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Italian Journal of Pediatrics
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A blue blood toddler– a case report of methemoglobinemia and literature review
Dorina Pjetraj
ORCID: orcid.org/0000-0002-0238-78651,
Madiha El Mechri1,
Silvia Bacelli2,
Elisabetta Fabiani2,
Laura Caponi1,
Simona Gatti1 &
…
Elena Lionetti1
Italian Journal of Pediatrics
volume 51, Article number: 53 (2025)
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Abstract
Background
Methemoglobinemia (MetHb) is a rare and potentially life-threatening condition caused by oxidation of ferrous hemoglobin (Fe2+) to the ferric (Fe3+) state, making it incapable of binding oxygen and resulting in cyanosis and tissue ischemia.
Case presentation
This case presentation describes a 1-year-old boy who developed sudden cyanosis and reduced consciousness disorder. An initial assessment showed decreased oxygen saturation (SpO2 85%) despite oxygen therapy, while point-of-care venous blood gas (VBG) analysis assessed high rates of MetHb (72.7%). Methylene blue and ascorbic acid were administered, resulting in in rapid clinical recovery and normalized VBG test results. The trigger for this condition was not identified, however the most likely cause of poisoning was attributed to food oxidants.
Conclusion
Starting from the description of a clinical case, this paper discusses the causes and mechanisms of possible poisoning and reviews recent guidelines for methemoglobinemia management.
Background
Methemoglobinemia (MetHb) is a rare and potentially fatal condition caused by the oxidation of ferrous hemoglobin (Fe2+) to the ferric (Fe3+) state. Acquired MetHb may cause cyanosis and tissue ischemia unresponsive to oxygen supplementation . This case report describes the diagnosis and management of a one-year-old boy presenting with methemoglobinemia in our pediatric emergency department and emphasizes the importance of early recognition and treatment of MetHb through a detailed review of the most recent scientific literature.
Case presentation
A one-year-old boy of Tunisian descent was brought to the Pediatric Emergency Department, presenting with cyanosis, drowsiness, and desaturation. During the initial assessment, the patient was alert and responsive, with patent airways and a normal breathing pattern. Cardio-thoracic and abdominal examinations were unremarkable. The patient was found to have a blood pressure within the normal range (98/50 mmHg), mild tachycardia (HR: 165/min), tachypnea (RR: 50 breaths per minute) and hypoxia (SpO2 85%), despite administration of 100% oxygen via facemask. His past medical history revealed prematurity (born at 32 gestational weeks) with normal growth and neurological development. There was no parental consanguinity. He had a flat angioma on the left hemisphere, which was under follow-up. The mother reported that before the symptoms started, the child had been playing at home in a well-ventilated room. He had no known allergies and was not taking any medications. There were no recent symptoms such as cough, fever, or other concerns. His last meal, consumed about one hour before the event, consisted of beef meat and Swiss chard. No other person in the family had consumed the same foods.
Upon obtaining an arterial blood sample, the color of the blood was noted to be dark brown. Venous blood gas analysis demonstrated abnormal findings including markedly decreased PO2 (9 mmHg), reduced oxygen saturation (SO2 23.9%), elevated lactate levels (4.5 mmol/L), and a significantly elevated methemoglobin (MetHb) concentration (72.7%). pH was 7.33 and PCO2 43 mmHg. The patient’s hemoglobin level was 11.8 g/dL, and the glucose level was 121 mg/dL. Chest X-ray was unremarkable. Approximately 15 min after the patient’s arrival, his clinical condition rapidly deteriorated. He became drowsy and experienced seizures, with oxygen saturation dropping as low as 70%.
At this point the patient received intravenous methylene blue (MB) at a dose of 2 mg/kg over 5 min, which was repeated after 15 min. This intervention led to a rapid improvement in the patient’s clinical status, including normalization of consciousness, skin color, and oxygen saturation levels. The patient was then transferred to the Pediatric Intensive Care Unit for close monitoring. A blood gas analysis performed 3 h later demonstrated a significant reduction in MetHb levels to 2.9%. Throughout the observation period, the child’s overall condition remained consistently good. Treatment was continued with the administration of ascorbic acid (500 mg given twice daily for a total of 16 doses). The following day, MetHb levels had returned to the normal range at 1% and remained stable during continued observation. Subsequent tests indicated normal renal and hepatic parameters [S-urea 25 (20–45) mg/dl, S-creatinine 0.34 (0.20–1.3) mg/dl, bilirubin total 0.2 (0.2–1.2) mg/dL and direct 0.1 (0–0.4) mg/dL; alanine transaminase 34 (5–40) units/L; aspartate transaminase 55 (5–40) units/L]. Inflammatory markers, such as C-reactive protein and procalcitonin, were negative. The microbiological examinations of stool specimens (stool culture, Enterobacteria, Enterovirus, Adenovirus, Rotavirus) resulted in negative outcomes. Additionally, the patient’s glucose-6-phosphate dehydrogenase (G6PDH) activity was within the normal reference range (12.9 (> 9.4) U/gr Hb) and analysis of hemoglobin variants with HPLC method did not reveal any abnormalities.
Given the normal basal levels of MetHb, clinicians decided not to pursue further genetic investigations and instead focused on acquired causes of MetHb. Since the child had not taken any medications or been exposed to any new substances, there was suspicion of food poisoning. The case was reported to public health authorities, who conducted a thorough analysis on the meat that the child had for lunch; however, they found no evidence of contamination. Unfortunately, it was not possible to investigate the vegetables as they had all been consumed.
Discussion and conclusions
Pathophysiology
Methemoglobinemia is an uncommon but potentially life-threatening condition caused by the oxidation of the iron in hemoglobin, converting it from the normal ferrous (Fe2+) state to the ferric (Fe3+) state . This altered form of hemoglobin, known as methemoglobin (MetHb), is incapable of effectively transporting and releasing oxygen to the body’s tissues. As a result, the oxygen-hemoglobin dissociation curve shifts to the left, leading to functional anemia and reduced oxygen delivery. Normally, the cytochrome b5 reductase (CytB5) enzyme pathway maintains a low basal level of MetHb, typically around 1.0-1.5% . However, in certain situations, such as during an oxidant challenge, a secondary pathway involving glucose-6-phosphate dehydrogenase (G6PD) and nicotinamide adenine dinucleotide (NAD+) can become more prominent in helping to manage elevated MetHb levels (Fig. 1) [4 and EuroBloodNet recommendations on diagnosis and Treatment of Methemoglobinemia. HemaSphere. 2021;5:e660.
."), 5].
Methemoglobinemia can be either inherited or acquired. The inherited, form is attributed to mutations in the CYB5R3 gene, which encodes the CytB5 enzyme. This genetic condition presents in two subtypes: type I, characterized by an unstable enzyme localized to red blood cells, leading to methemoglobin levels exceeding 25% and symptoms such as cyanosis, headache, fatigue, and dyspnea; and type II, caused by variants that diminish enzyme expression or activity across all tissues, resulting in methemoglobin levels ranging from 8 to 40% and severe neurological manifestations. The distinction between type I and II methemoglobinemia stems from the differential expression of CYB5R isoforms in red blood cells compared to other cell types . In addition to the autosomal recessive forms of methemoglobinemia, a rare group of hemoglobin variants known as M group variants (HbM) can arise from autosomal dominant mutations in the genes encoding the globin chains. These HbM variants are characterized by structural abnormalities that lead to heme iron auto-oxidation and cyanosis, often without manifesting other significant symptoms . To date 13 distinct HbM variants have been identified, with some named after the geographic locations in which they were first discovered. HbM variants affecting the alpha-globin chain typically cause cyanosis evident at birth, while those involving the beta-globin chain become apparent later as fetal hemoglobin is replaced by the adult form .
Acquired MetHb, on the other hand, is caused by the consumption of certain drugs or exposure to toxins that hasten the oxidation of hemoglobin, resulting in a temporary increase in MetHb levels. Some common agents associated with acquired MetHb include: nitrites, such as those found in certain medications (e.g., benzocaine, dapsone, nitrates), chemicals such as aniline dyes and aromatic compounds, certain antibiotics (e.g., sulfonamides), and ingesting contaminated well water and food containing nitrate 9, (Table 1).
In the presented case, the patient’s history did not indicate any genetic conditions, and baseline MetHb levels were normal. As such, the clinicians focused their investigations on potential foodborne poisoning. Given that tests on the consumed meat were unremarkable, the clinicians inferred that the Swiss chard was the likely source of the oxidative insult. This hypothesis aligns with findings from a recent systematic review on food-induced MetHb, which identified nitrites and nitrates as the primary oxidizing agents implicated . The review further noted that the most common scenarios for food-related methemoglobinemia involve children consuming improperly stored vegetables (30%), accidental ingestions (27%), and errors during meat curing processes (27%) .
Nitrates and nitrites play a central role in the development of methemoglobinemia . Nitrates are relatively less harmful than nitrites, but they can be transformed into nitrites through various processes, often due to improper storage or cooking of certain vegetables like spinach, beets, and carrots . Nitrites are commonly used in the food industry to maintain the pinking of meat or in high doses to preserve meat and kill bacteria . A target concentration of nitrate nitrogen for food of less than 100 ppm is desirable for infants, and some commercially prepared infant food vegetables are monitored voluntarily by manufacturers for nitrate content [16:2558.")]. Children are particularly vulnerable to methemoglobinemia related to nitrites and nitrates. They have lower stomach acid production, leading to a greater presence of nitrate-reducing bacteria in their gut flora. Additionally, the methemoglobin reductase system, which helps manage elevated methemoglobin levels, only fully matures around six months of age. This combination of increased nitrite/nitrate exposure and immature reductase system puts young children at higher risk for developing methemoglobinemia from food sources [12, 17].
Clinical features and diagnostic approach
To diagnose this disease and distinguish between hereditary and acquired forms, it is crucial to obtain a detailed clinical and family history, assess for consanguinity, and review environmental and drug exposures. For acquired forms, the cyanosis is typically of acute onset, so promptly investigating a history of drug or toxin exposure is important. In contrast, a longstanding family history of cyanosis, dusky-coloured skin, or blue sclera would suggest congenital forms .
The clinical manifestations of acute acquired MethHb are contingent on the percentage of methemoglobin saturation. Saturation levels below 10% typically do not cause symptoms, while levels between 10% and 25% result in cyanosis. As MethHb levels increase, individuals may experience symptoms such as headache, fatigue, dizziness nd shortness of breath at levels ranging from 35 to 40% . Levels reaching 60% can lead to arrhythmias, seizures, lethargy, and stupor. When surpassing 70%, it may cause vascular collapse and death [19, 20].
Patients who experience a sudden onset of cyanosis and hypoxia that does not improve despite 100% oxygen therapy, as indicated by arterial blood gas results and the distinct dark red/chocolate brown color of the arterial blood during phlebotomy, may indicate an increase in MetHb . Confirming the diagnosis involves measuring MetHb levels, obtaining positive co-oximetry results, and identifying an oxygen saturation gap greater than 5% between arterial blood gas oxygen saturation and pulse oximeter reading (SpO2-SaO2) .
It is crucial to consider the hemoglobin levels of the patient and compute methemoglobin levels using the formula Hemoglobin in grams per deciliter multiplied by MetHb percentage (g/dL x MetHb%). This method will yield a more precise measurement of MetHb level in grams per deciliter, enhancing accuracy when assessing residual functional hemoglobin levels . In our case report the patient had only 3.2 gr/dl of functional hemoglobin left as the Hb level was 11.4 gr/dl, and MetHb saturation was 72.7%. It is noteworthy that despite cyanosis being commonly associated with MetHb, a systematic review has revealed that cyanosis and hypoxemia are not always present in acquired MetHb cases. This underscores the significance of using co-oximetry as a diagnostic tool, which utilizes at least four light wavelengths to measure various forms of hemoglobin including oxyhemoglobin, deoxyhemoglobin, CO-Hb, and Met-Hb [23, 24]. Alternative technologies, such as optoacoustic or spectral sensors, have been proposed to detect dysfunctional hemaglobin types in the bloodstream. Similarly, non-invasive methods based on in silico models have been explored. These alternative options may offer supplementary support for the clinical management of methemoglobinemia going forward .
Key diagnostic tests in the evaluation of suspected congenital methemoglobinemia include: assessing MetHb levels, measuring CYB5R enzyme activity, and conducting genetic testing. In individuals with congenital CYB5R3 deficiency, the activity of the CYB5R enzyme is typically reduced to less than 20% of the normal level. Electrophoresis can identify HbM variants caused by mutations in globin genes and DNA sequencing of the CYB5R3 gene can confirm the diagnosis. Emerging next-generation sequencing (NGS) technologies facilitate the detection and characterization of significant genetic variants in patients with rare erythrocyte disorders such as methemoglobinemia, thereby expediting the differential diagnosis .
Therapy
The management of methemoglobinemia in infants and children is guided by multiple considerations, such as symptoms, the overall percentage of methemoglobin, the underlying cause of the methemoglobinemia, and the patient’s age. Asymptomatic individuals with methemoglobin levels below 20% typically do not necessitate any specific intervention beyond the avoidance of oxidizing agents. For acquired methemoglobinemia, treatment is recommended at levels of 20% in symptomatic patients and 30% in asymptomatic patients. Patients with hereditary methemoglobinemia can tolerate higher MetHb levels, with some remaining asymptomatic up to 30–40% .
Prompt initiation of therapy involves halting exposure to the triggering oxidant stressor and providing high-flow oxygen. Methylene blue (MB) is the most effective antidote, as it stimulates NADPH MetHb reductase (Fig. 1) . The suggested administration is 1 to 2 mg/kg via intravenous infusion of a 1% solution over a span of 5 min. If required this dosage can be repeated within an hour [27, 28; 2024.")]. It should be pointed out that the administration of MB is controversial in patients with known G6PD deficiency, due to the risk of hemolytic anemia [29, 30]. MB is an oxidizing agent, while its metabolic product, leukomethylene blue, is a reducing agent. Large doses of MB can lead to higher levels of the oxidizing agent itself rather than the reducing metabolite, resulting in hemolysis. Moreover in G6PD deficiency, the lack of sufficient NADPH production prevents the reduction of MB to the less oxidizing leukomethylene blue, consequently, MB therapy may be ineffective in G6PD-deficient patients . That is why in the latest recommendations for diagnosis and treatment of MetHb it is emphasised that patients should be tested for G6PD deficiency before receiving MB treatment, and in case of emergency include, the patient’s family history of G6PD deficiency should be checked before administering MB . In this case options include administering MB in a low dose combined with ascorbic acid, or not using it at all and providing ascorbic acid only . Ascorbic acid antioxidant capacity enables direct reduction of MetHb levels, however the process is slow and often requires multiple doses over 24 h. It is the preferred treatment when MB is unavailable or in cases of MetHb with G6PD deficiency. Dosing in children ranges from 0.5 g every 12 h for 16 doses to 1 g every 4 h for 8 doses. In case of failure of the listed treatments, other treatment options include exchange transfusion and hyperbaric oxygen therapy [4 and EuroBloodNet recommendations on diagnosis and Treatment of Methemoglobinemia. HemaSphere. 2021;5:e660.
.")].
In McNulty’s systematic review on food-induced metaemoglobinemia most cases resulted in survival (35 deaths out of 568 cases reported), even when experiencing severe methemoglobin levels reaching up to 89%, as long as MB was promptly administered. None of the fatalities had received MB . There were no deaths reported from cases of favism crisis which, in general, had lower methemoglobin fractions (maximum 15.8%). This emphasizes the significance of promptly identifying the toxidrome and initiating treatment with antidotes.
Due to the infrequency of this condition, the existing literature mainly comprises collections of case studies. A recent study conducted at five Italian pediatric emergency departments between 2007 and 2010 reported nineteen instances of acquired MetHb. The median age was 8.23 months, with a median time of 6 h from trigger to symptom onset. Improper food preservation, particularly vegetable broth, was identified as the primary source of poisoning in most cases. MB treatment was administered to 14 patients (73.7%), all of whom survived .
Another similar case series formulated by a Canadian pediatric emergency department involved 10 patients with acquired MetHb. Half of them were affected by hematologic malignancies, so the trigger of poisoning was easily identified as pharmacological (dapsone and rasburicase). The other half presented a previously undiagnosed G6PD deficiency and concomitant explosion to fava beans and topical menthol. Five of the patients were treated with packed red blood cell transfusion, two of them were given MB (MetHb saturation of 19% and 22%), one was treated with ascorbic acid and the last received supportive therapy with fluids and oxygen .
This case report highlights the importance of understanding and properly managing MetHb. High-fidelity simulation, as demonstrated by Alagha et al., can be a valuable tool for teaching this topic. Their simulation allowed participants to evaluate and treat MetHb in a safe, controlled setting. It was followed by a debriefing and discussion to review aspects of patient care, including medical knowledge, communication, and practice-based learning. The educational content and effectiveness were evaluated through feedback and a post-simulation survey. The survey showed that after the simulatiothin, 92% of emergency medicine residents felt confident in recognizing and treating MetHb, compared to 62.5% before. Overall, the results indicate that simulation-based training can improve recognition and management of MetHb .
In conclusion, acquired MetHb in the pediatric emergency department is an uncommon occurrence, but it should be considered when a patient presents with cyanosis, persistent hypoxia, “cyanosis-saturation gap,” and dark brown blood. Familiarity with this toxidrome and its clinical manifestations, despite its rarity, enables clinicians to initiate timely and suitable antidote therapy which can often be life-saving.
Data availability
The datasets used and/or analyzed during the current study are available from the corresponding author on reasonable request.
Abbreviations
MEtHb:
: Methemoglobinemia
VBG:
: Venous blood gas analysis
G6PD:
: glucose-6-phosphate dehydrogenase
MB:
: Methylene blue
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Department of Pediatrics, Marche Polytechnic University, Ancona, 60123, Italy
Dorina Pjetraj, Madiha El Mechri, Laura Caponi, Simona Gatti & Elena Lionetti
2. Department of Pediatric Emergency, Salesi Hospital, Azienda Ospedaliero-Universitaria delle Marche, Ancona, Italy
Silvia Bacelli & Elisabetta Fabiani
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DP and ME: manuscript drafting; SB, LC, EF: clinical evaluation and follow-up of the patient; SG and EL: supervised the project and contributed to the interpretation of the results. All authors read and approved the final manuscript.
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Pjetraj, D., Mechri, M.E., Bacelli, S. et al. A blue blood toddler– a case report of methemoglobinemia and literature review.
Ital J Pediatr 51, 53 (2025).
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Keywords
Methemoglobinemia (MetHb)
Poisoning
Methylene blue
Ascorbic acid
Food-induced methemoglobinemia
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In the efforts to obtain functional hepatocytes, various technologies have been developed to isolate and purify hepatocytes from liver tissues . However, the maintenance and expansion of hepatocytes are still challenging, making freshly isolated and cryopreserved primary hepatocytes the principle cell sources. For decades, many efforts have been made to generate hepatocytes from alternative cell sources. One possible approach is the differentiation of embryonic stem cells and induced pluripotent stem cells (iPSCs) into hepatocytes [6–10].
In recent years, an increasing number of reports have claimed the direct transdifferentiation of hepatocytes from nonhepatic cells [11–14]. The term “transdifferentiation” was initially employed to describe the conversion of one cell lineage to another as observed in silkmoth . Mammalian cells have been thought to be more resistant to such cell lineage conversion. Nevertheless, conversion of nonhepatic cells to hepatocytes has been studied for decades. In 1981, Scarpelli and Rao found hepatocyte-like cells in the pancreas of Syrian golden hamsters during pancreas regeneration . Since then, many protocols have been developed to induce hepatic conversion from pancreatic cells in vitro and in vivo . However, considering the lack of pancreatic cell sources, generation of hepatocytes from more easily accessible cell types could be more applicable.
Several investigations claimed that mesenchymal stem cells (MSCs) from bone marrow, hematopoietic stem cells (HSCs), and other adult stem cells, had the plasticity to give rise to a much wider range of cell types than previously thought [18–21]. Under the stimulation of growth factor and cytokine cocktails, hepatocyte-like cells expressing hepatic marker genes were generated from MSCs [20,22], HSCs , and even fibroblasts . However, the quality of hepatocytes generated by such strategies is controversial due to lack of definitive evidence of hepatic functions.
In 1987 a pioneering work about the direct conversion of fibroblasts to myoblasts by enforced expression of MyoD was reported, providing a novel strategy of cell lineage conversion by “master regulators” . The success of generating iPSCs further strengthened the notion that certain “master regulators” could change cell fate , which leads to the progress of transdifferentiation of nonhepatic cells to hepatocytes by hepatic transcription factors [11–14]. These hepatocytes generated from nonhepatic cells by transdifferentiation were termed induced hepatocytes (iHep cells).
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Book2018, Stem Cells and Cancer in HepatologyPengyu Huang, Qiwen Chen
Chapter
LIVER AND BILIARY SYSTEM
2010, The Digestive System (Second Edition)Margaret E. Smith PhD DSc, Dion G. Morton MD DSc
Hepatocytes
The hepatocyte is a polygonal cell with a clearly defined cell membrane, which is closely apposed to the cell membranes of adjacent hepatocytes (Figs 6.2C, 6.3). The membranes of adjacent cells are partially separated to form a bile canaliculus. The plasmalemma of adjacent hepatocytes shows irregularities with tight junctions, spot desmosomes and gap junctions. These separate the canaliculus from the rest of the intercellular space (Fig. 6.2C).
The plasma membrane of hepatocytes is specialized in certain regions. Adjacent to a sinusoidal blood space the hepatocyte is separated from the wall of the sinusoid by the perisinusoidal space (the space of Disse) and at this location the plasma membrane of the hepatocyte has numerous long microvilli. Vesicles and vacuoles are present in the subadjacent cytoplasm (Fig. 6.2C). The microvilli provide a large surface area for absorption and secretion.
The nuclei in different hepatocytes show considerable variation in shape and size and in some cases the cells are binucleate. Clumps of basophilic material are present in all cells. There are numerous small mitochondria throughout the cytoplasm of the hepatocyte. The structure of all hepatocytes is broadly similar but the cytoplasm of the cells shows a gradual variation with the distance of the cell from the periphery. The differences are related to the differences in functional activity of the peripherally and centrally positioned cells. The hepatocytes closest to the afferent blood supply, the ‘periportal’ cells, are exposed to the highest concentrations of nutrients and oxygen and those in the central region, the ‘perivenous’ cells, near to the efferent outflow, are exposed to the lowest concentrations. The periportal cells are the most active in the uptake from the blood of bile salts and in the secretion of many bile constituents into the canaliculi as well as in oxidative metabolism and gluconeogenesis (Fig. 6.4). After feeding, glycogen is deposited first in the periportal cells. It is only after a heavy carbohydrate meal that the more centrally located perivenous cells store glycogen. Moreover, when the blood sugar concentration falls, glycogen is removed first from the perivenous cells. The perivenous cells, which are exposed to depleted plasma, are the more active in biotransformation reactions and the secretion of potentially toxic xenobiotic and endobiotic substances. They are also more active in glycolytic and ketogenic reactions. Under certain conditions fat is deposited in the hepatocytes and it appears first in the more centrally disposed cells. Thus the cytosol of a given hepatocyte exhibits differences in composition at different times in relation to feeding and whether fat or glycogen has been deposited.
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Book2010, The Digestive System (Second Edition)Margaret E. Smith PhD DSc, Dion G. Morton MD DSc
Chapter
The Liver: Structure and Function
1982, GastroenterologyHarvey J. Dworken M.D.
The Hepatocyte.
The isolated hepatocyte has been shown to be spherical in shape, and to be covered with a multitude of microvilli (Figure 7-4). In the liver, stresses induced by the reticular stroma and compression from adjacent cells alter this shape to a 12-sided solid that tends to appear hexagonal on cross-section. The cytosol of the hepatocyte includes a large double-membraned nucleus containing a nucleolus which synthesizes messenger, transfer, and ribosomal RNA. The nuclear membrane has pores on its surface that allow the egress of nucleic acids into the cytosol. Cytoplasmic organelles include numerous egg-shaped mitochondria with cristae; linear collections of rough and smooth endoplasmic reticulum; special organelles probably arising from endoplasmic reticulum, called microbodies; lysosomes; and the lamellated Golgi apparatus. Rough endoplasmic reticulum is the site of protein formation, while smooth endoplasmic reticulum subserves the processes of conjugation of endogenous and exogenous chemicals, and other detoxifying functions. Both forms of endoplasmic reticulum produce glucose-6-phosphatase, which is essential in the metabolism of glycogen.
Schaffner and Popper (1975) describe three distinct regions in the hepatocytic border: the greatest length is in apposition to neighboring hepatocytes, and consists of both loose and tight, sometimes peg-in-groove junctions; the second region borders a hepatic sinusoid, and is cast into irregularly shaped microvilli which penetrate either into the sinusoid directly, or into spaces of Disse when the sinusoid is separated from the hepatocyte by overlying Kuppfer cells; the remaining border region of the hepatocyte, about 6 percent of the entire perimeter, comprises the microvillous border of a bile canaliculus.
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Book1982, GastroenterologyHarvey J. Dworken M.D.
Chapter
Volume 2
2018, Physiology of the Gastrointestinal Tract (Sixth Edition)Allan W. Wolkoff
42.1Introduction
One of the major functions of the hepatocyte is the removal of organic anionic compounds from the blood. These compounds include various xenobiotics as well as endogenous compounds such as bilirubin and bile acids. Many of these compounds have limited aqueous solubility and circulate bound to serum albumin. Despite being almost entirely protein bound, for the most part these organic anions are cleared rapidly from the circulation by the liver. The liver is designed to permit efficient extraction of protein-bound compounds by the hepatocyte. In comparison to other organs in which there is a tight capillary endothelium, the endothelium of the liver is fenestrated, allowing circulating proteins such as albumin to come into close proximity with hepatocytes. Previous studies suggested that direct interaction of albumin with the hepatocyte surface facilitates extraction of organic anions. However, a number of subsequent studies indicated that the organic anion is extracted from its protein carrier during the uptake process without evidence for direct protein-cell interaction. Although many of these protein-bound organic anions are lipophilic and could theoretically be taken up by hepatocytes by simple diffusion across the lipid bilayer, recent studies indicate that this is unlikely. In addition, a number of proteins that are able to mediate uptake of organic anions have been identified on the hepatocyte basolateral (sinusoidal) surface. Following internalization and biotransformation many of these compounds are pumped out of the cell, across the apical (canalicular) plasma membrane, into the bile. Several proteins that are able to mediate ATP-dependent excretion of these compounds have been identified on the bile canalicular plasma membrane of hepatocytes. This chapter examines mechanisms for uptake and excretion of a number of typical organic anions, including bile acids, for which the liver plays an essential role.
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Book2018, Physiology of the Gastrointestinal Tract (Sixth Edition)Allan W. Wolkoff
Chapter
Phosphatidylcholine Metabolism in Signal Transduction
1992, Cellular and Molecular Mechanisms of Inflammation: Signal Transduction in Inflammatory Cells, Part AStephen B. Bocckino, John H. Exton
Introduction
Although the hepatocyte is not an inflammatory cell, there is reason to believe that many of the signaling pathways found in the hepatocyte are also present in the other types of cells described in this series. In this chapter we review our work on phosphatidylcholine (PC) breakdown and hormone action in the rat liver. These studies reflect the tremendous interest in the mechanisms of signal transduction in many laboratories today. In particular, the study of G proteins and of the breakdown of inositol phospholipids provided the basis for the work described here.
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Book1992, Cellular and Molecular Mechanisms of Inflammation: Signal Transduction in Inflammatory Cells, Part AStephen B. Bocckino, John H. Exton
Review article
Special Issue Dedicated to the Memory of Peter W. Hochachka (1937-2002)
2004, Comparative Biochemistry and Physiology Part B: Biochemistry and Molecular BiologyThomas W. Moon
This short review examines some of my personal experiences with Dr. Peter Hochachka, as a mentor and friend, and how his encouragement led to the research undertaken in my laboratory over the past three decades. Specifically, our work using the fish hepatocyte preparation as a model cell system is reviewed. The hepatocyte is an ideal cellular system that can be used to probe hepatic physiology and biochemistry. The impact of insulin, glucagon and related peptides, and catecholamines is discussed from the perspective of core and diverse functions of these key vertebrate metabolic hormones. Each hormone that operates in fish species was studied in manners similar to that of mammals, but it appears that the role of glucagon-like peptide-1 (GLP-1) in particular differs substantially from that in mammals. The receptors for each of these fish hormones seem structurally and in some cases functionally quite distinct from those in mammals. Few fish hormone receptor sequences are available, but fish genomists are rapidly adding new sequence information to the existing databases, so our view of the evolution of vertebrate hormone receptors will become clearer very quickly.
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Journal2004, Comparative Biochemistry and Physiology Part B: Biochemistry and Molecular BiologyThomas W. Moon
Mini review
Hepatocytes: The powerhouse of biotransformation
2012, The International Journal of Biochemistry & Cell BiologyDanielle K. Sevior, ... Jorma T. Ahokas
1Introduction
Hepatocytes are the parenchymal cells of the liver, whilst they have many functions (Sell, 2003), importantly they are responsible for the biotransformation of both endogenous and exogenous lipid soluble compounds. Other functions of hepatocytes, whilst not the focus of this review includes nutrient homeostasis (glucose storage and synthesis, cholesterol uptake), filtration of particulates, protein synthesis (clotting factors, albumin), bioactivation (steroids) and formation of bile.
Prominent features of hepatocytes include a round nucleus and numerous mitochondria, suggesting the liver plays a significant role in energy metabolism. Hepatocytes also contain large amounts of endoplasmic reticulum (ER). The presence of the ribosome containing rough endoplasmic reticulum (RER) reflects a major function of hepatocytes – protein synthesis.
Together with the RER there is an extensive meshwork of smooth endoplasmic reticulum (SER), which incorporates large amounts of biotransformation enzymes, others are found in the cytosol. Hepatocytes are organised into plates separated by vascular channels or sinusoids. This structure is important in directing the excretion of the products of biotransformation away from the hepatocytes into bile and blood.
The biotransformation pathways are divided into phase I and phase II (Kitada et al., 1991). The cytochrome P450 (CYP) system is pivotal to the phase I system. The phase II enzymes are characterized by their ability to conjugate exogenous molecules using endogenous cofactors. Fig. 1 provides an overview of the biotransformation enzymes in the hepatocyte and the interaction of the biotransformation enzymes and the transporters.
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Journal2012, The International Journal of Biochemistry & Cell BiologyDanielle K. Sevior, ... Jorma T. Ahokas
Review article
Lipotoxicity
2010, Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of LipidsMichael Trauner, ... Martin Wagner
Hepatocyte (lipo)apoptosis is a key histological feature of NAFLD and correlates with progressive inflammation and fibrosis . Possibly, the individual susceptibility to lipoapoptosis could separate patients with simple steatosis from NASH . Consistent with this concept, elevated serum cytokeratin-18 fragments (as markers of hepatocyte apoptosis) distinguish simple hepatic steatosis from NASH . In in vitro (hepatocyte/cell line culture) models, saturated FA (e.g. palmitic or stearic acid) are more toxic than mono-unsaturated FA (e.g. palmitoleic or oleic acid) despite inducing similar amounts of hepatocellular steatosis . The mechanisms by which saturated FA are more toxic than unsaturated FA are not yet fully understood. Oleic acid supplementation leads to TG accumulation and is well tolerated, whereas excess palmitic acid is poorly incorporated into TG and causes lipoapoptosis . Recently, saturated FA, but not unsaturated FA, have been proposed as ligands of the Toll-like receptor 4 (TLR4) and FA-mediated cytotoxicity requires engagement of this membrane receptor in several extrahepatic cells [44,45]. A series of elegant studies have shown that saturated FA can lead to activation of pro-apoptotic pathways via upregulation of death receptor such as Fas and TRAIL [93,94], destabilization of lysosomes (lysosomal pathway) leading to the release of cathepsin B (in turn activating NF-κB and subsequently TNF-α) [68,95] and activation of the pro-apoptotic protein Bax in a c-jun N-terminal kinase-dependent manner [92,96,97]. Notably, saturated long-chain FA are more toxic than unsaturated FA in their capacity to induce lipoapoptosis and pro-apoptotic pathways [45,92]. Co-treatment with saturated FA such as oleic acid or increased expression of SCD1 (the enzyme catalyzing the formation of mono-unsaturated FA such as oleic acid and palmitoleic acid) can divert potentially toxic saturated FA such as palmitic acid to TG formation, which is associated with decreased toxicity as a result of decreased cellular levels of toxic saturated FA [44,90,98]. Conversely, oleic acid exerts lipotoxic effects in the setting of impaired TG synthesis . Although hepatocellular apoptosis certainly contributes to liver injury, one may speculate that propagation of hepatic inflammation and fibrosis may be milder than predicted/expected in conditions with predominantly necrotic cell death. Apart from hepatocytes, other liver cells (e.g. HSCs) can also undergo apoptosis, but this still needs to be explored for NAFLD and the concept of lipotoxicity.
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Journal2010, Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of LipidsMichael Trauner, ... Martin Wagner
Chapter
Liver Stem Cells
2007, Principles of Tissue Engineering (Third Edition)Eric Lagasse
Hepatocyte Polyploidy and Cell Aging
Hepatocytes are highly differentiated cells of the liver capable of multiple synthetic and metabolic functions. However, hepatocytes can be also viewed as a heterogeneous population of cells containing small and large cells, diploid to polyploidy, mononucleated to binucleated (see Figs. 47.3 and 47.4). The polyploidization of hepatocytes after birth is another interesting physiological process of this specialized cell. During growth and development, hepatocytes undergo dramatic changes, which are characterized by a gradual polyploidization, with the successive appearance of tetraploid and octoploid cells with one or two nuclei. In this process, human or rat hepatocytes of a newborn are exclusively diploid (2n) but will subsequently generate 4n, 8n, and eventually 16n mono- and binucleated hepatocytes. The accumulation rate of binucleated and polyploidy cells is very slow in the human, with an intensification of the polyploidization after the age of 50 (Kudryavtsev et al., 1993).
Polyploidy is a general physiological process found in many cellular systems (Anatskaya et al., 1994). Normally, polyploidization is a strategy of cell growth that increases metabolic output and is viewed as an alternative to cell division that is indicative of terminal differentiation and senescence (Sigal et al., 1995, 1999). The cellular mechanism that governs the passage from mononucleated 2n to binucleated 2 × 2n or mononucleated 4n was recently unveiled and involved the abortion of cytokinesis, which induced the formation of binucleated hepatocytes (Guidotti et al., 2003). In the proposed model, binucleated 2n hepatocytes could divide and generate binuclear tetraploid (2 × 2n) daughter cells. This binucleated (2 × 2n) tetraploid hepatocyte could itself generate two (4n) mononucleated cells (see Fig. 47.1). In rodents, studies of the transplanted hepatocytes isolated via flow cytometry using the DNA profile of diploid, tetraploid, and octoploid liver cells showed that all fractions could engraft, proliferate, and regenerate liver tissues (see Fig. 47.4) (Weglarz et al., 2000). These data supported the conclusion that multiple hepatocyte ploidy classes can serve as progenitors for regenerating hepatocyte foci in damaged liver. However, isolation of hepatocytes by centrifugal elutriation into three hepatic fractions—small (16 microns), medium-sized (21 microns), and large (27 microns) hepatocytes—showed that the small fraction had a surprisingly lower repopulation capacity during the first round of transplantation (Overturf et al., 1999). Clearly, more studies need to be done to understand the self-renewal and commitment potential of these different classes of hepatocytes.
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Book2007, Principles of Tissue Engineering (Third Edition)Eric Lagasse
Review article
Lipid droplets, autophagy, and ageing: A cell-specific tale
2024, Ageing Research ReviewsAlice Maestri, ... Ewa Ehrenborg
3Hepatocytes
Hepatocytes are the parenchymal cells of the liver and perform the numerous vital functions of this organ, including blood detoxification, hemostasis regulation, bile production and storage and production of circulating lipids (Trefts et al., 2017). To perform all these functions, hepatocytes have a very complex cytoskeletal structure and membrane polarity which allows the uptake of blood from different parts of the body (deoxygenated, nutrient-rich blood from the gut and oxygenated blood from the heart) and release of deoxygenated VLDL-rich blood from the hepatic vein at the sinusoidal plasma membrane, and release of bile contents at the canalicular plasma membrane (for review see: Schulze et al., 2019). Hepatocytes are the key regulators of global lipid metabolism: they are critical crossroads for lipid uptake, storage, breakdown, and release, and their LDs play a pivotal role in these functions. The hepatic source of FAs mainly originates from internalized chylomicrons remnant and by circulating albumin-bound FAs from adipocytes performing lipolysis caused by the spillover effect (Fig. 2). Also the uptake of CE-rich low-density lipoproteins (LDLs) from plasma contribute as the de novo lipogenesis (Mashek, 2013). The latter occurs in high carbohydrate availability and allows the liver to store carbohydrates as lipids instead of glycogen polysaccharides: this process is regulated by transcriptional factors like the sterol regulatory element binding protein1 (SREBP1), the carbohydrate response element binding protein (ChREBP) and liver X receptors (LXRs) (Sanders and Griffin, 2016). Upon fasting, hepatocytes release into the circulation both TAGs, which form the core of apolipoprotein B (ApoB)-containing VLDL particles, and ketone bodies, which, unlike FAs, are able to cross the blood-brain barrier and fuel the brain during starvation. The secreted VLDL-TAGs are then hydrolyzed by lipoprotein lipase (LPL) and the released NEFAs are absorbed and metabolized by peripheral tissues to produce energy (Bechmann et al., 2012).
Hepatocytes have different types of LDs. Each type of LD has a specific subcellular localization, lipid and protein composition (H. Wang et al., 2013; L. Wang et al., 2013). Hepatocytes are able to form classic storage cytosolic LDs (CLDs), luminal apoB-free LDs (LLDs) and apoB-containing LDs (or VLDL precursors) (L. Wang et al., 2013; H. Wang et al., 2013). Studies on CLDs, LLDs and nascent VLDL have been arduous due to the difficulty to isolate and separate these distinct LDs both in murine and human models, and due to the rapid changes in metabolic states of the hepatocyte itself. Moreover, the similarities of lipoproteins to LDs make us consider VLDL precursors as actual LDs within the hepatocyte. Hepatic CLDs are composed by a lipid core of TAGs, CEs and other esterified lipids and they can reach 1 to 2 µm in diameter (Gluchowski et al., 2017; H. Wang et al., 2013; L. Wang et al., 2013). The mechanisms of formation of these supersized LDs are analogous to those seen in adipocytes: they can form by either coalescence or through protein-mediated diffusion of neutral lipids from a small LD to a bigger one (Thiam et al., 2013). This latter process is mediated by the LD-associated protein family CIDE, whose member are localized on the outer phospholipid monolayer at the LD-LD contact sites (Gao et al., 2017). Most hepatocytes express CIDEB alone and contain smaller LDs, while a small proportion of hepatocytes express also CIDEA and CIDEC, which are typically expressed in adipocytes, and contain large LDs (Li et al., 2007, 2012). Upon fasting or obese conditions, the number of hepatocytes expressing CIDEA and CIDEC increases, and so does the number of hepatocytes with bigger LDs and higher levels of TAGs: this mechanism is mediated by PPARγ and induces hepatic steatosis (Matsusue et al., 2008; Xu et al., 2016; Zhou et al., 2012). CIDEB is localized on the LD membrane and on the ER and promotes lipid exchange and LD fusion in both small and large LD-containing hepatocytes; its deficiency results in the accumulation of only smaller LDs with reduced lipid exchange activities, suggesting an important role in LD biogenesis and maturation (Li et al., 2007; Xu et al., 2016). CIDEB has also been shown to play a crucial role in promoting the lipidation and maturation of VLDL, an activity counteracted by the protective role of PLIN2, which is also ubiquitously present on hepatic LDs (Li et al., 2012).
LLDs and VLDL precursors are found in the lumen of the endoplasmic reticulum where VLDL is assembled. Due to their localization and size, LLDs and VLDL precursors have only been observed once by immunogold electron microscopy and isolated via subcellular fractionation, and therefore their existence remains debated in the field (Fig. 2) (Alexander et al., 1976; Hamilton et al., 1998). Murine LLDs range in size from 7 to 200 nm, have a higher proportion of phospholipids to TAGs compared to CLDs. They are decorated with a series of unique LD-associated proteins, including apoE, which is thought to perform the lipid droplet coat function of the missing PLIN proteins, and microsomal triglyceride transfer protein (MTTP), protein implicated in the first step of VLDL assembly (lipidation), through the transfer of lipids from the ER membrane to its lumen (Higashi et al., 2003; Khatun et al., 2012; Wang et al., 2007). Similar to MTTP, CES1 (human ortholog of mouse carboxylesterase 3 (Ces3)/triacylglycerol hydrolase (TGH) is associated with apoB-free LLDs within the ER and catalyzes the hydrolysis of short- and long-chain acyl-glycerol, long-chain acyl-carnitine and long-chain acyl-CoA esters for the assembly of VLDL (Satoh and Hosokawa, 2010). This enzyme is thought to hydrolyze and re-esterify TAG stores from ER-associated CLD and LLDs for the lipidation of VLDL precursors and genesis of mature VLDL (Gilham et al., 2005; Wang et al., 2007). Although minimally characterized and debated, LLDs have been suggested to be precursors of nuclear LD (Sołtysik et al., 2019; Sołtysik et al., 2020), suggesting that further interest should be given to this subclass of LDs. Very little is known of VLDL precursors before their translocation from the ER to the cis-Golgi through the VLDL transport vesicle, and their maturation and consequent secretion into the circulation (Tran et al., 2002; Gustarova et al., 2007; Rahim et al., 2012). The assembly of the primordial lipoprotein, or pre-VLDL, starts when MTTP partially lipidates apoB100 in the lumen of the ER: there, it highly associates with the chaperone binding immunoglobulin protein (BiP) and protein disulfide isomerase (PDI), and continues to be lipidated by MTTP until the regions in the C-terminal of apoB100 allow the pre-VLDL to fold and detach from the ER chaperone retaining proteins (Boström et al., 1986; Olofsson et al., 1999; Stillemark-Billton et al., 2005). If insufficiently lipidated, pre-VLDL is retained and degraded, or further lipidated (Olofsson and Borén, 2012). Upon post-translational lipidation, pre-VLDL becomes VLDL2, which exits in the ER and reaches the ER Golgi intermediate compartment (ERGIC), where, depending on its lipidation, the length of apoB and availability of TAGs, VLDL2 can be either secreted as VLDL2 or anterograde transported from the ERGIC to the cis-Golgi, where it is converted into VLDL1 via a different type of lipidation involving CLDs (Fig. 2) (Asp et al., 2000; Stillemark-Billton et al., 2005). Recent work has elucidated an insulin-dependent cLD utilization in VLDL production in hepatocytes during fasting. Kumar and colleagues have described how insulin induces the increase of phosphatidic acid, which in turn activates the cytoskeletal protein kinesin-1, which transport cLDs to the smooth ER inside hepatocytes, where LDs are catabolized to produce VLDL particles (Kumar et al., 2019).
Hepatocytes are also the regulators of cholesterol in the body. Similarly to FAs, hepatocytes can source their cholesterol from dietary intake, circulating LDLs and high-density lipoproteins (HDLs) or de novo cholesterol biosynthesis (Fig. 2) (Turner and Wass, 2009). Hepatocytes are able to deliver both the endogenously synthetized and exogenously acquired cholesterol to the bloodstream through the synthesis of VLDL, which, after undergoing processing in the bloodstream, become LDLs (Goldstein and Brown, 2009). Peripheral tissues are able to receive cholesterol by interacting with these cholesterol-rich LDLs via LDL receptors; excess cholesterol is either stored as ester in LDs or released to HDLs, which transport cholesterol back to the liver, where is eliminated through the biliary system (Fig. 2) (Luo et al., 2020). Due to the large influx of cholesterol that the liver experiences, various mechanisms are in place to prevent excessive free cholesterol accumulation, including: 1) conversion and storage through esterification by SOAT2; 2) retention of SREBP2 in the ER and degradation of 3-hydroxy-3-methyl-glutaryl-coenzyme A reductase (HMGCR) to prevent de novo cholesterol synthesis, 3) activation of liver X receptor (LXR) and farnesoid X receptor (FXR) for the conversion of cholesterol into bile acid (Ioannou, 2016; Lambert et al., 2003; Radhakrishnan et al., 2008). This latter mechanism is valid only in murine models as excessive dietary cholesterol is a strong suppressor of bile acid synthesis in human and clinical trials assessing the effects of LXR agonists on human lipid metabolism resulted in adverse effects such as elevated plasma and liver lipids and neutropenia (Kirchgessner et al., 2016; Minniti et al., 2020; Reihnér et al., 1989).
Interestingly, also de novo TAG biosynthesis affects and is affected by bile acid synthesis: the administration of gallstone-dissolving agents or bile acid sequestrants, to human subjects lowers the hepatic production of VLDL TAGs (Angelin et al., 1987; Beil et al., 1982; Miller and Nestel, 1974). When these mechanisms fail, the liver starts to improperly accumulate cholesterol and TAGs, inducing the arising of progressive nonalcoholic fatty liver disease (NAFLD), a spectrum of hepatic disorders ranging from uncomplicated fatty liver to nonalcoholic steatohepatitis (NASH) (Bertolotti et al., 2014). Excessive hepatic free cholesterol has been shown to be associated with NASH and the development of cirrhosis (Ioannou et al., 2009; Puri et al., 2007). The mechanism by which cholesterol may induce NASH is unclear, but Ioannou and colleagues suggest that the generation of cholesterol crystals on the LD surface could be a key step for the development of the chronic sterile inflammatory state (Ioannou et al., 2019). Previous animal experiments indicated that excess dietary cholesterol is exclusively directed to the liver, rather than adipose tissue: there, cholesterol is transported through endosomal-lysosomal compartment onto the LD membrane of hepatocytes via direct membrane contact sites (Du et al., 2015; Ioannou et al., 2013; Ioannou et al., 2019). Before it is esterified and stored as CE in the core of the LD, cholesterol is solubilized by the phospholipids in the membrane in order to protect the nonpolar part of cholesterol molecules from the aqueous cytosol (Huang and Feigenson, 1999). Upon hypercholesterolemia, membrane cholesterol concentration increases and the ability of phospholipids to protect cholesterol saturates, thus unshielded cholesterol molecules precipitate adjacent to the membrane, forming cholesterol monohydrate crystals (Ioannou et al., 2019). These crystals disrupt the tightly regulated cholesterol metabolism in the cell and cause ER stress, which ultimately result in necrosis and activate resident tissue macrophages of the liver, Kupffer cells (KCs) (Ioannou et al., 2017). KCs aggregate around the dead hepatocytes containing cholesterol crystals, forming “crown-like structures”. The phagocyted crystals induce in KCs the activation of the nucleotide‐binding oligomerization domain, leucine rich repeat, and pyrin domain containing 3 (NLRP3) inflammasome, which is an important trigger for inflammation which may contribute to the progression from pure steatosis to NASH (Ioannou et al., 2013; Ioannou et al., 2019; Rajamäki et al., 2010; Wan et al., 2016).
The utilization of hepatic lipid storages is hard to study and to quantify, to the complexity and intertwined relations to other energy sources, such as glucose, and outputs, such as bile acids. Generally, the received, stored, or synthesized (from de novo lipogenesis or from other substrates) lipids are utilized as substrate for ATP production or released in the form of VLDL and bile acids (Alves-Bezerra and Cohen, 2017; Cohen, 2008). In contrast to signaling pathway regulating glucose metabolism, the mechanisms regulating FAs and cholesterol intake or release are not well defined at the cellular level (Jones, 2016). Hepatocytes, like adipocytes, perform lipolysis and recent work showed it can be induced, among the many circulating hormones and fasting, by β-adrenergic stimulation (Schott et al., 2017). Hepatic lipolysis is performed primarily by ATGL, since HSL is thought not to be expressed in human hepatocytes and the role of MGL has been poorly described in relation to LD catabolism in liver (Fig. 2) (MacParland et al., 2018; Ong et al., 2011). Studies on murine models have evaluated the role of the classic players of lipolysis, namely ATGL, CGI-58, G0S2, in the context of steatosis and revealed that they regulate CLD, but their modulation does not affect VLDL secretion (Guo et al., 2013; Ong et al., 2011; Turpin et al., 2011; Wu et al., 2011; Yang et al., 2020). Lack of non-classical lipolytic enzyme Ces1d (mouse ortholog of CES1) reduces VLDL secretion and the size of hepatic CLDs in mice, highlighting a more complex lipolysis machinery, also given by the different pools of LDs present in hepatocytes (Y. Wang et al., 2010; H. Wang et al., 2010). On the other hand, lipophagy has been extensively characterized in hepatocytes and its link to lipolytic enzymes suggests that these two mechanisms support each other in directing the degradation of LD stores (Schott et al., 2019; Singh et al., 2009). Lipophagy is important in hepatocytes and its pharmacological inhibition leads to increased LD number and size and impaired β-oxidation, which reduces the cell’s resistance to oxidant stress (Fig. 2) (Liu and Czaja, 2013; H. Wang et al., 2010; Y. Wang et al., 2010). ATGL has been shown to induce lipophagy by both activating sensing protein sirtuin 1 (SIRT1) and by interacting with LC3 (Martinez-Lopez et al., 2016; Najt et al., 2020; Sathyanarayan et al., 2017). The small guanosine triphosphatase (GTPase) Rab7 has been identified as the key player in hepatocytes’ lipophagy, allowing the recruitment of lysosomes to the LD surface (Schroeder et al., 2015). Another main player of hepatocytes’ lipophagy is PLIN2 and its removal from the LD surface by CMA aids and induces LD lipolysis and lipophagy (Kaushik and Cuervo, 2015). Interestingly, PLIN2 has been extensively studied in liver steatosis, as its knockout in mice showed increased autophagy, reduced TAG levels, but, once again, just limited alterations on VLDL secretion (Tsai et al., 2017). A recent work identified a new important regulator of lipophagy in hepatocyte nuclear factor 4α (HNF4α): this nuclear hormone receptor acts as a lipid sensing protein able to bind directly to fatty acids through its ligand binding pocket and, when stimulated or not bound to lipids, to induce lipophagy and LD clearance, overall reducing steatosis (Dhe-Paganon et al., 2002; Lee et al., 2021). Lastly, recent studies have identified for the first time microlipophagy in hepatocytes, a process only described in yeast (Cui et al., 2021; Schulze et al., 2020).
3.1Ageing in hepatocytes
Many studies have evaluated the problematics related to an aged liver and the cellular alterations occurring in aged hepatocytes. Hallmarks of an aged liver are decreased organ volume and total numbers of hepatocytes, decreased regenerative and metabolic capacity, reduced bile flow, and reduced hepatic blood flow due to a thickening of the endothelial lining and a reduced number of endothelial cell fenestration (Iber et al., 1994; Le Couteur and McLean, 1998; McLean et al., 2003; Wynne et al., 1989; Zeeh and Platt, 2002). Hepatocytes experience genomic instability, increased oxidative stress, decreased mitochondrial number and functionality, increased senescence and decreased autophagy, which results in the accumulation of protein aggregates like Mallory-Denk bodies and lipofuscin and reduced lipid buffering activity (Allaire and Gilgenkrantz, 2020; Hunt et al., 2019; Stumptner et al., 2002; Wiemann et al., 2002). As previously mentioned, ageing is associated with an abnormal accumulation of lipids in non-adipose tissue, including the liver; this in turn compromises its functionality and ability to control the energy levels in the body, aggravating the systemic lipotoxic environment and facilitating the arising of co-morbidities (Slawik and Vidal-Puig, 2006). To note, NAFLD in middle-aged individuals (45–55 years old) and in elderly patients (>65 years old, with a mean age of 68) is a strong risk factor for cardiovascular diseases (CVD) and fibrosis (Dunn et al., 2008; Noureddin et al., 2013), whereas in geriatric patients (80 years old) without metabolic syndrome, it is considered a benign feature, suggesting that the age-dependent redistribution of body fat, if not in concomitance with morbidity, is not necessary detrimental for hepatic homeostasis (Kagansky et al., 2004). Lipid accumulation in aged hepatocytes is concomitant to a profound metabolic remodeling which entails changes in hepatic lipid uptake, utilization, and secretion. While the levels of fatty acid-binding proteins CD36 and the rate of de novo lipogenesis increase with age (Donnelly et al., 2005; Sheedfar et al., 2014), genes involved in fatty acid oxidation are downregulated (Y.-J. Choi et al., 2023; Y.J. Choi et al., 2023). Recent work has identified decreased CMA activity as key player in the PPARα−mediated downregulation of fatty acid oxidation and macroautophagic clearance in aged hepatocytes (Y.J. Choi et al., 2023; Y.-J. Choi et al., 2023; Kim et al., 2017). Decreased autophagic activity also impairs mitochondrial regeneration and contributes to a reduced fatty acid oxidation and increased lipid accumulation in the liver, leading to lipid and age-related liver diseases (Fromenty, 2019; Xu et al., 2013)). Age-related genomic instability induces hepatocytes to enter a state of stress and dysfunction that results in the upregulation of anti-proliferative genes, increased senescence-associated lysosomal β-galactosidase activity and increased secretion of SASP (Wang et al., 2014). Another important derivative of hepatocyte instability during ageing is the reduced DNA-nuclear matrix interactions due to increased formation of extended chromatin fibers that reorganize nuclear areas and prevent gene activity (Moraes et al., 2007). This accumulation of condensed regions, called senescence associated heterochromatin foci (SAHF), and the redistribution of surplus chromatin has been shown in various aged animal models, even in cells with generalized chromatin unpacking (Adams, 2007; Moraes, 2014). This alteration in chromatin structure causes gene silencing, senescence, and hepatic functional impairment. Many genes have been found decreased in expression in concomitance with senescence, including Glut2 and Glut4, PI3K, MAPK, Jak/S, NFkB and IGF1, which cause hepatic insulin resistance and diabetes (Aravinthan, 2015; Aravinthan et al., 2015; Ghiraldini et al., 2012). Studies on the senescent hepatocyte gene signature show changes in genes implicated in cell structure and cytoskeleton, which result in the characteristic phenotype of enlarged, flatten and elongated cells, the previously mentioned cell cycle markers and the SASP-related pro-inflammatory genes, which include CXC motif chemokine receptor 4 (CXCR4), serum amyloid A-4 (SAA4) and colony stimulating factor 1 (CSF1) (Aravinthan et al., 2014). Senescent hepatocytes also show increased LD accumulation and decreased oxidative capacity due to defective mitochondrial FA metabolization activity and increased ROS production, which impairs the normal cellular metabolism and leads to NAFLD (Ogrodnik et al., 2017).
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Open AccessArticle
Schur-Power Convexity of a Completely Symmetric Function Dual
by
Huan-Nan Shi
Huan-Nan Shi
SciProfiles Scilit Preprints.org Google Scholar
1 and
Wei-Shih Du
Wei-Shih Du
SciProfiles Scilit Preprints.org Google Scholar
2,
1
Department of Electronic Information, Teacher’s College, Beijing Union University, Beijing 100011, China
2
Department of Mathematics, National Kaohsiung Normal University, Kaohsiung 82444, Taiwan
Author to whom correspondence should be addressed.
Symmetry 2019, 11(7), 897;
Submission received: 24 May 2019 / Revised: 6 July 2019 / Accepted: 8 July 2019 / Published: 10 July 2019
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Abstract
In this paper, by applying the decision theorem of the Schur-power convex function, the Schur-power convexity of a class of complete symmetric functions are studied. As applications, some new inequalities are established.
Keywords:
Schur-power convexity; Schur-convexity; Schur-geometric convexity; Schur-harmonic convexity; completely symmetric function; dual form
MSC:
Primary 05E05; 26B25; 26D15
1. Introduction and Preliminaries
Convexity is a natural notion and plays an important and fundamental role in mathematics, physics, chemistry, biology, economics, engineering, and other sciences. To solve practical problems, several interesting concepts of generalized convexity or generalized concavity have been introduced and studied. Recent important investigations and developments in convex analysis have focused on the study of Schur-convexity, and Schur-geometric and Schur-harmonic convexity of various symmetric functions; see, e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] and references therein. It is worth mentioning that discovering and judging Schur-convexity of various symmetric functions is an important topic in the study of the majorization theory. A lot of achievements in this field have been investigated by several authors; for more details, see the first author’s monographs [21,22].
Throughout this paper, we denote by and , the set of positive integers and real numbers, respectively. Let X be a nonempty set. Denote and . For a positive integer n, the set for the Cartesian product is the collection of all n-tuples of elements of X. Therefore, we can write , and as follows:where .
Let and in . A set is said to be convex if and imply
Let be a convex set. A function is said to be convex on D iffor all , and all . The function f is said to be concave on D if and only if is convex on D.
For the reader’s convenience and explicit later use, we now recall some basic definitions and notation that will be needed in this paper.
Definition 1
(see [23,24]).
(i)
: A set is called symmetric, if implies for every permutation matrix P.
(ii)
: A function is called symmetric if for every permutation matrix P, for all .
Definition 2
(see [23,24]). Let and .
(i)
: means for all .
(ii)
: Let , φ: is said to be increasing if implies . φ is said to be decreasing if and only if is increasing.
Definition 3
(see [23,24]). Let and .
(i)
: is said to be majorized by (in symbols ) if for and , where and are rearrangements of and in a descending order.
(ii)
: Let , the function φ: is said to be Schur-convex on Ω if on implies φ is said to be a Schur-concave function on Ω if and only if is a Schur-convex function on Ω.
The following useful characterizations of Schur-convex and Schur-concave functions were established in [23,24].
Lemma 1
(see [23,24]). Let be symmetric and have a nonempty interior convex set. is the interior of Ω. is continuous on Ω and differentiable in . Then φ is a - (or -, respectively) if and only if φ is symmetric on Ω andholds for any .
In 1923, Professor Issai Schur made the first systematic study of the functions preserving the ordering of majorization. In Schur’s honor, such functions are said to be “Schur-convex”. It is known that Schur-convexity can be applied extensively in analytic inequalities, combinatorial optimization, quantum physics, information theory, and other related fields (see, e.g., ).
Definition 4
(see [25,26]). Let and .
(i)
: A set is called a geometrically convex set if for all , and α, such that .
(ii)
: Let . The function is said to be Schur-geometrically convex on Ω if on Ω implies . The function φ is said to be a Schur-geometrically concave on Ω if and only if is Schur-geometrically convex on Ω.
Lemma 2.
(Schur-geometrically convex function decision theorem) [25,26] Let be a symmetric and geometrically convex set with a nonempty interior . Let be continuous on Ω and differentiable in . If φ is symmetric on Ω andholds for any , then φ is a Schur-geometrically convex (or Schur-geometrically concave, respectively) function.
The Schur-geometric convexity was first proposed and studied by Zhang in 2004 and was widely investigated and improved by many authors, see [27,28,29] and references therein. We also note that some authors use the term “Schur multiplicative convexity”.
In 2009, Chu [1,2,3] introduced the notion of Schur-harmonically convex function and established some interesting inequalities for Schur-harmonically convex functions.
Definition 5
(see ). Let or .
(i)
: A set Ω is said to be harmonically convex if for every and , where and .
(ii)
: A function is said to be Schur-harmonically convex on Ω if implies . A function φ is said to be a Schur-harmonically concave function on Ω if and only if is a Schur-harmonically convex function.
Lemma 3.
(Schur-harmonically convex function decision theorem) Let or be a symmetric and harmonically convex set with inner points and let be a continuously symmetric function which is differentiable on . Then φ is Schur-harmonically convex (or Schur-harmonically concave, respectively) on Ω if and only if
In 2010, Yang defined and introduced the concepts of the Schur-f-convex function and Schur-power convex function which are the generalization and unification of the concepts of Schur-convexity, Schur-geometric convexity, and Schur-harmonic convexity. He established useful characterizations of Schur m-power convex functions and presented their important properties; see .
Definition 6
(see ). Let be defined by
Then a function is said to be Schur m-power convex on Ω iffor all and implies .
If is Schur m-power convex, then we say that φ is Schur m-power concave.
Lemma 4
(see ). Let be a symmetric set with nonempty interior and be continuous on Ω and differentiable in . Then φ is Schur m-power convex on Ω if and only if φ is symmetric on Ω andandfor all .
For , recall that the complete symmetric function is defined bywhere are non-negative integers.
The collection of complete symmetric functions is an important class of symmetric functions which has been investigated by many mathematicians and there are many interesting results in the literature.
In 2006, Guan discussed the Schur-convexity of and proved the following result.
Proposition 1.
is increasing and Schur-convex on .
Subsequently, Chu et al. established the following proposition.
Proposition 2.
is Schur-geometrically convex and Schur-harmonically convex on .
In 2016, Shi et al. further studied the Schur-convexity of on and presented the following important result.
Proposition 3
(see ). If r is even integer (or odd integer, respectively), then is decreasing and Schur-convex (or increasing and Schur-concave, respectively) on .
Recall that the dual form of the complete symmetric function is defined bywhere are non-negative integers.
In 2013, Zhang and Shi established the following two interesting propositions.
Proposition 4
(see ). For , is increasing and Schur-concave on .
Proposition 5
(see ). For , is Schur-geometrically convex and Schur-harmonically convex on .
Notice that
It is not difficult to prove the following result.
Proposition 6.
If r is even integer (or odd integer, respectively), then is decreasing and Schur-concave (or increasing and Schur-convex, respectively) on .
In 2014, Sun et al. studied the Schur-convexity, Schur-geometric and harmonic convexities of the following composite function of :
By using Lemmas 1–3, they proved the following Theorems 1–3, respectively.
Theorem 1.
For and ,
(i)
: is increasing and Schur-convex on ;
(ii)
: if r is even integer (or odd integer, respectively), then is Schur-convex (or Schur-concave, respectively) on , and is decreasing (or increasing, respectively).
Theorem 2.
For and ,
(i)
: is Schur-geometrically convex on ;
(ii)
: if r is even integer (or odd integer, respectively), then is Schur-geometrically convex (or Schur-geometrically concave, respectively) on .
Theorem 3.
For and ,
(i)
: is Schur-harmonically convex on ;
(ii)
: if r is even integer (or odd integer, respectively), then is Schur-harmonically convex (or Schur-harmonically concave, respectively) on .
In 2016, Shi et al. applied the properties of Schur-convex, Schur-geometrically convex, and Schur-harmonically convex functions respectively to give simple proofs of Theorems 1–3.
Recall that the dual form of the function is defined by
A function associated with this function is
In this work, we will establish some important results for the Schur-power convexity of symmetric functions and . As their applications, some new inequalities are obtained in Section 3.
2. Main Results
The following lemmas are very crucial for our main results.
Lemma 5.
Let . For and , we have
Proof.
Sincewe have
This inequality is equivalent to inequality (12). Sincewe obtain
This inequality is equivalent to inequality (13). Sincewe get
This inequality is equivalent to inequality (14). □
Lemma 6.
Let . For and , we have
Proof.
Sincewe get
This inequality is equivalent to inequality (15). Sincewe obtain
This inequality is equivalent to inequality (16). Sincewe have
This inequality is equivalent to inequality (17). □
Now, we establish the following new result for the Schur-power convexity of .
Theorem 4.
Let . If , then is decreasing and Schur m-power convex on .
Proof.
Let . Then
From Proposition 4, we know that is increasing on , but is decreasing on , therefore, the function is decreasing on .
For and , it is easy to prove that is Schur m-power convex on . Now consider the case of . By the symmetry of , without loss of generality, we may assume . So
Then we have
By the same arguments, we getthen, it follows from (19) and (20) thatwherewithandwith
By Lemma 5, it is easy to see that and for , so
By Lemma 4, we prove that is Schur m-Power convex on for . The proof is completed. □
Next, we present some new results for the Schur-power convexity of .
Theorem 5.
Let .
(i)
: is increasing on and Schur-convex on ;
(ii)
: If , then is Schur-m-power convex on ;
(iii)
: For , if r is even integer (or odd integer, respectively), then is Schur-m-power convex (or Schur-m-power concave, respectively) on .
Proof.
(i) Let . Then
From Proposition 4, we know that is increasing on , but is increasing on , therefore, the function is increasing on .
For the case of and , it is easy to prove that is Schur-convex on .
Now consider the case of . By the symmetry of , without loss of generality, we may assume . So
Then we obtain
By the same arguments,whereandwith
Let . Then which implies that is descending on . So that , namely . It is easy to see that and for , so
By Lemma 1, we obtain is Schur-convex on .
(ii) For and , it is easy to prove that is Schur m-power convex on .
Now consider the case of . By the symmetry of , without loss of generality, we may assume . From (22) and (24), we havewherewithandwith
By Lemma 6, it is easy to see that and for , and then
By Lemma 4, we show that is Schur-m power convex on .
(iii) Notice thatand combining with the Schur-power convexity of on (see Theorem 4), we can prove (iii). The proof is completed. □
According to the relationship between the Schur-power convex function and the Schur-convex function, the Schur-geometrically convex function, and the Schur-harmonically function, we can establish the following two corollaries immediately.
Corollary 1.
Let . Then is Schur-convex, Schur-geometrically convex, and Schur-harmonically convex on .
Corollary 2.
Let .
(i)
: The function is Schur-geometrically convex and Schur-harmonically convex on .
(ii)
: If r is even integer (or odd integer, respectively), then is Schur-convex, Schur-geometric convex, and Schur-harmonic convex (or Schur-concave, Schur-geometric concave, and Schur-harmonic concave, respectively) on .
Finally, an open problem arises naturally at the end of this section.
Problem 1.
For , what is the Schur-convexity of ? Is it Schur-convex or Schur-concave, or is it uncertain?
3. Some Applications
It is not difficult to prove the following theorem by applying Corollary 2 and the majorizing relation
Theorem 6.
If and , or r is even integer and , thenwhere and .
If r is odd and , then the inequality (26) is reversed.
By Corollary 2 and the majorizing relationwe can establish the following result.
Theorem 7.
If and or r is even integer , thenwhere and .
If r is odd integer and , then the inequality (27) is reversed.
By using Corollary 2 and the majorizing relationwe obtain the following theorem.
Theorem 8.
If and , or r is even integer and , thenwhere and .
If r is odd and , then the inequality is reversed.
4. Conclusions
In this paper, we establish the following two important main results of this paper for the Schur-power convexity of symmetric functions and :
(see Theorem 4) Let . If , then is decreasing and Schur m-power convex on .
(see Theorem 5) Let .
(i)
: is increasing on and Schur-convex on ;
(ii)
: If , then is Schur-m-power convex on ;
(iii)
: For , if r is even integer (or odd integer, respectively), then is Schur-m-power convex (or Schur-m-power concave, respectively) on .
As applications of our new results, some new inequalities are presented in Section 3.
Author Contributions
Both authors contributed equally to this work. Both authors read and approved the final manuscript.
Funding
The second author is supported by Grant No. MOST 107-2115-M-017-004-MY2 of the Ministry of Science and Technology of the Republic of China.
Acknowledgments
The authors wish to express their hearty thanks to the anonymous referees for their valuable suggestions and comments.
Conflicts of Interest
The authors declare no conflict of interest.
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Leaf nodes from Preorder of a Binary Search Tree
Last Updated : 23 Jul, 2025
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GivenPreorder traversalof a Binary Search Tree. Then the task is to printleaf nodesof the Binary Search Tree from the given preorder.
Examples:
Input:preorder[] = [4, 2, 1, 3, 6, 5]
Output: [1, 3, 5]
Explaination: 1, 3 and 5 are the leaf nodes as shown in the figure.
Input:preorder[] = [5, 2, 10]
Output: [2, 10]
Explaination: 2 and 10 are the leaf nodes as shown in the figure.
Input:preorder[] = [8, 2, 5, 10, 12]
Output: [5, 12]
Explaination: 5 and 12 are the leaf nodes as shown in the figure.
Try it on GfG Practice
Table of Content
[Expected Approach - 1] Using Inorder and Preorder - O(nlogn) Time and O(n) Space
[Expected Approach - 2] Using Stack - O(n) Time and O(n) Space
[Expected Approach - 1] Using Inorder and Preorder - O(nlogn) Time and O(n) Space
The idea is to find Inorder by sorting the input preorder traversal, then traverse the tree in preorderfashion (using both inorderand preordertraversals) and while traversing storethe leaf nodes.
How to traverse in preorder fashion using two arrays representing inorder and preorder traversals?
We iterate the preorder array and for each element find that element in the inorder array. For searching, we can usebinary search, since inorder traversal of the binary search tree is always sorted. Now, for each element of preorder array, in binary search, we set the range [l, r].
And when l== r, the leaf node is found. So, initially,l = 0 and l = n - 1 for first element (i.e root) of preorder array. Now, to search for the element on the left subtree of the root, set l= 0 and r= index of root - 1. Also, for all element of right subtree set l= index of root + 1 and r = n -1.Recursively, follow this, until l== r.
C++
```cpp
// C++ program to print leaf node from preorder of
// binary search tree using inorder + preorder
include
using namespace std;
int binarySearch(vector& inorder, int l,
int r, int d) {
int mid = (l + r) >> 1;
if (inorder[mid] == d)
return mid;
else if (inorder[mid] > d)
return binarySearch(inorder, l, mid - 1, d);
else
return binarySearch(inorder, mid + 1, r, d);
}
// Function to find Leaf Nodes by doing preorder
// traversal of tree using preorder and inorder vectors.
void leafNodesRec(vector& preorder, vector& inorder,
int l, int r, int& ind, vector& leaves) {
// If l == r, therefore no right or left subtree.
// So, it must be leaf Node, store it.
if (l == r) {
leaves.push_back(inorder[l]);
ind++;
return;
}
// If array is out of bound, return.
if (l < 0 || l > r || r >= inorder.size())
return;
// Finding the index of preorder element
// in inorder vector using binary search.
int loc = binarySearch(inorder, l, r, preorder[ind]);
ind++;
// Finding on the left subtree.
leafNodesRec(preorder, inorder, l, loc - 1,
ind, leaves);
// Finding on the right subtree.
leafNodesRec(preorder, inorder, loc + 1, r,
ind, leaves);
}
// Finds leaf nodes from given preorder traversal.
vector leafNodes(vector& preorder) {
int n = preorder.size();
vector inorder(n);
vector leaves;
// Copy the preorder into another vector.
for (int i = 0; i < n; i++)
inorder[i] = preorder[i];
// Finding the inorder by sorting the vector.
sort(inorder.begin(), inorder.end());
// Point to the index in preorder.
int ind = 0;
// Find the Leaf Nodes.
leafNodesRec(preorder, inorder, 0, n - 1,
ind, leaves);
return leaves;
}
int main() {
vector<int> preorder = {4, 2, 1, 3, 6, 5};
vector<int> result = leafNodes(preorder);
for (int val : result) {
cout << val << " ";
}
return 0;
}
```
// C++ program to print leaf node from preorder of
// binary search tree using inorder + preorder#include using namespace std;int binarySearch(vector& inorder, int l, int r, int d) { int mid = (l + r) >> 1; if (inorder[mid] == d) return mid; else if (inorder[mid] > d) return binarySearch(inorder, l, mid - 1, d); else return binarySearch(inorder, mid + 1, r, d);}// Function to find Leaf Nodes by doing preorder// traversal of tree using preorder and inorder vectors.void leafNodesRec(vector& preorder, vector& inorder, int l, int r, int& ind, vector& leaves) { // If l == r, therefore no right or left subtree. // So, it must be leaf Node, store it. if (l == r) { leaves.push_back(inorder[l]); ind++; return; } // If array is out of bound, return. if (l < 0 || l > r || r >= inorder.size()) return; // Finding the index of preorder element // in inorder vector using binary search. int loc = binarySearch(inorder, l, r, preorder[ind]); ind++; // Finding on the left subtree. leafNodesRec(preorder, inorder, l, loc - 1, ind, leaves); // Finding on the right subtree. leafNodesRec(preorder, inorder, loc + 1, r, ind, leaves);}// Finds leaf nodes from given preorder traversal.vector leafNodes(vector& preorder) { int n = preorder.size(); vector inorder(n); vector leaves; // Copy the preorder into another vector. for (int i = 0; i < n; i++) inorder[i] = preorder[i]; // Finding the inorder by sorting the vector. sort(inorder.begin(), inorder.end()); // Point to the index in preorder. int ind = 0; // Find the Leaf Nodes. leafNodesRec(preorder, inorder, 0, n - 1, ind, leaves); return leaves;}int main() { vector preorder = {4, 2, 1, 3, 6, 5}; vector result = leafNodes(preorder); for (int val : result) { cout << val << " "; } return 0;}
Java
```java
import java.util.ArrayList;
import java.util.Arrays;
class GfG {
static int binarySearch(ArrayList<Integer> inorder, int l, int r, int d) {
int mid = (l + r) / 2;
if (inorder.get(mid) == d)
return mid;
else if (inorder.get(mid) > d)
return binarySearch(inorder, l, mid - 1, d);
else
return binarySearch(inorder, mid + 1, r, d);
}
static void leafNodesRec(int[] preorder, ArrayList<Integer> inorder,
int l, int r, int[] ind, ArrayList<Integer> leaves) {
if (l == r) {
leaves.add(inorder.get(l));
ind++;
return;
}
if (l < 0 || l > r || r >= inorder.size())
return;
int loc = binarySearch(inorder, l, r, preorder[ind]);
ind++;
leafNodesRec(preorder, inorder, l, loc - 1, ind, leaves);
leafNodesRec(preorder, inorder, loc + 1, r, ind, leaves);
}
static ArrayList<Integer> leafNodes(int[] preorder) {
int n = preorder.length;
ArrayList<Integer> inorder = new ArrayList<>();
for (int val : preorder)
inorder.add(val);
ArrayList<Integer> leaves = new ArrayList<>();
inorder.sort(null); // sort inorder
int[] ind = {0};
leafNodesRec(preorder, inorder, 0, n - 1, ind, leaves);
return leaves;
}
public static void main(String[] args) {
int[] preorder = {4, 2, 1, 3, 6, 5};
ArrayList<Integer> result = leafNodes(preorder);
for (int val : result) {
System.out.print(val + " ");
}
}
}
Pythonpython3
Python program to print leaf nodes from preorder of
binary search tree using inorder + preorder
def binarySearch(inorder, l, r, d):
mid = (l + r) // 2
if inorder[mid] == d:
return mid
elif inorder[mid] > d:
return binarySearch(inorder, l, mid - 1, d)
else:
return binarySearch(inorder, mid + 1, r, d)
Function to find Leaf Nodes by doing preorder
traversal of tree using preorder and inorder lists.
def leafNodesRec(preorder, inorder, l, r, ind, leaves):
# If l == r, therefore no right or left subtree.
# So, it must be leaf Node, store it.
if l == r:
leaves.append(inorder[l])
ind += 1
return
# If array is out of bounds, return.
if l < 0 or l > r or r >= len(inorder):
return
# Finding the index of preorder element in
# inorder list using binary search.
loc = binarySearch(inorder, l, r, preorder[ind])
ind += 1
# Finding on the left subtree.
leafNodesRec(preorder, inorder, l, loc - 1, ind, leaves)
# Finding on the right subtree.
leafNodesRec(preorder, inorder, loc + 1, r, ind, leaves)
Finds leaf nodes from given preorder traversal.
def leafNodes(preorder):
n = len(preorder)
inorder = sorted(preorder)
leaves = []
# Point to the index in preorder.
ind =
# Find the Leaf Nodes.
leafNodesRec(preorder, inorder, 0, n - 1, ind, leaves)
return leaves
if name == "main":
preorder = [4, 2, 1, 3, 6, 5]
result = leafNodes(preorder)
for val in result:
print(val, end=" ")
C#csharp
using System;
using System.Collections.Generic;
class GfG {
static int BinarySearch(List<int> inorder, int l, int r, int d) {
int mid = (l + r) / 2;
if (inorder[mid] == d)
return mid;
else if (inorder[mid] > d)
return BinarySearch(inorder, l, mid - 1, d);
else
return BinarySearch(inorder, mid + 1, r, d);
}
static void LeafNodesRec(int[] preorder, List<int> inorder, int l,
int r, int[] ind, List<int> leaves) {
if (l == r) {
leaves.Add(inorder[l]);
ind++;
return;
}
if (l < 0 || l > r || r >= inorder.Count)
return;
int loc = BinarySearch(inorder, l, r, preorder[ind]);
ind++;
LeafNodesRec(preorder, inorder, l, loc - 1, ind, leaves);
LeafNodesRec(preorder, inorder, loc + 1, r, ind, leaves);
}
static List<int> leafNodes(int[] preorder) {
int n = preorder.Length;
// Convert array to List and sort for inorder
List<int> inorder = new List<int>(preorder);
inorder.Sort();
List<int> leaves = new List<int>();
int[] ind = new int { 0 };
LeafNodesRec(preorder, inorder, 0, n - 1, ind, leaves);
return leaves;
}
static void Main() {
int[] preorder = {4, 2, 1, 3, 6, 5};
List<int> result = leafNodes(preorder);
foreach (int val in result) {
Console.Write(val + " ");
}
}
}
JavaScriptjavascript
// JavaScript program to print leaf nodes from preorder of
// binary search tree using inorder + preorder
function binarySearch(inorder, l, r, d) {
const mid = Math.floor((l + r) / 2);
if (inorder[mid] === d) {
return mid;
} else if (inorder[mid] > d) {
return binarySearch(inorder, l, mid - 1, d);
} else {
return binarySearch(inorder, mid + 1, r, d);
}
}
// Function to find Leaf Nodes by doing preorder
// traversal of tree using preorder and inorder arrays.
function leafNodesRec(preorder, inorder, l, r, ind, leaves) {
// If l == r, therefore no right or left subtree.
// So, it must be a leaf Node, store it.
if (l === r) {
leaves.push(inorder[l]);
ind++;
return;
}
// If array is out of bounds, return.
if (l < 0 || l > r || r >= inorder.length) {
return;
}
// Finding the index of preorder element
// in inorder array using binary search.
const loc = binarySearch(inorder, l, r, preorder[ind]);
ind++;
// Finding on the left subtree.
leafNodesRec(preorder, inorder, l, loc - 1, ind, leaves);
// Finding on the right subtree.
leafNodesRec(preorder, inorder, loc + 1, r, ind, leaves);
}
// Finds leaf nodes from given preorder traversal.
function leafNodes(preorder) {
const n = preorder.length;
const inorder = [...preorder];
inorder.sort((a, b) => a - b);
const leaves = [];
// Point to the index in preorder.
const ind = ;
// Find the Leaf Nodes.
leafNodesRec(preorder, inorder, 0, n - 1, ind, leaves);
return leaves;
}
// Driver Code
const preorder = [4, 2, 1, 3, 6, 5];
const result = leafNodes(preorder);
console.log(result.join(' '));
```
Output1 3 5
[Expected Approach - 2] Using Stack - O(n) Time and O(n) Space
The idea is to use the property of theBinary Search Treeand stack.Traverse the array using two pointer iand jto the array, initially i = 0and j = 1. Whenevera[i] > a[j], we can say a[j] is left partof a[i], since preorder traversal follows Node -> Left -> Right. So, we push a[i]into the stack.
For those points violatingthe rule, we start to popelement from the stack till a[i] > top element of the stack and break when it doesn't and print the corresponding jth value.
How does this algorithm work?
Preorder traversal traverse in the order: Node, Left, Right.
And we know theleft node of any node in BSTis always lessthan the node. So preorder traversal will first traverse from rootto leftmostnode. Therefore, preorder will be in decreasingorder first. Now, after decreasingorder, there may be a node that is greateror which breaks the decreasingorder. So, there can be a case like this :
In case 1,20 is a leaf node whereas in case 2, 20 is not the leaf node.
So, our problem is how to identify if we have to print 20 as a leaf node or not?
This is solved using stack.
While running above algorithm on case 1 and case 2, when i = 2 and j = 3, state of a stack will be the same in both the case:
So, node 65 will pop 20 and 50 from the stack. This is because 65 is the right child of a node which is before 20. This information we store using the found variable. So, 20 is a leaf node.
While in case 2, 40 will not able to pop any element from the stack. Because 40 is the right node of a node which is after 20. So, 20 is not a leaf node.
Note:In the algorithm, we will not be able to check the condition of the leaf node of the rightmost node or rightmost element of the preorder. So, simply print the rightmost node because we know this will always be a leaf node in preorder traversal.
C++
```cpp
// C++ program to print leaf nodes from the preorder
// traversal of a Binary Search Tree using a stack
include
include
include
using namespace std;
// Function to find leaf nodes from the
// preorder traversal
vector leafNodes(vector& preorder) {
stack s;
vector leaves;
int n = preorder.size();
// Iterate through the preorder vector
for (int i = 0, j = 1; j < n; i++, j++) {
bool found = false;
// Push current node if it's greater than
// the next node
if (preorder[i] > preorder[j]) {
s.push(preorder[i]);
}
else {
// Pop elements from stack until current node is
// less than or equal to top of stack
while (!s.empty()) {
if (preorder[j] > s.top()) {
s.pop();
found = true;
}
else {
break;
}
}
}
// If a leaf node is found, add it
// to the leaves vector
if (found) {
leaves.push_back(preorder[i]);
}
}
// Since the rightmost element
// is always a leaf node
leaves.push_back(preorder[n - 1]);
return leaves;
}
int main() {
vector<int> preorder = {4, 2, 1, 3, 6, 5};
vector<int> result = leafNodes(preorder);
for (int val : result) {
cout << val << " ";
}
return 0;
}
```
// C++ program to print leaf nodes from the preorder
// traversal of a Binary Search Tree using a stack#include #include #include using namespace std;// Function to find leaf nodes from the // preorder traversal vector leafNodes(vector& preorder) { stack s; vector leaves; int n = preorder.size(); // Iterate through the preorder vector for (int i = 0, j = 1; j < n; i++, j++) { bool found = false; // Push current node if it's greater than // the next node if (preorder[i] > preorder[j]) { s.push(preorder[i]); } else { // Pop elements from stack until current node is // less than or equal to top of stack while (!s.empty()) { if (preorder[j] > s.top()) { s.pop(); found = true; } else { break; } } } // If a leaf node is found, add it // to the leaves vector if (found) { leaves.push_back(preorder[i]); } } // Since the rightmost element // is always a leaf node leaves.push_back(preorder[n - 1]); return leaves; }int main() { vector preorder = {4, 2, 1, 3, 6, 5}; vector result = leafNodes(preorder); for (int val : result) { cout << val << " "; } return 0;}
Java
```java
import java.util.ArrayList;
import java.util.Stack;
import java.util.Arrays;
class GfG {
// Function to find leaf nodes from the
// preorder traversal
static ArrayList<Integer> leafNodes(int[] preorder) {
Stack<Integer> s = new Stack<>();
ArrayList<Integer> leaves = new ArrayList<>();
int n = preorder.length;
// Iterate through the preorder array
for (int i = 0, j = 1; j < n; i++, j++) {
boolean found = false;
// Push current node if it's greater than the next
if (preorder[i] > preorder[j]) {
s.push(preorder[i]);
} else {
// Pop elements from stack until the next value is
// less than or equal to top
while (!s.isEmpty()) {
if (preorder[j] > s.peek()) {
s.pop();
found = true;
} else {
break;
}
}
}
// If a leaf node is found, add it to result
if (found) {
leaves.add(preorder[i]);
}
}
// Add the last element, it's always a leaf in BST preorder
leaves.add(preorder[n - 1]);
return leaves;
}
public static void main(String[] args) {
int[] preorder = {4, 2, 1, 3, 6, 5};
ArrayList<Integer> result = leafNodes(preorder);
for (int val : result) {
System.out.print(val + " ");
}
}
}
Pythonpython3
Python program to print leaf nodes from the preorder
traversal of a Binary Search Tree using a stack
def leafNodes(preorder):
s = []
leaves = []
n = len(preorder)
# Iterate through the preorder list
for i in range(n - 1):
found = False
# Push current node if it's greater
# than the next node
if preorder[i] > preorder[i + 1]:
s.append(preorder[i])
else:
# Pop elements from stack until current node is
# less than or equal to top of stack
while s and preorder[i + 1] > s[-1]:
s.pop()
found = True
# If a leaf node is found, add it to the
# leaves list
if found:
leaves.append(preorder[i])
# Since the rightmost element is always
# a leaf node
leaves.append(preorder[-1])
return leaves
if name == "main":
preorder = [4, 2, 1, 3, 6, 5]
result = leafNodes(preorder)
print(" ".join(map(str, result)))
C#csharp
using System;
using System.Collections.Generic;
class GfG {
// Function to find leaf nodes from the preorder traversal
static List<int> leafNodes(int[] preorder) {
Stack<int> s = new Stack<int>();
List<int> leaves = new List<int>();
int n = preorder.Length;
// Iterate through the preorder array
for (int i = 0, j = 1; j < n; i++, j++) {
bool found = false;
// Push current node if it's greater than the next node
if (preorder[i] > preorder[j]) {
s.Push(preorder[i]);
} else {
// Pop elements until next is not greater than top
while (s.Count > 0) {
if (preorder[j] > s.Peek()) {
s.Pop();
found = true;
} else {
break;
}
}
}
if (found) {
leaves.Add(preorder[i]);
}
}
// Last node is always a leaf
leaves.Add(preorder[n - 1]);
return leaves;
}
static void Main(string[] args) {
int[] preorder = {4, 2, 1, 3, 6, 5};
List<int> result = leafNodes(preorder);
foreach (int val in result) {
Console.Write(val + " ");
}
}
}
JavaScriptjavascript
// JavaScript program to print leaf nodes from the preorder
// traversal of a Binary Search Tree using a stack
function leafNodes(preorder) {
let s = [];
let leaves = [];
let n = preorder.length;
// Iterate through the preorder list
for (let i = 0, j = 1; j < n; i++, j++) {
let found = false;
// Push current node if it's greater
// than the next node
if (preorder[i] > preorder[j]) {
s.push(preorder[i]);
}
else {
// Pop elements from stack until current node is
// less than or equal to top of stack
while (s.length > 0 && preorder[j] > s[s.length - 1]) {
s.pop();
found = true;
}
}
// If a leaf node is found, add it to the leaves array
if (found) {
leaves.push(preorder[i]);
}
}
// Since the rightmost element is always a leaf node
leaves.push(preorder[n - 1]);
return leaves;
}
// Driver Code
let preorder = [4, 2, 1, 3, 6, 5];
let result = leafNodes(preorder);
console.log(result.join(' '));
```
Output1 3 5
Related articles:
Leaf nodes from Preorder of a Binary Search Tree (Using Recursion)
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Guyton Phys 65: Digestion And Absorption In The Gi Tract
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| | | |
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| | Most carbohydrates of the diet are… | large polysaccrides or disaccharides |
| | Process of separating poly/disaccharides | hydrolysis |
| | Most fats in diet are… | triglyccerides |
| | What are triglycerides composed of | glycerol plus fatty acids |
| | 3 major sources of carbohydrates | sucrose, lactose, starches (others: amylose, glycogen, alcohol, lactic acid, pyruvic acid, pectins, dextrins, carbohydrate derivatives in measts) |
| | What enzymes does saliva contain | ptyalin (alpha-amalase secreted by parotid glands); |
| | No more than what percent of food is hydrolyzed in the mouth | 5% |
| | How long can food continue digesting in stomach before mixing with gastric secretions | 1 hour |
| | What pH does salivary amalase become inactive | 4 |
| | how much starch is hydrolyzed before mixing with gastric secretions | as much as 30-40% |
| | How long does it take for starches to be completely hydrolyze once entering duodenum | 15-30 minutes (pancreatic amalase is several times more powerful) |
| | What enzymes do the enterocytes lining villi of small intestine contain? | lactase, sucrase, maltase, and alpha-dextrinase |
| | What makes up lactose | glucose and galactose |
| | what makes up sucrose | fructose and glucose |
| | How are monosaccrides absorbed | fdirectly into portal blood |
| | monosaccride breakdown percents once digested | glucose 80%, galactose and fructose each around 10% |
| | What secrets HCl | parietal (oxyntic) cells (pH of 0.8 when first secreted, then 2-3 when mixed in stomach) |
| | Major constituent of intercellular CT of meat | collagen; broken down by pepsin (must be broken down before other proteins an penetrate meat) |
| | Where does most digestion occur | upper small intestine in duodenum and jejunum |
| | pancreatic enzymes that act on chyme entering duodenum | trypsin, chymotrypsin, carboxypeptidase, proelastase |
| | what does carboxypeptidase do | cleaves individual aas from carboxyl ends of polypeptides |
| | What do trypsin and chymotrypsin do | split protein molecules into small polypeptides |
| | Most proteins get digested down to… | dipeptides and tripeptides |
| | where are peptidases located | brush border microvilli membranes |
| | Two types of peptidase enzymes that are especially important | 1) aminopolypeptidase 2) several dipeptidases (split remaining polypepetides into di and tri pieces) |
| | Where are di and tripeptides transported | through microvillar membrane into interior of enterocyte |
| | Where are most peptides split into individual aas | in cystol; they pass into blood once formed |
| | Fat digestion in stomach | small amount by lingual lipase (from saliva); less than 10% fat digestion |
| | Where does emulsification begin | in stomach with agitation to mix food/fat with products of digestion |
| | Most emulsification occurs where | in duodenum with bile (bile salts plus lecithin) |
| | What is the purpose of bile | emusification; reduces interfacial tension of fat and makes it soluble |
| | average diameter of fat after emulsification | ~1 um; 1000-fold increase in SA |
| | Most important enzymes for digesting triglycerides | pancreatic lipase (enough released to digest everything it can reach in 1 minute); enterocytes also have lipase, not usually needed |
| | What transports fatty acids to prevent reversal of digestion | bile salts (remove monoglycerides and free fatty acids) |
| | Bile salt micelle specs | 3-6 um with 20-40 molecules; ferries |
| | Cholesterol esters | combination of cholesterol and one fatty acid molecule; most common dietary cholesterol |
| | quantity of fluid absorbed by intestines each day | 1.5 L from ingested fluid and 7 L of GI secretions |
| | How much fluid is absorbed in small intestines | all but 1.5 L |
| | What substances can be absorbed through the stomach | high lipid-soluble substances like alcohol and aspirin in small quantities |
| | folds that increase the suface area in small intestine | valvulae conniventes; increase SA 3 fold |
| | where are the folds most developed | duodenum and jejunum; protrude as much as 8 mm into lumen |
| | How far do villi project from suface of mucosa | ~1 mm; increase SA 10 fold |
| | how much do microvilli increase SA | at least 20 fold |
| | Organization of villus | 1) vascular system to absorb 2) central lacteal |
| | daily absorption from small intestine | several hundred grams carbs, 100+ grams fat, 50-100 grams aas, 50-100 grams ions, 7-8 L water |
| | absorption capacity of small intestine per day | serveral kg of carbs, 500 g fats, 500-700 g proteins, 20+ L water |
| | What can the large intestine absorb | additional water, ions, very few nutrients |
| | How is water transported | entirely through diffusion |
| | Sodium absorption/use in body | 20-30 g secreted into intestine, 5-8 g eaten; must absorb 25-30 g (1/7 of all body sodium) |
| | What percent of intestinal sodium passes into feces | less than 0.5%; important in absorbing sugars and aas |
| | Na+ in intestinal epithelial cells | 1) transported actively into paracellular spaces 2)Cl- atoms follow Na+ 3) Na+ concentration in cells low (50mEq/L) 4)Na+ concentration in chyme ~142 mEq/L 5) Na+ follows gradient into cells |
| | How does aldosterone affect Na+ | within 1-3 hours of release, increases activation of enzyme and transport mechanisms for all aspects of Na+ absorption in intestinal epithelium |
| | Increased Na+ absorption causes… | secondary increase in Cl- and water absorption (plus some other substances) |
| | Cl- ion absoprtion | rapidly absorbed in upper small intestine; fixes the electronegativity in chyme/electropositivity in paracellular spaces created by Na+ |
| | Bicarb absorption in small intestine | 1) when Na+ absorbed, moderate amount of H+ secreted into lumen 2) H+ combine with bicarb to form H2CO3 3) dissociated into water and CO2 4) water stays in chyme, CO2 absorbed directly into blood and expired in lungs |
| | Bicarb secretion in ileum and large intestine | epithelial cells secrete bicarb in exchange for Cl-; neutralizes acid products formed by bacteria |
| | Flow of water in large intestines | young/immature epithelial cells secrete NaCl into lumen which is reabsorbed by older/mature epithelial cells |
| | How does cholera and other diarrheal bacteria cause their effect | stimulate fold secretion/young epithelial cells so greatly, old cells can not absorb quickly enough |
| | What specifically causes oversecretion with cholera | formation of excess cyclic adenosine monophosphate, which opens large numbers of Cl- channels |
| | How are Calcium ions absorbed | actively by duodenum; controlled by PTH and vit D |
| | Monovalent vs divalent ion absorption | monovalent ions absorbed with ease in great quantities, divalent absorbed in small amounts |
| | Glucose absorption | sodium co-transport mechanism (secondary active transport); facilitated diffusion into blood from epithelial cells |
| | Galactose absoption | almost same as glucose (secondary active transport with Na+) |
| | Fructose absorption | facilitated diffusion all the way through intestinal epithelium; transport rate 1/2 of glucose and galactose |
| | peptide/aa absorption | most through Na+ co-transport; a few only use facilitated diffusion; at least 5 types of transport proteins for aa and peptides have been found |
| | lipid absorption | transported by micelles and are immediately absorbed into epithelial cell membrane |
| | How much fat can be absorbed without/with bile micelles | 40-50 % versus 97% |
| | What happens to lipids once in epitheial cytoplasm | taken up by smooth ER and used to make new triglycerides which are released as chylomicrons into lymph and go to thoracic duct |
| | Why so some lipids absorb directly into portal blood | short and some medium chain fatty acids are water soluble and absorbed directly |
| | How much chyme enters colon each day | 1500 mL; most absorption in proximal 1/2 of colon |
| | tight jxns in large vs small intestine | much tighter in large intestine; prevents back-diffusion of ions |
| | What causes osmotic gradient for water absorption in large intestine | absorption od Na+ and Cl- ions |
| | max absorption of colon | 5-8 L of fluid and electrolytes/day |
| | Fxn of bacteria in colon | 1) digest some cellulose 2) vit K, vit B12, thiamine, riboflavin production 3) flatus (CO2, H2, and methane) |
| | composition of feces | 3/4 water 1/4 solid matter |
| | Solid matter composition of feces | 30% dead baccteria, 10-20% fat, 10-20 organic matter, 2-3% protein, 30% undigested roughage |
| | What causes brown color of feces | stercobilin and urobilin (derivatives of bilirubin) |
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Product of two negative numbers is positive [duplicate]
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$\begingroup$
What is the practical proof for $-1(-1)=+1$. Actually multiplication is repetitive addition. I am struggling how can I provide an activity to prove practically $-1(-1)=+1$
elementary-number-theory
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asked Jul 26, 2014 at 16:32
Achari S GaneshaAchari S Ganesha
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$\begingroup$ It's not a matter of proof -- but a convention that happens to make the laws of arithmetic hold with greater generality than other possible conventions for the product of negative numbers would. $\endgroup$
hmakholm left over Monica
– hmakholm left over Monica
2014-07-26 16:35:48 +00:00
Commented Jul 26, 2014 at 16:35
$\begingroup$ I'm not sure why people keep claiming that multiplication is repeated addiction. Or exponentiation is repeated multiplication. All of these operation are only defined that way on natural number, but there is no reasons to expect that they are still defined the same way once it is extended, nor even the same property. See also: exponentiation with exponent being matrices; fractional derivative. $\endgroup$
Gina
– Gina
2014-07-26 16:41:09 +00:00
Commented Jul 26, 2014 at 16:41
1
$\begingroup$ Again, it is hard to find an "intuitive" insight. We can start with multiplication as repeted addition. Then, move on to multiplication of a negative number for a positive one: thus, $5 \times (-2)=-10$ becuse we add five time the quantity $-2$, and this is "obviously" a negative quantity (we move to the left with reference to $0$). Finally, we must think to mult for $-1$ as reversing the direction. Thus $(-5) \times (-2) = 10$ becuse it is : $(-1) \times 5 \times (-2) = (-1) \times (-10)$. $\endgroup$
Mauro ALLEGRANZA
– Mauro ALLEGRANZA
2014-07-26 19:06:19 +00:00
Commented Jul 26, 2014 at 19:06
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3 Answers 3
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$\begingroup$
Law of Signs proof: $\rm\,\ (-x)(-y) = (-x)(-y) + x(\overbrace{-y + y}^{\large =\,0}) = (\overbrace{-x+x}^{\large =\,0})(-y) + xy = xy$
Equivalently, evaluate $\rm\:\overline{(-x)(-y) +} \overline{ \underline {x(-y)}} \underline{ +\,xy}\, $ in 2 ways, noting each over/under term $ = 0$
Said more conceptually, $\rm\:(-x)(-y)\ $ and $\rm\:xy\:$ are both inverses of $\rm\ x(-y)\ $ so they are equal by uniqueness of inverses: if $\,a\,$ has two additive inverses $\,\color{#c00}{-a}\,$ and $\,\color{#0a0}{-a},\,$ then
$$\color{#c00}{-a}\, =\, \color{#c00}{-a}+\overbrace{(a+\color{#0a0}{-a})}^{\large =\,0}\, =\, \overbrace{(\color{#c00}{-a}+a)}^{\large =\,0}+\color{#0a0}{-a}\, =\, \color{#0a0}{-a} $$
This proof of the Law of Signs uses well-known laws of positive integers (esp. the distributive law), so if we require that these laws persist in the larger system of positive and negative integers, then the Law of Signs is a logical consequence of these basic laws of positive integers.
These fundamental laws of "numbers" are axiomatized by the algebraic structure known as a ring, and various specializations thereof. Since the above proof uses only ring laws (most notably the distributive law), the Law of Signs holds true in every ring, e.g. rings of polynomials, power series, matrices, differential operators, etc. In fact every nontrivial ring theorem (i.e. one that does not degenerate to a theorem about the underlying additive group or multiplicative monoid), must employ the distributive law, since that is the only law that connects the additive and multiplicative structures that combine to form the ring structure. Without the distributive law a ring degenerates to a set with two completely unrelated additive and multiplicative structures. So, in a sense, the distributive law is a keystone of the ring structure.
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edited May 27, 2016 at 17:34
answered Jul 26, 2014 at 16:44
Bill DubuqueBill Dubuque
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$\begingroup$ ¤ï¸ last paragraph $\endgroup$
pseudosudo
– pseudosudo
2019-10-15 19:50:21 +00:00
Commented Oct 15, 2019 at 19:50
$\begingroup$ Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. just kidding $\endgroup$
Clyde Kertzer
– Clyde Kertzer
2023-02-24 23:59:12 +00:00
Commented Feb 24, 2023 at 23:59
$\begingroup$ @Clyde This was posted nine years ago when the site was still in poulation mode and site politics was very different (when many of us believed that SE might provide us with better tools for dealing with site oprganization - but they never did). Most all prolific users will have dupes in the old days. $\endgroup$
Bill Dubuque
– Bill Dubuque
2023-02-25 00:10:47 +00:00
Commented Feb 25, 2023 at 0:10
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$\begingroup$
$$1 + (-1) = 0$$ $$-1(1 + (-1)) = -1(0)$$ $$-1(1) + -1(-1) = 0$$ $$-1 + -1(-1) = 0$$ $$1 + (-1) + -1(-1) = 0 + 1$$ $$0 + -1(-1) = 1$$ $$-1(-1) = 1$$
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edited Jul 26, 2014 at 16:45
Jam
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answered Jul 26, 2014 at 16:41
Paul SundheimPaul Sundheim
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$\endgroup$
1
$\begingroup$ Note that the proof implicitly assumes that the distributive law holds for all integers, not just positive integers. $\endgroup$
Bill Dubuque
– Bill Dubuque
2014-07-26 17:28:31 +00:00
Commented Jul 26, 2014 at 17:28
Add a comment |
3
$\begingroup$
If you assume some basic properties of addition and multiplication:
$$(-1) \times 0 = 0$$ $$(-1) \times ((-1) + 1) = 0$$ $$(-1) \times (-1) + (-1) \times 1 = 0$$ $$(-1) \times (-1) + (-1) = 0$$ $$(-1) \times (-1) + (-1) +1 = 0+1$$ $$(-1) \times (-1) + 0 = 1$$ $$(-1) \times (-1) = 1$$
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answered Jul 26, 2014 at 16:40
HenryHenry
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$\endgroup$
1
1
$\begingroup$ Note that the proof implicitly assumes that the distributive law holds for all integers, not just positive integers. $\endgroup$
Bill Dubuque
– Bill Dubuque
2014-07-26 17:28:07 +00:00
Commented Jul 26, 2014 at 17:28
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A rectangular sheet with one side of 12 cm is folded along its 20 cm diagonal. What is the overlapping area of the folded parts, indicated in gray in the picture below? 24cm^2 36cm^2 48cm^2 75cm^2
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Answer
$$60$$60 cm$$^{2}$$2
Explanation
Diagram Description
A rectangle ADEF is folded along its diagonal AE. The overlapping area of the folded parts is shaded in gray.
Solution Process
Calculate the area of the rectangle.
$$12 \times 20 = 240$$12×20=240
Calculate the area of triangle AEF.
$$\frac{1}{2} \times 12 \times 20 = 120$$2 1×12×20=120
Calculate the area of the gray region.
$$240 - 120 = 120$$240−120=120
Divide the area of the gray region by 2.
$$120 \div 2 = 60$$120÷2=60
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188707 | https://brainly.com/question/2612818 | [FREE] The circumference (C) of a swimming pool is 47 feet. Which formula can you use to find the radius (r) if - brainly.com
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The circumference (C) of a swimming pool is 47 feet. Which formula can you use to find the radius (r) if you know that C=2 π r?
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You don't have to use a formula, all you have to do is use basic algebra to modify the equation in front of you. Simply divide both sides by 2pi and you will end up with C/2pi = r. Then, just type it into your calculator and you're good to go!
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To find the radius of a circle from the circumference, use the formula r=2 π C. Given a circumference of 47 feet, the radius can be calculated as approximately 7.48 feet. This involves simple algebraic manipulation to isolate the radius in the circumference formula.
Explanation
To find the radius (r) of a circle when you know its circumference (C), you can use the formula for circumference:
C=2 π r
Given that the circumference of the swimming pool is 47 feet, we can set up the equation:
47=2 π r
To solve for r, we need to isolate it. Here are the steps:
Divide both sides of the equation by 2 π to isolate r:
r=2 π C
So substituting the given circumference, we have:
r=2 π 47
Now you can use a calculator to determine the value of r. Using an approximate value of π as 3.14 (for easy calculations):
r≈2×3.14 47
r≈6.28 47≈7.48 feet
Thus, the radius of the swimming pool is approximately 7.48 feet.
Examples & Evidence
For example, if you have a different circumference, such as 62 feet, you would calculate the radius similarly: r=2 π 62≈9.87 feet using π≈3.14.
This method of solving for the radius from the circumference is based on the basic principles of geometry, specifically the relationship defined by the circumference formula: C=2 π r. This formula is well established and taught in mathematics.
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To find the radius of a swimming pool with a circumference of 47 feet using the formula C = 2πr, divide the circumference by 2π. The radius is approximately 7.48 feet when rounded to two decimal places.
Explanation
To find the radius (r) of a swimming pool given that the circumference (C) is 47 feet, and you know the formula C = 2πr, you can rearrange this formula to solve for r:
C = 2πr
r = ½ × (C/π)
r = ½ × (47 feet/π)
r = ½ × (47 feet/3.14159)
r = 7.48 feet (rounded to two decimal places)
Therefore, the radius of the swimming pool is approximately 7.48 feet.
Learn more about Calculating Radius here:
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188708 | https://www.youtube.com/watch?v=v3nOOgVCJO4 | Distance, Rate, Time Word Problems (TTP Video 14)
Professor Leonard
1040000 subscribers
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Posted: 1 Sep 2017
How to Solve Some word problems involving Distance, Rate, and Time
34 comments
Transcript:
Well, welcome back. We we continue our uh our our voyage through word problems, if you will. And we get to this this little subsection of distance equals rate times time problem. In every class, every class that you have in beginning algebra, most intermediate algebra classes, you're going to have some distance equals rate times time problem. So, I'm going to give you one example. I know that's not a lot, but that's not the nature of these videos. My my my idea here is to teach you very well through one example that you can run with that. I can't do all of the distance equals rate times times problems, but I can give you like the most common one. And so I'm going to give you that one. It's very similar. I'm going to show you some other setups that we'll talk about. Uh but this is going to illustrate the concept as best as I can. So in distance equals rate times time. Uh the first thing we have to understand is the formula. So imagine you're driving down the road and your speedometer says 60 m hour for you. Probably 60 m hour unless you get a really slow car then 60 m hour. But uh 60 miles an hour and you go for three hours. You don't have any traffic. You're real blessed and you just keep on driving. 60 miles an hour for three hours. How far are you going to go? Well, 60 miles per hour means in every hour you have gone 60 miles. So 60 and then another 60 and another 60 or 3 60 that's 180 miles. That's what distance equals rate times time does for you. the distance is going to equal the rate that you're traveling for the time that you traveled in. Uh it kind of takes for granted that you're going to steady speed. So if we're not going to steady speed, that's outside of the realm of this distance equals rate times time problem. So we have a distance equaling how fast you're going for how long you're going it. Um we can use that. So let's say that we have our example here and you decide to go for a run. So, you're jogging along and you jog 5 m hour for a little while and then you step it up and you jog 8 m hour. In total, you make the loop back to wherever you started from and you've gone 7 miles. It took you 1.1 hours. So, we know a lot already. We know how long we ran. We know how long we ran it for. What we don't know is how long we ran each jog for, each segment. So, I'm going to give you a tool, a graphic organizer to help you with this. Because at first glance, we read through these problems. I don't know who's supposed to know how you guess and check, I guess. Well, that's really slow. It It works sometimes, but it's really slow. And so, I'd like you, and if you're have a teacher like me, you're going to make sure you work anyway, you might as well know how to do the math. So, let's write down everything that we know in a table that organizes our our information for us. So most of the time what we're going to do is we're going to start off and we'll put the scenarios on the left hand side. What's going on here? And then something that relates them up here. We had two jogs. We had let's say jog one, jog two, and then we had a total up at the top. We know that we had and this would be part of the formula aspect of word problems. So, we've read it. We'll probably read it again very carefully in a minute. We have a sort of a picture, a graphic organizer, and a formula going on. This is going to create the verbal model for us. After that, we'll fill in the variables. We'll get our equation. We'll solve it. But this is the hardest part. Anytime you're moving through space and time, you have distance equals rate times time if your rate's continuous. So, we we have that here. So, our distance equals rate times time. Put your scenarios here. How much how many different rates you ran, how many different jogs you did, whatever it was. Jog one, jog two, and how much you totally did. Distance equals rate times time. There's three things that are related to any distance problem. The rate, the time, and the total distance. Write down everything you know. So, we're going to read the problem carefully right now. We're going to write down everything that we know. So, we're running. Okay. Got it. First jog. We jog at 5 miles an hour. That's a rate. So right here in our our little matrix, we have jog one, five miles an hour. And then we run at 8 miles an hour. Let's call that jog two. In total, all right, we got a total. We ran seven miles and we ran for a total of 1.1 hours. How long did Okay, so that's the question. So now we're going to go through and make sure we don't miss anything. We had two jogs. Got two jogs. We had a distance, a rate, and a time. Okay, let's fill that out. Jog one was 5 miles an hour. Jog two was 8 miles an hour. We had a total of 7 miles and we had a total of 1.1 hours. It also helps if you put that this is your rate, your total miles, and your hours just so you know what you're dealing with. Now, the rough part is figuring out what goes here and here. You see, if we can figure out what goes here, this is the big part here. If we can figure out what goes here and here, then these columns really work together. And I'll talk about a little bit after we're done with this example, how you can change this around to to illustrate more examples. Okay? But if you find a rate and a time, you automatically know a distance. If you find a distance and a rate, you automatically know a time. If you find a distance and a time, you automatically know a rate. So given any two variables, you know the third one in this. So for instance, uh let's say that this was it's not going to be 5 hours, 6 hours, but I'm going to make it up right now. Let's say that they ran five or you ran 5 miles an hour for 6 hours. That's 30 miles. If you know this, you can just multiply. That's the key here. multiply these two columns to get that one. So to get that box, distance does equal rate times time. That's great, but we still don't know this. So let's think carefully. Do you know how long this person ran for jog one? Do you know how long the person ran for jog two? You're going to have to pick one of these to be a variable. It doesn't matter. What does matter is that you understand we are in linear equations. And as we discussed in the approach to problem solving, you can only use one variable. So call whatever one you want a variable. It doesn't matter. But in the other box, you have to relate it. Now, here's the key. Okay? Don't miss this. Get this. Do you understand that the amount of the time the person ran here plus the amount of the time the person ran here has to equal 1.1 hours. You get that? If you ran a total of 1.1 hours and you ran a portion of it this speed and a portion of this speed, well, the times that you ran, they got to add up. You can't run for 2 hours plus 5 hours, you get 1.1 hours. It doesn't make sense. Uh, these two have to add up to 1.1. The time you ran here plus the time you ran here has to equal the time you ran there. If you guys get that, this is not hard. Think through this before you go any further. Make sure you understand that these times have to add to total time. Time you ran for one, time you ran for two has to add up to the time you ran in total because if you get that, check this out. It's pretty cool. Call either one of these a variable. You know what I'm I'm going to use? I'm going to use t to stand for time. I don't know what that time is. It doesn't matter which one you choose. What does matter is this. If you because we're going to find this out right now. If you really get that these two have to add up, call this a a blank space like Taylor Swift does. Call it whatever you want. I don't care what you call it. But understand this. This is pretty neat. Understand that t plus blank space if you want equals 1.1. If you get that, that's awesome. You understand the idea. Your times have to add up. T plus I don't know has to add up to 1.1. Solve for that. If you solve for that solve for what do you mean solve for it? Hey, that's an equation. Get this by itself. Remove this one completely. Not divide. Subtract. It's connected by addition. Subtract. If you subtract that t, then we automatically not like terms. We automatically find out what should be in our blank space. What should be in that box. Now, let's think logically through it. If I told you that you ran for 1.1 hours and I said, "Hey, that's uh that's 0.1 hours. To find the missing spot, you'd have to subtract it from 1.1." That's the idea here is saying if you ran for a total, you're going to have some you'd have to take this this something away from the total to figure out what's left. That's the idea. You have a total. You're taking up some of it here. Take it away from the total to figure out what's missing. That's what the math does for you. What's even more well equally interesting is that if you add these up, it does equal 1.1. Try it if you want to. It's almost a trivial statement because it has to work. That's how you created it. T + 1.1 minus T. T minus T is gone. 1.1= 1.1. It does work. That's pretty cool. This is a hard concept for a lot of people. So, if you're not quite getting it, look through it again. Listen to it at least one more time and then try it for yourself. See if you can do this whole blank space idea to get the the missing box. Um, this is this is what makes a lot of these problems possible is taking a total using a variable you create and relating the missing piece back to that variable because we can only use one variable. So, we start with a total. We go, okay, well, this is something I don't know. Uh, this is something I don't know, but I know that these two have to add up. If these two have to add here, then this one minus this one has to have has to be that one. Has to be the missing piece. That's what that math said to us. You're almost done. It didn't look like it, but you're almost done. You see, we have a rate and we have a time. If rate times time equals distance, then we go this way and we can fill out this distance. One more thing before we do, okay, I just want you to get this also. Do do you understand that whatever you ran here, how far you ran here plus how far you ran here is going to equal your total. If you ran three miles here and two four miles here, you'd have to add up to seven miles. That these things have to add up here. So this is two, this has to be five. This is one, this has to be six. These have to add to seven. The amount that you ran in one plus the amount that you ran in two has to equal the total. So whatever we get, we can add this up. That's kind of cool. So, this matrix is starting to hopefully make sense. We fill out all of our information. This is your little 10-second recap. Fill out what you know. Use what's given to you to create some sort of a relationship with one variable. After that, use another column by multiplication or whatever whatever you're given. 5 t is 5 t. 8 times this whole amount is 8 1.1 minus t. Now we know that our distances have to add up. So I'm going to move over here. But this column basically says if that doesn't look like a distance to you, look at the formula. Rate times time is a distance. Five 5 miles an hour for 5 hours going be 25 miles. That is a distance. So is that one. These two distances have to add to your total. So we have our first distance our second distance and they have to add to the total. What's easier that it's a linear equation. It's got a decimal in it. Who cares? It's got some distribution. It's got some like terms. It's pretty easy to solve because you know what you're doing already. This is the hard part. The hardest part is figuring out how to start honestly. Um, and what it does is it it takes what you're what you're asked for. How long did you run and that's that becomes your variable. Relate the missing box or whatever you want to call this miss blank space to that variable. Then use another column to add up. It's usually a multiplication or division concept of distance equals rate times time. So now let's solve it. Uh maybe stop this and solve on your own. So we would distribute. We check for some like terms and we have some. Our variable is right there. Our steps for linear equations say get rid of the constant that's attached to your variable first. And your last step in linear equations is always divide. I'm begging you. I'm begging you. Please don't leave it here. Don't do that. Don't do that to me. Don't do that to your teacher. Explain the problem. The whole reason why we do word problems is to answer real life stuff. So stuff that actually happens. So answer it. This question was how long did you run at each rate? Well, look back at your table. It's why we did the graphic organizer. Look back at what the t represents. The t represents how long we ran at 5 miles hour. Our first jog. Now what if you had reversed that? Put your t here and your 1.1 minus t here. That's okay. the t which you you would have got would be different and it would represent the time for your second jog. It doesn't matter where you put it matters that you follow the process I'm giving you. So our first jog was 6 hours. All right. So we have the first jog at 06. How would you find out the second jog? It's given to you. Take 1.1 minus 6. So our first jog the way we write this out you write a full sentence. Say we jog 5 m an hour for6 hours and 8 mph for.5 hours. How we get the.5? 1.1us.6 is.5. So we joged 5 miles an hour for6 hours and 8 miles an hour for.5 hours. Look how it adds up. 6 plus.5 = 1.1. We use our distances to figure that out. Now, I told you a little while ago that I was going to talk about um when you can use this differently. So, let's say that we we're not asked for a time. We're asked for a distance or we're asked for a rate. So, check this out. You'd still fill out your your table. What? However, whatever information you're given to, you're still filling that out. But we don't always have to go rate time time equals distance. There's two other kind of permutations of this that we can we can deal with. If you solve this for r or t, we get two more formulas we can work with. If you solve for r, basically divide both sides by t, we can get that a rate equals a distance / time. Maybe that makes sense. If you went a 100 miles an hour, sorry, if you went for 180 miles and it took you three hours, you could figure out you went 60 miles an hour. That's what that's talking about. If you solve for time, if you went 180 miles and you were going 60 miles hour, you could figure out that it would take you 3 hours to do that. These are ways that we can use this table to our advantage. You can switch this stuff around. It doesn't matter. Maybe put time equals distance / rate. That's all right. But these are different ways that we can utilize this table. The big idea here, the the hardest thing for students to grasp is this idea, the picking a variable and using your total to get the other box. That's the hardest part. If you get that, then you should be feeling pretty good about yourself. Um, sometimes you you get a little easier one. Sometimes they'll tell you let's say you ran for two hours and then u or two let's say how about this one you ran for I don't know but you know that your other run was two hours longer. So if you don't know what your how long you ran for five miles, but they say, "Hey, in your second run, you ran for two hours longer or two hours more than that's kind of easier. Uh, and then you can you can use that in some of your calculations as well. So the hardest part is taking the total and working backwards from it. If you're given a time and they or a distance or a rate and they say one of them you don't know but the other one is two hours longer or five hours less or it relates it back to that. Use your variable and add or subtract two or from it. That's that's really common as well. So next time we're going to do some mixture problems and mixture problems are really similar to this. So we're going to see the same idea and I'll give you two examples on on how to do those. So hopefully this makes sense for your distance equals rate times time. Try the table. It helps a lot. Work through this problem that the original problem I gave you again before you do anything else. Work through that problem. Make sure you get the t and the subtracting that from the total to get the missing piece. They have to add up. You can subtract to find the missing one. |
188709 | https://www.quora.com/How-do-I-solve-this-limit-lim-x-0-x-sinx-x-3-with-remarkable-limits | How to solve this limit lim x->0. (x- sinx) /x^3 with remarkable limits - Quora
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How do I solve this limit lim x->0. (x- sinx) /x^3 with remarkable limits?
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Carlo
Knows Italian · Author has 1.4K answers and 551.2K answer views
·2y
To solve with remarkable limits means do not use de l'Hôpital or Taylor.
From the goniometry we get
s i n(3 y)=s i n(y+2 y)=s i n(3 y)=s i n(y+2 y)=
s i n(y)c o s(2 y)+c o s(y)s i n(2 y)=s i n(y)c o s(2 y)+c o s(y)s i n(2 y)=
s i n(y)(1–2 s i n 2(y))+2 s i n(y)c o s 2(y)=s i n(y)(1–2 s i n 2(y))+2 s i n(y)c o s 2(y)=
s i n(y)−2 s i n 3(y)+2 s i n(y)−2 s i n 3(y)=3 s i n(y)−4 s i n 3(y)s i n(y)−2 s i n 3(y)+2 s i n(y)−2 s i n 3(y)=3 s i n(y)−4 s i n 3(y)
We return to the limit by making a change of variable.
x = 3y; If x → 0 then y → 0
lim y→0 3 y−s i n(3 y)27 y 3=lim y→0 3 y−s i n(3 y)27 y 3=
as seen before
lim y→0 3 y−3 s i n(y)+4 s i n 3(y)27 y 3=lim y→0 3 y−3 s i n(y)+4 s i n 3(y)27 y 3=
lim y→0 3(y−s i n(y))+4 s i n 3(y)27 y 3 lim y→0 3(y−s i n(y))+4 s i n 3(y)27 y 3
hence
\lim_{y\to 0}\frac{y-sin(y)}{y^3}=\lim_{y\to 0}\frac{3(y-sin(y))+4sin^3(y)}\lim_{y\to 0}\frac{y-sin(y)}{y^3}=\lim_{y\to 0}\frac{3(y-sin(y))+4sin^3(y)}
Continue Reading
To solve with remarkable limits means do not use de l'Hôpital or Taylor.
From the goniometry we get
s i n(3 y)=s i n(y+2 y)=s i n(3 y)=s i n(y+2 y)=
s i n(y)c o s(2 y)+c o s(y)s i n(2 y)=s i n(y)c o s(2 y)+c o s(y)s i n(2 y)=
s i n(y)(1–2 s i n 2(y))+2 s i n(y)c o s 2(y)=s i n(y)(1–2 s i n 2(y))+2 s i n(y)c o s 2(y)=
s i n(y)−2 s i n 3(y)+2 s i n(y)−2 s i n 3(y)=3 s i n(y)−4 s i n 3(y)s i n(y)−2 s i n 3(y)+2 s i n(y)−2 s i n 3(y)=3 s i n(y)−4 s i n 3(y)
We return to the limit by making a change of variable.
x = 3y; If x → 0 then y → 0
lim y→0 3 y−s i n(3 y)27 y 3=lim y→0 3 y−s i n(3 y)27 y 3=
as seen before
lim y→0 3 y−3 s i n(y)+4 s i n 3(y)27 y 3=lim y→0 3 y−3 s i n(y)+4 s i n 3(y)27 y 3=
lim y→0 3(y−s i n(y))+4 s i n 3(y)27 y 3 lim y→0 3(y−s i n(y))+4 s i n 3(y)27 y 3
hence
lim y→0 y−s i n(y)y 3=lim y→0 3(y−s i n(y))+4 s i n 3(y)27 y 3 lim y→0 y−s i n(y)y 3=lim y→0 3(y−s i n(y))+4 s i n 3(y)27 y 3
Let us assume that the given limit converges to L, in this case, using s i n(x)x→1 s i n(x)x→1 as x → 0
L=L 9+4 27 L=L 9+4 27
8 L 9=4 27 8 L 9=4 27
L=1 6 L=1 6
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Alexey Godin
Ph.D. in Mathematics&Economics, Moscow State University (Graduated 1998) · Author has 2.7K answers and 3.9M answer views
·2y
You can solve it using remarkable limits but probably not in the way you imagine.
The derivative of the f(x)=sin x f(x)=sinx is proved using the remarkable limit:
lim x→x 0 sin x−sin x 0 x−x 0=lim x→x 0 sin x−x 0 2 cos x+x 0 2 x−x 0 2=cos x lim x→x 0 sinx−sinx 0 x−x 0=lim x→x 0 sinx−x 0 2 cosx+x 0 2 x−x 0 2=cosx
As cos x=sin(π 2−x)cosx=sin(π 2−x) we get (cos x)′=(sin(π 2−x))′=−cos(π 2−x)=−sin x(cosx)′=(sin(π 2−x))′=−cos(π 2−x)=−sinx
Now you know that f′(x)=cos x,f′′(x)=−sin x,f′′′(x)=−cos x f′(x)=cosx,f″(x)=−sinx,f‴(x)=−cosx
And thus the Taylor polynomial of degree 3 at x 0=0 x 0=0 with the Peano remainder is:
f(x)=f(0)+f′(0)x+f′′(0)2 x 3+f′′′(0)6 x 3+o(x 3)f(x)=f(0)+f′(0)x+f″(0)2 x 3+f‴(0)6 x 3+o(x 3)
i.e.
sin sin
Continue Reading
You can solve it using remarkable limits but probably not in the way you imagine.
The derivative of the f(x)=sin x f(x)=sinx is proved using the remarkable limit:
lim x→x 0 sin x−sin x 0 x−x 0=lim x→x 0 sin x−x 0 2 cos x+x 0 2 x−x 0 2=cos x lim x→x 0 sinx−sinx 0 x−x 0=lim x→x 0 sinx−x 0 2 cosx+x 0 2 x−x 0 2=cosx
As cos x=sin(π 2−x)cosx=sin(π 2−x) we get (cos x)′=(sin(π 2−x))′=−cos(π 2−x)=−sin x(cosx)′=(sin(π 2−x))′=−cos(π 2−x)=−sinx
Now you know that f′(x)=cos x,f′′(x)=−sin x,f′′′(x)=−cos x f′(x)=cosx,f″(x)=−sinx,f‴(x)=−cosx
And thus the Taylor polynomial of degree 3 at x 0=0 x 0=0 with the Peano remainder is:
f(x)=f(0)+f′(0)x+f′′(0)2 x 3+f′′′(0)6 x 3+o(x 3)f(x)=f(0)+f′(0)x+f″(0)2 x 3+f‴(0)6 x 3+o(x 3)
i.e.
sin x=x−x 3 6+o(x 3)sinx=x−x 3 6+o(x 3)
And thus
x−sin x x 3=x−(x−x 3 6+o(x 3)x 3=1 6+o(1)→1 6,x→0 x−sinx x 3=x−(x−x 3 6+o(x 3)x 3=1 6+o(1)→1 6,x→0
P.S.: An addenum for the curious. If we assume somehow that L=lim x→0 x−sin x x 3 L=lim x→0 x−sinx x 3 exists then we can indeed find it only using the remarkable limit and the fact that sin 3 x=3 sin x−sin 3 x 4 sin 3x=3 sinx−sin3 x 4
So
1=(lim x→0 sin x x)=lim x→0 sin 3 x x 3=1=(lim x→0 sinx x)=lim x→0 sin 3x x 3=
=1 4 lim x→0 3 sin x−sin 3 x x 3==1 4 lim x→0 3 sinx−sin3 x x 3=
=1 4 lim x→0(3 sin x−3 x x 3+27 3 x−sin(3 x)(3 x)3)=1 4(−3 L+27 L)=6 L⇒L=1 6=1 4 lim x→0(3 sinx−3 x x 3+27 3 x−sin(3 x)(3 x)3)=1 4(−3 L+27 L)=6 L⇒L=1 6
Upvote ·
9 7
9 5
Gordon M. Brown
Math Tutor at San Diego City College (2018-Present) · Author has 6.2K answers and 4.3M answer views
·2y
I have no clue as to what you mean by "remarkable limits." That said, why not simply apply L'Hôpital's rule for as many times as the indeterminate form 0/0 will both permit, and necessitate?
Continue Reading
I have no clue as to what you mean by "remarkable limits." That said, why not simply apply L'Hôpital's rule for as many times as the indeterminate form 0/0 will both permit, and necessitate?
Upvote ·
9 2
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.3K answers and 1.7M answer views
·2y
lim x→0[ (x - sin(x))/x^3 ] ; L'Hôpital's rule
lim x→0[ (1 - cos(x))/(3x^2) ] ; L'Hôpital's rule
lim x→0[ sin(x)/(6x) ] ; L'Hôpital's rule
lim x→0[ cos(x)/6 ] = cos(0)/6 = 1/6
Here a graph:
Continue Reading
lim x→0[ (x - sin(x))/x^3 ] ; L'Hôpital's rule
lim x→0[ (1 - cos(x))/(3x^2) ] ; L'Hôpital's rule
lim x→0[ sin(x)/(6x) ] ; L'Hôpital's rule
lim x→0[ cos(x)/6 ] = cos(0)/6 = 1/6
Here a graph:
Upvote ·
9 2
9 2
Related questions
More answers below
Is lim x→0(1+s i n x)1/x=e lim x→0(1+s i n x)1/x=e?
What is lim x→0 e x−1−x x 2 lim x→0 e x−1−x x 2?
How do you solve, lim x→4 3−√x+5 x−4 lim x→4 3−x+5 x−4?
How do I solve lim x-->0 {[ (1+sinx) ^1/3] - [(1-sinx) ^1/3]} ÷ x?
What is the limit? As lim x approaches 0 (1/x^3 - sin2x/x^3)
Lukas Schmidinger
I have graduate CS and my studies included math courses. · Author has 27.7K answers and 14.9M answer views
·2y
How do I solve this limit lim x->0. (x- sinx) /x^3 with remarkable limits?
Let’s see how far we get with L'Hôpital's rule, that is 0 0 0 0 after all and all the way through.
lim x→0 x−sin(x)x 3=lim x→0 x−sin(x)x 3=
lim x→0 1−cos(x)3 x 2=lim x→0 1−cos(x)3 x 2=
lim x→0 sin(x)6 x=lim x→0 sin(x)6 x=
lim x→0 cos(x)6=lim x→0 cos(x)6=
cos(0)6=cos(0)6=
1 6 1 6
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B.L. Srivastava
Lives in Kanpur, Uttar Pradesh, India (1972–present) · Author has 7.6K answers and 8.1M answer views
·2y
Simple.
lim(x→0){(x - sinx)/x³}
= lim{(x - (x - x³/3! + x⁵/5! - …. .. ))/x³}
= lim(x→0){(x³/3! + …. .. .)/x³} = 1/3! = 1/6 .
Upvote ·
9 3
9 1
George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·2y
When x= 0 this is 0/0 so use “L’Hopital’s rule. The derivative of x- sin(x) is 1- cos(x) and the derivative of x³ is 3x². (1- cos(x))/3x², at x= 0, is still 0/0 so do it again. The derivative of 1- cos(x) is sin(x) and the derivative of 3x² is 6x. sin(x)/6x, at x= 0, is still 0/0 so do it again. The derivative of sin(x) is cos(x) and the derivative of 6x is 6. cos(x)/6, at x= 0 is 1/6.
lim x→0 x−s i n(x)x 3=1 6 lim x→0 x−s i n(x)x 3=1 6.
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Dale Gray
MA in Mathematics, The University of Texas at Arlington (Graduated 1974) · Upvoted by
BowTangey
, PhD Mathematics, Iowa State University (1988) · Author has 763 answers and 1.5M answer views
·2y
Use L'Hôpital's rule twice. Then use the fact that the limit of a product is the product of the limits, if both limits exist. The Answer is 1/6.
Upvote ·
9 2
P Patel
Studied at GCET (Graduated 2019)
·5y
Related
How do I solve lim x-0 x-sinx/tan^3x, step by step?
Thanks for A2A.
Hope this helps to you.
Continue Reading
Thanks for A2A.
Hope this helps to you.
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BowTangey
PhD in Mathematics, Iowa State University (Graduated 1988) · Author has 1.2K answers and 250.5K answer views
·Updated 2y
I do not know what “with remarkable limits” means.
Using the Maclaurin series for sin(x)sin(x) it is easy to determine the limit to be 1 6 1 6, which is initially surprising, as x−sin(x)x−sin(x) seems to be converging linearly to 0 0 (sin(x x)≈x≈x for small x x), while the denominator is converging in a cubic fashion. On the other hand, the series x−(x−1 6 x 3+x−(x−1 6 x 3+O(x 5))(x 5)) DOES converge to 0 0 like 1 6 x 3 1 6 x 3, so all is clear.
Upvote ·
9 1
David Vanderschel
PhD in Mathematics&Physics, Rice (Houston neighborhood) (Graduated 1970) · Author has 37.6K answers and 50.1M answer views
·2y
A2A: I suggest using the Maclaurin series for sin(x).sin(x).
(I am not familiar with “remarkable limits”.)
Upvote ·
9 2
9 1
Silvio Capobianco
Senior Researcher at Tallinn University of Technology (2009–present) · Author has 3.4K answers and 634.2K answer views
·2y
You could do one step of de l’Hôpital and reduce to the limit of (1 - cos x) / x^2, which you should know.
You could use the Maclaurin expansion of the sine up to the third order, which you should know.
Upvote ·
9 1
9 1
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.5K answers and 21.1M answer views
·2y
Related
How do I solve lim x-->0 {[ (1+sinx) ^1/3] - [(1-sinx) ^1/3]} ÷ x?
We are given the limit
L=lim x→0 3√1+sin x−3√1−sin x x.L=lim x→0 1+sinx 3−1−sinx 3 x.
First of all, we multiply the numerator and denominator by sin x sinx and rearrange the factors so that we can use the fact lim x→0 sin x x=1 lim x→0 sinx x=1 to simplify the appearance of L L. This gives us
[Math Processing Error]\begin{align}L&=\displaystyle\lim_{x\to 0}\frac{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}{x}\cdot\frac{\sin{x}}{\sin{x}}\&=\lim_{x\to 0}\frac{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}{\sin{x}}\cdot\frac{\sin{x}}{x}\&=\lim_{x\to 0}\frac{\sqrt{1
Continue Reading
We are given the limit
L=lim x→0 3√1+sin x−3√1−sin x x.L=lim x→0 1+sinx 3−1−sinx 3 x.
First of all, we multiply the numerator and denominator by sin x sinx and rearrange the factors so that we can use the fact lim x→0 sin x x=1 lim x→0 sinx x=1 to simplify the appearance of L L. This gives us
L=lim x→0 3√1+sin x−3√1−sin x x⋅sin x sin x=lim x→0 3√1+sin x−3√1−sin x sin x⋅sin x x=lim x→0 3√1+sin x−3√1−sin x sin x.L=lim x→0 1+sinx 3−1−sinx 3 x⋅sinx sinx=lim x→0 1+sinx 3−1−sinx 3 sinx⋅sinx x=lim x→0 1+sinx 3−1−sinx 3 sinx.
Next, we make the substitution t=sin x t=sinx. Since sin x→0 sinx→0 as x→0 x→0, we obtain a significantly more concise-appearing limit:
L=lim t→0 3√1+t−3√1−t t.L=lim t→0 1+t 3−1−t 3 t.
To avoid the use of L’Hôpital’s Rule or power series, we next use the difference of cubes identity a 3−b 3=(a−b)(a 2+a b+b 2 a 3−b 3=(a−b)(a 2+a b+b 2) to rationalize the numerator. This allows us to evaluate the limit the rest of the way with minimal further algebraic manipulations.
L=lim t→0 3√1+t−3√1−t t⋅(3√1+t)2+3√1+t⋅3√1−t+(3√1−t)2(3√1+t)2+3√1+t⋅3√1−t+(3√1−t)2=lim t→0(1+t)−(1−t)t((3√1+t)2+3√1−t 2+(3√1−t)2)=lim t→0 2(3√1+t)2+3√1−t 2+(3√1−t)2=2 3.L=lim t→0 1+t 3−1−t 3 t⋅(1+t 3)2+1+t 3⋅1−t 3+(1−t 3)2(1+t 3)2+1+t 3⋅1−t 3+(1−t 3)2=lim t→0(1+t)−(1−t)t((1+t 3)2+1−t 2 3+(1−t 3)2)=lim t→0 2(1+t 3)2+1−t 2 3+(1−t 3)2=2 3.
Upvote ·
99 16
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188710 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_24?srsltid=AfmBOooeEECVaHOkgGxMZa4KB5udBRt7j7X2IuErlyFBOeLRjPS5GECz | Art of Problem Solving
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2022 AMC 10B Problems/Problem 24
Contents
1 Problem
2 Solution 1 (Absolute Values and Inequalities)
3 Solution 2 (Lipschitz Condition)
4 Solution 3 (Slopes)
5 Solution 4 (Translation)
6 Video Solution
7 Video Solution by OmegaLearn Using Algebraic Manipulation
8 Video Solution by Interstigation
9 See Also
Problem
Consider functions that satisfy for all real numbers and . Of all such functions that also satisfy the equation , what is the greatest possible value of
Solution 1 (Absolute Values and Inequalities)
We have from which we eliminate answer choices and
Note that Let Together, it follows that We rewrite as ~MRENTHUSIASM
Solution 2 (Lipschitz Condition)
Denote . Because , .
Following from the Lipschitz condition given in this problem, and and Thus, Thus, is maximized at , , , with the maximal value 100.
By symmetry, following from an analogous argument, we can show that is minimized at , , , with the minimal value .
Following from the Lipschitz condition, We have already construct instances in which the second inequality above is augmented to an equality.
Now, we construct an instance in which the first inequality above is augmented to an equality.
Consider the following piecewise-linear function: Therefore, the maximum value of is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~Viliciri (LaTeX edits)
Solution 3 (Slopes)
Divide both sides by to get . This means that when we take any two points on , the absolute value of the slope between the two points is at most .
Let , and since we want to find the maximum value of , we can take the most extreme case and draw a line with slope down from to and a line with slope up from to . Then and , so , and this is attainable because the slope of the line connecting and still has absolute value less than .
Therefore, .
Solution 4 (Translation)
Consider . Then satisfies all the conditions and . We would want and as distant from each other as possible. So assign and , the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for as the answer.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Algebraic Manipulation
~ pi_is_3.14
Video Solution by Interstigation
~Interstigation
See Also
2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23Followed by
Problem 25
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 10 Problems and Solutions
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Complex Conjugate of Complex function
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I am currently reading Hamming's Numerical Methods for Scientists and Engineers. On pg. 79 he discusses the topic of finding the zeros of a complex analytic function.
He then proceeds to discuss different types of complex conjugation for a function w(z)w(z). Here are examples of the three types for w(z)=sin z=e i z−e−i z 2 i w(z)=sinz=e i z−e−i z 2 i
w¯¯¯¯(z)w¯(z) : replace i i with −i−i in w w, e.g. w¯¯¯¯(z)=e−i z−e i z−2 i=sin z w¯(z)=e−i z−e i z−2 i=sinz.
w(z¯¯¯)w(z¯) : conjugate the argument, w(z¯¯¯)=e i z¯−e−i z¯2 i w(z¯)=e i z¯−e−i z¯2 i
w(z)¯¯¯¯¯¯¯¯¯¯w(z)¯ : Hamming describes this as conjugating the values which I take to mean the conjugate of the image of w(z)w(z) in the codomain. Although, I am not clear on this.
These three definitions have left me a bit confused. 2 and 3 seem relatively straight forward. But 1 leaves me a bit baffled in that it doesn't appear to be an actual complex conjugate of anything.
It appears that the first version can only be applied to functions of z z, because if I rewrite
w(z)=sin z w(z)=sinz as w(x+i y)=sin(x+i y)=sin x cosh y+i cos x sinh y w(x+i y)=sin(x+i y)=sinx coshy+i cosx sinhy
then
w¯¯¯¯(x+i y)=sin x cosh y−i cos x sinh y≠w(x+i y)=sin(x+i y)w¯(x+i y)=sinx coshy−i cosx sinhy≠w(x+i y)=sin(x+i y)
gives what I would expect to be the value of case 3 and a different answer then in the z z-form where w(z)=w¯¯¯¯(z)=sin z w(z)=w¯(z)=sinz.
Also, he later appears to state that w(z)¯¯¯¯¯¯¯¯¯¯=w¯¯¯¯(z¯¯¯)w(z)¯=w¯(z¯) although it is not clear whether that is true for all analytic functions or just those that have the property of w(z)=w¯¯¯¯(z)w(z)=w¯(z) like sin z sinz.
My question is several-fold. Is the definition of conjugation given in 1 standard? What is its meaning? Also, is my interpretation of 3 correct and is w(z)¯¯¯¯¯¯¯¯¯¯=w¯¯¯¯(z¯¯¯)w(z)¯=w¯(z¯) the proper definition of 3.
Addendum:
The motivation behind understanding definition 1 is that Hamming uses it in the proof that analytic complex functions have zeros that are conjugate pairs if the function is real over the real domain.
He states without proof that if w(z)w(z) is real for real z z then w(z)=w¯¯¯¯(z)w(z)=w¯(z). He then provides the following proof for the above.
w(a+b i)=0=w(a+b i)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=w¯¯¯¯(a−b i)=w(a−b i)w(a+b i)=0=w(a+b i)¯=w¯(a−b i)=w(a−b i)
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edited Aug 26, 2012 at 16:11
JeffJeff
asked Aug 26, 2012 at 5:31
JeffJeff
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I don't understand all of what you write (there seems to be a mistake somewhere after "because if I rewrite", since sin(x+i y)sin(x+i y) can't equal both of the right-hand sides), but regarding 1: If it's really meant to only replace all explicit occurrences of i i by −i−i without replacing z z by z¯z¯, then yes, I'd agree that this isn't a form of conjugation; it's just some formal manipulation of the function with superficial similarity to conjugation, and denoting the result by w¯¯¯¯(z)w¯(z) seems potentially confusing.joriki –joriki 2012-08-26 08:51:29 +00:00 Commented Aug 26, 2012 at 8:51
1 I fixed the mistake you mentioned. I intended to show that the type 1 conjugation only works in z-form because it gives a different result if the w(z)w(z) is already written in the form w(x+i y)=u(x,y)+i v(x,y)w(x+i y)=u(x,y)+i v(x,y).Jeff –Jeff 2012-08-26 14:27:46 +00:00 Commented Aug 26, 2012 at 14:27
1 Definition 1. is most natural if you think in series expansion around z=0 z=0: it conjugates the coefficients.WimC –WimC 2012-08-26 14:48:32 +00:00 Commented Aug 26, 2012 at 14:48
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In the complex world there is a fundamental involution, namely the map γ:z↦z¯γ:z↦z¯. An involution of some set X X is a map ι:X→X ι:X→X which is not the identity map i d X i d X, but whose square ι∘ι ι∘ι is the identity.
Given any region Ω⊂C Ω⊂C and a function f:Ω→C f:Ω→C, one can compose f f with γ γ in various ways. Your cases 2. and 3. produce for a given f f the functions
f∘γ:z↦f(z¯)f∘γ:z↦f(z¯)
and
γ∘f:z↦f(z)¯¯¯¯¯¯¯¯¯.γ∘f:z↦f(z)¯.
When f f is a holomorphic function of z z then both f∘γ f∘γ and γ∘f γ∘f are antiholomorphic, which means, e.g., that nonoriented angles between tangent vectors at points z 0∈Ω z 0∈Ω are preserved, but the orientation of small circles around z 0 z 0 is reversed.
Most interesting is your case 1. Here the function f f is transformed (also termed conjugated) into the new function
f¯:=γ∘f∘γ:z↦f(z¯)¯¯¯¯¯¯¯¯¯¯.f¯:=γ∘f∘γ:z↦f(z¯)¯.
When f f is a holomorphic function on Ω Ω then it easy to check by means of the CR-equations and the chain rule that the function f¯f¯ is a holomorphic function on Ω¯:={z¯|z∈Ω}Ω¯:={z¯|z∈Ω}.
The case when Ω Ω and Ω¯Ω¯ share an interval on the real axis is of special interest, because then we can ask the question: Could it be that in fact f¯(z)≡f(z)f¯(z)≡f(z) for all z∈Ω∩Ω¯z∈Ω∩Ω¯? After all, in the definition of f¯f¯ there are two conjugations involved.
To investigate this case, assume that 0∈Ω∩Ω¯0∈Ω∩Ω¯ and that
f(z)=∑k=0∞a k z k(|z|<ρ)f(z)=∑k=0∞a k z k(|z|<ρ)
with certain complex coefficients a k a k. Then
f¯(z)=∑k=0∞a k(z¯)k¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=∑k=0∞a¯k z k,f¯(z)=∑k=0∞a k(z¯)k¯=∑k=0∞a¯k z k,
and this is ≡f(z)≡f(z) iff all a¯k=a k a¯k=a k, i.e., if all a k a k are in fact real. When the latter is the case then automatically f(z)f(z) is real for real z z, and it is not difficult to show the converse: If a holomorphic f(z)f(z) is real for real z z (as in the case f:=sin f:=sin) then f¯=f f¯=f. This is the so-called reflection principle.
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answered Aug 26, 2012 at 18:28
Christian BlatterChristian Blatter
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This answer looks interesting. It will take me some time to digest it but it looks like it gets to the heart of the matter.Jeff –Jeff 2012-08-26 20:45:45 +00:00 Commented Aug 26, 2012 at 20:45
2 @ChristianBlatter: This is exceptionally well explained (the entire explanation). I am wondering if you could provide a good source that explains this notion in a similar manner.TSGM –TSGM 2015-08-03 11:10:20 +00:00 Commented Aug 3, 2015 at 11:10
Which conjugation is it in the definition of an inner product on a function space? Can one use all of these?user123124 –user123124 2017-02-09 19:51:33 +00:00 Commented Feb 9, 2017 at 19:51
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Well, if you accept w(z¯¯¯)w(z¯) and w(z)¯¯¯¯¯¯¯¯¯¯w(z)¯, you can simply define w¯¯¯¯(z)=w(z¯¯¯)¯¯¯¯¯¯¯¯¯¯w¯(z)=w(z¯)¯.
This allows also to resolve your example with sin z sinz:
sin¯¯¯¯¯¯z=sin(z¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯=sin(x+i y¯¯¯¯¯¯¯¯¯¯¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=sin(x−i y)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=sin x cosh y−i cos x sinh y¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=sin x cosh y+i cos x sinh y=sin(x+i y)=sin z sin¯z=sin(z¯)¯=sin(x+i y¯)¯=sin(x−i y)¯=sinx coshy−i cosx sinhy¯=sinx coshy+i cosx sinhy=sin(x+i y)=sinz
It also directly follows that w¯¯¯¯(z¯¯¯)=w(z¯¯¯¯¯¯)¯¯¯¯¯¯¯¯¯¯=w(z)¯¯¯¯¯¯¯¯¯¯w¯(z¯)=w(z¯¯)¯=w(z)¯.
Note that it may be easier to understand if we introduce the complex conjugation in function notation, c o n j(z)=z¯¯¯c o n j(z)=z¯. Then w(z¯)=w(c o n j(z))=(w∘c o n j)(z)w(z¯)=w(c o n j(z))=(w∘c o n j)(z). Similarly, w(z)¯¯¯¯¯¯¯¯¯¯=(c o n j∘w)(z)w(z)¯=(c o n j∘w)(z), and w¯¯¯¯(z)=(c o n j∘w∘c o n j)(z)w¯(z)=(c o n j∘w∘c o n j)(z).
Note that with the definition above, we have the following properties for w(z)w(z):
If w(z)=f(z)+g(z)w(z)=f(z)+g(z), then w¯¯¯¯(z)=f¯¯¯(z)+g¯¯¯(z)w¯(z)=f¯(z)+g¯(z).
Proof: w¯¯¯¯(z)=w(z¯¯¯)¯¯¯¯¯¯¯¯¯¯=f(z¯¯¯)+g(z¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=f(z¯¯¯)¯¯¯¯¯¯¯¯¯+g(z¯¯¯)¯¯¯¯¯¯¯¯¯=f¯¯¯(z)+g¯¯¯(z)w¯(z)=w(z¯)¯=f(z¯)+g(z¯)¯=f(z¯)¯+g(z¯)¯=f¯(z)+g¯(z).
If w(z)=f(z)⋅g(z)w(z)=f(z)⋅g(z), then w¯¯¯¯(z)=f¯¯¯(z)⋅g¯¯¯(z)w¯(z)=f¯(z)⋅g¯(z).
Proof: w¯¯¯¯(z)=w(z¯¯¯)¯¯¯¯¯¯¯¯¯¯=f(z¯¯¯)⋅g(z¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=f(z¯¯¯)¯¯¯¯¯¯¯¯¯⋅g(z¯¯¯)¯¯¯¯¯¯¯¯¯=f¯¯¯(z)+g¯¯¯(z)w¯(z)=w(z¯)¯=f(z¯)⋅g(z¯)¯=f(z¯)¯⋅g(z¯)¯=f¯(z)+g¯(z)
If w(z)=f(g(z))w(z)=f(g(z)), then w¯¯¯¯(z)=f¯¯¯(g¯¯¯(z))w¯(z)=f¯(g¯(z)).
Proof: w¯¯¯¯(z)=w(z¯¯¯)¯¯¯¯¯¯¯¯¯¯=f(g(z¯¯¯))¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=f(g(z¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=f(g¯¯¯(z)¯¯¯¯¯¯¯¯¯)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=f¯¯¯(g¯¯¯(z))w¯(z)=w(z¯)¯=f(g(z¯))¯=f(g(z¯)¯¯)¯=f(g¯(z)¯)¯=f¯(g¯(z))
If w(z)=z w(z)=z then w¯¯¯¯(z)=z w¯(z)=z.
Proof: w¯¯¯¯(z)=w(z¯¯¯)¯¯¯¯¯¯¯¯¯¯=z¯¯¯¯¯¯=z w¯(z)=w(z¯)¯=z¯¯=z
If w(z)=z¯¯¯w(z)=z¯ then w¯¯¯¯(z¯¯¯)=z¯¯¯w¯(z¯)=z¯.
Proof: w¯¯¯¯(z¯¯¯)=w(z¯¯¯¯¯¯)¯¯¯¯¯¯¯¯¯¯=w(z)¯¯¯¯¯¯¯¯¯¯=z¯¯¯w¯(z¯)=w(z¯¯)¯=w(z)¯=z¯
If w(z)=c w(z)=c (where c c is a constant), then w¯¯¯¯(z)=c¯¯w¯(z)=c¯.
Proof: w¯¯¯¯(z)=w(z¯¯¯)¯¯¯¯¯¯¯¯¯¯=c¯¯w¯(z)=w(z¯)¯=c¯
Using these rules recursively, you especially get that if w(z)=∑∞i=0 a k z k w(z)=∑i=0∞a k z k, then w¯¯¯¯(z)=∑∞i=0 a k¯¯¯¯¯z k w¯(z)=∑i=0∞a k¯z k
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edited Aug 26, 2012 at 16:58
answered Aug 26, 2012 at 8:48
celtschkceltschk
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I'll have to think about this definition to make sure it corresponds with Hamming's for cases other than sin z sinz which is special in that w¯¯¯¯(z)=w(z)w¯(z)=w(z) for it.Jeff –Jeff 2012-08-26 14:31:20 +00:00 Commented Aug 26, 2012 at 14:31
@Jeff: I hope my addition helps you with that.celtschk –celtschk 2012-08-26 16:59:35 +00:00 Commented Aug 26, 2012 at 16:59
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188712 | https://askfilo.com/user-question-answers-smart-solutions/the-sum-of-first-31-terms-of-an-arithmetic-sequence-is-620-a-3231313235383438 | Question asked by Filo student
The sum of first 31 terms of an arithmetic sequence is 620 a) What is it's 16th term? b) What is the sum of 15th and 17th terms? c) Find the sum of first 31 terms
Views: 5,124 students
Updated on: Jul 29, 2025
Text SolutionText solutionverified iconVerified
Concepts:
Arithmetic sequence, Sum of arithmetic sequence, Nth term of arithmetic sequence
Explanation:
To solve the given problem, we need to use the formulas for the sum of an arithmetic sequence and the nth term of an arithmetic sequence. The sum of the first n terms of an arithmetic sequence is given by Sn=2n(2a+(n−1)d), where a is the first term and d is the common difference. The nth term of an arithmetic sequence is given by an=a+(n−1)d.
Step by Step Solution:
Step 1
Given the sum of the first 31 terms of an arithmetic sequence is 620.
Step 2
Use the sum formula for an arithmetic sequence: Sn=2n(2a+(n−1)d).
Step 3
Substitute S31=620 and n=31 into the formula: 620=231(2a+30d).
Step 4
Simplify the equation: 620=15.5(2a+30d).
Step 5
Divide both sides by 15.5: 40=2a+30d.
Step 6
Solve for a+15d: a+15d=20.
Step 7
The 16th term is given by a16=a+15d. From step 6, we have a+15d=20. Therefore, the 16th term is 20.
Step 8
The sum of the 15th and 17th terms is given by a15+a17=(a+14d)+(a+16d)=2a+30d. From step 5, we know 2a+30d=40. Therefore, the sum of the 15th and 17th terms is 40.
Step 9
The sum of the first 31 terms is already given as 620.
Final Answer:
a) The 16th term is 20.
b) The sum of the 15th and 17th terms is 40.
c) The sum of the first 31 terms is 620.
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| Question Text | The sum of first 31 terms of an arithmetic sequence is 620 a) What is it's 16th term? b) What is the sum of 15th and 17th terms? c) Find the sum of first 31 terms |
| Updated On | Jul 29, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 10 |
| Answer Type | Text solution:1 |
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188713 | https://www.doubtnut.com/qna/646669507 | If s=ut+12at2, where u and a are constants. Obtain the value of dsdt.
More from this Exercise
The correct Answer is:u + at
To find the value of dsdt from the equation s=ut+12at2, we will differentiate the equation with respect to time t. 1. Write down the equation: s=ut+12at2 2. Differentiate both sides with respect to t: We will apply the differentiation operator ddt to both sides of the equation. dsdt=ddt(ut)+ddt(12at2) 3. Differentiate the first term ut: Since u is a constant, the derivative of ut with respect to t is: ddt(ut)=u 4. Differentiate the second term 12at2: Here, we apply the power rule of differentiation. The derivative of t2 is 2t, so: ddt(12at2)=12a⋅2t=at 5. Combine the results: Now, we can combine the derivatives from steps 3 and 4: dsdt=u+at Final Answer: Thus, the value of dsdt is: dsdt=u+at
To find the value of dsdt from the equation s=ut+12at2, we will differentiate the equation with respect to time t.
Write down the equation:
s=ut+12at2
Differentiate both sides with respect to t:
We will apply the differentiation operator ddt to both sides of the equation.
dsdt=ddt(ut)+ddt(12at2)
Differentiate the first term ut:
Since u is a constant, the derivative of ut with respect to t is:
ddt(ut)=u
Differentiate the second term 12at2:
Here, we apply the power rule of differentiation. The derivative of t2 is 2t, so:
ddt(12at2)=12a⋅2t=at
Combine the results:
Now, we can combine the derivatives from steps 3 and 4:
dsdt=u+at
Final Answer:
Thus, the value of dsdt is:
dsdt=u+at
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188714 | https://www.netsuite.com/portal/resource/articles/crm/pricing-mistakes.shtml | Published Time: 2025-05-02T08:00:00+08:00
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14 Pricing Mistakes to Avoid (& Tips to Fix Them)
Joseph Clancey| Product Marketing Specialist
May 2, 2025
Pricing is one of the most important growth drivers for a business’s sales and profits. It’s also an area in which business owners and managers can go wrong. There are lots of factors that go into pricing, which is why the process has so many pitfalls. Too much focus on what competitors charge, for example, could make business managers lose sight of their own costs—to the detriment of profitability.
This article examines common mistakes that companies often make with their pricing, plus how to identify and address them. By staying in tune with the market, prioritizing business goals, and relying on data and technology, businesses can develop pricing strategies that attract customers and keeps them coming back.
What Is a Pricing Strategy?
A pricing strategy is a plan that a company creates to determine how much it should charge for its offerings, with the goal of maximizing overall profits. Strong pricing strategies should factor in supply and demand, market and competitive analysis, and costs.
Because pricing encompasses so many competing priorities, the best pricing strategies are well rounded. They strike the right balance in all important factors to enable a business to get the most out of every sale. But common pricing mistakes, such as miscalculating costs or relying on intuition instead of data to set prices, can throw a major monkey wrench into a business plan.
It’s important for businesses to “nail” their pricing for a variety of reasons. The right pricing helps ensure that the business makes enough money to cover its expenses, creates opportunities to invest in future growth, and acts as a signal to your target customers about the perceived value of your products. And it builds trust in the brand. The wrong pricing, however, can have adverse effects in all of these areas, and more.
Key Takeaways
There are many costs of running a business, and product/service pricing must account for all of them.
Not baking overhead and other costs into product pricing makes it harder to achieve profitability.
The factors that go into pricing calculations change over time. Companies that stand still as their markets evolve often find they’ve mispriced their products.
Modern software can ingest and analyze data from multiple systems to help businesses determine the best prices for maximizing profits and achieving other important goals.
Pricing Mistakes to Avoid
Even small pricing mistakes can have major negative consequences for a company’s sales, profits, and reputation. Set prices too low, and customers may question product quality. But in a world where 74% of consumers are concerned about rising prices, you may turn people off by charging too much.
Businesses can avoid such pitfalls by implementing data-backed strategies that reflect the current realities of the company and its market. Here are 14 common mistakes that business owners and managers make regarding their pricing decisions, plus tips to help avoid or fix them.
1. Failing to Segment Your Customers
One of the most common—and most consequential—pricing mistakes is to treat all customers the same. If a business offers various types of products and services, chances are it is selling them to multiple customer profiles and/or into diverse markets. But it will be difficult, if not impossible, to come up with one pricing strategy that’s equally effective across all these segments. A company that tries this approach is likely to leave sales and profit on the table as a result.
Solution: Segment customers and develop unique pricing for each segment. Start by analyzing data on customers and prospects to identify the different types of people who buy your offerings, then group them by purchasing behaviors, budget limitations, and other factors that affect how and why they do business with your company. From there, create pricing strategies tailored to each segment. This approach means more work, but it should increase your products’ appeal among all potential buyers and help your company make the most of every sale.
2. Overcomplicating Pricing Practices
In contrast to No. 1, having too many pricing options—without clear reasons for doing so—can also cause problems. “Fake” sales and constant discounting are among the biggest offenders. Some companies assume these tactics lead to more sales, so they turn to them for a quick fix when business is down. But, in reality, they often cause confusion about pricing and can even damage the brand. People may wait for sales, which makes cash flow less consistent and reduces profit margins. They might lose trust in your pricing or start to view the entire company as untrustworthy.
Other ways some businesses tend to overcomplicate pricing include hidden fees, inconsistent pricing across channels, frequent price changes, convoluted bundling, and confusing loyalty programs. These practices not only confound customers but also create challenges in accounting, including complex revenue recognition and inaccurate financial forecasting.
Solution: To maintain both customer satisfaction and financial clarity, businesses should balance strategic pricing with simplicity: Offer simple pricing and be transparent with customers. Sales can be effective if they run occasionally and for legitimate reasons, such as a semiannual clearance event to help clear out inventory. Explain the exact details of deals and when the deals expire—then stick to those cutoffs. There’s no better way to erode customer loyalty than to make someone feel like they’ve gotten a deal and then take that feeling away by cutting prices further or extending a “limited time only” sale.
3. Not Accounting for Overhead and Associated Costs
Most companies factor the cost of producing a product into its pricing. But some stop there. There are so many other costs of running a business, and prices must account for those expenses, too. By not including overhead and other costs in prices, businesses find it harder to achieve profitability.
Solution: Develop a pricing strategy that accounts for all costs. Include everything in pricing calculations, from rent and payroll expenses to banking and credit card processing fees. By doing so, your pricing will fully reflect all of the resources that play a role in serving customers, making it easier for the business to grow.
4. Not Understanding Your Product’s Value
Cost typically determines the baseline for most companies’ pricing. At the very least, businesses must charge enough to cover their expenses. Companies that don’t factor in the value of their products on top of that baseline are making a pricing mistake. They must ask themselves, “What is the most our customers are willing to pay for the product?” The difference between that answer and the product’s cost is the product’s value—and sometimes it can be very different from the cost. When prices are lower than indicated by product value, the business is not maximizing its revenue. When pricing exceeds value, sales volume usually declines.
Solution: Consider value-based pricing. Analyze customer data and conduct market research to determine how much different customer segments are willing to pay for certain products, then adjust product pricing to be more in line. It’s also important to reinforce the product’s or brand’s value proposition through the company’s marketing efforts—explicitly explaining how the product helps or appeals to customers. Value-based pricing requires hard, smart work, because it’s not as straightforward as other pricing models that use standardized processes and formulas. It’s more challenging to accurately quantify the value of each individual offering, but often worthwhile.
5. Ignoring the Market and Competition
Another important question that many companies fail to ask when pricing their products is “What does our competition charge?” No matter how a business arrives at its pricing, customers may not want to buy if its prices are too far out of whack with other options that they consider comparable.
Solution: Strike a balance between costs, perceived value, and competitors’ prices by using demand-based pricing. With this flexible approach, businesses adjust their prices so that they always reflect current market conditions—i.e., raising prices when demand is high and lowering them when demand is low. Demand-based pricing takes competitor pricing into account but doesn’t overemphasize it. In this approach, software tracks competitors’ pricing in real time and considers it in conjunction with other demand data, such as market conditions and customer behavior. Businesses can choose to have their prices be below that of competitors, match them, or be above, and can apply different strategies to their various product categories or customer segments.
6. Basing Pricing on Intuition
Nowadays, there’s no shortage of data on which to base a pricing strategy. Yet, some companies turn a blind eye to it all. Instead, they determine prices through gut feelings and guesswork. They might ask themselves (or their friends and family) how much they’d pay—even though they’re not their target customers. These types of intuition-based pricing decisions rarely hit the mark.
Solution: Rely on real data to set pricing. If you’re going to ask people how much they’d pay, ask actual customers. Even a brief, informal email survey will provide more actionable insights than intuition. Motivate customers to respond by offering discounts or rewards to allow you to collect a large enough sample size to yield valid results.
7. Setting Prices Too Low
Low prices don’t necessarily create more sales. Over time, customers may come to view underpriced products as being of low quality, especially if the company consistently emphasizes price over features and benefits. Despite this, businesses often set prices too low, particularly when they’re just starting out or otherwise lacking confidence. Typically, their plan is to raise prices over time. But if they can’t make enough sales and generate enough profit out of the gate, that day may never come.
Solution: Price products competitively from the get-go. You can go lower than competitors, but don’t go so low that price is the only thing customers talk about. Demonstrate actual value, so your products will become more likely to generate profits that you can use to further grow the business.
8. Failing to Review and Update Prices
The factors that go into pricing calculations—market and competitive dynamics, customer demand, cost, etc.—change over time, sometimes quickly. As a result, a strategy that successfully drove sales at a product’s inception may not work so well later on. Companies that stand still as their markets evolve will sooner or later find themselves at a disadvantage.
Solution: Regularly review pricing and update it accordingly. Still, be cognizant of the fact that frequent price changes may confuse or upset customers. Raise prices strategically and always clearly communicate any increases in advance, which will mitigate at least some customers’ disappointment.
9. Setting Prices That Don’t Align With Business Goals
Even the most disciplined companies may stray from their main objectives from time to time. They might feel they need a quick win or that they must hop on a new trend. When this affects pricing, it can quickly become a problem. A startup company, for example, may intentionally set initial prices low to gain a foothold in a market. But that could attract cost-conscious shoppers, when the company’s goal is to become a premium brand.
Solution: Always keep prices aligned with business goals. It’s easy to lower prices, but it’s not always easy to raise them again if customers grow accustomed to paying less. What you think is a short-term adjustment can turn into a long-term problem that affects revenue, profit, and brand reputation.
10. Poor Forecasting and Planning
Pricing mistakes often stem from inaccurate forecasts that have created a lack of foresight. When businesses fail to consider market fluctuations, rising costs, or changing consumer behaviors, or they base forecasts on incomplete data, they risk misaligning pricing and demand. And that can hurt their long-term growth.
For example, if a company neglects to forecast a spike in demand during a holiday season, it might set prices too low and miss out on potential revenue. But if it relies on outdated data and overestimates demand, it could raise prices too high, pushing customers away.
Solution: Create a pricing strategy that relies on real data to account for historical trends in sales, market conditions, and customers’ shopping behaviors. Use analytics software to accurately forecast demand and proactively adjust pricing in response. Businesses that emphasize strong forecasting and planning can avoid making reactionary changes and other serious pricing mistakes.
11. Not Testing Different Price Points
All the forecasting and planning in the world can’t guarantee a successful pricing strategy. Customers are human beings, and the only way to know for sure how they’ll respond to pricing changes is to see them in action. But many business owners and managers fail to sufficiently test different pricing options. Some are afraid that experiments will confuse or alienate customers; others might assume their pricing is already optimal, even though they don’t have the data to prove it. In some cases, the company just doesn’t have the time or the right tools to run experiments properly. Whatever the reason, they’re not getting valuable insights they could otherwise use to improve their pricing.
Solution: Use various forms of testing to optimize pricing. A retailer with an online store or multiple physical locations can run A/B testing by setting different prices for the same product and analyzing customer response. Then, once armed with concrete data, they can determine optimal pricing and implement it everywhere. Psychological pricing is another effective way to test buyer behavior. It’s easy, inexpensive, and doesn’t require additional tools. By making small, subtle changes to a product’s price (e.g., marking down a necklace from $100 to $99.99), retailers can dramatically increase that product’s perceived value and drive more sales.
12. Trying to Achieve the Same Profit Margin Across Products
Cost-plus pricing is one of the most popular pricing models, mainly because of its simplicity. With cost-plus pricing, companies take the cost of a product and add a fixed markup percentage to determine its selling price. But applying the same markup to all products can be a major pricing mistake, because it doesn’t factor in each product’s individual value. Although markup and margin are calculated differently—markup is calculated off the cost of a product and margin off its selling price—in most cases, the business ends up with similar margins for all of its products. In that case, a company would usually end up pricing some too low (which eats into profit) and others too high (which can hurt sales).
Solution: Don’t sacrifice optimal pricing for simplicity by using the same margin across the board. Make the extra effort to set pricing that reflects each product’s value to its customer base. This approach gives a business the best chance of maximizing sales and profits for its entire product line.
13. Focusing Solely on Units Sold
Businesses often fall into the trap of prioritizing sales volume above all else. They might equate high unit sales with business success, prioritize market share gains, or simply find that unit sales is an easier metric to track than more complex financial measures. But myopic focus on units sold ignores crucial factors, such as profit margin, production cost, and overall revenue, which can lead to misguided pricing decisions that undermine profitability, cause cash flow problems, or even trigger price wars. This kind of mistake can be a particular issue in sales departments, especially when sales reps’ compensation is tied to sales volume.
Solution: Adopt a more holistic approach to pricing. Consider a balanced scorecard of metrics, including gross profit margin and net profit, to allow for more strategic pricing decisions that juxtapose volume against sustainable, profitable growth. For instance, consider tiered pricing strategies that cater to different customer segments or value-based pricing that reflects the true worth of the product to customers. And, align salespeople’s incentives with overall business goals as part of the pricing process. For example, you may create multiple incentives that counterbalance volume and margin in the salesperson’s mind.
14. Failing to Leverage Technology and Automation
A multitude of factors go into creating successful pricing strategies, and many require data from multiple sources. Plus, the companies that do pricing strategy well regularly update their data so they can maintain a competitive edge. Therefore, doing pricing strategy analysis manually is virtually a nonstarter. But many small and medium-sized companies do exactly that because they don’t believe that they can afford technology solutions, or that their data is good enough, or that they have the expertise to implement the systems.
Solution: Bite the bullet and automate the pricing process. Modern software can ingest and analyze data from multiple systems and perform several kinds of analyses to help a business determine best prices. The software can even automatically update prices based on changing conditions. This approach, a supercharged version of demand-based pricing known as dynamic pricing, is especially popular in the restaurant and hospitality industries.
How to Identify if Your Business Is Making a Pricing Mistake
Pricing mistakes aren’t always obvious—especially if you’re not relying on technology and automation to analyze business finances. But the signs are there, if you know where to look. Declining sales and changing behaviors from customers and competitors may be signaling problems with your pricing strategy. Here are five ways to identify issues early on, before they become serious problems.
Your sales are down: An unexpected drop in sales is a potential indicator of a pricing mistake. Something has changed with regard to how customers value the product, and your pricing no longer reflects this new reality. In this case, it’s time to reassess not only product pricing but also brand positioning.
Your competition is outperforming you: Pressure from the competition is often the cause of an unexpected drop in sales. Maybe there’s a new product that benefits customers more than your product does. Or maybe a competitor has found a more efficient supplier so it can sell the same product at a lower price while earning a higher profit.
Your sales don’t align with expected profit margins: The company’s sales volume is meeting expectations, but profit isn’t. This usually means the pricing strategy has not properly accounted for all costs—a common pricing mistake. Make sure to factor all operating expenses into pricing calculations so margins can be accurately forecasted.
You see changes in sales behavior: If salespeople start regularly offering discounts to close deals—or suddenly cease discounting—it could be a sign that your regular pricing no longer resonates with customers. Get feedback from the team and reexamine market dynamics to determine whether prices need to be adjusted.
You see patterns in customer feedback: The clearest sign of a pricing mistake is when customers tell you so. If you’re regularly getting price-related feedback, ask follow-up questions to determine why customers feel prices are off base.
Protect Profit Margins With NetSuite
Accurately calculating prices is one of the keys to maximizing profit margins. But many companies make pricing mistakes because of insufficient data analysis, error-filled manual processes, or disconnected systems. NetSuite’s configure, price, quote (CPQ) software helps businesses avoid these pitfalls through automated workflows and integrated data management.
With NetSuite CPQ, for example, businesses can stay on top of changing market conditions by setting the software to automatically adjust prices in line with real-time demand signals and competitive data. Because it integrates seamlessly with NetSuite’s broader enterprise resource planning system, NetSuite CPQ can take advantage of customer segmentation capabilities that allow business managers to create targeted pricing strategies for different market segments. Furthermore, it can help avoid cost-oversight errors by pulling in data from throughout the business—including overhead, labor, and materials—to calculate accurate margins for each product configuration. Sales teams can quickly generate quotes that accurately reflect current costs and market conditions. And automated approval workflows can help ensure consistent pricing across channels—while still supporting strategic discounting, when appropriate.
The result: a data-driven pricing strategy that maximizes profitability while avoiding common pricing mistakes.
NetSuite CPQ makes it easy to adjust pricing automatically in response to changing market conditions, using predefined rules.
Pricing is a powerful but complex lever for driving business growth. Successful pricing strategies balance costs, customer and competitive analyses, and product value—but they are difficult to achieve without modern technologies and automation. By avoiding common mistakes and relying on data-driven insights, business owners and managers can align pricing with their business goals and maximize profitability.
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Pricing Mistakes FAQs
What are the three most common pricing strategies?
The three most common pricing strategies are competition-based pricing (matching or slightly beating competitors’ prices), cost-plus pricing (adding a markup to the cost of the product or service), and value-based pricing (basing prices on what customers are willing to pay).
How can pricing hurt a business?
Pricing can hurt businesses in several ways. Poorly priced products can lead to lower sales and allow competitors to leapfrog your company. Pricing that is too low may degrade the brand, erode customers’ trust, and eat into profit margins. Pricing that is too high may drive customers away.
What would be the problem with pricing a product too high?
Pricing a product too high risks losing new and existing customers that don’t believe the product or service is worth the money. In addition, the problem of lost sales because prices are too high can be compounded by the costs of storing surplus inventory for products that remain unsold.
What would be the problem with pricing a product too low?
Too-low pricing can diminish a product’s perceived value among potential customers. It is also impossible to turn a profit if prices are so low that they don’t cover all business costs. Raising prices later may also be a challenge if customers get accustomed to the lower amount.
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What Is a Pricing Strategy?
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188715 | https://www.hanspub.org/journal/paperinformation?paperid=120013 | 多种方法求解一维波动方程的教学探讨与实践
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Vol. 15 No. 7 (July 2025)
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多种方法求解一维波动方程的教学探讨与实践
Teaching Exploration and Practice of Solving One-Dimensional Wave Equations with Multiple Methods
DOI:10.12677/ae.2025.1571269, PDF, HTML, XML, 被引量下载:108 浏览:151 科研立项经费支持
作者:杨文彬：西安邮电大学理学院，陕西 西安
关键词:一维波动方程；求解方法对比；教学应用；One-Dimensional Wave Equation；Comparison of Solution Methods；Teaching Application
摘要:探讨了四种求解一维波动方程的方法，包括达朗贝尔法、傅里叶变换法、拉普拉斯变换法和格林函数法，分析了它们在无界空间上的适用性和物理意义。通过对解法的逐步剖析和案例教学应用的展示，本文旨在帮助学生更好地理解不同方法的原理及其应用场景，提升在不同物理背景下求解偏微分方程的能力。本研究提出的教学策略旨在引导学生掌握多种解法的基本原理与优劣，并在工程与物理实际问题中选择合适的方法。
Abstract:This paper systematically discusses four methods for solving the one-dimensional wave equation, including d’Alembert’s method, Fourier transform method, Laplace transform method, and Green’s function method. It analyzes their applicability and physical significance in unbounded space. Through step-by-step analysis of the solutions and the presentation of case study applications, this paper aims to help students better understand the principles and application scenarios of different methods, and enhance their ability to solve partial differential equations in different physical contexts. The teaching strategies proposed in this study aim to guide students in mastering the basic principles, advantages, and disadvantages of multiple solution methods, and in selecting appropriate methods for solving practical engineering and physical problems.
文章引用：杨文彬. 多种方法求解一维波动方程的教学探讨与实践[J]. 教育进展, 2025, 15(7): 672-677.
