2/t + 3/t2 −1/t3 6.90195 1/t2 + 1/t4 −1/t5 + 1/t6 −1/t7 6.90196 1/t + 1/t3 −1/t4 6.90197 1/t + 1/t3 −1/t4 6.90198 1 6.90199 1 6.90200 1 6.90201 1/t + 1/t3 −1/t4 6.90202 1 6.90203 1 6.90204 1 6.90205 1 6.90206 1 6.90207 1 6.90208 −t4 + t3 + t 6.90209 t −1 + 2/t −2/t2 + 2/t3 −2/t4 + 1/t5 6.90210 t2 −t + 1 −1/t + 1/t2 6.90211 1 6.90212 1 6.90213 −t6 + t5 −t4 + 2t3 −t2 + t 6.90214 1/t −1/t2 + 2/t3 −1/t4 + 1/t5 −1/t6 6.90215 1/t + 1/t3 −1/t4 6.90216 1/t + 1/t3 −1/t4 6.90217 1 6.90218 1 6.90219 1 6.90220 t2 −t + 1 −1/t + 1/t2 6.90221 1 6.90222 1 6.90223 1 6.90224 1 6.90225 1 6.90226 1 6.90227 t2 −t + 2 −2/t + 1/t2 −1/t3 + 1/t4 6.90228 1/t −2/t2 + 3/t3 −1/t4 + 2/t5 −2/t6 6.90229 1/t + 1/t3 −1/t4 6.90230 1 6.90231 1 6.90232 2 −1/t −1/t3 + 1/t4 6.90233 t2 −t + 1 −1/t + 1/t2 6.90234 1 6.90235 1 Table C.1: Alternating virtual knots and their Jones polynomial. 95 Bibliography [AFLT02] Colin Adams, Thomas Fleming, Michael Levin, and Ari M. 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Doubled Khovanov homology, 2017. ArXiv/1704.07324. [Thi87] Morwen B. Thistlethwaite. A spanning tree expansion of the Jones polynomial. Topology, 26(3):297–309, 1987. [Thi88] Morwen B. Thistlethwaite. Kauﬀman’s polynomial and alternating links. Topology, 27(3):311–318, 1988. [Tub14] Daniel Tubbenhauer. Virtual Khovanov homology using cobor-disms. J. Knot Theory Ramiﬁcations, 23(9):1450046, 91, 2014. [Tur87] Vladimir G. Turaev. A simple proof of the Murasugi and Kauﬀman theorems on alternating links. Enseign. Math. (2), 33(3-4):203–225, 1987. [Tur17] Paul Turner. Five lectures on Khovanov homology. J. Knot Theory Ramiﬁcations, 26(3):1741009, 41, 2017. [Tut46] William T. Tutte. On Hamiltonian circuits. J. London Math. Soc., 21:98–101, 1946. [vAEdB51] T. van Aardenne-Ehrenfest and Nicolaas G. de Bruijn. Circuits and trees in oriented linear graphs. Simon Stevin, 28:203–217, 1951. [Wal68] Friedhelm Waldhausen. On irreducible 3-manifolds which are suﬃ-ciently large. Ann. of Math. (2), 87:56–88, 1968. [ZJ06] Paul Zinn-Justin. Alternating virtual link database, 2006. pzinn/virtlinks/. [ZJZ04] Paul Zinn-Justin and Jean-Bernard Zuber. Matrix integrals and the generation and counting of virtual tangles and links. J. Knot Theory Ramiﬁcations, 13(3):325–355, 2004. 102 Index 0-smoothing, 9 1-smoothing, 9 r-parallel, 21 abstract link diagram, 29 abstract R-move equivalent, 29 adequate, 20 Alexander determinant, 36 Alexander matrix, 12 Alexander module, 11 Alexander polynomial, 13 almost classical diagram, 34 almost classical knot, 34 alternate orientation, 55 alternating diagram, 17, 52 alternating link, 17, 52 alternating pattern, 82 alternating polynomial, 18 amphichiral knot, 49 articulation vertex, 57 band move, 14 birth, 14 black surface, 15 boundary property, 37 checkerboard colorable, 33, 44 checkerboard coloring, 33 checkerboard determinants, 43 checkerboard link, 33 circuit, 56 cobordant, 13 coloring matrix, 35 complement, 10 complete invariant, 31 crossing number, 8 death, 14 destabilisation, 28 determinant, 13 diagram category, 38 diﬀeomorphism, 28 elementary cobordism, 14 elementary ideal, 12 equivalent, 7 Eulerian circuit, 57 Eulerian graph, 57 even diagram, 38 ﬁssion type saddle, 14 ﬂat diagram, 33 Fox diﬀerentiation, 11 Fox Jacobian matrix, 11 fusion type saddle, 14 Gauss diagram, 26 Gauss pattern, 36 generalized Reidemeister moves, 26 homological grading, 70 horizontal grading, 70 horizontal grading shift, 70 horizontal mirror image, 48 horizontally amphichiral, 49 103 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics incidence number, 15 indegree, 56 index, 33 invariant, 8 inverse of a knot, 48 joining saddle, 14 Kauﬀman bracket, 9 knot, 7 knot determinant, 36 link, 7 link diagram, 8 long virtual knot, 43 minimal diagram, 30 minus-adequate, 20 mod p almost classical diagram, 34 mod p almost classical knot, 34 mod-2 almost classical, 38 negative link, 72 normal knots, 38 nugatory crossing, 57 nullity, 13 outdegree, 56 parity projection, 38 plus-adequate, 20 positive link, 72 proper crossing, 67 proper diagram, 67 quantum grading, 70 Rasmussen invariant, 71 realization, 29 reduced, 17 reduced diagram, 64 regular projection, 8 Reidemeister moves, 8 removable crossing, 17, 63 rooted tree, 54 saddle, 14 Seifert circuit, 12 Seifert genus, 12 Seifert matrix, 13 Seifert pairing, 12 Seifert surface, 12 Seifert’s algorithm, 12 separating saddle, 14 signature, 13 single cycle smoothing, 73 slice disk, 30 slice genus, 14 smoothly concordant, 13 smoothly slice, 13 source-sink orientation, 55 span, 10 spanning surface, 14 special diagram, 15 split diagram, 8 split link, 8, 30 stabilisation, 28 stable equivalence, 28 standard decoration, 75 state, 36 strongly alternating polynomial, 18 supporting genus, 30 Tait graph, 3 thickened surface, 28 topological quantum ﬁeld theory, 69 trail, 56 valuation, 54 vertical annulus, 28 104 Ph.D. Thesis - Homayun Karimi McMaster University - Mathematics vertical grading, 70 vertical grading shift, 70 vertical mirror image, 48 vertically amphichiral, 49 virtual Alexander module, 32 virtual crossing number, 31 virtual genus, 30 virtual knot, 25 virtual link, 25 virtual link diagram, 26 virtually concordant, 30 virtually slice, 30 walk, 56 welded knot, 32 white surface, 15 Wirtinger presentation, 10 writhe, 10 105
15206	https://www.wallstreetoasis.com/resources/skills/finance/exponential-growth	Published Time: 2021-12-13T09:51:52-05:00 Exponential Growth - Overview, How It Works, Compounding \| Wall Street Oasis Skip to main content Academy Academy Program Overview Team About us Testimonials View WSO Academy Courses Most popular program WSO Elite Modeling Program 6 Courses to Modeling Mastery WSO Elite Modeling Program View all courses Financial Modeling Courses Excel Modeling Course Foundation for All Finance Careers PowerPoint for Finance Course Pitchbook + Presentation Mastery Financial Statement Modeling 3 Statement Mastery Unlocked LBO Modeling Course Master Private Equity LBOs M&A Modeling Course M&A Mastery Unlocked Valuation Modeling Course Trading + Precedent Comps View all (+4) Interview Prep Courses IB Interview Course 7,548 Questions Across 469 IBs Private Equity Interview Course 9 LBO Modeling Tests + More View all (+2) Other Finance Courses Private Equity Deals Process PE Analyst + Associate Training Venture Capital Course Cap Table Modeling + Term Sheets View all (+2) Training Live Public Training Modeling & Valuation Bootcamp (2 Day) Excel, M&A, 3-stmt, DCF, Comps, LBO+ IB Interview Bootcamp (4 Hrs) IB Technical + Behavioral Questions Private Equity Interview Btcamp (1 day) Detailed PE LBO Tests + Case Venture Capital Bootcamp (4 Hrs) Cohorts, ARR, Cap Table Modeling+ View all (+2) Advanced Industry-Specific Private Equity Master Program PE Deals Process + Elite & Foundations Programs Venture Capital Master Program 11 Courses \| VC + Elite & Foundations Programs RE Master Program 11 Courses \| 50+ Hours Data Analytics Master Program 11 Courses \| 50+ Hours View all (+2) WSO Corporate Training Investment Banking Training Elite IB Pros \| Live + Online Private Equity Training Elite PE Pros \| Live + Online Venture Capital Training Elite VC Pros \| Live + Online Custom Training 1/2 Day to 10+ Days Available View All Upcoming Events (+2) Career WSO Academy High Finance Offer \| Guaranteed \| 12-week Program View WSO Academy Career Services Free WSO Application Tracker Apply and track 2k+ internships and FT roles WSO Resume Review Land More Interviews \| Detailed Bullet Edits \| Proven Process WSO Mock Interviews Land More Offers \| 1,000+ Mentors \| Global Team The Talent Oasis Find Elite Finance Talent for Less For Employers \| Flat Fee or Commission Available Access Elite Jobs First For Candidates \| Free to Apply WSO Finance Research Analyst Internship Build Your CV \| Earn Free Courses \| Join the WSO Team \| Remote/Flex Resources Excel Resources Excel Shortcuts PC & Mac Excel Self Study Best Excel Courses Excel Modeling Best Practices Index Match Advanced Excel Formulas View all (+334) Financial Modeling Resources What is Financial Modeling? 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Passion for sustainability. View Profile Last Updated:January 3, 2025 In This Article What Is Exponential Growth? Formula For Exponential Growth Applications Of Exponential Growth Exponential Growth Importance Advantages Of Exponential Growth Limitations Of Exponential Growth Uses Of Exponential Growth What Is Exponential Growth? What Is Exponential Growth? Formula For Exponential Growth Applications Of Exponential Growth Exponential Growth Importance Advantages Of Exponential Growth Limitations Of Exponential Growth Uses Of Exponential Growth What Is Exponential Growth? Exponential growth refers to a rapid increase in quantity over time, where the rate of growth is proportional to the current amount. In this type of growth, the quantity grows at an accelerating rate, creating a curve that steepens over time. Each interval, whether it's a fixed period or discrete steps, results in a larger increase than the previous one. Exponential growth is often observed in various natural phenomena, such as population growth, compound interest in finance, and the spread of diseases. It's characterized by its tendency to quickly amplify small changes, leading to significant and sometimes unforeseen outcomes. Growth in finance implies an increase in a company or a country’s earnings, revenue, GDP, or other monetary measures for a given period, like six months, one year, five years, or even ten years. Key Takeaways Exponential growth denotes a rapid increase in quantity over time, where the growth rate is proportional to the current amount, leading to an accelerating curve. Exponential growth finds applications in finance, population studies, and disease spread analysis and is mathematically represented by: V = S × (1 + R)T Exponential growth aids technological innovation, trend identification, and financial metric measurement. It's easy to calculate, enhances forecast accuracy, and emphasizes recent data observations. Despite its advantages, exponential growth forecasts may lag behind actual trends and struggle with seasonal variations. It finds utility in savings accounts, financial modeling, technology advancement, medical research, and environmental studies. Join the Waitlist High Finance Offer Guaranteed Be the first to know when applications open for WSO Academy's next cohort. WSO Academy's 12-week program has a 92% success rate First Name Email Use your personal email, not a .edu to ensure we can reach you. Target industry Select.. Leave this field blank Formula for Exponential Growth An exponential growth graph is an upward-sloping curve, and the equation for the curve can be depicted as follows: V = S × (1 + R)T Where V = Total Amount S = Principal R = Growth rate T = Number of time intervals Applications of Exponential Growth To understand the concept, let us consider a simple example of a $1,000 deposit with a return of 10%. If the account carries simple interest, the investment will yield an interest payment of $100 at the end of every year, and the amount of interest will only change if the principal amount is changed. If the account earns compound interest, the investor will earn compound interest on the cumulative account total, the sum of the principal amount, and the earned interest. Each year, interest will be added to the already earned interest on the principal. So, in this case, the amount after one year will be $1,100. However, the amount after two years will be $1,210 as the interest is calculated on the total amount of $1,100, which includes the initial principal and the interest earned in the first year. The compounding effects are not visible in the initial years but increase as time passes. This implies that as the investment matures, the rate of growth accelerates. After 40 years of depositing the initial $1,000, the investor will end up with $45,259.26, demonstrating the significant impact of compounding on investments. Exponential Growth Importance Some of the importance are: 1. Aids in the innovation of technology In industries like information technology and computer sciences, statistical information on population growth is crucial in practical applications. In such environments, data scientists may apply these methods to calculate output increases over a given period by using specific inputs when developing automated software, machine learning algorithms, and artificial intelligence processes. Note If a developer uses an algorithmic program as input to generate outputs over time, understanding exponential growth can help confirm whether the increase in information values follows an exponential function. 2. Helps to identify trends Exponential growth models are used to derive information about the growing trends of the population. They are used by professionals in different industries, such as health care, science and technology, and even business. For example, clinical researchers use exponential functions to chart the increases in positive health outcomes due to the introduction of new medicines. Business leaders might use such functions to identify changes in brand developments, customer segmentations, advertising costs, etc. 3. Helps to measure essential financial metrics Companies in the accounting and finance industry often depend heavily on growth functions to determine the expected returns from investment projects. One of the standard methods used is the compound return, a type of return that accrues on initial investment and growth in investment experiences. Note Compound returns play a significant role in investment accounts, leading to exponential increases in value with each increase in principal. This principle also applies to loans with adjustable interest rates that compound over time. Advantages of Exponential Growth The advantages of using the model can be explained as follows: 1. Easy to calculate and understand This growth model is easy to apply in real-life circumstances. Furthermore, the formula used for calculation is straightforward and understandable, even for individuals with limited knowledge of finance. Only three values are required to calculate the growth of an investment. These are a forecast for the most recent period, the actual value of the period, and the compounding factor. 2. Increases the accuracy of forecasts Analysts use such models to predict the results for future periods. Various trend projection techniques can be utilized to calculate forecasts for future periods. These calculations are accurate as they consider the difference between the actual projections and what happens. 3. Puts more significance on recent observations Observed data consists of multiple components, including random error, representing the difference between the observed and actual values. Using such models neglects such random variations, and thus, it is easier to grasp and understand the underlying phenomenon. Limitations Of Exponential Growth The cons of using such models for analysis are explained as follows: 1. Forecasts lagging behind actual trends One limitation is the forecast lag, as these models often overlook the fluctuations associated with random variable variations. As a result, the graph shows a smoother line or curve, but ignoring the random variables enables the analyst to see the underlying phenomenon, which helps present the data accurately and forecast future values. 2. Cannot factor the trends accurately Such models are best suited for forecasts that are short-term in nature and do not have any seasonal or cyclical variations. In such variations, exponential models do not yield accurate results. Exponential analysis methods are most accurate when there is a significant continuity between past and future data, making them suitable for short-term forecasts assuming that future patterns will resemble present patterns. Note Long-term analysis using these methods is not very accurate; however, specific different types of models can handle certain variations. 3. Limited use Growth models of physical phenomena can only be applied in limited cases, as exponential growth does not seem realistic in every case. Although investment growth may initially follow an exponential pattern, the model would eventually encounter previously disregarded factors that significantly affect predictions' accuracy. In such cases, other model assumptions, such as continuity or instantaneous feedback, will break down, rendering the model void and inaccurate. Uses Of Exponential Growth Exponential growth models are commonly utilized in high-interest savings account investments due to the stability of interest rates across economic cycles. Additionally, analysts employ these models for financial modeling purposes. Furthermore, analysts utilize exponential models to forecast future investment returns, particularly when the growth rate remains stable and consistent. Note In the fast-paced, technologically driven world, data is now being valued more by analysts and users of such data to draw inferences about different industry trends. Various firms across different industries utilize growth models to anticipate industry-specific trends and developments. Technology sectors leverage growth rates to drive software development and advancements in machine learning, while scientists and researchers utilize exponential growth rates and decay in various experiments. This data is helpful in the following ways: The exponential growth rate is beneficial for identifying trends and factors causing changes in financial metrics like compound interest rates, investments, revenues, expenses, assets, and liabilities. The data benefits researchers and economists as it helps them observe the effects of certain variables within the population, such as in clinical and medical trials and environmental studies. Exponential growth data also aids doctors and medical professionals diagnose conditions such as cancer, viral illnesses, and autoimmune deficiencies, providing valuable insights for treatment and research. ### Everything You Need To Build Your Accounting Skills To Help you Thrive in the Most Flexible Job in the World. 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15207	https://alteredstatesof.money/shell-money-as-ceremoney-in-png/	Altered States of Monetary Consciousness Economic anthropology 'Shell Money' isn't a commodity currency. It's cere-money What is 'shell money', and what we can learn from its resurgence in Papua New Guinea? Brett Scott • 21 min read In the early days of the pandemic media outlets were looking for uplifting novelty stories that would show people trying out new economic ideas. For example, there were a wave of stories about a US town 'printing its own wooden money'. In reality, this so-called 'wooden dollar' was really just a voucher system printed on thin pieces of plywood by the local government of Tenino in Washington. It was ‘scrip’, a legally-backed form of limited-purpose credit money. In such a system, the token material (and token production process) should take a conceptual back seat to the legal environment within which the token circulates. The Tenino voucher, like any voucher, could have been printed on paper, card, rubber, leather, ceramic tile or pretty much anything. Nevertheless, in my piece Wooden Promises for Digital Money, I showed how the media (and the public more generally) often fixates upon the visible appearance of money tokens rather than the system that activates them in the background. Our attention - and the visual imagery of money - is often focused on tokens as if they were some kind of autonomous force in themselves. Journalists were arriving in Tenino to snap shots of the voucher's wood, distracted by the somewhat irrelevant ‘commodity’ body. This is a very common practice. There is a long tradition of people believing that the body of the token is what matters most. This is partly due to the simple fact that physical tokens are the most visible element of any money system, and hence hog attention, but the fixation goes deeper: many people are pushed into this style of thinking by the difficulty they have in seeing the interdependencies within large-scale economic systems. In a vast capitalist economy, it is easier to imagine yourself as an autonomous individual using some commodity-like 'substance of value' to induce others to trade with you, than it is to see yourself as but one node enmeshed within an interdependent network structure held together by a monetary system that transcends you (for an introduction to this topic, see my piece Money Through the Eyes of Mowgli). This tendency we have towards fixating on the token body gets politicised in the hands of so-called 'sound money' promoters, who insist that any money token not made from some limited supply valuable material is some kind of fraud (for example, so-called goldbugs have this mentality, but so do many Bitcoiners, despite the fact that Bitcoin tokens have no body at all). Needless to say, there is a generalised tendency in society to have a commodity orientation to money, in which units of money, regardless of what type, are seen as some kind of substance (or a 'fictitious substance' that emulates a real substance). Even when people know that the money is not a commodity, they still often superimpose a mental model of 'commodity money' over it, seeking out its commodity elements, and using commodity-like imagery and language to represent it. So what is that mental model? We can identify three main elements, all of which can be contrasted to the competing paradigm of credit money, in which the body of the token is not particularly relevant. Commodity money is imagined to be ‘money from something’ (it is produced through labour), while credit money is ‘money from nothing’ (it takes the form of redeemable accounting records of promises, which can be printed on anything) Commodity money only increases (once it is made it is out there for good, until it disintegrates or gets lost), while credit money expands and contracts (it gets issued and redeemed) Commodity money is ‘one-sided’ (it is imagined to be a self-contained object in the world), while credit money is ‘two-sided’ (we might hold it, but it only has power insofar as there is an issuer on the other side) Want to receive these articles to you inbox? Sign up to my Substack here... One of the biggest areas of ambiguity in commodity money thinking, is whether or not the supposed 'value' of the commodity money is self-apparent or not. For example, goldbugs often assume that gold is 'obviously' valuable to everyone, despite the fact that it historically had very few immediate or practical uses for everyday people. Gold often features in the 'barter to binary code' narrative, the bog-standard story about the 'history of money' trotted out by standard economists, in which they insist that money emerged out of barter, and initially took the form of simple commodities like shells, before moving on to metal, paper, and finally, digital 'value'. It is a super dodgy story, but even in its own terms it is ambiguous: their shell story is particular fraught: Can't you just find shells by walking along the beach? Don't they break and fade? Why would those become 'commodity money'? This sense of incredulity that you can just wonder around a beach and ‘find money’ by picking up shells continues to be a source of fascination to this day. On the 21st August 2020, the Guardian ran a piece called “The return of shell money: PNG revives old ways after Covid's blow to economy”. Before I even began reading it I predicted that the story would fixate upon the shells, and ignore the systems that ‘activate’ their power in a particular context. The article was decent enough, but - as predicted - offered no account of what exactly 'shell money' is, and how it works. It carried that background hint of the 'barter to binary code' story: somehow, once upon a time, shells just were a crude form of money, and, in places like Papua New Guinea (PNG), some people still hold onto that memory, and - despite the 'upgrade' to a more advanced monetary system that surrounds them - they were now 'going back' to shells. Almost all goofy or popular references to shell money make the assumption that it is in the same class of phenomena as modern fiat money. They essentially assume that it facilitated a crude form of early capitalism. To grapple with shell money, however, requires dropping this assumption. The shells aren’t the money, and the ‘money’ doesn’t do what you think it does. I made a video about this. Ceremonial ‘commodity money’ (or ‘cere-money’) ‘Shell money’ tokens - also called Tabu by the Tolai people who use them in PNG- are not just shells. They are shells carefully and arduously manufactured into long strings of shells, a process governed by very strict production rules to produce set denominations (see this video for footage of the production). They are far more akin to jewellery than they are to some raw material. At first glance, this places ‘shell money’ into that broad paradigm of ‘commodity money’ - money that's apparently given its power by the physical body of its token. In this case, the body of the token is a kind of ornament (the token is only a token when the shells are manufactured into strings), and - unlike a voucher, which is retired when it is returned to its issuer - they remain in circulation ‘for all time’ (perpetually). They are somewhat like artistic creations that are slowly produced, and then left to circulate amongst collectors. By all appearances, then, Tabu are a ‘commodity money’, but here’s where the language politics kick in. In the imagination of someone who has been immersed in a large-scale capitalist economy, ‘money’ has a particular connotation. It is imagined to be general-purpose: money tokens give you access to all the other things, and are primarily transferred in private settings to obtain a huge range of goods and services. Furthermore, there are historically various taboos on what it can be spent on. For example, if you insulted your mother-in-law, handing her $20 in compensation is not going to heal that social wound. It will probably make it a lot worse. Tabu ‘shell money’ (and many similar things like Wampum) on the other hand, is - historically at least - limited purpose, and historically was used publically at elaborate events and rituals to ‘pay’ for existential things that normal money simply cannot buy. This includes ‘paying’ for ‘bride-price’ at weddings, compensating for injury and insult, commissioning cult members to raise spirits, paying tribute to invisible forces, and hosting elaborate memorial ceremonies to display power and status. Not only is the production of Tabu ritualistic, but the ‘exchange’ is too. There are set ‘prices’ (not truly set by supply and demand) for specific things - for example, a bride ‘costs’ 4 Mars, which are bound wheels of many strings of shells. The exchanges are more like performances, with almost comical affectations, with people feigning outrage and making jokes as if to reject the offer (see 2:14 below). Fetishistic tokens activated by cultural forcefields In the traditional ‘commodity money’ imagination found within capitalist societies, it is imagined that the token value is ‘self-apparent’ due to the material the token is made from (or - failing that - from some almost mystical property generated by its ‘scarcity’). But, the power of these shell strings is not self-apparent to any random stranger. Indeed, colonial officials used to see these as ‘primitive’ money, as if the tribal people were like silly children imagining that shells were valuable. But this inability to recognise the power of the shell pendants is because they only have power within a particular cultural field. An Australian officer saw a string of shells as a mere tricket, but within the Tolai cultural field these objects have deep political power, and are even seen as a political agents, or ‘power tokens’. When I use the term ‘cultural field’, I mean it literally, like a forcefield. We are all social creatures, and all of us frequently experience that feeling in social settings where you sense that you are part of a collective that extends beyond yourself. Indeed, your individual perceptions and judgements seem connected to some looming ‘entity’. Imagine a simple line from a novel, like ‘a heavy silence descended on the room as he ate before the others. Looking up, he sensed he’d done something wrong’. That ‘sense’ is generated by the social forcefield. (As an aside, in occult circles this is referred to this as an ‘egregore’, a looming feeling generated by the interconnections between individuals in a group that can be personified into a being.) These ‘fields’ exist around us all the time - because, as discussed in my tribute to David Graeber, we are never just individuals. We are normally enmeshed in many of these fields - some large, and some small. They can even be generated between just two people (many couples have private languages and symbols they only use around each other). Crucially, these fields have the power to ‘activate’ objects into more than just objects. Imagine, for example, a couple picked up a piece of driftwood on a beach where they fell in love. Now they keep it as a sentimental token imbued with deep meaning. Imagine then that a friend, house-sitting their house while they are away, decides to burn it to make a barbeque. Upon returning they are outraged. The friends says ‘come on, it was just some wood, I’ll get you more from the forest’. This obviously doesn’t cut it. It wasn’t ‘just a piece of wood’. Within their private social field, it was a material embodiment of their relationship. The friend, who is not part of that field, can’t see that. To him, it is just wood. Now imagine the social field generated by a much larger collective of people. It’s in this setting that the anthropological concept of ‘fetish objects’ emerges. Webs of human relationships are often tense and amorphous - we barely even understand ourselves, never mind the complex and subtle power dynamics with others, so it’s very common for people to take these vague semi-visible dynamics and focus them into objects that are concrete and easy to see. This then enables you to manipulate the objects as a way to concretely and visibly negotiate the invisible relationships. We have fetish objects all around us, and some of them span public and private settings. Consider, for example, a wedding ring. In reality it’s just a piece of metal, but within certain cultural fields it is publically recognised that it’s not just a piece of metal. It is - supposedly - a physical embodiment of a relationship between two people - complex and changing - manifested in an object, simple and unchanging. This public understanding permeates into a private one. Upon a divorce, the bearer might go to the ocean and ritualistically hurl the object into the sea, metaphorically giving up the relationship. Want to receive these articles to you inbox? Sign up to my Substack here... With this in mind, we can begin to look deeper at Tabu. It is very risky to see them as being ‘basically the same’ as normal money tokens (or, even worse, ‘a crude version of our money’). Rather, they are fetishistic power tokens used for fundamental rituals, and are activated by a cultural field within which they are used to negotiate considerable tensions. A cultural field is not a legal system - and many of our modern money tokens are activated by legal systems - but these fields are powerful enough to make it unthinkable to treat these shell pendants as ‘mere commercial money’. I will go into the ambiguities of the use of Tabu - because they have seeped into commercial settings -but the biggest amounts of tabu are used for all sorts of things that ‘money cannot buy’. For example, in this video they are used for the reconciliation of Japan and East New Britain (the region of PNG that was occupied by the Japanese). At point 5:20 in the video, the announcer says 'Japan, we forgive you' as Japanese delegates hand shell strings out in a gesture of ‘repaying’ some kind of existential debt that doesn’t really have a price. The complex history of Japanese invaders is being ‘resolved’ as if it were a quantifiable exchange of some sort, when everyone knows that it is not. Similarly, at Tolai marriages, the groom’s family is not literally 'buying’ a bride on a market’. Rather, they are ritualistically acknowledging the loss of labour the wedding brings to her family by offering a performative ‘exchange’. An alternative would have been for the family to, for example, collectively perform 10,000 pushups while the community watched. The limits of a normal commercial voucher are defined by its issuer, who guarantees to give you particular things if you present the voucher to them. Tabu ‘cere-money’ is also limited purpose, but these limits are defined by the cultural field, which will only ensure their acceptability for very specific spiritual and political services if their production process follows certain ritual procedures. Part of the fetishisation process is dependent upon the non-utilitarian labour involved in producing the cere-money tokens. Rather than spending a day weaving nets for fishing, a person exerts intense concentration stringing shells together. These shell strings will come to be seen as being brought to life, and thereby ‘containing’, some kind of ritualistic exertion, and this act is required before the shell tokens can be ‘activated’. Put simply, if a company mass produced shells and automated the process of stringing them, the cultural field would not recognise those as valid. Now, some readers may be thinking ‘Isn’t this a little bit like Bitcoin?’ After all, in Bitcoin, digital objects - tokens - are ‘produced’ through a process that requires large amounts of ‘pointless’ energy to be expended. Bitcoin, however, is explicitly intended to create a type of capitalist commodity money for ordinary market exchange, and the 'labour' is completely impersonal, abstract and automated. By contrast, Tabu requires highly personal, non-automated labour to create a limited-purpose power token for negotiating human relatioships. It’s not like any old person can find some shells, attach them to string, and then buy Coca Cola. The grey zones of cere-money Cultural fields, though, are not without internal tensions or contradictions. One of the most detailed accounts of Tabu I’ve read comes from the anthropologist Richard Frank Salisbury, who worked in PNG in the 50s and 60s. Salisbury explicitly explores the ambiguous nature of the tokens, and how there is a zone of ‘convertibility’ from sacred power token to non-sacred commercial object for everyday commerce. At the time he was observing, a small percentage of the Tabu was in fact being used as an everyday ‘commodity currency’, with standard prices for goods-to-Tabu. The largest amounts, however, were being used for ‘buying’ or commissioning dances, songs, spirit-raisings, and memorial ceremonies. The latter in particular are characterised by extensive competitive gifting, with political upstarts throwing Tabu around as a kind of ‘dick-swinging’ exercise. Salisbury is more ‘economisty’ than some anthropologists: he suggests there are ‘rate-of-return’ type calculations going on in these ceremonies, in which powerful ‘big men’ spend a lot and get a lot, acting like conduits via which people pool resources. Furthermore, he describes 'companies' formed by pooling shell-tokens, along with a coordinator that manages them, and who can end up capturing the surplus. He is, however, a lot less ‘economisty’ than most economists, painting a picture of a strange and complex mix of 'economic' and 'non-economic' logics, with Tabu-as-Political-Power-Token seaguing into Tabu-as-Economic-Commerce-Token. The blending of logics creates grey zones: for example, only a small fraction circulates for day-to-day commerce, with most of it being held in large bundles that - by convention - can only be broken open at non-commercial events. Furthermore, when people die, their unbroken tabu wheels are broken open and distributed equally to every clan member who comes to the funeral, forming a natural redistribution mechanism. As much as I love Anthropology, however, it also has a dark history of being associated with colonial power. A modern anthropologist would explicitly try to parse out the degree to which cultural fields are warped by colonial forces, whereas older anthropologists had a tendency to speak of cultures as unchanging. The key question to ask when reading Salisbury's account, is whether the increased commercial logics in Tabu had been induced by the presence of colonial capitalist systems, or whether they were ‘always there’. Colonialism, syncretic money and token possession Cultural fields are not static, and this is especially the case when they comes into contact with powerful external networks. Think of cultural fields as being somewhat like magnetic fields. If there is just one, you get drawn to its centre, but if a more powerful external one turns up, it can dilute and co-opt the weaker one, especially if it’s an imperial military force. This happened extensively during colonial times, where powerful nation states walked into much smaller and more fragmented societies, and set about dismantling or dissolving their internal structures in order to absorb them into a larger network in a subordinated position. Think about it as two cultural fields - or networks - colliding. The more powerful one can warp, or even break, the internal coherence of the weaker one. At the very least, what often ends up happening is the creation of interstitial zones of overlap - or syncretism. Syncretic religions are extremely common in the aftermath of colonialism, creating situations in which, for example, animist beliefs get practiced alongside Christianity. This is explicitly the case with Papua New Guinea, where local Christian pastors participate in ceremonies where Tabu is used to invoke cult spirits that are definitely not found in the Bible. PNG has a long history of colonialism, starting in 1828 when the Dutch claimed it, followed by the Germans, English and Australians, until it finally got independence in 1975. This man lays it out pretty succinctly. The arrival of anthropologists coincides with the colonial period, and this is where all our original accounts of shell money come from. But this means the anthropological literature - which has had an ongoing debate about the use of shell tokens for ‘high’ ritualistic purposes versus ‘low’ everyday commerce - cannot be separated from the political context which was creating extensive changes in PNG life. For example, at the time Salisbury was writing, colonialism had been in full swing for over a century, and the local PNG economy had changed to a ‘syncretic’ blend of traditional subsistence farming and cash cropping. There was extensive syncretic religion, and, a syncretic dual token system, with the traditional cere-money shell tokens existing alongside a much newer national money (which at that time was the Australian dollar) - used to access foreign manufactured goods. Also, there was now an official fixed ‘exchange rate’ between the Tabu and Australian currency. The economics discipline has left us with a legacy of depolitising money, describing it through the flat and vague ‘functions of money’ paradigm, in which money is also assumed to simply spontaneously emerge out of ‘natural’ human exchange and barter. The reality of capitalist money - as colonised people have experienced - is that it is anything but spontaneous. It arrives with imperialistic nation states and gets forced into your society. Indeed, the introduction of capitalist money is a core feature of all colonisation processes. In my country of South Africa, demanding taxation in state money was one way to ‘proletarianize’ the population, forcing them into markets to find the token required to meet the taxation obligation (this observation lies at the core of even progressive monetary movements like MMT today). General purpose money in the capitalist sense thrives when people have been thrown off their land and are forced into wage labour, but many pre-capitalist communities historically had basic subsistence. They were not - initially at least - precarious wage labourers trying to use their meagre general-purpose credits from the coal mine to buy tea from merchants importing from distant lands. This is why colonisation processes either have to pull at the cultural field from the outside, or - alternatively - seek to corrupt it from within. The first thing any good colonist does is look through their own tinted cultural lens for things that vaguely approximate behaviours they are used to. In this context, ritualistic fetish-like exchange which involves passing cere-money tokens around gets seens as ‘primitive monetary exchange’. If you are an official from a colonial nation, this offers you a route to co-opt the cere-money token like a trojan horse. It’s already embedded in the cultural field, so you start treating it as capitalist money - allowing it to be used to be handed in as tax, for example - as a way to pull on the threads of the society from the inside. This, for example, is exactly what happened with Wampum beads in North America. This is done in conjunction with the creation of official ‘bonding’ points between the two discrete token networks. Think of this as where two membranes meet. The official ‘exchange rate’ between Tabu and the colonial currency is an attempt to create a bridge between the two networks, and this opens a portal from the outside world of the colonial power into the inner one of the colonised. From here, the internal logics of the token get all disorientated, and a limited purpose form of cere-money can be parasited upon - or ‘possessed’ by - a militaristic all-purpose money. This is used as a mechanism via which those systems creep into previously clan-based societies to dissolve and rearrange economic relationships. For example, Salisbury arrived at a time when ‘cash cropping’ was present. This is not organic. ‘Cash cropping’ is a euphamism that accompanies colonialism, referring to small-scale farmers who will sell for a pittance to colonial traders who will resell the goods for major profits on international markets. It is how you turn an autonomous island nation into a subservient commodity-producing vassal of a much broader international system. And this is what the normal money system in PNG was originally for. The Bank of Papua New Guinea was originally part of the Reserve Bank of Australia, and is nowadays connected into vast international money networks. Modern Bonding We’re getting closer to understanding the Guardian article, but we need a few more pieces of the puzzle. In recent decades, as transnational globalisation has kicked in, the inter-network boundary between Tabu and ordinary money - already warped and blurred - has got even more grey, as new forms of inter-network ‘bonding’ emerge. For example, Tabu are accepted for modern taxation (see here for numerous examples), and other official payments associated with the state, such as school fees and local fines. A second form of bonding emerges from the fact that the raw material for the tokens - the shells - increasingly are bought from the Solomon Islands, which means the raw material for the tokens gets a money price. A third form of bonding is through tourist markets. For example, the World Bank has been encouraging women to treat shell money as tourist objects to be sold for normal money. This creates an explicit external market for the tokens (somewhat similar to the collectors market for novely vouchers like the Tenino Wooden Dollar). More than pulling on the internal cultural field, this literally pulls the tokens out of that field and into the vast realm of the global capitalist market, appearing as an image on Amazon to be sold in return for general purpose money, which in turn will be used to buy imported tinned goods produced from afar (and which promote diabetes and other issues, which in turn requires money to get medical help). In this context, Tabu begins to live an explicit double life, one as a kind of subservient object sold for normal money, and another as a proud token embedded in a community trying to hold onto some sense of unique place in the world. In other words, the token has much the same status as colonised people do. By day it does dances for tourists in the capitalist market, but on the weekends it has a place just for itself, with it’s own language and zone of power, as it were. And this small zone of power is protected. The Tolai people try to only allow the membrane between their token network and the outside world to work one way: while there are increasingly stories of people ‘selling their Tabu for money’ to buy stuff at shops, ordinary capitalist money is not used in the ritualistic settings that Tabu is used for. The spirits will not rise when dollars are thrown at them. Thus, despite the blending, and despite the cognitive dissonance that accompanies the blending, in their essence the two systems remain separate, conceptually at least. The limited purpose cere-money tokens activated by a cultural field are not the same as the general purpose fiat tokens that operate within a government-bank legal field. The present day story: Counter-trading cere-money So, finally we arrive to the Guardian article about the ‘resurgence’ of shell money in PNG during Covid-19. With our background knowledge, the story becomes clearer. Over the years the ‘syncretic’ dual money system has evolved, with limited-purpose cere-money eking out a life in the shadow of normal general-purpose money. But, as a result of Covid-19 transport disruptions, there is now a breakdown in the distribution system of national physical fiat money tokens (cash). Central banks and banks run an elaborate system for getting cash to ATMs and retailers and back again, and when that system breaks down there can be cash shortages in particular areas. This means that proletarianized, formerly colonised people that are partially integrated into global capitalist markets suddenly have reduced access to the general-purpose fiat money tokens that connect them into those markets. This means that locally the cere-money token - which has been increasingly taking on a mentality of a dollar-priced commodity - is stepping in to take on a ‘ordinary money’ role even more. It is however, more complex than this. The article notes that people initially turned to subsistence: Personal gardens were quickly depleted quickly, prompting residents to “go walkabout” within their village boundaries in search of a friendly exchange. The article also notes that 'barter was revived'. As soon as you see lines like this, know that you are in dubious territory. In de-politicised accounts of money, barter is always assumed to be the origin of money (hence the term ‘revived’), and, furthermore, is assumed to have operated much like capitalist exchange does. What is always missed out is that historically barter only really occurred on the boundaries between groups, in situations where trust was low. Subsistence mixed with reciprocity is a much more important mode of distribution internal to groups, where trust and interdependence is higher. Nevertheless, barter becomes far more likely after money is introduced, because money creates the distance between people that in turn induces barter in situations where the money is suddenly removed. To put this another way, capitalist money is a disassociative, or dissolving agent, which loosens the tight communal bonds that hold pre-capitalist societies together. This is in contrast to the original experience of non-capitalist cere-money, which is a bonding agent. Thus, it is only when communal bonds are loosened, and people have become accustomed to normal monetary exchange, that ‘barter’ temporarily emerges to fill the vacuum when the normal money system breaks down. It’s in this context that the Guardian article then talks about the revival of Tabu, but implies that it is object with a commodity-like 'value'. This places the article in broad alignment with ‘commodity money’ thinking, in which monetary exchange is basically assumed to be an elaborate form of barter. "Vanessa Mulas, a resident of Kuradui village, says those who were able to circumvent the cancellation of public buses brought back store goods and exchanged them with neighbours for tabu." What the article doesn’t talk about, though, is all the dynamics I went through earlier, and in particular the fact that the over the years Tabu have been ‘colonised’ by a much stronger monetary network, such that the tokens are now commoditised as an object sold for money. This means that what is actually occuring here is a form of countertrade. Yes, countertrade. This is a major concept I will keep returning to in this series. Countertrade is the situation in which a ‘barter-like’ transaction appears to be occuring, when in reality what is actually occuring is two superimposed fiat money transactions (incidentally, this is what Alfred Mitchell-Innes used to blow up some of Adam Smith’s bogus accounts of ‘commodity money’). In the global capitalist market, there is - nowadays - a fiat currency price for Tabu, and a fiat currency price for, say, rice. You then compare those two prices to create an exchange ratio for Tabu-to-rice. (As an aside, fiat countertrade is also how Bitcoin pricing works). From my perspective, what is most likely going on here is fiat currency countertrade, with people using the body of an old cere-money token that has been ‘bonded’ to fiat currency to countertrade for goods. The biggest subtlety we have to deal with, however, is to reconcile the fact that - according to people like Salisbury - there already was some internal commercial logic within the cere-money tokens in the pre-globalisation world, which in time must have been warped by the post-globalisation world. The soul of the token is in flux, like a tidal estuary that has one coherent but diverse upriver life that then mixes with a chaotic ocean that comes in and out. Sometimes it is itself. Sometimes it is inhabited by an external system. Despite this ambiguity, the Tabu cere-money/money/countertrade-object is undoubtedly local, and its use - like the Tenino Wooden Dollar - does signal a more general return to localism. This is a resurging theme in the age of Covid. In the case of PNG, the country is historically very fragmented, with colonial powers, and then development agencies like the World Bank trying to integrate and homogenise the people into bigger global economic networks. Big currency systems induce that, but now that integration is proving risky, creating a new impetus to create smaller networks that split off from bigger ones in order to localise. Insofar as they occur, returns to localisation, reciprocity, communal work and subsistence are a reversal of the dissolution, expansion and specialisation impulse that accompany mainstream money and the mass goods that accompany it. Ending the ideological abuse of cere-money One final theme I will bring up is the symbolic abuse of cere-money by entrepreneurs and economists who wish to assert that it operates by the same principles as any money. Many dubious 'history of money' accounts start by talking about various cere-money forms and then placing them in a ‘chain of progress’ leading up to modern money, as if they were basically the same thing. For example, I’ve seen the Bank of England speak about state central bank money alongside images of cowrie shells in their ‘What is money’ page. I’ve also seen Bitcoin entrepreneurs try to argue that the Bitcoin system is akin to the Rai stones of the Island of Yap. I wish they would back off, because you cannot understand things like cowrie shell money and Rai stones until you reverse - temporarily at least - the Pandora’s box of capitalist money thinking that has been opened in our minds. This returns us finally to David Graeber. Our conventional accounts of money often disguise deep politics. The desire to present shell tokens as being ‘basically the same’ as modern money, and operating on the same principles, is nothing else but a subliminal attempt to deny the vast power of a historically unique system of nation states, private banks, and vast depersonalised markets within which exploitation can be hidden very effectively, and explained away with pseudo-scientific models. In closing, while the Tabu might nowadays have a day job being used to buy tinned sardines, they moonlight as a portal into a pre-capitalist spirit world that very few of us will ever see. Want to receive these articles to you inbox? Sign up to my Substack here...
15208	https://www.matematica.pt/en/faq/calculate-lcm.php	How do you calculate the L.C.M? Languages Portuguese English Home Contact Toggle navigation Worksheet Grade 4 Grade 5 Grade 6 Grade 7 Cheatsheet Platonic Solid Triangle Classification Polygon Classification Important Points of Triangles Function Transformations Statistical Graphs Angle Names Angle Relationships Angles in Circles Types of Matrices Line and Plane Relationships Calculator Scientific Calculator Percentage Calculator BMI Calculator VAT Calculator Quadratic Equation Calculator Convert Units Useful FAQ Math Formulas Math Symbols Kangaroo Questions TED Conferences Famous Curves Free Software Study Tips ASCII Table Geometry Angles Complementary Angles Supplementary Angles Inscribed Angle Inscribed Angle on Diameter Points and Lines Midpoint of a Line Segment Slope of a Line Slope Formula Distance Between Points Functions Function Transformations Isometries: Reflection Isometries: Rotation Quadratic Functions Trigonometry Radian Measure Unit Circle Graph Sine Function Graph Cosine Function Graph Tangent Function Equations System of Linear Equation Circle Equation Vectors Vector Addition Vector Equation of a Line Statistics Box Plot Find Mean Find Median Find Mode Complex Complex Numbers Fun Sudoku Tic-Tac-Toe Magic Symbols Tower of Hanoi Numbers Curiosities Einstein's Riddle Palindrome Jokes Search Discover more math Mathematics mathematics MATEMATICA How do you calculate the L.C.M? short answers for big questions T here are several processes that allow us to calculate the L.C.M (Lowest Common Multiple) between two or more numbers. We could start by creating two lists containing all the multiples of both numbers and we would stop when we found the common number between them. But this is not a practical process! The best method is to start by decomposing each one of the numbers according to a product of prime numbers. This process is called to factorize and it is very practical, because it can be used not only to calculate the G.C.D but also the L.C.M. Let´s suppose we want to calculate the L.C.M between 168 and 180. After having decomposed the numbers in prime factors, we can notice that and . The following step is to calculate the product between all the common factors having the biggest exponent as well as the one of non-common factors. In this case we would have . Thus, we arrive to the conclusion that the Lowest Common Multiple between 168 and 180 is 2520. What is L.C.M useful for? There are several examples of real problems where the calculation of the L.C.M becomes useful. Let’s imagine that a circular running track in which two runners are competing and they run round the track several times. Now, let’s suppose that it takes one of them 168 seconds to complete a full lap of the track while it takes 180 to the other. We know they leave at the same time: when will they join together at the start? In this example, as the L.M.C of these two numbers is 2520, it allows us to say that they will join together at the start in 2520 seconds (42 minutes). But, do not forget that the faster one has already completed more laps. But, after all, what is the L.C.M of two numbers? We call your attention to the fact that the L.C.M is not only calculated between two numbers. We can calculate the L.C.M of 2, 3, 4 or more numbers! Having this in mind, if we have two natural numbers and we start writing multiples of each one of them, we can assure you that, at a certain point, we will reach a common multiple (the worst that can happens is that we can always multiply one number by the other). The first common multiple we will find is called the L.C.M. C heck out our List of Questions to get to know a little more about the most diverse topics related to mathematics. If you have any pertinent (math) question whose answer can not easily be found, send us an email on the Contact page with the question. We will be happy to respond. In the event that you detect any errors in our answers, do not hesitate to contact us! Discover more MATEMATICA mathematics Mathematics math Discover more math mathematics Mathematics MATEMATICA MATEMATICA.PT © 2025 - Vitor Nunes Se eliminar a causa, o efeito cessa. × Receber Novidades Subscreve a nossa newsletter para estar informado sobre as mais recentes novidades! Subscrever Fechar
15209	https://artofproblemsolving.com/community/c864h980706?srsltid=AfmBOorRQv4r07OeDal6HbNu9GixBdPKBKlx1aWdYlY3mJPAzrc7VxGv	the Search for Intelligence Continues. . . : Calculus Community » Blogs » the Search for Intelligence Continues. . . » Calculus Sign In • Join AoPS • Blog Info the Search for Intelligence Continues. . . ========================================== Calculus by rrusczyk, Jul 25, 2008, 3:22 PM As a lot of you know, we're working on a calculus book here at AoPS. We're also considering offering a calculus class this fall. So, a question: if you were to take such a class, why would you be taking it? Just for the AP test? To get a more rigorous look at calculus? For fun? AoPS 19 Comments (Post your comment) Comment 19 Comments The post below has been deleted. Click to close. This post has been deleted. Click here to see post. My daughter will be taking calculus in the 2009 school year. Since we homeschool, she will want an online class. If you offer one, she will be there! by lfm, Jul 25, 2008, 4:23 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Well, if I ever take it in the future, it will most likely be a combination of all 3 reasons. by Nerd_of_the_Ages, Jul 25, 2008, 4:40 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I am learning basic calculus and i am doing it do learn physics but mainly because its just extremely fun by Poincare, Jul 25, 2008, 5:56 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I don't care too much about the AP tests ( honestly, I've already taken one AP test and there not really all that great). My main reason for taking the class would be so that I would be able to do Putnam calculus problems, and so that I could finally see the proofs of certain theorems that require the use of calculus (e.g. Fundamental Theorem of Algebra, among others). This would be a great class. by mihail911, Jul 25, 2008, 7:15 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. ! I am pretty surprised by this, given your article on the subject. (I think it was yours -- might have been MCrawford's.) I guess I never followed the advice of that particular article though; this is a great idea. by MysticTerminator, Jul 25, 2008, 7:34 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I know Marcus (gauss202) is a strong supporter of incorporating physics into such an AoPS book/class. Personally, I agree that, if done well, this could make for a really awesome, interesting, useful class/book. It could also help you attract physics olympiad people (admittedly though, those people overlap with math olympiad people quite a bit already). With the AP prep route, I'm certain you can do that well but I guess the question is: do you really want to? There are already a ton of books and classes out there and the people you've attracted to AoPS thus far usually have very little trouble acing their AP's. by joml88, Jul 25, 2008, 8:25 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. In my next life, I'll write a book that integrates teaching physics and calculus with a problem solving approach. I've always been enamored of the idea, even though marketing such a beast would be nearly impossible (and I'd have to re-learn a lot of physics and get more experience teaching both calculus and physics). Mystic: my article, the main point of which was intended to be simply that calculus shouldn't be the goal of middle and high school education, and students who make it such are setting themselves up for a hard time. I've seen way too many 7th and 8th graders taking calculus, but not able to pass the AMCs. Those kids are almost always making a mistake. I'm guessing you were well beyond 110-120 on the AMC12 when you started in on calculus, even if it was in the 6th grade. You're the exception -- most of the middle school calculus takers I see are ones who memorized their way through the earlier classes. They memorize their way through calculus, too, because, well, most calculus classes are pretty simple. And then they think they are smart. Very smart. But they don't know anything yet, and when they hit a challenge, which everyone eventually does, they shut down -- math was always easy before, because they are smart (not because they work hard, etc). . . Then the challenge comes and they deduce they're not smart anymore, so it's time to stop. And the trap slams shut. by rrusczyk, Jul 25, 2008, 11:09 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. It seems like there are two groups this class could be aimed at: People who have not yet taken calculus in high school and want to get a head start, or people who have already taken AP Calculus and want to see it more in depth, and perhaps applied to contest problem-solving. I would prefer the latter, but I think the former would attract more people. Either way, I'm assuming the material covered would be single variable calculus only, similar to an AP Calculus BC class...? by mysmartmouth, Jul 26, 2008, 1:56 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I am also pretty surprised and delighted about this. I'm going to echo some sentiments, and say yeah, don't do something that is geared toward the AP exam exactly. I would personally love a more rigorous treatment of Calculus, because there is much left to be desired in all of the calculus courses that I have taken. Would this be a class that covers presumably through a standard Calc II course (i.e. BC Calculus?) or is it merely a Calc I class? I'd be interested in seeing either a putnam class, or like Calc III/Diffy Q/LinearAlgebra/Vector Calc related things. I'm sure there would be an okay market for a Putnam class, perhaps not for the latter. by Pakman2012, Jul 26, 2008, 2:19 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I'm curious about what you're envisioning by combining physics and calculus. I've seen the idea thrown about a bit and don't understand what's so entrancing about it. Certainly in my calculus texts, there was a fair bit of physics, like balls being thrown under the force of gravity and whatnot, so when I did take physics, I was fairly well-prepared from that, but is there some deeper connection you're looking for here? By the way, Joml, regarding physics books in general, there's so far only one book that I would recommend as enthusiastically as AoPS (well, vol 2 at least -- haven't had enough experience with vol 1 and these new ones), which would be Dave Morin's text on mechanics, so if you'd like to see a physics text with the same level of problem-solving emphasis, I'd recommend checking it out. by MysticTerminator, Jul 26, 2008, 6:05 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Isn't Div, Grad, Curl and all that about multivariable calculus and physics at the same time? Anyway, I think you should include nice connections to other areas of mathematics (the proof that diverges springs to mind), by Boy Soprano II, Jul 26, 2008, 12:11 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. yeah, and I wasn't overly impressed. the physics is nice to have around for examples, but it's not like it's at all central to the theory. by MysticTerminator, Jul 26, 2008, 1:20 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I expect we'd cover, at a minimum, through "Calc BC". I don't really understand "Calc AB" -- that stuff should only take at most a dozen weeks for decent students. I would also expect a much more rigorous treatment than you get in your typical calculus class. Part of the point of offering this class would be try out some of these more rigorous areas and more Putnam-esque problem solving problems before writing our calculus book. Physics and calculus: A great deal of calculus was essentially developed for physics. This is why integrating the two is appealing to me -- why teach these ideas twice; they were developed as one idea, more or less. If they can be understood this way, I could imagine that it would be much easier to learn. But again, I'm more or less thinking out loud about this, and there are plenty of areas of physics that came well after the mathematics on which they rely, and integrating these into the curriculum would be trickier. Ultimately, it would feel pretty synthetic and contrived in those places, which might offset the niceness of the others. (I have another similar "in another lifetime, I'll do it" idea -- teach history backwards. Forget the stone ages. Start from today and work backwards. I could imagine this approach being very successful with kids who ordinarily find history boring and pointless.) Putnam: I'm not as convinced there's a market for a Putnam class -- I don't think there are that many students that care so much about the Putnam, and many of the ones who do are clustered at schools that certainly don't need a Putnam class. That said, I'm probably unreasonably extrapolating from my own experience -- I never took the Putnam. by rrusczyk, Jul 26, 2008, 4:09 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I agree physics and calculus are essentially the same in many places, and I was bored seeing the material twice; in general, however, I don't think it's inherently bad to see some material over again. The connections between the relevant concepts in elementary physics and those in calculus and vector operations and other branches of math are usually fairly obvious and spelled out completely so students really shouldn't have any difficulty. In fact, more than that -- the students are probably beaten over the head with it at every turn. I will agree with your thinking completely, however, in more advanced physics -- probably my most deep-seated concern regarding my study of physics is that I don't really understand what's going on because too often physics texts assume their readers don't know the math involved so they simply obfuscate the mathematical details. (for example tensors in the two subjects, spinors, clifford algebras, and representation theory in particle, functional analysis in qmech and many many other things) So, I do spend a lot of time looking for good sources of mathematical physics that actually explains how the two threads intertwine and I am always grateful for such a source at higher levels. At just a basic entry level though, I gotta say I don't see the point. Putnam -- again, I think the first batch of kids y'all started reaching in a big way are just now starting to enter college so at this point, there's probably not a huge demand. I'd bet that in four or five years, though, all the middle-schoolers nowadays for whom AoPS really is their comprehensive universe of problem solving will start getting to college and really will totally want Putnam prep from you guys. by MysticTerminator, Jul 26, 2008, 7:53 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I'm wondering about how much analysis you're going to involve. I feel as if there's a market for a heavily heuristic text centered between the epsilon delta high-brow rigor of rudin and standard calculus texts (i think stewart and the like, maybe even apostol(i've never read it so i don't know)), involving the problem solving approach of your texts. Sort of a bottom up and top down approach that isn't coming all the way from the bottom or all the way at from top (hard-technical problems that suck out your soul). by jamiemorgan88, Jul 27, 2008, 3:47 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. The AP exams to me look like more memorization, not much of a problem solving aspect. I'd take a Physics/Calc course more for the fun of it, and to learn more (in a problem solving aspect, not just brute force mem). It's be a challenge, btu a fun challenge. if you ever do offer one though, I'd have to go finish Volume 1 and 2 though first. I'm not that smart for a Calc/Physics course yet (especially a rigorous AOPS one!) by themorninglighttt, Jul 27, 2008, 9:17 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. To be clear, we won't be integrating physics in this calculus class. That's just me thinking out loud. by rrusczyk, Jul 27, 2008, 11:25 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I've almost finished my online calculus BC class, but I'll get your book for a more rigorous understanding. My online homeschool calculus material is kinda boring, because the homework is like most other schoolwork: just checking if I know the formulas. Reading posts on the AoPS calculus subforum helps supplement that, though. Do you, by any chance, have a tentative publish date for your book? by undefined117, Jul 28, 2008, 3:01 AM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. 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15210	https://artofproblemsolving.com/community/c1059h986738?srsltid=AfmBOoqgNx4nPepfb46_TwBjhnq7-PcfY33-OyY-2ERzJuAIQx1PmJMo	Casework and Brute Force are Tools of Satan : Strategy Community » Blogs » Casework and Brute Force are Tools of Satan » Strategy Sign In • Join AoPS • Blog Info Casework and Brute Force are Tools of Satan =========================================== Strategy by PenguinIntegral, Jun 17, 2006, 3:21 PM OK, here is my training schedule thingy. First, I want you to know that saying "I solved a USAMO problem" sounds way cooler than "I went through USAMO problems and finally solved one". Or " I went through like 5 AIME's and got a double digit one." Ok, second. I need to go back to basics. I suck on geometry. Absolutly suck at it. Some AoPS 1 Geom stuff gives me a challenge. This is bad. Third. I cannot do Trig, logs, or complex numbers, and I haven't even memorized stuff like Binominal theorm and binominal coefficiants.At all. I know things like and basic things like that, but I still look up things like Decarte's rule of sign's when I recognize I need to use it. Or vectors. Or roots of unity. I mean, I know what they are, but knowing what they are and knowing how to munipulate them are different things. As a result of this, I waste so much time proving to myself things. For example, the formula for xCy. Now, before I develop some kind of training method, I need to know where this is all going to lead. The knee-jerk response is chanting IMO, but I don't kid myself. I'm not that good. I pretty sure that in 7th grade many of the site admins could do more than score double digits on an AIME. So I'm going to aim slightly lower. My goal is to be able to do any national(like USAMO, not international, like IMO) test in the kalva archives and get a socre of half the possible points consistantly. OK, so the first step will be to plug gaps in my knowledge. First phase-Backround (to be finished in a few weeks to 2 months) Attempt to do every Geom problem in volume 1 and read the solutions. Twice. Do everything in vol.2. Twice. Even the stuff I know. All the way through. Skipping nothing. Reading and understanding every solution. Expect annoying flurries of help postings in intermidiate. Second Phase-AIME prep!(Due to me not being able to take AMC tests this year, probably finished a little before summer.) Do all the availible AHSME tests from the KSU site and all AIME tests from kalva. Once again, twice. Mix in some AcoPS everyday. Expect annoying flurries of help postings in intermidiate of questions that have already been answered. Everything Else Lots of AcoPS. Lots. The USSR olympiad problem book. All of it. Twice. Problem solving strategies. All of this, reading ACoPS again in relevant sections. After all of that, the only thing left to do if refine my new skills in the AoPS fourms and do national olympiads. Oh, for you who have been nice enough to read my ramblings and have some rusty programming skills, MIT facuty lectures have been recorded and posted Here with the accompanying textbook provided free by MIT Here. Happy New Years! Well, off to the USAMO database again. On second thought, I need to program some probability simulations to help with tonights card game. We play a very cool variant of hearts with contracts and pointcards mixed in. 9 Comments (Post your comment here. I am lonely, talk to me) Comment 9 Comments The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I think your goal might be a little high. I do believe that half the possible points on an USAMO is 21; as far as I can tell some years that gets a <11th grader into MOP. Generally a 24 gets one in, from my guesses. Might as well say your goal is MOP! by solafidefarms, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. OK, I want to go to MOP. by PenguinIntegral, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Quote: Lots of AcoPS. Lots. The USSR olympiad problem book. All of it. Twice. Problem solving strategies. All of this, reading ACoPS again in relevant sections. Unless you can somehow spend every second of every day on math, it will take you a long time to go through USSR, PSS, and ACoPS. There are over 1300 problems in PSS alone and a good percentage of them appeared on Olympiad level tests (there are still a lot of them that aren't that hard, as well). Just thought you might want to take that into consideration. (I'm not trying to discourage you. I just want to help give you a good idea of how long it would take you to do all that). by joml88, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. I know. But if I am to start sometime, sooner is better than later. by PenguinIntegral, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. What grade are you in? by mna851, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. 7th. by PenguinIntegral, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. You definitely have a good shot at Red MOP in that case. Anyway, you certainly have an ambitious plan, but you are missing out on a CRUCIAL part of your preparation--my WIDELY SUCCESSFUL mock AIME. So therefore you must DOWNLOAD IT NOW if you are to have any chance at achieving your goal. OK seriously though, it's great to have all this stuff planned out, but really, are you going to be able to go through ALL of that? From my experience, math gets very tiring when you structure what you do in a strict way. Therefore, I just give myself some general milestones that I think are reasonable and hope to meet over 75% of them. That's what I think you should do. Plus download my mock AIME. by probability1.01, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. Where can I get your mock AIME? by PenguinIntegral, Jun 17, 2006, 3:22 PM Report The post below has been deleted. Click to close. This post has been deleted. Click here to see post. lol in the AMC forum by chess64, Jun 17, 2006, 3:22 PM Report I power my car on harvested angst PenguinIntegral Archives July 2007 ????? June 2007 Junggi May 2007 Hmm asy test OK CHAT CHAT omjf April 2007 avatar WOOT AMP Youtube AMP Hmm AMP The Indiana Forum is Whatever OMG Hmm this is bad Cars March 2007 Olympiads WOOT WOOT CARS AIME II AIME II WOOT Whatever OMG APOSTOL AoPS2 AIME Andreescu, you RULE funny February 2007 Nullify the vegetarian moral crusade Let's Blow up the Moon Ok, new plan How they really grade USAMTS ... AMCs, OH NOES I'm Stupid January 2007 Story :( AMC 12 So... AIMEs Tag line Titu Andreescu, why must you price your books so highly!? December 2006 AIME 1984 AIME 1983 I hate mathcounts. :( November 2006 AIME MC USAMO October 2006 Yayness September 2006 >.> Hmmmmm Ho Hum Test Iron Maiden Olympiad Problems August 2006 Classes ??? Ho hum AIME Oh no! v Post Count Arnav Birthday AoPS July 2006 AOPS Classes Letting others infringe on copyright so you don't have to! "Fun"-dementals Tetherball is fun Pwnanation Wooot Awesome Math Quote June 2006 ICP <(^.^)><(^.^<) (>^.^)> AIME 2005 I Conversation Stuff Stuff Yay No Woot:( AIME problem series A Calculus really isn't hard. AoPS classes Subject Advert USAMO stats Chess Yay Contest! :( I suck at life. MOP Math Leningrad Math Olympiads Geometry! Birthday No! Important Thread Regarding USAMO Hippie Math Mathcounts MC MC I can make funny symbols -> £ Role Model Yar Mathcounts No! I WILL fail mc! YEAH! Wait MC practice Haha Typing Chess GAME! Game-part duex :( !:(! Game-Part III Me! Game MC Mock up Mathcounts AMCs Math(duh!) Grr at CPIG Geometry No! AIME score Write in C :( 18th AIME1 2000 Goals Education ! Power of a point AMC tests science advice? ? Books! Advertisment! USAMO Short story Help!!! Hello! I see you! yay! Hello! Chuck Norris Flashbacks Norris J1 Grr :( Wow Idea Menelaus's Theorem! Guitar Violin Algebra AHSME Solving old AIME and USAMO questions. Strategy Trig Math Die vectors, Die!!! Im sending an SOS Conics Almost done!!! DONE!!! I am DONE!!!!! :( AIME scores ? F&G Critique my chess game Free chess software ...meh Sadness ... ... Click ... I feel better now I WIN! Bye Grr, im back f I hate pompous people Hello Jazz band typing Programming meh The first sane post in a long time ??? ERROR:Flying cows Grr GAHHHHH!!!!!! Part 2 Part 3 KITTENZ!!!!!!! SO excited! I'm a contributor The Inept Explorer Internet excrement Subject goes here Audio AIME! Convert Ye Heathens! meh yay! Sigh I C U No! Seeking help... Yay! YAY! Meh meh meh Meh Suprising, I'm a contributor. click me ROMANIA WILL DIE!!! Im happy SLeEp I feel bad I be thinking of math Penguins are Crunchy! Physics and AoPS classes im getting worried.... Yay! Sigs Grr Math, of course Im sad I am a MOD! Meh Summer camp was so fun. Im still here...wasting away Training Begins! Nooooooooooooooooooooooo! I would like to direct you to the AMC fourm marathon! Debate! Sorry for being gone Chess 64 should be a mod! Shouts Submit dead blog(10 years later) by fuegocaliente, Feb 1, 2019, 9:09 PM dead blog... by $LaTeX$., Aug 4, 2009, 9:53 PM Meh. I wonder if you remember me... by Nexmus, Dec 31, 2008, 8:51 PM zing by IntrepidMath, Dec 19, 2007, 5:05 PM It's 17 by my count. ? by PenguinIntegral, Oct 11, 2007, 9:36 PM I knew that Penguin would stop posting after AMSP. He's probs on a different account. BTW, penguin, if you ever see this, I'm on page 15 of the camper section of AMSP book. (AS) by IntrepidMath, Sep 30, 2007, 9:15 PM cool PenguinIntegral. This is your second shout!!! by #H34N1, Aug 5, 2007, 8:45 PM CHECK OUT MY BLOG!!!!!!!!!!!!! by shaggy75, Jul 29, 2007, 6:14 PM 8 shouts Contributors #H34N1 • 236factorial • bubala • chess64 • h_s_potter2002 • hwenterprise • junggi • math92 • MCrawford • MithsApprentice • MysticTerminator • nonie • PenguinIntegral • Sly Si • solafidefarms Tags About Owner Posts: 1794 Joined: May 14, 2005 Blog Stats Blog created: Aug 2, 2005 Total entries: 248 Total visits: 81322 Total comments: 548 Search Blog Something appears to not have loaded correctly. Click to refresh. a
15211	https://aclanthology.org/2018.gwc-1.52.pdf	WordNet Troponymy and Extraction of “Manner-Result” Relations Aliaksandr Huminski Institute of High Performance Computing, ASTAR, Singapore huminskia@iphc.a-star.edu.sg Zhang Hao Nanyang Technological University Singapore hao.zhang@ntu.edu.sg Abstract Commonsense knowledge bases need to have relations that allow to predict the consequenc-es of specific actions (say, if John stabbed Pe-ter, Peter might be killed) and to unfold the possible actions for the specific results (Peter was killed. It could happen because of poison-ing, stabbing, shooting, etc.) This kind of causal relations are established between man-ner verbs and result verbs: manner-result rela-tions. We offer a procedure on how to extract manner-result relations from WordNet through the analysis of the troponym glosses. The pro-cedure of extraction includes three steps and the results are based on the analysis of the whole set of verbs in WordNet. 1 Introduction WordNet (WN) as a database is widely used in variety of tasks related with extraction of seman-tic relations. Verbs in WN are organized hierar-chically as troponym-hypernym relations. Meanwhile, the definition of troponym has something in common with the definition of a manner verb suggested by B. Levin and M. Rap-paport Hovav. We consider in more details both types of rela-tions: troponym-hypernym and manner verb-result verb relations. 1.1 Troponym-Hypernym Relation Verbs in WN are linked through different types of relations – antonym, cause, entailment – but troponym-hypernym relation is a basic and the most frequently found relation among verb synsets (Fellbaum and Miller, 1990). If a hyper-nym is a verb of a more generalized meaning, a troponym replaces the hypernym by indicating more precisely the manner of doing something. The troponym-hypernym relations are hierar-chical (vertical). Therefore, it makes it possible to create a huge verb net with top synsets that represent the highest hypernyms and branches going down to the bottom with corresponding troponyms. The closer to the bottom, the more specific is the verb synset. There are no further clarifications between different types of tropon-ymy in WN. As a result, the manner relation is polysemic and many different semantic elements are hidden behind the label ‘manner’ (Fellbaum, 2010). It can be volume as in talk-whisper, speed as in jog-run, intensity of emotion as in love-adore-idolize, etc. The specific manner depends on the semantic field and corresponding dimension. 1.2 Manner Verbs and Result Verbs The definition of troponym has something in common with the definition of a manner verb suggested by Beth Levin and Malka Rappaport Hovav (2010). They pointed out that a study of the English verb lexicon reveals that within par-ticular semantic domains there can be verbs that describe carrying out activities – manners of do-ing; and there can be verbs that describe bringing about results. Manner verbs are walk, jog, stab, scrub, sweep, swim, wipe, yell, etc. Result verbs are break, clean, crush, destroy, shatter, etc. There are 3 features of manner-result relations that make extraction of them so important for commonsense knowledge bases. 1) Manner verbs and result verbs are in causal relations: stabbing causes killing; sweeping causes cleaning and etc. 2) It is an empirical, not a logical causality with probability less than 100%. Actions represented by manner verbs can fail in achievement of desirable results: I wiped the table, but it’s not clean. John shot Peter, but he survived. 3) It is a common situation when several manner verbs cause the same result verb: sweeping, wiping, blowing cause clean-ing. 2 Troponym-Hypernym and Manner-Result Relations In the WN glossary of terms1, a troponym is de-fined as a verb expressing a specific manner elaboration of another verb. X is a troponym of Y if to X is to Y in some manner. Having this definition, the obvious question arises: if tropo-nym is defined through the manner, can one state that troponym-hypernym relation equals in man-ner verb-result verb relation? In other words, is there any correlation between troponym-hypernym relation and manner verb-result verb? The general answer on this question is “no” since there are several types of correspondence that can be unfolded in WN: 1) troponym-hypernym relation can be equal “manner verb-manner verb” rela-tion. For example, the verb stroll (walk + slow + relaxed) is a troponym for the verb walk. But both of them are manner verbs. 2) troponym-hypernym relation can be equal “manner verb-underspecified verb” relation. For example, the verb walk (move + by steps) is a troponym for the verb move. The verb walk is a man-ner verb, the verb move is underspeci-fied: it is neither a path verb since it doesn’t encode direction, nor a manner verb since it doesn’t specify any particu-lar manner. So, it is an underspecified verb taking into consideration that man-ner-result dichotomy does not fully and exhaustively classify verbs. 3) troponym-hypernym relation can be equal “result verb-result verb” relation. For example, the verb fracture (break in-to pieces) is a troponym for the verb break (destroy the integrity). 4) troponym-hypernym relation can be equal “manner verb-result verb” relation. For example, the verbs stone, lapidate (kill by throwing stones at) and poison (kill with poison) are troponyms for the verb kill (cause to die; put to death). 1 Now, we need to find out the way how to ex-tract the 4th type of correspondence which repre-sents exactly what we are looking for. 3 General Procedure to Extract Man-ner-Result Relations from WordNet Manner-result relations are hidden in the WN verb hierarchy. We know for sure that this kind of relations is a subset of troponym-hypernym relations. However, there are not any explicit ways to extract them yet. Our idea is that manner-result relations can be extracted from the set of troponym-hypernym relations if two conditions, applied to troponym-hypernym relation are valid: 1) The hypernym is a result synset; 2) In the glosses of its troponyms one of the two templates can be found: “V + by” or “V + with”; where V = hypernym. For example, if we consider the result synset {clean, make clean} as a hypernym, some its troponyms have glosses that satisfy the patterns:  sweep (clean by sweeping)  brush (clean with a brush)  steam, steam clean (clean by means of steaming) In this case, it can be stated that sweep, brush, steam, steam clean are manner verbs for clean and the following causality can be constructed: sweep, brush, steam, steam clean  clean This idea is the basis of the general procedure for manner-result extraction. It includes 3 steps. 3.1 Extraction of Top Verb Synsets There are 13789 verb synsets in WN 3.1 ordered by troponym-hypernym hierarchical relation. At this stage, we need to extract synsets locat-ed on the top level of the hierarchy. This kind of synsets will be called further “top verb synsets”. The procedure of extraction is based on the following characteristic of the top verb synsets: they don’t have any hypernyms, only troponyms. Using this, all the extracted 13789 verb synsets have been tested whether they have a hypernym. As a result, 564 top verb synsets have been ex-tracted automatically. 3.2 Extraction of Top Result Verb Synsets Within 564 top synsets we made a manual classi-fication to extract only the result verb synsets. The classification revealed the following 5 clas-ses. 1) one-level top synsets. This type of top synsets has only one level: the top verb synset itself. It is a substantial portion of top synsets: 203. Example: admit (give access or entrance to). 2) manner and underspecified verb synsets. Total number: 105. Example of the top manner verb synset: splash (strike and dash about in a liquid). Example of the top underspecified verb synset: {travel, go, move, locomote}. 3) state verb synsets. Total number: 69. Ex-ample of the top state verb: lie (be lying, be prostrate; be in a horizontal position). 4) second order predicates. Total number: 60. Second order predicates govern the other predicate. Example: {begin, start} (have a beginning, in a temporal, spatial, or evaluative sense). 5) result and change-of-state verb synsets. Total number: 127. We combine these 2 classes of verbs since, as it turned out, change-of-state verbs have manner verbs as troponyms. For instance, the verb die has a troponym synset {suffocate, stifle, asphyxiate} which obviously contains manner verbs. Example of the result verb synset: {destroy, ruin}. We further analyze the 5th class only. Our as-sumption was that result verbs as hypernyms can have either result verbs or manner verbs as troponyms. But manner verbs as hypernyms can-not have result verbs as troponyms. They can only have manner verbs as troponyms. Following the assumption, the sequence of troponyms de-rived from the top result verb hypernym cannot have the subsequence of manner verb as a hyper-nym and result verb as a troponym. For example, the sequence of 4-level verbs with the top result verb and the bottom manner verb can have the following 3 possible distributions:  result-result-result-manner  result-result-manner-manner  result-manner-manner-manner The distribution of “result-manner-result-manner” is impossible. The next step is extraction of manner verbs from the tree with result verb synset on the top. 3.3 Extraction of Manner Verbs through the Patterns in Glosses At this stage, we look for the manner verbs for each result verb synset through the patterns “V + with” or “V + by” in the glosses of troponyms. If the synset doesn’t contain any patterns we mark it as “NONE”. If the synset contains at least one of the patterns we mark it with its gloss. As a result, we get a sequence of marked synsets from the top verb synset to the bottom verb synset. If the sequence of all synsets or only the tail of it contains “NONE” we exclude the whole sequence or the tail accordingly from the consideration since there is no manner verbs there. The purpose is to extract all lower synsets that contain the patterns. The procedure of the extraction is automatic. Following the assumption from 3.2 one can get different types of result-manner sequences. For example, for the top synset {change, alter, modify} we will get the following 3 sequences among many others: {change, alter, modify}-NONE {damage}-NONE {frost}-damage by frost The causality frost  damage can be made from this sequence, where frost is manner verb and damage is result verb. {change, alter, modify}-NONE {damage}-NONE {burn}-damage by burning with heat, fire, or radiation {scald}-burn with a hot liquid or steam The causality is scald, burn  damage {chang e, alter, modi-fy}-NONE {indis-pose}-NONE {hurt}-NONE {in-jure, wound }-NONE {tram-ple}-injure by trampling or as if by trampling The causality is trample  injure, wound, hurt. For each n-level synset one can get a restricted number of the valid sequences. For example, for each 3-level synset we can get only two valid sequences: “result-result-manner” as in {change, alter, modify}-NONE {sharpen}-NONE {whet}-sharpen by rubbing, as on a whetstone and “result-manner-manner” as in {damage}-NONE {burn}-damage by burning with heat, fire, or radiation {scald}-burn with a hot liquid or steam As a whole, for the different sublevels of the same top result synset, one can get full variety of valid n-level sequences: {change, alter, modi-fy}-NONE {shape, form}-NONE {tabulate}-Manner verb {change, alter, modi-fy}-NONE {shape, form}-NONE {roll}-Manner verb {change, alter, modi-fy}-NONE {shape, form}-NONE {draw}-Manner verb {change, alter, modi-fy}-NONE {shape, form}-NONE {fit}-NONE {dovetail}-Manner verb {change, alter, modi-fy}-NONE {shape, form}-NONE {flatten}-NONE {steamroll, steamroll-er}-Manner verb {change, alter, modi-fy}-NONE {shape, form}-NONE {flatten}-NONE {roll_out, roll}-Manner verb {change, alter, modi-fy}-NONE {shape, form}-NONE {flatten}-NONE {roll_out, roll}-Manner verb {mill}-Manner verb Table 1. Part of valid n-level sequences from {change, alter, modify} result synset. To make the table more compact we replaced the glosses that match the patterns to the phrase “Man-ner verb”. Figure 1. shows the Table 1. in the structural graphical form with glosses. Figure 1. Visualization of the valid n-level sequences. It is necessary to stress that each line in result-manner causal relation can contain both direct (frost  damage) and indirect (scald  damage) causality. Regardless of that, each line is consid-ered as one specific type of causal relations. After running all the top 127 result verb synsets and counting the lines we got the total number of 1541 lines. It means, 1541 manner-result causal relations have been extracted from WN. 4 Scope of the Results To evaluate what is the scope of the results we compare them with another type of causal rela-tions that is explicitly presented in WN 3.1: cause-relation. Cause-relation refers to the relation between two verbs V1 and V2 where V1 logically causes V2 (Fellbaum, 1998). For example, the verb kill causes the verb die. Running through 13789 verb synsets in WN 3.1 we automatically extracted 219 verb synsets that contain cause-relation. Among them there are 63 verb synsets that cause the same synset. In other words, there are 63 causal relations with absolutely identical left and right sides: {dry, dry_out} causes {dry, dry_out} {lengthen} causes {lengthen}, etc. It happened because of polysemy in verb mean-ing. Synsets here are formally identical but rep-resent different meanings of verbs. Since it is hard to use such kind of causality in applications, the real number of the verb synsets that contain cause-relation can be reduced to 156. Comparison of 156 verb synsets containing logical cause-relation with 1541 non-logical (empirical) causal relations shows that the scope of the latter relations is significant. 5 Conclusions and Future Work In this paper, we have described how to extract manner verb-result verb causal relations from WN. The procedure of extraction includes 3 steps: a) extraction of the top verb synsets (total 564), b) extraction of the result synsets and the change-of-state synsets among them manually (total 127), c) running automatically the algo-rithm “V + by” and “V + with” on 127 top synsets and getting 1541 types of manner-result causal relations. The results are considered as preliminary ones. As future work, the algorithm can be elaborat-ed by adding new patterns and tuning the original ones. For example, the change-of-state verb die has a troponym synset suffocate, stifle, asphyxi-ate (be asphyxiated; die from lack of oxygen) which clearly indicates the manner of dying but the gloss doesn’t contain the patterns we are working with. These types of extracted relations can be wide-ly used in commonsense knowledge bases for the prediction of action consequences and unfolding the possible reasons for the results. Com-monsense knowledge bases enriched by using this approach can be exploited in dialog systems and the other specific technologies and applica-tions. References Fellbaum C. and G. Miller. 1990. Folk psychology or semantic entailment? A reply to Rips and Conrad. The Psychological Review. 97, 565-570. Fellbaum, C. 2000. Autotroponymy. In Yael Ravin (ed). Polysemy: theoretical and computational ap-proaches. New York, Oxford University Press. Levin, B. and M. Rappaport Hovav. 2010. Lexicalized Meaning and Manner/Result Complementarity. Ms. Stanford University and The Hebrew Universi-ty of Jerusalem. Fellbaum C. 1998. WordNet: An Electronic Lexical Database. Cambridge, MA: MIT Press.
15212	https://www.sjsu.edu/faculty/watkins/clausius.htm	\| San José State University \| \| applet-magic.com Thayer Watkins Silicon Valley & Tornado Alley USA \| \| \| \| The Clausius-Clapeyron Equation: Its Derivation and Application in Meteorology \| The equilibrium between water and water vapor depends upon the temperature of the system. If the temperature increases the saturation pressure of the water vapor increases. The rate of increase in vapor pressure per unit increase in temperature is given by the Clausius-Clapeyron equation. Let p be the saturation vapor pressure and T the temperature. The Clausius-Clapeyron equation for the equilibrium between liquid and vapor is then dp/dT = L/(T(Vv-Vl)) where L is the latent heat of evaporation, and Vv and Vl are the specific volumes at temperature T of the vapor and liquid phases, respectively. More generally the Clausius-Clapeyron equation pertains to the relationship between the pressure and temperature for conditions of equilibrium between two phases. The two phases could be vapor and solid for sublimation or solid and liquid for melting. The material below is an examination of the application of the Clausius-Clapeyron equation to meteorology where the most relevant systems are water and water vapor and salt water and water vapor. But the equation also applies to ice and water vapor and ice and water. Finally there is a derivation of the Clausius-Clapeyron equation from thermodynamic principles. Note that Vv is much greater than Vl so that to a good approximation dp/dT = L/(TVv) Furthermore the ideal gas equation applies to the vapor; i.e., pVv = RT and hence Vv = RT/p where R is the universal gas constant. Thus dp/dT = L/(RT²/p) or, equivalently (1/p)(dp/dT) = L/(RT²) In differential form this is dp/p = (L/R)(dT/T²) or, equivalently d(ln(p)) = (L/R)d(-1/T) If L is independent of temperature then the solution to the differential equation is ln(p) = c0 - (L/RT) or, equivalently p = c1exp(-L/RT) where c1 is a constant. The shape of this function is given below: The empirical curves have the opposite curvature but clearly the equation dp/dT = L/(RT²/p) = Lp/T² indicates that as temperature increases the slope (dp/dT) decreases. Of course, the empirical curves take into account that the latent heat L can change with temperature, but the change in L with T is relative small. There is a fundamental discrepancy between the Clausius-Clapeyron Equation and the empirical relationships and the published derivations of the equation blithely show an empirical curve in which dp/dT increases with T and an equation in which dp/dT is inversely proportional to T. However this discrepancy will be left unresolved here and the implications of the Clausius-Clapeyron equation will will be further explored. For some purposes the density of the vapor is of more interest than the pressure. By the ideal gas equation, the molecular density D is given by D(T) = 1/Vv = p/RT which for the above pressure-temperature relationship is D(T) = c1exp(-L/RT)/RT The shape of this latter relationship for some arbitrary values of the parameters is shown below: It is a surprise that there would be a case such that as the temperature goes up the density of decreases. This needs to be checked for generality. The ideal gas equation indicates that density would go down as temperature increases if pressure remained constant. Even if pressure increases with temperature the density will decrease if the increase in pressure is not enough to offset the direct effect of the temperature increase on density. In differential terms the ideal gas equation is dD/D = dp/p - dT/T The Clausius Clapeyron equation, when the specific volume of the liquid is assumed to be zero, gives dp/p = (LD/p)(dT/T) = (L/RT)(dT/T) so dD/D = L/RT − 1 Thus the effect of a temperature increase on molecular density depends upon the magnitude of the latent heat of vaporization compared to RT. The latent heat of vaporization for water is 2.257×106 J/kg. The gas constant for water vapor in SI units is 461.5 J/(kg K) so at T=300 K, RT=1.3845×105 J/kg. Thus L/RT=16.3 and hence dD/D=15.3(dT/T) so if (dT/T)=1/300 then dD/D = 15.3/300=.051. Thus the relationship between density and temperature should look more like the following: (To be continued.) Derivation of the Clausius-Clapeyron Equation The derivation will be given for a liquid-vapor equilibrium interface but it equally well applies to the interface between any two phases. Let sv and sl be the specific entropy of the vapor and liquid phases, respectively. The pressure and temperature of the two phases are equal. The chemical potential μ is equal on either side of the phase boundary curve. Therefore the changes dμ in the chemical potential for movements along the phase boundary curve are also equal. This means that dμv = Vvdp - svdT = Vldp - sldT = dμl Solving for dp gives dp = (sv - sl)dT/(Vv - Vl) or, equivalently dp/dT = (sv - sl)/(Vv - Vl) But the change in entropy Δs for the phase change is just L/T so dp/dT = L/(T(Vv - Vl)) This is the Clausius-Clapeyron equation. \| \| \| HOME PAGE OF applet-magic HOME PAGE OF Thayer Watkins \|
15213	https://www.youtube.com/watch?v=16fFjAcxJSc	Arrhenius Equation Activation Energy and Rate Constant K Explained The Organic Chemistry Tutor 9730000 subscribers 9001 likes Description 705503 views Posted: 13 Jul 2016 This chemistry video tutorial focuses on the Arrhenius equation and how to derive it's many different forms within the subject of chemical kinetics. Chem Kinetics Formula Sheet: Activation Energy & Collision Theory - 1 Hour 30 Minute Video on Patreon: Printable PDF Worksheet With 13 Questions on Patreon: Direct Link to The Activation Energy & Collision Theory Video on Patreon: Here is a list of topics: 1. Arrhenius Equation and the Rate Constant K 2. The units of R and the activation energy 3. Rate law expression and the concentration of reactant A 4. Frequency Factor, Collision Frequency and Steric Factor 5. Rate of Reaction, Rate Constant K, and Activation Energy 6. The Effect of a Catalyst on Activation Energy and Reaction Rate 7. Slope, Ea, and R 8. Slope Intercept Form Linear Arrhenius Equation 9. Factors Affecting the rate of the reaction - concentration, temperature, catalyst, activation energy and rate constant K 10. Arrhenius Equation / Formula Graph Chemistry - Basic Introduction: Stoichiometry Practice Problems: Molarity, Molality, Density, & Mass %: Vapor Pressure & Clausius Equation: Raoult's Law - Vapor Pressure: Colligative Properties: General Chemistry 1 Final Exam Review: Chemical Kinetics - Initial Rate Method: Integrated Rate Laws - 1st & 2nd Order: Reaction Rate Factors: Collision Theory & Activation Energy: Potential Energy Diagrams: Elementary Rate Laws: Rate Laws of Reaction Mechanisms: Intermediates & Catalysts: Types of Catalysts: The Equilibrium Expression: Calculating Kp From Kc: Chemical Equilibrium & Ice Tables: Le Chatelier's Principle: Acids and Bases - Introduction: The 7 Strong Acids to Memorize: Conjugate Acid-Base Pairs: pH and pOH Calculations: Estimate The pH Without a Calculator: Autoionization of Water - Kw: Which Acid Is Stronger? Acidic, Basic, & Neutral Salts: pH of Weak Acids: Buffer Solutions: Acid Base Titration Curves: Acids and Bases - Practice Test: SAT Chemistry Subject Test Review: General Chemistry 2 Final Exam Review: 216 comments Transcript: in this video we're going to go over the AR ranous equation where K or the rate constant K is equal to a e raised to the E A / RT so what do these variables mean a is the frequency factor a is equal to Z times P where Z is the Collision frequency and P is the steric factor but for the most part for a typical chemical kinetics question you really don't have to worry about a or Z or P EA is the activation energy that's how much energy you need to get the reaction started we're going to talk a lot about that now R is the energy constant and it's equal to 8.3145 and the units are Jewels per mole per Kelvin so therefore temperature is measured in kelvin you don't want to plug in the Celsius temperature in this equation the activation energy has the units Jews per mole it has to match with the units of R now we need to understand the relationship between temperature and and K and activation energy but before we do that you need to know how K plays a role in the rate law expression let's say if we have this rate law expression rate is equal to K a is this A first order reaction second order or is it a zero order reaction what would you say now the exponent of a is one so it's a first order reaction which means that that if you double the concentration of reactant a the rate of the reaction will double if you quadruple the concentration of a the rate will quadruple so anytime you increase the concentration of a reactant if it's first order or second order the rate will increase now if it's a zero order reaction if you increase the concentration of that reactant the rate will not increase if it's zero order because let's say if you doubled it two to the0 power is one now if you double or if you increase the rate constant K what's going to happen to the rate of the reaction will it increase or decrease if you increase the rate constant K the rate of the reaction will go up now as you mentioned from the our heus equation K depends on the activation energy which is EA and it depends on the temperature so if you can increase the temperature the rate constant K will go up however if you decrease the activation energy the rate constant K will also go up now typically when you increase the numerator the value of the whole fraction goes up but since it's on the exponent and sits there's a negative sign it's kind of opposite to that usual Trend so it turns out when you decrease the activation energy the rate of the reaction goes up because K goes up now let's draw a potential energy diagram so here we have the reactants the products and this is the transition state also known as the activated complex the activation energy or at least the forward activation energy is the difference between the energy of the reactants and the activated complex the reverse activation energy is the difference between the activated complex and the energy of the products Delta H the enthropy of the reaction is the products minus the reactants now what's going to happen if we add a catalyst to this reaction what would a catalyst do now you know Catalyst speeds up a reaction but how exactly does it do that well what a catalyst does it lowers the activation energy and it does so by providing the reaction an alternative pathway another way to get to the products and so that's how it lowers the activation energy by the way that reaction is it exothermic or endothermic anytime the products have less energy than the reactants they're low in energy it's an exothermic reaction now anytime you add a catalyst the activation energy decreases when the activation energy decreases the rate con K goes up and whenever K goes up the rate of the reaction goes up as well when you increase the temperature the rate constant K goes up and therefore the rate of the reaction will go up as well now if you increase the concentration of a reactant if it's first order or second order and if it's not zero order then K will stay the same K is not affected by the concentration however the rate will increase if you increase the concentration of the reactant because the rate depends on k and a concentration of a if it's a first or second order reaction so if you increase the concentration or if you raise the temperature or if you add a catalyst you can increase the rate of the reaction so now let's say if we start from this equation the uranous equation we're going to do is we're going to take the natural log of both sides starting from this equation we're going to come up with a few other equations where if you're solving a problem it might be useful to know those equations so right now we're going to have the natural log of K is equal to uh the natural log of a e raised to the other stuff now a property of natural logs can allow us to separate a single log into two logs for example Ln a Ln or Ln a B is equal to Ln a plus Ln B so this is going to be like a and this is going to be like B so we're going to separate it into two um separate natural logs so on the right side we're going to have uh LNA a plus Ln e raised to the E A over RT now another property of logs is you're allowed to move the exponent to the front so this is equal to 2 natural log a so let's take this exponent and let's move it to the front so right now we have Ln K is equal to LNA a and then it's going to be minus E A over RT Ln e the natural log of e is one so that's going to disappear and I'm going to switch these two so what we now have is uh natural log of K is equal to EA over RT plus natural log of a now I'm going to put this in slope intercept form so Ln K is equal to E A over R Time 1 / T which is the same as EA over RT so this is like Y and this is m x plus b it's in slope intercept form so L and K is y the slope is negative EA over r m is the slope in the slope intercept form equation and one / T is X and the Y intercept which is B is l and a so if we were to graph this function since LM K represents y we can put that on the Y AIS and 1 / T represents X so we can put that on the x axis and the slope is negative so it should we should get a straight line but going down and so therefore as 1 /t increases Ln K decreases which means that K decreases if L and K Goes Down K goes down as well so if T increases T is the reciprocal of one/ T that means Ln K increases which mean k goes up that's why we can say that as you increase the temperature the rate constant K goes up and the rate of the reaction goes up as well now there's some other things to know um about this equation and one of those things that you want to know is the slope we said that the slope which is m is equal to E A over R so if you ever need to find the activation energy the activation energy is R the slope so if you can find a slope with this line by doing R over one or you know by using equation Y2 - y1 is equal to X2 - X1 well you can calculate the activation energy but using that equation we're going to get another equation so here's what we're going to do let's replace M with what it equals negative E A over R so e a/ r is equal to this thing R over run now Y is associated with Ln K so it's going to be Ln K2 minus Ln K1 which is like Y2 minus y1 now X is associated with 1 over T so it's X2 is going to be 1 over T2 and X1 is going to be one over T1 now let's uh make some space now keep in mind Ln a plus lnb was Ln a B now turns out that Ln a minus lnb is Ln a / B so using that fact Ln K2 minus Ln K1 is Ln K2 / K1 and if we so we have that and then 1 over T2 - 1 over T1 and if we multiply both sides by r or negative R these will cancel and so the activation energy is equal to R times this whole equation now you might have seen this equation differently perhaps you've seen this equation like this Ln K2 over K1 is equal to EA / R 1 / T2 - 1/ T1 if you rearrange the equation you can get this equation this is like the standard form of the arous equation now if you need to solve for K here's what we need to do let's say if you have Ln a equals B the base of Ln is e e ra to the B is equal to a so you can change it to this equation so using that fact this natural log has a base e so e raised to everything on the right side is equal to to everything inside so that means that K2 over K1 is equal to e raised to the EA over R 1 / T2 - 1/ T1 and if we multiply both sides by K1 this is how you could solve for the rate constant K2 so K2 is equal to K1 e raised to the netive EA / R 1 / T1 - 1/ T2 and make sure if you want to find a rate constant K the activation energy must be in Jews per mole don't plug in kles or you will get the answer wrong and keep in mind one KJ is a th000 Jew so to go from kles to Jews multiply by th000 if you needs to go from Jews to kles divide by th000 so now you know how to find the rate constant K and we use the other equation to find the activation energy so now sometimes you may need to find the temperature so let's see if we can rearrange the equation to get the temperature so let's start with this equation so in addition to the last form you also want to remember this form of the equation so let's multiply both sides sides by R over EA so on the left side we're going to have R Ln K2 over K1 and then it's going to be divided by EA on the right side well let's add the negative sign as well so negative R so the negative will cancel R will cancel and EA will cancel on the right side so we have 1 over T2 - 1 over T1 so what I'm going to do now is I'm going to take this term and move it to this side so one / T1 is negative on the right side but it's going to be positive on the left side if you add one over T1 to both sides so 1/ T1 minus r Ln K2 divid K1 / EA is equal to 1 / T2 so therefore if we raise both sides to the Nega one 1 over T2 is going to flip to T2 so here's the other form of the equation if you need to calculate the temperature use this it's going to be 1 / T1 minus r Ln K2 / K1 which is equal to well divided by EA and then raised to the minus one so that's what you need if you need to find the temperature let's put all the equations together and if you need to find a rate constant K use this equation it's going to be K1 e raised to the E A over R and then 1 / T2 - 1 T1 and in order to find the activation energy we said that the activation energy is equal to R Ln K2 / K1 there's a negative sign in front of the r and it's going to be divided by 1 1/ T2 - 1/ T1 so now you have the three forms of the equation so if you need to find the temperature use the first one if you need to find a rate constant K use the second and if you need to find the activation energy use the third and that is it for this video so thanks for watching and uh have a great day
15214	https://thewritepractice.com/story-arcs/	How to Shape a Story: The 6 Types of Story Arcs for Powerful Narratives Articles Resources Hire a Book Coach Join Our Writing Community Select Page How to Shape a Story: The 6 Types of Story Arcs for Powerful Narratives by Joe Bunting \| 2 comments In life, it can feel like things happen randomly, without causation, and with little or no meaning. The human brain, though, needs meaning. We need to understand why things are going badly for us so we can avoid it or why things are going well, so we can do more of whatever’s working. This is why humans love stories. Stories give us a sense of purpose, meaning, and shape, and they do that through story arcs. In stories, we get to see the cause-and-effect connections between otherwise random events. We get to experience the deeper meaning in life. We get to see through the chaos of real life and see the underlying pattern. The literary term for this pattern is story arc, and humans love story arcs. In this article, we’re going to talk about the definition of story arcs, look at the six most commonly found story arcs in literature, talk about how to use them in your writing, and, finally, study which story arcs are the most successful. Definition of Story Arc or Narrative Arc The arc of story, or narrative arc, describes the shape of the change in value, whether rise or fall, over the course of the story. That’s the definition, but what does it actually mean? Let’s break it down. “ If you want to master story arcs, you need to know the six story arcs in literature. Which one are you currently writing? Tweet this Tweet Story Arcs Rise and Fall Stories change. If there is no rise or fall in a narrative, it isn’t a story. It’s a series of events. Maintaining the status quo won't engage readers. It's the rise and fall of characters’ fortunes that interests us more than anything else. This change, the rise and fall in a story, can be plotted on a graph to form a curve shape line. And when you graph them, you begin to see patterns across all forms of story. Here is a simple graph of a dramatic arc that Kurt Vonnnegut describes as “Man in a Hole”: The x-axis of the graph describes the chronology of the narrative structure and the y-axis describes the positive or negative value the main character experiences. That means that story arcs can also be character arcs, illustrating the character development that occurs throughout the plot. However, while all character arcs are story arcs, not ALL story arcs are character arcs. In other words, some story arcs illustrate things separate from the main character's development, which we'll talk more about in the section “Story Arcs Measure Values” below. Want to write a story? Get The Write Plan Planner to write your way from idea to finished book. You'll start by creating a plan for your story (including drawing your story's arc!). Then, you'll track your writing each day until you reach the final page. It's the planner designed by writers,for writers, so you actually finish your book. Get The Write Plan planner today The 6 Primary Story Arcs Story arcs, of course, do not always follow such simple graphs. In fact, story arcs can often look more like this than a smooth curve: Yes, stories must change, but that doesn’t mean they all change in the same ways. But when you compare the story arcs of the best stories throughout history, patterns begin to emerge, and you find that these arcs are much more uniform than you might think. That’s what Andrew Reagan and his team of researchers from the University of Vermont found after analyzing over 4,000 of the best novels from the Project Gutenberg library. In fact, they found that stories fall into six primary arcs, which I’ll list below. You can find the full study, Toward a Science of Human Stories, here (the part we’re talking about begins on page 73). 1. Rags to Riches (rise) All stories move, but some stories only have one movement. In the “Rags to Riches” story arc, that movement is a continuous upward climb toward a happily ever after. The Rags to Riches story arc is one of the most common story types; however, these stories lag in popularity, according to Reagan, the researcher from the University of Vermont, who found that other arcs were more widely read. Examples of Rags to Riches story arcs: Disney’s Tangled A Winter’s Tale by William Shakespeare Pride and Prejudice by Jane Austen Matilda by Roald Dahl Holes by Louis Sachar The BFG by Roald Dahl My Fair Lady(film) / Pygmalion (play) by George Bernard Shaw The Great American Dream / Progress 2. Riches to Rags (fall) As with Rags to Riches, in a Riches to Rags story, there is just one movement. However, this movement is in the opposite direction—a fall, rather than a rise. In a Riches to Rags story, the protagonist begins the plot in a fairly high place, but slowly their life devolves until by the end, their life is a ruin of its former self. Examples of Riches to Rags story arcs: Catcher in the Rye by J.D. Salinger Animal Farm by George Orwell Catch-22 by Joseph Heller Love You Forever by Robert Munsch Picture of Dorian Gray by Oscar Wilde Addiction stories or stories about mental health often fit into this structure. 3. Man in a Hole (fall then rise) This is one of the most common and highly rated arcs, where the slope rises/falls and then rises/falls again, forming what looks like a hole. This is even an arc I used in my book Crowdsourcing Paris. Examples of Man in a Hole story arcs: The Hobbit by J.R.R. Tolkien Alice in Wonderland by Lewis Carroll Disney’s Monsters, Inc. Finding Nemo “Make America Great Again,” Donald Trump’s Campaign Slogan Many stories actually include two sequential Man in a Hole story arcs, as illustrated by this curve: According to Reagan and the researchers at the University of Vermont, this is one of the most popular structures, and the most popular arc with a happy ending. He says in his paper: We find “Icarus” (-SV 2), “Oedipus” (-SV 3), and two sequential “Man in a hole” arcs (SV 4), are the three most successful emotional arcs. Examples of the Double Man in a Hole arc include: Harry Potter and the Prisoner of Azkaban by J.K. Rowling Disney’s The Lion King And more Some stories even contain many Man in a Hole arcs—becoming Man in a Hole, Man in a Hole, Man in a Hole ad infinitum. Lord of the Rings and the 6,700-page online serialized novel Worm are examples of this. 4. Icarus / Freytag’s Pyramid (rise then fall) This is the plot structure Gustav Freytag was interested in when he coined the plot structure now known as Freytag’s Pyramid (contrary to popular belief, Freytag’s Pyramid is not a universal structure for plot, but a description of a single arc). For more on this literary concept (and how it’s since been misunderstood), see our full Freytag’s Pyramid guide here. The Icarus arc, named after the cautionary Greek story about a boy who escapes imprisonment on an island by constructing wings made of wax and ultimately flies too close to the sun, is one of the most popular story arcs. Examples of the Icarus story arc include: Hunger Games by Suzanne Collins The upcoming novel Pluck by J.H. Bunting (me!) Macbeth by William Shakespeare Disney’s Peter Pan The Old Man and the Sea/A Farewell to Arms by Ernest Hemingway The Fault in Our Stars by John Green Jurassic Park by Michael Crichton Titanic (film) Great Expectations by Charles Dickens The Great Gatsby by F. Scott Fitzgerald The Great Santini by Pat Conroy If the word “great” is in the title, you know you’re in for a sad ending! This is a popular story structure with literary writers, and tends to be a staple structure for many classics. 5. Cinderella (rise then fall then rise) The Cinderella arc, like Rags to Riches, is one of the most common arcs, often found in love stories, sports stories, Disney movies, and other stories with happy endings. If you’re writing a Disney movie, there’s a good chance it’s going to be Cinderella. This is also commonly the arc of stories that follow theHero's Journey. While the hero's journey can be more complicated than a single arc, most fit the Cinderella arc. Examples of Cinderella story arcs: Disney’s Frozen Disney’s Up Dreamwork's How to Train Your Dragon Jane Eyre by Emily Bronte Disney’s Pinocchio Disney’s Aladdin 6. Oedipus (fall then rise then fall) The Oedipus arc is one of the most difficult structures to pull off, but it’s also one of the most highly read structures. Examples of the Oedipus story arc include: Moby Dick by Herman Melville Frankenstein by Mary Shelley And Then There Were None by Agatha Christie Lolita by Vladimir Nabakov The Sun Also Rises by Ernest Hemingway Gone with the Wind by Margaret Mitchell The Godfather by Mario Puzo Gone Girl by Gillian Flynn Hamlet by William Shakespeare Top 3 Best-Selling Story Arcs and Character Arcs You now know the six story arcs, but there are three that researchers have shown to be the arcs of more bestselling stories. Watch the video to learn what they are: How Story Arcs Fit Dramatic Structure Dramatic structure describes the elements of a story's movement, and each of the above story arcs incorporates the dramatic structure. At The Write Practice, we identify six plot points, orelements of plot: Exposition Inciting Incident Rising Action/Progressive Complications Dilemma Climax Denouement Note that many people include the falling action in their dramatic structure. I don’t include it because I believe the term “falling action” is misleading and really only appropriate for Freytag’s narrow definition of the tragic, Icarus structure, not a modern three-act story structure. You can read more about this in my falling action article. “ Dramatic structure describes the elements of a story's movement, and each of the above story arcs incorporates the dramatic structure. Master story arc and these elements with this post. Tweet this Tweet Here’s how these elements of dramatic structure fit into the Rags to Riches Story Arc: In this arc, the exposition has little to no movement and is primarily to acclimate the reader to the world of the story and its characters. The inciting incident begins the upward movement. The rising action describes the upward motion of the movement. The combination of the dilemma, where the character must make a critical choice, and climax, the moment of highest conflict and action, is the point of the peak—the make-or-break moment when things could either continue to improve or reverse. Last, the denouementor resolution wraps up the plot at the end of the story with one or two scenes of relative stability. Denouement means untying in French, and in these final moments, the loose ends of the plot are tied up. These components of dramatic structure can be found in every arc, and are part of what gives each arc their structure. Story Arcs Can Also Fit In A Three-Act Structure The Greek philosopher Aristotle gave the first recorded writing tip, saying that stories should have a beginning, a middle, and an end. Not the most profound advice ever, but over time, this evolved into the three-act structure, the most commonly used structure today (as opposed to the five-act structure, which you can learn more about here). The three-act structure combines perfectly with story arcs, allowing flexibility in terms of the arc you're trying to create while also providing a structure that flows with the reader's expectations. While this is not a law, commonly twenty-five percent of the arc is in the first act, fifty-percent is in the second act, and the rest of the story fits into the final twenty-five percent of the third act. More complicated arcs may actually have nine acts; in other words, they have three three-act structures. Longer series or epics, stories with arcs that combine to form more complicated patterns, may even have twelve, eighteen, or even twenty-seven acts. Story Arcs Measure Values A story’s rise and fall in value can be expressed generally in terms of “fortune,” but you can also get more specific by measuring a story’s movement based on six different story values. You’ve heard that your story needs conflict, but what does that actually mean? Because the kind of conflict stories need is (probably) not more fistfights and loud arguments (although, depending on the story, that might not hurt!). No, the kind of conflict your story needs is between one value and its opposite. Which values? A good story rises and falls on the spectrum of one of six different values, according to Shawn Coyne, the author of Story Grid. The six values, which follow Maslow’s Hierarchy of Human Need, are as follows: Physiological. The value of food, water, air, warmth, and rest. Life vs. death. Safety. The value of personal and group security. In story terms, life vs. a fate worse than death. Love/Belonging.The value of intimate relationships and friendships. Love vs. hate. Esteem.The value of personal prestige and accomplishment. Accomplishment vs. failure. Self-Actualization. The value of reaching your potential. Maturity vs. naiveté. Transcendence. The value of becoming more than yourself. Right vs. wrong. The rise and fall of these values dictate the rise and fall of the arc. For example, in an adventure story(Man in the Hole arc) set in space like the film Gravity, where the core value is physiological survival, you would measure the arc based on this life vs. death metric. Let’s break this arc down, analyzing the rise and fall of the life vs. death value throughout the key moments in the story: Spoiler Alert Exposition Dr. Ryan Stone (Sandra Bullock) and astronaut Matt Kowalksi (George Clooney) are on a space walk on the Hubble Space Telescope. Life vs. Death value measure: stable. Inciting Incident A missile strike causes a chain reaction of space debris that threatens to destroy much of the spacecraft around the planet. Life vs. Death value measure: a threat of death appears. Rising Action The space debris field begins to destroy spacecraft, including Stone and Kowalski’s ship, and they have to escape to the International Space Station. But the spacecraft ISS has been damaged, and they have to travel to the Chinese space station. While en route, Kowalski sacrifices his life to save Stone. Other space shenanigans happen until Stone is out of options for survival. Life vs. Death value measure: inching closer and closer toward likely death. Dilemma The sole survivor of the debris field and trapped in a Soyuz capsule without fuel, Stone has to choose whether to end her life or keep working to survive. Initially, she decides to turn off life support, but as she is losing consciousness, a vision of Kowalski gives her a final solution to reach the working Chinese re-entry capsule. Life vs. Death value measure: near death. Climax Stone reaches the Chinese re-entry capsule just as the space station is about to crash into the atmosphere. She unlocks from the station and is descending to Earth when a fire starts. After she lands safely in a lake, she has to evacuate the capsule immediately because of the smoke and nearly drowns before finally swimming to shore. Life vs. Death value measure: near death but survival becoming a slim possibility. Denouement Stone takes her first steps on Earth, thanking Kowalski, and as she watches the debris burn in Earth’s atmosphere, a helicopter flies overhead, signaling her rescue. Life vs. Death value measure: survival by a small margin! End Spoiler Alert Notice how the story moves from virtually no chance of death to death being almost a certainty to the resolution, where survival seems even more precious because of how close the protagonist came to death. The story moves the value from the positive form to the negative, and, depending on the value, back again. The story’s arc is created through this rise and fall movement. This same arc can be used to tell a love story, a performance story, or even a coming of age story. The arc stays the same, but the value being represented by the arc changes. Can You Have Multiple Story Arcs? Yes! In fact, most stories have multiple arcs. Most novels and films are made by combining three plots, three different value scales like those listed above: Main plot Internal plot Subplot Here's the key point: Each plot must have its own arc That means if you're writing an adventure story with a coming of age internal plot and a love story subplot — like The Alchemist, a quarter of the Harry Potter series, A Tale of Two Cities, The Da Vinci Code — then you will have three different arcs, one for each plot. Short stories only have one arc Conversely, if you're writing a short story, it only has one arc. And, usually, that arc will only have one or two movements. Sometimes there are separate character arcs for each POV character If you have multiple point of view characters or protagonists, you may have multiple story arcs—one for each character. This is one reason to avoid having multiple protagonists, because it complicates the story, sometimes creating more arcs than you the writer (and your readers!) can keep track of, especially if you're relatively inexperienced. (If you really think your story benefits from multiple protagonists, make sure itchecks off these three qualifications.) Epics, novel or film series, and episodic stories have many arcs Some stories will have more arcs, especially series, epics, or episodic stories. Soap operas, for example, often have so many arcs going that if you find yourself in the middle of a random episode, it will feel extremely chaotic. Often, episodic TV series will have one or two long arcs that last the whole series, while each episode has two or three smaller arcs. Sitcoms, according to Noah Charney, often follow this structure: Teaser (exposition) – one to three minutes Trouble: Story A (inciting incident) – minute three Trouble: Story B (inciting incident) – minute six The Muddle: Story A (rising action, dilemma) – minute nine The Muddle: Story B: (rising action, dilemma) – minute twelve The Triumph/Failure: Story A (climax) – minute thirteen The Triumph/Failure: Story B (climax) – minute fifteen The Kicker: Story A + B (denouement) – minute nineteen In sitcoms, these two or three (when a C story is involved) arcs are usually independent from the overall arc of the series. However, occasionally, one or both of these arcs involve a longer running master arc of the show, carrying on plots that have been previously established. A big reason for this is because, unlike in novels, we don't want our favorite characters in sitcoms to change that much from episode to episode. While main characters like Sheldon in the Big Bang Theory do undergo character transformation in the series, his major quirks remain consistent episode to episode. This also provides more opportunities for main characters to frequently fail as they attempt to achieve their episode goal, which provides more opportunities to drop jokes that pull at the character's steadfast personality traits and flaws. Likewise, the Ross + Rachel arc from the sitcom Friends contains many movements over hundreds of episodes built up in ten years, but it was all done in one arc, usually using the same structure as the one above. “ Understanding the six main story arcs and how they interact with a story's core value is crucial to becoming a great writer. These six tips will help you use this information to write great stories. Tweet this Tweet How to Use Story Arcs in Your Writing: 6 Writing Tips Now that you understand the six main arcs and how the shapes of stories interact with a story’s core value, how do you actually use this information to write great stories? Here are five writing tips for using story arcs in your writing: 1. Above all, make sure your story moves. It can move up, it can move down, it can move up and then down. But it must move, and that movement must begin early. A narrative that stays the same is not a story but an account of events. 2. In the first draft, don’t worry about matching your story to a particular arc. You may know what your arc is when you start writing, and you may not. Don’t worry too much about it. Just tell your story (and make sure that story moves). Don't get me wrong: you can use these arcs as templates, especially if your story idea is somewhat formless right now. But if you have a clear idea of your story, don't worry too much about whether it matches the arcs above. 3. In the first draft, do worry about finding your core value. While you don’t need to worry about finding the right shape of your story when you start writing, you should try to discover the core value, the y-axis that your story will move on. If you can discover your core value (see the list of six values above for the options), you will be much more equipped to making sure it moves the way it needs to. And while you may choose more than one value—perhaps a value for a subplot or the internal genre—if you try to move your story on too many values, it will become muddied and will be very hard to work with in your second draft. Above all, keep it simple. You can always write another book, but a book that’s trying to do too much can easily become unworkable. 4. Know your genre and form and consider your arcs accordingly. Different forms have different arc conventions. As we explored above, most novels and films have three arcs, most sitcoms have two arcs, and most short stories, on the other hand, have just one arc. Genres have different conventions as well. Fantasy stories and romance stories often follow the Cinderella arc. Science fiction stories often use the double man in a hole arc. Literary novels often use the Oedipus arc. Study your genre and form to know which arcs are most common. If you find that they're commonly using one specific arc, it doesn't mean you're forced to use it too. But it should inform how you approach your arc choices, if you decide to use a different one. 5. Write toward the dilemma. When you’re writing a first draft, you don’t need to know everything that’s going to happen. If you’re a pantser rather than a plotter, you might not know anything that happens. But the best thing you can do is to write toward the dilemma. The dilemma is the primary turning point in a story. It is the moment when a character is presented with a difficult choice that will determine his or her fate. This moment is usually found at the very bottom of a dip in a story arc or the very top of a peak. It will be followed almost immediately by the climax. If you can find that dilemma, you will have found your story. Everything in a story builds to the dilemma. For more on how to discover the dilemma in your story, read my full guide on literary crisis here. 6. In your second draft, find each arc and enhance them. While you don’t need to know the shape of your main story arc or sub-arcs in your first draft, after you finish your first draft and before you start your second draft, find your arc. What is its shape? How does it rise and fall? Does it fall enough? Does it rise enough? Is there enough movement? The purpose of the second draft is to enhance your arc, to make it more pronounced, smoother, and more effective. All Good Stories Have an Arc Good stories are about change, and thus all good stories have an arc. By finding the arc in your story, and making that arc better, you can give your readers what they want: meaning. All humans need meaning. While the world often can feel confusing, chaotic, and meaningless, the role of the storyteller is to help people find meaning in their lives. This is why humans love stories. And soon, it’s why readers will love your story. Which of the six story arcs is your favorite? Which story arc do you want to use for your next book?Let me know in the comments section. Have a story arc you'd love to turn into a bestselling novel, but not sure how to make it to “The End”? We've helped thousands of writers finish their books, and we'd love to help you, too. In 100 Day Book, you'll start and finish your book in—you guessed it—100 days. Check it out and join our next semester: Discover 100 Day Book PRACTICE Let’s practice using story arcs with a creative writing exercise. Here’s what we’re going to do: Choose one of the six story arcs: Rags to Riches, Riches to Rags, Man in a Hole, Icarus, Cinderella, Oedipus. Write a six-sentence story based on that arc using the six elements of dramatic structure: exposition, inciting incident, rising action, crisis, climax, and resolution. Then, set your timer for fifteen minutes and expand your six-sentence story as much as you can. When you finish, Write Practice Pro members can post in the Pro Practice Workshop here. Not a member yet? Check out how to join here. There's even a free option! We'd love to have you join our community of practicing writers. About the author Related Posts Joe Bunting Website Joe Bunting is an author and the leader of The Write Practice community. He is also the author of the new book Crowdsourcing Paris, a real life adventure story set in France. It was a #1 New Release on Amazon. Follow him on Instagram (@jhbunting). Want best-seller coaching?Book Joe here. Joe Bunting The Write Practice Needs Your Help Joe Bunting The 9 Types of Stories and How to Master Them Joe Bunting What Is an Epilogue? And Is It Okay to Use One in YOUR Book? Joe Bunting Maybe vs. May Be: The Simple Trick to Always Keep Them Straight 16 X1.2k FacebookEmail120 PinterestPrint1 Reddit 1.4k SHARES About the author Related Posts Joe Bunting Website Joe Bunting is an author and the leader of The Write Practice community. He is also the author of the new book Crowdsourcing Paris, a real life adventure story set in France. It was a #1 New Release on Amazon. Follow him on Instagram (@jhbunting). Want best-seller coaching?Book Joe here. Joe Bunting How to Shape a Story: The 6 Types of Story Arcs for Powerful Narratives Joe Bunting The Write Practice Needs Your Help Joe Bunting The 9 Types of Stories and How to Master Them Joe Bunting What Is an Epilogue? And Is It Okay to Use One in YOUR Book? Work with Joe Bunting? WSJ Bestselling author, founder of The Write Practice, and book coach with 14+ years experience. Joe Bunting specializes in working with Action, Adventure, Fantasy, Historical Fiction, How To, Literary Fiction, Memoir, Mystery, Nonfiction, Science Fiction, and Self Help books. Sound like a good fit for you? View Joe Bunting's Profile 2 Comments Sherry Green on August 8, 2024 at 11:54 am Thank you so much for this article and video. I have been having a difficult time unstanding this concept, wrongly thinking there was just one arc. You have clarified the elements of “arc” very well, and now I will be able to make my story move more appropriately. Reply 2. abdessamed gtumsila on May 15, 2025 at 7:25 am Thank you for sharing. 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15215	https://en.wikipedia.org/wiki/Maximum_principle	Jump to content Maximum principle Català Deutsch Español Français 한국어 Italiano עברית 日本語 Português Edit links From Wikipedia, the free encyclopedia Theorem in complex analysis This article describes the maximum principle in the theory of partial differential equations. For the maximum principle in optimal control theory, see Pontryagin's maximum principle. For the theorem in complex analysis, see Maximum modulus principle. In the mathematical fields of differential equations and geometric analysis, the maximum principle is one of the most useful and best known tools of study. Solutions of a differential inequality in a domain D satisfy the maximum principle if they achieve their maxima at the boundary of D. The maximum principle enables one to obtain information about solutions of differential equations without any explicit knowledge of the solutions themselves. In particular, the maximum principle is a useful tool in the numerical approximation of solutions of ordinary and partial differential equations and in the determination of bounds for the errors in such approximations. In a simple two-dimensional case, consider a function of two variables u(x,y) such that The weak maximum principle, in this setting, says that for any open precompact subset M of the domain of u, the maximum of u on the closure of M is achieved on the boundary of M. The strong maximum principle says that, unless u is a constant function, the maximum cannot also be achieved anywhere on M itself. Such statements give a striking qualitative picture of solutions of the given differential equation. Such a qualitative picture can be extended to many kinds of differential equations. In many situations, one can also use such maximum principles to draw precise quantitative conclusions about solutions of differential equations, such as control over the size of their gradient. There is no single or most general maximum principle which applies to all situations at once. In the field of convex optimization, there is an analogous statement which asserts that the maximum of a convex function on a compact convex set is attained on the boundary. Intuition [edit] A partial formulation of the strong maximum principle [edit] Here we consider the simplest case, although the same thinking can be extended to more general scenarios. Let M be an open subset of Euclidean space and let u be a C2 function on M such that where for each i and j between 1 and n, aij is a function on M with aij = aji. Fix some choice of x in M. According to the spectral theorem of linear algebra, all eigenvalues of the matrix [aij(x)] are real, and there is an orthonormal basis of ℝn consisting of eigenvectors. Denote the eigenvalues by λi and the corresponding eigenvectors by vi, for i from 1 to n. Then the differential equation, at the point x, can be rephrased as The essence of the maximum principle is the simple observation that if each eigenvalue is positive (which amounts to a certain formulation of "ellipticity" of the differential equation) then the above equation imposes a certain balancing of the directional second derivatives of the solution. In particular, if one of the directional second derivatives is negative, then another must be positive. At a hypothetical point where u is maximized, all directional second derivatives are automatically nonpositive, and the "balancing" represented by the above equation then requires all directional second derivatives to be identically zero. This elementary reasoning could be argued to represent an infinitesimal formulation of the strong maximum principle, which states, under some extra assumptions (such as the continuity of a), that u must be constant if there is a point of M where u is maximized. Note that the above reasoning is unaffected if one considers the more general partial differential equation since the added term is automatically zero at any hypothetical maximum point. The reasoning is also unaffected if one considers the more general condition in which one can even note the extra phenomena of having an outright contradiction if there is a strict inequality (> rather than ≥) in this condition at the hypothetical maximum point. This phenomenon is important in the formal proof of the classical weak maximum principle. Non-applicability of the strong maximum principle [edit] However, the above reasoning no longer applies if one considers the condition since now the "balancing" condition, as evaluated at a hypothetical maximum point of u, only says that a weighted average of manifestly nonpositive quantities is nonpositive. This is trivially true, and so one cannot draw any nontrivial conclusion from it. This is reflected by any number of concrete examples, such as the fact that and on any open region containing the origin, the function −x2−y2 certainly has a maximum. The classical weak maximum principle for linear elliptic PDE [edit] The essential idea [edit] Let M denote an open subset of Euclidean space. If a smooth function is maximized at a point p, then one automatically has: as a matrix inequality. One can view a partial differential equation as the imposition of an algebraic relation between the various derivatives of a function. So, if u is the solution of a partial differential equation, then it is possible that the above conditions on the first and second derivatives of u form a contradiction to this algebraic relation. This is the essence of the maximum principle. Clearly, the applicability of this idea depends strongly on the particular partial differential equation in question. For instance, if u solves the differential equation then it is clearly impossible to have and at any point of the domain. So, following the above observation, it is impossible for u to take on a maximum value. If, instead u solved the differential equation then one would not have such a contradiction, and the analysis given so far does not imply anything interesting. If u solved the differential equation then the same analysis would show that u cannot take on a minimum value. The possibility of such analysis is not even limited to partial differential equations. For instance, if is a function such that which is a sort of "non-local" differential equation, then the automatic strict positivity of the right-hand side shows, by the same analysis as above, that u cannot attain a maximum value. There are many methods to extend the applicability of this kind of analysis in various ways. For instance, if u is a harmonic function, then the above sort of contradiction does not directly occur, since the existence of a point p where is not in contradiction to the requirement everywhere. However, one could consider, for an arbitrary real number s, the function us defined by It is straightforward to see that By the above analysis, if then us cannot attain a maximum value. One might wish to consider the limit as s to 0 in order to conclude that u also cannot attain a maximum value. However, it is possible for the pointwise limit of a sequence of functions without maxima to have a maxima. Nonetheless, if M has a boundary such that M together with its boundary is compact, then supposing that u can be continuously extended to the boundary, it follows immediately that both u and us attain a maximum value on Since we have shown that us, as a function on M, does not have a maximum, it follows that the maximum point of us, for any s, is on By the sequential compactness of it follows that the maximum of u is attained on This is the weak maximum principle for harmonic functions. This does not, by itself, rule out the possibility that the maximum of u is also attained somewhere on M. That is the content of the "strong maximum principle," which requires further analysis. The use of the specific function above was very inessential. All that mattered was to have a function which extends continuously to the boundary and whose Laplacian is strictly positive. So we could have used, for instance, with the same effect. The classical strong maximum principle for linear elliptic PDE [edit] Summary of proof [edit] Let M be an open subset of Euclidean space. Let be a twice-differentiable function which attains its maximum value C. Suppose that Suppose that one can find (or prove the existence of): a compact subset Ω of M, with nonempty interior, such that u(x) < C for all x in the interior of Ω, and such that there exists x0 on the boundary of Ω with u(x0) = C. a continuous function which is twice-differentiable on the interior of Ω and with : and such that one has u + h ≤ C on the boundary of Ω with h(x0) = 0 Then L(u + h − C) ≥ 0 on Ω with u + h − C ≤ 0 on the boundary of Ω; according to the weak maximum principle, one has u + h − C ≤ 0 on Ω. This can be reorganized to say for all x in Ω. If one can make the choice of h so that the right-hand side has a manifestly positive nature, then this will provide a contradiction to the fact that x0 is a maximum point of u on M, so that its gradient must vanish. Proof [edit] The above "program" can be carried out. Choose Ω to be a spherical annulus; one selects its center xc to be a point closer to the closed set u−1(C) than to the closed set ∂M, and the outer radius R is selected to be the distance from this center to u−1(C); let x0 be a point on this latter set which realizes the distance. The inner radius ρ is arbitrary. Define Now the boundary of Ω consists of two spheres; on the outer sphere, one has h = 0; due to the selection of R, one has u ≤ C on this sphere, and so u + h − C ≤ 0 holds on this part of the boundary, together with the requirement h(x0) = 0. On the inner sphere, one has u < C. Due to the continuity of u and the compactness of the inner sphere, one can select δ > 0 such that u + δ < C. Since h is constant on this inner sphere, one can select ε > 0 such that u + h ≤ C on the inner sphere, and hence on the entire boundary of Ω. Direct calculation shows There are various conditions under which the right-hand side can be guaranteed to be nonnegative; see the statement of the theorem below. Lastly, note that the directional derivative of h at x0 along the inward-pointing radial line of the annulus is strictly positive. As described in the above summary, this will ensure that a directional derivative of u at x0 is nonzero, in contradiction to x0 being a maximum point of u on the open set M. Statement of the theorem [edit] The following is the statement of the theorem in the books of Morrey and Smoller, following the original statement of Hopf (1927): Let M be an open subset of Euclidean space ℝn. For each i and j between 1 and n, let aij and bi be continuous functions on M with aij = aji. Suppose that for all x in M, the symmetric matrix [aij] is positive-definite. If u is a nonconstant C2 function on M such that on M, then u does not attain a maximum value on M. The point of the continuity assumption is that continuous functions are bounded on compact sets, the relevant compact set here being the spherical annulus appearing in the proof. Furthermore, by the same principle, there is a number λ such that for all x in the annulus, the matrix [aij(x)] has all eigenvalues greater than or equal to λ. One then takes α, as appearing in the proof, to be large relative to these bounds. Evans's book has a slightly weaker formulation, in which there is assumed to be a positive number λ which is a lower bound of the eigenvalues of [aij] for all x in M. These continuity assumptions are clearly not the most general possible in order for the proof to work. For instance, the following is Gilbarg and Trudinger's statement of the theorem, following the same proof: Let M be an open subset of Euclidean space ℝn. For each i and j between 1 and n, let aij and bi be functions on M with aij = aji. Suppose that for all x in M, the symmetric matrix [aij] is positive-definite, and let λ(x) denote its smallest eigenvalue. Suppose that and are bounded functions on M for each i between 1 and n. If u is a nonconstant C2 function on M such that on M, then u does not attain a maximum value on M. One cannot naively extend these statements to the general second-order linear elliptic equation, as already seen in the one-dimensional case. For instance, the ordinary differential equation y″ + 2y = 0 has sinusoidal solutions, which certainly have interior maxima. This extends to the higher-dimensional case, where one often has solutions to "eigenfunction" equations Δu + cu = 0 which have interior maxima. The sign of c is relevant, as also seen in the one-dimensional case; for instance the solutions to y″ - 2y = 0 are exponentials, and the character of the maxima of such functions is quite different from that of sinusoidal functions. See also [edit] Maximum modulus principle Hopf maximum principle Notes [edit] ^ Protter, Murray H.; Weinberger, Hans Felix (1984). Maximum principles in differential equations. New York Berlin Heidelberg [etc.]: Springer. ISBN 978-3-540-96068-3. ^ Chapter 32 of Rockafellar (1970). References [edit] Research articles [edit] Calabi, E. An extension of E. Hopf's maximum principle with an application to Riemannian geometry. Duke Math. J. 25 (1958), 45–56. Cheng, S.Y.; Yau, S.T. Differential equations on Riemannian manifolds and their geometric applications. Comm. Pure Appl. Math. 28 (1975), no. 3, 333–354. Gidas, B.; Ni, Wei Ming; Nirenberg, L. Symmetry and related properties via the maximum principle. Comm. Math. Phys. 68 (1979), no. 3, 209–243. Gidas, B.; Ni, Wei Ming; Nirenberg, L. Symmetry of positive solutions of nonlinear elliptic equations in Rn. Mathematical analysis and applications, Part A, pp. 369–402, Adv. in Math. Suppl. Stud., 7a, Academic Press, New York-London, 1981. Hamilton, Richard S. Four-manifolds with positive curvature operator. J. Differential Geom. 24 (1986), no. 2, 153–179. Hopf, Eberhard. Elementare Bemerkungen Über die Lösungen partieller Differentialgleichungen zweiter Ordnung vom elliptischen Typus. Sitber. Preuss. Akad. Wiss. Berlin 19 (1927), 147-152. Hopf, Eberhard. A remark on linear elliptic differential equations of second order. Proc. Amer. Math. Soc. 3 (1952), 791–793. Nirenberg, Louis. A strong maximum principle for parabolic equations. Comm. Pure Appl. Math. 6 (1953), 167–177. Omori, Hideki. Isometric immersions of Riemannian manifolds. J. Math. Soc. Jpn. 19 (1967), 205–214. Yau, Shing Tung. Harmonic functions on complete Riemannian manifolds. Comm. Pure Appl. Math. 28 (1975), 201–228. Kreyberg, H. J. A. On the maximum principle of optimal control in economic processes, 1969 (Trondheim, NTH, Sosialøkonomisk institutt Textbooks [edit] Caffarelli, Luis A.; Xavier Cabre (1995). Fully Nonlinear Elliptic Equations. Providence, Rhode Island: American Mathematical Society. pp. 31–41. ISBN 0-8218-0437-5. Evans, Lawrence C. Partial differential equations. Second edition. Graduate Studies in Mathematics, 19. American Mathematical Society, Providence, RI, 2010. xxii+749 pp. ISBN 978-0-8218-4974-3 Friedman, Avner. Partial differential equations of parabolic type. Prentice-Hall, Inc., Englewood Cliffs, N.J. 1964 xiv+347 pp. Gilbarg, David; Trudinger, Neil S. Elliptic partial differential equations of second order. Reprint of the 1998 edition. Classics in Mathematics. Springer-Verlag, Berlin, 2001. xiv+517 pp. ISBN 3-540-41160-7 Ladyženskaja, O. A.; Solonnikov, V. A.; Uralʹceva, N. N. Linear and quasilinear equations of parabolic type. Translated from the Russian by S. Smith. Translations of Mathematical Monographs, Vol. 23 American Mathematical Society, Providence, R.I. 1968 xi+648 pp. Ladyzhenskaya, Olga A.; Ural'tseva, Nina N. Linear and quasilinear elliptic equations. Translated from the Russian by Scripta Technica, Inc. Translation editor: Leon Ehrenpreis. Academic Press, New York-London 1968 xviii+495 pp. Lieberman, Gary M. Second order parabolic differential equations. World Scientific Publishing Co., Inc., River Edge, NJ, 1996. xii+439 pp. ISBN 981-02-2883-X Morrey, Charles B., Jr. Multiple integrals in the calculus of variations. Reprint of the 1966 edition. Classics in Mathematics. Springer-Verlag, Berlin, 2008. x+506 pp. ISBN 978-3-540-69915-6 Protter, Murray H.; Weinberger, Hans F. Maximum principles in differential equations. Corrected reprint of the 1967 original. Springer-Verlag, New York, 1984. x+261 pp. ISBN 0-387-96068-6 Rockafellar, R. T. (1970). Convex analysis. Princeton: Princeton University Press. Smoller, Joel. Shock waves and reaction-diffusion equations. Second edition. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 258. Springer-Verlag, New York, 1994. xxiv+632 pp. ISBN 0-387-94259-9 Retrieved from " Categories: Harmonic functions Partial differential equations Mathematical principles Hidden categories: Articles with short description Short description is different from Wikidata
15216	https://mathspace.co/textbooks/syllabuses/Syllabus-465/topics/Topic-8720/subtopics/Subtopic-115511/	Multiplicity of a Root \| Grade 12 Math \| Ontario 12 Advanced Functions (MHF4U) \| Mathspace Book a Demo Topics Log inSign upBook a Demo CanadaON Grade 12 Multiplicity of a Root LessonPractice Lesson Share Outcomes 12F.C.1.5 Make connections, through investigation using graphing technology, between a polynomial function given in factored form and the x-intercepts of its graph, and sketch the graph of a polynomial function given in factored form using its key features What is Mathspace About Mathspace
15217	https://pmc.ncbi.nlm.nih.gov/articles/PMC10045406/	ACTH Stimulation Test for the Diagnosis of Secondary Adrenal Insufficiency: Light and Shadow - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Biomedicines . 2023 Mar 15;11(3):904. doi: 10.3390/biomedicines11030904 Search in PMC Search in PubMed View in NLM Catalog Add to search ACTH Stimulation Test for the Diagnosis of Secondary Adrenal Insufficiency: Light and Shadow Maria Francesca Birtolo Maria Francesca Birtolo 1 Department of Biomedical Sciences, Humanitas University, 20072 Pieve Emanuele, Italy 2 Endocrinology, Diabetology and Andrology Unit, IRCCS Humanitas Research Hospital, 20089 Rozzano, Italy Find articles by Maria Francesca Birtolo 1,2, Simone Antonini Simone Antonini 1 Department of Biomedical Sciences, Humanitas University, 20072 Pieve Emanuele, Italy 2 Endocrinology, Diabetology and Andrology Unit, IRCCS Humanitas Research Hospital, 20089 Rozzano, Italy Find articles by Simone Antonini 1,2, Andrea Saladino Andrea Saladino 3 Fondazione IRCCS Istituto Neurologico Carlo Besta, Unit of Neurosurgery, 20133 Milan, Italy Find articles by Andrea Saladino 3, Benedetta Zampetti Benedetta Zampetti 4 Endocrinology Department, ASST Grande Ospedale Metropolitano Niguarda, 20162 Milan, Italy Find articles by Benedetta Zampetti 4, Elisabetta Lavezzi Elisabetta Lavezzi 1 Department of Biomedical Sciences, Humanitas University, 20072 Pieve Emanuele, Italy 2 Endocrinology, Diabetology and Andrology Unit, IRCCS Humanitas Research Hospital, 20089 Rozzano, Italy Find articles by Elisabetta Lavezzi 1,2, Iacopo Chiodini Iacopo Chiodini 4 Endocrinology Department, ASST Grande Ospedale Metropolitano Niguarda, 20162 Milan, Italy 5 Department of Medical Biotechnology and Translational Medicine, University of Milan, 20122 Milan, Italy Find articles by Iacopo Chiodini 4,5, Gherardo Mazziotti Gherardo Mazziotti 1 Department of Biomedical Sciences, Humanitas University, 20072 Pieve Emanuele, Italy 2 Endocrinology, Diabetology and Andrology Unit, IRCCS Humanitas Research Hospital, 20089 Rozzano, Italy Find articles by Gherardo Mazziotti 1,2, Andrea G A Lania Andrea G A Lania 1 Department of Biomedical Sciences, Humanitas University, 20072 Pieve Emanuele, Italy 2 Endocrinology, Diabetology and Andrology Unit, IRCCS Humanitas Research Hospital, 20089 Rozzano, Italy Find articles by Andrea G A Lania 1,2,, Renato Cozzi Renato Cozzi 4 Endocrinology Department, ASST Grande Ospedale Metropolitano Niguarda, 20162 Milan, Italy Find articles by Renato Cozzi 4 Editors: Elzbieta Skowronska-Jozwiak, Krzysztof C Lewandowski Author information Article notes Copyright and License information 1 Department of Biomedical Sciences, Humanitas University, 20072 Pieve Emanuele, Italy 2 Endocrinology, Diabetology and Andrology Unit, IRCCS Humanitas Research Hospital, 20089 Rozzano, Italy 3 Fondazione IRCCS Istituto Neurologico Carlo Besta, Unit of Neurosurgery, 20133 Milan, Italy 4 Endocrinology Department, ASST Grande Ospedale Metropolitano Niguarda, 20162 Milan, Italy 5 Department of Medical Biotechnology and Translational Medicine, University of Milan, 20122 Milan, Italy Correspondence: andrea.lania@hunimed.eu Roles Elzbieta Skowronska-Jozwiak: Academic Editor Krzysztof C Lewandowski: Academic Editor Received 2023 Feb 27; Revised 2023 Mar 10; Accepted 2023 Mar 13; Collection date 2023 Mar. © 2023 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: PMC10045406 PMID: 36979883 Abstract Secondary Adrenal Insufficiency (SAI) is a condition characterized by inappropriately low ACTH secretion due to a disease or injury to the hypothalamus or the pituitary. The evaluation when suspected is often challenging for the non-specific symptoms, the rarity of the disease, and the pitfalls associated with laboratory tests. A prompt and correct diagnosis of SAI is essential because although an adequate hormonal replacement therapy could be lifesaving, inappropriate life-long therapy with steroids can be harmful. The gold standard test for assessing the hypothalamus-pituitary-adrenal axis (HPA) is the insulin tolerance test (ITT), but due to safety issues is not widely used. Conversely, the ACTH stimulation test is a safer and well-tolerated tool for SAI diagnosis. However, data about its diagnostic accuracy show great variability due to both technical and interpretative aspects, such as dose, route of administration, the timing of the test, and assay used for cortisol measurements. Consequently, the clinical background of the patient and the pretest probability of HPA axis impairment become of paramount importance. We aimed to summarize the recent literature evidence in the conduction and interpretation of the ACTH stimulation test for the diagnosis of SAI to provide updated insights on its correct use in clinical practice. Keywords: ACTH Test, adrenal insufficiency, pituitary surgery, new cutoffs 1. Introduction Adrenal insufficiency (AI) can be defined as a defective function of the hypothalamus-pituitary-adrenal (HPA) axis, leading to a non-adequate response of cortisol levels to stress factors. It is a chronic disorder, with an overall prevalence estimated at 300 cases/million people/year . The HPA axis is composed of several components, starting from the secretion of corticotroph-releasing hormone (CRH) in the hypothalamus, which stimulates the transcription of pro-opio-melanocortin (POMC) in the anterior pituitary gland; one of the fragments of POMC is the adrenocorticotrophic hormone (ACTH), which has a direct effect on the adrenal glands, stimulating the secretion of cortisol. This hormone is fundamental for life both being the key element of the stress response in humans and having pleiotropic effects on other organs, such as the action as a counter-regulatory hormone on glycemic homeostasis. Based on its etiology AI can be classified as primary adrenal insufficiency (PAI), due to a reduced function of the adrenal glands, secondary adrenal insufficiency (SAI), caused by impaired production of adrenocorticotrophic hormone (ACTH) in the pituitary gland, and tertiary adrenal insufficiency (TAI), due to prolonged exposure to glucocorticoid (GC) therapy at a supraphysiological dosage which can lead to subsequent persistent suppression of the axis. All these patients are at risk of an adrenal crisis, which is a severe and acute episode of AI, characterized by weakness, hypotension, nausea, vomiting, confusion, and typical biochemical alterations (hyponatremia, hyperkalemia, low fasting glucose). It can be life-threatening and requires prompt treatment and urgent hospital care, with its pillar being the administration of intravenous hydrocortisone (HC), and monitoring. From a recent paper published in 2023 and based on Australian records, the admissions for an adrenal crisis are rising in the last two decades, with a stable amount of PAI and an increase in SAI; the authors theorized that the cause of the progressive overcome of adrenal crisis due to SAI rather than PAI are increasing the diagnosis of pituitary adenomas and the use of cancer immunotherapy, which seldom can trigger a hypophysitis . An adequate and precocious GC replacement therapy in AI is vital being the key element to preventing an adrenal crisis. Conversely, GC replacement as chronic therapy, in the long run, has also demonstrated detrimental effects, such as an increased incidence of arterial hypertension, type 2 diabetes mellitus, vertebral fractures, reduced bone mineral density (BMD), and increased body mass index (BMI) [3,4]. Considering the dramatic importance but also the risks of GC replacement therapy, the correct diagnosis of AI is pivotal. The diagnosis of PAI is usually overt and is based on clinical signs and symptoms (often the ones of an adrenal crisis), possibly combined with skin hyperpigmentation, very low serum cortisol in the morning, and high ACTH and since confirmatory stimulation tests are not always required, most centers do not routinely perform them. Conversely, in SAI the diagnosis is often challenging. SAI is mainly caused by pituitary diseases, such as macroadenomas and hypophysitis, or therapeutic procedures around the pituitary region, such as trans-nasal-sphenoidal (TNS) surgery and irradiation. The increased rate of diagnosis of pituitary adenomas in the last decades and the minor rate of complications of TNS surgeries compared to open surgeries have increased the amount of patients undergoing these procedures, hence increasing the amount of patients at risk for SAI. Among patients undergoing pituitary surgery in a retrospective study on 519 patients the incidence of new hypopituitarism after 2 months was 3.1%, all requiring hydrocortisone replacement . Oshino et al. demonstrated that the incidence of SAI in macroadenomas correlates with Knosp grade, male sex, and admission for pituitary apoplexy . In addition, cancer therapies, and in particular immunomodulating therapies, became important risk factors in SAI occurrence. In this concern, a recent meta-analysis, among a treated population of 30,014 individuals, 3.2% of the patients had hypophysitis and 0.42% had a following diagnosis of hypopituitarism, with the axes most affected being corticotropic, thyrotrophic, and gonadotrophic . The ACTH stimulation test has been validated against the “gold standard” insulin tolerance test (ITT) to be a reliable tool in the investigation of patients with suspected AI regardless of the underlying cause since it is safer, more convenient, and has no contraindications . The diagnostic value of the ACTH Test in SAI lies in the assumption that chronic ACTH deficiency leads to adrenal atrophy and a consequent impaired response to stimulation by corticotropin analogs . The aim of this paper is to collect and evaluate the most recent available literature on the diagnosis of SAI, with a focus on the role of the ACTH test, in order to provide updated insights on its correct use in clinical practice. We are going to discuss the indication for the test, the patients that should and should not undergo the procedure, the timing to plan it, the preparation of the compound, and how to perform and interpret it. We will also consider the differences between the gold standard test, standard dose ACTH test, and low dose ACTH test in terms of specificity and sensitivity. In conclusion, we will be discussing the new cortisol cut-offs to adopt when the laboratory has a monoclonal method for cortisol determination. 2. Methodology Using the PubMed database, we conducted a literature narrative review on the ACTH Stimulation Test for the diagnosis of SAI to detect the utility and limits of this tool in clinical practice. We identified relevant studies using the following search terms: “ACTH stimulation test”, “cosyntropin stimulation test”, or “short Synacthen Test (SST)” combined with “secondary adrenal insufficiency” or “secondary hypoadrenalism”, “pituitary surgery”, “pituitary disease”, or “trans-nasal-sphenoidal surgery”. We reviewed all the papers on the ACTH stimulation test written in English and then we selected only those on adrenal insufficiency secondary to pituitary surgery or pituitary disease. 3. Patients to Be Tested and Timing for Testing According to the Endocrine Society Clinical Practice Guideline for the management of hypopituitarism in adults, in patients undergoing pituitary surgery a morning cortisol sample < 3 μg/dL (82.7 nmol/L) is indicative of AI, while a cortisol level > 15 μg/dL (413 nmol/L) likely excludes an AI diagnosis. Hence, the ACTH stimulation test is suggested in patients with cortisol levels between 3 and 15 μg dL (82.7–413 nmol/L) to exclude an insufficiency of the HPA axis. The guidelines recommend performing the ACTH stimulation test 6 weeks after surgery and at least 18 h after the last HC dose has been administered . Despite data about the timing of postoperative testing are not uniform, with marked variability across centers, from 24 h to 48 h and from 1 to 6 weeks., the most reported timing for testing after pituitary surgery is 6 weeks, while further testing beyond this time point is not regarded as routine and performed in few centers . A recent study demonstrated that the basal cortisol during the ACTH test correlates with the values 30 min after the stimulus and cortisol at baseline ≤ 4.49 μg/dL (124 nmol/L) predicts the failure to reach the peak at the test in 100% of the patients, while a cortisol ≥ 11.38 μg/dL (314 nmol/L) predicts a successful response. In this study, an immunoassay using monoclonal antibodies was used to detect the cortisol levels, and the local laboratory established as a cutoff for the ACTH test failure a cortisol value ≤ 16.3 μg/dL (450 nmol/L) at 30 min after the injection of the corticotropin analog. As discussed in the following paragraph, the interpretation of cortisol value, both at the baseline and stimulated, depends on the assay used for the measurement. According to the results of this study, if the cortisol evaluation is made with immunoassays using monoclonal antibodies, the grey range of 2016 guidelines potentially tighten even more allowing to identify of a narrower population to test . In this context, the most recent studies stressed the importance of selecting patients to test according to the pretest probability of SAI that should be quantified by evaluating the patients’ clinical background and the morning cortisol levels [14,15]. Bioletto et al. proposed an integrated score for the prediction of SAI when morning cortisol is in the grey zone based on morning cortisol levels, sex, and the presence of at least three other pituitary deficits, trying to narrow even more the patients that need to be tested . The morning cortisol level is the most studied predictive factor of SAI, but clinicians should always bear in mind the diurnal variation of cortisol for its correct interpretation and that if the blood test is not performed early in the morning the cortisol value at the baseline has no more a role in the assessment of adrenal function. 4. How to Perform the Test Several different protocols for ACTH test are reported in the literature with a lack of consensus on the most reliable one, mainly in terms of the dose of corticotropin used, the route of administration, the duration of the test procedure, and the timing of blood sampling. According to the standard protocol, cortisol levels should be measured 30 min before and 60 min after intravenous (iv) administration of 250 μg of corticotropin as bolus injection . The low-dose ACTH test is a variation of the standard protocol based on the administration of 1 μg corticotropin for adrenal stimulation. Despite it is still a matter of debate, it seems that the low dose provides the same or even superior diagnostic accuracy compared to the standard one in the evaluation of SAI [9,17,18]. Among these studies, the meta-analysis conducted by Kazlauskaite et al. is relevant since the authors took into account only the published studies on SAI and constructed receiver operator characteristic (ROC) curves using patient-level data from each study and then merged results to create summary ROC curves, adjusting for study size and cortisol assay method. This patient-level meta-analysis found that the diagnostic value of the low-dose ACTH test was superior to the standard-dose test (AUC 0.94 and 0.85, respectively; p< 0.001) in the diagnosis of SAI . The same result was found by a more recent study, which concludes that the low-dose ACTH test should be preferred to the 250 μg ACTH test because has a better sensitivity than the standard-dose one since the latter test has more false positives due to supra-physiological adrenal stimulation . In this context, Daidoh et al. found that the minimal dose of ACTH-inducing peak cortisol is 0.5 μg of cosyntropin analog administrated iv, suggesting that even 1 μg leads to supra-physiological adrenal stimulations . To perform the ACTH test synthetic corticotropin analogs are used and the two commercially available are supplied as 250 μg/mL ampoules that must be diluted to obtain 1 μg . Therefore, technical remarks arose on the accuracy and reproducibility of making up low-dose tests in terms of dilution methods. In this context, a British survey in 2012 identified 14 different dilution methods of preparing the low-dose test which differed in the amount of the sample utilized for the initial dilution, the volume and the type of diluent, and the number of dilution steps . Based on these findings Cross et al. addressed this crucial clinical issue first identifying through an international survey the ten most used dilution methods and then reporting the accuracy, reproducibility, and reliability of each method. This study showed high variability in both inter-method and intra-method and mostly that the final actual delivered dose of corticotropin analog with the different methods was inadequate in all cases, ranging from 0.16 μg to 0.81 μg, up to seven-fold less than required in some cases. The least variable methods were two: the injection of 1 mL of the 250 μg/mL ampoules into a 500 mL bag of 5% dextrose and the administration of a volume of 2 mL of the resulting solution and the injection of 1 mL of the 250 μg/mL ampoules into 50 mL bag of 0.9% sodium chloride solution (saline), the transfer of 1 mL of this solution to 10 mL syringe containing 9 mL of saline and the administration of a volume of 2 mL. The first one was also the most accurate method with a final real delivered dose close to the desired 1 μg (0.79 to 0.84 μg), but it was the only one to use 5% dextrose as diluent instead of saline and therefore needs to be investigated further before making any recommendation . Being 1 μg an extremely little amount other concerns are about the losses of corticotropin analog when pushed through the devices used for the administration. It was demonstrated that the extent of the loss increases proportionally to the length of the device . A recent study reported that in the low-dose test administrating the corticotropin analog via 2.5-cm plastic tubes or directly intravenous ensures an adequate quantity of corticotropin analog and provides equal cortisol responses . As regards the route of administration most studies were conducted using the IV route, but corticotropin analogs requested for the IV test are not readily available in all countries and although data are limited and mostly obtained in pediatric cohorts the test conducted with intramuscular (IM) long-acting adrenocorticotropic hormone seems to be safe, effective, and reliable in cosintropin the diagnosis of SAI [26,27]. In clinical practice, the ACTH test is performed mostly in the morning , but very limited and conflicting data exist on the impact of the time of the day on the test outcome. Previous studies reported that the low-dose test is more prone to abnormal results if conducted in the afternoon [24,29], while more recent analyses showed for the standard dose test no difference in cortisol responses at different times of the day . It is not clear why in the low-dose test the time of the day affects the outcomes test and this aspect should be further investigated in large prospective studies since the interpretation of the ACTH test should be based on the peak stimulated serum cortisol regardless of the basal cortisol value which is influenced by diurnal variation. Finally, among the technical aspects, there is still some ongoing debate over the timing of blood sampling after the stimulus. Indeed, since several studies demonstrated that the cortisol peak occurs between 20 and 35 min after the administration of the low dose of corticotropin analog, in clinical practice the cortisol is mostly measured only at baseline and 30 min after the ACTH injection . Nevertheless, the latest evidence suggests that although the majority of patients peak 30 min after the stimulus, measuring cortisol both 30 and 60 min is necessary to reduce the risk of false positives and overdiagnosis of adrenal insufficiency. Indeed, from 10 to 24% of patients, according to several studies, classified as AI based on the 30-min cortisol level are reclassified as adrenally sufficient if the 60-min cortisol level is performed [31,32]. 5. Alternative Tests for the Assessment of HPA Axis Although less often used, other tests to assess the preserved function of the HPA axis and exclude secondary adrenal insufficiency are available. As an example, the glucagon stimulation test (GST) has been described in some papers as an effective option to evaluate this axis. A Russian article published in 2019 used the glucagon stimulation test in 28 patients that underwent craniospinal irradiation for neoplasms of the central nervous system without pituitary localization and in 10 healthy controls, performing both the glucagon and the insulin tolerance test to evaluate whether the HPA axis’ function was preserved or not. A peak of cortisol during the glucagon stimulation test above 18.1 μg/dL (500 nmol/L) ruled out secondary adrenal insufficiency (100% sensitivity), while a peak below 12.2 μg/dL (340 nmol/L) was diagnostic. However, several patients had discordant results (32.2%) between the glucagon stimulation test and insulin tolerance test, and 25% experienced adverse events during the procedure . Another more recent study investigated more deeply the safety profile of the test on a wider population, describing 43.2% of patients with adverse effects during the test, mainly characterized by nausea (29.6%), vomiting (27.1%), abdominal cramps (18.5%), but no patient needed to stop the procedure. However, these adverse events were significantly more common in elder patients (p = 0.01) . A very recent study, published in 2023 by Yalovitsky et al. compared the glucagon stimulation test with the low-dose ACTH test for the diagnosis of SAI in 120 pediatric patients with short stature from Israel. The data collected showed that the glucagon stimulation test had poorer performance than the low-dose ACTH test. The best performance was obtained using the cut-off of 11.18 μg/dL (320 nmol/L), but specificity and sensitivity were respectively 83% and 56% . The overnight single-dose metyrapone test has also been used to test patients for the HPA axis function and to rule out secondary adrenal insufficiency. It was very common mainly in the nineties and it consisted of the administration of a single dose administration of 30 mg/kg oral metyrapone and the evaluation of cortisol and 11-deoxycortisol using radioimmunoassay on blood samples obtained the next morning between 08:00 and 09:30 a.m. . This test is different from the stimulation tests (ITT, ACTH test, and GST) since it works by inducing a negative feedback stimulus instead of direct stimulation of the hypothalamus and/or pituitary. Some authors suggest its use in doubt cases. Indeed, overnight single-dose metyrapone was found to be more sensitive than the ITT and standard dose stimulation test in detecting subtle degrees of HPA axis insufficiency . This test has the advantage of not requiring parenteral administration of any drug, but data on its diagnostic accuracy are controversial, some side effects, such as nausea, vomiting, and dizziness, have also been reported and finally the need to have 11-deoxycortisol assays available, has progressively reduced its application . At the moment, no clear data can support the routinary use of the glucagon stimulation test or metyrapone test in place of the ITT and/or the ACTH stimulation test. No comparative studies are available to compare the performance of the glucagon stimulation test to the ACTH test in adults, while data on pediatric populations, although limited to few studies and small cohorts, do not suggest preferring it. 6. How to Interpret the Results: Old and New Cortisol Cutoffs According to the current guidelines for the diagnosis and management of both PAI and SAI, a peak cortisol level > 18.1 μg/dL (500 nmol/L) at 30 or 60 min after corticotropin analogs administration indicates an appropriate cortisol secretion and therefore excludes the diagnosis of AI [5,11]. This current cortisol cutoff threshold for the diagnosis of AI is based on the old immunological methods that are no longer in use in clinical practice. Indeed, traditionally cortisol cutoff levels across the ACTH test were established with immunoassays using polyclonal antibodies (Elecsys Cortisol generation I) , but these assays were characterized by low specificity because they had cross-reactivity with other serum steroids [39,40]. Over the last years, newer-generation assays with greater specificity for cortisol have been developed and have replaced the old assays in almost all the centers. The new tools available to detect cortisol are the immunoassays utilizing monoclonal antibodies, such as Elecsys Cortisol generation II and Beckman Access Cortisol, and the Liquid chromatography–tandem mass spectrometry (LC-MS/MS), which is a non-antibody structural assay highly specific for cortisol [41,42,43,44]. Serum cortisol concentrations measured with the newer assays when compared to those measured with the older ones are sharply lower by 20 up to 36%, according to different studies and assays [45,46,47]. Based on these findings several studies have been published with the aim of identifying the new ACTH-stimulated cortisol threshold values using new cortisol assays; we have summarized the main results of these studies in Table 1. Most of the studies were conducted using the standard dose ACTH test, in small cohorts and without differentiating the cause underlying the AI. Despite these limits, all the studies found a lower cutoff for the 30-min cortisol post-ACTH stimulus ranging from 12.6 μg/dL to 14.6 μg/dL, using the Elecsys Cortisol generation II assay, and from 13.3 μg/dL to 14.9 μg/dL, using the Beckman Access Cortisol assay. All these studies agree on the importance to use the new cutoffs in the presence of the new assays to accurately diagnose AI and to minimize the risk of lifelong glucocorticoid replacement therapy in patients with well-functioning HPA axis. Table 1. Summary of the available studies in the literature on new cortisol cutoffs. \| Author, Year \| ACTH Dose \| Cohort \| N. of Patients \| 30-Min Cortisol Post-ACTH Stimulus \| :---: :---: \| Elecsys I \| Elecsys II \| Access \| LC-MS/MS \| \| Raverot et al., 2016 \| NA \| PAI/SAI \| 109 \| 18.1 μg/dL (500 nmol/L \| 13.6 μg/dL (374 nmol/L) \| NA \| NA \| \| Kline et al., 2017 \| 250 μg/1 μg \| NA \| 56 \| 18.1 μg/dL (500 nmol/L) \| 12.6 μg/dL (350 nmol/L) \| NA \| NA \| \| Ueland et al., 2018 \| 250 μg \| PAI/SAI \| 94 \| 18.1 μg/dL (500 nmol/L) \| NA \| NA \| 14.9 μg/dL (412 nmol/L) \| \| Grassi et al., 2020 \| 250 μg/1 μg \| PAI/SAI \| 30 \| 18.1 μg/dL (500 nmol/L) \| 12.7 μg/dL (351 nmol/L) \| NA \| 13.3 μg/dL (368 nmol/L) \| \| Javorsky et al., 2021 \| 250 μg \| PAI/SAI/TAI \| 110 \| 18.1 μg/dL (500 nmol/L) \| 14.6 μg/dL (402 nmol/L) \| 14.8 μg/dL (408 nmol/L) \| 14.5 μg/dL (400 nmol/L) \| \| Zha et al., 2022 \| 250 μg \| NA \| 50 \| 18.1 μg/dL (500 nmol/L) \| NA \| 13.2 μg/dL (364 nmol/L) \| NA \| \| Husni et al., 2022 \| 250 μg \| healthy volunteers \| 63 \| 18.1 μg/dL (500 nmol/L) \| 15.7 μg/dL (433 nmol/) \| NA \| NA \| Open in a new tab NA: Not applicable; PAI: primary adrenal insufficiency; SAI: secondary adrenal insufficiency; TAI: tertiary adrenal insufficiency; In the study were included the results of 11-low dose ACTH tests and 45 standard dose ACTH tests, respectively. If cortisol values are detected with the new assays also the baseline cortisol cutoffs suggestive of AI should be revisited and validated in wide cohorts. Moreover, indications for the management of patients with discrepancies in the basal cortisol value and the stimulated one should be better addressed. 7. When to Repeat the Test? There are currently few studies on the appropriate frequency of repeat dynamic testing and the eventual likelihood of recovery of HPA axis function, both in patients undergoing pituitary surgery and in those exposed to supraphysiological doses of GC. Pofi et al. demonstrated that HPA axis recovery can occur as late as 9 to 12 months after TNS demonstrating the need for periodic reassessment of patients who initially failed to respond to the ACTH test . The same group found that a 30-min cortisol level above or below 12.7 μg /dL (350 nmol/L) across the first ACTH Test after TNS best predicts HPA axis recovery, enabling early identification of subgroups more likely to recover and hence patients to be prioritized for retesting . 8. Conclusions A summary of the latest evidence on the conduction and interpretation of the ACTH stimulation test for the diagnosis of SAI is provided in Figure 1. Both standard and low-dose ACTH tests have good diagnostic accuracy in the diagnosis of SAI. The low-dose ACTH test appears to have a better sensitivity than the standard dose one, which probably provides an overstimulation compared to the physiology. Nevertheless, the low-dose test should be performed only in tertiary centers where personnel have the expertise and knowledge of the multiple steps required for the correct preparation and administration. The main issue is the dilution of the 250 μg/mL ampoules, which often leads to a final real delivered dose lower than the desired 1 μg. For the correct interpretation of the results, clinicians must be aware of the assay used for cortisol assessment in their center. If the newer assays are used, the new cortisol cutoffs should be considered for diagnosis, but further studies on both standard and low-dose tests are urgently advocated to validate in a large cohort these redefined cutoffs. Considering the controversies and limits of the test clinicians should select patients to be tested according to the pre-test probability of AI and always correlate the results with clinical findings bearing in mind the dramatic importance but also the risks of the GC replacement therapy. Figure 1. Open in a new tab Summary of the latest evidence on the conduction and interpretation of the ACTH test for the diagnosis of SAI. The range of new cutoffs proposed in the figure are from studies conducted using Elecsys Cortisol generation II assay. Abbreviations Abbreviations used in the manuscript: PAI (primary adrenal insufficiency), SAI (secondary adrenal insufficiency), TAI (tertiary adrenal insufficiency), CRH (corticotroph-releasing hormone), POMC (pro-opio-melanocortin), ACTH (adrenocorticotrophic hormone), GC (glucocorticoid), TNS (trans-nasal-sphenoidal), ITT (insulin tolerance test), SST (short synacthen test), GST (glucagon stimulation test). Author Contributions Conceptualization, M.F.B. and S.A.; methodology, M.F.B. and S.A.; writing—original draft preparation, M.F.B. and S.A.; writing—review and editing, R.C and A.G.A.L.; supervision, A.S., B.Z., E.L., I.C., G.M., R.C. and A.G.A.L. All authors have read and agreed to the published version of the manuscript. Institutional Review Board Statement Not applicable. Informed Consent Statement Not applicable. 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Articles from Biomedicines are provided here courtesy of Multidisciplinary Digital Publishing Institute (MDPI) ACTIONS View on publisher site PDF (521.0 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Methodology 3. Patients to Be Tested and Timing for Testing 4. How to Perform the Test 5. Alternative Tests for the Assessment of HPA Axis 6. How to Interpret the Results: Old and New Cortisol Cutoffs 7. When to Repeat the Test? 8. Conclusions Abbreviations Author Contributions Institutional Review Board Statement Informed Consent Statement Data Availability Statement Conflicts of Interest Funding Statement Footnotes References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15218	https://www.haoduoyun.cc/book/hjb/shuxue/11s.shtml	沪教版高二数学上册电子课本\|教材\|教科书目录 - 好多电子课本网回顶部中小学古诗文\|电子课本\|汉语拼音\|加入主页中学文言文好多电子课本网好好学习，天天向上！--->少年强，则中国强好多课本人教版外研版译林版沪教版北京版古诗文文言文当前位置：好多电子课本网>沪教版> 高二数学上册教材目录(上海教育出版社) 深入探索科学教材数学教科书课本教育沪教版高二数学上册(义务教育教科书)课本目录如下：关注公众号免费下载PDF电子课本封面/目录 ...... 1 第7章数列与数学归纳法 ...... 5 一数列 ...... 6 7.1 数列 ...... 6 7.2 等差数列 ...... 11 等差数列及其通项公式 ...... 11 等差中项 ...... 12 等差数列的前n项和 ...... 17 7.3 等比数列 ...... 20 等比数列及其通项公式 ...... 20 等比中项 ...... 21 等比数列的前n项和 ...... 25 二数学归纳法 ...... 30 7.4 数学归纳法 ...... 30 7.5 数学归纳法的应用 ...... 33 7.6 归纳-猜想-论证 ...... 35 三数列的极限 ...... 38 7.7 数列的极限 ...... 38 数列的极限运算法则 ...... 41 7.8 无穷等比数列各项的和 ...... 45 阅读材料雪花曲线 ...... 50 第8章平面向量的坐标表示 ...... 55 8.1 向量的坐标表示及其运算 ...... 56 向量的定比分点和中点坐标公式 ...... 59 8.2 向量的数量积 ...... 60 向量的夹角 ...... 61 向量的数量积的坐标表示 ...... 64 8.3 平面向量的分解定理 ...... 66 8.4 向量的应用 ...... 68 第9章矩阵和行列式初步 ...... 73 一矩阵 ...... 74 9.1 矩阵的概念 ...... 74 9.2 矩阵的运算 ...... 77 二行列式 ...... 89 9.3 二阶行列式 ...... 89 9.4 三阶行列式 ...... 95 余子式 ...... 98 三元一次方程组的行列式解法 ...... 101 第10章算法初步 ...... 109 10.1 算法的概念 ...... 110 10.2 程序框图 ...... 114 10.3 计算机语句和算法程序 ...... 122 附录 ...... 138 1 向量 ...... 138 2 向量的加减法(平行四边形法则和三角形法则) ...... 141 3 实数与向量的乘积 ...... 144 古诗词推荐 1小学李白的古诗 2小学杜甫的古诗 3小学白居易的古诗 4小学苏轼的古诗 5小学王维的古诗 6初中李白的古诗 7初中杜甫的古诗词 8初中白居易的古诗词 9初中苏轼的古诗词 10初中王维的古诗词赞助商链接部编版语文上册电子课本 1一年级语文上册 2二年级语文上册 3三年级语文上册 4四年级语文上册 5五年级语文上册 6六年级语文上册 7七年级语文上册 8八年级语文上册 9九年级语文上册部编版道德与法治上册电子课本 1一年级道德与法治上册 2二年级道德与法治上册 3七年级道德与法治上册 4八年级道德与法治上册 5九年级道德与法治上册好多电子课本网、电子课本、电子教科书、电子课本下载、电子教材、教科书、教材网、小学英语课文翻译手机版Copyright © 2016-2020 www.haoduoyun.cc. All Rights Reserved. 苏ICP备16003350-2号好多电子课本网提供最新的在线沪教版电子课本阅读学习，涵盖小学、初中和高中电子教材，包括语文、数学、英语、物理、化学、生物、历史、政治、地理、科学等新版电子教科书。