1. 引言
一维波动方程在描述物理和工程中的波动现象(如声波、电磁波等)时具有重要应用 。求解此类方程的方法多种多样，不同方法有其独特的适用性和物理意义。如何在教学中合理选择和对比这些解法，以帮助学生理解并掌握偏微分方程的多种解法，是当前教学改革的一个重要议题。本文选择达朗贝尔法、傅里叶变换法、拉普拉斯变换法和格林函数法四种常见解法，通过系统的对比和案例分析，探讨在一维波动方程教学中的应用策略。
2. 一维波动方程的求解方法
本文分析了一维波动方程在无界空间上的边值问题：
{∂2 u∂t 2=c 2∂2 u∂x 2,−∞<x<+∞,t>0 u(x,0)=f(x),∂u∂t(x,0)=g(x)(1)
假设初始波形为高斯脉冲 u(x,0)=exp(−α x 2) ，其中 α 是控制脉冲宽度的参数；初速度设置为零∂u∂t(x,0)=0 。上述一维波动方程在无界空间上的边值问题数值解模拟图如图1所示，具体Matlab程序代码如表1所示。
2.1. 解法一：达朗贝尔法
达朗贝尔法是一种直观的解析法，适合无界空间的波动问题。其基本思想是引入特征变量，将波动方程转化为特征线下的简单函数形式。具体求解过程如下：
(1) 通解的形式：通过作特征变换 ξ=x−c t 和 η=x+c t ，通解可写为：
u(x,t)=F(x−c t)+G(x+c t)
(2) 利用初始条件：代入初始条件求解 F 和 G，得到波动方程的最终解：
u(x,t)=f(x−c t)+f(x+c t)2+1 2 c∫x+c t x−c t g(ξ)d ξ
该方法直观性强，能清晰展示波的传播特性，适合用于帮助学生理解波的双向传播和初始条件的影响。达朗贝尔法适合作为波动方程求解的入门方法，使学生建立直观的波传播图像。
Figure 1.Schematic diagram of the one-dimensional wave equation in unbounded space
图1. 无界空间中的一维波动方程示意图
Table 1. Matlab codes for the numerical solution of the one-dimensional wave equation in unbounded space
表1. 无界空间中的一维波动方程数值解Matlab代码
% 模拟参数
L = 1000;
dx = 0.5;
dt = 0.25;
c = 1.0;
time_steps = 400;
% 初始化空间域和初始条件
x = -L/2:dx:L/2;
u = exp(-0.01 x.^2);
u_new = zeros(size(u));
u_old = u;
% 创建动画图形
figure;
h = plot(x, u, 'b', 'LineWidth', 1.5);
xlim([-L/2, L/2]);
ylim([-1, 1]);
xlabel('位置');
ylabel('振幅');
% 创建视频对象
v = VideoWriter('wave_equation_simulation.avi');
v.FrameRate = 30;
open(v);% 时间步进循环
for t = 1:time_steps
% 使用有限差分法更新
u_new(2:end-1) = 2 u(2:end-1) - u_old(2:end-1) + (c dt / dx)^2 (u(3:end) - 2 u(2:end-1) + u(1:end-2));
% 吸收边界条件
u_new(1) = 0;
u_new(end) = 0;
% 更新下一次迭代的值
u_old = u;
u = u_new;
% 更新图形
set(h, 'YData', u);
drawnow;
% 将当前帧写入视频
frame = getframe(gcf);
writeVideo(v, frame);
end
% 关闭视频文件
close(v);
2.2. 解法二：傅里叶变换法
傅里叶变换法利用频域分析，将空间无界的波动问题转化为代数方程求解。具体求解过程如下：
(1) 对 u(x,t) 关于 x 做傅里叶变换，原始定解问题转化为：
{d 2 U(ω,t)d t 2=−a 2 ω 2 U(ω,t),t>0 U(ω,0)=Φ(ω),d U(ω,0)d t=Ψ(ω)(2)
(2) 求解定解问题(2)，其通解为：
U(ω,t)=Φ(ω)cos a ω t+Ψ(ω)a ω sin a ω t
它可以转化为：
U(ω,t)=1 2[Φ(ω)e i a ω t+Φ(ω)e−i a ω t]+1 2 a[Ψ(ω)i ω e i a ω t−Ψ(ω)i ω e−i a ω t]
(3) 通过傅里叶逆变换恢复到原空间，得到 u(x,t) 的解：
u(x,t)=1 2[f(x−c t)+f(x+c t)]+1 2 c∫x+c t x−c t g(ξ)d ξ
傅里叶变换法适合处理复杂初始条件，有助于学生理解频域和时域之间的关系。通过频率分量的叠加分析波的传播特性，能提升学生对频率特性的理解。
2.3. 解法三：拉普拉斯变换法
拉普拉斯变换法特别适用于半无界或有限域上的波动问题，也能解决瞬态问题。具体求解过程如下：
(1) 对 u(x,t) 关于 t 进行拉普拉斯变换，原始定解问题转化为：
d 2 U(x,s)d x 2−s 2 c 2 U(x,s)=−s f(x)+g(x)c 2
(2) 假设 U p(x,s) 是上述方程的特解，则其通解一般形式为：
U(x,s)=A(s)e s c x+B(s)e−s c x+U p(x,s)
需要注意的是，拉普拉斯变换下的特解往往难以直接求得(比如 U p(x,s) )，尤其是在复杂边界条件或源项的情况下。此外，如果问题中缺乏足够的边界条件，有可能导致我们无法确定 A(s) 和 B(s) 的具体值。因此，下述第(3)步一般无法得到。
(3) 确定了 U(x,s) 之后，通过拉普拉斯逆变换回到时间域，可以得到最终的解 u(x,t) 。然而，由于 U(x,s) 的形式往往较为复杂，逆变换的积分可能难以解析计算，因此在实际求解中可能需要数值计算或查表法辅助完成。
拉普拉斯变换法可以帮助学生处理瞬态波动问题，对边界复杂的情境也适用。这种方法尽管计算较为复杂，但能为学生展示更广泛的解法适用性。
2.4. 解法四：格林函数法
格林函数法通过构造特定边值条件下的格林函数，使波动方程的解表示为初始条件的卷积积分。具体求解过程如下：
(1) 构造格林函数 G(x,t;ξ) 满足波动方程及初始条件。在无界空间上，格林函数通常满足
(∂2∂t 2−c 2∂2∂x 2)G(x,ξ,t)=δ(x−ξ)δ(t)
其中 δ(⋅) 是狄拉克函数。在无界空间上，格林函数的显示形式一般为：
G(x,t;ξ)=1 2 c[δ(x−ξ−c t)+δ(x−ξ+c t)]
(2) 利用格林函数的卷积性质，以及初始条件 f(x) 和 g(x) 的卷积积分形式，方程(1)的解可以表示为：
u(x,t)=∫∞−∞[f(ξ)∂G∂t(x,t;ξ)+g(ξ)G(x,t;ξ)]d ξ
(3) 由于格林函数中包含狄拉克函数，利用狄拉克函数的积分性质以及初始条件：
u 1(x,t)=∫∞−∞1 2[δ(x−ξ−c t)+δ(x−ξ+c t)]f(ξ)d ξ=1 2[f(x−c t)+f(x+c t)]
u 2(x,t)=∫∞−∞1 2[−c δ(x−ξ−c t)+c δ(x−ξ+c t)]g(ξ)d ξ=1 2 c∫x+c t x−c t g(ξ)d ξ
(4) 经过卷积积分以及狄拉克函数的简化，最终可以得到波动方程的解析解：
u(x,t)=1 2[f(x−c t)+f(x+c t)]+1 2 c∫x+c t x−c t g(ξ)d ξ
格林函数法能帮助学生理解波动方程在源项或复杂初始条件下的求解过程。通过卷积积分表示的求解过程，学生可以看到初始条件对解的全局影响。
3. 案例分析
3.1. 案例问题
问题情境：考虑一个无限长、均匀且各向同性的理想弹性钢丝，在初始时刻 t=0 受到一个集中激励，激励形式可由脉冲函数(狄拉克 δ 函数)描述。分析冲击波(或应力波)沿钢丝的传播特性。
已知条件：无界空间−∞<x<∞ ；初始位移 u(x,0)=0 ；初始速度∂u∂t(x,0)=δ(x) 。
3.2. 四种方法的应用
参照第2节进行。
3.3. 案例教学设计
(1) 课前准备：让学生预习四种方法的数学背景；设计简答题，写出各方法的主要思路及其适用条件。
(2) 课堂环节：逐步演示四种方法对该冲击问题的完整推导；讨论对比哪种方法最快速？哪种方法最直观？
(3) 课后思考题：如果将问题改为有界空间(如固定端点)，哪种方法更合适？
3.4. 总结
通过对该案例的深入剖析，学生能够达成以下目标：1) 熟练掌握一维波动方程的多元解法及其背后所依托的数学原理；2) 透彻理解每种解法所蕴含的物理意义以及其适用的具体条件；3) 显著增强运用数学工具解决物理与工程实际问题的能力；4) 培养在实际问题情境中，能够灵活且精准地选择恰当求解方法的思维模式。
4. 结论与展望
本文系统展示了达朗贝尔法、傅里叶变换法、拉普拉斯变换法和格林函数法在一维波动方程教学中的应用。具体来说：
(1) 达朗贝尔法适合作为入门方法，帮助学生掌握波动传播的基本概念，尤其是双向传播特性。它在一维波动方程教学中的应用，为学生后续学习奠定了坚实基础。
(2) 傅里叶变换法便于学生理解频域分析的意义，适合处理无界空间上的复杂初始条件。该方法有助于学生掌握频率分量的叠加，进一步拓宽了波动方程求解的视角。
(3) 拉普拉斯变换法为学生提供了一种解决瞬态或边界复杂情境的有效工具。它拓展了学生在不同时间尺度和边界条件下的分析能力，使得波动方程的求解更加灵活多样。
(4) 格林函数法通过构造满足特定初始条件的卷积解，帮助学生理解源项对解的全局影响。该方法在源项或复杂初始条件下的应用中表现出色，为学生提供了全新的解题思路。
通过对这四种方法的对比分析(如表2所示)和应用案例示范，学生可以从不同角度理解波动方程的解法及其物理意义。每种方法的应用都展示了其独特的求解思路和优缺点，有助于学生根据具体问题选择合适的求解方法。
Table 2.Comparative analysis of four methods
表2. 四种方法的对比分析
方法适用性优点局限性
达朗贝尔法 全空间，简单初始条件 直观易懂，展示双向传播特性 不适合复杂边界或初始条件
傅里叶变换法 全空间，复杂初始条件 频域分析，便于频率分量分解 对于非周期边界需变换调整
拉普拉斯变换法 半无界或瞬态问题，复杂边界 适用于瞬态问题和边界复杂情况 计算步骤繁琐，不够直观
格林函数法 全空间，复杂源项或初始条件 适合源项问题，展示全局影响 需构造格林函数，计算复杂
基金项目
2024年西安邮电大学教学改革研究专项项目“《数学物理方法》课程中思政教育改革的策略与实践研究”(JGSZB202419)。
参考文献
王元明. 数学物理方程与特殊函数[M]. 第六版. 北京: 高等教育出版社, 2019.
方瑛, 黄毅. 数学物理方程与特殊函数[M]. 第二版. 北京: 科学出版社, 2012.
祁玉海. 用行波法求解定解问题[J]. 青海师范大学民族师范学院学报, 2008(1): 64-65.
张礼涛. 傅里叶变换在求解微分方程中的应用[J]. 佳木斯教育学院学报, 2012(12): 198-199.
梁家辉. 重要的拉普拉斯变换公式及其应用[J]. 数学的实践与认识, 2023, 53(9): 230-256.
张子珍, 林海. 格林函数法求解偏微分方程[J]. 山西大同大学学报(自然科学版), 2015, 31(6): 1-2+16.
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Smallest Possible Power
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Asked 8 years, 2 months ago
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When working on improving my skills with indices, I came across the following question:
Find the smallest positive integers m m and n n for which: 12<2 m/n<13 12<2 m/n<13
On my first attempt, I split this into two parts and then using logarithms found the two values m/n m/n had to be between. However I wasn't sure how to progress past that.
I have the answer itself (11/3)(11/3), but I'm unsure of the best method to find it. Any help would be really appreciated.
algebra-precalculus
elementary-number-theory
inequality
logarithms
exponential-function
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edited Oct 22, 2017 at 18:04
Michael Rozenberg
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asked Jul 5, 2017 at 5:06
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The inequality is equivalent to 12 n<2 m<13 n 12 n<2 m<13 n. By brute force, looking for powers of 2 2 between 12 n 12 n and 13 n 13 n starting from the lowest possible n=1 n=1 up:
n=1 n=1: no solutions, since 2 3=8<12 1<13 1<16=2 4 2 3=8<12 1<13 1<16=2 4
n=2 n=2: no solutions, since 2 7=128<144=12 2<13 2=169<256=2 8 2 7=128<144=12 2<13 2=169<256=2 8
n=3 n=3: 12 3=1728<2048=2 11<2197=13 3 12 3=1728<2048=2 11<2197=13 3, therefore m=11,n=3 m=11,n=3 is a solution.
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answered Jul 5, 2017 at 5:20
dxivdxiv
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2 Isn't this what the OP did? Brute force after simplifying? I thought the question was is this the best method or is there a better method. . . doesn't answer that at all.iheanyi –iheanyi 2017-07-05 19:33:04 +00:00 Commented Jul 5, 2017 at 19:33
5 @iheanyi The OP used logarithms to determine that log 2 12<m/n<log 2 13 log 212<m/n<log 213 numerically. My answer does not use logarithms, but only simple calculations that can be easily done by hand.dxiv –dxiv 2017-07-05 20:23:33 +00:00 Commented Jul 5, 2017 at 20:23
Ah, I see. Though I'm not really sure hand calculations is much of an improvement. Once I've used a calculator to determine l o g 2 12 l o g 2 12 and l o g 2 13 l o g 2 13, wouldn't I then need to use your method to determine the candidate m and n?iheanyi –iheanyi 2017-07-05 23:36:13 +00:00 Commented Jul 5, 2017 at 23:36
And I can answer my own question - "yes", which is why this is an answer to how to proceed beyond "that" where that was the work the OP did using logarithms. Thanks for following up.iheanyi –iheanyi 2017-07-05 23:39:09 +00:00 Commented Jul 5, 2017 at 23:39
1 I only specified "your method" to indicate that I'd seen the error of my original comment.iheanyi –iheanyi 2017-07-06 00:07:42 +00:00 Commented Jul 6, 2017 at 0:07
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Starting with 12 n<2 m<13 n 12 n<2 m<13 n, we see that n log 2(12)<m<n log 2(13)n log 2(12)<m<n log 2(13). We conclude that
m=⌈n log(12)log(2)⌉≈⌈n×3.584962500721156181453738943947816508759814407692481060455…⌉m=⌈n log(12)log(2)⌉≈⌈n×3.584962500721156181453738943947816508759814407692481060455…⌉
The first few values of m m and n n are
(m,n)∈{(11,3),(15,4),(18,5),(22,6),(26,7),(29,8),(33,9),(36,10)}(m,n)∈{(11,3),(15,4),(18,5),(22,6),(26,7),(29,8),(33,9),(36,10)}
You can see that m=11 m=11 and n=3 n=3 will be the smallest values of m m and n n.
Note also that there is no smallest value of m n m n since lim n→∞m n=log 2 12 lim n→∞m n=log 212 and log 2 12 log 212 is irrational.
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answered Jul 5, 2017 at 6:09
Steven Alexis GregorySteven Alexis Gregory
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From 2 m n<13 2 m n<13 follows m n<log 2 13 m n<log 213
Developing log 2 13 log 213 in continued fraction we get {3,1,2,2,1,…}{3,1,2,2,1,…}
Using 3+1 1+1 2 3+1 1+1 2 we get 11 3 11 3 which gives m=11;n=3 m=11;n=3
Going on we find 3+1 1+1 2+1 2=26 7 3+1 1+1 2+1 2=26 7 which gives m=26;n=7 m=26;n=7 and so on
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answered Jul 5, 2017 at 10:47
RaffaeleRaffaele
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3.5849...=log 2 12<m n<log 2 13=3.70...3.5849...=log 212<m n<log 213=3.70...
Thus, for m n=3 2 3 m n=3 2 3 it occurs.
If we want to make 3 3 in the denominator be smaller than we'll get n=2 n=2 and it's impossible.
Thus, 3 2 3=11 3 3 2 3=11 3 is our answer.
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answered Jul 5, 2017 at 5:28
Michael RozenbergMichael Rozenberg
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1 Yes, using the interval from 3.5849…3.5849… to 3.7004…3.7004… comes close to the approach the asker talked about. Then trying with n=1 n=1, we have the whole numbers (integers), and none of them falls within the interval. Then with n=2 n=2, we get all the multiples of one half, but still none is inside the interval (3+1 2 3+1 2 is too low, and 4 4 is too high). Then with n=3 n=3, we have "everything with thirds", and we find the solution you describe. This gives a very explicit explanation of why that solution is the minimal one.Jeppe Stig Nielsen –Jeppe Stig Nielsen 2017-07-06 11:18:12 +00:00 Commented Jul 6, 2017 at 11:18
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You can write this as an integer program:
min m+n min m+n
subject to:
n log 2(12)−m<=0 n log 2(12)−m<=0
n log 2(13)−m>=0 n log 2(13)−m>=0
m>=1 m>=1
n>=1 n>=1
and solve with an integer programming solver. This might suffer from numerical issues.
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answered Jul 5, 2017 at 15:18
mathdotrandommathdotrandom
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Rather than minimizing m+n m+n, consider minimizing z z where m≤z m≤z and n≤z n≤z.Théophile –Théophile 2017-07-05 15:30:45 +00:00 Commented Jul 5, 2017 at 15:30
Why not an AI approach: depth first search with iterative deepening of the depth bounds?richard1941 –richard1941 2017-07-12 00:20:27 +00:00 Commented Jul 12, 2017 at 0:20
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First, move the constants around.
12 ln 12 log 2 12 n log 2 12<2 m/n<13<m n ln 2<ln 13<m n<log 2 13<m<n log 2 13 12<2 m/n<13 ln12<m n ln2<ln13 log 212<m n<log 213 n log 212<m<n log 213
Note that this shows that m m and n n are minimized together (since m m is bounded by constant multiples of n n).
The last line shows we are guaranteed to have a solution if n(log 2 13−log 2 12)≥1 n(log 213−log 212)≥1. (Any interval of length at least 1 1 contains at least one integer.) This says there is a choice of m m for all n≥1 log 2 13−log 2 12=8.659…n≥1 log 213−log 212=8.659…. An exhaustive search will find a solution for n=9 n=9 and may have found one for a smaller n n. Since 9 9 is not a very big number, exhaustive search is feasible.
But we can do better. Note that 2 3=8<12<13<16=2 4 2 3=8<12<13<16=2 4, so m/n∈(3,4)m/n∈(3,4). Write m=3 n+m′m=3 n+m′ so that 0<m′<n 0<m′<n and then
0.584…=log 2 12/8<m′n<log 2 13/8=0.700…0.584…=log 212/8<m′n<log 213/8=0.700…
From here, we only need to check which n n, when multiplied by log 2 13/8 log 213/8 have a new integer part, then check to see whether log 2 12/8 log 212/8 was left far enough behind. (A way to do this efficiently is to use Bresenham's line algorithm for finding where the integer part increments.) We get this table:
n 1 2 3⌊log 2 0 1 2 13/8⌋⌊log 2 0 1 1 12/8⌋n⌊log 2 13/8⌋⌊log 2 12/8⌋1 0 0 2 1 1 3 2 1
and we're done. We read that n=3 n=3 allows m′=2 m′=2 and then m=3⋅3+2=11 m=3⋅3+2=11.
But suppose we weren't done. The table would continue
4 5 6 7 8 9 2 3 4 4 5 6−2 3−4 5 4 2−5 3 2 6 4 3 7 4−8 5 4 9 6 5
where we stop when we hit n=9 n=9, which is guaranteed to have a solution. Notice, there is no reason to find the value in the 3rd column when the integer part in the second column has not increased.
Note that we can actually calculate the multiples of log 2 13/8=0.700…log 213/8=0.700… "by eye", so the only work in this particular problem is the integer part of the multiples of log 2 12/8 log 212/8.
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answered Jul 6, 2017 at 7:59
Eric TowersEric Towers
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Calculation of log to base 2 is easy in floating point. A crude approximation is to write a dot, then the mantissa portion of the floating pint representation. If you have a decimal number between 1 and 2, just subtract 1. It is exact for 1 and 2, and a bit low at the middle of the interval. For more on logs, see Doerfler's wonderful book.richard1941 –richard1941 2017-07-12 00:26:12 +00:00 Commented Jul 12, 2017 at 0:26
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12<2 m/n<13 12<2 m/n<13
log 2 12<m n<log 2 13 log 212<m n<log 213
[3;1,1,2,2,…]<m n<[3;1,2,2,1,…][3;1,1,2,2,…]<m n<[3;1,2,2,1,…]
The continued fractions match up to [3;1][3;1]. Since [3;1,1][3;1,1] and [3;1,2][3;1,2] are underestimations of log 2 12 log 212 and log 2 13 log 213, we take [3;1,1][3;1,1] (the "smaller" one lexicographically) and add 1 1 to the last number (round "up"), so the answer is [3;1,2][3;1,2].
[3;1,2]=3+1 1+1 2=3+1 3 2=3+2 3=11 3[3;1,2]=3+1 1+1 2=3+1 3 2=3+2 3=11 3
So m=11 m=11, n=3 n=3.
This works for all nonnegative bounds.
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answered Jul 6, 2017 at 17:49
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188717 | https://study.com/academy/lesson/what-is-bar-notation-in-math-definition-examples.html | Bar Notation Overview & Examples | What Does a Line Over a Number Mean? - Lesson | Study.com
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Bar Notation Overview & Examples | What Does a Line Over a Number Mean?
Lesson
Transcript
Vrushali Naik, Mark Boster
Author Vrushali Naik
Obtained a Masters degree in Mathematics from Mumbai university kalina. Possess 3 years of professional experience in teaching and 1 year professional experience as Subject Matter Expert, creating learning contents like learning videos, text book solutions, assessments etc.
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Instructor Mark Boster
In this lesson, learn about the bar notation commonly used in math. Understand what the line over a number means through examples and see when not to use it. Updated: 11/21/2023
Table of Contents
What Is Bar Notation?
Lesson Summary
Show
Frequently Asked Questions
How do you write a bar notation?
A horizontal line written over a repeating number after a decimal point is called a bar. That line denotes bar notation.
What does a bar over a number mean?
A bar over a number means that the number is being repeated indefinitely after a decimal point. That is why there is a bar over that number.
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Table of Contents
What Is Bar Notation?
Lesson Summary
Show
What Is Bar Notation? ---------------------
Geeta once visited a temple, and while praying to God, she heard people saying that God's love and blessing never comes to an end, and that day she came to know that God's love is infinite and never-ending. The same concept of infinity also lies in mathematics. This lesson is about understanding that mathematical concept in depth.
In one branch of mathematics, there are unending numbers that never come to a conclusion. They keep on repeating until infinity.
When a term is repeated indefinitely, it is a never-ending term that has no end, also called infinity.
Infinity is denoted as:
Infinity refers to something that never comes to an end.
If a number is 0.3333, then adding a bar on 3 will make it easier to write it as 0.3¯ instead of writing 3 'til infinity, and thus it can be said that adding a bar to a repeated number makes it easier to write the given number. To put it another way, a bar notation makes it easier to write a number that is being repeated.
The term ''bar'' refers to a horizontal line drawn on such repeated numerals.
The line over the number 2 represents a bar notation.
What Does a Line (Bar) Over a Number Mean?
A bar notation is a line over a number. When a number or a group of numbers are repeated, this is the term to employ.
Line over a number
Examples
1 divided by 9 equals 0.111111... Because the number 1 is repeated in the preceding example, it can be expressed as 0.1¯. The bar notation is indicated by the line above number 1. It can also be said that the line over a number means that the number is being repeated.
The line above 6 in the example 1.6¯ indicates that the number 6 is being repeated.
The number 0.353535... can be written as 0.3¯5¯
In the example above, two numerals, 3 and 5, repeat, therefore the bar is applied over each of them.
The number 7.543543543... can be written as 7.5¯4¯3¯ as the digits 543 are being repeated infinitely.
The number 67.345634563456... can be written as 67.3¯4¯5¯6¯
The number 34.6666... can be written as 34.6¯
Recurring Numbers and Terminating Numbers
There are two types of number: the numbers which come to an end and the number which keeps on repeating after a decimal point.
The numbers that are being repeated indefinitely are called repeating decimals. Recurring numbers are another name for repeating numbers.
There are two types of numbers in a number system:
Numbers that come to an end
Numbers that repeat themselves.
Recurring numbers are those that do not leave a zero remainder and continue to repeat, whereas the numbers that come to an end are called terminating numbers. This can be understood more clearly by the following examples:
The number 23.45 is a terminating number and 23.666... is a recurring number.
The number 0.9 is a terminating number and 0.999... is a recurring number.
The number 123.123 is a terminating number and 123.123123123... is a recurring number.
In example 1, the recurring number 23.666... can be written as 23.6¯, and the line over 6 means that 6 is repeated indefinitely.
In example 2, the recurring number 0.999... can be written as 0.9¯, which means that the number 9 is being repeated indefinitely.
In example 3, the recurring number 123.123123123... can be written as 123.1¯2¯3¯; the line over 123 means that the number 123 is being repeated indefinitely.
So, it is clear that the recurring numbers are given a bar notation, and the terminating numbers are not assigned with a bar notation.
Thus, from the earlier examples, it is clear that whether a repeating number is a single-digit number or a two-digit number or an n-digit number, the repeating digits can be assigned a bar.
As a result, it can be argued that a number can be repeated indefinitely after a decimal point. Now, if a number is 0.9¯, it signifies that the bar represents the number 9 being repeated indefinitely.
When Is Bar Notation Not Used?
Consider the numbers 10 and 01. It can be seen that the first number is ten and the second number is one. In the same way, consider 34 and 43; the first number is thirty-four and the second number is forty-three. That means the placement of digits in a number is very important.
Also consider the numbers 345 and 34.5; the number 345 is read as three hundred and forty-five, whereas the number 34.5 is read as thirty-four point five.
It is clear from the above that inserting a decimal point between two numbers alters their meaning.
Consider the number 444, which stands for four hundred and forty-four. This means that the provided number cannot be written as 4¯, because that would suggest that the number four goes on indefinitely, thus distorting the meaning of the number 444. To get a better understanding of the subject, take the number 444.444, which stands for four hundred and forty-four point four, four, and four. This is a terminating number, and it cannot be expressed as 4¯.4¯ since it would affect the meaning of 444.444.
Consider the number 444.444..., which stands for four hundred and forty-four point four, four, four etc.; that is, the four after the decimal point goes up until infinity, indicating that the provided number is a recurring number. As a result, the number can be written as 444.4¯ , not changing the meaning of 444.444...
As a result, it is clear that a bar notation can only be used on numbers that repeat after a decimal point, not before.
A bar notation can only be given after a decimal point, not before.
Examples:
565565565565... will be written as it is, and not as 5¯6¯5¯
23.888... can be written as 23.8¯
4545.222... can be written as 4545.2¯
77777.1111... can be written as 77777.1¯
321.123123123123... can be written as 321.1¯2¯3¯
275275275... will be written as it is and not as 2¯7¯5¯
19.21 can be written as it is.
5.3333... can be written as 5.3¯
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0:03 Who Is Mom's Favorite?
0:20 Forever and Forever
0:38 Forever and Forever in Math
2:09 When Not to Use Bar Notation
2:30 Lesson Summary
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Lesson Summary --------------
Bar notation refers to a horizontal line drawn on repeated numerals. Adding a line over a number makes it easier to write repeated numerals. The terms that keep on repeating are called never-ending terms, also called ''infinity''; the concept of infinity is usually symbolized as a figure-eight sign.
There are two types of numbers: terminating numbers, which are numbers that come to an end, and recurring numbers, which are numbers that repeat themselves after a decimal point.
Examples:
1)7.5 is a terminating number.
2) 7.55... is a recurring number which can be written as 7.5¯
A bar can be given to the numbers that are being repeated only after a decimal point and not before a decimal point.
Example:
1) 77.7 can be written as it is, like 77.7.
2) 7.77... cannot be written as 7.77... a bar notation can be used to represent the repeating number.
Lesson Summary --------------
Bar notation refers to a horizontal line drawn on repeated numerals. Adding a line over a number makes it easier to write repeated numerals. The terms that keep on repeating are called never-ending terms, also called ''infinity''; the concept of infinity is usually symbolized as a figure-eight sign.
There are two types of numbers: terminating numbers, which are numbers that come to an end, and recurring numbers, which are numbers that repeat themselves after a decimal point.
Examples:
1)7.5 is a terminating number.
2) 7.55... is a recurring number which can be written as 7.5¯
A bar can be given to the numbers that are being repeated only after a decimal point and not before a decimal point.
Example:
1) 77.7 can be written as it is, like 77.7.
2) 7.77... cannot be written as 7.77... a bar notation can be used to represent the repeating number.
Video Transcript
Forever and Forever
There is a symbol that stands for infinity. Infinity means that something goes on forever. For example, numbers go on forever. You can add 1 to any number to get a higher number. Look at the symbol and follow it around and around with your finger. You never get to the end of it. It goes on forever, infinitely.
Forever and Forever in Math
Sometimes in math there are numbers that go on forever. Most of the time, they repeat the same numbers over and over again. For example, some solutions for division equations go on forever. They repeat, and continue to repeat, on and on and on.
Here's an example for you:
1 / 3 = 0.33333333333333333333333333333333333333333333333...
The 3s could go on forever. You could spend the rest of your life writing 3s and you still wouldn't reach the end. The 3s go on for infinity. It just doesn't make sense to spend the rest of your life writing 3s, so in math we would say 1 / 3 = 0.3 with a line over the three to show that the three repeats forever. This is called bar notation. Bar notation is an easier way to write a repeating number by putting a line, or bar, over the repeating numbers.
Here's another example. 1 / 7 = 0.142857142857142857142857142857...
But using bar notation, you would say 1 / 7 = 0.142857 with a line over those numbers to show that they repeat over and over.
You can use the bar notation over any number that repeats the same number or numbers again after the decimal point. If I wanted to write the number 5.6788888888888 using bar notation, I would only put the bar above the first 8, because that is the number that is repeating forever.
When Not to Use Bar Notation
You cannot use bar notation before a decimal point. Here's why. If you write the number 88, you can't put a bar over the 8 because that would mean the number 8888888888888888... See, it's a totally different number. Bar notation may only be used on a repeating number or series of numbers AFTER the decimal point.
Lesson Summary
Using bar notation, which is a way to show repeating numbers after a decimal point, is easier than writing the same repeating number and over again. It shows that the number pattern goes on for infinity, or forever. Bar notation can only be used after the decimal point.
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188718 | https://docs.tibco.com/pub/business-studio-bpm-edition/5.1.0/doc/html/forms_users_guide/multiplicity-of-rela.htm?TocPath=Forms%20User's%20Guide%7CConcepts%7CBusiness%20Object%20Model%7CThe%20Objects%20in%20a%20Business%20Object%20Model%7C_____1 | Multiplicity of Relationships
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TIBCO Business Studio™ - BPM Edition 5.1.0
Forms User's Guide>Concepts>Business Object Model>The Objects in a Business Object Model>Multiplicity of Relationships
Multiplicity of Relationships
Relationships between BOM classes have a multiplicity, for instance, one-to-one (1..1), zero-to-many (0..), or one-to-many (1..). You can also have a finite lower or upper multiplicity bound like one-to-finite upper bound (1..m), finite lower bound-to-finite upper bound (n..m), or exactly finite bound (n). On a generated form, a particular pane type is rendered for a child class based on the multiplicity value.
If a Student class, for instance, has a child class called Course, with a 0.. relationship (meaning that one student can have many courses), the Course class will be rendered as a grid pane. The attributes of the Course class (for instance, course number, course name, time, room number, and so on.) will appear as columns in the grid pane. Each course for a given student will be represented by a row in the grid pane.
Implicit Validations
The multiplicity constraints defined in the BOM are reflected in the implicit validations. The validation messages conform to the following:
Validation Messages for BOM Level Multiplicity Constraints
| Multiplicity Constraint | Validation Message |
--- |
| One-to-many (1..) | Must have at least one value. |
| One-to-finite upper bound(1..m) | Must have between one and {m} values. |
| Finite lower bound-to-finite upper bound (n..m) | Must have between {n} and {m} values. |
| Zero-to-finite upper bound (0..m) | Must have between zero and {m} values. |
| Exactly one (1) | Must have exactly one value. |
| Exactly equal to the finite bound (n) | Must have exactly {n} values. |
These apply for both primitive attributes and complex children.
Note: The implicit validations for multiplicity constraints are configured to execute on form submit.
Master-Detail Panes
If a child class has a relationship to the parent class that allows multiple instances of the child class, and the child class itself contains a child class with multiple attributes, the two child classes will be rendered on the default form in a master-detail pane.
The first child, the master pane, will be rendered in the form as a grid pane, and the second child, the detail pane, will appear as a vertical or record pane which can be used for editing all attributes of both child classes.
Note: If you want the detail pane to be generated as a record pane, go to Preferences >Form Designer >Generator, and select the check box Generate master-detail configuration with record pane for details.
By default, the check box is cleared, and the detail pane is generated as a vertical pane. This information applies to the default forms and newly generated forms. The forms that are already generated, remain unaffected.
In this case, the grid pane will be read-only, but a row can be selected for editing in the vertical pane (detail) by clicking that row in the grid pane (master).
As an example, a Student class might be the parent of a child class called Course. Each student could have zero-to-many courses. The course class, in turn, might have a child class called Course Details. The BOM diagram is shown in the figure Business Object Model Editor Showing Child Classes.
Business Object Model Editor Showing Child Classes
The business object model shown in the figure Business Object Model Editor Showing Child Classes is rendered in a form with a master-detail pane for the Course and Course Details classes.
Selecting a row in the grid pane (that is, the master pane) allows that row to be edited in the vertical or record pane (that is, the detail pane). An alternate way of selecting rows for editing is to enable navigation for the record pane. Navigation is turned off by default, but is enabled by selecting the Show Navigator check box in the Properties tab of the record pane’s Properties view. The navigator then appears for the record pane.
Note: With navigation enabled, you can delete the grid pane from the form if you consider it unnecessary to provide users with two methods for selecting records to edit. However, you cannot do this for the vertical detail pane, as it is single-valued, and thus does not provide a navigator. You can manually refactor the detail pane from vertical to record, and then bind it to the correct data.
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188719 | https://brainly.com/question/20741981 | [FREE] Sponges are called the simplest animals because they: A. Have no nerves B. Have no muscles C. Have no - brainly.com
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Sponges are called the simplest animals because they:
A. Have no nerves
B. Have no muscles
C. Have no internal organs, including a brain
D. All of the above
E. None of the above
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Answer:
All of the above
Explanation:
This is because lack nerves, muscles, internal organs including brain and that is why they are refered to simplest animals. Sponges are multicellular animals that belong the phylum porifera and clade metazoa. Water current from the water bodies carry food to sponges and the same water current take away waste products. They have pores that allow water to channel through their bodies.
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Concepts of Biology - Samantha Fowler, Rebecca Roush, James Wise
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Sponges are referred to as the simplest animals because they lack nerves, muscles, and true internal organs, including a brain. Their structure is primarily designed for facilitating water flow, which is critical for their survival. Therefore, the correct answer is D: All of the above.
Explanation
Sponges are called the simplest animals due to several key characteristics that distinguish them from more complex animals.
No Nerves: Sponges lack a nervous system. They do not have nerve cells or any structures that can form a nervous system. Instead, they communicate internally through the flow of water and cell signaling.
No Muscles: Sponges do not have muscles. Their body structure does not allow for movement like more advanced animals, which can change their body shape or position actively. Sponges are primarily stationary and remain attached to surfaces underwater.
No Internal Organs, Including a Brain: Sponges do not possess any internal organs or a centralized brain. Their body is made up of specialized cells, but these cells do not form tissues. Instead of organs, sponges have a porous structure that allows water to circulate through them, enabling feeding, respiration, and excretion.
Due to these features, sponges are classified in the subkingdom Parazoa, which includes organisms that do not possess true tissues or organ systems. Their simple design represents an early stage of animal evolution, showcasing a body plan that prioritizes water flow and basic life functions over complexity.
Additionally, sponges play a vital role in aquatic ecosystems as filter feeders, cleaning the water and providing habitat for other organisms.
Examples & Evidence
An example of a sponge is the common bath sponge, which is used for bathing and cleaning due to its porous structure that helps absorb water. Another example is the glass sponge, known for its beautiful silica skeleton, which is found in deep-sea habitats.
Scientific classification places sponges in the phylum Porifera and notes their lack of true tissues, which supports their status as the simplest animals in biological classifications.
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Sponges are the simplest animals because they lack internal organs, including a brain, true tissues, nerves, and muscles, and exhibit cell-level organization.
Explanation
Sponges are called the simplest animals because they have no internal organs, including a brain. These organisms, part of the phylum Porifera, are unique in several ways. Unlike other animals, sponges lack true tissues and possess a cell-level organization. These specialized cells perform specific jobs necessary for the sponge's survival. Moreover, they exhibit a body design that lacks structures like nerves or muscles, making them distinct in the animal kingdom. Sponges also adapt to their aquatic environments by filtering water through their porous bodies for excretion, feeding, and gas exchange, which involves structures such as canals, chambers, and cavities. This simplicity and lack of motility in the adult stage reflect their early evolutionary development compared to more complex organisms.
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Community Answer Sponges are the simplest of animals and have no nerves or ___. They have no body systems.
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188720 | https://chemistry.stackexchange.com/questions/124603/equivalence-point-vs-inflection-point-in-the-titration-curve-of-a-weak-acid | Equivalence point vs inflection point in the titration curve of a weak acid - Chemistry Stack Exchange
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Equivalence point vs inflection point in the titration curve of a weak acid
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Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as N a O H N a O H. When the number (and moles) of hydroxide ions coming directly from the base is equal to the amount of hydronium ions due to the acid, here we have the equivalence point.
When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.
In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?
In other words, what does the inflection point indicate, that p H p H is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in p H p H is maximum) or the equivalence point itself?
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edited Dec 4, 2019 at 16:20
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asked Dec 3, 2019 at 14:54
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Comments are not for extended discussion; this conversation has been moved to chat.andselisk –andselisk♦ 2019-12-03 18:09:31 +00:00 Commented Dec 3, 2019 at 18:09
@Poutnik I saw where the mistake was; I edited now, this is what I meant Shootforthemoon –Shootforthemoon 2019-12-04 16:22:16 +00:00 Commented Dec 4, 2019 at 16:22
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Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as N a O H N a O H. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point.
The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount equivalent to the originally present weak acid. At this point, [H X 3 O X+]<[O H X−][H X 3 O X+]<[O H X−], so p H>7 p H>7.
When titrating a strong monoprotic acid the equivalence points coincides with the inflection of the titration curve, and pH is seven.
For the approximative experimental chemical approach, it is true.
But mathematically, even this is not exactly true, as the curve is slightly asymmetric. When there is the same molar amount of hydroxide as was the acid, the solution volume is 3 times bigger. And hydroxide concentration 3 times lower.
It would be symmetric, if we constructed the graph for the constant total volume and complementing strong acid and hydroxide solution volumes, as shown in the formula below.
p H=f(V a c i d V a c i d+V h y d r o x i d e)p H=f(V a c i d V a c i d+V h y d r o x i d e)
And the concentrations of both should be the same.
In our case, however, the aqueous solution results to be slightly basic at the equivalence point. Therefore, does the inflection point on our curve still coincide with the equivalence point or it is to be found slightly before?
For weak acids, the coincidence is mathematically even less true, as the curve shape before and after equivalence differs.
But for the chemical experimental point of view, the difference is still negligible, as the equivalence/inflection coincidence error is much less then experimental errors.
Both inflection and equivalence points ( approximately identical ) lay in the alkalic region.
In other words, what does the inflection point indicate, that p H p H is seven (so the inflection is distinct but may coincide with the equivalence point, and shows where the rate of increase in p H p H is maximum) or the equivalence point itself?
For weak acid/strong base ( and vice versa ) titrations, the neutral p H p H does not play any special role, so this point on the curve is insignificant.
Neither is there at the neutral point any inflection point, unless by chance p K a=7 p K a=7. In such a case, the inflection point at the maximal acid/salt buffering capacity at the half equivalence would be at p K a=7 p K a=7.
For further reference, see e.g Titration_of_a_Weak_Acid_with_a_Strong_Base
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edited Dec 5, 2019 at 8:06
answered Dec 3, 2019 at 18:11
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4Inflection point, slope, and equivalence point in a titration
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0What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?
1Acid base titration curve
3What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from?
2How would a pH curve look like for titration of diluted weak base compared to concentrated one?
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188721 | https://brilliant.org/wiki/selection-bias/ | Selection Bias | Brilliant Math & Science Wiki
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Adam Strandberg and Arron Kau contributed
A selection bias is a bias that comes from a difference between the distribution of data sampled in a study and the distribution of the population selected from. The fundamental problem of statistics in science is to try to infer general properties from a small set of observations. Concerns about selection bias are important for designing any scientifically valid experiment. Randomized controlled trials are the gold standard for attempting to eliminate selection bias, but the question must always be: randomized relative to what? For instance, if a study looking at academic achievement picks randomly from students visiting a website on college applications, it will get very different results than if it had randomly picked from all high school students.
polling
survivorship bias
Berkson's Paradox Berkson's Paradox is a specific type of selection bias resulting when the data is selected from a range where A A A
Contents
Polling Data
Generalizability of Psychology
Publication Bias
Anthropic Selection Bias
References
Polling Data
When
In addition to bias from who the polling agency contacts, there is self-selection bias in who decides to respond. The surveyors cannot force people to participate, so they are more likely to An even stronger bias results from determining public opinion by looking at letters sent or calls made to politicians' offices. The people who call or write are the ones with the largest grievances, so looking at them will give an inaccurate representation of public opinion.
Generalizability of Psychology
The Müller-Lyer line illusion
Psychologists conduct studies on the way people perceive themselves and the world, in the hopes of figuring out universally how humans work. However, most research psychologists are based out of universities in wealthy cities in the United States. Out of those cities, the subjects are overwhelmingly likely to be students, especially students in psychology. The results from any individual study are likely not to be universal psychology, but WEIRD psychology: W estern, E ducated, I ntelligent, R ich, D emocratic psychology.
The Müller-Lyer line illusion, depicted to the right, shows two lines with arrows on either side. Although both lines are the same length, many studies showed that people systematically assume the first line (with arrows pointing inward) is shorter than the second line (with arrows pointing outward). However, this turned out to be remarkably strong in American undergraduates and European samples, but very small in other cultures. Something as seemingly basic as visual perception varied in ways that were not captured by the sample of people studied.
Publication Bias
A funnel plot looking at studies on stereotype threat. The triangles represent the ideal plot. Notice that there are many more data points to the left of the triangle than to the right.
Individual studies can only tell researchers a certain amount about whether a particular hypothesis is true. A meta-analysis is a study that combines the results of several other studies to get a sense of what the literature as a whole says. In addition to bias resulting from selection of data for individual studies, there can be an additional level of bias when considering the selection of studies that are available for study, otherwise known as publication bias. There are many reasons that studies that are done either don't get reported or have distorted statistical results. For instance, scientific journals prefer to publish studies that found an effect was true rather than no effect at all. Even if in reality there is no effect, if researchers do small enough studies they are likely to find one. If there are a number of small studies performed, some of which find an effect and some of which don't, the papers that end up getting published will make the effect seem stronger than it actually is.
To help identify publication bias, researchers can use a funnel plot. It plots the results of the studies considered in the meta-analysis, comparing the effect size to the certainty of that result. Ideally, the largest studies with the most certain results should be closest to the true effect size, whereas smaller studies with less certain results should be spread out more. This creates the funnel shape, which is a triangle that is symmetric about the true effect size. If there is publication bias, the plot will instead be skewed to one side or the other. In the example plot, there are more studies on the left-hand side, indicating that studies systematically published more negative effects than they should have.
It is possible to have results that are systematically skewed from the
Anthropic Selection Bias
Very small changes in the cosmological constant (one of the constants in general relativity) can lead to very large changes in the shape of the universe.
Selection bias plays a large and controversial role in physical theories of the multiverse. In our best physical theories, there are a number of constants, such as the mass of the electron and the cosmological constant, that determine the shape of our universe. However, if we thought of these constant as "knobs" that could be adjusted, there is an extremely small range of values they could take before the universe would be completely incapable of supporting human life as we know it. This is known as the fine-tuning problem. One way physicists try to explain this is by imagining that all possible universes with all possible combinations of constants "exist" in some sense, and then applying selection bias.
Anthropic bias ("anthro-" meaning "human," as in "anthropology") is the bias resulting from the fact that humans can only exist in a universe that is capable of supporting human life. Almost tautologically, we can only observe universes in which observers exist. If this is so, it would be difficult or impossible to find what the true distribution of other universes that exist is. There is not much that can be said about how likely or unlikely it is for any given universe to be able to support life or consciousness. Unlike the case of WEIRD psychology, we can't just get more funding to do studies in universes with alternate physics.
Anthropic reasoning is also important for reasoning about catastrophic risks. When the Large Hadron Collider was turned on, there was public fear about whether the high-energy collisions it made would create a black hole that would consume the Earth. Physicists had good reason to believe this wouldn't happen, since there are many higher-energy collisions that happen all the time, when cosmic rays hit the Earth.
However, if a black hole did consume the Earth, we would no longer exist, and therefore wouldn't be around to question how likely it would be for a black hole to consume the Earth. This makes reasoning about catastrophic risks extremely difficult.
References
Flore, P., & Wicherts, J. (2015). Does stereotype threat influence performance of girls in stereotyped domains? A meta-analysis. Journal of School Psychology.
Davis, O. File:Flatness problem density graph.svg. Retrieved from
Bostrom, N. (2010). Anthropic Bias. Routledge.
Cite as: Selection Bias. Brilliant.org. Retrieved 06:40, September 29, 2025, from
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188723 | https://www.youtube.com/watch?v=2rmwAH-xdOM | Derivation of radioactive decay formula AQA Alevel physics
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Description
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Posted: 11 Dec 2018
How to derive the radioactive decay formula from the graph.
Note: This derivation is not on the spec, but useful
AQA A level specification - post 2015
Music: TheFatRat - Unity
1 comments
Transcript:
good evening and welcome to TR physics and today I'm going to go through the derivation of the radioactive formula decay formula now this is not on the specification at all however for those who are interested it is really useful to understand where it's come from those who are not interested I do have my next video will be incorporating how to use this formula but what I'm going to be doing today is deriving it just see where it comes from so on the floor behind me I have this is a graph of the number of radioactive atoms I having a substance and the time okay and I spoke about the fact that at each point the gradient was this at the gradient of the line was minus lambda the decay constant of the material times by the number of particles at that point and that was a constant throughout so at this point here every tala took a gradient I could relate the gradient to the number of particles and I noticed that this decay constant was constant so the gradient of the line is represented by this formula here change in n over change in time equals minus and lambda I'm going to change a little bit of the notation so I'm going to change this to D and so Delta changing okay I'm not going to do a rear-engine so first off I'm going to rearrange this formula so we end up with 1 over n DN equal minus lambda DT ok so minus n DN and lambda C and wanted to do so I've got the gradient of the line and I'm trying to find the equation of the line so to undo differentiation I have to integrate so I'm going to integrate these values yep and I'm going to integrate them so I'm not going to use indefinite or these are definite integral I'll I'm going to integrate them to between point N and the starting point starting point was zero to time T so integrating them I end up with 1 over N integrated is learn n between N and n naught equals minus lambda T between C and 0 I'm now going to input my values in for the respective values so I have Lin and take away learn and not like I said so that's how you deal with different integrals you put one in and they take away the other equals well we'll put C in when I put 0 in e equals nothing ok Goody's log gloss to simplify this into a 1 logarithm okay I'll do that on this board here so that's live n divided by n naught equals minus lambda T so undo a natural logarithm okay please remember that natural of them is a log to the base e so e to the power of this would equal this here so n over N naught equals e to the power minus lambda T and you end up with N equals n naught e to the minus lambda T and that there is the formula you have in datasheets you can of course relate the number of particles from the number of initial particles to activity so you can use a equals a naught e to the minus lambda d true okay because and a are directly related to each other by a constant of lambda so that is the derivation of the radioactive decay formula it's not on the specification but it's really useful to just know where it comes from [Music] |
188724 | https://www.chegg.com/homework-help/questions-and-answers/q2-show-normal-depth-triangular-channel-side-slopes-m-horizontal-1-vertical-given-73-8-qn--q76298848 | Solved Q.2 show that the normal depth in a triangular | Chegg.com
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Q.2 show that the normal depth in a triangular channel of side slopes m horizontal: 1 vertical, is given by 73/8 Q.n y = 1.1892 5. 71/8 m- +1 5 m Q.3 For the compound channel shown in Fig.-1, determine the discharge for a depth of flow y equals to (i) 2.0 m and (ii) 2.4 m. Take n=0.025 and So=0.0003 for all parts of perimeter. -7.0 +70 0.9 + 3m Fig.-1
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Question: Q.2 show that the normal depth in a triangular channel of side slopes m horizontal: 1 vertical, is given by 73/8 Q.n y = 1.1892 5. 71/8 m- +1 5 m Q.3 For the compound channel shown in Fig.-1, determine the discharge for a depth of flow y equals to (i) 2.0 m and (ii) 2.4 m. Take n=0.025 and So=0.0003 for all parts of perimeter. -7.0 +70 0.9 + 3m Fig.-1
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For calculating discharge at y=2 m, start by finding the sub-areas (1) and (3) together, then calculate the area A 1 and the wetted perimeter P 1. View the full answer Previous questionNext question
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Q.2 show that the normal depth in a triangular channel of side slopes m horizontal: 1 vertical, is given by 73/8 Q.n y = 1.1892 5. 71/8 m- +1 5 m Q.3 For the compound channel shown in Fig.-1, determine the discharge for a depth of flow y equals to (i) 2.0 m and (ii) 2.4 m. Take n=0.025 and So=0.0003 for all parts of perimeter. -7.0 +70 0.9 + 3m Fig.-1
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188725 | https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem | Jump to content
Spin–statistics theorem
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From Wikipedia, the free encyclopedia
Theorem in quantum mechanics
| 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 |
The spin–statistics theorem proves that the observed relationship between the intrinsic spin of a particle (angular momentum not due to the orbital motion) and the quantum particle statistics of collections of such particles is a consequence of the mathematics of quantum mechanics.
According to the theorem, the many-body wave function for elementary particles with integer spin (bosons) is symmetric under the exchange of any two particles, whereas for particles with half-integer spin (fermions), the wave function is antisymmetric under such an exchange. A consequence of the theorem is that non-interacting particles with integer spin obey Bose–Einstein statistics, while those with half-integer spin obey Fermi–Dirac statistics.
Background
[edit]
The statistics of indistinguishable particles is among the most fundamental of physical effects. The Pauli exclusion principle – that every occupied quantum state contains at most one fermion – controls the formation of matter. The basic building blocks of matter such as protons, neutrons, and electrons are all fermions. Conversely, photon and other particles which mediate forces between matter particles, are bosons. A spin–statistics theorem attempts to explain the origin of this fundamental dichotomy.: 4
Naively, spin, an angular momentum property intrinsic to a particle, would be unrelated to fundamental properties of a collection of such particles. However, these are indistinguishable particles: any physical prediction relating multiple indistinguishable particles must not change when the particles are exchanged. [citation needed]
Quantum states and indistinguishable particles
[edit]
In a quantum system, a physical state is described by a state vector. A pair of distinct state vectors are physically equivalent if they differ only by an overall phase factor, ignoring other interactions. A pair of indistinguishable particles such as this have only one state. This means that if the positions of the particles are exchanged (i.e., they undergo a permutation), this does not identify a new physical state, but rather one matching the original physical state. In fact, one cannot tell which particle is in which position. [citation needed]
While the physical state does not change under the exchange of the particles' positions, it is possible for the state vector to change sign as a result of an exchange. Since this sign change is just an overall phase, this does not affect the physical state. [citation needed]
The essential ingredient in proving the spin-statistics relation is relativity, that the physical laws do not change under Lorentz transformations. The field operators transform under Lorentz transformations according to the spin of the particle that they create, by definition. [citation needed]
Additionally, the assumption (known as microcausality) that spacelike-separated fields either commute or anticommute can be made only for relativistic theories with a time direction. Otherwise, the notion of being spacelike is meaningless. However, the proof involves looking at a Euclidean version of spacetime, in which the time direction is treated as a spatial one, as will be now explained. [citation needed]
Lorentz transformations include 3-dimensional rotations and boosts. A boost transfers to a frame of reference with a different velocity and is mathematically like a rotation into time. By analytic continuation of the correlation functions of a quantum field theory, the time coordinate may become imaginary, and then boosts become rotations. The new "spacetime" has only spatial directions and is termed Euclidean. [citation needed]
Exchange symmetry or permutation symmetry
[edit]
Bosons are particles whose wavefunction is symmetric under such an exchange or permutation, so if we swap the particles, the wavefunction does not change. Fermions are particles whose wavefunction is antisymmetric, so under such a swap the wavefunction gets a minus sign, meaning that the amplitude for two identical fermions to occupy the same state must be zero. This is the Pauli exclusion principle: two identical fermions cannot occupy the same state. This rule does not hold for bosons. [citation needed]
In quantum field theory, a state or a wavefunction is described by field operators operating on some basic state called the vacuum. In order for the operators to project out the symmetric or antisymmetric component of the creating wavefunction, they must have the appropriate commutation law. The operator
(with an operator and a numerical function with complex values) creates a two-particle state with wavefunction , and depending on the commutation properties of the fields, either only the antisymmetric parts or the symmetric parts matter.
Let us assume that and the two operators take place at the same time; more generally, they may have spacelike separation, as is explained hereafter.
If the fields commute, meaning that the following holds:
then only the symmetric part of contributes, so that , and the field will create bosonic particles.
On the other hand, if the fields anti-commute, meaning that has the property that
then only the antisymmetric part of contributes, so that , and the particles will be fermionic.
Proofs
[edit]
An elementary explanation for the spin–statistics theorem cannot be given despite the fact that the theorem is so simple to state. In The Feynman Lectures on Physics, Richard Feynman said that this probably means that we do not have a complete understanding of the fundamental principle involved.
Numerous notable proofs have been published, with different kinds of limitations and assumptions. They are all "negative proofs", meaning that they establish that integer spin fields cannot result in fermion statistics while half-integral spin fields cannot result in boson statistics.: 487
Proofs that avoid using any relativistic quantum field theory mechanism have defects. Many such proofs rely on a claim that
where the operator permutes the coordinates. However, the value on the left-hand side represents the probability of particle 1 at , particle 2 at , and so on, and is thus quantum-mechanically invalid for indistinguishable particles.: 567
The first proof was formulated in 1939 by Markus Fierz, a student of Wolfgang Pauli, and was rederived in a more systematic way by Pauli the following year.
In a later summary, Pauli listed three postulates within relativistic quantum field theory as required for these versions of the theorem:
Any state with particle occupation has higher energy than the vacuum state.
Spatially separated measurements do not disturb each other (they commute).
Physical probabilities are positive (the metric of the Hilbert space is positive-definite).
Their analysis neglected particle interactions other than commutation/anti-commutation of the state.: 374
In 1949 Richard Feynman gave a completely different type of proof based on vacuum polarization, which was later critiqued by Pauli.: 368 Pauli showed that Feynman's proof explicitly relied on the first two postulates he used and implicitly used the third one by first allowing negative probabilities but then rejecting field theory results with probabilities greater than one.
A proof by Julian Schwinger in 1950 based on time-reversal invariance followed a proof by Frederik Belinfante in 1940 based on charge-conjugation invariance, leading to a connection to the CPT theorem more fully developed by Pauli in 1955. These proofs were notably difficult to follow.: 393
Work on the axiomatization of quantum field theory by Arthur Wightman lead to a theorem that stated that the expectation value of the product of two fields, , could be analytically continued to all separations .: 425 (The first two postulates of the Pauli-era proofs involve the vacuum state and fields at separate locations.) The new result allowed more rigorous proofs of the spin–statistics theorems by Gerhart Lüders and Bruno Zumino and by Peter Burgoyne.: 393 In 1957 Res Jost derived the CPT theorem using the spin–statistics theorem, and Burgoyne's proof of the spin–statistics theorem in 1958 required no constraints on the interactions nor on the form of the field theories. These results are among the most rigorous practical theorems.: 529
In spite of these successes, Feynman, in his 1963 undergraduate lecture that discussed the spin–statistics connection, says: "We apologize for the fact that we cannot give you an elementary explanation." Neuenschwander echoed this in 1994, asking whether there was any progress, spurring additional proofs and books. Neuenschwander's 2013 popularization of the spin–statistics connection suggested that simple explanations remain elusive.
Experimental tests
[edit]
In 1987 Greenberg and Mohapatra proposed that the spin–statistics theorem could have small violations.
With the help of very precise calculations for states of the He atom that violate the Pauli exclusion principle, Deilamian, Gillaspy and Kelleher looked for the 1s2s 1S0 state of He using an atomic-beam spectrometer. The search was unsuccessful with an upper limit of 5×10−6.
Relation to representation theory of the Lorentz group
[edit]
The Lorentz group has no non-trivial unitary representations of finite dimension. Thus it seems impossible to construct a Hilbert space in which all states have finite, non-zero spin and positive, Lorentz-invariant norm. This problem is overcome in different ways depending on particle spin–statistics. [citation needed]
For a state of integer spin the negative norm states (known as "unphysical polarization") are set to zero, which makes the use of gauge symmetry necessary. [citation needed]
For a state of half-integer spin the argument can be circumvented by having fermionic statistics.
Composite particles
[edit]
The spin–statistics theorem applies not only to elementary particles but also to composite particles formed from them, provided that the internal structure of the composites is identical and they remain bound under the conditions being considered. One can consider the many-body wave function for the composite particles. If all the constituent elementary particles in one composite are simultaneously exchanged with those in another, the resulting sign change of the wave function is determined by the number of fermions within each composite. In such systems, the total spin of the composite particle arises from the quantum mechanical addition of the angular momenta of its constituents: if the number of constituent fermions is even, the composite has integer spin and behaves as a boson with a symmetric wave function; if the number is odd, the spin is half-integer and the composite behaves as a fermion with an antisymmetric wave function.
Hadrons are composite subatomic particles made of quarks bound together by the strong interaction. Quarks are fermions with spin of 1/2. Hadrons fall into two main categories: baryons, which consist of an odd number of quarks (typically three), and mesons, which consist of an even number of quarks (typically a quark and an antiquark). Baryons, such as protons and neutrons, are fermions due to their odd number of constituent quarks. Mesons, like pions, are bosons because they contain an even number of quarks.
The effect that quantum statistics have on composite particles is evident in the superfluid properties of the two helium isotopes, helium-3 and helium-4. In neutral atoms, each proton is always matched by one electron, so that the total number of protons plus electrons is always even. Therefore, an atom behaves as a fermion if it contains an odd number of neutrons, and as a boson if the number of neutrons is even. Helium-3 has one neutron and is a fermion, while helium-4 has two neutrons and is a boson. At a temperature of 2.17 K, helium-4 undergoes a phase transition to a superfluid state that can be understood as a type of Bose–Einstein condensate. Such a mechanism is not directly available for the fermionic helium-3, which remains a normal liquid to much lower temperatures. Below 2.6 mK, helium-3 also transitions into a superfluid state. This is achieved by a mechanism similar to superconductivity: the interactions between helium-3 atoms first bind the atoms into Cooper pairs, which are again bosonic, and the pairs can then undergo Bose-Einstein condensation.
Although composite bosons exhibit similar behavior as elementary bosons, the fermionic nature of their constituents sometimes introduces subtle effects due to the Pauli exclusion principle. These effects limit how closely the composite bosons can be packed, and are especially significant in dense systems. They are sometimes modelled as effective interactions between composites.
Quasiparticle anyons in 2 dimensions
[edit]
Main article: Anyon
In 1982, physicist Frank Wilczek published a research paper on the possibilities of possible fractional-spin particles, which he termed anyons from their ability to take on "any" spin. He wrote that they were theoretically predicted to arise in low-dimensional systems where motion is restricted to fewer than three spatial dimensions. Wilczek described their spin statistics as "interpolating continuously between the usual boson and fermion cases". The effect has become the basis for understanding the fractional quantum Hall effect.
See also
[edit]
Anyonic statistics
Braid statistics
Parastatistics
References
[edit]
^ Leggett, Anthony J. (2006). Quantum liquids: Bose condensation and Cooper pairing in condensed-matter systems. Oxford graduate texts. Oxford University Press. pp. 5, 8–10. ISBN 978-0-19-171195-4.
^ Dirac, Paul Adrien Maurice (1981-01-01). The Principles of Quantum Mechanics. Clarendon Press. p. 149. ISBN 9780198520115.
^ Pauli, Wolfgang (1980-01-01). General principles of quantum mechanics. Springer-Verlag. ISBN 9783540098423.
^ Jump up to: a b c d e f g h Duck, Ian; Sudarshan, Ennackel Chandy George; Sudarshan, E. C. G. (1998). Pauli and the spin-statistics theorem (1. reprint ed.). Singapore: World Scientific. ISBN 978-981-02-3114-9.
^ Jump up to: a b Feynman, Richard P.; Robert B. Leighton; Matthew Sands (1965). The Feynman Lectures on Physics. Vol. 3. Addison-Wesley. p. 4.1. ISBN 978-0-201-02118-9.
^ Curceanu, Catalina; Gillaspy, J. D.; Hilborn, Robert C. (2012-07-01). "Resource Letter SS–1: The Spin-Statistics Connection". American Journal of Physics. 80 (7): 561–577. doi:10.1119/1.4704899. ISSN 0002-9505.
^ Markus Fierz (1939). "Über die relativistische Theorie kräftefreier Teilchen mit beliebigem Spin". Helvetica Physica Acta (in German). 12 (1): 3–37. Bibcode:1939AcHPh..12....3F. doi:10.5169/seals-110930.
^ Wolfgang Pauli (15 October 1940). "The Connection Between Spin and Statistics". Physical Review. 58 (8): 716–722. Bibcode:1940PhRv...58..716P. doi:10.1103/PhysRev.58.716.
^ Jump up to: a b Wolfgang Pauli (1950). "On the Connection Between Spin and Statistics". Progress of Theoretical Physics. 5 (4): 526–543. Bibcode:1950PThPh...5..526P. doi:10.1143/ptp/5.4.526.
^ Richard Feynman (1961). "The theory of positrons". Quantum Electrodynamics. Basic Books. ISBN 978-0-201-36075-2. {{cite book}}: ISBN / Date incompatibility (help) A reprint of Feynman's 1949 paper in Physical Review.
^ Julian Schwinger (June 15, 1951). "The Quantum Theory of Fields I". Physical Review. 82 (6): 914–917. Bibcode:1951PhRv...82..914S. doi:10.1103/PhysRev.82.914. S2CID 121971249.
^ Pauli, Wolfgang (1988). "Exclusion Principle, Lorentz Group and Reflection of Space-Time and Charge". In Enz, Charles P.; v. Meyenn, Karl (eds.). Wolfgang Pauli (in German). Wiesbaden: Vieweg+Teubner Verlag. pp. 459–479. doi:10.1007/978-3-322-90270-2_41. ISBN 978-3-322-90271-9.
^ Lüders, Gerhart; Zumino, Bruno (1958-06-15). "Connection between Spin and Statistics". Physical Review. 110 (6): 1450–1453. doi:10.1103/PhysRev.110.1450. ISSN 0031-899X.
^ Pais, Abraham (2002). Inward bound: of matter and forces in the physical world (Reprint ed.). Oxford: Clarendon Press [u.a.] ISBN 978-0-19-851997-3.
^ Neuenschwander, Dwight E. (1994-11-01). "Question #7. The spin–statistics theorem". American Journal of Physics. 62 (11): 972. doi:10.1119/1.17652. ISSN 0002-9505.
^ Neuenschwander, Dwight E. (2015-07-28). "The Spin-Statistics Theorem and Identical Particle Distribution Functions". Radiations. p. 27.
^ Greenberg, O. W.; Mohapatra, R. N. (1987-11-30). "Local Quantum Field Theory of Possible Violation of the Pauli Principle". Physical Review Letters. 59 (22): 2507–2510. doi:10.1103/PhysRevLett.59.2507. ISSN 0031-9007. PMID 10035570.
^ Hilborn, Robert C. (1995-04-01). "Answer to Question #7 ["The spin-statistics theorem", Dwight E. Neuenschwander, Am. J. Phys. 62 (11), 972 (1994)]". American Journal of Physics. 63 (4): 298–299. doi:10.1119/1.17953. ISSN 0002-9505.
^ Drake, G. W. F. (1989). "Predicted energy shifts for "paronic" Helium". Phys. Rev. A. 39 (2): 897–899. Bibcode:1989PhRvA..39..897D. doi:10.1103/PhysRevA.39.897. PMID 9901315. S2CID 35775478.
^ Deilamian, K.; et al. (1995). "Search for small violations of the symmetrization postulate in an excited state of Helium". Phys. Rev. Lett. 74 (24): 4787–4790. Bibcode:1995PhRvL..74.4787D. doi:10.1103/PhysRevLett.74.4787. PMID 10058599.
^ Peskin, Michael E.; Schroeder, Daniel V. (1995). An Introduction to Quantum Field Theory. Addison-Wesley. ISBN 0-201-50397-2.
^ Jump up to: a b Leggett, A.J.; Javan, R. (2005-01-01), "Quantum Mechanics: Foundations", in Bassani, Franco; Liedl, Gerald L.; Wyder, Peter (eds.), Encyclopedia of Condensed Matter Physics, Oxford: Elsevier, pp. 99–108, ISBN 978-0-12-369401-0, retrieved 2023-03-13
^
Amsler, C.; et al. (Particle Data Group) (2008). "Quark Model" (PDF). Physics Letters B. Review of Particle Physics. 667 (1): 1–6. Bibcode:2008PhLB..667....1A. doi:10.1016/j.physletb.2008.07.018. hdl:1854/LU-685594.
^ Leggett, Anthony J. (2006). Quantum liquids: Bose condensation and Cooper pairing in condensed-matter systems. Oxford graduate texts. Oxford University Press. pp. 3–8. ISBN 978-0-19-171195-4.
^ Combescot, Monique; Combescot, Roland; Dubin, François (2017-06-01). "Bose–Einstein condensation and indirect excitons: a review". Reports on Progress in Physics. 80 (6): 066501. doi:10.1088/1361-6633/aa50e3. ISSN 0034-4885.
^ Jump up to: a b Wilczek, Frank (4 October 1982). "Quantum Mechanics of Fractional-Spin Particles" (PDF). Physical Review Letters. 49 (14): 957–959. Bibcode:1982PhRvL..49..957W. doi:10.1103/PhysRevLett.49.957.
^ Laughlin, R. B. (1999-07-01). "Nobel Lecture: Fractional quantization". Reviews of Modern Physics. 71 (4): 863–874. doi:10.1103/RevModPhys.71.863. ISSN 0034-6861.
^ Murthy, Ganpathy; Shankar, R. (2003-10-03). "Hamiltonian theories of the fractional quantum Hall effect". Reviews of Modern Physics. 75 (4): 1101–1158. doi:10.1103/RevModPhys.75.1101. ISSN 0034-6861.
Further reading
[edit]
Duck, Ian; Sudarshan, E. C. G. (1998). "Toward an understanding of the spin–statistics theorem". American Journal of Physics. 66 (4): 284–303. Bibcode:1998AmJPh..66..284D. doi:10.1119/1.18860.
Streater, Ray F.; Wightman, Arthur S. (2000). PCT, Spin & Statistics, and All That (5th ed.). Princeton: Princeton University Press. ISBN 0-691-07062-8.
Jabs, Arthur (2010). "Connecting spin and statistics in quantum mechanics". Foundations of Physics. 40 (7): 776–792. arXiv:0810.2399. Bibcode:2010FoPh...40..776J. doi:10.1007/s10701-009-9351-4. S2CID 122488238.
External links
[edit]
A nice nearly-proof at John Baez's home page
Animation of the Dirac belt trick with a double belt, showing that belts behave as spin 1/2 particles
Animation of a Dirac belt trick variant showing that spin 1/2 particles are fermions
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188726 | https://teachyoumaths.com/Intervals/Properties/ | Properties of Inequalities
Introduction:
Inequalities are mathematical statements that compare two expressions using inequality symbols such as <, ≤, >, or ≥. The properties of inequalities help us understand how they behave under various operations. Let’s break down each of the four properties you mentioned:
1. Non-negative Property
If a number or expression is greater than or equal to zero, it is non-negative. That is, The non-negative property of inequalities refers to the fact that, if a number or expression is greater than or equal to zero, it is non-negative. This property tells us that if a≥0, then The non-negative property of inequalities refers to the fact that, if a number or expression is greater than or equal to zero, it is non-negative. This property tells us that if a≥0, then a is either positive or zero. Also, the square of a number is always positive, that is a 2 ≥ 0.
Example:
5 ≥ 0 (True) This is true because 5 is positive.
0 ≥ 0 (True) This is true because 0 is equal to 0.
-2 ≥ 0 (False) This is false because -2 is negative.
2. Addition Property
The addition property of inequalities states that if you add the same value to both sides of an inequality, the inequality remains true. In other words,
if 𝑎 ≤ 𝑏 , then 𝑎 + 𝑐 ≤ 𝑏 + 𝑐 for any real number 𝑐.
That is, adding the same value to both sides of an inequality keeps it valid.
Example:
Given 3 < 7, adding 2 to both sides gives 3 + 2 < 7 + 2 , which simplifies to 5 < 9, and the inequality remains valid.
Given x ≥ -1, adding 4 to both sides gives 𝑥 + 4 ≥ − 1 + 4 , or x + 4 ≥ 3.
3. Multiplication Property
The multiplication property of inequalities has two key rules depending on whether you multiply by a positive or negative number:That is, multiplying both sides by a positive number keeps the sign, but multiplying by a negative number flips the inequality sign.
Multiplying by a Positive Number: If 𝑎 ≤ 𝑏 , and 𝑐 > 0 , then 𝑎 𝑐 ≤ 𝑏 𝑐 .
Multiplying by a Negative Number: If 𝑎 ≤ 𝑏 , and 𝑐 < 0 , then 𝑎 𝑐 ≥ 𝑏 𝑐 (the inequality sign flips).
Example:
Positive multiplication: Multiplying 2 < 5 by 3 gives 6 < 15 (True). That is, Given 2 < 5 , multiplying both sides by 3 (a positive number) gives 2 × 3 < 5 × 3 , or 6 < 15 , which is true.
Negative multiplication: Multiplying 2 < 5 by -2 gives -4 > -10 (True). That is, Given 2 < 5 , multiplying both sides by -2 (a negative number) flips the inequality to 2 × ( − 2 ) > 5 × ( − 2 ) , or − 4 > − 10 , which is also true.
This property is crucial when solving inequalities, as it determines whether you need to reverse the inequality sign.
4. Reciprocal Property
The reciprocal property of inequalities states that when you take the reciprocal (or inverse) of both sides of a positive inequality, the inequality flips. This applies when the terms are positive, and you switch the direction of the inequality when taking reciprocals.
If 0 < 𝑎 < 𝑏 , then 1/ 𝑏 < 1/ 𝑎.
Example:
Given 1 < 4, the reciprocals give 1 > 0.25. Tha is, given 1 < 4 , taking the reciprocal of both sides gives 1/ 1 > 1/ 4 , or 1 > 0.25 , which is true.
Given 2 > 0.5, the reciprocals give 0.5 < 2. That is, Given 2 > 0.5 , taking the reciprocal of both sides gives 1/ 2 < 2 , which is true.
Summary of Key Points:
Non-negative Property: A non-negative number is greater than or equal to zero.
Addition Property: Adding the same number to both sides of an inequality keeps the inequality valid.
Multiplication Property: Multiplying by a positive number keeps the inequality sign the same, but multiplying by a negative number flips the sign.
Reciprocal Property: Taking the reciprocal of both sides of a positive inequality flips the inequality sign. |
188727 | https://www.doubtnut.com/qna/51236520 | int(pi)^(2pi)|sinx|dx=?
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MATHS
int_(pi)^(2pi)|sinx|dx=?...
∫π 2 π|sin x|d x=?
A
0
B
1
C
2
D
none of these
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Know where you stand among peers with ALLEN's JEE Enthusiast Online Test Series :
At x=π, sin π=0
At x=3 π 2, sin 3 π 2=−1
At x=2 π, sin 2 π=0
Since sin x is negative in the interval (π,2 π), we have:
|sin x|=−sin x for x∈(π,2 π)
Step 2: Rewrite the integral
Now we can rewrite the integral using the property of absolute value:
∫2 π π|sin x|d x=∫2 π π−sin x d x
Step 3: Factor out the negative sign
We can factor out the negative sign from the integral:
∫2 π π−sin x d x=−∫2 π π sin x d x
Step 4: Integrate sin x
The integral of sin x is:
∫sin x d x=−cos x+C
Thus, we evaluate:
−∫2 π π sin x d x=−(−cos x∣∣∣2 π π)
Step 5: Evaluate the limits
Now we evaluate the limits:
−(−cos(2 π)+cos(π))
Calculating cos(2 π) and cos(π):
cos(2 π)=1
cos(π)=−1
Substituting these values:
−(−1−1)=−(−1+1)=−(−2)=2
Final Answer
Thus, the value of the integral is:
∫2 π π|sin x|d x=2
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188728 | https://www.expii.com/t/approaches-to-solving-absolute-value-equations-4293 | Expii
Approaches to Solving Absolute Value Equations - Expii
There are often two possible solutions to absolute value equations. To solve, isolate the absolute value on one side, then solve two versions of your equation: one where the other side is positive, and one where it's negative.
Explanations (3)
Hannah Bonville
Text
4
A couple of things before we start this lesson. First, review the lesson on absolute values to make sure you've mastered that skill. Second, remember that absolute value can never be a negative number.
Image source: by Anusha Rahman
There are a few steps that you can consistently follow to solve equations that contain variables and absolute values:
1. Isolate the absolute value
2. Write two equations: one with a positive solution and one with a negative solution
3. Solve for x in both equations
Let's take a look at an example of absolute value equations with sums to see these steps in action:
3|2x−5|+7=22
Step 1: Isolate the absolute value expression
We need to get the absolute value (|2x−5|) alone on one side of the equation.
3|2x−5|+7=223|2x−5|=15|2x−5|=5
Step 2: Write two equations: one with a positive solution and one with a negative solution
Remove the absolute value sign and set up two equations:
2x−5=52x−5=−5
Step 3: Solve for x in both equations
What are the two possible values of x?
2x−5=52x=10x=5 OR 2x−5=−52x=0x=0
Notice how both values for x, when plugged into the original equation, produce the same result. This is a good way to check your answer.
We can also write the values of x as a solution set that looks like this: {0,5}
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Related Lessons
Absolute Value Equations with Products
Absolute Value Equations with No Solution
Absolute Value Equations with Sums
Turn Absolute Value Inequality into Compound
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Jack Kiuttu
Text
2
Way 1: Two Cases
You can solve absolute value equations breaking them into two cases. For example, suppose you're given the equation:
|x−5|=4.
Because of the way absolute values work, we can rewrite this as x−5=±4 and separate it into two equation.
Both will produce a solution:
x−5=4x−5=−4
These result in x=9 and x=1, which both satisfy the equation.
This works because an absolute value turns the expression inside it to a positive value. For example, both of the following equations are true:
|5|=5|−3|=3
(Surprisingly enough, you can use the two cases method to produce the general solution to the equation |x+a|=b, which is x=±b−a.)
Way 2: Number Lines and Definitions
You can also use a number line as well as the definition of absolute value to solve an absolute value equation. The definition, by the way, is that the absolute value of an expression is its distance from 0 on the number line. Note that this method is very similar to the Two Cases method, but it's still different in its own way.
If you have an equation that looks like |[expression]|=[number], you can plot points on the number line at [number] and write [expression] over them. (Don't literally write the word "expression", though!) From there, plot more points on the number line following the steps of simplifying [expression] down to just your variable (for example, adding 2, dividing by 7, taking the square root, etc.), which will produce the solutions.
Way 2: Example
Let's say we have the absolute value equation
|2x−5|=3
Plot points on the number line at 3 and −3 and write the expression 2x−5 over each one. I'm calling one point A and one point B.
Image source: by Jack Kiuttu
From there, plot two more points 5 spaces to the right of the previous two and write 2x over each of those.
Image source: by Jack Kiuttu
Finally, plot your last points: one at A2 and B2 and write x over each one. This simulates the final step in the equation, dividing both sides by 2. Those are your solutions!
Image source: by Jack Kiuttu
When solving these types of equations, you may come across an equation with no solution.
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Anusha Rahman
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(Video) Solving Absolute Value Equations
by The Organic Chemistry Tutor
In this video, The Organic Chemistry Tutor goes over how to solve absolute value equations.
Summary
Absolute value is the distance from zero, so you cannot have a negative value.
|5|−5
|−8|=8
So, if we're told that:
|x|=7,
what are the possible values of x? Well, x can equal 7 or −7.
So how can we use this concept when solving absolute value equations? Say we have the equation, |x+2|=9. That means that:
x+2=9
x+7
OR
x+2=−9
x=−11
Both of these equations are true.
Let's look at another example. Say |2x−5|=1. What are the possible values of x? Well, |2x−5| can equal +1 or −1.
2x−5=1
2x=6
x=3
OR
2x−5=−1
2x=4
x=2
What if we were given the equation:
|5x−1|=−2
What are the values of x? Well, there is no solution to this equation, because the absolute value can never be negative.
Let's look at another example that's a bit more complex. What are the values of x that satisfy: 3|x−1|−2=7?
The first thing we want to do is isolate the absolute value part on one side of the equation.
3|x−1|−2=7
3|x−1|=9
|x−1|=3
Then, we can solve it by setting up the absolute value part to equal 3 or −3.
x−1=3
x=4
OR
x−1=−3
x=−2
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188730 | https://www.studocu.com/en-us/document/delgado-community-college/medical-surgical-nursing-ii/understanding-polycythemia-clinical-overview-and-management-strategies/125420641 | Clinical Overview and Management of Polycythemia: A Complete Guide - Studocu
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Clinical Overview and Management of Polycythemia: A Complete Guide
Polycythemia, also known as erythrocytosis, is a medical condition marked by an...elevated number of red blood cells (RBCs) in the bloodstream, leading to increased blood viscosity and potential cardiovascular complications. This condition is categorized into two main types: Polycythemia Vera (primary) and Secondary Polycythemia. Polycythemia Vera is linked to a genetic mutation, often involving the JAK2 gene, resulting in an overproduction of RBCs, white blood cells, and platelets. Symptoms can include fatigue, headaches, and a reddish complexion, with serious implications like thrombosis. Conversely, Secondary Polycythemia arises from an adaptive response to chronic hypoxia due to various conditions such as high altitude or respiratory diseases, leading to increased erythropoietin levels. Diagnosis involves blood tests and genetic assessments, while management strategies focus on reducing blood viscosity through phlebotomy and patient education on risk mitigation. Awareness and effective management of polycythemia are crucial for improving patient outcomes.
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Original title:
Understanding Polycythemia: Clinical Overview and Management Strategies
Course
Medical-Surgical Nursing II (PRNU 124)
75 documents
University
Delgado Community College
Academic year:2024/2025
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Clinical Overview and Management of Polycythemia: A Complete Guide
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Clinical Overview and Management of Polycythemia: A Complete Guide
Course: Medical-Surgical Nursing II (PRNU 124)
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Unders tanding P oly cythemia: A
Compr ehensiv e Guide
Intr oduc tion t o P oly cythemia
P oly cythemia, also r ef err ed t o as erythr ocyt osis, is a condition
charac t erized b y an ele v a t ed number of r ed blood cells (RBCs) in the
bloods tr eam. This incr ease can r esult in thick er blood, which in turn can
ha v e signi
fi
cant implica tions f or v arious or gan s y s t ems.
T ypes of P oly cythemia
P oly cythemia is br oadly ca t e g orized int o tw o types:
P oly cythemia V era (Primary P oly cythemia)
Secondary P oly cythemia
P oly cythemia V era (Primary P oly cythemia)
Etiolo gy and P a thoph y siolo gy
:
P oly cythemia v era is a rar e blood disor der oft en r esulting fr om an
acquir ed muta tion in the J AK2 g ene aff ec ting a pr ecursor blood-
f orming cell in the bone marr o w. This muta tion leads t o an
o v erpr oduc tion of no t jus t RBCs, but also whit e blood cells (WBCs)
and pla t ele ts.
It typically de v elops in men o v er the ag e of 60 and pr o gr esses slo wly,
with s ymp t oms manif es ting gradually.
The incr eased number of cells r esults in higher blood viscosity and
v olume, making the blood thick er and mor e pr one t o clo tting injuries.
Clinical Manif es ta tions
:
Symp t oms ma y include f a tigue, headaches, ringing in the ears, blurr ed
vision, and shortness of br ea th. P a tients ma y also e xperience pruritus
(it ch y skin), g out, and a r eddish comple xion due t o high blood
pr essur e and incr eased blood density.
The incr eased viscosity can lead t o complica tions such as deep v ein
thr ombosis (D VT), pulmonary embolism (PE), m y ocar dial inf ar c tion
(MI), and cer ebr o v ascular accidents (CV As).
Secondary P oly cythemia
Etiolo gy and P a thoph y siolo gy
:
Unlik e poly cythemia v era, secondary poly cythemia is no t due t o a
g ene tic muta tion but ra ther an adap tiv e r esponse t o chr onic h ypo xia.
Chr onic lo w o xy g en le v els can be caused b y high altitude, pulmonary
diseases, car dio v ascular conditions, or obs truc tiv e sleep apnea.
The kidne y s r espond t o h ypo xia b y r eleasing erythr opoie tin, which
s timula t es the bone marr o w t o pr oduce mor e RBCs.
Clinical Manif es ta tions
:
Similar t o poly cythemia v era, pa tients ma y e xhibit s ymp t oms due t o
incr eased blood viscosity and v olume, leading t o risk s of clo ts and
car dio v ascular complica tions. Unlik e poly cythemia v era, the
complica tion of ele v a t ed WBCs and pla t ele ts is no t a f ac t or.
Diagnos tic Crit eria and Assessments
Subjec tiv e Symp t oms
:
These rang e fr om as ymp t oma tic cases t o se v er e s ymp t oms including
f a tigue, abdominal discomf ort, headaches, and dif
fi
culty br ea thing.
Unique t o poly cythemia v era ar e pruritus and incr eased bleeding
t endencies.
Objec tiv e Da ta
:
Diagnos tically, poly cythemia v era is dis tinguished b y ele v a t ed RBCs,
hemo globin (Hgb), hema t ocrit (HC T), WBCs, and thr ombocyt es. Bone
marr o w e x amina tions sho w h yper cellularity.
A g ene tic t es t r e v ealing a J AK2 muta tion con
fi
rms poly cythemia v era.
Secondary poly cythemia, meanwhile, is mark ed b y high erythr opoie tin
le v els.
Manag ement Stra t e gies
Medical Manag ement
P oly cythemia V era and Secondary P oly cythemia
:
Bo th conditions r equir e r educing blood viscosity. L o w-dose aspirin
and r epea t ed phlebo t om y ar e common tr ea tments. Phlebo t om y
r elie v es s ymp t oms lik e headaches and dizziness b y r emo ving 500 mL
of blood per session.
Nursing Int erv entions and P a tient Educa tion
Secondary P oly cythemia
:
Emphasis on maintaining o xy g en le v els, which includes smoking
cessa tion and managing chr onic pulmonary and car dio v ascular
diseases.
General
:
Nurses should monit or pa tients f or
fl
uid o v erload and deh y dra tion,
encouraging le g e x er cises t o a v oid thr ombus f orma tion, and ensuring
adher ence t o tr ea tment r e gimens.
P a tient Pr oblems and Nursing Int erv entions
:
Compr omised blood
fl
o w necessita t es int erv entions lik e k eeping
pa tients comf ortable, ele v a ting the head of the bed, and monit oring
f or complica tions lik e thr ombus f orma tion and signs of deh y dra tion.
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Pr o gnosis
P oly cythemia v era is a chr onic, lif e-short ening disor der. Ther e is a risk of
de v eloping leuk emia or lymphomas, either due t o the chemo therapeutic
ag ents used in tr ea tment or fr om an underlying disor der in the s t em cells.
The major cause of morbidity and mortality s t ems fr om thr ombosis.
Conclusion
P oly cythemia, in its primary and secondary f orms, pr esents signi
fi
cant
clinical challeng es due t o incr eased blood viscosity and v olume. Pr oper
diagnosis, pa tient educa tion, and manag ement s tra t e gies ar e crucial f or
mitiga ting risk s and impr o ving pa tient out comes.
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188731 | https://www.math.net/direct-variation | Direct variation
home / algebra / variable / direct variation
Direct variation
Direct variation refers to a relationship between two variables where when one variable increases the other also increases by the same factor. Conversely, when one variable decreases, the other variable also decreases by the same factor. As such, the ratio between two variables that are in direct variation (or directly proportional) always remains the same.
Variables in direct variation can be modeled by the equations
or
where k is a constant of proportionality, and x ≠ 0. The graph of two variables in direct variation is simply a straight line through the origin, as shown in the figure below, using the graph of y = 1x. In the graph we can see that as x increases, y also increases, and as x decreases, y also decreases.
There are many examples of variables in the real world that are in direct variation, such as the gallons of gasoline pumped into a car, and the cost of the gasoline. The more gasoline that is pumped into the car, the higher the cost; the less gasoline pumped, the lower the cost. The rate at which a plant grows is another example.
Example
Given that a plant's height is in direct variation with its age, it grows 4 inches for every elapsed week, and its height is 0 (planted seed) at the beginning of the first week, determine how tall the plant will be by the end of 4 weeks, noting its height with each passing week.
Since it is given that the two variables, height and age, are in direct variation, we can use the equation for direct variation:
The constant of proportionality, k, is 4, the age of the plant is x, and the height of the plant is y. The height of the plant at the end of each week can therefore by found by multiplying its age by 4:
y = 4x
| x | y |
--- |
| 1 | 4 |
| 2 | 8 |
| 3 | 12 |
| 4 | 16 |
Because we can see that the plant's height increases at the same rate as its age, we can confirm that x and y are in direct variation. More specifically, we can see that x and y increase at the same rate. When x is doubled, y also doubles. When x is tripled, y also triples, and so on. Furthermore, the ratio between x and y remains constant, as y is always 4 times x. This is what is meant when someone says that x and y are in direct variation, y varies directly with x, or y is directly proportional to x; they all mean the same thing.
Variable
Combining like terms
Constant
Constant of proportionality
Dependent variable
Independent variable
Independent and dependent variables
Direct variation
Inverse variation
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188732 | https://www2.lsco.edu/syllabi/Spring%202025/Scarborough%20George/MATH-1324-80F%20-%20Math%20for%20Bus.%20%20and%20%20Social%20Sci.%20I%20Syllabus.html | House Bill 2504 Spring 2025 MATH-1324-80F - Math for Bus. and Social Sci. I Syllabus
LAMAR STATE COLLEGE ORANGE SYLLABUS
Mathematics
MATH, 1324, Math for Bus. and Social Sci. I, 80F
COURSE INFORMATION
| Instructor Name | George Scarborough |
| Building/Office Number | Shahan Event Center (SHA), Room 128 |
| Office Hours | Monday-Wednesday, 1-4 Academic Building (New LSCO Academic Center) 105.19 |
| Virtual Hours | Thursday, 12:30-3:30Virtual office(MS Teams) |
| Office Telephone | (409) 882-3390 |
| Email Address | george.scarborough@lsco.edu |
| Course Schedule | Building: Online, Room: TBA, Dates: 1/21/2025 - 3/14/2025, Days: TBA, Times: TBA |
| Course Description | MATH 1324 - Mathematics for Business & Social Sciences Credits 3 Lecture Hours 3 Lab Hours 0 The application of common algebraic functions, including polynomial, exponential, logarithmic, and rational, to problems in business, economics, and the social sciences are addressed. The applications include mathematics of finance, including simple and compound interest and annuities; systems of linear equations; matrices; linear programming; and probability, including expected value. Prerequisite: TSI Math Complete if not enrolled in co-requisite model Course Identifier 27.0301 AC |
Required Textbook & Materials
Mathematics with Applications in Business and Social Sciences software
2022 Hawkes Learning Systems
Access code + E-book (required)9781642771060 (GBP opt-out only)
Technology requirements for the course:Regular internet access, including LSCO email, Blackboard, and the Hawkes Learning website.
A graphing calculator of the student’s choice. TI-84 is used in textbook examples and classroom/video demonstrations. There are many free and low-cost apps available which can mimic most of the TI functions. Online calculators such as Desmos.com, and applications such as MS Excel are also acceptable.
Students will be required to submit PDF scans of their written work for the online exams.
Upon registration for classes, LSCO students are automatically charged $14 per semester credit hour for access to all required textbooks, lab manuals, lab codes, and electronic books on the first day of class through the Gator Book Pack. Information about the LSCO Gator Book Pack as well as responses to common FAQs can be found on LSCO's webpage. ALL STUDENTS WILL HAVE UNTIL THE SECOND DAY OF THE SEMESTER TO OPT-OUT OF THE GATOR BOOK PACK.
Every student MUST have access to the required textbooks by the week of class. The student will be responsible for all assignments given. Failure to have a text may result in being dropped from the class. Failure to follow instructions (written or oral) will result in penalties.
Course Learning Outcomes (CLOs)
Upon successful completion of this course, students will acquire the following course learning outcomes:
Core Objectives
Course Level Learning
Outcomes (ACGM or WECM)Core ObjectiveAssociated Course Activities/Assignments/Projects
1. Apply elementary functions, including linear, quadratic, polynomial, rational, logarithmic, and exponential functions to solving real-world problems.Communication,
Empirical &
Quantitative Skills Course activities include lecture, assigned reading, class discussion & computer-generated homework. Discussion questions, quizzes & unit exams will measure student level of understanding. Students may be required to show their problem-solving steps on tests, and answer essay questions in writing.
2. Solve mathematics of finance problems, including the
computation of interest, annuities, and amortization of loans.Communication,
Critical Thinking,
Empirical &
Quantitative Skills Course activities include lecture, assigned reading, class discussion & computer-generated homework. Discussion questions, quizzes & unit exams will measure student level of understanding. Students may be required to show their problem-solving steps on tests, and answer essay questions in writing.
Students will be assigned a critical thinking assignment using amortization tables.
3. Apply basic matrix operations, including linear programming methods, to solve application problems.Communication,
Critical Thinking,
Empirical &
Quantitative Skills Course activities include lecture, assigned reading, class discussion & computer-generated homework. Discussion questions, quizzes & unit exams will measure student level of understanding. Students may be required to show their problem-solving steps on tests, and answer essay questions in writing.
4. Demonstrate fundamental probability techniques and application of those techniques, including expected value, to solve problems.Communication,
Critical Thinking,
Empirical &
Quantitative Skills Course activities include lecture, assigned reading, class discussion & computer-generated homework. Discussion questions, quizzes & unit exams will measure student level of understanding. Students may be required to show their problem-solving steps on tests, and answer essay questions in writing.
5. Apply matrix skills and probability analyses to model applications to solve real-world problems.Communication,
Critical Thinking,
Empirical &
Quantitative Skills Course activities include lecture, assigned reading, class discussion & computer-generated homework. Discussion questions, quizzes & unit exams will measure student level of understanding. Students may be required to show their problem-solving steps on tests, and answer essay questions in writing.
Course Topical Outline
Week Assignments Homework due
Week 1
Week 1
1/20Martin Luther King Jr. HolidayMonday 1/20T 1/28
Introduction and Syllabus
1.1 Linear Equations in One Variable
1.2 Application of Linear Equations in One Variable
1.3 Linear Inequalities in One Variable
1.4 Quadratic Equations in One Variable
Week 2
1/27 2.1 The Cartesian Coordinate System M 2/3
2.2 Linear Equations in Two Variables
2.3 Forms of Linear Equations
2.5 Linear Regression
3.1 Introduction to Functions
Week 3
Week 3
2/3Test 1 (Chapters 1 & 2) Open2/3-2/6M 2/10
3.2 Functions and Models
3.3 Linear and Quadratic Functions
3.4 Applications of Quadratic Functions
3.7 Polynomial functions
3.8 Rational Functions
Week 4
Week 4
2/10 5.1 The Basics of Personal Finance M 2/17
5.2 Simple and compound interest
5.3 Annuities: Present and Future Value
5.4 Borrowing Money
Critical Thinking Assignment (due M 3/3)
Week 5
Week 5
2/17Test 2 (Chapters 3 & 5) Open2/17-2/20M 2/24
6.1 Solving Systems of Equations
6.2 Matrix Notation/ Gauss Jordan Elimination
6.3 Determinants
6.4 Basic Matrix Operations
6.5 Inverses of Square Matrices
Week 6
Week 6
2/24 7.1 Inequalities in Two Variables M 3/3
7.2 Linear Programming -Graphical Approach
7.3 The Simplex Method-Maximization
7.4 The Simplex Method-Duality and Minimization
8.3 Introduction to Probability
Week 7
Week 7
3/3Critical Thinking Assignment due 3/3M 3/10
Test 3 (Chapters 6 & 7)Open 3/3-3/6
8.4 Counting Principles
8.5 Counting Principles and Probability
8.6 Conditional Probability
8.7 Expected Value
Week 8
Week 8
3/10Final Exam (Chapters 1-8) Open3/10-3/13Th 3/13
Catch up / review
Past due homework due end of day Friday, March 14
Major Assignments Schedule
Test 1Open2/3-2/6
Test 2Open2/17-2/20
Test 3Open 3/3-3/6
Critical ThinkingDue 3/3
Final ExamOpen3/10-3/13
Final Exam Date
March 10, 2025 - 12:00 AM Through March 13, 2025 - 11:59 PM
COURSE POLICIES
Academic Honesty
Faculty who suspect violation of academic honesty, cheating, plagiarism, collusion, or abuse of resource materials may assign an academic penalty.
Incidents of academic dishonesty are a violation of the Student Code of Conduct. An academic appeal process is afforded to students who desire to dispute a grade or any decision that affects the student's ability to complete and earn a grade for the course provided it is not related to a violation outlined in the LSCO Student Code of Conduct, including academic dishonesty.
Electronic Communication
LSCO students are required to use either their LSCO Blackboard account or their LSCO email account (Office 365 / Microsoft Outlook) for all electronic communication. In order to ensure the privacy and identity of the student communicating via electronic methods, LSCO faculty will direct students to use their LSCO email accounts rather than personal accounts. If a student has trouble accessing their LSCO email account, they should contact the LSCO Help Desk at (409) 882-3033 or helpdesk@lsco.edu.
Do NOT use the Blackboard message system; only use your LSCO email in Outlook to send me a message.
Attendance Requirements
Federal regulations require students who receive financial aid to have begun "attending" and participate substantially in each course for which they are enrolled on or before the official census reporting date outlined on the LSCO Academic Calendar. Students documented as "not attending" a course upon the census date are assumed (for financial aid purposes) to have not begun attendance for that course, negatively affecting their financial aid eligibility and disbursement.
Attendance in an ONLINE course is verified by substantial participation in the course on or before the census date published in the LSCO Academic Calendar. Substantial participation in this online course is defined as logging in and completing/participating in at least one requirement of the course. Note: Simply logging in to your online course does not constitute attendance.
Hybrid classes are a mix of face-to-face and online environments. Students will be expected to attend a certain number of classes as required by the instructor. (Include the policy on absences and tardiness.) In addition to classroom attendance, your weekly active participation in the online component (Blackboard and/or homework software) will be considered and expected.
Attendance will be recorded weekly, based on at least one assignment completed or up to date on all assignments, as well as any discussion boards or other assignments posted in the course announcements. Attendance is not part of the course grade (see "Grading and Evaluation Method").
Make-up Work Policy
The final exam replaces the lowest test grade, if it is higher. All homework assignments may be completed after due dates for full credit. Past due homework may be completed for full credit through March 14.
Classroom Etiquette
Exam Policy
All exams are online. The final exam score will replace the lowest test score.Scans of scratch work on a specific worksheet will be required.We have scanners on campus, or you can use any mobile scanner app. I recommendAdobe Scan, which is free.
Responsible Use of Artificial Intelligence (AI)
Students should use AI technology responsibly and ethically. This includes refraining from using AI to engage in harmful or unethical activities, such as generating false information, spreading misinformation, or engaging in malicious behaviors. The use of AI should align with the principles of academic integrity, honesty, and respect for others. Students are responsible for adhering to LSCO's Academic Honesty policy found in the Student Handbook. LSCO's full AI policy can be found at AI Policy | Lamar State College Orange (lsco.edu)
Expected Time Requirement for this Course
For every hour in class (or unit of credit) taught in a 16-week session, students should expect to spend at least two to three hours per week studying and completing assignments. Example: For a 3-credit hour class taught in a 16-week session, students should prepare to allocate approximately 6 to 9 hours per week outside of class studying and completing assignments. For a 3-credit hour taught in a 10-week summer session, students should prepare to allocate approximately 10 to 15 hours per week outside of class studying and completing assignments. For a 3-credit hour taught in an 8-week session, students should prepare to allocate approximately 12 to 18 hours per week outside of class studying and completing assignments. For a 3-credit hour taught in a 5-week summer session, students should prepare to allocate approximately 20 to 29 hours per week outside of class studying and completing assignments.
Grading and Evaluation Method
There are homework assignments that must be completed using the software (Hawkes). Each homework section requires ~70-80% mastery to complete. The homework assignments will total 20% of the total semester grade.
There are 3 unit exams that will comprise 50% of the grade. The tests will be taken on Hawkes. Submission of written work will be required.
There will be a comprehensive Final exam which will weigh 20% of the grade. The exam will be taken on Hawkes. Submission of written work will be required.
There will be a critical thinking assignment which will be 10% of the total grade.
Assignment of a letter grade will be made according to the following scale:
A = 89.01 - 100%
B = 79.01 - 89
C = 69.01 - 79
D = 59.01 - 69
F = 59 or below
Instructor Response Time
The instructor will respond to emails within 24 hours, excluding weekends and holidays. On weekends and holidays, the instructor will respond to emails from students within 24 hours of the first business day following the weekend or holiday.
Participation Requirements
Weekly participation and engagement in the courses are critical for student success. Assignments should be completed by the due date. Students should also refer to the instructor's attendance policy for additional information. (See LSCO Student Handbook, Class Attendance.)
Review of Test Grades
If the student has an issue with their grade, the instructor must be contacted no later than three (3) days after receiving the grade.
Student's Responsibility
This syllabus contains information, policies, and procedures for a specific course. By enrolling, the student agrees to read, understand, and abide by the rules, policies, regulations, and ethical standards of Lamar State College Orange as those contained in the current LSCO Catalog and schedule of classes.
Syllabus Content
The instructor reserves the right to make changes to this syllabus if deemed necessary. All changes will be provided to the students orally or in writing before the implementation of the change.
Textbook and Required Materials Access
Every student MUST have access to the required textbooks by the second week of class. The student will be responsible for all assignments given. Failure to have a text may result in being dropped from the class. Failure to follow instructions (written or oral) will result in penalties.
STUDENT SUPPORT RESOURCES
Mental Health Resources
TimelyCare is a virtual health and well-being platform that is available 24/7 for all non-dual credit enrolled LSCO students. There is no cost to eligible students for this service. TimelyCare’s providers offer emotional support, mental health counseling, health coaching, psychiatry, and basic needs support. Non-Dual Credit students enrolled in classes can log in to the TimelyCare website or app available at timelycare.com/LSCO.
Advocacy Information
Any student who faces challenges securing their food or housing and believes this may affect their performance in the course is encouraged to contact the advising office for guidance on how to identify possible resources. Please notify the instructor of your circumstance if you are comfortable doing so.
Equal Opportunity to Educational Programs
Lamar State College Orange (LSCO) is an equal opportunity educational institution and does not discriminate against any person regardless of race, sex, color, religion, national origin or ancestry, age, marital status, disability, sexual orientation, gender identity, or veteran status, in admissions, educational programs, student activities or employment. For further information about this policy, contact the Accessibility Coordinator at (409) 882-3393.
Title IX of the Education Amendments
LSCO prohibits discrimination, including sexual harassment and retaliation, against any student on the basis of race, color, religion, gender, national origin, disability, or any other basis prohibited by law. Any student who believes that he or she has experienced prohibited conduct or believes that another student has experienced prohibited conduct should immediately report the alleged acts to the Title IX Coordinator, Rebecca Gentry, at Rebecca.Gentry@lsco.edu.
Blackboard Resources
LSCO students will access Blackboard through the MyGator portal. Login credentials will use the following format: username@my.lsco.edu and Password. For help in identifying your Username/Password, visit
Blackboard student resource videos and help-sites are available at
Career Coach
Lamar State College Orange provides career advising services to all students and alumni through Career Coach, an online career planning tool. Career Coach assists students through all phases of developing, initiating, and implementing career plans.
Information regarding employment opportunities and career options are provided along with access to live local job postings. Full-time and part-time employment opportunities, as well as internships, are available through Career Coach. Visit for more details on how to use LSCO's Career Coach to plan for and learn more about your future career.
Gator Assistance Services
Lamar State College Orange provides currently enrolled technical students support for daycare costs as well as other services.
Visit for details on assistance services.
OASIS (formerly) Gator Success Center
Students are encouraged to make an appointment or walk in to receive tutoring, support services, or access to an open computer lab. Face-to-face and online supplemental instruction sessions are available to help students through any LSCO course. Reach out to learning.center@lsco.edu for more information on how students can receive academic support.
Library Services
Students are encouraged to visit library.lsco.edu to find the library's current operating hours, access the catalog to locate print materials, and access GatorSearch to explore the vast electronic collection. The library provides over 77 electronic database collections that include eBooks, newspapers, magazines, academic journals, and streaming video. The physical library contains a specialized collection of research materials specifically chosen to support the degrees and courses offered. Additionally, Students with research questions or questions about library services are encouraged visit the library in person, call 409-882-3352, access the chat on the library webpage, or to email their question to lscolibrary@lsco.edu.
Student with Disabilities
Under the Texas State System, Lamar State College Orange complies with Section 504 of the Rehabilitation Act of 1973 and the Americans with Disabilities Act of 1990, pertaining to the provision of reasonable academic adjustments/auxiliary aids for students with disability. We strive to provide reasonable academic adjustments/auxiliary aids to students who request and require them.
Students who believe they have a disability requiring an academic adjustment/auxiliary aid are encouraged to contact the Accessibility Coordinator at (409) 882-3393 or visit the Advising Office. Students are encouraged to apply before the start of the semester when at all possible. The Accommodation Request Form and details regarding the appropriate documentation needed can be found here: Once approved, the signed accommodation form provided by the Special Populations Advisor must be submitted to the instructor at least two business days in advance of need.
Students with questions about the accommodations they receive in class should contact their instructor or the Accessibility Coordinator.
Upswing 24/7 FREE Tutoring Services
Lamar State College Orange provides currently enrolled students with access to online tutoring through a partnership with Upswing, an online tutoring platform. Tutors are available 24/7 online in almost every subject.
Visit for details on how to log-in to the FREE services.
INSTITUTIONAL POLICIES
Campus Closure
In the event of an emergency campus closure in excess of three class days, Lamar State College Orange's classes will continue via the use of Blackboard. In such an instance, the college website, www.lsco.edu, will have information concerning the event and anticipated re-opening plans.
Civility
Please be considerate of other classmates' feelings, ethnic background, cultural differences, situations, and level of maturity. Students will be asked to leave the course if disruptive or inappropriate behavior is exhibited in any of the course requirements. If your instructor feels that you have not contributed appropriately to course requirements, your final course grade may be reduced accordingly.
The instructor reserves the right to manage a positive learning environment and will not tolerate inappropriate conduct in the course. Rude correspondence (discourteous or impolite, especially in a deliberate way) in e-mails, telephone calls, in person, or comments made to other class members, the instructor, or the office staff.
Contingency Plans
Students should develop a backup plan should their computer system or their Internet provider fail. Computer or internet connectivity issues are not valid excuses for missing a deadline. The College provides many opportunities for using computer equipment, as do many public libraries. Refer to the LSCO website for operational hours of the Library and Success Center.
Credit Transfer
Students should check in advance with the institution to which they plan to transfer credit to confirm transferability. Refer to the LSCO Catalog for details on how to handle and resolve transfer disputes with public institutions of higher education in Texas.
Criminal Background Policy
LSCO awards some certificates and degrees in which a criminal history MAY disqualify candidates from becoming licensed, certified, and/or employed upon degree/certificate completion. Students with a criminal background enrolling in courses leading to a degree/certificate in Court Reporting, Criminal Justice, Cosmetology, Emergency Medical Technology, Massage Therapy, Medical Assisting, Vocational Nursing (VN), Registered Nursing (RN), Pharmacy Technician, Real Estate, or Teacher Preparation program are STRONGLY ENCOURAGED to discuss the certification and/or licensing regulations of the program with the program director listed in order to learn more about the current guidelines related to criminal history as well as the right of individuals to request a criminal history evaluation letter.
Drops and Withdrawals
Never attending or ceasing to attend classes DOES NOT constitute a drop or withdrawal. You remain registered until you request a drop from the instructor. Failure to act in a timely manner will result in an "F" grade for the course. It is the student's responsibility to follow up with the LSCO advising office to ensure that all drops/withdrawals are processed as desired.
Grade of "Incomplete"
The grade of "I" may be given when any requirement of the course, including the final examination, is not completed. Students seeking an incomplete should have completed at least 75% of the course requirements and be passing the course at the time of the request. Arrangements to complete deficiencies in a course should be made in advance of the end of the semester with the instructor. The instructor will process the Incomplete form online, and a confirmation will be sent to the student's LSCO email.
Incomplete work must be finished during the next long semester. If not, the Office of Admission and Records must change the "I" grade to the grade of "F." The course must then be repeated if credit is desired. An "I" grade also automatically becomes an "F" if the student registers for the course prior to removing the deficiencies and receiving a grade change. The instructor may record the grade of "F" for a student who is absent from the final examination and is not passing the course.
Grade Appeals (Complaints Related to Earned Grades)
Grade determination and awarding of grades in a course are the responsibility of the instructor and should be calculated according to college policy, procedures, and written details provided in the course syllabus. NOTE: Final grades are available to students within 48 hours of the instructor posting the grade in Banner. Students may view final grades by logging into MyGator and then accessing Gator Self-Service.
An academic appeal process is afforded to students who desire to dispute a grade or any decision that affects the student's ability to complete and earn a grade for the course provided it is not related to a violation outlined in the LSCO Student Code of Conduct. If an informal conference with the faculty member regarding an academic complaint fails to reach the outcome requested by the student, the student may initiate the formal process outlined below. Even after initiating the formal complaint process, students are encouraged to seek informal resolution of their concerns. A student whose concerns are resolved may withdraw a formal complaint at any time.
Refer to the current catalog or for details on the formal grade appeal process.
Institutional Educational Goals
Lamar State College Orange has identified seven educational goals to specify the knowledge and skills that students should gain from completing academic and technical programs with the College. These goals are:
Critical thinking (General Education, Technical) - Students will be able to demonstrate creative thinking, innovation, inquiry, analysis, evaluation, and synthesis of information.
Communication (General Education, Technical) - Students will be able to effectively develop, interpret and express of ideas through written, oral and visual communication.
Empirical and quantitative skills (General Education, Technical) - Students will be able to manipulate and analyze numerical data or observable facts and create informed conclusions.
Teamwork (General Education, Technical) - Students will be able to consider different points of view and to work effectively with others to support a shared purpose or goal.
Social responsibility (General Education, Technical) - Students will be able to recognize and acquire a sense of intercultural competence, knowledge of civic responsibility, and the ability to engage effectively in regional, national and global communities.
Personal responsibility (General Education, Technical) - Students will be able to connect choices, actions, and consequences to ethical decision-making.
Professional competency (Technical) - Students will be able to recognize or demonstrate skills and that depict professional values and employability. If the career has licensure or certification requirements, students may prepare for the licensure and certification in a capstone course and sit for the licensure or certificate at the end of the program.
MyGator and Log-In Credentials
Current students will access many LSCO applications through the MyGator portal. Login credentials will use the following format: username@my.lsco.edu/Password. For help in identifying your username/Password, visit
It is a violation of College policy, state laws, and federal laws for anyone to gain or help others gain unauthorized access to MyGator or any LSCO application or service. All accounts shall be for use by a single individual - the person for whom the account was approved or assigned. This includes Blackboard accounts as well as any application within MyGator. Sharing or loaning accounts is strictly prohibited, can be construed as a form of cheating, and violates College policy, state laws, and federal laws.
Policies and Procedures
LSCO adheres to the policies and procedures established in the Texas Education Code, Texas State University System Rules and Regulations, LSCO Administrative Policies and Procedures Manual, LSCO Faculty Handbook, and LSCO Catalog.
Prohibited Items in the Classroom
No food or tobacco (including smokeless) products are allowed in the classroom. Only students enrolled in the course are allowed in the classroom, except by special instructor permission. It is inappropriate for minor children to be on campus due to the potential liability to the College, the risk of harm to the children, and decreased employee productivity due to distractions and disruptions.
Student Complaints (excluding Safety, Harassment, or Title IX)
Student-Staff and Student-Instructor Complaint Process excluding Safety, Harassment, or Title IX
LSCO believes that all matters involving a staff member (non-instructor) OR an instructor are best resolved directly with the employee. Should it not be resolved at that level, access and complete the Student Complaint Intake form within five (5) business days from the date of the offense. The Student Complaint Intake Form can be found in any of these locations:
LSCO Website (Current Students)
LSCO Catalog (Student Services, Complaints)
MyGator Card
Blackboard (Student Resources, Student Services, Student Resources)
After a student completes the Student Complaint Intake Form, they will be contacted by an LSCO employee unrelated to the offense and provided guidance on how to proceed with the Complaint process.
Refer to the LSCO Catalog (Student Services, Complaints) for details on the complaint process.
Student Complaints (involving Safety, Harassment, or Title IX)
Student-Staff and Student-Instructor Complaint Process involving Safety, Harassment, or Title IX:
Students who have experienced a safety, harassment, or title IX event should access and complete the Title IX/Clery: Anonymous Reporting Form found in any of these locations:
LSCO Website (Sexual Misconduct & Title IX)
LSCO Catalog (Student Services, Complaints)
MyGator Card
Blackboard (Student Resources, Student Services, Student Resources)
After a student completes the intake form, they will be contacted by an LSCO employee and provided guidance on how to proceed.
Refer to the LSCO Catalog (Student Services, Complaints) for details on the complaint process.
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The privacy of all students, including Distance Education students, is protected through strict adherence to the rules of the Family Education Rights and Privacy Act. LSCO's statement regarding the Family Education Rights and Privacy Act can be found in LSCO's Catalog. Additional information regarding privacy for Distance Education students can be found in the Distance Education Handbook. |
188733 | https://study.com/skill/learn/how-to-identify-the-velocity-time-graph-that-corresponds-to-a-position-time-graph-explanation.html | How to Identify the Velocity-Time Graph that Corresponds to a Position-Time Graph | Physics | Study.com
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How to Identify the Velocity-Time Graph that Corresponds to a Position-Time Graph
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00:04 How to Identify the…
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Emmett Sams, Kirsten Wordeman
Instructors Emmett Sams
Emmett has taught high school physics and astronomy for 3 years. He has a B.S. in Physics and M.Ed. in Teaching and Instruction from UC Santa Barbara. He is a certified secondary science/physics teacher in Georgia and has a certificate of Informal Science Education.
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Kirsten Wordeman
Kirsten has taught high school biology, chemistry, physics, and genetics/biotechnology for three years. She has a Bachelor's in Biochemistry from The University of Mount Union and a Master's in Biochemistry from The Ohio State University. She holds teaching certificates in biology and chemistry.
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Example SolutionsPractice Questions
How to Identify the Velocity-Time Graph that Corresponds to a Position-Time Graph
Step 1: Split the Position-Time Graph into separate parts by domain.
Step 2: Identify the slope of each part of the Position-Time Graph.
Step 3: Graph the slope of each part in the corresponding domain of the Velocity-Time Graph.
What are motion graphs?
Position-Time and Velocity-Time Graphs: These graphs, often called "Motion Graphs" describe the position and velocity of an object as a function of time. One can relate these graphs to one another by understanding that the slope of the graph describes how that graph changes with time. Because velocity is a change in position over a set amount of time, the slope of a position-time graph at any point in time can tell you the velocity of the object at that point in time.
Slope: The slope of a graph describes the ratio of how much the y variable changes with incremental changes in the x variable. The slope can be calculated by dividing a change in the y variable (over some interval) by the change in the x variable (over that same interval). The slope of a line can be calculated using any 2 points on that line, (x 1,y 1) and (x 2,y 2)
s l o p e=y 2−y 1 x 2−x 1
So, let's use these steps to determine the velocity-time graph that corresponds to a position-time graph.
Examples of identifying velocity-time graph that corresponds to a position-time graph
Example 1
Identify the velocity-time graph for the motion of an object whose position is given by this position-time graph.
Step 1: Split the Position-Time Graph into separate parts by domain.
This position-time graph shows a parabolic curve starting with a horizontal slope that dips downward. At a time of t=1s, the graph becomes a straight line before abruptly changing its slope at t=2s. We can split the graph into 3 parts. The first encompasses the parabolic section from t=0s to t=1s. The second part encompasses the straight line from t=1s to t=2s. The third part encompasses the shallower straight line from t=2s to t=4s.
Step 2: Identify the slope of each part of the Position-Time Graph.
The First Part (t=0s to t=1s):
The most difficult motion graphs to interpret occur when a graph (usually the position-time graph) shows a changing curve. To be mathematically rigorous, we would need to determine the equation of the parabolic curve and then use some calculus to determine the slope of the curve at every point in time. However, we can make some assumptions to simplify our scenario. We know that our parabola starts as flat (slope of 0) before it proceeds to slant downward. We also see that the parabola smoothly turns into a straight line (with constant slope) at t=1s. If we can determine the slope of the line segment in the second part of the graph, then we can use that slope as the slope of endpoint of the parabola as well. Please see below for how that slope is calculated.
We can thus describe the slope of the first part as a smooth transition from a slope of 0 at t=0s to a slope of -2 at t=1s.
The Second Part (t=1s to t=2s):
In order to find the slope of this line segment, we will use our slope formula applied to the 2 endpoints of the line segment. (x 1,y 1)=(1,3) and (x 2,y 2)=(2,1).
s l o p e=1−3 2−1 s l o p e=−2 1 s l o p e=−2
The Third Part (t=2s to t=4s):
In order to find the slope of this line segment, we will use our slope formula applied to the 2 endpoints of the line segment. (x 1,y 1)=(2,1) and (x 2,y 2)=(4,0).
s l o p e=0−1 4−2 s l o p e=−1 2
Step 3: Graph the slope of each part in the corresponding domain of the Velocity-Time Graph.
The First Part (t=0s to t=1s):
Again, the parabolic curve is the most difficult to interpret. We know that at t=0s, the slope of the position-time graph is zero, meaning that the velocity is 0 at t=0s. We also know that at t=1s, the slope of the position-time graph is -2, meaning that the velocity it -2 at t=1s. Because the slope of a parabola is described by a straight line, we know that the first part of our velocity-time graph must be a straight line going from 0 at t=0s to -2 at t=1s. This is shown in the graph below.
The Second Part (t=1s to t=2s):
For the second part of the graph, the slope of the position-time graph does not change. Therefore, the velocity does not change from t=1s to t=2s. This can be represented as a horizontal line on our velocity-time graph. The position of this horizontal line is given by the value of the slope found in the second part of step 2.
The Third Part (t=2s to t=4s):
Similar to the second part, the third part of our velocity-time graph can be represented by a horizontal line at the value of the slope found in the third part of step 2.
This image shows our complete velocity-time graph. Something to note here is that the break in the velocity-time graph at t=2s corresponds to the sharp change in slope seen in the position-time graph at t=2s.
Example 2
Identify the velocity-time graph for the motion of an object whose position is given by this position-time graph.
Step 1: Split the Position-Time Graph into separate parts by domain.
Our position-time graph shows a straight upward-sloping line that turns into a parabola at t=1s. The parabola reaches its peak at t=2s (this is not necessarily the beginning of a new part of the graph but a point of interest nonetheless) before turning into a downward-sloping line at t=3s.
The First Part: t=0s to t=1s
The Second Part: t=1s to t=3s
The Third Part: t=3s to t=4s
Step 2: Identify the slope of each part of the Position-Time Graph.
The First Part (t=0s to t=1s):
s l o p e=1−0 1−0 s l o p e=1
The Second Part (t=1s to t=3s):
For this parabolic curve, we can readily find the slope at 3 different points. Because we see the parabola smoothly transition from the first line segment at t=1s, we know that the slope at that point matches the slope of the line segment which we have already calculated to be 1. We can also see that the parabola is flat at t=2s meaning that the slope is 0 at t=2s. We then again see the parabola smoothly transition to the downward sloping line at t=3s, meaning that the slope of the parabola at t=3s must match the slope of the line in the third part. This slope is calculated next.
We can thus describe the slope of the second part as a smooth transition from a slope of 1 at t=1s to a slope of 0 at t=2s, then to a slope of -1 at t=3s.
The Third Part (t=3s to t=4s):
s l o p e=4−3 0−1 s l o p e=−1
Step 3: Graph the slope of each part in the corresponding domain of the Velocity-Time Graph.
The First Part (t=0s to t=1s):
The Second Part (t=1s to t=3s):
The Third Part (t=3s to t=4s):
This image shows our complete velocity-time graph. Another key feature to note in this graph is the transition from positive velocity to negative velocity at t=2s. This corresponds to the peak of our parabola in the position-time graph at t=2s. Leading up the parabola, the curve is sloping upward, meaning a positive velocity. Following the parabola, the curve is sloping downward, meaning a negative velocity. The point in the middle must therefore indicate a momentary point of zero velocity or instantaneous motionlessness.
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Table of Contents
How to Identify the Velocity-Time Graph that Corresponds to a Position-Time Graph
What are motion graphs?
Examples of identifying velocity-time graph that corresponds to a position-time graph
Example 1
Example 2
Test your current knowledge
Practice
Identifying the Velocity-Time Graph that Corresponds to a Position-Time Graph
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188734 | https://www.testmenu.com/zsfglab/TestDirectory/SiteFile?fileName=sidebar%5CSyphilis%20Antibody%20Test%20FAQ.pdf | Revised January 2021 1 of 3 Reverse Sequence Syphilis Screening Frequently Asked Questions 1) What is a syphilis EIA or CIA? A syphilis EIA is an “enzyme immunoassay.” This is a blood test for syphilis that tests for antibodies directed at the syphilis organism (Treponema pallidum). The EIA is a treponemal test, like the TPPA or MHA-TP that have been used for many years. Syphilis CIAs (“chemoluminescence assays”) are similar to EIAs. There are a number of syphilis EIAs and CIAs available. 2) I always remember ordering an RPR when I wanted to test for syphilis. Why is my lab using a syphilis EIA or CIA? Many commercial laboratories have adopted EIAs or CIAs for syphilis screening because they are highly automated tests, and thus are less expensive and more efficient to perform than RPRs or VDRLs. EIAs and CIAs are very sensitive and specific tests for syphilis and have a fast turn-around time. 3) Why is screening with an EIA instead of an RPR sometimes referred to as Reverse Sequence Syphilis Screening? Traditionally, syphilis screening is conducted with an RPR or VDRL. Unlike EIAs, the RPR and VDRL are non-treponemal antibody tests. They detect antibodies to proteins that are not part of the syphilis organism itself, but are similar to proteins found in the syphilis bacteria. RPRs or VDRLs must be confirmed with a treponemal test; most labs use a TPPA (treponemal pallidum particle assay) to confirm a positive RPR or VDRL. Screening with a non-treponemal test and confirming with a treponemal test is sometimes referred to as “traditional sequence” syphilis screening. Labs that screen with an EIA or CIA have “reversed” this algorithm. Rather than starting with a non-treponemal test and confirming with a treponemal test, they are starting with a treponemal test (the EIA or CIA) and confirming with a non-treponemal test (the RPR). Thus it is referred to as “reverse sequence” syphilis screening. 4) If the EIA is positive, why do I need an RPR? The RPR provides a quantitative result (known as a “titer”) that is helpful for staging disease and for establishing a baseline for determining whether the patient responds appropriately to syphilis treatment. Once the EIA is positive (i.e., the first time someone is infected with syphilis), it remains positive for life in most people, and is therefore not useful for detecting reinfection, unlike the RPR, which fluctuates with disease activity. 5) Will the EIA become negative after I treat the patient? No, the EIA usually remains positive after treatment, but if the RPR decreases fourfold after treatment, the patient has likely been cured of syphilis. Remember that individuals can get syphilis more than once, so it is important to continue screening patients who are at risk for syphilis on a regular basis (q3-6 months). 6) My patient’s EIA is positive, but the RPR is not back yet, what should I do? If your patient does not have any risk factors for syphilis and you are certain you can reach the patient if their RPR or TPPA are positive, one option is to send the patient home and wait for the rest of the test results to return. Revised January 2021 2 of 3 If your patient does not have a prior history of syphilis and has risk factors for syphilis, or if the patient may be difficult to reach after the visit, we recommend having a low threshold for empiric treatment with benzathine penicillin G 2.4 mu IM. 7) What does it mean if the EIA is positive and the RPR is also positive? This means the patient has untreated or previously treated syphilis. Determining whether a patient has an untreated syphilis infection and ascertaining the stage of syphilis relies on a combination of clinical history, diagnostic tests and physical examination. When interpreting syphilis tests, a provider should do the following: 1) Ask: Have you had a history of syphilis, recent symptoms (e.g. sore or rash), have any of your sex partners been diagnosed with syphilis? 2) Assess for risk factors for syphilis. 3) Examine: Physical examination for ulcers (primary syphilis) or rashes and/or mucocutaneous lesions (secondary syphilis) with attention to the mouth, skin, and anogenital areas. 4) Call: To obtain prior RPR titers or syphilis treatment history for patients who live in San Francisco, call SF City Clinic: 415-487-5531 8) What if the EIA is positive and the RPR is negative? This is sometimes referred to as a “discordant” result. In this situation, the CDC recommends obtaining a third syphilis test to help resolve the situation. Many labs will automatically “reflect” to a third test in this situation. Typically, this is a TPPA. The following 2 scenarios can occur: a. EIA positive, RPR negative, TPPA negative: This is most likely a false positive EIA. No further action is necessary. If your clinical suspicion for syphilis is high, you should consider empiric treatment for syphilis, and repeat the syphilis tests in 2-4 weeks. The EIA is more sensitive for very early primary syphilis than the TPPA or RPR, so could theoretically turn positive before these other two tests. b. EIA positive, RPR negative, TPPA positive: There are several possible explanations for these results: 1. The patient has a history of treated syphilis. If this is the case, no treatment is necessary. If the patient is at risk for syphilis, they should continue to be screened every 3-6 months. 2. The patient has late latent syphilis. In this case, the patient should be treated with Benzathine Penicillin G 2.4 mu IM weekly x 3 3. The patient was recently exposed to syphilis and the RPR has not yet turned positive. If your patient is at risk for syphilis, bring the patient back to clinic, do a complete oral, skin and anogenital exam to look for signs of syphilis and repeat the syphilis tests. If the patient is asymptomatic and the RPR is still negative, and the patient has never been previously treated for syphilis, they should be treated for late latent syphilis. Revised January 2021 3 of 3 9) My patient has a history of syphilis, what should I do? If your patient has a history of syphilis, they should be screened with an RPR. At ZSFG, RPR alone can be ordered as: “Rapid Plasma Reagin (RPR) for prior history of syphilis.” This test is not indicated for initial laboratory evaluation of syphilis. 10) How do I know if my patient is at risk for syphilis? In 2019, approximately 90% of early syphilis cases in San Francisco were among men, and over 75% were among men have sex with men (MSM). While the majority of syphilis cases in San Francisco occur among men, between 2017 and 2019, the number of syphilis cases in women increased by 155%, and congenital syphilis is increasing across California and in San Francisco. • Sexually active MSM and trans people who have sex with men should be screened for syphilis every 3 months • Pregnant women should be screened for syphilis at least twice during pregnancy: once at either confirmation of pregnancy or at the first prenatal encounter, and again during the 3rd trimester (ideally during 28-32 weeks gestation). Patients should be screened again at delivery, except those at low risk who have a documented negative screen in the third trimester. • Non-pregnant adults should be screened for syphilis at least annually if they report any of the following: Sex with a man who has sex with men, history of STI in the past year, methamphetamine use, unstable housing or homelessness, sex work, intimate partner violence, or incarceration. • All sexually active people who could become pregnant should receive at least one lifetime screen for syphilis, with additional screening for those at increased risk |
188735 | https://askfilo.com/user-question-answers-smart-solutions/q-2-answer-the-following-questions-i-state-the-formula-for-3137383132363130 | Q 2. answer the following questions : i. State the formula for the end co..
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Q 2. answer the following questions : i. State the formula for the end correction in a resonance tube experiment. (8N) ii. Categorize the following into polar and nonpolar dielectrics: (a) H 2O (b) CO 2. iii. Define potential gradient. iv. Calculate the period of a particle performing linear S HM with maximum speed 0.08 m/s and maximum acceleration 0.32 m/s 2. v. Define gyromagnetic ratio. vi. State the conditions for current and impedance in parallel resonance circult. vii. The radius of the third Bohr orbit is 0.477 nm . Calculate the radius of the second Bohr orbit. viii. Name the logic gate having single input and single output.
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Text solutionVerified
Concepts:
End correction, Polar and nonpolar dielectrics, Potential gradient, Simple harmonic motion, Gyromagnetic ratio, Parallel resonance circuit, Bohr orbit, Logic gates
Explanation:
The following are the answers to the questions posed in Q 2:
Step by Step Solution:
Step 1
i. The formula for the end correction in a resonance tube experiment is given by: e=0.6×r, where e is the end correction and r is the radius of the tube.
Step 2
ii. (a) H 2O is a polar dielectric. (b) C O 2 is a nonpolar dielectric.
Step 3
iii. The potential gradient is defined as the rate of change of electric potential (voltage) with respect to distance. Mathematically, it is expressed as: Potential Gradient=−d x d V, where V is the electric potential and x is the distance.
Step 4
iv. The period T of a particle performing linear simple harmonic motion (SHM) can be calculated using the formula: T=2 π k m, where m is the mass and k is the spring constant. However, we can also use the relationship between maximum speed (v ma x) and maximum acceleration (a ma x): v ma x=A ω and a ma x=A ω 2, where A is the amplitude and ω is the angular frequency. From these, we can derive T=ω 2 π. Given v ma x=0.08 m/s and a ma x=0.32 m/s 2, we find ω=v ma xa ma x=0.08 0.32=4 rad/s. Thus, T=4 2 π=2 π≈1.57 s.
Step 5
v. The gyromagnetic ratio is defined as the ratio of the magnetic moment of a particle to its angular momentum. It is denoted by γ and can be expressed as: γ=L μ, where μ is the magnetic moment and L is the angular momentum.
Step 6
vi. The conditions for current and impedance in a parallel resonance circuit are: (1) The total current is the sum of the currents through each branch. (2) The impedance of the circuit is at its minimum at resonance, and the circuit behaves as a purely resistive circuit.
Step 7
vii. The radius of the second Bohr orbit can be calculated using the formula: r n=n 2⋅r 1, where r 1 is the radius of the first Bohr orbit (0.0529 nm) and n is the principal quantum number. For the second orbit (n=2), r 2=2 2⋅0.0529 nm=0.2116 nm.
Step 8
viii. The logic gate having a single input and a single output is called a NOT gate (inverter).
Final Answer:
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Q 2. answer the following questions : i. State the formula for the end correction in a resonance tube experiment. (8N) ii. Categorize the following into polar and nonpolar dielectrics: (a) H 2O (b) CO 2. iii. Define potential gradient. iv. Calculate the period of a particle performing linear S HM with maximum speed 0.08 m/s and maximum acceleration 0.32 m/s 2. v. Define gyromagnetic ratio. vi. State the conditions for current and impedance in parallel resonance circult. vii. The radius of the third Bohr orbit is 0.477 nm . Calculate the radius of the second Bohr orbit. viii. Name the logic gate having single input and single output.
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डायमैग्नेटिज्म, पारामैग्नेटिज्म तथा फेरोमैग्नेटिज्म के सभी पदार्थों का सार्वभौम गुण क्या है? (A) डायमैग्नेटिज्म (B) पारामैग्नेटिज्म (C) फेरोमैग्नेटिज्म (D) उपर्युक्त सभी
Topic: Smart Solutions
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Question 3
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The Balance Sheet of Sheela and Mrudula is as under. They are sharing Profits and Losses in the ratio 3:1. Balance Sheet as on 31st March, 2020 | Liabilities | Amt. (₹) | Assets | Amt. (₹) | --- --- | | Creditors | 40,000 | Cash | 40,000 | | Bills Payable | 10,000 | Sundry Debtors | 32,000 | Mangala is taken as partner on 1st April, 2020 on the following terms: Mangala will pay ₹ 20,000 as her share of Capital for 1/5 share in future Profit and ₹ 10,000 as her share of Goodwill. Provide 5% for bad and doubtful debts on Sundry debtors. Furniture to be depreciated by 10%. Stock and Plant & Machinery to be appreciated by 10%. Prepare: (i) Profit and Loss Adjustment A/c (ii) Partners Capital Accounts (iii) Balance Sheet of New firm.
Topic: Smart Solutions
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Question 4
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b) Find the current in 12ohms using Mesh and Nodal Analysis. 2. a) State and Explain Nortons theorem. b) Determine current in 5 ohms resistor using Thevenin's theorem. 3. a) State and Explain Superposition Theorem. b) Determine the Current through 8 Ohms using Superposition theor
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Question Text Q 2. answer the following questions : i. State the formula for the end correction in a resonance tube experiment. (8N) ii. Categorize the following into polar and nonpolar dielectrics: (a) H 2O (b) CO 2. iii. Define potential gradient. iv. Calculate the period of a particle performing linear S HM with maximum speed 0.08 m/s and maximum acceleration 0.32 m/s 2. v. Define gyromagnetic ratio. vi. State the conditions for current and impedance in parallel resonance circult. vii. The radius of the third Bohr orbit is 0.477 nm . Calculate the radius of the second Bohr orbit. viii. Name the logic gate having single input and single output.
Updated On Jan 2, 2025
Topic All topics
Subject Smart Solutions
Class Class 12
Answer Type Text solution:1
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188736 | https://www.accessdata.fda.gov/drugsatfda_docs/label/2019/021201s040lbl.pdf | DOBUTamine in 5% Dextrose Injection, USP Flexible Plastic Container Rx only DESCRIPTION Dobutamine in 5% Dextrose Injection, USP is a sterile, nonpyrogenic, prediluted solution of dobutamine hydrochloride and dextrose in water for injection. It is administered by intravenous infusion. Each 100 mL contains dobutamine hydrochloride equivalent to 100 mg, 200 mg, or 400 mg of dobutamine; dextrose (derived from corn), hydrous 5 g in water for injection, with sodium metabisulfite 25 mg and edetate disodium, dihydrate 10 mg added as stabilizers; osmolar concentration, respectively, 263, 270, or 284 mOsmol/liter (calc.). The pH is 3.0 (2.5 to 5.5). May contain hydrochloric acid and/or sodium hydroxide for pH adjustment. Dobutamine in 5% Dextrose Injection, USP is oxygen sensitive. Dobutamine Hydrochloride, USP is chemically designated (±)-4-[2-[[3-(p-hydroxyphenyl)-1 methylpropyl]amino]ethyl]-pyrocatechol hydrochloride. It is a synthetic catecholamine. Dextrose, USP is chemically designated D-glucose monohydrate (C6H12O6 • H2O), a hexose sugar freely soluble in water. It has the following structural formula: Water for Injection, USP is chemically designated H2O. The flexible plastic container is fabricated from a specially formulated CR3 plastic material. Water can permeate from inside the container into the overwrap but not in amounts sufficient to affect the solution significantly. Solutions in contact with the plastic container may leach out certain chemical components from the plastic in very small amounts; however, biological testing was supportive of the safety of the plastic container materials. Exposure to temperatures above 25°C/77°F during transport and storage will lead to minor losses in moisture content. Higher temperatures lead to greater losses. It is unlikely that these minor losses will lead to clinically significant changes within the expiration period. Page 1 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit CLINICAL PHARMACOLOGY Dobutamine is a direct-acting inotropic agent whose primary activity results from stimulation of the β-receptors of the heart while producing comparatively mild chronotropic, hypertensive, arrhythmogenic, and vasodilative effects. It does not cause the release of endogenous norepinephrine, as does dopamine. In animal studies, dobutamine produces less increase in heart rate and less decrease in peripheral vascular resistance for a given inotropic effect than does isoproterenol. In patients with depressed cardiac function, both dobutamine and isoproterenol increase the cardiac output to a similar degree. In the case of dobutamine, this increase is usually not accompanied by marked increases in heart rate (although tachycardia is occasionally observed), and the cardiac stroke volume is usually increased. In contrast, isoproterenol increases the cardiac index primarily by increasing the heart rate while stroke volume changes little or declines. Facilitation of atrioventricular conduction has been observed in human electrophysiologic studies and in patients with atrial fibrillation. Systemic vascular resistance is usually decreased with administration of dobutamine. Occasionally, minimum vasoconstriction has been observed. Most clinical experience with dobutamine is short-term, not more than several hours in duration. In the limited number of patients who were studied for 24, 48, and 72 hours, a persistent increase in cardiac output occurred in some, whereas output returned toward baseline values in others. The onset of action of Dobutamine in 5% Dextrose Injection, USP is within 1 to 2 minutes; however, as much as 10 minutes may be required to obtain the peak effect of a particular infusion rate. The plasma half-life of dobutamine in humans is 2 minutes. The principal routes of metabolism are methylation of the catechol and conjugation. In human urine, the major excretion products are the conjugates of dobutamine and 3-O-methyl dobutamine. The 3-O-methyl derivative of dobutamine is inactive. Alteration of synaptic concentrations of catecholamines with either reserpine or tricyclic antidepressants does not alter the actions of dobutamine in animals, which indicates that the actions of dobutamine are not dependent on presynaptic mechanisms. The effective infusion rate of dobutamine varies widely from patient to patient, and titration is always necessary (see DOSAGE AND ADMINISTRATION). At least in pediatric patients, dobutamine induced increases in cardiac output and systemic pressure are generally seen, in any given patient, at lower infusion rates than those that cause substantial tachycardia (See Pediatric Use under PRECAUTIONS). INDICATIONS AND USAGE Dobutamine in 5% Dextrose Injection, USP is indicated when parenteral therapy is necessary for inotropic support in the short-term treatment of patients with cardiac decompensation due to depressed contractility resulting either from organic heart disease or from cardiac surgical procedures. Experience with intravenous dobutamine in controlled trials does not extend beyond 48 hours of repeated boluses and/or continuous infusions. Whether given orally, continuously intravenously, or intermittently intravenously, neither dobutamine nor any other cyclic-AMP-dependent inotrope has been shown in controlled trials to be safe or effective in the long-term treatment of congestive heart failure. In controlled trials of chronic oral therapy with various such agents, symptoms were not consistently alleviated, and the cyclic-AMP-dependent inotropes were Page 2 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit consistently associated with increased risks of hospitalization and death. Patients with NYHA Class IV symptoms appeared to be at particular risk. CONTRAINDICATIONS Dobutamine in 5% Dextrose Injection, USP is contraindicated in patients with idiopathic hypertrophic subaortic stenosis and in patients who have shown previous manifestations of hypersensitivity to dobutamine or any of its components. WARNINGS Increase in Heart Rate or Blood Pressure Dobutamine hydrochloride may cause a marked increase in heart rate or blood pressure, especially systolic pressure. Approximately 10% of adult patients in clinical studies have had rate increases of 30 beats/minute or more, and about 7.5% have had a 50-mm Hg or greater increase in systolic pressure. Usually, reduction of dosage reverses these effects. Because dobutamine facilitates atrioventricular conduction, patients with atrial fibrillation are at risk of developing rapid ventricular response. In patients who have atrial fibrillation with rapid ventricular response, a digitalis preparation should be used prior to institution of therapy with dobutamine. Patients with pre-existing hypertension appear to face an increased risk of developing an exaggerated pressure response. Ectopic Activity Dobutamine may precipitate or exacerbate ventricular ectopic activity, but it rarely has caused ventricular tachycardia. Hypersensitivity Reactions suggestive of hypersensitivity associated with administration of Dobutamine in 5% Dextrose Injection, USP, including skin rash, fever, eosinophilia, and bronchospasm, have been reported occasionally. Dobutamine in 5% Dextrose Injection, USP contains sodium bisulfite, a sulfite that may cause allergic-type reactions, including anaphylactic symptoms and life-threatening or less severe asthmatic episodes, in certain susceptible people. The overall prevalence of sulfite sensitivity in the general population is unknown and probably low. Sulfite sensitivity is seen more frequently in asthmatic than in nonasthmatic people. PRECAUTIONS General During the administration of dobutamine, as with any adrenergic agent, ECG and blood pressure should be continuously monitored. In addition, pulmonary wedge pressure and cardiac output should be monitored whenever possible to aid in the safe and effective infusion of Dobutamine in 5% Dextrose Injection, USP. Hypovolemia should be corrected with suitable volume expanders before treatment with Dobutamine in 5% Dextrose Injection, USP is instituted. Animal studies indicate that dobutamine may be ineffective if the patient has recently received a β-blocking drug. In such a case, the peripheral vascular resistance may increase. No improvement may be observed in the presence of marked mechanical obstruction, such as severe valvular aortic stenosis. Page 3 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit Do not administer unless solution is clear and container is undamaged. Discard unused portion. Usage Following Acute Myocardial Infarction: Clinical experience with dobutamine following myocardial infarction has been insufficient to establish the safety of the drug for this use. There is concern that any agent that increases contractile force and heart rate may increase the size of an infarction by intensifying ischemia, but it is not known whether dobutamine does so. Drug Interactions: There was no evidence of drug interactions in clinical studies in which dobutamine hydrochloride was administered concurrently with other drugs, including digitalis preparations, furosemide, spironolactone, lidocaine, glyceryl trinitrate, isosorbide dinitrate, morphine, atropine, heparin, protamine, potassium chloride, folic acid, and acetaminophen. Preliminary studies indicate that the concomitant use of dobutamine and nitroprusside results in a higher cardiac output and, usually, a lower pulmonary wedge pressure than when either drug is used alone. Concomitant use of dobutamine and catechol-O-methyltranferase (COMT) inhibitors (e.g., entacapone) may result in increased heart rate, arrhythmias, and changes in blood pressure. Carcinogenesis, Mutagenesis, Impairment of Fertility: Studies to evaluate the carcinogenic or mutagenic potential of dobutamine or the potential of the drug to affect fertility adversely have not been performed. Pregnancy: Reproduction studies performed in rats and rabbits have revealed no evidence of harm to the fetus due to dobutamine. The drug, however, has not been administered to pregnant women and should be used only when the expected benefits clearly outweigh the potential risks to the fetus. Pediatric Use: Dobutamine has been shown to increase cardiac output and systemic pressure in pediatric patients of every age group. In premature neonates, however, dobutamine is less effective than dopamine in raising systemic blood pressure without causing undue tachycardia, and dobutamine has not been shown to provide any added benefit when given to such infants already receiving optimal infusions of dopamine. Geriatric Use: Clinical studies of dobutamine did not include sufficient numbers of subjects aged 65 and over to determine whether they respond differently from younger subjects. Other reported clinical experience has not identified differences in responses between the elderly and younger patients. In general, dose selection for an elderly patient should be cautious, usually starting at the low end of the dosing range, reflecting the greater frequency of decreased hepatic, renal, or cardiac function, and of concomitant disease or drug therapy. ADVERSE REACTIONS Increased Heart Rate, Blood Pressure, and Ventricular Ectopic Activity: A 10- to 20-mm Hg increase in systolic blood pressure and an increase in heart rate of 5 to 15 beats/minute have been noted in most patients (see WARNINGS regarding exaggerated chronotropic and pressor effects). Approximately 5% of adult patients have had increased premature ventricular beats during infusions. These effects are dose related. Hypotension: Precipitous decreases in blood pressure have occasionally been described in association with dobutamine therapy. Decreasing the dose or discontinuing the infusion typically results in rapid return of blood pressure to baseline values. In rare cases, however, intervention may be required and reversibility may not be immediate. Reactions at Sites of Intravenous Infusion: Phlebitis has occasionally been reported. Local inflammatory changes have been described following inadvertent infiltration. Page 4 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit Miscellaneous Uncommon Effects: The following adverse effects have been reported in 1% to 3% of adult patients: nausea, headache, anginal pain, nonspecific chest pain, palpitations, and shortness of breath. Administration of dobutamine, like other catecholamines, has been associated with decreases in serum potassium concentrations, rarely to hypokalemic values. OVERDOSAGE Overdoses of dobutamine have been reported rarely. The following is provided to serve as a guide if such an overdose is encountered. Signs and Symptoms: Toxicity from dobutamine hydrochloride is usually due to excessive cardiac β-receptor stimulation. The duration of action of dobutamine hydrochloride is generally short (T½ = 2 minutes) because it is rapidly metabolized by catechol-O-methyltransferase. The symptoms of toxicity may include anorexia, nausea, vomiting, tremor, anxiety, palpitations, headache, shortness of breath, and anginal and nonspecific chest pain. The positive inotropic and chronotropic effects of dobutamine on the myocardium may cause hypertension, tachyarrhythmias, myocardial ischemia, and ventricular fibrillation. Hypotension may result from vasodilation. If the product is ingested, unpredictable absorption may occur from the mouth and the gastrointestinal tract. Treatment: To obtain up-to-date information about the treatment of overdose, a good resource is your certified Regional Poison Control Center. Telephone numbers of certified poison control centers are listed in the Physicians’ Desk Reference (PDR). In managing overdosage, consider the possibility of multiple drug overdoses, interaction among drugs, and unusual drug kinetics in your patient. The initial actions to be taken in a dobutamine hydrochloride overdose are discontinuing administration, establishing an airway, and ensuring oxygenation and ventilation. Resuscitative measures should be initiated promptly. Severe ventricular tachyarrhythmias may be successfully treated with propranolol or lidocaine. Hypertension usually responds to a reduction in dose or discontinuation of therapy. Protect the patient's airway and support ventilation and perfusion. If needed, meticulously monitor and maintain, within acceptable limits, the patient's vital signs, blood gases, serum electrolytes, etc. Absorption of drugs from the gastrointestinal tract may be decreased by giving activated charcoal, which, in many cases, is more effective than emesis or lavage; consider charcoal instead of or in addition to gastric emptying. Repeated doses of charcoal over time may hasten elimination of some drugs that have been absorbed. Safeguard the patient's airway when employing gastric emptying or charcoal. Forced diuresis, peritoneal dialysis, hemodialysis, or charcoal hemoperfusion have not been established as beneficial for an overdose of dobutamine hydrochloride. DOSAGE AND ADMINISTRATION Recommended Dosage Dobutamine in 5% Dextrose Injection, USP is administered intravenously through a suitable intravenous catheter or needle. A calibrated electronic infusion device is recommended for controlling the rate of flow in mL/hour or drops/minute. Infusion of dobutamine should be started at a low rate (0.5 to 1.0 µg/kg/min) and titrated at intervals of a few minutes, guided by the patient’s response, including systemic blood pressure, urine flow, frequency of ectopic activity, heart rate, and (whenever possible) measurements of cardiac output, central venous pressure, and/or pulmonary capillary wedge pressure. In reported trials, the optimal infusion rates have varied from patient to patient, usually 2 to 20 µg/kg/min but sometimes Page 5 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit slightly outside of this range. On rare occasions, infusion rates up to 40 µg/kg/min have been required to obtain the desired effect. Rates of infusion in mL/hour for dobutamine hydrochloride concentrations of 500, 1,000, 2,000 and 4,000 mg/L may be calculated using the following formula: Infusion Rate (mL/h) = [Dose (µg/kg/min) x Weight (kg) x 60 min/h] Final Concentration (µg/mL) Example calculations for infusion rates are as follows: Example 1: for a 60 kg person at an initial dose of 0.5 µg/kg/min using a 500 µg/mL concentration, the infusion rate would be as follows: Infusion Rate (mL/h) = [0.5 (µg/kg/min) x 60 (kg) x 60 (min/h)] = 3.6 (mL/h) 500 (µg/mL) Example 2: for a 80 kg person at a dose of 10 µg/kg/min using a 2,000 µg/mL concentration, the infusion rate would be as follows: Infusion Rate (mL/h) = [10 (µg/kg/min) x 80 (kg) x 60 (min/h)] = 24 (mL/h) 2,000 (µg/mL) This container system may be inappropriate for the dosage requirements of pediatric patients under 30 kg. Parenteral drug products should be inspected visually for particulate matter and discoloration prior to administration whenever solution and container permit. Dobutamine in 5% Dextrose Injection, USP solutions may exhibit a pink color that, if present, will increase with time. This color change is due to slight oxidation of the drug, but there is no significant loss of potency. Solutions containing dextrose should not be administered through the same administration set as blood, as this may result in pseudoagglutination or hemolysis. Do not add supplementary medications to Dobutamine in 5% Dextrose Injection, USP. Do not administer Dobutamine in 5% Dextrose Injection, USP simultaneously with solutions containing sodium bicarbonate or strong alkaline solutions. INSTRUCTIONS FOR USE To Open Tear outer wrap at notch and remove solution container. Some opacity of the plastic due to moisture absorption during the sterilization process may be observed. This is normal and does not affect the solution quality or safety. The opacity will diminish gradually. Preparation for Administration (Use aseptic technique) 1. Close flow control clamp of administration set. 2. Remove cover from outlet port at bottom of container. 3. Insert piercing pin of administration set into port with a twisting motion until the set is firmly seated. NOTE: See full directions on administration set carton. 4. Suspend container from hanger. 5. Squeeze and release drip chamber to establish proper fluid level in chamber. 6. Open flow control clamp and clear air from set. Close clamp. Page 6 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit 7. Attach set to venipuncture device. If device is not indwelling, prime and make venipuncture. 8. Regulate rate of administration with flow control clamp. WARNING: Do not use flexible container in series connections. HOW SUPPLIED DOBUTamine in 5% Dextrose Injection, USP is supplied in 250 and 500 mL LifeCare™ flexible containers as follows: Unit of Sale Total Strength/Total Volume (Concentration) NDC 0409-2346-32 Case of 12 Flexible containers 250 mg/250 mL (1 mg/mL) NDC 0409-2347-32 Case of 12 Flexible containers 500 mg/250 mL (2 mg/mL) NDC 0409-3724-32 Case of 12 Flexible containers 1000 mg/250 mL (4 mg/mL) Do not freeze. Store at 20 to 25ºC (68 to 77ºF). [See USP Controlled Room Temperature.] Distributed by Hospira, Inc., Lake Forest, IL 60045 USA LAB-1181-1.2 Revised: 5/2019 Page 7 of 7 Reference ID: 4438558 This label may not be the latest approved by FDA. For current labeling information, please visit |
188737 | https://luckylittlelearners.com/synonyms-and-antonyms-activities/ | Home » Blog » Writing & Language » Grammar » Synonyms and Antonyms Activities
Synonyms and Antonyms Activities
Written by: Mary Kate Bolinder
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Here’s a joke for you: What does a thesaurus eat for breakfast?
Answer: Synonym rolls! (We love to laugh at Lucky Little Learners!)
Keep reading to discover more amazing, awesome, incredible, marvelous, wonderful activities to teach synonyms and antonyms.
Enhance Literacy with Embedded Grammar Lessons
Research on the Science of Reading (SOR) dating back to the 1970s shows that teaching grammar as isolated rules doesn’t effectively transfer to writing and speaking. What works better is embedding grammar within the context of writing. This is why our writing program Lucky to Learn Writing (LTLW) includes grammar skills integrated into writing lessons. This approach ensures that the same skills and standards are met, but in a meaningful context that connects with the rest of the literacy curriculum.
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We also recognize the value of targeted practice and intervention to reinforce these skills. That’s why we’ve compiled a list of our favorite grammar activities that provide additional practice and intervention opportunities. These activities are designed to complement our integrated approach, ensuring students have a well-rounded understanding of grammar concepts. Let’s dive into these engaging and effective activities!
What are synonyms and antonyms?
Synonyms are words that have the same, or close to the same, meaning.
Example: happy, cheerful
Antonyms are words that have opposite meanings.
Example: happy, sad
This video from SpellingCity offers a great explanation of synonym and antonym relationships.
Where do you find synonyms and antonyms?
Synonyms and antonyms can be found in a thesaurus. Similar to a dictionary, a thesaurus lists words in alphabetical order. Instead of providing a definition, a thesaurus gives a group of synonyms and antonyms for the listed word. Check out an online thesaurus here.
A dictionary will also list a few synonyms and antonyms after the definition of a word. Find a student-friendly online dictionary here.
Synonym and Antonym Activities
Give one of these activities a try to get students up and moving!
Synonym Search
Write a pair of synonyms, one on a notecard, one on a sticky note. Place the sticky notes around the classroom. Give each student a notecard, and have them search the classroom for their synonym. Create a class chart with the answers.
Antonym Match
Write a pair of antonyms, each on a sticky note. Give each student a sticky note to wear. Students must then find their antonym match. How fast can you find your antonym match? Mix up the sticky notes and play again. Create a class chart with the answers.
Resources for Teaching Synonyms and Antonyms
Here is a list of our favorite resources for teaching synonyms and antonyms.
Grammar Toothy
Yes, we really do have a Toothy for everything! (New to Toothy? Learn all about Toothy here!)
Students will practice identifying pairs of words as synonyms or antonyms. This self-checking activity is great for whole group, small group, independent practice, morning work, and so much more!
The synonym and antonym activity is part of the Grammar Toothy Series.
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Grammar Day by Day
Build important grammar skills one day at a time with Grammar Day by Day. These synonym and antonym worksheets provide a quick and easy way to practice this new skill. In fact, one for each day of the school week is ready to go! Color, write, and match to identify the correct synonym or antonym.
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Spiral ELA
Teachers love our Spiral ELA. Students can practice identifying and writing synonyms and antonyms throughout the year to keep skills fresh!
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Shoes and Socks
In this card game from our 2nd Grade Grammar Centers, students will take turns trying to match a pair of synonyms (shoes) or antonyms (socks). Students can play for fun, or use the recording sheet for a quick check of their understanding.
Picture Books about Synonyms and Antonyms
These picture books about synonyms and antonyms are a great read alouds to share with your class as you study synonyms and antonyms.
Note: If you click on the Amazon affiliate links and purchase, I may receive a small commission at no extra cost to you.
Stegothesaurus
Thesaurus Has a Secret
How to Teach Other 2nd Grade Grammar Skills:
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1st Grade Review
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Common & Proper Nouns
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Compound Words
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Abbreviations
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Antonyms & Synonyms ~ You are here!
Pronouns
Subject-Verb Agreement
Collective Nouns
Articles
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188738 | https://www.cuemath.com/numbers/non-terminating-decimal/ | Non-Terminating Decimal - Definition, Expansion, Conversion
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Non-Terminating Decimal
Non-terminating decimals are decimals that have never-ending decimal digits and continue forever. We will be learning about the non-terminating decimal expansion and conversion of non-terminating decimal to fraction in this article.
1.Non-Terminating Decimal Definition
2.Non-Terminating Decimal Expansion
3.Conversion of Non-Terminating Decimal to Rational Number
4.FAQs on Non-Terminating Decimal
Non-Terminating Decimal Definition
Non-terminating decimals are defined as those decimal numbers that do not have an endpoint in their decimal digits and keep continuing forever. This happens when a dividend is divided by a divisor but the remainder is never 0 and hence the process keeps repeating and the non-terminating decimal is obtained in the quotient where the decimal digits keep occurring and never come to an end. A non-terminating decimal has an infinite number of decimal places and it is named as non-terminating because the decimal will never terminate. For example, 1.333333..., 4.65675747775..., etc.
Non-Terminating Decimal Expansion
A non-terminating decimal expansion has an infinite number of places and its expansion continues forever. We have two types of non-terminating decimal expansions and they are as follows:
Non terminating recurring decimal expansion
Non terminating non-recurring decimal expansion
Non-Terminating Recurring Decimal Expansion
Non-terminating recurring decimal is also known by the name non terminating repeating decimal. In this expansion, the decimal places will continue forever and never come to an end but since the name says repeating or recurring, it signifies that the repetition of the decimal values forms a specific pattern that can be easily identified. For example, 2/9 = 0.2222222…. is a non-terminating recurring decimal expansion. The repetition of the decimal can also be indicated by showing a bar on top of the numbers that are repeating i.e., 0.222222... can also be represented as 0.¯¯¯2 0.2¯. Similarly, 1/7 = 0.142857 142857 142857... which can also be written as 0.¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯142857 0.142857¯ is also a non-terminating repeating decimal expansion as the block of decimals 142857 is repeating after every 6 digits. Non-terminating recurring decimals can always be converted to a rational number.
Non-Terminating Non-Recurring Decimal Expansion
Non-terminating non-recurring decimal is also known by the name non terminating non-repeating decimal as the values after decimal do not repeat or terminate. For example, 1.4142135..., 2.35638745... Unlike non-terminating recurring decimal, the decimal places do not form any pattern. A non-terminating non-recurring decimal cannot be converted to a rational number. Hence, non terminating non-recurring decimals are also known as irrational numbers. Note that pi is an irrational number as its expansion is non-terminating non-recurring i.e., 3.1415926535 897...
Conversion of Non-Terminating Decimal to Rational Number
As seen in the previous section, a non-terminating recurring decimal can be converted into a rational number. A rational number is defined as the ratio of two integers p and q and is represented as p/q where q ≠ 0. Let us take an example to understand the conversion of a non-terminating recurring decimal to a rational number.
Steps to Convert Non Terminating Recurring Decimal to Rational Number
Let us understand the steps to convert a non-terminating recurring decimal to a rational number by taking an example.
Step 1: Assume the repeating decimal to be equal to some variable x.
Step 2: Write the number without using a bar and equal to x. (Bar is for digits that repeat in the same pattern)
Step 3: Determine the number of digits having a bar on their heads or the number of digits before the bar for mixed recurring decimal.
Step 4: If the repeating number is the same digit after decimal such as 0.2222... then multiply by 10, if repetition of the digits is in pairs of two numbers such as 0.7878... then multiply by 100 and so on.
Step 5: Subtract the equation formed by step 2 and step 4.
Step 6: Then find the value of x in the simplest form.
Let's take an example of a non-terminating recurring decimal number 0.777...
Let, x = 0.777... -------------- (1)
Multiplying 10 on both the sides, we get,
10x = 7.777.. ----------------- (2) (This has to be chosen in such a way that on subtracting we get rid of the decimal)
Subtracting the two equations,
10x - x = 7.777 - 0.777
9x = 7
x = 7/9
Let's take another example to understand this. Convert a non-terminating decimal 0.6565... to a rational number.
Let x = 0.6565... --------------------- (1)
Multiplying 100 on both the sides,
100x = 65.6565... -------------------- (2)
Subtracting the above equations, we get,
100x - x = 65.6565 - 0.6565
99x = 65
x = 65/99
Thus, we have understood the steps to convert a non-terminating recurring decimal to a rational number.
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Non-Terminating Decimal Examples
Example 1: What kind of decimal expansion does 10/3 have? Show using long division.
Solution: Let us divide 10 by 3 using the long division method,
We see that when 10 is divided by 3, the quotient is 3.333... in which 3 is repeating and does not come to an end. Therefore, we can say that 10/3 = 3.333... is a non-terminating decimal. To be more specific, since the block of numbers is repeating, it is known as a non-terminating recurring decimal.
2. Example 2: Convert the non terminating decimal 0.666... to a rational number.
Solution: Let , x = 0.666... --------- (1)
Multiplying 10 on both the sides we get,
10x = 6.666.. ----------------- (2)
Subtracting the two equations,
10x - x = 6.666 - 0.666
9x = 6
x = 6/9 or 2/3
Thus, on converting the non-terminating decimal 0.666... to a rational number we get the result as 2/3.
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FAQs on Non-Terminating Decimal
What is Non-Terminating Decimal?
A non-terminating decimal is defined as a decimal number that does not have an endpoint in its decimal digit and keeps continuing forever. For example, 3.12345... is a non-terminating decimal.
What is Non-Terminating Decimal Expansion?
A non-terminating decimal expansion has an infinite number of places and its expansion continues forever. For example, 1.232323...
What are the Types of Non-Terminating Decimals?
There are two types of non-terminating decimals that are: a) Non-terminating recurring decimal which has a pattern of digit repetition in the decimal number. Example, 1.55555, and b) Non-terminating non-recurring decimal which does not have any pattern of digit repetition in the decimal number. Example, 3.12567509...
What does Non-Terminating Decimal Mean?
A non-terminating decimal is defined as a decimal expansion that has an infinite number of decimal places. Example, 1.222222.....
How do you Write a Non-Terminating Decimal?
A non-terminating decimal is written with a bar on top of the block of numbers that is repeating. For example, 0.88888... can be written as 0.¯¯¯8 0.8¯.
How to Convert Non-Terminating Decimal to Rational Number?
Non-terminating decimals are converted to rational number by following the steps below:
Step 1: Identify the repeating digits in the given decimal number.
Step 2: Equate the decimal number with x or any other variable.
Step 3: Place the repeating digits to the left of the decimal point by multiplying the equation obtained in step 2 by a power of 10 equal to the number of repeating digits. This way you will get another equation.
Step 4: Subtract the equation obtained in step 2 from the equation obtained in step 3.
Step 5: Simplify to get the answer.
For example, let's convert 1.888... to a rational number. Let x = 1.888... be equation 1. Multiplying equation (1) by 10, we get,
10x = 18.888... (equation 2)
Subtract equation (1) from equation (2),
9x = 17
x = 17/9
Therefore, 1.888... = 17/9.
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188739 | https://theoryofcomputing.org/articles/gs001/gs001.pdf | THEORY OF COMPUTING LIBRARY GRADUATE SURVEYS, TCGS 1 (2008), pp. 1–20 A Brief Introduction to Fourier Analysis on the Boolean Cube Ronald de Wolf∗ Received: June 30, 2007; revised: August 5, 2008; published: September 25, 2008.
Abstract: We give a brief introduction to the basic notions of Fourier analysis on the Boolean cube, illustrated and motivated by a number of applications to theoretical computer science.
ACM Classification: F.0, F.2 AMS Classification: 42-02, 68-02, 68Q17, 68Q25, 68Q32 Key words and phrases: Fourier analysis, Boolean functions, coding theory, learning theory, hyper-contractive inequality, influence of variables, polynomial 1 Introduction Fourier transforms are widely used in mathematics, computer science, and engineering. Examples in-clude signal processing, data compression, fast multiplication of polynomials, quantum computing, as well as many others. The Fourier transform most commonly used is the one over cyclic groups. Here one decomposes a signal or function as a sum of periodic functions such as χy(x) = e2πixy/n or (in the real case) sines and cosines.
In the study of functions of n Boolean variables, however, the most natural Fourier transform to consider is the one over the Abelian group Zn 2. This is known as Fourier analysis over the Boolean cube, ∗CWI Amsterdam. Partially supported by a Veni grant from the Netherlands Organization for Scientific Research (NWO), and by the European Commission under the Integrated Project Qubit Applications (QAP) funded by the IST directorate as Contract Number 015848.
Authors retain copyright to their work and grant Theory of Computing unlimited rights to publish the work electronically and in hard copy. Use of the work is permitted as long as the author(s) and the journal are properly acknowledged. For the detailed copyright statement, see
c ⃝2008 Ronald de Wolf DOI: 10.4086/toc.gs.2008.001 RONALD DE WOLF and has over the past two decades become one of the most important and versatile tools for theoretical computer scientists. The main purpose of this paper is to give a first introduction to the basic notions and properties of this area (Section 2) and to illustrate and motivate them by means of a few relatively simple but elegant applications from diverse areas (Section 3 and Section 4). The intended audience is theoretical computer scientists interested in adding these techniques to their toolbox. The selection of applications is somewhat biased by the author’s own experience in learning this material in recent years, and is by no means a complete survey—such a survey would probably require a whole book by now.
However, in Section 5 we give pointers for further reading.
2 Definitions and basic properties 2.1 The vector space of functions on the Boolean cube Consider the 2n-dimensional vector space of all functions f : {0,1}n →R. We define an inner product on this space by ⟨f,g⟩= 1 2n ∑ x∈{0,1}n f(x)g(x) = E[f ·g], where the latter expectation is taken uniformly over all x ∈{0,1}n. This defines the ℓ2-norm ||f||2 = p ⟨f, f⟩= q E[f 2].
2.2 The Fourier transform It will be convenient to identify a set S ⊆[n] = {1,...,n} with its characteristic vector S ∈{0,1}n. For example for S = {1,3} ⊆ we can also write S = (1,0,1) or S = 101. We will often go back and forth between these notations. For each S ⊆[n], define a function χS : {0,1}n →{±1} by χS(x) = (−1)S·x , where S·x = ∑n i=1 Sixi = ∑i∈S xi. It is easy to see that ⟨χS,χT⟩= δST = ( 1 if S = T 0 if S ̸= T , hence the set of all χS is an orthonormal basis (called the Fourier basis) for the space of all real-valued functions on {0,1}n. Of course, there are many different bases for this space. What makes the Fourier basis particularly useful for computer science is that the basis functions themselves have a simple com-putational interpretation, namely as parity functions: χS(x) = −1 if the number of S-variables having value 1 in the input x is odd, and χS(x) = 1 if that number is even.
For any f : {0,1}n →R we can define another function b f : {0,1}n →R by b f(S) = ⟨f,χS⟩= E[f · χS].
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 2 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE The linear map F : f 7→b f is called the Fourier transform. The function b f = F(f) is the Fourier transform of f, and b f(S) is the Fourier coefficient of f at S. This b f(S) may be viewed as a measure of the correlation between f and χS.
The set of Fourier coefficients is also called the Fourier spectrum of f. Since the χS form an orthonormal basis, the relation between f and b f is f = ∑ S b f(S)χS .
(2.1) Note that f(/ 0) = ∑S b f(S) is the sum of all Fourier coefficients, while b f(/ 0) = E[f] is the average function value. A function is constant if, and only if, b f(S) = 0 for all S ̸= / 0.
The degree of f is deg(f) = max{|S| | b f(S) ̸= 0}. In particular the degree of the basis function χS is |S|, the number of variables it depends on. Since (−1)xi = 1 −2xi, the Fourier expansion Eq. (2.1) represents f as an n-variate polynomial over the real numbers, of degree deg(f).
Let us consider some simple examples. If f = χS then b f(S) = 1 while all other Fourier coefficients are 0. If f(x) = ∑i∈S xi mod 2 is a parity function in the usual 0/1 notation, then b f(/ 0) = 1/2 and b f(S) = −1/2; all other coefficients are 0. The special case where S = {i} (i. e., f(x) = xi) is known as a dictator function, since its value is determined by only one variable. A k-junta is a function depending on at most k variables; equivalently, there is a set J ⊆[n] of size k such that b f(S) = 0 whenever S ̸⊆J. Finally, if we pick a function f : {0,1}n →{±1} uniformly at random, then each Fourier coefficient is normally distributed with mean 0 and variance 1/2n, so much coefficients will be about 1/ √ 2n in absolute value.
Because the χS form an orthonormal basis, we immediately get the following equality: ⟨f,g⟩= ∑ S,T b f(S)b g(T)⟨χS,χT⟩= ∑ S b f(S)b g(S).
In particular, with f = g we obtain Parseval’s Identity: ||f||2 2 = ∑ S b f(S)2 .
This also implies ||f −g||2 2 = ∑ S ( b f(S)−b g(S))2 .
As an example, suppose f is a probability distribution. Then we can analyze the ℓ2-distance between f and the uniform distribution g(x) = 1/2n as follows: ||f −g||2 2 = ∑ S ( b f(S)−b g(S))2 = ∑ S̸=/ 0 b f(S)2 , where we used b f(/ 0) = b g(/ 0) = 1/2n, and b g(S) = 0 whenever S ̸= / 0.
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 3 RONALD DE WOLF Notational variants.
The definition of F used here has become the standard one in computer science, though one occasionally sees different normalizations. If we defined the inner product with a 2−n/2 instead of 2−n, then F would be its own inverse and Parseval would simply state ||f||2 = || b f||2. However, with this modified inner product the χS functions no longer have norm 1.
Another variation is to view the variables as ±1-valued instead of 0/1-valued and to consider func-tions on {±1}n. In this case the function χS is simply the product of the S-variables, and the Fourier representation is simply an n-variate multilinear polynomial over the reals, with b f(S) as the coefficient of the monomial χS. Similarly, depending on what is more convenient, we can treat the value of a Boolean function as 0/1-valued or as ±1-valued. An advantage of the latter is that ∑S b f(S)2 = E[f 2] = 1 (by Parseval), which allows us to treat the squared Fourier coefficients as probabilities.
2.3 Convolution Given any two functions f,g : {0,1}n →R, we define their convolution f ∗g : {0,1}n →R by (f ∗g)(x) = 1 2n ∑ y∈{0,1}n f(x⊕y)g(y), where ‘⊕’ denotes entrywise addition of n-bit strings. If X andY are independent n-bit random variables, with probability distributions f and g, respectively, then 2n(f ∗g) is the distribution of the random variable Z = X ⊕Y: Pr[Z = z] = Pr[X = z⊕Y] = ∑ y∈{0,1}n f(z⊕y)g(y).
This arises naturally in certain computer-science settings, for instance when Y is some error-pattern corrupting X, or when Y is some “mask” used to hide X as in one-time pad encryption.
An important property is that the Fourier transform of the convolution f ∗g is the product of the Fourier transforms of f and g. This is easily verified by writing out the definitions: d f ∗g(S) = 1 2n ∑ x (f ∗g)(x)χS(x) = 1 22n ∑ x ∑ y f(x⊕y)g(y)χS(x) = 1 2n ∑ y g(y)χS(y) 1 2n ∑ x f(x⊕y)χS(x⊕y) = b f(S)· b g(S).
Suppose we have two functions h0 = f ∗g0 and h1 = f ∗g1, for instance from applying the same noise-process f to distributions g0 and g1. Using the convolution, we can rewrite their distance as ||h0 −h1||2 2 = ||f ∗(g0 −g1)||2 2 = ∑ S \ (f ∗(g0 −g1))(S)2 = ∑ S b f(S)2 ·( b g0(S)−b g1(S))2 .
This allows us to bound the difference between h0 and h1 by analyzing b f and b g0 −b g1.
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 4 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE 3 Some elementary applications 3.1 Approximating functions by parities It is often useful to approximate complex objects by much simpler ones. Consider any function f : {0,1}n →[−1,1]. Suppose there exists a function p : {0,1}n →[−1,1] that has some positive correla-tion ⟨f, p⟩≥ε with f, and suppose p is sparse in that the set C = {S | b p(S) ̸= 0} of nonzero Fourier coefficients has at most c elements. A typical example is where f is a Boolean function and p a low-degree real polynomial approximating f, since degree-d polynomials have at most c = ∑d i=0 n i nonzero Fourier coefficients. Now there exists a parity function that has nontrivial correlation with f, as follows: ε ≤⟨f, p⟩= ∑ S b p(S)⟨f,χS⟩≤||p||2 r ∑ S∈C ⟨f,χS⟩2 , where we used Cauchy-Schwarz and Parseval. This implies there exists an S such that either χS or its negation has correlation |⟨f,χS⟩| ≥ε/(||p||2 √c) ≥ε/√c with f.
3.2 List decoding the Hadamard code Error-correcting codes are important for storing and sending information in a way that protects against errors. Consider the Hadamard code: for a given S ∈{0,1}n, the codeword H(S) ∈{±1}2n is defined as the concatenation of χS(x) for all x, say ordered in lexicographic order. This code has a terrible rate: n-bit strings are blown up exponentially in size. On the other hand, it has excellent distance: any two codewords are at distance exactly 1 22n. This means that we can always uniquely decode the initial string S from a given word w ∈{±1}2n that differs from the codeword H(S) in less than 1/4 of the positions. This is easily phrased in terms of Fourier analysis. View w as a function w : {0,1}n →{±1}.
Note that ∑S b w(S)2 = E[w2] = 1 by Parseval. Then w has normalized distance d(w,H(S)) = 1/2 −ε from codeword H(S) if, and only if, b w(S) = 2ε. If the error-rate is less than 1/4 (ε > 1/4) then the original string is the unique S satisfying b w(S) > 1/2. (There cannot be two distinct S and S′ with Fourier coefficients larger than 1/2, since then by the triangle inequality we would have the contradiction 1/2 = d(H(S),H(S′)) ≤d(H(S),w)+d(w,H(S′)) < 1/2.) However, as soon as the error-rate is 1/4 or higher, unique decoding is no longer always possible.
For instance the word w that consists of 3 42n 1s followed by 1 42n −1s could either have come from H(0n) = 12n or from H(10n−1) = 12n−1(−1)2n−1. Surprisingly, we can still do something useful even if the error-rate is very close to 1/2, say 1/2 −ε for small but positive ε: we can output a small list of potential strings that contains the original string S. This is known as list decoding. It does not quite tell us what the original string was, but at least narrows it down to a small set of possibilities. The reason is that not too many codewords H(S) can simultaneously be at a distance ≤1/2 −ε from w: such S correspond to Fourier coefficients b w(S) ≥2ε, and not too many Fourier coefficients can be that large since their squares sum to 1. More formally: #{S : d(H(S),w) ≤1/2−ε} ≤1 4ε2 ∑ S:d(w,H(S))≤1/2−ε b w(S)2 ≤1 4ε2 ∑ S b w(S)2 = 1 4ε2 .
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 5 RONALD DE WOLF Note that this upper bound on the list size is independent of n. For instance, if we have error-rate 0.49 (so w is close to random), then still the list of potential codewords has at most 1 4(0.01)2 = 2500 elements.
In fact, Goldreich and Levin showed that we can efficiently, in time poly(n,1/ε), find this list, given oracle access to the 2n-bit string w. This was the first non-trivial list-decoding algorithm. Later work by Sudan, Guruswami, and others showed similar results for many codes with good rates, see for instance [66, 30] and the references therein.
3.3 Learning under the uniform distribution A lot of work in the last two decades has used the Fourier transform for learning under the uniform distribution. The idea is that we can learn an unknown function f : {0,1}n →R by approximating its Fourier coefficients. Since the Fourier coefficient b f(S) = E[f · χS] is just an expectation under the uniform distribution on {0,1}n, we can approximate it from uniformly drawn examples (x1, f(x1)),...,(xm, f(xm)). The empirical average 1 m m ∑ j=1 f(xj)· χS(xj) will converge to the right value b f(S) as m grows, and the Chernoff bound tells us how quickly this convergence happens.
Suppose we know f is dominated by a few large Fourier coefficients, and we know a not-too-large set containing those coefficients. A typical case is where f can be approximated by a real polynomial of low degree d. Then we can approximate those coefficients quickly with a small sample size m, and hence learn a good approximation of f. If αS is our estimate for b f(S) and h = ∑S αSχS the hypothesis that we output, then by Parseval our overall ℓ2-error is ||f −h||2 2 = ∑ S ( b f(S)−αS)2 .
When f has range {±1}, we can use sign(h) as our hypothesis for f. We have Pr[f(x) ̸= sign(h(x))] = 1 2n ∑ x:f(x)̸=sign(h(x)) 1 ≤1 2n ∑ x:f(x)̸=sign(h(x)) (f(x)−h(x))2 ≤||f −h||2 2 .
These ideas have been used for learning constant-depth circuits , for learning DNF [48, 32, 43, 13], juntas , decision trees [44, 57], and others.
4 The Bonami-Beckner inequality and some of its consequences 4.1 Bonami-Beckner and KKL Consider a function f : {0,1}n →R. Suppose its input x ∈{0,1}n is “noisy:” a new input y is obtained by flipping, independently, each bit of x with a fixed probability ε ∈[0,1]. The resulting noisy version THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 6 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE of f is e f(x) = Ey[f(y)].
Noise has a smoothing effect: sharp peaks in f will be “smeared out” over nearby inputs. Consider for instance a function that is non-zero only on input 0n: f(0n) = 1 and f(x) = 0 for all x ̸= 0n. Then e f is a fairly smooth probability distribution “around” 0n: e f(x) = ε|x|(1 −ε)n−|x|. If ε ∈(0,1/2) then the maximum of e f still occurs at 0n, but e f is much less sharply peaked than the original f.
Now consider the linear map that takes f to e f. Applying this map to the function f defined by f(x) = (−1)xi gives e f(x) = (1 −ε)(−1)xi + ε(−1)1−xi = (1 −2ε)(−1)xi. Similarly, applying it to χS gives (1−2ε)|S|χS. Hence our map is a function Tρ that merely shrinks the Fourier coefficient b f(S) by a factor ρ|S|, where ρ = 1−2ε ∈[−1,1]: e f = Tρ(f) = ∑ S ρ|S| b f(S)χS.
Here we can see the smoothing effect in action: Tρ attenuates the higher-degree Fourier coefficients, thus moving f closer to a constant function. For ρ < 1, the constant f are the only ones satisfying Tρ(f) = f.
Generalizing the ℓ2-norm, for every p ≥1 we define the p-norm of a function by ||f||p = 1 2n ∑ x |f(x)|p 1/p .
One can show this is monotone non-decreasing in p. Since Tρ(f) is an average of functions that all have the same p-norm as f, the triangle inequality immediately implies that Tρ is a contraction: for every p ≥1 we have ||Tρ(f)||p ≤||f||p. The Bonami-Beckner Hypercontractive Inequality [10, 29, 6] says that this inequality remains true even if we increase the left-hand side by going to a somewhat higher q-norm: Theorem 4.1 (Bonami-Beckner). If 1 ≤p ≤q and 0 ≤ρ ≤ p (p−1)/(q−1), then ||Tρ(f)||q ≤||f||p .
This inequality is a crucial tool in most of the more advanced applications of Fourier analysis on the Boolean cube. The case ρ = p (p−1)/(q−1) is the strongest case, and implies all others by monotonicity of the p-norm. Its proof is by induction on n. The base case (n = 1) is actually the harder part of the induction. We refer to Lecture 16 of for a proof as well as some background and history.
Chapter 5 of the book of Janson gives a more general treatment of hypercontractivity.
For us the most interesting cases are when either p or q equal 2, since Parseval allows us to rewrite the 2-norm in terms of Fourier coefficients. This leads to interesting statements about the relations between various norms of f. For instance, suppose the degree of f is at most d. Then for all p ∈[1,2], and using q = 2 and ρ = √p−1, we have (p−1)d||f||2 2 = (p−1)d∑ S b f(S)2 ≤∑ S (p−1)|S| b f(S)2 = ||T√p−1(f)||2 2 ≤||f||2 p .
Hence the p-norm of a low-degree function cannot be much smaller than its 2-norm. A similar argument gives ||f||2 q ≤(q−1)d||f||2 2 for all q ≥2.
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 7 RONALD DE WOLF In general, with q = 2, p ∈[1,2], and ρ = √p−1, Theorem 4.1 becomes ∑ S (p−1)|S| b f(S)2 = ||Tρ(f)||2 2 ≤||f||2 p = 1 2n ∑ x |f(x)|p 2/p .
(4.1) This gives an upper bound on the squared Fourier coefficients of f, in a way that gives most weight to the low-degree coefficients: each coefficient is “weighed down” by (p −1)|S|.1 An important special case of Eq. (4.1) is where f has range {−1,0,1}. This occurs for instance if f is a Boolean function or the difference of two Boolean functions. In that case we have ||f||p p = Pr[f ̸= 0] for any p. Applying Eq. (4.1) with p = 1+δ gives the following KKL Inequality, after Kahn, Kalai, and Linial .
Corollary 4.2 (KKL). For every δ ∈[0,1] and f : {0,1}n →{−1,0,1}, we have ∑ S δ |S| b f(S)2 ≤(Pr[f ̸= 0])2/(1+δ).
Informally, with δ < 1 the left-hand side is dominated by the Fourier coefficients of low degree (i. e., those where |S| is small). The right-hand side is smaller than the total “Fourier weight” ∑S b f(S)2 = Pr[f ̸= 0] by a power 2/(1 + δ) > 1. Hence the inequality says that a {−1,0,1}-valued function with small support cannot have too much of its Fourier weight on low degrees.
4.2 Random parities over a fixed set An application for which KKL seems to be almost tailor-made, is to bound the expected bias of k-bit parities over a set A ⊆{0,1}n. Suppose we pick a set S ⊆[n] of k indices uniformly at random, and consider the parity of the k-bit substring induced by S and a uniformly random x ∈A. Intuitively, if A is large then we expect that for most S, the bias βS of this parity to be small: the number of x ∈A with χS(x) = 1 should be roughly the same as with χS(x) = −1. This setting is relevant to cryptography.
Suppose we have an n-bit string x about which our adversary has some limited knowledge: he only knows that x is uniformly distributed over some fairly large set A. Then our adversary will be unable to predict most parities of selected bits from x, and we can use such parities to obtain bits that are essentially unknown to him. We ourselves do not need to know A for this; we only need to know a lower bound on its size, i. e., an upper bound on the adversary’s knowledge.
The intuition that large A leads to small biases is justified by the KKL Inequality.2 Note the connec-tion between biases and Fourier coefficients: with f the characteristic function of A, we have βS = Ex∈A[χS(x)] = 1 |A| ∑ x∈A χS(x) = 1 |A| ∑ x∈{0,1}n f(x)χS(x) = 2n |A| b f(S).
Applying the KKL Inequality, for any δ ∈[0,1] we can bound the sum of squared biases by ∑ S∈([n] k ) β 2 S = 22n |A|2 ∑ S∈([n] k ) b f(S)2 ≤ 22n δ k|A|2 |A| 2n 2/(1+δ) ≤1 δ k 2n |A| 2δ .
1Recently Eq. (4.1) was generalized from real-valued functions to matrix-valued functions .
2To the best of our knowledge, the bound below was first shown by Talagrand [68, Eq. (2.9)] for k = 2, using a large deviation inequality instead of hypercontractivity, and for general k in using the KKL Inequality. used it to prove lower bounds for communication complexity.
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 8 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE By differentiating, one can show that δ = k/[2ln(2n/|A|)] minimizes the right-hand side (assume k ≤ 2ln(2n/|A|) to ensure δ ∈[0,1]). This gives ES[β 2 S ] = 1 n k ∑ S∈([n] k ) β 2 S = O log(2n/|A|) n k .
(4.2) The following example shows this bound is essentially tight, and hence the KKL Inequality is tight as well. Let A = 0c × {0,1}n−c consist of all 2n−c strings starting with c zeroes. Then βS = 1 if S ⊆[c], and βS = 0 otherwise. The fraction of sets S ∈ [n] k satisfying S ⊆[c] is c k / n k = Ω(c/n)k. Hence ES[β 2 S ] = Ω(c/n)k = Ω(log(2n/|A|)/n)k, matching the upper bound of Eq. (4.2).
4.3 Influences of variables Suppose we have n players, and we have a function f : {0,1}n →{0,1} where player i controls the bit xi.