15219	https://www.kembaraxtra.com/law/kembaraxtra-case-law-nash-v-inman-1908-ca	KembaraXtra- Case Law -Nash v Inman (1908) CA - kembara Xtra kembara Xtra Introduction Earth Gemstones Medicine Finance Law Psychology Technology Travel The World Who We Are Get In Touch more... Introduction Earth Gemstones Medicine Finance Law Psychology Technology Travel The World Who We Are Get In Touch LAW KembaraXtra- Case Law -Nash v Inman (1908) CA 4/26/2025 0 Comments KembaraXtra- Case Law -Nash v Inman (1908) CA This case concerns the legal definition of "necessaries" in the context of a minor's contract. Understanding this case hinges on the concept of a minor's capacity to enter into legally binding contracts. I. Key Facts: Plaintiff: A Saville Row tailor. Defendant: An undergraduate at Cambridge University (a minor). Contract: Sale of clothing described as "extravagant and ridiculous" and "an extravagant number of waistcoats". Price: £145 (a substantial sum at the time). Evidence: The defendant's father testified that the son already had sufficient clothing. II. Legal Issue: Was the clothing supplied "necessaries" for which the minor is liable? This is the central question. The law recognizes that minors can be held liable for contracts for goods or services considered "necessary" for their well-being. However, the definition of "necessaries" is flexible and fact-specific. III. Key Arguments & Reasoning: Plaintiff's Argument: The plaintiff argued that the clothing, while perhaps extravagant in style, was still necessary in that it provided the defendant with clothing. Defendant's Argument: The defendant, through his father's testimony, argued that the clothing was not necessary, as he already possessed sufficient clothing for his needs. The extravagance of the clothing was a key factor here. The court emphasized that the issue wasn't simply about whether clothing was necessary, but whether this specific clothing was necessary. Court's Holding: The Court of Appeal held that the clothing supplied was not considered "necessaries." The extravagance and excess clearly exceeded what was reasonably necessary for a Cambridge undergraduate. The court stressed that "necessaries" are judged based on the minor's station in life and their existing provisions. IV. Legal Principle Established: This case clarifies the limitations of a minor's liability for "necessaries." Simply because something is generally considered necessary (e.g., clothing) does not automatically make a specific item or quantity of that item "necessary" for a minor. The court will consider: The minor's station in life: A wealthy minor may have different needs than a less wealthy minor. Existing provisions: The minor's current supply of the goods or services in question. Reasonableness: Whether the quantity and quality of the goods or services are reasonable in light of the minor's needs and circumstances. Extreme extravagance or excess will disqualify the goods as "necessaries." V. Study Questions: 1. Define "necessaries" in the context of minors' contracts. How is this definition applied in Nash v Inman? 2. What factors did the court consider in determining whether the clothing was "necessary"? Why were these factors important? 3. How does this case illustrate the flexible nature of the "necessaries" doctrine? 4. What would the outcome likely be if the defendant had been a homeless minor with no existing clothing? By carefully considering these points, you should achieve a solid understanding of Nash v Inman and its implications for the law surrounding minors and contracts. Remember that this case highlights the need for a contextual approach when determining whether goods or services constitute "necessaries" for a minor. 0 Comments Leave a Reply. --------------Author Learn the fact about law Archives August 2025 July 2025 June 2025 May 2025 April 2025 February 2025 January 2025 December 2024 November 2024 October 2024 September 2024 August 2024 July 2024 June 2024 May 2024 March 2024 December 2023 November 2023 Categories All Case Law Contract Law Intellectual Property International Law Land Law Legal Terms Medical Law RSS Feed Psychology Technology Travel The World Who We Are Get In Touch Powered by Create your own unique website with customizable templates.Get Started
15220	https://math.stackexchange.com/questions/1295138/what-is-the-mathematical-distinction-between-closed-and-open-sets	general topology - What is the mathematical distinction between closed and open sets? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What is the mathematical distinction between closed and open sets? Ask Question Asked 10 years, 4 months ago Modified4 years, 7 months ago Viewed 46k times This question shows research effort; it is useful and clear 31 Save this question. Show activity on this post. If you wanted me to spell out the difference between closed and open sets, the best I could do is to draw you a circle one with dotted circumference the other with continuous circumference. Or I would give you an example with a set of numbers (1,2) vs [1,2] and tell you which bracket signifies open or closed. But in many theorems the author is dead set about using either closed or open sets. What is the strict mathematical difference that distinguish between the two sets and signifies the importance for such distinction? Can someone demonstrate with an example where using closed set for a theorem associated with open set would cause some sort of a problem? general-topology Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited May 23, 2015 at 9:15 Your neighbor TodorovichYour neighbor Todorovich asked May 23, 2015 at 9:09 Your neighbor TodorovichYour neighbor Todorovich 8,629 8 8 gold badges 56 56 silver badges 108 108 bronze badges 8 3 Important question. But a finite set of numbers should be written {1,2,3,4,5}. Round and square brackets are for intervals like (1,5) or [1,5].Stanley –Stanley 2015-05-23 09:12:15 +00:00 Commented May 23, 2015 at 9:12 6 Closed sets contain their supremum and infimum but open sets do not contain them alkabary –alkabary 2015-05-23 09:12:34 +00:00 Commented May 23, 2015 at 9:12 3 Open sets have a little bit of space around each point; one reason they're important is because differentiation is usually defined only for functions defined on an open set. On the other hand, closed sets are closed under limits of nets, which is why they are important. Miraculously, open sets are precisely the complements of closed sets, and vice versa. Does that answer your question?goblin GONE –goblin GONE 2015-05-23 09:15:46 +00:00 Commented May 23, 2015 at 9:15 4 Closed sets contain all their boundary points, open sets contain none of theirs. In a metric space, closed sets can be so sparse they contain no metric balls at all, unlike open sets which have "enough space" that there is a ball around every point. As for facts that use the adjectives open/closed - they're almost surely using the adjectives for a reason. One wouldn't e.g. state a fact for open sets if it were also true for closed sets for the same reasons. Literally pick any such claim and consider it an exercise to see what goes wrong if you alter its adjectives "open" and "closed."anon –anon 2015-05-23 09:36:04 +00:00 Commented May 23, 2015 at 9:36 2 I am not sure I understand your comment Ms. Tank.anon –anon 2015-05-23 09:52:06 +00:00 Commented May 23, 2015 at 9:52 \|Show 3 more comments 8 Answers 8 Sorted by: Reset to default This answer is useful 24 Save this answer. Show activity on this post. Let's talk about real numbers here, rather than general metric or topological spaces. This way we don't need notions of Cauchy sequences or open balls, and can talk in more familiar terms. We define that a set X⊂R is open if for every x∈X there exists some interval (x−ϵ,x+ϵ) with ϵ>0 such that this interval is also fully contained in X. An example is the inverval (0,1)={x∈R:0<x<1}. Note that this is an infinite set, because there are infinitely many points in it. If you choose a number a∈(0,1) and let ϵ=min{a−0,1−a} then we can guarantee that (a−ϵ,a+ϵ)⊂X. The set X is open. A set X is defined to be closed if and only if its complement R−X is open. For example, [0,1] is closed because R−[0,1]=(−∞,0)∪(1,∞) is open. It gets interesting when you realise that sets can be both open and closed, or neither. This is a case where strict adherence to the definition is important. The empty set ∅ is both open and closed and so is R. Why? The set [1,2) is neither open nor closed. Why? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2015 at 20:57 answered May 23, 2015 at 9:42 StanleyStanley 3,254 19 19 silver badges 24 24 bronze badges 3 2 Sorry to revive this, but it might be better to do this than open a new question! I wanted to ask if choosing ϵ=min{a−0,1−a} is valid to show the openness(?) of the set X=(0,1). If we use that in the definition of open set, then (a−(1−a),a+1−a)=(2 a−1,1)⊈X, right?OFRBG –OFRBG 2016-01-19 00:29:16 +00:00 Commented Jan 19, 2016 at 0:29 No problem OFRBG! But why do you think (2 a−1,1) is not contained in (0,1)? Just have to show that if 1−a<a−0 then 0<2 a−1<1.Stanley –Stanley 2016-01-19 04:05:06 +00:00 Commented Jan 19, 2016 at 4:05 1 Sorry! I misused the notation. I meant to type the set [2 a−1,1]. It seems as if the definition of the epsilon is including the bounds. Am I missing something?OFRBG –OFRBG 2016-01-19 04:34:18 +00:00 Commented Jan 19, 2016 at 4:34 Add a comment\| This answer is useful 21 Save this answer. Show activity on this post. The question you asked has answers at various levels of sophistication. You really want to be thinking about intervals rather than finite sets of points, and think also about regions in the plane. Here is one way of looking at things. A closed set is one which contains all its limit points - so when you are working with closed sets you can be confident abut taking limits because you know that the limit points exist. An open set can be thought of as one in which every point is an interior point, at least that is a useful guide when you are thinking about sets on the line or in the plane. So the idea is that if you pick a point in an open set, you have enough points close to it in the set to work with, and that there are no points close to it which are outside the set (every point in an open set has a neighbourhood of points wholly within the open set). This cashes out particularly when thinking about continuous functions - the classic epsilon-delta definition is essentially talking about the relationships between nearby points. These ideas can be considerably generalised and made precise as part of the machinery of topology. Note it is possible to have a set which is both open and closed -- the whole of the real line for example -- or to have a set that is neither open nor closed, such as the set of all rational numbers. The basic properties of closed and open sets are not the only useful things about them. For example a closed and bounded subset of the real line has a useful property called "compactness", which enables us to reduce some infinite problems to finite ones, and hence get better results. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2015 at 14:20 hmakholm left over Monica 292k 25 25 gold badges 441 441 silver badges 706 706 bronze badges answered May 23, 2015 at 9:29 Mark BennetMark Bennet 102k 14 14 gold badges 119 119 silver badges 232 232 bronze badges 1 1 @Henning Makholm - if I could give votes for edits, you'd get one!Mark Bennet –Mark Bennet 2015-05-23 14:28:32 +00:00 Commented May 23, 2015 at 14:28 Add a comment\| This answer is useful 10 Save this answer. Show activity on this post. A real-valued continuous function on a closed bounded subset of R n always has a maximum and a minimum. This is not true for open subsets. Example: the function 1 x(1−x) has no maximum on (0,1). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Feb 10, 2021 at 22:27 answered May 23, 2015 at 9:25 BernardBernard 180k 10 10 gold badges 75 75 silver badges 182 182 bronze badges 1 1 Yay for the EVT.bjb568 –bjb568 2015-05-23 14:55:42 +00:00 Commented May 23, 2015 at 14:55 Add a comment\| This answer is useful 6 Save this answer. Show activity on this post. Can someone demonstrate with an example where using closed set for a theorem associated with open set would cause some sort of a problem? An arbitrary intersection of closed sets is closed, but you can't say the same about open sets: for an intersection of open sets to be open, it has to be finite (counterexample, ⋂n∈N(−1 n,1 n)={0} is not open). Similarly, an arbitrary union of open sets is open, but an arbitrary union of closed sets is not necessarily closed (counterexample, for any non-closed X, ⋃x∈X{x}=X is not closed). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 11, 2018 at 9:00 answered May 23, 2015 at 9:37 fkraiemfkraiem 3,179 1 1 gold badge 14 14 silver badges 12 12 bronze badges 1 1 Can't believe this incorrect answer got five upvotes. :D Fixed.fkraiem –fkraiem 2018-12-11 09:01:21 +00:00 Commented Dec 11, 2018 at 9:01 Add a comment\| This answer is useful 5 Save this answer. Show activity on this post. The question you've asked is difficult to answer because it is broad. An open set is a lot more concrete and intuitive in a metric space, where it is defined as some set U so that for every point x in U, there is a neighborhood (or an open ball) around x that is completely contained in U. Think about picking any number in (0,1). I assure you, that no matter what number you pick, I can find some teeny tiny neighborhood around that number that is still inside (0,1). However, in [0,1], if you pick x=0, then I simply cannot find a neighborhood around 0 that is completely contained in [0,1] (since I'd always have to include some teeny tiny negative number in that neighborhood). An open set in a general topological space is a little bit harder to grasp for a beginner, though. Let X be a set, and τ is a set of sets. Then τ is a topology if: X and ∅ are in τ, Any union of sets in τ are also in τ, Any finite intersection of sets in τ are also in τ. Then we define everything in τ to be open sets. In either event, a closed set is a set whose complement is open. (A much simpler definition :) It's also important to note that sets can be open, closed, neither, or both! (0,1), [0,1], [0,1), are open, closed, and neither (respectively). For an example that is both open and closed, consider the set of complex numbers. Its complement is the empty set, which is open (see (1)), and so the complex numbers are closed. But we also know C is open in τ by (1). So it is both open and closed. The idea to take home here is that the concept of an open set can be anything. It's a relative term. It's like how time is relative, and depends on your frame of reference. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2015 at 10:31 answered May 23, 2015 at 9:49 Sultan of SwingSultan of Swing 1,540 1 1 gold badge 14 14 silver badges 25 25 bronze badges 1 If a downvoter notices any errors or mistakes, or has something to add, letting me know in the comments is much appreciated :)Sultan of Swing –Sultan of Swing 2015-05-23 11:05:18 +00:00 Commented May 23, 2015 at 11:05 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The easiest way to think of it is that in an open set you can go arbitrarily "close" to some point but never quite get the point in question without leaving the set itself. In a closed set there is no such point and all points you can go arbitrarily close to is within the set hence you can reach the point without leaving the set. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 23, 2015 at 9:29 Zelos MalumZelos Malum 6,686 2 2 gold badges 16 16 silver badges 36 36 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Try to define a closed interval by using open sets. This refers to the way we are normally defining open and closed intervals. So consider that we normally define B(R)=σ(K), with K={[a,b),a,b∈R,−∞<a,b<∞}. Now note that B(R) is a σ-algebra (by normal way of verifying this) which contains all open and closed intervals, cause it is closed under countably many intersections and unions. This follows cause we can write (a,b)=⋃∞n=1[a+1 n,b) for all open intervals with a,b∈R,a<b and [a,b]=⋂∞n=1[a,b+1 n) for all closed intervals with a,b∈R,a≤b. Now try to define the same set where you are only using open intervals and try to construct closed intervals from there. The problem then is that you have boundary points which are not inside this set. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 23, 2015 at 9:33 AlwinAlwin 125 3 3 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. (I am assuming everything inside a metric space) consider any cauchy-sequence inside a closed set...then this will always converge where this is not true for any open set.. same time for any point in the open set there always exist a neighbour hood which is properly contained in the open set..which is in general not true for any closed set... theses two things are very important for proving a lots of theorem. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited May 23, 2015 at 9:41 answered May 23, 2015 at 9:16 Anubhav MukherjeeAnubhav Mukherjee 6,748 1 1 gold badge 19 19 silver badges 38 38 bronze badges 1 Intuitively thé second thing given by Anubhav can be viewed as follows: closed set contains points such that if you are on one of them there is some direction such that if you move towards this direction by a so small movement (so small as you want) you Will fall down outside the set. However, for an open set, if you are on any of its points and you take any direction you can always move by a so small movement such you remains inside the set during the movement as well as on the destination point!Idris Addou –Idris Addou 2015-05-23 09:30:05 +00:00 Commented May 23, 2015 at 9:30 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions general-topology See similar questions with these tags. 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15221	https://www.mathplanet.com/education/algebra-1/formulating-linear-equations/parallel-and-perpendicular-lines	Parallel & perpendicular lines - Mathplanet Mathplanet A free service from Mattecentrum Other tools Mathplanet Matteboken Matteboken (Arabic) Webmatematik (Danish) MenuAlgebra 1/Formulating linear equations/ Parallel and perpendicular lines Do excercisesShow all 2 exercises Choose the parallel line Perpendicular line If two non-vertical lines that are in the same plane has the same slope, then they are said to be parallel. Two parallel lines won't ever intersect. If two non-vertical lines in the same plane intersect at a right angle then they are said to be perpendicular. Horizontal and vertical lines are perpendicular to each other i.e. the axes of the coordinate plane. Example Compare the slope of the perpendicular lines The slope of the red line: m 1=−3−1 2−(−2)=−4 4=−1 m 1=−3−1 2−(−2)=−4 4=−1 The slope of the blue line m 2=2−(−2)3−(−1)=4 4=1 m 2=2−(−2)3−(−1)=4 4=1 The slopes of two perpendicular lines are negative reciprocals. The product of the slopes of two perpendicular lines is -1 since m⋅−1 m=−1,w h e r e m 1=m a n d m 2=−1 m m⋅−1 m=−1,w h e r e m 1=m a n d m 2=−1 m Video lesson Are these two line parallel? You need to accept marketing cookies to play the video. Accept marketing cookies Do excercisesShow all 2 exercises Choose the parallel line Perpendicular line More classes on this subject Algebra 1 Formulating linear equations: Writing linear equations using the slope-intercept formAlgebra 1 Formulating linear equations: Writing linear equations using the point-slope form and the standard form Next Chapter: FORMULATING LINEAR EQUATIONS –Scatter plots and linear models Search Math Playground All courses Math PlaygroundOverview Programming All courses ProgrammingOverview Pre-Algebra All courses Pre-AlgebraOverview Algebra 1 All courses Algebra 1Overview Discovering expressions, equations and functions Algebra 1 Discovering expressions, equations and functions Overview Expressions and variables Operations in the right order Composing expressions Composing equations and inequalities Representing functions as rules and graphs Exploring real numbers Algebra 1 Exploring real numbers Overview Integers and rational numbers Calculating with real numbers The Distributive property Square roots How to solve linear equations Algebra 1 How to solve linear equations Overview Properties of equalities Fundamentals in solving equations in one or more steps Ratios and proportions and how to solve them Similar figures Calculating with percentages Visualizing linear functions Algebra 1 Visualizing linear functions Overview The coordinate plane Linear equations in the coordinate plane The slope of a linear function The slope-intercept form of a linear equation Formulating linear equations Algebra 1 Formulating linear equations Overview Writing linear equations using the slope-intercept form Writing linear equations using the point-slope form and the standard form Parallel and perpendicular lines Scatter plots and linear models About Mathplanet Cookie settings) Linear inequalities Algebra 1 Linear inequalities Overview Solving linear inequalities Solving compound inequalities Solving absolute value equations and inequalities Linear inequalities in two variables Systems of linear equations and inequalities Algebra 1 Systems of linear equations and inequalities Overview Graphing linear systems The substitution method for solving linear systems The elimination method for solving linear systems Systems of linear inequalities Exponents and exponential functions Algebra 1 Exponents and exponential functions Overview Properties of exponents Scientific notation Exponential growth functions Factoring and polynomials Algebra 1 Factoring and polynomials Overview Monomials and polynomials Special products of polynomials Polynomial equations in factored form Quadratic equations Algebra 1 Quadratic equations Overview Use graphing to solve quadratic equations Completing the square The quadratic formula Radical expressions Algebra 1 Radical expressions Overview The graph of a radical function Simplify radical expressions Radical equations The Pythagorean Theorem The distance and midpoint formulas Rational expressions Algebra 1 Rational expressions Overview Simplify rational expression Multiply rational expressions Division of polynomials Add and subtract rational expressions Solving rational equations Algebra 2 All courses Algebra 2Overview Geometry All courses GeometryOverview SAT All courses SATOverview ACT All courses ACTOverview Mathplanet is licensed byCreative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. 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15222	https://kaybeebio.com/blog/gibberellic-acid-ga-unveiling-the-growth-elixir-for-plants/?srsltid=AfmBOoqfAEzE277QJ7gFL3fy69H2QQWYfPefS4JQOFaLVhQJXJHaEGno	Gibberellic Acid (GA) - Unveiling the Growth Elixir for Plants Home Products Bio Pesticides Products Neem Based Products Granule Based Products Stickers & Spreaders Plant Growth Regulators About Us Why Us Blogs Careers Contact Us Powered by Translate 0 No products in the cart. Search × Home\|Blog\|Gibberellic Acid (GA) – Unveiling the Growth Elixir for Plants By Admin September 23, 2023 Gibberellic Acid (GA) – Unveiling the Growth Elixir for Plants Gibberellic acids are naturally occurring plant hormones that are used as plant growth regulators to stimulate both cell division and elongation that affect leaves and stems. These hormones are synthesized within plants, primarily in young and actively growing tissues. Plant growth regulator plays a vital role in regulating various aspects of plant growth and development. It’s an essential compound that influences processes such as seed germination, stem elongation, flowering, and fruit development. In this blog, we’ll delve into the history, origin, importance, advantages, disadvantages, and recommended usage of GA in the world of agriculture. History of GA The history of GA dates back to the 1920s when Japanese plant pathologists discovered its existence while studying a fungal disease that caused excessive stem elongation in rice plants. This disease, known as “bakanae” or “foolish seedling,” was eventually traced back to a fungus that produced Gibberellic Acid. This revelation marked the beginning of our understanding of this remarkable plant growth regulator. Origin of GA GA is naturally produced by certain fungi, bacteria, and plants. It’s present in various parts of plants, including seeds, stems, leaves, and roots. While some plants can synthesize GA on their own, others rely on external sources of this growth regulator to stimulate specific growth processes. Why Is Gibberellic Acid Plant Growth Regulator (GA) Important? Gibberellic Acid for plants is of paramount importance in the world of agriculture and horticulture for several reasons: Promotes Germination: GA plays a crucial role in breaking seed dormancy, kickstarting the germination process, and ensuring that seeds sprout and grow into healthy plants. Enhances Stem Elongation: It helps plants elongate their stems, a crucial factor for crops such as wheat and rice, which need to reach a certain height to maximize yield. Facilitates Flowering and Fruit Development: GA induces flowering in many plant species and contributes to the development of larger and more bountiful fruits. Improves Fruit Quality: It can enhance the size, color, and overall quality of fruits, making them more appealing to consumers. Advantages Of Gibberellic Acid (Plant Growth Regulator) In the realm of agriculture, Gibberellic Acid (GA) is like a secret ingredient that boosts plant growth and enhances crop yield. While we’ve touched upon its significance, let’s delve deeper into the advantages of using GA and explore why it’s a staple for many growers. Improved Germination Rate and Uniformity: One of GA’s primary roles is breaking seed dormancy. When seeds remain dormant, they do not sprout, and farmers face uneven germination. GA acts as a wake-up call for these dormant seeds, encouraging them to sprout uniformly. This uniformity is critical for ensuring a strong and consistent stand of plants in the field. Boosted Stem Elongation: For certain crops, particularly grains like wheat and rice, stem elongation is crucial. Taller plants often mean higher yields. GA can be employed strategically to induce stem elongation, helping crops reach the desired height for optimal sunlight exposure and easier harvesting. Enhanced Flowering: In many horticultural crops, flowering is directly linked to fruit production. GA serves as a floral promoter, encouraging plants to blossom profusely. This not only increases the visual appeal of ornamental plants but also results in higher fruit sets for crops like apples, citrus fruits, and grapes. Quality Fruit Development: The size, color, and overall quality of fruits can significantly impact their market value. GA steps in as a quality enhancer, ensuring that fruits develop to their full potential. Larger, more vibrant fruits not only fetch better prices but also attract consumers looking for top-quality produce. Among the myriad of GA products available, Nova GA stands out as an exceptional choice for environmentally conscious growers. This organic solution offers several benefits that align perfectly with sustainable farming practices Disadvantages of Overusing Gibberellic Acid (Plant Growth Regulator) While GA offers a plethora of benefits, it’s essential to recognize that more isn’t always better. Overusing GA can have adverse effects on plants and the environment, as well as impact your bottom line: Excessive Stem Elongation: An overdose of GA can lead to plants growing excessively tall and becoming structurally weak. To prevent these towering plants from collapsing under their weight, additional support may be necessary, incurring additional costs. Inhibited Flowering: Paradoxically, using too much GA can sometimes hinder flowering, especially in certain plant species. This can lead to reduced fruit production, defeating the purpose of using GA to enhance yield. Also Read:An Overview Of Best Plant Growth Regulators In India Careful Considerations When Using GA (Plant Growth Promoter Products) To maximize the benefits of GA plant growth promoter products while avoiding their potential downsides, consider these careful considerations: Dosage Control: The appropriate dosage of GA depends on its concentration and the specific crop you’re cultivating. Always follow recommended guidelines and calculate the required amount accurately to prevent overuse. Application Timing: Timing is everything when it comes to applying GA. Different crops and growth stages require precise timing for optimal results. Ensure that you apply GA when it will have the most significant positive impact on your plants. Environmental Factors: Environmental conditions such as temperature, humidity, and soil type can influence the effectiveness of GA. Be aware of these factors and make necessary adjustments in your application strategy. Dosage Guidelines for (GA) Plant Growth Promoter Products While dosage varies depending on the concentration of GA and the crop, here’s a general guideline for using GA at a concentration of 10 ppm: Mix approximately 2 ml of GA per liter of water for application. However, always refer to the specific product instructions and consult with agricultural experts for crop-specific recommendations. Conclusion In conclusion, Gibberellic Acid (GA) is an invaluable tool in modern agriculture and horticulture. Its capacity to enhance germination, promote stem elongation, encourage flowering and fruit development, and improve fruit quality makes it a go-to solution for growers seeking higher yields and better-quality crops. Among GA products, Nova GA stands out as an organic and sustainable choice. Its eco-friendly nature, support for robust germination, enhanced fruit set, and versatility make it a trusted companion for farmers dedicated to sustainable and high-yield crop production. As we navigate the complex landscape of plant growth regulators, it’s vital to strike a balance between reaping the benefits of GA and avoiding its potential pitfalls. By adhering to recommended guidelines, considering timing and environmental factors, and carefully controlling dosage, growers can harness the growth-enhancing properties of GA while ensuring the health and vitality of their crops and the environment they thrive. In the ever-evolving field of agriculture, GA remains a key player, unlocking the full growth potential of plants and ensuring a bountiful harvest for generations to come. Share: ##### Leave a Comment Cancel reply Your email address will not be published.Required fields are marked Your Name Email (will not be shared) [x] Save my name, email, and website in this browser for the next time I comment. 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15223	https://www.mathworks.com/help/optim/ug/constrained-nonlinear-optimization-algorithms.html	Skip to content Main Content Constrained Nonlinear Optimization Algorithms Constrained Optimization Definition Constrained minimization is the problem of finding a vector x that is a local minimum to a scalar function f(x) subject to constraints on the allowable x: such that one or more of the following holds: c(x) ≤ 0, ceq(x) = 0, A·x ≤ b, Aeq·x = beq, l ≤ x ≤ u. There are even more constraints used in semi-infinite programming; see fseminf Problem Formulation and Algorithm. fmincon Trust Region Reflective Algorithm Trust-Region Methods for Nonlinear Minimization Many of the methods used in Optimization Toolbox™ solvers are based on trust regions, a simple yet powerful concept in optimization. To understand the trust-region approach to optimization, consider the unconstrained minimization problem, minimize f(x), where the function takes vector arguments and returns scalars. Suppose you are at a point x in n-space and you want to improve, i.e., move to a point with a lower function value. The basic idea is to approximate f with a simpler function q, which reasonably reflects the behavior of function f in a neighborhood N around the point x. This neighborhood is the trust region. A trial step s is computed by minimizing (or approximately minimizing) over N. This is the trust-region subproblem, \| \| \| --- \| \| \| (1) \| The current point is updated to be x + s if f(x + s) < f(x); otherwise, the current point remains unchanged and N, the region of trust, is shrunk and the trial step computation is repeated. The key questions in defining a specific trust-region approach to minimizing f(x) are how to choose and compute the approximation q (defined at the current point x), how to choose and modify the trust region N, and how accurately to solve the trust-region subproblem. This section focuses on the unconstrained problem. Later sections discuss additional complications due to the presence of constraints on the variables. In the standard trust-region method (), the quadratic approximation q is defined by the first two terms of the Taylor approximation to F at x; the neighborhood N is usually spherical or ellipsoidal in shape. Mathematically the trust-region subproblem is typically stated \| \| \| --- \| \| \| (2) \| where g is the gradient of f at the current point x, H is the Hessian matrix (the symmetric matrix of second derivatives), D is a diagonal scaling matrix, Δ is a positive scalar, and ‖ . ‖ is the 2-norm. Good algorithms exist for solving Equation 2 (see ); such algorithms typically involve the computation of all eigenvalues of H and a Newton process applied to the secular equation Such algorithms provide an accurate solution to Equation 2. However, they require time proportional to several factorizations of H. Therefore, for large-scale problems a different approach is needed. Several approximation and heuristic strategies, based on Equation 2, have been proposed in the literature ( and ). The approximation approach followed in Optimization Toolbox solvers is to restrict the trust-region subproblem to a two-dimensional subspace S ( and ). Once the subspace S has been computed, the work to solve Equation 2 is trivial even if full eigenvalue/eigenvector information is needed (since in the subspace, the problem is only two-dimensional). The dominant work has now shifted to the determination of the subspace. The two-dimensional subspace S is determined with the aid of a preconditioned conjugate gradient process described below. The solver defines S as the linear space spanned by s1 and s2, where s1 is in the direction of the gradient g, and s2 is either an approximate Newton direction, i.e., a solution to \| \| \| --- \| \| \| (3) \| or a direction of negative curvature, \| \| \| --- \| \| \| (4) \| The philosophy behind this choice of S is to force global convergence (via the steepest descent direction or negative curvature direction) and achieve fast local convergence (via the Newton step, when it exists). A sketch of unconstrained minimization using trust-region ideas is now easy to give: Formulate the two-dimensional trust-region subproblem. Solve Equation 2 to determine the trial step s. If f(x + s) < f(x), then x = x + s. Adjust Δ. These four steps are repeated until convergence. The trust-region dimension Δ is adjusted according to standard rules. In particular, it is decreased if the trial step is not accepted, i.e., f(x + s) ≥ f(x). See and for a discussion of this aspect. Optimization Toolbox solvers treat a few important special cases of f with specialized functions: nonlinear least-squares, quadratic functions, and linear least-squares. However, the underlying algorithmic ideas are the same as for the general case. These special cases are discussed in later sections. Preconditioned Conjugate Gradient Method A popular way to solve large, symmetric, positive definite systems of linear equations Hp = –g is the method of Preconditioned Conjugate Gradients (PCG). This iterative approach requires the ability to calculate matrix-vector products of the form H·v where v is an arbitrary vector. The symmetric positive definite matrix M is a preconditioner for H. That is, M = C2, where C–1HC–1 is a well-conditioned matrix or a matrix with clustered eigenvalues. In a minimization context, you can assume that the Hessian matrix H is symmetric. However, H is guaranteed to be positive definite only in the neighborhood of a strong minimizer. Algorithm PCG exits when it encounters a direction of negative (or zero) curvature, that is, dTHd ≤ 0. The PCG output direction p is either a direction of negative curvature or an approximate solution to the Newton system Hp = –g. In either case, p helps to define the two-dimensional subspace used in the trust-region approach discussed in Trust-Region Methods for Nonlinear Minimization. Linear Equality Constraints Linear constraints complicate the situation described for unconstrained minimization. However, the underlying ideas described previously can be carried through in a clean and efficient way. The trust-region methods in Optimization Toolbox solvers generate strictly feasible iterates. The general linear equality constrained minimization problem can be written \| \| \| --- \| \| \| (5) \| where A is an m-by-n matrix (m ≤ n). Some Optimization Toolbox solvers preprocess A to remove strict linear dependencies using a technique based on the LU factorization of AT . Here A is assumed to be of rank m. The method used to solve Equation 5 differs from the unconstrained approach in two significant ways. First, an initial feasible point x0 is computed, using a sparse least-squares step, so that Ax0 = b. Second, Algorithm PCG is replaced with Reduced Preconditioned Conjugate Gradients (RPCG), see , in order to compute an approximate reduced Newton step (or a direction of negative curvature in the null space of A). The key linear algebra step involves solving systems of the form \| \| \| --- \| \| \| (6) \| where approximates A (small nonzeros of A are set to zero provided rank is not lost) and C is a sparse symmetric positive-definite approximation to H, i.e., C = H. See for more details. Box Constraints The box constrained problem is of the form \| \| \| --- \| \| \| (7) \| where l is a vector of lower bounds, and u is a vector of upper bounds. Some (or all) of the components of l can be equal to –∞ and some (or all) of the components of u can be equal to ∞. The method generates a sequence of strictly feasible points. Two techniques are used to maintain feasibility while achieving robust convergence behavior. First, a scaled modified Newton step replaces the unconstrained Newton step (to define the two-dimensional subspace S). Second, reflections are used to increase the step size. The scaled modified Newton step arises from examining the Kuhn-Tucker necessary conditions for Equation 7, \| \| \| --- \| \| \| (8) \| where and the vector v(x) is defined below, for each 1 ≤ i ≤ n: If gi < 0 and ui < ∞ then vi = xi – ui If gi ≥ 0 and li > –∞ then vi = xi – li If gi < 0 and ui = ∞ then vi = –1 If gi ≥ 0 and li = –∞ then vi = 1 The nonlinear system Equation 8 is not differentiable everywhere. Nondifferentiability occurs when vi = 0. You can avoid such points by maintaining strict feasibility, i.e., restricting l < x < u. The scaled modified Newton step sk for the nonlinear system of equations given by Equation 8 is defined as the solution to the linear system \| \| \| --- \| \| \| (9) \| at the kth iteration, where \| \| \| --- \| \| \| (10) \| and \| \| \| --- \| \| \| (11) \| Here Jv plays the role of the Jacobian of \|v\|. Each diagonal component of the diagonal matrix Jv equals 0, –1, or 1. If all the components of l and u are finite, Jv = diag(sign(g)). At a point where gi = 0, vi might not be differentiable. is defined at such a point. Nondifferentiability of this type is not a cause for concern because, for such a component, it is not significant which value vi takes. Further, \|vi\| will still be discontinuous at this point, but the function \|vi\|·gi is continuous. Second, reflections are used to increase the step size. A (single) reflection step is defined as follows. Given a step p that intersects a bound constraint, consider the first bound constraint crossed by p; assume it is the ith bound constraint (either the ith upper or ith lower bound). Then the reflection step pR = p except in the ith component, where pRi = –pi. fmincon Active Set Algorithm Introduction In constrained optimization, the general aim is to transform the problem into an easier subproblem that can then be solved and used as the basis of an iterative process. A characteristic of a large class of early methods is the translation of the constrained problem to a basic unconstrained problem by using a penalty function for constraints that are near or beyond the constraint boundary. In this way the constrained problem is solved using a sequence of parametrized unconstrained optimizations, which in the limit (of the sequence) converge to the constrained problem. These methods are now considered relatively inefficient and have been replaced by methods that have focused on the solution of the Karush-Kuhn-Tucker (KKT) equations. The KKT equations are necessary conditions for optimality for a constrained optimization problem. If the problem is a so-called convex programming problem, that is, f(x) and Gi(x), i = 1,...,m, are convex functions, then the KKT equations are both necessary and sufficient for a global solution point. Referring to GP (Equation 1), the Kuhn-Tucker equations can be stated as \| \| \| --- \| \| \| (12) \| in addition to the original constraints in Equation 1. The first equation describes a canceling of the gradients between the objective function and the active constraints at the solution point. For the gradients to be canceled, Lagrange multipliers (λi, i = 1,...,m) are necessary to balance the deviations in magnitude of the objective function and constraint gradients. Because only active constraints are included in this canceling operation, constraints that are not active must not be included in this operation and so are given Lagrange multipliers equal to 0. This is stated implicitly in the last two Kuhn-Tucker equations. The solution of the KKT equations forms the basis to many nonlinear programming algorithms. These algorithms attempt to compute the Lagrange multipliers directly. Constrained quasi-Newton methods guarantee superlinear convergence by accumulating second-order information regarding the KKT equations using a quasi-Newton updating procedure. These methods are commonly referred to as Sequential Quadratic Programming (SQP) methods, since a QP subproblem is solved at each major iteration (also known as Iterative Quadratic Programming, Recursive Quadratic Programming, and Constrained Variable Metric methods). The 'active-set' algorithm cannot use sparse data; see Sparsity in Optimization Algorithms. Sequential Quadratic Programming (SQP) SQP methods represent the state of the art in nonlinear programming methods. Schittkowski , for example, has implemented and tested a version that outperforms every other tested method in terms of efficiency, accuracy, and percentage of successful solutions, over a large number of test problems. Based on the work of Biggs , Han , and Powell ( and ), the method allows you to closely mimic Newton's method for constrained optimization just as is done for unconstrained optimization. At each major iteration, an approximation is made of the Hessian of the Lagrangian function using a quasi-Newton updating method. This is then used to generate a QP subproblem whose solution is used to form a search direction for a line search procedure. An overview of SQP is found in Fletcher , Gill et al. , Powell , and Schittkowski . The general method, however, is stated here. Given the problem description in GP (Equation 1) the principal idea is the formulation of a QP subproblem based on a quadratic approximation of the Lagrangian function. \| \| \| --- \| \| \| (13) \| Here you simplify Equation 1 by assuming that bound constraints have been expressed as inequality constraints. You obtain the QP subproblem by linearizing the nonlinear constraints. Quadratic Programming (QP) Subproblem \| \| \| --- \| \| \| (14) \| This subproblem can be solved using any QP algorithm (see, for instance, Quadratic Programming Solution). The solution is used to form a new iterate xk + 1 = xk + αkdk. The step length parameter αk is determined by an appropriate line search procedure so that a sufficient decrease in a merit function is obtained (see Updating the Hessian Matrix). The matrix Hk is a positive definite approximation of the Hessian matrix of the Lagrangian function (Equation 13). Hk can be updated by any of the quasi-Newton methods, although the BFGS method (see Updating the Hessian Matrix) appears to be the most popular. A nonlinearly constrained problem can often be solved in fewer iterations than an unconstrained problem using SQP. One of the reasons for this is that, because of limits on the feasible area, the optimizer can make informed decisions regarding directions of search and step length. Consider Rosenbrock's function with an additional nonlinear inequality constraint, g(x), \| \| \| --- \| \| \| (15) \| This was solved by an SQP implementation in 31 iterations compared to 37 for the unconstrained case. The figure shows the path to the solution point x = [0.9072,0.8228] starting at x = [–1.9,2.0]. Code for Creating the Figure "fmincon" "sqp" function end function persistent switch case 'init' % log for cleaner contours on equal "ko" case 'iter' "kx" % Plot iterative points'--k' case 'done' ' Start point' ' Solution' "SQP Method on Nonlinearly Constrained Rosenbrock's Function" off end end SQP Implementation The SQP implementation consists of three main stages, which are discussed briefly in the following subsections: Updating the Hessian Matrix Quadratic Programming Solution Initialization Line Search and Merit Function Updating the Hessian Matrix.At each major iteration a positive definite quasi-Newton approximation of the Hessian of the Lagrangian function, H, is calculated using the BFGS method, where λi, i = 1,...,m, is an estimate of the Lagrange multipliers. \| \| \| --- \| \| \| (16) \| where Powell recommends keeping the Hessian positive definite even though it might be positive indefinite at the solution point. A positive definite Hessian is maintained providing is positive at each update and that H is initialized with a positive definite matrix. When is not positive, qk is modified on an element-by-element basis so that . The general aim of this modification is to distort the elements of qk, which contribute to a positive definite update, as little as possible. Therefore, in the initial phase of the modification, the most negative element of qksk is repeatedly halved. This procedure is continued until is greater than or equal to a small negative tolerance. If, after this procedure, is still not positive, modify qk by adding a vector v multiplied by a constant scalar w, that is, \| \| \| --- \| \| \| (17) \| where and increase w systematically until becomes positive. The functions fmincon, fminimax, fgoalattain, and fseminf all use SQP. If Display is set to 'iter' in options, then various information is given such as function values and the maximum constraint violation. When the Hessian has to be modified using the first phase of the preceding procedure to keep it positive definite, then Hessian modified is displayed. If the Hessian has to be modified again using the second phase of the approach described above, then Hessian modified twice is displayed. When the QP subproblem is infeasible, then infeasible is displayed. Such displays are usually not a cause for concern but indicate that the problem is highly nonlinear and that convergence might take longer than usual. Sometimes the message no update is displayed, indicating that is nearly zero. This can be an indication that the problem setup is wrong or you are trying to minimize a noncontinuous function. Quadratic Programming Solution.At each major iteration of the SQP method, a QP problem of the following form is solved, where Ai refers to the ith row of the m-by-n matrix A. \| \| \| --- \| \| \| (18) \| The method used in Optimization Toolbox functions is an active set strategy (also known as a projection method) similar to that of Gill et al., described in and . It has been modified for both Linear Programming (LP) and Quadratic Programming (QP) problems. The solution procedure involves two phases. The first phase involves the calculation of a feasible point (if one exists). The second phase involves the generation of an iterative sequence of feasible points that converge to the solution. In this method an active set, , is maintained that is an estimate of the active constraints (i.e., those that are on the constraint boundaries) at the solution point. Virtually all QP algorithms are active set methods. This point is emphasized because there exist many different methods that are very similar in structure but that are described in widely different terms. is updated at each iteration k, and this is used to form a basis for a search direction . Equality constraints always remain in the active set . The notation for the variable is used here to distinguish it from dk in the major iterations of the SQP method. The search direction is calculated and minimizes the objective function while remaining on any active constraint boundaries. The feasible subspace for is formed from a basis Zk whose columns are orthogonal to the estimate of the active set (i.e., ). Thus a search direction, which is formed from a linear summation of any combination of the columns of Zk, is guaranteed to remain on the boundaries of the active constraints. The matrix Zk is formed from the last m – l columns of the QR decomposition of the matrix , where l is the number of active constraints and l < m. That is, Zk is given by \| \| \| --- \| \| \| (19) \| where Once Zk is found, a new search direction is sought that minimizes q(d) where is in the null space of the active constraints. That is, is a linear combination of the columns of Zk: for some vector p. Then if you view the quadratic as a function of p, by substituting for , you have \| \| \| --- \| \| \| (20) \| Differentiating this with respect to p yields \| \| \| --- \| \| \| (21) \| ∇q(p) is referred to as the projected gradient of the quadratic function because it is the gradient projected in the subspace defined by Zk. The term is called the projected Hessian. Assuming the Hessian matrix H is positive definite (which is the case in this implementation of SQP), then the minimum of the function q(p) in the subspace defined by Zk occurs when ∇q(p) = 0, which is the solution of the system of linear equations \| \| \| --- \| \| \| (22) \| A step is then taken of the form \| \| \| --- \| \| \| (23) \| At each iteration, because of the quadratic nature of the objective function, there are only two choices of step length α. A step of unity along is the exact step to the minimum of the function restricted to the null space of . If such a step can be taken, without violation of the constraints, then this is the solution to QP (Equation 18). Otherwise, the step along to the nearest constraint is less than unity and a new constraint is included in the active set at the next iteration. The distance to the constraint boundaries in any direction is given by \| \| \| --- \| \| \| (24) \| which is defined for constraints not in the active set, and where the direction is towards the constraint boundary, i.e., . When n independent constraints are included in the active set, without location of the minimum, Lagrange multipliers, λk, are calculated that satisfy the nonsingular set of linear equations \| \| \| --- \| \| \| (25) \| If all elements of λk are positive, xk is the optimal solution of QP (Equation 18). However, if any component of λk is negative, and the component does not correspond to an equality constraint, then the corresponding element is deleted from the active set and a new iterate is sought. Initialization.The algorithm requires a feasible point to start. If the current point from the SQP method is not feasible, then you can find a point by solving the linear programming problem \| \| \| --- \| \| \| (26) \| The notation Ai indicates the ith row of the matrix A. You can find a feasible point (if one exists) to Equation 26 by setting x to a value that satisfies the equality constraints. You can determine this value by solving an under- or overdetermined set of linear equations formed from the set of equality constraints. If there is a solution to this problem, then the slack variable γ is set to the maximum inequality constraint at this point. You can modify the preceding QP algorithm for LP problems by setting the search direction to the steepest descent direction at each iteration, where gk is the gradient of the objective function (equal to the coefficients of the linear objective function). \| \| \| --- \| \| \| (27) \| If a feasible point is found using the preceding LP method, the main QP phase is entered. The search direction is initialized with a search direction found from solving the set of linear equations \| \| \| --- \| \| \| (28) \| where gk is the gradient of the objective function at the current iterate xk (i.e., Hxk + c). If a feasible solution is not found for the QP problem, the direction of search for the main SQP routine is taken as one that minimizes γ. Line Search and Merit Function.The solution to the QP subproblem produces a vector dk, which is used to form a new iterate \| \| \| --- \| \| \| (29) \| The step length parameter αk is determined in order to produce a sufficient decrease in a merit function. The merit function used by Han and Powell of the following form is used in this implementation. \| \| \| --- \| \| \| (30) \| Powell recommends setting the penalty parameter \| \| \| --- \| \| \| (31) \| This allows positive contribution from constraints that are inactive in the QP solution but were recently active. In this implementation, the penalty parameter ri is initially set to \| \| \| --- \| \| \| (32) \| where represents the Euclidean norm. This ensures larger contributions to the penalty parameter from constraints with smaller gradients, which would be the case for active constraints at the solution point. fmincon SQP Algorithm The sqp algorithm (and nearly identical sqp-legacy algorithm) is similar to the active-set algorithm (for a description, see fmincon Active Set Algorithm). The basic sqp algorithm is described in Chapter 18 of Nocedal and Wright . The sqp algorithm is essentially the same as the sqp-legacy algorithm, but has a different implementation. Usually, sqp has faster execution time and less memory usage than sqp-legacy. The most important differences between the sqp and the active-set algorithms are: Strict Feasibility With Respect to Bounds The sqp algorithm takes every iterative step in the region constrained by bounds. Furthermore, finite difference steps also respect bounds. Bounds are not strict; a step can be exactly on a boundary. This strict feasibility can be beneficial when your objective function or nonlinear constraint functions are undefined or are complex outside the region constrained by bounds. Robustness to Non-Double Results During its iterations, the sqp algorithm can attempt to take a step that fails. This means an objective function or nonlinear constraint function you supply returns a value of Inf, NaN, or a complex value. In this case, the algorithm attempts to take a smaller step. Refactored Linear Algebra Routines The sqp algorithm uses a different set of linear algebra routines to solve the quadratic programming subproblem, Equation 14. These routines are more efficient in both memory usage and speed than the active-set routines. Reformulated Feasibility Routines The sqp algorithm has two new approaches to the solution of Equation 14 when constraints are not satisfied. The sqp algorithm combines the objective and constraint functions into a merit function. The algorithm attempts to minimize the merit function subject to relaxed constraints. This modified problem can lead to a feasible solution. However, this approach has more variables than the original problem, so the problem size in Equation 14 increases. The increased size can slow the solution of the subproblem. These routines are based on the articles by Spellucci and Tone . The sqp algorithm sets the penalty parameter for the merit function Equation 30 according to the suggestion in . Suppose nonlinear constraints are not satisfied, and an attempted step causes the constraint violation to grow. The sqp algorithm attempts to obtain feasibility using a second-order approximation to the constraints. The second-order technique can lead to a feasible solution. However, this technique can slow the solution by requiring more evaluations of the nonlinear constraint functions. fmincon Interior Point Algorithm Barrier Function The interior-point approach to constrained minimization is to solve a sequence of approximate minimization problems. The original problem is \| \| \| --- \| \| \| (33) \| For each μ > 0, the approximate problem is \| \| \| --- \| \| \| (34) \| There are as many slack variables si as there are inequality constraints g. The si are restricted to be positive to keep the iterates in the interior of the feasible region. As μ decreases to zero, the minimum of fμ should approach the minimum of f. The added logarithmic term is called a barrier function. This method is described in , , and . The approximate problem Equation 34 is a sequence of equality constrained problems. These are easier to solve than the original inequality-constrained problem Equation 33. To solve the approximate problem, the algorithm uses one of two main types of steps at each iteration: A direct step in (x, s). This step attempts to solve the KKT equations, Equation 2 and Equation 3, for the approximate problem via a linear approximation. This is also called a Newton step. A CG (conjugate gradient) step, using a trust region. By default, the algorithm first attempts to take a direct step. If it cannot, it attempts a CG step. One case where it does not take a direct step is when the approximate problem is not locally convex near the current iterate. At each iteration the algorithm decreases a merit function, such as \| \| \| --- \| \| \| (35) \| The parameter may increase with iteration number in order to force the solution towards feasibility. If an attempted step does not decrease the merit function, the algorithm rejects the attempted step, and attempts a new step. If either the objective function or a nonlinear constraint function returns a complex value, NaN, Inf, or an error at an iterate xj, the algorithm rejects xj. The rejection has the same effect as if the merit function did not decrease sufficiently: the algorithm then attempts a different, shorter step. Wrap any code that can error in try-catch: function val = userFcn(x) try val = ... % code that can error catch val = NaN; end The objective and constraints must yield proper (double) values at the initial point. Direct Step The following variables are used in defining the direct step: H denotes the Hessian of the Lagrangian of fμ: \| \| \| --- \| \| \| (36) \| Jg denotes the Jacobian of the constraint function g. Jh denotes the Jacobian of the constraint function h. S = diag(s). λ denotes the Lagrange multiplier vector associated with constraints g Λ = diag(λ). y denotes the Lagrange multiplier vector associated with h. e denote the vector of ones the same size as g. Equation 38 defines the direct step (Δx, Δs): \| \| \| --- \| \| \| (37) \| This equation comes directly from attempting to solve Equation 2 and Equation 3 using a linearized Lagrangian. You can symmetrize the equation by premultiplying the second variable Δs by S–1: \| \| \| --- \| \| \| (38) \| In order to solve this equation for (Δx, Δs), the algorithm makes an LDL factorization of the matrix; see ldl. This is the most computationally expensive step. One result of this factorization is a determination of whether the projected Hessian is positive definite or not; if not, the algorithm uses a conjugate gradient step, described in Conjugate Gradient Step. Update Barrier Parameter For the approximate problem Equation 34 to approach the original problem, the barrier parameter μ needs to decrease toward 0 as the iterations proceed. The algorithm has two barrier parameter update options, which you specify using the BarrierParamUpdate option: 'monotone' (default) and 'predictor-corrector'. The 'monotone' option decreases the parameter μ by a factor of 1/100 or 1/5 when the approximate problem is solved with sufficient accuracy in the previous iteration. The option uses a factor of 1/100 when the algorithm takes only one or two iterations to achieve sufficient accuracy, and uses 1/5 otherwise. The measure of accuracy is the following test, which determines if the size of all terms on the right side of Equation 38 is less than μ: Note fmincon overrides the BarrierParamUpdate setting to 'monotone' in either of these cases: The problem has no inequality constraints, including bound constraints. The SubproblemAlgorithm option is 'cg'. The 'predictor-corrector' algorithm for updating the barrier parameter μ is similar to the linear programming Predictor-Corrector algorithm. Predictor-corrector steps can accelerate the existing Fiacco-McCormick (monotone) approach by adjusting for the linearization error in the Newton steps. The effects of the predictor-corrector algorithm are twofold: it often improves step directions and simultaneously updates the barrier parameter adaptively with the centering parameter σ to encourage iterates to follow the central path. See Nocedal and Wright’s discussion of predictor-corrector steps for linear programs to understand why the central path allows larger step sizes and, consequently, faster convergence. The predictor step uses the linearized step with μ = 0, meaning without a barrier function: Define ɑs and ɑλ to be the largest step sizes that do not violate the nonnegativity constraints. Now compute the complementarity from the predictor step. \| \| \| --- \| \| \| (39) \| where m is the number of constraints. The first corrector step adjusts for the quadratic term neglected in the Newton root-finding linearization To correct the quadratic error, solve the linear system for the corrector step direction. The second corrector step is a centering step. The centering correction is based on the variable σ on the right side of the equation Here, σ is defined as where μP is defined in equation Equation 39, and . To prevent the barrier parameter from decreasing too quickly, potentially destabilizing the algorithm, the algorithm keeps the centering parameter σ above 1/100. This action causes the barrier parameter μ to decrease by no more than a factor of 1/100. Algorithmically, the first correction and centering steps are independent of each other, so they are computed together. Furthermore, the matrix on the left for the predictor and both corrector steps is the same. So, algorithmically, the matrix is factorized once, and this factorization is used for all these steps. The algorithm can reject the proposed predictor-corrector step when the step increases the merit function value Equation 35, increases the complementarity by at least a factor of two, or the computed inertia is incorrect (the problem looks nonconvex). In these cases, the algorithm attempts to take a different step or a conjugate gradient step. Conjugate Gradient Step The conjugate gradient approach to solving the approximate problem Equation 34 is similar to other conjugate gradient calculations. In this case, the algorithm adjusts both x and s, keeping the slacks s positive. The approach is to minimize a quadratic approximation to the approximate problem in a trust region, subject to linearized constraints. Specifically, let R denote the radius of the trust region, and let other variables be defined as in Direct Step. The algorithm obtains Lagrange multipliers by approximately solving the KKT equations in the least-squares sense, subject to λ being positive. Then it takes a step (Δx, Δs) to approximately solve \| \| \| --- \| \| \| (40) \| subject to the linearized constraints \| \| \| --- \| \| \| (41) \| To solve Equation 41, the algorithm tries to minimize a norm of the linearized constraints inside a region with radius scaled by R. Then Equation 40 is solved with the constraints being to match the residual from solving Equation 41, staying within the trust region of radius R, and keeping s strictly positive. For details of the algorithm and the derivation, see , , and . For another description of conjugate gradients, see Preconditioned Conjugate Gradient Method. Feasibility Mode When the EnableFeasibilityMode option is true and the iterations do not decrease the infeasibility quickly enough, the algorithm switches to feasibility mode. This switch happens after the algorithm fails to decrease the infeasibility in normal mode, and then fails again after switching to conjugate gradient mode. Therefore, for best performance when the solver fails to find a feasible solution without feasibility mode, set the SubproblemAlgorithm to 'cg' when using feasibility mode. Doing so avoids fruitless searching in normal mode. The feasibility mode algorithm is based on Nocedal, Öztoprak, and Waltz . The algorithm ignores the objective function and instead tries to minimize the infeasibility, defined as the sum of the positive parts of the inequality constraint functions and the absolute value of the equality constraint functions. In terms of the relaxation variables , which correspond to inequalities, positive parts of equalities, and negative parts of equalities, respectively, the problem is subject to the constraints To solve the relaxed problem, the software uses an interior-point formulation with a logarithmic barrier function and the slacks to minimize subject to the constraints The solution process for the relaxed problem begins with μ initialized to the current barrier parameter value. The slack variable sI is initialized to the current inequality slack value, inherited from the main mode. The r variables are initialized to The remaining slacks are initialized to Starting at this initial point, the feasibility mode algorithm reuses the code for the normal interior-point algorithm. This process requires special step computations because the r variables are linear and, therefore, their associated second derivatives are zero. In other words, the objective function Hessian for the feasibility problem is rank-deficient. Therefore, the algorithm cannot take a Newton step. Instead, the algorithm takes a steepest-descent direction step. The algorithm starts with the gradient of the objective with respect to the variables, projects the gradient onto the null space of the Jacobian of the constraints, and rescales the resulting vector so that it has an appropriate step length. This step can be effective at reducing the infeasibility. The feasibility mode algorithm ends when it reduces the infeasibility by a factor of 10. When feasibility mode ends, the algorithm passes the variables x and sI to the main algorithm, and discards the other slack variables and relaxation variables r. References Nocedal, Jorge, Figen Öztoprak, and Richard A. Waltz. An Interior Point Method for Nonlinear Programming with Infeasibility Detection Capabilities. Optimization Methods & Software 29(4), July 2014, pp. 837–854. Interior-Point Algorithm Options Here are the meanings and effects of several options in the interior-point algorithm. HonorBounds — When set to true, every iterate satisfies the bound constraints you have set. When set to false, the algorithm may violate bounds during intermediate iterations. HessianApproximation — When set to: 'bfgs', fmincon calculates the Hessian by a dense quasi-Newton approximation. 'lbfgs', fmincon calculates the Hessian by a limited-memory, large-scale quasi-Newton approximation. 'fin-diff-grads', fmincon calculates a Hessian-times-vector product by finite differences of the gradient(s); other options need to be set appropriately. HessianFcn — fmincon uses the function handle you specify in HessianFcn to compute the Hessian. See Including Hessians. HessianMultiplyFcn — Give a separate function for Hessian-times-vector evaluation. For details, see Including Hessians and Hessian Multiply Function. SubproblemAlgorithm — Determines whether or not to attempt the direct Newton step. The default setting 'factorization' allows this type of step to be attempted. The setting 'cg' allows only conjugate gradient steps. For a complete list of options see Interior-Point Algorithm in fmincon options. fminbnd Algorithm fminbnd is a solver available in any MATLAB® installation. It solves for a local minimum in one dimension within a bounded interval. It is not based on derivatives. Instead, it uses golden-section search and parabolic interpolation. fseminf Problem Formulation and Algorithm fseminf Problem Formulation fseminf addresses optimization problems with additional types of constraints compared to those addressed by fmincon. The formulation of fmincon is such that c(x) ≤ 0, ceq(x) = 0, A·x ≤ b, Aeq·x = beq, and l ≤ x ≤ u. fseminf adds the following set of semi-infinite constraints to those already given. For wj in a one- or two-dimensional bounded interval or rectangle Ij, for a vector of continuous functions K(x, w), the constraints are Kj(x, wj) ≤ 0 for all wj∈Ij. The term “dimension” of an fseminf problem means the maximal dimension of the constraint set I: 1 if all Ij are intervals, and 2 if at least one Ij is a rectangle. The size of the vector of K does not enter into this concept of dimension. The reason this is called semi-infinite programming is that there are a finite number of variables (x and wj), but an infinite number of constraints. This is because the constraints on x are over a set of continuous intervals or rectangles Ij, which contains an infinite number of points, so there are an infinite number of constraints: Kj(x, wj) ≤ 0 for an infinite number of points wj. You might think a problem with an infinite number of constraints is impossible to solve. fseminf addresses this by reformulating the problem to an equivalent one that has two stages: a maximization and a minimization. The semi-infinite constraints are reformulated as \| \| \| --- \| \| \| (42) \| where \|K\| is the number of components of the vector K; i.e., the number of semi-infinite constraint functions. For fixed x, this is an ordinary maximization over bounded intervals or rectangles. fseminf further simplifies the problem by making piecewise quadratic or cubic approximations κj(x, wj) to the functions Kj(x, wj), for each x that the solver visits. fseminf considers only the maxima of the interpolation function κj(x, wj), instead of Kj(x, wj), in Equation 42. This reduces the original problem, minimizing a semi-infinitely constrained function, to a problem with a finite number of constraints. Sampling Points.Your semi-infinite constraint function must provide a set of sampling points, points used in making the quadratic or cubic approximations. To accomplish this, it should contain: The initial spacing s between sampling points w A way of generating the set of sampling points w from s The initial spacing s is a \|K\|-by-2 matrix. The jth row of s represents the spacing for neighboring sampling points for the constraint function Kj. If Kj depends on a one-dimensional wj, set s(j,2) = 0. fseminf updates the matrix s in subsequent iterations. fseminf uses the matrix s to generate the sampling points w, which it then uses to create the approximation κj(x, wj). Your procedure for generating w from s should keep the same intervals or rectangles Ij during the optimization. Example of Creating Sampling Points.Consider a problem with two semi-infinite constraints, K1 and K2. K1 has one-dimensional w1, and K2 has two-dimensional w2. The following code generates a sampling set from w1 = 2 to 12: % Initial sampling interval if isnan(s(1,1)) s(1,1) = .2; s(1,2) = 0; end % Sampling set w1 = 2:s(1,1):12; fseminf specifies s as NaN when it first calls your constraint function. Checking for this allows you to set the initial sampling interval. The following code generates a sampling set from w2 in a square, with each component going from 1 to 100, initially sampled more often in the first component than the second: % Initial sampling interval if isnan(s(1,1)) s(2,1) = 0.2; s(2,2) = 0.5; end % Sampling set w2x = 1:s(2,1):100; w2y = 1:s(2,2):100; [wx,wy] = meshgrid(w2x,w2y); The preceding code snippets can be simplified as follows: % Initial sampling interval if isnan(s(1,1)) s = [0.2 0;0.2 0.5]; end % Sampling set w1 = 2:s(1,1):12; w2x = 1:s(2,1):100; w2y = 1:s(2,2):100; [wx,wy] = meshgrid(w2x,w2y); fseminf Algorithm fseminf essentially reduces the problem of semi-infinite programming to a problem addressed by fmincon. fseminf takes the following steps to solve semi-infinite programming problems: At the current value of x, fseminf identifies all the wj,i such that the interpolation κj(x, wj,i) is a local maximum. (The maximum refers to varying w for fixed x.) fseminf takes one iteration step in the solution of the fmincon problem: such that c(x) ≤ 0, ceq(x) = 0, A·x ≤ b, Aeq·x = beq, and l ≤ x ≤ u, where c(x) is augmented with all the maxima of κj(x, wj) taken over all wj∈Ij, which is equal to the maxima over j and i of κj(x, wj,i). 3. fseminf checks if any stopping criterion is met at the new point x (to halt the iterations); if not, it continues to step 4. 4. fseminf checks if the discretization of the semi-infinite constraints needs updating, and updates the sampling points appropriately. This provides an updated approximation κj(x, wj). Then it continues at step 1. MATLAB Command You clicked a link that corresponds to this MATLAB command: Run the command by entering it in the MATLAB Command Window. Web browsers do not support MATLAB commands. Select a Web Site Choose a web site to get translated content where available and see local events and offers. 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15224	https://math.uchicago.edu/~may/REU2017/REUPapers/Xu,Yifan.pdf	KURATOWSKI’S THEOREM YIFAN XU Abstract. This paper introduces basic concepts and theorems in graph the-ory, with a focus on planar graphs. On the foundation of the basics, we state and present a rigorous proof of Kuratowski’s theorem, a necessary and suﬃ-cient condition for planarity. Contents 1. Introduction 1 2. Basic Graph Theory 1 3. Planar Graphs 3 4. Kuratowski’s Theorem 7 Acknowledgments 12 References 12 1. Introduction The planarity of a graph, whether a graph can be drawn on a plane in a way that no edges intersect, is an interesting property to investigate. With a few simple theorems it can be seen that K5 (see ﬁgure 1) and K3,3 (see ﬁgure 2) are nonpla-nar. Kuratowski pushes this nearly eﬀortless observation into a powerful theorem exposing the suﬃcient and necessary condition of planarity. Simple as the theorem appears to be, to prove this we need a signiﬁcant amount of preparations. In this paper, we start with basic graph theory and proceed into concepts and theorems related to planar graphs. In the last section we will give a proof of Kuratowski’s theorem, which in general corresponds with that in Graph Theory with Applica-tions (see in the list of references) but provides more details and hopefully more clarity. 2. Basic Graph Theory We need several deﬁnitions as a start. Deﬁnition 2.1. A graph G is an ordered pair (V (G), E(G)), consisting of a nonempty set V (G) of vertices and a set E(G) of edges, each edge a two-element subset of V . Denote \|V (G)\|, the number of vertices in the vertex set, by ν and \|E(G)\| by ϵ. Note that (i) E(G) can be empty and (ii) an edge can link a vertex to itself. Date: August 28, 2017. 1 2 YIFAN XU Figure 1. a diagram of K5 Deﬁnition 2.2. An edge whose ends are identical is a loop. Otherwise, the edge is a link. A graph is simple if it has no loops and no two links have the same pair of unordered vertices. Deﬁnition 2.3. A vertex is incident to an edge if the vertex is one of the ends of the edge; two vertices are adjacent to each other if they are connected by an edge. The degree of a vertex v, denoted by deg(v), is the number of edges incident to v, with a loop counted as two edges. We let δ := minv∈V (deg v) and ∆:= maxv∈V (deg v). Deﬁnition 2.4. A walk W is a set of alternating vertices and edges, denoted by W=v0e1v1e2...ekvk where ei (i ∈[1, k], i ∈N) links vi−1 with vi. A walk where all ei’s are distinct is a trail. In addition, if all vertices are distinct, then W is a path. A graph G is connected if there exists a path between every pair of vertices in G. A walk is closed if it has positive length and identical ends. A closed trail with distinct vertices is a cycle. A family of walks is internally disjoint if no vertex is an internal vertex of more than one walk in the family. Deﬁnition 2.5. H is a subgraph of G if V (H) ⊆V (G), E(H) ⊆E(G), and end-points of all edges in E(H) are included in V (H). In addition, if H is a maximally connected subgraph, H is a component of G. The number of components in G is denoted by ω(G). An edge e is a cut edge if ω(G −e) > ω(G). Deﬁnition 2.6. A complete graph is a graph where each pair of vertices is con-nected by a unique edge. We let Km denote the complete graph on m vertices. A graph G is bipartite if its vertex set V can be divided into two nonempty subsets X and Y such that every edge in G connects one vertex in X to another one in Y . A graph G is complete bipartite if for all x ∈X, y ∈Y , x is connected to y by a unique edge. When X contains m vertices and Y contains n vertices, G is denoted by Km,n. Theorem 2.7. For a graph G with vertex set V and edge set E, X v∈V deg(v) = 2ϵ KURATOWSKI’S THEOREM 3 s X1 s X2 s X3 s Y1 s Y2 s Y3 cc ccccc c aaaaaaaaaaaaaa a # # # # # # # # cc ccccc c ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! # # # # # # # s X1 s Y1 s Y3 s X2 s X3 s Y2 S S S S S S S S S S T T T T T T T T T T Figure 2. two diagrams of K3,3 Proof. For v ∈V and e ∈E, we let n(v) denote the set of edges incident to v and m(e) the set of endpoints of e. Then X v∈V deg(v) = X v∈V X e∈n(v) 1 = X e∈E X v∈m(e) 1 = 2ϵ. □ 3. Planar Graphs Deﬁnition 3.1. A way to draw the graph, representing vertices by points and edges by lines connecting points, is called a diagram of the graph. A diagram is embedded in the plane. A graph that has a diagram whose edges do not intersect anywhere besides their ends (i.e., vertices) is called a planar graph. The diagram is then called the planar embedding of a planar graph, or simply, a plane graph. Deﬁnition 3.2. Closures of regions partitioned by a plane graph are faces, the number of which is denoted by φ. In a plane graph, the degree of a face f, denoted by deg(f), is the number of edges incident to f, with cut edges being counted twice. Deﬁnition 3.3. A dual of a plane graph G, denoted by G∗, can be constructed as follows: every vertex v∗in G∗corresponds to a face f in G and every edge e∗in G∗ corresponds to an edge e in G. Two vertices in G∗, v∗and w∗, are joined by the edge e∗if and only if their corresponding faces in G, f and g, are separated by e. 4 YIFAN XU Theorem 3.4. For G a plane graph, let F(G) denote the set of faces in G. The following holds: X f∈F (G) deg(f) = 2ϵ. Proof. Consider the dual, G∗, of G. By Theorem 2.13, X v∗∈V ∗ deg(v∗) = 2ϵ∗. By deﬁnition of a dual graph, ∀f ∈F(G), deg(f)=deg(v∗), and ϵ=ϵ∗.Then X f∈F (G) deg(f) = X v∗∈V ∗ deg(v∗) = 2ϵ∗= 2ϵ. □ Theorem 3.5. (Euler’s Formula) For G a connected plane graph, the following relationship holds: ν −ϵ + φ = 2. Proof. We prove this by induction on ϵ. (1) Basic step: when ϵ = 0, since G is connected, G must be trivial. Then ν = 1, ϵ = 0, φ = 1. ν −ϵ + φ = 1 −0 + 1 = 2. Clearly the formula holds. (2) Inductive step: suppose for G with ϵ(G) = k, ν(G) −ϵ(G) + φ(G) = 2. Let H be a planar graph such that G is a subgraph and ϵ(H) = k + 1. There are 3 cases: Case(a): e is a loop added on some v ∈V . Then ν(H) = ν(G). Since H needs to remain a plane graph, the loop does not intersect with any other edge in G. Therefore the loop constitutes a new face, φ(H) = φ(G) + 1, which then gives us ν(H) −ϵ(H) + φ(H) = ν(G) −(ϵ(G) + 1) + (φ(G) + 1) = ν(G) −ϵ(G) + φ(G) = 2. Case(b): e adds a link between two vertices v1, v2 in G. Since G is connected, there is a path between v1 and v2. The addition of e creates a new cycle, and hence a new face (if there are more than one path between v1 and v2, there is necessarily one that borders the new face). Therefore ν(H) = ν(G), φ(H) = φ(G) + 1, ϵ(H) = ϵ(G) + 1. Similar to (a), the formula holds. Case(c): e links some v1 ∈V (G) to a new vertex v2. Then ν(H) = ν(G) + 1. Since v2 is a new vertex, e lies in some face bordering v1, without creating any new face. Then φ(H) = φ(G). In this case, ν(H) −ϵ(H) + φ(H) = (ν(G) + 1) −(ϵ(G) + 1) + φ(G) = ν(G) −ϵ(G) + φ(G) = 2. Therefore in all 3 cases, the formula holds in H. By the principle of induction, the formula holds for all connected plane graphs. □ KURATOWSKI’S THEOREM 5 Corollary 3.6. For G a simple planar graph with ν ≥3, ϵ ≤3ν −6. Proof. Since G is a simple graph with ν(G) ≥3, the planar embedding of G, G′, is also simple with ν(G′) ≥3. For f ∈F(G′), there exist 3 cases: (1) if f is bounded by a cycle, since G′ is simple, the minimum size of a cycle is 3 and degG′(f)≥3; (2) if f is incident to some edges in addition to a cycle, degG′(f)>3; (3) if f is incident to no cycle at all, then there exists at least two cut edges on the boundary of f—if not, the rest of the boundary must be a cycle, which contradicts the condition—which means degG′(f) ≥4 > 3. Therefore, for every f ∈F(G′), X degG′(f) ≥3φ(G′) then by Theorem 3.7, X f∈F (G′) deg(f) = 2ϵ(G′) ≥3φ(G′) which yields φ(G′) ≤2 3ϵ(G′). Then by Theorem 3.8, ν(G′) −ϵ(G′) + φ(G′) = 2 ≤ν(G′) −ϵ(G′) + 2 3ϵ(G′) = ν(G′) −1 3ϵ(G′) = ν(G) −1 3ϵ(G). Therefore, ν(G) −1 3ϵ(G) ≥2 and ϵ ≤3ν −6. □ Corollary 3.7. K5 is nonplanar. Proof. We have ϵ(K5) = 5 2 = 10 > 3ν(K5) −6 = 9, by Corollary 3.9, K5 is nonplanar. □ Corollary 3.8. K3,3 is nonplanar. Proof. We are going to prove this by contradiction. Suppose there exists a planar embedding of K3,3. In a simple bipartite graph, the minimum size of a cycle is 4, which means for f ∈F(K3,3), deg(f) ≥4 6 YIFAN XU Figure 3. an example of 3 bridges on a cycle (the red one is skew to the blue one; the cyan one avoids the other two; red and cyan outer bridges, blue an inner bridge) (the argument similar to that in Corollary 3.9). By Theorem 3.7, 4φ ≤ X f∈F (G) deg(f) = 2ϵ = 18 then since φ ∈N, φ ≤4. By Theorem 3.8, ν −ϵ + φ = 2 ≤6 −9 + 4 = 1, which is impossible. Therefore, K3,3 is nonplanar. □ Deﬁnition 3.9. Let H be a subgraph of a graph G. Deﬁne an equivalence relation ∼on E(G) \ E(H) as follows: a ∼b if there is a walk W such that a and b are the ﬁrst and last edge in W respectively, and that no internal vertex of W is in V (H). A bridge of H in G is a subgraph of G −E(H) induced by an equivalent class of ∼(a bridge containing e is the subgraph containing every e′, e′ ∼e, e and e′ edges of G). For a bridge B of H, we deﬁne vertices of attachment of B to H as the vertices in the set V (B) ∩V (H). Let C be a cycle. Then two bridges of C, B1 and B2, are skew if two vertices of attachment of B1, say u1 and v1, and two of B2, u2 and v2, appear in the order of u1, u2, v1, v2 on C. Deﬁnition 3.10. Suppose C is a cycle in a planar embedding of a planar graph G. Then for some bridge B of C, B is contained entirely in either Int(C) (the region inside C) or Ext(C) (the region outside C). A bridge in Int(C) is an inner bridge, while one in Ext(C) is an outer bridge. In the planar embedding, inner (or outer) bridges avoid each other: for all B1, B2 two inner(outer) bridges, all vertices of attachment in B1 lie on the arc uv of C which contains no vertices of attachment of B2 other than u and v. KURATOWSKI’S THEOREM 7 Deﬁnition 3.11. In some planar embedding G1 of a planar graph G, an inner bridge B of C (a cycle in G) is transferrable if there exists another planar embedding G2 of G, where B is an outer bridge but everything else remains the same as in G1. Theorem 3.12. Let G be a plane graph and C a cycle in G. An inner bridge B of C is transferrable if B avoids every outer bridge of C. Proof. Find an inner bridge B that avoids every outer bridge. Then we can ﬁnd a face in Ext(C) whose boundary contains all vertices of attachment of B. Drawing B on the new face gives us another planar embedding, which means B is transferrable. □ 4. Kuratowski’s Theorem In 1930, Kuratowski published the theorem giving a necessary and suﬃcient condition for planarity. Kuratowski’s Theorem states that a graph is planar if and only if it contains no subdivision of K5 or K3,3. To prove this theorem, we ﬁrst need some simple lemmas. Lemma 4.1. Every subgraph of a planar graph is planar. Proof. If G is planar, then there exists a planar embedding of G. For every subgraph H of G, we can ﬁnd the vertices and edges of H in the planar embedding of G. This is how we can construct a planar embedding of H. □ Deﬁnition 4.2. A subdivision of an edge is the operation where the edge is replaced by a path of length 2, the internal vertex added to the original graph. A subdivision of a graph G is a graph achieved by a sequence of edge-subdivisions on G. Lemma 4.3. Every subdivision of a nonplanar graph is nonplanar. Proof. Suppose for G, there exists a planar embedding of its subdivision, G′. When we remove the vertices created in edge-subdivisions, and reconstruct the original edge (without changing the shape and position of the path), we get a planar em-bedding of G and ﬁnd G planar. Therefore, if G is nonplanar, every subdivision of G is nonplanar. □ From the two lemmas above the necessity easily follows. Then it suﬃces to prove that the condition is also suﬃcient. In order to show if G contains no subdivisions of K5 or K3,3, G is planar, it is equivalent to show that if G is nonplanar, G has to contain some subdivision of K5 or K3,3. Two deﬁnitions are necessary before we get to the strategy of proving the equivalent statement. Deﬁnition 4.4. For a graph G, a vertex cut V ′ is a subset of V whose removal renders G −V ′ disconnected (when we remove a vertex, we remove the vertex as well as all edges incident to it). The connectivity of G, denoted by κ(G), is the minimum size of the vertex cut V ′. A graph G is said to be k-connected if k ≤κ(G). Deﬁnition 4.5. For a graph G, H is a proper subgraph of G if V (H)⊊V (G) and E(H)⊊E(G). A minimal nonplanar graph is a nonplanar graph that does not have any nonplanar proper subgraph. Clearly, it suﬃces to prove the statement for all G some minimal nonplanar graph. The strategy is as follows: (1) Show that if minimal nonplanar graphs without any subdivision of K5 or K3,3 8 YIFAN XU as subgraphs did exist, they would be 3-connected and simple. (2) Show that every 3-connected graph with no subdivision of K5 or K3,3 as sub-graphs is in fact planar. This is how we arrive at a contradiction, forcing the original statement to be true. To show (1), we need to establish a few more lemmas. Lemma 4.6. A minimal nonplanar graph is 2-connected. Proof. First show that a minimal nonplanar graph is 1-connected (connected). Sup-pose G is disconnected and nonplanar, but all of its components are planar. With-out loss of generality, suppose G has two components, G1 and G2. Since G1 and G2 are both planar, we can add a planar embedding of G1 to one of the faces of a planar embedding of G2 (the inﬁnite face for example), which yields a planar embedding of G, a contradiction. Then we show that it is 2-connected. Suppose G is nonplanar, with κ(G) = 1. By deﬁnition of connectivity, there exists a vertex v such that G−v is disconnected. Without loss of generality, suppose G −v has two components, H1 and H2. We know that H1 ∪v and H2 ∪v are both planar. In the planar embedding of each, we can ﬁnd a face f whose boundary contains v. With stereographic projection, we can get a planar embedding for each of H1 ∪v and H2 ∪v where v lies on the boundary of the unbounded face, by placing the point at inﬁnity on the sphere inside f. Then we can combine H1 ∪v and H2 ∪v by merging v and get a planar embedding of G, a contradiction. Therefore if G is a minimal nonplanar graph, G is 2-connected. □ Lemma 4.7. If G is a graph that has the fewest edges possible among all connected nonplanar graphs with no subdivision of K5 or K3,3, then G is 3-connected. Proof. The hypothesis suggests G is a minimal nonplanar graph. By the previous lemma, G is 2-connected. Suppose κ(G) = 2. Then there exists a vertex cut {u, v} such that G−{u, v} is disconnected. Name the components of G−{u, v} H1, H2,..., Hk. Construct M1, M2,..., Mk, where Mi is Hi ∪{u, v} with the addition of a new edge uv. We claim here that among Mi, 1 ≤i ≤k, there exists at least one Mi that is nonplanar. Below is a proof for the claim: Suppose all Mi’s are planar for 1 ≤i ≤k. Then there is a planar embedding for each. Since {u, v} and the edge uv are the only part Mi’s share, we can merge the planar embeddings of Mi’s and get a planar embedding of G + uv (G ∪{uv}), which means G + uv is planar. Then by Lemma 4.1, G is planar, a contradiction. Therefore, there exists some Mj where 1 ≤j ≤k that is nonplanar. It is clear that ϵ(Mj) < ϵ(G). But since G, by the original hypothesis, is the smallest connected nonplanar graphs with no subdivision of K5 or K3,3, Mj must have some subdivision of K5 or K3,3. Moreover, since G contains no such subdivi-sion, Mj is not a subgraph of G, which means G does not have an edge uv. Now we combine Mj −uv with Mp −uv where p ̸= j, 1 ≤p ≤k by merging the vertices u and v and get a subgraph of G. Since Mp −uv is connected, there exists a path between u and v. When we combine this path with Mj −uv, we get a subdivision of K5 or K3,3. This means G contains such a subdivision, a contradiction. Therefore, G has to be 3-connected. □ Now (1) has been shown. To complete (2), we again need some lemmas. KURATOWSKI’S THEOREM 9 Lemma 4.8. (Whitney’s Theorem) Let G be a graph with ν ≥3. Then G is 2-connected if and only if for all u, v ∈V (G), there are at least two internally-disjoint paths between them. Proof. (⇐) If any two vertices in G are connected by at least two internally-disjoint paths, then clearly there exists no 1-vertex cut (since no matter which vertex is removed, between every two vertices that remain, there still exists at least one path between them). Hence G is 2-connected. (⇒) Suppose G is 2-connected. We shall prove this direction by induction. Take two vertices u, v ∈V (G). Denote the number of edges in the shortest paths between them by d(u, v). (a) Basic step: when d(u, v) = 1. Since G is 2-connected, there exists another path connecting u, v, which does not contain the edge uv. (b) Inductive step: Suppose there exist at least two internally-disjoint paths for all u, v with d(u, v) ≤k. For x, y with d(x, y) = k + 1, ﬁnd a path P0 of length d(x,y) between x, y and a vertex z that is closest to y (d(y, z) = 1) in P0. Then d(x, z) = d(x, y)−1. By the inductive hypothesis, there exist two internally-disjoint paths P1 and P2 between x, z. Since G is 2-connected, there exists another path Q between x, y that does not contain z (or {w} would be a 1-vertex cut). Let w be the vertex in (Q ∩(P1 ∪P2)) that is closest to y on Q. Without loss of generality we suppose w is contained in P1. Then we can ﬁnd two internally-disjoint paths between x, y: the ﬁrst one would be the part from x to w on P1 combined with the part from w to y on Q; the second one would be P2 combined with the edge zy. □ Lemma 4.9. If G is simple and 3-connected and uv is an edge in G, then G −uv is 2-connected. Proof. We want to show that for all a, b ∈V (G −uv), there exist at least two internally-disjoint paths between them. In other words, we want to show for every two vertices of G −uv, there exists a cycle they both lie on. We prove this by discussing 3 cases. (1) {a, b} = {u, v}. Clearly ν(G) ≥4. Pick another two vertices c and d in G −uv. Without loss of generality, assume u = a. Now consider u and c. Since G is 3-connected, G does not contain any 2-vertex cut, which means when v and d are removed, u and c are still connected. In other words, there exists a path P1 between u and c which does not contain v and d. Similarly, there exists a path P2 between c and v that avoids u and d, a P3 between v and d that avoids u and c, and ﬁnally a P4 between d and u that avoids c and v. But then u and v lie on the same cycle u-P1-c-P2-v-P3-d-P4-u. (2) One and only one of {a, b} is u or v. Without loss of generality, let a = u and b ̸= v. Find c ̸= b that is neither u nor v. Then following the similar argument as in (1), we can ﬁnd a path P1 avoiding c and v between u and b, a P2 avoiding u and v between c and b, and a P3 avoiding v between c and u. Again u-P1-b-P2-c-P3-u is a cycle containing u, b. (3) Neither of {a, b} equals u or v. Once again following the same argument, we can ﬁnd a path P1 avoiding u, v between a, b, a P2 avoiding u, a between v, b, and a P3 avoiding u, b between a, v. Then a-P1-b-P2-v-P3-a is a cycle containing a, b. Since in all 3 cases, we construct a cycle without the edge uv where both a, b lie, the same cycles can be constructed in G −uv, which means G −uv must be 2-connected. □ 10 YIFAN XU s y1 s u s x2 sy2 s v s x1 l l l l l l l l l s x1 s u s y1 s x2 s v s y2 Figure 4. Case (1) and (2), with colors indicating bipartition Now we are ready for the actual proof. Theorem 4.10. (Kuratowski’s Theorem) A graph is planar if and only if it does not contain any subdivision of K5 or K3,3. Proof. (⇒) It is true by the ﬁrst two lemmas in this section. (⇐) Suppose there exists a nonplanar graph that does not contain any subdi-vision of K5 or K3,3. Without loss of generality, let G be a nonplanar graph that contains no subdivision K5 or K3,3 and has the fewest edges possible. Then G is a minimal nonplanar graph. By Lemma 4.7, G is 3-connected (and simple). Take two adjacent vertices u, v ∈V (G). Consider the subgraph G −uv. By minimality, G −uv is planar. By Lemma 4.9, G −uv is 2-connected. By Lemma 4.8, there are at least two internally-disjoint paths between u and v. In other words, u and v lie on some common cycle. Among all cycles containing u and v in a planar embedding of G −uv, ﬁnd C0 with the most edges in Int(C0). Now consider the bridges of C0 in G −uv (if G −uv does not contain any bridge of C0, then it is clear that with the addition of edge uv, the graph is still planar, which means G is planar, a contradiction). Suppose there exists a bridge with only one vertex of attachment v1. Then v1 is a one-vertex cut of G −uv, which KURATOWSKI’S THEOREM 11 s x1(y1) s u(y3) sx2(y2) s v(y4) s z s x1(y1) s u(y3) s z2 sx2(y2) s v(y4) s z1 Figure 5. Case (3) and (4) contradicts the condition that G −uv is 2-connected. Therefore, all bridges of C0 in G −uv have at least two vertices of attachment. Moreover, if an outer bridge of C0 has more than 2 vertices of attachment, we can always ﬁnd a new cycle that contains parts of the outer bridge and has more edges in its interior. Therefore all outer bridges of C0 have exactly 2 vertices of attachment. Following the same argument, if an outer bridge avoids the arc uv, then there would be another cycle with more edges in the interior. Hence all outer bridges overlap the arc uv, that is, for any outer bridge, not all vertices of attachment lie on the same arc uv. Also, if the size of an outer bridge is more than one, there exists a vertex that is not on C0 in the bridge. Then the two vertices of attachment form a 2-vertex cut of both G −uv and G, which contradicts the condition of G being 3-connected. Therefore, we can conclude that all outer bridges of C0 have 2 vertices of attachment, have the size 1, and overlap the arc uv. Find an outer bridge B1 and an inner bridge B2 that overlap. Justiﬁcation for ﬁnding such B1, B2 is as follows. If all bridges of C0 are inner (outer) bridges, then we can draw the edge uv in the exterior (interior) of C0 and achieve a planar 12 YIFAN XU embedding of G, which contradicts the hypothesis. Hence C0 has to have both inner and outer bridges. The reason why there exists a pair that overlap is that if not, then every inner bridge of C0 avoids every outer bridge, and by Theorem 3.16, all inner bridges of C0 are transferrable. We can then ﬁnd a planar embedding of G −uv where C0 have only outer bridges, which again contradicts the hypothesis. Let the vertices of attachment of B1 be x1, x2, and those of B2 be y1, y2, y3, .... We know that B2 overlaps the arc uv, and is skew to B1. We consider 4 cases in terms of the relative position of B1 and B2. Without loss of generality, we assume that u, x2, v, x1 lie on the cycle in a clockwise order. (1) Among all vertices of attachment of B2, there exist y1, y2 such that y1 lies between x1 and v, y2 between u and x2. Then G contains a subdivision of K3,3, which is against our assumption. (2) There exist y1, y2 such that y1 lies between x2 and v, y2 between x1 and u. Still G contains a subdivision of K3,3, a contradiction. (3) There exist {y1, y2, y3, y4} = {x1, x2, u, v} such that the u-v path P1 and the x1-x2 path P2 have one and only one vertex z in common (P1 and P2 must have some vertices in common because of the planarity of G −uv). Then G contains a subdivision of K5, a contradiction. (4) There exist {y1, y2, y3, y4} = {x1, x2, u, v} such that P1 and P2 have more than one vertex in common. Then G again contains a subdivision of K3,3. By now we have covered every possible case and derived a contradiction from each of them. Therefore, the theorem is true. □ Acknowledgments. It is a pleasure to thank my mentor, Reid Harris, for his help-ful guidance and advice. I would also like to thank Professor Babai for introducing me to graph theory and Professor May for organizing the REU. References J. A. Bondy and U. S. R. Murty. Graph Theory with Applications. The Macmillan Press Ltd., 1982, page 1-15, 143-156. Walter Klotz. ”A constructive proof of Kuratowski’s theorem.” publication/256078009.
15225	https://math.stackexchange.com/questions/2010063/trigonometric-identify-1-cosx-sinx-cotx-2	trigonometry - Trigonometric Identify: $(1+\cos{x})/\sin{x}=\cot{(x/2)}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Trigonometric Identify: (1+cos x)/sin x=cot(x/2)(1+cos⁡x)/sin⁡x=cot⁡(x/2) Ask Question Asked 8 years, 10 months ago Modified8 years, 10 months ago Viewed 7k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I am having trouble with this. I have tried many times already. I just can not figure out how to deal with the x/2 x/2. 1+cos x sin x=cot x 2 1+cos⁡x sin⁡x=cot⁡x 2 trigonometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 12, 2016 at 3:50 Mc Cheng 2,314 15 15 silver badges 27 27 bronze badges asked Nov 12, 2016 at 3:29 jon kjon k 13 1 1 silver badge 4 4 bronze badges Add a comment\| 4 Answers 4 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Notice cos x=cos 2(x/2)−sin 2(x/2)cos⁡x=cos 2⁡(x/2)−sin 2⁡(x/2) and sin x=2 sin(x/2)cos(x/2)sin⁡x=2 sin⁡(x/2)cos⁡(x/2). Thus, 1+cos x sin x=1+cos 2(x/2)−sin 2(x/2)2 sin(x/2)cos(x/2)=2 cos 2(x/2)2 sin(x/2)cos(x/2)=cos(x/2)sin(x/2)=cot(x/2)1+cos⁡x sin⁡x=1+cos 2⁡(x/2)−sin 2⁡(x/2)2 sin⁡(x/2)cos⁡(x/2)=2 cos 2⁡(x/2)2 sin⁡(x/2)cos⁡(x/2)=cos⁡(x/2)sin⁡(x/2)=cot⁡(x/2) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 12, 2016 at 3:33 ILoveMathILoveMath 11.1k 9 9 gold badges 57 57 silver badges 128 128 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. cot(x/2)=cos(x/2)sin(x/2)=2 cos 2(x/2)sin(x)=cos(x)+1 sin(x)cot⁡(x/2)=cos⁡(x/2)sin⁡(x/2)=2 cos 2⁡(x/2)sin⁡(x)=cos⁡(x)+1 sin⁡(x) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 12, 2016 at 4:02 robjohn♦robjohn 355k 38 38 gold badges 499 499 silver badges 892 892 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. (1+cos x sin x)2=(1+cos x)2 1−cos 2 x(1+cos⁡x sin⁡x)2=(1+cos⁡x)2 1−cos 2⁡x =1+cos x 1−cos x=1+cos x 2 1−cos x 2=1+cos⁡x 1−cos⁡x=1+cos⁡x 2 1−cos⁡x 2 =cos 2(x/2)sin 2(x/2)=cot 2(x 2)=cos 2⁡(x/2)sin 2⁡(x/2)=cot 2⁡(x 2) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 12, 2016 at 4:12 answered Nov 12, 2016 at 3:31 RJMRJM 1,127 9 9 silver badges 21 21 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. 1+c o s x s i n x=c o t x 2 1+c o s x s i n x=c o t x 2 1+c o s x=2 c o s 2 x 2 1+c o s x=2 c o s 2 x 2 s i n x=2 s i n x 2 c o s x 2 s i n x=2 s i n x 2 c o s x 2 ⇒2 c o s 2 x 2 2 s i n x 2 c o s x 2=c o s x 2 s i n x 2⇒2 c o s 2 x 2 2 s i n x 2 c o s x 2=c o s x 2 s i n x 2 c o s x s i n x=c o t x c o s x s i n x=c o t x ⇒c o s x 2 s i n x 2=c o t x 2⇒c o s x 2 s i n x 2=c o t x 2 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 12, 2016 at 6:11 Deepak SuwalkaDeepak Suwalka 2,372 2 2 gold badges 15 15 silver badges 23 23 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1Trigonometric formula simplifies to sin x cos x[tan x+cot x]sin⁡x cos⁡x[tan⁡x+cot⁡x] 5Solve trigonometric equation cot x+cos x=1+cot x cos x cot⁡x+cos⁡x=1+cot⁡x cos⁡x 0Trig identity cos x sec x+sin x csc x=csc 2 x−cot 2 x cos⁡x sec⁡x+sin⁡x csc⁡x=csc 2⁡x−cot 2⁡x 0Prove (1+sin x+cos x−cos 2 x sin x+sin 2 x cos x)/(cos x sin x)=(1+sin 3 x+cos 3 x)/(cos x sin x)(1+sin⁡x+cos⁡x−cos 2⁡x sin⁡x+sin 2⁡x cos⁡x)/(cos⁡x sin⁡x)=(1+sin 3⁡x+cos 3⁡x)/(cos⁡x sin⁡x) 5Prove that (sin 20∘+cos 20∘)2 cos 40∘=cot 25∘(sin⁡20∘+cos⁡20∘)2 cos⁡40∘=cot⁡25∘ 4Solve this trigonometric equation tan 2 x+cot 2 x−3(tan x−cot x)=0 tan 2 x+cot 2 x−3(tan⁡x−cot⁡x)=0 0Prove cos(x)1−tan(x)+sin(x)1−cot(x)=sin(x)+cos(x)cos⁡(x)1−tan⁡(x)+sin⁡(x)1−cot⁡(x)=sin⁡(x)+cos⁡(x) 2Proving (sec 2 x+tan 2 x)(csc 2 x+cot 2 x)=1+2 sec 2 x csc 2 x(sec 2⁡x+tan 2⁡x)(csc 2⁡x+cot 2⁡x)=1+2 sec 2⁡x csc 2⁡x and cos x 1−tan x+sin x 1−cot x=sin x+cos x cos⁡x 1−tan⁡x+sin⁡x 1−cot⁡x=sin⁡x+cos⁡x 1Limit lim x→0(cot x−1 sin x)lim x→0(cot⁡x−1 sin⁡x) 0Solving the inequality (tan(4 x)+cot(4 x)−5 2)(cos(2 x)+\|sin(x)\|)>0(tan⁡(4 x)+cot⁡(4 x)−5 2)(cos⁡(2 x)+\|sin⁡(x)\|)>0 Hot Network Questions On being a Maître de conférence (France): Importance of Postdoc How exactly are random assignments of cases to US Federal Judges implemented? 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15226	https://www.britannica.com/science/conditional-probability	SUBSCRIBE Ask the Chatbot Games & Quizzes History & Society Science & Tech Biographies Animals & Nature Geography & Travel Arts & Culture ProCon Money Videos conditional probability mathematics Print Written by Stephen Eldridge Stephen Eldridge is a writer and editor of fiction and nonfiction for all ages. Stephen Eldridge Fact-checked by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Last Updated: • Article History Related Topics: : Monty Hall problem : logistic regression : stationary transition probability : transition probability : a priori distribution On the Web: : IOPscience - Journal of Physics A: Mathematical and Theoretical - Conditional probability framework for entanglement and its decoupling from tensor product structure (PDF) (Aug. 23, 2025) See all related content conditional probability, the probability that an event occurs given the knowledge that another event has occurred. Understanding conditional probability is necessary to accurately calculate probability when dealing with dependent events. Dependent events can be contrasted with independent events. A dependent event is one where the probability of the event occurring is affected by whether or not another event occurred. In contrast, an independent event is one where the probability of the event occurring is the same regardless of the outcome of any other events. Suppose one draws two cards from a standard deck. If the deck is well shuffled, the chance of the first card being red would be 26 out of 52, or 50 percent. However, when one draws the second card, the odds have changed because there is now one less card in the deck. The probability of the second card being red is dependent on the first card being red. This second draw is a dependent event and so is a scenario in which conditional probability would be used. More From Britannica probability theory: Conditional probability However, if one replaces the first card and reshuffles before drawing another card, both draws are made with the full deck of cards. The second draw is no longer affected by the result of the first draw, and so these events are independent. Conditional probability would not apply in this scenario. The probability that an event A will occur given another event B occurs is written as P(A\|B), meaning “the probability of A given B.” Assuming that the probability of B is not zero, this can be calculated using the formulaP(A\|B) = P(A ∩ B)/P(B).Here P(A ∩ B) is the probability of A and B, meaning A and B both occur. This is called the intersection of A and B. P(B) is the probability of B. For example, imaging someone playing a video game against a computer opponent. The human player wants to know if going first (event B) affects the probability that one will win the game (event A). Doing some observation, one can construct a probability distribution table for the two events, using 1 as the true condition for the event and 0 as the false condition. \| \| A = 0 (computer wins) \| A = 1 (human wins) \| P(B) \| --- --- \| \| B = 0 (computer goes first) \| 0.25 \| 0.25 \| 0.5 \| \| B = 1 (human goes first) \| 0.15 \| 0.35 \| 0.5 \| \| P(A) \| 0.4 \| 0.6 \| 1 \| In 35 percent of games, it is true both that the human player goes first (B = 1) and wins the game (A = 1). This is expressed as P(A ∩ B) = 0.35. To know the conditional probability P(A\|B), the probability of the human player’s victory given the human player goes first, one also needs to know P(B), or the probability of the human player going first (B = 1). In the table, P(B) = 0.5. Access for the whole family! Bundle Britannica Premium and Kids for the ultimate resource destination. Subscribe Dividing 0.35 by 0.5 results in P(A\|B) = 0.7. Given the player goes first, the probability of the human player winning the game is 70 percent. Because that is higher than the overall probability of the human player winning, P(A) = 0.6, going first improves the chances of the human player winning the game. Note that P(A\|B) (the probability of A given B) and P(B\|A) (the probability of B given A) are rarely the same. To find the relationship between the two, one uses Bayes’s theorem, named after 18th-century clergyman Thomas Bayes. Bayes’s theorem allows one to find “reverse” probability, meaning it allows you to calculate the probability of an event having occurred given a later dependent event having occurred. Bayes’s theorem is an extension of the equation above, often represented asP(A\|B) = P(A ∩ B)/P(B) = P(A)P(B\|A)/P(B).For example, suppose that a doctor performs a test to determine if a patient has a particular genetic condition. The prevalence P(A) of the condition in the population is 0.01, or 1 percent, and thus the probability of not having the condition P(not-A) is 0.99 or 99 percent. The chance that someone with the condition gets a positive test result B when they have the condition is P(B\|A) = 0.95, or 95 percent. The chance of someone without it getting a false positive test result, P(B\|not-A) is 0.02, or 2 percent. Given a positive test result B, the doctor wants to know the probability that the patient really has the condition, P(A\|B). To use Bayes’s theorem, one needs P(A), P(B\|A), and P(B). The first two items have already been stated; P(A) = 0.01, and P(B\|A) = 0.95. To find P(B), the probability of getting a positive test result, one must consider that people both with and without the condition get a positive result. Therefore one must find the number of people with the condition who get a positive by multiplying P(B\|A) by P(A), then add that result to P(B\|not-A) multiplied by P(not-A). That is,P(B) = P(B\|A)P(A) + P(B\|not-A)P(not-A),P(B) = 0.95(0.01) + 0.02(0.99) = 0.0293. Inserting this result into Bayes’s theorem to find P(A\|B), 0.01(0.95)/0.0293 = 0.0095/0.0293 = 0.3242.The chance of a patient who receives a positive test result actually having the condition is about 32 percent. Stephen Eldridge inference Introduction References & Edit History Quick Facts & Related Topics Quizzes Numbers and Mathematics inference statistics Print Also known as: statistical inference Written by The Editors of Encyclopaedia Britannica Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree.... The Editors of Encyclopaedia Britannica Last Updated: • Article History Related Topics: : statistics : classical inference On the Web: : OpenStax - Principles of Data Science - Statistical Inference and Confidence Intervals (Aug. 29, 2025) See all related content inference, in statistics, the process of drawing conclusions about a parameter one is seeking to measure or estimate. Often scientists have many measurements of an object—say, the mass of an electron—and wish to choose the best measure. One principal approach of statistical inference is Bayesian estimation, which incorporates reasonable expectations or prior judgments (perhaps based on previous studies), as well as new observations or experimental results. Another method is the likelihood approach, in which “prior probabilities” are eschewed in favour of calculating a value of the parameter that would be most “likely” to produce the observed distribution of experimental outcomes. In parametric inference, a particular mathematical form of the distribution function is assumed. Nonparametric inference avoids this assumption and is used to estimate parameter values of an unknown distribution having an unknown functional form. Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Eldridge, Stephen. "conditional probability". Encyclopedia Britannica, 23 Aug. 2025, Accessed 15 September 2025. Share Share to social media Facebook X External Websites PennState - Eberly College of Science - Conditional Probability LOUIS Pressbooks - Finite Mathematics - Conditional Probability and Bayes Theory Open Library Publishing Platform - Introduction to Statistics - Conditional Probability University of Connecticut - Conditional probability Stanford University - Conditional Probability IOPscience - Journal of Physics A: Mathematical and Theoretical - Conditional probability framework for entanglement and its decoupling from tensor product structure (PDF) Massachusetts Institute of Technology - Conditional Probability (PDF) Corporate Finance Institiute - Conditional Probability Khan Academy - Conditional probability and independence Toronto Metropolitan University Pressbooks - Mathematics for Public and Occupational Health Professionals - Conditional Probability Mathematics LibreTexts - Conditional Probability Feedback Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. verifiedCite While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style The Editors of Encyclopaedia Britannica. "inference". Encyclopedia Britannica, 29 Aug. 2025, Accessed 15 September 2025. Share Share to social media Facebook X URL External Websites University of Toronto - Department of Statistical Sciences - On Some Principles of Statistical Inference University of Wisconsin Pressbooks - Process of Science Companion: Data Analysis, Statistics and Experimental Design - Statistical Inference ï¿½ Basic Concepts National Center for Biotechnology Information - PubMed Central - Statistical inference through estimation: Recommendations from the International Society of Physiotherapy Journal Editors OpenStax - Principles of Data Science - Statistical Inference and Confidence Intervals University of Bristol - 4 Ideas of statistical inference Corporate Finance Institute - Inferential Statistics Statistics LibreTexts - Foundations for Inference
15227	https://community.monogame.net/t/can-someone-explain-override-abstract-and-virtual/19698	Can someone explain: override,abstract and virtual? Hi so I’m having trouble understanding what each bit does?? Also if someone could tell me what would be the best for a base class??? C# Keywords - C# C# Keywords thx <3 About the C# Player’s Guide (csharpplayersguide.com) This might be the book for you @Shawty I recently completed it, highly recommend joining their Discord. Its very hard to read lots of dry documentation - but its also very hard to simplify it to the point of getting the concepts. But I’ll give it a shot. (20 years of C# speaking) classes used as a base-class All it means, is I create a class that inherit from some other class. For example, you might create a class to represent sprites in your game. Then you find that you need a more specialised kind of sprite to represent the main player on screen, but the main player is also a sprite. This means when you have “class PlayerSprite : Sprite” you also put a player instance in a collection of Sprite instances, such as a List because the PlayerSprite class IS A Sprite instance, just a more specialised kind. classes as abstract When you have methods, and/or properties that you want to share, but the class itself doesn’t make sense, as it may be incomplete - and needs more to make it useful. For example, you may never want to create an instance of an Animal, as its too abstract, where as a class like Dog:Animal makes more sense. So adding ‘abstract’ to the Animal definition - “abstract class Animal” means you enforce the fact no one can make an instance of that type, but are free to inherit from it. methods as abstract The biggest benefit of abstract methods, is that you can write code in your abstract class that calls the abstract method - more on that in a second. You then inherit from your abstract class, and are forced to implement the abstract methods. Which means you’re not providing any base functionality unlike a virtual method. The inheriting class must provide the real code. Back in the base class then, you can write methods that call the abstract method, and in reality, it will call the implementation provided by the inheriting class. Therefore a Cat:Animal could provide a different implementation for an abstract method called “MakeANoise()” that what a Dog:Animal will provide. This is a powerful concept to grasp. methods as virtual / override Unlike an abstract method, a virtual method is optional. You can provide code in a base-class, and optionally override it. Simplest example is “.ToString()” method. By default that will return the name of your type. But you can optionally choose to override this method, and represent some value of the class, for example “I am a Dog, and I make a Bark! noise”. Often I use this to provide some debugging information. When debugging in Visual Studio, it uses the “.ToString()” when you hover over a collection of a type, such as Animal. In this case, it would give me more useful information. But this technique has many many possibilities, and is incredibly powerful in altering the behaviour of a class. (side effects). I hope this gives a little insight - and if you’re hungry for more, hit the Microsoft docs! class keyword - C# Reference - C# \| Microsoft Learn // In C#, override, abstract and virtual are keywords that are used to define the behavior of methods in classes and interfaces. Let’s see what they mean and how they work. // A method is a block of code that performs a specific task, such as adding two numbers or printing a message. A method can have a name, parameters and a return value. For example: public int Add(int x, int y) // This is a method named Add that takes two integers as parameters and returns their sum { return x + y; // This is the body of the method that contains the code to execute } // A class is a blueprint for creating objects, which are instances of the class. A class can have fields, properties and methods that define the state and behavior of the objects. For example: public class Calculator // This is a class named Calculator { public int Add(int x, int y) // This is a method of the class that can be called by the objects { return x + y; } } // An interface is a contract that specifies what methods a class must implement if it wants to conform to the interface. An interface does not provide any implementation for the methods, it only declares their names, parameters and return values. For example: public interface ICalculator // This is an interface named ICalculator { int Add(int x, int y); // This is a method declaration that does not have a body } // A class can implement an interface by using the colon ( operator and providing the implementation for all the methods declared in the interface. For example: public class SimpleCalculator : ICalculator // This is a class named SimpleCalculator that implements the ICalculator interface { public int Add(int x, int y) // This is the implementation of the Add method declared in the interface { return x + y; } } // Now, let’s see what override, abstract and virtual mean. // Override is a keyword that is used to modify a method in a derived class that has the same name and signature as a method in a base class. A derived class is a class that inherits from another class, which is called the base class. For example: public class Animal // This is a base class named Animal { public virtual void MakeSound() // This is a method named MakeSound that can be overridden by derived classes { Console.WriteLine("Generic animal sound"); } } public class Dog : Animal // This is a derived class named Dog that inherits from Animal { public override void MakeSound() // This is a method named MakeSound that overrides the method in the base class { Console.WriteLine("Woof"); } } // In this example, the Animal class has a method named MakeSound that prints “Generic animal sound”. The Dog class inherits from Animal and overrides the MakeSound method to print “Woof”. The override keyword indicates that this method replaces the behavior of the method in the base class. // Abstract is a keyword that is used to declare a method or a class that cannot be instantiated or implemented directly. An abstract method is a method that has no body and must be overridden by derived classes. An abstract class is a class that has at least one abstract method and cannot be used to create objects. For example: public abstract class Shape // This is an abstract class named Shape { public abstract double Area(); // This is an abstract method named Area that must be overridden by derived classes } public class Circle : Shape // This is a derived class named Circle that inherits from Shape { private double radius; // This is a field of the class public Circle(double radius) // This is a constructor of the class that takes the radius as a parameter { this.radius = radius; // This assigns the parameter value to the field } public override double Area() // This is a method named Area that overrides the abstract method in the base class { return Math.PI radius radius; // This returns the area of the circle using the formula pi r^2 } } // In this example, the Shape class is an abstract class that has an abstract method named Area. The Circle class inherits from Shape and overrides the Area method to provide the implementation for calculating the area of a circle. The abstract keyword indicates that this class or method cannot be used directly and must be inherited or overridden. // Virtual is a keyword that is used to declare a method that can be overridden by derived classes, but also has a default implementation in the base class. A virtual method is similar to an abstract method, but it provides a body that can be executed if no derived class overrides it. For example: public class Vehicle // This is a base class named Vehicle { public virtual void Start() // This is a virtual method named Start that can be overridden by derived classes { Console.WriteLine("Starting the vehicle"); } } public class Car : Vehicle // This is a derived class named Car that inherits from Vehicle { public override void Start() // This is a method named Start that overrides the virtual method in the base class { Console.WriteLine("Starting the car"); } } public class Bicycle : Vehicle // This is a derived class named Bicycle that inherits from Vehicle { // This class does not override the Start method } // In this example, the Vehicle class has a virtual method named Start that prints “Starting the vehicle”. The Car class inherits from Vehicle and overrides the Start method to print “Starting the car”. The Bicycle class also inherits from Vehicle, but does not override the Start method, so it uses the default implementation from the base class. The virtual keyword indicates that this method can be overridden, but also has a fallback behavior. // I hope this explanation helps you understand the meaning and usage of override, abstract and virtual keywords in C# code. If you have any questions, feel free to ask me. Hope that helps… Powered by Discourse, best viewed with JavaScript enabled
15228	https://brainly.com/question/40965681	[FREE] The word "advocate" contains the root "voc." What does the root "voc" mean? A. own or self B. hear or - brainly.com 7 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +40,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +22,1k Ace exams faster, with practice that adapts to you Practice Worksheets +8,7k Guided help for every grade, topic or textbook Complete See more / English Textbook & Expert-Verified Textbook & Expert-Verified The word "advocate" contains the root "voc." What does the root "voc" mean? A. own or self B. hear or listen C. call or speak 1 See answer Explain with Learning Companion NEW Asked by patientNight95 • 10/28/2023 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 11128 people 11K 0.0 0 Upload your school material for a more relevant answer The root 'voc' in the word 'advocate' means 'call or speak'. Explanation The root 'voc' in the word 'advocate' means 'call or speak'. It comes from the Latin word 'vocare', which means 'to call' or 'to speak'. In the word 'advocate', the 'voc' root signifies someone who speaks or calls in support of a cause or person. Learn more about Meaning of the root 'voc' in the word 'advocate' here: brainly.com/question/30367002 Answered by wildField49 •181 answers•11.1K people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 11128 people 11K 0.0 0 Threshold Concepts in Biochemistry - Julian Pakay; Hendrika Duivenvoorden; Thomas Shafee; and Kaitlin Clarke Simplified Signs A Manual Sign-Communication System for Special Populations, Volume 1 - John D. Bonvillian; Nicole Kissane Lee; Tracy T. Dooley; Filip T. Loncke Communication in the Real World Upload your school material for a more relevant answer The root 'voc' in 'advocate' means 'call or speak', deriving from the Latin 'vocare'. It indicates someone who speaks in favor of a cause. Understanding this root helps clarify similar terms and their meanings. Explanation The root 'voc' in the word 'advocate' means 'call or speak'. This root comes from the Latin word 'vocare', which translates to 'to call' or 'to speak'. In the context of the word 'advocate', the 'voc' root signifies someone who speaks up or calls for support of a cause or person. Understanding this root can help clarify the meaning of not only 'advocate' but also related words such as 'vocal' (pertaining to the voice) and 'invoke' (to call upon). In advocacy, it is the act of publicly recommending or supporting a particular cause or policy, which ties back to the meaning of 'call'. Thus, knowing the meaning behind the root 'voc' can greatly enhance your understanding of various terms in both everyday language and specific fields such as law and public service. Examples & Evidence An example of using 'voc' can be found in the term 'vocal', which refers to using the voice to express oneself, such as singing or speaking in public. Another example is 'invoke', which means to call upon a deity or a greater power for assistance or support. The definition of 'voc' is supported by etymological dictionaries that trace many English words back to their Latin origins, confirming its meaning connected to calling or speaking. Thanks 0 0.0 (0 votes) Advertisement patientNight95 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free English solutions and answers Community Answer the word advocate contains the root voc. what does the root voc mean? Community Answer 9 The criteria retailer must meet to receive a reduced penalty and/or protect the license/permit if an illegal alcohol sale takes place at the establishment is often referred to Community Answer 4.1 240 Based on the passage you just read, what conclusion can you draw about the cultural values of the iroquois? “The right-handed twin accused his brother of murdering their mother, and their quarrels continued until it was time to bury their mother. With the help of their grandmother, they made her a grave. From her head grew the three sister plants: corn, beans, and squash. From her heart grew tobacco, which people still use to give thanks in ceremony. She is called ‘our mother’ and the people dance and sing to her to make the plants grow.” Community Answer 2 Making them clean the floors would be a(n) because it would be outside their usual duties, Community Answer 5 a letter to the Ambassador highlighting hardship faced by Nigerian students and soliciting for assistance Community Answer 19 What does Stitch say in the beginning of "Lilo and Stitch" in Alien to make the robot throw up Community Answer 2 Then, from heaven, the voice of the god called to Gilgamesh: "Hurry, attack, attack Humbaba while the time is right, before he enters the depths of the forest, before he can hide there and wrap himself in his seven auras with their paralyzing glare. He is wearing just one now. Attack him! Now!" –Gilgamesh: A New English Translation, Stephen Mitchell Identify a feature of epic poetry in the passage. In three to four sentences, explain its impact on the epic's plot. Community Answer 4.5 487 Write two to three sentences explaining how Gilgamesh demonstrates the characteristics of an epic hero. Use evidence from the text to support your answer New questions in English An alphabetical list of topics, along with the page numbers where each topic appears is the A. glossary B. appendix C. index D. table of contents Motivation to read faster is not a factor in your reading speed. A. True B. False If the body of the speech is well developed, you don't always need to prepare an introduction and conclusion ahead of time. Perspective is A. a version or adaptation of a story. B. a way of thinking about or seeing a situation. C. the way a character is presented and developed. D. the position from which a story is narrated. For many listeners, an intriguing feature of podcasts is which of the following? A. Biased recommendations generated by endorsements B. Live platforms allowing for real-time conversation with other listeners C. Interviews with celebrities, scientists, and successful businesspeople Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
15229	https://www.ncbi.nlm.nih.gov/books/NBK513060/	Vitamin A - Drugs and Lactation Database (LactMed®) - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log in Show account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. Drugs and Lactation Database (LactMed®) [Internet]. Bethesda (MD): National Institute of Child Health and Human Development; 2006-. Drugs and Lactation Database (LactMed®) [Internet]. Show details Bethesda (MD): National Institute of Child Health and Human Development; 2006-. Contents Search term < PrevNext > Vitamin A Last Revision: September 15, 2024. Estimated reading time: 15 minutes CASRN: 68-26-8; 79-81-2 Go to: Drug Levels and Effects Summary of Use during Lactation Vitamin A (retinol) and provitamin A carotenoids are normal components of human milk. Vitamin A in milk has antioxidant properties. The recommend dietary intake in lactating women is 1300 mcg retinol daily, compared to 770 mcg daily during pregnancy. The recommended daily intake for infants aged 6 months or less is 400 mcg. Vitamin A deficiency is not uncommon, and maternal supplementation may be needed during lactation to achieve the recommended daily vitamin A intake. Maternal single doses up to 120,000 mcg (400,000 IU) or daily doses up to 1500 mcg (5,000 IU) are not expected to harm the breastfed infant. Higher maternal doses have not been studied. Daily maternal doses above 3,000 mcg (10,000 IU) should be avoided. Drug Levels Vitamin A (retinol) measurements are expressed in the literature as mcg, IU, or micromoles. One mcg is equal to 3.33 IU (0.3 mcg = 1 IU), and also to 0.0035 micromoles (286 mcg = 1 micromole). Beta-carotene from vitamin supplements is one-half as potent as retinol. One mcg of beta-carotene from vitamin supplements is therefore equal to 0.5 mcg retinol (see the LactMed beta-carotene record). A serum retinol concentration less than 0.7 micromoles/L (<200 mcg/L) indicates biochemical vitamin A deficiency, and 0.7 to 1.05 micromoles/L (200 to 300 mcg/L) indicates marginal deficiency. Milk levels less than 1.05 micromoles/L (<300 mcg/L) indicates low milk vitamin A content. Most retinol present in milk are retinyl esters (e.g., retinyl acetate), which can be de-esterified in the infant's digestive tract. Mature milk retinol levels correlate with maternal serum retinol levels, and levels of milk retinol and beta-carotene are correlated with one another.[6-11] Milk retinol and carotenoid levels in colostrum are higher than in mature milk. Intra-individual colostrum levels vary widely, but are consistently higher than and not correlated with maternal serum levels, suggesting active colostrum enrichment. Levels decrease over the first 1 to 2 weeks postpartum and then stabilize.[2,13,14] Vitamin A partitions into milk fat and milk retinol levels are positively correlated with milk fat.[9,15] Hindmilk levels are higher than foremilk.[16,17] Milk vitamin A content increases as maternal dietary vitamin A increases. Nutritionally deficient mothers have lower milk retinol and provitamin A carotenoid levels than those of well-nourished mothers.[6,19,20] Premature birth may also be associated with lower milk retinol levels.[21,22] Maternal Levels. Healthy mothers not taking a supplement have retinol levels averaging 800 to 1400 mcg/L in colostrum and 300 to 800 mcg/L in mature milk.[10,12,14,17,20,23,24] Beta-carotene, lutein and lycopene are among the most abundant carotenoids in milk, totaling 300 to 400 mcg/L in the first week postpartum and 50 to 200 mcg/L in mature milk.[1,10,13,14,20] Numerous studies targeting lactating women living in regions with endemic vitamin A deficiency have evaluated the effects of a single vitamin A dose of 200,000 to 400,000 IU given in the early postpartum period. Most of these studies demonstrate small increases in milk retinol levels compared to the control or placebo group during the first 3 months postpartum. In a meta-analysis of six such studies, the average increase in milk retinol was 57 mcg/L in the mothers treated with vitamin A compared to placebo. Studies that controlled for milk fat content had similar outcomes. In two of the analyzed studies, the frequencies of low milk levels (<300 mcg/L) were 25% and 50% lower in supplemented mothers compared to control. Based on the average increase in milk retinol reported in this meta-analysis, the single, high-dose vitamin A postpartum maternal supplementation strategy would provide breastfed infants with an extra 3.5 to 13 mcg/kg of retinol daily after the first postpartum week. A study involving 102 mothers from Hyderabad, India given 200,000 IU of vitamin A within 24 hours of birth reported higher average milk retinol levels in the vitamin A group at postpartum day 10 (1100 mcg/L vs. 800 mcg/L), and day 30 (750 mcg/L vs 600 mcg/L), but not on or after day 45. The increase in milk retinol intake from maternal supplementation during the first month after birth for a breastfed infant based on these results would be about 30 mcg/kg daily. In a similar study conducted in northeastern Brazil, average milk retinol was higher in the colostrum of the supplemented group 24 hours after the dose (2360 mcg/L vs 1010 mcg/L), but was not significantly different at 30 days (525 mcg/L vs. 478 mcg/L). A large study conducted in Ghana, Peru, and India gave vitamin A 200,000 IU or placebo to 2,990 breastfeeding mothers at 18 to 42 days postpartum. The average milk retinol level at baseline was similar in both groups (52.2 vs 51.6 nmol/gram of fat), but was higher at 2 months postpartum in the vitamin A group (49.8 vs. 42.7 nmol/gram of fat). The percentage of mothers at 2 months with milk retinol of 28 nanomoles/gram of fat or less, indicating low milk vitamin A content, was 49% in the treatment group compared to 79% in the placebo group. At 6 months postpartum, there were no differences between the groups. Studies comparing a 400,000 IU dose to a 200,000 IU dose of vitamin A have found no differences in milk vitamin A levels between the two dosage groups.[28-32] One hundred forty healthy, low-income mothers in Rabat, Morocco were given 200,000 IU of vitamin A as a one-time dose beginning at 2 to 3 weeks after birth. In addition, the mothers were randomized to receive vegetable oil fortified with retinyl palmitate 30 IU per gram of oil, or non-fortified oil. The oil was to be used in the home during routine food preparation, not as a daily required study dose. Both groups had milk retinol levels near 515 mcg/L at baseline. After three months, average milk levels were 386 mcg/L in the nonfortified group and 592 mcg/L in the fortified group; 20% of mothers in the nonfortified group had low milk levels (<300 mcg/L) compared to none in the fortified group. After 6 months, the differences were 172 mcg/L vs. 398 mcg/L, and 100% vs. 2.7%, respectively. Based on the average milk levels reported, exclusively breastfed infants would receive approximately an extra 30 mcg/kg retinol daily from the regimen used in this study. Daily supplementation of 2165 IU vitamin A was compared to no supplementation in 119 Gambian pregnant and postpartum women. Milk retinol levels collected once weekly between postpartum weeks 3 and 15 were higher in the supplemented mothers, ranging from 800 to 1000 mcg/L, compared to 500 to 700 mcg/L in the non-supplemented mothers. Adjusting for milk fat content did not change the results. Based on the milk levels reported, breastfed infants would receive an extra 45 mcg/kg retinol daily as a result of this long-term, pregnancy through postpartum extended supplementation approach. Two hundred forty-five healthy, exclusively breastfeeding mothers in Nanjing, China were randomized to receive a daily vitamin A supplement of 1800 IU or placebo for 2 months beginning at 1 to 2 months postpartum. Both groups had similar average milk retinol levels of approximately 450 mcg/L at baseline. After two months, average milk level in the supplemented group was 290 mcg/L compared to 240 mcg/L in the control group. The difference amounts to an extra 7.5 mcg/kg retinol daily received by the breastfed infant from the short period of maternal supplementation. HIV-infected pregnant women with vitamin A deficiency or marginal deficiency in Tanzania were enrolled in a study to receive 1 of 4 supplements during pregnancy and lactation. Groups received either multivitamins (thiamine, riboflavin, vitamin B6, niacin, vitamin B12, vitamin C, vitamin E, and folic acid), multivitamins plus vitamin A 5000 IU and beta-carotene 30 mg, the same doses of vitamin A and beta-carotene alone, or placebo once daily beginning in their second trimester and continuing for two years. Breastmilk samples were collected at delivery and at 3-month intervals thereafter up to 1 year. Breastmilk retinol levels in the vitamin A and beta-carotene supplemented groups were consistently about 2 to 3 micromoles/L (570 to 860 mcg/L) higher during the entire study period than in the groups that received no supplement. The average milk levels were 1400 mcg/L vs 640 mcg/L at 3 months, 1350 mcg/L vs 640 mcg/L at 6 months, and 1300 mcg/L vs 630 mcg/L at 12 months. Based on the maximum average retinol levels reported, an exclusively breastfed infant would receive an extra 114 mcg/kg daily of retinol from the maternal supplementation used in this study. Premature birth may be associated with lower milk retinol levels. A southeastern Brazilian study of non-supplemented mothers reported average levels of 286 mcg/L at around 30 days postpartum in those who delivered preterm babies compared to 534 mcg/L in those delivering at term. Similarly, in a study from Mexico of non-supplemented mothers who experienced high-risk pregnancies (e.g., preeclampsia, anemia, advanced maternal age or gestational diabetes), premature birth increased the odds of low milk retinol levels compared to term birth, and compared to mothers with normal pregnancies. Infant Levels. One hundred nine Indian mothers were given a single dose of vitamin A 200,000 IU or no vitamin A within 48 hours of delivery. At 3 months postpartum the average infant serum retinol level was higher (1.06 micromoles/L vs 0.77 micromoles/L) and the percentage of infants with serum levels <0.7 micromoles/L was lower (2.6% vs 44.8%) in the vitamin A group. One-half of all infants had umbilical cord retinol levels <0.7 micromoles/L, but differences between the two groups for this baseline measurement were not reported. One hundred forty Indonesian mothers were given a single oral dose of 300,000 IU vitamin A as retinyl palmitate or placebo between 7 and 21 days postpartum. Baseline infant serum retinol was not measured. At 6 months postpartum, average infant serum retinol levels did not differ between the two groups. However, the proportion of infants with levels <0.52 micromoles/L was lower in the treatment group, 15% vs 36%. Breastfeeding rates and infant diet were the same in both groups. One hundred forty mothers in Bangladesh were given 200,000 IU vitamin A or placebo between 1 and 3 weeks postpartum. Baseline infant serum retinol was not measured. Infant vitamin A dietary and supplemental intakes during the first 6 months of age were not different between the two groups. Infants were partially breastfed for 6 months, at which time the average infant serum retinol level was higher in the vitamin A group, 0.84 micromoles/L compared to 0.77 micromoles/L in the placebo group. Biochemical indicators of adequate infant vitamin A storage were also better in the vitamin A group. One hundred ten Gambian women were given 200,000 IU vitamin A shortly after delivery. Breastfeeding rates were not given, but >90% of participants were able to provide milk for study analysis during the 6 months postpartum milk collection period. Average infant serum retinol increased from 0.67 micromoles/L in cord blood to 0.85 micromoles/L at 9 months. The percentage of infants with levels <0.7 micromoles/L decreased from 61% at birth to 31% at 9 months. To compare a 200,000 IU to a 400,000 IU maternal dose, 173 Brazilian mothers were given 200,000 IU vitamin A once after delivery, or once after delivery and again at 10 days postpartum. Their infants were more than 75% breastfed for 6 months, but not supplemented with vitamin A directly. The average infant serum levels in umbilical cord blood and in blood at 2, 4 and 6 months postpartum were similar in both groups: 1.1, 1.2, 1.5, and 1.55 micromoles/L, respectively. The increase compared to baseline was statistically significant in both groups. This study suggests that a single high maternal dose of vitamin A can improve infant vitamin A status through breastmilk, but a 400,000 IU dose offered no added improvement in status over a 200,000 IU maternal postpartum dose. One hundred exclusively breastfeeding Indian mothers were given 200,000 IU vitamin A or placebo as a single oral dose within 24 hours after delivery. Average infant serum retinol levels increased from approximately 0.5 micromoles/L in cord blood to 0.8 to 0.9 micromoles/L at 6 weeks to 6 months in both groups. In Kenya, 564 mothers were given 400,000 IU vitamin A or placebo within 24 hours of birth. About 30% of the mothers were vitamin A deficient at study entry. Their infants were also randomized to receive 100,000 IU vitamin A or placebo at 14 weeks of age. The average serum retinol level was similar in all infants at 14 weeks of age, about 0.9 micromoles/L, and again at 26 weeks among all infants who received placebo, about 1.0 micromoles/L. The lack of a difference did not change when controlling for maternal baseline serum retinol. Breastfeeding rates were not reported. In Brazil, 61 breastfeeding mothers received 200,000 IU vitamin A or placebo between 20 and 30 days postpartum. At 3 months postpartum the average infant serum retinol levels were similar: 0.69 micromoles/L in the vitamin A group and 0.64 micromoles/L in the placebo group. Seventy-five percent of infants were exclusively breastfed throughout the study period. Effects in Breastfed Infants Maternal vitamin A supplementation during pregnancy and lactation has not reduced mortality or the risk of anemia in breastfed infants living in developing regions, but there may be health benefits related to improved antioxidant capacity and immune function.[25,41,42] One hundred exclusively breastfeeding mothers in India were given 200,000 IU vitamin A or placebo as a single oral dose within 24 hours after delivery. There were no differences in weight gain of their infants during postpartum hospitalization between the groups, and no infants in the vitamin A group had symptoms of hypervitaminosis A, such as excessive crying, raised fontanelle and vomiting. All the infants received a dose of oral polio vaccine (OPV) between 48 and 72 hours after birth. There were no differences between the two groups in OPV seroconversion rates or antibody titers at 6 weeks. Similarly, 1,085 exclusively breastfeeding mothers in Ghana were given 200,000 IU vitamin A or placebo at 3 to 4 weeks postpartum. There were no differences in infant immune response to polio or tetanus vaccination at 6 weeks and 6 months of age. The infants of 197 mothers in Gambia given either 200,000 or 400,000 IU vitamin A in the first week after delivery showed no signs of adverse reactions to vitamin A during the 12-month follow-up period. Two hundred forty-five healthy, exclusively breastfeeding mothers in Nanjing, China were randomized to receive a daily vitamin A 1800 IU supplement or placebo for 2 months beginning at 1 to 2 months postpartum. By the end of the 2 month study, infants in both groups had similar rates of febrile illnesses, respiratory tract infections, diarrhea, and eczema, although it was not stated how these data were collected. Effects on Lactation and Breastmilk HIV-infected mothers in Tanzania had a higher rate of severe subclinical mastitis (36%), determined by measuring serial milk electrolyte concentrations, when taking a daily vitamin A supplement during pregnancy and postpartum compared to placebo (23%). Mothers taking a multivitamin without vitamin A also had a higher rate (38%) than placebo, suggesting a vitamin A-independent effect. Given the many other benefits of multivitamin supplementation in this specific patient population, the authors did not recommend avoiding supplementation due to a potential risk of mastitis. In a Brazilian study, 57 mothers received either 200,000 IU vitamin A or no vitamin A within 2 days after birth. At 24 hours after administration, the average level of alpha-tocopherol (vitamin E) was 16.4% lower in the colostrum of mothers who received in the vitamin A (28 micromoles/L compared to 24.5 micromoles/L), but it was not significantly different at 30 days after administration, at about 6 micromoles/L. The reported vitamin E colostrum levels in this study are within normal limits, thus the small reduction in the vitamin A-supplemented mothers may not be clinically important. A different Brazilian research group conducting the same experiment using a vitamin A supplement with low-dose vitamin E 110 IU added as a preservative found no effect of vitamin A on colostrum alpha-tocopherol. Alternate Drugs to Consider Beta-Carotene References 1. Khachik F, Spangler CJ, Smith JC, Jr, et al. Identification, quantification, and relative concentrations of carotenoids and their metabolites in human milk and serum. Anal Chem 1997;69:1873-81. [PubMed: 9164160] 2. Szlagatys-Sidorkiewicz A, Zagierski M, Jankowska A, et al. Longitudinal study of vitamins A, E and lipid oxidative damage in human milk throughout lactation. Early Hum Dev 2012;88:421-4. [PubMed: 22085741] 3. National Institutes of Health Office of Dietary Supplements. Vitamin A fact sheet for health professionals. 2022. 4. Brown E, Akre J. Indicators for assessing vitamin A deficiency and their application in monitoring and evaluating intervention programmes. Geneva: World Health Organization 1996;WHO/NUT/96.10. 5. Lammi-Keefe CJ, Jensen RG. Fat-soluble vitamins in human milk. Nutr Rev 1984;42:365-71. [PubMed: 6396543] 6. Butte NF, Calloway DH. Evaluation of lactational performance of Navajo women. Am J Clin Nutr 1981;34:2210-5. [PubMed: 7293949] 7. Mello-Neto J, Rondo PH, Oshiiwa M, et al. The influence of maternal factors on the concentration of vitamin A in mature breast milk. Clin Nutr 2009;28:178-81. [PubMed: 19249141] 8. Martins TM, Ferraz IS, Daneluzzi JC, et al. Impact of maternal vitamin A supplementation on the mother-infant pair in Brazil. Eur J Clin Nutr 2010;64:1302-7. [PubMed: 20842169] 9. Dror DK, Allen LH. Retinol-to-fat ratio and retinol concentration in human milk show similar time trends and associations with maternal factors at the population level: A systematic review and meta-analysis. Adv Nutr 2018;9 (Suppl 1):332S-46S. [PMC free article: PMC6008956] [PubMed: 29846525] 10. Gebre-Medhin M, Vahlquist A, Hofvander Y, et al. Breast milk composition in Ethiopian and Swedish mothers. I. Vitamin A and beta-carotene. Am J Clin Nutr 1976;29:441-51. [PubMed: 944526] 11. Green MH, Lopez-Teros V, Avila-Prado J, et al. Use of theoretical women and model-based compartmental analysis to evaluate the impact of vitamin A intake with or without a daily vitamin A supplement on vitamin A total body stores and balance during lactation. J Nutr 2024;154:2374-80. [PubMed: 38857673] 12. de Vries JY, Pundir S, Mckenzie E, et al. Maternal circulating vitamin status and colostrum vitamin composition in healthy lactating women-a systematic approach. Nutrients 2018;10:E687. [PMC free article: PMC6024806] [PubMed: 29843443] 13. Gossage CP, Deyhim M, Yamini S, et al. Carotenoid composition of human milk during the first month postpartum and the response to beta-carotene supplementation. Am J Clin Nutr 2002;76:193-7. [PubMed: 12081834] 14. Sakurai T, Furukawa M, Asoh M, et al. Fat-soluble and water-soluble vitamin contents of breast milk from Japanese women. J Nutr Sci Vitaminol (Tokyo) 2005;51:239-47. [PubMed: 16261995] 15. Stoltzfus RJ, Underwood BA. Breast-milk vitamin A as an indicator of the vitamin A status of women and infants. Bull World Health Organ 1995;73:703-11. [PMC free article: PMC2486808] [PubMed: 8846497] 16. Hampel D, Shahab-Ferdows S, Islam MM, et al. Vitamin concentrations in human milk vary with time within feed, circadian rhythm, and single-dose supplementation. J Nutr 2017;147:603-11. [PMC free article: PMC5368580] [PubMed: 28202638] 17. Nimmannun K, Davis CR, Srisakda P, et al. Breast milk retinol concentrations reflect total liver vitamin a reserves and dietary exposure in thai lactating women from urban and rural areas. J Nutr 2023;152:2689-98. [PubMed: 36170963] 18. Kodentsova VM, Vrzhesinskaya OA. Evaluation of the vitamin status in nursing women by vitamin content in breast milk. Bull Exp Biol Med 2006;141:323-7. [PubMed: 17073150] 19. Bates CJ, Prentice A. Breast milk as a source of vitamins, essential minerals and trace elements. Pharmacol Ther 1994;62:193-220. [PubMed: 7991643] 20. Lu Z, Chan YT, Lo KK, et al. Carotenoids and vitamin A in breastmilk of Hong Kong lactating mothers and their relationships with maternal diet. Nutrients 2022;14:2031. [PMC free article: PMC9148123] [PubMed: 35631170] 21. Souza G, Dolinsky M, Matos A, et al. Vitamin A concentration in human milk and its relationship with liver reserve formation and compliance with the recommended daily intake of vitamin A in pre-term and term infants in exclusive breastfeeding. Arch Gynecol Obstet 2015;291:319-25. [PubMed: 25118833] 22. Sámano R, Martínez-Rojano H, Hernández RM, et al. Retinol and alpha-tocopherol in the breast milk of women after a high-risk pregnancy. Nutrients 2017;9:E14. [PMC free article: PMC5295058] [PubMed: 28045436] 23. Bhaskaram P, Balakrishna N, Nair M, Sivakumar, B. Vitamin A deficiency in infants: Effects of postnatal maternal vitamin A supplementation on the growth and vitamin A status. Nutr Res 2000;20:769–78. doi:10.1016/S0271-5317(00)00176-7 [CrossRef] 24. Ding Y, Hu P, Yang Y, et al. Impact of maternal daily oral low-dose vitamin a supplementation on the mother–infant pair: A randomised placebo-controlled trial in China. Nutrients 2021;13:2370. [PMC free article: PMC8308679] [PubMed: 34371880] 25. Oliveira JM, Allert R, East CE. Vitamin A supplementation for postpartum women. Cochrane Database Syst Rev 2016;3:CD005944. [PMC free article: PMC8407451] [PubMed: 27012320] 26. Grilo EC, Medeiros WF, Silva AG, et al. Maternal supplementation with a megadose of vitamin A reduces colostrum level of alpha-tocopherol: A randomised controlled trial. J Hum Nutr Diet 2016;29:652-61. [PubMed: 27231056] 27. Bahl R, Bhandari N, Wahed MA, et al. Vitamin A supplementation of women postpartum and of their infants at immunization alters breast milk retinol and infant vitamin A status. J Nutr 2002;132:3243-8. [PubMed: 12421835] 28. Darboe MK, Thurnham DI, Morgan G, et al. Effectiveness of an early supplementation scheme of high-dose vitamin A versus standard WHO protocol in Gambian mothers and infants: a randomised controlled trial. Lancet 2007;369:2088-96. [PubMed: 17586304] 29. Idindili B, Masanja H, Urassa H, et al. Randomized controlled safety and efficacy trial of 2 vitamin A supplementation schedules in Tanzanian infants. Am J Clin Nutr 2007;85:1312-9. [PubMed: 17490968] 30. Bezerra DS, de Araujo KF, Azevedo GM, et al. A randomized trial evaluating the effect of 2 regimens of maternal vitamin A supplementation on breast milk retinol levels. J Hum Lact 2010;26:148-56. [PubMed: 20110563] 31. Tomiya MTO, de Arruda IKG, da Silva Diniz A, et al. The effect of vitamin A supplementation with 400 000 IU vs 200 000 IU on retinol concentrations in the breast milk: A randomized clinical trial. Clin Nutr 2017;36:100-6. [PubMed: 26725194] 32. Bezerra DS, de Melo ATA, de Oliveira Kcan, et al. Breast milk retinol levels after vitamin A supplementation at different postpartum amounts and intervals. Nutrients 2022;14:3570. [PMC free article: PMC9460635] [PubMed: 36079825] 33. Atalhi N, El Hamdouchi A, Barkat A, et al. Combined consumption of a single high-dose vitamin A supplement with provision of vitamin A fortified oil to households maintains adequate milk retinol concentrations for 6 months in lactating Moroccan women. Appl Physiol Nutr Metab 2020;45:275-82. [PubMed: 31365834] 34. Villard L, Bates CJ. Effect of vitamin A supplementation on plasma and breast milk vitamin A levels in poorly nourished Gambian women. Hum Nutr Clin Nutr 1987;41:47-58. [PubMed: 3570862] 35. Webb AL, Aboud S, Furtado J, et al. Effect of vitamin supplementation on breast milk concentrations of retinol, carotenoids and tocopherols in HIV-infected Tanzanian women. Eur J Clin Nutr 2009;63:332-9. [PMC free article: PMC3095494] [PubMed: 17940544] 36. Vinutha B, Mehta MN, Shanbag P. Vitamin A status of pregnant women and effect of post partum vitamin A supplementation. Indian Pediatr 2000;37:1188-93. [PubMed: 11086300] 37. Stoltzfus RJ, Hakimi M, Miller KW, et al. High dose vitamin A supplementation of breast-feeding Indonesian mothers: Effects on the vitamin A status of mother and infant. J Nutr 1993;123:666-75. [PubMed: 8463867] 38. Rice AL, Stoltzfus RJ, de Francisco A, et al. Maternal vitamin A or beta-carotene supplementation in lactating Bangladeshi women benefits mothers and infants but does not prevent subclinical deficiency. J Nutr 1999;129:356-65. [PubMed: 10024613] 39. Dos Santos CS, Kruze I, Fernándes T, et al. The effect of a maternal double megadose of vitamin A supplement on serum levels of retinol in children aged under six months. J Nutr Metab 2013;2013:876308. [PMC free article: PMC3886490] [PubMed: 24455219] 40. Ayah RA, Mwaniki DL, Magnussen P, et al. The effects of maternal and infant vitamin A supplementation on vitamin A status: A randomised trial in Kenya. Br J Nutr 2007;98:422-30. [PubMed: 17391562] 41. Miller MF, Stoltzfus RJ, Iliff PJ, et al. Effect of maternal and neonatal vitamin A supplementation and other postnatal factors on anemia in Zimbabwean infants: a prospective, randomized study. Am J Clin Nutr 2006;84:212-22. [PubMed: 16825698] 42. Cox SE, Arthur P, Kirkwood BR, et al. Vitamin A supplementation increases ratios of proinflammatory to anti-inflammatory cytokine responses in pregnancy and lactation. Clin Exp Immunol 2006;144:392-400. [PMC free article: PMC1941972] [PubMed: 16734607] 43. Bhaskaram P, Balakrishna N. Effect of administration of 200,000 IU of vitamin A to women within 24 hrs after delivery on response to PPV administered to the newborn. Indian Pediatr 1998;35:217-22. [PubMed: 9707874] 44. Newton S, Cousens S, Owusu-Agyei S, et al. Vitamin A supplementation does not affect infants' immune responses to polio and tetanus vaccines. J Nutr 2005;135:2669-73. [PubMed: 16251628] 45. Arsenault JE, Aboud S, Manji KP, et al. Vitamin supplementation increases risk of subclinical mastitis in HIV-infected women. J Nutr 2010;140:1788-92. [PMC free article: PMC2937574] [PubMed: 20739447] 46. García L, Ribeiro K, Araujo K, et al. Alpha-tocopherol concentration in the colostrum of nursing women supplemented with retinyl palmitate and alpha-tocopherol. J Hum Nutr Diet 2010;23:529-34. [PubMed: 20831709] Go to: Substance Identification Substance Name Vitamin A CAS Registry Number 68-26-8; 79-81-2 Drug Class Breast Feeding Lactation Milk, Human Vitamins Disclaimer: Information presented in this database is not meant as a substitute for professional judgment. You should consult your healthcare provider for breastfeeding advice related to your particular situation. The U.S. government does not warrant or assume any liability or responsibility for the accuracy or completeness of the information on this Site. Drug Levels and Effects Substance Identification Copyright NoticeAttribution Statement: LactMed is a registered trademark of the U.S. Department of Health and Human Services. 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15230	https://cdn.wou.edu/chemistry/files/2021/08/Lecture-2-Physical-foundations-PDF.pdf	Foundations of Biochemistry 1.2 Physical Foundations In the second section of Chapter 1, we will focus on key elements of physical chemistry that provide a foundation understanding cellular reactions and enzyme kinetics. 1 1.2 Physical Foundations • Reactions and Energy Changes • Reversible Reactions • Irreversible Reactions • Reaction Equilibrium • Change in Free Energy These include the major types of chemical reactions and the energy changes inherent within them. 2 Reversible Reactions • Reactions and Energy Changes Consider the reaction: A + B P + Q Scenario 1: Reversible reaction in which the reactants and products are equally favored First let’s take a look at reversible reactions. If we consider the reaction: A + B being converted to the products, P + Q over time, we can imagine a few different scenarios. In the first scenario, shown here, this is a reversible reaction, where the reactants and products are equally favored. Shown on the graph is the concentration of one of the reactants [A] and one of the products [P]. You can see that overtime, as the reaction proceeds, that the concentration of the reactants decreases and the concentration of the products increases, until they reach the same level and stabilize at equilibrium. Recall, that at equilibrium the reaction is still proceeding and dynamic…there is just no net gain or loss in reactants or products as the rate of the forward and reverse reactions is the same. 3 1.2 Physical Foundations • Reactions and Energy Changes Consider the reaction: A + B P + Q Scenario 2: Reversible reaction in which the reactants are favored (reverse reaction is favored… In scenario 2, the reverse reaction is favored over the forward reaction, causing an accumulation of reactants over time. This is seen in the reaction graph showing that some product is formed, but that when equilibrium is reached, the concentrations of reactant are higher than that of the product, and 4 1.2 Physical Foundations • Reactions and Energy Changes Consider the reaction: A + B P + Q Scenario 3: Reversible reaction in which the products are favored (forward reaction is favored… In scenario 3, the forward reaction is favored causing an accumulation of more product than reactant at the point that equilibrium is reached. 5 1.2 Physical Foundations • Reactions and Energy Changes Consider the reaction: A + B P + Q Scenario 3: Irreversible reaction where the reaction is driven to completion (forward products are not shown on the graph) We will also find that some reactions are irreversible in nature and are driven to completion. Note that forward products are not shown on this graph, only the complete consumption of the reactants during the reaction. 6 Equilibrium Constants Equilibrium Constants (Keq), as its name implies, is constant and independent of the concentration of the reactants and products. • A Keq > 1 implies that the products are favored. • A Keq < 1 implies that reactants are favored. • When Keq = 1, both reactants and products are equally favored. For reactions that are reversible and reach an equilibrium state, they can be defined mathematically by an equilibrium constant called K-eq or K-equilibrium, due to the stability and nature of the equilibrium state. For this, we need to also take into account a fully balanced equation where the correct molar equivalents are shown. This is indicated in our equation here by the lower case letters in front of the reactants and products, such that we can now define Keq as the concentration of the products (raised to the power of the molar equivalent) over the concentration of the reactants (also raised to the power of the molar equivalent). This helps us define the dynamic nature of the reaction and the static nature of equilibrium. Regardless of whether we perturb the system by adding in more reactants or more products (or do the opposite! Remove products or remove reactants), there must be a shift in the concentration of the other amounts, such that they maintain the equilibrium state at the characteristic constant level. Because Keq is defined by the products in the numerator and the reactants in the denominator, a Keq that is > 1 indicates that the products are favored, whereas a Keq that is < 1 implied that the reactants are favored. If it is exactly 1, then the products and reactants are equally favored. 7 Change in Gibbs Free Energy (DG) • Controls the extent of a reaction • Two different pairs of factors influence the ΔG • One pair is concentration and inherent reactivity of reactants compared to products • The other pair is enthalpy and entropy changes Recall from general chemistry lessons, that the spontaneity of a reaction is measured by the change in the Gibbs Free Energy (termed Delta G). Delta G in influenced by two different pairs of factors. The first pair is the concentration of reactants and products and the inherent reactivity of the reactants compared to the products (these are the chemical characteristics of the compounds themselves). The second pair of factors include enthalpy and entropy changes. 8 HCl(aq) + H2O(l) –> H3O+(aq) + Cl–(aq) CH3CO2H(aq) + H2O(l) –> H3O+(aq) + CH3CO2 –(aq) For example, consider the two reactions shown. The top reaction showing hydrochloric acid reacting with water and the lower reaction showing acetic acid reacting with water. At t=0, there will be no product formed yet, and we will just be placing 0.1 moles of each acid into the water. When equilibrium is reached, there is essentially no HCl left in solution. This reaction has gone to completion. Whereas, 99% of the acetic acid remains. Why is that? All we can say is that there is something about the structure of HCl that makes it intrinsically less stable in water than the acetic acid. This is reflected in the Keq for each reaction ( >>1 for HCl and <<1 for acetic acid). Thus, intrinsic stability of the molecule is one factor that contributes to Delta G. The other factor in the first pair is concentration. For example, if we added 0.25 mol/L of acetic acid to a solution of water, this solution will not conduct electricity, suggesting that very little ionization of the acid has occurred. However, if we add more concentrated acid to the solution, this will drive the formation of more product and the solution will conduct electricity. This is an example of Le Chatelier’s Principle, which states that if a reaction at equilibrium is perturbed, the reaction will be driven in the direction that will relieve the perturbation. 9 Effects of Le Chatelier’s Principle • if more reactant is added, the reaction shifts to form more products • if more product is added, the reaction shifts to form more reactants • if products are selectively removed (by distillation, crystallization, or further reaction to produce another species), the reaction shifts to form more product. • if reactants are removed (as above), the reaction shifts to form more reactants. • if heat is added to an exothermic reaction, the reaction shifts to get rid of the excess heat by shifting to form more reactants. (opposite for an endothermic reaction). So, conlcusions we can make from Le Chatlier’s Principle are: 10 Change in Gibbs Free Energy (DG) So we can think about this first pair of factors in a more mathematical way to help us define Delta G of the reaction. We can first say that it is equal to the change in free energy associated with the intrinsic stability of the reactants, as well as the change in free energy associated with the concentrations of reactants present. Delta G knot is used to reflect the contribution from the intrinsic stability of the reactants and products (and differentiate it from the overall delta G of the rection). The component of delta G that is dependent on the concentration of reactants and products can be further defined by RTlnQrx, where R is the gas constant, T is temperature in Kelvin, and Qrx is the reaction quotient (which reflects the concentrations of the products over the reactants. 11 Change in Gibbs Free Energy (DG) • if ΔG < 0, the reaction goes toward products P and Q The reaction is EXERGONIC (Spontaneous) • if ΔG = 0, the reaction is at equilibrium and no further change occurs in the concentration of reactants and products. The reaction is at EQUILIBRIUM • if ΔG > 0, the reaction goes toward reactants A and B. The reaction is ENDERGONIC (Not Spontaneous) Thus the change in Gibbs Free Energy can be simplified to this equation. Recall that if delta G is < 0, the reaction is spontaneous (or Exergonic). If delta G is 0, the reaction is at equilibrium, and if delta G is > 0, the reaction is Endergonic or Not spontaneous. 12 HCl(aq) + H2O(l) –> H3O+(aq) + Cl–(aq) CH3CO2H(aq) + H2O(l) –> H3O+(aq) + CH3CO2 –(aq) So if we look at our two acids from our previous reactions, we can characterize the reactions according to these following characteristics. We know that for HCl that the intrinsic stability of the molecule favors the products, whereas for acetic acid, it favors the reactants. For both reactions starting at time = 0, there will be a shift in the reaction such that the reaction will move to the right towards the formation of the products (as no products will have been formed yet at t=0; thus, products will form until equilibrium is established). The combination of these two factors, leads to the overall delta G of the reaction, which in the case of HCl is much larger than 1, and in the case of acetic acid is much smaller than 1. 13 HCl(aq) + H2O(l) –> H3O+(aq) + Cl–(aq) CH3CO2H(aq) + H2O(l) –> H3O+(aq) + CH3CO2 –(aq) Once equilibrium is reached, you can see that the two pairs of factors have reached equivalency, such that neither factor favors a shift to the reactant or product side (they essentially cancel eachother out and the reaction remains at equilibrium). 14 What is the significance of DGo? So what is the significance of delta G knot? One thing that we know, is that it is independent of concentration and does not change… 15 What is the significance of DGo? Think of the reaction at equilibrium Think about the reaction at equilibrium….What do we know about the value of Delta G at equilibrium? It is zero, right? 16 What is the significance of DGo? Think of the reaction at equilibrium Conclusions? At equilibrium, delta G knot will equal the negative RTln[P][Q]/[A][B]. We then know that this reaction is at equilibrium and can substitute in Keq. Thus, our equation reduces to delta G knot equals negative RT, natural log Keq. And we can convert this to a standard log equation if needed. Thus, depending on a given problem, we can potentially calculate a value for delta G knot that can help us further evaluate the delta G of a reaction under different conditions (ie not at equilibrium). 17 What is the significance of DGo? Think of the reaction at standard state (all concentrations are at 1 M) What happens if in our equation, all of the concentrations are at 1 M? 18 What is the significance of DGo? Well…the concentration factor reduces to zero, and we find that under this circumstance that the delta G of the reaction will be equal to the delta G knot value (ie the reaction will be independent of concentration and only be affected by the intrinsic reactivity of the molecules. This is defined as the standard state…when the delta G of the reaction equals delta G knot. 19 New symbol = DGo’ • when all reactants are at this concentration, defined as the standard state (1 M for solutes), the ΔG at that particular moment just happens to equal the ΔGo for the reaction. The ΔGo’(delta G naught prime) is defined as the free energy change of a reaction under standard conditions. Note that standard conditions also define temperature and pressure constraints for the system. The following are true for ΔGo’: •all the reactants and products are at an initial concentration of 1.0 M •Pressure is at 1.0 atm •Temperature is at 25oC So at this point, we will introduce a new symbol for the steady state which is delta G knot prime. Standard conditions also define the temperature and pressure conditions of the reaction to a pressure of 1.0 atm and a tempterature of 25 degrees C. 20 So What? Energy flow. What is the direction of each of the following reactions when the reactants are initially present in equimolar amounts? Use the data given in the table DGo’ This helps us to evaluate energy flow within a system. For example, if we look at reaction (a), we can see that conversion of ATP to ADP is going to yield a delta G knot prime of -7.3 kcal/mol (that is in the forward direction for our reaction). If we look at our table, we can also see that the breakdown of creatine phosphate will yield -10.3 kcal/mol (that is in the reverse direction for our equation). So for the forward reaction of creatine to creatine phosphate, our delta G knot prime will be +10.3 kcal/mol. Thus our overall delta G knot prime will be a sum of these values going in the forward direction or (-7.3) + (+10.3) = +3.0. So we know that reaction (a) under steady state conditions is Not Spontaneous! 21 So What? Energy flow. What is the direction of each of the following reactions when the reactants are initially present in equimolar amounts? Use the data given in the table DGo’ The overall DGo’ for the reactions above can be taken as the sum/difference of those listed in the table for the individual hydrolysis steps Practice calculating delta G knot prime for reactions (b) – (d). What did you get for (c)? [ -7.3 for the ATP ADP & +14.8 for the pyruvate phosphoenolpyruvate = +7.5 for the overall reaction and Non spontaneous]. 22 Calculate ΔG°′ and K′eq at 25°C for this reaction, by using the data given in the table DGo’ R = 0.0019872 kcal/molK We calculated delta G knot prime on the last slide…can you use that information to calculate the Keq?? [Hint: Think about the reaction at equilibrium] 23 Solving the Problem…. Fill in the values (make sure units match!) and isolate the log. Then exponentiate the equation to solve for Keq +7.5 kcal/mol = -2.303(0.0019872 kcal/molK)(25+273K)logKeq +7.5 kcal/mol = (-1.364kcal/mol)logKeq -5.94 = logKeq 10-5.94 = Keq 0.000001148 or 1.1 x 10-6 = Keq Did you use this equation to come up with your final answer? First fill in the known values, checking to be sure that your units all match and will cancel out. Solve the problem down to the log, and then exponentiate. Once in exponential form, you can solve for Keq. This number turns out to be much << 1 indicating that the reaction favors the reactant side. Note that reactions in the body are not typically at equilibrium or at the steady state. These are conditions that we use in the lab to help us understand the nature of a reaction. The two fluctuating components then, in the reaction are temperature and the concentration of the reactants and the products. In vivo, it is not likely that you can vary the temperature very much to significantly impact the rate of reaction. In vivo, the main driver will be concentration of the reactants and products. Thus, if you have a highly unfavorable reaction, such as the one that we’ve been discussing, removing the product from the area as it is formed can drive the reaction in the forward direction (ie Le Chatelier’s Principle is important in vivo!!). 24 What about Enthalpy and Entropy?? • Enthalpy relates to the heat of the reaction • Entropy relates to the state of order/disorder in a reaction What is the equation that relates DG to these terms? Enthalpy and Entropy are also important components that can affect the overall change in Gibb’s free energy within a reaction. Do you remember the equation that relates these terms from general chemistry? 25 What about Enthalpy and Entropy?? • Enthalpy relates to the heat of the reaction • Entropy relates to the state of order/disorder in a reaction What is the equation that relates DG to these terms? ΔG = ΔH – TΔS Yes, the change in Gibb’s free energy is equal to the change in enthalpy within the system minus the temperature (in Kelvin) times the change in entropy. This equation is more useful in the lab when we are studying isolated systems. 26 - Breaking Bonds Requires Energy - Forming New Bonds Releases Energy The final concept that I want to remind you of before we begin our journey into biochemistry, is that bonded elements that are sharing electrons exist in an optimal energy state and are highly attracted to each other. Thus, it always takes energy to break a bond, or pull the atoms apart from one another. When new bonds form, it releases energy. This amount of energy is called the bond energy, and it will be different for different atoms pairs involved in the bond and depending on what type of bond is formed (ie a sigma bond vs a pi bond or a hybridized mixture of the two). 27 ATP Hydolysis A good example of where this concept is often misused or misunderstood is the hydrolysis of ATP. Many people erroneously state that breaking the high energy phosphate bond in ATP releases lots of energy. This is an incorrect way of thinking about this reaction. It is the formation of the hydrolysis product (the inorganic phosphate) that releases the energy! We will see that this mechanism is used over and over again to create energy for biological reactions in vivo. 28
15231	https://math.stackexchange.com/questions/4743978/simplifying-integral-based-on-symmetry	calculus - Simplifying Integral based on Symmetry - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Simplifying Integral based on Symmetry Ask Question Asked 2 years, 2 months ago Modified2 years, 2 months ago Viewed 93 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. The following integral is to be evaluated over the disc consisting of the interior R of the unit circle, centered at the origin. I'm trying to simplify the integral using the symmetries of R and the integrand: ∬R e x d A∬R e x d A simplifies to 2∬S e x d A 2∬S e x d A where S S is the upper half of R R. I'm looking for a reason to understand this simplification. It is not that e^x is particularly even or odd. calculus integration Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 28, 2023 at 17:02 David G. Stork 30.5k 5 5 gold badges 33 33 silver badges 60 60 bronze badges asked Jul 28, 2023 at 16:06 OrpheusOrpheus 924 9 9 silver badges 23 23 bronze badges 1 ∫x=−1 1∫y=0 1−x 2√e x d x d y=π I 1(1)∫x=−1 1∫y=0 1−x 2 e x d x d y=π I 1(1) where I 1 I 1 is the Bessel I function.David G. Stork –David G. Stork 2023-07-28 17:08:26 +00:00 Commented Jul 28, 2023 at 17:08 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. You are correct, e x e x is not even or odd as a function of x x, but the bivariate function f(x,y):=e x f(x,y):=e x is even as a function of y y: f(x,−y)=f(x,y)f(x,−y)=f(x,y) for all x x and y y. One way to argue is to apply Fubini's theorem to replace the double integral with an iterated integral, integrating over y y first for each x x, then integrating over x x. For the inner integral over y y, you can appeal to even symmetry of both the integrand and the domain (as noted in the comment by @ThomasAndrews). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jul 28, 2023 at 16:18 answered Jul 28, 2023 at 16:10 grand_chatgrand_chat 41k 1 1 gold badge 46 46 silver badges 86 86 bronze badges 4 And the domain is also invariant under (x,y)↦(x,−y).(x,y)↦(x,−y).Thomas Andrews –Thomas Andrews 2023-07-28 16:14:05 +00:00 Commented Jul 28, 2023 at 16:14 So you are telling me in a situation where f(x,y)=f(−x,−y)f(x,y)=f(−x,−y) I can replace the 2 by 4?Orpheus –Orpheus 2023-07-28 17:00:35 +00:00 Commented Jul 28, 2023 at 17:00 If both f(x,y)=f(x,−y)f(x,y)=f(x,−y) and f(x,y)=f(−x,y)f(x,y)=f(−x,y) are true, you can replace integrating over the unit circle with 4 4 copies of an integral over the first quadrant of the unit circle.grand_chat –grand_chat 2023-07-28 17:23:10 +00:00 Commented Jul 28, 2023 at 17:23 Well thanks a lot grand_chat Orpheus –Orpheus 2023-07-28 17:44:02 +00:00 Commented Jul 28, 2023 at 17:44 Add a comment\| You must log in to answer this question. 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15232	https://www.youtube.com/watch?v=05HvMLOfgPU	1.9 \| Review Of Vector Algebra \| Volume of a Tetrahedron Using Scalar Triple Product MathandBolt 1390 subscribers Description 950 views Posted: 29 Mar 2023 1.9 \| Review Of Vector Algebra \| Volume of a Tetrahedron Using Scalar Triple Product mathematicalphysics #electromagnetictheory In this video, we'll show you how to use the scalar triple product to determine a tetrahedron's volume. We will begin by defining the terms tetrahedron and scalar triple product. The formula for determining a tetrahedron's volume using the scalar triple product will next be discussed, along with an example of how to use it. You will have a better knowledge of how to use the scalar triple product to determine the volume of a tetrahedron by the end of the video. About The mission of MathandBolt is to create a library of lectures that provide simple explanations for many complex mathematical and scientific topics. We believe that any difficult topic can be explained in simple terms. You can browse and watch our library of videos, and work on tons of practice problems. Whether broadening your knowledge or trying to ace a college course, MnB is there with you. 👉 There are multiple ways you can contribute to the support of my channel 😊 ➡️ Website: ➡️ Subscribe to us on YouTube: ➡️Instagram: 1 comments Transcript: foreign we need to determine the volume of a tetrahedron whose vertex have the coordinate given by a b c n d let's try to sketch this tetrahedron the volume of the tetrahedron will be 1 6 of the scalar triple product between AV vector dotted width aec Vector cross a d vector now let's try to find out the vector a b a UC and AED will start with Vector a b to compute the vector a b we consider deposition Vector OB and from it we subtract the position Vector o a vector now our B Vector so this is given by four one three minus a vector so that's to negative 1 negative 3. so this gives us that we have 4 minus 2 so that will be 2. 1 minus minus 1 so that's positive two and three minus minus 3 so that's three plus three six so we have a b vector next we'll determine the vector AC so to find out AC Vector we consider the position Vector OC and from it we subtract the position Vector o a so our OC Vector will be 3 to negative 1. minus a wave Vector so a vector will be 2 negative 1 negative 3 and that will be equals to next we have a d Vector so a d Vector will be OD Vector minus o a vector the OD Vector will be 1 4 2. minus or a vector so a vector will be 2 negative 1 negative 3 and this will be equals to so now we have the three vectors so let's try to find out the volume so a volume is 1 6 times so we have a b Vector so our av Vector is to I cap plus 2 J cap plus 6 K cap this is dotted width the cross product between AC Vector which is I cap plus 3 J cap plus 2K cap cross width a d Vector so that's negative I cap plus 5 J cap plus 5 K cap so since this is a scalar triple product we can easily compute it by considering the three by three determinant so in the first row we'll consider the components of a v Vector that will be 2 2 6. then we have the components of AC vectors are that will be 1 3 2 and finally we have the component a d Vector so that will be negative 1 5 5. so we'll expand it along the first row so we have two times multiplied two so we're 3 times 5 so that's 15 minus 5 times 2 so that is 10. minus 2 times 1 times 5 so that's 5 then we have minus 2 times negative 1 so that's negative 2 so this will become positive 2 plus 6 times 1 times 5 so that is positive 5 minus 3 times negative 1 so that's negative 3 and this will become positive 3. so we will have 1 6 of 2 times 15 minus 10 so that's 5 so this will be 10 minus so 5 plus 2 is 7 so 7 times 2 is 14 5 plus 3 so that's eight eight times six forty eight so this will be 44 or simply 22 over 3. so this is going to be the volume of the tetrahedron foreign [Music]
15233	https://note.nkmk.me/en/python-numpy-broadcasting/	Published Time: 2021-10-13 NumPy: Broadcasting rules and examples \| note.nkmk.me note.nkmk.me Home Python NumPy NumPy: Broadcasting rules and examples Posted: 2021-10-13 \| Tags: Python, NumPy In operations between NumPy arrays (ndarray), each shape is automatically converted to be the same by broadcasting. This article describes the following contents. Broadcasting rules in NumPy Broadcasting examples in NumPy Examples of 2D array Examples of 3D array Cases that cannot broadcast Functions to get the broadcasted array Broadcast an array to a specified shape.: np.broadcast_to() Broadcast multiple arrays: np.broadcast_arrays() The official documentation explaining broadcast is below. Broadcasting — NumPy v1.16 Manual Use reshape() or np.newaxis if you want to reshape ndarray to any shape you want. NumPy: How to use reshape() and the meaning of -1 NumPy: Add new dimensions to ndarray (np.newaxis, np.expand_dims) Broadcasting rules in NumPy There are the following two rules for broadcasting in NumPy. Make the two arrays have the same number of dimensions. If the numbers of dimensions of the two arrays are different, add new dimensions with size 1 to the head of the array with the smaller dimension. Make each dimension of the two arrays the same size. If the sizes of each dimension of the two arrays do not match, dimensions with size 1 are stretched to the size of the other array. If there is a dimension whose size is not 1 in either of the two arrays, it cannot be broadcasted, and an error is raised. Note that the number of dimensions of ndarray can be obtained with the ndim attribute and the shape with the shape attribute. NumPy: Get the number of dimensions, shape, and size of ndarray Broadcasting examples in NumPy Examples of 2D array 2D array and 1D array The following 2D and 1D arrays are used as examples. To make it easier to understand the result of the broadcast, one of them uses zeros() to set all the elements to 0. NumPy: Create an ndarray with all elements initialized with the same value ``` import numpy as np a = np.zeros((3, 3), np.int) print(a) [[0 0 0] [0 0 0] [0 0 0]] print(a.shape) (3, 3) b = np.arange(3) print(b) [0 1 2] print(b.shape) (3,) ``` source: numpy_broadcasting.py The shape of 1D array is (3,) instead of (3) because tuples with one element have a comma at the end. A tuple with one element requires a comma in Python The result of the addition of these two ndarray is as follows. ``` print(a + b) [[0 1 2] [0 1 2] [0 1 2]] ``` source: numpy_broadcasting.py Let's transform the array with a smaller number of dimensions (1D array b) according to the rules described above. First, according to rule 1, the array is transformed from shape (3,) to (1, 3) by adding a new dimension of size 1 at the head. The reshape() method is used. NumPy: How to use reshape() and the meaning of -1 ``` b_1_3 = b.reshape(1, 3) print(b_1_3) print(b_1_3.shape) (1, 3) ``` source: numpy_broadcasting.py Next, the size of each dimension is stretched according to rule 2. The array is stretched from (1, 3) to (3, 3). The stretched part is a copy of the original part. np.tile() is used. numpy.tile — NumPy v1.16 Manual ``` print(np.tile(b_1_3, (3, 1))) [[0 1 2] [0 1 2] [0 1 2]] ``` source: numpy_broadcasting.py Note that reshape() and np.tile() are used here for the sake of explanation, but if you want to get the broadcasted array, there are functions np.broadcast_to() and np.broadcast_arrays() for that purpose. See below. 2D array and 2D array The result of addition with the 2D array of (3, 1) is as follows. ``` b_3_1 = b.reshape(3, 1) print(b_3_1) [ ] print(b_3_1.shape) (3, 1) print(a + b_3_1) [[0 0 0] [1 1 1] [2 2 2]] ``` source: numpy_broadcasting.py In this case, since the number of dimensions is already the same, the array is stretched from (3, 1) to (3, 3) according to rule 2. ``` print(np.tile(b_3_1, (1, 3))) [[0 0 0] [1 1 1] [2 2 2]] ``` source: numpy_broadcasting.py In the previous examples, only one of the arrays is converted, but there are cases where both are converted by broadcasting. The following is the result of adding arrays whose shapes are (1, 3) and (3, 1). ``` print(b_1_3) print(b_1_3.shape) (1, 3) print(b_3_1) [ ] print(b_3_1.shape) (3, 1) print(b_1_3 + b_3_1) [[0 1 2] [1 2 3] [2 3 4]] ``` source: numpy_broadcasting.py Both (1, 3) and (3, 1) are stretched to (3, 3). ``` print(np.tile(b_1_3, (3, 1))) [[0 1 2] [0 1 2] [0 1 2]] print(np.tile(b_3_1, (1, 3))) [[0 0 0] [1 1 1] [2 2 2]] print(np.tile(b_1_3, (3, 1)) + np.tile(b_3_1, (1, 3))) [[0 1 2] [1 2 3] [2 3 4]] ``` source: numpy_broadcasting.py The same applies if one of them is 1D array. ``` c = np.arange(4) print(c) [0 1 2 3] print(c.shape) (4,) print(b_3_1) [ ] print(b_3_1.shape) (3, 1) print(c + b_3_1) [[0 1 2 3] [1 2 3 4] [2 3 4 5]] ``` source: numpy_broadcasting.py 1D array is converted like (4,) ->(1, 4) ->(3, 4), and 2D array like (3, 1) ->(3, 4). ``` print(np.tile(c.reshape(1, 4), (3, 1))) [[0 1 2 3] [0 1 2 3] [0 1 2 3]] print(np.tile(b_3_1, (1, 4))) [[0 0 0 0] [1 1 1 1] [2 2 2 2]] print(np.tile(c.reshape(1, 4), (3, 1)) + np.tile(b_3_1, (1, 4))) [[0 1 2 3] [1 2 3 4] [2 3 4 5]] ``` source: numpy_broadcasting.py Note that the dimension is stretched only when the original size is 1. Otherwise, it cannot be broadcasted, and an error is raised, as described below. Examples of 3D array Rule 1 applies even if the difference in the number of dimensions is two or more. Using 3D and 1D arrays as examples, the addition results are as follows: ``` a = np.zeros((2, 3, 4), dtype=np.int) print(a) [[[0 0 0 0] [0 0 0 0] [0 0 0 0]] [[0 0 0 0] [0 0 0 0] [0 0 0 0]]] print(a.shape) (2, 3, 4) b = np.arange(4) print(b) [0 1 2 3] print(b.shape) (4,) print(a + b) [[[0 1 2 3] [0 1 2 3] [0 1 2 3]] [[0 1 2 3] [0 1 2 3] [0 1 2 3]]] ``` source: numpy_broadcasting_3d.py The shape is changed as (4, ) ->(1, 1, 4) ->(2, 3, 4). ``` b_1_1_4 = b.reshape(1, 1, 4) print(b_1_1_4) ] print(np.tile(b_1_1_4, (2, 3, 1))) [[[0 1 2 3] [0 1 2 3] [0 1 2 3]] [[0 1 2 3] [0 1 2 3] [0 1 2 3]]] ``` source: numpy_broadcasting_3d.py Cases that cannot broadcast As mentioned above, the dimension is stretched only if the original size is 1. If the sizes of the dimensions are different and the sizes of both arrays are not 1, it cannot be broadcasted, and an error is raised. ``` a = np.zeros((4, 3), dtype=np.int) print(a) [[0 0 0] [0 0 0] [0 0 0] [0 0 0]] print(a.shape) (4, 3) b = np.arange(6).reshape(2, 3) print(b) [[0 1 2] [3 4 5]] print(b.shape) (2, 3) print(a + b) ValueError: operands could not be broadcast together with shapes (4,3) (2,3) ``` source: numpy_broadcasting_error.py The same applies to the following case. ``` a = np.zeros((2, 3, 4), dtype=np.int) print(a) [[[0 0 0 0] [0 0 0 0] [0 0 0 0]] [[0 0 0 0] [0 0 0 0] [0 0 0 0]]] print(a.shape) (2, 3, 4) b = np.arange(3) print(b) [0 1 2] print(b.shape) (3,) print(a + b) ValueError: operands could not be broadcast together with shapes (2,3,4) (3,) ``` source: numpy_broadcasting_error.py In this example, if a new dimension is added at the end, the array can be broadcasted. ``` b_3_1 = b.reshape(3, 1) print(b_3_1) [ ] print(b_3_1.shape) (3, 1) print(a + b_3_1) [[[0 0 0 0] [1 1 1 1] [2 2 2 2]] [[0 0 0 0] [1 1 1 1] [2 2 2 2]]] ``` source: numpy_broadcasting_error.py It is easy to understand whether it can be broadcasted or not by right-aligned shape. ``` Not broadcastable: (2, 3, 4) ( 3) Broadcastable: (2, 3, 4) ( 3, 1) -> (1, 3, 1) -> (2, 3, 4) ``` If the sizes are different when right-aligned and compared vertically, one of them must be 1 to be broadcasted. For example, in the case of images, a color image is a 3D array whose shape is (height, width, 3) (3 means red, green, and blue), while a grayscale image is a 2D array whose shape is (height, width). Image processing with Python, NumPy In the case of computing the value of each color in a color image and the value of a grayscale image, it is impossible to broadcast even if the height and width are the same. You need to add a dimension to the end of the grayscale image with np.newaxis, np.expand_dims(), and so on. NumPy: Add new dimensions to ndarray (np.newaxis, np.expand_dims) ``` Not broadcastable: (h, w, 3) ( h, w) Broadcastable: (h, w, 3) (h, w, 1) -> (h, w, 3) ``` Functions to get the broadcasted array Broadcast an array to a specified shape.: np.broadcast_to() Use np.broadcast_to() to broadcast ndarray with the specified shape. numpy.broadcast_to — NumPy v1.16 Manual The first argument is the original ndarray, and the second is a tuple or list indicating shape. The broadcasted ndarray is returned. ``` a = np.arange(3) print(a) [0 1 2] print(a.shape) (3,) print(np.broadcast_to(a, (3, 3))) [[0 1 2] [0 1 2] [0 1 2]] print(type(np.broadcast_to(a, (3, 3)))) ``` source: numpy_broadcast_to.py An error occurs when specifying a shape that cannot be broadcasted. ``` print(np.broadcast_to(a, (2, 2))) ValueError: operands could not be broadcast together with remapped shapes [original->remapped]: (3,) and requested shape (2,2) ``` source: numpy_broadcast_to.py Broadcast multiple arrays: np.broadcast_arrays() Use np.broadcast_arrays() to broadcast multiple ndarray. numpy.broadcast_arrays — NumPy v1.16 Manual Specify multiple arrays separated by commas. A list of ndarray is returned. ``` a = np.arange(3) print(a) [0 1 2] print(a.shape) (3,) b = np.arange(3).reshape(3, 1) print(b) [ ] print(b.shape) (3, 1) arrays = np.broadcast_arrays(a, b) print(type(arrays)) print(len(arrays)) 2 print(arrays) [[0 1 2] [0 1 2] [0 1 2]] print(arrays) [[0 0 0] [1 1 1] [2 2 2]] print(type(arrays)) ``` source: numpy_broadcast_arrays.py An error occurs when specifying a combination of arrays that cannot be broadcasted. ``` c = np.zeros((2, 2)) print(c) [[0. 0.] [0. 0.]] print(c.shape) (2, 2) arrays = np.broadcast_arrays(a, c) ValueError: shape mismatch: objects cannot be broadcast to a single shape ``` source: numpy_broadcast_arrays.py Related Categories Python NumPy Related Articles How to fix "ValueError: The truth value ... is ambiguous" in NumPy, pandas Matrix operations with NumPy in Python NumPy: Arrange ndarray in tiles with np.tile() NumPy: np.sign(), np.signbit(), np.copysign() NumPy: Get the dimensions, shape, and size of an array NumPy: Transpose ndarray (swap rows and columns, rearrange axes) NumPy: arange() and linspace() to generate evenly spaced values NumPy: Slicing ndarray NumPy: astype() to change dtype of an array NumPy: Calculate the absolute value element-wise (np.abs, np.fabs) NumPy: Count values in an array with conditions NumPy: Create an array with the same value (np.zeros, np.ones, np.full) NumPy: clip() to limit array values to min and max NumPy: Calculate cumulative sum and product (np.cumsum, np.cumprod) Check NumPy version: np.version Search × Categories Python NumPy OpenCV pandas Pillow pip scikit-image Git Jupyter Notebook Mac Windows Image Processing File CSV JSON Date and time String Regex Numeric Dictionary List Error handling Mathematics Summary About GitHub: nkmk Related Articles How to fix "ValueError: The truth value ... is ambiguous" in NumPy, pandas Matrix operations with NumPy in Python NumPy: Arrange ndarray in tiles with np.tile() NumPy: np.sign(), np.signbit(), np.copysign() NumPy: Get the dimensions, shape, and size of an array English / Japanese \| Disclaimer Privacy policy GitHub ©nkmk.me
15234	https://en.wikipedia.org/wiki/CycleBeads	Jump to content Search Contents (Top) 1 How to Use 2 Main Drawbacks 3 Considerations for Family Planning Programs 4 Additional Evidence 5 Origin 6 References 7 External links CycleBeads العربية Deutsch Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Visual tool used for family planning CycleBeads is a visual tool that was developed by the Institute for Reproductive Health at Georgetown University. This device helps women use the Standard Days Method, a fertility awareness-based family planning method. The sole manufacturer is a US for-profit company, Cycle Technologies. The Standard Days Method is based on the fact that there is a fertile window during a woman's menstrual cycle which begins several days before ovulation and ends a few hours after ovulation. During this time a woman can become pregnant. The Standard Days Method identifies days 8-19 of cycle for women with cycles between 26 and 32 days long, as the potential fertile window. This formula is based on computer analysis of 7,500 menstrual cycles and takes into account cycle length, the timing of ovulation, the variation of the timing of ovulation from one cycle to the next, as well as the lifespan of the sperm and ovum. To prevent pregnancy using the Standard Days Method and CycleBeads, users avoid unprotected sex by using a condom or abstaining during days 8-19 of the cycle. How to Use [edit] CycleBeads, a color-coded string of beads that represents the days of a woman's cycle, helps an individual use the Standard Days Method, by helping her track her cycle days. Starting the first day of her period, she moves a band to the red bead then to a new bead every day. The color of the bead lets her know if today is a day she is highly likely to be fertile or not. Couples use condoms or barrier methods to prevent pregnancy on fertile days. An efficacy trial found that CycleBeads was more than 95% effective at preventing pregnancy with correct use and approximately 88% effective with typical use among women who reported recent cycles of 26–32 days. While this is similar or better than the efficacy of most other Natural Family Planning methods. its perfect use is significantly less effective than many other modern birth control methods and Long-acting reversible contraceptives. Women can use CycleBeads to plan pregnancy as well. Couples target those days where the bead colour indicates fertility is highest for intercourse to conceive a child. Main Drawbacks [edit] This method is not as effective for women who have cycles outside of the 26- to 32-day range. Women who are breastfeeding or have recently used contraceptive injections must wait before using CycleBeads. Many natural family planning methods require male involvement. Efficacy, like all birth control, is highly dependent on continuing correct use. Considerations for Family Planning Programs [edit] CycleBeads can be included in a wide variety of programs and offered by different levels of providers without significant additional resources. It has the potential to expand contraceptive prevalence by bringing new users to family planning. The method involves men and which gives programs an opportunity to develop strategies for reaching men with a variety of reproductive health messages. CycleBeads are a low-cost, one-time purchase which can help address concerns about increasing costs for contraceptives. Additional Evidence [edit] According to research studies, most women can learn to use CycleBeads in a single counseling session of about 20–30 minutes. Almost all women who choose to use CycleBeads do so because it is "natural" and does not have side effects. Follow up interviews with users and their partners found a high levels of satisfaction with the method among women and their partners. Origin [edit] The original birth control necklace was invented in 1989 by the Austrian gynecologist Maria Hengstberger who developed this necklace together with local Ethiopian women and adapted it to their wishes and needs. The newest and certainly most understandable model of this necklace is called "Baby Necklace", since the beads of the days of highest fertility have the shape of a baby. In 1995 the Necklace was taken over by the "Pastoral of the Child" in Brazil for joint use and further development and passed on to the Institute for Reproductive Health (IRH) at Georgetown University, for scientific research. In the form of "CycleBeads", developed and patented by IRH, the necklace is the basis of the "Standard Days Method", which has been spread worldwide by IRH. References [edit] ^ Weis, Julianne; Festin, Mario (30 March 2020). "Implementation and Scale-Up of the Standard Days Method of Family Planning: A Landscape Analysis". Global Health: Science and Practice. 8 (1): 114–124. doi:10.9745/GHSP-D-19-00287. PMC 7108942. PMID 32033980. ^ M. Arevalo et al. "A fixed formula to define the fertile window of the menstrual cycle as the basis of a simple method of Natural Family Planning," Contraception 60 (1999);357-60. ^ M. Arevalo et al. "Efficacy of a new method of family planning: the Standard Days Method," Contraception 65 (2002) 333-338. ^ R. Hatcher, ed. et al. Contraceptive Technology, 18th edition, 2004 ^ J. Cachan and R. Lundgren. "Reference Guide for Counseling Clients in the SDM," 2003. ^ [bare URL image file] ^ a b c J. Gribble, et al. "Being strategic about Contraceptive Introduction: The Experience of the Standard Days Method", Contraception 77 (2008); 147-154. ^ J. Gribble. "The Standard Days Method of Family Planning: A Response to Cairo," International Family Planning Perspectives 29 (2003). ^ "Being Strategic about Contraceptive Introduction: The Experience of the Standard Days Method", Contraception 2008 ^ J. Gribble et al. "Mind the Gap: responding to the funding crisis in family planning," Journal of Family Planning and Reproductive Care 30 (2004); 155-157. ^ E. Pinto, et al. "Introducing the Standard Days Method in CEMOPLAF," Final Report, 2003. ^ "Rain Tools". aktionregen.at (in German). Retrieved 2018-07-12. ^ "Rain Tools". aktionregen.at. Retrieved 2018-07-12. External links [edit] Official website Retrieved from " Category: Fertility awareness Hidden categories: All articles with bare URLs for citations Articles with bare URLs for citations from March 2022 Articles with image file bare URLs for citations CS1 German-language sources (de) Articles with short description Short description is different from Wikidata Add topic
15235	https://www.studocu.com/en-us/messages/question/6418159/what-is-the-probability-that-while-rolling-two-six-sided-dice-you-roll-a-sum-of-7	[Solved] What is the probability that while rolling two sixsided dice you - Math for the Real World (MATH108X) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent What is the probability that, while rolling two six-sided dice, you roll a sum of 7?Group of answer choices19.4%16.7%25.0%22.2%27.8% Math for the Real World (MATH108X) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist Brigham Young University-Idaho Math for the Real World Question What is the probability that while rolling two sixsided dice you Brigham Young University-Idaho Math for the Real World Question ### Momina 1 year ago What is the probability that, while rolling two six-sided dice, you roll a sum of 7? Group of answer choices 19.4% 16.7% 25.0% 22.2% 27.8% Like 0 Related documents W12 Homework Assignment B Math for the Real World Assignments 0%(1) Probability and Confidence intervals Math for the Real World Lecture notes 100%(1) SAT Math Assessment Questions and Solutions - Practice Set Math for the Real World Practice materials 100%(1) 04 HW Assignment Social Science Statistics Assignments 100%(5) Math108x W12A Probability Concepts and Homework Assignment Math For The Real World Assignments 100%(1) lesson 3 assignment part 8 Biostatistics Assignments None W12 Group Quiz: Understanding Probability Concepts & Applications Math For The Real World Coursework None Answer Created with AI 1 year ago To calculate the probability of rolling a sum of 7 with two six-sided dice, we can use the total number of favorable outcomes divided by the total number of possible outcomes. The favorable outcomes for rolling a sum of 7 are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). This gives us a total of 6 favorable outcomes. The total number of possible outcomes when rolling two six-sided dice is 6 6 = 36. So, the probability of rolling a sum of 7 is: js 6 favorable outcomes / 36 total outcomes = 1/6 ≈ 16.7% Therefore, the correct answer choice is 16.7%. Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: Math for the Real World MATH108X Brigham Young University-Idaho 34 Documents Go to course 5 Math108x Loan Case Study: Analyzing Interest Rates and Terms Math for the Real World 100%(3) 216 MAT 108X - Craig's Background Case Study on Charts & Graphs Math for the Real World 100%(2) 244 FDMAT 108 - Craig's Report Case Study: Sales Data Analysis and Graphs Math for the Real World 100%(2) 4 Calculus (MATH101) Formula Sheet for Derivatives & Integrals Math for the Real World 100%(2) Discover more from: Math for the Real World MATH108XBrigham Young University-Idaho34 Documents Go to course 5 Math108x Loan Case Study: Analyzing Interest Rates and Terms Math for the Real World 100%(3) 216 MAT 108X - Craig's Background Case Study on Charts & Graphs Math for the Real World 100%(2) 244 FDMAT 108 - Craig's Report Case Study: Sales Data Analysis and Graphs Math for the Real World 100%(2) 4 Calculus (MATH101) Formula Sheet for Derivatives & Integrals Math for the Real World 100%(2) 2 Math108x Checkpoint: Analyzing Kenji's Loan for Business Startup Math for the Real World 100%(2) 156 SAT Math Assessment Questions and Solutions - Practice Set Math for the Real World 100%(1) Related Answered Questions 10 days ago Why could this graph be misleading?Graph representing the national average cost of gas. The cost of gas is increasing significantly over time. Three time points are represented evenly spaced along the x-axis: Last year, last, week, and current. Group of answer choicesThe x-axis doesn't have equally spaced time units.The data comes from too small of a sampleMore labels needed to be added.A time series is not an appropriate graph for this data. Math for the Real World (MATH108X) 23 days ago Solve With his new company, Ruben has a two-year assignment living in a large city. The company will reimburse most of his expenses. Ruben is using the Quantitative Reasoning Process to build a budget. Calculate Ruben's estimated living expenses for the next two years using the following:Rent: $1891 per monthGym membership: $167 per year with a monthly fee of $22Renters' insurance: $75 every 6 monthsGroceries: $410 per monthEating out & entertaining: $126 per weekAssume 52 weeks in 1 year and 12 months in 1 year.(Round your answer to the nearest whole dollar) Math for the Real World (MATH108X) 23 days ago Solve In Tasheena's Anthropology class Quizzes are worth 15% of the final grade, Exams are worth 55%, Projects are worth 25%, and Attendance is worth 5%.At mid-semester Tasheena scored 144 out of 150 points on quizzes, 75, 83, and 96 on the first three exams, each worth 100 points. She got extra credit on her project with a score of 27 out of 25 possible points, and she had perfect attendance to class. Compute Tasheena's grade percentage in the class so far. (Make sure to change your answer to a percent and then round your percent to a whole number.) Math for the Real World (MATH108X) 23 days ago With his new company, Ruben has a two-year assignment living in a large city. The company will reimburse most of his expenses. Ruben is using the Quantitative Reasoning Process to build a budget. Calculate Ruben's estimated living expenses for the next two years using the following:Rent: $1891 per monthGym membership: $167 per year with a monthly fee of $22Renters' insurance: $75 every 6 monthsGroceries: $410 per monthEating out & entertaining: $126 per weekAssume 52 weeks in 1 year and 12 months in 1 year.(Round your answer to the nearest whole dollar) Math for the Real World (MATH108X) 23 days ago Tenielle currently walks to school from her apartment, which is 1.3 miles away from her first class. She typically walks at a speed of 3 miles per hour. She is considering buying a used bicycle to ride to campus. Tenielle assumes that if she were riding a bike, she could go about 6 miles per hour.How many minutes could Tenielle save getting to class each morning if she were to ride the bike?(Round your answer to 2 decimal places) Math for the Real World (MATH108X) 23 days ago Tenielle currently walks to school from her apartment, which is 1.3 miles away from her first class. She typically walks at a speed of 3 miles per hour. She is considering buying a used bicycle to ride to campus. Tenielle assumes that if she were riding a bike, she could go about 6 miles per hour.How many minutes could Tenielle save getting to class each morning if she were to ride the bike?(Round your answer to 2 decimal places) Math for the Real World (MATH108X) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. 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15236	https://www.scribd.com/document/843380454/11-1-Worksheet	11.1 Worksheet \| PDF Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 117 views 2 pages 11.1 Worksheet The document is a worksheet for Algebra 1 focused on simplifying radicals. It contains a series of problems that require students to simplify various radical expressions. 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15237	https://www.todaysrdh.com/a-hygienists-review-of-obtaining-medical-clearance-before-dental-treatment/	A Hygienists’ Review of Obtaining Medical Clearance before Dental Treatment During my clinical career in dental hygiene, I had the opportunity to work in five different dental practices across three states. Some were private practices, and some were DSOs. Although many protocols were similar or, in some cases, precisely the same, the one thing that was always different in every practice was the protocols for medical clearance. Differing protocols for medical clearance from practice to practice appear to be the case for other dental professionals, as the topic often arises in many social media groups. This begs the question: What is the proper protocol for acquiring medical clearance before dental treatment for pregnant patients and conditions such as myocardial infarction, cerebrovascular accident, or patients on anticoagulant or antiplatelet therapy? Pregnancy Too often, dental professionals treat pregnancy as a disease – pregnancy is not a disease. Though there are complications that can occur during pregnancy, a healthy pregnant patient should be treated like any other healthy patient. The American Dental Association (ADA), the American College of Obstetrics and Gynecology (ACOG), and the Health Resources and Services Administration’s Maternal and Child Health Bureau have issued a consensus statement and offered guidance for treating pregnant patients.1 The consensus statement states, “Reassure women that oral health care, including use of radiographs, pain medication, and local anesthesia, is safe throughout pregnancy.”1 The only time medical clearance/consultation is needed before dental care is in the following scenarios:1 Pharmacological considerations for the use of analgesics, antibiotics, anesthetics, and antimicrobials for pregnant patients include:1 \| \| \| --- \| \| Analgesic \| Indications, Contraindications, and Special Considerations \| \| Acetaminophen Acetaminophen with codeine Hydrocodone Oxycodone Codeine Meperidine Morphine \| -May be used during pregnancy -Oral pain can often be managed with non-opioid medications -If opioids are used, prescribe the lowest dose for the shortest duration (usually less than 3 days) \| \| Aspirin Ibuprofen Naproxen \| -First trimester: Avoid use -Second trimester, 13 up to 20 weeks: May use for short duration, 48 to 72 hours -Second trimester, 20 up to 27 weeks: Limit use -Third trimester: Avoid use \| Acetaminophen Acetaminophen with codeine Hydrocodone Oxycodone Codeine Meperidine Morphine -May be used during pregnancy -Oral pain can often be managed with non-opioid medications -If opioids are used, prescribe the lowest dose for the shortest duration (usually less than 3 days) Aspirin Ibuprofen Naproxen -First trimester: Avoid use -Second trimester, 13 up to 20 weeks: May use for short duration, 48 to 72 hours -Second trimester, 20 up to 27 weeks: Limit use -Third trimester: Avoid use Analgesic considerations for pregnant patientsAdapted from Oral Health Care During Pregnancy Expert Workgroup1 \| \| \| --- \| \| Antibiotic \| Indications, Contraindications, and Special Considerations \| \| Amoxicillin Cephalosporins Clindamycin Metronidazole Penicillin \| May be used during pregnancy \| \| Ciprofloxacin Clarithromycin Levofloxacin Moxifloxacin \| Avoid during pregnancy \| \| Tetracycline \| Never use during pregnancy \| Amoxicillin Cephalosporins Clindamycin Metronidazole Penicillin Ciprofloxacin Clarithromycin Levofloxacin Moxifloxacin Tetracycline Antibiotic considerations for pregnant patientsAdapted from Oral Health Care During Pregnancy Expert Workgroup1 \| \| \| --- \| \| Anesthetic \| Indications, Contraindications, and Special Considerations \| \| Local anesthetic with epinephrine: bupivacaine, lidocaine, mepivacaine \| May be use during pregnancy \| \| Nitrous oxide 30% \| May be used during pregnancy when topical or local anesthetics are inadequate. Pregnant women require lower levels of nitrous oxide to achieve sedation; consult with prenatal care health professional before administration. \| Local anesthetic with epinephrine: bupivacaine, lidocaine, mepivacaine Nitrous oxide 30% May be used during pregnancy when topical or local anesthetics are inadequate. Pregnant women require lower levels of nitrous oxide to achieve sedation; consult with prenatal care health professional before administration. Anesthetic considerations for pregnant patientsAdapted from Oral Health Care During Pregnancy Expert Workgroup1 \| \| \| --- \| \| Antimicrobial \| Indications, Contraindications, and Special Considerations \| \| Cetylpyridinium chloride mouth rinse Chlorhexidine mouth rinse Xylitol \| May be used during pregnancy \| Cetylpyridinium chloride mouth rinse Chlorhexidine mouth rinse Xylitol Antimicrobial considerations for pregnant patientsAdapted from Oral Health Care During Pregnancy Expert Workgroup1 Dental care is not only safe during pregnancy, but it is also encouraged. Rest assured that providing treatments, including prophylaxis, non-surgical periodontal therapy (NSPT), radiographs, and restorative care, is safe. There is no need to acquire medical clearance to provide care for pregnant patients. Studies show that pregnant patients are at an increased risk of dental caries, gingivitis, and periodontal disease. Let us stop delaying care for these patients by requiring medical clearance for standard dental care.1,2 Myocardial Infarction There is clear evidence of an association between cardiovascular disease and oral health. Therefore, providing dental care to patients with cardiovascular disease is imperative to help prevent or manage disease progression. Irrespective of dental treatment, over 70% of myocardial infarction relapses occur within the first month after the initial incident.3,4 In the past, the American Heart Association (AHA) guidelines recommended avoiding dental care for six months after the initial incident. This was based on the recovery through the creation of collateral circulation and restoration of contractility of the damaged areas of the myocardium.However, over the past two decades, advancements in cardiac management have made these limitations no longer necessary.3 The priority for patients recovering from myocardial infarction is to manage inflammation, including inflammation in the oral cavity. Avoiding dental treatment can lead to an increased production of endogenous catecholamines, increasing the burden of repairing the damaged heart muscle. Therefore, there may be better solutions than an over-cautious approach.3 The current protocol is providing dental care four to six weeks post-incident. Tooth extraction and surgical procedures are an exception. All other preventative and complex dental treatments can be completed four to six weeks after myocardial infarction unless otherwise contraindicated by the cardiologist.3 Acquiring medical clearance from the patient’s cardiologist post-myocardial infarction is a reasonable precaution. The cardiologist can better determine if the patient is stable through tests and symptoms reported by the patient. However, the six-month waiting period is no longer standard. Newer guidelines indicate dental treatment is safe after four to six weeks.3 Anticoagulant or Antiplatelet Therapy Patients often start antiplatelet or anticoagulation therapy after a myocardial infarction or, in some cases, when the patient is at increased risk of having a myocardial infarction. Increased risks include, but are not limited to, type 2 diabetes, high blood pressure, and obesity.5 You can expect increased bleeding with antiplatelet therapy, such as aspirin. However, modifications to the patient’s medications are unnecessary for dental treatment. Anticoagulation therapy (i.e., warfarin) is often used after myocardial infarction but may also be utilized in patients who are unable to adhere to antiplatelet therapy, as well as patients with atrial fibrillation, left ventricular thromboses, or systemic or pulmonary emboli.4 Dental treatment and elective dental surgery can be safely performed on patients on anticoagulation therapy whose international normalization ratio (INR) is 2.0–3.0 without bleeding problems. Therefore, dental care need not be postponed in most cases if the patient knows their INR and treatment is noninvasive, such as prophylaxis, NSPT, and restorative treatment.4 Nonetheless, precautions should be taken for dental surgery, such as placing a hemostatic dressing, multiple sutures, and intraoral pressure packs. If the patient reports a history of excessive bleeding after dental surgery, a medical clearance is reasonable in those specific cases.4 Cerebrovascular Accident Cerebrovascular accident (CVA), also referred to as stroke, is classified by the cause. There are two major classes of CVA: ischemic and hemorrhagic. An ischemic stroke occurs when there is a blockage in the cerebral blood vessels. A hemorrhagic stroke occurs when a cerebral blood vessel ruptures.4 Following a CVA, patients are at an increased risk of recurrence during dental treatment. As a general rule, patients should not undergo elective dental care within six months of a CVA, as the risk of recurrence is more significant during this period. Emergency care during this timeframe should be managed noninvasively through medication if possible. It should be completed in a hospital setting if invasive treatment is required.4 Patients who experience a transient ischemic attack (TIA), also referred to as a “mini-stroke,” are also at increased risk of recurrence during the first six months after the incident. Therefore, elective dental treatment should also be delayed for these patients.4 Medical clearance and/or consultation is recommended for patients who have experienced any type of CVA.4 \| \| \| \| \| \| \| --- --- --- \| \| Cause \| Approximate % of All Strokes \| Mortality Rate within 30 Days \| Mortality Rate within 1 Year \| Recurrence Rate within 30 Days \| Recurrence Rate within a Year \| \| All strokes \| \| \| 23% \| 2% \| 8% \| \| Ischemic stroke \| 87% \| 3%-20% \| 20%-35% \| 1%-6% \| 5%-25% \| \| Atherothrombotic brain infarction \| 60% \| 8%-12% \| \| 8% \| 12% \| \| Cardioembolic stroke \| 25.1% \| 30.3% \| 50.3% \| \| \| \| Hemorrhagic stroke \| 13% \| 35%-47% \| \| \| \| \| Intracerebral hemorrhage \| 10%-15% \| 35%-52% \| 80% \| \| \| \| Subarachnoid hemorrhage \| 5.4% \| 28.7%-40% \| 50%(within 6 months) \| \| \| All strokes Ischemic stroke Atherothrombotic brain infarction Cardioembolic stroke Hemorrhagic stroke Intracerebral hemorrhage Subarachnoid hemorrhage Types of strokes and mortality and recurrence ratesAdapted from Malamed, S.F.4 In Closing Although some conditions and situations require medical clearance, in many cases, there is clear guidance that certain conditions no longer require medical clearance. For pregnant patients, obtaining medical clearance or postponing treatment is not necessary as long as the patient is healthy. Delaying treatment for pregnant patients is more detrimental to the patient and the developing fetus than completing dental treatment.1 However, in the case of post-CVA, delaying treatment and obtaining medical clearance is the best practice.4 Though it is difficult to stay informed with newer guidance, it is in the patients’ and the clinicians’ best interest to ensure they stay abreast of changes and follow current guidance. This will improve patient care and outcomes as well as greatly reduce medical emergencies in the dental setting. Before you leave, check out the Today’s RDH self-study CE courses. All courses are peer-reviewed and non-sponsored to focus solely on high-quality education. Click here now. Listen to the Today’s RDH Dental Hygiene Podcast Below: References RELATED ARTICLESMORE FROM AUTHOR Understanding and Addressing Patient Concerns Regarding Dental Radiographs Patient Education: Why Dental Hygienists Should Stop Being Minimizers Trismus: A Potential Complication of Administering Local Anesthesia Like and Follow on Facebook Trending Now Categories Most Recent Understanding and Addressing Patient Concerns Regarding Dental Radiographs An Evaluation of the FDA’s Proposed Removal of Fluoride Prescription Drugs... Scoping Review Evaluates Parental Oral Health Literacy’s Influence on Children’s Oral... Curiosity Killed the Plaque Ep. 36: Fentanyl Exposure, Overdose Signs, and... Don't Miss Be Prepared to Provide Dental Hygiene Care for Pregnant Patients How Dental Hygienists Can Help Patients with Dental Phobias Dental Practitioners Address Serious COVID-19 Issues in Northern Italy The Importance of Screening Pediatric Patients for Sleep Related Breathing Disorders
15238	https://www.endocrine-abstracts.org/ea/0029/ea0029s52.3	Treatment of hyperthyroidism in pregnancy \| ICEECE2012 \| 15th International & 14th European Congress of Endocrinology \| Endocrine Abstracts Searchable abstracts of presentations at key conferences in endocrinology ISSN 1470-3947 (print) \| ISSN 1479-6848 (online) Endocrine Abstracts Menu SearchIssues/ConferencesCiteAboutOur ServicesPoliciesContactDisclaimer Search Issues/Conferences Cite About Our Services Policies Contact Disclaimer S52.3 PrevNext SectionContentsCite Endocrine Abstracts (2012) 29 S52.3 ICEECE2012SymposiaThyroid & Pregnancy (3 abstracts) Treatment of hyperthyroidism in pregnancy Fereidoun. Azizi Treatment of hyperthyroidism in pregnancy Author affiliations Research Institute for Endocrine Sciences, Shahid Beheshti University of Medical Sciences, Tehran, Iran. Graves’ disease is the most common cause of autoimmune hyperthyroidism in pregnancy accounting for 0.1 to 1% (0.4% clinical and 0.6% subclinical) of all pregnancies. It may be diagnosed for the first time in pregnancy, may present as a recurrent episode in a woman with past history of hyperthyroidism, or in a women on antithyroid drugs (ATD). Less common non autoimmune causes include multinodular goiter, toxic adenoma, and factitious hyperthyroidism. More frequent than Graves’ disease as the cause of hyperthyroidism is the syndrome of gestational hyperthyroidism defined as ‘transient hyperthyroidism, limited to the first half of pregnancy characterized by elevated FT 4 or adjusted TT 4 and suppressed or undetectable serum TSH, in the absence of serum markers of thyroid autoimmunity’. Poor control of thyrotoxicosis is associated with miscarriages, PIH, prematurity, low birth weight, intrauterine growth restriction, stillbirth, thyroid storm and maternal congestive heart failure. Antithyroid drugs are the mainstay of treatment for hyperthyroidism during pregnancy. Side effects occur in a 3–5%, of patients taking thionamide drugs, mostly allergic reactions such as skin rash. The greatest concern in the use of ATD in pregnancy is related to teratogenic effects. Exposure to MMI may produce several congenital malformations, mainly aplasia cutis and the syndrome of ‘methimazole embryopathy’ that includes choanal or esophageal atresia, and dysmorphic faces. Although very rare complications, they have not been reported with the use of PTU. Report from the Adverse Event Reporting System of the Food and Drug Administration (FDA) has called attention to the risk of hepatotoxicity in patients exposed to PTU and an advisory committee recommended to limit the use of PTU to the first trimester of pregnancy. Hepatotoxicity may occur at any time during PTU treatment, without warning; however it may be appropriate to monitor liver function tests regularly while PTU is administered. In women with hyperthyroidism due to Graves’ disease or a toxic nodule goiter, ATDs are initiated, or adjusted in those women on treatment, at conception. PTU is preferred in the first trimester. Following the first trimester, PTU should be switched to methimazole. The combination of ATD plus levothyroxine is not recommended in the management of hyperthyroidism in pregnancy. Adrenergic β blocking agents, such as propranolol 20–40 mg every 6–8 h may be used for controlling hyper-metabolic symptoms. In the vast majority of cases the drug could be discontinued in 2 to 6 weeks. Long term treatment with propranolol has been associated with intrauterine growth restriction, fetal bradycardia and neonatal hypoglycemia. Thyroidectomy should be considered in cases of allergies to both ATDs, women requiring large doses of ATDs and the occasional patient who is not adherent to drug therapy. If surgery is indicated, second trimester is the optimal time. A determination of serum TRAb titers is mandatory at the time of surgery in order to assess the potential risk of fetal hyperthyroidism. Preparation with β-blocking agents and a short course of potassium iodine solution (50–100 mg a day) are recommended. Volume 29 PrevNext 15th International & 14th European Congress of Endocrinology European Society of Endocrinology Browse other volumes SummaryProgrammeVolume EditorsAbstractsAbstract Book Article tools ▼ \| Disclaimer My recent searches No recent searches My recently viewed abstracts Treatment of hyperthyroidism in pregnancy (<1 min ago) Authors Azizi Fereidoun. Endocrine AbstractsGoogle ScholarPub Med Endocrine Abstracts ISSN 1470-3947 (print) \| ISSN 1479-6848 (online) © Bioscientifica 2025 \| Privacy policy \| Cookie settings BiosciAbstracts Biosci Abstracts Bioscientifica Abstracts is the gateway to a series of products that provide a permanent, citable record of abstracts for biomedical and life science conferences. Find out more This site uses cookies × Essential cookies Some cookies are essential and the website may not work correctly without them. These cookies can only be removed by changing your browser preferences. Other cookies We use tracking cookies and other non-essential cookies to help improve our services. On Off More information can be found on our privacy policy Close Original text Rate this translation Your feedback will be used to help improve Google Translate We place cookies on your device to give you the best experience of this website. If you don't change your cookie settings, we'll assume you're happy with this. 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15239	https://artofproblemsolving.com/wiki/index.php/Radical_axis?srsltid=AfmBOophbsqoIroR8maC9jJTA_Ff80Yi7SbtlGIjjvtanVTrL0Gb7bW-	Art of Problem Solving Radical axis - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Radical axis Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Radical axis Contents [hide] 1 Introduction 2 Definitions 3 Results 4 Proofs 5 Exercises 6 Problems 6.1 Simple 6.2 Intermediate 6.3 Olympiad 7 See Also Introduction The theory of radical axis is a priceless geometric tool that can solve formidable geometric problems fairly readily. Problems involving it can be found on many major math olympiad competitions, including the prestigious USAMO. Therefore, any aspiring math olympian should peruse this material carefully, as it may contain the keys to one's future success. Not all theorems will be fully proven in this text. The objective of this document is to introduce you to some key concepts, and then give you a chance to derive some of the beautiful results on your own. In that way, you will understand and retain the information in here much more solidly. Finally, your newfound knowledge will be tested on a few challenging problems that are exemplary examples on how radical axis theory can be used and why it pertains to that situation. I hope after you read this text, you will become a better math student, armed with another tool to solve difficult problems. But, anyway, good luck. i Definitions The power of point with respect to circle (with radius and center ), which shall thereafter be dubbed , is defined to equal . Note that the power of a point is negative if the point is inside the circle. The radical axis of two non-concentric circles is defined as the locus of the points such that the power of with respect to and are equal. In other words, if are the center and radius of , then a point is on the radical axis if and only if (i.e., the radical axis is the line that one gets when you subtract the equations of two circles). Results Theorem 1: (Power of a Point) If a line drawn through point P intersects circle at points A and B, then . Theorem 2: (Radical Axis Theorem) a. The radical axis is a line perpendicular to the line connecting the circles' centers (line ). b. If the two circles intersect at two common points, their radical axis is the line through these two points. c. If they intersect at one point, their radical axis is the common internal tangent. d. If the circles do not intersect, and if one does not fully contain the other, their radical axis is the perpendicular to through point A, the unique point on such that . Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center. Theorem 4: (Radical Centre of Intersecting Circles) (EGMO Theorem 2.9) Let and be two circles with centers and . Select two points A and B on and C and D on . Then the following are equivalent lie on a circle with center not on line . Lines and intersect on the radical axis of and . Theorem 5: (EGMO Lemma 2.11) Let ABC be a triangle and consider a point in its interior. Suppose that is tangent to and , ray bisects Proofs Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.) Theorem 2 shall be proved here. Assume the circles are and with centers and and radii and , respectively. (It may be a good idea for you to draw some circles here.) First, we tackle part (b). Suppose the circles intersect at points and and point P lies on . Then by Theorem 1 the powers of P with respect to both circles are equal to , and hence by transitive . Thus, if point P lies on , then the powers of P with respect to both circles are equal. Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on . In particular, line does not intersect X. Then intersects circles and a second time at distinct points and , respectively. (If is tangent to , for example, we adopt the convention that ; similar conventions hold for . Power of a Point still holds in this case. Also, notice that and cannot both equal , as cannot be tangent to both circles.) Because is not equal to , so does not equal , and thus by Theorem 1 is not congruent to , as desired. This completes part (b). For the remaining parts, we employ a lemma: Lemma 1: Let be a point in the plane, and let be the foot of the perpendicular from to . Then . The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise. Lemma 2: There is an unique point P on line such that . Proof: First show that P lies between and via proof by contradiction, by using a bit of inequality theory and the fact that . Then, use the fact that (a constant) to prove the lemma. Lemma 1 shows that every point on the plane can be equivalently mapped to a line on . Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to , completing part (a). Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.) Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axes.)The proof of this theorem along with the proof of Theorem 4are given in EGMO as Example 2.7 and Theorem 2.9. Theorem 5 Let intersect at at and let , . Clearly, is a radical axis of , . We see thatas desired. Exercises If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. Note: No solutions will be provided to the following problems(laziness). If you are stuck, ask on the forum. Problem 1. Two circles and intersect at and . Point is located on such that and ~~15~~. If the radius of is , find the radius of . Note: An error in this problem previously rendered it unsolvable. Problem 2. Solve 2009 USAMO Problem 1. If you already know how to solve it. Problem 3. Two circles P and Q with radii 1 and 2, respectively, intersect at X and Y. Circle P is to the left of circle Q. Prove that point A is to the left of if and only if . Problem 4. Solve 2012 USAJMO Problem 1. Problem 5. Does Theorem 2 apply to circles in which one is contained inside the other? How about internally tangent circles? Concentric circles? Problem 6. Construct the radical axis of two circles. What happens if one circle encloses the other? Problem 7. Solve 1995 IMO Problem 1 in two different ways. Compare your solutions with the solutions provided. Problems Simple Let triangle points and be given. Denote Prove that Proof WLOG, the order of the points is as shown on diagram. Intermediate 2021 AIME Problem 13. Circles and with radii and , respectively, intersect at distinct points and . A third circle is externally tangent to both and . Suppose line intersects at two points and such that the measure of minor arc is . Find the distance between the centers of and . Olympiad 2014 USAMO Problem 5. Let be a triangle with orthocenter and let be the second intersection of the circumcircle of triangle with the internal bisector of the angle . Let be the circumcenter of triangle and the orthocenter of triangle . Prove that the length of segment is equal to the circumradius of triangle . 2017 USAMO Problem 3. Let be a scalene triangle with circumcircle and incenter Ray meets at and again at the circle with diameter cuts again at Lines and meet at and is the midpoint of The circumcircles of and intersect at points and Prove that passes through the midpoint of either or 2012 IMO Problem 5. Let be a triangle with , and let be the foot of the altitude from . Let be a point in the interior of the segment . Let be the point on the segment such that . Similarly, let be the point on the segment such that . Let . Prove that . See Also This article is a stub. Help us out by expanding it. 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15240	https://www.chemguide.co.uk/inorganic/group7/testing.html	\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- \| \| TESTING FOR HALIDE IONS This page describes and explains the tests for halide ions (fluoride, chloride, bromide and iodide) using silver nitrate solution followed by ammonia solution. Using silver nitrate solution Carrying out the test This test has to be done in solution. If you start from a solid, it must first be dissolved in pure water. The solution is acidified by adding dilute nitric acid. (Remember: silver nitrate + dilute nitric acid.) The nitric acid reacts with, and removes, other ions that might also give a confusing precipitate with silver nitrate. Silver nitrate solution is then added to give: \| ion present \| observation \| --- \| \| F- \| no precipitate \| \| Cl- \| white precipitate \| \| Br- \| very pale cream precipitate \| \| I- \| very pale yellow precipitate \| The chloride, bromide and iodide precipitates are shown in the photograph: The chloride precipitate is obviously white, but the other two aren't really very different from each other. You couldn't be sure which you had unless you compared them side-by-side. All of the precipitates change colour if they are exposed to light - taking on grey or purplish tints. The absence of a precipitate with fluoride ions doesn't prove anything unless you already know that you must have a halogen present and are simply trying to find out which one. All the absence of a precipitate shows is that you haven't got chloride, bromide or iodide ions present. The chemistry of the test The precipitates are the insoluble silver halides - silver chloride, silver bromide or silver iodide. Ag+(aq) + Cl-(aq) AgCl(s) Ag+(aq) + Br-(aq) AgBr(s) Ag+(aq) + I-(aq) AgI(s) Silver fluoride is soluble, and so you don't get a precipitate. Confirming the precipitate using ammonia solution Carrying out the confirmation Ammonia solution is added to the precipitates. \| original precipitate \| observation \| --- \| \| AgCl \| precipitate dissolves to give a colourless solution \| \| AgBr \| precipitate is almost unchanged using dilute ammonia solution, but dissolves in concentrated ammonia solution to give a colourless solution \| \| AgI \| precipitate is insoluble in ammonia solution of any concentration \| Explaining what happens Background There is no such thing as an absolutely insoluble ionic compound. A precipitate will only form if the concentrations of the ions in solution in water exceed a certain value - different for every different compound. This value can be quoted as a solubility product. For the silver halides, the solubility product is given by the expression: Ksp = [Ag+(aq)][X-(aq)] The square brackets have their normal meaning, showing concentrations in mol dm-3. If the actual concentrations of the ions in solution produce a value less than the solubility product, you don't get a precipitate. If the product of the concentrations would exceed this value, you do get a precipitate. Essentially, the product of the ionic concentrations can never be greater than the solubility product value. Enough of the solid is precipitated so that the ionic product is lowered to the value of the solubility product. \| \| \| \| --- Note:If your syllabus says that you need to know about solubility product calculations, you might be interested in my chemistry calculations book where they are explained in detail. --- \| \| Look at the way the solubility products vary from silver chloride to silver iodide. (You can't quote a solubility product value for silver fluoride because it is too soluble. Solubility products only work with compounds which are very, very sparingly soluble.) \| \| Ksp (mol2dm-6) \| --- \| \| AgCl \| 1.8 x 10-10 \| \| AgBr \| 7.7 x 10-13 \| \| AgI \| 8.3 x 10-17 \| \| \| \| \| --- Note:These figures come from the Chemistry Data Book by Stark and Wallace. --- \| \| You can see that the compounds are all pretty insoluble, but become even less soluble as you go from the chloride to the bromide to the iodide. What is the ammonia doing? The ammonia combines with silver ions to produce a complex ion called the diamminesilver(I) ion, [Ag(NH3)2]+. This is a reversible reaction, but the complex is very stable, and the position of equilibrium lies well to the right. A solution in contact with one of the silver halide precipitates will contain a very small concentration of dissolved silver ions. The effect of adding the ammonia is to lower this concentration still further. What happens if you multiply this new silver ion concentration by the halide ion concentration? If the answer is less than the solubility product, the precipitate will dissolve. That happens with the silver chloride, and with the silver bromide if concentrated ammonia is used. The more concentrated ammonia tips the equilibrium even further to the right, lowering the silver ion concentration even more. The silver iodide is so insoluble that the ammonia won't lower the silver ion concentration enough for the precipitate to dissolve. An alternative test using concentrated sulphuric acid If you add concentrated sulphuric acid to a solid sample of one of the halides you get these results: \| ion present \| observation \| --- \| \| F- \| steamy acidic fumes (of HF) \| \| Cl- \| steamy acidic fumes (of HCl) \| \| Br- \| steamy acidic fumes (of HBr) contaminated with brown bromine vapour \| \| I- \| Some steamy fumes (of HI), but lots of purple iodine vapour (plus various red colours in the tube) \| \| \| \| \| --- Note:The chemistry of this test is explained in detail on another page. --- \| \| The only possible confusion is between a fluoride and a chloride - they would behave identically. You could distinguish between them by dissolving the original solid in water and then testing with silver nitrate solution. The chloride gives a white precipitate; the fluoride doesn't give a precipitate. \| \| \| Questions to test your understanding If this is the first set of questions you have done, please read the introductory page before you start. You will need to use the BACK BUTTON on your browser to come back here afterwards. questions on testing for halide ions \| Where would you like to go now? To the Group 7 menu . . . To the Inorganic Chemistry menu . . . To Main Menu . . . --- © Jim Clark 2002 (last modified March 2022) \| \|
15241	https://www.youtube.com/watch?v=drZX6i8YpEw	AP Physics 2 Lecture 10-1 Charge and the Electric Force KcoolScience 3280 subscribers 16 likes Description 1636 views Posted: 5 Nov 2022 An lecture designed for AP Physics 2 students in Unit 3 Electrostatic. Covers the basics of charge, methods of charging, coulomb's law, and the law of superposition (vector math with charges). Covers AP Topics 10.1 Electric Charge and Electric Force and 10.2 Conservation of Electric Charge and the Process of Charging as described by Collegeboard's AP Physics 2 Course and Exam Description- (Previously topics 3.1-3.8) Want the lecture notes and a practice that goes along with this lecture? Check out my shop on TPT! 5 comments Transcript: hey guys miss Peterson here and welcome to AP Physics 2 lecture 3-1 all about charge and the electric force this is our first lecture of the unit on electrostatics where we're going to be covering a ton of AP topics um that kind of all go together so best to do them all in one uh we're going to be talking about charges and how they interact go through Coulomb's law which is the quantitative way that we figure out the charge and go through conductors and insulators and different methods of charging Okay so before we get into all of that what even is charge okay quick review atoms okay the things that make up matter have protons neutrons and electrons electrons are on the outside and they are negative okay protons are on the inside and they are positive the neutrons are neutral so when we're talking about charge we don't need to worry about the neutrons okay if an atom gains an electron and has more electrons that it has protons it is going to be negatively charged where if you have more protons then you have electrons you will be positively charged okay now note because the electrons are on the outside okay matter can only be charged by the transfer of those electrons okay that is what moves to make things charged and then of course just like we have conservation of matter and conservation of energy we have a conservation of electric charge okay and the classic rules of electric force are the Opposites Attract so positive and negative charges will attract and light charges will repel okay so how does an object even become charged well if you wanted to charge up a balloon to stick it on the wall you would just rub it on your head okay rub it on your hair um that is called charging by friction which is also sometimes called triboelectric charging and it's called the triboelectric charging because it has to do with something called the triboelectric series okay so things like wool okay or cotton and the human body then they tend to become positive meaning they tend to lose electrons okay where things like silicone and uh PVC that's what polyethylene vinyl is um and sulfur and rubber like the balloons tend to become negative okay they tend to gain electrons in order to become negative meaning they have a higher electron affinity they like electrons more than some other atoms our materials might okay now that is one way of charging another way that we charge things is by conduction so when two objects that are conductors touch they transfer that charge okay charge likes to spread out okay picture if you have a bunch of negative charge on a sphere it's going to want to be as far away from each other as possible because it's all going to repel so if say we have this positive plate with all that positive charge close to it and we touch it to a neutral sphere okay those positive charges are going to attract electrons in this sphere and electrons are going to add on to there which will remove some of the positive charge from the plate but because that's fear loss electrons okay it hasn't gained the positive charge okay so while triboelectric charging becomes oppositely charged when you charge by conduction the objects have the same charge okay now another way that we will see things is charging by induction so when a charged object is brought near a neutral object it polarizes it AKA moves that charge to one side Okay then if that polarized object say it was one object or maybe it was two spheres touching each other okay they're separated during the polarization you can be left with two oppositely charged objects okay and here's a little diagram of how that happens we have neutral spheres okay they are neutral now if you bring a charged object toward it then we get that polarization okay that charge object repels the electrons so gay they gather over here and attracts the positive charge so that gathers over there and because they are touching they are conducting okay because they're touching there where if then you separated them during the polarization they have two opposite charges okay cool so let's go ahead and practice okay uh charged rubber rods okay so we got charged rods are placed near a neutral conducting sphere causing a redistribution of charge which of these depict the proper distribution what do you think yep well if you said diagram a okay a lot checks out these little things are the positives charges sorry they're not that clear in the picture but the negative will repel the negative and attract the positive okay here we have positive attracting positive that's not gonna happen Okay negative attracting negative not going to happen positive attracting negative yeah and now diagram e we do have the positive attracting the negative but why is that one not correct okay because this sphere is not neutral okay it says neutral conducting sphere so we would need a balance of those positive charges okay we would need those three positive over here for that to be the case but since they are not there it is not neutral okay and then a triboelectric charging one Saran Wrap has a larger electron affinity than nylon okay so that means Saran Wrap likes electrons more if nylon is rubbed against Saran Wrap which would end up with the excess negative charge well electrons are negative Saran Wrap has a larger electron affinity meaning it is that it likes the electrons more so Saran Wrap would be negative okay and the nylon would be positive okay cool which brings us to conductors versus insulators okay when I say conductors that think metals and if you remember from a chemistry class okay metals have these positive ions flowing in a sea of negative electrons okay and they are free to flow okay they flow freely from particle to particle and the electrons can go wherever they want within the substance okay for example in this metal okay the charge will spread out all along the surface because it's a conductor and it can move now because everything repels each other that charge will repel and only reside on the surface okay so if you have a big metal sphere all of the charge is going to be on the outside of that sphere meaning no net electric field in the middle and now when two conductors touch that charge gets spread out between it like it was just one conductor um and yeah just think Metals tap water is very conductive things like that now insulators are a little different insulators impede the flow of electrons but this doesn't mean that they can't become charged okay if you have a really charged object and you touch it to an insulator it will transfer some of that charge the only difference is it will stay in the location it was transferred to okay it would look kind of like this image okay where all that charge stays in one place it's stuck on whatever atoms or molecules were there to begin with okay rubber is a good example of an insulator glass dry air is a very good insulator okay so that's the general properties and of course this is a range where everything falls within the range okay now a couple other ways that charges move is through grounding okay so grounding is the removal of a charge okay so this means that taking it to neutral and that's because we think of the ground as an infinite reservoir of electrons okay and since it is an infinite Reservoir there's tons of electrons there and available it can add or remove okay if you touch a grounding object to a positively charged object it's going to add electrons okay so if positive okay it will add electrons okay where if it's negative it will remove it okay for example we can also use this to charge things here we have an Ebonite Rod um that is at negatively charged being brought toward a metal sphere okay and right now this is neutral but polarized okay now if we touch the grounding wire on it okay that is a wire it is a conductor okay meaning those charges are free to move so those negative charges are going to get repelled by that negative rod and go to the ground okay and then if we remove the grounding wire we could be left with a net positive charge yep just by using the grounding while it's polarized and just a little note on being polarized both conductors and insulators can become polarized it's just the scale to which they do it okay so conductors are going to have a lot more separation of charge because those charges can move freely okay as you can see in oops okay so as you can see in this picture both conductors and insulators can become polarized but in a conductor the whole charges can move okay they can move between the atoms and molecules that are in the substance creating an entire net negative charge over here and net positive charge over here where for insulators the polarization happens on the like molecular level okay polarized on molecular level meaning within the molecules themselves we have that separation of charge which results in a net difference in the charge on the two sides of that insulator but it's not as dramatic as with the conductors at bagel now let's get to the math stuff okay Coulomb's law so Coulomb's law is the force between two charged objects and when we look at it we're going to be talking about the charges on electrons and protons okay we use the letter e for that where e is the elementary charge it's the basic unit of electron charges there are no charges smaller than it and our all charges have to be a multiply multiple of e okay charge is quantized you can't have half an electrons worth of charge okay now it's in units of coulombs this is the variable Q for charge a lowercase Q not to be confused with the capital Q that we use for heat energy okay now one coulomb is a lot of charge you need 6.24 times 10 to the 18 electrons to have one coulomb most of our measurements of uh coulombs will be on the order of like 10 to the negative um something okay now on your equation sheet on your AP Physics 2 equation sheet Coulomb's a lot is written like this okay the absolute value of the electric force vector equals 1 over 4 Pi e naught okay now e naught is the permittivity of free space okay and we often will not use that at all we'll use the coulombs constant K which is equal to 1 over 4 Pi e naught okay so we have that and then they actually have it in absolute values where you have q1 times Q2 over the distance between them squared okay they set it as an absolute value because uh the tendency is to think that if it's negative if you have a negative force that means repel but if we look at the equation okay if we had a negative charge at a positive charge that would result in a negative number and we know that a negative charge and a positive charge are opposite they will attract each other okay where if both of these charges were positive or if both of them were negative we would get a positive force so that means repel okay and I think that is kind of confusing because I think negative normally means repel so that's why I say use the absolute value okay use the absolute value and the idea that Opposites attract to figure out the direction rather than trying to use at something like this okay and the form that I write Coulomb's law in most often okay is just that Force as K q1 Q2 over r squared okay oh and R of course is uh not really the radius but it is the separation distance okay between those two charges okay and again this K is a constant it is Coulomb's law constant 9 times 10 to the ninth noon meter squared coulomb squared but on your equation sheet it is written as 1 over 4 Pi e naught in terms of the permittivity of the free space which basically is how well the space can hold an electric field um and just so you know I've never seen a value for it that is that not this number okay so just use that one for all purposes let's look at an example okay so we are considering the proton and the electron in a hydrogen atom okay so uh they're separated by a Bohr radius which is 5 times 10 to the negative 14 which is cool because then like you know we actually have the radius um in what we want yep or it's it's the radius actually R times that 10 to the negative 11 meters now for the charge on a proton and electron they're equal to each other so Q proton equals the negative charge on the electron negative Q electron and they are both equal to that fundamental charge 1.6 times 10 to the negative 19 coulombs okay so what is the electric force between them we have our equation our equation is that the electric force equals K times q1 times Q2 over r squared and like I said um ignore these signs K this is going to be an attractive Force yep so that's why I'm not going to put the signs of those charges into here I know that it's attractive okay the sign of it just tells me that direction of that Force um with the type of charge but yeah okay so plug in my numbers okay is 9.0 times that 10 to the ninth newton meters squared per coulomb squared times q1 which is 1.60 times 10 to the negative 19 coulombs the charge on a proton times the charge on an electron which is negative 1.6 but again if I'm just calculating the force easier to just do it with the absolute values divided by the distance between them squared okay and then we got to be able to plug that all into our calculator okay now calculators get very confused when doing scientific notation and order of operations so I highly recommend that you use your exponential notation button or your ee button your exp button so that your calculator knows to treat each of these as one whole number and then you can do the calculation all at once so I have nine times 10 to the ninth times 1.6 times 10 to the negative 19. times 1.6 times 10 to the negative 19. I went ahead and put all of that in parentheses again I am using that ee button okay so and then divided by 5.0 oops times 10 to the negative 11. so squared okay I highly recommend putting that one in parentheses okay if you guys can see that okay so that they know that the square applies to the 5 times 10 to the negative 11 not just the 10 to the negative 11. okay and I press enter and I got 9.216 okay so I'll just say 9.22 times that 10 to the negative 8 Newtons okay and the units work out coulomb squared coulombs coulombs meters squared with the meter squared we're left with units of Newtons now I'm asking you about the gravitational force which did this equation look a little familiar okay it is almost identical to the gravitational force equation okay which is just a different constant the gravitational constant G times the two masses over the radius okay they have very similar equations what is the difference between these two forces there um there are quick side there are four fundamental forces in physics nuclear strong force nuclear weak Force we only deal with those when we're dealing with nucleuses so we can ignore them our main two are gravitational and the electromagnetic these two equations okay but there is a fundamental difference between them does gravity ever repel the electric force can attract or repel but in our experience with gravity gravity only ever attracts we've never seen enough instance where two masses will repel each other um um now for the masses and G uh those can be found on your AP Physics 2 reference sheet okay so I'm just going to look at that reference sheet to plug them in okay we have G which is 6.67 times that 10 to the negative 11. okay Newton kilogram squared oh oh nope sorry Newton meter squared over kilogram squared times the mass on a proton which is 1.67 times 10 to the negative 27 kilograms times the mass on an electron which is 9.11 times that 10 to the negative 31 kilograms divided by the distance between them squared it and we plug that into our calculators times 9.11 times 10 to the negative 31 divided by 5.0 times 10 to the negative 11. squared and just by looking at the constants the K versus G and those relative masses we can tell that this number is going to be much smaller and in fact it is much much much smaller 4.06 times 10 to the negative 47 Newtons okay so a negative order of negative 8 versus order of negative 47 okay if we look at the ratio between those two forces oops okay so if we have that 9.22 times 10 to the negative 8 Newtons over hey just to see how dramatic this is okay the electric force is 2.27 times 10 to the 39 times larger then the gravitational force okay to the 39 times that is huge so okay we when considering the electric force between two objects we do not need to consider the gravitational force between them okay when considering the electric force between two masses we can ignore the gravitational gravitational force between them yep because you know it is so so small now this doesn't mean we ignore the gravity acting on the object okay gravity is still exerting a force on that object and we will do problems where you have to balance out the force of gravity down on an object between Earth and the object with an electric force between an object above it that's how the Millikin oil drop experiment worked okay but if you're balancing out those two forces you're not considering the gravitational force between that mass and the mass above it okay you would consider the gravitational force between the object and Earth but not the gravitational force between those two objects okay because really it's not going to affect it at all if I want to solve for the acceleration due to that electric force okay acceleration would be equal to the electric force divided by the mass of the electron where here we'll do it as that force of gravity divided by the mass of the electron so on the acceleration due to the electric force would be 9.22 times 10 to the negative 8 Newtons divided by the electrons Mass which is 9.11 times 10 to the negative 31 kilograms and we'll compare that to the gravitational acceleration which is 4.06 times 10 to the negative 47 Newtons divided by 9.11 times 10 to the negative 31 kilograms so let's go ahead and calculate those out and for due to the electric force we have 1.01 times 10 to the 23rd meters per second squared so very very fast uh versus okay 4.46 times that 10 to the negative 17. okay so we can tell that this number is a um the acceleration due to the electric force is much much much greater than the acceleration due to gravity okay that is why we can ignore the acceleration due to gravity okay cool okay cool now let's get into Force Edition okay anytime we have more than two charges we're gonna do this is basically Vector Mass okay a little review of vectors but if we um yeah but just using the electric force we're going to do one here where it's just charges in one dimension and then we'll do one here where we have to break the vector down into its components to solve it in three dimensions okay so first one we'll do this one right here we have three charges lie along the x-axis okay so we got the x-axis here the positive charge at q1 of 15 micro coulombs is at x equals two okay so x equals 2.0 meters and we have on there okay q1 which has a charge of 15 micro coulombs remember what micro is okay micro is at times at 10 to the negative 6. so q1 you can write it as 15 times 10 to the negative 6 coulombs yep um and then the positive charge Q2 is at the origin okay so over here at zero we have Q2 which has a is another positive charge it is six times that 10 to the negative 6 coulombs and then it's asking where must a negative Q3 be placed on the x-axis so the resultant net force is zero okay so we're wanting to know where to put this on the axis so that the net force is zero well first let's think about how it's going to interact with both of these charges it's negative so it is going to be attracted to both of these now if we want it to be attracted to both and we want the resultant net force to be zero we're going to need to put it in between them okay so we know that this charge must be somewhere in between it but we can do even better than that okay since q1 is stronger it's going to attract it stronger if it was closer to this charge it would feel a stronger attraction to this and a weaker one to this due to both its charge and its distance so we know that Q3 actually needs to be closer to the smaller charge okay so that it can receive an equal force from that one as it does from this larger charge so let's go ahead and narrow oops and narrow that down draw that line back in okay we'll put 1.0 meters here as that middle Mark and we know that it must be closer to Q2 okay so for right now I'm just going to put Q3 right there so we can look at the forces on it it's going to be attract this one and attracted to that one and we want those forces to be equal okay actually yep there we go so for our Force we want the force of one on Three to be equal and opposite to the force of two on three okay so when we set up that equation what we're going to do is say this distance here is X to charge 3. how could we talk about this distance here so that we have less variables well we know Q3 is at 2. so this would be equal to 2 minus X okay then that distance X would give us where Q3 needs to be placed on the x-axis so that resulted in net force is zero okay let's go ahead and start setting up our equation uh force of two on three will be equal to K times q1 times Q3 over the distance between them x squared where for f one on Three we'll have K times it oops this was two on three two on three this one is Force One on three so Q2 Q3 over 2 minus X squared okay does it matter that we don't know what Q3 is nope not at all it's going to affect the force on both of them equally so we can cancel out Q3 and we can also cancel out okay because it'll just divide out so our equation simplifies to Q2 over X2 equals q1 over 2 minus X let's go ahead and put in the actual numbers and start doing that algebra so for Q2 we have six times ten to the negative 6 coulombs over X squared equals q1 15 times 10 to the negative 6 coulombs over 2 minus X squared okay let's make our math even easier we can cancel out the times 10 to the negative 6. okay so and I'm also going to cross multiply so I don't have a fraction so my equation becomes 6 times 2 minus x squared equals 15 times x squared okay now let's work smarter not harder let's go ahead and combine our like terms so I have um I'm gonna put the uh x's on one side and divide the 15 to the other side so we have 6 divided by 15. which is 0.4 equals x squared over 2 minus x squared okay easiest way to solve this is going to be the take the square root of the whole thing so the square root of 0.4 is a I thought it was 0.2 it is not 0.2 okay it is 0.632 equals x over 2 minus X okay so then do more algebra 0.62 times 2 gives us 1.26 minus 0.632 x equals x we add the 0.632 X to both sides uh which gives us 1.26 equals 2.632 X so x equals that 1.26 divided by 2.632 which is 0.48 and I feel like I'm not gonna lie I feel like I messed up on my math somewhere but 0.48 meters that is in our range so it is what we expect but I feel like I got 0.78 last time foreign I figured out why I went wrong okay um 1 plus 0.632 is a 1.632 X so the answer is 1.26 divided by 1.632 there we go which is 0.772 so 0.77 meters okay so this should be placed at 0.77 meters okay okay cool now for the trigonometry problem okay consider three point charges of the right triangle as shown where Q one is six times that 10 to the negative 9 coulombs Q2 is negative two times that 10 to the negative nine coulombs and Q3 is at 5 times 10 to the negative nine coulombs the distance between q1 and Q2 is 3 meters and between two three and four is four meters and we're going to find the force of one on Three as well as there's a force on two on three as well as the resultant Force okay so basically we got this charge here it is going to be repelled by this charge that's the force of one on Three which we see in that direction and it's going to be attracted to this one okay so that's going to be in that direction so we can picture the dot net force is going to be somewhere in this direction okay but we need to find what it's actually going to be so first let's go ahead and just find those forces okay starting with the force of one on Three and we got k nine times that ten to the ninth Newton meter squared per coulomb squared times at q1 6.00 times 10 to the negative 9 coulombs times Q3 5 times 10 to the negative 9 coulombs divided by the distance between them squared so when we solve that out we get 1.08 times 10 to the negative 8. so the force of one on Three is one point zero eight times that 10 to the negative 8. and then let's go ahead and do the force of two on three okay Newtons which is a 9 times 10 to the ninth times that Q what or Q2 and again I'm just going to use the absolute values here um and then do the directions at the end because I think that's easier times 10 to the negative 9 coulombs divided by 4.00 meters squared and we plug that into our calculators divide by 16 and we get 5.625 okay so I'll say 5.63 times that 10 to the negative 9 Newtons okay so we have our forces and then I'm just going to go ahead and draw or we'll just zoom in on that here so you guys can kind of see how that works okay the force of two on three is already in the horizontal component but the force of one on Three we are going to break down into its components okay so we know this angle here is 36.9 okay because we know this angle here is 36.9 um and again that comes from uh setting the tangent or yeah we'll use the tangent of that angle equal to three over four so then it's the inverse tangent of three over four yep which gives you uh 36.9 okay 36.9 using those complementary angles so we're going to break it down into its components okay so I'm going to call this first one okay the force of X for one on Three okay that one in the X direction that is the adjacent side so it's going to be equal to the force 1.08 times that 10 to the negative 8 Newtons times it's the adjacent so the cosine of 36.9 okay and again cosine of this comes from cosine of 36.9 degrees is equal to X over that force of one on three so if we're solving for that X component you can see how that math works out so that comes out too make sure your calculator is in degrees mode times the cosine of 36.9 yep and I got 8.64 times that 10 to the negative 9 Newtons now we'll do the Y components FY of one on Three 1.08 times 10 to the negative 9 Newtons and now it'll be the sine okay because now it's the opposite angle that we're talking about 36.9 degrees we plug that into our calculator and we get 6.48 times that 10 to the negative 9 Newtons okay um so let me go ahead and zoom in and draw what we have so far on Q3 okay so we have Q3 and we broke it down into its components so we have the X component of force or sorry yeah the X component of force one on Three which is 8.64 times 10 to the negative 9 Newtons okay we have the force due to two on three which is 5.63 times 10 to the negative 9 Newtons and then we figured out that upward or Y component of this to be 6.48 times 10 to the negative 9 nuance so in order to get the resultant Force first I'm going to combine the horizontal components okay and then set up a new triangle with that so they're in opposite directions so I know this one is bigger okay the force due to X um on three the X component of the force of one on Three which makes sense the force on two on three is a lot smaller because it's a much smaller charge so we're going to subtract that 8.64 times 10 to the negative 9 minus 5.63 times 10 to the negative 9 and we get 3.01 times 10 to the negative 9. since these forces bigger it's going to be in that direction okay so let me go ahead and draw that a little clear yep there this is our Q3 and now we figured out that the X component of that force was three times 10 to the negative 9 Newtons so these are the X and Y components of that Force to figure out the resultant okay we would line them up tail to tip okay so if I take this one and I'm going to move it over there okay and then we will be solving for this angle right there okay and in order to do that we use Pythagorean theorem a squared plus b squared equals c squared okay so our resultant Force okay is going to be equal to the square root of 3.01 times 10 to the negative 9 squared plus six point four eight times ten to the negative 9 squared okay and we take the square root of that to get our resultant Force times 10 to the negative 9 squared Plus 6.48 times 10 to the negative 9 squared and then we take the square root of that answer to get 7.14 times 10 to the negative 9 Newtons okay so the force of one on Three the net force would be that okay and now if we wanted to find the direction of that Force okay let's go ahead and look back here on the picture okay we know the force is going to be somewhere in that direction okay or actually no probably not because we found that this component was stronger than this one okay it had more in that so actually we expect that net force to be somewhere in this Direction okay now to figure out what it actually is we need to find this angle here so to find the angle I'm going to use tangent so the inverse tangent of the opposite which is 6.48 times that 10 to the negative 9 Newtons over the adjacent TOA times 10 negative 9 Newtons to get okay 6.48 times 10 to the negative 9 divided by 3.01 times 10 to the negative 9 okay and then I'm going to take the inverse tangent of that number to get 65 degrees okay so that based on how I drew it would be 65 degrees above horizontal yep so this angle here between it and the force it's probably up in a this direction then somehow yep okay that does make sense and this angle would be 65 degrees okay cool that is a big heavy complicated Vector math problem using electrostatic charge okay now you're unlikely to see something that complex on the AP exam but you might see pieces of it so it's helpful to understand the concepts okay cool okay cool
15242	https://www.geeksforgeeks.org/competitive-programming/dynamic-programming-in-game-theory-for-competitive-programming/	Dynamic Programming in Game Theory for Competitive Programming Last Updated : 23 Jul, 2025 Suggest changes 1 Like In the fast-paced world of competitive programming, mastering dynamic programming in game theory is the key to solving complex strategic challenges. This article explores how dynamic programming in game theory can enhance your problem-solving skills and strategic insights, giving you a competitive edge. Whether you're a seasoned coder or a newcomer, this article uncover the power of Dynamic Programming in Game Theory for Competitive Programming. Problem Identification of Dynamic Programming in Game Theory: In these type of problems you are basically given 2 players(first and second), set of operations, and a win condition. Given that both the players play optimally and they take turn one after the other we have to determine the winner of the game. Winning State and Loosing State: If a player is standing at a particular state 'S' and it can send its opponent to one of the K different states T1, T2...TK then: Condition for 'S' to be a winning State: If there exists atleast one state from T1 to TK which is a Loosing States. Condition for 'S' to be a Loosing State: All the states from T1 to TK are Winning States. Obviously, in order to win the player will try to send its opponent to any loosing state and if it is not possible to do so then that player itself looses the game. follow the below image for better understanding. Lets's take a look over some problems to understand this concept more thoroughly: Problem 1: Two players (First and Second) are playing a game on a 2-D plane in which a Token has been place at coordinate (0,0). Players can perform two types of operations on the token: Increase the X coordinate by exactly K distance Increase the Y coordinate by exactly K distance You are given an integer D, In order to perform above operations the player must ensure that the token stays within Euclidean distance D from (0,0) i.e. after each operation, X2 + Y2 <= D2 where X and Y are the coordinates of the token. Given the value of K and D, determine the Winner of the Game assuming both the players play optimally and First player plays first. Example: Input: D=2, K=1 Output: Second Explanation: First player moves token from (0,0) to (0,1) Second player moves token from (0,1) to (0,2) Now, whatever move First player makes, the distance will exceed the Euclidean Distance Input: D=5, K=2 Output: Second Approach: Let DP[i][j] denote whether the coordinate (i,j) are a winning state or a loosing state for any player that is: DP[i][j] = 0 means a loosing state DP[i][j] = 1 means a winning state Recurrence Relation: On the basis of given operations we can either increment x coordinate by K or y coordinate by K, hence DP[i][j] depends on two values: DP[i+k][j] DP[i][j+k] Condition for (i, j) to be winning state i.e. DP[i][j] = 1 : DP[i+k][j] = 0 OR DP[i][j+k] = 0 (Why? read the definition of winning state) Condition for (i, j) to be loosing state i.e. DP[i][j] = 0: DP[i+k][j] = 1 AND DP[i][j+k] = 1 (Why? read the definition of Loosing state) Below is the implementation of the above approach: C++ ```` include define ll long long using namespace std; ll d, k; ll findWinner(ll i, ll j, vector >& dp) { // ans varible determines whether one of dp[i+k][j] or // dp[i][j+k] is 0 or not ll ans = 1; if (dp[i][j] != -1) return dp[i][j]; // x and y stores the possible eucledian distances from // current coordinates i and j ll x = (i + k) (i + k) + j j; ll y = (i) (i) + (j + k) (j + k); // if x is valid eucledian distance if (x <= d d) { ans = ans & (findWinner(i + k, j, dp)); } // if y is valid eucledian distance if (y <= d d) { ans = ans & (findWinner(i, j + k, dp)); } // ans=0 means current state is a winning state if (ans == 0) return dp[i][j] = 1; return dp[i][j] = 0; } // driver code int main() { d = 5; k = 2; vector<vector<int> > dp(d d + 1, vector<int>(d d + 1, -1)); int ans = findWinner(0, 0, dp); if (ans == 1) { cout << "First player wins"; } else cout << "Second player wins"; cout << endl; } ```` ``` include #include ``` ``` define ll long long #define ll long long ``` using namespace std; using namespace std ll d, k; ll d k ll findWinner(ll i, ll j, vector<vector<int> >& dp) ll findWinner ll i ll j vector< vector< int>>& dp { // ans varible determines whether one of dp[i+k][j] or // ans varible determines whether one of dp[i+k][j] or // dp[i][j+k] is 0 or not // dp[i][j+k] is 0 or not ll ans = 1; ll ans = 1 if (dp[i][j] != -1) if dp i j!= - 1 return dp[i][j]; return dp i j // x and y stores the possible eucledian distances from // x and y stores the possible eucledian distances from // current coordinates i and j // current coordinates i and j ll x = (i + k) (i + k) + j j; ll x = i + k i + k + j j ll y = (i) (i) + (j + k) (j + k); ll y = i i + j + k j + k // if x is valid eucledian distance // if x is valid eucledian distance if (x <= d d) {if x<= d d ans = ans & (findWinner(i + k, j, dp)); ans = ans& findWinner i + k j dp } // if y is valid eucledian distance // if y is valid eucledian distance if (y <= d d) {if y<= d d ans = ans & (findWinner(i, j + k, dp)); ans = ans& findWinner i j + k dp } // ans=0 means current state is a winning state // ans=0 means current state is a winning state if (ans == 0) if ans == 0 return dp[i][j] = 1; return dp i j = 1 return dp[i][j] = 0; return dp i j = 0 } // driver code // driver code int main() int main { d = 5; d = 5 k = 2; k = 2 vector<vector<int> > dp(d d + 1, vector< vector< int>> dp d d + 1 vector<int>(d d + 1, -1)); vector< int> d d + 1 - 1 int ans = findWinner(0, 0, dp); int ans = findWinner 0 0 dp if (ans == 1) {if ans == 1 cout << "First player wins"; cout<< "First player wins" } else else cout << "Second player wins"; cout<< "Second player wins" cout << endl; cout<< endl } Java ```` // Java Implementation : import java.util.Arrays; public class GameWinner { private static int d, k; public static int findWinner(int i, int j, int[][] dp) { int ans = 1; if (dp[i][j] != -1) { return dp[i][j]; } int x = (i + k) (i + k) + j j; int y = i i + (j + k) (j + k); if (x <= d d) { ans &= findWinner(i + k, j, dp); } if (y <= d d) { ans &= findWinner(i, j + k, dp); } if (ans == 0) { return dp[i][j] = 1; } return dp[i][j] = 0; } public static void main(String[] args) { d = 5; k = 2; int[][] dp = new int[d d + 1][d d + 1]; for (int[] row : dp) { Arrays.fill(row, -1); } int ans = findWinner(0, 0, dp); if (ans == 1) { System.out.println("First player wins"); } else { System.out.println("Second player wins"); } } } // This code is contributed by Sakshi ```` Python3 ```` Function to find the winner def findWinner(i, j, dp): # ans variable determines whether one of dp[i+k][j] or # dp[i][j+k] is 0 or not ans = 1 if dp[i][j] != -1: return dp[i][j] # x and y store the possible Euclidean distances from # current coordinates i and j x = (i + k) (i + k) + j j y = i i + (j + k) (j + k) # If x is a valid Euclidean distance if x <= d d: ans = ans & findWinner(i + k, j, dp) # If y is a valid Euclidean distance if y <= d d: ans = ans & findWinner(i, j + k, dp) # ans=0 means the current state is a winning state if ans == 0: dp[i][j] = 1 else: dp[i][j] = 0 return dp[i][j] Driver code d = 5 k = 2 dp = [[-1 for _ in range(d d + 1)] for _ in range(d d + 1)] ans = findWinner(0, 0, dp) if ans == 1: print("First player wins") else: print("Second player wins") ```` C# ```` // C# program for the above approach using System; public class GFG { static long d, k; static long FindWinner(long i, long j, long[, ] dp) { // ans variable determines whether one of dp[i+k,j] // or dp[i,j+k] is 0 or not long ans = 1; if (dp[i, j] != -1) return dp[i, j]; // x and y store the possible Euclidean distances // from current coordinates i and j long x = (i + k) (i + k) + j j; long y = i i + (j + k) (j + k); // if x is a valid Euclidean distance if (x <= d d) { ans = ans & (FindWinner(i + k, j, dp)); } // if y is a valid Euclidean distance if (y <= d d) { ans = ans & (FindWinner(i, j + k, dp)); } // ans=0 means the current state is a winning state if (ans == 0) return dp[i, j] = 1; return dp[i, j] = 0; } static void Main() { d = 5; k = 2; // Initialize a 2D array to store the intermediate // results long[, ] dp = new long[d d + 1, d d + 1]; // Initialize the array with -1 (indicating that the // result is not calculated yet) for (int i = 0; i <= d d; i++) { for (int j = 0; j <= d d; j++) { dp[i, j] = -1; } } // Call the function to find the winner and convert // the result to an integer int ans = Convert.ToInt32(FindWinner(0, 0, dp)); // Output the result based on the calculated winner if (ans == 1) { Console.WriteLine("First player wins"); } else { Console.WriteLine("Second player wins"); } } } // This code is contributed by Susobhan Akhuli ```` JavaScript ```` // Function to find the winner function findWinner(i, j, dp) { // ans varible determines whether one of dp[i+k][j] or // dp[i][j+k] is 0 or not let ans = 1; if (dp[i][j] !== -1) return dp[i][j]; // x and y stores the possible eucledian distances from // current coordinates i and j let x = (i + k) (i + k) + j j; let y = i i + (j + k) (j + k); // if x is valid eucledian distance if (x <= d d) ans = ans & findWinner(i + k, j, dp); // if y is valid eucledian distance if (y <= d d) ans = ans & findWinner(i, j + k, dp); // ans=0 means current state is a winning state if (ans === 0) return dp[i][j] = 1; return dp[i][j] = 0; } // driver code let d = 5; let k = 2; let dp = new Array(d d + 1).fill().map(() => new Array(d d + 1).fill(-1)); let ans = findWinner(0, 0, dp); if (ans === 1) { console.log("First player wins"); } else { console.log("Second player wins"); } ```` Output First player wins Problem 2: Two players (First and Second) are playing a game on array arr[] of size N. The players are building a sequence together, initially the sequence is empty. In one turn the player can perform either of the following operations: Remove the rightmost array element and append it to the sequence. Remove the leftmost array element and append it to the sequence. The rule is that the sequence must always be strictly increasing, the winner is the player that makes the last move. The task is to determine the winner. Example: Input: arr=[5, 4, 5] Output: First Player Wins Explanation: After the first player append 5 into the sequence the array would look like either [4,5] or [5,4] and second player won't be able to make any move. Input: arr=[5, 8, 2, 1, 10, 9] Output: Second Player Wins Explanation: For any element the first player append to the sequence, the second player can always append a strictly greater elements. Solution: Let DP[L][R] denote whether the subarray from L to R is a Loosing state or a Winning state for any player that is: DP[L][R] = 0 means L to R forms a Loosing subarray. DP[L][R] = 1 means L to R forms a Winning subarray. Recurrence Relation: As the operations allow a player to only remove from the rightmost end or the leftmost end therefore DP[L][R] will depend on two ranges based on two conditions: DP[L+1][R] if we can append Arr[L] into our strictly increasing sequence. DP[L][R-1] if we can append Arr[R] into our strictly increasing sequence. Condition for (L, R) to be winning state i.e. DP[L][R] = 1 : DP[L+1][R] =0 OR DP[L][R-1] = 0 (Why? read the definition of winning state) Condition for (L, R) to be loosing state i.e. DP[L][R] = 0: DP[L+1][R] =1 AND DP[L][R-1] = 1 (Why? read the definition of Loosing state) Below is the implementation of the above approach: C++ ```` include define ll long long using namespace std; // function to determine the winner ll findWinner(ll l, ll r, ll last, vector& arr, vector >& dp) { // if l>r means current state is a loosing state as we can // not make any move if (l > r) { return 0; } // x = leftmost element of current subarray l to r ll x = arr[l]; // y = rightmost element of current subarray l to r ll y = arr[r]; // ans variable is used to know whether either of the // next transitions are loosing state or not ll ans = 1; if (dp[l][r] != -1) return dp[l][r]; // if we can take the leftmost element in the sequence if (x > last) { ans = ans & (findWinner(l + 1, r, x, arr, dp)); } // if we can take rightmost element in the sequence if (y > last) { ans = ans & (findWinner(l, r - 1, y, arr, dp)); } // ans=0 means we found a loosing state in some next // transistions hence current state is winning if (ans == 0) return dp[l][r] = 1; else return dp[l][r] = 0; } // driver code int main() { vector arr = { 5, 8, 2, 1, 10, 9 }; ll n = arr.size(); // dp of size nn to store all the L to R ranges vector > dp(n, vector(n, -1)); // recursive call ll ans = findWinner(0, n - 1, INT_MIN, arr, dp); if (ans) { cout << "First player wins"; } else cout << "Second player wins"; } ```` ``` include #include ``` ``` define ll long long #define ll long long ``` using namespace std; using namespace std // function to determine the winner // function to determine the winner ll findWinner(ll l, ll r, ll last, vector<ll>& arr, ll findWinner ll l ll r ll last vector< ll>& arr vector<vector<ll> >& dp) vector< vector< ll>>& dp { // if l>r means current state is a loosing state as we can // if l>r means current state is a loosing state as we can // not make any move // not make any move if (l > r) {if l> r return 0; return 0 } // x = leftmost element of current subarray l to r // x = leftmost element of current subarray l to r ll x = arr[l]; ll x = arr l // y = rightmost element of current subarray l to r // y = rightmost element of current subarray l to r ll y = arr[r]; ll y = arr r // ans variable is used to know whether either of the // ans variable is used to know whether either of the // next transitions are loosing state or not // next transitions are loosing state or not ll ans = 1; ll ans = 1 if (dp[l][r] != -1) if dp l r!= - 1 return dp[l][r]; return dp l r // if we can take the leftmost element in the sequence // if we can take the leftmost element in the sequence if (x > last) {if x> last ans = ans & (findWinner(l + 1, r, x, arr, dp)); ans = ans& findWinner l + 1 r x arr dp } // if we can take rightmost element in the sequence // if we can take rightmost element in the sequence if (y > last) {if y> last ans = ans & (findWinner(l, r - 1, y, arr, dp)); ans = ans& findWinner l r - 1 y arr dp } // ans=0 means we found a loosing state in some next // ans=0 means we found a loosing state in some next // transistions hence current state is winning // transistions hence current state is winning if (ans == 0) if ans == 0 return dp[l][r] = 1; return dp l r = 1 else else return dp[l][r] = 0; return dp l r = 0 } // driver code // driver code int main() int main { vector<ll> arr = { 5, 8, 2, 1, 10, 9 }; vector< ll> arr = 5 8 2 1 10 9 ll n = arr.size(); ll n = arr size // dp of size nn to store all the L to R ranges // dp of size nn to store all the L to R ranges vector<vector<ll> > dp(n, vector<ll>(n, -1)); vector< vector< ll>> dp n vector< ll> n - 1 // recursive call // recursive call ll ans = findWinner(0, n - 1, INT_MIN, arr, dp); ll ans = findWinner 0 n - 1 INT_MIN arr dp if (ans) {if ans cout << "First player wins"; cout<< "First player wins" } else else cout << "Second player wins"; cout<< "Second player wins" } Java ```` // Java program for the above approach import java.util.Arrays; public class GFG { // Function to determine the winner static long findWinner(int l, int r, long last, long[] arr, long[][] dp) { // If l > r means the current state is a losing // state as we cannot make any move if (l > r) { return 0; } // x = leftmost element of the current subarray l to // r long x = arr[l]; // y = rightmost element of the current subarray l // to r long y = arr[r]; // ans variable is used to know whether either of // the next transitions are losing states or not long ans = 1; if (dp[l][r] != -1) return dp[l][r]; // If we can take the leftmost element in the // sequence if (x > last) { ans &= (findWinner(l + 1, r, x, arr, dp)); } // If we can take the rightmost element in the // sequence if (y > last) { ans &= (findWinner(l, r - 1, y, arr, dp)); } // ans = 0 means we found a losing state in some // next transitions, hence the current state is // winning if (ans == 0) return dp[l][r] = 1; else return dp[l][r] = 0; } // Driver code public static void main(String[] args) { long[] arr = { 5, 8, 2, 1, 10, 9 }; int n = arr.length; // DP array of size nn to store all the L to R // ranges long[][] dp = new long[n][n]; for (long[] row : dp) Arrays.fill(row, -1); // Recursive call long ans = findWinner(0, n - 1, Integer.MIN_VALUE, arr, dp); if (ans == 1) { System.out.println("First player wins"); } else { System.out.println("Second player wins"); } } } // This code is contributed by Susobhan Akhuli ```` Python3 ```` Python Implementation def find_winner(l, r, last, arr, dp): if l > r: return 0 x = arr[l] y = arr[r] ans = 1 if dp[l][r] != -1: return dp[l][r] if x > last: ans = ans & find_winner(l + 1, r, x, arr, dp) if y > last: ans = ans & find_winner(l, r - 1, y, arr, dp) if ans == 0: dp[l][r] = 1 else: dp[l][r] = 0 return dp[l][r] arr = [5, 8, 2, 1, 10, 9] n = len(arr) dp = [[-1] n for _ in range(n)] ans = find_winner(0, n - 1, float('-inf'), arr, dp) if ans: print("First player wins") else: print("Second player wins") This code is contributed by Tapesh(tapeshdua420) ```` C# ```` // C# program for the above approach using System; using System.Collections.Generic; public class GFG { static long FindWinner(long l, long r, long last, List arr, List > dp) { // If l > r means the current state is a losing // state as we cannot make any move if (l > r) { return 0; } // x = leftmost element of the current subarray l to // r long x = arr[(int)l]; // y = rightmost element of the current subarray l // to r long y = arr[(int)r]; // ans variable is used to know whether either of // the next transitions is a losing state or not long ans = 1; if (dp[(int)l][(int)r] != -1) return dp[(int)l][(int)r]; // If we can take the leftmost element in the // sequence if (x > last) { ans = ans & (FindWinner(l + 1, r, x, arr, dp)); } // If we can take the rightmost element in the // sequence if (y > last) { ans = ans & (FindWinner(l, r - 1, y, arr, dp)); } // ans = 0 means we found a losing state in some // next transitions hence the current state is // winning if (ans == 0) return dp[(int)l][(int)r] = 1; else return dp[(int)l][(int)r] = 0; } static void Main() { List<long> arr = new List<long>{ 5, 8, 2, 1, 10, 9 }; long n = arr.Count; // dp of size nn to store all the L to R ranges List<List<long> > dp = new List<List<long> >(); for (int i = 0; i < n; i++) { dp.Add(new List<long>(new long[n])); for (int j = 0; j < n; j++) { dp[i][j] = -1; } } // Recursive call long ans = FindWinner(0, n - 1, int.MinValue, arr, dp); if (ans == 1) { Console.WriteLine("First player wins"); } else { Console.WriteLine("Second player wins"); } } } // This code is contributed by Susobhan Akhuli ```` JavaScript ```` // function to determine the winner function findWinner(l, r, last, arr, dp) { // if l>r means current state is a loosing state as we can // not make any move if (l > r) { return 0; } // x = leftmost element of current subarray l to r let x = arr[l]; // y = rightmost element of current subarray l to r let y = arr[r]; // ans variable is used to know whether either of the // next transitions are loosing state or not let ans = 1; if (dp[l][r] !== -1) { return dp[l][r]; } // if we can take the leftmost element in the sequence if (x > last) { ans = ans & (findWinner(l + 1, r, x, arr, dp)); } // if we can take rightmost element in the sequence if (y > last) { ans = ans & (findWinner(l, r - 1, y, arr, dp)); } // ans=0 means we found a loosing state in some next // transitions hence current state is winning if (ans === 0) { return dp[l][r] = 1; } else { return dp[l][r] = 0; } } // driver code function main() { let arr = [5, 8, 2, 1, 10, 9]; let n = arr.length; // dp of size nn to store all the L to R ranges let dp = Array.from({ length: n }, () => Array(n).fill(-1)); // recursive call let ans = findWinner(0, n - 1, Number.MIN_SAFE_INTEGER, arr, dp); if (ans) { console.log("First player wins"); } else { console.log("Second player wins"); } } main(); ```` Output Second player wins Practice Problems on Dynamic Programming in Game Theory: \| Optimal Strategy for a Game \| \| Optimal Strategy for a Game \| Set 2 \| \| Optimal Strategy for a Game \| Set 3 \| \| Optimal Strategy for the Divisor game using Dynamic Programming \| \| Game of N stones where each player can remove 1, 3 or 4 \| \| Find the player who will win by choosing a number in range [1, K] with sum total N \| \| Find the winner of the game with N piles of boxes \| \| Coin game of two corners (Greedy Approach) \| V vaibhav_gfg Improve Article Tags : Competitive Programming Algorithms-Dynamic Programming Explore Complete CP Guide Competitive Programming - A Complete Guide 5 min read Basics DSA Tutorial - Learn Data Structures and Algorithms 6 min readMaths for DSA 15+ min readMathematical Algorithms 5 min read Bit manipulation Bit Manipulation for Competitive Programming 15+ min readBit Tricks for Competitive Programming 7 min readBitwise Hacks for Competitive Programming 14 min read DP for CP Dynamic Programming (DP) Introduction 15+ min readDynamic Programming or DP 3 min readDP on Trees for Competitive Programming 15+ min readDynamic Programming in Game Theory for Competitive Programming 15+ min read Advanced Graph Algorithms 3 min readSegment Tree 2 min readBinary Indexed Tree or Fenwick Tree 15 min readArray Range Queries 3 min read Improvement Suggest Changes Help us improve. 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15243	https://physics.nist.gov/PhysRefData/ASD/ionEnergy.html	Published Time: Thu, 07 Nov 2024 07:39:40 GMT NIST: Atomic Spectra Database - Ionization Energies Form NIST Atomic Spectra Database Ionization Energies Form Best viewed with the latest versions of Web browsers and JavaScript enabled This form provides access to NIST critically evaluated data on ground states and ionization energies of atoms and atomic ions. Spectra:e.g., Fe I or Na or H-Ds I or Mg+ or Al3+ or mg iv,vi-VIII;S V-xii or Fe ne-like-S-like or Ne-Fe I-III or Ni-like or H-like-Ne-like Units: Format output:- [x] No JavaScript Ordered by Z Ordered by sequence Output Data:- [x] Atomic number [x] Spectrum name [x] Ion charge [x] Element name [x] Isoelectronic sequence- [x] Ground-state electronic shells [x] Ground-state configuration [x] Ground-state level [x] Ionized configuration Ionization energy Total binding energy [x] Uncertainty Bibliographic references:- [x]
15244	https://mathoverflow.net/questions/18094/polynomial-with-the-primes-as-coefficients-irreducible	nt.number theory - Polynomial with the primes as coefficients irreducible? - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Polynomial with the primes as coefficients irreducible? Ask Question Asked 15 years, 6 months ago Modified14 years, 3 months ago Viewed 5k times This question shows research effort; it is useful and clear 52 Save this question. Show activity on this post. If p n p n is the n n'th prime, let A n(x)=x n+p 1 x n−1+⋯+p n−1 x+p n A n(x)=x n+p 1 x n−1+⋯+p n−1 x+p n. Is A n A n then irreducible in Z[x]Z[x] for any natural number n n? I checked the first couple of hundred cases using Maple, and unless I made an error in the code those were all irreducible. I have thought about this for a long time now, and asked many others, with no answer yet. polynomials nt.number-theory Share Share a link to this question Copy linkCC BY-SA 2.5 Cite Improve this question Follow Follow this question to receive notifications edited Mar 13, 2010 at 20:19 Yemon Choi 26.4k 9 9 gold badges 72 72 silver badges 163 163 bronze badges asked Mar 13, 2010 at 20:15 Tobias KildetoftTobias Kildetoft 2,518 2 2 gold badges 24 24 silver badges 31 31 bronze badges 11 6 You might want to entertain the possibility that this is a very hard problem. It might not be, but if it's not a "problem from a book", if it's just a "made-up" problem, then it might be hard. Reducibility of a polynomial is a very subtle fact about the coefficients, and we can't even prove if there are infinitely many primes of the form p+2, or n^2+1, or 2p+1, or...Kevin Buzzard –Kevin Buzzard 2010-03-13 21:16:01 +00:00 Commented Mar 13, 2010 at 21:16 3 Assuming one can rule out the rational root −p n−p n, I am trying to see the consequences of your coefficients being strictly positive and strictly increasing, ignoring primality. I just have this sense that reducible polynomial coefficients a j>0 a j>0 "ought to" either exhibit faster growth themselves, as in (x+B)n(x+B)n for large fixed B B, or have maximum value a j a j in the middle, not at either end. It worked for n=4 n=4, when I get access to Maple again I will experiment with this. Meanwhile, I like Ram Murty's article, through en.wikipedia.org/wiki/Cohn%27s_irreducibility_criterionWill Jagy –Will Jagy 2010-03-13 21:38:28 +00:00 Commented Mar 13, 2010 at 21:38 9 A reducible monic polynomial with prime constant coefficient has to split into monic factors whose constant coefficients are either \pm 1 or \pm p, and the latter case occurs exactly once. In particular, such a polynomial has to have at least one root on or inside the unit circle and at least one root outside. Can anyone rule this out?Qiaochu Yuan –Qiaochu Yuan 2010-03-13 23:27:40 +00:00 Commented Mar 13, 2010 at 23:27 7 @Kevin, Douglas: after multiplying this new polynomial by (x - 1), it's not hard to see by the triangle inequality that it can't have any roots inside or on the unit circle. Qiaochu Yuan –Qiaochu Yuan 2010-03-14 01:38:37 +00:00 Commented Mar 14, 2010 at 1:38 5 @Kevin: For your last question, it has been proven so for almost all integers n. It was a conjecture of Filaseta. Reference: "Classes of polynomials having only one non-cyclotomic irreducible factor", by Borisov, Filaseta, Lam, Trifonov.Gjergji Zaimi –Gjergji Zaimi 2010-03-14 01:47:55 +00:00 Commented Mar 14, 2010 at 1:47 \|Show 6 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 77 Save this answer. Show activity on this post. I will prove that A n A n is irreducible for all n n, but most of the credit goes to Qiaochu. We have (x−1)A n=b n+1 x n+1+b n x n+⋯+b 1 x−p n(x−1)A n=b n+1 x n+1+b n x n+⋯+b 1 x−p n for some positive integers b n+1,…,b 1 b n+1,…,b 1 summing to p n p n. If \|x\|≤1\|x\|≤1, then \|b n+1 x n+1+b n x n+⋯+b 1 x\|≤b n+1+⋯+b 1=p n\|b n+1 x n+1+b n x n+⋯+b 1 x\|≤b n+1+⋯+b 1=p n with equality if and only x=1 x=1, so the only zero of (x−1)A n(x−1)A n inside or on the unit circle is x=1 x=1. Moreover, A n(1)>0 A n(1)>0, so x=1 x=1 is not a zero of A n A n, so every zero of A n A n has absolute value greater than 1 1. If A n A n factors as B C B C, then B(0)C(0)=A n(0)=p n B(0)C(0)=A n(0)=p n, so either B(0)B(0) or C(0)C(0) is ±1±1. Suppose that it is B(0)B(0) that is ±1±1. On the other hand, ±B(0)±B(0) is the product of the zeros of B B, which are complex numbers of absolute value greater than 1 1, so it must be an empty product, i.e., deg B=0 deg⁡B=0. Thus the factorization is trivial. Hence A n A n is irreducible. Share Share a link to this answer Copy linkCC BY-SA 2.5 Cite Improve this answer Follow Follow this answer to receive notifications answered Mar 14, 2010 at 7:25 Bjorn PoonenBjorn Poonen 24k 7 7 gold badges 92 92 silver badges 110 110 bronze badges 6 7 Fantastic! So the statement is actually stronger, right? The coefficients can be any sequence of non-decreasing positive integers as long as the constant term is prime.Qiaochu Yuan –Qiaochu Yuan 2010-03-14 07:58:02 +00:00 Commented Mar 14, 2010 at 7:58 3 All you use is that the coefficients are strictly increasing and the constant term is prime, right? Will Jagy also pointed out that strictly increasing was a funny property for a reducible polynomial to have.Kevin Buzzard –Kevin Buzzard 2010-03-14 08:04:28 +00:00 Commented Mar 14, 2010 at 8:04 I think Qiaochu said it right, and they need only be non-decreasing rather than strictly increasing. (You just need the b_k's to be nonnegative.)Jonas Meyer –Jonas Meyer 2010-03-14 08:08:30 +00:00 Commented Mar 14, 2010 at 8:08 3 @Kevin: Yes, that's right. @Qiaochu and Jonas: Nondecreasing is not quite strong enough; consider x 3+x 2+2 x+2 x 3+x 2+2 x+2. But you can say that if f(x)f(x) is a monic irreducible polynomial with nondecreasing coefficients and f(0)f(0) is prime and f′(0)<f(0)f′(0)<f(0), then f(x)f(x) is irreducible.Bjorn Poonen –Bjorn Poonen 2010-03-14 08:28:17 +00:00 Commented Mar 14, 2010 at 8:28 Thank you for the correction. I see one of my oversights: if some of the b_k's can be zero, equality in the inequality doesn't necessarily imply that x=1, e.g. if the odd indexed coefficients are zero and n is odd, like in your example. It seems to generalize to the case where the set of k k such that b k≠0 b k≠0 has gcd 1. In particular, as you mentioned it works if b 1≠0 b 1≠0.Jonas Meyer –Jonas Meyer 2010-03-14 08:57:48 +00:00 Commented Mar 14, 2010 at 8:57 \|Show 1 more comment This answer is useful 14 Save this answer. Show activity on this post. Since I don't have enough "reputation" to comment on Bjorn's answer, I will write this in an answer. The remark about the location of the zeros of A n A n goes back at least to Kakeya, On the Limits of the Roots of an Algebraic Equation with Positive Coefficients, Tôhoku Math. J., vol. 2, 140-142, 1912. It also appears in the nice book by E. Landau, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, Springer 1916, (Hilfssatz p.20). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Jul 2, 2011 at 13:44 Francesco SicaFrancesco Sica 313 3 3 silver badges 6 6 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polynomials nt.number-theory See similar questions with these tags. 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15245	https://www.thermopedia.com/content/1053/?utm_source=TrendMD&utm_medium=cpc&utm_campaign=Thermopedia_TrendMD_0&_gl=161xc58_gaMTc5MjYwNDcyLjE2Nzg3NzE2MDI._ga_1L8Z33S10Z*MTY3ODc3MTYwMi4xLjEuMTY3ODc3MTY0Ny4wLjAuMA..	PRANDTL NUMBER Shires, G. L. DOI: 10.1615/AtoZ.p.prandtl_number Article added: 2 February 2011 Article last modified: 7 February 2011 Share article View in A-Z Index Number of views: 146672 Prandtl number, Pr, is a dimensionless parameter representing the ratio of diffusion of momentum to diffusion of heat in a fluid. where ν is kinematic viscosity and κ thermal diffusivity. Prandtl number is a characteristic of the fluid only. For air at room temperature Pr is 0.71 and most common gases have similar values. The Prandtl number of water at 17°C is 7.56. Liquids in general have high Prandtl numbers, with values as high as 105 for some oils. REFERENCES Hewitt, G. F., Shires, G. L., and Bott, T. R. (1994) Process Heat Transfer, CRC Press, Boca Raton, FL. References Hewitt, G. F., Shires, G. L., and Bott, T. R. (1994) Process Heat Transfer, CRC Press, Boca Raton, FL. Back to top © Copyright 2008-2025 Related content in other products Begell House Inc. Annual Review of Heat Transfer AIHTC ICHMT
15246	https://mathematicalmysteries.org/k-12-commutative-associative-distributive-identity-and-inverse-laws/	Skip to content Search K-12 – Commutative, Associative, Distributive, Identity and Inverse Laws Contents Commutative Law Commutative Property of Addition Commutative Property of Multiplication Associative Law Associative Property of Addition Associative Property of Multiplication Distributive Law Distributive Property of Multiplication Identity Law Identity Property of Addition Identity Property of Multiplication Important Notes on Additive Identity and Multiplicative Identity Inverse Law Inverse Property of Addition Inverse Property of Multiplication Summary References Additional Reading The big four math operations — addition, subtraction, multiplication, and division — let you combine numbers and perform calculations. Certain operations possess properties that enable you to manipulate the numbers in the problem, which comes in handy, especially when you get into higher math like algebra. The important properties you need to know are the commutative property, the associative property, and the distributive property. Understanding an identity and inverse operation is is also helpful. \| Law / Property \| Addition \| Multiplication \| --- \| Commutative \| a + b = b + a \| a ⋅ b = b ⋅ a \| \| Associative \| (a + b) + c = a + (b + c) \| (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) \| \| Distributive \| \| a(b + c) = ab + ac \| \| Identity \| a + 0 = a \| a ⋅ 1 = a \| \| Inverse \| a + (-a) = 0 \| a ⋅ (1/a) = 1 \| Properties of Real Numbers Commutative Law Commutative Property of Addition The commutative property of addition states that when two numbers are being added, their order can be changed without affecting the sum. For any real numbers a and b, a + b = b + a. Subtraction is not commutative. Commutative Property of Multiplication The commutative property of multiplication states that when two numbers are being multiplied, their order can be changed without affecting the product. For any real numbers a and b, a ⋅ b = b ⋅ a. Just as subtraction is not commutative, neither is division commutative. Associative Law Associative Property of Addition The associative property of addition states that numbers in an addition expression can be grouped in different ways without changing the sum. You can remember the meaning of the associative property by remembering that when you associate with family members, friends, and co-workers, you end up forming groups with them. For any real numbers a, b, and c, (a + b) + c = a + (b + c). Associative Property of Multiplication Multiplication has an associative property that works exactly the same as the one for addition. The associative property of multiplication states that numbers in a multiplication expression can be regrouped using parentheses. For any real numbers a, b, and c, (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c). Distributive Law Distributive Property of Multiplication The product of a sum (or a difference) and a is the same as the sum (or difference) of the product of each addend (or each being subtracted) and the number. For any real numbers a, b, and c, a(b + c) = ab + ac. Identity Law Additive Identity and Multiplicative Identityare the two basic algebraic identities widely used in mathematics. Additive identity refers to a unique number which when added to any other number, it returns the same number. Multiplicative identity refers to the number which when multiplied by any number, gives back the original number. Identity Property of Addition The identity property of addition is a property of real numbers that states that the sum of 0 and any number is equal to that number. For any real number a, a + 0 = a. Additive identity,denoted as 0, is a unique element in arithmetic operations that, when added to any number, leaves the number unchanged. Identity Property of Multiplication The identity property of multiplication is a property of real numbers that states that a number retains the same value when multiplied by 1. For any real number a, a ⋅ 1 = a. Multiplicative identity, denoted as 1, is a special element in arithmetic operations such that when any number is multiplied by 1, the result is the number itself. Formally, for any real number a, a × 1 = a. Important Notes on Additive Identity and Multiplicative Identity Following are few important points related to additive and multiplicative identity: 0 is the additive identity. 1 is the multiplicative identity. For any real number a, a + 0 = a and a × 1 = a. -1 is not a multiplicative identity. Inverse Law Inverse Property of Addition The inverse property of addition is a property of real numbers that states that the sum of a number and its negative (the “additive inverse”) is always zero. For any real number a, a ⋅ (-a) = 0. Inverse Property of Multiplication The inverse property of multiplication is a property of real numbers that states that the multiplication of a number and its reciprocal (the “multiplicative inverse”) is always one. For any real number a, a ⋅ (1/a) = 1. Summary The commutative, associative, and distributive properties help you rewrite a complicated algebraic expression into one that is easier to deal with. When you rewrite an expression by a commutative property, you change the order of the numbers being added or multiplied. When you rewrite an expression using an associative property, you group a different pair of numbers together using parentheses. You can use the commutative and associative properties to regroup and reorder any number in an expression as long as the expression is made up entirely of addends or factors (and not a combination of them). The distributive property can be used to rewrite expressions for a variety of purposes. When you are multiplying a number by a sum, you can add and then multiply. You can also multiply each addend first and then add the products together. The same principle applies if you are multiplying a number by a difference. References “9.3.1: Associative, Commutative, and Distributive Properties.” 2021. Mathematics LibreTexts. September 5. “Additive Identity Vs Multiplicative Identity: Difference, Properties and Examples.” 2024. GeeksforGeeks. GeeksforGeeks. May 24. Additional Reading Remick, Karen. 2022. “A(B+C) = AB + AC.” Medium. Math Simplified. May 23. When I first was shown A(B+C) = AB + AC (the distributive property) , I was like “OK, that’s obvious. Why are they bothering to show us this? What use is it?” I thought it was yet more of the useless bits of information taught to us to use in mathematical proofs. Little did I know how critical it is to doing math in your head. Stapel, Elizabeth. 2023. “How to Tell the Basic Number Properties Apart.” Purplemath. Accessed October 27. ⭐ “What Are the Number Properties? (Commutative, Distributive, Associative & Identity) – BYJUS.” 2022. BYJU’S. July 24. Number properties lay down some rules that we can follow while performing mathematical operations. There are four number properties: commutative property, associative property, distributive property and identity property. Number properties are only associated with algebraic operations that are addition, subtraction, multiplication and division. However, some of these properties are not applicable to subtraction and division operations. ⭐ I suggest that you read the entire reference. Other references can be read in their entirety but I leave that up to you. Medium Member Only The featured image on this page is from the YouTube video Commutative, Associative & Distributive Law. Subscribe Subscribed Mathematical Mysteries Already have a WordPress.com account? Log in now. Mathematical Mysteries Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar
15247	https://eaglepubs.erau.edu/introductiontoaerospaceflightvehicles/chapter/worked-examples-aircraft-rockets/	Skip to content 77 Worked Examples: Airplane, Rocket, & Spacecraft Performance These worked examples have been fielded as homework problems or exam questions. Worked Example #1 A propeller-driven airplane has a fuel flow to the engines of the form where is the weight of fuel burned per unit time, is the true airspeed, is the brake specific fuel consumption. The values of and depend on the characteristics of the airplane, its weight, and the density altitude at which it is flying, i.e., Based on this fuel flow equation, show that the airspeed for the best endurance for the airplane will be obtained when the true airspeed is and the corresponding airspeed for the best range for the airplane will be obtained when the true airspeed is Explain the operational significance of flying at and , and give an example of a flight profile (or a part of one) when such airspeeds might be specifically used. The first term in the fuel flow equation is the contribution from the profile/parasitic (non-lifting) part of the drag, which grows with the cube of the airspeed. The second part is the contribution from the induced drag, which is inversely proportional to airspeed. Starting from then the lowest fuel burn (hence maximum flight endurance) can be determined by finding when is a minimum. Differentiating the fuel flow equation with respect to gives which is zero for a minimum, i.e., and so The best range is obtained when the ratio is a minimum. In this case so that which is zero for a minimum, i.e., and so There are various missions where flying at the optimal speed for the best range or endurance is crucial. For example, an airplane’s long-range ferry mission (such as over water) may require the airplane to be flown at or near its optimal airspeed for maximum range, even though this airspeed is typically lower than the airplane’s best cruise speed. Likewise, flying at an airspeed for best endurance might be necessary for a reconnaissance mission where the airplane must remain at or near a particular location for an extended period. Worked Example #2 Starting from the steady, level-flight flight assumption of and , and assuming the engine thrust specific fuel consumption TSFC () is constant, then show that for a jet-propelled airplane flying at a constant altitude that the fuel flow to the engines in terms of weight of fuel burned per unit time can be expressed in the form where you should evaluate and in terms of the the wing area, , wing aspect ratio, , flight weight, , parasitic drag coefficient & , Oswald’s efficiency factor, , engine TSFC, . The standard formula gives the lift on the airplane. which is equal to the weight . The drag on the airplane is where the drag coefficient is given by i.e., the sum of non-lifting and lifting (induced) parts. is the wing’s aspect ratio, and is Oswald’s efficiency factor. Using the equation for lift (as well as the assumption that ), the lift coefficient is Also, because the thrust equals drag, then and after rearrangement, then For a jet engine, the fuel burn rate will be the product of the thrust and the thrust-specific fuel consumption, i.e., Substituting for gives or which is of the form where and Both constants are based on the stated assumptions of all the values involved for a constant weight and altitude, i.e., = constant. Worked Example #3 Using the fuel flow equation from the previous worked example, i.e., show that the airspeed for the best endurance for a jet airplane will be obtained when In this problem, we are asked to start from The lowest fuel burn (hence maximum flight endurance) can be determined by differentiating the fuel flow equation with respect to , i.e., which is zero for a minimum, i.e., and so where and Therefore, is the airspeed at which the airplane should fly to obtain the minimum fuel flow. This speed depends on the airplane’s weight and altitude. Worked Example #4 An ERAU Cessna 172 Skyhawk airplane is flying along at an estimated in-flight weight of 2,100 lb and an airspeed of 120 knots at 2,000 ft, where the air density is 0.00216 slugs/ft. The student pilot is flying solo on a long cross-country flight. When flying over Jacksonville airport on the way back to ERAU (125 nautical miles left to fly), the pilot notices from the fuel gauge that only 8 gallons of usable AVGAS fuel remain in the tanks. Assume that the airplane’s weight for this analysis is the initial in-flight weight minus half the remaining fuel weight. Assume also that AVGAS fuel weighs 6.0 lb per gallon. The engineering characteristics of the airplane are given below: Wing span, = 36 ft Wing area, = 174 ft Non-lifting drag coefficient, = 0.02 Average propeller efficiency, = 0.85 Oswald’s efficiency factor, = 0.81 Engine BSFC, = 0.45 lb hp hr Calculate the following: (a) The operating lift coefficient of the wing, . (b) The induced drag coefficient, and the total drag coefficient, . (c) The lift-to-drag ratio of the airplane. (d) The propeller thrust and engine power (in hp) required for the airplane to fly. (e) The fuel flow in gallons per hour. Will the pilot be able to return to ERAU using the remaining fuel? We are instructed to assume that the airplane’s weight for this analysis is the initial in-flight weight, denoted as , minus half the remaining fuel weight. We have only 8 gallons of AVGAS fuel, and we are told to assume AVGAS weighs 6.0 lb per gallon, so = 48 lb of fuel. Therefore, the weight to perform the analysis is An airspeed of 120 kts is equivalent to 120 1.688 = 202.54 ft/s = . (a) The operating lift coefficient of the wing, , is (b) To find the induced drag coefficient, we need the aspect ratio of the wing, , i.e., The induced drag coefficient will be The total drag coefficient, , is (c) The lift-to-drag ratio of the airplane at the given conditions of flight is (d) The thrust from the propeller required for the airplane to fly is The power required for flight will be where the conversion factor 550 has been used to convert the base USC units to standard units of horsepower (hp). (e) The fuel flow now follows directly, i.e., so the airplane is using about 6 gallons per hour. (f) The time to burn off all of the relatively small amount of available fuel is So, the potential approximate range of the airplane is In conclusion, the airplane has enough fuel to return to ERAU. But, in practice, the FAA requires a minimum fuel reserve of 30 minutes of flight time (visual rules) and 45 minutes of flight time (instrument rules), so legally, the student pilot is probably going to have to land and pick up more fuel before completing the final leg back to ERAU. Worked Example #5 Consider a small jet-powered airplane with an initial in-flight weight of 21,000 lb flying in unaccelerated level flight at an airspeed of 250 knots where the air density is 0.0015 slugs/ft. The airplane weighs 850 lb of usable fuel remaining in its tanks. The airplane’s wingspan is 48.0 ft, and the wing panels have a trapezoidal shape with a root chord = 9 ft and a tip chord = 4 ft. The other characteristics of the airplane are: Non-lifting drag coefficient, = 0.02; Oswald’s efficiency factor, = 0.81; Engine TSFC = = 0.5 lb lb-1 hr-1. Assume for the following analysis that the airplane’s weight from fuel burning is the initial weight minus half the remaining fuel weight. Calculate the following: (a) The operating lift coefficient of the wing, . (b) The induced drag coefficient, and the total drag coefficient, . (c) The lift-to-drag ratio of the airplane in the given flight conditions. (d) The thrust required from the engines for the airplane to fly. (e) The approximate maximum potential remaining flight range of the airplane. (f) The approximate flight time to reach the remaining flight range. We need initial information, including the wing area and aspect ratio, as well as the weight at which to run the calculations. Calculating the area of the wing gives The aspect ratio can be calculated using We are told to use the weight of the airplane at a point which is its initial weight minus half the remaining fuel weight, so (a) An airspeed of 250 kts is equal to 422 ft/s. The operating lift coefficient of the wing, , is (b) The total drag coefficient, , is (c) The lift-to-drag ratio of the airplane at the given conditions of flight is (d) The thrust required for the airplane to fly is (e) The time to burn off all of the relatively small amount of available fuel is So, the potential approximate range of the airplane is Worked Example #6 Consider an airplane with a wing of lifting planform area 60 m, an aspect ratio 12, and Oswald’s efficiency factor 0.90. The wing has a non-lifting profile drag coefficient of 0.01. The remainder of the airplane has a non-lifting drag coefficient of 0.03. All force coefficients are based on wing area . The mass of the airplane is 16,000 kg. If the airplane is flying at a density altitude of 10,000 ft and its true airspeed is 253 kts, then calculate (a) The lift force produced by the wing; (b) The lift coefficient of the wing; (c) The drag force on the wing; (d) The lift-to-drag ratio of the wing; (e) The total drag force on the airplane; (f) The lift-to-drag ratio of the airplane. (a) With the assumption that the airplane is flying along in steady, unaccelerated flight, the lift force produced by the wing will equal the weight of the airplane, i.e., (b) The lift coefficient of the wing is given by To find , we need the density of the air (in this case, at 10,000 ft) and the true airspeed in units of m/s. The density can be found from the ISA (assuming standard temperature), so and converting from nautical miles per hour (kts) to m/s gives Inserting the numbers gives (c) The drag force on the wing will be given by where is the drag coefficient of the wing, which will be given by where the second part is the induced drag (i.e., drag due to lift). Inserting the numbers gives Therefore, the drag force on the wing is (d) Now that the lift and drag on the wing are known, the lift-to-drag ratio of the wing is (e) The total drag force on the airplane is where is the net drag coefficient of the airplane, which will be given by where we are given that for the remainder of the airplane, then (the non-lifting part), and the second part will be from the wing (which has already been calculated). Notice that all drag coefficients are defined using wing area as a reference. For the entire airplane, then Therefore, the drag force on the airplane is (f) Now the lift and drag are known then, the lift-to-drag ratio of the entire airplane is Worked Example #7 The goal is to estimate the endurance and range of a general aviation airplane similar to a Cessna 182 Skylane. The parameters describing this propeller-driven airplane are listed in the table below. \| \| \| \| --- \| Wing span \| \| 35.8 ft \| \| Wing area \| \| 174 ft \| \| Wing aspect ratio \| \| 7.37 \| \| Gross takeoff weight \| \| 2,950 lb \| \| Fuel capacity (tankage) \| \| 65 U.S. gals AVGAS \| \| Engine rated power \| \| 230 hp @ MSL ISA \| \| Engine BSFC \| \| 0.45 lb hp hr \| \| Parasitic drag coefficient \| \| 0.025 \| \| Oswald’s efficiency factor \| \| 0.8 \| \| Average propeller efficiency \| \| 0.8 \| Based on the provided information, several key performance characteristics of the airplane can be determined, including the power requirements for flight. It can be assumed that the weight of the airplane is the gross takeoff weight, so and so Rearranging for the lift coefficient gives Also, the drag coefficient for the airplane (profile drag plus induced drag) is and so the total drag is For level flight, the brake power required is remembering to factor in the propeller efficiency; the power curve is shown in the figure below for a pressure altitude of 5,000 ft, where = 0.002048. Finally, assuming a constant BSFC gives the fuel flow rate as which is shown graphically in the figure below. The airplane’s endurance and range can be estimated using the appropriate Breguet formulas. This airplane’s fuel capacity is 65 gallons of AVGAS, and at 6.01 lb per gallon, the maximum fuel weight that could be carried would be 390.65 lb. This is about 13% of the gross weight, so the assumption of constant weight over the flight time is reasonable for evaluating the maximum values of and . While these values and their corresponding airspeeds will change with weight and altitude, a representative weight for analysis is the in-flight weight, which equals the Gross Takeoff Weight (GTOW) minus half of the total fuel weight. Not all of this fuel would be usable, however, because at least some quantity of fuel would be needed for startup, taxi, takeoff, and climb, as well as for the descent and landing, plus an allowance of 30 minutes of VFR flight or 45 minutes of IFR flight must be included to comply with FAA regulations. If an average fuel burn allowance of 120 lb is assumed, which seems reasonable for takeoff and landing, this leaves 270 lb of fuel for actual flight use. So, for the endurance and range estimates in this case, then = 2,950-60 = 2,890 lb and = 2,890 – 270 = 2,620 lb. The estimated maximum endurance can be found using From the graph below, the maximum value of is 12.79 at 81 mph or 118.8 ft/s. Notice that the BSFC value must be converted into appropriate engineering units so = 0.45/550/3600 = 2.27 in units of (lb) (lb-ft s) (s). Substituting the appropriate values gives So, if we loiter, the airplane at or around an airspeed of 81 mph would remain airborne for nearly 10 hours. The estimated maximum range can be found using The maximum value of is 13.6 at 106 mph or 155 ft/s. Substituting the appropriate values gives which would only be obtained if this airplane were to be flown at or around 106 mph. Of course, this is rather slow compared to the airplane’s top speed, which is around 170 mph; however, at that airspeed, the range would be reduced by nearly half. Remember that these latter results are only estimates of the maximum range and endurance, but they are typically within 10% of the actual values that can be demonstrated in flight. Worked Example #8 It is desired to estimate the endurance and range of a jet-powered airplane in the form of a small jet similar to the Cessna Citation. The parameters describing this airplane are listed in the table below. \| \| \| \| --- \| Wing span \| \| 53.3 ft \| \| Wing area \| \| 318 ft \| \| Wing aspect ratio \| \| 8.93 \| \| Gross takeoff weight \| \| 19,815 lb \| \| Fuel capacity (tankage) \| \| 1,119 U.S. gals JET-A \| \| Engine rated thrust ( 2) \| \| 3,650 lb @ MSL ISA \| \| Engine TSFC \| \| 0.6 lb lb hr \| \| Parasitic drag coefficient \| \| 0.02 \| \| Oswald’s efficiency factor \| \| 0.81 \| We can assume for the following calculations that the airplane’s weight is the gross takeoff weight, so , although this assumption will tend to overpredict the fuel burn. Alternatively, one can use the less half the total fuel weight, which is usually considered the preferred approach. The drag coefficient for the airplane is assuming no transonic wave drag. For level flight, then, the total thrust required is where Substituting the known values for our exemplar jet airplane gives the thrust required curve shown below for a pressure altitude of 15,000 ft, where = 0.0014963. Because this is a twin-engine airplane, each engine would produce half of this required thrust. Assuming a constant TSFC gives the total fuel flow rate as which is shown in the figure below. Notice that the fuel flow mimics the thrust requirements. For flight at higher airspeeds, the fuel flow increases rapidly, reflecting that for any airplane, there is a high fuel cost associated with flying fast. We can estimate the endurance and range of this jet airplane using the appropriate Breguet formulas. We should always use the proper formulas for jet airplanes and distinguish them from propeller airplanes. The fuel capacity of this jet airplane is 1,119 gallons of JET-A, and at 6.8 pounds per gallon, the maximum fuel weight would be 7,609.2 pounds, which is approximately 38% of the airplane’s gross weight. This is why a more accurate estimate of endurance and range would be obtained by using minus half the total fuel weight. But we will continue here with . Not all this fuel would be usable, and allowances are needed for previously discussed reasons. We can proceed by assuming that = 19,815 -1,000 (the 1,000 lb being the allowance) = 18,815 lb and = 18,815 – 5,600 = 13,215 lb. The endurance is evaluated by using According to the results provided below, the airplane’s best ratio of 16.85 occurs at an airspeed of 208 kts (351 ft/s). The needed ratios can also be calculated because we have formulas for calculating both and . Substituting the numbers and remembering to convert into units of s by dividing by 3,600 gives where the final result has been converted back from seconds to hours. The range is evaluated by using The best ratio of 23.4 occurs at an airspeed of 274 kts or 462 ft/s. Substituting the actual numbers gives which seems fairly reasonable for this class of jet-powered airplanes. Of course, these results would depend on the payload, which may need to be traded for fuel. Note: For all aircraft, a maximum useful load can be carried, which is the sum of the payload and the fuel load. More payload (e.g., passengers) may, and usually does, allow for less fuel load. Worked Example #9 A small jet airplane weighs 10,000 lb, has a wing area of 200 ft, and a drag polar given by . Because of a fuel leak, the airplane runs out of fuel, and the engines stop at an altitude = 20,000 ft. Estimate the best glide range from this altitude. The drag polar is given by Therefore, Differentiating with respect to gives which will be zero for a maximum or minimum, i.e., or The drag coefficient at this lift coefficient is Therefore, the lift-to-drag ratio is The gliding distance is given by Worked Example #10 – Flight performance of a drone A reconnaissance drone aircraft, as shown in the figure below, is cruising in trim in steady flight at an in-flight weight of 3,100 lb at a Mach number of 0.30 at an altitude of 10,000 ft. The aircraft’s drag polar is given by: The wing has an area of 122.5 ft² and an aspect ratio of 19.6. Assume the following ambient atmospheric conditions: and . Explain the balance of forces and moments acting on the airplane. Determine the operating lift coefficient of the wing. Determine the induced drag and total drag coefficients. Calculate Oswald’s efficiency factor. Calculate the total drag force and the lift-to-drag ratio. Calculate the brake power required for flight if the propeller efficiency is 0.8. If the BSFC of its piston engine is , approximately how much fuel will be burned in 15 minutes of flying time? Assume that the net change in the aircraft’s weight from burning fuel is negligible. The figure below explains the balance of forces, where lift equals weight and thrust equals drag. All the moments are balanced in trim, so the net moments about all axes are zero. The lift equation is given by: Because , the operating lift coefficient is The density of air is The airspeed is Substituting all the values gives the lift coefficient as The induced drag coefficient is In this case , so The total drag coefficient is The induced drag factor is Rearranging for gives The total drag force is given by Substituting the values gives Therefore, the lift-to-drag ratio is The brake power (in hp) required for flight is Substituting values gives The fuel flow rate is Substituting gives For 15 minutes (), then Worked Example #11 The Nemeth Parasol was an early aircraft designed for short takeoff and landing capabilities. It used a circular wing, which the inventors claimed would operate at very low airspeeds and act as a parachute during landing. In early flight demonstrations, it was claimed that the aircraft came down “almost vertically with a gentle landing.” The diameter of the circular wing was 15 ft, and the aircraft had a total weight of 1,000 lbs. Calculate the aspect ratio of the circular wing and discuss the implications on induced drag. Why do aircraft typically avoid using wings with such low aspect ratios? Assuming the aircraft stalls at a maximum lift coefficient of = 1.8, estimate the stall speed of the aircraft at sea level standard conditions. Is this value reasonable? Calculate the terminal velocity in feet per second at which the aircraft will “parachute” to a landing, assuming purely vertical motion. Hint: The drag coefficient of a circular disk with the flow perpendicular to the disk is 1.4. Compare your answers from parts (2) and (3). Would the transition from horizontal forward flight to vertical descent happen smoothly or abruptly? What flight characteristics would help ensure a “gentle” landing? You may decide on the outcome from part (c) that a more comfortable vertical landing speed of 12 ft/s would be appropriate. What would the diameter of the circular wing need to be for this scenario? What do you think of the inventor’s original claims? The aspect ratio, , of a wing is defined as where is the wing span and is the planform area. For a circular wing of diameter = 15 ft, the span and the area is so that A low aspect ratio leads to high induced drag because the wing generates strong trailing vortices and inefficient lift distribution. Aircraft typically avoid low aspect ratios to minimize induced drag and achieve improved aerodynamic efficiency, especially during cruise. 2. The stall speed, , can be found from the lift equation at maximum lift, i.e., Solving for gives The air density at sea level standard conditions is = 0.002378 slug/ft, so that This stall speed is very low and reasonable for an aircraft designed for short takeoff and landing. 3. In a vertical descent, the weight of the aircraft is balanced by drag, i.e., Solving for gives where = 1.4. Therefore, 4. From parts (2) and (3), 51.4 ft/s and 58.3 ft/s. The terminal velocity is slightly higher than the stall speed, suggesting that once the aircraft stalls, it would transition into a somewhat faster vertical descent. The transition would likely be abrupt without careful control. 5. Given a desired terminal velocity = 12 ft/s, solve for required circular wing area, , using Substituting values gives The corresponding diameter is found from and substituting values gives Therefore, the diameter of the circular wing would need to be approximately 73 ft to achieve a vertical landing speed of 12 ft/s. 6. The inventor’s claims are partially credible. The aircraft likely exhibited low forward stall speeds and significant drag during descent, leading to relatively low and survivable landing speeds. However, the transition to vertical descent might not have been perfectly smooth without careful pilot control, and truly “parachuting” down almost vertically would require a much larger wing to slow the descent to safer speeds. Worked Example #12 – Propellant needed for a single-stage rocket A single-stage rocket must provide a speed of 6,000 m/s (6 km/s) to a payload mass, , of 12,000 kg. The vehicle’s structural mass coefficient, , is 0.06. The propellant used in the engine has a specific impulse, , of 325 s. What must be the rocket’s initial mass, , and its propellant mass, , to meet these requirements? The equivalent exhaust velocity is (1) The initial mass-to-burnout mass ratio, , is (2) and the payload ratio, , is (3) where is the structural mass coefficient. Therefore, the initial mass of the rocket, , is (4) and the burnout mass, , is (5) Finally, the propellant mass, , is (6) which is a fairly large rocket. Worked Example #13 A small rocket is launched vertically from the Earth’s surface with an initial mass of 1,000 kg. If the rocket engine has a thrust of 10 kN and a specific impulse of 250 s, calculate: (a) The rocket’s acceleration at liftoff. (b) The flight velocity of the missile after 30 seconds. (a) The acceleration at liftoff will be (b) The equivalent exhaust velocity is and the thrust from the engine is The mass flow rate is Therefore, using the rocket equation, the flight velocity after seconds will be and inserting the numerical values for = 30 seconds gives which is just below the speed of sound. Worked Example #14 Consider a small single-stage rocket with the following design characteristics: propellant mass = 7,200 kg; structural mass = 800 kg; payload mass = 50 kg. The specific impulse for this rocket is 275 s. The rocket blasts off from Earth and climbs vertically. The fuel burns steadily, and the burnout time is 60 seconds. The aerodynamic drag can be neglected, but the effects of gravity should be included. (a) What will be the value of the burnout velocity? (b) Find the thrust generated by the rocket. (c) What is the initial acceleration of the rocket? (d) What will the speed and acceleration of the rocket be 30 seconds into the flight? (a) The equivalent velocity from the rocket engine will be The burnout mass has no fuel left, so that this mass will be the sum of the structural mass and payload mass, i.e., The initial mass of the rocket has all the unburned fuel, so With the gravity loss term included, the burnout velocity of the rocket will be (b) The thrust generated by the rocket will be (c) With the consideration of gravity, the initial acceleration will be so about 3. (d) After 30 seconds from liftoff, the mass of the rocket will be and so the acceleration at 30 seconds will be so about 6. Of course, including aerodynamic drag on the rocket would reduce this value slightly. Worked Example #15 A single-stage rocket has a total mass of kg and a burnout mass of kg, including engines, structural shell, and payload. The rocket blasts off from Earth and climbs vertically, exhausting its propellant in 2 minutes and 20 seconds. The propellant burning occurs at a steady rate, and the specific impulse of the propulsion system is 240 seconds. (a) If air resistance and gravity are neglected, what will the rocket’s velocity be at burnout conditions? (b) What thrust does the rocket engine develop at liftoff? (c) What is the initial acceleration of the rocket if gravity is not neglected? (d) What is the acceleration of the rocket at 60 seconds into the flight? (a) We have that specific impulse is 240 seconds and the initial mass is 1.14 kg and the burnout mass is 1.11 kg. Therefore, the propellant mass will be The rocket equation (we are told to ignore the gravity loss term in this equation) gives us the change in the velocity of the vehicle , i.e., The equivalent exhaust velocity is given in terms of the specific impulse, i.e., The burnout velocity is if we do not include the gravity loss term, as stated. (b) We need to find the net mass flow rate to the engines. Assuming the rate of propellant consumption is constant, then the mass of the rocket varies over time as where is the burnout time and is the rate of burning propellant. The thrust produced, , will be (c) The initial acceleration will be (d) The mass of the rocket at 60 seconds into the flight will be So, the acceleration will be Worked Example #16 Use the rocket equation to determine its burnout velocity and maximum achievable height, assuming it was launched vertically. Neglect the aerodynamic drag forces. Solve for the burnout velocity and maximum altitude if the burnout time is 60 seconds. The specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. We are given that the specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. The rocket equation gives us the change in the velocity of the vehicle , i.e., where is the initial mass of the vehicle and is the final or burnout mass. If gravity is included (but no aerodynamic drag), then where is the burnout time, which is 60 seconds in this case. The equivalent exhaust velocity is given in terms of the specific impulse, i.e., The burnout mass is given by So now we have the burnout velocity, which is so if we do not include the gravity loss term. With the gravity loss term included, then So at the burnout is reduced to Assuming the propellant consumption rate is constant, then the mass of the rocket varies over time as The velocity then is The height achieved at the burnout time is then. which, after the application of mathematics, gives Inserting the values gives so The final additional coasting height of the rocket can then be determined by equating the kinetic energy of the rocket at its burnout time with its change in potential energy between that point and the maximum obtained height. Worked Example #17 A rocket must provide a speed of 6,000 m/s to a payload of 12,000 kg. The vehicle’s structural coefficient is 0.06. The propellants used in the engines have a specific impulse of 325 s. What must the initial mass and propellant mass be to meet these requirements? The equivalent exhaust velocity so and Therefore, the initial mass is and the burnout mass is so the propellant mass is Worked Example #18 Use the rocket equation to determine a rocket’s burnout velocity and maximum achievable height if its burnout time is 60 seconds. The specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. Assume the rocket is launched vertically. The specific impulse is 250 seconds, the initial mass is 12,700 kg, and the propellant mass is 8,610 kg. The rocket equation gives us the change in the velocity of the vehicle , i.e., where is the initial mass of the vehicle and is the final or burnout mass. If gravity is included (but no aerodynamic drag), then where is the burnout time, which is 60 seconds in this case. The equivalent exhaust velocity is given in terms of the specific impulse, i.e., The burnout mass is given by So now we have the burnout velocity, which is so if we do not include the gravity loss term. With the gravity loss term included, then So at the burnout is reduced to Assuming the propellant consumption rate is constant, then the mass of the rocket varies over time as The velocity is The height achieved at the burnout time is then. which, after some mathematics, gives Inserting the values gives so The final additional coasting height of the rocket can then be determined by equating the kinetic energy of the rocket at its burnout time with its change in potential energy between that point and the maximum obtained height. Worked Example #19 Consider a two-stage rocket with a payload mass = 60 kg. The first stage has two solid rocket boosters attached. First stage: propellant mass = 7,200 kg, structural mass = 800 kg, and mass flow rate = 80.0 kg s. For each booster, the propellant mass is 1,400 kg, the structural mass is 200 kg, and the burn time is 45 seconds. For the second stage: propellant mass = 5,400 kg, structural mass = 600 kg. The specific impulse for the first stage is 250 and 290s for the boosters. Find the extra burnout velocity of the rocket at first-stage separation when using boosters compared to when not using them. Here is a summary of the information provided: Propellant mass for stage 1, = 7,200 kg Propellant mass for stage 2, = 5,400 kg Propellant mass for each booster, = 1,400 kg Structural mass for stage 1, = 800 kg Structural mass for stage 2, = 600 kg Structural mass for each booster, = 200 kg Payload mass, = 60 kg Specific impulse for stage 1, = 250.0 s Specific impulse for boosters, = 290.0 s Burnout time for each booster, = 45 s Mass flow rate for stage 1, = 80 kg/s The initial mass, , of the rocket without the boosters is and inserting the values gives The burnout mass, , just before the first stage separation is which gives the burnout mass, , as Therefore, the increment for the first stage without the boosters is and inserting the values gives Therefore, for the first stage, the burnout velocity is The equivalent exhaust velocity for each booster, , is and the net thrust, , produced is The mass flow rate for each booster is so the net thrust (to augment the thrust of stage 1) will be remembering that there are two boosters. Notice that the boosters nearly double the thrust at launch. To obtain the burnout velocity at booster burnout at 45 seconds into the flight, we first need the mean equivalent exhaust velocity for the stage 1 core and the two boosters, which is given by where . The mass flow rates and the equivalent velocities of the stage 1 core and the boosters have already been calculated, so The burnout velocity of the rocket at booster burnout, , at = 45 seconds into the flight is given by In this case, the initial mass, , of the rocket with the two boosters is and inserting the values gives The burnout mass, , at booster burnout at 45 seconds is noting that the first stage core has burned, in this case, half of its propellant. Therefore, the burnout mass at booster burnout, , is Therefore, the increment for the first stage at the burnout time is and inserting the values gives Therefore, at booster burnout, the velocity of the rocket is The first stage then burns for another 45 seconds, over which it burns the other half of its propellant. After booster separation, the new initial weight is and the burnout mass just before stage 1 separation is then Therefore, the burnout mass is and so the for this part of the launch is Inserting the values gives Therefore, at stage 1 burnout, the velocity of the rocket is This means the rocket travels nearly twice the velocity it would have traveled without the boosters. Worked Example #20 – Space Shuttle launch Consider a Space Shuttle, which is a parallel rocket system launch vehicle. The Shuttle used LH2/LOX for the “core” main engines on the Orbiter, with two solid rocket boosters, or SRBs. It is required to estimate the final burnout velocity. Neglect the gravity loss term in the rocket equation. The information available includes the following: Orbiter: Structural mass = 110,000 kg Payload mass = 24,000 kg Specific impulse = 454 seconds Burnout time = 480 seconds External tank Structural mass = 30,000 kg Propellant mass = 720,000 kg SRB (for each): Structural mass = 86,000 kg Propellant mass = 500,000 kg Specific impulse = 269 seconds Burnout time = 124 seconds For the Orbiter or the “core,” the equivalent exhaust velocity, , is For the SRBs, the equivalent exhaust velocity, , is The fraction of propellant mass remaining in the core at booster burnout will be The mean equivalent exhaust velocity, , is and putting in the values gives The initial mass at the point of launch is and putting in the values gives The final mass of the vehicle at SRB burnout is and putting in the values gives Therefore, the at SRB burnout is The initial vehicle mass after SRB separation is and putting in the values gives The final mass at the depletion of the propellant in the external tank is The extra at complete burnout is Therefore, the final burnout velocity of the Orbiter is which is fairly close to the values quoted by NASA, considering that the gravitational has been neglected. Worked Example #21 – Change in kinetic energy of a spacecraft A spacecraft with a mass of 10,000 kg travels with a velocity of 5,000 m/s. If a rocket engine with a specific impulse of 250 seconds is turned on and produces a constant thrust of 50 kN for 100 seconds, then calculate (a) the final velocity of the spacecraft. (b) The change in kinetic energy of the spacecraft. (a) The final velocity of the spacecraft is given by (b) The change in kinetic energy of the spacecraft is given by The initial mass of the spacecraft is 10,000 kg, and the final mass can be determined based on the mass flow rate, i.e., giving the final mass as Therefore, the change in kinetic energy of the spacecraft is Worked Example #22 A single-stage rocket is launched vertically, carrying a payload of 200 kg. It uses 3,000 kg of propellant and has a structural mass of 400 kg. The engine has a specific impulse of 300s. Compute the rocket’s burnout velocity. A design proposal reduces the structural mass to 300 kg while keeping everything else the same. What is the new burnout velocity? Based on your answers, briefly discuss whether it is more effective to reduce dry mass by 100 kg or increase propellant mass by 100 kg while keeping the same structure. You may assume all else remains constant. The burnout velocity for a vertical launch, neglecting the gravity term on (we are not given the burnout time) and drag, can be found from the ideal rocket equation, i.e., where and and Using the rocket equation gives 2. With the structural mass reduced to 300 kg, the new masses are and so that Reducing the structural mass by 100 kg increased the burnout velocity from approximately 5,271 m/s to 5,725 m/s, a gain of about 454 m/s. 3. If instead the propellant mass were increased by 100 kg (to 3,100 kg), keeping the structural mass at 400 kg, the new masses would be and so that Increasing the propellant mass by 100 kg results in a increase from 5,271 m/s to 5,355 m/s, a gain of approximately 84 m/s. Therefore, reducing structural (dry) mass is much more effective than increasing propellant mass for improving burnout velocity.
15248	https://www.tiger-algebra.com/zh-CN/%E8%A7%A3%E5%86%B3%E6%96%B9%E6%A1%88/%E4%BA%8C%E6%AC%A1%E4%B8%8D%E7%AD%89%E5%BC%8F/x%5E2-4%3E%3D0/	版权 Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. 解答 - 使用二次公式解决二次不等式其他解决方法逐步解答 1. 确定二次不等式的系数 a，b和c 我们的不等式系数，即x2+0x−4≥0，是： a = 1 b = 0 c = -4 2. 将这些系数插入到二次公式中要找到二次方程的根，将其系数（a，b和c）插入到二次公式中: x=(-b±sqrt(b2-4ac))/(2a) a=1 b=0 c=−4 x=(-0±sqrt(02-41-4))/(21) 简化指数和平方根 x=(-0±sqrt(0-41-4))/(21) 从左到右进行任何乘法或除法操作： x=(-0±sqrt(0-4-4))/(21) x=(-0±sqrt(0--16))/(21) 按照从左到右的顺序，计算任何加法或者减法。 x=(-0±sqrt(0+16))/(21) x=(-0±sqrt(16))/(21) 从左到右进行任何乘法或除法操作： x=(-0±sqrt(16))/(2) 得到结果： x=(-0±sqrt(16))/2 3. 简化根号下的 (16) 通过找出其质因数来简化16： 16的质因数分解是24 写出素因数： 16=2·2·2·2 将素因数分成对并以指数形式重写它们： 2·2·2·2=22·22 使用规则(x2)=x进一步简化： 22·22=2·2 从左到右进行任何乘法或除法操作： 2·2=4 4. 解出 x的方程 x=(-0±4)/2 ±表示有两个可能的根。分离这两个方程： x1=(-0+4)/2 和 x2=(-0-4)/2 x1=(-0+4)/2 按照从左到右的顺序，计算任何加法或者减法。 x1=(-0+4)/2 x1=(4)/2 从左到右进行任何乘法或除法操作： x1=42 x1=2 x2=(-0-4)/2 按照从左到右的顺序，计算任何加法或者减法。 x2=(-0-4)/2 x2=(-4)/2 从左到右进行任何乘法或除法操作： x2=−42 x2=−2 5. 求得区间我们首先通过找出其抛物线来寻找二次不等式的区间。抛物线的根（即抛物线穿过x轴的点）是：-2, 2。既然 a 系数是正的 (a=1)，那么这是一个"正"的二次不等式，抛物线向上，像一个笑脸！若不等式符号是≤或≥，则区间包括根，我们使用实线。若不等式符号是<或>，则区间不包括根，我们使用虚线。 6. 选择正确的区间（解决方案）由于x2+0x−4≥0具有≥的不等号，我们寻找抛物线间隔位于x轴上方。解决方案：区间记号：我们做得怎么样？为什么学习这个二次方程表达了弧线的路径以及沿线的点，而二次不等式表达了这些弧线内外的区域和覆盖的范围。换句话说，如果二次方程告诉我们边界在哪里，那么二次不等式则帮助我们理解相对于该边界，我们应该关注哪些内容。更实际地说，二次不等式被用来创建强大软件的复杂算法，并追踪随时间变化的情况，例如杂货店的价格。术语和主题相关链接最新相关的解决方案版权 Ⓒ 2013-2025 tiger-algebra.com
15249	https://brainly.com/question/62684766	[FREE] In N_2O and NO_2 , the ratio of masses of oxygen that combines with a fixed mass of nitrogen is: 1. 1 : - brainly.com 4 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +26,4k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +19k Ace exams faster, with practice that adapts to you Practice Worksheets +5,9k Guided help for every grade, topic or textbook Complete See more / Chemistry Expert-Verified Expert-Verified In N 2O and N O 2, the ratio of masses of oxygen that combines with a fixed mass of nitrogen is: 1 : 2 2 : 3 1 : 4 1 : 8 1 See answer Explain with Learning Companion NEW Asked by JkgamingFTW4843 • 06/17/2025 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 854840 people 854K 0.0 0 Upload your school material for a more relevant answer To determine the ratio of the masses of oxygen that combines with a fixed mass of nitrogen in N₂O and NO₂, we need to use the concept from chemistry known as the Law of Multiple Proportions. Explanation N₂O (Nitrous oxide) consists of 2 nitrogen atoms and 1 oxygen atom. Molar mass of nitrogen (N) = 14 g/mol Molar mass of oxygen (O) = 16 g/mol Total molar mass of N₂O = 2(14) + 1(16) = 28 + 16 = 44 g/mol Oxygen's contribution = 16 g/mol of the 44 g/mol NO₂ (Nitrogen dioxide) consists of 1 nitrogen atom and 2 oxygen atoms. Total molar mass of NO₂ = 1(14) + 2(16) = 14 + 32 = 46 g/mol Oxygen's contribution = 32 g/mol of the 46 g/mol Ratio of the masses of oxygen: Fix the mass of nitrogen at 28 g (from 2 moles of nitrogen in N₂O): In N₂O, using 28 g of nitrogen will result in 16 g of oxygen. In NO₂ with the same 28 g of nitrogen: We would need twice the ratio of moles of N to keep nitrogen constant, so for 1 mole of NO₂, we consider NO₄ for comparison: 14 g of N combines with 32 g of O. So, 28 g of nitrogen will combine with 2 times 32 g of oxygen = 64 g. Now, the oxygen mass ratio can be calculated: Ratio of oxygen in N₂O and NO₂=64 16=4 1 Therefore, the ratio of masses of oxygen that combines with a fixed mass of nitrogen in N₂O and NO₂ is 1:4. Correct Option: 3. 1 : 4 Answered by OliviaLunaGracy •68.1K answers•854.8K people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) This answer helped 854840 people 854K 0.0 0 Upload your school material for a more relevant answer The ratio of the masses of oxygen that combines with a fixed mass of nitrogen in N₂O and NO₂ is determined to be 1:4. This is calculated by comparing the amounts of oxygen that react with the same mass of nitrogen in each compound. Thus, the correct answer is option 3. Explanation To determine the ratio of the masses of oxygen that combines with a fixed mass of nitrogen in the compounds N₂O (nitrous oxide) and NO₂ (nitrogen dioxide), we use the concept from the Law of Multiple Proportions. Step-by-step Explanation: Analyzing N₂O: N₂O consists of 2 nitrogen atoms and 1 oxygen atom. Molar mass of nitrogen (N) = 14 g/mol. Molar mass of oxygen (O) = 16 g/mol. Total molar mass of N₂O = 2(14) + 1(16) = 28 + 16 = 44 g/mol. The mass of oxygen in N₂O = 16 g (from 1 mole of O). Analyzing NO₂: NO₂ consists of 1 nitrogen atom and 2 oxygen atoms. Total molar mass of NO₂ = 1(14) + 2(16) = 14 + 32 = 46 g/mol. The mass of oxygen in NO₂ = 32 g (from 2 moles of O). Fixing the mass of nitrogen: Let’s fix the mass of nitrogen at 28 g, which corresponds to 2 moles of nitrogen in N₂O. In N₂O, 28 g of nitrogen uses 16 g of oxygen. In NO₂, 28 g of nitrogen (from 2 moles of nitrogen) would need twice as much oxygen: 1 mole (14 g) of N combines with 32 g of O for 2 moles (28 g of N), so we have: 2 x 32 g of O = 64 g of O combined. Calculating the ratio: The ratio of the oxygen mass in N₂O to NO₂ for the same nitrogen mass: Thus, the ratio is: Ratio of oxygen in N₂O and NO₂=64 16=4 1 Therefore, the ratio of the masses of oxygen that combines with a fixed mass of nitrogen in N₂O and NO₂ is 1:4. Correct Option: 3. 1 : 4 Examples & Evidence For example, if we consider 28 grams of nitrogen (which corresponds to 2 moles in N₂O), it combines with 16 grams of oxygen in N₂O. In NO₂, the same 28 grams of nitrogen combines with 64 grams of oxygen, demonstrating the 1:4 ratio. The calculations for molar mass and the ratios used are based on established chemical formulas and principles from the Law of Multiple Proportions, which states that when elements combine, they do so in fixed and simple whole-number ratios. Thanks 0 0.0 (0 votes) Advertisement JkgamingFTW4843 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer Calculate the ratio by mass of nitrogen that combines with a fixed amount of oxygen during the formation of NO and N₂O₃. a) 1:2 b) 2:3 c) 2:1 d) 3:2 Community Answer Show that for fixed mass of nitrogen the masses of oxygen in four oxides of Nitrogen : N₂O, NO, NO₂, N₂O₅ are in the ratio of small integers. Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. New questions in Chemistry Predict the products of the following reaction. If no reaction will occur, use the NO REACTION button. Be sure your chemical equation is balanced! C H 3(C H)2C H 3(g)+O 2(g)→ Predict the products of the following reaction: C H 3CC H(g)+O 2(g)→□ Predict the products of the following reaction. If no reaction will occur, use the NO REACTION button. Be sure your chemical equation is balanced! C H 3(C H 2)2O H(l)+O 2(g)→□ The Haber process can be used to produce ammonia (N H 3) from hydrogen gas (H 2) and nitrogen gas (N 2). The balanced equation for this process is shown below. 3 H 2+N 2→2 N H 3 The molar mass of N H 3 is 17.03 g/m o l. The molar mass of H 2 is 2.0158 g/m o l. In a particular reaction, 0.575 g of N H 3 forms. What is the mass, in grams, of H 2 that must have reacted, to the correct number of significant figures? A. 0.1 grams B. 0.102 grams C. 0.10209 grams D. 0.1021 grams What is the hybridization of the C atoms in C 2C l 4? 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15250	https://live.poshenloh.com/past-contests/amc8/2024	2024 AMC 8 Exam Problems Scroll down and press Start to try the exam! Or, go to theprintable PDF, answer key, or professional video solutions and written solutions curated by LIVE, by Po-Shen Loh. All of thereal AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA). Want to learn professionally through interactive video classes? Learn LIVE Time Left: 40:00 What is the ones digit of 222,222−22,222−2,222−222−22−2? 0 2 4 6 8 Answer: B Video solution: Click to load, then click again to play Written solution: We only need to consider the ones digits of each number (except for the first one so we avoid getting a negative answer): 22−2−2−2−2−2=12 which has a ones digit of 2. Thus, B is the correct answer. What is the value of this expression in decimal form? 1144+44110+110044 6.4 6.504 6.54 6.9 6.94 Answer: C Video solution: Click to load, then click again to play Written solution: We can simplify the fractions by taking out the common factor 11: 1144 simplifies to 4, 44110 simplifies to 410=25=2.5, and 110044 simplifies to 1004=0.04. Therefore, we have 4+2.5+0.04=6.54. Thus, C is the correct answer. Four squares of side length 4, 7, 9, and 10 units are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color, as shown in the figure. What is the area of the visible colored region in square units? 42 45 49 50 52 Answer: E Video solution: Click to load, then click again to play Written solution: The visible colored region includes the area of the square with side length 10 subtracted by the area of the square with side length 9 plus the area of the square with side length 7 minus the area of the square with side length 4. Hence, we get 100−81+49−16=52. Thus, E is the correct answer. 4. When Yunji added all the integers from 1 to 9, she mistakenly left out a number. Her incorrect sum turned out to be a square number. Which number did Yunji leave out? 5 6 7 8 9 Answer: E Video solution: Click to load, then click again to play Written solution: To find the number that Yunji left out, we need to find the sum of the integers from 1 to 9 and find its difference with the largest perfect square below the sum. We can calculate the sum of the integers from 1 to 9 as follows: 1+…+9=29(9+1)=45 The largest perfect square less than 45 would be 36 and 45−36=9. Thus, E is the correct answer. 5. Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of 6. Which of the following integers cannot be the sum of the two numbers? 5 6 7 8 9 Answer: B Video solution: Click to load, then click again to play Written solution: can draw a 2x2 chart of possible rolls We can simply list down all possible combinations of the two rolls: (1,6),(2,3),(2,6),(3,6), (4,6),(5,6),(6,6) Respectively, the sums would be 7,5,8,9,10,11,12. Among the choices, only 6 is not a possible sum. Thus, B is the correct answer. 6. Sergei skated around an ice rink, gliding along different paths. The gray lines in the figures below show four of the paths labeled P, Q, R, and S. What is the sorted order of the four paths from shortest to longest? P, Q, R, S P, R, S, Q Q, S, P, R R, P, S, Q R, S, P, Q Answer: D Video solution: Click to load, then click again to play Written solution: From inspection, we can notice that Path R is the shortest path since it avoids going through the entire loop by taking straight-line shortcuts instead of going through the arc portions of the rink. This leaves us with two possible answer choices, D and E, which only differ in how they sort Path P and S. To determine which path is longer between Path P and S, notice that the difference between the two paths can be related by a right triangle as shown here: The hypotenuse of the triangle was traversed by Sergei in Path S while one of the legs is part of Path P. By the Pythagorean Theorem, we know that the hypotenuse will always be longer than the legs, and so Path S will be longer than Path P. Thus, B is the correct answer. 7. A 3×7 rectangle is covered without overlap by 3 shapes of tiles: 2×2, 1×4, and 1×1, shown below. What is the minimum possible number of 1×1 tiles used? 1 2 3 4 5 Answer: E Video solution: Click to load, then click again to play Written solution: Notice that if we only fill the 3×7 rectangle using 2×2 and 1×4 tiles, then we will always be filling the rectangle in multiples of 4. From the answer choices, it will suffice to consider the cases where we are able to fill 16 or 20 tiles out of the 21 tiles of the rectangle, since the former will leave us 5 spaces for the 1×1 tiles while the latter will leave us with one. The other options are not possible since those numbers of tiles cannot be arrived by subtracting any multiple of 4 from 21. By attempting to place the 2×2 and 1×4 tiles, we can immediately notice that it's not possible to fill 20 tiles in the rectangle, and we can easily find cases where we are left with five tiles. Thus, E is the correct answer. 8. On Monday Taye has $2. Every day, he either gains $3 or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, 3 days later? 3 4 5 6 7 Answer: D Video solution: Click to load, then click again to play Written solution: We can systematically exhaust the cases using a tree diagram where the upper branch covers the case where the amount gains $3 while the other branch represents the case where the amount doubles. From the diagram, we can see that we end up with six unique dollar amounts after three days. Thus, D is the correct answer. 9. All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection? 24 25 26 27 28 Answer: E Video solution: Click to load, then click again to play Written solution: We can let r be the number of red marbles that Maria has. Since Maria has half as many red marbles as green, then we know that she has 2r green marbles. Moreover, since she has twice as many blue marbles as green, then she will have 2(2r)=4r blue marbles. Adding these together gives us r+2r+4r=7r, and so the answer must be a multiple of 7. Among the answer choices, only 28 is a multiple of 7. Thus, E is the correct answer. 10. In January 1980 the Mauna Loa Observatory recorded carbon dioxide CO2 levels of 338 ppm (parts per million). Over the years the average CO2 reading has increased by about 1.515 ppm each year. What is the expected CO2 level in ppm in January 2030? Round your answer to the nearest integer. 399 414 420 444 459 Answer: B Video solution: Click to load, then click again to play Written solution: There are 50 years between 1980 and 2030, so we can expect the CO2 reading to increase by 50×1.515=75.75≈76 ppm by 2030. Since the CO2 reading in 1980 was 338 ppm, then we will have 338+76=414 ppm by 2030. Thus, B is the correct answer. 11. The coordinates of △ABC are A(5,7), B(11,7), and C(3,y), with y>7. The area of △ABC is 12. What is the value of y? 8 9 10 11 12 Answer: D Video solution: Click to load, then click again to play Written solution: Consider the base of the triangle to be AB which has length 11−5=6. Given that the area of the triangle is 12, its height must be of length 62(12)=4. Since y>7, then y must be 7+4=11. Thus, D is the correct answer. 12. Rohan keeps a total of 90 guppies in 4 fish tanks. • There is 1 more guppy in the 2nd tank than in the 1st tank. • There are 2 more guppies in the 3rd tank than in the 2nd tank. • There are 3 more guppies in the 4th tank than in the 3rd tank. How many guppies are in the 4th tank? 20 21 23 24 26 Answer: E Video solution: Click to load, then click again to play Written solution: Let x be the number of guppies in the 1st tank. Hence, there are x+1 guppies in the 2nd tank, x+3 guppies in the 3rd tank, and x+6 guppies in the 4th tank. We then use the fact that there are a total of 90 guppies in the 4 tanks to find x: x+x+1+x+3+x+6=90 4x+10=90 x=20. Note that we are not yet done since we are asked for the number of guppies in the 4th tank and not the 1st. There are x+6=20+6=26 guppies in the 4th tank. Thus, E is the correct answer. 13. Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of 6 hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.) 4 5 6 8 12 Answer: B Video solution: Click to load, then click again to play Written solution: We can deduce from the choices that it is possible to exhaust all possible cases for this problem. Note that all sequences must start with up (U) and end with down (D), and that it should not be possible to go down more times than Buzz has gone up so far. Keeping this in mind, we can arrive at the following possible cases: UUUDDD UUDUDD UUDDUD UDUDUD UDUUDD which is a total of five possible sequences. Thus, B is the correct answer. 14. The one-way routes connecting towns A, M, C, X, Y, and Z are shown in the figure below (not drawn to scale). The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers? 28 29 30 31 32 Answer: A Video solution: Click to load, then click again to play Written solution: A systematic way of tracking the shortest overall distance to Z is to consider the shortest distance to get to each town from A. For instance, the shortest distance to get to town X from A is 5 km, trivially. Then, for town M, going to town X first will be shorter compared to going directly from A, so the shortest path to town M has a length of 7 km. For town Y, it will take us 15 km if we come from town X and only 13 km coming from M, so 13 km is the length of shortest path to Y from A. Doing the same for town C will give us 18 km as the shortest distance by coming from town Y. Finally, for town Z, we can either come from town Y, C, or M. The total distance if we come from each three towns respectively would be 30, 28, and 32. Hence, 28 km is the shortest distance from A to Z. The diagram below summarizes our process. Here, the town name is replaced by the shortest distance to get to that town from town A. Thus, A is the correct answer. 15. Let the letters F, L, Y, B, U, G represent distinct digits. Suppose FLYFLY is the greatest number that satisfies the equation 8⋅FLYFLY=BUGBUG. What is the value of FLY+BUG? 1089 1098 1107 1116 1125 Answer: C Video solution: Click to load, then click again to play Written solution: Firstly, note that FLYFLY=1001(FLY) and, similarly, BUGBUG=1001(BUG) so the equation can be simplified to 8⋅FLY=BUG. For BUG to remain three digits, F must be 1. Moreover, L must also be less than 3 to avoid carrying over 2 to the hundreds digit and making the product 4 digits. Since we need FLY to be the greatest number, L must be 2. To identify the possible values for Y, we note that so far we have 8(12)=96, so we must avoid carrying 4 to the tens digit to keep the resulting product three digits. Hence, Y≤5. We can try 4 and verify that the resulting product has unique digits that haven't been used yet: 8(124)=992, which does not have unique digits. Trying 3, we get 8(123)=984, which satisfies our criteria. Hence, FLY+BUG=123+984=1107. Thus, C is the correct answer. 16. Minh enters the numbers 1 through 81 into the cells of a 9×9 grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by 3? 8 9 10 11 12 Answer: D Video solution: Click to load, then click again to play Written solution: We first need to note that there are 27 multiples of 3 from 1 through 81. If any row or column has a multiple of 3, then the product of the numbers in the row or column will be divisible by 3. From this, it is evident that we must try to place all 27 multiples of 3 together in some corner of the grid so the least number of products will be divisible by 3. We can place 25 multiples of 3 in a 5×5 grid so only 5 rows and 5 columns will have products that are divisible by 3. The remaining 2 multiples of 3 can then be placed in the 6th column along the 1st and 2nd row of the 5×5 grid so in total we will only have 6 columns and 5 rows that have products that are divisible by 3, for a total of 6+5=11 products divisible by 3. The diagram below illustrates our process. The highlighted cells contain the multiples of 3 while the dotted lines mark the rows or columns that have products which are divisible by 3. Thus, D is the correct answer. 17. A chess king is said to attack all the squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a 3×3 grid attacks all 8 other squares, as shown below. Suppose a white king and a black king are placed on different squares of a 3×3 grid so that they do not attack each other. In how many ways can this be done? 20 24 27 28 32 Answer: E Solution: Firstly, we note that a king cannot be on the center square as it will attack any other piece on the 3×3 grid. To solve this problem, we can simply consider some possible cases: one where a king is in the corner and another when one king is on an edge (but not a corner). When one king is placed in any corner, then the other king can be placed in 5 other squares without them attacking each other. Since there are four corners, we have 5×4=20 possibilities for this case. Another possible case is when one king is on an edge that is not a corner. In this scenario, the other king can be in 3 other squares without the two kings attacking each other. There are 4 edges in the grid so we have 4(3)=12 possibilities for this case. Adding the number of possibilities, we get 20+12=32 total number of ways. Thus, E is the correct answer. 18. Three concentric circles centered at O have radii of 1, 2, and 3. Points B and C lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle BOC, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of ∠BOC in degrees? 108 120 135 144 150 Answer: A Solution: Let θ be the measure of ∠BOC. One component of the shaded region is the area of the circle with radius 2 minus the area of the circle with radius 1. This part has area 4π−π=3π. The remaining area is a sector of the biggest circle minus the area of the circle with radius 2. This has area 360θ(9π−4π)=360θ(5π). Hence, the total area of the shaded region is 3π+360θ(5π). Next, we note that the unshaded region is composed of the smallest circle and the unshaded portion of the outer ring. This will have a total area of π+360360−θ(5π) Lastly, we equate the area of both regions and solve for θ: 3π+360θ(5π)=π+360360−θ(5π) 2π=360360−θ−θ(5π) 52=1−3602θ 2θ=53(360) θ=108. Thus, A is the correct answer. 19. Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction? 0 51 154 31 52 Answer: C Solution: Jordan has 53×15=9 pairs of red sneakers and 6 pairs of white sneakers. Moreover, 32×15=10 are high-top and 5 are low-top. If we want to minimize the number of red high-top sneakers, then we can set all 6 white sneakers to be high-top, leaving 10−6=4 red sneakers as high-top. Hence, the fraction of red high-top sneakers would be 154. Thus, C is the correct answer. 20. Any three vertices of the cube PQRSTUVW, shown in the figure below, can be connected to form a triangle. (For example, vertices P, Q, and R can be connected to form isosceles △PQR.) How many of these triangles are equilateral and contain P as a vertex? 0 1 2 3 6 Answer: D Solution: We first note that we can only form equilateral triangles if we go through the diagonals of the square faces, otherwise at least one angle of the triangle will be different. Afterwards, it is easy to exhaust all possible equilateral triangles that can be formed: △PVT, △PRT, and △PRV. Thus, D is the correct answer. 21. A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was 3:1. Then 3 green frogs moved to the sunny side and 5 yellow frogs moved to the shady side. Now the ratio is 4:1. What is the difference between the number of green frogs and yellow frogs now? 10 12 16 20 24 Answer: E Video solution: Click to load, then click again to play Written solution: We can let g be the number of green frogs and y be the number of yellow frogs. Initially, we have g=3y. Then, after some frogs moved, we have the following proportion: y+3−5g+5−3=14. Substituting 3y for g will allow us to determine the number of yellow frogs originally (y): y−23y+2=4 3y+2=4(y−2)=4y−8 y=10 Hence, there were 10 yellow frogs and 3×10=30 green frogs initially. After some frogs moved, we now have 10+3−5=8 yellow frogs and 30+5−3=32 green frogs, giving us a difference of 32−8=24 between the number of green and yellow frogs. Thus, E is the correct answer. 22. A roll of tape is 4 inches in diameter and is wrapped around a ring that is 2 inches in diameter. A cross section of the tape is shown in the figure below. The tape is 0.015 inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest 100 inches. 300 600 1200 1500 1800 Answer: B Video solution: Click to load, then click again to play Written solution: We have a huge margin of error for this problem so we can freely estimate values. Firstly, we note that the entire roll of tape is 1-inch thick, and given that a single tape is 0.015 inches thick, then there are 0.0151=0.032=3200 layers of tape in the entire roll. Then, we must identify an estimate for the circumference of one layer of tape. Near the center, one layer of tape will have a circumference of 2π while the layers near the outer section will have a circumference of about 4π. A good estimate would be to take the average circumference which is 3π. With this, we can estimate the total length for the entire roll: 3200(3π)=200π≈200(3)≈600. Thus, B is the correct answer. 23. Rodrigo has a very large piece of graph paper. First he draws a line segment connecting point (0,4) to point (2,0) and colors the 4 cells whose interiors intersect the segment, as shown below. Next, Rodrigo draws a line segment connecting point (2000,3000) to point (5000,8000). Again he colors the cells whose interiors intersect the segment. How many cells will he color this time? 6000 6500 7000 7500 8000 Answer: C Video solution: Click to load, then click again to play Written solution: From the given example, observe that if we consider a different line segment with the same slope, for instance the line connecting points (0,2) and (1,0), then the number of colored cells will be halved. In general, it is possible to scale down the problem as long as we still have the same slope for the line. Next, note that the line segment passing through points (2000,3000) and (5000,8000) has a slope of 35. Hence, a scaled down version that we can consider is a segment connecting the points (0,0) and (3,5). From the diagram, it is evident that the line passes through 7 cells. We know that this will happen 1000 times as the segment passes through points (2000,3000) and (5000,8000). Hence, Rodrigo will need to color 7(1000)=7000 cells. Thus, C is the correct answer. Jean made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is 8 feet high and the other peak is 12 feet high. Each peak forms a 90∘ angle, and the straight sides of the mountains form 45∘ angles with the ground. The artwork has an area of 183 square feet. The sides of the mountains meet at an intersection point near the center of the artwork, h feet above the ground. What is the value of h? 4 5 42 6 52 Answer: B Video solution: Click to load, then click again to play Written solution: Observe that the we can extend the two mountains at the intersection point to make three triangles as shown in the diagram below. All triangles formed are 45∘-45∘-90∘ triangles and so based on the given height information, we can determine that the side lengths of the triangles, from left to right, are 82,h2, and 122. Hence, we have areas of 64,h2, and 144, respectively. To find h, we use the total area of the diagram which we can obtain by adding the area of the two large triangles and subtracting the area of the triangle with height h since this area was counted twice. Doing this, we get 64+144−h2=183 h2=144+64−183=25 h=5 Thus, B is the correct answer. A small airplane has 4 rows of seats with 3 seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be 2 adjacent seats in the same row for the couple? 158 5532 3320 5534 118 Answer: C Video solution: Click to load, then click again to play Written solution: For simplicity, we can disregard the order in which passengers are seated so we only consider the number of ways that the 12 seats can be filled by 8 passengers for the total number of possibilities, which is given by (812)=495. Out of these 495 possibilities, we consider the number of ways where no adjacent seats are available, then subtract this from 495. This scenario can happen when two passengers are occupying the edge seats of one row or one passenger is seated in the middle seat of a row. Hence, we consider the following cases: • No passenger is in the middle seat (all 8 passengers are on the edge seats): (88)=1 way. • Exactly 1 passenger is in the middle seat (7 are seated on the edge seats): There are (14)=4 ways where one row can contain a passenger in the middle seat and (12)=2 ways for the eighth passenger to be seated on the row where the one passenger is sat on the middle seat. Hence, the total for this case is 4(2)=8 ways. • Exactly 2 rows have a passenger in the middle seat: (24)=6 ways to select rows with occupied middle seat and another (24)=6 ways for the remaining 2 passengers to be seated in the rows with occupied middle seats. Thus, this case has a total of 6(6)=36 possibilities. • Exactly 3 rows have occupied middle seats: (34)=4 ways to select rows with occupied middle seats and (36)=20 ways for the remaining 3 passengers to be seated on the rows with occupied middle seats, for a total of 4(20)=80 ways. • All 4 rows have occupied middle seats: (48)=70 ways for the remaining 4 passengers to be seated. Adding all the possibilities in each case, we get 1+8+36+80+70=195 ways for there to be no two adjacent seats available. Hence, the probability that the couple will be seated together would be 495495−195=3320. Thus, C is the correct answer.