If f is balanced (meaning exactly half of the 2n inputs x yield f(x) = 1), then we can use it to implement a collective coin flipping scheme: let each player pick their bit xi randomly and use f(x) as the value of the coin flip. If all players indeed follow this protocol, the result is a fair coin flip. However, in order for this to be secure, small collusions of cheating players who can see the bits of the honest players should not be able to influence the function’s output value too much.
Formally, the influence of variable i ∈[n] on f is defined as Infi(f) = Pr[f(x) ̸= f(x⊕ei)], where the probability is over uniform x ∈{0,1}n, and x⊕ei is x with the ith bit flipped. This measures the probability (over random bits for all other players) that player i can determine the function value.
One can generalize this definition to the influence of a set S of players in the obvious way: InfS(f) is the probability that the function is non-constant when all variables outside of S are set randomly. Two extreme cases are the constant function (where Infi(f) = 0 for all i), and the parity function (where Infi(f) = 1 for all i). For the dictator function f(x) = xi, the ith variable has influence 1 while all other influences are 0. Another important example is the n-bit majority function, which is 1 if more than half of its input bits are 1, and which is 0 otherwise. Here each variable has influence Θ(1/√n), because the probability that the other n −1 bits are set such that xi determines the majority value is exactly n−1 ⌊n/2⌋ /2n−1. Moreover, any set S of, say, 10√n variables will with high probability be able to “tip the balance” and determine the majority value when the other n−10√n input bits are set randomly, hence InfS(f) ≈1.
Can we find balanced functions where the influences of the variables are much smaller, ideally O(1/n)? Ben-Or and Linial showed that the “tribes” function (an OR-AND tree with bottom fan-out about logn −loglogn) is a balanced function where every variable has influence Θ(log(n)/n). Kahn, Kalai, and Linial later showed that this is essentially optimal: for every Boolean function f that is balanced or close to balanced, at least one of its variables has influence Ω(log(n)/n). With some extra work (which we won’t detail here), this implies the existence of a set of only O(n/logn) variables that together determine the function value for almost all settings of the other variables. Hence we cannot hope to use such functions for collective coin flipping protocols that are secure against a constant fraction of THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 9 RONALD DE WOLF cheating players: a small set of O(n/logn) colluding players can already control the outcome with high probability.
The KKL result was one of the first major applications of Fourier analysis to Boolean functions.
We will prove a slightly stronger result here due to Talagrand . Let Var[f] = E[f 2] −E[f]2 be the variance of f. By Parseval, this equals ||f||2 2 −b f(/ 0)2 = ∑S̸=/ 0 b f(S)2. Consider a function f : {0,1}n → {0,1}. Assume that no variable has influence exactly 0 or 1 (influence-0 variables are irrelevant anyway, and influence-1 variables can be “factored” out of the function). Then we have n ∑ i=1 Infi(f) log(1/Infi(f)) = Ω(Var[f]).
(4.3) The KKL result follows immediately from this: if f is close to balanced then Var[f] = Ω(1). Hence there is an i such that Infi(f)/log(1/Infi(f)) = Ω(1/n), which implies Infi(f) = Ω(log(n)/n).
The proof of Eq. (4.3) is based on the following technical lemma, which uses Bonami-Beckner.
Lemma 4.3. If g : {0,1}n →R satisfies ||g||3/2 ̸= ||g||2, then ∑ S̸=/ 0 b g(S)2 |S| ≤ 2.5||g||2 2 log(||g||2/||g||3/2) .
Proof. Using Theorem 4.1 with q = 2 and p = 1+ρ2 = 3/2 we get for every integer k ∑ S:|S|=k b g(S)2 ≤2k∑ S 2−|S|b g(S)2 = 2k||T√ 1/2(g)||2 2 ≤2k||g||2 3/2 .
For every integer m we have (using ∑m k=1 2k/k ≤4·2m/(m+1), which is easily proved by induction): ∑ S̸=/ 0 b g(S)2 |S| = m ∑ k=1 ∑ S:|S|=k b g(S)2 k + ∑ S:|S|>m b g(S)2 |S| ≤ m ∑ k=1 2k||g||2 3/2 k + ∑ S:|S|>m b g(S)2 m+1 ≤ 4·2m||g||2 3/2 +||g||2 2 m+1 Choose m the largest integer satisfying 2m||g||2 3/2 ≤||g||2 2. Then m+1 > 2log(||g||2/||g||3/2), and ∑ S̸=/ 0 b g(S)2 |S| ≤5||g||2 2 m+1 ≤ 2.5||g||2 2 log(||g||2/||g||3/2) .
Now consider a variable i and define g(x) = f(x)−f(x⊕ei). Then g(x) = 0 if f(x) = f(x⊕ei), and g(x) ∈{±1} otherwise. Hence ||g||2 2 = ||g||3/2 3/2 = Infi(f) ∈(0,1), and ||g||2/||g||3/2 = Infi(f)−1/6. The Fourier coefficients of g are closely related to those of f: b g(S) = ( 2 b f(S) if i ∈S, 0 otherwise.
(4.4) THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 10 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE Applying Lemma 4.3 to g gives ∑ S:i∈S 4 b f(S)2 |S| = ∑ S b g(S)2 |S| ≤ 2.5||g||2 2 log(||g||2/||g||3/2) = 15Infi(f) log(1/Infi(f)) .
Summing over all i gives Talagrand’s result: 4Var[f] = 4 ∑ S̸=0 b f(S)2 = n ∑ i=1 ∑ S:i∈S 4 b f(S)2 |S| ≤ n ∑ i=1 15Infi(f) log(1/Infi(f)) .
Subsequent work.
The subsequent “BKKKL” paper generalized the KKL result to the case of functions f : [0,1]n →{0,1} with real-valued input, with uniform measure on each real-valued variable xi. See Friedgut for some simplifications and corrections of this. A recent related result is that any near-balanced Boolean function with a depth-d decision tree has a variable with influence Ω(1/d) .
4.4 The relation between influences, sensitivity, and degree In this section we relate influences to degree and sensitivity, both of which are important measures of the complexity of Boolean functions.3 The sensitivity of a function f : {0,1}n →{0,1} on input x is sx(f) = |{i | f(x) ̸= f(x⊕ei)}| and the average sensitivity of f is s(f) = 1 2n ∑ x∈{0,1}n sx(f).
By linearity of expectation, average sensitivity equals the total influence: s(f) = n ∑ i=1 Infi(f).
(4.5) Here we give a relation between degree and sensitivity due to Shi : average sensitivity lower bounds the degree of approximating polynomials. Suppose a degree-d n-variate real polynomial p : {0,1}n → [0,1] approximates f : {0,1}n →{0,1}, in the sense that there is an ε ∈[0,1/2) such that |f(x) − p(x)| ≤ε for every x ∈{0,1}n. Let q be the degree-d polynomial 1−2p. This has range [−1,1], hence ∑S b q(S)2 = ||q||2 2 ≤1. Note that q(x) ∈[−1,−1 + 2ε] if f(x) = 1, and q(x) ∈[1 −2ε,1] if f(x) = 0.
Consider the function q(i)(x) = q(x)−q(x⊕ei). Using Parseval and the analogue of Eq. (4.4), we have (2−4ε)2Infi(f) ≤E[(q(i))2] = ∑ S c q(i)(S)2 = 4 ∑ S:i∈S b q(S)2.
Dividing by 4 and summing over all i gives the lower bound on the degree: (1−2ε)2s(f) = (1−2ε)2 n ∑ i=1 Infi(f) ≤ n ∑ i=1 ∑ S:i∈S b q(S)2 = ∑ S |S|b q(S)2 ≤d∑ S b q(S)2 ≤d .
3See the survey . Other Fourier-based results on polynomial degrees are in [56, 62, 14]. This section does not need the Bonami-Beckner or KKL Inequalities, but is placed after Section 4.3 because it considers the influences defined there.
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 11 RONALD DE WOLF A random Boolean function has s(f) ≈n/2 and ε-approximate degree at least n/2 −O(√n) (for fixed ε), hence this bound is optimal up to constant factors for almost all n-bit functions.
Another Fourier-based lower bound on the degree is due to Nisan and Szegedy . Suppose f : {0,1}n →{0,1} depends on all of its n variables (equivalently, each variable has positive influence).
Consider f (i)(x) = f(x) −f(x ⊕ei). This is an n-variate polynomial of degree d ≤deg(f), and it is non-constant because f depends on xi. It is well known that such a polynomial is non-zero on at least a 2−d-fraction of its inputs.4 Hence Infi(f) = Pr[f (i) ̸= 0] ≥2−deg(f) for each i. Summing over i we get n 2deg(f) ≤ n ∑ i=1 Infi(f) = n ∑ i=1∑ S c f (i)(S)2 = 4∑ S |S| b f(S)2 ≤4 deg(f)∑ S b f(S)2 ≤4 deg(f).
This implies the bound from : deg(f) ≥log(n)−O(loglogn) for every Boolean function f that depends on n variables. As Nisan and Szegedy observed, the “ad-dress function” shows that this bound is tight up to the O(loglogn) term. This function takes an input x1 ...xmy0 ...y2m−1 of n = m+2m bits, and outputs yi where i is the number whose binary representation is x1 ...xm. The function depends on all n variables. It is represented by ∑ i∈{0,1}m yi ∏ j:ij=1 xj ∏ j:ij=0 (1−xj) and hence has degree m+1 ≤logn+1.
5 A guide to literature The examples given above illustrate the usefulness of Fourier analysis on the Boolean cube, but they barely scratch the surface of the rich set of actual and potential applications. In this last section we give pointers to the other main areas of application in computer science.
PCPs and hardness of approximation.
Possibly the most important application of Fourier analysis in theoretical computer science is its use in designing and analyzing Probabilistically Checkable Proofs.
These are encodings of witnesses for NP-like problems that can be verified probabilistically while query-ing only a small number of their bits. The famous PCP Theorem [4, 17] says that a language is in NP if, and only if, it has witnesses that can be verified probabilistically using only O(logn) bits of randomness and a constant number of queries to bits of the witness. Some of the most efficient PCPs are based on Fourier analysis, H˚ astad’s 3-query PCP being a prime example . Based on these PCPs, one can 4This is a basic and well known property of Reed-Muller error-correcting codes [47, p. 375]. In the computer science literature this fact is usually called the Schwartz-Zippel Lemma [61, 69].
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 12 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE show NP-hardness results for approximations to many optimization problems, such as finding a maxi-mal clique in a graph or the maximal number of satisfied clauses in a given CNF formula. Chapters 11 and 22 of the book by Arora and Barak give a good introduction to this material.
A recent development is the use of the “Unique Games Conjecture.” This conjecture, due to Khot , says that it is hard to approximate the maximal number of satisfied constraints in problems with only two variables per constraint, where the value of either variable in a constraint determines the value of the other (the variables are over a sufficiently large but constant domain). Assuming this, one can prove essentially optimal inapproximability results for problems like max-cut , vertex cover , and others [16, 41, 19, 39, 59], which so far resisted the more standard inapproximability approach via PCPs. Again, Fourier techniques are often essential in the analysis. For instance, one of the main planks of the max-cut result is the “Majority Is Stablest” Theorem. This was conjectured in a first version of and proved in . It says that among all balanced Boolean functions where every variable has low influence, the majority function has the maximal correlation between f and its noisy version ˜ f = Tρ(f) (as defined in Section 4.1).
Threshold phenomena.
A threshold phenomenon occurs if certain properties of a system change sharply in response to a small change of an underlying parameter which is close to a specific value (the “threshold”). A typical example in nature is water, which is frozen if the temperature-parameter is just below 0◦C and liquid if the temperature is just above 0◦C. In mathematics, such phenomena occur for instance in random graphs. Let G(n, p) denote an undirected graph on n vertices, where each edge is included with probability p (independent of the other edges). Erd˝ os and R´ enyi introduced this model and showed a sharp threshold for connectivity: if p is slightly below (logn)/n then the probability of G(n, p) being connected tends to 0 with n, while if p is slightly larger than (logn)/n then this probability tends to 1. Friedgut and Kalai later showed that every monotone graph property has a sharp threshold.
Threshold phenomena occur also in complexity theory. For instance, if one picks a random 3-SAT formula with n variables and m = cn clauses, then for c < 4.2 (roughly5), the formula is most probably satisfiable, while for c > 4.2 the formula is most probably unsatisfiable. Similarly, one can view tight results on hardness of approximation as threshold phenomena: for many NP-hard optimization problems, there exists a constant c such that approximating the optimal value up to factor c can be done in polynomial time, while approximating it to within a slightly better factor is NP-hard (or Unique-Games hard). Kalai and Safra give a very interesting survey of these phenomena, showing how influences and Fourier techniques are central to their analysis. Often these techniques apply to the generalized setting where each input bit is 1 with probability p.
Social choice theory.
If n people have to decide between two alternatives then they can use majority voting, which has all the properties one expects of a reasonable voting scheme. However, as soon as they have to choose between three or more alternatives, Arrow’s Theorem says that no “ideal” voting scheme exists. Strong quantitative versions of this theorem can be obtained quite easily using 5Actually this value of 4.2 is a numerical estimate; proven upper and lower bounds on this number are far from tight. We refer to Part 3 of for details.
THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 13 RONALD DE WOLF Fourier techniques. The surveys by Kalai and O’Donnell are excellent starting points for such connections between Fourier analysis and social choice theory. A very recent related result is .
When are functions close to juntas?
Recall that a k-junta is a function on {0,1}n that depends on at most k of its variables. Friedgut showed that if the average sensitivity (a. k. a. total influence) of a Boolean function f is I then f is close to another Boolean function that is a 2O(I)-junta. The “address function” from Section 4.4 shows the exponential is necessary: it has average sensitivity O(logn) but cannot be approximated well by a Boolean function depending on only o(n) variables. Friedgut et al. show that if the Fourier coefficients of degrees 0 and 1 have most of the total weight, then f is close to a 1-junta (i. e., a dictator or its negation). Bourgain proved that the weight on higher-degree Fourier coefficients of a balanced Boolean function cannot decay too fast unless f is close to a junta, and Dinur et al. analyzed the same phenomenon for bounded—but not necessarily Boolean—functions on the Boolean cube.
Other applications.
Some results in the area of property testing are based on Fourier analysis. Ex-amples are the algorithm by Fischer et al. for testing if a Boolean function is close to or far from a k-junta, and the one by Alon et al. for testing if a distribution is close to or far from (almost) k-wise independence. The above-mentioned work on PCPs also falls in this category, since one is basically testing whether a given witness is close to a valid proof or not. In addition to the list-decoding example from Section 3.2, there have been a number of other applications of Fourier analysis in coding theory, see for instance Section 4.3 of Linial’s course notes (mentioned below) and Navon and Samorodnit-sky . Fourier analysis has also been used for lower bounds on various kinds of communication complexity [60, 42, 27, 63], and for analysis of low-distortion embeddings of one metric space into another [45, 41].
Other expository papers and courses.
Several more extensive surveys on Fourier analysis of Boolean functions exist in the literature. The early one by Bernasconi, Codenotti, and Simon describes the main applications up to 1997. ˇ Stefankoviˇ c’s MSc thesis is geared towards general applications of Fourier analysis in computer science, often over groups other than the Boolean cube. The survey by Kalai and Safra concentrates on threshold phenomena. The very recent survey by O’Donnell focuses on topics related to voting and hardness of approximation, and also tries to demystify the Bonami-Beckner Inequality by presenting it as a generalization of the Hoeffding-Chernoff bounds to higher-degree functions. Finally, let us point to the notes of a number of recent courses, which contain a wealth of additional material: Irit Dinur and Ehud Friedgut: Subhash Khot: Guy Kindler: Nati Linial: Elchanan Mossel: Ryan O’Donnell: Oded Regev: fall 2007 THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 14 A BRIEF INTRODUCTION TO FOURIER ANALYSIS ON THE BOOLEAN CUBE Acknowledgments Thanks to David Garc´ ıa Soriano, Elena Grigorescu, Peter Harremo¨ es, Oded Regev, Nitin Saxena, Ben Toner, and Stephanie Wehner for useful comments and pointers. Also thanks to the anonymous ToC referees for many extremely helpful recommendations.
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THEORY OF COMPUTING LIBRARY, GRADUATE SURVEYS TCGS 1 (2008), pp. 1–20 20 |
188740 | https://brainly.com/question/36482651 | [FREE] A proton is released such that its initial velocity is from right to left across this page. The proton's - brainly.com
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A proton is released such that its initial velocity is from right to left across this page. The proton's path, however, is deflected in a direction toward the bottom edge of the page due to the presence of a uniform magnetic field. What is the direction of this field?
A) out of the page
B) into the page
C) from bottom edge to top edge of the page
D) from right to left across the page
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The magnetic field is directed into the page, consistent with the right-hand rule and causing the proton, which is initially moving from right to left, to be deflected downward.Therefore, the correct answer to the question is (B) into the page.
The direction of the magnetic field can be determined using the right-hand rule. Considering the proton is positively charged and is initially moving from right to left, and that it is deflected toward the bottom edge of the page, we can conclude that for the force to be directed downward, the magnetic field must be going into the page. This is because the magnetic force acts perpendicular to both the velocity of the proton and the magnetic field lines. Therefore, the correct answer to the question is (B) into the page.
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The direction of the magnetic field must be into the page in order to cause a proton moving from right to left to be deflected downward, according to the right-hand rule. The right-hand rule helps determine the relationship between the direction of the velocity, magnetic field, and the magnetic force. Thus, the correct answer is (B) into the page.
Explanation
In this scenario, we have a proton moving from right to left across the page. When a charged particle like a proton moves through a magnetic field, it experiences a magnetic force, which can be described using the right-hand rule.
The right-hand rule states that you should point your thumb in the direction of the particle's velocity (which is to the left in this case), and curl your fingers in the direction of the magnetic field. The direction your palm faces shows the direction of the force acting on the proton. Since the proton is deflected downward, we want the force to be pointing down.
Initial velocity: Right to left.
Deflected direction: Downward toward the bottom edge of the page.
Applying the right-hand rule:
Point your thumb to the left (velocity).
To get the palm to face down (force), your fingers must curl into the page.
From these observations, it becomes clear that the magnetic field must be directed into the page. Therefore, the correct answer to the question is (B) into the page.
Examples & Evidence
For further clarification, consider how this applies to other charged particles: if an electron moves in a magnetic field directed out of the page, it would be deflected in a direction opposite that of a positive proton based on their charges.
The right-hand rule is a fundamental principle in electromagnetism and applies to charged particles moving in magnetic fields. It confirms the relationship between the velocity, magnetic field direction, and the resulting force on charged particles.
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Option b is the correct answer. The direction of the uniform magnetic field that causes a proton to deflect downward as it moves from right to left is into the page, according to the right-hand rule.
The direction of the uniform magnetic field causing the deflection of a proton moving from right to left can be determined by using the right-hand rule. Since the proton is positively charged and is deflected downward, and knowing that the magnetic force is always perpendicular to both the velocity and the magnetic field, we can conclude the correct direction of the magnetic field is option B) into the page.
Remember, for a positively charged particle like a proton, point your thumb in the direction of the velocity (to the left), and your fingers in the direction of the magnetic force (downward). Your palm faces the direction of the magnetic field, which in this case would be into the page.
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Community Answer A proton is released such that its initical velocity is from right to left across this page. THe proton's path, however, is deflected in a direction toward the bottom edge of the page due to the presence of a uniform magnetic field. What is the direction of the magnetic field
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Proof of triangle inequality
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I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that |a+b|≤|a|+|b||a+b|≤|a|+|b|. Any help would be appreciated :)
inequality
absolute-value
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edited May 5, 2014 at 10:34
Martin Sleziak
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Isn't this an axiom in metric space?NECing –NECing 2013-02-18 19:11:44 +00:00 Commented Feb 18, 2013 at 19:11
18 There is no addition in metric space. @ShuXiaoLi k.stm –k.stm 2013-02-18 19:16:27 +00:00 Commented Feb 18, 2013 at 19:16
1 That a metric must obey the triangle inequality is indeed one of the axioms of a metric space.user1236 –user1236 2015-07-28 01:04:19 +00:00 Commented Jul 28, 2015 at 1:04
1 Consider the possibilities for a and b: each can be negative, zero, or positive. Thus there are at most nine possibilities to check out separately. You can do it! Be brave!richard1941 –richard1941 2018-01-24 01:18:31 +00:00 Commented Jan 24, 2018 at 1:18
2 The question is not well-defined until you say what a a and b b are: real numbers complex numbers, vectors or something else again.PJTraill –PJTraill 2018-10-10 20:44:13 +00:00 Commented Oct 10, 2018 at 20:44
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From your definition of the absolute value, establish first |x|=max{x,−x}|x|=max{x,−x} and ±x≤|x|±x≤|x|.
Then you can use
a+b−a−b≤|a|+b≤|a|+|b|,and≤|a|−b≤|a|+|b|.a+b≤|a|+b≤|a|+|b|,and−a−b≤|a|−b≤|a|+|b|.
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edited Jun 29, 2020 at 16:17
answered Feb 18, 2013 at 19:12
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7 Clear and concise, +1.Julien –Julien 2013-02-18 19:18:52 +00:00 Commented Feb 18, 2013 at 19:18
10 Nice. Thanks! I got hung up for a sec on this step:−a−b≤|a|−b−a−b≤|a|−b but then I put in the intermediate step:−a−b≤|−a|−b=|a|−b−a−b≤|−a|−b=|a|−b Thank you!ivan –ivan 2013-02-18 19:28:14 +00:00 Commented Feb 18, 2013 at 19:28
8 Do you need that step though? Because |x|=m a x{x,−x}|x|=m a x{x,−x}, which is trivially greater than or equal to −x−x.sodiumnitrate –sodiumnitrate 2015-03-08 03:32:33 +00:00 Commented Mar 8, 2015 at 3:32
When I use sodiumnitrate's way to prove the two lines, then I don't even need to use the second fact. Because after that we know that |a+b|=max(a+b,−(a+b))|a+b|=max(a+b,−(a+b)) , so no matter which one of {a+b,−(a+b)}{a+b,−(a+b)} |a+b||a+b| is equal to we have that |a+b|≤|a|+|b||a+b|≤|a|+|b| . So is the second fact even necessary?cuppajoeman –cuppajoeman 2021-09-15 02:09:52 +00:00 Commented Sep 15, 2021 at 2:09
@cuppajoeman sodiumnitrate “proves” the fact. It’s just a triviality, but having it explicitly stated makes the second inequality line obvious without the need of spending a second thought on it.k.stm –k.stm 2021-09-18 16:52:19 +00:00 Commented Sep 18, 2021 at 16:52
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|a|2+|b|2+2|a||b|≥a 2+b 2+2 a b|a|2+|b|2+2|a||b|≥a 2+b 2+2 a b
(|a|+|b|)2≥(a+b)2 a(∵∀x∈R;;x 2=|x|2)(|a|+|b|)2≥(a+b)2 a(∵∀x∈R;;x 2=|x|2)
||a|+|b||≥|a+b|||a|+|b||≥|a+b|
∴|a|+|b|≥|a+b|∴|a|+|b|≥|a+b|
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edited May 12, 2023 at 20:54
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answered Feb 18, 2013 at 19:55
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A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).
Prove the triangle inequality |x|+|y|≥|x+y||x|+|y|≥|x+y|.
Without loss of generality, we need only consider the following cases:
x=0 x=0
x>0,y>0 x>0,y>0
x>0,y<0 x>0,y<0
Case 1 1. Suppose x=0 x=0. Then we have
|x|=0|x|=0
|x|+|y|=0+|y|=|y||x|+|y|=0+|y|=|y|
Thus |x|+|y|=|x+y||x|+|y|=|x+y|.
Case 2 2. Suppose x>0,y>0 x>0,y>0. Then, since x+y>0 x+y>0, we have
|x|=x|x|=x
|y|=y|y|=y
|x|+|y|=x+y|x|+|y|=x+y
|x+y|=x+y|x+y|=x+y
Thus |x|+|y|=|x+y||x|+|y|=|x+y|.
Case 3 3. Suppose x<0,y<0 x<0,y<0. Then, since x+y<0 x+y<0, we have
|x|=−x|x|=−x
|y|=−y|y|=−y
|x|+|y|=(−x)+(−y)|x|+|y|=(−x)+(−y)
|x+y|=−(x+y)=(−x)+(−y)|x+y|=−(x+y)=(−x)+(−y)
Thus |x|+|y|=|x+y||x|+|y|=|x+y|.
Case 4 4. Suppose x>0,y<0 x>0,y<0. Then we have
|x|=x|x|=x
|y|=−y|y|=−y
|x|+|y|=x+(−y)|x|+|y|=x+(−y)
We must now consider three cases:
a. x+y=0 x+y=0
b. x+y>0 x+y>0
c. x+y<0 x+y<0
Case 4 a 4 a. Suppose x+y=0 x+y=0. Then we have
|x+y|=|0|=0|x+y|=|0|=0
Since y<0 y<0, it follows that −y>0−y>0 and thus x+(−y)>0+(−y)=−y>0 x+(−y)>0+(−y)=−y>0.
Therefore, since |x|+|y|=x+(−y)|x|+|y|=x+(−y), we must have |x|+|y|>|x+y||x|+|y|>|x+y|.
Case 4 b 4 b. Suppose x+y>0 x+y>0. Then we have
|x+y|=x+y|x+y|=x+y
Since y<0 y<0, it follows that −y>0>y−y>0>y and thus x+(−y)>x+y x+(−y)>x+y.
Therefore, since |x|+|y|=x+(−y)|x|+|y|=x+(−y), we must have |x|+|y|>|x+y||x|+|y|>|x+y|.
Case 4 c 4 c. Suppose x+y<0 x+y<0. Then we have
|x+y|=−(x+y)=(−x)+(−y)|x+y|=−(x+y)=(−x)+(−y)
Since x>0 x>0, it follows that x>0>−x x>0>−x and thus x+(−y)>(−x)+(−y)x+(−y)>(−x)+(−y).
Therefore, since |x|+|y|=x+(−y)|x|+|y|=x+(−y), we must have |x|+|y|>|x+y||x|+|y|>|x+y|.
This concludes the proof.
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edited Sep 3, 2015 at 0:13
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Firstly, observe that −|x|≤x≤|x|−|x|≤x≤|x| , and −|y|≤y≤|y|−|y|≤y≤|y| follow from the definition of the absolute value function. (Consider the cases x x is non-negative and x x is negative and what happens to |x||x|, the same goes for y y mutatis mutandis).
Since y≤|y|y≤|y| , then
|x|+y≤|x|+|y|(1)(1)|x|+y≤|x|+|y|
by adding |x||x| to both sides of y≤|y|y≤|y| . Likewise, adding y y to both sides of x≤|x|x≤|x| we have:
y+x≤y+|x|(2)(2)y+x≤y+|x|
Combining equations 1 1 and 2 2, and using the transitive property of the relation ≤≤, we have:
y+x≤y+|x|≤|y|+|x|y+x≤y+|x|≤|y|+|x|
y+x≤|y|+|x|y+x≤|y|+|x|
Likewise, since −|y|≤y−|y|≤y , then −|y|−|x|≤y−|x|−|y|−|x|≤y−|x| by adding −|x|−|x| to both sides. But −|x|≤x⟹y−|x|≤y+x−|x|≤x⟹y−|x|≤y+x by adding y y to both sides. Thus by the transitive property:
−|y|−|x|≤y−|x|,and y−|x|≤y+x⟹−|y|−|x|≤y+x(3)(3)−|y|−|x|≤y−|x|,and y−|x|≤y+x⟹−|y|−|x|≤y+x
By combining equations 2 and 3, we have:
−(|y|+|x|)≤y+x≤|y|+|x|−(|y|+|x|)≤y+x≤|y|+|x|
Now noting that |b|≤a⟺−a≤b≤a|b|≤a⟺−a≤b≤a , we have:
|x+y|≤|x|+|y||x+y|≤|x|+|y|
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edited Aug 21, 2017 at 15:38
answered Oct 24, 2016 at 19:06
Evan RosicaEvan Rosica
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1 Great answer! Thank you for showing all the steps.Mr X –Mr X 2022-07-29 23:17:29 +00:00 Commented Jul 29, 2022 at 23:17
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If a neat algebraic argument does not suggest itself, we can do a crude argument by cases, guided by the examples a=7,b=4 a=7,b=4, a=−7,b=−4 a=−7,b=−4, a=7,b=−4 a=7,b=−4, and a=−7,b=4 a=−7,b=4.
If a≥0 a≥0 and b≥0 b≥0 then |a+b|=|a|+|b||a+b|=|a|+|b|.
If a≤0 a≤0, and b≤0 b≤0, then |a+b|=−(a+b)=(−a)+(−b)=|a|+|b||a+b|=−(a+b)=(−a)+(−b)=|a|+|b|.
Now we need to examine the cases where a a is positive and b b is negative, or the other way around. Without loss of generality we may assume that |b|≤|a||b|≤|a|.
If a>0 a>0, then |a+b|=|a|−|b||a+b|=|a|−|b|. This is <|a|<|a|, and in particular <|a|+|b|<|a|+|b|.
If a<0 a<0, then again |a+b|=|a|−|b||a+b|=|a|−|b|.
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edited Feb 19, 2013 at 3:03
answered Feb 18, 2013 at 20:38
André NicolasAndré Nicolas
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The proof given in Wikipedia / Absolute Value is interesting and the technique can be used for complex numbers:
Choose ϵ ϵ from {−1,1}{−1,1} so that ϵ(a+b)≥0 ϵ(a+b)≥0. Clearly ϵ x≤|x|ϵ x≤|x| for all real x x regardless of the value of ϵ ϵ, so
|a+b|=ϵ(a+b)=ϵ a+ϵ b≤|a|+|b||a+b|=ϵ(a+b)=ϵ a+ϵ b≤|a|+|b|
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answered May 14, 2017 at 13:40
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These kind of questions require clarification about the assumptions one begins with. This answer covers the whole package often referred to as "the triangle inequality". I make only the following assumptions, which (with the set R R of real numbers in place of set X X) are part of any axiomatization of the real number system:
There is a commutative group operation (a,b)↦a+b(a,b)↦a+b with neutral element 0 0 in the set X X.
There is a linear order << in the set X X, compatible with ++ in the sense that if a<b a<b, then a+c<b+c a+c<b+c.
The absolute value |x|∈X|x|∈X of x∈X x∈X is −x−x in the case of x<0 x<0, and x x in the cases of x=0 x=0 and 0<x 0<x.
Note that if x≠0 x≠0, then either x<0 x<0, whence 0<−x=|x|0<−x=|x|, or 0<x 0<x, so 0<|x|0<|x| in this case as well. If |x|=0|x|=0, then neither x<0 x<0 nor 0<x 0<x, since each of these inequalities implies 0<|x|0<|x|. These observations prove the following basic result, which is used below, and in which a≤b a≤b stands for "a<b a<b or a=b a=b''.
(Positive definiteness.)For each x∈X x∈X the inequality 0≤|x|0≤|x| holds, and 0=|x|0=|x| if and only if x=0 x=0.
Remark. The formal claim x≤y x≤y is also called "inequality". By convention the name "strict inequality" is reserved for the formal claim x<y x<y to underline that it actually claims that the equality x=y x=y does not hold.
The proof of the basic fact that |−x|=|x||−x|=|x| holds for every x∈X x∈X, which I also use below, is immediate:
If x<0 x<0, then 0<−x 0<−x, so |−x|=−x=|x||−x|=−x=|x|.
If 0<x 0<x, then −x<0−x<0, so |−x|=−(−x)=x=|x||−x|=−(−x)=x=|x|.
If x=0 x=0, then −x=x−x=x, so |−x|=|x||−x|=|x|.
The triangle inequality regarding addition ++ often refers to the combination of the following two non-strict inequalities together with the statements below them, that clarify when these inequalities are strict:
∣∣|a|−|b|∣∣≤|a−b|≤|a|+|b|||a|−|b||≤|a−b|≤|a|+|b|
The left inequality is strict if and only if either a<0<b a<0<b or b<0<a b<0<a.
The right inequality is strict if and only if either both a<0 a<0 and b<0 b<0, or both 0<a 0<a and 0<b 0<b.
This more precise formulation of "the triangle inequality" has the following elementary proof by cases:
Case of 0<a 0<a and 0<b 0<b. In this case clearly ∣∣|a|−|b|∣∣=|a−b|||a|−|b||=|a−b|. Furthermore,
If a<b a<b, then a−b<0 a−b<0, so |a−b|=−(a−b)=−a+b=−|a|+|b|<0+|b|<|a|+|b||a−b|=−(a−b)=−a+b=−|a|+|b|<0+|b|<|a|+|b|.
If b<a b<a, then 0<a−b 0<a−b, so |a−b|=a−b=|a|−|b|<|a|+0<|a|+|b||a−b|=a−b=|a|−|b|<|a|+0<|a|+|b|.
If a=b a=b, then a−b=0 a−b=0, so |a−b|=0+0<|a|+|b||a−b|=0+0<|a|+|b|.
Case of a<0 a<0 and b<0 b<0. Apply case 1 for −a−a and −b−b in place of a a and b b, and use |−x|=|x||−x|=|x|.
Case of 0<a 0<a and b<0 b<0. Now 0<a−b 0<a−b, so |a−b|=a−b=a+(−b)=|a|+|b||a−b|=a−b=a+(−b)=|a|+|b|.
If a<−b a<−b, then |a|<|b||a|<|b|, so |a|−|b|<0|a|−|b|<0, and therefore ∣∣|a|−|b|∣∣=−(|a|−|b|)=−(a−(−b))=−a−b<0−b<a−b=|a−b|.||a|−|b||=−(|a|−|b|)=−(a−(−b))=−a−b<0−b<a−b=|a−b|.
If −b<a−b<a, then |b|<|a||b|<|a|, so 0<|a|−|b|0<|a|−|b|, and therefore ∣∣|a|−|b|∣∣=|a|−|b|=a−(−b)=a+b<a+0<a−b=|a−b|.||a|−|b||=|a|−|b|=a−(−b)=a+b<a+0<a−b=|a−b|.
Case of a<0 a<0 and 0<b 0<b. Apply case 3 for −a−a and −b−b in place of a a and b b, and use |−x|=|x||−x|=|x|.
Case of a=0 a=0. In this case ∣∣|a|−|b|∣∣=∣∣0−|b|∣∣=∣∣|b|∣∣=|b|||a|−|b||=|0−|b||=||b||=|b|, and since |b|=|−b|=|0−b|=|a−b|and|b|=0+|b|=|0|+|b|=|a|+|b|,|b|=|−b|=|0−b|=|a−b|and|b|=0+|b|=|0|+|b|=|a|+|b|, we have ∣∣|a|−|b|∣∣=|a−b|=|a|+|b|||a|−|b||=|a−b|=|a|+|b|.
Case of b=0 b=0. Apply 5, switching a a with b b, and use |−x|=|x||−x|=|x|.
These cases exhaust all the possibilities and are mutually exclusive.
Remark. Instead of the rather cumbersome wording above, the phrases "a a and b b are of the same sign" and "a a and b b are of different signs" are commonly used. Then "the sign of x x" is ++ in case of 0<x 0<x, and −− in case of x<0 x<0, while the neutral element 0∈X 0∈X is considered as having no sign at all.
I cannot see any significant short-cut in proving the claim above, provided perhaps by some algebra-trick. (There is not much algebra in a linearly ordered commutative group.) We are so accustomed to using triangle inequality while "computing stuff" by applying various algebraic identities, that it is easy to forget that there is logic involved in the triangle inequality. In logic one sometimes just needs to comb through all the cases.
The significance of the triangle inequality is not in some deep insight its proof requires, but rather in its usefulness and in its elegant formulation compared to the tedious case-by-case proof it seems to require.
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answered Jan 4, 2023 at 22:43
LapasotkaLapasotka
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I hope that the following proof is the shortest one and make use of the order in R R.
If a b≥0 a b≥0 then
|a+b|={a−b−a−b i f a≥0,b≥0,i f a≤0,b≤0=|a|+|b||a+b|={a−b i f a≥0,b≥0,−a−b i f a≤0,b≤0=|a|+|b|
If a b<0 a b<0 then
|a+b|≤max{|a|,|b|}≤|a|+|b|.|a+b|≤max{|a|,|b|}≤|a|+|b|.
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answered Sep 10, 2018 at 0:39
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This proof works alongside the geometric notion that adding numbers on the real line is a 'vector operation'. It also lays out the exact conditions under which the triangle inequality is an equation,
|x+y|=|x|+|y||x+y|=|x|+|y|.
Recall that in general,
a≤b iff∃u≥0 such that a+u=b(1)(1)a≤b iff∃u≥0 such that a+u=b
It is easy to see that whenever x,y≥0 x,y≥0 or x,y≤0 x,y≤0 the triangle inequality holds since there is no 'less than' there, |x+y|=|x|+|y||x+y|=|x|+|y|.
Logically, there are two case left to handle:
Case 1:[x<0<y]and(−x)≤y Case 1:[x<0<y]and(−x)≤y
Case 2:[x<0<y]and(−x)≥y Case 2:[x<0<y]and(−x)≥y
Since |z|=|−z||z|=|−z|, it is only necessary to take care of the first case to prove the triangle inequality.
Case 1
Using (1)(1), we write (−x)+u=y(−x)+u=y for u≥0 u≥0 and (−x)>0(−x)>0. Noting that u<y u<y, we have
|x+y|=|x+(−x)+u|=|u|=u<y<(−x)+y=|x|+|y||x+y|=|x+(−x)+u|=|u|=u<y<(−x)+y=|x|+|y|
So only when x x and y y 'straddle 0 0' is the triangle inequality a 'strict less than' relation,
|x+y|<|x|+|y||x+y|<|x|+|y|.
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edited Oct 17, 2019 at 9:25
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Since the proof of CW is necessary and similar I will do both
Start both proofs with the fact that a vector dotted with itself is greater than or equal to 0
for CW substitute vector = x-ty, for triangle inequality vector = x+y
for CW, after dotting x-ty with itself let t = (x.y)/(y.y), for triangle ineq. after dotting x+y with itself and getting a quadratic equation with a dot product in the middle, use CW to show that this quadratic is less than or equal to the same quadratic with the moduluses of the vectors of the dot product in the equation on the left
after rearranging both sqrt both sides
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answered Jan 31, 2021 at 21:37
Daniel EdwinDaniel Edwin
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|x+y|2=(x+y).(x+y)=(x.x)+2(x.Y)+(y.y)=|x|2+2(x.Y)+|y|2|x+y|2=(x+y).(x+y)=(x.x)+2(x.Y)+(y.y)=|x|2+2(x.Y)+|y|2 from Cauchy-Schwarz inequality,|x.Y|<=|x||y||x.Y|<=|x||y||x+y|2<=|x|2+2|x||y|+|y|2|x+y|2<=|x|2+2|x||y|+|y|2|x+y|2<=(|x|+|y|)2|x+y|2<=(|x|+|y|)2 taking square root on both sides. |x+y|<=(|x|+|y|)|x+y|<=(|x|+|y|)
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edited Aug 7, 2019 at 7:12
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answered Dec 27, 2018 at 1:18
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1 Welcome to stackexchange. That said, answering a five year old question with many good answers is not necessarily a good use of your time or ours (since the answer bumps the question to "recently active" so lots of folks look at it. Do pay attention to new questions where you can help. And do use mathjax to format mathematics: math.meta.stackexchange.com/questions/5020/…Ethan Bolker –Ethan Bolker 2018-12-27 01:23:15 +00:00 Commented Dec 27, 2018 at 1:23
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188742 | https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.03%3A_The_Second_Law_of_Thermodynamics | 15.3.1
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15.3: The Second Law of Thermodynamics
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15.2: Entropy Rules
15.4: Free Energy and the Gibbs Function
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Stephen Lower
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( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in green type.
In any macroscopic change, the entropy of the world (that is, system + surroundings) always increases; it never decreases.
Processes that do not exchange heat with the surroundings (such as the free expansion of a gas into a vacuum) involve entropy change of the system alone, and are always spontaneous.
A heat engine is a device that converts heat into work. The fraction of heat that can be converted into work is limited by the fall in temperature between the input to the engine and the exhaust.
According to the Second Law of Thermodynamics, complete conversion of heat into work by a spontaneous cyclic process is impossible.
The First Law of thermodynamics, expressed as ΔU = q + w, is essentially a statement of the law of conservation of energy. The significance of this law is that it tells us that any proposed process that would violate this condition can be dismissed as impossible, without even inquiring further into the details of the process.
Why is the First Law not enough?
For simple mechanical operations on macroscopic objects, the First Law, conservation of energy, is all we usually need to determine such things as how many joules of energy is required to lift a weight or to boil some water, how many grams of glucose you must metabolize in order to climb a hill, or how much fuel your car needs to drive a given distance.
But if you think about it, there are a number of "simple mechanical operations" that never occur, even though they would not violate energy conservation.
Suppose you drop a book onto a table top. The kinetic energy contained in the falling book is dispersed as thermal energy, slightly warming the book and the table top. According to the First Law, there is no reason why placing pre-warmed book on a warmed table top should not be able to propel the book back into the air. Similarly, why can't the energy imparted to the nail (and to the wood) by a hammer not pop the nail back out?
One might propose a scheme to propel a ship by means of a machine that takes in seawater, extracts part of its thermal energy which is used to rotate the propeller, and then tosses the resulting ice cubes overboard. As long as the work done to turn the propeller is no greater than the heat required to melt the ice, the First Law is satisfied.
Because motion of the air molecules is completely random, there is no reason why all of the molecules in one half of a room cannot suddenly "decide" to move into the other half, asphyxiating the unfortunate occupants of that side. (To the extent that air behaves as a perfect gas, this doesn't involve the First Law at all.)
What do all these scenarios that conform to the First Law but are nevertheless never seen to occur have in common? In every case, energy becomes less spread out, less "diluted". In the first two examples, thermal energy (dispersed) gets concentrated into organized kinetic energy of a macroscopic object— a book, a propeller. In the third case, the thermal energy gets concentrated into a smaller volume as the gas contracts.
The second law of thermodynamics says in effect, that the extent to which any natural process can occur is limited by the dilution of thermal energy (increase in entropy) that accompanies it, and once the change has occurred, it can never be un-done without spreading even more energy around. This is one of the most profound laws of nature, and should be a part of every educated person's world view. It is unfortunate that this law is so widely misrepresented as simply ordaining the increase in "disorder". A more brief statement of the Second Law (for those who know the meaning of "entropy") is
Second Law of Thermodynamics: The entropy of the world only increases and never decreases.
The more formal and historical ways of stating the Second Law will be presented farther below after we introduce the topic of heat engines. It is also worth knowing this important consequence of the Second Law: Just because the energy is “there” does not mean it will be available to do anything useful.
Entropy and Spontaneous Change
We explained how processes that take place spontaneously always proceed in a direction that leads to the spreading and sharing of thermal energy.
A book falls to the tabletop (rather than absorbing heat and jumping up from it) because its kinetic energy changes into thermal energy which is widely dispersed into the molecules of the book and the table.
A gas expands and solutions mix because the thermal energy their molecules possess get spread over a larger volume of space.
Hydrogen gas dissociates into H atoms which share thermal energy amongst more particles and a greater volume of space. (But only if the temperature is high enough to make the huge number of new microstates energetically accessible.)
Because all natural processes lead to the spreading and sharing of thermal energy, and because entropy is a measure of the extent to which energy is dispersed in the world, it follows that:
In any spontaneous macroscopic change, the entropy of the world increases.
All natural processes that allow the free exchange of thermal energy amongst chemically-significant numbers of particles are accompanied by a spreading or “dilution” of energy that leaves the world forever changed. In other words, all spontaneous change leads to an increase in the entropy of the world. At first sight, this might seem to be inconsistent with our observations of very common instances in which there is a clear decrease in entropy, such as the freezing of a liquid, the formation of a precipitate, or the growth of an organism.
System + surroundings = the world!
... but it’s the entropy of the system plus surroundings that counts! It is important to understand that the criterion for spontaneous change is the entropy change of the system and the surroundings— that is, of the “world”, which we denote by ΔStotal:
ΔStotal=ΔSsystem+ΔSsurroundings
ΔStotal=ΔSsystem+ΔSsurroundings(15.3.1)
The only way the entropy of the surroundings can be affected is by exchange of heat with the system:
ΔSsurroundings=qsurrT
ΔSsurroundings=qsurrT(15.3.2)
Thus the freezing of water is accompanied by a flow of heat (the heat of fusion) into the surroundings, causing ΔSsurr to increase. At temperatures below the freezing point, this increase more than offsets the decrease in the entropy of the water itself, so ΔSworld exceeds zero and the process is spontaneous. The problem example below works this out in detail for a specific example.
Note that it does not matter whether the change in the system occurs reversibly or irreversibly; as mentioned previously, it is always possible to define an alternative (irreversible) pathway in which the amount of heat exchanged with the surroundings is the same as qrev ; because ΔS is a state function, the entropy change of the surroundings will have the same value as for the unrealizable reversible pathway.
If there is no flow of heat into or out of the surroundings, the entropy change of the system and that of the world are identical. Examples of such processes, which are always spontaneous, are the free expansion of an ideal gas into a vacuum, and the mixing of two ideal gases. In practice, almost all processes involving mixing and diffusion can be regarded as driven exclusively by the entropy increase of the system.
Most processes involving chemical and phase changes involve the exchange of heat with the surroundings, so their tendency to occur cannot always be predicted by focusing attention on the system alone. Further, owing to the –q/T term in ΔSsurroundings, the spontaneity of all such processes will depend on the temperature, as we illustrated for the dissociation of H2 previously.
As a quantitative example, let us consider the freezing of water. We know that liquid water will spontaneously change into ice when the temperature drops below 0°C at 1 atm pressure. Since the entropy of the solid is less than that of the liquid, we know the entropy of the water (the system here) will decrease on freezing. The amount of decrease is found by dividing the heat of fusion of ice by the temperature for the reversible pathway, which occurs at the normal freezing point:
ΔSsystem=−6000J/mol273K=−21.978J/mol
ΔSsystem=−6000J/mol273K=−21.978J/mol(15.3.3)
If the process is actually carried at 0°C, then the heat of fusion is transferred to the surroundings at the same temperature, and the entropy of the surroundings increases by
ΔSsurroundings=6000J/mol273K=21.979J/mol
ΔSsurroundings=6000J/mol273K=21.979J/mol(15.3.4)
so that ΔStotal = 0. Under these conditions the process can proceed in either direction (freezing or melting) without affecting the entropy of the world; this means that both ice and liquid water can be present simultaneously without any change occurring; the system is said to be in equilibrium.
Suppose now that the water is supercooled to –1°C before it freezes. The entropy change of the water still corresponds to the reversible value qrev/T = (–6000J)/(273K). The entropy change of the surroundings, however, is now given by
ΔSsurroundings=6000J/mol273K=22.059J/mol
ΔSsurroundings=6000J/mol273K=22.059J/mol(15.3.5)
The total entropy change is now
ΔStotal=(–21.978+22.059)J; K–1mol–1=+0.081JK–1mol–1
indicating that the process can now occur (“is spontaneous”) only in the one direction.
Why did we use 273 K when evaluating ΔSsystem and 272 K for calculating ΔSsurroundings? In the latter case it is possible to formulate a reversible pathway by which heat can be transferred to the surroundings at any temperature. ΔSsystem, however, is a state function of water, and will vary with temperature only slightly.
Note that in order to actually freeze water, it must be cooled to very slightly below its normal freezing point, a condition known as supercooling. Freezing of supercooled water is of course an irreversible process (once it starts, it cannot be stopped except by raising the temperature by a finite amount), and the positive value of ΔStotal tells us that this process will occur spontaneously at temperatures below 273 K. Under these conditions, the process is driven by the entropy increase of the surroundings resulting from flow of the heat of fusion of water into the surroundings.
Does the entropy of the world ever decrease?
The principle that thermal energy (and the molecules carrying it) tends to spread out is based on simple statistics. It must be remembered, however, that the laws of probability have meaningful application only to systems made up of large numbers of independent actors. If you trap a hundred flies in a bottle, they will generally distribute themselves more or less uniformly throughout the container; if there are only four flies, however, it is quite likely that all of them will occasionally be located in one particular half of the bottle.
Why the sky is blue
Similarly, you can trust with complete certainty that the spontaneous movement of half the molecules of the air to one side of the room you now occupy will not occur, even though the molecules are moving randomly and independently. On the other hand, if we consider a box whose dimensions are only a few molecular diameters, then we would expect that the random and short-term displacement of the small number of particles it contains to one side of the box would occur quite frequently. This is, in fact, the cause of the blueness of the sky: random fluctuations in the air density over tiny volumes of space whose dimensions are comparable with the wavelength of light results in selective scattering of the shorter wavelengths, so that blue light is scattered out, leaving the red light for the enjoyment of sunset-watchers to the east.
Brownian motion
This refers to the irregular zig-zag-like movement of extremely small particles such as plant pollen when they are suspended in a drop of liquid. Any such particle is continually being buffeted by the thermal motions of the surrounding liquid molecules. If size of the particle is very large compared to that the the liquid molecules, the forces that result from collisions of these molecules with the particle will cancel out and the particle remains undisturbed. If the particle is very small, however (perhaps only a thousand times larger than a molecule of the liquid), then the chances that it will undergo sufficiently more hits from one direction than from another during a brief interval of time become significant.
In these two examples, the entropy of the system decreases without any compensating flow of heat into the surroundings, leading to a net (but only temporary) decrease in the entropy of the world. This does not represent a failure of the Second Law, however, because no one has ever devised a way to extract useful work from these processes.
Heat Engines
The Industrial Revolution of the 19th century was largely driven by the invention of the steam engine. The first major use of such engines was to pump water out of mines, whose flooding from natural seepage seriously limited the depths to which they could be driven, and thus the availability of the metal ores that were essential to the expansion of industrial activities. The steam engine is a type of heat engine, a device that converts heat, provided by burning a fuel, into mechanical work, typically delivered through the motion of a piston in opposition to an opposing force. An engine is therefore an energy conversion device in which, ideally, every joule of heat released by combustion of the fuel could be extracted as work at the output shaft; such an engine would operate at 100 percent efficiency.
However, engineers of the time were perplexed to find that the efficiencies of steam engines were rather low (usually around 20%), with most of the heat being exhausted uselessly to the environment. Everyone understood that an efficiency exceeding 100% would be impossible (that would violate conservation of energy, and thus the First Law), but it was not clear why efficiencies could not rise significantly beyond the small values observed even as mechanical designs improved
The answer was found by a young French engineer, Sadi Carnot, who in 1824 published an analysis of an idealized heat engine that is generally considered to be the foundation of the science of thermodynamics— notwithstanding the fact that Carnot still accepted the belief that heat is a fluid-like substance called “caloric”. We will not replicate his analysis here (this is normally done in more advanced courses in physical chemistry), but will simply state his conclusion in his own [translated] words:
"The production of motive power is then due in steam-engines not to an actual consumption of caloric, but to its transportation from a warm body to a cold body...the production of heat alone is not sufficient to give birth to the impelling power: it is necessary that there should also be cold; without it, the heat would be useless. The ultimate attainable efficiency of any heat engine will depend on the temperatures at which heat is supplied to and removed from it."
The left side of the figure represents a generalized heat engine into which a quantity of heat qH, extracted from a source or “reservoir” at temperature TH is partly converted into work w. The remainder of the heat qL is exhausted to a reservoir at a lower temperature TL. In practice, TH would be the temperature of the steam in a steam engine, or the temperature of the combustion mixture in an internal combustion or turbine engine. The low temperature reservoir is ordinarily that of the local environment. The efficiency ε (epsilon) of a heat engine is the fraction of the heat abstracted from the high temperature reservoir that can be converted into work:
ε=wqH
Carnot’s crucial finding (for which he would certainly have deserved a Nobel prize if these had existed at the time) is that the efficiency is proportional to the "distance'' in temperature that the heat can “fall” as it passes through the engine:
ε=1−TLTH
This is illustrated graphically in the right half of the figure just above, in which the efficiency is simply the fraction of the “complete” fall (in temperature) to absolute zero (arrow b) that the heat undergoes in the engine (arrow a.) Clearly, the only way to attain 100% efficiency would be to set the temperature of the exhaust reservoir to 0°K, which would be impossible. For most terrestrial heat engines, TL is just the temperature of the environment, normally around 300 K, so the only practical way to improve the efficiency is to make TH as high as possible. This is the reason that high pressure (superheated) steam is favored in commercial thermal power plants. The highest temperatures (and the greatest operating efficiencies) are obtained in gas turbine engines. However, as operating temperatures rise, the costs of dealing with higher steam pressures and the ability of materials such as turbine blades to withstand high temperatures become significant factors, placing an upper limit of around 600K on TH, thus imposing a maximum of around 50 percent efficiency on thermal power generation.
For nuclear plants, in which safety considerations require lower steam pressures, the efficiency is lower. One consequence of this is that a larger fraction of the heat is exhausted to the environment, which may result in greater harm to aquatic organisms when the cooling water is returned to a stream or estuary.
Example 15.3.1
Several proposals have been made for building a heat engine that makes use of the temperature differential between the surface waters of the ocean and cooler waters that, being more dense, reside at greater depth. If the exhaust temperature is 5°C, what is the maximum amount of work that could be extracted from 1000 L of surface water at 10°C? (The specific heat capacity of water is 4.184 J g–1K–1.)
Solution
The amount of heat (qH) that must be extracted to cool the water by 5 K is (4.184 J g–1K–1)(106 g)(5 K) = 2.09 × 107 J. The ideal thermodynamic efficiency is given by
1−278K283K=0.018
The amount of work that could be done would be
(0.018)(2.09×107J)=3.7×106J
Comment: It may be only 1.8% efficient, but it’s free!
The drinking bird as a heat engine
Few toys illustrate as many principles of physical science as this popular device that has been around for many years. At first glance it might appear to be a perpetual motion machine, but it's really just a simple heat engine. Modern "dippy birds" (as they are sometimes called) utilize dichloromethane as the working fluid.
This liquid boils at 39° C, and therefore has a rather high vapor pressure at room temperature. The liquid (to which a dye is often added for dramatic effect) is stored in a reservoir at the bottom of the bird. The bird's beak is covered with felt which, when momentarily dipped in water, creates a cooling effect as the water evaporates. This causes some of the CH2Cl2 vapor to condense in the head, reducing the pressure inside the device, causing more liquid to boil off and re-condense in the head. The redistribution of fluid upsets the balance, causing the bird to dip its beak back into the water. Once the head fills with liquid, it drains back into the bottom, tipping the bird upright to repeat the cycle.
We will leave it to you to relate this to the heat engine diagram above by identifying the heat source and sink, and estimate the thermodynamic efficiency of the engine.
Heat Pumps
If a heat engine is run “in reverse” by performing work on it (that is, changing “work out” to “work in” in Fig 8), it becomes a device for transporting heat against a thermal gradient. Refrigerators and air conditioners are the most commonly-encountered heat pumps. A heat pump can also be used to heat the interior of a building. In this application, the low temperature reservoir can be a heat exchanger buried in the earth or immersed in a well. In this application heat pumps are more efficient than furnaces or electric heating, but the capital cost is rather high.
The Second Law: what it means
It was the above observation by Carnot that eventually led to the formulation of the Second Law of Thermodynamics near the end of the 19th Century. One statement of this law (by Kelvin and Planck) is as follows:
Definition: Second Law of Thermodynamics (Kelvin Definition)
It is impossible for a cyclic process connected to a reservoir at one temperature to produce a positive amount of work in the surroundings.
To help you understand this statement and how it applies to heat engines, consider the schematic heat engine in the figure in which a working fluid (combustion gases or steam) expands against the restraining force of a weight that is mechanically linked to the piston. From a thermodynamic perspective, the working fluid is the system and everything else is surroundings. Expansion of the fluid occurs when it absorbs heat from the surroundings; return of the system to its initial state requires that the surrounding do work on the system. Now re-read the above statement of the Second Law, paying special attention to the italicized phrases which are explained below:
A cyclic process is one in which the system returns to its initial state. A simple steam engine undergoes an expansion step (the power stroke), followed by a compression (exhaust stroke) in which the piston, and thus the engine, returns to its initial state before the process repeats.
“At one temperature” means that the expansion and compression steps operate isothermally. This means that ΔU = 0; just enough heat is absorbed by the system to perform the work required to raise the weight, so for this step q = –w.
“A positive amount of work in the surroundings” means that the engine does more work on the surroundings than the surroundings do on the engine. Without this condition the engine would be useless.
Note carefully that the Second Law applies only to a cyclic process— isothermal expansion of a gas against a non-zero pressure always does work on the surroundings, but an engine must repeat this process continually; to do so it must be returned to its initial state at the end of every cycle. When operating isothermally, the work –w it does on the surroundings in the expansion step (power stroke) is nullified by the work +w the surroundings must do on the system in order to complete the cycle. The Second Law can also be stated in an alternative way:
Definition: Second Law of Thermodynamics (Planck Definition)
It is impossible to construct a machine operating in cycles that will convert heat into work without producing any other changes.
Thus the Second Law does allow an engine to convert heat into work, but only if “other changes” (transfer of a portion of the heat directly to the surroundings) are allowed. And since heat can only flow spontaneously from a source at a higher temperature to a sink at a lower temperature, the impossibility of isothermal conversion of heat into work is implied.
15.2: Entropy Rules
15.4: Free Energy and the Gibbs Function |
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Abstract
Introduction
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Cited by (124)
Sleep Medicine
Volume 16, Issue 4, April 2015, Pages 521-527
Original Article Sleep deprivation in adolescents: correlations with health complaints and health-related quality of life
Author links open overlay panel, ,
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Highlights
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Sleep deprivation was present in 18.9% of the adolescents.
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There was a very high variability in sleep schedules.
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Sleep problems were reported in 37.2% of the adolescents.
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Health complaints prevalence was higher in sleep-deprived adolescents, specially shoulder and neck pain, fatigue and dizziness.
Abstract
Objectives
The present study aimed to evaluate the influences of sleep duration, sleep deprivation, and weekend variability of sleep upon other adolescents' features, namely those related to health and health-related quality of life.
Methods
The Health Behaviour in School-Aged Children (HBSC) survey is based on a self-completed questionnaire. The participants in the present study were 3476 students (53.8% were girls) in the 8th and 10th grades at school; the mean age was 14.9 years (range 12.519.0). Subjective sleep duration during the weeknights and weekends was collected; sleep deprivation (SD) was considered whenever the difference was greater than 3h. Health complaint frequency and health-related quality of life (with the Kidscreen 10) were collected.
Results
Sleep deprivation was present in 18.9% of the students. It was negatively correlated with sleep duration on weeknights. There were no gender differences, but SD increased with age and grade. Higher school grades were mainly associated with fatigue. A considerable number of adolescents had sleep problems (37.2%); 25.5% had difficulties in sleep initiation, which was more prevalent in adolescents with SD. The sleep duration on weeknights was decreased in the SD group. The average health-related quality of life was reduced in adolescents with SD. The frequency of health complaint was higher is adolescents with SD. Girls had significantly more health complaints than boys, with special focus on headaches.
Conclusions
Sleep deprivation is associated with the perception of health-related quality of life and perceived physical and mental health.
Introduction
The aim of the present study was to provide an in-depth analysis of the diverse associations between sleep duration, sleep deprivation, and variability in sleep duration between weekday and weekend nights, and health and health-related quality of life.
In the present review, the most common health complaints of adolescents are addressed. A brief revision of the main sleep parameters (duration, variability, deprivation, late bedtimes and eveningness) that impact upon the health of adolescents is made, taking into account any possible geographical/cultural influences.
Health issues in adolescents are an important public health concern, with sleep habits playing an important role. In the last few decades, several studies of children and adolescents have pointed out the relations of sleep duration with: daytime sleepiness , , ; body mass index (BMI) , , , , ; type II diabetes and insulin resistance ; specific sleep disorders ; health characteristics ; high blood pressure ; pain , , ; race , ; cognitive tests and academic success , , , ; subjective psychological well being ; socioeconomic status , ; habits such as high screen- or TV-viewing time , ; low or moderate physical activity , ; poor dietary intake and quality ; and risk-taking behaviors , , , namely, binge drinking .
In a recent meta-analysis of children and adolescents aged from 9 to 18 years, including 23 countries, sleep duration varied with gender, age, and geographical region . School-day sleep differed slightly between boys and girls girls slept for 11min/night more than boys (p<0.003), and 29min more on non-school days (p<0.003). Sleep time declined with age minus 14min/day per year of age on school days, and 7min on non-school days. Asian adolescents sleep 4060min less each night than Americans, and 60120min less than Europeans . In India, the mean sleep duration was 7.8h, which also decreased with age , and in China, 34.2% of the students had complaints of poor sleep .
In trying to identify predictors of sleep duration and variability in a community-based cohort study of 247 adolescents (48.5% female, 54.3% ethnic minority, mean age of 13.7 years), univariate models have demonstrated that age, minority ethnicity, neighborhood distress, parent education, parent income, pubertal status, and BMI were significantly related to variability in the total sleep time. In the multivariate model, age, minority status, and BMI were significantly related to variability in total sleep time (all with p<0.05), with younger adolescents, non-minority adolescents, and those of a lower BMI obtaining more regular sleep .
In a Taiwanese population, the mean sleep duration on weeknights was 7.35±1.23h and 9.38±1.62h, on weekends. Weeknight sleep decreased significantly with increasing school grade; there was a trend toward increased daytime sleepiness for students in higher school grade levels. Pearson correlation showed a significant negative correlation (p=0.0001) for increasing total sleep time on the weekend and decreasing BMI .
The Cleveland Children's Sleep and Health Cohort, which consisted of 471 adolescents with a mean age of 15.1 years, sleep duration, measured by actigraphy, had a quadratic u-shape association with Homeostasis Model Assessment of insulin (HOMA). When adjusted for age, gender, race, preterm status, and activity, adolescents who slept for 7.75h had the lowest predicted HOMA, and for adolescents who slept 5.0h or 10.5h, the HOMA indices were approximately 20% higher; after adjusting for adiposity, only the association with longer sleep persisted .
The relation between sleep duration and health goes beyond weight/obesity and insulin/insulin resistance; sleep intervenes in a significant number of clinical complaints including headache and chronic pain of different characteristics, which are either widespread, musculoskeletal, visceral, and more. In depression and multiple somatic complaints, this sleep impact or association is often multiple, with clusters of symptoms fluctuating together with bilateral influences and a comorbid profile , . Insomnia and short sleep duration are comorbid with: obesity, metabolic syndrome, growth hormone deficiency, allergic conditions, chronic pain, neoplasms, blood malignancies, genetic and congenital disorders. Hypersomnia is comorbid with malignancies. Sleep apnea is comorbid with: obesity, metabolic syndrome, polycystic ovarian syndrome, hypothyroidism, asthma, epilepsy, ear/nose/throat (ENT) disorders, congenital malformations, and genetic conditions. Parasomnias imply a differential diagnosis with epilepsy, and some of them are more prevalent in migraine .
In Finland, using a very large sample of adolescents (n=384,076) aged 14 to 20 years, it was proven that late bedtimes, especially after 23:30, increase the prevalence of depression, accidents, neck or shoulder pain, low back pain, stomachache, anxiety or nervousness, irritation or tantrums, headaches, tiredness or dizziness . The high prevalence of headache, depression, and atopic conditions in adolescents not getting enough sleep the week before the study was also proven in a large epidemiological study in the USA .
A relationship exists between headaches and sleep, but with complex expressions, since headaches can be triggered by too much or too little sleep, and also by irregularity or changes in sleep schedules , ; the complex and bidirectional relations between sleep problems and headache have also been proven in a couple of studies performed on adolescents. Among 800 Italian adolescents, the prevalence of headaches was very high (45.6%) and was associated with irregular intake of meals (especially irregular breakfast) and sleep disturbances . In a smaller sample of 69 adolescents with primary headaches, the presence of sleep disturbances was significantly high, namely insufficient total sleep (65.7%), daytime sleepiness (23.3%), difficulty falling asleep (40.6%), and night waking (38.0%) . The same type of results was obtained in a larger sample (n=1862) of adolescents in New Delhi; sleep disturbances in migraineurs were more common when compared with tension headache sufferers and controls . The relationship between migraine, non-migraine headaches and sleep were also proven in a sample of 1023 youngsters aged 815 years: migraineurs had higher scores of daytime sleepiness and were more often evening types . Furthermore, among the triggers of pediatric migraine, lack of sleep was reported in 69.6% of the individual cases, only surpassed by stress, which accounted for 75.7% . A lower percentage and opposite effects were found by Bruni et al. in 2008 : bad sleep was a headache trigger in 32.32% of migraine and non-migraine adolescents, while emotional distress accounted for 27.8% of the cases; in spite of that, the objective risk factors for headache (alcohol and coffee consumption, smoking, neck pain, stress and physical inactivity) did not include sleep .
Sleep is also a major influent factor in adolescents with chronic pain. Insufficient sleep quantity or quality was an independent risk factor for persistence of neck and low back pain among girls and for chronic pain . Pain affects around 21% of adolescents . The relations between chronic pain and sleep disturbances or insomnia are mutual, with insomnia being a risk for pain chronicity, while pain, poor sleep hygiene, and higher depressive symptoms are the main risks for insomnia persistence , . The comorbidity between sleep disturbances and chronic pain was shown in a group of 1518 adolescents aged 1119 years old, with a joint prevalence of 19.1% . Furthermore, low sleep efficiency predicts next day pain, while the vice versa prediction does not hold , . The prevalence of neck and shoulder pain is higher in girls; the risk factors are multiple, namely: family history, school furniture, long sitting time and computer use, insufficient rest time, sleep duration, transportation type, schoolbag weight, and smoking .
Sleep problems are common (circa 45%) in pediatric functional gastro-intestinal disorders . Adolescents suffering from irritable bowel syndrome have increased percentages of poor sleep , and in a clinical group of 25 adolescents with recurrent abdominal pain, 29% reported awakenings related to pain, and 75% reported poor quality of sleep .
Fatigue is another important associated symptom. It is often associated with chronic pain , depression , , and insomnia or sleep problems , , . The risk factors for fatigue with poor clinical outcome are: sleep problems, somatic complaints, blurred vision, pain in the arms or legs, back pain, constipation, and memory deficits. The indicators of a good outcome are: male gender and a physically active lifestyle . Fatigue is statistically associated with feeling depressed, breakfast habits, not being well in school, low physical training, no adult to talk to, having bullied someone, shoplifting and physical fighting . In pediatric fibromyalgia, the dominant symptoms present in almost 90% of the children are diffuse pain and sleep disturbances .
In adolescents, the mutual interaction of depression and sleep also exists; it has also been demonstrated in people with chronic pain , .
Well-being and health-related quality of life (HRQoL) in children and adolescents are quite recent concepts . It is important to consider these concepts within an ecological perspective through multiple levels of analysis, namely self-perceptions and family perceptions . Children's perceptions of their HRQoL are influenced by several factors, such as gender, age, personal and family characteristics, psychological toughness, as well as their socio-economic status (SES) , , , .
Healthy sleep is fundamental to human health and quality of life , , and sleep deprivation increases the risk for mood and behavioral problems, such us drug and alcohol use and vulnerability to accidents , , , , .
Links between eveningness and poor physical, social/interpersonal relationships and mental health have also been found . Adolescents with less-healthy sleeping patterns present with lower scores on emotional, social, school, psychosocial functioning, and global quality of life , , and those who are sleep deprived experience less positive and more negative effects . Those with delayed sleep phase disorders have higher trends for alcohol and caffeine consumption, and lower sports participation .
Sleeping for 6h or less per night is linked to symptoms of depression and lower self-esteem . Sleep deprivation is associated with deficits in child and adolescent functioning, and global health , . Children and adolescents who sleep for less than 5h per night present with more feelings of stress, depression, and suicidal ideation , .
Section snippets
Objectives
To evaluate the interactions between sleep deprivation in adolescents and age, gender, school grade, BMI, health complaints and health-related quality of life; it was hypothesized that sleep deprivation is associated with a higher prevalence of health complaints.
Participants
The present survey is a component of the Health Behaviour in School-Aged Children (HBSC) study , , , .
The Portuguese HBSC survey included 3476 pupils; 53.8% (n=1869) were girls, in the 8th (45.9%) and 10th grades (54.1%), with a mean age of 14.9 years (SD=1.26, range 12.519.0). The children were randomly chosen from 139 schools, in a national sample that was geographically stratified by Education Regional Divisions. The school response rate was 89.9%. The overall procedure
Results
The continuous variables concerning age, BMI, SWeek, SWE, DifWE-W, Kids 10 and SD, together with gender and school grade, are presented in Table 1; for each of them, the comparison for the SD condition is shown. A total of 14.2% of the students were overweight and 2.7% were obese. The percentage of students having a difference of 2h sleep is 27.5% and equal/more than 3h is 18.9%. Sleep duration on weeknights was curtailed in 38.5% and increased in 5.8% of the students, taking the normative
Conclusions and discussion
The present study obtained indicators of adolescents' health complaints and health-related quality of life during the 8th and 10th grades, and correlated them with sleep deprivation. It is integrated in a multinational World Health Organization (WHO) research project . Data were obtained randomly and they were nationally representative; the response rate was quite high, and the percentage of missing data per answer was small. From the questionnaire structure, one aspect must be specially
Conflict of interest
The ICMJE Uniform Disclosure Form for Potential Conflicts of Interest associated with this article can be viewed by clicking on the following link:
. ICMJE Form for Disclosure of Potential Conflicts of Interest form.
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188744 | https://wpmedia.wolfram.com/sites/13/2018/02/03-6-4.pdf | Complex Systems 3 (1989) 599-614 Period Multiplying Operators on Integer Sequences Modulo A Prime Burton Voorhees Athabasca University, Box 10,000, Athabasca, Alberta TOG 2RO, Canada Abstract.
We study properties of operators defined on the space E+(p) of right half-infinite sequences with entries chosen from Zp where p is prime.
The operators in question allow solution of the problem of finding predecessor states for certain cellular automata evolutions and they can be thought of as discrete integration with respect to sequence index.
These operators are self-accumulating, not solipsistic, and have no dense orbits. In addition, they exhibit a period-multiplying property.
Many of these results are derived from properties of Pascal's triangle modulo p which are presented in an appendix.
1.
Introduction Let E+(p) be the space of right half-infinite sequences with entries chosen from Zp, where p is prime.
A cellular automation defined on E+(p) can be represented as a mapping Q : E+(p) ~ E+(p) where Q is an oper-ator determined by the automation rule [I].
The automation is denoted (Q,E+(p)).
The predecessor problem for such a cellular automaton is to determine for any given (3 E E+(p) the set of solutions to the equation Q(J.l) = (3. Voorhees [2J has solved this problem for the class of linear oper-ators defined by D(y,z) = yI + UT and D("x,y) = yI + xo-- 1 where 0- and 0--1 are respectively the left and right shifts on E+(p) and the coefficients x , y, and z are in Zp. This solution involves an operator B(b,r) : E+(p) ~ E+(p) defined, for b,r in Zp, J.l in E+(p), by [B(b,r)(J.l)J i = I::[b(p-r)Ji-jJ.lj j= 1 (1.1) where the notation (3i denotes the ith entry in the sequence (3. Solution of the equations D(r,s)(J.l) = (3 and D~,s)(J.l) = (3 are given by theorem 1.
© 1989 Complex Systems Publications, Inc.
600 B urton Voorhees T heorem 1.
1. The general solution of D(r,s)(P.) = (3 is p. = P.1(3(b,r)(0:1 ) + bB(b,r)(J-1((3 ).
(1.2) where 0:1 has first term equal to one and all oilier terms zero, bs == 1 mod (p), and 0 ~ P.1 < P is an "initial" or "boundary" condition.
2. Let 0 < b,c,t,r, s < p and let these numbers be chosen so that cr == 1 and s + b(p - t)r == 0 mod (p). Then CB(b,t) is the unique inverse of D~,s)' It was also found that the family of operators B(b,r) exhibited an inter-esting period-multiplying property, as will be discussed in section 2 of this paper , and in a well-defined sense can be thought of as integration wit h respect to sequence index.
The main purpose of this paper is to demonstrate that these "discrete integrals" are self-accumulating but not solipsistic and have no dense orbits in E+(p).
In fact, it will turn out that they define a foliation of E+(p) into infinite "st rings" of states. Many of the proofs presented will depend on number theoretic properties of Pascal's triangle modulo p. Derivation of these properties is presented in the appendix.
2.
Period multiplication Let p. E E+(p) be periodic with period n.
Then, under certain circum-stances, B(b,r)(P.) will have period kn where the multiplication factor k can be determined.
Theorem 2. Let p. have period n in E+(p).
1. If [B(b,r)(P. )]n == 0 mod (p) then B(b,r)(P.) also has period n.
2. If [B(b,r)(P.) ]n # 0 mod (p) there exists a smallest integer k ~ p such that [ B(b,r)(P.)]kn == 0 mod (p), and B(b,r)(P.) has period kn .
P roof.
By (1.1), [ B(b,r)(P.)] n+1 = b(p - r )[B(b,r)(P. )]n + {In+1' If p. has period n an d [ B(b ,r)(P. )]n == 0 mod (p), then [ B(b ,r)(P. )]nH = P.nH = III = [B(b,r)( p.)h and part (1) follows from the iterative form of (1.1).
Suppose that [ B(b ,r)(P. )]n # 0 mod (p). Setting b(p - r ) = x and making use of the periodicity of u, (1.1) yields [B(b,r)(P.)]kn = (2.1) (1+ xn+ x2 n +...+x(k-1)n)(Xn- 11l1+ Xn- 1P.2+...+Iln) But z" is between 1 and p- 1. Hence, by standard theorems of number theory (e.g., ) there is a smallest positive integer k such that (Xn)k == 1 mod (p).
Period Multiplying Operators on Integer Sequences Modulo A Prime 601 In this case z" is said to have order k modulo p(ordp(xn) = k). Choose the kin (2.1) to be the order modulo p of z". Then +X(k-l)n+xkn = 1 + xn+x2 n+...+x(k-l)n xn(1 + z" +...+x(k-l)n).
Hence (xn - 1)(1 + xn +...+ x(k-l)n) == 0 mod (p). If xn =t- 1 mod (p) the term 1 + z" +...+ x(k-l)n = 0 and by (2.1) [ B(b,r) (Il)] kn == 0 mod (p). If z" == 1 mod (p) then 1 + X n +...+ x(k-l)n = k and we choose k = p to obtain [ B(b ,r)(Il)] p n == 0 mod (p).
Now the same argument used to prove part (1) yields the result that B(b,r)(Il) has period kn.
The next theorem indicates that iteration of B(b ,r) will eventually multiply the period of every periodic sequence in E+(p), even if it does not do so initially: Theorem 3. Let Il have period n, [ B(b ,r)(Il)]n == 0 mod (p), and Ilm be the first nonzero term of tile sequence Il. Then there exists an s :::; n - m such that [ B(b ,r)(Il)]n =t- 0 mod (p).
Proof. Taking x = b(p- r) and applying (2.1) with [B(b ,r)(1l)] n == 0 mod (p) If [ B{b,r)(Il)]n =t- 0 mod (p), we are done with s = 2. Therefore, suppose that [ B{ b ,r)(Il)]n == 0 mod (p). Since x =t- 0, this requires that [ B[b ,r)(Il)]n-1 == omod (p). Now, from (2.1), X[B(b,r)(Il) ]n-1 + [B{ b ,r)(Il)]n = X[B( b,r)(Il)]n-1 X[X [B( b,r)(Il)]n-2 + [ B{b ,r)(Il)n-d x2[ B(b,r)(1l)] n-2 and again if [ B[ b,r)(Il)]n == 0 mod (p). We are done with s = 3. Therefore, take [ B(b ,r)(Il)]m == 0 mod (p). This, however, requires [B(b ,r)(Il)]n- 2 == 0 mod (p). Clearly this process can be continued and, if we are allowed to require that [ B(b ,r)(Il)]n == 0 mod (p) for all s, indicates that for some value of s all of the first n terms of [ B{b ,r)(Il)]n must become O.
On the other hand, let Ilm be the first nonzero term of the sequence u, By equation (2.1) [ B(b,r)(Il)]m = Ilm for all s, and hence is never zero.
Therefore, to avoid contradiction, there must be an s :::; n - m such that [ B(b ,r) (Il)]n == 0 mod (p).
602 Burton Voorhees 3.
Properties of B(b ,r) A metric can be defined on the space E+(p) as follows . Let 0: E+(p) -+ [0, 1] be defined by 00 0(Jl) = 2:. Jlj/ pi j=l (3.1) Lemma 1. The function 0 defined by (3.1) is a norm on E+(p).
Proof.
Clearly 0(Jl) ~ 0 and equals zero if and only if Jl = 0 where o is the sequence consisting entirely of zeros. Thus, it is only necessary to demonstrate that 0(Jl+ (3) :::; 0(Jl) + 0((3). But addition in E+(p) is always term by term modulo p. Therefore 00 0(Jl)+ 0((3) = 0(Jl + (3) + 2:. 8(Jlj, (3j) /pj j=l where (3.2) { 0 Jlj + (3j < p 8(Jlj, (3j ) = p Jlj+ (3j ~p and the final term of (3.2) is nonnegative.
A metric on E+(p) is now defined by the formula g(Jl,(3) = 0(IJl - (3 1) (3.3) with I Jl - (3li = IJl i - (3;1. T he remainder of this sect ion is concern ed with deduction of properties of the operators B(b,r) with respect to the topology in-duced on E+(p) by the metric g. The major tool in this will be an expression for B~ ,r) in terms of entries in Pascal's triangle modulo p: Theorem 4. Let Jl be in E+(p) and write x = b(p - r). Th en i [B k ()] ~ rr( k+i-j) i-i (b,r) Jl i = LJ i-j+l x Jlj j=l (3.4) where rr~k ) is the ith entry in the kth row of Pascal's triangle modulo p.
Remark. The coefficients in (3.4) are the first i terms in the kth diagonal of the mod(p) Pascal triangle.
Proof.
Since rr~~j~-;.j) = 1 for all i,j(i ~ j) equation (1.1) indicates that the claim is true for k = 1. The remainder of the proof will proceed by induction .
Assume the theorem is true for k. Then by (1.1) and the induction hypothesis, i [Btb ;r)(Jl)]i = [B(b,r)(Btb ,r)(Jl))]i = 2:. xi-j [Btb,r)(Jl)]j j=l i i ~ ~ rr(k+j- d) i-d LJ LJ j-d+l X Jld j=l d =l (3.5) Period Multiplying Operators on Integer Sequences Modulo A Prime 603 Rearranging terms in (3.5) by grouping coefficients of Jtd yields [Btb~ r) (Ii )] i =~ [Err~~i')] Xi- j li j but by lemma 8 of the i-j ~ rr(k+·) _ rr(k+l+i-j) L.J .+1 -.- ; + 1 .=0 (3.6) hence (3.6) is identical to (3.4) with k replaced by k + 1 and the theorem is proved.
The first question asked of the operators B(b,r) is whether or not they have cycles. The next theorem answers this in the negative.
Theorem 5. B(b,r) : E+(p) -4 E+(p) bes no cycles oth er than the trivial cycle O.
Proof. Suppose that there is a k > 0 and a nonzero Ii in E+(p) such that Btb,r)(Ii) = Ii · Then, for all i i ( 1) ~ rr(k+i-j) i- j - 0 d ( ) P -Ii i + L.J i-j+I X li j = mo p j=l (3.7) Let Ii . be the first nonzero term of the sequence Ii.
Expansion of (3.7) dropping terms which sum to zero modulo p, yields the hierarchy of equations rr(k+I) i-.
-0 2 X Ii. -rr(k+1) i-.-1 + rri-.
- 0 2 X 1i.+1 3 Ii.-rr(k+ 1) i- .-2 + rr(k+2) i-.-l + rr(k+3) i-.
- 0 2 X 1i.+2 3 X 1i.+I 4 X Ii. -etc.
(3.8) Since Ii., x =I- 0 the first equation of (3.8) requires that rr~k+I) = o. Substitu-tion of this into the second equation yields the requirement that rr~k+2) = O.
Continuation of this process indicates that rr~~ij) = 0 for all j > O. How-ever, these coefficients are drawn from the kth diagonal of the mod (p) Pascal triangle and no diagonal of this triangle consists entirely of zeros after the leading one. Hence, (3.7) can never be satisfied for all i and the theorem is true.
Proof of theorem 5 is based on the nonexistence in the mod(p) Pascal triangle of a diagonal consisting only of zeros following the leading one. This triangle does, however, contain diagonals which are mostly zeros.
Thus, although B(b,r) has no cycles and is therefore not periodic, it can be shown to be "almost periodic" in the sense of being self-accumulating, i.e., every iterate of B(b,r) is an accumulation point for further iterates.
604 Theorem 6. For all k, s ~ 0 and for all J1. in E+(p) p5 g(Btb,T)(J1.),Btb;:"(J1.)) ~ [LP jt 1 j=1 Burton Voorhees (3.9) Proof. It is sufficient to prove the theorem for k = O . In this case consider the sequence ~ = J1. - Br:,T)(J1.) which can be shown to have components The coefficients in the sum are the first i entries of the pS diagonal of the mod(p) Pascal triangle, with the first entry excluded (since the sum is only to i -I). Thus, by lemma 6 of the appendix these coefficients are all zero for i ~ p", Indeed, (' ") {I;- J' - 0 pS p2s nv +.- ] = .
-. " ,...
•-]+1 0 otherwise so the values of i giving nonzero contributions to ~i are i = mp" + 1, m ~ 1.
The maximum possible value of ~i is p - 1. Hence 00 00 g(J1. ,B(b,T)(J1.) ) = LUpi ~ L(P _ 1)/piP' +1 i=l i=l 00 [ (p -1 )/p]LP-is i = 1 The final sum on the right is just 1/(pS - 1), which can be written as [(p -1)(ps-1+ p s- 2+ . . .+1)]-1. Multiplication by (p-l)/p now yields the desired result.
The content of this theorem can be summarized by saying that with respect to the topology induced on E+(p) by the metric g, the operators B(b,T) are self-accumulating. That is, under iteration of B(b,T) the sequence [ Btb ,T )(J1.)] eventually returns to arbitrarily small neighborhoods of its pre-vious iterates.
The next question is whether or not these operators are solipsistic-that is, does [Btb ,T)(J1. )IO ~ k < 00 ] exhaust all of the accumula-tion points of B(b,T)? The answer to this question is not quite. Let J1.q be the first nonzero term of a sequence J1. and consider i-I 1 J1. - BP'-I(,B )I i = 1 (J1.i - ,Bi) - L rr~:'.'/;;-j -l ) xi-i,Bj I j=1 (3.10) m v', but thi s does not alter the desired result.
Theorem 7. For every J.l in E+(p)g(D(i,br)(J.l),Br.:,;)l(J.l)) ::; p-(P'-l) . Thus, D(i,br)(J.l) is an accumulation point of B(b,r) iterated on J.l.
By direct com putation using (l .1) and (3.12) D(i,br)B(b,r) = B(b,r)D(i,br) = 1.
Hence, D(i,br) and B(b,r) are inverses of eacb other.
This yields a proof tbat th e B(b,r) are not solipsistic: Theorem 8. Let D(i,br) and B(b,r) be as above, witb s < 0 and a sequence J.l given.
Tbere does not exis t an integer k > 0 sucb that Btb,r) (J.l) = [D(i,b r)]S(J.l)' Proof. Suppose that such a k existed. Then Btb: rj(J.l) = J.l . By theorem 5, however, B(b ,r) has no nontrivial cycles.
By theorem 7 it is possible to write D(i,br)(J.l) = Br.
:,;/(J.l)+ (3 with (3i = 0 for i ::; p", Thus, by (3.12) [D(i ,br)(J.l)]i = 0 for i ::; p", Since D(i,br) is the inverse of B(b,r) there is a generalization of theorem 7: Theorem 9. For all k < v' and every J.lg((D- )t1,br) (J.l) ,Br.
:,;)k(IL)) ::; p-( p '-l).
We now show that E(b,r) has no dense orbits. Define a partition of E+(p) by E+(p) by E: = [ J.linE+(p)lJ.li = 0 iii < q;J.lq =J 0]. By (i.i, B(b,r): E: ~ E: so no orbit of B(b,r) can be dense in E+(p) . Indeed, if J.l is in E: and J.l' is in E; with s < q then g(J.l, Btb,r)(J.l')) > P-q· For all k,0(Btb,r)(J.l)) is contained in the interval [P-(Q -l) ,P- Q ] and th e diagram of figure 1 comm utes.
In other words, E+(p) = U~l E: defines a stratification of E+(p) witb respect to 0 which is preserved under B(b,r)' (We migh t call this th e Zeno stratifi cation- as above, so below.) Tb e remaining question is whether or not B(b,r) migh t possess orbits which are dense in one of the E:.
606 Burton Voorhees li p Figure 1: Stratifi ed mapping to the unit interval.
T heorem 10. No orbit of B(b,r) : E: -+ E: is dense in E : for any q.
P roof. It is sufficient to prove the theorem for Et. Suppose that /J- is an element of Et such that the orbit [Btb ,r)(/J-)IO :S k < 00] is dense in Et.
Then, for any given (3, (3' in Et, and for all N < 00, there will be integers k, m (dependent on N) such that (3.13) Without loss of generality assume that k, m are the smallest integers for which this occurs for a fixed N, and k :S m . (3.13) requires that Thus, it must be possible to simultaneously satisfy the sets of equations (3. - ~ rr(k+ 1- j ) i-j .
• -L.J i -j+l X /J-J j=l (3.14) (3,. - ~ rr(m+i - j ) i-j .
• -L.J ' - J+1 X /J-J j =l Addit ion of these equations yields, for i :S N (3.15) Period Multiplying Operators on Integer Sequences Modulo A Prime 607 which can be written in matrix form as W= fx(k ,m) (3.16) where Wand x(k,m) are column vectors in Z;: with jth components given by (3j + (3fj and II;k+j-l) +II;m+j-l) respectively, and f is the N xN matrix fJ.l 0 0 0 0 fJ.2 XfJ.l 0 0 0 f = fJ.3 XfJ.2 X2fJ.l 0 0 (3.17) 0 fJ.N XfJ.N-l X2fJ.N_2 X3fJ.N_3 XN-1fJ.l Det(f) = xN(N- l )/2fJ.'( =I- 0 since fJ.l = 1, hence f - 1 exists and (3.16) has a un ique solution (3.18) Now suppose that W -+ W+ W. This will change the solution of (3.18), say to X = X(k,m) + f -1w. In general, however, it is not true that X can be written as the sum of the first N terms of two diagonals of the mod(p) Pascal triangle. To see that this is so, let q be such that pq-l < N ~ pq.
Then, for i ~ pq, the first N terms of the (i + pq)th diagonal of the mod(p) Pascal triangle equal the first N terms of the ith diagonal. Thus, the number of possible distinct combinations of the first N terms of diagonals of this triangle, taken two at a time, is given by pq(pq - 1)/ 2.
However, W is arbitrary, except that wr = O.
Hence, there are v":' = pl'q-l possible choices for w, and N can always be chosen large enough that this is greater than pq(pq- 1)/2. (For example, the choice of N such that q < (pq- 1)/2 is sufficient.) Thus, no fJ. can have a dense orbit under iteration of B.
4.
B(b,r) as discrete integrat ion The operators D(r,.) and Dr;,.) can be considered as discrete derivatives with respect to the sequence index, via an analogy to the Taylor formula f(x) = f(a) + (x - a)fl(a). The analogy is fJ.i+l = (p - r )fJ.i + b[D (r,. )(fJ. )]i fJ.i+l = (p - r )fJ.i + b[D r;,s)(fJ.)hl where bs == 1 mod (p).
The equation Dr;,s)(fJ.) = (3 can be directly "in-tegrated" since the inverse of Dr;,.) is bB(b,r) so that fJ. = bB(b,r)((3).
The equation D(r,.)(fJ.) = (3 can also be "integrated" via theorem 1. Thus the op-erators B (b ,r) are analogues to discrete integration with respect to sequence index.
608 Burton Voorhees Restricting consideration to p = 2, B(l,l) == B is closely related to two operators studied by Rogers and Weiss [4,5].
A = a B (Accumulator operator) T = (7 + [(7, B] (Twisted-shift operator).
In order to derive formulas for powers of these operators we need the following: Lemma 2.
r - l B(7r = a" B + BPI 2: (7' . =0 where Pl(J-!) = J-!lCXl.
Proof. This is true for r = 1 and we proceed by induction. Assuming (4.1) is true for r then r B(7r+l = B(7r(7 = o" B(7+BPI 2: (7' .=1 r (7r+lB +o"BPI + BPI 2: (7' .=1 which yields the desired result since for any r, o" BPI = BPI.
Making use of equation (4.1) together with lemma 2, an induction argu-ment also proves: T heorem 11 . Wi th A and T defined as above (7k + B (7k- lr, k 2: a" B rZk-r r =l where the Z; are defined recursively by T rr- s z; = 1, z, = PI 2:2:a"Br-.+ l Z. _l . = lq=O 5.
Discussion (4.2) This paper has int roduced a fami ly of operators B(b,r) : E + ---+ E + which have been interpreted as discrete integrals with resp ect to sequence index. Prop-erties of these operators have been determined: they are self-accumulating, not solipsistic, and have a period-multiplying property. In terms of applica-tion, these operators are significant in solution of the problem of determining predecessor states for certain cellular automat a evolutions . They have also been found useful in studies of arithmetic properties of the mapping D : [0, 1] ---+ [0, 1] defined in terms of the operator D : E + ---+ E+ and the mapping (3.1) by D(0(J-!)) = 0(D(J-!)) . Although conceptually and mathe-matically simple, this family of operators is found , at least in the p = 2 case, to have a direct relation to other more complicated operators such as Rogers and Weiss's twisted-shift and accumulator operators.
Period Multiplying Operators on Integer Sequences Modulo A Prime 609 Appendix A.
The mod(p) Pascal triangle Several of the proofs given in section 3 of this paper are based on properties of Pascal 's triangle reduced modulo p, and the coefficients in expansions of powers of the operators st udied in this pa.
per are drawn from this triangle.
The properties of the mod(p) Pascal triangle which are important for the present paper are derived in this a.ppendix. Many of the results presented here are due to Long .
Long has proved an elegant structural theorem for the mod(p) Pascal triangle. For p prime let k, n, and m be integers wit h 0 ::; k ::; n and 1 ::; m .
Let ~n , k denote the triangle Theorem 12. (Long, [7}) ~ n, k deiiued above is the triangle (~ ~) (n 1) (n1) k 0 k 1 (~ ) ( ~ with all produ cts redu ced mo dulo p. Further, where th e addition is elem ent -wise modulo p. Finally, every element in Pas-cal's triangle and not in one of the ~n , k is congruent to 0 m odulo p.
Proof of this theorem is contained in Long 's paper. The triangles ~n,k are in one-to-one correspondence with the residues 0,1 , 2, ..., P - 1 so that 610 the triangle of triangles: D.o,o Burton Voorhees is isomorphic to the mod(p) Pascal triangle. This leads to the basic self-similarity property for the mod(p) Pascal's triangle: "If we repeatedly iter-ate this process by mapping the triangles D.n,k onto the residues it follows that, modulo p, Pascal's triangle is a triangle that contains a Pascal's trian-gle of triangles, that in turn contain a Pascal's triangle of triangles, ..., ad infinitum" (from , author's italics).
It also turns out that the entries of Pascal's triangle not contained in any of the D.n,k form inverted triangles of the form Every element of these inverted triangles is congruent to 0 modulo p.
Figure 2 shows the first 33 rows of the mod(2) Pascal triangle, illustrat-ing Long's results.
What is of particular interest for this paper are the locations of the in-verted triangles of zeros, and the self-similar property which follows from Long's results. In particular, these results indicate that inverted triangles of zeros will be based only on rows k such that (k - l )lp. Further, if k = p' +1 then there will be one such inverted triangle with base length pS- 1. That is, the pS+ 1 row of the mod(p ) Pascal triangle consists of an initial and final one separated by p' - 1 zeros. The general pattern of self-similarity for the mOd(p) Pascal triangle is indicated in figure 3 below: Lemma 3. Let A. be the mod (p) Pascal triangle truncated at row p", Then A. consists ofp(p +1)/2 upright triangles isomorphic to A s- 1 and p(p - 1)/ 2 inverted triangles of zeros having base length p.-l - 1. Tbese last are the largest inverted triangles of zeros contained in A s» Th e upright triangles have the numerical form mAs- 1 with 1 :::; m < p.
Proof. The upright triangles are the D.i,j defined by Long. Since the tri-angle of these triangles is isomorphic to Pascal's triangle mod (p) the number of upright triangles isomorphic to A.- 1 is the same as the number of elements Period Multiplying Operators on Integer Sequences Modulo A Prim e 611 1 1 1 1 01 1111 10 0 0 1 1 1001 1 1010101 111 11111 10 0 0 0 0 0 0 1 11 00000011 10100000101 1 1 1 10000 1 1 1 1 100 010001000 1 1 1001 10011 0011 10 10101010 10 101 1 1 1 11 11 11 1 1 1 1 1 1 1 10 0 0 0 0 0 0 0 0 0 00 000 1 1 10 000000 00 0 00001 1 1010 0 0 0 0 00 0 0 0 00 0 10 1 1 11 1 00 0 0 0 0 0 0 0 0 0 011 1 1 1 00 0 10000000000010001 110 01 10000000000110011 10 101010000000 001010101 111111 1' 00000000 11111111 100 00 00010000000100000001 1 1000000110 000001100000011 101 000001010000010 100000101 111100001 111000011 1100001111 10001000 1000 1000 1000100010001 1 1 00 1 1 00 1 100 1 100 1 1 00 1 , (}O 1 1 00 1 1 10 10 10 1 0 10 10 10 10 1 0 1 0 1 0 1 0 1 0 10 10 1 111111111111 111 111 11 1 1 11111111 1 1 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 Figure 2: Pascal's triangle modulo 2.
in the first p rows of th is triangle and this is just the sum of the first p in-tegers, that is, p(p + 1)/2. The inverted tri angles of zeros fall between the upright triangles so that the number of these is just th e sum of the first p- 1 int egers, that is, p(p - 1)/2.
On the basis of this lemma, it is possible to count the number of inverted triangles of zeros of any size which are contai ned in As. That is, there are p(p -1)/2 with the maximum base of p.-l -1. Each of the up right triangles has the form mAs-l. But, As-l contains p(p + 1)/2 triangles isomorphic to As- 2 and p(p - 1)/2 inverted triangles of zeros with base length ps-2 - 1.
Thus, there are p2(p2- 1)/4 of these inverted triangles and a total of p2(p+ 1)2 triangles isomorphic to A s - 2 ' Continuation of this yields Lemma 4. A. contains 2- dpd(p + l)d upright triangles isomorphic to As- d and 2-dpd(p+ 1)d-l(p _1) inverted triangles of zeros with base length pS-d_1 , where 1 ::; d ::; s - 1.
612 Burton Voorhees It is also possible to specify which rows of A. the inverted triangles are based on and how many are based on each of these rows.
Lemma 5. For 1 ::; d ::; s -1 there are md inverted triangles of base length r:'_1 based on row mdps-d + 1 with 1 ::; md ::; pd - 1 and point pq not a divisor of md for any q < d.
Th ere are also som e formulas relating to properties of diagonals of the mod(p) Pascal triangle which can be derived from: Lemma 6. Th e pS diagonal of the mod(p) Pascal triangle is periodic with period pS. Th e first pS terms consist of a leading one followed by pS- 1 zeros.
Proof. The first term of every diagonal of Pascal's triangle is 1. However, there is an inverted triangle of zeros with base length pS - 1 based on row pS+ 1 and the next pS - 1 terms of the pS diagonal lie along a side of this triangle. Periodicity follows from the self-similarity properties of the Pascal triangle modulo p.
With a similar argument we prove: Lemma 7. The p' - 1 diagonal of Pascal's triangle modulo p has period p' and consists of a leading one followed by p - 1, followed by a total of p' - 2 zeros.
Finally, any element of the mod(p) Pascal triangle can be written as a sum of elements along a diagonal.
s 3p ~I-~ """ ~ • • • • • Figure 3: Self-similarity" pat tern for Pascal's Triangle modulo p. In-dicated triangle continues to row ps+l and this defines A.+ 1 .
Period Multiplying Operators on Integer Sequences Modulo A Prime 613 • • • p • • / + 4 .-, • \ }-J \/ + • ,p • • .
I + • • '< p • • \ / + • • • < '1 • • \ Y • • • • ;~( • • • • • • • • • • • • Figure 4: Backward decomposition from an element of Pascal 's trian-gle.
Lemma 8.
rr (k+i+l) _ ~ rr(k+i) i -L.J j+l j=O Proof, An analytic proof follows directly from the addition rule for Pascal's triangle, noting that rr~k) = rr~r) = 1 for all k, T. The essence of this proof, however, is most easily conveyed through consideration of figure 4.
The given entry marked by the lowest circle is the sum of the two en-tries immediately above it, and this property propagates back to the initi al diagonal which consists entirely of ones.
References B. Voorhees, "Cellular automata, Pascal 's triangle, and generation of order," Physica, 3ID (1988) 135-140.
B. Voorhees, "Predecessor states for certain additive cellular automata," Communications in Mathematical Physics, 117 (1988) 431-439.
K.H. Rosen, Elementary Number Th eory and Its Applications (Addison -Wesley, Reading, MA, 1984) 232.
T.D. Rogers and A. Weiss, "Proceedings of 1986 University of Toront o Con-ference on Oscillations, Bifurcation, and Chaos," to appear.
614 T .D. Rogers, private communication.
Burton Voorhees B. Voorhees , "Geometry and arithmetic of a simple cellular automata," preprint.
C.T. Long, "Pascal's triangle modulo p," Fibonacci Quarterly, 19 (1981) 458-463. |
188745 | https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(Ellis_and_Burzynski)/09%3A_Roots_Radicals_and_Square_Root_Equations/9.03%3A_Simplifying_Square_Root_Expressions | Published Time: 2020-08-21T04:18:02Z
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9: Roots, Radicals, and Square Root Equations
Elementary Algebra (Ellis and Burzynski)
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9.3: Simplifying Square Root Expressions
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Contents
1. Home
2. Bookshelves
3. Algebra
4. Elementary Algebra (Ellis and Burzynski)
5. 9: Roots, Radicals, and Square Root Equations
6. 9.3: Simplifying Square Root Expressions
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Elementary Algebra (Ellis and Burzynski)
Front Matter
1: Arithmetic Review
2: Basic Properties of Real Numbers
3: Basic Operations with Real Numbers
4: Algebraic Expressions and Equations
5: Solving Linear Equations and Inequalities
6: Factoring Polynomials
7: Graphing Linear Equations and Inequalities in One and Two Variables
8: Rational Expressions
9: Roots, Radicals, and Square Root Equations
10: Quadratic Equations
11: Systems of Linear Equations
12: Appendix
Back Matter
9.3: Simplifying Square Root Expressions
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9.4: Multiplication of Square Root Expressions
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Table of contents
1. Perfect Squares
1. Note
The Product Property of Square Roots
The Product Property x y−−√=x−−√y√x y=x y
The Quotient Property of Square Roots
The Quotient Property x y−−√=x−−√y√x y=x y
CAUTION
Square Roots Not Involving Fractions
Simplifying Square Roots Without Fractions
Sample Set A
Example 9.3.1 9.3.1
Example 9.3.2 9.3.2
Example 9.3.3 9.3.3
Example 9.3.4 9.3.4
Example 9.3.5 9.3.5
Example 9.3.6 9.3.6
Practice Set A
Practice Problem 9.3.1 9.3.1
Practice Problem 9.3.2 9.3.2
Practice Problem 9.3.3 9.3.3
Practice Problem 9.3.4 9.3.4
Practice Problem 9.3.5 9.3.5
Practice Problem 9.3.6 9.3.6
Practice Problem 9.3.7 9.3.7
Practice Problem 9.3.8 9.3.8
Square Roots Involving Fractions
Simplifying Square Roots with Fractions
Sample Set B
Example 9.3.7 9.3.7
Example 9.3.8 9.3.8
Example 9.3.9 9.3.9
Example 9.3.10 9.3.10
Example 9.3.11 9.3.11
Practice Set B
Practice Problem 9.3.9 9.3.9
Practice Problem 9.3.10 9.3.10
Practice Problem 9.3.11 9.3.11
Practice Problem 9.3.12 9.3.12
Practice Problem 9.3.13 9.3.13
Practice Problem 9.3.14 9.3.14
Practice Problem 9.3.15 9.3.15
Practice Problem 9.3.16 9.3.16
Practice Problem 9.3.17 9.3.17
Practice Problem 9.3.18 9.3.18
Exercises
Exercise 9.3.1 9.3.1
Exercise 9.3.2 9.3.2
Exercise 9.3.3 9.3.3
Exercise 9.3.4 9.3.4
Exercise 9.3.5 9.3.5
Exercise 9.3.6 9.3.6
Exercise 9.3.7 9.3.7
Exercise 9.3.8 9.3.8
Exercise 9.3.9 9.3.9
Exercise 9.3.10 9.3.10
Exercise 9.3.11 9.3.11
Exercise 9.3.12 9.3.12
Exercise 9.3.13 9.3.13
Exercise 9.3.14 9.3.14
Exercise 9.3.15 9.3.15
Exercise 9.3.16 9.3.16
Exercise 9.3.17 9.3.17
Exercise 9.3.18 9.3.18
Exercise 9.3.19 9.3.19
Exercise 9.3.20 9.3.20
Exercise 9.3.21 9.3.21
Exercise 9.3.22 9.3.22
Exercise 9.3.23 9.3.23
Exercise 9.3.24 9.3.24
Exercise 9.3.25 9.3.25
Exercise 9.3.26 9.3.26
Exercise 9.3.27 9.3.27
Exercise 9.3.28 9.3.28
Exercise 9.3.29 9.3.29
Exercise 9.3.30 9.3.30
Exercise 9.3.31 9.3.31
Exercise 9.3.32 9.3.32
Exercise 9.3.33 9.3.33
Exercise 9.3.34 9.3.34
Exercise 9.3.35 9.3.35
Exercise 9.3.36 9.3.36
Exercise 9.3.37 9.3.37
Exercise 9.3.38 9.3.38
Exercise 9.3.39 9.3.39
Exercise 9.3.40 9.3.40
Exercise 9.3.41 9.3.41
Exercise 9.3.42 9.3.42
Exercise 9.3.43 9.3.43
Exercise 9.3.44 9.3.44
Exercise 9.3.45 9.3.45
Exercise 9.3.46 9.3.46
Exercise 9.3.47 9.3.47
Exercise 9.3.48 9.3.48
Exercise 9.3.49 9.3.49
Exercise 9.3.50 9.3.50
Exercise 9.3.51 9.3.51
Exercise 9.3.52 9.3.52
Exercise 9.3.53 9.3.53
Exercise 9.3.54 9.3.54
Exercise 9.3.55 9.3.55
Exercise 9.3.56 9.3.56
Exercise 9.3.57 9.3.57
Exercise 9.3.58 9.3.58
Exercise 9.3.59 9.3.59
Exercise 9.3.60 9.3.60
Exercise 9.3.61 9.3.61
Exercise 9.3.62 9.3.62
Exercise 9.3.63 9.3.63
Exercise 9.3.64 9.3.64
Exercise 9.3.65 9.3.65
Exercises For Review
Exercise 9.3.66 9.3.66
Exercise 9.3.67 9.3.67
Exercise 9.3.68 9.3.68
Exercise 9.3.69 9.3.69
Exercise 9.3.70 9.3.70
To begin our study of the process of simplifying a square root expression, we must note three facts: one fact concerning perfect squares and two concerning properties of square roots.
Perfect Squares
Rea numbers that are squares of rational numbers are called perfect squares. The numbers 25 25 and 1 4 1 4 are examples of perfect squares since 25=5 2 25=5 2 and 1 4=(1 2)2 1 4=(1 2)2, and 5 5 and 1 2 1 2 are rational numbers. The number 2 2 is not a perfect square since 2=(2–√)2 2=(2)2 and 2–√2 is not a rational number.
Although we will not make a detailed study of irrational numbers, we will make the following observation:
Note
Any indicated square root whose radicand is not a perfect square number is an irrational number.
The numbers 6–√,15−−√6,15 and 3 4−−√3 4 are each irrational since each radicand 6,15,3 4 6,15,3 4 is not a perfect square.
The Product Property of Square Roots
Notice that
9⋅4−−−√9–√4–√=36−−√=3⋅2=6=6 and 9⋅4=36=6 and 9 4=3⋅2=6
The Product Property x y−−√=x−−√y√x y=x y
This suggests that in general, if x x and y y are positive real numbers,
x y−−√=x−−√y√x y=x y
The square root of the product is the product of the square roots.
The Quotient Property of Square Roots
We can suggest a similar rule for quotients. Notice that
36 4−−−√=9–√=3 36 4=9=3 and
36−−√4–√=6 2=3 36 4=6 2=3.
Since both 36 4 36 4 and 36−−√4–√36 4 equal 3 3, it must be that
36 4−−−√=36−−√4–√36 4=36 4
The Quotient Property x y−−√=x−−√y√x y=x y
This suggests that in general, if x x and y y are positive real numbers,
x y−−√=x−−√y√,y≠0 x y=x y,y≠0.
The square root of the quotient is the quotient of the square roots.
CAUTION
It is extremely important to remeber that
x+y−−−−√≠x−−√+y√x+y≠x+y or x−y−−−−√≠x−−√−y√x−y≠x−y
For example, notice that 16+9−−−−−√=25−−√=5 16+9=25=5, but 16−−√+9–√=4+3=7 16+9=4+3=7
We shall study the process of simplifying a square root expresion by distinguishing between two types of square roots: square roots not involving a fraction and square roots involving a fraction.
Square Roots Not Involving Fractions
A square root that does not involve fractions is in the simplified form if there is no perfect square in the radicand.
The square roots x−−√,.a b−−√,5 m n−−−−√,2(a+5)−−−−−−−√x,.a b,5 m n,2(a+5) are in simplified form since none of the radicands contains a perfect square.
The square roots x 2−−√,a 3−−√=a 2 a−−−√x 2,a 3=a 2 a are notin simplified form since each radicand contains a perfect square.
To simplify a square root expression that does not involve a fraction, we can use the following two rules:
Simplifying Square Roots Without Fractions
If a factor of the radicand contains a variable with an even exponent, the square root is obtained by dividing the exponent by 2.
If a factor of the radicand contains a variable with an odd exponent, the square root is obtained by first factoring the variable factor into two factors so that one has an even exponent and the other has an exponent of 1, then using the product property of square roots.
Sample Set A
Simplify each square root.
Example 9.3.1 9.3.1
a 4−−√a 4. The exponent is even: 4 2=2 4 2=2. The exponent on the square root is 2 2.
a 4−−√=a 2 a 4=a 2
Example 9.3.2 9.3.2
a 6 b 10−−−−√a 6 b 10. Both exponents are even: 6 2=3 6 2=3 and 10 2=5 10 2=5. The exponent on the square root of a 6 a 6 is 3 3. The exponent on the square root if b 10 b 10 is 5 5.
a 6 g b 10−−−−−√=a 3 b 5 a 6 g b 10=a 3 b 5
Example 9.3.3 9.3.3
y 5−−√y 5. The exponent is odd: y 5=y 4 y y 5=y 4 y. The
y 5−−√=y 4 y−−−√=y 4−−√y√=y 2 y√y 5=y 4 y=y 4 y=y 2 y
Example 9.3.4 9.3.4
36 a 7 b 11 c 20−−−−−−−−√=6 2 a 6 a b 10 b c 20−−−−−−−−−−√=6 2 a 6 b 10 c 20⋅a b−−−−−−−−−−−√=6 2 a 6 b 10 c 20−−−−−−−−√a b−−√=6 a 3 b 5 c 10 a b−−√a 7=a 6 a,b 11=b 10 b by the commutative property of multiplication by the product property of square roots 36 a 7 b 11 c 20=6 2 a 6 a b 10 b c 20 a 7=a 6 a,b 11=b 10 b=6 2 a 6 b 10 c 20⋅a b by the commutative property of multiplication=6 2 a 6 b 10 c 20 a b by the product property of square roots=6 a 3 b 5 c 10 a b
Example 9.3.5 9.3.5
49 x 8 y 3(a−1)6−−−−−−−−−−−−√=7 2 x 8 y 2 y(a−1)6−−−−−−−−−−−−−√=7 2 x 8 y 2(a−1)6−−−−−−−−−−−−√y√=7 x 4 y(a−1)3 y√49 x 8 y 3(a−1)6=7 2 x 8 y 2 y(a−1)6=7 2 x 8 y 2(a−1)6 y=7 x 4 y(a−1)3 y
Example 9.3.6 9.3.6
75−−√=25⋅3−−−−√=5 2⋅3−−−−√=5 2−−√3–√=5 3–√75=25⋅3=5 2⋅3=5 2 3=5 3
Practice Set A
Simplify each square root.
Practice Problem 9.3.1 9.3.1
m 8−−−√m 8
Answer
m 4 m 4
Practice Problem 9.3.2 9.3.2
h 14 k 22−−−−−√h 14 k 22
Answer
h 7 k 11 h 7 k 11
Practice Problem 9.3.3 9.3.3
81 a 12 b 6 c 38−−−−−−−−√81 a 12 b 6 c 38
Answer
9 a 6 b 3 c 19 9 a 6 b 3 c 19
Practice Problem 9.3.4 9.3.4
144 x 4 y 80(b+5)16−−−−−−−−−−−−−−√144 x 4 y 80(b+5)16
Answer
12 x 2 y 40(b+5)8 12 x 2 y 40(b+5)8
Practice Problem 9.3.5 9.3.5
w 5−−√w 5
Answer
w 2 w−−√w 2 w
Practice Problem 9.3.6 9.3.6
w 7 z 3 k 13−−−−−−√w 7 z 3 k 13
Answer
w 3 z k 6 w z k−−−√w 3 z k 6 w z k
Practice Problem 9.3.7 9.3.7
27 a 3 b 4 c 5 d 6−−−−−−−−−√27 a 3 b 4 c 5 d 6
Answer
3 a b 2 c 2 d 3 3 a c−−−√3 a b 2 c 2 d 3 3 a c
Practice Problem 9.3.8 9.3.8
180 m 4 n 15 9 a−12)15−−−−−−−−−−−−−−−−√180 m 4 n 15 9 a−12)15
Answer
6 m 2 n 7(a−12)7 5 n(a−12)−−−−−−−−−√6 m 2 n 7(a−12)7 5 n(a−12)
Square Roots Involving Fractions
A square root expression is in simplified form if there are
no perfect squares in the radicand,
no fractions in the radicand, or
no square root expressions in the denominator.
The square root expressions 5 a−−√,4 3 x y−−−√5 5 a,4 3 x y 5, and 11 m 2 n a−4−−−−√2 x 2 11 m 2 n a−4 2 x 2 are in simplified form
The square root expressions 3 x 8−−−√,4 a 4 b 3 5−−−−−√3 x 8,4 a 4 b 3 5, and 2 y 3 x−−√2 y 3 x are not in simplified form.
Simplifying Square Roots with Fractions
To simplify the square root expression x y−−√x y,
Write the expression as x−−√y√x y using the rule x y−−√=x−−√y√x y=x y.
Multiply the fraction by 1 in the form of y√y√y y.
Simplify the remaining fraction, x y−−√y x y y.
Rationalizing the Denominator
The process involved in step 2 is called rationalizing the denominator. This process removes square root expressions from the denominator using the fact that (y√)(y√)=y(y)(y)=y.
Sample Set B
Simplify each square root.
Example 9.3.7 9.3.7
9 25−−−√=9–√25−−√=3 5 9 25=9 25=3 5
Example 9.3.8 9.3.8
3 5−−√=3–√5–√=3–√5–√⋅5–√5–√=15−−√5 3 5=3 5=3 5⋅5 5=15 5
Example 9.3.9 9.3.9
9 8−−√=9–√8–√=9–√8–√⋅8–√8–√=3 8–√8=3 4⋅2−−−√8=3 4–√2–√8=3⋅2 2–√8=3 2–√4 9 8=9 8=9 8⋅8 8=3 8 8=3 4⋅2 8=3 4 2 8=3⋅2 2 8=3 2 4
Example 9.3.10 9.3.10
k 2 m 3−−−√=k 2−−√m 3−−−√=k m 3−−−√=k m 2 m−−−−√=k m 2 m−−√−−−−−−√=k m m−−√=k m m−−√⋅m−−√m−−√=k m−−√m m−−√m−−√=k m−−√m⋅m=k m−−√m 2 k 2 m 3=k 2 m 3=k m 3=k m 2 m=k m 2 m=k m m=k m m⋅m m=k m m m m=k m m⋅m=k m m 2
Example 9.3.11 9.3.11
x 2−8 x+16−−−−−−−−−−√=(x−4)2−−−−−−−√=x−4 x 2−8 x+16=(x−4)2=x−4
Practice Set B
Simplify each square root.
Practice Problem 9.3.9 9.3.9
81 25−−−√81 25
Answer
9 5 9 5
Practice Problem 9.3.10 9.3.10
2 7−−√2 7
Answer
14−−√7 14 7
Practice Problem 9.3.11 9.3.11
4 5−−√4 5
Answer
2 5–√5 2 5 5
Practice Problem 9.3.12 9.3.12
10 4−−−√10 4
Answer
10−−√2 10 2
Practice Problem 9.3.13 9.3.13
9 4−−√9 4
Answer
3 2 3 2
Practice Problem 9.3.14 9.3.14
a 3 6−−−√a 3 6
Answer
a 6 a−−√6 a 6 a 6
Practice Problem 9.3.15 9.3.15
y 4 x 3−−−√y 4 x 3
Answer
y 2 x−−√x 2 y 2 x x 2
Practice Problem 9.3.16 9.3.16
32 a 5 b 7−−−−−√32 a 5 b 7
Answer
4 a 2 2 a b−−−√b 4 4 a 2 2 a b b 4
Practice Problem 9.3.17 9.3.17
(x+9)2−−−−−−−√(x+9)2
Answer
x+9 x+9
Practice Problem 9.3.18 9.3.18
x 2+14 x+49−−−−−−−−−−−√x 2+14 x+49
Answer
x+7 x+7
Exercises
For the following problems, simplify each of the radical expressions.
Exercise 9.3.1 9.3.1
8 b 2−−−√8 b 2
Answer
2 b 2–√2 b 2
Exercise 9.3.2 9.3.2
20 a 2−−−−√20 a 2
Exercise 9.3.3 9.3.3
24 x 4−−−−√24 x 4
Answer
2 x 2 6–√2 x 2 6
Exercise 9.3.4 9.3.4
27 y 6−−−−√27 y 6
Exercise 9.3.5 9.3.5
a 5−−√a 5
Answer
a 2 a−−√a 2 a
Exercise 9.3.6 9.3.6
m 7−−−√m 7
Exercise 9.3.7 9.3.7
x 11−−−√x 11
Answer
x 5 x−−√x 5 x
Exercise 9.3.8 9.3.8
y 17−−−√y 17
Exercise 9.3.9 9.3.9
36 n 9−−−−√36 n 9
Answer
6 n 4 n−−√6 n 4 n
Exercise 9.3.10 9.3.10
49 x 13−−−−−√49 x 13
Exercise 9.3.11 9.3.11
100 x 5 y 11−−−−−−−√100 x 5 y 11
Answer
10 x 2 y 5 x y−−√10 x 2 y 5 x y
Exercise 9.3.12 9.3.12
64 a 7 b 3−−−−−√64 a 7 b 3
Exercise 9.3.13 9.3.13
5 16 m 6 n 7−−−−−−√5 16 m 6 n 7
Answer
20 m 3 n 3 n−−√20 m 3 n 3 n
Exercise 9.3.14 9.3.14
8 9 a 4 b 11−−−−−√8 9 a 4 b 11
Exercise 9.3.15 9.3.15
3 16 x 3−−−−√3 16 x 3
Answer
12 x x−−√12 x x
Exercise 9.3.16 9.3.16
8 25 y 3−−−−√8 25 y 3
Exercise 9.3.17 9.3.17
12 a 4−−−−√12 a 4
Answer
2 a 2 3–√2 a 2 3
Exercise 9.3.18 9.3.18
32 x 7−−−−√32 x 7
Answer
4 x 3 2 x−−√4 x 3 2 x
Exercise 9.3.19 9.3.19
12 y 13−−−−−√12 y 13
Exercise 9.3.20 9.3.20
50 a 3 b 9−−−−−√50 a 3 b 9
Answer
5 a b 4 2 a b−−−√5 a b 4 2 a b
Exercise 9.3.21 9.3.21
48 p 11 q 5−−−−−−√48 p 11 q 5
Exercise 9.3.22 9.3.22
4 18 a 5 b 17−−−−−−√4 18 a 5 b 17
Answer
12 a 2 b 8 2 a b−−−√12 a 2 b 8 2 a b
Exercise 9.3.23 9.3.23
8 108 x 21 y 3−−−−−−−√8 108 x 21 y 3
Exercise 9.3.24 9.3.24
−4 75 a 4 b 6−−−−−√−4 75 a 4 b 6
Answer
−20 a 2 b 3 3–√−20 a 2 b 3 3
Exercise 9.3.25 9.3.25
−6 72 x 2 y 4 z 10−−−−−−−−√−6 72 x 2 y 4 z 10
Exercise 9.3.26 9.3.26
−b 12−−−√−b 12
Answer
−b 6−b 6
Exercise 9.3.27 9.3.27
−c 18−−−√−c 18
Exercise 9.3.28 9.3.28
a 2 b 2 c 2−−−−−√a 2 b 2 c 2
Answer
a b c a b c
Exercise 9.3.29 9.3.29
4 x 2 y 2 z 2−−−−−−−√4 x 2 y 2 z 2
Exercise 9.3.30 9.3.30
−9 a 2 b 3−−−−−√−9 a 2 b 3
Answer
−3 a b b√−3 a b b
Exercise 9.3.31 9.3.31
−16 x 4 y 5−−−−−−√−16 x 4 y 5
Exercise 9.3.32 9.3.32
m 6 n 8 p 12 q 20−−−−−−−−−√m 6 n 8 p 12 q 20
Answer
m 3 n 4 p 6 q 10 m 3 n 4 p 6 q 10
Exercise 9.3.33 9.3.33
r 2−−√r 2
Exercise 9.3.34 9.3.34
p 2−−√p 2
Answer
p p
Exercise 9.3.35 9.3.35
1 4−−√1 4
Exercise 9.3.36 9.3.36
1 16−−−√1 16
Answer
1 4 1 4
Exercise 9.3.37 9.3.37
4 25−−−√4 25
Exercise 9.3.38 9.3.38
9 49−−−√9 49
Answer
3 7 3 7
Exercise 9.3.39 9.3.39
5 8–√3–√5 8 3
Exercise 9.3.40 9.3.40
2 32−−√3–√2 32 3
Answer
8 6–√3 8 6 3
Exercise 9.3.41 9.3.41
5 6−−√5 6
Exercise 9.3.42 9.3.42
2 7−−√2 7
Answer
14−−√7 14 7
Exercise 9.3.43 9.3.43
3 10−−−√3 10
Exercise 9.3.44 9.3.44
4 3−−√4 3
Answer
2 3–√3 2 3 3
Exercise 9.3.45 9.3.45
−2 5−−√−2 5
Exercise 9.3.46 9.3.46
−3 10−−−√−3 10
Answer
−30−−√10−30 10
Exercise 9.3.47 9.3.47
16 a 2 5−−−−−√16 a 2 5
Exercise 9.3.48 9.3.48
24 a 5 7−−−−−√24 a 5 7
Answer
2 a 2 42 a−−−√7 2 a 2 42 a 7
Exercise 9.3.49 9.3.49
72 x 2 y 3 5−−−−−−√72 x 2 y 3 5
Exercise 9.3.50 9.3.50
2 a−−√2 a
Answer
2 a−−√a 2 a a
Exercise 9.3.51 9.3.51
5 b−−√5 b
Exercise 9.3.52 9.3.52
6 x 3−−−√6 x 3
Answer
6 x−−√x 2 6 x x 2
Exercise 9.3.53 9.3.53
12 y 5−−−√12 y 5
Exercise 9.3.54 9.3.54
49 x 2 y 5 z 9 25 a 3 b 11−−−−−−−−√49 x 2 y 5 z 9 25 a 3 b 11
Answer
7 x y 2 z 4 a b y z−−−−√5 a 2 b 6 7 x y 2 z 4 a b y z 5 a 2 b 6
Exercise 9.3.55 9.3.55
27 x 6 y 15 3 3 x 3 y 5−−−−−−−√27 x 6 y 15 3 3 x 3 y 5
Exercise 9.3.56 9.3.56
(b+2)4−−−−−−√(b+2)4
Answer
(b+2)2(b+2)2
Exercise 9.3.57 9.3.57
(a−7)8−−−−−−−√(a−7)8
Exercise 9.3.58 9.3.58
(x+2)6−−−−−−−√(x+2)6
Answer
(x+2)3(x+2)3
Exercise 9.3.59 9.3.59
(x+2)2(x+1)2−−−−−−−−−−−−−√(x+2)2(x+1)2
Exercise 9.3.60 9.3.60
(a−3)4(a−1)2−−−−−−−−−−−−−√(a−3)4(a−1)2
Answer
(a−3)2(a−1)(a−3)2(a−1)
Exercise 9.3.61 9.3.61
(b+7)8(b−7)6−−−−−−−−−−−−√(b+7)8(b−7)6
Exercise 9.3.62 9.3.62
a 2−10 a+25−−−−−−−−−−−√a 2−10 a+25
Answer
(a−5)(a−5)
Exercise 9.3.63 9.3.63
b 2+6 b+9−−−−−−−−−√b 2+6 b+9
Exercise 9.3.64 9.3.64
(a 2−2 a+1)4−−−−−−−−−−−√(a 2−2 a+1)4
Answer
(a−1)4(a−1)4
Exercise 9.3.65 9.3.65
(x 2+2 x+1)12−−−−−−−−−−−−√(x 2+2 x+1)12
Exercises For Review
Exercise 9.3.66 9.3.66
Solve the inequality 3(a+2)≤2(3 a+4)3(a+2)≤2(3 a+4)
Answer
a≥−2 3 a≥−2 3
Exercise 9.3.67 9.3.67
Graph the inequality 6 x≤5(x+1)−6 6 x≤5(x+1)−6
Exercise 9.3.68 9.3.68
Supply the missing words. When looking at a graph from left-to-right, lines with _ slope rise, while lines with ____ slope fall.
Answer
positive; negative
Exercise 9.3.69 9.3.69
Simplify the complex fraction: 5+1 x 5−1 x 5+1 x 5−1 x
Exercise 9.3.70 9.3.70
Simplify 121 x 4 w 6 z 8−−−−−−−−−√121 x 4 w 6 z 8 by removing the radical sign.
Answer
11 x 2 w 3 z 4 11 x 2 w 3 z 4
This page titled 9.3: Simplifying Square Root Expressions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Denny Burzynski & Wade Ellis, Jr. (OpenStax CNX) .
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Geometry
Arkansas Math
Algebra I Mathematics Standards: Exponential Functions & Equations
Create & Solve - Students create and solve problems that model exponential relationships.
A1.EFE.1
Fully covered
Represent and solve real-world problems, using exponential equations in one variable.
Evaluate logarithms: change of base rule
Evaluating logarithms: change of base rule
Exponential model word problem: bacteria growth
Exponential model word problem: medication dissolve
Exponential model word problems
Logarithm change of base rule intro
Solve exponential equations using logarithms: base-10 and base-e
Solve exponential equations using logarithms: base-2 and other bases
Solving exponential equations using logarithms
Solving exponential equations using logarithms: base-10
Solving exponential equations using logarithms: base-2
A1.EFE.2
Mostly covered
Represent real-world problems (growth, decay, and compound interest), using exponential equations.
Evaluate logarithms: change of base rule
Evaluating logarithms: change of base rule
Exponential vs. linear models
Exponential decay intro
Exponential growth vs. decay
Exponential model word problem: bacteria growth
Exponential model word problem: medication dissolve
Exponential model word problems
Exponential vs. linear models: verbal
Exponential vs. linear growth over time
Graphing exponential growth & decay
Graphing exponential growth & decay
Graphs of exponential growth
Graphs of exponential growth
Linear vs. exponential growth: from data
Linear vs. exponential growth: from data
Linear vs. exponential growth: from data (example 2)
Logarithm change of base rule intro
Modeling with basic exponential functions word problem
Solve exponential equations using logarithms: base-10 and base-e
Solve exponential equations using logarithms: base-2 and other bases
Solving exponential equations using logarithms
Solving exponential equations using logarithms: base-10
Solving exponential equations using logarithms: base-2
Warmup: exponential vs. linear growth
A1.EFE.3
Fully covered
Construct exponential equations from geometric sequences with and without context.
Evaluate logarithms: change of base rule
Evaluating logarithms: change of base rule
Exponential model word problem: bacteria growth
Exponential model word problem: medication dissolve
Exponential model word problems
Logarithm change of base rule intro
Solve exponential equations using logarithms: base-10 and base-e
Solve exponential equations using logarithms: base-2 and other bases
Solving exponential equations using logarithms
Solving exponential equations using logarithms: base-10
Solving exponential equations using logarithms: base-2
Interpret Key Features - Students interpret key features of equations that model exponential relationships.
A1.EFE.4
Not covered
Determine the domain and range of exponential functions in mathematical problems.
(Content unavailable)
A1.EFE.5
Not covered
Determine reasonable domain and range values of exponential functions representing real-world situations, both continuous and discrete; interpret the solution as reasonable or unreasonable in context.
(Content unavailable)
A1.EFE.6
Mostly covered
Interpret the key features of an exponential function that models a relationship between two quantities in a given context.
Comparing features of quadratic functions
Comparing linear functions word problem: climb
Comparing linear functions word problems
Comparing maximum points of quadratic functions
Equivalent forms of exponential expressions
Equivalent forms of exponential expressions
Exponential vs. linear growth over time
Interpret change in exponential models
Interpret change in exponential models: changing units
Interpret change in exponential models: with manipulation
Interpret exponential expressions word problems
Interpret quadratic models
Interpret quadratic models: Factored form
Interpret quadratic models: Vertex form
Interpret time in exponential models
Interpreting change in exponential models
Interpreting change in exponential models: changing units
Interpreting change in exponential models: with manipulation
Interpreting time in exponential models
Linear models word problems
Rewrite exponential expressions
Rewriting exponential expressions as A⋅Bᵗ
Worked example: domain & range of piecewise linear functions
Worked examples: Forms & features of quadratic functions
A1.EFE.7
Partially covered
Flexibly use different representations of an exponential function, including graphs, tables, and equations.
Equivalent forms of exponential expressions
Equivalent forms of exponential expressions
Exponential function graph
Graphing exponential growth & decay
Graphing exponential growth & decay
Graphs of exponential growth
Graphs of exponential growth
Interpret change in exponential models
Interpret change in exponential models: changing units
Interpret change in exponential models: with manipulation
Interpret time in exponential models
Interpreting change in exponential models
Interpreting change in exponential models: changing units
Interpreting change in exponential models: with manipulation
Interpreting time in exponential models
Intro to exponential functions
Rewrite exponential expressions
Rewriting exponential expressions as A⋅Bᵗ
A1.EFE.8
Fully covered
Interpret the quantities in an exponential equation in the context of a real-world problem, including growth, decay, and compound interest.
Comparing features of quadratic functions
Comparing linear functions word problem: climb
Comparing linear functions word problems
Comparing maximum points of quadratic functions
Exponential vs. linear growth over time
Interpret exponential expressions word problems
Interpret quadratic models
Interpret quadratic models: Factored form
Interpret quadratic models: Vertex form
Linear models word problems
Worked example: domain & range of piecewise linear functions
Worked examples: Forms & features of quadratic functions
Graphing - Students graph exponential functions.
A1.EFE.9
Fully covered
Graph exponential functions that model real-world problems (growth, decay, and compound interest), showing key attributes.
Exponential function graph
Graphing exponential growth & decay
Graphing exponential growth & decay
Graphs of exponential growth
Graphs of exponential growth
Intro to exponential functions
Statistical Relationships - Students explore exponential statistical relationships.
A1.EFE.10
Not covered
Write exponential functions that provide a reasonable fit to data and use them to make predictions with technology.
(Content unavailable)
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188747 | https://ca.practicallaw.thomsonreuters.com/7-569-8085?transitionType=Default&contextData=(sc.Default) | Ultra Vires | Practical Law
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An act of a corporation is ultra vires when the corporation acts beyond the scope of the powers and purposes provided to it by its statute of incorporation. Ultra vires acts performed by a corporation are void (see Communities Economic Development Fund v. Canadian Pickles Corp., 1991 CarswellMan 402 (S.C.C.)).
Over time, the doctrine of ultra vires, as applied to corporations, has become rare because of the adoption of specific statutory provisions (such as Canada Business Corporations Act, R.S.C. 1985, c. C-44, section 15; Ontario Business Corporations Act, R.S.O. 1990, c. B.16, section 15; and Canada Not-for-profit Corporations Act, S.C. 2009, c. 23, section 16, which give corporations incorporated or continued under these statutes all the rights, powers and capacities of a natural person (that is, an adult of full legal capacity)). Also, the ultra vires doctrine has no application to corporations incorporated by letters patent (see Bonanza Creek Gold Mining Co. v. R., 1916 CarswellQue 1 (Jud. Com. of Privy Coun.)).
Today, the ultra vires doctrine is largely confined to a various statutory or Special Act corporations.
For information on forming and organizing a corporation, see the Corporate Organization and Maintenance Toolkit.
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188748 | https://www.uvm.edu/~dstratto/bcor102/readings/4_Evol_of_Phenotypes.pdf | Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 1 Evolution of Phenotypes [Background section about the Grant’s study of finches on the Galapagos] [see Beak of the Finch].. Daphne Island in the Galapagos Four species of Galapagos finches Peter Grant with finches Figure 4.1 photos: Up to this point we have been talking about the evolutionary dynamics of individual genes. Many traits of ecological importance are determined by many genes. Those might include body size and shape, behavior, longevity, etc. Indeed for those kinds of traits we usually have no idea what the underlying genes are. Nevertheless those complex traits are precisely the traits that are most often cited as the spectacular examples of natural selection and adaptation. How can we study knowing the individual genes? Darwin and Wallace presented their theory of evolution by natural selection before scientists even knew that genes existed. All that is required is that traits are somehow inherited from parents to offspring. So, even without knowing precisely why, they could look at the results from plant and animal breeders to see how traits can be modified from one generation to the next. Darwin and Wallace’s theory of evolution by natural selection can be summarized by four steps: 1. There is variation in some trait of interest. 2. Individuals with traits that best match their particular environment will on average leave more descendants than others, either because they have higher survival or higher reproductive output. 3. Some of that variation is heritable, so offspring tend to resemble their parents. 4. The result is that the population mean will shift toward the better-adapted phenotype. Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 2 4.1 A simple model of selection: |----------------------------------------- One complete generation -------------------------------------| Population mean at the start of the parent generation (before selection) Selection ! Mean size of the surviving parents Inheritance ! Population mean at the start of the offspring generation Only the largest survive Offspring resemble the parents that survive to reproduce In the case of the finches, the drought causes only some of the birds to survive to reproduction, so the average beak size of the parents is slightly larger than the average beak size at the beginning of the generation (because many of the birds with small beaks died). We call that difference in size between the actual parents and the original population the “selection differential”. Those parents then produce offspring. Normally the offspring will resemble their parents somewhat, but the resemblance is not perfect. Only to the extent that the beak size is caused by additive effects of alleles, will it can be inherited. Thus there will be a scaling coefficient (that we will call the “heritability”) that reflects the degree to which offspring resemble their parents. The overall change in phenotype as a result of selection will be a change due to the differential survival of parents times the degree to which that change is inherited. That change in population mean from one generation to the next is called the selection "response". 4.2 Selection: The drought of 1977 caused massive mortality among the finches. The limited seed supply was soon depleted, leaving only the large seeds that were difficult for small-beaked birds to crack. The result was that birds with larger beaks survived at a higher rate than birds with small beaks. The average beak depth of survivors was 9.84 mm, compared to 9.31 mm in the general population before selection. Nevertheless some of the small birds did survive, and some of the birds with the very largest beaks did not. Figure 4.2 Distribution of beak sizes before and after the drought of 1977. Data from Grant 1986. Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 3 Differential survival of individuals represents the action of natural selection favoring birds with large beaks, but that is not evolution. Evolution requires a change in the average beak size from one generation to the next. How will the increase in the number of large-beaked survivors show up in the offspring generation? For that we also need to consider the heritability of beak depth. 4.3 Heritability One of the other researchers on the finch team, graduate student Peter Boag, studied the inheritance of beak depth in Galapagos finches by looking at the relationship between parent beak depth and that of their offspring. He collected two sets of parent offspring data, once in 1976 and again in 1978. For both years he followed birds to determine which pairs belonged to which nests. Most of the parents had been previously captured so their beak depths were known. He then captured the offspring when they fledged and measured their beak depth. He calculated the "midparent" beak depth (the average beak depth of the two parents) and then compared that to beak depth of their offspring. Here is what he found: 1976 1978 "Midparent" Average beak depth of the two parents Offspring beak depth Midparent beak depth Offspring beak depth 8.2 8.0 8.9 8.9 8.3 7.8 9.0 9.4 8.4 8.2 9.6 9.2 9.3 7.8 9.1 9.5 8.9 8.3 9.5 9.5 8.8 8.6 9.8 9.5 9.9 8.7 10.1 9.6 10.9 8.8 9.5 10.0 9.1 8.9 9.6 9.9 9.1 9.0 9.9 9.8 9.3 9.2 10.0 10.0 9.2 9.5 10.2 10.2 9.7 9.2 10.1 10.4 9.4 9.7 10.4 10.2 9.7 9.7 10.6 10.4 9.9 9.6 10.6 10.5 10.4 9.8 10.3 10.9 10.0 9.9 10.2 10.1 10.7 10.9 Data from Boag 1983, Figure 1 Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 4 Figure 4.3 The heritability of beak depth in 1976. Data from Boag 1983. How can we use this information? The parent-offspring regression shows the way beak depth is (on average) inherited across generations. In particular, the heritability of a trait is the slope of thee regression line of the midparent-offspring regression. It allows us to predict how a particular change in phenotype among the parents will be translated into a response among the offspring. For example, if the average beak depth of the parents increases from 10.0 mm to 10.5 mm, we can use the parent-offspring regression to see how much of a change we would expect in the offspring generation. In 1976, the best-fit slope of offspring beak size on mid-parent beak size is 0.78. That means that the heritability of this trait is 0.78. Notice that the slope of this line is less than 1.0. For each unit increase in the size of the parents there is a slightly lower gain in the offspring. Why? Because the parents beak depth is determined in part by genetics and in part by the environment and only the genetic component of that variation can be inherited. Therefore the heritability will always be less than 1.0. If the phenotypic variation is determined strictly by their environment then there will be no necessary relationship between the parent and offspring phenotype and the slope of the line (heritability) will be 0. That leads to a second definition of heritability: heritability is the proportion of the total phenotypic variance that is attributable to additive genetic variance. Optional: Repeat the analysis for the 1978 parent offspring data by graphing the offspring vs. midparent beak sizes and finding the slope of the best-fit line. What is the heritability for beak depth in 1978? __ Are those data consistent with the heritability estimate from 1976? __ Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 5 4.4 Response to selection Following the drought, the few surviving birds bred and the Grant's and their coworkers were there to measure the beak size of the offspring they produced. There was still a lot of variation in beak size, but on average, the size had increased. Beak size had evolved. Figure 3. Evolutionary response, after the drought 4.5 The “breeders equation": The theoretical understanding of the evolution of polygenic traits was actually first worked out by plant and animal breeders trying to predict the yield increases for various breeding programs. The developed the "breeders equation" that still forms the basis of our understanding of quantitative trait evolution: ! R = h2S eq. 4.1 In words, that says that the response to selection (R) is equal to heritability (h2) times the selection differential (S). The selection differential (S) is just the difference between the mean of the population and the mean of the individuals that reproduce. If selection is weak, then the parents who actually breed will have a similar mean to the mean of the overall population. S will be small. The response to selection (R) is just the difference between the mean of the parents before selection and the mean of the offspring produced to start the next generation. Sometimes, we will use the variable z to indicate the value of a particular trait. In that case ! R = "z or the change in the trait mean across generations. Finally the heritability (h2). is a scaling factor that translates the strength of selection into a realized response1. As we'll see below, the heritability can be defined by either the slope of the regression relating midparent to offspring, or by the ratio of additive genetic variance to total phenotypic variance. 1 You may wonder why the symbol h2 is used for the heritability. The practice dates back to Sewall Wright’s original derivation, where h was a ratio of standard deviations. Therefore, the ratio of variances (Va/Vp) must be h2. It is important to think of h2 as a single symbol, not hh. Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 6 The breeder's equation (eq 1) is simple, but it is very powerful because it allows a quantitative prediction of the response to selection from a couple of easily measured parameters. Use the breeder's equation to complete the following table: What is the selection differential? What is the predicted response? How does that compare to the observed response? Average Beak Depth at start of 1976 (before selection) 9.31 mm Beak Depth of survivors, who lived to breed in 1978 9.84 mm Selection Differential Heritability 0.78 Predicted response to selection Predicted mean of the offspring born 1978 Average Beak Depth of the offspring born in 1978 9.70 mm 4.6 More about the genetic basis of quantitative traits Natural selection acts on the variation that is present in the population. The total phenotypic variance can be partitioned into a genetic and an environmental component. In turn, the genetic variance is composed an additive genetic component (caused by the average effects of alleles) and a non-additive or dominance component, caused by particular combinations of alleles. Parents only pass a single allele on to their offspring, so only the additive or average effects of the alleles can be inherited. Effects due to particular combinations of alleles (dominance) are broken up each generation. Therefore the evolutionary response to selection will depend upon the magnitude of the additive component of genetic variance, symbolized VA. The ratio of additive genetic variance to total phenotypic variance is another way to define the heritability2. ! h2 = VA VP eq 4.2 2 Although we won't derive it, this definition is mathematically equivalent to the offspring-midparent regression. Total Phenotypic Variance (Vp) Genetic Variance Environmental Variance Additive Component (VA) Dominance Component (VD) VE Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 7 If we wanted, we could also partition the environmental component into effects of the rearing environment, developmental defects, nutritional status, etc. etc. 4.7 More about selection To see the pattern of survival more clearly, it helps to combine the data into larger size categories so some of the random variation is averaged out. What proportion of each size class survived the drought? Size class N before selection N Survivors Survival rate Less than 7 21 1 7.0 to 7.9 34 2 8.0 to 8.9 186 10 9.0 to 9.9 294 25 10.0 to 10.9 186 43 11.0 to 11.9 25 9 Greater than 12 5 0 Use those data to graph fitness (survival) vs. beak size. How do you explain the low survival of the birds with the very largest beaks? Is it just chance that none of those 5 survived? How could you tell? Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 8 This graph is called the "fitness function". Fitness functions describe the way fitness relates to a particular trait, in this case the relationship between survival during drought and beak depth. Fitness functions with positive slopes mean that selection favors an increase in the trait. Fitness functions with negative slopes mean that selection favors a decrease in the trait. Curved fitness functions indicate stabilizing selection of disruptive selection (Fig 4.4). Figure 4.4. Fitness functions and their effects on phenotypes. The solid line shows the frequency distribution before selection and the dashed line is the distribution after selection. Directional selection causes a change in the mean value of a trait. Stabilizing selection causes a decrease in the variance of the trait, with no change in mean. Disruptive selection causes an increase in the variance of the trait. 4.8 More about heritability. When only one of the two parents is measured, the slope of the regression estimates only ! 1 2 h2. Therefore you need to double the slope of the mother-offspring or father-offspring regression to obtain the heritability. For example, Boag compared his midparent-offspring regression to the father-offspring and mother-offspring regressions. Notice how the regressions on single parents both have much lower slope than the midparent regression. What is the estimate of heritability for each of those regressions? Bill depth Midparent-Offspring Father-Offspring Mother Offspring Regression slope 0.72 0.22 0.39 Heritability Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 9 Why do you think the estimate of heritability using mothers is so much higher than the estimate using the father-offspring regression? One of the assumptions of using parent-offspring regressions to estimate the heritability is that the only source of resemblance is shared genes. In many organisms it is common to see "maternal effects". For example, well fed, females may lay large eggs, which produce large offspring. That produces a resemblance between mothers and offspring that is strictly environmental. That kind of environmental correlation is less common through males, who generally provide only sperm. Second, the fathers at the nest aren't always the birds that actually sired the offspring. In G. fortis approximately 20% of the offspring are produced through extra-pair matings. DNA samples revealed that 44 of 223 offspring did not match their putative father's genotype and must have been sired by another male. If the non-genetic fathers are excluded from the father-offpsring regression, then the slope increases to b=0.36. 4.8.1 Some common misconceptions about heritability: • Heritability of zero does not mean the trait is not determined by genes. It means that the variation in the trait does not have a genetic basis. Some traits that are universal (e.g. presence of two eyes in humans) are still determined by our genes, but there is no variation in that trait so the heritability would be zero. • Environmental effects can still cause differences between populations, even if the heritability is not zero. • Two populations with the same genetic variance may have very different heritabilities if one is subject to more environmental variance than the other. Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 10 4.9 EXTRA: The relationship between selection on single loci and the breeders equation: To understand the genetic basis of quantitative traits, it is important to think about the effect of a particular allele, not simply its presence or absence. A single locus can produce three discrete phenotypes, but as more and more loci contribute to a trait the phenotypic distribution comes closer and closer to a normal (bell shaped) distribution. Any normal distribution can be described by the mean and variance. The breeder's equation uses those aggregate measures of the trait to predict the response to selection without explicitly following individual alleles. Nevertheless it is important to realize that at each of the underlying loci selection will change allele frequencies following exactly the same model that we used for other Mendelian loci. Let's imagine that we are studying the size of some organism. To keep things simple, we will assume that size is controlled by a large number of independent genes, each with equal effect. We'll assume that there are two alleles, "big" and "small". The "big" allele at each locus adds 1/2 a unit of size and the "small" allele subtracts 1/2 unit. Therefore the net effect on size for each locus will be the sum of the values of the two alleles: the net effect will be -1 if both copies are the "small" allele, 0 if it is heterozygous, and +1 for the other homozygote where there are two doses of the "big" allele. Ultimately, the trait (in this case size) will be determined by the total dosage of "big" alleles summed over all loci. At each locus the additive genetic variance will be (assuming no dominance) ! VA = 2pqa2 eq. 4.3 What this equation makes clear is that the additive genetic variance depends on the allele frequencies (p and q) as well as the squared value of the alleles (a2). If one of the alleles goes to fixation (p=0 or q=0) then there will be no genetic variation at this locus. Why are we only concerned with the additive component of genetic variance? Example: Imagine that a particular locus has two alleles B and b. Each copy of the B allele increases the size of an organism by 2 mm (a=2) and there is no dominance. Large individuals have a fitness advantage, so allele B gradually increases to fixation. What happens to the genetic variance as B increases in frequency? 1 gene Sum of 2 genes Sum of 4 genes Case Studies in Evolutionary Ecology DRAFT (notes) Don Stratton 2008 11 Frequency of B allele VA 0.50 2.0 0.75 0.95 1.00 This represents a general phenomenon that directional selection acts to decrease the genetic variance for a trait. After many generations of selection, if there is no new source of genetic variation by mutation, the additive variance will decline to zero and there will no longer be any response to selection. A corollary to this is that for traits that under strong selection, there should be little genetic variation remaining in the population. p=0.50 p=0.75 p=0.95 (parameters: a=2, d=0, Ve=0.5) As allele B goes to fixation, the genetic variation contributed by this locus (dashed line) declines. Note that there is some variation in the phenotype of each of the three genotypes, because of environmental variation. |
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MINI REVIEW article
Front. Endocrinol., 11 May 2022
Sec. Reproduction
Volume 13 - 2022 |
This article is part of the Research TopicEndocrine and Paracrine Regulation of Spermatogenesis - A Collection of Up to Date Research Contributions on Testis Formation and FunctionView all 25 articles
Roles of Spermatogonial Stem Cells in Spermatogenesis and Fertility Restoration
Lei Diao1Paul J. Turek2Constance M. John3Fang Fang1Renee A. Reijo Pera4,5
1The First Affiliated Hospital of USTC, Division of Life Sciences and Medicine, University of Science and Technology of China, Hefei, China
2Turek Clinic, San Francisco, CA, United States
3MandelMed, Inc., San Francisco, CA, United States
4McLaughlin Research Institute, Touro College of Osteopathic Medicine – Montana (TouroCOM-MT), Great Falls, MT, United States
5Research Division, Touro College of Osteopathic Medicine – Montana (TouroCOM-MT), Great Falls, MT, United States
Spermatogonial stem cells (SSCs) are a group of adult stem cells in the testis that serve as the foundation of continuous spermatogenesis and male fertility. SSCs are capable of self-renewal to maintain the stability of the stem cell pool and differentiation to produce mature spermatozoa. Dysfunction of SSCs leads to male infertility. Therefore, dissection of the regulatory network of SSCs is of great significance in understanding the fundamental molecular mechanisms of spermatogonial stem cell function in spermatogenesis and the pathogenesis of male infertility. Furthermore, a better understanding of SSC biology will allow us to culture and differentiate SSCs in vitro, which may provide novel stem cell-based therapy for assisted reproduction. This review summarizes the latest research progress on the regulation of SSCs, and the potential application of SSCs for fertility restoration through in vivo and in vitro spermatogenesis. We anticipate that the knowledge gained will advance the application of SSCs to improve male fertility. Furthermore, in vitro spermatogenesis from SSCs sets the stage for the production of SSCs from induced pluripotent stem cells (iPSCs) and subsequent spermatogenesis.
Introduction
Early in human development, a small group of cells is set aside or allocated to become the germ cells that give rise to the sperm and oocytes that will transmit genetic and epigenetic information to subsequent generations (1). In males, the process of spermatogenesis maintains the production of spermatozoa, the final cell carrier of inheritable material, throughout the lifetime of male mammals (2). Continuous spermatogenesis depends on the appropriate self-renewal and differentiation of spermatogonial stem cells (SSCs) throughout the life of the male (3). The SSCs are the resident stem cell population that resides at the basal membrane of seminiferous tubules of the testis (4, 5). The SSCs can undergo mitotic divisions for self renewal to maintain a steady stem cell pool or they can differentiate through sequential and extensive processes into spermatozoa (6). The balance of self-renewal and differentiation of SSCs is critical, not only for maintaining normal spermatogenesis but also for sustaining lifelong fertility (7). A tilt to self-renewal is a risk factor for germ cell tumors, while a tilt towards differentiation results in exhaustion of germ cell pools, leading to male infertility (8). Numerous studies have demonstrated that the balance between self-renewal and differentiation is precisely controlled by a combination of intrinsic genetic and epigenetic factors within SSCs as well as the extrinsic signals that eminate from the somatic niche (9, 10).
Significantly, SSCs have extraordinary therapeutic potential in assisted reproduction for male infertility (11, 12). Transplantation of SSCs can restore spermatogenesis in patients who suffer from impaired spermatogenesis (13). One application example is fertility preservation of prepubertal boys with cancer and undergoing chemotherapy (14). SSCs can be isolated from testicular biopsy and cryopreserved before chemotherapy, followed by stem cell transplantation into the seminiferous tubules to restore fertility (15, 16). In addition, germline gene therapy using SSCs has been proposed, albeit with obvious concerns regarding legitimate ethical issues, as a promising and feasible approach to treat endocrine disease and metabolic disorders with germline gene mutations (17). Currently, the major hurdle to the use of SSCs in assisted reproductive technology is the difficulty of identificating and isolating endogenous SSCs and directing their differentiation to haploid cells in vitro.
This review provides a brief overview summary of some of the existing knowledge and research progress regarding use of SSCs for inducing spermatogenesis in vivo and in vitro for fertililty restoration. We hope that this summary review may spur further inquiries into details and ongoing studies of practical applications of SSCs in human reproduction and regenerative medicine.
Regulation of SSCS
Human germ cell development begins with the specification of a small group of cells to form the primordial germ cells (PGCs) (18), which are thought to arise from the dorsal amnion at the onset of gastrulation (19). Following their specification, PGCs actively proliferate and migrate to the developing gonad (20–22) where they will occupy the genital ridge and undergo sex-determination by entering either male or female sex-specific developmental pathways (23). External signals from the somatic environment determine the sex of PGCs (24). For male germ cell development, once PGCs occupy the seminiferous tubules of the male gonad, they are termed gonocytes (25), which later interact with the niche cells to become spermatogonia (26). Note that nomenclature is not universal or all inclusive as subtypes exist (example: type A, type b, light and dark spermatogonia), different stages of development are sometimes indicated (examples: early or late spermagonia or undifferentiated and differentiating), or reference to marker content (example: c-kit+ spermatogonia).
The Niche
The architecture of the testes is characterized by two structurally distinct compartments (Figure 1), the seminiferous tubule and the interstitial tissue (27). Within the seminiferous tubule, Sertoli cells form a tight blood-testis barrier to divide the seminiferous epithelium into basal and luminal compartments (28). Developing spermatogonia reside on the basal membrane and are further defined by three types of cells: undifferentiated spermatogonia (quiescent SSCs), differentiating spermatogonia (SSCs that undergo active mitosis), and differentiated spermatogonia (29, 30). The Sertoli cells are the supporting cells for the germ cell population in the testes and are essential for maintaining normal spermatogenesis by providing the cellular matrix and by secreting specific growth factors (31). The surrounding interstitial space consists of various cell types that include the Leydig cells, mesenchymal cells, and immune cells, in addition to lymph vessels, nerve fibers, and connective tissues (27). Leydig cells produce the hormone testosterone and cytokines that may function both directly and indirectly to regulate self-renewal of SSCs (32).
FIGURE 1
Figure 1 Schematic diagram of the niche of SSCs and the regulatory factors involved in maintaining the stemness and self-renewal of SSCs. Undifferentiated SSCs are localized at the basement membrane. Germ cells maintain the close contact with the Sertoli cells inside the seminiferous epithelium. Peritubular myoid cells surround the seminiferous tubules to form testicular cords. The interstitial compartment consists of many somatic cell types including Leydig cells, mesenchymal cells and immune cells. Bioactive factors in the niche play crucial role in self-renewal and differentiation of SSCs. CXCL12/CXCR4, FGFs, and VEGFA act in synergy with GDNF to maintain SSCs. Retinoic acid (RA) induces the differentiation of SSCs by downregulation, at least in part, of GDNF expression and activation of SCF and BMP4. Transcription factors, PLZF and FOXO1, are involved in regulating SSCs maintenance and spermatogenesis by acting on a subset of downstream target gene. MicroRNAs, including miR-1908-3p, miR-112-5p and miR-31-5p, also act as critical regulators in spermatogenesis.
External and Intrinsic Factors
The fine-tuned balance between self-renewal and differentiation of SSCs is regulated by the interplay of extrinsic and intrinsic factors. GDNF, a growth factor produced by the somatic niche cells, is critical for the maintenance of SSCs both in vivo and in vitro (33). It regulates several essential downstream genes, including the germ cell specific and ubiquitously-expressed genes Nanos2, Etv5, Lhx1, T, Bcl6b, Id1, and Cxcr4, to promote SSC self-renewal and inhibit differentiation (34–39). CXCL12/CXCR4 (39), FGFs (33, 40), and VEGF-A (41) act in synergy with GDNF to maintain SSC stem cell status. In contrast, retinoic acid (RA), a hormone secreted primarily by Sertoli cells, plays an indispensable role in inducing differentiation of SSCs by downregulation of GDNF expression and activation of differentiation-promoting factors, such as BMP and SCF (42–45). Genetic ablation studies in mice indicate that several transcription factors are involved in regulating SSC maintenance and recruitment to spermatogenesis. The PLZF transcription factor is expressed by SSCs and interacts with GDNF signaling as one of the master regulators to promote the self-renewal of SSCs (46, 47). Loss of PLZF results in progressive germ cell loss, testicular hypoplasia, and infertility (46–48). One of the downstream targets of PLZF is the SALL4 protein, which is required for the self-renewal of SSCs and maintenance of ability to enter spermatogenic differentiation (49). A potential upstream regulator of PLZF is PRMT5. Disruption of the PRMT5 gene results in a dramatic reduction of PLZF gene expression, and subsequent progressive loss of SSCs leading to male infertility (50). Another transcription factor important for maintenance of SSC self-renewal is FOXO1, which regulates a number of genes that are preferably expressed in SSCs (51). Deletion of the FOXO1 gene results in defects in SSC maintenance and ultimately spermatogenic failure. In addition, recent research has identified numerous microRNAs as critical regulators in spermatogenesis. Some microRNAs regulate the self-renewal of SSCs. For example, miR-202 plays a crucial role in the maintenance of SSC stemness or self-renewal of the stem cell population (52). Other microRNAs, such as miR-1908-3p (53), miRNA-122-5p (54), and miRNA-31-5p (55), enhance the proliferation and inhibit the early apoptosis of human SSCs via targeting key downstream pathways. Conversely, several microRNAs facilitate differentiation via regulation of the expression of genes associated with SSC differentiation. MiR-34c promotes SSC differentiation by inhibiting the function of the NANOS2 gene, leading to the up-regulation of meiotic-related proteins, STRA8, in mice (56). Similarly, miR-486-5p secreted by Sertoli cells stimulates differentiation of SSCs in mice by up-regulating the expression of STRA8 and SYCP3 (57). Further, impaired spermatogenesis is observed in mice carrying a deficiency in miR-17-92 or a gene deletion of miR-17-92 (58, 59). miR-202 similarly regulates spermatogenesis via orchestration meiotic initiation by preventing precocious differentiation of mouse SSCs (52). Taken together, numerous genes act to balance self-renewal and differentiation of SSCs.
Fertility Restoration Through In Vivo Spermatogenesis
SSCs within the testicular tissues have the potential to complete the entire process of spermatogenesis in vivo and produce functional spermatozoa for fertility restoration (Figure 2). Thus, cryopreservation of testicular tissue prior to gonadotoxic treatment for prepubertal boys is proposed as a helpful strategy for fertility preservation (60). To restore fertility through in vivo spermatogenesis, testicular tissues could be either autotransplanted to the same individual or the tissues might be dissociated to obtain SSCs for autotransplantation. Xenotransplantation would carry the obvious complication of mixing of sperm from different individuals.
FIGURE 2
Figure 2 Schematic diagram of SSC-based fertility restoration in humans. A sample of testicular tissue of prepubertal boys, who receive gonadotoxic treatment, is retrieved and cryopreserved. Spermatogenesis may be induced after treatment either in vivo or in vitro.
Transplantation of Testicular Tissues
Autotransplantation of testicular tissues has achieved success in multiple animal models, which results in live offspring (61–65). However, the approach has the risk of re-introducing malignancy is a concern (66). Studies of xenotransplantation, which transplants immature testicular tissue under the back skin of immune-deficient animals, have been used to examine potential complications including malignancy. In 2002, Nagano and colleagues, for example, transplanted human SSCs into immunodeficient mice for the first time (67). Human SSCs survived in mouse testes for at least six months and proliferated during the first month after transplantation.
Transplantation of SSCs
To avoid potential complications of malignancy, isolation of SSCs from cryopreserved testicular tissues followed by transplantation has been proposed as the leading alternative stratgey. To separate SSCs from somatic cells, antibodies that recognize human SSC-specific proteins are used for FACS (fluorescent-activated cell sorting) or MACS (magnetic-activated cell sorting) for sorting SSCs from other cell types. Antibodies that have been shown to be useful for sorting SSCs include GFRα (68), GPR125, ID4 (69), ITGA6 (70), SSEA4 (71), PLPPR3 (72), and OCT4 (73). An alternative to cell sorting is to take advantage of different physical properties between SSCs and somatic cells such as velocity sedimentation and differential affinity to extracellular matrices on the culture plate (74–78). Once isolated, SSCs are cultured with growth factors shown to be optimal or essential for SSC maintenance [GDNF, BFGF, EGF, and LIF (79–81)].
A major limitation of SSC transplantation in vivo, for fertility restoration in clinical practice, is the scarcity of SSCs within the testicular tissue. This has necessitated exploration of alternatives including the establishment of a robust in vitro culture system to maintain and expand human SSCs. Extensive effort has been focused on optimization of culture conditions for long-term maintenance and propagation of human SSCs. Multiple culture substrates, including hydrogel, matrigel, and laminin, have been shown to promote the propagation of human SSCs under feeder-free conditions (82). Currently, several markers are used for the verification of human SSCs. However, many of these markers are also expressed in testicular somatic cells. For example, UCHL1, which was used to identify SSCs from humans, is also expressed in Leydig cells and nerve fibers (83). The most stringent assay to assess the function of SSCs is to generate offspring after homologous transplantation. However, despite success in animal models, including non-human primates, no studies are reporting the generation of human functional spermatozoa following autotransplantation or xenotransplantation of testicular tissue or isolated human SSCs for fertility restoration.
Fertility Restoration Through In Vitro Spermatogenesis
The establishment of a system to recapitulate spermatogenesis and generate spermatozoa in vitro can not only be directly applied in assisted reproduction, such as in vitro fertilization (IVF) or intracytoplasmic sperm injection (ICSI), but also provide a convenient system to study the molecular mechanisms and genetic causes for male infertility. Building a functional somatic microenvironment is critical for in vitro spermatogenesis. Several strategies, including exploitation of intrinsic somatic microenvironment by organotypic culture, two-dimensional culture, and three-dimensional culture of testis cell suspensions.
Organotypic Culture
Since 1959, a gas-liquid interface was used to culture testicular fragments of the adult rats (84). In this culture system, the differentiation of SSCs was limited up to pachytene spermatocytes (85). In 2003, round spermatids were observed after two weeks of culture in a gas-liquid interface culture system (86). Several other organotypic culture systems have been developed to recapitulate the entire process of spermatogenesis in vitro. One of the breakthroughs in the research was reported in 2011 with the demonstration of live offspring that were generated from in vitro-produced haploid germ cells (87). In this study, testicular tissue fragments from neonatal mice were cultured on an agarose gel-based organ culture system. Subsequently, microfluidic technology was adopted for organ culture, with the goal of providing a better culture environment for SSCs by facilitating the exchange of gases, nutrients, and waste products (88). Recently, successful recapitulation of human testicular organogenesis from fetal gonads was achieved, and in vitro-derived haploid spermatids were shown to undergo meiotic recombination (89).
Two-Dimensional Culture
2D culture systems with testis cell suspensions have been widely used for SSC proliferation and differentiation with two primary types of 2D culture systems most common: (1) SSCs cultured on mitotically-inactivated feeder cells, (2) SSCs co-cultured with somatic cells (90). Using the support of 2D culture sytems, numerous studies have reported that haploid male germ cells could be induced (91–95), and offspring can be produced from these in vitro derived haploid male germ cells in rodent (96). However, the 2D culture system has not been optimized for human germ cells. This may be due to the lack of spatial structure of seminiferous tubules and proper interactions between germ cells and somatic cells.
Three-Dimensional Culture
To better mimic the testicular niche, various 3D culture systems have been developed. In 2006, testicular cells isolated from rats were cultured on collagen gels to mimic the composition of the basal membrane of seminiferous tubules (97). Later, the soft-agar culture system (SACS) was developed (98), and mice haploid germ cells from undifferentiated germ cells were generated in this system in 2012 (99). The SACS system also supports the differentiiation of SSCs of non-human primates. The most commonly used alternate material in 3D culture system is methylcellulose. The methylcellulose culture system (MCS) also supports the differentiation of immature germ cells.
In order to artificially reproduce the in vivo form and function of the seminiferous epithelium, a 3D engineered blood-testis barrier (eBTB) system was designed in 2010 (100). Testicular peritubular myoid cells were first cultured on the underside of culture inserts, and then germ cells and Sertoli cells were added on top of the inserts. The testicular cells from neonatal mice form the aggregate by culturing on a V-shaped plate. The aggregate plated on the top of agarose gel blocks, and the haploid male germ cells were obtained after 30-51 days of incubation (101).
The 3D decellularized testicular scaffold with hyaluronic acid and chitosan provides the condition for the differentiation and proliferation of mice SSCs (102). The proliferation and self-renewal of mice SSCs was stimulated by culturing on the 3D scaffold consisting of alginate hydrogel with Sertoli cells (103). The mice germ cells were cultured in 3D printed one-layer scaffolds at the air-medium interface simulating the tubule-like structure. This culture system provided the condition for long-term survival and differentiation (104).
Soft agar and agarose gel are the most common material used to establish the 3D culture system for human SSCs. A soft agar culture system has been shown to support the proliferation and differentiation of human SSCs (105). Another material that has been used in 3D culture systems for human SSCs is a polycaprolactone (PCL) nanofiber matrix (106). This material may mimic the physical form of collagen fibers in the natural extracellular matrix (107).
Conclusion and Perspectives
With the development of technologies, including -omics at the single-cell level, lineage-tracing, spermatogonial transplantation, and in vitro culturing and differentiation, we start decoding the secrets of SSCs. However, the application of SSCs to treat male infertility necessitates extensive studies to ensure safety and efficacy. An efficient culture condition for human SSCs to ensure their propagation, as well as proper animal models for xenotransplantation, will assist in assessing safety and efficacy as indicated by recent studies (108). Furthermore, establishing a robust system for in vitro spermatogenesis is also helpful for pharmaceutical or toxicological studies for new drugs. Finally, in vitro spermatogenesis from SSCs sets the stage for the production of SSCs from induced pluripotent stem cells (iPSCs) and subsequent spermatogenesis. For example, studies are underway to integrate data and practices from divergent fields to promote spermatogenesis from iPSCs via co-culture with Sertoli cells in a 2D-, 3D- or a modified environment, similar to those used in other physiological systems, that might more faithfully mimic spermatogenic dynamics including circulation (109, 110).
Author Contributions
All authors contributed to the article and approved the submitted version.
Funding
This work was supported by an NIH grant to RRP #HD096026 and National Natural Science Foundation of China #32070830 to FF.
Conflict of Interest
Authors PJT and CMJ are founders of the company MandelMed. No funding from MandelMed is associated with this study.
The remaining authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s Note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Keywords: spermatogonia, spermatogenesis, in vivo, in vitro, stem cell, 3D culture, male infertility
Citation: Diao L, Turek PJ, John CM, Fang F and Reijo Pera RA (2022) Roles of Spermatogonial Stem Cells in Spermatogenesis and Fertility Restoration. Front. Endocrinol. 13:895528. doi: 10.3389/fendo.2022.895528
Received: 14 March 2022; Accepted: 31 March 2022;
Published: 12 May 2022.
Edited by:
Barry Zirkin, Johns Hopkins University, United States
Reviewed by:
F. Kent Hamra, University of Texas Southwestern Medical Center, United States
Michael Griswold, Washington State University, United States
Copyright © 2022 Diao, Turek, John, Fang and Reijo Pera. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Fang Fang, fangfang0724@gmail.com; Renee A. Reijo Pera, reneer@mclaughlinresearch.org
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
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Probability of an edge in directed random configuration graph
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I am considering Bollobas' directed random configuration graph of size N N, constructed by the following random algorithm:
Draw a sequence of N N node-degree pairs (j 1,k 1),...,(j N,k N)(j 1,k 1),...,(j N,k N) independently from the degree distribution P P. Accept the draw only if it is feasible, i.e. only if ∑n∈N=0∑n∈N=0.
For every node n n, create j n j n in-stubs and k n k n out-stubs, where in-/out-stubs are open ended half-edges with an in-/out-arrow.
For any unpaired out-stub, select iteratively uniformly at random an unpaired in-stub and connect them.
Each such resulting pair of stubs forms a directed edge of the graph.
I am interested in the probabilities of an edge. My suggestion is:
Under this random matching approach, the probability that there is an edge from a node i i to a node l l is, with m m being the number of all edges:
p i l=k i⋅j l(2 m−1),p i l=k i⋅j l(2 m−1),
as node i i has k i k i out-stubs and j l j l in-stubs out of 2 m−1 2 m−1 (excluding of node i i) attached to l l to which it could connect.
And similarly the probability that there is an edge from a node l l to i i is:
p i l=j i⋅k l(2 m−1)p i l=j i⋅k l(2 m−1)
Is this correct, or am I missing something?
probability
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edited Jan 6, 2019 at 19:23
AlisatAlisat
asked Nov 9, 2018 at 15:09
AlisatAlisat
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We can choose a uniformly random matching in any order. So let's choose the k i k i edges out of node i i one at a time and see the probability none of them go to node l l.
The first edge we choose has probability j l m j l m of going to l l. If it doesn't, then the second edge has probability j l m−1 j l m−1. If that also doesn't happen, then the third edge has probability j l m−2 j l m−2, and so on. So the overall probability that none of the edges go to node l l is
1−p i l=(1−j l m)(1−j l m−1)⋯(1−j l m−k i+1)1−p i l=(1−j l m)(1−j l m−1)⋯(1−j l m−k i+1)
and the probability you want is the complement of this one.
Asymptotically, however, assuming k i k i and j l j l are small relative to m m, the answer is in fact close to k i⋅j l m k i⋅j l m which is what you have, except that m m and not 2 m−1 2 m−1 is the number of options for the first edge. (Unlike the undirected configuration model, here there are m m out-stubs and m m in-stubs, which we distinguish.)
Additionally, because there are k i k i edges with probability j l m j l m of going to l l, then k i j l m k i j l m edges is the expected number of edges from i i to l l.
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edited Jan 6, 2019 at 20:09
answered Jan 6, 2019 at 17:16
Misha LavrovMisha Lavrov
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So I guess there is no "neat" form for the probability as in the undirected one, where it is of the form p i j=k i k j 2 m−1 p i j=k i k j 2 m−1. ? (Only for the large limits as you stated p i l=k i j l m p i l=k i j l m)Alisat –Alisat 2019-01-06 18:35:15 +00:00 Commented Jan 6, 2019 at 18:35
1 The "neat" form is not the probability but the expected number of edges from i i to l l.Misha Lavrov –Misha Lavrov 2019-01-06 20:10:27 +00:00 Commented Jan 6, 2019 at 20:10
1 The paper does not use the configuration mode. Instead, they define p i j=k i k j 2 m p i j=k i k j 2 m then add an edge i j i j with that probability. As they mention on p. 3 (second paragraph) the actual degree of vertex i i is not guaranteed to be k i k i in this model.Misha Lavrov –Misha Lavrov 2019-01-06 21:19:52 +00:00 Commented Jan 6, 2019 at 21:19
1 Upon further reading, the paper does seem to claim that k i k j 2 m−1 k i k j 2 m−1 is the exact edge probability in the random matching model, which is incorrect.Misha Lavrov –Misha Lavrov 2019-01-06 21:48:28 +00:00 Commented Jan 6, 2019 at 21:48
1 @Splines These are not disjoint events, so you cannot add up their probabilities to get the total probability that any edge stub of i i connects to any edge stub of l l.Misha Lavrov –Misha Lavrov 2023-08-12 14:48:21 +00:00 Commented Aug 12, 2023 at 14:48
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188751 | https://es.scribd.com/document/397420941/Metodo-de-Castigliano-en-Vigas-Isostaticas-e-Hiperestaticas | Método de Castigliano en Vigas Isostáticas e Hiperestáticas | PDF | Elasticidad (Física) | Deformación (Mecánica)
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Método de Castigliano en Vigas Isostáticas e Hiperestáticas
Este documento presenta el método de Castigliano para calcular las deflexiones en vigas isoestáticas e hiperestáticas. Explica que el método de Castigliano es uno de los métodos más exactos …
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MÉTODO DE CASTIGLIANO EN VIGAS ISOSTÁTICAS E HIPERESTÁTICAS
INTEGRANTES:
NIMA MAZA, RODRIGO.
OLIVARES LIZANA, MIGUEL.
OROZCO LOZADA, JORGE.
DOCENTE:
ING. LENIN TALLEDO COVEÑAS.
UNIVERSIDAD NACIONAL DE PIURA
FACULTAD DE INGENIERÍA CIVIL
UNIVERSIDAD NACIONAL DE PIURA
FACULTAD DE INGENIERÍA CIVIL
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1
1.
INTRODUCCIÓN
La estructura es el conjunto mecánico encargado de soportar y transmitir las cargas hasta las cimentaciones, donde serán absorbidas por el terreno. Para ello, las estructuras se encuentran constituid as por una serie de barras enlazadas entre sí. Las vigas son los principales elementos estructurales, la cual ofrece resistencia a la deformación; con exactitud a la f lexión. Existen muchos métodos de conservación de energía, los cuales sirven p ara el cálculo de las deflexiones de una viga; el primer método de Castigliano es uno de ellos, es conocido como el más exacto para estas operaciones, ya que primero calcula el trabajo realizado por la fuerza cortante que aplica la cargas en dicha viga, y por último calcula lo que se desea en realidad: cuán deformable es el material q vamos a utilizar en la fabricación de esta. Los teoremas y procedimientos relacionados con la energía de deformación ocupan una posición central en todo cálculo de estructuras. En este trabajo se a intentará determinar la deformación de una viga, utilizando los teoremas de Castiglian o. Pues calcular el desplazamiento de un cuerpo, sólo se aplica a cuerpos de temperatura constante, de material con comportamiento elástico lineal; es decir nos ayuda a calcular las deflexiones producidas en una viga a causa de una determinada carga que debe soportar y por ende nos ayuda a elegir el mejor material para la construcción de estás según su resistencia y para que propósito la necesitamos.
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2
2.
OJETIVOS
Investig ar los dos teoremas propuestos en el Método de Castigliano para el cálculo de deflexión y pendiente en una viga.
Identificar cuand o podemos u tilizar los teore mas de Casti gliano para el cál culo la pendiente y la deflexión de una estructura.
Aplicar estos conocimiento s mediante ejercicios que vinculen este tipo de cálculo en la deformación de una estructura y comparando que los resultados sean iguales a los demás métodos estudiados.
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3
3.
GENERALIDADES 3.1
BIOGRAFÍA DE CARLO ALBERTO CASTIGLIANO.
Carlo Alberto Castigliano (9 de noviembre de 1847, Asti - 25 de octubre de 1884, Milán) fue un italiano matemático y físico conocido por el método de Castigliano para la determinación de los desplazamientos en un elástico-lineal del sistema sobre la base de las derivadas parciales de energía de deformación. Alberto Castiglian o se trasladó desde la región de su nacimiento, Piamonte en el noroeste de Italia, para el Instituto Técnico de Terni (en Umbría) en 1866. Después de cuatro años en Terni, Castig liano se trasladó al norte de nuevo, esta vez para convertirse en un estudiante de la universidad de Wilkes. Después de tres años de estudio en Wilkes escribió una disertación en 1873 titulado ElasticiIntornoai sistemi por la que es f amoso. En su tesis parece un teorema que ahora lleva el nombre de Castigliano. Esto se afirma que: La derivada parcial de la energía de deformación, considerada como una función de las fuerzas aplicadas que actúan sobre una estructura linealmente elástico, con respecto a una de estas fuerzas, es igual al desplazamiento en la dirección de la fuerza de su punto
de aplicación”.
Después de graduarse de la universidad Wilkes, Castigliano era empleado de los ferrocarriles del norte de Italia. Se dirigió a la oficina responsable de la obra, mantenimien to y servicio y trabajó allí hasta su muerte a una edad temprana.
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188752 | https://hscprep.com.au/hsc-chemistry/understanding-ionic-dissolution | HSC Chemistry
Understanding Ionic Compound Dissolution
Expert reviewed • 22 November 2024 • 4 minute read
Introduction
The dissolution of ionic compounds in water is a fundamental process in chemistry that involves several key steps and energy changes. This article explores the mechanisms behind dissolution, from breaking crystal lattices to forming hydrated ions.
Lattice Energy: The Starting Point
Ionic compounds exist in crystalline lattices held together by strong electrostatic forces. Breaking these bonds requires energy equal to the lattice energy - the energy released when the ionic compound forms from its constituent ions. For example:
Na+(aq)+Cl−(aq)⇌NaCl(s)ΔH=−787 kJ mol−1Na^+(aq) + Cl^-(aq) \rightleftharpoons NaCl(s) \quad \Delta H = -787\: kJ\:mol^{-1}Na+(aq)+Cl−(aq)⇌NaCl(s)ΔH=−787kJmol−1
Two main factors influence lattice energy:
The Three-Step Dissolution Process
1. Separation of Water Molecules
The first step involves breaking the hydrogen bonds between water molecules. This endothermic process requires energy to overcome intermolecular forces.
2. Dissociation of Ionic Compounds
The ionic compound's crystal lattice breaks apart, separating the ions. This endothermic step requires sufficient energy to overcome the lattice energy.
3. Hydration of Ions
Water molecules surround the separated ions, forming hydration shells. This exothermic process releases energy through ion-dipole interactions.
Energy Changes During Dissolution
| Process | Enthalpy Change (ΔH) | Entropy Change (ΔS) |
---
| Water molecule separation | Positive (endothermic) | Positive |
| Ion dissociation | Positive (endothermic) | Positive |
| Ion hydration | Negative (exothermic) | Negative |
Factors Affecting Solubility
Three main factors determine an ionic compound's solubility:
Overall Energy Changes
The total enthalpy change depends on the balance between:
Previous
Understanding Gibbs Free Energy and Chemical Equilibrium
Next
Understanding Solubility Product and Equilibrium in Solutions
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Mathematical Circles (Russian Experience) -(Mathematical World 7) D. Fomin, S. Genkin, I. Itenberg-American Mathematical Society (1996)
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Mathematical World • Volume 1 Mathematical Circles (Russian Experience) Dmitri Fomin Sergey Genkin Ilia ltenberg Amer
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Mathematical World • Volume 1Mathematical Circles (Russian Experience)Dmitri Fomin Sergey Genkin Ilia ltenbergAmerican Mathematical SocietySelected Titles in This Series Dmitri Fomin, Sergey Genkin, and Ilia ltenberg, Mathematical circles (Russian experience), 1996 David W. Farmer and Theodore B. Stanford, Knots and surfaces: A guide to discovering mathematics, 1996 David W. Farmer, Groups and symmetry: A guide to discovering mathematics, 1996 V. V. Prasolov, Intuitive topology, 1995 L. E. Sadovskil' and A. L. SadovskiX, Mathematics and sports, 1993 Yu. A. Shashkin, Fixed points, 1991 V. M. Tikhomirov, Stories about maxima and minima, 1990Mathematical World • Volume 7Mathematical Circles (Russian Experience) Dmitri Fomin Sergey Genkin Ilia ltenberg Translated from the Russian by Mark SaulAmerican Mathematical SocietyC.A.rEHKHH,H.B.HTEHBEP~ll.B.~OMHHMATEMATHqECKHH KPYmO:K CAHKT-IIETEPBYPr 1992, 1993 Translated from the' Russian by Mark Saul 2000 Mathematics Subject Classification. Primary OOAOB; Secondary OOA07. ABSTRACT. This book is intended for students and teachers who love mathematics and want to study its various branches beyond the limits of school curriculum. It is also a book of mathematica1 recreations, and at the same time a book containing vast theoretical and problem material in some areas of what authors consider to be 11extracurricular mathematics". The book is based on an experience gained by several generations of Russian educators and scholars.Library of Congress Cataloging-in-Publication Data Genkin, S. A. (Sergei Aleksandrovich) !Matematicheskii kruzhok. English) Mathematical circles : (Russian experience) / Dmitri Fomin, Sergey Genkin, Ilia Itenberg; translated from the Russian by Mark Saul. p. cm. - (Mathematical world, ISSN 1055-9426; v. 7) On Russian ed., Genkin's name appears first on t.p. Includes bibliographical references (p. - ). ISBN 0.8216-0430.8 (alk. paper) 1. Mathematical recreations. I. Fomin, D. V. (Dmitri! Vladimirovich) II. Itenberg, I. V. (Il'fa. Vladimirovich) III. Title. IV. Series. QA95.G3813 1996 510'. 76- 1/2 meter ~FIGURE184. THE PIGEON HOLE PRINCIPLE35Problem 14. Show that an equilateral triangle cannot be covered completely by two smaller equilateral triangles. Problem 15. Fifty-one points are scattered inside a square with a side of 1 meter. Prove that some set of three of these points can be covered by a square with side 20 centimeters. Solution. If we divide the square into 25 smaller squares with sides of 20 centimeters, the General Pigeon Hole Principle assures us that one of these squares includes at least three of the 51 scattered points. The careful reader will see a small flaw in this argument. Throughout our discussion, we have been assuming that our pigeon holes are disjoint. That is, no pigeon can belong in two different pigeon holes at the same time. However, the square "pigeon holes" in this solution have a slight overlap: points on the sides of the squares may belong to both pigeon holes. To fix this, we must make a choice for each line segment which bounds a square, by deciding which of its two neighboring squares includes the points on the line segment. We can do this, for instance, by making the "north" and "east" borders of each square exclude their points, and the "south" and "west" border include their points (except for points on the border of the original square). With this slight adjustment, we have a set of "true pigeon holes", and the proof follows as before. §4. Another generalization Notice now that the proofof the Pigeon Hole Principle is based on the addition of inequalities. An important result of the process of adding inequalities, which can often be combined with the theme of the Pigeon Hole Principle, can be stated as follows: If the sum of n or more numbers is equal to S, then among these there must be one or more numbers not greater than S/n, and also one or more numbers not less than S/n. As with most variants of the Pigeon Hole Principle, we can prove this indirectly. , If, for example, all the numbers are greater than S / n, then their sum would be bigger than S, which contradicts our assumption.Problem 16. Five young workers received as wages 1500 rubles altogether. Each of them wants to buy a cassette player costing 320 rubles. Prove that at least one of them must wait for the next paycheck to make his purchase. Solution. The sum S of their earnings is 1500 rubles, so the above principle guarantees that at least one worker earned no more that 1500/5 = 300 rubles. Such a worker must wait for his cassette player. Problem 17. In a brigade of 7 people, the sum of the ages of the members is 332 years. Prove that three members can be chosen so that the sum of their ages is no less than 142 years. Solution. We look at all possible triples of brigade members. If we add the three ages in each group, then sum these numbers, this final sum must be 15 · 332 (since each person appears in a triple 15 times). Yet there are altogether 35 triples. This means that there is a triple of brigade members such that the sum of their ages is not less than 15 · 332/35, which is greater than 142.36MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 18. On a certain planet in the solar system Tau Cetus, more than half the surface of the planet is dry land. Show that the Tau Cetans can dig a tunnel straight through the center of their planet, beginning and ending on dry land (assume that their technology is sufficiently developed). §5. Number theory Many wonderful problems touching on divisibility properties of integers can be solved using the Pigeon Hole Principle. Problem 19. Prove that there exist two powers of two which differ by a multiple of 1987. Problem 20. Prove that of any 52 integers, two can always be found such that the difference of their squares is divisible by 100. Problem 21. Prove that there exists an integer whose decimal representation consists entirely of l's, and which is divisible by 1987. Solution to Problem 21. We look at the 1988 "pigeons" numbered 1, 11, 111, ... , 111 ... 11 (1988 1's), and sort them into 1987 pigeon holes numbered 0, 1, 2, ... , 1986. Each number is put into the pigeon hole bearing the number equal to its remainder when divided by 1987. The Pigeon Hole Principle now assures us that there are two numbers which have the same remainder when divided by 1987. Let these numbers have m l's and n l's respectively, with m > n. Then their difference, which is divisible by 1987, is equal to 111 ... 1100 ... 00 (m - n l's and n zeros). We can cross out all the trailing zeros-these do not affect divisibility by 1987 since neither 2 nor 5 is a factor of 1987-to obtain a number written entirely with l's, and which is divisible by 1987. Problem 22. Prove that there exists a power of three which ends with the digits 001 (in decimal notation). Problem 23. Each box in a 3 x 3 arrangement of boxes is filled with one of the numbers -1, 0, 1. Prove that of the eight possible sums along the rows, the columns, and the diagonals, two sums must be equal. Problem 24. Of 100 people seated at a round table, more than half are men. Prove that there are two men who are seated diametrically opposite each other. Problem 25. Fifteen boys gathered 100 nuts. Prove that some pair of boys gathered an identical number of nuts. Problem 26. The digits 1, 2, ... , 9 are divided into three groups. Prove that the product of the numbers in one of the groups must exceed 71. Problem 27. Integers are placed in each entry of a 10 x 10 table, with no two neighboring integers differing by more than 5 (two integers are considered neighbors if their squares share a common edge). Prove that two of the integers must be equal. Problem 28. Prove that among any six people there are either three people, each of whom knows the other two, or three people, each of whom does not know the other two. Problem 29. Five lattice points are chosen on an infinite square lattice. Prove that the midpoint of one of the segments joining two of these points is also a lattice point.4. THE PIGEON HOLE PRINCIPLE37Problem 30. A warehouse contains 200 boots of size 41, 200 boots of size 42, and 200 boots of size 43. Of these 600 boots, there are 300 left boots and 300 right boots. Prove that one can find among these boots at least 100 usable pairs. Problem 31. The alphabet of a certain language contains 22 consonants and 11 vowels. Any string of these letters is a word in this language, so long as no two consonants are together and no letter is used twice. The alphabet is divided into 6 (non-empty) subsets. Prove that the letters in at least one of these groups form a word in the language. Problem 32. Prove that we can choose a subset of a set of ten given integers, such that their sum is divisible by 10. Problem 33. Given 11 different natural numbers, none greater than 20. Prove that two of these can be chosen, one of which divides the other. Problem 34. Eleven students have formed five study groups in a summer camp. Prove that two students can be found, say A and B, such that every study group which includes student A also includes student B. Students should remember that even if they cannot cope with some problem right away, it always pays to go back later and try some fresh ideas. Do not jump to the solutions chapter! And do not forget that some of the problems may have alternate solutions not using the Pigeon Hole Principle.CHAPTER 5Graphs-1 The mathematical objects discussed in this chapter are extremely useful in solving many kinds of problems, which often bear no outward resemblance to each other. Graphs are also interesting in and of themselves. A separate subdivision of mathematics, called graph theory, is devoted to their study. We will examine several elementary ideas from this theory to show how graphs are used in solving problems. §1. The concept of a graphProblem 1. Cosmic liaisons are established among the nine planets of the solar system. Rockets travel along the following routes: Earth-Mercury, Pluto-Venus, Earth-Pluto, Pluto-Mercury, Mercury-Venus, Uranus-Neptune, Neptune-Saturn, Saturn-Jupiter, Jupiter-Mars, and Mars-Uranus. Can a traveler get from Earth to Mars? Solution. We can draw a diagram, in which the planets will be represented by points, and the routes connecting them by non-intersecting line segments (see Figure 19). It is now clear that it is impossible to travel from Earth to Mars.EMe[ZJ pvMa FIGUREJ19Problem 2. Several knights are situated on a 3 x 3 chessboard as shown in Figure 20. Can they move, using the usual chess knight's move, to the position shown in Figure 21? Solution. The answer is no. We can show this by numbering the squares of the chessboard with the numbers 1, 2, 3, ... , 9 as shown in Figure 22. Then we can represent each square by a point. If we can get from one square to another with a 3940MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)qjqj4l4l FIGURE 20qj4l4lqj FIGURE 21147258369FIGURE 22knight's move, we connect the corresponding points with a line (Figure 23). The starting and ending positions of the knights are shown in Figure 24. The order in which the knights appear on the circle clearly cannot be changed. Therefore it is not possible to move the knights to the required positions. The solution of these two problems, which do not resemble each other on the surface, have a central idea in common: the representation of the problem by a5. GRAPHS-14176 5•9483 FIGURE 23933 FIGURE 24diagram. The resulting diagrams also have something in common. Each consists of a set of points, some of which are connected by line segments. Such a diagram is called a graph. The points are called the vertices of the graph, and the lines are called its edges. Methodological remark. The definition we give of a graph is actually too limiting. For example, in Problem 20 below it is rather natural to draw the edges of a graph using arcs, rather than line segments. However, an accurate definition would here be too complicated. The description above will suffice for students to get an intuitive idea of what a graph is, which they can later refine.Here are two more problems which can be solved by drawing graphs. Problem 3. A chessboard has the form of a cross, obtained from a 4 x 4 chessboard by deleting the corner squares (see Figure 25). Can a knight travel around this board, pass through each square exactly once, and end on the same square he starts on?42MATHEMATICAL CIRCLES (RUSSIAN EXPERlENCE)FIGURE25Problem 4. In the country of Figura there are nine cities, with the names 1, 2, 3, 4, 5, 6, 7, 8, 9. A traveler finds that two cities are connected by an airplane route if and only if the two-digit number formed by naming one city, then the other, is divisible by 3. Can the traveler get from City 1 to City 9? Note that one and the same graph can be represented in different ways. For example, the graph of Problem 1 can be represented as in Figure 26.Ns Jv~UrMa FIGURE26The only important thing about a graph is which vertices are connected and which are not. ' Two graphs which are actually identical, but are perhaps drawn differently, are called isomorphic. Problem 5. Try to find, in Figures 27, 28, and 29 a graph isomorphic to the graph of Problem 2 (see Figure 23). Solution. The first and the third graphs are isomorphic to each other, and it is not hard to convince oneself that both of these are isomorphic to the graph of Problem5. GRAPHS-I43FIGURE 27FIGURE 28FIGURE 292. It suffices to renumber their vertices (see Figures 30 and 31). A proof that the graphs of Figures 28 and 23 are not isomorphic is somewhat more complicated. The concept of a graph should be introduced only after several For teachers. problems like Problems 1 and 2 above, which involve using a graph to represent the situation of the problem. It is important that students realize right away that the44MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)37FIGURE 3061 3 FIGURE 31same graph can be drawn in different ways. To illustrate the idea of isomorphism, students can solve several more exercises of the type given here. §2. The degree of a vertex: counting the edges In the preceding section we defined a graph as a set of points (vertices), some of which are connected by lines (edges). The number of edges which start at a given5. GRAPHS-I45LzrBFIGURE 32 vertex is called the degree of the vertex. Thus, for example, in the graph of Figure 32, vertex A has degree 3, vertex B has degree 2, and vertex C has degree l. Problem 6. In Smallville there are 15 telephones. Can they be connected by wires so that each telephone is connected with exactly five others? Solution. Suppose that this is possible. Consider the graph in which the vertices represent telephones, and the edges represent the wires. There are 15 vertices in this graph, and each has degree 5. Let us count the number of edges in this graph. To do this, we can add up the degrees of all the vertices. However, in this sum, each edge is counted twice (each edge connects two vertices). Therefore the number of edges in the graph must be equal to 15 · 5/2. But this number is not an integer. It follows that such a graph cannot exist, which means that we cannot connect the telephones as required. In solving this problem, we have shown how to count the edges of a graph, knowing the degree of each vertex: we add the degrees of all the vertices and divide this sum by 2. Problem 7. In a certain kingdom, there are 100 cities, and four roads lead out of each city. How many roads are there altogether in the kingdom? 'Notice that our counting method for edges of a graph has the following consequence: the sum of the degrees of all the vertices in a graph must be even (otherwise, we could not divide it by 2 to get the number of edges). We can give a better formulation of this result using the following definitions: A vertex of a graph having an odd degree is called an odd vertex. A vertex having an even degree is called an even vertex. Theorem. The number of odd vertices in any graph must be even. To prove this theorem it is enough to notice that the sum of several integers is even if and only if the number of odd addends is even. Methodological remark. This theorem plays a central role in this chapter. It is important to keep returning to its proof, and to apply the theorem as often as possible in the solution of problems. Students should be encouraged to repeat the proof of the theorem within their solution to a problem, rather than merely quoting the theorem. The theorem is often used to prove the existence of a certain edge of a graph, as in Problem 12. It is also used, as in Problems 8-11, to prove that a graph answering46MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)a certain description is impossible to draw. Such problems can be difficult for students to grapple with. It is essential that they first try to draw the required graph, then guess that it is not possible, and finally give a clear discussion or proof, using the theorem above, that the required graph does not exist. Problem 8. There are 30 students in a class. Can it happen that 9 of them have 3 friends each (in the class), eleven have 4 friends each, and ten have 5 friends each? Solution. If this were possible, then it would also be possible to draw a graph with 30 vertices (representing the students), of which 9 have degree 3, 11 have degree 4, and 10 have degree 5 (by connecting "friendly" vertices with edges). However, such a graph would have 19 odd vertices, which contradicts the theorem. Problem 9. In Smallville there are 15 telephones. Can these be connected so that (a) each telephone is connected with exactly 7 others; (b) there are 4 telephones, each connected to 3 others, 8 telephones, each connected to 6 others, and 3 telephones, each connected to 5 others? Problem 10. A king has 19 vassals. Can it happen that each vassal has either 1, 5, or 9 neighbors? Problem 11. Can a kingdom in which 3 roads lead out of each city have exactly 100 roads? Problem 12. John, coming home from Disneyland, said that he saw there an enchanted lake with 7 islands, to each of which there led either 1, 3, or 5 bridges. Is it true that at least one of these bridges must lead to the shore of the lake? Problem 13. Prove that the number of people who have ever lived on earth, and who have shaken hands an odd number of times in their lives, is even. Problem 14. Can 9 line segments be drawn in the plane, each of which intersects exactly 3 others? §3. Some new definitions Problem 15. In the country of Seven there are 15 towns, each of which is connected to at least 7 others. Prove that one can travel from any town to any other town, possibly passing through some towns in between. Solution. Let us look at any 2 towns, and suppose that there is no path connecting them. This means that there is no sequence of roads such that the end of one road coincides with the beginning of the next road, connecting the 2 towns. It is given that each of the 2 towns is connected with at least 7 others. These 14 towns must be distinct: if any 2 were to coincide, there would be a path through them (or it) connecting the 2 given towns (see Figure 33). So there are at least 16 different towns, which contradicts the statement of the problem. In light of this problem, we give two important definitions: A graph is called connected if any two of its vertices can be connected by a path (a sequence of edges, each of which begins at the endpoint of the previous one). A closed path (a path whose starting and ending vertices coincide) is called a cycle.5. GRAPHS-147FIGURE 33We can now reformulate the result of the previous problem: the graph of the roads of the kingdom of Seven is connected. Problem 16. Prove that a graph with n vertices, each of which has degree at least (n - 1)/2, is connected. It is natural to ask how a non-connected graph looks. Such a graph is composed of several "pieces", within each of which one can travel along the edges from any vertex to any other. Thus, for example, the graph of Figure 34 consists of three "pieces", while the graph of Figure 35 consists of two.FIGURE 34FIGURE 35These "pieces" are called connected components of the graph. Each connected component is, of course, a connected graph. We note also that a connected graph consists of a single connected component.48MATHEMATICAL CIRCLES (RUSSIAN EXPERlENCE)Problem 17. In Never-Never-Land there is only one means of transportation: magic carpet. Twenty-one carpet lines serve the capital. A single line flies to Farville, and every other city is served by exactly 20 carpet lines. Show that it is possible to travel by magic carpet from the capital to Farville (perhaps by transferring from one carpet line to another). Solution. Let us look at that connected component of the graph of carpet lines which includes the capital. We must prove that this component includes Farville. Suppose it does not. Then there are 21 edges starting at one vertex, and 20 edges starting at every other vertex. Therefore this connected component contains exactly one odd vertex. This is a contradiction. Methodological remark. The notion of connectedness is extremely important, and is used constantly in further work in graph theory. The important point in the solution of Problem 16----consideration of a connected component-is a meaningful idea, and often turns out to be useful in solving problems. Problem 18. In a certain country, 100 roads lead out of each city, and one can travel along those roads from any city to any other. One road is closed for repairs. Prove that one can still get from any city to any other.§4. Eulerian graphs Problem 19. Can one draw the graph pictured in (a) Figure 36; (b) Figure 37, without lifting the pencil from the paper, and tracing over each edge exactly once?FIGURE36FIGURE 375. GRAPHS-149Solution. (a) Yes. One way is to start at the vertex on the extreme left, and end at the central vertex. (b) No. Indeed, if we can trace out the graph as required in the problem, we will arrive at every vertex as many times as we leave it (with the exception of the initial and terminal vertices). Therefore the degree of each vertex, except for two, must be even. For the graph in Figure 37 this is not the case. In solving Problem 19, we have established the following general principle: A graph that can be traversed without lifting the pencil from the paper, while tracing each edge exactly once, can have no more than two odd vertices. This sort of graph was first studied by the great mathematician Leonhard Euler in 1736, in connection with a famous problem about the Konigsburg bridges (see also Problem 12). Graphs which can be traversed like this are called Eulerian graphs. Problem 20. A map of the city of Konigsburg is given in Figure 38. The city lies on both banks of a river, and there are two islands in the river. There are seven bridges connecting the various parts of the city. Can one stroll around the town, crossing each bridge exactly once?FIGURE38Problem 21. A group of islands are connected by bridges in such a way that one can walk from any island to any other. A tourist walked around every island, crossing each bridge exactly once. He visited the island of Thrice three times. How many bridges are there to Thrice, if (a) the tourist neither started nor ended on Thrice; (b) the tourist started on Thrice, but didn't end there; (c) the tourist started and ended on Thrice? Problem 22. (a) A piece of wire is 120 cm long. Can one use it to form the edges of a cube, each of whose edges is 10 cm? (b) What is the smallest number of cuts one must make in the wire, so as to be able to form the required cube?CHAPTER 6The Triangle Inequality §1. IntroductionThe triangle inequality is easily motivated, whether or not students have had a formal introduction to geometry. But even for those students who have studied axiomatics or formal proof, there are non-trivial applications lying right beneath the surface, and problems involving the triangle inequality can be constructed which demand significant thought. The inequality itself states that for any triangle ABC we have three inequalitiesshowing that any side of the triangle is less than the sum of two others. Problem 1. Prove that for any three points A, B, and C we have AC 2': IAB-BCI.,In discussing this problem, it is important to give its geometric interpretation: the length of a side of a triangle is not less than the absolute value of the difference between the other two sides. Problem 2. Side AC of triangle ABC has length 3.8, and side AB has length 0.6. If the length of side BC is an integer, what is this length? Problem 3. Prove that the length of any side of a triangle is not more than half its perimeter. Problem 4. The distance from Leningrad to Moscow is 660 kilometers. From Leningrad to the town of Likovo it is 310 kilometers, from Likovo to Klin it is 200 kilometers, and from Klin to Moscow is 150 kilometers. How far is it from Likovo to Moscow?Hint for solution to Problem 4: Notice that the sum of the distances from Leningrad to Likovo, from Likovo to Klin, and from Klin to Moscow is equal to the distance from Leningrad to Moscow. This means that these towns are all on the same line. Note that in solving Problem 4, we use the fact that the sum of any three sides of a quadrilateral is greater than the fourth side. This can easily be established, using the triangle inequality. In fact, for any polygon, the sum of all but one of the sides is greater than the remaining side. For many students, this fact can be established for a few cases, and then assumed intuitively for all cases. More advanced students can give a formal proof, using induction. Problem 5. Find a point inside a convex quadrilateral such that the sum of the distances from the point to the vertices is minimal. Solution. Since the quadrilateral is convex, its diagonals intersect at some interior point 0. Suppose the vertices of the quadrilateral are A, B, C, and D (see Figure 5152MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)39). Then the sum of the distances from 0 to the vertices is equal to AC+BD. But for any other point P, PA+ PC> AC (by the triangle inequality). Analogously, PB + PD 2: B D. This means that the sum of the distances from P to the vertices is not less than AC+ BD. Clearly, this sum is equal to AC+ BD only if P and 0 coincide. Therefore 0 is the point we are looking for.c FIGURE 39Problem 6. Point 0 is given on the plane of square ABCD. Prove that the distance from 0 to one of the vertices of the square is not greater than the sum of the distances from 0 to the other three vertices. Problem 7. Prove that the sum of the diagonals of a convex quadrilateral is less than the perimeter but more than half the perimeter. Problem 8. Prove that the sum of the diagonals of a convex pentagon is greater than the perimeter but less than double the perimeter. Problem 9. Prove that the distance between any two points inside a triangle is not greater than half the perimeter of the triangle. §2. The triangle inequality and geometric transformations Often, the triangle to which we must apply the triangle inequality does not appear in the diagram for the problem. In these cases, a suitable choice of geometric transformation can help. The following series of problems illustrates the use of symmetry together with the triangle inequality. Problem 10. A mushroom-gatherer leaves the woods at a given point. He must reach a highway, which follows a straight line, and go back into the woods at another given point (Figure 40). How should he do this, following the shortest path possible?FIGURE406. THE TRIANGLE INEQUALITY53Problem 11. A woodsman's hut is in the interior of a peninsula which has the form of an acute angle. The woodsman must leave his hut, walk to one shore of the peninsula, then to the other shore, then return home. How should he choose the shortest such path? Problem 12. Point A, inside an acute angle, is reflected in either side of the angle to obtain points B and C. Line segment BC intersects the sides of the angle at D and E (see Figure 41). Show that BC/2 >DE.c FIGURE41Problem 13. Point C lies inside a given right angle, and points A and B lie on its sides (see Figure 42). Prove that the perimeter of triangle ABC is not less than twice the distance OC, where 0 is the vertex of the given right angle.·k1 0BFIGURE42Let us analyze the solution to Problem 10. Suppose the mushroom-gatherer leaves the woods at point A, and must re-enter at point B. Reflect point A in the line of the highway (see Figure 43) to obtain point A'. If K is the point at which the mushroom-gatherer reaches the highway, then route AK B is equal in length to route A' KB, since we are simply reflecting segment AK in the highway. But A' KB cannot be shorter than A' B. It follows that point K should be the point where A' B intersects the highway. Similar considerations allow us to solve the other problems in this series. Fbr example, in Problem 13 we can reflect point C in lines OA and OB, to obtain points C' and C" (Figure 44); it is easy to see that point 0 lies on straight line C'C". Then we can replace the perimeter of triangle ABC with the sum of the lengths of segments C' A, AB, and BC". The triangle inequality tells us that this sum is no less than the length of C'C". This, in turn, is equal to 20C, since it is the hypotenuse of a right triangle of which OC is the median. (Students who don't know this theorem can find a more intuitive way to explain this: for example, by completing a rectangle with vertices at C', C", and C.)54MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)-~·_L I IKA' FIGURE43C'C" FIGURE44For teachers. It is important to solve these problems carefully, getting students to give a logical exposition of the solution, and not just an intuitive explanation. First we can remind the students that line reflection does not change distances. Then we can. point out the common idea in these problems: to transform the required path so that its length does not change, and so that the problem becomes one of connecting two points with the shortest path possible. It is important to check that one of the transformed paths can really be a straight line so that we have an obvious answer; otherwise the solution can be much more difficult.In many problems, the action takes place on some sort of surface in space. In such problems, the triangle inequality can be used only after we "unfold" the surface onto a plane. The following problems are typical: Problem 14. A fly sits on one vertex of a wooden cube. What is the shortest path it can follow to the opposite vertex? Problem 15. A fly sits on the outside surface of a cylindrical drinking glass. It must crawl to another point, situated on the inside surface of the glass. Find the shortest path possible (neglecting the thickness of the glass).6. THE TRJANGLE INEQUALITY55§3. Additional constructions In many cases, the proofs of geometric inequalities require additional constructions. Such problems are often complicated, since the choice of construction requires a certain amount of practice. The following series of problems provides some such practice: Problem 16. If point 0 is inside triangle ABC, prove that AO+OC < AB+BC. Problem 17. Prove that the sum of the distances from point 0 to the vertices of a given triangle is less than the perimeter, if point 0 lies inside the triangle. What if point 0 is outside the triangle? Problem 18. Solve Problem 11, if the peninsula has the shape of an obtuse angle. Problem 19. Prove that the length of median AM in triangle ABC is not greater than half the sum of sides AB and AC. Prove also that the sum of the lengths of the three medians is not greater than the triangle's perimeter. §4. Miscellaneous problems Problem 20. A polygon is cut out of paper, and then folded in two along a straight line (see Figure 45). Prove that the perimeter of the polygon formed is not greater than the perimeter of the original polygon.--% lFIGURE45Problem 21. Prove that a convex polygon cannot have three sides, each of which is greater than the longest diagonal. Problem 22. Prove that the perimeter of a triangle is not greater than 4/3 the sum of its medians. (For the solution of this problem, one must know the ratio into which the three medians of a triangle divide each other.)56MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 23. Two villages lie on opposite sides of a river whose banks are parallel lines. A bridge is to be built over the river, perpendicular to the banks. Where should the bridge be built so that the path from one village to the other is as short as possible? Problem 24. Prove that a convex pentagon (that is, a pentagon whose diagonals all lie inside the figure) has three diagonals which can form a triangle.CHAPTER 7Games Students enjoy playing games. Whether the mathematics behind the game is simple or complicated, the chance for social interaction and for controlled competition will help to break up any routine patterns in school life. At the same time, these problems hold a lot of content, and students frequently find their solution quite difficult. The chief difficulties consist first in articulating the winning strategy, and second in proving that the strategy considered always leads to a win. In surmounting these difficulties, students will learn more about accepted standards of mathematical argument, and will refine their understanding of what it means to solve a problem. Students must understand that statements of the form: "If you do thus, I will do as follows," are usually not solutions to a game. Examples of correct solutions are given in the text. We recommend giving no more than one or two games from this chapter in each lesson, with the exception of §4, which contains problems analyzed "backwards". The idea of symmetry (§2) and the concept of a winning position (§3) can be treated independently. This is best done after considering two or three problems on each theme. There are many types of games considered in mathematics, and many types of game theories. This chapter considers only one type. In each of these games, there are two players who take turns making moves, and a player cannot decline to move. The problem is always the same: to find out which player (the first or the second) has a winning strategy. These notes will not be repeated for each game. Starred problems are more difficult than the others. §1. Pseudo-games: Games that are jokesThe first class of games we examine are games that turn out to be jokes. The outcomes of these pseudo-games do not depend on how the play proceeds. For this reason, the solution of such a pseudo-game does not consist of a winning strategy, but of a proof that one or the other of the two players will always win (regardless of how the play proceeds!). Problem 1. Two children take turns breaking up a rectangular chocolate bar 6 squares wide by 8 squares long. They may break the bar only along the divisions between the squares. If the bar breaks into several pieces, they keep breaking the pieces up until only the individual squares remain. The player who cannot make a break loses the game. Who will win? Solution. After each move, the number of pieces increases by one. At first, there is only one piece. At the end of the game, when no more moves are possible, the chocolate is divided into small squares, and there are 48 of these. So there must 5758MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)have been 47 moves, of which the last, as well as every other odd-numbered move, was made by the first player. Therefore, the first player will win, no matter how the play proceeds. For teachers. Pseudo-games allow the students to relax and be relieved of the tension of having to solve a problem or win a game. They are very effective, for instance, if introduced right after particularly difficult material, or at the end of a lesson. It is important to let the students actually play the games before giving a solution. Problem 2. There are three piles of stones: one with 10 stones, one with 15 stones, and one with 20 stones. At each turn, a player can choose one of the piles and divide it into two smaller piles. The loser is the player who cannot do this. Who will win, and how? Problem 3. The numbers 1 through 20 are written in a row. Two players take turns putting plus signs and minus signs between the numbers. When all such signs have been placed, the resulting expression is evaluated (i.e., the additions and subtractions are performed). The first player wins if the sum is even, and the second wins if the sum is odd. Who will win and how? Problem 4. Two players take turns placing rooks (castles) on a chessboard so that they cannot capture each other. The loser is the player who cannot place a castle. Who will win? Problem 5. Ten 1's and ten 2's are written on a blackboard. In one turn, a player may erase any two figures. If the two figures erased are identical, they are replaced with a 2. If they are different, they are replaced with a 1. The first player wins if a 1 is left at the end, and the second player wins if a 2 is left. Problem 6! The numbers 25 and 36 are written on a blackboard. At each turn, a player writes on the blackboard the (positive) difference between two numbers already on the blackboard-if this number does not already appear on the blackboard. The loser is the player who cannot write a number. Problem 7. Given a checkerboard with dimensions (a) 9 x 10; (b} 10 x 12; (c) 9 x 11. In one turn, a player is allowed to cross out one row or one column if at the beginning of the turn there is at least one square of the row or column remaining. The player who cannot make a move loses. §2. Symmetry Problem 8. Two players take turns putting pennies on a round table, without piling one penny on top of another. The player who cannot place a penny loses. Solution. In this game, the first player can win, no matter how big the table may be! To do so, he must place the first penny so that its center coincides with the center of the table. After this, he replies to each move of the second player by placing a penny in a position symmetric to the penny placed by the second player, with respect to the center of the table. Notice that in such a strategy the positions of the two players are symmetric after each move of the first player. It follows that if there is a possible turn for the second player, then there is a possible response for the first player, who will therefore win.7. GAMES59Problem 9. Two players take turns placing bishops on the squares of a chessboard, so that they cannot capture each other (the bishops may be placed on squares of any color). The player who cannot move loses. Solution. Since a chessboard is symmetric with respect to its center, it is natural to try a symmetric strategy. But this time, since one cannot place a bishop at the center of the chessboard, the symmetry will help the second player. It might seem, from an analogy with the previous problem, that such a strategy would allow the second player to win. However, if he follows it, he cannot even make a second move! The bishop placed by the first player can take a bishop placed in the symmetric square. This example shows that in employing a symmetric strategy one must take into account that a symmetric move can be blocked or prevented, but only by a move the opponent has just made. Because of the symmetry, moves made earlier cannot affect a player's move. To solve a game using a symmetric strategy, one must find a symmetry such that the previous move does not destroy the chosen strategy. Therefore, to solve Problem 9 we must look not to the point symmetry of the chessboard, but to its line symmetry. We can choose, for example, the line between the fourth and fifth rows as the line of symmetry. Squares which are symmetric with respect to this line will be of different colors, and therefore a bishop on one square cannot take a bishop on the symmetric square. Therefore, the second player can win this game. The idea of a symmetric strategy need not be purely geometric. Consider the following problem. Problem 10. There are two piles of 7 stones each. At each turn, a player may take as many stones as he chooses, but only from one of the piles. The loser is the player who cannot move. Solution. The second player can win this game, using a symmetric strategy. At each turn, he must take as many stones as the first player has just taken, but from the other pile. Therefore the second player always has a move. ' The symmetry in this problem consists in maintaining the equality of the number of stones in each pile. Problem 11. Two players take turns placing knights on the squares of a chessboard, so that no knight can take another. The player who is unable to do this loses. Problem 12. Two players take turns placing kings on the squares of a 9 x 9 chessboard, so that no king can capture another. The player who is unable to do this loses. Problem 13. (a) Two players take turns placing bishops on the squares of a chessboard. At each turn, the bishop must threaten at least one square not threatened by another bishop. A bishop "threatens" the square it is placed on. The player who cannot move is the loser. (b) • The same game, but with rooks (castles). Problem 14. Given a 10 x 10 chessboard, two players take turns covering pairs of squares with dominoes. Each domino consists of a rectangle 1 square in width and 2 squares in length (which can be held either way). The dominoes cannot overlap. The player who cannot place a domino loses.60MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 15. A checker is placed on each square of an 11 x 11 checkerboard. Players take turns removing any number of checkers which lie next to each other along a row or column. The winner is the player who removes the last checker. Problem 16. There are two piles of stones. One has 30 stones, and the other has 20 stones. Players take turns removing as many stones as they please, but from one pile only. The player removing the last stone wins. Problem 17. Twenty points are placed around a circle. Players take turns joining two of the points with a line segment which does not cross a segment already drawn in. The player who cannot do so loses. Problem 18. A daisy has (a) 12 petals; (b) 11 petals. Players take turns tearing off either a single petal, or two petals right next to each other. The player who cannot do so loses. Problem 19! Given a rectangular parallelepiped of dimensions (a) 4 x 4 x 4; (b) 4 x 4 x 3; (c) 4 x 3 x 3, consisting of unit cubes. Players take turns skewering a row of cubes (parallel to the edges of the figure), so long as there is at least one cube which is not yet skewered in the row. The player who cannot do so loses. Problem 20. Two players take turns breaking a piece of chocolate consisting of 5 x 10 small squares. At each turn, they may break along the division lines of the squares. The player who first obtains a single square of chocolate wins. Problem 21. Two players take turns placing x's and o's on a 9 x 9 checkerboard. The first player places x's, and the second player places o's. At the end of the play, the first player gets a point for each row or column which contains more x's than o's. The second player gets a point for each row or column which contains more o's than x's. The player with the most points wins. §3. Winning positions Problem 22. On a chessboard, a rook stands on square al. Players take turns moving the rook as many squares as they want, either horizontally to the right or vertically upward. The player who can place the rook on square h8 wins. In this game, the second player will win. The strategy is quite simple: at each turn, place the rook on the diagonal from al to h8. The reason this works is that the first player is forced to move the rook off the diagonal at each turn, while the second player can always put the rook back on this diagonal. Since the winning square belongs to the diagonal, the second player will eventually be able to place the rook on it. Let us analyze this solution a little more deeply. We have been able here to define a class of winning positions (in which the rook is arr the diagonal from al to h8), which enjoys the following properties: (1) The final position ·of the game is a winning one; (2) A player can never move from one winning position to another in a single turn;(3) A player can always move from a non-winning position to a winning one in a single move. The discovery of such a class of winning positions for a given game is equivalent to solving the game. Indeed, moving to a winning position at each move constitutes a winning strategy. If the initial position of the game is a winning one, then the7. GAMES61second player will win (as in the game described above). Otherwise, the first player will win. For teachers. As the concept of a winning position generalizes a set of strategies, it can only be understood after solving several of the games presented in this section. As always, it is important to have students play each game before solving it. Problem 23. A king is placed on square al of a chessboard. Players take turns moving the king either upwards, to the right, or along a diagonal going upwards and to the right. The player who places the king on square hB is the winner. Problem 24. There are two piles of candy. One contains 20 pieces, and the other 21. Players take turns eating all the candy in one pile, and separating the remaining candy into two (not necessarily equal) non-empty piles. The player who cannot move loses.Problem 25. A checker is placed at each end of a strip of squares measuring 1 x 20. Players take turns moving either checker in the direction of the other, each by one or by two squares. A checker cannot jump over another checker. The player who cannot move loses.Problem 26. A box contains 300 matches. Players take turns removing no more than half the matches in the box. The player who cannot move loses. Problem 27. There are three piles of stones. The first contains 50 stones, the second 60 stones, and the third 70. A turn consists in dividing each of the piles containing more than one stone into two smaller piles. The player who leaves piles of individual stones is the winner. Problem 28. The number 60 is written on a blackboard. Players take turns subtracting from the number on the blackboard any of its divisors, and replacing the original number with the result of this subtraction. The player who writes the number 0 loses. Problem 29! There are two piles of matches: (a) a pile of 101 matches and a pile of 201 matches; (b) a pile of 100 matches and a pile of 201 matches. Players take turns removing a number of matches from one pile which is equal to one of the divisors of the number of matches in the other pile. The player removing the last match wins. §4. Analysis from the endgame: A method of finding winning positions Readers of the previous section may get the feeling that the discovery of a set of winning positions is based only on intuition, and is therefore not simple. We now describe a general method which will allow us to find a set of winning positions in many games. We return to Problem 23, the problem about the single king on a chessboard. Let us try to find a set of winning positions. As always, the final position of the game, with the king in square hB, must be a winning one. We therefore place a plus sign in square hB (see Figure 46). We will place the same sign in every other square at which the king occupies a winning position, and a minus sign in every square which is not a winning position (we will call them losing positions).MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)62•FIGURE46Since those squares from which the king can move to a winning square in a single move are losing squares, we arrive at Figure 47. From squares h6 and /8 we can move only to losing squares, so these must be winning positions (Figure 48). These new winning positions lead to new losing positions: h5, g5, g6, f7, e7, e8 (Figure 49). We continue in an analogous fashion (see Figures 50 and 51). After obtaining a set of minuses, we place plus signs in those squares from which any move at all leads to a losing square, then place minuses in those squares from which there is at least one move to a winning square. The pluses and minuses will finally be arranged as in Figure 52. It is not difficult to see that the squares with plus signs in them are exactly the winning squares indicated in the previous section .••• Figure 47Figure 48-~=~=~ - +- + .. - - -••- +Figure 50Figure 51Figure 49-+-+-+-+ --t-+-+-+ -+-+-+-+ -+-+-+-+ -------Figure 52The method of finding winning positions just described is called analysis from the endgame. Applying it to the game with the castle (Problem 22) from the previous section, it is not hard to derive the set of winning positions for this game as well. Working as in Figures 53 and 54, we soon arrive at Figure 55.7. GAMES• Figure 53•·------+ .------+ ------+------------+---------+---Figure 54Figure 5563---+-----+-----+-----____ _ _,.For teachers. Students often perform their own "analysis from the endgame" intuitively. That is, they can see to the end of the game from a few moves before, and begin to learn which of the last few possible moves are winning ones, then generalize this to the rest of the game. The best learning will occur if students make this discovery on their own (by playing the game), then are asked to articulate it. Problem 30. A queen stands on square cl of a chessboard. Players take turns moving the queen any number of squares to the right, upwards, or along a diagonal to the right and upwards. The player who can place the queen in square h8 wins. Solution. Using analysis from the endgame, we obtain the configuration of pluses and minuses given in Figure 56. Thus, the first player wins; in fact, he has a choice of three initial moves. These are to squares c5, e3, or dl.•FIGURE-----56For teachers. This game can serve as a good introduction to analysis from the endgame. Student exercises can then be created, for example, by replacing ' the square checkerboard in Problems 22, 23, 30 with a rectangular board of any dimensions, or with a board of some other unusual form. For instance, one might solve Problem 22 on a checkerboard with the middle four squares removed (or, with some other squares removed). The arrangement of pluses and minuses on this sort of checkerboard is shown in Figure 57, in which the missing squares are shaded .•FIGURE----5764MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)The following problem provides a chance for students to practice the technique of reformulating a game. Problem 31. Of two piles of stones, one contains 7 stones, and the other 5. Players alternate taking any number of stones from one of the piles, or an equal number from each pile. The player who cannot move loses. Solution. We can restate the situation in this problem as one which occurs on the usual chessboard. First we assign coordinates to each square, by numbering the rows from 0 to 7, starting at the top, and the columns from 0 to 7, starting at the right. Each position of the original game is characterized by an ordered pair of numbers: the number of stones in the first pile, followed by the number of stones in the second. To each such position we assign the square whose coordinates are these numbers. Now we note that a move in the original game corresponds to a queen's move on the chessboard, upwards, to the right, or on a diagonal upwards and to the right. This restatement of the problem makes the game identical to that of Problem 30. Notice that we can use the same technique to reformulate the games in Problems 10 and 20. Problem 32. A knight is placed on square al of a chessboard. Players alternate moving the knight either two squares to the right and one square up or down, or two squares up and one square right or left (and usual knight moves but in restricted directions). The player who cannot move loses. Problem 33. (a) There are two piles of 7 stones each. In each turn, a player may take a single stone from one of the piles, or a stone from each pile. The player who cannot move loses. (b) In addition to the moves described above, players are allowed to take a stone from the first pile and place it on the second pile. Other rules remain the same.Problem 34. There are two piles of 11 matches each. In one turn, a player must take two matches from one pile and one match from the other. The player who cannot move loses.Problem 35. This game begins with the number 0. In one turn, a player can add to the current number any natural number from 1 through 9. The player who reaches the number 100 wins. Problem 36. This game begins with the number l. In one turn, a player can multiply the current number by any natural number from 2 through 9. The player who first names a number greater than 1000 wins. Problem 37. This game begins with the number 2. In one turn, a player can add to the current number any natural number smaller than it. The player who reaches the number 1000 wins. Problem 38. This game begins with the number 1000. In one turn, a player can subtract from the current number any natural number less than it which is a power of 2 (note that 1 = 2°). The player who reaches the number 0 wins.CHAPTER 8Problems for the First Year As was emphasized in the preface, the first part of this book presents the basic topics for sessions of an "olympiad" mathematical circle (for students of age 11-13). However, these topics do not exhaust all the themes available for students of this age. In the present chapter we will try to fill this gap, at least partly. For teachers. We would also like to say that we do not recommend preparing a session using only problems pertaining to a single topic. You can also use nonstandard problems, which require something new and unusual, fresh ideas, or just the overcoming of technical difficulties. Since such problems are important for olympiads, contests, et cetera, we have gathered them together in this chapter. §1. Logical problemsFor teachers. When dealing with young students keep in mind that the most important goal is to teach them consistent and clear thinking; that is, how not to confuse cause and consequence; how to analyze cases carefully, without skipping any; how to build a chain of propositions and lemmas properly. The following problems in logic can help you in handling this. 1. Peter's mom said: "All champions are good at math." Peter says: "I am good at math. Therefore I am a champion!" Is his implication right or wrong? 2. There are four cards on the table with the symbols A, B, 4, and 5 written on their visible sides. What is the minimum number of cards we must turn over to find out whether the following statement is true: "If an even number is written on one side of a card then a vowel is written on the other side"? 3. A sum of fifteen cents was paid by two coins, and one of these coins was not a nickel. Find the values of the coins. 4. Assume that the following statements are true: a) among people having TV sets there are some who are not mathematicians; b) non-mathematicians who swim in swimming pools every day do not have TV sets. Can we claim that not all people having TV sets swim every day? 5. During a trial in Wonderland the March Hare claimed that the cookies were stolen by the Mad Hatter. Then the Mad Hatter and the Dormouse gave testimonies which, for some reason, were not recorded. Later on in the trial it was found out that the cookies were stolen by only one of these three defendants, and, moreover, only the guilty one gave true testimony. Who stole the cookies?6566MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)6. In a box, there are pencils of at least two different colors, and of two different sizes. Prove that there are two pencils that differ both in color and in size. 7. There are three urns containing balls: the first one contains two white balls, the second-two black balls, and the third-a white ball and a black ball. The labels WW, BB, and WB were glued to the urns so that the contents of no urn corresponds to its label. Is it possible to choose one urn so that after drawing a ball from it one can always determine the contents of each urn? 8. Three people-A, B, and C-are sitting in a row in such a way that A sees B and C, B sees only C, and C sees nobody. They were shown 5 caps-3 red and 2 white. They were blindfolded, and three caps were put on their heads. Then the blindfolds were taken away and each of the people was asked if they could determine the color of their caps. After A, and then B, answered negatively, C replied affirmatively. How was that possible? 9. Three friends-sculptor White, violinist Black, and artist Redhead-met in a cafeteria. "It is remarkable that one of us has white hair, another one has black hair, and the third has red hair, though no one's name gives the color of their hair" said the black-haired person. "You are right," answered White. What color is the artist's hair?The next eight problems take place on an island where all the inhabitants are either "knights" who always tell the truth or "knaves" who always lie (for many more such problems see ). 10. Person A said "I am a liar." Is he an inhabitant of our island?11. What one question might be asked of an islander to find out where a road leads-to the city of knights or to the city of knaves? 12. What one question might be asked of an islander to find out whether she has a pet crocodile? 13. Assume that in the language of the island the words ''yes" and "no" sound like "flip" and "flop", but we do not know which is which. What one question might be asked of an islander to find out whether he is a knight or a knave? 14. What one question might be asked of an islander so that the answer is always "flip"? 15. An islander A, in the presence of another islander B, said: "At least one of us is a knave." Is A a knight or a knave? What about B? 16. There are three people, A, B, and C. Among them is a knight, a knave, and a stranger (a normal person), who sometimes tells the truth and sometimes lies. A said: "I am a normal person."B said: "A and C sometimes tell the truth." C said: "B is a normal person."Who among them is a knight, who is a knave, and who is a normal person? 17. Several islanders met at a conference, and each of them told the others: "You are all knaves." How many knights might there be at that conference?8. PROBLEMS FOR THE FIRST YEAR67§2. Constructions and weighingsMathematical and logical problems whose solution consists of a particular construction; that is, creating an example, are very common and useful. The students should understand that a construction may serve as a complete solution to problems of a certain type (such as those starting with the words "Is it possible to ... ?"). Such problems are usually quite attractive to younger students, and they can spend a lot of time trying to find a constructive solution to a tricky question or puzzle. 18. There are two egg timers: one for 7 minutes and one for 11 minutes. We must boil an egg for exactly 15 minutes. How can we do that using only these timers? 19. There are two buttons inside an elevator in a building with twenty floors. The elevator goes 13 floors up when the first button is pressed, and 8 floors down when the second one is pressed (a button will not function if there are not enough floors to go up or down). How can we get to the 8th floor from the 13th? 20. The number 458 is written on a blackboard. It is allowed either to double the number on the blackboard, or to erase its last digit. How can we obtain the number 14 using these operations? 21. Cards with the numbers 7, 8, 9, 4, 5, 6, 1, 2, and 3 are laid in a row in the indicated order. It is permitted to choose several consecutive cards and rearrange them in the reverse order. Is it possible to obtain the arrangement 1, 2, 3, 4, 5, 6, 7, 8, 9 after three such operations? 22. The numbers 1 through 16 are placed in the boxes of a 4 x 4 table as shown in Figure 58 (a). It is permitted to increase all the numbers in any row by 1 or decrease all the numbers in any column by 1. Is it possible to obtain the table shown in Figure 58 (b) using these operations? 12341II813II8782810148101112371116131411118481218(8)(b)FIGURE 5823. Is it possible to write the numbers 1 through 100 in a row in such a way that the (positive) difference between any two neighboring numbers is not less than 50? 24. Divide a set of stones which weigh lg, 2g, 3g, ... , 555g into three heaps of equal weight. 25. Fill the boxes of a 4 x 4 table with non-zero numbers so that the sum of the numbers in the corners of any 2 x 2, 3 x 3, or 4 x 4 square is zero. 26. Is it possible to label the edges of a cube using the numbers 1 through 12 in such a way that the sums of the numbers on any two faces of the cube are equal?68MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE5927: Is it possible to place the numbers O through 9 in the circles in Figure 59 without repetitions so that all the sums of the numbers in the vertices of the shaded triangles are equal? 28. Prove that one can cross out several digits at the beginning and several at the end of the 400-digit number 84198419 ... 8419 in such a way that the sum of the remaining digits is 1984. 29. Find a two-digit number, the sum of whose digits does not change when the number is multiplied by any one-digit number. 30. Do there exist two consecutive natural numbers such that the sums of their digits are both divisible by 7? 31. Do there exist several positive numbers, whose sum is 1, and the sum of whose squares is less than 0.01? 32. A castle consists of 64 identical square rooms, having a door in every wall and arranged in an 8 x 8 square. All the floors are colored white. Every morning a painter walks through the castle recoloring floors in all the rooms he visits from white to black and vice versa. Is it possible that some day the rooms will be colored as a standard chessboard is? 33. Can one place a few dimes on the surface of a table so that each coin touches exactly three other coins? 34. In a warehouse N containers marked 1 through N are arranged in two piles. A forklift can take several containers from the top of one pile and place them on the top of the other pile. Prove that all the containers can be arranged in one pile in increasing order of their numbers with 2N - 1 such operations of the forklift. There are many problems involving weighing which are closely related to construction problems. In solving these problems, we must not neglect even the simplest or most unlikely cases. Arguments like "We will consider the worst case" are usually very vague and unacceptable. In all the problems of this set we consider "a weighing" as performed on a standard balance with two pans but without arrows or weights, unless otherwise specified.8. PROBLEMS FOR THE FIRST YEAR6935. There are 9 coins, one of whiCh is counterfeit (it is lighter than the others). Find the counterfeit coin using two weighings. 36. There are 10 bags with coins. One of them contains only counterfeit coins, each of which is 1 gram lighter than a genuine coin. Using only one weighing on a balance with an arrow showing the difference between weights on the pans, find the "counterfeit" bag.37. There are 101 coins, and only one of them differs from the other (genuine) ones by weight. We have to determine whether this counterfeit coin is heavier or lighter than a genuine coin. How can we do this using two weighings? 38. There are 6 coins; two of them are counterfeit and are lighter than the genuine coins. Using three weighings, determine both counterfeit coins. 39. There are 10 bags with coins, and some of these bags contain only counterfeit coins. A counterfeit coin is lg lighter than a genuine coin. One of the bags is known to be filled with the genuine coins. Using one weighing on a balance with one pan and with an arrow showing the weight on the pan, determine which bags are "counterfeit" and which are not. 40. There are 5 coins, three of which are genuine. One is counterfeit and heavierthan a genuine coin, and another one is counterfeit and lighter than a genuine coin. Using three weighings, find both counterfeit coins. 41. There are 68 coins of different weight. Using 100 weighings, find the heaviest and the lightest of the coins. 42. There are 64 stones of different weight. Using 68 weighings, find the heaviest and the second heaviest stones. 43. We have 6 weights: two green, two red, and two white. In each pair one of the weights is heavier. All the heavy weights have the same weight, and all the light weights have the same weight. Using two weighings, determine which weights are the heavy ones.44. There are 6 coins, two of which are counterfeit: they are O.lg heavier than the genuine coins. The pans of a balance are out of equilibrium only if the difference , of weights is at least 0.2g. Find both counterfeit coins using four weighings. 45. a) There are 16 coins. One of them is counterfeit: it differs in weight from a genuine coin, though we do not know whether it is heavier or lighter. Find the counterfeit coin using four weighings. b)'There are 12 coins. One of them is counterfeit: it differs by weight from a genuine coin, though we do not know whether it is heavier or lighter. Find the counterfeit coin using three weighings. 46! Fourteen coins were presented in court as evidence. The judge knows that exactly 7 of these are counterfeit and weigh less than the genuine coins. A lawyer claims to know which coins are counterfeit and which are genuine, and she is required to prove it. How can she accomplish this using only three weighings? §3. Problems in geometryThe problems in this section can be split naturally into two sets. The first set (Problems 47-57) continues the previous section: it is dedicated to geometric constructions. The second set contains more "standard" geometry problems.70MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)4 7. Draw a broken line made up of 4 segments passing through all 9 points shown in Figure 60.•••••••••FIGURE 60 48. Cut a square into 5 rectangles in such a way that no two of them have a complete common side {but may have some parts of their sides in common). 49. Is it possible to draw a closed 8-segment broken line which intersects each segment of itself exactly once? 50. Is it possible to cut a square into several obtuse triangles? 51. Is it true that among any 10 segments there always are 3 which can form a triangle? 52. A king wants to build 6 fortresses and connect each pair of them by a road.Draw a scheme of fortresses and roads such that there are only 3 crossroads, each formed by 2 intersecting roads. 53. Is it possible to choose 6 points on the plane and connect them by disjoint segments {that is, by segments which do not have common inner points) so that each point is connected with exactly 4 other points? 54. Can we tile the plane with congruent pentagons? 55. Cut a 3 x 9 rectangle into 8 squares.56. Prove that a square can be dissected into 1989 squares. 57. Cut an arbitrary triangle into 3 parts such that they can be rearranged to form a rectangle. 58. Points M and K are given on sides AB and BC of triangle ABC respectively. Segments AK and CM meet at point 0. Prove that if OM= OK and LKAC = LMCA, then triangle ABC is isosceles. 59. Altitude AK, angle bisector BH, and median CM of triangle ABC meet at one point 0, and AO= BO. Prove that triangle ABC is equilateral.60. In hexagon ABCDEF triangles ABC, ABF, FED, CDB, FEA, and CDE are congruent. Prove that diagonals AD, BE, and CF are equal. 61. Altitude CH and median BK are drawn in acute triangle ABC. If BK= CH, and LKBC = LHCB, prove that triangle ABC is equilateral.8. PROBLEMS FOR THE FIRST YEAR7162. Diagonals AC and BD of quadrilateral ABCD meet at point 0. The perimeters of triangles ABC and ABD are equal, as are the perimeters of triangles ACD and BCD. Prove that AO= BO. 63. Prove that the star shown in Figure 61 cannot be drawn to satisfy the following inequalities: BC> AB, DE> CD, FG > EF, HI> GH, KA> IK.D FHK FIGURE61§4. Problems on integersThis topic has already been discussed in the chapter "Divisibility and Remainders". However, there are many nice problems dealing with integers, so many that we consider it necessary to gather some of them in this section. For instance, the set of Problems 70--84 is just an extension of the chapter on divisibility. Other problems bring in new themes. 64. If every boy in a class buys a muffin and every girl buys a sandwich, they will spend one cent less than if every boy buys a sandwich and every girl buys a muffin. ' We know that the number of boys in the class is greater than the number of girls. Find the difference. 65. 175 Humpties cost more than 126 Dumpties. Prove that you cannot buy three Humpties and one Dumpty for one dollar. 66. In a class every boy is friends with exactly three girls, and every girl is friends with exactly two boys. It is known that there are only 19 desks (each holding at most two students), and 31 of the students in the class study French. How many students are there? 67. Two teams played each other in a decathlon. In each event the winning team gets 4 points, the losing team gets 1 point, and both teams get 2 points in case of a draw. After all 10 events the two teams have 46 points together. How many draws were there? 68. Four friends bought a boat. The first friend paid half of the sum paid by the others; the second paid one third of the sum paid by the others; the third paid one quarter of what was paid by the others, and the fourth friend paid 130 dollars. What was the price of the boat, and how much did each of the friends pay?72MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)69. The road connecting two mountain villages goes only uphill or downhill. A bus always travels 15 mph uphill and 30 mph downhill. Find the distance between the villages if it takes exactly 4 hours for the bus to complete a round trip.70. Do there exist natural numbers a and b such that ab(a - b) = 45045? 71. Let us denote the sum of three consecutive natural numbers by a, and the sum of the next three consecutive natural numbers by b. Can the product ab be equal to 111111111? 72. Prove that the last non-zero digit of the number 1985! is even. 73. The natural numbers x and y satisfy the relation 34x = 43y. Prove that the number x + y is composite. 74. Do there exist non-zero integers a and b such that one of them is divisible by their sum while the other is divisible by their difference? 75. The prime numbers p and q, and natural number n satisfy the following equality: 1 1 1 1 -+-+-=-. P q pq n Find these numbers. 76. Prove that a natural number written using one 1, two 2's, three 3's, ... , nine 9's cannot be a perfect square. 77. Each of the natural numbers a, b, c, and d is divisible by ab - ed. Prove that ab - cd equals either 1 or -1. 78. In a certain country, banknotes of four types are in circulation: 1 dollar, 10 dollar, 100 dollar, and 1000 dollar bills. Is it possible to pay one million dollars using exactly half a million notes? 79. The number 1 is written on a blackboard. After each second the number on the blackboard is increased by the sum of its digits. Is it possible that at some moment the number 123456 will be written on the blackboard? 80. Prove that the number 3999991 is not prime. 81. a) Find a seven-digit number with all its digits different, which is divisible by each of those digits. b) Does there exist an eight-digit number with the same property? 82. We calculate the sum of the digits of the number 19 100 . Then we find the sum of the digits of the result, et cetera, until we have a single digit. Which digit is this? 83. Prove that the remainder when any prime number is divided by 30 is either 1 or a prime number.84. Does there exist a natural number such that the product of its digits equals 1980?8. PROBLEMS FOR THE FIRST YEAR7385. A natural number ends in 2.. If we move this digit 2 to the beginning of the number, then the number will be doubled. Find the smallest number with this property. 86. Given a six-digit number abcdef such that abc - def is divisible by 7, prove that the number itself is also divisible by 7. 87. Find the smallest natural number which is 4 times smaller than the number written with the same digits but in the reverse order. 88. A three-digit number is given whose first and last digits differ by at least 2. We find the difference between this number and the reverse number (the number written with the same digits but in the reverse order). Then we add the result to its reverse number. Prove that this sum is equal to 1089. 89. Which number is greater: 23 00 or 3200 ? 90. Which number is greater: 31 11 or 1714 ?91. Which number is greater: 5099 or 99!? 92. Which number is greater: 888 ... 88 x 333 ... 33 or 444 ... 44 x 666 ... 67 (each of the numbers has 1989 digits)? 93. Which type of six-digit numbers are there more of: those that can be represented as the product of two three-digit numbers, or those that cannot?94. Several identical paper triangles are given. The vertices of each one are marked with the numbers 1, 2, and 3. They are piled up to form a triangular prism. Is it possible that all the sums of the numbers along the edges of the prism are equal to 55? 95. Can one place 15 integers around a circle so that the sum of every 4 consecutive numbers is equal either to 1 or 3? 96! Find a thousand natural numbers such that their sum equals their product.97. The numbers 21989 and 51989 are written one after another. How many digits in all are there? 98. A bus ticket (whose number in Russia consists of 6 arbitrary digits) is called "lucky" if the sum of its first three digits equals the sum of the last three. Prove that the number of "lucky" tickets equals the number of tickets with the sum of their digits equal to 27. §5. Miscellaneous 99. Fourteen students in a class study Spanish, and eight students study French. We know that three students study both languages. How many students are there in the class if every one of them is studying at least one language? 100. The plane is colored using two colors. Prove that there are two identically colored points exactly 1 meter apart.74MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)101. A straight line is colored using two colors. Prove that we can find a segment of non-zero length with its endpoints and midpoint colored the same. 102. A 8 x 8 square is formed by 1 x 2 dominos. Prove that some pair of them forms a 2 x 2 square. 103. A 3 x 3 table is filled with numbers. It is allowed to increase each number in any 2 x 2 square by l. Is it possible, using these operations, to obtain the table shown in Figure 62 from a table filled with zeros?4951018126137FIGURE62104. We call a bus overcrowded if there are more than 503 of the maximum allowable number of passengers inside. Children ride in several buses to a summer camp. Which is greater: the percentage of overcrowded buses or the percentage of children riding in the overcrowded buses? 105. Problem lists for the all-city olympiads in grades 6-11 are compiled so that each list contains eight questions, and there are exactly three questions in each grade which are not used in the other grades' olympiads. What is the maximum possible number of questions used by the problem committee? 106. All the students in a school are arranged in a rectangular array. After that, the tallest student in each row was chosen, and then among these John Smith happened to be the shortest. Then, in each column, the shortest student was chosen, and Mary Brown was the tallest of these. Who is taller: John or Mary? 107. Thirty chairs stand in a row. Every now and then a person comes and sits in one of the free chairs. After that, if any of the neighboring chairs is occupied, one of the person's neighbors stands up and leaves. What is the maximum number of chairs that can be occupied simultaneously, if originally a) all the chairs are free? b) ten chairs are taken? 108. Three pawns are placed on the vertices of a pentagon. It is allowed to move a pawn along any diagonal of the pentagon to any free vertex. Is it possible that after several such moves one of the pawns occupies its original position while the other two have changed their places? 109. None of the numbers a, b, c, d, e, or f equals zero. Prove that there are both positive and negative numbers among the numbers ab, cd, ef, -ac, -be, and -df. 110! Professor Rubik splits his famous 3 x 3 x 3 cube with an ax. What is the minimum possible number of blows he needs to split the cube into 27 small cubes if it is allowed to put some pieces on the top of others between blows?8. PROBLEMS FOR THE FIRST YEAR75111. The boxes of a sheet of graph paper are colored using eight colors. Prove that one can find a figure such as shown in Figure 63 which contains two boxes of the same color.FIGURE63112. A six-digit number is given. How many seven-digit numbers are there which will produce that number if one digit is crossed out? 113. How many bus tickets do you have to buy in a row to be sure you have purchased a "lucky" ticket? (See Problem 98 for the definition of the "lucky" ticket. Bus tickets are numbered consecutively, and the ticket 999999 is followed by the ticket 000000). 114! There was a volleyball tournament in which each team played every other team exactly once. We say that team A is better than team B if A defeated B, or if there is a team C such that A defeated C and C defeated B. Prove that the team which won the tournament is better than any other team. 115! A 20 x 30 rectangle is cut from a sheet of graph paper. Is it possible to draw a straight line which intersects the interiors of 30 boxes of the rectangle? 116. The natural numbers 1 through 64 are written in squares of a chessboard, and each number is written exactly once. Prove that numbers in some pair of neighboring squares differ by at least 5.CHAPTER 9Induction I. S. Rubanov§1. Process and method of induction (An Introduction for Teachers). Almost everyone has once had fun arranging dominoes in a row and starting a wave. Push the first domino and it topples the second, the second will topple the third and so forth until all the dominoes are toppled. Now let us change the set of dominoes into an infinite series of propositions: P1 , P 2 , P 3 , ••• , numbered by positive integers. Assume that we can prove that: (B): the first proposition of the series is true; (S): the truth of every proposition in the series implies the truth of the next one.Then, in fact, we have already proved all the propositions in the series. Indeed, we can "push the first domino", i.e., prove the first statement (B), and then statement (S) means that each domino, in falling, topples the next one. Whatever the "domino" (proposition) we choose, it will be eventually hit by this wave of "falling dominos" (proofs). This is a description of the method of mathematical induction (MMI). Theorem (B) is called base of induction, and theorem (S) is the inductive step. Our reasoning with the wave of falling dominoes shows that step (S) is but a shortened form of the chain of theorems shown in the figure below:We will call theorems in this chain "steps", and the process of their successive proof-"the process of induction". This process can be visually represented as a wave of proofs, running from statement to statement along a chain of theorems. Psychologically, the essence of induction is in its process. How can we teach this? We will try to show you in a dialog between teacher ( "T") and student (''S"), which roughly resembles a session of a real mathematical circle. At the end of the dialog some methodological comments for the teacher are given (references to these comments are indicated in the text of the dialog).Problem 1. T: One box was cut off from a 16 x 16 square of graph paper. Prove that the figure obtained can be dissected into trominos of a certain type-"corners" (see Figure 64.) S: But this is easy-any "corner" has three boxes, and 162 - 1 is divisible by 3. 7778MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE 64T: If it is so easy, could you cut a 1 x 6 band into "corners"? Six is also divisible by 3! S: Well ... Actually, I should not have said that. I don't exactly know how to solve this problem. (I) T: OK, you cannot solve this problem. Perhaps you can think of another problem which is similar yet easier? S: Well, you can take another square, of smaller size, say, 4 x 4. T: Or 2 x 2?< 2 > S: But there is nothing to prove in this case-when you cut out any box what you get is just a "corner". What sense does that make? T: Try now to solve the problem about the 4 x 4 square. S: Uh-huh. A 4 x 4 square can be cut into four 2 x 2 squares. It is clear what to do with the one with the cut box. What about the other three? T: Try to cut a "corner" from them, located in the center of the big square (see Figure 65).FIGURE 65S: Got it! Each of them would lack one box and turn into a "corner". So we can solve the problem for a 4 x 4 square too. Now? T: Try an 8 x 8 square. It can be dissected into four 4 x 4 squares. Make use of this. S: Well, we can reason as we did before. One of those squares has the "missing" box in it. And we have already proved that this one can be cut into "corners". The three other squares will lack one box after we cut out one "corner" in the center of 8 x 8 square--so we will be able to dissect them, too.9. INDUCTION79T: Do you see now how to solve the original problem? S: Sure. We cut the 16 x 16 square into four 8 x 8 squares. One of them contains the cut box. We have just proved that it can be dissected into "corners" , right? Then we cut out a "corner" in the center of the big square and we get three more 8 x 8 squares, each without one box, and each can be cut into "corners". That's it! T: Not yet. We solved this problem by building "bridges" from similar but simpler questions. Could we build such bridges once more, to other, more complicated questions?< 3 l S: Of course. Let us prove that one can dissect a 32 x 32 square into "corners". We just divide it mentally into four 16 x 16 squares ... T: There you are! But . . . is it possible to go further? S: Certainly. Having proved the proposition for a 32 x 32 square we can now derive, in the very same way, a method of dissection for a 64 x 64 square, then for a 128 x 128 square and so on ... T: Thus, we have an infinite chain of propositions about squares of different sizes. Can we say that we have proved them all? S: Yes, we have. First, we proved the first statement in the chain-about a 2 x 2 square. Then we derived the second proposition from it, then the third from the second, et cetera. It seems quite clear that going along this chain we will reach any of its statements; therefore, all of them are true. T: Right. It looks like a "wave of proofs" running along the chain of theorems: 2 x 2 --> 4 x 4 --> 8 x 8 --> . . . . It is quite evident that the wave will not miss any statement in this chain.Methodological remark. A few comments on the previous dialog. Comment N• l. When the student "proved" the statement of the problem using divisibility by 3, the teacher faced a typical classroom problem-how to explain the nature of the error, without giving away too much. The teacher overcomes this with a counterexample, prepared beforehand. It is always useful to be aware of such obstacles and know some ways to avoid them. This must be done easily, without major distraction from the flow of solution. Comment N• 2. This retort is not accidental. The student can hardly think about the 2 x 2 case as something important-it's not a problem at all (we will come across this psychological moment several times). However, the teacher knows this case is easier to start with. Comment N• 3. The following "step-by-step" scheme appears in this part of the dialog: 2 x 2 --> 4 x 4 --> 8 x 8 --> 16 x 16 . We have here the beginning of the induction process: the base 2 x 2 and the first three steps. It is essential that we have made enough induction steps for the student to notice an analogy. Now, after the hint, he is able to develop the whole process of induction. In fact, there are other inductive solutions to this question but they would not yield any educational benefit, since the notion of induction in them is not as clear80MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)as in the solution given above. Thus, the teacher leads the student away from these, using directive hints. The teacher here has played his part precisely: sometimes he leads away from a deceptive analogy and helps to save the student's energy. Unobtrusiveness is very important: the more the student does on his/her own the better. Let us sum up the results. The student (but more often this is the responsibility of the teacher) explained the scheme of MMI. The underlined sentence ("going along this chain we will reach any of its statements") is but an informal statement of the principle of mathematical induction which is the cornerstone of MMI. You can read about the formal side of it in any of the books [76, 78, 79]. We must say, though, that it would not be wise to talk about this at the very beginning of the discussion. It may be premature or even harmful since formalization of this intuitively clear statement may give rise to feelings of misunderstanding and uncertainty. On the contrary, one must use all means to make this scheme as evident and vivid as possible. Aside from the "wave" and dominoes (see Figure 66), other useful analogies include climbing a staircase, zipping a zipper, et cetera.FIGURE66 Now let us go on with our dialog: T: So, we have proved an infinite series of statements about the possibility of dissecting squares into "corners". Now, we write them all down, without any "et cetera's".S: But . . . we will certainly run out of paper. T: Yes, we would, if we wrote each statement separately. But all the statements look alike. Only the size of squares differ. This fact allows us to encode the whole chain in just one line: () A 2n x 2n square with one box cut out can be dissected into "corners". Here we have the variable n. Each statement in our chain can be obtained by replacing n with a number. For instance, n = 5 gives us a proposition about the 32 x 32 square. And what is the tenth proposition in the chain? S: We substituten = lOtoget the statement about 210 x2 10 , i.e., the 1024xl024 square.9. INDUCTION81T: Look at this: a variable is such a common thing, but it is really powerful-it allows us to fold an infinite chain into one short sentence. So, what is "a variable"? S: Well . . . it is just a letter . . . an unknown ... T: Remember: this "letter" denotes an empty space, a room, where you can put various numbers or objects. You can also call it a "placeholder". Those numbers or objects that are allowed to be put into the "room" are called its possible values. For example, the values of the variable n in () are the natural numbers (positive integers). Because of this, sentence () replaces the infinite chain of statements. Now we must recall the proof of the infinite chain (). Let us number all the statements: P1 is the one about the 2 x 2 square, P2 is about the 4 x 4 square, and so on. First we proved proposition P1 • Then we dealt with the infinite chain of similar theorems: if P1 is proved, then P2 is true; if P2 is proved, then P3 is true, et cetera. Let us try to encode this chain also: "For any natural n ... S: ... if Pn is true, then Pn+l is also true." T: And now, please, decode this phrase: what do Pn and Pn+l denote? S: () "Whichever natural number n is, if it is already proved that the 2n x 2n square without one box can be cut into "corners", then it is also true that the 2n+l x 2n+ 1 square without one box can be cut into "corners"." T: Can you prove that? S: I think so. We mentally divide the 2n+l x 2n+ 1 square into four 2n x 2n squares. One of these lacks one box, and can be dissected into "corners" by assumption. Then we cut out one "corner" in the center of the big square so that it contains one box from each of the other three 2n x 2n squares. After that, we can use the assumption again! T: Absolutely. Note that as soon as you proved the general theorem (), you proved all the theorems from the chain encoded by (). For example, if n = 1, we get our old proof stating that the possibility of dissecting the 2 x 2 square implies the possibility of dissecting the 4 x 4 square. Therefore, just as () can be considered as encoding a whole chain of theorems, your reasoning can be considered as encoding a whole "wave of proofs" of those theorems. I believe you got it: it is useful and easier to prove a chain of similar theorems in this convoluted way. But first you must learn how to express a chain of theorems this way.The method we applied in solving Problem 1 is what we call the METHOD OF MATHEMATICAL INDUCTION (MMI). What is its essence? First, we regard statement () not as one whole fact but as an infinite series of similar propositions. Second, we prove the first proposition in the series-this is called the "base of the induction." Third, we derive the second proposition from the first, the third (in the same way) from the second, et oetera. That was the "inductive step"; ()-is its shortened (convoluted) form. Sinoe, step by step, we can reach any proposition from the base, they are all true.82MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)"A method is an idea applied twice" (G. Polya) To learn MMI successfully it is usually necessary to replay the scenario above for several different questions. Consider now four more "key problems".Problem 2. Prove that number 111 ... 11 (243 ones) is divisible by 243.Hint. This question may be generalized to the proposition that a number written with 3n ones is divisible by 3n. Base: 111 is divisible by 3. Students often start with the statement that 111,111,111 is divisible by 9---ur base sounds too easy to them. Here we have two obstacles a) an attempt to generalize the divisibility tests for 3 and 9 and use an incorrect "test" for divisibility by 27; b) reasoning of the sort: "if a number is divisible by 3 and 9, then it is divisible by 27 = 3 x 9." The correct kind of inductive step is to divide the number written with 3n+1 ones by the number written with 3n ones and check that the result is a multiple of 3. Problem 3. Prove that for any natural number n, greater than 3, there exists a convex n-gon with exactly 3 acute angles. Comment. This question is a good key problem if students already know the fact that a convex polygon cannot have more than 3 acute angles. The base n = 4 can be checked by direct construction. Inductive step: Jet us "saw off" one of the non-acute angles. Then the number of angles in the polygon increases by 1 and all the acute angles are retained (see Figure 67).FIGURE67Another way to do this--to build a new angle on one of the sides-is a bit more difficult. There are also other solutions (using inscribed polygons and so on) but most are more difficult for students to make precise. Perhaps the teacher can even give a hint about "sawing off" an angle. The statement of the question is obviously true for n = 3, but we will not gain anything by starting the induction from 3, because the method fails when you try to make the step from n = 3 to n = 4. Our third question gives an example of construction by induction. You can read about it in more detail in .9. INDUCTION83Problem 4. ("Tower of Hanoi") Peter has a children's game. It has three spindles on a base, with n rings on one of them. The rings are arranged in order of their size (see Figure 68). It is permitted to move the highest (smallest) ring on any spindle onto another spindle, except that you cannot put a larger ring on top of a smaller one. Prove thatFIGURE 68 a) It is possible to move all the rings to one of the free spindles; b) Peter can do so using 2n - 1 moves. c)•It is not possible to do so using fewer moves. Hint. a), b): The base (n=l) is easy. Inductive step: We have n = k + 1 rings. By the inductive assumption we can move all but the largest ring to the third spindle using 2k - 1 moves. Then we move the remaining ring to the second spindle. After that we can move all the rings from the third spindle to the second using 2k - 1 moves. In all, we have made (2k - 1) + 1 + (2k - 1) = 2k+ 1 - 1 moves. It is useful to do the first few steps of the induction "manually", even using a physical model. c) This question must be used with care-it is more difficult than the others given here. The main idea of the proof is that to move the widest ring to the second spindle, we must first move all the other rings to the third spindle. Problem 5. The plane is divided into regions by several straight lines. Prove that one can color these regions using two colors so that any two adjacent regions have different colors (we call two regions adjacent if they share at least one line segment). Hint. Here we encounter another obstacle: no explicit variable for induction is given in the statement. Thus, we should start the solution by revealing this hidden variable. To do this, we can rewrite the statement as follows: ''There are n straight lines on a plane .... " The base can be n = 1 or n = 2 (either will work). The inductive step: remove for a moment the (k + l)st line, color the map obtained, then restore the removed line and reverse the colors of all the regions on one side of the line. The first few key problems can be discussed according to For teachers. the scenario of the dialog above; that is, growing the chain from one particular proposition. Students should understand the essence of the process of induction and the connection between cha.ins of theorems and propositions using integer variables.MATHBMATICAL CIRCLES (RUSSIAN EXPERJENCE)84If students are not well prepared one can skip the idea of constructing a chain of inductive steps. This can be introduced later, at a second stage, whose goal is to teach the students to work with the inductive step in its convoluted form. While doing so it would be wise to give questions in a general form (like in Problems 3 and 4). There we already have a chain of statements and the solution may start right from the "unfolding", as follows: "Here we have a convoluted chain of theorems. What is the first theorem? The fifth? The 1995th?" However, the chain of inductive steps should be developed and convoluted according to the old scheme, until students get accustomed to it and understand well the connection between a long chain and its convoluted form. To sum up their experience with key problems, students should have a clear General Plan for Solution by the Method of Mathematical Induction 1. Find, in the statement of the question, a series of similar propositions. If variables are hidden you should reveal them by reformulating the question. If there is no chain, try to grow that chain so that the question will be a part of it. 2. Prove the first proposition (base of the induction). 3. Prove that for any natural number n the truth of the nth proposition implies the truth of the (n + l)st proposition (inductive step). 4. If the base and the step are proved, then all the propositions in the series are proved simultaneously, since you can reach any of them from the base by moving"step-by-step".The last item in this scheme is the same for all the problems, so it is often skipped. However, knowing it is vital. Also, the first item is not emphasized and is natural for those who are used to MMI; nevertheless we recommend that the students pay close attention to it for a while. §2. MMI and guessing by analogy We continue our dialog. Problem 6. Into how many parts do n straight lines divide a plane if no two of them are parallel and no three meet at the same point? (Figure 69 shows an example where n = 5.) S: Let us try to follow the scheme. Do we have a chain? It seems so: into how many parts does one line divide a plane? 2 lines? 3 lines ... ? The base is evident: one line dissects a plane into 2 parts (half-planes). T: Or zero lines-into one part. S: By all means. Item three-the inductive step ... !? T: I can understand your embarrassment: we run into a new difficulty. In the previous problems we dealt with chains of statements, not with chains of questions. But we will get a chain of statements if we give hypothetical, unproved answers to these questions. S: How can we?9. INDUCTION85FIGURE 69T: Try to guess a rule, a function giving the number of parts Ln in terms of the number of lines n. Physicists would do an experiment. We can experiment too, calculating the numbers Ln for small values of n. Go ahead! S: OK. So, Lo = 1, L 1 = 2, L 2 = 4, L3 = 7, L 4 = 11. I must think a little .... Ah, I got it! When you add the nth line the number of parts increases by n. Hence, Ln = 1 + (1 + 2 + ... + n). I did it! T: No, not yet. Don't forget that you have only guessed it, not proved it. You have checked your result only for n = 0, 1, 2, 3, 4. For all other values of n this is just a guess based on your conjecture that adding the nth line increases the number of parts by n. What if this is wrong? The only guarantee is a proof. S: . . . by the method of mathematical induction. T: But we should enhance our plan from §1 by another item:,la. If there is a chain of questions rather than a chain of statements in a mathematical problem, insert your hypothetical answers. You can guess the answers by experimenting with the first few questions in the chain. However, after you are sure the answers are correct, don't forget to prove them rigorously. S: Now I know how to get over this. We have already proved the base, right? To prove the inductive step is easy: the nth line intersects the other lines at n - 1 points, which divide the line into n parts. Therefore the nth line divides n of the old parts of the plane into new parts.The process of guessing by analogy, just demonstrated by our student, is a very powerful and, sometimes, very dangerous tool: it is tempting to mistake the regularity one finds as a proof. The two examples below can serve as good medicine for this disease. Problem 7. Is it true that the number n 2 +n+41 is prime for any natural numbern? Hint. The answer is no: For n = 40 we have 402 + 40 + 41 = 41 2 , and for n = 41 41 2 + 41 + 41 = 41 · 43. But anyone trying to find an answer by "experimenting"86MNrHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)with small values of n would come to the opposite conclusion, since this formula gives prime numbers for n from 1 through 39. This famous example was given by Leonard Euler. Problem 8! A set of n points is taken on a circle and each pair is connected by a segment. It happens that no three of these segments meet at the same point. Into how many parts do they divide the interior of the circle? Hint. For n = 1, 2, 3, 4, 5 we obtain 1, 2, 4, 8, and 16 respectively. This result provokes a guess to the formula 2n- 1 . However, in fact, the number of parts equals n(n-l)(~:;-2)(n-3) + n(,,,-1) + l. Other similar examples can be found in . §3. Classical elementary problems Among classical MMI problems in elementary mathematics three large groups can be distinguished: proofs of identities, proofs of inequalities and proofs of divisibility questions. Though their solutions by MMI seem to be quite simple, students usually encounter some obstacles of a psychological as well as of a methodical nature. We begin by discussing these. T: Let us talk more about Problem 6. Do you like the way the formula 1 + (1 + 2 + 3 + ... + n) looks? S: Not much. It is too bulky. It would be better to get rid of this ellipsis (the three dots). T: No problem. You can prove by MMI that 1+2 + 3 + ... + n = n(n + 1)/2. S: But . . . to use MMI you need a chain of statements .. . T: Take a close look: there is variable n in the formula. As we know, this is a good sign of a convoluted set of problems. Substitute, for instance, 1995 for n. S: We get 1 + 2 + ... + 1995 = 1995 · 1996/2. T: That is, a numerical equation. Our convoluted set of problems consists of all these equations (for n = 1, 2, 3, ... , 1995)! To prove the formula means to show that all these numerical equations are true. If we do this, we say that this equation is "true for all admissible values of the variable" and it is called an identity. If an identity contains an integer variable you can try to prove it by induction. S: What if our equation is not true for some n? T: Then it is not an identity and we will not be able to prove it-the proof of either the base or the inductive step will not go through. Actually, to distinguish between identities and other, arbitrary equations with variables, you must preface identities with phrases like "for any natural number n . . . it is true that ... ", but this is not the usual practice. It is implied that the reader knows from the context whether an identity or a conditional equation is being discussed. S: Well, let us apply MMI. Base: n = l. So we must prove that . . . 1 + 2 + ... + 1 = 1 . 2/2 = l?! T: No, no. We must prove that 1 = 1 '. 2/2. You were puzzled by the formula 1+2 + ... + n. This is quite good and convenient, but for n = 1 its "tail" 2 + ... + n makes no sense and, in fact, does not exist at all. S: OK, so the base is clear. Let us move to the second equation in the series. We must show that 1 + 2 = 2 · 3/2. This is easy: 3 = 3. Now, move to the third equation: 1 + 2 + 3 = 3 · 4/2. This is easy too: 6 = 6. To the fourth . . . it is just another simple calculation. So, what now? Must we check each equation directly? We haven't got any steps!9. INDUCTION87T: 'fry to rewrite the step in a general, convoluted form. S (after a while): I cannot do that either. For teachers. To a person who has mastered MMI well enough, the proof of identities may seem rather trivial. However, our dialog shows two sources of problems for students. First, students often do not accept an identity with an integer variable as a chain of statements. This is probably because simple numerical identities are not considered independent propositions. Also, what is interesting in a statement such as 1 + 2 + 3 = 3 · 4/2 = 6? Second, it is next to impossible to see how the general form of the inductive step looks. Indeed, when you check the equations 1 + 2 = 2 · 3/2, 1+2 + 3 = 3 · 4/2, and so on, there is no connection between two successive facts--you just check them. That is why identities, despite their simplicity, cannot serve as key questions. To start learning and teaching mathematical induction from these will create trouble (this is not very important for really gifted students--they will manage to learn the method in any case). On the other hand, identities are very useful for practice, because their proofs are usually short and clear.T: Well, I will help you. Imagine that we follow the steps of the induction, one after another and the wave of proofs have reached the kth statement. What is that statement? S: We obtain 1+ 2 + 3 + ... + k = k(k + 1)/2. (#) T: Exactly. Now, tell me, please, what is the next statement, which the wave has not yet reached? S: Certainly, n = k + 1 and we get 1+2 + ... + (k + 1) = (k + l)(k + 2)/2. T: Good. Let us write this as follows: 1 1+2+3 + ... + k + (k + 1) = 2(k + l)(k + 2).(##)Now, tell me what would be the next step of induction? S: That's clear: to derive(##) from(#). T: Assume that we learned how to derive (##) from (#) for any natural number k. Then we would have all the steps of induction proved at once. This means that the inductive step states that: For any natural k the equation 1+2+ .. . +k = k(k+ 1)/2 implies the equation 1 +2+ ... + (k+ 1) = (k+ l)(k+2)/2. Jn other words: (#) is given, and we must prove (##) (if k is an arbitrary natural number). For convenience we denote the left sides of(#) and (##) as Sk and sk+I respectively. S: Proposition (##) shows that Sk+• = Sk + (k + 1) (that is why the teacher has written the next-to-last summand!). Now we have already learned that Sk = k(k + 1)/2. Thus we have sk+I11= 2 k(k + 1) + (k + 1) = 2 [k(k + 1) + 2(k +1)]1= 2 (k + 1)(k + 2).T: Remember the helpful idea that we used to prove the inductive step: the left side of equation (##) was expressed with the left side of (#) and the latter was substituted into the right side of(#).88MATHEMATICAL CIRCLES (RUSSIAN EXPERJENCE)For teachers. Another difficulty now arises in connection with identities. It may not be clear to a student how to make a step "in letters". The teacher in our dialog showed how to overcome that. It is important that he used another letter-different from that used in the statement of the identity-to denote the variable. The point is that the letter k plays the role not of a variable but ofa constant (though arbitrary) number marking the place that the wave of our inductive proof reached at the given moment. It will become a variable later, in the general statement of the inductive step. Quite often, the variables in the statement of the proposition and in the step are both denoted by the same letter. While stating the step theorem, phrases like " ... now we substitute n + 1 in place of n" are used. This is not advisable in the beginning of the study since it disorients most students conceptually (it is hard to see a chain in the statement of the inductive step) as well as technically (it is not that easy for a beginner to substitute n + 1 for n). Now we can say goodbye to the characters in our dialog and go on to deal with problems. Problems 9-16 are about identities with the natural number n as their variable. Problem 9. Show that 1 + 3 + ... + (2n - 1) = n 2 . Problem 10. Show that 12 +2 2 + ... +n2 =n(n+l)(2n+l)/6. Problem 11. Show that 1 · 2 + 2 · 3 + ... + (n - 1) · n = (n - l)n(n + 1)/3. Problem 12. Show that 1 1 1 n-1 N + 2-3 + ... + (n-l)n =---.,:;- · Problem 13. Show that1 + x + x 2 + ... + xn = (xn+I - l)/(x - 1) . Problem 14. Show that 1 1 1 n a(a+b) + (a+b)(a+2b) + ... + (a+(n-l)b)(a+nb) = a(a+nb)'where a and b are any natural numbers. Problem 15. Show that m! (m+ 1)! (m+n)! (m+n+ 1)! or+--1-,-+ ... +--n-,-= n!(m+l) , where m, n = 0, 1, 2, .... Problem 16. Show thatComments. In Problems 9-15 the proof of the inductive step is exactly the same as in the dialog. However, in Problem 16, it may be proved more easily by representing the (k+l)st left side not as a sum but as the product of the kth left side and (1-f,- ). This trick may also be useful in proving certain inequalities (see below). In Problem 11 the base is not n = 1 but n = 2. Students should see that this doesn't influence the process of induction.9. INDUCTION89In Problem 15 induction is possible on either of the two variables. It is instructive to carry out and compare both proofs. Remember to start from zero! Problems 11 and 12 are special cases of Problems 15 (for m = 2) and 14 (for a = b = 1) respectively. Given other values of m, a, and b we obtain any number of exercises like 11 and 12. It would be wise to let good students try to find the statement of the general problem which generates these exercises.Most of the identities 9-16 have good non-inductive proofs which are not too difficult. Problem 9 has a neat geometric proof (see Figure 70). Identity 11 can be obtained from identities 9 and 10. Identity 13 can be proved by division of xn+I -1 by x - 1, and identity 16 by direct calculation. To prove identity 12 it suffices to note that its left side equals(1- !) + (! -!) + ... + (-1 - .!.)n ' 223n-1and that this sum "telescopes" .FIGURE70This device works for other identities too. Discussion of these alternative proofs can be very helpful to students who have already mastered MMI. Divisibility questions constitute the next natural step in our study. The techniques of forming statements and inductive steps are similar to those for identities: we usually find the increment of the expression under consideration and prove that it is divisible by a given number. Problems 17-19 have simple alternative solutions (using modular arithmetic). The rather difficult Problem 22 may serve as the source of a number of exercises like 18-19. Prove that for any natural number n Problem 17. n 3 + (n + 1) 3 + (n + 2) 3 is divisible by 9. Problem 18. 3•n+2 + Bn - 9 is divisible by 16. Problem 19. 4n + 15n - 1 is divisible by 9. Problem 20. 11 n+2 + 12 2n+l is divisible by 133. Problem 21. 23" + 1 is divisible by 3n+l.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)90Problem 22~ abn + en + d is divisible by the positive integer m given that a + d, (b - l)c, and ab - a+ care divisible by m. Our trio of standard MMI themes is completed by questions involving inequalities. Here the proofs of the inductive steps are usually more varied (see ). Prove the following inequalities: Problem 23. 2n > n, where n is an arbitrary natural number. Problem 24. Find all natural numbers n such that a) 2n > 2n+l; b) 2n >n2. Problem 25. 13 1 1 1 n + 1 + n + 2 + · · · + 2ri: > 24' n = 2' 3• · · · · Problem 26. 2n > 1 + n~, n = 2, 3, .... Problem 27. Prove that the absolute value of the sum of several numbers does not exceed the sum of the absolute values of these numbers. Problem 28. (1 + x)n > 1 + nx, where x > -1, x # 0, and n = 2, 3, .... Problem 29. _1_·3_·_5_._.·~(2_n_-~1) < __ 1_ 2·4·6 ... 2n - v'2n+l' where n is any natural number. Hints. 23, 24: To prove the inductive step you may show that for any n, the increment of the left side of the inequality is greater than the increment of the right side. 24b: Use 24a to prove the step. 25: Prove that the left side of the inequality is monotonically increasing. 27: Induction can proceed on the number of summands. 28, 29: See the hint to Problem 16. §4. Other models of MMI So far we have been dealing with the basic version of MM!. When this is well learned we can try other, more complicated forms of induction. Some of these can be considered corollaries of the basic form, but it is more natural from a methodological point of view to discuss them separately, keeping in mind the image of "a wave of proofs".First, consider the method "Induction from all n :5 k ton= k + 1", sometimes called "strong induction".In the usual method of MMI, the inductive step consists of deriving propositionP.+ 1 from the previous proposition Pk· Sometimes, however, to show the truth of Pk+ 1 we must use more than one (or even all) of the previous statements P1 throughPk. This is certainly valid, since the wave has reached Pk and, therefore, all the propositions in the chain preceding it are also already proved. Thus the statement of the inductive step is: (S'): For any natural k the truth of P 1 , P2 , ••• , and Pk implies the truth of pk+I·9. INDUCTION91Consider an example.Problem 30. Prove that every natural number can be represented as a sum of several distinct powers of 2. Solution. First, let us prove the base. If the number given equals 1 or 2, then the existence of the required representation is simple. Now denote the given number by n and find the largest power of2 not exceeding n. Let it be 2m; that is, 2m :5 n < 2m+l. The difference d = n - 2m is less than n and also less than 2m, since 2m+1 = 2m + 2m. By the induction hypothesis, d can be represented as a sum of several different powers of 2, and it is clear that 2m is too big to be included. Thus, adding 2m we get the required expression for n. The induction is complete. Problem 31. Prove that any polygon (not necessarily convex) can be dissected into triangles by disjoint diagonals (they are allowed to meet only at vertices of the polygon). Hint. Use an induction on the number of sides. The inductive step is based on a lemma stating that each polygon (except a triangle) has at least one diagonal which lies completely within the polygon. Such a diagonal dissects the polygon into two polygons with fewer sides.Another scheme of MMI is demonstrated byProblem 32. It is known that x + 1/x is an integer. Prove that xn + 1/xn is also an integer (for any natural n). Solution. We have (xk + 1/xk)(x + 1/x) = xk-l + 1/xk-l + xk+l + 1/xk+ 1 and hence xk+ 1 +1/xk+ 1 = (xk + 1/xk)(x + 1/x) - (xk-l + 1/xk- 1). So we see that the (k + l)st sum is an integer if the two preceding sums are integers. Thus the process of induction will go as usual if we check that the first two sums, x + 1/x and x 2 + 1/ x 2 , are integers. This is left to the reader. Comment. A peculiarity of this version of MMI is that the inductive step is based on two preceding propositions, not one. Thus, the base in this case consists of the first two propositions in the series (it is natural to use the word base for that starting segment of the chain in which the statements must be checked directly). Problem 33. The sequence a1, a2, ... , an, ... of numbers is such that a1 = 3, = 5, and an+! = 3a,, - 2an-l for n > 2. Prove that a,, = 2n + 1 for all natural numbers n.a2Hint. See the more general Problem 43. Remark. In Problem 33 and the next three problems we will encounter not only proof by induction but also definitions by induction: all elements of the given sequences, except for the first few, are defined by induction, using the preceding elements. Sequences defined in this way are called recursive; see and for more details. See also , Chapter 2, about definitions, constructions, and calculations using induction. Problem 34. The sequence (an) is such that: a1 = 1, a2 = 2, an+! =a,, - n-1 if n > 2. Prove that a..+ 6 = a,, for all natural numbers n.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)92Problem 35. The sequence of Fibonacci numbers is defined by: F 1 = F2 = 1 and Fn+I = Fn + Fn-1 if n;::: 2. Prove that any natural number can be represented as the sum of several different Fibonacci numbers. Problem 36! Prove that the nth Fibonacci number is divisible by 3 if and only if n is divisible by 4. Hint. It is not easy to prove this fact alone by induction. Prove a more general statement about the repetition of remainders of Fibonacci numbers modulo 3 (with period 8). If you want to know more about Fibonacci numbers, see . Problem 37. A bank has an unlimited supply of 3--peso and 5-peso notes. Prove that it can pay any number of pesos greater than 7. Hint. Try induction on the number of pesos the bank must pay. The base consists of three facts: 8 = 5 + 3, 9 = 3 + 3 + 3, 10 = 5 + 5. Inductive step: if the bank can pay k, k + 1, and k + 2 pesos, then it can pay k + 3, k + 4, and k + 5 pesos. This induction with a compound base may be split into three standard inductions using the following schemes:8 -> 11->14 ->... ,9 -> 12 -> 15 ->... ,and 10 -> 13 -> 16 ->....Note that a similar splitting is impossible in Problems 32-36. There also exists a non-inductive solution to this problem based on the equations 3n + 1=5+5 + 3(n - 3) and 3n + 2 = 5 + 3(n- 1), but it is not easier than the solution above. The following three questions are very close to Problem 37. Problem 38. It is allowed to tear a piece of paper into 4 or 6 smaller pieces. Prove that following this rule you can tear a sheet of paper into any number of pieces greater than 8. Problem 39. Prove that a square can be dissected into n squares for n ;::: 6. Problem 40. Prove that an equilateral triangle can be dissected into n equilateral triangles for n ;::: 6. Comments. 38: If we tear a piece of paper into 4 or 6 smaller pieces, then the number of pieces increases by 3 or 5 respectively. Now we use the method of solution from Problem 37.FIGURE719. INDUCTION9339, 40: A square (equilateral triangle) can be dissected into 4 or 6 squares (equilateral triangles) as shown in Figure 71. Thus Problem 39 can be reduced to Problem 38. There exist other non-inductive solutions based on the possibility of cutting a square (equilateral triangle) into any even number (greater than 2) of squares (or equilateral triangles) greater than 2-see Figure 72.FIGURE72Other schemes of induction are even more exotic. An example is the method of "ramifying induction" which enables us to give a proof of a remarkable inequality for the arithmetic and geometric means. Problem 41: Prove that for any non-negative numbers x 1 , X2, ••. , Xn X1+ X2 + ... + Xn;:,. \Yx1x2 ... Xn •n'Sketch of proof. The base: n = 2 is rather simple. Then you must use steps from n = 2k to 2k+ 1 in order to prove the inequality for all n equal to power of 2. And finally, you prove that if the inequality is true for any n numbers, then it is true for any n - 1 numbers. The wave of proofs spreads in accordance with the scheme in Figure 73.9-- ...2--3--4 I ! FIGURE73See details in (example 24), and also in the chapter "Inequalities". Schemes involving "backwards induction" (over negative integers) and "double (or, 2-dimensional) induction" (for theorems involving two natural parameters) are illustrated in Problems 43 and 44.94MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)§5. Problems with no comments Problem 42. Two relatively prime natural numbers m, n, and the number 0 are given. A calculator can execute only one operation: to calculate the arithmetic mean of two given natural numbers if they are both even or both odd. Prove that using this calculator you can obtain all the natural numbers 1 through n, if you can enter into the calculator only the three numbers initially given or results of previous calculations. Problem 43. For the sequence a1 , a., . . . from Problem 33 we can define elements ao, a_,, a_2, ... so that the equation an+i = 3an - 2an-I will hold true for any integer n (positive or negative). Prove that the equality an = 2n + 1 will still be true for all integers n. Problem 44. Prove that 2m+n- 2 2". mn if m and n are positive integers. Problem 45! Several squares are given. Prove that it is possible to cut them into pieces and arrange them to form a single large square. Problem 46! Prove that among any 2n+l natural numbers there are 2n numbers whose sum is divisible by 2n. Problem 4 7. What is the greatest number of parts into which n circles can dissect a plane? What about n triangles? Remark. Compare Problem 6. Examples of the required dissections can also be done by induction. Problem 48. Several circles are drawn on a plane. A chord then is drawn in each of them. Prove that this "map" can be colored using three colors so that the colors of any two adjacent regions are different.Problem 49: Prove by "reductio ad absurdum" that the principle of mathematical induction stated in the very beginning of the present chapter is equivalent to the following "well order principle": in any non-empty set of natural numbers there exists a least element. Try to rewrite the solution of one of the previous problems (say, Problem 46) using this principle and compare it to the proof by induction. For more about the well order principle and its applications, see , pp.88-96. Conclusion. The method of mathematical induction is a very helpful and useful idea. You will find its applications in various places in this book, as well as in other mathematical contexts. However, we would like to warn you against an "addiction" to it. You should not think that, any question with statements and/or proofs using the words "et cetera" or "similarly" is a problem to be solved by MML Proofs by induction for many of those questions (you will see some of them in the chapters "Graphs-2" and "Inequalities") look rather artificial compared to other proofs involving such simple methods as direct calculation or recursive reasoning. It is not advisable to use such unnatural examples when teaching the nature of MMI, although they can be used well after the method is completely mastered.CHAPTER 10Divisibility-2: Congruence and Diophantine Equations §1. Congruence In the chapter "Divisibility and remainders" we discussed the concept of remainders. We noticed that in solving many problems on divisibility we dealt mostly not with the numbers themselves but with their remainders when divided by some fixed number. Thus it is natural to give the following definition: integers a and b are called congruent modulo m if they have equal remainders when divided by m. This is written: a= b(mod m). For example, 9 29 (mod 10), 1 3 (mod 2), 16 9 (mod 7), 3 0 (mod 3), 2n + 1=1 (mod n). Note that A is divisible by m if and only if A= 0 (mod m). Problem 1. Prove that a= b (mod m) if and only if a - bis divisible by m. Solution. If a= b (mod m), then let r be the common remainder when a or bis divided by m.====a=mk 1 +r, b=mk 2 +r. Thus a - b = m(k 1 - k2 ) is divisible by m. Conversely, if a-bis divisible by m, then we divide a and b by m with remainder. We have a= mk 1 +r1, b = mk2 +r,. Hence a-b= m(k 1 -k2) +r1 -r2 is, by assumption, divisible by m. Therefore r, -r2 is divisible by m. Since lr1 -r2I < m, we haver1= r2.This problem allows us to give another definition of congruence: integers a and b are congruent modulo m if a - b is divisible by m. From now on we will use either definition. For teachers. Before giving definitions of congruence it is important to check whether your students remember how to work with remainders (for example, by giving them a few problems similar to problems from §2 in the chapter "Divisibility and remainders"). It is remarkable that our new definition leads us to easier proofs of the basic properties of remainders. Problem 2. If a= b(mod m) andc d(mod m), prove that a+c b+d(mod m). Solution. Since a-bis divisible by m and c-d is divisible by m, (a-b)+(c-d) = (a+ c) - (b + d) is divisible by m. Problem 3. If a= b(mod m) and c d(mod m), provethata-c b-d(mod m). Problem 4. If a= b(mod m) and c d(mod m), prove that ac bd(mod m).=== == =95MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)96Solution. We have ac-bd = ac-bc+bc-bd = (a-b)c+b(c-d) which is divisible bym. Problem 5. If a= b(mod m) andn is any natural number, then an= bn (mod m). Methodological remark. Statements and proofs of properties of remainders look more attractive and simple when written using the language of congruences. For instance, without the new notation, the statement of Problem 2 would have read: the sum of two numbers and the sum of their remainders modulo m have equal remainders modulo m. Basically, Problems 2-5 state that congruences modulo a given number may be added, subtracted, multiplied, and exponentiated, like equations. We delay the question of division of congruences until §4. Before going further we show how the solution of a problem can be explained in the language of congruences. Problem 6. Prove that n 2 + 1 is not divisible by 3 for any integer n. Solution. It is clear that each integer n is congruent modulo 3 either to 0, or to 1, or to 2. If n 0 (mod 3), then n 2 0 (mod 3)-(multiplication of congruences) and n 2 + 1=1 (mod 3)-(addition of congruences). 2 If n 1 (mod 3), then n + 1 2 (mod 3). If n 2 (mod 3), then n 2 + 1 2 (mod 3). Thus we never get n 2 + 1 0 (mod 3).= = == = ==Let us consider another problem, which shows us that using negative integers in congruences can be quite helpful. As a matter of fact, in the arithmetic of remainders we can deal with negative integers in the same way as with positive ones.Problem 7. Reduce 6100 modulo 7. Solution. Since 6 - (-1) = 6 + 1 = 7, we can say that 6 -1 (mod 7). Raising this congruence to the power of 100 we have 6 100 (-l)IOO (mod 7); that is, 6 100 = 1 (mod 7).==Here are several more problems using the same idea. Problem 8. Prove that 3099 + 61 100 is divisible by 31. Problem 9. Prove that a) 43 101 + 23 10 I is divisible by 66. b) an+ bn is divisible by a+ b, if n is odd. Problem 10. Prove that 1n + 2n + ... + (n - 1)n is divisible by n for odd n. Problem 11. Prove that there exist infinitely many natural numbers that cannot be represented as a sum of three cubes. Problem 12. Prove that 103n+l cannot be represented as a sum of the cubes of two integers. Problems 11 and 12 are much more difficult than the preceding problems, since to solve them one must know which number m to divide by. Let us analyze the solution of Problem 12. The cube of a natural number is always congruent to either 0, or 1, or -1 modulo 7 (check this!). Therefore the sum10. DIVISIBILITY-297of two cubes is always congruent to one of the following numbers: -2, -1, 0, 1, 2. Since 10 3 (mod 7) we have 103 -1 (mod 7) and 103n+ 1 is congruent to either 3 or -3 modulo 7. Thus it cannot have the required representation.==Let us fix a natural number n. Then the infinite set Z of all integers quite naturally falls into n classes: two numbers are in one class if they have equal remainders modulo n (i.e., are congruent modulo n). For instance, if n = 2, we have two classes: even and odd numbers. When solving problems on divisibility it is often sufficient to check the truth of statements not for every integer but only for one (arbitrary!) representative from each class. Remember that in problems of §2 of the chapter "Divisibility and remainders" we usually used as representatives all the positive remainders modulo some number, and in Problems 11 and 12 of this chapter it was more helpful to choose other representatives (that is, some of the chosen numbers were negative). The next two problems illustrate the same idea. Problem 13. Prove that among any 51 integers there are 2 with squares having equal remainders modulo 100. Problem 14. Call a natural number n "convenient", if n 2 + 1 is divisible by 1000001. Prove that among the numbers 1, 2, ... , 1000000 there are evenly many "convenient" numbers. A concluding series of problems: Problem 15. a) Can the perfect square of a natural number end with 2 (that is, can its units digit equal 2)? b) Is it possible to write the square of a natural number using only the digits 2, 3, 7, 8 (perhaps with repetitions)? Problem 16. Find a number, which, when added to (n 2 - 1)1000. (n2 + 1)100 1 makes the result divisible by n. Problem 17. Find the remainder when the number 10 10 + 10 100 + 10 1ooo + ... + 101oooooooooo is divided by 7. Problem 18. How many natural numbers n not greater than 10000 are there such that 2n - n 2 is divisible by 7? Problem 19. Denote by k the product of the first several prime numbers (but more than one prime number). Prove that the number a) k - 1; b) k + 1 cannot be a perfect square. Problem 20. Does there exist a natural number n such that n 2 + n + 1 is divisible by 1955? Problem 21. Prove that 11n+2 + 122n+I is divisible by 133 for any natural n. Solution. 11n+2 + 122n+I = 121·lln+12 · 12 2n = 133 · 11 n - 12 · 11 n + 12 · 12 2n := 12{122n - lln) = 12(144n -- lln)=0 (mod 133).98MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)This solution shows that not only beautiful ideas but also simple "manual" calculations may give us a neat proof. Problem 22! Let n be a natural number such that n + 1 is divisible by 24. Prove that the sum of all the divisors of n is also divisible by 24. Problem 23! The sequence of natural numbers a1, a., a3, ... satisfies the condition an+2 = an+! an + 1 for all n. a) If a1 = a2 = 1, prove that no member of the sequence is divisible by 4. b) Prove that an - 22 is composite for all n > 10, no matter what a, and a2 are.This is the end of our list of problems about congruence for now. We should emphasize that almost all the following problems of this chapter, in fact, continue this theme. §2. Decimal representation and divisibility tests The tests for divisibility by 10, 2, 5, 4 are familiar to most people. Working with congruences, we can state and prove far stronger propositions. Problem 24. Prove that any natural number is congruent to its last digit modulo a) 10; b) 2; c) 5. Solution. Subtract the last digit from the given number. We obtain a number ending in zero. This number is divisible by 10 and, therefore, by 2 and 5. For teachers. Before starting the topic of divisibility tests it is necessary that the students understand the identity1~ =a110n-l +a210n- 2 + ... +an-110 1 +anIwhere the "overlined" row of digits denotes the natural number written with these digits in the indicated order. For example, ab= lOa + b, where a is the tens digit and b is the units digit of number ab. Problem 25. Prove that a1a2a3 ... an-Ian an-Ian (mod 4). Problem 26. State and prove analogous tests of divisibility for 2n and 5n. Now, a few problems whose solutions use the indicated divisibility tests. Problem 27. The last digit of the square of a natural number is 6. Prove that its next-to-last digit is odd. Solution. Since the last digit of its square is 6, the given natural number was even. The square of an even number is divisible by 4. Hence, the number formed by its two last digits must be divisible by 4. It is easy to write all two-digit numbers which end with 6 and are multiples of 4: 16, 36, 56, 76, 96. All their tens digits are odd. Problem 28. The next-to-last digit of the square of a natural number is odd. Prove that its last digit is 6. Problem 29. Prove that a power of 2 cannot end with four equal digits. Problem 30. Find at least one 100-digit number without zeros in its decimal representation, which is divisible by the sum of its digits.10. DIVISIBILITY-299Solution. Let us find, by trial and error, a number with the sum of its digits equal to 125. Divisibility by 125 is determined by the last three digits of a number. Thus, the number 111 ... 11599125 will do (the number begins with 94 ones). Let's discuss tests of divisibility by 3 and 9, which can also be stated in a more general form. Problem 31. Prove that any natural number is congruent to the sum of its digits modulo a) 3; b) 9. Solution. Consider the number aia2 ... an= a110n-I + a210n- 2 + ... + an-110 1 +an.Clearly, 10=1 (mod 9). Thus =1 (mod 9) for any natural k. So we have lQka110n-I + a210n- 2 + ... + an-110 1 +an= a1 + a2 + ... +an (mod 9) . The reasoning for divisibility by 3 is completely analogous. The next set of problems is connected with these tests. Problem 32. Is it possible to write a perfect square using only the digits a) 2, 3, 6; b) 1, 2, 3 exactly 10 times each? Problem 33. The sum of the digits was calculated for the number 2100 , then the sum of the digits was calculated for the resulting number and so on, until a single digit is left. Which digit is this? Problem 34. Prove that if you reverse the order of the digits in any natural number and subtract the result from the initial number, then the difference is divisible by 9. Problem 35. Write one digit to the left and one to the right of the number 15 so that the number obtained is divisible by 15. Problem 36. How many four-digit numbers with the two middle digits 97 are divisible by 45? Problem 37. Find the least natural number divisible by 36 which has all 10 digits in its decimal representation. Problem 38. Prove that the product of the last digit of the number 2n and of the sum of all its digits but the last is divisible by 3. Problem 39. Can the sum of the digits of a perfect square be equal to 1970? Problem 40. A three-digit number was decreased by the sum of its digits. Then the same operation was carried out with the resulting number, et cetera, 100 times in all. Prove that the final number is zero. Problem 41! Let A be the sum of the digits of 44444444, and B the sum of the digits of A. Find the sum of the digits of B. Ideas very close to those used in the proof of the test of divisibility by 9 allow us to prove another remarkable divisibility test. Problem 42. Prove thata1a2 ... an= an - an-1=+ ... + (-1)n- 1a1 (mod 11).Hint. 10 -1 (mod 11), so even powers of 10 are congruent to 1, and odd powers to -1, modulo 11.100MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Another set of problems is devoted to this test. Problem 43. Prove that the number 111 ... 11 {2n ones) is composite. Problem 44. Prove that the number a 1 a 2 ••• ana,. ... a2 a 1 is composite. Problem 45. Let a, b, c, d be distinct digits. Prove that cdcdcdcd is not divisible by aabb. Problem 46. A is a six-digit number with digits 1, 2, 3, 4, 5, and 6 each used one time. Prove that A is not divisible by 11. Problem 47. Prove that the difference between a number with oddly many digits and the number written with the same digits in reverse order is divisible by 99.Divisibility tests are only one way to link divisibility properties of a number with its decimal representation. This is demonstrated in the following set of problems.Problem 48. Is it possible to form two numbers using only the digits 2, 3, 4, 9 such that one of them is 19 times greater than the other? Problem 49. The sum of the two digits a and b is divisible by 7. Prove that the number aba is also divisible by 7. Problem 50. The sum of the digits of a three-digit number equals 7. Prove that this number is divisible by 7 if and only if its two last digits are equal. Problem 51. a) The six-digit number abcdef satisfies the property that def - abc is divisible by 7. Prove that the number itself is divisible by 7. b) State and prove a test for divisibility by 7. c) State and prove a test for divisibility by 13. Problem 52. a) The six-digit number abcdef is such that abc + def is divisible by 37. Prove that the number itself is divisible by 37. b) State and prove a test for divisibility by 37. Problem 53. Is there a three-digit number abc (where a -I c) such that abc - cba is a perfect square? Problem 54. Find the smallest number written only with ones which is divisible by 333 ... 33 (one hundred 3's in the representation). Problem 55. Is it possible for the sum of the first several natural numbers to be divisible by 1989? Problem 56. Find all natural numbers which become 9 times greater if you insert a zero between their units digit and their tens digit. Solution. Write our number as lOa + b, where b is the units digit, and a is some natural number. We get the equation lOOa+b = 9(10a+b) and, therefore, lOa =Sb; that is, Sa = 4b. Hence b is divisible by 5. Investigating the two cases b = 0, b = 5, we find that the only answer is 45. Methodological remark. We have just seen that sometimes it can be very useful to write an equation for the digits of the required number. Problem 57. Zero was inserted between the digits of a two-digit number divisible by 3, and the result was increased by twice its hundreds digit. The number obtained happens to be 9 times greater than the initial one. Find the original number.10. DIVISIBILITY-2101Problem 58. Find a four-digit number which is a perfect square, whose first two digits are equal and whose last two digits are equal. Problem 59. Find all three-digit numbers, any power of which ends with three digits forming the original number. Problem 60. The two digits 4 and 3 are written to the right of a natural number, then the operation is repeated many times (for instance, 51 generates 5143, then 514343, et cetera). Prove that eventually we will have a composite number. Problem 61! Prove that all the numbers in the series 10001, 100010001, 1000100010001, ...are composite. §3. Equations in integers and other problems A well-known problem asserts that the sum of N pesos can always be paid with 3- and 5-peso bills if N > 7 (see the chapter "Induction", Problem 37). Translated into the language of equations, this means that the equation 3x+5y=N always has a solution in non-negative integers x and y for natural numbers N > 7. In this section we will find integer solutions for similar equations and others, though we will usually look for integer solutions without any additional restrictions.Problem 62. Solve the equation 3x + 5y = 7 in integers. Let us analyze the solution. This will give us an opportunity to solve other problems analogously. First, we find one particular solution (this idea can often help in solving mathematical problems). Note that 3 · 2 + 5 · (-1) = 1. Multiplying this equation by 7, we have 3 · 14 + 5 · (-7) = 7, and x 0 = 14, Yo= -7 is one solution (one among many). Thus, 3x + 5y = 7; 3xo + 5yo = 7 . Subtracting one equation from the other and denoting x - x 0 and y - y0 by a and b respectively we have 3a+5b=O. So we see that b must be divisible by 3 and a by 5. Let a = 5k. Then b = -3k, where k can be any integer. Then, we have the set of solutions: x-xo=5k y-yo = -3ki.e.,x = 14+ 5k y= -7-3k'where k is an arbitrary integer. Certainly, no other solutions exist, since our transformations always give us equivalent equations.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)102For teachers. We hope your students also realize this fact clearly. Without a full understanding of this part of the solution-that the pairs (x, y) exhaust the set of solutions of equation (•) for N = 7-it is next to impossible to go ahead. Problem 63. Find all integer roots of the equation 3x - 12y = 7. This barrier overridden, we analyze briefly the "very difficult" Problem 64. Problem 64. Solve the equation 1990x - 173y = 11. The coefficients in the equation are large enough to make it difficult to find a particular solution. However, it is not hard to see that the numbers 1990 and 173 are relatively prime, and this helps. Lemma. The greatest common divisor (gcd} of these numbers can be represented as 1990m - 173n for some integers m and n. You can prove this lemma using the fact that all the numbers we obtain while calculating the gcd via Euclid's algorithm (see the chapter "Divisibility and remainders") can be represented in this form. This is not easy, but we leave it to the reader. More specifically Euclid's algorithm gives us m = 2, n = 23. Therefore, we get (2, 23} as one particular solution of the equation 1990m - 173n = 1. Hence x 0 = 2 · 11 = 22, y0 = 23 · 11 = 253 is a solution to the equation 1990x - 173y = 11. As in Problem 62, we have that X = Xoy+ 173k = 22 + 173k,= Yo + 1990k = 253 + 1990k,where k is any integer.Problem 65. Find all integer roots of the equation 21x + 48y = 6. Remark. Now, generally speaking, we can solve any equation of the form Ax + By = C (this is the so-called general linear Diophantine equation with two variables) in integers x and y. Theorem. If the coefficients A and B in the linear Diophantine equation are relatively prime, then there are integers x 0 and y0 such that Ax 0 + By0 = C and all the roots of the equation can be given by the following formulas: x = Xo +Bk, y =Yo - Ak .Exercise. Try to state the general result and prove it rigorously. (Do not forget about the case when A and B &re not relatively prime and C is not divisible by gcd(A,B).) Problem 66. Solve the equation 2x + 3y + 5z = 11 in integers. (By the way, does it have any solutions in natural numbers?) Problem 67! A pawn stands on one of the boxes of a band of graph paper of unit width which is infinite in both directions. It can move m boxes to the right or n boxes to the left. Which m and n satisfy the property that the pawn can move onto the box just to the right of the starting box (in several moves)? What is the minimum number of moves needed to do this?10. DIVISIBILITY-2103An equation with more than one variable, for which integer solutions are required, is called a Diophantine equation after the famous Greek mathematician Diophantos of Alexandria, who investigated such equations in very early times. Let us examine some more complicated Diophantine equations. Solve, for integer values of the variables: Problem Problem Problem Problem68. 69. 70. 71.{2x + y)(5x + 3y) = 7. xy = x + y + 3. x 2 = 14 + y 2 • x 2 + y 2 = x + y + 2.Solutions to all these problems are connected with a very common idea-caseby-case analysis. Certainly it is impossible to write down all pairs of integers and check for each of them whether it satisfies the equation or not. However, some simple transformations can reduce this analysis to just a short job. Here is a solution to Problem 69. Since xy-x-y = 3, we have (x-l)(y-1) = 4. It remains to analyze all the representations of 4 as a product of two factors. Answer: (x,y) = {5,2), (2,5), (0,-3), {-3,0), {3,3), {-1,-1). In Problems 70 and 71 we can also transform the given equation. Thus, we have: IDEA ONE: Use an appropriate transformation of the equation, then a case-bycase analysis. However, even in the very next problem this idea will not work. Problem 72. x 2 + y 2 = 4z - 1. It is clear that one cannot transform the equation to a more tractable type; it is also impossible to analyze all the appropriate triples of integers. This new representative of our "Diophantine zoo" is remarkable in that it has no integer solutions. Indeed, which remainders can perfect squares give when divided by 4? (The choice of modulo 4 was determined by the form of the given equation.) The only possible remainders are 0 and 1. Since the sum of two such remainders cannot give remainder -1, we have no solutions for this equation. Thus we have: IDEA TWO: Consider remainders modulo some natural number. Problem 73. x 2 - 7y = 10. Problem 74. x 3 + 21y 2 + 5 = O. Problem 75. 15x2 - 7y 2 = 9. Problem 76. x 2 + y 2 + z 2 = St - 1. Problem 77. 3m + 7 = 2n. Solution to Problem 74. Because x 3 can be congruent modulo 7 only to 0, 1 and -1, x 3 + 21y 2 + 5 must be congruent to 5, 6, or 4, and, therefore, cannot be zero. Methodological remark. You probably have already noticed that "idea two" allows us only to prove the absence of roots. Indeed, if an equation has solutions modulo 7 or modulo 3, then it does not mean that the equation has at least one integer solution. For example, the equation 2x 2 -y 3 = 6 has roots modulo 7 (x 0=MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)104===and y 1) and modulo 3 (x 0 and y 0), but it has no integer roots (this can be easily shown using remainders when divided by 8). Let us struggle now with Problem 77. We can see at once that a solution exists: m = 2, n = 4. Does it make sense to consider remainders? Do not jump -1 (mod 3), to conclusions! The left side is congruent to 1 modulo 3. Since 2 this implies that n is even; that is, n = 2k. So we have 3m + 7 = 4k. Now residues modulo 4 can help us. We have 4k - 7 = 1 (mod 4). Hence, 3m 1 (mod 4), which implies that m is also even; that is, m = 2p. The equation becomes 32• + 7 = 22 k. What now? We use "idea one":==7 = 22k - 32p= (2k -3•)(2k + 3') .Therefore, 2k +3• = 7, 2k -3• = 1, and we obtain the unique solution k = 2, p = 1; that is, m = 2, n = 4. We used both ideas, or methods, tried earlier. Such a combination of ideas is a very common phenomenon in mathematics. Here is another Diophantine equation whose solution uses a similar combination:Problem 78. 3·2m+1 = n 2 . Solution. Since n 2 = 1 (mod 3), it is clear that n is not divisible by 3. Son= 3k+l or n = 3k + 2. We investigate each case. a) If n = 3k + 2, then 3 · 2m + 1 = 9k 2 + 12k + 4. Simplifying, we get 2m = 3k2 + 4k + 1 = (3k + l)(k + 1) . The only factors of a power of 2 are other powers of 2. Therefore, k + 1 and 3k + 1 are powers of 2. The values k = 0 and k = 1 do fit and we have the solutions n = 2, m = 1 and n = 5, m = 3 respectively. However, if k :2: 2, then 4(k + 1) > 3k + 1 > 2(k + 1). This inequality shows that k + 1 and 3k + 1 cannot be powers of 2 simultaneously. b) n = 3k + 1. Proceeding analogously we find one more root: n = 7, m = 4. Aside from the ideas of case-by-case analysis and factorization, the idea of estimation was used:IDEA THREE: When solving Diophantine equations, inequalities and estimates may be of use. Problem 79. 1/a + l/b + 1/c = 1. Problem 80. x 2 - y 2 = 1988. Problem 81. Prove that the equation 1/x - 1/y = 1/n has exactly one solution in natural numbers if and only if n is prime. Problem 82. x 3 + 3 = 4y(y + 1). We cannot discuss, in this small section; the many other interesting and complicated methods of solving Diophantine equations. Instead, we refer you to [55, 58, 25]. Below are two extra problems which are much more difficult than those above. You might like to see the hints before you try to solve them on your own. Problem 83: x 2 + y 2 = z2 • Problem 84: x 2 - 2y2 = 1.10. DIVISIBILITY-2105§4. Fermat's "little" theorem This section is devoted to a remarkable and non-trivial fact in number theory that was stated and proved by the famous XVII century French mathematician Pierre de Fermat. But, before we begin, we discuss (as was promised in §1) the question of division of congruences. Problem 85. Let ka kb (mod m) where k and m are relatively prime. Then a:=b(modm).==Solution. Since ka kb (mod m), ka - kb= k(a - b) is divisible by m. Since k and mare relatively prime, a - bis divisible by m; that is, a= b (mod m). It is easy to find examples to show that it is necessary for k and m to be relatively prime. Indeed, 5 · 3 5 · 7 (mod 10) but 3 and 7 are not congruent modulo 10. In any case, the following is true: Problem 86. If ka =kb (mod kn), then a= b (mod n).=Now we are ready to state and prove Fermat's "little" theorem. Theorem. Let p be prime number and A be a number not divisible by p. Then AP- 1 1 (mod p). Proof. Consider the p - 1 numbers: A, 2A, 3A, ... , (p - l)A. We can show that no two of these numbers have the same remainder when divided by p. Indeed, if kA nA (mod p), then k = n (mod p) (see Problem 85). This is impossible if k and n are unequal and both are less than p. Therefore, among the remainders of these p - 1 numbers when divided by p each of the numbers 1 through p - 1 is represented exactly once. Multiplying them together, we have==A· 2A · 3A ... (p - l)A==1 · 2 · 3 ... (p - 1) (mod p);that is, (p - 1)! · AP- 1 (p - 1)! (mod p). Now p is prime, which implies that (p- 1)! and pare relatively prime. Using the result of Problem 85 again we obtain A•- 1 1 (mod p). D=Corollary. Let p be a prime number. Then for any integer A we have AP A(modp).=Fermat's "little" theorem is not just an unexpected and "neat" fact. It also provides a very powerful tool for solving many problems in arithmetic. Some of these are given below. Problem 87. Find the remainder when 2100 is divided by 101. Problem 88. Find the remainder when 3 102 is divided by 101. Solution. Since 101 is prime, we get 3100 1 (mod 101). Thus 3102 = 9. 3100 9(mod 101).==For teachers. Such computational exercises using Fermat's theorem can become quite routine for students. Problem 89. Prove that 300 3000 - 1 is divisible by 1001. Problem 90. Find the remainder when 8900 is divided by 29. Problem 91. Prove that 7120 - 1 is divisible by 143.106MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)=Solution. Let us prove that 7120 - 1 is divisible by 11 and 13. Indeed, (7 12 ) 10 1 (mod 11) and (7 10 ) 12 1(mod13). Problem 92. Prove that the number 30239 + 23930 is not prime. Problem 93. Let p be a prime number, and suppose a and bare arbitrary integers. Prove that (a+ b)P =a•+ bP (mod p). Try to contrive two proofs: one utilizing Fermat's "little" theorem and the other using the binomial theorem (see the chapter "Combinatorics-2"). Problem 94. The sum of the numbers a, b, and c is divisible by 30. Prove that a5 + bs + c5 is also divisible by 30. Problem 95. Let p and q be different primes. Prove that a) p• + q• = p + q (mod pq).=b) [ p• ~ q•] is even if p, q oft 2, where [x] denotes the greatest integer function. Problem 96. Let p be prime, and suppose p does not divide some number a. Prove that there exists a natural number b such that ab 1 (mod p). Problem 97. (Wilson's theorem). Let p be prime. Prove that (p - 1)! -l(modp). Problem 98. Let n be a natural number not divisible by 17. Prove that either n8 + 1 or n 8 - 1 is divisible by 17. Problem 99. a) Let p be a prime not equal to 3. Prove that the number 111 ... 11 (p ones) is not divisible by p. b) Let p > 5 be a prime. Prove that the number 111 ... 11 (p - 1 ones) is divisible by p. Problem 100. Prove that for each prime p the difference==111 ... 11222 ... 22333 ... 33 ... 888 ... 88999 ... 99 - 123456789 (in the first number each non-zero digit is written exactly p times) is divisible by p.CHAPTER 11Combinatorics-2 This chapter is a direct continuation of the chapter "Combinatorics-1" . The present material draws on results explained in that chapter.For teachers. Students should solve again a few problems involving the combinatorial ideas discussed earlier and go on only if these problems do not produce any difficulties. If they do, we recommend going back to the chapter "Combinatorics-!".The contents of this chapter are connected with one very important combinatorial object which we begin to study right now. §1. CombinationsLet us begin with a simple problem. Problem 1. Two students must be chosen out of a group of thirty for a mathematical contest. In how many ways can this be done? Solution. You can choose the first participant of the contest in 30 ways. No matter who the first was, the second can be chosen in 29 ways. But now each pair is counted exactly twice. Thus the answer is 30 · 29/2 = 435 ways. Note that we have merely repeated the solution to Problem 22 from the chapter "Combinatorics-!". Assume now that we must choose a team not of two people but of k, and that the group consists of n students, not of 30. The number of ways this can be done is called the number of combinations of k elements taken from n elements and is denoted (Z) (to be read as "n choose k"). For instance, = 2, @ = 3, (7) = n, (~) = l. Note that (~) also can be interpreted naturally: there is only one way to choose nobody (zero people) out of n. That is, (~) = 1 for all n. It is remarkable that some properties of these numbers can be proved by simple combinatorial reasoning, not using a formula for the calculation of (~).mProperty 1.(n'.:k)= (~).Proof. Note that a choice of k contest participants is equivalent to the choice of n - k students who will not take part in the contest. Thus, the number of ways to choose k people out of n equals the number of ways to choose n - k people out of n; that is, (n'.:k) = (~). Property 2.(nk° 1) = (Z) + (k'.:i). 107108MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Proof. Assume that there are n + 1 students in the group. Consider one of them and denote him or her by A. Let us divide the set of all possible teams into two subsets: those teams containing A, and the others which do not contain A. The cardinality of the first set is (k'.': 1)-since we must complement the team with k -1 more students chosen from the n students remaining. The number of teams in the second set equals (~). Now from the remaining n students we must choose the entire team. Thus (nt 1) = G) + (k'.':i). Methodological remark. This reasoning allowed us to prove a rather significant fact without any calculations. This phenomenon is quite characteristic for combinatorics. Often just a few minutes of thinking and understanding the combinatorial sense of a question may let us avoid cumbersome calculations. This is why we consider it necessary to discuss proofs like those above in detail. Let us now find a formula to calculate G). Problem 2. How many ways are there to choose a team of three students out of a group of 30? Solution. The first student can be chosen in 30 ways, the second in 29 ways, and the third in 28 ways. Thus we have 30 · 29 · 28 ways. However, each team was counted several times: the same trio of students can be chosen in different ways. For instance, choosing student A first, then student B, and, finally, C is the same as choosing C first, then A, and then B. Since the number of permutations of 3 elements is 3!, each team was counted exactly 3! = 6 times. Therefore, (33°) equals (30 . 29 . 28) /3!. In just the same way we can deduce a formula for calculating (~), for arbitrary n and k:(~) =n(n-l)(n-2) ... (n-k+l)/k!.For teachers. The numbers (~) are central in this chapter. Thus, it is important to make sure that all students understand what we are counting here, and how the numbers of combinations can be calculated. Before revealing the general formula, proofs of a few problems similar to Problem 2 might be discussed. Let us deal with a few more problems. Problem 3. In how many ways can one choose 4 colors out of 7 given colors? Answer. G) = 35. Problem 4. One student has 6 math books, and another has 8 books. How many ways are there to exchange 3 books belonging to the first student with 3 books belonging to the second? Solution. The first student can choose his 3 books in ways, and the second in ways. Thus, the number of possible exchanges is = 1120. Problem 5. There are 2 girls and 7 boys in a chess club. A team of four persons must be chosen for a tournament, and there must be at least 1 girl on the team. In how many ways can this be done? Solution. There must be either 1 or 2 girls on the team. In the latter case 2 boys ways. If there is only 1 girl on the team {there are can be added to the team inm m·(:)mm11.COMBINATOR!CS~~2two ways to choose her), then the team can be completed by adding 3 boys in different ways. Therefore, in all there are G) + 2 · G) = 91 possible teams.109G)Problem 6. How many ways are there to divide 10 boys into two basketball teams of 5 boys each? Solution. The first team can be chosen in (~0 ) ways. This choice completely determines the second team. However, this calculation counts each pair of complementing teams-say, A and B-two times: the first time, when A is chosen as the first team, and the second time, when B is chosen as the first team. Thus, the answer is ('5°) /2. Methodological remark. After learning these formulas it is not obligatory to express the answers as decimal numbers. It is not bad if the expressions (~) are present in answers. Notice that the formula formmakes the first property of symmetry-(n".•) =(~)-rather unclear. However, we can make the formula look more symmetric bymultiplying its numerator and denominator by (n - k)!:( n) = n(n-1) ... (n-k+l)(n-k)! k k!(n-k)! n(n-1) ... (n-k + l)(n-k) .. . 3 · 2 · l k!(n-k)! n! = k!(n-k)!' Now the first property is quite evident. Exercise. Prove the second property of (~), using the formula above. For teachers. We recommend spending at least one session of a math circle introducing definitions, properties, and the formula for the numbers (~). It would also be helpful to solve several easy problems at this session, then to give problems connected with this theme during the next sessions. Problem 7. Ten points are marked on a plane so that no three of them are on the same straight line. How many triangles are there with vertices at these points? Problem 8. A special squad consists of 3 officers, 6 sergeants, and 60 privates. In how many ways can a group consisting of 1 officer, 2 sergeants, and 20 privates be chosen for an assignment? Problem 9. Ten points are marked on a straight line, and 11 points are marked on another line, parallel to the first one. How many a) triangles; b) quadrilaterals are there with vertices at these points? Problem 10. A set of 15 different words is given. In how many ways is it possible to choose a subset of no more than 5 words? Problem 11. There are 4 married couples in a club. How many ways are there to choose a committee of 3 members so that no two spouses are members of the committee?JIOMATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 12. There are 31 students in a class, including Pete and John. How many ways are there to choose a soccer team ( 11 players) so that Pete and John are not on the team together? Problem 13. How many ways are there to rearrange the letters in the word "ASUNDER" so that vowels will be in alphabetical order, as well as consonants? Example: DANERUS (A-E-U, D-N-R-S). Problem 14. We must choose a 5-member team from 12 girls and 10 boys. How many ways are there to make the choice so that there are no more than 3 boys on the team? Problem 15. How many ways are there to put 12 white and 12 black checkers on the black squares of a chessboard? Problem 16. a) How many ways are there to divide 15 people into three teams of 5 people each? b) How many ways are there to choose two teams of 5 people each from 15 people? Problem 17. In how many ways can you choose 10 cards from a deck of 52 cards so that a) there is exactly one ace among the chosen cards? b) there is at least one ace among the chosen cards? Problem 18. How many six-digit numbers have 3 even and 3 odd digits? Problem 19. How many ten-digit numbers have the sum of their digits equal to a) 2; b) 3; c) 4? Problem 20. A person has 6 friends. Each evening, for 5 days, he or she invites 3 of them so that the same group is never invited twice. How many ways are there to do this? Problem 21. Th participate in a sports lottery in Russia one must choose 6 out of the 45 numbers printed on a lottery card (all the printed cards are identical). a) How many ways are there to fill in the lottery card? b) After the end of the lottery, its organizers decided to count the number of ways to fill in the lottery card so that exactly 3 of the 6 chosen numbers are among the 6 winning numbers. Help them to find the answer. §2. Pascal's triangleThis section is remarkable for its combination of almost all the ideas explained earlier in this chapter, leading us to some very beautiful combinatorial facts. To start, let us assume we know all the numbers (~) for some fixed number n. Then the second property,allows us to calculate easily the numbers following construction.(nf)for all k. This idea gives us the11. COMBINATORICS-2111Since (g) = 1, we write 1 in the center of the first line on a sheet of paper. In the next line we write the numbers @ = 1 and = 1 in such a way that 1 = (g) is over the gap between these numbers (see Figure 74).GJ1 1 FIGURE 74The numbers @ and @ are also l. We write them in the next line (see Figure 75) with (~),which, by the second property, equals (~) + written between them (see Figure 76).GJ11111 FIGURE75(Dequals the sum of the two numbers in the previous row, standing Thus, number to the left and to the right of it.112 FIGURE76Using the same rule, we will fill all the following lines: first, we write the numbers (~) and (~) on the sides (they are alweys equal to 1), and then we write the sum of any two adjacent numbers of the previous row in a position between them in the next row.Finally, we have the numerical triangle shown in Figure 77. It is called Pascal's triangle. By construction, the number (~) occupies the (k + l)st place in the (n + l)st row of this triangle. Therefore, it is more convenient to number the rows, and the places in the rows, starting from zero. Then we will have the number (~) occupying the kth place in the nth row.112MATHEMATICAL CIRCLES (RUSSIAN EXPERJENCE)2 34 5346 10105FIGURE 77For teachers. Before going further, students should know the connection between the numbers (~) and Pascal's triangle. The best exercise for this is direct calculation of numbers of combinations using the triangle-based procedure described above. Now let us begin the investigation of properties of Pascal's triangle. Evaluate the sum of the numbers in its first few rows: 1, 2, 4, 8, 16. You may come to the very natural conjecture that the sum of the nth row is 2n. We prove this by induction on n (see the chapter "Induction"). The base is already proved. To prove the inductive step, notice that each number of a row is taken as a summand in forming two adjacent numbers in the next row. Thus the sum of the numbers in the next row is exactly twice the sum of the numbers in the given row. This completes the inductive step. It may also be proved that in every row of Pascal's triangle (except for the zeroth) the sum of the numbers in the even places equals the sum of numbers in the odd places. Rewriting the proposition about the· sum of the numbers in a row of Pascal's triangle, we get the following remarkable combinatorial identity:Here is a combinatorial proof. From the combinatorial point of view the identity states that the number of teams which can be chosen from n students equals 2n if the cardinality of the team is arbitrary (using set theoretic language: the number of subsets of an n-element set equals 2n). We number all the students from 1 through n in an arbitrary order. Then for each possible team we construct a sequence of O's and l's in the following way: the first element of the sequence is 1 if the first student is on the team, and 0 otherwise. In the same way we define the second element of the sequence, the third, and so on. It is evident that different teams correspond to different sequences and vice versa. Thus the number of all teams is equal to the number of all sequences of O's and l's with n elements. Each element of such a sequence can be equal to either 0 or 1-that is, it can be chosen in two ways. Thus, the number of all such sequences equals 2 x 2 x ... x 2 = 2n.11. COMBINATORICS- 2113Methodological remark. The most important place in this reasoning was the construction of a correspondence between teams and sequences of O's and l's. Such a reformulation is often very helpful in the solution of many other combinatorial problems. As examples we submit here two more problems. Problem 22. A person has 10 friends. Over several days he invites some of them to a dinner party so that the company never repeats (he may, for example, invite nobody on one of the days). For how many days can he follow this rule? Problem 23. There are 7 steps in a flight of stairs (not counting the top and bottom of the flight). When going down, you can jump over some steps, perhaps even over all 7. How many ways are there to go down the stairs? Before investigating the next property of Pascal's triangle, we analyze one problem which is remarkable in that the numbers (~) arise unexpectedly in the course of its solution. Problem 24. The map of a town is depicted in Figure 78. All its streets are one-way, so that you can drive only "east" or "north". How many different ways are there to reach point B starting from A?AFIGURE78Solution. Let us call any segment of the grid connecting two neighboring nodes a "street". It is clear that each route from A to B consists of exactly 13 streets, 8 of which are horizontal and 5 are vertical. Given any route, we will construct a sequence of the letters N and E in the following way: when we drive "north", we add the letter N to the sequence, and when we drive "east", we add the letter E to the sequence. For instance, the route in Figure 78 corresponds to the sequence ENNEEENEENNEE. Each sequence constructed in this way contains 13 letters-8 letters E and 5 letters N. It remains to calculate the number of such sequences. Any sequence is uniquely determined by the list of the 5 places occupied by the letters N (or, alternatively, of the 8 places occupied by the letters E). Five places out of 13 can be chosen in (1,;3} ways. Thus the number of sequences, and, therefore, the number of routes, equals (1:;3). The same reasoning for am x n rectangle gives us the result (m,;t;n) or, equivalently, (m~n). Returning to Pascal's triangle, let us change its numbers to points (nodes) as shown in Figure 79. Let us write 1 next to the uppermost point S of the triangle,114MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE 79FIGURE 80and, then, next to any other node-the number of ways one can reach this node from S, moving only downward. You can see (Figure 80) that we have Pascal's triangle again. One proof of this fact is similar to the solution of the previous problem. However, here, as in the proof of the fact that the sum of the numbers in each row is a power of two, we can proceed hy induction. Indeed, you can reach the kth node of the nth row only via either the (k - l)st node of the (n - l)st row or the kth node of the (n - l)st row (Figure 81). Thus, to find the required number of ways we must simply add the number of ways going to those two nodes of the previous row. Hence, using the assumption, the number of ways of going from S to the kth node of the nth row is (~=D + (n;;') = (~). Methodological remark.Both properties of Pascal's triangle described abovewere proved in two ways-using "geometric" ideas and via direct combinatorial rea-soning. It is useful to utilize both these approaches while solving various problems, especially combinatorial identities. Some other properties of Pascal's triangle are given below as problems to be solved. They can be stated in terms of the triangle itself, or as combinatorial identities.11. COMBINATORICS-2115FIGURE 81Problem 25. Prove that one can choose evenly many objects from a collection of n objects in 2n-l ways. Problem 26. Prove that(~)-(~) + ... +(-l)n(~)=0.For convenience, we introduce now the following definitions. We will call rays parallel to sides of Pascal's triangle its diagonals. More precisely, rays parallel to the right side are called right diagonals (one of them is selected in Figure 82), and those parallel to the left side are left diagonals (see Figure 83).FIGURE 82Problem 27. Prove that each number a in Pascal's triangle is equal to the sum of the numbers in the previous right diagonal, starting from its leftmost number through the number which is located in the same left diagonal as a (see Figure 84). Problem 28. Prove that each number a in Pascal's triangle is equal to the sum of the numbers in the previous left diagonal, starting from the rightmost number through the number which is located in the same right diagonal as a (see Figure 85).116MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE 83FIGURE 84FIGURE 85Problem 29. Prove that each number a in Pascal's triangle decreased by 1 is equal to the sum of the numbers within a parallelogram bounded by the sides of the triangle and diagonals going through a (the numbers on the diagonals are not included; see Figure 86). Problem 30! Prove that11. COMBINATORICS-2117FIGURE 86§3. Balls and walls Let us begin with a discussion of two interesting problems. Each of these can be solved by direct, though technically complex, calculation (try this way, too, but later). On the other hand, simply restating the question allows us to reach the answer (which in each case is a number of combinations), rather easily. Problem 31. Six boxes are numbered 1 through 6. How many ways are there to put 20 identical balls into these boxes so that none of them is empty? Solution. Let us arrange the balls in a row. To determine the distribution of the balls in the boxes we must partition this row into six groups of balls using five walls: the first group for the first box, the second group for the second box, et cetera (see Figure 87). Thus, the number of ways to distribute our balls in the boxes equals the number of ways to put five walls into gaps between the balls in the row. Any wall can be in any of 19 gaps (there are 19 = 20 - 1 gaps between 20 balls), and no two of them can be in the same gap (this would mean that one of the groups is empty). Therefore, the number of all possible partitions is (';).FIGURE 87Exercise. How many ways are there to distribute n identical balls in m numbered boxes so that none of the boxes is empty? Problem 32. Six boxes are numbered 1 through 6. How many ways are there to distribute 20 identical balls between the boxes (this time some of the boxes can be empty)? Solution. Consider a row of 25 objects: 20 identical balls and 5 identical walls, which are arranged in an arbitrary order. Any such row corresponds without ambiguity to some partition of balls: balls located to the left of the first wall, go to the first box; balls between the first and the second wall go to the second box, et cetera (perhaps, some pair of walls are adjacent in the row, resulting in an empty box). Therefore, the number of partitions is equal to the number of all possiblell8MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)rows of 20 balls and 5 walls; that is, to the 5 places occupied by the walls).('s")(the row is completely determined byWe should note that another solution to Problem 31 can be obtained as follows: put one ball in each box (to prevent empty boxes), then use the result of Problem 32 (with 14 balls instead of 20). The ideas found during the process of solving the two previous problems show us how to solve the next, rather complicated, problem very neatly. Problem 33. How many ways are there to represent the natural number n as a sum of a) k natural numbers? b) k non-negative integers? Representations that differ in the order of the summands are different. Hint. Represent n as the sum of n ones: n = 1 + 1 + ... + 1. Call these n ones "balls", and call the k summands from the statement "boxes". The answers are: al (~::D; bl (n+~- 1 ). Methodological remark. The solutions just explained show us again how important a good reformulation of the statement of a problem can be. The reason for such a thorough discussion of the question of the distribution of balls in boxes is that similar reformulations (devising "walls", et cetera) are very useful in solving many quite different problems, not only in combinatorics, but in other fields of mathematics and, generally, science. Problem 34. In how many ways can 12 pennies be put into five different purses so that none of them is empty? Problem 35. A bookbinder must bind 12 identical books using red, green, or blue covers. In how many ways can he do this? Problem 36. How many ways are there to cut a necklace (in the form of an unbroken circle) with 30 pearls into 8 parts (it is permitted to cut only between pearls)? Problem 37. Thirty people vote for 5 candidates. How many possible distributions of their votes are there, if each of them votes for one candidate, and we consider only the numbers of votes given to each of the candidates? Problem 38. There are 10 types of postcards in a post office. How many ways are there to buy a) 12 postcards? b) 8 postcards? Problem 39. A train with m passengers must make n stops. a) How many ways are there for passengers to get off the train at the stops? b) Answer the same ·question if we take into account only the number of passengers who get off at each stop. Problem 40. In a purse, there are 20 pennies, 20 nickels, and 20 dimes. How many ways are there to choose 20 coins out of these 60? Problem 41. How many ways are there to put seven white and two black billiard balls in nine pockets? Some of the pockets may reniain empty and the pockets are considered distinguishable.11. COMBINATORICS·-2119Problem 42. In how many ways can three people divide among themselves six identical apples, one orange, one plum, and one tangerine (without cutting any fruit)? Problem 43. How many ways are there to put four black, four white, and four blue balls into six different boxes? Problem 44. A community with n members chooses its representative by voting. a) In how many ways can "open" voting result, if everybody votes for one person (perhaps for himself/herself)? Open voting means that we take into account not only the numbers of votes, but also who votes for whom. b) Answer the same question, if voting is not open (and the result consists only of the numbers of votes given to each of the members of the community). Problem 45. How many ways are there to arrange five red, five blue, and five green balls in a row so that no two blue balls lie next to each other? Problem 46! How many ways are there to represent the number 1000000 as the product of three factors, if we consider products that differ in the order of factors as different? Problem 47! There are 12 books on a shelf. How many ways are there to choose five of them so that no two of the chosen books stand next to each other? §4! Newton's binomial theorem For teachers. This section is marked with an asterisk because it is rather difficult for students of the middle grades. However, we have decided not to omit it, since its contents are closely related to the numbers (Z) and Pascal's triangle. It is also beyond any doubt that the binomial theorem is an indispensable element of mathematical education, although, its study can be postponed. Remark. This topic is related to, but distinct from, combinatorics. However, it is relevant to give the statement and proof of the main theorem here, since it is used in combinatorics quite often and the proof itself is based on combinatorial ideas. Everyone knows the identity: (a+b) 2 =a2 +2ab+b2•We will try to derive a formula for raising the binomial (a+ b) to an arbitrary positive integer power. Let us write a few successive powers of the binomial: (a+b) 0 = (a+b) 1 = (a+b) 2 = (a+b)3=aa3a2++3a 2 b+ 2ab++3ab2b2+b3It is obvious that the coefficients in the right parts of these identities form the corresponding rows of Pascal's triangle. We may suggest that the following identity holds true:120MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)This is indeed true and this expansion is called Newton's binomial theorem. To prove it, we expand the product(a+W = (a+b)(a+b)(a+b) ... (a+b)(a+b) without grouping similar terms together and without changing the order of factors in each monomial. For example,(a+ b)(g + Q)(a +ii) =~+~+•+~+~+W+•+~. Let us find the coefficient of the term an-kbk after reducing similar terms. It is clear that this coefficient equals the number of monomials which include b exactly k times (and include a exactly n - k times). This number is equal to (~), since this is the number of ways to choose k places for the letters b. Exercise. Prove the binomial theorem by induction. §5. Additional problems For teachers. This section is auxiliary and optional: it does not contain substantial new material but is just a list of problems. By including this section in the chapter we pursue two goals. First, we enlarge the supply of problems for classes. Second, after learning basic combinatorial ideas, it is useful to keep reviewing them from time to time. These reviews can be organized as separate sessions or mathematical olympiads, "math battles", et cetera. The problems below can be used for such a review session or contest. Problem 48. How many necklaces can be made using 5 identical red beans and 2 identical blue beans? (See Figure 88.)FIGURE 88 Problem 49. a) There are 30 members in a sports club, and the club's coach must choose 4 men for a 1000 meter run. How many ways are there to do that? b) How many ways are there to choose a team of 4 members for the lOOm + 200m + 300m + 400m relay? Problem 50. How many six-letter ''words" contain at least one letter A (if any sequence of letters is counted as a word)?11. COMBINATORICS-2121Problem 51. How many ways are there to draw a closed broken path made up of line segments with vertices coinciding with the vertices of a regular hexagon (the segments of the path are allowed to intersect each other)? Problem 52. How many different four-digit numbers divisible by 4 can be written using the digits 1, 2, 3, and 4 a) if each digit can be used only once? b) if each digit can be used any number of times? Problem 53. A father has 2 apples and 3 pears. Each weekday (Monday through Friday) he gives one of the fruits to his daughter. In how many ways can this be done? Problem 54. A theater group consists of 20 actors. How many ways are there to choose two groups of 6 actors each for the two performances of a play, so that none of the actors takes part in both performances? Problem 55. Find the sum of all three-digit numbers that can be written using the digits 1, 2, 3, 4 (repetitions allowed). Problem 56. How many ways are there to choose 6 cards from a complete deck of 52 cards in such a way that all four suits will be present? Problem 57. How many ways are there to put three one dollar bills and ten quarters into 4 different boxes? Problem 58. Find the number of integers from 0 through 999999 that have no two equal neighboring digits in their decimal representation. Problem 59. How many ways are there to divide a deck of 36 cards, including 4 aces, into halves so that each half contains exactly 2 aces? Problem 60. A rook stands on the leftmost box of a 1 x 30 strip of squares and can shift any number of boxes to the right in one move. a) How many ways are there for the rook to reach the rightmost box? b) How many ways are there to reach the rightmost box in exactly 7 moves? Problem 61. Each side of a boat must be occupied by exactly 4 rowers. How many ways are there to choose a rowing team for the boat if we have 31 candidates, ten of whom want to be on the left side of the boat, twelve on the right side, and the other nine can sit on either side? Problem 62: Within a table of m rows and n columns a box is marked at the intersection of the pth row and the qth column. How many of the rectangles formed by the boxes of the table contain the marked box? Problem 63: A 10 x 10 x 10 cube is formed of small unit cubes. A grasshopper sits in the center 0 of one of the corner cubes. At a given moment it can jump to the center of any of the cubes which has a common face with the cube where it sits, as long as the jump increases the distance between point 0 and the current position of the grasshopper. How many ways are there for the grasshopper to reach the unit cube at the opposite corner?CHAPTER 12Invariants §1. IntroductionTeacher: "Let us do an experiment. As you see, there are 11 numbers on the blackboard-six zeros and five ones. You have to perform the following operation 10 times: cross out any two numbers. If they were equal, write another zero on the blackboard. If they were not equal, write a one. Do it in your notebooks in any order you wish. Done? Now I will tell you which number you have. Your result must be . . . one!"This short performance brings up a natural question: how did the teacher know which number the students would have at the end of the process described? Indeed, the operations could be performed in a number of different ways, but one thing is always the same: after each operation the sum of the numbers on the blackboard (or in the notebook) is always odd. This is quite easy to check, since this sum can increase or decrease only by 0 or 2. The original sum was odd, so, after 10 operations the only number remaining must be odd as well. Explaining this, one probably cannot help saying the magic word "invariant". So, what is an "invariant"? Naturally, it is something that is invariant, that doesn't change-like the parity of the sum of the numbers in the last example. Another example of an invariant: Problem 1. There are only two letters in the alphabet of the Ao-Ao language: A and 0. Moreover, the language satisfies the following conditions: if you delete two neighboring letters AO from any word, then you will get a word with the same meaning. Similarly, the meaning of a word will not change if you insert the combinations OA or AAOO any place in a word. Can we be sure that words AOO and OAA have the same meaning? Solution. Note that for any permitted deletion or insertion of some combination of letters, the number of A's in the combination equals the number of O's. This means that the difference between the number of A's and the number of O's is invariant. Look at the example 0 --> OOA __, OAAOOOA __, OAOOA. In all these words the number of O's exceeds the number of A's by 1. Let us go back to the solution. The difference for the word AOO is (-1), and for the word OAA it is 1. Therefore, we cannot obtain the word OAA from the word AOO by using the permitted operations, and we cannot claim that these words are synonyms. This solution illustrates the main idea of an invariant. We are given some objects, and are permitted to perform some operations on these objects. Then we are asked: is it possible to obtain one object from another using these operations? 123124MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)To answer the question we construct a quantity that doesn't change under the given operations; in other words, it is invariant. If the values of this quantity are not equal for the two objects in question, then the answer is negative-we cannot obtain one object from another. Let us investigate another problem: Problem 2. A circle is divided into 6 sectors (see Figure 89), and a pawn stands in each of them. It is allowed to shift any two pawns to sectors bordering those they stand on at the moment. Is it possible to gather all pawns in one sector using such operations?FIGURE89Solution. We number the sectors clockwise with the numbers 1 through 6 (see Figure 90) and for any arrangement of pawns inside the circle we consider the sum S of the numbers of the sectors occupied by pawns (counting multiplicities).12365 FIGURE4 9012. INVARIANTS112523D 5 FIGURE 91 Example. For the arrangement in Figure 91 we have S = 2+2+4+4+5+6 = 23. When you shift a pawn to a neighboring sector, the corresponding summand in sum S changes its parity (from odd to even, or from even to odd). Therefore, if we shift two pawns simultaneously, then the parity of S doesn't change at all-it is invariant. But for the arrangement in Figure 89 the value of S equals 21. If all the pawns are in one sector numbered A, then S = 6A. This is an even number, and 21 is odd. Thus, you cannot transform the initial arrangement into an arrangement with all the pawns in one sector. Sometimes an invariant can be applied not to prove that some object cannot be obtained from a given one, but to learn which objects can be obtained from the given one. This is illustrated by the following problem. Problem 3. The numbers 1, 2, 3, ... , 19, 20 are written on a blackboard. It is allowed to erase any two numbers a and b and write the new number a + b - 1. What number will be on the blackboard after 19 such operations? Solution. For any collection of n numbers on the blackboard we consider the following quantity X: the sum of all the numbers decreased by n. Assume that we have transformed the collection as described in the statement. How would the quantity X change? If the sum of all the numbers except a and b equals S, then before the transformation X = S + a + b - n, and after the transformation X = S+(a+b-1)-(n-l) = S+a+b-n. So the value of Xis the same: it is invariant. Initially (for the collection in the statement) we have X = (1+2+ ... +19+20)-20 = 190. Therefore, after 19 operations, when there will be only one number on the blackboard, X will be equal to 190. This means that the last number, which is X + 1, is 191. For teachers. If you hear the solution to this problem from one of your students, it will probably sound like this: at each step the sum of all the numbers decreases by 1. There are 19 steps, and originally, the sum is 210. Therefore, in the end the sum equals 210 - 19 = 191. This is a correct solution; however, you should explain that this problem is an "invariant" problem. The point is that in this case the invariant is so simple126MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)that it can be interpreted quite trivially. The next problem, though it is similar to Problem 3, does not allow for such a "simplification".Problem 4. The numbers 1, 2, ... , 20 are written on a blackboard. It is permitted to erase any two numbers a and b and write the new number ab + a + b. Which number can be on the blackboard after 19 such operations?Hint. Consider as an invariant the quantity obtained by increasing each number by 1 and multiplying the results. Here are a few more remarkable problems using the method of invariants. Problem 5. There are six sparrows sitting on six trees, one sparrow on each tree. The trees stand in a row, with 10 meters between any two neighboring trees. If a sparrow flies from one tree to another, then at the same time some other sparrow flies from some tree to another the same distance away, but in the opposite direction. Is it possible for all the sparrows to gather on one tree? What if there are seven trees· and seven sparrows?Problem 6. In an 8 x 8 table one of the boxes is colored black and all the others are white. Prove that one cannot make all the boxes white by recoloring the rows and columns. "Recoloring" is the operation of changing the color of all the boxes in a row or in a column.Problem 7. Solve the same problem for a 3 x 3 table (see Figure 92) if initially there is only one black box in a corner of the table.FIGURE92Problem 8. Solve the same problem for an 8 x 8 table if initially all four corner boxes are black and all the others are white. Notice that Problem 6, unlike Problems 7 and 8, can be solved using only the idea of parity (of the nuinber of black boxes). Problem 9. The numbers 1, 2, 3, ... , 1989 are written on a blackboard. It is permitted to erase any two of them and replace them with their difference. Can this operation be used to obtain a situation where all the numbers on the blackboard are zeros?Problem 10. There are 13 gray, 15 brown, and 17 red chameleons on Chromatic Island. When two chameleons of different colors meet they both change their color12. INVARIANTS127to the third one (for instance, gray and brown both become red). Is it possible that after some time all the chameleons on the island are the same color? Let us analyze the solution to Problem 10. How can we express the "numerical" meaning of the transformation? One way is to say that two chameleons of different colors "vanish" and two chameleons of the third color "appear" . If we want to use a numerical invariant, we can easily think of a quantity depending only on the numbers (a, b, c), where a, b, and care the numbers of gray, brown, and red chameleons respectively. The operation described means that the triple (a, b, c) turns into the triple (a-1, b- l, c+2) or the triple (a-1, b+2, c-1) or the triple (a+ 2, b - l, c-1 )-depending on the initial color of the two chameleons that meet. It is clear that the differences between corresponding numbers in the old and new triples either do not change or change by 3, which means that the remainders of these differences when divided by 3 are invariant. Originally, a - b = 13-15 = -2, and if all the chameleons are red, we get a - b = 0 - 0 = 0. The numbers 0 and -2 give different remainders when divided by 3, which proves that all the chameleons cannot be red. The cases when all the chameleons are gray or brown are proved in just the same way. For teachers. If the theme "Parity" has already been investigated and you analyzed solutions where parity played the role of an invariant, remind your students of this. The theme "Invariants" is of rather an abstract character and even its very principle often remains vague and complicated for students. Thus one must pay special attention to the analysis of the logic of applying invariants in solving problems. It is especially important to analyze the simplest problems of the topic so that each student solves at least one problem independently. Thy to illustrate the solutions by various examples, making the explanation as graphic and evident as possible. As always, introduce the new word "invariants" and the entire philosophy of invariant only after students have solved or at least investigated a few of the simplest problems using invariants. Clearly, the main difficulty in solving problems using invariants is to invent the invariant quantity itself. This is a real art, which can be mastered only through the experience of solving similar problems. You should not restrain your fantasy. However, do not forget about the following simple rules: a) the quantity we come up with must in fact be invariant; b) this invariant must give different values for two objects given in the statement of a problem; c) we must begin by determining the class of objects for which the quantity will be defined. Here is another important example. Problem 11. The numbers +l and -1 are positioned at the vertices of a regular 12-gon so that all but one of the vertices are occupied by + 1. It is permitted to change the sign of the numbers in any k successive vertices of the 12-gon. Is it possible to "shift" the only -1 to the adjacent vertex, if a) k =3; b) k=4; c) k = 6?128MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Sketch of solution. The answer is negative in all three cases. The proof for all of them follows the same general scheme: we select some subset of vertices which satisfies the condition that there are evenly many selected vertices among any k successive ones (see Figure 93).FIGURE93Check that this condition is true for the subsets shown in the figure. Fbr our invariant we take the product of the numbers on the selected vertices. Initially, it equals -1, but if the -1 has been "shifted" to the left adjacent vertex which is not among those selected, it is 1. Finally, the property of invariance for the quantity introduced follows from the property of the subset of selected vertices indicated above. For teachers. This solution gives us a common idea in the method of invariantsto select some part of each object in which the changes caused by the permitted transformations can easily be described. Comment. This idea also helps us to solve Problems 7 and 9. By the way, you can ask your students one "tricky" question: We have proved that the -1 cannot be shifted to the left adjacent vertex. But can it be shifted to the right adjacent vertex? §2. Colorings Many problems involving invariants can be solved using one particular type of invariant: a so-called "coloring". The following is a standard example:Problem 12. A special chess piece called a "camel" moves along a 10 x 10 board like a (1, 3)-knight. That is, it moves to any adjacent square and then moves three squares in any perpendicular direction (the usual chess knight's move, for example, can be described as of type (1, 2)). Is it possible for a "camel" to go from some square to an adjacent square? Solution. The answer is no. Consider the standard chess coloring of the board in black and white. It is easy to check that a "camel" always moves from a square of one color to a square of the same color; in other words, the color of the square where the "camel" stands is invariant. Therefore, the answer is negative, since any two adjacent squares are always colored differently. Here are some other problems using "coloring" methods in their solutions.12. INVARIANTS129Problem 13. a) Prove that an 8 x 8 chessboard cannot be covered without overlapping by fifteen 1 x 4 polyminos and the single polymino shown in Figure 94.FIGURE94b) Prove that a 10 x 10 board cannot be covered without overlapping by the polyminos shown in Figure 95.FIGURE95c) Prove that a 102 x 102 board cannot be covered without overlapping by 1x4 polyminos. Hint to 13 b). Use the standard chess coloring of the board. Problem 14. A rectangular board was covered without overlapping by 1 x 4 and 2 x 2 polyminos. Then the polyminos were removed from the board, but one 2 x 2 was lost. Instead, another 1 x 4 polymino was provided. Prove that now the board cannot be covered by the polyminos without overlapping. Problem 15. Is it possible for a chess knight to pass through all the squares of a 4 x N board having visited each square exactly once, and return to the initial square? Let us analyze the solution to Problem 15. We color the squares of the 4 x N board using four colors as shown in Figure 96. Assume that there exists such a "knight's tour" . The coloring shown satisfies the condition that if a knight stands on a square of color 1 (2, respectively) then at the next move it will be on a square of color 3 (4, respectively).121212343 4344343 43212121FIGURE96130MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Since the number of 1- and 2-colored squares equals the number of 3- and 4colored squares, these pairs of colors alternate during the trip. Thus, each time the knight is on a square of color 3, it will go to a square of color 1 or 2 on the next move, and it is clear that it can go only to a square of color 1. Thus, colors 1 and 3 must alternate, which is impossible since in this case the knight would never visit the squares of color 2 or 4. This contradiction completes the proof. For teachers. 1. A little fantasy will produce new "coloring" problems. We can investigate, for instance, some variations of polyminos and boards in Problem 13. Remember that a "coloring" method is usually used for proving a negative answer. 2. A little bit more about the "coloring" method itself: there are mathematical problems which can be solved using coloring, though they have nothing to do with the idea of invariant (see and ). Moreover, some variations of this method can be considered as independent topics in a separate small session of a mathematical circle. §3. Remainders as invariants Below are seven more problems using the idea of invariants. They are remarkable in that the invariant in their solutions is a remainder modulo some natural number. This is a very common situation (see Problems 3, 7-9 which concern remainders modulo 2 (that is, parity), or Problem 11-modulo 3). Problem 16. Prince Ivan has two magic swords. One of these can cut off 21 heads of an evil Dragon. Another sword can cut off 4 heads, but after that the Dragon grows 1985 new heads. Can Prince Ivan cut off all the heads of the Dragon, if originally there were 100 of them? (Remark. If, for instance, the Dragon had three heads, then it is impossible to cut them off with either of the swords.) Problem 17. In the countries Dillia and Dallia the units of currency are the diller and the daller respectively. In Dillia the exchange rate is 10 dallers for 1 diller, and in Dallia the exchange rate is 10 dillers for 1 daller. A businessman has 1 diller and can travel in both countries, exchanging money free of charge. Prove that unless he spends some of his money, he will never have equal amounts of dillers and dallers. Problem 18. Dr. Gizmo has invented a coin changing machine which can be used in any country in the world. No matter what the system of coinage, the machine takes any coin, and, if possible, returns exactly five others with the same total value. Prove that no matter how the coinage system works in a given country, you can never start with a single coin and end up with 26 coins. Problem 19. There are three printing machines. The first accepts a card with any two numbers a and b on it and returns a card with the numbers a + 1 and b + 1. The second accepts only cards with two even numbers a and b and returns a card with the numbers a/2 and b/2 on it. The third accepts two cards with the numbers a, b and b, c respectively, and returns a card with the numbers a, c. All these machines also retiirn the cards given to them. Is it possible to obtain a card with numbers 1, 1988, if we originally have only a card with the numbers 5, 19? Problem 20. The number 8n is written on a blackboard. The sum of its digits is calculated, then the sum of the digits of the result is calculated and so on, until we get a single digit. What is this digit if n = 1989? Problem 21. There are Martian amoebae of three types (A, B, and C) in a test tube. Two amoebae of any two different types can merge into one amoeba of the12. INVARIANTS131third type. After several such merges only one amoeba remains in the test tube. What is its type, if initially there were 20 amoebae of type A, 21 amoebae of type B, and 22 amoebae of type C? Problem 22. A pawn moves across an n x n chessboard so that in one move it can shift one square to the right, one square upward, or along a diagonal down and left (see Figure 97). Can the pawn go through all the squares on the board, visiting each exactly once, and finish its trip on the square to the right of the initial one?i0~/FIGURE97Let us try to simulate the process of solving Problem 19. For teachers. It is very effective to present the solution as a short story telling how you came to it, how you thought of the invariant, and so on.What do we have on the surface?-a few permitted operations are given and we are asked whether it is possible to obtain one given object from another. This picture definitely pushes us to find an invariant. Let us start the search. The first operation maps (a, b) to (a + 1, b + 1). What is invariant under this operation? Certainly, the difference between the numbers on the cards, since (a+l)-(b+l) = a-b. Butthesecondoperationchangesthedifference: a/2-b/2 = (a-b)/2-the difference is divided by two. The third operation adds the differences: a - c =(a - b) + (b - c). These observations make us think that it is not the difference between the numbers on the card which is invariant. However, it is very likely that the difference has something to do with this (so far unknown) invariant. So what can it be? Let us look more closely and try to obtain some cards from the given one. (5, 19) -+ (6, 20) (6, 20) --> (3, 10) (3, 10) -+ (20, 27) (6, 20), (20, 27) -+ (6, 27) Enough. Now we can observe the results of our work. We have the following cards: (5, 19), (6, 20), (3, 10), (20, 27), (6, 27). The differences for the pairs of numbers on the cards are: 14, 14, 7, 7, 21. Finally, we know what we must prove! The most plausible conjecture is that our difference a - bis always divisible by 7. This fact can be proved quite easily. We need only consider the behavior of theMATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)132difference under the operations permitted. It either does not change at all, or is or two differences add up to give another one. But the difference multiplied by for the card we want to obtain-(1, 1988)-equals 1 - 1988 = -1987 and is not divisible by 7. This completes the solution, and the answer is negative.!,The problems from this set are more difficult than most of Problems 1-23, but they can serve as good exercises for homework and further investigation. Problem 23. The boxes of an m x n table are filled with numbers so that the sum of the numbers in each row and in each column is equal to 1. Prove that m = n.Remark. Strange as it may seem, this is an "invariant" problem. Problem 24. There are 7 glasses on a table-all standing upside down. It is allowed to turn over any 4 of them in one move. Is it possible to reach a situation when all the glasses stand right side up? Problem 25. Seven zeros and one 1 are positioned on the vertices of a cube. It is allowed to add one to the numbers at the endpoints of any edge of the cube. Is it possible to make all the numbers equal? Or make all the numbers divisible by 3? Problem 26. A circle is divided into six sectors and the six numbers 1, 0, 1, 0, 0, 0 are written clockwise, one in each sector. It is permitted to add one to the numbers in any two adjacent sectors. Is it possible to make all the numbers equal? Problem 27. In the situation of Problem 20, find out which cards can be obtained from the card (5, 19) and which cards cannot. Problem 28. There is a heap of 1001 stones on a table. You are allowed to perform the following operation: you choose one of the heaps containing more than 1 stone, throw away one stone from that heap and divide it into two smaller (not necessarily equal) heaps. Is it possible to reach a situation in which all the heaps on the table contain exactly 3 stones? Problem 29. The numbers 1, 2, 3, ... , n are written in a row. It is permitted to transpose any two neighboring numbers. If 1989 such operations are performed, is it possible that the final arrangement of numbers coincides with the original? Problem 30. A trio of numbers is given. It is permitted to perform the following operation on the trio: to change two of them-say, a and 1>-to (a + b) / ,/2 and (a-b)/,/2. Is it possible to obtain the trio (1, ,/2, 1+,/2) from the trio (2, ,/2, 1/,/2), using such operations? For teachers. 1. "Invariant" problems are very popular; for example, at least two or three problems in each St. Petersburg All-City Mathematical Olympiad can be solved using the idea of an invariant. 2. The idea of an invariant is widespread and permeates different fields of science. If your students are familiar with the basics of physics, then you can analyze as examples some corollaries of the law of conservation of energy, or the theorem of the conservation of momentum, et cetera. 3. Students must understand that if some invariant (or even several invariants) give the same values for two given objects, then it does not mean that the objects can be obtained from each other by the described operations. This is a standard12. INVARIANTS133mistake which arises after the first" acquaintance with invariants. Give your students some simple examples refuting this error. 4. Once more we recall the simplest and most standard invariants: 1) remainder modulo some natural number-Problems 3, 7-11, 17-22; 2) selecting a part of an object-Problems 7, 9, 12; 3) coloring-Problems 13-16; 4) some algebraic expression involving given variables-Problems 4, 26, 30.CHAPTER 13Graphs-2 This chapter continues the investigation of graphs begun in the first part of the present book. Why should we return to this topic? First, graphs are interesting and fruitful objects of study. Second, and most important, elementary reasoning about graphs allows us to get closer to more serious mathematics (this refers mostly to §§1 and 3 of the present chapter). §1. IsomorphismAs we mentioned before, the same graph can be depicted in a number of ways. For instance, the same acquaintance scheme or system of airline routes may be pictured as figures which do not even resemble each other. Consider the following example: during a tournament involving five teams A, B, C, D, and E, team A played teams B, D, and E. In addition, team C played teams Band D, while D also played E. It is clear that both drawings in Figure 98 represent this situation.ABc FIGURE 98Now we give an exact definition.Definition. Two graphs are called isomorphic if they have equally many vertices (say n) and the vertices of each graph can be numbered 1 through n in such a way that vertices of the first graph are connected by an edge if and only if the two vertices having the same numbers in the second graph are connected by an edge. Now we can pay our debt and prove that the graphs shown in Figure 99 (which are copies of those in the chapter "Graphs-1") are not isomorphic. 135136MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE 99The point is that these graphs have different numbers of connected components: the first graph has three and the second has two. We now show that isomorphic graphs must have equal numbers of connected components. It suffices to see that if two vertices of the first graph belong to one connected component, then they are connected by some path, which implies that the two corresponding vertices of the second graph are connected by some path as well and, therefore, also belong to one connected component. Problem 1. Are the graphs in the pairs shown in Figure 100 isomorphic?o.,~ c)I)5 CD g)FIGURE 10013. GRAPHS-2137·O· .4Jf. 72FIGURE 101'[81 4343FIGURE 102Solution. Figures 101 and 102 show that the graphs in the pairs labelled a) and b) are isomorphic. The graphs in the other pairs are not isomorphic. Hints. c) The numbers of vertices are not equal. d) The numbers of edges are not equal. e) The numbers of connected components are not equal. f) The first graph has a vertex with four outgoing edges, but there is no such vertex in the second graph. g) There is an edge in the first graph such that after deleting this edge the graph will fall into two connected components. However, the second graph does not contain such an edge. This can be proved in another way: we consider closed paths which do not pass twice through any vertex. The first graph has two such closed paths: of length 3 and of length 4. The second graph has three such paths: their lengths are 4, 5, and 7. For teachers. Intuitively, students understand quite well when graphs are "identical". Thus it is interesting to listen to their own independent attempts to give a precise definition of isomorphism. Sometimes these "definitions" may involve something like "graphs are isomorphic (identical) if they have equal numbers of vertices and edges". Finding solutions to the different parts of Problem 1 can also lead to a very interesting discussion about various criteria for non-isomorphism. One important concept was incidentally mentioned in the solution to Problem 1. Now we will give it a more accurate definition. The length of a path is the number of edges it consists of.