15251	https://arxiv.org/abs/2201.12573	[2201.12573] High-precision electron-capture $Q$ value measurement of $^{111}$In for electron-neutrino mass determination Skip to main content We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate >nucl-ex> arXiv:2201.12573 Help \| Advanced Search Search GO quick links Login Help Pages About Nuclear Experiment arXiv:2201.12573 (nucl-ex) [Submitted on 29 Jan 2022] Title:High-precision electron-capture Q value measurement of ^{111}In for electron-neutrino mass determination Authors:Z.Ge, T.Eronen, A.deRoubin, K.S.Tyrin, L.Canete, S.Geldhof, A.Jokinen, A. Kankainen, J. Kostensalo, J. Kotila, M. I. Krivoruchenko, I. D. Moore, D. A. Nesterenko, J. Suhonen, M. Vilén View a PDF of the paper titled High-precision electron-capture $Q$ value measurement of $^{111}$In for electron-neutrino mass determination, by Z.Ge and 13 other authors View PDF Abstract:A precise determination of the ground state ^{111}In (9/2^+) electron capture to ground state of ^{111}Cd (1/2^+) Qvalue has been performed utilizing the double Penning trap mass spectrometer, JYFLTRAP. A value of 857.63(17) keV was obtained, which is nearly a factor of 20 more precise than the value extracted from the Atomic Mass Evaluation 2020 (AME2020). The high-precision electron-capture Qvalue measurement along with the nuclear energy level data of 866.60(6) keV, 864.8(3) keV, 855.6(10) keV, and 853.94(7) keV for ^{111}Cd was used to determine whether the four states are energetically allowed for a potential ultra-low Q-value \beta^{}decay or electron-capture decay. Our results confirm that the excited states of 866.60(6) keV with spin-parity (J^\pi) of 3/2^{+}and 864.8(3) keV with J^\pi= 3/2^{+}are ruled out due to their deduced electron-capture Qvalue being smaller than 0 keV at the level of around 20\sigmaand 50\sigma, respectively. Electron-capture decays to the excited states at 853.94(7) keV (J^\pi= 7/2^+) and 855.6(10) keV (J^\pi= 3/2^+), are energetically allowed with Qvalues of 3.69(19) keV and 2.0(10) keV, respectively. The allowed decay transition ^{111}In (9/2^{+}) \rightarrow^{111}Cd (7/2^{+}), with a Qvalue of 3.69(19) keV, is a potential a new candidate for neutrino-mass measurements by future EC experiments featuring new powerful detection technologies. The results show that the indium level 2p_{1/2}for this decay branch leads to a significant increase in the number of EC events in the energy region sensitive to the electron neutrino mass. Comments:12 pages, 4 figures Subjects:Nuclear Experiment (nucl-ex) Cite as:arXiv:2201.12573 [nucl-ex] (or arXiv:2201.12573v1 [nucl-ex] for this version) Focus to learn more arXiv-issued DOI via DataCite Related DOI: Focus to learn more DOI(s) linking to related resources Submission history From: Zhuang Ge [view email] [v1] Sat, 29 Jan 2022 12:14:26 UTC (1,523 KB) Full-text links: Access Paper: View a PDF of the paper titled High-precision electron-capture $Q$ value measurement of $^{111}$In for electron-neutrino mass determination, by Z.Ge and 13 other authors View PDF TeX Source Other Formats view license Current browse context: nucl-ex <prev \| next> new \| recent \| 2022-01 References & Citations INSPIRE HEP NASA ADS Google Scholar Semantic Scholar export BibTeX citation Loading... BibTeX formatted citation × Data provided by: Bookmark Bibliographic Tools Bibliographic and Citation Tools [x] Bibliographic Explorer Toggle Bibliographic Explorer (What is the Explorer?) 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15252	https://infinitalab.com/astm/coefficient-of-linear-thermal-expansion-astm-e831-astm-d696/?srsltid=AfmBOorxerdxNGg4hEKfirQkK7lkJXvA-TyyDi_G88TQ89eJ1IFZVyy6	ASTM E831, ASTM D696 Coefficient of Linear Thermal Expansion Lab In US - Infinita Lab Services Metrology Testing Service Scanning Electron Microscope (SEM) Electron Energy Loss Spectroscopy (EELS) Secondary Ion Mass Spectroscopy – SIMS Electron Backscatter Diffraction (EBSD) Liquid Chromatography-Mass Spectroscopy – LC-MS Gas Chromatography Mass Spectroscopy – GCMS FTIR Emissivity Inductively Coupled Plasma Mass Spectrometry – ICP-MS Atomic Force Microscopy – AFM Inductively Coupled Plasma Optical Emission Spectroscopy (ICP-OES) Automotive Volatile Organic Compound (VOC) Testing Inductively Coupled Plasma Atomic Emission Spectroscopy Rutherford Backscattering Spectrometry – RBS X-Ray Diffraction Analysis – XRD X-ray Photoelectron Spectroscopy – XPS Raman Spectrometry (Raman) Material Testing Service Ash Testing Coefficient of Friction (COF) ASTM D4603 ASTM C326 ASTM D8280 ASTM D950 ASTM E1921 ASTM E2714 Electrostatic Discharge (ESD) Testing Heat Deflection Temperature (HDT) Highly Accelerated Life (HALT) ASTM A754 ASTM D2294 ASTM E 18 / ISO 6508 / ASTM D 785 / ISO 2039 Shear Strength Test Product Testing Refractive Index (RI) Bearing Failure Analysis Spraying Water and Splashing Water (IPX3 & IPX4) Altitude Testing Altitude Testing ASTM Ceramics ASTM C1161 ASTM E903 ASTM C368 ASTM C373 ASTM C1211 ASTM C1341 ASTM C1499 ASTM C1505 ASTM C1292 ASTM C1259 Chemical Analysis ASTM D543 ASTM E34 MIL-STD-810 Mil-Std-810 ASTM C146 ASTM D543 ASTM D 1239 ASTM E1232 Corrosion ASTM B117 ASTM G154, ASTM D4329.. 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CLTE measures the rate at which a given material expands as a function of temperature. This test is important to select thermally stable material for designing parts of different equipment and machines. Start Testing- [x] You are agreeing to receive customer care text messages from Infinita Lab inc. Message frequency may vary. Standard Message and Data Rates may apply. Reply STOP to opt out. Reply Help for help. View our privacy policy. Δ TRUSTED BY ENGINEERS FROM Scope: ASTM E831 and ASTM D696 test methods change the ambient temperature with a change in the size of an object . The coefficient of Linear Thermal Expansion (CLTE) measures the rate at which a given material expands as a function of temperature. ASTM E831, ASTM D696 test is important to select thermally stable material for designing parts of different equipment and machines. Knowledge about relative expansion/contraction features of two materials in contact is utilized to use the tested materials successfully. Test Procedure: For the ASTM E831 and ASTM D696 test method a thermomechanical analyzer (TMA) is employed. The material specimen is placed in the holder at ambient temperature, and the specimen’s length is measured. Next, the specimen is cooled to a specific temperature and then heated again by increasing the temperature inside the furnace. The ASTM E831 and ASTM D696 test specimen is heated at a particular rate over the desired temperature range. CLTE versus temperature graph is drawn to obtain a curve. Dimensional changes in a specimen as a function of temperature can also be measured by using a dilatometer. The specimen is placed in the holder inside the dilatometer tube and is then heated. The overhead measuring compartment gauges change in dimension of the specimen. Specimen size: The test specimen to be used for TMA can be anywhere between 2 and 10 mm in length and a maximum of 10 mm in breadth and should be flat at both ends. 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15253	https://math.stackexchange.com/questions/26037/intuitive-understanding-of-the-constant-e	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Intuitive Understanding of the constant "$e$" Ask Question Asked Modified 1 year, 8 months ago Viewed 11k times $\begingroup$ Potentially-related questions, shown before posting, didn't have anything like this, so I apologize in advance if this is a duplicate. I know there are many ways of calculating (or should I say "ending up at") the constant e. How would you explain e concisely? It's a rather beautiful number, but when friends have asked me "what is e?" I'm usually at a loss for words -- I always figured the math explains it, but I would really like to know how others conceptualize it, especially in common-language (say, "English"). related but not the same: exponential-function intuition big-list constants eulers-number-e edited Dec 25, 2023 at 19:27 J. W. Tanner 64k44 gold badges4343 silver badges8989 bronze badges asked Mar 9, 2011 at 19:14 sovasova 1,79733 gold badges1717 silver badges1111 bronze badges $\endgroup$ 10 $\begingroup$ Can you justify why this is not the same question? My answer to this question is the same as my answer to that question. $\endgroup$ Qiaochu Yuan – Qiaochu Yuan 2011-03-09 19:18:29 +00:00 Commented Mar 9, 2011 at 19:18 $\begingroup$ There are any number of "semi-natural" ways to arrive at $e$ (what you call "ending up at $e$"), such as compound interest, a function that is equal to its own rate of growth, etc. The compound interest one is particularly succint ("$e$ is the amount of interest you would have at the end of one year if you deposit one dollar, at 100% annual compound interest, compounded each and every instant."), though it may take some motivation to explain to the lay public (who has enough trouble grasping financial matters, it would seem), why the frequency of compounding matters. $\endgroup$ Arturo Magidin – Arturo Magidin 2011-03-09 19:25:03 +00:00 Commented Mar 9, 2011 at 19:25 $\begingroup$ @Qiaochu: IMHO the questions are not at all the same; they are complementary and so it is natural that they share the same basis for an answer. $\endgroup$ – Eelvex 2011-03-09 19:28:45 +00:00 Commented Mar 9, 2011 at 19:28 $\begingroup$ @Arturo Magidin: The number $e$ is what everybody wants to have (at least on his/her bankaccount). Imagine a world where the banks would pay interest continously... $\endgroup$ – Fabian 2011-03-09 19:29:28 +00:00 Commented Mar 9, 2011 at 19:29 $\begingroup$ @Fabian: Well, keep in mind that they would likely also charge interest continuously... $\endgroup$ Arturo Magidin – Arturo Magidin 2011-03-09 19:30:09 +00:00 Commented Mar 9, 2011 at 19:30 \| Show 5 more comments 23 Answers 23 Reset to default 54 $\begingroup$ If someone asks me, "what is $e$?" I sketch the graph of $y=1/x$, draw a line segment from $(1,1)$ on the curve down to $(1,0)$ on the $x$-axis, and ask, how far to the right do I have to draw another vertical segment to rope off an area of 1? Anyone who is familiar with the idea of graphing a function can appreciate that definition, and it's not surprising that something with such a down-to-earth definition is going to turn up in lots of other places in Mathematics. And anyone who knows Calculus can be shown that all the other properties of $e$ and of $e^x$ and of $\log x$ can be derived from this one property of $e$. The yellow area equals the red one: Share edited Sep 8, 2017 at 9:31 answered Jul 29, 2011 at 1:32 Gerry MyersonGerry Myerson 187k1313 gold badges232232 silver badges406406 bronze badges $\endgroup$ 4 1 $\begingroup$ I don't understand. So simply you want to find an area under the curve? Can you illustrate the idea? $\endgroup$ Ooker – Ooker 2017-09-08 09:17:59 +00:00 Commented Sep 8, 2017 at 9:17 1 $\begingroup$ Someone made an illustrate and I'd added into your post. $\endgroup$ Ooker – Ooker 2017-09-08 09:29:46 +00:00 Commented Sep 8, 2017 at 9:29 $\begingroup$ Also, would this still be true with different functions? $\endgroup$ Ooker – Ooker 2017-09-08 09:30:23 +00:00 Commented Sep 8, 2017 at 9:30 2 $\begingroup$ Would what still be true with other functions, @Ooker? It's certainly not true that for every function $f$ the area under the graph from $x=1$ to $x=e$ is 1 I'm sure you can come up with very simple examples for which that area isn't 1. $\endgroup$ Gerry Myerson – Gerry Myerson 2017-09-08 09:33:24 +00:00 Commented Sep 8, 2017 at 9:33 Add a comment \| 29 $\begingroup$ For the somewhat-calculus-literate, your "related but not the same" question is what I'd go for: $e$ is the number for which the exponential function with that base is its own derivative. Without calculus, I'd go for the notion of compound interest: With a rate $r$ per period, compounded $n$ times per period, $A$ grows to $A(1+\frac{r}{n})^n$ after 1 period; as $n\to\infty$, $A(1+\frac{r}{n})^n\to Ae^r$. Share answered Mar 9, 2011 at 19:23 IsaacIsaac 37.1k1515 gold badges112112 silver badges142142 bronze badges $\endgroup$ 1 8 $\begingroup$ The interest method is exactly how I heard Sir Michael Atiyah explain it at a symposium. He explained it as investing $1 with an interest rate of 100%, both of which are easy numbers to wrap your head around. $\endgroup$ JavaMan – JavaMan 2011-03-09 19:31:04 +00:00 Commented Mar 9, 2011 at 19:31 Add a comment \| 23 $\begingroup$ Nobody has yet given a combinatorial way of visualizing $e,$ so I thought I'd add one. The constant $\pi$ is, of course, a ratio, that of the circumference of a circle to its diameter. If the corresponding ratios for regular polygons are used as a starting point, then $\pi$ is defined by taking the limit as the number of sides of the polygon goes to infinity. In a similar way, the constant $e$ is the limit of a ratio as a certain parameter is taken to infinity. Imagine a lottery in which, to enter, you must submit a sequence of $20$ numbers in the range $1$ to $20.$ Order matters, and numbers can be used multiple times. So, for example, you can submit all $1\text{s}$ if you like. The winning number could be any of $20^{20}$ possibilities. Now imagine that a similar lottery has been set up in a different locale, except that, because of local superstition, use of the number $13$ has been banned. Lottery entries still consist of $20$ numbers in the range $1$ to $20$, but only $19$ number choices are available, giving $19^{20}$ possible winning numbers. Clearly it's easier to win the second lottery, but by how much? The answer is the ratio $(1/19^{20})/(1/20^{20})=20^{20}/19^{20}\approx2.79.$ You are $2.79$ times as likely to win the second lottery as the first. To generalize, imagine comparing a lottery in which submissions consist of $n$ numbers in the range $1$ to $n$ with a lottery in which one of the numbers in the range $1$ to $n$ is unlucky and may not be used. The probability of winning the second lottery is higher by a factor of $n^n/(n-1)^n.$ The limiting value of this ratio as $n$ goes to infinity is $e\approx2.718281828$. What does this have to do with rates of growth of exponential processes? Imagine a quantity that goes up by a fixed ratio every year, say $72\%.$ Suppose we preferred to use months as units rather than years. It would be incorrect to simply divide the yearly rate by $12$ to get $72\%/12=6\%$ monthly growth, or, if we did so, we'd be describing a different process. The reason it's different is compounding. Growth of $6\%$ per month means growth by a factor of $1.06$ each month. In two months this translates to growth by a factor of $1.06\times1.06=1.1236$ or an increase of $12.36\%,$ which is bigger than $2\times6\%=12\%$ since the $6\%$ of quantity gained in the first month itself grows by $6\%$ in the second month. Similarly, over the course of a year, the factor of increase is $1.06^{12}\approx2.0122,$ which translates to $101.22\%$ growth per year instead of $72\%.$ More generally, if we prefer to use time units of $1/n$ years, an increase of $72\%/n$ every $1/n$ years is bigger than an increase of $72\%$ every year because of compounding. Specifically, it corresponds to growth by a factor of $(1+0.72/n)^n$ every year, which, if you compute it for particular $n,$ is bigger than $72\%$ per year. Moreover, if we make the time unit smaller by making $n$ bigger, the annual growth gets bigger because the compounding effect gets enhanced. In calculus we like talking about instantaneous rates of change, which involves make the time interval over which the change is measured arbitrarily small. If we do this in our problem, by taking the limit as $n$ goes to infinity of $(1+0.72/n)^n,$ we maximize the compounding effect, obtaining a growth factor of $2.05443,$ or $105.443\%$ per year. The limiting growth factor of $2.05443$ turns out to equal to $e^{0.72}.$ To relate this to the combinatorics problem, imagine a quantity that doubles every year, that is, that increases by $100\%$ per year. If we instead split this $100\%$ over $n$ equal time intervals, we get a growth factor of $(1+1/n)^n=(n+1)^n/n^n.$ As $n$ goes to infinity, this has the same limiting value as $n^n/(n-1)^n,$ namely $e.$ Hence, because we are compounding on arbitrarily short time intervals, we grow by a factor of approximately $2.718281828$ per year rather than by a factor of $2.$ Interestingly, there's a related combinatorics problem in which $e$ appears. Imagine, once again, a lottery in which $20$ numbers in the range $1$ to $20$ must be chosen. But in this lottery all $20$ numbers must be used. So our number selection boils down to choosing a permutation of the sequence $1,2,3,\ldots,20.$ Now imagine a similar lottery with the extra stipulation that the first number may not be $1,$ the second number may not be $2,$ the third number may not be $3,$ and so on. Such a permutation is called a derangement. Clearly there are fewer derangements than permutations, so the second lottery is easier to win. Once again, the probability of winning the second lottery is approximately $e$ times bigger than the probability of winning the first. Share answered Mar 31, 2013 at 10:20 Will OrrickWill Orrick 19.2k22 gold badges5757 silver badges9090 bronze badges $\endgroup$ Add a comment \| 15 $\begingroup$ It is my opinion there is no "intuitive understanding of the number $e$". Presumably, what you want to explain to your friends is not some mythical intuitive content of the number but some actual, concrete property it has which makes you appreciate it. Explain that. Share edited Mar 9, 2011 at 19:41 answered Mar 9, 2011 at 19:33 Mariano Suárez-ÁlvarezMariano Suárez-Álvarez 140k1414 gold badges260260 silver badges395395 bronze badges $\endgroup$ 6 1 $\begingroup$ Well put indeed! $\endgroup$ Phonon – Phonon 2011-07-29 00:49:19 +00:00 Commented Jul 29, 2011 at 0:49 11 $\begingroup$ -1. I fully agree that there is no "intuitive understanding of a number," and that a more meaningful question would be, "How can I intuit the properties of the number $e$?" Now, we also seem to agree that the OP really meant to ask the latter question. So if that's the question, what's the answer? The response you gave does not provide one. I feel that this should be a comment rather than an answer. $\endgroup$ Jesse Madnick – Jesse Madnick 2011-08-05 01:06:28 +00:00 Commented Aug 5, 2011 at 1:06 $\begingroup$ @JesseMadnick, I find the idea that one can intuit properties of a number as ununderstandable as that of reaching an intuitive understanding of the number $e$. $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2013-04-01 03:33:32 +00:00 Commented Apr 1, 2013 at 3:33 1 $\begingroup$ @MarianoSuárez-Alvarez: Then what are the things that we can intuit, if not mathematical properties? $\endgroup$ Jesse Madnick – Jesse Madnick 2013-04-01 04:07:24 +00:00 Commented Apr 1, 2013 at 4:07 $\begingroup$ I doubt I intuit any property of $e$ whatsoever... what propertes of $e$ do you intuit? (hopefully this verb does not exist...) $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2013-04-01 04:14:37 +00:00 Commented Apr 1, 2013 at 4:14 \| Show 1 more comment 14 $\begingroup$ Just to illustrate Gerry Myerson's answer. The yellow and red zones have the same area. Share edited May 19, 2018 at 3:54 answered Jun 19, 2013 at 1:51 leonbloyleonbloy 66.4k1010 gold badges7878 silver badges167167 bronze badges $\endgroup$ Add a comment \| 12 $\begingroup$ Geometric interpretations help with the intuition, and I liked Gerry Myerson's explanation for that reason. Here's another geometric explanation you might give, using exponential decay. Start by imagining a process, such as radioactive decay, where at the end of every hour you have half the amount of material you started with at the begining of the hour. So if you start with 1 unit of material, then the amount of material remaining at hours 0, 1, 2, 3 is 1, 1/2, 1/4, 1/8. This process is described by $(1/2)^t$ or $2^{-t}$. Or, you can imagine faster decay where you have only 1/3 of the material left at the end of each hour, so that the amounts are 1, 1/3, 1/9, 1/27 and the process is described by $(1/3)^t$ or $3^{-t}$. You can then sketch these two functions; both asymptotically approach 0, with $3^{-t}$ getting there faster than $2^{-t}$. You can then mention the amazing fact that the area bounded by either curve and the horizontal and vertical axes, although infinite in extent, has a definite finite area. This is highly plausible since the curve is approaching the horizontal so quickly. Tell them that if you calculate this area, you find that for $(1/2)^t$ the area is bigger than 1, while for $(1/3)^t$, the area is smaller than 1. Then ask how can you adjust the decay rate, or equivalently, the fraction remaining after each hour, so that the area exactly equals 1? The answer turns out to be that $1/e\approx1/2.718$ of the material should remain after each hour - a process described by $e^{-t}$. Not surprisingly, this is a number between 2 and 3. For the very curious and dedicated listener who knows about geometric series, you can justify the assertions that the areas are finite, and that the area is greater than 1 for $2^{-t}$ and less than 1 for $3^{-t}$. For example, for $2^{-t}$ you can get an overestimate of the area using rectangles: $1+1/2+1/4+\ldots=2$. So the area under the curve is finite. To show that for $2^{-t}$ the area is bigger than 1, you can do an underestimate using rectangles: $1/2+1/4+1/8+\ldots=1$.To show that $3^{-t}$ is smaller than 1, you can do an overestimate using trapezoids. If you break each trapezoid into a rectangle and a triangle, you get the overestimate $$ \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}\right)+\left(\frac{1}{9}+\frac{1}{2}\cdot\frac{2}{9}\right)+\ldots=2\left(\frac{1}{3}+\frac{1}{9}+\ldots\right)=1. $$ This provides some explanation for the magnitude of $e$. Share edited Jul 29, 2011 at 19:03 answered Jul 29, 2011 at 14:46 Will OrrickWill Orrick 19.2k22 gold badges5757 silver badges9090 bronze badges $\endgroup$ 2 2 $\begingroup$ thanks for a very cool answer, Will. Would it be more fair to say that the area under e^-t approaches 1, rather than is equal to it? $\endgroup$ sova – sova 2011-08-03 02:21:21 +00:00 Commented Aug 3, 2011 at 2:21 2 $\begingroup$ @sova : That's a matter of definitions. I say that the proper integral $\int_0^b e^{-t}\,dt$ approaches 1 as $b$ approaches infinity or, in other words, that $\lim_{b\rightarrow\infty}\int_0^b e^{-t}\,dt=1$. I define "area under the curve" to be the value of the improper integral $\int_0^\infty e^{-t}\,dt$, which in turn is defined to be this limit, assuming it exists. $\endgroup$ Will Orrick – Will Orrick 2011-08-04 02:21:07 +00:00 Commented Aug 4, 2011 at 2:21 Add a comment \| 12 $\begingroup$ Suppose you have a particle with the following property: starting at time zero (in seconds, s), the magnitude of its velocity $v = \|\mathbf{v}\|$ (in meters per second, m/s) at time $t$ is exactly equal to its distance $d$ (in meters, m) from its starting place at the same time. At $1$ m its speed is $1$ m/s, at $2$ m, $2$ m/s and so on. What is the distance function $d = d(t)$ of this particle as a function of time, you might ask? Since $v(t) = d^{\prime}(t) = d(t)$, then $d(t) = d_{0} \ e^{t}$, where $d_{0}$ is some multiplicative constant. Since we haven't yet specified any initial data, we can (without loss of generality) simply take $d_0 = 1 \ \text{m}$ and $v_{0} = 1 \ \text{m/s}$ both at $t = 0 \ \text{s}$. The constant $e$ is the magnitude of this particle's distance or velocity at time $t = 1$ s. Share edited Aug 5, 2011 at 0:17 answered Mar 9, 2011 at 19:38 user02138user02138 17.3k55 gold badges6060 silver badges8989 bronze badges $\endgroup$ 2 1 $\begingroup$ it's unfortunate that you cannot hug someone or high five someone through a web browser yet, but I would at least like to say thanks for this wonderful answer :D $\endgroup$ sova – sova 2011-08-04 17:06:13 +00:00 Commented Aug 4, 2011 at 17:06 1 $\begingroup$ When I was younger, I would often wonder about this while on road trips (except I would be driving toward my destination, not away from it). I since figured out what you have written here, but thanks for writing it! (+1) $\endgroup$ The Chaz 2.0 – The Chaz 2.0 2011-12-06 15:25:14 +00:00 Commented Dec 6, 2011 at 15:25 Add a comment \| 11 $\begingroup$ One way of understanding what is $e$, is to see it as a rate of growth. This article explains it very well. Share answered Mar 9, 2011 at 19:19 EelvexEelvex 1,4241111 silver badges2424 bronze badges $\endgroup$ 2 2 $\begingroup$ what a wonderful link! many thanks for this. $\endgroup$ sova – sova 2011-03-09 19:45:04 +00:00 Commented Mar 9, 2011 at 19:45 9 $\begingroup$ Thanks for the mention (I'm the author :)). You might want to check out betterexplained.com/articles/developing-your-intuition-for-math I cover 4 common definitions of e and how you can see the intuition behind them. $\endgroup$ Kalid – Kalid 2011-03-09 21:47:14 +00:00 Commented Mar 9, 2011 at 21:47 Add a comment \| 6 $\begingroup$ I also believe this to be equivalent to the cited question about e^x, but will answer the explicit question for the sake of argument. To be concise, I remark to friends and students that e is the most important number in calculus, just as pi is the most important constant in geometry. Certainly these claims are arguable, but my friends/students don't argue! Share answered Mar 9, 2011 at 19:47 The Chaz 2.0The Chaz 2.0 10.6k77 gold badges5050 silver badges8585 bronze badges $\endgroup$ 4 $\begingroup$ Well, with that argument, you could also claim that $e$ is home to a herd of unicorns... $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2011-03-09 19:54:37 +00:00 Commented Mar 9, 2011 at 19:54 $\begingroup$ @mar With what argument?!? I usually don't justify the statement unless the listener has timefor something less-than concise. $\endgroup$ The Chaz 2.0 – The Chaz 2.0 2011-03-09 20:07:16 +00:00 Commented Mar 9, 2011 at 20:07 2 $\begingroup$ That is the argument I had in mind... the lack thereof. $\endgroup$ Mariano Suárez-Álvarez – Mariano Suárez-Álvarez 2011-03-09 20:09:12 +00:00 Commented Mar 9, 2011 at 20:09 $\begingroup$ @MarianoSuárez-Alvarez - I think I get it now! Will leave the answer, for posterity's sake :) $\endgroup$ The Chaz 2.0 – The Chaz 2.0 2015-06-10 13:49:35 +00:00 Commented Jun 10, 2015 at 13:49 Add a comment \| 6 $\begingroup$ The derivative of an exponential function is $$\left(a^x\right)'=\lim_{h\to0}\frac{a^{x+h}-a^x}h=\lim_{h\to0}\frac{a^h-1}ha^x=\ln(a)\ a^x,$$ where $\ln(a)$ is some function of $a$. The number $e$ is by definition the one such that $\ln(a)=1$, so that the derivative is the most "natural": $$\color{green}{\left(e^x\right)'=e^x}.$$ Informally, we can "invert" the limit and get: $$\ln(a)=\lim_{h\to0}\frac{a^h-1}h\to a=\lim_{h\to0}(1+h\ln(a))^{1/h}.$$ This has two useful consequences: $\color{green}{e=\lim_{h\to0}(1+h)^{1/h}}$ allows to compute $e$. $1.1^{10}=2.593\cdots\1.01^{100}=2.7048\cdots\1.001^{1000}=2.71692\cdots\1.0001^{10000}=2.718145\cdots\\cdots\1.00000\cdots1^{100000\cdots0}=2.718281828459045235\cdots$ $a=\lim_{h\to0}(1+h\ln(a))^{1/h}=\lim_{h'\to0}(1+h')^{\ln(a)/h'}=\lim_{h'\to0}\left((1+h')^{1/h'}\right)^{\ln(a)}=e^{\ln(a)}$, and this shows that the logarithm function is the inverse of the exponential. Share edited Jun 10, 2015 at 10:03 answered Jun 10, 2015 at 9:49 user65203user65203 $\endgroup$ 2 $\begingroup$ Thank you for this answer. I wish to ask you how it can be proven that their exists an $a$ such that $\ln(a) = \lim_{h\to0}\frac{a^h-1}h = 1$ since that is already assumed in the first step of your answer. More generally, how would you demonstrate that the function $f(a) = \lim_{h\to0}\frac{a^h-1}h$ is surjective onto $\mathbb{R}$? This I feel is worth considering because then it's clear that for any $k$ there exists a function of the form $a^x$ such that (differentiating with respect to $x$) $(a^x)'$ = $k(a^x)$. $\endgroup$ Insert Pseudonym – Insert Pseudonym 2015-08-07 23:10:57 +00:00 Commented Aug 7, 2015 at 23:10 $\begingroup$ @InsertPseudonym: the question is about intuition for the constant $e$. The key point is: $(a^x)'=a^x$, informally. $\endgroup$ user65203 – user65203 2015-08-08 07:25:14 +00:00 Commented Aug 8, 2015 at 7:25 Add a comment \| 5 $\begingroup$ If $a > 0$, draw the curve $y = a^x$. You will notice it has a tangent line when it strikes the $y$-axis. If $a$ is much larger than 1, the tangent line is steep. If $a$ is small the slope is small. This slope depends continuously on $a$. The unique value making it 1 is $e$. This unlocks all of the magic if you think carefully. Share answered Aug 5, 2011 at 2:18 ncmathsadistncmathsadist 50.2k33 gold badges8585 silver badges134134 bronze badges $\endgroup$ Add a comment \| 4 $\begingroup$ There is a phenomenon called "exponential growth". We all have a feeling for it. Exponential growth of some quantity $Q$ with time is characterized by the property that in equal time intervals the quantity $Q$ increases by the same factor. Exponential growth can be slow or fast; therefore we have to have a "unit" in order to measure and compare the speed of growth in individual cases. Mathematicians tell us that the natural unit for this speed is encoded in a certain number $e\doteq 2.718$, in the same way that the unit of length was encoded in a rod of platinum and stored 1799 in a vault somewhere in Paris. Share answered Jan 29, 2013 at 14:32 Christian BlatterChristian Blatter 233k1414 gold badges204204 silver badges490490 bronze badges $\endgroup$ 1 $\begingroup$ very elegant explanation. exponential rates (multiplicative increases) need a measuring stick and e is it. $\endgroup$ sova – sova 2023-02-17 07:17:57 +00:00 Commented Feb 17, 2023 at 7:17 Add a comment \| 4 $\begingroup$ Professor Ghrist of University of Pennsylvania would say that e^x is the sum of the infinite series with k going from zero to infinity of (x^k)/k!. If you are interested in Euler's number then you should not miss his Calculus of a Single Variable Course on Coursera Share answered Jun 19, 2013 at 0:42 mark ptakmark ptak 4111 bronze badge $\endgroup$ Add a comment \| 4 $\begingroup$ Preface: The best first definition of e is one that nobody has yet given in this thread, namely, that it is the base such that the corresponding exponential function has a slope of unity at 0. If it is not immediately clear what this means, then sketch two or three exponential functions with different bases (say, 2, 3, and 1/2), note that they all pass through the point (0,1), and then note that they can be distinguished by what slope they have at that point (ie, what slope their tangent line has at that point). The most convenient one is the one that has a slope of unity. This base can be taken as the definition of e. However, once the edifice of one-variable calculus has been erected, it turns out to be convenient to define e in another manner, and this first definition is obtained as a theorem. Now, on to the answer: There is a joke, from the old West, about how to sell a covered wagon. As best as I recall, it goes something like this: Say that the price is one hundred dollars. If the buyer doesnt wince, add just for the wheels. If the buyer still doesnt wince, add for each one¦ You get the idea. In a similar vein, the answer to this question ought to be layered. The first best answer is that given by The Chaz (whose answer I have therefore up-voted), that is, just state that it is the most important number in calculus. Now, the audience mentioned by The Chaz is very limited, namely, his friends and students, however, the OP, naturally, wants us to consider the public at large, such as a casual acquantance encountered in an elevator who, stuck for something to say to fill the silence, asks you, whom they know to be an adept of mathematics, about this mysterious number e. The answer that The Chaz gives his friends and students is the best first answer for anyone, the best, as they say, elevator pitch (ie, something intelligible that you can say in the time that it takes an elevator to go between floors). If the questioner does not wince, that is, does not inquire further, then just leave it at that. If the questioner is not satisfied, then the next answer to give is the exponential base whose tangent at 0 is 1 one (that I mentioned above). If the questioner does not inquire further, then just leave it at that. If the questioner is not satisfied, then the next answer to give is the compound-interst one. If the questioner does not wince, then leave it at that. If the questioner is still not satisfied, then give the calculus rate-of-change explanation the velocity version by user02138 is especially good (and so I have up-voted it too). If the questioner is still not satisfied, then say, You have reached the limit, no pun intended, of my anyones ability to explain this to you, short of you yourself learning calculus. The questioner may then say, I appreciate all you have said, but what I want to know is why e has the value it does, rathter than some other value. You should then say, Thats a very good question, and no one has ever answered that, but for that matter, no one has ever answered the corresponding question for Ï. For example, why is the third decimal digit of Ï 1, rather than, say, 2? But remember that mathematics is not unique in this regard. In Physics, there is a number, approximately equal to 1/137, the intuitive understanding of which is a genuine mystery. This number is called the fine structure constant. The big mystery is exactly what you are asking regarding e, namely, why it has the value that it does. There is a long-standing brouhaha about it, nothing like the placid acceptance of e in mathematics. After reading up on it, you just might come back to mathematics and look upon e as an old friend. Share edited Sep 1, 2013 at 18:58 answered Jul 28, 2011 at 22:37 Mike JonesMike Jones 4,56811 gold badge4040 silver badges4242 bronze badges $\endgroup$ 9 2 $\begingroup$ I don't know why your post was downvoted. I think it's pretty witty! I voted it up. $\endgroup$ Bruno Joyal – Bruno Joyal 2011-07-28 23:36:05 +00:00 Commented Jul 28, 2011 at 23:36 5 $\begingroup$ I had no strong feelings one way or the other about this answer until I read the final paragraph. Questions about mathematical constants like "Why is $\pi$ what it is, rather than something else?" are not deep, they are trivial (or meaningless, depending on how you far you want to keep pushing on the "why" aspect), and I don't think they should be portrayed otherwise to the layperson, who won't know better. Thus, I have downvoted. $\endgroup$ Zev Chonoles – Zev Chonoles 2011-07-29 00:24:32 +00:00 Commented Jul 29, 2011 at 0:24 1 $\begingroup$ I enjoyed this answer very much, and I'm sorry it was downvoted. Although I don't think I would ever be satisfied giving curt replies in an elevator to someone genuinely interested in Mathematics, you pointed out user02138's answer (the velocity one) to me -- [mind blown] meets [mental satisfaction] $\endgroup$ sova – sova 2011-08-04 17:10:49 +00:00 Commented Aug 4, 2011 at 17:10 1 $\begingroup$ @Zev Chonoles: If it is true that the question of why mathematical constants are what they are is trivial, or even meaningless, then it seems like this metamathematical principle ought to appear somewhere as part of the ground rules. Im willing to admit that there is at least some truth to this, but I think one could in good faith disagree. Would you also consider the question Why is e irrational, rather than rational? a trivial question? My guess is that many would not consider it trivial, but if it is not trivial, then can the question of its choice of digits be entirely trivial? $\endgroup$ Mike Jones – Mike Jones 2011-08-16 04:47:52 +00:00 Commented Aug 16, 2011 at 4:47 1 $\begingroup$ @Mike Jones: I made a perfectly reasonable complaint about your original post, seeing as how it contained no mathematics and no keywords that led to anything relevant; in other words, it was useless. You have since fixed the latter - and less important - of the two issues I raised. So, I never left anyone with any impressions, other than that your original post was unhelpful, to say the least. $\endgroup$ Zev Chonoles – Zev Chonoles 2011-08-16 05:23:48 +00:00 Commented Aug 16, 2011 at 5:23 \| Show 4 more comments 3 $\begingroup$ It's easy to show that $\dfrac{d}{dx} 2^x = (2^x\cdot\text{constant})$. And $\left.\dfrac{d}{dx} 2^x\right\|_{x=0} = \text{that constant}$. Since the graph of $y=2^x$ gets steeper as $x$ grows, the slope at $x=0$ must be less than the slope of the secant line involving $x=0$ and $x=1$. That latter slope is 1. Therefore the "constant" is less than 1. By thinking about $y=4^x$ and considering the secant line involving $x=-1/2$ and $x=0$, one sees that that "constant" is more than 1. Therefore 2 is too small, and 4 is too big, to be $e$. For $y=e^x$, the "constant" is exactly 1. (One can show that 3 is too big via the secant line at $x=-1/6$ and $x=0$, but the arithmetic is a bit messy.) Similarly $2.5$ is too small, via $x=0$ and $x={}$ . . . . I don't remember which number I used here. A positive number, obviously, and less than 1. Messy arithmetic again. Share answered Aug 5, 2011 at 1:39 Michael HardyMichael Hardy 1 $\endgroup$ Add a comment \| 3 $\begingroup$ The thing that is special is the exponential function $\mathrm{exp}$, which satisfies $$\mathrm{exp}(0)=1,\quad \mathrm{exp'}=\mathrm{exp}.$$ Then of course $e:=\mathrm{exp}(1)$, and because of $\mathrm{exp}(x+y)=\mathrm{exp}(x)\mathrm{exp}(y)$ it makes sense to write $\exp(x)=e^x$. Share answered Oct 29, 2014 at 16:08 Carsten SCarsten S 8,91311 gold badge2424 silver badges4141 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ If you want to determine a function which equals its own derivative, you can try a numerical approximation $$f(x)=f'(x)\approx\frac{f(x+h)-f(x)}h,$$ which can be rewritten $$f(x+h)=(1+h)f(x).$$ By induction, $$f(x+nh)=(1+h)^nf(x).$$ This means that the function $f$ follows a geometric progression. Then setting $x=0,nh=t$, and choosing the solution such that $f(0)=1$, $$f(t)=(1+h)^{t/h}.$$ Now if you want an exact result, you have to let $h$ tend to $0$, which yields $$f(t)=e^t$$ and in particular $$f(1)=e.$$ A few values of $(1+h)^{1/h}$: $$ 1\to2\ 0.5\to2.25\ 0.25\to2.44140625\ 0.125\to2.565784513950347900390625\ 0.0625\to2.6379284973665998587631122129782\cdots\ \vdots\ 0.000\cdots\to2.7182818284590452353602874713527\cdots $$ Share edited May 25, 2017 at 15:43 Michael Hardy 1 answered Dec 31, 2016 at 17:49 user65203user65203 $\endgroup$ Add a comment \| 2 $\begingroup$ Personally I like the engineering way of explaining the number e: $$e = 3$$ $$\pi = 3$$ $$4 = 3$$ Share answered Feb 11, 2020 at 0:25 Tomer WolbergTomer Wolberg 48633 silver badges1212 bronze badges $\endgroup$ 1 $\begingroup$ en.wikipedia.org/wiki/Euler%27s_identity =D $\endgroup$ sova – sova 2020-02-11 22:26:38 +00:00 Commented Feb 11, 2020 at 22:26 Add a comment \| 2 $\begingroup$ Maybe not really intuitive... but if you can view things in higher dimensions this could help? Share answered Mar 13, 2020 at 20:37 marco trevimarco trevi 3,35811 gold badge2020 silver badges4040 bronze badges $\endgroup$ 1 $\begingroup$ this is amazing $\endgroup$ sova – sova 2023-02-17 07:12:02 +00:00 Commented Feb 17, 2023 at 7:12 Add a comment \| 0 $\begingroup$ A verbal (aphorism like sounding !?) definition may be interesting. A function made by raising a constant to a variable power remains invariant by differentiation... such a unique constant is $e.$ Share answered Dec 31, 2016 at 16:40 NarasimhamNarasimham 42.4k77 gold badges4545 silver badges111111 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ One spring afternoon you set out for a stroll. You're enjoying the warmth and the new blooms when suddenly you realize the sky has darkened. One heavy raindrop splatters on the ground near your feet, then another, and you hastily take cover under a nearby shelter. Mesmerized by the roar of the now steady spring shower, your gaze falls on a slick square paving stone a short distance away. You absently begin counting raindrops falling on the stone. With nowhere to go, you decide to get more systematic. Your watch and a bit of arithmetic tell you that one raindrop hits the square every five seconds, on average. A question suggests itself. Set a five-second timer and count raindrops hitting the stone. Sometimes your one expected raindrop will hit; sometimes two, three, or more will fall within the square; sometimes none at all. How often will each eventuality occur? Repeating a number of times, you discover that exactly one raindrop hits the square during the five-second experiment in less than half of trials but in somewhat more than a third of them. Aiming for precision, you keep experimenting for a while longer, and estimate that exactly one raindrop hits in about $\frac{1}{2.7}$ of trials. Remarkably, the fraction of experiments in which no raindrop falls in the square is also about $\frac{1}{2.7}$, but the fraction in which precisely two raindrops hit is half that: about $\frac{1}{5.4}$. Now intensely curious, you hold out for three raindrops. Scribbling tally marks in the little notebook you carry with you on walks, you eventually satisfy yourself that this happens one third as often as two raindrops, or one sixth as often as one raindrop---in about $\frac{1}{6\times2.7}$ of experiments. Excited, you realize that if this pattern continues, and you account for all possibilities, you should get $$ \frac{1}{2.7\ldots}\times\left(1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\ldots\right)=1, $$ which, doing the rough numbers, seems about right. But wait---you started with the observation that the average number of raindrops hitting the square per five-second trial was $1$. How does that work out with your formula? Let's see...you get the average by multiplying each number of raindrops by its probability of occurrence and doing the sum, which works out to \begin{align} &\frac{1}{2.7\ldots}\times\left(0\cdot1+1\cdot1+2\cdot\frac{1}{2}+3\cdot\frac{1}{6}+4\cdot\frac{1}{24}+5\cdot\frac{1}{120}+\ldots\right)\ &\qquad=\frac{1}{2.7\ldots}\times\left(1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\ldots\right). \end{align} Where did you see that before? You begin to feel that you might be onto something... Share answered Mar 31, 2023 at 11:02 Will OrrickWill Orrick 19.2k22 gold badges5757 silver badges9090 bronze badges $\endgroup$ Add a comment \| 0 $\begingroup$ The differential equation ${ \frac{df}{dx} = \lambda f }$ is basic and occurs often in physics applications. Assuming there is a power series solution ${ f(x) = \sum _{k=0} ^{\infty} a _k x ^k }$ (for ${ x }$ within some radius of convergence), we see the coefficients satisfy ${ (k+1) a _{k+1} = \lambda a _k }.$ Hence coefficients ${ a _k = \frac{\lambda ^k a _0}{k!} },$ that is the power series solution is ${ f(x) = a _0 \left( \sum _{k=0} ^{\infty} \frac{(\lambda x) ^k}{k!} \right) .}$ Note series ${ \sum _{k=0} ^{\infty} \frac{x ^k}{k!} }$ has infinite radius of convergence. So defining $${ \exp(x) := \sum _{k=0} ^{\infty} \frac{x ^k}{k!} },$$ the solution to ${ \frac{df}{dx} = \lambda f }$ is ${ f(x) = f(0) \exp(\lambda x) }.$ One can show ${ \exp(x) = \exp(1) ^x }$ for any rational ${ x }.$ First show ${ \exp(x+y) = \exp(x)\exp(y) }$ for any real ${ x, y }.$ Now ${ \exp(p) = \exp(1) ^{p} }$ for any integer ${ p },$ and ${ \exp(\frac{p}{q}) ^{q} = \exp(p) }$ for any integer ${ p }$ and integer ${ q > 0 }.$ Hence ${ \exp(\frac{p}{q}) ^q = \exp(1) ^p }$ that is ${ \exp(\frac{p}{q}) = \exp(1) ^{\frac{p}{q}} }$ for any integer ${ p }$ and integer ${ q > 0 }.$ So ${ \exp(x) = \exp(1) ^x }$ for any real ${ x }.$ The base ${ \exp(1) }$ is written ${ e }.$ Share edited Dec 15, 2023 at 10:31 answered Dec 15, 2023 at 4:19 Venkata Karthik BandaruVenkata Karthik Bandaru 1,65422 gold badges1414 silver badges1616 bronze badges $\endgroup$ Add a comment \| $\begingroup$ To actually explain it to somebody not so math-literate: e is the constant of growth. Whenever you want to relate the future value / quantity of something to now, thus define the future as a function of the present or vice versa, the only way to do this constantly and continuously is with e. That's why it shows up in Finance (interest), calculus integrals/derivatives, population growth etc. Share answered Feb 16, 2023 at 5:43 AlecAlec 35444 silver badges1212 bronze badges $\endgroup$ 3 1 $\begingroup$ a number that reflects "the natural rate of growth" is very intriguing indeed $\endgroup$ sova – sova 2023-02-17 07:11:08 +00:00 Commented Feb 17, 2023 at 7:11 $\begingroup$ Downvoters should explain. There's a reason the top posts here are not intuitive at all to someone without math education $\endgroup$ – Alec 2023-03-28 21:38:50 +00:00 Commented Mar 28, 2023 at 21:38 $\begingroup$ The issue is that growth can occur at different rates; there is no single constant of growth. Saying that $e$ is the constant of growth sounds mysterious and intriguing. Saying that $e$, like any positive number, can be used as the base when describing (exponential) growth and is a choice that makes some formulas easier to remember sounds...boring. But the latter is much closer to the truth. Some of those people without math education will later take a course. Prior conceptions have a way of sticking around and getting in the way, e.g. use of $e$ as base $\Leftrightarrow$ continuous growth. $\endgroup$ Will Orrick – Will Orrick 2023-04-01 01:35:27 +00:00 Commented Apr 1, 2023 at 1:35 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions exponential-function intuition big-list constants eulers-number-e See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Visit chat Linked 7 Why is $e$ the number that it is? 6 Why is $e$ so special? 8 How would you explain why "e" is important? (And when it applies?) 2 Can someone please explain $e$ in layman's term? 2 What's so special about $e$? 0 what is $e$ really? what is its meaning? 3 What is the constant $e$, fundamentally? 2 Practical significance of $e$ Is $e$ arbitrary? If not, how is it derived? 2 What is actually $e$ ? Why $e$ is so significant over other real numbers ? See more linked questions Related 3 Intuitive Understanding of the First Isomorphism Theorem 4 Intuitive understanding of the "Multiplication Rule"? 2 Intuitive understanding of integration 2 Intuitive understanding of probability puzzle 4 Any intuitive way to understand the graph for $x^x$? 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15254	https://www.khanacademy.org/science/organic-chemistry/organic-structures/acid-base-review/v/ka-and-pka-review	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
15255	https://books.google.com/books/about/Genetics.html?id=jxhH8sWRn_oC	Genetics - Daniel Hartl, Maryellen Ruvolo - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» ### Get Textbooks on Google Play Rent and save from the world's largest eBookstore. Read, highlight, and take notes, across web, tablet, and phone. Go to Google Play Now » My library My History Genetics ======== Daniel Hartl, Maryellen Ruvolo Jones & Bartlett Publishers, 2012 - Science - 800 pages Thoroughly revised and updated with the latest data from this every changing field, the Eighth Edition of Genetics: Analysis of Genes and Genomes provides a clear, balanced, and comprehensive introduction to genetics and genomics at the college level. Expanding upon the key elements that have made this text a success, Hartl has included updates throughout, as well as a new chapter dedicated to genetic evolution. He continues to treat transmission genetics, molecular genetics, and evolutionary genetics as fully integrated subjects and provide students with an unprecedented understanding of the basic process of gene transmission, mutation, expression, and regulation. New chapter openers include a new section highlighting scientific competencies, while end-of-chapter Guide to Problem-Solving sections demonstrate the concepts needed to efficiently solve problems and understand the reasoning behind the correct answer. More » Preview this book » Selected pages Title Page Table of Contents Index Contents Genes Genomes and Genetic Analysis 1 DNA Structure and Genetic Variation 39 Transmission Genetics The Principle of Segregation 79 Chromosomes and SexChromosome Inheritance 116 Genetic Linkage and Chromosome Mapping 154 Molecular Biology of DNA Replication and Recombination 194 Molecular Organization of Chromosomes 226 Human Karyotypes and Chromosome Behavior 256 Genetic Control of Development 480 Molecular Mechanisms of Mutation and DNA Repair 520 Molecular Genetics of the Cell Cycle and Cancer 563 Mitochondrial DNA and Extranuclear Inheritance 597 Molecular Evolution and Population Genetics 621 The Genetic Basis of Complex Traits 662 Human Evolutionary Genetics 694 Answers to EvenNumbered Problems 726 More Genetics of Bacteria and Their Viruses 302 Molecular Biology of Gene Expression 346 Molecular Mechanisms of Gene Regulation 387 Genomics Proteomics and Transgenics 439 Prefixes Suffixes and Combining Forms 745 Concise Dictionary of Genetics and Genomics 748 Index774 Copyright Less Other editions - View all ‹ › Common terms and phrases activityaffectedallele frequencyamino acidanalysisbacterialbandsbase pairsbindingcancercell cyclecentromereChapterchrochromatidschromochromosomecleavagecodingcodoncolicomplexcontainscopiescrosscrossovercyclincytoplasmdeletiondiploidDNA fragmentsDNA moleculeDNA sequenceDNA stranddouble-strandedDrosophiladuplexEcoRIembryoencodedeukaryotesexampleexpectedexpressionF₁femalesFIGUREfunctiongametesgenesgenetic mapgenomegenotypeheterozygoushomologoushomozygoushumanhybridincludesindividualinheritanceintronslineagemalemarkersmatingmeiosismitochondrialmolecularmosomemRNAmutationnormalnucleotideoperonorganismsparentalpathwaypedigreepercentphagephenotypeplantsplasmidpolymerasepolymorphismspolypeptide chainpopulationprimerPROBLEMproducedprogenyproteinrandomratiorecessive allelerecombinationregionrepressorrestriction enzymeresultribosomesegmentsegregationshownspeciessplicingstrainstructuresynthesistelomeretemplatetiontraittranscriptiontranslocationtRNAtypesundergowildtypeWw GgX chromosomeyeastyield About the author(2012) Daniel L. Hartl is Higgins Professor of Biology at Harvard University, a Professor of Immunology and Infectious Diseases at the Harvard T. H. Chan School of Public Health, and a Senior Associate Member of the Broad Institute of M.I.T. and Harvard. He is a member of the National Academy of Sciences and the American Academy of Arts and Sciences. He received his B.S. degree and Ph.D. from the University of Wisconsin and carried out postdoctoral research at the University of California at Berkeley. His research interests include molecular genetics, genomics, molecular evolution, and population genetics Bibliographic information Title Genetics AuthorsDaniel Hartl, Maryellen Ruvolo Edition illustrated Publisher Jones & Bartlett Publishers, 2012 ISBN 1449626106, 9781449626105 Length 800 pages SubjectsScience › Life Sciences › Biology Science / Life Sciences / Biology Science / Life Sciences / Genetics & Genomics Science / Life Sciences / Molecular Biology Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
15256	https://adebray.github.io/lecture_notes/108notes.pdf	MATH 108 NOTES: COMBINATORICS ARUN DEBRAY JUNE 11, 2013 These notes were taken in Stanford’s Math 108 class in Spring 2013, taught by Professor Kannan Soundararajan. I live-T EXed them using vim, and as such there may be typos; please send questions, comments, complaints, and corrections to a.debray@math.utexas.edu. Contents 1. Introduction to Graph Theory: 4/1/13 1 2. More Graph Theory: 4/3/13 3 3. Cayley’s Theorem: 4/8/13 5 4. Another Proof of Cayley’s Theorem: 4/10/13 7 5. Ramsey Theory: 4/15/13 8 6. More Ramsey Theory: 4/17/13 10 7. Tur´ an’s Theorem: 4/22/13 12 8. Coloring the Vertices of a Graph: 4/24/13 14 9. Probabilistic Constructions: The Legacy of Paul Erd¨ os: 4/25/13 15 10. Chromatic Polynomials: 4/29/13 17 11. Planar Graphs and Matchings: 5/1/13 18 12. The Gale-Shapley Theorem and Network Flow: 5/6/13 20 13. Network Flow II: 5/13/13 22 14. Network Flow III: 5/15/13 22 15. Sperner’s Theorem: 5/20/13 24 16. The Principle of Inclusion and Exclusion: 5/22/13 25 17. The Twelvefold Way: 5/29/13 26 18. Combinatorial Functions: 6/3/13 28 19. More Techniques for Asymptotics: 6/5/13 30 1. Introduction to Graph Theory: 4/1/13 This course will cover graph theory and enumerative combinatorics. Graph theory will be covered in the ﬁrst 9 or 10 chapters of the book (and perhaps some supplementary material) and enumerative combinatorics will be in chapters 14 and 15, as well as the ﬁrst chapter of Stanley’s book. If time permits, other topics will be discussed, such as coding theory or Hadamard matrices. Deﬁnition 1.1. A graph is an object consisting of vertices V and a set of edges E such that each edge e is a pair of vertices {x, y}, such that edge e is thought to connect edges x and y. Alternatively, one could think of sets V and E and a function φ : E →V × V . One says that x and y are on the edge e, or that they are incident on it. Note that x = y is allowed. 1 Example 1.2. If G1 = (V, E1) with V = {1, 2, 3, 4} and E1 = {{1, 3}, {3, 4}, {2, 4}, {1, 4}} and G2 = (V, E2) with E2 = E1 ∪{2, 2}, the graphs are G1 = 1 2 4 3 G2 = 1 2 4 3. ( Deﬁnition 1.3. A simple graph is a graph that contains no loops and no multiple edges, such as G1 in Example 1.2 above, but not G2. Deﬁnition 1.4. A ﬁnite graph G = (E, V ) is a graph in which both E and V are ﬁnite sets. Deﬁnition 1.5. If vi, vj ∈V are joined by an edge, they are called adjacent. Then, the adjacency matrix of a graph G is a matrix A of size \|V \| such that aij is equal to the number of edges joining vi and vj. If the graph is undirected, then the matrix is symmetric. In a simple graph, the adjacency matrix consists of only zeroes and ones. A symmetric matrix with real entries has nice properties, so the subject of spectral graph theory attempts to understand a graph by performing linear algebra on its adjacency matrix. Deﬁnition 1.6. The degree of a vertex is the number of edges coming out of it. Note that for loops on an undirected graph, each loops counts twice. The degree of a vertex is also referred to as its valency. Thus, deg(vi) = Pn j=1 aij (summing a column of the adjacency matrix), which does require the convention given above. Deﬁnition 1.7. A graph is called regular if all vertices have the same degree. If A is the adjacency matrix of a k-regular graph, then k is an eigenvalue, because A    1 . . . 1   =    k . . . k   . In particular, since the matrix is symmetric for an undirected graph, the eigenvalues are all real. Understanding the adjacency matrix is extremely helpful for applications such as random walks on graphs. Lemma 1.8 (Handshaking1). In any graph, the number of vertices of odd degree is even. Proof. It can be assumed that there are no loops, since removing the loops of a graph doesn’t change the parity of the degree of any vertex. Consider every ordered pair (x, y) which forms an edge {x, y}, so that (x, y) is counted if (y, x) is. Thus, there are 2\|E\| such pairs.2 However, this is also equal to the sum X x∈V  X {x,y}∈E 1  = X x∈V deg(x), which is also equal to the sum of the elements in the adjacency matrix. Thus, P x∈V deg(x) = 2\|E\|. Notice that throwing out the loops isn’t necessary to get this result. Thus, the number of vertices with odd degree must be even; otherwise, the sum would be odd. ⊠ The lemma owes its name to a scenario where a bunch of people at a party exchange some handshakes, and then count the sum of the number of handshakes that each person made. Some more examples of graphs: (1) The complete graph on n vertices, denoted Kn, is a simple graph on vertices V = {1, . . . , n} with E = {{i, j}, i, j ∈V } (i.e. all vertices are connected). For example, K3 = 1 2 3 K4 = 1 2 3 4. 1This is of course not to be confused with the Handwaving Lemma. 2This is a common trick in combinatorics — double-counting something, or counting in two diﬀerent ways, can show that two things must be equal. 2 Kn has n 2 = n(n −1)/2 edges. Thus, if G is any ﬁnite simple graph with \|V \| = n, then \|E\| ≤ n 2 , since there are no more edges to add to Kn. (2) A bipartite graph is a graph such that V = A ∪B, with A and B disjoint, such that the edges join vertices in A to vertices in B, and there are no edges between two elements of A or two elements of B. The complete bipartite graph Km,n, where \|A\| = m and \|B\| = n, has \|V \| = m + n and \|E\| = mn. (3) Using probability, one can construct some interesting classes of graphs, such as the Erd¨ os-Renye random graph: if \|V \| = n and there is some probability p ∈[0, 1], then for each vi, vj ∈V (considering vj, vi and vi, vj in the same trial), add an edge between them with probability p. This leads to questions such as how this sort of graph might look. The expected number of edges is \|E\| = p n 2 , since there are n 2 trials. One could examine triangles in a graph: when does a graph G contain vertices 1, 2, 3 and edges {1, 2}, {2, 3}, {1, 3}? Graphs without triangles certainly exist, such as bipartite graphs. Example 1.9. The complete bipartite graph Km,n has n2 edges and 2n vertices, so it has half of the “maximum” number of edges. Nonetheless, it doesn’t looks like a random graph (with p = .5) might look at all, since triangles would be expected for large n. More precisely, the number of possible triangles is \|V \| 3 , so the probability is p3 over \|V \| 3 trials. Thus, it is expected to happen fairly often. This can be generalized to an important topic of research: what kinds of structures tend to exist in large, seemingly random graphs? ( The idea of investigating graphs probabilistically is called the probabilistic method, due to Erd¨ os. Szemeredi also did a lot in this area (and won the Abel prize in 2012 for his work). Exercise 1.10. If G is a k-regular bipartite graph (which already implies that there aren’t any loops), then show that −k is an eigenvalue of the adjacency matrix. (Hint: try to guess an eigenvector, such as indexing 1 if xi ∈A and −1 if xi ∈B.) Deﬁnition 1.11. A walk is an alternating sequence of vertices and edges v1, e1, v2, e2, . . . , en−1, vn, such that edge ej connects vk and vk+1. This is exactly what you would think it is. Notice that a walk can visit an edge or vertex many times. Deﬁnition 1.12. A walk is called a trail if no edge is repeated in it.3 Example 1.13. v1, e1, v2, e2, v3, e3, v4, e4, v5, e5, v6, e5, v3, e2, v2, e1, v1 is a walk that isn’t a trail. ( Deﬁnition 1.14. A closed walk is (again) exactly what you might think it to be: a walk in which the starting and ending vertices are the same. One can similarly deﬁne a closed trail. Deﬁnition 1.15. A path is a walk in which all vertices are distinct.4 Deﬁnition 1.16. An Euler trail is a trail through a graph that traverses every edge exactly once. Not every graph has an Euler trail: look at K4. Theorem 1.17 (Euler). A connected graph has an Euler trail iﬀthere are at most two vertices with odd degree. Connectedness will be deﬁned in the next lecture, and its deﬁnition will not come as a surprise. The proof will also be given in the next lecture. Note also that by Lemma 1.8, the condition of at most 2 is equivalent to specifying either 0 or 2 such edges. 2. More Graph Theory: 4/3/13 Deﬁnition 2.1. A graph is connected if for any two vertices in the graph, there is a walk (equiv. path) from one to the other. Deﬁnition 2.2. Similarly to how one deﬁnes a closed walk and a closed trail, one has a closed path. This requires the ﬁrst and last vertices to be identical (which strictly implies that it isn’t a path). This is also a cycle. 3Note that the textbook calls this sort of walk a path. 4The textbook calls this a simple path. 3 Last lecture, we also saw Euler trails. A closed Euler trail is also called an Euler tour. Euler was interested in a question about the bridges of K¨ onigsberg: speciﬁcally, if the graph created from the bridges and islands admitted an Euler tour. By Theorem 1.17, this isn’t possible. So let’s prove the theorem. Figure 1. Schematic of the Bridges of K¨ onigsberg. Source Proof of Theorem 1.17. Let G be a connected graph. If it has an Euler tour, then every edge going into a vertex must come out of it, so the degree of every vertex is even. If it only has an Euler trail, then the starting and ending point don’t necessarily meet this, so they might have odd degree, but that makes for at most 2 such edges. If G has only edges with even degree, start at an arbitrary point and build a trail by choosing distinct edges. Then, the trail won’t get stuck anywhere, but it might not get all of the edges. Thus, remove the edges contained in this trail, leaving a smaller graph. This graph isn’t necessarily connected, but it has several connected components, each of which has vertices of only even degree. By induction, we can assume that each of these components has an Euler tour, since they all have fewer edges than G.5 Then, an Euler tour can be made by joining the edges in the original trail with each of these tours at some vertex. If G has instead two vertices of odd degree, take the graph G′ to be G with a new edge e between them. Then, G′ has an Eulerian tour by above, so taking that tour and removing e gives an Eulerian trail on G. ⊠ Deﬁnition 2.3. A tree is a connected graph that has no cycles. These are called trees because many of them can be made to look like trees with trunks and branches. A tree with n vertices must have n −1 edges, which can be proven straightforwardly by induction. Additionally, any two vertices are connected by a unique path: a tree is connected, so such a path exists, but if there were two distinct paths p1 and p2, one could obtain a cycle from them, by taking p1 ∪p2 \ (p1 ∩p2). If one adds an edge to a tree without adding any vertices, then it connects two vertices v1 and v2. However, there was already a path between these vertices and this edge gives another path, so the result is not a tree. If one removes a path from a tree, the result is disconnected. Thus, if one removes an edge, it splits into two trees, so the formula for the number of edges (i.e. \|V \| −1) follows. For any connected graph, deﬁne the distance between two edges x and y to be d(x, y), the smallest number of edges in any path between x and y. This can be used to show that trees are bipartite: take some vertex v and let A be the set of vertices of even distance from v, and B be the set of vertices of odd distance from v. Then, having edges from A to be is possible, but there will never be an edge from A to A, or B to B, since this would cause a nontrivial cycle (since it’s of odd length). This is a special case of a more general theorem: Theorem 2.4. A graph that has no cycles of odd length is bipartite. This can be proven by a modiﬁcation of the above argument, as well as reasoning that a bipartite graph cannot have cycles of odd length by trying to sort the vertices in the cycle into sets. 5This would need to be formalized for the graph with one edge and then making an inductive assumption, but it’s just as valid. 4 Deﬁnition 2.5. An isomorphism of graphs G21 = (V1, E1) and G2 = (V2, E2) called f : G1 →G2 is a bijection V1 →V2 that preserves edges. For example, one can use an isomorphism to make K4 planar, by moving one of the edges around the graph. There’s a website called planarity.net, which allows one to try and make a graph planar. Deﬁnition 2.6. An automorphism of a graph is an isomorphism with itself. One can try to enumerate all trees of a given size. For example, the trees with 3 edges aren’t very interesting; they’re just lines of two edges. Thus, one might consider labelled trees, where the vertices are chosen among {1, . . . , n} and the order of the labelling matters. There aren’t 6 labelled trees with 3 vertices, though, because 1 −2 −3 is isomorphic to 3 −2 −1. Thus, there are 3 labelled trees with 3 vertices. Theorem 2.7 (Cayley). There are nn−2 trees with n labelled vertices. The proof of this theorem will require the following lemma: Lemma 2.8. Every tree has at least two vertices of degree 1. Proof. Let di be the degree of vertex i; then, Pn i=1 di = 2\|E\| = 2n −2. Since di ≥1, then at most n −2 have degree at least 2, or else one vertex would have to have degree 0. ⊠ 3. Cayley’s Theorem: 4/8/13 Returning to the problem of labelled trees, which is a diﬀerent condition than isomorphism, we will prove Theorem 2.7. Deﬁnition 3.1. A subgraph H of a graph G is a subset of the vertices and edges of the graph: H = (V (H), E(H)), such that the edges in E(H) only connect the vertices in V (H), with E(H) ⊆E(G) and V (H) ⊆V (G). This is exactly what you would think it would be. Deﬁnition 3.2. A subgraph of G is called spanning if it contains all of the vertices of G. The most interesting spanning subgraphs are those whose edge sets are a proper subset of the edge set of G. Then, Cayley’s Theorem can be reformulated in terms of the number of spanning trees contained in Kn. First, it is useful to observe that every connected graph has a spanning tree. There are many ways to do this; for example, one could order the edges somehow and add the edges in order, adding only the edges that don’t produce a cycle. Alternatively, you could add vertices in one per step; since the graph is connected, there is always another vertex to add (until all of them have been used). Claim. If a connected graph has n vertices and n −1 edges, then it is a tree. Proof. Using Lemma 1.8, there is a vertex with degree 1. Then, remove it, and apply induction. ⊠ There are plenty of other ways to prove this. Now, it will be possible to prove Cayley’s Theorem. Two proofs will be given: Proof of Theorem 2.7. Deﬁnition 3.3. A Pr¨ ufer code is a sequence of n −2 numbers [y1, . . . , yn−2] with 1 ≤y1, . . . , yn−2 ≤n. Thus, there are nn−2 distinct Pr¨ ufer codes, and the goal is to show that there is a bijective correspondence between labeled trees and Pr¨ ufer codes. Generate a Pr¨ ufer code from a tree in the following way: of all of the vertices with degree 1, let x1 be the vertex with the smallest value. Then, let y1 be the unique vertex to which x1 is connected. Then, remove the edge connecting them, and repeat; the next entry in the Pr¨ ufer code is the next yi found by this algorithm. Then, continue until n −2 numbers are found. Claim. If [y1, . . . , yn−2] is the Pr¨ ufer code for a tree, then yn−1 = n. Proof. A vertex will be removed from the game if it is the least vertex in the tree, which will never be true for vertex n, and at the last step, there are only two vertices, n and a for some a < n, which are connected, so yn−1 is whatever is connected to a, which is n. ⊠ 5 3 1 2 9 5 4 6 7 8 Figure 2. This tree’s Pr¨ ufer code is [1, 2, 2, 1, 9, 9, 9]. This is why the Pr¨ ufer code is written as an (n −2)-tuple, rather than an (n −1)-tuple; if it were of size (n −1), the last entry wouldn’t have any meaning. Now, given a Pr¨ ufer code, it is possible to obtain a tree. First observe that each vertex v of the tree will appear in its Pr¨ ufer code exactly deg(v) −1 times, since it is removed once (in which case it isn’t added to the Pr¨ ufer code), but the remaining deg(v) −1 times the other vertex on the considered edge is removed, and v is added to the code. Thus, given a Pr¨ ufer code, such as [1, 2, 2, 1, 9, 9, 9] as in Figure 2, one can reconstruct the tree according to the following algorithm: • First, place vertex y1 on the graph and add the smallest vertex with degree 0, connecting them. • Remove y1 from the Pr¨ ufer code and repeat, updating the degrees left to add based on the new Pr¨ ufer code. Then, repeat these steps. Though this seems a little fuzzy, this is a bijective correspondence between a tree and its Pr¨ ufer code, since each one can be used to obtain the other, and a more rigorous proof of this can be given by induction. Then, there are clearly nn−2 Pr¨ ufer codes, so there are nn−2 labelled trees. ⊠ The binomial coeﬃcient is the number of ways to choose k elements out of an n-element set, n k = n!/(k!(n −k)!). This obeys the recurrence n k = n−1 k−1 + n−1 k , which can be seen in Pascal’s triangle: pick a set S of k elements and a favorite element t; either t ∈S, in which case you get the ﬁrst term, or t ̸∈S, which gives the second term. Additionally, there is the following theorem: Theorem 3.4 (Binomial6). (x + y)n = n X k=0 n k xky n−k. There are several ways to prove it: induction on n is straightforward, but one can also directly observe that when multiplying out (x + y) and looking for a particular coeﬃcient, one chooses k xs and n −k ys. The binomial coeﬃcient can be generalized to the multinomial coeﬃcient: ℓ X i=1 xi !n = X k1+···+kℓ=n n k1, . . . , kℓ ℓ Y i=1 xki i . (3.5) Here, n k1,...,kℓ = n!/(k1! · · · kℓ!) is called the multinomial coeﬃcient. This can be thought of as assigning n people to ℓteams of sizes k1, . . . , kℓ, so the recursive formulation is n k1, . . . , kℓ = ℓ X j=1 n −1 k1, . . . , kj−1, kj −1, kj+1, . . . , kℓ . (3.6) A complete deﬁnition does also require that n k1,...,kℓ = 0 if some kj < 0. The multinomial coeﬃcient reduces the the binomial coeﬃcient of n with k and n −k. Then, (3.5) will be used in a second proof of Cayley’s Theorem to show that (1 + · · · + 1)n−2 = X k1,...,kn≥0 k1+···+kn=(n−2) n −2 k1, . . . , kn . (3.7) 6“About the Binomial Theorem I am teeming with a lot o’ news. . . ” 6 4. Another Proof of Cayley’s Theorem: 4/10/13 Another proof of Cayley’s Theorem can be given by enumerating all trees with given properties. This will be an example of enumerative combinatorics. Second proof of Theorem 2.7. Let t(n; d1, . . . , dn) be the number of trees on n vertices such that vertex i has degree di. Then, all of the di ≥1 and Pn i=1 di = 2n −2. Thus, Pn i=1(di −1) = n −2, since n terms have been taken away. Claim. t(n; d1, . . . , dn) = n −2 d1 −1, . . . , dn −1 . Supposing this claim, the total number of trees is X d1,...,dn t(n; d1, . . . , dn) = X d1,...,dn n −2 d1 −1, . . . , dn −1 = X k1,...,kn≥0 P ki=n−2 n −2 k1, . . . , kn = (1 + · · · + 1 \| {z } n times )n−2 = nn−2 by (3.7). Proof of the claim. It’s probably worth checking this claim for n = 2 or n = 3 to build intuition, since these cases are pretty simple. The general case will be given by a recursive formula that is similar to the one given for multinomial coeﬃcients in (3.6). Of the n vertices, some vertex has degree 1: di = 1. Then, vertex i can be connected to any of the vertices 1, . . . , i −1, i + 1, . . . , n. If it’s connected to vertex 1, the number of options is t(n −1; d1 −1, d2, d3, . . . , di−1, di+1, . . . , dn) = n −3 d1 −2, d2 −1, . . . , di−1 −1, di+1 −1, . . . , dn −1 by the inductive assumption,7 so one can take the sum over all such vertices t(n; d1, . . . , dn) = n X i=1 n −3 d1 −2, d2 −1, . . . , di−1 −1, di+1 −1, . . . , dn −1 = n −2 d1 −1, . . . , dn −1 by the multinomial theorem. ⊠ Then, the theorem follows as above. ⊠ Here are two applications of this: (1) Take a connected graph and make it a weighted graph, or a graph in which every edge e has a weight w(e) ≥0. These can be thought of as costs of traversal, or distances between the points, or travel times, etc. Then, if one starts at some vertex A, what is the shortest path to some other vertex? Here, “shortest” means the most economical: if a path uses edges e1, . . . , ek, then its weight is w(e1) + · · · + w(ek), and the goal is to ﬁnd the path that minimizes the sum of the weights of the edges. There is an algorithm for this, which produces a spanning tree for G such that the path on this tree connecting A to any other point on the tree is the shortest. Algorithm 4.1 (Dijkstra). 7Alternatively, t(n; d1, . . . , dn) can be thought of as a symmetric function of the degree. Thus, it can be assumed they are in order: d1 ≥d2 ≥· · · ≥dn, so dn = 1. Then, repeat the argument as above: vertex n is connected to some other vertex, so t(n, d1, . . . , dn) = n−1 X i=1 t(n −1; d1, . . . , di−1, di −1, di+1, . . . , dn), since removing that vertex and edge leaves behind a tree with the given degrees of its vertices. 7 • Start at A, or letting S = {A}. • At every stage, there is a set S of vertices such that the shortest path from A to these vertices is known. Here, ﬁnd a v ∈V \ S such that the distance from A to some vertex u ∈S plus the distance from u to v is minimized, over all u ∈S adjacent to v, over all v ∈V \ S. Since G is connected, there will always be such v and u, so this algorithm is correct. Of course, there’s a bit more to think about, but this is not so bad. If the weights aren’t assumed to be nonnegative, there is no unique shortest path, so the problem wouldn’t be well-formed, and the problem wouldn’t have as much physical meaning. (2) Given some weighted, connected graph G, one problem is to ﬁnd a minimal spanning tree for G: to ﬁnd a spanning tree T such that P e∈T w(e) is minimum; such a tree is called a minimal spanning tree. This has applications such as laying pipe eﬃciently in order to cover every city in an area. In the previous lecture, it was shown that every connected graph has a spanning tree, so there must be a minimal spanning tree as well. Kruskal’s algorithm is the standard solution here, but there are others, such as reverse-delete. They are all of the same family, referred to as greedy algorithms, and do the following: at each step, ﬁnd the cheapest edge that hasn’t been used and doesn’t form a cycle, and add it to the tree, or from G remove the most expensive edge that doesn’t disconnect G.8 Another solution is Prim’s Algorithm. This builds up a sequence of vertices S such that for each step, one chooses a vertex from V \ S, and connected it in the cheapest way to a vertex in S. This diﬀers from Kruskal’s algorithm in that it requires the tree to always be connected as it is built, whereas Kruskal’s doesn’t. Claim. Prim’s Algorithm produces an MST.9 5. Ramsey Theory: 4/15/13 Ramsey theory can be summarized as: in any suﬃciently large system, total disorder is not possible. For example, any suﬃciently large graph will have some sort of pattern in it. Color the complete graph Kn with two colors, red and blue. Is it always possible to ﬁnd a completely blue Kp or a red Kq for some given p, q? Theorem 5.1 (Ramsey). If n is suﬃciently large (in terms of p and q), then there always exists either a completely red Kp or a completely blue Kq. Example 5.2. Take p = 2. This asks if there is a blue edge in the graph. If there is, that’s okay, but if not, then the entire graph is red, and all edges are red, so a red Kq can be found if n ≥q. ( Let R(p, q) be the smallest value of n such that the theorem is true. Then, it was just shown that R(2, q) = q. Example 5.3. p = 3 and q = 3 is the basic example. The question thus becomes about ﬁnding a red triangle or a blue triangle. Claim. R(3, 3) = 6. Proof. First, it will be necessary to show that ﬁve vertices aren’t enough. Take the complete graph K5 and color some pentagon red and every other edge blue, as in Figure 3. Additionally, six vertices are enough, which will be shown by the pidgeonhole principle (like most things in this theory). Choose some vertex: at least three of the vertices coming out of it are the same color (without loss of generality, red). Then, look at the edges between the vertices those edges connect to: if any of these edges is red, there is a red triangle. But if none of them are, then they form a blue triangle. ⊠ ( Another easy argument is that R(p, q) = R(q, p) for reasons of symmetry. Now, the general theorem can be tackled: 8This seems like it would be more expensive in terms of running time, but this is a math class, so nobody cares. 9There could be many such MSTs, though if the weights are distinct, then there is exactly one. 8 Figure 3. A coloring of K5 which contains no triangles. Source. Proof of Theorem 5.1. Proceed by induction on p + q, and bound the number of vertices that are needed. Assume the theorem is false for n = p + q. Pick some vertex v, which is connected to n −1 vertices. Of these edges, suppose b of them are blue and r are red. The b blue edges connect to b vertices forming a complete graph Kb. This can’t contain any blue Kp−1 or red Kq, since these would prove the theorem. Similarly, looking at the r vertices connected to v by red edges, they cannot contain a red Kq−1 or a blue Kp for the same reasons. Thus, b ≤R(p −1, q) −1 and r ≤R(p, q −1) −1. Thus, n = b + r + 1 ≤R(p −1, q) + R(p, q −1) −1. In other words, this is an upper bound for R(p, q): R(p, q) ≤R(p −1, q) + R(p, q −1). ⊠ This proof looks locally at a graph, since nobody knows how to look globally at a graph to make these sorts of proofs. Notice that this recurrence looks just like the one from the Binomial Theorem, and thus: Corollary 5.4. R(p, q) ≤ p+q−2 p−1 . Proof. R(2, q) = q 1 = q and R(3, 3) ≤R(2, 3) + R(3, 2) = 6, so by induction, R(p, q) ≤R(p −1, q) + R(p, q −1) ≤ p + q −3 p −2 + p + q −3 p −1 = p + q −2 p −1 . ⊠ Beyond these bounds, very little is known about R(p, q) for general p and q: • R(3, k) is known exactly for k ≤9. For example, R(3, 9) = 36. • R(4, 4) = 18, and R(4, 5) = 25. That’s it, though lower and upper bounds for other values are known. “Imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, we should marshal all our computers and all our mathematicians and attempt to ﬁnd the value. But suppose, instead, that they ask for R(6, 6). In that case, we should attempt to destroy the aliens.” –Paul Erd¨ os These are both ﬁnite problems, just incredible large ones. But a brute-force calculation of all two-colored graphs of K100 would require 2( 100 2 ) calculations, which is impossible. Looking at the diagonal Ramsey numbers (p = q), the upper bound can be given as R(p, p) ≤ 2p−2 p−1 ≤2p−2.10 For lower bounds, there is a probabilistic method due to Erd¨ os, which manages to provide a proof without an example. In this ﬁnite case, though, probability is just counting. The goal is to ﬁnd the largest possible n for which there exists some coloring with no red Kp or blue Kp. Do this by summing over all possible 2-colorings of edges of Kn: X 2-colorngs of Kn (# blue Kp + # red Kp). (5.5) 10With a little more calculation, a better bound can be obtained. 9 The goal is to show this is strictly less than the number of possible two-colorings of Kn, which is 2( n 2), since this would imply there is some coloring for which there are no such monochromatic subgraphs Kp. This bound can be found by double-counting, calculating the sum in two ways to obtain two diﬀerent kinds of information. For a given red or blue Kp, count the number of 2-colorings of Kn in which it is monochromatic. Choose p vertices from the n, which can be done in n p ways, and pick either red or blue. Then, there are 2( n 2)−( p 2) ways to color the rest of the graph, so the sum (5.5) is equal to 2 2 p 2( n 2)−( p 2). Thus, we want 2 n p 2( n 2)−( p 2) < 2( n 2). Rearranging, this is n p < 2( p 2)−1. Since n p = n(n−1)···(n−p+1) p! ≤np/p!, it is suﬃcient to check if np/p! ≤2( p 2)−1, or that n < 2p(p−1)/2−1p! 1 p . The factorial term dominates, so it’s suﬃcient to have n < 2p2/2+p/2−21/p . It’s possible to do better with more care, again, but it’s enough to have n < 2p/2. Thus, we have the following theorem: Theorem 5.6. There exists a coloring of K⌊2p/2⌋with no monochromatic Kp. In other words, R(p, p) ≥2p/2. Thus, the bounds are ( √ 2)p ≤R(p, p) ≤4p, so is R(p, p) ≈cp for some c? Maybe. Nobody knows. A big open problem is to improve these bounds; even changing the 4 to a 3.999 would be a major improvement. Using Stirling’s Formula, which approximates n! ≈ √ 2πn(n/e)n, the actual bound for large numbers is something like R(p, p) ≥p2p/2/(e √ 2). If you’re willing to work a lot harder, there’s a better result (Spencer 1980) that gives a factor of two, which is the best known: R(p, p) ≥p √ 22p/2/e. The best known upper bound is due to Conlon in 2011, and shows that R(p, p) ≤4p/pA for any A for suﬃciently large p. Nonetheless, this is still nowhere near 3.999p. 6. More Ramsey Theory: 4/17/13 Recall that Ramsey’s Theorem stated that if one chooses p and q, then there exists a complete graph on n vertices for some n such that if every vertex of Kn is colored either red or blue, then there exists either an entirely red Kp or entirely blue Kq as a subgraph of Kn. The minimal such n is called R(p, q); for example, R(2, q) = q and R(3, 3) = 6. We showed an upper bound R(p, q) ≤ p+1−2 p−1 , and a lower bound was obtained by the probabilistic method: R(p, p) ≥2p/2. There are plenty of related questions and generalizations of Ramsey’s Theorem: Theorem 6.1. For given p1, . . . , pr, then if n is suﬃciently large, then every r-coloring of Kn will contain some Kpi that is monochromatic of color i. The smallest n for which this is true can be denoted R(p1, . . . , pi). Proof of Theorem 6.1. The proof will not be much diﬀerent than the proof of Theorem 5.1: suppose Kn is r-colored but there is no Kpi of color i, where the proof uses induction on n. Pick some vertex v. Then, there are strictly less than R(p1 −1, p2 . . . , pr) vertices connected to v such that the connecting edge has color 1, since that would satisfy the theorem (since if there were p −1 of them of color 1, then there would be a monochromatic Kp1 created by adjoining v). Similarly, the number of edges of color i must be R(p1, . . . , pi−1, pi −1, pi+1, . . . , pr). Thus, summing up, n ≤1 −r + r X i=1 (p1, . . . , pi−1, pi −1, pi+1, . . . , pr), so adding one more vertex is suﬃcient to prove the theorem. ⊠ Thus, there is an upper bound on this generalized Ramsey number: R(p1, . . . , pn) ≤(2 −r) + r X i=1 (p1, . . . , pi−1, pi −1, pi+1, . . . , pr). Here, 2 −r is nonpositive, so it can be ignored if one wishes. 10 Exercise 6.2. Prove a generalization of Corollary 5.4: that R(p1, . . . , pn) ≤ (p1 −1) + · · · + (pr −1) p1 −1, . . . , pr −1 . Consider R(3, 3, 3): if one takes a vertex v, then there must be at most 5 vertices connected to v by an edge of color 1, and similarly for colors 2 and 3, or there would be a triangle because R(3, 3) = 6. Thus, R(3, 3, 3) ≤17, and it’s known that R(3, 3, 3) = 17 as well. Philosophically, in a suﬃciently large set of data, there will be small patterns. This is why there are constellations that look like something in the stars, or something which might look like a secret message if one takes some sequence of letters in a book. In the last thirty years, there has been a lot of activity in the ﬁeld of Ramsey theory on natural numbers. Here, the goal is to color N with r colors. Theorem 6.3 (Schur). Given r, if N is suﬃciently large and [1, N] is colored with r colors, then there is a monochromatic solution to x + y = z. Proof. Color KN and color the edge (i, j) between vertices i and j with the color of \|i −j\| on [1, N]. Then, there is a monochromatic triangle if N is suﬃciently large. Label its vertices a, b, and c in descending order. Thus, a −b, b −c, and a −c are monochromatic, so the equation a −c = (a −b) + (b −c) is monochromatic. Let z = a −c, x = a −b, and y = b −c to explicitly give the equation. ⊠ There are a lot of complicated versions of this result. One of the more beautiful examples: Theorem 6.4 (Van der Waerden). In any r-coloring of the integers, there exist arbitrarly long monochromatic arithmetic progressions (i.e. sequences of the form a, a + d, a + 2d, . . . , a + (k + 1)d). There is a ﬁnite version of this theorem, as with Schur’s theorem: there is some number N(r, k) such that an arithmetic progression of length k always exists if [1, N(r, k)] is r-colored. Theorem 6.5 (Gowers). An upper bound for N(r, k) is N(n, k) ≤22r22k+9 . Gowers got a Fields medal in part because of this theorem. It leads to lots of useful things, such as the theorem that the primes contain arbitrarily long arithmetic progressions. Another example involves playing Tic-Tac-Toe in high dimensions. Suppose one has a d-dimensional board of size k × k × · · · × k. Denote the set {1, . . . , k}d as k-tuples of coordinates (k1, . . . , kd). Then, deﬁne a combinatorial line to be a function of these k-tuples where each coordinate is either constant or varies with the same value (e.g. (1, 1, ∗, ∗, 3, 5), where ∗is always the same in both components; ∗can be thought of as a wildcard). This sort of line is more restrictive than the lines allowed in Tic-Tac-Toe, and is sometimes known as a Hales-Jewett line. Theorem 6.6 (Hales-Jewett). If d is large enough (depending on k and r)and {1, . . . , k}d is r-colored, then there exists a monochromatic combinatorial line. This means that for suﬃciently large board, there is always a winner in Tic-Tac-Toe of many dimensions. Note that this is explicitly not true when r = 2 and d = 2. A statement called the density version of this theorem also deals with how much of {1, . . . , k}d is colored before the statement holds. It is possible to restate Ramsey’s theorem in a diﬀerent way: for any graph G on n vertices, one can color Kn by taking an edge to be red if it’s in G, and blue if it’s not. Then, Deﬁnition 6.7. An independent set on a simple graph G is a set of vertices such that no edge connects any of the vertices. A clique is the opposite: it is a set of vertices where all possible edges between them exist. Thus, Ramsey’s theorem states that if n ≥R(p, q), then any simple graph G on n vertices either contains p independent vertices or a q-clique. Thus, one can ask how many edges a simple graph on n vertices contains given that it has no triangles. For example, one can have a bipartite graph, with the two sets of sizes a and (n −a). Thus, one can obtain a graph 11 of a(n −a) vertices without triangles, which can be maximized as about a2/4, when a ≈n/2. Speciﬁcally, one has ⌊n/2⌋(n −⌊n/2⌋) edges, which is almost half as many as in a complete graph. However, this is a highly nonrandom graph, since it’s so structured, so one could have some theory as to whether a graph looks random or looks structured, or whether these patterns are true in random graphs as well. Theorem 6.8 (Tur´ an). If G is a simple graph that has more than ⌊n/2⌋(n −⌊n/2⌋) edges, then G contains a triangle. Proof. Let K(n) be the maximum example of edges that a simple, triangle-free graph can have. A bipartite graph gives a lower bound of K(n) ≥⌊n/2⌋(n −⌊n/2⌋) edges. For some small values: K(1) = 1, K(2) = 1, K(3) = 2, etc. By induction, an upper bound can be discovered. Choose an edge e = {1, 2} and remove it and its vertices. Then, there are at most K(n−2) edges. Then, back in the original graph G, each vertex can be connected to at most on of vertices 1 or 2, so K(n) ≤K(n−2)+n−1. Applying this repeatedly, one can check by induction that K(n) ≤⌊n/2⌋(n −⌊n/2⌋). ⊠ It’s actually possible to prove a stronger result, which is that the only graph with K(n) edges is the bipartite one mentioned above. 7. Tur´ an’s Theorem: 4/22/13 Recall Tur´ an’s theorem from the previous lecture: it shows that if a graph has no triangles, then it can have the maximum number of edges if it is a complete bipartite graph. More generally, let G be a simple graph on n vertices, such that G contains no Kp. Let M(n, p) denote the maximum number of edges that G can have. Then, if G has M(n, p) edges, then adding an edge to G means it contains a Kp, so G must contain a Kp−1, and there are n −(p −1) vertices not in this Kp−1. Thus, the maximum number of edges such that G doesn’t contain a Kp can be found: among the n −(p −1) vertices, there must be at most M(n −(p −1), p) edges to ensure there is no Kp. There are p−1 2 edges connecting the vertices in the Kp, and in order to avoid a Kp coming from the Kp−1, the number of vertices joining the two parts must be at most (p −2)(n −(p −1)), since for each vertex in n −(p −1), there must be at most p −2 connections to the Kp−1. Thus, one obtains the bound of M(n, p) ≤M(n −(p −1), p) + p −1 2 + (p −2)(n −(p −1)). If n ≤p −1, then M(n, p) = n 2 , since there aren’t enough vertices to have a Kp. Thus, write n = t(p −1) + r, where t is the quotient and r < p −1 is the remainder. Thus, M(n, p) ≤M((t −1)(p −1) + r, p) + p −1 2 + (p −2)((t −1)(p −1) + r). Exercise 7.1. Check that M(t(p −1) + r, p) ≤(p −2)r 2 2(p −1) −r(p −1 −r) 2(p −1) . In fact, this bound is optimal, which can be shown using a (p −1)-partite graph: divide the vertex set “evenly” into p −1 sets V1, . . . , Vp−1, such that V1, . . . , Vr each contain t + 1 points, and Vr+1, . . . , Vp−1 each contain t points. Draw edges from any v ∈Vi to all vertices not in Vi. This doesn’t contain a Kp−1, since that would imply there are connections within one of the Vi, and one can check that the number of edges matches the upper bound. Of course, there’s a variant on the problem: if instead of 3-cycles on considers 4-cycles, how many edges can such a graph have? This recalls the notion of girth from the ﬁrst week, which is the length of the smallest cycle in a graph. If a graph is triangle-free and quadrilateral-free, then its girth is at least 5. Theorem 7.2. If G is simple and has no 3- or 4-cycles, then G has at most O(n3/2) edges. This is substantially smaller than the triangle case, and as the girth increases, the exponent on n decreases. 12 Proof of Theorem 7.2. Proof by double-counting: pick a vertex u and look at the set {{v, w} \| uv ∈E, uw ∈E}. If d(u) = deg u, then the size of this set is d(u) 2 = d(u)(d(u) −1)/2. Then, for any edge {v, w} there is at most one u that they come from, or else there would be a 4-cycle. Thus, the number of “tents” of edges v −u −w is P u∈V d(u) 2 ≤ n 2 by above, so by the Handshaking Lemma, so X u∈V d(u)2 −d(u) 2 = 1 2 X u∈V d(u)2 −\|E\|. This requires a little bit of analysis; speciﬁcally, take the Cauchy-Schwarz Inequality n X i=1 xiyi 2 ≤ n X i=1 x2 i ! n X i=1 y 2 i ! . Then, 4\|E\|2 = X u∈V d(u) · 1 !2 ≤ X u∈V 1 ! X u∈V d(u)2 ! = n X u∈V d(u)2 ! . Thus, 1/2 P d(u)2 ≥2\|E\|2/n, so 2\|E\|2 n −\|E\| ≤n(n −1) 2 = ⇒4\|E\|2 −2n\|E\| ≤n2(n −1). Intuitively, the use of Cauchy-Schwarz is due to considering the most evenly distributed case, or a pidgeonhole argument. But now, the quadratic formula can be used to ﬁnd the maximum number of edges, giving \|E\| ≤2n + p 4n2 + 16n2(n −1) 8 = n + n√4n −3 4 . ⊠ This also gives a more concrete upper bound: G must have at most (n(1 + n√4n −3)/4 edges. This is not a tight bound, but no tight bound is known. In some sense, the Cauchy-Schwarz Inequality is a statement that the largest number of edges occurs when they are evenly distributed. Deﬁnition 7.3. If G is a simple graph, a Hamiltonian circuit on G is a closed path that travels through each vertex in the graph exactly once. This is diﬀerent from the Eulerian circuit, which travels every edge exactly once. Theorem 7.4 (G. Dirac11). Suppose that G is a simple graph on n vertices and each vertex has degree at least n/2. Then, G contains a Hamiltonian cycle. Proof. Take the maximal counterexample (i.e. one such that if any edge is added, then there is a Hamiltonian cycle). Thus, G doesn’t contain any edge uv such that G ∪uv contains a Hamiltonian cycle (that contains uv). Suppose u = u1 is connected to ui+1 and v = un is connected to ui, where u1, . . . , un is the cycle. Then, there exist such i such that u = u1 is connected to at least n/2 vertices ui+1 and v = un is connected to at least n/2 vertices ui by the pidgeonhole principle. ⊠ Deﬁnition 7.5. The chromatic number of a graph G is the minimum number of colors necessary to color the vertices of G such that no two adjacent vertices share the same color. To think about the deﬁnition, here’s a nice toy unsolved problem: color all of the points on the plane with χ colors. Are there exactly two points exactly distance 1 apart that have the same color? Somewhere between χ = 4 and χ = 7, inclusive, there aren’t. For χ = 3 with colors R, B, and W, take some equilateral triangle and color each vertex a diﬀerent color. Then, take the reﬂection of that triangle through the line between the red and blue vertices; the remaining vertex must be white. Thus, all points in a circle of radius √ 3 must be white, and take any two points on that circle which are one inch apart, so any 3-coloring of the plane has points of the same color of distance 1 apart. 11Not the famous Dirac. 13 8. Coloring the Vertices of a Graph: 4/24/13 A lot of the problems in this class can be diﬃcult: they tend to require some sort of clever trick. There are various ways to think about this, but one good one is due to P´ olya: if you can’t solve, then there is a smaller problem you can’t solve. Solving the smaller problem (or applying the same heuristic) will give insight into the larger one. A big question in vertex coloring is: given a graph G, color the vertices such that no adjacent vertices have the same color. How many colors are necessary? As discussed in the previous lecture, take the set of points in the plane; then, if they are 3-colored, then there always exist 2 points of distance 1 apart with the same color. Thus, at least four colors are necessary to ensure there are no two points of distance 1 with the same color, but not much more than that is known. However, it is possible to prove that an upper bound exists? Imagine tiling the plane with squares with diagonal distance slightly smaller than 1. Take each 3 × 3 set of these squares and color each one with a diﬀerent color, and then repeat this. Thus, any two points with the same color are either within the same square (so they have distance less than 1) or diﬀerent squares of more than distance 1 away. Thus, the problem has an upper bound of 9, and there’s a more clever way of using hexagons to show that 7 colors is also suﬃcient. Thus, the answer to the problem is one of 4, 5, 6, or 7, but nobody knows which. Deﬁnition 8.1. The chromatic number χ(G) of a ﬁnite graph G is the smallest number r such that the vertices of G can be r-colored without two adjacent vertices being the same color. It can be assumed that G is simple: the existence of loops would cause issues with the statement of the problem, so they won’t be considered, and any additional edges don’t change the adjacency of the problem, and therefore don’t change the chromatic number. χ(Kn) = n, since all pairs of vertices are adjacent. Thus, for any G with n vertices, χ(G) ≤n, since each vertex can be given a diﬀerent color. If χ(G) = 1, then G has no edges. If χ(G) = 2, then let V1 be the set of vertices of color 1 and V2 be the set of vertices of color 2. Then, there are no edges in V1 or in V2, so G is bipartite. More generally, if χ(G) = r, then the set V of vertices of G can be written as V = Sr i=1 Vi, where the Vi are disjoint, and there are no vertices entirely within any given Vi. In general, if there are more edges, the chromatic number increases. Thanks to Tur´ an’s theorem, there exist triangle-free graphs with arbitrarily large chromatic number. Is it possible to do better than this? Proposition 8.2. Suppose ∆= maxv∈V deg v. Then, χ(G) ≤∆+ 1. Proof. Use a greedy algorithm: build up the graph vertex-by-vertex. When any particular vertex is added, it must be connected to at most ∆vertices, so it can be given a color that is distinct from the color of each of these vertices, since there are ∆+ 1 colors. ⊠ In general, this is only a modest strengthening of the theorem, and it’s possible to do a bit better. Theorem 8.3 (Brooks). If G is a connected simple graph, then χ(G) ≤∆except in two cases: • G = K∆+1, or • ∆= 2 and G is an odd-length cycle. It’s also possible to state a lower bound. Deﬁnition 8.4. Deﬁne the clique number of G to be the largest ℓsuch that G contains an ℓ-clique (i.e. a Kℓ). All of the vertices in a clique must be colored diﬀerent colors, so χ is at least the clique number. Additionally, if G can be split into independent sets V1, . . . , Vr, then there is some Vj such that \|Vj\| > n/r. Thus, another lower bound is found: χ(G) > n/\|Vj\|. This may be better or worse than the previous lower bound. Combining, χ(G) ≥max(ℓ, n/\|Vj\|). Deﬁnition 8.5. The chromatic function χ(G, k) is the number of diﬀerent ways in which one can k-color the graph G. 14 In this scheme, two colorings are the same if each vertex has the same color in the colorings. For example, a triangle can be k-colored in k(k −1)(k −2) ways. In general, χ(G, k) = 0 for 0 ≤k < χ(G). If G = Kn, then there are k ways to color the ﬁrst vertex, k −1 ways to color the second, etc. Thus, χ(Kn, k) = k n , which also neatly implies that χ(Kn, k) = 0 when 0 ≤k < n. Theorem 8.6. For any graph G, the chromatic function is an nth-degree polynomial in k. Proof. Divide V into r independent sets V1, . . . , Vr. Then, given such a partition, one can color all of the points in each Vi with the same color, and use diﬀerent colors for diﬀerent regions. There are k r such ways of doing this. Then, this must be summed over all ways of partitioning V into indepdent sets: χ(G, k) = X partions of G into independent sets k(k −1) · · · (k −r + 1). Thus, it is a ﬁnite linear combination of polynomials, so it is a polynomial. Its degree comes from the partition into sets of single points, since otherwise there are fewer permutations. This gives k(k −1) · · · (k −n + 1), so the degree is at most n. If the degree were n −1, then there is a doubleton V1 and some singletons V2, . . . , Vn−2, so the number of options is k(k −1) · · · (k −n + 2) n 2 , so the product is k(k −1) · · · (k −n + 1) + n 2 −\|E\| (k(k −1) · · · (k −n + 2)) + · · · , where everything not written has degree at most n−2. Thus, the coeﬃcient of kn = 1, of kn−1 is − n 2 + n 2 −\|E\| = −\|E\|. This proof will be continued in the next lecture, but notice that the constant term is 0 (since a no-coloring of a graph has problems), the leading coeﬃcient is 1, the degree is n, and the second coeﬃcient is −\|E\| in a simple graph. Additionally, this polynomial is 0 for 0 ≤k < χ. It will also be possible to show that the coeﬃcients of the polynomial alternate in sign. 9. Probabilistic Constructions: The Legacy of Paul Erd¨ os: 4/25/13 This lecture wasn’t technically part of the class, as it was part of this quarter’s SUMO Speaker Series, given by Professor Amir Dembo. However, it was relevant to Math 108, so it has been included in these notes. Paul Erd¨ os initiated the idea of the probabilistic method in 1947, and gave lots of nice examples and uses of it as some of his over 1500 papers. This is such a large number of papers that mathematicians are often measured based on their graph distance to Erd¨ os, where connectedness is given by coauthorship of a paper. Professor Dembo’s Erd¨ os number is 2, and they tend to be small natural numbers. A relevant book on this subject is Alon and Spencer’s The Probabilistic Method. This lecture will touch only on the ﬁrst two chapters. Sometimes, a course called Math 159 is oﬀered, which discusses this in more detail. The general goal is to use probability in ways that don’t seem like they would need probability. On some ﬁnite set S let f : S →{0, 1} (corresponding to some property: f (x) = 1 if x has the property, and f (x) = 0 if it doesn’t). Then, the goal is to ﬁnd an x ∈S such that f (x) = 1. To do so, put a probability distribution P on S, X ∼P, such that E[f (X) > 0] > 0 (i.e. the expected value that f (x) is positive is itself positive). If this is the case, there must be an x such that f (x) = 1. This isn’t constructive, but this is precisely its power: it turns some nightmarish constructions into simple proofs of existence. However, it will require some cleverness. Example 9.1 (Erd¨ os, 1947). Take the complete graph Kn on n vertices (so that there are n 2 edges) and color its edges with 2 colors, red and blue. The goal is to ﬁnd a monochromatic k-clique (i.e. a Kk ⊂Kn whose edges are either all red or all blue). Then, the Ramsey number R(k, k) is the smallest number n such that this holds true for any coloring of Kn. The key here is that of course such a Kk can be created, but the condition is that it is present in all colorings. ( Theorem 9.2 (Erd¨ os). If n 2 21−( k 2) < 1, then R(k, k) > n. Proof. One way to ﬁnd a lower bound is to ﬁnd a coloring of Kn without any monochromatic Kk. But this is hard, so colorings will be chosen at random. Let S be the set of colorings of Kn, so that \|S\| = 2( n 2), and let 15 P be the uniform distribution on S. Let f (x) = 1 is there exists no monochromatic Kk in the coloring and f (x) = 0 otherwise. Then, the goal is to show that P(f (x) = 1) > 0, or equivalently that P(f (x) = 0) < 1. There are n k choices of k vertices in Kn, and each such choice is equivalent to a subgraph Kk of Kn. The union bound of probability states that P(A ∪B) ≤P(A) + P(B), so P(f (x) = 0) ≤ n k P(a speciﬁc Kk is monochromatic) = n k 21−( k 2), since there are k 2 ways to color Kk and exactly two of them are monochromatic. ⊠ Thus, R(k, k) ≥2k/2 for all k ≥3, and thus the Ramsey numbers grow at least exponentially. Notice how easy this is compared to a construction. If one wants an algorithm, there’s the possibility of an exhaustive search. which takes exponential time and is unrealistic for large n, or a smarter way: if n = 2k/2 , then n k 21−( k 2) ≪1, so there is a very small probability that any given graph doesn’t have the property. Since it’s easy to check, one can just randomly guess a graph, and guess again in the unlikely event that the graph is bad. Thus, this nonconstructive proof provides an algorithm, albeit a probabilistic one. In some ways, this is elementary, but not trivial: the proof is very easy to follow but hard to create. In some sense, it’s really nice: someone else does all of the hard work. Deﬁnition 9.3. A tournament on V = {1, . . . , n} is a directed graph T = (V, E) of n vertices such that for any distinct x, y ∈V , either (x, y) ∈E or (y, x) ∈E.12 Tournament graphs can be used to represent some competition, where the direction of the edge between x and y indicates the winner of a match between them. A tournament has a property Sk such that if for any subset of k players there exists some other player who beat all of them. Clarly, this requires k < n, and the “for any subset” aspect makes it diﬃcult unless n ≫k. What is the minimum value of n such that there exists a tournament on T with Sk? The probabilistic method can provide an upper bound by showing that a randomly chosen tournament T will have Sk with positive probability. Claim (Erd¨ os, 1963). If n k (1 −2−k)n−k < 1, then there exists a tournament on n vertices with Sk. Proof. It will be shown that a random tournament will have probability of not satisfying Sk with probability strictly less than 1. For all ﬁxed subsets K of size k, let Ak be the complement to Sk. Thus, P(Ak) = (1 −2−k)n−k: iterate over all j ̸∈K (there are n −k such j) and check if each j beat every k ∈K. Since the tournament was chosen uniformly at random, this has probability 1 −2−k, so P(T doesn’t have Sk) = P     [ K⊆V \|K\|=k AK    = X K⊆V \|K\|=k P(Ak) = n k (1 −2−k)n−k, and if this is less than 1, then there must be a T with probability Sk. ⊠ This bound was chosen to be easy to prove, as with the probability distribution. A better bound can certainly be found, but it requires much more work and is more elaborate. Remark. (1) The bound guaranteed by the theorem is a chore to calculate, but ends up being O k2k ). (2) There is a construction of such a T when n ≥(1 + δ)k for some δ > 0, which is a little better. ( Deﬁnition 9.4. A family of sets F is called intersecting (also a family of intersecting sets) if for any A, B ∈F, A ∩B ̸= ø. Theorem 9.5 (Erd¨ os-Ko-Rado). If F is a family of intersecting k-subsets13, of {0, 1, . . . , n −1} and n ≥2k, then \|F\| ≤ n−1 k−1 . 12When this graph is taken as an undirected graph, one obtains a complete graph. 13i.e. they are all of size k. 16 Proof. This proof is due to Kentona in 1972. Let As = {s, s + 1 mod n, . . . , (s + k −1) mod n}, so that there are n of them. Lemma 9.6. F contains at most k of these sets. Assuming the lemma (whose proof is skipped, but very easy), let σ be a uniformly chosen permutation of {0, . . . , n −1} and i uniformly chosen in {0, . . . , n −1} independently of σ. Then, let A = {σ(i), σ(i + 1 mod n), . . . , σ(i + k −1 mod n)}. Clearly, P(A ∈F \| σ) ≤k/n since there are n choices for i, and only k can be present by the lemma, so uncondition: P(A ∈F) ≤k/n. But since σ and i were chosen at random, A is uniformly distributed among all k-subsets of {0, . . . , n −1}, so P(A ∈F) = \|F\|/ n k . ⊠ This upper bound is actually tight: take F = {A ⊂{0, . . . , n −1} \| 0 ∈A}, so there are n−1 k−1 such sets that all intersect. The beauty of the probabilistic method is that there is no probability in the questions it tackles. Thus, it can be used to say a lot of very deep things about how probability relates to the rest of mathematics, and in more ways than just counting. This relates probability to group theory, topology, etc., and is the most far-reaching use of probability in mathematics. 10. Chromatic Polynomials: 4/29/13 In the proof of Theorem 8.6, the key was to write the vertex set V of a graph G as V = V1 ∪· · · ∪Vr, were the sets V1, . . . , Vr are disjoint and independent. Then, enumerating the possible colorings was fairly easy: there are k(k −1) · · · (k −r + 1) of them. Thus, the chromatic function is a polynomial of degree n and its second-order term is −\|E\|, where E is the edge set of G. Some other things were shown; see the previous lecture for details. A recurrence can be used to compute χ(G, k): let e ∈E, and denote G −e be the graph obtained by removing e and leaving the vertices unchanged. Clearly, χ(G −e, k) ≥χ(G, k), since every coloring still works when an edge is removed. If the endpoints of e have diﬀerent colors, then a coloring of G −e implies a coloring of G, and the leftover case is when the endpoints have the same color. Let G · e be the graph formed by removing e from G and identifying its endpoints. This has one fewer vertex and one fewer edge than G. Note that G · e might not be simple, even if G is. Then, χ(G −e, k) = χ(G, k) + χ(G · e, k), where χ(G · e, k) is calculated by removing the extraneous edges (since they have no eﬀect on the coloring). This can be rewritten as χ(G, k) = χ(G −e, k) −χ(G · e, k), which is useful because both graphs on the right-hand side are smaller. It can be used to compute the chromatic polynomial of a graph, though at some base case it is necessary to just compute it straightforwardly. This argument can be used to show that if T is a tree, then χ(T, k) = k(k −1)n−1. Remark. Using a similar contraction-deletion argument, one can obtain a formula for τ(G), the number of spanning trees of G. From Theorem 2.7, we know τ(Kn) = nn−2, but more generally τ(G) = τ(G −e)+τ(G ·e): choose an edge and ask whether it belongs in the spanning tree (in which case we go to the right) or not (left, or τ(G −e)). ( Claim. The coeﬃcients of χ(G, k) alternate in sign. Proof. This can be seen by the recurrence: if G has no edges, then χ(G, k) = kn, which trivially alternates in sign. Then, using induction, χ(G, k) = χ(G −e, k) −χ(G · e, k); the ﬁrst term on the right-hand side has degree n, so its (n −1)st term is negative and its terms alternate, and deg χ(G · e, k)) has degree n −1, so its terms alternate in the opposite way. Thus, taking −χ(G · e, k)) makes all of the signs align, so to speak. ⊠ If G is not connected, so that G = G1 ∪G2, where G1, G2 are its connected components, then χ(G, k) = χ(G1, k)χ(G2, k). Theorem 10.1 (Erd¨ os). There exist simple connected graphs with arbitrarily large girth and arbitrary large chromatic number. The probabilistic method is also useful here. Deﬁnition 10.2. A planar graph is a graph that can be drawn in the plane such that no edges cross (they only meet at vertices). 17 There are many ways of drawing a graph; K4 is planar, even though the intuitive representation of it is not planar, since there exists a drawing of it that is planar. Technically, in order to speak of the “inside” and “outside” of curves (the edges), one needs to prove something complicated like the Jordan Curve Theorem, but here, any edge can be represented by a piecewise linear curve, for which the theorem isn’t necessary. Thus, (1) there are no topological problems, (2) and even if there were, we wouldn’t care, anyway. A planar graph thus separates the plane into regions, called faces. The number of faces isn’t obviously well-deﬁned, since it depends on the way that the graph is drawn. However, the following is known. Theorem 10.3 (Euler). Suppose a planar graph G has V vertices, E edges, and divides the plane into F faces. Then, V −E + F = 2. Example 10.4. If G is a tree, then there is one face (since there are no cycles), so V −E + 1 = 2, or V = E + 1. ( There are hundreds of proofs of Euler’s theorem, and here’s one: Proof of Theorem 10.3. Proceed by induction on F. The base case is due to Example 10.4, so suppose that G is not a tree, or that there exists a cycle in G. Take some edge in the cycle and delete it; then, the nuber of faces is smaller, so by induction V −(E −1) + (F −1) = 2, so V −E + F = 2. ⊠ Thus, the number of faces for a planar graph is actually well-deﬁned. The 2 in Euler’s theorem relates to the surface, and is called the Euler characteristic of a surface: for example, on a torus there would be a diﬀerent constant. Another useful fact is that a connected, simple planar graph doensn’t have “too many” edges, which allows for a nice bound on the chromatic number. Theorem 10.5. Every planar graph can be colored using ﬁve colors. Actually, only four are necessary: Theorem 10.6 (Four-Color Conjecture). Every planar graph can be colored using four colors. This was proven using a computer, which sounds scary until one realizes that the 1970s didn’t have a lot of computing power compared to today. There were philosophical concerns about using a computer-generated proof, but 40 years later most people accept it. However, a better proof was given in 1997 by Seymour et. al., which only required 5 minutes of computing time. After all, we have computers, so why not use them? A computer-aided proof was also used to verify the Kepler conjecture, which stated that a hexagonal arrangement is optimal for packing spheres. The statement that a planar graph doesn’t have too many edges can be clariﬁed: each face gives rise to at least 3 edges, but each edge could be counted twice (or not at all for some edges), so E ≥3F/2. Using Euler’s formula, 2 = V −E + F ≤V −E + 2E/3, so E/3 ≤V −2, or E ≤3V −6. This is relatively low given how many edges are possible. 11. Planar Graphs and Matchings: 5/1/13 As was shown last lecture, for a planar graph, E ≤3(V −2) and 2 = V −E + F. Thus, K5 is not planar: it has 5 vertices and 10 edges, but to be planar it would need E ≤3(5 −2) = 9. Similarly, K3,3 isn’t planar: it has 6 vertices and 9 edges. In a bipartite graph, each face must be bounded by at least four edges (since there are no triangles), so E ≥4F/2 = 2F, so 2 = V −E + F ≤V −E/2, so E ≤2(V −2) if a bipartite graph is planar, which is a problem. Claim. If G is a planar graph, then G has a vertex of degree at most 5. Proof. If deg(v) ≥6 for all v ∈V , then 2E = P deg v ≥6V , so E ≥3V , but it is necessary that E ≤ 3(V −2). ⊠ This leads to a (not strict) upper bound on the chromatic number of a planar graph: 18 Theorem 11.1 (Six-Color Theorem). Any planar graph can be 6-colored, or if G is a planar graph, then χ(G) ≤6. Proof. Proceed by induction on the number of vertices. If V < 6, then of course G is 6-colorable, because each vertex can be assigned a diﬀerent color. In general, let x be a vertex of degree at most 5, as per the claim. Then, by induction G −x is 6-colored, because it is also planar and has strictly fewer vertices. Then, bring that coloring to G; at most ﬁve diﬀerent colors are adjacent to x, so it can be given the remaining color. ⊠ Six colors are actually too many: Proof of Theorem 10.5. Proceed by induction again, with the base case essentially the same: a graph with fewer than 5 vertices can of course be 5-colored. In a general planar graph, pick a vertex x of degree at most 5 and by induction, 5-color G −x. It can be assumed that deg(x) = 5, and furthermore that each of its neighbors x1, . . . , x5 have diﬀerent colors (so color xi with color i), because otherwise we would be done. Thus, it will be necessary to recolor some of the neighbors of x. Look at the subgraph of vertices colored only with colors 1 and 3, called H1,3. This is a planar bipartite graph, and isn’t necessarily connected. • If x1 and x3 are in diﬀerent connected components of H1,3, take the connected component of x1 in H1,3 and switch all coloings of 1 and 3 in that component. This is still a valid coloring of G, since there wasn’t any conﬂict beforehand and switching the color in a connected component doesn’t create one. Thus, x1 can be given color 3 while x3 still has color 3, so x can be given color 1 and G has a valid 5-coloring. • If x1 and x3 lie in the same connected component of H1,3, then take the same argument with colors 2 and 4 and the analogously deﬁned subgraph H2,4. If they’re in diﬀerent connected components, then the previous item allows G −x to be recolored and G to be 5-colored, or they are in the same connected component. Then, there is a path from x1 to x3 containing vertices only of vertices colored with colors 1 and 3, and there is a path from x2 to x4 in the same manner, but they must intersect, because G is planar.14 Thus, this can’t be possible. ⊠ This proof doesn’t use x5 at all, which made people think there was some clever trick that would allow it to be reduced to four colors. In fact, the discoverer of the proof, Kempe, published it as a proof of the Four-Color theorem, and it was accepted as such for ten years. This was not the only such false proof, so much that when the real proof was reported, people were wary of reporting it! Moving to matchings of graphs, Gale and Shapley’s solution to the stable marriage problem recently won a Nobel Prize in Ecomonics.15 The paper itself was called College Admissions and the Stability of Marriage, one of the most ﬂavorful names for a math paper in a long tine, and one that illustrates some of the many applications of this problem (there are lots of them, in ﬁelds as diverse as organ transplants). Deﬁnition 11.2. A matching of a graph G is a subset of the edge set such that no vertex appears in more than one edge of the matching. A perfect matching is when all vertices appear exactly once. One can consider a bipartite graph of a set M of men and a set W of women, where it is guaranteed that there exists a matching such that every vertex in M is in an edge (so that \|W\| ≥\|M\|).16 Additionally, for any m ∈M, deg(m) ≥1, or else no suitable matching would exist. More generally, if I ⊆M, then \| S i∈I W(i)\| ≥\|I\|, where W(i) is the set of vertices neighboring I. Clearly, these are necessary conditions. Theorem 11.3 (Hall’s Marriage Theorem). These conditions are suﬃcient; if for any I ⊆M, \| S i∈I W(i)\| ≥\|I\|, then G admits a matching in which all elements of M are paired up. Corollary 11.4. If G = M ∪W is a k-regular bipartite graph and \|M\| = \|W\|, then G has a perfect matching. Proof. Exercise; it is necessary to check that this condition implies the one in the theorem. 14This requires that x2 or x4 be between x1 and x3 in the plane, but strictly speaking this isn’t always the case. However, the colors can be rearranged such that x2 does lie between x1 and x3, since some colors must lie between other colors around x. 15Since Gale was dead, Shapley won the prize, along with Roth, a Stanford business professor. 16If one wishes to have a perfect matching, it is necessary that \|M\| = \|W \|. 19 Theorem 11.3 can be reformulated as: if A1, . . . , AN are some ﬁnite sets such that for any I ⊆{1, . . . , N} we have \| S i∈I Ai\| ≥\|I\|, then it is possible to ﬁnd distinct elements a1, . . . , aN with ai ∈Ai. This is called a system of distinct representatives (SDR). This has applications to group theory: suppose G is a ﬁnite group and H ≤G. Then, there exist g1, . . . , gm ∈G (where m = [G : H]) such that G = Sm i=1 g1H = Sm i=1 Hg1. Proof of Theorem 11.3. Proceed by induction on N. If N = 1, then of course an SDR exists for one set; just choose any element of the set. Call a collection of sets {Ai : i ∈I} for 1 ≤\|I\| ≤N critical if S i∈I Ai = \|I\| (e.g. ten men who among them know exactly ten women). Thus, there are two cases: • Suppose there are no critical collections. Then, pick some aN ∈AN, and let ˜ Ai = Ai −aN, and each such set is nonempty. Thus, for any collection {Ai}, S i∈I Ai ≥\|I\| + 1 ≥\|I\|, so ˜ A1, . . . , ˜ AN−1 satisﬁes the hypothesis by induction. Thus, there is an SDR ˜ a1, . . . , ˜ aN−1, and each element is distinct from aN, Thus, ˜ a1, . . . , ˜ aN−1, aN is an SDR for the whole collection. 12. The Gale-Shapley Theorem and Network Flow: 5/6/13 • Continuing with the proof of Theorem 11.3, suppose there exists a critical collection A1, . . . , Ak for 1 ≤k ≤N −1 (by relabeling if necessary). Then, Sn i=1 Ai = k. By the inductive hypothesis, there are distinct a1 ∈A1, . . . , ak ∈Ak. Let ˜ Aj = Aj −{a1, . . . , ak}, for k + 1 ≤j ≤n. These sets are nonempty: if one was, then (re)label it Ak+1, so that \|A1 ∪· · · ∪Ak+1\| = k, which contradicts the initial assumption. Claim. For any J ⊆{k + 1, . . . , n}, S j∈J ˜ Aj ≥\|J\|. Proof. k [ i=1 Ai ∪ [ j∈J Aj = [ j∈J ˜ Aj + k ≥k + \|J\| by the original assumption. ⊠ Thus, one can ﬁnd ak+1 ∈˜ Ak+1, . . . , an ∈˜ AN distinct from each other and the a1, . . . , ak. ⊠ This is useful in lots of places; for example, the 2012 Putnam B3 can be solved by using this, and the trick is seeing exactly how. A related problem is that of stable marriage, in which each man and woman has a list of preferences, and each person prefers to be married to someone than to be unmarried. Thus, the goal is to pair the men and women, but to avoid unstable marriages, deﬁned as pairs m1, w1 and m2, w2 if m1 prefers w2 to w1 and w2 prefers m1 to m2. A stable pairing is a pairing such that there are none of these sorts of marriages. Theorem 12.1 (Gale-Shapley). For any such choices of preferences, there exists a stable matching. There are lots and lots of variations, such as ﬁnding stable pairs of roomates (in which case the graph isn’t bipartite), ﬁnding optimal preferences for colleges, etc. This can even help in terms of organ transplants. Apparently, this is used by Stanford Housing to assign staﬀto dorms. Proof of Theorem 12.1. The proof is essentially the “old-fashioned marriage proposal.” To wit: (1) Each man proposes to the woman he likes most.17 Each man proposes to only one woman, and keeps the proposal until it is rejected. (2) If all women get a proposal, they all accept, and a stable matching is found. (3) Otherwise, some women get more than one proposal. Each such woman accepts the one she prefers most, and rejects the rest. A woman who has only one proposal still waits; after all, someone better might come along. (4) The men who were rejected propose to the woman they most prefer who have not yet turned them down. Then, return to step 2. 17Of course, you can switch the women and the men in this scenario, which is better for the women and worse for the men. . . 20 This algorithm terminates, because each time a man is rejected, the number of choices he has dwindles, but since everyone would prefer to be married, then eventually someone accepts. The matching found is stable, because if it weren’t, then suppose m1, w1 and m2, w2 were unstable with m1 and w2 preferring each other. Then, m1 must have proposed to w2 before w1, and therefore rejected by her, which means that w2 prefers m2 to m1, which generates stability.18 ⊠ Note that this stable matching is not necessarily unique: one might obtain a diﬀerent pairing by letting the woman propose and the men choose. It turns out, though, that the algorithm discussed above is most optimal for the men: each man gets the best of all possible women in a stable pairing, and each woman gets the worst of all possible men such that the pairing is still stable. The last graph-theoretic topic of this class will be network ﬂow. Here, one takes a directed graph and two distinguished nodes s and t to try to get something from s to t. Deﬁnition 12.2. A ﬂow on a directed graph is a real-valued positive function on the edges of the graph, such that the net ﬂow of all values at a vertex (signed by whether they’re coming in or going out) is zero. Formally, if v ∈V , then P x f (v, x) −P y f (y, v) = 0 (where the sums are over all x, y where those edges exist, or such that the ﬂow is deﬁned). Note that negative weights could be used instead of directed edges, but one is more intuitive than the other. Then, there can be a capacity on each edge c(x, y) ≥0, representing the maximum possible ﬂow through the pipe x to y, so the only ﬂows considered are those for which f (x, y) ≤c(x, y). Deﬁne the volume of a ﬂow f to be \|f \| = P v f (s, v) −P u f (u, s). Then, what is the ﬂow with the maximum volume? Claim. Then, since there are no points of accumulation, the amount ﬂowing out of s must be equal to the amount ﬂowing into t. Proof. X v̸=s,t X w f (v, w) − X u f (u, v) ! = 0, and X v X w f (v, w) − X u f (u, v) ! = 0, so X v=s,t X w f (v, w) − X u f (u, v) ! = 0, which is just the diﬀerence in the net ﬂow out of s and the net ﬂow into t. ⊠ More generally, one could take some set S ⊆V , and look at the ﬂow from things in S to things not in S. Claim. X u∈S v̸∈S f (u, v) − X u̸∈S v∈S f (u, v) = X u∈S X v f (u, v) − X v∈S X u f (u, v). That is, if the things in S are included in the ﬂow calculation, the result doesn’t change: this is just the sum of the net ﬂows at u for each u ∈S. If S ⊆V −{s, t}, then this is zero, and if s ∈S but t ̸∈S, then this is vol(f ) = \|f \|. Deﬁnition 12.3. A cut is a collection of vertices S with s ∈S but t ∈Sc, or a partition such that s is in one half, but t is in the other. Thus, for any cut, we have \|f \| = X u∈S v∈Sc f (u, v) − X u∈Sc v∈S f (u, v). 18Notice the lack of formalism in this proof. This is okay, because the original paper didn’t have any either! 21 13. Network Flow II: 5/13/13 Recall that on a directed graph G, a ﬂow is a nonnegative function f (x, y) on the edges of G (i.e. f (x, y) = 0 if there is no edge from x to y), such that for any v ̸= s, t, P w f (v, w) −P w f (w, v) = 0 (intuitively, there are no leaks in the graph). If the maximum possible ﬂow is c(x, y), which is some other nonnegative function, one considers only ﬂows such that 0 ≤f (x, y) ≤c(x, y). In some situations, it will be assumed that the capacity is ﬁnite, which doesn’t make too much of a diﬀerence if it’s suﬃciently large. Suppose S is any subset of the vertex set V , and S = V −S. Then, X x∈S y∈S f (x, y) − X x∈S y∈S f (y, x) = X x∈S f (x, y) − X x∈S f (x, y). This is zero if S ⊆V −{s, t}, and if s ∈S but t ̸∈S, then this becomes \|f \|. Recall that a cut is a partition of G such that s is on one side and t is on the other. Then, the capacity of a cut is P x∈S,y∈S c(x, y), which represents the amount that can ﬂow at the boundary. Then, since f (x, y) ≤c(x, y) for all x and y, then \|f \| = X x∈S y∈S f (x, y) − X x∈S y∈S f (y, x) ≤ X x∈S y∈S c(x, y). Thus, the capacity of any cut is at least the volume of the ﬂow. This leads to something a little more interesting: if the goal is to ﬁnd a ﬂow of maximum volume, then the maximum volume is less than the capacity of any cut, or the maximum volume is at most the minimum capacity of a cut. Since there are a ﬁnite number of options, then clearly the minimum capacity exists, but the maximum ﬂow is more nuanced: just because it is bounded above doesn’t mean it has a maximum, as in the sequence {x ∈Q, x2 < 2}. This requies a little bit of elementary analysis to address fully: speciﬁcally. using the Bolzano-Weierstrauss theorem, it is possible to show there is a maximal ﬂow on each edge and therefore on the entire graph by taking subsequences that converge on each edge. Theorem 13.1. In fact, these quantities are equal: the maximum ﬂow is equal to the minimum cut. This leads to some number of algorithms for ﬁnding the maximum ﬂow given the capacities. The maximum ﬂow might not be unique, however; several diﬀerent ﬂows could lead to the same total ﬂow. The following proof is due to Ford and Fullerson. Proof of Theorem 13.1. Take a maximal ﬂow f . Then, the goal is to construct a cut S such that the capacity of S is equal to \|f \|. First, put s ∈S. Then, pick any v ̸∈S, and check if there is an x ∈S such that f (x, v) < c(x, v) or f (v, x) > 0. If so, then add v to S. Then, repeat as long as there are vertices to add. Then, there are two things to show: ﬁrst, that S is a cut (so that t ̸∈S) and that the capacity of S is \|f \|. Suppose t ∈S. Then, there is a sequence x0 = s, x1, . . . , xn = t such that for all j, either c(xj, xj+1) − f (xj, xj+1) > 0 (so there’s more room) or f (xj+1, xj) > 0 (so there is backﬂow). Let εj = max(c(xj, xj+1) − f (xj, xj+1), f (xj+1, xj)), so that εj > 0, and ε = min εj. Then, adjust the ﬂow: if ε is the ﬁrst case, then increase the ﬂow through that point by ε, and if the second case happens, reduce the backﬂow by ε. This is a valid ﬂow that satisﬁes the capacity bounds, and its volume is \|f \| + ε. This is very much like an Euler trail, with something going in and something coming out. Then, since the volume is increased, f isn’t a maximal ﬂow, which is a problem. Thus, it is necessary that t ̸∈S, so S is a cut. The capacity of this cut is X x∈S y∈S c(x, y) = X x∈S y∈S f (x, y) − X x∈S y∈S f (x, y) = \|f \|. ⊠ 14. Network Flow III: 5/15/13 An algorithm for ﬁnding the maximum ﬂow can be given if it is assumed that the capacities are all nonnegative integers: (1) Start with a ﬂow of 0 on every edge. 22 (2) Construct S as in the proof of Theorem 13.1, and in particular, ﬁnd some ε > 0. Then, it is possible to increase the ﬂow by 1 (since all of the capacities are integers, so this won’t overﬂow it). (3) Repeat until S is a cut. This proof clearly also works for rational numbers, because the capacities can just be scaled to some common denominator. However, for irrational numbers, this is not so easy, even though the theorem guarantees that there at least is a solution. In the real world, this doesn’t make too much of a diﬀerence, since everything can be approximated by a rational number. This has a nice application to the proof of Hall’s theorem: let G be a bipartite graph G = V1 ∪V2, with \|V1\| = \|V2\| = n. Proof of Theorem 11.3. Add to G two vertices s, connected to every vertex in V1, and t, connected to every vertex in t2. Then, give capacity 1 to all edges adjacent to s or t, and some integer large capacity to every other edge, such as n. Then, the minimum ﬂow must be equal to the maximum cut, so suppose S is a cut that contains s but not t. Then, if S ∩V1 and y ∈S ∩V2 and (x, y) is an edge in G, then this is ﬁne, because c(x, y) = n. Then, the ﬂow from s to t passes through the neighbors of S ∩V1 in V2, of which there are at least \|S ∩V1\| by the precondition, so each they are also in S. Thus, the capacity of the ﬂow is at least n, so the theorem follows. ⊠ There are lots of variants on Menger’s theorem, and here’s one related to network ﬂow and edge connectivity: Theorem 14.1 (Menger). Suppose G is an undirected graph and s, t ∈V (G). Then, let p be the maximum number of paths from s to t such that all used edges are disjoint, and let p′ be the minimum number of edges required to disconnect s and t. Then, p = p′. Proof. Consider a directed graph in which every edge {x, y} in G corresponds to two edges − → xy and − → yx, and give each edge a capacity of 1. Then, the max ﬂow with f (x, y) = 0 or 1 is the number of edge-disjoint paths from s to t, which makes sense by thinking about how the ﬂow works, and the min cut is the smallest number of edges necessary to disconnect s and t, so since the max-ﬂow and min-cut are equal, then p = p′. ⊠ Turning to enumerative combinatorics, consider some famous theorems: ﬁrst, Hall’s theorem, Theorem 11.3 on SDRs, can be considered enumerative. Deﬁnition 14.2. A poset, or partially ordered set, is a set S with an order relation < such that not all elements are comparable, but if x, y ∈S are comparable, then x < y, y < x, or x = y, and if x < y and y < z, then x < z. Example 14.3. Let X be a ﬁnite set and S be its power set, ordered by inclusion, where if A, B ⊆S, then A < B iﬀA ⊊B (equivalently, A ≤B iﬀA ⊂B). This is clearly a partial order, but there exist A ̸= B such that A ̸< B and B ̸< A. ( Deﬁnition 14.4. A chain in a poset is a sequence x1 < · · · < xn, and an anti-chain is a set {x1, . . . , xn} such that no two elements are comparable. Theorem 14.5 (Sperner). If X is a ﬁnite set of set n and S is its power set ordered by inclusion, then the size of a maximal antichain in S is n ⌊n/2⌋ . Some examples of antichains are sets of singleton sets, such as {x1}, . . . , {xn}, and note that maxk n k = n ⌊n/2⌋ . Proof of Theorem 14.5. Consider chains of n elements A1 ⊆A2 ⊆· · · ⊆An = X, where \|Ai\| = i. Each Ai corresponds to adding one element to Ai−1, so this chain corresponds to a permutation of {1, . . . , n} specifying the order the numbers are added in, so there are n! such chains. Each chain contains at most one element of a given maximal antichain, so it has at most n! elements. However, they can be double-counted: look at the antichains. If \|A\| = k, then A is in k!(n −k)! chains, because the ﬁrst k elements can be permuted while preserving A, and similarly the last n −k ordered can be permuted. Then, P \|A\|=k k!(n −k)! ≤n! 23 15. Sperner’s Theorem: 5/20/13 Recall that a partial order on a set S (a poset) is an order < such that exactly one of the following must hold: a < b, b < a, a = b. or a and b are incomparable. If the fourth option doesn’t happen for any a, b ∈S, then < is a total order on S (e.g. (R, <)). Continuing the proof of Theorem 14.5, since P \|A\|=k n!/ n k = P \|A\|=k k!(n −k)! ≤n!, so if A is any antichain, then X A∈A 1 ≤ X \|A\|=k n ⌊n/2⌋ n k ≤ n ⌊n/2⌋ , so \|A\| ≤ n ⌊n/2⌋ . ⊠ Theorem 15.1 (Dilworth). If S is any ﬁnite poset, then the minimal number of chains in a partition of S into a union over all such partitions is equal to the size of the maximal antichain. Showing ≥is considerably easier than ≤, but that is beyond the scope of this class. In a set X, a family of k-subsets A = {A1, . . . , Aℓ} is called intersecting if Ai ∩Aj ̸= ø for all 1 ≤i, j ≤ℓ. If \|X\| = n, then the size of the largest intersecting family is at most n k , and if k > n/2 then all k-element subsets intersect by the pigeonhole principle. Theorem 15.2 (Erd¨ os-Ko-Rado). Suppose X = {1, . . . , n}. Then, if n ≥2k, then any intersecting family of k-subsets has at most n−1 k−1 elements. Proof. Arrange {1, . . . , n} around a circle in some order. Since the circle can be rotated, there are (n −1)! ways of arranging the elements. Given some k-family of intersecting sets A, count the number of sets in A that appear as k consecutive elements as the circle is traversed clockwise (out of n possible). It happens that there are at most k such sets, which we will go back and prove in a moment. Assuming it, there are at most k(n −1)! sets over all such arrangements on the circle. Thus, we have k(n −1)! = X circular arrangements #{k-consecutive sets in A} = X A∈A #{circular arrangements in which A appears k-consecutively}, so \|A\|k!(n −k)! ≤k(n −1)!, so \|A\| ≤(n −1)!/(k −1)!(n −k)! = n−1 k−1 . This is good, but now the fact still needs to be proven: suppose A, B ∈A are k-consective. Then, they must intersect, and since n ≥2k, then A and B must intersect, so each element of A must give rise to at most one element in A; therefore, there are at most k of them in this arrangement. ⊠ This leads into various problems in enumeration. For example, one might count functions from n-element sets to k-element sets. Let S be a set of n elements and A1, . . . , Ak be a collection of subsets of S. Then, one can use the principle of inclusion and exclusion to calculate the number of elements of Sk i=1 Ai = Tk i=1(S −Ai) given T i∈I Ai for I ⊆{1, . . . , k}. Theorem 15.3 (Principle of Inclusion and Exclusion). S − k [ i=1 Ai = n − k X i=1 \|Ai\| + X 1≤i<j≤k \|Ai ∩Aj\| − X 1≤i<j<ℓ≤k \|Ai ∩Aj ∩Aℓ\| + · · · = n + X I⊆{1,...,k} I̸=ø (−1)\|I\| \ i∈I Ai . (15.4) In some sense, one starts by overcounting, then undercounting, then overcounting, and so on. A neat application of this is that if S = {1, . . . , N} and p ≤√n is prime, then let Ap = {n ≤N \| p \| n}. If π(x) represents the number of primes less than x, then then S −S p≤ √ N Ap is the set of primes between √ N 24 and N. Thus, the principle says that 1 + π(N) −π( √ N) = N − X p≤ √ N \|Ap\| + X p,q \|Ap ∩Aq\| −· · · More generally, for d not necessarily prime, let Ad = {n ≤N \| d \| n}, so that \|Ad\| = ⌊N/d⌋. The numbers that are interesting here are products of distinct primes d = p1 · · · pℓ(i.e. are square-free). This number d appears in the formula with sign (−1)ℓ. In number theory, the M¨ obius function is an important concept: µ(d) =    1, d = 1 (−1)ℓ, d = p1 · · · pℓare distinct 0, p2 \| d for some prime p The formula then becomes 1 + π(N) −π( √ N) = X 1≤d≤N p\|d = ⇒p≤ √ N µ(d)\|Ad\|. This is an example of a sieve identity, called the Sieve of Eratosthenes.19 16. The Principle of Inclusion and Exclusion: 5/22/13 First, it will be helpful to prove the principle of inclusion and exclusion: Proof of Theorem 15.3. The formula for Sk i=1 Ak will be found: pick some a ∈S and suppose that it appears in r of the A1, . . . , Ak, so that 0 ≤r ≤k. Then, the left-hand side of the formula (15.4) counts a once if r ≥1 and zero times if r = 0. The right-hand side counts a zero times if r = 0, and if r ≥1, then suppose a ∈Ai1, . . . , Air , and in the right-hand side of (15.4) a is counted X ø̸=I⊆{i1,...,ir } (−1)\|I\|−1 · 1 = r X ℓ=1 (−1)ℓ−1 r ℓ , times, where \|I\| = ℓ. By the Binomial theorem, 0 = r X ℓ=0 r ℓ (−1)r−ℓ= 1 − r X ℓ=1 r ℓ (−1)ℓ−1, so the quantity calculated must be 1. The second formula is the complement of this one, and its formula follows as a result. ⊠ Consider the set of permutations on {1, . . . , n}, which are required to be bijections.20 There are n! permuta-tions, but consider the number of derangements: permutations which have no ﬁxed points. Imagine putting n letters in n envelopes such that no letter goes in the correct envelope.21 How many derangements are there? Let Aℓbe the set of permutations that ﬁx ℓ; then, none of the derangements belong to any of these sets, so the total number of derangements is n \ ℓ=1 (Sn −Aℓ) = n! + X I⊆{1,...,n} I̸=ø \ i∈I Ai . 19There is a sculpture near the Cantor Arts Center that is relevant to this sieve. 20Recall that a function f : A →B is injective if f (a) = f (b) for a, b ∈A, then a = b. A function is surjective if it is onto: f : A →B is surjective if every b ∈B has an a ∈A such that f (a) = b. A bijection is a function that is both injective and surjective. 21Apparently this is a homework question in one of the professor’s other classes. Oops. 25 In T i∈I Ai , the elements of I are ﬁxed and the rest are permuted, so there are (n −\|I\|)! such permutations, so this becomes n n \ ℓ=1 (Sn −Aℓ) = n! + n X \|I\|=ℓ ℓ=1 (−1)ℓ(n −ℓ)! n ℓ = n! + n X ℓ=1 n! ℓ! = n! 1 −1 1! + 1 2! −1 3! + · · · + (−1)n n! , which approaches n!/e as n →∞. Thus, most permutations are derangements for large n. There is literature on what a random permutation might look like (which has applications, such as shuﬄing cards). The probability of having k ﬁxed points is approximately 1/ek!, which forms a Poisson distribution. There are lots of similar problems: imagine a set of 100 prisoners who are given numbers on their backs (which they cannot see). There are 100 boxes, and each prisoner can look at 50. If all of them see their numbers, they can go free. There’s a reasonably good strategy that they can employ; can you ﬁnd it? A probem called the twelve-fold way asks for the number of functions f : N →K, where N = {1, . . . , n} and K = {1, . . . , k}. One may wish to count injections or surjections, or arbitrary functions, giving bijections. Then, there are four ways of thinking of the functions as the same (which leads to the number twelve): there may be no restricton, or one may wish to consider them up to a permutation of N, or of K, or of both. This can be imagined as placing n balls in k boxes, such that there are no restrictions on the number of balls in each box, at most one ball in each box, or at least one ball in each box. One could have the balls colore with n colors, or that the balls have the same color and the boxes have diﬀerent labels, or the balls have diﬀerent colors and the boxes look identical, or the balls and boxes are indistinguishable among themselves. This forms a natural collection of enumeration problems, ﬁve of which are diﬃcut and seven of which are easy. Case i. There are kn functions N →K, where all balls and boxes are distinct. Case ii. How many injective functions are there? There are k choices for the location for the ﬁrst ball, and k −1 for the second, k −2 choices for the third, etc. If k < n, then there are zero choices, so the total number of functions is k(k −1)(k −2) · · · (k −n + 1), a number called the falling factorial. Case iii. How many surjective functions are there? We will return to this next week, using the principle of inclusion or exclusion or setting up a recurrence. This would be a good thing to think about. Let Ai be the set of balls in box i; since the function is surjective, then there are all nonempty sets. Thus, the goal is to partition {1, . . . , n} into k nonempty sets. This is diﬀerent than writing n as the sum of k numbers because the elements of the set are distinct. Case iv. Suppose the balls are indistinguishable, but the boxes aren’t. Then, the goal is to consider all such functions, so if box i contains ai balls, then a1 + · · · + ak = n, and it is possible to permute the ai. There are two ways to approach this: Imagine all of the balls are red. Put some partitions, represented by k −1 green balls, between the red ones, such that green balls may be adjacent. Then, everything to the left of the ﬁrst green ball goes in box 1, everything between the ﬁrst and the second goes to box 2, etc. Thus, the goal is to choose locations for the k −1 green balls among the total of n + k −1 slots (after which the red balls can be added), giving a total of n+k−1 k−1 . The second proof uses a very useful idea called a generating function. For each box i, take a sequence (1+x +x2 +x3 +· · · ). Then, the number of balls in box i is one of these coeﬃcients, and the coeﬃcient of xn in the product of all of these terms is the product we want: it includes all distinguishable partitions of the n balls into k boxes. The geometric series converges to 1/(1 −x), so if F(x) = 1/(1 −x)k represents the product, one can compute the nth derivative at zero, which once divided by n! is the coeﬃcient of xn. This becomes n+k−1 k−1 . 17. The Twelvefold Way: 5/29/13 Case v. Consider the set of injective functions in which the boxes, but not the balls, are distinct. Then, each box contains at most one ball, so choose n boxes out of the k, giving k n options. This is 0 if k > n. Case vi. If the functions are requred to be surjective, the goal is to put n identical balls into k boxes such that each box has at least one ball. After removing one ball from each box, the goal is to put n −k balls 26 in k boxes with no conditions, which reduces to Case iv, giving n−k+k−1 k−1 = n−1 k−1 , and there are of course no ways if k < n. This illustrates another way to approach Case iv: one chooses ℓboxes for some 1 ≤ℓ≤k that have at least one ball, so the total number is k X ℓ=1 k ℓ n −1 ℓ−1 = n + k −1 k −1 . This itself is a useful combinatorial identity. This can again be done with generating functions, by taking (x + x2 + · · · )k = xk/(1 −x)k and looking at the coeﬃcient of xn. Case vii. So now we suppose that the balls are distinguishable, but the boxes aren’t. Then, the balls go in ℓboxes for some ℓ, leading to the sum Pk ℓ=1 S(n, ℓ), where S(n, ℓ) is as given below. Case viii. For the injective functions, if k ≥n then there is exactly one way to do this (since the boxes can be shuﬄed around), and otherwise there are none. Case ix. For the surjective case, this asks the number of ways to partition {1, . . . , n} into exactly k nonempty subsets (provided n ≥k; otherwise, it is zero). This number is known as the Stirling number of the second kind, S(n, k). This leads to a solution for Case iii: since there are k! ways to permute the boxes, the number is k!S(n, k). Case x. Here, neither the balls nor the boxes have labels on them so the goal is to determine the number of ways of writing n = a1 + · · · + ak, where a1 ≥a2 ≥· · · ≥ak. Case xi. If these functions are injective, then each of the ai above must be at most 1 and therefore there is one such way to order if k ≥n and none if k < n. Using the notation deﬁned below, suppose the balls end up in ℓboxes, and after the summation one has Pk ℓ=1 pℓ(n). Case xii. If they must be surjective, then each of the ai must be strictly positive. Thus, one obtains the number of ways of partitioning the number n into exactly k parts, which is similar to the Stirling number, but not the same, and it is denoted pk(n). Of course, this only makes sense if n ≥k; otherwise, it is zero. Thus, it will be useful to obtain a formula for the Stirling number. There are several ways to obtain it. First, consider the number of cases where {n} appears by itself. Thus, there are S(n −1, k −1) ways for this to happen. Alternatively, if n appears in some set with at least one other element, then there are kS(n −1, k) choices, so S(n, k) = S(n −1, k −1) + kS(n −1, k). Then, one has the base cases S(n, 1) = 1 for n ≥1 and S(k, k) = 1. One often adopts the convention that S(0, 0) = 1. One could also use the principle of inclusion and exclusion, since we know there are k!S(n, k) functions in one of the above cases: there are kn functions from {1, . . . , n} to {1, . . . , k}, so splitting up based on the subset of {1, . . . , k} is the image, one obtains k!S(n, k) = k X ℓ=0 (−1)ℓ k ℓ (k −ℓ)n, which isn’t very illuminating, but is at least explicit. One could also consider coloring the graph on n vertices with no edges and x colors, so that there are xn options. Then, the chromatic polynomial can be obtained by partitioning the vertices into k sets, and multiply by the number of ways to color these sets: x(x −1)(x −2) · · · (x −k + 1) = (x)k, which is the falling factorial function. Thus, xn = n X k=1 S(n, k)(x)k. It will also be nice to know pk(n), the number of partitions of n into k parts. A recursive formula is given by noticing that if one wishes to write n = a1 + · · · + an such that a1 ≥· · · ≥ak ≥1, then either ak = 1, whih gives pk−1(n −1) options, or ak ≥2, in which case all of the ai are, so there is an expression for n−k = (a1 −1)+· · ·+(ak −1) ≥1, contributing a factor of pk(n−k). Thus, pk(n) = pk−1(n−1)+pk(n−k). 27 18. Combinatorial Functions: 6/3/13 Recall the deﬁnitions of some functions given in the Twelvefold Way: the binomial coeﬃcients are probably familiar, but also S(n, k), the Stirling number, which is the number of set partitions of {1, . . . , n} into k nonempty sets, which obeys the relation S(n, k) = kS(n −1, k) + S(n −1, k −1). Additionally, there is the function pk(n), which is the number of partitions of n as n = a1 + · · · + ak such that a1 ≥a2 ≥· · · ≥ak ≥1. This obeys pk(n) = pk−1(n −1) + pk(n −k), and there are other relations given in the textbook. Consider bijections of {1, . . . , n}. Each bijection π can be written as a cycle decomposition by seeing where 1 goes, then π(1), etc., until 1 is reached again. This is written as (1 π(1) π(π(1)) · · · ). Then, repeat with a number that wasn’t in the ﬁrst cycle, and so on. Some numbers are ﬁxed points: if π(i) = i, then the cycle is (i). The Stirling number of the ﬁrst kind is the number of permutations of {1, . . . , n} which have exactly k cycles. Then, summing over k, one would obtain all n! permutations. Notice that the 100 prisoners problem presented earlier has a nice solution in terms of cycles: what is the probability there is a cycle of size at least 50? This comes up in “nature:” 52! is the number of posible shuﬄes for a deck of cards. One could use this to determine whether a deck has been shuﬄed correctly: does it “look like” a random permutation? Markers indicating too many cards in the same position are a red ﬂag. In all of the n! permutations, how many n-cycles are there? There are (n −1)!, because some cycle decompositions are identical, such as (1 2 3) and (2 3 1). One can also use the principle of inclusion and exclusion: for example, how many permutations of {1, . . . , n} have no cycle of length greater than n/2? By size constraints, each permutation must have at most one such cycle, so the answer is n! − X n≥k>n/2 π has a cycle of length k 1 = n! − X n≥k>n/2 n k k Y i=1 (k −i)! = n! − X n≥k>n/2 n! k . Notice that if k ≥n/2 this wouldn’t work, because some things would be double-counted. Recall the prisoners problem: one solution is to take prisoner n to look at box n and get a number, then look at the box with that number, and so on. The goal is to look up the cycles, and the prisoners can go free if all cycles are of length at most 50 (otherwise, the prisoners might not see the boxes with their own numbers). This is optimal, though it’s not easy to show it. As shown above, the number of permutations for which this holds is 100! 1 −P100 k=51 1 k . An approximation can be given as 100 X i=51 Z i+1 i dt t < 100 X k=51 1 k < 100 X i=51 Z i i−1 dt t , showing that the chance is at least log(101/51) ≈1/3 and at most log 2 ≈0.693, which is pretty nice. Notice that if this fails, then at least 50 people fail to see their number. Further questions can be asked: what happens if ﬁfty boxes are replaced by 35? In general, if they must inspect n/u out of n boxes, then as n gets large one sees the Dickman-de Bruijn function ρ(u): we saw that ρ(2) = ln 2, and this number is always positive. This actually has to do with unpublished work of Ramanujan (of course). This function ocurs in lots of ways: if one factors a random n (whatever random means here) into primes n = pα1 1 · · · pαk k , then what is the chance that all of these prime factors are less than √n? This ends up being 1 −ln 2, and there are all sorts of interesting connections between what a random number looks like, a random permutation looks like, and a random polynomial looks like, and there’s a lot of structure here (Don Knuth did some work here, too). This function satisﬁes a diﬀerential diﬀerence equation, funnily enough: uρ′(u) = −ρ(u −1), and ρ(u) = 1 for 0 ≤u ≤1 (the prisoners always succeed). Ramanujan wrote down what the principle of inclusion and exclusion would give for this in cases for u ≤6, and then said “and so on.” These sorts of things are useful in case one doesn’t care as much about exact formulas, but instead the asympotics of how these combinatorial functions behave over the long run. Generally, one would use tools of analysis or calculus to determine this. This also relates to the philsophical idea of a good formula, which is an expression for a function that is ideally easy to compute or easy to understand. One of the most basic and useful identities is Stirling’s formula for n! We don’t know how to compute factorials exactly faster than O(n) (just multiplying things together). Since there doesn’t seem to be a great 28 formula, there’s at least one for an approximation: n! ∼ √ 2πn n e n , where ∼means that lim n→∞ n! √ 2πn(n/e)n = 1. This can be used to understand how pk(n) or S(n, k) behave as n →∞. One could also consider the horrible Bell numbers B(n) = Pn k=1 S(n, k) (all possible partitions of {1, . . . , n} into nonempty sets), which have no good formula, but asymptotically are better to understand, and similarly p(n) = Pn k=1 pk(n). One can count trees on n vertices up to isomorphism, for example, but this is a headache for general n, and the best case is an asymptotic understanding. Now for some formulas: one useful technique is to compare a sum with an integral of the expression being summed. log N! = N X k=1 log k ≤ N X k=1 Z k+1 k log t dt = Z N+1 1 log t dt. For a lower bound, we have log N! = N X k=2 log k ≥ N X k=2 Z k k−1 log t dt = [t log t −t]N 1 = N log N −N + 1. Expontntiating, N! ≥(N/e)Ne, which is nice to know. The integral for the upper bound can also be evaluated, giving (N + 1) log(N + 1) −N. Since log(N + 1) −log N = log(1 + 1/N), this can be expanded out using a Taylor series: log(1 + x) = Z x 0 dt 1 + t = Z x 0 (1 −t + t2 −t3 + · · · ) dt = x −x2 2 + x3 3 −· · · , so the diﬀerence can be found and simpliﬁed to log N! ≤(N + 1)(log N + 1/N) −N. In summary, N e N e ≤N! ≤NN+1 eN e1+1/N, and the diﬀerence is a factor of Ne1/N. Stirling’s formula indicates that this diﬀerence is split, up to constant, as the √x term. In general, converting to an integral is a good way to approach a solution; for example, the harmonic series PN k=1 1/k should be approximately log n. Similarly, N X k=1 nk ≈ Z N 1 xk dx ≈Nk+1 k + 1, and this approximation can be made precise with the upper and lower bounds, as seen in the previous examples. Another approximation can be made: since log k ≈ R k k−1 log t dt, then for k ≥2, Z k k−1 log k dt − Z k k−1 log t dt = Z k k−1 log k t dt, If t = k −y, then 0 ≤y ≤1 is an appropriate change of variables, yielding = Z 1 0 log 1 1 −y/k dy. The Taylor series log(1/(1 −x)) = x + x2/2 + x3/3 + · · · gives = Z 1 0 ∞ X ℓ=1 y k ℓ1 ℓdy = ∞ X ℓ=1 1 kℓ· 1 ℓ Z 1 0 y ℓdy = ∞ X ℓ=1 1 kℓ(ℓ)(ℓ+ 1). Thus, log N! = N X k=2 log k = Z N 1 log t dt + N X k=2 log k − Z k k−1 log t dt . 29 A lot of things cancel out, as shown above, with the remainder N X k=2 ∞ X ℓ=1 1 ℓ(ℓ+ 1)kℓ ! = 1 2k = X ℓ≥2 1 ℓ(ℓ+ 1)kℓ, where the sum is now constant as N →∞, and PN k=2 1/(2k) →(1/2) log N + C for some constant C. Thus, the square root occurs in Stirling’s formula, and the constant will lead to the rest of the term. The point is to see just how much can be done with single-variable calculus and some elbow grease. 19. More Techniques for Asymptotics: 6/5/13 Recall that Stirling’s formula claims that n! ∼ √ 2πn(n/e)n, where ∼means that the limit as n →∞of their ratio is 1. We showed that n! ∼C√n(n/e)n last time, and can be more speciﬁc. We know that 2n n is the largest of all binomial coeﬃcients 2n k where n is ﬁxed. One can ask how big it is, and how close nearby coeﬃcients (e.g. 2n n+1 , or more generally 2n n+ℓ for a “small” ℓ) are to it, at least in orders of magnitude. Using Stirling’s formula, 2n n + ℓ = (2n)! (n + ℓ)!(n −ℓ)! ∼ C√n C √ n + ℓC √ n −ℓ (2n/e)2n ((n + ℓ)/e)n+ℓ((n −ℓ)/e)n−ℓ A couple of things quickly cancel out, showing 22n/(2n + 1) ≤ 2n n ≤22n: ∼ √ 2 C 1 √n (2n)2n (n + ℓ)n+ℓ(n −ℓ)n−ℓ ∼ √ 2 C 22n √n nn+ℓnn−ℓ (n + ℓ)n+ℓ(n −ℓ)n−ℓ= √ 2 C 22n √n n n + ℓ n+ℓ n n −ℓ n−ℓ . We can take the logarithm of this quantity: log n n + ℓ n+ℓ n n −ℓ n−ℓ! = (n + ℓ) log n n + ℓ + (n −ℓ) log n n −ℓ . When x is small, log(1 + x) = x −x2/2 + x3/3 −· · · and log(1 −x)−1 = x + x2/2 + x3/3 + · · · , so (n + ℓ) log 1 1 + ℓ/n = −(n + ℓ) log 1 + ℓ n = −(n + ℓ) ℓ n −1 2 ℓ n 2 + · · · ! (n −ℓ) log 1 1 −ℓ/n = (n −ℓ) ℓ n + 1 2 ℓ n 2 + · · · ! = ⇒(n + ℓ) log 1 1 + ℓ/n + (n −ℓ) log 1 1 −ℓ/n = −2ℓ ℓ n + 1 2 ℓ n 2 (2n) + smaller terms = −ℓ2 n + smaller terms. Thus, the approximation for the binomial coeﬃcient is 2n n + ℓ ≈ √ 2 C√n22ne−ℓ2/n + smaller terms. Thus, this is largest when ℓ= 0, which we already knew, but the nearby coeﬃcients are almoat as large when ℓ2/n is small (i.e. ℓis on the scale of √n). This can be applied to the binomial distribution in probability, where one ﬂips n coins and receives ℓmore heads than tails. Then, it’s most likely that the number of heads and tails will be about the same (to about √n diﬀerence). Thus, we can approximate something else: n X ℓ=−n 2n n + ℓ = 22n ∼ √ 2 C√n22n n X ℓ=−n e−ℓ2/n. 30 Thus, we can approximate with an integral: n X ℓ=−n e−ℓ2/n ≈ Z n −n e−x2/n dx ≈ Z ∞ −∞ e−x2/n dx, because when x > n, the value of the function is very small. Setting y = x/√n, 22n ∼ √ 2 C 22n Z ∞ −∞ e−y 2 dy, so C = √ 2 R ∞ −∞e−y 2 dy. This integral is a common joke, if you’re into that kind of humor, and it evaluates to √π, so C = √ 2π. Moving to a diﬀerent function, recall pk(n) is the number of ways to write n = a1 + · · · + ak such that a1 ≥a2 ≥· · · ≥ak ≥1. We want to understand this when k is ﬁxed and n is large. Since some of the ai may be the same, then computing all of the permutations of them would be overcounting them, so k!pk(n) ≥#{n = a1 + · · · + ak, ai ≥1} = n−1 k−1 (i.e. the number of compositions of n into k parts). Thus, pk(n) ≥1 k! n −1 k −1 ≈(n −1)(n −2) · · · (n −(k −1)) k!(k −1)! ≈ nk−1 k!(k −1)! if k is ﬁxed as n →∞. Notice that this approximation isn’t valid if k also grows with n. Then, there is a nice trick to obtain the upper bound: suppose n = a1 + · · · + ak, written in descending order, and let bi = ai + (k −i). Then, n + k(k −1)/2 = (a1 + (k −1)) + · · · + (ak + (k −k)) = b1 + · · · + bk. In fact, every permutation of the bi gives a set of ai that works, so the number of such bi provides an upper bound: k!pk(n) ≤#{compositions of n + k(k −1)/2 into k parts}. Thus, pk(n) ≤1 k! n + k(k −1)/2 −1 k −1 . If k is ﬁxed as n →∞, then this is approximately nk−1/(k!(k −1)!). Thus, we have shown the following theorem: Theorem 19.1. pk(n) ∼1 k! nk−1 (k −1)!. This means there are lots and lots of such partitions: it grows faster than any polynomial. One can also use generting functions to analyze these: the convergence of a generating function yields information about the coeﬃcients of a recurrence. Example 19.2. Consider the Fibonacci numbers: F0 = 0, 1 = 1, and Fn+2 = Fn + fn+1. The generating function is f (x) = P∞ n=0 Fnxn, so xf (x) = ∞ X n=0 Fnxn+1 = ∞ X n=1 Fn−1xn, and x2f (x) = ∞ X n=0 Fnxn+2 = ∞ X n=2 Fn−2xn. Thus, f (x) −xf (x) −x2f (x) = x, using the recurrence Fn = Fn−1 = Fn−2. Thus, one can solve to get f (x) = x/(1 −x −x2). No discussion of convergence has been mentioned here; these are just formal power series. Using a partial fraction decomposition, 1 −x −x2 = − x − √ 5 −1 2 x + 1 + √ 5 2 = √ 5 + 1 2 1 + x ( √ 5 + 1)/2 √ 5 −1 2 1 − x ( √ 5 −1)/2 = 1 + x ( √ 5 + 1)/2 1 + x ( √ 5 −1)/2 . Thus, we have x 1 −x −x2 = A 1 + x(1 + √ 5)/2 + B 1 −x( √ 5 −1)/2, 31 with 1 1 −x(1 + √ 5)/2 = ∞ X n=0 1 + √ 5 2 n xn, and 1 1 −x(1 − √ 5)/2 = ∞ X n=0 1 − √ 5 2 n xn. Thus, Fn = A √ 5 −1 2 n + B √ 5 + 1 2 n , so asympotically, they are exponential, behaving like a constant times the exponential in the Golden Ratio. This is nice because it’s not obvious from the recurrence. ( Example 19.3. The generating function for p(n) is P(x) = P∞ n=0 p(n)xn. The geometric series gives us that ∞ Y n=1 (1 −xn)−1 = ∞ Y n=1 ∞ X j=0 xjn. In this inﬁnite product, the coeﬃcient of xn is the number of ways to write n as a sum of other numbers, or p(n).22 This product Q(1 −xn)−1 converges when P log(1 −xn)−1 does, and after a little bit of work this works when \|x\| < 1. The general technique here is for a generating function P∞ n=0 a(n)xn = A(x), assume a(n) ≥0, so that a(n) ≤x−nA(x). Then, the minimum value of x−nA(x), which is just a single-variable calculus problem, is a reasonable choice for an upper bound of the function. Thus, in this speciﬁc case, p(n) ≤x−nP(x) for all 0 < x < 1, so we want p(n) ≤min0<x<1(x−nP(x)). Thus, use logarithms: log(x−nP(x)) = log P(x) −n log x = ∞ X k=1 log(1 −xk)−1 −n log x, so diﬀerentiating, ∞ X k=1 kxk−1 1 −xk = n k , so n = ∞ X k=1 kxk 1 −xk .23 Of course, it is a great pain to ﬁgure out what this exactly means. One will need to solve for x0 such that this is maximized, so the bound is obtained. This looked complicated, but of course that’s because it is. A rough idea for x0 is that it should go to 1, but how can this be quantiﬁed in terms of n? Crudely, if x →1, then 1 −xk →0, but a little more precisely, 1 −xk = (1 −x)(1 + x + · · · + xk−1) ≈k(1 −x). Thus, n ≈x/(1 −x)2, so 1 −x ≈1/√n. This calculation can be made more precise, of course, and once the details are worked out, then one sees that p(n) < Ceπ√ 2n/3 √n . ( The true bound is a beautiful theorem: Theorem 19.4 (Hardy-Ramanujan). p(n) ∼eπ√ 2n/3 2n √ 3 . 22A lot of this may feel handwavy, especially in terms of convergence, but it’s possible to work in the ring of formal power series over a ﬁeld, in which the goal is to only consider formal expressions, and ignore questions of convergence. Then, two power series are the same if their coeﬃcients are termwise the same. However, one could also argue convergence on some interval, which is an equally valid, but alternate, approach. 23At this point, you’re supposed to ﬁgure out if it is a minimum or maximum, but we know here it must be a minimum. 32
15257	https://www.quora.com/How-do-I-prove-a-function-is-increasing	How to prove a function is increasing - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Increasing Intervals Derivatives (finance) Monotonic Function Functions (general) Calculus (Mathematics) Increasing/Decreasing Fun... Derivatives and Different... Continuous Functions 5 How do I prove a function is increasing? All related (37) Sort Recommended Sebastian Bozlee MA in Mathematics, University of Colorado Boulder (Graduated 2016) · Author has 168 answers and 332.5K answer views ·8y To show that f(x)f(x) is increasing, the goal is always to show that if x<y x<y then f(x)0 f′(x)>0. You can’t always do it this way, though. Let’s do an example where the derivative method works. Consider f(x)=x 3+x f(x)=x 3+x. This is an increasing function since f′(x)=3 x 2+1 f′(x)=3 x 2+1 is always greater than or equal to 1, since x 2≥0 x 2≥0. (Graph f(x)f(x) and f′(x)f′(x)!) However, just checking that the derivative is positive doesn’t work if your domain is not an int Continue Reading To show that f(x)f(x) is increasing, the goal is always to show that if x<y x<y then f(x)0 f′(x)>0. You can’t always do it this way, though. Let’s do an example where the derivative method works. Consider f(x)=x 3+x f(x)=x 3+x. This is an increasing function since f′(x)=3 x 2+1 f′(x)=3 x 2+1 is always greater than or equal to 1, since x 2≥0 x 2≥0. (Graph f(x)f(x) and f′(x)f′(x)!) However, just checking that the derivative is positive doesn’t work if your domain is not an interval. For example, consider the function f(x):(−1,1)∪(2,4)f(x):(−1,1)∪(2,4) defined by f(x)=x f(x)=x on (−1,1)(−1,1) and f(x)=x−4 f(x)=x−4 on (2,4)(2,4). The derivative is always positive: f′(x)=1 f′(x)=1 for x x in (−1,1)∪(2,4)(−1,1)∪(2,4). Despite this, f(x)f(x) is not an increasing function, since 0<3 0<3 but f(0)=0 f(0)=0 is not less than f(3)=−1 f(3)=−1. (Graph f(x)f(x) and f′(x)f′(x)!) Conversely, not every increasing function has f′(x)>0 f′(x)>0! For example, take f(x)=x 3 f(x)=x 3. Since x<y x<y implies x 3<y 3 x 3<y 3, f f is an increasing function, but its derivative is not always positive. In fact, f′(0)=0 f′(0)=0. The situation gets worse when you have functions that don’t really have derivatives. Take, for example, the function f(n)=n!f(n)=n!, defined for the positive integers. f(x)f(x) is increasing, but it doesn’t even have a derivative. Upvote · 99 14 9 1 Related questions More answers below How can you prove that a function is increasing or decreasing without using the definition? How can you prove a function is increasing without using calculus? How do I prove a function is decreasing without using calculus? What is an increasing function? How do I prove erf(x) is an increasing function? David Joyce Professor Emeritus of Mathematics at Clark University · Upvoted by Allan Steinhardt , PhD EE, published in various math journals, inventor, hyperbolic Householder and Luís Sequeira , PhD Mathematics, University of Lisbon (2001) · Author has 9.9K answers and 68.4M answer views ·6y Related How do I prove that a function is increasing/decreasing on an interval without differentiating? For example, the functions x 2,1/x,x 2,1/x, or √x.x. You can directly prove a function is increasing or decreasing from the definition. Here are the definitions for strictly increasing and strictly decreasing, sometimes just called increasing and decreasing, respectively. A function f f is increasing on an interval I I if for all a,b∈I a,b∈I with a<b a<b it is the case that f(a)<f(b),f(a)<f(b), equivalently, f(b)−f(a)f(b)−f(a) is positive. It is decreasing if a<b a<b implies f(b)<f(a),f(b)<f(a), equivalently, f(b)−f(a)f(b)−f(a) is negative. Each function will require a different proof. Example. The function f(x)=x 2 f(x)=x 2 is increasing on the interval [0,∞).[0,∞). Suppose a a and b b are two non-negative numbers Continue Reading You can directly prove a function is increasing or decreasing from the definition. Here are the definitions for strictly increasing and strictly decreasing, sometimes just called increasing and decreasing, respectively. A function f f is increasing on an interval I I if for all a,b∈I a,b∈I with a<b a<b it is the case that f(a)<f(b),f(a)<f(b), equivalently, f(b)−f(a)f(b)−f(a) is positive. It is decreasing if a<b a<b implies f(b)<f(a),f(b)<f(a), equivalently, f(b)−f(a)f(b)−f(a) is negative. Each function will require a different proof. Example. The function f(x)=x 2 f(x)=x 2 is increasing on the interval [0,∞).[0,∞). Suppose a a and b b are two non-negative numbers with a<b.a<b. Then f(b)−f(a)=b 2−a 2=(b−a)(b+a).f(b)−f(a)=b 2−a 2=(b−a)(b+a). That’s positive since b>a b>a and both a a and b b are positive. Example. The function f(x)=1/x f(x)=1/x is decreasing on the interval (0,∞).(0,∞). Let a a and b b are two non-negative numbers with a<b.a<b. Then f(b)−f(a)=1 b−1 a=a−b a b.f(b)−f(a)=1 b−1 a=a−b a b. That’s negative since b>a b>a and both a a and b b are positive. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 99 29 Assistant Bot · 1y To prove that a function f(x)f(x) is increasing on an interval, you can use the following approaches: Using the Derivative A common method involves the derivative of the function: Step 1: Compute the derivative f′(x)f′(x). Step 2: Show that f′(x)≥0 f′(x)≥0 for all x x in the interval of interest. Step 3: If f′(x)>0 f′(x)>0 for all x x in the interval, then f(x)f(x) is strictly increasing. If f′(x)=0 f′(x)=0 at some points but is non-negative elsewhere, f(x)f(x) is non-decreasing. Using the Definition of Increasing Functions You can also use the formal definition: Step 1: For any two points a a and b b in the interval where a<b a<b, show Continue Reading To prove that a function f(x)f(x) is increasing on an interval, you can use the following approaches: Using the Derivative A common method involves the derivative of the function: Step 1: Compute the derivative f′(x)f′(x). Step 2: Show that f′(x)≥0 f′(x)≥0 for all x x in the interval of interest. Step 3: If f′(x)>0 f′(x)>0 for all x x in the interval, then f(x)f(x) is strictly increasing. If f′(x)=0 f′(x)=0 at some points but is non-negative elsewhere, f(x)f(x) is non-decreasing. Using the Definition of Increasing Functions You can also use the formal definition: Step 1: For any two points a a and b b in the interval where a<b a<b, show that f(a)≤f(b)f(a)≤f(b). Step 2: If this condition holds for all a a and b b in the interval, then the function is increasing on that interval. Using the Mean Value Theorem If f(x)f(x) is continuous on the closed interval [a,b][a,b] and differentiable on the open interval (a,b)(a,b): Step 1: Apply the Mean Value Theorem, which states that there exists some c∈(a,b)c∈(a,b) such that: f′(c)=f(b)−f(a)b−a f′(c)=f(b)−f(a)b−a Step 2: If you can show that f′(c)>0 f′(c)>0, then f(b)>f(a)f(b)>f(a) and thus f(x)f(x) is increasing on [a,b][a,b]. Example Consider the function f(x)=x 2 f(x)=x 2 on the interval [0,∞)[0,∞): Derivative method: Compute f′(x)=2 x f′(x)=2 x. For x≥0 x≥0, f′(x)≥0 f′(x)≥0. Therefore, f(x)f(x) is increasing on [0,∞)[0,∞). Definition method: For any a,b∈[0,∞)a,b∈[0,∞) with a<b a<b, f(a)=a 2 f(a)=a 2 and f(b)=b 2 f(b)=b 2. Since a<b a<b implies a 2<b 2 a 2<b 2 (for a,b≥0 a,b≥0), it follows that f(a)<f(b)f(a)<f(b). By using these methods, you can effectively prove that a function is increasing on a specified interval. Upvote · Vincent Cavallo I try sometimes to answer some maths issues · Author has 65 answers and 152.5K answer views ·8y A few ways of doing it : Prove that for all x, y, x>y => f(x)>f(y) If your function is differentiable, find its derivative : your function is increasing whenever it’s derivative is positive. Chose the most appropriate way, but these 2 should cover most of your cases. Upvote · 9 2 9 2 Related questions More answers below What is the minimum value for increasing a function? How do I increase of pancreas function? How does one prove that e x e x is an increasing function on R? How can I prove that the following function is increasing in x: ∞∑i=1 x(1−x)i−1 log(1+μ(1−x)i−1)∑i=1∞x(1−x)i−1 log⁡(1+μ(1−x)i−1)? How do you prove that the exponential function e^x is an increasing function? Muhammad Farhan Azmine 8y After taking the differentiation if u see the f`(x) differetiated fuction is increasing for a certain range of x value, the function is increasing for those values of x Upvote · Evan De Vries 3y Originally Answered: What are the ways I can prove that a particular function grows? · By finding the derivative of a function, you can see how the function grows over time, and thus find if the function grows. Alternatively you can just graph the function and observe which direction it grows Upvote · Ashutosh Studied Joint Entrance Examination&Indian Institutes of Technology at FIITJEE ·4y Related What is the difference between increasing and strictly increasing function? A strictly increasing function can be simply understood as a function that is always increasing, Mathematically, We say a function is strictly increasing on the interval I I(closed, open, semiclosed) if f(x 1)<f(x 2)f(x 1)<f(x 2) for all x 1<x 2 x 1<x 2 in I I. If the function is differentiable then f′(x)>0 f′(x)>0, & in case f′(x)=0 f′(x)=0 then it will happen only for some discrete value of x only not for an interval. A increasing function can be simply understood as a function that is always non decreasing, Mathematically, We say a function is increasing on the interval I I(closed, open, semiclosed) if f(x 1)≤f(x 2)f(x 1)≤f(x 2) for Continue Reading A strictly increasing function can be simply understood as a function that is always increasing, Mathematically, We say a function is strictly increasing on the interval I I(closed, open, semiclosed) if f(x 1)<f(x 2)f(x 1)<f(x 2) for all x 1<x 2 x 1<x 2 in I I. If the function is differentiable then f′(x)>0 f′(x)>0, & in case f′(x)=0 f′(x)=0 then it will happen only for some discrete value of x only not for an interval. A increasing function can be simply understood as a function that is always non decreasing, Mathematically, We say a function is increasing on the interval I I(closed, open, semiclosed) if f(x 1)≤f(x 2)f(x 1)≤f(x 2) for all x 1<x 2 x 1<x 2 in I I. If the function is differentiable then f′(x)>0 f′(x)>0, & in case f′(x)=0 f′(x)=0 then it will happen for an interval. Since, Every strictly increasing function is also an increasing function so in case f′(x)=0 f′(x)=0 then it will happen also for some discrete value of x. So, we came to the conclusion that in case f′(x)=0 f′(x)=0, it can happen for both descrete point as well as for an interval. So, It doesnot matter whether f′(x)=0 f′(x)=0 or not, In either cases f(x) is increasing. All we need is just f′(x)>0 f′(x)>0 for increasing function. Upvote · 99 19 9 1 Mavuso Dingani BS in Physics, University at Albany ·8y essay prove that its derivative is always positive> for example, x^3 is an increasing function because its derivative is 3x^2 and x^2 will always be positive in R^1 Upvote · 9 1 Reuven Harmelin Lecturer at Technion - Israel Institute of Technology (1982–present) · Author has 2.3K answers and 1.9M answer views ·4y Related How do you prove $a^x$ is an increasing function (calculus, functions, exponential function, math)? It is wrong, unless you add the requirement that a>1. It is quite a basic consequence of 2 of the most elementary properties of exponential functions on base a>1, namely (i) (a^x)(a^y)=a^(x+y) (ii) a^x>=1+kx where k=ln(a)>0 (ln(a)=natural logarithm of a) The last inequality demonstrated by the following picture in which the black curve is the graph of y=a^x for a>1, and it is lying above the red line which is the graph of y=1+kx: From the 2 properties (i-ii) you may deduce that: (a^0)(a^0)=a^(0+0)=a^0>=1+0=1 ==> a^0=1 and therefore (a^(-y))(a^y)=a^(-y+y)=a^0=1. Hence: a^(-y)=1/(a^y) ==> (a^x)/(a^ Continue Reading It is wrong, unless you add the requirement that a>1. It is quite a basic consequence of 2 of the most elementary properties of exponential functions on base a>1, namely (i) (a^x)(a^y)=a^(x+y) (ii) a^x>=1+kx where k=ln(a)>0 (ln(a)=natural logarithm of a) The last inequality demonstrated by the following picture in which the black curve is the graph of y=a^x for a>1, and it is lying above the red line which is the graph of y=1+kx: From the 2 properties (i-ii) you may deduce that: (a^0)(a^0)=a^(0+0)=a^0>=1+0=1 ==> a^0=1 and therefore (a^(-y))(a^y)=a^(-y+y)=a^0=1. Hence: a^(-y)=1/(a^y) ==> (a^x)/(a^y)=(a^x)(a^(-y))=a^(x-y) Thus, whenever x>y, so that x-y>0, it readily follows that (a^x)/(a^y)=a^(x-y)>=1+k(x-y)>1 ==> a^x>a^y Q.E.D Upvote · Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views ·1y Related How can it be proven that the function f(x) = (1/x) √(x+1) is strictly increasing? The logical idea is that if we can show the gradient is always positive then the function must always be increasing! So let’s find the gradient… Continue Reading The logical idea is that if we can show the gradient is always positive then the function must always be increasing! So let’s find the gradient… Upvote · 99 10 Awnon Bhowmik Studied at University of Dhaka · Author has 3.7K answers and 11.2M answer views ·8y Related What are the best ways to show that a function f(x) is increasing/decreasing? A2A I would give you the simplest answer that pops up in my head. Take a function f(x)f(x), take the first derivative f′(x)f′(x). The interval(s) where f′(x)>0⟺f(x)↗f′(x)>0⟺f(x)↗ The interval(s) where f′(x)<0⟺f(x)↘f′(x)<0⟺f(x)↘ For the case of monotonic functions, we have either f′(x)≥0 f′(x)≥0 or f′(x)≤0 f′(x)≤0 And for the case of strictly monotonic functions, we have either f′(x)>0 f′(x)>0 or f′(x)<0 f′(x)<0 Upvote · 9 9 Paul K. Young Former PhD candidate in Mathematics, lecturer in Calculus at Cornell · Author has 1.1K answers and 18.4M answer views ·9y Related How do I prove a function is decreasing without using calculus? There are many non-calculus techniques that can be applied to show a function is increasing. One approach is to show for k > 0 that f(x + k) is larger than f(x). Here’s a simple example. Suppose f(x)f(x) is linear. That is f(x)=m x+b f(x)=m x+b. Now take k>0 k>0 and compare f(x)f(x) to f(x+k)f(x+k). f(x+k)=m(x+k)+b=m x+b+m k f(x+k)=m(x+k)+b=m x+b+m k f(x+k)−f(x)=m k f(x+k)−f(x)=m k Since k>0 k>0 we know m k m k is positive iff m m is positive. So f(x)f(x) is increasing iff m m is positive. Here’s a slightly more involved example. Suppose f(x)=x 3.f(x)=x 3. f(x+k)=(x+k)3=x 3+3 k x 2+3 k 2 x+k 3.f(x+k)=(x+k)3=x 3+3 k x 2+3 k 2 x+k 3. f(x+k)−f(x)=3 k x 2+3 k 2 x+k 3 f(x+k)−f(x)=3 k x 2+3 k 2 x+k 3 For x>0 x>0 all terms Continue Reading There are many non-calculus techniques that can be applied to show a function is increasing. One approach is to show for k > 0 that f(x + k) is larger than f(x). Here’s a simple example. Suppose f(x)f(x) is linear. That is f(x)=m x+b f(x)=m x+b. Now take k>0 k>0 and compare f(x)f(x) to f(x+k)f(x+k). f(x+k)=m(x+k)+b=m x+b+m k f(x+k)=m(x+k)+b=m x+b+m k f(x+k)−f(x)=m k f(x+k)−f(x)=m k Since k>0 k>0 we know m k m k is positive iff m m is positive. So f(x)f(x) is increasing iff m m is positive. Here’s a slightly more involved example. Suppose f(x)=x 3.f(x)=x 3. f(x+k)=(x+k)3=x 3+3 k x 2+3 k 2 x+k 3.f(x+k)=(x+k)3=x 3+3 k x 2+3 k 2 x+k 3. f(x+k)−f(x)=3 k x 2+3 k 2 x+k 3 f(x+k)−f(x)=3 k x 2+3 k 2 x+k 3 For x>0 x>0 all terms are positive hence the function is increasing. At x=0 x=0 the difference is k 3 k 3 which is also positive. Now assume x<0 x<0. Note that we can take k to be arbitrarily small (but still positive) to show that f(x)f(x) is increasing. So let k<\|x\|k<\|x\|. Then we get f(x+k)−f(x)f(x+k)−f(x) =3 k x 2+3 k 2 x+k 3=3 k x 2+3 k 2 x+k 3 =k(3 x 2+3 k x)+k 3=k(3 x 2+3 k x)+k 3 k(3 x 2+3 k x)>k(3 x 2+3 k x) k(3 x 2+3\|x\|x)>k(3 x 2+3\|x\|x) - this works because x is negative. =k(3 x 2−3 x 2)=k(3 x 2−3 x 2) =0=0 So the difference is positive hence the function is increasing. In both cases the calculus approach is simpler. f′(x)=m f′(x)=m. So f f is increasing if m>0 m>0. f′(x)=3 x 2 f′(x)=3 x 2 which is positive except at the critical point x=0 x=0. But that means the function is increasing both to the left and to the right of x=0 x=0 so x=0 x=0 is an inflection point. Therefore the function is increasing. Upvote · 99 10 9 1 Jay Wacker theoretical physicist · Author has 4.4K answers and 26.9M answer views ·8y Related How do I prove erf(x) is an increasing function? What it is asking for is the derivative of the error function and asking whether it is always positve. The definition of the error function is \erf(x)=2√π∫x 0 exp(−t 2)d t\erf(x)=2 π∫0 x exp⁡(−t 2)d t This can easily be differentiated by the fundamental theorem of calculus d d x\erf(x)=2√π d d x∫x 0 exp(−t 2)d t=2√π exp(−x 2)d d x\erf(x)=2 π d d x∫0 x exp⁡(−t 2)d t=2 π exp⁡(−x 2) Thus the derivative of the error function is a Gaussian, which is a strictly positive function (assuming you’re only considering real arguments). Upvote · 9 7 Related questions How can you prove that a function is increasing or decreasing without using the definition? How can you prove a function is increasing without using calculus? How do I prove a function is decreasing without using calculus? What is an increasing function? How do I prove erf(x) is an increasing function? What is the minimum value for increasing a function? How do I increase of pancreas function? How does one prove that e x e x is an increasing function on R? How can I prove that the following function is increasing in x: ∞∑i=1 x(1−x)i−1 log(1+μ(1−x)i−1)∑i=1∞x(1−x)i−1 log⁡(1+μ(1−x)i−1)? How do you prove that the exponential function e^x is an increasing function? How do you prove a function is decreasing (calculus, math)? What is the definition of an increasing function? Does an increasing function always have a discontinuity? How can I prove that x^3 is strictly increasing, where x>=0? How do I prove that a function f is a one-to-one function? How do you prove that if f(x) is both increasing and decreasing then it must be one-to-one (or onto)? Related questions How can you prove that a function is increasing or decreasing without using the definition? How can you prove a function is increasing without using calculus? How do I prove a function is decreasing without using calculus? What is an increasing function? How do I prove erf(x) is an increasing function? What is the minimum value for increasing a function? How do I increase of pancreas function? How does one prove that e x e x is an increasing function on R? How can I prove that the following function is increasing in x: ∞∑i=1 x(1−x)i−1 log(1+μ(1−x)i−1)∑i=1∞x(1−x)i−1 log⁡(1+μ(1−x)i−1)? How do you prove that the exponential function e^x is an increasing function? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15258	https://www.superteacherworksheets.com/pictograph/elm-street-picto.pdf	Name: _____ Using a Pictograph Title: Number of Students at Elm Street School Table Pictograph Grade Number of Students Grade Number of students Kindergarten 35 Kindergarten     1st 40 1st 2nd 25 2nd 3rd 35 3rd 4th 30 4th Key Each  = 10 students 1. Use the data in the table to complete the pictograph. 2. How are the pictograph and the table alike? _________ 3. How are the pictograph and the table different? _________ 4. What is the purpose of the pictograph's key? _________ 5. Describe how you would find the total number of students at Elm Street School. _________ 6. What is the total number of students at Elm Street School? Show your work in the space below. Super Teacher Worksheets - www.superteacherworksheets.com Name: _____ Using a Pictograph- ANSWER KEY Title: Number of Students at Elm Street School Table Pictograph Grade Number of Students Grade Number of students Kindergarten 35 Kindergarten     1st 40 1st     2nd 25 2nd    3rd 35 3rd     4th 30 4th    Key Each  = 10 students 1. Use the data in the table to complete the pictograph. 2. How are the pictograph and the table alike? They both show the same data. 3. How are the pictograph and the table different? The table uses numbers to show data while the pictograph uses symbols. 4. What is the purpose of the pictograph's key? To show how many items/people is represented by each symbol. 5. Describe how you would find the total number of students at Elm Street School. Add up the number of students in each grade. 6. What is the total number of students at Elm Street School? Show your work in the space below. 35 + 40 + 25 + 35 + 30 = 165 students Super Teacher Worksheets - www.superteacherworksheets.com