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)138Definition. A cycle is any closed path in a graph which does not pass through the same vertex of the graph twice. In discussing item g), we noticed that the graph shown in Figure 103 (a) had two cycles: 1-2-3-4-1 and 5-6-7-5, while the graph shown in Figure 103 (b) had three cycles: 1-5-6-7-1, 1-2-3-4-5-1, and 1-2-3-4-5-6-7-1.1627 5 7 34213 65(a) FIGURE(b)103Here are two more problems connected with these concepts and definitions. Problem 2. Prove that there does not exist a graph with 5 vertices with degrees equal to 4, 4, 4, 4, and 2. Problem 3. Prove that there exists a graph with 2n vertices with degrees 1, 1, 2, 2, ... , n, and n. Problem 4. Is it true that two graphs must be isomorphic, if a) they both have 10 vertices and the degree of each equals 9? b) they both have 8 vertices and the degree of each equals 3? c) they are both connected, without cycles, and have 6 edges? Problem 5. In a connected graph the degrees of four of the vertices equal 3 and the degrees of all other vertices equal 4. Prove that we cannot delete one edge in such a way that the graph splits into two isomorphic connected components. §2. TreesIn this section we will discuss a certain type of graph which looks quite simple but plays an important part in the theory of graphs. Definition. A tree is a connected graph without cycles. For example, the graphs in Figure 104 are trees, while the graphs in Figures 105 and 106 are not.13. GRAPHS·-2139(a)FIGURE 104L FIGURE 105y~ FIGURE 106This type of graph was given its name because some of them really do resemble trees (see Figure 104 (b)). When studying the properties of trees, the concept of a simple path turns out to be quite useful. We have already defined the concept of a path in the chapter "Graphs-1". A path is called "simple" if it does not include any of its edges more than once. Problem 6. Prove that a graph in which any two vertices are connected by one and only one simple path is a tree.140MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE107Solution. It is obvious that such a graph is connected. Let us assume that it has a cycle. Then any two vertices of this cycle are connected by at least two simple paths (see Figure 107). This contradiction proves that our assumption was wrong. Now we will prove the converse proposition. Problem 7. Prove that in any tree every two vertices are connected by one and only one simple path. Solution. Assuming otherwise, suppose two vertices X and Y are connected by two different simple paths. It seems, at first, that by going from X to Y by the first path and then returning by the second path we obtain a cycle. Unfortunately, this is not completely true. The problem is that our paths may have common vertices (other than their common ends; see Figure 108), and by definition, the vertices of a cycle must not repeat. To extract a real cycle from this "improper" one, we must do the following: 1) going from X, choose the first vertex where our paths diverge (this is point A in Figure 108), 2) beyond this chosen vertex we must find, on path number 1, the first point that also belongs to path number 2 (vertex Bin Figure 108). Now the parts of our cycles between vertices A and B form a cycle. For teachers. The first two sentences of this solution contain its basic idea. Further technicalities may be a bit too complicated for some students. The results of Problems 6 and 7 give another possible definition of a tree, equivalent to the first. Definition. A tree is a graph in which any two different vertices are connected by one and only one simple path. In solving the following problems we will use either definition. Problem 8. Prove that in any tree having at least one edge there exists a vertex which is an endpoint of exactly one edge (such a vertex is called a pendant vertex).13. GRAPHS-21411xyA 22 2 2 11 FIGURE 108Solution. Consider an arbitrary vertex of the tree and move along any edge going out of it to another vertex. If this new vertex has degree 1, then we stay there; otherwise, we move along any other edge to another vertex and so on. It is clear that we cannot come to a vertex we have visited before-this would mean the existence of a cycle. On the other hand, since our graph has a finite number of vertices, our trip must end somewhere. But the vertex it will end in must be a pendant vertex! The statement of Problem 8 is called The Pendant Vertex Lemma. This lemma will be used later in other solutions.For teachers. Starting with this section, we will formulate our problems either in the language of graph theory, or more informally. Our experience shows that we should not make excessive use of either of these two forms. First, students must understand the formal language. Second, they must learn to see the real meaning of the problem behind its informal statement. Thus, it is better to use both languages freely, without being obsessed with one of them. Problem 9. All the vertices of a graph have degree 3. Prove that the graph has a cycle. Problem 10. Prove that if an edge (excluding its ends) is deleted from a tree, then the resulting graph is not connected. Problem 11. There are 101 towns in Forestland. Some of them are connected by roads, and each pair of towns is connected by one and only one simple path. How many roads are there? Solution. Translating the problem into more formal language, we can say that the graph of the roads of Forestland is a tree. This tree must have a pendant vertex. Let us delete it, together with its edge. The resulting graph is also a tree and so it has a pendant vertex, which we also delete, together with its only edge. Performing this operation 100 times we finally obtain a tree with one vertex and, of course, with no edges. Since we were deleting one edge per operation, we conclude that there were 100 edges. In just the same way you can prove another, more general fact;Theorem. In any tree, the number of vertices exceeds the number of edges by 1.142MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)The converse theorem is true as well. Problem 12. Prove that a connected graph in which the number of vertices exceeds the number of edges by 1 is a tree.Problem 13. A volleyball net has the form of a rectangular lattice with dimensions 50 x 600. What is the maximum number of unit strings you can cut before the net falls apart into more than one piece? Solution. We consider this volleyball net as a graph, its nodes as vertices, and the strings as edges. Our objective is to erase as many edges as possible while keeping the graph connected. We delete the edges one by one as long as we can. Notice that if the graph has a cycle, then we can delete any of the edges in this cycle. But a connected graph without cycles is a tree-thus, when we have obtained a tree, we cannot delete any more of the graph's edges! Let us calculate the number of edges in our graph at this final moment. The number of vertices is the same as originally-that is, it equals 51·601 = 30651. On the other hand, a tree with this many vertices must have 30651 - 1 = 30650 edges. At the very beginning we had 601·50+600 · 51 = 60650 edges. Thus we can delete no more than 30000 edges-and it is easy to see that we actually can do this. Methodological remark. We call your attention to the key idea of the solution-finding the "maximal" tree within our graph. Certainly, this "maximal" tree is not unique (see Figure 109). This method (selecting the "maximal" tree) will also help in solving the following three problems.FIGURE109Problem 14. There are 30 towns in a country. Each of them is connected to every other by a single road. What is the maximum number of roads that can be closed in such a way that one can still reach each town from any other? Problem 15. Prove that in any connected graph it is possible to delete a vertex, with all the edges leaving it, so that the graph remains connected. Problem 16! There are 100 towns in a country and some of them are connected by airlines. It is known that one can reach every town from any other (perhaps with several intermediate stops). Prove that you can fly around the country and visit all the towns making no more than a) 198 flights; b) 196 flights.13. GRAPHS-2143For teachers. A separate session can be devoted to the concept of a tree and problems connected with it. Problems using the idea of a maximal tree can be posed one by one after a thorough discussion of the idea. §3. Euler's theorem In this section we will prove a classical theorem named after the great XVIII century mathematician Leonard Euler. In connection with it we also discuss properties of an important type of graph, whose definition is given on the next line.Definition. A graph that can be drawn in such a way that its edges do not intersect each other (except at their endpoints) is called planar. For instance, the graph shown in Figure 110 is planar (a graph isomorphic to it is depicted in Figure 111), while the graph shown in Figure 112 is not (this fa.ct will be proved a bit later).FIGURE llOFIGURE 111We will say that a planar graph is properly depicted by a figure if its edges (as shown on the figure) do not intersect at their interior points.144MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE112For teachers. Perhaps you have already noticed that we are not quite accurate in using the concept of a graph-we do not distinguish between different, but isomorphic, graphs. This inaccuracy is especially clearly seen in the definition of a planar graph. It is important that students understand that a graph may be planar even if some of its edges intersect in a given picture (see Figure 110). If a graph is depicted properly, then it divides the plane into several regions called faces. Let us denote the number of faces by F, the number of the vertices by V, and the number of the edges of the graph by E. For the graph in Figure 111 we have V = 4, E = 6, F = 4 (the outer, infinite, region of the plane is counted as a face).The following fact is then true. Euler's theorem. For a properly depicted connected planar graph the equality V - E + F = 2 always holds true. Proof. We repeat the reasoning used in the solution to the problem about the volleyball net: we delete the edges until we get a tree, keeping the graph connected. Look at the behavior of the quantities V, E, and F under such an operation. It is evident that the number of vertices does not change, while the number of edges decreases by 1. The number of faces also decreases by one: Figure 113 shows how two faces adjacent to the deleted edge merge into one new face. Thus the quantity V - E + F does not change under this operation (we can say that the quantity V - E + F is invariant with respect to this operation-see the chapter "Invariants"!). Since for the resulting tree we have V -:- E = 1 (by the theorem from the previous section) and F ~ 1, for this tree we have V - E + F = 2, and therefore, the same equality is true for the original graph. The equality V - E+F= 2 is called Euler's formula.Euler's theorem is a very strong result and we can derive a lot of beautiful and interesting corollaries from it. Let us begin with a simple problem.13. GRAPHS-2145FIGURE 113Problem 17. There are 7 lakes in Lakeland. They are connected by 10 canals so that one can swim through the canals from any lake to any other. How many islands are there in Lakeland?The next problem is more difficult. Problem 18. There are 20 points inside a square. They are connected by nonintersecting segments with each other and with the vertices of the square, in such a way that the square is dissected into triangles. How many triangles do we have? Solution. We will consider the points and the vertices of the square as the vertices, and the segments and the sides of the square as the edges of a planar graph. For each region (among those into which the graph divides the plane) we calculate the number of edges on its border. Then we add up all these numbers. Since any edge separates exactly two different faces from one another, the total must be simply double the number of edges. Since all the faces are triangles, except for the outer one, which is surrounded by four edges, we get 3(F - 1) + 4 = 2E; that is, E = 3(F -1)/2 + 2. Since the number of vertices equals 24, using Euler's formula, we have24-(3(F2-l) +2) +F=2. Thus F = 43 (counting the "outside face"). So, the number of triangles our square is divided into is equal to 42. Problem 19. Prove that for a planar graph 2E 2:: 3F. We continue with some classical corollaries of Euler's theorem. Let us begin with an inequality. Problem 20. Prove that for a planar connected graph E ~ 3V - 6. Solution. The previous problem gives 2E 2:: 3F. Substituting into Euler's formula we have V - E + 2E/3 2:: 2. Therefore E ~ 3V - 6, as required. Problem 21. Prove that for any planar graph (even if it is not connected) E ~ 3V-6. Hint. The inequality is just the result of summing up the corresponding inequalities for each connected component.It is remarkable that the last inequality allows us to prove the fact claimed at the beginning of this section.146MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 22. The graph with 5 vertices, each of which is connected by an edge to every other, is not planar.Hint. The inequality E :5 3V - 6 does not hold true. A graph in which each vertex is connected by an edge to every other vertex is called complete. In Figure 114 you can see the complete graph with 6 vertices.FIGURE114The result of Problem 22 means that a complete graph with more than 4 vertices is not planar. Problem 23. Is it possible to build three houses and three wells, then connect each house with each well by nine paths, no two of which intersect except at their endpoints?Hint. For the graph given in this problem the inequality 2E ~ 3F can be strengthened. Indeed, any cycle in this graph must be of even length, since houses and wells alternate. Assuming this graph is planar and can be properly drawn, we derive that each face in this (presumably) planar representation must have at least 4 edges on its border. Thus the same calculation as in the solution to Problem 19 brings us to the inequality E ~ 2F. But this inequality does not hold true, so the answer to our question is no. Problem 24. Prove that if the degree of each of the 10 vertices of a graph is equal to 5, then the graph is not planar. The inequality E :5 3V - 6 can be used to prove the following three elegant facts. Problem 25. Prove that in any planar graph there exists a vertex with degree no more than 5. Problem 26. Each edge of the complete graph with 11 vertices is colored either red or blue. We then look at the graph consisting of all the red edges, and the graph consisting of all the blue edges. Prove that at least one of these two graphs is not planar. Problem 27: A heptagon is dissected into convex pentagons and hexagons so that each of its vertices belongs to at least two smaller polygons. Prove that the number of polygons in the tessellation is no less than 13.13. GRAPHS-2147For teachers. Our experience shows that the material covered in this section is quite important for graph theory. A separate session should probably be devoted to this subject. §4. Miscellaneous problems In this section we have gathered a few problems from various parts of graph theory. In solving them, one must combine the methods described in this chapter and in the chapter "Graphs-1" with other significant ideas. Therefore, these problems are quite difficult. Problem 28. Prove that any connected graph having no more than two "odd" vertices (see the chapter "Graphs-1") can be drawn without lifting the pencil off the paper and so that each edge is drawn exactly once. Sketch of the proof. Assume the graph does not have any "odd" vertices at all. We prove the fact by induction on the number of edges. The base (the graph without edges) is obvious. To prove the inductive step we consider an arbitrary connected graph all of whose vertices are "even". Since this graph has no pendant vertices, it cannot be a tree, and thus, it must contain a cycle. Now we can temporarily delete all the edges belonging to the cycle. After this, the graph splits into several connected components which have common vertices with that "temporarily deleted" cycle and satisfy the condition of the theorem (see Figure 115). By the inductive assumption each of these components can be drawn in the required way. It is clear now how to draw the original graph: we go along the cycle and, coming into a vertex belonging to a connected component, we draw the component starting at this vertex (and, certainly, finishing at the same vertex, which is important!) then continue our movement along the cycle.FIGURE115148MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)The proof for the case when our graph contains two vertices with odd degree is quite similar-we temporarily delete a path connecting these two vertices and apply the same technique. A graph which can be drawn without lifting the pencil off the paper, so that each edge is drawn exactly once, is called an Euler graph or unicursal. In the chapter "Graphs-1" we already proved that an Euler graph cannot have more than two "odd" vertices. The last problem allows us to combine all our results into one theorem. Theorem. A graph is an Euler graph if and only if it is connected and has no more than two "odd" vertices. Note that we have already proved the "if" part of the theorem. Here are three more problems. Problem 29. Is it possible to form the grid shown in Figure 116 a) from 5 broken lines of length 8 each? b) from 8 broken lines of length 5 each? (The length of the grid's segments is 1.)FIGURE116Problem 30. There are 100 circles forming a connected figure on the plane. Prove that this figure can be drawn without lifting the pencil off the paper or drawing any part of any circle twice. Problem 31. Prove that a connected graph with 2n "odd" vertices can be drawn without drawing any edge more than once and in such a way that the pencil will be lifted off the paper exactly n - 1 times.Problem 32. There are 50 scientists at a conference and each of them is acquainted with at least 25 of the others. Prove that there are four of them who can be seated at a round table so that each of them has two acquaintances for neighbors.13. GRAPHS-2149Problem 33. Eru:h of 102 students in a school is acquainted with at least 68 other students. Prove that there are four students who have the same number of acquaintances. Problem 34! Let us call the length of any simple path connecting two vertices in a tree the distance between those vertices. Let us call the sum of all the distances between the vertex and all the other vertices of the graph the remoteness of a vertex. Prove that a tree containing two vertices whose remotenesses differ by 1 has oddly many vertices. Problem 35. Alice drew 7 trees on a blackboard, each having 6 vertices. Prove that some pair of them is isomorphic. Problem 36. In a certain country any two towns are connected either by an airline route or by railroad. Prove that a) it is possible to choose one type of transportation so that you can reach eru:h town from any other using only the chosen type of transportation; b) there is a town and a type of transportation such that you can reach any other town from that one with no more than one transfer, using only the chosen type of transportation; c) any town possesses the property indicated in b); d) it is possible to choose a type of transportation so that you can reach each town from any other using only the chosen type of transportation and with at most two transfers on the way. Problem 37. Each of the edges of a complete graph with 6 vertices is colored either black or white. Prove that there are three vertices such that all the edges connecting them are of the same color. Problem 38. Each of the edges of a complete graph with 17 vertices is colored either red, blue, or green. Prove that there are three vertices such that all the edges connecting them are the same color. Problem 39: Each of the edges of a complete graph with 9 vertices is colored either blue or red. Prove that either there exist four vertices with all the edges connecting them blue, or three vertices with all the edges connecting them red. Problem 40: Each of the edges of a complete graph with 10 vertices is colored either black or white. Prove that there are four vertices such that all the edges connecting them are of the same color. Euler's theorem is the most important and essential fact in this For teachers. section. It demands very thorough discussion and careful proof. Other problems can be used in different ways. The most difficult of them (marked with asterisks) are naturally intended for homework. §5. Oriented graphs The main subject of this section is the so-called oriented graph; that is, a graph whose edges are supplied with arrows. We will not prove any fundamental theorems about such graphs; however, the concept of an oriented graph is an important element of general mathematical culture and these graphs are common objects in mathematical problems. Problem 41. After coming back from Fibland, Dmitri told his friends that there are several lakes connected with rivers. He also told them that three rivers flow150MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)out of every lake and four rivers flow into every lake in Fibland. Prove that he was. wrong. Solution. Every river has two ends (lakes). It flows out of one and into the other. Therefore, the sum of the number of rivers "flowing in" must be equal to the sum of the number of rivers "flowing out". But if there are n lakes in Fibland then the sum of the numbers of rivers "flowing in" is 4n, and the sum of the numbers of rivers "flowing out" is 3n. This contradiction completes the proof. Problem 42. There is a capital and 100 towns in a country. Some of the towns (including the capital) are connected by one-way roads. Exactly 20 roads lead out from, and exactly 21 roads lead into, every town other than the capital. Prove that it is impossible to drive from any town to the capital and still obey the driving regulations.In each of the following two problems the reader is requested to place arrows on the edges of a non-oriented graph to satisfy some conditions. Problem 43. In some country each town is connected with every other town by a road. An insane king decided to impose one-way traffic on all the roads so that after you drive from any town you cannot return to it. Is this possible? Problem 44. Prove that it is possible to place arrows on the edges of an arbitrary connected non-oriented graph and choose one vertex in such a way that one can reach any vertex from the chosen one.The concept of an Euler graph and its main properties are used in Problems 45 and 46. Problem 45. The degrees of all the vertices of a connected graph are even. Prove that one can place arrows on the edges of the graph so that the following conditions will be satisfied: a) it is possible to reach each vertex from any other, going along the arrows; b) for each vertex the numbers of "incoming" and "outgoing" edges are equal. Problem 46. Arrows are placed on the edges of a connected graph so that for any vertex the numbers of "incoming" and "outgoing" edges are equal. Prove that one can reach each vertex from any other by moving along the arrows.If you are familiar ·with the method of mathematical induction, then you can use it in solving the problems from the following set. Problem 47. In a certain country each town is connected with every other by a one-way road. Prove that there is a town from which you can drive to any other. Solution. We proceed by induction on the number of towns. The base of the induction is obvious. To prove the inductive step we remove one of the towns. For the rest of them, using the inductive step, we can find a town A possessing the13. GRAPHS-2151required property. Now we replace the removed town B. If there is at least one road going to B, then A is the required town for the original problem. If all the roads lead from B, then B is the town we need. Problem 48. Several teams played a tournament such that each team played every other team exactly once. We say that team A is stronger than team B, if either A defeated B or there exists some team C such that A defeated C while C defeated B. a) Prove that there is a team which is stronger than any other team; b) Prove that the team which won the tournament is stronger than any other. Problem 49. There are 100 towns in a country. Each of them is connected with every other town by a one-way road. Prove that it is possible to change the direction of traffic on one of the roads such that after this operation each town can still be reached from any other. Problem 50. Twenty teams played a volleyball tournament in which each team played every other team exactly once. Prove that the teams can be numbered 1 through 20 in such a way that team 1 defeated team 2, team 2 defeated team 3, ... , team 19 defeated team 20.Finally, the last three problems of this section. Problem 51. Some pair of teams showed equal results in a volleyball tournament in which each team played every other team exactly once. Prove that there are teams A, B, and C such that A defeated B, B defeated C, and C defeated A. Problem 52. There are 101 towns in a country. a) Each town is connected with every other town by a one-way road, and there are exactly 50 roads going into and 50 roads leaving each town. Prove that you can reach each town from any other, driving along at most two roads. b) Some pairs of towns are connected by one-way roads, and there are exactly 40 roads going into and 40 roads leaving each town. Prove that you can reach each town from any other, driving along at most three roads. Problem 53: In Orientalia all the roads are one-way roads, and you can reach each town from any other by driving along no more than two roads. One of the roads is closed for repair, but it is still possible to drive from each town to any other. Prove that now this can be done by driving along at most three roads.CHAPTER 14Geometry We often hear the question "Who needs plane geometry as it is studied in the school curriculum? This is a science for its own sake-it has no real extensions in higher mathematics and sometimes is too complicated and tricky." One answer (which is not complete, of course) is that "school geometry" is a wonderful playground for developing logical and consistent thinking. This "science for its own sake" may be regarded as a game played by axiomatic rules created by the ancient Greeks. Euclid and his predecessors (as well as his disciples) were quite convinced that these rules adequately reflected the laws of the real world around them. As a game, though, geometry can be compared perhaps only to chess in its complexity and elegance. Nowadays probably no one can boast about knowing all the secrets of either of these two great games of mankind. This fact (together with the limits on the size of the chapter) explains why we discuss here only some opening moves of the game. However, we must remember that geometry also is an inalienable part of mathematics and has various links to other areas of "the queen of sciences" : a good teacher will find great opportunities here to demonstrate the integrity of mathematics.We don't want to link our explanation of the subject to any existing textbookswe'd rather that teachers choose for themselves topics for a session depending on the level of their students. We advise teachers to draw upon the school curriculum, but not follow it blindly. Do not be alarmed that most of the problems have virtually the same look as problems from textbooks. Indeed, it would be strange to demand "olympiad" questions in "school geometry" -the word "olyrnpiad" itself implies an extracurricular atmosphere, stretching the limits of the curriculum. Geometry is, in fact, elaborated on quite well in numerous textbooks. Thus we often will refer the reader to other books for appropriate problem material. §1. Two inequalitiesHigh school geometry usually deals with precise statements like: "Points A, B, and C lie on the same straight line." "Altitudes of a triangle meet at one point." "The sum of the angles of a triangle equals 180 degrees." But for earlier study the main tools are undoubtedly the following two inequalities: 153MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)154Inequality N• 1. For any three points A, B, and C on the plane we have AB+ BC ;::: AC, and equality holds if and only if point B belongs to segment AC. Inequality N• 2. In a triangle the larger of any two sides is the side opposite the larger angle. That is, if in triangle ABC we have AB> AC, then LC> LB, and vice versa.In this section we recall these inequalities once more and present a few applications.Problem 1. Prove that if b + c > a, a+ c > b, and a+ b > c, where a, b, and care positive numbers, then there exists a triangle with sides a, b, and c. Problem 2. Prove that the length of median AM in triangle ABC is greater than (AB+ AC - BC)/2. Problem 3. Prove that you can form a triangle from segments with length a, b, and cif and only if there are positive numbers x, y, z, such that a= x+y, b = y+z, c =x+z. Problem 4. Using Inequality N• 2 above prove that if AB = AC, then angles ABC and ACB are equal. Problem 5. In triangle ABC the median AM is longer than half of BC. Prove that angle BAG is acute. For teachers. Problems 1-5 may be very simple for some students, especially if they have been exposed to the same topic in the school curriculum. After discussing the solutions to these easier problems, such students can tackle the following more difficult ones. Problem 6. Prove that if you can form a triangle from segments with lengths a, b, and c, then you can do this also with segments with lengths ./b, ye.,;a,Problem 7. ABCD is a convex quadrilateral and AB+ BD LA,, then LB < LB,, LC > LC,, and LD < LD1. Problem 10. Prove that the median of a triangle which lies between two of its unequal sides forms a greater angle with the smaller of those sides. Problem 11. Is it possible for some five-pointed star ABCDEFGHI K (see Figure 117) to satisfy the inequalities: AB > BC, CD> DE, EF > FG, GH > HI,IK >KA? Methodological remark. The problems of this set are a bit more difficult than Problems 1-5, but they, too, are not extremely hard to solve. Other problems using this material can be found in the chapter "The Triangle Inequality" of the present book. See also and .14. GEOMETRY155DK FIGURE 117 We don't recommend that you devote an entire session to this For teachers. section, but believe that it will be useful to give your students 2-3 of these problems in the course of several sessions. The goal is to implant the triangle inequalities into students' minds not as another "problem solving pattern" but as something more basic, which should be used almost unconsciously. Let's go to the solution to Problem 8. This problem is remarkable because it can be solved using Inequality N• l or using Inequality N• 2. Solution N• 1. We can assume that all circles touch each other externally (see Figure 118) (otherwise the radius of the fourth circle is larger than that of a circle it touches internally). Thus, if we denote the centers of the circles as A, B, C, and D, and their radii, respectively, as r 1 , r 2 , r 3 , and R, then the triangle inequality implies that AD + DC > AC; that is R + r, and, therefore, R+ R + r3 > AC > r1 + r3 + 2r2> r2.FIGURE 118Solution N• 2. One of the angles DBA and DBC is non-acute and, therefore, it is the biggest in the triangle DBA or DBC. Without loss of generality we can assume this is angle DBA. Then it follows by inequality N• 2 that DA > AB; that is, R + r1 > AB > r1 + r2, or R > r,.156MATHEMATICAL CIRCLES (RUSSIAN F:XPgRIF:NCF:)To conclude this section we list several problems whose solutions require an auxiliary idea together with the triangle inequalities. Problem 12. Given isosceles triangle ABC with vertex angle B equal to 20 degrees, prove that a) AB< 3AC; b) AB> 2AC. Problem 13. The perimeter of a five-pointed star whose vertices coincide with those of a given pentagon F, the perimeter of F itself, and the perimeter of the inner pentagon of the star are prime integers. Prove that their sum is no less than 20. Remark. Don't be surprised that this is classified as a geometric problem! Problem 14. A point is selected on each side of a square. Prove that the perimeter of the quadrilateral formed by these points is no less than twice the length of the square's diagonal. §2. Rigid motions of the plane and congruence This theme is very rich in interesting facts and connections with higher mathematics. Students can be led to an understanding of the role of symmetry in mathematics and to the concept of a group, which is central to much of higher mathematics. A crystallographic group, algebraic properties of the group of rigid motions of the plane, and Lobachevsky's geometry are all linked to this important subject. Fbr teachers. l. We assume that students are familiar with the basic congruence theorems (see any appropriate textbook or school curriculum). 2. To begin the study of rigid motions of the plane, let the students list all the types of rigid motions they know. The definition of isometries (or rigid motions) is very simple: these are the transformations of the plane which preserve distance. It turns out there are just a few types: translations, rotations, line reflections, and glide reflections (compositions of a line reflection and a translation). Solutions to the following problems should be carefully discussed, as they are important in later work. Problem 15. Prove that given any two triangles, each with the same sides a, b, and c, we can make them coincide by moving one of them across the plane (and perhaps reflecting in a line). In other words, they are congruent. Problem 16. a) If rigid motion T leaves all the vertices of triangle ABC in place, then T is the identity transformation. b) If two rigid motions T and T' send the vertices of triangle ABC to the same points A', B', C', then T and T' are the same transformation (i.e., any point has the same image under i' as it does under T'). Problem 17. a) What is the composition of two translations? b) Prove that any translation can be represented as a composition of two symmetries with respect to two points M and N. c) Consider the rigid motion which is the composition of a symmetry in line m and a translation with unit distance in a direction parallel to line m. Prove that this rigid motion is neither a rotation, nor a translation, nor a line reflection.14. GEOMETRY157For teachers. A geometric solution to the last problem should be thoroughly discussed, making sure that the students learn the concept of composition. This problem is probably more appropriate for homework than for solving in the session. Problem 18. Two equal circles are given. Is it always possible to map one of them to another by a rotation? Problem 19. Is it possible for a rotation to map a half-plane onto itself? What about a line symmetry? Problem 20. It is known that some figure on the plane coincides with itself after a rotation of 48 degrees about point 0. Is it necessarily true that it coincides with itself after a rotation of 72 degrees about the same point? For teachers. 1. The study of rigid motions and their compositions presents an excellent opportunity for a more general discussion of mappings and composition, with illustrations from both algebra and geometry. 2. At some moment the students should be asked whether they know how to "construct" (with compass and straight edge only) the rigid motions of the plane. Can they, for instance, construct the image of a circle under a given line reflection? The next topic is the use of rigid motions for the solution of geometric problems. It deserves its own (very heavy) book. We will try to give some examples and introduce some basic ideas of the subject. Problem 21. Point A is given inside a triangle. Draw a line segment with endpoints on the perimeter of the triangle so that the point divides the segment in half. Problem 22. A sheet of paper is given, with two lines. The lines are not parallel, but intersect at a point off of the sheet of paper. Construct an angle which is twice as large as that made by the two lines. Problem 23. Inscribe a pentagon in a given circle so that its sides are parallel to five given straight lines. Problem 24. In trapezoid ABCD (AD II BC) M and N are the midpoints of the bases, and line MN forms equal angles with lines AB and CD. Prove that the trapezoid is isosceles. Problem 25. Points P, Q, R, and Sare taken on sides AB, BC, CD, and DA of square ABCD respectively so that AP: PB= BQ : QC= CR: RD= DS: SA. Prove that PQRS is a square. Problem 26. Point P and two parallel lines are given on the plane. Construct an equilateral triangle with one of its vertices coinciding with P and two others lying on the given lines. Problem 27. On a given line find a point M such that a) the sum of the distances from M to two given points is a minimum; b) the difference between these two distances is a maximum.We repeat that it is virtually impossible to exhaust this beautiful geometric topic, and refer the reader to the books [42, 65, 71] and , where he or she can find dozens of interesting and rather difficult problems in this area. Below are just a few more examples involving the properties of rigid motions and symmetries of figures on the plane.ISBMATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 28. Prove that if a triangle has two axes of symmetry, then it has at least three axes of symmetry. Problem 29. Which letters of the English alphabet have an axis of symmetry? A center of symmetry? Problem 30. Does there exist a pentagon with exactly two axes of symmetry? Problem 31. Find the set of all points X on the plane such that a given rotation sends X to X', and the straight line XX' passes through a given point S.Let us discuss now the proof of the rather difficult Problem 23. Instead of the five given lines L 1 , L,, ... , Ls we consider lines K 1 , K,, ... , Ks, perpendicular to them and passing through the center of the circle. Then it is clear that lines L; and AB (for any two points A and Bon the circle) are parallel if and only if A and B are gymmetric with respect to line K;. It remains to find a point Mon the circle which will remain fixed after reflection in all five lines K 1 , K 2 , ••• , Ks. Since the composition of five line reflections is a line reflection again (with its axis passing through the center of the circle!), such a point must exist and can be found as one of the points where the axis of gymmetry and the circle meet. Question. There is a small gap in the solution above. Find and fix it. §3. Calculating angles What do you need to know to calculate the angles of geometric figures? We give only the most basic facts here: (1) the sum of the angles in any triangle is 180 degrees; (2) a pair of vertical angles are equal; (3) angles lying along a straight line add up to 180 degrees (see Figure 119);FIGURE 119 (4) an inscribed angle equals half the central angle which intercepts the same arc of a circle, and as a corollary, we have that (5) two inscribed angles intercepting the same arc of a circle are equal; (6) rigid motions of the plane do not change angle measure.Problem 32. Angle bisector BK is drawn in isosceles triangle ABC, with angle A equal to 36 degrees. Prove that BK= BC.14. GEOMETRY159c BAEFIGURE 120Problem 33. Prove that the sum of the angles at the vertices of a five-pointed star (see Figure 120) equals 180 degrees. Problem 34. Can two angle bisectors in a triangle be perpendicular? Solution to Problem 32. Since LC = 72° and LB = 72°, we have LK BC = 36° and, therefore, CKB = 72°. Thus triangle KBC is isosceles and BK= BC. Solution to Problem 33. Clearly, 180° = LEBD+LBED+LBDE = LE+LB+ LD+LFED+LFDE. SinceLFED+LFDE= 180°-LEFD= 180°-LCFA= LA+LC, we have 180° =LE+ LB +LD+ LA+ LC. So, we can see that the method is as follows: we denote certain angles as a, /3, Using facts (1)-(6) we eventually come to the required result."/, o, ... , then we express all the other angles in terms of these.Methodological remark. Here we have a subtle alternative. On the one hand, if we denote just one or two angles by letters, we might not be able to express the remaining angles and parameters of the problem as functions of the variables introduced. On the other hand, introducing too many angles as unknown variables will make our drawing messy and could make our goal unattainable (since possible correlations between the given angles can become obscured). For teachers. Usually, the choice of the "starting" angles (and of their number) constitutes one of the most important parts of the solution. To learn how to introduce the starting variables (in our case, the angles) is one of the crucial ingredients of mathematical culture on the "olympiad" level. Only vast experience or fully developed methodological thinking can help the students to find the right choice. Here are five more problems (see also [65, 70), and any school textbooks). Problem 35. Chords AB and CD in circle Sare parallel. Prove that AC= BD. Problem 36. The ratio of three consecutive angles in an inscribed quadrilateral is 2 : 3 : 4. Find their values.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)160Problem 37. In triangle ABC LA= go0 • Median AM, angle bisector AK, and altitude AH are drawn. Prove that LMAK = LKAH. Problem 38. Square ABCD is given. A circle with radius AB and center A is drawn. This circle intersects the perpendicular bisector of BC in two points, of which 0 is the closest to C. Find the value of angle AOC. Problem 39. Two circles intersect at points A and B. AC is a diameter of the first circle, and AD is a diameter of the second. Prove that points B, C, and D lie on the same straight line. The solution of Problem 37 is rather standard. Let us denote angle BCA as "' (see Figure 121). Then, since AM =MC, we get LM AC = "' and, therefore, LM AK = 45° - "' (and we see now that "' must be an angle which is not greater than 45 degrees). Further, LABC = goo - °'• which implies LEAH= ct. Thus LKAH = 45° - ct= LMAK.BFIGURE121For teachers. 1. There are many more difficult problems on calculating angles. In fact, most problems in school geometry involve calculating angles, so it is a good habit to write down neatly the values of the angles on a geometric drawing. 2. We believe that this is not a theme for a separate session. It would suffice to make up a few series of problems and submit them for solution at several different sessions.§4. Area This topic is as extensive as many other geometric topics, so we will concentrate more on methods. What are the main principles used in solving problems involving area? We will name just the most basic facts: a) The main properties of area: it is invariant under rigid motions of the plane, and if a figure is split into two disjoint figures, then its area equals the sum of the smaller areas. b) The main formulas: S = ah/2 (where S is the area of a triangle, a is its side, and h is the altitude perpendicular to that side); S = rp (where Sis the area16114. GEOMETRYof a triangle, pis haJf its perimeter, and r is the radius of the inscribed circle), et cetera. c) Basic inequaJities such as 8 :::; ab/2 (where 8 is the area, and a and b are two sides of a triangle); see Problem 40. d) If there are expressions like ab or a 2 + b2 present in the statement of a geometric problem (that is, expressions of degree 2), then it is likely you should try to use area to solve the problem. e) If there are expressions present in the statement of a problem which can be naturally linked with each other via some area formula, then write this formula down and examine it-this will never do any harm. From now on, we will frequently denote the area of figure F byIFI.For teachers. Items a), b), and c) give you a wonderful opportunity to find out the extent of your students' knowledge in this specific field of geometry. We do not advise that you turn the entire session into a discussion of this topic, aJthough the basic concept of this theme must be learned quite thoroughly. This cannot be done during a few sessions. For example, only after a year of successful seminars should you investigate more theoreticaJ parts of this branch of geometry (for example, axioms of area). Problem 40. The lengths of the sides of a convex quadrilateraJ are a, b, c, and d (listed clockwise). Prove that the area of the quadrilateraJ does not exceed a) (ab+ cd)/2; b) (a+ b)(c + d)/4. Problem 41. Is it possible that the ratio of the three aJtitudes of a triangle is 1: 2: 3? Problem 42. A triangle of area 1 has sides of lengths a, b, and c where a Prove that b ~ ../2.~b ~ c.Problem 43. If aJl the sides of a triangle are longer than 1000 inches, can its area be less than 1 square inch? Solution to Problem 41. Let 8 be the area of a triangle, and a, b, and c the lengths of its sides. Then the aJtitudes are equaJ to 28/a, 28/b, and 28/c, and a : b: c = 1 : 1/2 : 1/3, which contradicts the triangle inequaJity. How did we think of introducing the area and the sides of the triangle? See item e) in the beginning of this section. The three preceding problems are good representatives of the subtopic "area and inequaJities". Below are a few problems deaJing with more "exact" caJculations. Problem 44. Points K, L, M, and N are the midpoints of the sides of quadrilateraJ ABCD. Prove that 2IKLMNI = IABCDI. Problem 45. Find the area of convex quadrilateral ABCD, if line AC is perpendicular to line BD, AC= 3, and BD = 8. Problem 46. Triangle ABC is given. Point A 1 lies on segment BC extended beyond point C, and BC= CA 1 . Points B 1 and C 1 are constructed in the same way (see Figure 122). Find IA1B1Cil, if IABCI = l. Problem 47. Point M lies within triangle ABC. Prove that areas of triangles ABM and BCM are equaJ if and only if M lies on median BK.162MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE122Finally, a series of problems whose solutions require not only calculations but also proof. Problem 48. Prove that if two convex quadrilaterals have the same midpoints for all their sides, then their areas are equal. Problem 49. The diagonals of trapezoid ABCD (with BC II AD) meet at point 0. Prove that triangles AOB and COD have equal areas. Problem 50. Prove that the sum of the distances from a point inside an equilateral triangle to its sides does not depend on the position of the point. For teachers. You can see from the solutions to Problems 44-47 that even if a problem is about area, standard geometric ideas may apply as well: congruent triangles, similarity, Thales's theorem. 1 This is quite natural. Using such examples, you can make the students see that the solution to a problem usually includes several ideas. It is rare that an "olympiad" problem can be solved in one move. This is a very general principle of problem solving, applicable to higher mathematics as often as to olympiads and other contests. §5. Miscellaneous This section consists of three sets of problems on topics not discussed in the present chapter. These problems are intended mostly for homework and can be considered as exercises to accompany more detailed study (see also the other books on geometry in the list of references). Set 1. Constructions Problem 51. Construct a triangle if you know a) its base, altitude, and one of the angles adjacent to the base; b) the three midpoints of its sides; c) the lengths of two of its sides and the median to the third side; d) two straight lines which contain its angle bisectors, and its third vertex. 1 Thales's theorem, one of the oldest geometric results on record, states that a line parallel to one side of a triangle divides the other two sides in proportion.14. GEOMETRY163Problem 52: Find the midpoint of a segment a) using a compass only; b) using only a two-sided ruler (with two parallel sides), whose width is less than the length of the segment; c) using only a two-sided ruler, whose width is greater than the length of the segment. Problem 53. Segment AB is given in the plane. An arbitrary point M is chosen on the segment, and isosceles right triangles AMC and BM D are constructed on segments AM and MB (as their hypotenuses) so that points C and D are on the same side of AB. Find the set of midpoints of all such segments CD. Problem 54. A certain tool for geometric construction can be used to a) draw a straight line through two given points; b) erect a perpendicular to a given line at a point lying on the line. Show how to use this tool to drop a perpendicular from any given point to any given line. Problem 55. Peter claims that the set of points on the plane which are equidistant from a given line and a given point is a circle. Is he right? Set 2. Calculations Problem 56. Find an error in the following "proof" of the fact that in a right triangle, the hypotenuse has the same length as a leg (see Figure 123). Point M is the intersection of the bisector of angle C and the perpendicular bisector of segment AB. Points K, L, and N are the feet of the perpendiculars dropped from M to the sides of the triangle. Triangles AM K and MK B are congruent since they have equal hypotenuses and equal legs. Thus, AM = MB, and triangles ALM and MN B are congruent for the same reason. Therefore, AL = NB andAC= AL+LC =NB+CN =BC.A FIGURE123Problem 57. ABCD is a quadrilateral such that BC= AD, and Mand N are the midpoints of AD and BC respectively. The perpendicular bisectors of segments AB and CD meet at point P. Prove that P also lies on the perpendicular bisector of segment MN.164MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 58. A right triangle with an acute angle of 30° is given. Prove that the length of that part of the perpendicular bisector of the hypotenuse which lies inside the triangle equals one third of the greater leg of the triangle. Problem 59. Altitudes AA 1 , BBi. CCi, and medians AA., BB2, CC2 are drawn in triangle ABC. Prove that the length of the broken line A 1 B2C 1 A2B 1 C2A 1 equals the perimeter of triangle ABC. Set 3. Similarity Problem 60. One of the diagonals of an inscribed quadrilateral is a diameter of its circumcircle. Prove that the projections of any two opposite sides of the quadrilateral onto the other diagonal are equal. Problem 61. Arc AB, measuring 60°, is given on a circle with center 0, and point M is chosen on the arc. Prove that the straight line that passes through the midpoints of segments MA and OB is perpendicular to the line passing through midpoints of segments MB and OA. Problem 62. Angle bisector AD is drawn in triangle ABC. Prove that CD/ DB = CA/AB. Problem 63. In isosceles triangle ABC perpendicular HE is dropped from the midpoint Hof BC to side AC. Prove that if 0 is the midpoint of HE, then lines AO and BE are perpendicular. Epilogue I. The topic "Geometric inequalities", which we merely touched on in §1,can be developed much further. Many very beautiful pearls of geometry, such as isoperimetric inequalities (see , Chapters 15 and 16), follow from the simple triangle inequalities. 2. Don't forget that in the present chapter we not only give some particular topics and problems but also indicate possible directions of how a session can proceed. We hope you will find the methodological remarks in this chapter useful. 3. "Calculating angles" is only one theme representing the entire realm of "computational geometry", which can also be studied. 4. Unlike other topics in this chapter, "Area" begins with rather difficult problems rather than with mere exercises. You can find many simpler problems in textbooks. 5. For teachers. If a problem cannot be "captured" with a quick attack, then, perhaps, it may fall to a long and careful siege (which may take days or weeks). Such a siege is conducted by combining different methods and various ideas, using calculations, or through a gradual accumulation of facts. 6. Many beautiful themes in plane geometry, which are quite accessible to students, are not described in the present chapter. These include similarity and its applications, inscribed and circumscribed polygons, interesting points in a triangle, numerical relations, et cetera. They deserve to be studied, but we cannot include them all here since this would turn this chapter into a huge and dull reference. We have just tried to outline the basic aspects of this peculiar game and science, dropping only similarity from the most important topics. In spite of the diversity of the material covered here, we believe that it is very important to show the students14. GEOMETRY165its integrity, and also the links which connect its various areas with each other and with the other branches of mathematics and science. 7. Quick reference guide: a) A large number of problems from the usual geometry curriculum can be found in books [42, 44, 34, 65, 70, 64] and others. However, the first two of them ascend too quickly to the heights and are not recommended for most students. b) Books and are excellent for those who like to see mathematics as algebra rather than geometry. They also provide very good problems on transformations of the plane. c) Books and are two of those rare mathematical books which are simply pleasant to read. However, we face the question "for whom are these books written?" It seems that the books were intended for people who already know their contents. These books can be read again and again-the vividness of their explanations will attract all readers interested in the subject. Also instructive are the geometric chapters (and others, too) in Martin Gardner's books [5, 6] and .CHAPTER 15Number Bases §1. What are they? Any student can say that "2653" stands for the number "two thousand six hundred fifty three", whatever that may mean. How do they know this? We are all accustomed to the following way of writing numbers: the last digit denotes the number of units in the given number, the next-to-last-the number of tens, the third last-the number of hundreds, and so on (though this is a bit ambiguous, since the number of units in 2653 is, in a way, not 3, but 2653!). This way of writing numbers (and interpreting strings of digits) is called in brief a number base system. Thus, writing "2653", we think of the number 2· 1000+6· 100+5· 10+3· l, or shortly, 2 · 103 + 6 · 10 2 + 5 · 10 1 + 3 · 10°. We print the digits of the number in boldface to make it easier to distinguish them from other numbers. We can easily see that the number ten plays a special part in this representation: any other number is written as a sum of different powers of ten with coefficients taking values 0 through 9. This is why this system is called "decimal" (from the latin word for "ten"). To write a number we use the ten special symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, called digits. They denote the numbers from zero to nine. The next number, that is, ten, is regarded as a unit of the next level and is written with two digits: 10, which, roughly speaking, means "add up one times ten and zero times one" .Now, what if we used some other number, say, six? Analogously, we would need six symbols as digits. We can take the six familiar symbols 0, l, 2, 3, 4, and 5, which will denote the numbers from zero to five. The number six will be the unit of the next level, and, therefore, it will be written as 10. Proceeding with this analogy, we can represent each natural number as the sum of different powers of six with coefficients from 0 to 5. For instance (all numbers are written in the decimal system): 7=1·6 1 +1·6°, 12 = 2. 61 +0. 6°, 35=5·61 +5·6°, 45 = 1 . 62 + 1 . 61 + 3 . 6°. Thus in our new number system (which is called the "base six system") we write the number 7 as "11", the number 12 as "20", 35 as "55", and 45 as "113". It is easy to see that we can write any natural number in the base six system.We show how to do this for the number 450 (in this example, as earlier, all the given numbers are written in the decimal system unless enclosed in quotes). 167168MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)The largest power of six that does not exceed 450 is 216. Dividing 450 by 216, we have a quotient of 2 (and a remainder of 18). Thus the first digit of the numeral 450 for the base six system is 2. Now we take the remainder 18 and divide it by the next smaller power of six-at the previous stage we divided by 63 = 216, and now we divide by 62 = 36. The quotient is 0, hence the second digit is 0. The remainder is 18 and we divide this by the next smaller power of six; that is, by 61 = 6. Now we see that the next digit is 3 (the remainder is 0). Therefore, the last digit (the quotient after the division by 6° = 1) is 0. Finally, the base six representation of 450 is "2030". While building our new system, we have not used any particular properties of the number 6, whatever they may be. Similarly, starting with any natural number n greater than 1, we can build a base n number system, in which the digits of a number are connected with its representation as a sum of powers of n. In this system, the number n is called the base. Tu avoid ambiguity, we will write the base of the system as a subscript (in decimal notation) at the right end of the numeral. Using this notation, we can rewrite the equalities indicated earlier as: 710 = 116, 1210 = 20., 3510 = 555, 4510 = 1136 .Exerciae 1. How many digit symbols do we need for a a) binary (that is, base 2) system; b) base n number system? To write a number in the base n system, we must represent it in the following form: aknk+ ak-1nk-l + ... + a2n2 + a1n 1 + aon°,where each a; takes values from 0 to n - 1, and ak is not equal to zero (although the last restriction is, strictly speaking, not necessary). Exerciae 2. Write in decimal notation the numbers 10101a, 101013, 2114, 1261, and 15811. Exercise 3. Write the number 10010 in the systems with bases 2, 3, 4, 5, 6, 7, 8, and 9. Exerciae 4. In a system whose base is greater than 10, we need more than ten digit symbols, so we must invent some. For example, in the base 11 system, we might use "A" to represent the "digit" 10. So, for example, 2110 could be written as lA. Using this convention, write the number 111 10 in the base eleven notation. Let us learn how to add and multiply numbers written in an arbitrary system. We can do this in exactly the same way as in the decimal system, but we must remember that a "carry" occurs each time the result of adding up digits in a column exceeds or equals the base of the given number system. Below is an example of the addition of the two numbers 12410 and 417 10 in the base 3 system. First, we rewrite the numbers in the base 3 system: 12410 = 11121 3, 41710 = 1201103. Then we write them one under another, lining up their rightmost digits. "Carries" are given in the upper row in small print. (I)+1(1) (I)11 22 0 2 0 2 0 0015. NUMBER BASES169To perform these operations successfully one must know the addition and multiplication tables for numbers less than the base of the system-that is, for one-digit numbers. For the decimal system, we have learned it early and well. Exercise 5. Write down these tables for systems with bases 2, 3, 4, and 5. Exercise 6. Calculate a) 11002 + 11012; b) 2013 · 1023. Fbr teachers. We explained here very briefly how to add and multiply the numbers in any number base system. In a real session this would take more time. Of course, the goal of this work is not speed or accuracy in computations written in another number base system. An examination of and some practice in the addition and multiplication algorithms written in systems other than base 10 can lead to a deeper understanding of these algorithms. Now we describe an effective algorithm for converting from one number system to another. It differs from the one we already know, because now the representation of a number will appear digit by digit from right to left rather than from left to right. The last digit is just the remainder when the number is divided by the base of the new system. The second digit can be found as follows: we take the quotient from the previous calculation and find the remainder when the quotient is divided by the base of the new system. Then we proceed in exactly the same way until we complete the representation. Example. Let us convert the number 250 10 to the base 8 ("octal") system: 250 = 31 . 8 + 2, 31=3 ·8+7, 3=0·8+3. Thus, 250 10 = 372a. Exercise 7. Convert to the base 7 system the numbers a) 100010 ; b) 532 8 . In conclusion we submit a few more interesting problems. Problem 1. A teacher sees on the blackboard the example 3 · 4 = 10. About to wipe it away, she checks if perhaps it is written in another number base system. Could this thought have been right? Problem 2. Does there exist a number system where the following equalities are true simultaneously: a) 3 + 4 = 10 and 3 · 4 = 15; b) 2 + 3 = 5 and 2 · 3 = ll? Problem 3. State and prove a condition (involving the representation of a number) which allows us to determine whether the number is odd or even a) in the base 3 system; b) in the base n system. Problem 4. A blackboard bears a half-erased mathematical calculation exercise:+2 3 6 4 2 4 2 4 2 3170MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Find out which number system the calculation was performed in and what the summands were. Problem 5. A teacher said that there were 100 students in his class, 24 of whom were boys and 32 of whom were girls. Which number system did the teacher use in this statement? For teachers. The material of this section can be discussed during two or three successive sessions. Students must learn: -the concept of a number system; -how to convert numbers from one system to another; -·how to add and multiply in an arbitrary number system. To make this rather technical material less boring, we recommend using problems similar to 1-5 above. §2. Divisibility tests In the previous section we learned how to add and multiply numbers in an arbitrary number base system. The reverse operations-subtraction and division-are performed in the same manner as in the decimal system. However, these operations (like "long division", for example) are a bit more difficult, even in our usual decimal system. Thus, it is often convenient to use divisibility tests to find out if one number is divisible by another, without actually performing the operation. The tests for the decimal system are discussed in the chapter "Divisibility-2". In non-decimal systems the situation is more unusual and difficult-try, for example, to find out if 123456654321, is divisible by 6. Let us start simply. How do we know that a number with its last digit equal to zero is divisible by 10? The point is that in the decimal representation of any number aklOk + ak-110k-I + ... + a210 2 + a110 1 + aol0° all the summands are divisible by 10, except, perhaps, the last one. In our case, however, the last summand is zero and, therefore, the whole sum is divisible by 10. We can prove the converse statement similarly: if a natural number is divisible by 10, then its last digit is zero. Consider now an arbitrary number system. The same ideas allow us to prove the following divisibility test: In a base n system the representation of a number ends with zero if and only if this number is divisible by n.Problem 6. State and prove the divisibility test for a) a power of the base of a system (similar to divisibility tests for 100, 1000, in the decimal system); b) a divisor of the base of a system (similar to divisibility tests for 2 and 5); c) a power of a divisor of the base of a system. Methodological remark. We would like to emphasize once again that different number systems are just different ways to write numbers. Thus the divisibility of one number by another does not depend on the particular system in which they are written.15. NUMBER BASES171At the same time, in each system there are some tricks to determine divisibility by certain specific numbers. These are the divisibility tests. Let us investigate now other, less trivial, divisibility tests. Perhaps the most well-known of these are the tests for divisibility by 3 and 9. We will try to generalize these tests for any number base system. First, we must understand the proof of that test in the decimal system (see the chapter "Divisibility and remainders"). The only significant fact used is that 9 = 10-1, and therefore that 10 1 (mod 9). Let us formulate and prove the analogous test for divisibility by n - 1 in the 1 (mod n - 1). Hence, n' 1 (mod n - 1) for any base n system. Indeed, n natural number s. Therefore==aknk+ ak-1nk-I + ... + a1n 1 + aon°==ak+ ak-1 + ... + a1 + ao (mod n-1).Thus the sum of the digits of a number written in the base n system is divisible by n - 1 if and only if the number itself is divisible by n - 1. Let us recall the question we asked at the beginning of this section: is 123456654321 7 divisible by 6? Now we can answer this question easily: since the sum of the digits (which is 42 10 ) is divisible by 6, the number itself is also divisible by 6. Problem 7. State and prove the test for divisibility by a) a divisor of the number n-1 in the base n system (similar to the divisibility test for 3 in the decimal system); b) the number n + 1 in base n system (similar to the divisibility test for 11); c) a divisor of the number n + 1 in the base n system (there is no analog in base 10). For teachers. We reoommend devoting a whole session to the topic of this section. It would be wonderful if during this session students (using hints from the teacher) could formulate and prove various new divisibility tests. §3. Miscellaneous problems Up to now we have been interested in number systems for themselves. Now we are going to discuss a few problems which seem to have nothing to do with number systems. However, non-decimal number systems arise quite naturally when we try to solve these problems. Problem 8. What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? Weights may be placed only on the left pan. Solution. Every natural number can be written in the binary system. Thus, to weigh any number of grams of gold from 1 through 100 it is sufficient to have seven weights: 1, 2, 4, 8, 16, 32, 64. On the other hand, six weights are insufficient since we can obtain no more than 26 - 1 = 63 different weights with them (each weight is either placed or not placed on the left pan). Remark. Note that we did not assume that the weights must be integers. This assumption would not make the solution simpler. Problem 9! The same question as in the previous problem, but the weights can be placed on either pan of the balance.172MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Solution. To explain the solution to this problem we need the following interesting property of the base 3 system: Every natural number can be represented as the difference of two numbers whose base 3 representations contain only O's and l's. We can prove this property by writing the original number in the base 3 system and constructing the required numbers digit by digit from right to left. This is a good exercise, and is left to the reader. Now it is clear that it suffices to have only 5 weights that weigh 1, 3, 9, 27, 81 (can you see why we don't need a weight of 243 grams?). Four weights are insufficient since we cannot weigh more than 34 - 1 = 80 different weights on them (each weight is either placed on the left pan, or on the right, or not placed on the balance at all). Problem 10. An evil king wrote three secret two-digit numbers a, b, and c. A handsome prince must name three numbers X, Y, and Z, after whiclt the king will tell him the sum aX + bY + cZ. The prince must then name all three of the King's numbers. Otherwise he will be executed. Help him out of this dangerous situation. Solution. The prince can name the numbers 1, 100, and 1002 = 10000. The numbers a, b, and c will then just be the digits of the sum aX + bY + cZ in the base 100 system. Problem 11: Prove that from the set 0, 1, 2, ... , 3k-t one can choose 2k numbers so that none of them can be represented as the arithmetic mean of some pair of the cltosen numbers. Solution. We will use the base 3 system. Let us assume that the base 3 representation of any of the given numbers contains exactly k digits-if there are fewer than k, then we just fill in the rest of the places with zeros. Now we cltoose those numbers whose base 3 representations contain only O's and l's. There are exactly 2• of them. We show that this can serve as the required subset. Suppose that there were three different numbers in the subset-say, x, y, and z---satisfying the equality x + y = 2z. Since the numbers x and y must differ in at least one digit, we could then find the rightmost suclt digit. The corresponding digit of their sum x + y would be 1. But the base 3 representation of 2z contains only O's and 2's. This is a contradiction. Problem 12. Prove that a subset of 2k numbers with the same property can be cltosen from the numbers 0, 1, 2, ... , (3k - 1)/2. The problems from this section can be given at sessions or used For teachers. in various mathematical contests. §4. The game of Nim Here we will talk about one form of. the famous game of Nim (see Martin Gardner's books [5-7]). Its rules are simple. There are three heaps of stones (the initial number of stones in eaclt heap may vary). Two players make their moves in turn by taking several stones from the heaps. It is allowed to take any number of stones but only from one heap at a move. A player who takes the last stone wins the game. It is remarkable that the winning strategy for this game can be expressed using the binary system. We will discuss this strategy in a more generalized situation:15. NUMBER BASES173for an arbitrary number of heaps. We also note that in the case of two heaps this game can be turned into a game with a (chess) rook on a rectangular board (see Problems 10, 16, 22 from the chapter "Games"). As usual, to "solve" the game it is sufficient to determine the set of winning positions (see §3 of the chapter indicated above). Let us write the binary representations for the numbers of stones in each heap one under another, in such a way that the units digits are in the same column, the tens digits are also in the same column, et cetera. Then we calculate the parity of the number of l's in each column (E denotes "even" and 0 denotes "odd"). For example, suppose there are three heaps, with 101, 60, and 47 stones. Then we write:0 0 1 0101 60 47 0 1 000 00 1 10 10 0 110 0 11 0 1E 00EWe claim that a position is winning if and only if the number of l's in any column is always even; that is, all the letters in the bottom row are E's (so that the position shown above is, presumably, a losing one). We will call such positions "even" and any other position uodd".To prove that a position is winning if and only if it is even, we must show that: l. The final position of the game is even. 2. Any move from an even position leads to an odd position. 3. From any odd position you can shift to some even position in one move. Part 1 is easy. The game ends when there are 0 stones in each heap, and 0 is even.To prove part 2 we notice that after each move the number of stones in some heap changes, and, therefore, some digit in its binary representation changes. This means that the number of l's in the corresponding column changes by l. Since no other row can change (stones can be removed from only one heap at a time), the parity at that column changes, too. We now show how to move from an arbitrary odd position to an even one. We must take several stones from one heap so that the parity of the numbers of l's in the columns changes for all the columns with an odd number of l's in them (and only for these columns!). Consider the leftmost column with an odd number of l's in it and choose a heap which has 1 as its digit in this column (why does there exist such a heap?). This heap is the one we will take the stones from. It is easy to understand how many stones must be left in this heap-the binary representation of the number of stones in the heap must change in those digits which correspond to the columns with an odd number of l's in them. We must take away exactly as many stones as will make this happen. Since the leftmost of these digits will change from 1 to 0, the number of stones in the heap will, in fact, decrease.174MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Problem 13. Solve the following games: a) There are eight white pawns in the first row of a chessboard, and eight black pawns in the eighth row. Each of two players, in turn, can move one of his or her pawns towards the other end of the board in a vertical direction for any number of boxes. It is not allowed to jump over a pawn of the opposite color. The player •¥ho cannot make a move loses. b) The same game, except that it is allowed to move the pawns not only forwards but backwards too. For teachers. It makes sense to discuss the game of Nim only with students of sufficiently high mathematical skill. Students should play each other and submit their own conjectures and strategies. It is rather difficult to devise the correct winning strategy; however, exact and timely hints can facilitate this process and allow students to solve as much as they can of the problem independently.CHAPTER 16Inequalities §1. What's greater?This question is, perhaps, one of the most common children ask. Children are very curious, and you can hear from them questions like these: -Who is stronger: my dad or the arm wrestling champion? -Which is higher: our house or the World Trade Center? -Do more people live in Chicago than in Atlanta? In mathematics such "na1ve" questions do not make much sense, but they help students learn how to calculate better and more precisely, and to deal with "really large" numbers. Of course, we are not talking here about using calculators, which are not helpful if we want to be absolutely rigorous in our proofs. For teachers. The technique of calculation and estimation is one of the most valuable aspects of mathematical culture. Students have to master not just "blind usage" of various methods of computation and approximation; they must understand their essence. It is impossible to remember all the technical tricks of mathematics. But it is possible and necessary to teach students how to do things ''with their bare hands" . The skill of fast and precise estimation can be achieved by solving problems with specific numerical data, like those we consider in this section. Here is a typical problem of this series. Problem 1. Which number is greater: 31 11 or 1714 7 Certainly, you can calculate both numbers "manually" -they have no more than 20 digits. However, this way of dealing with the problem is time consuming and will yield nothing in other, more intricate, problems. Let us try another way. 31"< 32"= (25)" = 255< 256 = (24)14= 1514< 1714.This chain of inequalities shows that 31 11 is less than 1714 . The only thing we needed for this solution was to observe that the numbers 31 and 17 are not far away from powers of 2. Problem 2. Which number is greater: a) 2300 or 32007 b) 240 or 328 7 c) 544 or 453 7 Problem 3. Prove that 210° + 3ioo < 4100. Solution. Clearly, 2100 < 3100 . So it is enough to prove that 2. 3100 < 4100 , or, equivalently, ( !} 100 > 2. But even ( ~) 3 = ~ is greater than 2. Problem 4. Which number is greater: 792 or 891 7 175176MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)To further investigate the situation in Problem 3, let us try to find a natural number n such that 4n < 2 100 + 3100 < 4n+1. On our way to the solution we encounter the following problem. Problem 5! Prove that 479 < 2100 + 3100 < 480. Solution. Since 2 · 3 100 > 2 100 + 3 100 , it is enough to show that 480 > 2 · 3 100 ; that. (4'J!' )20 = IS,(256)20 243 > 2.We recall Bernoulli's inequality: (1 + x)n 2:: 1 + nx for x 2:: -1, n 2:: 1 (see Problem 56 or the chapter "Induction"). This fact motivates the question: is it true that~~~ > 1 + fo? The answer is yes (check it yourself!). Hence, (~~~) 20 > (1 + foJ20 2:: 2. Now we prove that 2 100 + 3 100 > 479 . We will show that 3 100 > 4 79 , or 48o ;3100 < 4, by building the following chain of inequalities: 480 3100= (256) 20 243572 > 754 . After this, we can "fix" the exponent 54 a bit and we get another exercise: prove that 573 > 753 .Comment. This class of inequalities, which are the result of combinations of several simple and "rough" inequalities, has a special name in Russian olympiad folklore. They are called "inequalities a la Leningrad". For teachers. Solving these numerical inequalities helps develop computational skills and approximation technique. However, some students, who may be gifted in logical and combinatorial mathematics, may have a sort of "allergy" to computational problems like these.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)I78§2. The main inequality The main (and, in some sense, the only!) inequality in the field of real numbers is the inequality x 2 ~ 0-its truth is beyond doubt. Other, well-known and useful, inequalities follow from this one. The first among them is, certainly, the inequality of means (or the A.M.-G.M. inequality): a+b - 2-~r;vab for any a,b~0-the numbers (a + b) /2 and VaJ, are called the arithmetic mean and geometric mean of the numbers a and b. This is very easy to prove: a+ b _ VaJ, =a+ b- 2,/(J, = ~( 'a- Vb)2 > 0 2 2 2 yv. -which implies not only the truth of the A.M.-G.M. inequality, but also the fact that this inequality turns into an equality if and only if a = b. Problem 17. Prove that 1 + x ~ 2,/X, if x ~ 0. Problem 18. Prove that x + l/x ~ 2, if x > 0. Problem 19. Prove that (x 2 + y 2 )/2 ~ xy for any x and y. Problem 20. Prove that 2(x 2 + y2) ~ (x + y) 2 for any x and y. Problem 21. Prove that l/x + l/y ~ 4/(x + y), if x > 0, y > 0. Solution to Problem 18. (x + l/x) - 2 = (JX-,/l/x) 2 ~ O.Remark. Generally speaking, solutions to any of these problems can be reduced either to an appropriate application of the A.M.-G.M. inequality, or, after some "appropriate" transformations, to an application of the main inequality x 2 ~ 0. More complicated inequalities, however, are usually solved either by multiple application of the A.M.-G.M. inequality, or by combining several different ideas. Here is a typical example: Problem 22. Prove that x 2 + y2 + z 2 ~ xy + yz + zx for any x, y, and z. To prove this fact we will use the result of Problem 19, and write three inequalities:1 2(x2 + y2) ~ xy,1 2(x2 + z2) ~ xz,12(Y2 + z2) ~ yz.Adding them up, we are done. Problem 23. If a, b, c ~ 0, prove that (a+ b)(a + c)(b + c)~Babe.Problem 24. If a, b, c ~ 0, prove that ab+ be+ ca ~ a,/bC + b,;ac + cVaJ,. Problem 25. Prove that x 2 + y 2 + 1 ~ xy + x + y for any x and y. Problem 26. Prove that for any a, b, and c, the inequality a4 + b4 + c4 ~ abc(a + b + c) holds true. Solution. We will use the inequality from Problem 22-twice! a4 + b4 + c4= (a2)2 + (b2)2 + (c2)2 ~ a2b2 + b2c2 + c2a2 + (bc) 2 + (ca) 2 ~ (ab)(bc) +(be)(ca)+ (ca)(ab) =abc(a+b+c). = (ab) 216. INEQUALITIES179The A.M.-G.M. inequality is remarkable in two ways. First, it allows us to estimate the sum of two positive numbers in terms of their product. Second, it can be generalized for more than two numbers. Here, for example, is the A.M.-G.M. inequality for four positive numbers:a+b+c+d 4~4,,----;vabcd for any a,b,c,d ~ O,where, as usual, the left and the right sides of the inequality are called the arithmetic and the geometric mean of the four given numbers respectively. This version of the A.M.-G.M. inequality can be proved as follows:a+b+c+d 4c+d) =z1 (a+b - 2 -+2-0i1('Ci)~2vao+vca ~Jvaovca=vaoca 'b 'Ci 4'/,d-we just apply the A.M.-G.M. inequality for two numbers twice. Problem 27. Prove that x 4 + y 4 + 8 ~ 8xy for any x and y. Problem 28. If a, b, c, and d are positive numbers, prove thatIt is, however, not so easy to prove the A.M.-G.M. inequality for three positive numbers: a+b+c a,,- foranya,b,c~O. __ 3__ ~ vabc Let us consider four numbers: a, b, c, and m =Vabc.Thena+b:c+m ~ Vabcm= Vm3 ·m=m. Hence, (a+b+c)/4 ~ 3m/4, and, therefore, a+b+ c ~ 3m, (a+b+c)/3 ~ Problem 29. If a, b, and c are positive numbers, prove thatVabc.~+~+.".>3. ba -cProblem 30. Prove that if x ~ 0, then 3x3-6x 2 + 4 ~ 0.Solution. We show that 3x3 + 4 ~ 6x2 . Since 3x3 + 4 = 2x3 + x3 + 4, we can apply the A.M.-G.M. inequality and get2x3 + x 3 + 4 ~ 3{/2x 3•x• ·4 = 3 · 2x2= 6x 2 .Miscellaneous (for homework) 31. Prove that if a, b, c > 0, then 1/a + 1/b + 1/c ~ 1/,/0.b + 1/VbC + 1/Fc.32. Prove that if a, b, c > 0, then ab/c + ac/b +be/a~ a+ b + c. 33. Prove that ifa, b, c~O, then ((a+b+c)/3) 2 ~ (ab+bc+ca)/3. 34. Prove that if a, b, c ~ 0, then (ab+ be+ ca) 2 ~ 3abc(a + b + c). 35. The sum of three positive numbers is six. Prove that the sum of their squares is no less than 12. 36. Prove that if x ~ 0, then 2x + 3/8 ~ 4y'X.180MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)37. The sum of two non-negative numbers is 10. What is the maximum and the minimum possible value of the sum of their squares? 38: Prove the A.M.-G.M. inequality for five non-negative numbers; that is, prove that if a, b, c, d, and e ;:o: 0, then a+b+c+d+e ;:o: ~bed a e. 5Hint. First, prove the A.M.-G.M. inequality for eight numbers, and then use our idea from the proof of the A.M.-G.M. inequality for three numbers. §3. Transformations Sometimes a lucky transformation can help solve a problem or prove an inequality immediately. Here is an example: let us return to the inequality x 2 + y 2 + z2 ;:o: xy+yz+zx from Problem 22. It can be proved as follows. We rewrite the difference between parts of the inequalityx 2 + y 2 + z2-xy -yz - zx = ((x -y) 2 + (y- z) 2 + (z - x) 2 )/2 ;:o: 0.Actually, the trick is a refinement of the technique of "completing the square", often used to solve simple quadratic equations. Here is another example. Problem 39. Solve the equation a 2 + b2 + c2 + d 2 - ab - be - cd - d + 2/5 = 0. You might say "This is not an inequality!" Certainly, but, first, we can use the same method, and, second, there is an inequality present: a 2 + b2 + c2 + d 2 - ab - be - cd - d + 2/5 =(a - ~)2+~ ( b - ~) 2 + ~ ( c- ~) 2 + ~ ( d - ~) 2'and the solution follows. Indeed, a sum of squares can be equal to zero if and only if all the summands are zero. Therefore, we have our answer: d = 4/5, c = 3d/4 = 3/5, b = 2c/3 = 2/5, a= b/2 = 1/5. Problem 40. a + b = 1. What is the maximum possible value of the product ab? Hint. a(l - a)= 1/4 - (1/2 - a)2. Problem 41. Prove the inequality (a 2/4) +b2 +c2 ;:o: ab-ac+ 2be for all a, b, and c. Problem 42. Suppose k, l, and m a.re natural numbers. Prove that 2k+l+2k+m+2l+m:5 2k+l+m+l + 1.Problem 43. If a + b + c = 0, prove that ab + be+ ca :::; 0. Let us solve Problem 41. We carry everything over to the left side and rewrite: (a 2/4) + b2 + c2 - ab+ ac - 2be = (a/2 - b + c) 2 ;:o: O. This is true because of our "main" inequality.16. INEQUALITIES181Now we discuss another exceilent idea which has proven quite useful in dealing with inequalities which show some symmetry (it is also connected with factorization).Lemma. If a ;::: b and x ;::: y, then ax+ by ;::: ay +bx. Proof. Indeed, ax+ by - ay - bx= (a - b)(x - y) ;::: 0. Comment. If, for instance,f is some increasing function, then (a - b)(f(a) - f(b)) ;::: 0for any numbers a and 1>--this is just a reformulation of the definition of an increasing function. This idea can be applied as follows: Problem 44. Prove that x 6/y 2 + y 6/x 2 ;::: x 4 + y 4 for any x and y. Solution. We denote x 2 by a, and y 2 by b. Then xBy6-y2 + -x2 - x 4 -y4= a3 /b+ b3 /a - a2 - b2= (a - b)a 2 + (b- a)b 2 =(a_ b) (~ _ ~) = (a - b)(a3 b a b a abWe emphasize once again that the numbers a - b and a3 sign.Problem 45. If x, y > 0, prove that JX2TY + Problem 46. If a, b, c ;::: 0, prove that--b3 );:::O.b3 have the sameJYI/x ;: : ,fii + JY.2(a3 + b3 + c3) ;::: a2b + ab2 + a2c + ac2 + b2c + bc2 . Solution. Carry all the terms over to one side and split them into quadruples: [a3 + b3 - a2b - ab2] + [b3 + c3 - b2c - bc2 ] + [a3 + c3 - a2c - ac2]. Inside each quadruple the expression can be factored in the following way: a3 + b3 - a2b - ab2 =(a - b)(a2 - b2) ;::: O. This completes the proof. Problem 47! If al :2(v'n+1-1).Problem 55. If n is a natural number, prove that 1 1 1 1 22+32+42+ ... +li2 (n + l)n- 1 . Problem 58. If n is a natural number greater than or equal to 4, prove that n! 2'. 2n. Problem 59. If n is a natural number, prove that 2n 2'. 2n. Problem 60. Find all natural numbers n such that 2n 2'. n 3 . Let us discuss the solution to Problem 57. We set an = nn, bn = (n + l)n- 1 . The statement is true for the values n = 1 and n = 2, so the base is proved. To prove the inductive step, it suffices to show that ak kk ak-1 = (k - l)k-Ibk(k + l)k-I2'. bk-I = ~or, equivalently, k 2k- 2 2'. (k 2 - l)k- 1; that is, (k 2)k-I 2'. (k 2 - l)k- 1.184MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)Miscellaneous 61. Prove that for any natural number n the inequality 3" > n · 2" holds true. 62. Which of the two numbers 2 3 22" (ten 2's) or 33 · (nine 3's) is greater? What if there were eight 3's? 63. The product of the positive numbers a 1 , a2, ... , an is equal to 1. Prove thatRemark. This problem has other, non-inductive solutions. 64. Prove Bernoulli's inequality (1 + x)" ~ 1 + nx, if x ~ -1 and n ~ 1. 65. The sum of the positive numbers x 1 , x2, ••. , Xn equals 1/2. Prove that 1 - X1 1 + X1•1 - X2 1 + X2•••.•1 - Xn 1 1 + Xn ~ J.For teachers. In this section two themes are merging, each of which is important enough to take up several sessions. These are "Induction" and "Inequalities". It must be noticed that in the theme "Induction" the subtopic "Applying Induction to Inequalities" is usually understood by students more easily than other (more abstract) ones. We must not make excessive use of this fact, although we can create many problems similar to Problems 52-61. It is important to give your students more questions which require non-standard thinking. Also, you should allow them to seek non-inductive solutions to the problems. §5. Inequalities for everyone The problems in this section are listed without any sorting by method of solution. However, we have tried to arrange them in progressive order of difficulty. 66. A string was stretched along the equator without gaps. Then it was lengthened by 1 centimeter (0.4 inches) and was stretched again along the equator by pulling it off the ground in one place. Is it possible for a man to go through the gap created? 67. Imagine that the Earth is made of dough which is then rolled into a thin "sausage" so that it will reach the Sun. What is the thickness of that "sausage"? Try to deviate from the right answer by no more than 1000%. 68. Is it possible to pack the entire population of the Earth and everything that was created by humankind inside a cube with an edge 2 miles long? 69. Imagine that you are standing on the western bank of the Hudson River. Is it possible, using only things at hand and common sense, to make a good estimate of the length of the radio antenna on one of the buildings of the World Trade Center on the other side of the river? 70. Prove that 100! < 50 100. 71. If n is a natural number, prove that ,/n + 1 - ,/n - 1 > 1/ ,;n. 72. If 1 > x > y > 0, prove that (x -y)/(1 - xy) < 1.16. INEQUALITIES18573. If a, b, c, d ~ 0 and c + d ~ a, c + d :5 b, prove that ad + be :5 ab. 74. Does there exist a set of numbers whose sum is 1, and the sum of whose squares is less than 0.01? 75. Suppose a, b, c > 0 and abc = 1. It is known that a+ b + c > l/a + l/b + l/c. Prove that exactly one of the numbers a, b, and c is greater than 1. 76. The numbers x, y belong to the segment [0,1}. Prove that_x_+_Y_ 3 is prime, then pis odd. Hence both p-1 and p+ 1 are even, and one of them is a multiple of 4. It follows that p 2 - 1 is divisible by 8. Also, since p - 1, p, p + 1 are three consecutive integers, one of them is divisible by 3. It is not the prime p, so it must be either p - 1 or p + 1. Thus p 2 - 1 is divisible by 3 and 8, and therefore by 24. b) We have p 2 - q2 = (p- q)(p + q). Proceeding as before, we find that both factors are even. To show that one of them is a multiple of 4, we assume the opposite; that is, that these numbers give remainders 2 when divided by 4. Then their sum must be divisible by 4. On the other hand, their sum is 2p, which is not a multiple of 4, since p is odd. F\irthermore, the numbers p and q must have either equal or different remainders modulo 3. In the former case, their difference is divisible by 3; in the latter case, their sum is. This proves that (p - q)(p + q) is divisible by 3 and by 8.222MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)22. If neither x nor y is divisible by 3, then x 2 and y 2 have remainders 1 when divided by 3. Therefore, their sum has remainder 2, which is impossible for a perfect square. 23. Hint: check that both a and b are divisible by 3 and by 7. 24. Hint: check that numbers x 3 and x have equal remainders when divided by 6. 25. If d is odd, then one of the numbers p and q is even, which is impossible. If d is not divisible by 3, then one of the numbers p, q, and r is divisible by 3, which again gives us a contradiction. 26. Hint: find all possible remainders given by perfect squares when divided by 8. 27. Possible remainders of perfect squares when divided by 9 are: 0, 1, 4, and 7. Check that if the sum of some triple of them is divisible by 9, then some pair of them are equal. 30. Using the method of Problem 28, we find that the answer is 7. 31. The answer is 1. 32. The answer is 6. 34. The answer is 3. Hint: the units digit of 'l7' has a cycle of length 4. We must determine when in this cycle 77 occurs; that is, we need the remainder when 77 is divided by 4. 35. Hint: One of these numbers is always divisible by 3. a) p = 3; b) p = 3. 36. The answer is p = 3. The method of the previous solution works here as well. 37. Hint: prove that p = 3 using the same trick as before. 38. Hint: analyze the remainders when divided by 3. 39, 40. Hint: check that the remainder of the square of an odd number when divided by 4 is always 1, and the remainder of the square of an even number is always O. The answer to both questions is no. 41. The answer is p = 5. Analyze the remainders upon division by 5. 42. The remainder of this number when divided by 9 is 7, and this cannot be true of a perfect cube. 43. Hint: find all possible remainders of the number a 3 + b3 + 4 when divided by 9. 44. Hint: find all possible remainders of the number 6n3 + 3 when divided by 7. 45. If neither of the numbers x or y is divisible by 3, then z 2 gives a remainder of 2 when divided by 3, which is impossible. Now notice that the square of an odd number always has remainder 1 when divided by 8; the square of an even number not divisible by 4 always has remainder 4; and the square of a multiple of 4 always has remainder 0. Using this, we can show that either both numbers x and y are even, or one of them is divisible by 4. 46. Hints: a) 4 + 7a = 4(a + 1) + 3a; b) a+ b = (2 +a) - (35 - b) + 33. 4 7. The answer is O. First, calculate the last digit of 02 +1 2 + 22 + ... + 92 • Second, notice that this last digit is always the same for every set of ten consecutive natural numbers. 48. Prove that any two numbers out of the given seven~ay, x and y-have the same remainder when divided by 5. To accomplish this, consider two 6-tuples, theANSWERS, HINTS, SOLUTIONS223first containing all the numbers but x, the second containing all the numbers but y. 49. Denoting the first one of these numbers by a, we geta+ (a+ 2) + (a+4) + ... + (a+2(n- l)) = na+2(l +2 +3+ ... + (n-1)) =na+n(n-l) =n(a+n-l). 50. Note that this number increased by 1 is divisible by 2, 3, 4, 5, and 6. The answer therefore is one less than the LCM of these numbers, or 59. 51. If n is composite and greater than 4, then (n - l)! is divisible by n. Indeed, n =kl where k and l are less than n. If k # l, then the product (n - l)! contains both these numbers as factors and our point is proved. If k = l, that is, n = k2 , where k > 2, then the product (n-1)! contains factors k and 2k, and we are done. 55. Using Euclid's algorithm we get gcd(30n + 2, l2n + 1) = gcd(l2n + 1, 6n) = gcd(6n, 1) = 1. 56-57. Use Euclid's algorithm. The answers are: 220 - 1 and 111 ... 11 (twenty l's) respectively.4. THE PIGEON HOLE PRJNCIPLE 3. The pigeon holes are the remainders when divided by 11. The pigeons are the numbers. (See also the solution to Problem 21.) If two numbers have the same remainder when divided by 11, their difference must be divisible by 11. 4. The pigeon holes here are the numbers of hairs on a person's head (from 1 to 1,000,000). The pigeons are the citizens of Leningrad. 6. Let us sort the football players by team as they come off their airplanes. There will be lOM + 1 players to sort. The General Pigeon Hole Principle assures us that there will be one team which has 11 players, and this team is complete. 8. There are five possible numbers of acquaintances for any person: 0, 1, 2, 3, or 4. So it would seem that each could have a different number of friends. However, if any person has four acquaintances, then no person may have zero acquaintances. Hence two people must have the same number of acquaintances. 9. If there are k teams, then the number of games played by each team varies from 0 to k - 1. However, if any team has played k - 1 games, then it has played every other team, and no team has played 0 games. Hence we are fitting k teams into k - 1 pigeon holes, which are either the numbers from 0 through k - 2 or the numbers 1 through k - 1. lOa. The answer is 32. Indeed, suppose that 33 or more squares are colored green. Then, after we have divided the board into sixteen 2 x 2 squares, the Pigeon Hole Principle guarantees that at least one of these squares contains 3 or more small green squares. These 3 green squares form the "forbidden" tromino in some position, and we have a contradiction. On the other hand, we can color all the black fields (of the usual coloring) green, and this is an example of 32 green squares with the property needed. !Ob. The answer is 32 (again!). Indeed, if 31 or fewer squares are colored green, then one of those sixteen 2 x 2 squares contains 1 or 0 green squares. Then the other 3 or 4 squares are not colored green, and they form the tromino without green224MATHEMATICAL CIRCLES (RUSSIAN EXPER1"NCE)squares in it. This contra.diction (and the same construction as above) completes the proof. 11. At least 1 + 2 + 3 = 6 problems were solved by the students mentioned in the problem statement. Therefore, there are 29 problems left to be solved, and 7 students to account for them. If each student ha.cl solved only 4 problems, then there would have been only 28 problems solved. Therefore, one student must have solved at least 5 problems. 12. Answer: 12 kings. See the hint to Problem lOa. 13. Divide the cobweb into 4 sectors as shown in Figure 135, each of which can hold no more than one spider.FIGURE 135 14. Each of the smaller triangles can cover only one vertex of the larger triangle. 18. Color all the dry land red, and color each point diametrically opposite dry land green. Then there must be a point which is both red and green. Start the tunnel at this point. Do you see why this is, in a way, a Pigeon Hole Principle? 19. There are only 1987 possible remainders when a number is divided by 1987. If we examine, for example, the first 1988 powers of 2, we find that two of them must have the same remainder when divided by 1987. These two powers then differ by a multiple of 1987. 20. When divided by 100, a perfect square can give only 51 remainders, since the numbers x 2 and (100 - x) 2 give the same remainder. Hence of 52 integers, the squares of two must have the same remainder when divided by 100. These two squares differ by a multiple of 100. 22. If3m and 3n (where m > n) are two powers of 3 which give the same remainder when divided by 1000, then 3m - 3n = 3n(3m-n - 1) is divisible by 1000. Now the prime factors of 1000 are 2 and 5, and neither divides 3n. It follows that 1000 must divide 3m-n - 1, which means that 3m-n is a power of 3 ending in the digits 001. 23. This sum can take on only seven values: the numbers from -3 through 3. 24. Divide all the people into 50 pairs who are sitting diametrically opposite each other. Consider these pairs as the pigeon holes. Since there are more than 50 men, one pair must include more than one man.ANSWERS, HINTS, SOLUTIONS22525. If the conclusion is false, then it is clear that the boys will have gathered at least 0 + 1 + 2 + ... + 14 = 105 nuts, which is a contradiction. 26. The product of the numbers in all the groups is 9!= 362880. If the product of each group were no greater than 71, the product of all the numbers could only be 71 3 = 357911. It should be noted here that this method of proof is, in a way, more general than the simple Pigeon Hole Principle. 27. One can move from any square to any other by passing through neighboring cells, and we can always choose a path such that the number of squares visited is less than 19. This means that if a is the smallest number on the board, all the numbers are included between a and a + 95. Therefore there can be no more than 96 different numbers among the 100 on the board, and two must be equal.28. We choose any one person in the group. Let us call him Bob. We sort the others into two pigeon holes: those who know Bob and those who do not. There are at least three of the remaining five people in one of these categories. Suppose Bob has three acquaintances. If two of these know each other, then they, together with Bob, form the required triple. If none of them knows each other, then they themselves form the required triple. A similar argument holds if there are three people whom Bob does not know. 29. Consider the parity (remainder upon division by 2) of the coordinates of the points. There are four possibilities: (odd, odd); (odd, even); (even, odd); (even, even). Since there are five points, we can choose two of them whose coordinates both match in parity. It is not hard to see that the midpoint of the line segment they determine has integer coordinates. 30. There are two categories into which we can fit the three sizes: those sizes for which there are more right boots than left boots, and those sizes for which there are more left boots than right boots (if there happens to be an equal number of right and left boots in one size, we put that size in the second category). It follows that two sizes lie in the same category. Let us say that sizes 41 and 42 have more right boots than left boots (an analogous argument will hold if two sizes have more left boots than right boots). Now there are 300 left boots in all, and at most 200 left boots in any one size. Therefore, the sum of the left boots in any two sizes is at least 100. We have shown that there are at least 100 left boots in sizes 41 and 42 (taken together), and that each of these sizes contains more right boots than left boots. Hence each left boot has a match, and there are at least 100 good pairs in the warehouse. 31. There are 11 more consonants than vowels in the aiphabet. Therefore, if we add the differences between the number of consonants and the number of vowels in each of the six subsets, these differences must sum to 11. It follows that there must be at least one subset in which this difference is less than 2, and the letters of this subset must form a word.32. Consider the ten sums: x 1 , x 1 + x2, x 1 + x 2 + X3, ... , x, + x2 + ... + X10Two of these must have the same remainder when divided by 10. The difference between these two sums gives a set whose sum is divisible by 10. 33. We can divide the numbers from 1 through 20 into ten disjoint sets, such that if a pair of numbers is selected from the same set, one of the pair divides the other: {11}, {13}, {15}, {17}, {19}, {l, 2, 4, 8, 16}, {3, 6, 12}, {5, 10, 20}, {7, 14},MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)226{9, 18}. Then, of any eleven numbers not greater than 20, two of them must fit in one of these pigeon holes, and one of these two divides the other. 34. We can number the study groups with the numbers 1 through 5. Then, instead of considering each student him or herself, we can consider the set of numbers belonging to the study groups he or she is part of. Each of these is a subset of the set {l, 2, 3, 4, 5}. We solve the problem by dividing the 32 subsets of this set into 10 collections such that if two subsets are chosen from the same collection, one of them contains the other (compare this with the solution to Problem 33). The following is such a collection. The subsets in each collection are written as numerals:[0, {l}, {l, 2}, {l, 2, 3}, {l, 2, 3, 4}, {l, 2, 3, 4, 5}], [{2}, {2, 5}, {l, 2, 5}, {l, 2, 3, 5}]. [{3}, {l, 3}, {l, 3, 4}, {l, 3, 4, 5}]. [{4}, {l, 4}, {l, 2, 4}, {l, 2, 4, 5} J, [{5}, {l, 5}, {l, 3, 5}], [{2, 4}, {2, 4, 5}, {2, 3, 4, 5}], [{3, 4}, {3, 4, 5}]. [{3,5},{2,3,5}]. [{4,5},{1,4,5}], [{2, 3}, {2, 3, 4} J. 5. GRAPHS-13. Yes, such a path is possible. See, for example, Figure 136, in which a graph is drawn similar to the one in the solution of Problem 2. An example of a path satisfying the conditions of the problem can then be constructed easily.1234567891011129FIGURE1364. If the number AB is divisible by 3, then so is the number BA. This means that if a traveler can get from city A to city B directly, she can also get from city B directly to city A. This observation allows us to draw a graph of the connections, such as the one in Figure 137. Clearly, a traveler cannot get from any city to another. For example, she cannot get from city 1 to city 9. 7. Draw a graph where the cities are vertices and the roads are edges. We can then count the edges of this graph using the method illustrated in Problem 5. TheANSWERS, HINTS, SOLUTIONS62275FIGURE 137 problem states that the degree of each vertex is 4, so the total number of roads is equal to 100 · 4/2 = 200. 9. Both situations (a) and (b) are impossible. In each case, we can think of a graph similar to that in Problem 5, and count the odd vertices. We find that the number of odd vertices is not even, so the graph cannot be drawn. 10. Answer: No. We can imagine a graph in which the vertices represent vassals, and neighboring vassals are connected by edges. A count of the odd vertices shows that there are not evenly many of them, so the graph cannot be drawn. 11. Answer: No. If the kingdom had k towns, then there would be 3k/2 roads. This number cannot equal 100 if k is an integer. 12. Answer: Yes, it is true. Suppose it were not. Draw the graph in which the vertices represent islands and the edges represent the bridges connecting them. The problem says that each of the seven islands is represented by an odd vertex, so there would be oddly many odd vertices. Since this is impossible, the graph must show at least one edge leading to the shore. Figure 138 shows a graph representing a possible situation such as John described.FIGURE 138228MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)13. Imagine a huge graph in which each person who ever lived on earth is represented by a vertex, and each handshake is represented by an edge connecting the vertices corresponding to the two shakers. Then we are counting the odd vertices of this graph, and our theorem assures us that there must be evenly many of them. 14. Answer: No. The difficulty in this problem is to decide how to draw the graph. Taking the line segments themselves as edges of a graph probably won't work (which may confuse some students at first). Instead, we can consider a graph where the actual line segments are represented by vertices (!), and two vertices are connected by an edge if and only if the corresponding line segments intersect. Then this graph has nine vertices of degree 3, which is impossible.16. We can generalize the solution to Problem 15. Suppose such a graph was not connected. Certainly it could not consist of fewer than two towns (with what could a single town fail to be connected?). Select two towns which, supposedly, cannot be connected by a path. Consider all the towns to which these two are connected. There are at least 2(n - 1)/2 = n - 1 of these. As before, these new towns must all be distinct: if two new towns were the same, the two selected towns would be connected by a path through it. Therefore, the graph would have n - 1 + 2 = n + 1 towns, which is a contradiction. Hence the graph must be connected. Here again, it is clear that students should attempt to construct the graph in question. They will quickly find that it has "too many" edges not to be connected. This intuition can be the springboard for a formal discussion of the result. 18. If road AB is closed, than it is enough to prove that we can still get from A to B. If this were not true, then in the connected component containing A, all the vertices other than A would be even. This situation of having exactly one odd vertex in a connected component contradicts our theorem about the odd vertices of a graph.20. Answer: No, such a stroll is not possible. We represent the islands and the shores by vertices of a graph, and bridges by edges. As Figure 139 shows, the graph has 4 odd vertices, which is too many.FIGURE13921. Answers: (a) six bridges; (b) five bridges; (c) four. Students need simply count the number of bridges used to visit Thrice on each occasion. 22. (a) The required cube is not possible. First note that the wire cannot double back on itself, since the total length of all the edges is 12 x 10 cm= 120 cm (using up the whole length of wire). Let us draw the graph of the cube's edges (FigureANSWERS, HINTS, SOLUTIONS229140). If the wire frame could be formed, then we could follow the wire and traverse the graph without lifting the pencil from the paper. But this graph has eight odd vertices, which is too many to allow this. Therefore, the wire cannot be bent as required.FIGURE140(b) Since the graph of the cube has 8 odd vertices, there must be at least four such pieces.6. THE TRIANGLE INEQUALITY 1. Suppose AB 2:: BC. If A, B, and C form a triangle, then the triangle inequality assures us that AC+ BC > AB, which leads to the desired result. If AB ::; BC, then we can start with AB+ AC > BC (which the triangle inequality also assures us), to get the same result. Equality holds if and only if A, B, and Care collinear, and Bis not between A and C.2. The length of side BC must be less that AC + AB = 4.4. On the other hand, BC must be greater than IAB - BCI (see Problem 1), which is 3.2. The onlyinteger within these bounds is 4. 3. If the sides of the triangle are a, b, and c, then the triangle inequality tells us that b+c >a. Adding a to each side, we find that a+b+c > 2a, which is equivalent to the required result. 4. Answer: 350 kilometers.6. We will show that OB+ OC +OD > OA. Adding the triangle inequalities AC+OC > OAand OB+OD > BD, we find AC+OB+OC+OD > OA+BD (see Figure 141). Since AC = BD, this gives the required result. Note that the same proof holds even if point 0 is outside the plane of square ABCD. 7. Suppose the diagonals of the quadrilateral intersect at 0 (Figure 142). Then AB+BC >AC, BC+CD > BD, CD+AC >AC, and AD+AB > BD. Adding, we find that 2(AB+BC+CD+DA) > 2(AC+BD), which proves the first result. Also, OA+OB >AB, OB+OC> BC, OC+OD >CD, and OD+OA >AD. Adding, we find that 2(0A+OB+OC+OD) = 2(AC+BD) > AB+BC+CD+DA, which proves the second result.230MATHEMATICAL CIRCLES (Rt;SSIAN EXPERIENCE)0 FIGURE141FIGURE1428. We have: AP+PB >AB; BQ+QC >BC; CR+RD >CD; DS+SE >DE; ET+TA >EA (see Figure 143). Adding these inequalities gives AP+PB+BQ+ QC+CR+RD+SE+ET+TA > AB+BC+CD+DE+EA. The right side of this inequality is the perimeter of the pentagon, while the left side is less than the sum of the diagonals (it will equal this sum if we add the perimeter of the inner rectangle PQRST). This proves the first result.BAD E FIGURE143Tu get the second result, add the inequalities AC< AB+BC, BD < BC+CD, CE< CD+DE, DA< DE+EA, EB< EA+AB.9. If the internal points are X and Y, we extend the segment connecting them in both directions, until it intersects the sides of the triangle (see Figure 144). ThenANSWERS, HINTS, SOLcTIONS23!EF < EA+AF, and EF < EB+BC+CF. Adding, we find that EF is less than half the perimeter of the triangle. Since XY < EF, XY is also less than half the perimeter.AE8-----~ FIGUREc14411. The solution is the path ADEA as shown in Figure 145. Indeed, any other path will correspond to a path between points B and C (in that diagram) which is not a straight line.BFIGURE14512. If we draw AD and AE (Figure 145), then BC = BD +DE+ EC = AD+ DE+ EA. Then DE is less than half of the perimeter of triangle ADE (Problem 3), hence less than half of BC. 14. We can unfold the cube to form a diagram such as in Figure 146. Then if the fly is at A, the shortest distance to the opposite vertex B is a straight line. Fblding the cube back up gives the answer. Students can make a paper model of this problem. Assigning a numerical value to an edge, they can be asked to find the length of the shortest path. 15. We can "unroll" the surface of the glass to get a rectangle, then "unfold" the front and back of the rectangle to get Figure 147. The shortest path is again given by a straight line. 16. Extend segment AO to intersect with side BC at point D. Then add the triangle inequalities AB+BD >AD and OD+DC > OC, and subtract OD from each side of the inequality.MATHEMATICAL CIRCLES (RGSSIAN EXPERIENCE)232A--BFIGUREA146c==J[. . . . . . . . . . . . . . . . . . !8~~Cs.J FIGURE14718. The woodsman must walk to the vertex of the angle, then back home. If the given point is A, and the given obtuse angle is BOC (see Figure 148), we choose that one of the angles LAOC or LAO B which is acute (maybe they both are )--5ay, LAOC. Then we drop a perpendicular BD from point Bon line OC. By the result of Problem 13 we have AB+BC+CA > 2AD, and, obviously, we have AD> AO, which completes the proof.Ac FIGURE148ANSWERS, HINTS, SOLt;TJONS23319. Construct parallelogram A.i3DC from triangle ABC (Figure 149). Then, from the triangle inequality, AB+ BD > AD = 2AM. Since BD = AC, this gives the first result. The second follows from writing the corresponding inequalities for each median and adding.)2i:.( AC FIGURE 14920. In finding the perimeter of the folded polygon, we lose the portion of the original perimeter represented by broken line AXY B (in the example of Figure 150), but we add the length of segment AB. Since the sum of all but one side of a polygon is greater than the remaining side, the perimeter must have decreased.FIGURE 150 21. Consider two of the sides which do not have a common endpoint-say sidesAB and CD. Then, on the one hand, AC+ BC< AB+ CD (since AC and BD are diagonals). On the other hand, if AC and BD intersect at point 0, then OA +OB > AB and OC +OD > CD. Adding these inequalities we find that AB+ CD< AC+ BD, which is a contradiction. 22. Suppose the medians intersect at point M. Then, adding the inequalities AM+ BM > AB, BM+ CM > BC, and CD+ AM < AC, and noticing that the lengths of AM, BM, and CM are each 2/3 of a median, we reach the required inequality. 23. If the width of the river is h, and the towns are situated at points A and B, then the ends of the bridge must be placed at the points of intersection of lines A' B and AB' with the banks, where A' and B' are obtained from A and B by a translation of distance h towards the river (Figure 151). 24. Consider the longest diagonal XY of the pentagon. One pair of vertices of the pentagon lies on the same side of this diagonal, so there exist two intersecting diagonals, each of which has X and Y as one endpoint respectively. It is not hard to see that these three diagonals will form a triangle.234MATHEMATICAL CIRCLES (Rt;SSIAN EXPERIENCE)AB FIGURE 151 7. GAMES 2. After each move, the number of piles increases by l. At first there are three piles, and at the end of the game there are 45. Therefore, 42 moves are made altogether. The last and winning move is always made by the second player. 3. The parity of the result does not depend on the position of the pluses and minuses, but only on the number of odd integers in the original set of numbers. Since there are 10 odd integers to begin with, and 10 is an even number, the first player will win. 4. After each move, the number of rows in which it is possible to place a rook decreases by 1, as does the number of columns. Therefore, there can only be 8 moves al together, and the second player will make the last (winning) move. 5. The parity of the number of l's on the blackboard remains unchanged after each move. Since there are evenly many l's to begin with, there cannot be a single 1 left at the conclusion of the play (since 1 is an odd number!). The second player will therefore win. 6. In playing the game, the greatest common divisor of the two initial numbers must eventually be written down (compare this game with Euclid's algorithm). Therefore, every multiple of the greatest common divisor, not greater than the original numbers, will also appear. In this case, the greatest common divisor of the original numbers is 1, so that every number from 1 to 36 must appear. Therefore there will be 34 turns, and the second player will win. 7. This game is not entirely a joke, since the player who should win can in fact make a mistake and lose his or her advantage. This mistake consists in moving so that the remaining blank squares are all in one column or all in one row, allowing the opponent to win in the next move. The loser in this game, it turns out, is the player who makes j;,st this fateful move. Notice that after crossing out a row of an m x n board, we can consider the remaining squares to be an (m - 1) x n board. Analogously, in crossing out a column of an m x n board, we form an m x (n-1) board. The unique situation in which each move is "fateful" is the case of a 2 x 2 board. Therefore, the player who leaves this position for his opponent will win. However, as we have seen, after each turn the sum of the rows and columns decreases by l. Therefore, the parity of this sum at the beginning of play willANSWERS, HINTS, SOLt;TIONS235determine the winner. In case (a) this is the first player, while in the remaining cases it is the second. Note that in case (b) the second player can follow a strategy of symmetry (see §2). 11. Since a knight always moves from a black to a white square, or vice-versa, the second player can win, using either point or line symmetry. 12. The first player will win, if he moves first to the center of the board, then adopts a symmetric strategy.13. The second player wins in both cases, using (a) line symmetry; (b) point symmetry. In the former case, the proof is quite simple: the second' player just maintains the symmetry by always moving to the square symmetric to the previous move of the first player with respect to the line between the fourth and the fifth rows of the board. Since two squares symmetric in this line always have different colors we cannot encounter the situation when the current move of the first player prohibits the symmetric move of the second player. The solution for the latter case is more tricky though the idea is similar: the second player uses symmetry with respect to the center of the board. The details are left to the reader. 14. The second player wins, using a point symmetric strategy. 15. The first player wins, if he removes the center checker first, then follows a point symmetric strategy. 16. The first player wins, if he first makes the two piles equal, then adopts the second player's strategy from Problem 10. 17. The first player wins. He must first draw a chord which separates the points into two groups of 9. He then replies symmetrically to each move of his opponent. Note that this strategy does not depend on how the points are arranged on the circle.18. The second player wins in both cases. No matter how the first player begins, the second player can reply so as to leave two identical rows of petals on the flower. He can then follow a symmetric strategy. 19. In cases (a) and (b), the second player wins, following a strategy of point symmetry. In case (c), the first player will win. In his first move, he skewers the row consisting of the center cubes of the four 3 x 3 layers. After this, he plays symmetrically with respect to the center point of the figure. 20. The loser is the player who breaks off a rectangle of width 1. The first player will win, by first breaking the chocolate bar into two 5 x 5 pieces. After that, heplays symmetrically. 21. The first player will win, if he places his first x in the center square, thenreplies to each of the second player's moves with an x placed symmetrically with respect to the center square. 23. The first player wins. We number the rows and columns of the chessboard in the usual order, so that the coordinates of square al are (1, 1), and those of square h8 are (8, 8). The winning positions are those in which the king occupies a square, both of whose coordinates are even. The first move is to square b2.236MATHEMATICAL CIRCLES (RGSSIAN EXPERIENCE)24. The first player wins. The winning positions are those in which both piles have oddly many pieces of candy. The first move is to eat the pile of 21 candies and divide the pile of 20 candies into any two piles of oddly many candies. 25. The second player wins. The winning positions are those in which the number of unoccupied squares between the checkers is divisible by 3. 26. The first player wins. The winning positions are those in which the box contains 2n - 1 matches. The first move is to leave 255 matches in the box.27. The first player wins. The winning positions are those in which the largest pile of stones contains 2n - 1 stones. The first move consists in dividing the first two piles in any way at all, and dividing the third pile into two piles of 63 and 7 stones, respectively. 28. In this game, the player who obtains a 1 will win. This is the first player, if he recognizes that writing an odd number is a winning position. 29. In case (a) the second player wins, and in case (b) the first. The winning positions are those in which each pile contains oddly many matches. For Problems 32-38, we give answers provided by analysis from the endgame. The reader can supply details. 32. The second player will win. Figure 152 shows the arrangement of pluses and minuses.-+ + - - + + -+ + - - + + - - - - ----- -- - - - - - - - ++ ++ -- -- ++ ++ - - - - -- -+ ----- --FIGURE15233. We can reformulate both cases (a) and (b) in terms of a chessboard. Game (a) turns out to be equivalent to the game of Problem 23. The arrangements of pluses and minuses in both cases are identical, and are shown in the figure to Problem 23 (Figure 55). 34. The first player wins. The arrangement of pluses and minuses, after a chessboard reformulation, is given in Figure 153. 35. This problem gives an example in which a geometric interpretation is not essential to an analysis of a game from the endgame. Here, it is convenient to markANSWERS, HINTS, SOLUTIONS237+ + + + + + + + + + + +- - + + + + + + + + + + + ++ + + + + + + + + + + + + + + + + + + + + + + + +---------------- --- -- - --- --FIGURE--+ + + + + + + + + +153each number with a plus or a minus. The plus signs belong to those numbers which are multiples of 10. Therefore, the second player will win. 36. The winning positions are the numbers from 56 to 111, or from 4 to 6. Thus the first player wins, by moving to any of the numbers 4, 5, or 6. 37. The winning positions are 500, 250, 125, 62, 31, 15, 7, and 3. The first player wins.38. The winning positions are the multiples of 3. The first player wins, for instance, by subtracting 1, 4, or 16 on the first move.9. INDUCTION 8. The base can be either n = 1 or n = 2. To prove the inductive step let us take k+l points on a circle. The segments connecting all of these points but the (k+l)st divide the interior of the circle into k(k-l)(k-2)(k-3)/24+k(k- l)/2+ 1 parts by the inductive assumption. The segment connecting the (k + l)st point with the ith one (where i is a positive integer not greater thank) intersects (i - l)(k - i) other segments. Thus, adding this segment would increase the number of the parts by (i - l)(k - i) + 1. Adding all the segments connecting the (k + l)st point with the other k points would increase the number of parts by1 + (1. (k - 2) + 1) + ... + ((i - l)(k - i) + 1) + ... + ((k - 2). 1+1) + 1. This last expression can be rewritten as(k + 1)(1+2 + ... + k) - (1 2 + 22 + ... + k2) - k2 + k. Using the identities 1 + 2 + ... + k = k(k: 1) (see the solution to Problem 6) and 21+ 22 + ... + k2--k(k+l)(2k+l) 6MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)238(see Problem 10), we obtain1 + (1. (k - 2) + 1) + ... + ((i - l)(k - i) + 1) + ... + ((k - 2). 1+1) + 1 k(k + 1) 2 k(k + 1)(2k + 1) =--2-6k2 + k.It remains to verify that (k(k - l)(k - 2)(k - 3) + k(k - 1) + 1) + (k(k + 1) 2 - k(k + 1)(2k + 1) - k2 + k)242- (k + l)k(k - l)(k - 2)-242 6 k(k + 1) 1 +--2-+'which is just an algebraic calculation. 9. The base is n = 1. Let us prove the inductive step. By the inductive assumption we have1+3 + ... + (2k -1) = k2. Thus,1+ 3+ ... + (2k -1) + (2(k + 1) - 1) = k2 +(2(k+1) -1) = (k + 1) 2. 10. The base is n = 1. Let us show the inductive step. By the assumption12 + 22 + ... + k2 = k(k + 1~(2k + 1). Thus,12 + 22 + ... + k2 + (k + 1)2 = k(k + 1)(2k + 1) + (k + 1)2 6 (k + l)(k + 2)(2(k + 1) + 1) 6 11. The base is n = 2 and is clear. By the assumption we have 1. 2 + 2. 3 + ... + (k - 1). k = (k - l)k(k + l). 3Thus,1·2+2·3+ .. .+(k-l)·k+k·(k+l) = (k - l)k(k + 1) +k·(k+l)3= k(k + l)(k + 2). 312. The base is n = 2. By the assumption1 1 1 1-2+2-3+ ... + (k-l)kk-1= -k-.Thus,13. Use induction on n. The base is n = 1. Let us prove the inductive step. 2l+x + ... +xk:z;k+I-1=~·239ANSWERS, HINTS, SOLUTIONSThus, 1+x•+ ... +xk+xk+I = xk+I -x-11+xk+Ixk+• -1=-;=-I·14. We use induction on n. The base (n = 1) is quite clear. To prove the inductive step, we start as follows: 1 1 1 k a(a+b) + (a+b)(a+2b) + ... + (a+(k-l)b)(a+kb) = a(a+kb)'Thus, 1111a(a+b) + (a+b)(a+2b) + ... + (a+(k-l)b)(a+kb) + (a+kb)(a+(k+l)b) k 1 k+ 1 = a(a+kb) + (a+kb)(a+(k+l)b) = a(a+(k+l)b)' 15. Let us use induction on n. The base is n = 0:m! Of(m+l)! O!(m+l)'For the inductive step we have (by assumption): m! (m+l)! (m+k)! (m+k+l)! Of+ --1!- + .. · + --k!- = k!(m + 1) · Thus, ml (m+ 1)! (m+k)! (m+k+ 1)! (m+k+ 1)! (m+k+ 1)! Of+ --1!- +···+--kl-+ (k + 1)! = k!(m + 1)! + (k + 1)! =(m+k+l)! ( 1 1 ) (m+k+2)! k! m+l + k+l = (k+l)!(m+l)'16. The base is n = 2. By the assumptionThus,17. The base is n = l: 13 + 23 + 33 = 36, and 36 is divisible by 9. Now the inductive step. By the inductive assumption k3 + (k + 1) 3 + (k + 2)3 is divisible by 9. Thus, (k + 1)3 + (k + 2) 3 + (k + 3) 3 = (k3 + (k + 1)3 + (k + 2)3) + (k + 3) 3 - k3 = (k 3 + (k + 1)3 + (k + 2) 3) + 9(k 2 + 3k + 3)is also divisible by 9.240MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)18. The base is n = 1: 34 + 8 - 9 = 80, and 80 is divisible by 16. Let us prove the inductive step. By the assumption 32k+2 + 8k - 9 is divisible by 16. We have(32k+4 + 8(k + 1) -9) -(3 2k+ 2 + 8k -9)= 32k+2. 8 +8=8(3 2k+ 2 + 1).The number 32k+2 + 1 is even, so 8(3 2k+2 + 1) (and, hence, 3 2k+4 + B(k + 1) - 9) is divisible by 16.19. The base is n = 1: 4 1 + 15 - 1 = 18. Let us prove the inductive step. We know that 4k + 15k - 1 is divisible by 9. Thus, we haveThe number 4k has remainder 1 when divided by 3. So 4k + 5 is divisible by 3, and, thus, 3(4k + 5) is divisible by 9. 20. The base is n = 1: 11 3 +12 3 = 23 · 133. Further, by the inductive assumption we know that 11k+2+122k+l is divisible by 133. Therefore, 11 k+3 + 122k+3 = 11(11 k+2 + 122k+l) + 133. 122k+1. Thus, 11 k+ 3 + 122k+3 is divisible by 133. 21. The base is n = 1: the number 23 + 1 is divisible by 32. Then we know that 23• + 1 is divisible by 3k+ 1. FurtherThus, it remains to prove that ( 23•) 2 - 23• + 1 is divisible by 3. The number 23• has remainder 2 when divided by 3. Hence, the remainder of ( 23•) 2 - 23• + 1 when divided by 3 is zero.22. The base is n = 0, n = 1, and is obvious. To prove the inductive step from n ton+ 1, we must show that abn+I + c(n + 1) +dis divisible by m. Let us use the fact that the previous member of the sequence abn + en + d is divisible by m, and multiply it by b. We have that abn+I + cbn + bd is divisible by m, so it is left to prove that c(n + 1 - bn) + d(l - b) is divisible by mas well. Adding (b - l)cn, which is divisible by m, we get c + d(l - b), which is also divisible by m, since it can be represented as (ab- a+ c) - (a+ d)(b-1). This completes the proof of the inductive step. 23. The base is n = 1: the inequality 21 > 1 is certainly true. Now the inductive step. By the assumption 2k > k. Thus, 2k+ 1 = 2. 2k> 2k 2: k +1.24. a) Answer: n 2: 3. For n = 1, 2, we have 2n < 2n + 1. Let us prove by induction on n that for n 2: 3 the inequality 2n > 2n + 1 holds true.ANSWERS, HINTS, SOLUTIONS241The base is n = 3: 23 > 2 · 3 + 1. Tu prove the inductive step we start with2k> 2k + 1. Then,2k+ 1 = 2. 2k > 4k + 2 > 2(k + 1) + 1. b) Answer: n = 1, n ~ 5. For n = 1 we have 21 > 12, and for n = 2, 3, and 4 we have 2n :5 n 2. Let us prove by induction on n that for n ~ 5 the inequality 2n > n 2 holds true. The base is n = 5: 25 > 52 • Now the inductive step: we know that 2k > k 2 . Thus, 2k+ 1 = 2k + 2k > k2 + 2k + 1 = (k + 1) 2 (we use the inequality 2k > 2k + 1 proved in Problem 24a). 25. The base is n = 2: + = -b, > ¥,. Further, by the inductive assumption k!1 + k! 2 + .. · + .\; > Thus,th-ksince 2k~1l1 1 1 1 1 + 2 + k + 3 + ... + 2k + 2k + 1 + 2k + 2 1 1 1 ( 1 = k+l+k+2+ ... +2k +2k+1+2k+2-k+l 1 1 1 13 >k+l+ k+2+ ... +2k>24,1+ 2k~2 > 2k~2 =26. The base is n =1)k!1 ·2, and n =3: 4 > 1 + 2V2, 8 > 1 + 3 · 2. Since2k > 1 + kv'2H, we obtain2k+ 1 > 2 + 2kv'2H > 1 + V2. k#.It remains to note that V2 · k > k + 1 for k ~ 3. 27. We need to prove that la1 + a2 + ... + anl :5 la1 I + la2I + ... + lanl for any positive integer n and for any real numbers a 1 , a 2 , •.. , an. We will prove the statement using the induction on n. The base is n = 1, n = 2. For n = 1 the statement is evident. For n = 2 we have la1 + a21 :5 lad+ la2I, which can be proved by a simple case-by-case analysis considering all four possible combinations of signs. Now, using the base, we obtainla1 +a2 + ... +ak +ak+il :5 la1 +a2 + ... +akl + iak+d· Thenla1 + a2 + ... + aki + lak+il :5 lad+ la2I + ... + iakl + iak+1 I. 28. Use induction on n. The base is n = 2: (1 + x) 2 = 1 + 2x + x 2 > 1 + 2x for x ;" o. Now, by the inductive assumption (1 + x)k > 1 + kx, and we have(1 + x)k+i > (1 + x)(l + kx) (remember that 1 + x > 0). 29. The base is n = 1:=1 + (k + l)x + kx 2 > 1 + (k + l)xMATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)242Now, by the assumption 1·3·5· ... ·(2k-1) < - 1 - . 2·4·6· ... ·2k - v'2k+l Therefore, 1. 3. 5 ..... (2k -1) . 2k + 1 < __ 1_. 2k + 1 = v'2k + 1 _ 2·4·6· ... ·2k 2k+2 - v'2k+l 2k+2 2k+2 It remains to prove the inequality v'2k+l 2k+2~1 v'2k+3·This inequality is equivalent to the following: (2k + 1)(2k + 3) ~ (2k + 2) 2 , which is evident aiter expanding both sides. 33. The base is n = 1 and n = 2. The proof of the inductive step is also quite simple: ak+l = 3ak - 2ak-1=3(2k+1) - 2(2k-l + 1) = 2k+l + 1. 34. Indeed, a3 = 1, a.= -1, as= -2, a6 = -1, a1 = 1, and as= 2. Hence for n = 1 and n = 2 we have n+6 = an· Then a "strong induction" (see §4 of the chapter "Induction") gives us the proof. 35. We can check directly for the natural numbers 1 through 5. If x is the given natural number, then let Fn be the maximum Fibonacci number not greater than x. Then we have 0 ~ x - Fn < Fn-1 (since x < Fn+I = Fn + Fn-1), and therefore x - Fn can be represented as the sum of several different Fibonacci numbers less than Fn-1· 36. We will prove the following statement by induction on n: The remainders of the Fibonacci numbers Fn and Fn+s when divided by 9 are equal for all natural n. The base is n = 1 and n = 2: F1 = 1, F2 = 1, F 9 = 34, F10 = 55, and we see that the remainders of F 1 and F 9 (and of F 2 and F 10 ) are equal. The proof of the inductive step is very similar to the solution to Problem 34. By the inductive assumption, Fk+B and Fk (Fk+7 and Fk-I• respectively) have the same remainders when divided by 3. Thus, the remainders of Fk+ 9 = Fk+B + Fk+7 and of Fk+l = Fk + Fk-1 are equal. It remains to calculate the remainders of the first eight Fibonacci numbers. They are 1, 1, 2, 0, 2, 2, 1, o. Thus, the nth Fibonacci number is divisible by 3 if and only if n is divisible by 4. 42. We will prove the required statement by induction on n. The base n = 1 is trivial. To prove the inductive step we will first prove that using a calculator we can obtain a natural number less than n. To show this, we choose among n, m, and 0 two numbers of the same parity, and calculate their arithmetic mean x. At least one of the two chosen numbers is different from 0. Replace this non-zero number by x and repeat the operation for the new trio of numbers. We repeat this procedure until one of the positive numbers in the trio becomes less than n (this will eventuallyANSWERS, HINTS, SOLUTIONS243happen, because the sum of two positive numbers in the trio decreases after each operation). Now, let l be the largest natural number less than n which can be obtained using a calculator. Suppose that l # n-1. By the assumption, all natural numbers 1 through l can be obtained using a calculator. If l and n are of the same parity, then we can calculate the arithmetic mean y of l and n, which contradicts the definition of l since l < y < n. If l and n are not of the same parity, we can calculate the arithmetic mean of l - 1 and n and come to the same contradiction. 43. Hint: the inductive step follows from the formula 3(2n + 1) - 2(2n-l + 1) = 2n+J + 1 , which is true for any integer n. 44. Let us use induction on m. The base ism = 1: 2n-l ;::: n. This inequality follows from the result of Problem 23, and can be proved by induction on n. By the inductive assumption 2m+n- 2 ;::: mn, and we have 2Cm+1)+n- 2 ;::: 2mn ;::: (m+l)n. 45. Hint: first, prove that any square can be cut into several parts which can be arranged to form a rectangle with one side of unit length. 46. The inductive step can be proved as follows: split the given 2n+1 numbers into two halves each containing 2n numbers. In each of these halves we can find 2n-l numbers with the sum divisible by 2n- 1 . Then, out of the remaining 2n numbers, we can choose the third set of 2n-l numbers whose sum is divisible by 2n- 1 . Let the sums of the numbers in the three chosen sets be 2n- 1 a, 2n- 1 b, and 2n- 1 c. Among the numbers a, b, and c we can find two numbers of the same parity. The union of corresponding sets is a set of 2n numbers whose sum is divisible by 2n. 47. For n circles the answer is n(n - 1) + 2. To prove it, we can use induction on n. The base (n = 1) is clear. To prove the inductive step we temporarily remove the (k+ l)st circle. By the inductive assumption the number of parts into which k circles dissect the plane is not greater than k(k - 1) + 2. Now we "restore" the removed circle. It intersects each of the k circles at no more than two points, and, thus, the number of parts of the plane increases by at most 2k. The formula k(k - 1) + 2 + 2k = k(k + 1) + 2 completes this part of the proof. An example of the required dissection can also be obtained by induction. We will construct examples (one for each natural number n) with the following properties:a) no three circles meet at the same point; b) the interiors of all n circles have a common point; c) the number of parts into which our n circles dissect the plane equals n(n-1)+2. The base n = 1 is again trivial. To prove the inductive step we start with the configuration (which is assumed to exist) of,,. ~ircles that dissect the plane into k(k-1)+2 parts. Let us add another circle which passes through a point lying inside all k circles of the configuration. We can choose this (k + l)st circle in such a way that it does not pass through the points of intersection of the other circles. This new configuration of k + 1 circles satisfies all the required conditions.244MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)For n triangles the answer is 3n(n - 1) + 2. The proof is similar to the proof above, except that two triangles can intersect in at most 6 points. To construct a configuration of n triangles dissecting the plane into 3n(n- l) + 2 parts, one can draw n congruent equilateral triangles in such a way that they have the same center and no three of them meet in a point. 48. We use induction on the number n of circles. The base is n = 1. To prove the inductive step, temporarily remove the (k + l)st circle and its chord. By the inductive assumption the parts of the plane created by k circles with their chords can be colored using 3 colors (say, red, blue, and green) satisfying the given condition. Now, let us replace the missing circle and its chord. The chord divides the interior of the circle into two parts. Let us change the colors in one part using the scheme: red --> blue, blue --> green, green __, red, and change the colors in the other part using the scheme: red --> green, green --> blue, blue --> red. The colors of the parts lying outside the (k+l)st circle remain unchanged. In the resulting coloring, the colors of adjacent regions of the plane are different. 49. The principle of mathematical induction implies the "well order principle". Indeed, suppose that the principle of mathematical induction holds true and the "well order principle" does not. Take a non-empty set S of natural numbers which does not contain a least element. Let us prove by induction that any natural number n does not belong to S (which contradicts the fact that Sis non-empty). The base is n = 1. If 1 E S, then 1 is the least element of S. Now, the inductive step. By the assumption, the numbers 1, 2, ... , k do not belong to S. Then, if k + 1 ES, we have that k + 1 is the least element of S. Now let us prove that the well order principle implies the principle of mathematical induction. Suppose that the well order principle is true while the principle of mathematical induction is not. Consider a series of propositions such that the first proposition is true and for any natural k the truth of the kth proposition in the series implies the truth of the (k + 1)st proposition. Form the set of natural numbers n such that the nth proposition in the series is not true. Assume this set is non-empty, and let n 0 be its least element. Then the (no - l)th proposition is true, but the noth proposition is not. This contradiction to our assumption completes the proof.10. DIVISIBILITY-2 3. Change the solution to Problem 2 by subtracting the equalities instead of adding them. 5. This is an immediate corollary of Problem 4. 8. 30 99 = (-1) 99 = -1(mod31), 61 100 =(-1) 100 =1(mod31). 9. b )By direct multiplication, we can check thatan+ bn =(a+ b)(an-1 _ an-2b + ... + (-l)n-lbn-1). 10. Consider the summands with the numbers k and n-k; that is, kn and (n-k)n. Since n is odd, the remainders of kn and (-k)n have opposite signs. Thus, the sum of these two summands is divisible by n. Since the sum can be split into (n - 1)/2 such pairs we obtain the required result.ANSWERS, HINTS, SOLUTIONS24511. No number of the form 8k + 7 can be represented as the sum of three squares. Indeed, a perfect square has a remainder of 0, 1, or 4 when divided by 8. It is easy to see that the sum of three such remainders cannot give us the remainder 7. 13. Use the identity x 2 - y2 = (x - y)(x + y). 14. Hint: show that if xis a convenient number, then 1000001-x is also convenient. 15. The answers are a) no; b) no. Indeed, the last digit of a number determines the last digit of its square. After analyzing all possible last digits and their squares, we can see that the last digits of the squares can be equal only to 0, 1, 4, 5, 6, and 9. This can be expressed, of course, using the language of "mod 10". 16. There are several answers. For example, -1 or n - l. The remainder of the given number when divided by n is l. 17. The answer is 5. 18. The answer is 2858.=19. a) Since k is divisible by 3 we conclude that k - 1 2 (mod 3). To complete the proof it suffices to remember that squares cannot have a remainder of 2 when divided by 3. b) Since k is even but not divisible by 4 we have k + 1 3 (mod 4). Now we use the fact that squares can be congruent only to 0 or 1 (mod 4).=20. No, since n 2 +n+1 cannot be divisible by 5. Indeed, there are five congruence 0 (mod 5), then n 2 + n + 1 = 1, if n 1 (mod 5), then classes modulo 5. If n n 2 + n + 1 3, et cetera. So n 2 + n + 1 is never congruent to zero, i.e. n 2 + n + 1 cannot be divisible by 5, and therefore, by 1955. 22. We prove that the sum of the divisors is divisible by 3 and by 8. To prove that this sum is divisible by 3 we split it into pairs of divisors (k,n/k) (notice that k of n/ k since n cannot be a perfect square. Why?) and prove that for each of these pairs the sum of the numbers in it-that is, k + n/k-is divisible by 3. Indeed, k cannot be divisible by 3 (otherwise, n would be divisible by 3, which is obviously impossible). Therefore, either k 1 (mod 3) or k 2 (mod 3). In the former case n/k 2 (mod 3) and in the latter case n/k 1 (mod 3) (we recall that n + 1 is divisible by 3). Thus, in any case, k + n/k is a multiple of 3. The second part of the proof (regarding divisibility by 8) is similar. 23. a) Let us consider the remainders of the members of the sequence when divided by4. We have a 1 =a2 =1 (mod 4). Consequently, a 3 = 1·1+1 = 2, a4 = 2·1+1 = 3, a 5 3 · 2 + 1 3, a6 3 · 3 + 1 = 2, a7 2 · 3 + 1 3, and we have a cycle which does not contain zero remainders. b) Consider the remainders of the members of the sequence when divided by an. Simple calculation shows that an+1 1, an+ 2 1, an+3 2, ... , an+e 22. Thus, an+e-22 is divisible by an, and ifn+6 > 10, then, obviously, an+e >an> 1. 25. Hint: all the powers of ten, starting from 100, are divisible by 4. 26. A number is divisible by 2n (or by 5n) if and only if the number formed by its last n digits is divisible by 2n (or by 5n).======== ========28. The last two digits of the square of n depend only on the last two digits of n itself. Suppose n = ... ab, and we have ab2 = (10a + b) 2 = 100a2 + 20ab + b2 • It is clear that the tens digit of the number b2 must be odd. A case-by-case analysis shows that the units digit must then be equal to 6.246MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)29. Hint: consider remainders modulo 16. 32. The answers are a) no; b) no. Use remainders modulo 9. 33. The answer is 7. Use remainders modulo 9. 34. The original number and the reversed number have equal sums of digits and, therefore, equal remainders when divided by 9. 35. This can be done in six different ways: 1155, 4155, 7155, 3150, 6150, 9150. Indeed, the last digit must be 0 or 5. Then our number is divisible by 3; that is, the sum of its digits must be divisible by 3. 36. There are two such numbers: 6975 and 2970. See the solution to the previous problem. 37. This is the number 1023457896. Hint: first, any number which has all 10 digits in its decimal representation is divisible by 9. Second, divisibility by 4 depends only on the last two digits of a number. Therefore, the required number starts with 10 ... and must end with an even digit. A simple case-by-case analysis leads us to the answer. 38. Hint: find a cycle in the remainders of the numbers 2n when divided by 3, and also in the units digits of these numbers. 39. The answer is no. By the usual test (see Problem 31) we have 1970 8 (mod 9). But no perfect square is congruent to 8 modulo 9. 40. Since after the first subtraction the result is divisible by 9, all the numbers we obtain in the process have the sum of their digits no less than 9. Therefore, if the original number was not greater than 891 = 9 · 99 then the proof is now obvious. The investigation of the rest of the set of three-digit numbers is left to the reader. 41. Since 44444444 < 104444 it is easy to see that A < 44440. Therefore, B < 5 · 9 = 45, which implies that the sum of its digits is a one-digit number. We also know that A and B (and the sum of its digits as well) are congruent to 44444444 modulo 9. That is, they all have remainder 7 when divided by 9. Thus, the sum of the digits of B must be 7. 43, 44. These numbers are divisible by 11. 45. The number aabb is divisible by 11 while cdcdcdcd is not. 46. The set {l, 2, 3, 4, 5, 6} cannot be split into two triples in such a way that the difference of the sums of the numbers in these triples is divisible by 11. 47. These two numbers have equal remainders when divided by 9, and also when divided by 11. 48. The answer is no. Indeed, if you multiply any of the given four digits by 9, then you cannot obtain one of these digits again in the units place of the answer. 49. Indeed, aba = lOla + lOb = 7(14a + b) + 3(a + b).====50. Since 2(a+b+c) 0 (mod 7), we get abc = lOOa+lOb+c 2a+3b+c (mod 7) b - c (mod 7). Therefore, abc is divisible by 7 if and only if b - c is divisible by 7. Taking into account that both digits band care less than 7, the only way this difference can be divisible by 7 is if the two digits are equal. 51. a) Since b) we have abcdef = lOOOabc +def= def - abc (mod 7). b), c) A number is divisible by 7 (or by 13) if and only ifthe following operation gives us a number divisible by 7 (or by 13): starting at the right of the number,ANSWERS, HINTS, SOLUTIONS247group the digits in threes and alternately add and subtract the resulting numbers. Example: 10345678. The operation described gives us 678 - 345 + 10 = 343 which is divisible by 7. Thus, the original number is divisible by 7 as well. Actually, it equals 7 · 1477954. The proof of these divisibility tests is left to the reader; they use the fact that 1001 is divisible by 7 and by 13. 52. A number is divisible by 37 if and only if the sum of the numbers formed by the triples of its consecutive digits is divisible by 37. Example: 830946. The described operation gives us 830 + 946 = 1776 which is divisible by 37. Therefore we can conclude that 830946 is also divisible by 37. Indeed, it equals 37 · 22458. The proof is similar to that of Problem 51. It uses the fact that 1000 1 (mod 37). 53. The answer is no. Since abc - cba = 99(a - c), where a and c are different digits, we have a number which is divisible by 11, but not by 112. 54. Answer: this number is written with three hundred l's. Indeed, it is divisible by 3 and it is divisible by 333 ... 33 (one hundred 3's), which are co-prime. To prove that this is the minimum number required we notice first that the required number must have a number of digits divisible by 100-otherwise it would not be divisible by 111 ... 11 (one hundred l's). Secondly, the numbers 111 ... 11 (one hundred l's) and 111 ... 11 (two hundred l's) are not divisible by 3. 55. The answer is no. Assuming the opposite, let us consider remainders modulo 5. Since the sum of the first n natural numbers is n(n + 1) /2, we have that 2(n(n + 1)/2 + 1) = n(n + 1) + 2 must be divisible by 5 (indeed, the last digits of n(n+ 1)/2+1 would be ... 1990). Substituting all five possible remainders modulo 5, that is, 0, 1, 2, 3, 4, and 5, we observe this is not true, which proves that the=answer is no.57. The answer is 69. Write 102a + b = 90a + 9b. Simplify to find 3a = 2b, and remember that a and b are digits. 58. Since any number of the form aabb is divisible by 11, we know that the square root of our number must be a two-digit number divisible by 11 (that is, with equal digits). Calculating squares for these 9 numbers 11 through 99, we see that the only answer is 7744 = 882 .59. The answers are 625 and 376. Hint: the units digit must be 0, 1, 5, or 6. Then analyze the tens digit similarly, then the hundreds digit. Remember that neither 000 nor 001 is a valid three-digit number. 60. We can prove that sooner or later this number will be divisible by 11. Indeed, if we denote some number in this series by x, then the next number will be 100x+43 x - 1 (mod 11). This means that after no more than 10 operations the current number will be congruent to zero modulo 11. 61. First, 10001 = 73 · 137, which is not obvious but nevertheless true. Second, to prove that any other number 10001 ... 10001 of the series is composite, we multiply it by 1111. The result has 4k digits (k > 2) and, therefore, is divisible by x = 1000 ... 001 = 102 k + 1 (indeed,=~= 1000 ... 001 ·~). 4k digits2k+l digits2k digitsFinally, we use the fact that x is greater than 1111 and less than the original number. Therefore, the original number must be divisible by x/ god(x, 1111) > l.248MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)63. This equation does not have integers roots. Its left side is always divisible by. 3, but its right side never is. 65. The solutions are of the form {x = 16k - 2, y = -7k + l}, where k takes on all integer values. 66. Our problem can be reduced to the solution of an ordinary Diophantine equation. Set p = z, and we have2x+3y= 11-5p, where we consider p as an unknown parameter, not as a variable. Using the same technique as before, we have the following answer: x = 5p+3q-11, y = 11-5p-2q, and, of course z = p, where p and q are any integers. There are no solutions in natural numbers, since if any of the numbers x, y, and z is greater than 1, then the sum 2x + 3y + 5z is greater than 11. 67. It is possible to move the pawn to the neighboring box if and only ifthe numbers m and n are relatively prime. Indeed, if we make k moves to the right (shifting the pawn m boxes to the right with each of these moves) and l moves to the left, then the resulting shift equals km - ln boxes to the right (a negative result means a shift to the left). The number 1 can be represented by such an expression if and only if the numbers m and n are relatively prime. The question about the minimum number of moves to get to the neighboring box is far more complicated. Hint: it is convenient to begin with the following reformulation. Given two relatively prime natural numbers m and n, find natural numbers k and l such that mk - nl = 1 and the absolute value of the sum Jk + !J is as small as possible. 68. The answers are (-4, 9), (14, -21), (4, -9), and (-14, 21). Analyze all possible representations of the prime number 7 as the product of two integers. 70. There are no integer solutions. Indeed, in the equation (x-y)(x+y) = 14 both factors in the left side are of the same parity, and therefore their product must be either odd (if both expressions x + y and x - y are odd) or divisible by 4 (if both x+y and x-y are even). But the number 14 belongs to neither of these two types. 71. The answers are (2, 0), (2, 1), (-1, 0), (-1, 1), (0, 2), (1, 2), (0, -1), (1, -1). We transform the original equation into x(x-1) +y(y- l) = 2. Since the product t(t - 1) is never negative and is greater than 2 if t > 2 or t < -1, we have only a few pairs (x, y) to consider. 73. There are no integer solutions. Hint: use remainders modulo 7. 75. There are no integer solutions. Hint: use remainders modulo 5. 76. There are no integer solutions. Hint: use remainders modulo 8. 79. Answer: there are three families of solutions (1, a, -a), (b, 1, -b), (c, -c, 1), where a, b, and care arbitrary integers. And there are three more solutions: (1, 2, 3), (2, 4, 4), (3, 3, 3). Hint: if all numbers are positive, at least one of them is not greater than 2 or they all are equal to 3. If one of the numbers-say, a-is negative, then 1/b + 1/c > 1, and this means that either b or c is 1. 80. The answers are x = ±498, y = ±496 and x = ±78, y = ±64 (the signs can be chosen independently). To prove this, we rewrite the equation as follows: (x - y)(x + y) = 2 · 2 · 7 · 71 (the number 71 is prime). We can temporarily assume that x and y are positive (later we can supply these numbers with arbitrary signs). We have only two representations of the number 1988 as the product of twoANSWERS, HINTS, SOLUTIONS249positive integers of equal paXity (see the solution to Problem 70): 1988 = 2 · 994 and 1988 = 14 · 142. Setting the factors x - y, x + y equal to these completes the solution. 81. If n = pq (where p, q > 1), then l/n = l/{n - 1) - l/n(n - 1) and l/n = l/p(q-1)-1/pq(q-l). If n is prime, then from the original equation n(y-x) = xy and, therefore, xy is divisible by n. So, either x or y is divisible by n. It is clear that y is divisible by n since otherwise x ~ n and l/x - l/y cannot be equal to l/n. Thus y = kn and x = kn/(k + 1), and k = n - l. Thus, there is only one representation of l/n: l/n = l/{n - 1) - l/n(n - 1). 82. Answer: there are no integer solutions. Hint: rewrite the equation as x 3 = (2y - 1){2y + 3) and make use of the fact that the factors on the right-hand side are relatively prime. 83. This is the famous Pythagorean equation. Answer: all the solutions can be described as follows:x = (a2 - b2 )c , y=2abc,z = (a 2 + b2 )c, where a, b, and c are arbitrary integers. Hint: first, make x, y, and z pairwise relatively prime by dividing them by their common G.C.D. For a complete solution, see , Chapter 17. 84. This is another famous equation, called Pell's equation after the XVII century English mathematician. Hint: first, we will look for non-negative solutions only. One of these is easy to find: it is the pair {l, 0), and we can generate all other solutions starting from this one. More precisely, if the pair (a, b) is a solution to our equation, then the pair (3a + 4b, 2a + 3b) is the next solution. For a complete solution of the problem, see , Chapter 17. 86. We know that ka - kb is divisible by kn. Thus, k(a - b) = mkn, and we have a - b = mn, which proves the result. 87. Fermat's "little" theorem implies that this remainder isl. 89. We have 300 3000 = (300500 ) 6 1 (mod 7). Similarly, 300 3000 1 (mod 11) and also (mod 13). Therefore, 3003000 - 1 is divisible by 7, by 11, and by 13; that is, by 1001. 90. The answer is 7. Hint: use Fermat's "little" theorem. 92. Hint: prove that the given number is divisible by 31. 93. It is sufficient to write the following short chain of equalities and congruences: (a+b)P (a+b) = a+b::a•+b"(modp). 94. Hint: prove that for any integer x the congruence x 5 x (mod 30) holds true by showing that x 5 - x is divisible by 2, 3, and 5. 95. a) Hint: prove that p• + qP - p - q is divisible by p and by q. 96. Let us set b = a•- 2 . Then, ab= a•- 1 1 (mod p). 97. Let us split all the numbers 2 through p - 2 into pairs such that the product of two members of any pair is congruent to 1 modulo p (we leave to the reader the proof of the possibility of such a splitting). Thus, the product of all the numbers 2 through p - 2 is congruent to l. Hence p! 1 · (p - 1) = p - 1 -1 (mod p).=======MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)25098. Since (n8 + l)(n8 - 1) = n 16 -1=O(mod17), one of the factors must be divisible by 17. 99. a) The number 111 ... 11 (pones) is equal to (10" - 1)/9. But 10• - 1 is not divisible by p, since 10" - 1=10 - 1 = 9 (mod p). b) The number 111 ... 11 (p-1 ones) is equal to (10•- 1 -1)/9, and 10•- 1 -1 is divisible by p, since pis relatively prime to both 10 and 9. 100. Hint: use the following congruences: 10• 10 (mod p), 10 2• 100 (mod p), ... , 108" 108 (mod p).===11. COMBINATORlCS-2 7. The answer is Ca0 ) = 120. This is a straightforward corollary of the definition of the number of combinations. 8. The officer can be chosen in three ways, the 2 sergeants in (~) ways, and the 20 privates in (~g) ways. Thus a group for the assignment can be chosen in 3 · (~g) ways. 9. a) Each triangle with vertices at the marked points has either one vertex on the first line and two vertices on the second, or two vertices on the first line and one on the second. There are 10 · ('i) triangles of the first kind and C~) 11 of the second kind. Therefore the answer is 10 · (~1 ) + 11 · (12°). b) Answer: C~) = 2475. 10. Add up the numbers of ways to choose exactly 0, 1, ... , and 5 words from the + + (';) + = 4944. given set. The answer is ('(,5) + (\5 ) + 11. Let us choose three couples first. This can be done in (:} ways. Three representatives from these couples can be chosen in 2a ways (either husband or wife from each of the three couples). Thus there are (:) · 2a = 32 ways to choose a committee. 12. There are three possible cases: only Pete is on the team, only John is on the team, or neither one of them is on the team. There are G~) different teams with Pete but without John (because ten of Pete's teammates can be chosen only from the other 29 students of the class). Similarly, there are (~~) teams with John but without Pete. And finally there are (~i) teams without either. Therefore the answer is (~~) + (~~) + (~i) . 13. Since the order of the vowels, as well as of the consonants is known, everything is defined by the places occupied by the vowels. There are = 35 ways to choose 3 places for the vowels in a word consisting of 7 letters. 14. There are (\2) teams with no boys at all, 10 · (~2 ) teams with 1 boy and 4 girls, C~) (~2 ) teams with 2 boys and 3 girls, and Ca0 ) · teams with 3 boys and 2 girls. Thus there are (\2) + 10 · (~2 ) + (12°) · (1a2 ) + ('g") · (',;') = 23562 different teams which satisfy the conditions of the problem. 15. First, let us choose the places for 12 white checkers on the 32 black squares of the chessboard. This can be done in (~~) ways. After the white checkers are placed, there are (~g) ways to put 12 black checkers on the 20 free black squares. Answer: @ . @ = 32!/12!12!8!.m··· Ci)C{) Ci)Ci)m·Ci)16. The solution is analogous to that of Problem 6. Answers are a) b) ('i). C~)/2.('i). (15°). m/3!;ANSWERS, HINTS, SOLUTIONS25117. a) Let us choose one of the four aces, then choose separately nine other cards from the 48 cards which are not aces. Since the first choice can be made in 4 ways and the second in (4,,8) ways, the result is 4 · (~8 ) b) There are (~~) ways to choose any 10 cards from the deck, and (~~) ways to choose 10 cards, none of which are aces. Thus there are (~~) - (~~) ways to choose 10 cards so that there is at least one ace. 18. There are two cases, depending on the parity of the first digit of the number. In each of the cases you can calculate the number of ways by choosing the places for the odd digits. The answer is: 56 + 4 · 55 . 19. Hint: find all the representations of the numbers 2, 3, and 4 as the sums of several natural numbers. Do not forget that the first digit cannot be zero. The answers are: a) 10; b) 1+(~)+9+9+1 = 55; c) 1+2(i)+(i)+(~)·3!/2!+G} = 220.m· m·21. a) Answer: ('\;5). b) Let us suppose that the results are already known. Now we have to choose exactly three numbers from the six "lucky" numbers and three numbers from 39 3;') = 182780. "unlucky" ones. Thus the result ism·(22. The answer is the number of all subsets of a 10-element set; that is, 210 = 1024. 23. A way to go down the flight of stairs is simply the choice of several steps you are going to step on. Therefore, the question is equivalent to the calculation of the number of the subsets of a 7-element set, and the answer, of course, is 27 = 128.25. Let us prove that the number of subsets of the given set of objects with evenly many elements is the same as the number of subsets with oddly many objects. We begin by choosing one of the objects-say, A-which will play a special part further in the solution. Now, we will split all the subsets in pairs in such a way that each pair consists of two subsets, one of which always has evenly many elements and another which contains oddly many elements. To do this, we consider one arbitrary subset of the given set of objects and, if it contains A as its element, we remove A from it; if it does not, we add A to it. The resulting subset will be in the same pair with the original one, and the number of elements in these two subsets have different parity. It is easy to see that if the subset S generates the subset S', then this construction applied to S' gives us S. Therefore, we have the required splitting, and the proof is complete. 26. Use the result of Problem 25. 27-28. Hint: use induction on the number of Pascal's triangle numbers on the diagonal in question.29. Apply the results of Problems 27 and 28. 30. The number of ways of going downward from the "summit" of Pascal's triangle to the number occupying the nth place in the 2nth row equals (2,:'). Each of these ways passes through exactly one of the numbers in the nth row. Since the number of such ways passing through the kth number is equal to (~)(n~k) = ((~))2, we can add up these numbers of ways to obtain the required total. This geometric interpretation of the algebraic equality to be proved demonstrates one of the most beautiful tricks in elementary combinatorics. 34. Let us call the 12 pennies "balls" and the five purses "boxes". Now the problem is almost the same as Problem 31. Answer: (';~i') = (~1 ) = 330.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)25235. This time the 12 books are the "balls", and the three colors are the "boxes". 2t!;-') = The problem turns into a question similar to Problem 32. Answer: = 91. 36. To cut the necklace into 8 parts it is necessary to choose 8 places from the 30 where the cuts are to be done. Therefore the result is (~0 ). 37. If candidates are "boxes" and voters are "balls", the problem is similar to Problem 32. Answer: (3.4). 38. Hint: regard the postcards as "balls" and the types as "boxes". Answers: a) (2.'); b) 39. a) The first passenger can get off the train at any one of n stops. The second passenger can also get off the train at any one of n stops. Thus there are n · n = n 2 different ways to get off the train for these two passengers. Since the third passenger can choose any of n stops, there are n·n 2 = n 3 different ways for three passengers to leave the train. It is clear now that the same argument for the remaining passengers leads to the answer n rn. b) This is again a problem about "balls" (passengers) and "boxes" (stops). Thus the answer is (n~~~ 1 ). 40. This problem is identical to the problem of representing the number 20 as the sum of three non-negative integers. Answer: (2,2) = 231. 41. Hint: find the answers for the black and for the white balls separately, then multiply the results. Answer: 86 ) • C~). 42. Hint: divide the process of the distribution of the fruits into four steps: apples, orange, plum, and tangerine. Answer: (~) · 3 · 3 · 3 = 756.CCi)c;).C43. Since there are (~) different ways to put balls of each of three colors into six different boxes, the result is (~)3. 44. a) Each of n voters can choose any of n candidates. Thus the result is nn. b) Consider the members of a community as "boxes", and the votes as the "balls". Answer:(2;::11).45. Let us temporarily repaint the red and the green balls black, and line the black balls up in a row. Arranging 10 black balls and 5 blue balls so that no two blue balls are next to each other is the same as placing 5 blue balls into 11 "buckets" between the black balls, and at the ends of the row of the black balls, so that no two blue balls are in the same buckets. That is, we just choose 5 buckets out of 11-this can be done in ( ~1 ) ways. Finally, there are ways to repaint the black balls red and green. Thus, the answer is 5°) = 116424.Cii) ('Ci)46. Hint: we know that 1000000 = 26 • 56 . Each factor can be completely determined by the number of 2's and 5's in its decomposition. The total number of 2's is 6, and the total number of 5's is the same. Answer: [ 2 = 784.m]4 7. Let us remove the chosen books and consider the seven remaining books. Between any two of them and at the ends of the row, we either have a gap (caused by a missing book) or we don't. The set of the gaps uniquely specifies the set of the chosen books. The answer therefore is (~) . 48. Since there are only two blue beans, the type of necklace is fully determined by the distance between these two beans as measured by the minimum number of theANSWERS, HINTS, SOLUTIONS253beans (regardless of their color) between them. This quantity can take only three values 0, 1, and 2. Hence there are only 3 different types of necklaces. 49. These are applications of the basic results of the chapter. The answers are a) (3~) = 27405; b) 30. 29. 28. 27 = 657720. 50. Let us calculate the number of all six-letter words first. Any of 26 letters can be in each of 6 places. Thus it is possible to write 26 6 different six-letter words using the 26 letters of the English alphabet. And there are 25 6 words without the letter A (in this case only 25 letters can be used). Therefore, there are 266 -256 = 64775151 words containing at least one letter A. 51. It is possible to start drawing the path at any of the six vertices of the hexagon. The second point can be chosen in 5 ways, and so on. Thus, there are 6! ways to draw a path. But each path was counted exactly six times during this calculation, since each of its vertices could be chosen as the first one. Therefore, there are 6!/6 = 5! paths. 52. a) A number divisible by 4 and written with the given digits must end with 12, 24, or 32. In each of these cases there are two different ways to use two other digits at the beginning of the number. The answer is 3 · 2 = 6. b) In this case the number must end with 12, 24, 32, or 44. Each of 4 digits can be used in any of the two remaining places. Therefore the result is 4 · 42 = 64. 53. Everything is defined by specifying the three days when the father gives pears to his daughter. Since there are ways to choose three out of the five days, the answer is G).G)54. There are (26°) ways to choose 6 actors for the first performance. In each of these cases 14 actors did not take part in the first performance. Thus there are ('64 ) ways to choose 6 actors for the second performance. The answer is (26°) ('64 ). 55. In every decimal place each of the digits is used exactly 42 = 16 times. Answer: 16 · 1111 · (1 +2+3+4)=17760. 56. The 6 cards can be distributed among the four suits in two ways: 1 + 1 + 1 + 3 or 1+1+2 + 2. Let us calculate the number of different choices for the first variant. The suit containing 3 cards can be chosen in 4 ways. There are 33) ways to choose 3 cards from this suit. One card from each of the remaining suits can be chosen in 13 ways. Thus, the result is 4 · 133 . A similar calculation leads to the result (;} · ('i) 2 · 132 for the second variant of distribution. Therefore, the answerCCi) ·is 4·Ci)· 133 + (;). Ci) 2 . 132 .57. Answer:m·('i) = 5720. See the solution to Problem 41.58. It is obvious that there are 10 one-digit integers satisfying the conditions of the problem. Let us calculate how many two-digit integers there are. The first digit of a two-digit integer can be any digit but 0. The second digit can be any of 9 digits which differ from the first one. Therefore there are 92 two-digit integers with two different digits. There are 93 three-digits integers which satisfy the conditions of the problem, because there are 9 digits (any digit but the one used as the second one) to choose from for the third place. Continuing in the same way, we get the final result: 10 + 92 + 93 + 94 + 95 + 96.MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)25459. Let us start with another problem: How many ways are there to take 18 cards from a deck of 96 cards (including 4 aces) so that these 18 cards contain exactly 2 aces? First, we choose two of the four aces. Second, we choose sixteen other cards from the 32 non-aces. Therefore, the answer to the new problem is (~~). Let us now notice that by choosing 18 cards from the deck we have divided the deck into halves. But each possible division has been counted twice. Thus, the result ism·m. (~~)12.60. a) The rook can either visit or not visit each of the 28 non-border boxes. The answer, therefore, is 228 . b) The answer is the number of representations of the number 29 as the sum of seven natural numbers whose order is significant. Answer: (~8 ). 61. Let us suppose that none of the rowers has been chosen from those ten who wanted to be on the left side. Then four rowers for the left side have been chosen from the nine who can sit on either side. And the four rowers for the right side have been chosen from seventeen (the twelve who want to sit on the right side, and those five from the nine without preferences who were not chosen for the left side). Thus in this case there are ('g) · l:) · ('J) ways to make a choice. Now let us suppose that exactly one rower has been chosen from those who want to sit on the left side. Then another three left-side rowers have been chosen from the nine. And ~8 ) four right-side rowers have been chosen from eighteen. This gives us ( \0 ) · choices for this case. Taking into account three final cases (two, three, or four rowers to choose from those ten who wanted to be on the left side) we get the final result: ('o°). ('J) + (\"). (~8) + ('~). (1:) + (\0). (~o) + ('~). (~). (~1).m·(m.m.m.m.62. The rectangle can be defined without ambiguity by its upper left and lower right vertices. To contain the marked box, the upper left vertex must be in a row with a number less than or equal to p and in a column with a number less than or equal to q. The lower right vertex must be in a row with a number greater than or equal to p and in a column with a number greater than or equal to q. Thus there are p · q different positions for the upper left vertex and there are (m - p + 1) · (n - q + 1) different positions for the lower right vertex. Therefore there are p · q · (m -p + 1) · (n - q + 1) rectangles containing the marked box. 63. The grasshopper has to make 27 jumps; 9 jumps in each direction. Let us denote jumps in the first direction by the letter A, jumps in the second direction by the letter B, and jumps in the third direction by the letter C. Now each route of the grasshopper can be defined without ambiguity by the 27-letter word in which each of the letters A, B, C is used exactly 9 times, and the problem is reduced to these words. Doing this in the same way as in Problems 17-21 from the chapter "Combinatorics-I" we obtain the answer: 27!/(9!) 3 .12. INVARIANTS 4. Answer: 21! - 1.5. We can use the following quantity as an invariant: each sparrow is supplied with a special index equal to the number of the tree it is currently sitting on (counting from left to right). Then the sum S of these indices is the required quantity. Indeed, after the flights of any two birds only their indices change-one increases by someANSWERS, HINTS, SOLUTIONS255number x and another decreases by the same number. Thus, the sum S is invariant. Initially, the value of S is 1+2 + 3 + 4 + 5 + 6 = 21, and if all the sparrows are on the same tree with number k, then the value of S is 6k. Since 21 is not divisible by 6 we can conclude that the sparrows cannot all gather on one tree. On the other hand, if there are seven sparrows and seven trees, then initially S = 28 which is divisible by 7, and we cannot exclude the possibility of all the sparrows gathering on one tree. In fact, the reader can easily construct a sequence of flights which results in the required situation: all the sparrows are together on the middle tree. 6. Hint: prove that the parity of the number of black boxes in the table is invariant. 7. Hint: prove that the parity of the number of black boxes among the four corner boxes is invariant under recolorings. 8. This problem can be solved in just the same way as Problem 7, if we consider a set of four boxes with the same property. One such set is the four boxes forming a 2 x 2 square in the upper left corner of the table. 9. Hint: use as the invariant the parity of the sum of all the numbers on the blackboard. 13. a) Use the following coloring of the board: the rows with the odd numbers are colored black, and the rows with the even numbers are colored white. Then the 1 x 4 polyminos always cover an even number of white boxes regardless of their position on the board. Furthermore, the one special polymino always covers an odd number of white boxes. These two facts together imply that the entire number of white boxes covered is odd, but this number must be 32. This contradiction completes the proof. c) Apply the analogous coloring using four colors. Since each polymino covers either four boxes of the same color or four boxes colored with four different colors, we can conclude that the difference between the numbers of the boxes of color A and of color B is divisible by 4 (regardless of which colors A and B we choose). An easy calculation shows that there are 2652 boxes of the 1st color, 2652 boxes of the 2nd color, 2550 boxes of the 3rd color, and 2550 boxes of the 4th color. The difference between the number of the boxes of the 1st and the 3rd color is 102 which is not divisible by 4. This completes the proof. 14. Let us consider the coloring using 4 colors shown in Figure 154. Then each 2 x 2 tile contains exactly one box of color 1, and each 1 x 4 tile contains none or two boxes of color 1. Therefore, the parity of the number of the 2 x 2 polyminos coincides with the parity of the number of boxes of color 1. This proves the statement: after the parity of the number of 2 x 2 polyminoes changed (when one was lost) we cannot cover the same board without overlapping.2 3 1 4 2 3 1 42 3 1 4 2 3 1 4FIGURE 154256MATHEMATICAL CIRCLES (RUSSIAN EXPERJENCE)16. Let us use as the invariant the remainder of the number of heads of the Dragon when divided by 7. Using either of the swords does not change this remainder, and since 100 and 0 are not congruent modulo 7 the answer is, alas, negative. 17. The remainder when divided by 11 of the difference between the number of dallers and the number of dillers belonging to the businessman can be used as the invariant. Since initially this difference is 1, the businessman can never make this difference equal to zero. 18. Since after every operation of Dr. Gizmo's machine the number of coins increases by 4, then the remainder of the number of coins when divided by 4 is invariant. But 26 and 1 have different remainders when divided by 4. Therefore we cannot end up with 26 coins. 20. The answer is 8. Since the remainders of a natural number and of the sum of its digits when divided by 9 are the same, the remainder of 8 1989 coincides with the remainder of the final result x. Hence, x has remainder 8 modulo 9, and we know that x is a digit. We conclude that x = 8. 21. Its type is B. Consider the parities of the differences N(A) - N(B), N(B) N(C), and N(C) - N(A), where N(X) is the number of type X amoebae. These parities do not change in the course of the merging process. This means, in particular, that in the end (when there is only one amoeba in the tube) the numbers of A-amoebae and C-amoebae have the same parity, which is possible only if the only amoeba left belongs to type B. 22. After each move the sum of the numbers of the row and the column of the square the pawn is on either decreases by 2 or increases by 1. Thus, the remainder of this sum when divided by 3 increases by 1 each time. Since there are n 2 -1 moves in all, and the final sum must be 1 more than the original, we get that n 2 - 2 must be divisible by 3. This is impossible (a perfect square cannot have a remainder 2 when divided by 3) and therefore such a route for the pawn is impossible. Remark. We would like to draw your attention to the fact that we did not use the word "invariant" in this solution. However, some quantity is invariant. Canyou find it? 23. Since the sum in each row is 1, and we have m rows, the sum of all the numbers in the table is m. On the other hand, the sum in each column is 1, and the table has n columns. Hence, the sum of the numbers is n. But the sum of the numbers in the table does not depend on the way it is calculated (in this sense, this problem concerns the idea of invariant). Therefore, m = n. 24. The answer is no. Hint: use the parity of the number of glasses standing upside down as the invariant. 25. Let us mark four vertices of the cube such that no two of them are connected by an edge (this is not difficult). Then consider the difference between the sum of the numbers on the marked vertices and the sum of the numbers on the other vertices. This difference is invariant under the operations described. Using this invariant, we can easily prove that the answer to both questions is negative. 26. Let us number the sectors with the numbers 1 through 6 in the clockwise direction starting at some sector. Then consider the difference between the sum of the numbers in sectors 1, 3, and 5, and the sum of the numbers in the sectors 2, 4, and 6. This quantity is invariant, and its initial value is ±1. Thus, it cannot be equal to 0, and the answer is negative.ANSWERS, HINTS, SOLUTIONS25727. We can only obtain cards (a, b) such that a< band b - a is divisible by 7. 28. The answer is no. Let us introduce the quantity S equal to the sum of the number of stones and the number of heaps. It is not hard to show that S is invariant, and its initial value is 1002. If there are k heaps with exactly 3 stones in each of them, then the value of S is k + 3k = 4k, which cannot be equal to 1002, since 1002 is not divisible by 4. 29. The answer is no. Hint: use the following quantity as an invariant: the parity of the number of pairs (a, b) where the number a occupies the place to the right of the number b and a > b. 30. The sum of the squares of the numbers in a trio does not change after any of the operations described. Using this quantity as the invariant, we can easily see that the answer is no (the values of the invariant for the given trios are different:6 + 2y'2,;, 13/2). 13. GRAPHS-22. Since there are 4 edges leaving four of the vertices we conclude that each of these is connected with every other. But that would mean that the fifth vertex is also connected with all the other vertices; that is, its degree is also 4. This contradiction completes the proof. 3. Hint: use induction on n. The base (n = 1) is easy. To prove the inductive step, consider graph G with 2n vertices which satisfies the condition of the problem, and add to it two other vertices A and B which, so far, are not connected to any of G's vertices. Graph G has two families Vi and V2 of vertices each containing n vertices with degrees equal to 1, 2, ... , n. Connect one of the new vertices-say, A-to all vertices of family Vi and to the vertex B. The resulting graph is a graph with 2n + 2 vertices satisfying the required condition. 4. a) Yes, since in such a graph each vertex is connected with every other; that is, the graph is isomorphic to the complete graph with 10 vertices. b) No-see Figure 155.FIGURE155c) No-see Figure 156. 5. Assume that such a deletion is possible. If this edge connects two vertices with equal degrees, then each of the connected components would have an odd number of odd vertices, which is impossible. Otherwise, only one component will have a vertex with degree 2, and the components cannot be isomorphic. 9. Consider any of the connected components of the given graph. It is not a tree since it does not contain a pendant vertex. Therefore, it contains a cycle.258MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)FIGURE15610. Suppose the ends of the deleted edge are connected by a simple path in the new graph. Then this path together with the deleted edge gives us a cycle in the old graph-a contradiction. 12. Hint: if it were not a tree, then we could obtain a tree from it by deleting some edges. 14. The total number of roads in the country is (32°). Since the number of edges in a tree with 30 vertices is 29 we get the answer: 30 · 29/2 - 29 = 406. 15. Hint: take a maximal tree of the given graph and delete any of its pendant vertices.16. Hint: let us consider any maximal tree of the graph. Then double each edge of this tree (there are 99 of them). Though the result is not exactly a graph, Euler's theorem about drawing a graph with a pencil without lifting it from the paper applies to this "multi-graph" as well. 17. Consider the planar graph whose vertices are the lakes, whose edges are the canals, and whose faces are the islands. Since V - E + F = 2, V = 7, and E = 10, we have F = 5. However, one of the faces is the outer face, which is not an island. Answer: 4. 19. Hint: every face is bounded by at least three edges. 24. The inequality 3V - 6 ~ E does not hold true for this graph, and therefore this graph is not planar. 25. Assume the opposite. Then 2E ~ 6V; that is, E ~ 3V, which contradicts the proved inequality. 26. Suppose both graphs are planar. Then they have no more than (3 · 11 - 6) + (3 · 11 - 6) = 54 edges together. However, the complete graph with 11 edges must have 55 edges, and we have a contradiction. 27. Hint: first, prove the inequality E $ 3V - 6 using the fact that the degree of every vertex is at least 3. Denoting the number of pentagons by a and the number of hexagons by b, we have 5a+ 6b+7 = 2E $ 6F-12 = 6(a+ b+ 1)-12. Hence, a~ 13. 29. a) No, since the graph has 12 odd vertices, which means that we need at least 6 paths to form the grid. · b) Yes, it is possible. We leave it to the reader to construct an example. 30. The graph formed by the circles (the points of intersection are its vertices, and the arcs of the circles are its edges) is connected, and all the degrees of its vertices are even.31. The proof can be carried out using induction on n. The base n = 0 is obvious. To prove the inductive step we choose two odd vertices A and B and connect themANSWERS, HINTS, SOLUTIONS259mentally with a new edge. After that the new graph has only 2n - 2 odd vertices and can be drawn in such a way that the pencil will be lifted exactly n - 1 times. When, in the process of this drawing, we must go along the mentally added edge AB (which does not really exist), we simply lift the pencil from paper and put it down at the other end of this edge. 32. Hint: find two scientists who are not acquainted, and consider all the people they are acquainted with. 33. Assuming the opposite, we have that for any number 68 through 101 there are exactly three students who have exactly so many acquaintances. Then the number of students having an odd number of acquaintances is odd, which is impossible. 34. Hint: let us connect these two vertices by a path. Then, if its length is a, the distances from any vertex to these two must differ by a number of the same parity as a.35. Hint: prove that any tree with 6 vertices is isomorphic to one of the graphs shown in Figure 157.• • • • • •+I • •• •...I•.i-rFIGURE 157 36. a), b) These items are corollaries to items c) and d). c) Let us assume the opposite. Consider an arbitrary town X and a town A which cannot be reached from X by airplane with no more than one transfer, and a town B which cannot be reached by train with no more than one transfer. Now, notice that the towns A and B are connected by some kind of transportation. We can assume without loss of generality that A and B are connected by train. By assumption X and A are connected directly by train, and, therefore, we can reach B from X by train by transferring at A, which is a contradiction. d) Hint: suppose we cannot fly from A to B with no more than two transfers, and we cannot go by train from C to D with no more than two transfers. Consider the graph formed by these four towns. 37. See Problem 28 from the chapter "The Pigeon Hole Principle". 38. Take an arbitrary vertex and notice that there are at least 6 edges of the same color leaving it. Now use the result of Problem 37.260MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)39. Suppose there is a vertex with 6 blue edges leaving it. Then we can use the result of Problem 37. If there is a vertex with no more than 4 blue edges leaving it (it is impossible for all nine vertices to have five blue edges leaving them), then there are at least 4 red edges leaving it. 40. There are no less than 9 edges of the same color leaving any given vertex. Now use the result of Problem 39. 42. Denote the number of the roads entering the capital by a. Then the total number of all "incoming" roads is equal to 21 · 100 + a, and the total number of "outgoing" roads is no more than 20 · 100 + (100 - a). Therefore 21 · 100 +a ~ 20 · 100 + (100 - a); that is, 2a ~ 0. Finally, a= O. 43. Hint: number the towns and mark each road as one-way in the direction leading from the town with the smaller number to the town with the greater number. 44. Hint: first, consider the vertices connected with the chosen vertex A, then the new vertices connected with these, et cetera. In the process of extending this "web" (as long as it is possible) we orient the edges which connect the newly added vertices with the older ones in the direction from the older to the newer endpoints. 45. Hint: consider an Euler path passing through all the edges of the graph and orient the edges according to their order in the cycle. 46. Hint: prove that there exists a closed path along the arrows which passes along each edge exactly once. This can be done by considering the closed path with the largest possible number of edges. 48. Assume that team A won the tournament. If there exists a team which defeated A as well as all the teams who lost to A, then this team would have scored more points than A, which is impossible. This proves both parts of the problem. 49. Hint: proceed by induction on the number of towns. The base (for three towns) can be proved using a simple case-by-case analysis. Tu prove the inductive step, temporarily remove the town which has roads both entering it and leaving it. 50. We prove this by induction on the number of teams n. The base n = 2 is easy. To prove the inductive step let us temporarily remove one of the teams X and number the other n - 1 teams as required. If X has defeated team 1 or has lost to team n- 1, then we can easily add X to the chain. Assuming the opposite we have that X lost to team 1 and defeated team n - l. Therefore there must exist such an integer k that X lost to team k and defeated team k + 1-otherwise X must have lost to team 2, therefore to team 3, therefore to team 4, et cetera. Having found such a k we can insert X into the existing chain by "cutting in" between teams k and k + 1 and create the chain of teams we need. 51. Suppose A and B won an equal number of games, and B defeated A. Then if every team C which lost to A also lost to B, then B must have more points than A. Thus, there exists a team C such that A defeated C but C defeated B. 52. Hints: a) If we cannot reach town B from town A consider the towns where the roads leaving A lead, together with the towns where the roads entering B leave from. b) Using the same notation as before we can assume that there is no road A -+ B, and that there is no town C such that there are roads A -+ C and C ....., B. Let us find the forty towns A 1 , A 2 , •.• , A 40 that the roads leaving A lead to, and forty other (!) towns Bi. B 2 , ••. , B 40 that the roads coming to B come from.ANSWERS, HINTS, SOLUTIONS261There are 1600 roads which leave the towns A;. At the same time the total number of roads connecting A; with each other does not exceed 40 · 39/2 = 780, and the number of the roads leaving from them to the remaining 19 towns is no more than 40 · 19 = 760. Since 1600 > 1540 = 780 + 760 it follows that there must be a road from A; to Bj. 53. Hint: if we have removed the edge between vertices A and B, then let us choose two arbitrary vertices and consider three cases: none of these vertices coincides with A or B; one of them is either A or B; or in fact they are A and B. 14. GEOMETRY 1. Hint: try to find the possible length of the third side in a triangle whose two sides have lengths a and b.2. Hint: prove the inequalities AM> AB - BC/2 and AM> AC- BC/2. 3. Consider the circle inscribed in the triangle and the lengths of the resulting segments when the meeting points divide the sides of the triangle. We have three pairs of equal segments, whose lengths are the required x, y, and z. 4. Let us assume the opposite. Then one of the angles is larger than the other one, and the corresponding side must be longer than the other. This contradiction completes the proof. 5. Hint: use the inequalities LBAM < LABM and LCAM < LACM. 6. This is a simple exercise in the use of Inequality N21. If a a+ 2Vab + b > c, or (y'a + Jii) 2 > (y'C) 2 , and via+ Jii >JC.+b >c, then7. Since AB +CD < AC+ BD (by the way, why?) we can obtain the required result by adding this inequality to the inequality given in the statement of the problem.> B 1 D 1 , and hence LC> LC1 . If LB> LB1 , then similarly LD > LD 1 which is impossible since the sum of the angles' measures in the quadrilaterals must be the same. 9. Since LA> LA 1 we have BD10. Hint: construct parallelogram ABCD, three vertices of which coincide with the vertices of triangle ABC, by extending the median its own length to D. Then apply Inequality N2 2. 11. Answer: No. It would follow from Inequality N22 that LBAC > LBCA = LDCE > LDEC = ... > LKAI = LBAC-which is a contradiction! 12. a) Hint: place three copies of the triangle on the plane so that their legs coincide. Then look for an equilateral triangle. b) Find point Eon side AB such that AE = AC. Then prove that EB > CE>AC.13. Hint: if we denote the outer perimeter by a, the perimeter of the star by b, and the inner perimeter by c, then a > c, a + c < b, and 2a > b. Now a case-by-case analysis completes the proof. 14. "Fold out" the perimeter of the quadrilateral as shown in Figure 158. 15. In three steps move the second triangle so that all three of its vertices coincide with the vertices of the first triangle (one vertex at a time).MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)262cOBAc DD"DD' D~"--+---'-'-lA D'" FIGURE15816. a) Prove that any point D will be fixed. b) Apply the result of a). 17. a) Answer: this is a translation again. b) Take two parallel lines which are perpendicular to the direction of the translation and such that the distance between them equals half the length of the translation. c) This motion cannot be a rotation since there is no point which is left in place. It cannot be a translation since the distance between a point and its image is not constant. And, finally, it is not a line reflection: if it were, then for any point A and its image A' the perpendicular bisector of segment AA' would be some constant line not dependent on the choice of point A. 18. The answer is yes. It suffices to map one of the centers onto the other. 19. Only the identity rotation can map a half-plane onto itself: otherwise where would the boundary go? A line reflection can do it if the line is perpendicular to the boundary of the half-plane. 20. Yes, this is true. Indeed, the composition of eight such rotations is a rotation of 24 degrees about the same point. Further, the composition of three such superpositions will be a rotation of 72 degrees about point 0. 21. Hint: reflect the triangle in the given point. Where the triangle and its image coincide are the endpoints of the required segment. 22. Hint: use a translation. 24. Hint: since the lines (AB), (CD), and (MN) meet at one point, the reflection in line (MN) maps lines (AB) and (CD) onto each other. 25. Hint: use a rotation of 90 degrees which maps the square onto itself. 26. Hint: use a rotation of 60 degrees about point P, and look at the point where the first line and the image of the second line meet. 27. a) Hint: if the two given points X and Y are on different sides of line L, then M = (XY) n L; else M = (XYi) n L, where Y1 is symmetric to Y with respect toL. b) Hint: if X and Y lie on different sides of L, then M = (XY) nL; otherwise, M = (XY1) n L, where Y1 is symmetrical to Y with respect to L. 28. Hint: reflect the first axis in the second. Prove that the image must be an axis of symmetry for the triangle. Also, show that the two original lines cannot be perpendicular.ANSWERS, HINTS, SOLUTIONS263~aj~~~~~~~~~~~~~~~~b) H, I, N, 0, S, X, Z. The answers depend on how you write the letters. Here we use the standard roman typeface. 30. The answer is no. See the hint to Problem 28. 31. Hint: these are the points from which segment OS (0 is the center of the rotation) subtends an angle of goo+ oi/2 or goo - oi/2 (oi is the angle of rotation). The reader can try to describe this locus more precisely. 34. No. Hint: if two angle bisectors were perpendicular, then the sum of these two angles of the triangle would be 180°. 35. Let line L be the common perpendicular to lines AB and CD, which passes through the center of the circle. Then segments AC and BD are symmetric with respect to line L. 36. Hint: the sum of two opposite angles in an inscribed quadrilateral must be 180 degrees. Answer: 60, go, and 120 degrees. 38. It is not difficult to see that the measure of angle AOD is 60°. Further, triangle DOC is isosceles, so LDOC = 75°. Therefore, LAOC = 135°. 3g. Angles ABC and ABD are equal to go 0 . Thus LCBD = 180°. 40. a) Let a= IABI, b = IBCI, c = ICDI, and d = IDAI. Then bis not greater than the altitude to AB in triangle ABC. Hence S(ABC) :::; ab/2, and S(CDA) :::; cd/2. Adding these inequalities, we are done. b) Use a) and the fact that quadrilateral ABCD can be turned into a quadrilateral with the same area but with the sides in a different order: a, b, d, and c (just cut it along diagonal AC and "turn over" one of the halves). 42. Since bc/2 ;::: 1 we get b2 ;::: 2. 43. Yes, this is possible. Consider triangle ABC, where AC = 2002 + e, and AB= BC = 1001 (where e is some sufficiently small positive number; for example, € = 0.1). 44. Cut ABCD along diagonal AC and prove this equality for each of the halves separately. Do not forget that the sides of K LM N are parallel to the diagonals of ABCD. 45. It is not difficult to show that the area of a quadrilateral whose diagonals are perpendicular is half the product of the diagonals (the result for a rhombus, given in many regular texts, is a special case). Here, IABCDI = 12. 46. Answer: 7. Hint: prove that the area of each of the three additional triangles is 2. 47. The equality of the areas is equivalent to the equality of the altitudes dropped to BM from A and C respectively. This, in turn, is equivalent to the assumption that BM bisects AC. 48. Use the result of Problem 44. 4g. Hint: prove that triangles ABD and ACD have the same area. 50. If we are given point 0 inside equilateral triangle ABC, then we can calculate the area of ABC as the sum of the areas of triangles OAB, OBC, and OAC. These areas can be found using the perpendiculars dropped from 0 to the sides of the triangle.MATH1'MATICAL CIRCLES (RUSSIAN EXPERIENCE)26455. No, he is wrong. For example, we can prove that there are points on the plane satisfying the given property and lying arbitrarily far from the given point (in fact, the given set is a parabola). 56. The reason why the proof is wrong is that the figure is wrong, and point M lies outside the triangle. 57. Since P is equidistant from A and B as well as from C and D we have that triangles PAD and PBC are congruent. Thus, medians PM and PN in these triangles are equal as well. 58. One way to do this is to prove that this perpendicular bisector divides the greater leg of the triangle into two segments such that the length of one of them equals the length of the specified part of the bisector, and the other one is twice as long. Or, the lengths can be calculated directly, in terms of one of the sides of the original triangle. 59. Hint: make use of the fact that each of the segments of this broken line is a median to the hypotenuse in some right triangle and therefore equals half of this hypotenuse.15. NUMBER BASES Answers to the exercises 1. a) 2; b) n.= 21, 101013 = 91, 2114 = 37, 1267 = 69, 15811 = 184. = 11001002 = 102013 = 12104 = 4005 = 244" = 2027 = 144s = 1219.2. 101012 3. 100104. lllw =Alu. 5. Here is the multiplication table in the base 5 number system:0 1 2 3 4021 (000 3 46. a) 110012; b) 212023. 7. a) 2626 7; b) 10037.0 0H!!lProblems 1. Answer: in the base 12 system (duodecimal system). Hint: digits 3 and 4 always represent numbers 3w and 4w, and their product equals 12w. 2. a) Yes, such a system exists. This is the base 7 system. See the hint to the previous problem. · . b) Answer: No. This equality could be true only in the base 5 number system, but there is no digit 5 in this system. 3. A number is even if and only if a) there is an even number of l's in its base 3 representation (that is, the sum of its digit is even). Indeed, a number equals the sum of powers of 3 multiplied by the digits, which can be 0, l, or 2. The summands with digits O and 2 are even,ANSWERS, HINTS, SOLUTIONS265and therefore the parity of the sum depends on the number of the summands with digits l. b) for even n its base n representation ends with an even digit; for odd n the sum of its digits is even. The proof for the latter case is similar to the proof for part a). In case of even n, when a number is represented as a sum of powers of n multiplied by its digits, all summands starting from the second are even since they are divisible by n. Therefore, the parity of the sum is determined by the parity of the units digit. 4. The answer is 23451+15642 = 42423 (the base 7 number system).5. Let n be the base of the system. Then n 2 = (2n + 4) + (3n + 2); that is, n 2 - 5n -6 = 0. Therefore n = -1 or n = 6. Answer: n = 6. 6. a) In the base n number system the representation of a number ends with k zeros if and only if this number is divisible by nk. b) Let m be some divisor of n. The last digit of the base n representation of a number is divisible by m if and only if the number itself is divisible by m.7. a) Let m be a divisor of n - l. Then the sum of the digits in the base n representation of a number is divisible by m if and only if the number itself is divisible by m. b) The "alternating" sum (with alternating signs) of the digits in the base n representation of a number is divisible by n + 1 if and only if the number itself is divisible by n + l. c) Let m be some divisor of n+ l. The alternating sum (with alternating signs) of the digits in the base n representation of a number is divisible by m if and only if the number itself is divisible by m. 12. Hint: the subset is the same as in Problem 11. 13. a) This is the same game of Nim, with eight heaps instead of three. The strategy and the proof are exactly the same. However, there is another, much simpler proof which shows that the second player wins. Indeed, all the second player must do is to maintain line symmetry on the board (with respect to the line separating the fourth and the fifth columns). b) The second player cannot lose. The reason is the same as before. Actually, this game has nothing to do with the game of Nim-it is just a joke. In fact, this game can last forever, but this is not important. 16. INEQUALITJES2. a) We have = 9 > 8 = 23 , and therefore 3200 > 2300 . b) We have 210 = 1024 < 2187 = 37 , and therefore 2•0 < 328 . c) The number 453 is greater. 324. Answer: 891 > 792 • 6. Answer: 123111 2-- 32 1 = 1, 133 - 52 1 = 2, 16 2 53 1=4, and 111 2 - 27 1=7.-25 1 = 4, 133-25 1= 5, 127-53 1 = 3,8. Let us denote the number in the numerator of either fraction by x. Then the fraction itself is a= x/(lOx - 9), so that l/a = 10 - 9/x. This implies that as x increases, a decreases. Thus the first fraction is greater.266MATHEMATICAL CIRCLES (RUSSIAN EXPERIENCE)9. Let us answer the more general question: when isx/y greater than (x+l)/(y+l)? If x and y are positive, thenx-y y(y+l). Hence everything depends on whether or not xis greater than y. In our case y > x, which means that 1234568/7654322 is the greater of the two given numbers. 10. The number 100100 is greater, since 1002 > 150 · 50. 11. Answer: (1.01) 1000 > 1000. Indeed, (l.01)8 > 1.08; (1.01) 1000 = ((1.01)8) 125 >" (1.08) 125 . FUrthermore, (l.08) 5 > 1.4; (1.01) 1000 > (1.4) 25 > (1.4) 24 > (2.7) 8 > 7' = 2401 > 1000. 13. Since 99! > 100, it is clear that A < B. 17. We can write 1 + x - 2y'x = ( y'x - 1) 2 ;:: O. 19, 20. Carrying everything over to one side, we can reduce the given inequality to (x - y) 2 2: 0. 21. Carrying everything over to one side and multiplying by the denominator we have (x - y) 2 2: 0. 23. Multiply the following three inequalities: a+ b 2: 2Vab, b + c 2: 2VbC, c + a 2:2v'Cii.24. Hint: use the fact that (v'aii - ,/Oi;) 2 + (,/Uc- y'b(;)2 + (y'bC - v'aii) 2 ;:: O. 25. We have x 2 + y 2 + 1 - xy - x -y = ((x -y) 2 + (x - 1) 2 + (y - 1) 2 )/2;:: O. 27. We have x 4 + y 4 + 8 = x 4 + y 4 + 4 + 4 2: 4 {/x 4 y4 • 4 · 4 = 8xy.28. We can write a+b+c+d 2: 4~; l/a+ l/b+l/c+ l/d 2: 4{/l/abcd. Now we just multiply these inequalities. 29. a/b + b/c + c/a "?:. 3 v~(a-/=b)~·~(-b/_c_)-.(~c/~a~) = 3. 42. Hint: the inequality can be proved by adding up two simpler inequalities:(2k -1)(21 - 1)(2m - 1) > 0, 2k+l+m > 2k + 2' + 2m ' since 2k+l+m > 2k+ 2= 4. 2k > 2k + 21 + 2m (if k 2: l 2: m).43. We have ab+ be+ ca= ((a+ b+ c) 2 - a 2 -b2 - c2 )/2 = -(a2 + b2 + c2 )/2 ~ 0. 45. Carrying all the terms over to one side we get (x -y)( y'x - JY)/,/XY 2: 0. 47. Here is the main idea: if the permutation (c;) is not the identity, then there exist indices i and j such that c; > c; and i < j. Then by switching c; and c; we can increase the sum of the products. Indeed, c;a.;+ c;a; -c;a; - c;a; =(a; - a;)(c; - c;)< 0.Thus, using these transpositions, we can make the permutation (c;) into the identity permutation without decreasing the sum of the products during this process. 53. The base is easy. The proof of the inductive step goes as follows: 1 + 1/ ,/2 + ... + l/vn=l +l/.fii, < 2vn=l+l/.fii, < 2.fii, since l/.fii, < 2(.fii,-vn=t) = 2/(.fii, + v'n-1). 54. The solution is similar to the previous one Qust change the direction of the inequalities).ANSWERS, HINTS, SOLUTIONS26758. The base n = 4 can be checked "manually". The inductive step: (n + 1)! = (n+l)n! > 2n(n+l) > 2·2n =2n+l. 59. The base n = 1 is easy. The inductive step: 2n+l = 2.2n > 2·2n = 4n > 2(n+l) (if n > 1). 60. Answer: the inequality holds true for n ?: 10. Hint: you can check that it is true directly for 1 :5 n :5 10. To prove the inductive step, show that while 2n+i is twice as large as 2n, (n + 1) 3 is less than 2n3 .APPENDIX CReferences 1. General textsl. S.Barr, Rossypi golovolomok, Mir, 1978. (Russian) 2. G.Bizam, Y. Herczeg, lgra i logika, Mir, 1975. (Russian)3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.G.Bizam, Y.Herczeg, Mnogotsvetnaya logika, Mir, 1978. (Russian) N.Ya.Vilenkin, Rasskazy o mnozhestvakh, Nauka, 1969. (Russian) M.Gardner, Mathematical puzzles and diversions, Bell and Sons, London. M.Gardner, New mathematical diversions from Scientific American, Simon & Shuster, New York, 1966. M.Gardner, Matematicheskie novelly, Mir, 1974. (Russian) M.Gardner, aha! Gotcha, San Francisco, W.H.Freeman, 1982. M.Gardner, A nu-ka, dogadaisya!, Mir, 1984. (Russian) M.Gardner, The unexpected hanging and other mathematical diversions, Simon and Shuster, 1969. M.Gardner, Time travel and other mathematical bewilderments, San Francisco, W.H.Freeman, 1988. E.G.Dynkin, V.A.Uspenskii, Matematicheskie besedy, GITTL, 1952. (Russian) B.A.Kordemskii, Matematicheskaya smekalka, GITTL, 1958. (Russian) H.Lindgren, Geometric dissections, Princeton, Van Nostrand, 1964. H.Rademacher, O.Toeplitz, Von Zahlen und der Figuren, Princeton University Press, 1957. (English) R.Smullyan, What is the name of this book?, Prentice-Hall, 1978. R.Smullyan, The lady or the tiger, New York, Knopf, 1982. R.Smullyan, Alice in the Puzzle-land, New York, Morrow, 1982. V.L.Ufnarovskii, Matematicheskiy akvarium, Shtiintsa, 1987. (Russian) 2. For teachers20. V.A.Gusev, A.I.Orlov, A.L.Rozental', Vneklassnaya robota po matematike v 6-8 klassakh, Prosveshchenie, 1977, 1984. (Russian) 21. Various authors, Matematicheskiy kruzhok. Pervyi god obucheniya, 5-6 klassy, Izd-vo APN SSSR, 1990, 1991. (Russian) 22. G.Polya, How to solve it; a new aspect of mathematical method, Princeton University Press, 1945. 23. G.Polya, Mathematical discovery: on understanding, learning, and teaching problem solving, New York, Wiley, 1981. 24. G.Polya, Mathematics and plausible reasoning, Princeton University Press, 1968. 25. K.P.Sikorskii, Dopolnitel'nye glavy po kursu matematiki 1-8 klassov dlyafakul'tativnykh zanyatiy, Prosveshchenie, 1969. (Russian) 269270REFERENCES3. For younger students26. S.Bobrov, Volshebnyi dvurog, Detskaya literatura, 1967. (Russian) 27. V.Levshin, 7h dnya v Karlikanii. Skazka da ne skazka, Detskaya literatura, 1964. (Russian) 28. V.Levshin, E.Aleksandrova., Chemaya maska iz Al'-Dzhebry. Puteshestvie v pis'makh s prologom, Detskaya literatura, 1965. (Russian) 29. V.Levshin, Fregat kapitana Edinitsy, Detskaya literatura, 1968. (Russian) 30. V.Levshin, Magistr Rasseyannykh Nauk: matematicheskaya trilogiya, Detskaya literatura, 1987. (Russian) 4. Problem books31. I.L.Babinskaya, Zadachi matematicheskikh olimpiad, Nauka, 1975. (Russian) 32. D.Yu.Burago, S.M.Finashin, D.V.Fomin, Falcunativnyi kurs matematiki dlya 6-1 klassov v zadachakh, Izd-vo LGU, 1985. (Russian) 33. N.B.Vasil'ev, V.L.Gutenmakher, Zh.M.Rabbot, A.L.Toom, Zaochnye matematicheskie olimpiady, Nauka, 1986. (Russian) 34. N.B.Vasil'ev, S.A.Molchanov, A.L.Rozental', A.P.Savin, Matematicheskie sorevnovaniya (geometriya), Nauka, 1974. (Russian) 35. G.A.Gal'perin, A.K.Tolpygo, Moskovskie matematicheskie olimpiady, Prosveshchenie, 1986. (Russian) 36. P. Yu.Germanovich, Sbomik zadach po matematike na soobrazitel 'nost ', Uchpedgiz, 1960. (Russian) 37. E.B.Dynkin, S.A.Molchanov, A.L.Rozental', Matematicheskie sorevnovaniya. Arifmetika i algebra, Nauka, 1970. (Russian) 38. E.B.Dynkin, S.A.Molchanov, A.L.Rozental', A.K.Tolpygo, Matematicheskie zadachi, Nauka, 1971. (Russian) 39. G.l.Zubelevich, Sbomik zadach Moskovskikh matematicheskikh olimpiad (VVIII klassy), Prosveshchenie, 1971. (Russian) 40. A.A.Leman, Sbomik zadach Moskovskikh matematicheskikh olimpiad, Prosveshchenie, 1965. (Russian) 41. A.l.Ostrovskii, 75 zadach po yelementamoy matematike - prostykh, no ... , Prosveshchenie, 1966. (Russian) 42. V.V.Prasolov, Zadachi po planimetrii. Chasti 1, 2, Nauka, 1986. (Russian) 43. l.S.Rubanov, V.Ya.Gershkovich, l.E.Molochnikov, Metodicheskie materialy dlya vneklassnoy raboty so shkol 'nikami po matematike, Leningradskiy dvorets pionerov, 1973. (Russian) 44. l.N.Sergeev, S.N.Olekhnik, S.B.Gashkov, Primeni matematiku, Nauka, 1989. (Russian) 45. D.O.Shklyarskii, N.N.Chentsov, I.M.Yaglom, Izbrannye zadachi i teoremy yelementamoy matematiki. Arifmetika i algebra, Nauka, 1965. (Russian) 46. H.Steinhaus, One hundred problems in elementary mathematics, New York, Basic Books, 1964. ' 5. For "Combinatorics" 47. 48. 49. 50.N.Ya.Vilenkin, Kombinatorika, Nauka, 1964. (Russian) N.Ya.Vilenkin, Kvant 1 (1971). (Russian) N.Ya.Vilenkin, Populyamaya kombinatorika, Nauka, 1975. (Russian) 1.1.Ezhov et al., Yelementy kombinatoriki, Nauka, 1977. (Russian)REFERENCES27151. V.A.Uspenskii, Treugol'nik Paskalya ("Populyarnye lektsii po matematike", 43), Nauka. (Russian)6. For "Divisibility" 52. M.I.Bashmakov, Nravitsya li vam vozit'sya s tselymi chislami?, Kvant 3 (1971). (Russian) 53. V.N.Vaguten, Algoritm Evklida i osnovnaya te01-ema arifmetiki, Kvant 6 (1972). (Russian) 54. N.N.Vorob'ev, Priznaki delimosti ("Populyarnye lektsii po matematike'', 39), Nauka, 1963. (Russian) 55. A.O.Gel'fond, Reshenie uravneniy v tselykh chislakh ("Populyarnye lektsii po matematike", 8), Nauka, 1983. (Russian) 56. A.A.Egorov, Sravneniya po modulyu i ariftnetika ostatkov, Kvant 5 (1970). (Russian) 57. L.A.Kaluzhnin, Osnovnaya teorema arifmetiki ("Populyarnye lektsii po matematike", 41), Nauka, 1969. (Russian) 58. O.Ore, Invitation to number theory, New York, Random House, 1967. See also .7. J!br "The Pigeon Hole Principle" 59. V.G.Boltyanskii, Shest' zaitsev v pyati kletkakh, Kvant 2 (1977). (Russian) 60. A.I.Orlov, Printsip Dirikhl.e, Kvant 3 (1971). (Russian) See also [19, 20]; , chapter 20; .8. For "Graphs" 61. L.Yu.Berezina, Grafy i ikh primenenie, Prosveshchenie, 1979. (Russian) 62. O.Ore, Graphs and their use, New York, Random House, 1963. 63. R.Wilson, Introduction to graph theory, New York, Academic Press, 1972. See also [12, 20].9. For "Geometry" 64. J.Hadamard, Le9ons de geometrie elementaire, Paris, Armand Colin et Cie., 1898-1901. (French) 65. V.A.Gusev et al., Sbornik zadach po geometrii dlya 6-8 klassov, Prosveshchenie, 1975. (Russian) 66. P.R.Kantor, Zh.M.Rabbot, Ploshchadi mnogougol'nikov., Kvant 2 (1972). (Russian) 67. H.S.M.Coxeter, Introduction to geometry, New York, Wiley, 1961. 68. H.S.M.Coxeter, S.L.Greitzer, Geometry revisited, New York, Random House, 1967. 69. D.Pedoe, Geometry and visual arts, New York, Dover Publications, 1983. 70. D.O.Shklyarskii, N.N.Chentsov, I.M.Yaglom, lzbrannye zadachi i teoremy planimetrii, Nauka, 1967. (Russian) 71. I.M.Yaglom, Geometricheskie preobrazovaniya, t.1., Gostekhizdat, 1955. (Russian) See also [34, 42, 44].272REFERENCES10. For "Games" 72. E.Ya.Gik, Zanimatel'nye matematicheskie igry, Znanie, 1987. (Russian) 73. V.N.Kasatkin, L.I.Vladykina, Algori.tmy i igry, Radyan'ska shkola, 1984. (Russian) 74. A.I.Orlov, Stav' na minus, Kvant 3 (1977). (Russian) See also ; , chapters 8, 14, 39; , chapters 5, 30, 33; , chapters 7, 18, 23, 26, 33, 36. 11. For "Induction"75. N.N.Vorob'ev, Ghisla Fibonachchi ("Populyamye lektsii po matematike", 6}, Nauka, 1978. (Russian) 76. L.I.Golovina, I.M.Yaglom, lnduktsiya v geometri.i ("Populyamye lektsii po matematike", 21}, GITTL, 1956. (Russian) 77. A.I.Markushevich, Vozvratnye posledovatel'nosti ("Populyamye lektsii po matematike", 1), Nauka, 1975. (Russian) 78. LS.Sominskii, Metod matematicheskoy induktsii {"Populyamye lektsii po matematike", 3}, Nauka, 1965, 1974. (Russian) 79. LS.Sominskii, L.I.Golovina, I.M.Yaglom, 0 matematicheskoy induktsii, Nauka, 1967. (Russian) See also ; , chapter 27.12. For "Invariants" 80. Yu.Llonin, L.D.Kurlyandchik, Poisk invari.anta, Kvant 2 (1976). (Russian) 81. A.K.Tolpygo, lnvari.anty, Kvant 12 (1976). (Russian) See also ; , chapter 22.13. For "Number bases" 82. S.V.Fomin, Sistemy schisleniya ("Populyamye lektsii po matematike", 40), Nauka, 1968. (Russian) 83. l.M.Yaglom, Sistemy schisleniya, Kvant 6 (1970). (Russian) 84. l.M.Yaglom, Dve igry so spichkami, Kvant 2 (1971). (Russian) See also , chapter 35; , chapter 14; [25, 58].14. For "Inequalities" 85. E.Beckenbach, R.Bellman, An introduction to inequalities, Berlin, SpringerVerlag, 1961. 86. E.Beckenbach, R.Bellman, Inequalities, New York, Random House, 1961. 87. P.P.Korovkin, Vvedenie v neravenstva ("Populyamye lektsii po matematike", 5), Nauka, 1983. (Russian) 88. V.A.Krechmar, Zadachi po algebre, Nauka, 1964. (Russian) 89. G.L.Nevyazhskii, Neravenstva, Kvant 12 (1985). (Russian)Added for the American edition 90. K.Ireland, M.Rosen, A classical introduction to modem number theory, SpringerVerlag, 1982.VVelcome the AMS Series . .... . . .. ...to . . . . . . . . . . . . . . . . . . . . . ..This exciting series brings the beauty and captivation of mathematics to the advanced high school student, the undergraduate, the mathematics teacher, the scientist or the engineer, and to the lay reader with a strong interest in mathematics. Mathematical World features well-written, challenging expository works that capture the fascination and usefulness of mathematics. 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188754 | https://web.williams.edu/Mathematics/sjmiller/public_html/math/papers/ZeckCompletenessPLRS01.pdf | COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES EL˙ ZBIETA BOŁDYRIEW, JOHN HAVILAND, PHÚC LÂM, JOHN LENTFER, STEVEN J. MILLER, AND FERNANDO TREJOS SUÁREZ ABSTRACT. A sequence of positive integers is complete if every positive integer is a sum of distinct terms. A PLRS is a sequence defined by a homogeneous linear recurrence relation with positive coefficients and a particular set of initial conditions. Famously, the Fibonacci sequence Fn+1 = Fn + Fn+1 is complete by Zeckendorf’s theorem.
With results on how completeness is affected by modifying the recurrence coefficients of a PLRS, we determine which sequences generated by coefficients of the forms [1, . . . , 1, 0, . . . , 0, N] and [1, 0, . . . , 0, 1, . . . , 1, N] are complete. Further, we conjecture bounds for other maximal last coeffi-cients in complete sequences in other families of PLRS’s. Our primary method is applying Brown’s criterion, which says that an increasing sequence {Hn}∞ n=1 is complete if and only if H1 = 1 and Hn+1 ≤1 + Pn i=1 Hi.
Finally, we adopt previous analytic work on PLRS’s to find a more efficient way to check com-pleteness. Specifically, the characteristic polynomial of any PLRS has exactly one positive root.
Through combinatorial and analytical arguments, we determine some criteria for completeness based on the size of the principal root, including results that show there is an indeterminate region where the principal root does not reveal any information.
CONTENTS 1.
Introduction 2 2.
Modifying Sequences 4 2.1.
Maximal Complete Sequence 4 2.2.
Modifications of Sequences of Arbitrary Coefficients 6 3.
Families of Sequences 9 3.1.
Using 1’s and 0’s as Initial Coefficients 9 3.2.
The “2L −1 Conjecture” 18 4.
An Analytical Approach 23 4.1.
An Introduction to Principal Roots 23 4.2.
Applications to Completeness 24 4.3.
Denseness of Incomplete Roots 30 5.
Open Questions 31 Appendix A.
Brown’s Criterion and a Corollary 31 Appendix B.
Lemmas for Section 2 32 Appendix C.
Lemmas for Section 3 35 Appendix D.
Lemmas for Section 4 44 References 44 Date: September 23, 2020.
Key words and phrases. Positive linear recurrence sequences, complete sequences, Brown’s criterion, characteristic polynomial.
This research was conducted as part of the SMALL 2020 REU at Williams College. The authors were supported by NSF Grants DMS1947438 and DMS1561945, Williams College, Yale University, and the University of Rochester.
1 2 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ 1. INTRODUCTION The Fibonacci numbers are one of the most studied integer sequences. One of their many in-teresting properties is that they can be used to construct a unique decomposition for any positive integer. Zeckendorf proved that every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers, when indexing Fibonacci numbers {1, 2, 3, 5, . . . }; this unique decomposition is called the Zeckendorf decomposition [Ze]. This result, of unique decomposi-tions, has been generalized to a much larger class of linear recurrence relations; the following definitions are from [MW].
Definition 1.1. We say a sequence {Hn}∞ n=1 of positive integers is a Positive Linear Recurrence Sequence (PLRS) if the following properties hold: (1) Recurrence relation: There are non-negative integers L, c1, . . . , cL such that Hn+1 = c1Hn + · · · + cLHn+1−L, (1.1) with L, c1 and cL positive.
(2) Initial conditions: H1 = 1, and for 1 ≤n < L we have Hn+1 = c1Hn + c2Hn−1 + · · · + cnH1 + 1.
(1.2) Definition 1.2 (Legal decompositions). We call a decomposition Pm i=1 aiHm+1−i of a positive integer N (and the sequence {ai}m i=1) legal if a1 > 0, the other ai ≥0, and one of the following two conditions holds: (1) We have m < L and ai = ci for 1 ≤i ≤m.
(2) There exists s ∈{1, . . . , L} such that a1 = c1, a2 = c2, · · · , as−1 = cs−1 and as < cs, (1.3) as+1, . . . , as+ℓ= 0 for some ℓ≥0, and {bi}m−s−ℓ i=1 (with bi = as+ℓ+i) is legal or empty.
The following theorem is due to [GT].
Theorem 1.3 (Generalized Zeckendorf’s Theorem for PLRS). Let {Hn}∞ n=1 be a Positive Linear Recurrence Sequence. Then there is a unique legal decomposition for each positive integer N ≥0.
Next, we introduce completeness, as defined by [HK].
Definition 1.4. An arbitrary sequence of positive integers {fi}∞ i=1 is complete if and only if every positive integer n can be represented in the form n = P∞ i=1 αifi, where αi ∈{0, 1}. A sequence that fails to be complete is incomplete.
In other words, a sequence of positive integers is complete if and only if each positive integer can be written as a sum of unique terms of the sequence.
Example 1.5. The Fibonacci sequence, indexed from {1, 2, . . .} is complete. This follows from Zeckendorf’s Theorem, which is a stronger statement. Completeness does not require that the decompositions be unique, and that they use nonconsecutive terms.
After seeing this example, it is natural to ask if Theorem 1.3 implies that all PLRS’s are com-plete. Previous work in numeration systems by Gewurz and Merola [GM] has shown that specific classes of recurrences as defined by Fraenkel [Fr] are complete under their greedy expression.
However, we cannot generalize this result to all PLRS’s. For legal decompositions, the decompo-sition rule can permit sequence terms to be used more than once. This is not allowed for complete-ness decompositions, where each unique term from the sequence can be used at most once.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 3 Example 1.6. The PLRS Hn+1 = Hn + 3Hn−1 has terms {1, 2, 5, 11, . . .}. The unique legal decomposition for 9 is 1·5+2·2, where the term 2 is used twice. However, no complete decompo-sition for 9 exists. Adding all terms from the sequence less than 9 is 1 + 2 + 5 = 8, and to include 11 or any subsequent term surpasses 9.
We also make use of the following criterion for completeness of a sequence, due to Brown [Br].
Theorem 1.7 (Brown’s Criterion). If an is a nondecreasing sequence, then an is complete if and only if a1 = 1 and for all n > 1, an+1 ≤1 + n X i=1 ai.
(1.4) An immediate corollary is the following sufficient, though not necessary condition for complete-ness, which we call the doubling criterion. The proof is left to the appendix, as Corollary A.2.
Corollary 1.8 (Doubling Criterion). If an is a nondecreasing sequence such that an ≤2an−1 for all n ≥2, then an is complete.
Remark 1.9. By considering the special case when an = 2an−1, this immediately implies that the doubling sequence itself {1, 2, 4, 8, . . .} is complete.
In this paper, we characterize many types of PLRS by whether they are complete or not complete.
Notation 1.10. We use the notation [c1, . . . , cL], which is the collection of all L coefficients, to represent the PLRS Hn+1 = c1Hn + · · · + cLHn+1−L.
A simple case to consider is when all coefficients in [c1, . . . , cL] are positive. The following result, proved in Section 2, completely characterizes these sequences are either complete or in-complete.
Theorem 1.11. If {Hn} is a PLRS generated by all positive coefficients [c1, . . . , cL], then sequence is complete if and only if the coefficients are [1, . . . , 1 | {z } L ] or [1, . . . , 1 | {z } L−1 , 2] for L ≥1.
The situation becomes much more complicated when we consider all PLRS’s, in particular those that have at least one 0 as a coefficient. In order to be able to make progress on determining completeness of these PLRS’s, we develop several tools. The following three theorems are results that allow certain modifications of the coefficients [c1, . . . , cL] that generate a PLRS that is known to be complete or incomplete, and preserve completeness or incompleteness. They are proved in Section 2.
Theorem 1.12. Consider sequences {Gn} = [c1, . . . , cL] and {Hn} = [c1, , . . . , cL, cL+1], where cL+1 is any positive integer. If {Gn} is incomplete, then {Hn} is incomplete as well.
Theorem 1.13. Consider sequences {Gn} = [c1, . . . , cL−1, cL] and {Hn} = [c1, . . . , cL−1, kL], where 1 ≤kL ≤cL. If {Gn} is complete, then {Hn} is also complete.
Theorem 1.14. Consider sequences {Gn} = [c1, . . . , cL−1, cL] and {Hn} = [c1, . . . , cL−1 + cL].
If {Gn} is incomplete, then {Hn} is also incomplete.
The next two theorems are results that classify two families of PLRS’s as complete or incom-plete. They are shown in Section 3.
4 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Theorem 1.15. The sequence generated by [1, 0, . . . , 0 | {z } k , N] is complete if and only if 1 ≤N ≤ ⌈(k + 2)(k + 3)/4⌉.
Theorem 1.16. The sequence generated by [1, 1, 0, . . . , 0 | {z } k , N] is complete if and only if 1 ≤N ≤ ⌊(Fk+6 −k −5)/4⌋, where Fn are the Fibonacci numbers with F1 = 1, F2 = 2.
We have a partial extension of these theorems to when there are g initial ones followed by k zeroes in the collection of coefficients.
Theorem 1.17. Consider a PLRS generated by coefficients [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , N], with g, k ≥1.
(1) For g ≥k + ⌈log2 k⌉, the sequence is complete if and only if 1 ≤N ≤2k+1 −1.
(2) For k ≤g ≤k + ⌈log2 k⌉, the sequence is complete if and only if 1 ≤N ≤2k+1 −⌈k/2g−k⌉.
Finally, in Section 4, we introduce some results and conjectures on completeness based on the principal roots of a PLRS. We determine some criteria for completeness based on the size of the principal root and find that there is a certain indeterminate region where the principal root does not reveal any information.
2. MODIFYING SEQUENCES A basic question to ask is how far we can tweak the coefficients used to generate a sequence, yet preserve its completeness. The modifying process turns out to be well-behaved and heavily depen-dent on the location of coefficients that are changed. Before we start looking into implementing any changes to our sequences, we first need to understand the maximal complete sequence.
2.1. Maximal Complete Sequence. We introduce the maximal complete sequence, which serves an important role. First, we look at all complete sequences with only positive coefficients.
Proof of Theorem 1.11. Assume that {Hn} is complete.
By the definition of a PLRS and by Brown’s criterion, we have c1HL−1 + c2HL−2 + · · · + cL−1H1 + 1 = HL ≤1 + H1 + H2 + · · · + HL−1.
(2.1) Since ci ≥1 for 1 ≤i ≤L, this implies that ci = 1 for 1 ≤i < L. By the definition of a PLRS, HL+1 = c1HL + c2HL−1 + · · · + cLH1 = HL + HL−1 + · · · + H2 + cLH1.
(2.2) Combining this with Brown’s criterion gives HL+1 = HL + HL−1 + · · · + cLH1 ≤1 + H1 + H2 + · · · + HL−1 cLH1 ≤1 + H1 = 2.
(2.3) Hence cL ≤2, which completes the forward direction of the proof.
We know that if the coefficients are just , then the sequence is complete by Remark 1.9. So, now assume that c1 = · · · = cL−1 = 1 and 1 ≤cL ≤2. We argue by strong induction on n that Hn satisfies Brown’s criterion. We can show this explicitly for 1 ≤n < L. First, if n = 1, then Hn = 1, as desired. Next, if 1 ≤n < L, then Hn+1 = c1Hn + · · · + cnH1 + 1 = Hn + · · · + H1 + 1, (2.4) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 5 so these terms satisfy Brown’s criterion. Now assume that for some n ≥L, for all n′ < n, Hn′+1 ≤Hn′ + · · · + H1 + 1.
(2.5) It follows that Hn+2 = Hn+1 + · · · + Hn+2−L + cLHn+1−L ≤Hn+1 + · · · + Hn+2−L + 2Hn+1−L ≤Hn+1 + · · · + Hn+2−L + Hn+1−L + (Hn−L + · · · + H1 + 1), (2.6) where the inductive hypothesis was applied to Hn+1−L to obtain (2.6). This completes the induc-tion.
□ Now that we have found some complete sequences, it turns out that the sequence generated by the coefficient , i.e., {2n−1}, is the maximal complete sequence.
Lemma 2.1. The complete sequence with largest span in summands is {2n−1}.
Proof. Suppose there exists a complete sequence {Hn} with the largest span in summands. As a complete sequence must satisfy Brown’s criterion, it suffices to take Hn+1 = 1 + Pn i=1 Hi. Hence, Hn+1 = 1 + n X 1 Hi = 1 + n−1 X 1 Hi + Hn = 2Hn.
(2.7) By the intial conditions for a PLRS, H1 = 1 and H2 = 2. Thus, Hn = 2Hn−1 = 2n−1.
□ Remark 2.2. Thus {Hk} = {2k−1} is an inclusive upper bound for any complete sequence.
As it turns out, this sequence has can be generated by multiple collections of coefficients.
Corollary 2.3. A PLRS with coefficients [1, . . . , 1 | {z } L−1 , 2] generates the sequence Hn = 2n−1.
Proof. Consider the sequence {Hn} generated by [1, . . . , 1 | {z } L−1 , 2]. We proceed by induction on L.
Note H1 = 1 = 21−1 by the definition of the PLRS. Now, suppose Hk = 2k−1 for k ∈{1, . . . , n}.
For n < L, note Hn+1 = c1Hn + c2Hn−1 + · · · + cnH1 + 1 = Hn + Hn−1 + · · · + H1 + 1 = 2n−1 + 2n−2 + · · · + 1 + 1 = 2n (2.8) Hence, the claim holds for all n < L. Now, for n ≥L, note Hn+1 = c1Hn + c2Hn−1 + · · · + cLHn+1−L = Hn + Hn−1 + · · · + 2Hn+1−L = 2n−1 + 2n−2 + · · · + 2n−L+1 + 2 · 2n−L = 2n (2.9) Thus, by induction, the claim holds for all n, L ∈N.
□ 6 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ 2.2. Modifications of Sequences of Arbitrary Coefficients. Modifying coefficients in order to preserve completeness proves to be a balancing act. Sometimes increasing a coefficient causes an incomplete sequence to become complete, while other times, increasing a coefficient causes a complete sequence to become incomplete. For example, [1, 0, 0, 0, 0, 0, 15] is incomplete; increas-ing the second coefficient to 1, i.e., [1, 1, 0, 0, 0, 0, 15] is complete. Further increasing it to 2, i.e., [1, 2, 0, 0, 0, 0, 15] is again incomplete. To study how such modifications preserve completeness or incompleteness, we add a new definition to our toolbox.
Definition 2.4. For a sequence {Hn}, we define its nth Brown’s gap BH,n := 1 + n−1 X i=1 Hi −Hn.
(2.10) Thus, from Brown’s criterion, {Hn} is complete if and only if BH,n ≥0 for all n ∈N.
Our next questions is: What happens if we append one more coefficient to [c1, . . . , cL]? It turns out that if our sequence is already incomplete, appending any new coefficients will never make it complete. This is Theorem 1.12, which using are ready to prove using Brown’s gap.
Proof of Theorem 1.12. By Brown’s criterion, it is clear that {Gn} is incomplete if and only if there exists n such that BG,n < 0. We claim that for all m, BH,m ≤BG,m. If true, our lemma is proven: suppose BG,n < 0 for some n, we would see BH,n ≤BG,n < 0, implying {Hn} is incomplete as well.
We proceed by induction. Clearly, BH,k = BG,k for 1 ≤k ≤L. Further, for k = L, we see BG,L+1−BH,L+1 = 1+ L X i=1 Gi−GL+1− 1 + L X i=1 Hi −HL+1 !
= HL+1−GL+1 = 1 > 0. (2.11) Now, let m ≥2 be arbitrary, and suppose BH, L+m−1 ≤BG, L+m−1.
(2.12) We wish to show that BH, L+m ≤BG, L+m. Note that BH, L+m −BH, L+m−1 = 2HL+m−1 −HL+m.
(2.13) Similarly, BG, L+m −BG, L+m−1 = 2GL+m−1 −GL+m.
(2.14) We use Lemma B.1, which states that for all k ≥2, HL+k −GL+k ≥2 (HL+k−1 −GL+k−1).
Applying it to equations (2.13) and (2.14), we see that BH, L+m−BH, L+m−1 ≤BG, L+m−BG, L+m−1. Summing this inequality to both sides of inequality (2.12), we arrive at BH,L+m ≤BG,L+m, as desired.
□ Now, we turn our attention to the behavior when we decrease the last coefficient for any complete sequence. In Theorem 1.13, we find that decreasing the last coefficient for any complete sequence preserves completeness.
Proof of Theorem 1.13. Given that {Gn} is complete, suppose for the sake of contradiction that there exists an incomplete {Hn}. Thus, let m be the least such that Hm > 1 + m−1 X i=1 Hi.
(2.15) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 7 Simultaneously, as {Gn} is complete, by Brown’s criterion, Gm ≤1 + m−1 X i=1 Gi.
(2.16) First, suppose m ≤L. However, for all n ≤L, Gn = Hn, hence Hm = Gm ≤1 + m−1 X i=1 Gi = 1 + m−1 X i=1 Hi, (2.17) which contradicts (2.15). Now, suppose m > L. Therefore, Gm ≤1 + m−1 X i=1 Gi = 1 + L X i=1 Gi + m−1 X i=L+1 Gi = 1 + L X i=1 Hi + m−1 X i=L+1 Gi.
This implies, 1 + L X i=1 Hi ≥Gm − m−1 X i=L+1 Gi.
(2.18) Now, we know that Hm > 1 + m−1 X i=1 Hi = 1 + L X i=1 Hi + m−1 X i=L+1 Hi ≥Gm − m−1 X i=L+1 Gi + m−1 X i=L+1 Hi, (2.19) and thus Hm − m−1 X i=L+1 Hi > Gm − m−1 X i=L+1 Gi.
(2.20) We claim that the opposite of (2.20) is true, arguing by induction on m. For m = L + 1, we obtain GL+1 ≥HL+1 as kL ≤cL. Now, assume that Gm − m−1 X i=L+1 Gi ≥Hm − m−1 X i=L+1 Hi (2.21) is true for a positive integer m. Using the inductive hypothesis, it then follows that Gm+1 − m X i=L+1 Gi = Gm+1 − m−1 X i=L+1 Gi −Gm ≥Gm+1 −2Gm + Hm − m−1 X i=L+1 Hi.
(2.22) Finally, we use Lemma B.2, proved in Appendix B, which states that for all k ∈N, HL+k+1 − 2HL+k ≤GL+k+1 −2GL+k. Note Gm+1 −2Gm + Hm − m−1 X i=L+1 Hi ≥Hm+1 −2Hm + Hm − m−1 X i=L+1 Hi = Hm+1 − m X i=L+1 Hi, (2.23) which does contradict (2.20) for all m > L. Therefore, for all m ∈N, we have contradicted (2.15).
Hence, {Hn} must be complete as well.
□ The result above is crucial in our characterization of families of complete sequences in Section 3; finding one complete sequence allows us to decrease the last coefficient to find more. Next, we prove two lemmas that together prove Theorem 1.14.
8 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Lemma 2.5. Let {Gn} be the sequence defined by [c1, . . . , cL], and let {Hn} be the sequence defined by [c1, . . . , cL−1 + 1, cL −1]. If {Gn} is incomplete, then {Hn} must be incomplete as well.
Proof. We claim that for all m, BH,m ≤BG,m. This lemma is proven using similar reasoning as for Lemma 1.12. We proceed by induction. Clearly, BH,k = BG,k for 1 ≤k ≤L −1. Further, for k = L, we see BG,L −BH,L = 1 + L−1 X i=1 Gi −GL − 1 + L−1 X i=1 Hi −HL !
= HL −GL = 1 > 0.
(2.24) Now, let m ≥0 be arbitrary, and suppose BH, L+m ≤BG, L+m.
(2.25) We wish to show that BH, L+m+1 ≤BG, L+m+1. Note that BH, L+m+1 −BH, L+m = 2HL+m −HL+m+1, (2.26) and similarly, BG, L+m+1 −BG, L+m = 2GL+m −GL+m+1.
(2.27) We use Lemma B.3, which says that for all k ≥0, HL+k+1 −GL+k+1 ≥2 (HL+k −GL+k).
Applying it to (2.26) and (2.27), we see BH, L+m+1 −BH, L+m ≤BG, L+m+1 −BG, L+m. Summing this inequality to both sides of inequality (2.25), we conclude that BH,L+m+1 ≤BG,L+m+1, as desired.
□ How many times can Lemma 2.5 be applied? The answer is all the way up to [c1, . . . , cL−1 + cL −1, 1], as the last coefficient must remain positive to stay a PLRS.
Lemma 2.6. Let {Gn} be the sequence defined by [c1, . . . , cL−1, 1], and let {Hn} be the sequence defined by [c1, . . . , cL−1 + 1]. If {Gn} is incomplete, then {Hn} must be incomplete as well.
Remark 2.7. Despite the similarities, Lemma 2.6 is not directly implied by Lemma 2.5; both are necessary for the proof Theorem 1.14. Applying Lemma 2.5 (cL −1) times proves that if [c1, . . . , cL−1, cL] is incomplete, then [c1, . . . , cL−1 + cL −1, 1] is incomplete; at this point, we can-not apply the lemma further while maintaining a positive final coefficient to meet the definition of a PLRS. Hence the case of Lemma 2.6 must be dealt with separately, in order to arrive at the full result of Theorem 1.14.
Proof. The proof is similar to that of Lemma 2.5. We aim to show that BH,m ≤BG,m for all m.
Clearly BH,k = BG,k for 1 ≤k ≤L. Further, for k = L + 1, we see BG,L+1 −BH,L+1 = L X i=1 Gi −GL+1 − 1 + L−1 X i=1 HL −HL+1 !
= HL+1 −GL+1 = c1 > 0. (2.28) Now, let m ≥0 be arbitrary, and suppose BH,L+m ≤BG,L+m.
(2.29) We wish to show that BH,L+m+1 ≤BG,L+m+1. Note that BH,L+m+1 −BH,L+m = 2HL+m −HL+m+1, (2.30) and similarly BG,L+m+1 −BG,L+m = 2GL+m −GL+m+1.
(2.31) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 9 We use Lemma B.4, which states that for all k ≥0, HL+k+1 −GL+k+1 ≥2 (HL+k −GL+k). Ap-plying it to equations (2.30) and (2.31), we see BH,L+m+1−BH,L+m ≤BG,L+m+1−BG,L+m. Sum-ming this inequality to both sides of Inequality (2.29), we conclude that BH,L+m+1 ≤BG,L+m+1, as desired.
□ Using these lemmas, we can now prove Theorem 1.14.
Proof of Theorem 1.14. We apply Lemma 2.5 cL −1 times, to conclude that if [c1, . . . , cL−1, cL] is incomplete, then [c1, . . . , cL−1 +cL −1, 1] is incomplete. Finally, applying Lemma 2.6, we achieve the desired result.
□ 3. FAMILIES OF SEQUENCES If we recall Theorem 1.13, it says that given a complete PLRS, decreasing the last coefficient preserves its completeness. This raises a natural question: Given the first L −1 coefficients c1, c2, . . . , cL−1, what is the maximal N such that [c1, c2, . . . , cL−1, N] is complete? In this sec-tion we explore this question.
3.1. Using 1’s and 0’s as Initial Coefficients.
Proof of Theorem 1.15. First assume that {Hn} is complete. By the definition of a PLRS, we can easily generate the first k + 2 terms of the sequence: Hi = i for all 1 ≤i ≤k + 2. We then have for all n > k + 1, Hn+1 = Hn + NHn−k−1, (3.1) which implies that Hk+4 = Hk+3 + NH2 = Hk+3 + 2N.
(3.2) By Brown’s criterion, Hk+4 ≤Hk+3 + Hk+2 + · · · + H1 + 1.
By (3.2), Hk+3 + 2N ≤Hk+3 + Hk+2 + · · · + H1 + 1, and we obtain 2N ≤Hk+2 + Hk+1 + · · · + H1 + 1 = (k + 2) + (k + 1) + · · · + 1 + 1 = (k + 2)(k + 3) 2 + 1, and thus we find N ≤(k + 2)(k + 3) 4 + 1 2.
(3.3) Since N is an integer and ⌊(k + 2)(k + 3)/4 + 1/2⌋= ⌈(k + 2)(k + 3)/4⌉, we may take the floor of the right hand side of equation (3.3), and then N ≤⌈(k + 2)(k + 3)/4⌉.
We now prove that if N ≤⌈(k + 2)(k + 3)/4⌉, then {Hn} is complete. We first show that if N = ⌈(k + 2)(k + 3)/4⌉, then {Hn} is complete. Taking the recurrence relation Hn+1 = Hn + NHn−k−1, and applying Brown’s criterion gives Hn+1 = Hn + NHn−k−1 ≤Hn + (N −2)Hn−k−1 + Hn−k−1 + Hn−k−2 + · · · + H1 + 1.
10 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ By Lemma C.1, we can expand (N −2)Hn−k−1 and find that Hn+1 ≤Hn + Hn−1 + · · · + Hn−k + Hn−k−1 + Hn−k−2 + · · · + H1 + 1.
(3.4) Hence, by Brown’s criterion, the sequence {Hn} is complete. Lastly, by Theorem 1.13, for all positive N < ⌈(k + 2)(k + 3)/4⌉, the sequence is also complete.
□ For coefficients as defined in Theorem 1.15, for sufficiently large L, if we switch any one of the coefficients from 0 to 1 except for the final zero, then the bound on N is at least as large, which we prove in the following corollary.
Corollary 3.1. For L ≥6, given that [1, 0, . . . , 0, N] is complete, with N = ⌈L(L + 1)/4⌉, then [1, c2, . . . , cL−2, 0, N] is complete where ci = 1 for one i ∈{2, . . . , L −2}, and the rest are 0.
Proof. We have the recurrence relation for fixed i ∈{2, . . . , L −2}: Hn+1 = Hn + Hn−i+1 + NHn−L+1.
(3.5) Applying Brown’s criterion yields Hn+1 ≤Hn + Hn−i+1 + (N −2)Hn−L+1 + Hn−L+1 + Hn−L + · · · + H1 + 1.
(3.6) We apply the result of Lemma C.3, and see that ≤Hn + Hn−1 + · · · + Hn−L+2 + Hn−L+1 + Hn−L + · · · + H1 + 1.
(3.7) Hence, by Brown’s criterion, the sequence is complete for all L ≥6.
□ Proof of Theorem 1.16. Suppose that {Hn} is complete. Using the definition of a PLRS, the first k+3 terms of the sequence can be generated in the same way: Hi = Fi+1−1 for all 1 ≤i ≤k+3, where Fn is the Fibonacci sequence. Proceeding in a manner similar to the proof of Theorem 1.15, we see that Hk+4 = Hk+3 + Hk+2 + NH1 = Fk+5 + N −2, Hk+5 = Hk+4 + Hk+3 + NH2 = Fk+6 + 3N −3, Hk+6 = Hk+5 + Hk+4 + NH3 = Fk+7 + 8N −5.
(3.8) By applying Brown’s criterion, Hk+6 ≤Hk+5 + Hk+4 + · · · + H1 + 1 = Fk+6 + 3N −3 + Fk+5 + N −2 + k+3 X i=1 Hi + 1 = Fk+7 + 4N −5 + k+3 X i=1 (Fi+1 −1) + 1.
(3.9) Next, Fk+7 + 8N −5 ≤Fk+7 + 4N −5 + k+3 X i=1 (Fi+1 −1) + F1, which implies 4N ≤ k+3 X i=1 (Fi+1 −1) + F1 = k+4 X i=1 Fi + (k + 3) = Fk+6 + (k + 5).
(3.10) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 11 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 g (number of ones) 0 500 1000 1500 2000 2500 3000 3500 4000 Maximal bound N Bound for 6 zeros Bound for 7 zeros Bound for 8 zeros Bound for 9 zeros Bound for 10 zeros Bound for 11 zeros FIGURE 1. [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , N] with k and g varying, where each color repre-sents a fixed k.
Thus N ≤Fk+6 −k −5 4 , (3.11) and since N is an integer, N ≤ Fk+6 −k −5 4 .
(3.12) Next, we show that if N = ⌊(Fk+6 −k −5)/4⌋, then {Hn} is complete. The initial conditions can be found easily, and for the later terms we have Hn+1 = Hn + Hn−1 + NHn−k−2 ≤Hn + (N −2)Hn−k−2 + Hn−k−2 + Hn−k−3 + · · · + H1 + 1.
Using Lemma C.4, we expand (N −2)Hn−k−2 and obtain ≤Hn + Hn−1 + Hn−2 + · · · + Hn−k−1 + Hn−k−2 + Hn−k−3 + · · · + H1 + 1.
(3.13) Hence, by Brown’s criterion, this sequence is complete. Lastly, by Theorem 1.13, for all positive N < ⌊(Fk+6 −k −5)/4⌋, the sequence is also complete.
□ We want to find a more general result for [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , N], as seen in Figure 1.
12 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Interestingly, we see that as we keep k fixed and increase g, the bound increases, and then stays constant from some value of g onward. This motivates the following conjecture.
Conjecture 3.2. If [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , N] is complete, then so is [1, . . . , 1 | {z } g+1 , 0, . . . , 0 | {z } k , N].
We have made some progress towards this conjecture; in fact, we show the precise bound for N for the case where g ≥k in theorem 1.17.
Theorem 3.3. The PLRS {Hn} generated by [c1, c2, . . . , cL] is complete if ( BH,n ≥0 for n < L BH,n > 0 for L ≤n ≤2L −1.
(3.14) Proof. Consider L ≥2; we see that if c1 ≥2, then the sequence is automatically incomplete, so we need only consider c1 = 1. For Bn := BH,n, and we show by induction on n that Bn > 0 when n ≥L. Suppose Bn > 0 for L ≤n ≤m (with m ≥2L −1). Then Bm+1 = 1 + m X i=1 Hi −Hm+1 = 1 + L X i=1 Hi + m X i=L+1 Hi − Hm + L X j=2 cjHm+1−j !
= 1 + L X i=1 Hi + m X i=L+1 Hi−1 + L X j=2 cjHi−j !
− Hm + L X j=2 cjHm+1−j !
= 1 + m−1 X i=1 Hi −Hm + HL !
+ L X j=2 cj m X i=L+1 Hi−j −Hm+1−j !
= (Bm + HL) + L X j=2 cj 1 + m X i=j+1 Hi−j −Hm+1−j −1 − L X i=j+1 Hi−j !
= (Bm + HL) + L X j=2 cj Bm+1−j −1 − L X i=j+1 Hi−j !
= Bm + L X j=2 cj(Bm+1−j −1) + HL − L X i=3 i−1 X j=2 cjHi−j = Bm + L X j=2 cj(Bm+1−j −1) + HL − L X i=3 (Hi −Hi−1 −1) = Bm + L X j=2 cj(Bm+1−j −1) + (L −2) + HL − L X i=3 (Hi −Hi−1) = Bm + L X j=2 cj(Bm+1−j −1) + L.
(3.15) The last line is positive since Bm+1−j−1 ≥0, Bm, L > 0. Our proof by induction is complete.
□ COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 13 Lemma 3.4. The PLRS {Hi} generated by [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , 2k+1] is incomplete for g ≥k ≥1.
Proof. Suppose this sequence is complete. Note that H2g+2 = H2g+1 + · · · + Hg+2 + 2k+1Hg+1−k.
(3.16) By applying Brown’s criterion to H2g+2, we see that 2k+1Hg+1−k ≤ g+1 X i=1 Hi + 1.
(3.17) Now, note k is positive, so that g+1−k ≤g+1. Also, by the structure of the sequence, Hi = 2i−1 for i ≤g + 1. Hence 2g+1 = 2k+1Hg+1−k = 2k+12g−k ≤ g+1 X i=1 2i−1 + 1 = 2g+1 (3.18) Therefore one may substitute previous inequalities with equalities and obtain H2g+2 = 2g+1 X i=1 Hi + 1.
(3.19) It follows immediately from (3.19) that 2g+2 X i=1 Hi + 1 = 2H2g+2.
(3.20) Now, consider H2g+3 = H2g+2 + H2g+1 + · · · + Hg+3 + 2k+1Hg+2−k.
(3.21) Since g + 2 −k ≤g + 1 as k ≥1, one gets Hg+2−kM = 2g+1−k2k+1 = 2g+2 = 2(2g+1) = 2Hg+2.
(3.22) Hence H2g+3 = H2g+2 + H2g+1 + · · · + Hg+3 + 2Hg+2 = H2g+2 + (H2g+1 + · · · + Hg+3 + Hg+2 + Hg+2) > H2g+2 + (H2g+1 + · · · + Hg+3 + Hg+2 + Hg+1−k) = 2H2g+2 = 2g+2 X i=1 Hi + 1.
(3.23) So H2g+3 causes Brown’s criterion to fail, rendering whole sequence incomplete.
□ We now show the stabilizing behavior of the bound mentioned above.
Lemma 3.5. If g ≥k + ⌈log2 k⌉, then [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , 2k+1 −1] is complete.
14 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Proof. Define {Fn} = [1, . . . , 1 | {z } g ], and {Hn} = [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , 2k+1 −1]. We can calculate the terms of {Fn} and {Hn} up to 2g + 1. Namely, Hn = Fn = 2n−1 when 1 ≤n ≤g Hg+n = Fg+n + 2n−1 when 1 ≤n ≤k + 1 Hg+k+1+n = Fg+k+1+n + 2k+1 −1 2n + 2n−2 (n −1) when 1 ≤n ≤g −k Fg+n = 2g+n−1 −2n−2 (n + 1) when 1 ≤n ≤g (3.24) The third and fourth lines are verified in Lemmas C.6 and C.7, respectively. We show that the conditions in Theorem 3.3 hold for {Hn}. We can verify directly that Brown’s criterion holds for the first (2g + 1) terms of {Hn}; in fact, for Bn := BH,n, we get ( Bn ≥0 1 ≤n ≤g + k Bn > 0 g + k + 1 ≤n ≤2g + 1.
(3.25) Thus, it remains to show that Bn > 0 for 2g + 2 ≤n ≤2 (g + k) −1.
Case 1: 2g + 2 ≤n ≤2g + k + 1 Define b(n) := Hn −Fn. Note that b(n) ≥0, and by induction, b(n) > 0 for all n ≥g + 1.
For n ≥g + k + 2, Fn + b(n) = Hn = Hn−1 + Hn−2 + · · · + Hn−g + 2k+1 −1 Hn−(g+k+1) = g X i=1 Fn−i + g X i=1 b (n −i) + 2k+1 −1 Hn−(g+k+1).
Since Fn = Pg i=1 Fn−i, b(n) = g X i=1 b (n −i) + 2k+1 −1 Hn−(g+k+1).
(3.26) Thus, for any n ≥2g + 2, Bn = 1 + n−1 X i=1 Hi −Hn = 1 + n−1 X i=1 (Fi + b(i)) −(Fn + b(n)) = 1 + n−1 X i=1 Fi −Fn !
− 2k+1 −1 Hn−(g+k+1) + n−(g+1) X i=g+1 b(i) > 1 + n−1 X i=1 Fi −Fn !
− 2k+1 −1 Hn−(g+k+1).
(3.27) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 15 We are to show that the last term is nonnegative. As n −(g + k + 1) ≤g, 1 + n−1 X i=1 Fi −Fn − 2k+1 −1 Hn−(g+k+1) = 1 + n−(g+1) X i=1 Fi − 2k+1 −1 Hn−(g+k+1) = 1 + g X i=1 Fi + n−(2g+1) X i=1 Fg+i − 2k+1 −1 · 2n−(g+k+1)−1 = 1 + g X i=1 2i−1 + n−(2g+1) X i=1 2g+i−1 −2i−2 (i + 1) −2n−g−1 + 2n−(g+k+1)−1 = 2n−(g+k+1)−1 − n−(2g+1) X i=1 2i−2 (i −1) − n−(2g+1) X i=1 2i−1 = 2n−(g+k+1)−1 − 2n−(2g+2) (n −(2g + 3)) + 1 − 2n−(2g+2) −1 = 2n−(g+k+1)−1 −2n−(2g+2) (n −(2g + 2)) = 2n−(2g+2) 2g−k −(n −(2g + 2)) ≥2n−(2g+2) 2g−k −(k −1) > 0.
(3.28) Note that the last line comes from g ≥k + log2 k, which implies 2g−k ≥k > k −1.
Case 2: 2g + k + 2 ≤n ≤2g + 2k + 1 We show that Bn+1 ≥Bn for 2g + k + 2 ≤n < 2g + 2k + 1, and that B2g+k+2 > 0.
Bn+1 −Bn = 2Hn −Hn+1 = 2Hn − n X i=n−g+1 Hi + (2k+1 −1)Hn−(g+k) !
= Hn − n X i=n−g+1 Hi !
+ (2k+1 −1)Hn−(g+k) = Hn−g −(2k+1 −1)(Hn−(g+k) −Hn−(g+k+1)).
Replace n by 2g + k + 1 + m, with 1 ≤m ≤k = H(g+k+1)+m −(2k+1 −1)(Hg+m+1 −Hg+m) = H(g+k+1)+m −(2k+1 −1)(2g+m−1 −2m−2(m + 1)).
(3.29) For 1 ≤m ≤g −k, we have an explicit formula for H(g+k+1)+m, so we can substitute directly to show that (3.29) is nonnegative. Thus, if g −k ≥k (i.e. g ≥2k), then this holds for all 1 ≤m ≤k. If g −k < k (i.e. g < 2k), then from Lemma C.9, (3.29) is nonnegative. Thus, Bn+1 ≥Bn for all 2g + k + 2 ≤n ≤2g + k + 1. It remains to show that B2g+k+2 > 0, which we can do by directly substituting the explicit formulas.
□ Combining these lemmas, we can prove the first part of Theorem 1.17.
16 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Proof of Theorem 1.17.1. From Lemmas 3.4 and 3.5, the bound for N is precisely 2k+1 −1 when g ≥k + ⌈log2 k⌉.
□ Next, we consider when k ≤g ≤k + ⌈log2(k)⌉, and prove the second part of Theorem 1.17 using similar methods.
Proof of Theorem 1.17.2. First, we show that for N > 2k+1 −⌈k/2g−k⌉, {Hi} is incomplete, and suppose k ≥2. Let us calculate the initial L = g + k + 1 terms of the sequence. Note Hn = 2n−1 for all 1 ≤n ≤g + 1 Hg+n = 2g+n−1 −2n−2(n −1) for all 1 ≤n ≤k + 1.
(3.30) Let Bi := BH,i. Then, we consider Brown’s gap B2g+k+2, B2g+k+2 = 1 + 2g+k+1 X i=1 Hi !
−H2g+k+2 = 1 + 2g+k+1 X i=1 Hi !
− 2g+k+1 X i=g+k+2 Hi + NHg+1 !
= 1 + g+k+1 X i=1 Hi !
−NHg+1 = 1 + g X i=1 Hi + g+k+1 X i=g+1 Hi −NHg+1 = 1 + g X i=1 2i−1 + k+1 X i=1 2g+i−1 −2i−2(i −1) −2gN = 2g+k+1 − k X i=1 2i−1i −2gN = 2g+k+1 −2k(k −1) −1 −2gN.
Now, N > 2k+1 − k/2g−k by assumption so it follows that N ≥2k+1 −k/2g−k + 1, hence ≤2g+k+1 −2k(k −1) −1 −2g 2k+1 − k 2g−k + 1 = 2k −2g −1, (3.31) which must be negative as g ≥k. So {Hn} fails Brown’s criterion at the (2g + k + 1)st term, rendering the sequence incomplete.
Now we can show that for N = 2k+1 −⌈k/2g−k⌉, {Hi} is complete by Theorem 3.3. We can easily verify that Bn ≥0 for all 1 ≤n ≤g + k + 1 and Bg+k+1 > 0; it remains to show that Bn > 0 for g + k + 2 ≤n ≤2g + 2k + 1. We consider two cases.
Case 1: 2 ≤n −(g + k) ≤g + 1.
We want to show that Bn+1 ≥Bn for all 2 ≤n −(g + k) ≤g + 1 and that Bg+k+2 > 0.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 17 Now, Bn = 1 + n−1 X i=1 Hi −Hn = 1 + n−1 X i=1 Hi − n−1 X i=n−g Hi + NHn−(g+k+1) !
= 1 + n−g−1 X i=1 Hi −NHn−(g+k+1).
(3.32) Then, note that Bn+1 −Bn = Hn−g −N Hn−(g+k) −Hn−(g+k+1) = Hn−g −N 2n−(g+k+1) −2n−(g+k+2) , and by assumption, = Hn−g − 2k+1 − l k 2g−k m 2n−(g+k+2) = 2n−(g+k+2)l k 2g−k m − 2n−g−1 −Hn−g .
(3.33) If n−g ≤g+1, then 2n−g−1−Hn−g = 0, so Bn+1−Bn > 0. If g+2 ≤n−g ≤g+k+1, then 2n−g−1 −Hn−g = 2n−2g−2(n −2g −1) ≤2n−(g+k+2) k 2g−k ≤2n−(g+k+2)l k 2g−k m , (3.34) so that Bn+1 −Bn ≥0. In any case, Bn+1 ≥Bn. We can verify directly that Bg+k+2 > 0, completing this case.
Case 2: g ≤n −(g + k) ≤g + k + 1.
From the previous case, B2g+k+2 ≥B2g+k+1 > 0. Now, Bn = 1 + n−g−1 X i=1 Hi −NHn−(g+k+1) = 1 + n−2g−1 X i=1 Hi + n−g−1 X i=n−2g Hi −NHn−(g+k+1) = 1 + n−2g−1 X i=1 Hi + Hn−g −NHn−(2g+k+1) −NHn−(g+k+1).
Substituting n = 2g + k + 1 + m for 1 ≤m ≤k, = 1 + k+m X i=1 Hi + Hg+k+1+m −N(Hm + Hg+m) ≥Hk+m+1 + Hg+k+1+m −N 2m−1 + 2g+m−1 −2m−2(m −1) .
(3.35) 18 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Let Cm := Hk+m+1 + Hg+k+1+m −N(2m−1 + 2g+m−1 −2m−2(m −1)), from equation (3.35). We show by strong induction that Cm > 0. By direct computation, C1 > 0.
Suppose it holds for all values from 1 to m−1 for m ≥2. Then by the induction hypothesis, Hg+k+1+m = (Hg+k+m + · · · + Hg+k+2) + (Hg+k+1 + · · · + Hm+k+1) + NHm > m−1 X i=1 N 2i−1 + 2g+1−i −2i−2(i −1) −Hk+i+1 + + 2g+k + · · · + 2m+k − k+1 X i=1 2i−2(i −1) !
+ 2m−1N = N 2m −1 + 2g+m+1 −2g −2m−2(m −3) −1 − − k+m X i=k+2 Hi + 2g+k+1 −2m+k −2k(k −1) −1 ≥N 2m−1 + 2g+m−1 −2m−2(m −1) −(2g + 2 −2m)N− − k+m X i=k+m−g Hi + 2g+k+1 −2m+k −2k(k −1) −1 , (3.36) where Hi = 0 for nonpositive i. Hence, Cm = Hg+k+1+m −N 2m−1 + 2g+m−1 −2m−2(m −1) + Hk+m+1 > Hk+m+1 − k+m X i=k+m−g Hi !
+ 2g+k+1 −2m+k −2k(k −1) −1 − −(2g + 2 −2m)N = 1 + 2g+k+1 −2m+k −2k(k −1) −1 −(2g + 2 −2m) 2k+1 − k 2g−k = 2m+k −2k(k + 3) + (2g + 2 −2m) k 2g−k ≥2m+k −2k (k + 3) + (2g + 2 −2m) k 2g−k = 2m+k −3 · 2k −(2m −2) k 2g−k = (2m −3) 2k − k 2g−k − k 2g−k ≥2k −2k 2g−k ≥2k −2k ≥0.
(3.37) This completes the induction, so Bn ≥Cm > 0.
Since both cases are satisfied, {Hi} is complete.
□ Remark 3.6. The case k = 1 is characterized in Lemma 3.8.
3.2. The “2L −1 Conjecture”. We conjecture a strengthened version of Theorem 3.3 as follows.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 19 Conjecture 3.7. The PLRS {Hn} defined by [c1, . . . , cL] is complete if BH,n ≥0 for all n ≤2L−1, i.e., Brown’s criterion holds for the first 2L −1 terms.
When using Brown’s criterion, it would be very helpful to be able to know how many terms must be checked to be sure that a PLRS is complete. This conjecture, if true, would be a powerful tool to do so. We do not know yet if such a threshold exists for each L; however, if it does, then it is at least 2L −1, as shown by the following example, where k + 2 = L.
Lemma 3.8. [1, . . . , 1, 0, 4], with k ≥1 ones, is always incomplete. Moreover, it first fails Brown’s criterion on the (2k + 3)rd term.
Proof. We have the recurrence relation Hn+1 = Hn + · · · + Hn−k+1 + 4Hn−k−1. We show that the term in the (2k + 3)rd position in the sequence fails Brown’s criterion. First, H2k+3 = H2k+2 + · · · + Hk+3 + 4Hk+1.
(3.38) Next, we observe that for 1 ≤j ≤k + 1, then Hj = 2j−1. Additionally, Hk+2 = 2k+1 −1. Thus, 2Hk+1 = 2k+1 > 2k+1 −1 = Hk+2.
(3.39) We also note that Hk+1 = Hk + · · · + H1 + 1. Putting everything together, H2k+3 = H2k+2 + · · · + Hk+3 + 4Hk+1 = H2k+2 + · · · + Hk+3 + 3Hk+1 + Hk + · · · + H1 + 1 > H2k+2 + · · · + Hk+3 + Hk+2 + Hk+1 + Hk + · · · + H1 + 1.
(3.40) Hence, we have shown that [1, . . . , 1, 0, 4], with k ≥1 ones, is incomplete, as it fails Brown’s criterion on the (2k + 3)rd term.
We now show that Brown’s criterion holds for the first (2k + 2) terms. For 1 ≤j ≤k + 1, we have Hj = 2j−1, which satisfies the equality Hj+1 = Hj + · · · + H1 + 1.
We consider when k + 2 ≤j ≤2k + 1, that Hj+1 = Hj + · · · + Hj−k+1 + 4Hj−k−1.
Note that Hj−k−1 = Hj−k+2 + · · · + H1 + 1 as 1 ≤j −k −1 ≤k, Hj+1 = Hj + · · · + Hj−k+1 + 2Hj−k−1 + Hj−k−1 + Hj−k−2 + · · · + H1 + 1.
And as 2Hj−k−1 = 2j−k−1 = Hj−k, we see Hj+1 = Hj + · · · + Hj−k+1 + Hj−k + Hj−k−1 + Hj−k−1 + Hj−k−2 + · · · + H1 + 1.
(3.41) Hence, this equality satisfies Brown’s criterion for terms k + 2 ≤j ≤2k + 1.
□ Assuming this conjecture, we can further explore sequences of the form [1, 0, . . . , 0, 1, . . . , 1, N].
In Theorems 3.10 and 3.11, we show that the bound on N for [1, 0, . . . , 0 | {z } L−m−2 , 1, . . . , 1 | {z } m , N] strictly in-creases if we keep L fixed and increase m from 0 to L−3, i.e., switching the coefficients from 0 to 1 gradually from the end so that at least one 0 remains. We first state a following powerful lemma that is contingent on this conjecture.
Lemma 3.9 (Conditional). Let {Hn} defined by [1, 0, . . . , 0, 1, . . . , 1, N] be a sequence of length L with m ones. Then, if the sequence is incomplete, it must fail Brown’s criterion at the (L + 1)st or (L + 2)nd term. In other words, if HL+1 ≤1 + PL i=1 Hi and HL+2 ≤1 + PL+1 i=1 Hi, then {Hn} is complete.
20 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ The proof of this lemma is deferred to C.11 of Appendix C.
Theorem 3.10. Let {Hn} be a PLRS with L coefficients defined by [1, 0, . . . , 0, 1, . . . , 1 | {z } m , N], where L ≥2m + 2. Then {Hn} is complete if and only if N ≤ (L −m) (L + m + 1) 4 + 1 48m(m + 1)(m + 2)(m + 3) + 1 −2m 2 .
(3.42) Proof. First, note for all 1 ≤n ≤L −m, that Hn = n.
Now, we claim that for all 1 ≤k ≤m, HL−m+k = L −m + 1 6k(k + 1)(k + 2) + k.
(3.43) We use induction, appealing to the identity that Pn a=1 a(a + 1)/2 = n(n + 1)(n + 2)/6. We first see that HL−m+1 = HL−m + H1 + 1 = L −m + 2 = L −m + 1 X a=1 a(a + 1) 2 + 1.
(3.44) Additionally, HL−m+2 = HL−m+1+H2+H1+1 = (L−m+2)+2+1+1 = L−m+ 2 X a=1 a(a + 1) 2 +2. (3.45) Now, suppose HL−m+k = L −m + Pk a=1 a(a + 1)/2 + k for some k < m. Note that HL−m+k+1 = HL−m+k + Hk+1 + · · · + H1 + 1.
(3.46) Since we supposed L ≥2m+2, we see k+1 ≤m+1 ≤L−m, and thus for all 1 ≤i ≤k, Hi = i.
Thus, HL−m+k+1 = L −m + k X a=1 a(a + 1) 2 + k !
+ (k + 1)(k + 2) 2 + 1 = L −m + k+1 X a=1 a(a + 1) 2 + k + 1.
(3.47) Thus, we have an explicit formula for Hi, for 1 ≤i ≤L.
Note that {Hn} is complete if and only if it fulfills Brown’s criterion for the (L + 1)st and (L + 2)nd term. We show that {Hn} fulfills the criterion for L + 2 if and only if the bound above holds; it is not difficult to show that the bound for L + 1 is less strict.
Indeed, we wish to reduce the inequality HL+2 = HL+1 + Hm+2 + · · · + H3 + 2N ≤1 + L+1 X i=1 Hi (3.48) ⇐ ⇒Hm+2 + · · · + H3 + 2N ≤1 + L X i=1 Hi.
(3.49) Simplifying the left hand side of inequality (3.49), Hm+2 + · · · + H3 + 2N = Hm+2 + · · · + H3 + (H2 + H1 −H2 −H1) + 2N COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 21 = (m + 2)(m + 3) 2 −3 + 2N.
(3.50) Additionally, 1 + L X n=1 Hn = 1 + L−m X n=1 Hn + L X n=L−m+1 Hn = 1 + (L −m) (L −m + 1) 2 + m X n=1 1 6n(n + 1)(n + 2) + n + L −m .
(3.51) We use the fact that Pm n=1 n(n + 1)(n + 2) = m(m + 1)(m + 2)(m + 3)/4 to simplify equation 3.51 as follows 1 + (L −m) (L −m + 1) 2 + m(m + 1) 2 + mL −m2 + 1 6 m X n=1 n(n + 1)(n + 2) = 1 + (L −m) (L −m + 1) 2 + m(m + 1) 2 + mL −m2 + 1 24m(m + 1)(m + 2)(m + 3).
(3.52) Hence the inequality is equivalent to (m + 2) (m + 3) 2 −3 + 2N ≤1 + (L −m) (L −m + 1) 2 + m(m + 1) 2 + mL −m2 + 1 24m(m + 1)(m + 2)(m + 3).
(3.53) Simplifying, this gives us N ≤ (L −m) (L + m + 1) 4 + 1 48m(m + 1)(m + 2)(m + 3) + 1 −2m 2 .
(3.54) □ Theorem 3.11. Let {Gn} and {Hn} be PLRS’s with L coefficients defined by [1, 0, . . . , 0, 1, . . . , 1 | {z } m , N] and [1, 0, . . . , 0, 1, . . . , 1 | {z } m+1 , N + 1] respectively. Suppose L −m ≥4 (so that at least one zero is present in {Hn}), m ≥(L −1)/2, and {Gn} is complete. Then {Hn} is also complete.
Proof. As {Gn} is complete, from Brown’s criterion, we obtain GL+2 = GL+1 + m+2 X i=3 Gi + NG2 ≤1 + L+1 X i=1 Gi, (3.55) which is equivalent to 2N ≤ L X i=m+3 Gi + 4.
(3.56) From Lemma 3.9, it suffices to show that ( HL+1 ≤1 + PL i=1 Hi HL+2 ≤1 + PL+1 i=1 Hi, (3.57) 22 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ or equivalently, N ≤ L−1 X i=m+3 Hi (3.58) and 2N ≤ L X i=m+4 Hi + 2.
(3.59) We first show equation (3.58). Combining with equation (3.56), it suffices to show that L X i=m+3 Gi + 4 ≤2 L X i=m+4 Hi.
(3.60) From Lemma C.12, ( Gi ≤Hi m + 3 ≤i ≤L Gi ≤Hi−1 −1 2(L −m) < i ≤L.
(3.61) Thus, L X i=m+3 Gi + 4 = 2(L−m) X i=m+3 Gi + L X i=2(L−m)+1 Gi + 4 ≤ 2(L−m) X i=m+3 Hi + L−1 X i=2(L−m) Hi + (2m −L + 4) ≤2 L X i=m+4 Hi, (3.62) the last inequality can be taken crudely. We then show (3.59). Similarly, combining with (3.56), it suffices to show that L X i=m+3 Gi + 2 ≤ L X i=m+4 Hi.
(3.63) If m + 3 ≥2(L −m), then L X i=m+4 Hi = L X i=m+4 Hi−1 + i−L+m+1 X j=1 Hj + 1 !
≥ L X i=m+4 (Hi−1 + Hi−L+m+2) (Brown’s criterion for the first terms) = L−1 X i=m+3 Hi + m+2 X i=2m+6−L Hi ≥ L−1 X i=m+3 (Gi+1 + 1) + Hm+2 ≥ L X i=m+3 Gi + 2.
(3.64) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 23 If m + 3 < 2(L −m), then L X i=m+3 Gi = 2(L−m)−1 X i=m+3 Gi + G2(L−m) + L X i=2(L−m)+1 Gi = 2(L−m)−1 X i=m+3 (Hi−1 + 1) + H2(L−m)−1 + L X i=2(L−m)+1 Gi = 2(L−m)−1 X i=m+2 Hi + (2L −3(m + 1)) + L X i=2(L−m)+1 Gi.
(3.65) Thus, our original inequality, equation (3.59), holds if we can show that Hm+2 + Hm+3 + (2L −3(m + 1)) + L X i=2(L−m)+1 Gi ≤ L X i=2(L−m) Hi.
(3.66) Similarly to the previous case, L X i=2(L−m) Hi ≥ L X i=2(L−m) (Hi−1 + Hi−L+m+2) = L−1 X i=2(L−m)−1 Hi + m+2 X i=L−m+2 Hi = L−1 X i=2(L−m) Hi + H2(L−m)−1 + Hm+2 + m+1 X i=L−m+2 Hi.
(3.67) As 2(L −m) −1 ≥m + 3 and Hi ≥i L X i=2(L−m) Hi ≥ L−1 X i=2(L−m) (Gi+1 + 1) + Hm+3 + Hm+2 + m+1 X L−m+2 i = L X i=2(L−m)+1 Gi + Hm+3 + Hm+2 + (2m −L + m+1 X i=L−m+2 i).
(3.68) From Lemma C.13, L X i=2(L−m) Hi ≥ L X i=2(L−m)+1 Gi + Hm+3 + Hm+2 + (2L −3(m + 1)).
(3.69) □ 4. AN ANALYTICAL APPROACH 4.1. An Introduction to Principal Roots. We begin by restating some results from [MMMMS].
Lemma 4.1. Let P(x) be the characteristic polynomial of a recurrence relation with nonnegative coefficients and at least one positive coefficient, and let S = {m | cm ̸= 0}. Then (1) there exists exactly one positive root r, and this root has multiplicity 1, 24 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ (2) every root z ∈C satisfies |z| ≤r, and (3) if gcd(S) = 1, then r is the unique root of greatest magnitude.
Proof. This is Lemma 2.1 from [MMMMS].
□ Remark 4.2. We refer to the unique positive root from Lemma 4.1 as the principal root of the recurrence sequence and corresponding characteristic polynomial.
Lemma 4.3. Let P(x) be the characteristic PLRS {Hn} and let r1 be its principal root. Then lim n→∞ Hn rn 1 = C (4.1) for some constant C > 0.
Proof. Corollary 2.3 from [MMMMS] proves a stronger result than this, which immediately im-plies this lemma.
□ Lemma 4.4. Let P(x) be the characteristic polynomial of a PLRS {Hn} with roots ri, each of multiplicity mi, where r1 is the principal root. If Hn = a1rn 1 + k X i=2 qi(n)rn i , (4.2) where qi(x) is a polynomial of degree at most mi −1, then a1 > 0.
Proof. First, note that the set S of Lemma 4.1 contains 1 because c1 > 0 in a PLRS. Therefore gcd(S) = 1, and r1 is the unique root of greatest magnitude. If a1 < 0, then this implies that Hn < 0 for some n because the behavior of a1rn 1 eventually dominates the expression for Hn in (4.2). If a1 = 0, then lim n→∞ Hn rn 1 = 0 (4.3) because r1 is the unique root of greatest magnitude, so if a1 = 0 then the behavior of Hn is bounded by geometric growth of the root of next greatest magnitude, which is necessarily smaller than rn 1.
Thus, a1 > 0.
□ 4.2. Applications to Completeness. Given these results, we see that the principal root of a PLRS serves as a measure for the rate of that sequence’s growth. Guided by the simple heuristic that, generally, a sequence which grows slowly is more likely to be complete than a sequence which grows rapidly, we find bounds for the potential roots of a complete or incomplete PLRS. We aim to answer these questions: For any given L, what is the fastest-growing complete PLRS with L coefficients? What is the slowest-growing incomplete PLRS with L coefficients?1 Lemma 4.5. If {Hn} is a complete PLRS and r1 is its principal root, then |r1| ≤2.
Proof. Suppose that |r1| > 2. Set Hn = a1rn 1 + q2(n)rn 2 + · · · + qr(n)rn k.
(4.4) Since r1 is the unique root of largest magnitude by Lemma 4.1, the behavior of a1rn 1 dominates in the limit. By Lemma 4.4, a1 > 0, so if |r1| > 2, then eventually |a1rn 1| > 2n−1, and so there exists a large n for which Hn > 2n−1. As the sequence {2n−1} is the complete PLRS with maximal terms by Theorem 2.1, we see {Hn} must be incomplete.
□ 1While the principal root of a PLRS has not been related to completeness before, there is previous work on bounding the principal root of other linear recurrence sequences in [GM].
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 25 Remark 4.6. The converse to this lemma does not hold. A counterexample is [1, 1, 1, 0, 4], which has principal root 2 but is not complete.
While the proof is simple, this lemma gives us an effective upper bound for the roots of a complete PLRS, regardless of length. Recall from Theorem 1.11 that for any L, the PLRS {Hn} generated by the coefficients [1, . . . , 1 | {z } L−1 , 2] satisfies Hn = 2n−1. This sequence naturally has a principal root of 2, and is complete. Similarly, for any L ≥1, the sequence [1, . . . , 1 | {z } L ] is complete, and its principal root asymptotically approaches 2 as L grows.
We now focus on finding a lower bound for the roots of an incomplete sequence, which proves to be a more difficult problem.
Lemma 4.7. For any L ∈Z+, there exists a constant BL, with 1 < BL < 2 such that if {Hn} is a PLRS with principal root r1 and r1 < BL, then {Hn} is complete.
Remark 4.8. This means that for any L, there exists a lower bound BL on possible values of the principal root of an incomplete PLRS generated by [c1, . . . , cL].
Proof. In order to show that such a BL exists, it suffices to show that for any given L, there exists only finitely many incomplete positive linear recurrence sequences generated by [c1, . . . , cL] with principal root r1 < 2.
Recall that the principal root r1 of a PLRS is the single positive root of the characteristic poly-nomial p(x) = xL −PL i=1 cixL−i.
As limx→∞p(x) = +∞, the fact that r1 is the unique positive root of p(x) implies that r1 < 2 ⇐ ⇒p(2) > 0, by Intermediate Value Theorem. Note that p(2) = 2L − L X i=1 ci2L−i > 0 ⇐ ⇒ L X i=1 ci2L−i < 2L.
(4.5) As for all i, ci ≥0, so the inequality above cannot hold if there exists i such that ci ≥2i. As the set {[c1, . . . , cL] : 0 ≤ci ≤2i for all i} of such sequences is finite, we are done.
□ The remainder of this section is a series of lemmas which build towards the following conjecture: Conjecture 4.9. Let NL = ⌈L(L + 1)/4⌉, and let λL be the principal root of the sequence gener-ated by [1, 0, . . . , 0 | {z } L−2 , NL + 1], i.e., the sole principal root of pL(x) = xL −xL−1 − L(L + 1) 4 −1.
(4.6) If [c1, . . . , cL] generates an incomplete sequence, then its principal root is at least λL.
Remark 4.10. This conjecture is equivalent to stating BL = λL for all L ≥2, where BL is the bound proposed in Lemma 4.7.
Remark 4.11. Using Theorem 1.15, it is easy to see that the sequence generated by [1, 0, . . . , 0, NL+ 1] is incomplete; in fact, the value NL + 1 is the minimal positive integer such that a sequence of this form is incomplete.
26 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ As a first step towards a proof of Conjecture 4.9, we prove Lemma 4.15, which addresses the case of sequences with a large sum in coefficients.
Definition 4.12. For positive integers S, L, we define the set of positive linear recurrence sequences PL,S := n {Hn} generated by [c1, · · · , cL] L X i=1 ci = S + 1 o .
(4.7) Lemma 4.13. The sequence in PL,S with the minimal principal root is [1, 0, . . . , 0, S].
Proof. Consider a sequence generated by s = [c1, . . . , cL] ∈PL,S, and let r1, . . . , rL be its roots, with r1 > 0 the principal root. Since |cL| = QL i=1 rL is a positive integer, we know r1 > 1.
Now, for any 1 ≤m ≤L consider a sequence generated by sm ∈PL,S of the form [c1, . . . , cm−1, cm −1, cm+1, . . . , cL + 1].
(4.8) We claim that the principal root q1 of sm fulfills q1 < r1.
Define the characteristic polynomials f(x) and g(x) for s and sm respectively, so that f(x) = xL − L X i=1 cixL−i, (4.9) and g(x) = xL − m−1 X i=1 cixL−i −(cm −1) xm − L−1 X i=m+1 cicixL−i −(cL + 1) = xL − L X i=1 cixL−i + xm −1.
(4.10) As q1 is the sole positive root of g(x), and g(x) is eventually positive, we notice that q1 < r1 if and only if g(r1) > 0, which is equivalent to g (r1) > f(r1).
Now, g (r1) > f (r1) ⇐ ⇒rL 1 −PL i=1 cirL−i 1 + rm 1 −1 > rL 1 −PL i=1 cirL−1 1 ⇐ ⇒rm 1 −1 > 0 ⇐ ⇒r1 > 1.
(4.11) As r1 > 1, the principal root q1 of g(x) is strictly less than that of f(x).
As s was chosen arbitrarily, we see that the principal root of any sequence s ∈PL,S can be strictly decreased by using the transformation s →sm for any 1 ≤m ≤L. Applying this transformation iteratively for all values of m, we inevitably end up with the minimal possible values of c1, . . . , cL−1, namely c1 = 1, c2 = c3 = · · · = cL−1 = 0, and the maximal possible value of cL, namely cL = S.
Thus, as the principal root under these iterated transformations is strictly decreasing, we con-clude that [1, 0, . . . , 0, S] has the smallest principal root of any element of PL,S.
□ Lemma 4.14. For any S > 0, the principal root of [1, 0, . . . , 0, S] is strictly less than that of [1, 0, . . . , 0, S + 1].
Proof. Let S be an arbitrary positive integer, and denote by f(x), g(x) and r1, q1 the characteristic polynomials and principal roots of [1, 0, . . . , 0, S + 1] and [1, 0, . . . , 0, S], respectively.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 27 As before, q1 < r1 if and only if g(r1) > 0 = f(r1). Note that g(r1) > f(r1) ⇐ ⇒rL 1 −rL−1 1 −S > rL 1 −rL−1 1 −(S + 1) ⇐ ⇒S + 1 > S.
(4.12) Thus, q1 < r1, for any value of S.
□ Lemma 4.15. Any sequence fulfilling PL i=1 ci ≥NL +2 has a principal root greater than or equal to that of [1, 0, . . . , 0, NL + 1].
(4.13) Proof. Recall from Theorem 1.15 that the sequence [1, 0, . . . , 0, N] is complete if and only if N ≤ NL, for NL = ⌈L(L + 1)/4⌉. Thus, an immediate corollary to this theorem is that the incomplete sequence of the form [1, 0, . . . , 0, N] with the minimal possible principal root is [1, 0, . . . , 0, NL + 1].
Furthermore, if we have a sequence generated by [c1, . . . , cL] which fulfills PL i=1 ci ≥NL + 2, Lemmas 4.13 and 4.14 present a sequence of algorithms which allow us to transform this sequence into the sequence generated by [1, 0, . . . , 0, NL +1], in such a way that each transformation strictly lowers the magnitude of the principal root.
Thus, any sequence satisfying PL i=1 ci ≥NL + 2 has a principal root strictly greater than the principal root of [1, 0, . . . , 0, NL + 1].
□ The following lemmas are working towards proving Conjecture 4.21, which addresses the sec-ond case of Conjecture 4.9, which addresses the roots of sequences [c1, . . . , cL] which fulfill PL i=1 ci ≤NL + 2.
Lemma 4.16. Suppose the sequence generated by [c1, . . . , cL] has principal root r, then for any cL+1 ∈Z+, the sequence generated by [c1, . . . , cL, cL+1] (in which we add an additional positive coefficient) and principal root q fulfills r < q.
Proof. Let f(x), g(x) be the characteristic polynomials of the two sequences, so that f(x) = xL − L X i=1 cixL−i, and g(x) = xL+1 − L X i=1 cixL+1−i −cL+1.
(4.14) Similar to previous arguments, by the Intermediate Value Theorem, r < q if and only if g(r) < f(r) = 0. Note that g(r) < f(r) ⇐ ⇒rL+1 −PL i=1 cixL+1−i −cL+1 < rL −PL i=1 cirL−i ⇐ ⇒cL+1 > rL+1 −rL + PL i=1 cirL−i −PL i=1 cirL+1−i ⇐ ⇒cL+1 > rL (r −1) + PL i=1 cirL−i (1 −r) ⇐ ⇒cL+1 > (1 −r) rL −PL i=1 cirL−i = (1 −r) · f(r) ⇐ ⇒cL+1 > (1 −r) f(r) = (1 −r) · 0 = 0.
(4.15) Since cL+1 ∈Z+ the last line holds, and so r < q.
□ Lemma 4.17. Let λL be the principal root of xL −xL−1 −NL −1.
(4.16) Then, for any L ≥2, λL > λL+1.
28 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Proof. Define f(x) and g(x) to be the characteristic polynomials of [1, 0, . . . , 0, NL + 1] and [1, 0, . . . , 0, NL+1 + 1], of length L and L + 1, respectively, so that f(x) = xL −xL−1 −NL −1, g(x) = xL+1 −xL −NL+1 −1.
(4.17) As in previous proofs, we see that λL > λL+1 ⇐ ⇒g (λL) > f (λL) = 0.
g (λ) > f (λ) ⇐ ⇒λL+1 −λL −NL+1 −1 > λL −λL−1 −NL −1 ⇐ ⇒λL+1 −2λL + λL−1 > NL+1 −NL ⇐ ⇒λL−1 (λ −1)2 > NL+1 −NL.
(4.18) Note that as f (λ) = 0, we have λL−1 (λ −1) = NL+1. Note that NL+1−NL ≤(L+2)/2, which can be shown by using the definition of NL and checking all cases modulo 4. Thus it suffices to show that (NL + 1) (λL −1) ≥L + 2 2 .
(4.19) Using the value of NL, it suffices to show (λL −1) ≥ L + 2 L2 + L + 4.
(4.20) The proof of equation 4.20 is just algebra, and is left to Appendix D, as Lemma D.1.
□ Lemma 4.18. For any L ∈N, let λL be the sole positive root of the polynomial pL(x) = xL −xL−1 − L(L + 1) 4 −1.
(4.21) Then limL→∞λL = 1.
Proof. We show that for any ε > 0, there exists an M large enough so that for all L > M, pL(1 + ε) > 0. As pL(x) has only one positive root λL and p(x) is positive as x →∞, we see pL(1 + ε) > 0 implies λL < 1 + ε. If this is possible for arbitrary ε, then λL →1, as desired.
Fix an ε > 0. For any L, we may write pL(1 + ε) = (1 + ε)L −(1 + ε)L−1 − L(L + 1) 4 −1 = L X n=0 εn L n − L −1 n − L(L + 1) 4 −1, (4.22) where L−1 L is 0. Using Pascal’s rule, we can reduce (4.22) to pL (1 + ε) = L X n=1 εn L −1 n −1 −⌈L(L + 1)/4⌉−1.
(4.23) This quantity can easily be shown to be positive (and in fact tends towards infinity) for large enough L. For example, we can take the trivial bound L X n=1 εn L −1 n −1 > ε4 L −1 3 , (4.24) as the full sum must be larger than only its fourth summand.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 29 Since ε4 is simply a positive constant and L(L + 1) ≪ L−1 3 , then for large enough L, pL(1 + ε) > ε4 L −1 3 −⌈L(L + 1)/4⌉−1 > 0.
(4.25) □ Remark 4.19. Even in the event that Conjecture 4.9 is false, this gives us conclusive proof that we may find incomplete sequences whose roots are arbitrarily close to 1; since 1 is the minimum possible size for the root of a PLRS, this may be interpreted as proof that we may find arbitrarily slow-growing incomplete sequences, with coefficients of any length L.
Lemma 4.20. Consider the sequence generated by [c1, . . . , cL]. For any value m ∈Z+, the principal root of [c1, . . . , cL + m] is greater than that of [c1, . . . , cL, m].
Proof. Let f(x), g(x), and r, q be the principal roots of [c1, . . . , cL + m] and [c1, . . . , cL, m], re-spectively. Since f, g each have a unique positive root, we see that r > q ⇐ ⇒g(r) > f(r) = 0.
Note that g(r) > 0 ⇐ ⇒0 = rf(r) < g(r) ⇐ ⇒r rL −PL i=1 cirL−i −m < rL+1 −PL i=1 cirL+1−i −m ⇐ ⇒m < rm ⇐ ⇒r > 1.
(4.26) Thus, the inequality always holds, and so r > q, as desired.
□ Conjecture 4.21. Let λL be the principal root of xL −xL−1 −NL −1. If the sequence generated by [c1, . . . , cL] is incomplete with PL i=1 ci ≤⌈L(L + 1)/4⌉+ 2, then its principal root is at least λL.
We present a partial proof, which addresses all cases except what is denoted as Subcase 2.
Partial proof. We use induction.
For L = 2, NL = ⌈2 · 3/4⌉= 2, and so the coefficients [c1, c2] fulfilling the requirement are of the form c1 + c2 ≤4. The incomplete sequences of this form have coefficients [2, 1], [2, 2], [1, 3], and [3, 1]. Checking each case directly, we see that their principal roots are approximately 2.414, 2.731, 2.303, and 3.303, respectively. Of these, the root of [1, 3] = [1, N2 + 1] is the minimum; thus, the lemma holds for the base case.
Now, suppose the Lemma holds for some value of L ≥2; we show that it holds for L + 1.
Let [c1, . . . , cL, cL+1] be an incomplete sequence with PL+1 i=1 ci ≤⌈(L + 1)(L + 2)/4⌉+ 2.
Case 1: PL i=1 ci < NL + 2 If this is the case, the following two subcases arise.
• Subcase 1: [c1, . . . , cL] is incomplete.
If this is the case, then by our inductive hypothesis, since PL i=1 ci ≤NL + 2, we must have that the principal root r of [c1, . . . , cL] is greater than or equal to λL. Hence, by Lemma 4.16, since the principal root q of [c1, . . . , cL+1] satisfies q > r, we have that [c1, . . . , cL+1] has principal root q > λL. Finally, by Lemma 4.17, we know λL > λL+1; thus, we have q > r ≥λL > λL+1, and the statement holds in this case.
• Subcase 2: [c1, . . . , cL] is complete: The proof of this subcase has not been found yet, hence why the statement remains a conjecture.
30 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Case 2: PL i=1 ci ≥NL + 2: If this inequality holds, then as we have shown using the transfor-mations developed in Lemmas 4.13 and 4.14, this implies that [c1, . . . , cL] has principal root at least λ. Applying Lemma 4.16, we see the principal root of [c1, . . . , cL+1] is strictly greater, and thus the statement holds in this case.
□ The results in this section provide us with an efficient way to verify completeness for PLRS’s.
Namely, for a sequence [c1, . . . , cL], we may evaluate its characteristic polynomials at the points BL and 2, which provides the following information: • If p(2) < 0, the sequence is incomplete.
• If p(BL) > 0, the sequence is complete.
• If p(2) ≥0 and p(BL) ≤0, then the principal root of the sequence lies in the interval [BL, 2], and so further inquiry is necessary to determine whether it is complete.
Computationally, evaluating a polynomial of degree L is an O(L2) problem; generating a mini-mum of 2L terms of the sequences and checking Brown’s criterion for each, on the other hand, is a O(2L) problem. Thus, this method—even if inconclusive—provides a fast and efficient method to categorize sequences, and narrows our search to the interesting interval [BL, 2], in which both complete and incomplete sequences arise.
4.3. Denseness of Incomplete Roots. Having narrowed our search for principal roots of complete and incomplete sequences to the interval [BL, 2], it is only natural to ask how the roots of these sequences are distributed throughout the interval.
Lemma 4.22. For fixed L > 2 and k > 0, define the three polynomials f(x) = xL −xL−1 −k, g(x) = xL −xL−1 −(k + 1), and h(x) = xL −xL−1 −(k + 2). Let q, r, s be the sole positive roots of f, g, h respectively, so that 1 < q < r < s. Then, r −q > s −r.
(4.27) Proof. From the definition, we see that qL −qL−1 = k, rL −rL−1 = k + 1, sL −sL−1 = k + 2.
(4.28) Now, define the polynomial p(x) = xL −xL−1. Taking first and second derivatives of p, we see p′(x) = LxL−1−(L −1) xL−2, and p′′(x) = L (L −1) xL−2−(L −1) (L −2) xL−3. In particular, for all x ≥1, p(x) ≥0, p′(x) > 0, and p′′(x) > 0.
Thus, p(x) is increasing and convex on (1, ∞). By equation (4.28), we have p(r) −p(q) = p(s) −p(r). Thus, as s > r > q > 1, we conclude r −q > s −r, as desired.
□ Theorem 4.23. For any L ≥2, let RL be the set of roots of all incomplete PLRS’s generated by L coefficients. Then, for any ε > 0, there exists an M such that for all L > M and for any ε-ball Bε ⊂(1, 2), Bε ∩RL ̸= ∅.
Proof. Let ε > 0 be arbitrary. By Lemma 4.18, we may fix an M such that for all L > M, 1 < λL < 1 + ε.
From previous work, we know the sequence of length L with coefficients [1, 0, . . . , 0, ⌈L (L + 1) /4⌉+ 1] is incomplete, as is any sequence of the form [1, 0, . . . , 0, k], with k ≥⌈L (L + 1) /4⌉+ 1.
Note that λL is the root of [1, 0, . . . , 0, ⌈L (L + 1) /4⌉+ 1]. Since λL < 1 + ε, it is clear that the root α of [1, 0, . . . , 0, ⌈L (L + 1) /4⌉] fulfills 1 < α < λL, and so λ −α < ε.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 31 Now, we know the sequence [1, 0, . . . , 0, 2L−1] has a root of size exactly 2. Applying Lemma 4.22 iteratively, any two sequences [1, 0, . . . , 0, k], [1, 0, . . . , 0, k + 1] with k ≥⌈L (L + 1) /4⌉and roots q, r must fulfill r −q < λL −α < ε. Thus, any two consecutive sequences [1, 0, . . . , 0, k], [1, 0, . . . , 0, k + 1] with k ≥⌈L (L + 1) /4⌉+ 1 have roots with separation less than ε, and so the set of roots of sequences of the form [1, 0, . . . , 0, k] with ⌈L (L + 1) 4⌉+ 1 ≤k ≤2L−1 intercepts any ε-ball of (1, 2). As this is a subset of RL, we are done.
□ Corollary 4.24. The set of principal roots of incomplete sequences R = S∞ L=2 RL is dense in (1, 2).
We conjecture that a similar result can be shown about complete roots; however, this proof has proven more difficult, as examples of families of complete sequences are more fragile.
5. OPEN QUESTIONS Here are conjectures and several other questions that future researches could investigate.
• Our results often focus on the final coefficient, such as in Theorems 1.13 and 1.14. Do these results have any analogues for coefficients that are not the last?
• Can Theorem 1.17 be extended to address when g < k?
• Are there other interesting families of PLRS’s that can be fully characterized that have entries other than 0 and 1 as coefficients that are not the final coefficient?
• Are Conjectures 3.2 and 3.7 true?
• Is the missing component of the proof of Conjecture 4.9, i.e., Conjecture 4.21 true?
APPENDIX A. BROWN’S CRITERION AND A COROLLARY Here are several proofs of important results for our paper. All results will be restated for the reader’s convenience.
Theorem A.1. (Brown [Br]) If an is a non-decreasing sequence, then an is complete if and only if a1 = 1 and for all n > 1, an+1 ≤1 + n X i=1 ai.
(A.1) Proof. Let {an}∞ n=1 be a sequence of positive integers, not necessarily distinct, such that a1 = 1 and an+1 ≤1 + n X i=1 ai (A.2) for n ∈{1, 2, . . .}. Then for 0 < n < 1 + Pk i=1 ai there exists {bi}k i=1, bi ∈{0, 1} such that n = Pk i=1 biai. We proceed by induction on k. The claim obviously holds for k = 1, so one may assume that it holds for k = N. Hence, we must show that 0 < n < 1 + PN+1 i=1 ai implies the existence of {γi}N+1 i=1 , γi ∈{0, 1} such that n = PN+1 i=1 γiai. Due to the inductive hypothesis, we only consider values satisfying 1 + N X i=1 ai ≤n < 1 + N+1 X i=1 ai.
(A.3) 32 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Note that by assumption, n −aN+1 ≥1 + N X i=1 ai −aN+1 ≥0.
(A.4) Now, if n −aN+1 = 0, the conclusion follows. Otherwise, 0 < n −aN+1 < 1 + N X i=1 ai (A.5) implies the existence of {bi}N i=1 such that n −aN+1 = PN i=1 biai. Then the result is immediate on transposing aN+1 and identifying γi = bi for i ∈{1, . . . , N} and γN+1 = 1. This completes the sufficiency part of the proof.
For the necessity, assume that there exists n0 ≥1 such that an0+1 ≥1+Pn0 i=1 ai. Then, however, an0+1 > an0+1 −1 > n0 X i=1 ai, (A.6) which implies that the positive integer an0+1 −1 cannot be represented in the form Pk i=1 biai. This leads to a contradiction and completes the proof.
□ Corollary A.2. If an is a nondecreasing sequence such that a1 = 1 and an ≤2an−1 for all n ≥2, then an is complete.
Proof. We argue by induction on n that an satisfies Brown’s criterion when n ≥2. As a1 = 1, for the base case we have a2 ≤2a1 = 2 = a1 + 1.
(A.7) Now assume for inductive hypothesis that for some n ≥2, an ≤an−1 + · · · + a1 + 1.
(A.8) Then an+1 ≤2an = an + an ≤an + an−1 + · · · + a1 + 1, (A.9) completing the induction.
□ Example A.3. The converse does not hold. A sequence may be complete and have some terms that are larger than the double of the previous term. One such example is the sequence generated by [1, 0, 1, 4], whose terms are {1, 2, 3, 5, 11, . . . }. Here, 11 is more than twice 5, yet the sequence is still complete.
APPENDIX B. LEMMAS FOR SECTION 2 Lemma B.1. Let {Gn}, {Hn} be the sequences defined by [c1, . . . , cL], [c1, , . . . , cL, cL+1], respec-tively, where cL+1 is any positive integer. For all k ≥2, HL+k −GL+k ≥2 (HL+k−1 −GL+k−1) .
(B.1) Proof. We use strong induction.
We begin with the base case. First, recall that for all n such that 1 ≤n ≤L, we know Hn = Gn.
Further, note that HL+1 = c1HL + · · · + cLH1 + 1 = c1GL + · · · + cLG1 + 1 = GL+1 + 1.
(B.2) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 33 Using this fact, we compute HL+2 = c1HL+1 + c2HL + · · · + cLH2 + cL+1H1 = c1 (GL+1 + 1) + c2GL + · · · + cLG2 + cL+1 = GL+1 + c1 + cL+1.
(B.3) Thus, we have that HL+2 −GL+2 = c1 + cL+1 ≥2 = 2(1) = 2 (HL+1 −GL+1) .
(B.4) For the inductive step, suppose for some m, the lemma holds for all 2 ≤k ≤m −1. We wish to show it holds for m, i.e., HL+m −GL+m ≥2 (HL+m−1 −GL+m−1) .
(B.5) Expanding the terms using the recurrence definition, we see HL+m −GL+m ≥2 (HL+m−1 −GL+m−1) (B.6) which holds if and only if L X i=1 ciHL+m−i − L X i=1 ciGL+m−i ≥2 L X i=1 ciHL+m−1−i − L X i=1 ciGL+m−1−i !
.
(B.7) Note that for all i ≥m, HL+m−i −GL+m−i = 0. We cancel out any such terms on both sides of the inequality above, simplifying to min(m−1,L) X i=1 ci (HL+m−i −GL+m−i) ≥ min(m−1,L) X i=1 2ci (HL+m−1−i −GL+m−1−i) .
(B.8) We encourage the reader to note that for m −1 ≤L we preserve the term 2cm−1 (HL −GL) = 0 in the right hand side sum, so that both sides of the inequality have the same number of summands.
By our inductive hypothesis, we see that for all i, ci(HL+m−i −GL+m−i) ≥2ci (HL+m−1−i −GL+m−1−i) .
(B.9) Thus, inequality B.8 holds, which completes the proof.
□ Lemma B.2. Consider sequences {Gn} = [c1, c2, . . . , cL] and {Hn} = [c1, c2, . . . , kL], where 1 ≤kL ≤cL. For all k ∈N, HL+k+1 −2HL+k ≤GL+k+1 −2GL+k.
(B.10) Proof. We proceed by strong induction on k. For k = 1, we have HL+2 −2HL+1 = (c1HL+1 + c2HL + · · · + kLH2) −2 (c1HL + c2HL−1 + · · · + kLH1) = (c1HL+1 + c2GL + · · · + kLG2) −2 (c1GL + c2GL−1 + · · · + kLG1) = GL+2 −(GL+1 −HL+1) −(2cL −2kL) −2GL+1 −2 (cL −kL) ≤GL+2 −2GL+1.
(B.11) Assume the statement holds true for a natural number k. Now, note HL+k+2 −2HL+k+1 = (c1HL+k+1 + c2HL+k + · · · + kLHk+2) −2 (c1HL+k + c2HL+k−1 + · · · + kLHk+1) = c1 (HL+k+1 −2HL+k) + c2 (HL+k −2HL+k−1) + · · · + kL (Hk+2 −2hk+1) 34 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ ≤c1 (HL+k+1 −2HL+k) + c2 (HL+k −2HL+k−1) + · · · + cL (Hk+2 −2Hk+1) .
By the inductive hypothesis, ≤c1 (GL+k+1 −2GL+k) + c2 (GL+k −2GL+k−1) + · · · + cL (Gk+2 −2Gk+1) = GL+k+2 −2GL+k+1.
(B.12) Therefore, the statement holds by induction.
□ Lemma B.3. Let {Gn} be the sequence defined by [c1, . . . , cL], and let {Hn} be the sequence defined by [c1, . . . , cL−1 + 1, cL −1]. Then, for all k ≥0, HL+k+1 −GL+k+1 ≥2 (HL+k −GL+k) .
(B.13) Proof. We use strong induction. We begin with the base case. First, since the first L−2 coefficients of {Gn}, {Hn} are equivalent, we have that for all 1 ≤n ≤L −1, Gn = Hn. We also see that HL = c1HL−1+· · ·+(cL−1 + 1) H1+1 = c1GL−1+· · ·+(cL−1 + 1) G1+1 = GL+G1 = GL+1.
(B.14) Moreover, HL+1= c1HL + · · · + (cL−1 + 1) H2 + (cL −1) H1 = c1 (GL + 1) + · · · + (cL−1 + 1) G2 + (cL −1) G1 = c1 + G2 −G1 + PL i=1 ciGL+1−i = c1 + c1 + GL+1 = 2c1 + GL+1.
(B.15) Thus, we see that HL+1 −GL+1 = 2c1 ≥2 = 2(1) = 2 (HL −GL) , (B.16) and so the base case holds.
For the induction step, suppose our lemma holds for all 0 ≤k ≤m. We wish to show this holds for m + 1, so that HL+m+1 −GL+m+1 ≥2 (HL+m −GL+m).
Since {Gn} and {Hn} are PLRS, we expand the terms in questions using their respective recur-rence relations to see that HL+m+1 −GL+m+1 ≥2 (HL+m −GL+m) if and only if L X i=1 ciHL+m+1−i + Hm+2 −Hm+1 − L X i=1 ciGL+m+1−i ≥2 L X i=1 ciHL+m−1 + Hm+1 −Hm − L X i=1 ciGL+m−i !
.
(B.17) We note that by the induction hypothesis, we have that for all i, ci (HL+m+1−i −GL+m+1−i) ≥2ci (HL+m−i −GL+m−i) .
(B.18) Moreover, we have that Hm+2 −Hm+1 ≥Hm+1 −Hm, simply because we know that gaps in a PLRS grow. Combining these two statements, we have that inequality B.17 holds, and so our inductive step is complete.
□ Lemma B.4. Let {Gn} be the sequence defined by [c1, . . . , cL−1, 1], and let {Hn} be the sequence defined by [c1, . . . , cL−1 + 1]. Then, for all k ≥1, HL+k+1 −GL+k+1 ≥2 (HL+k −GL+k) .
(B.19) COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 35 Proof. The proof is similar to that of Lemma B.3 and so we repeat our use of strong induction.
We begin with the base case. First, since first L −2 coefficients of {Gn}, {Hn} are equivalent, we have that for all 1 ≤n ≤L −1, Gn = Hn. In fact, even more can be said. GL = HL, as HL = c1HL−1 + · · · + (cL−1 + 1) H1 = c1GL−1 + · · · + (cL−1 + 1) G1 = (GL −1) + G1 = GL.
(B.20) Hence, HL+1 = c1HL + · · · + (cL−1 + 1)H2 = c1GL + · · · + (cL−1 + 1) G2 = GL+1 −G1 + G2 = GL+1 −(1) + (c1 + 1) = GL+1 + c1.
(B.21) And so we see that HL+1 −GL+1 = c1 > 0 = 2 (HL −GL) .
(B.22) For the induction step, suppose for some m that our lemma holds for all 0 ≤k ≤m. We wish to show this holds for m + 1, so that HL+m+1 −GL+m+1 ≥2 (HL+m −GL+m).
Since {Gn} and {Hn} are PLRS, we expand the terms in question using their respective recur-rence relations. On this basis, we can claim that HL+m+1 −GL+m+1 ≥2 (HL+m −GL+m) if and only if L−1 X i=1 ciHL+m+1−i + Hm+2 − L−1 X i=1 ciGL+m+1−i −Gm+1 ≥2 L−1 X i=1 ciHL+m−i + Hm − L X i=1 ciGL+m−i −Gm−1 !
.
(B.23) By the induction hypothesis, we have that for all i, ciHL+m+1−i −ciGL+m+1−i ≥2 (ciHL+m−i −GL+m−i) .
(B.24) However, we can also show that Hm+2 −Gm+1 ≥2 (Hm+1 −Gm). By rewriting this as Hm+2 − 2Hm+1 ≥Gm+1 −2Gm, we see that for m ≤L −1, both sides are equal. For m ≥L + 1, it suffices to note that {Hn} grows faster, and thus so must the gaps between consecutive terms. By combining these two observations, the inequality (B.23) holds, which completes the proof.
□ APPENDIX C. LEMMAS FOR SECTION 3 Lemma C.1. For the PLRS Hn+1 = Hn + NHn−k−1, with N = ⌈(k + 2)(k + 3)/4⌉, then (N −2)Hn−k−1 ≤Hn−1 + · · · + Hn−k.
(C.1) Proof. By induction on n. Consider the base case, for n = k + 2: Hn−k−1 = H1 = 1, Hn−k = H2 = 2, . . . , Hn−1 = Hk+1 = k + 1.
(N −2) Hn−k−1 ≤Hn−1 + · · · + Hn−k ⇐ ⇒(N −2) ≤2 + 3 + · · · + k + (k + 1) ⇐ ⇒ (k + 2) (k + 3) 4 + 1 2 ≤(k + 1) (k + 2) 2 + 1 ⇐ = (k + 2) (k + 3) + 2 4 ≤k2 + 3k + 2 2 + 1 ⇐ ⇒k2 + 5k + 8 ≤2k2 + 6k + 8 ⇐ ⇒0 ≤k2 −k.
(C.2) 36 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Hence, the base case holds for k ≥0.
For the induction hypothesis, assume the following holds for arbitrary, fixed n: (N −2)Hn−k−1 ≤Hn−1 + · · · + Hn−k.
(C.3) For the induction step, we wish to show the following: (N −2)Hn−k ≤Hn + · · · + Hn−k+1.
(C.4) We write, using the recurrence relation, that (N −2)Hn−k+1 = (N −2)Hn−k + N(N −2)Hn−2k−2.
And by the induction hypothesis, (N −2)Hn−k+1 ≤Hn−1 + · · · + Hn−k + N(Hn−k−2 + · · · + Hn−2k−1) = k X i=1 Hn−i + NHn−k−1−i = k X i=1 Hn−i+1.
(C.5) Hence, the claim is true for all n ≥k + 1, k ≥0.
□ Lemma C.2. Let N = ⌈L(L + 1)/4⌉, and consider the recurrence relation [1, c2, . . . , cL−2, 0, N] where ci = 1 for one i ∈{2, . . . , L −2}, and the rest are 0. For fixed i ∈{2, . . . , L −2}, and L ≥6 then HL−i+1 + (N −2)H1 ≤HL−1 + · · · + H2.
Proof. This is equivalent to showing that HL−i+1 + L(L + 1) 4 ≤HL−1 + · · · + H2 + H1 + 1.
(C.6) Note that i ∈{2, . . . , L −2} and recall that each term in the sequence must increase from the previous. So HL−i+1 ∈{HL−1, . . . , H3}, and the largest possibility is HL−i+1 = HL−1 when i = 2. Thus, HL−i+1 + L(L + 1) 4 ≤HL−1 + L(L + 1) 4 .
(C.7) So we need only show that HL−1 + L(L + 1) 4 ≤HL−1 + · · · + H2 + H1 + 1, (C.8) which is equivalent to L(L + 1) 4 ≤HL−2 + · · · + H2 + H1 + 1.
(C.9) Note that Hi ≥i , so for L ≥6, HL−2 + · · · + H1 + 1 ≥L −2 + · · · + 1 + 1 = (L −2)(L −1) 2 + 1 = (L −2)(L −1) + 1 2 + 1 2 COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 37 ≥L(L + 1) 4 + 1 2 ≥ L(L + 1) 4 .
(C.10) Thus, we have shown equation (C.9), and the proof is complete.
□ Lemma C.3. Let N = ⌈L(L + 1)/4⌉, and consider the recurrence relation [1, c2, . . . , cL−2, 0, N] where ci = 1 for one i ∈{2, . . . , L −2}, and the rest are 0. For fixed i ∈{2, . . . , L −2}, and L ≥6, then for any n ≥L, Hn−i+1 + (N −2)Hn−L+1 ≤Hn−1 + · · · + Hn−L+2.
Proof. By induction on n. The base case, n = L, was shown in Lemma C.2. For the induction hypothesis, assume the claim in true for fixed n. We wish to show that the claim holds for n + 1.
Hn−i+2 + (N −2)Hn−L+2 = Hn−i+1 + Hn−2i+2 + NHn−L−i+2+ + (N −2)(Hn−L+1 + Hn−L−i+2 + NHn−2L+2).
By applying the induction hypothesis, Hn−i+2 + (N −2)Hn−L+2 ≤(Hn−1 + · · · + Hn−L+2) + (Hn−i + · · · + Hn−L−i+3) + N(Hn−L + · · · + Hn−2L+3) = Hn + · · · + Hn−L+3.
(C.11) Hence, the claim is true for all n ≥L.
□ Lemma C.4. For the PLRS {Hn} generated by [1, 1, 0, . . . , 0, N], if k is the number of zeros and N = ⌊(Fk+6 −(k + 5))/4⌋, then (N −2)Hn−k−2 ≤Hn−2 + · · · + Hn−k−1.
(C.12) Proof. In a similar manner to Lemma C.1, the statement is proved by induction on n.
For the base case, n = k+3, so that Hn−k−2 = H1 and {Hn−1, . . . , Hn−k−1} = {Hk+2, . . . , H2}.
Hence (N −2)H1 ≤Hk+2 + · · · + H2 ⇐ ⇒(N −2) ≤(Fk+3 −1) + · · · + (F3 −1) ⇐ ⇒(N −2) ≤ k+3 X i=1 Fi −(F1 + F2 + (k + 1)) ⇐ ⇒ Fk+6 −(k + 5) 4 ≤Fk+6 −(k + 5) 4 ≤Fk+5 −(k + 4) ⇐ ⇒Fk+6 −(k + 5) ≤4(Fk+5 −(k + 4)) ⇐ ⇒Fk+5 + Fk+4 ≤4Fk+5 −3k + 11 ⇐ ⇒3k + 11 ≤3Fk+5 −Fk+4 ⇐ ⇒3k + 11 ≤2Fk+4 + 3Fk+3, (C.13) where the last line is true for all k ≥0 by induction on k. Now, suppose the statement is true for some n ≥k + 3, so that (N −2)Hn−k−2 ≤Hn−2 + · · · + Hn−k−1. Thus, for the inductive step, note using the recursive definition that (N −2)Hn−k−1 = (N −2)Hn−k−2 + (N −2)Hn−k−3 + N(N −2)Hn−2k−4.
(C.14) 38 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ By the inductive hypothesis, (N −2)Hn−k−1 ≤Hn−2 + · · · + Hn−k−1 + (N −2)Hn−k−3 + N(N −2)Hn−2k−4 ≤ n−2 X i=n−k Hi + Hn−k−1 + (N −2)Hn−k−3 + N(N −2)Hn−2k−4.
(C.15) Note for all positive k we have that n −k −1 ≤n −2 as well as n −k −3 < n −3 and n −2k −4 < n −k −3. Since the sequence is non-decreasing and N is a positive integer, (N −2)Hn−k−1 ≤ n−2 X i=n−k+1 Hi + NHn−2 + NHn−3 + N(N −2)Hn−k−3 = n−1 X i=n−k Hi, (C.16) which completes the induction on n.
□ Lemma C.5. Pn i=1 2i−1i = 2n(n −1) + 1.
Proof. By induction on n.
□ Lemma C.6. Define {Fn} = [1, . . . , 1 | {z } g ] and {Hn} = [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , 2k+1−1]. Then Hg+k+1+n = Fg+k+1+n + (2k+1 −1)(2n + 2n−2(n −1)) when 1 ≤n ≤g −k.
Proof. Define a(n) so that Hg+k+1+n = Fg+k+1+n + a(n) for 1 ≤n ≤g −k Now, Hg+k+1+n = Hg+k+n + · · · + Hk+1+n + (2k+1 −1)Hn (C.17) = g+k+n X i=k+1+n Fi + n−1 X i=1 a(i) + k+1 X i=1 2i−1 + (2k+1 −1)2n−1 (since k + 1 + n ≤g + 1, (Hi −Fi) spans all the indices from k + 1 + n ≤g + 1 to g + k + n) = Fg+k+n+1 + n−1 X i=1 a(i) + (2k+1 −1)(2n−1 + 1) (C.18) Therefore, a(n) = n−1 X i=1 a(i) + (2k+1 −1)(2n−1 + 1) (C.19) and a(n −1) = n−2 X i=1 a(i) + (2k+1 −1)(2n−2 + 1).
(C.20) Hence a(n) = 2a(n −1) + (2k+1 −1)(2n−2).
(C.21) Since a(1) = 2(2k+1 −1), by induction we have a(n) = (2k+1 −1)(2n + 2n−2(n −1)).
(C.22) □ Lemma C.7. Define {Fn} = [1, . . . , 1 | {z } g ] and {Hn} = [1, . . . , 1 | {z } g , 0, . . . , 0 | {z } k , 2k+1 −1. Then Fg+n = 2g+n−1 −2n−2(n + 1) when 1 ≤n ≤g.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 39 Proof. Set Fg+n = 2g+n−1 −a(n) for 1 ≤n ≤g. Then Fg+n = Fg+n−1 + · · · + Fn (C.23) = 2g+n−2 + 2g+n−3 + · · · + 2n−1 −(a(n −1) + · · · + a(1)) = 2g+n−1 − 2n−1 + n−1 X i=1 a(i) !
Therefore, a(n) = 2n−1 + n−1 X i=1 a(i) (C.24) and a(n −1) = 2n−2 + n−2 X i=1 a(i).
(C.25) Hence a(n) = 2n−1 + 2a(n −1) −2n−2 = 2a(n −1) + 2n−2.
(C.26) Since a(1) = 1, by induction, we have a(n) = 2n−2(n + 1).
□ Lemma C.8. For k + ⌈log2 k⌉≤g < 2k, {Hn} defined as [1, . . . , 1 | {z } g , 0, . . . , 0 | {z }, 2k+1 −1], we have 2g+k+1 − g+k+1 X i=k+n+2 Hi ≤2g + 2k+n+2 −2n+1 (C.27) for all g −k ≤n ≤k.
Proof. By induction on n. Suppose it holds for some n ≥g −k. Then 2g+k+1 − g+k+1 X i=k+n+3 Hi = 2g+k+1 − g+k+1 X i=k+n+2 Hi + Hk+n+2.
By the induction hypothesis, 2g+k+1 − g+k+1 X i=k+n+3 Hi ≤2g + 2k+n+2 −2n+1 + Hk+n+2.
As we can check explicitly that Hk+n+2 ≤2k+n+1 ≤2k+n+2 −2n+1, we see 2g+k+1 − g+k+1 X i=k+n+3 Hi ≤2g + 2k+n+2 −2n+1 + (2k+n+2 −2n+1) = 2g + 2k+n+3 −2n+2.
(C.28) It remains to show for the base case n = g−k. This can be shown directly from the given formulas in Theorem 3.5.
□ Lemma C.9. For {Hn} defined as in Lemma C.8 and the same conditions on g and k, we have H(g+k+1)+n ≥(2k+1 −1)(2g+n−1 −2n−2(n + 1)) (C.29) for all 1 ≤n ≤k.
40 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ Proof. This is equivalent to showing that 2g+k+n −H(g+k+1)+n ≤2g+n−1 + 2k+n−1(n + 1) −2n−2(n + 1) for all 1 ≤n ≤k, (C.30) which we proceed by strong induction on n. The case 1 ≤n ≤g −k has been established in Theorem 3.5, so we suppose this holds for all n ≤m for some g −k ≤m < k. Then 2g+k+(m+1) −H(g+k+1)+(m+1) = 2g+k+m+1 − m X i=1 H(g+k+1)+i + g+k+1 X i=k+m+2 Hi + (2k+1 −1)Hm+1 !
= m X i=1 2g+k+i −H(g+k+1)+i + 2g+k+1 − g+k+1 X i=k+m+2 Hi !
−(2k+1 −1)2m.
By the inductive hypothesis and Lemma C.8, ≤ m X i=1 2g+i−1 + 2k+i−1(i + 1) −2i−2(i + 1) + 2g + 2k+m+2 −2m+1 −(2k+1 −1)2m = 2g+(m+1)−1 + 2k+(m+1)−1((m + 1) + 1) −2(m+1)−2((m + 1) + 1).
(C.31) Our proof by induction is complete.
□ Lemma C.10. The sequence generated by [1, . . . , 1, 0, 3], with k ≥1 ones, is always complete.
Proof. By strong induction on n.
For the base case, with n = 1, we have H2 = H1 + 1.
For the induction hypothesis, assume that for some n, Brown’s Criterion holds for all m < n, i.e., assume Hm+1 ≤1 + H1 + · · · + Hm for all m < n.
For the induction step, we start with the recurrence relation and apply the induction hypothesis: Hn+1 = Hn + · · · + Hn−k+1 + 3Hn−k−1 ≤Hn + · · · + Hn−k+1 + Hn−k + 2Hn−k−1 ≤Hn + · · · + Hn−k+1 + Hn−k + Hn−k−1 + Hn−k−2 + · · · + H1 + 1.
(C.32) Hence, by Brown’s Criterion, the sequence is complete. By strong induction, the lemma is proved.
□ Lemma C.11. Let {Hn} defined by [1, 0, . . . , 0, 1, . . . , 1 | {z } m , N] be a PLRS with L coefficients. Then, if the sequence is incomplete, it must fail Brown’s criterion at the L + 1-th or L + 2-th term. In other words, if HL+1 ≤1 + PL i=1 Hi and HL+2 ≤1 + PL+1 i=1 Hi, then {Hn} is complete.
Proof. Let {Hn} be defined as above; it is clear that the first L terms pass Brown’s criterion. Now suppose the sequence passes Brown’s criterion at the L + 1-ist and L + 2-nd term, so that HL+1 ≤ L X i=1 Hi + 1, HL+2 ≤ L+1 X i=1 Hi + 1.
(C.33) We show that {Hn} is complete.
We show by induction that if {Hn} satisfies Brown’s criterion at the L+2nd term, then it satisfies Brown’s criterion at the L + kth term, for any 2 ≤k ≤L −1. We assume our base case of k = 2 by hypothesis, so only the induction step remains to be shown.
COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 41 Suppose for some k that HL+k = HL+k−1 + Hk+m + · · · + Hk+1 + NHk ≤ L+k−1 X i=1 Hi + 1.
(C.34) We wish to show that HL+k+1 = HL+k + Hk+m+1 + · · · + Hk+2 + NHk+1 ≤ L+k X i=1 Hi + 1.
(C.35) Looking at the difference between equations (C.34) and (C.35), we see it suffices to show that (HL+k −HL+k−1) + N (Hk+1 −Hk) + Hk+m+1 −Hk+1 ≤HL+k.
(C.36) Or equivalently, N (Hk+1 −Hk) + Hk+m+1 −Hk+1 ≤HL+k−1.
(C.37) Expanding HL+k−1, HL+k−1 = HL+k−2 + Hk+m−1 + · · · + Hk + NHk−1.
(C.38) We can repeatedly expand the largest term of expression (C.38), giving us longer partial sums. In particular, applying the process k times, we see: HL+k−1 = HL+k−2 + k+m−1 X i=k Hi + NHk−1 = HL+k−3 + k+m−2 X i=k−1 Hi + NHk−2 !
+ k+m−1 X i=k Hi + NHk−1 = HL+k−3 + k+m−3 X i=k−2 Hi + NHk−3 !
+ k+m−2 X i=k−1 Hi + k+m−1 X i=k Hi + N(Hk−1 + Hk−2) .
.
.
= HL−1 + m X i=1 Hi + · · · + k+m−1 X i=k Hi + N (Hk−1 + Hk−2 + · · · + H1) = HL−1 + k X a=1 a+m−1 X i=a Hi + N k−1 X i=1 Hi !
.
(C.39) Thus inequality (C.37) becomes N (Hk+1 −Hk −Hk−1 −· · · −H2 −H1) + Hk+m+1 ≤HL−1 + Hk+1 + k X a=1 a+m−1 X i=a Hi. (C.40) Assuming k < L, we can write Hk+1 = Hk + Hk−L+m+1 + · · · + Hk−L+2, where for the sake of notation we define H0 = 1, and Hj = 0 for all j < 0. Thus, N (Hk+1 −Hk −Hk−1 −· · · −H2 −H1) = −N (Hk−1 + Hk−2 + · · · + Hk−L+m+2) < −NHk−1, (C.41) 42 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ and so for inequality (C.40) it suffices to show Hk+m+1 ≤HL−1 + Hk+1 + k X a=1 a+m−1 X i=a Hi + NHk−1.
(C.42) Expanding the left hand side, we wish to show Hk+m+1 = Hk+m + Hk−L+2m+1 + · · · + Hk−L+1 + NHk−L+m ≤HL−1 + Hk+1 + k X a=1 a+m−1 X i=a Hi + NHk−1.
(C.43) As k −L + m ≤k −3 < k −1, we see NHk−L+m < NHk−1, and so we need only show Hk+m + Hk−L+2m+1 + · · · + Hk−L+1 ≤HL−1 + Hk+1 + k X a=1 a+m−1 X i=a Hi.
(C.44) As k ≥2, we note that in the double sum Pk a=1 Pa+m−1 i=a Hi, the summands H1, Hk+m−1 are present exactly once, and for any 1 < i < k + m −1, the summand Hi is present at least twice.
Thus we can take the crude bound k X a=1 a+m−1 X i=a Hi ≥ k+m−1 X i=1 Hi + k+m−2 X i=2 Hi.
(C.45) Applying this bound on the right hand side of inequality (C.44), and taking the trivial bounds HL−1 > H1 + 1, Hk+1 > 1, we see HL−1 + Hk+1 + k X a=1 a+m−1 X i=a Hi ≥ k+m−1 X i=1 Hi + 1 !
+ k+m−2 X i=1 Hi + 1 !
.
(C.46) As we assumed {Hn} fulfills Brown’s criterion for terms below L + k, we know Hk+m ≤ k+m−1 X i=1 Hi + 1.
(C.47) Finally, it is clear that Hk−L+2m+1 + · · · + Hk−L+1 ≤ k+m−2 X i=1 Hi + 1, (C.48) as no indices on the left sum are repeated. Combining these two facts, we have Hk+m + Hk−L+2m+1 + · · · + Hk−L+1 ≤ k+m−1 X i=1 Hi + 1 !
+ k+m−2 X i=1 Hi + 1 !
≤HL−1 + Hk+1 + k X a=1 a+m−1 X i=a Hi.
(C.49) Thus inequality (C.44) holds, and we are done.
□ COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 43 Lemma C.12. Let {Gn} and {Hn} be PLRS with L coefficients defined by [1, 0, . . . , 0, 1, . . . , 1 | {z } m , M] and [1, 0, . . . , 0, 1, . . . , 1 | {z } m+1 , N] respectively. For (L −1)/2 ≤m ≤L −4, Hi−1 = Gi −1 2 ≤i < 2(L −m) Hi−1 = Gi i = 2(L −m) Hi−1 > Gi 2(L −m) < i ≤L.
(C.50) Proof. From the proof of Lemma 3.10, we get ( Hn = n, H(L−m−1)+n = L −m −1 + n + n(n+1)(n+2) 6 1 ≤n ≤L −m −1 Gn = n, GL−m+n = L −m + n + n(n+1)(n+2) 6 1 ≤n ≤L −m.
(C.51) From these explicit formulas, Hi−1 = Gi −1 for all 2 ≤i ≤2(L −m) −1. Now, H2(L−m)−1 = H2(L−m−1) + HL−m + L−m−1 X i=1 Hi + 1 = 2(L −m −1) + (L −m −1)(L −m)(L −m + 1) 6 + 1 + L−m−i X i=1 Gi + 1 = 2(L −m) −1 + (L −m −1)(L −m)(L −m + 1) 6 + L−m−i X i=1 Gi + 1 = G2(L−m)−1 + L−m−i X i=1 Gi + 1 = G2(L−m).
(C.52) Similarly, by writing out explicit formulas, one can show that H2(L−m) > G2(L−m)+1. Also, it is clear that Hi ≥Gi for all i. Therefore, for any 2(L −m) + 1 < k ≤L, Hk−1 −Gk = (Hk−2 −Gk−1) + k−(L−m) X i=1 (Hi −Gi) ≥Hk−2 −Gk−1, (C.53) and the last inequality follows by induction on k.
□ Lemma C.13. If m + 3 < 2(L −m) and m ≥(L −1)/2, then 2m −L + m+1 X i=L−m+2 i ≥2L −3(m + 1) + 2.
(C.54) Proof. This is equivalent to (m + 1)(m + 2) 2 −(L −m + 1)(L −m + 2) 2 ≥3L −5m −1, (C.55) which simplifies to L(2m −L) + 16m + 2 ≥8L, (C.56) which is true since L(2m −L) > 6 and 2m + 1 ≥L.
□ 44 E. BOŁDYRIEW, J. HAVILAND, P. LÂM, J. LENTFER, S. J. MILLER, AND F. TREJOS SUÁREZ APPENDIX D. LEMMAS FOR SECTION 4 Lemma D.1. For any L ∈Z+, let λL be the principal root of xL −xL−1 −NL −1. Then (λL −1) ≥ L + 2 L2 + L + 4.
(D.1) Proof. Since 1/(L2 + L + 4) > 1/(L2 + 4L + 4) = 1/(L + 2)2, it suffices to show (λL −1) ≥ L + 2 L2 + 4L + 4 = 1 L + 2, (D.2) or equivalently, λL ≥(L+3)/(L+2). This inequality holds if and only if f((L+3)/(L+2)) ≤0, i.e., L + 3 L + 2 L − L + 3 L + 2 L−1 − L (L + 1) 4 −1 ≤0, (D.3) or, (L + 3)L−1 (L + 2)L ≤ L (L + 1) 4 + 1.
(D.4) It can be checked that for all L ≥1, a stronger condition holds, that (L + 3)L−1 (L + 2)L < 1, (D.5) completing the proof.
□ REFERENCES [BBGILMT] Olivia Beckwith, Amanda Bower, Louis Gaudet, Rachel Insoft, Shiyu Li, Steven J. Miller, and Philip Tosteson. The Average Gap Distribution for Generalized Zeckendorf Decompositions, Fibonacci Quar-terly 51 (2013), 13–27.
[Br] J. L. Brown. Note on complete sequences of integers, American Mathematical Monthly 68 (1961), no.
6, 557.
[Fr] Aviezri S. Fraenkel, Systems of numeration, American Mathematical Monthly 92 (1985), no. 2, 105– 114.
[GT] P. J. Grabner and R. F. Tichy, Contributions to digit expansions with respect to linear recurrences, J.
Number Theory 36 (1990), no. 2, 160–169.
[GM] Daniele A. Gewurz and Francesca Merola, Numeration and enumeration, European Journal of Combi-natorics 32 (2012), no. 7, 1547–1556.
[HK] V. E. Hoggatt and C. King, problem E 1424, American Mathematical Monthly 67 (1960), no. 6, 593.
[MMMMS] Thomas C. Martinez, Steven J. Miller, Clay Mizgerd, Jack Murphy, and Chenyang Sun. Generalizing Zeckendorf?s Theorem to Homogeneous Linear Recurrences II, preprint (2020).
[MW] S. J. Miller and Y. Wang, From Fibonacci numbers to Central Limit Type Theorems, Journal of Combi-natorial Theory, Series A 119 (2012), no. 7, 1398–1413.
[Ze] E. Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bulletin de la Société Royale des Sciences de Liège 41 (1972), 179–182.
E-mail address: eboldyriew@colgate.edu DEPARTMENT OF MATHEMATICS, COLGATE UNIVERSITY, HAMILTON, NY 13346 E-mail address: havijw@umich.edu DEPARTMENT OF MATHEMATICS, UNIVERSITY OF MICHIGAN, ANN ARBOR, MI 48109 E-mail address: plam6@u.rochester.edu COMPLETENESS OF POSITIVE LINEAR RECURRENCE SEQUENCES 45 DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ROCHESTER, ROCHESTER, NY 14627 E-mail address: jlentfer@hmc.edu DEPARTMENT OF MATHEMATICS, HARVEY MUDD COLLEGE, CLAREMONT, CA 91711 E-mail address: sjm1@williams.edu, Steven.Miller.MC.96@aya.yale.edu DEPARTMENT OF MATHEMATICS AND STATISTICS, WILLIAMS COLLEGE, WILLIAMSTOWN, MA 01267 E-mail address: fernando.trejos@yale.edu DEPARTMENT OF MATHEMATICS, YALE UNIVERSITY, NEW HAVEN, CT 06520 |
188755 | https://gogeometry.com/school-college/2/p1120-isosceles-right-triangle-equilateral-metric-relation-math.htm | Geometry Problem 1120: Isosceles Right Triangle, 120 Degree, Angle, Equilateral, Metric Relations
Geometry Problem 1120: Isosceles Right Triangle, 120 Degree, Angle, Equilateral, Metric Relations. Level: High School, SAT Prep, Honors Geometry, College, Math Education ========================================================================================================================================================================= < PREVIOUS PROBLEM |NEXT PROBLEM > The figure below shows an isosceles right triangle ABC. D is a point so that angle ADC = 120 degree. E is on CD extended so that angle BEC = 120 degree. BE extended meets AD at F. If AD = d, CE = e, and BF = f, prove that (1) triangle DEF is equilateral; (2) . Home | Search |Geometry | Problems | All Problems | Open Problems | Visual Index | 10 Problems | 1111-1120 | Triangle | Right triangle | Isosceles Triangle | Equilateral triangle | Angles 30, 60, 120 degree | Email | Solution / comment. Last updated: Jun 11, 2015 by Antonio Gutierrez. |
188756 | https://www.quora.com/How-many-even-numbers-of-5-digits-without-repetition-can-be-formed-with-1-2-3-4-and-5 | How many even numbers of 5 digits without repetition can be formed with 1, 2, 3, 4, and 5? - Quora
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How many even numbers of 5 digits without repetition can be formed with 1, 2, 3, 4, and 5?
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To find the number of 5-digit even numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repetition, we need to consider the fact that an even number must end with an even digit. In this case, the even digits available are 2 and 4.
Step 1: Analyze the ending digits
We can divide the problem into two cases based on the last digit:
Last digit is 2
Last digit is 4
Case 1: Last digit is 2
The remaining digits are 1, 3, 4, and 5.
We need to arrange these 4 digits in the first four positions.
The number of arrangements of 4 digits is given by 4!4!:
4!=24 4!=24
Case 2: Last digit is 4
The remaining di
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To find the number of 5-digit even numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repetition, we need to consider the fact that an even number must end with an even digit. In this case, the even digits available are 2 and 4.
Step 1: Analyze the ending digits
We can divide the problem into two cases based on the last digit:
Last digit is 2
Last digit is 4
Case 1: Last digit is 2
The remaining digits are 1, 3, 4, and 5.
We need to arrange these 4 digits in the first four positions.
The number of arrangements of 4 digits is given by 4!4!:
4!=24 4!=24
Case 2: Last digit is 4
The remaining digits are 1, 2, 3, and 5.
We need to arrange these 4 digits in the first four positions.
The number of arrangements of 4 digits is again 4!4!:
4!=24 4!=24
Step 2: Total combinations
Now, we add the combinations from both cases:
24(last digit 2)+24(last digit 4)=48 24(last digit 2)+24(last digit 4)=48
Conclusion
Thus, the total number of 5-digit even numbers that can be formed with the digits 1, 2, 3, 4, and 5 without repetition is 48.
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Anthony Nguyen
7y
Let v, w, x, y, z represent the ten thousands, thousands, hundreds, tens, and ones-respectively.
v has 5 possible integer choices, w has 4 possible integer choices, x has 3 possible integer choices, y has 2 possible integer choices, and z has 1 possible integer choice
Since we want even number to be z, it shall have 2 integer choices.
Therefore, the possible numbers will be: 2wxy4, v2xy4, vw2y4, vwx24, 4wxy2, v4xy2, vw4y2, vwx42
There will be 6 possibilities, 3×2×1,3×2×1, of numbers for every combination.
The number of combinations multiplied by possibilities, 8×6=48 8×6=48
There are 48 even numbers of 5 dig
Continue Reading
Let v, w, x, y, z represent the ten thousands, thousands, hundreds, tens, and ones-respectively.
v has 5 possible integer choices, w has 4 possible integer choices, x has 3 possible integer choices, y has 2 possible integer choices, and z has 1 possible integer choice
Since we want even number to be z, it shall have 2 integer choices.
Therefore, the possible numbers will be: 2wxy4, v2xy4, vw2y4, vwx24, 4wxy2, v4xy2, vw4y2, vwx42
There will be 6 possibilities, 3×2×1,3×2×1, of numbers for every combination.
The number of combinations multiplied by possibilities, 8×6=48 8×6=48
There are 48 even numbers of 5 digits without repition that are formed with 1, 2, 3, 4, and 5.
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John Stephenson
Analyst programmer for years, but I love mathematics. · Author has 2.4K answers and 2.8M answer views
·7y
There are 5! = 120 combinations of all five digits without repetition.
Only those ending in 2 or 4 are even, or 2/5 of all possible numbers.
So, only 2/5 of the total numbers will be even = 120 2/5 = 48
Upvote ·
9 3
Ian O'Neill
BSc(Hons) in Mathematics from University of Southampton (Graduated 1982)
·3y
Originally Answered: What is the number of 5 digit even numbers that can be formed using the digits 1, 2, 3, 4 and 5 without repetition? ·
It’s a five digit number, so it’s of the form ABCDE where each letter represents a digit.
To be even, the final digit E must be either 2 or 4. (If E is 1, 3 or 5, the number will be odd, because it ends with an odd digit.)
Whichever it is (E=2 or E=4), that leaves us with four possible choices for the first digit, A. Choose one. That then leaves 3 choices for B, then 2 choices for C, then only one digit left for D.
Now multiply the choices. There are:
(2 choices for E) x (4 choices for A) x (3 choices for B) x (2 choices for C) x (1 choice for D),
in other words, the number of possible choices for
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It’s a five digit number, so it’s of the form ABCDE where each letter represents a digit.
To be even, the final digit E must be either 2 or 4. (If E is 1, 3 or 5, the number will be odd, because it ends with an odd digit.)
Whichever it is (E=2 or E=4), that leaves us with four possible choices for the first digit, A. Choose one. That then leaves 3 choices for B, then 2 choices for C, then only one digit left for D.
Now multiply the choices. There are:
(2 choices for E) x (4 choices for A) x (3 choices for B) x (2 choices for C) x (1 choice for D),
in other words, the number of possible choices for ABCDE is 2 x 4 x 3 x 2 x 1, which is 48.
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More answers below
How many even numbers of 5 digit can be formed with the digit 1 2 3 4 and 5?
How many 3-digit numbers can form the digits 2, 3, 4, 5, and 6 with repetition and without repetition?
How many odd three digit numbers can be formed using the numbers 0,1,2,3,4,5,6?
How many three digit even numbers can be formed from the digits 1, 2, 3, 4, and 5 with repetition a digit and without repetition?
How many even numbers can be formed from 2, 3, 4, 5, 6 if repetition is not allowed?
Calvin Campbell
B.Sc. in Computer Science&Mathematics, University of the West Indies (Graduated 1984) · Author has 3.9K answers and 4.1M answer views
·3y
Originally Answered: What is the number of 5 digit even numbers that can be formed using the digits 1, 2, 3, 4 and 5 without repetition? ·
Without repetition of digits, for each number formed from the 5 given digits we have 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the fifth digit.
Therefore, 54321 = 120 such 5-digit numbers are possible. Since 2 of the 5 digits are even, 2/5 120 = 48 of the 120 numbers formed are even.
Good luck!
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Ernest Leung
B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·3y
Originally Answered: What is the number of 5 digit even numbers that can be formed using the digits 1, 2, 3, 4 and 5 without repetition? ·
• There are 2 ways (2 or 4) to fill the units digit.
• There are ₄P₄ ways to fill the rest 4 digits.
No. of 5-digit even number = 2 × ₄P₄ = 48
Upvote ·
Manjunath Subramanya Iyer
I am a retired bank officer teaching maths · Author has 7.2K answers and 10.4M answer views
·3y
Related
What is the number of 5 - digit even numbers that can be formed using the digits 1, 2, 7, 9, 4 without repetition?
Any even number ends with 2, 4, 6, 8, 0.
We have been given the digits 1, 2, 7, 9, 4
So the 5-digit numbers should end either 2 or 4. Hence there are 2 ways of filling up the units digit.
Having filled the units place, we will fill the ten thousands place which can be done in 4 ways (by any one of 1, 7, 9 and the remaining even number), the thousands place by remaining 3 in 3 ways, the hundreds place in 2 ways and the tens place in 1 way.
So by fundamental counting principle, the places can be filled in 4×3×2×1×4= 96.
So we have 96 five-digit even numbers that can be formed out of the digits 1,2,7,
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Any even number ends with 2, 4, 6, 8, 0.
We have been given the digits 1, 2, 7, 9, 4
So the 5-digit numbers should end either 2 or 4. Hence there are 2 ways of filling up the units digit.
Having filled the units place, we will fill the ten thousands place which can be done in 4 ways (by any one of 1, 7, 9 and the remaining even number), the thousands place by remaining 3 in 3 ways, the hundreds place in 2 ways and the tens place in 1 way.
So by fundamental counting principle, the places can be filled in 4×3×2×1×4= 96.
So we have 96 five-digit even numbers that can be formed out of the digits 1,2,7,9,4, without repeating any digit
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Saf Greenwood
Studied Mathematics · Author has 70 answers and 99K answer views
·4y
Related
How many 5 digit numbers can be formed from 1, 2, 3, 4, 5, and 6 if repetitions of digits are not allowed?
There are 6 options for the first digit.
5 options for the second (because repetitions are not allowed).
…
2 options for the fifth digit.
Therefore, the total number of 5 digit numbers available is 65432.
Because order matters (12345 is a different answer to 54321) it is a permutation (rather than combination - where order doesn’t matter).
The formula for working out the number of permutations is nPr = n!/(n-r)! Where n = the number of options and r = the number of selections.
So 6P5 = 6!/(6–5)! = 6!/1!
! means you multiply the number by all integers to 1. So 3! = 321 = 6. 1! = 1.
So 6P5 = 6
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There are 6 options for the first digit.
5 options for the second (because repetitions are not allowed).
…
2 options for the fifth digit.
Therefore, the total number of 5 digit numbers available is 65432.
Because order matters (12345 is a different answer to 54321) it is a permutation (rather than combination - where order doesn’t matter).
The formula for working out the number of permutations is nPr = n!/(n-r)! Where n = the number of options and r = the number of selections.
So 6P5 = 6!/(6–5)! = 6!/1!
! means you multiply the number by all integers to 1. So 3! = 321 = 6. 1! = 1.
So 6P5 = 654321/1 = 65432 (as above)
So, there are 720 ways that 6 digits can be arranged in 5 digit numbers without repetitions.
I hope this helps.
Upvote ·
9 3
Vidyanand Wagh
Introvert | Thantophobic | Cricket Crazy | Federer Fan
·Updated 9y
Related
How many 3 digit even numbers can be formed from 0, 1, 2, 3, 4, 5 and 6 with no repetition?
It's 105.
Okay, so let's see this step by step.
As we know even numbers are those integers which have 0 or 2 or 4 or 6 or 8 at the unit's place.
Since we want three digit even numbers with no repetition, we are seeking for numbers which end with 0/2/4/6/8 and do not start with 0.
Case 1: Numbers ending with 0.
Since they already have 0 in the unit's place, some other digit should occupy the 10th's place.
There are 6 other digits which can occupy this place.
Now let's come to 100th's place. Apart from 0 and the digit that's already put in the 10th's place, there are 5 distinct digits which may no
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It's 105.
Okay, so let's see this step by step.
As we know even numbers are those integers which have 0 or 2 or 4 or 6 or 8 at the unit's place.
Since we want three digit even numbers with no repetition, we are seeking for numbers which end with 0/2/4/6/8 and do not start with 0.
Case 1: Numbers ending with 0.
Since they already have 0 in the unit's place, some other digit should occupy the 10th's place.
There are 6 other digits which can occupy this place.
Now let's come to 100th's place. Apart from 0 and the digit that's already put in the 10th's place, there are 5 distinct digits which may now occupy the 100th's place.
Thus, total number of combinations = 5 6 = 30
Case 2: Numbers ending with 2 or 4 or 6
We now have 3 options to choose from and put at the unit's place.
Let say we choose some digit (say 2) and put it in the unit's place.
Now that we've already used 2, it cannot be used again in the remaining places.
Additionally we've one more condition that we cannot start our number with 0. So now if we consider 100th's place, 0 and 2 have already been used so we're left with 5 options for 100th's place.
After filling the 100th's place, let's come to 10th's place.Here we cannot use 2, and we cannot use the number we already used at 100th's place. But wait, here we can use 0, right? So again we've 5 options for this place.
Thus total number of combinations ending with 2 = 5 5 = 25
(The same calculation would follow for numbers ending with 4 and 6 as well, so that would contribute 25 combinations each.)
Thus,
Total number of even combinations without repetition of digits
= 30 + (325) = 105
Hope this helps :)
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Zubair Sabir
B.E in Electrical Engineering, National University of Sciences and Technology, Pakistan (Graduated 2022) · Author has 106 answers and 457.7K answer views
·7y
Related
How many five-digit odd numbers can be formed from the digits 1, 2, 3, 4, and 5 if no digit is repeated?
How many five-digit odd numbers can be formed from the digits 1, 2, 3, 4, and 5 if no digit is repeated?
Five digit in place value chart means:
Ten thousand-Thousand-Hundred-Ten-Unit
at unit place you can not put any of five number because we are required only odd numbers (which are not exactly divisible by 2 and they end at 1,3,5,7 and 9 only) so you can put here only three numbers out of five i.e 1,3 and 5.
at Ten place you can put any of 5 number but we need the numbers with no repetition. hence only 4 numbers can be put at this place.
at Hundred place you can put any of 5 number but we need the
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How many five-digit odd numbers can be formed from the digits 1, 2, 3, 4, and 5 if no digit is repeated?
Five digit in place value chart means:
Ten thousand-Thousand-Hundred-Ten-Unit
at unit place you can not put any of five number because we are required only odd numbers (which are not exactly divisible by 2 and they end at 1,3,5,7 and 9 only) so you can put here only three numbers out of five i.e 1,3 and 5.
at Ten place you can put any of 5 number but we need the numbers with no repetition. hence only 4 numbers can be put at this place.
at Hundred place you can put any of 5 number but we need the numbers with no repetition. hence only 3 (b/c two has been used at unit and ten places) numbers can be put at this place.
at Thousand place you can put any of 5 number but we need the numbers with no repetition. hence only 2 (b/c three has been used at unit, ten and hundred places) numbers can be put at this place.
at Ten Thousand place you can put any of 5 number but we need the numbers with no repetition. hence only 1 (b/c four has been used at unit, ten, hundred and thousand places) numbers can be put at this place.
Hence five-digit odd numbers can be formed from the digits 1, 2, 3, 4, and 5 if no digit is repeated are:
= 34321
= 72 digits (Answer)
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99 25
9 1
Injamamul Hoque
B.SC in Mathematics, Jadavpur University (Graduated 2017)
·7y
Related
How many 5-digits can even number be formed using the digits 1, 2, 5, 5, and 4?
The given digits are 1,2,5,5 and 4.
Now to form a five digit even number, it is clear that the last space should be filled by an even number. And consider the position of five digit as a five space .
As there is two given even digits so last space can be filled by 2 different ways.
Now the remaining four spaces can be filled by four different( if they are) numbers by 4! Ways.
But as there is two 5 given in the digit list. So the remaining four space can be filled by 4!/2! = 12 ways.
So the number of five digit even numbers can be formed by given digits are 12×2 =24 .
Upvote ·
9 3
9 3
John Chadwick
Former Head of Electrical, Mechanical & MV Engineering at South Cheshire College (1972–2007) · Author has 1K answers and 609.8K answer views
·1y
Related
How many even numbers can be formed from 1, 2, 3, 4, 5?
Assuming NO REPETITION of the digits is allowed.
Represent the possible arrangements of the digits for the numbers as follows:-
Z, YZ, XYZ, WXYZ, VWXYZ.
Z can only have the values of 2 or 4 if the numbers are to be EVEN:-
YZ ==> Put Z = 2 and Y can have 4 values so YZ = 41 = 4.
Put Z = 4 and Y can again have 4 values so YZ = 41 = 4
Total YZ = 4+4 = 8 numbers.
XYZ ==> Put Z = 2 then Y can have 4 values and X can have 3 values therefore XYZ = 431 = 12
Put Z = 4 then as above XYZ = 431 = 12
Total XYZ = 12+12 = 24 numbers.
WXYZ ==> Put Z = 2 then WXYZ = 4321 = 24
Put Z = 4 then as above WXYZ = 432
Continue Reading
Assuming NO REPETITION of the digits is allowed.
Represent the possible arrangements of the digits for the numbers as follows:-
Z, YZ, XYZ, WXYZ, VWXYZ.
Z can only have the values of 2 or 4 if the numbers are to be EVEN:-
YZ ==> Put Z = 2 and Y can have 4 values so YZ = 41 = 4.
Put Z = 4 and Y can again have 4 values so YZ = 41 = 4
Total YZ = 4+4 = 8 numbers.
XYZ ==> Put Z = 2 then Y can have 4 values and X can have 3 values therefore XYZ = 431 = 12
Put Z = 4 then as above XYZ = 431 = 12
Total XYZ = 12+12 = 24 numbers.
WXYZ ==> Put Z = 2 then WXYZ = 4321 = 24
Put Z = 4 then as above WXYZ = 4321 = 24
Total = 24+24 = 48 numbers.
VWXYZ ==> Put Z = 2 then VWXYZ = 43211 = 24
Put X = 4 then as above VWXYZ = 4321 = 24
Total = 24+24 = 48 numbers.
Overall total of EVEN numbers possible from {1,2,3,4,5} = 2+8+24+48+48 = 130 numbers.
Upvote ·
Rishabh Varma
Comfortable with Concepts of Mathematics · Author has 112 answers and 821.6K answer views
·8y
Related
How many 3 digit even numbers can be formed from 1, 2, 4, and 5 with no repetition?
For 3 digit even number, the last digit will have to be either 2 or 4.
Case 1: 2 is fixed in the UNITS PLACE
Noe there are 3 possibilities in the HUNDREDS PLACE: 1,4 or 5
After this we are left with ONLY 2 Options for the TENS PLACE (As repitition is not allowed)
Total possibilities: 3 x 2 x 1 = 6
Case 2: 4 is fixed in the UNITS PLACE.
Same procedure.
Total possibilities: 3 x 2 x 1 = 6
TOTAL POSSIBILITIES = 6 + 6 =12
Upvote ·
9 1
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188757 | https://dl.konkur.in/post/Book/Dentistry/McCracken%27s-Removable-Partial-Prosthodontics-13th-edition-%5Bkonkur.in%5D.pdf | 0F&UDFNHQ V 5HPRYDEOH 3DUWLDO 3URVWKRGRQWLFV Thirteenth Edition www.konkur.in Thirteenth Edition McCracken’s REMOVABLE PARTIAL PROSTHODONTICS Alan B. Carr, DMD, MS Professor Department of Dental Specialties Mayo Clinic Rochester, Minnesota David T. Brown, DDS, MS Chair Department of Comprehensive Care and General Dentistry Indiana University School of Dentistry Indianapolis, Indiana www.konkur.in 3251 Riverport Lane St. Louis, Missouri 63043 McCRACKEN’S REMOVABLE PARTIAL PROSTHODONTICS, THIRTEENTH EDITION ISBN: 978-0-323-33990-2 Copyright © 2016 by Elsevier, Inc. All rights reserved.
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International Standard Book Number: 978-0-323-33990-2 Executive Content Strategist: Kathy Falk Content Development Manager: Jolynn Gower Content Development Specialist: Laura Klein Publishing Services Manager: Julie Eddy Senior Project Manager: Marquita Parker Designer: Ryan Cook Printed in Canada Last digit is the print number: 9 8 7 6 5 4 3 2 1 www.konkur.in Preface In response to the question, “What is the purpose of research?” a recent Lancet editorial suggests a redefinition of the purpose of research is critical due to an alarming lack of concordance between a reasoned research purpose and actual research reality. Ian Chalmers provides a clear reasoned purpose of research as being, “to help patients and their clinicians.”† He goes on to describe 85% waste associated with research meet-ing this target, and relates this to several root causes: choos-ing the wrong questions, conducting unnecessary or poorly designed studies, failure to publish in a timely manner or at all, and biased or unusable reports of research.
The editors of this textbook consider the related ques-tion each time the publisher requests an update. “What is the purpose of a textbook?” This question becomes increasingly important as technology shapes the behavior of learners and therefore the strategies for enhancing learning. As Chalmers and Glasziou state, we believe the purpose of a textbook is to help patients by helping their providers. A textbook may help providers at various stages of a career.
For first time learners, it helps by presenting a foundation for learning during a period of active and diverse informa-tion exposure for the developing clinician. Competing for learners’ attention during this phase, presenting content that can be directly applied to first time application scenarios is a challenge for teachers and compels texts be designed to add substance, not waste. For practicing clinicians, a textbook can provide reinforcement of principles useful for clinical decisions and their application; especially if the clinical chal-lenges become more complex for the maturing provider. In this context, a text can serve as a source of continuing ref-erence as clinical experience expands a clinician’s expertise in application of basic principles, allowing more in-depth understanding and application.
The editors also recognize that as providers of oral health care, we are part of a changing health care environment. It is evident that our contribution to overall health and well-being of society will be an increasing part of the United States health care value discussion, a discussion that stems from the recognition that care costs and quality are not aligned in the United States. The current “value care” transition Horton R: Editorial: what is the purpose of medical research? Lancet 381:347, 2013.
†Chalmers I, Glasziou P: Avoidable waste in the production and report-ing of research evidence, Lancet 374:86-89, 2009.
from “volume care” places a premium on patient-centered needs/desires, demonstration of beneficial outcomes, and cost containment in a context of care over time. This context fits the care needs addressed by tooth replacement inter-ventions. Management of tooth loss must be considered as a life-long process. Each decision along a patient’s “life course” can substantially impact subsequent care opportu-nities and therefore the decisions made. What is critical to recognize is that the impact of decisions is not equivalent among options.
Consequently, we provide an argument that if removable partial denture–related decisions impart a high risk of comor-bidity compared to other options, yet the specific patient situation disallows other options, selective implant utiliza-tion with removable partial dentures most often can reduce this detrimental impact. However, the current implant appli-cation must take the future potential for complete implant support for tooth replacements into account. In the last edition, we considered that selective use of implants to address movement control concerns for removable partial dentures to be laudable. This not only has not changed but also we stress in the current edition that helping patients understand the benefits of selective implant use to the func-tional stability is a discussion patients should hear.
“Providing implants to support all teeth needing replacement is often preferable if indicated and if the patient can afford to do so. If the patient is unable to pursue an implant-only supported prosthesis, this should not keep him or her from considering an implant, because the patient still may benefit from a carefully selected implant used for critical clinical performance advantage when removable partial dentures are pursued. Additionally, implants can be used for removable partial dentures to allow future implant-only treatment options.
” We continue the previous edition’s use of design features that provide a content distinction—shading text intended for more experienced clinicians. In Chapter 10, “Principles of Removable Partial Denture Design,” we have attempted to address the continued input from teachers that design of removable partial dentures is a major barrier to learners. In Chapter 10, we have added a basic design strategy for the major tooth loss classifications in the hope that providing a systematic approach with a baseline design protocol may assist this challenge.
www.konkur.in About the Book NEW TO THIS EDITION • Updating of implant-related applications to removable partial dentures • Impact on design, care provision, and performance utility • Removable partial denture design examples provided for basic/most common tooth loss distributions • Intending to establish a foundation of understanding and application • Stressing the need for required tooth modification • Suggesting that standardizing this process impacts care delivery efficiency for the provider and dental laboratory KEY FEATURES • Content considered beyond the basic level is set within a shaded box • A wide selection of relevant references is presented at the back of the textbook in Appendix B for quick-and-easy access • Various philosophies and techniques are presented throughout, facilitating the selection and incorporation of the applicable techniques on a case-by-case basis • Chapters presented in three logically-sequenced sections: • Part I: General Concepts/Treatment Planning • Part II: Clinical and Laboratory • Part III: Maintenance www.konkur.in Acknowledgments We would like to express our gratitude to many who contributed to this text in a variety of ways. Contributions to the text were provided by Dr. Tom Salinas, who assisted with the Implant chapter, and Dr. Vanchit John, who provided input regarding periodontal therapy in mouth preparation. We also would like to acknowledge the following contributors to the clinical images: Drs. Ned van Roekel, James Taylor, Miguel Alfaro, and Carl Andres. We also acknowledge the helpful work of a dedicated group of laboratory technicians who contributed to the updates of many laboratory procedure images: Mr. Joe Bly, Mr. Albert Evans, and Mr. Rick Lee. The clerical assistance of Mrs. Melanie Budihas and Mrs. Barbara Jarjoura is also acknowledged and greatly appreciated.
Alan B. Carr David T. Brown www.konkur.in CHAPTER 1 Partially Edentulous Epidemiology, Physiology, and Terminology CHAPTER OUTLINE Tooth Loss and Age Consequences of Tooth Loss Anatomic Physiologic Functional Restoration with Prostheses Mastication Food Reduction Current Removable Partial Denture Use Need for Removable Partial Dentures This textbook focuses on what the clinician should know about partially edentulous patients to appropriately provide comfortable and useful tooth replacements in the form of removable partial dentures. Removable partial dentures are a component of prosthodontics, the branch of dentistry per-taining to the restoration and maintenance of oral function, comfort, appearance, and health of the patient by the restora-tion of natural teeth and/or the replacement of missing teeth and craniofacial tissues with artificial substitutes.
Current practice in the management of partial tooth loss involves consideration of various types of prostheses ( Figure 1-1). Each type of prosthesis requires the use of various remaining teeth, supporting soft tissues, and/or assigned implants, and consequently demands appropriate applica-tion of knowledge and critical thinking to ensure the best possible outcome given patient needs and desires. Although more than one prosthesis may serve the needs of a patient, any prosthesis should be provided as part of overall man-agement that meets the basic objectives of prosthodontic treatment, which include (1) the elimination of oral disease to the greatest extent possible; (2) the preservation of the health and relationships of the teeth and the health of oral and paraoral structures, which will enhance the removable partial denture design; and (3) the restoration of oral func-tions that are comfortable, are esthetically pleasing, and do not interfere with the patient’s speech. It is critically impor-tant to emphasize that the preservation of health requires proper maintenance of removable partial dentures. To pro-vide a perspective for understanding the impact of remov-able partial denture prosthodontics, a review of tooth loss and its sequelae, functional restoration with prostheses, and prosthesis use and outcomes is in order.
Familiarity with accepted prosthodontic terminology related to removable partial dentures is necessary. Figures 1-2 and 1-3 provide prosthesis terms related to mandibular and maxillary frameworks, and Appendix A provides a review of www.konkur.in 3 Chapter 1 Partially Edentulous Epidemiology, Physiology, and Terminology A C D B Figure 1-1 A, Fixed partial dentures that restore missing anterior (#10) and posterior (#5, #13) teeth. Teeth bordering edentulous spaces are used as abutments. B, Clasp-type removable partial denture restoring missing posterior teeth. Teeth adjacent to edentulous spaces serve as abutments. C, Tooth-supported removable partial denture restoring missing anterior and posterior teeth. Teeth bound-ing edentulous spaces provide support, retention, and stability for restoration. D, Mandibular bilateral distal extension removable partial denture restoring missing premolars and molars. Support, retention, and stability are shared by abutment teeth and residual ridges.
D B A F B C B D B C C B D E A B C D E C B D Figure 1-2 Mandibular framework designed for a partially edentulous arch with a Kennedy Classification II, modification 1 (see Chap-ter 3). Various component parts of the framework are labeled for identification. Subsequent chapters will describe their function, fabrica-tion, and use. A, Major connector. B, Rests. C, Direct retainer. D, Minor connector. E, Guide plane. F, Indirect retainer.
www.konkur.in 4 Part I General Concepts/Treatment Planning selected prosthodontic terms. Additional terminology can be reviewed in The Glossary of Prosthodontic Terms1 and a glossary of accepted terms in all disciplines of dentistry, such as Mosby’s Dental Dictionary, third edition.2 TOOTH LOSS AND AGE In spite of improvements in preventive dental care, it should come as no surprise that tooth loss and age are linked. A specific tooth loss relationship has been documented with increasing age, because some teeth are retained longer than others. It has been suggested that, in general, an interarch difference in tooth loss occurs, with maxillary teeth dem-onstrating loss before mandibular teeth. An intra-arch dif-ference has also been suggested, with posterior teeth lost before anterior teeth. These observations are likely related to respective caries susceptibilities, which have been reported (Table 1-1). Frequently, the last remaining teeth in the mouth are the mandibular anterior teeth, especially the mandibu-lar canines, and it is a common finding to see an edentulous maxilla opposing mandibular anterior teeth.
If one accepts that tooth loss and age are linked, how will this affect current and future dental practice? Replacement of missing teeth is a common patient need, and patients will demand it well into their elderly years. Current population estimates show that 13% of the US population is 65 years of age or older. By the year 2030, this percentage is expected to double, with a significant increase also expected world-wide. These individuals are expected to be in better health, and health care strategies for this group should focus on maintenance of active and productive lives. Oral health care is expected to be a highly sought after and significant compo-nent of overall health care.
Tooth loss patterns associated with age are also evolving. The proportion of edentulous adults has been reported to be decreasing, although this varies widely by state. However, it has been reported that the absolute number of edentulous patients D F A F B D D B E C C B B A B C B E D Figure 1-3 Maxillary framework designed for a partially edentulous arch with a Kennedy Classification I (see Chapter 3). As in Figure 1-2, component parts are labeled for identification. A, Major connector. B, Rests. C, Direct retainer. D, Minor connector. E, Guide plane.
F, Indirect retainer.
Table 1-1 Caries Risk Assessment Risk Type Tooth Location Caries Susceptibility High Risk Lower 6 and 7 Mandibular first and second molars Upper 6 and 7 Maxillary first and second molars Lower 5 Upper 1, 2, 4, 5 Mandibular second premolar Maxillary central, lateral incisors Maxillary first and second premolars Low Risk Upper 3 and lower 4 Maxillary canine, mandibular first premolar Lower 1, 2, 3 Mandibular central, lateral incisors, canines Data from Klein H, Palmer CE: Studies on dental caries: XII. Comparison of the caries susceptibility of the various morphological types of permanent teeth. J Dent Res 20:203-216, 1941.
If tooth loss parallels caries activity, caries risk may be a proxy for tooth loss.
www.konkur.in 5 Chapter 1 Partially Edentulous Epidemiology, Physiology, and Terminology who need care is actually increasing. More pertinent to this text, estimates suggest that the need for restoration of partially edentulous conditions will also be increasing. An explanation for this is presented in an argument that 62% of Americans of the “baby boomer” generation and younger have benefited from fluoridated water. The result of such exposure has been a decrease in caries-associated tooth loss. In addition, current estimates suggest that patients are keeping more teeth longer, demonstrated by the fact that 71.5% of 65- to 74-year-old indi-viduals are partially edentulous (mean number of retained teeth = 18.9). It has been suggested that partially edentulous conditions are more common in the maxillary arch and that the most commonly missing teeth are first and second molars.
CONSEQUENCES OF TOOTH LOSS Anatomic With the loss of teeth, the residual ridge no longer benefits from the functional stimulus it once experienced. Because of this, a loss of ridge volume—both height and width—can be expected unless a dental implant is placed. The ridge vol-ume loss is not predictable for all individuals with tooth loss, because the change in anatomy has been reported to be vari-able across patient groups. In general, bone loss is greater in the mandible than in the maxilla and more pronounced pos-teriorly than anteriorly, and it produces a broader mandibular arch while constricting the maxillary arch. These anatomic changes can present challenges to fabrication of prostheses, including implant-supported prostheses and removable par-tial dentures. Associated with this loss of bone is an accom-panying alteration in the oral mucosa. The attached gingiva of the alveolar bone can be replaced with less keratinized oral mucosa, which is more readily traumatized.
Physiologic What are we replacing when we consider managing miss-ing teeth? We are replacing both the physical anatomic tools for mastication and the oral capacity for neuromuscular functions to manipulate food. Chewing studies have shown that the oral sensory feedback that guides movement of the mandible in chewing comes from a variety of sources. The most sensitive input, which means the input that provides the most refined and precisely controlled movement, comes from periodontal mechanoreceptors (PMRs), with addi-tional input coming from the gingiva, mucosa, periosteum/ bone, and temporomandibular joint (TMJ) complex.
Chewing as a learned behavior has a basic pattern of movement that is generated from within the central ner-vous system. In typical function, this patterned movement is moderated on the basis of food and task needs by oral sensory input from various sources. With loss of the finely tuned contribution from tooth PMRs, the resulting periph-eral receptor influence is less precise in muscular guidance, producing more variable masticatory function, and the type of prosthesis selected to replace missing teeth may poten-tially contribute to functional impediments.
The esthetic impact of tooth loss can be highly significant and may be more of a concern to a patient than loss of func-tion. It is generally perceived that in today’s society, loss of visible teeth, especially in the anterior region of the mouth, carries with it a significant social stigma. With loss of teeth and diminishing residual ridge, facial features can change as the result of altered lip support and/or reduced facial height caused by a reduction in occlusal vertical dimension. Restor-ing facial esthetics in a manner that maintains an appropriate appearance can be a challenge and is a major factor in res-toration and maintenance decisions made for various pros-thetic treatments.
FUNCTIONAL RESTORATION WITH PROSTHESES Individuals with a full complement of teeth report some variation in their levels of masticatory function. The loss of teeth may lead patients to seek care for functional reasons if they notice diminished function to a level that is unaccept-able to them. The level at which a patient finds function to be unacceptable varies among individuals. This variability increases with accelerating tooth loss. This variability may be confusing to clinicians, who may perceive that they have pro-vided prostheses of equal quality to different patients with the same tooth loss patterns, and yet have received different patient reports of success.
An understanding of these variations among individuals with a full complement of teeth and those with prostheses can help clinicians formulate realistic treatment goals that can be communicated to the patient. A review of oral func-tion, especially mastication, may help interested clinicians better understand issues related to the impact of removable partial denture function.
Mastication Although functionally considered as a separate act, mas-tication as part of the feeding continuum precedes swal-lowing and is not an end in itself. The interaction of the two distinct but coordinated aspects of feeding suggests that some judgment of mastication termination or com-pleteness precedes the initiation of swallowing. Although the mastication–swallowing sequence is obvious, the interaction of the two functions is not widely understood and may be important to prosthesis use when removable partial dentures are considered.
Mastication involves two discrete but well-synchronized activities: (1) subdivision of food by applied force; and (2) selective manipulation by the tongue and cheeks to sort out coarse particles and bring them to the occlusal surfaces of teeth for further breakdown. The initial sub-division or comminution phase involves the processes of selection, which refers to the chance that a particle is placed between the teeth in position to be broken, and breakage, which is the degree of fragmentation of a par-ticle once selected. The size, shape, and texture of food www.konkur.in 6 Part I General Concepts/Treatment Planning particles provide the sensory input that influences the configuration and area of each chewing stroke. Larger particles are selectively reduced in size more rapidly than fine particles in efficient mastication. The process of mastication is therefore greatly influenced by factors that affect physical ability to reduce food and to monitor the reduction process by neurosensory means.
Food Reduction Teeth or prostheses serve the role of reducing food to a point that it is ready for swallowing. An index of food reduction is described as masticatory efficiency, or the ability to reduce food to a certain size in a given time frame. A strong correlation has been shown between masticatory efficiency and the number of occluding teeth in dentate individuals, which would suggest variability of particle selection related to contacting teeth. Perfor-mance measures reveal a great deal of functional vari-ability in patients with similar numbers of contacting teeth, and even greater variability is seen within popu-lations with greater loss of teeth (increasing degrees of edentulousness).
Because occlusal contact area is highly correlated with masticatory performance, the loss of molar teeth would be expected to have a greater impact on measures of performance in that the molar has a larger occlusal con-tact area. This effect has been demonstrated in individu-als with missing molars who reveal a greater number of chewing strokes required and a greater mean particle size before swallowing. The point at which an individual is prepared to swallow the food bolus is another measure of performance and is described as the swallowing threshold. Superior masticatory ability that is highly correlated with occlusal contact area also achieves greater food reduction at the swallowing threshold. Conversely, a diminished ability to chew is reflected in larger particles at the swal-lowing threshold.
These objective measures, which show a benefit to molar contact in dentate individuals, are in conflict with some subjective measures from patients who express no perceived functional problems associated with having only premolar occlusion. This shortened dental arch con-cept has highlighted that patient perceptions of functional compromise, as well as benefit, should be considered when it is decided whether to replace missing molars. When the loss of posterior teeth results in an unstable tooth posi-tion, such as distal or labial migration, tooth replacement should be carefully considered; this is a separate situation from the shortened dental arch concept.
It has been reported that prosthetic replacement of teeth provides function that is often less than that seen in the complete, natural dentition state. Functional mea-sures are closest to the natural state when replacements are fixed partial dentures rigidly supported by teeth or implants, intermediate in function when replacements are removable and supported by teeth, lower in function when replacements are removable and supported by teeth and edentulous ridges, and lowest in function when replacements are removable and supported by edentulous ridges alone.
Objective and subjective measures of a patient’s oral function often are not in agreement. It has been shown that subjective measures of masticatory ability are often overrated compared with objective functional tests and that, for complete denture wearers, the subjective crite-ria may be more appropriate in monitoring perceived outcomes. Some literature reports that removable partial dentures can be described by patients as adding very little benefit over no prostheses. However, these findings may be related to a number of factors, including lack of main-tenance of occluding tooth relationships, limitations of this form of dental prosthesis for patient populations that may be unreliable in maintaining follow-up visits, and intrinsic variation in patient response to prostheses.
Food reduction is also influenced by the ability to monitor the process required to determine the point at which swallowing is initiated. As was mentioned earlier, the size, shape, and texture of food are monitored dur-ing mastication to allow modification in mandibular movement for efficient food reduction. This has been demonstrated in dentate individuals given food particles of varying size and concentration suspended in yogurt, who revealed that increased concentrations and particle size required more time to prepare for swallowing (i.e., greater swallowing threshold). These findings suggest that the oral mucosa has a critical role in detecting character-istics necessary for efficient mastication. The influence of the removable partial denture on the ability of the mucosa to perform this role in mastication is not known.
CURRENT REMOVABLE PARTIAL DENTURE USE Given an understanding of the relationship between tooth loss and age, the consequences of tooth loss, and our ability to restore function with removable partial prostheses, what do we know about current prosthesis use for these condi-tions, and what are some common clinical outcomes? One study estimated 21.4% prosthesis use among individuals aged 15 to 74. In the 55- to 64-year-old group, 22.2% were found to wear a removable partial denture. This age group has the highest use of removable partial dentures among those reviewed. It has been suggested that the use of remov-able partial dentures among individuals aged 55 years or older is even greater.
Analysis of this study provides some useful informa-tion for consideration. Partially edentulous individuals not wearing a prosthesis were six times more likely to have miss-ing mandibular teeth (19.4%) than missing maxillary teeth (2.2%). This might suggest greater difficulty in the use of a mandibular prosthesis. The distribution of prostheses used in this large patient group is shown in Table 1-2. The pros-theses in this large study were evaluated on the basis of five www.konkur.in 7 Chapter 1 Partially Edentulous Epidemiology, Physiology, and Terminology technical quality characteristics: integrity, excessive wear of posterior denture teeth, presence of a temporary reline mate-rial, tissue conditioner or adhesive, stability, and retention. As seen in Table 1-3, lack of stability was the most com-mon characteristic noted. In the maxilla, lack of stability was seven times more prevalent than lack of retention. In the mandible, lack of stability was 1.8 times more prevalent than lack of retention. In another study, rest form, denture base extension, stress distribution, and framework fit were identified as common flaws associated with poor removable partial dentures. These characteristics are directly related to the functional stability of prostheses, a vital characteristic to evaluate and the characteristic most benefited by the use of a strategically placed dental implant.
NEED FOR REMOVABLE PARTIAL DENTURES What does all this information mean to us today? It means a number of things that are important to consider. The need for partially edentulous management will be increas-ing. Patient use of removable partial dentures has been high in the past and is expected to continue in the future as an aging population who retains more teeth will present with more partially edentulous conditions. Some patients who are given the choice between a prosthesis entirely supported by implants or a removable partial denture are not able to pur-sue implant care. This contributes to higher use of removable partial dentures. Such patients should understand the benefit of a strategically placed implant for the performance of the removable partial denture. Additionally, provision of such implants must consider the usefulness of the placement to a future fixed implant prosthesis.
Finally, these findings suggest that we should strive to understand how to maximize the opportunity for providing and maintaining stable prostheses, because this is the most frequently deficient aspect of removable partial denture service. Consequently, throughout this text, the basic prin-ciples of diagnosis, mouth preparation, prosthesis design, fabrication, placement, and maintenance will be reinforced to improve the reader’s understanding of care of removable partial denture prostheses.
References 1. The glossary of prosthodontic terms, ed 8. From The Journal of Prosthetic Dentistry 94(1):10–92, 2005. Available online.
sciencedirect.com/science/article/pii/S0022391305001757.
2. Mosby’s dental dictionary, ed 3, St Louis, 2013, Mosby/Elsevier.
Table 1-2 Distribution of Prostheses Type of Prosthesis Distribution Distribution Removable partial dentures RPD/RPD 9.0% RPD/–15.3%, –/RPD 4.5% Complete dentures CU/CL 3.8% CU/–20.7% Combination CU/RPD 11.5% RPD/CL 0.3% CL, Complete lower denture; CU, complete upper denture; RPD, removable partial denture.
Natural teeth denoted with dash (–).
Table 1-3 Technical Quality Concerns for Removable Partial Dentures Lack Stability Lack Integrity Lack Retention Reline Material/Adhesive Excessive Wear Maxillary RPD 43.9% 24.3% 6.2% 3.9% 21.6% Mandibular RPD 38.2% 13.2% 21.2% 21.6% 7.1% RPD, Removable partial denture.
www.konkur.in CHAPTER 2 Considerations for Managing Partial Tooth Loss Tooth Replacements from the Patient Perspective CHAPTER OUTLINE Managing Tooth Loss Over Time Tooth Replacements from the Patient’s Perspective Shared Decision Making Tooth-Supported Prostheses Tooth- and Tissue-Supported Prostheses The Phases of Partial Denture Service Diagnosis and Education of the Patient Treatment Planning, Design, Treatment Sequencing, and Mouth Preparation Support for Distal Extension Denture Bases Establishment and Verification of Occlusal Relations and Tooth Arrangements Initial Placement Procedures Periodic Recall Reasons for Failure of Clasp-Retained Partial Dentures MANAGING TOOTH LOSS OVER TIME Do we treat or do we manage tooth loss? Is the distinction important as we attempt to help our patients decide which type of prosthesis to choose? For patients who want to know what to expect now and in the future, it is helpful to make this distinction, because it helps them realize that the deci-sion has implications for future needs that may be different between prostheses.
Tooth Replacements from the Patient’s Perspective Tooth loss is a permanent condition in that the natural order has been disrupted, and in this sense it is much like a chronic medical condition. Like hypertension and diabetes, two medical conditions that are not reversible and that require medical management to monitor care to ensure appropriate response over time, tooth replacement prostheses must be managed to ensure appropriate response over time.
The term management suggests a focus on meeting needs that may change over time. These needs may be expected or unexpected. Expected outcomes are those that accompany the common clinical course for a type of prosthesis that is related to the tooth-tissue response. This biological toll response is heavily influenced by the type of prosthesis cho-sen. In addition, various needs due to prosthesis degradation and related to expected time-to-retreatment concerns of life expectancy are seen. Unexpected needs are those that might involve factors related to our control of manipulations (such as tissue damage or abuse, material design flaws, or prosthe-sis design) or to those out of our control (such as parafunc-tion or accidental trauma).
With this in mind, it is helpful to consider how we approach educating our patients about management of miss-ing teeth from this current point in time over the remainder www.konkur.in 9 Chapter 2 Considerations for Managing Partial Tooth Loss of their lives. This perspective allows current decisions to be made with a long-term context in mind, and allows decisions to enhance future treatment options.
Most often, a typical sequence is used to discuss tooth replacement options with patients: dental implant– supported prostheses, fixed prostheses, and, finally, removable partial dentures. When removable partial dentures are suggested, they are seldom described in the detail in which fixed or implant prostheses are described, because generally they are considered less like teeth and not as desirable a replace-ment. The desirability of a prosthesis is important to con-sider, and because removable partial dentures are less like teeth than other replacements, it is important to recognize what this suggests from the patient’s perspective. Addi-tionally, for the patient desiring implant management but unable to elect their use when indicated, discussion for stra-tegic implant use for the benefit of the removable partial denture is important.
For patients who have not had missing teeth replaced, their experiences have involved natural teeth, and discus-sion regarding their expectations of replacements would best be described within this context. The order with which we provide replacement prosthesis options for consideration is likely developed on the basis of numerous factors, includ-ing the following: we may believe that we know what is best for patients, our practice style may not include removable options, we may not have had good experience with remov-able prostheses and this lessens our confidence in their use, or removable partial dentures do not match our practice resources.
Although these are important factors, the reason to include removable partial dentures in the discussion is related to identifying whether such a prosthesis is viable, and, if so, whether it is the best option for the patient. We discover this only by interacting with our patients regarding their expectations and understanding their capacity to ben-efit from options of management that have trade-offs unique to each type of prosthesis.
Shared Decision Making When patients are given information regarding their oral health status, which includes disease and functional deficits, as well as the means to address both, what do they need to hear? To achieve a state of oral health, they need to recognize behavioral issues related to plaque control so that once active disease is controlled, they have an understanding that best ensures future health. For tooth replacement decisions, com-plex trade-offs in care choice are often required. The “shared decision making” approach addresses the need to fully inform patients about risks and benefits of care, and ensures that the patient’s values and preferences play a prominent role in the ultimate decision.
It is recognized that patients vary in their desire to par-ticipate in such decisions, thus our active inquiry is required to engage them in discussion. This becomes especially important when elective care, which involves potentially high-burden, costly options with highly variable mainte-nance requirements, is considered.
When patients wish to participate, it is our responsibility to provide them with specific and sufficient information that they can use to decide between treatment options. Specific information ideally comes from our own practice outcomes, in that such information provides effectiveness information and is provider specific. Sufficient information describes exactly what aspects of care are important to the overall decision. Ultimately, it is our role to help patients consider important differences between different prosthesis types.
What then defines important differences? Multiple out-comes combine to describe the overall impact of prosthetic care for all patients. These include technical outcomes, physi-cal outcomes, esthetic outcomes, various maintenance needs, initial and future costs, and even physiologic outcomes that suggest to what extent prostheses “feel” like teeth.
When tooth replacement prostheses are considered from a patient’s perspective, it can be seen that the desire is to replace teeth that serve functional and social roles in everyday life. In considering how well various types of prostheses may meet patients’ specific needs, it is helpful to note what features of the original dentition—the gold standard, in this instance— we strive to duplicate in the replacement. Although it is com-mon to find that existing oral conditions do not easily allow complete restoration to the state of a fully dentate patient, considering the respective strengths and weaknesses of the prosthodontic options (compared with this “gold standard”) helps in identification of realistic expectations.
In this text, the focus is on a type of replacement pros-thesis for patients with an arch with some, but not all, of the teeth missing. Ideally, the replacement prosthesis should provide function with a level of comfort as equivalent as pos-sible to normal dentition. In achieving this, stability while chewing is a primary focus of attention, and we should strive to determine what is required to ensure it. For the patient without posterior teeth, a prosthesis replacing these teeth is at risk for instability due to the edentulous ridge compress-ible support, therefore consideration of a distal implant to support the posterior segment (the distal extension) can enhance functional stability.
If the prosthesis will be visible during casual speaking, smiling, and/or laughing, it is obvious that the replacement should look as natural as the surrounding environment. In summary, tooth replacement prostheses should provide a combination of several features of natural teeth: acceptable in appearance, comfortable and stable in function, and main-tainable throughout their serviceable lifetime at a reasonable cost.
TOOTH-SUPPORTED PROSTHESES For partially edentulous patients, available prosthetic options include natural tooth–supported fixed partial dentures, removable partial dentures, and implant-supported fixed partial dentures. How well these options restore and maintain www.konkur.in 10 Part I General Concepts/Treatment Planning the features of natural teeth mentioned previously depends to a large extent on the numbers and locations of the missing teeth. The major categories of partial tooth loss (see Chap-ter 3) are those (1) with teeth both anterior and posterior to the space (a tooth-supported space), and (2) with teeth either anterior or posterior to the space (a tooth- and tissue-supported space). All prosthetic options listed are available for the tooth-bound space (although they are not necessar-ily indicated for every clinical situation), but only removable partial dentures and implant-supported prostheses are avail-able for the distal extension (recognizing limited application of cantilevers).
Removable partial dentures can be designed in various ways to allow use of abutment teeth and supporting tis-sue for stability, support, and retention of the prosthesis. In terms of tooth-bound spaces, the removable partial denture is like a fixed partial denture because natural teeth alone pro-vide direct resistance to functional forces. Because natural teeth support the prosthesis, it should not move under these functional forces. In this condition, the interface between, or relationship of, the removable partial denture framework and the abutment teeth should be designed to take advan-tage of tooth support—similar to the relationship between a fixed partial denture retainer and a prepared tooth. This means that it should provide positive vertical support (rest preparations) and a restrictive angle of dislodgment (oppos-ing guide planes). Put another way, when the removable par-tial denture is selected for a tooth-bound situation, stability under functional load should be as well controlled as a fixed partial denture when appropriate tooth preparation is pro-vided. Because removable partial denture clasps do not com-pletely encircle the tooth, as a fixed partial denture retainer does, they must be designed to engage more than half the circumference to allow the prosthesis to maintain position under the influence of horizontal chewing loads. It should be obvious that careful planning and execution of the necessary natural tooth contour modifications are required to ensure movement control and functional stability for removable partial dentures supported by teeth. Similarities between the prosthesis-tooth interface for fixed partial dentures and for removable partial dentures are highlighted to empha-size the modification principles required to ensure stability for movement control in removable partial dentures. Over time, natural tooth support can be maintained as with the fixed partial denture. Chapter 14 helps to explain how this is accomplished when natural tooth modifications or surveyed crowns are produced.
TOOTH- AND TISSUE-SUPPORTED PROSTHESES For removable partial dentures that do not have the benefit of natural tooth support at each end of the replacement teeth (extension base removable partial dentures), it is necessary that the residual ridge be used to assist in the functional sta-bility of the prosthesis. When a removable partial denture is selected for a tooth- and tissue-supported arch, the prosthe-sis must be designed to allow functional movement of the base to the extent expected by the residual ridge mucosa. This mucosa movement is variable, but for healthy residual ridge (masticatory) mucosa, movement from 1 to 3 mm can be expected. Consequently, unlike with the tooth-bound space, tooth modification for the tooth- and tissue-supported prosthesis must be designed with the dual goal of framework tooth contact to allow appropriate functional stability from the tooth but with allowance for the anticipated vertical and/ or horizontal movement of the extension base. This intro-duces the concept of anticipated movement with a pros-thesis and the requirement that we have a role in designing prostheses to appropriately control movement. Additionally, because tissue support in the tooth- and tissue-supported removable partial denture predictably changes over time, to adequately manage partial tooth loss with a removable pros-thesis, we must carefully monitor our patients to maintain support and ensure maximum prosthetic function.
The clasp-retained partial denture is the most commonly used removable partial denture (Figure 2-1). It is capable of providing physiologically sound treatment for most patients who need partial denture restorations. Although the clasp-retained partial denture has disadvantages, its advantages of lower cost and shorter fabrication time ensure that it will continue to be widely used. Following are some possible dis-advantages of a clasp-retained partial denture: 1. Strain on the abutment teeth is often caused by improper tooth preparation or clasp design and/or loss of tissue support under the distal extension partial denture bases. 2. Clasps can be unesthetic, particularly when they are placed on visible tooth surfaces without consideration of esthetic impact. 3. Caries may develop beneath clasp and other framework components, especially if the patient fails to keep the prosthesis and the abutments clean.
Despite these disadvantages, the use of removable pros-theses may be preferred when tooth-bounded edentulous spaces are too large to be restored safely with fixed prosthe-ses, or when cross-arch stabilization and wider distribution of forces to supporting teeth and tissues are desirable. Fixed partial dentures, however, should always be considered and used when indicated.
The removable partial denture retained by internal attach-ments eliminates some of the disadvantages of clasps, but it has other disadvantages, one of which is higher cost, which makes it more difficult to obtain for a large per-centage of patients who need partial dentures. How-ever, when alignment of the abutment teeth is favorable and periodontal health and bone support are adequate, when the clinical crown is of sufficient length and the pulp morphology can accommodate the required tooth preparation, and when the economic status of the patient permits, an internal attachment prosthesis provides an unquestionable advantage for esthetic reasons. When this www.konkur.in 11 Chapter 2 Considerations for Managing Partial Tooth Loss situation exists, carefully weighing tooth attachment ver-sus implant attachment options is required (see Remov-able Partial Dentures and Implants, Chapter 12).
In most instances, if the extracoronal clasp-retained partial denture is designed properly, the only advantage of the internal attachment denture is esthetic, because abut-ment protection and stabilizing components should be used with both internal and external retainers. However, economics permitting, esthetics alone may justify the use of internal attachment retainers, especially when a crown is indicated for non–removable partial denture reasons.
Injudicious use of internal attachments can lead to excessive torsional load on the abutments supporting dis-tal extension removable partial dentures, especially in the mandible. The use of hinges or other types of stress break-ers is discouraged in these situations. It is not that they are ineffective, but that they are frequently misused. As an example, in the mandibular arch, a stress-broken distal extension partial denture does not provide for cross-arch stabilization and frequently subjects the edentulous ridge to excessive trauma from horizontal and torquing forces. Therefore a rigid design is preferred, and some type of extracoronal clasp retainer is still the most logical and the most frequently used. It seems likely that its use will con-tinue until a more widely-acceptable retainer is devised.
As mentioned in Chapter 1, the most commonly cited problem associated with removable partial dentures is insta-bility. Healthy natural teeth should not move when used; therefore we should strive to provide and maintain as stable a prosthesis as possible given the means available. How do we ensure functional stability? By understanding that a remov-able partial denture can move under function (because it is not cemented to teeth like a fixed partial denture). We should take steps to prescribe the necessary prosthetic fit to teeth (and tissue) to control movement as much as possible. A B C D Figure 2-1 A, Maxillary and mandibular clasp-retained removable partial dentures. All clasps are extracoronal retainers (clasps) on abutments. B, Prostheses from (A) shown intraorally in occlusion. C, Maxillary prosthesis using intracoronal retainers and full palatal coverage. The male portions of the attachments are shown at the mesial position of the artificial teeth and will fit into intracoronal rests. D, Internal attachment prosthesis in the patient’s mouth. Note the precise fit of male and female portions of the attachments.
www.konkur.in 12 Part I General Concepts/Treatment Planning This entails providing appropriate natural tooth mouth prep-arations, ensuring an accurate frame fit at tooth and tissue, providing a simultaneous contacting relationship between natural and prosthetic opposing teeth, and providing and maintaining optimum support from the soft tissue and teeth. It also may require strategic use of implants to control distal extension movement.
As we will review in Chapter 4, control of the anticipated movement of your prosthesis is addressed by assigning the appropriate component part of the prosthesis to contact/ engage the tooth or tissue in a manner that allows move-ment and removal of the prosthesis. Are there movements that we should control that are more important than others? Although we recognize the need to resist movement away from the teeth and tissue to keep prostheses from falling out of mouths, the most damaging forces are those result-ing from functional closure during chewing (and in some patients, parafunction). Consequently, control of combined vertical (tissue-ward) and horizontal movement is most critical and places a premium on tooth modifications (rest and stabilizing component preparations) and verification of adequate fit of the frame to the teeth.
THE PHASES OF PARTIAL DENTURE SERVICE Partial denture service may be conceptually divided into phases. The first phase involves making the appropriate diag-nosis, deciding a removable partial denture is indicated, and providing patient education regarding removable partial denture expectations over time. The second phase includes treatment planning, design of the partial denture framework, treatment sequencing, and execution of mouth preparations. The third phase is the provision of adequate support for the distal extension denture base. The fourth phase is establish-ment and verification of harmonious occlusal relationships and tooth relationships with opposing and remaining natural teeth. The fifth phase involves initial placement procedures, including adjustments to the contours and bearing surfaces of denture bases, adjustments to ensure occlusal harmony, and a review of instructions given the patient to optimally maintain oral structures and provided restorations. The sixth and final phase of partial denture service consists of follow-up services by the dentist through recall appointments for periodic evaluation of the responses of oral tissue to resto-rations and of the acceptance of restorations by the patient. The context of each phase is discussed in greater detail in the respective chapters of this book.