15259	https://www.youtube.com/watch?v=mBr3c54Vhgk	Curious case of 0 \| Permutations and Combinations \| Grade 11 \| Math \| Khan Academy Khan Academy India - English 541000 subscribers 11 likes Description 504 views Posted: 19 Feb 2025 This video tackles the case where a digit is torn between two conflicting constraints. We try to make two digit numbers using a set of digits that include 0. We first solve for a case where repetition is allowed. We run into problems where repetition is not allowed - even though we are dealing with the last digit constraint. We quickly realise that the issue is not with the boxes but with the digit zero. We arrive at a neat solution involving cases and use it to solve another similar practice problem. Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now! ( Timestamps: 00:00 Problem 1 (Find all even numbers, 0 is one of the digits) 00:21 Solution 1 (repetition is allowed) 01:53 Problem 2 (Same digits, repetition not allowed) 02:03 Attempt 1 (taking no even values) 02:34 Comparing result with actual 02:57 Attempt 2 (taking one even value) 03:28 Comparing result with actual 03:50 Breaking down the issue with 0 04:18 Separating the lists through cases 05:12 Solving case 1 (last digit zero) 05:33 Solving case 2 (last digit non zero) 06:15 Practice problem 2 (3 digit even numbers) 06:44 Solving case 1 (last digit zero) 07:17 Solving case 2 (last digit non zero) 08:33 Adding results from both cases Khan Academy India is a nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. We have videos and exercises that have been translated into multiple Indian languages, and 15 million people around the globe learn on Khan Academy every month. Support Us: Created by Ashish Gupta 1 comments Transcript: Problem 1 (Find all even numbers, 0 is one of the digits) here's another problem we have 0 1 2 and 3 we have these four digits and we have to make two digigit even numbers and the digits are allowed to repeat so two digit numbers will have these two digits we have to fill this digit and we have to fill this one let's set this one up we can pick from these four digits we're making a two-digit even number and the Solution 1 (repetition is allowed) repetition is allowed here are the boxes and we have to make sure that this is even so pause the video think about it okay so because we have to make sure this is even this is a constraint and because this is a constraint we'll deal with this first so let's do that so ping box can't have one or three because we want to make it even so it can pick anything from zero and two so there are two options so we'll put two ways and let's say it picks zero so we'll put zero here let's move to the next one because repetition is allowed all options are available but are all options really available think about it we're making a two-digit number is 01 a two-digit number is 02 a two-digit number they're not they're one-digit numbers this means there is a natural constraint here we cannot start with zero this means we cannot have zero in this box so when we picking digits for the yellow box we cannot have zero here so zero is out we can pick from one 2 or three which means we have three ways so let's say we pick one we got our example as 10 and the number of ways is 3 2 that's six ways and because 6 is such a small number let's actually list down these numbers we have 1 0 2 0 3 0 1 2 22 and 32 that's it these are your six two-digit even numbers when you allow for repetition and what happens if you don't allow for repetition let's see let's pick the same numbers but this time repetition is not allowed here are Problem 2 (Same digits, repetition not allowed) your boxes you can pause the video and start having fun all right let's do this together this this is a constraint which means we have to take care of this first Attempt 1 (taking no even values) which means we'll start with a pink box so it can't have one it can't have three it can have anything from zero and two so we have two ways to do this let's pick a number let's say we pick two so two's here so for this yellow box how many options do we have we can't start with zero so zero is off and we also can't pick two so we have one and we have three so this box also has two ways let's put a number here let's put one all right so how many total numbers we have 2 2 that's 4 let's check our Comparing result with actual answer we had 10 20 30 12 22 and 32 the only number that actually has repetition is 22 let's strike that out how many numbers are left we have 1 2 3 4 and five we actually have five such numbers and the answer that we're getting is four is there something wrong with our approach let's try it again for the pink Attempt 2 (taking one even value) box we can't have one we can't have three we can pick from 0 and two so it has two ways last time we picked two let's say this time we pick zero so let's put 0er here what about the yellow one we can't startop with zero but we don't have to worry about it because zero is taken care of zero has been already picked we're left with these three digits 1 2 and three this means there are three ways to fill this let's say we pick in one so one goes here and the total number of ways are 3 2 that's 6 wait 6 is also not the right Comparing result with actual answer we just saw that we had five such numbers this means we again got the wrong answer what's actually happening here pause the video think about it what's wrong with our approach and what can we do to fix it all right let's resolve this mystery together we have 0 1 2 and 3 and we have to fill these two boxes the problem here is not in the boxes the problem is in zero this box Breaking down the issue with 0 says we never want to see Zero here and this pink box says well sometimes you can have zero here but but not always we would also want to have two here and this is what's Happening Here zero is torn between these two boxes let's not let zero move around a lot instead let's tell zero where it can and can't be the way to do this will be to think about the situation in terms of cases here's what I mean you have these nice digits 1 Separating the lists through cases 2 and three and you have this zero which is causing us some problems we'll solve this problem twice we'll write 0 1 2 3 here and we'll write 0 1 2 3 here again and we'll separate these two worlds so here are boxes for the left world and here are boxes for the right world now in this left World we're calling it case one we're telling zero to go to the last digit and stay there and in the right World which we're calling case two we're saying the exact opposite we're saying never go to the last digit the last digit is non zero which means zero can never be here think about it on the left we'll count the numbers which have last digit zero and on the right we'll count the numbers where the last digit is never zero which means two things first there's not going to be any overlap between these two worlds and second together we'll cover all possible cases so let's get started if you have to put the last St zero you can do that in only one way and because zero is already Solving case 1 (last digit zero) taken care of we have these three digits for this yellow box so this box can be filled in three ways this means in the left world we have 3 1 that's three numbers those numbers are actually 10 20 and 30 all three of them have the last digit zero let's move to case two here we're saying the last digit can never be zero which means we can't use Solving case 2 (last digit non zero) zero here so if you have to make this number even the only other digit that we can use is two which means this box can be filled in only one way which is by using two and now for this yellow box we can't use zero because we want to make a two-digit number and we can't use two because two is already taken the only ones left are one and three so there are two ways to fill the yellow box which means in case two we have 2 1 that's two numbers and those numbers are 12 and 32 and we can pick a number from either left world or the right world so we'll add the numbers so 3 + 2 that means five and finally we get the right answer let's quickly practice what we have just learned here are six numbers 0 1 2 3 4 5 Practice problem 2 (3 digit even numbers) and 6 and using these digits we going to make three digigit even numbers and the repetition is not allowed and let me help you set things up for case one we're making the last digit zero we're fixing it and for case two we're making sure that that never happens we're making the last digit non zero here are the boxes for case one now you can pause the video and start solving it on your own all right fixing zero here means that you can fill this box in only one Solving case 1 (last digit zero) way that's by putting in zero so zero is gone next we have this box and we have these six digits so this box can be filled in six ways let's say we fill it with three so we are left with 1 2 3 4 and five five other digits which means the yellow box can be filled in five ways let's put a number here so for case one we have 5 6 that's 30 ways we have 30 such numbers here let's clean this up and move to case two we have our boxes and we're going to make sure that the last digit is not zero so how do we do Solving case 2 (last digit non zero) this we make this digit even without using zero which means we can use either two or four or six which means we can fill the blue box in three ways let's let's pick a number let's say we pick six all right so this digit is even now which box should we pick next if we pick Green we're going to run into some problems of these two green is the nice one green has no problems with any of these digits and yellow actually has a problem with this digit zero if you put in zero here this will no longer be a three-digit number because of these two this one is a constraint let's start with the yellow one we need to first take care of the boxes that have problems and then move to boxes that don't have any so for the yellow box how many options do we have we can't have zero but we can have anything from 1 to 3 4 and five six is already taken so five options for the yellow box let's pick one let's say we pick four now how many options for the green one well green one can take zero so it has 0 1 2 3 and five again five options let's pick one let's say we pick zero so for case 2 we have 5 5 3 that's 25 3 that's 75 cases so we have 30 numbers with last Adding results from both cases digit zero and we have 75 numbers with last digit not zero and together they will make up all the possible cases where we are having a three-digit even number and the repetition is not allowed so if we add both of them up we get 30 + 75 that's 105 105 ways
15260	https://flexbooks.ck12.org/cbook/ck-12-algebra-ii-with-trigonometry-concepts/section/10.5/primary/lesson/ellipses-centered-at-the-origin-alg-ii/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 10.5 Ellipses Centered at the Origin Written by:Lori Jordan \| Kate Dirga Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Your homework assignment is to draw the ellipse @$\begin{align}16x^2 + 4y^2 = 144\end{align}@$. Where will the foci of your graph be located? Ellipses Centered at the Origin The third conic section is an ellipse. Recall that a circle is when a plane sliced through a cone and that plane is parallel to the base of the cone. An ellipse is formed when that plane is not parallel to the base. Therefore, a circle is actually a more specific version of an ellipse. By definition, an ellipse is the set of all points such that the sum of the distances from two fixed points, called foci (the plural of focus), is constant. Drawing an Ellipse We will use the definition of an ellipse to draw an ellipse. Step 1: On a piece of graph paper, draw a set of axes and plot @$\begin{align}(-2, 0)\end{align}@$ and @$\begin{align}(2, 0)\end{align}@$. These will be the foci. Step 2: From the definition, we can conclude a point @$\begin{align}(x, y)\end{align}@$ is on an ellipse if the sum of the distances is always constant. In the picture, @$\begin{align}d_1+d_2=r\end{align}@$ and @$\begin{align}g_1+g_2=r\end{align}@$. Step 3: Determine how far apart the foci are. Then, find @$\begin{align}d_1\end{align}@$ and @$\begin{align}d_2\end{align}@$. Step 4: Determine if the point @$\begin{align}(-2, 3)\end{align}@$ is on the ellipse. In this concept, the center of an ellipse will be @$\begin{align}(0, 0)\end{align}@$. An ellipse can have either a vertical or horizontal orientation (see below). There are always two foci and they are on the major axis. The major axis is the longer of the two axes that pass through the center of an ellipse. Also on the major axis are the vertices, which its endpoints and are the furthest two points away from each other on an ellipse. The shorter axis that passes through the center is called the minor axis, with endpoints called co-vertices. The midpoint of both axes is the center. Equation of an Ellipse, Centered at the Origin \| @$\begin{align}\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1\end{align}@$ \| HORIZONTAL major axis is the @$\begin{align}x\end{align}@$-axis with length @$\begin{align}2a\end{align}@$. minor axis is the @$\begin{align}y\end{align}@$-axis with length @$\begin{align}2b\end{align}@$. \| \| @$\begin{align}\frac{x^2}{b^2}+ \frac{y^2}{a^2}=1\end{align}@$ \| VERTICAL major axis is the @$\begin{align}y\end{align}@$-axis with length @$\begin{align}2a\end{align}@$. minor axis is the @$\begin{align}x\end{align}@$-axis with length @$\begin{align}2b\end{align}@$. \| Other Important Facts @$\begin{align}a\end{align}@$ is ALWAYS greater than @$\begin{align}b\end{align}@$. If they are equal, we have a circle. The foci, vertices, and co-vertices relate through a version of the Pythagorean Theorem: @$\begin{align}c^2=a^2-b^2\end{align}@$ Let's find the vertices, co-vertices, and foci of @$\begin{align}\frac{x^2}{64}+ \frac{y^2}{25}=1\end{align}@$. Then, let's graph the ellipse. First, we need to determine if this is a horizontal or vertical ellipse. Because @$\begin{align}64 > 25\end{align}@$, we know that the ellipse will be horizontal. Therefore, @$\begin{align}a^2=64\end{align}@$ making @$\begin{align}a=\sqrt{64}=8\end{align}@$ and @$\begin{align}b^2=25\end{align}@$, making @$\begin{align}b=\sqrt{25}=5\end{align}@$. Using the pictures above, the vertices will be @$\begin{align}(8, 0)\end{align}@$ and @$\begin{align}(-8, 0)\end{align}@$ and the co-vertices will be @$\begin{align}(0, 5)\end{align}@$ and @$\begin{align}(0, -5)\end{align}@$. To find the foci, we need to use the equation @$\begin{align}c^2=a^2-b^2\end{align}@$ and solve for @$\begin{align}c\end{align}@$. @$$\begin{align}c^2&=64-25=39 \ c&=\sqrt{39}\end{align}@$$ The foci are @$\begin{align}\left(\sqrt{39},0\right)\end{align}@$ and @$\begin{align}\left(- \sqrt{39},0\right)\end{align}@$. To graph the ellipse, plot the vertices and co-vertices and connect the four points to make the closed curve. Now, let's graph @$\begin{align}49x^2+9y^2=441\end{align}@$ and identify the foci. This equation is not in standard form. To rewrite it in standard form, the right side of the equation must be 1. Divide everything by 441. @$$\begin{align}\frac{49x^2}{441}+ \frac{9y^2}{441}&= \frac{441}{441} \ \frac{x^2}{9}+ \frac{y^2}{49}&=1\end{align}@$$ Now, we can see that this is a vertical ellipse, where @$\begin{align}b=3\end{align}@$ and @$\begin{align}a=7\end{align}@$. To find the foci, use @$\begin{align}c^2=a^2-b^2\end{align}@$. @$$\begin{align}c^2&=49-9=40 \ c&=\sqrt{40}=2\sqrt{10}\end{align}@$$ The foci are @$\begin{align}\left(0,2\sqrt{10}\right)\end{align}@$ and @$\begin{align}\left(0,-2\sqrt{10}\right)\end{align}@$. Finally, let's write equations for the ellipses with the given characteristics below and centered at the origin. In either part, you may wish to draw the ellipse to help with the orientation. vertex: @$\begin{align}(-6, 0)\end{align}@$, co-vertex: @$\begin{align}(0, 4)\end{align}@$ We can conclude that @$\begin{align}a=6\end{align}@$ and @$\begin{align}b=4\end{align}@$. The ellipse is horizontal, because the larger value, @$\begin{align}a\end{align}@$, is the @$\begin{align}x\end{align}@$-value of the vertex. The equation is @$\begin{align}\frac{x^2}{36}+ \frac{y^2}{16}=1\end{align}@$. vertex: @$\begin{align}(0, 9)\end{align}@$, focus: @$\begin{align}(0, -5)\end{align}@$ We know that @$\begin{align}a=9\end{align}@$ and @$\begin{align}c=5\end{align}@$ and that the ellipse is vertical. Solve for @$\begin{align}b\end{align}@$ using @$\begin{align}c^2=a^2-b^2\end{align}@$ @$$\begin{align}5^2&=9^2-b^2 \ 25&=81-b^2 \ b^2&=56 \rightarrow b=2\sqrt{14}\end{align}@$$ The equation is @$\begin{align}\frac{x^2}{56}+ \frac{y^2}{81}=1\end{align}@$ Examples Example 1 Earlier, you were asked to determine where the foci of your graph would be located. This equation is not in standard form. To rewrite it in standard form, the right side of the equation must be 1. Divide everything by 144. @$$\begin{align}\frac{16x^2}{144}+ \frac{4y^2}{144}&= \frac{144}{144} \ \frac{x^2}{9}+ \frac{y^2}{36}&=1\end{align}@$$ Now, we can see that this is a vertical ellipse, where @$\begin{align}b=3\end{align}@$ and @$\begin{align}a=6\end{align}@$. To find the foci, use @$\begin{align}c^2=a^2-b^2\end{align}@$. @$$\begin{align}c^2&=36-9=27 \ c&=\sqrt{27}=3\sqrt{3}\end{align}@$$ The foci are therefore @$\begin{align}\left(0,3\sqrt{3}\right)\end{align}@$ and @$\begin{align}\left(0,-3\sqrt{3}\right)\end{align}@$. Example 2 Find the vertices, co-vertices, and foci of @$\begin{align}\frac{x^2}{4}+ \frac{y^2}{36}=1\end{align}@$. Then, graph the equation. Because the larger number is under @$\begin{align}y^2\end{align}@$, the ellipse is vertical. Therefore, @$\begin{align}a=6\end{align}@$ and @$\begin{align}b^2\end{align}@$. Use @$\begin{align}c^2=a^2-b^2\end{align}@$ to find @$\begin{align}c\end{align}@$. @$$\begin{align}c^2&=6^2-2^2=36-4=32 \ c&=\sqrt{32}=4\sqrt{2}\end{align}@$$ vertices: @$\begin{align}(0, 6)\end{align}@$ and @$\begin{align}(0, -6)\end{align}@$ co-vertices: @$\begin{align}(2, 0)\end{align}@$ and @$\begin{align}(-2, 0)\end{align}@$ foci: @$\begin{align}\left(0,4\sqrt{2}\right)\end{align}@$ and @$\begin{align}\left(0,-4\sqrt{2}\right)\end{align}@$ Example 3 Graph @$\begin{align}49x^2+64y^2=3136\end{align}@$ and find the foci. Rewrite @$\begin{align}49x^2+64y^2=3136\end{align}@$ in standard form. @$$\begin{align}\frac{49x^2}{3136}+ \frac{64y^2}{3136}&=\frac{3136}{3136} \ \frac{x^2}{64}+ \frac{y^2}{49}&=1\end{align}@$$ This ellipse is horizontal with @$\begin{align}a=8\end{align}@$ and @$\begin{align}b=7\end{align}@$. Find @$\begin{align}c\end{align}@$. @$$\begin{align}c^2&=64-49=15 \ c&=\sqrt{15}\end{align}@$$ The foci are @$\begin{align}\left(- \sqrt{15},0\right)\end{align}@$ and @$\begin{align}\left(\sqrt{15},0\right)\end{align}@$. Example 4 Find the equation of the ellipse with co-vertex @$\begin{align}(0, -7)\end{align}@$, focus @$\begin{align}(15, 0)\end{align}@$ and centered at the origin. Because the co-vertex is @$\begin{align}(0, -7), b = 7\end{align}@$ and the ellipse is horizontal. From the foci, we know that @$\begin{align}c = 15\end{align}@$. Find @$\begin{align}a\end{align}@$. @$$\begin{align}15^2&=a^2-7^2 \ a^2&=225+49=274 && \text{The equation is} \ \frac{x^2}{274}+ \frac{y^2}{49}=1. \ a&=\sqrt{274}\end{align}@$$ Review Find the vertices, co-vertices, and foci of each ellipse below. Then, graph. @$\begin{align}\frac{x^2}{9}+ \frac{y^2}{16}=1\end{align}@$ @$\begin{align}4x^2+25y^2=100\end{align}@$ @$\begin{align}\frac{x^2}{64}+y^2=1\end{align}@$ @$\begin{align}81x^2+100y^2=8100\end{align}@$ @$\begin{align}\frac{x^2}{49}+ \frac{y^2}{16}=1\end{align}@$ @$\begin{align}121x^2+9y^2=1089\end{align}@$ Find the equation of the ellipse, centered at the origin, with the given information. vertex: @$\begin{align}(-3, 0)\end{align}@$ co-vertex: @$\begin{align}(0, 1)\end{align}@$ co-vertex: @$\begin{align}(7, 0)\end{align}@$ major axis: 18 units vertex: @$\begin{align}(0, 5)\end{align}@$ minor axis: 4 units vertex: @$\begin{align}(0, 6)\end{align}@$ co-vertex: @$\begin{align}(-2, 0)\end{align}@$ co-vertex: @$\begin{align}(17, 0)\end{align}@$ focus: @$\begin{align}(0, 17)\end{align}@$ vertex: @$\begin{align}(4, 0)\end{align}@$ focus: @$\begin{align}(-3, 0)\end{align}@$ co-vertex: @$\begin{align}(-6, 0)\end{align}@$ focus: @$\begin{align}(0, 5)\end{align}@$ focus: @$\begin{align}(0, -9)\end{align}@$ minor axis: 16 units Real Life Application A portion of the backyard of the White House is called The Ellipse. The major axis is 1058 feet and the minor axis is 903 feet. Find the equation of the horizontal ellipse, assuming it is centered at the origin. Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- \| \| \| Credit: Ad Meskens Source: \| \| \| \| Credit: Ad Meskens Source: \| \| \| \| Credit: Ad Meskens Source: \| \| \| \| Credit: Ad Meskens Source: \| \| \| \| Credit: Ad Meskens Source: \| \| \| \| Credit: Ad Meskens Source: License: CC BY-NC 3.0 \| \| \| \| Credit: Ad Meskens Source: \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. 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15262	https://www.quora.com/Given-matrices-A-and-B-how-do-I-solve-for-C-in-the-equation-A-BC-using-a-non-numerical-method	Given matrices A and B how do I solve for C in the equation A=BC using a non-numerical method? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Given matrices A and B how do I solve for C in the equation A=BC using a non-numerical method? All related (38) Sort Recommended Charles H Martin Calculation Consulting; we predict things · Upvoted by Daniel McLaury , Ph.D. Student in Mathematics at University of Illinois at Chicago · Author has 1.2K answers and 4.1M answer views ·13y More generally, we can define a family of solutions parameterized by lambda This is called Tikhonhov Regularization: and has been studied recently and extensively by Smale (the Fields Medal winner) Obviously this is a formal solution requires a numerical solver to obtain values for C . This can be generalized by adding more adjustable parameters (i.e. a full matrix instead of just lambda), or by performing matrix factorization (i.e. SVD) on This can also be solved via effective operator theor Continue Reading More generally, we can define a family of solutions parameterized by lambda This is called Tikhonhov Regularization: and has been studied recently and extensively by Smale (the Fields Medal winner) Obviously this is a formal solution requires a numerical solver to obtain values for C . This can be generalized by adding more adjustable parameters (i.e. a full matrix instead of just lambda), or by performing matrix factorization (i.e. SVD) on This can also be solved via effective operator theory, Bayesian linear regression, etc. Upvote · 99 13 Tracy Chou Founder & CEO, Block Party · Author has 579 answers and 7.5M answer views ·15y Originally Answered: Given matrices A and B how do I solve for C in the equation A=BC using a non numerical method? · If B B has full rank / is invertible, you can just solve for C=B−1 A C=B−1 A. Otherwise, the least-squares approximation is given by left-multiplying A A with the pseudoinverse: C=(B T B)−1 B T A C=(B T B)−1 B T A. Upvote · 99 13 9 2 Assistant Bot · 10mo To solve for matrix C C in the equation A=B C A=B C using a non-numerical method, you can follow these steps: Step 1: Understanding the Equation The equation A=B C A=B C implies that you want to express C C in terms of A A and B B. Here, A A is a matrix of dimensions m×n m×n, B B is a matrix of dimensions m×p m×p, and C C will be a matrix of dimensions p×n p×n. Step 2: Rearranging the Equation To isolate C C, you can rearrange the equation as follows: C=B−1 A C=B−1 A This requires that B B is a square matrix (i.e., m=p m=p) and invertible (i.e., det(B)≠0 det(B)≠0). If B B is not square or not invertible, you will need to use a Continue Reading To solve for matrix C C in the equation A=B C A=B C using a non-numerical method, you can follow these steps: Step 1: Understanding the Equation The equation A=B C A=B C implies that you want to express C C in terms of A A and B B. Here, A A is a matrix of dimensions m×n m×n, B B is a matrix of dimensions m×p m×p, and C C will be a matrix of dimensions p×n p×n. Step 2: Rearranging the Equation To isolate C C, you can rearrange the equation as follows: C=B−1 A C=B−1 A This requires that B B is a square matrix (i.e., m=p m=p) and invertible (i.e., det(B)≠0 det(B)≠0). If B B is not square or not invertible, you will need to use a different method. Step 3: Using Pseudoinverse (if B is not square or invertible) If B B is not square or is singular (not invertible), you can use the Moore-Penrose pseudoinverse B+B+ to find a solution for C C: C=B+A C=B+A The pseudoinverse can be computed using Singular Value Decomposition (SVD) or other numerical methods. Step 4: Solving for C If B B is invertible: Calculate B−1 B−1 and then multiply it by A A: C=B−1 A C=B−1 A 2. If B B is not invertible: Calculate the pseudoinverse B+B+: C=B+A C=B+A Step 5: General Solution For cases where there are multiple solutions (e.g., when A A is not in the column space of B B), you might also express C C in terms of a particular solution plus a homogeneous solution: C=C p+C h C=C p+C h where C p C p is a particular solution and C h C h is any matrix that satisfies B C h=0 B C h=0. Conclusion The method you choose depends on the properties of matrix B B. If B B is square and invertible, use its inverse. If not, use the pseudoinverse. This approach allows you to derive C C without needing specific numerical values for the matrices involved. Upvote · Related questions More answers below If A B=C A B=C, where A A, B B and C C are matrices, then how can I obtain matrix A A given matrices B B and C C? How can I solve for A and B, given AB=C and BA=D? How do I solve matrix equation where I have AX=B, A and B have 3x3, I have to find X? Can someone link a video or give me advice about how I can solve it? Why do we use numerical iterative methods for solving equations? In the system of equations AX=B, what is the matrix (A, B) called as? Marc Millstone PhD in Numerical Analysis, Numerical Linear Algebra and Optimization. · Author has 73 answers and 314.9K answer views ·14y Just to extend on answers above, remember, that in this case the matrices are just a convenient shorthand for writing down a system of equations. Let A A be an n by n matrix and assume C C is (s by n) (So B B is n by s). The equation A=B C A=B C is simply shorthand for the n 2 n 2 linear equations A i,j=∑k B i,k C k,j A i,j=∑k B i,k C k,j in the n s n s variables corresponding to elements of C C. A little bit of algebra will then allow you to convert this into the standard form for linear equations a=M x a=M x (see below). Now as others have said, you can just apply Gaussian Elimination to M M to see if you Continue Reading Just to extend on answers above, remember, that in this case the matrices are just a convenient shorthand for writing down a system of equations. Let A A be an n by n matrix and assume C C is (s by n) (So B B is n by s). The equation A=B C A=B C is simply shorthand for the n 2 n 2 linear equations A i,j=∑k B i,k C k,j A i,j=∑k B i,k C k,j in the n s n s variables corresponding to elements of C C. A little bit of algebra will then allow you to convert this into the standard form for linear equations a=M x a=M x (see below). Now as others have said, you can just apply Gaussian Elimination to M M to see if you can compute the solution. To compute the matrix M M, you can use the following identity: v e c(X Y Z)=(Z T⊗X)v e c(Y)v e c(X Y Z)=(Z T⊗X)v e c(Y) where v e c v e c maps a matrix to the vector with stacked columns and ⊗⊗ is the Kronecker product. Setting X=B,Y=C,Z=I X=B,Y=C,Z=I you see that M=I⊗B M=I⊗B. So the linear system becomes v e c(A)=M v e c(C)v e c(A)=M v e c(C) Upvote · Neal Harris Works at Salesforce.com (product) ·14y Originally Answered: Given matrices A and B how do I solve for C in the equation A=BC using a non numerical method? · Several people have discussed the possibility of B not being invertible. Note that in this case, there is not necessarily a unique solution to A = BC. A silly example is where both A and B are the n x n 0 matrix. In this case, C can be any n x n matrix. Upvote · 9 3 Bill Bell Lives in Canada · Author has 4.1K answers and 6.2M answer views ·Updated 10y Is it possible that you simply want to do exact, formal calculations? As mentioned by one or two other respondents, you can solve for C by left-multiplying A by the inverse of B, if its inverse exists. There are a number of software systems capable of making exact calculations. The Python-language sympy library is one of them. Here is how you can use it for this situation. Clearly, life gets more complicated if the matrices involved are fairly big or if B is nearly singular. EDIT: I absent-mindedly omitted why this approach might be useful. It is that sympy calculates the exact inverse. Therefor Continue Reading Is it possible that you simply want to do exact, formal calculations? As mentioned by one or two other respondents, you can solve for C by left-multiplying A by the inverse of B, if its inverse exists. There are a number of software systems capable of making exact calculations. The Python-language sympy library is one of them. Here is how you can use it for this situation. Clearly, life gets more complicated if the matrices involved are fairly big or if B is nearly singular. EDIT: I absent-mindedly omitted why this approach might be useful. It is that sympy calculates the exact inverse. Therefore, the result is exact as well. Upvote · Related questions More answers below Why are partial differential equations hard to solve? Can irreducible algebraic quintic equations be solved using online graphers or other numerical methods? What are the advantages and disadvantages of using numerical methods for solving ordinary differential equations (ODEs)? How can the equation [(a-b) (a-c)] / [(b-c) +(c-a)] = 0 be solved? What is the Value a, b and c in equation, a b - c = a + b + c? Erik Frey Author has 120 answers and 690.5K answer views ·15y Originally Answered: Given matrices A and B how do I solve for C in the equation A=BC using a non numerical method? · By algebraic solution, do you mean something you can do by hand? If so, to supplement Mike Kross's answer, you can find the inverse of matrix B by performing a Gaussian elimination (just like high school!) of B and the identity matrix: Upvote · 9 2 Michael Harris Software Engineer, Palantir Technologies ·15y Originally Answered: Given matrices A and B how do I solve for C in the equation A=BC using a non numerical method? · Probably the best general way to do this is with the SVD. If you're using matlab, this should be easy. If the SVD of B is B = USV, then the pseudoinverse of B is VS'U, where S' is just S with the non-zero diagonal values inverted. The SVD I guess is a numerical method, but it's something that everyone implements, so it's barely harder to code than matrix multiplication. This will give you an easy way to compute C = B'A where B' is the pseudoinverse of B. Upvote · Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.7K answers and 52.2M answer views ·2y Related How do you solve Ax=b using matrices? Suppose the equations are: 2x + 5y = 16 x + 3y = 9 We write them in matrix form like this… Continue Reading Suppose the equations are: 2x + 5y = 16 x + 3y = 9 We write them in matrix form like this… Upvote · 99 17 Nikesh Ghimire Author has 133 answers and 235.4K answer views ·7y Related If A,B and C are matrices, and if AB=C, how would I find B if A and C are given? A B=C A−1 AI B=A−1 C[Take A−1 on both sides]I B=A−1 C[I B=B]B=A−1 C A B=C A−1 A⏟I B=A−1 C[Take A−1 on both sides]I B=A−1 C[I B=B]B=A−1 C Upvote · 99 13 9 1 Sean Owen Data Science Lead at Databricks (2018–present) · Author has 708 answers and 4.7M answer views ·6y Related If A,B and C are matrices, and if AB=C, how would I find B if A and C are given? I’d like to add to these answers that this is still solvable even if A is not invertible. You only need a left inverse of A to exist. This is sometimes called the pseudo-inverse of A. For example, in Octave, you can get it with pinv(A). You should find that B = pinv(A)C. Upvote · 9 6 9 2 Christian Claudel Assistant Professor at The University of Texas at Austin (2015–present) · Author has 138 answers and 821.2K answer views ·8y Related If A,B and C are matrices, and if AB=C, how would I find B if A and C are given? This is a classical linear algebra problem. If A, B and C can be anything, then A may not be invertible (it may not even be square). So you have to look at the map f such that f(X)= AX. This is a linear map from some vector space to some other vector space (the domain of f is the space of matrices that have the same dimensions as B, and the codomain of f is the space of matrices that have the same dimensions as C). Being a linear map between finite dimensional vector spaces, f has some matrix associated to it. Let s call this matrix M (it depends on the bases you choose for the domain and codo Continue Reading This is a classical linear algebra problem. If A, B and C can be anything, then A may not be invertible (it may not even be square). So you have to look at the map f such that f(X)= AX. This is a linear map from some vector space to some other vector space (the domain of f is the space of matrices that have the same dimensions as B, and the codomain of f is the space of matrices that have the same dimensions as C). Being a linear map between finite dimensional vector spaces, f has some matrix associated to it. Let s call this matrix M (it depends on the bases you choose for the domain and codomain). We then have a classical linear system of equations of the form Mx=c, where x corresponds to the coordinates of the matrix X you are looking for, and c is the vector of coordinates corresponding to C (in the codomain basis). You then just have to solve this linear system of equations. Upvote · 9 6 9 1 Deb P. Choudhury Former Professor at University of Allahabad · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 10K answers and 7.9M answer views ·4y Related If A, B and C are non-square matrices, and if A.B = C is possible (is true), and we know A and C, what is the value of B? If inverse(B) is not possible, then how is A. B = C possible? Suppose that the orders of the matrices A, B and C are m×n, n×p and m×p respectively. The problem of finding B is equivalent to solving p separate systems of linear equations given as A(B)_k=(C)_k where (B)_k and (C)_k are the k_th columns of the matrices B and C respectively, for k= 1, 2, …., p.Here the column vectors of C are given and those of B are unknowns. We can check the solvability of these p different systems of linear equations by checking whether the rank of A is equal to each of the augumented m×(n+1) matrices [A; (C_k)]. Hence the given matrix equation A B = C can be solved with t Continue Reading Suppose that the orders of the matrices A, B and C are m×n, n×p and m×p respectively. The problem of finding B is equivalent to solving p separate systems of linear equations given as A(B)_k=(C)_k where (B)_k and (C)_k are the k_th columns of the matrices B and C respectively, for k= 1, 2, …., p.Here the column vectors of C are given and those of B are unknowns. We can check the solvability of these p different systems of linear equations by checking whether the rank of A is equal to each of the augumented m×(n+1) matrices [A; (C_k)]. Hence the given matrix equation A B = C can be solved with the help of usual methods of linear algebra and a solution may or may not exist and may not be unique. Edit: Note that such a matrix B will exist if and only if each column of C is a linear combination of the column vectors of A, i.e. the column space of C is a subspace of the column space of A. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · 9 2 9 1 Related questions If A B=C A B=C, where A A, B B and C C are matrices, then how can I obtain matrix A A given matrices B B and C C? How can I solve for A and B, given AB=C and BA=D? How do I solve matrix equation where I have AX=B, A and B have 3x3, I have to find X? Can someone link a video or give me advice about how I can solve it? Why do we use numerical iterative methods for solving equations? In the system of equations AX=B, what is the matrix (A, B) called as? Why are partial differential equations hard to solve? Can irreducible algebraic quintic equations be solved using online graphers or other numerical methods? What are the advantages and disadvantages of using numerical methods for solving ordinary differential equations (ODEs)? How can the equation [(a-b) (a-c)] / [(b-c) +(c-a)] = 0 be solved? What is the Value a, b and c in equation, a b - c = a + b + c? What is the method called for solving systems of equations using identity matrices? How can I solve a.b.c+a.b.c+a.b.c.=c.c.c? How can we find the characteristic equation for a rectangular matrix? How do I achieve a numerical solution of equation with free parameters? What are the non-trivial solutions to the matrix equation A²=-A? Related questions If A B=C A B=C, where A A, B B and C C are matrices, then how can I obtain matrix A A given matrices B B and C C? How can I solve for A and B, given AB=C and BA=D? How do I solve matrix equation where I have AX=B, A and B have 3x3, I have to find X? Can someone link a video or give me advice about how I can solve it? Why do we use numerical iterative methods for solving equations? In the system of equations AX=B, what is the matrix (A, B) called as? Why are partial differential equations hard to solve? Can irreducible algebraic quintic equations be solved using online graphers or other numerical methods? What are the advantages and disadvantages of using numerical methods for solving ordinary differential equations (ODEs)? How can the equation [(a-b) (a-c)] / [(b-c) +(c-a)] = 0 be solved? What is the Value a, b and c in equation, a b - c = a + b + c? 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15263	https://math.stackexchange.com/questions/5003153/count-triangles-of-a-shape	puzzle - Count triangles of a shape - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Count triangles of a shape Ask Question Asked 10 months ago Modified10 months ago Viewed 347 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Assume you are supposed to count the number of triangles of this shape: What would be your strategy to count them as fast as possible, accurately? It just looks mind-boggling to me. What I have tried: Upper-right section has 3: 2 single-shape triangles 1 double-share triangles Upper-left section has 4: 2 single-shape triangles 2 double-shape triangles Upper section has 2 triangles by combining right & left sections So, upper section has 9 triangles. Lovely symmetry gives us another 9 triangles at the bottom. Triangles by combining up & bottom sections 2+4: 2 triangles on the left 4 triangles on the right So, in total, we will have 9+9+2+4 triangles i.e. 24. Am I right? I am looking for a convenient trick to count such things easily. triangles puzzle triangulation Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Nov 25, 2024 at 13:32 MegiddMegidd asked Nov 25, 2024 at 11:34 MegiddMegidd 321 4 4 silver badges 11 11 bronze badges 4 2 Have you tried any work yourself? You might try to "divide and conquer" to get most of the triangles.Oscar Lanzi –Oscar Lanzi 2024-11-25 11:43:40 +00:00 Commented Nov 25, 2024 at 11:43 1 How many sets of 3 points are there? How many sets do not form a triangle? Is there any symmetry you can use?Paul –Paul 2024-11-25 11:50:30 +00:00 Commented Nov 25, 2024 at 11:50 1 I'd count the ones that contain the rightmost vertex (using symmetry) and then erase that vertex and the associated edges and keep counting.John Hughes –John Hughes 2024-11-25 13:14:10 +00:00 Commented Nov 25, 2024 at 13:14 @OscarLanzi Right. I added what I have tried so far.Megidd –Megidd 2024-11-25 13:32:27 +00:00 Commented Nov 25, 2024 at 13:32 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Typically counting triangles is error prone because one can potentially count some triangles multiple times not count some triangles There are two ways to reduce or even eliminate these errors. First order all line segments and all corner points. Perhaps naming numerically and/or alphabetically. This is tedious but feasible. Sequentially, for every line segment, count triangles it participates to. To do so: sequentially, go over all possible opposite corner points. Sequentially, for every corner point, count triangles it participates to. To do so: sequentially, go over all possible opposite line segments. Because line segments and corner points are ordered there is less or even no chance to not count triangles or to count triangles multiple times. Of course in both procedures one counts triangles 3 times so one needs to divide by 3. see for example my answer in PSE: Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Nov 25, 2024 at 16:07 answered Nov 25, 2024 at 14:39 FirstName LastNameFirstName LastName 334 1 1 gold badge 3 3 silver badges 19 19 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. 1) How many triangles consist of one single shape? => eight 2) How many triangles consist of two single shapes? => seven (I think) ... 10) How many triangles consist of ten single shapes? => zero Just add everything together :-) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Nov 25, 2024 at 12:33 DominiqueDominique 3,429 7 7 gold badges 25 25 silver badges 33 33 bronze badges Add a comment\| You must log in to answer this question. 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15264	https://www.reddit.com/r/AskPhysics/comments/jx1ws1/electrostatics_question_in_electrical_field/	Electrostatics Question, In electrical field strength calculation do you use the sign of the charge? : r/AskPhysics Skip to main contentElectrostatics Question, In electrical field strength calculation do you use the sign of the charge? : r/AskPhysics Open menu Open navigationGo to Reddit Home r/AskPhysics A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to AskPhysics r/AskPhysics r/AskPhysics 1.5M Members Online •5 yr. ago Chongzzy Electrostatics Question, In electrical field strength calculation do you use the sign of the charge? If you use the typical convention of a positive test charge, would you still use the either +ve or -ve associated with a charge. For example, if something had a charge of -5C, would we use -5 in the calculation, or just +5 because we used the typical convention? Read more Share Related Answers Section Related Answers Applications of quantum entanglement Physics behind black hole event horizons Role of symmetry in particle physics Explaining time dilation with relativity Physics principles in everyday technology New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of November 19, 2020 Reddit reReddit: Top posts of November 2020 Reddit reReddit: Top posts of 2020 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
15265	http://www.mat.uniroma3.it/users/chierchia/PREPRINTS/Chierchia_Pinzari_NBP_10.pdf	The planetary N–body problem: Symplectic foliation, reductions and invariant tori∗ Luigi Chierchia Dipartimento di Matematica Universita “Roma Tre” Largo S. L. Murialdo 1, I-00146 Roma (Italy) luigi@mat.uniroma3.it Gabriella Pinzari Dipartimento di Matematica ed Applicazioni “R. Caccioppoli” Universit a di Napoli “Federico II” Monte Sant’Angelo – Via Cinthia I-80126 Napoli (Italy) pinzari@mat.uniroma3.it November 25, 2010 Revised version Abstract The 6n–dimensional phase space of the planetary (1+n)–body problem (after the classical reduction of the total linear momentum) is shown to be foliated by symplectic leaves of dimension (6n −2) invariant for the planetary Hamiltonian H. Such foliation is described by means of a new global set of Darboux coordinates related to a symplectic (partial) reduction of rotations. On each symplectic leaf H has the same form and it is shown to preserve classical symmetries. Further sets of Darboux coordinates may be introduced on the symplectic leaves so as to achieve a complete (total) reduc-tion of rotations. Next, by explicit computations, it is shown that, in the reduced settings, certain degeneracies are removed. In particular, full torsion is checked both in the partially and totally reduced settings. As a consequence, a new direct proof of Arnold’s theorem on the stability of planetary system (both in the partially and in the totally reduced setting) is easily deduced, pro-ducing Diophantine Lagrangian invariant tori of dimension (3n −1) and (3n −2). Finally, elliptic lower dimensional tori bifurcating from the secular equilibrium are easily obtained. Keywords: Planetary system. Symplectic invariants. Darboux coordinates. Deprit’s reduction of the nodes. Birkhoﬀ normal form. KAM tori. Kolmogorov set. N–body problem. MSC2000 numbers: 70H08, 70K43, 70F10, 70H12, 70K45, 70F15, 70E55, 34C20, 34C27, 34C29, 34D10 ∗Acknowledgments. Discussing with J. Fej´ oz, since 2004, on the N–body problem has been a continuous source of learning and inspiration; we are deeply indebted with him. We thank M. Berti and L. Biasco for their interest. This paper is based on the PhD thesis . Partially supported by European Research Council under F P 7 project “New connections between dynamical systems and Hamiltonian PDEs with small divisors phenomena”. 1 1 Introduction A major breakthrough in the mathematical treatment of the 3–body problem was the “reduction of the nodes” introduced by Jacobi in 1842 . Jacobi’s reduction allows to lower the number of diﬀerential equations which describe the dynamics, frees the system from extra integrals of motion (related to invariance by rotation) and, in general, clariﬁes the structure of the phase space1. Applications of the reduction of the nodes in the general theory of the 3–body problem are countless. Jacobi’s reduction was fully generalized to the general spatial N–body problem only in 1983 by A. Deprit , but, strangely enough, Deprit’s reduction did not have a similar mathematical success2. This fact might be partly due to the pessimistic attitude of Deprit himself, who at p. 194 of , writes: “Whether the new phase variables (34) and (35) are practical in the General Theory of Perturbations is an open question. At least for planetary theories, the answer is likely to be negative [...]”. In this paper we show how, starting from Deprit’s reduction of the nodes, one can give a new description of the analytic structure of the phase space of the general N–body system, which not only might have some theoretical interest in itself, but can be useful in practical computations. For example, we will explicitly compute the second order Birkhoﬀinvariants of the (reduced) secular planetary3 perturbation showing, in particular, full torsion of the secular planetary system. This fact leads immediately to a new “direct” proof of a celebrated theorem by V.I. Arnold on the existence of a positive measure set of phase points evolving in relatively bounded motions in the planetary (1 + n)–body problem for small values of the planet masses. Before describing our results, let us briﬂy discuss the history of Arnold’s theorem, which illustrates quite well the fundamental rˆ ole of having “proper” analytic symplectic variables for general N–body systems. In 1963 V.I. Arnold claimed that, in the general spatial planetary (1 + n)–body prob-lem, there is a positive Liouville measure set of phase space points whose evolution lies on invariant (3n−1)–dimensional Lagrangian Diophantine tori, solving a more than centennial problem. Arnold gave a complete proof for the planar 3–body case4, giving some indications on how to generalize his approach. Very roughly, Arnold’s scheme of proof consists in: averaging the planetary Hamiltonian over the mean anomalies (“fast angles”); put the averaged system (the “secular system”) in Birkhoﬀnormal form up to order four5; introduce polar symplectic variables for the secular system and check Kolmogorov’s non– degeneracy, i.e., the non–degeneracy of the matrix of the second order Birkhoﬀinvariants (“torsion”), so as to apply a KAM theorem for properly–degenerate systems6. For many years it was believed that the problem was completely settled by Arnold, also in view of authoritative endorsements, compare, e.g., Siegel–Moser’s Lectures on Celestial Mechanics [28, p. 277]. However, M. Herman in the 1990’s realized that there were serious hindrances preventing the extension of Arnold’s approach to the general case. Incidentally, by using Jacobi’s reduction of the nodes, in 1995, Robutel extended Arnold’s proof to the spatial 3–body case. The ﬁrst problem is related to the presence of secular resonances, i.e, resonances among the ﬁrst order Birkhoﬀinvariants Ωi of the secular perturbation. In particular, Herman discovered a strange resonance that seemed not to have been fully noticed before, namely that the sum of Ωi’s vanishes7. These reso-nances prevent standard applications of Birkhoﬀnormal form theory8 and, what is worst for Herman’s 1For a symplectic version of Jacobi’s reduction of the nodes, see, e.g., [6, §4.4]. 2At the present date, the MathSciNet database (MR0682462) shows only one citation of Deprit’s paper. 3In the the planetary (1 + n)–body problem one considers one star and n planets – all regarded as point masses – interacting only through gravity. 4In this case the degrees of freedom are four and the tori are 4–dimensional. 5Actually, Arnold requires the Birkhoﬀnormal form up to order six, but this is not necessary, compare . 6We recall that a properly–degenerate Hamiltonian system is a system for which the unperturbed Hamiltonian does not depend on all the action–variables. In [2, Ch. 4], Arnold worked out a general KAM theory for properly–degenerate Hamiltonian systems for which the averaged perturbation admits an elliptic equilibrium. Such theory has been revisited and strengthened in . We also recall that in properly–degenerate systems there appear two perturbative parameters: one (say µ) measures the size of the perturbation and a second one (say ϵ) measures the distance from the reference elliptic equilibrium of the averaged perturbation. 7This resonance is now known as “Herman resonance”; for more information, compare , and § 7.1 below. 8For generalities on Birkhoﬀnormal forms and Birkhoﬀinvariants, see . 2 approach (which we will shortly recall), they imply that the frequency map lies on a plane. A second problem in trying to pursue Arnold’s strategy is the allegedly lack of torsion of the secular perturba-tion. Herman, using Poincare variables9, tried to compute the torsion in the 3–body case (in a suitable asymptotics) but did not succeed and computed, instead, the torsion for a modiﬁed system10. Indeed, it is a fact that the torsion computed in Poincar e variables vanishes at all orders () . To overcome such problems, Herman (unpublished), and later Fejoz in 2004 (, ), introduced two ideas: 1) use a weaker KAM theory based on non–degeneracy conditions involving only the ﬁrst order Birkhoﬀinvariants Ωi, i.e., one requires that the map a →Ω(a) is non planar (a being the vector of the n semimajor–axes and “non planar” means that Ωhas not to be contained in any hyperplane); 2) use a trick by Poincar e, consisting in modifying the Hamiltonian by adding a commuting Hamiltonian, so as to remove the degeneracy: by a Lagrangian intersection theory argument, commuting Hamiltonians have the same maximal transitive invariant tori, so that the KAM tori constructed for the modiﬁed Hamilto-nian are indeed invariant tori also for the original system. Incidentally, we mention that Herman’s KAM theory in yields smooth tori also if the Hamitlonian system is analytic; for a real–analytic version of Herman’s KAM theory, see or . Let us now describe the results obtained in this paper. First, we recall that the phase space of the planetary (1 + n)–body problem, after the classical Poincare reduction of the linear momentum, is 6n–dimensional and is endowed with standard (heliocentric) symplectic variables. On such phase space we introduce a new set of Darboux coordinates, which we call Regularized Planetary Symplectic (RPS) variables: such variables are analytic in a neighborhood of the secular elliptic equilibrium, corresponding to co–planar and co–circular motions, of the secular Hamiltonian. The RPS variables are obtained, ﬁrst, by considering an action–angle version of Deprit’s variables and then performing a Poincar e regularization to remove the singularity due to the vanishing of the eccentricities and mutual inclinations. The RPS variables (Λ, λ, z) := (Λ, λ, η, ξ, p, q) ∈Rn×Tn×B4n, Tn being the standard ﬂat n–dimensional torus and B4n denoting a ball around the origin in R4n, share with the spatial Poincar´ e variables several nice features related to symmetries. In fact, the averaged perturbation, expressed in RPS variables, is even in the secular variables z and commutes with rotations; the D’Alembert relations in the quadratic forms describing the linearized secular system are retained. But unlike Poincar´ e variables, RPS variables may be used to perform a partial and total symplectic reduction of the (1 + n)–body problem11. In particular, the partial reduction leads to a remarkable symplectic foliation of the phase space into invariant symplectic submanifold (“symplectic leaves”) with a prescribed orientation of the total angular momentum. The partial symplectic reduction (3n −1 degrees of freedom) may be described as follows. The region M6n of the phase space corresponding to bounded motions in the integrable limit is foliated by (6n−2)– dimensional symplectic leaves M6n−2 p∗ n,q∗ n, which are invariant for the planetary ﬂow. In fact, two conjugated variables pn and qn, depending only upon the total angular momentum, are both cyclic for the planetary Hamiltonian H, and the invariant leaves M6n−2 p∗ n,q∗ n are sections obtained by ﬁxing pn = p∗ n and qn = q∗ n. The induced symplectic form on any leaf is simply dΛ ∧dλ + dη ∧dξ + d¯ p ∧d¯ q where ¯ p and ¯ q denote the ﬁrst (n −1) components of p and q. Furthermore, the restriction of H to each symplectic leaf is the same and is given by hKep(Λ) + µf(Λ, λ, η, ξ, ¯ p, ¯ q), where hKep coincide with the classical expression of n–decoupled two–body systems in Delaunay variables. We then study the secular Hamiltonian fav (i.e. , the average over the λ’s of f) on M6n−2 p∗ n,q∗ n. The ﬁrst order Birkhoﬀinvariants of fav satisfy identically only Herman’s resonance, while the well known “rotational resonance” (one of the ﬁrst order Birkhoﬀinvariant vanishing identically) present in the unreduced setting, disappears. 9See, e.g. , [15, § 6,1] and references therein. 10Compare ed in particular the remark at the end of p. 24 where Herman says: “Jignore si det T2(a) est identiquement nulle!” ,where T2 is the (4 × 4)–matrix of the second order Birkhoﬀinvariants for the 3–body case and a the ratio of the semimajor–axes. 11For formal series reductions based on Poincar´ e variables, see . 3 Now, it is a general fact (discussed in § 7.2 below), that, for rotational invariant Hamiltonian systems, the construction of the Birkhoﬀnormal form is simpler: indeed the only dangerous resonances (leading to zero divisors) are those non–vanishing integer vectors k for which P ki = 0. Since fav is rotation invariant, one sees immediately that Herman’s resonance does not aﬀect the construction of Birkhoﬀ normal form, which is explicitly carried out up to order four in the limit of well separated semi–major axes. In particular, we shown that the matrix formed by the second order Birkhoﬀinvariant is non– singular, proving full torsion of the secular Hamiltonian12. The total symplectic reduction (3n −2 degrees of freedom) may be described as follows. Since the planetary Hamiltonian is the same on any symplectic leaf M6n−2 p∗ n,q∗ n, without loss of generality, one can consider only the “vertical leaf” M6n−2 0 = M6n−2 0,0 . In an open region avoiding certain “conic singular-ities” (compare (9.7)), we introduce a new set of Darboux coordinates, which includes the symplectic couple (G, g) ∈R+ × T, where G denotes the Euclidean length of the total angular momentum. Since G is an integral, the periodic variable g is cyclic and the motion is described by a (3n −2)–degrees of freedom system ˆ M6n−4 G , ˆ HG where the phase space ˆ M6n−4 G is endowed with the standard symplectic form dΛ∧dˆ λ+dˆ η ∧dˆ ξ +dˆ p∧dˆ q, with (ˆ p, ˆ q) ∈R2(n−2). Symmetries are now broken and in particular the secular origin is no longer an equilibrium for the averaged perturbation. However, (essentially through the standard Implicit Function Theorem) one can perform a symplectic re–centering of the variables and still compute the Birkhoﬀnormal form up to order four and check again full torsion. In particular, we see that in the partially reduced setting there still is an independent commuting integral (namely, G), while in the totally reduced system the are no other integral related to invariance by rotations. Similarly, in the partially reduced setting, Herman’s resonance is the only resonance among the ﬁrst order Birkhoﬀinvariants, while in the totally reduced setting no secular resonance exist. As an application of the above theory, we consider the persistence of quasi–periodic motions for small values of the planetary masses. We show how to construct the Birkhoﬀnormal form up to order four in both the partially and totally reduced settings and then check the non–vanishing of the torsion in all symplectic leaves. This fact allow to resume Arnold’s direct approach and give a complete proof of Arnold’s theorem. Furthermore, having Kolmogorov’s non–degeneracy allows for explicit (and sharp) measure estimates: in the partially reduced case, if ϵ denotes the distance from the elliptic equilibrium, the Kolmogorov’s set (i.e., the union of (3n−1)–dimensional Diophantine tori) ﬁll asymptotically an open set of measure ϵ4n−2 with a density of order 1 −√ϵ; compare Eq. (11.2) below. In the fully reduced setting one obtains a positive measure set of invariant Lagrangian KAM tori in ˆ M6n−4 G with (3n−2) Diophantine frequencies; indeed, the estimate on the measure of the Kolmogorov’s set in ˆ M6n−4 G improves, being proportional to ϵ4n−4. Lifting such tori in M6n−2 amounts to a trivial rotation, leading to (3n −1)–dimensional tori; however such tori may now be Diophantine, Liouvillean or resonant according to the value of the quasi–periodic average of ∂G ˆ HG (integrated along the mo-tion). Incidentally, we notice that, in the 6n–dimensional “ambient” phase space all the invariant tori constructed are resonant since the RPS variables pn and qn do not move. We ﬁnally turn to study the bifurcation of the linear elliptic equilibrium corresponding to Keplerian motions in the fast variables. We prove that such linear equilibrium bifurcates, in the fully non–linear set-ting, into Cantor families of n–dimensional elliptic Diophantine tori; for more comments and references on this topic, see § 11.2 below. The organization of the paper is reﬂected by the table of contents reported below. As the reader will see the paper contains also a lengthy appendix (Appendix B), where the averaged secular Hamiltonian is expanded in RPS variables up to order 4. This computation is clearly central for the computation of the Birkhoﬀnormal forms and for checking the full torsion of the planetary system, which is the main hypothesis for constructing KAM tori, both in partially and totally reduced settings. On the other 12On the other hand, as mentioned above, the torsion evaluated in the unreduced (6n)–dimensional phase space (using Poincare variables) is identically zero. 4 hand, once the symmetries are established and the form of the expansion is therefore derived (§ 6), these computations are in a sense “straightforward” and this is the reason why they are relegated to the appendix. Another reason to spell out these computations is to convince the reader that they can be done by pencil and paper. Contents 2 The planetary Hamiltonian in Cartesian variables 5 3 Deprit’s action–angle variables 6 4 Regularized Planetary Symplectic (RPS) variables (Λ, λ, z) 9 5 Partial symplectic reduction of the planetary system 11 6 Symmetries of the partially reduced secular Hamiltonian 12 7 Birkhoﬀnormal form of the partially reduced secular Hamiltonian 15 7.1 First order Birkhoﬀinvariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 7.2 Birkhoﬀnormal forms for rotation invariant functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 7.3 Second order Birkhoﬀinvariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 8 Full torsion of the partially reduced planetary system 21 8.1 Full torsion in the three–Body case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 8.2 Asymptotic full torsion in the general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 9 The totally reduced planetary system 25 10 Birkhoﬀnormal form and full torsion of the totally reduced secular Hamiltonian 27 10.1 Fourth order Birkohﬀnormal form of the totally reduced planetary Hamiltonian . . . . . . . . . . . . . . . 27 10.2 Full torsion of the totally reduced planetary system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 11 Quasi–periodic motions in the planetary problem 33 11.1 Lagrangian Diophantine tori in the planetary system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 11.2 n–dimensional elliptic KAM tori in the planetary system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 A Explicit formulae for the RPS map 36 A.1 Deprit map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 A.2 RPS map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 B Expansion of the secular Hamiltonian up to order four in RPS variables 40 References 51 2 The planetary Hamiltonian in Cartesian variables Let us consider the planetary (1 + n)–body system, i.e., a system of n + 1 bodies (point masses) with masses m0 (corresponding to “the star” or the Sun) and µmi (corresponding to n “planets”), subject only to the mutual gravitation attraction13. By translation invariance of the Newton equations governing this system, we can restrict our attention to the invariant, symplectic submanifold of vanishing total linear momentum and zero center of mass. On such manifold, following Poincar´ e, one can introduce 13Eventually, µ will be taken 0 < µ ≪1. 5 symplectic heliocentric coordinates, eliminating the coordinates of the Sun and lowering the number of degrees of freedom by 3 units. In such coordinates the planetary system is governed by the Hamiltonian Hplt = X 1≤i≤n \|y(i)\|2 2Mi −Mi ¯ mi \|x(i)\| + µ X 1≤i<j≤n y(i) · y(j) m0 − mimj \|x(i) −x(j)\| =: hplt + µ fplt (2.1) where x(i) = (x(i) 1 , x(i) 2 , x(i) 3 ) ∈R3, y(i) = (y(i) 1 , y(i) 2 , y(i) 3 ) ∈R3 are standard symplectic conjugate vari-ables, x · y = P 1≤i≤3 xiyi and \|x\| := (x · x)1/2 denote, respectively, the standard inner product in R3 and the Euclidean norm; Mi := m0mi m0 + µmi , ¯ mi := m0 + µmi ; the phase space is the collisionless domain P6n := (y, x) = (y(1), . . . , y(n)), (x(1), . . . , x(n)) ∈R3n × R3n : 0 ̸= x(i) ̸= x(j) ∀i ̸= j , (2.2) endowed with the standard symplectic form X 1≤i≤n dy(i) ∧dx(i) = X 1≤i≤n X 1≤j≤3 dy(i) j ∧dx(i) j . Physically, the coordinates x(i) represent the diﬀerence between the position of the ith planet and the position of the Sun, while y(i) are the associated symplectic momenta rescaled by µ; the position and velocity of the Sun is recovered by recalling that the center of mass and the total linear momentum are zero. For details, see, e.g., [15, § 6.1] and references therein14. The Hamiltonian hplt, as well known, describes the integrable limit given by n decoupled 2–body systems formed by the Sun and the ith planet. In this way symplectic reduction of the total linear momentum has been achieved, but the three com-ponents C1, C2, C3 of the total angular momentum C := X 1≤i≤n x(i) × y(i) are still integrals for the Hamiltonian (2.1). Such integrals are independent but do not commute, as the well known cyclic Poisson commutation rules hold. Nevertheless one can form, out of them, two independent commuting integrals, for example: C3 = X 1≤i≤n (x(i) 1 y(i) 2 −x(i) 2 y(i) 1 ) and G := \|C\| = X 1≤i≤n x(i) × y(i) . (2.3) The conservation of C3 and G along the Hplt–trajectories is equivalent to the invariance of Hplt under rotations around the k(3)–axis15 and around the C–axis, and will be at the basis of the further symplectic reductions described in § 5 and § 9 below. 3 Deprit’s action–angle variables Following Deprit , we introduce, in an open subset of P6n avoiding co–circular and co–planar phase points, a remarkable set of action–angle variables. In the next two sections, after regularizing such 14The parameters µ, Mj, ¯ mj here correspond to, respectively, ϵ, µj, Mj in . 15(k(1), k(2), k(3)) denotes the standard orthonormal basis in R3, i.e. , k(i) j = δij, where δij denotes the usual Kronecker delta. 6 variables (allowing again for co–circular and co–planar points), we will achieve the symplectic reduction of the inclination of the total angular momentum C and ﬁnd a global symplectic chart for the reduced symplectic submanifolds of dimension (6n −2), which foliate the phase space. Consider the the ﬂow (y(i), x(i)) →φt hi(y(i), x(i)) generated by the two–body problem Hamiltonian hi(y(i), x(i)) := \|y(i)\|2 2Mi −Mi ¯ mi \|x(i)\| with y(i) ∈R3 , x(i) ∈R3 \ {0} . (3.1) As well known, initial data (y(i), x(i)) with stricltly negative energies hi(y(i), x(i)) = Ei < 0 give rise to bounded motions, evolving on Keplerian ellipses Ei = Ei(y(i), x(i)), having one focus in the origin. Let ai = ai(y(i), x(i)), ei = ei(y(i), x(i)) denote, respectively, the semi–major axis and the eccentricity of Ei(y(i), x(i)), C(i) := y(i) × x(i) the angular momentum of the ith the planet, orthogonal to the plane of Ei and, ﬁnally, S(i) := i X j=1 C(j) the “partial” angular momentum of the ﬁrst i planets, for16 i = 2, . . . , n. We consider the following phase spaces. (d) D6n := the set of phase points17 (y, x) ∈P6n such that: Ei < 0; ei < 1; S(i) is not anti–parallel18 to C(i) for 2 ≤i ≤n; C is not anti–parallel to k(3). (d∗) D6n ∗ := the subset of D6n where the following inequalities also hold: D6n ∗ ⊂D6n :    ei(y(i), x(i)) > 0 , S(i) × C(i) ̸= 0 , ∀2 ≤i ≤n , k(3) × C ̸= 0 . (3.2) Deprit action–angle variables19 (L, Γ, Ψ), (ℓ, γ, ψ) are deﬁned through a symplectic map φ∗: (y, x) ∈D6n ∗ → (L, Γ, Ψ), (ℓ, γ, ψ) ∈(0, +∞)3n−1 × R × T3n (3.3) which we now proceed to describe. (d1) The variables L = (L1, . . . , Ln), ℓ= (ℓ1, . . . , ℓn) and Γ = (Γ1, . . . , Γn) are standard planar Delau-nay variables20, namely: – Li := Mi √¯ miai, where ai is semi–major axis of Ei; – the angle conjugated to Li is the mean anomaly ℓi of x(i), i.e. , the area spanned by the orbital radius starting from the ith perihelion Pi (:= the point of Ei at minimum distance from a focus) and ending at x(i), normalized to 2π; – the action variable Γi = \|C(i)\| is the Euclidean length of the ith angular momentum21 C(i). 16In , Deprit uses the diﬀerent (but equivalent) convention S(i) = Pn j=i+1 C(j). 17Taking the phase points in the collisionless phase space P6n (2.2) is actually only needed in considering the planetary Hamiltonian, but could be disregarded in the general symplectic–geometry arguments. 18We say that u and v in R3 are anti–parallel if u × v = 0 and u · v < 0. 19Actually, the variables in are slightly diﬀerent from the variables in (3.3), compare Remark 3.1–(i) below. 20For an analytical description of planar Delaunay variables see, e.g., [7, § 3.2]. 21Recall also the classical relation Γi := \|C(i)\| = Li q 1 −e2 i . 7 (d2) For 1 ≤i ≤n −1, Ψi := \|S(i+1)\|, while Ψn := C3 is the vertical component of the total angular momentum as in (2.3). Notice that Ψn−1 = \|S(n)\| = \|C(1) + · · · + C(n)\| = \|C\| =: G. To describe the remaining Deprit angles γi and ψi, we introduce the following notations. Given three non vanishing vectors u, v and w of R3, with w orthogonal to u and v, we denote as αw(u, v) the positively oriented angle between u and v in the plane πw orthogonal to22 w. We also denote by Rw(g) the positive (3 × 3)–matrix rotation by an angle g around the axis oriented as23 w. In view of (3.2), on D6n ∗ the following “nodes” are well deﬁned24. ¯ ν := k(3) × C , νi := C(1) × C(2) i = 1 S(i) × C(i) 2 ≤i ≤n . (3.4) (d3) γi := αC(i)(νi, Pi), i.e. , γi is (in the plane orthogonal C(i)) the “argument of the perihelion with respect to the node the νi”. Thus, γi diﬀers form the Delaunay’s angle gi, which is deﬁned as the argument of the perihelion with respect to the node ¯ νi := k(3) × C(i), simply by a shift: γi = αC(i)(νi, Pi) = αC(i)(νi, ¯ νi) + αC(i)(¯ νi, Pi) = αC(i)(νi, ¯ νi) + gi , ¯ νi := k(3) × C(i) . (3.5) (d4) ψn−1 := αC(¯ ν, νn) and25 ψn := αk(3)(k(1), ¯ ν). (d5) When n > 2, the angles ψ1,...,ψn−2 are deﬁned as26 ψi := αS(i+1)(νi+2, νi+1), 1 ≤i ≤n −2. Remark 3.1 (i) In Appendix A.1 it is given the analytic expression of the map φ−1 ∗. Here, we just remark that the variables (L, Γ, ℓ, γ) are “horizontal variables”: each quadruple (Li, Γi, ℓi, γi) describes the position of the ith planet on its orbital plane, while the variables (Ψ, ψ) play the rˆ ole of “vertical variables”, since they are related to the orientations of the diﬀerent orbital planes. In the original paper by Deprit we ﬁnd a diﬀerent set of horizontal variables, namely, the set (R, Φ, r, φ) = (R1, . . . , Rn), (Φ1, . . . , Φn), (r1, . . . , rn), (φ1, . . . , φn) , (3.6) where (ri, φi) are the usual polar coordinates of x(i) on πC(i) with respect to νi (as polar axis); (Ri, Φi) are their respective conjugated actions. In Deprit proved that the variables (R ,Φ, Ψ, r, φ, ψ) are symplectic. Since the map (Ri, Φi, ri, φi) →(Li, Γi, ℓi, γi) is well known to be symplectic27, it follows that the variables d1÷d5 are symplectic28; for a diﬀerent (inductive) proof of the symplecticity of the variables d1÷d5, see . (ii) A remarkable property of Deprit’s variables is the presence, among the action variables Ψ, of the two commuting Hplt–integrals G = Ψn−1 and C3 = Ψn. The variables Ψ1, . . . , Ψn−2 complete Ψn−1, Ψn so as to obtain a commuting set of action variables Ψ = (Ψ1, . . . , Ψn). (iii) Besides G and C3, also the angle ζ (the longitude of the node among the planes πk(3) and πC) is an integral of motion: giving G, C3 and ζ corresponds to giving the three components of the angular momentum C. This fact will be crucial in the forthcoming reduction. 22For example, if u = k(1), v = k(2) and w = k(3), then αk(3)(k(1), k(2)) = π/2. 23For example, Rk(3)(π/2)k(1) = k(2). 24I.e., do not vanish. Notice that (for later convenience) the node ν1 is deﬁned as ν1 = ν2 (and that assumption (3.2) with i = 2 implies ν1 ̸= 0). In ν1 is chosen with the opposite sign, possibly to recover, for n = 2, the so–called “Jacobi’s opposition of the nodes” (namely, the relation C × C(1) = −C × C(2)). Notice also that ¯ ν is orthogonal to k(3) and νi is orthogonal to C(i) and S(i). 25Such angles will be often denoted, respectively, by g and ζ. 26Note that νi+2 := (S(i+2) × C(i+2)) = (S(i+1) × C(i+2)) so that also νi+2 ∈πS(i+1). 27Compare, e.g., § 3.2 of . 28Note that the variables (Ψ, ψ) correspond, in Deprit’s notation (up to an unessential reordering of C(1), . . . , C(n)), to: Ψi = Θ∗ n−1−i for 1 ≤i ≤n −1; Ψn = N∗ 0 ; ψ1 = θ∗ n−2 + π, ψi = θ∗ n−1−i for 2 ≤i ≤n −2; ψn−1 = θ∗ 0 + π; ψn = ν∗ 0. Notice also that the names in of the variables (Φ, φ) in (3.6) are (Θ, θ). 8 (iv) As in Delaunay variables, the integrable part hplt of the planetary Hamiltonian (2.1) takes the well known “Keplerian form”: hKep(L) = − X 1≤i≤n M 3 i ¯ m2 i 2L2 i . (3.7) 4 Regularized Planetary Symplectic (RPS) variables (Λ, λ, z) The action–angle Deprit variables deﬁned in § 3 become singular when some of the inequalities in (3.2) fails. We describe now a regularization procedure which allows for ei = 0 and for C(j+1) parallel to S(j), and C parallel to k(3), including, in particular, the case of co–circular and co–planar motions29. This procedure is analogous to the Poincar´ e regularization of the Delaunay variables. The new symplectic variables (Λ, λ, z) with Λ = (Λ1, . . . , Λn) , λ = (λ1, . . . , λn) , z = (η, ξ, p, q) η = (η1, . . . , ηn) , ξ = (ξ1, . . . , ξn) , p = (p1, . . . , pn) , q = (q1, . . . , qn) will be called Regularized Planetary Symplectic (RPS) variables and are deﬁned as follows. Let (L, Γ, Ψ, ℓ, γ, ψ) = φ∗(y, x); let ψ0 := 0, Ψ0 := Γ1 and Γn+1 := 0. Then, for 1 ≤i ≤n, we deﬁne: Λ = L , λi = ℓi + γi + ψn i−1 , where ψn i = X i≤j≤n ψj , ηi = p 2(Λi −Γi) cos γi + ψn i−1 ξi = − p 2(Λi −Γi) sin γi + ψn i−1 ( pi = p 2(Γi+1 + Ψi−1 −Ψi) cos ψn i qi = − p 2(Γi+1 + Ψi−1 −Ψi) sin ψn i (4.1) Remark 4.1 Writing out (pn, qn) explicitly one ﬁnds    pn = p 2(Ψn−1 −Ψn) cos ψn = p 2(G −C3) cos ζ qn = − p 2(Ψn−1 −Ψn) sin ψn = − p 2(G −C3) sin ζ (4.2) showing that the conjugated variables pn and qn are both integrals for Hplt. This fact is at the basis of the partial symplectic reduction described in next section. The map from the cartesian heliocentric variables (y, x) to the RPS variables (Λ, λ, z) will be denoted by φ : (y, x) ∈D6n ∗ →(Λ, λ, z). Its inverse map φ−1 can be described as follows (see § A.2 for full details): φ−1 :      y(i) = Ri(Λ, z)y(i) pl (Λi, λi, ηi, ξi) x(i) = Ri(Λ, z)x(i) pl (Λi, λi, ηi, ξi) (4.3) where: • (Λi, λi, ηi, ξi) →(y(i) pl , x(i) pl ) denotes the planar Poincar´ e map, i.e. , the map which sends a point (Λi, λi, ηi, ξi) ∈(0, +∞)×T×{ η2 i +ξ2 i 2Λi < 1} to the point (y(i) pl , x(i) pl ) ∈R2 ×R2 ⊂R3 ×R3 recovering the Cartesian coordinates on the “instantaneous orbital plane” (i.e. the Ei plane). The planar Poincar´ e map is explicitly given by30 x(i) pl = x(i) 1 , x(i) 2 , 0 , y(i) pl = βi ∂λix(i) pl (4.4) 29Actually, the regularization procedure depends on whether one wants to allow parallel or antiparallel angular momenta and C parallel or antiparallel to k(3). For deﬁniteness, we shall describe only the case corresponding to parallel angular momenta, and C parallel to k(3) (the other cases being analogous). 30Compare, [3, Lemma 2.1]. 9 where          x(i) 1 := 1 ¯ mi Λi Mi 2 cos ui − ξi 2Λi (ηi sin ui + ξi cos ui) − ηi √Λi q 1 −ηi2+ξi2 4Λi x(i) 2 := 1 ¯ mi Λi Mi 2 sin ui − ηi 2Λi (ηi sin ui + ξi cos ui) + ξi √Λi q 1 −ηi2+ξi2 4Λi βi := ¯ m2 i M 4 i Λ3 i (4.5) and ui = ui(Λi, λi, ηi, ξi) = λi + O(\|(ηi, ξi)\|) is the unique solution of the (regularized) Kepler equation ui − 1 √Λi s 1 −ηi2 + ξi 2 4Λi (ηi sin ui + ξi cos ui) = λi ; (4.6) • for 1 ≤i ≤n, Ri are products of matrices: Ri = R∗ nR∗ n−1 · · · R∗ i Ri (4.7) where R1 = id and Ri, R∗ j are 3 × 3 unitary matrices depending on (Λ, z) but with Ri, R∗ j, for 1 ≤i ≤n and 2 ≤j ≤n −1, independent of the cyclic couple (pn, qn). The matrices Ri, R∗ j have the form R∗ i =   1 −q2 i c∗ i −piqic∗ i −qis∗ i −piqic∗ i 1 −p2 i c∗ i −pis∗ i qis∗ i pis∗ i 1 −(p2 i + q2 i )c∗ i  , 1 ≤i ≤n Ri =   1 −q2 i−1ci −pi−1qi−1ci −qi−1si −pi−1qi−1ci 1 −p2 i−1ci −pi−1si qi−1si pi−1si 1 −(p2 i−1 + q2 i−1)ci  , 2 ≤i ≤n (4.8) where ci, si, c∗ i , s∗ i , explicitly computed in Appendix A (compare (A.22), (A.23) and (A.27)), are analytic functions of ρi = η2 i +ξ2 i 2 , rj = p2 j+q2 i 2 for 1 ≤i ≤n and 1 ≤j ≤n −1. Remark 4.2 (i) By (4.8), when p = 0 = q, one has Ri = id = R∗ i for all i’s. In this case, by (4.3) and (4.7), the remaining variables (Λ, λ, η, ξ) are seen to coincide with the planar Poincar´ e variables. (ii) By (4.1), (ηi, ξi) = 0 corresponds to ei = 0 and (pj, qj) = 0 to S(j) parallel to C(j+1) (j ̸= n). By (4.2), (pn, qn) = 0 corresponds to G = C3, i.e. , to C parallel to k(3). (iii) From (4.3)÷(4.8) it follows at once that the map φ : (y, x) ∈D6n ∗ →(Λ, λ, z) ∈φ(D6n ∗) deﬁned in (4.1) can be extended to a real–analytic diﬀeomorphism φ : (y, x) ∈D6n →M6n := φ(D6n) (4.9) where D6n is the set deﬁned in § 3–(d) . (iv) The variables (Λ, λ, z) are symplectic. In fact, let, on D6n ∗, (ρi, ϕi), (ri, χi) denote the sym-plectic polar coordinates associated to31 (ηi, ξi), (pi, qi), and let ρ = (ρ1, . . . , ρn), r = (r1, . . . , rn), ϕ = (ϕ1, . . . , ϕn), χ = (χ1, . . . , χn). In terms of Λ, ρ, r, λ, ϕ, χ, equations (4.1) become   Λ ρ r  = M   L Γ Ψ  ,   λ ϕ χ  = ˆ M   ℓ γ ψ   31I.e., in complex notation, ηj + iξj = p2ρjeiϕj and pj + iqj = p2rjeiχj , where i:=√−1. 10 where M and ˆ M denote the matrices of order 3n uniquely deﬁned by the equations (1 ≤i ≤n −1)    Λi = Li ρi = Li −Γi ri = Γi+1 + Ψi−1 −Ψi    λi = ℓi + γi + ψn i−1 ϕi = −γi −ψn i−1 χi = −ψn i One easily recognizes that the matrices M and ˆ M are related by ˆ M=(Mt)−1, where (·)t denotes matrix transpostion. This relation implies that φ is symplectic on D6n ∗, hence, by regularity, on D6n. The properties in (iii) and (iv) explain the name given to the variables (Λ, λ, z). (v) The following relations are immediately checked X 1≤i≤n ρi = X 1≤i≤n η2 i + ξ2 i 2 = X 1≤i≤n Λi − X 1≤i≤n Γi (4.10) X 1≤j≤n−1 rj = X 1≤j≤n−1 p2 j + q2 j 2 = X 1≤j≤n Γj −Ψn−1 = X 1≤j≤n Γj −G (4.11) rn = p2 n + q2 n 2 = Ψn−1 −Ψn = G −C3 . (4.12) 5 Partial symplectic reduction of the planetary system We now go back to the planetary many–body problem showing, in particular, that its phase space D6n (deﬁned in § 3–(d)) is foliated by Hplt–invariant symplectic submanifolds of dimension (6n −2) with a natural “global” symplectic structure. Let H(Λ, λ, z) := Hplt ◦φ−1 = hKep(Λ) + µf denote the planetary Hamiltonian expressed in RPS variables with phase space given by M6n as in (4.9), endowed with the standard symplectic form dΛ ∧dλ + dη ∧dξ + dp ∧dq. As mentioned in Remark 4.1, the variables pn, qn in (4.2) are both integrals and cyclic for H. This means that the perturbation function f does not depend upon (pn, qn), i.e. , f = f(Λ, λ, ¯ z) with ¯ z := (η, ξ, ¯ p, ¯ q) := (η1, . . . , ηn), (ξ1, . . . , ξn), (p1, . . . , pn−1), (q1, . . . , qn−1) . The upshot is that the phase space M6n is foliated by symplectic H–invariant submanifolds M6n−2 p∗ n,q∗ n := {(Λ, λ, z) ∈M6n : pn = p∗ n, qn = q∗ n} , (5.1) p∗ n and q∗ n being ﬁxed constants. A global symplectic chart is given simply by the ﬁrst (6n−2) variables (Λ, λ, ¯ z), the restriction of the symplectic form on each “symplectic leaf ” M6n−2 p∗ n,q∗ n being dΛ ∧dλ + dη ∧dξ + d¯ p ∧d¯ q . Finally, the restriction of the planetary Hamiltonian to each leaf M6n−2 p∗ n,q∗ n is the same, as the perturbation f is independent from the couple (pn, qn): we shall keep denoting H the partially reduced Hamiltonian and f the planetary perturbation: H(Λ, λ, ¯ z) = Hplt ◦φ−1 = hKep(Λ) + µf(Λ, λ, ¯ z) . (5.2) Remark 5.1 (i) In view of this last observation, without loss of generality, we can restrict our atten-tion to the symplectic leaf M6n−2 0 := M6n−2 0,0 with p∗ n = 0 = q∗ n, which corresponds to the “vertical submanifold” {C1 = 0 = C2} where the total angular momentum is oriented in the k(3)–direction. (ii) When the “vertical variables” p and q are set to be zero, that is, when the secular set z is taken z = zpl = (η, ξ, 0, 0), the map φ−1 reduces to the planar Poincar´ e map. 11 (iii) The expression “partial reduction” refers to the fact that, on each leaf M6n−2 p∗ n,q∗ n, H admits another independent integral, namely, the restriction of G = Ψn−1 = \|C\| onto M6n−2 p∗ n,q∗ n. From (4.10) and (4.11) there follows G = X 1≤i≤n Λi −1 2\|¯ z\|2 . (5.3) We also note that the expression of C3 on M6n is given32 by C3 = G −p2 n + q2 n 2 = X 1≤i≤n Λi −1 2\|z\|2 . (5.4) Notice that the formula for C3 on M6n coincides with the well known one in spatial Poincar´ e variables. (iv) The conservation of G along H–trajectories induces into the averaged perturbation33 fav some symmetries (discussed in detail in § 6 below), which imply, in particular, that fav is an even function of ¯ z and that its quadratic part splits into the sum of two separated terms34: Qh(Λ) · η2 + ξ2 2 + ¯ Qv(Λ) · ¯ p2 + ¯ q2 2 (5.5) where the “horizontal part” Qh is a quadratic form of order n coinciding with that of the planar problem, while the “vertical part” ¯ Qv is a quadratic form of order n −1. As we will see below, Qh and ¯ Qv are non–degenerate35, hence, the point ¯ z = (η, ξ, ¯ p, ¯ q) = 0, corresponding to zero eccentricities and zero relative inclinations, is a non–degenerate elliptic equilibrium for the secular Hamiltonian fav. (v) In Poincar´ e variables36 a splitting similar to (5.5) holds with the same horizontal quadratic form and with the vertical quadratic form replaced by a quadratic form Qv of order n. One can show that the eigenvalues of the quadratic form ¯ Qv into (5.5) coincide with the non identically vanishing eigenvalues of Qv (compare ). As shown in and Proposition 7.1 below, the trace tr (Qh+Qv) = tr (Qh+Qv) = 0, where Qv is the (n × n) matrix Qv := ¯ Qv 0 0 0 ; this relation is known as “Herman’s resonance”. 6 Symmetries of the partially reduced secular Hamiltonian The planetary perturbation function f in (5.2) enjoys several symmetry properties, which, in particular, simplify the Taylor expansion of the secular Hamiltonian37 fav(Λ, z) = (2π)−n R Tn fdλ. In view of Remark 5.1–(i), we will consider only the symplectic “vertical leaf” M6n−2 0 . The angular momentum integral G Poisson commutes with the unperturbed Keplerian Hamiltonian hKep, whence G commutes with f. This fact implies that f is invariant under the Hamiltonian ﬂow g →Rg at time g generated by G, i.e. : f Rg(Λ, λ, ¯ z) = f(Λ, λ, ¯ z) for any g ∈T, (Λ, λ, ¯ z) ∈M6n−2 0 . (6.1) 32Recall (4.12) and observe that, clearly, C3 = G on M6n−2 0 . 33Here and in what follows, the index “av” denotes the average over the fast angles conjugated to Λ; the function fav is usually called “secular Hamiltonian”. 34For exact notation, see footnote 39 below. The symmetries expressed by (5.5) is sometimes called “D’Alembert relations”. 35Actually, they are, respectively, negative and positive deﬁnite. The positive–deﬁniteness of Qh has been proved in [15, proposition 73]. From the analytical expression of ¯ Qv there follows that, since C1 < 0, its eigenvalues ¯ ςi’s are (maybe not strictly) positive. From their asymprotic evaluation (compare the proof of Proposition 7.2 below) there follows that they are actually strictly positive on an open set A of semimajor axes described below (see Eq. (7.2) ). By standard arguments of complex analysis there follows that ¯ Qv is strictly positive on an open dense set. 36For the analytic deﬁnition of spatial Poincar´ e variables, see or . Notice that secular “vertical” variables of the Poincar´ e set (i.e. , the p’s and the q’s) are in Rn. 37By (4.4) and (4.3), the λ–average of the perturbation (2.1) expressed in RPS variables (Λ, λ, z) reduces to the averaging the newtonian potential V (Λ, λ, z) := P 1≤i<j≤n − mimj \|x(i)−x(j)\| ´ . 12 The action of Rg corresponds, in Cartesian variables (y, x), to a positive rotation of all the y(i)’s and the x(i)’s by an angle g around the C–axis, which coincide with the k(3)–axis on M6n−2 0 . By the expression (5.3) of G in RPS variables, such ﬂow is given by Rg : Λ′ = Λ , λ′ i = λi + g , ¯ z′ = Sg ¯ z (6.2) where Sg acts as synchronous rotation in the symplectic (ηi, ξi) and (pi, qi)–planes: Sg : η′ i ξ′ i = R(−g) ηi ξi , p′ j q′ j = R(−g) pj qj , 1 ≤i ≤n , 1 ≤j ≤n −1 (6.3) R(g) being the plane rotation by g R(g) := cos g −sin g sin g cos g . (6.4) As one checks immediately, the planetary perturbation fplt (2.1) is invariant also under the following transformations (x(i), y(i)) →(x′(i), y′(i)): R− 1 : x′(i) = −x(i) 1 , x(i) 2 , x(i) 3 , y′(i) = y(i) 1 , −y(i) 2 , −y(i) 3 R− 2 : x′(i) = x(i) 1 , −x(i) 2 , x(i) 3 , y′(i) = −y(i) 1 , y(i) 2 , −y(i) 3 R− 3 : x′(i) = x(i) 1 , x(i) 2 , −x(i) 3 , y′(i) = y(i) 1 , y(i) 2 , −y(i) 3 R1↔2 : x′(i) = x(i) 2 , x(i) 1 , x(i) 3 , y′(i) = −y(i) 2 , −y(i) 1 , −y(i) 3 . (6.5) Such transformations correspond, respectively, to the reﬂections with respect to the planes {x1 = 0}, {x3 = 0}, {x2 = 0} and {x1 = x2}. Notice that the invariance by R− 2 and R1↔2 is implied by the invariance by R− 1 and Rg. The actions of the reﬂections (6.5) in the variables (Λ, λ, z) are found using (4.3)÷(4.8) and their expressions are given by R− 1 : Λ′ i = Λi, λ′ i = π −λi, ¯ z′ = (−η, ξ, ¯ p, −¯ q) := S− 14¯ z R− 2 : Λ′ i = Λi, λ′ i = −λi, ¯ z′ = (η, −ξ, −¯ p, ¯ q) := S− 23¯ z R− 3 : Λ′ i = Λi , λ′ i = λi , ¯ z′ = (η, ξ, −¯ p, −¯ q) := S− 34¯ z R1↔2 : Λ′ i = Λi , λ′ i = π 2 −λi , ¯ z′ = (ξ, η, ¯ q, ¯ p) := S1↔2 ¯ z . (6.6) Let us check, for instance, the expression of R− 1 , the others being similar. From Kepler’s equation (4.6), it follows that u′ i := ui(Λi, π −λi, −ηi, ξi) = π −ui(Λi, λi, ηi, ξi) := π −ui, hence, cos u′ i = −cos ui and sin u′ i = sin ui. By (4.5), if x(i)(Λi, λi, ηi, ξi)=(x(i) 1 , x(i) 2 , 0), then (x′ 1 (i), x′ 2 (i), 0) := x(i)(Λi, π −λi, −ηi, ξi) = (−x(i) 1 , x(i) 2 , 0) . If one changes simultaneously (¯ p, ¯ q) →(¯ p, −¯ q), then, by (4.3) and (4.8), one obtains that, if x(i)(Λ, λ, η, ξ, ¯ p, ¯ q) = x(i) 1 (Λ, λ, η, ξ, ¯ p, ¯ q), x(i) 2 (Λ, λ, η, ξ, ¯ p, ¯ q), x(i) 3 (Λ, λ, η, ξ, ¯ p, ¯ q) := (x(i) 1 , x(i) 2 , x(i) 3 ) then38 x′(i) := x(i)(Λ, π −λ, −η, ξ, ¯ p, −¯ q) = (−x(i) 1 , x(i) 2 , x(i) 3 ) . (6.7) 38π −λ denotes, for short, the vector of components (π −λ1, π −λ2, . . . , π −λn). Notice that R∗ n = id since we have assumed pn = 0 = qn (compare (4.8)). This relation easily implies (6.7). 13 This is the ﬁrst identity into the ﬁrst line of (6.5). The second one is then obtained taking the derivative of (6.7) with respect to λi, multiplied by βi. Remark 6.1 (i) The fast angles λ’s are averaged out when considering the secular Hamiltonian fav. (ii) The invariance of f under the reﬂections (6.6) yields the following symmetries for fav: fav(Λ, ¯ z) = fav(Λ, Sg¯ z) = fav(Λ, S¯ z) for any S ∈{S− 14, S− 23, S− 34, S1↔2} . (6.8) (iii) By the symmetries S− 14, S− 23, and S− 34, it follows that fav is an even function of the variables (η, ¯ q), (ξ, ¯ p) and (¯ p, ¯ q). Thus, the parity holds also in the variables (η, ξ) and (η, ¯ p). By the S1↔2–invariance, fav does not change when (η, ξ) is changed to (ξ, η) and, simultaneously, (¯ p, ¯ q) to (¯ q, ¯ p). We are now ready to discuss the form of the Taylor expansion of the secular Hamiltonian fav around the elliptic equilibrium ¯ z = 0 up to the fourth order. By parity, in second order term, only the monomials39 η2, ξ2 will appear and, in the fourth order term, only the monomials η4, η2ξ2, ξ4, ¯ p4, ¯ p2¯ q2, ¯ q4, η2¯ q2, ξ2¯ p2, ξ2¯ q2, ξ2¯ p2, ηξ¯ p¯ q. By the symmetry S1↔2, the tensors in front of each monomial of the couples (η2, ξ2), (¯ p2, ¯ q2), (η4, ξ4), (¯ p4, ¯ q4), (η2¯ q2, ξ2¯ p2), (η2¯ p2, ξ2¯ q2) will be pairwise equal. Let us denote Qh · η2 2 , ¯ Qv · ¯ p2 2 the quadratic forms associated to the monomials η2, ¯ p2 and Fh · η2ξ2 , Fv · ¯ p2¯ q2 , Fhv · η2¯ q2 , F′ hv · η2¯ p2 the quartic forms associated to, respectively, η2ξ2, ¯ p2¯ q2, η2¯ q2, η2¯ p2. By the invariance for the transfor-mation ηi →(ηi −ξi)/ √ 2, ξi →(ηi + ξi)/ √ 2, ¯ pi →(¯ pi −¯ qi)/ √ 2, ¯ qi →(¯ pi + ¯ qi)/ √ 2 (which corresponds to a rotation by π/4 around the C–axis: compare (6.3)), the quartic tensors associated to monomials η4, ¯ p4 coincide, respectively, with one half the quartic tensors Fh, Fv associated to η2ξ2, ¯ p2¯ q2. By the same reason, the tensor associated to ηξ¯ p¯ q coincides with one half the diﬀerence F′ hv −Fhv, provided (as it is always possible) the entries (Fhv)ijkl, (F′ hv)ijkl of such tensors are chosen so as to satisfy (F′ hv)ijkl = (F′ hv)jilk, (Fhv)ijkl = (Fhv)jilk. By the previous considerations, the expansion of fav has the form fav(Λ, ¯ z) = C0(Λ) + Qh(Λ) · η2 + ξ2 2 + ¯ Qv(Λ) · ¯ p2 + ¯ q2 2 + 1 2F(Λ) · ¯ z4 + P(Λ, ¯ z) (6.9) where P has a zero of order 6 (due to the parity of fav as a function of ¯ z) in ¯ z = 0 and 1 2F(Λ) · ¯ z4 denotes the fourth order term 1 2F(Λ) · ¯ z4 := Fh(Λ) · η4 + ξ4 + 2η2ξ2 2 + Fv(Λ) · ¯ p4 + ¯ q4 + 2¯ p2¯ q2 2 + Fhv · η2¯ q2 + ξ2¯ p2 −2ηξ¯ p¯ q 2 + F′ hv · η2¯ p2 + ξ2¯ q2 + 2ηξ¯ p¯ q 2 . (6.10) Remark 6.2 The invariance by R− 3 actually implies that the whole perturbation (and not only its average) is even in (¯ p, ¯ q). The explicit values of the tensors appearing in (6.9), (6.10) will be given in Appendix B. 39For tensors, we use the same notation as in . Thus, if a = (ai,j)i∈{1,...,n}r1 ,j∈{1,...,n−1}r2 is a tensor with r = r1 + r2 indices and ki is an integer between 0 and ri, then, a · ηk1ξr1−k1 ¯ pk2 ¯ qr2−k2 denotes X i∈{1,...,n}r1 ,j∈{1,...,n−1}r2 ai,j ηi1 · · · ηik1 ξik1+1 · · · ξir1 pj1 · · · pjk2 qjk2+1 · · · qjr2 . In the Taylor expansion are present only those monomials ηaξbpcqd which are left unvaried by the symmetries S− 14, S− 34, S− 23, S1↔2 in (6.6) (and by their compositions). 14 7 Birkhoﬀnormal form of the partially reduced secular Hamil-tonian In this section we shall discuss non–resonance properties of the ﬁrst order Birkhoﬀinvariants of the partially reduced secular planetary Hamiltonian and its full torsion (i.e. , the non–vanishing of the determinant of the second order Birkhoﬀinvariants). In particular, we will show that: (i) the ﬁrst order Birkhoﬀinvariants do not verify, at any order, any other resonance besides Herman’s resonance (which, actually, as follows from § 7.2 below, does not aﬀect the construction of Birkhoﬀnormal forms); (ii) the matrix of the second order Birkhoﬀinvariants is not singular (“full torsion”). Furthermore, we will construct explicitly the Birkhoﬀnormal form up to order four for any n, in the well–spaced regime. The secular planetary Hamiltonian is rotation invariant40 and this simpliﬁes the structure of the “dangerous resonances” in constructing Birkhoﬀnormal forms: this is a general fact that will be explained in § 7.2 below. The computations in this section are based upon the Taylor expansion up to order four of the secular planetary perturbation, as given in Appendix B. 7.1 First order Birkhoﬀinvariants Proposition 7.1 (Herman’s resonance) tr (Qh + Qv) = n X i=1 σi + n−1 X i=1 ςi = 0 , Qv := ¯ Qv 0 0 0 , (7.1) where (σ, ¯ ς) ∈Rn × Rn−1 are the eigenvalues of the two quadratic forms Qh and ¯ Qv in (6.9). Proof By (B.1) and (B.3), the diagonal entries of the matrices Qh and ¯ Qv are given by (Qh)ii = P 1≤k̸=i≤n mimk C1(ai,ak) Λi and (¯ Qv(Λ))ii = −P 1≤j<k≤n mjmkC1(aj, ak)(Lji −Lki)2. Using (B.4), one readily ﬁnds (Lji −Lki)2 =          1 Λi+1 + 1 Li , k = i + 1 , 1 ≤j ≤i 1 Λi+1 − 1 Li+1 , i + 1 < k ≤n, j = i + 1 1 Li − 1 Li+1 , i + 1 < k ≤n , 1 ≤j < i + 1 (when j ≤k −2) 0 , otherwise. This implies that P 1≤i≤n−1(Lji −Lki)2 = ( 1 Λj + 1 Λk ) and hence n−1 X i=1 ςi = tr ¯ Qv = X 1≤i≤n−1 (¯ Qv(Λ))ii = − X 1≤i≤n−1 X 1≤j<k≤n mjmkC1(aj, ak)(Lji −Lki)2 = − X 1≤j<k≤n mjmkC1(aj, ak) X 1≤i≤n−1 (Lji −Lki)2 = − X 1≤j<k≤n mjmkC1(aj, ak)( 1 Λj + 1 Λk ) = −tr Qh = − n X i=1 σi . We shall now prove that σ(Λ) and ¯ ς(Λ) do not satisfy any other resonance, up to a preﬁxed order, on a suitable open set of Λ’s. More precisely, we consider the following subset of {a1 < · · · < an} A := Λ : aj < aj < aj for any 1 ≤j ≤n (7.2) where a1, · · · , an, a1, · · · , an, are positive numbers verifying aj < aj < aj+1 for any 1 ≤j ≤n, an+1 := ∞(“well–spaced regime”). 40I.e., it is invariant under Sg; compare (6.8) and (6.3). 15 Proposition 7.2 For any n ≥2, s ∈N, there exist aj, aj and d such that \|(σ, ¯ ς) · k\| ≥d > 0 for any Λ ∈A , k ∈Z2n−1 : 0 < \|k\| ≤s with ki ̸= kj for some i ̸= j . (7.3) In particular, the ﬁrst order Birkhoﬀinvariants σ, ¯ ς ∈Rn × Rn−1 of the partially reduced planetary system verify, identically, only Herman’s resonance (7.1). To prove Proposition 7.2, we need the following simple41 Lemma 7.1 Let M ∗∈Mat(m × m), M∗∈Mat(r × r), M # ∈Mat(m × r), M# ∈Mat(r × m). Let Mδ := M ∗+ O(δ) δM # δM# δ1+tM∗+ O(δ2) , 0 ≤t < 1, (7.4) where δ is a small parameter. Then, (i) if λ∗̸= 0 is a simple eigenvalue of M ∗, then, for \|δ\| small enough, Mδ has an eigenvalue of the form λ∗ δ = λ∗+ O(δ); (ii) if λ∗is a simple eigenvalue of M∗and det M ∗̸= 0, then, for \|δ\| small enough, Mδ has an eigenvalue of the form λδ ∗= δ1+tλ∗+ O(δ2); (iii) if r = 1 and (U∗)tM ∗U∗= diag [λ∗ 1, · · · , λ∗ m] , with U∗∈SO(m)and 0 ̸= λ∗ i ̸= λ∗ j, then there exists Uδ = diag [U∗, 1] + O(δ) such that (Uδ)tMδUδ is diagonal. Proof Statement (i) follows applying the IFT (Implicit Function Theorem) to the function F1(λ, δ) := det(Mδ −λ id m+r), noticing that F1(λ, 0) = (−λ)r det(M ∗−λ id m), For statement (ii), apply the IFT to F2(λ, δ) = det M ∗−δ1+tλ id m δ1−tM # M# M∗−λ id r noticing that det(Mδ −δ1+tλ id m+r) = δ(1+t)rF2(λ, δ). Finally, for statement (iii), apply the IFT the m + 1 functions (w, δ) →(F(i), 1 −\|w\|2), with 1 ≤i ≤m + 1, where F(i)(w, δ) = (M(i) δ −λ(i) δ id m)w, λ(1) δ , · · · , λ(m+1) δ denote the eigenvalues of Mδ which are obtained by continuation of λ∗ 1, · · · , λ∗ m and λ := M∗; M(i) δ the matrix which is obtained by Mδ dropping its ith row and column. Proof of Proposition 7.2 Fix s ≥1. We shall prove by induction that, for any n ≥2, the eigenvalues σ2, · · · , σn, ς1, · · · , ςn−1 do not satisfy any non trivial resonance up to order s in A and thus (7.3) follows. For n = 2, the assertion follows by the direct computation: σ2 = −3 4m1m2 a1 a2 2Λ2 a1 a2 + O(a1 a2 )2 , ς = +3 4m1m2 a1 a2 2 1 Λ1 + 1 Λ2 a1 a2 + O(a1 a2 )3 . (7.5) By (7.5), the functions a2 →\|k2σ2 + κς\| with a1 < a1 < a1 and (k2, κ) ̸= 0, have a positive inﬁmum on a suitable neighborhood of a2 = +∞. Assume now that, when n −1 ≥2, ˆ σ = (ˆ σ1, · · · , ˆ σn−1), ˆ ς = (ˆ ς1, · · · , ˆ ςn−2), ˆ A replace n, σ, ¯ ς, A, where ˆ σ = (ˆ σ1, · · · , ˆ σn−1), ˆ ς = (ˆ ς1, · · · , ˆ ςn−2) denote the eigenvalues of the matrices ˆ Qh, ˆ Qv at rank n −1 and ˆ A is a suitable set of the form (7.2), with n−1 replacing n, then, ˆ σ2, · · · , ˆ σn−1, ˆ ς1,· · · , ˆ ςn−2 do not satisfy any non trivial linear combination on ˆ A. Then, (7.3) holds with n −1, ˆ A, ˆ σ, ˆ ς. In particular, on ˆ A, 0 ̸= ˆ σi ̸= ˆ σj and 0 ̸= ˆ ςh ̸= ˆ ςk , ∀1 ≤i < j ≤n −1 , ∀1 ≤h < k ≤n −2 . (7.6) As it follows from formulae in (B.1)÷(B.4), at rank n, the matrices Qh and ¯ Qv are given by Qh = ˆ Qh + O(δ) O(δ) O(δ) αδ 1 + O(δ2/3) , ¯ Qv = ˆ Qv + O(δ) O(δ) O(δ) βδ 1 + O(δ2/3) (7.7) 41Compare also Lemma B.2 of . 16 where42 δ := a−3 n , α = −3mn 4Λn X 1≤i 0 and integer s, \|Ω(I) · k\| ≥a > 0, ∀k ∈Zm : m X i=1 ki = 0 , 0 < \|k\|1 := m X i=1 \|kj\| ≤2s , I ∈B . (7.10) Then, there exists 0 < ˘ r ≤r and a symplectic transformation ˘ φ : (I, ˘ ϕ, ˘ w) ∈˘ D := B × Tn × B2m ˘ r → (I, ϕ, w) ∈D which puts f into Birkhoﬀnormal form up to the order44 2s. Furthermore, ˘ φ leaves the I–variables unvaried, acts as a ˘ ϕ–indepenent shift on ˘ ϕ, is ˘ ϕ–independent on the remaining variables and is such that ˘ φ ◦Rg = Rg ◦˘ φ . (7.11) 42From (B.1)÷(B.4) it follows that (Qh)nn = −P 1≤i≤n−1 mimn 2Λn ai a2 n b3/2,1(ai/an) and (B.4), that ( ¯ Qv)n−1,n−1 = P 1≤i≤n−1 mimn 2 ( 1 Ln−1 + 1 Λn ) ai a2 n b3/2,1(ai/an). In fact that, by (B.4), the diﬀerences Lj,n−1 −Lk,n−1 vanish unless k = n, in which case they take the value − q 1 Ln−1 + 1 Λn for any 1 ≤j ≤n −1. 43In such set–up there appear a “vertical” invariant ςn = 0, which in the reduced setting is absent. 44I.e., such that f ◦˘ φ = f0+Ω·˘ r+P 2≤h≤s Ph(˘ r; I)+o(\| ˘ w\|2s), where Ph are homogeneous polynomials in ˘ rj = \| ˘ wj\|2/2 := (˘ u2 j + ˘ v2 j )/2 of degree h. 17 Proof As customary, we pass to complex symplectic variables45 (t, t∗) = (t1, . . . , tm), (t∗ 1, . . . , t∗ m) : ( tj = uj−ivj √ 2 t∗ j = uj+ivj √ 2 i (7.12) and write f in such variables46 f(I, t, t∗) = X 0≤k<+∞ X \|α\|1+\|α∗\|1=k cα,α∗(I) Y 1≤i≤m tαi i t∗ i α∗ i . (7.13) In terms of the coordinates t, t∗the rotations Rg in (7.9) becomes Rg : I′ i = Ii, ϕ′ i = ϕi + g , 1 ≤i ≤n ; (t′, t′∗) = Sg(t, t∗) (7.14) where Sg : (t, t∗) →(t′, t′∗) with t′ j := tjeig and t′ j ∗:= t∗ je−ig. Thus, one sees immediately that f being rotation invariance is equivalent to have cα,α∗(I) = 0 , ∀\|α\|1 ̸= \|α∗\|1 . (7.15) In particular, a rotation invariant function is even in (t, t∗). Thus, the ﬁrst non vanishing (and not in normal form) polynomial of the Taylor expansion of f with respect to the w–variables has degree 4, i.e. , f(I, t, t∗) = f0 + X 1≤j≤m Ωj\|tj\|2 + P4(I, t, t∗) + O(\|t\|6), P4 = X \|α\|1+\|α∗\|1=4 c(4) α,α∗ Y 1≤j≤m tαj j t∗ j α∗ j ; (7.16) f0, Ωj and c(4) α,α∗depend on I and we denote \|tj\|2 := itjt∗ j. Because of rotation invariance, c(4) α,α∗= 0 for \|α\|1 ̸= \|α∗\|1. From Birkhoﬀnormal form theory (see, e.g., ) one knows that the symplectic transformation putting (7.16) into Birkhoﬀnormal form of order 2 can be obtained as the time–1 ﬂow φ2 generated by the Hamiltonian function K4 = X α̸=α∗ c(4) α,α∗ iΩ· (α −α∗) Y 1≤j≤n tαj j t∗ j α∗ j = X \|α\|1=\|α∗\|1 c(4) α,α∗ iΩ· (α −α∗) Y 1≤j≤n tαj j t∗ j α∗ j . (7.17) Notice that, since P 1≤i≤m(αi −α∗ i ) = \|α\|1 −\|α∗\|1, P 1≤i≤m(αi −α∗ i ) = 0 in (7.17). Hence, in view of (7.10), we have that \|Ω(I) · (α −α∗)\| ≥a > 0 so that K4 is well–deﬁned and analytic in a neighborhood of t = 0 = t∗. Clearly, K4 is rotation invariant (and ϕ–independent). Let (I, ϕ, t, t∗) →Φθ K4(I, ϕ, t, t∗) the Hamiltonian ﬂow (which leaves I unvaried) generated by K4 at time θ. Since the transformations Rg in (7.9) are symplectic and K4 is invariant by Rg, the ﬂow Φθ K4 commutes with Rg: Φθ K4 ◦Rg(I, ϕ, t, t∗) = Rg ◦Φθ K4(I, ϕ, t, t∗) . Taking θ = 1, we ﬁnd that the transformation φ2 := Φ1 K4 commutes with Rg. Furthermore, φ2 puts f in Birkhoﬀnormal form up to the order 4, acting on the ϕ–variables as a ϕ–independent shift and as the identity on the I–variables and being ϕ–independent on the remaining variables. This implies that the function f2 := f ◦φ2 is ϕ–independent, too. To check that f2 is again rotation invariant, we denote I, (t, t∗) →(tφ2, t∗ φ2) the projection of φ2 on the (t, t∗)–variables and φg := Rg ◦φ2 = φ2 ◦Rg. Then, the projection of φg the (t, t∗)–variables is given by Sg(tφ2, t∗ φ2), thus f2 I, Sg(t, t∗) = f ◦φ2 I, Sg(t, t∗) = f ◦φ2 ◦RgI, (t, t∗) = f ◦φg I, (t, t∗) = f I, Sg(tφ2, t∗ φ2) = f I, (tφ2, t∗ φ2) = f ◦φ2 I, (t, t∗) = f2 I, (t, t∗) . 45As above, i denotes the imaginary unit √−1. 46With abuse of notation, we denote with the same symbol the function f in complex symplectic coordinates. 18 The argument can now be iterated, with f2 = f ◦φ2 replacing f. After s −1 steps, we have the thesis of Proposition 7.3. Remark 7.2 Birkhoﬀnormal form up to the order 2s for the secular Hamiltonian is achieved in s −1 steps, rather than 2s −2, because of parity. 7.3 Second order Birkhoﬀinvariants From now on we shall consider, in phase space, a neighborhood of the secular origin (which corresponds to co–planar and co–circular motions) with well–spaced semi–major axes. In such neighborhood we will construct the Birkhoﬀnormal form. Therefore, we let47 ˜ M6n−2 := A × Tn × B2(2n−1) ϵ0 ⊂M6n−2 0 (7.18) where A is as in (7.2) and ϵ0 is some positive number. Now, let Uh = Uh(Λ) ∈SO(n) a matrix which diagonalizes Qh(Λ) and ¯ Uv = ¯ Uv(Λ) ∈SO(n −1) a matrix which diagonalizes ¯ Qv(Λ): Ut hQhUh = diag[σ1, · · · , σn] , (¯ Uv)t ¯ Qv ¯ Uv = diag[ς1, · · · , ςn−1] . (7.19) Put ¯ U := diag [Uh, Uh, ¯ Uv, ¯ Uv] and consider the unitary transformation of B2(2n−1) ϵ0 into itself ¯ z = ¯ U(Λ) ˜ z = ¯ U(Λ) (˜ η, ˜ ξ, ˜ p, ˜ q), namely, the transformation η = Uh˜ η , ξ = Uh ˜ ξ , ¯ p = ¯ Uv˜ p , ¯ q = ¯ Uv˜ q . (7.20) In the coordinates ˜ z, the quadratic part of fav in (6.9) is in diagonal form X 1≤i≤n σi(Λ) ˜ η2 i + ˜ ξ2 i 2 + X 1≤i≤n−1 ςi(Λ) ˜ p2 i + ˜ q2 i 2 , (7.21) with the ﬁrst order Birkhoﬀinvariants (σ, ¯ ς) satisfying in the set A Herman’s resonance and only that. The change of coordinates (7.20) is lifted to a symplectic transformation ˜ φ of the domain ˜ M6n−2 into itself which, leaving the Λ’s unvaried, acts as ˜ φ : (Λ, ˜ λ, ˜ z) →(Λ, ˜ λ −λ1(Λ, ˜ z), ¯ U(Λ) ˜ z) (7.22) where λ1(Λ, ˜ z) is a suitable shift making ˜ φ symplectic48. Let us denote ˜ H(Λ, λ, ˜ z) := H ◦˜ φ(Λ, ˜ λ, ˜ z) = hKep(Λ) + µ ˜ f(Λ, ˜ λ, ˜ z), (Λ, ˜ λ, ˜ z) ∈˜ M6n−2 . (7.23) It is easy to see that the transformation ˜ φ in (7.22) preserves G. Indeed, since it acts as the identity on Λ and as a unitary transformation ¯ z = ¯ U(Λ)˜ z on ˜ z, we have G ◦˜ φ(Λ, ˜ λ, ˜ z) = G(Λ, ¯ U(Λ)˜ z) = X 1≤i≤n Λi −1 2\|¯ U(Λ)˜ z\|2 = X 1≤i≤n Λi −1 2\|˜ z\|2 = G(Λ, ˜ z) . (7.24) This fact implies that ˜ φ commutes with the the G–ﬂow Rg (6.2) and hence, since the Hamiltonian (5.2) is Rg–invariant, also the Hamiltonian (7.23) is Rg–invariant. Sice ˜ φ commutes also with the reﬂections 47Actually such neighborhood can be lifted to any leaf M6n−2 p∗,q∗. 48The generating function of ˜ φ is S(˜ Λ, ˜ η, ˜ p, λ, ξ, ¯ q) := ˜ Λ · λ + ξ · Uh(˜ Λ)˜ η + ¯ q · ¯ Uv(˜ Λ)˜ p so that ˜ λ1(Λ, ˜ z) = ξ · ∂ΛUh(Λ)˜ η + ¯ q · ∂Λ ¯ Uv(Λ)˜ p evaluated at ξ = Uh(Λ)˜ ξ, ¯ p = ¯ Uv(Λ)˜ p. 19 R− 1 , R− 3 , in (6.6) and H is invariant by them, we have also that ˜ H is invariant by R− 1 , R− 3 (and hence also by R− 2 and R1↔2). Thus, the averaged perturbation ˜ fav admits the same symmetries as fav, namely, for any (Λ, ˜ z) ∈ A × B2(2n−1) ϵ0 one has ˜ fav(Λ, ˜ z) = ˜ fav(Λ, Sg˜ z) = ˜ fav(Λ, S˜ z) , ∀S ∈{S− 14, S− 23, S− 34, S1↔2} (7.25) where S− 14, S− 23, S− 34, S1↔2 are deﬁned in (6.6). As discussed in § 6 above, such symmetries imply that ˜ fav has the form49 ˜ fav(Λ, ˜ z) := (f ◦˜ φ)av(Λ, ˜ z) = fav ◦˜ φ(Λ, ˜ z) = C0(Λ) + X 1≤i≤n σi(Λ) ˜ η2 i + ˜ ξ2 i 2 + X 1≤i≤n−1 ςi(Λ) ˜ p2 i + ˜ q2 i 2 + 1 2 ˜ F(Λ) · ˜ z4 + ˜ P(Λ, ˜ z) (7.26) where ˜ P(Λ, ˜ z) is of order \|˜ z\|6 and (compare (6.9) and (6.10)) 1 2 ˜ F(Λ) · ˜ z4 := ˜ Fh(Λ) · ˜ η4 + ˜ ξ4 + 2˜ η2 ˜ ξ2 2 + ˜ Fv(Λ) · ˜ p4 + ˜ q4 + 2˜ p2˜ q2 2 + ˜ Fhv · ˜ η2˜ q2 + ˜ ξ2˜ p2 −2˜ η˜ ξ˜ p˜ q 2 + ˜ F′ hv · ˜ η2˜ p2 + ˜ ξ2˜ q2 + 2˜ η˜ ξ˜ p˜ q 2 . The new quartic tensors ˜ Fh, ˜ Fv, ˜ Fhv, ˜ F′ hv are easily identiﬁed observing that ˜ F(Λ) · ˜ z4 = F(Λ) · (¯ U(Λ)˜ z)4 (7.27) with F(Λ) as in (6.10). Expression (7.26) is the starting point for the construction of the normal form up to fourth order for the averaged perturbation of the planetary problem: we make use of Proposition 7.3 with m = 2n −1 , B = A , r = ϵ0 , s = 2 , and f = ˜ fav . Herman’s resonance, indeed, does not violate assumption (7.10) (at any order 2s of the Birkhoﬀtrans-formation), since it holds with k = (1, · · · , 1), hence, P 1≤i≤2n−1 ki = 2n −1 > 0. Let ϵ0 > ϵ1 := ˘ r > 0 , ˘ φ : (Λ, ˘ λ, ˘ z) ∈˘ M6n−2 := A × Tn × B2(2n−1) ϵ1 →(Λ, ˜ λ, ˜ z) ∈˜ M6n−2 (7.28) be the Birkhoﬀtransformation given by Proposition 7.3. Denote ˘ H := ˜ H ◦˘ φ(Λ, ˘ λ, ˘ z) = hkep(Λ) + µ ˘ f(Λ, ˘ λ, ˘ z) (7.29) the normalized Hamiltonian, with50 ˘ fav(Λ, ˘ z) := ( ˜ f ◦˘ φ)av(Λ, ˘ z) = ˜ fav ◦˘ φ(Λ, ˘ z) = C0(Λ) + Ω· ˘ R + 1 2τ ˘ R · ˘ R + ˘ P(Λ, ˘ z) (7.30) where ˘ z := (˘ η, ˘ ξ, ˘ p, ˘ q), ˘ P(Λ, ˘ z) is of order \|˘ z\|6 (compare (7.26)), ˘ R = (˘ ρ, ˘ r), ˘ ρ = (˘ ρ1, · · · , ˘ ρn), ˘ r = (˘ r1, · · · , ˘ rn−1), ˘ ρi := ˘ η2 i +˘ ξ2 i 2 , ˘ ri = ˘ p2 i +˘ q2 i 2 , and, ﬁnally, Ω= (σ, ¯ ς) (compare Proposition 7.2). 49Since ˜ φ acts as a ˜ λ–independent shift on the ˜ λ variables and is ˜ λ–independent on the remaining variables, then, the actions of averaging and applying ˜ φ can be interchanged, i.e. , (f ◦˜ φ)av = fav ◦˜ φ, whence, (7.26). 50Again, ( ˜ f ◦˘ φ)av(Λ, ˘ z) = ˜ fav ◦˘ φ(Λ, ˘ z), since ˘ φ acts as a ˘ λ–independent shift on the ˘ λ–variables and is ˘ λ–independent on the remaining variables. 20 Remark 7.3 (i) Notice that, since ˜ f is invariant by Rg, by (7.11) the perturbation ˘ f in (7.29) is again invariant by Rg: ˘ f ◦Rg(Λ, ˘ λ, ˘ z) = ˜ f ◦˘ φ ◦Rg(Λ, ˘ λ, ˘ z) = ˜ f ◦Rg ◦˘ φ(Λ, ˘ λ, ˘ z) = ˜ f ◦˘ φ(Λ, ˘ λ, ˘ z) = ˘ f(Λ, ˘ λ, ˘ z) . (ii) Since ˜ fav is even in ˜ z, to exlicitly evaluate the second order Birkhoﬀinvariants, we can use the procedure described in the proof of Proposition 7.3, with f = ˜ fav and m = 2n −1. In the three–body case (n = 2) one can compute the Birkhoﬀinvariants explicitly (§ 8), while for arbitrary n, we shall evaluate the Birkhoﬀinvariants in the asymptotic limit of well–spaced semi–major axes, i.e. , on the set A in (7.2) (§ 8.2). (iii) As remarked in , since the Birkhoﬀinvariants are real–analytic functions of the semi–major axes, they deﬁne complex holomorphic functions. Now, since in § 8–8.2 we will see that the determinant of the second order Birkhoﬀinvariants (which is also an holomorphic function of the semi–major axes) does not vanish on the set A, by complex function theory51, it then follows that the determinant does not vanish on an open dense set. 8 Full torsion of the partially reduced planetary system In this section we shall check the full torsion of the partially reduced planetary (1 + n)–body problem, i.e. , we shall show that the matrix of the second order Birkhoﬀinvariants τ computed above is non singular. In the case n = 2 we will evaluate exactly τ, while for the general case we shall deduce a simple inductive formula in the asymptotics of well–spaced semi–major axes. 8.1 Full torsion in the three–Body case In the case n = 2, we can compute explicitely τ, re–obtaining, in the planar limit, the computation performed by Arnold in . In this case, we have two couples of horizontal variables (η1, ξ1), (η2, ξ2) and one couple (p, q) of vertical variables. Writing down the matrices Qh and ¯ Qv (of order 2, 1 respectively), the quartic tensors Fh, Fv, Fhv, F′ hv appearing in the expansion (6.9) (compare Appendix B, deﬁnitions in (B.1)÷ (B.17)), computing the matrix Uh which diagonalizes Qh (the matrix ¯ Uv = 1 being trivial) as in (7.19) and computing ﬁnally the normal form of the polynomial (7.27) with ¯ U = diag [Uh, Uh, 1, 1] as described in proof of Proposition 7.3, one ﬁnds the following expressions of the second order Birkhoﬀinvariants: τ11 = 4m1m2 (1 + d2)2 hr1(a1, a2) Λ2 1 + d4r1(a2, a1) Λ2 2 + 2d2r2(a1, a2) Λ1Λ2 (8.31) −2dr3(a1, a2) Λ1 √Λ1Λ2 −2d3r3(a2, a1) Λ2 √Λ1Λ2 + d2r4(a1, a2) Λ1Λ2 i τ12 = 4m1m2 (1 + d2)2 h(1 −d2)2r2(a1, a2) Λ1Λ2 + 2d2r1(a1, a2) Λ2 1 + 2d2r1(a2, a1) Λ2 2 + 2d(1 −d2)r3(a1, a2) Λ1 √Λ1Λ2 −2d(1 −d2)r3(a2, a1) Λ2 √Λ1Λ2 + (1 −6d2 + d4)r4(a1, a2) 4Λ1Λ2 i τ22 = 4m1m2 (1 + d2)2 hr1(a2, a1) Λ2 2 + d4r1(a1, a2) Λ2 1 + 2d2r2(a1, a2) Λ1Λ2 −2dr3(a2, a1) Λ2 √Λ1Λ2 −2d3r3(a1, a2) Λ1 √Λ1Λ2 + d2r4(a1, a2) Λ1Λ2 i 51Compare, also, with the argument used in Proposition 74, p. 1573 of . 21 τ13 = m1m2 (1 + d2)2 h 1 Λ1 1 Λ1 + 1 Λ2 (s1(a1, a2) + s∗ 1(a1, a2)) − 1 Λ2 1 + d2 Λ2 2 C1(a1, a2) − d √Λ1Λ2 1 Λ1 + 1 Λ2 (s2(a1, a2) + s∗ 2(a1, a2)) + d2 Λ2 1 Λ1 + 1 Λ2 (s1(a2, a1) + s∗ 1(a2, a1)) i τ23 = m1m2 (1 + d2)2 h 1 Λ2 1 Λ1 + 1 Λ2 (s1(a2, a1) + s∗ 1(a2, a1)) − 1 Λ2 2 + d2 Λ2 1 C1(a1, a2) + d √Λ1Λ2 1 Λ1 + 1 Λ2 (s2(a1, a2) + s∗ 2(a1, a2)) + d2 Λ1 1 Λ1 + 1 Λ2 (s1(a1, a2) + s∗ 1(a1, a2)) i τ33 = 4m1m2 h r∗ 1(a1, a2) 1 Λ1 + 1 Λ2 2 + C1(a1, a2) 4Λ1Λ2 i where d := \|a −b\| 2c s 1 + 4c2 (a −b)2 −1 with a := C1(a1, a2) Λ1 , b := C1(a1, a2) Λ2 , c := C2(a1, a2) √Λ1, Λ2 and C1, C2, r1, · · · , r4, r∗ 1, r∗ 2, , s1, s2, s∗ 1, s∗ 2 are deﬁned in terms of the Laplace coeﬃcients β(r) k (α) := br/2,k(α) in Appendix B (compare (B.2), (B.6), and (B.8)). The expression (8.31) generalizes the normal form of the planar Three–Body problem (the only known), computed in52 [2, Arnold, 1963]. And in fact, if we consider Arnold’s case (n = 2 and p = 0 = q) and use the following asymptotics (where we regard a1 = O(1) and 1/a2 small) r1(a1, a2) = 3 16a2 a2 1 a2 2 + O a4 1 a4 2 , r1(a2, a1) = −3 4a2 a2 1 a2 2 + O a4 1 a4 2 r2(a1, a2) = r2(a2, a1) = − 9 16a2 a2 1 a2 2 + O a4 1 a4 2 , r3(a1, a2) = O a3 1 a4 2 = r3(a2, a1) r4(a1, a2) = O a4 1 a5 2 , d = O(a−5/4 2 ) (8.32) then, asymptotically, the matrix τ of the coeﬃcients of (7.30), which is reduced to “horizontal” submatrix of (8.31) τpl = τ11 τ12 τ12 τ22 takes the value τpl = m1m2 a2 1 a3 2 3 4Λ2 1 − 9 4Λ1Λ2 − 9 4Λ1Λ2 −3 Λ2 2 ! (1 + O(a−5/4 2 )) . (8.33) We, then, recover det τpl = − 117 16Λ2 1Λ2 2 m1m2 a2 1 a3 2 2 (1 + O(a−5/4 2 )) . (8.34) This is Arnold’s check53 of non–degeneracy (or “full torsion”) of the frequency–map of the planar three– Body problem. 52Recall that when p = 0 = q, the RPS map coincides with the planar Poincar´ e map. 53Compare (8.34) with [2, p. 138, eq. (3.4.31)]. Notice that, in , the second order Birkhoﬀinvariants are deﬁned as one half the τij’s and that, in , a4 2 should be corrected into a7 2. 22 The ﬁrst non–trivial case after that considered by Arnold, is the (spatial) Three–Body problem. As Arnold, we consider well–spaced semimajor axes. Then, using the expansions r∗ 1(a1, a2) = − 3 16a2 a2 1 a2 2 + O a4 1 a4 2 , s1(a1, a2) = 3 a2 a2 1 a2 2 + O a4 1 a4 2 s1(a2, a1) = 9 8a2 a2 1 a2 2 + O a4 1 a4 2 , s∗ 1(a1, a2) = −3 4a2 a2 1 a2 2 + O a4 1 a4 2 s∗ 1(a2, a1) = 9 8a2 a2 1 a2 2 + O a4 1 a4 2 , s2(a1, a2), s∗ 2(a1, a2) = O a3 1 a4 2 C1(a1, a2) = −3 4a2 a2 1 a2 2 + O a4 1 a4 2 , (8.35) we ﬁnd, from (8.31), that the planar matrix (8.33) is completed, in the spatial case, to τ = m1m2 a2 1 a3 2    3 4Λ2 1 − 9 4Λ1Λ2 3 Λ2 1 − 9 4Λ1Λ2 −3 Λ2 2 9 4Λ1Λ2 3 Λ2 1 9 4Λ1Λ2 − 3 4Λ2 1   (1 + O(a−5/4 2 )) (8.36) For small 1/a2, this matrix is non singular: det τ = − 27 16Λ4 1Λ2 2 m1m2 a2 1 a3 2 3 (1 + O(a−5/4 2 )) = −27 16 m2 m1m3 0 a4 1 a7 2 (1 + O(µ) + O(a−5/4 2 )) (8.37) having used Λ2 i = m2 i m0ai + O(µ). 8.2 Asymptotic full torsion in the general case The asymptotic evaluation (8.37) can be extended to the general case: Proposition 8.1 (Full torsion of the planetary frequency map) For n ≥2 and 0 < δ⋆< 1 there exist54 ¯ µ > 0, 0 < a1 < a1 < · · · < an < an such that, on the set A deﬁned in (7.2) and for 0 < µ < ¯ µ, the matrix τ = (τij) is non–singular: det τ = dn(1 + δn), where \|δn\| < δ⋆with dn = (−1)n−1 3 5 45 16m2 0 n−1 m2 m1m0 a1 a1 an 3 Y 2≤k≤n 1 ak 4 . (8.38) Proof The proof is by induction on n. The case n = 2 has been proved by explicit computation, compare (8.37). Assume now that, when, in the statement of Proposition 8.1, n −1 ≥2, ˆ A and ˆ τ replace n, A, τ, then, det ˆ τ = dn−1(1 + O(µ) + O(1/√a2) (8.39) and let us prove this equality at rank n. Proceeding as in the proof of Proposition 7.2 and using Lemma 7.1–(iii), one sees that, at rank n, matrices which diagonalize Qh, ¯ Qv may be taken of the form Uh = U(0) h + O(δ) , ¯ Uv = ¯ U(0) v + O(δ) (8.40) where U(0) h = diag [ˆ Uh, 1], ¯ U(0) v = diag [ˆ Uv, 1] and ˆ Uh, ˆ Uv denote matrices which diagonalize the second oder matrices ˆ Qh, ˆ Qv at the previous rank. Using (B.5)÷(B.17), one also sees that the polynomial of degree 4 in (6.9) has the form 54¯ µ is taken small only to simplify (8.38), but a similar evaluation hold with ¯ µ = 1. 23 F(Λ) · ¯ z4 = (ˆ F(ˆ Λ) + F0(Λ))(z(n−1))4 + X 1≤i≤3 Fi(Λ) · (z(n−1))4−izi + F4(Λ) · z4 , (8.41) where Fi are O(δ), ˆ Λ, z(n−1), z denote the sets of variables ˆ Λ = (Λ1, · · · , Λn−1) , z(n−1) = (η1, · · · , ηn−1, ξ1, · · · , ξn−1, p1, · · · , pn−2, q1, · · · , qn−2) z := (ηn, ξn, pn−1, qn−1) and ˆ F · (z(n−1))4 is the fourth order polynomial in (6.9) associated to rank n −1. Later, we will need the explicit expression of the polynomial F4 · z4 in (8.41), which, as we will shortly see, is given by F4 · z4 = ¯ Fh(Λ)(η2 n + ξ2 n)2 + ¯ Fv(Λ)(p2 n−1 + q2 n−1)2 + ¯ Fhv(ηnqn−1 −ξnpn−1)2 + ¯ F′ hv(ηnpn−1 + ξnqn−1)2 , (8.42) where ¯ Fh(Λ) = X 1≤j<n mjmn r1(an, aj) Λ2 n ¯ Fv(Λ) = X 1≤j<n mjmn h (r∗ 1(aj, an) 1 Ln−1 + 1 Λn 2 + C1(aj, an) 4Ln−1Λn i ¯ Fhv = X 1≤j<n mjmn h 1 Ln−1 + 1 Λn s1(an, aj) Λn −C1(aj, an) 2Λ2 n i ¯ F′ hv = X 1≤j<n mjmn h 1 Ln−1 + 1 Λn s∗ 1(an, aj) Λn −C1(aj, an) 2Λ2 n i . (8.43) In fact, the ﬁrst line in (8.43) is found using (B.5). Next, by (B.7), the action Fv1, Fhv1, F′ hv1 on z4, is found selecting in the formula of Fv1 only the indices j = k = n −1 and, in the formula of Fhv, F′ hv, the indices j = n and h = n −1. Using, as in the proof of Proposition 7.2, that Li,n−1 −Lj,n−1 = − q 1 Λn + 1 Ln−1 δj,n, one ﬁnds the contribution to ¯ Fv, ¯ Fhv and ¯ F′ hv proportional to, respectively, r∗ 1, s1 and s∗ 1 in (8.43). Finally, the actions of the tensors Fv2, Fhv2, F′ hv2 on z4 are found observing that the matrices Ti+1, T∗ i , ¯ Ti deﬁned in (B.9)÷(B.17) contain only (ηj, ξj)–variables with indices j ≤i + 1 and (pj, qj)–variables with indices j ≤i. Hence, the monomials with literal part z4 in the polynomial (B.17) are only those appearing in P 1≤i<n mimnC1(ai, an) tr ¯ Tn−1. Since the quartic part of the trace of ¯ Tn−1 is −¯ Qn−1(p2 n−1 + q2 n−1), with ¯ Qn−1 = −Ln−1Λn(p2 n−1 + q2 n−1) −2(Ln−1)2(η2 n + ξ2 n) 4L2 n−1Λ2 n (up to monomials which do not depend on z: compare (B.9)), it follows that the sum of the actions of Fv2, Fhv2, F′ hv2 on z4 is identiﬁed with with the polynomial X 1≤i<n mimnC1(ai, an)(p2 n−1 + q2 n−1)Ln−1Λn(p2 n−1 + q2 n−1) −2L2 n−1(η2 n + ξ2 n) 4L2 n−1Λ2 n and hence (8.43) follow. By (8.40) and (8.41), letting, for short, ¯ U(0) := diag [U(0) h , U(0) h , ¯ U(0) v , ¯ U(0) v ] and ˆ U := diag [ˆ Uh, ˆ Uh, ˆ Uv, ˆ Uv], one readily ﬁnds that the polynomial (7.27) is given by ˜ F(Λ) · ˜ z4 = F(Λ) · (¯ U˜ z)4 = F(Λ) · (¯ U(0) + O(δ))˜ z 4 = (ˆ F(ˆ Λ) + ˜ F0(Λ))(ˆ U˜ z(n−1))4 + X 1≤i≤3 ˜ Fi(Λ) · (˜ z(n−1))4−i˜ zi + (F4(Λ) + O(δ2)) · ˜ z4 , (8.44) 24 where ˜ Fi are O(δ) and ˜ z(n−1) := (˜ η1, · · · , ˜ ηn−1, ˜ ξ1, · · · , ˜ ξn−1, ˜ p1, · · · , ˜ pn−2, ˜ q1, · · · , ˜ qn−2) , ˜ z := (˜ ηn, ˜ ξn, ˜ pn−1, ˜ qn−1) . Then, computing the second order Birkhoﬀinvariants τ = (τij) from (8.44), one ﬁnds τ = ˆ τ + O(δ) O(δ) O(δ) ¯ τ + O(δ2) (8.45) where ˆ τ is the second order Birkhoﬀinvariants matrix at rank n −1 and ¯ τ is the matrix of the Birkhoﬀ invariants of order 2 associated to the polynomial in (8.42), i.e. , ¯ τ11 = 4¯ Fh = 4mn Λ2 n X 1≤j<n mjr1(an, aj) ¯ τ12 = ¯ τ21 = ¯ Fhv + ¯ F′ hv = mn X 1≤j<n mj h 1 Ln−1 + 1 Λn s1(an, aj) + s∗ 1(an, aj) Λn −C1(aj, an) Λ2 n i ¯ τ22 = 4¯ Fv = 4mn X 1≤j<n mj h (r∗ 1(aj, an) 1 Ln−1 + 1 Λn 2 + C1(aj, an) 4Ln−1Λn i . Using now Ln−1 = Λn−1(1 + O(Λn−2/Λn−1), Λ2 i = m2 i m0ai(1 + O(µ)) and the asymptotics expressions in (8.32) and (8.35), one ﬁnds ¯ τ11 = −3mn−1a2 n−1 m0mna4 n (1 + O(1/Λn−1) + O(µ)) , ¯ τ12 = 9 4 a2 n−1 m0√an−1ana3 n (1 + O(1/Λn−1) + O(µ)) ¯ τ22 = −3 4 mnan−1 m0mn−1a3 n (1 + O(1/Λn−1) + O(µ)) . (8.46) Finally in view of (8.45), det τ = det ˆ τ det ¯ τ(1 + O(δ)) = −45 16m2 0 a3 n−1 a7 n dn−1(1 + O(1/√a2) + O(µ)) relation which immediately implies (8.38) at step n, because of the inductive assumption (8.39). 9 The totally reduced planetary system We turn now to the total reduction of the planetary system. Indeed, the partially reduced Hamiltonian (in Birkhoﬀnormalized variables (7.28)÷(7.30)) ˘ H = hkep+µ ˘ f still commutes with the length of the total angular momentum G = \|C\|, or, equivalently, is invariant under the Rg action; compare Remark 7.3. This allows to reduce completely the total angular momentum by introducing symplectic variables, which include an action–angle couple (G, g), with g cyclic for the reduced planetary Hamiltonian. However, to achieve the total reduction one has, in general, to exclude certain conical singularities, compare (9.7) below55. Recall (7.28) and let, at ﬁrst, ˘ M6n−2 ∗ := A × Tn × B∗ (9.1) where B∗denotes the open set B∗:= B2(2n−1) ϵ1 ∩ ˘ z : (˘ ηi, ˘ ξi) ̸= 0 , (˘ pj, ˘ qj) ̸= 0 for any 1 ≤i ≤n , 1 ≤j ≤n −1 . 55These singularities include the secular origin (i.e., co–circular and co–planar phase points); but, say, all co–circular and all–but–one co–planar are allowed; compare Remark 9.1–(iii) below. In the case n = 2, these singularities are removable; see . 25 For ˘ z ∈B∗, let ( ˘ ρi := ˘ η2 i +˘ ξ2 i 2 ˘ ϕi := arg(˘ ξi, ˘ qi) , ( ˘ rj := ˘ p2 j+˘ q2 j 2 ˘ χj := arg(˘ pj, ˘ qj) (9.2) be the symplectic polar coordinates associated to (˘ ηi, ˘ ξi), (˘ pj, ˘ qj). The integral G becomes then a linear function of Λ, ˘ ρ1, . . . , ˘ ρn, ˘ r1, . . . , ˘ rn−1, G = X 1≤i≤n Λi − X 1≤i≤n ˘ ρi − X 1≤j≤n−1 ˘ rj . (9.3) Denote ˆ r := (˘ r1, . . . ˘ rn−2), ˆ χ := (˘ χ1, . . . ˘ χn−2), 1k := (1, . . . , 1) ∈Rk and observe that Λ · d˘ λ + ˘ η · d˘ ξ + ˘ p · d˘ q = Λ · d˘ λ + ˘ ρ · d ˘ ϕ + ˘ r · d˘ χ (9.4) = Λ · d˘ λ + ˘ ρ · d ˘ ϕ + ˆ r · dˆ χ + ˘ rn−1 · d˘ χn−1 = Λ · d(˘ λ + ˘ χn−11n) + ˘ ρ · d( ˘ ϕ −˘ χn−11n) + ˆ r · d(ˆ χ −˘ χn−11n−2) + Gd(−˘ χn−1) . Deﬁne, now, the symplectic map ˆ φ −1 : (Λ, ˘ λ, ˘ z) →(Λ, G, ˆ λ, g, ˆ z) by letting, for 1 ≤i ≤n and 1 ≤j ≤ n −2: ˆ λ = (ˆ λ1, . . . ˆ λn) ∈Tn , (G, g) ∈R+ × T , ˆ z = (ˆ η, ˆ ξ, ˆ p, ˆ q) with ˆ η = (ˆ η1, . . . , ˆ ηn) , ˆ ξ = (ˆ ξ1, . . . , ˆ ξn) ∈Rn ; ˆ p = (ˆ p1, . . . , ˆ pn−2) , ˆ q = (ˆ q1, . . . , ˆ qn−2) ∈Rn−2 Λi ˆ λi := ˘ λi + ˘ χn−1 , G = G(Λ, ˘ z) g := −˘ χn−1 = −arg(˘ pn−1, ˘ qn−1) ˆ ηi = √2˘ ρi cos ( ˘ ϕi −˘ χn−1) ˆ ξi = √2˘ ρi sin ( ˘ ϕi −˘ χn−1) , ˆ pj = p 2˘ rj cos (˘ χj −˘ χn−1) ˆ qj = p 2˘ rj sin (˘ χj −˘ χn−1) . (9.5) Conclusions are drawn in the following Remark 9.1 (i) In view of (9.4) the map ˆ φ −1 : ˘ M6n−2 ∗ →ˆ M 6n−2 ∗ := ˆ φ −1( ˘ M6n−2 ∗ ) is symplectic. (ii) Inverting formulae (9.5) we ﬁnd that the symlpectic map ˆ φ is explicitly given by ˘ ηi ˘ ξi = R(−g) ˆ ηi ˆ ξi , ˘ λi = ˆ λi + g , 1 ≤i ≤n ˘ pj ˘ qj = R(−g) ˆ pj ˆ qj , ˘ pn−1 ˘ qn−1 = R(−g) p ϱ2 −\|ˆ z\|2 0 , 1 ≤j ≤n −2 (9.6) where ϱ2 := 2 P 1≤i≤n Λi −G and R(g) is the matrix (6.4). Thus, ˆ φ can be extended to a real–analytic map ˆ φ : ˆ M 6n−2 →˘ M6n−2 with56: ˆ M 6n−2 := Λ ∈A , G ∈R+ , \|ˆ z\| < ϱ(Λ, G) := s 2 X 1≤i≤n Λi −G < ϵ1 , ˆ λ ∈Tn , g ∈T . (9.7) (iii) In particular, points on the hyperplanes (ˆ ηi, ˆ ξi) = 0 or (ˆ pj, ˆ qj) = 0, corresponding to co–circular or co–planar motions for n −1 planets (and one possibly co–circular with the others but not co–planar) are regular points for ˆ φ. (iv) Consider the Hamiltonian ˆ H := ˘ H ◦ˆ φ on the phase space ˆ M 6n−2. As already mentioned, the invariance of ˘ H by Rg implies that the variable g is cyclic for ˘ H ◦ˆ φ. We then set ˆ HG(Λ, ˆ λ, ˆ z) := ˘ H ◦ˆ φ(Λ, ˆ λ, ˆ z; G) = hKep(Λ) + µ ˆ f G(Λ, ˆ λ, ˆ z) . (9.8) 56Recall (9.1) and notice that by (9.2) and (9.3) P 1≤i≤n Λi −G = \|˘ z\|2/2. Notice also that ˆ φ is well deﬁned for any Λ ∈Rn +. 26 For ﬁxed values of the parameter G ∈R+, the “totally reduced planetary Hamiltonian” ˆ HG governs the motions of the planetary system on the (6n −4)–dimensional phase space ˆ M 6n−4 G := ˆ AG × Tn × B2(2n−2) ϱ(Λ,G) , ˆ AG := {Λ ∈A : 0 < ϱ(Λ, G) < ϵ1} , (9.9) endowed with the standard symplectic form dΛ ∧dˆ λ + dˆ η ∧dˆ ξ + dˆ p ∧dˆ q. (v) The full motion on ˆ M 6n−2 is then simply obtained by integrating the cyclic variable g, which amounts to a rotation around the direction of C with instant speed given by ∂G ˆ HG. 10 Birkhoﬀnormal form and full torsion of the totally reduced secular Hamiltonian In this section, by means of the standard IFT, we cast the totally reduced secular Hamiltonian ˆ f G,av in Birkhoﬀnormal form up to order four on ˆ M 6n−4 G . In this context, at contrast with the partially reduced case, there are no secular resonances (see Remark 10.1 below); on the other hand, as in the partially reduced case, the torsion matrix is nonsingular. 10.1 Fourth order Birkohﬀnormal form of the totally reduced planetary Hamiltonian We ﬁrst consider the “ambient phase space” ˆ M 6n−2. Using the expression of ˆ φ in (9.6) and the inde-pendence of ˆ f G on g, we have ˆ f G = ˘ f ◦ˆ φ = ( ˘ f ◦ˆ φ)\|g=0 = ˘ f ◦ˆ φ0 (10.1) where ˆ φ0 := ˆ φ (ϱ) 0 := ˆ φ\|g=0 has the simple expression ˆ φ (ϱ) 0 : ˘ λ = ˆ λ , ˘ η = ˆ η ˘ ξ = ˆ ξ , ˘ pi = ˆ pi ˘ qi = ˆ qi , ˘ pn−1 = p ϱ2 −\|ˆ z\|2 ˘ qn−1 = 0 , 1 ≤i ≤n −2 . (10.2) Taking the average of (10.1) with respect to ˆ λ and using (7.30) one readily ﬁnds the following expression ˆ f G,av(Λ, ˆ z) = ˘ fav ◦ˆ φ (ϱ) 0 = ˆ C(Λ, ϱ) + ˆ Ω(Λ, ϱ) · ˆ R + 1 2 ˆ τ(Λ)ˆ R · ˆ R + ˆ F(Λ, ˆ z, ϱ) (10.3) where ˆ R := (ˆ ρ, ˆ r) , ˆ ρi := ˆ η2 i + ˆ ξ 2 i 2 , ˆ rj := ˆ p2 j + ˆ q2 j 2 1 ≤i ≤n , 1 ≤j ≤n −2 ; the constants ˆ C(Λ, ϱ) and ˆ Ω= (ˆ Ω1, . . . , ˆ Ω2n−2) are deﬁned as: ˆ C(Λ, ϱ) := C0(Λ) + ϱ2 2 ςn−1(Λ) + ϱ4 4 τmm(Λ), m := 2n −1 (10.4) ˆ Ωi(Λ, ϱ) = Ω0 i (Λ) + ϱ2(τim(Λ) −τmm(Λ)), 1 ≤i ≤2n −2 (10.5) with Ω0 := (Ω0 1, . . . , Ω0 2n−2) given by Ω0 i (Λ) := σ0 i (Λ) := σi(Λ) −ςn−1(Λ), 1 ≤i ≤n ς0 i−n(Λ) := ςi−n(Λ) −ςn−1(Λ), n + 1 ≤i ≤2n −2 ; (10.6) and the matrix ˆ τ = (ˆ τ ij) by ˆ τ ij(Λ) := τij(Λ) −τim(Λ) −τjm(Λ) + τmm(Λ), 1 ≤i, j ≤2n −2 ; (10.7) 27 ﬁnally, ˆ F(Λ, ˆ z, ϱ) := ˘ P ◦ˆ φ (ϱ) 0 (Λ, ˆ z) . (10.8) In general, for n ≥3, the function ˆ F(Λ, ˆ z, ϱ) has non vanishing derivatives in ˆ z = 0. But the existence of a suitable elliptic equilibrium for ˆ f G(Λ, ˆ z) can be estabilished by a simple IFT argument and the relative normal form around such equilibrium next restored by classical symplectic diagonalization and Birkhoﬀnormal form. Remark 10.1 The functions (10.6) do not exhibit any resonance of order 4 or lower in the range of well separated semi–major axes. In fact, the existence of a non–trivial linear relation among the Ω0 i ’s, would imply a linear relation among σ1, . . . , σn, ς1, . . . , ςn−1 diﬀerent from Herman’s resonance (since the coeﬃcient of ςn−1 should be opposite to the sum of the coeﬃcients of σ1, . . . , σn, ς1, . . . , ςn−2), which is in contraddiction with Proposition 7.2. Let I := (Λ, G) , B∗= n I = (Λ, G) ∈A × R+ : X 1≤i≤n Λi −G > 0 o ⊂Rn + × R+ ˆ ϕ := (ˆ λ, ˆ g) ∈Tn+1 , ˆ u := (ˆ η, ˆ p) , ˆ v := (ˆ ξ, ˆ q) , ˆ z := (ˆ u, ˆ v). Proposition 10.1 There exists ϵ2 ∈(0, ϵ1) such that, on the domain ˇ M 6n−2 := n I ∈B∗, ˇ ϕ ∈Tn+1 , 0 < ϱ(I) < ϵ2 , \|ˇ z\| < ϱ(I) 4 o , (10.9) one can ﬁnd a real–analytic and symplectic transformation with respect to dI ∧dϕ + dˆ u ∧dˆ v, ˇ φ : (I, ˇ ϕ, ˇ z) ∈ˇ M 6n−2 →(I, ˆ ϕ, ˆ z) ∈ˆ M 6n−2, with the following properties. The map ˇ φ leaves the I–variables unvaried, maps ˇ ϕ →ϕ = ˇ ϕ + ˇ ϕ1(I, ˇ z), and is ˇ ϕ–independent on the remaining variables; furthermore, it puts ˆ HG into the form ˇ HG = ˇ HG ◦ˇ φ = hKep(Λ) + µ ˇ f G(Λ, ˇ λ, ˇ z) (10.10) with: ˇ f G,av(Λ, ˇ z) = ˇ CG(Λ) + ˇ ΩG(Λ) · ˇ R + 1 2 ˇ R · ˇ τ G(Λ)ˇ R + ˇ P G(Λ, ˇ z) , (10.11) ˇ z = (ˇ u, ˇ v) = (ˇ u1, . . . , ˇ u2n−2), (ˇ v1, . . . , ˇ v2n−2) , ˇ R = (ˇ R1, . . . , ˇ R2n−2) , ˇ Ri = ˇ u2 i + ˇ v2 i 2 ; with ˇ P G(I, ˇ z) having a zero of order 5 in ˇ z = 0. The ﬁrst and second Birkhoﬀinvariants ˇ ΩG, ˇ τ G verify \|ˇ ΩG(Λ) −Ω0(Λ)\| ≤Cϱ(I)2 , ∥ˇ τ G(Λ) −ˆ τ(Λ)∥≤Cϱ(I)2 , (10.12) where Ω0 and ˆ τ are as in (10.6) and (10.7). Finally, ˇ ΩG is non–resonant up to order four57. Remark 10.2 (i) Clearly, since the Hamiltonian ˆ HG does not depend on g, it projects to a Hamiltonian (which we still call ˆ HG) on the totally reduced phase space Λ ∈A , ˇ λ ∈Tn , \|ˇ z\| < ϱ(I) 4 < ϵ2 4 , (10.13) 57And, in fact, by possibly reducing ϵ2, at any ﬁnite order. 28 endowed with the standard symplectic form dΛ ∧dˇ λ + dˇ η ∧ˇ dξ + dˇ p ∧dˇ q, (ˇ p, ˇ q ∈Rn−2 and G is a parameter). Notice that, now, the secular center ˇ z = 0 is an elliptic equilibrium for the totally reduced Hamiltonian ˆ HG. (ii) For the applications in § 11 below, we will consider, for simplicity, a subset of (10.13) of the form ˇ M 6n−4 G := Λ ∈ˇ AG , ˇ λ ∈Tn , \|ˇ z\| < ϵ3 , with ˇ AG := {Λ ∈A : 4ϵ3 < ϱ(Λ, G) < ϵ2} , (10.14) where ϵ3 is an arbitrary positive number smaller than ϵ2/4. Proof of Proposition 10.1. We divide the proof of into four steps. Step 1: The new elliptic equilibrium For n = 2, the point ˆ ze := 0 in an equilibrium for ˆ f G,av for any I ∈B∗. When n ≥3, to ﬁnd an equilibrium I →ˆ ze(I), we use the IFT (Proposition 10.2 below). Fix a number 0 < θ < 1. For ˇ z = (ˇ u, ˇ v) = (ˇ u1, . . . , ˇ u2n−2), (ˇ v1, . . . , ˇ v2n−2 ∈B 2(2n−2) θ , (10.15) let ˇ φ (ϱ) the map ˇ φ (ϱ) : (Λ, ˆ λ, ˇ z) →ˆ φ (ϱ) 0 (Λ, ˆ λ, ϱˇ z) := (Λ, ˆ λ, ˇ φ (ϱ) ˘ z ) = (Λ, ˆ λ, ϱˇ φ (1) ˘ z ) , where ˆ φ (ϱ) 0 is the map (10.2) and hence ˇ φ (1) ˘ z : ˇ z = (ˇ u, ˇ v) →˘ z = (˘ u, ˘ v) : ˘ u = ˇ u , ˘ vi = ˇ vi 1 ≤i ≤n −2 , ˘ vn−1 = p 1 −\|ˇ z\|2 . (10.16) Let us also put f (ϱ) := ˆ f G,av\|ˆ z=ϱˇ z = ˆ C0(Λ, ϱ) + ϱ2 ˆ Ω(Λ, ϱ) · ˇ R + ϱ2 2 ˇ R · ˆ τ(Λ)ˇ R + ϱ4Q(Λ, ˇ z, ϱ) where ˇ Ri := ˇ u2 i + ˇ v2 i 2 ( 1 ≤i ≤2n −2) and, by (10.8) and (10.16), the function Q(Λ, ˇ z, α) is deﬁned by α6Q(Λ, ˇ z, α) = ˆ F(Λ, αˇ z, α) = ˘ P ◦ˆ φ (α) 0 (Λ, αˇ z) = ˘ P(Λ, ˇ φ (α) ˘ z (ˇ z)) = ˘ P(Λ, αˇ φ (1) ˘ z (ˇ z)) . (10.17) Since the function ˘ z →˘ P(Λ, ˘ z) is regular on the closed ball B 2(2n−1) ϵ1/2 ⊂B2(2n−1) ϵ1 , has a zero of order 6 in 0, ˇ φ (1) ˘ z is regular on B 2(2n−2) θ and \|ˇ φ (1) ˘ z (ˇ z)\| = 1 for any ˇ z ∈B 2(2n−2) θ , by (10.17), for any ﬁxed Λ, the function (ˇ z, α) →Q(Λ, ˇ z, α) is a regular function of ˇ z, α on the domain \|ˇ z\| ∈B 2(2n−2) θ and 0 ≤α ≤ϵ1 2 . We now need a quantitative formulation of the standard IFT58: Proposition 10.2 (Quantitative IFT) Let w0 ∈Rh, and let A be a compact set of Rk. Let F : (w, α) ∈BR(w0) × A →F(w, α) ∈R (BR(w0) denoting the closed ball of radius R and center w0) be a continuous function with invertible and C1 Jacobian matrix ∂wF(w0, α), for any α ∈A. Denote by M(α) := ∂wF(w0, α) −1 , by m an upper bound on supA ∥M∥(∥· ∥denoting the standard “operator norm” on matrices). If 4m2 sup A \|F(w0, α)\| sup BR(w0)×A ∥∂2 wF∥≤1 (10.18) then, there exists a unique continuous function α ∈A →w(α) ∈Bρ(w0) such that F(w(α), α) ≡0 for any α ∈A, where ρ := 2m supA \|F(w0, α)\|. 58The elementary proof can be found in [5, Theorem 1]. 29 We apply Proposition 10.2, with59 h = 2n −2, k = 1, A = {α}, w0 = 0, R = θ and F(ˇ z, α) := ∂ˇ z(f (α) −ˆ C0(Λ, α))α−2 = ∂ˇ z ˆ Ω(Λ, α) · ˇ R + α2 2 ˇ R · ˆ τ(Λ)ˇ R + α4Q(Λ, ˇ z, α) , having omitted to explicitate in F the dependence on Λ. We have in fact      F(0, α) = α4∂ˇ zQ(Λ, 0, α) ∂ˇ zF(0, α) = diag [ˆ Ω, ˆ Ω] + α4∂2 ˇ zQ(Λ, 0, α) ∂2 ˇ zF(ˇ z, α) = α2∂3 ˇ z 1 2 ˇ R · ˆ τ ˇ R + α2Q(Λ, ˇ z, α) easily implying the existence of three constants60 c1, c2, c3 (independent on Λ) for which the following bounds hold    \|F(0, α)\| ≤c1α4 ∥(∂ˇ zF(0, α))−1∥≤2 supA ∥( diag [ˆ Ω, ˆ Ω])−1∥=: m when c2α4 ≤1 supˇ z∈B 2(2n−2) θ ∥∂2 ˇ zF(ˇ z, α)∥≤c3α2 . (10.19) Notice that the constant m is well deﬁned by the non resonance assumption of the Ω′s. Let c4 := 2mc1 , c5 := 2c4 θ , c6 := 4m2c1c3 and let ¯ ϱ such that the inequalities max n c6α6, c5α5o ≤1 (10.20) hold for α ≤¯ ϱ . The assumption (10.18) of Proposition 10.2 is easily met, since, in view of (10.19), (10.20) 4m2\|F(0, α)\| sup B 2(2n−2) θ ∥∂2 ˇ zF∥≤c6α6 ≤1 . Then, by Proposition 10.2, a root ˇ ze = ˇ ze(Λ, α) of equation F(ˇ z, α) = 0 can be found, satisfying \|ˇ ze\| ≤2m\|F(0, α)\| ≤c4α4 (10.21) for any α ≤¯ ϱ. Thus, the point ˆ ze = ˆ ze(Λ, G) ∈B 2(2n−2) c4ϱ5 ⊂B 2(2n−2) θϱ/2 deﬁned as ˆ ze := ϱˇ ze(Λ, ϱ) is an equilibrium for fG,av for any (Λ, G), with Λ ∈A and G such that ϱ(Λ, G) ≤ϵ2, where ϵ2 ≤¯ ϱ. Step 2: Symplectic shift of the equilibrium into the origin The equilibrium point ˆ z = ˆ ze can be then shifted into the origin with the change of variables ˆ z = z⋆+ ˆ ze , i.e. , ˆ u = u⋆+ ˆ ue(I) , ˆ v = v⋆+ ˆ ue(I) , (10.22) where z⋆:= (u⋆, v⋆) is taken varying into the closed ball B 2(2n−2) θϱ/2 , so to have \|ˆ z\| = \|z⋆+ ˆ ze\| ≤\|z⋆\| + \|ˆ ze\| ≤ϱθ 2 + ϱθ 2 = ϱθ < ϱ . (10.23) 59In particular, Proposition 10.2 holds also when the compact A is a set of one point only. 60The ci’s can be taken 8 > < > : c1 := supΛ∈A,0≤α≤ϵ1/2 \|∂ˇ zQ(Λ, 0, α)\| c2 := 2 supΛ∈A,0≤α≤ϵ1/2 ∥( diag [ˆ Ω, ˆ Ω])−1∂2 ˇ zQ(Λ, 0, α)∥ c3 := supΛ∈A,ˇ z∈B2n−2 θ ,0≤α≤ϵ1/2 ∥∂3 ˇ z( 1 2 R · ˆ τR + α2Q(Λ, ˇ z, α))∥ 30 By construction, the function f ⋆(I, z⋆) := ˆ f G,av\|ˆ z=z⋆+ˆ ze has vanishing linear part61 in z⋆= 0. Using this fact and (10.3), we can write f ⋆(I, z⋆) := ˆ f G,av\|ˆ z=z⋆+ˆ ze = C⋆(I) + Ω⋆· R⋆+ 1 2 R⋆· ˆ τR⋆+ Q⋆(I, z⋆) , (10.24) where62          C⋆(I) := ˆ C(I) + ˆ Ω· ˆ Re + 1 2 ˆ Re · ˆ τ ˆ Re , where ˆ R (i) e := (ˆ u(i) e )2+(ˆ v(i) e )2 2 R⋆ i := (u⋆ i )2+(v⋆ i )2 2 Ω⋆:= ˆ Ω+ ˆ τ ˆ Re Q⋆(I, z⋆) := P⋆(I, z⋆) −Lz⋆(P⋆(I, z⋆)) (10.25) where P⋆(I, z⋆) := ˆ F(Λ, z⋆+ ˆ ze, ϱ) = ˘ P ◦ˆ φ (ϱ) 0 (Λ, z⋆+ ˆ ze) (10.26) and Lz⋆(P⋆(I, z⋆)) denotes the linear part of P⋆(I, z⋆) with respect to z⋆in z⋆= 0. The shift (10.22) is next lifted to a symplectic transformation φ⋆which, leaving the I′s unvaried, acts on their respective conjugated angles as63 ˆ ϕ = ϕ⋆+ u⋆· ∂Iˆ ve(I) −(v⋆+ ˆ ue(I))∂I ˆ ue(I) . Step 3: Symplectic diagonalization around the equilibrium Let Q⋆, C⋆, F⋆denote, respectively, the coeﬃcients of the expansion Q⋆(I, z⋆) = P⋆(I, z⋆) −Lz⋆(P⋆(I, z⋆)) = Q⋆ 0(I) + z⋆· Q⋆(I)z⋆+ C⋆(I) · z⋆3 + 1 2F⋆(I) · z⋆4 + Q⋆ 5(I, z⋆) , (10.27) with Q⋆ 5(I, z⋆) ≤C\|z⋆\|5. Using (10.23) and (10.26), one easily ﬁnds suitable constants c7 ÷c9 such that sup A ∥Q⋆(I)∥≤c7ϱ4 , sup A ∥C⋆(I)∥≤c8ϱ3 , sup A ∥F⋆(I)∥≤c9ϱ2 . (10.28) In particular, when ϱ is suﬃciently small, the quadratic form Q∗:= diag [ˆ Ω, ˆ Ω] + Q⋆ (10.29) associated to f ⋆(I, z⋆) has purely imaginary eigenvalues (hence, the equilibrium z⋆= 0 is elliptic). Let them be denoted as ˇ Ω= (ˇ Ω1, . . . , ˇ Ωm−1) Then, ˇ Ωsatisfy \|ˇ Ω−ˆ Ω\| ≤c10ϱ4 . (10.30) Let ˇ Ω· R∗, where R∗ i = (u∗ i )2+(v∗ i )2 2 , the diagonal form of the quadratic form Q∗as in (10.29) and let z∗→L(I)z∗, where z∗= (u∗, v∗), the symplectic tansformation64 which transforms Q∗in such diagonal form. By (10.28), L is ϱ4–close to the identity and hence we can assume that is well deﬁned on the domain B2(2n−2) ϱ/3 . Being linear and symplectic, L can be lifted to a symplectic transformation φ∗: (Λ, λ∗, z∗) →(Λ, λ⋆, z⋆) 61If z →g(a, z) is analytic on z = 0, its linear part in z = 0 is ∂zg(a, 0) · z. 62If v ∈R2n−2, v(i) or vi denotes its ith component. 63The generating function of this transformation is S(I′, u′, ϕ, v) = I′ · ϕ + u′ · (v −ve(I′) + v · ue(I′). 64With respect to du∗∧dv∗. 31 on the domain D∗: I ∈B∗, ϕ ∈Tn+1 , 0 < ϱ(I) < ϱ∗, \|z∗\| < ϱ(I) 3 , such in a way to leave the I′s unvaried and shifting the angles ϕ∗of a ϕ∗–independent quantity and is ϕ∗–independent on the remaining variables. Let f ∗ av := f ⋆ av ◦φ∗= C⋆ 0 + ˇ Ω· R∗+ C∗· (z∗)3 + 1 2R∗· τ ∗R∗+ F∗· (z∗)4 + O(\|z∗\|5) , where τ ∗, C∗and F∗are ϱ4–close to ˆ τ, C⋆and F ⋆, respectively. Step 4: Construction of the Birkhoﬀnormal form By (10.30) and (10.4), the ﬁrst invariants ˇ Ωof f ∗ av are ϱ2–close to the functions Ω0 i into (10.6) which, as already observed, do not satisfy any resonance of order 4 or lower in A. Then, when ϱ is suﬃciently small, Birkhoﬀtheory can be applied, proving the thesis of Proposition 10.1. The ﬁnal claim follows at once by the non–resonance of the Ω0 i ’s (Remark 10.1) and by taking ϵ2 small enough (compare (10.12)). 10.2 Full torsion of the totally reduced planetary system We are now ready to check full torsion in the fully reduced setting. Corollary 10.1 Fix n ≥2 and 0 < δ⋆< 1. Then, there exist ¯ µ > 0, 0 < a1 < a1 < · · · < an < an such that for any µ < ¯ µ and for any Λ ∈ˇ AG, where ˇ AG is the set in (10.14), the matrix ˇ τ G is non–singular: det ˇ τ G = ˆ dn(1 + δn), with \|δn\| < δ⋆and ˆ dn = 4 5 m0mn mn−1 a4 n a2 n−1 dn (10.31) where dn is as in (8.38). Proof By the second inequality in (10.12) (eventually, taking ϵ2 smaller), it suﬃcies to prove that the matrix ˆ τ deﬁned (in terms of the unreduced matrix τ) in (10.7) is non–singular. We distinguish two cases. Case n = 2 By the computation of the τij’s performed in § 7.3, (compare the exact expression in (8.31), or, better, the asymptotics for small 1/a2 in (8.36)), when 1/a2 is small enough, one has ˆ τ = m1m2 a2 1 a3 2 −6 Λ2 1 −15 4Λ2 1 −15 4Λ2 1 − 3 4Λ2 1 ! (1 + o(1)) , easily proving non–singularity: det ˆ τ = − m1m2 a2 1 a3 2 2 153 16Λ4 1 (1 + o(1)) ̸= 0 . Case n ≥3. Since τi,2n−1 = τ2n−1,i = O(δ) for any 1 ≤i ≤2n −1, one easily ﬁnds that ˆ τ = ˇ τ + O(δ) O(δ) O(δ) ˆ τ 2n−2,2n−2 + O(δ2) (10.32) where δ = 1/a3 n, ˇ τ denotes the unreduced matrix relatively to n −1 bodies (of order 1 in δ) and ˆ τ 2n−2,2n−2 = ¯ τ11 −2¯ τ12 + ¯ τ22 = −3 4 mn L2 n−1 δ X 1≤j 0 and let H(I, ϕ, p, q; µ) = H0(I) + µ P(I, ϕ, p, q; µ) be a real–analytic Hamiltonian on Pϵ0, endowed with the standard symplectic form dI ∧dϕ + dp ∧dq. Assume that H verify the following non–degeneracy assumptions: (A1) I ∈V →∂IH0 is a diﬀeomorphism; (A2) P av(p, q; I) = P0(I)+ n2 X i=1 Ωi(I)ri + 1 2 n2 X i,j=1 βij(I)rirj +o4 with ri := p2 i +q2 i 2 and lim (p,q)→0 o4 \|(p, q)\|4 = 0; (A3) \| det β(I)\| ≥const > 0 for all I ∈V . Then, there exist positive numbers ϵ∗< ϵ0, C∗and c∗such that, for 0 < ϵ < ϵ∗, 0 < µ < ϵ6 (log ϵ−1)c∗, (11.1) one can ﬁnd a set K ⊂Pϵ formed by the union of H–invariant n–dimensional Lagrangian tori, on which the H–motion is analytically conjugated to linear Diophantine quasi–periodic motions with frequencies (ω1, ω2) ∈Rn1 ×Rn2 with ω1 = O(1) and ω2 = O(µ). The set K has positive Liouville–Lebesgue measure and satisﬁes meas Pϵ > meas K > 1 −C∗ √ϵ meas Pϵ . (11.2) This is Theorem 1.3 in , to which we refer for the proof65. • Consider the partially reduced planetary (1 + n)–body system in normalized Birkhoﬀcoordinates, i.e. , consider the Hamiltonian system ( ˘ H, ˘ M6n−2 ϵ ) where ˘ H is as in (7.29) and ˘ M6n−2 ϵ is deﬁned as in 65Actually, Theorem 11.1 is a corollary of a more general result holding under much milder conditions than (11.1); compare Theorem 1.2 in . 33 (7.28) with B2(2n−1) ϵ1 replaced by B2(2n−1) ϵ with ϵ ≤ϵ1. Consider also the “well separated regime”, i.e. , let µ < ¯ µ and let the semi–major axes ai be as in (7.2) with aj and aj as in Proposition 8.1, so that full torsion holds (det τ ̸= 0). Then, ˘ M6n−2 ϵ has exactly the form of Pϵ with V = A, n1 = n , n2 = 2n −1. Furthermore, by the form of the Keplerian part, by (7.30) and by Proposition 8.1, one sees immediately that the Hamiltonian ˘ H satisﬁes assumptions (A1)÷(A3) of Theorem 11.1. Thus, the following result follows at once. Theorem 11.2 If µ < ¯ µ and ϵ < ϵ1 verify condition (11.1), then each symplectic leaves M6n−2 p∗ n,q∗ n (5.1) contains a positive measure H–invariant Kolmogorov set Kp∗ n,q∗ n, which is actually the suspension of the same Kolmogorov’s set K, which in normalized Birkhoﬀsymplectic variables (Λ, ˘ λ, ˘ z) is ˘ H–invariant. Furthermore, K is formed by the union of (3n −1)–dimensional Lagrangian, real–analytic tori on which the ˘ H–motion is analytically conjugated to linear Diophantine quasi–periodic motions with frequencies (ω1, ω2) ∈Rn × R2n−1 with ω1 = O(1) and ω2 = O(µ). Finally, K satisﬁes the bound in (11.2) with Pϵ replaced by ˘ M6n−2 ϵ ; in particular meas K ≃ϵ2(2n−1). Remark 11.1 (i) In fact, from Remark 7.3–(iii) it follows (giving up the constructive approach) that the same result holds in an open dense set of M6n. (ii) Since the “secular variables” ˘ z vary in a ball around the origin, which correspond to co–circular and co–planar motions, we recover and strengthen Arnold’s results in ; compare [2, p. 125 and p. 142]. The approach followed here extends to the general situation Arnold’s proof, which was given only for the planar n = 2 case. (iii) The proof in of Arnold’s result is rather diﬀerent from the one presented here. Indeed, the proof in is based on the following: (a) one works in the unreduced phase space endowed with Poincar´ e spatial symplectic variables; (b) the KAM non–degeneracy condition used in involves only the ﬁrst order Birkhoﬀinvariants66, which requires the frequency–map (formed by ∂ΛhKep and by the ﬁrst order Birkhoﬀinvariants of the secular Hamiltonian) to be non–planar (i.e. , not to lie in any hyperplane); (c) in the unreduced phase space, however, two secular resonances are present (Herman resonance and the vanishing of one of the spatial eigenvalues, ςn); (d) to overcome the problem of the secular resonances, F´ ejoz, following an idea of Herman (based, in turn, on a Poincar´ e trick), modiﬁes the system by adding a term proportional to C2 3, since, by an abstract Lagrangian intersection theory argument, two commuting Hamiltonians have the same transitive tori; (e) to obtain (3n −1)–dimensional tori one restricts to the vertical symplectic submanifold67 {C1 = 0 = C2}. As side remarks, we point out also that: in the unreduced setting the full torsion is false (i.e. , one can show that the determinant of the second order Birkhoﬀinvariant vanish identically68); the KAM theory developed in the invariant tori constructed are C∞even if the starting Hamiltonian is analytic69; it is not clear what kind of measure estimates one would obtain from the scheme in . • We turn now to the construction of Lagrangian tori in the fully reduced setting and well–spaced regime. Let ˇ M 6n−4 G,ϵ be as in (10.14) with ϵ3 replaced by a generic ϵ ≤ϵ3. Then, also in this case, ˇ M 6n−4 G,ϵ has the form Pϵ with V = AG, n1 = n, n2 = 2n −2 and the totally reduced Hamiltonian ˇ HG in (10.10)–(10.11), by Proposition 10.1, veriﬁes assumptions (A1)÷(A3) of Theorem 11.1. Thus, a statement parallel to Theorem 11.2 holds also in the totally reduced case: Theorem 11.3 If µ < ¯ µ and ϵ < ϵ3 verify condition (11.1), then, ˇ M 6n−4 G,ϵ contains a positive measure Kolmogorov set KG, which is ˇ HG–invariant and is formed by the union of 3n−2–dimensional Lagrangian, real–analytic tori on which the ˇ HG–motion is analytically conjugated to linear Diophantine quasi–periodic motions with frequencies (ω1, ω2) ∈Rn×R2n−2 with ω1 = O(1) and ω2 = O(µ); furthermore KG satisﬁes the bound in (11.2) with Pϵ replaced by ˇ M 6n−4 G,ϵ . In particular meas KG ≃ϵ2(2n−2). 66Such condition is called “Arnold–Pyarti’s condition” in ; elsewhere is also called “R¨ ussmann condition”. 67In , however, no explicit symplectic variables are available on the vertical submanifold. 68Compare . 69For a real–analytic version of see . 34 Remark 11.2 (i) The tori found in this theorem were not mentioned in . (ii) The (3n−1)–dimensional invariant tori in the partially reduced phase space obtained by integrating (rotating) the angle g in the Hamilton equation ˙ g = ∂G ˆ HG, may be resonant according to whether the quasi–periodic average of ∂G ˆ HG is rationally independent or not with the (3n−2) Diophantine frequencies of the invariant tori belonging to KG. In general one expects to have all kind of tori (resonant, Liouvillean and Diophantine). Clearly all the 3n–dimensional tori lifted in the unreduced space will be resonant, since the pn and qn variables are, in fact, always constant. 11.2 n–dimensional elliptic KAM tori in the planetary system In this ﬁnal section we discuss brieﬂy elliptic lower dimensional tori in the planetary reduced (and fully reduced) system, generalizing to the spatial case the results in . The Lagrangian tori found in § 11 have fast Keplerian rotations and slow secular quasi–periodic varia-tions around the elliptic linear equilibrium. It is natural to ask whether the linear secular equilibrium bifurcates in the full nonlinear dynamics into lower dimensional elliptic tori of dimension n. This is indeed the case, as we will shortly prove. Results in this direction were obtained in in the planar 3–body case, in for the spatial 3–body case and in in the planar (1+n)–body case. All these results are based on the application of the lower dimensional KAM theory as developed by Melnikov , Kuksin , Eliasson , R¨ ussmann and P¨ oschel . The extension of the previous results to the general spatial case in the partially and fully reduced setting is possible, because of absence of low–order resonances (in the well–spaced region or in an open dense set). We also remark that in this discussion one only needs non degeneracy of the ﬁrst order Birkhoﬀinvariants (“Melnikov’s conditions”) and no information is needed on the second order Birkhoﬀinvariants. The existence of lower dimensional tori will be based upon the following Theorem 11.4 Let Pϵ and H be as in Theorem 11.1 with (A2) and (A3) replaced, respectively, by (A′ 2) P av(p, q; I) = P0(I) + n2 X i=1 Ωi(I)ri + o2 with ri := p2 i +q2 i 2 and lim (p,q)→0 o2 \|(p, q)\|2 = 0; (A′ 3) \|Ωi\| ≥const , \|Ωi −Ωj\| ≥const , ∀i ̸= j, ∀I ∈V . Then, if ϵ and µ are small enough, there exists a Cantor set of actions I in V of positive n1–measure, which parameterizes a family of n1–dimensional elliptic H–invariant tori on which the H–motion is analytically conjugated to linear quasi–periodic motions with n1 Diophantine frequencies close to the unperturbed frequencies ∂IH0. The proof of Theorem 11.4 is given is § 4 of ; compare 4.1÷4.4, which describe the steps necessary to prove Theorem 11.4. Taking Pϵ = ˘ M6n−2 (as in (7.28) with ϵ1 replaced by ϵ ≤ϵ1) and ˘ H = hkep(Λ) + µ ˘ f(Λ, ˘ λ, ˘ z) (as in (7.29)–(7.30)), and observing that assumption (A′ 3) follows from Proposition 7.2, one obtains the following corollary: Theorem 11.5 For ϵ, µ small enough, the well–spaced, partially reduced planetary system ˘ H, ˘ M6n−2 possesses a Cantor family of n–dimensional elliptic ˘ H–invariant tori with frequencies close to the un-perturbed Keplerian frequencies. Such tori are “surrounded” by the Lagrangian tori constructed in The-orem 11.2. An analogous discussion can be done in the totally reduced planetary system, leading to the existence of n–dimensional elliptic tori also in ˇ M 6n−4 G (compare (10.14)). 35 A Explicit formulae for the RPS map A.1 Deprit map In this section we describe the map which relates the action–angle Deprit variables (Λ, Γ, Ψ, λ, γ, ψ) deﬁned in § 3 to the standard Cartesian variables (y, x) ∈R3n × R3n. To describe properly the rotations occurring in such map, we introduce the following notation. Given two frames F= (k(1), k(2), k(3)) and G= (κ(1), κ(2), κ(3)), we say that G is (i, ψ)–rotated with respect to F if κ(1) ⊥k(3) , ακ(1)(k(3), κ(3)) = i and αk(3)(k(1), κ(1)) = ψ i, ψ ∈T . (A.1) So, if (z1, z2, z3), (ζ1, ζ2, ζ3) denote, respectively, the respective sets of Cartesian coordinates of a point P with respect to F, G, i.e. , P = z1k(1) + z2k(2) + z3k(3) = ζ1κ(1) + ζ2κ(2) + ζ3κ(3), the matrix RFG of the change of coordinates from G to F, i.e. , such that   z1 z2 z3  = RFG   ζ1 ζ2 ζ3   is given by RFG = R31(ψ, i) := R3(ψ)R1(i) , with R1(i), R3(ψ) denoting the matrices R1(i) =   1 0 0 0 cos i −sin i 0 sin i cos i  , R3(ψ) =   cos ψ −sin ψ 0 sin ψ cos ψ 0 0 0 1  . (A.2) Let F=(k(1), k(2), k(3)) a preﬁxed positively oriented orthonormal frame. With the same notations as in § 3, we introduce the following positively oriented frames. • The “orbital frames” Fj, for 1 ≤j ≤n, deﬁned by the orthonormal triples (k(1,j), k(2,j), k(3,j)), where k(3,j) is in the direction of C(j) and k(1,j) in the direction of the node νj (compare the deﬁnition (3.4)); • the frames F∗ j, for 1 ≤j ≤n, deﬁned as follows. F∗ 1 is the orbital frame F1, deﬁned above. For 2 ≤j ≤n, the frames F∗ j are deﬁned by the orthonormal triples (k(1,j) ∗ , k(2,j) ∗ , k(3,j) ∗ ), where k(3,j) ∗ is in the direction of S(j) and k(1,j) ∗ in the direction of νj+1, with νn+1 := ¯ ν. The planes (k(1,j), k(2,j)) of the frames Fj = (k(1,j), k(3,j), k(3,j)) contain the osculating ellipses Ej (orthogonal to C(j) ∥k(3,j)). We recall that they are deﬁned as the ellipses with the perihelia Pj forming an angle γj with the k(1,j) = νj–axis, semimajor axis aj = 1 ¯ mj ( Lj Mj )2 and eccentricity ej = q 1 −( Γj Lj )2. We also recall that the symplectic Cartesian coordinates (y(j) pl , x(j) pl ) with respect to the orbital frame Fj of a point on the osculating ellipse Ej are recovered by the so–called Kepler map, analytically deﬁned by equations x(j) pl = R3(γj)x(j) orb , y(j) pl = βj ∂ℓjx(j) pl (A.3) where βj is as in (4.5) and x(j) orb = aj    cos uj −ej q 1 −e2 j sin uj 0   , (A.4) uj being the unique solution of Kepler’s equation uj −ej sin uj = ℓj. 36 By (A.3) and the deﬁnition of the frames Fj, the Cartesian coordinates (y(j), x(j)) with respect to the preﬁxed frame F=(k(1), k(2), k(3)) are recovered by the following formulae      y(j) = (y(j) 1 , y(j) 2 , y(j) 3 ) = RFFj(Γ, Ψ, ψ) y(i) pl (Lj, Γj, ℓj, γj) = βj∂ℓjx(j) x(j) = (x(j) 1 , x(j) 2 , x(j) 3 ) = RFFj(Γ, Ψ, ψ) x(i) pl (Lj, Γj, ℓj, γj) 1 ≤j ≤n , (A.5) where RFFj denotes the matrix which describes the change of coordinates from Fj to F. To describe RFFj in terms of the Deprit variables deﬁned in § 3, we consider the following tree of frames F → F∗ n → F∗ n−1 → · · · → F∗ j → · · · → F∗ 2 →F∗ 1 = F1 ↓ ↓ ↓ ↓ Fn Fn−1 Fj F2 (A.6) and we decompose RFFj as RFFj =    RFF∗ nRF∗ nF∗ n−1 · · · RF∗ j+1F∗ j RF∗ 2F1 , j = 1 RFF∗ nRF∗ nF∗ n−1 · · · RF∗ j+1F∗ j RF∗ j Fj , 2 ≤j ≤n (A.7) with RFG generically denoting the matrix associated to the change of coordinates from G to F. We have thus to express the matrices appearing at the right hand side of (A.7) in terms of the Deprit variables. To do it,we deﬁne the fhe following “absolute” angles ij, i∗ j, i.e. , the angles in (0, π) deﬁned by cos i∗ j = Ψ2 j + Ψ2 j−1 −Γ2 j+1 2ΨjΨj−1 , cos i∗ n = C3 G , i1 := i∗ 1 , cos ij+1 = Ψ2 j + Γ2 j+1 −Ψ2 j−1 2ΨjΓj+1 (A.8) where Ψ0 := Γ1 and 1 ≤j ≤n −1. Notice that i∗ n has the meaning of the absolute angle between C and k(3) and, considering the triangle with sides Ψj−2 = \|S(j−1)\|, Γj = \|C(j)\| and Ψj−1 = \|S(j)\| = \|S(j−1)+C(j)\|, for 1 ≤j−1 ≤n−1, i∗ j−1 is the absolute angle between S(j−1) and S(j) (with S(1) := C(1)), while, for 2 ≤j ≤n, ij is the absolute angle between C(j) and S(j). Notice also that values 0 or π are never reached because of the assumptions70 (3.2). By the above deﬁnitions, one has that • F∗ n is (i∗ n, ζ)-rotated with respect to F and hence, as noticed at the beginning of this section, RFF∗ n = R31(ζ, i∗ n) . (A.9) In fact, k(1,n) ∗ ∥νn+1 := ¯ ν = k(3) × C ⊥k(3); by the deﬁnition of i∗ n, αk(1,n) ∗ (k(3), k(3,n) ∗ ) = α¯ ν(k(3), C) = αk(3)×C(k(3), C) = i∗ n and, by deﬁnition of ζ, αk(3)(k(1), k(1,n) ∗ ) = αk(3)(k(1), ¯ ν) = ζ. • When n ≥3, RF∗ j F∗ j−1 = R31(ψj−1, −i∗ j−1) for 2 ≤j −1 ≤n −1 , (A.10) since F∗ j−1 is (−i∗ j−1, ψj−1)–rotated with respect to F∗ j. In fact, one has that k(1,j−1) ∗ ∥νj = S(j)×C(j) ⊥S(j) ∥k(3,j) ∗ ; furthermore, αk(1,j−1) ∗ (k(3,j) ∗ , k(3,j−1) ∗ ) = ανj(S(j), S(j−1)) = αS(j)×C(j)(S(j), S(j−1)) = α−S(j)×S(j−1)(S(j), S(j−1)) = −i∗ j−1, by deﬁnition of i∗ j−1. Finally, by deﬁnition of ψj−1, αk(3,j) ∗ (k(1,j) ∗ , k(1,j−1) ∗ ) = αS(j)(νj+1, νj) = ψj−1. 70S(j) × C(j) ̸= 0 implies that also S(j−1) × S(j) ̸= 0, since C(j) = S(j) −S(j−1). 37 • For 2 ≤j ≤n, Fj is (ij, ψj−1)–rotated with respect to F∗ j, hence, RF∗ j Fj = R31(ψj−1, ij), 2 ≤j ≤n . (A.11) In fact, k(1,j) ∥νj = S(j) × C(j) ⊥S(i) ∥k(3,j) ∗ ; furthermore, by deﬁnition of ij, αk(1,j)(k(3,j) ∗ , k(3,j)) = ανj(S(j), C(j)) = αS(j)×C(j)(S(j), C(j)) = ij and ﬁnally, by deﬁnition of ψj−1, αk(3,j) ∗ (k(1,j) ∗ , k(1,j)) = αS(j)(νj+1, νj) = ψj−1 . • For j = 1, F1 is (−i∗ 1, ψ1)–rotated with respect to F∗ 2, implying RF∗ 2F1 = R31(ψ1, −i∗ 1) . (A.12) In fact, similarly to the previous case, we have that k(1,1) ⊥k(3,2) ∗ and that αk(1,1)(k(3,2) ∗ , k(3,1)) = αν2(S(2), C(1)) = −αν2(C(1), S(2)) = −i1 = −i∗ 1. Furthermore, since ν1 = ν2, we have that αk(3,2) ∗ (k(1,2) ∗ , k(1,1)) = αS(2)(ν3, ν1) = αS(2)(ν3, ν2) = ψ1. Then, in view of (A.7), (A.9), (A.10), (A.11) and (A.12), the description (A.5) of the Deprit map is completed by the following formulae RFFj(Γ, Ψ) =    R31(ψn, i∗ n)R31(ψn−1, −i∗ n−1) · · · R31(ψ2, −i∗ 2)R31(ψ1, −i∗ 1), j = 1 R31(ψn, i∗ n)R31(ψn−1, −i∗ n−1) · · · R31(ψj, −i∗ j)R31(ψj−1, ij), 2 ≤j ≤n . (A.13) where i∗ j, ij are as in (A.8). A.2 RPS map By (4.1) and (4.2), the angles ζ, ψn j = P j≤k≤n ψk and γj are related to regularized variables by71 ζ = ψn = −arg(pn, qn) , ψn j = −arg(pj, qj), γ1 −arg(p1, q1) = −arg(η1, ξ1), γj −arg(pj−1, qj−1) = −arg(ηj, ξj) 2 ≤j ≤n . (A.14) From the formula of ψn j , we ﬁnd the angles ψ1, . . . , ψn−2, ψn−1 = g: ψj = ψn j −ψn j+1 = arg(pj+1, qj+1) −arg(pj, qj) for 1 ≤j ≤n −1 . (A.15) We substitute such expressions of into the matrices RFFj deﬁned in (A.13), and we ﬁnd RFF1 = R1R3 −arg(p1, q1) , RFFj = Rj R3 −arg(pj−1, qj−1) for 2 ≤j ≤n (A.16) where the matrices Rj are the products (4.7), with R∗ j :=    R313(arg(pj, qj), −i∗ j), 1 ≤j ≤n −1 R313(arg(pn, qn), i∗ n), j = n (A.17) and Rj := R313(arg(pj−1, qj−1), ij), 2 ≤j ≤n with R313(ψ, i) := R3(−ψ)R1(i)R3(ψ) . (A.18) 71Recall the choice ψ0 = 0. As usual, if (u, v) ∈R2 \ {0}, arg(u, v) denotes the unque t ∈T such that cos t = u √ u2+v2 and sin t = v √ u2+v2 . Recall also that assumptions (3.2) imply (ηj, ξj) ̸= 0, (pj, qj) ̸= 0 for any 1 ≤j ≤n. 38 Remark A.1 Notice that, if (Λ, λ, z) →Rg(Λ, λ, z), where Rg is as in (7.9), then, Ri →R(g)RiR(−g) (since the same holds for the matrices Ri, R∗ i , as it follows by their deﬁnitions). Substituting (A.16) into the expression of y(i), x(i) in (A.5), with x(i) pl as in (A.4) and using ﬁnally the expressions in the second line of (A.14), one easily ﬁnds the expression in (4.3), with Ri as in (4.7) and x(j) pl := R3 −arg(ηj, ξj) x(j) pl easily recognized to be the planar Poincar´ e map deﬁned in (4.4) and (4.5). To complete the analytical expression of the RPS map (4.3), we have to express the matrices R∗ i , Ri in (A.17) and (A.18) in terms of the variables (Λ, λ, z). To this end, we let ρi := η2 i + ξ2 i 2 , ri := p2 i + q2 i 2 , Li := X 1≤j≤i Λj (A.19) z0 := (η1, ξ1) , zi = (η1, . . . , ηi+1, ξ1, . . . , ξi+1, p1, . . . , pi, q1, . . . , qi), 1 ≤i ≤n −1 so that zn−1 = ¯ z, (¯ z, (pn, qn)) = (zn−1, (pn, qn)) = z. By (4.1), (4.2), Γi = Λi −ρi , 1 ≤i ≤n Ψi = i+1 X j=1 Λj − i+1 X j=1 (Λj −Γj) − i X j=1 (Ψj−1 + Γj+1 −Ψj) = Li+1 −1 2\|zi\|2 , 1 ≤i ≤n −1 C3 = G −p2 n + q2 n 2 = Ψn−1 −p2 n + q2 n 2 = Ln −1 2\|z2\| (A.20) Then, by (A.8), we have 1 −cos i∗ n = G −C3 G = p2 n + q2 n 2Ln −\|¯ z\|2 = (p2 n + q2 n)c∗ n (A.21) where c∗ n := 1 2Ln −\|¯ z\|2 (A.22) Similarly deﬁne c∗ 1, . . . , c∗ n−1, c2, . . . , cn by c∗ j := 2Λj+1 −2ρj+1 −rj (2Lj+1 −\|zj\|2)(2Lj −\|zj−1\|2) 1 ≤j ≤n −1 cj := 2Lj−1 −\|zj−2\|2 −rj−1 2(Λj −ρj)(2Lj −\|zj−1\|2) 2 ≤j ≤n (A.23) such that, by (A.8) and (A.20), for 1 ≤j ≤n −1 1 −cos i∗ j = 1 −Ψ2 j + Ψ2 j−1 −Γ2 j+1 2ΨjΨj−1 = (Γj+1 −Ψj + Ψj−1)(Γj+1 + Ψj −Ψj−1) 2ΨjΨj−1 = (p2 j + q2 j )c∗ j (\|z0\|2 := 2ρ1) (A.24) and, for 2 ≤j ≤n 1 −cos ij = 1 −Ψ2 j−1 + Γ2 j −Ψ2 j−2 2Ψj−1Γj = (Ψj−2 −Ψj−1 + Γj)(Ψj−2 + Ψj−1 −Γj) 2Ψj−1Γj = (p2 j−1 + q2 j−1)cj . (A.25) 39 Now, if R1, R3 are as in (A.2) then, the matrix R313(ψ, i) = R3(−ψ)R1(i)R3(ψ) has the expression R313(ψ, i) =   1 −sin2 ψ(1 −cos i) −sin ψ cos ψ(1 −cos i) −sin ψ sin i −sin ψ cos ψ(1 −cos i) 1 −cos2 ψ(1 −cos i) −cos ψ sin i sin ψ sin i cos ψ sin i cos i   (A.26) Then, by (A.21), (A.24), (A.25) and (A.26), we ﬁnd that the matrices R∗ j, Rj into (A.17) and (A.18) have the expressions (4.8), with cj, c∗ j as in (A.22) and (A.23) and s∗ j := sin(−i∗ j) q p2 j + q2 j = − r c∗ j 2 −(p2 j + q2 j )c∗ j 1 ≤j ≤n −1 s∗ n := sin i p p2 n + q2 n = r c∗ n 2 −(p2 n + q2 n)cn sj := sin ij q p2 j−1 + q2 j−1 = r cj 2 −(p2 j−1 + q2 j−1)cj 2 ≤j ≤n . (A.27) Notice that the matrices R∗ j, Rj , hence the matrices Ri in (4.7), are regular also when some of (ηj, ξj) or (pj, qj) vanishes. Since, as it is known, also the planar Poincar´ e map (y(j) pl , x(j) pl ) (compare (4.5), (4.6)) is regular, the RPS (4.3) map is so. B Expansion of the secular Hamiltonian up to order four in RPS variables Here, we will prove that the explicit form of the constant C0, the quadratic tensors Qh, ¯ Qv and the quartic tensors Fh, Fv, Fhv, F′ hv appearing into the expansion (6.9) and (6.10) are given by the following explicit formulae. • The constant C0 is trivial and is given by C0(Λ) := − X 1≤j<k≤n mjmk ak b1/2,0(aj/ak) where ak = ak(Λk) = ¯ m−1 k (Λk/Mk)2 is the kth semi–major axis and bh,k’ s denote the Laplace coeﬃcients72. Let us, now, denote by β(r) k := br/2,k and, for two any positive numbers a ̸= b, α = α(a, b) := a/b. Then: • the horizontal quadratic form Qh is73 Qh(Λ) · η2 := X 1≤j 0, k ∈Z, α ∈C, with α ̸= ±1, the Laplace coeﬃcient bh,k(α) are the Fourier coeﬃcient of the function t →(1 + α2 −2α cos t)−h, i.e. , bh,k(α) = 1 2π R 2π 0 cos kt (1+α2−2α cos t)h dt. Recall that we are in a region of phase where a1 < a2 < · · · < an. Notice that, when ¯ z = (η, ξ, ¯ p, ¯ q) = 0, the x(i)–projection of the RPS map reduces to x(i) = ai(cos λi, sin λi, 0), whence the expression of C0 follows. 73Formulae (B.1), (B.2) are as in [15, (36), (37)], where Poincar´ e variables are used. This is due to the fact that when ¯ p = ¯ q = 0 and pn = qn = 0, the two sets of variables (Poincar´ e and RPS) coincide: see Remark 4.2–(i). The “vertical” quadratic form in (10.29), on the other hand, diﬀers from that in . 40 ¯ Qv(Λ) · ¯ p2 := − X 1≤j<k≤n mjmkC1(aj, ak) X 1≤h≤n−1 (Ljh −Lkh)ph 2 (B.3) where C1 is as in (B.2) and L is the n × (n −1) matrix given by Lij :=        − q Λj+1 Lj+1Lj for i = 1 and 1 ≤j ≤n −1 , or 2 ≤i ≤j ≤n −1 q Li−1 LiΛi for 2 ≤i ≤n and j = i −1 0 otherwise. (B.4) where Lj := P 1≤k≤j Λk. • the horizontal quartic tensor Fh(Λ) is given by Fh(Λ) · η2ξ2 := X 1≤i<j≤n mimj r1(ai, aj)η2 i ξ2 i Λ2 i + r1(aj, ai)η2 j ξ2 j Λ2 j + r2(ai, aj) η2 i ξ2 j ΛiΛj + r2(aj, ai) η2 j ξ2 i ΛiΛj + r3(ai, aj) ηiηjξ2 i Λi p ΛiΛj + r3(aj, ai) ηiηjξ2 j Λj p ΛiΛj + r3(ai, aj) η2 i ξiξj Λi p ΛiΛj + r3(aj, ai) η2 j ξiξj Λj p ΛiΛj + r4(ai, aj)ηiηjξiξj ΛiΛj (B.5) where, for a ̸= b, r1(a, b) := −α 256b [(−60α5 + 4311α3 −300α) β(9) 0 (α) + 8 · (7α6 −252α4 −222α2 + 7) β(9) 1 (α) + 4 · (75α5 −503α3 + 135α) β(9) 2 (α) + 24 · (23α4 + 13α2) β(9) 3 + 37α4 β(9) 4 (α)] r2(a, b) := 3α 512b [(84α5 −8832α3 + 84α) β(9) 0 (α) −8 · (5α6 −652α4 −652α2 + 5) β(9) 1 (α) −5 · (328α5 −561α3 + 328α) β(9) 2 (α) + (216α6 −1020α4 −1020α2 + 216) β(9) 3 (α) + (116α5 + 200α3 + 116α) β(9) 4 (α) −(20α4 + 20α2) β(9) 5 (α) + 3α3 β(9) 6 (α)] r3(a, b) := α 256b [(1146α4 + 1266α2) β(9) 0 (α) + (−744α5 + 2014α3 −864α) β(9) 1 (α) + 8 · (28α6 −321α4 −321α2 + 28) β(9) 2 (α) + (552α5 + 423α3 + 672α) β(9) 3 (α) + 6(29α4 + 9α2) β(9) 4 (α) −5α3 β(9) 5 (α)] r4(a, b) := −α 128b [(−36α5 −7956α3 −36α) β(9) 0 (α) + 8 · (α6 + 828α4 + 828α2 + 1) β(9) 1 (α) + (−3096α5 + 1039α3 −3096α) β(9) 2 (α) + (648α6 −1332α4 −1332α2 + 648) β(9) 3 (α) + (348α5 + 700α3 + 348α) β(9) 4 (α) −60 · (α4 + α2) β(9) 5 (α) + 9α3 β(9) 6 (α)] . (B.6) • the vertical quartic tensor Fv(Λ) and the mixed quartic tensors Fhv(Λ), F′ hv(Λ) can be splitted as Fv(Λ) = Fv1(Λ) + Fv2(Λ) , Fhv(Λ) = Fhv1(Λ) + Fhv2(Λ) F′ hv(Λ) = F′ hv1(Λ) + F′ hv2(Λ) where • Fv1(Λ) · ¯ p2¯ q2, Fhv1(Λ) · η2¯ q2 and F′ hv1(Λ) · η2¯ p2 are respectively given by 41 X 1≤i<j≤n mimj r∗ 1(ai, aj) X 1≤h≤n−1 (Lih −Ljh)ph 2 X 1≤k≤n−1 (Lik −Ljk)qk 2 + r∗ 2(aj, ak) X∗ 1≤h<k≤n−1(Lih −Ljh)(Lik −Ljk)(phqk −pkqh) 2 (B.7) X 1≤i<j≤n mimj s1(ai, aj)ηi2 Λi + s2(ai, aj) ηiηj p ΛiΛj + s1(aj, ai)ηj2 Λj X 1≤h≤n−1 (Lih −Ljh)qh 2 X 1≤i<j≤n mimj s∗ 1(ai, aj)ηi2 Λi + s∗ 2(ai, aj) ηiηj p ΛiΛj + s∗ 1(aj, ai)ηj2 Λj X 1≤h≤n−1 (Lih −Ljh)ph 2 with r∗ 1(a, b) := −3 32 α2 b (2β(5) 0 (α) + β(5) 2 (α)) (B.8) r∗ 2(a, b) := −3 16 α2 b (β(5) 0 (α) + β(5) 2 (α)) s1(a, b) := α 64b[2α(57α2 + 117)β(7) 0 (α) + (−12α4 −291α2 −12)β(7) 1 (α) + 2α(15α2 −45)β(7) 2 (α) + 27α2β(7) 3 (α)] s2(a, b) := α 64b[554α2β(7) 0 (α) + α(−376 −376α2)β(7) 1 (α) + 2(16α4 + 10α2 + 16)β(7) 2 (α) + 56α(α2 + 1)β(7) 3 (α) + 2α2β(7) 4 (α)] s∗ 1(a, b) := α 64b[2α(71α2 + 11)β(7) 0 (α) + (−20α4 + 119α2 −20)β(7) 1 (α) + 2α(−79α2 −19)β(7) 2 (α) −47α2β(7) 3 (α)] s∗ 2(a, b) := α 64b[−430α2β(7) 0 (α) + α(8 + 8α2)β(7) 1 + 2(16α4 + 118α2 + 16)β(7) 2 (α) + 56α(α2 + 1)β(7) 3 (α) + 2α2β(7) 4 (α)] . • The deﬁnition of Fv2(Λ) · ¯ p2¯ q2, Fhv2(Λ) · η2¯ q2 and F′ hv2(Λ) · η2¯ p2 is more involved (reﬂecting the products (4.7) appearing in the deﬁnition of the rotation matrices Ri). Let c2, . . . , cn, c∗ 1, . . . , c∗ n−1, ¯ c1, . . . , ¯ cn−1 be deﬁned by    c∗ 1 := L11 i = 1 ci := Li,i−1 2 ≤i ≤n      c∗ i := Lki 2 ≤i ≤n −1 , 1 ≤k ≤i ¯ ci = ci+1 −c∗ i = q Li+1 LiΛi+1 1 ≤i ≤n −1 (B.9) where Lij is the matrix (B.4). Notice, for later convenience, that the above deﬁnitions imply that the diﬀerences Li −Lj of the rows of L are related to c2, . . . , cn, c∗ 1, . . . , c∗ n−1, ¯ c1, . . . , ¯ cn−1 by Li −Lj =    (−αij, 0n−j) , i = 1, 2 (0i−2, −αij, 0n−j) , i ≥3 αij :=    (−c∗ 1, −c∗ 2, . . . , −c∗ j−2, ¯ cj−1) , i = 1 (−ci, −c∗ i , . . . , −c∗ j−2, ¯ cj−1) , i ≥2 (B.10) where 0r denotes the null vector of dimension r. 42 Deﬁne the homogeneous polynomials, of degree 2 in ¯ z, Q2, . . . , Qn, Q∗ 1, . . . , Q∗ n−1, ¯ Q1, . . . , ¯ Qn−1 by74 Qi := 2Li−1(Li−1 + 2Λi)ρi + Λi(Li−1 −Λi)ri−1 −Λ2 i \|zi−2\|2 4L2 i Λ2 i , 2 ≤i ≤n Q∗ 1 := 2Λ2(2Λ1 + Λ2)ρ1 −2Λ2 1ρ2 + Λ1(Λ2 −Λ1)r1 4Λ2 1L2 2 , Q∗ i := −2L2 i ρi+1 + Li(Λi+1 −Li)ri + Λi+1(2Li + Λi+1)\|zi−1\|2 4L2 i L2 i+1 , 2 ≤i ≤n −1 ¯ Qi := Λ2 i+1\|zi−1\|2 + 2L2 i ρi+1 −LiΛi+1ri 4L2 i Λ2 i+1 , 1 ≤i ≤n −1 (B.11) where zi := (η1, . . . , ηi+1, ξ1, . . . , ξi+1, p1, . . . , pi, q1, . . . , qi) for 1 ≤i ≤n −1 . (B.12) Next, for arbitrary numbers Q, c ̸= 0, r, p, q, denote T(c, Q, r, p, q) :=      1 −q2 c2 2 + Q −pq c2 2 + Q −qS(c, Q, r) −pq c2 2 + Q 1 −p2 c2 2 + Q −pS(c, Q, r) qS(c, Q, r) pS(c, Q, r) 1 −(p2 + q2) c2 2 + Q     ; (B.13) where S(c, Q, r) := c + 1 c Q −c3 4 r (B.14) Now, deﬁne T1 := id , Ti := T(ci, Qi, ri−1, pi−1, qi−1) for 2 ≤i ≤n T∗ i := T(c∗ i , Q∗ i , ri, pi, qi) , ¯ Ti := T(¯ ci, ¯ Qi, ri, pi, qi)) for 1 ≤i ≤n −1 ; (B.15) and let Tij := (Ti)t(T∗ i )t · · · (T∗ j−2)t ¯ Tj−1 , (B.16) where Tt denotes the transpose of T and the product (T∗ i )t · · · (T∗ j−2)t in front of ¯ Tj−1 is absent when j = i + 1. Let, ﬁnally, Qij denote the quartic part (i.e. , the homogeneous part of order 4 in z) of the upper left (2 × 2) submatrix of Tij and deﬁne the polynomial X 1≤i<j≤n mimjC1(ai, aj) tr Qij , (B.17) where tr Qij denotes the trace of Qij. Then, Fv2(Λ)· ¯ p2¯ q2, Fhv2(Λ)·η2¯ q2 and F′ hv2(Λ)·η2¯ p2 are identiﬁed as the monomials associated respec-tively to the literal parts ¯ p2¯ q2, η2¯ q2, η2¯ p2 of the polynomial (B.17). Proof of (B.1)÷(B.17) Using (4.3), we write the Eucledean distance of x(i) and x(j) as \|x(i) −x(j)\| = \|Rix(i) pl −Rjx(j) pl \| = \|x(i) pl −Rijx(j) pl \| , where Rij := Rt iRj , 74Recall (A.19). 43 so as to write fav as75 fav = − X 1≤i<j≤n mimj 4π2 Z T2 dλidλj \|x(i) pl −Rijx(j) pl \| . (B.18) Since x(i) pl and x(j) pl have vanishing third components, we can substitute, into (B.18), the matrix Rij with R(2) ij , deﬁned as its upper left submatrix of order 2. It is easy to see, using (4.7),(A.17) and (A.18) that R(2) ij even in z. We therefore denote R(2) ij = id + qij + Qij + O(\|z\|6) (B.19) its expansion up to the fourth order, where qij and Qij are respectively matrices of order 2 of homoge-neous polynomials in z of degree 2, 4 respectively. We interrupt for a while the description of the expansion, just to point out how the 2 × 2 matrices qij, Qij may be computed. This is done in the following Remark B.1 At ﬁrst, by (4.7), write Rij as Rij = Rt iRj = (Ri)t(R∗ i )t · · · (R∗ j−1)tRj = (Ri)t(R∗ i )t · · · (R∗ j−2)t ¯ Rj−1 for 1 ≤i < j ≤n (B.20) where Ri := id , ¯ Rj−1 := (R∗ j−1)tRj for 1 ≤j −1 ≤n −1 (B.21) and the productories (R∗ i )t · · · (R∗ j−1) do not appear for j = i + 1. Notice that the matrix ¯ Rj−1 = (R∗ j−1)tRj has the expression (compare (A.17) and (A.18)) ¯ Rj−1 = (R∗ j−1)tRj = R313 arg(pj−1, qj−1), i∗ j−1 + ij = R313 arg(pj−1, qj−1),¯ ij−1 (B.22) where ¯ ij−1 := i∗ j−1 + ij for 1 ≤j −1 ≤n −1 . (B.23) By the deﬁnition of i∗ j−1 and ij and (B.23), the inclination ¯ ij−1 corresponds to be the outer angle of Ψj−2 and Γj in the triangle of Ψj−2, Γj, Ψj−1. Hence, with the same notations as in § A.2, using (A.20) 1 −cos¯ ij−1 = 1 −Ψ2 j−1 −Ψ2 j−2 −Γ2 j 2ΓjΨj−2 = (Γj + Ψj−2)2 −Ψ2 j−1 2ΓjΨj−2 = (Γj + Ψj−2 −Ψj−1)(Γj + Ψj−2 + Ψj−1) 2ΓjΨj−2 = (p2 j−1 + q2 j−1)¯ cj−1 where ¯ cj−1 := 2Lj −\|zj−2\|2 −2ρj −rj−1 (2Λj −2ρj)(2Lj−1 −\|z2 j−2\|) (j ≥2) (B.24) Then, by (B.22) and (A.26), ¯ Rj−1 has the expression ¯ Rj−1 =   1 −q2 j−1¯ cj−1 −pj−1qj−1¯ cj−1 −qj−1¯ sj−1 −pj−1qj−1¯ cj−1 1 −p2 j−1¯ cj−1 −pj−1¯ sj−1 qj−1¯ sj−1 pj−1¯ sj−1 1 −(p2 j−1 + q2 j−1)¯ cj−1   (B.25) where ¯ sj−1 := sin¯ ij−1 q p2 j−1 + q2 j−1) = r ¯ cj−1 2 −(p2 j−1 + q2 j−1)¯ cj−1 . (B.26) 75As we have already observed, also in the variables (Λ, λ, z) the averaged perturbation coincides with the average of the averaged Newtonian potential (B.18). 44 (i) To compute the second order term qij of R(2) ij , where R(2) ij is the submatrix of order 2 of the matrix Rij in (B.20), we truncate the matrices Ri, R∗ i , ¯ Ri to the second order and denote as Si, S∗ i , ¯ Si the respective truncations. In fact, by (B.20), the matrix qij can be obtained taking the term of degree 2 of the submatrix of order 2 of the products (Si)t(S∗ i )t · · · (S∗ j−2)t¯ Sj−1 for 1 ≤i < j ≤n (S1 := id ) (B.27) of the respective truncated matrices. By (4.8), (B.25) and the deﬁnitions (A.23), (A.27), (B.24) and (B.26) of the functions ci, c∗ i , ¯ ci, si, s∗ i , ¯ si, one sees that the expressions of the respective truncated matrices are Si+1 = S(ci+1, pi, qi) , S∗ i = S(c∗ i , pi, qi) , ¯ Si = S(¯ ci, pi, qi) 1 ≤i ≤n −1 where ci+1, c∗ i , ¯ ci, are deﬁned in as in (B.9) and S(c, p, q) denotes S(c, p, q) :=   1 −1 2c2q2 −1 2c2pq −cq −1 2c2pq 1 −1 2c2p2 −cp cq cp 1 −1 2c2(p2 + q2)  . It is quite immediate to check that, if α = (α1, . . . , αm) , y = (y1, . . . , ym) , x = (x1, . . . , xm) ∈Rm , then, the second order term of the pricipal submatrix of order 2 of the productry S(α1, y1, x1) · · · S(αm, ym, xm) is given by the matrix q(α, y, x) = −1 2 (α · x)2 (α · y)(α · x) −∆(α, y, x) (α · y)(α · x) + ∆(α, y, x) (α · y)2 (B.28) where ∆(α, y, x) denotes ∆(α, y, x) := X 1≤h<k≤m αhαk(yhxk −ykxh) (when m ≥2) . Then, in the products in (B.27), the entries aij, bij, cij, dij of the matrix qij = aij bij cij dij are found using (B.28) with m =    j −1 i = 1 j −i + 1 i ≥2 , α = αij y =    (p1, . . . pj−1) i = 1 (pi−1, . . . , pj−1) i ≥2 , x =    (q1, . . . , qj−1) i = 1 (qi−1, . . . , qj−1) i ≥2 (B.29) 45 with αij as in (B.10). Therefore, one ﬁnds aij = −1 2 X 1≤k≤n−1 (Lik −Ljk)qk 2 bij = −1 2 X 1≤h,k≤n−1 (Lik −Ljk)(Lih −Ljh)phqk + 1 2 X∗ 1≤h<k≤n−1(Lih −Ljh)(Lik −Ljk)(phqk −pkqh) cij = −1 2 X 1≤k,h≤n−1 (Lik −Ljk)(Lih −Ljh)phqk −1 2 X∗ 1≤h<k≤n−1(Lih −Ljh)(Lik −Ljk)(phqk −pkqh) dij = −1 2 X 1≤k≤n−1 (Lik −Ljk)pk 2 (B.30) where the asterisc in P∗means that such sums exist only when n ≥3; we have also used the ﬁrst relation in (B.10). (ii) The fourth order matrix Qij has a less explicit espression. To compute it, we have to consider the fourth order truncations of the matrices Ri, R∗ i , ¯ Ri. By (A.23), (A.27), (B.24), (B.26) one easily sees that the functions ci, c∗ i , ¯ ci, si, s∗ i , ¯ si verify ci = c2 i 2 + Qi + O(\|¯ z\|4) , si = S(ci, Qi, ri−1) + O(\|¯ z\|4) c∗ i = (c∗ i )2 2 + Q∗ i + O(\|¯ z\|4) , s∗ i = S(c∗ i , Q∗ i , ri) + O(\|¯ z\|4) ¯ ci = ¯ c2 i 2 + ¯ Qi + O(\|¯ z\|4) , ¯ si = S(¯ ci, ¯ Qi, ri) + O(\|¯ z\|4) where ci, c∗ i ¯ ci, Qi, Q∗ i , ¯ Qi and S(c, Q, r) are deﬁned in (B.9), (B.11) and (B.14). Then, since Qi, Q∗ i , ¯ Qi, Si, S∗ i , ¯ Si are O(\|¯ z\|2), the fourth order truncations of the matrices Ri, R∗ i , ¯ Ri are given by the matrices Ti, T∗ i , ¯ Ti into (B.15). Hence, The matrix Qij into (B.19) of R(2) ij is uniquely identiﬁed as the principal submatrix of order 2 of the fourth order term of the matrix Tij into (B.16). We proceed with the expansion of the function (B.18). Using (B.19), we write the squared Euclidean distance Dij := \|x(i) pl −Rijx(j) pl \|2 as Dij = \|x(i) pl −Rijx(j) pl \|2 = \|x(i) pl −R(2) ij x(j) pl \|2 = \|x(i) pl \|2 + \|x(j) pl \|2 −2x(i) pl · R(2) ij x(j) pl = \|x(i) pl −x(j) pl \|2 −2x(i) pl · qijx(j) pl −2x(i) pl · Qijx(j) pl + O(\|¯ z\|6) We have then the following expansion for the inverse Euclidean distance 1 \|x(i) pl −Rijx(j) pl \| = 1 \|x(i) pl −x(j) pl \| + x(i) pl · qijx(j) pl \|x(i) pl −x(j) pl \|3 + x(i) pl · Qijx(j) pl \|x(i) pl −x(j) pl \|3 + 3 2 (x(i) pl · qijx(j) pl )2 \|x(i) pl −x(j) pl \|5 + O(\|¯ z\|6). (B.31) 46 Then, multiplying this expression by −mimi, taking the average over λi, λj ∈T and the sum for 1 ≤i < j ≤n, we split fav as fav(Λ, ¯ z) = fh Λ, (η, ξ) + f (1) hv Λ, ¯ z + f (2) hv Λ, ¯ z + fv Λ, ¯ z + O(\|¯ z\|6) where fh Λ, (η, ξ) := − X 1≤i<j≤n mimj 4π2 Z T2 dλidλj \|x(i) pl −x(j) pl \| f (1) hv Λ, ¯ z := − X 1≤i<j≤n mimj 4π2 Z T2 x(i) pl · qijx(j) pl \|x(i) pl −x(j) pl \|3 dλidλj f (2) hv Λ, ¯ z := − X 1≤i<j≤n mimj 4π2 Z T2 x(i) pl · Qijx(j) pl \|x(i) pl −x(j) pl \|3 dλidλj fv Λ, ¯ z := − X 1≤i<j≤n mimj 4π2 Z T2 3 2 (x(i) pl · qijx(j) pl )2 \|x(i) pl −x(j) pl \|5 dλidλj (B.32) Remark B.2 Notice that each of the previous functions is rotation invariant, i.e. , it is invariant under the transformations (Λ, λ, ¯ z) →Rg(Λ, λ, ¯ z) deﬁned in (6.2). Indeed, when (Λ, λ, ¯ z) →Rg(Λ, λ, ¯ z), the matrices Rij = Rt iRj transform as Rij →R(g)RijR(−g) since the same holds for the matrices Ri (compare Remark A.1) and the planar Poincar´ e map as x(i) pl →R(g)x(i) pl . Then, the scalar products x(i) pl ·R(2) ij x(j) pl = x(i) pl ·Rijx(j) pl are rotation invariant and hence (since they are so term by term) the scalar products x(i) pl · qijx(j) pl and x(i) pl · Qijx(j) pl are. The rotation invariance of fh, f (1) hv , f (2) hv , fv then follows. Also invariance by the reﬂections (6.6) can easily be checked. Noticing ﬁnally that • fh depends only on (η, ξ); • f (1) hv is O(\|(¯ p, ¯ q)\|2) and its quartic part vanishes for (η, ξ) = 0; • f (2) hv is O(\|¯ z\|4) and vanishes for (¯ p, ¯ q) = 0; • fv is O(\|(¯ p, ¯ q)\|4); the following expansions hold fh = C0(Λ) + Qh · η2 + ξ2 2 + Fh · η4 + ξ4 + 2η2ξ2 2 + O(\|(η, ξ)\|6) (B.33) f (1) hv = ¯ Qv · ¯ p2 + ¯ q2 2 + Fhv1 · η2¯ q2 + ξ2¯ p2 −2ηξ¯ p¯ q 2 + F′ hv1 · η2¯ p2 + ξ2¯ q2 + 2ηξ¯ p¯ q 2 + O(\|¯ z\|6) f (2) hv = Fhv2 · η2¯ q2 + ξ2¯ p2 −2ηξ¯ p¯ q 2 + F′ hv2 · η2¯ p2 + ξ2¯ q2 + 2ηξ¯ p¯ q 2 + Fv2 · ¯ p4 + ¯ q4 + 2¯ p2¯ q2 2 + O(\|¯ z\|6) fv = Fv1 · ¯ p4 + ¯ q4 + 2¯ p2¯ q2 2 + O(\|¯ z\|6) • The tensor Fv1 47 This tensor deﬁnes the quartic part of the function fv in (B.32). Since qij is a matrix of degree 2 in ¯ z, to compute Fv1, it suﬃcies to truncate the planar Poincar´ e map x(i) pl to its zero order, i.e. , to substitute x(i) pl with x(i) tr0 = ai(cos λi, sin λi) where ai = 1 ¯ mi Λi Mi 2 . (B.34) If aij, bij, cij, dij are the entries of the matrix qij as in (B.30), we have Fv1 · ¯ p4 + ¯ q4 + 2¯ p2¯ q2 2 = − X 1≤i<j≤n mimj 4π2 Z T2 3 2 a2 i a2 j aij cos λi cos λj + bij cos λi sin λj + cij sin λi cos λj + dij sin λi sin λj 2 (a2 i + a2 j −2aiaj cos (λi −λj))5/2 dλidλj Computing the integral easily gives the quartic form Fv1 · ¯ p4 + ¯ q4 + 2¯ p2¯ q2 2 = − X 1≤i<j≤n 3 16 a2 i a3 j (a2 ij + b2 ij + c2 ij + d2 ij)(2β(5) 0 + β(5) 2 ) + 2(aijdij −bijcij)β(5) 2 mimj Using the expressions (B.30) and selecting the monomial with literal part ¯ p2¯ q2, we ﬁnd the result in (B.7) and (B.8). • The tensors Fhv2 and F′ hv2 and Fv2 Such tensors deﬁne the quartic part of the function f (2) hv into (B.32). Since Qij is of degree 4 in ¯ z, to compute them we substitute, as before, x(i) pl with its 0–order truncation (B.34). The computation of the integral is immediate and gives the quartic form Fhv2 · η2¯ q2 + ξ2¯ p2 −2ηξ¯ p¯ q 2 + F′ hv2 · η2¯ p2 + ξ2¯ q2 + 2ηξ¯ p¯ q 2 + Fv2 · ¯ p4 + ¯ q4 + 2¯ p2¯ q2 2 = − X 1≤i<j≤n mimj 4π2 Z T2 x(i) tr0 · Qijx(j) tr0 \|x(i) tr0 −x(j) tr0\|3/2 = − X 1≤i<j≤n mimj 4π2 × Z T2 aiaj Aij cos λi cos λj + Bij cos λi sin λj + Cij sin λi cos λj + Dij sin λi sin λj a2 i + a2 j −2aiaj cos (λi −λj) 3/2 dλidλj = − X 1≤i<j≤n mimj ai 2a2 j (Aij + Dij)β(3) 1 = X 1≤i<j≤n mimjC1(ai, aj) tr Qij (B.35) where Aij, Bij, Cij, Dij denote the entries of Qij = Aij Bij Cij Dij and C1(ai, aj) is as in (B.2). • The tensor Fh For the computation of the tensor Fh, deﬁning the quartic part of fh (see (B.32)), we have to consider the truncation of x(i) pl = (x(i) 1 , x(i) 2 , 0) up to degree 2 in η and ξ separately. Let such truncation be denoted as x(i) tr4 = (x(i) 1,tr4, x(i) 2,tr4, 0). Proceeding as in , we write the regularized Kepler equation (4.6) in the form vi = si sin vi + ti cos vi , where vi := ui −λi (B.36) 48 and si := 1 √Λi s 1 −η2 i + ξ2 i 4Λi (ηi cos λi −ξi sin λi) = 1 √Λi 1 −η2 i + ξ2 i 8Λi (ηi cos λi −ξi sin λi) + O(\|(ηi, ξi)\|5) ti := 1 √Λi s 1 −η2 i + ξ2 i 4Λi (ηi sin λi + ξi cos λi) = 1 √Λi 1 −η2 i + ξ2 i 8Λi (ηi sin λi + ξi cos λi) + O(\|(ηi, ξi)\|5) (B.37) We then expand, from (B.36), the variable vi = ui −λi in powers of (si, ti) vi = ti + si ti + s2 i ti −1 2 t3 i + s3 i ti −5 3 si t3 i + O(\|(si, ti)\|5) (B.38) Using (B.37) and (B.38) into (4.5), we ﬁnd x(i) 1,tr4 = 1 ¯ mi Λi Mi 2 cos λi + ηi 2√Λi (cos 2λi −3) − ξi 2√Λi sin 2λi + 3 8Λi η2 i (cos 3λi −cos λi) −ηiξi 4√Λi (3 sin 3λi + sin λi) −ξ2 i 8Λi (3 cos 3λi + 5 cos λi) − η2 i ξi 16Λi √Λi (16 sin 4λi −5 sin 2λi) − ηiξ2 i 16Λi √Λi (16 cos 4λi + 9 cos 2λi −3) −η2 i ξ2 i 64Λ2 i (125 cos 5λi + 9 cos 3λi −14 cos λi) (B.39) The expresson of x(i) 2,tr4 can be obtained from the right–hand–side of (B.39), letting (λi, ηi, ξi) →( π 2 − λi, ξi, ηi). Using ﬁnally such truncated espressions into the deﬁnition of fh in (B.32) will provide, through the computation of the (αi, αj, βi, βj)–derivatives with respect to (ηi, ηj, ξi, ξj) over αi!αj!βi!βj!, the quartic form Fh · η2ξ2 as in (B.5) and (B.6). We omit further details. • The tensors ¯ Qv, Fhv1, F′ hv1 The tensors ¯ Qv, Fhv1, F′ hv1 deﬁne the expansion of the function f (1) hv . For their computation (in particular, for the computation of the quartic tensors Fhv1 and F′ hv1), we shall put ξ = 0 and then shall select into the expansion of f (1) hv \|ξ=0 the monomials with literal part ¯ p2, η2¯ q2, η2¯ p2 respectively. By the previous paragraph, we ﬁnd the following truncations of the Kepler map up to degree 2 in η and with ξ = 0 x(i) 1,tr2 = 1 ¯ mi Λi Mi 2 cos λi + ηi 2√Λi (cos 2λi −3) + η2 i 8Λi (3 cos 3λi −3 cos λi) x(i) 2,tr2 = 1 ¯ mi Λi Mi 2 sin λi + ηi 2√Λi sin 2λi + η2 i 8Λi (3 sin 3λi −5 sin λi) (B.40) Denoting x(i) tr2 := (x(i) 1,tr2, x(i) 2,tr2), by the expression of f (1) hv in (B.32), since the matrix qij is O(\|(p, q)\|2) and x(i) 1,tr2, x(i) 2,tr2 are, respectively, even, odd in λi, f (1) hv = − X 1≤i<j≤n mimj 4π2 Z T2 x(i) tr2 · qijx(j) tr2 \|x(i) tr2 −x(j) tr2\|3 dλidλj + O(\|(¯ p, ¯ q)\|2η4) = − X 1≤i<j≤n mimj 4π2 Z T2 aijx(i) 1,tr2x(j) 1,tr2 + dijx(i) 2,tr2x(j) 2,tr2 \|x(i) tr2 −x(j) tr2\|3 dλidλj + O(\|(¯ p, ¯ q)\|2η4) (B.41) 49 where aij, dij are the diagonal entries of qij (given in (B.30)). In particular, if we put also η = 0, so to have x(i) tr2 = x(i) tr0 = ai(cos λi, sin λi), by (B.33) and (B.41), we can identify ¯ Qv · ¯ p2 + ¯ q2 2 = X 1≤i<j≤n mimjC1(ai, aj) tr qij (the computation being the same as in (B.35)) with tr qij = aij + dij the trace of qij. Using the expressions for aij, bij in (B.30) and selecting the monomial in ¯ p2, we ﬁnd the result in (B.3). If we use the whole expression for x(i) 1,tr2, x(i) 2,tr2 as in (B.40), computing the derivatives of order two with respect to (ηi, ηj), for (ηi, ηj) = 0, we then have the second order in η of f (1) hv , which we identify with Fhv1 · η2¯ q2 + F′ hv1 · η2¯ p2. We omit the details of this straightforward computation. References K. Abdullah and A. Albouy. On a strange resonance noticed by M. Herman. Regul. Chaotic Dyn., 6(4):421–432, 2001. V.I. Arnold. Small denominators and problems of stability of motion in classical and celestial mechanics. Russian Math. Surveys, 18(6):85–191, 1963. L. Biasco, L. Chierchia, and E. Valdinoci. Elliptic two-dimensional invariant tori for the planetary three-body problem. Arch. Rational Mech. Anal., 170:91–135, 2003. See also: Corrigendum. Arch. Ration. Mech. Anal.. 180: 507–509, 2006. L. Biasco, L. Chierchia, and E. Valdinoci. n-dimensional elliptic invariant tori for the planar (n + 1)-body problem. SIAM Journal on Mathematical Analysis, 37(5):1560–1588, 2006. A. Celletti and L. Chierchia. Construction of stable periodic orbits for the spin-orbit problem of celestial mechanics. Regul. Chaotic Dyn., 3(3):107–121, 1998. J. Moser at 70 (Russian). A. Celletti and L. Chierchia. KAM tori for N-body problems: a brief history. Celestial Mech. Dynam. Astronom., 95(1-4):117–139, 2006. A. Celletti and L. Chierchia. KAM stability and celestial mechanics. Mem. Amer. Math. Soc., 187(878):viii+134, 2007. L. Chierchia and G. Pinzari. Deprit’s reduction of the nodes revised, 2010. Preprint, users/chierchia. L. Chierchia and G. Pinzari. Planetary Birkhoﬀnormal forms, 2010. Preprint, chierchia. L. Chierchia and G. Pinzari. Properly–degenerate KAM theory (following V.I. Arnold). Discrete Contin. Dyn. Syst. S, 3(4):545–578, 2010. L. Chierchia and F. Pusateri. Analytic Lagrangian tori for the planetary many-body problem. Ergodic Theory Dynam. Systems, 29(3):849–873, 2009. A. Deprit. Elimination of the nodes in problems of n bodies. Celestial Mech., 30(2):181–195, 1983. L. H. Eliasson. Perturbations of stable invariant tori for Hamiltonian systems. Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 15(1):115–147 (1989), 1988. J. F´ ejoz. Quasiperiodic motions in the planar three-body problem. J. Diﬀerential Equations, 183(2):303–341, 2002. J. F´ ejoz. D´ emonstration du ‘th´ eoreme d’Arnold’ sur la stabilit´ e du syst eme plan´ etaire (d’apres Herman). Ergodic Theory Dynam. Systems, 24(5):1521–1582, 2004. J. F´ ejoz. D´ emonstration du ‘th´ eor eme d’Arnold’ sur la stabilit´ e du systeme plan´ etaire (d’apr es Herman). Version r´ evis´ ee de l’article paru dans le Michael Herman Memorial Issue Ergodic Theory Dyn. Sys. 24(5) : 1521–1582 (2004), 2007. Available at J. F´ ejoz. A simple proof of invariant tori theorems, 2009. Preprint, Articles/invariantTori.pdf. M.R. Herman. Torsion du probleme plan etaire, edited by J. Fej´ oz in 2009. Available in the electronic ‘Archives Michel Herman’ at H. Hofer, E. Zehnder. Symplectic Invariants and Hamiltonian Dynamics. Birkh¨ auser Verlag, Basel, 1994. C. G. J. Jacobi. Sur l’´ elimination des noeuds dans le probleme des trois corps. Astronomische Nachrichten, Bd XX:81–102, 1842. 50 S. B. Kuksin. Perturbation theory of conditionally periodic solutions of inﬁnite-dimensional Hamiltonian systems and its applications to the Korteweg-de Vries equation. Mat. Sb. (N.S.), 136(178)(3):396–412, 431, 1988. F. Malige, P. Robutel, and J. Laskar. Partial reduction in the n-body planetary problem using the angular momentum integral. Celestial Mech. Dynam. Astronom., 84(3):283–316, 2002. V. K. Melnikov. On certain cases of conservation of almost periodic motions with a small change of the Hamiltonian function. Dokl. Akad. Nauk SSSR, 165:1245–1248, 1965. G. Pinzari. On the Kolmogorov set for many–body problems. PhD thesis, Universit a Roma Tre, April 2009. http: //www.mat.uniroma3.it/users/chierchia/TESI/PhD_Thesis_GPinzari.pdf. J. P¨ oschel. On elliptic lower-dimensional tori in Hamiltonian systems. Math. Z., 202(4):559–608, 1989. P. Robutel. Stability of the planetary three-body problem. II. KAM theory and existence of quasiperiodic motions. Celestial Mech. Dynam. Astronom., 62(3):219–261, 1995. See also: Erratum, Celestial Mech. Dynam. Astronom., 84(3):317, 2002. H. R¨ ussmann. Nondegeneracy in the perturbation theory of integrable dynamical systems. In Stochastics, algebra and analysis in classical and quantum dynamics (Marseille, 1988), volume 59 of Math. Appl., pages 211–223. Kluwer Acad. Publ., Dordrecht, 1990. C. L. Siegel, and J. K. Moser. Lectures on Celestial Mechanics. Springer Verlag, 1995. Reprint of the 1971 Edition. 51