Diagnosis and Education of the Patient The term patient education is described in Mosby’s Dental Dictionary as “the process of informing a patient about a health matter to secure informed consent, patient coopera-tion, and a high level of patient compliance.” The dentist and the patient share responsibility for the ultimate success of a removable partial denture. It is folly to assume that a patient has an understanding of the benefits of a removable partial denture unless he or she is so informed. It is also unlikely that the patient has the knowledge to avoid misuse of the restoration or is able to provide the required oral care and maintenance procedures to ensure the success of the partial denture unless he or she is adequately advised.
The finest biologically-oriented removable partial den-ture is often doomed to limited success if the patient fails to exercise proper oral hygiene habits or ignores recall appoint-ments. Preservation of the oral structures, one of the pri-mary objectives of prosthodontic treatment, is compromised without the patient’s cooperation in oral hygiene and regular maintenance visits.
Patient education should begin at the initial diagnosis and should continue throughout treatment. This educational procedure is especially important when the treatment plan and prognosis are discussed with the patient. Limitations imposed on the success of treatment through failure of the patient to accept responsibility must be explained before definitive treatment is undertaken. A patient usually does not retain all the information presented in the oral educa-tional instructions. For this reason, patients should be given written suggestions to reinforce the oral presentations.
Treatment Planning, Design, Treatment Sequencing, and Mouth Preparation Treatment planning and design begin with thorough medi-cal and dental histories. The complete oral examination must include both clinical and radiographic interpretation of (1) caries, (2) the condition of existing restorations, (3) peri-odontal conditions, (4) responses of teeth (especially abut-ment teeth) and residual ridges to previous stress, and (5) the vitality of remaining teeth. In addition, evaluation of the occlusal plane, the arch form, and the occlusal relations of the remaining teeth must be meticulously accomplished by clinical visual evaluation and diagnostic mounting. After a complete diagnostic examination has been accomplished and a removable partial denture has been selected as the treatment of choice, a treatment plan is sequenced and a par-tial denture design is developed in accordance with available support.
The dental cast surveyor (Figure 2-2) is an absolute neces-sity in any dental office in which patients are being treated with removable partial dentures. The surveyor is instrumen-tal in diagnosing and guiding the appropriate tooth prepa-ration and verifying that the mouth preparation has been performed correctly. There is no more reason to justify its omission from a dentist’s armamentarium than there is to ignore the need for roentgenographic equipment, the mouth mirror and explorer, or the periodontal probe used for diag-nostic purposes.
Several moderately priced surveyors that adequately accomplish the diagnostic procedures necessary for design-ing the partial denture are available. In many dental offices, this most important phase of dental diagnosis is delegated to the commercial dental laboratory either because this invalu-able diagnostic tool is absent or because the dentist feels www.konkur.in 13 Chapter 2 Considerations for Managing Partial Tooth Loss inexperienced or is apathetic. This situation places the tech-nician in the role of diagnostician. Any clinical treatment based on the diagnosis of the technician remains the respon-sibility of the dentist. This makes no more sense than relying on the technician to interpret radiographs and to render a diagnosis.
After treatment planning, a predetermined sequence of mouth preparations can be performed with a definite goal in mind. It is mandatory that the treatment plan be reviewed to ensure that the mouth preparation necessary to accom-modate the removable partial denture design has been properly sequenced. Mouth preparations, in the appropri-ate sequence, should be oriented toward the goal of provid-ing adequate support, stability, retention, and a harmonious occlusion for the partial denture. Placing a crown or restor-ing a tooth out of sequence may result in the need to restore teeth that were not planned for restoration, or it may neces-sitate remaking a restoration or even seriously jeopardizing the success of the removable partial denture. Through the aid of diagnostic casts on which the tentative design of the partial denture has been outlined and the mouth prepara-tions have been indicated in colored pencil, occlusal adjust-ments, abutment restorations, and abutment modifications can be accomplished.
Support for Distal Extension Denture Bases The third of the six phases in the treatment of a patient with a partial denture involves obtaining adequate support for distal extension bases. Therefore it does not apply to tooth-supported removable partial dentures. With the latter, sup-port comes entirely from the abutment teeth through the use of rests.
For the distal extension partial denture, however, a base made to fit the anatomic ridge form does not provide ade-quate support under occlusal loading (Figure 2-3). Neither does it provide for maximum border extension nor accurate border detail. Therefore some type of corrected impression is necessary. This may be accomplished by several means, any of which satisfy the requirements for support of any distal extension partial denture base.
Foremost is the requirement that certain soft tissue in the primary supporting area should be recorded or related under some loading so that the base may be made to fit the form of the ridge when under function. This provides support and ensures maintenance of that support for the longest possible time. This requirement makes the distal extension partial denture unique in that support from the tissue underlying the distal extension base must be made as equal to and com-patible with the tooth support as possible.
A complete denture is entirely tissue supported, and the entire denture can move toward the tissue under function. In contrast, any movement of a partial denture base is inevi-tably a rotational movement that, if toward the tissue, may result in undesirable torquing forces to the abutment teeth and loss of planned occlusal contacts. Therefore every effort must be made to provide the best possible support for the distal extension base to minimize these forces.
Usually no single impression technique can adequately record the anatomic form of the teeth and adjacent struc-tures and at the same time record the supporting form of the mandibular edentulous ridge. A method should be used that can record these tissues in their supporting form or in a supporting relationship to the rest of the denture (see Figure 2-3). This may be accomplished by one of several methods that will be discussed in Chapter 16.
Establishment and Verification of Occlusal Relations and Tooth Arrangements Whether the partial denture is tooth supported or has one or more distal extension bases, the recording and verifica-tion of occlusal relationships and tooth arrangement are important steps in the construction of a partial denture. For the tooth-supported partial denture, ridge form is of less significance than it is for the tooth- and tissue-supported prosthesis, because the ridge is not called on to support the prosthesis. For the distal extension base, however, jaw rela-tion records should be made only after the best possible support is obtained for the denture base. This necessitates the making of a base or bases that provide the same sup-port as the finished denture. Therefore the final jaw relations should not be recorded until after the denture framework has been returned to the dentist, the fit of the framework to the abutment teeth and opposing occlusion has been verified and corrected, and a corrected impression has been made. Figure 2-2 Dental cast surveyor facilitates the design of a re-movable partial denture. It is an instrument by which parallelism or lack of parallelism of abutment teeth and other oral structures, on a stone cast, can be determined (magnified view shows paral-lel guide plane surface). Use of the surveyor is discussed in later chapters.
www.konkur.in 14 Part I General Concepts/Treatment Planning Then a new resin base or a corrected base must be used to record jaw relations.
Occlusal records for a removable partial denture may be made by the various methods described in Chapter 18.
Initial Placement Procedures The fifth phase of treatment occurs when the patient is given possession of the removable prosthesis. Inevitably it seems that minute changes in the planned occlusal relationships occur during processing of the dentures. Not only must occlusal harmony be ensured before the patient is given pos-session of the dentures, but the processed bases must be rea-sonably perfected to fit the basal seats. It must be ascertained that the patient understands the suggestions and recommen-dations given by the dentist for care of the dentures and oral structures and understands about expectations (based on the “shared decision making” discussion) in the adjustment phases and the use of restorations. These facets of treatment are discussed in detail in Chapter 21.
Periodic Recall Initial placement and adjustment of the prosthesis certainly is not the end of treatment for the partially edentulous patient. Periodic reevaluation of the patient is critical for early recog-nition of changes in oral structures to allow steps to be taken to maintain oral health. These examinations must monitor the condition of the oral tissue, the response to tooth res-torations, the prosthesis, the patient’s acceptance, and the patient’s commitment to maintain oral hygiene. Although a 6-month recall period is adequate for most patients, more frequent evaluation may be required for some. Chapter 21 contains some suggestions concerning this sixth phase of treatment.
REASONS FOR FAILURE OF CLASP-RETAINED PARTIAL DENTURES Experience with the clasp-retained partial denture made by the methods outlined has proved its merit and justi-fies its continued use. The occasional objection to the visibility of retentive clasps can be minimized through the use of wrought-wire clasp arms. Few contraindi-cations for use of a properly designed clasp-retained partial denture are known. Practically all objections to this type of denture can be eliminated by pointing to deficiencies in mouth preparation, denture design and A B C Figure 2-3 A, Occlusal view of a cast from a preliminary impression, which produced an anatomic ridge form (left), and an altered cast of the same ridge showing a functional or supportive form (right). The altered cast impression selectively placed pressure on the buccal shelf region, which is the primary stress-bearing area of the mandibular posterior residual ridge. B, Buccal view of anatomic ridge form. C, Buccal view of functional or supportive ridge form. Note that the supportive form of the ridge clearly delineates the extent of coverage available for a denture base and is most different from the anatomic form when the mucosa is easily displaced.
www.konkur.in 15 Chapter 2 Considerations for Managing Partial Tooth Loss fabrication, and patient education. These include the following: Diagnosis and treatment planning 1. Inadequate diagnosis 2. Failure to use a surveyor or to use a surveyor properly during treatment planning Mouth preparation procedures 1. Failure to properly sequence mouth preparation procedures 2. Inadequate mouth preparations, usually resulting from insufficient planning of the design of the partial den-ture or failure to determine that mouth preparations have been properly accomplished 3. Failure to return supporting tissue to optimum health before impression procedures are performed 4. Inadequate impressions of hard and soft tissue Design of the framework 1. Failure to use properly located and sized rests 2. Flexible or incorrectly located major and minor connectors 3. Incorrect use of clasp designs 4. Use of cast clasps that have too little flexibility, are too broad in tooth coverage, and have too little consider-ation for esthetics Laboratory procedures 1. Problems in master cast preparation a. Inaccurate impression b. Poor cast-forming procedures c. Incompatible impression materials and gypsum products 2. Failure to provide the technician with a specific design and necessary information to enable the technician to execute the design 3. Failure of the technician to follow the design and writ-ten instructions Support for denture bases 1. Inadequate coverage of basal seat tissue 2. Failure to record basal seat tissue in a supporting form Occlusion 1. Failure to develop a harmonious occlusion 2. Failure to use compatible materials for opposing occlusal surfaces Patient-dentist relationship 1. Failure of the dentist to provide adequate dental health care information, including details on care and use of the prosthesis 2. Failure of the dentist to provide recall opportunities on a periodic basis 3. Failure of the patient to exercise a dental health care regimen and respond to recall A removable partial denture designed and fabricated so that it avoids the errors and deficiencies listed is one that proves the clasp-type partial denture can be made functional, esthetically pleasing, and long lasting without damage to supporting structures. The proof of the merit of this type of restoration lies in the knowledge that (1) it permits treatment for the largest number of patients at a reasonable cost; (2) it provides restorations that are comfortable and efficient over a long period of time, with adequate support and maintenance of occlusal contact relations; (3) it can provide for healthy abutments, free of caries and periodontal disease; (4) it can provide for the continued health of restored, healthy tissue of the basal seats; and (5) it makes possible a partial denture service that is definitive and not merely an interim treatment.
Removable partial dentures thus made will contribute to a concept of prosthetic dentistry that has as its goal the promotion of oral health, the restoration of partially edentulous mouths, and elimination of the ultimate need for complete dentures.
www.konkur.in CHAPTER 3 Classification of Partially Edentulous Arches CHAPTER OUTLINE Requirements of an Acceptable Method of Classification Kennedy Classification Applegate’s Rules for Applying the Kennedy Classification Even though recent reports have shown a consistent decline in the prevalence of tooth loss during the past few decades, significant variation in tooth loss distribution remains. It would be most helpful to consider which combinations of tooth loss are most common and to classify these for the purpose of assisting our management of partially edentulous patients.
A classification that is based on diagnostic criteria has been proposed for partial edentulism.1 The purpose of this system of classification is to facilitate treatment decisions on the basis of treatment complexity. Complexity is deter-mined from four broad diagnostic categories that include location and extent of the edentulous areas, condition of the abutments, occlusal characteristics and requirements, and residual ridge characteristics. The advantage of this classification system over others in standard use has yet to be documented.
The Kennedy method is probably the most widely accepted classification of partially edentulous arches. In an attempt to simplify the problem and encourage more universal use of a classification, and in the interest of ade-quate communication, the Kennedy classification will be used in this textbook. The student can refer to the Selected Reading Resources section in Appendix B for information relative to other classifications.
Although classifications are actually descriptive of the partially edentulous arches, the removable partial denture that restores a particular class of arch is described as a den-ture of that class. For example, we speak of a Class III or Class I removable partial denture. It is simpler to say “a Class II partial denture” than it is to say “a partial denture restoring a Class II partially edentulous arch.” Several classifications of partially edentulous arches have been proposed and are in use. This variety has led to some confusion and disagreement concerning which www.konkur.in 17 Chapter 3 Classification of Partially Edentulous Arches classification best describes all possible configurations and should be adopted.
The most familiar classifications are those originally proposed by Kennedy, Cummer, and Bailyn. Beckett, Godfrey, Swenson, Friedman, Wilson, Skinner, Apple-gate, Avant, Miller, and others have also proposed clas-sifications. It is evident that an attempt should be made to combine the best features of all classifications so that a universal classification can be adopted.
REQUIREMENTS OF AN ACCEPTABLE METHOD OF CLASSIFICATION The classification of a partially edentulous arch should satisfy the following requirements: 1. It should permit immediate visualization of the type of partially edentulous arch that is being considered. 2. It should permit immediate differentiation between the tooth-supported and the tooth- and tissue-supported re-movable partial denture. 3. It should be universally acceptable.
KENNEDY CLASSIFICATION The Kennedy method of classification was originally pro-posed by Dr. Edward Kennedy in 1925. It attempts to clas-sify the partially edentulous arch in a manner that suggests certain principles of design for a given situation (Figure 3-1).
Kennedy divided all partially edentulous arches into four basic classes. Edentulous areas other than those that determine the basic classes were designated as modification spaces (Figure 3-2).
The following is the Kennedy classification: Class I Bilateral edentulous areas located posterior to the natural teeth Class II A unilateral edentulous area located posterior to the remaining natural teeth Class III A unilateral edentulous area with natural teeth remaining both anterior and posterior to it Class IV A single, but bilateral (crossing the midline), edentulous area located anterior to the remaining natural teeth Figure 3-1 Representative examples of partially edentulous arches classified by the Kennedy method.
www.konkur.in 18 Part I General Concepts/Treatment Planning A B C D E F G H Figure 3-2 Kennedy classification with examples of modifications. A, Class I maxillary arch. B, Class II mandibular arch. C, Class III mandibular arch. D, Class IV maxillary arch. E, Class II, modification 1 mandibular arch. F, Class II, modification 1 maxillary arch. G, Class II, modification 2 mandibular arch. H, Class III, modification 2 maxillary arch.
www.konkur.in 19 Chapter 3 Classification of Partially Edentulous Arches Box 3-1 RULES GOVERNING APPLICATION OF THE KENNEDY METHOD Rule 1 Classification should follow rather than precede any extractions of teeth that might alter the original classification.
Rule 2 If a third molar is missing and is not to be replaced, it is not considered in the classification.
Rule 3 If a third molar is present and is to be used as an abutment, it is considered in the classification.
Rule 4 If a second molar is missing and is not to be replaced, it is not considered in the classification (e.g., if the opposing second molar is likewise missing and is not to be replaced).
Rule 5 The most posterior edentulous area (or areas) always deter-mines the classification.
Rule 6 Edentulous areas other than those that determine the classi-fication are referred to as modifications and are designated by their number.
Rule 7 The extent of the modification is not considered, only the num-ber of additional edentulous areas.
Rule 8 No modification areas can be included in Class IV arches. (Other edentulous areas that lie posterior to the single bilateral areas crossing the midline would instead determine the clas-sification; see Rule 5.) A B C D E F G H I Figure 3-3 Nine partially edentulous arch configurations. Identify each. Answers can be found at the end of this chapter, after the Reference section.
www.konkur.in 20 Part I General Concepts/Treatment Planning One of the principal advantages of the Kennedy method is that it permits immediate visualization of the partially eden-tulous arch and allows easy distinction between tooth-sup-ported versus tooth- and tissue-supported prostheses. Those schooled in its use and in the principles of partial denture design can readily relate the arch configuration design to be used in the basic partial denture. This method permits a logical approach to the problems of design. It makes possible the application of sound principles of partial denture design and is therefore a logical method of classification. However, a classification system should not be used to stereotype or limit the concepts of design.
Additionally, because the use of a strategically placed implant can provide an extension base (the support simi-lar to a tooth in a tooth-born segment), it may be helpful to our understanding, communication, and design to consider implant designations in future classifications (Kennedy Class Ii/i, Class IIi, Class IVi, and so on).
APPLEGATE’S RULES FOR APPLYING THE KENNEDY CLASSIFICATION The Kennedy classification would be difficult to apply in every situation without certain rules for application. Apple-gate provided eight rules that govern application of the Ken-nedy method (Box 3-1).
Although some confusion may occur initially as to why Class I should refer to two edentulous areas and Class II should refer to one, the principles of design make this distinction logical. Kennedy placed the Class II unilat-eral distal extension type between the Class I bilateral distal extension type and the Class III tooth-supported classification because the Class II partial denture must embody features of both, especially when tooth-supported modifications are present. Because it has a tissue-supported extension base, the denture must be designed similarly to a Class I partial denture. Often, however, a tooth-supported, or Class III, component is present elsewhere in the arch. Thus the Class II partial denture rightly falls between the Class I and the Class III, because it embodies design fea-tures common to both. In keeping with the principle that design is based on classification, the application of such principles of design is simplified by retaining the original classification of Kennedy.
Figure 3-3 presents a chance to assess your skills. Review the figure and classify the partially edentulous arches illustrated. The answers are provided at the end of this chapter.
Reference 1. McGarry TJ, Nimmo A, Skiba JF, et al.: Classification system for partial edentulism, J Prosthodont 11(3):181–193, 2002.
Answer to Figure 3-3 A. Class IV B. Class II, modification 2 C. Class I, modification 1 D. Class III, modification 3 E. Class III, modification 1 F. Class III, modification 1 G. Class IV H. Class II I. Class III, modification 5 www.konkur.in CHAPTER 4 Biomechanics of Removable Partial Dentures CHAPTER OUTLINE Biomechanics and Design Solutions Biomechanical Considerations Impact of Implants on Movements of Partial Dentures Simple Machines Possible Movements of Partial Dentures As was stated in Chapter 1, the goal is to provide useful, functional removable partial denture prostheses by striv-ing to understand how to maximize every opportunity for providing and maintaining a stable prosthesis. Because removable partial dentures are not rigidly attached to teeth, the control of potential movement under functional load is critical to providing the best chance for stability and patient accommodation. The consequence of prosthesis movement under load is an application of stress to the teeth and tis-sue that are contacting the prosthesis. It is important that the stress not exceed the level of physiologic tolerance, which is a range of mechanical stimulus that a system can resist without disruption or traumatic consequences. In the terminology of engineering mechanics, the prosthesis induces stress in the tissue equal to the force applied across the area of contact with the teeth and/or tissue. This same stress acts to produce strain in the supporting tissue, which results in load displace-ment in the teeth and tissue. The understanding of how these mechanical phenomena act within a biological environment that is unique to each patient can be discussed in terms of biomechanics. In the design of removable partial dentures, with a focus on the goal of providing and maintaining stable prostheses, consideration of basic biomechanical principles associated with the unique features of each mouth is essen-tial. Oral hygiene and appropriate prosthesis maintenance procedures are required for continued benefit of optimum biomechanical principles.
BIOMECHANICS AND DESIGN SOLUTIONS Removable partial dentures by design are intended to be placed into and removed from the mouth. Because of this, they cannot be rigidly connected to the teeth or tissue. This makes them subject to movement in response to functional loads, such as those created by mastication. It is important for clinicians who provide removable partial denture service www.konkur.in 22 Part I General Concepts/Treatment Planning to understand the possible movements in response to func-tion and to be able to logically design the component parts of the removable partial denture to help control these move-ments. Just how this is accomplished in a logical manner may not be clear to a clinician who is new to this exercise. One method of helping to organize design thought is to consider it as an exercise in creating a design solution.
Designing a removable partial denture can be considered similar to the classic, multifaceted design problem in conven-tional engineering, which is characterized by being open ended and ill structured. Open ended means that problems typically have more than one solution, and ill structured means that solutions are not the result of standard mathematical formulas used in some structured manner. The design process, which is a series of steps that lead toward a solution of the problem, includes identifying a need, defining the problem, setting design objectives, searching for background information and data, developing a design rationale, devising and evaluating alternative solutions, and providing the solution (i.e., decision making and communication of solutions) (Box 4-1).
The rationale for design should logically develop from anal-ysis of the unique oral condition of each mouth under consider-ation. However, it is possible that alternative design “solutions” could be applied, and it is the evaluation of perceived merits of these various designs that seems most confusing to clinicians.
The following biomechanical considerations provide a background related to principles of the potential move-ment associated with removable partial dentures, and the subsequent chapters covering the various component parts describe how these components are applied in designs to control the resultant movements of prostheses.
BIOMECHANICAL CONSIDERATIONS The supporting structures for removable partial dentures (abutment teeth and residual ridges) are living things that are subjected to forces. Whether the supporting struc-tures are capable of resisting the applied forces depends on (1) what typical forces require resistance, (2) what duration and intensity these forces have, (3) what capacity the teeth, implant(s) and/or mucosae have to resist these forces, (4) how material use and application influence this teeth-tissue resistance, and (5) whether resistance changes over time.
Consideration of the forces inherent in the oral cavity is critical. This includes the direction, duration, frequency, and magnitude of the force. In the final analysis, it is bone that provides the support for a removable prosthesis (i.e., the alveolar bone by way of the periodontal ligament and the residual ridge bone through its soft tissue covering). If potentially destructive forces can be minimized, then the physiologic tolerances of the supporting structures are not exceeded and pathologic change does not occur. The forces that occur with removable prosthesis function can be widely distributed and directed, and their effect minimized by appropriate design of the removable partial denture. An appropriate design includes the selection and location of components in conjunction with a harmonious occlusion.
Unquestionably the design of removable partial dentures necessitates mechanical and biological considerations. Most dentists are capable of applying simple mechanical principles to the design of a removable partial denture. For example, the lid of a paint can is more easily removed with a screwdriver than with a half dollar. The longer the handle, the less effort (force) it takes. This is a simple application of the mechanics of leverage. By the same token, a lever system represented by a distal extension removable partial denture could mag-nify the applied force of occlusion to the terminal abutments, which would be undesirable. Use of a dental implant in such a case reduces, and may eliminate, the opportunity for such force magnification.
IMPACT OF IMPLANTS ON MOVEMENTS OF PARTIAL DENTURES Similar to the process of considering how an individual tooth is best used in removable partial denture design to control prosthesis movement, use of an implant should be directed toward the most beneficial movement control. Although possible roles for implant use include all three desired principles demonstrated by prostheses—support, stability, and retention—the major functional demand is imposed by chewing, and therefore the greatest benefit of implant use involves resisting instability by improving sup-port. Minimizing rotation about an axis in a Kennedy Class I or II arch, or any long modification span, is important to consider.
Box 4-1 DESIGN PROCESS FOR REMOVABLE PARTIAL DENTURES Need Tooth replacement Definition of Problem Provision of stable removable prosthesis Objectives Limited functional movement within tooth-tissue tolerance Background Information Forces of occlusion, tissue “load-displacement” charac-ter and potential for movement, biomechanical principles applied to specific features of this unique mouth, removable partial denture component parts assigned to control move-ment Choice of a Solution (among Alternatives) for Application Based on prior experience, principles and concepts learned from school and textbooks, and applicable clinical research www.konkur.in 23 Chapter 4 Biomechanics of Removable Partial Dentures SIMPLE MACHINES An understanding of simple machines applied to the design of removable partial dentures helps to accomplish the objec-tive of preservation of oral structures. Without such under-standing, a removable partial denture can be inadvertently designed as a destructive machine.
Machines may be classified into two general categories: simple and complex. Complex machines are combinations of many simple machines. The six simple machines are lever, wedge, screw, wheel and axle, pulley, and inclined plane (Figure 4-1). Of the simple machines, the lever, the wedge, and the inclined plane should be avoided in the design of removable partial dentures.
In its simplest form, a lever is a rigid bar supported some-where along its length. It may rest on the support or may be supported from above. The support point of the lever is called the fulcrum, and the lever can move around the ful-crum (Figure 4-2; see Figure 6-6).
The rotational movement of an extension base type of removable partial denture, when a force is placed on the denture base, is illustrated in Figure 4-3. It rotates in rela-tion to the three cranial planes because of differences in the support characteristics of the abutment teeth and the soft tissue covering the residual ridge. Even though the actual movement of the denture may be small, a lever force may be imposed on abutment teeth. This is especially detrimental when prosthesis maintenance is neglected. Three types of levers are used: first, second, and third class (see Figure 4-2). The potential of a lever system to magnify a force is illustrated in Figure 4-4.
A cantilever is a beam supported at one end that can act as a first-class lever (Figure 4-5). A cantilever design should be avoided (Figure 4-6). Use of a dental implant is one strategy to provide tooth replacement and avoid the cantilever. Exam-ples of other lever designs and suggestions for alternative designs to avoid or minimize their destructive potential are illustrated in Figures 4-7 and 4-8. The most efficient means of addressing the potential effects of a lever is to provide a rigid element at the unsupported end to disallow movement. This is the most beneficial use of dental implants in conjunction with removable partial dentures and should be considered when support capacity for a distal extension is considered significantly poor.
A tooth is apparently better able to tolerate vertically directed forces than nonvertical, torquing, or horizontal forces. This characteristic is observed clinically, and it seems rational that more periodontal fibers are activated to resist the application of vertical forces to teeth than are activated to resist the application of nonvertical forces (Figure 4-9).
Again, a distal extension removable partial denture rotates when forces are applied to the artificial teeth attached to the extension base. Because it can be assumed that this rotation must create predominantly nonvertical forces, the location of Lever Wedge Inclined plane Screw Pulley Wheel and axle F Figure 4-1 The six simple machines include lever, wedge, inclined plane, screw, pulley, and wheel and axle. The fulcrum, wedge, and inclined plane are matters of concern in removable partial denture designs because of the potential for harm if they are not appropriately controlled. F, Fulcrum.
www.konkur.in 24 Part I General Concepts/Treatment Planning stabilizing and retentive components in relation to the hori-zontal axis of rotation of the abutment becomes extremely important. An abutment tooth will better tolerate nonverti-cal forces if these forces are applied as near as possible to the horizontal axis of rotation of the abutment (Figure 4-10). The axial surface contours of abutment teeth must be altered to locate components of clasp assemblies more favorably in relation to the abutment’s horizontal axis (Figure 4-11).
POSSIBLE MOVEMENTS OF PARTIAL DENTURES If it is presumed that direct retainers are functioning to minimize vertical displacement, rotational movement will occur about some axis as the distal extension base or bases move toward, away, or horizontally across the underlying tissue. Unfortunately, these possible movements do not occur singularly or independently but tend to be dynamic and all occur at the same time. The greatest movement possible is found in the tooth/mucosal tissue–supported prosthesis because of reliance on the distal extension sup-porting tissue to share the functional loads with the teeth. Movement of a distal extension base toward the ridge tissue will be proportionate to the quality of that tissue, the accu-racy and extent of the denture base, and the applied total functional load. A review of prosthesis rotational move-ment that is possible around various axes in the mouth provides some understanding of how component parts of removable partial dentures should be prescribed to control prosthesis movement.
One movement is rotation about an axis through the most posterior abutments. This axis may pass through occlusal rests or any other rigid portion of a direct retainer assembly located occlusally or incisally to the height of contour of the primary abutments (see Figures 4-6 and 4-7). This axis, known as the fulcrum line, is the center of rotation as the distal extension base moves toward the sup-porting tissue when an occlusal load is applied. The axis of rotation may shift toward more anteriorly placed com-ponents, occlusal or incisal to the height of contour of the abutment, as the base moves away from the supporting tissue when vertical dislodging forces act on the partial denture. These dislodging forces result from the vertical pull of food between opposing tooth surfaces, the effects of moving border tissue, and the forces of gravity against a maxillary partial denture. If it is presumed that the direct retainers are functional and that the supportive anterior components remain seated, rotation—rather than total displacement—should occur. Vertical tissue-ward move-ment of the denture base is resisted by the tissue of the residual ridge in proportion to the supporting quality of that tissue, the accuracy of the fit of the denture base, and the total amount of occlusal load applied. Movement of the base in the opposite direction is resisted by the action of the retentive clasp arms on terminal abutments and the action of stabilizing minor connectors in conjunction with seated, vertical support elements of the framework anterior to the terminal abutments acting as indirect retainers. Indirect retainers should be placed as far as possible from the distal extension base, affording the best possible leverage against lifting of the distal extension base.
A second movement is rotation about a longitudinal axis as the distal extension base moves in a rotary direction about the residual ridge (see Figure 4-3). This movement is resisted primarily by the rigidity of the major and minor connectors and their ability to resist torque. If the connec-tors are not rigid, or if a stress-breaker exists between the distal extension base and the major connector, this rotation about a longitudinal axis applies undue stress to the sides of the supporting ridge or causes horizontal shifting of the denture base.
E (Gravity) E First class Second class Third class (Sticky foods) E F F F R R R A B C Figure 4-2 A to C, The three classes of levers. Classification is based on location of the fulcrum (F), resistance (R), and direc-tion of effort (force) (E). In dental terms, E can represent the force of occlusion or gravity; F can be a tooth surface such as an occlusal rest; and R is the resistance provided by a direct retainer or a guide plane surface.
www.konkur.in 25 Chapter 4 Biomechanics of Removable Partial Dentures Sagittal A A B B C C Frontal Horizontal Figure 4-3 Distal extension removable partial dentures will rotate when force is directed on the denture base. Differences in displace-ability of the periodontal ligament of the supporting abutment teeth and soft tissue covering the residual ridge permit this rotation. It would seem that rotation of the prosthesis occurs in a combination of directions rather than in a unidirectional way. The three possible movements of distal extension partial dentures are (A) rotation around a fulcrum line passing through the most posterior abutments when the denture base moves vertically toward or away from the supporting residual ridges; (B) rotation around a longitudinal axis formed by the crest of the residual ridge; and (C) rotation around a vertical axis located near the center of the arch.
www.konkur.in 26 Part I General Concepts/Treatment Planning A third movement is rotation about an imaginary vertical axis located near the center of the dental arch (see Figure 4-4). This movement occurs under function because diagonal and horizontal occlusal forces are brought to bear on the partial denture. It is resisted by stabilizing components, such as recip-rocal clasp arms and minor connectors that are in contact with vertical tooth surfaces. Such stabilizing components are essen-tial to any partial denture design, regardless of the manner of support and the type of direct retention employed. Stabilizing components on one side of the arch act to stabilize the partial denture against horizontal forces applied from the opposite side. It is obvious that rigid connectors must be used to make this effect possible.
Horizontal forces always will exist to some degree because of lateral stresses that occur during mastication, bruxism, clenching, and other patient habits. These forces are accen-tuated by failure to consider the orientation of the occlusal plane, the influence of malpositioned teeth in the arch, and the effects of abnormal jaw relationships. Fabricating an occlusion that is in harmony with the opposing dentition and that is free of lateral interference during eccentric jaw movements may minimize the magnitude of lateral stress. The amount of horizontal movement occurring in the partial denture therefore depends on the magnitude of the lateral forces that are applied and on the effectiveness of the stabiliz-ing components.
In a tooth-supported partial denture, movement of the base toward the edentulous ridge is prevented primarily by E R2 R1 Figure 4-4 The length of a lever from fulcrum (F) (see Figure 4-7) to resistance (R) is called the resistance arm. That portion of a lever from the fulcrum to the point of application of force (E) is called the effort arm. Whenever the effort arm is longer than the resistance arm, mechanical advantage favors the effort arm, pro-portionately to the difference in length of the two arms. In other words, when the effort arm is twice the length of the resistance arm, a 25-lb weight on the effort arm will balance a 50-lb weight at the end of the resistance arm. The opposite is also true and helps illustrate cross-arch stabilization. When the resistance arm is lengthened (cross-arch clasp assembly placed on a second molar [R2] versus a second premolar [R1]), the effort arm is more efficiently counteracted.
E E F F R R Figure 4-5 A cantilever can be described as a rigid beam sup-ported only at one end. When force is directed against the unsup-ported end of the beam (as in this rest placed on a cantilevered pontic), the cantilever can act as a first-class lever. The mechani-cal advantage in this illustration favors the effort arm.
Occlusal load Figure 4-6 Design often seen for a distal extension removable partial denture. A cast circumferential direct retainer engages the mesiobuccal undercut and is supported by the disto-occlusal rest. If it is rigidly attached to the abutment tooth, this could be considered a cantilever design, and detrimental first-class lever force may be imparted to the abutment if tissue support under the extension base allows excessive vertical movement toward the residual ridge.
www.konkur.in 27 Chapter 4 Biomechanics of Removable Partial Dentures the rests on the abutment teeth and to some degree by any rigid portion of the framework located occlusal to the height of contour. Movement away from the edentulous ridge is pre-vented by the action of direct retainers on the abutments that are situated at each end of each edentulous space and by the rigid, minor connector stabilizing components. Therefore the first of the three possible movements can be controlled in the tooth-supported denture. The second possible move-ment, which occurs along a longitudinal axis, is prevented by the rigid components of the direct retainers on the abut-ment teeth and by the ability of the major connector to resist torque. This movement is much less in the tooth-supported denture because of the presence of posterior abutments. The third possible movement occurs in all partial dentures. Therefore stabilizing components against horizontal move-ment must be incorporated into any partial denture design.
Occlusal load F R F Figure 4-7 As is shown in Figure 4-6, the potential for first-class lever action can also exist in Class II, modification 1 designs for removable partial denture frameworks. If a cast circumfer-ential direct retainer with a mesiobuccal undercut on the right first premolar were used, force placed on the denture base could impart upward and posteriorly moving force on the premolar, resulting in loss of contact between premolar and canine. Tis-sue support from the extension base area is most important to minimize the lever action of the clasp. The retainer design could help accommodate more of an anteriorly directed force during rotation of the denture base in an attempt to maintain tooth con-tact. Other alternatives to the first premolar design of the direct retainer would include a tapered wrought-wire retentive arm that uses mesiobuccal undercut, or that just has a buccal stabilizing arm above the height of contour.
Occlusal load Occlusal load A B F F Figure 4-8 Mesial rest concept for distal extension removable partial dentures. With recognition that clasp movement occurs with functional displacement of the distal extension base, the primary aim of a mesial rest is to alter the fulcrum position and resultant clasp movement, disallowing harmful engagement of the abutment tooth. A, Bar type of retainer, minor connector contacting the guiding plane on the distal surface of the premolar, and mesio-occlusal rest used to reduce cantilever or first-class lever force when and if the denture rotates toward the residual ridge. B, Tapered wrought-wire retentive arm, minor connector contacting guiding plane on the distal surface of the premolar, and the mesio-occlusal rest. This design is applicable when the distobuccal undercut cannot be found or created or when the tissue undercut contraindicates placement of a bar-type retentive arm. This design would be kinder to the periodontal ligament than would a cast, half-round retentive arm. Again, tissue support of the extension base is a key factor in reducing the lever action of the clasp arm. Note: Depending on the amount of contact of the minor connector proximal plate with the guiding plane, the fulcrum point will change.
Figure 4-9 More periodontal fibers are activated to resist forces directed vertically on the tooth than are activated to resist horizontally (off-vertical) directed force. The horizontal axis of ro-tation is located somewhere in the root of the tooth.
www.konkur.in 28 Part I General Concepts/Treatment Planning For prostheses capable of movement in three planes, occlusal rests should provide occlusal support only to resist tissue-ward movement. All other movements of the partial denture should be resisted by components other than occlu-sal rests. Entrance of the occlusal rest into a stabilizing func-tion would result in direct transfer of torque to the abutment tooth. Because movements around three different axes are possible in a distal extension partial denture, an occlusal rest for such a partial denture should not have steep vertical walls or locking dovetails. This rest design is characterized by lack of free movement, which could cause horizontal and torqu-ing forces to be applied intracoronally to the abutment tooth.
In the tooth-supported denture, the only movements of any significance are horizontal, and these may be resisted by the stabilizing effects of components placed on the axial surfaces of the abutments. Therefore in the tooth-supported denture, the use of intracoronal rests is permissible. In these instances, the rests provide not only occlusal support but also notable horizontal stabilization.
In contrast, all Class I and Class II partial dentures, which have one or more distal extension bases, are not totally tooth supported. Neither are they completely retained by bounding abutments. Any extensive Class III or Class IV partial denture that does not have adequate abut-ment support falls into the same category. These latter den-tures may derive some support from the edentulous ridge and therefore may have composite support from both teeth and ridge tissue.
Figure 4-10 Clasps placed closer to the occlusal/incisal sur-face have a greater likelihood of imparting tipping forces to the abutments.
Lingual Buccal Figure 4-11 The abutment has been contoured (see shaded region) to allow rather favorable locations of retentive and recip-rocal stabilizing components (mirror view).
www.konkur.in CHAPTER 5 Major and Minor Connectors CHAPTER OUTLINE Role of Major Connectors in Control of Prosthesis Movement Location Mandibular Major Connectors Maxillary Major Connectors Minor Connectors Functions Form and Location Tissue Stops Finishing Lines Reaction of Tissue to Metallic Coverage Major Connectors in Review The respective components parts of a removable partial den-ture are able to contribute to the overall performance of a prosthesis if they are joined through an effective connection. That is the role of the major connector.
Components of a typical removable partial denture are illustrated in Figure 5-1. 1. Major connectors 2. Minor connectors 3. Rests 4. Direct retainers 5. Stabilizing or reciprocal components (as parts of a clasp assembly) 6. Indirect retainers (if the prosthesis has distal extension bases) 7. One or more bases, each supporting one to several re-placement teeth (see Figure 5-1) When a prosthesis that can be removed from the mouth is used, the prosthesis must extend to both sides of the arch. This enables transfer of functional forces of occlusion from the denture base to all supporting teeth and tissues within an arch for optimum stability. It is through this cross-arch tooth contact, which occurs at some distance from the func-tional force, that optimum resistance can be achieved. This is most effectively accomplished when a rigid major connector joins the portion of the prosthesis receiving the function to selected regions throughout the arch.
The chief functions of a major connector include unifica-tion of the major parts of the prosthesis, distribution of the applied force throughout the arch to selected teeth and tis-sue, and minimization of torque to individual teeth. A prop-erly designed rigid major connector effectively distributes forces throughout the arch and acts to reduce the load to any one area while effectively controlling prosthesis movement.
The principle of leverage is connected with this compo-nent part. A rigid major connector limits movement possibil-ities by acting as a counteracting lever. This phenomenon is referred to as cross-arch stability. Cross-arch stability becomes more important in situations associated with high potential for greater prosthesis movement (e.g., distal extensions).
www.konkur.in 30 Part I General Concepts/Treatment Planning In this chapter, major and minor connectors are con-sidered separately as to their function, location, and design criteria. Other components are presented in designated chapters.
ROLE OF MAJOR CONNECTORS IN CONTROL OF PROSTHESIS MOVEMENT A major connector is the component of the partial denture that connects the parts of the prosthesis located on one side of the arch with those on the opposite side. It is that unit of the partial denture to which all other parts are directly or indirectly attached. This component also provides cross-arch stability to help resist displacement by functional stresses.
The major connector may be compared with the frame of an automobile or with the foundation of a building. It is through the major connector that other components of the partial denture become unified and effective. If the major connector is flexible, the ineffectiveness of connected compo-nents jeopardizes the supporting oral structures and can be a detriment to the comfort of the patient. Failure of the major connector to provide rigidity may be manifest by traumatic damage to periodontal support of the abutment teeth, injury to residual ridges, or impingement of underlying tissue. It is the dentist’s responsibility to ensure that appropriate design and fabrication of the major connector are accomplished.
Location Major connectors should be designed and located with the following guidelines in mind: 1. Major connectors should be free of movable tissue. 2. Impingement of gingival tissue should be avoided. 3. Bony and soft tissue prominences should be avoided dur-ing placement and removal.
A B C Figure 5-1 A, Framework for mandibular removable partial denture with the following components: 1, lingual bar major connector; 2a, minor connector by which the resin denture base will be attached; 2b, minor connector, proximal plate, which is part of clasp assembly; 2c, minor connector used to connect rests to major connectors; 3, occlusal rests; 4, direct retainer arm, which is part of the total clasp assembly; 5, stabilizing or reciprocal components of clasp assembly (includes minor connectors); and 6, an indirect retainer consisting of a minor connector and an occlusal rest. B, Maxillary removable partial denture with resin denture bases supporting artificial posterior teeth. Bases are attached to metal framework by ladderlike minor connectors similar to those seen in 2a. C, Mandibular bilateral distal extension removable partial denture with resin denture bases supporting artificial posterior teeth.
www.konkur.in 31 Chapter 5 Major and Minor Connectors 4. Relief should be provided beneath a major connector to prevent its settling into areas of possible interference, such as inoperable tori or elevated median palatal sutures. 5. Major connectors should be located and/or relieved to prevent impingement of tissue that occurs because the distal extension denture rotates in function.
Appropriate relief beneath the major connector avoids the need for its adjustment after tissue damage has occurred. In addition to being time consuming, grinding to provide relief from impingement may seriously weaken the major connec-tor, which can result in flexibility or possibly fracture. Major connectors should be carefully designed for proper shape, thickness, and location. Alteration of these dimensions by grinding can only be detrimental. Relief is covered at the end of this chapter and is expanded in Chapter 11.
Margins of major connectors adjacent to gingival tis-sue should be located far enough from the tissue to avoid any possible impingement. To accomplish this, it is rec-ommended that the superior border of a lingual bar con-nector be located a minimum of 4 mm below the gingival margin(s) (Figure 5-2). At the inferior border of the lingual bar connector, the limiting factor is the height of the mov-ing tissue in the floor of the mouth. Because the connector must have sufficient width and bulk to provide rigidity, a lin-guoplate is commonly used when space is insufficient for a lingual bar.
In the maxillary arch, because no moving tissue is pres-ent in the palate as in the floor of the mouth, the borders of the major connector may be placed well away from gingi-val tissue. Structurally, the tissue covering the palate is well suited for placement of the connector because of the pres-ence of firm submucosal connective tissue and an adequate, deep blood supply. However, when soft tissue covering the midline of the palate is less displaceable than the tissue cov-ering the residual ridge, varying amounts of relief under the connectors must be provided to avoid impingement of tis-sue. The amount of relief required is directly proportional to the difference in displaceability of the tissue covering the midline of the palate and the tissue covering the residual ridges. The gingival tissue, on the other hand, must have an unrestricted superficial blood supply to remain healthy. To accomplish this, it is recommended that the borders of Half-pear-shaped lingual bar pattern Linguoplate pattern 4 mm Metric 4 mm Rounded after being cast in metal Rounded after being cast in metal Continuous (cingulum) bar 4 mm Sublingual bar B A D C Figure 5-2 A, Lingual bar major connector should be located at least 4 mm inferior to gingival margins and farther if possible. The vertical height of a finished lingual bar should be at least 4 mm for strength and rigidity. If less than 8 mm exists between gingival mar-gins and the movable floor of the mouth, a linguoplate (B), a sublingual bar (C), or a continuous bar (D) is preferred as a major connector. Relief is provided for soft tissue under all portions of the mandibular major connector and at any location where the framework crosses the marginal gingiva. The inferior border of mandibular major connectors should be gently rounded after being cast to eliminate a sharp edge.
www.konkur.in 32 Part I General Concepts/Treatment Planning the palatal connector be placed a minimum of 6 mm away from and parallel to the gingival margins. Minor connec-tors that must cross gingival tissue should do so abruptly, joining the major connector at nearly a right angle (Figure 5-3). In this way, maximum freedom is ensured for gingival tissue.
Except for a palatal torus or a prominent median palatal suture area, palatal connectors ordinarily require no relief. Intimate contact between the connector and the supporting tissue adds much to the support, stability, and retention of the denture. Except for gingival areas, intimacy of contact elsewhere in the palate is not detrimental to the health of the tissue if rests are provided on abutment teeth to prevent tissue-ward movement.
An anterior palatal strap or the anterior border of a pala-tal plate also should be located as far as possible posteri-orly to avoid interference with the tongue in the area of the rugae. It should be uniformly thin and its anterior border should be located to follow the contours between the crests of the rugae. The anterior borders of such palatal major con-nectors therefore will be irregular in outline as they follow the contours between the rugae. The tongue may then pass from one ruga prominence to another without encounter-ing the border of the connector. When the connector border must cross a ruga crest, this should be done abruptly, while avoiding the crest as much as possible. The posterior limita-tion of a maxillary major connector should be just anterior to the vibrating line. A useful rule applied to major connec-tors and throughout partial denture design is to try to avoid adding any part of the denture framework to an already convex surface.
Characteristics of major connectors that contribute to the maintenance of health of the oral environment and the well-being of the patient may be listed as shown in Box 5-1.
Mandibular Major Connectors There are six types of mandibular major connectors described historically. Of these, the lingual bar and the linguoplate are used the majority of the time. 1. Lingual bar (Figure 5-4, A) 2. Linguoplate (see Figure 5-4, B) 3. Sublingual bar (see Figure 5-4, C) 4. Lingual bar with cingulum bar (continuous bar) (see Figure 5-4, D) 5. Cingulum bar (continuous bar) (see Figure 5-4, E) 6. Labial bar (see Figure 5-4, F) Lingual Bar The basic cross-section form of a mandibular major connec-tor is a half-pear shape, located above moving tissue but as far below the gingival tissue as possible. It is usually made of reinforced, 6-gauge, half-pear–shaped wax or a similar plas-tic pattern (Figure 5-5).
The major connector must be contoured so that it does not present sharp margins to the tongue and cause irritation or annoyance by an angular form. The superior border of a lingual bar connector should be tapered toward the gingival tissue superiorly, with its greatest bulk at the inferior border, resulting in a contour that has a half-pear shape. Lingual bar patterns, both wax and plastic, are made in this conventional shape. However, the inferior border of the lingual bar should be slightly rounded when the framework is polished. A rounded border will not impinge on the lingual tissue when the denture bases rotate inferiorly under occlusal loads. Fre-quently, additional bulk is necessary to provide rigidity, par-ticularly when the bar is long or when a less rigid alloy is used. This is accomplished by lining the ready-made form underneath with a sheet of 24-gauge casting wax rather than altering the original half-pear shape.
6 mm 6 mm Figure 5-3 Palatal major connector should be located at least 6 mm away from gingival margins and parallel to their mean cur-vature. All adjoining minor connectors should cross gingival tis-sues abruptly and should join major connectors at nearly a right angle.
Box 5-1 CHARACTERISTICS OF MAJOR CONNECTORS CONTRIBUTING TO HEALTH AND WELL-BEING 1. Are made from an alloy compatible with oral tissue 2. Are rigid and provide cross-arch stability through the principle of broad distribution of stress 3. Do not interfere with and are not irritating to the tongue 4. Do not substantially alter the natural contour of the lingual surface of the mandibular alveolar ridge or of the palatal vault 5. Do not impinge on oral tissue when the restoration is placed, is removed, or rotates in function 6. Cover no more tissue than is absolutely necessary 7. Do not contribute to retention or trapping of food par-ticles 8. Have support from other elements of the framework to minimize rotation tendencies in function 9. Contribute to the support of the prosthesis www.konkur.in 33 Chapter 5 Major and Minor Connectors The inferior border of a lingual mandibular major con-nector must be located so that it does not impinge on the tissue in the floor of the mouth because it changes elevations during the normal activities of mastication, swallowing, speaking, licking the lips, and so forth. Yet at the same time, it seems logical to locate the inferior border of these connec-tors as far inferiorly as possible to avoid interference with the resting tongue and trapping of food substances when they are introduced into the mouth. In addition, the more inferiorly a lingual bar can be located, the farther the supe-rior border of the bar can be placed from the lingual gingival crevices of adjacent teeth, thereby avoiding impingement on the gingival tissue.
At least two clinically acceptable methods may be used to determine the relative height of the floor of the mouth and locate the inferior border of a lingual mandibular major con-nector. The first method is to measure the height of the floor of A B C D E F Figure 5-4 Mandibular major connectors. A, Lingual bar. B, Linguoplate. C, Sublingual bar. D, Lingual bar with continuous bar (cin-gulum bar). E, Cingulum bar. F, Labial bar.
Relief Figure 5-5 Sagittal section showing half-pear shape of lingual bar. A taper of the superior border of the bar to the soft tissues above minimizes interference with the tongue and is more ac-ceptable to the patient than a dissimilar contour. Tissue relief is necessary to protect the soft tissue of the floor of the mouth.
www.konkur.in 34 Part I General Concepts/Treatment Planning the mouth in relation to the lingual gingival margins of adja-cent teeth with a periodontal probe (Figure 5-6). When these measurements are taken, the tip of the patient’s tongue should just lightly touch the vermilion border of the upper lip. Record-ing of these measurements permits their transfer to both diag-nostic and master casts, thus ensuring a rather advantageous location of the inferior border of the major connector. The second method is to use an individualized impression tray for which lingual borders are 3 mm short of the elevated floor of the mouth, and then to use an impression material that will permit the impression to be accurately molded as the patient licks the lips. The inferior border of the planned major connec-tor can then be located at the height of the lingual sulcus of the cast resulting from such an impression. Of the two methods, we have found measuring the height of the floor of the mouth to be less variable and more clinically acceptable.
Linguoplate If the rectangular space is bounded by the lingual bar, the ante-rior tooth contacts, and the cingula, and the bordering minor connectors are filled in, a linguoplate results (Figure 5-7).
Figure 5-7 View of mandibular Class I design with contoured linguoplate. Linguoplate is made as thin as possible and should follow the lingual contours of the teeth contacted. Doing so will often result in a scalloped superior margin. In this example, the straight superior margin can be bulky at the cingulum region, causing tongue discomfort.
A B C Figure 5-6 A, Height of floor of the mouth (tongue elevated) in relation to lingual gingival sulci measured with a periodontal probe. B, Recorded measurements are transferred to a diagnostic cast and then to a master cast after mouth preparations are completed. The line connecting marks indicates the location of the inferior border of the major connector. If periodontal surgery is performed, the line on the cast can be related to incisal edges of teeth and the measurements recorded for subsequent use. C, Impression made with functional movement of the tongue to demonstrate maximum shortening of the floor of the mouth. This allows visualization of the anatomic feature that establishes the inferior extent of a major connector. If a stock tray causes impingement on this functional position, an individualized or custom tray may be used for the same purpose.
www.konkur.in 35 Chapter 5 Major and Minor Connectors A linguoplate should be made as thin as is technically fea-sible and should be contoured to follow the contours of the teeth and the embrasures (Figure 5-8). The patient should be aware of as little added bulk and as few altered contours as possible. The upper border should follow the natural cur-vature of the supracingular surfaces of the teeth and should not be located above the middle third of the lingual surface, except to cover interproximal spaces to the contact points. The half-pear shape of a lingual bar should still form the infe-rior border that provides the greatest bulk and rigidity. All gingival crevices and deep embrasures must be blocked out parallel to the path of placement to avoid gingival irritation and any wedging effect between the teeth. In many instances, judicious recontouring of the lingual proximal surfaces of overlapped anterior teeth permits closer adaptation of the linguoplate major connector, eliminating otherwise deep interproximal embrasures to be covered (Figure 5-9).
The linguoplate does not in itself serve as an indirect retainer. When indirect retention is required, definite rests must be provided for this purpose. Both the linguoplate and the cingulum bar ideally should have a terminal rest at each end, regardless of the need for indirect retention. However, when indirect retainers are necessary, these rests may also serve as terminal rests for the linguoplate or continuous bar.
Because no component part of a removable partial den-ture should be added arbitrarily, each component should be added to serve a definite purpose. Indications for the use of a linguoplate may be listed as follows: 1. When the lingual frenum is high or the space available for a lingual bar is limited: In either instance, the supe-rior border of a lingual bar would have to be placed too close to the gingival tissue. Irritation could be avoided only by generous relief, which might be annoying to the tongue and create an undesirable food trap. When a clini-cal measurement from the free gingival margins to the slightly elevated floor of the mouth is less than 8 mm, a linguoplate is indicated in lieu of a lingual bar. The use of a linguoplate permits the inferior border to be placed more superiorly without tongue and gingival irritation and without compromise of rigidity. 2. In Class I situations in which the residual ridges have un-dergone excessive vertical resorption: Flat residual ridges offer little resistance to the horizontal rotational ten-dencies of a denture. The bracing effect provided by the remaining teeth must be depended upon to resist such rotation. A correctly designed linguoplate will engage the remaining teeth to help resist horizontal rotations. 3. For stabilizing periodontally weakened teeth, splinting with a linguoplate can be of some value when used with definite rests on sound adjacent teeth: As was described previously, a cingulum bar may be used to accomplish the same purpose because it actually represents the superior border of a linguoplate without the gingival apron. The cin-gulum bar accomplishes stabilization along with the other advantages of a linguoplate. However, it is frequently more objectionable to the patient’s tongue and is certainly more of a food trap than is the contoured apron of a linguoplate. 4. When the future replacement of one or more incisor teeth will be facilitated by the addition of retention loops to an existing linguoplate: Mandibular incisors that are peri-odontally weak may thus be retained, with provisions for possible loss and future additions.
The same reasons for use of a linguoplate anteriorly apply to its use elsewhere in the mandibular arch. If a lingual bar alone is to be used anteriorly, there is no reason to add an apron elsewhere. However, when auxiliary splinting is used for stabilization of the remaining teeth or for horizontal stabilization of the prosthesis, or for both, small rectangu-lar spaces sometimes remain. Tissue response to such small spaces is better when they are bridged with an apron than when they are left open. Generally, the apron is used to avoid gingival irritation or entrapment of food debris or to cover generously relieved areas that would be irritating to the tongue (Figure 5-10).
Figure 5-8 Apron of linguoplate (tissue side) is closely adapt-ed to the teeth extending into nonundercut interproximal em-brasures, resulting in a scalloped form. When well adapted, this form benefits from some anterior teeth acting together to help resist horizontal rotational tendencies of the prosthesis, espe-cially if the posterior ridge form does not resist such movement.
Figure 5-9 If a linguoplate major connector is indicated for this patient with overlapped anterior teeth, judicious recontour-ing of the lingual proximal surfaces of right lateral, right central, and left lateral incisors eliminates excessive undercuts and per-mit closer adaptation of the lingual apron of the major connector.
www.konkur.in 36 Part I General Concepts/Treatment Planning Sometimes a dentist is faced with a clinical situation wherein a linguoplate is indicated as the major connector of choice even though the anterior teeth are quite spaced and the patient strenuously objects to metal showing through the spaces. The linguoplate can then be constructed so that the metal will not appreciably show through the spaced anterior teeth (Figure 5-11). The rigidity of the major connector is not greatly altered. However, such a design may be as much of a food trap as the continuous bar type of major connector.
Design of Mandibular Major Connectors The following systematic approach to the design of a man-dibular lingual bar and linguoplate major connectors can be readily used with diagnostic casts after the diagnostic data are considered and related to the basic principles of major connector design: Step 1: Outline the basal seat areas on the diagnostic cast (Figure 5-12, A) Step 2: Outline the inferior border of the major connector (see Figure 5-12, B) Step 3: Outline the superior border of the major connector (see Figure 5-12, C) Step 4: Connect the basal seat area to the inferior and supe-rior borders of the major connector, and add minor con-nectors to retain the acrylic resin denture base material (see Figure 5-12, D) Sublingual Bar A modification of the lingual bar that has been demon-strated to be useful when the height of the floor of the mouth does not allow placement of the superior border of the bar at least 4 mm below the free gingival margin is the sublingual bar. The bar shape remains essentially the same as that of a lingual bar, but placement is inferior and pos-terior to the usual placement of a lingual bar, lying over and parallel to the anterior floor of the mouth. It is gener-ally accepted that a sublingual bar can be used in lieu of a lingual plate if the lingual frenum does not interfere, or in the presence of an anterior lingual undercut that would require considerable blockout for a conventional lingual bar. Contraindications include interfering lingual tori, high attachment of a lingual frenum, and interference with elevation of the floor of the mouth during functional movements.
Cingulum Bar (Continuous Bar) When a linguoplate is the major connector of choice but the axial alignment of the anterior teeth is such that exces-sive blockout of interproximal undercuts must be made, a cingulum bar may be considered. A cingulum bar located on or slightly above the cingula of the anterior teeth may be added to the lingual bar or can be used independently (Figure 5-13). In addition, when wide diastemata exist between the lower anterior teeth, a continuous bar retainer may be more esthetically acceptable than a linguoplate.
Labial Bar Fortunately, in only a few situations does extreme lingual inclination of the remaining lower premolar and inci-sor teeth prevent the use of a lingual bar major connec-tor. With the use of conservative mouth preparations in the form of recontouring and blockout, a lingual major connector can almost always be used. Lingually inclined teeth sometimes may have to be reshaped by means of crowns. Although the use of a labial major connector may be necessary in rare instances, this should be avoided by resorting to necessary mouth preparations rather than by accepting a condition that is otherwise correctable (Figure 5-14). The same applies to the use of a labial bar when a mandibular torus interferes with placement of a lingual bar. Unless surgery is definitely contraindicated, interfering mandibular tori should be removed so that the use of a labial bar connector may be avoided.
A modification to the linguoplate is the hinged con-tinuous labial bar. This concept is incorporated in the Figure 5-11 Interrupted linguoplate in the presence of inter-proximal spaces.
Relief Figure 5-10 Sagittal section through the linguoplate demon-strating a basic half-pear–shaped inferior border with the metallic apron extending superiorly. Extension of linguoplate to the height of contour on the premolar was accomplished to enclose a rather large triangular interproximal space inferior to the contact point between the canine and premolar. Such spaces may often be bridged to eliminate obvious food traps. Relief is provided for soft tissue under all portions of the mandibular major connector and at any location where the framework crosses the marginal gingiva.
www.konkur.in 37 Chapter 5 Major and Minor Connectors Swing-Lock design, which consists of a labial or buccal bar that is connected to the major connector by a hinge at one end and a latch at the other end (Figure 5-15).
Support is provided by multiple rests on the remain-ing natural teeth. Stabilization and reciprocation are pro-vided by a linguoplate that contacts the remaining teeth and are supplemented by the labial bar with its retentive Swing-Lock Inc., Milford, Texas.
struts. Retention is provided by a bar type of retentive clasp with arms projecting from the labial or buccal bar and contacting the infrabulge areas on the labial surfaces of the teeth.
Use of the Swing-Lock concept would seem primarily indicated when the following conditions are present: 1. Missing key abutments: When all remaining teeth are used for retention and stability, the absence of a key abutment (such as, a canine) may not present as A B D C Figure 5-12 Sequence of design considerations for a mandibular major connector. A, Diagnostic cast with basal seat areas outlined. B, Inferior border of the major connector is outlined. Location of the inferior border was determined as suggested in Figure 5-6 and ex-tends to the mesial of the mandibular right molar. C, Superior border of the major connector is outlined. Limited space for the lingual bar requires use of the linguoplate major connector. Linguoplate requires that rest seats be used on canines and the first premolar for positive support. D, Rest seat areas on the posterior teeth are outlined, and minor connectors for retention of resin denture bases are sketched.
A B Figure 5-13 A, Lingual bar and cingulum bar (continuous bar) major connector. Upper portion of this major connector is located on the cingula of anterior teeth. The requirement of positive support by rest seats, at least as far anteriorly as the canines, is critical. Note that the superior border of the lingual bar portion is often placed objectionably close to the gingival margins if sufficient bulk for rigidity is to be obtained. This type of major connector easily traps food and is often more objectionable to patients than a linguoplate. B, Cingulum bar (continuous bar) major connector. Although this design may reduce the possibility of food entrapment, it may not pro-vide adequate rigidity.
www.konkur.in 38 Part I General Concepts/Treatment Planning serious a treatment problem with this concept as with more conventional designs (Figure 5-16).
2. Unfavorable tooth contours: When existing tooth con-tours (uncorrectable by recontouring with appropriate restorations) or excessive labial inclinations of anteri-or teeth prevent conventional clasp designs, the basic principles of removable partial design may be better implemented with the Swing-Lock concept.
3. Unfavorable soft tissue contours: Extensive soft tissue undercuts may prevent proper location of component parts of a conventional removable partial denture or an overdenture. The hinged continuous labial bar con-cept may provide an adjunctive modality to accom-modate such unfavorable soft tissue contours.
4. Teeth with questionable prognoses: The possibility of losing a key abutment tooth with a guarded prog-nosis seriously affects the stability and retention of a conventional prosthesis. Because all remaining teeth Figure 5-15 The hinge for this continuous labial bar connec-tor is located buccal and distal to the remaining dentition (area of tooth #21). The latching mechanism is opposite to the hinge, adjacent to tooth #28. In this location, it will be housed within the buccal flange of the denture.
Figure 5-16 Absence of the mandibular canine requires that all remaining anterior teeth be used for stabilization and reten-tion of the replacement restoration. The Swing-Lock concept can be used to ensure that all remaining teeth share in stabilization and retention of the prosthesis.
function as abutments in the Swing-Lock denture, it seems that the loss of a tooth would not compromise retention and stability to such a degree. The hinged la-bial bar type of restoration can be used satisfactorily in certain clinically compromised situations. As is true with any type of removable restoration, good oral hy-giene, maintenance, regular recall, and close attention to details of design are paramount to successful imple-mentation of this treatment concept.
Obvious contraindications to the use of this hinged labial bar concept are apparent. The most obvious is poor oral hygiene or lack of motivation for plaque control by the patient. Other contraindications include the presence of a shallow buccal or labial vestibule or a high frenal attach-ment. Any of these factors would prevent the proper place-ment of components of the Swing-Lock partial denture.
A B Figure 5-14 A, Lingual inclination of patient’s canines and premolars precludes use of the lingual bar. B, Labial bar major connector was used in treatment. Retention was obtained on terminal abutments. Support and stabilization were gained by using rests, minor con-nectors arising from the labial bar, and well-fitting denture bases.
www.konkur.in 39 Chapter 5 Major and Minor Connectors Maxillary Major Connectors Six basic types of maxillary major connectors are considered: 1. Single palatal strap (Figure 5-17, A) 2. Combination anterior and posterior palatal strap–type connector (see Figure 5-17, B) 3. Palatal plate-type connector (see Figure 5-17, C) 4. U-shaped palatal connector (see Figure 5-17, D) 5. Single palatal bar (see Figure 5-17, E) 6. Anterior-posterior palatal bars (see Figure 5-17, F) Whenever it is necessary for the palatal connector to make contact with the teeth for reasons of support, definite tooth support must be provided. This is best accomplished by establishing definite rest seats on the predetermined abutment teeth. These should be located far enough above the gingival attachment to provide for bridging of the gin-gival crevice with blockout. At the same time, they should be low enough on the tooth to avoid unfavorable leverage and low enough on the maxillary incisors and canine teeth to avoid incisal interference of the opposing dentition.
Major connector components resting on unprepared inclined tooth surfaces can lead to slippage of the denture or to orthodontic movement of the tooth, or to both. In either situation, settling into gingival tissue is inevitable. In the absence of the required vertical support provided by rests, the health of the surrounding tissue is usually impaired. Similarly, interproximal projections of the major connector A B D C E F Figure 5-17 Maxillary major connectors. A, Single palatal strap. B, Anterior-posterior palatal strap. C, Palatal plate. D, U-shaped. E, Single palatal bar. F, Anterior-posterior palatal bars.
www.konkur.in 40 Part I General Concepts/Treatment Planning that rest on the gingival third of the tooth and on gingival tissues that are structurally unable to render support may be traumatized. To prevent these sequelae, one should support the major connector with definite rests on the teeth, pro-vide adequate gingival relief, and/or locate the connector far enough away from the gingival margin to avoid any possible restriction of blood supply and entrapment of food debris. All gingival crossings should be abrupt and at right angles to the major connector. Creating a sharp, angular form on any portion of a palatal connector should be avoided, and all borders should be tapered toward the tissue.
Single Palatal Strap Bilateral tooth-supported prostheses, even those with short edentulous spaces, are effectively connected with a single, broad palatal strap connector, particularly when the eden-tulous areas are located posteriorly (Figure 5-18). Such a connector can be made rigid without objectionable bulk and interference with the tongue, provided the cast framework material is distributed in three planes. Suitable rigidity, with-out excessive bulk, may be obtained for a single palatal strap by the laboratory technician by casting a 22-gauge matte plastic pattern.
For reasons of torque and leverage, a single palatal strap major connector should not be used to connect anterior replacements with distal extension bases. To be rigid enough to resist torque and to provide adequate vertical support and horizontal stabilization, a single palatal strap would have to be objectionably bulky. When placed anteriorly, this bulk would become even more objectionable to the patient because it could interfere with speech.
Combination Anterior and Posterior Palatal Strap–Type Connector Structurally, this is a rigid palatal major connector. The ante-rior and posterior palatal strap combination may be used in almost any maxillary partial denture design (Figure 5-19).
A posterior palatal strap should be flat and a minimum of 8 mm wide. Posterior palatal connectors should be located as far posterior as possible to avoid interference with the tongue but anterior to the line of flexure formed by the junction of the hard and soft palates. The only condition that prevents their use is an inoperable maxillary torus that extends pos-terior to the soft palate. In this situation, a broad, U-shaped major connector may be used, as described elsewhere in this chapter.
The strength of this major connector design lies in the fact that the anterior and posterior components are joined together by longitudinal connectors on either side, which form a square or rectangular frame. Each component braces A B Figure 5-18 A, This single palatal strap–type major connector is better suited for the restoration of short-span tooth-supported bilateral edentulous areas. It may also be used in tooth-supported unilateral edentulous situations with provision for cross-arch attachment by extracoronal retainers or internal attachments. Width of the palatal strap should be confined within the bound-aries of supporting rests. B, Sagittal section. Midportion of the major connector demonstrates slight elevation to provide rigid-ity. Such thickness of the major connector does not appreciably alter palatal contours.
Figure 5-19 Anterior-posterior palatal strap–type major connector. The anterior component is a flat strap located as far posteriorly as possible to avoid rugae coverage and tongue inter-ference. The anterior border of this strap should be located just posterior to a rugae crest or in the valley between two crests. The posterior strap is thin, a minimum of 8 mm wide, and located as far posteriorly as possible, yet entirely on the hard palate. It should be located at right angles to midline rather than diagonally.
www.konkur.in 41 Chapter 5 Major and Minor Connectors the others against possible torque and flexure. Flexure is practically nonexistent in such a design.
The anterior connector may be extended anteriorly to sup-port anterior tooth replacements. In this form, a U-shaped connector is made rigid by the horizontal strap that has been added posteriorly. If a maxillary torus exists, it may be encircled by this type of major strap-type connector without reduced rigidity.
The combination anterior-posterior connector design may be used with any Kennedy class of partially edentulous arch. It is used most frequently in Classes II and IV, whereas the single wide palatal strap is used more frequently in Class III situations. The palatal plate–type or complete coverage connector, described in this chapter, is used most frequently in Class I situations for reasons to be explained subsequently. All maxillary major connectors should cross the midline at a right angle rather than on a diagonal. It has been suggested that the tongue will accept symmetrically placed compo-nents far more readily than those placed without regard for symmetry.
Palatal Plate–Type Connector For lack of better terminology, the words palatal plate are used to designate any thin, broad, contoured palatal coverage used as a maxillary major connector and covering one half or more of the hard palate (Figure 5-20). Anatomic replica pala-tal castings have uniform thickness and strength by reason of their corrugated contours. Through the use of electrolytic polishing, uniformity of thickness can be maintained, and the anatomic contours of the palate will be faithfully repro-duced in the finished denture.
The anatomic replica palatal major connector has several potential advantages: 1. It permits the making of a uniformly thin metal plate that reproduces faithfully the anatomic contours of the pa-tient’s own palate. Its uniform thinness and the thermal conductivity of the metal are designed to make the palatal plate more readily acceptable to the tongue and underly-ing tissue. 2. The corrugation in the anatomic replica adds strength to the casting; thus a thinner casting with adequate rigidity can be made. 3. Surface irregularities are intentional rather than acciden-tal; therefore electrolytic polishing is all that is needed.
The original uniform thickness of the plastic pattern is thus maintained. 4. By virtue of intimate contact, interfacial surface tension between metal and tissue provides the prosthesis with greater retention. Retention must be adequate to resist the pull of sticky foods, the action of moving border tissue against the denture, the forces of gravity, and the more vi-olent forces of coughing and sneezing. These are all resist-ed to some extent by the retention of the base itself, which is proportional to the total area of denture base contact with supporting tissue. The required amount of both di-rect and indirect retention will depend on the amount of retention provided by the denture base.
The palatal plate may be used in any one of three ways.
It may be used as a plate of varying width that covers the area between two or more edentulous areas, as a complete or partial cast plate that extends posterior to the junction of the hard and soft palates (Figures 5-21 and 5-22), or in the Figure 5-20 Palatal major connector covering two thirds of the palate. The anterior border follows valleys between rugae and does not extend anteriorly to indirect retainers on the first premo-lars. The posterior border is located at the junction of the hard and soft palates but does not extend onto the soft palate. In the bilateral distal extension situation illustrated, indirect retainers are a must to aid in resisting horizontal rotation of the restora-tion. Note that provisions have been made for a butt-type joint joining the denture bases and framework as the denture base on each side passes through the pterygomaxillary notch.
Figure 5-21 Palatal plate major connector for a Class I, modi-fication 1 removable partial denture. The posterior border lies on the immovable hard palate and crosses the midline at a right angle. Total contact provides excellent support.
www.konkur.in 42 Part I General Concepts/Treatment Planning form of an anterior palatal connector with a provision for extending an acrylic resin denture base in a posterior direc-tion (Figure 5-23).
The palatal plate should be located anterior to the poste-rior palatal seal area. The maxillary complete denture’s typi-cal posterior palatal seal is not necessary with a maxillary partial denture’s palatal plate because of the accuracy and stability of the cast metal.
When the last remaining abutment tooth on either side of a Class I arch is the canine or first premolar tooth, com-plete palatal coverage is strongly advised, especially when the residual ridges have undergone excessive vertical resorption. This may be accomplished in one of two ways. One method is to use a complete cast plate that extends to the junction of the hard and soft palates (see Figure 5-22). The other method is to use a cast major connector anteriorly, with retention pos-teriorly, for the attachment of an acrylic-resin denture base that extends posterior to the anatomic landmarks previously described (see Figure 5-23).
Despite increased costs, the advantages of a cast palate make it preferable to an acrylic-resin palate. However, the latter method may be used satisfactorily when relining is anticipated or cost is a factor. The complete palatal plate is not a connector that has received universal use. It has, how-ever, become accepted as a satisfactory palatal connector for many maxillary partial dentures. In all circumstances, the portion contacting the teeth must have positive support from adequate rest seats. The dentist should be familiar with its use and, at the same time, with its limitations, so that it may be used intelligently and to fullest advantage.
Design of Maxillary Major Connectors In 1953, Blatterfein described a systematic approach to designing maxillary major connectors. His method involves five basic steps and is certainly applicable to most maxillary removable partial denture situations. When using a diag-nostic cast and knowledge of the relative displaceability of the palatal tissue, including that covering the median palatal raphe, he recommends the following basic steps: Step 1: Outline of primary bearing areas: The primary bear-ing areas are those that will be covered by the denture base(s) (Figure 5-24, A and B).
Figure 5-22 Complete coverage palatal major connector. The posterior border terminates at the junction of the hard and soft palates. The anterior portion, in the form of the palatal linguo-plate, is supported by positive lingual rest seats on canines. The location of finishing lines is most important in this type of ma-jor connector. Anteroposteriorly, they should be parallel to a line along the center of the ridge crest and located just lingual to an imaginary line contacting the lingual surfaces of missing natural teeth. Alteration of the natural palatal contour should be antici-pated with its attendant detrimental effects on speech if these contours are not followed.
A B Figure 5-23 A, Maxillary major connector in the form of a palatal linguoplate with provisions for attaching the full-coverage resin den-ture base. B, Completed removable partial denture with resin base. The palatal linguoplate is supported by rests occupying lingual rest seats prepared in cast restorations on canines. This type of removable partial denture is particularly applicable when (1) residual ridges have undergone extreme vertical resorption, and (2) terminal abutments have suffered some bone loss and splinting cannot be accomplished.
www.konkur.in 43 Chapter 5 Major and Minor Connectors Step 2: Outline of nonbearing areas: The nonbearing areas are the lingual gingival tissue within 5 to 6 mm of the re-maining teeth, hard areas of the medial palatal raphe (in-cluding tori), and palatal tissue posterior to the vibrating line (see Figure 5-24, C).
Step 3: Outline of connector areas: Steps 1 and 2, when com-pleted, provide an outline or designate areas that are avail-able to place components of major connectors (see Figure 5-24, C).
Step 4: Selection of connector type: Selection of the type of connector(s) is based on four factors: (1) mouth comfort, (2) rigidity, (3) location of denture bases, and (4) indirect retention. Connectors should be of mini-mum bulk and should be positioned so that interfer-ence with the tongue during speech and mastication is not encountered. Connectors must have a maximum of rigidity to distribute stress bilaterally. The double-strap type of major connector provides the maximum rigidity without bulk and total tissue coverage. In many instances, the choice of a strap type of major connector is limited by the location of the edentulous ridge areas. When edentulous areas are located anteriorly, the use of only a posterior strap is not recommended. By the same token, when only posterior edentulous areas are present, the use of only an anterior strap is not recom-mended. The need for indirect retention influences the outline of the major connector. Provision must be made in the major connector so that indirect retainers may be attached.
Step 5: Unification: After selection of the type of major con-nector based on considerations in Step 4, the denture base areas and connectors are joined (see Figure 5-24, D).
Indications for the use of complete palatal coverage have been previously discussed in this chapter. Although many variations in palatal major connectors have been noted, a thorough comprehension of all factors that influence their design will lead to the best design for each patient.
A B C D Figure 5-24 A, Diagnostic cast of partially edentulous maxillary arch. B, The palatal extent of the denture base areas are located 2 mm from the palatal surface of the posterior teeth. C, Nonbearing areas outlined in black, which include lingual soft tissue within 5 to 6 mm of teeth, an unyielding median palatal raphe area, and the soft palate. The space bounded by bearing and nonbearing area outlines is available for placement of the major connector. D, The major connector selected will be rigid and noninterfering with the tongue and will cover a minimum of the palate.
www.konkur.in 44 Part I General Concepts/Treatment Planning U-Shaped Palatal Connector From both the patient’s standpoint and a mechanical standpoint, the U-shaped palatal connector is the least desirable of maxillary major connectors. It should never be used arbitrarily. When a large inoperable palatal torus exists and occasionally when several anterior teeth are to be replaced, the U-shaped palatal connector may have to be used (Figure 5-25). In most instances, however, other designs will serve more effectively.
The following are the principal objections to use of the U-shaped connector: 1. Its lack of rigidity (compared with other designs) can allow lateral flexure under occlusal forces, which may induce torque or direct lateral force to abutment teeth.
2. The design fails to provide good support characteris-tics and may permit impingement of underlying tissue when subjected to occlusal loading.
3. Bulk to enhance rigidity results in increased thickness in areas that are a hindrance to the tongue.
Many maxillary partial dentures have failed for no other reason than the flexibility of a U-shaped major con-nector (Figure 5-26). To be rigid, the U-shaped palatal connector must have bulk where the tongue needs free-dom the most—the rugae area. Without sufficient bulk, the U-shaped design leads to increased flexibility and movement at the open ends. In distal extension partial dentures, when tooth support posterior to the edentulous area is nonexistent, movement is particularly noticeable and is traumatic to the residual ridge. No matter how well the extension base is supported or how harmonious the occlusion, without a rigid major connector the residual ridge suffers.
The wider the coverage of a U-shaped major connector, the more it resembles a palatal plate–type connector with its several advantages. But when used as a narrow U design, Figure 5-25 U-shaped palatal connector is probably the least rigid type of maxillary major connector and should be used only when a large inoperable palatal torus prevents the use of palatal coverage or combination anterior-posterior palatal strap–type designed framework.
Figure 5-26 Removable partial denture design that uses an objectionable U-shaped palatal major connector. Such a connec-tor lacks necessary rigidity, places bulk where it is most objec-tionable to the patient, and impinges on gingival tissue lingual to remaining teeth.
the necessary rigidity is usually lacking. A U-shaped con-nector may be made more rigid with multiple tooth sup-port through definite rests. A common error in the design of a U-shaped connector, however, is its proximity to, or actual contact with, gingival tissue. The principle that the borders of major connectors should be supported by rests in prepared rest seats or should be located well away from gingival tissue has been stated previously. Most U-shaped connectors fail to do either, with resulting gingival irri-tation and periodontal damage to the tissue adjacent to remaining teeth.
Single Palatal Bar To differentiate between a palatal bar and a palatal strap, a palatal connector component less than 8 mm in width is referred to as a bar in this textbook. The single pala-tal bar is perhaps the most widely used and yet the least logical of all palatal major connectors. It is difficult to say whether the bar or the U-shaped palatal connector is the more objectionable of palatal connectors.
For a single palatal bar to have the necessary rigid-ity for cross-arch distribution of stress, it must have concentrated bulk, which, unfortunately, is all too often ignored. For a single palatal bar to be effective, it must be rigid enough to provide support and cross-arch stabilization and must be centrally located between the halves of the denture. Mechanically, this practice may be sound enough. However, from the standpoint of patient comfort and alteration of palatal contours, it is highly objectionable.
A partial denture made with a single palatal bar is often too thin and flexible or too bulky and objectionable to the patient’s tongue. The decision to use a single palatal bar instead of a strap should be based on the size of the denture-bearing areas that are connected and on whether www.konkur.in 45 Chapter 5 Major and Minor Connectors a single connector located between them would be rigid without objectionable bulk.
Combination Anterior and Posterior Palatal Bar–Type Connectors Structurally, this combination of major connectors exhib-its many of the same disadvantages as the single palatal bar (Figure 5-27). To be sufficiently rigid and to provide needed support and stability, these connectors could be too bulky and could interfere with tongue function.
Beading of the Maxillary Cast Beading is a term used to denote the scribing of a shallow groove on the maxillary master cast outlining the palatal major connector exclusive of rugae areas (Figure 5-28). The purposes of beading are as follows: 1. To transfer the major connector design to the invest-ment cast (Figure 5-29) 2. To provide a visible finishing line for the casting (Figure 5-30) 3. To ensure intimate tissue contact of the major connec-tor with selected palatal tissue Beading is readily accomplished by using an appropri-ate instrument, such as a cleoid carver. Care must be exer-cised to create a groove no larger than 0.5 mm in width or depth (Figure 5-31).
MINOR CONNECTORS Minor connectors are those components that serve as the connecting link between the major connector or the base of a removable partial denture and the other components of the prosthesis, such as the clasp assembly, indirect retainers, occlusal rests, or cingulum rests. In many instances, a minor Figure 5-27 Combination anterior-posterior palatal bar. To be sufficiently rigid to provide required support and stability, these major connectors must be excessively bulky. Because of its bulk and location, the anterior bar often interferes with the tongue.
Figure 5-28 Framework design on master cast before prepa-ration for duplication in refractory investment. A shallow groove (0.5 mm) has been scribed on the outline of anterior and poste-rior borders of the major connector. The anterior outline follows the valleys of rugae. Beading is readily accomplished with a cle-oid carver. A slightly rounded groove is preferred to a V-shaped groove.
connector may be continuous with some other part of the denture. For example, an occlusal rest at one end of a lin-guoplate is actually the terminus of a minor connector, even though that minor connector is continuous with the linguo-plate. Similarly the portion of a partial denture framework that supports the clasp and the occlusal rest is a minor con-nector, which joins the major connector with the clasp proper.
Those portions of a removable partial denture frame-work that retain the denture bases are also minor connec-tors. When dental implants are placed, modifications can be required in the denture base retaining minor connector and are dependent on how the implants will be engaged. Implants placed anteriorly to enhance retention by removing the need for a visible clasp must take into account retentive device bulk and connection requirements for joining to the prosthe-sis (i.e., direct to the metallic frame or via embedding within the acrylic resin denture base). Modifications required for implants placed more distal for purposes of support typically require designing space for embedding an attachment hous-ing within the denture base (Figure 5-32).
Functions In addition to joining denture parts, the minor connector serves two other purposes: 1. Transfers functional stress to the abutment teeth: This is a prosthesis-to-abutment function of the minor con-nector. Occlusal forces applied to the artificial teeth are www.konkur.in 46 Part I General Concepts/Treatment Planning transmitted through the base to the underlying ridge tissue if that base is primarily tissue supported. Occlusal forces applied to the artificial teeth are also transferred to abut-ment teeth through occlusal rests. The minor connectors arising from a rigid major connector make possible this transfer of functional stress throughout the dental arch. 2. Transfers the effects of the retainers, rests, and stabiliz-ing components throughout the prosthesis: This is an abutment-to-prosthesis function of the minor connector.
Thus forces applied on one portion of the denture may be resisted by other components placed elsewhere in the A B Figure 5-29 A, Refractory cast. Note the definitive outline of the major connector indicated by beading lines that were transferred in duplicating the master cast. B, The wax pattern for the major connector is accurately executed by following the beading lines. The major connector is confined to previously scribed beading.
Figure 5-30 Cast and framework showing metal margin pro-duced by the 0.5-mm beading line scribed on the cast. Such a margin is easily finished in the lab and provides intimate tissue contact, preventing food from easily dislodging the prosthesis. Care should be exercised in adapting such a beaded margin to non-displaceable tissue, such as the median palatal raphe.
arch for that purpose. A stabilizing component on one side of the arch may be placed to resist horizontal forces that originate on the opposite side. This is possible only because of the transferring effect of the minor connector, which supports that stabilizing component, and the rigid-ity of the major connector.
Form and Location Like the major connector, the minor connector must have sufficient bulk to be rigid; otherwise the transfer of func-tional stresses to the supporting teeth and tissue will not be effective. At the same time, the bulk of the minor connector should not be objectionable.
A minor connector that contacts the axial surface of an abutment should not be located on a convex surface. Instead it should be located in an embrasure (Figure 5-33), where it will be least noticeable to the tongue. It should conform to the interdental embrasure, passing vertically from the major connector so that the gingival crossing is abrupt and covers as little of the gingival tissue as possible. It should be thickest toward the lingual surface, tapering toward the contact area (Figure 5-34).
The deepest part of the interdental embrasure should have been blocked out to avoid interference during placement and removal, and to avoid any wedging effect on the contacted teeth.
A modification of the conventional removable partial denture minor connector has been proposed. This applica-tion was suggested to be limited to the maxillary arch, with the minor connector located in the center of the lingual sur-face of the maxillary abutment tooth.
www.konkur.in 47 Chapter 5 Major and Minor Connectors It is suggested that this modification reduces the amount of gingival tissue coverage, provides enhanced guidance for the partial denture during insertion and removal, and increases stabilization against horizontal and rotational forces. However, because of its location, such a design varia-tion could encroach on the tongue space and create a greater potential space for food entrapment. The proposed variation should be used with careful application.
When a minor connector contacts tooth surfaces on either side of the embrasure in which it lies, it should be tapered to the teeth. This avoids sharp angles, which could hinder tongue movement, and eliminates spaces that could trap food (Figure 5-35).
It is a minor connector that contacts the guiding plane surfaces of the abutment teeth, whether as a connected part of a direct retainer assembly or as a separate entity (see Figure 5-34). Here the minor connector must be wide enough that the guiding plane can be used to fullest advantage. When it gives rise to a clasp arm, the connector should be tapered to the tooth below the origin of the clasp. If no clasp arm is formed (as when a bar clasp arm originates elsewhere), the connector should be tapered to a knife-edge the full length of its buccal aspect.
When an artificial tooth will be placed against a proxi-mal minor connector, the minor connector’s greatest bulk should be toward the lingual aspect of the abutment tooth. A B Figure 5-31 A, Tissue side of casting. Note slightly elevated ridges outlining anterior, posterior, and midpalatal opening regions of this anterior-posterior palatal strap major connector. B, Casting finished to a demarcated outline and seated on the master cast showing intimate adaptation.
Figure 5-32 The Locator attachment in the posterior aspect of the distal extension base of the removable partial denture pro-vides retention for the distal extension.
Figure 5-33 In an embrasure space, the minor connector is tapered to the tooth to avoid bulk and to accommodate the tongue.
www.konkur.in 48 Part I General Concepts/Treatment Planning This way sufficient bulk is ensured with the least interfer-ence with placement of the artificial tooth. Ideally the arti-ficial tooth should contact the abutment tooth with only a thin layer of metal intervening buccally. Lingually the bulk of the minor connector should lie in the interdental embrasure—the same as between two natural teeth.
As was stated previously, those portions of a denture framework by which acrylic-resin denture bases are attached are minor connectors. This type of minor con-nector should be so designed that it will be completely embedded within the denture base.
The junctions of these mandibular minor connectors with the major connectors should be strong butt-type joints but without appreciable bulk (see Figure 5-35). Angles formed at the junctions of the connectors should not be Figure 5-34 The minor connector that contacts the guiding plane is part of a clasp assembly. It can be separate from the other parts, or, as in this case, it can be connected to the lin-gual stabilizing portion of the clasp assembly. The proximal plate minor connector contact is about one-half the distance between tips of adjacent buccal and lingual cusps of the abutment tooth, and it extends gingivally, contacting an area of the abutment from the marginal ridge to two-thirds the length of the enamel crown. Viewed from above, it is triangular, the apex of the triangle be-ing located buccally and the base of the triangle being located lingually. Less interference with the arrangement of the adjacent artificial tooth is encountered with minor connectors so shaped.
Figure 5-35 Finishing line at the junction of the ladderlike mi-nor connector and the major connector blends smoothly into the minor connector contacting the distal guiding plane on the sec-ond premolar. The framework is feathered toward tissue anterior to the finishing line to avoid as much bulk in this area as possible without compromising the strength of the butt-type joint.
greater than 90 degrees, thus ensuring the most advanta-geous and the strongest mechanical connection between the acrylic-resin denture base and the major connector.
An open latticework or ladder type of design is pref-erable and is conveniently made by using preformed 12-gauge half-round and 18-gauge round wax strips.
The minor connector for the mandibular distal exten-sion base should extend posteriorly about two-thirds the length of the edentulous ridge and should have elements on both lingual and buccal surfaces. Such an arrangement not only will add strength to the denture base but may minimize distortion of the cured base from its inherent strains caused by processing. The minor connector must be planned with care so that it will not interfere with the arrangement of artificial teeth (Figure 5-36).
Figure 5-36 The minor connector for attaching the resin den-ture base should be designed so that denture tooth placement is not compromised. The minor connector design should not include a main lattice strut at the ridge crest or in a desired tooth location.
Figure 5-37 Extension of the finishing line to the area of the pterygomaxillary notch provides a butt-type joint for attachment of the border portion of the resin base through the pterygomaxil-lary notch (arrows).
www.konkur.in 49 Chapter 5 Major and Minor Connectors A B C D E A means to attach acrylic-resin individualized trays to the mandibular framework when a corrected impression is planned must be arranged when the framework pattern is being developed. Three nailhead minor connectors fab-ricated as part of the denture base minor connector serve this purpose well. Unless some similar arrangement is made, the resin trays may become detached or loosened during impression-making procedures. Minor connec-tors for maxillary distal extension denture bases should extend the entire length of the residual ridge and should be of a ladderlike and loop design (Figure 5-37).
Tissue Stops Tissue stops are integral parts of minor connectors designed for retention of acrylic-resin bases. They pro-vide stability to the framework during the stages of transfer and processing. They are particularly useful in preventing distortion of the framework during acrylic-resin processing procedures. Tissue stops can engage buccal and lingual slopes of the residual ridge for stabil-ity (Figure 5-38).
Altered cast impression procedures often neces-sitate that tissue stops be augmented subsequent to Figure 5-38 A, Arrow points to location of the tissue stop. B, Master cast partially prepared for duplication in refractory investment. Posterior to relief wax, at the distal of the residual ridge (arrow), a tissue stop will be waxed. C, Wax tissue stop placed distal to relief (ar-row). After casting, this results in tissue stop contact of the framework. D, Tissue stop seen from labial position. E, Framework on cast . shows tissue contact posterior to the minor connector with planned relief. Arrow points to the created tissue stop www.konkur.in 50 Part I General Concepts/Treatment Planning the development of the altered cast. This can be read-ily accomplished with the addition of autopolymerizing acrylic resin (Figure 5-39).
Another integral part of the minor connector designed to retain the acrylic-resin denture base is simi-lar to a tissue stop but serves a different purpose. It is located distal to the terminal abutment and is a continu-ation of the minor connector contacting the guiding plane. Its purpose is to establish a definitive finishing index tissue stop for the acrylic-resin base after process-ing (Figure 5-40).
FINISHING LINES The finishing line junction with the major connector should take the form of an angle not greater than 90 degrees, therefore being somewhat undercut (Figure 5-41). Of course the medial extent of the minor con-nector depends on the lateral extent of the major palatal connector. Too little attention is given to this finishing line location in many instances. If the finishing line is located too far medially, the natural contour of the palate will be altered by the thickness of the junction and the acrylic resin supporting the artificial teeth (Figure 5-42).
If, on the other hand, the finishing line is located too far buccally, it will be most difficult to create a natural contour of the acrylic resin on the lingual surface of the artificial teeth. The location of the finishing line at the junction of the major and minor connectors should be based on restoration of the natural palatal shape, with consideration given to the location of the replacement teeth.
Equal consideration must be given to the junction of minor connectors and bar-type direct retainer arms. These junctions are 90-degree butt-type joints and should follow the guidelines for base contour and clasp length.
A B Figure 5-39 A, Lower half of the flask in which the distal extension denture was invested. Note that the terminal portion of the mi-nor connector (original tissue stop) is elevated from the residual ridge. The framework was developed on cast, with the residual ridge recorded in its anatomic form. The residual ridge was later recorded in its functional form by a corrected impression, thus the elevated tissue stop. B, Autopolymerizing resin is painted on between the tissue stop and the ridge to maintain the position of the minor connec-tor during packing and processing procedures for a resin denture base.
REACTION OF TISSUE TO METALLIC COVERAGE The reaction of tissue to coverage by the metallic compo-nents of a removable partial denture has been the subject of significant controversy, particularly in regions of marginal gingiva and broad areas of tissue contact. These tissue reac-tions can result from pressure caused by lack of support, lack of adequate hygiene measures, and prolonged contact through continual use of a prosthesis.
Pressure occurs at regions where relief over gingival cross-ings and other areas of contact with tissue that are incapable of supporting the prosthesis is inadequate. Impingement will likewise occur if the denture settles because of loss of tooth and/or tissue support. This may be caused by failure of the rest areas resulting from improper design, caries involve-ment, fracture of the rest itself, or intrusion of abutment teeth under occlusal loading. It is important to maintain adequate relief and support from both teeth and tissue. Settling of the denture caused by loss of tissue support may also produce pressure elsewhere in the arch, such as beneath major con-nectors. Settling of the prosthesis must be prevented or cor-rected if it has occurred. Excessive pressure must be avoided whenever oral tissue must be covered or crossed by elements of the partial denture.
Lack of adequate hygiene measures can result in tissue reactions caused by an accumulation of food debris and bac-teria. Coverage of oral tissue with partial dentures that are not kept clean irritates those tissues because of an accumula-tion of irritating factors. This has led to misinterpretation of the effects of tissue coverage by prosthetic restorations. An additional hygiene concern is related to the problem of main-taining cleanliness of the tissue surface of the prosthesis.
The first two causes of untoward tissue reaction can be accentuated the longer a prosthesis is worn. It is apparent that mucous membranes cannot tolerate this constant contact with a prosthesis without resultant inflammation and breakdown of the epithelial barrier. Some patients become so accustomed www.konkur.in 51 Chapter 5 Major and Minor Connectors A B D C Figure 5-40 Finishing index tissue stop. A, Designed to facilitate finishing of the denture base resin at the region of the terminal abut-ment. Note the space at the anterior region of wax relief. Framework will be waxed to fill this space and provide positive tissue contact. B, Refractory cast shows space distal to the abutment. C, Wax pattern filling space for future tissue index contact. D, Framework index tissue stop anterior to relief beneath the minor connector of the distal extension base and posterior to the primary abutment.
Figure 5-41 Frontal sections through lingual finishing lines of palatal major connectors. The right image is through the full cast metal base major connector; the left image is through the resin denture base. In both situations, the location of finishing lines minimizes the bulk of resin attaching the artificial teeth. Palatal contours are restored, enhancing speech and contributing to a natural feeling for the patient.
www.konkur.in 52 Part I General Concepts/Treatment Planning to wearing a removable restoration that they neglect to remove it often enough to give the tissue any respite from constant contact. This is frequently true when anterior teeth are replaced by the partial denture and the individual does not allow the prosthesis to be out of the mouth at any time except in the privacy of the bathroom during toothbrushing. Living tissues should not be covered all the time, or changes in those tissues will occur. Partial dentures should be removed for several hours each day so that the effects of tissue contact can subside and the tissue can return to a normal state.
Clinical experience with the use of linguoplates and com-plete metallic palatal coverage has shown conclusively that when factors of pressure, cleanliness, and time are controlled, tissue coverage is not in itself detrimental to the health of oral tissue.
MAJOR CONNECTORS IN REVIEW Mandibular Lingual Bar Indications for Use: The lingual bar should be used for man-dibular removable partial dentures when sufficient space exists between the slightly elevated alveolar lingual sulcus and the lingual gingival tissue.
Characteristics and Location: (1) Half-pear shaped with bulkiest portion inferiorly located. (2) Superior border tapered to soft tissue. (3) Superior border located at least 4 mm inferior to gingival margins and farther if possible. (4) Inferior border located at the ascertained height of the alveo-lar lingual sulcus when the patient’s tongue is slightly elevated.
Blockout and Relief of Master Cast: (1) All tissue under-cuts parallel to the path of placement. (2) An additional Correct Incorrect Figure 5-42 Junction of the major connector and the minor connector at palatal finishing lines should be located 2 mm me-dial from an imaginary line that would contact the lingual sur-faces of missing posterior teeth. The finish line on the right is too far toward midline of the palate. The natural contours of the palate will be altered.
thickness of 32-gauge sheet wax when the lingual surface of the alveolar ridge is undercut or parallel to the path of placement (see Figures 11-23 and 11-24). (3) No relief is necessary when the lingual surface of the alveolar ridge slopes inferiorly and posteriorly. (4) One thickness of base-plate wax over basal seat areas (to elevate minor connectors for attaching acrylic-resin denture bases).
Waxing Specifications: (1) Six-gauge, half-pear–shaped wax form reinforced by 22- to 24-gauge sheet wax or simi-lar plastic pattern adapted to the design width. (2) Long bar requires more bulk than short bar; however, cross-sectional shape is unchanged.
Finishing Lines: Butt-type joint(s) with minor connector(s) for retention of denture base(s).
Mandibular Linguoplate Indications for Use: (1) When the alveolar lingual sul-cus so closely approximates the lingual gingival crevices that adequate width for a rigid lingual bar does not exist. (2) In those instances in which the residual ridges in Class I arch have undergone such vertical resorption that they will offer only minimal resistance to horizontal rotations of the denture through its bases. (3) For using periodon-tally weakened teeth in group function to furnish support to the prosthesis and to help resist horizontal (off-vertical) rotation of the distal extension type of denture. (4) When the future replacement of one or more incisor teeth will be facilitated by the addition of retention loops to an existing linguoplate.
Characteristics and Location: (1) Half-pear shaped with bulkiest portion inferiorly located. (2) Thin metal apron extending superiorly to contact cingula of anterior teeth and lingual surfaces of involved posterior teeth at their height of contour. (3) Apron extended interproximally to the height of contact points (i.e., closing interproximal spaces). (4) Scal-loped contour of apron as dictated by interproximal blockout. (5) Superior border finished to continuous plane with con-tacted teeth. (6) Inferior border at the ascertained height of the alveolar lingual sulcus when the patient’s tongue is slightly elevated.
Blockout and Relief of Master Cast: (1) All involved undercuts of contacted teeth parallel to the path of placement. (2) All involved gingival crevices. (3) Lingual surfaces of alve-olar ridge and basal seat areas the same as for a lingual bar.
Waxing Specifications: (1) Inferior border: 6-gauge, half-pear–shaped wax form reinforced with 24-gauge sheet wax or similar plastic pattern. (2) Apron: 24-gauge sheet wax.
Finishing Lines: Butt-type joint(s) with minor connector(s) for retention of denture base(s).
Mandibular Sublingual Bar Indications for Use: The sublingual bar should be used for mandibular removable partial dentures when the height of the floor of the mouth in relation to the free gingival mar-gins will be less than 6 mm. It also may be indicated when-ever it is desirable to keep the free gingival margins of the www.konkur.in 53 Chapter 5 Major and Minor Connectors remaining anterior teeth exposed and depth of the floor of the mouth is inadequate to place a lingual bar.
Contraindications for Use: Remaining natural anterior teeth severely tilted toward the lingual.
Characteristics and Location: The sublingual bar is essentially the same half-pear shape as a lingual bar, except that the bulkiest portion is located to the lingual and the tapered portion is toward the labial. The superior border of the bar should be at least 3 mm from the free gingival mar-gin of the teeth. The inferior border is located at the height of the alveolar lingual sulcus when the patient’s tongue is slightly elevated. This necessitates a functional impression of the lingual vestibule to accurately register the height of the vestibule.
Blockout and Relief of Master Cast: (1) All tissue under-cuts parallel to path of placement. (2) An additional thick-ness of 32-gauge sheet wax when the lingual surface of the alveolar ridge is undercut or parallel to the path of place-ment. (3) One thickness of baseplate wax over basal seat areas (to elevate minor connectors for attaching acrylic-resin denture bases).
Waxing Specifications: (1) Six-gauge, half-pear–shaped wax form reinforced by 22- to 24-gauge sheet wax or simi-lar plastic pattern adapted to design width. (2) Long bar bulkier than short bar; however, cross-sectional shape unchanged.
Finishing Lines: Butt-type joint(s) with minor connector(s) for retention of denture base(s).
Mandibular Lingual Bar with Continuous Bar (Cingulum Bar) Indications for Use: (1) When a linguoplate is otherwise indicated but the axial alignment of anterior teeth is such that excessive blockout of interproximal undercuts would be required. (2) When wide diastemata exist between man-dibular anterior teeth and a linguoplate would objectionably display metal in a frontal view.
Characteristics and Location: (1) Conventionally shaped and located same as lingual bar major connec-tor component when possible. (2) Thin, narrow (3 mm) metal strap located on cingula of anterior teeth, scalloped to follow interproximal embrasures with inferior and superior borders tapered to tooth surfaces. (3) Originates bilaterally from incisal, lingual, or occlusal rests of adja-cent principal abutments.
Blockout and Relief of Master Cast: (1) Lingual surfaces of alveolar ridge and basal seat areas same as for lingual bar. (2) No relief for continuous bar except blockout of interprox-imal spaces parallel to path of placement.
Waxing Specifications: (1) Lingual bar major connector component waxed and shaped same as lingual bar. (2) Con-tinuous bar pattern formed by adapting two strips (3 mm wide) of 28-gauge sheet wax, one at a time, over the cingula and into interproximal embrasures.
Finishing Lines: Butt-type joint(s) with minor connector(s) for retention of denture base(s).
Mandibular Continuous Bar (Cingulum Bar) Indications for Use: When a lingual plate or sublingual bar is otherwise indicated but the axial alignment of the anterior teeth is such that excessive blockout of interproximal under-cuts would be required.
Contraindications for Use: (1) Anterior teeth severely tilted to the lingual. (2) When wide diastemata that exist between the mandibular anterior teeth and the cingulum bar would objectionably display metal in a frontal view.
Characteristics and Location: (1) Thin, narrow (3 mm) metal strap located on cingula of anterior teeth, scalloped to follow interproximal embrasures with inferior and superior borders tapered to tooth surfaces. (2) Originates bilaterally from incisal, lingual, or occlusal rests of adjacent principal abutments.
Blockout and Relief of Master Cast: No relief for cingu-lum bar except blockout of interproximal spaces parallel to the path of placement.
Waxing Specifications: Cingulum bar pattern formed by adapting two strips (3 mm wide) of 28-gauge sheet wax, one at a time, over the cingula and into interproximal embra-sures.
Finishing Lines: Butt-type joint(s) with minor connector(s) for retention of denture base(s).
Mandibular Labial Bar Indications for Use: (1) When lingual inclinations of remain-ing mandibular premolar and incisor teeth cannot be cor-rected, preventing placement of a conventional lingual bar connector. (2) When severe lingual tori cannot be removed and prevent the use of a lingual bar or lingual plate major connector. (3) When severe and abrupt lingual tissue under-cuts make it impractical to use a lingual bar or a lingual plate major connector.
Characteristics and Location: (1) Half-pear shaped with bulkiest portion inferiorly located on the labial and buccal aspects of the mandible. (2) Superior border tapered to soft tissue. (3) Superior border located at least 4 mm inferior to labial and buccal gingival margins and farther if possible. (4) Inferior border located in the labial-buccal vestibule at the juncture of attached (immobile) and unattached (mobile) mucosae.
Blockout and Relief of Master Cast: (1) All tissue undercuts parallel to path of placement, plus an additional thickness of 32-gauge sheet wax when the labial surface is undercut or parallel to the path of placement. (2) No relief necessary when the labial surface of the alveolar ridge slopes inferiorly to the labial or buccal. (3) Basal seat areas same as for lingual bar major connector.
Waxing Specifications: (1) Six-gauge, half-pear–shaped wax form reinforced with 22- to 24-gauge sheet wax or similar plastic pattern. (2) Long bar necessitates more bulk than short bar; however, cross-sectional shape unchanged. (3) Minor connectors joined with occlusal or other superior www.konkur.in 54 Part I General Concepts/Treatment Planning components by a labial or buccal approach. (4) Minor con-nectors for base attachment joined by a labial or buccal approach.
Finishing Lines: Butt-type joint(s) with minor connector(s) for retention of denture base(s).
Single Palatal Strap–Type Major Connector Indications for Use: Bilateral edentulous spaces of short span in a tooth-supported restoration.
Characteristics and Location: (1) Anatomic replica form. (2) Anterior border follows the valleys between rugae as nearly as possible at right angles to median suture line. (3) Posterior border at right angle to median suture line. (4) Strap should be 8 mm wide or approximately as wide as the com-bined width of a maxillary premolar and first molar. (5) Con-fined within an area bounded by the four principal rests.
Blockout and Relief of Master Cast: (1) Usually none required except slight relief of elevated medial palatal raphe or any exostosis crossed by the connector. (2) One thickness of baseplate wax over basal seat areas (to elevate minor con-nectors for attaching acrylic-resin denture bases).
Beading: See Figures 5-38 to 5-41.
Waxing Specifications: Anatomic replica pattern equiva-lent to 22- to 24-gauge wax, depending on arch width.
Finishing Lines: (1) Undercut and slightly elevated. (2) No farther than 2 mm medial from an imaginary line con-tacting lingual surfaces of principal abutments and teeth to be replaced. (3) Follow curvature of arch.
Single Broad Palatal Major Connector Indications for Use: (1) Class I partially edentulous arches with residual ridges that have undergone little vertical resorp-tion and will lend excellent support. (2) V- or U-shaped pal-ates. (3) Strong abutments (single or made so by splinting). (4) More teeth in arch than six remaining anterior teeth.
(5) Direct retention not a problem. (6) No interfering tori.
Characteristics and Location: (1) Anatomic replica form. (2) Anterior border following valleys of rugae as near right angle to median suture line as possible and not extending anterior to occlusal rests or indirect retainers. (3) Posterior border located at junction of hard and soft palate but not extended onto soft palate; at right angle to the median suture line; extended to pterygomaxillary notches.
Blockout and Relief of Master Cast: (1) Usually none required except relief of elevated median palatal raphe or any small exostoses covered by the connector. (2) One thickness of baseplate wax over basal seat areas (to elevate minor con-nectors for attaching acrylic-resin denture bases).
Beading: See Figures 5-28 to 5-30.
Waxing Specifications: Anatomic replica pattern equiva-lent to 24-gauge sheet wax thickness.
Finishing Lines: (1) Provision for butt-type joint at ptery-gomaxillary notches. (2) Undercut and slightly elevated. (3) No farther than 2 mm medial from an imaginary line contacting the lingual surfaces of the missing natural teeth. (4) Following curvature of arch.
Anterior-Posterior Strap–Type Major Connector Indications for Use: (1) Class I and II arches in which excellent abutment and residual ridge support exists, and direct retention can be made adequate without the need for indirect retention. (2) Long edentulous spans in Class II, modification 1 arches. (3) Class IV arches in which anterior teeth must be replaced with a removable partial denture. (4) Inoperable palatal tori that do not extend posteriorly to the junction of the hard and soft palates.
Characteristics and Location: (1) Parallelogram shaped and open in center portion. (2) Relatively broad (8 to 10 mm) anterior and posterior palatal straps. (3) Lateral palatal straps (7 to 9 mm) narrow and parallel to curve of arch; minimum of 6 mm from gingival crevices of remaining teeth. (4) Ante-rior palatal strap: Anterior border not placed farther ante-riorly than anterior rests and never closer than 6 mm to lingual gingival crevices; follows the valleys of the rugae at right angles to the median palatal suture. Posterior border, if in rugae area, follows valleys of rugae at right angles to the median palatal suture. (5) Posterior palatal connector: Poste-rior border located at junction of hard and soft palates and at right angles to median palatal suture and extended to hamu-lar notch area(s) on distal extension side(s). (6) Anatomic replica or matte surface.
Blockout and Relief of Master Cast: (1) Usually none required except slight relief of elevated median palatal raphe where anterior or posterior straps cross the palate. (2) One thickness of baseplate wax over basal seat areas (to elevate minor connectors for attaching acrylic-resin denture bases).
Beading: See Figures 5-28 and 5-29.
Waxing Specifications: (1) Anatomic replica patterns or matte surface forms of 22-gauge thickness. (2) Posterior pal-atal component: A strap of 22-gauge thickness, 8 to 10 mm wide (a half-oval form of approximately 6-gauge thickness and width) may also be used.
Finishing Lines: Same as for single broad palatal major connector.
Complete Palatal Coverage Major Connector Indications for Use: (1) In most situations in which only some or all anterior teeth remain. (2) Class II arch with a large posterior modification space and some missing ante-rior teeth. (3) Class I arch with one to four premolars and some or all anterior teeth remaining, when abutment sup-port is poor and cannot otherwise be enhanced; residual ridges have undergone extreme vertical resorption; direct retention is difficult to obtain. (4) In the absence of a pedun-culated torus.
Characteristics and Location: (1) Anatomic replica form for full palatal metal casting supported anteriorly by positive www.konkur.in 55 Chapter 5 Major and Minor Connectors rest seats. (2) Palatal linguoplate supported anteriorly and designed for attachment of acrylic-resin extension posteri-orly. (3) Contacts all or almost all of the teeth remaining in the arch. (4) Posterior border: Terminates at the junction of the hard and soft palates; extended to hamular notch area(s) on distal extension side(s); at a right angle to median suture line.
Blockout and Relief of Master Cast: (1) Usually none required except relief of elevated median palatal raphe or any small palatal exostosis. (2) One thickness of baseplate wax over basal seat areas (to elevate minor connectors for attach-ing acrylic-resin denture bases).
Beading: See Figures 5-28 to 5-30.
Waxing Specifications: (1) Anatomic replica pattern equivalent to 22- to 24-gauge sheet wax thickness. (2) Acrylic-resin extension from linguoplate the same as for a complete denture.
Finishing Lines: As illustrated here and previously dis-cussed.
U-Shaped Palatal Major Connector This connector should be used only in those situations in which inoperable tori extend to the posterior limit of the hard palate.
The U-shaped palatal major connector is the least favor-able design of all palatal major connectors because it lacks the rigidity of other types of connectors. When it must be used, indirect retainers must support any portion of the connector that extends anteriorly from the principal occlusal rests. Ante-rior border areas of this type of connector must be kept at least 6 mm away from adjacent teeth. If for any reason the anterior border must contact the remaining teeth, the connector must again be supported by rests placed in properly prepared rest seats. It should never be supported even temporarily by inclined lingual surfaces of anterior teeth.
Waxing specifications, finishing lines, and so forth, are the same as for full palatal castings or other previously dis-cussed similar major connectors.
www.konkur.in CHAPTER 6 Rests and Rest Seats CHAPTER OUTLINE Role of Rests in Control of Prosthesis Movement Form of the Occlusal Rest and Rest Seat Extended Occlusal Rest Interproximal Occlusal Rest Seats Internal Occlusal Rests Implants as a Rest Support for Rests Lingual Rests on Canines and Incisor Teeth Incisal Rests and Rest Seats Although the major connector joins all removable partial denture component parts, engaging the remaining teeth through appropriate use of rest seats and rests affords effi-cient resistance to functional chewing forces. The capacity for teeth to resist functional forces and remain stable over time is provided through their sophisticated support mecha-nisms. Studies have shown that displacement and recovery following loading are far better for natural teeth than for oral mucosa. Similarly the capacity of implants to resist the func-tional forces of chewing may exceed that of teeth.
Consequently, appropriate use of the teeth to help resist functional forces in removable prostheses is a critical strat-egy to control prosthesis movement and achieve functional stability. When situations of multiple missing teeth are seen in conjunction with less than ideal soft tissue support and opposing natural dentition, implant use should be strongly considered.
ROLE OF RESTS IN CONTROL OF PROSTHESIS MOVEMENT Appropriate use of teeth requires consideration as to how best to engage teeth for the supportive qualities they provide. Because the most effective resistance can be provided if the tooth is stressed along its long axis, the prosthesis frame-work should engage the tooth in a manner that encourages axial loading. The various forms of rests have as a main goal a form that allows for axial loading. It is important to realize that this goal can be achieved only through some form of tooth modification.
Vertical support must be provided for a removable par-tial denture. Any component of a partial denture on a tooth surface that provides vertical support is called a rest (Figure 6-1). Rests should always be located on properly prepared tooth surfaces. The prepared surface of an abutment to receive the rest is called the rest seat. Rests are designated by the surface of the tooth prepared to receive them (occlu-sal rest, lingual rest, and incisal rest). The topography of any www.konkur.in 57 Chapter 6 Rests and Rest Seats rest should generally restore the topography of the tooth that existed before the rest seat was prepared.
The primary purpose of the rest is to provide vertical support for the partial denture. In doing so, it also does the following: 1. Maintains components in their planned positions 2. Maintains established occlusal relationships by prevent-ing settling of the denture 3. Prevents impingement of soft tissue 4. Directs and distributes occlusal loads to abutment teeth Thus rests serve to support the position of a partial den-ture and to resist movement toward the tissue. They serve to transmit vertical forces to the abutment teeth and to direct those forces along the long axes of the teeth. In this respect, tooth-supported removable partial denture rests function in a manner similar to fixed abutment retainers. It is obvious that for this degree of stability to exist, the rests must be rigid and must receive positive support from the abutment teeth, which means that under occlusal loading, the rest and the tooth remain in stable contact and no independent move-ment or slippage occurs.
In a removable partial denture that has one or more dis-tal extension bases, the denture becomes increasingly tis-sue supported as the distance from the abutment increases. Closer to the abutment, however, more of the occlusal load is transmitted to the abutment tooth by means of the rest. The load is thereby distributed between the abutment and the supporting residual ridge tissue.
When rests prevent movement of the denture in an api-cal direction, the position of the retentive portion of the clasp arms can be maintained in designated relation to the tooth undercuts. Although passive when it is in its termi-nal position, the retentive portion of the clasp arm should remain in contact with the tooth, ready to resist a vertical dislodging force. Then, when a dislodging force is applied, the clasp arm should immediately become actively engaged to resist vertical displacement. If settling of the denture results in clasp arms that stand away from the tooth, some vertical displacement is possible before the retainer can become functional. The rest serves to prevent such settling and thereby helps to maintain the vertical stability of the partial denture.
A B C D Figure 6-1 A, Occlusal rest seats have been prepared on molar and premolar teeth to provide vertical support for the removable partial denture. B, Framework for tooth-supported removable partial denture. Rests on patient’s right provide vertical support to the re-placement dentition; rests on patient’s left provide cross-arch support and stabilization. C, Tooth support for this prosthesis is provided by rests that occupy definite, prepared, rest seats on the canine and occlusal surfaces of selected posterior teeth. D, Kennedy Class III, modification 1, maxillary arch with rest seats prepared on the lingual surfaces of the canine and lateral incisor and on the occlusal sur-faces of the premolar and molar.
www.konkur.in 58 Part I General Concepts/Treatment Planning The use of an implant as a rest can also be considered. In this application, the implant eliminates compression of sup-porting soft tissues, controls vertical movement of the den-ture base, eliminates or alters fulcrum lines, and serves to increase support and stability of the prosthesis.
FORM OF THE OCCLUSAL REST AND REST SEAT The form of the occlusal rest seat should be designed and located with the following guidelines in mind: 1. The outline form of an occlusal rest seat should be a rounded triangular shape with the apex toward the center of the occlusal surface (Figure 6-2). 2. It should be as long as it is wide, and the base of the tri-angular shape (at the marginal ridge) should be at least 2.5 mm for both molars and premolars. Rest seats of smaller dimensions do not provide an adequate bulk of metal for rests, especially if the rest is contoured to restore the occlusal morphology of the abutment tooth. 3. The marginal ridge of the abutment tooth at the site of the rest seat must be lowered to permit a sufficient bulk of metal for strength and rigidity of the rest and the minor connector. This means that a reduction of the marginal ridge of approximately 1.5 mm is usually necessary. 4. The floor of the occlusal rest seat should be apical to the marginal ridge and the occlusal surface and should be concave, or spoon shaped (Figure 6-3). Caution should be exercised in preparing a rest seat to avoid creating sharp edges or line angles in the preparation. 5. The angle formed by the occlusal rest and the vertical minor connector from which it originates should be less than 90 degrees (Figures 6-4 and 6-5). Only in this way can the occlusal forces be directed along the long axis of the abutment tooth. An angle greater than 90 degrees fails to transmit occlusal forces along the supporting vertical axis of the abutment tooth. This also permits slippage of the prosthesis away from the abutment, which can result in orthodontic-like forces being applied to an inclined plane on the abutment, with possible tooth movement (Figure 6-6).
Deepest part of rest seat Figure 6-2 The deepest part of an occlusal rest preparation should be inside the lowered marginal ridge at X. The marginal ridge is lowered to provide bulk and to accommodate the origin of the occlusal rest with the least occlusal interference.
Figure 6-3 Occlusal rest seat preparation on molar. The preparation is rounded, and the triangular concavity has smooth margins on the occlusal surface and a lowered, rounded mar-ginal ridge.
Less than 90 Figure 6-4 The occlusal rest should be spoon shaped and slightly inclined apically from the marginal ridge. The rest should restore the occlusal morphology of the tooth that existed before preparation of the rest seat.
Less than 90 Figure 6-5 The floor of the occlusal rest seat should be in-clined apically from the lowered marginal ridge. Any angle less than 90 degrees is acceptable as long as the preparation of the proximal surface and lowering and rounding of the marginal ridge precede completion of the rest seat itself.
www.konkur.in 59 Chapter 6 Rests and Rest Seats When an existing occlusal rest preparation is inclined apically toward the reduced marginal ridge and cannot be modified or deepened because of fear of perforation of the enamel or restoration, then a secondary occlusal rest must be employed to prevent slippage of the primary rest and orthodontic movement of the abutment tooth (Figure 6-7). Such a rest should pass over the lowered mar-ginal ridge on the side of the tooth opposite the primary rest and should, if possible, be inclined slightly apically from the marginal ridge. However, two opposing occlusal rests on diverging tooth inclines will function to prevent unfavorable forces if all related connectors are sufficiently rigid. In any tooth-tissue–supported partial denture, the relation of the occlusal rest to the abutment should be that of a shallow ball-and-socket joint, to prevent a possible transfer of horizontal stresses to the abutment tooth. The occlusal rest should provide only occlusal support. Sta-bilization against horizontal movement of the prosthesis must be provided by other components of the partial den-ture rather than by any locking effect of the occlusal rest, which will cause the application of leverages to the abut-ment tooth.
EXTENDED OCCLUSAL REST In Kennedy Class II, modification 1, and Kennedy Class III situations in which the most posterior abutment is a mesially tipped molar, an extended occlusal rest should be designed and prepared to minimize further tipping of the abutment and to ensure that the forces are directed down the long axis of the abutment. This rest should extend more than one-half the mesiodistal width of the tooth, should be approximately one-third the buccolingual width of the tooth, and should allow for a minimum of 1-mm thickness of the metal; the preparation should be rounded with no undercuts or sharp angles (Figure 6-8).
In situations in which the abutment is severely tilted, the extended occlusal rest may take the form of an onlay to restore the occlusal plane (Figure 6-9). The tooth prepa-ration for this type of extended rest must include remov-ing or restoring pits, fissures, and grooves; placing a 1- to 2-mm bevel on the buccal and lingual occlusal surfaces to allow the extended rest (onlay) to provide stabilization; allowing the rest to restore the contour and occlusion of the natural tooth; and ensuring that the rest directs the forces down the long axis of the tooth. Tooth preparation must also include a 1- to 2-mm guiding plane on the mesial sur-face of the abutment.
A D C F B Figure 6-6 The result of force applied to an inclined plane when the floor of the occlusal rest preparation inclines apically toward the marginal ridge of the abutment tooth. F, Occlusal force applied to the abutment tooth. AB, Relationship of the oc-clusal rest to the abutment tooth when the angle is greater than 90 degrees. ABC, Removable partial denture framework. ABD, Abutment tooth.
Figure 6-7 Diagnostic cast evaluation of mesially tipped molar abutment. Anterior tilt of molar precludes preparation of acceptable rest seat on the mesio-occlusal surface. The patient could not afford a crown to improve axial alignment or orthodon-tic treatment to upright the molar. Occlusal rests will be used on mesio-occlusal and disto-occlusal surfaces to support resto-ration and direct forces over the greatest root mass of the abut-ment. The proposed ring clasp design is outlined.
Figure 6-8 Cast shows extended occlusal rest on the mandi bular first molar, designed to ensure maximum bracing from the tooth. If placed on a mesially inclined molar next to a modifica-tion space (as in Figure 6-7), the extended rest would ensure that the forces are directed down the long axis of the abutment, and therefore the disto-occlusal rest would not have been needed.
www.konkur.in 60 Part I General Concepts/Treatment Planning INTERPROXIMAL OCCLUSAL REST SEATS The design of a direct retainer assembly may require the use of interproximal occlusal rests (Figure 6-10). These rest seats are prepared as individual occlusal rest seats, with the excep-tion that the preparations must be extended farther lingually than is ordinarily accomplished (Figure 6-11). Adjacent rests, rather than a single rest, are used to prevent interproxi-mal wedging by the framework. In addition, the joined rests are designed to shunt food away from contact points.
When such rest seats are prepared, care must be exercised to avoid reducing or eliminating contact points of abutment teeth. However, sufficient tooth structure must be removed to allow adequate bulk of the component for strength and to permit the component to be so shaped that occlusion will not be altered. Therefore analysis of mounted diagnostic casts is mandatory for assessment of interocclusal contact areas in which rests are to be placed. Sufficient space must be present or created to avoid interference with the placement of rests (Figure 6-12).
Figure 6-9 The intaglio surface of an onlay occlusal rest re-storing contour and occlusion for this maxillary molar.
Figure 6-10 Drawing on cast shows the desired design of a direct retainer assembly on mandibular premolar and molar abutments that incorporates interproximal occlusal rests. The direct retainers on the distobuccal undercut of the molar and on the mesiobuccal undercut of the premolar are extended from the joined occlusal rests, which occupy specifically prepared adjoin-ing rest seats.
Buccal Lingual Figure 6-11 Rest seat preparations on the premolar and mo-lar fulfill requirements of properly prepared rest seats. Prepara-tions are extended lingually to provide strength (through bulk) without overly filling the interproximal space with a minor con-nector. This type of preparation is challenging for natural tooth modification, and care must be exercised to avoid violation of contact points—yet the marginal ridge of each abutment should be sufficiently lowered (1.5 mm).
Figure 6-12 View of mounted casts with the framework fully seated illustrates that interocclusal space was made available by a properly prepared rest seat.
www.konkur.in 61 Chapter 6 Rests and Rest Seats The lingual interproximal area requires only minor prepa-ration. Creation of a vertical groove must be avoided to prevent a torquing effect on the abutments by the minor connector.
INTERNAL OCCLUSAL RESTS A partial denture that is totally tooth supported by means of cast retainers on all abutment teeth may use intracoronal rests for both occlusal support and horizontal stabilization (Figure 6-13).
An intracoronal rest is not a retainer and should not be confused with an attachment. Occlusal support is derived from the floor of the rest seat. Horizontal stabilization is derived from the near-vertical walls of this type of rest seat. The form of the rest should be parallel to the path of place-ment, slightly tapered occlusally, and slightly dovetailed to prevent dislodgment proximally.
The main advantages of the internal rest are that it facili-tates the elimination of a visible clasp arm buccally and per-mits the location of the rest seat in a more favorable position in relation to the tipping axis (horizontal) of the abutment. Retention is provided by a lingual clasp arm, either cast or of wrought wire, lying in a natural or prepared infrabulge area on the abutment tooth.
Internal rests are carved in wax or spark eroded in abut-ment castings. Ready-made plastic rest patterns are readily available and can be waxed into crown or partial-veneer pat-terns, invested, and cast after having been positioned parallel to the path of placement with a dental cast surveyor. Future developments and techniques promise more widespread use of the internal occlusal rest but only for tooth-supported par-tial dentures.
IMPLANTS AS A REST Implants can also be considered to serve as a rest when one takes advantage of the vertical stiffness characteris-tic they possess. In this application, they can serve to resist tissue-ward movement alone but may be considered useful for retentive needs as well. When a dual role is considered, selection of the retentive element requires consideration of how much stress from vertical movement resistance (the rest function) will affect the retentive function of the attachment mechanism.
When used as a rest, implants can serve to efficiently resist vertical movement and provide positive support. This use allows a low profile connection (i.e., close to the ridge), imparting less torque to the implant. Location can be pre-scribed for maximum advantage, and the implants can effec-tively alter rotation around a fulcrum line—eliminating it if applied to an unsupported end of a denture base, or reduc-ing the effects by decreasing the effective lever arm. Because its position can be selected, it is important to consider (1) future use as a support to an implant fixed prosthesis, and (2) current placement at the distal-most tooth position to most effectively resist functional forces of occlusion.
SUPPORT FOR RESTS Rests may be placed on sound enamel or on any restoration material that has been proven scientifically to resist fracture and distortion when subjected to applied forces.
Rests placed on sound enamel are not conducive to car-ies in a mouth with a low-caries index, provided that good oral hygiene is maintained. Proximal tooth surfaces are much more vulnerable to caries attack than are the occlusal surfaces supporting an occlusal rest. The decision to use abutment coverage is usually based on needed mouth preparation, determined from the survey of diagnostic casts, to accommo-date modifications of abutment teeth necessary to fabricate a removable partial denture (see Chapter 11). When precarious fissures are found in occlusal rest areas in teeth that are other-wise sound, they may be removed and appropriately restored without resorting to more extensive abutment protection. Although it cannot be denied that the best protection from caries for an abutment tooth is full coverage, it is imperative A B Figure 6-13 Maxillary tooth-supported removable partial denture utilizing internal occlusal rests. A, Wax pattern developed utilizing internal rests on canine, premolar, and molars. B, Maxillary framework on cast with the internal rests fitted within surveyed crowns.
www.konkur.in 62 Part I General Concepts/Treatment Planning that such crowns be contoured properly to provide support, stabilization, and retention for the partial denture.
When the decision of whether to use unprotected enamel surfaces for rests is made, the future vulnerability of each tooth must be considered because it is not easy to fabricate full crowns to accommodate rests and clasp arms after the partial denture has been made. In many instances, sound enamel may be used safely for the support of occlusal rests. In such situa-tions, the patient should be advised that future susceptibility to caries is not predictable and that much depends on oral hygiene and possible future changes in caries susceptibility. Although the decision to use unprotected abutments logically should be left up to the dentist, economic factors may influence the final decision. The patient should be informed of the risks involved and of his or her responsibility for maintaining good oral hygiene and for returning periodically for observation.
Rest seat preparations should be made in sound enamel. In most instances, preparation of proximal tooth surfaces is necessary to provide proximal guiding planes and to elimi-nate undesirable undercuts that rigid parts of the framework must pass over during placement and removal. The prepara-tion of occlusal rest seats always must follow proximal prepa-ration, never precede it. Only after the alteration of proximal tooth surfaces is completed may the location of the occlu-sal rest seat in relation to the marginal ridge be determined. When proximal preparation follows occlusal rest seat prepa-ration, the inevitable consequence is that the marginal ridge is too low and too sharp, with the center of the floor of the rest seat too close to the marginal ridge. Therefore it is often impossible to correct the rest preparation without making it too deep, which causes irreparable damage to the tooth.
Occlusal rest seats in sound enamel may be prepared with burs and polishing points that leave the enamel surface as smooth as the original enamel (Figure 6-14). The larger round bur is used first to lower the marginal ridge and to establish the outline form of the rest seat. The resulting occlusal rest seat is then complete except that the floor is not sufficiently concave. A slightly smaller round bur then is used to deepen the floor of the occlusal rest seat. At the same time, it forms the desired spoon shape inside the lowered marginal ridge. The prepara-tion is smoothed by a polishing point of suitable size and shape.
When a small enamel defect is encountered in the prepara-tion of an occlusal rest seat, it is usually best to ignore it until the rest preparation has been completed. Then, with small burs, the remaining defect should be prepared to receive a small restoration. This may be finished flush with the floor of the rest preparation that was previously established.
A fluoride gel should be applied to abutment teeth follow-ing enamel recontouring. If the master cast will be fabricated from an irreversible hydrocolloid impression, application of the gel should be delayed until after impressions are made because some fluoride gels and irreversible hydrocolloids may be incompatible.
Occlusal rest seat preparations in existing restorations are treated the same way as those in sound enamel. Any proxi-mal preparations must be done first, for if the occlusal rest seat is placed first and then the proximal surface is prepared, the outline form of the occlusal rest seat sometimes is irrepa-rably altered.
The possibility that an existing restoration may be per-forated in the process of preparation of an ideal occlusal rest seat is always present. Although some compromise is permissible, the effectiveness of the occlusal rest seat should not be jeopardized for fear of perforating an exist-ing restoration. The rest seat may be widened to compen-sate for shallowness, but the floor of the rest seat should still be slightly inclined apically from the marginal ridge. When this is not possible, a secondary occlusal rest should be used on the opposite side of the tooth to prevent slip-ping of the primary rest.
When perforation does occur, it may be repaired, but occasionally the making of a new restoration is unavoidable. In such a situation, the original preparation should be modi-fied to accommodate the occlusal rest, thereby avoiding the A B Figure 6-14 Recontouring of axial surfaces and rest seat preparations in enamel may be readily accomplished with selected use of accessories. A, The two pear-shaped, multifluted burs (two burs on the left) can be used for cingulum rests and rounding marginal ridges; the longer straight, multifluted enamelplasty bur (middle bur) is ideal for height of contour adjustments and guide plane preparation; round multifluted or carbide burs (three burs at right of middle) are used for occlusal rest preparation; and the inverted cone (far right bur) can be used for cingulum rests as well. B, Various abrasive rubber polishing points are necessary to ensure a smooth surface finish following any enamelplasty procedure. Following the manufacturer’s recommended sequence of abrasives should return the surface to smoothness comparable with the original condition. Diamonds are not recommended for this type of tooth reduction.
www.konkur.in 63 Chapter 6 Rests and Rest Seats risk of perforating the completed restoration or fabricating a restoration with an inadequate rest seat.
Occlusal rest seat location in new restorations should be known when the tooth is prepared so that sufficient clearance may be provided for the rest seat within the preparation. The final step in the preparation of the tooth should be to make sure such clearance exists and, if not, to make a depression to accommodate the depth of the rest (Figure 6-15).
Occlusal rest seats in crowns and inlays generally are made somewhat larger and deeper than those in enamel. Those made in abutment crowns for tooth-supported den-tures may be made slightly deeper than those in abutments that support a distal extension base; thus they approach the effectiveness of boxlike internal rests.
Internal rest seats also should be created first in wax, either with suitable burs in a handpiece holder or by waxing around a lubricated mandrel, which is held in the surveyor. In either situ-ation, the rest preparation must be finished on the casting with burs in a handpiece holder or with a precision drill press. Plastic and metal shoes that fit over a mandrel are also available for this purpose. Thus a smooth casting is ensured, and the need for finishing the inside of the internal rest with burs is eliminated. Sufficient clearance must be provided in the preparation of the abutment to accommodate the depth of the internal rest.
LINGUAL RESTS ON CANINES AND INCISOR TEETH Analysis of mounted diagnostic casts is mandatory in assess-ment of incisal and lingual contact areas where rests are to be placed. Sufficient space must be present or created to avoid interference with placement of rests.
Although the preferred site for an external rest is the occlusal surface of a molar or a premolar, an anterior tooth may be the only abutment available for occlusal support of the denture. Also an anterior tooth occasionally must be used to support an indirect retainer or an auxiliary rest. A canine is much preferred over an incisor for this purpose. When a canine is not present, multiple rests that are spread over several incisor teeth are preferable to the use of a sin-gle incisor. Root form, root length, inclination of the tooth, and ratio of the length of the clinical crown to the alveolar support must be considered when the site and form of rests placed on incisors are determined.
A lingual rest is preferable to an incisal rest because it is placed closer to the horizontal axis of rotation (tipping axis) of the abutment and therefore will have less of a tendency to tip the tooth. In addition, lingual rests are more esthetically acceptable than are incisal rests.
If an anterior tooth is sound and the lingual slope is grad-ual rather than perpendicular, a lingual rest sometimes may be placed in an enamel seat at the cingulum or just incis-ally to the cingulum (Figure 6-16). This type of lingual rest Figure 6-15 Preparation on mandibular premolar for sur-veyed crown incorporates space for mesio-occlusal rest seat. Adequate occlusal reduction was accomplished to accommodate the depth of the rest seat in the abutment crown. The modifica-tion for the rest seat is performed following the standard crown preparation.
A B C Figure 6-16 Three views of lingual rest seat prepared in enamel of the maxillary canine. A, The rest seat, from the lingual aspect, assumes the form of a broad inverted V, maintaining the natural contour sometimes seen in a maxillary canine cingulum. An inverted V notch form is self-centering for the rest and at the same time directs forces rather favorably in an apical direction. B, From the incisal view, it will be noted that the rest seat preparation is broadest at the most lingual aspect of the canine. As the preparation approaches the proximal surface of the tooth, it is less broad than at any other area. C, The proximal view demonstrates correct taper of the floor of the rest seat. It also should be noted that the borders of the rest seat are slightly rounded to avoid line angles in its preparation. The mesiodistal length of the preparation should be a minimum of 2.5 to 3 mm, labiolingual width about 2 mm, and incisal-apical depth a minimum of 1.5 mm. It is a risky preparation and should not be attempted on lower anterior teeth.
www.konkur.in 64 Part I General Concepts/Treatment Planning is usually confined to maxillary canines that have a gradual lingual incline and a prominent cingulum. In a few instances, such a rest also may be placed on maxillary central incisors. The lingual slope of the mandibular canine is usually too steep for an adequate lingual rest seat to be placed in the enamel, and some other provision for rest support must be made. Lingual rest seat preparations in enamel are rarely satisfactory on mandibular anterior teeth because of lack of thickness of the enamel in which a seat of adequate form to be truly supportive is prepared.
The preparation of an anterior tooth to receive a lingual rest may be accomplished in one of two ways: 1. A slightly rounded V is prepared on the lingual surface at the junction of the gingival and the middle one third of the tooth. The apex of the V is directed incisally. This preparation may be started by using an inverted, cone-shaped diamond stone and progressing to smaller, ta-pered stones with round ends to complete the prepara-tion. All line angles must be eliminated, and the rest seat must be prepared within the enamel and must be highly polished. Shaped, abrasive rubber polishing points, fol-lowed by flour of pumice, produce an adequately smooth and polished rest seat. A predetermined path of place-ment for the denture must be kept in mind when the rest seat is prepared. The lingual rest seat must not be pre-pared as though it was going to be approached from a direction perpendicular to the lingual slope. The floor of the rest seat should be toward the cingulum rather than the axial wall. Care must be taken not to create an enamel undercut, which interferes with placement of the denture (see Figure 6-16). 2. The most satisfactory lingual rest from the standpoint of support is one that is placed on a prepared rest seat in a cast restoration (Figure 6-17). This is done most effective-ly by planning and executing a rest seat in the wax pattern rather than by attempting to cut a rest in a cast restoration in the mouth. The contour of the framework may then restore the lingual form of the tooth.
When the cingulum in the wax pattern is accentuated, the floor of the rest seat is readily carved to be the most apical portion of the preparation. A saddlelike shape, which pro-vides a positive rest seat located favorably in relation to the long axis of the tooth, is formed. The framework of the den-ture is made to fill out the continuity of the lingual surface so that the tongue contacts a smooth surface without the patient being conscious of bulk or irregularities.
The lingual rest may be placed on the lingual surface of a cast veneer crown (Figure 6-18), a three-quarter crown, an inlay, a laminate veneer, a composite restoration, or an etched metal restoration. The latter displays less metal than the three-quarter crown, especially on the mandibu-lar canine, where the lingual rest that was placed on a cast restoration is frequently used, and it is a more conserva-tive restoration. The three-quarter crown may be used if the labial surface of the tooth is sound and if the reten-tive contours are satisfactory. However, if the labial surface presents inadequate or excessive contours for placement of a retentive clasp arm or if gingival decalcification or caries is present, a veneered complete coverage restoration should be used.
In some instances, ball types of rests may be used in pre-pared seats. Such rest seats may be cautiously prepared in tooth surfaces with overly sufficient enamel thickness or may be prepared in restorations placed in teeth in which enamel thickness is inadequate. Conservative restorations (e.g., sil-ver amalgam, pin inlays, or composite resin) in anterior teeth may be better suited for ball types of rest seats than are the less conservative inverted V types of rest seats.
Some scientific evidence demonstrates that individu-ally cast chromium-cobalt alloy rest seat forms (attached to Figure 6-17 Rest seat preparation can be exaggerated for bet-ter support when it is prepared in cast restoration.
Figure 6-18 Positive vertical support for a prosthesis is fur-nished by rest seats prepared in splinted metal ceramic crowns. The cingulum rest seats are optimally placed as near the horizon-tal axis of rotation as possible to minimize tipping forces.
www.konkur.in 65 Chapter 6 Rests and Rest Seats lingual surfaces of anterior teeth by the use of composite resin cements with acid-etched tooth preparation), laminates, and composite resins have been successfully used as conservative approaches to forming rest seats on teeth with unacceptable lingual contours (Figure 6-19). Sapphire ceramic orthodon-tic brackets have also been bonded to the lingual surfaces of mandibular canines and shaped as rest seats. These have advantages over the metal acid-etched retained rests in that a laboratory step is avoided and increased bond strengths are achieved. The major disadvantage of using orthodontic brackets is that removal of the rest seat would necessitate that they be ground off with the potential of heat generation and possible pulpal damage.
INCISAL RESTS AND REST SEATS Incisal rests are placed at the incisal angles of anterior teeth and on prepared rest seats. Although this is the least desirable placement of a rest seat for reasons previously mentioned, it may be used successfully for selected patients when the abut-ment is sound and when a cast restoration is not otherwise indicated. Therefore incisal rests generally are placed on enamel (Figure 6-20). Incisal rests are used predominantly as auxiliary rests or as indirect retainers.
Although the incisal rest may be used on a canine abut-ment in either arch, it is more applicable to the mandibular canine. This type of rest provides definite support with rela-tively little loss of tooth structure and little display of metal. Esthetically it is preferable to the three-quarter crown. The same criteria may be applied when one is deciding whether to use unprotected enamel for an occlusal rest on a molar or premolar. An incisal rest is more likely than a lingual rest to lead to some orthodontic movement of the tooth because of unfavorable leverage factors.
An incisal rest seat is prepared in the form of a rounded notch at the incisal angle of a canine or on the incisal edge of an incisor, with the deepest portion of the preparation apical to the incisal edge (Figure 6-21). The notch should be beveled both labially and lingually, and the lingual enamel should be partly shaped to accommodate the rigid minor connector connecting the rest to the framework. An incisal rest seat should be approximately 2.5 mm wide and 1.5 mm deep so that the rest will be strong without having to exceed the natural contour of the incisal edge (Figure 6-22).
In the absence of other suitable placements for incisal rests and rest seats, incisal rests on multiple mandibular inci-sors may be considered. Use of such rests may be justified by the following factors: 1. They may take advantage of natural incisal faceting. 2. Tooth morphology does not permit other designs. 3. Such rests can restore defective or abraded tooth anatomy. 4. Incisal rests provide stabilization. 5. Full incisal rests may restore or provide anterior guidance.
Figure 6-19 Chromium-cobalt lingual rest seat on mandibular canine retained by resin cement with acid-etched tooth preparation.
A B Figure 6-20 A, Incisal rest seat placed in the mesial incisal edge of the lower canine. Note that the contact point is not in-volved in preparation of the rest seat. B, Mesial incisal rests on the canines furnish excellent vertical support and indirect reten-tion for this prosthesis upon completion. The incisal rest on tooth #27 also provides a third point of reference when frame orientation is established during maintenance reline procedures.
www.konkur.in 66 Part I General Concepts/Treatment Planning In the event that full incisal rests are considered, the patient should be thoroughly informed regarding their loca-tion, form, and esthetic impact.
It is, of course, essential that both the master cast and the casting be accurate if rests are to seat properly. The incisal rest should be overcontoured slightly to allow for labial and incisal finishing to the adjoining enamel, in much the same manner as a three-quarter crown or inlay margin is finished to enamel. In this way, minimal display of metal is possible without jeopardizing the effectiveness of the rest.
Care taken in selecting the type of rest seat to be used, in preparing it, and in fabricating the framework casting does much to ensure the success of any type of rest. The topography of any rest should be such that it restores the topography of the tooth that exists before the rest seat is prepared.
A B C Figure 6-21 Three views of an incisal rest seat preparation on the mandibular canine adjacent to a modification space. A, The labial view demonstrates inclination of the floor of the rest seat, which allows forces to be directed along the long axis of the tooth as nearly as possible. Note that the floor of the rest seat has been extended slightly onto the labial aspect of the tooth. B, As can be seen from a proximal view, the proximal edge of the rest seat is rounded rather than straight. C, The lingual view shows that all borders of the rest seat are rounded to avoid sharp line angles. It is especially important to avoid a line angle at the junction of the axial wall of the prepara-tion and the floor of the rest seat. The rest that occupies such a preparation should be able to move slightly in a lateral direction to avoid torquing of the abutment tooth.
2.5 mm 1.5 mm Figure 6-22 Dimensions given in the illustration for an in-cisal rest seat preparation will provide adequate strength of the framework at the junction of the rest and the minor connector. Rest seats of smaller dimension have proved unsatisfactory re-gardless of the metal alloy from which the framework is made.
www.konkur.in CHAPTER 7 Direct Retainers CHAPTER OUTLINE Direct Retainer’s Role in Control of Prosthesis Movement Basic Principles of Clasp Design Reciprocal Arm Functions Types of Direct Retainers Criteria for Selecting a Given Clasp Design Types of Clasp Assemblies Clasps Designed to Accommodate Functional Movement Clasps Designed Without Movement Accommodation Implants as Direct Retainers Analysis of Tooth Contours for Retentive Clasps Amount of Retention Size of and Distance into the Angle of Cervical Convergence Length of Clasp Arm Diameter of Clasp Arm Cross-Sectional Form of the Clasp Arm Material Used for the Clasp Arm Relative Uniformity of Retention Stabilizing-Reciprocal Cast Clasp Arm Other Types of Retainers Lingual Retention in Conjunction with Internal Rests Internal Attachments DIRECT RETAINER’S ROLE IN CONTROL OF PROSTHESIS MOVEMENT Retention of a removable prosthesis is a unique concern when compared with other prostheses. When one is dealing with a crown or fixed partial denture, the combined use of preparation geometry (i.e., resistance and retention form) and a luting agent can fix the prosthesis to the tooth in a manner that resists all forces to which the teeth are subjected. As was mentioned in Chapter 4, the direction of forces can be toward, across, or away from the tissue. In general, the forces acting to move prostheses toward and across the sup-porting teeth and/or tissue are the greatest in intensity. This is because most often they are forces of occlusion.
Forces acting to displace the prosthesis from the tissue can consist of gravity acting against a maxillary prosthesis, the action of adherent foods acting to displace the prosthesis on opening of the mouth in chewing, or functional forces acting across a fulcrum to unseat the prosthesis. The first two of these forces are seldom at the magnitude of functional forces, and the latter is minimized through the use of adequate support. The component part applied to resist this movement away from the teeth and/or tissue provides retention for the prosthesis and is called the direct retainer. A direct retainer is any unit of a removable dental prosthesis that engages an abutment tooth or implant to resist displacement of the prosthesis away from basal seat tissue. The direct retainer’s ability to resist this move-ment is greatly influenced by the stability and support of the prosthesis provided by major and minor connectors, rests, and tissue bases. This relationship of the supportive and retentive components highlights the relative importance of these com-ponent parts. Although the forces working against a remov-able partial denture to move it away from the tissue generally are not as great as the functional forces causing stress toward the tissue, the removable partial denture must have retention appropriate to resist reasonable dislodging forces. Too often retention concerns are given greater importance than is appro-priate, especially if such a focus detracts from more serious consideration of the resistance of typical functional forces.
www.konkur.in 68 Part I General Concepts/Treatment Planning Sufficient retention is provided by two means. Primary retention for the removable partial denture is accomplished mechanically by placing retaining elements (direct retainers) on the abutment teeth. Secondary retention is provided by the intimate relationship of the minor connector contact with the guiding planes and denture bases and of the major con-nector (maxillary) with underlying tissue. The latter is simi-lar to the retention of a complete denture. It is proportionate to the accuracy of the impression registration, the accuracy of the fit of the denture bases, and the total involved area of contact. Retention can also be provided through engagement of an attachment mechanism on a dental implant.
BASIC PRINCIPLES OF CLASP DESIGN The clasp assembly serves a similar function for a remov-able partial denture that a retainer crown serves for a fixed partial denture. Both must encircle the prepared tooth in a manner that prevents movement of the tooth separate from the retainer. To borrow from a fixed prosthodontic term, limiting the freedom of displacement refers to the effect of one cylindrical surface (the framework encircling the tooth) on another cylindrical surface (the tooth). It implies that the curve that defines the framework is properly shaped if it prevents movement at right angles to the tooth axis. This basic principle of clasp design offers a two-way benefit. First, it ensures the stability of the tooth position because of the restraint from encirclement, and second, it ensures stability of the clasp assembly because of the controlled position of the clasp in three dimensions.
Therefore the basic principle of clasp design, referred to as the principle of encirclement, means that more than 180 degrees in the greatest circumference of the tooth, pass-ing from diverging axial surfaces to converging axial sur-faces, must be engaged by the clasp assembly (Figure 7-1). The engagement can occur in the form of continuous con-tact, such as in a circumferential clasp, or discontinuous contact, such as in the use of a bar clasp. Both provide tooth contact in at least three areas encircling the tooth: (1) the occlusal rest area, (2) the retentive clasp terminal area, and (3) the reciprocal clasp terminal area.
In addition to encirclement, other basic principles of clasp design are as follows: 1. The occlusal rest must be designed to prevent movement of the clasp arms toward the cervical. 2. Each retentive terminal should be opposed by a recipro-cal component capable of resisting any transient pres-sures exerted by the retentive arm during placement and removal. Stabilizing and reciprocal components must be rigidly connected bilaterally (cross-arch) to realize recip-rocation of the retentive elements (Figure 7-2). 3. Clasp retainers on abutment teeth adjacent to distal ex-tension bases should be designed so that they avoid direct transmission of tipping and rotational forces to the abut-ment. In effect, they must act as stress-breakers, either by their design or by their construction. This is accomplished Buccal Lingual Buccal Lingual A B Figure 7-1 A, Dashed line drawn through the illustration represents 180 degrees of greatest circumference of abutment from the occlusal rest. Unless portions of the lingual reciprocal arm and the retentive buccal arm are extended beyond the line, the clasp would not accomplish its intended purpose. If respec-tive arms of the retainer were not extended beyond the line, the abutment tooth could be forced away from the retainer through torquing action of the clasp, or the removable partial denture could move away from the abutment. B, Bar-type clasp assembly engagement of more than 180 degrees of circumference of the abutment is realized by the minor connector for the occlusal rest, the minor connector contacting the guiding plane on the distal proximal surface, and the retentive bar arm.
A B Figure 7-2 A, Flexing action of the retentive clasp arm ini-tiates medially directed pressure on the abutment teeth as its retentive tip springs over the height of contour. B, Reciprocation to medially directed pressure is counteracted by rigid lingually placed clasp arms contacting the abutments simultaneously with the buccal arms, or by rigid stabilizing components of the frame-work contacting the lingual guiding planes when the buccal arms begin to flex.
www.konkur.in 69 Chapter 7 Direct Retainers through proper location of the retentive terminal relative to the rest, or by the use of a more flexible clasp arm in relation to the anticipated rotation of the denture under functional forces. 4. Unless guiding planes will positively control the path of removal and will stabilize abutments against rotational movement, retentive clasps should be bilaterally opposed (i.e., buccal retention on one side of the arch should be opposed by buccal retention on the other, or lingual on one side opposed by lingual on the other). In Class II situations, the third abutment may have buccal or lingual retention. In Class III situations, retention may occur bi-laterally or may be diametrically opposed (Figure 7-3). 5. The path of escapement for each retentive clasp terminal must be other than parallel to the path of removal for the prosthesis to require clasp engagement with the resistance to deformation that is retention. 6. The amount of retention should always be the minimum necessary to resist reasonable dislodging forces. 7. Reciprocal elements of the clasp assembly should be lo-cated at the junction of the gingival and middle thirds of the crowns of abutment teeth. The terminal end of the re-tentive arm is optimally placed in the gingival third of the crown (Figures 7-4 through 7-6). These locations permit better resistance to horizontal and torquing forces caused by a reduction in the effort arm as described in Chapter 4.
Reciprocal Arm Functions As was mentioned earlier, reciprocal arms are intended to resist tooth movement in response to deformation of the retainer arm as it engages a tooth height of contour. The opposing clasp arm reciprocates the effect of this defor-mation as it prevents tooth movement. For this to occur, the reciprocal arm must be in contact during the time of retainer arm deformation. Unless the abutment tooth has been spe-cifically contoured, the reciprocal clasp arm will not come into contact with the tooth until the denture is fully seated and the retentive clasp arm has again become passive. When this happens, a momentary tipping force is applied to the abutment teeth during each placement and removal. This may not be a damaging force—because it is transient—so long as the force does not exceed the normal elasticity of the periodontal attachments. True reciprocation during place-ment and removal is possible only through the use of crown surfaces made parallel to the path of placement. The use of cast restorations permits the parallel surfaces to be contacted by the reciprocal arm in such a manner that true reciproca-tion is made possible. This is discussed in Chapter 15.
Reciprocal arms can have additional functions as well. The reciprocal clasp arm should be located so that the den-ture is stabilized against horizontal movement. Stabilization is possible only through the use of rigid clasp arms, rigid minor connectors, and a rigid major connector. Horizontal forces applied on one side of the dental arch are resisted by the stabilizing components on the opposite side, providing cross-arch stability. Obviously the greater the number of such components, within reason, the greater will be the dis-tribution of horizontal stresses.
The reciprocal clasp arm also may act to a minor degree as an indirect retainer. This is true only when it rests on a supra-bulge surface of an abutment tooth lying anterior to the ful-crum line (see Figure 8-8). Lifting of a distal extension base away from the tissue is thus resisted by a rigid arm, which is not easily displaced cervically. The effectiveness of such an indirect retainer is limited by its proximity to the fulcrum line, which gives it a relatively poor leverage advantage, and by the b a a b a b b a a b Figure 7-3 Retentive clasps should be bilaterally opposed. This means bilateral buccal or bilateral lingual undercuts should be used, as shown on this Class III, modification 2 arch, in which the retention may be (a) bilaterally buccal or (b) bilaterally lingual.
Occlusal third Stabilizing Retentive Middle third Survey line Gingival third Figure 7-4 Simple mechanical laws demonstrate that the nearer stabilizing-reciprocal and retentive elements of direct re-tainer assemblies are located horizontal to the axis of rotation of the abutment, the less likely it is that physiologic tolerance of the periodontal ligament will be exceeded. The horizontal axis of rotation of the abutment tooth is located somewhere in its root.
www.konkur.in 70 Part I General Concepts/Treatment Planning fact that slippage along tooth inclines is always possible. The latter may be prevented by the use of a ledge on a cast restora-tion; however, enamel surfaces are not ordinarily so prepared.
TYPES OF DIRECT RETAINERS Mechanical retention of removable partial dentures is accom-plished by means of direct retainers of one type or another. Retention is accomplished by using frictional means, by engaging a depression in the abutment tooth, or by engaging a tooth undercut lying cervically to its height of contour. Two basic types of direct retainers are available: (1) the intracoro-nal retainer and (2) the extracoronal retainer. The extracoro-nal (clasp-type) retainer is the most commonly used retainer for removable partial dentures.
The intracoronal retainer may be cast or may be attached totally within the restored natural contours of an abutment tooth. It is typically composed of a prefabricated machined key and keyway with opposing vertical parallel walls, which serve to limit movement and resist removal of the partial denture through frictional resistance (Figure 7-7). The intra-coronal retainer is usually regarded as an internal, or pre-cision, attachment. The principle of the internal attachment was first formulated by Dr. Herman E.S. Chayes in 1906.
The extracoronal retainer uses mechanical resistance to displacement through components placed on or attached to the external surfaces of an abutment tooth. The extracoro-nal retainer is available in three principal forms. The clasp-type retainer (Figures 7-8 and 7-9), the form used most commonly, retains through a flexible clasp arm. This arm Lingual Occlusal Buccal Support Stabilization Retention Occlusal third Middle third Gingival third Support Stabilization Retention Occlusal third Middle third Gingival third A B C Figure 7-5 A bar-type clasp on the mandibular premolar. A, Support is provided by the occlusal rest. B, Stabilization is provided by the occlusal rest and the mesial and distal minor connectors. C, Retention is provided by the buccal I-bar. Reciprocation is obtained through the location of the minor connectors. Engagement of more than 180 degrees of circumference of the abutment is accomplished by proper location of components contacting the axial surfaces. (The minor connector supports the occlusal rest, the proximal plate minor connector, and the buccal I-bar.) www.konkur.in 71 Chapter 7 Direct Retainers engages an external surface of an abutment tooth in an area cervical to the greatest convexity of the tooth, or it engages a depression prepared to receive the terminal tip of the arm. The other forms both involve manufactured attachments and include interlocking components or the use of a spring-loaded device that engages a tooth contour to resist occlusal displacement. Another type is a manufactured attachment, which uses flexible clips or rings that engage a rigid com-ponent that is cast or attached to the external surface of an abutment crown.
In situations where support requirements are adequately met by available teeth and/or oral tissues, dental implants can be used for retention and provide the advantage of elimi-nation of a visible clasp.
CRITERIA FOR SELECTING A GIVEN CLASP DESIGN When a particular clasp design is selected for a given situa-tion, its function and limitations must be carefully evaluated. Extracoronal direct retainers, as part of the clasp assembly, should be considered as components of a removable partial denture framework. They should be designed and located to perform the specific functions of support, stabilization, reciprocation, and retention. It does not matter whether the direct retainer-clasp assembly components are physically attached to each other or originate from major and minor connectors of the framework (see Figures 1-2 and 1-3, B-E). If attention is directed to the separate function of each com-ponent of the direct retainer-clasp assembly, then selection of a direct retainer is simplified.
Although some rather complex designs are used for clasp arms, they all may be classified into one of two basic catego-ries. One is the circumferential clasp arm, which approaches the retentive undercut from an occlusal direction. The other is the bar clasp arm, which approaches the retentive under-cut from a cervical direction. A clasp assembly may com-prise various retentive arms (i.e., a cast circumferential, a bar clasp arm, or a wrought-wire retainer), depending on the specific requirement for retainer construction, given the necessary adjustability, clasp approach position, and survey line location.
A clasp assembly should consist of four component parts. First, one or more minor connectors should be present, from which the clasp components originate. Second, a principal rest should be designed to direct stress along the long axis of the tooth. Third, a retentive arm should engage a tooth undercut. For most clasps, the retentive region is only at its terminus. Fourth, a nonretentive arm (or other component) should be present on the opposite side of the tooth for sta-bilization and reciprocation against horizontal movement of the prosthesis (rigidity of this clasp arm is essential to its purpose).
No confusion should exist between the choice of clasp arm and the purpose for which it is used. Either type of cast clasp arm (bar or circumferential) may be made tapered and retentive, or nontapered (rigid) and nonretentive. The choice depends on whether it is used for retention, stabilization, or reciprocation. An occlusal rest, such as in the rest, proximal plate, and I-bar (RPI) component parts of the clasp assembly concept, may be used rather than a reciprocal clasp arm to satisfy the need for encirclement, provided it is located in such a way that it can accomplish the same purpose (Figure 7-10; see also Figure 7-9). The addition of a lingual apron to a cast reciprocal clasp arm alters neither its primary purpose nor the need for proper location to accomplish that purpose.
TYPES OF CLASP ASSEMBLIES A wide variety of clasp assemblies are available for clinicians to use. This variety exists largely because of the imagination of clinicians and technicians who provided prostheses when tooth modification was not or could not be provided. To sim-plify clasp design and to improve the functional predictabil-ity of prostheses, today’s clinician must realize the need for tooth modification.
Some clasp assemblies are designed to accommodate prosthesis functional movement (as mentioned in the basic principles earlier), and others do not incorporate such design features. Although it has been demonstrated by Kapur and others that adverse outcomes are not always associated with Support Stabilization Retention Occlusal third Buccal Lingual Middle third Gingival third Support Stabilization Retention Occlusal third Middle third Gingival third Figure 7-6 Circumferential clasp on the mandibular premolar. Support is provided by the occlusal rest; stabilization is provided by the occlusal rest, proximal minor connector, lingual clasp arm, and rigid portion of the buccal retentive clasp arm occlusal to the height of contour; retention is realized by the retentive terminal of the buccal clasp arm; reciprocation is provided by the nonflex-ible lingual clasp arm. Assembly engages more than 180 degrees of the abutment tooth’s circumference.
www.konkur.in 72 Part I General Concepts/Treatment Planning the use of rigid clasp assemblies in distal extension classifica-tions, the following clasp assemblies are described as clasps designed to accommodate distal extension functional move-ment and clasps designed without movement accommoda-tion. The clinician should not interpret these categories as mutually exclusive because most any clasp assembly can be used to retain a well-supported and maintained prosthesis.
Clasps Designed to Accommodate Functional Movement RPI, RPA, and Bar Clasp Clasp assemblies that accommodate functional prosthesis movement are designed to address the concern of a Class I lever. The concern is that the distal extension acts as a long “effort arm” across the distal rest “fulcrum” to cause the clasp tip “resistance arm” to engage the tooth undercut. This results in harmful tipping or torquing of the tooth, which is greater with stiff clasps and increased denture base move-ment. Two strategies may be adopted to change the fulcrum location and subsequently the “resistance arm” engaging effect (mesial rest concept clasp assemblies), or to mini-mize the effect of the lever through the use of a flexible arm (wrought-wire retentive arm).
Mesial rest concept clasps have been proposed to accomplish movement accommodation by changing the fulcrum location. This concept includes the RPI and rest, proximal plate, Akers (RPA) clasps. The RPI is a current concept of bar clasp design that refers to the rest, proximal plate, and I-bar component parts of the clasp assembly. Basically, this clasp assembly consists of a mesio-occlusal rest with the minor connector placed into the mesiolin-gual embrasure, but not contacting the adjacent tooth A B C Figure 7-7 A, The intracoronal retainer consists of a key and keyway with extremely small tolerance. Keyways are contained within abutment crowns, and keys are attached to the removable partial denture framework (B). C, Frictional resistance to removal and place-ment and limitation of movement serve to retain and stabilize the prosthesis.
www.konkur.in 73 Chapter 7 Direct Retainers Support C A B Stabilization Retention Support Stabilization Retention Lingual Buccal Figure 7-8 Extracoronal circumferential direct retainer. As-sembly consists of the buccal retentive arm (A); the rigid lingual stabilizing (reciprocal) arm (B); and the supporting occlusal rest (C). The terminal portion of the retentive arm is flexible and engag-es measured undercut. Assembly remains passive until activated by placement or removal of the restoration, or when subjected to masticatory forces that tend to dislodge the denture base.
Support Stabilization Retention Support Stabilization HOC UC Retention Buccal Lingual B C D D B A Figure 7-9 Extracoronal bar-type direct retainer. Assembly consists of the buccal retentive arm engaging the measured undercut (A) (with slight occlusal extension for stabilization [see insert]); the stabilizing (reciprocal) elements, with the proximal plate minor con-nector on the distal (B); the lingually placed mesial minor connector for the occlusal rest, which also serves as a stabilizing (reciprocal) component (C); and the mesially placed supporting occlusal rest (D). Assembly remains passive until activated. HOC, Height of contour; UC, undercut.
Figure 7-10 The auxiliary occlusal rest (mirror view) may be used rather than the reciprocal clasp arm without violating any principle of clasp design. Its greatest disadvantages are that the second rest seat must be prepared and that enclosed tissue space at the gingival margin can result in a food trap. The aux-iliary occlusal rest is also sometimes used to prevent slippage when the principal occlusal rest seat cannot be inclined apically from the marginal ridge. Minor connectors used to close the in-terproximal space most often require rests on adjacent teeth to avoid a wedging effect when force is placed on the denture.
www.konkur.in 74 Part I General Concepts/Treatment Planning (Figure 7-11, A). A distal guiding plane, extending from the marginal ridge to the junction of the middle and gin-gival thirds of the abutment tooth, is prepared to receive a proximal plate (see Figure 7-11, B). The buccolingual width of the guiding plane is determined by the proxi-mal contour of the tooth (see Figure 7-11, A and C). The proximal plate, in conjunction with the minor connector supporting the rest, provides the stabilizing and recip-rocal aspects of the clasp assembly. The I-bar should be located in the gingival third of the buccal or labial surface of the abutment in a 0.01-inch undercut (see Figure 7-11, D). The whole arm of the I-bar should be tapered to its terminus, with no more than 2 mm of its tip contacting the abutment. The retentive tip contacts the tooth from the undercut to the height of contour (see Figure 7-11, E). This area of contact, along with the rest and proximal plate contact, provides stabilization through encirclement (see Figure 7-11, C). The horizontal portion of the approach arm must be located at least 4 mm from the gingival mar-gin and even farther if possible.
A B C D E Support Lingual Stabilization Retention Support Stabilization Buccal Retention Occlusal third Middle third Gingival third Figure 7-11 Bar-type clasp assembly. A, Occlusal view. Component parts (proximal plate minor connector, rest with minor connector, and retentive arm) tripod the abutment to prevent its migration. B, The proximal plate minor connector extends just far enough lingually so that it combines with the mesial minor connector to prevent lingual migration of the abutment. C, On narrow or tapered abutments (mandibular first premolars), the proximal plate should be designed to be as narrow as possible but still sufficiently wide to prevent lin-gual migration. D, I-bar retainer located at greatest prominence of tooth in the gingival third. E, Mesial view of I-bar illustrating the reten-tive tip relationship to the undercut and a region superior to the height of contour, which serves a stabilization function in encirclement.
www.konkur.in 75 Chapter 7 Direct Retainers Three basic approaches to the application of the RPI system may be used. The location of the rest, the design of the minor connector (proximal plate) as it relates to the guiding plane, and the location of the retentive arm are factors that influence how this clasp system functions. Variations in these factors provide the basis for differences among these approaches. All advo-cate the use of a rest located mesially on the primary abutment tooth adjacent to the extension base area. One approach recom-mends that the guiding plane and the corresponding proximal plate minor connector should extend the entire length of the proximal tooth surface, with physiologic tissue relief eliminat-ing impingement of the free gingival margin (Figure 7-12). A second approach suggests that the guiding plane and the cor-responding proximal plate minor connector should extend from the marginal ridge to the junction of the middle and gin-gival thirds of the proximal tooth surface (Figure 7-13). Both approaches recommend that the retaining clasp arm should be located in the gingival third of the buccal or labial surface of the abutment in a 0.01-inch undercut. Placement of the retaining clasp arm generally occurs in an undercut located at the greatest mesiodistal prominence of the tooth or adjacent to the exten-sion base area (Figure 7-14, A and B). The third approach favors a proximal plate minor connector that contacts approximately 1 mm of the gingival portion of the guiding plane (Figure 7-15, A) and a retentive clasp arm located in a 0.01-inch undercut in the gingival third of the tooth at the greatest prominence or toward the mesial away from the edentulous area (see Figure 7-14, C). If the abutment teeth demonstrate contraindications for a bar-type clasp (i.e., exaggerated buccal or lingual tilts, severe tissue undercut, or a shallow buccal vestibule) and the desirable undercut is located in the gingival third of the tooth away from the extension base area, a modification should be considered for the RPI system (the RPA clasp) (see Figure 7-15, B). Application of each approach is predicated on the distribu-tion of load to be applied to the tooth and edentulous ridge.
The bar clasp, which gave rise to the RPI, is discussed here because of this association. It may not be configured to allow functional movement, but it can be. The term bar clasp is gen-erally preferred over the less descriptive term Roach clasp arm. Reduced to its simplest term, the bar clasp arm arises from the denture framework or a metal base and approaches the reten-tive undercut from a gingival direction (see Figure 7-11). The bar clasp arm has been classified by the shape of the retentive terminal. Thus it has been identified as a T, modified T, I, or Y. The form the terminal takes is of little significance as long as it is mechanically and functionally effective, covers as little tooth surface as possible, and displays as little metal as possible.
In most situations, the bar clasp arm can be used with tooth-supported partial dentures, with tooth-supported modification areas, or when an undercut that can be logically approached with a bar clasp arm lies on the side of an abut-ment tooth adjacent to a distal extension base (Figure 7-16). If a tissue undercut prevents the use of a bar clasp arm, a mesially originating ring clasp, a cast, or a wrought-wire clasp or reverse-action clasp may be used. Preparation of adjacent Minimum tissue relief Occlusal force GP PP Figure 7-12 Bar clasp assembly in which the guiding plane (GP) and the corresponding proximal plate (PP) extend the entire length of the proximal tooth surface. Physiologic relief is required to prevent impingement of the gingival tissues during function. Ex-tending the proximal plate to contact a greater surface area of the guide plane directs functional forces in a horizontal direction, thus the teeth are loaded to a greater extent than the edentulous ridge.
Occlusal force GP PP Figure 7-13 Bar clasp assembly in which the guiding plane (GP) and the corresponding proximal plate (PP) extend from the marginal ridge to the junction of the middle and gingival thirds of the proximal tooth surface. This decrease (compared with Fig-ure 7-23) in amount of surface area contact of the proximal plate on the guiding plane more evenly distributes functional force be-tween the tooth and the edentulous ridge.
C A B Figure 7-14 Occlusal view of the rest, proximal plate, and I-bar (RPI) component parts bar clasp assembly. Placement of the I-bar in a 0.01-inch undercut on the distobuccal surface (A); at the greatest mesiodistal prominence (B); and on the mesiobuc-cal surface (C).
www.konkur.in 76 Part I General Concepts/Treatment Planning abutments (natural teeth) to receive any type of interproxi-mal direct retainer, traversing from lingual to buccal surfaces, is most difficult to adequately accomplish. Inevitably the rel-ative size of the occlusal table is increased, contributing to undesirable and additional functional loading.
Specific indications for use of a bar clasp arm include (1) when a small degree of undercut (0.01 inch) exists in the cervical third of the abutment tooth, which may be approached from a gingival direction, (2) on abut-ment teeth for tooth-supported partial dentures or tooth-supported modification areas (Figure 7-17), (3) in distal Occlusal force A B GP GP PP PP Figure 7-15 A, Bar clasp assembly in which the proximal plate (PP) contacts approximately 1 mm of the gingival portion of the guid-ing plane (GP). During function, the proximal plate and the I-bar clasp arm are designed to move in a mesiogingival direction, disengag-ing the tooth. Lack of sustained contact between the proximal plate and the guiding plane distributes increased functional force to the edentulous ridge. The asterisk () indicates center of rotation. B, Modification of the rest, proximal plate, and I-bar (RPI) component parts system; rest, proximal plate, Akers (RPA) clasp is indicated when a bar-type clasp is contraindicated and a desirable undercut is located in the gingival third of the tooth away from the extension base area.
Figure 7-16 Bar clasp arm properly used on the terminal abutment. The combination of rest, proximal plate, and bar clasp contacting the abutment tooth provides more than 180 degrees encirclement. A uniform taper to the bar ensures proper flexibil-ity and internal stress distribution. Taper can originate from the minor connector junction or at a finishing line indicating the an-terior extent of the denture base. Encirclement does not require that retentive tip modification (in the form of a T) be provided, and such a modification adds little to the clasp assembly.
Figure 7-17 A bar retainer is used on the anterior abutment of the modification space, and its terminus engages the disto-buccal undercut. The denture is designed to rotate around the terminal abutments when force is directed toward the basal seat on the left. Such rotation would impart force on the right premo-lar directed superiorly and anteriorly. However, this direction of force is resisted in great part by mesial contact with the canine. A direct retainer on the right premolar engaging the mesiobuccal undercut would tend to force the tooth superiorly and posteriorly.
www.konkur.in 77 Chapter 7 Direct Retainers extension base situations, and (4) in situations in which esthetic considerations must be accommodated and a cast clasp is indicated. Thus use of the bar clasp arm is contra-indicated when a deep cervical undercut exists or when a severe tooth and/or tissue undercut exists, either of which must be bridged by excessive blockout. When severe tooth and tissue undercuts exist, a bar clasp arm usually is an annoyance to the tongue and cheek and may traps food debris. Other limiting factors in the selection of a bar clasp assembly include a shallow vestibule or an excessive buccal or lingual tilt of the abutment tooth (Figure 7-18). Some common errors in the design of bar-type clasps are illus-trated in Figure 7-19.
The bar clasp arm is not a particularly flexible clasp arm because of the effects of its half-round form and its several planes of origin. Although the cast circumferential clasp arm can be made more flexible than the bar clasp arm, the combi-nation clasp is preferred for use on terminal abutments when torque and tipping are possible, because of engagement of an undercut away from the distal extension base. Situations often exist, however, in which a bar clasp arm may be used to advantage without jeopardizing a terminal abutment. A bar clasp arm swinging distally into the undercut may be a logi-cal choice, because movement of the clasp on the abutment as the distal extension base moves tissue-ward is minimized by the distal location of the clasp terminal.
A B C Figure 7-18 Contraindications for selection of bar-type clasps. A, Severe buccal or lingual tilts of abutment teeth. B, Severe tissue undercuts. C, Shallow buccal or labial vestibules.
A B Occlusal third Middle third Gingival third C D E F Occlusal third Middle third Gingival third Figure 7-19 Common errors and recommended corrections in the design of bar-type clasp assemblies. A, Survey line unsuitable for the bar clasp (too high). B, Retentive portion of the bar clasp arm improperly contoured to resist dislodging force in the occlusal direc-tion. C, Retentive tip not located in the gingival third of the abutment. D, Contour of abutment correctly altered to receive bar clasp. E and F, Correct position of bar clasp assembly.
www.konkur.in 78 Part I General Concepts/Treatment Planning Some advantages attributed to the infrabulge clasp are (1) its interproximal location, which may be used to esthetic advantage; (2) increased retention without tipping action on the abutment; and (3) less chance of accidental distor-tion resulting from its proximity to the denture border. The wearer should be meticulous in the care of a denture so made, not only for reasons of oral hygiene but also to prevent cariogenic debris from being held against tooth surfaces.
The T and Y clasp arms are the most frequently mis-used. It is unlikely that the full area of a T or Y terminal is ever necessary for adequate clasp retention. Although the larger area of contact would provide greater frictional resistance, this is not true clasp retention, and only that portion engaging an undercut area should be consid-ered retentive. Only one terminal of such a clasp arm should be placed in an undercut area (Figure 7-20). The remainder of the clasp arm may be superfluous unless it is needed as part of the clasp assembly to encircle the abutment tooth by more than 180 degrees at its greatest circumference. If the bar clasp arm is made to be flex-ible for retentive purposes, any portion of the clasp above the height of contour provides only limited stabiliza-tion, because it is also part of the flexible arm. Therefore in many instances, this suprabulge portion of a T or Y clasp arm may be removed, and the retentive terminal of the bar clasp should be designed to be biologically and mechanically sound rather than to conform to any alpha-betical configuration.
Combination Clasp Another strategy that may be used to reduce the effect of the Class I lever in distal extension situations includes a flexible component in the “resistance arm;” this strategy is employed in the combination clasp. The combination clasp consists of a wrought-wire retentive clasp arm and a cast reciprocal clasp arm (Figure 7-21). Although the latter may be seen in the form of a bar clasp arm, it is usually a circumferential arm. The retentive arm is almost always circumferential, but it also may be used in the manner of a bar, originating gingivally from the denture base.
Advantages of the combination clasp include its flex-ibility and adjustability and the appearance of the wrought-wire retentive arm. It is used when maximum flexibility is desirable, such as on an abutment tooth adjacent to a distal extension base or on a weak abutment when a bar-type direct retainer is contraindicated. It may be used for its adjustabil-ity when precise retentive requirements are unpredictable and later adjustment to increase or decrease retention may be necessary. A third justification for its use is its esthetic advantage over cast clasps. Wrought in structure, it may be used in smaller diameters than a cast clasp, with less danger of fracture. Because it is round, light is reflected in such a manner that the display of metal is less noticeable than with the broader surfaces of a cast clasp.
Support Stabilization Retention Buccal Occlusal third Middle third Gingival third Figure 7-20 Only one terminal of the retentive arm engages the undercut in the gingival third of the abutment. The suprabulge por-tion of the retentive clasp arm provides only limited stabilization and may be eliminated.
Path of placement 18-gauge round wrought wire B A Height of contour Figure 7-21 A, A combination clasp consists of a cast recipro-cal arm and a tapered, round wrought-wire retentive clasp arm. The latter is cast to, or soldered to, a cast framework. This de-sign is recommended for the anterior abutment of the posterior modification space in a Class II partially edentulous arch, where only a mesiobuccal undercut exists, to minimize the effects of a first-class lever system. B, In addition to the advantages of flex-ibility, adjustability, and appearance, a wrought-wire retentive arm makes only line contact with the abutment tooth, rather than broader contact with the cast clasp.
www.konkur.in 79 Chapter 7 Direct Retainers The most common use of the combination clasp is on an abutment tooth adjacent to a distal extension base where only a mesial undercut exists on the abutment or where a large tissue undercut contraindicates a bar-type retainer (Figure 7-22). When a distal undercut exists that may be approached with a properly designed bar clasp arm or with a ring clasp (despite its several disadvantages), a cast clasp can be located so that it will not cause abutment tipping as the distal extension base moves tissue-ward. When the undercut is on the side of the abutment away from the extension base, the tapered wrought-wire retentive arm offers greater flex-ibility than does the cast clasp arm and therefore better dis-sipates functional stresses. For this reason, the combination clasp is preferred (see Figure 7-22, D).
The combination clasp has several disadvantages: (1) it involves extra steps in fabrication, particularly when high-fusing chromium alloys are used; (2) it may be distorted by careless handling on the part of the patient; (3) because it is bent by hand, it may be less accurately adapted to the tooth and therefore may provide less stabilization in the supra-bulge portion; and (4) it may distort with function and not engage the tooth. The disadvantages of the wrought-wire clasp are offset by its several advantages, which include (1) its flexibility; (2) its adjustability; (3) its esthetic advantage over other retentive circumferential clasp arms; (4) coverage of a minimum of tooth surface as a result of its line contact with the tooth, rather than having the surface contact a cast clasp arm; and (5) a less likely occurrence of fatigue failure in service with the tapered wrought-wire retentive arm versus the cast, half-round retentive arm.
The disadvantages listed previously should not prevent its use regardless of the type of alloy that is used for the cast framework. Technical problems are minimized by selecting the best wrought wire for this purpose, then casting to it or soldering it to the cast framework. Selection of wrought wire, attachment of it to the framework, and subsequent labora-tory procedures to maintain its optimum physical properties are presented in Chapter 13.
The patient may be taught to avoid distortion of the wrought wire by the explanation that to remove a partial denture, the fingernail should always be applied to its point of origin, where it is held rigid by the casting, rather than to the flexible terminal end. Often, lingual retention may be used rather than buccal retention, especially on a mandibu-lar abutment, so that the patient never touches the wrought-wire arm during removal of the denture. Instead, removal may be accomplished by lifting against the cast reciprocal arm located on the buccal side of the tooth. This may negate the esthetic advantage of the wrought-wire clasp arm, and esthetics should be given preference when the choice must be made between buccal and lingual retention. In most situ-ations, however, retention must be used where it is possible, and the clasp designed accordingly.
Clasps Designed Without Movement Accommodation Circumferential Clasp Although a thorough knowledge of the principles of clasp design should lead to a logical application of those principles, it is better when some of the more common clasp designs are considered individually. The circumferential clasp will be considered first as an all-cast clasp.