15266	https://books.google.com/books/about/Craig_s_Restorative_Dental_Materials.html?id=EDFqAAAAMAAJ	Craig's Restorative Dental Materials - Google Books Sign in Hidden fields Try the new Google Books Books Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» ### Get Textbooks on Google Play Rent and save from the world's largest eBookstore. Read, highlight, and take notes, across web, tablet, and phone. Go to Google Play Now » My library My History Craig's Restorative Dental Materials John M. Powers, Ronald L. Sakaguchi Mosby Elsevier, 2006 - Medical - 632 pages This comprehensive exploration of restorative dental materials presents everything readers need to know to correctly use dental materials in the clinic and dental laboratory, from fundamental concepts to advanced skills. The scientific basis for technical procedures and manipulation of materials is provided, and the book's problem-solving approach focuses on applying new information to practical situations. At the end of each chapter, a case-based scenario presents the opportunity to work through problems and verify solutions. Extensive figures and tables of data throughout the book clarify the text. Comprehensive information on restorative dental materials provides insight into both the fundamentals and the latest practical knowledge in the field. Problem-solving approach presents cases with solutions at the end of each chapter, providing an opportunity to apply and test important principles. Thoroughly revised content includes the latest information on: optical, thermal, and electrical properties; mechanical properties; preventive materials; composites; and impression materials. The latest standards in the field are covered, including biocompatibility and standards of dental materials. Easy-to-follow organization progresses according to the way in which material is most often taught in dental schools. An appendix contains useful references such as weights and measurements, conversion tables, metric weights, and more. Each chapter concludes with a list of references for further reading, divided by topic sections. A new chapter on Dental Implants thoroughly discusses implant materials, coatings, and textures, and cell and bone interaction. A new chapter on Tissue Engineering presents the history of tissue engineering and addresses the different methods of engineering tissue, as well as clinical applications for dentistry and other disciplines. Mind Maps - key words and images from each chapter - are provided in the appendix and on the accompanying Evolve site to help readers remember and review material. More » From inside the book Contents Scope and History of Restorative 1 MOSBY 12 Applied Surface Phenomena 13 Copyright 27 other sections not shown Other editions - View all Craig's Restorative Dental Materials John M. Powers,Ronald L. Sakaguchi Snippet view - 2006 Craig's Restorative Dental Materials John M. Powers,Ronald L. Sakaguchi Snippet view - 2006 Common terms and phrases acidacrylicadhesivealginateANSI/ADA Specificationappliedbiocompatibilitybond strengthbonding agentsbonecalciumcalcium sulfatecasting alloyscavitycellsceramicceramic-metalchemicalclinicalcolorcompomerscompressive strengthcorrosionCraig RGcrowndeformationDent AssocDent MaterDent Resdental alloysdental amalgamdental implantsdental materialsdental restorationsdental stonedentindentistrydenture baseeffectelasticelastic modulusenameletchingfillerfluoridefractureglass ionomer cementsgypsumheatincreaseinlayinvestmentlayerlight-curedlinersliquidmechanical propertiesmeltingmercurymetalmethacrylatemixingmodulusmoldmonomermouth protectorocclusalorthodonticoxidepartial denturesparticlesphasephosphateplasticpolymerpolymerizationporcelainpowderPowers JMPROBLEMproduceProsthet DentProsthodontreactionresin cementsresin compositesresistancerestorative materialsresultsealantsettingshown in FigshrinkagesolderSolutionspruestresssurfaceTabletechniqueteethtemperaturetensileteststhermal expansiontiontissuetitaniumviscosityvitrowax patternzinc About the author(2006) Dr. Sakaguchi integrates a systems perspective, design tools, and strategic innovation and entrepreneurship with a focus on execution in collaborating with graduate students in healthcare management and executive leaders. Dr. Sakaguchi has a clinical and biomaterials science research background, with experience in business development, design for strategic initiatives, facilitation of strategy development, and coaching. In the Division of Management, Dr. Sakaguchi is the course director for an MBA course in healthcare innovation and the graduate capstone business consulting program. Bibliographic information Title Craig's Restorative Dental Materials Craig's Restorative Dental Materials, Ronald L. Sakaguchi EditorsJohn M. Powers, Ronald L. Sakaguchi Edition 12, illustrated Publisher Mosby Elsevier, 2006 Original from the University of Michigan Digitized Jul 19, 2008 ISBN 0323036066, 9780323036061 Length 632 pages SubjectsMedical › Dentistry › General Medical / Dentistry / General Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
15267	https://mathematica.stackexchange.com/questions/83568/print-all-numbers-from-a-sequence-which-dont-contain-the-digit-7	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Print all numbers from a sequence which don't contain the digit 7 Ask Question Asked Modified 10 years, 4 months ago Viewed 604 times 3 $\begingroup$ so i have this: Cases[RecurrenceTable[{a[n + 1] == 7 a[n] + n, a == 1}, a, {n, 7}], Except] But i noticed that it will only exclude numbers that are 7, and not numbers which contain 7. If anyone can tell me the answer, i will be very grateful. Any help is appreciated. Thanks! expression-manipulation homework integer-sequence Share Improve this question edited May 15, 2015 at 23:54 ciao 26k22 gold badges6262 silver badges145145 bronze badges asked May 15, 2015 at 22:45 laralara 3911 bronze badge $\endgroup$ 12 1 $\begingroup$ Select[RecurrenceTable[{a[n + 1] == 7 a[n] + n, a == 1}, a, {n, 7}], FreeQ[IntegerDigits@#, 7] &] $\endgroup$ ciao – ciao 2015-05-15 22:47:57 +00:00 Commented May 15, 2015 at 22:47 2 $\begingroup$ The last condition in Cases[] should probably be n_ /; FreeQ[IntegerDigits[n], 7]¦ $\endgroup$ J. M.'s missing motivation – J. M.'s missing motivation 2015-05-15 22:48:50 +00:00 Commented May 15, 2015 at 22:48 $\begingroup$ @ciao, this works great, however it affected the number of terms in the list .. it removed all numbers containing 7, but now i have less than 10 terms. Suggestion? $\endgroup$ lara – lara 2015-05-15 23:02:22 +00:00 Commented May 15, 2015 at 23:02 $\begingroup$ Same goes for @J. M.'s answer! $\endgroup$ lara – lara 2015-05-15 23:04:38 +00:00 Commented May 15, 2015 at 23:04 1 $\begingroup$ This is some homework exercise, say the prof. decides he wants 20 terms printed. He will have to calculate how many more to add to that {n, }. He won't bother with that, i believe. And that's why i'm looking for a more efficient solution. $\endgroup$ lara – lara 2015-05-15 23:39:59 +00:00 Commented May 15, 2015 at 23:39 \| Show 7 more comments 4 Answers 4 Reset to default 4 $\begingroup$ An interesting problem which at first sight looks innocent. But then ... My (bold) conjecture is that the maximum number lg of terms you can get of this recurrence if a == 1 is lg = 11. First we solve the recurrence explicitly sol = RSolve[a[n + 1] == 7 a[n] + n && a == 1, a[n], n]; a[n] /. First[sol] ( Out= 1/252 (-7 + 43 7^n - 42 n) ) This can be simplified to c[n_] = 7^(-1 + n) + 1/36 (-1 + 7^n - 6 n) Remark: I did it by hand because Mathematica was reluctant to do so as almost always in simple cases like this. Next we calculate nn terms of the series. Let's start with nn = 50 nn = 50; t = Table[c[n], {n, 1, nn}]; and select the terms which do not contain any digit 7. s = Select[t, FreeQ[IntegerDigits[#], d] &] ( Out= {1, 8, 58, 409, 140524, 48200140, 2361806941, 16532648599, 39694889291465, \ 549428363095106064500063, 306880939816820326605841486684503658800800} ) The length of this list is lg = Length[s] ( Out= 11 ) nn = 50 is "critical" in the sense that nn = 49 only leads to lg = 10. The last term c ( Out= 306880939816820326605841486684503658800800 ) has Length[IntegerDigits[%]] ( Out= 42 ) digits. Now we increase nn in order to find longer lists. I tested up to nn = 10^5 without finding new terms. The maximum term tested had Length[IntegerDigits[c[10^5]]] ( Out= 84510 ) digits. Extension 1: Considering different values for a we observe that the lengths of the series remain rather small. With nn = 10^4 and a between 0 and 200 the lengths and their multiplicities, respectively, are tallylga = {{4, 2}, {5, 2}, {6, 7}, {7, 21}, {8, 28}, {9, 31}, {10, 44}, {11, 26}, {12, 20}, {13, 13}, {14, 5}, {15, 1}, {16, 1}} The maximum of 16 is reached for a = 86. Extension 2: Replacing 7 by any other (decimal) digit. But wait, there's still a hole to be fixed: the proof that no more terms exist or the falseness of my conjecture. But this requires sufficient theoretical knowledge in number theory - which I don't have at the moment. EDIT #1 21.05.15 This is not a proof of the conjecture but a plausibility argument that for a large integer the probability of lacking a specific decimal digit goes to zero. In fact, let d be the number of decimal digits of n. The number of nubers with d digits in which one specific digit is missing is less than 9^d. Hence the density of theses numbers within all numbers of d digits is (9/10)^d which goes to zero with increasing d i.e. increasing n. However, this does not help in our specific problem with expressions of the form b^n for which we conjecture the much stronger statement that above a certain n = n_crit there is no (!) decimal digit missing. Share Improve this answer edited May 21, 2015 at 10:33 answered May 18, 2015 at 16:43 Dr. Wolfgang HintzeDr. Wolfgang Hintze 13.2k2121 silver badges5252 bronze badges $\endgroup$ Add a comment \| 3 $\begingroup$ This is mostly a copy and paste of the brilliant answer by @WReach regarding implementation of lazy lists in Mathematica. I'll refer you to that answer for a detailed explanation of his concept of a stream and only detail the modifications I made for this particular problem. ClearAll[stream] SetAttributes[stream, {HoldAll, Protected}] sEmptyError[] := (Message[stream::empty]; Abort[]) stream::empty = "Attempt to access beyond the end of a stream."; ClearAll[sEmptyQ, sHead, sTail, sTake, sList, sMap, sFilter, sIntegers] sEmptyQ[stream[]] := True sEmptyQ[stream[_, _]] = False; sHead[stream[]] := sEmptyError[] sHead[stream[h_, _]] := h sTail[stream[]] := sEmptyError[] sTail[stream[_, t_]] := t sTake[s_stream, 0] := stream[] sTake[s_stream, n_] /; n > 0 := With[{nn = n - 1}, stream[sHead[s], sTake[sTail[s], nn]]] sList[s_stream] := Module[{tag}, Reap[NestWhile[(Sow[sHead[#], tag]; sTail[#]) &, s, ! sEmptyQ[#] &], tag] /. {l_} :> l] sMap[stream[], _] := stream[] sMap[s_stream, fn_] := stream[fn[sHead[s]], sMap[sTail[s], fn]] sFilter[s_, pred_] := NestWhile[sTail, s, (! sEmptyQ[#] && ! pred[sHead[#]]) &] /. stream[h_, t_] :> stream[h, sFilter[t, pred]] The above code block defines operations on expressions with head stream. Explanations are given in the linked answer. sSequence[v_: 1, n_: 1] := With[{nn = n + 1, vv = 7 v + n}, stream[v, sSequence[vv, nn]]] I create a function called sSequence which generates a lazy list conforming to the rule given by OP: next element is seven times previous element plus number of element. As you can see, it takes two arguments - value of element, plus number of element and creates a lazy list generating all subsequent elements in accordance with the rule. Then define a filter to get rid of numbers with the digit 7: noSevenQ = FreeQ[IntegerDigits[#], 7] & and take the first nine elements. Ten is also fine. Mathematica fails to find an 11th element without 7 quickly enough. sSequence[]~sFilter~noSevenQ~sTake~9 // sList Share Improve this answer edited Apr 13, 2017 at 12:56 CommunityBot 1 answered May 16, 2015 at 11:47 LLlAMnYPLLlAMnYP 11.6k2828 silver badges6767 bronze badges $\endgroup$ 1 $\begingroup$ Now that I think about it, DigitCount[] could also be used: no7Q = DigitCount[#, 10, 7] == 0 &. $\endgroup$ J. M.'s missing motivation – J. M.'s missing motivation 2015-05-16 12:03:15 +00:00 Commented May 16, 2015 at 12:03 Add a comment \| 2 $\begingroup$ I don't have Mathematica to test this, but I think something like this ought to work: Block[{n = 10, k= 0}, NestList[NestWhile[(k++; 7 # + k) &, #, DigitCount[#, 10, 7] != 0 &, {2, 1}] &, 1, n]] Share Improve this answer answered May 16, 2015 at 12:29 community wiki J. M.'s missing motivation $\endgroup$ 9 $\begingroup$ It is working :) $\endgroup$ Sektor – Sektor 2015-05-24 13:22:59 +00:00 Commented May 24, 2015 at 13:22 $\begingroup$ Thanks. :) I forgot to test the blasted thing a few hours ago, when I had Mathematica access. Oh well¦ $\endgroup$ J. M.'s missing motivation – J. M.'s missing motivation 2015-05-24 13:33:50 +00:00 Commented May 24, 2015 at 13:33 $\begingroup$ And I just saw it :D BTW you can access the Programming Cloud and still use (somewhat) Mathematica. $\endgroup$ Sektor – Sektor 2015-05-24 13:53:10 +00:00 Commented May 24, 2015 at 13:53 $\begingroup$ I know a number of you guys are exhorting me to try the thing out, but somehow I feel I won't fully appreciate the thing on a smartphone¦ $\endgroup$ J. M.'s missing motivation – J. M.'s missing motivation 2015-05-24 13:59:56 +00:00 Commented May 24, 2015 at 13:59 $\begingroup$ Have to set-up a Mathematica session over SSH for you then :D BTW If it is not private aren't you working with computers ? How come you are left without one ? $\endgroup$ Sektor – Sektor 2015-05-24 14:41:06 +00:00 Commented May 24, 2015 at 14:41 \| Show 4 more comments 2 $\begingroup$ s[n_] := Reap[ NestList[ Sow[{# + 1, 7 # + #}, FreeQ[IntegerDigits[7 # + #], 7]] &, Sow[{1, 1}, True], n], True]; So for the first 100 members of sequence 11 comply: Grid[Prepend[s, {"n", "a[n]"}]] Share Improve this answer answered May 21, 2015 at 10:55 ubpdqnubpdqn 67.8k33 gold badges6666 silver badges164164 bronze badges $\endgroup$ 2 $\begingroup$ you might be interested that in my solution I already had tested up to n = 10^5 $\endgroup$ Dr. Wolfgang Hintze – Dr. Wolfgang Hintze 2015-05-24 13:59:25 +00:00 Commented May 24, 2015 at 13:59 $\begingroup$ @Dr.WolfgangHintze...thank you, NestList...will get unwieldy, so like a constructive method $\endgroup$ ubpdqn – ubpdqn 2015-05-24 23:24:57 +00:00 Commented May 24, 2015 at 23:24 Add a comment \| Start asking to get answers Find the answer to your question by asking. 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15268	https://www.teatrosancassiano.it/en/the-project/	Skip to main content en it ico en it News Press Education About us Newsletter Support us THE PROJECT A LIVING LEGACY The project has three clear objectives To reconstructthe Teatro San Cassiano of 1637 as faithfully as modern scholarship and traditional craftsmanship will allow to deliver a fully functioning, dedicated Baroque opera house, complete with its own fully operational Baroque stage machinery, moving scene-sets and special effects. To restorehistorically-informed Baroque opera to Venice. To establishthe Teatro San Cassiano as a world-renowned centre for the research, the exploration and staging of Historically Informed Performance, literally studying Baroque opera through its recital on stage and in the orchestra pit. THE CONCEPT The Teatro San Cassiano is conspicuous in history as the first opera house to open its doors to a wider ticket-buying public and to move opera away from the preserve of private patronage and forward to a commercially independent entertainment accessible to all. This momentous act sparked a global opera boom with Venice as its celebrated capital and ensured that forevermore the Teatro San Cassiano would be revered as the world’s first public opera house. In fact, one can go further: it is the birthplace of modern-day opera. It is undoubtedly the most culturally significant theatre in the history of music, and yet…it no longer exists. While the demand for Baroque opera continues to grow, in Venice and in Italy, no single, fully-functioning, commercially active Baroque theatre survives. All have perished with time. Venice has lost its ability to stage historically-informed Baroque opera. The Teatro San Cassiano itself was demolished under Napoleon’s instruction in 1812. But it is not too late. We have the means to recover the theatre from the clutches of time and Venice can once again lead the world in Baroque opera production; better still, all can be achieved within a modern vision of a forward looking and sustainable Serenissima. WIDER AIMS AND OBJECTIVES Having rebuilt the theatre, our aim will then be to preserve for the benefit of the people of Venice, Italy and the world, the musicological, historical, and architectural heritage of one of Venice’s greatest achievements and contributions to world culture. Our goal will be to join with Venice in celebrating its history in a form that is relevant to the world today, by expanding its music industry and by increasing investment and employment opportunities within the city. Our plan is to build upon the theatre’s heritage as the first ‘public’ opera house to re-interpret in a modern context how the theatre can reach out to its modern-day public in new and imaginative ways. We are passionate about creating a legacy for future generations of Venetians through our museum and school outreach projects, bringing children into the theatre and into close interactive contact with all aspects of opera performance and production. We also want to go further by promoting, maintaining and advancing public awareness of the benefits of music therapy by initiating a programme of music therapy services—related where possible to opera—that reaches out to Venice and allows opera to play an intrinsic role in the well-being of its local community. We will deliver the project using an independent hybrid charitable/commercial model that seeks to pursue charitable objectives for the greater good, but finances its activities through commercial enterprise rather than public funding. CURRENT PREDICAMENT: ITALY AND VENICE Italy is arguably the only country in the world where you can visit theatres from the Renaissance through to the modern day. Sadly, however, due to a process in which its earlier treasures have either ceased to exist or have seen their interiors modernised, enlarged or adapted to serve other genres, Italy has very quietly lost the ability to stage Baroque opera in its original context. Today, the country that gave the world opera finds itself without a single commercially active Baroque opera theatre with functioning period stage machinery and scene-sets. Venice is home to the Gran Teatro La Fenice, the Teatro Goldoni and the Teatro Malibran, but none offer the small intimate experience of a Baroque theatre built in 1637. In noting this, we of course salute La Fenice (built 1792) for its excellence in staging Baroque opera and its huge contribution to preserving the genre. It sets the standard to which we will shall aspire. Our project and the need to restore the world’s first public opera house to Venice should not detract from its successes. Indeed, our project is only possible because it alone has made Venice the world’s preeminent centre for opera. For our part, the addition of the Teatro San Cassiano will make Venice the only city in the world able to stage opera in theatres appropriate to every era of operatic production: from its birth through to today. THE SOLUTION: A VENETIAN BAROQUE OPERA HOUSE A rebuilt Teatro San Cassiano offers the perfect solution to the current predicament, and at once restores Venice as arguably the world’s foremost centre for Baroque opera. The reconstructed 1637 theatre will go to the heart of addressing the need in Venice (and indeed Italy) for precisely the type of small intimate auditorium that shaped the development of opera (and Venetian opera especially) in the 17th and 18th centuries. The Teatro San Cassiano will be the only fully active Baroque theatre in the world. Its iconic status will provide the perfect conditions for it to become the world’s foremost centre for of Baroque opera, attracting the world’s greatest conductors, singers and musicians. It will establish at its core (both on and off stage) an unprecedented collaboration to bring the worlds of musicology and music performance into a practical union which will go to the very heart of the methodology and practical experimentation necessary to truly explore Historically Informed Performance within opera production. Singers, musicians, the theatre and the public will be able to engage and to interact as part of a uniquely Baroque experience, which has until now been lost to opera. REGENERATION, IMPACT AND SUSTAINABILITY This project is as much about the future as it is the past. By putting Venice and Venetians at the centre of this project, the theatre will offer Venice a sympathetic contribution to a commercially secure and environmentally sustainable future. Both in build and in operation, the Teatro San Cassiano will create new jobs, new businesses, new opportunities, and all driven out of an ethos for a sustainable regeneration of Venice. By embracing its shared cultural heritage, but with a modern global outlook, the Teatro San Cassiano will deliver the perfect blend of employment, investment and regeneration to Venice that is simply unprecedented in modern times. This is an Impact project delivering long-term sustainability. Since its launch, the Group has targeted contracts to over 35 Venetian companies and the completed theatre will employ c. 200 direct and indirect staff. The theatre will create and maintain its own historically-informed scenery and costumes, offering apprenticeships and employment to Italy’s talented artisans. The Teatro San Cassiano’s outreach programme will reimagine the meaning of “public opera” in the 21st century. In addition to an “open-access” policy for all the community to make it “their” theatre, the planned outreach projects will inspire the next generation and all parts of the community to place the theatre at the heart of a shared Venetian identity. Singers, musicians and production staff will take opera scenes into schools and to the wider community to educate, to rehearse and ultimately to perform in the San Cassiano before their own families. Music therapy will be a continuous and core element therein. The theatre itself will form the perfect example of “Historical New Build” using traditional methods blending with modern technology with the objective of a carbon negative contribution to the Venetian environment, flora and fauna both in its build and operation.
15269	https://www.tec-science.com/mechanics/gases-and-liquids/reynolds-number-laminar-and-turbulent-flow/	Reynolds number (laminar and turbulent flow) \| tec-science Youtube Home Mechanics Gases and liquids Chemistry Structure of matter Atomic models Chemical bonds Material science Structure of metals Ductility of metals Solidification of metals Alloys Steelmaking Iron-carbon phase diagram Heat treatment of steels Material testing Mechanical power transmission Basics Gear types Belt drive Planetary gear Involute gear Cycloidal gear Thermodynamics Temperature Kinetic theory of gases Heat Thermodynamic processes in closed systems Thermodynamic processes in open systems Optics Geometrical optics Search Sign in Welcome! Log into your account your username your password Forgot your password? Get help Privacy Policy Password recovery Recover your password your email A password will be e-mailed to you. tec-science Home Mechanics Gases and liquids Chemistry Structure of matter Atomic models Chemical bonds Material science Structure of metals Ductility of metals Solidification of metals Alloys Steelmaking Iron-carbon phase diagram Heat treatment of steels Material testing Mechanical power transmission Basics Gear types Belt drive Planetary gear Involute gear Cycloidal gear Thermodynamics Temperature Kinetic theory of gases Heat Thermodynamic processes in closed systems Thermodynamic processes in open systems Optics Geometrical optics Youtube HomeMechanicsGases and liquidsReynolds number (laminar and turbulent flow) Mechanics Gases and liquids Reynolds number (laminar and turbulent flow) By tec-science 04/03/2020 26364 The Reynolds number is a dimensionless similarity parameter for describing a forced flow, e.g. whether it is an alminar or turbulent flow. Laminar and turbulent flow The definition of viscosity implies that the motion of the fluid can be divided into individual layers that shift against each other. Such a layered flow is also called laminar flow. If one imagines in thought massless particles that one introduces into such a flow, then these would move along straight paths with the flow. These imaginary flow paths are also called streamlines. Streamlines are imaginary flow paths on which massless particles would move in a fluid! Figure: Pathlines in a laminar and a turbulent flow At high flow speeds, however, turbulence occurs in fluids, so that laminar flow no longer occurs. In this case one speaks of a turbulent flow. The turbulent flow is caused by disturbances in the well-ordered flow, which are always present. However, these disturbances can be compensated to a certain degree by a relatively strong internal cohesion of the fluid, so that the flow remains laminar. Animation: Laminar and turbulent flow in a pipe At high flow speeds, however, the inertial forces of the fluid particles are so great that the disturbances can no longer be compensated for by the cohesion forces. Crossflows are formed, which interfere with the main flow and thus lead to the formation of vortices. The flow velocity at which such vortices or turbulences are generated is determined by the kinematic viscosity. After all, a high kinematic viscosity means a relatively strong internal cohesion of the fluid, which is able to compensate for disturbances. Reynolds number The flow type (i.e. whether laminar or turbulent) is thus determined by the ratio of inertia and viscosity of the fluid. This ratio is expressed by the so-called Reynolds number Re. It is determined by the (mean) flow velocity v and the kinematic viscosity ν (Greek small letter Nu) of the fluid. On the other hand, the Reynolds number is determined by the spatial dimension of the flow. In the case of a pipe this is the pipe diameter d. In this context one speaks generally of the so-called characteristic length. Since kinematic viscosity is related to dynamic viscosity by density, the Reynolds number can also be expressed in terms of dynamic viscosity η: R e:=v⋅d ν=v⋅d⋅ρ η Reynolds number[R e]=1(1)(1)R e:=v⋅d ν=v⋅d⋅ρ η Reynolds number[R e]=1 The Reynolds number is a dimensionless similarity parameter for describing the flow processes for forced flows. Only if the Reynolds numbers are identical, physically similar flow processes are obtained regardless of the size of the system. The Reynolds number is very important for all kinds of flows. In the chemical industry, for example, gaseous and liquid substances are very often pumped through pipelines. However, before chemical plants are built on a real scale, they are first tested or researched on a smaller scale (e.g. in a laboratory or pilot plant). In order to obtain the same or “similar” flow behavior as later on in the real scale, the Reynolds number must be the same on all scales. The Reynolds number is therefore determined on a small scale and then applied to the real scale. The Reynolds number is also very important for model tests in wind tunnels or water channels. Here, too, the following applies: only if the Reynolds numbers in the model experiment correspond to the real Reynolds numbers can valid results be obtained in the model experiment that can be transferred to reality. In the case of objects around which flow occurs, the characteristic length L for calculating the Reynolds number corresponds to the length of the object in the direction of flow: R e=v⋅L ν=v⋅L⋅ρ η(2)(2)R e=v⋅L ν=v⋅L⋅ρ η Figure: Aircraft model in a wind tunnel to study the flow around the aircraft Reynolds number for stirred vessels In chemistry, the flows in stirred tanks, which are generated when mixing liquids with a paddle, are also of great importance. The type of flow that occurs depends on the speed with which the paddle stirs through the liquid. Figure: Determination of the Reynolds number of a flow in a stirred vessel The reference point for the speed is the outermost part of the paddle. This speed thus depends on the diameter D and the frequency f of the rotating paddle v~D⋅f). Even if this is not the actual flow velocity of the fluid, for practical reasons this velocity is still used as flow velocity to define a Reynolds number. In this particular case of stirred vessels, the Reynolds number Re v is determined as follows (the frequency is to be given in the unit of revolution per second): R e v=f⋅D 2 ν=f⋅D 2⋅ρ η Reynolds number for stirred vessels(3)(3)R e v=f⋅D 2 ν=f⋅D 2⋅ρ η Reynolds number for stirred vessels Critical Reynolds numbers (transition from laminar to turbulent flow) The transition from laminar flow to turbulent flow has been empirically studied for different kinds of flows. For flows in pipes, a transition from laminar to turbulent flow takes place at Reynolds numbers around 2300. This is also called the critical Reynolds number. The transition from laminar to turbulent flow can range up to Reynolds numbers of 10,000. Animation: Laminar and turbulent flow in a pipe The critical Reynolds number is the Reynolds number at which a laminar flow is expected to change into a turbulent flow! When a fluid flows over a flat plate, a turbulent flow is to be expected if the Reynolds numbers are greater than 100,000. In stirred vessels, the critical Reynolds numbers are around 10,000. In this case, turbulent flows need not be a disadvantage, but contribute essentially to rapid mixing! However, in the case of vehicles or airplanes, turbulent flows are generally disadvantageous, as they ultimately mean that energy is dissipated. That is why these objects should be designed streamlined, so no turbulences come up. Typical Reynolds numbers for pipe flows In engineering, we are often dealing with flows through pipes. Think for example of water pipes or gas pipes in buildings. In such pipes the flow velocities in the case of water are in the order of 1 m/s. The inner diameter of the water pipes is about 20 mm. With a dynamic viscosity of water of 1 mPas (millipascal second) and a density of 1000 kg/m³, one already obtains Reynolds numbers in the order of 20,000! Similar results are obtained for natural gas pipelines with a diameter of e.g. 50 mm and a flow velocity of 5 m/s. With a density of 0.7 kg/m³ and a dynamic viscosity of 11 µPas, Reynolds numbers of 15,000 are obtained. These examples show that turbulent pipe flows occur far more frequently in technical practice than laminar flows! Previous articleViscosity of liquids and gases Next articleEnergetic analysis of the Hagen-Poiseuille law tec-science RELATED ARTICLESMORE FROM AUTHOR Gases and liquids Archimedes’ principle of buoyancy (crown of Archimedes) Gases and liquids Why does water boil faster at high altitudes? Gases and liquids Derivation of the Navier-Stokes equations Gases and liquids Derivation of the Euler equation of motion (conservation of momentum) Gases and liquids Derivation of the continuity equation (conservation of mass) Gases and liquids Viscosity of an ideal gas Youtube Legal notice Privacy Policy © Copyright 2025 tec-science This website uses cookies. 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15270	https://pmc.ncbi.nlm.nih.gov/articles/PMC9006834/	Successful management of life-threatening cardiac tamponade by pericardial aspiration in a boy following blunt trauma to the chest - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. 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Learn more: PMC Disclaimer \| PMC Copyright Notice BMJ Case Rep . 2022 Apr 12;15(4):e247761. doi: 10.1136/bcr-2021-247761 Search in PMC Search in PubMed View in NLM Catalog Add to search Successful management of life-threatening cardiac tamponade by pericardial aspiration in a boy following blunt trauma to the chest Iruthayanathan Reginald Ragunathan Iruthayanathan Reginald Ragunathan 1 Department of Cardiology, Teaching Hospital Karapitiya, Galle, Southern Province, Sri Lanka Find articles by Iruthayanathan Reginald Ragunathan 1, Charitha Munasinghe Charitha Munasinghe 1 Department of Cardiology, Teaching Hospital Karapitiya, Galle, Southern Province, Sri Lanka Find articles by Charitha Munasinghe 1, Chathuranga Lakmal Fonseka Chathuranga Lakmal Fonseka 2 Department of Medical Sciences, Weatherall Institute of Molecular Medicine, University of Oxford, Oxford, Oxfordshire, UK 3 Department of Clinical Medicine, Faculty of Medicine, University of Ruhuna, Galle, Southern Province, Sri Lanka Find articles by Chathuranga Lakmal Fonseka 2,3,✉ Author information Article notes Copyright and License information 1 Department of Cardiology, Teaching Hospital Karapitiya, Galle, Southern Province, Sri Lanka 2 Department of Medical Sciences, Weatherall Institute of Molecular Medicine, University of Oxford, Oxford, Oxfordshire, UK 3 Department of Clinical Medicine, Faculty of Medicine, University of Ruhuna, Galle, Southern Province, Sri Lanka ✉ Correspondence to Dr Chathuranga Lakmal Fonseka; chathuranga.fonseka@rdm.ox.ac.uk ✉ Corresponding author. Accepted 2022 Mar 29; Collection date 2022. © BMJ Publishing Group Limited 2022. No commercial re-use. See rights and permissions. Published by BMJ. PMC Copyright notice PMCID: PMC9006834 PMID: 35414577 Abstract A 7-year-old boy was presented with significant chest pain, reduced consciousness with haemodynamic instability following a minor blunt trauma to the chest. He was diagnosed to have a life-threatening pericardial effusion in FAST (Focused Assessment with Sonography for Trauma scan) ultrasound examination which was confirmed as haemopericardium causing cardiac tamponade in 2D echocardiogram. Emergency cardiac catheterisation ruled out active bleeding and prompt pericardiocentesis under fluoroscopy guidance rapidly restored patients’ haemodynamic parameters. He was successfully discharged without complications after a few days. This case report highlights uncommon presentation of cardiac tamponade without major cardiac injury after a minor blunt trauma in a paediatric patient which was detected early and successfully managed without complications. Keywords: Interventional cardiology, Emergency medicine Background Pericardial tamponade, a life-threatening condition caused by the accumulation of fluid in the pericardial sac, is treated by drainage.1 The main cause of cardiac tamponade is a penetrating chest injury (80%–90%). Comparatively, blunt chest trauma injuries occur in only 10% of the cases and are often associated with structural cardiac and/or great vessel injuries.2 Large pericardial effusions and cardiac tamponade are rare in childhood.3 Clinically, any patient with severe chest trauma, hypotension disproportionate to estimated loss of blood or with an inadequate response to fluid administration should be suspected of having a cardiac cause of shock. For patients with severe hypotension, the treatment of choice is resuscitative thoracotomy with pericardotomy. Blunt traumatic cardiac injury presenting with shock is associated with a poor prognosis.2 Here, we report a paediatric patient who presented haemodynamic instability caused by cardiac tamponade without major cardiac injury after a seemingly trivial trauma to the chest. Case presentation A 7-year-old boy was presented to the emergency department following a fall from a parked bicycle while trying to climb on top of it. He slipped and fell off the bicycle and had hit his right side of the chest on the bicycle pedal. He had no external injury but described a significant pain. He was taken to the nearest general hospital within an hour due to the pain. During transfer, the boy became drowsy. On arrival to the hospital, he had a Glasgow Coma Scale of 10/15, tachycardia of 158 beats per minute (BPM) with a rapid regular low volume pulse and blood pressure (BP) of 97/61 mm Hg (height adjusted 50th centile, BP 92/55 mm Hg). Pulsus paradoxus was not detected at the emergency department. His haemoglobin level was 152 g/L on admission and ECG showed normal voltage complexes. Arterial blood gas showed significant acidosis pH of 7.21 and HCO 3- of 16. FAST ultrasound examination at the emergency department revealed a moderate right haemothorax and a moderate pericardial effusion without intra-abdominal fluid accumulation. He was referred to the on-call surgical team and a right chest tube was placed after securing the airway by placing an endotracheal tube. The patient’s perfusion and haemodynamic status improved temporarily following the chest tube placement and blood transfusion. Subsequently, he was assessed by an adult cardiologist and noted a large pericardial effusion (6.5×3.0 cm) with a possible blood clot with early right atrial (RA)/right ventricle (RV) collapse in 2D echocardiogram signifying cardiac tamponade. Approximately 3 hours after initial presentation, he was transferred to a tertiary care hospital for further management of pericardial effusion after discussing with the cardiothoracic and paediatric cardiology team. It took a time exceeding 2 hours to reach a tertiary care centre. After admission, the patient continued to deteriorate with hypotension (BP 78/51 mm Hg) and persistent tachycardia (170 BPM). The intercostal tube was functioning and has drained 150 mL of blood. Repeat echocardiogram showed a large pericardial effusion (6.5×3.0 cm) with RA and RV collapse (figure 1A, B). Large homogenous collection of blood, which was initially detected as a clot, measuring 5×2 cm was noted in the pericardial space. As he kept deteriorating, he was taken to the cardiac catheterisation laboratory with the plan of pericardial aspiration after excluding cardiac damage under fluoroscopy guidance. After discussing the preferences and consent on the type of procedure with the parents, the patient was taken directly for cardiac catheterisation. A cardiac CT scan and open thoracotomy was considered as a second option due to the expected time to arrange the resources and parents’ preferences. Figure 1. Open in a new tab 2D echocardiography views of cardiac tamponade. (A) Apical four-chamber echo view of the heart showing pericardial effusion (arrow) with collapsed right atrium (RA). (B) Foreshortened parasternal long-axis echo view of the heart showing a large pericardial effusion (PE) with compressed right ventricle (RV). Postaspiration echocardiography images showing resolution of PE with expansion of cardiac chambers in (C) apical four-chamber view and (D) foreshortened parasternal long-axis view. Ao, aorta; CS, coronary sinus; LA, left atrium; LV, left ventricle; RVOT, right ventricular outflow tract. Differential diagnosis Due to the initial presentation of drowsiness after the fall, intracranial haemorrhage was considered and was promptly ruled out. Subsequently, an encephalitic illness and sepsis was considered due to the presence of haemodynamic instability and metabolic acidosis. Although fever and other signs of infection were not present on this presentation this was excluded due to the possibility of atypical presentation of infectious illnesses at this age group. Due to the presence of normal sepsis markers, lack of response to fluid resuscitation, a FAST scan was planned due to the fact that the boy had a fall although the trauma seemed trivial. Additionally, the boy did not have any external injuries. FAST ultrasound screening led to the detection of pericardial effusion and haemothorax. Due to the nature of the trauma and disproportionate pericardial and pleural involvement, basic clotting studies were done which were detected to be normal. Treatment In catheterisation laboratory, right femoral access was obtained. An angiogram was performed in left ventricle, aortic root, superior vena cava (SVC), inferior vena cava, right atrium (RA) and RV to see any significant bleeding into the pericardial space. Both coronary arteries were visualised and found to be intact. In evaluating the haemodynamic parameters, aortic pressure was 80/56 mm Hg and other haemodynamic cath studies were not performed in the interest of time. Total cath time was 30 min and injected contrast dose was 5 mL/kg). After ruling out any active bleeding to the pericardial space, pericardial aspiration was performed immediately under direct fluoroscopic guidance. A 5F pigtail catheter was inserted and approximately 110 mL of blood was aspirated from the pericardial space. Outcome and follow-up Postprocedure echocardiogram confirmed no residual pericardial effusion with good cardiac function (figure 1C, D). His haemodynamics improved dramatically with a blood pressure of 92/55 mm Hg) and heart rate reduced from 150 to 120 BPM. He has adequately hydrated and his urine output was maintained throughout. A repeat blood gas analysis was within normal parameters. He was sent to the cardiothoracic intensive care unit (ICU) for further care. He was extubated on the following day after confirming no reaccumulation of pericardial effusion. The boy had normal cognition and complete examination was normal. His chest drain and pericardial sheath was removed on day 2 of his ICU stay. The drains were kept to observe for further accumulation of blood/fluid which did not occur. He was observed in the ICU and discharged on day 5. Basic clotting studies were performed and they were found to be normal. He was reviewed in the paediatric cardiology clinic after 2 weeks and 6 months where he was found to be completely normal in cognition, development. Repeat echocardiogram showed no pericardial or pleural effusion with normal cardiac function. Discussion We present a case of cardiac tamponade to a seemingly inapparent trauma to the chest. In instances where history of trauma is subtle and external injuries are inapparent life-threatening emergencies could go unnoticed. This would make early detection challenging.4 Cardiac tamponade is considered to be rare following blunt chest trauma. Young children present subtle haemodynamic signs and mental status findings which can be especially difficult to assess in children.3 Seven prior cases of cardiac tamponade in children after blunt injury have been reported elsewhere.3–9 These cases included ages 21 months to 17 years. Four of the seven patients presented associated cardiac injuries, three with ventricular rupture and one with an SVC tear and two of the seven patients did not survive their injuries. Apart from two case reports,8 9 all other cases involved high speed motor vehicle accidents without proper restraints where the patients were subjected to significant acceleration/deceleration forces even without external injuries. Tabansi and Otaigbe described a case where a child developed a delayed pericardial effusion after an impact with a bed rail about 3 weeks prior to presentation.9 Another 21-month-old girl was reported to have late haemopericardium with cardiac tamponade after an initially unrecognised blunt chest wall injury.8 In the absence of major structural cardiac damage, usual site of bleeding is from lacerated pericardiophernic artery which supplies pericardium which may either resolve spontaneously or result in a fatal outcome.10 Our patient presented tachycardia, low blood pressure and altered mental status. Although inclusion of FAST exams in routine trauma care may or may not have benefit for stable patients, it remains to be essential in the evaluation of unstable children.11 This case highlights the importance of actively investigating for cardiac or pulmonary causes of shock even when the trauma to the chest is subtle and when external injuries are absent. Despite an absence of classic clinical features, cardiac tamponade was successfully and appropriately treated with a good outcome. Standard recommendation is to go for open pericardial drainage when the patient presents following trauma.2 Additionally, guidelines suggest to perform a cardiac CT to rule out bleeding in to the pericardial space. Our case highlights the value of angiography and fluoroscopy guided pericardiocentesis in a resource-limited setting, as an additional benefit, this boy did not have to undergo a major invasive cardiothoracic procedure. Although, open pericardial drainage is appropriate in children having significant chest trauma, the evidence is sparse in managing children with subtle chest trauma without any external injuries. In the absence of a detectable bleeding source or reaccumulation, guided pericardiocentesis was comparatively less invasive; hence, the patient could be discharged early from the hospital without significant morbidity. Additionally, we used a 5F pig tail catheter to facilitate aspiration of the effusion due to the suspicion of possible clot formation. We think that this may have facilitated the aspiration of the clot. Additionally, open cardiac surgery causes considerable amount of anxiety and surgical scarring. Though this patient had evidence of cardiac tamponade, he was relatively stable after resuscitation and the pericardial effusion was not rapidly expanding which led to the current management pathway. In an event with cardiac tamponade with deteriorating clinical parameters, open cardiac drainage with surgical exploration will be appropriate.2 The final decision on definitive treatment was taken after discussing with the family and the cath lab resource was accessible immediately on admission. In this situation cardiac cath offered the benefit of monitoring, resuscitation and therapeutic intervention in the acute setting. In our case, the bleeding source was evaluated through an angiogram, while the effusion was aspirated concurrently, and the team could complete the procedure within 30 min. The total contrast dose was around 5 mL/kg which is higher than the average dose necessary for a contrast CT imaging. Therefore, the risk contrast induced nephropathy (CIN) was attempted to be reduced through adequate preprocedure and postprocedure hydration. Additionally, CIN is mostly reversible. While the usual prevalence of CIN was considered to be higher, approximately 4%–10% in children, certain studies have not detected a higher incidence of acute kidney injury in the CIN groups in a cohort of patients who did not have a prior predisposition.12 However, it is imperative to adequately treat with available management strategies and we went ahead with this management pathway as a life-saving procedure. Learning points. Cardiac tamponade can occur with seemingly trivial trauma to the chest; hence, high degree of suspicion to investigate is vital. Even with the absence of external chest wall injuries, cardiac tamponade should be considered as a differential diagnosis of shock. Poor responses to fluid resuscitation should prompt bedside ultrasound investigation to exclude cardiac and pulmonary causes of shock in children. Subtle changes in consciousness could be a warning sign of poor cerebral perfusion and indicate the rapid need for a complete clinical and sonographic assessment. In case of emergencies, in a resource-limited setting, pericardiocentesis may be life saving in a patient presenting cardiac tamponade. Footnotes Contributors: IRR and CMM provided care for this patient. IRR, CMM and CLF conceived the idea for the case report. IRR, CMM and CLF wrote the initial draft and did the editing. All authors accepted the final version. Funding: The authors have not declared a specific grant for this research from any funding agency in the public, commercial or not-for-profit sectors. Case reports provide a valuable learning resource for the scientific community and can indicate areas of interest for future research. They should not be used in isolation to guide treatment choices or public health policy. Competing interests: None declared. Provenance and peer review: Not commissioned; externally peer reviewed. Ethics statements Patient consent for publication Consent obtained from the parent(s)/guardian(s). References 1.Salem K, Mulji A, Lonn E. Echocardiographically guided pericardiocentesis - the gold standard for the management of pericardial effusion and cardiac tamponade. Can J Cardiol 1999;15:1251–5. [PubMed] [Google Scholar] 2.Fitzgerald M, Spencer J, Johnson F, et al. Definitive management of acute cardiac tamponade secondary to blunt trauma. 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Childhood Kidney Diseases 2019;23:77–85. 10.3339/jkspn.2019.23.2.77 [DOI] [Google Scholar] Articles from BMJ Case Reports are provided here courtesy of BMJ Publishing Group ACTIONS View on publisher site PDF (377.2 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Background Case presentation Differential diagnosis Treatment Outcome and follow-up Discussion Footnotes Ethics statements References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15271	https://brainly.com/question/14407218	[FREE] Use formal charge arguments to explain why CO has a much smaller dipole moment than would be expected based - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +84,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +42,2k Ace exams faster, with practice that adapts to you Practice Worksheets +6,2k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Use formal charge arguments to explain why CO has a much smaller dipole moment than would be expected based on electronegativity. 2 See answers Explain with Learning Companion NEW Asked by scavalieri2142 • 01/14/2020 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 600631 people 600K 3.0 1 Upload your school material for a more relevant answer Explanation Analizing only the difference of electronegativity between C and O the dipole moment is important. But why is it smaller than expected? In the figure we can see the Lewis structure of the CO applying Formal charge, in which the oxygen shares one pair of lone electrons with the carbon in addition to the double bond they have, resulting on a triple bond (following the formal charge arguments). This acummulation of electrons in the triple decreases the effect of electronegativity anf therefore the magnitude of the dipole Answered by Giangiglia •263 answers•600.6K people helped Thanks 1 3.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 600631 people 600K 3.0 1 The Basics of General, Organic, and Biological Chemistry - David W Ball Organic Chemistry - Morsch et al. Principles of Chemistry I - Michael Nelson Upload your school material for a more relevant answer Carbon monoxide (CO) has a smaller dipole moment than expected due to the strong polarization of electron density from a triple bond between carbon and oxygen. The formal charges on the atoms indicate no significant charge separation despite the electronegativity difference, leading to an unexpected distribution of electron density. Thus, the overall structure of CO results in a near-zero dipole moment despite polar bonds. Explanation To understand why carbon monoxide (CO) has a much smaller dipole moment than expected based on electronegativity, let's begin by exploring the concept of formal charge and molecular structure. Lewis Structure and Formal Charge: In the Lewis structure of CO, carbon (C) and oxygen (O) create a triple bond, consisting of one sigma bond and two pi bonds. This triple bond is formed through the sharing of six electrons (three pairs) between the two atoms. The reasonable distribution of these electrons results in a formal charge of 0 for both carbon and oxygen, indicating that the atom's electronegativity differences do not lead to significant charge separation within the molecule. Electronegativity Comparison: Oxygen is more electronegative than carbon, which typically suggests that oxygen should attract the bonding electrons more strongly, leading to a dipole moment pointing toward the oxygen atom. However, due to the nature of the triple bond and the electron-sharing, the electron density is not unevenly distributed as might be expected based on simple electronegativity arguments. Electron Density Distribution: Despite the electronegativity difference, the electron density in the triple bond is more concentrated due to the strong bonding interactions present. In CO, the electron density is greater at carbon than a single-bonded carbon atom would typically have because of the electron delocalization effect in the triple bond configuration and the contributions from lone pairs of electrons. Contributions from Atomic Polarization: The involvement of the carbon atom's p-orbitals contributes to a phenomenon known as atomic polarization. This means that the effective charge distribution in the molecule does not align with the expected dipole moment based solely on electronegativity. The resultant charge on carbon in CO has been calculated to be slightly negative, contradicting the usual assumptions derived from electronegativity values alone. In conclusion, while CO features polar bonds, the overall molecular structure and the presence of a strong triple bond lead to a smaller than anticipated dipole moment. The formal charge analysis helps clarify how electronegativity and electron distribution interact in the CO molecule, resulting in these unexpected characteristics. Examples & Evidence For instance, in the case of formaldehyde (H₂C=O), the dipole moment is larger compared to CO because the carbon atom's bonding density decreases as it forms single bonds with hydrogen, which allows the charge transfer to be more pronounced towards oxygen. Another example is CO₂, which has two polar C=O bonds that exactly offset each other due to its linear shape, leading to no overall dipole moment even though the individual bonds are polar. The evidence lies in the calculations of partial charges based on formal charge arrangements and molecular orbital theory where the distribution of electron density around carbon actually results in a negative formal charge in CO, highlighting how electronegativity differences do not straightforwardly translate into dipole moments. Thanks 1 3.0 (1 vote) Advertisement Community Answer This answer helped 11625522 people 11M 0.0 0 The smaller dipole moment in CO compared to what would be expected based on electronegativity is due to the partial cancellation of charges resulting from the formal charges on carbon and oxygen. Explanation Formal charge is a way to determine the distribution of electrons in a molecule. It is calculated by assigning electrons to the atoms in a molecule based on their electronegativity. In the case of CO, the oxygen atom is more electronegative than the carbon atom, so it is assigned 6 electrons and carbon is assigned only 2 electrons. This results in a negative formal charge on oxygen and a positive formal charge on carbon. The dipole moment of a molecule is a measure of the separation of positive and negative charges within the molecule. In CO, due to the difference in electronegativity between carbon and oxygen, there is a separation of charges. However, the magnitude of the dipole moment is smaller than expected based on electronegativity because the formal charges on carbon and oxygen partially cancel each other out. The positive charge on carbon partially compensates for the negative charge on oxygen, resulting in a smaller dipole moment. Overall, the smaller dipole moment in CO compared to what would be expected based on electronegativity is due to the partial cancellation of charges resulting from the formal charges on carbon and oxygen. Learn more about dipole moment in CO here: brainly.com/question/33540618 SPJ3 Answered by Qwseed •24.2K answers•11.6M people helped Thanks 0 0.0 (0 votes) Advertisement ### Free Chemistry solutions and answers Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. Community Answer 8 For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval mc015-1. Jpg? mc015-2. Jpg mc015-3. Jpg mc015-4. Jpg mc015-5. Jpg. Community Answer 4.8 268 How did Mendeleev come up with the first periodic table of the elements? (1 point) A He determined the mass of atoms of each element. B He estimated the number of electrons in atoms of each element. C He arranged the elements by different properties to find a pattern. D He organized the elements by their atomic number. New questions in Chemistry Which equation represents the combined gas law? P₁V₁ = P₂V₂ Create your own example of a science experiment that involves salt. Be sure to follow the steps of the scientific method. Place the following transitions of the hydrogen atom from smallest to largest frequency of light absorbed: (a) n=3 to n=6 (b) n=4 to n=9 (c) n=2 to n=3 (d) n=1 to n=2 How many molecules are in 0.400 moles of N O 3 ? What type of chemical bond holds C a 2+ and O 2− together in CaO? A. metallic bond B. covalent bond C. ionic bond D. hydrogen bond Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
15272	https://doctorabad.com/app/uptodate/d/topic.htm?path=clinical-manifestations-diagnosis-and-treatment-of-osteomalacia	Clinical manifestations, diagnosis, and treatment of osteomalacia Clinical manifestations, diagnosis, and treatment of osteomalacia Drug Interactions Calculators PCUs What's New Patient Education Contents Clinical manifestations, diagnosis, and treatment of osteomalacia Topic Outline SUMMARY & RECOMMENDATIONS INTRODUCTION CLINICAL FEATURES Clinical manifestations Laboratory findings Radiographic findings - Changes in vertebral bodies - Looser zones - Secondary hyperparathyroidism - Other radiograph findings Bone mineral density DIAGNOSIS AND EVALUATION Interpretation of laboratory abnormalities Bone biopsy DIFFERENTIAL DIAGNOSIS TREATMENT Vitamin D deficiency Other causes PREGNANCY SUMMARY AND RECOMMENDATIONS ACKNOWLEDGMENT REFERENCES GRAPHICS View All DIAGNOSTIC IMAGES - Blurred spine in osteomalacia - Osteomalacia of the lumbar spine CT - Pseudofractures in osteomalacia - Tibial bowing in osteomalacia - Coxa profunda in osteomalacia - Cephalopelvic disproportion in osteomalacia radiographic finding PICTURES - Low-turnover osteomalacia TABLES - Lab findings in osteomalacia RELATED TOPICS Causes of hypophosphatemia Clinical features, laboratory manifestations, and diagnosis of multiple myeloma Clinical manifestations and diagnosis of Paget disease of bone Clinical manifestations, diagnosis, and evaluation of osteoporosis in postmenopausal women Epidemiology and etiology of osteomalacia Etiology and diagnosis of distal (type 1) and proximal (type 2) renal tubular acidosis Etiology and treatment of calcipenic rickets in children Hereditary hypophosphatemic rickets and tumor-induced osteomalacia Management of secondary hyperparathyroidism in adult nondialysis patients with chronic kidney disease Management of secondary hyperparathyroidism in dialysis patients Osteoporotic fracture risk assessment Overview of rickets in children Periodontal disease in children: Associated systemic conditions Primary hyperparathyroidism: Clinical manifestations Primary hyperparathyroidism: Diagnosis, differential diagnosis, and evaluation Treatment of distal (type 1) and proximal (type 2) renal tubular acidosis Vitamin D deficiency in adults: Definition, clinical manifestations, and treatment Clinical manifestations, diagnosis, and treatment of osteomalacia Authors:Adi Cohen, MD, MHSMatthew T Drake, MD, PhDSection Editor:Peter J Snyder, MDDeputy Editor:Jean E Mulder, MD Contributor Disclosures All topics are updated as new evidence becomes available and our peer review process is complete. Literature review current through:Feb 2018.\|This topic last updated:Aug 03, 2017. INTRODUCTION—Osteomalacia is a disorder of bone, characterized by decreased mineralization of newly formed osteoid at sites of bone turnover. Several different disorders cause osteomalacia via mechanisms that result in hypocalcemia, hypophosphatemia, or direct inhibition of the mineralization process. (See "Epidemiology and etiology of osteomalacia".) This topic review will present an overview of the clinical manifestations, diagnosis, and treatment of adults with osteomalacia. The treatment of nutritional, hereditary vitamin D resistant, and pseudovitamin D-deficient rickets in children is discussed separately. (See "Etiology and treatment of calcipenic rickets in children" and "Hereditary hypophosphatemic rickets and tumor-induced osteomalacia".) CLINICAL FEATURES Clinical manifestations—Osteomalacia may be asymptomatic and present radiologically as osteopenia. It can also produce characteristic symptoms, independently of the underlying cause, including diffuse bone and joint pain, muscle weakness, and difficulty walking [1-3]. In a report of 17 patients with osteomalacia on bone biopsy, the following findings were observed : ●Bone pain and muscle weakness in 16 (94 percent) ●Bone tenderness in 15 (88 percent) ●Fracture in 13 (76 percent) ●Difficulty walking and waddling gait in four (24 percent) ●Muscle spasms, cramps, a positive Chvostek's sign, tingling/numbness, and inability to ambulate in one to two (6 to 12 percent) These symptoms may be insidious in onset. Bone pain is usually most pronounced in the lower spine, pelvis, and lower extremities, where fractures have taken place, and may be associated with tenderness to palpation. The pain is characterized as dull and aching and is aggravated by activity and weight bearing. Fractures may occur with little or no trauma, typically involving the ribs, vertebrae, and long bones. Abnormal spinal curvature or deformity of the thorax or pelvis appears only in severe osteomalacia of long duration . The muscle weakness characteristically is proximal and may be associated with muscle wasting, hypotonia, and discomfort with movement . There may also be a waddling gait. It is likely that high levels of parathyroid hormone (PTH) and low levels of phosphate and calcitriol all contribute to the myopathy since similar findings occur in severe primary hyperparathyroidism. Osteomalacia secondary to hypophosphatasia is associated with poorly healing metatarsal or other fractures, chondrocalcinosis, and premature loss of teeth during childhood. Laboratory findings—Laboratory abnormalities in osteomalacia are largely dependent upon the cause of the osteomalacia (table 1). (See "Epidemiology and etiology of osteomalacia", section on 'Etiologic diagnosis'.) In retrospective reviews of patients with biopsy-proven nutritional osteomalacia, the following laboratory abnormalities were observed [4,5]: ●Alkaline phosphatase elevated in 95 to 100 percent ●Serum calcium and phosphorus reduced in 27 to 38 percent ●Urinary calcium low in 87 percent ●25-hydroxyvitamin D (25[OH]D, calcidiol) <15 ng/mL in 100 percent ●PTH elevated in 100 percent The majority of patients (40 of 43) in this review had nutritional osteomalacia from either a gastrointestinal disorder or suboptimal nutrition and inadequate sun exposure. In these cases, 25(OH)D levels were very low (<10 ng/mL [25 nmol/L]), which differentiates vitamin D deficiency from the other causes of osteomalacia, such as the renal phosphate wasting syndromes. Radiographic findings—The radiological abnormalities in adults who develop osteomalacia are less striking than those seen in children with rickets (see "Overview of rickets in children"). Reduced bone mineral density (BMD) with thinning of the cortex is the most common finding, but it is very nonspecific. More specific are changes in vertebral bodies and Looser zones. Infrequently, radiologic evidence of secondary hyperparathyroidism can be seen. Changes in vertebral bodies—Inadequate mineralization of osteoid and loss of secondary trabeculae lead to a loss of radiologic distinctness of vertebral body trabeculae, making the radiograph appear of poor quality (image 1 and image 2). With more advanced disease, softening leads to a concavity of the vertebral bodies called codfish vertebrae. The vertebral disks appear large and biconvex. There may be spinal compression fractures, but these are more common in osteoporosis. Looser zones—Looser pseudofractures, fissures, or narrow radiolucent lines, 2 to 5 mm in width with sclerotic borders, are a characteristic radiologic finding in osteomalacia (image 3) . They often are bilateral, symmetric, and lie perpendicular to the cortical margins of bones. They are most commonly found at the femoral neck, on the medial part of the femoral shaft, immediately under the lesser trochanter or a few centimeters beneath, and in the pubic and ischial rami. They may also occur at the ulna, scapula, clavicle, rib, and metatarsal bones. Pseudofractures can also be seen with bone scans where they appear as hot spots . The term "Milkman syndrome" refers to the combination of multiple bilateral and symmetric pseudofractures in a patient with osteomalacia . Looser zones have been postulated to represent either [9,10]: ●Stress fractures that have been repaired by the laying down of inadequately mineralized osteoid, or ●Erosion of bone by arterial pulsations, since they often lie in apposition to arteries Secondary hyperparathyroidism—Skeletal changes induced by longstanding secondary hyperparathyroidism are less frequent than the above abnormalities. When they do occur, they include subperiosteal resorption of the phalanges, bone cysts, and resorption of the distal ends of long bones such as the clavicle and humerus. (See "Primary hyperparathyroidism: Clinical manifestations", section on 'Skeletal'.) Other radiograph findings—More severe osteomalacia can lead to shortening and bowing of the tibia, pathologic fractures, coxa profunda hip deformity (image 4A-B), and cephalopelvic disproportion (image 5). Bone mineral density—Several studies have demonstrated markedly reduced spine, hip, and forearm BMD (as measured by dual-energy x-ray absorptiometry [DXA]) in patients with osteomalacia related to vitamin D deficiency [4,5]. However, BMD is not required for the diagnosis of osteomalacia, and BMD (DXA) findings are unable to differentiate osteomalacia and osteoporosis. In contrast, BMD tends to be normal or increased (especially at the lumbar spine) in adults with X-linked hypophosphatemic rickets (XLH), axial osteomalacia, fibrogenesis imperfecta, and skeletal fluorosis [11-13]. This discrepancy may be related in part to differences in PTH values in the two groups (high in vitamin D deficiency and normal in the others). (See "Hereditary hypophosphatemic rickets and tumor-induced osteomalacia", section on 'X-linked hypophosphatemia' and "Epidemiology and etiology of osteomalacia", section on 'Defective bone matrix'.) DIAGNOSIS AND EVALUATION—Osteomalacia should be suspected in cases of bone pain associated with malabsorption, gastric bypass surgery, celiac sprue, chronic hepatic disease, or chronic kidney disease. The diagnosis is based on a combination of clinical features (which may include bone pain and tenderness, fractures, and/or muscle weakness), laboratory results, radiologic findings, and, rarely, transiliac bone histomorphometry. The majority of patients have nutritional osteomalacia and will have a very low serum 25-hydroxyvitramin D (25[OH]D) (<10 ng/mL [25 nmol/L]), low to low-normal serum calcium and phosphate, and high parathyroid hormone (PTH) and alkaline phosphatase (both total and bone-specific) levels. Clinical evaluation, including history of gastrointestinal diseases or procedures, sun exposure, dietary habits, and onset (insidious or acute) and duration of symptoms, may help determine the etiology of osteomalacia. Laboratory abnormalities in osteomalacia are largely dependent on its cause (table 1), and therefore, laboratory findings are used to diagnose and determine the etiology of osteomalacia. The initial laboratory evaluation should include measurement of serum concentrations of: ●Calcium ●Phosphate ●Alkaline phosphatase ●25(OH)D ●PTH ●Electrolytes, blood urea nitrogen (BUN), and creatinine Radiographs may be helpful in certain settings (eg, severe bone pain), to distinguish osteomalacia from multiple myeloma or Paget disease of bone. Bone biopsy and histomorphometry should be performed only when the diagnosis of osteomalacia is in doubt or the cause of osteomalacia is not determined by noninvasive testing (eg, to assess for one of the rare disorders of defective bone matrix, such as axial osteomalacia or fibrogenesis imperfecta). (See "Epidemiology and etiology of osteomalacia", section on 'Defective bone matrix'.) A delay in diagnosis of osteomalacia is commonly reported [4,14,15]. In one study of 33 women with osteomalacia, the mean duration of symptoms before diagnosis was 2.5 years . Diagnoses considered prior to confirmation of osteomalacia included osteoporosis, Paget disease, malignancy, pseudohypoparathyroidism, osteoarthritis, malabsorption, irritable bowel syndrome with depression, fibromyalgia, and somatization disorders. Interpretation of laboratory abnormalities—The goal of the laboratory evaluation is to distinguish vitamin D deficiency or resistance from the phosphate wasting syndromes and other less common causes of osteomalacia (table 1): ●In nutritional osteomalacia, 25(OH)D (calcidiol) is typically very low (<10 ng/mL [25 nmol/L]), calcium and phosphate low to low-normal, and PTH and alkaline phosphatase high . The serum concentration of 1,25-dihydroxyvitamin D may be normal, low, or high, depending upon the severity and duration of vitamin D deficiency, and is therefore not helpful in making the diagnosis . ●In primary renal phosphate wasting, serum phosphate is low and phosphate clearance is elevated. Serum 1,25-dihydroxyvitamin D levels are frequently inappropriately normal for the degree of hypophosphatemia (1,25-dihydroxyvitamin D should increase in response to severe hypophosphatemia). Serum calcium and 25(OH)D levels are normal, and PTH and alkaline phosphatase levels are normal or mildly elevated. Patients may also have other tubular defects (hypouricemia, aminoaciduria, and glucosuria) if the phosphate wasting is part of a generalized Fanconi syndrome. Heavy metals in the urine may be increased if they are the cause of the Fanconi syndrome. In patients with tumor-induced osteomalacia and hereditary hypophosphatemic rickets, serum fibroblast growth factor (FGF) 23 levels may be elevated . (See "Causes of hypophosphatemia" and "Hereditary hypophosphatemic rickets and tumor-induced osteomalacia".) ●In type 2 (proximal) renal tubular acidosis, there is a hyperchloremic metabolic acidosis and hypophosphatemia. The latter reflects both proximal renal tubule phosphate wasting and secondary hyperparathyroidism due to acidosis-induced hypercalciuria. (See "Etiology and diagnosis of distal (type 1) and proximal (type 2) renal tubular acidosis", section on 'Proximal (type 2) RTA'.) ●In hypophosphatasia, the alkaline phosphatase level is low while the serum calcium and phosphate concentrations are normal. Reduced activity of the alkaline phosphatase enzyme results in accumulation of substrates, including phosphoethanolamine, inorganic pyrophosphate, and pyridoxal 5'-phosphate (PLP), in blood and urine. In patients not taking a vitamin B6 (pyridoxine) supplement, elevated plasma PLP is a marker of hypophosphatasia. Mutational analysis of the tissue-nonspecific isoenzyme of alkaline phosphatase is available commercially, but mutations may not always correlate with disease severity [18,19]. (See "Epidemiology and etiology of osteomalacia", section on 'Hypophosphatasia'.) ●In fibrogenesis imperfecta and axial osteomalacia, alkaline phosphatase, calcium, phosphate, and vitamin D are usually normal. (See "Epidemiology and etiology of osteomalacia", section on 'Defective bone matrix'.) ●In skeletal fluorosis, serum calcium and phosphate are usually normal and alkaline phosphate is elevated. Serum, urine, and bone fluoride content is increased . Bone biopsy—Bone biopsy with double tetracycline labeling is the most accurate way to diagnose osteomalacia. However, it is infrequently performed clinically because it is invasive and because the diagnosis can usually be made from a combination of clinical and laboratory findings. The histomorphometric characteristics of osteomalacia include (picture 1) : ●Prolonged mineralization lag time ●Widened osteoid seams ●Increased osteoid volume All of these features are necessary for the diagnosis because other disorders may show one of these findings. Wide osteoid seams reflecting high turnover, for example, can be seen with hyperthyroidism, Paget disease, and hyperparathyroidism. However, the mineral apposition rate is elevated in these disorders in contrast to the low values in osteomalacia. (See "Epidemiology and etiology of osteomalacia", section on 'Pathogenesis'.) DIFFERENTIAL DIAGNOSIS—Other causes of bone fractures, bone pain, and reduced bone mineral density (BMD) include osteoporosis, malignancy, Paget disease, and hyperparathyroidism. Most of these diagnoses can be distinguished from osteomalacia by the clinical history, physical examination, and a combination of laboratory and radiologic studies. Bone biopsy using double tetracycline labeling (assuming that the patient is not allergic to tetracycline) is performed rarely and is reserved for cases that are difficult to diagnose using noninvasive methods . ●Osteoporosis occurs in different settings from osteomalacia, particularly postmenopausal women, older adult subjects, and patients treated with chronic corticosteroid therapy. (See "Osteoporotic fracture risk assessment" and "Clinical manifestations, diagnosis, and evaluation of osteoporosis in postmenopausal women".) Osteoporosis is characterized by normal serum levels of calcium, phosphate, and alkaline phosphatase. This is in contrast to the frequent findings of one or more of the following in the different causes of osteomalacia: hypophosphatemia, hypocalcemia, low levels of 25-hydroxyvitamin D (25[OH]D) (<10 ng/mL [25 nmol/L]), and increased parathyroid hormone (PTH) and alkaline phosphatase levels (table 1) . Although 25(OH)D levels may be low in patients with osteoporosis and a subset of these patients may also have secondary elevations of PTH, 25(OH)D levels rarely are below 10 ng/mL (25 nmol/L). (See 'Laboratory findings' above.) Reduced BMD does not distinguish osteoporosis from osteomalacia. Patients with osteomalacia due to vitamin D deficiency may have markedly reduced spine, hip, and forearm BMD. In such patients, treatment with bisphosphonates, teriparatide, or other osteoporosis medications is not appropriate and may exacerbate hypocalcemia. Osteomalacia related to vitamin D deficiency should be treated with vitamin D and calcium, which often results in marked improvement in BMD. (See 'Vitamin D deficiency' below.) ●In patients with Paget disease of bone, alkaline phosphatase is elevated, but bone scan and radiographic findings are unique. Plain radiographs of involved areas reveal cortical thickening, expansion, coarsening of trabecular markings and mixed areas of lucency and sclerosis. (See "Clinical manifestations and diagnosis of Paget disease of bone".) ●In patients with multiple myeloma, weakness, fatigue, and bone pain are common. Conventional radiographs often reveal lytic lesions, as well as diffuse osteopenia and vertebral fractures. Many patients have anemia and abnormal renal function at diagnosis, whereas patients with osteomalacia generally have normal renal function. Alkaline phosphatase is usually not elevated in multiple myeloma, and hypercalcemia may be present in some patients. Multiple myeloma is a common cause of type 2 renal tubular acidosis in adults . (See "Clinical features, laboratory manifestations, and diagnosis of multiple myeloma".) ●In patients with hyperparathyroidism, both PTH and calcium are elevated, whereas calcium levels are either low or normal in osteomalacia. (See "Primary hyperparathyroidism: Diagnosis, differential diagnosis, and evaluation".) TREATMENT—The treatment of osteomalacia should be directed at reversal of the underlying disorder, if possible, and correction of hypophosphatemia, hypocalcemia, and vitamin D deficiency. Vitamin D deficiency—Vitamin D supplementation in patients who are deficient in this hormone leads to a dramatic improvement in muscle strength and bone tenderness within weeks. Effects are typically most dramatic when adequate calcium intake occurs simultaneously. Bone mineral density (BMD) may improve within three to six months . Multiple preparations of vitamin D and its metabolites are available. Vitamin D, rather than its metabolites, is used when possible since the cost is modest. Vitamin D metabolites are required when there is abnormal vitamin D metabolism (significant liver or renal disease). The recommended preparation and dose vary with the clinical condition, as described below. With each regimen, the serum calcium concentration and urinary calcium excretion are monitored, initially after one month and three months, and then less frequently (every 6 to 12 months), until 24-hour urinary calcium excretion is normal. The serum calcium concentration is monitored to permit early detection of hypercalcemia from excessive dosing. Serum 25-hydroxyvitamin D (25[OH]D) should be measured approximately three to four months after initiating therapy. The dose should be adjusted to prevent hypercalciuria or hypercalcemia. In most cases, serum calcium and phosphate are normal after a few weeks of treatment, but alkaline phosphatase remains elevated for several months. Healing of osteomalacia is considered to have occurred when there are increases in urinary calcium excretion and BMD. Healing of the osteomalacia may take many months to a year or more and varies with the degree and duration of the deficiency . In addition to vitamin D supplementation, all patients should maintain a calcium intake of at least 1000 mg per day since inadequate intake of calcium may contribute to the development of osteomalacia [25,26]. The combination of calcium and vitamin D is more likely to produce radiographic evidence of nearly complete healing of rickets (58 versus 19 percent in Nigerian children with nutritional rickets) . A higher calcium dose (up to 4 g/day) may be necessary in patients with malabsorption (such as occurs following gastric bypass). The treatment of vitamin D deficiency is reviewed in detail separately and briefly summarized here. (See "Vitamin D deficiency in adults: Definition, clinical manifestations, and treatment", section on 'Vitamin D repletion'.) ●For patients with severe vitamin D deficiency (25[OH]D <10 ng/mL [25 nmol/L]), one common approach is to treat with 50,000 international units of vitamin D2 or D3 orally once per week for six to eight weeks, and then 800 international units of vitamin D3 daily thereafter. However, the efficacy of this practice compared with daily, weekly, or monthly dosing has not been rigorously established. ●In malabsorptive states, oral dosing and duration of treatment depend upon the vitamin D absorptive capacity of the individual patient. Doses of vitamin D of 10,000 to 50,000 international units daily may be necessary to replete patients with gastric bypass or malabsorption. Patients who remain deficient or insufficient on such doses will need to be treated with hydroxylated vitamin D metabolites (calcidiol or calcitriol), which are more readily absorbed, or with sun or sunlamp exposure. ●In liver disease, the vitamin D metabolite calcidiol (25[OH]D) should be used because it does not require hepatic 25-hydroxylation. The onset of action is more rapid and the half-life of two to three weeks is shorter than that of vitamin D3 and similar to that of vitamin D2. The dose in this situation is 50 to 200 micrograms/day . Calcidiol is not readily available in the United States, so calcitriol may be used in patients with severe liver disease who remain deficient after treatment with vitamin D2 or vitamin D3. ●Calcitriol (1,25-dihydroxyvitamin D) is a vitamin D metabolite available in capsules of 0.25 and 0.5 micrograms. It has a rapid onset of action and the half-life is only six hours. It is associated with a fairly high incidence of hypercalcemia, and patients should be followed carefully. It is most useful in those with decreased synthesis of calcitriol, as occurs in chronic renal failure or in type 1 vitamin D-dependent rickets (due to an inactivating mutation in the 1-hydroxylase gene). (See "Etiology and treatment of calcipenic rickets in children", section on '1-alpha-hydroxylase deficiency'.) ●Dihydrotachysterol (DHT) is available as tablets of 0.125, 0.2, and 0.5 mg. It is functionally equivalent to 1alpha-hydroxyvitamin D. It requires hepatic 25-hydroxylation prior to becoming therapeutically active. DHT can be used in the disorders for which calcitriol is used. It has a rapid onset of action and a relatively short duration of action. Other causes—The treatment of hereditary and acquired renal phosphate wasting syndromes, renal osteodystrophy, and renal tubular acidosis are discussed in greater detail elsewhere: ●Hereditary hypophosphatemic rickets is treated with a combination of phosphate supplementation and calcitriol. Tumor-induced osteomalacia is treated similarly until the causative tumor can be removed, or indefinitely if tumor removal is not possible. (See "Hereditary hypophosphatemic rickets and tumor-induced osteomalacia".) ●In renal insufficiency, both oral and intravenous calcitriol can be used. (See "Management of secondary hyperparathyroidism in dialysis patients" and "Management of secondary hyperparathyroidism in adult nondialysis patients with chronic kidney disease".) ●Osteomalacia of renal tubular acidosis is treated initially with 5000 to 10,000 international units per day of vitamin D. The acidosis is corrected with sodium and/or potassium citrate. Once the bone heals, supraphysiologic doses of vitamin D (above 800 international units/day) should not be necessary. (See "Treatment of distal (type 1) and proximal (type 2) renal tubular acidosis".) ●For hypophosphatasia, there have been few treatment options. However, enzyme replacement therapy (asfotase alfa) for perinatal, infantile, and juvenile-onset hypophosphatasia became available in October 2015 . In a preliminary report, infusion of recombinant human tissue-nonspecific isoenzyme of alkaline phosphatase was associated with improvement in skeletal radiographs and in pulmonary and physical function in infants and young children . In subsequent open-label, prospective studies including a total of 99 patients with perinatal, infantile, or juvenile-onset hypophosphatasia, enzyme replacement therapy was associated with improved overall survival, ventilator-free survival, growth, and bone mineralization compared with a historic cohort . Studies in adult hypophosphatasia are underway . (See "Epidemiology and etiology of osteomalacia", section on 'Hypophosphatasia' and "Periodontal disease in children: Associated systemic conditions", section on 'Hypophosphatasia'.) ●For patients with the rare skeletal disorder of defective bone matrix (axial osteomalacia and fibrogenesis imperfecta), there are no established therapies, although a report of two brothers with fibrogenesis imperfecta reported clinical, radiographic, and histologic improvement following treatment with recombinant human growth hormone . Patients with axial osteomalacia do not appear to deteriorate over time. In contrast, patients with fibrogenesis imperfecta develop severe skeletal pain, debilitating fractures, and progressive immobility . (See "Epidemiology and etiology of osteomalacia", section on 'Defective bone matrix'.) PREGNANCY—Severe vitamin D deficiency resulting in osteomalacia has been described in dark-skinned pregnant women who have emigrated from warm to cold climates in North America and Europe [31-33]. Other risk factors for osteomalacia during pregnancy include limited sun exposure due to protective clothing, malabsorption (eg, celiac disease, dietary intake of phytates that prevent calcium absorption), and malnutrition. Pregnant women with osteomalacia have similar symptoms as nonpregnant adults (see 'Clinical manifestations' above). They may present with persistent and nonspecific musculoskeletal pain and inability to bear weight. Evaluation reveals similar biochemical findings as in nonpregnant adults. In some case reports, multiple fractures were present [33,34]. Severe osteomalacia can be associated with cephalopelvic disproportion (image 5), necessitating cesarean delivery . In case reports, pregnant women with severe osteomalacia (diagnosed at the time of delivery) were successfully treated with high-dose vitamin D (600,000 international units intramuscularly as a single dose, after delivery) and calcium supplementation (up to 1.5 g daily) [33,34]. The administration during pregnancy of such a high dose of vitamin D has not been adequately studied. In case series from the 1960s, pregnant women with osteomalacia were safely treated with calcium and 3000 to 6000 international units of vitamin D daily . There are modern day trials evaluating vitamin D dosing in pregnant women with vitamin D deficiency, some of whom had secondary elevations in serum parathyroid hormone (PTH), but none of whom had vitamin D deficiency of sufficient severity and duration to cause osteomalacia [36,37]. In these trials, vitamin D supplementation at 12 to 27 weeks gestation with 400, 800, 2000, or 4000 international units D3 daily or 200,000 international units as a single oral dose was safe and successfully increased serum vitamin D concentrations. These trials are reviewed in detail separately. (See "Vitamin D deficiency in adults: Definition, clinical manifestations, and treatment", section on 'Pregnancy'.) Women who are diagnosed with osteomalacia during pregnancy should receive adequate calcium (approximately 1000 to 1500 mg daily) and vitamin D. We typically start with 2000 to 4000 international units daily and measure the serum 25-hydroxyvitamin D (25[OH]D) and calcium and urinary calcium excretion after one month and three months, then less frequently (every 6 to 12 months), until 24-hour urinary calcium excretion is normal. If the initial dose does not improve serum 25(OH)D after three to four months, the dose of vitamin D can be increased by 1000 to 2000 international units/day, with continued monitoring of serum vitamin D and calcium and urinary calcium. The serum calcium concentration is monitored to permit early detection of hypercalcemia from excessive dosing. The dose of vitamin D should be decreased, as needed, to prevent hypercalciuria or hypercalcemia. SUMMARY AND RECOMMENDATIONS ●The clinical manifestations of osteomalacia may include bone pain and tenderness, muscle weakness, difficulty walking, and a waddling gait. Laboratory findings depend upon the underlying cause of osteomalacia (table 1). Typical laboratory features of nutritional osteomalacia include elevations in alkaline phosphatase and parathyroid hormone (PTH) and decreases in calcium, phosphate, and 25-hydroxyvitamin D (25[OH]D) concentrations. (See 'Clinical features' above and 'Laboratory findings' above.) ●The characteristic radiologic findings are Looser pseudofractures, fissures, or narrow radiolucent lines. In addition, inadequate mineralization of osteoid and loss of secondary trabeculae lead to a loss of radiologic distinctness of the vertebral body trabeculae and concavity of the vertebral bodies (codfish vertebrae). Bone mineral density (BMD) is not required for the diagnosis of osteomalacia, and reduced BMD does not distinguish osteoporosis from osteomalacia. However, several studies have demonstrated markedly reduced spine, hip, and forearm BMD (as measured by dual-energy x-ray absorptiometry [DXA]) in patients with osteomalacia related to vitamin D deficiency. (See 'Radiographic findings' above.) ●Osteomalacia should be suspected in cases of bone pain associated with malabsorption, gastric bypass surgery, celiac sprue, chronic hepatic disease, or chronic kidney disease. The diagnosis is based upon a combination of clinical features (bone pain, tenderness, and fractures; muscle weakness), laboratory results (table 1), radiologic findings, and, rarely, bone histomorphometry. (See 'Diagnosis and evaluation' above.) ●Laboratory abnormalities in osteomalacia are largely dependent on its cause (table 1), and therefore, laboratory findings are used to diagnose and determine the etiology of osteomalacia. Laboratory evaluation should include measurement of serum concentrations of calcium, phosphate, alkaline phosphatase, 25(OH)D, PTH, electrolytes, blood urea nitrogen (BUN), and creatinine. Radiographs may be helpful in certain settings (severe bone pain), to distinguish osteomalacia from multiple myeloma or Paget disease of bone. Bone biopsy and histomorphometry should be performed only when the diagnosis of osteomalacia is in doubt or the cause of osteomalacia is not determined with noninvasive testing, eg, to assess for one of the rare disorders of defective bone matrix. (See 'Diagnosis and evaluation' above.) ●Treatment of osteomalacia depends upon the underlying etiology. For patients with osteomalacia secondary to severe vitamin D deficiency, vitamin D repletion leads to a dramatic improvement in muscle strength and bone tenderness within weeks. Many clinicians treat nutritional deficiency (25[OH]D <10 ng/mL [25 nmol/L]) with 50,000 international units of vitamin D2 or D3 orally once per week for six to eight weeks, followed by a maintenance dose (eg, 800 international units of vitamin D3 daily) thereafter. (See 'Vitamin D deficiency' above and "Vitamin D deficiency in adults: Definition, clinical manifestations, and treatment", section on 'Dosing'.) ●After initiation of vitamin D treatment, the serum calcium concentration and urinary calcium excretion are monitored, first after one month and three months, and then less frequently (every 6 to 12 months). The serum calcium concentration is monitored to permit early detection of hypercalcemia from excessive dosing. Serum 25(OH)D should be measured approximately three to four months after initiating therapy. The dose should be adjusted to prevent adverse effects of hypercalciuria or hypercalcemia. Biological and radiological abnormalities may take up to one year or more to disappear. (See 'Vitamin D deficiency' above.) ●Hereditary hypophosphatemic rickets is treated with a combination of phosphate supplementation and calcitriol. Tumor-induced osteomalacia is treated similarly until the causative tumor can be removed, or indefinitely if tumor removal is not possible. (See "Hereditary hypophosphatemic rickets and tumor-induced osteomalacia".) ACKNOWLEDGMENT—The editorial staff at UpToDate would like to acknowledge the late Charles J Menkes, MD, who contributed to earlier versions of this topic review. Use of UpToDate is subject to the Subscription and License Agreement. REFERENCES Lips P, van Schoor NM, Bravenboer N. Vitamin D-related disorders. In: Primer on the Metabolic Bone Diseases and Disorders of Mineral Metabolism, 8th ed, Rosen CJ (Ed), John Wiley & Sons, Inc, Ames 2013. Gifre L, Peris P, Monegal A, et al. Osteomalacia revisited : a report on 28 cases. Clin Rheumatol 2011; 30:639. Bhan A, Rao AD, Rao DS. Osteomalacia as a result of vitamin D deficiency. Endocrinol Metab Clin North Am 2010; 39:321. Basha B, Rao DS, Han ZH, Parfitt AM. Osteomalacia due to vitamin D depletion: a neglected consequence of intestinal malabsorption. Am J Med 2000; 108:296. Bhambri R, Naik V, Malhotra N, et al. Changes in bone mineral density following treatment of osteomalacia. J Clin Densitom 2006; 9:120. Frame B, Parfitt AM. Osteomalacia: current concepts. Ann Intern Med 1978; 89:966. Kim S, Park CH, Chung YS. Hypophosphatemic osteomalacia demonstrated by Tc-99m MDP bone scan: a case report. Clin Nucl Med 2000; 25:337. Milkman LA. Multiple spontaneous idiopathic symmetrical fractures. Am J Roentgenol 1934; 32:622. Steinbach HL, Kolb FO, Crane JT. Unusual roentgen manifestations of osteomalacia. Am J Roentgenol 1959; 82:863. Goldring SR, Krane SM. Disorders of calcification: Osteomalacia and rickets. In: Endocrinology, DeGroot LJ (Ed), Grune & Stratton, New York 1979. Rosenthall L. DEXA bone densitometry measurements in adults with X-linked hypophosphatemia. Clin Nucl Med 1993; 18:564. Whyte MP. Sclerosing bone disorders. In: Primer on the Metabolic Bone Diseases and Disorders of Mineral Metabolism, 8th ed, Rosen CJ (Ed), John Wiley & Sons, Inc, Ames 2013. Kurland ES, Schulman RC, Zerwekh JE, et al. Recovery from skeletal fluorosis (an enigmatic, American case). J Bone Miner Res 2007; 22:163. de Torrenté de la Jara G, Pécoud A, Favrat B. Female asylum seekers with musculoskeletal pain: the importance of diagnosis and treatment of hypovitaminosis D. BMC Fam Pract 2006; 7:4. Jin J, Sun F, Wang G, et al. [The clinical characteristics of 26 cases of hypophosphatemia osteomalacia misdiagnosed as spondyloarthritis]. Zhonghua Nei Ke Za Zhi 2014; 53:847. Lips P. Relative value of 25(OH)D and 1,25(OH)2D measurements. J Bone Miner Res 2007; 22:1668. Endo I, Fukumoto S, Ozono K, et al. Clinical usefulness of measurement of fibroblast growth factor 23 (FGF23) in hypophosphatemic patients: proposal of diagnostic criteria using FGF23 measurement. Bone 2008; 42:1235. Whyte MP. Enzyme defects and the skeleton. In: Primer on the Metabolic Bone Diseases and Disorders of Mineral Metabolism, 8th ed, Rosen CJ (Ed), John Wiley & Sons, Inc, Ames 2013. Mornet E. Molecular Genetics of Hypophosphatasia and Phenotype-Genotype Correlations. Subcell Biochem 2015; 76:25. Parfitt AM. Vitamin D and the pathogenesis of rickets and osteomalacia. In: Vitamin D, second edition, Feldman D, Pike JW, Glorieux FH (Eds), Elsevier Academic Press, San Diego 2005. p.1029. Recker RR. Bone biopsy and histomorphometry in clinical practice. In: Primer on the Metabolic Bone Diseases and Disorders of Mineral Metabolism, 8th ed, Rosen CJ (Ed), John Wiley & Sons, Inc, Ames 2013. Bingham CT, Fitzpatrick LA. Noninvasive testing in the diagnosis of osteomalacia. Am J Med 1993; 95:519. Clarke BL, Wynne AG, Wilson DM, Fitzpatrick LA. Osteomalacia associated with adult Fanconi's syndrome: clinical and diagnostic features. Clin Endocrinol (Oxf) 1995; 43:479. Allen SC, Raut S. Biochemical recovery time scales in elderly patients with osteomalacia. J R Soc Med 2004; 97:527. Thacher TD, Fischer PR, Pettifor JM, et al. A comparison of calcium, vitamin D, or both for nutritional rickets in Nigerian children. N Engl J Med 1999; 341:563. Bishop N. Rickets today--children still need milk and sunshine. N Engl J Med 1999; 341:602. (Accessed on October 29, 2015). Whyte MP, Greenberg CR, Salman NJ, et al. Enzyme-replacement therapy in life-threatening hypophosphatasia. N Engl J Med 2012; 366:904. (Accessed on July 24, 2017). Bhadada SK, Dhiman V, Mukherjee S, et al. Fibrogenesis Imperfecta Ossium and Response to Human Growth Hormone: A Potential Therapy. J Clin Endocrinol Metab 2017; 102:1750. Lee C, Lashari S. Pseudofracture of the neck of femur secondary to osteomalacia. J Bone Joint Surg Br 2007; 89:956. Docker C, Starks I, Wade R, Wynn-Jones C. Delayed fixation of displaced bilateral, atraumatic, femoral neck fractures in a patient with pregnancy related osteomalacia. Acta Orthop Belg 2011; 77:402. Henry A, Bowyer L. Fracture of the neck of the femur and osteomalacia in pregnancy. BJOG 2003; 110:329. Radhika AG, Goel M, Radhakrishnan G, et al. Severe osteomalacia presenting as numerous fractures in late pregnancy. Int J Gynaecol Obstet 2008; 100:92. Felton DJ, Stone WD. Osteomalacia in asian immigrants during pregnancy. Br Med J 1966; 1:1521. Yu CK, Sykes L, Sethi M, et al. Vitamin D deficiency and supplementation during pregnancy. Clin Endocrinol (Oxf) 2009; 70:685. Dawodu A, Saadi HF, Bekdache G, et al. Randomized controlled trial (RCT) of vitamin D supplementation in pregnancy in a population with endemic vitamin D deficiency. J Clin Endocrinol Metab 2013; 98:2337. Topic 2040 Version 19.0
15273	https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9e81a4f98389efdf:hyperb-foci/v/proof-hyperbola-foci	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
15274	https://isaacscience.org/concepts/cp_electric_field	Electric Field — Isaac Science Skip to main content About Isaac Question finder Concepts News Events Books Help Problem-solving guide Student FAQ Teacher FAQ Glossary Contact Us Sign up / log in My IsaacYou're not currently logged in. Log in or sign up for free below! Sign up / log in Explore by learning stage 11-14 11-14 Physics GCSE GCSE Physics GCSE Maths GCSE Chemistry A Level A Level Physics A Level Maths A Level Chemistry A Level Biology University University Physics University Maths University Chemistry Explore by subject Physics 11-14 Physics GCSE Physics A Level Physics University Physics Maths GCSE Maths A Level Maths University Maths Chemistry GCSE Chemistry A Level Chemistry University Chemistry Biology A Level Biology Search Search Isaac Science is the new home of Isaac Physics. If you haven't already, please read more about how the change to Isaac Science might affect you. 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Got it Concepts Concept Related concepts Electromagnetic Waves Field Potential Flux and Gauss's Law Inverse Square Laws in Physics Related questions Accelerating Particles A Level Challenge 2 Charged Massive Shell Further A Challenge 1 Charged Stars A Level Challenge 2 Charges in a Square A Level Challenge 2 Charges in Three Corners A Level Challenge 2 Current of an Orbiting Electron A Level Challenge 2 Current of an Orbiting Electron A Level Challenge 2 Inkjet Printing A Level Challenge 3 Oil Drop Experiment A Level Challenge 2 Oil Drop Experiment A Level Challenge 2 Question key Not started In progress All attempted (some errors) All correct Electric Field Coulomb's law, potential, vector fields Electric charges experience forces from other charges, for example two positive charges push apart, or repel. We explain this by saying that an electric charge sets up (or induces) an electric field around it. This electric field then affects other charges nearby, causing forces on them. For a given electric field, larger charges will experience larger forces. The force is proportional to the charge. We define the electric field strength as E=q F and as a result, electric field strength is measured in N C−1, that is newtons on each coulomb of charge. Electric field strength is a vector, and its direction is the direction in which a positive charge would be forced. The direction is shown using field lines as in the diagram below. The closer together the field lines, the stronger the field. Figure 1: Electric field patterns for charges, and between two charged surfaces. The electric field at a distance r from a small charge Q is given by the equation E=4 π ϵ 0r 2 Q. This means that the force on a charge q at this point will be F=qE=4 π ϵ 0r 2 qQ. A Level Part A A Level Basic concepts The value of the electric field at a point is the electric force per unit charge exerted at that point. It is a vector field, and points in the direction of the force that a small positive charge would feel at that point. The force acting on a particle with a charge of q at a point where the electric field is E is given by: F=q E The electric field can be visualised by drawing a series of lines going from a positive charge to a negative charge, or to or from infinity if only one charge is present. The electric field is stronger where electric field lines are closer together, and weaker where they are further apart. For a point charge the electric field gets weaker as you move further away from the charge. Figure 2: Electric field lines around a positive and negative point charge and inside a parallel plate capacitor. Part B A Level Uniform electric field Consider two identical flat conducting plates placed parallel to each other, so that the distance d between the plates is much smaller than the size of each plate. If the plates have opposite charges, then a voltage V is created between the plates. The electric field in the region between the plates is uniform, which means that the strength and direction of the field is the same everywhere between the plates. Figure 3: Electric field pattern between two surfaces with a potential difference V applied between them. The dotted lines are lines of equal electric potential. The force F on any small charge q is also the same everywhere between the plates: F=qE If the charge moves from one plate to the other, the work done by the force is: W=F×d=qE d But this work done is equal to the change in electric potential energy U, which is related to the change in electric potential V by the equation U=q V. This leads us to an expression for the electric field: V=q U=E d or E=d V From the units of the quantities in this equation, you can see that the unit of electric field strength, N C−1 is equivalent to the volt per metre, V m−1. Quick Q1 Quick Q2 Quick Q1 Quick Q2 Two parallel, charged conducting plates have a voltage of 35 V between them. There is a uniform electric field of 25 kN C−1 between the plates. What is the distance between the plates? Show answer Quick Q1 Quick Q2 Two parallel, conducting plates are separated by 1.0 cm. One plate is connected to ground with a potential of 0 V, and the other plate has a positive potential of +600 V. What is the potential between the plates, 2.5 mm away from the grounded plate? Show answer Part C A Level Coulomb's Law Coulomb's Law states that the force acting between two point charges q 1 and q 2, which are separated by a distance r in a vacuum, has magnitude F∝r 2 q 1q 2 and is directed along the line joining their centres. The constant of proportionality in the SI unit system is given by 4 π ε 01, (4 π appears due to spherical symmetry - the field is identical at any point the same distance from the charge).The numerical value of this constant is 8.99×1 0 9 N m 2 C−2, implying that ε 0=8.85×1 0−12 C 2 N−1 m−2. The magnitude of the force between two point charges becomes F=4 π ϵ 0r 2 q 1q 2 A positive F corresponds to a repulsive force, and a negative F corresponds to an attractive force. In other words, like charges repel and opposite charges attract. Figure 4 shows several separate pairs of particles and the Coulomb forces for each pair. Forces F1 and F2 form an action-reaction pair as described by Newton's 3rd Law of Motion and so have the same magnitude and opposite direction, as expected from considering Coulomb's Law. Figure 4: Coulomb Force between pairs of similar and opposite charges. Since the magnitude of the electric field at a point is equal to the force exerted per unit charge, we can derive the equation for the field strength stated earlier by dividing the equation given in Coulomb's Law by the charge of the particle. For the field made by the charge q 1, the resultant electric field acts radially and, at the location of the charge q 2, is given by E=q 2F=4 π ε 0r 2 q 1. Quick Q1 Quick Q2 Quick Q1 Quick Q2 What is the size of the electric field due to an electron at a distance of 1.0 μ m away from it? Show answer Quick Q1 Quick Q2 If another electron was 1.0 μ m away from the first electron, what force would it experience due to the first electron? Show answer Part D A Level Visualizing and adding electric fields Electric fields are vector fields. They can be visualized in two ways - either by drawing an arrow representing the electric field vector at that point, or by drawing field lines. For more details, see Vector Fields. When several charges are present, the resultant electric field at any point is the vector sum of the electric fields at that point due to each of the charges. For more detail see Adding Vector Fields. Figure 5 contains a plot of the field lines of the field arising from two equal and opposite charges. This is called a dipole. Figure 5: Electric field due to a nearby positive and negative charge. Figure adapted from original by Geek3 under CC BY-SA 3.0 Part E A Level Electric PE and electric potential The electric potential energy of a configuration of charged particles is the energy stored in the configuration due to the electrostatic interactions between the particles. For two point charges of charge q 1 and q 2, separated by a distance r, the electric potential energy is U elec=4 π ε 01r q 1q 2, It is zero when the charges are infinitely far apart. This quantity equals minus the work done by the electrostatic force exerted by q 1 on q 2. Equivalently, it equals the work that has to be done to assemble this charge configuration, which is "stored" as the potential energy of the final charge configuration. The electric field potential due to a charge q 1, at a point a distance r from q 1, is therefore V elec=q 2U elec=4 π ε 01r q 1. Part F A Level Relating electric field and electric potential The electric field at a point is related to the electric potential at that same point through the equation: E=−d rd V eleci.e.E x=−δ x δ V elec,E y=−δy δ V elec,E z=−δz δ V elec as explained in more detail here. Part G A Level Electric fields in conductors In the absence of an electromotive force (e.m.f) provided by a power supply or a changing magnetic field inside the conductor, free charges will flow inside conductors. This happens until the electric field caused by these charges inside the conductor cancels out the external electric field. Consequently, in electrostatics, there is no resultant electric field inside a conductor. If there was a field inside the conductor, the free charges would be continually moving, generating thermal energy from zero energy input. Funded by University of Cambridge. Supported by Department for Education and The Ogden Trust. About Isaac News STEM SMART Events Books Contact us Explore by subject Physics Maths Chemistry Biology Computer science Follow us All materials on this site are licensed under the Creative Commons license unless stated otherwise. Accessibility statementPrivacy policyCookie policyTerms of use
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Preferenze Statistiche Marketing Incertezza di misura: meno errori, risultati migliori Incertezza di misura: meno errori, risultati migliori \| HBM Toggle navigation +39 02 45471616 Home Contatti HBMshop \| Language \| Country \| --- \| \| English \| International Version \| \| \| \| Deutsch \| \| \| \| \| Español \| \| \| \| \| Français \| \| \| \| \| Italiano \| \| \| \| \| Português \| \| \| \| \| 中文 \| \| \| \| \| 日本語 \| \| \| \| \| 한국어 \| \| Sensori Deformazione Estensimetri Estensimetri OEM Trasduttori Ottici di Deformazione Accessori per Estensimetria Carico Celle di Carico Celle di Carico da Piattaforma Celle di Carico a Flessione Celle di Carico Digitali Elettronica industriale Forza Trasduttori di Forza Trasduttori Multi Asse Sensori piezoelettrici Coppia Torsiometri TrasduttoriMulti Asse Sensori OEM Sensori OEM personalizzati Altri sensori Sensori di pressione Sensori di spostamento Trasduttori di corrente Sonde ad alta tensione TEDS Microfoni Accelerometri Strumenti Sistemi DAQ QuantumX – universale SomatXR – robusto MGCplus – elevato numero di canali Interrogatori ottici Genesis HighSpeed Software e Driver catman – universale catman Enterprise – elevato numero di canali Perception – alta velocità Driver e APIper sistemi DAQ Driver e APIper Genesis HighSpeed e Perception Automazione aperta– Software industriale, API e soluzioni basate sul web Analizzatori di potenza eDrive and eGrid Analizzatori di potenza Alta tensione Sistemi di prova e isolamento Knowledge base sull’analisi di potenza Elettronica industriale PMX DAQ industriale ClipX Condizionatore di segnale DSE Elettronica per trasduttori digitali Elettronica di pesatura e industriale Amplificatori e trasmettitori Strumenti di precisione DMP41 Amplificatore di precisione Standard ponte Strumenti di taratura Soluzioni Servizi & Supporto Chi siamo Servizi & Supporto Consigli e Suggerimenti Misurazione della deformazione: concetti base Estensimetri: principi di base Meno errori, risultati migliori Meno errori, risultati migliori Questo articolo affronta l'argomento delle possibili fonti di errore nell'uso degli estensimetri durante le analisi sperimentali delle sollecitazioni e dimostra come valutare correttamente l'incertezza della misura già in fase di progettazione La precisione della misura nell'analisi sperimentale delle sollecitazioni La tecnologia degli estensimetri, con le sue vaste opportunità di compensazione degli errori, è stata oggetto di ottimizzazione per decenni. Eppure nelle estensimetrie, le fonti di disturbo continuano a essere numerose. Lo scopo del presente articolo è individuare le svariate (e spesso evitabili) fonti d'errore durante l'uso degli estensimetri nell'analisi sperimentale delle sollecitazioni e fornire un'assistenza che consenta di individuare l'incertezza di misura già in fase di progettazione. Questioni fondamentali per la configurazione della misurazione Prima di effettuare le misurazioni nell'analisi sperimentale delle sollecitazioni con estensimetro può essere utile riflettere sulle seguenti osservazioni, che riassumono le esperienze dell'autore. Le seguenti domande sono fondamentali per individuare gli interventi necessari (ad es. la protezione del punto di misura) e l'incertezza di misura ottenibile: Quando si raggiunge il termine della vita utile del punto di misura? Quanto saranno alti i valori di sollecitazione? Ci saranno variazioni di temperatura? Se sì, di che entità e a quale velocità? Le condizioni ambientali (acqua, umidità, ecc.) influiscono sul punto di misura? Su quali materiali è installato l'estensimetro (disomogenei, anisotropi, altamente igroscopici, ecc.)? Esiste la possibilità di tarare il punto zero, se necessario? Un ingegnere esperto cercherà le risposte già in fase di analisi del compito di misura (molto prima di installare il primo estensimetro). La risposta all’ultima domanda è ciò che determina se la misura è riferita al punto zero oppure non riferita al punto zero. Misurazioni riferite al punto zero Le misurazioni riferite al punto zero sono normalmente intese come misurazioni che implicano un confronto dei valori attualmente misurati con i valori misurati all'inizio della misurazione, nel corso di più settimane, mesi o anni. In questo arco di tempo non si prevede alcun "bilanciamento del punto zero" della catena di misura. Le misurazioni riferite al punto zero presentano molte più criticità rispetto a quelle non riferite al punto zero, in quanto le derive dello zero (per via della temperatura e altri fattori ambientali) sono comprese integralmente nel risultato della misurazione Gli errori di taratura sono particolarmente rischiosi quando i valori di sollecitazione sono ridotti, in quanto comportano derivemolto grandi rispetto al valore misurato. Le sollecitazioni nei componenti e nelle strutture di macchinari spesso non arrivano a 100 µm/m, in quanto hanno un fattore di sicurezza "integrato". In questo caso, con uno scostamento dello zero di 100 µm/m, si avrebbe un errore di misurazione pari al 100 %. Dato che la misurazione continua per il monitoraggio strutturale è quasi sempre una misurazione riferita al punto zero, occorre prestare particolare attenzione a proteggere gli estensimetri dai fattori ambientali. È fondamentale che il punto di misurazione offra una stabilità sufficiente a lungo termine. Prevedendo ampie variazioni di temperatura, i coefficienti della temperatura devono essere contenuti. Le ampiezze di segnale ridotte nelle misurazioni su componenti di grandi dimensioni probabilmente saranno sovrapposte dagli effetti derivanti da un'errata installazione dell'estensimetro. L'elettronica di misurazione risponde ad ogni variazione di resistenza con un cambiamento sul display. Ciò può essere dovuto ad una variazione delle quantità da misurare, oppure all'ingresso di molecole di acqua. Il valore effettivamente misurato, come il segnale aggregato di tutte le misure della sollecitazione sull'estensimetro, non consente di distinguere le misure desiderate da quelle indesiderate. Misurazioni non riferite al punto zero Le misurazioni non riferite al punto zero sono considerate compiti di misura che consentono il bilanciamento dello zero senza perdite di informazioni in determinati punti nel tempo. In questo caso è rilevante soltanto la variazione della quantità misurata dopo il "bilanciamento dello zero". (Le moderne bilance pesapersone vengono tarate automaticamente ad ogni accensione, senza perdite d'informazione). Il "bilanciamento dello zero" spesso è possibile con le prove di carico one-off (spesso in forma di misurazioni a breve termine), mentre la deriva dello zero è del tutto insignificante. È nelle prove distruttive che si verificano sollecitazioni fortissime, pertanto è necessario impiegare estensimetri con intervalli di misurazione adeguati. Sarebbe infatti imbarazzante, oltre che costoso, accorgersi dopo settimane di preparativi che gli estensimetri installati sui componenti non eseguano correttamente il lavoro. Le misurazioni in laboratorio e celle di prova sono considerate piuttosto affidabili, in quanto le condizioni ambientali (temperatura, umidità) sono moderate. Le misurazioni in campo e in camere ambientali con elevata umidità e ampi gradienti termici presentano invece criticità. I componenti della catena di misura Per motivi di chiarezza e comprensibilità, qui di seguito si prenderà in considerazione solamente lo stato di sollecitazione monoassiale. Il diagramma a blocchi (Fig. 1) illustra il flusso del segnale di misura, le grandezze di influenza e l’effetto delle stesse in relazione alle caratteristiche principali della catena di misura. Caratteristiche ed effetti vengono evidenziati in blu se possono influenzare il punto zero. L’oggetto della misura (DUT) Quando l’oggetto della misura in esame viene caricato, nel materiale si produce la sollecitazione σ. Ciò provoca nel materiale una deformazione che si comporta in modo inversamente proporzionale al modulo di elasticità. Questa sollecitazione del materiale può essere determinata come deformazione superficiale utilizzando un estensimetro. Il modulo di elasticità presenta un’incertezza (tolleranza del modulo di elasticità). Analisi approfondite condotte su acciai strutturali hanno evidenziato la presenza di un coefficiente di variazione del 4,5%. Il modulo di elasticità dipende anche dalla temperatura (che in questo caso rappresenta una grandezza di influenza) e dal coefficiente di temperatura del modulo di elasticità. Se l’estensimetro è incollato a una superficie (come una barra di flessione) ad elevata convessità, la deformazione sulla griglia di misura è maggiore di quella sulla superficie del componente. Ciò dipende dalla distanza dalla fibra neutra: all’aumentare della distanza tra la griglia di misura e la fibra neutra e al diminuire dello spessore del componente, il valore misurato aumenta. Lo spessore dell’adesivo e la struttura dell’estensimetro hanno un’influenza minore. La variazione di temperatura (∆t) che agisce insieme al coefficiente di espansione termica del materiale causa anche una dilatazione termica significativa per le misurazioni relative al punto zero. Le conseguenze a livello elastico (causate dai processi di rilassamento nella microstruttura del materiale) provocano una diminuzione della deformazione del materiale dopo il carico spontaneo. La formula indicata nel grafico evidenzia la presenza di numerose incertezze. Indice delle formule L'installazione La grandezza di input richiesta è la deformazione del materiale che, in condizioni ideali, è uguale alla deformazione effettiva della griglia di misura sull’estensimetro: Nella pratica, tuttavia, anche nel caso in cui si presti la massima attenzione, si verificano errori di allineamento e altri errori di installazione. L’estensimetro è un elemento elastico soggetto a sollecitazione meccanica; per questo motivo ritorna spontaneamente al suo stato iniziale dopo l’applicazione del carico di deformazione, ma anche delle proprietà reologiche dell’adesivo e del supporto dell’estensimetro stesso. Mostra inoltre una leggera isteresi. L’effetto dello scorrimento dell’estensimetro viene utilizzato nella costruzione del trasduttore per ridurre al minimo gli effetti del materiale che producono una deformazione supplementare indesiderata, regolando le lunghezze delle anse trasversali non sottoposti alla tensione esercitata sull’estensimetro. Questa compensazione può essere implementata solo con grande sforzo nell’analisi sperimentale delle sollecitazioni. Una sollecitazione eccessiva può essere dovuta anche alla presenza di una superficie di installazione curva (vedi sopra). Se i punti di misura non sono protetti in maniera adeguata da umidità e condensa, l’adesivo e il supporto possono assorbire l’umidità e gonfiarsi. Questa eventualità verrà espressa negli estensimetri come frazione di errore sotto forma di deformazione non intenzionale specifica dell’attività in corso. Il contenuto di umidità influenza anche la stabilità dei valori misurati in tutti i metodi di misurazione (vedi “estensimetro: resistenza all’isolamento”). In particolare, nel caso di misurazioni relative al punto zero, il tecnico responsabile della prova potrebbe non capire con certezza se sta osservando la deformazione del materiale rilevante o se si tratta solamente di uno degli effetti descritti in precedenza. Per questo motivo, la protezione del punto di misura è una condizione preliminare essenziale per ottenere risultati affidabili, in particolar modo nel caso di misurazioni relative al punto zero. Di conseguenza, la deformazione della griglia di misura non corrisponderà perfettamente alla deformazione del materiale nella direzione della sollecitazione. L'estensimetro L’estensimetro converte la deformazione nella griglia di misura in una variazione relativa della resistenza proporzionale alla deformazione. La tolleranza del fattore K e la sua sensibilità alla temperatura contribuiscono all’incertezza. Si prega di notare che se la deformazione non è distribuita in maniera omogenea, l’estensimetro converte il valore medio della deformazione nella griglia di misura nella variazione relativa della resistenza. Di conseguenza, se si seleziona una lunghezza attiva sbagliata dell’estensimetro, i valori misurati di deformazione e sollecitazione del materiale risulteranno troppo piccoli o troppo grandi. Ciò è di fondamentale importanza per la determinazione a livello metrologico dei valori massimi dei picchi di sollecitazione meccanica. La risposta a livello di temperatura dell’estensimetro influisce sul punto zero nel caso di grandi differenze di temperatura, in particolare se si utilizzano estensimetri che non sono stati adeguatamente adattati al coefficiente di espansione termica del materiale (DUT), in quanto tali estensimetri interferiscono con gli effetti di compensazione. L’auto-riscaldamento (dovuto all’energia elettrica trasformata all’interno dell’estensimetro) provoca effetti simili, poiché causa una differenza di temperatura tra il materiale e l’estensimetro. Per questo motivo, negli amplificatori di misura moderni è possibile impostare tensioni di eccitazione molto basse. I dispositivi sono in grado di amplificare in maniera accurata anche tensioni di uscita dal ponte molto basse. Si raccomanda tuttavia di prestare la massima attenzione nel caso di materiali molto sottili o che non dissipano facilmente il calore. Nel caso di deformazione alternata frequente con notevole ampiezza (> 1500 µm/m), il materiale della griglia di misura può essere sottoposto a fatica e provocare una deriva dello zero. Esiste una sensibilità trasversale dell’estensimetro che non produce però variazioni significative. Nello stato di sollecitazione non assiale, la sensibilità trasversale viene presa in considerazione nella determinazione sperimentale del fattore K. Per le deformazioni, una deviazione di linearità fino a 1000 µm/m è considerata trascurabile. La penetrazione di umidità e condensa riduce la resistenza di isolamento, il che a sua volta provoca una deviazione di resistenza verso le connessioni dell’estensimetro che in genere si riflette in un’instabilità nella visualizzazione dei valori misurati. Gli estensimetri a bassa resistenza sono meno sensibili all’influenza di umidità e condensa. L'Amplificatore di Misura La grandezza di input nell’amplificatore di misura è la variazione relativa della resistenza dell’estensimetro. Poiché si tratta di un valore molto piccolo (a 1000 µm/m e con un fattore K pari a 2 equivale soltanto allo 0,2 % o a 0,24 Ω su 120 Ω), nell’analisi sperimentale delle sollecitazioni si effettua un’aggiunta al ponte di Wheatstone (circuito a quarto di ponte) pari a tre resistori fissi (solitamente nell’amplificatore di misura). In questa sede non verranno analizzati i vantaggi dei circuiti a mezzo ponte o ponte intero e le diverse modalità in cui essi possono essere utilizzati per ridurre le incertezze. Verrà considerato soltanto il collegamento di un singolo estensimetro all’interno di un circuito a quarto di ponte. In genere, il rapporto tra sbilanciamento del ponte e variazione di resistenza relativa viene descritto con Il rapporto reale evidenzia un basso grado di non linearità che verrà analizzato più approfonditamente in seguito. L’amplificatore di misura fornisce tensione al circuito a ponte, amplifica la tensione in uscita dal ponte e genera il valore misurato. Si è deciso di non prendere in considerazione in questa sede gli errori di misurazione che possono verificarsi a causa di resistenze dei cavi di alimentazione, campi di interferenza, tensioni termoelettriche e componenti elettronici di misura. Tali errori possono essere completamente evitati utilizzando alcune tecnologie molto diffuse (tecnologie multi-cavo, circuiti Kreuzer estesi, metodi di schermatura, amplificatori di misura TF moderni). Tolleranza del modulo di elasticità Il modulo di elasticità (specifiche del fabbricante) presenta un’incertezza (tolleranza del modulo di elasticità) di diversi punti percentuali. La determinazione precisa del modulo di elasticità in un laboratorio idoneo è costosa e spesso non può essere applicata. Nelle misurazioni sperimentali delle sollecitazioni, oppure in quella che noi a volte chiamiamo analisi sperimentale delle sollecitazioni (ESA), l’incertezza relativa del modulo di elasticità produce un’incertezza relativa di pari entità nella sollecitazione meccanica. Ciò significa che se il materiale ha un modulo di elasticità con valore noto entro un’incertezza del 5%, solo questo produce un’incertezza del 5% nella sollecitazione meccanica. Il modulo di elasticità dipende anche dalla temperatura come quantità d’incidenza e dal coefficiente di temperatura (TC) del modulo di elasticità (per l'acciaio ≈ -2 • 10^-4/K). Il cambiamento relativo nel modulo di elasticità deriva dal prodotto: Equivalente all’incertezza aggiuntiva delle sollecitazioni meccaniche. Esempio: Se il modulo di elasticità dell'acciaio viene dato per una temperatura di 23 °C e la misurazione viene eseguita a 33 °C, il modulo di elasticità cala dello 0,2%. Se l’effetto non è compensato da calcoli, continua ad esserci una deviazione dello 0,2% oltre alla tolleranza specificata dal modulo di elasticità. Si noti che TC del modulo di elasticità dipende di per sé dalla temperatura, il che significa che questo effetto non può essere mai interamente compensato. Index of formulas Stimare l’incertezza di misura per le misurazioni non riferite al punto zero Un elemento importante di questa procedura di misurazione è che il punto zero, per analizzare i risultati di misurazione, non è necessario. Questo perché soltanto i cambiamenti nella quantità misurata sono interessanti e il punto zero non si sposta durante la misura (tipicamente per test di misurazione brevi). Ne sono un esempio i crash test, le prove di trazione e i test di carico rapido. Gli effetti a posteriori sul materiale e lo scorrimento degli estensimetri possono avere una certa importanza nelle misurazioni non riferite al punto zero, per questo motivo sono trattati in questa sezione. Dall’altro lato, i fenomeni quali l’espansione termica, il rigonfiamento dell'adesivo, la riduzione della resistenza dell’isolamento, la reazione alla temperatura dell’estensimetro e la fatica degli estensimetri nelle misurazioni non riferite al punto zero sono quasi del tutto irrilevanti. Ovviamente, la resistenza durante un breve test di carico della resistenza all’isolamento non si ridurrà in modo così drastico da permettere un guasto del punto di misurazione. Raggio di misura per gli oggetti misurati soggetti a sollecitazioni flessionali (aumento della sollecitazione) Se si colloca l’estensimetro su un componente che si flette longitudinalmente rispetto alla griglia di misura, la sollecitazione della griglia di misurazione devia dalla sollecitazione superficiale del componente (Fig. 2). I valori misurati ottenuti sono troppo grandi. Minore è il raggio di curvatura e maggiore lè a distanza della griglia di misurazione dalla superficie dei componenti, tanto maggiore è l’effetto. Se l'estensimetro viene posizionato nell’area concava, i valori misurati saranno troppo grandi semplicemente in termini di quantità. Il fattore che descrive l’errore di misurazione sarebbe lo stesso. Il risultato è una deviazione moltiplicativa in relazione al valore misurato. L’equazione di calcolo è: Per una distanza media di 100 μm dalla griglia di misurazione alla superficie del componente e un raggio di flessione di 100 mm, l’aumento risultante della sollecitazione è 1/1000 in relazione al valore reale della sollecitazione. La sollecitazione reale del componente in questo esempio è dello 0,1% inferiore alla sollecitazione misurata. Ciò significa che la deformazione misurata è troppo grande dello 0,1%. Questo errore di misurazione chiaramente è rilevante soltanto per raggi di flessione ridotti. Post-effetti elastici In molti materiali, la deformazione aumenta ulteriormente dopo un carico meccanico spontaneo. Questo fenomeno è di gran lunga completato dopo circa 30 minuti (acciaio a 23 °C) e avviene anche quando si rimuove il carico. Il quoziente della quantità di questa deformazione aggiuntiva e la deformazione spontanea dipende in grande misura dal materiale. I post-effetti del materiale producono un errore di misurazione aggiuntivo (positivo). Questo succede soltanto quando si acquisiscono i valori della deformazione. Questa deviazione può pertanto essere completamente evitata in molte operazioni di misurazione. Tuttavia, se il valore misurato viene acquisito molto tempo dopo aver applicato il carico e la deformazione del materiale è aumentata dell’1% (rispetto alla deformazione spontanea), il risultato sarà un valore misurato per la deformazione del materiale maggiore dell’1%. Disallineamento dell'estensimetro Se l’estensimetro non è perfettamente allineato in direzione della sollecitazione del materiale (stato di sollecitazione monoassiale), si produce un errore di misurazione negativo. La sollecitazione misurata sarà quindi inferiore alla sollecitazione del materiale. L’errore di sollecitazione relativo si determina come segue: Un errore di allineamento di 5 gradi e un coefficiente di Poisson di 0,3 (acciaio) producono un errore di sollecitazione di -1%. Pertanto, la sollecitazione effettiva e la sollecitazione del materiale sono maggiori dell’1%. Scorrimento dell'estensimetro Dopo aver ridotto spontaneamente la sollecitazione del materiale, la griglia di misurazione dell'estensimetro scorre un poco indietro. Il processo, determinato in modo primario dalle proprietà dell’adesivo e dalla geometria dell'estensimetro (le griglie di misurazione brevi sono critiche, mentre quelle con lunghezze di inversione molto lunghe non scorrono) dipende anche dalla temperatura. Dopo lo scorrimento di ritorno, la sollecitazione della griglia è inferiore alla sollecitazione del materiale. L’estensimetro spesso usato in ESA (HBM tipo LY11-6/120 con lunghezza della griglia di misura attiva di 6 mm), se usato con adesivo Z70 (HBM) ad una temperatura di 23 °C, ha uno scorrimento di ritorno di circa 0,1% in un’ora. Ciò equivale ad un errore di misurazione pari a -0,1% in relazione alla deformazione misurata. Ovviamente la deviazione sarà inferiore se viene determinata subito dopo il carico spontaneo. In presenza di segno negativo, lo scorrimento dell'estensimetro compensa almeno in parte i post-effetti elastici, pertanto può essere spesso ignorato completamente in ESA. Tuttavia, si raccomanda cautela quando si usano altri adesivi a temperature più alte. Ad esempio, con l’adesivo X60 (HBM) applicato a 70 °C con sollecitazione di 2000 μm/m, la deviazione risultante dopo appena un’ora è di -5%. Isteresi dell'estensimetro Lo stesso si applica all’isteresi: griglie di misurazione brevi tendono ad essere critiche e l’adesivo ha alcuni effetti. L’isteresi dell’estensimetro LY11-6/120 è di appena 0,1% con sollecitazione di ±1000 μm/m se si utilizza Z70 come adesivo. Pertanto è trascurabile. Se invece deve essere usato un estensimetro molto piccolo (LY11-0,6/120) con lunghezza della griglia di misurazione attiva di 0,6 mm, l’isteresi aumenta, e con essa anche l’incertezza dimisura della sollecitazione o deformazione a 1%. Il fattore estensimetro Tolleranza del fattore estensimetro Si assume una regolazione precisa della catena di misura al valore nominale del fattore estensimetro (come specificato dal fabbricante sulla confezione dell'estensimetro). Questo fattore descrive il rapporto tra il cambiamento di sollecitazione e il cambiamento della resistenza relativa. È stato determinato in modo sperimentale dal produttore. L’incertezza del fattore estensimetro è solitamente dell’1%. Il fattore estensimetro è anche specificato sulla confezione. Produce lo stesso grado di incertezza relativa sia nelle misure delle sollecitazioni che della deformazione. Coefficiente di temperatura (TC) del fattore estensimetro Il fattore estensimetro dipende dalla temperatura. Il segno e la quantità della dipendenza sono determinati dalla lega della griglia di misurazione. Il fatto che TC del fattore estensimetro sia di per sé dipendente dalla temperatura può essere ignorato allo scopo dell’ESA. Il TC per una griglia di misurazione fatta di costantana è di circa 0,01% per Kelvin. Pertanto, il fattore estensimetro diminuisce dello 0,1% con un aumento della temperatura di 10 K, un dato in linea di massima trascurabile. Eseguendo le misurazioni a 33 °C, i valori di sollecitazione o deformazione subirebbero una deviazione verso l’alto di appena 0,1%. Ad ogni modo, a 120 °C sarebbe di 1%, un valore degno di nota. Lunghezza della griglia di misurazione Secondo la consueta definizione, un estensimetro integra le sollecitazioni sotto la superficie attiva. Se il campo di deformazione sotto quella superficie non è omogeneo, il cambiamento relativo della resistenza non corrisponderà alla maggiore sollecitazione locale, ma piuttosto alla sollecitazione media sotto la griglia di misurazione attiva. Si tratta di un fattore cruciale, perché sono proprio le sollecitazioni maggiori a destare interesse. I valori misurati dunque deviano verso il basso dai valori massimi desiderati, comportando deviazioni negative. Dato che è un fenomeno ben noto ed esistono contromisure adatte (griglia di misurazione corta), nelle applicazioni pratiche è raro che si verifichino grandi errori. Ad ogni modo, vediamo un esempio: si applica la misurazione ad una deformazione flettente all’inizio del paranco. L’estensimetro acquisisce la sollecitazione media con la propria griglia di misurazione (fig. 3). Le sollecitazioni si comportano come le deformazioni: Il massimo valore di deformazione effettivamente ricercato potrebbe essere trovato in questo semplice caso esemplificativo con un calcolo di correzione. In caso contrario si avrà una deviazione del risultato di misurazione dalla deformazione massima. La deviazione relativa è: Se si utilizza una griglia di misurazione con lunghezza attiva inferiore al 2% di l2 nell'esempio sopra, la deviazione scende a meno dell’1% del valore misurato. In sostanza, il rapporto della sollecitazione massima e della sollecitazione misurata dipende sempre dalla distribuzione della sollecitazione sotto la griglia di misurazione. Se questo dato è noto grazie a un calcolo a elementi finiti, è possibile calcolare il valore massimo desiderato a partire dalla media aritmetica della deformazione. Ovviamente ci saranno deviazioni nei casi in cui l'estensimetro non sia posizionato correttamente. Si tratta di una situazione evitabile, che va pertanto evitata. Deviazioni di linearità Deviazione di linearità dell'estensimetro Gli estensimetri con materiali della griglia di misurazione adatti (costantana, karma, cromo-nichel V, platino-tungsteno) presentano una linearità eccellente. Ciononostante, per grandi sollecitazioni è possibile dimostrare deviazioni apprezzabili nelle griglie di misurazione in costantana. La curva caratteristica statica effettiva può essere descritta in modo molto preciso (empiricamente) con un’equazione al quadrato: Se le sollecitazioni fossero individuate con il rapporto non ci sarebbe alcuna deviazione di linearità. Tuttavia, dato che il componente quadratico è semplicemente trascurato nelle applicazioni pratiche, l’errore risultante dovrebbe essere indicato qui. La deviazione relativa del valore della sollecitazione individuato dal valore effettivo è ampia come la sollecitazione stessa: per sollecitazioni fino a 1000 μm/m, il valore della deviazione relativa della sollecitazione non supera lo 0,1%. Questo equivale a 1 μm/m, un valore trascurabile. La deviazione dalla linearità diventa percettibile solo con sollecitazioni maggiori: con 10.000 μm/m risulta dell’1% con 100.000 μm/m risulta del 10% In gran misura fortunatamente questo comportamento è compensato dalla deviazione di linearità del circuito con quattro resistori. Deviazione di linearità del circuito con quattro resistori I piccoli cambiamenti relativi nella resistenza sono solitamente analizzati con un circuito con ponte di Wheatstone. Come indicato sopra, solitamente si usa un solo estensimetro per punto di misurazione in ESA. Pertanto, le altre resistenze del ponte sono indipendenti dalla sollecitazione. Il rapporto corretto per il rapporto di deformazione in questo caso è: Nonostante il rapporto sia non lineare, nelle misurazioni pratiche (che sia nota o meno)si presuppone la linearità e viene utilizzata l’equazione di approssimazione La deviazione relativa derivante da questa semplificazione può essere calcolata con eq. Una sollecitazione di 1000 μm/m (con k = 2) determinaun cambiamento pari a 0,2% nella resistenza relativa. L’errore di misurazione relativo secondo quanto determinato con eq. 17 è -0,1%. Questo equivale ad una deviazione assoluta di -1 μm/m. La deviazione dal valore reale è trascurabile. A sollecitazioni maggiori si ottengono deviazioni di linearità sensibili, come indicato sopra: 10.000 μm/m determina una deviazione di -1% 100.000 μm/m determina una deviazione di -9.1%. Quando si usano estensimetri in costantana (non linearità simile in termini di grandezza, ma con il segno opposto), le due deviazioni si annullano reciprocamente, quindi non devono essere più considerate. Si noti tuttavia che non ci sono compensazioni sempre del tutto riuscite, soprattutto per il fatto che il fattore estensimetro devia leggermente dal 2 e che la curva caratteristica attuale statica non corrisponde con esattezza all’eq. empirica 12. Riepilogo delle incertezze parziali Le incertezze individuali sono difficili da collegare tra loro. Tuttavia, nella misura in cui possono essere collegate (post-effetti del materiale e scorrimento estensimetro, deviazione della linearità dell’estensimetro e ponte a quattro resistori), i loro effetti entro una certa misura si annullano. Pertanto, è ammissibile combinare le incertezze individuali con il metodo root sum square. I valori indicati sopra in grassetto sono usati per ottenere un risultato per l'esempio. L’incertezza della misura della sollecitazione è inferiore al 3%. La misura della sollecitazione raggiunge circa il 6% del valore misurato. Questa percentuale, moltiplicata per il valore misurato, fornisce la deviazione in μm/m o N/mm2. L’incertezza del modulo di elasticità è solitamente responsabile della maggior parte degli errori nelle misurazioni non riferite al punto zero in ESA. Per le misurazioni riferite al punto zero occorre considerare incertezze aggiuntive. La valutazione dell’incertezza di misura per le misurazioni relative al punto zero In queste misurazioni, il punto zero è importante. Si tratta essenzialmente di misurazioni a lungo termine su edifici e di prove di fatica su componenti. Se il punto zero varia durante compiti di misura di questo tipo, il risultato è un ulteriore errore di misura. Alle incertezze di misura evidenziate nella presente sezione devono essere aggiunte quelle già trattate nell’ultima parte della presente serie. Rigonfiamento dell’adesivo e del supporto della griglia Il rigonfiamento è dovuto principalmente all’elevata mobilità delle molecole dell’acqua e alle proprietà igroscopiche degli adesivi e dei materiali del supporto. L’effetto ottenuto è una deriva zero non chiaramente identificabile (o distinguibile dalle sollecitazioni del materiale). Potrebbe raggiungere valori elevati. Viene misurata una sollecitazione che non esiste, perlomeno nel componente analizzato. Questa sollecitazione parassita è solo parzialmente reversibile. Purtroppo, non è possibile “soffiare via” le molecole d’acqua. La velocità di deriva del valore misurato dipende dalla protezione del punto di misura e dalle condizioni ambientali. La costante temporale può rientrare in un intervallo di molte ore. Due fattori particolarmente critici sono la temperatura elevata e l’umidità relativa elevata. Purtroppo in questa sede non è possibile fornire formule o dati concreti. Resistenza di isolamento Anche i residui di materiale decapante possono assorbire le molecole d’acqua. Nelle applicazioni pratiche questo appare come una “visualizzazione pulsante” spesso distinguibile dai valori misurati fluttuanti per via di una corrente d'aria o di una causa simile. Un addetto alle prove esperto riconoscerà il segnale di allarme e provvederà a pulire meticolosamente tutti i punti di contatto. In alcuni casi è anche possibile “asciugare” il residuo. Tuttavia, tutte queste contromisure possono essere effettuate solamente se le parti umide non sono già chiuse sotto al coperchio di protezione del punto di misura, come spesso avviene per motivi più che validi. Se il punto di misura è stato predisposto per essere coperto, può essere utile scaldarlo di alcuni gradi Kelvin rispetto alla temperatura ambiente prevalente e poi coprirlo immediatamente. Ciò escluderà la possibilità che in seguito si formi condensa sotto al coperchio. Se le resistenze di isolamento sono troppo basse, si verificherà la deriva a zero dei valori misurati. Le resistenze di isolamento all’interno del circuito a ponte rivestono un’importanza critica in questo caso. L’isolamento elettrico non efficace dei contatti dell’estensimetro tra loro ha un effetto simile a quello di un derivatore di resistenza. Non può essere misurato direttamente, ma è per natura simile in ampiezza a quello della resistenza di isolamento. Il rapporto tra deformazione apparente e derivatore di resistenza è espresso come segue: Questa equazione mostra che l’effetto è inferiore se si utilizzano estensimetri a resistenza elevata. Gli errori di misura seguenti sono definiti per estensimetri da 120 Ω (fattore di scala = 2): In circostanze “normali”, è possibile raggiungere resistenze di isolamento maggiori di 50 MΩ; gli scostamenti inferiori a1,2 μm/msono considerati trascurabili. A 500 kΩ e con un valore misurato pari a 1000 μm/m, l’errore zero sarebbe già pari a -12%! Ciò mostra chiaramente che un calo significativo delle resistenze di isolamento potrebbe causare un errore di misurazione. I trasduttori a estensimetri hanno resistenze di isolamento pari a molti GΩ. Valori elevati di umidità relativa combinati con valori di temperatura elevati (ad esempio, vapore saturo) causano l’aumento della pressione del vapore acqueo. Le minuscole molecole di acqua vengono spinte in avanti e superano gradualmente la protezione del punto di misura. È impossibile prevedere senza effettuare un test quando avverrà l’errore di misurazione nel punto di misura (se entro pochi giorni o molti anni). Fatica Durante il carico dinamico del componente compaiono alcuni segni di fatica che vengono espressi in una deriva dello zero (deformazione apparente nel materiale). Maggiore è l’ampiezza delle deformazioni che si alternano, maggiore è il numero di cicli di carica e di conseguenza anche l’effetto (Fig. 5). Anche l’installazione e la media aritmetica influenzano la deriva dello zero. Se la media è negativa, la durata a fatica migliora; se è positiva, peggiora. Per deformazioni alternate con ampiezza fino a 1000 μm/m, non dovrebbe verificarsi alcuna deriva dello zero. Per ampiezze maggiori, la situazione è più critica. Ci si può aspettare un errore zero pari a 10 μm/m nei seguenti casi: 1500 μm/m e circa 2 milioni di cicli di carico 2000 μm/m e circa 100.000 cicli di carico 2500 μm/m e circa 4000 cicli di carico 3000 μm/m e circa 100 cicli di carico Anche il campione usato per il test è sottoposto a fatica. Se la sua resistenza alle deformazioni alternate è maggiore di quella della lamina dell’estensimetro, potrebbe essere utile utilizzare estensimetri ottici (con reticolo a fibra di Bragg). Fig. 5: Dipendenza della deriva dello zero dall’ampiezza della deformazione e dal numero di cicli di carico. Riepilogo di tutte le incertezze parziali Mentre gli scostamenti trattati nellaparte 3 parte 3 della presente seriesi moltiplicano e sono indicati come percentuale del valore misurato, gli scostamenti trattati in questa sezione si sommano. La loro unità di misura è μm/m e sono praticamente indipendenti dal valore misurato. Se lo scostamento relativo viene calcolato con l’equazione, l valore è paragonabile a quelli della parte 3. Se i valori del tipo in grassetto qui sopra vengono combinati con la somma pitagorica, il risultato è pari a 16,01 μm/m. Dal momento che le incertezze di misura non possono essere arrotondate, l’incertezza per il punto zero è pari a 17 μm/m. Con una deformazione di 1000 μm/m, lo scostamento espresso in percentuale è pari a 1,7%, il che è sicuramente ragionevole. È evidente che questo valore assume un’importanza critica nel caso di deformazioni ridotte: 17 μm/m di 100 μm/m è già pari al 17%. Ora l’incertezza del punto zero (1,7% o 17%) deve ancora essere aggiunta all’incertezza indicata nella parte 3 (3% per la misurazione della deformazione). Il risultato della somma pitagorica è: 4% con un valore misurato di 1000 μm/m, 18% con un valore misurato di 100 μm/m. Solitamente la sollecitazione meccanica rappresenta la grandezza effettivamente misurata, quindi la sua incertezza deve essere stimata. L’incertezza della misura delle sollecitazioni calcolata nella parte 3 è pari al 6%. Se aggiungiamo l’incertezza del punto zero (1,7% o 17%) con la somma pitagorica, il risultato sarà: 7% con una deformazione di 1000 μm/m, 19% con una deformazione di 100 μm/m. Con compiti di misura relativi al punto zero, in particolare con deformazioni minime, si verificano errori di misurazione relativa consistenti. Estensimero installato su una struttura in acciaio. Estensimetro installato sulla testa del rotore di un elicottero. Possibili installazioni degli estensimetri Estensimetro installato su un binario. I punti di misura dell’estensimetro presente sulla piattaforma di ricerca FINO 1 vengono preparati per uso subacqueo nel Mare del Nord. Estensimetro installato su materiale composito (scheda di circuito stampato). L’importanza di un’installazione corretta Finora abbiamo sempre dato per scontato che l’installazione del punto di misura fosse pianificata correttamente ed eseguita ad arte. Per questo motivo, solamente alcuni degli scostamenti singoli presenti negli esempi precedenti superavano l’intervallo stabilito. È però purtroppo necessario precisare che se l’installazione viene eseguita in modo non corretto, gli errori di misurazione possono assumere valori estremamente alti. Provate a immaginare di utilizzare un estensimetro molto lungo per misurare le sollecitazioni all’intaglio oppure di avere resistenze di contatto all’estensimetro variabili dello 0,24 Ω (equivalenti a un errore di deformazione di 1000 μm/m per un estensimetro da 120 Ω). L’importanza della protezione del punto di misura non può essere trascurata, in particolare nel caso di misurazioni relative al punto zero. Un ottimo esempio a questo riguardo è costituito dai 44 punti di misura dell’estensimetro sulla piattaforma di ricerca FINO 1 (altezza globale 129 m) nel Mare del Nord (45 km a nord dell’Isola di Borkum). Gli estensimetri sono situati da 5 a 25 m sotto alla superficie dell’oceano. La loro funzione era quella di misurare le deformazioni da carico sulla struttura di supporto della piattaforma causate dai battipali e dall’effetto delle onde e del vento. Dopo due anni nelle acque del Mare del Nord, 42 punti di misura erano ancora perfettamente funzionanti. Un altro errore grossolano si verifica se l’estensimetro ha solamente una connessione interna parziale con la superficie o con il componente esaminato. Le possibili cause sono: scarsa pulizia o movimentazione non corretta della superficie di applicazione e dell’adesivo sovrapposto. Queste cause devono e possono essere evitate. In genere, il test della gomma da cancellare chiarifica la situazione. Anche se è possibile evitare la protezione del punto di misura per misurazioni a breve termine (prova di trazione), l’installazione degli estensimetri richiede un approccio attento e spesso una buona dose di esperienza. In nessun altro metodo di misurazione, la conoscenza e l’esperienza dell’installatore giocano un ruolo così importante. Per questo motivo, le aziende e gli istituti sono sempre più impegnati nelle attività di certificazione del proprio personale in accordo alle norme VDI/VDE/GESA 2636 relative a diversi livelli di qualificazione. Fotografia e disegno della piattaforma di ricerca FINO 1, per gentile concessione di GL Garrad Hassan. Prodotti correlati Strain Gauges Riferimento per analisi sperimentale delle sollecitazioni Karl Hoffmann descrive le nozioni fondamentali delle misurazioni con estensimetro, dalla scelta del sensore all’analisi e valutazione dei dati acquisiti. Estensimetri: principi di base Principi di base della misurazione delle deformazioni in riferimento a estensimetri e analisi sperimentale delle sollecitazioni. QuantumX MX1615B/MX1616B L’amplificatore per estensimetri ideale misurare tutte le grandezze meccaniche: per tutti i sensori con configurazione full-bridge o quarter-bridge Chi siamo Dal 1950 HBM è considerata leader nella produzione di strumenti per test e misurazioni, precisi e affidabili. Con filiali in 30 paesi nel mondo, i nostri clienti ottengono risultati di cui possono fidarsi. Per saperne di più Prodotti Estensimetri Celle di Carico Trasduttori di Coppia Trasduttori di Forza Strumenti di Misurazione Software di Acquisizione Dati Supporto & Assistenza Accedi / Il mio account Supporto Download Know-How Webinar & Formazione Opportunità di Lavoro Contatti Contatta HBM Ufficio commerciale: +1 800-578-4260 HBK USA Scopri le ultime tendenze nel settore della misurazione e accedi ai nostri esclusivi eventi e promozioni email Dettagli della Società P.IVA 03737590962 Privacy Notice Termini per l'uso del Sito Web Cookie Notice Anti-Slavery Statement Incertezza di misura: meno errori, risultati migliori - © HBK X $h2 $hl X
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MEMBERS ONLY Recipes Mark Bittman’s Summer Menu for Two MEMBERS ONLY MEMBERS ONLY Declutter What to Do With Your Unwanted Stuff MEMBERS ONLY Back Auto Close Menu Car Buying Driver Safety Maintenance & Safety Trends & Technology MEMBERS ONLY Maintenance Get More Mileage Out of Your Car MEMBERS ONLY We Need To Talk Assess Your Loved One's Driving Skills Driver Safety AARP Smart Driver Course Back Staying Sharp Close Menu Assessment Challenges Articles Videos Recipes Building Resilience in Difficult Times Tips for Finding Your Calm Weight Loss After 50 Challenge Back Podcasts Close Menu Cautionary Tales of Today's Biggest Scams 7 Top Podcasts for Armchair Travelers Jean Chatzky: ‘Closing the Savings Gap’ Back Videos Close Menu Quick Digest of Today's Top News AARP Top Tips for Navigating Life Get Moving With Our Workout Series AARP AARP States New York Money New York State Taxes: What You’ll Pay in 2025 By Elissa Chudwin, Rocky Mengle, February 14, 2025 08:33 AM The top income tax rate in New York is among the highest in the nation. New York also has some of the highest sales and property tax rates in the country, though rates vary depending on where you live within the state. The big picture: Income tax: 4 percent to 10.9 percentNew York has nine tax brackets that range from 4 to 10.9 percent. Residents of New York City and Yonkers also pay local income tax. Property tax: 1.54 percent of a home’s assessed value (average) Real estate taxes vary widely by county and municipality across New York, with an average tax rate of 1.54 percent of a home’s assessed value in 2022, according to the Tax Foundation. Sales tax: 8.53 percent (average combined state and local)A 4 -percent state sales tax is levied across New York, in addition to local state sales tax as high as 4.875 percent in local state sales tax. The state’s average combined sales tax rate is 8.53 percent, according to the Tax Foundation. How is income taxed in New York? The state’s nine tax brackets are listed below. New York City and Yonkers both add local taxes on income on top of state income tax.Note that all of your income is not taxed at the same rate. For example, if you’re a single filer who earned $80,000, the first $8,500 would be taxed at 4 percent. The next $8,500 to $11,700 would be taxed at 4.5 percent and so on.Single filers and married couples filing separately: \| \| \| --- \| \| Income \| Tax Rate \| \| $0 to $8,500 \| 4% \| \| Over $8,500 to $11,700 \| $340, and 4.5% of income over $8,500 \| \| Over $11,700 to $13,900 \| $484, and 5.25% of income over $11,700 \| \| Over $13,900 to $80,650 \| $600, and 5.5% of income over $13,900 \| \| Over $80,650 to $215,400 \| $4,271, and 6% of income over $80,650 \| \| Over $215,400 to $1,077,550 \| $12,356, and 6.85% of income over $215,400 \| \| Over $1,077,550 to $5,000,000 \| $71,413, and 9.65% of income over $1,077,550 \| \| Over $5,000,000 to $25,000,000 \| $449,929, and 10.3% of income over $5,000,000 \| \| Over $25,000,000 \| $2,509,929, and 10.9% of income over $25,000,000 \| Source:New York State Department of Taxation and FinanceJoint filers and surviving spouses: \| \| \| --- \| \| Income \| Tax Rate \| \| $0 to $17,150 \| 4% \| \| Over $17,150 to $23,600 \| $686, and 4.5% of income over $17,150 \| \| Over $23,600 to $27,900 \| $976, and 5.25% of income over $23,600 \| \| Over $27,900 to $161,550 \| $1,202, and 5.5% of income over $27,900 \| \| Over $161,550 to $323,200 \| $8,553, and 6% of income over $161,550 \| \| Over $323,200 to $2,155,350 \| $18,252, and 6.85% of income over $323,200 \| \| Over $2,155,350 to $5,000,000 \| $143,754, and 9.65% of income over $2,155,350 \| \| Over $5,000,000 to $25,000,000 \| $418,263, and 10.3% of income over $5,000,000 \| \| Over $25,000,000 \| $2,478,263, and 10.9% of income over $25,000,000 \| Source: New York State Department of Taxation and FinanceHead-of-household filers: \| \| \| --- \| \| Income \| Tax Rate \| \| $0 to $12,800 \| 4% \| \| Over $12,800 to $17,650 \| $512, and 4.5% of income over $12,800 \| \| Over $17,650 to $20,900 \| $730, and 5.25% of income over $17,650 \| \| Over $20,900 to $107,650 \| $901, and 5.5% of income over $20,900 \| \| Over $107,650 to $269,300 \| $5,672, and 6% of income over $107,650 \| \| Over $269,300 to $1,616,450 \| $15,371, and 6.85% of income over $269,300 \| \| Over $1,616,450 to $5,000,000 \| $107,651, and 9.65% of income over $1,616,450 \| \| Over $5,000,000 to $25,000,000 \| $434,163, and 10.3% of income over $5,000,000 \| \| Over $25,000,000 \| $2,494,163, and 10.9% of income over $25,000,000 \| Source: New York State Department of Taxation and FinanceAll residents, regardless of your filing status, who have an adjusted gross income over $107,650 also pay a supplemental tax. This amount is calculated based on your earnings and your tax-filing status (single, married filing jointly, married filing separately, head of household, or qualifying surviving spouse). Check New York Form IT-201 (for residents) and Form IT-203 (for nonresidents/part-year residents) for filing details.Watch the video below to learn how to identify your 2024 federal income tax brackets. Understanding Your 2024 Income Tax Are pensions or retirement income taxed in New York? Yes, money withdrawn from pensions and 401(k)s, 403(b)s and IRAs are combined and generally taxed as regular income to the same extent they’re taxed at the federal level. Tax rates run from 4 percent to 10.9 percent.But federal and New York state government pensions and military retirement pay are tax-exempt.For those 59½ or older, the first $20,000 of retirement income (from a corporate pension, an IRA, a 401(k) account or another retirement plan) is tax-exempt. If you are married, each spouse is eligible for the $20,000 exclusion, for a total of $40,000.AARP’s retirement calculator can help you determine if you are saving enough to retire when — and how — you want. What about investment income? Capital gains from investments (including proceeds from property sales) are treated as ordinary personal income and are taxed at the same rates. Does New York tax Social Security benefits? No, but you may payfederal taxes on a portion of your Social Security benefits, depending on your "provisional income." In most cases, provisional income is equal to the combined total of half your Social Security benefits, your adjusted gross income (not including any Social Security benefits) and any tax-exempt interest for the year.Up to 50 percent of your benefits will be taxed if your provisional income is $25,001 to $34,000--or if you file jointly and your provisional income is $32,001 to $44,000.Up to 85 percent of your benefits will be taxed if your provisional income is more than $34,000 individually or more than $44,000 as a couple.AARP’s Social Security calculator can assist you in determining when to claim your Social Security benefits and how to maximize them. How is property taxed in New York? Property tax in New York is a local tax that’s based on the value of your home. The average rate is 1.54 percent of the assessed value of your home, according to 2022 data from the Tax Foundation.Note that property taxes vary widely by county and municipality across the state. The median property taxes paid exceed $10,000 in six New York counties — Nassau, New York, Putnam, Rockland, Suffolk, and Westchester — according to Tax Foundation data. Hamilton County has the lowest, at $2,197.Learn how your home value is assessed on the New York Department of Taxation and Finance website — and even contest it. New York City residents can learn about how their property is assessed and calculate their property taxes at the New York City Department of Finance website. The state does not collect tax on personal property, such as cars, boats or jewelry. What about sales tax and other taxes? Sales tax: Many, but not all, consumer goods and services are taxed at 4 percent statewide in addition to local sales tax as high as 4.875 percent.The average combined rate is 8.53 percent, according to the Tax Foundation. Find your local tax rate at the New York Department of Taxation and Finance’s website. Groceries: The state doesn’t tax most groceries, clothing valued under $110, prescription or nonprescription drugs, medical equipment and certain medical care services. Find a list of taxed and tax-exempt goods and services on the state website. Gas: There is a 8.05 cent-per-gallon tax on motor fuel in addition to state sales tax. Alcohol: Beer is taxed at 14 cents per gallon. Wine is taxed at 30 cents to $6.44 per gallon, depending on the percentage of alcohol. The tax on liquor is either 67 cents or $1.70 per liter, depending on the percentage of alcohol. New York City charges an additional 12 cents per gallon of beer and an additional 26 cents per liter of alcohol or wine with higher percentages of alcohol. State excise taxes on alcohol are paid by the vendor, but some or all may be included in the retail price. Commuter tax: A tax is levied on certain employers and self-employed people conducting business in the New York City metro area, which includes these boroughs and counties: the Bronx, Brooklyn, Dutchess, Manhattan, Nassau, Orange, Putnam, Queens, Rockland, Staten Island, Suffolk and Westchester. The tax rate for employers ranges from .11 percent to .6 percent. For self-employed people, the rate is either .34 percent or .6 percent. Lottery: Lottery winnings are subject to local, state and federal taxes, but the amount varies depending on where you live and if you've moved to another state. More information is available at Publication 140-W. Will I or my heirs have to pay inheritance and estate tax in New York? While there is no inheritance tax in New York, estates are taxed at rates ranging from 3.06 percent to 16 percent after the current exemption of $7,160,000. Are there any tax breaks for older New York residents? In addition to the Social Security exemption mentioned earlier, New York offers a few other tax breaks to help older residents. Pension income exclusion: As noted above, the state offers an income tax exemption on the first $20,000 of pension and annuity income — up to $40,000 for married couples — for those 59½ or older. Real property tax credit: An income tax credit is available to qualified residents who have a household income of $18,000 or less and pay real property taxes or rent for their residences. For most people, the annual credit tops out at $75. However, if at least one member of your household is 65 or older, the credit can be as much as $375. Enhanced STAR benefits: Under the School Tax Relief (STAR) program, eligible homeowners can get either an exemption or credit for school district property taxes. However, residents age 65 or older with incomes below a certain level qualify for increased STAR benefits. The income threshold is $98,700 for the 2024-2025 school year ($107,300 for the 2025-2026 school year). Senior citizens exemption: Eligible residents can reduce the assessed value of their home by as much as 50 percent to lower their property taxes. You must be 65 or older and meet certain income limitations as well as other requirements. Note that each county, city, town, village or school district also sets a maximum income limit between $3,000 and $50,000 for that 50 percent exemption. Municipalities may also use a sliding scale for partial tax relief for seniors with incomes higher than the local limit. Senior citizen rent increase exemption: Local governments can adopt property tax exemptions that essentially freeze rent payments for low-income residents who are at least 62 years old and live in a rent-controlled apartment. If there’s a rent increase, the older adult’s rent stays the same and the landlord receives a property tax credit equal to the rent increase. To qualify, the older person’s income can’t exceed a maximum amount set by local law (not to exceed $50,000). This exemption expires on June 30, 2026. Are military benefits taxed in New York? New York does not tax military pensions. Active-duty pay is taxed like normal income if you are a resident of the state. You are not considered a resident and do not pay taxes on military pay if you weren’t living in New York full-time when you entered the military.Military spouses may be eligible for certain tax benefits under the federal Service members Civil Relief Act, including income tax exemption and an option to use the same state of residency as the service member. What is the deadline for filing New York state taxes in 2025? The deadline to file a New York state tax return is April 15, which is also the deadline for federal tax returns. For help estimating your annual income taxes, use AARP’s tax calculator.New Yorkers who are required to file a state income tax return but need more time should apply for an extension on or before the April 15 deadline. Details on how to get a six-month extension (until Oct. 15) are available on the state taxation department’s website.Even with an extension, New York filers will still be subject to late tax payment penalties if they don’t pay any tax due by April 15. Filing for an extension by April 15 only eliminates late filing penalties. If you fail to file your tax return by Oct. 15, you will be subject to both penalties in addition to your initial tax payment.Editor's note: This guide was originally published on Dec. 16, 2022, and has been updated to reflect new information.Elissa Chudwin covers federal and state policy and writes the podcastToday’s Tips from AARP. She previously worked as a digital producer forThe Press Democratin Santa Rosa, California, and as an editor forAdvocatemagazines in Dallas.Rocky Mengle has more than 25 years of experience covering federal and state tax developments for CPAs, tax attorneys, and ordinary taxpayers, including stints at Kiplinger Personal Finance and Wolters Kluwer Tax & Accounting. Also of Interest: 10 Things You Need to Know Before Filing Your 2024 Tax Return 9 States With No Income Tax How to Get Free Help With Your Taxes taxnew yorkretirement income taxSale and Property Tax About AARP New York Contact information and more from your state office. Learn what we are doing to champion social change and help you live your best life.
15277	https://math.stackexchange.com/questions/4782581/how-do-derivatives-even-work	implicit differentiation - How do derivatives even work? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How do derivatives even work? Ask Question Asked 1 year, 11 months ago Modified1 year, 11 months ago Viewed 138 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I'm a student currently looking into partial derivatives and was writing notes for a simple example. find the derivative with respect to x:x 2+x y−y 2=8 find the derivative with respect to x:x 2+x y−y 2=8 I got to the point where I was differentiating the −y 2−y 2 term, I wrote d(−y 2)d y d y d x d(−y 2)d y d y d x. This got me thinking, with my low level knowledge of what derivatives even are. Why does it not work out when I take out a factor of y and cancel terms like this −y d y d y d y d x−y d y d y d y d x Any knowledge and deep understanding is really appreciated! derivatives implicit-differentiation learning Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Oct 7, 2023 at 16:22 Nathan WilliamsNathan Williams 1 4 You can't take out a factor of y y since d(y 2)=(y+d y)2−y 2=2 y d y+d y 2=2 y d y d(y 2)=(y+d y)2−y 2=2 y d y+d y 2=2 y d y Mason –Mason 2023-10-07 16:30:32 +00:00 Commented Oct 7, 2023 at 16:30 Because (f g)′=f′g+f g′(f g)′=f′g+f g′, and not f(g)′f(g)′. I.e. you can't take a factor out of a derivative when that factor depends on the thing you're differentiating with respect to. Also, with the notation you're using, you're not doing partial derivatives - you're doing implicit differentiation, and that assumes that y y is a function of x x.JonathanZ –JonathanZ 2023-10-07 16:30:48 +00:00 Commented Oct 7, 2023 at 16:30 1 Are you asking what is the correct method to find d y d x d y d x in this problem? Or if not, what are you asking? In general, if you want to know whether some alternate method works, you try it out on an example (such as this one), and you compare to the answer obtained from the correct method; if the answers don't match, then you know that the alternate method does not work.Lee Mosher –Lee Mosher 2023-10-07 16:31:07 +00:00 Commented Oct 7, 2023 at 16:31 d(−y 2)/d y d(−y 2)/d y isn't −y⋅d y/d y−y⋅d y/d y. It's close, but it's not quite right. As for what derivatives "really" are, there are a few different perspectives. Don't worry; your confusion with Leibniz notation is extremely common. Keep challenging and trying to understand how to prove calculus manipulations are valid, don't just push symbols around and hope for the best FShrike –FShrike 2023-10-07 16:34:11 +00:00 Commented Oct 7, 2023 at 16:34 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. (This is more a comment than an answer, as it's asking for clarification, but it's too long for a comment.) Is that the full text and context of an example that was assigned to you? Because I would describe that as "missing a boat-load of context". See, there are (at least) two ways to view that formula. Define f(x,y)=x 2+x y−y 2 f(x,y)=x 2+x y−y 2. Differentiate f f w.r.t. x x. This version would use the notation for partial derivatives (which you mentioned), but that's ∂f∂x∂f∂x, not d y d x d y d x, which is what you've written. Although just what the "=8=8" would be doing in there in this case I'm not sure. And in this case, ∂(y 2)∂x∂(y 2)∂x is actually equal to 0 0, as y y doesn't depend on x x! Assume the equation x 2+x y−y 2=8 x 2+x y−y 2=8 defines y y implicitly in terms of x x. Compute d y d x d y d x. In this version you would think of y y as a function of x x, i.e. y(x)y(x), and it should be clear that in this case you can't "factor y y out" anymore than your could, if, say, g(x)g(x) were a function of x x, write d(g⋅g)d x=g⋅d g d x d(g⋅g)d x=g⋅d g d x Distinguishing between versions 1 and 2 depends on making clear whether we are in a situation where x x and y y can vary freely and independently, or one where a change in x x induces a change in y y. The instruction "find the derivative with respect to x x" does nothing to help us understand which of the situations we are in, and the use of the description "partial derivatives" and the notation d y d x d y d x point in completely opposite directions. This example needs more details and more context to be able to provide an answer. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 7, 2023 at 17:41 answered Oct 7, 2023 at 17:02 JonathanZJonathanZ 13k 2 2 gold badges 24 24 silver badges 39 39 bronze badges Add a comment\| This answer is useful -3 Save this answer. Show activity on this post. There's several important facts about derivatives that might help untangle the mess: df/dx is linear operator. f is not a function, but polynomial to transform derivatives to your ordinary linear algebra, you just need to represent your polynomial in basis [1,x,x²,x³,x⁴,... ], and the coefficients need to be separable. polynomial representation in matrix form is just: ⎡⎣⎢⎢⎢⎢⎢⎢a f k p u b g l q v c h m r w d i n s x e j o t y⎤⎦⎥⎥⎥⎥⎥⎥⎡⎣⎢⎢⎢⎢⎢⎢1 x x ² x ³ x ⁴⎤⎦⎥⎥⎥⎥⎥⎥[a b c d e f g h i j k l m n o p q r s t u v w x y][1 x x ² x ³ x ⁴] Derivative in matrix form is just(for size 5): ⎡⎣⎢⎢⎢⎢⎢⎢0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 0⎤⎦⎥⎥⎥⎥⎥⎥[0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0] I.e. it's an identity matrix which has been "shifted" a little with coefficients correctly setup. This uses the normal polynomial derivative calculation: f(x)=x n⇒f′(x)=n x n−1 f(x)=x n⇒f′(x)=n x n−1 and ability to move constants outside of derivative f(x)=c g(x)⇒f′(x)=c g′(x)f(x)=c g(x)⇒f′(x)=c g′(x) and of course the sum rule f(x)=g(x)+h(x)⇒f′(x)=g′(x)+h′(x)f(x)=g(x)+h(x)⇒f′(x)=g′(x)+h′(x) To handle multiple variables like x and y, you might need to change the basis from [1,x,x²,x³,x⁴] to [1,y,y²,y³,y⁴]. Linear algebra has tons of information how the basis change can be done. Choosing correct basis for your derivative makes the derivative matrix look reasonable/easier to remember. derivative matrix anyway only works if your calculations are in correct basis. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Oct 7, 2023 at 18:36 answered Oct 7, 2023 at 17:57 tp1tp1 686 9 9 silver badges 15 15 bronze badges 2 This nonanswer is at best helpful when studying linear algebra (not calculus). At worst it's just plain wrong. (2) is particularly confusing. "Separable" in (3) makes no sense.Ethan Bolker –Ethan Bolker 2023-10-07 18:02:58 +00:00 Commented Oct 7, 2023 at 18:02 separable means that if your matrix number slots a-y have x x in it, then the separable property does not hold, and this approach doesn't work. All x's need be inside the basis and not in the left side.tp1 –tp1 2023-10-07 18:11:46 +00:00 Commented Oct 7, 2023 at 18:11 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions derivatives implicit-differentiation learning See similar questions with these tags. 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15278	https://flexbooks.ck12.org/cbook/ck-12-interactive-algebra-2/section/4.2/primary/lesson/factoring-higher-degree-polynomials/	Factoring Higher Degree Polynomials \| CK-12 Foundation \n \n\n\n First, let's simplify the expressions: \n\n The divisor is 2(8!) - 21(6!). We can factor out 6! to get 6!(2 \cdot 7 \cdot 8 - 21), which simplifies to 6!(112 - 21) or 6!(91). \n\n The dividend is 14(7!) + 14(13!). We can factor out 7! to get 7!(14 + 14 \cdot 8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13), which simplifies to 7!(14 + 14 \cdot 1235520) or 7!(17297280). \n\n Now, we can divide the dividend by the divisor: \n \frac{7!(17297280)}{6!(91)} \n This simplifies to 7 \cdot 17297280 / 91. \n\n The remainder when 7 \cdot 17297280 is divided by 91 is 0. \n\n So, the remainder when 2(8!) - 21(6!) divides 14(7!) + 14(13!) is 0. \n\n
15279	https://k12.libretexts.org/Bookshelves/Mathematics/Analysis/01:_Analyzing_Functions/1.05:_Transformations/1.5.01:_Vertical_and_Horizontal_Transformations	1.5.1: Vertical and Horizontal Transformations - K12 LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 1.5: Transformations Unit 1: Analyzing Functions { } { "1.5.01:_Vertical_and_Horizontal_Transformations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.5.02:_Stretching_and_Reflecting_Transformations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.5.03:_Combining_Transformations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.5.04:_Notation_for_Transformations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "1.01:_Relations_and_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.02:_Average_Rate_of_Change" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.03:_Limits" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.04:_Function_Families" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.05:_Transformations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.06:_Operations_on_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "1.07:_Models_of_Functions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Sun, 27 Mar 2022 16:56:48 GMT 1.5.1: Vertical and Horizontal Transformations 14139 14139 admin { } Anonymous Anonymous User 2 false false [ "article:topic", "program:ck12", "authorname:ck12", "license:ck12", "source@ ] [ "article:topic", "program:ck12", "authorname:ck12", "license:ck12", "source@ ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Bookshelves 3. Mathematics 4. Analysis 5. Unit 1: Analyzing Functions 6. 1.5: Transformations 7. 1.5.1: Vertical and Horizontal Transformations Expand/collapse global location Analysis Front Matter Unit 1: Analyzing Functions Unit 2: Polynomial and Rational Functions Unit 3: Exponential and Logarithmic Functions Unit 4: Polar Equations and Complex Numbers Unit 5: Vector Analysis Unit 6: Conic Sections Unit 7: Sequences, Series, and Mathematical Induction Unit 8: Introduction to Calculus Back Matter 1.5.1: Vertical and Horizontal Transformations Last updated Mar 27, 2022 Save as PDF 1.5: Transformations 1.5.2: Stretching and Reflecting Transformations picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 14139 ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Vertical and Horizontal Transformations 2. Vertical and Horizontal Transformations 3. Examples 1. Example 1 2. Example 2 3. Example 3 4. Example 4 5. Example 5 6. Example 6 Review Review (Answers) Vocabulary Vertical and Horizontal Transformations Horizontal and vertical transformations are two of the many ways to convert the basic parent functions in a function family into their more complex counterparts. What vertical and/or horizontal shifts must be applied to the parent function of y=x 2 in order to graph g⁡(x)=(x−3)2+4 ? Vertical and Horizontal Transformations Have you ever tried to draw a picture of a rabbit, or cat, or dog? Unless you are talented, even the most common animals can be a bit of a challenge to draw accurately (or even recognizably!). One trick that can help even the most "artistically challenged" to create a clearly recognizable basic sketch is demonstrated in nearly all "learn to draw" courses: start with basic shapes. By starting your sketch with simple circles, ellipses, rectangles, etc., the basic outline of the more complex figure is easily arrived at, then details can be added as necessary, but the figure is already recognizable for what it is. The same trick works when graphing equations. By learning the basic shapes of different types of function graphs, and then adjusting the graphs with different types of transformations, even complex graphs can be sketched rather easily. This lesson will focus on two particular types of transformations: vertical shifts and horizontal shifts. We can express the application of vertical shifts this way: Formally: For any function f(x), the function g(x) = f(x) + c has a graph that is the same as f(x), shifted c units vertically. If c is positive, the graph is shifted up. If c is negative, the graph is shifted down. Informally: Adding a positive number after the x outside the parentheses shifts the graph up, adding a negative (or subtracting) shifts the graph down. We can express the application of horizontal shifts this way: Formally: given a function f(x), and a constant a > 0, the function g(x) = f(x - a) represents a horizontal shift a units to the right from f(x). The function h(x) = f(x + a) represents a horizontal shift a units to the left. Informally: Adding a positive number after the x inside the parentheses shifts the graph left, adding a negative (or subtracting) shifts the graph right. " frameborder="0" height="450px" name="91933" src=" thumbnailurl="" title="VideoObject?hash=0b9a7343fab39ccd5979563a60978084" uploaddate="2016-07-01 16:36:12" width="95%"> Examples Example 1 Solution Earlier, you were given a question about applying vertical and/or horizontal shifts to a parent function in order to graph a different function in the same function family. What transformations must be applied to y=x 2, in order to graph g⁡(x)=(x−3)2+4 ? The graph of g⁡(x)=(x−3)2+4 is the graph of y=x 2 shifted 3 units to the right, and 4 units up. Example 2 What must be done to the graph of y=x 2 to convert it into the graphs of y=x 2−3 and y=x 2+4 ? Solution At first glance, it may seem that the graphs have different widths. For example, it might look like y = x 2 + 4, the uppermost of the three parabolas, is thinner than the other two parabolas. However, this is not the case. The parabolas are congruent. If we shifted the graph of y = x 2 up four units, we would have the exact same graph as y = x 2 + 4. If we shifted y = x 2 down three units, we would have the graph of y = x 2 - 3. Example 3 Identify the transformation(s) involved in converting the graph of _f_(_x_) = \|_x_\| into _g_(_x_) = \|_x_ - 3\|. Solution From the examples of vertical shifts above, you might think that the graph of _g_(_x_) is the graph of f(x), shifted 3 units to the left. However, this is not the case. The graph of _g_(_x_) is the graph of _f_(_x_), shifted 3 units to the right. The direction of the shift makes sense if we look at specific function values. \| _x_ \| _g_(_x_) = abs(_x_ - 3) \| --- \| \| 0 \| 3 \| \| 1 \| 2 \| \| 2 \| 1 \| \| 3 \| 0 \| \| 4 \| 1 \| \| 5 \| 2 \| \| 6 \| 3 \| From the table we can see that the vertex of the graph is the point (3, 0). The function values on either side of _x_ = 3 are symmetric, and greater than 0. Example 4 What transformations must be applied to y=x 2, in order to graph g(x)=(x+2)2−2? Solution The graph of g(x)=(x+2)2−2 is the graph of y=x 2 shifted 2 units to the left, and 2 units down. Example 5 Use the parent function f(x) = x 2 to graph f(x) = x 2 + 3. Solution The function f(x) = x 2 is a parabola with the vertex at (0, 0). Adding outside the parenthesis shifts the graph vertically. Therefore, f(x) = x 2 + 3 will be a parabola with the vertex 3 units up.g Example 6 Use the parent function f(x) = \|x\| to graph f(x) = \|x - 4\|. Solution The graph of the absolute value function family parent function f(x) = \|x\| is a large "V" with the vertex at the origin. Adding or subtracting inside the parenthesis results in horizontal movement. Recall that the horizontal shift is right for negative numbers, and left for positive numbers. Therefore f(x) = \|x - 4\| is a large "V" with the vertex 4 units to the right of the origin. Review Graph the function f(x)=2\|x−1\|−3 without a calculator. What is the vertex of the graph and how do you know? Does it open up or down and how do you know? For the function: f(x)=\|x\|+c if c is positive, the graph shifts in what direction? For the function: f(x)=\|x\|+c if c is negative, the graph shifts in what direction? The function g(x)=\|x−a\| represents a shift to the right or the left? The function h(x)=\|x+a\| represents a shift to the right or the left? If a graph is in the form a⋅f(x). What is the effect of changing the _a_? Describe the transformation that has taken place for the parent function f(x)=\|x\|. f(x)=\|x\|−5 f(x)=5\|x+7\| Write an equation that reflects the transformation that has taken place for the parent function g⁡(x)=1 x, for it to move in the following ways: Move two spaces up Move four spaces to the right Stretch it by 2 in the y-direction Write an equation for each described transformation. a V-shape shifted down 4 units. a V-shape shifted left 6 units a V-shape shifted right 2 units and up 1 unit. The following graphs are transformations of the parent function f(x)=\|x\| in the form of f(x)=a\|x−h\|=k. Graph or sketch each to observe the type of transformation. f(x)=\|x\|+2. What happens to the graph when you add a number to the function? (i.e. f(x) + k). f(x)=\|x\|−4. What happens to the graph when you subtract a number from the function? (i.e. f(x) - k). f(x)=\|x−4\|. What happens to the graph when you subtract a number in the function? (i.e. f(x - h)). f(x)=\|x+2\|. What happens to the graph when you add a number in the function? (i.e. f(x + h)). Practice: Graph the following. f(x)=2\|x\| f⁡(x)=5 2⁢\|x\| f⁡(x)=1 2⁢\|x\| f⁡(x)=2 5⁢\|x\| Let f(x)=x 2. Let g(x) be the function obtained by shifting the graph of f(x) two units to the right and then up three units. Find a formula for g(x) and then draw its graph Suppose H(t) gives the height of high tide in Hawaii(H) on a Tuesday, (t) of the year. Use shifts of the function H(t) to find formulas of each of the following functions: F(t), the height of high tide on Fiji on Tuesday (t), given that high tide in Fiji is always one foot higher than high tide in Hawaii. S(d), the height of high tide in Saint Thomas on Tuesday (t), given that high tide in Saint Thomas is the same height as the previous day's height in Hawaii. Review (Answers) To see the Review answers, open this PDF file and look for section 1.12. Vocabulary \| Term \| Definition \| --- \| \| Absolute value function \| An absolute value function graphs a V shape, and is in the form y=\|x\|. \| \| Function families \| Function families are groups of functions with similarities that make them easier to graph when you are familiar with the parent function, the most basic example of the form. \| \| Function Family \| Function families are groups of functions with similarities that make them easier to graph when you are familiar with the parent function, the most basic example of the form. \| \| Horizontal shift \| A horizontal shift is the result of adding a constant term to the function inside the parentheses. A positive term results in a shift to the left and a negative term in a shift to the right. \| \| parent function \| A parent function is the simplest form of a particular type of function. All other functions of this type are usually compared to the parent function. \| \| shift \| A shift, also known as a translation or a slide, is a transformation applied to the graph of a function that does not change the shape or orientation of the graph, only the location of the graph. \| \| shifts \| A shift, also known as a translation or a slide, is a transformation applied to the graph of a function that does not change the shape or orientation of the graph, only the location of the graph. \| \| Transformations \| Transformations are used to change the graph of a parent function into the graph of a more complex function. \| \| Translation \| A translation is a transformation that slides a figure on the coordinate plane without changing its shape, size, or orientation. \| \| Vertical shift \| A vertical shift is the result of adding a constant term to the value of a function. A positive term results in an upward shift, and a negative term in a downward shift. \| This page titled 1.5.1: Vertical and Horizontal Transformations is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform. LICENSED UNDER Back to top 1.5: Transformations 1.5.2: Stretching and Reflecting Transformations Was this article helpful? Yes No Recommended articles 1.5.2: Stretching and Reflecting Transformations 1.5.3: Combining Transformations 1.5.4: Notation for Transformations 1.7: Models of Functions Article typeSection or PageAuthorCK12LicenseCK-12OER program or PublisherCK-12 Tags source@ © Copyright 2025 K12 LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 1.5: Transformations 1.5.2: Stretching and Reflecting Transformations Complete your gift to make an impact
15280	https://chem.libretexts.org/Courses/University_of_Arkansas_Little_Rock/Chem_1300%3A_Preparatory_Chemistry/Learning_Modules/04%3A_Compounds_and_Molecules/4.03%3A_Ionic_Compounds_and_Formulas	Skip to main content 4.3: Ionic Compounds and Formulas Last updated : Mar 11, 2022 Save as PDF 4.2: Molecular Compounds 4.4: Polyatomic Ions Page ID : 279971 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning objectives Recognize that metals lose electrons to form cations and that nonmetals gain electrons to form anions Predict the charge of monatomic main group elements based on their group number. Write formulas for ionic compounds using monatomic and polyatomic ions by applying the principle of charge neutrality. Ionic Bonds and Ionic Compounds If the number of protons in an atom or molecule is not equal to the number of electrons it has a net charge and is called an ion. Note, if it is an atom, it is called a monatomic ion, and if it is a molecule, it is called a polyatomic ion. There are obviously two types of ions, positive and negative, which are called cation and anion respectively. The charge of the ion is the number of protons minus the number of electrons. Cation: Net positive charge, the number of protons is greater than the number of electrons Anion: Net negative charge, the number of protons is less than the number of electrons. Formula for Ionic Compounds. When writing the formula of an ionic compound we use the lowest whole number ratio of cations ([+] ions) to anions ([-] ions). If you look at a salt crystal you will see that each positive ion is surrounded on all 6 sides by negative ions, and that a crystal lattice forms. So it makes no sense to have a "molecular formula" for an ionic compound, because the number of atoms will grow as the crystal grows. \| \| \| --- \| \| \| \| Figure On the left is a crystal lattice of NaCl table salt, which is also shown on the right, where it forms Halite crystals. Because the number of ions depends on the size of the crystal, we use the lowest ratio of positive to negative ions that results in a neutral structure. The concept of ionic bonds will be further developed in the next section. Introduction In the preceding section we discussed covalent bonds of molecules and introduced the concept of ionic bonds. In this section we will develop a deeper understanding of ionic compounds. We will see that to be stable an ionic compound must be neutral, and that there is not a discrete formula as there is with covalent compounds, in fact we use the lowest ratio of positive to negative charged ions that results in a neutral formula. In the previous section we did introduce the concept of cations (positive ions) and anions (negative ions). In the next section we will go over nomenclature, the naming of compounds. Crystal Lattice and Principle of Charge Neutrality Why must ionic compounds be neutral? An ionic bond is not between two ions, but the result of the interaction of multiple cations ([+] ions) and anions ([-] ions) as the coordinate with each other and form an ionic crysta. Salts are typical ionic compounds. That is, a salt crystal has +/+ and -/- repulsions (increasing the energy), and +/- attractions (decreasing the energy, and making the compound more stable), and if the attractive forces are greater than the repulsive, an ionic compound is formed. In the following diagram we see how a crystal forms which maximizes the attractive interactions of the opposite charges while minimizing the repulsive interactions of the like charges. If the compound was not neutral, but had more cations, or more anions, the like/like repulsive forces would be larger and the material would become unstable and break apart. Therefore, ionic compounds must be neutral. \| \| \| --- \| \| \| \| Figure : In the above image the sodium (a metal) loses and electron (becoming smaller) while the Chlorine atom (nonmetal) gains an electron (becoming bigger). A crystal lattice results where the cations and anions pack in three dimensional space that can result in a macroscopic crystal that is visible to the naked eye (CC BY-SA-NC; anonymous). The salt crystals seen to the naked eye are the result of the way cations and anions pack to form a stable crystal structure that minimizes like/like repulsions and maximizes opposite charge attractions as defined by Coulomb's Law. The following video from YouTube describes the formation of sodium chloride. Properties of Ionic Compounds Because of the arrangement of cations and anions in a the crystal lattice, these ions are held tightly into their position. Although the ions can vibrate and oscillate about their position, it take much more energy than room temperature for one ion to break free of the lattice structure. This results in very high melting points and boiling points for ionic compounds. Therefore, ionic compounds are crystal solids at room temperature. They are not malleable or soft. When struck sharply, they do shatter cleanly, with layers of ions breaking off at once. Monatomic Ions Metals tend to form cations and nonmetals tend to form anions. The following periodic table shows some of the common ions formed by common elements. Note, the noble gasses (turquoise) do not tend to form ions. Figure should not be considered to be a comprehensive Figure and this topic will be brought up again in the next section. There are several things you can note from this table though. First, metals tend to from cations and nonmetals tend to form anions, while the noble gasses do not tend to form ions. Second, the nonmetals have only one charge state, which represents the number of electrons that need to be added for them to have the same number as a noble gas. Third, there are two types of metals, those that have only one charge state (like the nonmetals), and those that have multiple charge states. These different types of ion forming metals will be discussed more in the next section on naming compounds. Monatomic Cations: When atoms (typically metals) lose one or more electrons the become positively charged cations. For example, when a neutral sodium atom loses one electron, it becomes the sodium ion, Na+ as shown in the following shorthand notation: Another example shows the formation of the magnesium ion from the neutral magnesium atom losing 2 electrons to pick up a +2 charge. Can we predict the stable charge of a monoatomic cation? Yes, it is easy for non-transition metals that have only one charged state (alkali metals, alkaline earth - see Figure 2.4.2). The trick is to remove enough electrons so that the cation has the same number of outer shell electrons as a noble gas, that is, it is isoelectronic to the noble gas. This is easy if we use the atom's group number's in terms of the "A" designation (remember there are two conventions for numbering the groups), as the group number effectively tells you how many electrons are needed to be removed. So Sodium (Group 1A) has 11 electrons and so if it looses one it has 10, like Neon (a noble gas). Magnesium has 12, so it looses 2 to become like Neon, and Aluminum has 13 and so looses 3 to become like Neon. Note: Why does hydrogen not form a +1 cation? Although hydrogen has one electron in its outer shell (like all the alkali metals), it only has one electron and so completely removing it would leave behind a bare nucleus, a subatomic particle the proton, which would not be stable. Consequently hydrogen does not form a cation, but can result in an acid, which is a polar covalent bond. In the above video sodium with 11 electrons gave an electron to chlorine, forming the cation and the anion, which then formed the crystal table salt NaCl. If chlorine was to try and remove the electron from hydrogen, it would not be able to as the resulting nucleus would pull it back, this results in the molecular acid HCl, where the chlorine can not completely remove the electron from the hydrogen. This is called a "polar covalent bond" and will be discussed later this semester. Monatomic Anions: When atoms (typically non-metals) gain one or more electrons they become negatively charged anions. For example, when a neutral fluorine atom gains one electrons, it becomes the fluoride ion as shown by the following shorthand notation: Another example is the formation of the sulfide ion when a neutral sulfur atom gains two electrons picking up a -2 charge. Can we predict the stable charge of a monoatomic anion? Yes, just as with the monatomic cations, stable monatomic anions are isoelectronic to the nearest noble gas, but in this case you add electrons instead of subtract. Once again, if we use the "A" designations for numbering the groups, we simply add the number of electrons required to be the same as a noble gas. So chlorine gains 1, oxygen gains 2 and nitrogen gains 3. This number can be determined by taking the group number 5, 6 or 7 from groups 5A, 6A and 7A and subtracting from 8 (for group 8A). Exercise Based on the position of the periodic table, predict the charge of the following elements if they are to produce ions. (do not look at periodic table above to come up with your answer) Sr Ga P Answer : A) . Sr is a group 2A main group metal. So, +2 charge. B) . Ga is a group 3A main group metal. So, +3 charge. C) . P is a group 5A main group nonmetal. So, 5-8 = -3 charge. Ionic Compound Formula and the Principle of Charge Neutrality The Principle of Charge Neutrality simply states that for matter to be stable it must be neutral. If an ionic compound had a formula of Na2F, and you had 10,000 of those units, then there would be 20,000 positive things and 10,000 negative things, meaning an excess of 10,000 positive things repelling each other, and so it would not be stable, and so would not form (and realize that real substances on macrosocopic scales that we can measure have in the order of 1023 entities. You might ask, if something is neutral, why are the ion-ion attractions greater than the repulsion, that is, why isn't the energy of interaction neutral? The reason that the attractions are greater than the repulsions is because opposite ions tend to be closer together (they are attracted) than the ions of same sign (they are repulsed), and according to Coulomb's law, the closer they are the greater the force, and so a neutral compound has a net attractive force and forms a stable compound. The formula of an ionic compound represents the lowest whole number ratio of cations to anions, it is as simple as that. Most cations have charges of [+1] through [+6] while most ions have charges of [-1] through [-3]. Trick: Set # of Anions = Charge of Cation and set # of Cations = Charge of Anion Beware that this must be the Lowest Whole # Ratio of cation to anion. Strategy: Identify charge of each ion multiply charge of each ion by absolute value of the charge of the counter ion. Make sure resulting ratio is the lowest whole number ratio, if not, divide by common denominator so all values are integers. Watch video to see how to perform this calculation Exercise What is the formula for the ionic compound that forms when the calcium ion reacts with the oxide ion? Answer : . and . You need one calcium ion for every one oxide ion for the compound to be charge neutral. s Exercise What is the formula for the ionic that forms when the lead(IV) () ion reacts with the sulfate () ion? Answer : Add . and . For every one Lead(IV) ion, you need two sulfate ions to be charge neutral. 1(+4) + 2(-2)=0. You may use the criss-cross rule, but you must make sure that you reduce to the simplest whole number ratio. Determining Ion Charges from Formula If given a formula like NaCl or FeO or Fe2O3 you need to be able to calculate the charge of the ions. In the case of NaCl it is easy as Na is always [+1] and Cl is always [-1]. But iron can have multiple charges and so needs to be Figured out. Fortunately, the anion is always of known charge and so you can use that to algebraically reverse engineer the principle of charge neutrality. that is for a general formula CxAy where C is the Cation with charge [+m] and there are x of them, and A is the Anion with charge of [-n] and there are y of them, we know from the principle of charge neutrality that So for FeO we know the charge of the oxygen is -2, so 1(charge of Fe) +1(charge of O) = 0, so Fe=+2 as )=-1 for Fe2O3 we have 2(charge of Fe) + 3(charge of Oxygen) = 0 so that is: 3x2 -2x3=0, which makes sense. Exercise What is the the charge for the cation in each of the following comounds? a. b. Answer : a. because oxygen is minus 2. b. because nitrogen is minus 3. Test Yourself Homework: Section 4.3 Graded Assignment: Section 4.3 Query Contributors Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to: Elena Lisitsyna (H5P interactive modules) 4.2: Molecular Compounds 4.4: Polyatomic Ions
15281	https://ellapizrojo.wordpress.com/2017/10/17/palabras-cuyo-significado-cambia-segun-el-genero/	Palabras cuyo significado cambia según el género ¿Sabías que el género puede cambiar el significado de una palabra? En español tenemos algunos sustantivos que tienen diferentes significados dependiendo del artículo utilizado, es decir, en los que un cambio de género va asociado a un cambio de significado. Son casos en que el sustantivo no cambia de forma al cambiar de género, por lo que lo importante es fijarse en el artículo que le acompaña para saber a qué nos estamos refiriendo. Esto es lo que ocurre, por ejemplo, con el orden (‘colocación, arreglo’) y la orden (‘mandato’). No es lo mismo el orden del día (la relación de asuntos que se han de tratar en una reunión), que la orden del día (las instrucciones de un superior que habremos de cumplir durante la jornada). Veamos unos cuantos más: —El margen es el espacio que dejamos en blanco a los lados de una hoja, mientras que la margen es la orilla, normalmente de un río, aunque también puede serlo de un campo o de un camino. Ejemplos: Deja un margen en el folio de 3 centímetros. Fui dando un paseo por la margen del río. —El pez es un animal que vive en el agua, mientras que la pez es una sustancia oscura y viscosa que se utiliza para impermeabilizar. Ejemplos: Tengo un pez en la pecera. Pinté la barca con la pez que me prestaste. —Un terminal es un teléfono o algo por el estilo; pero una terminal se refiere a un aeropuerto. Ejemplos: Me dieron un terminal nuevo en la tienda al cambiar el contrato. No conseguíamos llegar a la terminal a tiempo de coger el avión. —El cólera es una enfermedad, pero la cólera es una pasión. Ejemplos: Descubrió una cura para el cólera. Montó en cólera cuando se lo dijimos. —La coma es un signo ortográfico. El coma es una situación clínica. Ejemplos: Te falta una coma en esa oración. Todavía no salió del coma. —El editorial es el artículo de un diario en el que la redacción de este fija su posición respecto de algún asunto de actualidad. La editorial, por su parte, es la empresa que se dedica a publicar. Ejemplos: Me gustó mucho el editorial de hoy del periódico. Juan trabaja en la editorial de su padre. —Los curas son sacerdotes. Las curas se refieren a los cuidados que se nos prodigan para que sanemos. Ejemplo: El cura del pueblo ayuda a los necesitados. Voy al ambulatorio a que me hagan la cura de la herida. —Los cometas son cuerpos celestes que surcan el cielo, mientras que las cometas son juguetes que vuelan atados con una cuerda con ayuda del viento. Ejemplos: Vimos pasar el cometa Halley. Vamos a volar la cometa a la playa. —El frente es un lugar donde no conviene dejarse ver si estamos en medio de una guerra. La frente se refiere a una parte de la cara que va desde las cejas hasta el nacimiento del pelo. Ejemplos: Murieron muchos soldados en el frente de África. Tiene la frente muy ancha, debería dejarse flequillo. —La pendiente es una zona inclinada. El pendiente es un adorno que se pone en la oreja. Ejemplo: Se deslizó por la pendiente helada. Perdió el pendiente de su madre. —El parte es un informe, pero la parte es un trozo. Ejemplo: Transmitieron por la radio el parte diario. Dame la parte de tarta que me corresponde. En otras ocasiones, encontrarás algunas palabras, con las mismas características, usadas coloquialmente. Ejemplos: Había una rata (animal) en la cocina. Un rata (ladrón) fue detenido por la Policía. Juan Luis Guerra presenta su nuevo disco(música). ¿Vamos a la disco (discoteca)? Comparte esto: Relacionado Una respuesta a “Palabras cuyo significado cambia según el género” el pendiente y la pendiente Perdí el pendiente. La carretera tiene mucha pendiente Me gustaMe gusta Deja un comentario Cancelar la respuesta Δ Visitantes Entradas recientes Archivos Lo que más os ha gustado Categorías Nube de etiquetas Encuentra lo que buscas: Encuéntralo: Mis amigos seguidores Sígueme por correo electrónico Introduce tu dirección de correo electrónico para seguir este Blog y recibir las notificaciones de las nuevas publicaciones en tu buzón de correo electrónico. Dirección de correo electrónico: Sígueme por Email Sitios de interés sobre lectura y escritura Me encanta trabajar con vosotros vballesteros Me apasiona el mundo de la literatura y todo lo que le rodea. Soy lectora empedernida y, por ello, no me gusta encontrar textos con incorrecciones. Soy meticulosa en mi trabajo, responsable y de trato cercano. Me gusta pensar que colaboro con mis clientes en que su texto salga adelante en las mejores condiciones. Su satisfacción con mi trabajo es mi mejor gratificación. Ver perfil completo →
15282	https://chemistry.stackexchange.com/questions/69568/hybridization-of-nitrogen-in-trisilylamine-sih33n	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hybridization of nitrogen in trisilylamine, (SiH3)3N? Ask Question Asked Modified 4 years, 3 months ago Viewed 17k times 17 $\begingroup$ I want to know the hybridization of the central atom in $\ce{(SiH3)3N}$. I think it should be $\mathrm{sp^3}$, because $\ce{N}$ is attached to three silicon atoms and one lone pair. But actually it is supposedly $\mathrm{sp^2}$. How is this so? inorganic-chemistry hybridization Share edited Mar 2, 2017 at 12:44 orthocresol 72.7k1212 gold badges259259 silver badges438438 bronze badges asked Mar 1, 2017 at 17:14 heyhey 19311 gold badge33 silver badges1111 bronze badges $\endgroup$ 12 1 $\begingroup$ I don't see how's that related. Probably in this compound N is more planar than usual because of steric reasons, hence the higher s character of the lone pair. $\endgroup$ Ivan Neretin – Ivan Neretin 2017-03-01 17:58:42 +00:00 Commented Mar 1, 2017 at 17:58 3 $\begingroup$ @IvanNeretin More planar would mean lower s character of the lone pair. I think the question is related (but far from a dupe) because of the inversion mechanism. However, after reading up on the structure of $\ce{N(SiH3)3}$, I realise that it is far more interesting. I like that molecule! $\endgroup$ Jan – Jan 2017-03-01 18:02:55 +00:00 Commented Mar 1, 2017 at 18:02 1 $\begingroup$ @Jan Sorry, of course you're right: it is higher p character of the lone pair. $\endgroup$ Ivan Neretin – Ivan Neretin 2017-03-01 20:03:40 +00:00 Commented Mar 1, 2017 at 20:03 1 $\begingroup$ Also @IvanNeretin I dont think steric reasons are the key; see my answer ;) $\endgroup$ Jan – Jan 2017-03-01 20:09:42 +00:00 Commented Mar 1, 2017 at 20:09 2 $\begingroup$ Yes, I see. Indeed, the molecule turns out to be more interesting than it seemed. $\endgroup$ Ivan Neretin – Ivan Neretin 2017-03-01 20:11:14 +00:00 Commented Mar 1, 2017 at 20:11 \| Show 7 more comments 4 Answers 4 Reset to default 22 $\begingroup$ Ordinarily and according to Bents rule, we would expect nitrogens lone pair to be in an $\mathrm s$ orbital and nitrogen using its three $\mathrm p$ orbitals to form three bonds to the three silicon atoms. This configuration would allow for the greatest stabilisation. However, due to nitrogens small size, this perfect world already falls apart for ammonia ($\ce{NH3}$), where nitrogen is bound to three otherwise tiny hydrogen atoms. Because an electronically perfect angle of $90^\circ$ would generate much too much steric strain between the hydrogen atoms, $\mathrm s$ contribution is mixed into the bonding $\mathrm p$ orbitals to a certain extent; for ammonia, this extent happens to be almost perfect $\mathrm{sp^3}$ results for other amines will vary. This electronic situation is not ideal, however it is clearly better than having $\mathrm{sp^2}$ hybridisation and the lone pair in a $\mathrm p$ type orbital. An $\mathrm{sp^2}$ hybridisation of nitrogen in ammonia can be reached, but only as the transition state of nitrogen inversion. Carrying on to the compound $\ce{N(SiH3)3}$, we would be inclined to again assume a hybridisation of $\mathrm{sp^3}$ in line with the previous paragraph. However, Beagley and Conrad performed electron diffraction studies on $\ce{N(SiH3)3}$ and found the molecule to be practically planar within experimental error.[1,2] A planar molecule without doubt means that nitrogen is $\mathrm{sp^2}$-configured in $\ce{N(SiH3)3}$. The question remains why. There must be some kind of stabilising interaction of nitrogens remaining $\mathrm p$ orbital with something else to keep that molecule planar. Beagley and Conrad suggest in line with what was thought at the time that this be due to Ï bonds with silicons remote $\mathrm d$ orbitals.. Numerous evidence, much of which is collected on this site, speaks the opposite (namely that $\mathrm d$ orbitals do not play any role in the bonding situation of main group metals). Instead, I think we are dealing with something you may call inverse hyperconjugation. Remember that $\chi(\ce{Si}) = 1.9$ which is less than hydrogen, meaning that the $\ce{Si-H}$ bonds are polarised towards hydrogen. This in turn means that $\sigma^_{\ce{Si-H}}$ is a silicon-centred orbital with its primary lobe pointing towards nitrogen. Therefore, nitrogens $\mathrm p$ orbital can favourably interact with the antibonding $\sigma_{\ce{Si-H}}^$ orbital, increasing the $\ce{Si-N}$ bond order and decreasing the $\ce{Si-H}$ bond order. The effects are the same as with hyperconjugation stabilisation of secondary or tertiary carbocations but the electronic demand is reversed. We could attempt to draw the following resonance structures in Lewis formalism to explain this: $$\ce{H-SiH2-N(SiH3)2 <-> \overset{-}{H}\bond{...}SiH2=\overset{+}{N}(SiH3)2}\tag{1}$$ In this Lewis formalism, the double bond would be generated from a $\mathrm p$ orbital on both silicon and nitrogen. Note and reference: : B. Beagley, A. R. Conrad, Trans. Faraday Soc. 1970, 66, 27402744. DOI: 10.1039/TF9706602740. : Actually, $\angle(\ce{Si-N-Si}) \approx 119.5^\circ < 120^\circ$. The authors state: The apparant slight deviation from planarity is associated with a shrinkage effect11 on the $\ce{Si\dots Si}$ distance of about $\require{mediawiki-texvc}\pu{0.007 \AA}$ (see [table]). Spectroscopic results12 are entirely in agreement that the molecule is planar. 11 A. Allmenningen, O. Bastiansen and T. Munthe-Kaas, Acta Chem. Scand., 1956, 10, 261. [sic!] 12 E. A. V. Ebsworth, J. R. Hall, M. J. Mackillop, D. C. McKean, N. Sheppard and L. A. Woodward, Spectrochim Acta, 1958, 13, 202. [sic!] Share edited Jun 11, 2020 at 10:20 CommunityBot 1 answered Mar 1, 2017 at 18:01 JanJan 69.4k1212 gold badges211211 silver badges396396 bronze badges $\endgroup$ 2 1 $\begingroup$ I still frequently hear d orbitals being implicated in main group bonding situations in books and classes. Can you point me to a good example of evidence against this? $\endgroup$ gannex – gannex 2017-03-05 05:08:24 +00:00 Commented Mar 5, 2017 at 5:08 1 $\begingroup$ @gannex I can forward you an answer of Orthos, cant I? ;) $\endgroup$ Jan – Jan 2017-03-06 19:41:10 +00:00 Commented Mar 6, 2017 at 19:41 Add a comment \| 19 $\begingroup$ To test Jan's argument, I did an NBO analysis of your structure (optimized at PBE-D3/def2-SVP with NWChem 6.6 using a conformational search with MMFF94s and Avogadro as the starting point; a frequency calculation determined it was a true minimum). I used NBO version 5.9. Figure 1: optimized geometry (angles in degrees and distances in angstrom) The obtained geometry is in perfect agreement with Jan's answer, showing a $\ce{Si-N-Si}$ angle of 120°. The five most significative NBO second order stabilisation energies are: \| \| E(2) \| E(j)-E(i) \| F(i,j) Donor NBO (i) \| Acceptor NBO (j) \| kcal/mol \| a.u. \| a.u. ==============\|==================\|==========\|===========\|======== LP ( 1) N 1 \| BD( 1)Si 2- H 6 \| 5.08 \| 0.43 \| 0.043 LP ( 1) N 1 \| BD( 1)Si 2- H 7 \| 5.08 \| 0.43 \| 0.043 LP ( 1) N 1 \| BD( 1)Si 3- H 8 \| 5.08 \| 0.43 \| 0.043 LP ( 1) N 1 \| BD( 1)Si 3- H 9 \| 5.08 \| 0.43 \| 0.043 LP ( 1) N 1 \| BD( 1)Si 4- H12 \| 5.08 \| 0.43 \| 0.043 LP ( 1) N 1 \| BD( 1)Si 4- H13 \| 5.08 \| 0.43 \| 0.043 That is, a six-fold $\ce{n_\ce{N}} \rightarrow \sigma^(\ce{Si-H})$ donation, worth of 5.08 kcal/mol each, seems to be the most significant delocalisation. Figure 2: $\ce{n_\ce{N}} \rightarrow \sigma^(\ce{Si-H})$ delocalisation scheme On the other hand, the natural electron configurations are as follows: Atom \| Natural Electron Configuration ------\|---------------------------------- N \| [core]2s( 1.53)2p( 5.12) Si \| [core]3s( 1.01)3p( 1.96)3d( 0.02) H \| 1s( 1.15) Thus, the electron configuration of $\ce{N}$, according to NBO analysis, is $\ce{1s^{2} 2s^{1.53} 2p^{5.12}}$. Furthermore, the nitrogen lone pair is of pure $\pi$ character. Looking closer to the $\ce{N-Si}$ bond we see: (Occupancy) Bond orbital/ Coefficients/ Hybrids ------------------------------------------------------------------------------- 1. (1.97614) BD ( 1) N 1-Si 2 ( 81.00%) 0.9000 N 1 s( 33.32%)p 2.00( 66.66%)d 0.00( 0.02%) ( 19.00%) 0.4359Si 2 s( 23.84%)p 3.16( 75.42%)d 0.03( 0.74%) Thus, The $\ce{N-Si}$ bond is polarised towards $\ce{N}$ ($0.9000 h_\ce{N} + 0.4359 h_\ce{Si}$), as expected due to electronegativity; The $\ce{Si}$ hybrid ($h_\ce{Si}$) is of $\ce{sp^{3.16}}$ character, as expected; and The $\ce{N}$ hybrid ($h_\ce{N}$) is of $\ce{sp^{2.00}}$ character. In agreement with Jan's answer. Share edited May 8, 2020 at 12:47 answered Mar 2, 2017 at 0:43 schneiderfelipeschneiderfelipe 2,88322 gold badges2323 silver badges4242 bronze badges $\endgroup$ 5 1 $\begingroup$ I will now study this answer carefully so as to shamelessly rip off its technique. Very nice! Can you suggest any good reading for learning more about how to best carry out and/or interpret NBO analyses? $\endgroup$ hBy2Py – hBy2Py 2017-03-02 12:42:01 +00:00 Commented Mar 2, 2017 at 12:42 1 $\begingroup$ I really like Discovering Chemistry With Natural Bond Orbitals, a book by NBO author Frank Weinhold $\endgroup$ schneiderfelipe – schneiderfelipe 2017-03-02 12:54:48 +00:00 Commented Mar 2, 2017 at 12:54 3 $\begingroup$ @FelipeSchneider I've also been wanting to learn about techniques like this. Thanks for the reference!! $\endgroup$ gannex – gannex 2017-03-02 17:40:05 +00:00 Commented Mar 2, 2017 at 17:40 1 $\begingroup$ By any chance, do you still know which NBO version you have used? I don't think it'll make a huge difference as major improvements in the theory were towards the use with transition metals, but in general it would be nice to know. $\endgroup$ Martin - マーチン – Martin - マーチン ♦ 2020-05-07 21:56:46 +00:00 Commented May 7, 2020 at 21:56 $\begingroup$ @Martin-ãã¼ãã³ Excellent suggestion, it was version 5.9. I added it to the answer :) $\endgroup$ schneiderfelipe – schneiderfelipe 2020-05-08 12:47:50 +00:00 Commented May 8, 2020 at 12:47 Add a comment \| -6 $\begingroup$ We call this weird thing as back bonding. The lone pair kinda delocalises or seeks refuge in the empty d orbital of Si, basically providing each N-Si, bond, on an average, a one third of an extra bond. Share answered Jul 29, 2019 at 11:01 TheScholarAMTheScholarAM 3511 bronze badge $\endgroup$ 1 8 $\begingroup$ No d-orbitals, please... $\endgroup$ orthocresol – orthocresol 2019-07-29 11:26:24 +00:00 Commented Jul 29, 2019 at 11:26 Add a comment \| -7 $\begingroup$ It's actually due to the back bonding basically silicon is an element in the 3rd period and nitrogen is an element in the 2nd period so here silicon will contain empty orbitals of 3d and whereas nitrogen has lone pair of electrons so therefore back bonding occurs so while calculating the steric number in order to calculate the hybridization the lone pair on central atom nitrogen isn't considered which makes that the steric number is 3 and the hybridization is sp2. Share answered Jun 26, 2021 at 6:24 Vasudha PathedaVasudha Patheda 7 $\endgroup$ 1 4 $\begingroup$ Did you not read the answer given by @TheScholarAM and the comment about the lack of d-orbital participation? $\endgroup$ Safdar Faisal – Safdar Faisal 2021-06-26 09:09:26 +00:00 Commented Jun 26, 2021 at 9:09 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions inorganic-chemistry hybridization See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't? What is Drago's rule? Does it really exist? 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15283	https://www.youtube.com/watch?v=j511hg7Hlbg	Integrating factors 1 \| First order differential equations \| Khan Academy Khan Academy 9090000 subscribers 2992 likes Description 914997 views Posted: 31 Aug 2008 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Using an integrating factor to make a differential equation exact Watch the next lesson: Missed the previous lesson? Differential Equations on Khan Academy: Differential equations, separable equations, exact equations, integrating factors, homogeneous equations. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to KhanAcademy’s Differential Equations channel:: Subscribe to KhanAcademy: 179 comments Transcript: A lot of what you'll learn in differential equations is really just different bags of tricks. And in this video I'll show you one of those tricks. And it's useful beyond this. Because it's always good when, if maybe one day, you become a mathematician or a physicist, and you have an unsolved problem. Some of these tricks that solved simpler problems back in your education might be a useful trick that solves some unsolved problems. So it's good to see it. And if you're taking differential equations, it might be on an exam. So it's good to learn. So we'll learn about integrating factors. So let's say, we have an equation that has this form. Let's say this is my differential equation. 3xy-- I'm trying to write it neatly as possible-- plus y squared plus x squared plus xy times y prime is equal to 0. So, especially since we've covered this in recent videos, whenever you see an equation of this form where you have some function of xy, then you have another function of x and y times y prime equals 0, you said, oh, this looks like this could be an exact differential equation. And how do we test that? Well, we can take the partial derivative of this with respect to y, and we could call this function of x and y, M. So the partial of that, with respect to y, so M partial with respect to y, would be 3x plus 2y. And if this function right here, that expression right there, that's our function N, which is a function of x and y. We take the partial with respect to x, and we get that is equal to 2x plus y. And in order for this to have been an exact differential equation, the partial of this with respect to y would have to equal the partial of this with respect to x. But we see here, just by looking at these two, they don't equal each other. They're not equal. So, at least superficially, the way we looked at it just now, this is not an exact differential equation. But what if there were some factor, or I guess some function that we could multiply both sides of this equation by, that would make it an exact differential equation? So let's call that mu. So what I want to do is I want to multiply both sides of this equation by some function mu, and then see if I can solve for that function mu that would make it exact. So let's try to do that. So let's multiply both sides by mu. And just as a simplification, mu could be a function of x and y. It could be a function of x. It could be a function of just x. It could be function of just y. I'll assume it's just a function of x. You could assume it's just a function of y and try to solve it. Or you could just assume it's a function of x and y. If you assume it's a function of x and y, it becomes a lot harder to solve for. But that doesn't mean that there isn't one. So let's say that mu is a function of x. And I want to multiply it by both of these equations. So I get mu of x times 3xy plus y squared plus mu of x times x squared plus xy times y prime. And then, what's 0 times any function? Well, it's just going to be 0, right? 0 times mu of x is just going to be 0. But I did multiply the right hand side times mu of x. And remember what we're doing. This mu of x is-- when we multiply it, the goal is, after multiplying both sides of the equation by it, we should have an exact equation. So now, if we consider this whole thing our new M, the partial derivative of this with respect to y should be equal to the partial derivative of this with respect to x. So what's the partial derivative of this with respect to y? Well, if we're taking the partial with respect to y here, mu of x, which is only a function of x, it's not a function of y, it's just a constant term, right? We take a partial with respect to y. x is just a constant, or a function of x can be viewed just as a constant. So the partial of this with respect to y is going to be equal to mu of x, you could just say, times 3x plus 2y. That's the partial of this with respect to y. And what's the partial of this with respect to x? Well, here, we'll use the product rule. So we'll take the derivative of the first expression with respect to x. mu of x is no longer a constant anymore, since we're taking the partial with respect to x. So the derivative of mu of x with respect to x. Well, that's just mu prime of x, mu prime, not U. mu prime of x. mu is the Greek letter. It's for the muh sound, but it looks a lot like a U. So mu prime of x times a second expression, x squared plus xy, plus just the first expression. This is just the product rule, mu of x. Times the derivative of the second expression with respect to x. So times-- ran out of space on that line-- 2x plus y. And now for this new equation, where I multiplied both sides by mu. In order for this to be exact, these two things have to be equal to each other. So let's just remember the big picture. We're kind of saying, this is going to be exact. And now, we're going to try to solve for mu. So let's see if we can do that. So let's see, on this side, we have mu of x times 3x plus 2y. And let's subtract this expression from both sides. So it's minus mu of x times 2x plus y. You'll see a lot of these differential equation problems that get kind of hairy. They're really just a lot of algebra. And that equals-- what do we have left? I'll write it in yellow. That equals-- I'm going to run out of space. I'm going to do it a little bit lower. That equals, just this term right here. That equals mu prime of x times x squared plus xy. And let's see, if we factor out a mu of x here, we get mu of x times 3x plus 2y minus 2x minus y is equal to mu prime of x, the derivative of mu with respect to x, times x squared plus xy. Now, we can simplify this. So we get mu of x times-- what is this-- 3x minus 2x is x. 2y minus y, so x plus y, is equal to-- and I'm just going to simplify this side a little bit-- is equal to mu prime of x. Let's factor out an x here. And the reason why I'm doing that is because it seems like if I factor out an x here, I'll get an x plus y. So this is mu prime of x times x times x plus y. x times x plus y is x squared plus xy. So that's why I did it, and I have this x plus y on both sides equation, which I will now divide both sides by. So if you divide both sides by x plus y, we could maybe assume that it's not 0. That simplifies things pretty dramatically. We get mu of x is equal to mu prime of x times x. And now, just the way my brain works, I like to rewrite this expression just in our operator form, where instead of writing it mu prime of x, we could write that as d mu dx. So let's do that. So we could write mu of x is equal to d, the derivative of mu with respect to x, times x. And this is actually a separable differential equation in and of itself. It's kind of a sub-differential equation to solve our broader one. We're just trying to figure out the integrating factor right here. So let's divide both sides by x. So we get mu over x, this is just a separable equation now, is equal to d mu dx. And then, let's divide both sides by mu of x, and we get 1 over x is equal to 1 over mu. That's mu of x, I'll just write 1 over mu right now, for simplicity, times d mu dx. I'm actually going to go horizontal right here. Multiply both sides by dx, you get 1 over x dx is equal to 1 over mu of x d mu. Now, you could integrate both sides of this, and you'll get the natural log of the absolute value of x is equal to the natural log of the absolute value of mu, et cetera, et cetera. But it should be pretty clear from this that x is equal to mu, or mu is equal to x, right? They're identical. If you look at both sides of this equation there, you can just change x for mu, and it becomes the other side. So, this is obviously telling us that mu of x is equal to x. Or mu is equal to x. So we have our integrating factor. And if you want, you can take the antiderivative of both sides with the natural logs, and all of that. And you'll get the same answer. But this is just, by looking at it, by inspection, you know that mu is equal to x. Because both sides of this equation are completely the same. Anyway, we now have our integrating factor. But I am running out of time. So in the next video, we're now going to use this integrating factor. Multiply it times our original differential equation. Make it exact. And then solve it as an exact equation. I'll see you in the next video.
15284	https://www.antenna-theory.com/tutorial/smith/chart.php	The Smith Chart Back: Antenna Theory (Home) \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| Intro to Smith Charts Smith Charts Video Constant Resistance Circles Constant Reactance Curves Impedance Transformations Impedance Matching: Series L and C Impedance Matching: Tx Lines The Admittance Smith Chart Impedance Matching: Parallel L and C Impedance Matching: Parallel Tx Lines Immittance Charts Impedance Matching on Immittance Charts Dual-Band Impedance Matching \| \| \| \| \| \| \| \| \| \| --- --- --- \| In this tutorial, we will introduce and explain Smith Charts, and then given an introduction to impedance matching. We will then use the Smith Chart to perform impedance matching with transmission lines and lumped components (capacitors and inductors). --- Introduction to Smith Charts Up: Smith Chart and Impedance Matching Table of Contents The Smith Chart is a fantastic tool for visualizing the impedance of a transmission line and antenna system as a function of frequency. Smith Charts can be used to increase understanding of transmission lines and how they behave from an impedance viewpoint. Smith Charts are also extremely helpful for impedance matching, as we will see. The Smith Chart is used to display an actual (physical) antenna's impedance when measured on a Vector Network Analyzer (VNA). Smith Charts were originally developed around 1940 by Phillip Smith as a useful tool for making the equations involved in transmission lines easier to manipulate. See, for instance, the input impedance equation for a load attached to a transmission line of length L and characteristic impedance Z0. With modern computers, the Smith Chart is no longer used to the simplify the calculation of transmission line equatons; however, their value in visualizing the impedance of an antenna or a transmission line has not decreased. The Smith Chart is shown in Figure 1. A larger version is shown here. Figure 1. The basic Smith Chart. Figure 1 should look a little intimidating, as it appears to be lines going everywhere. There is nothing to fear though. We will build up the Smith Chart from scratch, so that you can understand exactly what all of the lines mean. In fact, we are going to learn an even more complicated version of the Smith Chart known as the immitance Smith Chart, which is twice as complicated, but also twice as useful. But for now, just admire the Smith Chart and its curvy elegance. This section of the antenna theory site will present an introduction to Smith Chart basics. Smith Chart Tutorial We'll now begin to explain the fundamentals of the Smith Chart. The Smith Chart displays the complex reflection coefficient [Equation 1, below], in polar form, for an arbitrary impedance (we'll call the impedance ZL or the load impedance). The reflection coefficient is completely determined by the impedance ZL and the "reference" impedance Z0. Note that Z0 can be viewed as the impedance of the transmitter, or what is trying to deliver power to the antenna. Hence, the Smith Chart is a graphical method of displaying the impedance of an antenna, which can be a single point or a range of points to display the impedance as a function of frequency. For a primer on complex math, click here. Recall that the complex reflection coefficient () for an impedance ZL attached to a transmission line with characteristic impedance Z0 is given by: \| \| \| [Equation 1] \| --- For this Smith Chart tutorial, we will assume Z0 is 50 Ohms, which is often, but not always the case. Note that the Smith Chart can be used with any value of Z0. The complex reflection coefficient, or , must have a magnitude between 0 and 1. As such, the set of all possible values for must lie within the unit circle: Figure 2. The Complex Reflection Coefficient must lie somewhere within the unit circle. In Figure 2, we are plotting the set of all values for the complex reflection coefficient, along the real and imaginary axis. The center of the Smith Chart is the point where the reflection coefficient is zero. That is, this is the only point on the Smith Chart where no power is reflected by the load impedance. The outter ring of the Smith Chart is where the magnitude of is equal to 1. This is the black circle in Figure 1. Along this curve, all of the power is reflected by the load impedance. Let's look at a few examples. Smith Chart Example 1. Suppose =0.5. From equation , we can solve for ZL to be: \| \| \| [Equation 2] \| --- From equation , with Z0=50 Ohms, a reflection coefficient of 0.5 corresponds to a load impedance ZL=150 Ohms. We can plot gamma_1 on the smith chart: . Figure 3. plotted on the Smith Chart. Since is entirely real, the point lies along the real gamma axis (x-axis) in Figure 3, and the imaginary axis value (y-axis) location is 0. Smith Chart Example 2. Suppose = -0.3 + i0.4 is plotted on the Smith Chart in Figure 4: Figure 4. plotted on the Smith Chart. From Equation and using Z0=50, we note that corresponds to a load impedance ZL = 20.27 + i21.62 [Ohms]. Smith Chart Example 3. = -i. is plotted on the Smith Chart in Figure 5: Figure 5. =-i plotted on the Smith Chart. From Equation and with Z0=50, corresponds to a load impedance ZL = -i50 Ohms. That is, the load impedance here is purely imaginary and negative, which indicates a purely capacitive load. VSWR on the Smith Chart Since VSWR is only a function of the absolute value of , we can get the VSWR for a load from the Smith Chart as well. That is, a VSWR = 1 would be the center of the Smith Chart, and VSWR=3 would be a circle centered around the center of the Smith Chart, with magnitude =0.5. Circles centered at the origin of the Smith Chart are constant-VSWR circles. Note that the outer boundary of the Smith Chart (where =1) corresponds to a VSWR of infinity. --- Smith Chart Lecture Here we present an introduction video to Smith Charts. This video can be skipped if you like; the material is covered in the remaining sections. However, if you like videos, here is another medium of presenting the Smith Chart that may be beneficial. Particularly, the Immittance Smith Chart is discussed in terms of the reflection coefficient. An example antenna impedance is plotted on the Smith Chart and explained. The goal of this movie is to present the information using a different method (video instead of webpage), which will hopefully increase your understanding of Smith Charts. To sum up this page so far, the Smith Chart is a plot of the reflection coefficient. Since the reflection coefficient corresponds directly to an impedance, we are actually plotting the impedance of a device on the Smith Chart. In the next section, we'll start filling in some of the crazy lines on the Smith Chart visible in Figure 1. --- Next: Constant Resistance Circles Smith Chart Table of Contents Topics Related To Antenna Theory Antenna Tutorial (Home) \| \| \| an ad
15285	https://www.ck12.org/flexi/cbse-math/integration-by-partial-fractions/how-to-convert-reciprocals-of-quadratic-equations-into-special-integrals/	How to convert reciprocals of quadratic equations into special integrals? - Steps \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Integration by Partial Fractions Question How to convert reciprocals of quadratic equations into special integrals? Flexi Says: To convert reciprocals of quadratic equations into special integrals, we can use the following steps: Step 1:Write the reciprocal of the quadratic equation in the form of an integral. For example, if you have a quadratic equation a x 2+b x+c, its reciprocal can be written as 1 a x 2+b x+c. This can be written as an integral as ∫1 a x 2+b x+c d x. Step 2: Complete the square in the denominator. This involves rewriting the quadratic equation in the form (d x+e)2+f, where d, e, and f are constants. This step makes the integral easier to solve. Step 3: Use the formula for the integral of a function divided by a quadratic. Analogy / Example Try Asking: What is the sum up to infinity in the series 1/4, 1/8, 1/16, 1/32, 1/64, ...?Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants. Unexpected text node: '\frac{6x^2 - 4x + 2}{(x - 7)(x^2 + 9)}'Unexpected text node: '\frac{x+1}{4}=\frac{x-2}{3}' How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
15286	https://zhuanlan.zhihu.com/p/674741143	统计学入门（七）：抽样方法的原理与实践（终章） - 知乎首页知乎直答焕新知乎知学堂等你来答切换模式登录/注册统计学入门（七）：抽样方法的原理与实践（终章）首发于数学切换模式登录/注册统计学入门（七）：抽样方法的原理与实践（终章）张人大目前对AI和数学感兴趣，也会分享我的其他想法收录于 · 数学 7 人赞同了该文章目录在过去的几篇文章中，我们一起探索了统计学的许多重要概念与方法，从理解样本与总体的基础出发，到深入探讨统计量、参数估计、假设检验、置信区间、方差分析，直至上一篇关于回归分析的精彩讨论。这一系列文章旨在为读者揭示统计学的核心原理和广泛应用，使之成为理解数据的强大工具。今天，我们将进入这个系列的最后一篇文章，专注于统计学中一个至关重要却经常被忽视的领域：抽样方法（Sampling Methods）。抽样方法是统计学的基石之一，它关系到如何从一个大的群体（即总体）中选取代表性的小群体（即样本）来进行研究。在实际应用中，由于成本、时间或其他资源的限制，我们很少有机会对整个总体进行完整的调查或实验。因此，如何通过抽样来获得能够准确反映总体特性的样本，成为了一项挑战性的任务。正确的抽样方法不仅能提高研究的效率，还能保证数据分析的有效性和可靠性。在本文中，我们将深入探讨不同类型的抽样技术，包括简单随机抽样、分层抽样、整群抽样、系统抽样等，每种方法都有其独特的适用场景和优缺点。我们还会讨论抽样误差的概念以及如何确定合适的样本量。通过实际案例的分析，我们将展示这些抽样方法在实际研究中的应用。此外，虽然本系列文章将以抽样方法的讨论告一段落，但统计学的世界远远不止于此。在文章的最后，我们还会简要提及一些未在本系列中深入探讨的相关主题，为对统计学有更深入兴趣的读者提供未来的学习方向。现在，让我们开始探索抽样方法的奥秘，理解它在统计学研究中的不可替代性和实际操作中的多样性。抽样方法的基础抽样（Sampling）是指从一个较大的集合（总体）中选择部分个体或项目作为样本的过程。在统计学中，这个过程不仅是一种实用的必要性，更是一种科学的艺术。总体（Population）通常是庞大且复杂的，例如一个国家的所有居民、某个工厂生产的全部产品，或者一个时期内所有的交易记录。由于研究或调查整个总体通常不现实，因此，我们通过抽样来获得一个较小但代表性强的样本（Sample），以此来估计或推断总体的特性。正确的抽样方法对于确保研究结果的准确性和可靠性至关重要。一个好的抽样设计能够减少偏差，提高样本的代表性，从而使得从样本得出的结论更有可能接近真实的总体情况。在预算、时间和其他资源有限的情况下，有效的抽样策略还可以大幅提高研究的效率。理解总体与样本之间的关系对于掌握抽样方法至关重要。总体是指我们想要研究或推断的整个群体。它可以是有限的，如一个城市里的所有居民，也可以是无限的，如无限期间内的所有交易记录。样本则是从总体中选出的一部分个体或项目，其目的是通过对样本的研究来推断总体的特性。在进行抽样时，最重要的原则之一是确保样本的代表性。这意味着样本中的个体应该在重要的特征上尽可能地反映总体的分布。例如，在研究一个国家的居民健康状况时，样本中的个体应该在年龄、性别、居住地等方面与总体相似。代表性的缺失会导致抽样偏差（Sampling Bias），从而影响研究的准确性和可靠性。主要的抽样技术简单随机抽样（Simple Random Sampling）简单随机抽样是最基本也是最常见的抽样方法。在这种方法中，每个总体成员被选中的概率都是相等的。简单地说，它就像是从一个装满号码的箱子里随机抽取一定数量的号码。这种抽样方式的主要优点在于它的公平性和简易性。然而，它的一个缺点是在总体很大时，可能无法保证样本的代表性，特别是当总体中的群体存在显著差异时。分层抽样（Stratified Sampling）分层抽样是一种更精细化的抽样方法，它首先将总体分成几个相互独立且在某些特征上内部相似的层次或子群体，然后从每个层次中独立地进行简单随机抽样。这种方法的优势在于能更好地保证样本在关键特征上的代表性，尤其是在总体内部差异较大时。分层抽样的挑战在于需要准确地定义和识别各个层次。整群抽样（Cluster Sampling）整群抽样是另一种节省资源的抽样方法。在这种方法中，总体首先被分成含多个成员的“群”或“簇”。然后，随机选择一些群体，对其内所有成员进行调查。整群抽样适合于当总体分布广泛且难以直接访问每个成员时。其主要缺点是，如果群体内部差异较小而群体间差异较大，可能导致较高的抽样误差。系统抽样（Systematic Sampling）系统抽样是一种在给定间隔内选择样本的方法。例如，你可能从名单的第一个人开始，然后每隔10个人选择一个。这种方法的实施相对简单，特别是在有序列表的情况下。然而，如果列表中存在周期性的模式，那么系统抽样可能会引入偏差。方便抽样（Convenience Sampling）方便抽样，顾名思义，是基于方便和可达性选择样本的方法。例如，一个研究者可能选择靠近他们的个体作为样本。虽然这种方法在某些情况下实用，但它通常不被认为是科学的，因为样本的选择可能高度偏见，不具代表性。抽样误差与样本量确定抽样误差（Sampling Error）抽样误差是指由于从总体中抽取样本而产生的结果与总体实际情况之间的差异。这种差异是不可避免的，因为样本只是总体的一个部分，无法完全复制总体的所有特性。抽样误差的大小可以作为样本结果准确性的一个指标。减少抽样误差的一种方法是增加样本量，另外，采用更精确的抽样方法也可以帮助降低误差。确定样本量的原则确定合适的样本量是抽样设计中的一个关键步骤。样本量过小会增加抽样误差，导致结果不够准确。而样本量过大则会增加成本和工作量。确定样本量时需要考虑的因素包括总体的大小、预期的抽样误差、置信水平和变异性。通常，统计学家会使用特定的公式或软件来帮助确定适当的样本量。在实践中，决定样本量的选择往往是一种平衡。研究者需要在理想的统计准确性和实际的资源限制之间找到平衡点。例如，在市场研究中，可能需要较大的样本量来确保结果的代表性，而在一项探索性研究中，较小的样本量可能就足够了。样本量与总体大小的关系虽然总体的大小是影响样本量决策的一个因素，但在很多情况下，即使总体非常大，所需的样本量并不会显著增加。这是因为一旦样本量达到一定程度，增加更多样本对于减少误差的效果逐渐变小。因此，即使是对非常大的总体进行研究，合理的样本量也可以在保持准确性的同时控制成本。抽样方法的实际应用案例为了更好地理解抽样方法的实际应用，本部分将通过几个案例来展示不同抽样技术的选择和实施。国家健康调查：使用分层抽样一个国家的卫生部门想要进行一项全国性的健康调查。由于国家范围广泛，人口分布不均，因此采用了分层抽样。他们首先根据地理区域、年龄和性别将总体分成多个层。然后，在每个层中使用简单随机抽样选择个体。这种方法确保了各个区域和不同人群在样本中都得到了代表，从而提高了调查结果的准确性和可靠性。市场研究：使用整群抽样一家大型零售公司想要研究不同地区消费者的购物习惯。由于公司的门店遍布全国，他们选择使用整群抽样。在这个案例中，每个门店被视为一个群体。公司随机选择了一些门店，并调查了这些门店的所有顾客。这种方法简化了数据收集过程，尤其是在有大量潜在调查对象且分布广泛时。教育研究：使用简单随机抽样一个教育研究小组想要研究某个城市中学生的学习习惯。由于总体规模（该城市中的中学生）相对较小，且易于获取完整的名单，研究小组采用了简单随机抽样。他们随机选择了一定数量的学生进行问卷调查。这种方法在操作上简单，且因其随机性，结果具有较高的可靠性。企业内部审计：使用系统抽样一家公司进行内部财务审计时，选择使用系统抽样来检查交易记录。假设公司每个月有成千上万的交易，他们可能每50个交易检查一次。这种抽样方法在大量数据中提供了一种高效的方式来进行审计，同时保持了一定程度的随机性。社会学研究：使用方便抽样一个社会学研究者可能对某个特定社区的意见感兴趣。由于资源有限，他们可能选择在该社区的一个活动中进行方便抽样，邀请到场的人参与调查。虽然这种方法的代表性可能受限，但它提供了快速且成本效益高的方式来获取初步数据。这些案例显示了抽样方法如何根据研究的具体目的、可用资源和总体的特点来选择和应用。每种方法都有其独特的优势和局限性，理解这些可以帮助研究者在实际工作中做出更明智的选择。结语随着我们对抽样方法的探讨接近尾声，这个统计学入门系列文章也随之画上句点。从最初探讨样本与总体的基础概念，到深入分析各种统计量、假设检验、置信区间、方差分析和回归分析，再到本篇文章对抽样方法的全面解读，我们共同走过了一段充满知识和发现的旅程。通过这些文章，我们希望读者能够更好地理解统计学的基本原理和应用，从而在面对数据和研究时做出更明智的决策。统计学是一个广博且不断发展的领域，每一种方法和技术都是为了更好地理解数据和现实世界。虽然本系列已经涵盖了许多重要主题，但仍有无数的知识等待我们去探索和学习。在结束这个系列之际，我们想简要提及一些本系列未能深入探讨的相关统计学主题，以供感兴趣的读者进一步探索：非概率抽样（Non-probability Sampling）这类抽样技术不依赖于随机选择的原则，包括方便抽样、判断抽样等。它们在某些情况下很有用，但也存在较大的偏差风险。时间序列分析（Time Series Analysis）时间序列分析关注于对随时间变化的数据进行分析，如股票价格、气候变化等。多元统计分析（Multivariate Statistical Analysis）多元统计分析涉及同时分析多个变量之间的关系，是理解复杂数据结构的关键。贝叶斯统计（Bayesian Statistics）贝叶斯统计是一种不同于传统频率统计的方法，它在更新证据的基础上进行概率的推断。实验设计（Experimental Design）实验设计涉及如何有效地规划和执行实验，以确保可靠和有效的结论。统计学的学习是一个不断探索和深入的过程。每个主题都是通往更深层次理解数据世界的一扇窗。希望这些文章能激发您的好奇心，引领您继续在统计学的道路上探索前行。感谢您跟随这个系列走到了这里。愿您的统计学之旅充满发现和启迪。编辑于 2023-12-28 02:13・中国香港抽样技术统计学统计赞同 7添加评论分享喜欢收藏申请转载写下你的评论... 还没有评论，发表第一个评论吧关于作者张人大目前对AI和数学感兴趣，也会分享我的其他想法回答 169文章 400关注者 2,466 关注他发私信推荐阅读统计学回顾 - 抽样分布 ============ Harol...发表于想用数据搞...抽样技术理论-名词解释（二） ============== 1. 不等概率抽样是概率样本，每个单元被选入样本的概率是不相等的，样本单元权数也不相等，因此数据处理中的权数就可以忽略，从而计算更简洁方便。 2. 有限总体修正系数1-f $f=n/N$为抽样… burx 发表于统计学笔记抽样技术理论-名词解释（一） ============== 非概率抽样抽取样本时并不依据随机原则，单元入样概率不知，无法计算抽样误差，不具备统计推断意义。概率抽样依据随机原则，按照某种实现设定的程序，从总体中抽取部分单元的抽样方法。… burx 推论统计分析——抽样分布 ============ 简简单单想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
15287	https://www.icicibank.com/blogs/mutual-fund/what-is-a-sinking-fund	Published Time: 2025-06-23T00:00:00.000+0530 What is a Sinking Fund? Meaning, Types, Advantages & Formula Try our new website for a faster, fresher, and smarter banking experience. Old Old VersionNew New Version Open/Close Website Toggle Widget Explore the new ICICI Bank website! Designed for a simple, fast and personalised banking experience. 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Meaning, Types, Advantages & Formula ORANGE HUB PLAN RETIREMENT DIY Stack GET A CALL BACK Want us to help you with anything? Request a Call back This field is required Only alphabetes are allowed This field is required Only alphabetes are allowed Please enter valid number Please enter valid email Select product type [x] Auto Loan [x] Personal Loan [x] Home Loan [x] Credit Card [x] Insta SIP [x] Mutual Funds [x] Money Coach [x] ELSS [x] Savings Account [x] Two Wheeler Loan Please select product type Please enter valid pincode Submit Thank you for your request. Your reference number is CRM Our executive will contact you shortly THE ORANGE HUB Blog 2 mins Read \| 1 Month Ago What is a Sinking Fund? Meaning, Types, Advantages & Formula Investment Share Copy Know MoreInvest now Financial planning for the future is important as it helps keep financial stress at bay. An important strategy used for future planning is the ‘Sinking Fund’. This is an amount of money kept aside to meet a defined future financial requirement. In this blog post, we will learn in detail about the nature, types and benefits of Sinking Funds. What is a Sinking Fund? A Sinking Fund is a dedicated pool of money, accumulated over a period of time to cover a significant future expense or to repay debt in the future. For example, companies or governments set aside money over time to repay debt or bond issues, typically when they mature. The money in a Sinking Fund is invested in low-risk securities to grow steadily and accumulate enough to cover the future debt repayment. Sinking Funds are sometimes also used for other long-term financial obligations, such as replacing equipment or maintaining capital projects. Let us understand this with an example: Imagine XYZ Company issues bonds worth ₹ 150 crore for a tenure of 5 years. Instead of waiting until the maturity date to gather funds to repay the entire amount, the company sets up a Sinking Fund. Every year, it contributes ₹ 30 crore to this Fund and by the end of 5 years, ₹ 150 crore will have been accumulated. This process ensures smooth debt repayment. Types of Sinking Funds There are different types of Sinking Funds maintained by businesses, each serving a specific purpose. Let’s look at some common types: 1. Callable Bond Sinking Fund Companies create Callable Bond Sinking Funds to save money specifically for repurchasing callable bonds early (before maturity). 2. Specific Purpose Sinking Fund This type of Sinking Fund is created for a particular financial goal. For instance, if a company plans to buy new machinery in five years, it can set up a Sinking Fund to systematically save money for this purchase. 3. Regular Payment Sinking Fund A company may have periodic financial commitments such as interest payments to bondholders or trustee fees. A regular payment Sinking Fund ensures that these recurring payments are made on time. 4. Purchase Back Sinking Fund If a company decides to buy back its issued bonds from the market before maturity, it can set up a Purchase Back Sinking Fund. Advantages of a Sinking Fund Some benefits of creating a Sinking Fund are: 1. Timely Debt Repayment By regularly contributing to a Sinking Fund, businesses ensure they have sufficient funds to repay their debts when due, reducing sudden financial strain. 2. Financial Stability A Sinking Fund helps companies maintain stable finances by spreading out large expenses over time rather than facing a sudden cash crunch. 3. Enhances Investor Confidence Investors feel more secure knowing that a company has a structured repayment plan in place. This increases investor trust and makes it easier for companies to raise funds in the future. 4. Flexibility in Redemption Companies with a Sinking Fund can redeem bonds before maturity, which can sometimes be beneficial when interest rates fluctuate. 5. Helps in Capital Planning Organisations planning for capital investments such as acquiring land, buildings or equipment, can benefit from Sinking Funds to avoid sudden financial strain. How to Calculate the Contribution to a Sinking Fund? To determine the periodic contribution required for a Sinking Fund, you can use the following formula: C = A × [ r / ((1 + r)^(n·t) − 1) ] where: C = Amount contributed in each period A = Total amount needed at maturity r = Interest rate (in decimal form) n = Number of contributions per year t = Total number of years Guide to Start a Sinking Fund 1. Know the Purpose It is very important to have a clear purpose for starting a Sinking Fund. Make a detailed plan about the expense or liability that you wish to cover with a Sinking Fund. 2. Set a Target Amount Based on the purpose of the Fund and other factors like market conditions, estimated projections etc., decide a target amount. 3. Choose a Timeframe Finalise a timeframe to accumulate money in the Sinking Fund such that the Fund is ready well in time for fulfilling its intended requirement. 4. Calculate Contribution Amount Use an online calculator or manually use the formula to get an insight into your contribution amount per year / per quarter etc. 5. Open a Dedicated Account Remember to keep the Sinking Fund separate from regular funds. 6. Make Regular Contributions Ensure that you make timely contributions to reach the target amount that you have set. Conclusion Now that you know the meaning of a Sinking Fund and its advantages, you also know how important it is for better financial management. Sinking Funds are useful for companies because they avoid sudden financial stress by systematically setting aside money for significant expenses. By making adequate provisions to have an effective Sinking Fund, you can ensure smooth debt repayment, boost investor trust in your business and maintain long-term financial stability. Know MoreInvest now Want us to help you with anything? Request a Call back This field is required Only alphabetes are allowed This field is required Only alphabetes are allowed Please select a country Please enter valid number Please enter valid email Select options Please select product type Select options Please select product type Please enter valid pincode Thank you for your request. Your reference number is CRM 786578956 Our executive will contact you shortly Sorry! Please check back in a few minutes as an error has occurred. Share this Blog Copy Link Get social with us. 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15288	https://study.com/skill/learn/how-to-solve-absolute-value-inequalities-explanation.html	How to Solve Absolute Value Inequalities \| Algebra \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up How to Solve Absolute Value Inequalities Algebra 1 Skills Practice Click for sound 2:49 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 Example 1 Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Allison Cooper, Amy McKenney Instructors Allison Cooper Allison earned her PhD from the University of Oregon and has taught mathematics for over ten years, both as a graduate student and as a professor at Indiana University South Bend. She has taught developmental algebra, mathematics for elementary school teachers, college algebra, geometry, trigonometry, and calculus. She loves making seemingly difficult math topics more clear and understandable to students. When she's not tutoring, she enjoys hiking, gardening, cooking healthy meals, yoga, and cuddling with her cats! View bio Amy McKenney Amy has taught high school mathematics for over 14 years. She has a master's degree in education from Plymouth State University and her undergraduate degree in mathematics. She is certified to teach grades 7-12 mathematics. View bio Example SolutionsPractice Questions How to Solve Absolute Value Inequalities Step 1: To solve an absolute value inequality algebraically, first rewrite the inequality without absolute value symbols. \|a x+b\|<c⇒−c<a x+b<c \|a x+b\|≤c⇒−c≤a x+b≤c \|a x+b\|>c⇒a x+b<−c or a x+b>c \|a x+b\|≥c⇒a x+b≤−c or a x+b≥c Step 2: Solve the resulting linear inequalities by isolating the variable. Step 3: To solve the absolute value inequality graphically, set y=left-hand side and y=right-hand side and use a graphing calculator to graph the two equations. Step 4: Use the graphs from Step 3 to determine for which values of x the first graph is above (in case of > or ≥) or below (in case of < or ≤) the second graph. Include endpoints if necessary. Step 5: Compare the solutions from each method and verify they are same. How to Solve Absolute Value Inequalities: Vocabulary Absolute value: The absolute value of A, denoted \|A\|, is the distance between A and 0 on a number line. Absolute value inequality: If \|A\|<d, then A is within d of 0 on the number line. Therefore, A must be between −d and d: −d<A<d. The same holds for \|A\|≤d except that the endpoints are included: −d≤A≤d. If \|A\|>d, then the distance between A and 0 is greater than d. Therefore, A is either greater than d or less than -d: A<−d or A>d. The same holds for \|A\|≥d except that the endpoints are included: \|A\|≤−d or\|A\|≥d. Interval notation: Interval notation is used to express solutions to inequalities. These are the four possible solutions sets we can encounter when working with absolute value inqualities of the type in this lesson: a≤x≤b⇔[a,b]a<x<b⇔(a,b)x≤a or x≥b⇔(−∞,a]∪[b,∞)xb⇔(−∞,a)∪(b,∞) Basic absolute value graph: Now let's practice solving two absolute value inequalities, one with a bounded solution and one with an unbounded solution. How to Solve Absolute Value Inequalities: Example with Bounded Solution Solve for x both algebraically and graphically: \|2 x−5\|≤7. Step 1: To solve algebraically, we rewrite the absolute value inequality as −7≤2 x−5≤7. Step 2: The above compound inequality is equivalent to −7≤2 x−5 and 2 x−5≤7. For x to be a solution, it must satisfy both inequalities. We can solve the two inequalities simultaneously. To isolate x in the middle of the compound inequality, first add 5 to all three parts of the compound inequality: −2≤2 x≤12. Next, divide all three parts of the compound inequality by 2 to get the final result: −1≤x≤6. We can express the solution set using interval notation as [−1,6]. We use square braces to include −1 and 6 because the inequality symbol used in the solution set includes equality. Step 3: To solve the absolute value inequality graphically, we first set y equal to each side of the inequality y=\|2 x−5\|y=7. Step 4: Next we graph these two equations: From the graphs, we see that y=\|2 x−5\| is below (less than or equal to) y=7 for x between -1 and 6: −1≤x≤6 or[−1,6]. Again, we include the endpoints because the inequality symbol includes equality. We see that the solution set obtained algebraically matches the solution set obtained graphically, as desired. How to Solve Absolute Value Inequalities: Example with Unbounded Solution Solve for x both algebraically and graphically: \|x+4\|>2. Step 1: Rewriting the inequality without the absolute value symbols, we get x+4<−2 or x+4>2. Step 2: Again, we can solve these two inequalities simultaneously, but we must keep the word "or" between them. In other words, x is a solution if it satisfies either the first inequality or the second inequality. Subtracting 4, we get x<−6 or x>−2. We can express the solution set using interval notation as (−∞,−6)∪(−2,∞). We use parentheses next to −6 and −2 because those values are not included in the solution set. Step 3: To solve graphically, we graph y=\|x+4\|y=2 Step 4: We see that y=\|x+4\| is above (greater than) y=2 for values of x satisfying x<−6 or x>−2. We see that this solution set matches the solution set we obtained algebraically. Get access to thousands of practice questions and explanations! Create an account Table of Contents How to Solve Absolute Value Inequalities Vocabulary Example with Bounded Solution Example with Unbounded Solution Test your current knowledge Practice Solving Absolute Value Inequalities Related Courses Math 101: College Algebra Precalculus: Homework Help Resource Study.com ACT® Test Prep: Help and Review Glencoe Pre-Algebra: Online Textbook Help Study.com SAT Study Guide and Test Prep Related Lessons Domain & Range of a Function \| Definition, Equation & Examples Algebraic Function \| Definition, Types & Examples Relation in Math \| Definition, Representations & Examples Function Notation Definition, Evaluation & Examples Representation of Function \| Overview & Equations Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources Solving Problems Involving Systems of Equations Ethnic Groups in Indonesia \| Demographics & People Gods of the Winter Solstice Holes by Louis Sachar \| Themes, Quotes & Analysis Aurangzeb \| Empire, Achievements & Failures Libya Ethnic Groups \| Demographics, Population & Cultures Cold War Lesson for Kids: Facts & Timeline The Chronicles of Narnia Series by C.S. 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15289	https://www.youtube.com/watch?v=Jjou2hU6xnk	Math-e-Magic: Sum of hidden faces on a stack of dice D1 Station 70 subscribers 4 likes Description 137 views Posted: 16 Apr 2021 How to find the sum of the hidden faces on a stack of dice - play this on your friends and amaze them! Transcript: hello and today we're going to learn how to do the dice prediction trick first you need a partner just roll the dice then stack them and then i'm going to guess some of the hidden faces should i say it it's 11. you're wondering how i guessed that as you can see it's four plus one is five five plus six is 11. so my prediction was correct now we're going to see how to do the trick but first we'll see with three dice so again same rules roll the dice stack them up and i say that the sum of the hidden faces is 15. wondering how i guess that let's check one and another one two and a six eight eight plus six fourteen fourteen plus one fifteen again so so this trick always works and let's see how to do the trick first now first you're going to need a partner to roll the dice in any other and you should not be looking um and then ask your partner to stack up the die and as you can see in one die one side has a six on it and the other side has a one that's still seven it works for every side so what you're gonna do is roll them stack them up and look you have three eye with seven on each face we have to do three times seven which is twenty one if you have two guys do seven times two or two times seven which will have fourteen next you have twenty one and you have seen this top number when you're not gonna be looking take a sneak peek at it and you subtract that number from the um from the 21 you had it it depends if it's a 14 or 21 or maybe even a seven so um it depends on which numbers you have but you have to subtract the top number from your sum and then the remaining num the remaining hidden faces the sum of them will add up to your um subtracted values so 21 minus 6 equals 15 let's check 1 plus five six six plus two is six plus two is eight eight eight plus one is nine and nine plus six is fifteen so this trick always works dry eyes on your parents or whoever you have and you'll be amazed bye
15290	https://www.nationalarchives.gov.uk/latin/stage-2-latin/lessons/lesson-22-deponent-and-semi-deponent-verbs/	This website uses cookies We place some essential cookies on your device to make this website work. We'd like to use additional cookies to remember your settings and understand how you use our services. This information will help us make improvements to the website. Set cookie preferences Home > Latin > Stage 2 > Lessons > Lesson 10 – Deponent and semi-deponent verbs Lesson 10 – Deponent and semi-deponent verbs When a Latin verb is passive in form, but has an active meaning, it is called a deponent verb. For example: sequor, sequi, secutus sum (3) means ‘to follow’ and not ‘to be followed’. Even though it appears to be passive, it is translated with an active meaning and can have an object following it. For example: Robertus Willelmum sequitur – Robert follows William. Examples of deponent verbs \| Latin \| English \| --- \| \| conor, conari, conatus sum (1) \| to try \| \| ingredior, ingredi, ingressus sum (3) \| to enter \| \| loquor, loqui, locutus sum (3) \| to speak \| \| morior, mori, mortuus sum (3) \| to die \| \| ordior, ordiri, orsus sum (4) \| to begin \| \| orior, oriri, orsus sum (4) \| to rise \| \| potior, potiri, potitus sum (4) \| to gain mastery of \| \| queror, queri, questus sum (3) \| to complain \| \| sequor, sequi, secutus sum (3) \| to follow \| \| testor, testari, testatus sum (1) \| to witness \| \| utor, uti, usus sum (3) \| to use \| \| vereor, vereri, veritus sum (2) \| to fear \| Participles of deponent verbs Deponent verbs have participles, formed in the same way as for normal verbs and the meaning is always active. For example: The present participle for sequor, sequi, secutus sum (3) ‘to follow’ is sequens, sequentis ‘following’. The past participle for sequor, sequi, secutus sum (3) ‘to follow’ is secutus, -a, -um ‘having followed’. The future participle for sequor, sequi, secutus sum (3) ‘to follow’ is secuturus, -a, -um ‘about to follow’. Semi-deponent verbs For these verbs only the perfect tense, and the tenses formed from it, are in the passive form, but just as with deponent verbs, the meaning is always active. Examples of semi-deponent verbs \| Latin \| English \| --- \| \| audeo, audere, ausus sum (2) \| to dare \| \| confido, confidere, confisus sum (3) \| to trust \| \| diffido, diffidere, diffisus sum (3) \| to distrust \| \| fido, fidere, fisus sum (3) \| to trust \| \| gaudeo, gaudere, gavisus sum (2) \| to be glad \| \| soleo, solere, solitus sum (2) \| to be accustomed \| Handy hint gaudere and solere(the last two verbs, in bold) are those which you have the greatest chance of encountering. However, it is worth noting that the past participles of these two verbs can also be used in a passive way, meaning ‘enjoyed’ and ‘accustomed’. Fortunately, in most cases, the overall sense of the sentence you will be working with will help you to translate this correctly. Checklist Are you confident with the meaning of a deponent and semi-deponent verb? the form of a deponent and semi-deponent verb? What next? Go to Lesson 23
15291	https://polytope.miraheze.org/wiki/Alternation	Alternation - Polytope Wiki Alternation From Polytope Wiki Jump to navigationJump to search A non-uniformsquare antiprism, created by alternation from an octagonal prism. Alternation (also known as alternated faceting) is a procedure by which half of some elements of a polytope are removed, thus creating a new one. Most often, and unless otherwise specified, the elements taken are vertices, but edges and other elements may sometimes be alternated too. The process of vertex alternation is closely related to snubbing. It applies to any polytope whose vertex adjacency graph is bipartite. To alternate a polytope, one first 2-colors the chosen elements, and removes all of those of a certain color, say black. The facets of the polytope then either are alternations of the former facets in turn or are so called sefas (sectioning facets underneath the being removed element of the original polytope). In case of vertex alternations those would just be the according vertex figures. If this process creates any degenerate facets, such as digons, these usually are removed. For instance, the alternation of a cube is considered to be the tetrahedron, but could well be treated as a tetrahedron with extra digons at each edge. Generally, any given polytope with an alternation of some of its elements has in fact two different alternations, resulting from either color choice of elements to remove. For instance, a rhombic dodecahedron can either be vertex alternated into a cube or into an octahedron. In the case where the polytope is uniform, however, both vertex alternations result in congruent polytopes (albeit sometimes in an enantiomorph pair). Thus, in this special case, alternation can be regarded as giving a unique result. Though having faces with an even amount of sides is a necessary condition for a polytope to be globally alternatable, this turns out not to be sufficient in the general case. Nevertheless, all convexpolyhedra whose faces have an even amount of sides can be alternated. Relation to snubbing [edit \| edit source] Main article: Snubbing Snubbing however adds to the process of mere vertex alternation usually also the secondary process of edge resizement back to all unit edges. It is this secondary process, which might or might not be applicable. The mere alternation however always is - at least locally, cf. the theorem below. In its oldest use of the word, snubbing was applied to omnitruncates only, but later became applied more generally. Alternation, when taken as a local process in fact alternatingly maintains respectively rejects a vertex (say, or any other element). Whenever all 2D faces have an even count of vertices, then that local process returns to the starting point again within the right parity whichever path you went. As soon as there are 2D faces encountered with an odd count of vertices, then the same local operation might return within the wrong parity to the starting point. That then would require to maintain as well as to reject that vertex. In fact, when this local application then being continued nonetheless is just what a holoalternation (holosnubbing) is. I.e. every vertex then gets maintained as such as well as replaced by its vertex figure (or the respective sectioning facet underneath). Accordingly a snub always has half the vertex count of its starting figure. But a holosnub always has the same vertex count as its starting figure. Alternatively one could also think of holosnubbing when one replaces the starting figure with the according Grünbaumian double cover, i.e. the one which uses 2 fully incident copies of every even 2D face and uses the doubly wound doublecover for every odd 2D face. If one then would apply usual snubbing to that replacement, the outcome surely is nothing else than the process of holosnubbing being applied to the starting figure instead. Examples [edit \| edit source] The following are examples of polytopes resulting from alternation. The vertex alternation of the hexagon results in the triangle. The alternation of one of the 2 edge types of a ditetragon results in a rectangle. The vertex alternation of the cube results in the tetrahedron. The alternation of the great rhombicuboctahedron results in a (non-uniform) snub cube. For every n, the vertex alternation of the 2n-gonal prism results in a (non-uniform) n-gonal antiprism. For every n, the alternation of the lacing edges of a 2n-gonal prism results in a (non-uniform) n-gonal prism. The alternation of the triangles of a small rhombicuboctahedron results in a (non-uniform) truncated tetrahedron. For every n, the alternation of the n-hypercube results in the n-demihypercube. The rhombic dodecahedron can be vertex-alternated in two different ways. One produces the cube while the other produces the octahedron. Alternating a square tiling produces another square tiling, with edges sqrt(2) times longer than the original. Which polytopes are alternatable? [edit \| edit source] A petrial tetrahedron cannot be alternated. A polytope is vertex alternatable iff its vertex adjacency graph is bipartite. Particularly, each one of its faces must have an even amount of sides. It might be tempting to declare that conversely, every polytope whose faces all have an even amount of sides is alternatable, but this turns out not to be the case. A simple counterexample is the Petrial tetrahedron, whose faces are all blended squares, but whose vertex adjacency graph is that of the tetrahedron, and therefore is not bipartite. There are also convex counterexamples with infinitely many faces and/or vertices. For instance, if an infinite amount of triangular prisms are joined by their triangles, the resulting apeirohedron will not be vertex alternatable, even though all of its faces will be squares. Nevertheless, the following result can be established for convex polyhedra. Theorem—Every convex polyhedron whose faces all have evenly many sides can be vertex alternated. Proof— A convex polyhedron's vertices, edges and faces can all be put in correspondence with those of a planar graph. Thus, it suffices to prove that any finite planar graph G whose faces all have evenly many sides is bipartite. Suppose for sake of contradiction that G contains some cycle with an odd amount of vertices. Since G is finite, there are finitely many such cycles, and we may thus take the one that encloses the least area. Call this cycle C. If any two vertices in C were linked by an edge interior to C, then C could be subdivided into two cycles enclosing a lesser area, one of which would necessarily have an odd amount of vertices, thus contradicting the minimality of the area of C. Thus, C must enclose a connected area, and is thus a face of G with an odd amount of sides, contradicting the assumption that G contains no such face. Reaching a contradiction in either case, we conclude our theorem. Retrieved from " Category: Polytope operations Cookies help us deliver our services. By using our services, you agree to our use of cookies. 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15292	https://elsevier-elibrary.com/contents/fullcontent/83273/epubcontent_v2/OPS/xhtml/chp00078.xhtml	Guyton and Hall Textbook of Medical Physiology Chapter 78 Adrenocortical Hormones The two adrenal glands, each of which weighs about 4 grams, lie at the superior poles of the two kidneys. As shown in Figure 78-1, each gland is composed of two major parts, the adrenal medulla and the adrenal cortex. The adrenal medulla, the central 20 percent of the gland, is functionally related to the sympathetic nervous system; it secretes the hormones epinephrine and norepinephrine in response to sympathetic stimulation. In turn, these hormones cause almost the same effects as direct stimulation of the sympathetic nerves in all parts of the body. These hormones and their effects are discussed in detail in Chapter 61 in relation to the sympathetic nervous system. Figure 78-1Secretion of adrenocortical hormones by the different zones of the adrenal cortex and secretion of catecholamines by the adrenal medulla. The adrenal cortex secretes an entirely different group of hormones, called corticosteroids. These hormones are all synthesized from the steroid cholesterol, and they all have similar chemical formulas. However, slight differences in their molecular structures give them several different but very important functions. Corticosteroids: Mineralocorticoids, Glucocorticoids, and Androgens Two major types of adrenocortical hormones, the mineralocorticoids and the glucocorticoids, are secreted by the adrenal cortex. In addition to these hormones, small amounts of sex hormones are secreted, especially androgenic hormones, which exhibit about the same effects in the body as the male sex hormone testosterone. They are normally of only slight importance, although in certain abnormalities of the adrenal cortices, extreme quantities can be secreted (which is discussed later in the chapter) and can result in masculinizing effects. The mineralocorticoids have gained this name because they especially affect the electrolytes (the “minerals”) of the extracellular fluids, especially sodium and potassium. The glucocorticoids have gained their name because they exhibit important effects that increase blood glucose concentration. They have additional effects on both protein and fat metabolism that are equally as important to body function as their effects on carbohydrate metabolism. More than 30 steroids have been isolated from the adrenal cortex, but two are of exceptional importance to the normal endocrine function of the human body: aldosterone, which is the principal mineralocorticoid, and cortisol, which is the principal glucocorticoid. Synthesis and Secretion of Adrenocortical Hormones The Adrenal Cortex Has Three Distinct Layers Figure 78-1 shows that the adrenal cortex is composed of three relatively distinct layers: The zona glomerulosa, a thin layer of cells that lies just underneath the capsule, constitutes about 15 percent of the adrenal cortex. These cells are the only ones in the adrenal gland capable of secreting significant amounts of aldosterone because they contain the enzyme aldosterone synthase, which is necessary for synthesis of aldosterone. The secretion of these cells is controlled mainly by the extracellular fluid concentrations of angiotensin II and potassium, both of which stimulate aldosterone secretion. The zona fasciculata, the middle and widest zone, constitutes about 75 percent of the adrenal cortex and secretes the glucocorticoids cortisol and corticosterone, as well as small amounts of adrenal androgens and estrogens. The secretion of these cells is controlled in large part by the hypothalamic-pituitary axis via adrenocorticotropic hormone (ACTH). The zona reticularis, the inner zone of the cortex, secretes the adrenal androgens dehydroepiandrosterone and androstenedione, as well as small amounts of estrogens and some glucocorticoids. ACTH also regulates secretion of these cells, although other factors such as cortical androgen-stimulating hormone, released from the pituitary, may also be involved. The mechanisms for controlling adrenal androgen production, however, are not nearly as well understood as those for glucocorticoids and mineralocorticoids. Aldosterone and cortisol secretion are regulated by independent mechanisms. Factors such as angiotensin II that specifically increase the output of aldosterone and cause hypertrophy of the zona glomerulosa have no effect on the other two zones. Similarly, factors such as ACTH that increase secretion of cortisol and adrenal androgens and cause hypertrophy of the zona fasciculata and zona reticularis have little effect on the zona glomerulosa. Adrenocortical Hormones Are Steroids Derived from Cholesterol. All human steroid hormones, including those produced by the adrenal cortex, are synthesized from cholesterol. Although the cells of the adrenal cortex can synthesize de novo small amounts of cholesterol from acetate, approximately 80 percent of the cholesterol used for steroid synthesis is provided by low-density lipoproteins (LDLs) in the circulating plasma. The LDLs, which have high concentrations of cholesterol, diffuse from the plasma into the interstitial fluid and attach to specific receptors contained in structures called coated pits on the adrenocortical cell membranes. The coated pits are then internalized by endocytosis, forming vesicles that eventually fuse with cell lysosomes and release cholesterol that can be used to synthesize adrenal steroid hormones. Transport of cholesterol into the adrenal cells is regulated by feedback mechanisms that can markedly alter the amount available for steroid synthesis. For example, ACTH, which stimulates adrenal steroid synthesis, increases the number of adrenocortical cell receptors for LDL, as well as the activity of enzymes that liberate cholesterol from LDL. Once the cholesterol enters the cell, it is delivered to the mitochondria, where it is cleaved by the enzyme cholesterol desmolase to form pregnenolone; this is the rate-limiting step in the eventual formation of adrenal steroids (Figure 78-2). In all three zones of the adrenal cortex, this initial step in steroid synthesis is stimulated by the different factors that control secretion of the major hormone products aldosterone and cortisol. For example, both ACTH, which stimulates cortisol secretion, and angiotensin II, which stimulates aldosterone secretion, increase the conversion of cholesterol to pregnenolone. Figure 78-2Pathways for synthesis of steroid hormones by the adrenal cortex. The enzymes are shown in italics. Synthetic Pathways for Adrenal Steroids. Figure 78-2 gives the principal steps in the formation of the important steroid products of the adrenal cortex: aldosterone, cortisol, and the androgens. Essentially all these steps occur in two of the organelles of the cell, the mitochondria and the endoplasmic reticulum, with some steps occurring in one of these organelles and some in the other. Each step is catalyzed by a specific enzyme system. A change in even a single enzyme in the schema can cause vastly different types and relative proportions of hormones to be formed. For example, very large quantities of masculinizing sex hormones or other steroid compounds not normally present in the blood can occur with altered activity of only one of the enzymes in this pathway. The chemical formulas of aldosterone and cortisol, which are the major mineralocorticoid and glucocorticoid hormones, respectively, are shown in Figure 78-2. Cortisol has a keto-oxygen on carbon number 3 and is hydroxylated at carbon numbers 11 and 21. The mineralocorticoid aldosterone has an oxygen atom bound at the number 18 carbon. In addition to aldosterone and cortisol, other steroids having glucocorticoid or mineralocorticoid activities, or both, are normally secreted in small amounts by the adrenal cortex. Furthermore, several additional potent steroid hormones not normally formed in the adrenal glands have been synthesized and are used in various forms of therapy. Some of the more important of the corticosteroid hormones, including the synthetic ones, are the following, as summarized in Table 78-1. Table 78-1 Adrenal Steroid Hormones in Adults; Synthetic Steroids and Their Relative Glucocorticoid and Mineralocorticoid Activities Steroids Average Plasma Concentration (free and bound, µg/100 ml)Average Amount Secreted (mg/24 hr)Glucocorticoid Activity Mineralocorticoid Activity Adrenal steroids Cortisol 12 15 1.0 1.0 Corticosterone 0.4 3 0.3 15.0 Aldosterone 0.006 0.15 0.3 3000 Deoxycorticosterone 0.006 0.2 0.2 100 Dehydroepiandrosterone 175 20—— Synthetic steroids Cortisone——0.7 0.5 Prednisolone——4 0.8 Methylprednisone——5— Dexamethasone——30— 9α-fluorocortisol——10 125 Glucocorticoid and mineralocorticoid activities of the steroids are relative to cortisol, with cortisol being 1.0. Mineralocorticoids • Aldosterone (very potent; accounts for about 90 percent of all mineralocorticoid activity) • Deoxycorticosterone (1/30 as potent as aldosterone, but very small quantities are secreted) • Corticosterone (slight mineralocorticoid activity) • 9α-Fluorocortisol (synthetic; slightly more potent than aldosterone) • Cortisol (very slight mineralocorticoid activity, but a large quantity is secreted) • Cortisone (slight mineralocorticoid activity) Glucocorticoids • Cortisol (very potent; accounts for about 95 percent of all glucocorticoid activity) • Corticosterone (provides about 4 percent of total glucocorticoid activity, but is much less potent than cortisol) • Cortisone (almost as potent as cortisol) • Prednisone (synthetic; four times as potent as cortisol) • Methylprednisone (synthetic; five times as potent as cortisol) • Dexamethasone (synthetic; 30 times as potent as cortisol) It is clear from this list that some of these hormones and synthetic steroids have both glucocorticoid and mineralocorticoid activities. It is especially significant that cortisol normally has some mineralocorticoid activity, because some syndromes of excess cortisol secretion can cause significant mineralocorticoid effects, along with its much more potent glucocorticoid effects. The intense glucocorticoid activity of the synthetic hormone dexamethasone, which has almost zero mineralocorticoid activity, makes it an especially important drug for stimulating specific glucocorticoid activity. Adrenocortical Hormones Are Bound to Plasma Proteins. Approximately 90 to 95 percent of the cortisol in the plasma binds to plasma proteins, especially a globulin called cortisol-binding globulin or transcortin and, to a lesser extent, to albumin. This high degree of binding to plasma proteins slows the elimination of cortisol from the plasma; therefore, cortisol has a relatively long half-life of 60 to 90 minutes. Only about 60 percent of circulating aldosterone combines with the plasma proteins, so about 40 percent is in the free form; as a result, aldosterone has a relatively short half-life of about 20 minutes. These hormones are transported throughout the extracellular fluid compartment in both the combined and free forms. Binding of adrenal steroids to the plasma proteins may serve as a reservoir to lessen rapid fluctuations in free hormone concentrations, as would occur, for example, with cortisol during brief periods of stress and episodic secretion of ACTH. This reservoir function may also help to ensure a relatively uniform distribution of the adrenal hormones to the tissues. Adrenocortical Hormones Are Metabolized in the Liver. The adrenal steroids are degraded mainly in the liver and are conjugated especially to glucuronic acid and, to a lesser extent, to sulfates. These substances are inactive and do not have mineralocorticoid or glucocorticoid activity. About 25 percent of these conjugates are excreted in the bile and then in the feces. The remaining conjugates formed by the liver enter the circulation but are not bound to plasma proteins, are highly soluble in the plasma, and are therefore filtered readily by the kidneys and excreted in the urine. Diseases of the liver markedly depress the rate of inactivation of adrenocortical hormones, and kidney diseases reduce the excretion of the inactive conjugates. The normal concentration of aldosterone in blood is about 6 nanograms (6 billionths of a gram) per 100 milliliters, and the average secretory rate is approximately 150 µg/day (0.15 mg/day). The blood concentration of aldosterone, however, depends greatly on several factors, including dietary intake of sodium and potassium. The concentration of cortisol in the blood averages 12 µg/100 ml, and the secretory rate averages 15 to 20 mg/day. However, blood concentration and secretion rate of cortisol fluctuate throughout the day, rising in the early morning and declining in the evening, as discussed later. Functions of the Mineralocorticoids—Aldosterone Mineralocorticoid Deficiency Causes Severe Renal Sodium Chloride Wasting and Hyperkalemia. Total loss of adrenocortical secretion may cause death within 3 days to 2 weeks unless the person receives extensive salt therapy or injection of mineralocorticoids. Without mineralocorticoids, potassium ion concentration of the extracellular fluid rises markedly, sodium and chloride are rapidly lost from the body, and the total extracellular fluid volume and blood volume become greatly reduced. Diminished cardiac output soon develops, which progresses to a shock-like state, followed by death. This entire sequence can be prevented by the administration of aldosterone or some other mineralocorticoid. Therefore, the mineralocorticoids are said to be the acute “lifesaving” portion of the adrenocortical hormones. The glucocorticoids are equally necessary, however, because they allow the person to resist the destructive effects of life's intermittent physical and mental “stresses,” as discussed later in the chapter. Aldosterone Is the Major Mineralocorticoid Secreted by the Adrenals. In humans, aldosterone exerts nearly 90 percent of the mineralocorticoid activity of the adrenocortical secretions, but cortisol, the major glucocorticoid secreted by the adrenal cortex, also provides a significant amount of mineralocorticoid activity. The mineralocorticoid activity of aldosterone is about 3000 times greater than that of cortisol, but the plasma concentration of cortisol is nearly 2000 times that of aldosterone. Cortisol can also bind to mineralocorticoid receptors with high affinity. However, the renal epithelial cells express the enzyme 11β-hydroxysteroid dehydrogenase type 2 (11β-HSD2), which has actions that prevent cortisol from activating mineralocorticoid receptors. One action of 11β-HSD2 is to convert cortisol to cortisone, which does not avidly bind mineralocorticoid receptors. There is also evidence that 11β-HSD2 may have effects on the intracellular redox (reduction and oxidation) state that prevent cortisol from activating the mineralocorticoid receptors. In patients with genetic deficiency of 11β-HSD2 activity, cortisol may have substantial mineralocorticoid effects. This condition is called apparent mineralocorticoid excess syndrome (AME) because the patient has essentially the same pathophysiological changes as a patient with excess aldosterone secretion, except that plasma aldosterone levels are very low in the patient with AME. Ingestion of large amounts of licorice, which contains glycyrrhetinic acid, may also cause AME because of its ability to block 11β-HSD2 enzyme activity. Renal and Circulatory Effects of Aldosterone Aldosterone Increases Renal Tubular Reabsorption of Sodium and Secretion of Potassium. It will be recalled from Chapter 28 that aldosterone increases reabsorption of sodium and simultaneously increases secretion of potassium by the renal tubular epithelial cells, especially in the principal cells of the collecting tubules and, to a lesser extent, in the distal tubules and collecting ducts. Therefore, aldosterone causes sodium to be conserved in the extracellular fluid while increasing potassium excretion in the urine. A high concentration of aldosterone in the plasma can transiently decrease the sodium loss into the urine to as little as a few milliequivalents a day. At the same time, potassium loss into the urine transiently increases severalfold. Therefore, the net effect of excess aldosterone in the plasma is to increase the total quantity of sodium in the extracellular fluid while decreasing the potassium. Conversely, total lack of aldosterone secretion can cause transient loss of 10 to 20 grams of sodium in the urine a day, an amount equal to one tenth to one fifth of all the sodium in the body. At the same time, potassium is conserved tenaciously in the extracellular fluid. Excess Aldosterone Increases Extracellular Fluid Volume and Arterial Pressure But Has Only a Small Effect on Plasma Sodium Concentration. Although aldosterone has a potent effect to decrease the rate of sodium excretion by the kidneys, the concentration of sodium in the extracellular fluid often rises only a few milliequivalents. The reason for this is that when sodium is reabsorbed by the tubules, simultaneous osmotic absorption of almost equivalent amounts of water occurs. Also, small increases in extracellular fluid sodium concentration stimulate thirst and increased water intake, if water is available, and increase secretion of antidiuretic hormone, which enhances water reabsorption by the distal and collecting tubules of the kidneys. Therefore, the extracellular fluid volume increases almost as much as the retained sodium, but without much change in sodium concentration. Even though aldosterone is one of the body's most powerful sodium-retaining hormones, only transient sodium retention occurs when excess amounts are secreted. An aldosterone-mediated increase in extracellular fluid volume lasting more than 1 to 2 days also leads to an increase in arterial pressure, as explained in Chapter 19. The rise in arterial pressure then increases kidney excretion of both sodium and water, called pressure natriuresis and pressure diuresis, respectively. Thus, after the extracellular fluid volume increases 5 to 15 percent above normal, arterial pressure also increases 15 to 25 mm Hg, and this elevated blood pressure returns the renal output of sodium and water to normal despite the excess aldosterone (Figure 78-3). Figure 78-3Effect of aldosterone infusion on arterial pressure, extracellular fluid volume, and sodium excretion in dogs. Although aldosterone was infused at a rate that raised plasma concentrations to about 20 times normal, note the “escape” from sodium retention on the second day of infusion as arterial pressure increased and urinary sodium excretion returned to normal.(Data from Hall JE, Granger JP, Smith MJ Jr, et al: Role of hemodynamics and arterial pressure in aldosterone “escape.” Hypertension 6[suppl I]:I183-I192, 1984.) This return to normal of sodium and water excretion by the kidneys as a result of pressure natriuresis and diuresis is called aldosterone escape. Thereafter, the rate of gain of sodium and water by the body is zero, and balance is maintained between sodium and water intake and output by the kidneys despite continued excess aldosterone. In the meantime, however, hypertension has developed, which lasts as long as the person remains exposed to high levels of aldosterone. Conversely, when aldosterone secretion becomes zero, large amounts of sodium are lost in the urine, not only diminishing the amount of sodium chloride in the extracellular fluid but also decreasing the extracellular fluid volume. The result is severe extracellular fluid dehydration and low blood volume, leading to circulatory shock. Without therapy, this usually causes death within a few days after the adrenal glands suddenly stop secreting aldosterone. Excess Aldosterone Causes Hypokalemia and Muscle Weakness; Aldosterone Deficiency Causes Hyperkalemia and Cardiac Toxicity. Excess aldosterone not only causes loss of potassium ions from the extracellular fluid into the urine but also stimulates transport of potassium from the extracellular fluid into most cells of the body. Therefore, excessive secretion of aldosterone, as occurs with some types of adrenal tumors, may cause a serious decrease in the plasma potassium concentration, sometimes from the normal value of 4.5 mEq/L to as low as 2 mEq/L. This condition is called hypokalemia. When the potassium ion concentration falls below about one-half normal, severe muscle weakness often develops. This muscle weakness is caused by alteration of the electrical excitability of the nerve and muscle fiber membranes (see Chapter 5), which prevents transmission of normal action potentials. Conversely, when aldosterone is deficient, the extracellular fluid potassium ion concentration can rise far above normal. When it rises to 60 to 100 percent above normal, serious cardiac toxicity, including weakness of heart contraction and development of arrhythmia, becomes evident, and progressively higher concentrations of potassium lead inevitably to heart failure. Excess Aldosterone Increases Tubular Hydrogen Ion Secretion and Causes Alkalosis. Aldosterone not only causes potassium to be secreted into the tubules in exchange for sodium reabsorption in the principal cells of the renal collecting tubules but also causes secretion of hydrogen ions in exchange for potassium in the intercalated cells of the cortical collecting tubules, as discussed in Chapters 28 and 31. This decreases the hydrogen ion concentration in the extracellular fluid, causing metabolic alkalosis. Aldosterone Stimulates Sodium and Potassium Transport in Sweat Glands, Salivary Glands, and Intestinal Epithelial Cells Aldosterone has almost the same effects on sweat glands and salivary glands as it has on the renal tubules. Both these glands form a primary secretion that contains large quantities of sodium chloride, but much of the sodium chloride, upon passing through the excretory ducts, is reabsorbed, whereas potassium and bicarbonate ions are secreted. Aldosterone greatly increases the reabsorption of sodium chloride and the secretion of potassium by the ducts. The effect on the sweat glands is important to conserve body salt in hot environments, and the effect on the salivary glands is necessary to conserve salt when excessive quantities of saliva are lost. Aldosterone also greatly enhances sodium absorption by the intestines, especially in the colon, which prevents loss of sodium in the stools. Conversely, in the absence of aldosterone, sodium absorption can be poor, leading to failure to absorb chloride and other anions and water as well. The unabsorbed sodium chloride and water then lead to diarrhea, with further loss of salt from the body. Cellular Mechanism of Aldosterone Action Although for many years we have known the overall effects of mineralocorticoids on the body, the molecular mechanisms of the actions of aldosterone on the tubular cells to increase transport of sodium are still not fully understood. However, the cellular sequence of events that leads to increased sodium reabsorption seems to unfold as follows. First, because of its lipid solubility in the cellular membranes, aldosterone diffuses readily to the interior of the tubular epithelial cells. Second, in the cytoplasm of the tubular cells, aldosterone combines with a highly specific cytoplasmic mineralocorticoid receptor (MR) protein (Figure 78-4), which has a stereomolecular configuration that allows only aldosterone or similar compounds to combine with it. Although renal tubular epithelial cell MR receptors also have a high affinity for cortisol, the enzyme 11β-HSD2 normally converts most of the cortisol to cortisone, which does not readily bind to MR receptors, as discussed previously. Figure 78-4Aldosterone-responsive epithelial cell signaling pathways. Activation of the mineralocorticoid receptor (MR) by aldosterone can be antagonized with spironolactone. Amiloride is a drug that can be used to block epithelial sodium channel proteins (ENaC). Third, the aldosterone-receptor complex or a product of this complex diffuses into the nucleus, where it may undergo further alterations, finally inducing one or more specific portions of the DNA to form one or more types of messenger RNA (mRNA) related to the process of sodium and potassium transport. Fourth, the mRNA diffuses back into the cytoplasm, where, operating in conjunction with the ribosomes, it causes protein formation. The proteins formed are a mixture of (1) one or more enzymes and (2) membrane transport proteins that, all acting together, are required for sodium, potassium, and hydrogen transport through the cell membrane (see Figure 78-4). One of the enzymes especially increased is sodium-potassium adenosine triphosphatase, which serves as the principal part of the pump for sodium and potassium exchange at the basolateral membranes of the renal tubular cells. Additional proteins, perhaps equally important, are epithelial sodium channel proteins inserted into the luminal membrane of the same tubular cells that allow rapid diffusion of sodium ions from the tubular lumen into the cell; then the sodium is pumped the rest of the way by the sodium-potassium pump located in the basolateral membranes of the cell. Thus, aldosterone does not have a major immediate effect on sodium transport; rather, this effect must await the sequence of events that leads to the formation of the specific intracellular substances required for sodium transport. About 30 minutes is required before new RNA appears in the cells, and about 45 minutes is required before the rate of sodium transport begins to increase; the effect reaches maximum only after several hours. Possible Nongenomic Actions of Aldosterone and Other Steroid Hormones Some studies suggest that many steroids, including aldosterone, elicit not only slowly developing genomic effects that have a latency of 45 to 60 minutes and require gene transcription and synthesis of new proteins, but also more rapid nongenomic effects that take place in a few seconds or minutes. These nongenomic actions are believed to be mediated by binding of steroids to cell membrane receptors that are coupled to second messenger systems, similar to those used for peptide hormone signal transduction. For example, aldosterone has been shown to increase formation of cyclic adenosine monophosphate (cAMP) in vascular smooth muscle cells and in epithelial cells of the renal collecting tubules in less than 2 minutes, a period that is far too short for gene transcription and synthesis of new proteins. In other cell types, aldosterone has been shown to rapidly stimulate the phosphatidylinositol second messenger system. However, the precise structure of receptors responsible for the rapid effects of aldosterone has not been determined, nor is the physiological significance of these nongenomic actions of steroids well understood. Regulation of Aldosterone Secretion Regulation of aldosterone secretion is so deeply intertwined with regulation of extracellular fluid electrolyte concentrations, extracellular fluid volume, blood volume, arterial pressure, and many special aspects of renal function that it is difficult to discuss control of aldosterone secretion independently of all these other factors. This subject is presented in more detail in Chapters 29 and 30, to which the reader is referred. However, it is important to list here some of the more important points of aldosterone secretion control. Regulation of aldosterone secretion by the zona glomerulosa cells is almost entirely independent of regulation of cortisol and androgens by the zona fasciculata and zona reticularis. The following four factors are known to play essential roles in regulation of aldosterone: Increased potassium ion concentration in the extracellular fluid greatly increases aldosterone secretion. Increased angiotensin II concentration in the extracellular fluid also greatly increases aldosterone secretion. Increased sodium ion concentration in the extracellular fluid very slightly decreases aldosterone secretion. ACTH from the anterior pituitary gland is necessary for aldosterone secretion but has little effect in controlling the rate of secretion in most physiological conditions. Of these factors, potassium ion concentration and the renin-angiotensin system are by far the most potent in regulating aldosterone secretion. A small percentage increase in potassium concentration can cause a severalfold increase in aldosterone secretion. Likewise, activation of the renin-angiotensin system, usually in response to diminished blood flow to the kidneys or to sodium loss, can increase aldosterone secretion severalfold. In turn, the aldosterone acts on the kidneys (1) to help them excrete the excess potassium ions and (2) to increase the blood volume and arterial pressure, thus returning the renin-angiotensin system toward its normal level of activity. These feedback control mechanisms are essential for maintaining life, and the reader is referred again to Chapters 28 and 30 for a more complete description of their functions. Figure 78-5 shows the effects on plasma aldosterone concentration caused by blocking formation of angiotensin II with an angiotensin-converting enzyme inhibitor after several weeks of a low-sodium diet that increases plasma aldosterone concentration. Note that blockade of angiotensin II formation markedly decreases plasma aldosterone concentration without significantly changing cortisol concentration, which indicates the important role of angiotensin II in stimulating aldosterone secretion when sodium intake and extracellular fluid volume are reduced. Figure 78-5Effects of treating sodium-depleted dogs with an angiotensin-converting enzyme (ACE) inhibitor for 7 days to block formation of angiotensin II (Ang II) and of infusing exogenous Ang II to restore plasma Ang II levels after ACE inhibition. Note that blocking Ang II formation reduced plasma aldosterone concentration with little effect on cortisol, demonstrating the important role of Ang II in stimulating aldosterone secretion during sodium depletion.(Data from Hall JE, Guyton AC, Smith MJ Jr, et al: Chronic blockade of angiotensin II formation during sodium deprivation. Am J Physiol 237:F424, 1979.) By contrast, the effects of sodium ion concentration per se and of ACTH in controlling aldosterone secretion are usually minor. Nevertheless, a 10 to 20 percent decrease in extracellular fluid sodium ion concentration, which occurs on rare occasions, can perhaps increase aldosterone secretion by about 50 percent. In the case of ACTH, if even a small amount is secreted by the anterior pituitary gland, it is usually enough to permit the adrenal glands to secrete whatever amount of aldosterone is required, but total absence of ACTH can significantly reduce aldosterone secretion. Therefore, ACTH appears to play a “permissive” role in regulation of aldosterone secretion. Functions of Glucocorticoids Even though mineralocorticoids can save the life of an acutely adrenalectomized animal, the animal still is far from normal. Instead, the animal's metabolic systems for utilization of proteins, carbohydrates, and fats remain considerably deranged. Furthermore, the animal cannot resist different types of physical or even mental stress, and minor illnesses such as respiratory tract infections can lead to death. Therefore, the glucocorticoids have functions just as important to the long-continued life of the animal as those of the mineralocorticoids. These functions are explained in the following sections. At least 95 percent of the glucocorticoid activity of the adrenocortical secretions results from the secretion of cortisol, known also as hydrocortisone. In addition, a small but significant amount of glucocorticoid activity is provided by corticosterone. Effects of Cortisol on Carbohydrate Metabolism Stimulation of Gluconeogenesis. The best-known metabolic effect of cortisol and other glucocorticoids on metabolism is the ability to stimulate gluconeogenesis (i.e., the formation of carbohydrate from proteins and some other substances) by the liver, often increasing the rate of gluconeogenesis as much as 6- to 10-fold. This increased rate of gluconeogenesis results mainly from direct effects of cortisol on the liver, as well as by antagonizing the effects of insulin. Cortisol increases the enzymes required to convert amino acids into glucose in liver cells. This results from the effect of glucocorticoids to activate DNA transcription in the liver cell nuclei in the same way that aldosterone functions in renal tubular cells, with formation of mRNAs that in turn lead to the array of enzymes required for gluconeogenesis. Cortisol causes mobilization of amino acids from the extrahepatic tissues, mainly from muscle. As a result, more amino acids become available in the plasma to enter into the gluconeogenesis process of the liver and thereby to promote the formation of glucose. Cortisol antagonizes insulin's effects to inhibit gluconeogenesis in the liver. As discussed in Chapter 79, insulin stimulates glycogen synthesis in the liver and inhibits enzymes involved in glucose production by the liver. The net effect of cortisol is to increase glucose production by the liver. The marked increase in glycogen storage in liver cells that accompanies increased gluconeogenesis potentiates the effects of other glycolytic hormones, such as epinephrine and glucagon, to mobilize glucose in times of need, such as between meals. Decreased Glucose Utilization by Cells. Cortisol also causes a moderate decrease in glucose utilization by most cells in the body. Although the precise cause of this decrease is unclear, one important effect of cortisol is to decrease translocation of the glucose transporters GLUT 4 to the cell membrane, especially in skeletal muscle cells, leading to insulin resistance. Glucocorticoids may also depress the expression and phosphorylation of other signaling cascades that influence glucose utilization directly or indirectly by affecting protein and lipid metabolism. For example, glucocorticoids have been reported to reduce the expression of insulin receptor substrate–1 and phosphatidylinositol 3 kinase, both of which are involved in mediating the actions of insulin, as well as oxidation of nicotinamide-adenine dinucleotide (NADH) to form NAD+. Because NADH must be oxidized to allow glycolysis, this effect could also contribute to diminished utilization of glucose by the cells. Elevated Blood Glucose Concentration and “Adrenal Diabetes.” Both the increased rate of gluconeogenesis and the moderate reduction in the rate of glucose utilization by the cells cause the blood glucose concentrations to rise. The rise in blood glucose in turn stimulates secretion of insulin. The increased plasma levels of insulin, however, are not as effective in maintaining plasma glucose as they are under normal conditions. For reasons that were discussed previously, high levels of glucocorticoid reduce the sensitivity of many tissues, especially skeletal muscle and adipose tissue, to the stimulatory effects of insulin on glucose uptake and utilization. Besides potential direct effects of cortisol on expression of glucose transporters and enzymes involved in glucose regulation, the high levels of fatty acids, caused by the effect of glucocorticoids to mobilize lipids from fat depots, may impair the actions of insulin on the tissues. In this way, excess secretion of glucocorticoids may produce disturbances of carbohydrate metabolism similar to those found in patients with excess levels of growth hormone. The increase in blood glucose concentration is occasionally great enough (50 percent or more above normal) that the condition is called adrenal diabetes. Administration of insulin lowers the blood glucose concentration only a moderate amount in adrenal diabetes—not nearly as much as it does in pancreatic diabetes—because the tissues are resistant to the effects of insulin. Effects of Cortisol on Protein Metabolism Reduction in Cellular Protein. One of the principal effects of cortisol on the metabolic systems of the body is reduction of the protein stores in essentially all cells of the body except those of the liver. This reduction is caused by both decreased protein synthesis and increased catabolism of protein already in the cells. Both these effects may result partly from decreased amino acid transport into extrahepatic tissues, as discussed later, but this is probably not the major cause because cortisol also depresses the formation of RNA and subsequent protein synthesis in many extrahepatic tissues, especially in muscle and lymphoid tissue. In the presence of great excesses of cortisol, the muscles can become so weak that the person cannot rise from the squatting position. In addition, the immunity functions of the lymphoid tissue can be decreased to a small fraction of normal. Cortisol Increases Liver and Plasma Proteins. Coincidentally with the effect of glucocorticoids to reduce proteins elsewhere in the body, the liver proteins are increased. Furthermore, the plasma proteins (which are produced by the liver and then released into the blood) are also increased. These increases are exceptions to the protein depletion that occurs elsewhere in the body. It is believed that this difference results from a possible effect of cortisol to enhance amino acid transport into liver cells (but not into most other cells) and to enhance the liver enzymes required for protein synthesis. Increased Blood Amino Acids, Diminished Transport of Amino Acids Into Extrahepatic Cells, and Enhanced Transport Into Hepatic Cells. Studies in isolated tissues have demonstrated that cortisol depresses amino acid transport into muscle cells and perhaps into other extrahepatic cells. The decreased transport of amino acids into extrahepatic cells decreases their intracellular amino acid concentrations and consequently decreases the synthesis of protein. Yet, catabolism of proteins in the cells continues to release amino acids that diffuse out of the cells to increase the plasma amino acid concentration. Therefore, cortisol mobilizes amino acids from the nonhepatic tissues and in doing so diminishes the tissue stores of protein. The increased plasma concentration of amino acids and enhanced transport of amino acids into the hepatic cells by cortisol could also account for enhanced utilization of amino acids by the liver to cause such effects as (1) increased rate of deamination of amino acids by the liver, (2) increased protein synthesis in the liver, (3) increased formation of plasma proteins by the liver, and (4) increased conversion of amino acids to glucose—that is, enhanced gluconeogenesis. Thus, it is possible that many of the effects of cortisol on the metabolic systems of the body result mainly from this ability of cortisol to mobilize amino acids from the peripheral tissues while at the same time increasing the liver enzymes required for the hepatic effects. Effects of Cortisol on Fat Metabolism Mobilization of Fatty Acids. In much the same manner that cortisol promotes amino acid mobilization from muscle, it also promotes mobilization of fatty acids from adipose tissue. This mobilization increases the concentration of free fatty acids in the plasma, which also increases their utilization for energy. Cortisol also seems to have a direct effect to enhance the oxidation of fatty acids in the cells. The mechanism by which cortisol promotes fatty acid mobilization is not completely understood. However, part of the effect probably results from diminished transport of glucose into the fat cells. Recall that α-glycerophosphate, which is derived from glucose, is required for both deposition and maintenance of triglycerides in these cells. In its absence, the fat cells begin to release fatty acids. The increased mobilization of fats by cortisol, combined with increased oxidation of fatty acids in the cells, helps shift the metabolic systems of the cells from utilization of glucose for energy to utilization of fatty acids in times of starvation or other stresses. This cortisol mechanism, however, requires several hours to become fully developed—not nearly so rapid or so powerful an effect as a similar shift elicited by a decrease in insulin, as we discuss in Chapter 79. Nevertheless, the increased use of fatty acids for metabolic energy is an important factor for long-term conservation of body glucose and glycogen. Excess Cortisol Causes Obesity. Despite the fact that cortisol can cause a moderate degree of fatty acid mobilization from adipose tissue, a peculiar type of obesity develops in many people with excess cortisol secretion, with excess deposition of fat in the chest and head regions of the body, giving a buffalo-like torso and a rounded “moon face.” Although the cause is unclear, it has been suggested that this obesity results from excess stimulation of food intake, with fat being generated in some tissues of the body more rapidly than it is mobilized and oxidized. Cortisol is Important in Resisting Stress and Inflammation Almost any type of stress, whether physical or neurogenic, causes an immediate and marked increase in ACTH secretion by the anterior pituitary gland, followed within minutes by greatly increased adrenocortical secretion of cortisol. This effect is demonstrated dramatically by the experiment shown in Figure 78-6, in which corticosteroid formation and secretion increased sixfold in a rat within 4 to 20 minutes after fracture of two leg bones. Figure 78-6Rapid reaction of the adrenal cortex of a rat to stress caused by fracture of the tibia and fibula at time zero. (In the rat, corticosterone is secreted in place of cortisol.) The following list details some of the different types of stress that increase cortisol release: Trauma Infection Intense heat or cold Injection of norepinephrine and other sympathomimetic drugs Surgery Injection of necrotizing substances beneath the skin Restraining an animal so it cannot move Debilitating diseases Even though cortisol secretion often increases greatly in stressful situations, we are not sure why this is of significant benefit to the animal. One possibility is that the glucocorticoids cause rapid mobilization of amino acids and fats from their cellular stores, making them immediately available both for energy and for synthesis of other compounds, including glucose, needed by the different tissues of the body. Indeed, it has been shown in a few instances that damaged tissues that are momentarily depleted of proteins can use the newly available amino acids to form new proteins that are essential to the lives of the cells. Also, the amino acids are perhaps used to synthesize other essential intracellular substances, such as purines, pyrimidines, and creatine phosphate, which are necessary for maintenance of cellular life and reproduction of new cells. All this is mainly supposition and is supported only by the fact that cortisol usually does not mobilize the basic functional proteins of the cells, such as the muscle contractile proteins and the proteins of neurons, until almost all other proteins have been released. This preferential effect of cortisol in mobilizing labile proteins could make amino acids available to needy cells to synthesize substances essential to life. Anti-inflammatory Effects of High Levels of Cortisol When tissues are damaged by trauma, by infection with bacteria, or in other ways, they almost always become “inflamed.” In some conditions, such as in rheumatoid arthritis, the inflammation is more damaging than the trauma or disease itself. Administration of large amounts of cortisol can usually block this inflammation or even reverse many of its effects once it has begun. Before attempting to explain the way in which cortisol functions to block inflammation, let us review the basic steps in the inflammation process, which are discussed in more detail in Chapter 34. Five main stages of inflammation occur: (1) release from the damaged tissue cells of chemicals such as histamine, bradykinin, proteolytic enzymes, prostaglandins, and leukotrienes that activate the inflammation process; (2) an increase in blood flow in the inflamed area caused by some of the released products from the tissues, an effect called erythema; (3) leakage of large quantities of almost pure plasma out of the capillaries into the damaged areas because of increased capillary permeability, followed by clotting of the tissue fluid, thus causing a nonpitting type of edema; (4) infiltration of the area by leukocytes; and (5) after days or weeks, ingrowth of fibrous tissue that often helps in the healing process. When large amounts of cortisol are secreted or injected into a person, the glucocorticoid has two basic anti-inflammatory effects: (1) it can block the early stages of the inflammation process before noticeable inflammation even begins, or (2) if inflammation has already begun, it causes rapid resolution of the inflammation and increased rapidity of healing. These effects are explained further in the following sections. Cortisol Prevents the Development of Inflammation by Stabilizing Lysosomes and by Other Effects. Cortisol has the following effects in preventing inflammation: Cortisol stabilizes lysosomal membranes. This stabilization is one of its most important anti-inflammatory effects because it is much more difficult than normal for the membranes of the intracellular lysosomes to rupture. Therefore, most of the proteolytic enzymes that are released by damaged cells to cause inflammation, which are mainly stored in the lysosomes, are released in greatly decreased quantities. Cortisol decreases permeability of the capillaries, probably as a secondary effect of the reduced release of proteolytic enzymes. This decrease in permeability prevents loss of plasma into the tissues. Cortisol decreases both migration of white blood cells into the inflamed area and phagocytosis of the damaged cells. These effects probably result from the fact that cortisol diminishes formation of prostaglandins and leukotrienes that otherwise would increase vasodilation, capillary permeability, and mobility of white blood cells. Cortisol suppresses the immune system, causing lymphocyte reproduction to decrease markedly. The T lymphocytes are especially suppressed. In turn, reduced amounts of T cells and antibodies in the inflamed area lessen tissue reactions that would otherwise promote inflammation. Cortisol attenuates fever mainly because it reduces release of interleukin-1 from white blood cells, which is one of the principal excitants to the hypothalamic temperature control system. The decreased temperature in turn reduces the degree of vasodilation. Thus, cortisol has an almost global effect in reducing all aspects of the inflammatory process. It is unclear how much of this reduction results from the simple effect of cortisol in stabilizing lysosomal and cell membranes versus its effect in reducing the formation of prostaglandins and leukotrienes from arachidonic acid in damaged cell membranes and other effects of cortisol. Cortisol Causes Resolution of Inflammation. Even after inflammation has become well established, administration of cortisol can often reduce inflammation within hours to a few days. The immediate effect is to block most of the factors that promote the inflammation. In addition, however, the rate of healing is enhanced. This probably results from the same, mainly undefined, factors that allow the body to resist many other types of physical stress when large quantities of cortisol are secreted. Perhaps this results from (1) the mobilization of amino acids and use of these acids to repair the damaged tissues; (2) the increased glucogenesis that makes extra glucose available in critical metabolic systems; (3) increased amounts of fatty acids available for cellular energy; or (4) some effect of cortisol for inactivating or removing inflammatory products. Regardless of the precise mechanisms by which the anti-inflammatory effect occurs, this effect of cortisol plays a major role in combating certain types of diseases, such as rheumatoid arthritis, rheumatic fever, and acute glomerulonephritis. All these diseases are characterized by severe local inflammation, and the harmful effects on the body are caused mainly by the inflammation and not by other aspects of the disease. When cortisol or other glucocorticoids are administered to patients with these diseases, almost invariably the inflammation begins to subside within 24 hours. Even though the cortisol does not correct the basic disease condition, preventing the damaging effects of the inflammatory response can often be a lifesaving measure. Other Effects of Cortisol Cortisol Blocks the Inflammatory Response to Allergic Reactions. The basic allergic reaction between antigen and antibody is not affected by cortisol, and even some of the secondary effects of the allergic reaction still occur. However, because the inflammatory response is responsible for many of the serious and sometimes lethal effects of allergic reactions, administration of cortisol, followed by its effect in reducing inflammation and the release of inflammatory products, can be lifesaving. For instance, cortisol effectively prevents shock or death as a result of anaphylaxis, a condition that otherwise kills many people, as explained in Chapter 35. Effect on Blood Cells and on Immunity in Infectious Diseases. Cortisol decreases the number of eosinophils and lymphocytes in the blood; this effect begins within a few minutes after the injection of cortisol and becomes marked within a few hours. Indeed, a finding of lymphocytopenia or eosinopenia is an important diagnostic criterion for overproduction of cortisol by the adrenal gland. Likewise, administration of large doses of cortisol causes significant atrophy of lymphoid tissue throughout the body, which in turn decreases the output of T cells and antibodies from the lymphoid tissue. As a result, the level of immunity for almost all foreign invaders of the body is decreased. This decrease occasionally can lead to fulminating infection and death from diseases that would otherwise not be lethal, such as fulminating tuberculosis in a person whose disease had previously been arrested. However, this ability of cortisol and other glucocorticoids to suppress immunity makes them useful drugs in preventing immunological rejection of transplanted hearts, kidneys, and other tissues. Cortisol increases the production of red blood cells by mechanisms that are unclear. When excess cortisol is secreted by the adrenal glands, polycythemia often results, and conversely, when the adrenal glands secrete no cortisol, anemia often results. Cellular Mechanism of Cortisol Action Cortisol, like other steroid hormones, exerts its effects by first interacting with intracellular receptors in target cells. Because cortisol is lipid soluble, it can easily diffuse through the cell membrane. Once inside the cell, cortisol binds with its protein receptor in the cytoplasm, and the hormone-receptor complex then interacts with specific regulatory DNA sequences, called glucocorticoid response elements, to induce or repress gene transcription. Other proteins in the cell, called transcription factors, are also necessary for the hormone-receptor complex to interact appropriately with the glucocorticoid response elements. Glucocorticoids increase or decrease transcription of many genes to alter synthesis of mRNA for the proteins that mediate their multiple physiological effects. Thus, most of the metabolic effects of cortisol are not immediate but require 45 to 60 minutes for proteins to be synthesized, and up to several hours or days to fully develop. Recent evidence suggests that glucocorticoids, especially at high concentrations, may also have some rapid nongenomic effects on cell membrane ion transport that may contribute to their therapeutic benefits. Regulation of Cortisol Secretion by Adrenocorticotropic Hormone from the Pituitary Gland ACTH Stimulates Cortisol Secretion. Unlike aldosterone secretion by the zona glomerulosa, which is controlled mainly by potassium and angiotensin II acting directly on the adrenocortical cells, secretion of cortisol is controlled almost entirely by ACTH that is secreted by the anterior pituitary gland. This hormone, also called corticotropin or adrenocorticotropin, also enhances the production of adrenal androgens. Chemistry of ACTH. ACTH has been isolated in pure form from the anterior pituitary. It is a large polypeptide, having a chain length of 39 amino acids. A smaller polypeptide, a digested product of ACTH having a chain length of 24 amino acids, has all the effects of the total molecule. ACTH Secretion Is Controlled by Corticotropin-Releasing Factor From the Hypothalamus. In the same way that other pituitary hormones are controlled by releasing factors from the hypothalamus, an important releasing factor also controls ACTH secretion. This factor is called corticotropin-releasing factor (CRF). It is secreted into the primary capillary plexus of the hypophysial portal system in the median eminence of the hypothalamus and then carried to the anterior pituitary gland, where it induces ACTH secretion. CRF is a peptide composed of 41 amino acids. The cell bodies of the neurons that secrete CRF are located mainly in the paraventricular nucleus of the hypothalamus. This nucleus in turn receives many nervous connections from the limbic system and lower brain stem. The anterior pituitary gland can secrete only minute quantities of ACTH in the absence of CRF. Instead, most conditions that cause high ACTH secretory rates initiate this secretion by signals that begin in the basal regions of the brain, including the hypothalamus, and are then transmitted by CRF to the anterior pituitary gland. ACTH Activates Adrenocortical Cells to Produce Steroids by Increasing cAMP. The principal effect of ACTH on the adrenocortical cells is to activate adenylyl cyclase in the cell membrane. This activation then induces the formation of cAMP in the cell cytoplasm, reaching its maximal effect in about 3 minutes. The cAMP in turn activates the intracellular enzymes that cause formation of the adrenocortical hormones, which is another example of cAMP as a second messenger signal system. The most important of all the ACTH-stimulated steps for controlling adrenocortical secretion is activation of the enzyme protein kinase A, which causes initial conversion of cholesterol to pregnenolone. This initial conversion is the “rate-limiting” step for all the adrenocortical hormones, which explains why ACTH is normally necessary for any adrenocortical hormones to be formed. Long-term stimulation of the adrenal cortex by ACTH not only increases secretory activity but also causes hypertrophy and proliferation of the adrenocortical cells, especially in the zona fasciculata and zona reticularis, where cortisol and the androgens are secreted. Physiological Stress Increases ACTH and Adrenocortical Secretion. As pointed out earlier in the chapter, almost any type of physical or mental stress can lead within minutes to greatly enhanced secretion of ACTH and consequently cortisol as well, often increasing cortisol secretion as much as 20-fold. This effect was demonstrated by the rapid and strong adrenocortical secretory responses after trauma shown in Figure 78-6. Pain stimuli caused by physical stress or tissue damage are transmitted first upward through the brain stem and eventually to the median eminence of the hypothalamus, as shown in Figure 78-7. Here CRF is secreted into the hypophysial portal system. Within minutes the entire control sequence leads to large quantities of cortisol in the blood. Figure 78-7Mechanism for regulation of glucocorticoid secretion. ACTH, adrenocorticotropic hormone; CRF, corticotropin-releasing factor. Mental stress can cause an equally rapid increase in ACTH secretion. This increase is believed to result from increased activity in the limbic system, especially in the region of the amygdala and hippocampus, both of which then transmit signals to the posterior medial hypothalamus. Inhibitory Effect of Cortisol on the Hypothalamus and on the Anterior Pituitary to Decrease ACTH Secretion. Cortisol has direct negative feedback effects on (1) the hypothalamus to decrease the formation of CRF and (2) the anterior pituitary gland to decrease the formation of ACTH. Both of these feedbacks help regulate the plasma concentration of cortisol. That is, whenever the cortisol concentration becomes too great, the feedbacks automatically reduce the ACTH toward a normal control level. Summary of the Cortisol Control System Figure 78-7 shows the overall system for control of cortisol secretion. The key to this control is the excitation of the hypothalamus by different types of stress. Stress stimuli activate the entire system to cause rapid release of cortisol, and the cortisol in turn initiates a series of metabolic effects directed toward relieving the damaging nature of the stressful state. Direct feedback of the cortisol to both the hypothalamus and the anterior pituitary gland also occurs to decrease the concentration of cortisol in the plasma at times when the body is not experiencing stress. However, the stress stimuli are the most potent ones; they can always break through this direct inhibitory feedback of cortisol, causing either periodic exacerbations of cortisol secretion at multiple times during the day (Figure 78-8) or prolonged cortisol secretion in times of chronic stress. Figure 78-8Typical pattern of cortisol concentration during the day. Note the oscillations in secretion, as well as a daily secretory surge an hour or so after waking in the morning. Circadian Rhythm of Glucocorticoid Secretion. The secretory rates of CRF, ACTH, and cortisol are high in the early morning but low in the late evening, as shown in Figure 78-8; the plasma cortisol level ranges between a high of about 20 µg/dl an hour before arising in the morning and a low of about 5 µg/dl around midnight. This effect results from a 24-hour cyclical alteration in the signals from the hypothalamus that cause cortisol secretion. When a person changes daily sleeping habits, the cycle changes correspondingly. Therefore, measurements of blood cortisol levels are meaningful only when expressed in terms of the time in the cycle at which the measurements are made. Synthesis and Secretion of ACTH in Association with Melanocyte-Stimulating Hormone, Lipotropin, and Endorphin When ACTH is secreted by the anterior pituitary gland, several other hormones that have similar chemical structures are secreted simultaneously. The reason for this secretion is that the gene that is transcribed to form the RNA molecule that causes ACTH synthesis initially causes the formation of a considerably larger protein, a preprohormone called pro-opiomelanocortin (POMC), which is the precursor of ACTH and several other peptides, including melanocyte-stimulating hormone (MSH), β-lipotropin, β-endorphin, and a few others (Figure 78-9). Under normal conditions, most of these hormones are not secreted in enough quantity by the pituitary to have a major effect on the human body, but when the rate of secretion of ACTH is high, as may occur in persons with Addison's disease, formation of some of the other POMC-derived hormones may also be increased. Figure 78-9Pro-opiomelanocortin processing by prohormone convertase 1 (PC1, red arrows) and PC2 (blue arrows). Tissue-specific expression of these two enzymes results in different peptides produced in various tissues. ACTH, adrenocorticotropic hormone; CLIP, corticotropin-like intermediate peptide; MSH, melanocyte-stimulating hormone. The POMC gene is actively transcribed in several tissues, including the corticotroph cells of the anterior pituitary, POMC neurons in the arcuate nucleus of the hypothalamus, cells of the dermis, and lymphoid tissue. In all of these cell types, POMC is processed to form a series of smaller peptides. The precise type of POMC-derived products from a particular tissue depends on the type of processing enzymes present in the tissue. Thus, pituitary corticotroph cells express prohormone convertase 1 (PC1) but not PC2, resulting in the production of N-terminal peptide, joining peptide, ACTH, and β-lipotropin. In the hypothalamus, the expression of PC2 leads to the production of α-MSH, β-MSH, γ-MSH, and β-endorphin, but not ACTH. As discussed in Chapter 72, α-MSH formed by neurons of the hypothalamus plays a major role in appetite regulation. In melanocytes located in abundance between the dermis and epidermis of the skin, MSH stimulates formation of the black pigment melanin and disperses it to the epidermis. Injection of MSH into a person over 8 to 10 days can greatly increase darkening of the skin. The effect is much greater in people who have genetically dark skins than in light-skinned people. In some animals, an intermediate “lobe” of the pituitary gland, called the pars intermedia, is highly developed, lying between the anterior and posterior pituitary lobes. This lobe secretes an especially large amount of MSH. Furthermore, this secretion is independently controlled by the hypothalamus in response to the amount of light to which the animal is exposed or in response to other environmental factors. For instance, some arctic animals develop darkened fur in the summer and yet have entirely white fur in the winter. ACTH, because it contains an MSH sequence, has about 1/30 as much melanocyte-stimulating effect as MSH. Furthermore, because the quantities of pure MSH secreted in humans are extremely small, whereas those of ACTH are large, it is likely that ACTH is normally more important than MSH in determining the amount of melanin in the skin. Adrenal Androgens Several moderately active male sex hormones called adrenal androgens (the most important of which is dehydroepiandrosterone) are continually secreted by the adrenal cortex, especially during fetal life, as discussed in Chapter 84. Also, progesterone and estrogens, which are female sex hormones, are secreted in minute quantities. Normally, the adrenal androgens have only weak effects in humans. It is possible that part of the early development of the male sex organs results from childhood secretion of adrenal androgens. The adrenal androgens also exert mild effects in the female, not only before puberty but also throughout life. Much of the growth of the pubic and axillary hair in the female results from the action of these hormones. In extra-adrenal tissues, some of the adrenal androgens are converted to testosterone, the primary male sex hormone, which probably accounts for much of their androgenic activity. The physiological effects of androgens are discussed in Chapter 81 in relation to male sexual function. Abnormalities of Adrenocortical Secretion Hypoadrenalism (Adrenal Insufficiency)—Addison's Disease Addison's disease results from an inability of the adrenal cortices to produce sufficient adrenocortical hormones, and this in turn is most frequently caused by primary atrophy or injury of the adrenal cortices. In about 80 percent of the cases, the atrophy is caused by autoimmunity against the cortices. Adrenal gland hypofunction may also be caused by tuberculous destruction of the adrenal glands or invasion of the adrenal cortices by cancer. In some cases, adrenal insufficiency is secondary to impaired function of the pituitary gland, which fails to produce sufficient ACTH. When ACTH output is too low, cortisol and aldosterone production decrease, and eventually the adrenal glands may atrophy because of a lack of ACTH stimulation. Secondary adrenal insufficiency is much more common than Addison's disease, which is sometimes called primary adrenal insufficiency. Disturbances in severe adrenal insufficiency are described in the following sections. Mineralocorticoid Deficiency. Lack of aldosterone secretion greatly decreases renal tubular sodium reabsorption and consequently allows sodium ions, chloride ions, and water to be lost into urine in great profusion. The net result is a greatly decreased extracellular fluid volume. Furthermore, hyponatremia, hyperkalemia, and mild acidosis develop because of failure of potassium and hydrogen ions to be secreted in exchange for sodium reabsorption. As the extracellular fluid becomes depleted, plasma volume falls, red blood cell concentration rises markedly, cardiac output and blood pressure decrease, and the patient dies in shock, with death usually occurring in the untreated patient 4 days to 2 weeks after complete cessation of mineralocorticoid secretion. Glucocorticoid Deficiency. Loss of cortisol secretion makes it impossible for a person with Addison's disease to maintain normal blood glucose concentration between meals because he or she cannot synthesize significant quantities of glucose by gluconeogenesis. Furthermore, lack of cortisol reduces the mobilization of both proteins and fats from the tissues, thereby depressing many other metabolic functions of the body. This sluggishness of energy mobilization when cortisol is not available is one of the major detrimental effects of glucocorticoid deficiency. Even when excess quantities of glucose and other nutrients are available, the person's muscles are weak, indicating that glucocorticoids are necessary to maintain other metabolic functions of the tissues in addition to energy metabolism. Lack of adequate glucocorticoid secretion also makes a person with Addison's disease highly susceptible to the deteriorating effects of different types of stress, and even a mild respiratory infection can cause death. Melanin Pigmentation. Another characteristic of most people with Addison's disease is melanin pigmentation of the mucous membranes and skin. This melanin is not always deposited evenly but occasionally is deposited in blotches, and it is deposited especially in the thin skin areas, such as the mucous membranes of the lips and the thin skin of the nipples. The melanin deposition is believed to have the following cause: When cortisol secretion is depressed, the normal negative feedback to the hypothalamus and anterior pituitary gland is also depressed, therefore allowing tremendous rates of ACTH secretion, as well as simultaneous secretion of increased amounts of MSH. The large amounts of ACTH probably cause most of the pigmenting effect because they can stimulate formation of melanin by the melanocytes in the same way that MSH does. Treatment of People with Addison's Disease. An untreated person with total adrenal destruction dies within a few days to a few weeks because of weakness and, usually, circulatory shock. Yet, such a person can live for years if small quantities of mineralocorticoids and glucocorticoids are administered daily. Addisonian Crisis. As noted earlier in the chapter, great quantities of glucocorticoids are occasionally secreted in response to different types of physical or mental stress. In a person with Addison's disease, the output of glucocorticoids does not increase during stress. Yet, during different types of trauma, disease, or other stresses, such as surgical operations, a person is likely to have an acute need for excessive amounts of glucocorticoids and often must be given 10 or more times the normal quantities of glucocorticoids to prevent death. This critical need for extra glucocorticoids and the associated severe debility in times of stress is called an addisonian crisis. Hyperadrenalism—Cushing's Syndrome Hypersecretion by the adrenal cortex causes a complex cascade of hormone effects called Cushing's syndrome. Many of the abnormalities of Cushing's syndrome can be ascribed to abnormal amounts of cortisol, but excess secretion of androgens may also cause important effects. Hypercortisolism can occur from multiple causes, including (1) adenomas of the anterior pituitary that secrete large amounts of ACTH, which then causes adrenal hyperplasia and excess cortisol secretion; (2) abnormal function of the hypothalamus that causes high levels of corticotropin-releasing hormone, which stimulates excess ACTH release; (3) “ectopic secretion” of ACTH by a tumor elsewhere in the body, such as an abdominal carcinoma; and (4) adenomas of the adrenal cortex. When Cushing's syndrome is secondary to excess secretion of ACTH by the anterior pituitary, this condition is referred to as Cushing's disease. Excess ACTH secretion is the most common cause of Cushing's syndrome and is characterized by high plasma levels of ACTH and cortisol. Primary overproduction of cortisol by the adrenal glands accounts for about 20 to 25 percent of clinical cases of Cushing's syndrome and is usually associated with reduced ACTH levels due to cortisol feedback inhibition of ACTH secretion by the anterior pituitary gland. Administration of large doses of dexamethasone, a synthetic glucocorticoid, can be used to distinguish between ACTH-dependent and ACTH-independent Cushing's syndrome. In patients who have overproduction of ACTH due to an ACTH-secreting pituitary adenoma or to hypothalamic-pituitary dysfunction, low doses of dexamethasone usually do not suppress ACTH secretion normally. By increasing the dose of dexamethasone to very high levels, ACTH eventually can be suppressed in most patients with Cushing's disease. In contrast, patients with primary adrenal overproduction of cortisol (ACTH-independent Cushing's syndrome) usually have low or undetectable levels of ACTH. The dexamethasone test, although widely used, can sometimes result in an incorrect diagnosis because some ACTH-secreting pituitary tumors do respond to dexamethasone with suppressed ACTH secretion. Also, nonpituitary malignant tumors that produce ACTH ectopically, such as some lung carcinomas, are not responsive to glucocorticoid-negative feedback. Therefore, the dexamethasone test is usually considered to be a first step in the differential diagnosis of Cushing's syndrome. Cushing's syndrome can also occur when large amounts of glucocorticoids are administered over prolonged periods for therapeutic purposes. For example, patients with chronic inflammation associated with diseases such as rheumatoid arthritis are often treated with glucocorticoids and may experience some of the clinical symptoms of Cushing syndrome. A special characteristic of Cushing's syndrome is mobilization of fat from the lower part of the body, with concomitant extra deposition of fat in the thoracic and upper abdominal regions, giving rise to a buffalo-like torso. The excess secretion of steroids also leads to an edematous appearance of the face, and the androgenic potency of some of the hormones sometimes causes acne and hirsutism (excess growth of facial hair). The appearance of the face is frequently described as a “moon face,” as demonstrated in the untreated patient with Cushing's syndrome to the left in Figure 78-10. About 80 percent of patients have hypertension, presumably because of the mineralocorticoid effects of cortisol. Figure 78-10A person with Cushing's syndrome before (left) and after (right) a subtotal adrenalectomy.(Courtesy Dr. Leonard Posey.) Effects on Carbohydrate and Protein Metabolism The abundance of cortisol secreted in Cushing's syndrome can increase blood glucose concentration, sometimes to values as high as 200 mg/dl after meals—as much as twice normal. This increase results mainly from enhanced gluconeogenesis and decreased glucose utilization by the tissues. The effects of glucocorticoids on protein catabolism are often profound in Cushing's syndrome, causing greatly decreased tissue proteins almost everywhere in the body with the exception of the liver; the plasma proteins also remain unaffected. The loss of protein from the muscles in particular causes severe weakness. The loss of protein synthesis in the lymphoid tissues leads to a suppressed immune system, and thus many of these patients die of infections. Even the protein collagen fibers in the subcutaneous tissue are diminished so that the subcutaneous tissues tear easily, resulting in development of large purplish striae where they have torn apart. In addition, severely diminished protein deposition in the bones often causes severe osteoporosis with consequent weakness of the bones. Treatment of Cushing's Syndrome. Treatment of Cushing's syndrome consists of removing an adrenal tumor if this is the cause or decreasing the secretion of ACTH, if possible. Hypertrophied pituitary glands or even small tumors in the pituitary that oversecrete ACTH can sometimes be surgically removed or destroyed by radiation. Drugs that block steroidogenesis, such as metyrapone, ketoconazole, and aminoglutethimide, or that inhibit ACTH secretion, such as serotonin antagonists and GABA-transaminase inhibitors, can also be used when surgery is not feasible. If ACTH secretion cannot easily be decreased, the only satisfactory treatment is usually bilateral partial (or even total) adrenalectomy, followed by administration of adrenal steroids to make up for any insufficiency that develops. Primary Aldosteronism (Conn's Syndrome) Occasionally a small tumor of the zona glomerulosa cells occurs and secretes large amounts of aldosterone; the resulting condition is called primary aldosteronism or Conn's syndrome. Also, in a few instances, hyperplastic adrenal cortices secrete aldosterone rather than cortisol. The effects of the excess aldosterone are discussed in detail earlier in the chapter. The most important effects are hypokalemia, mild metabolic alkalosis, a slight increase in extracellular fluid volume and blood volume, a modest increase in plasma sodium concentration (usually <4 to 6 mEq/L increase), and, almost always, hypertension. Especially interesting in persons with primary aldosteronism are occasional periods of muscle paralysis caused by the hypokalemia. The paralysis is caused by a depressant effect of low extracellular potassium concentration on action potential transmission by the nerve fibers, as explained in Chapter 5. One of the diagnostic criteria of primary aldosteronism is a decreased plasma renin concentration. This decrease results from feedback suppression of renin secretion caused by the excess aldosterone or by the excess extracellular fluid volume and arterial pressure resulting from the aldosteronism. Treatment of primary aldosteronism may include surgical removal of the tumor or of most of the adrenal tissue when hyperplasia is the cause. Another option for treatment is pharmacological antagonism of the mineralocorticoid receptor with spironolactone or eplerenone. Adrenogenital Syndrome Occasionally an adrenocortical tumor secretes excessive quantities of androgens that cause intense masculinizing effects throughout the body. If this phenomenon occurs in a female, virile characteristics develop, including growth of a beard, a much deeper voice, occasionally baldness if she also has the genetic trait for baldness, masculine distribution of hair on the body and the pubis, growth of the clitoris to resemble a penis, and deposition of proteins in the skin and especially in the muscles to give typical masculine characteristics. In the prepubertal male, a virilizing adrenal tumor causes the same characteristics as in the female plus rapid development of the male sexual organs, as shown in Figure 78-11, which depicts a 4-year-old boy with adrenogenital syndrome. In the adult male, the virilizing characteristics of adrenogenital syndrome are usually obscured by the normal virilizing characteristics of the testosterone secreted by the testes. It is often difficult to make a diagnosis of adrenogenital syndrome in the adult male. In adrenogenital syndrome, the excretion of 17-ketosteroids (which are derived from androgens) in the urine may be 10 to 15 times normal. This finding can be used in diagnosing the disease. Figure 78-11Adrenogenital syndrome in a 4-year-old boy.(Courtesy Dr. Leonard Posey.) Bibliography Baker ME, Funder JW, Kattoula SR. Evolution of hormone selectivity in glucocorticoid and mineralocorticoid receptors. J Steroid Biochem Mol Biol. 2013;137:57. Biller BM, Grossman AB, Stewart PM, et al. Treatment of adrenocorticotropin-dependent Cushing's syndrome: a consensus statement. J Clin Endocrinol Metab. 2008;93:2454. Bornstein SR. Predisposing factors for adrenal insufficiency. N Engl J Med. 2009;360:2328. Boscaro M, Arnaldi G. Approach to the patient with possible Cushing's syndrome. J Clin Endocrinol Metab. 2009;94:3121. Chapman K, Holmes M, Seckl J. 11β-hydroxysteroid dehydrogenases: intracellular gate-keepers of tissue glucocorticoid action. Physiol Rev. 2013;93:1139. Charmandari E, Nicolaides NC, Chrousos GP. Adrenal insufficiency. Lancet. 2014;383:2152. Feelders RA, Hofland LJ. Medical treatment of Cushing disease. J Clin Endocrinol Metab. 2013;98:425. Fuller PJ. Adrenal diagnostics: an endocrinologist's perspective focused on hyperaldosteronism. Clin Biochem Rev. 2013;34:111. Fuller PJ, Young MJ. Mechanisms of mineralocorticoid action. Hypertension. 2005;46:1227. Funder JW. Aldosterone and the cardiovascular system: genomic and nongenomic effects. Endocrinology. 2006;147:5564. Funder JW. The genetic basis of primary aldosteronism. Curr Hypertens Rep. 2012;14:120. Gomez-Sanchez CE, Oki K. Minireview: potassium channels and aldosterone dysregulation: is primary aldosteronism a potassium channelopathy? Endocrinology. 2014;155:47. Hall JE, Granger JP, Smith MJ Jr, Premen AJ. Role of renal hemodynamics and arterial pressure in aldosterone “escape. Hypertension. 1984;6:I183. Hammes SR, Levin ER. Minireview: recent advances in extranuclear steroid receptor actions. Endocrinology. 2011;152:4489. Mazziotti G, Giustina A. Glucocorticoids and the regulation of growth hormone secretion. Nat Rev Endocrinol. 2013;9:265. Pimenta E, Wolley M, Stowasser M. Adverse cardiovascular outcomes of corticosteroid excess. Endocrinology. 2012;153:5137. Prague JK, May S, Whitelaw BC. Cushing's syndrome. BMJ. 2013;346:f945. Spat A, Hunyady L. Control of aldosterone secretion: a model for convergence in cellular signaling pathways. Physiol Rev. 2004;84:489. Speiser PW, White PC. Congenital adrenal hyperplasia. N Engl J Med. 2003;349:776. Tritos NA, Biller BM. Advances in medical therapies for Cushing's syndrome. Discov Med. 2012;13:171. Vinson GP. The adrenal cortex and life. Mol Cell Endocrinol. 2009;300:2. Wendler A, Albrecht C, Wehling M. Nongenomic actions of aldosterone and progesterone revisited. Steroids. 2012;77:1002.
15293	https://pmc.ncbi.nlm.nih.gov/articles/PMC11014425/	Lung cancer in patients who have never smoked — an emerging disease - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Nat Rev Clin Oncol . Author manuscript; available in PMC: 2024 Apr 12. Published in final edited form as: Nat Rev Clin Oncol. 2024 Jan 9;21(2):121–146. doi: 10.1038/s41571-023-00844-0 Search in PMC Search in PubMed View in NLM Catalog Add to search Lung cancer in patients who have never smoked — an emerging disease Jaclyn LoPiccolo Jaclyn LoPiccolo 1 Department of Medical Oncology, Dana-Farber Cancer Institute, Boston, MA, USA 2 The Lowe Center for Thoracic Oncology, Dana-Farber Cancer Institute, Boston, MA, USA Find articles by Jaclyn LoPiccolo 1,2,✉, Alexander Gusev Alexander Gusev 1 Department of Medical Oncology, Dana-Farber Cancer Institute, Boston, MA, USA 3 The Eli and Edythe L. Broad Institute, Cambridge, MA, USA Find articles by Alexander Gusev 1,3, David C Christiani David C Christiani 4 Department of Environmental Health, Harvard T. H. Chan School of Public Health, Boston, MA, USA 5 Massachusetts General Hospital, Boston, MA, USA Find articles by David C Christiani 4,5, Pasi A Jänne Pasi A Jänne 1 Department of Medical Oncology, Dana-Farber Cancer Institute, Boston, MA, USA 2 The Lowe Center for Thoracic Oncology, Dana-Farber Cancer Institute, Boston, MA, USA Find articles by Pasi A Jänne 1,2 Author information Article notes Copyright and License information 1 Department of Medical Oncology, Dana-Farber Cancer Institute, Boston, MA, USA 2 The Lowe Center for Thoracic Oncology, Dana-Farber Cancer Institute, Boston, MA, USA 3 The Eli and Edythe L. Broad Institute, Cambridge, MA, USA 4 Department of Environmental Health, Harvard T. H. Chan School of Public Health, Boston, MA, USA 5 Massachusetts General Hospital, Boston, MA, USA Author contributions All authors researched data for the article and contributed substantially to the discussion of content. J.L. wrote the article. All authors reviewed and/or edited the manuscript before submission. ✉ Email: jaclyn_lopiccolo@dfci.harvard.edu Issue date 2024 Feb. PMC Copyright notice PMCID: PMC11014425 NIHMSID: NIHMS1977779 PMID: 38195910 The publisher's version of this article is available at Nat Rev Clin Oncol Abstract Lung cancer is the most common cause of cancer-related deaths globally. Although smoking-related lung cancers continue to account for the majority of diagnoses, smoking rates have been decreasing for several decades. Lung cancer in individuals who have never smoked (LCINS) is estimated to be the fifth most common cause of cancer-related deaths worldwide in 2023, preferentially occurring in women and Asian populations. As smoking rates continue to decline, understanding the aetiology and features of this disease, which necessitate unique diagnostic and treatment paradigms, will be imperative. New data have provided important insights into the molecular and genomic characteristics of LCINS, which are distinct from those of smoking-associated lung cancers and directly affect treatment decisions and outcomes. Herein, we review the emerging data regarding the aetiology and features of LCINS, particularly the genetic and environmental underpinnings of this disease as well as their implications for treatment. In addition, we outline the unique diagnostic and therapeutic paradigms of LCINS and discuss future directions in identifying individuals at high risk of this disease for potential screening efforts. Introduction Lung cancer is the leading cause of cancer-related death worldwide (see GLOBOCAN) and in every ethnic group in the USA1. Smoking-related lung cancers account for the majority of lung cancer diagnoses and continue to claim ~100,000 lives in the USA each year1; however, smoking rates have been decreasing for several decades. Although varying widely across the USA, cigarette smoking prevalence among adults reached an all-time low at 11.5% in 2021 (ref. 2), down by more than two-thirds (from ~42%) since the first Surgeon General’s Report linking cigarette smoking to lung cancer and heart disease in 1964 (ref. 3) (Fig. 1a) and is projected to fall further to 7.5% overall by 2065 (ref. 4). As smoking prevalence has declined, some reports indicate that the proportion of lung cancers occurring in individuals who have never smoked (LCINS) has increased, particularly among women and in younger age groups5. In 2023, >20,000 lung cancer-related deaths in the USA were projected to occur in people who have never smoked1, making LCINS the eighth leading cause of cancer-related mortality in the USA; data suggest that LCINS is currently the fifth most common cause of cancer-related deaths worldwide6 (Fig. 1b). Fig. 1 \|. Smoking rates and lung cancer-related deaths. Open in a new tab a, Smoking prevalence in the USA from 1965 to 2022 as the percentage of adults aged ≥18 years who reported currently smoking cigarettes, overall and by sex. The data are derived from the National Health Interview Survey (NHIS) 1965–2018 and 2019–2022 (refs. 2,3,385,386) and are not available for all years. b, Estimated number of global deaths attributed to 37 cancer types in 2020 (ref. 6). Smoking-related lung cancer and lung cancer in individuals who have never smoked (LCINS) are shown in red as the first and fifth leading cause of cancer deaths, respectively. LCINS have histological and epidemiological distinctions from smoking-related lung cancers given that they are almost exclusively adenocarcinomas and most commonly occur in women and individuals of Asian ancestry7,8. Moreover, several studies have revealed that LCINS are also genomically and molecularly distinct from smoking-related lung cancers, highly enriched for targetable oncogenic alterations (such as EGFR mutations or ALK rearrangements as well as less common alterations), and thus often require different diagnostic and therapeutic strategies. Over the past decade, new data have emerged regarding the genetic risk of LCINS, conferred by both common and rare germline variants, as well as environmental risk factors and potential interactions between the two. Conceivably, LCINS could eventually become the most common form of lung cancer, necessitating a thorough understanding of its pathogenesis and risk factors. Herein, we review the epidemiological, clinical and genomic features of LCINS, along with data from studies examining both genetic and environmental risk factors, as well as diagnostic and treatment strategies. Definitions In an effort to mitigate the stigmatization of patients with lung cancer, the International Association for the Study of Lung Cancer released a language guide in 2021 that includes replacing the term ‘smoker’ with language such as ‘person who has smoked’, and ‘never-smoker’ with ‘person who has never smoked’9. Definitions of smoking status have varied; however, the Centers for Disease Control and Prevention and others have previously used the term ‘never-smoker’ (or sometimes ‘non-smoker’) to refer to individuals who have smoked <100 cigarettes in their lifetime, and LCINS is defined as lung cancer arising in such individuals. The previously used terms ‘former smoker’ or ‘ex-smoker’ are defined as people who have smoked >100 cigarettes in their lifetime but quit smoking ≥12 months prior to a lung cancer diagnosis. The term ‘long-term former smoker’ has been used to refer to those who have smoked >100 cigarettes in their lifetime but quit smoking ≥15 years prior to a lung cancer diagnosis, and individuals with ‘remote’ smoking histories include those who smoked infrequently and/or socially as adolescents and/or young adults but still smoked >100 cigarettes in their lifetime; both of these categories might share certain characteristics with those who have never smoked. A current smoking status refers to individuals who have smoked >100 cigarettes in their lifetime and who still report smoking every day or some days. Nevertheless, data from several studies suggest that smoking history should be quantified as a continuous rather than discrete variable, supported by findings such as the likelihood of a lung adenocarcinoma (LUAD) harbouring an EGFR mutation decreasing as number of pack-years increases10. As discussed below, the discovery of tumour mutational signatures corresponding to tobacco smoking might eventually help to delineate what constitutes a clinically significant level of exposure from a genomic perspective. Epidemiology Approximately two-thirds of LCINS cases occur in women, making women who have not smoked more than twice as likely to develop lung cancer than men who have not smoked5,11. The proportion of lung cancers attributable to tobacco smoking varies across countries, at >80% in men and women in the USA and the UK12–14, and 57.5% in men and 13% in women in China15. A never-smoking status is more frequent among female patients in Asia than those in other regions and, interestingly, >55% of female Asian patients and >30% of Hispanic female patients diagnosed with non-small-cell lung cancer (NSCLC) in the USA have never smoked16. However, even when compared specifically with non-smoking women in other geographical regions, lung cancer incidence rates have been observed to be higher among women in East Asia17, suggesting that genetic and/or environmental factors other than tobacco smoke exposure contribute to the global variation in the prevalence of LCINS. The average age at LCINS diagnosis is similar to that reported for smoking-related lung cancers5 (median age at diagnosis of 67 years versus 65 years; P = 0.1)7, although younger patients with lung cancer are more likely to have never smoked. In a study including 121 patients <40 years of age at diagnosis and with a documented smoking history, 73% had never smoked; 90% of these patients had LUAD, >80% of which harboured a targetable oncogenic alteration18. Patients with ALK-rearranged NSCLC, which mostly occurs in individuals who have never smoked, are younger on average at diagnosis (median age at diagnosis 50–52 years)19–22, as are patients with other LUADs harbouring oncogenic fusions, for example, involving RET23,24, ROS1 (refs. 22,25–27) or NTRK1–3 (ref. 28). In economically developed countries, smoking prevalence has decreased among all age groups (in the USA, most steeply amongst those <30 years old in the USA), reflected in an overall decreasing incidence of lung cancer1,29. Lung cancer incidence is decreasing twice as fast in men than in women, and perhaps four times as fast, based on an analysis of United States Cancer Statistics data30 examining annual percentage change in NSCLC incidence31,32 over 2001–2019 (J.L. et al., unpublished data), a difference that is only partially explained by differences in smoking trends33–35. Notably, the incidence of lung cancer is now higher in women than in men in both Hispanic and non-Hispanic white individuals born during or after the mid-1960s whereas, prior to this time, the incidence was higher in men than in women35. This pattern is also expected to reverse (that is, lung cancer incidence will become higher in women than in men) in the remaining birth cohorts by 2045 if current smoking trends continue4. Trends in NSCLC histology also reflect the decreasing smoking prevalence, with a relative increase in LUAD and decrease in lung squamous cell carcinoma (LSCC) over time36,37. Precise national and global trends in LCINS epidemiology are unavailable owing to a lack of individual-level data on smoking status, which historically has not been collected in cancer registries or on death certificates. However, a retrospective study including >10,000 patients from three independent cancer centres revealed that the proportion of LCINS relative to total NSCLC diagnoses increased from 8.0% in 1990–1995 to 14.9% in 2011–2013 (P< 0.001)5. This trend was independent of sex, disease stage at diagnosis and ethnicity, and no statistically significant increase in the proportion of LCINS was observed among small-cell lung carcinoma (SCLC) or LSCC diagnoses5. However, additional and contemporary data are needed to determine true trends in LCINS epidemiology. Features of LCINS In addition to histological characteristics shared among LCINS, genomic and molecular features have been more recently characterized in LCINS as compared to lung cancers occurring in patients with a history of smoking. The findings underscore the distinct biology of LCINS, with important diagnostic and treatment implications (Table 1). Table 1 \|. Somatic features reported for LCINS compared to tobacco-related lung cancers \| Feature \| LCINS \| Tobacco-related lung cancer \| :--- \| Histology \| NSCLC; mostly LUAD1,2 \| LUAD, LSCC or SCLC; strongest association with LSCC and SCLC334 \| \| Targetable driver alterations \| Present in 78–92% of LUADs3–5 \| Present in 49.5% of LUADs (mainly KRAS mutations)7 \| \| PD-L1 expression level \| Low36–8,a \| Highest with current smoking; PD-L1 staining intensity is positively correlated with pack-years of smoking history136 \| \| TMB \| 0–3mut/Mb (refs. 3–5,9–13) (median 1.1 mut/Mb) \| Up to tenfold higher than in LCINS131, with a dose–response relationship between pack-years of smoking history and TMB132 \| \| Genomic signatures \| Devoid of tobacco-associated mutational signatures, including LCINS with SHS exposure3–5 \| SBS signature 4b (mainly C>A transversions), attributable to misrepair of DNA damage40,135; less strongly associated with indel-based signature 3c and doublet-base substitution signature 2 (ref. 335); total number of SBS nearly fivefold higher in smoking-related versus non-smoking-related LUADs (mean 12.09 vs 2.65; P = 2.7 × 10−13); total number of indels higher in smoking-related versus non-smoking-related LUADs (mean 0.39 vs 0.14; P = 7.9 × 10−14)40 \| Open in a new tab LCINS, Lung cancers in individuals who have never smoked; LUAD, lung adenocarcinoma; mut/Mb, mutations per megabase; NSCLC, non-small-cell lung cancer; SBS, single-base substitution; LSCC, lung squamous cell carcinoma; SCLC, small-cell lung cancer; SHS, secondhand smoke; TMB, tumour mutational burden. a Moderate to high levels of PD-L1 expression have been observed in LCINS with MET exon 14 mutations14,15. b Reliably detectable with targeted panel-based next-generation sequencing assays. c Reliable detection restricted to whole-exome or whole-genome sequencing analyses336,337; at present, these mutational signatures are mainly used for investigational purposes with limited use in clinical diagnostics. Histology As noted above, LCINS are near-exclusively LUADs5,16 and largely driven by oncogenic alterations in key pro-survival signalling pathways. Although LUAD accounts for a substantial number of smoking-related lung cancers, both LSCC and SCLC are more strongly associated with smoking and typically arise in the larger, central airways that are more readily accessible to tobacco smoke. An estimated 6–8% of LSCCs and SCLCs occur in patients who have never smoked16, and the age-adjusted odds ratios (OR) for LUAD, LSCC and SCLC development in men who currently smoke are 21.9, 103.5 and 111.3, respectively38–40. An analysis of 11 cases of SCLC diagnosed in individuals with a history of former light smoking or never smoking revealed that most tumours (8 of 11; 73%) were of mixed histology or non-pulmonary origin41. Furthermore, driver mutations were detected in EGFR, NRAS, KRAS, BRCA1 and ATM, and one tumour had a TMPRSS2–ERG fusion41. These findings suggest that SCLC arising in those who have never smoked constitutes a distinct disease entity, requiring different therapeutic approaches than smoking-related SCLC. LUADs occurring in the presence or absence of tobacco smoke exposure are largely similar in histological appearance and typically cannot be distinguished based on histological subtype or features when adjusted for pathological stage; differences in adverse prognostic features, such as lymphovascular invasion, visceral pleural invasion and spread of tumour through airways, have not been reported. However, cigarette smoking is strongly associated with the presence of a solid component within the tumour (prevalence of 53% versus 20% in stage I LUADs from individuals who had never smoked; multivariate HR 3.32, 95% CI 1.78–6.19; P< 0.001), and EGFR-mutant LUADs less frequently contain solid components than EGFR-wild-type LUADs (17% versus 51%; P = 0.033)42. Most studies have not characterized histology beyond LUAD and LSCC, although one study of 320 stage I LUADs revealed that those in patients with a never-smoking status are more frequently bronchioalveolar carcinoma (85% versus 58% in ever-smoking patients; P< 0.001) and more commonly have papillary components (81% versus 68%; P = 0.01)42. No histological differences in precursor lesions between never-smoking and ever-smoking patients have been reported, such that adenocarcinoma in situ in a never-smoking patient is indistinguishable from that occurring within the context of a field effect in a patient who has smoked; however, some evidence indicates that adenocarcinoma in situ and minimally invasive adenocarcinoma in patients with a history of smoking tend to be larger than those in patients who have never smoked43. Features including a signet ring cell morphology, cribriform formation and solid or acinar growth patterns have been associated with ALK-rearranged LUAD44–49; similar features have been identified in LUADs harbouring ROS1 or RET fusions50. Imaging characteristics Several studies have examined various radiographical features of molecular subtypes of LUAD and, in general, radiographical differences correlate with the molecular driver and not with smoking status. Fusion-positive LUADs often have striking solid components, corresponding to areas of invasiveness on histology51–53. In one study, ALK-rearranged LUADs tended to be centrally located, associated with large pleural effusions and lacking a pleural tail54. No characteristic imaging findings have been consistently identified in KRAS-mutant or EGFR-mutant LUADs, and conclusions have been confounded by differences in image acquisition, cancer stage and geographical location (that is, East Asian versus non-Asian cohorts)55–57. However, characteristic volumetric tumour response dynamics have been observed following treatment with EGFR tyrosine kinase inhibitors (TKIs), consisting of an initial marked decrease in tumour burden followed by a slower decrease until the point of maximal response58–62. The magnitude of the initial, and thus the maximal, response is predictive of survival duration in patients receiving EGFR TKIs58, and slower rates of tumour regrowth following maximal response are associated with longer overall survival (OS)63. Although not clinically useful at this time, radiomic machine learning algorithms using multimodal imaging features might eventually aid in determining the molecular driver present in an individual NSCLC57,64,65. Patterns of metastatic disease Certain patterns of metastatic spread also correlate with the molecular driver. Among patients with NSCLCs harbouring EGFR mutations or ALK rearrangements, 50–60% present with or eventually develop brain metastases, compared with 16–20% of unselected patients with NSCLC66–74. ROS1-rearranged LUADs are associated with roughly half this frequency of brain metastasis (~36%), although estimates vary widely owing to the relative rarity of this disease subtype22,25,75. With the advent of osimertinib – a third-generation, brain-penetrant EGFR TKI – the risk of developing brain metastasis while on treatment has been observed to be three times less than with earlier-generation TKIs76. Oncogenic driver alterations Constitutive activation of a growing number of oncogenes through mutation, rearrangement and/or amplification accounts for 78–92% of LCINS versus 49.5% of smoking-related LUADs7,40 (Fig. 2). The most common genetic alterations include mutations in EGFR, KRAS, HER2, MET and BRAF, rearrangements involving ALK, ROS1, RET and NTRK1–3, and MET amplification, and prevalences of these alterations vary according to age and genetic ancestry (for example, East Asian versus non-East Asian)77 as well as according to heterogeneity of the molecular testing assays used. Somatic sequencing studies include those at the whole-exome7 (Fig. 2a) and whole-genome78 (Fig. 2b) levels, as well as panel-based next-generation sequencing (NGS) data derived from the AACR Project GENIE cohort, which is the largest collective somatic sequencing study in patients with cancer and comprises clinical data from multiple institutions. A subset of the AACR Project GENIE cohort (Biopharma Collaborative NSCLC cohort79, n = 1,846) has been annotated for outcomes and clinicopathological variables, including smoking status, providing insights on the prevalence of oncogenic alterations in LUADs from patients who have never smoked (Fig. 2c). Rare molecular subtypes of lung cancer, most of which are more commonly found in LCINS, have been recently reviewed in this journal80. Fig. 2 \|. Prevalence of oncogenic somatic driver alterations in LCINS. Open in a new tab Data from somatic profiling studies of lung adenocarcinoma (LUAD) in individuals who have never smoked (LCINS). a, Whole-exome sequencing (WES) and RNA sequencing study7 of 160 samples, at an average depth of 20–30× for WES. The prevalence of each somatic driver alteration was calculated as a composite average using three individual cohorts included in the study, with values weighted proportionally to the number of samples in each cohort. Genetic ancestry was inferred by the authors based on germline sequencing data from matched peripheral blood. b, Whole-genome sequencing (WGS) study78 of 189 samples at an average depth of 85×. Genetic ancestry was inferred by the authors based on germline sequencing data from matched peripheral blood. c, Clinical panel-based next-generation sequencing (NGS) data from 1,508 LUADs, collected as part of the non-small-cell lung cancer (NSCLC) cohort of the AACR Project GENIE Biopharma Collaborative79, across four large academic cancer cancers in North America (Dana-Farber Cancer Institute, Memorial Sloan Kettering Cancer Center, Vanderbilt-Ingram Cancer Center, and Princess Margaret Cancer Centre-University Health Network). Clinical data available for this cohort include tumour histology, patient self-reported primary ethnicity and cigarette use at time of diagnosis: never smoked; quit smoking >1 year prior to diagnosis (‘former smoking (remote)’); quit smoking <1 year prior to diagnosis (‘former smoking (recent)’); and current smoking. Self-reported primary ethnicity is shown for samples with never-smoking status only. Included for analysis were somatic mutations (EGFR, KRAS, ERBB2, BRAF and MET) and gene fusions (ALK, ROS1 and RET) annotated as ‘putative drivers’ based on prior knowledge (OncoKB387,388) and statistical recurrence (Cancer Hotspots389,390). The GENIE Biopharma Collaborative Public release is available for download ( and can be visualized and analysed using the cBioPortal interface ( Note, wide ranges in the prevalence of particular alterations across studies probably reflect differences in cohort ascertainment (for example, 25.0% versus only 1.7% of patients were of East Asian ancestry in the WES and WGS studies, respectively) and heterogeneity of testing assays used. ND, none of the above alterations detected. EGFR. EGFR-mutant LUAD constitutes the largest proportion of LCINS, with rates as high as 60–74% in non-smoking East Asian women with lung cancer77. RNA sequencing (RNA-seq) and whole-exome sequencing (WES) data from a large cohort of never-smoking patients with LUAD (n = 160) found EGFR to be mutated in up to 52.5% of participants (versus ~10.4% of smoking-related LUADs)7. Canonical mutations in exons 19 and 21 (that is, exon 19 deletions and the L858R point mutation, respectively) account for the vast majority of EGFR alterations in LCINS (>85%), without a preponderance of one alteration relative to the other7, although the L858R mutation and exon 19 deletions seem to occur more frequently in older and younger patients, respectively18,81. EGFR mutations occur in a small fraction of smoking-related lung cancers, with the likelihood of EGFR mutation decreasing with increasing pack-years10. Regardless of pack-years smoked, the likelihood that a lung cancer will harbour an EGFR mutation increases with the number of smoke-free years prior to diagnosis, particularly in those who have stopped smoking for >25 years10,82. Uncommon sensitizing and non-sensitizing EGFR mutations might be more likely to occur in smoking-related compared with non-smoking-related lung cancers83,84; however, ~60% of patients with uncommon EGFR exon 20 insertions have never smoked and most are women, in contrast to those with EGFR-wild-type cancers (but similar to those with classical sensitizing EGFR mutations)84–87. In a study of 102 NSCLC samples harbouring uncommon EGFR mutations, 85% of those with exon 18 alterations and 43% of those with exon 20 alterations occurred in patients with a history of smoking84. Rarely, oncogenic EGFR fusions occur, most commonly with RAD51 as the fusion partner, and reports of EGFR–RAD51 fusions in LUAD most often involve patients with a history of never or light (<5 pack-year) smoking, who are mostly young (<40 years of age)88–90. These fusions are targetable with several generations of EGFR TKIs and sustained clinical responses have been reported88–90. ALK. Rearrangement of ALK is the second most common oncogenic driver alteration in LCINS and is found in up to 14% of these cancers91, although this prevalence varies widely according to study design and population as well as between different age groups7,78,92. Overall, ALK rearrangements are found in ~5% of patients with NSCLC, with a roughly equal incidence in Asian and European populations93. Most patients with ALK rearrangements (70–80%) have never smoked19, and up to 23% of EGFR-wild-type NSCLCs occurring in individuals <50 years of age with a history of never or light smoking harbour ALK rearrangements94. Similar to other molecular subtypes of LCINS, ALK-rearranged lung cancers are almost exclusively adenocarcinomas, although very rare cases of ALK-rearranged LSCC have been reported95,96. Other oncogenic fusions. Novel gene fusions have been identified as oncogenic drivers in NSCLC using DNA-based and/or RNA-based NGS80. These include a variety of fusions involving ROS1, RET, NTRK1–3, FGFR1, FGFR3 or NRG1 as well as more recently identified CLIP1–LTK fusions97,98. ROS1 and RET fusions each occur in approximately 1–4% of all LUADs, with the prevalence of other fusions estimated at <1%; detection sensitivities might be increased when using both DNA-based and RNA-based sequencing panels, particularly for ROS1 and RET fusions80. Many of these fusions are enriched in LCINS as well as in younger patients and are targetable with various FDA-approved TKIs97,99,100. KRAS. KRAS mutations comprise the largest molecularly defined subset of LUAD owing to the large percentage of smoking-related LUADs harbouring mutant KRAS. Activating point mutations in KRAS most commonly occur at codon 12 (G12C, G12D, G12V and G12A) and less commonly at codons 13, 10 or 61. KRAS mutations are found in only 5–15% of lung cancers occurring in white never-smoking patients7,82,91,101, with the G12D variant being most common in this context (accounting for ~56% of KRAS mutations)82. By contrast, KRAS mutations are present in up to 47% of smoking-related LUADs7, in which the G12C variant predominates (accounting for ~41% of these mutations)82. Indeed, transversions leading to the KRAS G12C, G12V, G12A and G12R variants are part of a mutational signature associated with tobacco carcinogens82. Nevertheless, smoking has the potential to cause transition mutations, albeit at a relatively lower frequency, thus accounting for the limited occurrence of mutations such as KRAS G12D in smoking-related LUADs. Furthermore, transversion mutations are also associated with non-tobacco exposures (for example, reactive oxygen species102 and prior chemotherapy103), which might explain the occasional observation of KRAS G12C mutations in LCINS. A study comparing KRAS G12D-mutant versus non-G12D-mutant NSCLC revealed that KRAS G12D-mutant tumours have lower PD-L1 expression, a lower tumour mutational burden (TMB), and lower intratumoural and total numbers of CD8+PD-1+ T cells. In keeping with these findings, KRAS G12D-mutant disease was associated with a worse objective response rate (ORR; 15.8% versus 28.4%; P = 0.03), progression-free survival (HR 1.51, 95% CI 1.44–2.00; P = 0.003) and OS (HR 1.45, 95% CI 1.05–1.99; P = 0.02) when treated with anti-PD-(L)1 antibody monotherapy104. Moreover, stratification of KRAS G12D-mutant tumours by smoking status revealed that PD-L1 expression and TMB were markedly lower in those with a history of light or never smoking (<10 pack-years), and these patients (n = 12) had an ORR of 0% with single-agent PD-(L)1 blockade. MET alterations. MET alterations are present in a small subset of NSCLCs and consist of mutations resulting in MET exon 14 skipping (METex14; estimated prevalence of 1–4%)80,105–108 and/or MET amplifications (prevalence of 1–6%, varying based on MET copy number)109–111. The percentage of patients with METex14-mutant NSCLC who have never smoked varies across studies (36–64%)108,112. One study found that METex14 mutations are significantly more likely to be identified in those who have never smoked compared to KRAS mutations (P< 0.001) but are more likely to be associated with a smoking history than EGFR mutations (P = 0.03)108. Patients with METex14-mutant NSCLC tend to be older at diagnosis (median age >70 years)108,112, whereas patients with MET amplifications are slightly younger (median age at diagnosis 60–64 years) and less often have a history of never smoking (7–34%)112–114. MET amplification also occurs as a mechanism of resistance to therapies targeting EGFR and ALK115–117. A high level of MET amplification (MET-to-CEP7 ratio of ≥5 on fluorescence in situ hybridization (FISH) or a ≥5–10-fold increase in MET copy number detected via NGS) has been used to distinguish tumours that are more likely to be MET-driven, based on the absence of co-occurring oncogenic drivers and response to MET-directed TKIs (highest in tumours with MET copy number ≥10)112,114,118; high-level MET amplification is considered an emerging biomarker in the National Comprehensive Cancer Network (NCCN) Guidelines for NSCLC. Case reports have identified the presence of MET fusions (specifically KIF5B–MET and STARD3NL–MET) in LUADs from patients with a history of never or light smoking, which were targetable with crizotinib (a TKI with activity against MET)119–121. HER2. Activating mutations in ERBB2 (also known as HER2) occur in 1–3% of NSCLCs and up to 5% of LUADs122–124, most commonly in patients who have never smoked and in women124,125. These mutations are typically small in-frame insertions in exon 20 (residues 770 to 783, most commonly the A775_G776insYVMA insertion or duplication) that result in constitutive HER2 kinase activity. Point mutations in the extracellular domain, transmembrane domain or kinase domain of HER2 have also been identified (for example, G660D, R678Q, E693K and Q709L)126,127. Notably, germline HER2 G660D mutations have been identified in patients with familial lung cancer127. The relevance of HER2 amplification and/or overexpression in NSCLC is less clear, and efforts to target HER2 in patients with NSCLC have been focused mostly on those harbouring activating HER2 mutations, approximately 54–58% of which occur in patients who have never smoked128,129. Unknown drivers. Large-scale genomic analyses of LCINS show that a small percentage of these tumours have no detectable oncogenic driver alterations (Fig. 2). In one of the largest aggregated studies of LUADs from never-smoking patients7, samples deemed oncogene negative by WES had a lower mean tumour cellularity than those in which driver alterations were readily identified. Furthermore, among 13 tumours deemed oncogene negative based on standard-coverage WES, two were found to harbour METex14 mutations after additional deep WES (~400×)7. Exploratory WES and/or whole-genome sequencing (WGS) might uncover new variants in known driver oncogenes in regions not covered in current NGS panels, particularly in the case of fusions. Additionally, current variants of uncertain significance might be reclassified as oncogenic drivers (for example, the EGFR exon 18–25 kinase domain duplication130 and kinase domain A955R7 variants identified in 2021) in this population with a high pre-test probability of driver alterations. Moreover, ongoing genomic, epigenomic and proteomic analyses of LCINS are likely to elucidate new drivers and/or therapeutic targets. Tumour mutational burden LCINS generally have a markedly lower TMB (measured as non-synonymous mutations per megabase (mut/Mb)) in coding and non-coding regions compared to smoking-related lung cancers, with early studies suggesting as much as a tenfold difference131 and a dose–response relationship between TMB and pack-years132 (Table 1). Genomic profiling of >15,000 NSCLC samples has also revealed that KRAS and BRAF driver mutations, which are more commonly associated with a history of smoking, are associated with substantially higher TMB when compared to alterations in EGFR, ALK or ROS1 (refs. 133,134). Responses to immune-checkpoint inhibitors (ICIs) in patients with NSCLC correlate with a high TMB as well as with a high number of transversions as part of a smoking-related mutational signature135. Indeed, the lack of ICI response associated with LCINS, and with NSCLCs harbouring alterations in EGFR or ALK more generally, is thought to partly reflect their lower TMB. PD-L1 expression PD-L1 expression is typically low or absent in LCINS7,136 as well as in NSCLCs with non-KRAS oncogenic drivers94,137. Furthermore, PD-L1 positivity among KRAS-mutant lung cancers is lowest in patients who have never smoked, higher in those who formerly smoked and highest in those who currently smoke, with the intensity of staining positively correlating with pack-years of smoking history136. An exception is METex14-altered NSCLC, in which moderate to high levels of PD-L1 expression have been observed, although with a median TMB still substantially lower than that of unselected NSCLCs138. ICIs have been associated with low ORRs in patients with METex14-altered NSCLC (17–36% with ICI monotherapy139), and responses do not seem to correlate with PD-L1 expression or TMB138. ALK-rearranged and ROS1-rearranged NSCLCs are more frequently PD-L1 positive compared with EGFR-mutant NSCLCs94,137,140–142 (70.1% and 72.7%, respectively, versus 50.3% in those with classical EGFR mutations)142; however, PD-L1 positivity in these tumours is thought to reflect differential intrinsic oncogene-driven activation of downstream signalling effectors that transcriptionally upregulate PD-L1 expression (such as STAT3 and HIF1α)143,144 rather than reflecting true T cell-mediated immunogenicity through IFNγ signalling that correlates with a response to ICIs. Genomic mutational signatures Analysis of somatic alterations in lung cancers has enabled the identification of genomic mutational signatures that can emerge in tissues directly exposed to tobacco smoke. An analysis of WES or WGS data from 5,243 cancers of various types often associated with smoking demonstrated an increased burden of somatic mutations and several distinct mutational signatures in tumours from patients with a history of tobacco smoking, with total base substitutions nearly fivefold higher in LUADs from such patients compared with LUADs from patients who have never smoked40. The smoking-associated single-base substitution signature 4, or simply ‘signature 4’, consists mainly of C>A transversions with lesser contributions from other base substitutions40 and is very similar to the mutational signature that results from exposing cells to benzo[a]pyrene145, a carcinogen found in tobacco smoke. This signature can also be found in cells derived from the non-malignant bronchial epithelium in individuals without cancer but with a former or current smoking status and is not present in non-malignant and tumour tissues from people who have never smoked146. Interestingly, most lung cancers occurring in patients with reported passive exposure to tobacco smoke also do not contain these signatures7,78. WES and WGS profiling of LCINS Two large-scale genomic studies of LCINS were reported in 2021 (refs. 7,78). One study was a WGS analysis of 232 LCINS from patients of mostly European ancestry (97.4% European and 1.7% East Asian by inferred genetic ancestry) and described three different tumour subtypes according to somatic copy number alterations78. Approximately 60% of samples had alterations in EGFR, KRAS, ALK, MET, HER2, RET or ROS1. As expected, median TMB was sevenfold less than that of smoking-related lung cancers at 1.1 mut/Mb, and no tobacco smoking signatures were observed, even in cases with reported passive smoking exposure78. The second study was the previously discussed WES and RNA-seq analysis of 160 tumour and matched non-malignant tissue samples from never-smoking patients with LUAD, with a substantially higher percentage of East Asian patients (62% European and 25% East Asian by inferred genetic ancestry), and including 40 and 36 samples from The Cancer Genome Atlas (TCGA) and Clinical Proteomic Tumour Analysis Consortium cohorts, respectively7. As noted above, 78–92% of these LCINS harboured clinically actionable driver alterations, compared with 49.5% of smoking-related LUADs (P< 0.001). Only 6% of LCINS contained a smoking-related mutational signature potentially indicative of passive exposure to tobacco smoke, and the median TMB ranged from 1.25 to 2.93 mut/Mb across the internal and external (TCGA and Clinical Proteomic Tumour Analysis Consortium) cohorts. The immune landscape of these LUADs was examined using RNA-seq and consensus clustering of immune and stromal cell type markers. On the basis of PD-L1 and immune cell marker expression, three clusters were identified as relatively ‘immune cold’, ‘immune hot’ or intermediate tumours, with each cluster having a similar TMB and frequency of EGFR and KRAS mutations – suggesting the potential to identify subsets of LCINS that might be more likely to respond to immunotherapy. Risk factors for LCINS Genetic risk Of the 20,000–40,000 LCINS diagnosed in the USA each year, second-hand smoke (SHS) and radon exposure are estimated to account for approximately 3,500 (ref. 13) and 2,900 cases147, respectively. In the remainder, few consistent environmental associations can be found, posing the question of underlying genetic predisposition. Overall lung cancer heritability has been estimated at 18%148 and might be even greater in those who have not smoked. Several large-scale genome-wide association studies (GWAS) have investigated common polymorphisms associated with lung cancer risk, mainly for smoking-associated cancers. Collectively, >50 loci mediating a small to moderate amount of lung cancer risk have been identified149. Additionally, data on rare germline pathogenic variants, for example, in DNA damage repair or tumour-suppressor genes, are emerging from somatic mutation profiling studies (WES and/or WGS analyses) that use matched non-malignant tissues for comparison150. The implications of germline variants for lung cancer therapy and familial genetic screening are largely unexplored. Indeed, a study testing a panel of 76–88 cancer predisposition genes to evaluate the prevalence of therapeutically actionable germline variants in almost 12,000 patients across >50 different cancer types did not include lung cancer151. Family history of cancer can be used as a simplified surrogate for inherited susceptibility, particularly in the context of LCINS given that family members do not always share a history of tobacco smoking. This surrogacy is especially robust for rare, monogenic variants with large effect sizes, which are more likely to exhibit Mendelian patterns of inheritance within families than common variants with individually small effect sizes. A systematic review published in 2005 found that having a first-degree relative with lung cancer was associated with a twofold increased lung cancer risk; this was even greater when the first-degree relative was diagnosed at a young age and/or in individuals with multiple affected family members152. A subsequent study of epidemiological risk factors in people who had never smoked revealed that a family history of any cancer diagnosed before 50 years of age in a first-degree relative was a significant predictor of increased lung cancer risk (OR 1.87, 95% CI 1.13–3.10)153. Other proxies for genetic predisposition to lung cancer might include a lack of lifetime exposure to tobacco smoke and a young age at diagnosis, particularly for LCINS (which predominate lung cancer diagnoses among individuals <40 years of age)16,18. Nevertheless, such patients might not have a family history of lung cancer owing to rare pathogenic variants with incomplete penetrance or a polygenic risk architecture that obscures an inheritance pattern. Insights from GWAS. GWAS and candidate gene studies conducted over the past decade have identified common genetic variants, typically defined as single-nucleotide polymorphisms (SNPs) with a minor allele frequency of ≥1–5%, that are associated with lung cancer149. Common variant heritability is defined as the proportion of heritability that can be explained by common SNPs and has been estimated on an observed scale at 8.3% for lung cancer and 7.1% after removing genomic loci known to be associated with smoking behaviour154. Many studies using polygenic risk scores (PRS) to further estimate this attributable risk are under way. The GWAS have been focused mainly on smoking-related lung cancer (particularly a susceptibility locus at 15q25, corresponding to a cluster of nicotinic acetylcholine receptor subunit genes that mediate nicotine metabolism and smoking behaviour)155–158, with several additional studies of LUAD in non-smoking Asian women. Reported susceptibility loci generally have a low to moderate effect size (OR 1–2) and, for many loci, suspected causal genes have been identified. To avoid spurious associations owing to population stratification (that is, systematic differences in allele frequencies between subpopulations), GWAS are often restricted to individuals of a single ethnicity and/or ancestry. For this reason and because the landscape of NSCLC differs considerably between European and East Asian populations159, most GWAS of lung cancer have considered these populations separately. A comprehensive review of all GWAS in lung cancer is beyond the scope of this article. Instead, key GWAS that have identified loci contributing to the risk of LCINS are summarized (Table 2). Table 2 \|. Key genome-wide association studies identifying LCINS risk loci, stratified by population ancestry \| Study \| Study cohorts: sample size (cases/controls) \| Chromosomal region \| Reference SNP cluster ID (rsID) \| Associated genetic loci \| OR (discovery; 95% CI) \| :--- :--- :--- \| \| European \| \| Li et al. (2010)160 \| Mayo Clinic: 377/377 MDACC: 328/407 Harvard University: 92/161 UCLA: 91/439 \| 13q31.3 \| rs2352028 \| GPC5 \| 1.46 (1.26–1.70; P = 5.94 × 10−6) \| \| Wang et al. (2010)338,a \| Candidate gene study: 259/553 \| 5p15.33 \| rs4975616 \| CLPTM1L–TERT \| 0.69 (0.55–0.85; P = 7.95 × 10−4) \| \| Sptiz et al. (2011)252,a \| Pathway-based (inflammatory) association study: 451/508 \| 12q13.13 \| rs12809597 \| ACVR1B \| 0.72 (0.59–0.88; P = 0.0012) \| \| Hung et al. (2019)339 \| ILCCO: 3,636/6,295 \| 5p15.33 \| rs380286; rs31490; rs4975616 \| CLPTM1L–TERT \| rs380286: 0.77 (0.72–0.82; P=5.31×10−16) rs31490: 0.77 (0.72–0.82; P=4.32×10−16) rs4975616: 0.78 (0.73–0.83; P=1.04×10−14) \| \| East Asian \| \| Hsiung et al. (2010)162,b \| GELAC (Han Chinese): 584/585 GELAC (replication set): 610/560 CAMSCH: 287/287 SNU: 259/293 SWHS: 209/213 WHLCS: 207/207 KNUH: 121/119 KUMC: 95/87 GEL-S: 194/546 NJLCS: 203/203 \| 5p15.33 \| rs2736100 \| CLPTM1L–TERT \| 1.54 (1.41–1.68; P=2.60×10−20); 1.62 (1.40–1.87; P=8.51×10−11) when heterozygous; 2.54 (1.95–2.83; P=3.05×10−19) when homozygous \| \| Hosgood et al. (2012)340,b \| GELAC, CAMSCH, SNU, KUMC, KNUH, SWHS, GEL-S, SLCS, FLCS and TLCS: 3,467 (2,557 LUAD, 309 SCC)/3,787 in total \| 3q28 \| rs10937405; rs4488809 \| TP63 \| rs10937405: 0.80 (LUAD: 0.74–0.87; P=7.1×10−8); 0.82 (SCC: 0.67–0.99; P=0.037) rs4488809: 1.16 (LUAD: 1.08–1.24; P=7.4×10−5) \| \| Shiraishi et al. (2012)341 \| Japanese population study: 1,722/5,846 \| 6p21.3 \| rs3817963 \| BTNL2 \| 1.18 (1.12–1.24; P=2.7×10−10) \| \| 17q24.3 \| rs7216064 \| BPTF \| 1.20 (1.13–1.26; P=7.4×10−11) \| \| Lan et al. (2012)163,b \| FLCCA: 5,510/4,544 (from 14 studies) and 1,099/2,913 (replication set) \| 10q25.2 \| rs7086803 \| VTI1A \| 1.32 (1.24–1.41; P=5.04×10−17) \| \| 6q22.2 \| rs9387478 \| ROS1, DCBLD1 \| 0.85 (0.81–0.90; P=7.79×10−8) \| \| 6p21.32 \| rs2395185/rs28366298 \| HLA class II region \| 1.16 (1.09–1.23; P=2.60×10−6) \| \| Ahn et al. (2012)171 \| Korean population study: 446/497 (discovery set) and 434/1,000 (replication set) \| 18p11.22 \| rs11080466; rs11663246 \| FAM38B (PIEZO2), APCDD1, NAPG \| rs11080466: 0.61 (0.44–0.77; P=2.68×10−5) rs11663246: 0.60 (0.48–0.76; P=1.74×10−5) \| \| Kim et al. (2013)175,b \| Korean population study: 285/1,455 (discovery set), 294/495 (replication set 1) 546/733 (replication set 2) \| 2p16.3 \| rs10187911 \| NRXN1 \| 1.47 (1.22–1.78; P<0.001) \| \| Wang et al. (2016)174,b \| FLCCA: 6,877/6,277 (from 4 studies) and 5,878/7,046 (replication set) \| 6p21.1 \| rs7741164 \| FOXP4, FOXP4–AS1 \| 1.18 (1.10–1.26; P=2.05×10−6) \| \| 9p21.3 \| rs72658409 \| CDKN2B, CDKN2B–AS1 \| 0.75 (0.67–0.83; P=1.37×10−7) \| \| 12q13.13 \| rs116101143 \| ACVR1B \| 0.88 (0.83–0.93; P=2.21×10−6) \| \| Shi et al. (2023)159 \| FLCCA: 4,438/4,544 NJLCS: 1,923/3,544 NCC: 3,291/19,910 ACC: 1,471/2,564 \| 2p23.3 \| rs682888 \| DTNB \| 0.89 (0.86–0.93; P=4.94×10−10) \| \| 3q22.2 \| rs137884934 \| PIK3CB \| 0.81 (0.74–0.89; P=6.33×10−6) \| \| 4p13 \| rs117715768 \| KCTD8 \| 1.24 (1.14–1.34; P=4.48×10−7) \| \| 4q32.1 \| rs1373058 \| PDGFC \| 1.10 (1.05–1.15; P=8.55×10−6) \| \| 6p12.1 \| rs531557 \| GCLC \| 0.90 (0.87–0.94; P=7.73×10−7) \| \| 7q31.33 \| rs4268071 \| GPR37 \| 1.39 (1.25–1.54; P=7.27×10−10) \| \| 10q26.13 \| rs10901793 \| FAM53B, METTL10 \| 1.10 (1.06–1.14; P=3.14×10−7) \| \| 11q12.2 \| rs174559 \| FADS1 \| 0.91 (0.88–0.94; P=6.10×10−7) \| \| 15q21.2 \| rs764014 \| RFX7 \| 0.91 (0.88–0.95; P=5.75×10−7) \| \| 15q21.3 \| rs71467682 \| FGF7, SECISBP2L \| 0.91 (0.87–0.95; P=2.46×10−6) \| \| 19p13.1 \| rs116863980 \| PALM \| 1.31 (1.16–1.47; P=7.94×10−6) \| \| East Asian and European \| \| Shi et al. (2023)159 \| ILCCO multi-ancestry meta-analysis (similar effect sizes observed in both groups): 11,753/30,562 (East Asian) and 11,273/55,483 (European) \| 2p11.2 \| rs1130866 \| SFTPB \| 1.08 (P=1.56×10−8)c \| \| 4q32.2 \| rs2320614 \| NAF1 \| 1.08 (P=6.51×10−9)c \| \| 16q23.3 \| rs34638657 \| MPHOSPH6 \| 1.09 (P=2.19×10−9)c \| \| 18q12.1 \| rs638868 \| GAREM1 \| 1.08 (P=3.60×10−8)c \| Open in a new tab The table lists the novel risk loci identified in each study. ACC, Aichi Cancer Center (Japan); CAMSCH, Chinese Academy of Medical Sciences Cancer Hospital; FLCCA, Female Lung Consortium in Asia; FLCS, Fudan Lung Cancer Study (China); GELAC, Genetic Epidemiological Study of Lung Adenocarcinoma (Taiwan); GEL-S, Genes and Environment in Lung Cancer, Singapore study; ILCCO, International Lung Cancer Consortium; KNUH, Kyungpook National University Hospital (South Korea); KUMC, Korea University Medical Center; LCINS, lung cancer in individuals who have never smoked; LUAD, lung adenocarcinoma; MDACC, MD Anderson Cancer Center; NCC, National Cancer Center of Japan Research Institute; NJLCS, Nanjing Lung Cancer Study (China); OR, odds ratio; SCC, squamous cell carcinoma; SLCS, Shenyang Lung Cancer Study (China); SNP, single-nucleotide polymorphism; SNU, Seoul National University (South Korea); SWHS, Shanghai Women’s Health Cohort Study (China); TLCS, Tianjin Lung Cancer Study (China); WHLCS, Wuhan Lung Cancer Study (China); UCLA, University of California Los Angeles (USA). a These studies are candidate gene and pathway-based analyses, which interrogate predefined biological pathways or gene sets; these are distinct from traditional genome-wide association studies and do not require genome-wide significance; therefore, the results should be interpreted with caution. b These studies were focused specifically on women. c No confidence intervals provided in summary statistics. One of the first GWAS to assess the genetic risk of lung cancer in individuals who have never smoked was conducted in Europeans, and the findings published in 2010 identified a single locus at 13q31.3 where variants resulting in lower transcription of GPC5 were associated with increased susceptibility (OR 1.46, 95% CI 1.26–1.70; P = 5.94 × 10−6)160. GPC5 encodes glypican 5, a proteoglycan with poorly understood roles in physiology and tumorigenesis. Interestingly, GPC5 expression is reduced in LUAD but not in other histological subtypes of lung cancer compared with the surrounding non-malignant lung tissues160. The 5p15.33 locus containing TERT and CLPTM1L has been identified as a lung cancer susceptibility region in both smoking and non-smoking populations, including never-smoking women of East Asian descent161–163 (Table 2). Although specific pathogenetic mechanisms have not been elaborated, this locus is associated with the risk of lung, bladder, prostate and cervical cancers in both European and Asian populations162–164. Results of a meta-analysis of GWAS performed in two independent cohorts to identify variants associated with OS in never-smoking European individuals with NSCLC were reported in 2013 (ref. 165). Out of the top 25 SNPs (combined P< 1 × 10−6), 6 variants showed a genotype–expression association upon expression quantitative trait loci analysis, none of which were within genes previously found to be associated with overall lung cancer risk in people who have not smoked nor within genes associated with OS in patients with smoking-related lung cancer166–168. GWAS of LCINS have been more extensive in patients of East Asian ancestry (Table 2), with risk variants at the aforementioned 5p15.33 locus recurrently identified in never-smoking women with LUAD162,169,170. Aside from this locus, the loci identified in never-smoking Asian women are generally distinct from those identified in European people with a history of smoking160,163,171. A LUAD susceptibility locus at 3q28 corresponding to TP63 was first identified in Japanese and Korean populations (OR 1.31; P = 7.26 × 10−12), with a trend towards higher ORs in women but no clear differences by smoking behaviour172, and was later confirmed specifically in never-smoking Asian women163. This locus was also subsequently associated with LUAD risk in European populations (OR 1.13; P = 7.22 × 10−10)173. A later study imputing data from four prior GWAS of lung cancer in never-smoking Asian women using the 1000 Genomes Project identified three new risk loci with small effect sizes at 6p21.1, 9p21.3 and 12q13.13, which map near FOXP4, CDKN2B and ACVR1B, respectively174 (Table 2). Subsequent studies in never-smoking Korean populations have identified additional LCINS susceptibility regions on chromosomes 2 and 18171,175 (Table 2). Most recently, a large, two-stage GWAS of LUAD occurring in individuals of East Asian ancestry identified 12 novel susceptibility variants159 and identified novel alveolar lineage-specific candidate genes, including FADS1 and ELF5, via expression quantitative trait loci colocalization analyses. A multi-ancestry meta-analysis performed as part of the same study revealed four additional novel risk loci shared among East Asian and European individuals (Table 2), although the majority of associations identified in the East Asian cohorts did not extend to the European population. Historically, few studies have identified gene–environment interactions in lung cancer risk176. However, a PRS generated from the top 25 independent susceptibility variants that achieved genome-wide significance in the East Asian population stratified individuals in the highest risk quintile from those in the middle quintile (corresponding to average risk in the general population) to a greater extent in the non-smoking versus smoking population (OR 2.07 versus 1.80; P Interaction = 0.0058)159. Taken together, the results from GWAS of LCINS are largely non-overlapping and have not converged on a set of robust associations. An analysis by the Transdisciplinary Research In Cancer of the Lung (TRICL) consortium using a custom SNP array restricted to rare variants in known cancer-associated genes identified a large effect association between germline ATM L2307F mutations and LUAD (OR 2.93–8.82 across discovery and replication cohorts)177. This effect seemed to be independent of smoking status, although never-smoking women carrying the L2307F variant were approximately four to seven times more likely to develop LUAD than non-carriers. Notably, this variant is much more prevalent in the Ashkenazi Jewish population (~4%) and might constitute a founder mutation. Variant-associated clinicopathological variables reaching high levels of statistical significance included never-smoking status and female sex but also adenocarcinoma histology and the presence of a somatic EGFR mutation177. Interestingly, never-smoking patients with lung cancers harbouring somatic EGFR mutations are more likely than those with EGFR-wild-type tumours to have a family history of lung cancer178,179. Moreover, a lung cancer genomics and ancestry analysis of admixed Latin American populations demonstrated that the frequency of somatic EGFR mutations varies by ethnicity, suggesting a germline component to EGFR mutation status180. Polygenic risk scores. PRS integrate the risk associated with GWAS-identified, disease-associated common SNPs present in the genome of an individual into weighted averages to produce a ‘personalized genetic susceptibility profile’ as a single measure of risk. Using data generated from previous GWAS conducted predominantly in people with a history of smoking, researchers developed PRS specific for 16 different cancer types; the investigators then compared the ability of the PRS versus self-reported family history of cancer in first-degree relatives to predict the risk of cancer within 5 years among 413,870 individuals included in the UK Biobank (with 22,755 incident cancer cases)181. Interestingly, lung cancer was unique in that family history was a significantly better predictor of risk than PRS, such that individuals with a positive family history and a low PRS had a higher 5-year risk of lung cancer than those with a high PRS and negative family history (0.54% versus 0.46%)181. This trend was not observed for several other common cancers, including prostate, breast and colorectal cancers. Assuming that affected relatives from different families did not share common exposures, this finding indicates that lung cancer heritability might be largely attributable to rare, high-effect alleles that mediate familial risk, although a possible alternative explanation is that a positive family history better predicted the risk in individuals who had not smoked given that the PRS was generated from GWAS of predominantly smoking-related NSCLCs. Another lung cancer-specific PRS was generated by combining genetic and epidemiological data from approximately 13,000 patients and 10,000 control individuals and subsequently validated in almost 336,000 individuals included in the UK Biobank182. This PRS was generated based on a population in which >90% of patients had a history of smoking; therefore, a de novo model to predict the risk of LCINS was derived based on age, sex, body mass index, family and personal history of cancer, impaired lung function, and exposure to ambient air pollution and SHS, in addition to PRS182. Interestingly, including environmental exposures did not improve the predictive performance of the model for LCINS182, which is in contrast to the previously mentioned data from East Asian populations showing an interaction of PRS with smoking status159, probably because smoking is a stronger risk factor. Ultimately, family history and PRS are not mutually exclusive but rather complementary factors that can provide different insights into the cancer risk of an individual. Indeed, combining family history and PRS has been shown to enhance risk assessment for breast and prostate cancers183,184, although a PRS for these malignancies has not yet been integrated into clinical practice. Whether PRS will have meaningful utility in predicting lung cancer risk remains unclear, and further efforts in this area will probably need to distinguish between smoking-related versus non-smoking-related lung cancers. Insights from WES and WGS. Susceptibility loci identified in GWAS account for only a small portion of the variation in lung cancer incidence, mainly because GWAS typically detect common variants with small to moderate effect sizes. Studies at the exome and genome levels are therefore needed to identify rare, large-effect/high-risk germline variants; only a handful of germline analyses in lung cancer have been conducted at these levels, either using data gleaned from matched non-malignant tissues in somatic tumour-profiling studies or as dedicated germline-directed efforts. The largest widely available datasets are derived from TCGA185 (containing WES data from ~580 LUADs and paired non-malignant tissues and WGS data from ~100 tumours) and the Hartwig Medical Foundation186,187. Additional datasets come from the Pan-Cancer Analysis of Whole Genomes (PCAWG) project188 as well as a WES study of LUAD in patients of East Asian ancestry189. In the Hartwig Medical Foundation cohort, which included 178 patients with lung cancer186, various nonsense, frameshift or splice site-altering germline variants were found in CHEK2 and Fanconi anaemia complementation group (FANC) genes (FANCI, FANCL, FANCM), involved in DNA repair, as well as DOCK8 and GJB2 (encoding dedicator of cytokinesis protein 8 and gap junction β2 protein, respectively). A germline WES study focused on SCLC or extrapulmonary small-cell carcinoma identified 42 deleterious germline variants across 35 cancer predisposition genes in 38 (44%) of 87 patients, although 90% had a current or former smoking status190. WES and/or WGS studies had not distinguished between smoking and non-smoking individuals until 2021, when results emerged from the two formerly discussed somatic alteration profiling studies of never-smoking LUAD, which included data from matched non-malignant tissue analyses,7,78. In the WGS study78, 8 of 232 patients carried pathogenic germline variants (PGVs) in CYP21A2, which encodes the 21-hydroxylase enzyme involved in the synthesis of cortisol and aldosterone, 6 patients carried the same PGV in GLUD2, which encodes the mitochondrial enzyme glutamate dehydrogenase 2, and 5 patients had PGVs in the AR gene (encoding the androgen receptor); PGVs in BRCA1, ATM and RAD51 were each found in 2 or 3 patients. Interestingly, both CYP21A2 and AR are involved in hormone production and signalling, and might thus contribute to sex differences in the incidence of LCINS. In the WES study7, rare pathogenic or likely pathogenic (P/LP) mutations in known cancer predisposition genes were identified in patients who had never smoked; although the prevalence of P/LP mutations was similar to that in those who had smoked (6.9% and 6.4%, respectively), variants in BRCA1, BRCA2, FANCG, FANCM, HMBS, MSH6, NF1,POLD1, TMEM127 and WRN were exclusive to never-smoking individuals. Using a cancer-free control cohort derived from the Genome Aggregation Database (gnomAD), the never-smoking LUAD group was enriched for P/LP variants in FANCG (encoding a component of the Fanconi anaemia DNA repair complex) and TMEM127 (encoding a negative regulator of mTOR signalling). Larger sample sizes along with functional validation will be needed to further implicate these germline variants in lung cancer pathogenesis, although these early studies in individuals who have never smoked have highlighted the presence of pathogenic alterations with large effect sizes in DNA repair-related genes. This finding is underscored by data from a broader study of common diseases showing that patients with a low common variant PRS are more likely to carry rare disease-specific pathogenic variants191, suggesting that these individuals could be prioritized for rare variant screening. Lastly, in a study using WES data from participants in the UK Biobank and Mass General Brigham Biobank, the presence of clonal haematopoiesis was found to be associated with increased risk of lung cancer (meta-analysed OR 1.35, 95% CI 1.08–1.68)192, specifically for LUAD (OR 1.68, 95% CI 1.23–2.29) and LSCC (OR 1.59, 95% CI 1.51–1.68) but not for SCLC. Clonal haematopoiesis was also associated with a 36% increase in lung cancer risk among the UK Biobank participants after adjusting for major risk factors, including pack-years of smoking, age at sequencing, family history of lung cancer and lung cancer PRS192. Whether clonal haematopoiesis is a surrogate of currently unknown shared risk factors, including inherited genetic risk (for example, mediated by rare variants), or has a more causal role in lung cancer remains unclear. Germline EGFR mutations in familial lung cancer. Germline EGFR mutations have been identified in familial lung cancers193. Most common is the T790M mutation, which, in its somatic form, can be present at diagnosis or develop as a mechanism of resistance to earlier-generation EGFR TKIs. The germline EGFR T790M mutation was identified in 2005 among a family of European descent with several members across multiple generations developing LUAD193. Germline EGFR T790M has since been found in multiple unrelated kindreds, predominantly comprising white individuals in the USA, Europe and Australia, more frequently in never-smoking women, and in association with multiple primary lung lesions (either nodules or invasive adenocarcinomas)193–196. A cluster of families has been identified in the southeastern USA, suggesting a possible founder effect197. Notably, germline EGFR T790M has not been reported in patients of East Asian ethnicity with lung cancer despite the high prevalence of somatic EGFR mutations in this population. Germline EGFR T790M mutations have been estimated to occur in 0.5–1.0% of patients with NSCLC195,198 and in roughly 1 in 100,000 individuals in the general population194 (allele frequency of 9.9 × 10−6 in the gnomAD v4.0 database)199. Additional studies with larger sample sizes are needed to determine more precise frequencies of germline EGFR T790M mutations in patients with lung cancer as well as the magnitude of their effect on lung cancer risk. Virtually all lung cancers that develop in the context of a germline EGFR mutation harbour a secondary somatic activating mutation in EGFR in cis193, most commonly L858R197,200. EGFR T790M carriers without known lung cancer are often found to have multiple groundglass nodules on CT imaging, suggestive of premalignant lesions that can develop into invasive adenocarcinoma over time194, and are likely to benefit from surveillance CT-based screening. EGFR T790M carriers do not seem to have an increased incidence of any other cancer type; the mechanism underlying LUAD specificity is unknown. Less common germline EGFR mutations include R776G/H201,202 and V769M in exon 20 (ref. 203) and V834L and V843I in exon 21, the latter identified specifically in Asian and Surinamese families204–206. Even rarer is the EGFR R831H germline variant reported in Chinese patients with NSCLC207,208, which was also described as co-segregating with prostate cancers harbouring somatic biallelic inactivation of CDK12 in two brothers within one family; prostate cancer cells derived from these patients were responsive to the EGFR TKI afatinib in vitro209. To our knowledge, EGFR R831H is the only germline EGFR mutation associated with a cancer type outside of the lung, although genotype–phenotype relationships among various germline EGFR mutations have not been extensively studied. Additional genetic or environmental modifiers might affect the lung cancer risk associated with germline EGFR mutations and account for phenotypic differences between individuals within families, despite carrying the same mutation. Non-EGFR germline mutations in familial lung cancer. Germline mutations in familial lung cancer pedigrees have also been found in HER2 (refs. 210-212), BRCA2, CHEK2 (ref. 173), MET213 and YAP1 (ref. 214), predominantly in never-smoking patients with LUAD and more commonly in female patients. The germline G660D mutation in the transmembrane domain of HER2 was initially reported in a Japanese family, identified in a female proband with a 1.2 pack-year smoking history and multifocal NSCLC210. Following lobectomy, pathology demonstrated that this woman had innumerable pre-invasive lesions; similar lesions were seen bilaterally on imaging in her 30-year-old daughter, also a HER2 G660D carrier212. No additional somatic alterations were identified (including HER2 and EGFR), and the proband was treated with second-line afatinib following disease progression on chemotherapy, with a partial response in the lung and stable disease for ≥16 months in the bone211. Additionally, a candidate gene study using WGS in a never-smoking Taiwanese family with high frequency of LUAD identified a germline variant (R331W) in the transactivation domain of YAP1 (ref. 214), a transcriptional regulator in the Hippo signalling pathway that has been implicated in resistance of EGFR-mutant NSCLC to EGFR TKIs215,216. In a validation cohort derived from the Genetic Epidemiology Study of Lung Adenocarcinoma in Taiwan, the YAP1 R331W allele frequency was 1.1% in patients with LUAD versus 0.18% in individuals without cancer, translating into an OR of 5.9 after adjusting for age, sex and smoking status214. All chest CT-screened YAP1 R331W carriers had LUAD or groundglass lesions (40% and 60%, respectively)214. These studies support the existence of rare, highly penetrant PGVs that contribute to a subset of LCINS. Hereditary cancer predisposition syndromes. LUAD is a principal malignancy associated with several multi-organ cancer predisposition syndromes resulting from germline mutation of tumour-suppressor genes (for example, TP53, PTEN or LKB1)217. Several case reports have demonstrated oncogene-driven NSCLC in patients with Li–Fraumeni syndrome218–221. In one study, 21 (91%) of 23 NSCLC tumours in patients with Li–Fraumeni syndrome harboured an oncogenic alteration, 20 (87%) of which were EGFR mutations, most commonly exon 19 deletions220. EGFR-mutant LUAD has also been described in Cowden syndrome, caused by germline PTEN mutations, in case reports of younger patients with a history of light or never smoking222. An increased risk of lung cancer has been reported in survivors of hereditary retinoblastoma with germline RB1 mutations, with lung cancer diagnosis tending to occur before 40 years of age223, and there are case reports of lung cancers associated with Bloom224, Werner225 and Birt–Hogg–Dube syndromes217. Environmental risk factors Various environmental exposures have been implicated or hypothesized to contribute to the risk of LCINS (Table 3). An exhaustive discussion of these risk factors is beyond the scope of this Review; here, we focus on key exposures with established associations and/or those with contemporary evidence of an effect on lung cancer risk. Table 3 \|. Environmental risk factors for LCINS \| Risk factor \| Effect summary \| Refs. \| :--- \| Radon \| Second-leading environmental cause of lung cancer, after smoking, and estimated to account for ~12% of lung cancers in the USA annually; the association of radon with lung cancer was first noted in uranium miners in the 1980s228,229; numerous large, global studies have demonstrated an association between prolonged residential radon exposure and lung cancer in the general population; the risk is greatest when combined with smoking230,234,242 but applies to both smoking and non-smoking populations244; the strongest histological association is with SCLC, followed by LUAD245–247 \| 226,228–239, 242–247,342 \| \| SHS \| SHS increases the risk of lung cancer development in individuals who do not smoke by 20–25% and is responsible for ~3,500 lung cancer deaths among non-smoking people in the USA annually1,13; other studies are limited but the reported effect sizes are moderate (OR range 1.2–2.1)153,253–256,343; no differences in the prevalence of various targetable alterations found in LCINS have been reported based on exposure to SHS258; tobacco-related mutational signatures have not been detected in SHS-related LCINS7,78 \| 1,7,13,78,153,249–256,258,343 \| \| Occupational carcinogens \| Elements, such as sulfur, arsenic, beryllium, cadmium, chromium, nickel, plutonium and radon, fumes and particulate matter with a diameter of ≤10μm (PM 10) or ≤2.5μm (PM 2.5), diesel engine exhaust emissions, silica dust, X-rays, and γ-radiation have each been associated with an increased risk of LCINS281, as have bis(chloromethyl)ether, chloromethyl methyl ether (technical grade), coal-tar pitch and asbestos (all forms, including actinolite, amosite, anthophyllite, chrysotile, crocidolite and tremolite); asbestos is definitively associated with an increased risk of bronchogenic carcinoma and pleural mesothelioma285,286, with the risk being several-fold higher for lung cancer than for mesothelioma285,286, although for both, the risk is greatly multiplied when combined with cigarette smoking286,290–292,344; silica, diesel exhaust emissions and welding fumes increase the risk of lung cancer independently of smoking and co-exposures282–284 \| 281–286,289–294,344,345 \| \| Air pollution \| Air pollution and fine particles (PM 2.5) are classified as group 1 carcinogens by the WHO IARC268,310; WHO has estimated that ~6% of outdoor air pollution-related premature deaths in 2016 were due to lung cancer268; associations with lung cancer risk and mortality have been demonstrated in individuals who have smoked and those who have not (HR < 2)265,267,271,272, and air pollution-associated lung cancer risk is probably influenced by underlying genetic susceptibility265,278; PM 2.5 levels positively correlated with rates of EGFR-mutant lung cancer in England, South Korea and Taiwan279 \| 261–265,268,276–279,346 \| \| Electronic cigarettes/vaping \| The carcinogenic potential of electronic cigarette/vaping particles is not currently known and is probably dependent on the composition of aerosolized liquid, the device and user habits302–304; long-term follow-up studies are required to determine any lung cancer risk \| 302–308,347–349 \| \| Household use of solid fuels and high-temperature frying \| Fumes and particulate matter generated from cooking and heating contain group 1 (carcinogenic) or group 2 (likely carcinogenic) substances as defined by the WHO IARC; dose–response relationships between reported exposure to cooking oil fumes and lung cancer risk have been demonstrated; in Chinese non-smoking women, the ORs are as high as 6.15 in the highest exposure groups (higher with deep frying versus stir-frying and in homes with poor ventilation)311–313; indoor burning of coal, wood and other biomass is associated with an increased risk of lung cancer with ORs <2, with greater levels of risk in women than in men315,316,350 \| 310–316,350 \| \| Diet \| No consistent data exist that implicate dietary differences as being associated with lung cancer risk; primary chemoprevention studies focused on lung cancer have not demonstrated benefit from dietary supplements, although some studies have shown an association between β-carotene supplementation and a reduced risk of lung cancer in men who have smoked351–353 \| 351–355 \| \| HRT \| Findings regarding the association between HRT and lung cancer risk are mixed; data from the Women’s Health Initiative trial demonstrated no statistically significant increase in the risk of lung cancer with either oestrogen–progestin or oestrogen-only HRT356–358; however, results of the Vitamins and Lifestyle study indicate an increased risk of lung cancer with oestrogen–progestin HRT after adjusting for smoking status (HR 1.48 if exposed to ≥10 years oestrogen plus progestin)359 \| 78,356–364 \| \| Prior and/or chronic lung disease \| A history of lung disease (COPD, pneumonia, tuberculosis or interstitial lung disease) is associated with an increased risk of lung cancer365–370, although the associations are confounded by tobacco smoking and other environmental exposures; an elevated lung cancer risk associated with emphysema, pneumonia and tuberculosis has been observed consistently in white men367; the underlying mechanisms are presumed to be related to chronic inflammation and parenchymal lung injury; whether autoimmune diseases are associated with an increased risk of developing lung cancer in individuals who do not smoke remains unclear \| 365–375 \| \| Recreational and illicit drugs \| Whether habitual smoking of marijuana or cocaine use increases the risk of lung cancer is unclear owing to confounding by tobacco smoking and/or insufficient follow-up time; opium has been classified as a group 1 carcinogen after smoking or ingestion of this drug was associated with the development of lung cancer among individuals in Golestan province, Iran (HR 2.2)376; the lung cancer risks associated with derivatives, such as heroin, morphine, codeine and fentanyl, have not been evaluated \| 376–382 \| \| Prior radiotherapy to the chest \| Chest irradiation increases the risk of second primary lung cancer, with the greatest risk in combination with tobacco smoking; for example, a high risk of secondary lung cancer has been demonstrated in individuals who received prior radiotherapy for Hodgkin lymphoma (RR 2.6–7.0), particularly those treated after ≥45 years of age, with chemotherapy having an additive effect383; lung cancer risk increases over time up to 20–25 years after radiotherapy; similar observations have been reported in breast cancer survivors, with a tenfold higher lung cancer risk in patients who received radiation for invasive breast cancer versus those who did not384 \| 383,384 \| Open in a new tab COPD, chronic obstructive pulmonary disease; HRT, hormone replacement therapy; IARC, International Agency for Research on Cancer; LCINS, lung cancer in individuals who have never smoked; LUAD, lung adenocarcinoma; SCLC, small-cell lung cancer; SHS, secondhand smoke. Radon. Radon has been identified as the second-leading environmental cause of lung cancer (after tobacco smoking) and is estimated to contribute to ~21,000 lung cancer-related deaths annually in the USA, with roughly 2,900 of these deaths occurring in individuals who have never smoked147,226,227. The increased lung cancer incidence associated with radon was first noted in uranium miners in the 1980s228–230, and numerous large, international epidemiological case–control studies have also demonstrated an association between prolonged residential radon exposure and lung cancer in the general public231–239. These and additional studies led the US Environmental Protection Agency (EPA) to classify radon as a carcinogen owing to its causal association with lung cancer, and both the EPA and the National Radon Safety Board advocate radon screening and mitigation in homes across the USA240,241. Radon exposure more than additively increases the lung cancer risk conferred by smoking234,242. In comparison with non-smoking individuals, lung cancer risk associated with radon exposure is eightfold to ninefold greater in individuals who also smoke, and >85% of radon-associated lung cancers occur in those who formerly smoked or currently smoke230,243, although the risk is still substantial in non-smoking groups244. Subsequent studies and meta-analyses have shown the strongest histological association of residential radon exposure with SCLC, followed by LUAD245–247. A population-based, case–control study performed among ~530 women with LCINS did not find an association between lung cancer and domestic levels of radon exposure, potentially owing to a relatively low level of exposure or differences in underlying genetic susceptibility. Interestingly, a meta-analysis has revealed a higher adjusted excess relative risk of lung cancer from residential radon exposure for men than for women among those who had never smoked (0.46 versus 0.09; P = 0.027)248. Secondhand smoke. SHS has long been studied as a potential causative factor in LCINS. SHS exposure is estimated to increase lung cancer risk by 20–25% in individuals who do not smoke and, in 2023, a projected 3,560 lung cancer-related deaths among non-smoking individuals in the USA were attributable to SHS1,13. In addition to the duration and intensity of exposure, genetic modifiers influencing carcinogen metabolism, DNA repair and inflammation all probably affect the lung cancer risk conferred by SHS249–252. Many epidemiological studies performed in the 1990s to early 2000s established SHS as a risk factor for LCINS153,253–256, although the effect sizes are moderate. Most of these studies have compared individuals who have a spouse who smokes with those who do not. In a meta-analysis of 37 such studies (n = 4,600), the pooled relative risk of lung cancer in women who were exposed to SHS versus those who were not was 1.24 (95% CI 1.13–1.36; P< 0.001)253. A subsequent and much smaller case–control study including 280 patients with LCINS found a larger effect (OR 2.08, 95% CI 1.25–3.43 across both sexes) but lower than that associated with exposure to environmental dust (OR 2.43, 95% CI 1.53–3.88)153. Interestingly, a large prospective cohort study involving >75,000 women revealed a 13-fold higher incidence of lung cancer in individuals who currently smoke (HR 13.44, 95% CI 10.80–16.75) and fourfold higher incidence in those who formerly smoked (HR 4.20, 95% CI 3.48–5.08) compared with those who had never smoked, but no increase was observed among never-smoking women with passive smoke exposure (HR 0.88, 95% CI 0.52–1.49)257. Despite the limited number of lung cancers in never-smoking women with SHS exposure, a borderline significant trend towards an increased risk of lung cancer was observed in those who cohabitated for ≥30 years with someone who smoked (HR 1.61, 95% CI 1.00–2.58)257. The molecular characteristics of SHS-associated lung cancers are similar to those of tumours in never-smoking patients, with no overall difference in rates of EGFR, ALK, KRAS, HER2, BRAF and PIK3CA alterations258. Although tobacco carcinogen metabolites have been found in the urine and blood of never-smoking individuals exposed to SHS253, large-scale genomic studies of LCINS have shown that reported SHS exposure often does not correlate with smoking-related somatic mutational signatures,7,78. Indeed, the small group in which such signatures were detected might consist of individuals with the highest SHS exposure and/or underlying genetic susceptibility to carcinogenesis. The WHO International Agency for Research in Cancer (IARC)259 and US National Institutes of Health260 have designated SHS as a human carcinogen, although SHS is generally accepted to confer only a modest risk of cancer and is not considered the sole causative factor for the majority of LCINS. Nonetheless, household exposure is probably more relevant than public exposure, and the continued declines in smoking rates as well as health regulations outlawing smoking in indoor public spaces will further mitigate the effects of SHS. Air pollution. Air quality can be measured by quantifying the amounts of ozone, particulate matter and chemical pollutants in ambient air. Particle pollution is a combination of small liquid and solid particles that can comprise dust, metals, soil, acids and organic chemicals. These small, inhaled particles emitted from vehicle exhaust, forest fires, coal-fired power plants and other industrial sources are thought to pass through the central airways and lodge in the more peripheral airways where LUADs typically form. The association between outdoor air pollution and lung cancer risk has been reported in several studies261–267, and both air pollution and particulate matter have been officially classified as group 1 carcinogens by the IARC265,268. Particulate matter with a diameter of ≤2.5 μm (PM 2.5) can reach the alveoli and has been associated with heart disease and lung cancer263,266,267,269. According to the IARC, the largest and most important studies consist of combined analyses spanning 17 cohorts from 9 countries in Europe263 and a multi-cohort study among large cities in the USA266, reporting HRs of 1.18 and 1.55 per 5 μg/m 3 of PM 2.5 exposure for lung cancer and LUAD, respectively. Additionally, a large study in those who had never smoked showed a 15–27% increase in lung cancer mortality per 10 μg/m 3 increase in PM 2.5 concentrations267. In 2016, the IARC estimated that ~6% of outdoor air pollution-related premature deaths were attributable to lung cancer268. Unsurprisingly, risk is proportional to the extent of exposure270 and is greater than additive when combined with cigarette smoking (risk of lung cancer mortality 2.2 times greater than additive for joint exposure)271, although evidence indicates an increased risk even after accounting for smoking263,272 as well as when restricting the analysis to those who have never smoked267. The updated 2021 WHO global air quality guidelines273 recommend annual mean PM 2.5 concentrations of ≤5 μg/m 3, while the EPA primary standard274 is ≤12 μg/m 3, with a proposal in January 2023 to revise this standard to 9.0–10.0 μg/m 3 (ref. 275). Modelled annual mean PM 2.5 concentrations for the USA276 and globally277 show the highest levels in portions of the Midwest, Southeast and California within the USA, and in northern Africa, India, China and Middle Eastern countries internationally (Fig. 3). Fig. 3 \|. Annual average PM 2.5 concentrations across the world. Open in a new tab a, Mean annual average concentrations of particulate matter measuring ≤2.5 μm in diameter (PM 2.5) across the contiguous United States, plotted in 1-km grids, spanning years 2000–2016 (ref. 276). The annual estimates are averages of daily predictions for each year in each grid cell. The current US Environmental Protection Agency annual average primary, health-based national ambient air quality standard for PM 2.5 is ≤12.0 μg/m 3, with a proposed revision to within 9.0–10.0 μg/m 3 announced in January 2023 (ref. 275). b, Mean annual average PM 2.5 concentrations across the globe, spanning years 2003–2022, according to European Centre for Medium-range Weather Forecasts Atmospheric Composition Reanalysis 4. Performed by the Copernicus Atmosphere Monitoring Service, the reanalysis combines data from modelling studies with observations from across the world into a globally complete dataset277. Annual estimates are averages of monthly predictions. The WHO air quality guidelines273 recommend annual mean PM 2.5 concentrations of ≤5 μg/m 3. As with other environmental exposures, the lung cancer risk conferred by air pollution is probably influenced by underlying genetic susceptibility265,278 (Fig. 4a). In 2021, a study analysed SNP data from >450,000 individuals included in the UK Biobank combined with estimated particulate matter exposure265. After controlling for confounders, such as obesity and smoking, the investigators generated a PRS to model genetic risk and found an additive interaction between genetic susceptibility and air pollution, such that individuals with high genetic risk and high levels of exposure to air pollution were at greatest risk of developing lung cancer (PM 2.5: HR 1.71, 95% CI 1.45–2.02) relative to those with low genetic risk and low air pollution exposure. Fig. 4 \|. Germline–somatic and gene–environment interactions in the pathogenesis of LCINS. Open in a new tab a, Model of germline–somatic interactions in lung cancer in individuals who have never smoked (LCINS). Dashed lines and arrows indicate interactions, whereas solid lines and arrows indicate direct processes. Germline variants and acquired somatic alterations can interact to promote cancer development and progression (potential mechanisms are detailed in the main text). Moreover, germline variants affecting carcinogen metabolism, DNA repair and inflammatory processes can interact with environmental exposures and thereby influence the acquisition of somatic alterations and tumour development, and polymorphisms in genes underlying drug metabolism can affect responses to therapy or the development of treatment-related toxicities (the concept of pharmacogenomics). b, Risk factors promoting LCINS include environmental exposures such as radon, secondhand smoke, outdoor (air) and indoor (household) pollution, occupational carcinogens, and prior radiotherapy to the chest. Patient-level risk factors include female sex b, Asian ethnicity and germline risk, the genetic architecture of which consists of common variants of small to moderate effect as well as rare and ultra-rare variants of large effect. Familial lung cancer syndromes are mediated by rare germline mutations in oncogenes such as EGFR. Interaction between genetics and environment can occur across any and/or multiple risk factors. GWAS, genome-wide association study; PRS, polygenic risk score; SNP, single-nucleotide polymorphism. a Copy number neutral loss of heterozygosity events can occur at loci harbouring either tumour-suppressor genes or oncogenes. b Note, environmental risks related to the socioeconomic and cultural aspects of female gender (such as historically greater exposure to household cooking fumes among women), and/or biological factors related to female sex, might contribute to the effect of female sex. More recently, another study used UK Biobank data to examine exposure to PM 2.5 in >400,000 individuals residing in the UK279. Increasing levels of PM 2.5 exposure correlated with increased risk of lung cancer (HR 1.08; 95% CI 1.04–1.12); nominally significant (P< 0.05 and false discovery rate > 0.05) associations were also reported for mesothelioma (HR 1.11, 95% CI 1.00–1.24) and lip and oropharyngeal cancers (HR 1.10, 95% CI 1.01–1.19). Analyses of the interaction between PM 2.5 exposure and ever-smoking status indicated that current or previous smoking and exposure to high levels of PM 2.5 might act in combination to increase lung cancer risk (P = 0.049). Interestingly, PM 2.5 levels correlated with the incidence of EGFR-mutant lung cancer in England as well as in South Korea and Taiwan, with the relative rates per 100,000 individuals increasing by 0.63 (P = 0.0028), 0.71 (P = 0.0091) and 1.82 (P = 4.01 × 10−6), respectively, with each additional 1 μg/m 3 PM 2.5 increment. Interestingly, DNA sequencing of non-malignant lung tissue revealed EGFR and KRAS mutations in 18% (54 out of 295) and 53% (43 out of 81) of samples, respectively, leading to the hypothesis that these mutations accumulate as part of the ageing process and that PM 2.5 promotes subsequent tumour formation in the ‘at-risk epithelium’ harbouring these driver mutations279. Further studies are necessary to elucidate the degree of lung cancer risk conferred by PM 2.5, particularly with regards to the amount and duration of exposure, as well as the direct mechanistic link between PM 2.5 and EGFR and KRAS mutation and/or tumorigenesis driven by these alterations. Importantly, inequities in exposure to air pollution disproportionately affect lower socioeconomic status groups and non-white populations280 as well as those with less education or who live closer to major sources of pollution, necessitating a global effort to improve air quality for the most vulnerable populations. Occupational carcinogens. In an extraction of data from the IARC monographs (spanning years 1971–2017), 20 different agents and their related compounds were found to be causally associated with lung cancer281 (Table 3). Several of these exposures (for example, silica, diesel exhaust and welding fumes) increase lung cancer risk independent of smoking status and co-exposures282–284. A large pooled analysis assessing the risk of lung cancer in men exposed to diesel exhaust fumes (with 16,901 cases and 20,965 controls) found an exposure–response relationship regardless of smoking history (OR 1.41, 95% CI 1.30–1.52 in never-smoking individuals at the highest exposure level), particularly for LSCC and SCLC (OR 1.38 for both cancer types at any exposure level, 95% CI 0.98–1.94 and 0.81–2.36, respectively)283. Similar exposure–response relationships were demonstrated for all histological subtypes of lung cancer using the same dataset to assess the risk associated with respirable crystalline silica282. Exposure to asbestos is definitively associated with a high risk of both bronchogenic carcinoma and pleural mesothelioma285,286, with the risk being several-fold higher for lung cancer than mesothelioma. Nevertheless, approximately 80% of patients with pleural mesothelioma report asbestos exposure287, and 15–20% of all asbestos-related deaths in the USA result from mesothelioma288. Although these proportions are decreasing owing to asbestos abatement laws effected in the latter half of the twentieth century, they are likely to remain considerable for at least a decade owing to a long latency period to lung cancer development (15–35 years). Individuals employed in mining, construction, shipbuilding and firefighting as well as veterans exposed to military asbestos products continue to constitute the groups at highest risk. The risk of lung cancer is greatly multiplied when asbestos exposure is combined with cigarette smoking286,289–292 (OR 8.70 versus 1.70 in those exposed to asbestos who never smoked)292. In men with LUAD, an association has been found between occupational asbestos exposure and an increased prevalence of KRAS codon 12 mutations, independent of smoking status and age (adjusted OR 6.9, 95% CI 1.7–28.6)293. In addition to occupational exposure, lung cancer risk might be higher in individuals who live near environmental sources of asbestos (OR 1.48, 95% CI 1.18–1.86)294. Electronic cigarettes and vaping. Whether the use of electronic cigarettes (e-cigarettes) and similar devices is associated with an increased incidence or risk of lung cancer remains unclear owing to their relatively recent development and the tendency for users to also smoke tobacco. Nicotine exposure in the absence of tobacco smoking is not thought to increase cancer risk based on studies involving users of nicotine replacement therapy295,296, although these studies have had relatively short surveillance times (for example, 5–7 years). In fact, when used as tools for smoking cessation, nicotine replacement therapy has been correlated with a decrease in lung cancer risk and mortality297–300. Several thousand cases of severe vaping-related lung injury were reported in 2019 (ref. 301), thereafter known as e-cigarette or vaping product use-associated lung injury, largely owing to vitamin E acetate contamination of e-cigarette cartridges with either manufacturer-derived or illicit substances. However, the carcinogenic potential of e-cigarette vapours is less clear, and their acute and chronic effects probably depend on the composition of the liquid being aerosolized, the specific device used and user habits302–304. When heated, common nicotine solvents can yield byproducts, such as propylene oxide and acrolein, that are carcinogenic305–307. Depending on the aerosol, e-cigarette vapour can contain ultrafine particles (<0.3 μm in diameter) capable of reaching pulmonary alveolar regions, the distribution of which is also dependent on liquid solvent content (for example, vegetable glycerin-to-propylene glycol ratio) and inhalation practices303,308,309. Although similar in size to those found in tobacco and diesel engine smoke, the carcinogenic potential of vaping particles is unknown. Ultimately, long-term follow-up studies are needed to determine any association between e-cigarette use and lung cancer. Household use of solid fuels and high-temperature frying. Fumes and particulate matter generated from burning cooking oils and solid fuels used for heating (consisting of ‘smoky’ or bituminous coal and ‘biomass’ such as wood, charcoal and crop residue) contribute to indoor air pollution in homes across the world, particularly in developing nations and parts of Africa and Southeast Asia. These emissions contain known carcinogen substances, mainly polyaromatic hydrocarbons and aldehydes as well as particulate matter with a diameter of ≤0.25 μm, that have been classified as group 1 (carcinogenic) or group 2 (likely carcinogenic) by the IARC310 based on numerous studies worldwide. Traditional cooking in many Asian countries involves heating cooking oils to very high temperatures during the process of frying, which has historically exposed women to the resulting carcinogens to a greater extent than men. A dose–response relationship has been observed between reported exposure to cooking oil fumes and lung cancer risk, which is generally higher with deep frying versus stir-frying and in homes with poor ventilation311–313. With respect to indoor burning of coal, a large retrospective cohort study among smoky coal users in China (n = 27,310) found that the absolute risk of death from lung cancer before 70 years of age was 18% and 20% for men and women, respectively, compared with <0.5% for both sexes among smokeless coal users (n = 9,962), regardless of tobacco smoking status314. The lung cancer risk associated with indoor burning of wood and other biomass is similar to that attributed to coal smoke, with ORs of <2; the risk is greater in women than in men315–317, which might reflect a greater amount of time spent in the household by women and might therefore be associated with the social construct of female gender rather than being biologically related to female sex. Household pollutants are predominantly associated with LUAD, although cases of LSCC and SCLC have been reported. Whether an underlying genetic susceptibility contributes to the lung cancer risk associated with indoor pollutants remains unclear. Germline–somatic–environment interactions Cancers arise in the context of a complex interplay between germline and somatic genomes and the environment (Fig. 4a). With regard to lung cancer, the architecture of germline risk includes many small-effect common variants and larger-effect rare and ultra-rare variants. Established rare risk variants include loss-of-function mutations in tumour-suppressor genes as well as gain-of-function alterations in oncogenes, the latter of which underlie familial lung cancer syndromes associated with germline mutations in EGFR and HER2. Interactions between inherited genetic variants and the acquisition of somatic alterations as well as environmental exposures can affect the risk and management of LCINS (Fig. 4), underscoring the importance of studies evaluating matched tumour and non-malignant tissue samples. An example of such an interaction comes from the previously discussed study in which germline ATM L2307F mutations were found to be associated with LUAD arising in individuals with a history of light versus heavy smoking, later age of onset, and broad and local somatic features (that is, EGFR mutation and biallelic ATM inactivation, respectively)177. Potential mechanisms for this interaction include (1) cooperation between germline and somatic variants in different pathways to promote a clonal advantage318, (2) germline variants acting downstream of somatic variants to modulate clonal advantage (and vice versa), (3) germline variants promoting the formation of larger-scale somatic events such as chromosomal abnormalities177, and (4) germline variants (for example, in DNA repair pathways) contributing to clonal advantage, agnostic to the nature of the somatic mutation driving the clone. Treatments for lung cancer typically target somatic alterations irrespective of germline genetic background, although a study in patients with SCLC demonstrated that PGVs in DNA repair genes can influence recurrence-free survival and response to agents targeting defective DNA repair190. Considering ICI treatment and toxicity, a germline PRS enriched for variants regulating gene expression in macrophages and dendritic cells has been shown to predict the nature of the tumour immune microenvironment and ICI response318. Additionally, a germline PRS for hypothyroidism is predictive of thyroid-specific immune-related adverse events in patients with NSCLC319. Moreover, a cross-cancer GWAS among patients receiving ICIs identified three common variants that are associated with all-grade immune-related adverse events, the most consistently replicated of which was an intronic variant in IL7 that is predictive of increased lymphocyte stability after ICI initiation as well as improved OS320. Thus, underlying germline variation can affect both tumour evolution and the development of targetable somatic alterations, and consequently influence treatment response and clinical outcomes. Diagnostic and management considerations Given the high likelihood of targetable somatic alterations, several diagnostic and management principles can be applied to LCINS (Fig. 5). The NCCN Guidelines for NSCLC321 recommend testing for mutations or fusions involving the following 11 oncogenes: EGFR, KRAS, ALK, ROS1, RET, HER2, N7RK1–3, BRAF and MET (exon 14 skipping), with high-level MET amplification as an emerging target. Testing methods vary, although panel-based NGS is preferable to maximize target coverage. Blood-based NGS might not detect driver alterations owing to insufficient shedding of tumour DNA; therefore, samples cannot be deemed oncogene negative based on liquid biopsy results alone. Most fusions (for example, those involving ALK, ROS1 or RET) are detectable with break-apart FISH assays or hybrid capture-based NGS, and DNA-based NGS has been used to identify novel fusion partners not detectable with FISH and reverse transcription PCR80. However, genomic breakpoints for rarer fusions (for example, involving NTRK1–3 or NRG1 as well as some ROS1 fusions) can occur within long intronic sequences that are not adequately covered in targeted DNA-based panels. For this reason, RNA-based NGS has been used, and can also enable the identification of fusions when low tumour purity limits DNA-based detection owing to the high expression of fusion oncogenes. Broader sequencing efforts, such as WES, can reveal therapeutic targets, particularly in the case of novel fusions or other genomic alterations in regions with low coverage in standard NGS panels. Fig. 5 \|. Molecular diagnostic algorithm for advanced-stage LCINS. Open in a new tab The algorithm outlines a sequential approach to evaluating the presence of targetable oncogenic driver alterations in lung cancers in individuals who have never smoked (LCINS) or who have a limited smoking history, which are almost exclusively non-small-cell lung cancer (NSCLC) of adenocarcinoma histology. The National Comprehensive Cancer Network (NCCN) Guidelines321 recommend that molecular analysis of advanced-stage NSCLC should include assays to identify actionable alterations in EGFR, KRAS, ALK, ROS1, RET, ERBB2 (also known as HER2), NTRK1–3, BRAF (V600E) and MET exon 14, for which FDA-approved targeted therapies are available. Emerging therapeutic biomarkers include high-level MET amplification, with the current NCCN Guidelines (version 5.2023) noting that thresholds constituting high-level MET amplification are evolving; when next-generation sequencing (NGS) is used, a MET copy number of ≥10 is consistent with high-level MET amplification and is associated with clinical benefit from MET tyrosine kinase inhibitors. Although NGS is the preferred analytical modality, molecular profiling is defined as broad testing that identifies all relevant clinically actionable biomarkers in either a single assay or a combination of a limited number of assays and, ideally, also identifies emerging biomarkers. WES, whole-exome sequencing; WGS, whole-genome sequencing. a Simultaneous blood-based and tissue-based NGS testing recommended, with ideal turnaround times of 7–10 days and 2–3 weeks, respectively. b If a matched targeted therapy is not approved in the first-line setting, may consider clinical trials offering first-line targeted therapy versus standard-of-care treatment (for example, NCT05048797 for HER2-mutant NSCLC). c Clinical decision-making should include consideration of clinicopathological variables such as smoking history, histology, PD-L1 positivity and tumour mutational burden, noting that the majority of LCINS are associated with poor response to immunotherapy. d If available, to maximize detection of fusion proteins and emerging targets. The management of LCINS is largely driven by genomic findings as outlined in NCCN guidelines for NSCLC321. Discussion of genotype-directed therapy is beyond the scope of this Review but, for patients with advanced-stage disease, targeted therapy is generally preferable when available321. Newer strategies involving TKI treatment to the point of maximal response followed by local consolidative treatment are under investigation322–325 as are upfront combinations of targeted agents with chemotherapy in patients with advanced-stage disease326. For patients with central nervous system metastases at diagnosis but without severe mass effects and in whom a targetable alteration is found for which a central nervous system-penetrant TKI is available, potential therapeutic strategies include TKI treatment prior to radiotherapy or surgical resection of brain lesions given the high intracranial efficacy of these drugs in clinical trials; however, comparisons with upfront radiotherapy and/or surgery have not yet been made. Ongoing or completed clinical trials have also incorporated targeted therapies in the neoadjuvant and adjuvant settings327–330. Indeed, adjuvant osimertinib is now approved for patients with completely resected stage IB–IIIA EGFR-mutant NSCLC, with a final analysis of the phase III ADAURA trial showing 5-year OS of 88% with osimertinib versus 78% with placebo (HR 0.49, 95% CI 0.33–0.73)328,331. In individuals with an elevated germline risk of cancer, screening has an important role in disease prevention. Low-dose CT (LDCT) screening has been shown to reduce lung cancer mortality332,333 but is only approved for individuals ≥50 years of age with a current or recent heavy smoking status. Future efforts in the LCINS space include identifying a population with elevated germline and/or exposure-mediated risk to prioritize for LDCT screening (for example, individuals carrying large-effect rare variants associated with familial EGFR-mutant lung cancer), with studies already under way (such as NCT05587439 and NCT05265429). These individuals might benefit from germline genetic testing and carriers of pathogenic variants might also benefit from LDCT screening as part of future changes to clinical care. Conclusions LCINS is an evolving and complex disease with several risk factors and open questions. With smoking rates declining, LCINS might eventually predominate lung cancer diagnoses. Investigations with specific attention to smoking history are therefore paramount both for determining the global and national trends in LCINS epidemiology, and establishing registries with detailed exposure histories. Advances have been made in understanding LCINS biology at the somatic exome and genome levels, yet the germline contribution remains largely unexplored and large-scale sequencing studies in diverse populations are needed to define germline genetic risk. Furthermore, our understanding of environmental carcinogens continues to evolve, and future directions for research should include the mechanisms driving carcinogenesis mediated by non-tobacco-related exposures, such as environmental pollution, and their interaction with underlying germline variation. The biological distinctions between LCINS and smoking-related lung cancers necessitate a unique approach to diagnosis and treatment, with integration of environmental exposures and both germline and somatic genetics to deliver precision oncology strategies for the treatment and prevention of this increasingly important disease. Key points. The global incidence of lung cancer is decreasing in parallel with declining smoking rates in developed countries; however, the incidence of lung cancer in individuals who have never smoked (LCINS) is stable or increasing. LCINS is the eighth leading cause of cancer-related mortality in the USA and the fifth most common cause of cancer-related deaths worldwide. LCINS has histological and epidemiological distinctions from smoking-related lung cancers, occurring almost exclusively as adenocarcinomas and most commonly in women and individuals of Asian ancestry. LCINS are highly enriched for targetable oncogenic alterations, have low tumour mutational burden and low rates of PD-L1 positivity, and lack mutational signatures, even in patients who report passive, secondhand smoke exposure. LCINS development probably involves interactions between genetic risk, mediated by common and rare germline variants, and environmental exposures, including air pollution and particulate matter, with potential opportunities for broader lung cancer screening. In the era of precision oncology, the biological underpinnings of LCINS necessitate unique diagnostic and treatment paradigms and warrant consideration of this disease as an important and distinct clinical entity. Acknowledgements The authors gratefully acknowledge M. Makarem, R. Collins, I. Odintsov and B. Johnson (at the Dana-Farber Cancer Institute) for their careful review of the manuscript and/or thoughtful discussions as well as Z. Wei and A. Tramontano (at the Dana-Farber Cancer Institute) and A. Shtein and Y. Wei (at the Harvard School of Public Health) for their assistance with data gathering and visualization. Footnotes Competing interests P.A.J. declares consultancy roles for Abbvie, Accutar Biotech, Allorion Therapeutics, AstraZeneca, Bayer, Biocartis, Boehringer Ingelheim, Chugai Pharmaceuticals, Daiichi Sankyo, DualityBio, Eisai, Eli Lilly, Frontier Medicines, Hongyun Biotechnology, Merus, Mirati Therapeutics, Monte Rosa, Novartis, Nuvalent, Pfizer, Roche/Genentech, Sanofi, Scorpion Therapeutics, SFJ Pharmaceuticals, Silicon Therapeutics, Syndax, Takeda Oncology, Transcenta and Voronoi; stock ownership in Gatekeeper Pharmaceuticals; research funding from AstraZenenca, Boehringer Ingelheim, Daiichi Sankyo, Eli Lilly, PUMA Biotechnology, Revolution Medicines and Takeda Oncology; and post-marketing royalties from Dana-Farber Cancer Institute-owned intellectual property relating to EGFR mutations that has been licensed to Lab Corp. The other authors declare no competing interests. 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[DOI] [PMC free article] [PubMed] [Google Scholar] ACTIONS View on publisher site PDF (1.5 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Definitions Epidemiology Features of LCINS Risk factors for LCINS Diagnostic and management considerations Conclusions Acknowledgements Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15294	https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/b5258beb40175f7ff9259d48959942dd_MIT18_03SCF11_s21_1text.pdf	Periodic Functions Periodic functions are functions which repeat: f (t + P) = f (t) for all t. For example, if f (t) is the amount of time between sunrise and sunset at a certain lattitude, as a function of time t, and P is the length of the year, then f (t + P) = f (t) for all t, since the Earth and Sun are in the same position after one full revolution of the Earth around the Sun. We state this explicitly as the following deﬁntion: a function f (t) is periodic with period P > 0 if f (t + P) = f (t) for all t. Example. f (t) = sin(2t) is periodic with period P = π. This is true because, for all t, f (t + π) = sin(2(t + π) = sin(2t + 2π) = sin(2t) = f (t). Notice, though, that in the example above f (t) = sin(2t) also has period P = 2π and period P = 3π. In fact, it has period P = nπ for any integer n = 1, 2, 3 . . . . Graphically, a function with period P is one whose graph stays the same if it is shifted P to the left or right. Base Period Most periodic functions have a minimal period, which is often called either the period or the base period. For example, sin t has minimal period is 2π. It follows from this that the minimal period for sin(2t) is π. The only exception is the constant function. Every value of P > 0 is a period and so it has no minimal period. (We don’t allow P = 0 to be a period because then every function would be periodic with period P = 0.) Windows To fully describe a periodic function you only need to specify the period and the value of the function over one full period. We call an interval containing one full period a window. Typical choices for windows are [−P/2, P/2) and [0, P), but any interval of length P will work. Frequency Terminology Angular frequency, also called circular frequency has units of radians/unit time. Frequency has units of cycles/unit time. Since one cycle is 2π radians the relationship is Periodic Functions OCW 18.03SC angular frequency = 2π × frequency. The above is the ofﬁcial terminology, but in actual practice many people say frequency when they mean angular frequency. In fact, that has been the general usage earlier in this course where we have called ω the frequency of cos(ωt). You will have to use the context to decide exactly which frequency is being used. For a function with period P the base angular frequency ω (also called the fundamental angular frequency) means the angular frequency corre sponding to the base (or minimal) period P that is 2π ω = . P Fourier Series We will see that a periodic function with base frequency ω can be written as a sum of sines and cosines whose frequencies are integer multiples of ω. This is called the Fourier series for the function. That is, sines and cosines, the simplest periodic functions, are the “building blocks" for more general periodic functions. Later in this session we will see exactly how to compute the Fourier series for a periodic function. 2 MIT OpenCourseWare 18.03SC Differential Equations Fall 2011 For information about citing these materials or our Terms of Use, visit:
15295	https://allendowney.github.io/ThinkBayes2/chap06.html	Odds and Addends — Think Bayes Skip to main content Back to top- [x] - [x] Ctrl+K Think Bayes Search Ctrl+K Think Bayes 2 Front Matter Preface Chapters 1. Probability 2. Bayes’s Theorem 3. Distributions 4. Estimating Proportions 5. Estimating Counts 6. Odds and Addends 7. Minimum, Maximum, and Mixture 8. Poisson Processes 9. Decision Analysis 10. Testing 11. Comparison 12. Classification 13. Inference 14. Survival Analysis 15. Mark and Recapture 16. Logistic Regression 17. Regression 18. Conjugate Priors 19. MCMC 20. Approximate Bayesian Computation Examples The Red Line Problem The Red Line Problem Estimating vaccine efficacy Flipping USB Connectors The Left Handed Sister Problem Bayesian Dice The Emitter-Detector Problem Grid algorithms for hierarchical models Comparing birth rates How Many Typos? How Many Books? The All-Knowing Cube of Probability What’s a chartist? The Poincaré Problem .ipynb .pdf Odds and Addends Contents 6.1. Odds 6.2. Bayes’s Rule 6.3. Oliver’s Blood 6.4. Addends 6.5. Gluten Sensitivity 6.6. The Forward Problem 6.7. The Inverse Problem 6.8. Summary 6.9. Exercises You can order print and ebook versions of Think Bayes 2e from Bookshop.org and Amazon. 6. Odds and Addends# This chapter presents a new way to represent a degree of certainty, odds, and a new form of Bayes’s Theorem, called Bayes’s Rule. Bayes’s Rule is convenient if you want to do a Bayesian update on paper or in your head. It also sheds light on the important idea of evidence and how we can quantify the strength of evidence. The second part of the chapter is about “addends”, that is, quantities being added, and how we can compute their distributions. We’ll define functions that compute the distribution of sums, differences, products, and other operations. Then we’ll use those distributions as part of a Bayesian update. 6.1. Odds# One way to represent a probability is with a number between 0 and 1, but that’s not the only way. If you have ever bet on a football game or a horse race, you have probably encountered another representation of probability, called odds. You might have heard expressions like “the odds are three to one”, but you might not know what that means. The odds in favor of an event are the ratio of the probability it will occur to the probability that it will not. The following function does this calculation. def odds(p): return p / (1-p) For example, if my team has a 75% chance of winning, the odds in their favor are three to one, because the chance of winning is three times the chance of losing. odds(0.75) 3.0 You can write odds in decimal form, but it is also common to write them as a ratio of integers. So “three to one” is sometimes written 3:1. When probabilities are low, it is more common to report the odds against rather than the odds in favor. For example, if my horse has a 10% chance of winning, the odds in favor are 1:9. odds(0.1) 0.11111111111111112 But in that case it would be more common I to say that the odds against are 9:1. odds(0.9) 9.000000000000002 Given the odds in favor, in decimal form, you can convert to probability like this: def prob(o): return o / (o+1) For example, if the odds are 3/2, the corresponding probability is 3/5: prob(3/2) 0.6 Or if you represent odds with a numerator and denominator, you can convert to probability like this: def prob2(yes, no): return yes / (yes + no) prob2(3, 2) 0.6 Probabilities and odds are different representations of the same information; given either one, you can compute the other. But some computations are easier when we work with odds, as we’ll see in the next section, and some computations are even easier with log odds, which we’ll see later. 6.2. Bayes’s Rule# So far we have worked with Bayes’s theorem in the “probability form”: P(H\|D)=P(H)P(D\|H)P(D) Writing odds(A) for odds in favor of A, we can express Bayes’s Theorem in “odds form”: odds(A\|D)=odds(A)P(D\|A)P(D\|B) This is Bayes’s Rule, which says that the posterior odds are the prior odds times the likelihood ratio. Bayes’s Rule is convenient for computing a Bayesian update on paper or in your head. For example, let’s go back to the cookie problem: Suppose there are two bowls of cookies. Bowl 1 contains 30 vanilla cookies and 10 chocolate cookies. Bowl 2 contains 20 of each. Now suppose you choose one of the bowls at random and, without looking, select a cookie at random. The cookie is vanilla. What is the probability that it came from Bowl 1? The prior probability is 50%, so the prior odds are 1. The likelihood ratio is 3 4/1 2, or 3/2. So the posterior odds are 3/2, which corresponds to probability 3/5. prior_odds = 1 likelihood_ratio = (3/4) / (1/2) post_odds = prior_odds likelihood_ratio post_odds 1.5 post_prob = prob(post_odds) post_prob 0.6 If we draw another cookie and it’s chocolate, we can do another update: likelihood_ratio = (1/4) / (1/2) post_odds = likelihood_ratio post_odds 0.75 And convert back to probability. post_prob = prob(post_odds) post_prob 0.42857142857142855 6.3. Oliver’s Blood# I’ll use Bayes’s Rule to solve another problem from MacKay’s Information Theory, Inference, and Learning Algorithms: Two people have left traces of their own blood at the scene of a crime. A suspect, Oliver, is tested and found to have type ‘O’ blood. The blood groups of the two traces are found to be of type ‘O’ (a common type in the local population, having frequency 60%) and of type ‘AB’ (a rare type, with frequency 1%). Do these data [the traces found at the scene] give evidence in favor of the proposition that Oliver was one of the people [who left blood at the scene]? To answer this question, we need to think about what it means for data to give evidence in favor of (or against) a hypothesis. Intuitively, we might say that data favor a hypothesis if the hypothesis is more likely in light of the data than it was before. In the cookie problem, the prior odds are 1, which corresponds to probability 50%. The posterior odds are 3/2, or probability 60%. So the vanilla cookie is evidence in favor of Bowl 1. Bayes’s Rule provides a way to make this intuition more precise. Again odds(A\|D)=odds(A)P(D\|A)P(D\|B) Dividing through by odds(A), we get: odds(A\|D)odds(A)=P(D\|A)P(D\|B) The term on the left is the ratio of the posterior and prior odds. The term on the right is the likelihood ratio, also called the Bayes factor. If the Bayes factor is greater than 1, that means that the data were more likely under A than under B. And that means that the odds are greater, in light of the data, than they were before. If the Bayes factor is less than 1, that means the data were less likely under A than under B, so the odds in favor of A go down. Finally, if the Bayes factor is exactly 1, the data are equally likely under either hypothesis, so the odds do not change. Let’s apply that to the problem at hand. If Oliver is one of the people who left blood at the crime scene, he accounts for the ‘O’ sample; in that case, the probability of the data is the probability that a random member of the population has type ‘AB’ blood, which is 1%. If Oliver did not leave blood at the scene, we have two samples to account for. If we choose two random people from the population, what is the chance of finding one with type ‘O’ and one with type ‘AB’? Well, there are two ways it might happen: The first person might have ‘O’ and the second ‘AB’, Or the first person might have ‘AB’ and the second ‘O’. The probability of either combination is (0.6)(0.01), which is 0.6%, so the total probability is twice that, or 1.2%. So the data are a little more likely if Oliver is not one of the people who left blood at the scene. We can use these probabilities to compute the likelihood ratio: like1 = 0.01 like2 = 2 0.6 0.01 likelihood_ratio = like1 / like2 likelihood_ratio 0.8333333333333334 Since the likelihood ratio is less than 1, the blood tests are evidence against the hypothesis that Oliver left blood at the scence. But it is weak evidence. For example, if the prior odds were 1 (that is, 50% probability), the posterior odds would be 0.83, which corresponds to a probability of 45%: post_odds = 1 like1 / like2 prob(post_odds) 0.45454545454545453 So this evidence doesn’t “move the needle” very much. This example is a little contrived, but it demonstrates the counterintuitive result that data consistent with a hypothesis are not necessarily in favor of the hypothesis. If this result still bothers you, this way of thinking might help: the data consist of a common event, type ‘O’ blood, and a rare event, type ‘AB’ blood. If Oliver accounts for the common event, that leaves the rare event unexplained. If Oliver doesn’t account for the ‘O’ blood, we have two chances to find someone in the population with ‘AB’ blood. And that factor of two makes the difference. Exercise: Suppose that based on other evidence, you prior belief in Oliver’s guilt is 90%. How much would the blood evidence in this section change your beliefs? What if you initially thought there was only a 10% chance of his guilt? Show code cell content Hide code cell content Solution post_odds = odds(0.9) like1 / like2 prob(post_odds) 0.8823529411764706 Show code cell content Hide code cell content Solution post_odds = odds(0.1) like1 / like2 prob(post_odds) 0.0847457627118644 6.4. Addends# The second half of this chapter is about distributions of sums and results of other operations. We’ll start with a forward problem, where we are given the inputs and compute the distribution of the output. Then we’ll work on inverse problems, where we are given the outputs and we compute the distribution of the inputs. As a first example, suppose you roll two dice and add them up. What is the distribution of the sum? I’ll use the following function to create a Pmf that represents the possible outcomes of a die: import numpy as np from empiricaldist import Pmf def make_die(sides): outcomes = np.arange(1, sides+1) die = Pmf(1/sides, outcomes) return die On a six-sided die, the outcomes are 1 through 6, all equally likely. die = make_die(6) Show code cell source Hide code cell source from utils import decorate die.bar(alpha=0.4) decorate(xlabel='Outcome', ylabel='PMF') If we roll two dice and add them up, there are 11 possible outcomes, 2 through 12, but they are not equally likely. To compute the distribution of the sum, we have to enumerate the possible outcomes. And that’s how this function works: def add_dist(pmf1, pmf2): """Compute the distribution of a sum.""" res = Pmf() for q1, p1 in pmf1.items(): for q2, p2 in pmf2.items(): q = q1 + q2 p = p1 p2 res[q] = res(q) + p return res The parameters are Pmf objects representing distributions. The loops iterate though the quantities and probabilities in the Pmf objects. Each time through the loop q gets the sum of a pair of quantities, and p gets the probability of the pair. Because the same sum might appear more than once, we have to add up the total probability for each sum. Notice a subtle element of this line: res[q] = res(q) + p I use parentheses on the right side of the assignment, which returns 0 if q does not appear yet in res. I use brackets on the left side of the assignment to create or update an element in res; using parentheses on the left side would not work. Pmf provides add_dist, which does the same thing. You can call it as a method, like this: twice = die.add_dist(die) Or as a function, like this: twice = Pmf.add_dist(die, die) And here’s what the result looks like: Show code cell content Hide code cell content from utils import decorate def decorate_dice(title=''): decorate(xlabel='Outcome', ylabel='PMF', title=title) Show code cell source Hide code cell source twice = add_dist(die, die) twice.bar(color='C1', alpha=0.5) decorate_dice() If we have a sequence of Pmf objects that represent dice, we can compute the distribution of the sum like this: def add_dist_seq(seq): """Compute Pmf of the sum of values from seq.""" total = seq for other in seq[1:]: total = total.add_dist(other) return total As an example, we can make a list of three dice like this: dice = [die] 3 And we can compute the distribution of their sum like this. thrice = add_dist_seq(dice) The following figure shows what these three distributions look like: The distribution of a single die is uniform from 1 to 6. The sum of two dice has a triangle distribution between 2 and 12. The sum of three dice has a bell-shaped distribution between 3 and 18. Show code cell source Hide code cell source import matplotlib.pyplot as plt die.plot(label='once') twice.plot(label='twice', ls='--') thrice.plot(label='thrice', ls=':') plt.xticks([0,3,6,9,12,15,18]) decorate_dice(title='Distributions of sums') As an aside, this example demonstrates the Central Limit Theorem, which says that the distribution of a sum converges on a bell-shaped normal distribution, at least under some conditions. 6.5. Gluten Sensitivity# In 2015 I read a paper that tested whether people diagnosed with gluten sensitivity (but not celiac disease) were able to distinguish gluten flour from non-gluten flour in a blind challenge (you can read the paper here). Out of 35 subjects, 12 correctly identified the gluten flour based on resumption of symptoms while they were eating it. Another 17 wrongly identified the gluten-free flour based on their symptoms, and 6 were unable to distinguish. The authors conclude, “Double-blind gluten challenge induces symptom recurrence in just one-third of patients.” This conclusion seems odd to me, because if none of the patients were sensitive to gluten, we would expect some of them to identify the gluten flour by chance. So here’s the question: based on this data, how many of the subjects are sensitive to gluten and how many are guessing? We can use Bayes’s Theorem to answer this question, but first we have to make some modeling decisions. I’ll assume: People who are sensitive to gluten have a 95% chance of correctly identifying gluten flour under the challenge conditions, and People who are not sensitive have a 40% chance of identifying the gluten flour by chance (and a 60% chance of either choosing the other flour or failing to distinguish). These particular values are arbitrary, but the results are not sensitive to these choices. I will solve this problem in two steps. First, assuming that we know how many subjects are sensitive, I will compute the distribution of the data. Then, using the likelihood of the data, I will compute the posterior distribution of the number of sensitive patients. The first is the forward problem; the second is the inverse problem. 6.6. The Forward Problem# Suppose we know that 10 of the 35 subjects are sensitive to gluten. That means that 25 are not: n = 35 num_sensitive = 10 num_insensitive = n - num_sensitive Each sensitive subject has a 95% chance of identifying the gluten flour, so the number of correct identifications follows a binomial distribution. I’ll use make_binomial, which we defined in <<_TheBinomialDistribution>>, to make a Pmf that represents the binomial distribution. from utils import make_binomial dist_sensitive = make_binomial(num_sensitive, 0.95) dist_insensitive = make_binomial(num_insensitive, 0.40) The results are the distributions for the number of correct identifications in each group. Now we can use add_dist to compute the distribution of the total number of correct identifications: dist_total = Pmf.add_dist(dist_sensitive, dist_insensitive) Here are the results: Show code cell source Hide code cell source dist_sensitive.plot(label='sensitive', ls=':') dist_insensitive.plot(label='insensitive', ls='--') dist_total.plot(label='total') decorate(xlabel='Number of correct identifications', ylabel='PMF', title='Gluten sensitivity') We expect most of the sensitive subjects to identify the gluten flour correctly. Of the 25 insensitive subjects, we expect about 10 to identify the gluten flour by chance. So we expect about 20 correct identifications in total. This is the answer to the forward problem: given the number of sensitive subjects, we can compute the distribution of the data. 6.7. The Inverse Problem# Now let’s solve the inverse problem: given the data, we’ll compute the posterior distribution of the number of sensitive subjects. Here’s how. I’ll loop through the possible values of num_sensitive and compute the distribution of the data for each: import pandas as pd table = pd.DataFrame() for num_sensitive in range(0, n+1): num_insensitive = n - num_sensitive dist_sensitive = make_binomial(num_sensitive, 0.95) dist_insensitive = make_binomial(num_insensitive, 0.4) dist_total = Pmf.add_dist(dist_sensitive, dist_insensitive) table[num_sensitive] = dist_total The loop enumerates the possible values of num_sensitive. For each value, it computes the distribution of the total number of correct identifications, and stores the result as a column in a Pandas DataFrame. Show code cell content Hide code cell content table.head(3) \| \| 0 \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| ... \| 26 \| 27 \| 28 \| 29 \| 30 \| 31 \| 32 \| 33 \| 34 \| 35 \| --- --- --- --- --- --- --- --- --- --- --- \| \| 0 \| 1.719071e-08 \| 1.432559e-09 \| 1.193799e-10 \| 9.948326e-12 \| 8.290272e-13 \| 6.908560e-14 \| 5.757133e-15 \| 4.797611e-16 \| 3.998009e-17 \| 3.331674e-18 \| ... \| 1.501694e-36 \| 1.251411e-37 \| 1.042843e-38 \| 8.690357e-40 \| 7.241964e-41 \| 6.034970e-42 \| 5.029142e-43 \| 4.190952e-44 \| 3.492460e-45 \| 2.910383e-46 \| \| 1 \| 4.011165e-07 \| 5.968996e-08 \| 7.162795e-09 \| 7.792856e-10 \| 8.013930e-11 \| 7.944844e-12 \| 7.676178e-13 \| 7.276377e-14 \| 6.796616e-15 \| 6.274653e-16 \| ... \| 7.508469e-34 \| 6.486483e-35 \| 5.596590e-36 \| 4.823148e-37 \| 4.152060e-38 \| 3.570691e-39 \| 3.067777e-40 \| 2.633315e-41 \| 2.258457e-42 \| 1.935405e-43 \| \| 2 \| 4.545987e-06 \| 9.741401e-07 \| 1.709122e-07 \| 2.506426e-08 \| 3.269131e-09 \| 3.940182e-10 \| 4.490244e-11 \| 4.908756e-12 \| 5.197412e-13 \| 5.365476e-14 \| ... \| 1.806613e-31 \| 1.620070e-32 \| 1.449030e-33 \| 1.292922e-34 \| 1.151034e-35 \| 1.022555e-36 \| 9.066202e-38 \| 8.023344e-39 \| 7.088005e-40 \| 6.251357e-41 \| 3 rows × 36 columns The following figure shows selected columns from the DataFrame, corresponding to different hypothetical values of num_sensitive: Show code cell source Hide code cell source table.plot(label='num_sensitive = 0') table.plot(label='num_sensitive = 10') table.plot(label='num_sensitive = 20', ls='--') table.plot(label='num_sensitive = 30', ls=':') decorate(xlabel='Number of correct identifications', ylabel='PMF', title='Gluten sensitivity') Now we can use this table to compute the likelihood of the data: likelihood1 = table.loc loc selects a row from the DataFrame. The row with index 12 contains the probability of 12 correct identifications for each hypothetical value of num_sensitive. And that’s exactly the likelihood we need to do a Bayesian update. I’ll use a uniform prior, which implies that I would be equally surprised by any value of num_sensitive: hypos = np.arange(n+1) prior = Pmf(1, hypos) And here’s the update: posterior1 = prior likelihood1 posterior1.normalize() Show code cell output Hide code cell output 0.4754741648615131 For comparison, I also compute the posterior for another possible outcome, 20 correct identifications. likelihood2 = table.loc posterior2 = prior likelihood2 posterior2.normalize() Show code cell output Hide code cell output 1.7818649765887378 The following figure shows posterior distributions of num_sensitive based on the actual data, 12 correct identifications, and the other possible outcome, 20 correct identifications. Show code cell source Hide code cell source posterior1.plot(label='posterior with 12 correct', color='C4') posterior2.plot(label='posterior with 20 correct', color='C1') decorate(xlabel='Number of sensitive subjects', ylabel='PMF', title='Posterior distributions') With 12 correct identifications, the most likely conclusion is that none of the subjects are sensitive to gluten. If there had been 20 correct identifications, the most likely conclusion would be that 11-12 of the subjects were sensitive. posterior1.max_prob() 0 posterior2.max_prob() 11 6.8. Summary# This chapter presents two topics that are almost unrelated except that they make the title of the chapter catchy. The first part of the chapter is about Bayes’s Rule, evidence, and how we can quantify the strength of evidence using a likelihood ratio or Bayes factor. The second part is about add_dist, which computes the distribution of a sum. We can use this function to solve forward and inverse problems; that is, given the parameters of a system, we can compute the distribution of the data or, given the data, we can compute the distribution of the parameters. In the next chapter, we’ll compute distributions for minimums and maximums, and use them to solve more Bayesian problems. But first you might want to work on these exercises. 6.9. Exercises# Exercise: Let’s use Bayes’s Rule to solve the Elvis problem from <<_Distributions>>: Elvis Presley had a twin brother who died at birth. What is the probability that Elvis was an identical twin? In 1935, about 2/3 of twins were fraternal and 1/3 were identical. The question contains two pieces of information we can use to update this prior. First, Elvis’s twin was also male, which is more likely if they were identical twins, with a likelihood ratio of 2. Also, Elvis’s twin died at birth, which is more likely if they were identical twins, with a likelihood ratio of 1.25. If you are curious about where those numbers come from, I wrote a blog post about it. Show code cell content Hide code cell content Solution prior_odds = odds(1/3) Show code cell content Hide code cell content Solution post_odds = prior_odds 2 1.25 Show code cell content Hide code cell content Solution prob(post_odds) 0.5555555555555555 Exercise: The following is an interview question that appeared on glassdoor.com, attributed to Facebook: You’re about to get on a plane to Seattle. You want to know if you should bring an umbrella. You call 3 random friends of yours who live there and ask each independently if it’s raining. Each of your friends has a 2/3 chance of telling you the truth and a 1/3 chance of messing with you by lying. All 3 friends tell you that “Yes” it is raining. What is the probability that it’s actually raining in Seattle? Use Bayes’s Rule to solve this problem. As a prior you can assume that it rains in Seattle about 10% of the time. This question causes some confusion about the differences between Bayesian and frequentist interpretations of probability; if you are curious about this point, I wrote a blog article about it. Show code cell content Hide code cell content Solution prior_odds = odds(0.1) Show code cell content Hide code cell content Solution post_odds = prior_odds 2 2 2 Show code cell content Hide code cell content Solution prob(post_odds) 0.4705882352941177 Exercise:According to the CDC, people who smoke are about 25 times more likely to develop lung cancer than nonsmokers. Also according to the CDC, about 14% of adults in the U.S. are smokers. If you learn that someone has lung cancer, what is the probability they are a smoker? Show code cell content Hide code cell content Solution prior_odds = odds(0.14) Show code cell content Hide code cell content Solution post_odds = prior_odds 25 Show code cell content Hide code cell content Solution prob(post_odds) 0.8027522935779816 Exercise: In Dungeons & Dragons, the amount of damage a goblin can withstand is the sum of two six-sided dice. The amount of damage you inflict with a short sword is determined by rolling one six-sided die. A goblin is defeated if the total damage you inflict is greater than or equal to the amount it can withstand. Suppose you are fighting a goblin and you have already inflicted 3 points of damage. What is your probability of defeating the goblin with your next successful attack? Hint: You can use Pmf.sub_dist to subtract a constant amount, like 3, from a Pmf. Show code cell content Hide code cell content Solution d6 = make_die(6) Show code cell content Hide code cell content Solution The amount the goblin started with is the sum of two d6 hp_before = Pmf.add_dist(d6, d6) Show code cell content Hide code cell content Solution Here's the number of hit points after the first attack hp_after = Pmf.sub_dist(hp_before, 3) hp_after \| \| probs \| --- \| \| -1 \| 0.027778 \| \| 0 \| 0.055556 \| \| 1 \| 0.083333 \| \| 2 \| 0.111111 \| \| 3 \| 0.138889 \| \| 4 \| 0.166667 \| \| 5 \| 0.138889 \| \| 6 \| 0.111111 \| \| 7 \| 0.083333 \| \| 8 \| 0.055556 \| \| 9 \| 0.027778 \| Show code cell content Hide code cell content Solution But -1 and 0 are not possible, because in that case the goblin would be defeated. So we have to zero them out and renormalize hp_after = 0 hp_after.normalize() hp_after \| \| probs \| --- \| \| -1 \| 0.000000 \| \| 0 \| 0.000000 \| \| 1 \| 0.090909 \| \| 2 \| 0.121212 \| \| 3 \| 0.151515 \| \| 4 \| 0.181818 \| \| 5 \| 0.151515 \| \| 6 \| 0.121212 \| \| 7 \| 0.090909 \| \| 8 \| 0.060606 \| \| 9 \| 0.030303 \| Show code cell content Hide code cell content Solution The damage from the second attack is one d6 damage = d6 Show code cell content Hide code cell content Solution Here's what the distributions look like hp_after.bar(label='Hit points') damage.plot(label='Damage', color='C1') decorate_dice('The Goblin Problem') Show code cell content Hide code cell content Solution Here's the distribution of points the goblin has left points_left = Pmf.sub_dist(hp_after, damage) Show code cell content Hide code cell content Solution And here's the probability the goblin is dead points_left.prob_le(0) 0.4545454545454545 Exercise: Suppose I have a box with a 6-sided die, an 8-sided die, and a 12-sided die. I choose one of the dice at random, roll it twice, multiply the outcomes, and report that the product is 12. What is the probability that I chose the 8-sided die? Hint: Pmf provides a function called mul_dist that takes two Pmf objects and returns a Pmf that represents the distribution of the product. Show code cell content Hide code cell content Solution hypos = [6, 8, 12] prior = Pmf(1, hypos) Show code cell content Hide code cell content Solution Here's the distribution of the product for the 4-sided die d4 = make_die(4) Pmf.mul_dist(d4, d4) \| \| probs \| --- \| \| 1 \| 0.0625 \| \| 2 \| 0.1250 \| \| 3 \| 0.1250 \| \| 4 \| 0.1875 \| \| 6 \| 0.1250 \| \| 8 \| 0.1250 \| \| 9 \| 0.0625 \| \| 12 \| 0.1250 \| \| 16 \| 0.0625 \| Show code cell content Hide code cell content Solution Here's the likelihood of getting a 12 for each die likelihood = [] for sides in hypos: die = make_die(sides) pmf = Pmf.mul_dist(die, die) likelihood.append(pmf) likelihood [0.1111111111111111, 0.0625, 0.041666666666666664] Show code cell content Hide code cell content Solution And here's the update posterior = prior likelihood posterior.normalize() posterior \| \| probs \| --- \| \| 6 \| 0.516129 \| \| 8 \| 0.290323 \| \| 12 \| 0.193548 \| Exercise:Betrayal at House on the Hill is a strategy game in which characters with different attributes explore a haunted house. Depending on their attributes, the characters roll different numbers of dice. For example, if attempting a task that depends on knowledge, Professor Longfellow rolls 5 dice, Madame Zostra rolls 4, and Ox Bellows rolls 3. Each die yields 0, 1, or 2 with equal probability. If a randomly chosen character attempts a task three times and rolls a total of 3 on the first attempt, 4 on the second, and 5 on the third, which character do you think it was? Show code cell content Hide code cell content Solution die = Pmf(1/3, [0,1,2]) die \| \| probs \| --- \| \| 0 \| 0.333333 \| \| 1 \| 0.333333 \| \| 2 \| 0.333333 \| Show code cell content Hide code cell content Solution pmfs = {} pmfs['Bellows'] = add_dist_seq([die]3) pmfs['Zostra'] = add_dist_seq([die]4) pmfs['Longfellow'] = add_dist_seq([die]5) Show code cell content Hide code cell content Solution pmfs'Zostra' 0.2345679012345679 Show code cell content Hide code cell content Solution pmfs'Zostra'.prod() 0.00915247412224499 Show code cell content Hide code cell content Solution hypos = pmfs.keys() prior = Pmf(1/3, hypos) prior \| \| probs \| --- \| \| Bellows \| 0.333333 \| \| Zostra \| 0.333333 \| \| Longfellow \| 0.333333 \| Show code cell content Hide code cell content Solution likelihood = prior.copy() for hypo in hypos: likelihood[hypo] = pmfshypo.prod() likelihood \| \| probs \| --- \| \| Bellows \| 0.006401 \| \| Zostra \| 0.009152 \| \| Longfellow \| 0.004798 \| Show code cell content Hide code cell content Solution posterior = (prior likelihood) posterior.normalize() posterior \| \| probs \| --- \| \| Bellows \| 0.314534 \| \| Zostra \| 0.449704 \| \| Longfellow \| 0.235762 \| Exercise: There are 538 members of the United States Congress. Suppose we audit their investment portfolios and find that 312 of them out-perform the market. Let’s assume that an honest member of Congress has only a 50% chance of out-performing the market, but a dishonest member who trades on inside information has a 90% chance. How many members of Congress are honest? Show code cell content Hide code cell content Solution n = 538 ns = range(0, n+1) table = pd.DataFrame(index=ns, columns=ns, dtype=float) for n_honest in ns: n_dishonest = n - n_honest dist_honest = make_binomial(n_honest, 0.5) dist_dishonest = make_binomial(n_dishonest, 0.9) dist_total = Pmf.add_dist(dist_honest, dist_dishonest) table[n_honest] = dist_total table.shape (539, 539) Show code cell content Hide code cell content Solution data = 312 likelihood = table.loc len(likelihood) 539 Show code cell content Hide code cell content Solution hypos = np.arange(n+1) prior = Pmf(1, hypos) len(prior) 539 Show code cell content Hide code cell content Solution posterior = prior likelihood posterior.normalize() posterior.mean() 431.4882114501996 Show code cell content Hide code cell content Solution posterior.plot(label='posterior') decorate(xlabel='Number of honest members of Congress', ylabel='PMF') Show code cell content Hide code cell content Solution posterior.max_prob() 430 Show code cell content Hide code cell content Solution posterior.credible_interval(0.9) array([388., 477.]) Think Bayes, Second Edition Copyright 2020 Allen B. Downey License: Attribution-NonCommercial-ShareAlike 4.0 International (CC BY-NC-SA 4.0) previous 5. Estimating Countsnext 7. Minimum, Maximum, and Mixture Contents 6.1. Odds 6.2. Bayes’s Rule 6.3. Oliver’s Blood 6.4. Addends 6.5. Gluten Sensitivity 6.6. The Forward Problem 6.7. The Inverse Problem 6.8. Summary 6.9. Exercises By Allen B. Downey © Copyright 2023.
15296	https://bio.libretexts.org/Courses/Norco_College/OpenStax_Biology_2e_for_Norco_College/10%3A_Cell_Reproduction	10: Cell Reproduction - Biology LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode OpenStax Biology 2e for Norco College Norco College 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MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Principles_of_Ecology : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Supplement_to_OpenStax_Biology_2e_for_Norco_College : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Tue, 15 Aug 2023 19:29:14 GMT 10: Cell Reproduction 103602 103602 Teresa Friedrich Finnern { } Anonymous Anonymous 2 false false [ "article:topic-guide", "authorname:openstax", "Cell Reproduction", "showtoc:no", "license:ccby", "printoptions:no-header-title", "cssprint:dense", "source-bio-1799", "licenseversion:40", "program:openstax", "source@ "source-bio-97033" ] [ "article:topic-guide", "authorname:openstax", "Cell Reproduction", "showtoc:no", "license:ccby", "printoptions:no-header-title", "cssprint:dense", "source-bio-1799", "licenseversion:40", "program:openstax", "source@ "source-bio-97033" ] Search site Search Search Go 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OpenStax Biology 2e for Norco College 5. 10: Cell Reproduction Expand/collapse global location OpenStax Biology 2e for Norco College Front Matter 1: The Study of Life 2: The Chemical Foundation of Life 3: Biological Macromolecules 4: Cell Structure 5: Structure and Function of Plasma Membranes 6: Metabolism 7: Plant Form and Function 8: Photosynthesis 9: Cellular Respiration 10: Cell Reproduction 11: Meiosis and Sexual Reproduction 12: Mendel's Experiments and Heredity 13: Modern Understandings of Inheritance 14: DNA Structure and Function 15: Genes and Proteins 16: Gene Expression 17: Biotechnology and Genomics 18: Evolution and the Origin of Species 19: The Evolution of Populations 20: Phylogenies and the History of Life 21: Viruses 22: Prokaryotes - Bacteria and Archaea 23: Protists 24: Fungi 25: Seedless Plants 26: Seed Plants 27: Introduction to Animal Diversity 28: Invertebrates 29: Vertebrates 30: Ecology and the Biosphere 31: Population and Community Ecology 32: Ecosystems 33: Conservation Biology and Biodiversity 34: Appendix 10: Cell Reproduction Last updated Aug 15, 2023 Save as PDF 9.9: Chapter Summary 10.1: Introduction Page ID 103602 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Contributors 10.1: IntroductionA human, as well as every sexually reproducing organism, begins life as a fertilized egg (embryo) or zygote. Trillions of cell divisions subsequently occur in a controlled manner to produce a complex, multicellular human. In other words, that original single cell is the ancestor of every other cell in the body. Once a being is fully grown, cell reproduction is still necessary to repair or regenerate tissues. 10.2: Cell DivisionThe continuity of life from one cell to another has its foundation in the reproduction of cells by way of the cell cycle. The cell cycle is an orderly sequence of events that describes the stages of a cell’s life from the division of a single parent cell to the production of two new daughter cells. The mechanisms involved in the cell cycle are highly regulated. 10.3: The Cell CycleThe cell cycle is an ordered series of events involving cell growth and cell division that produces two new daughter cells. Cells on the path to cell division proceed through a series of precisely timed and carefully regulated stages of growth, DNA replication, and division that produces two identical (clone) cells. The cell cycle has two major phases: interphase and the mitotic phase. 10.4: Control of the Cell CycleThe length of the cell cycle is highly variable, even within the cells of a single organism. In humans, the frequency of cell turnover ranges from a few hours in early embryonic development, to an average of two to five days for epithelial cells, and to an entire human lifetime spent in G0 by specialized cells, such as cortical neurons or cardiac muscle cells. There is also variation in the time that a cell spends in each phase of the cell cycle. 10.5: Cancer and the Cell CycleCancer is the result of unchecked cell division caused by a breakdown of the mechanisms that regulate the cell cycle. The loss of control begins with a change in the DNA sequence of a gene that codes for one of the regulatory molecules. Faulty instructions lead to a protein that does not function as it should. Any disruption of the monitoring system can allow other mistakes to be passed on to the daughter cells. Each successive cell division will give rise to daughter cells with even more damage 10.6: Prokaryotic Cell DivisionIn both prokaryotic and eukaryotic cell division, the genomic DNA is replicated and then each copy is allocated into a daughter cell. In addition, the cytoplasmic contents are divided evenly and distributed to the new cells. However, there are many differences between prokaryotic and eukaryotic cell division. Bacteria have a single, circular DNA chromosome but no nucleus. Therefore, mitosis is not necessary in bacterial cell division. 10.7: Chapter Summary Thumbnail: Life cycle of the cell. (CC BY-SA 4.0; BruceBlaus). Contributors Connie Rye (East Mississippi Community College),Robert Wise (University of Wisconsin, Oshkosh),Vladimir Jurukovski (Suffolk County Community College),Jean DeSaix (University of North Carolina at Chapel Hill),Jung Choi (Georgia Institute of Technology),Yael Avissar (Rhode Island College) among other contributing authors. Original content by OpenStax(CC BY 4.0;Download for free at This page titled 10: Cell Reproduction is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. Back to top 9.9: Chapter Summary 10.1: Introduction Was this article helpful? 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15297	https://www.youtube.com/watch?v=sBvRJUwXJPo	Multiply and simplify a radical expression 2 \| Algebra I \| Khan Academy Khan Academy 9090000 subscribers 234 likes Description 295333 views Posted: 24 Jun 2011 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Multiply and Simply a Radical Expression Watch the next lesson: Missed the previous lesson? Algebra I on Khan Academy: Algebra is the language through which we describe patterns. Think of it as a shorthand, of sorts. As opposed to having to do something over and over again, algebra gives you a simple way to express that repetitive process. It's also seen as a "gatekeeper" subject. Once you achieve an understanding of algebra, the higher-level math subjects become accessible to you. Without it, it's impossible to move forward. It's used by people with lots of different jobs, like carpentry, engineering, and fashion design. In these tutorials, we'll cover a lot of ground. Some of the topics include linear equations, linear inequalities, linear functions, systems of equations, factoring expressions, quadratic expressions, exponents, functions, and ratios. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy’s Algebra channel: Subscribe to Khan Academy: 17 comments Transcript: We're asked to divide and simplify. And we have one radical expression over another radical expression. The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. And it really just comes out of the exponent properties. If I have two things that I take to some power-- and taking the principal root is the same thing as taking it to the 1/2 power-- if I'm raising each of them to some power and then dividing, that's the same thing as dividing first and then raising them to that power. So let's apply that over here. This expression over here is going to be the same thing as the principal root-- it's hard to write a radical sign that big-- the principal root of 60x squared y over 48x. And then we can first look at the coefficients of each of these expressions and try to simplify that. Both the numerator and the denominator is divisible by 12. 60 divided by 12 is 5. 48 divided by 12 is 4. Both the numerator and the denominator are divisible by x. x squared divided by x is just x. x divided by x is 1. Anything we divide the numerator by, we have to divide the denominator by. And that's all we have left. So if we wanted to simplify this, this is equal to the-- make a radical sign-- and then we have 5/4. And actually, we can write it in a slightly different way, but I'll write it this way-- 5/4. And we have nothing left in the denominator other than that 4. And in the numerator, we have an x and we have a y. And now we could leave it just like that, but we might want to take more things out of the radical sign. And so one possibility that you can do is you could say that this is really the same thing as-- this is equal to 1/4 times 5xy, all of that under the radical sign. And this is the same thing as the square root of or the principal root of 1/4 times the principal root of 5xy. And the square root of 1/4, if you think about it, that's just 1/2 times 1/2. Or another way you could think about it is that this right here is the same thing as-- so you could just say, hey, this is 1/2. 1/2 times 1/2 is 1/4. Or if you don't realize it's 1/2, you say, hey, this is the same thing as the square root of 1 over the square root of 4, and the square root of 1 is 1 and the principal root of 4 is 2, so you get 1/2 once again. And so if you simplify this right here to 1/2, then the whole thing can simplify to 1/2 times the principal root-- I'll just write it all in orange-- times the principal root of 5xy. And there's nothing else that you can really take out of the radical sign here. Nothing else here is a perfect square.
15298	https://en.wikipedia.org/wiki/Debye_model	Jump to content Search Contents 1 Derivation 2 Debye's derivation 3 Another derivation 4 Temperature limits 5 Debye versus Einstein 6 Debye temperature table 7 Extension to other quasi-particles 8 Extension to liquids 9 Debye frequency 9.1 Definition 9.2 Relation to Debye's temperature 9.3 Debye's derivation 9.3.1 Three-dimensional crystal 9.3.2 One-dimensional chain in 3D space 9.3.3 Two-dimensional crystal 9.4 Polarization dependence 9.4.1 One-dimensional chain in 3D space 9.4.2 Two-dimensional crystal 9.4.3 Three-dimensional crystal 9.5 Derivation with the actual dispersion relation 9.6 Alternative derivation 10 See also 11 References 12 Further reading 13 External links Debye model العربية Català Deutsch Español Esperanto Français 한국어 Italiano עברית Nederlands 日本語 Oʻzbekcha / ўзбекча Polski Русский Slovenčina Svenska Türkçe Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikibooks Wikidata item Appearance From Wikipedia, the free encyclopedia Method in physics \| \| \| \| --- \| This article has multiple issues. Please help improve it or discuss these issues on the talk page. (Learn how and when to remove these messages) \| \| \| This article is written like a textbook. Please help rewrite it in a neutral, encyclopedic style. Relevant discussion may be found on the talk page. (January 2024) (Learn how and when to remove this message) \| \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Debye model" – news · newspapers · books · scholar · JSTOR (January 2024) (Learn how and when to remove this message) \| (Learn how and when to remove this message) \| \| Statistical mechanics \| \| Thermodynamics Kinetic theory \| \| Particle statistics Spin–statistics theorem Indistinguishable particles Maxwell–Boltzmann Bose–Einstein Fermi–Dirac Parastatistics Anyonic statistics Braid statistics \| \| Thermodynamic ensembles NVE Microcanonical NVT Canonical µVT Grand canonical NPH Isoenthalpic–isobaric NPT Isothermal–isobaric \| \| Models Debye Einstein Ising Potts \| \| Potentials Internal energy Enthalpy Helmholtz free energy Gibbs free energy Grand potential / Landau free energy \| \| Scientists Maxwell Boltzmann Helmholtz Bose Gibbs Einstein Dirac Ehrenfest von Neumann Tolman Debye Fermi Synge Ising Landau \| \| v t e \| In thermodynamics and solid-state physics, the Debye model is a method developed by Peter Debye in 1912 to estimate phonon contribution to the specific heat (heat capacity) in a solid. It treats the vibrations of the atomic lattice (heat) as phonons in a box in contrast to the Einstein photoelectron model, which treats the solid as many individual, non-interacting quantum harmonic oscillators. The Debye model correctly predicts the low-temperature dependence of the heat capacity of solids, which is proportional to the cube of temperature – the Debye T 3 law. Similarly to the Einstein photoelectron model, it recovers the Dulong–Petit law at high temperatures. Due to simplifying assumptions, its accuracy suffers at intermediate temperatures.[clarification needed] Derivation [edit] The Debye model treats atomic vibrations as phonons confined in the solid's volume. It is analogous to Planck's law of black body radiation, which treats electromagnetic radiation as a photon gas confined in a vacuum space. Most of the calculation steps are identical, as both are examples of a massless Bose gas with a linear dispersion relation. For a cube of side-length , the resonating modes of the sonic disturbances (considering for now only those aligned with one axis), treated as particles in a box, have wavelengths given as where is an integer. The energy of a phonon is given as where is the Planck constant and is the frequency of the phonon. Making the approximation that the frequency is inversely proportional to the wavelength, in which is the speed of sound inside the solid. In three dimensions, energy can be generalized to in which is the magnitude of the three-dimensional momentum of the phonon, and , , and are the components of the resonating mode along each of the three axes. The approximation that the frequency is inversely proportional to the wavelength (giving a constant speed of sound) is good for low-energy phonons but not for high-energy phonons, which is a limitation of the Debye model. This approximation leads to incorrect results at intermediate temperatures, whereas the results are exact at the low and high temperature limits. The total energy in the box, , is given by where is the number of phonons in the box with energy ; the total energy is equal to the sum of energies over all energy levels, and the energy at a given level is found by multiplying its energy by the number of phonons with that energy. In three dimensions, each combination of modes in each of the three axes corresponds to an energy level, giving the total energy as: The Debye model and Planck's law of black body radiation differ here with respect to this sum. Unlike electromagnetic photon radiation in a box, there are a finite number of phonon energy states because a phonon cannot have an arbitrarily high frequency. Its frequency is bounded by its propagation medium—the atomic lattice of the solid. The following illustration describes transverse phonons in a cubic solid at varying frequencies: It is reasonable to assume that the minimum wavelength of a phonon is twice the atomic separation, as shown in the lowest example. With atoms in a cubic solid, each axis of the cube measures as being atoms long. Atomic separation is then given by , and the minimum wavelength is making the maximum mode number : This contrasts with photons, for which the maximum mode number is infinite. This number bounds the upper limit of the triple energy sum If is a function that is slowly varying with respect to , the sums can be approximated with integrals: To evaluate this integral, the function , the number of phonons with energy must also be known. Phonons obey Bose–Einstein statistics, and their distribution is given by the Bose–Einstein statistics formula: Because a phonon has three possible polarization states (one longitudinal, and two transverse, which approximately do not affect its energy) the formula above must be multiplied by 3, Considering all three polarization states together also means that an effective sonic velocity must be determined and used as the value of the standard sonic velocity The Debye temperature defined below is proportional to ; more precisely, , where longitudinal and transversal sound-wave velocities are averaged, weighted by the number of polarization states. The Debye temperature or the effective sonic velocity is a measure of the hardness of the crystal. Substituting into the energy integral yields These integrals are evaluated for photons easily because their frequency, at least semi-classically, is unbound. The same is not true for phonons, so in order to approximate this triple integral, Peter Debye used spherical coordinates, and approximated the cube with an eighth of a sphere, where is the radius of this sphere. As the energy function does not depend on either of the angles, the equation can be simplified to The number of particles in the original cube and in the eighth of a sphere should be equivalent. The volume of the cube is unit cell volumes, such that the radius must be The substitution of integration over a sphere for the correct integral over a cube introduces another source of inaccuracy into the resulting model. After making the spherical substitution and substituting in the function , the energy integral becomes : . Changing the integration variable to , To simplify the appearance of this expression, define the Debye temperature where is the volume of the cubic box of side-length . Some authors describe the Debye temperature as shorthand for some constants and material-dependent variables. However, is roughly equal to the phonon energy of the minimum wavelength mode, and so we can interpret the Debye temperature as the temperature at which the highest-frequency mode is excited. Additionally, since all other modes are of a lower energy than the highest-frequency mode, all modes are excited at this temperature. From the total energy, the specific internal energy can be calculated: where is the third Debye function. Differentiating this function with respect to produces the dimensionless heat capacity: These formulae treat the Debye model at all temperatures. The more elementary formulae given further down give the asymptotic behavior in the limit of low and high temperatures. The essential reason for the exactness at low and high energies is, respectively, that the Debye model gives the exact dispersion relation at low frequencies, and corresponds to the exact density of states at high temperatures, concerning the number of vibrations per frequency interval.[original research?] Debye's derivation [edit] Debye derived his equation differently and more simply. Using continuum mechanics, he found that the number of vibrational states with a frequency less than a particular value was asymptotic to in which is the volume and is a factor that he calculated from elasticity coefficients and density. Combining this formula with the expected energy of a harmonic oscillator at temperature (already used by Einstein in his model) would give an energy of if the vibrational frequencies continued to infinity. This form gives the behaviour which is correct at low temperatures. But Debye realized that there could not be more than vibrational states for N atoms. He made the assumption that in an atomic solid, the spectrum of frequencies of the vibrational states would continue to follow the above rule, up to a maximum frequency chosen so that the total number of states is Debye knew that this assumption was not really correct (the higher frequencies are more closely spaced than assumed), but it guarantees the proper behaviour at high temperature (the Dulong–Petit law). The energy is then given by Substituting for , where is the function later given the name of third-order Debye function. Another derivation [edit] First the vibrational frequency distribution is derived from Appendix VI of Terrell L. Hill's An Introduction to Statistical Mechanics. Consider a three-dimensional isotropic elastic solid with N atoms in the shape of a rectangular parallelepiped with side-lengths . The elastic wave will obey the wave equation and will be plane waves; consider the wave vector and define , such that \| \| \| --- \| \| \| 1 \| Solutions to the wave equation are and with the boundary conditions at , \| \| \| --- \| \| \| 2 \| where are positive integers. Substituting (2) into (1) and also using the dispersion relation , The above equation, for fixed frequency , describes an eighth of an ellipse in "mode space" (an eighth because are positive). The number of modes with frequency less than is thus the number of integral points inside the ellipse, which, in the limit of (i.e. for a very large parallelepiped) can be approximated to the volume of the ellipse. Hence, the number of modes with frequency in the range is \| \| \| --- \| \| \| 3 \| where is the volume of the parallelepiped. The wave speed in the longitudinal direction is different from the transverse direction and that the waves can be polarised one way in the longitudinal direction and two ways in the transverse direction and ca be defined as . Following the derivation from A First Course in Thermodynamics, an upper limit to the frequency of vibration is defined ; since there are atoms in the solid, there are quantum harmonic oscillators (3 for each x-, y-, z- direction) oscillating over the range of frequencies . can be determined using \| \| \| --- \| \| . \| 4 \| By defining , where k is the Boltzmann constant and h is the Planck constant, and substituting (4) into (3), \| \| \| --- \| \| \| 5 \| this definition is more standard; the energy contribution for all oscillators oscillating at frequency can be found. Quantum harmonic oscillators can have energies where and using Maxwell-Boltzmann statistics, the number of particles with energy is The energy contribution for oscillators with frequency is then \| \| \| --- \| \| . \| 6 \| By noting that (because there are modes oscillating with frequency ), From above, we can get an expression for 1/A; substituting it into (6), Integrating with respect to ν yields Temperature limits [edit] The temperature of a Debye solid is said to be low if , leading to This definite integral can be evaluated exactly: In the low-temperature limit, the limitations of the Debye model mentioned above do not apply, and it gives a correct relationship between (phononic) heat capacity, temperature, the elastic coefficients, and the volume per atom (the latter quantities being contained in the Debye temperature). The temperature of a Debye solid is said to be high if . Using if leads to which upon integration gives This is the Dulong–Petit law, and is fairly accurate although it does not take into account anharmonicity, which causes the heat capacity to rise further. The total heat capacity of the solid, if it is a conductor or semiconductor, may also contain a non-negligible contribution from the electrons. Debye versus Einstein [edit] The Debye and Einstein models correspond closely to experimental data, but the Debye model is correct at low temperatures whereas the Einstein model is not. To visualize the difference between the models, one would naturally plot the two on the same set of axes, but this is not immediately possible as both the Einstein model and the Debye model provide a functional form for the heat capacity. As models, they require scales to relate them to their real-world counterparts. One can see that the scale of the Einstein model is given by : The scale of the Debye model is , the Debye temperature. Both are usually found by fitting the models to the experimental data. (The Debye temperature can theoretically be calculated from the speed of sound and crystal dimensions.) Because the two methods approach the problem from different directions and different geometries, Einstein and Debye scales are not the same, that is to say which means that plotting them on the same set of axes makes no sense. They are two models of the same thing, but of different scales. If one defines the Einstein condensation temperature as then one can say and, to relate the two, the ratio is used. The Einstein solid is composed of single-frequency quantum harmonic oscillators, . That frequency, if it indeed existed, would be related to the speed of sound in the solid. If one imagines the propagation of sound as a sequence of atoms hitting one another, then the frequency of oscillation must correspond to the minimum wavelength sustainable by the atomic lattice, , where : , which makes the Einstein temperature and the sought ratio is therefore Using the ratio, both models can be plotted on the same graph. It is the cube root of the ratio of the volume of one octant of a three-dimensional sphere to the volume of the cube that contains it, which is just the correction factor used by Debye when approximating the energy integral above. Alternatively, the ratio of the two temperatures can be seen to be the ratio of Einstein's single frequency at which all oscillators oscillate and Debye's maximum frequency. Einstein's single frequency can then be seen to be a mean of the frequencies available to the Debye model. Debye temperature table [edit] Even though the Debye model is not completely correct, it gives a good approximation for the low temperature heat capacity of insulating, crystalline solids where other contributions (such as highly mobile conduction electrons) are negligible. For metals, the electron contribution to the heat is proportional to , which at low temperatures dominates the Debye result for lattice vibrations. In this case, the Debye model can only be said to approximate the lattice contribution to the specific heat. The following table lists Debye temperatures for several pure elements and sapphire: \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| --- \| \| Aluminium \| 0428 K \| \| Beryllium \| 1440 K \| \| Cadmium \| 0209 K \| \| Caesium \| 0038 K \| \| Carbon (diamond) \| 2230 K \| \| Chromium \| 0630 K \| \| \| \| \| --- \| \| Copper \| 0343 K \| \| Germanium \| 0374 K \| \| Gold \| 0170 K \| \| Iron \| 0470 K \| \| Lead \| 0105 K \| \| Manganese \| 0410 K \| \| \| \| \| --- \| \| Nickel \| 0450 K \| \| Platinum \| 0240 K \| \| Rubidium \| 0056 K \| \| Sapphire \| 1047 K \| \| Selenium \| 0090 K \| \| Silicon \| 0645 K \| \| \| \| \| --- \| \| Silver \| 0215 K \| \| Tantalum \| 0240 K \| \| Tin (white) \| 0200 K \| \| Titanium \| 0420 K \| \| Tungsten \| 0400 K \| \| Zinc \| 0327 K \| \| \| The Debye model's fit to experimental data is often phenomenologically improved by allowing the Debye temperature to become temperature dependent; for example, the value for ice increases from about 222 K to 300 K as the temperature goes from absolute zero to about 100 K. Extension to other quasi-particles [edit] For other bosonic quasi-particles, e.g., magnons (quantized spin waves) in ferromagnets instead of the phonons (quantized sound waves), one can derive analogous results. In this case at low frequencies one has different dispersion relations of momentum and energy, e.g., in the case of magnons, instead of for phonons (with ). One also has different density of states (e.g., ). As a consequence, in ferromagnets one gets a magnon contribution to the heat capacity, , which dominates at sufficiently low temperatures the phonon contribution, . In metals, in contrast, the main low-temperature contribution to the heat capacity, , comes from the electrons. It is fermionic, and is calculated by different methods going back to Sommerfeld's free electron model.[citation needed] Extension to liquids [edit] It was long thought that phonon theory is not able to explain the heat capacity of liquids, since liquids only sustain longitudinal, but not transverse phonons, which in solids are responsible for 2/3 of the heat capacity. However, Brillouin scattering experiments with neutrons and with X-rays, confirming an intuition of Yakov Frenkel, have shown that transverse phonons do exist in liquids, albeit restricted to frequencies above a threshold called the Frenkel frequency. Since most energy is contained in these high-frequency modes, a simple modification of the Debye model is sufficient to yield a good approximation to experimental heat capacities of simple liquids. More recently, it has been shown that instantaneous normal modes associated with relaxations from saddle points in the liquid energy landscape, which dominate the frequency spectrum of liquids at low frequencies, may determine the specific heat of liquids as a function of temperature over a broad range. Debye frequency [edit] The Debye frequency (Symbol: or ) is a parameter in the Debye model that refers to a cut-off angular frequency for waves of a harmonic chain of masses, used to describe the movement of ions in a crystal lattice and more specifically, to correctly predict that the heat capacity in such crystals is constant at high temperatures (Dulong–Petit law). The concept was first introduced by Peter Debye in 1912. Throughout this section, periodic boundary conditions are assumed. Definition [edit] Assuming the dispersion relation is with the speed of sound in the crystal and k the wave vector, the value of the Debye frequency is as follows: For a one-dimensional monatomic chain, the Debye frequency is equal to with as the distance between two neighbouring atoms in the chain when the system is in its ground state of energy, here being that none of the atoms are moving with respect to one another; the total number of atoms in the chain; the size of the system, which is the length of the chain; and the linear number density. For , , and , the relation holds. For a two-dimensional monatomic square lattice, the Debye frequency is equal to with is the size (area) of the surface, and the surface number density. For a three-dimensional monatomic primitive cubic crystal, the Debye frequency is equal to with the size of the system, and the volume number density. The general formula for the Debye frequency as a function of , the number of dimensions for a (hyper)cubic lattice is with being the gamma function. The speed of sound in the crystal depends on the mass of the atoms, the strength of their interaction, the pressure on the system, and the polarisation of the spin wave (longitudinal or transverse), among others. For the following, the speed of sound is assumed to be the same for any polarisation, although this limits the applicability of the result. The assumed dispersion relation is easily proven inaccurate for a one-dimensional chain of masses, but in Debye's model, this does not prove to be problematic.[citation needed] Relation to Debye's temperature [edit] The Debye temperature , another parameter in Debye model, is related to the Debye frequency by the relation where is the reduced Planck constant and is the Boltzmann constant. Debye's derivation [edit] Three-dimensional crystal [edit] In Debye's derivation of the heat capacity, he sums over all possible modes of the system, accounting for different directions and polarisations. He assumed the total number of modes per polarization to be , the amount of masses in the system, and the total to be with three polarizations per mode. The sum runs over all modes without differentiating between different polarizations, and then counts the total number of polarization-mode combinations. Debye made this assumption based on an assumption from classical mechanics that the number of modes per polarization in a chain of masses should always be equal to the number of masses in the chain. The left hand side can be made explicit to show how it depends on the Debye frequency, introduced first as a cut-off frequency beyond which no frequencies exist. By relating the cut-off frequency to the maximum number of modes, an expression for the cut-off frequency can be derived. First of all, by assuming to be very large ( ≫ 1, with the size of the system in any of the three directions) the smallest wave vector in any direction could be approximated by: , with . Smaller wave vectors cannot exist because of the periodic boundary conditions. Thus the summation would become where ; is the size of the system; and the integral is (as the summation) over all possible modes, which is assumed to be a finite region (bounded by the cut-off frequency). The triple integral could be rewritten as a single integral over all possible values of the absolute value of (see Jacobian for spherical coordinates). The result is with the absolute value of the wave vector corresponding with the Debye frequency, so . Since the dispersion relation is , it can be written as an integral over all possible : After solving the integral it is again equated to to find It can be rearranged into One-dimensional chain in 3D space [edit] The same derivation could be done for a one-dimensional chain of atoms. The number of modes remains unchanged, because there are still three polarizations, so The rest of the derivation is analogous to the previous, so the left hand side is rewritten with respect to the Debye frequency: The last step is multiplied by two is because the integrand in the first integral is even and the bounds of integration are symmetric about the origin, so the integral can be rewritten as from 0 to after scaling by a factor of 2. This is also equivalent to the statement that the volume of a one-dimensional ball is twice its radius. Applying a change a substitution of , our bounds are now 0 to , which gives us our rightmost integral. We continue; Conclusion: Two-dimensional crystal [edit] The same derivation could be done for a two-dimensional crystal. The number of modes remains unchanged, because there are still three polarizations. The derivation is analogous to the previous two. We start with the same equation, And then the left hand side is rewritten and equated to where is the size of the system. It can be rewritten as Polarization dependence [edit] In reality, longitudinal waves often have a different wave velocity from that of transverse waves. Making the assumption that the velocities are equal simplified the final result, but reintroducing the distinction improves the accuracy of the final result. The dispersion relation becomes , with , each corresponding to one of the three polarizations. The cut-off frequency , however, does not depend on . We can write the total number of modes as , which is again equal to . Here the summation over the modes is now dependent on . One-dimensional chain in 3D space [edit] The summation over the modes is rewritten The result is Thus the Debye frequency is found The calculated effective velocity is the harmonic mean of the velocities for each polarization. By assuming the two transverse polarizations to have the same phase speed and frequency, Setting recovers the expression previously derived under the assumption that velocity is the same for all polarization modes. Two-dimensional crystal [edit] The same derivation can be done for a two-dimensional crystal to find The calculated effective velocity is the square root of the harmonic mean of the squares of velocities. By assuming the two transverse polarizations to be the same, Setting recovers the expression previously derived under the assumption that velocity is the same for all polarization modes. Three-dimensional crystal [edit] The same derivation can be done for a three-dimensional crystal to find (the derivation is analogous to previous derivations) The calculated effective velocity is the cube root of the harmonic mean of the cubes of velocities. By assuming the two transverse polarizations to be the same, Setting recovers the expression previously derived under the assumption that velocity is the same for all polarization modes. Derivation with the actual dispersion relation [edit] This problem could be made more applicable by relaxing the assumption of linearity of the dispersion relation. Instead of using the dispersion relation , a more accurate dispersion relation can be used. In classical mechanics, it is known that for an equidistant chain of masses which interact harmonically with each other, the dispersion relation is with being the mass of each atom, the spring constant for the harmonic oscillator, and still being the spacing between atoms in the ground state. After plotting this relation, Debye's estimation of the cut-off wavelength based on the linear assumption remains accurate, because for every wavenumber bigger than (that is, for is smaller than ), a wavenumber that is smaller than could be found with the same angular frequency. This means the resulting physical manifestation for the mode with the larger wavenumber is indistinguishable from the one with the smaller wavenumber. Therefore, the study of the dispersion relation can be limited to the first Brillouin zone without any loss of accuracy or information. This is possible because the system consists of discretized points, as is demonstrated in the animated picture. Dividing the dispersion relation by and inserting for , we find the speed of a wave with to be By simply inserting in the original dispersion relation we find Combining these results the same result is once again found However, for any chain with greater complexity, including diatomic chains, the associated cut-off frequency and wavelength are not very accurate, since the cut-off wavelength is twice as big and the dispersion relation consists of additional branches, two total for a diatomic chain. It is also not certain from this result whether for higher-dimensional systems the cut-off frequency was accurately predicted by Debye when taking into account the more accurate dispersion relation. Alternative derivation [edit] For a one-dimensional chain, the formula for the Debye frequency can also be reproduced using a theorem for describing aliasing. The Nyquist–Shannon sampling theorem is used for this derivation, the main difference being that in the case of a one-dimensional chain, the discretization is not in time, but in space. The cut-off frequency can be determined from the cut-off wavelength. From the sampling theorem, we know that for wavelengths smaller than , or twice the sampling distance, every mode is a repeat of a mode with wavelength larger than , so the cut-off wavelength should be at . This results again in , rendering It does not matter which dispersion relation is used, as the same cut-off frequency would be calculated. See also [edit] Bose gas Gas in a box Grüneisen parameter Bloch–Grüneisen temperature Electrical resistivity and conductivity#Temperature dependence References [edit] ^ Pohl, R. O.; Love, W. F.; Stephens, R. B. (1973-08-01). Lattice vibrations in noncrystalline solids (Report). Cornell Univ., Ithaca, N.Y. (USA). Lab. of Atomic and Solid State Physics. OSTI 4410557. ^ Debye, Peter (1912). "Zur Theorie der spezifischen Waerme". Annalen der Physik (in German). 39 (4): 789–839. Bibcode:1912AnP...344..789D. doi:10.1002/andp.19123441404. ^ a b Kittel, Charles (2004). Introduction to Solid State Physics (8 ed.). John Wiley & Sons. ISBN 978-0471415268. ^ Schroeder, Daniel V. "An Introduction to Thermal Physics" Addison-Wesley, San Francisco (2000). Section 7.5 ^ Hill, Terrell L. (1960). An Introduction to Statistical Mechanics. Reading, Massachusetts, U.S.A.: Addison-Wesley Publishing Company, Inc. ISBN 9780486652429. {{cite book}}: ISBN / Date incompatibility (help) ^ Oberai, M. M.; Srikantiah, G (1974). A First Course in Thermodynamics. New Delhi, India: Prentice-Hall of India Private Limited. ISBN 9780876920183. ^ Patterson, James D; Bailey, Bernard C. (2007). Solid-State Physics: Introduction to the Theory. Springer. pp. 96–97. ISBN 978-3-540-34933-4. ^ Shulman, L. M. (2004). "The heat capacity of water ice in interstellar or interplanetary conditions". Astronomy and Astrophysics. 416: 187–190. Bibcode:2004A&A...416..187S. doi:10.1051/0004-6361:20031746. ^ Flubacher, P.; Leadbetter, A. J.; Morrison, J. A. (1960). "Heat Capacity of Ice at Low Temperatures". The Journal of Chemical Physics. 33 (6): 1751. Bibcode:1960JChPh..33.1751F. doi:10.1063/1.1731497. ^ In his textbook Kinetic Theory of Liquids (engl. 1947) ^ Bolmatov, D.; Brazhkin, V. V.; Trachenko, K. (2012). "The phonon theory of liquid thermodynamics". Scientific Reports. 2: 421. arXiv:1202.0459. Bibcode:2012NatSR...2..421B. doi:10.1038/srep00421. PMC 3359528. PMID 22639729. ^ Baggioli, M.; Zaccone, A. (2021). "Explaining the specific heat of liquids based on instantaneous normal modes". Physical Review E. 104 (1): 014103. arXiv:2101.07585. Bibcode:2021PhRvE.104a4103B. doi:10.1103/PhysRevE.104.014103. PMID 34412350.{{cite journal}}: CS1 maint: article number as page number (link) ^ Debye, P. (1912). "Zur Theorie der spezifischen Wärmen". Annalen der Physik. 344 (14): 789–839. Bibcode:1912AnP...344..789D. doi:10.1002/andp.19123441404. ISSN 1521-3889. ^ "The one dimensional monatomic solid" (PDF). Retrieved 2018-04-27. ^ Fitzpatrick, Richard (2006). "Specific heats of solids". Richard Fitzpatrick University of Texas at Austin. Retrieved 2018-04-27. ^ a b c Simon, Steven H. (2013-06-20). The Oxford Solid State Basics (First ed.). Oxford: Oxford University Press. ISBN 9780199680764. OCLC 859577633. ^ "The Oxford Solid State Basics". podcasts.ox.ac.uk. Retrieved 2024-01-12. ^ Srivastava, G. P. (2019-07-16). The Physics of Phonons. Routledge. ISBN 978-1-351-40955-1. Further reading [edit] CRC Handbook of Chemistry and Physics, 56th Edition (1975–1976) Schroeder, Daniel V. An Introduction to Thermal Physics. Addison-Wesley, San Francisco (2000). Section 7.5. External links [edit] Experimental determination of specific heat, thermal and heat conductivity of quartz using a cryostat. Simon, Steven H. (2014) The Oxford Solid State Basics (most relevant ones: 1, 2 and 6) Retrieved from " Categories: Condensed matter physics Thermodynamic models American inventions Hidden categories: CS1 German-language sources (de) CS1 errors: ISBN date CS1 maint: article number as page number Articles with short description Short description matches Wikidata Wikipedia articles with style issues from January 2024 All articles with style issues Articles needing additional references from January 2024 All articles needing additional references Articles with multiple maintenance issues Wikipedia articles needing clarification from January 2024 All articles that may contain original research Articles that may contain original research from January 2024 All articles with unsourced statements Articles with unsourced statements from January 2024 Debye model Add topic
15299	https://www.youtube.com/watch?v=80YhRNQ6cM4	Venn Diagrams and Set Operations (Union, Intersection, Complement, etc) Math Problems Solved Craig Faulhaber 6540 subscribers Description 8222 views Posted: 6 May 2020 We first introduce the complement of a set. We show this set in a Venn Diagram. We then introduce a second set and draw a new Venn Diagram that includes some overlap between sets. This leads us to discuss the union and intersection of sets. We then look at subtracting sets and write a law for the cardinal numbers of (or number of elements in) sets involving unions and intersections. 8 comments Transcript: for the next couple of examples let's just consider all natural numbers from 1 to 10. in other words let's make our universal set 1 through 10. and let's say we have a set A and A includes the following numbers. for our first new concept let's talk about the complement of A. this is denoted as A with a prime on it. sometimes we might denote it as an A with a C for complement on it. either way what the complement of A is is the set of all elements that are not in A. so for this example A complement is going to be all of the numbers between 1 and 10 except 4 3 4 5 and 8. so a complement includes 1 2 6 7 9 and 10. we can visualize sets by drawing a venn diagram in a venn diagram typically is a rectangle representing our universal set U and circles representing our sets. in this example A contains the elements 3 4 5 and 8. outside of a somewhere the elements 1 2 6 7 9 and 10. now i kind of scattered my elements all over the place but you could have just as easily drawn a more ordered looking Venn diagram that would mean the exact same thing. for example something like this would be the exact same Venn diagram. now as we always do in math why don't we complicate this situation a little bit. let's start with the same universal set in the same set for A but let's add in a set B. let's say the elements in B are 1 3 5 7 and 9. now you'll notice that there are elements in a that are not in B there are elements in B that are not in A and there is some overlap between these two sets. there are some elements that are in both A and B, so if this is the case our Venn diagram is going to look something like this we have a set A and we have a set B and there is some overlap between those two sets the overlap between these two sets is the elements three and five. also in set A are the elements four and eight and in set B but not in set A are the elements 1 7 and 9. now still out here in the universal set outside of both A and B are the elements 2 6 and 10. now earlier we introduced the concept of A complement and decided that the complement of A was this right here. we could also find the complement of B. all the elements that are not in B are... now let's introduce this concept of an intersection down here we can see that the sets A and B intersect and that intersection between those two sets includes the elements 3 and 5. we have notation for this intersection. the intersection of A and B is denoted with this intersection symbol right here in between A and B. and this new set includes all of the elements that are both in A and B in our example the intersection of A and B includes the elements three and five. and as a note it is pretty common for us to think of this symbol right here as meaning AND so we define the intersection operation between sets now let's define the union. the union of A and B is denoted with this right here basically a U in between A and B and the union of A and B includes all elements that are in A or in B or in both sets. in our example A union B, I'll try to write them in order, is going to include the elements 1 3 4 5 7 8 and 9. those are all of the elements that are in either A or B or in both. and as another note this union sign here we often think of as an OR. now why don't I zoom out a little bit to make a little bit of space and we can do a few more examples. what if i asked you to find A union B complement. what this notation means is we need to find the union of A and B first and then take the complement of that. well we found A union with B right here so the complement of this set is everything that's not in this set which is just 2 6 and 10. let's make up another example let's say i wanted you to find a complement intersected with B well what we would need to do is we would need to find a complement, we would need to make sure we have B, and then we would look at the intersection of those two sets. fortunately above we already have A complement written out. it includes the elements 1 2 6 7 9 and 10. we also have B written out and that includes the elements 1 3 5 7 and nine. now the intersection of those two sets is all of the elements that are in both sets. that would be one seven and nine okay and you could imagine i could probably come up with any number of increasingly complicated looking intersections and unions of sets but i do want to introduce one or two more concepts to you the difference between two sets is given by A minus B this would be the set of all elements that belong to A but don't belong to B well we could do this problem just by looking at sets A and B, or we could look at the Venn diagram, it's easier for me to visualize this by looking at the Venn diagram. all of the elements that belong to A but don't belong to B are included in this piece of the Venn diagram right here. those are the elements four and eight. so our set A minus B is the set four eight. okay just a couple more thoughts and I'll let you go. let's recall the notation for the cardinal number of a set. n of A that just means the number of elements in A. and if we look over here at a we see that A has four elements. the number of elements in B you can see is five. let's go down here and look at the number of elements in A and B or A intersected with B. and we see from this set that we have two elements. one more let's look at the number of elements in A union B. we can see from this set that there are seven elements in the union of A and B. now if we look at this Venn diagram the intersection of A and B is in this little area right here. the union of A and B can be given by all of the elements in this whole region right here. so this formula is going to look a little bit complicated but it's probably not too surprising that the number of elements in A unioned with B, that's the number of elements in this region up here, should be the number of elements in A plus the number of elements in B but notice when we add the number of elements in A and the elements in B together we get some overlap in the middle we've counted these middle two elements or these intersection elements twice. so what we need to do to get the number of elements in this whole region A unioned with B is we need to subtract off the number of elements in the intersection of A and B. and if you look at the numbers for our particular example here you see that we get 7 equals 4 plus 5 minus 2 and that is a true statement. so this formula checks out okay i think that's all I have for you for an introduction to complements and unions and intersections. i hope that that helps you out and i will see you next time!

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