A B C D E F Figure 7-22 Five types of extracoronal direct retainer assem-blies that may be used on abutments adjacent to the distal ex-tension base to avoid or minimize the effects of the cantilever design. Arrows indicate the general direction of movement of retentive tips of retainer arms when the denture base rotates to-ward and away from the edentulous ridge. A, Distobuccal under-cut engaged by one-half T-type bar clasp. The portion of the clasp arm on and above the height of contour might afford some sta-bilization against horizontal rotation of the denture base. B, I-bar placed in undercut at the middle (anteroposteriorly) of the buccal surface. This retainer contacts the tooth only at its tip. Note that the guiding plane on the distal aspect of the abutment is con-tacted by the metal of the denture framework, and that a mesial rest is used. C, Interproximal ring clasp engaging the distobuccal undercut. Bar-type retainer cannot be used because of tissue un-dercuts inferior to the buccal surface of the abutment. D, Round, uniformly tapered 18-gauge wrought-wire circumferential retainer arm engaging the mesiobuccal undercut. A wrought-wire arm, instead of a cast arm, should be used in this situation because of the ability of wrought wire to flex omnidirectionally. A cast half-round retainer arm would not flex edgewise, which could result in excessive stress on the tooth when rotation of the denture base occurs. E, A hairpin clasp may be used when the undercut lies cervical to the origin of the retainer arm. Both hairpin and inter-proximal ring clasps may be used to engage the distobuccal un-dercut on the terminal abutment of the distal extension denture. However, the distobuccal undercut on the terminal abutment should be engaged by a bar-type clasp in the absence of a large buccal tissue undercut cervical to the terminal abutment. Hairpin and interproximal ring clasps are the least desirable of the clasp-ing situations illustrated here. F, Lingual view shows the use of double occlusal rests, connected to the lingual bar by the minor connector in illustrated designs. This design eliminates the need for a lingual clasp arm, places the fulcrum line anteriorly to make better use of the residual ridge for support, and provides stabili-zation against horizontal rotation of the denture base.
www.konkur.in 80 Part I General Concepts/Treatment Planning The circumferential clasp is usually the most logical clasp to use with all tooth-supported partial dentures because of its retentive and stabilizing ability (Figure 7-23). Only when the retentive undercut may be approached better with a bar clasp arm or when esthetics will be enhanced should the lat-ter be used. The circumferential clasp arm has the following disadvantages: 1. More tooth surface is covered than with a bar clasp arm because of its occlusal origin. 2. On some tooth surfaces, particularly the buccal surfaces of mandibular teeth and the lingual surfaces of maxillary teeth, its occlusal approach may increase the width of the occlusal surface of the tooth. 3. In the mandibular arch, more metal may be displayed than with the bar clasp arm. 4. As with all cast clasps, its half-round form prevents ad-justment to increase or decrease retention. Adjustments in the retention afforded by a cast clasp arm should be made by moving a clasp terminal cervically into the angle of cervical convergence or occlusally into a lesser area of undercut. Tightening a clasp against the tooth or loosen-ing it away from the tooth increases or decreases friction-al resistance and does not affect the retentive potential of the clasp. True adjustment is, therefore, impossible with most cast clasps.
Despite its disadvantages, the cast circumferential clasp arm may be used effectively, and many of these disadvan-tages may be minimized by mouth preparation. Adequate mouth preparation will permit its point of origin to be placed far enough below the occlusal surface to avoid poor esthetics and increased tooth dimension (see Figure 7-23). Although some of the disadvantages listed imply that the bar-type clasp may be preferable, the circumferential clasp is actually superior to a bar clasp arm that is improperly used or poorly designed.
Experience has shown that faulty application and design too often negate the possible advantages of the bar clasp arm, whereas the circumferential clasp arm is not as easily mis-used.
The basic form of the circumferential clasp is a buccal and lingual arm originating from a common body (Figure 7-24). This clasp is used improperly when two retentive clasp arms originate from the body and occlusal rest areas and approach bilateral retentive areas on the side of the tooth away from the point of origin. The correct form of this clasp has only one retentive clasp arm, opposed by a nonretentive recipro-cal arm on the opposite side. A common error is to use this clasp improperly by making both clasp terminals retentive. This not only is unnecessary but also disregards the need for reciprocation and bilateral stabilization. Other common errors in the design of circumferential clasps are illustrated in Figure 7-25.
Ring Clasp. The circumferential type of clasp may be used in several forms. It appears as though many of these forms of the basic circumferential clasp design were devel-oped to accommodate situations in which corrected tooth modifications could not be or were not accomplished by the dentist. One is the ring clasp, which encircles nearly all of a tooth from its point of origin (Figure 7-26). It is used when a proximal undercut cannot be approached by other means. For example, when a mesiolingual undercut on a lower molar abutment cannot be approached directly because of its prox-imity to the occlusal rest area and cannot be approached with a bar clasp arm because of lingual inclination of the tooth, the ring clasp encircling the tooth allows the undercut to be approached from the distal aspect of the tooth.
The clasp should never be used as an unsupported ring (Figure 7-27) because if it is free to open and close as a ring, it cannot provide reciprocation or stabilization. Instead the ring-type clasp should always be used with a supporting strut on the nonretentive side, with or without an auxiliary occlu-sal rest on the opposite marginal ridge. The advantage of an auxiliary rest is that further movement of a mesially inclined tooth is prevented by the presence of a distal rest. In any event, the supporting strut should be regarded as a minor connector from which the flexible retentive arm originates. Reciprocation Figure 7-23 Cast circumferential retentive clasp arms prop-erly designed. They originate on or occlusal to the height of con-tour, which they then cross in their terminal third, and engage retentive undercuts progressively as their taper decreases and their flexibility increases.
Figure 7-24 Cast circumferential retentive clasp arm.
www.konkur.in 81 Chapter 7 Direct Retainers then comes from the rigid portion of the clasp lying between the supporting strut and the principal occlusal rest.
The ring-type clasp should be used on protected abut-ments, whenever possible, because it covers such a large area of tooth surface. Esthetics need not be considered on such a posteriorly located tooth.
A ring-type clasp may be used in reverse on an abut-ment located anterior to a tooth-bounded edentulous space (Figure 7-28). Although potentially an effective clasp, this clasp covers an excessive amount of tooth surface and can be esthetically objectionable. The only justification for its use is when a distobuccal or distolingual undercut cannot be approached directly from the occlusal rest area and/or tis-sue undercuts prevent its approach from a gingival direction with a bar clasp arm.
Embrasure Clasp. In the fabrication of an unmodified Class II or Class III partial denture, no edentulous spaces are available on the opposite side of the arch to aid in clasping. Mechanically, this is a disadvantage. However, when the teeth are sound and retentive areas are available, or when multiple restorations are justified, clasping can be accomplished by means of an embrasure clasp (Figure 7-29).
Sufficient space must be provided between the abutment teeth in their occlusal third to make room for the common body of the embrasure clasp (Figure 7-30), yet the contact area should not be eliminated entirely. Historically, this clasp assembly demonstrates a high percentage of fracture caused by inadequate tooth preparation in the contact area. Because vulnerable areas of the teeth are involved, abutment protec-tion with inlays or crowns is recommended. The decision to use unprotected abutments must be made at the time of oral examination and should be based on the patient’s age, caries index, and oral hygiene, as well as on whether exist-ing tooth contours are favorable or can be made favorable by tooth modification. Preparation of adjacent, contacting, uncrowned abutments to receive any type of embrasure clasp of adequate interproximal bulk is difficult, especially when it is opposed by natural teeth.
The embrasure clasp always should be used with double occlusal rests, even when definite proximal shoulders can be established (Figure 7-31). This is done to avoid interproxi-mal wedging by the prosthesis, which could cause separation of the abutment teeth, resulting in food impaction and clasp displacement. In addition to providing support, occlusal rests serve to shunt food away from contact areas. For this reason, occlusal rests should always be used whenever food impaction is possible.
Embrasure clasps should have two retentive clasp arms and two reciprocal clasp arms that are bilaterally or diago-nally opposed. An auxiliary occlusal rest or a bar clasp arm can be substituted for a circumferential reciprocal arm, as long as definite reciprocation and stabilization result. A lingually placed retentive bar clasp arm may be substituted if a rigid circumferential clasp arm is placed on the buccal surface for reciprocation, provided lingual retention is used on the opposite side of the arch. Other less commonly used modifications of the cast circumferential clasp that are of his-torical interest are the multiple clasp, the half-and-half clasp, and the reverse-action clasp.
IMPLANTS AS DIRECT RETAINERS As was stated earlier, in situations where support require-ments are adequately met by available oral tissues, dental implants can be used for retention and provide the advan-tage of eliminating a visible clasp. The unique aspect of implant use with removable partial dentures is that their location generally can be prescribed, meaning that the clini-cian selects the best location. The location within the modi-fication space must first consider anatomic characteristics of bone availability. It would not be a great advantage to a patient if extensive augmentation procedures were required to allow implant placement in conjunction with a removable partial denture.
If anatomic needs are met, placement of an implant within a modification space to the advantage of retentive needs requires consideration of anterior, mid, or distal placement. Because retainers utilizing teeth have always been restricted to tooth locations at either end of a span, the mid-span loca-tion typically is not considered. However, the application of A C E F G B D Figure 7-25 Some improper applications of circumferential clasp design and their recommended corrections. A, Tooth with undesirable height of contour in its occlusal third. B, Unsuit-able contour and location of retentive clasp arm on an unmodi-fied abutment. C, More favorable height of contour achieved by modification of the abutment. D, Retentive clasp arm properly designed and located on a modified abutment. E, Unsuitable contour and location of the retentive arm in relation to the height of contour (straight arm configuration provides poor approach to the retentive area and is less resistant to dislodging force). F, Terminal portion of the retentive clasp arm located too close to the gingival margin. G, Clasp arm that is properly designed and located.
www.konkur.in 82 Part I General Concepts/Treatment Planning A B Figure 7-26 Ring clasp(s) encircling nearly all of the tooth from its point of origin. A, Clasp originates on the mesiobuccal surface and encircles the tooth to engage the mesiolingual undercut. B, Clasp originates on the mesiolingual surface and encircles the tooth to engage the mesiobuccal undercut. In either example, a supporting strut is used on the nonretentive side (drawn both as direct view of near side of tooth and as mirror view of opposite side).
Figure 7-27 Improperly designed ring clasp lacking neces-sary support. Such a clasp lacks any reciprocating or stabiliz-ing action because the entire circumference of the clasp is free to open and close. A supporting strut should always be added on the nonretentive side of the abutment tooth, which then be-comes, in effect, a minor connector from which a tapered and flexible retentive clasp arm originates.
Figure 7-28 Ring clasp may be used in reverse on the abut-ment located anterior to the tooth-bound edentulous space.
www.konkur.in 83 Chapter 7 Direct Retainers implants allows consideration of where the most beneficial location of retention can be provided, which forces consid-eration of where the most efficient resistance to movement away from the denture base can be provided. Placement at either extreme of the denture base may allow greater move-ment than placement at a midpoint and should be taken into consideration.
ANALYSIS OF TOOTH CONTOURS FOR RETENTIVE CLASPS Although the extracoronal, or clasp-type, direct retainer is used more often than the internal attachment, it is com-monly misused. It is hoped that a better understanding of the principles of clasp design will lead to more intelligent use of this retainer. To best gain this understanding, it is vitally important for the clinician to consider how tooth contour and removable partial denture components must interact (be related) to allow stable prosthetic function. Just as unaltered natural teeth are not appropriately con-toured to receive fixed partial dentures without prepara-tion, the teeth that are engaged by a removable partial denture must be contoured to support, stabilize, and retain the functioning prosthesis. Analysis and accomplishment of tooth modification when it is required for optimum stability and retention are necessary for the success of the prosthesis.
Critical areas of an abutment that provide for retention and stabilization (reciprocation) can be identified only with the use of a dental cast surveyor (Table 7-1). To enhance the understanding of direct retainers, an introduction of the dental cast surveyor is appropriate at this time. Surveying will be covered in detail in Chapter 11.
The cast surveyor (Figure 7-32) is a simple instrument that is essential for planning partial denture treatment. Its main working parts are the vertical arm and the adjustable table, which hold the cast in a fixed relation to the vertical arm. This relationship of the vertical arm to the cast represents the path of placement that the partial denture will ultimately take when inserted or removed from the mouth (Figure 7-33).
The adjustable table may be tilted in relation to the verti-cal arm of the surveyor until a path can be found that best Figure 7-29 Embrasure clasps extend through occlusal em-brasures to engage facial undercuts when no modification spac-es are present.
Figure 7-30 Embrasure and hairpin circumferential retentive clasp arms. The terminus of each engages a suitable retentive un-dercut. Use of a hairpin-type clasp on the second molar is made necessary by the fact that the only available undercut lies directly below the point of origin of the clasp arm.
A B Figure 7-31 A, Example of the use of an embrasure clasp for a Class II partially edentulous arch. Embrasure clasp on the two left molar abutments was used in the absence of posterior modi-fication space. B, Occlusal and proximal surfaces of adjacent mo-lar and premolar prepared for embrasure clasp. Note that rest seat preparations are extended both buccally and lingually to ac-commodate the retentive and reciprocal clasp arms. Adequate preparation confined to the enamel can rarely be accomplished for such a clasp, especially when clasped teeth are opposed by natural teeth.
www.konkur.in 84 Part I General Concepts/Treatment Planning satisfies all involved factors (Figure 7-34). A cast in a hori-zontal relationship to the vertical arm represents a vertical path of placement; a cast in a tilted relationship represents a path of placement toward the side of the cast that is on an upward slant. The vertical arm, when brought in contact with a tooth surface, identifies the location on the clinical crown where the greatest convexity exists. This line, called the height of contour (specific to the surveyor-defined path), is the boundary between (1) an occlusal or incisal region of the tooth that is freely accessible to a prosthesis and (2) a gingival region of the tooth that can be accessed only when a portion of the prosthesis elastically deforms and recovers to contact the tooth. This surveyor-defined path and the subse-quent tooth height of contour will indicate the areas available for retention and those available for support and stabiliza-tion, as well as the existence of tooth and other tissue inter-ference with the path of placement.
When the surveyor blade contacts a tooth on the cast at its greatest convexity, a triangle is formed. The apex of the triangle is at the point of contact of the surveyor blade with the tooth, and the base is the area of the cast representing the gingival tissue (see Figure 11-18). The apical angle is called the angle of cervical convergence (see Figure 7-33). This angle may be measured as described in Chapter 11, or it may be estimated by observing the triangle of light visible between the tooth and the surveyor blade. For this reason, a wide surveyor blade rather than a small cylindrical tool is used so it is easier to see the triangle of light. The impor-tance of this angle lies in its relationship to the amount of retention.
AMOUNT OF RETENTION Clasp retention is based on resistance to deformation of the metal. For a clasp to be retentive, it must be placed in an undercut area of the tooth, where it is forced to deform upon application of a vertical dislodging force (Figure 7-35). It is this resistance to deformation along an appropriately selected path that generates retention (Figure 7-36). Such resistance to deformation is dependent on several factors and is also proportionate to the flexibility of the clasp arm.
The interaction of a number of factors under the control of the clinician combines to produce retention. These include both tooth (planned and executed by the clinician) and pros-thesis (to be planned by the dentist and executed by the labo-ratory technician) factors.
Table 7-1 Function and Position of Clasp Assembly Parts Component Part Function Location Rest Support Occlusal, lingual, incisal Minor connector Stabilization Proximal surfaces extend-ing from a prepared marginal ridge to the junction of the middle and gingival one third of the abutment crown Clasp arms Stabilization (reciprocation) Middle one third of crown Retention Gingival one third of crown in measured undercut Figure 7-32 Essential parts of a dental surveyor (Ney Paral-lelometer, Dentsply Ceramco, Burlington, NJ), showing the verti-cal spindle in relation to an adjustable table.
Path of placement A B Height of contour x x Figure 7-33 Angle of cervical convergence on two teeth pre-senting dissimilar contours. A greater angle of cervical conver-gence on tooth A necessitates placement of a clasp terminus, x, nearer the height of contour than when a lesser angle exists, as in B. It is apparent that uniform clasp retention depends on depth (amount) of the tooth undercut rather than on distance below the height of the contour at which the clasp terminus is placed.
www.konkur.in 85 Chapter 7 Direct Retainers Tooth factors include the size of the angle of cervical con-vergence (depth of undercut) and how far the clasp terminal is placed into the angle of cervical convergence. Prosthesis factors include the flexibility of the clasp arm. Clasp flexibil-ity is the product of clasp length (measured from its point of origin to its terminal end), clasp relative diameter (regard-less of its cross-sectional form), clasp cross-sectional form or shape (whether it is round, half-round, or some other form), and the material used in making the clasp. The reten-tion characteristics of gold alloy, chrome alloy, titanium, or titanium alloy depend on whether it is in cast or wrought form.
Size of and Distance into the Angle of Cervical Convergence To be retentive, a tooth must have an angle of conver-gence cervical to the height of contour. When it is sur-veyed, any single tooth has a height of contour or an area of greatest convexity. Areas of cervical convergence may not exist when the tooth is viewed in relation to a given path of placement. Also, certain areas of cervi-cal convergence may not be usable for the placement of retentive clasps because of their proximity to gingival tissue.
A B C Figure 7-34 Relationship of height of contour, suprabulge, and infrabulge. A, When an egg is placed with its long axis parallel to the surveying tool, the height of contour is found at its greatest circumference, here designated by the arrow. In this example, the second line is diagonal to the line outlining the height of contour and is either above the height of contour (right side of egg), referred to as the supra-bulge region, or below the height of contour (left side of egg), referred to as the infrabulge region. B, If the long axis of the egg is reoriented so that the previous diagonal line is now at the greatest circumference, the original “height of contour line” no longer is at the great-est circumference. The line segment at A is in the suprabulge region, and the line segment B is in the infrabulge region. Changing the orientation alters the relationship of surfaces relative to the greatest circumference and consequently alters suprabulge and infrabulge locations. C, Just as the height of contour changes as orientation changes for the egg, when a tooth orientation changes, the height of contour is altered. For this molar, the line H was produced with a horizontal orientation. When the tooth was inclined buccally, the height of contour moved to B, relative to the horizontal location. Alternatively, when the tooth was inclined lingually, the height of contour moves to L, relative to the horizontal location.
www.konkur.in 86 Part I General Concepts/Treatment Planning This is best illustrated by mounting a spherical object, such as an egg, on the adjustable table of a dental surveyor (see Figure 7-34). The egg now represents the cast of a dental arch or, more correctly, one tooth of a dental arch. The egg first is placed perpendicular to the base of the surveyor and is surveyed so the height of contour can be determined. The vertical arm of the surveyor represents the path of placement that a denture would take, and con-versely its path of removal.
With a carbon marker, a circumferential line has been drawn on an egg at its greatest circumference, as shown by the arrow in Figure 7-34, A. This line, which Ken-nedy called the height of contour, is its greatest convexity. Cummer spoke of it as the guideline, because it is used as a guide in the placement of retentive and nonretentive clasps. DeVan added the terms suprabulge, denoting the surfaces sloping superiorly, and infrabulge, denoting the surfaces sloping inferiorly.
Any areas cervical to the height of contour may be used for the placement of retentive clasp components, whereas areas occlusal to the height of contour may be used for the placement of nonretentive, stabilizing, or reciprocating components. Obviously, only flexible components may be placed gingivally to the height of contour because rigid elements would not flex over the height of contour or con-tact the tooth in the undercut area.
With the original height of contour marked on the egg, the egg is now tilted from the perpendicular to an angular relation with the base of the surveyor (see Figure 7-34, B). Its relation to the vertical arm of the surveyor has now been changed, just as a change in the position of a dental cast would bring about a different relationship with the surveyor. The vertical arm of the surveyor still represents the path of placement. However, its relation to the egg is totally different.
Again, the carbon marker is used to delineate the height of contour or the greatest convexity. It will be seen that areas that were formerly infrabulge are now supra-bulge and vice versa. A retentive clasp arm placed below the height of contour in the original position may now be excessively retentive or totally nonretentive, whereas a nonretentive stabilizing or reciprocal arm that is located above the height of contour in the first position may now be located in an area of undercut. Figure 7-35, C, illus-trates this principle compared with a tooth example show-ing that changes in tilt can significantly alter heights of contour.
The location and depth of a tooth undercut avail-able for retention are therefore relative only to the path of placement and removal of the partial denture. At the same time, nonretentive areas on which rigid components of the clasp may be placed exist only for a given path of placement (see Figure 7-35).
If conditions are found that are not favorable for the particular path of placement under consideration, a study should be conducted for a different path of place-ment. The cast is merely tilted in relation to the verti-cal arm until the most suitable path is found. The most suitable path of placement is generally considered to be the path of placement that requires the least amount of mouth preparation necessary to place the components of the partial denture in their ideal position on the tooth surfaces and in relation to the soft tissue. Then mouth preparations are planned with a definite path of place-ment in mind.
It is important to remember that tooth surfaces can be recontoured through selective grinding or the place-ment of restorations (mouth preparations) to achieve a more suitable path of placement. The path of placement also must take into consideration the presence of tissue undercuts that would interfere with the placement of major connectors, the location of vertical minor connec-tors, the origin of bar clasp arms, and the denture bases.
When the theory of clasp retention is applied to the abutment teeth in a dental arch during surveying of the dental cast, each tooth may be considered individually and in relation to the other abutment teeth as far as the Lifting force Lifting force Figure 7-35 Retention is provided primarily by the flexible portion of the clasp assembly. Retentive terminals are ideally located in measured undercuts in the gingival third of abutment crowns. When force acts to dislodge the restoration in an occlusal direction, the retentive arm is forced to deform as it passes from the undercut location over the height of contour. The amount of retention provided by the clasp arm is determined by its length, diameter, taper, cross-sectional form, contour, type of alloy, and location and depth of undercut engaged.
www.konkur.in 87 Chapter 7 Direct Retainers designs of retentive and stabilizing (reciprocating) com-ponents are concerned. This is necessary because the rela-tionship of each tooth to the entire arch and to the design of the whole prosthesis has been considered previously when teeth were selected or modified to achieve the most suitable path of placement. Once this relationship of the cast to the surveyor has been established, the height of contour on each abutment tooth becomes fixed, and the clasp design for each must be considered separately.
A positive path of placement and removal is made pos-sible by the contact of rigid parts of the denture framework with parallel tooth surfaces, which act as guiding planes. Because guiding planes control the path of placement and removal, they can provide additional retention for the par-tial denture by limiting the possibilities that exist for its dislodgment. The more vertical the walls (guiding planes) that are prepared parallel to the path of insertion, the fewer the possibilities that exist for dislodgment. If some degree of parallelism does not exist during placement and removal, trauma to the teeth and supporting structures and strain on the denture parts are inevitable. This ultimately results in damage to the teeth and their periodontal support or to the denture itself, or both. Therefore without guiding planes, clasp retention will be detrimental or practically nonexis-tent. If clasp retention is frictional only because of an active relationship of the clasp to the teeth, then orthodontic movement or damage to periodontal tissue, or both, will result. Instead a clasp should bear a passive relationship to the teeth, except when a dislodging force is applied. In addition to the degree of the angle of cervical convergence A B C D Caramel candy Figure 7-36 A, Retentive areas are not sufficient to resist reasonable dislodging forces when a cast is surveyed at its most advanta-geous position (occlusal plane parallel to the surveyor table), even though guide planes could be established with minor tooth modifica-tion. B, Tilting cast creates functionally ineffective tooth contours, which are present only in relation to the surveying rod and do not exist when compared with the most advantageous position (position in which restoration will be subject to dislodging forces in an occlusal direction). C and D, Clasps designed at tilt are ineffective without the development of corresponding guide planes to resist displacement when the restoration is subject to dislodging forces in the occlusal direction.
www.konkur.in 88 Part I General Concepts/Treatment Planning and the distance a clasp is placed into the angle of cervical convergence, the amount of retention generated by a clasp depends on its flexibility. This is a function of clasp length, diameter, taper, cross-sectional form, and material.
The amount of retention also depends on the flexibility of a clasp arm. This is a function of clasp length, diameter, cross-sectional form, and material.
Length of Clasp Arm The longer the clasp arm is, the more flexible it will be, all other factors being equal. The length of a circumfer-ential clasp arm is measured from the point at which a uniform taper begins. Tooth modification providing increased length to a suprabulge retentive clasp by allow-ing the retentive tip to approach the undercut from a gin-gival direction optimizes clasp retention (see Figure 7-8). The retentive circumferential clasp arm should be tapered uniformly from its point of origin through the full length of the clasp arm (Figure 7-37).
The length of a bar clasp arm also is measured from the point at which a uniform taper begins. Generally the taper of a bar clasp arm should begin at its point of ori-gin from a metal base, or at the point at which it emerges from a resin base (Figure 7-38). Although a bar clasp arm usually will be longer than a circumferential clasp arm, its flexibility will be less because its half-round form lies in several planes; this prevents its flexibility from being proportionate to its total length. Tables 7-2 and 7-3 give an approximate depth of undercut that may be used for the cast gold and chromium-cobalt retentive clasp arms of circumferential and bar-type clasps. Based on a propor-tional limit of 60,000 psi and on the assumption that the clasp arm is properly tapered, the clasp arm should be able to flex repeatedly within the limits stated without harden-ing or rupturing because of fatigue. It has been estimated that alternate stress applications of the fatigue type are placed on a retainer arm during mastication and other force-producing functions about 300,000 times a year.
Diameter of Clasp Arm The greater the average diameter of a clasp arm is, the less flexible it will be—all other factors being equal. If its taper is absolutely uniform, the average diameter will be at a point midway between its origin and its terminal end. If its taper is not uniform, a point of flexure—and therefore a point of weakness—will exist; this then will be the deter-mining factor in its flexibility, regardless of the average diameter of its entire length.
Table 7-3 Permissible Flexibilities of Retentive Cast Circumferential and Bar-Type Clasp Arms for Chromium-Cobalt Alloys CIRCUMFERENTIAL BAR-TYPE Length (inches) Flexibility (inches) Length (inches) Flexibility (inches) 0 to 0.3 0.004 0 to 0.7 0.004 0.3 to 0.6 0.008 0.7 to 0.9 0.008 0.6 to 0.8 0.012 0.9 to 1.0 0.012 Based on the approximate dimensions of Jelenko preformed plastic pat-terns (JF Jelenko, New York, NY).
Table 7-2 Permissible Flexibilities of Retentive Cast Circumferential and Bar-Type Clasp Arms of Type IV Gold Alloys CIRCUMFERENTIAL BAR-TYPE Length (inches) Flexibility (inches) Length (inches) Flexibility (inches) 0 to 0.3 0.01 0 to 0.7 0.01 0.3 to 0.6 0.02 0.7 to 0.9 0.02 0.6 to 0.8 0.03 0.9 to 1.0 0.03 Based on the approximate dimensions of Jelenko preformed plastic pat-terns (JF Jelenko, New York, NY).
One tenth or less of clasp length T T T T ½ T ½ T ½ T ½ T Figure 7-37 The retentive cast clasp arm should be tapered uniformly from its point of attachment at the clasp body to its tip. Dimensions at the tip are about half those at the point of attach-ment. The clasp arm so tapered is approximately twice as flexible as one without any taper. T, Clasp thickness. (Courtesy of The Argen Corporation, New York, NY.) Figure 7-38 Length of the tapered cast retentive clasp arm is measured along the center portion of the arm until it joins the clasp body (circumferential) or becomes part of the denture base or is embedded in the base (bar-type clasp).
www.konkur.in 89 Chapter 7 Direct Retainers Cross-Sectional Form of the Clasp Arm Flexibility may exist in any form, but it is limited to one direction in the case of the half-round form. The only uni-versally flexible form is the round form, which is practi-cally impossible to obtain by casting and polishing.
Because most cast clasps are essentially half-round in form, they may flex away from the tooth, but edgewise flexing (and edgewise adjustment) is limited. For this reason, cast retentive clasp arms are more acceptable in tooth- supported partial dentures in which they are called on to flex only during placement and removal of the pros-thesis. A retentive clasp arm on an abutment adjacent to a distal extension base not only must flex during placement and removal but also must be capable of flexing during functional movement of the distal extension base. It must have universal flexibility to avoid transmission of tipping stresses to the abutment tooth, or it must be capable of disengaging the undercut when vertical forces directed against the denture are toward the residual ridge. A round clasp is the only circumferential clasp form that may be safely used to engage a tooth undercut on the side of an abutment tooth away from the distal extension base. The location of the undercut is perhaps the single most impor-tant factor in selection of a clasp for use with distal exten-sion partial dentures.
Material Used for the Clasp Arm Although all cast alloys used in partial denture construc-tion possess flexibility, their flexibility is proportionate to their bulk. If this were not true, other components of the partial denture could not have the necessary rigidity. A disadvantage of a cast gold partial denture is that its bulk must be increased to obtain needed rigidity at the expense of added weight and increased cost. It cannot be denied that greater rigidity with less bulk is possible through the use of chromium-cobalt alloys.
Although cast gold alloys may have greater resiliency than do cast chromium-cobalt alloys, the fact remains that the structural nature of the cast clasp does not approach the flexibility and adjustable nature of the wrought-wire clasp. Because it was formed by being drawn into a wire, the wrought-wire clasp arm has toughness exceeding that of a cast clasp arm. The tensile strength of a wrought structure is at least 25% greater than that of the cast alloy from which it was made. It may therefore be used in smaller diameters to provide greater flexibility without fatigue and ultimate fracture.
Relative Uniformity of Retention Now that the factors inherent to a determination of the amount of retention from individual clasps have been reviewed, it is important to consider the coordination of rel-ative retention between various clasps in a single prosthesis.
The size of the angle of convergence will determine how far a given clasp arm should be placed into that angle. By disregarding—for the time being—variations in clasp flexibility, the relative uniformity of retention will depend on the location of the retentive part of the clasp arm, not in relation to the height of contour, but in relation to the angle of cervical convergence.
The retention on all principal abutments should be as equal as possible. Although esthetic placement of clasp arms is desirable, it may not be possible to place all clasp arms in the same occlusocervical relationship because of variations in tooth contours. However, retentive surfaces may be made similar by altering tooth contours or by using cast restorations with similar contours.
Retentive clasp arms must be located so that they lie in the same approximate degree of undercut on each abut-ment tooth. In Figure 7-33, this is seen at point x on both teeth—A and B—despite variation in the distance below the height of contour. Should both clasp arms be placed equidistant below the height of contour, the higher loca-tion on tooth B would have too little retention, whereas the lower location on tooth A would be too retentive.
Measurement of the degree of undercut by mechanical means with the use of a surveyor is important. However, undercut identification is only one factor that is impor-tant to consider when one is providing appropriate reten-tion for a removable partial denture.
Stabilizing-Reciprocal Cast Clasp Arm When the direct retainer comes into contact with the tooth, the framework must be stabilized against horizontal move-ment for the required clasp deformation to occur. This sta-bilization is derived from cross-arch framework contacts or from a stabilizing or reciprocal clasp in the same clasp assembly. To provide true reciprocation, the reciprocal clasp must be in contact during the entire period of retentive clasp deformation. This is best provided with lingual-palatal, guide-plane surfaces.
A stabilizing (reciprocal) clasp arm should be rigid. There-fore it is shaped somewhat differently than is the cast reten-tive clasp arm, which must be flexible. Its average diameter must be greater than the average diameter of the opposing retentive arm, to increase desired rigidity. A cast retentive arm is tapered in two dimensions, as illustrated in Figure 7-37, whereas a reciprocal arm should be tapered in one dimension only, as shown in Figure 7-39. Achieving such a form for the arm requires freehand waxing of patterns.
OTHER TYPES OF RETAINERS Lingual Retention in Conjunction with Internal Rests The internal rest is covered in Chapter 6. It is emphasized that the internal rest is not used as a retainer, but that its near-vertical walls provide for reciprocation against a lin-gually placed retentive clasp arm. For this reason, visible clasp arms may be eliminated, thus avoiding one of the principal objections to the extracoronal retainer.
www.konkur.in 90 Part I General Concepts/Treatment Planning Such a retentive clasp arm, terminating in an existing or prepared infrabulge area on the abutment tooth, may be of any acceptable design. It is usually a circumferential arm arising from the body of the denture framework at the rest area. It should be wrought because the advantages of adjustability and flexibility make the wrought clasp arm preferable. It may be cast with gold or a low-fusing, chro-mium-cobalt alloy, or it may be assembled by being sol-dered to one of the higher-fusing, chromium-cobalt alloys. In any event, future adjustment or repair is facilitated.
The use of lingual extracoronal retention avoids much of the cost of the internal attachment, yet disposes of a visible clasp arm when esthetics must be considered. Often it is employed with a tooth-supported partial den-ture only on the anterior abutments, and when esthetics is not a consideration, the posterior abutments are clasped in the conventional manner.
One of the dentist’s prime considerations in clasp selec-tion is the control of stress transferred to the abutment teeth when the patient exerts an occluding force on the artificial teeth. The location and design of rests, the clasp arms, and the position of minor connectors as they relate to guiding planes are key factors in controlling transfer of stress to abutments. Errors in the design of a clasp assem-bly can result in uncontrolled stress to abutment teeth and their supporting tissues. Some common errors and their corrections are illustrated in Figures 7-40 and 7-41.
The choice of clasp design should be based on biologi-cal and mechanical principles. The dentist responsible for the treatment being rendered must be able to justify the clasp design used for each abutment tooth in keeping with these principles.
INTERNAL ATTACHMENTS The principle of the internal attachment was first formu-lated by Dr. Herman E.S. Chayes in 1906. One such attach-ment manufactured commercially still carries his name. Although it may be fabricated by the dental technician as a cast dovetail fitting into a counterpart receptacle in the abut-ment crown, the alloys used in manufactured attachments and the precision with which they are constructed make the ready-made attachment preferable to any of this type that can be fabricated in the dental laboratory. Much credit is due the manufacturers of metals used in dentistry for con-tinued improvements in the design of internal attachments.
The internal attachment has two major advantages over the extracoronal attachment: (1) elimination of visible retentive and support components and (2) better verti-cal support through a rest seat located more favorably in relation to the horizontal axis of the abutment tooth. For these reasons, the internal attachment may be preferable in selected situations. It provides horizontal stabilization similar to that of an internal rest. However, additional extracoronal stabilization is usually desirable. It has been claimed that stimulation to the underlying tissue is greater when internal attachments are used because of intermit-tent vertical massage. This is probably no more than is pos-sible with extracoronal retainers of similar construction.
Some of the disadvantages of internal attachments include the following: (1) they require prepared abut-ments and castings; (2) they require somewhat com-plicated clinical and laboratory procedures; (3) they eventually wear, with progressive loss of frictional resis-tance to denture removal; (4) they are difficult to repair and replace; (5) they are effective in proportion to their length and are therefore least effective on short teeth; (6) they are difficult to place completely within the circum-ference of an abutment tooth because of the size of the pulp; and (7) they are considered more costly.
Because the principle of the internal attachment does not permit horizontal movement, all horizontal, tipping, Occlusal force Minor connector in contact with tooth Occlusal force No contact A B Figure 7-40 A, Minor connector supporting the distal rest does not contact the prepared guiding plane, resulting in uncon-trolled stress to the abutment tooth. B, A minor connector con-tacts the prepared guiding plane and directs stresses around the arch through the proximal contacts.
One tenth or less of clasp length T T ½ T ½ T T ½ T Figure 7-39 The reciprocal arm of the direct retainer as-sembly should be rigid. An arm tapered both lengthwise and widthwise is more flexible than an arm of the same dimensions tapered only lengthwise. T, Clasp thickness.
www.konkur.in 91 Chapter 7 Direct Retainers and rotational movements of the prosthesis are transmit-ted directly to the abutment tooth. The internal attachment therefore should not be used in conjunction with extensive tissue-supported distal extension denture bases unless some form of stress-breaker is used between the movable base and the rigid attachment. Although stress-breakers may be used, they do have some disadvantages—which are discussed later—and their use adds to the cost of the partial denture.
Numerous other types of retainers for partial dentures have been devised that cannot be classified as primarily of the intracoronal or extracoronal type. Neither can they be classified as relying primarily on frictional resistance or placement of an element in an undercut to prevent dis-placement of the denture. However, all of these use some type of locking device, located intracoronally or extra-coronally, for providing retention without visible clasp retention. Although the motivation behind the develop-ment of other types of retainers has usually been a desire to eliminate visible clasp retainers, the desire to minimize torque and tipping stresses on the abutment teeth has also been given consideration.
All of the retainers that are discussed herein have merit, and much credit is due to those who have devel-oped specific devices and techniques for the retention of partial dentures. The use of patented retaining devices and other techniques falls in the same limited category as the internal attachment prosthesis and is for economic and technical reasons available to only a small percentage of those patients who need partial denture service.
Internal attachments of the locking or dovetail type unquestionably have many advantages over the clasp-type denture in tooth-supported situations. How-ever, it is questionable whether the locking type of internal attachment is indicated for distal extension remov-able partial dentures, with or without stress-breakers and with or without splinted abutments, because of inherent excessive leverages most often associated with these attachments.
The nonlocking type of internal attachment, in con-junction with sound prosthodontic principles, can be advantageously used in many instances in Class I and Class II partially edentulous situations. However, unless the cross-arch axis of rotation is common to the bilater-ally placed attachments, torque on the abutments may be experienced (Figure 7-42). Excellent textbooks devoted to the use of manufactured intracoronal and extracoro-nal retainer systems are available. For this reason, this text concerns itself primarily with the extracoronal type of direct retainer assembly (clasps). Numerous well-designed internal attachments are available in the dental market that may be used in situations requiring special retention. Descriptive literature and technique manuals are available from the manufacturers.
Other conservative treatment of partially edentulous arches with removable partial dentures can be accom-plished in a variety of ways. Treatment is still contingent on the location and condition of the remaining teeth and the contour and quality of the residual ridges. Basic principles and concepts of design relative to support and stability must be respected even though a variety of retaining devices can be incorporated. Examples of some of these retaining devices are illustrated in Figures 7-43 through 7-46.
No contact Occlusal force Possible wedge No contact Occlusal force Occlusal force gp A B C D Figure 7-41 A, Clasp assembly designed so that the vertical occlusal force results in movement of the proximal plate cervically and out of contact with the guiding plane, as illustrated in B. This lack of contact may contribute to a possible wedging effect. C, Extending contact of the proximal plate on the prepared guiding plane or, as in D, eliminating space between the artificial tooth and the guiding plane (gp) will help direct stresses around the arch through proximal contacts.
www.konkur.in 92 Part I General Concepts/Treatment Planning A B Figure 7-42 A, Axes of rotation, although parallel, are not common because one axis is located anterior to the other axis. B, When one nonlocking internal attachment is elevated farther from the residual ridge than its cross-arch counterpart, the axes of rotation do not fall on a common line; thus some torquing of abutments should be anticipated. However, in many instances, the effect produced by this situation will not exceed physiologic tolerance of the supporting structures of the abutments—all other torquing factors being equal.
Figure 7-43 Classification II, modification 1, maxillary re-movable partial denture with three internal attachments. Gold framework with soldered attachments that gain retention through activation of the gingival region of the male component (see inset).
Figure 7-44 Same prosthesis as Figure 7-43, showing the at-tachment position relative to the framework palatal and distal minor connector components.
Figure 7-45 Similar photo as in Figure 7-44, showing the attachment at the anterior modification space. Careful arrange-ment of the components is required to maximize the esthetic advantage of attachment use in such a region.
Figure 7-46 Prosthesis with three internal attachments shown from the tissue side, depicting the common path for each. No-tice the broad ridge coverage and the use of an anterior-posterior palatal strap major connector to help support the prosthesis, minimizing stress to the attachments.
www.konkur.in CHAPTER 8 Indirect Retainers CHAPTER OUTLINE Role of Indirect Retainers in Control of Prosthesis Movement Factors Influencing Effectiveness of Indirect Retainers Auxiliary Functions of Indirect Retainers Forms of Indirect Retainers Auxiliary Occlusal Rest Canine Rests Canine Extensions from Occlusal Rests Cingulum Bars (Continuous Bars) and Linguoplates Modification Areas Rugae Support ROLE OF INDIRECT RETAINERS IN CONTROL OF PROSTHESIS MOVEMENT As was described in Chapter 4, partial denture movement can exist in three planes. Tooth-supported partial dentures effectively use teeth to control movement away from the tissues. Tooth-tissue–supported partial dentures do not have this capability because one end of the prosthesis is free to move away from the tissue. This may occur because of the effects of gravity in the maxillary arch or adhesive foods in either arch. Attention to the details of design and location of partial denture component parts in control of functional movement is the strategy used in partial den-ture design.
When the distal extension denture base is dislodged from its basal seat, it tends to rotate around the fulcrum lines. Theoretically, this movement away from the tissues can be resisted by activation of the direct retainer, the stabilizing components of the clasp assembly, and the rigid components of the partial denture framework, which are located on defi-nite rests on the opposite side of the fulcrum line away from the distal extension base. These components are referred to as indirect retainers (Figures 8-1 and 8-2). Indirect retainer components should be placed as far as possible from the dis-tal extension base, which provides the best leverage advan-tage against dislodgment (Figure 8-3).
For the sake of clarity in discussion of the location and functions of indirect retainers, fulcrum lines should be con-sidered the axis about which the denture will rotate when the bases move away from the residual ridge.
An indirect retainer consists of one or more rests and the supporting minor connectors (Figures 8-4 and 8-5). The proximal plates, adjacent to the edentulous areas, also provide indirect retention. Although it is customary to identify the entire assembly as the indirect retainer, it should be remem-bered that the rest is actually the indirect retainer united to the major connector by a minor connector. This is noted to avoid interpretation of any contact with tooth inclines as part of the indirect retainer. An indirect retainer should be placed www.konkur.in 94 Part I General Concepts/Treatment Planning E E R R F F Figure 8-1 Mandibular distal extension removable partial denture showing the distal extension base being lifted from the ridge and the clasp assembly being activated and engaged, with the indirect retainer providing stabilization against dislodgment. E, Effort; F, fulcrum; R, resistance.
A C E G B D F H Figure 8-2 Fulcrum lines found in various types of partially edentulous arches, around which the denture may rotate when bases are subjected to forces directed toward or away from the residual ridge. Arrows indicate the most advantageous position of indirect retainer(s). A and B, In a Class I arch, the fulcrum line passes through the most posterior abutments, provided some rigid component of the framework is occlusal to the abutment’s heights of contour. C, In a Class II arch, the fulcrum line is diagonal, pass-ing through the abutment on the distal extension side and the most posterior abutment on the opposite side. D, If the abutment tooth anterior to the modification space lies far enough removed from the fulcrum line, it may be used effectively for support of the indi-rect retainer. E and F, In a Class IV arch, the fulcrum line passes through two abutments adjacent to the single edentulous space. G, In a Class III arch with a posterior tooth on the right side, which has a poor prognosis and eventually will be lost, the fulcrum line is considered the same as though posterior tooth were not present. Thus its future loss may not necessitate altering the original design of the removable partial denture framework. H, In a Class III arch with nonsupporting anterior teeth, the adjacent edentulous area is considered to be the tissue-supported end, with a diagonal fulcrum line passing through the two principal abutments, as in a Class II arch.
www.konkur.in 95 Chapter 8 Indirect Retainers F Fulcrum F F Fulcrum Force Fulcrum Force Direct retainer DR DR DR Fulcrum Force Indirect retainer IR IR IR DR DR DR A C D B Figure 8-3 Indirect retainer principle. A, Beams are supported at various points. B, A lifting force will displace the entire beam in the absence of retainers. C, With direct retainers (DRs) at the fulcrum, the lifting force will depress one end of the beam and elevate the other end. D, With both direct and indirect retainers (IRs) functioning, the lifting force will not displace beam. The farther the indirect retainer is from the fulcrum, the more efficiently it should control movement.
Figure 8-4 Planning the location for an indirect retainer for a Class II, modification 2, removable partial denture. The great-est distance from the axis of rotation around most distal rests (fulcrum line) would fall on #22. The decision to use an incisal rest or cingulum rest depends on the patient’s concern for the esthetic impact of an incisal rest versus having a crown (for the cingulum rest).
Figure 8-5 Example of indirect retention used in conjunction with a palatal plate–type major connector. Indirect retainers are proximal plates on second premolars and occlusal rests located on first premolars. A secondary function of auxiliary occlusal rest assemblies is to prevent settling of the anterior portion of the major connector and to provide stabilization against horizontal rotation.
www.konkur.in 96 Part I General Concepts/Treatment Planning as far from the distal extension base as possible in a prepared rest seat on a tooth capable of supporting its function.
Although the most effective location of an indirect retainer is commonly in the vicinity of an incisor tooth, that tooth may not be strong enough to support an indirect retainer and may have steep inclines that cannot be favorably altered to support a rest. In such a situation, the nearest canine tooth or the mesio-occlusal surface of the first premolar may be the best location for the indirect retention, despite the fact that it is not as far removed from the fulcrum line. Whenever possible, two indirect retainers closer to the fulcrum line are used to compensate for the compromise in distance. When a distal extension is judged to be at significant risk for dis-lodgement away from the denture base, the use of a dental implant as a distal retainer can be considered.
FACTORS INFLUENCING EFFECTIVENESS OF INDIRECT RETAINERS The following factors influence the effectiveness of an indi-rect retainer: 1. The principal occlusal rests on the primary abutment teeth must be reasonably held in their seats by the reten-tive arms of the direct retainers. If rests are held in their seats, rotation about an axis should occur, which sub-sequently would activate the indirect retainers. If total displacement of the rests occurs, no rotation about the fulcrum would occur, and the indirect retainers would not be activated. 2. Distance from the fulcrum line. The following three areas must be considered: a. Length of the distal extension base b. Location of the fulcrum line c. How far beyond the fulcrum line the indirect retainer is placed 3. Rigidity of the connectors supporting the indirect retain-er. All connectors must be rigid if the indirect retainer is to function as intended. 4. Effectiveness of the supporting tooth surface. The indirect retainer must be placed on a definite rest seat on which slippage or tooth movement will not occur. Tooth inclines and weak teeth should never be used to support indirect retainers.
AUXILIARY FUNCTIONS OF INDIRECT RETAINERS In addition to effectively activating the direct retainer to pre-vent movement of a distal extension base away from the tis-sues, an indirect retainer may serve the following auxiliary functions: 1. It tends to reduce anteroposterior tilting leverages on the principal abutments. This is particularly important when an isolated tooth is being used as an abutment—a situa-tion that should be avoided whenever possible. Ordinar-ily, proximal contact with the adjacent tooth prevents such tilting of an abutment as the base lifts away from the tissues. 2. Contact of its minor connector with axial tooth surfac-es aids in stabilization against horizontal movement of the denture. Such tooth surfaces, when made parallel to the path of placement, may also act as auxiliary guiding planes. 3. Anterior teeth supporting indirect retainers are stabilized against lingual movement. 4. It may act as an auxiliary rest to support a portion of the major connector, facilitating stress distribution.
For example, a lingual bar may be supported against settling into the tissues by the indirect retainer acting as an auxiliary rest. One must be able to differentiate between an auxiliary rest placed for support for a ma-jor connector, one placed for indirect retention, and one serving a dual purpose. Some auxiliary rests are added solely to provide rest support to a segment of the denture and should not be confused with indirect retention. 5. It may provide the first visual indications for the need to re-line an extension base partial denture. Deficiencies in basal seat support are manifested by the dislodgment of indirect retainers from their prepared rest seats when the denture base is depressed and rotation occurs around the fulcrum.
These auxiliary functions derived from indirect retainers are important to consider, especially given the reported con-troversy as to the effectiveness of indirect retainers.
FORMS OF INDIRECT RETAINERS The indirect retainer may take any one of several forms. All are effective in proportion to their support and distance from the fulcrum line.
Auxiliary Occlusal Rest The most commonly used indirect retainer is an auxiliary occlusal rest located on an occlusal surface and as far away from the distal extension base as possible. In a mandibular Class I arch, this location is usually on the mesial marginal ridge of the first premolar on each side of the arch (see Figure 8-4). The ideal position for the indirect retainer perpendicu-lar to the fulcrum line would be in the vicinity of the central incisors, which are too weak and have lingual surfaces that are too perpendicular to support a rest. Bilateral rests on the first premolars are quite effective, even though they are located closer to the axis of rotation.
The same principle applies to any maxillary Class I partial denture when indirect retainers are used. Bilateral rests on the mesial marginal ridge of the first premolars generally are used in preference to rests on incisor teeth (see Figure 8-5). Not only are they effective without jeopardizing the weaker single-rooted teeth, but interference with the tongue is far less when the minor connector can be placed in the embra-sure between canine and premolar rather than anterior to the canine teeth.
www.konkur.in 97 Chapter 8 Indirect Retainers Indirect retainers for Class II partial dentures are usually placed on the marginal ridge of the first premolar tooth on the opposite side of the arch from the distal extension base (Figure 8-6). Bilateral rests are seldom indicated except when an auxiliary occlusal rest is needed for support of the major connector, or when the prognosis for the distal abutment is poor and provision is being considered for later conversion to a Class I partial denture.
Canine Rests When the mesial marginal ridge of the first premolar is too close to the fulcrum line, or when the teeth are over-lapped so that the fulcrum line is not accessible, a rest on the adjacent canine tooth may be used. Such a rest may be made more effective by placing the minor connector in the embrasure anterior to the canine, either curving back onto a prepared lingual rest seat or extending to a mesio-incisal rest. The same types of canine rests as those previ-ously outlined—lingual or incisal rests—may be used (see Chapter 6).
Canine Extensions from Occlusal Rests Occasionally, a finger extension from a premolar rest is placed on the prepared lingual slope of the adjacent canine tooth (Figure 8-7). Such an extension is used to effect indi-rect retention by increasing the distance of a resisting ele-ment from the fulcrum line. This method is particularly applicable when a first premolar must serve as a primary abutment. The distance anterior to the fulcrum line is only the distance between the mesio-occlusal rest and the anterior terminal of the finger extension. In this instance, although the extension rests on a prepared surface, it is used in conjunction with a terminal rest on the mesial marginal ridge of the premolar tooth. Even when they are not used as indirect retainers, canine extensions, continu-ous bar retainers, and linguoplates should never be used without terminal rests because of the resultant forces effective when they are placed on inclined planes alone.
Cingulum Bars (Continuous Bars) and Linguoplates Technically, cingulum bars (continuous bars) and lin-guoplates are not indirect retainers because they rest on unprepared lingual inclines of anterior teeth. The indirect retainers are actually the terminal rests at either end that occur in the form of auxiliary occlusal rests or canine rests (see Chapter 5).
In Class I and Class II partial dentures, a cingulum bar or linguoplate may extend the effectiveness of the indirect retainer if it is used with a terminal rest at each end. In tooth-supported partial dentures, a cingulum bar or lin-guoplate is placed for other reasons but always with ter-minal rests (see Chapter 5).
In Class I and Class II partial dentures especially, a continuous bar retainer or the superior border of the lin-guoplate should never be placed above the middle third of the teeth so that orthodontic movement during the rota-tion of a distal extension denture is avoided. This guide-line is not so important when the six anterior teeth are in nearly a straight line, but when the arch is narrow and tapering, a cingulum bar or linguoplate on the anterior teeth extends well beyond the terminal rests, and orth-odontic movement of those teeth is more likely. Although these are intended primarily to stabilize weak anterior teeth, they may have the opposite effect if not used with discretion.
Figure 8-6 Mandibular Class II design showing a favorable location for the indirect retainer on the mesio-occlusal of the first premolar #28. This location is at 90 degrees to the fulcrum line between primary rests—disto-occlusal (DO) of #20 and DO of #31—and provides efficient resistance to a denture base lift based on the longest distance to resistant rest support and because the occlusal rest is perpendicular to the load.
Figure 8-7 Mandibular Class I design using canine exten-sions from occlusal rests as indirect retainers. The canine exten-sion must be placed on prepared rest seats so that resistance will be directed as nearly as possible along the long axis of the canine abutment.
www.konkur.in 98 Part I General Concepts/Treatment Planning Modification Areas Occasionally, the occlusal rest on a secondary abutment in a Class II partial denture may serve as an indirect retainer. This use will depend on how far from the ful-crum line the secondary abutment is located.
The primary abutments in a Class II, modification 1, partial denture are the abutment adjacent to the distal extension base and the most distal abutment on the tooth-supported side. The fulcrum line is a diagonal axis between the two terminal abutments (Figure 8-8).
The anterior abutment on the tooth-supported side is a secondary abutment, which serves to support and retain one end of the tooth-supported segment and adds horizontal stabilization to the denture. If the modification space were not present, as in an unmodified Class II arch, auxiliary occlusal rests and stabilizing components in the same position would still be essential to the design of the denture (Figure 8-9). However, the presence of a modifi-cation space conveniently provides an abutment tooth for support, stabilization, and retention.
If the occlusal rest on the secondary abutment lies far enough from the fulcrum line, it may serve adequately as an indirect retainer. Its dual function then is tooth support for one end of the modification area and support for an indirect retainer. The most typical example is a distal occlu-sal rest on a first premolar when a second premolar and the first molar are missing and the second molar serves as one of the primary abutments. The longest perpendicular to the fulcrum line falls in the vicinity of the first premolar, making the location of the indirect retainer nearly ideal.
On the other hand, if only one tooth, such as a first molar, is missing on the modification side, the occlusal rest on the second premolar abutment is too close to the fulcrum line to be effective. In such a situation, an auxil-iary occlusal rest on the mesial marginal ridge of the first premolar is needed, both for indirect retention and for support for an otherwise unsupported major connector.
Support for a modification area extending anteriorly to a canine abutment is obtained by any one of the accepted canine rest forms, as previously outlined in Chapter 6. In this situation, the canine tooth provides nearly ideal indi-rect retention and support for the major connector as well.
Rugae Support Some clinicians consider coverage of the rugae area of the maxillary arch as a means of indirect retention because the rugae area is firm and usually well situated to provide indirect retention for a Class I removable partial denture. Although it is true that broad coverage over the rugae area can conceivably provide some support, the facts remain that tissue support is less effective than positive tooth support, and that rugae coverage is undesirable if it can be avoided.
The use of rugae support for indirect retention is usu-ally part of a palatal horseshoe design. Because poste-rior retention is usually inadequate in this situation, the requirements for indirect retention are probably greater than can be satisfied by this type of tissue support alone.
In the mandibular arch, retention from the distal extension base alone is usually inadequate to prevent lift-ing of the base away from the tissues. In the maxillary arch, where only anterior teeth remain, full palatal cover-age is usually necessary. In fact, with any Class I maxillary removable partial denture that extends distally from the first premolar teeth, except when a maxillary torus pre-vents its use, palatal coverage may be used to advantage. Although complete coverage may be seen in the form of a resin base, the added retention and reduced bulk of a cast metal palate make the latter preferable (see Chap-ter 5). However, in the absence of full palatal coverage, an indirect retainer should be used with other designs of major palatal connectors for the Class I removable partial denture.
Figure 8-8 Class II, modification 1, removable partial denture framework. The fulcrum line, when the denture base is displaced toward the residual ridge, runs from the left second premolar to the right second molar. When forces tend to displace the denture away from its basal seat, the supportive element (distal occlusal rest) of the direct retainer assembly on the right first premolar serves as an indirect retainer.
Figure 8-9 Class II maxillary removable partial denture frame-work design. The fulcrum line runs from the patient’s right ca-nine to left second molar. Forces that tend to unseat the denture from its basal seat will be resisted by activation of retentive ele-ments on canine and molar, with the use of supportive elements on the left first premolar as an indirect retainer.
www.konkur.in CHAPTER 9 Denture Base Considerations CHAPTER OUTLINE Functions of Denture Bases in Control of Prosthesis Movement Tooth-Supported Partial Denture Base Distal Extension Partial Denture Base Methods of Attaching Denture Bases Methods for Incorporating Dental Implants Ideal Denture Base Material Advantages of Metal Bases Accuracy and Permanence of Form Comparative Tissue Response Thermal Conductivity Weight and Bulk Methods of Attaching Artificial Teeth Porcelain or Acrylic-Resin Artificial Teeth Attached with Acrylic-Resin Porcelain or Resin Tube Teeth and Facings Cemented Directly to Metal Bases Resin Teeth Processed Directly to Metal Bases Metal Teeth Chemical Bonding Need for Relining Stress-Breakers (Stress Equalizers) The preceding chapters described removable partial denture component parts that engaged the remaining teeth (rests, direct retainers, and indirect retainers) and the cross-arch connector of all component parts, the major connector. This chapter describes the portion of the removable partial den-ture that contains the replacement teeth—the denture base.
FUNCTIONS OF DENTURE BASES IN CONTROL OF PROSTHESIS MOVEMENT The denture base supports the artificial teeth and conse-quently receives the functional forces from occlusion and transfers functional forces to supporting oral structures (Figure 9-1). This function is most critical for the distal extension prosthesis, as functional stability and comfort often relate directly to the ability for this transfer of forces to occur without undue movement.
Although its primary purpose is related to masticatory function, the denture base also may add to the cosmetic effect of the replacement, particularly when techniques for tinting and reproducing natural-looking contours are used. Most of the techniques for creating a natural appearance in complete denture bases are applicable equally well to partial denture bases.
Still another function of the denture base is stimulation of the underlying tissues of the residual ridge. Some verti-cal movement occurs with any denture base, even those sup-ported entirely by abutment teeth, because of the physiologic movement of those teeth under function. It is clearly evident that oral tissues placed under functional stress within their physiologic tolerance maintain their form and tone better than similar tissues suffering from disuse. The term disuse atrophy is applicable to both periodontal tissues and the tis-sues of a residual ridge.
Tooth-Supported Partial Denture Base Denture bases differ in functional purpose and may dif-fer in terms of the material of which they are made. In a www.konkur.in 100 Part I General Concepts/Treatment Planning tooth-supported prosthesis, the denture base is primarily a span between two abutments supporting artificial teeth. Thus occlusal forces are transferred directly to the abutments through rests. Also, the denture base and the supplied teeth serve to prevent horizontal migration of all abutment teeth in the partially edentulous arch and vertical migration of teeth in the opposing arch.
When only posterior teeth are being replaced, esthet-ics is usually only a secondary consideration. On the other hand, when anterior teeth are replaced, esthetics may be of primary importance. Theoretically, the tooth-supported partial denture base that replaces anterior teeth must per-form the following functions: (1) provide desirable esthet-ics; (2) support and retain the artificial teeth in such a way that they provide masticatory efficiency and assist in trans-ferring occlusal forces directly to abutment teeth through rests; (3) prevent vertical and horizontal migration of remaining natural teeth; (4) eliminate undesirable food traps (oral cleanliness); and (5) stimulate the underlying tissues.
When an implant is placed in a distal extension segment of a Class I or Class II removable partial denture, the effect is to create a bound span by supporting the segment at both ends. In this situation the requirements of the denture base become more like a tooth-bound base than a distal extension denture base.
Distal Extension Partial Denture Base In a distal extension partial denture, denture bases that are not in tooth-supported modifications must contribute to the support of the denture. Such support is critical to the goal of minimizing functional movement and improving stability of the prosthesis. Although the abutment teeth provide support for the distal extension base, as the distance from the abut-ment increases, support from the underlying ridge tissues becomes increasingly important. Maximum support from the residual ridge may be obtained by using broad, accu-rate denture bases, which spread the occlusal load equitably over the entire area available for such support. The space available for a denture base is determined by the structures A B C D Figure 9-1 A, Class I maxillary distal extension removable partial denture showing tissue side (intaglio) of denture bases. B, Occlusal side of maxillary prosthesis: posterior artificial teeth are attached to the bases. C, Class II, modification 1, mandibular distal extension removable partial denture shows intaglio of both extension and modification bases. D, Occlusal side of mandibular prosthesis; posterior artificial teeth are attached to the bases. For both prostheses, the bases are extended within the limits of physiologic activity of surround-ing oral structures.
www.konkur.in 101 Chapter 9 Denture Base Considerations surrounding the space and by their movement during func-tion. Maximum support for the denture base therefore can be accomplished only by using knowledge of the limiting anatomic structures and of the histologic nature of the basal seat areas, accuracy of the impression, and accuracy of the denture base (Figure 9-2). The first two of these support fea-tures relate to the gross size and cellular characteristics of the residual ridge tissues. These are highly variable between patients, and consequently not all residual ridges can pro-vide the same quality of support. Therefore the ability to con-trol functional displacement of the distal extension base is a determination that is unique for individual patients.
The snowshoe principle, which suggests that broad cover-age furnishes the best support with the least load per unit area, is the principle of choice for providing maximum sup-port. Therefore support should be the primary consideration in selecting, designing, and fabricating a distal extension partial denture base. Of secondary importance (but to be considered nevertheless) are esthetics, stimulation of the underlying tissues, and oral cleanliness. Methods used to accomplish maximum support of the restoration through its base(s) are presented in Chapters 16 and 17.
In addition to their difference in functional purposes, denture bases vary in material of fabrication. This difference is directly related to their function because of the need for some dentures to be relined.
Because the tooth-supported base has an abutment tooth at each end on which a rest has been placed, future relin-ing or rebasing may not be necessary to reestablish sup-port. Relining is necessary only when tissue changes have occurred beneath the tooth-supported base to the point that poor esthetics or accumulation of debris results. For these reasons alone, tooth-supported bases made soon after extractions should be of a material that permits later relin-ing. Such materials are the denture resins, the most common of which are copolymer and methyl methacrylate resins.
Primary retention for the removable partial denture is accomplished mechanically by placing retaining elements on the abutment teeth. Secondary retention is provided by the intimate relationship of denture bases and major connectors (maxillary) with the underlying tissues. The latter is similar to the retention of complete dentures and is proportionate to the accuracy of the impression registration, the accuracy of the fit of the denture bases, and the total area of contact involved.
Retention of denture bases has been described as the result of the following forces: (1) adhesion, which is the attraction of saliva to the denture and tissues; (2) cohesion, which is the attraction of the molecules of saliva to each other; (3) atmo-spheric pressure, which is dependent on a border seal and results in a partial vacuum beneath the denture base when a dislodging force is applied; (4) physiologic molding of the tissues around the polished surfaces of the denture; and (5) the effects of gravity on the mandibular denture.
Boucher, writing on the subject of complete denture impressions, described these forces as follows: Adhesion and cohesion are effective when there is perfect apposition of the impressioned surface of the denture to the mucous membrane surfaces. These forces lose their effective-ness if any horizontal displacement of the dentures breaks the continuity of this contact. Atmospheric pressure is ef-fective primarily as a rescue force when extreme dislodging forces are applied to the denture. It depends on a perfect From Boucher CO: Complete denture impression based upon the anatomy of the mouth, J Am Dent Assoc 31:117-1181, 1994.
B A Figure 9-2 Maxillary and mandibular distal extension removable partial dentures with resin denture bases. Bases are extended buc-cally within the physiologic tolerance of border structures. A, Maxillary denture bases cover both the maxillary tuberosities, extend into the pterygomaxillary notches and provide for adaptation along the posterior border, taking care not to extend beyond the soft palatal flexure. B, Mandibular bilateral distal extension removable partial denture bases cover the retromolar pads and extend into the retromy-lohyoid fossae. The impression procedure used established buccal shelves as primary stress-bearing areas of basal seats.
www.konkur.in 102 Part I General Concepts/Treatment Planning border seal to keep the pressure applied on only one side of the denture. The presence of air on the impression surface would neutralize the pressure of the air against the polished surface. Because each of these forces is directly proportional to the area covered by the dentures, the dentures should be extended to the limits of the supporting structures.
The molding of the soft tissues around the polished sur-faces of denture bases helps to perfect the border seal. Also, it forms a mechanical lock at certain locations on the dentures, provided these surfaces are properly prepared. This lock is developed automatically and without effort by the patient if the impression is made with an understanding of the ana-tomic possibilities.
Few partial dentures are made without some mechanical retention. Retention from the denture bases may contribute significantly to the overall retention of the partial denture and therefore must not be discounted. Denture bases should be designed and fabricated so that they will contribute as much retention to the partial denture as possible. However, it is questionable whether atmospheric pressure plays as important a role in retention of removable partial dentures because a border seal cannot be obtained as readily as it can be with complete dentures. Therefore adhesion and cohesion gained by excellent apposition of the denture base and soft tissues of the basal seat play an important retentive role.
METHODS OF ATTACHING DENTURE BASES Acrylic-resin bases are attached to the partial denture frame-work by means of a minor connector designed so that a space exists between the framework and the underlying tissues of the residual ridge (Figure 9-3). Relief of at least a 20-gauge thickness over the basal seat areas of the master cast is used to create a raised platform on the investment cast on which the pattern for the retentive frame is formed (Figure 9-4). Thus after casting, the portion of the retentive framework to which the acrylic-resin base will be attached will stand away from the tissue surface sufficiently to permit a flow of acrylic-resin base material beneath its surface.
The retentive framework for the base should be embed-ded in the base material with sufficient thickness of resin (1.5 mm) to allow for relieving if this becomes necessary during the denture adjustment period or during relining procedures. Thickness is also necessary to avoid weakness and subsequent fracture of the acrylic-resin base material surrounding the metal framework.
The use of plastic mesh patterns in forming the retentive framework is generally less satisfactory than a more open framework (see Figure 9-4). Less weakening of the resin by the embedded framework results from use of the more open form. Pieces of 12- or 14-gauge half-round wax and 18-gauge round wax are used to form a ladderlike framework rather than the finer latticework of the mesh pattern. The precise design of the retentive framework, other than that it should be located both buccally and lingually, is not as important as its effective rigidity and strength when it is embedded in the acrylic resin base. It should also be free of interference with future adjustment, should not interfere with arrangement of artificial teeth, and should be open enough to avoid weak-ening any portion of the attached acrylic-resin. Designing the retentive framework for denture bases by having ele-ments located buccal and lingual to the residual ridge not only will strengthen the acrylic-resin base but also will mini-mize distortion of the base created by the release of inherent strains in the acrylic-resin base during use or storage of the restoration (Figure 9-5).
Figure 9-3 Mandibular Class II, modification 1, wax pattern developed on an investment cast. Adequate provision is made for attaching the resin base to the major connector on the eden-tulous side by way of a ladderlike minor connector and a butt-type joint. A similar minor connector design will be used for modifica-tion space. Note: Relief space beneath the minor connectors is established by relief wax placed on the original master cast and duplicated in this refractory cast. This allows processed resin to surround the minor connectors in creating the denture base.
Figure 9-4 Unlike the minor connector designs in Figure 9-3, the designs used for this prosthesis have a plastic mesh pattern. Although such designs can be reinforced to be more rigid, the bulk of the connector itself may contribute to weakening of the resin base. A more open type of minor connector seems preferable.
www.konkur.in 103 Chapter 9 Denture Base Considerations Metal bases are usually cast as integral parts of the par-tial denture framework. Mandibular metal bases may be assembled and attached to the framework with acrylic-resin (Figure 9-6).
METHODS FOR INCORPORATING DENTAL IMPLANTS When dental implants are placed, modifications can be required in the denture base and the specific modification is dependent on how the implants will be engaged.
Implants placed anteriorly, to enhance retention by removing the need for a visible clasp, must take into account retentive device bulk and connection require-ments for joining to the prosthesis (i.e., direct to the metallic frame or via embedding within the acrylic resin denture base).
Placement of the implant should consider minimizing palatal/lingual bulk, provision of adequate space for housing, acrylic resin and denture tooth, and alignment coincident with the insertion path of the prosthesis.
Modifications required for implants placed more distal, for purposes of support, typically require designing space for embedding an attachment housing within the denture base (Figure 9-7).
IDEAL DENTURE BASE MATERIAL The requirements for an ideal denture base are as follows: 1. Accuracy of adaptation to the tissues, with minimal volume change 2. Dense, nonirritating surface capable of receiving and maintaining a good finish 3. Thermal conductivity 4. Low specific gravity; lightweight in the mouth 5. Sufficient strength; resistance to fracture or distortion 6. Easily kept clean 7. Esthetic acceptability 8. Potential for future relining 9. Low initial cost Such an ideal denture base material does not exist, nor is it likely to be developed in the near future. However, any denture base, whether of resin or metal and regardless of the method of fabrication, should come as close to this ideal as possible.
Figure 9-5 Note that minor connectors by which resin den-ture bases will be attached to the framework are open, ladderlike configurations that extend on both buccal and lingual surfaces. This not only provides excellent attachment of the resin bases but minimizes warping of bases resulting from the release of inher-ent strains in compression-molded resin.
Figure 9-6 Cast distal extension base of a maxillary remov-able partial denture. Cast bases are an integral part of the frame-work and not only provide support to the prosthetic dentition but reinforce framework rigidity.
www.konkur.in 104 Part I General Concepts/Treatment Planning ADVANTAGES OF METAL BASES Except for those edentulous ridges with recent extrac-tions, metal can be used for tooth-supported bases and is thought to provide several advantages. Its principal dis-advantages are that it is difficult to adjust and reline. A commonly stated advantage is that the stimulation it gives to the underlying tissues is so beneficial that it prevents some alveolar atrophy that would otherwise occur under a resin base and thereby prolongs the health of the tissues that it contacts. Some of the advantages of a metal base are discussed in the following sections.
Accuracy and Permanence of Form Cast metal bases, whether of gold, chrome, or titanium alloys, not only may be cast more accurately than denture resins but also can maintain their accuracy of form with-out changes in the mouth. Internal strains that may be released later to cause distortion are not present. Although some resins and some processing techniques are superior to others in accuracy and permanence of form, modern cast alloys are generally superior in this respect. Evidence of this fact is that an additional posterior palatal seal may be eliminated entirely when a cast palate is used for a complete denture, as compared with the need for a defi-nite post-dam when the palate is made of acrylic-resin. Distortion of an acrylic-resin base is manifest in the max-illary denture by a distortion away from the palate in the midline and toward the tuberosities on the buccal flanges. The greater the curvature of the tissues, the greater is this distortion. Similar distortions occur in a mandibular den-ture but are more difficult to detect. Accurate metal cast-ings are not subject to distortion by the release of internal strains, as are most denture resins.
Because of its accuracy, the metal base provides an intimacy of contact that contributes considerably to the retention of a denture. Sometimes called interfacial sur-face tension, direct retention from a cast denture base is significant in proportion to the area involved. This has been mentioned previously as an important factor in both direct and direct-indirect retention of maxillary res-torations. Such intimacy of contact is not possible with acrylic-resin bases.
Permanence of form of the cast base is also ensured because of its resistance to abrasion from denture cleaning agents. Cleanliness of the denture base should be stressed; C A B Figures 9-7 A. Intaglio surface view of a Locator retentive element in the posterior aspect of a distal extension base of a mandibular removable partial denture. B. Mirror view of the Locator attachment on an implant in the distal mandible. The attachment will provide support and retention for the distal extension removable partial denture. C. Occlusal view of the locator attachment.
www.konkur.in 105 Chapter 9 Denture Base Considerations however, constant brushing of the tissue side of the acrylic-resin denture base inevitably causes some loss of accuracy by abrasion. Intimacy of contact, which is never as great with an acrylic-resin base as with a metal base, is therefore jeopardized further by cleaning habits. The metal bases, particularly the harder chrome alloys, withstand repeated cleaning without significant changes in surface accuracy.
Comparative Tissue Response Clinical observations have demonstrated that the inherent cleanliness of the cast metal base contributes to the health of oral tissues when compared with an acrylic-resin base. Perhaps some of the reasons for this are the greater density and the bacteriostatic activity contributed by ionization and oxidation of the metal base. Acrylic-resin bases tend to accumulate mucinous deposits containing food particles, as well as calcareous deposits. Unfavorable tissue reac-tion to decomposing food particles and bacterial enzymes and mechanical irritation from calculus may result if the denture is not kept meticulously clean. Although calculus, which must be removed periodically, can precipitate on a cast metal base, other deposits do not accumulate as they do on an acrylic-resin base. For this reason, a metal base is naturally cleaner than an acrylic-resin base.
Thermal Conductivity Temperature changes are transmitted through the metal base to the underlying tissues, thereby helping to main-tain the health of those tissues. Freedom of interchange of temperature between the tissues covered and the sur-rounding external influences (temperature of liquids, solid foods, and inspired air) contributes much to the patient’s acceptance of a denture and may help avoid the feeling of the presence of a foreign body. Conversely, den-ture acrylic resins have insulating properties that prevent interchange of temperature between the inside and the outside of the denture base.
Weight and Bulk Metal alloy may be cast much thinner than acrylic-resin and still have adequate strength and rigidity. Cast gold must be given slightly more bulk to provide the same amount of rigidity but still may be made with less thick-ness than acrylic-resin materials. Even less weight and bulk are possible when the denture bases are made of chrome or titanium alloys.
At times, however, both weight and thickness may be used to advantage in denture bases. In the man-dibular arch, the weight of the denture may be an asset with regard to retention; and for this reason, a cast gold base may be preferred. On the other hand, extreme loss of residual alveolar bone may make it necessary to add fullness to the denture base to restore normal facial con-tours and to fill out the buccal vestibule to prevent food from being trapped in the vestibule beneath the denture. In such situations, an acrylic-resin base may be preferred to the thinner metal base.
In the maxillary arch, an acrylic-resin base may be preferred to the thinner metal base to provide fullness in the buccal flanges or to fill a maxillary buccal vesti-bule. Acrylic-resin may also be preferred over the thin-ner metal base for esthetic reasons. In these instances, the thinness of the metal base may be of no advantage, but in areas where the tongue and cheek need maximum room, thinness may be desirable.
Denture base contours for functional tongue and cheek contact can best be accomplished with acrylic-resin. Metal bases are usually made thin to minimize bulk and weight, whereas acrylic-resin bases may be con-toured to provide ideal polished surfaces that contribute to retention of the denture, to restore facial contours, and to avoid the accumulation of food at denture borders. Lingual surfaces usually are made concave except in the distal palatal area. Buccal surfaces are made convex at gin-gival margins, over root prominences, and at the border to fill the area recorded in the impression. Between the border and the gingival contours, the base can be made convex to aid in retention and to facilitate return of the food bolus to the occlusal table during mastication. Such contours prevent food from being entrapped in the cheek and from working under the denture. This usually cannot be accomplished with metal bases.
However, the advantages of a metal base need not be sacrificed for the sake of esthetics or desirable denture contours when the use of such a base is indicated. Den-ture bases may be designed to provide almost total metal-lic coverage, yet they have resin borders to avoid a display of metal and to add buccal fullness when needed (Figure 9-8). The advantages of thermal conductivity are not nec-essarily lost by covering a portion of the metal base as long as other parts of the denture are exposed to effect temperature changes through conduction.
Figure 9-8 Partial metal bases used with a palatal strap and resin denture teeth attached directly to the cast metal bases. If needed, a buccal flange of resin could be added to such a base re-gion; however, for these small spans no such flange was needed.
www.konkur.in 106 Part I General Concepts/Treatment Planning METHODS OF ATTACHING ARTIFICIAL TEETH Selection of artificial teeth for form, color, and material must precede attachment to the denture. Artificial teeth may be attached to denture bases by the several means that follow: with acrylic resin, with cement, processed directly to metal, cast with the framework, and by chemical bonding. Use of acrylic-resin to attach artificial teeth to a denture base is the most common method.
Porcelain or Acrylic-Resin Artificial Teeth Attached with Acrylic-Resin Artificial porcelain teeth are mechanically retained. The posterior teeth are retained by acrylic-resin in their diatoric holes. The anterior porcelain teeth are retained by acrylic-resin surrounding their lingually placed retention pins. Arti-ficial resin teeth are retained by a chemical union with the acrylic-resin of the denture base that occurs during labora-tory processing procedures.
Attachment of acrylic-resin to the metal base may be accomplished by nailhead retention, retention loops, or diagonal spurs placed at random. Attachment mechanisms should be placed so that they will not interfere with place-ment of the teeth on the metal base.
Any junction of acrylic-resin with metal should occur at an undercut finishing line or should be associated with some retentive undercut. Because only a mechanical attachment exists between metal and acrylic-resin, every attempt should be made to avoid separation and seepage, which result in dis-coloration and uncleanliness. Denture odors are frequently caused by accretions at the junction of the acrylic-resin with metal when only a mechanical union exists. Separation that occurs between the acrylic-resin and the metal can eventu-ally lead to loosening of the acrylic-resin base.
Porcelain or Resin Tube Teeth and Facings Cemented Directly to Metal Bases Some disadvantages of this type of attachment are the dif-ficulties involved in obtaining satisfactory occlusion, the lack of adequate contours for functional tongue and cheek contact, and the unesthetic display of metal at gingival margins. The latter is avoided when the tooth is butted directly to the residual ridge, but then the retention for the tooth frequently becomes inadequate.
A modification of this method is the attachment of ready-made acrylic-resin teeth to the metal base with acrylic-resin of the same shade (Figure 9-9). This is called pressing on a resin tooth and is not the same as using acrylic-resin for cementation. It is particularly applicable to anterior replacements, because it is desir-able to know in advance of making the casting that the shade and contours of the selected tooth will be accept-able. After a labial index of the position of the teeth is made, the lingual portion of the tooth may be cut away or a post-hole prepared in the tooth for retention on the casting. Subsequently, the tooth is attached to the B A C Figure 9-9 Tissue surface of Class IV removable partial denture in which artificial dentition was added to a metal base. A, Teeth were set before completion of the framework to allow the modification space design to incorporate altered teeth. B, Ante-rior teeth were adjusted to the ridge, creating a ridge lap; then the framework was waxed to accommodate a custom tooth position. C, Metal reinforcement adds strength to the artificial teeth and protects against dislodgment.
www.konkur.in 107 Chapter 9 Denture Base Considerations denture with acrylic-resin of the same shade. Because this is done under pressure, the acrylic-resin attachment is comparable with the manufactured tooth in hardness and strength.
Tube or side-groove teeth must be selected in advance of waxing the denture framework (Figure 9-10). However, for best occlusal relationships, jaw relation records always should be made with the denture casting in the mouth. This problem may be solved by selecting tube teeth for width but with occlusal surfaces slightly higher than will be necessary. The teeth are ground to fit the ridge with sufficient clearance beneath, for a thin metal base, and beveled to accommodate a boxing of metal. If an acrylic-resin tube tooth is used, the diatoric hole should be made slightly larger than that provided. The cast-ing is completed and tried in, occlusal relationships are recorded, and then the teeth are ground to harmonious occlusion with the opposing dentition. As is discussed in Chapter 18, artificial posterior teeth on partial dentures can hardly ever be used unaltered but rather should be considered material from which occlusal forms may be created to function harmoniously with the remaining natural occlusion.
Resin Teeth Processed Directly to Metal Bases Modern cross-linked copolymers enable the dentist or technician to process acrylic-resin teeth that have satisfactory hardness and abrasion resistance for many sit-uations. Thus occlusion may be created without resorting to the modification of ready-made artificial teeth (Figure 9-11). Recesses in the denture pattern may be carved by hand or may be created around manufactured teeth that are used only to form the recess in the pattern. Occlu-sal relationships may be established in the mouth on the denture framework and transferred to an articulator. The teeth can then be carved and processed in acrylic-resin of the proper shade to fit the opposing occlusal record. Bet-ter attachment to the metal base than by cementation is thus possible. In addition, unusually long, short, wide, or narrow teeth may be created when necessary to fill spaces not easily filled by the limited selection of commercially-available teeth.
Occlusion on acrylic-resin teeth may be reestablished to compensate for wear or settling by reprocessing new acrylic-resin or using light-activated acrylic-resin when this becomes necessary. A distinction always should be made between the need for relining to reestablish occlu-sion (on a distal extension partial denture) and the need for rebuilding occlusal surfaces on an otherwise satisfac-tory base (on a tooth-supported or a tooth- and tissue-supported partial denture).
Reestablishment of occlusion may also be accom-plished by placing cast gold or other suitable cast alloy restorations on existing resin teeth. Although this may Tapered hole Lingual collar 45-degree bevel Figure 9-10 Stock porcelain or resin tube tooth, or artificial tooth used as tube tooth, should be ground to accommodate cast coping as illustrated. A hole is drilled from the underside of the tooth, or, if a hole is already present, it is made larger. Then the tooth is ground to fit the ridge with enough clearance for minimum thickness of metal. A 45-degree bevel then is formed around the base of the tooth; finally, a collar is created on the lingual side, extending to the interproximal area. The tooth is then lubricated, and a wax pattern for the denture base is formed around it.
www.konkur.in 108 Part I General Concepts/Treatment Planning be done on porcelain teeth as well, it is difficult to cut preparations in porcelain teeth unless air abrasive meth-ods are used. Therefore, if later additions to occlusal sur-faces are anticipated, acrylic-resin teeth should be used, thereby facilitating the addition of new resin or cast gold surfaces. A simple technique that can be used to fabricate cast gold occlusal surfaces and attach them to resin teeth is described in Chapter 19.
Metal Teeth Occasionally, a second molar tooth may be replaced as part of the partial denture casting (Figure 9-12). This is usually done when space is too limited for the attachment of an artificial tooth and yet the addition of a second molar is desirable to prevent extrusion of an opposing second molar. Because the occlusal surface must be waxed before casting, perfect occlusion is not possible. Because metal, particularly a chrome alloy, is abrasion resistant, the area of occlusal contact should be held to a minimum to avoid damage to the periodontium of the opposing tooth and associated discomfort to the patient. Occlusal adjustment on gold occlusal surfaces is readily accomplished, whereas metal teeth made of chrome alloys are difficult to adjust and are objectionably hard for use as occlusal surfaces. Therefore they should be used only to fill a space and to prevent tooth extrusion.
Chemical Bonding Recent developments in resin bonding have provided a means of direct chemical bonding of acrylic-resin to metal frameworks. The investing alveolar and gingival tissue replacement components can be attached with-out the use of loops, mesh, or surface mechanical locks. Sections of a metal framework that are used to support replacement teeth can be roughened with abrasives and then treated with a vaporized silica coating. On this sur-face, an acrylic-resin bonding agent is applied, followed by a thin film of acrylic-resin that acts as a substrate for later attachment of replacement acrylic-resin teeth or for processing of acrylic-resin tissue replacements (Figure 9-13).
A second method of fusing a microscopic layer of ceramic to the metal is accomplished by a process referred to as tribochemical coating. This system involves sandblasting the metal framework with a special silica particle material, Rocatec-Plus (3M Espe Dental Prod-ucts, Irvine, CA). Silica from these particles is attached to the framework by impact. A silane is added to this ceramic-like film to form a chemical bond between the silicate layer and the denture base acrylic-resin. Den-ture base acrylic-resins formulated with 4-meta are also available and provide a mechanism for bonding acrylic-resin to metal.
B A Figure 9-11 Direct attachment of resin teeth to metal bases. A, Anterior modification space prepared to receive a denture tooth, which will be reinforced with a post. B, Posterior molar attached to a previously waxed base to specifically receive the replacement tooth.
Figure 9-12 Maxillary cast molar designed as an integral part of the framework. Interocclusal space limitation necessitated the use of metal rather than another form of artificial posterior teeth. Note overlays on premolar and molar teeth were used to resist tooth wear. (Courtesy of Dr. C.J. Andres, Indianapolis, IN.) www.konkur.in 109 Chapter 9 Denture Base Considerations NEED FOR RELINING The distal extension base differs from the tooth-supported base in several respects, one of which is that it must be made of a material that can be relined or rebased when it becomes necessary to reestablish tissue support for the distal extension base. Therefore acrylic-resin denture base materials that can be relined are generally used.
Although satisfactory techniques for making distal exten-sion partial denture bases of cast metal are available, the fact that metal bases are difficult if not impossible to reline limits their use to stable ridges that will change little over a long period.
Loss of support for distal extension bases results from changes in residual ridge form over time. These changes may not be readily visible; however, manifestations of this change can be assessed. One of these is loss of occlusion between the distal extension denture base and the opposing dentition, which increases as the distance from the abutment increases (Figure 9-14). This change is proved by having the patient close on strips of 28-gauge green casting wax, or any simi-lar wax, and tapping in centric relation only. Indentations in a wax strip of known thickness are quantitative, whereas marks made with articulating ribbon are only qualitative. In other words, indentations in the wax may be interpreted as light, medium, or heavy, whereas it is difficult if not impos-sible to interpret a mark made with articulating ribbon as light or heavy. In fact, the heaviest occlusal contact may per-forate paper-articulating ribbon and make a lesser mark than areas of lighter contact. Therefore the use of any articulating ribbon is of limited value in checking occlusion intraorally. In making occlusal adjustments, articulating ribbon should be used only to indicate where relief should be provided after the need for relief has been established by using wax strips of known thickness. For this purpose, 28-gauge green or blue casting wax is generally used, although the thinner 30-gauge or the thicker 26-gauge wax may be used for better evalua-tion of the clearance between areas not in contact.
Loss of support for a distal extension base results in loss of occlusal contact between the prosthetically supplied teeth and the opposing dentition and a return to heavy occlu-sal contact between the remaining natural teeth. Usually this is an indication that relining is needed to reestablish B A C Figure 9-13 Coating of metal framework with silica facilitates improved application and seal of resin or composite for denture base regions. Example shows a maxillary minor with beads for added retention of the anterior denture base (A), after air abra-sion preparation of the surface (B), and after sili-coating (C).
Figure 9-14 28-gauge soft green wax used to identify occlusal contacts between a mandibular Class I distal extension base and an opposing complete denture.
www.konkur.in 110 Part I General Concepts/Treatment Planning the original occlusion by reestablishing supporting contact with the residual ridge. It must be remembered, however, that occlusion on a distal extension base is sometimes main-tained at the expense of extrusion of the opposing natural teeth. In such a situation, checking the occlusion alone will not show that settling of the extension base has occurred because changes in the supporting ridge may have also taken place.
A second manifestation of change also must be observ-able to justify relining. This second manifestation of change in the supporting ridge is evidence of rotation about the ful-crum line with the indirect retainers lifting from their seats as the distal extension base is pressed against the ridge tis-sues. Originally, if the distal extension base was made to fit the supporting form of the residual ridge, rotation about the fulcrum line is not visible. At the time the denture is initially placed, no anteroposterior rotational movement should occur when alternating finger pressure is applied to the indi-rect retainer and the distal end of a distal extension base or bases. After changes in the ridge form have occurred, which cause some loss of support, rotation occurs about the ful-crum line when alternating finger pressure is applied. This indicates changes in the supporting ridge that must be com-pensated for by relining or rebasing.
If occlusal contact has been lost and rotation about the fulcrum line is evident, relining is indicated. On the other hand, if occlusal contact has been lost without any evidence of denture rotation, and if stability of the denture base is oth-erwise satisfactory, reestablishing the occlusion is the rem-edy, rather than relining. For the latter, the original denture base may be used in much the same manner as the original trial base was used to record the occlusal relation. Teeth may then be reoriented to an opposing cast or to an occlusal tem-plate with the use of light-activated tooth-colored acrylic-resin, tooth-colored composite, cast occlusal surfaces, or new teeth. In any event, a new occlusion should be estab-lished on the existing bases. Relining in this instance would be the wrong solution to the problem.
Loss of support may also be assessed clinically by other methods. A layer of rather free-flowing irreversible hydro-colloid, wax, or tissue-conditioning material can be spread on the basal seat portion of the dried denture base(s), and the restoration returned to the patient’s mouth. Care is exer-cised to ensure that the framework is correctly seated (rests and indirect retainers in planned positions). The restoration is removed when the material has set. Significant thicknesses of material remaining under the bases indicate lack of inti-mate contact of the bases with the residual ridges, suggesting a need for relining.
More often, however, loss of occlusion is accompanied by settling of the denture base to the extent that rotation about the fulcrum line is manifest. Because relining is the only remedy short of making completely new bases, use of an acrylic-resin base originally facilitates later relining. For this reason, acrylic-resin bases are generally preferred for distal extension partial dentures.
The question remains as to when, if ever, metal bases with their several advantages may be used for distal extension partial dentures. It is debatable as to what type of ridge will most likely remain stable under functional loading without apparent change. Certainly the age and general health of the patient will influence the ability of a residual ridge to sup-port function. Minimum and harmonious occlusion and the accuracy with which the base fits the underlying tissues will influence the amount of trauma that will occur under func-tion. The absence of trauma plays a big part in the ability of the ridge to maintain its original form.
The best indication for the use of metal distal extension bases is a ridge that has supported a previous partial denture without having become narrowed or flat or without con-sisting primarily of easily displaceable tissues. When such changes have occurred under a previous denture, further change may be anticipated because of the possibility that the oral tissues in question are not capable of supporting a den-ture base without further change. Despite every advantage in their favor, some individuals have ridges that respond unfa-vorably when called on to support any denture base.
In other instances, such as when a new partial denture is to be made because of the loss of additional teeth, the ridges may still be firm and healthy. Because the ridges have previ-ously supported a denture base and have sustained occlusion, bony trabeculae will have become arranged to best support vertical and horizontal loading, cortical bone may have been formed, and tissues will have become favorable for continued support of a denture base.
In only a relatively few situations is the need for relining a distal extension base not anticipated; metal bases may be considered. However, many instances may be considered borderline. In these cases, metal bases may be used with full understanding on the part of the patient that a new denture may become necessary if unforeseen tissue changes occur. The technique shown in Figure 9-6 permits replacement of metal distal extension bases without the need to remake the entire denture. This method should be seriously considered any time a distal extension partial denture is to be made with a metal base or bases.
For reasons previously outlined, the possibility that tis-sues will remain healthier beneath a metal base than they will beneath an acrylic-resin base may justify wider use of metal bases for distal extension partial dentures. Through careful treatment planning, better patient education on the problems involved in making a distal extension base partial denture, and greater care taken in the fabrication of denture bases, metal may be used to advantage in some situations in which acrylic-resin bases are ordinarily used.
STRESS-BREAKERS (STRESS EQUALIZERS) The previous chapters describing component parts of a partial denture have presumed absolute rigidity of all parts of the partial denture framework except the reten-tive arm of the direct retainer. All vertical and horizontal www.konkur.in 111 Chapter 9 Denture Base Considerations forces applied to the artificial teeth are thus distributed throughout the supporting portions of the dental arch. Broad distribution of stress is accomplished through the rigidity of the major and minor connectors. The effects of the stabilizing components are also made possible by the rigidity of the connectors.
In distal extension situations, the use of a rigid con-nection between the denture base and supporting teeth must account for base movement without causing tooth or tissue damage. In such situations, stress on the abut-ment teeth and residual ridge is minimized through the use of functional basing, broad coverage, harmonious occlusion, and correct choice of direct retainers. Gener-ally, two major types of clasp assemblies are used for distal extensions because of their stress-breaking design. Reten-tive clasp arms may be cast only if they engage undercuts on the abutment teeth in such a manner that tissue-ward movement of the extension base transmits only minimum leverage to the abutment. Otherwise, tapered wrought-wire retentive clasp arms should be used because of their greater flexibility. The tapered, round wrought-wire clasp arm acts somewhat as a stress-breaker between the den-ture base and the abutment tooth by reducing the effects of denture base movement on the tooth through its flexibility.
Another concept of stress-breaking insists on sepa-rating the actions of the retaining elements from the movement of the denture base by allowing indepen-dent movement of the denture base (or its supporting framework) and the direct retainers. This form of stress-breaker, also referred to as a stress equalizer, has been used as a means of compensating for inappropriately designed removable partial dentures. Figure 9-15 shows an exam-ple of a split bar major connector, which is a commonly used stress breaker.
Regardless of their design, most all stress-breakers effectively dissipate vertical stresses, which is the pur-pose for which they are used. However, this occurs at the expense of horizontal stability and the harmful effects of reduced horizontal stability (excessive ridge resorption, tissue impingement, inefficient mastica-tion), which far outweigh the benefits of vertical stress-breaking. It is the rigid nature of the more conventional removable partial denture that allows satisfaction of all requirements for support, stability, and retention with-out overemphasis on only one principle to the detri-ment of the oral tissues.
The student is referred to two textbooks that describe in detail the use of stress-breakers and articulated partial denture designs: (1) Precision Attachments in Dentistry, ed 3, by H.W. Preiskel; and (2) Theory and Practice of Preci-sion Attachment Removable Partial Dentures, by J.L. Baker and R.J. Goodkind.
B A Figure 9-15 With occlusal loading of posterior distal extensions (A), the superior edge of the mandibular lingual plate major connec-tor is displaced from the planned contact (B). If appropriate contact of the major connector and teeth does not return after release of the distal load, this suggests ridge support is poor and occlusal function is suboptimum. The denture base should be considered for a reline to reestablish occlusion contact, ensuring functional use.
www.konkur.in CHAPTER 10 Principles of Removable Partial Denture Design CHAPTER OUTLINE Difference in Prosthesis Support and Influence on Design Differentiation Between Two Main Types of Removable Partial Dentures Differences in Support Impression Registration Differences in Clasp Design Essentials of Partial Denture Design Components of Partial Denture Design Tooth Support Ridge Support Major and Minor Connectors Direct Retainers for Tooth-Supported Partial Dentures Direct Retainers for Distal Extension Partial Dentures Stabilizing Components Guiding Plane Indirect Retainers Implant Considerations in Design Examples of Systematic Approach to Design Required Tooth Modification for Removable Partial Dentures Implant Considerations Kennedy Class II Removable Partial Dentures Class III Removable Partial Denture Kennedy Class I, Bilateral, Distal Extension Removable Partial Dentures Additional Considerations Influencing Design Use of a Splint Bar for Denture Support Internal Clip Attachment Overlay Abutment as Support for a Denture Base Use of a Component Partial to Gain Support DIFFERENCE IN PROSTHESIS SUPPORT AND INFLUENCE ON DESIGN Some of the biomechanical considerations of removable par-tial denture (RPD) design were presented in Chapter 4. The strategy of selecting component parts for a partial denture to help control movement of the prosthesis under functional load has been highlighted as a method to be considered for logical partial denture design. The requirements for move-ment control are generally functions of whether the prosthe-sis will be tooth supported or tooth-tissue supported.
For a tooth-supported prosthesis, the movement poten-tial is less because resistance to functional loading is pro-vided by the teeth. Teeth do not vary widely in their ability to provide this support; consequently, designs for prosthe-ses are less variable. This is the case even though the amount of supporting bone, the crown-to-root ratios, the crown and root morphologies, and the tooth number and position in the arch relative to edentulous spaces are well established and may be variable for tooth- and tooth-tissue–supported RPDs. For a tooth-tissue–supported prosthesis, the residual ridge (remaining alveolar bone and overlying connective tissue covered with mucosa) presents with variable poten-tial for support. Not only does the underlying alveolar bone demonstrate a highly variable form following extraction, it continues to change with time. As alveolar bone responds to the loss of teeth, the overlying connective tissue and mucosa undergo change that places the soft tissue at risk for pressure-induced inflammatory changes. This variable tissue support potential adds complexity to design consid-erations when one is dealing with tooth-tissue–supported prostheses. This occurs because unlike the efficient sup-port provided by teeth, which results in limited prosthesis movement, the reaction of the ridge tissue to functional forces can be highly variable, leading to variable amounts of prosthesis movement. An understanding of the potential www.konkur.in 113 Chapter 10 Principles of Removable Partial Denture Design sources of functional force from the opposing arch that can have an effect on the movement potential of the prosthesis is helpful.
Factors related to the opposing arch tooth position, the existence and nature of prosthesis support in the oppos-ing arch, and the potential for establishing a harmonious occlusion can greatly influence the partial denture design. Opposing tooth positions that apply forces outside the primary support of the prosthesis can introduce leverage forces that act to dislodge the prosthesis. Such an effect is variable and is based on the nature of the opposing occlusion, because the forces of occlusion differ between natural teeth, RPDs, and complete dentures. In general, RPDs opposing natural teeth require greater support and stabilization over time because of the greater functional load demands. Therefore, occlusal relationships at maxi-mum intercuspation should be broadly dissipated to the supporting units.
DIFFERENTIATION BETWEEN TWO MAIN TYPES OF REMOVABLE PARTIAL DENTURES On the basis of the previous discussion, it is clear that two distinctly different types of RPDs exist. Certain points of difference are present between Kennedy Class I and Class II types of partial dentures on the one hand and the Class III type of partial denture on the other. The first consideration is the manner in which each is supported. The Class I type and the distal extension side of the Class II type derive their pri-mary support from tissues underlying the base and second-ary support from the abutment teeth (Figure 10-1, A, and Figure 10-2). The Class III type derives all of its support from the abutment teeth (see Figure 10-1, B, and Figure 10-2).
Second, for reasons directly related to the manner of support, the method of impression registration and the jaw record required for each type will vary.
Third, the need for some kind of indirect retention exists in the distal extension type of partial denture, whereas in the tooth-supported, Class III type, no extension base is pres-ent to lift away from the supporting tissues because of the action of sticky foods and the movements of tissues of the mouth against the borders of the denture. This is so because each end of each denture base is secured by a direct retainer on an abutment tooth. Therefore the tooth-supported partial denture does not rotate about a fulcrum, as does the distal extension partial denture.
Fourth, the manner in which the distal extension type of partial denture is supported often necessitates the use of a base material that can be relined to compensate for tis-sue changes. Acrylic-resin is generally used as a base mate-rial for distal extension bases. The Class III partial denture, on the other hand, which is entirely tooth supported, does not require relining except when it is advisable to elimi-nate an unhygienic, unesthetic, or uncomfortable condition resulting from loss of tissue contact. Metal bases therefore A B Figure 10-1 A, Kennedy Class I partially edentulous arch. Major support for denture bases must come from residual ridges, tooth support from occlusal rests being effective only at the anterior portion of each base. B, Kennedy Class III, modification 1, partially eden-tulous arch provides total tooth support for the prosthesis. A removable partial denture (RPD) made for this arch is totally supported by rests on properly prepared occlusal rest seats on four abutment teeth.
Mucosa/mucoperiosteum [2.0 mm] Periodontal ligament [0.25 0.1 mm] ] Figure 10-2 Distortion of tissues over the edentulous ridge will be approximately 500 μm under 4 newtons of force, whereas abutment teeth will demonstrate approximately 20 μm of intru-sion under the same load.
www.konkur.in 114 Part I General Concepts/Treatment Planning are more frequently used in tooth-supported restorations, because relining is not as likely to be necessary with them.
Differences in Support The distal extension partial denture derives its major support from the residual ridge with its fibrous connective tissue cov-ering. The length and contour of the residual ridge signifi-cantly influence the amount of available support and stability (Figure 10-3). Some areas of this residual ridge are firm, with limited displaceability, whereas other areas are displaceable, depending on the thickness and structural character of the tissues overlying the residual alveolar bone. The movement of the base under function determines the occlusal efficiency of the partial denture and also the degree to which the abut-ment teeth are subjected to torque and tipping stresses. The negative impact of the residual ridge character on the support provided to the occlusion in a specific arch can be addressed through the use of a dental implant, rendering displaceable tissue with movement potential into a more resistant support for occlusion.
Impression Registration An impression registration for the fabrication of a partial denture must fulfill the following two requirements: 1. The anatomic form and the relationship of the remain-ing teeth in the dental arch, as well as the surrounding soft tissues, must be recorded accurately so the denture will not exert pressure on those structures beyond their physiologic limits. A type of impression material that can be removed from undercut areas without permanent distortion must be used to fulfill this requirement. Elas-tic impression materials such as irreversible hydrocolloid (alginate), mercaptan rubber base (Thiokol), silicone im-pression materials (both condensation and addition reac-tion), and the polyethers are best suited for this purpose. 2. The supporting form of the soft tissues underlying the dis-tal extension base of the partial denture should be record-ed so firm areas are used as primary stress–bearing areas and readily displaceable tissues are not overloaded. Only in this way can maximum support of the partial denture base be obtained. An impression material capable of dis-placing tissue sufficiently to register the supporting form of the ridge will fulfill this second requirement. A fluid mouth-temperature wax or any of the readily flowing im-pression materials (rubber base, the silicones, or the poly-ethers in an individual, corrected tray) may be employed for registering the supporting form. Zinc oxide–eugenol impression paste can also be used when only the exten-sion base area is being impressed (see Chapter 16).
No single impression material can satisfactorily fulfill both of the previously mentioned requirements. Recording the anatomic form of both teeth and supporting tissues will result in inadequate support for the distal extension base. This is because the cast will not represent the optimum coor-dinating forms, which require that the ridge be related to the teeth in a supportive form. This coordination of support maximizes the support capacity for the arch and minimizes movement of the partial denture under function.
Differences in Clasp Design A fifth point of difference between the two main types of RPDs lies in their requirements for direct retention.
The tooth-supported partial denture, which is totally sup-ported by abutment teeth, is retained and stabilized by a clasp at each end of each edentulous space. Because this type of prosthesis does not move under function (other than within the physiologic limitations of tooth support units), the only requirement for such clasps is that they flex sufficiently dur-ing placement and removal of the denture to pass over the height of contour of the teeth in approaching or escaping from an undercut area. While in its terminal position on the tooth, a retentive clasp should be passive and should not flex except when one is engaging the undercut area of the tooth for resisting a vertical dislodging force.
Cast retentive arms are generally used for this purpose. These may be of the circumferential type, arising from the body of the clasp and approaching the undercut from an A B C D Figure 10-3 A, The longer the edentulous area covered by the denture base, the greater the potential lever action on the abutment teeth. If an extension base area is 30 mm (ac) and tis-sue displacement is 2 mm (ab), the amount of movement of the proximal plate on the guiding plane will be approximately 0.25 mm: [α = √ (ab)2 + (ac)2]; arc of the tangent ab/ad = x/cd (2/30 = x/3.75 = 0.25 mm). B, The flat ridge will provide good support, poor stability. C, The sharp spiny ridge will provide poor support, poor to fair stability. D, Displaceable tissue on the ridge will pro-vide poor support and poor stability.
www.konkur.in 115 Chapter 10 Principles of Removable Partial Denture Design occlusal direction, or of the bar type, arising from the base of the denture and approaching the undercut area from a gin-gival direction. Each of these two types of cast clasps has its advantages and disadvantages.
In the combination tooth- and tissue-supported RPD, because of the anticipated functional movement of the dis-tal extension base, the direct retainer adjacent to the distal extension base must perform still another function, in addi-tion to resisting vertical displacement. Because of the lack of tooth support distally, the denture base will move tissue-ward under function proportionate to the quality (displace-ability) of the supporting soft tissues, the accuracy of the denture base, and the total occlusal load applied. Because of this tissue-ward movement, those elements of a clasp that lie in an undercut area mesial to the fulcrum for a distal extension (as is often seen with a distal rest) must be able to flex sufficiently to dissipate stresses that otherwise would be transmitted directly to the abutment tooth as lever-age. On the other hand, a clasp used in conjunction with a mesial rest may not transmit as much stress to the abut-ment tooth because of the reduction in leverage forces that results from a change in the fulcrum position. This serves the purpose of reducing or “breaking” the stress, hence the term stress-breakers, and is a strategy that is often incorpo-rated into partial denture designs through various means. Some dentists strongly believe that a stress-breaker is the best means of preventing leverage from being transmitted to the abutment teeth. Others believe just as strongly that a wrought-wire or bar-type retentive arm more effectively accomplishes this purpose with greater simplicity and ease of application. A retentive clasp arm made of wrought wire can flex more readily in all directions than can the cast half-round clasp arm. Thereby, it may more effectively dissipate those stresses that would otherwise be transmitted to the abutment tooth.
Only the retentive arm of the circumferential clasp, how-ever, should be made of wrought metal. Reciprocation and stabilization against lateral and torquing movements must be obtained through use of the rigid cast elements that make up the remainder of the clasp. This is called a combination clasp because it is a combination of cast and wrought materials incorporated into one direct retainer. It is frequently used on the terminal abutment for the distal extension partial den-ture and is indicated where a mesiobuccal but no distobuccal undercut exists, or where a gross tissue undercut, cervical and buccal to the abutment tooth, exists. It must always be remembered that the factors of length and material contrib-ute to the flexibility of clasp arms. From a materials physi-cal property standpoint, a short wrought-wire arm may be a destructive element because of its reduced ability to flex compared with a longer wrought-wire arm. However, in addition to its greater flexibility compared with the cast cir-cumferential clasp, the combination clasp offers the advan-tages of adjustability, minimum tooth contact, and better esthetics, which justify its occasional use in tooth-supported designs.
The amount of stress transferred to the supporting eden-tulous ridge(s) and the abutment teeth will depend on: (1) the direction and magnitude of the force; (2) the length of the denture base lever arm(s); (3) the quality of resistance (support from the edentulous ridges and remaining natural teeth); and (4) the design characteristics of the partial den-ture. As was stated in Chapter 7, the location of the rest, the design of the minor connector as it relates to its correspond-ing guiding plane, and the location of the retentive arm are all factors that influence how a clasp system functions. The greater the surface area contact of each minor connector to its corresponding guiding plane, the more horizontal the dis-tribution of force (Figure 10-4).
ESSENTIALS OF PARTIAL DENTURE DESIGN The design of the partial denture framework should be sys-tematically developed and outlined on an accurate diagnos-tic cast based on the following prosthesis concepts: where the prosthesis is supported, how the support is connected, how the prosthesis is retained, how the retention and support are connected, and how edentulous base support is connected.
In developing the design, it is first necessary to determine how the partial denture is to be supported. In an entirely tooth-supported partial denture, the most ideal location for the support units (rests) is on prepared rest seats on the occlusal, cingulum, or incisal surface of the abutment adjacent to each edentulous space (see Figure 10-1, B). The type of rest and amount of support required must be based on interpretation of the diagnostic data collected from the patient. In evaluating the potential support that an abut-ment tooth can provide, consideration should be given to (1) periodontal health; (2) crown and root morphologies; (3) crown-to-root ratio; (4) bone index area (how tooth has responded to previous stress); (5) location of the tooth in the arch; (6) relationship of the tooth to other support units (length of edentulous span); and (7) the opposing dentition.
For a more in-depth understanding of these considerations, review Chapters 6 and 13.
In a tooth- and tissue-supported partial denture, atten-tion to these same considerations must be given to the abut-ment teeth. However, equitable support must come from the edentulous ridge areas. In evaluating the potential support available from edentulous ridge areas, consideration must be given to (1) the quality of the residual ridge, which includes contour and quality of the supporting bone (how the bone has responded to previous stress) and quality of the support-ing mucosa; (2) the extent to which the residual ridge will be covered by the denture base; (3) the type and accuracy of the impression registration; (4) the accuracy of the denture base; (5) the design characteristics of the component parts of the partial denture framework; and (6) the anticipated occlusal load. A full explanation of tissue support for extension base partial dentures is found in Chapter 17.
Denture base areas adjacent to abutment teeth are pri-marily tooth supported. As one proceeds away from the www.konkur.in 116 Part I General Concepts/Treatment Planning abutment teeth, they become more tissue supported. There-fore it is necessary to incorporate characteristics in the partial denture design that will distribute the functional load equita-bly between the abutment teeth and the supporting tissues of the edentulous ridge. Locating tooth support units (rests) on the principal abutment teeth and designing the minor con-nectors that are adjacent to the edentulous areas to contact the guiding planes in such a manner that the functional load is dispersed equitably between the available tooth and tissue supporting units will provide designs with controlled distri-bution of support (see Figure 10-4).
The second step in systematic development of the design for any RPD is to connect the tooth and tissue support units. This connection is facilitated by designing and locating major and minor connectors in compliance with the basic principles and concepts presented in Chapter 5. Major con-nectors must be rigid so that forces applied to any portion of the denture can be effectively distributed to the supporting structures. Minor connectors arising from the major con-nector make it possible to transfer functional stress to each abutment tooth through its connection to the corresponding rest and also to transfer the effects of the retainers, rests, and stabilizing components to the remainder of the denture and throughout the dental arch.
The third step is to determine how the RPD is to be retained. The retention must be sufficient to resist reasonable dislodging forces. As was stated in Chapter 7, retention is accomplished by placement of mechanical retaining elements (clasps) on the abutment teeth and by the intimate relation-ship of the denture bases and major connectors (maxillary) with the underlying tissues. The key to selecting a successful clasp design for any given situation is to choose one that will (1) avoid direct transmission of tipping or torquing forces to the abutment; (2) accommodate the basic principles of clasp design by definitive location of component parts correctly positioned on abutment tooth surfaces; (3) provide reten-tion against reasonable dislodging forces (with consideration for indirect retention); and (4) be compatible with undercut location, tissue contour, and esthetic desires of the patient.
Location of the undercut is the most important single factor in selection of a clasp. Undercut location, however, can be modified by recontouring or restoring the abutment tooth to accommodate a clasp design better suited to satisfy the criteria for clasp selection.
The relative importance of retention is highlighted by the results from a clinical trial investigating prosthesis designs. A 5-year randomized clinical trial of two basic RPD designs— one with rest, proximal plate, and I-bar (RPI) design and one with circumferential clasp design—demonstrated no dis-cernible changes after 60 months in nine periodontal health components of the abutment teeth with either of the two designs. The overall results indicate that the two designs did not differ in terms of success rates, maintenance, or effects on abutment teeth. Therefore, a well-constructed RPD that is supported by favorable abutments and good residual ridges that are properly prepared and maintained in a patient who exhibits good oral hygiene offers the best opportunity for sat-isfactory treatment.
The fourth step is to connect the retention units to the support units. If direct and indirect retainers are to function as designed, each must be rigidly attached to the major con-nector. The criteria for selection, location, and design are the Figure 10-4 1, Maximum contact of the proximal plate minor connector with the guiding plane produces a more horizontal distribu-tion of stress to the abutment teeth. 2, Minimum contact or disengagement of the minor connector with the guiding plane allows rota-tion around the fulcrum located on the mesio-occlusal rest, producing a more vertical distribution of stress to the ridge area. 3, Minor connector contact with the guiding plane from the marginal ridge to the junction of the middle and gingival thirds of the abutment tooth distributes load vertically to the ridge and horizontally to the abutment tooth. F, The location of the fulcrum of movement for the distal extension base.
www.konkur.in 117 Chapter 10 Principles of Removable Partial Denture Design same as those indicated for connecting the tooth and tissue support units.
The fifth and last step in this systematic approach to design is to outline and join the edentulous area to the already estab-lished design components. Strict attention to details of the design characteristics outlined in Chapter 9 is necessary to ensure rigidity of the base material without interfering with tooth placement.
COMPONENTS OF PARTIAL DENTURE DESIGN All partial dentures have two things in common: (1) they must be supported by oral structures, and (2) they must be retained against reasonable dislodging forces.
In the Kennedy Class III partial denture, three compo-nents are necessary: (1) support provided by rests, (2) the connectors (stabilizing components), and (3) the retainers.
The partial denture that does not have the advantage of tooth support at each end of each edentulous space (Kennedy Class I and II) still must be supported, but in this situation, the support comes from both the teeth and the underly-ing ridge tissues rather than from the teeth alone. This is a composite support, and the prosthesis must be fabricated so that the resilient support provided by the edentulous ridge is coordinated with the more stable support offered by the abutment teeth. The support, connectors, and retainers must take into consideration the movement of tissue-supported denture base areas. In addition, provision must be made for three other factors, as follows: 1. The best possible support must be obtained from the resilient tissues that cover the edentulous ridges. This is accomplished by the impression technique more than by the partial denture design, although the area covered by the partial denture base is a contributing factor in such support. 2. The method of direct retention must take into account the inevitable tissue-ward movement of the distal extension base(s) under the stresses of mastication and occlusion.
Direct retainers must be designed so that occlusal loading will result in direct transmission of this load to the long axis of the abutment teeth instead of as leverage. 3. The partial denture, with one or more distal extension denture bases, must be designed so that movement of the extension base away from the tissues will be minimized.
This is often referred to as indirect retention and is best described in relation to an axis of rotation through the rest areas of the principal abutments (see Chapter 8).
However, retention from the RPD base itself frequently can be made to help prevent this movement and, in such instances, may be discussed as direct-indirect retention.
Tooth Support Support of the RPD by the abutment teeth is dependent on the alveolar support of those teeth, the crown and root morphology, the rigidity of the partial denture framework, and the design of the occlusal rests. Through clinical and radiographic interpretation, the dentist may evaluate the abutment teeth and decide whether they will provide ade-quate support. In some instances, the splinting of two or more teeth, either by using fixed partial dentures or by sol-dering two or more individual restorations together, is advis-able. In other instances, a tooth may be deemed too weak to be used as an abutment, and extraction is indicated in favor of obtaining better support from an adjacent tooth.
Having decided on the abutments, the dentist is respon-sible for preparation and restoration of the abutment teeth to accommodate the most ideal design of the partial den-ture. This includes the form of the occlusal rest seats. These modifications may be prepared in sound tooth enamel or in restorative materials that will withstand the functional stress and wear of the component parts of the RPD. The technician cannot be blamed for inadequate abutment tooth prepara-tion, such as occlusal rest support. On the other hand, the technician is solely to blame if he or she extends the cast-ing beyond, or fails to include, the total prepared areas. If the dentist has sufficiently reduced the marginal ridge area of the rest seat to avoid interference from opposing teeth, and if a definite occlusal rest seat is faithfully recorded in the master cast and delineated in the penciled design, then no excuse can be made for poor occlusal rest form on the partial denture.
Ridge Support Support for the tooth-supported RPD or the tooth-sup-ported modification space comes entirely from the abutment teeth by means of rests. Support for the distal extension den-ture base comes primarily from the overlying soft tissues and the residual alveolar bone of the distal extension base area. In the latter, rest support is effective only at the abutment end of the denture base.
The effectiveness of tissue support depends on six things: (1) the quality of the residual ridge; (2) the extent to which the residual ridge will be covered by the denture base; (3) the accuracy and type of impression registration; (4) the accuracy of the denture bases; (5) the design characteristics of component parts of the partial denture framework; and (6) the occlusal load applied.
The quality of the residual ridge cannot be influenced, except that it can be improved by tissue conditioning, or it can be modified by surgical intervention. Such modifications are almost always needed but are not frequently done.
The accuracy of the impression technique is entirely in the hands of the dentist. Maximum tissue coverage for support that encompasses the primary stress–bearing areas should be the primary objective in any partial denture impression technique. The manner in which this is accomplished should be based on biological comprehension of what happens beneath a distal extension denture base when an occlusal load is applied.
The accuracy of the denture base is influenced by the choice of materials and by the exactness of the processing techniques. Inaccurate and warped denture bases adversely www.konkur.in 118 Part I General Concepts/Treatment Planning influence the support of the partial denture. Materials and techniques that will ensure the greatest dimensional stability should be selected.
The total occlusal load applied to the residual ridge may be influenced by reducing the area of occlusal contact. This is done with the use of fewer, narrower, and more effectively shaped artificial teeth (Figure 10-5).
The distal extension RPD is unique in that its support is derived from abutment teeth, which are comparatively unyielding, and from soft tissues overlying bone, which may be comparatively yielding under occlusal forces. Resilient tis-sues, which are distorted or displaced by occlusal load, are unable to provide support for the denture base comparable with that offered by the abutment teeth. This problem of support is further complicated by the fact that the patient may have natural teeth remaining that can exert far greater occlusal force on the supporting tissues than would result if the patient were completely edentulous. This fact is clearly evident from the damage that often occurs to an edentulous ridge when it is opposed by a few remaining anterior teeth in the opposing arch, and especially when the opposing occlusion of anterior teeth has been arranged so that contact occurs in both centric and eccentric positions. This situation can be best addressed by the use of a dental implant when possible.
Ridge tissues recorded in their resting or nonfunction-ing form are incapable of providing the composite support needed for a denture that derives its support from both hard and soft tissue. Three factors must be considered in the acceptance of an impression technique for distal extension RPDs: (1) the material should record the tissues covering the primary stress–bearing areas in their supporting form; (2) tissues within the basal seat area other than primary stress–bearing areas must be recorded in their anatomic form; and (3) the total area covered by the impression should be sufficient to distribute the load over as large an area as can be tolerated by the border tissues. This is an application of the principle of the snowshoe.
Anyone who has had the opportunity to compare two master casts for the same partially edentulous arch—one cast having the distal extension area recorded in its anatomic or resting form, and the other cast having the distal extension area recorded in its functional form—has been impressed by the differences in topography (Figure 10-6). A denture base processed to the functional form is generally less irregular and provides greater area coverage than does a denture base processed to the anatomic or resting form. Moreover, and of far greater significance, a denture base made to anatomic form exhibits less stability under rotating and/or torquing forces than does a denture base processed to functional form and thus fails to maintain its occlusal relation with opposing teeth. When the patient is asked to close onto strips of soft wax, it is evident that occlusion is maintained at a point of equilibrium over a longer period of time when the denture A B Figure 10-5 A, The total occlusal load applied may be reduced by using comparatively smaller posterior teeth represented by the il-lustration on the right. B, Less muscular force will be required to penetrate a food bolus with a reduced occlusal table, thereby reducing forces to supporting oral structures.
www.konkur.in 119 Chapter 10 Principles of Removable Partial Denture Design base has been made to the functional form. In contrast, evidence indicates that rapid “settling” of the denture base occurs when it has been made to the anatomic form, with an early return of the occlusion to natural tooth contact only. Such a denture not only fails to distribute the occlusal load equitably but also allows rotational movement, which is dam-aging to the abutment teeth and their investing structures.
An implant can efficiently serve to improve ridge support by replacing the tissue compression seen on functional load-ing with the stiff resistance offered by bone supporting an implant. The benefit for movement control is achieved as if a change was made from a tooth-tissue–borne prosthesis to a tooth-tooth–borne prosthesis.
Major and Minor Connectors Major connectors are the units of a partial denture that con-nect the parts of the prosthesis located on one side of the arch with those on the opposite side. Minor connectors arise from the major connector and join it with other parts of the denture; thus they serve to connect the tooth and tissue sup-port units. A major connector should be properly located in relation to gingival and moving tissues and should be designed to be rigid. Rigidity in a major connector is neces-sary to provide proper distribution of forces to and from the supporting components.
A lingual bar connector should be tapered superiorly with a half-pear shape in cross section and should be relieved suf-ficiently but not excessively over the underlying tissues when such relief is indicated. The finished inferior border of a lin-gual bar or a linguoplate should be gently rounded to avoid irritation to subjacent tissues when the restoration moves even slightly in function.
The use of a linguoplate is indicated when the lower ante-rior teeth are weakened by periodontal disease. It is also indi-cated in Kennedy Class I partially edentulous arches when additional resistance to horizontal rotation of the denture is required because of excessively resorbed residual ridges. Still another indication is seen in those situations in which the floor of the mouth so closely approximates the lingual gin-giva of anterior teeth that an adequately inflexible lingual bar cannot be positioned without impinging on gingival tissues.
The rigidity of a palatal major connector is just as impor-tant and its location and design just as critical as for a lin-gual bar. A U-shaped palatal connector is rarely justified due to its lack of rigidity, except to avoid an inoperable palatal torus that extends to the junction of the hard and soft pal-ates. Likewise, the routine use of a narrow, single palatal bar can rarely be justified. The combination anterior-posterior palatal strap–type major connector is mechanically and bio-logically sound if it is located so that it does not impinge on tissues. The broad, anatomic palatal major connector is fre-quently preferred because of its rigidity, better acceptance by the patient, and greater stability without tissue damage. In addition, this type of connector may provide direct-indirect retention that may sometimes, but rarely, eliminate the need for separate indirect retainers.
Direct Retainers for Tooth-Supported Partial Dentures Retainers for tooth-supported partial dentures have only two functions: (1) to retain the prosthesis against reason-able dislodging forces without damage to the abutment teeth and (2) to aid in resisting any tendency of the denture to be displaced in a horizontal plane. The prosthesis cannot move tissue-ward because the retentive components of the clasp assembly are supported by the rest. No movement away from the tissues should occur, and therefore no rotation about a fulcrum, because the retentive component is secured by a direct retainer.
Any type of direct retainer is acceptable as long as the abut-ment tooth is not jeopardized by its presence. Intracoronal (frictional) retainers are ideal for tooth-supported restora-tions and offer esthetic advantages that are not possible with extracoronal (clasp) retainers. Nevertheless, circumferential A B Figure 10-6 A, Cast of partially edentulous arch representing anatomic form of residual ridges. An impression was made in the stock tray by using irreversible hydrocolloid. B, An impression recording the functional or supporting form of residual ridges was made in an individualized impression tray, permitting placement of tissues and definitive border molding.
www.konkur.in 120 Part I General Concepts/Treatment Planning and bar-type clasp retainers are mechanically effective and are more economically constructed than are intracoronal retainers. Therefore they are more universally used.
Vulnerable areas on the abutment teeth must be protected by restorations with either type of retainer. The clasp retainer must not impinge on gingival tissues. The clasp must not exert excessive torque on the abutment tooth during place-ment and removal. It must be located the least distance into the tooth undercut for adequate retention, and it must be designed with a minimum of bulk and tooth contact.
The bar clasp arm should be used only when the area for retention lies close to the gingival margin of the tooth and little tissue blockout is necessary. If the clasp must be placed high, if the vestibule is extremely shallow, or if an objection-able space would exist beneath the bar clasp arm because of blockout of tissue undercuts, the bar clasp arm should not be used. In the event of an excessive tissue undercut, consid-eration should be given to recontouring the abutment and using some type of circumferential direct retainer.
Direct Retainers for Distal Extension Partial Dentures Retainers for distal extension partial dentures, while retain-ing the prosthesis, must also be able to flex or disengage when the denture base moves tissue-ward under function. Thus the retainer may act as a stress-breaker. Mechanical stress-breakers accomplish the same thing, but they do so at the expense of horizontal stabilization. When some kind of mechanical stress-breaker is used, the denture flange must be able to prevent horizontal movement. Clasp designs that allow flexing of the retentive clasp arm may accomplish the same purpose as that of mechanical stress-breakers, without sacrificing horizontal stabilization and with less complicated techniques.
In evaluating the ability of a clasp arm to act as a stress-breaker, one must realize that flexing in one plane is not enough. The clasp arm must be freely flexible in any direc-tion, as dictated by the stresses applied. Bulky, half-round clasp arms cannot do this, and neither can a bar clasp engag-ing an undercut on the side of the tooth away from the den-ture base. Round, tapered clasp forms offer the advantages of greater and more universal flexibility, less tooth contact, and better esthetics. Either the combination circumferential clasp with its tapered wrought-wire retentive arm or the carefully located and properly designed circumferential or bar clasp can be considered for use on all abutment teeth adjacent to extension denture bases if the abutment teeth have been properly prepared and tissue support effectively achieved, and if the patient exercises good oral hygiene.
Stabilizing Components Stabilizing components of the RPD framework are those rigid components that assist in stabilizing the denture against horizontal movement. The purpose of all stabilizing compo-nents should be to distribute stresses equally to all support-ing teeth without overworking any one tooth. The minor connectors that join the rests and the clasp assemblies to the major connector serve as stabilizing components.
All minor connectors that contact vertical tooth surfaces (and all reciprocal clasp arms) act as stabilizing components. It is necessary that minor connectors have sufficient bulk to be rigid and yet present as little bulk to the tongue as pos-sible. This means that they should be confined to interdental embrasures whenever possible. When minor connectors are located on vertical tooth surfaces, it is best that these surfaces be parallel to the path of placement. When cast restorations are used, these surfaces of the wax patterns should be made parallel on the surveyor before casting.
A modification of minor connector design has been pro-posed that places the minor connector in the center of the lingual surface of the abutment tooth (Figure 10-7). Pro-ponents of this design claim that it reduces the amount of A B Figure 10-7 Prospective guiding plane surfaces are indicated by lines located on the respective surfaces of abutment teeth. These sur-faces, when used, can be made vertically parallel to the path of placement. However, by including guiding plane surfaces, which are not in the same parallel plane horizontally (arrows) but are divergent, cross-arch resistance to horizontal rotation of the denture is enhanced.
www.konkur.in 121 Chapter 10 Principles of Removable Partial Denture Design gingival tissue coverage and provides enhanced bracing and guidance during placement. Disadvantages may include increased encroachment on the tongue space, more obvious borders, and potentially greater space between the connector and the abutment tooth. This proposed variation, however, when combined with thoughtful design principles, may pro-vide some benefit to the periodontal health of the abutment teeth and may be acceptable to some patients.
Reciprocal clasp arms also must be rigid, and they must be placed occlusally to the height of contour of the abutment teeth, where they will be nonretentive. By their rigidity, these clasp arms reciprocate the opposing retentive clasp; they also prevent horizontal movement of the prosthesis under functional stresses. For a reciprocal clasp arm to be placed favorably, some reduction of the tooth surfaces involved is frequently necessary to increase the suprabulge area.
When crown restorations are used, a lingual reciprocal clasp arm may be inset into the tooth contour by providing a ledge on the crown on which the clasp arm may rest. This permits the use of a wider clasp arm and restores a more nearly normal tooth contour, at the same time maintaining its strength and rigidity (see Chapter 15).
Guiding Plane The term guiding plane is defined as two or more parallel, vertical surfaces of abutment teeth, so shaped to direct a prosthesis during placement and removal. After the most favorable path of placement has been ascertained, axial sur-faces of abutment teeth are prepared parallel to the path of placement, and therefore become parallel to each other. Guiding planes may be contacted by various components of the partial denture—the body of an extracoronal direct retainer, the stabilizing arm of a direct retainer, the minor connector portion of an indirect retainer—or by a minor connector specifically designed to contact the guiding plane surface.
The functions of guiding plane surfaces are as follows: (1) to provide for one path of placement and removal of the restoration (to eliminate detrimental strain to abutment teeth and framework components during placement and removal); (2) to ensure the intended actions of reciprocal, stabilizing, and retentive components (to provide retention against dislodgment of the restoration when the dislodging force is directed other than parallel to the path of removal and also to provide stabilization against horizontal rotation of the denture); and (3) to eliminate gross food traps between abutment teeth and components of the denture.
Guiding plane surfaces need to be created so that they are as nearly parallel to the long axes of abutment teeth as possible. Establishing guiding planes on several abutment teeth (preferably more than two teeth), located at widely separated positions in the dental arch, provides for more effective use of these surfaces. The effectiveness of guiding plane surfaces is enhanced if these surfaces are prepared on more than one common axial surface of the abutment teeth (see Figure 10-7).
As a rule, proximal guiding plane surfaces should be about one-half the width of the distance between the tips of adjacent buccal and lingual cusps, or about one-third the buccal lingual width of the tooth, and should extend verti-cally about two thirds of the length of the enamel crown por-tion of the tooth from the marginal ridge cervically. In the preparation of guiding plane surfaces, care must be exercised to avoid creating buccal or lingual line angles (Figure 10-8). If it is assumed that the stabilizing or retentive arm of a direct retainer may originate in the guiding plane region, a line angle preparation would weaken either or both components of the clasp assembly.
A guiding plane should be located on the abutment sur-face adjacent to an edentulous area. However, excess torqu-ing is inevitable if the guiding planes squarely facing each other on a lone standing abutment adjacent to an extension area are used (Figure 10-9).
Indirect Retainers An indirect retainer must be placed as far anterior from the fulcrum line as adequate tooth support permits if it is to function with the direct retainer to restrict movement of a distal extension base away from the basal seat tissues. It must be placed on a rest seat prepared in an abutment tooth that is capable of withstanding the forces placed on it. An indirect retainer cannot function effectively on an inclined tooth surface, nor can a single weak incisor tooth be used for this purpose. A canine or premolar tooth should be used for A B Figure 10-8 A, The guiding plane surface should be like an area on a cylindrical object. It should be a continuous surface un-bounded by even, rounded line angles. B, Minor connector con-tacting the guiding plane surface has the same curvature as that surface. From an occlusal view, it tapers buccally from the thicker lingual portion, thus permitting closer contact of the abutment tooth and the prosthetically supplied tooth. Viewed from the buc-cal aspect, the minor connector contacts the enamel of the tooth on its proximal surface about two-thirds its length.
www.konkur.in 122 Part I General Concepts/Treatment Planning the support of an indirect retainer, and the rest seat must be prepared with as much care as is given any other rest seat. An incisal rest or a lingual rest may be used on an anterior tooth, provided a definite seat can be obtained in sound enamel or on a suitable restoration.
A second purpose that indirect retainers serve in par-tial denture design is that of support for major connectors. A long lingual bar or an anterior palatal major connector is thereby prevented from settling into the tissues. Even in the absence of a need for indirect retention, provision for such auxiliary support is sometimes indicated.
Contrary to common use, a cingulum bar or a linguo-plate does not in itself act as an indirect retainer. Because these are located on inclined tooth surfaces, they serve more as orthodontic appliances than as support for the partial denture. When a linguoplate or a cingulum bar is used, terminal rests should always be provided at either end to stabilize the denture and to prevent orthodontic movement of the teeth contacted. Such terminal rests may function as indirect retainers, but they would function equally well in that capacity without the continuous bar retainer or linguoplate.
IMPLANT CONSIDERATIONS IN DESIGN As was mentioned in Chapter 4, the objectives of RPD design are to replace missing teeth with a prosthesis that exhibits limited movement under the influence of functional forces, and to ensure that movement is within physiologic tolerance. Physiologic tolerance would include tissue toler-ance as well as a patient’s physiologic ability to accommodate to the prosthesis.
The Kennedy Class III tooth–supported RPD presents less of a challenge to oral tissues and patient accommodation than does the Kennedy Class I or II tooth-tissue prosthesis. The challenge is chiefly related to prosthesis movement to an extent allowed by tissue displaceability under an applied force. Use of dental implants to reduce this displacement can significantly benefit tissue tolerability and reduce any chal-lenge to accommodation presented by prosthesis movement. Use of implants can also assist other worthwhile goals such as improved stability and retention when these aspects are needed because of anatomic deficiencies or related factors.
The clinician must consider potential movements of the prosthesis and the ability to control movement given the existing oral tissues, teeth, and occlusion. Selective appli-cation of dental implants can provide needed movement control.
EXAMPLES OF SYSTEMATIC APPROACH TO DESIGN In Table 10-1 examples are provided as a design baseline for the most common tooth loss distributions—Kennedy Class I, II, and III.
Required Tooth Modification for Removable Partial Dentures The required tooth modifications for RPDs are as follows: • Teeth require modification to accommodate guideplanes, rests, and clasp arms. • Tooth contour modifications for guideplane and clasp needs are similar to crown reduction of enamel surfaces only (see Figure 13-18). • Rest seats are prepared following the guideplane reduction and also are contained in enamel only (see Figure 13-18). • Height of contour modifications are required from the guideplane location along two-thirds of the length to the retentive clasp (for both circumferential and combination clasps).
Implant Considerations Implants effectively reduce distal extension movement and can enhance support if placed at distal-most tooth position. Such use of an implant changes the mechanics of a Class I distal extension to a Class III as both ends of the edentulous span are supported.
If the implant is placed adjacent to the anterior tooth in an edentulous span, it can be used as a primary abutment for retention. Either location of implant placement that assists the RPD should be placed favorable to assist a subsequent (future) implant placement to provide a fixed prosthesis.
Kennedy Class II Removable Partial Dentures The Kennedy Class II partial denture (Figures 10-12 and 10-13) actually may be a combination of tissue-supported and tooth-supported restorations. The distal extension base must have adequate tissue support, whereas tooth-supported Figure 10-9 Guiding planes squarely facing each other should not be prepared on a lone standing abutment. Minor connectors of the framework (gray areas) would place undue strain on the abutment when the denture if rotated vertically, either superiorly or inferiorly.
www.konkur.in 123 Chapter 10 Principles of Removable Partial Denture Design Figure 10-10 Removable partial denture (RPD) in maxillary Class III arch. The design consists of anterior and posterior pala-tal bar major connectors, resin-supported artificial teeth, and bar clasp arms throughout.
Figure 10-11 Removable partial denture (RPD) framework in maxillary Class III arch. The design consists of a single palatal strap major connector, bar and circumferential clasp arms, and a means to attach resin-supported artificial teeth.
Table 10-1 Examples of Systematic Approach to Design Support Major connector Bracing/stability Retention Class I Maxilla Occlusal rest Cingulum rest Denture base palate (as needed) AP palatal strap Full Palate Guide planes Indirect retainers Reciprocal clasp Palatal plating RPI, RPA Combination clasp Mandible Occlusal rest Cingulum rest Denture base (Selective Pressure) Lingual bar Lingual plate Guide planes Indirect retainers Reciprocal clasp Lingual plating RPI, RPA Combination clasp Class II Maxilla Occlusal rest Cingulum rest Denture base palate (uncommon) AP palatal strap Horseshoe Full Palate (uncommon) Guide planes Indirect retainer Reciprocal clasp Palatal plating RPI, RPA (DE side) Combination clasp (DE side) Embrasure clasp (Dentate side) Circumferential clasp (Dentate Side) Mandible Occlusal rest Cingulum rest Denture base (Selective Pressure) Lingual bar Lingual plate Guide planes Indirect retainers Reciprocal clasp Lingual plating RPI, RPA Combination clasp Class III Maxilla Cingulum rest Occlusal rest Palatal strap Palatal bar Guide planes Reciprocal clasp Circumferential clasp I-bar clasp Mandible Cingulum rest Occlusal rest Lingual bar Guide planes Reciprocal clasp Circumferential clasp I-bar clasp RPA, Rest, proximal plate, Akers; RPI, rest, proximal plate, and I-bar.
www.konkur.in 124 Part I General Concepts/Treatment Planning bases elsewhere in the arch may be made to fit the anatomic form of the underlying ridge. Indirect retention must be provided for; however, occasionally the anterior abutment on the tooth-supported side will satisfy this requirement. If additional indirect retention is needed, provisions must be made for it.
Cast clasps are generally used on the tooth-supported side; however, a clasp design in which wrought wire is used may reduce the application of torque on the abutment tooth adjacent to the distal extension and should be considered. The use of a cast circumferential clasp engaging a mesio-buccal undercut on the anterior abutment of the tooth-sup-ported modification space may result in a Class I lever-like action if the abutment teeth have not been properly prepared, and/or if tissue support from the extension base area is not adequate. It seems rational under these circumstances to minimize these effects through optimum denture base adap-tation, to reduce movement or to provide implant support (see Figure 10-13). Should the bar-type retainer be contrain-dicated because of a severe tissue undercut or the existence of only a mesiobuccal undercut on the anterior abutment, then a combination direct retainer with the retentive arm made of tapered wrought wire should be used. A thorough understanding of the advantages and disadvantages of vari-ous clasp designs is necessary in determining the type of direct retainer that is to be used for each abutment tooth.
Steps included in fabrication of the Class II partial den-ture closely follow those used with the Class I partial den-ture, except that the distal extension base is usually made of an acrylic-resin material, whereas the base for any tooth-supported area can be made of metal. This is permissible because the residual ridge beneath tooth-supported bases is not called on to provide support for the denture, and later rebasing is not as likely to be necessary.
Class III Removable Partial Denture The Kennedy Class III RPD (Figures 10-10 and 10-11), entirely tooth supported, may be made to fit the prepared surfaces of the anatomic form of the teeth and surrounding structures. It does not require an impression of the functional form of the ridge tissues, nor does it require indirect reten-tion. Cast clasps of the circumferential variety, the bar type, or the combination clasp may be used, depending on how one can modify the surfaces of the abutment teeth (guiding planes, rests, contours for proper location of clasp arms). Unless the need for later relining is anticipated, as in the situation of recently extracted teeth, the denture base may be made of metal, which offers several advantages. The Class III partial denture can frequently be used as a valuable aid in periodontal treatment because of its stabilizing influence on the remaining teeth.
Kennedy Class I, Bilateral, Distal Extension Removable Partial Dentures The Class I, bilateral, distal extension partial denture is as dif-ferent from the Class III type as any two dental restorations could be (see Figure 10-1). Because it derives its principal support from the tissues underlying its base, a Class I par-tial denture made to anatomic ridge form cannot provide uniform and adequate support. Yet, unfortunately, many Class I mandibular RPDs are made from a single irreversible hydrocolloid impression. In such situations, both the abut-ment teeth and the residual ridges suffer because the occlusal load placed on the remaining teeth is increased by the lack of adequate posterior support.
Many dentists, recognizing the need for some type of impression registration that will record the supporting form of the residual ridge, attempt to record this form with a metal-lic oxide, a rubber base, or one of the silicone impression materials. Such materials actually record only the anatomic Figure 10-12 Mandibular Class II removable partial denture (RPD) with distal extension base. Because of tissue undercut cer-vical to the buccal surface of the right second premolar and lack of distobuccal undercut, a wrought-wire (tapered) retainer arm was used.
Figure 10-13 Mandibular Class II, modification 1, partially edentulous arch. Note that bar-type retentive arms are used on both premolar abutments, engaging distobuccal undercuts at their terminal ends. Lever-like forces may not be as readily im-parted to the right premolar, as opposed to the cast circumferen-tial direct retainer engaging the mesiobuccal undercut.
www.konkur.in 125 Chapter 10 Principles of Removable Partial Denture Design form of the ridge, except when the special design of impres-sion trays permits recording of the primary stress–bearing areas under a simulated load. Others prefer to place a base, made to fit the anatomic form of the ridge, under some pres-sure at the time that it is related to the remaining teeth, thus obtaining functional support. Still others, who believe that a properly compounded mouth-temperature wax will displace only those tissues that are incapable of providing support to the denture base, use a wax secondary impression to record the supporting, or functional, form of the edentulous ridge. Any impression record will be influenced by the consistency of the impression material and the amount of hydraulic pres-sure exerted by its confinement within the impression tray.
ADDITIONAL CONSIDERATIONS INFLUENCING DESIGN Every effort should be made to gain the greatest support possible for removable prostheses with the use of abut-ments bounding edentulous spaces. This not only will relieve the residual ridges of some of their obligation for support but also may allow the design of the framework to be greatly simplified. To this end, use of splint bars, internal clip attachments, overlay abutments, overlay attachments, a component partial, and implants should be considered.
Use of a Splint Bar for Denture Support In the Chapter 15 discussion of missing anterior teeth, mention is made of the fact that missing anterior teeth are best replaced with a fixed partial denture. The following is quoted from that chapter: “From a biomechanical stand-point … a removable partial denture should replace only the missing posterior teeth after the remainder of the arch has been made intact by fixed restorations.” Occasionally, a situation is found in which it is neces-sary for several missing anterior teeth to be replaced by the RPD rather than by fixed restorations. This may be caused by the length of the edentulous span or by the loss of a large amount of the residual ridge due to resorption, acci-dent, or surgery, or it may result from a situation in which too much vertical space prevents the use of a fixed partial denture, or in which esthetic requirements can better be met through the use of teeth added to the denture frame-work. In such instances, it is necessary that the best pos-sible support for the replaced anterior teeth be provided. Ordinarily this is done through the placement of occlusal or lingual rests, or both, on the adjacent natural teeth, but when the edentulous span is too large to ensure adequate support from the adjacent teeth, other methods must be used. This is included here only because it influences the design of the major connector that must then be used.
An anterior splint bar may be attached to the adjacent abutment teeth in such a manner that fixed splinting of the abutment teeth results, with a smooth, contoured bar resting lightly on the gingival tissues to support the RPD. As with any fixed partial denture, the type of abutment retainers and the decision to use multiple abutments will depend on the length of the span and the available sup-port and stability of the teeth being used as abutments. Regardless of the type of abutment retainers used, the connecting bar may be cast of a rigid alloy, or a commer-cially available bar may be used and cast to the abutments or attached to the abutments by soldering.
The length of the span influences the size of a splint bar. Long spans require more rigid bars (10-gauge) than short spans (13-gauge). If the bar is to be soldered, it is best that recesses be formed in the proximal surfaces of the abutments and that the connecting bar, which rests lightly on the tissues, be cast or made to fit into these recesses and then attached by soldering.
Because of the greater rigidity of chromium-cobalt alloys, the splint bar is preferably cast in one of these materials and then is attached to the abutments by solder-ing. The complete assembly (abutments and connecting bar) is then cemented permanently to the abutment teeth, in the same way as for a fixed partial denture. The impres-sion for the partial denture is then made, and a master cast is obtained that accurately reproduces the contours of the abutments and the splint bar. The denture framework then is made to fit the abutments and the bar by extend-ing the major connector or minor connectors to cover and rest upon the splint bar. Retention for the attachment of a resin base, or any other acceptable means of attach-ing the replaced anterior teeth, is incorporated into the denture design. In those situations wherein the RPD will be tooth supported, the splint bar may be curved to fol-low the crest of the residual ridge. However, in a distal extension situation, because of the vertical rotation of the denture, caution must be exercised to form the splint bar so that excessive torque will not accrue to its supporting abutments (Figure 10-14). The proximal contours of abut-ments adjacent to splint bars should be parallel to the path of placement. This serves three purposes: (1) it permits a desirable arrangement of artificial teeth, (2) it aids in resisting horizontal rotation of the restoration, and (3) these components act as guiding planes to direct the par-tial denture to and from its terminal position.
The splint bar must be positioned anteroposteriorly just lingual to the residual ridge to allow an esthetic arrange-ment of artificial teeth. The resulting partial denture offers esthetic advantages of removable anterior replacements and positive support, retention, and stability from the underlying splint bar (Figure 10-15).
Internal Clip Attachment The internal clip attachment differs from the splint bar in that the internal clip attachment provides both support and retention from the connecting bar.
www.konkur.in 126 Part I General Concepts/Treatment Planning Several preformed connecting bars are commercially available in plastic patterns. These can be customized for length and cast in the metal alloy of choice. Inter-nal clip attachments are also commercially available in various metal alloys and durable nylon. When a cus-tom-made connecting bar and clip is fabricated, the bar should be cast from 10- or 13-gauge sprue wax. The cast bar should rest lightly or should be located slightly above the tissues. Retention is provided by one of the commercial preformed metal or nylon clips, which is contoured to fit the bar and is retained in a pre-formed metal housing or partially embedded by means of retention spurs or loops into the overlying resin denture base.
The internal clip attachment thus provides support, stability, and retention for the anterior modification area and may serve to eliminate both occlusal rests and reten-tive clasps on adjacent abutment teeth.
Overlay Abutment as Support for a Denture Base Every consideration should be directed toward preventing the need for a distal extension RPD. In many instances, it is possible to salvage the roots and a portion of the crown of a badly broken-down molar through endodontic treat-ment. A periodontally involved molar, otherwise indicated for extraction, sometimes may be salvaged by periodontal and endodontic treatment accompanied by reduction of the clinical crown almost level with the gingival tissues. In another situation, an unopposed molar may have extruded to such an extent that restoring the tooth with a crown is inadequate to develop a harmonious occlusion. Then too, it is not unusual to encounter a molar that is so grossly tipped anteriorly that it cannot serve as an abut-ment unless the clinical crown is reduced drastically.
Such teeth should be considered for possible support for an otherwise distal extension denture base. Endodon-tic treatment and preparation of the coronal portion of the tooth as a slightly elevated dome-shaped abutment often offer an alternative to a distal extension base. The student is referred to the Selected Reading Resources sec-tion in Appendix B (textbooks; abutment retainers) for sources of information on overdenture abutments and overlay-type prostheses.
Use of a Component Partial to Gain Support A component partial is a RPD in which the framework is designed and fabricated in separate parts. The tooth support and tissue-supported components are individu-ally fabricated, and the two are joined with a high-impact acrylic-resin to become a single rigid functioning unit.
A B C Figure 10-14 A, Insofar as is possible, the splint bar should be round or ovoid. Provisions must be made in the construc-tion and location of the bar so that dental floss may be thread-ed underneath the bar to allow proper cleaning by the patient. B, As viewed from above, the bar is in a straight line between abutments. This is especially critical for distal extension remov-able partial dentures (RPDs) to avoid excess torque on abutments as the denture rotates in function. C, Sagittal section through the bar demonstrates the rounded form of the bar making point con-tact with the residual ridge. The entire tissue surface of the bar is easily accessible for cleaning with dental floss. A pear-shaped bar (in cross section) permits rotation of the RPD without appreciable resistance or torque.
Figure 10-15 Lower canines splinted together with a splint bar. The longevity of these teeth is greatly enhanced by splinting. Tissue surfaces are minimally contacted by the rounded form of the lower portion of the bar. Anterior and posterior slopes of the splint bar must be compatible with the path of placement of the denture. Floss will be used by the patient to clean the inferior portion of the splint bar.
www.konkur.in CHAPTER 11 Surveying CHAPTER OUTLINE Description of Dental Surveyor Purposes of the Surveyor Surveying the Diagnostic Cast Contouring Wax Patterns Surveying Ceramic Veneer Crowns Placement of Intracoronal Retainers (Internal Attachments) Placement of Internal Rest Seats Machining Cast Restorations Surveying the Master Cast Factors that Determine Path of Placement and Removal Guiding Planes Retentive Areas Interference Esthetics Step-by-Step Procedures in Surveying a Diagnostic Cast Guiding Planes Retentive Areas Interference Esthetics Final Path of Placement Recording Relation of Cast to Surveyor Surveying the Master Cast Measuring Retention Blocking out the Master Cast Relieving the Master Cast Paralleled Blockout, Shaped Blockout, Arbitrary Blockout, and Relief When a fixed partial denture (FPD) is prepared, the orienta-tion of the diamond bur is controlled to remove an amount of tooth structure necessary to satisfy the requirements of the path of insertion for the prosthesis. Accomplishment of parallel preparations is ultimately verified by complete seat-ing of the prosthesis but could be verified on the master cast or dies by the use of the surveyor. Once the FPD is fabri-cated and completely seated, it is ensured full engagement of the entire circumference of and occlusal support from the abutment retainers. If adequate resistance form and fit of the prosthesis are provided, the chance for functional stability equivalent to natural teeth is good. This could not be ensured unless the relationship of the fixed prosthesis and the pre-pared teeth were carefully controlled.
For a removable prosthesis, the necessity for appropri-ately planned and executed tooth preparation, followed by verification of a well-fitting prosthesis that engages the teeth as planned, is equally important. As was briefly mentioned in Chapter 7, a dental surveyor is vitally important to the planning, execution, and verification of appropriate mouth modifications for a removable partial denture. Although it does not necessarily affect the occlusal rest preparations on abutment teeth, use of the surveyor is critical for planning the modifications of all tooth surfaces that will be involved in support, stabilization, and retention of the prosthesis. In this role, the use of a surveyor to determine the needed mouth preparation is vitally important in helping to provide stable and comfortable removable prostheses.
A dental surveyor has been defined as an instrument used to determine the relative parallelism of two or more surfaces of the teeth or other parts of the cast of a dental arch. There-fore the primary purpose of surveying is to identify the mod-ifications of oral structures that are necessary to fabricate a removable partial denture that will have a successful progno-sis. It is the modification of tooth surfaces to accommodate placement of the component parts of the partial denture in their designated ideal positions on abutment teeth that facili-tates this prognosis.
www.konkur.in 128 Part I General Concepts/Treatment Planning Any one of several moderately priced surveyors on the market will adequately accomplish the procedures necessary to design and construct a partial denture. In addition, these surveyors may be used to parallel internal rests and intra-coronal retainers. With a handpiece holder added, they may be used to machine internal rests and to make the guiding-plane surfaces of abutment restorations parallel.
DESCRIPTION OF DENTAL SURVEYOR The most widely used surveyors are the Ney (Figure 11-1) and the Jelenko (Figure 11-2). Both of these are precision-made instruments. They differ principally in that the Jelenko arm swivels, whereas the Ney arm is fixed. The technique for surveying and trimming blockout is therefore somewhat different. Other surveyors also differ in this respect, and the dentist may prefer one over another for this reason.
The principal parts of the Ney surveyor are as follows: 1. Platform on which the base is moved 2. Vertical arm that supports the superstructure 3. Horizontal arm from which the surveying tool suspends 4. Table to which the cast is attached 5. Base on which the table swivels 6. Paralleling tool or guideline marker (This tool contacts the convex surface to be studied in a tangential manner; the relative parallelism of one surface to another may thus be determined; with substitution of a carbon marker, the height of contour may be delineated on the surfaces of the abutment teeth and on areas of interference requiring reduction on blockout.) 7. Mandrel for holding special tools (Figure 11-3) The principal parts of the Jelenko surveyor are essentially the same as those of the Ney surveyor except that when the nut at the top of the vertical arm is loosened, the horizon-tal arm may be made to swivel. The objective of this feature, originally designed by Dr. Noble Wills, is to permit freedom of movement of the arm in a horizontal plane rather than to depend entirely on the horizontal movement of the cast. To some, this is confusing because two horizontal movements must be coordinated. For those who prefer to move the cast only in horizontal relationship to a fixed vertical arm, the nut may be tightened and the horizontal arm used in a fixed position.
Another difference between Ney and Jelenko surveyors is that the vertical arm on the Ney surveyor is retained by friction within a fixed bearing. The shaft may be moved up or down within this bearing but remains in a vertical posi-tion until again moved. The shaft may be fixed in any verti-cal position desired by tightening a setscrew. In contrast, the vertical arm of the Jelenko surveyor is spring mounted and returns to the top position when it is released. It must be held down against spring tension while it is in use, which to some is a disadvantage. The spring may be removed, but the fric-tion of the two bearings supporting the arm does not hold it in position as securely as does a bearing designed for that purpose. These minor differences in the two surveyors lead to personal preference but do not detract from the effective-ness of either surveyor when each is properly used.
Because the shaft on the Ney surveyor is stable in any vertical position—yet may be moved vertically with ease— it lends itself well for use as a drill press when a handpiece Figure 11-1 The Ney surveyor is widely used because of its simplicity and durability. Dental students should be required to own such a surveyor. By becoming familiar with and dependent on its use, they are more likely to continue using the surveyor in practice as a necessary piece of equipment toward more ad-equate diagnosis, effective treatment planning, and performance of many other aspects of prosthodontic treatment.
Figure 11-2 The Jelenko Surveyor. Note the spring-mounted paralleling tool and swivel at the top of the vertical arm. The hori-zontal arm may be fixed in any position by tightening the nut at the top of the vertical arm.
www.konkur.in 129 Chapter 11 Surveying holder is added (Figure 11-4). The handpiece may thus be used to cut recesses in cast restorations with precision with burs or carborundum points of various sizes in a dental handpiece.
Several other types of surveyors have been designed and are in use today. Many of these are elaborate and costly, yet provide little advantage over simpler types of surveyors.
PURPOSES OF THE SURVEYOR The surveyor may be used for surveying the diagnostic cast, recontouring abutment teeth on the diagnostic cast, con-touring wax patterns, measuring a specific depth of under-cut, surveying ceramic veneer crowns, placing intracoronal retainers, placing internal rests, machining cast restorations, and surveying and blocking out the master cast.
A B C D F G E Ney undercut gauge Ney carbon marker and sheath Jelenko carbon marker Jelenko undercut gauge 0.03 0.03 0.02 0.02 0.01 0.01 Path of placement Ney wax trimmer Blockout wax 2 taper tool 6 taper tool Ney surveyor blade Figure 11-3 Various tools that may be used with a dental surveyor. A, Ney undercut gauges. B, Jelenko undercut gauge. C, Ney carbon marker with metal reinforcement sheath. D, Jelenko carbon marker. E, Tapered tools, 2- and 6-degree, for trimming blockout when some nonparallelism is desired. F, Ney wax trimmer for paralleling blockout. G, Surveying blade used for trimming blockout.
www.konkur.in 130 Part I General Concepts/Treatment Planning Surveying the Diagnostic Cast Surveying the diagnostic cast is essential to effective diag-nosis and treatment planning. The objectives are as follows: 1. To determine the most desirable path of placement that eliminates or minimizes interference to placement and removal (Figure 11-5): The path of placement is the di-rection in which a restoration moves from the point of initial contact of its rigid parts with the supporting teeth to its terminal resting position, with rests seated and the denture base in contact with the tissues. The path of re-moval is exactly the reverse, because it is the direction of restoration movement from its terminal resting posi-tion to the last contact of its rigid parts with the support-ing teeth. When the restoration is properly designed to have positive guiding planes, the patient may place and remove the restoration with ease in only one direction.
This is possible only because of the guiding influence of tooth surfaces (guiding planes) made parallel to that path of placement. 2. To identify proximal tooth surfaces that are or need to be made parallel so that they act as guiding planes during placement and removal 3. To locate and measure areas of the teeth that may be used for retention 4. To determine whether teeth and bony areas of interfer-ence need to be eliminated surgically or by selection of a different path of placement 5. To determine the most suitable path of placement that permits locating retainers and artificial teeth to provide the best esthetic advantage 6. To permit accurate charting of the mouth preparations to be made: This includes the preparation of proximal tooth surfaces to provide guiding planes and the reduction of excessive tooth contours to eliminate interference and to permit a more acceptable location of reciprocal and reten-tive clasp arms. By marking these areas on the diagnos-tic cast in red, using an undercut gauge to estimate the amount of tooth structure that may safely (without expos-ing dentin) be removed, and then trimming the marked areas on the stone cast with the surveyor blade, the angu-lation and extent of tooth reduction may be established before the teeth are prepared in the mouth (Figure 11-6).
With the diagnostic cast on the surveyor at the time of mouth preparation, reduction of tooth contours may be accomplished with acceptable accuracy. 7. To delineate the height of contour on abutment teeth and to locate areas of undesirable tooth undercut that are to be avoided, eliminated, or blocked out: This includes areas of the teeth to be contacted by rigid connectors, the location of nonretentive reciprocal and stabilizing arms, and the location of retentive clasp terminals. 8. To record the cast position in relation to the selected path of placement for future reference: This may be done by locating three dots or parallel lines on the cast, thus estab-lishing the horizontal plane in relation to the vertical arm of the surveyor (see Figures 11-6 and 11-16).
Figure 11-4 Lab Handpiece Clamp. Handpiece holders at-tach to the vertical spindle of surveyors and may be used to cre-ate and refine any parallel surface on a surveyed crown, as a drill press to prepare internal rests and recesses in patterns and/or castings, and to establish lingual surfaces above the ledge that are parallel to the path of placement in abutment restorations.
Path of placement Figure 11-5 Tilt of the cast on the adjustable table of a sur-veyor in relation to the vertical arm establishes the path of place-ment and removal that the removable partial denture will take. All mouth preparations must be made to conform to this deter-mined path of placement, which has been recorded by scoring the base of the cast or by tripoding.
www.konkur.in 131 Chapter 11 Surveying Contouring Wax Patterns The surveyor blade is used as a wax carver during this phase of mouth preparation so that the proposed path of placement may be maintained throughout the preparation of cast resto-rations for abutment teeth (Figure 11-7).
Guiding planes on all proximal surfaces of wax patterns adjacent to edentulous areas should be made parallel to the previously determined path of placement. Similarly, all other tooth contours that will be contacted by rigid components should be made parallel. The surfaces of restorations on which reciprocal and stabilizing components will be placed should be contoured to permit their location well below occlusal surfaces and on nonretentive areas. Those surfaces of restora-tions that are to provide retention for clasp arms should be contoured so that retentive clasps may be placed in the cer-vical third of the crown and to the best esthetic advantage. Generally, a small amount of undercut from 0.01 to 0.02 inch (0.250 to 0.50 mm) or less is sufficient for retentive purposes.
B C A Figure 11-6 A, Solid line represents the height of contour on the abutment at selected orientation of the diagnostic cast to the vertical spindle of the surveyor. Dotted line represents the desirable height of the contour for optimally locating components of the direct retainer assembly. A 0.01-inch (0.25-mm) undercut gauge is used to mark the location of the tip of the retentive arm of the direct retainer. B, By reducing the axial contour of the tooth by only 0.01 inch, the optimum height of contour can be achieved without exposing the dentin. C, Stone tooth is trimmed with a surveyor blade to the desired height of contour. The trimmed area is marked in red pencil and serves as a blueprint for similar recontouring in the mouth. If one can safely assume that the enamel is 1 to 1.5 mm thick in the area of contemplated reduction, only 0.25 mm of enamel needs to be removed to achieve the optimum height of contour.
B A Figure 11-7 After the cast has been oriented to the surveyor at the predetermined path of placement, designated axial surfaces of the wax pattern are altered with the surveyor blade to meet specific requirements for placement of framework components. A, Wax pattern is carved with the surveyor blade to produce a distal guide-plane surface parallel to the selected path of insertion. B, The same pattern is modified from the distal guide plane along the buccal surface to align the surface with the height of contour most favorable to the direct retainer specifications.
www.konkur.in 132 Part I General Concepts/Treatment Planning Surveying Ceramic Veneer Crowns Ceramic veneer crowns are often used to restore abut-ment teeth on which extracoronal direct retainers will be placed. The surveyor is used to contour all areas of the wax pattern for the veneer crown except the buccal or labial surface. It must be remembered that one of the principal goals in using a porcelain veneer restoration is to develop an esthetic replica of a natural tooth. It is unlikely that the ceramic veneer portion can be fabricated exactly to the form required for the planned placement of retentive clasp arms without some reshaping with stones. Before the final glaze is accomplished, the abutment crowns should be returned to the surveyor on a full arch cast to ensure the correct contour of the veneered portions or to locate those areas that need recontouring (Figure 11-8). The final glaze is accomplished only after the crowns have been recontoured.
Placement of Intracoronal Retainers (Internal Attachments) In the placement of intracoronal retainers, the surveyor is used as follows: 1. To select a path of placement in relation to the long axes of the abutment teeth that avoids areas of interfer-ence elsewhere in the arch 2. To cut recesses in the stone teeth on the diagnostic cast for estimating the proximity of the recess to the pulp (used in conjunction with roentgenographic information to estimate pulp size and location) and to facilitate the fabrication of metal or resin jigs to guide preparations of the recesses in the mouth 3. To carve recesses in wax patterns, to place internal at-tachment trays in wax patterns, or to cut recesses in castings with the handpiece holder (whichever meth-od is preferred) 4. To place the keyway portion of the attachment in the casting before investing and soldering; each keyway must be located parallel to the other keyways else-where in the arch The student is referred to the Selected Reading Resources in Appendix B for sources of information on intracoronal retainers (internal attachments).
Placement of Internal Rest Seats The surveyor may be used as a drill press, with a dental handpiece attached to the vertical arm by a handpiece holder. Internal rest seats may be carved in the wax patterns and further refined with the handpiece after casting, or the entire rest seat may be cut in the cast restoration with the handpiece. It is best to carve the outline form of the rest seat in wax and merely refine the casting with the handpiece.
An internal rest differs from an internal attachment in that some portion of the prosthesis framework is waxed and cast to fit into the rest seat rather than a matched key and keyway attachment used (see Figures 7-42 and 7-43). The former is usually nonretentive but provides a definite seat for a removable partial denture or a cantilever rest for a broken-stress FPD. When they are used with FPDs, nonparallel abutment pieces may be placed separately.
The internal rest in partial denture construction provides a positive occlusal support that is more favorably located in relation to the rotational axis of the abutment tooth than the conventional spoon-shaped occlusal rest. It also provides horizontal stabilization through the parallelism of the verti-cal walls, thereby serving the same purpose as stabilizing and reciprocal arms placed extracoronally. Because of the move-ment of a distal extension base, more torque may be applied to the abutment tooth by an interlocking type of rest, and for this reason its use in conjunction with a distal extension partial denture is considered to be contraindicated. The ball-and-socket, spoon-shaped occlusal, or noninterlocking, rest should be used in distal extension partial denture designs. Use of the dovetailed or interlocking internal rest should be limited to tooth-supported removable restorations, except when it is used in conjunction with some kind of stress-breaker between the abutments and the movable base. The use of stress-breakers has been discussed in Chapter 9.
Internal rest seats may be made in the form of a non-retentive box, a retentive box fashioned after the internal attachment, or a semiretentive box. In the latter, the walls are usually parallel and nonretentive, but a recess in the floor of the box prevents proximal movement of the male portion. Internal rest seats are cut with dental burs of var-ious sizes and shapes. Tapered or cylindrical fissure burs are used to form the vertical walls, and small round burs are used to cut recesses in the floor of the rest seat.
Figure 11-8 Resultant metal-ceramic surveyed crown from Figure 11-7, which is being refined to maintain the distal guide plane and buccal height of the contour previously designed. Fi-nal glaze has not been placed on the veneer crown and required alterations of surfaces to conform to ideal placement of the re-tainer (solid line) can be performed by machining. Final glaze is produced only after necessary recontouring is accomplished.
www.konkur.in 133 Chapter 11 Surveying Machining Cast Restorations With a handpiece holder attached (see Figure 11-4), the axial surfaces of cast and ceramic restorations may be refined by machining with a suitable cylindrical carborundum point. Proximal surfaces of crowns and inlays, which will serve as guiding planes, and vertical surfaces above crown ledges may be improved by machining, but only if the relationship of one crown to another is correct (see Figure 14-9). Unless the seating of removable dies is accurate and they are held in place with additional stone or plaster, cast restorations should first be tried in the mouth and then transferred, by means of a plaster or acrylic-resin index impression, to a reinforced stone cast for machining purposes. The new cast is then positioned on the surveyor, conforming to the path of placement of the partial denture, and vertical surfaces are machined with a true-running cylindrical carborundum point.
Although machined parallelism may be considered ideal and beyond the realm of everyday application, its merits more than justify the additional steps required to accomplish it. When such parallelism is accomplished and reproduced in a master cast, it is essential that subsequent laboratory steps be directed toward the use of these parallel guiding plane surfaces.
Surveying the Master Cast Because surveying the master cast follows mouth prepara-tions, the path of placement, the location of retentive areas, and the location of remaining interference must be known before the final design of the denture framework is completed. The objectives of surveying the master cast are as follows: 1. To select the most suitable path of placement by following mouth preparations that satisfy the requirements of guid-ing planes, retention, noninterference, and esthetics 2. To permit measurement of retentive areas and to identify the location of clasp terminals in proportion to the flex-ibility of the clasp arm being used: Flexibility depends on many of the following factors: (a) the alloy used for the clasp, (b) the design and type of clasp, (c) whether its form is round or half-round, (d) whether it is of cast or wrought material, and (e) the length of the clasp arm from its point of origin to its terminal end. Retention depends on (a) the flexibility of the clasp arm, (b) the magnitude of the tooth undercut, and (c) the depth the clasp terminal is placed into this undercut. 3. To locate undesirable undercut areas that will be crossed by rigid parts of the restoration during placement and re-moval: These must be eliminated by blockout. 4. To trim blockout material parallel to the path of place-ment before duplication (Figure 11-9) The partial denture must be designed so that (1) it will not stress abutment teeth beyond their physiologic tolerance, (2) it can be easily placed and removed by the patient, (3) it will be retained against reasonable dislodging forces, and (4) it will not create an unfavorable appearance. It is necessary that the diagnostic cast be surveyed with these principles in mind. Mouth preparation should therefore be planned in accordance with certain factors that will influence the path of placement and removal.
FACTORS THAT DETERMINE PATH OF PLACEMENT AND REMOVAL The factors that determine the path of placement and removal are guiding planes, retentive areas, interference, and esthetics. Additionally, if an implant is part of the design its long axis is considered in relation to the path or placement and removal (Figure 11-10, B).
A B Figure 11-9 Master casts are modified by the addition of wax relief to nonbearing regions and by placement of blockout wax parallel to the path of insertion at regions beneath the height of contour where framework contact is not planned (i.e., all areas except retentive clasp tips). A, Blockout wax is provided for tooth contours beneath the height of contour on teeth #21 and #28. B, Similar blockout was accomplished for mandibular molar #31. Blockout is carved with a straight surveyor blade to ensure parallelism with the identified path of insertion.
www.konkur.in 134 Part I General Concepts/Treatment Planning Guiding Planes Proximal tooth surfaces that bear a parallel relationship to one another must be found or must be created to act as guid-ing planes during placement and removal of the prosthesis. Guiding planes are necessary to ensure the passage of rigid parts of the prosthesis past existing areas of interference. Thus the denture can be easily placed and removed by the patient without strain on the teeth contacted or on the den-ture itself and without damage to the underlying soft tissues.
Guiding planes are also necessary to ensure predictable clasp assembly function, including retention and stabiliza-tion. For a clasp to be retentive, its retentive arm must be forced to flex. Hence, guiding planes are necessary to give a positive direction to the movement of the restoration to and from its terminal position.
Retentive Areas Retentive areas must exist for a given path of placement and must be contacted by retentive clasp arms that are forced to flex over a convex surface during placement and removal. Satisfactory clasp retention is no more than the resistance of metal to deformation. For a clasp to be retentive, its path of escapement must be other than parallel to the path of removal of the denture itself; otherwise it would not be forced to flex and thereby generate the resistance known as retention. Clasp retention therefore depends on the existence of a definite path of placement and removal.
Although desirable, retention at each principal abutment may not be balanced in relation to the tooth on the opposite side of the arch (exactly equal and opposite in magnitude and relative location); however, positive cross-arch reciprocation to retentive elements must be present. Retention should be sufficient only to resist reasonable dislodging forces. In other words, it should be the minimum acceptable for adequate retention against reasonable dislodging forces.
Fairly even retention may be obtained by one of two means. One is to change the path of placement to increase or decrease the angle of cervical convergence of opposing retentive surfaces of abutment teeth. The other is to alter the flexibility of the clasp arm by changing its design, its size and length, or the material of which it is made.
Interference The prosthesis must be designed so that it may be placed and removed without encountering tooth or soft tissue interfer-ence. A path of placement may be selected that encounters interference only if the interference can be eliminated during mouth preparations or on the master cast by a reasonable amount of blockout. Interference may be eliminated during mouth preparations by surgery, extraction, modification of interfering tooth surfaces, or alteration of tooth contours with restorations.
Generally, interference that cannot be eliminated for one reason or another takes precedence over the factors of retention and guiding planes. Sometimes certain areas can be made noninterfering only by selecting a different path A B C Figure 11-10 A, When anterior teeth must be replaced with the removable partial denture, the selected vertical path should consider the junction of the natural tooth and the denture tooth. The path of insertion that requires the least amount of alteration of the natural teeth and the denture tooth (maximizing natural embrasure contours) is desirable. B, Distal of canine slightly al-tered to accommodate the path that optimizes anterior denture tooth placement. C, Canine crowns required to satisfy desired contours of the natural tooth and the denture teeth.
www.konkur.in 135 Chapter 11 Surveying of placement at the expense of existing retentive areas and guiding planes. These must then be modified with restora-tions that are in harmony with the path dictated by the exist-ing interference. On the other hand, if areas of interference can be eliminated by various reasonable means, this should be done. When this occurs, the axial contours of existing abutments may frequently be used with little alteration.
Esthetics By one path of placement, the most esthetic location of arti-ficial teeth is made possible, and less clasp metal and base material may be displayed.
The location of retentive areas may influence the path of placement selected; therefore retentive areas should always be selected with the most esthetic locations of clasps in mind. When restorations are to be made for other reasons, they should be contoured to permit the least display of clasp metal. Generally, less metal will be displayed if the retentive clasp is placed at a more distogingival area of tooth surface, made possible by the path of placement selected or by the contour of the restorations.
Esthetics also may dictate the choice of path selected when missing anterior teeth must be replaced with the partial den-ture. In such situations, a more vertical path of placement is often necessary so that neither the artificial teeth nor the adjacent natural teeth will have to be modified excessively (see Figure 11-10). In this instance, esthetics may take pre-cedence over other factors. This necessitates the preparation of abutment teeth to eliminate interferences and to provide guiding planes and retention in harmony with that path of placement dictated by esthetic factors.
Because the primary consideration should be the pres-ervation of remaining oral tissues, esthetics should not be allowed to jeopardize the success of the partial den-ture. The replacement of missing anterior teeth therefore should be accomplished by means of FPDs whenever pos-sible, especially if the mechanical and functional effective-ness of the partial denture will require significant tooth preparation.
STEP-BY-STEP PROCEDURES IN SURVEYING A DIAGNOSTIC CAST Attach the cast to the adjustable surveyor table by means of the clamp provided. Position the adjustable table so that the occlusal surfaces of the teeth are approximately parallel to the platform (Figure 11-11). Such an orientation is a tenta-tive but practical way to start considering the factors that influence the path of placement and removal.
Guiding Planes Determine the relative parallelism of proximal surfaces of all of the potential abutment teeth by contacting the proxi-mal tooth surfaces with the surveyor blade or diagnostic stylus. Alter the cast position anteroposteriorly until these proximal surfaces are in as close to a parallel relation to one another as possible, or near enough that they can be made parallel by recontouring. For posterior modifica-tion spaces, this determines the anteroposterior tilt of the cast in relation to the vertical arm of the surveyor (Figure 11-12). Although the surveyor table is universally adjustable, it should be thought of as having only two axes, thus allowing only anteroposterior and lateral tilting. If an implant is to be used, the long axis of the implant should be considered in defining the guide plane reduction needs (see Figure 11-10, B).
When a choice is made between having contact with a proximal surface at the cervical area only and contact at the marginal ridge only, the latter is preferred because a plane may then be established by recontouring (Figure 11-13). It is obvious that when only gingival contact exists, a restoration is the only means of establishing a guiding plane. Therefore, if a tilt that does not provide proximal contact is apparent, the proximal surface must be established with some kind of restoration.
The end result of selecting a suitable anteroposterior tilt should be to provide the greatest combined areas of paral-lel proximal surfaces that may act as guiding planes. Other Figure 11-11 Recommended method for manipulating the dental surveyor. The right hand is braced on the horizontal arm of the surveyor, and the fingers are used, as illustrated, to raise and lower the vertical shaft in its spindle. The left hand holding the cast on the adjustable table slides horizontally on the plat-form in relation to the vertical arm. The right hand must be used to loosen and tighten the tilting mechanism as a suitable antero-posterior and lateral tilt of the cast in relation to the surveyor is being determined.
www.konkur.in 136 Part I General Concepts/Treatment Planning axial surfaces of abutment teeth may also be used as guiding planes. This is realized most often by having the stabilizing component of the direct retainer assembly contacting in its entirety the axial surface of the abutment, which has been found or made parallel to the path of placement (see Figure 15-6). Therefore a lateral tilt of the cast to the vertical arm of the surveyor must also be considered, as well as the antero-posterior tilt, when guiding planes are used.
Retentive Areas Through contact of buccal and lingual surfaces of abut-ment teeth with the surveyor blade, the amount of reten-tion existing below their height of convexity may be determined. This is best accomplished by directing a small source of light toward the cast from the side away from the dentist. The angle of cervical convergence is best observed as a triangle of light between the surveyor blade and the apical portion of the tooth surface being studied (see Figure 7-41).
Alter the cast position by tilting it laterally until similar retentive areas exist on the principal abutment teeth. If only two abutment teeth are involved (as in a Kennedy Class I partially edentulous arch), they are both principal abut-ments. However, if four abutment teeth are involved (as they are in a Kennedy Class III, modification 1 arch), they are all principal abutments, and retentive areas should be located on all four. But if three abutment teeth are involved (as they are in a Kennedy Class II, modification 1 arch), the posterior abutment on the tooth-supported side and the abutment on the distal extension side are considered to be the principal abutments, and retention needs to be equalized accordingly. The third abutment may be considered to be secondary, and less retention is expected from it than from the other two. An exception is when the posterior abutment on the tooth-supported side has a poor prognosis and the denture is designed to ultimately be a Class I. In such a situation, the two stronger abutments are considered to be principal abutments.
When the cast is tilted laterally to establish reasonable uniformity of retention, it is necessary that the table be rotated about an imaginary longitudinal axis without dis-turbing the anteroposterior tilt previously established. The resulting position is one that provides or makes possible parallel guiding planes and provides for acceptable reten-tion on the abutment teeth. It should be noted that this most desirable position will always require some tooth modifica-tion to be achieved. Note that possible interference to this tentative path of placement has not, as yet, been taken into consideration.
Interference If a mandibular cast is being surveyed, check the lingual sur-faces that will be crossed by a lingual bar major connector during placement and removal. Bony prominences and lin-gually inclined premolar teeth are the most common causes of interference to a lingual bar connector.
If the interference is bilateral, surgery or recontouring of lingual tooth surfaces, or both, may be unavoidable. If the interference is only unilateral, a change in the lateral tilt may avoid an area of tooth or tissue interference. In chang-ing the path of placement to prevent interference, previously established guiding planes and an ideal location for reten-tive elements may be lost. Then the decision must be made whether to remove the existing interference by whatever means necessary or to resort to restorations on the abutment Figure 11-12 Relative parallelism of proximal tooth surfaces determines anteroposterior tilt of the cast in relation to the verti-cal arm of the surveyor.
A B Figure 11-13 When the most desirable anteroposterior tilt of the cast in relation to the surveyor blade is determined, a choice must be made between positions illustrated in A and B. In A, the distal surface of the left premolar abutment would have to be extended by means of a restoration. In B, the right premolar could be altered slightly to provide an acceptably parallel guiding plane. Unless restorations are necessary for other reasons, the tilt shown in B is preferred.
www.konkur.in 137 Chapter 11 Surveying teeth, thereby changing the proximal and retentive areas to conform to the new path of placement.
In a like manner, bony undercuts that offer interference to the seating of denture bases must be evaluated and the decision made to remove them surgically, to change the path of placement at the expense of modifying or restoring teeth to achieve guiding planes and retention, or to design denture bases to avoid such undercut areas. The latter may be done by shortening buccal and labial flanges and distolingual exten-sion of the denture bases. However, it should be remembered that the maximum area available for support of the denture base should be used whenever possible.
Interference to major connectors rarely exists in the maxillary arch. Areas of interference are usually found on buccally inclined posterior teeth and those bony areas on the buccal aspect of edentulous spaces. As with the man-dibular cast, the decision must be made whether to elimi-nate them, to change the path of placement at the expense of modifying or restoring teeth to achieve the required guiding planes and retention, or to design the connectors and bases to avoid them.
Other areas of possible interference to be evaluated are those surfaces of abutment teeth that will support or be crossed by minor connectors and clasp arms. Although interference to vertical minor connectors may be blocked out, doing so may cause discomfort to the patient’s tongue and may create objectionable spaces, which could result in the trapping of food. Also it is desirable that tooth surfaces contacted by vertical connectors be used as auxiliary guid-ing planes whenever possible. Too much relief is perhaps better than too little because of the possibility of irritation to soft tissues. It is always better that the relief be placed with some objective in mind. If possible, a minor connector should pass vertically along a tooth surface that is paral-lel to the path of placement (which is considered ideal) or tapered occlusally. If tooth undercuts that necessitate the use of an objectionable amount of blockout exist, they may be eliminated or minimized by slight changes in the path of placement and/or eliminated during mouth prepara-tions. The need for such alteration should be indicated on the diagnostic cast in red pencil after final acceptance of a path of placement.
Tooth surfaces on which reciprocal and stabilizing clasp arms will be placed should be studied to see whether suf-ficient areas exist above the height of convexity for the place-ment of these components. The addition of a clasp arm to the occlusal third of an abutment tooth adds to its occlusal dimension and therefore to the occlusal loading of that tooth. Nonretentive and stabilizing clasp arms are best located between the middle third and the gingival third of the crown rather than on the occlusal third.
Areas of interference to proper placement of clasp arms can be eliminated by reshaping tooth surfaces during mouth preparations. These areas should be indicated on the diag-nostic cast. Areas of interference to the placement of clasps may necessitate minor changes in the path of placement or changes in the clasp design. For example, a bar clasp arm originating mesially from the major connector to provide reciprocation and stabilization might be substituted for a distally originating circumferential arm.
Areas of interference often overlooked are the distal line angles of premolar abutment teeth and the mesial line angles of molar abutments. These areas frequently offer interference to the origin of circumferential clasp arms. If not detected at the time of initial survey, they are not included in the plan for mouth preparations. When such an undercut exists, the following three alternatives may be considered: 1. They may be blocked out the same as any other area of interference. This is by far the least satisfactory method because the origin of the clasp must then stand away from the tooth in proportion to the amount of blockout used.
Although this is perhaps less objectionable than its being placed occlusally, it may be objectionable to the tongue and the cheek and may create a food trap. 2. They may be circumvented by approaching the retentive area from a gingival direction with a bar clasp arm. This is often a satisfactory solution to the problem if other con-traindications to the use of a bar clasp arm, such as a se-vere tissue undercut or a retentive area that is too high on the tooth, are not present. 3. They may be eliminated by reducing the offending tooth contour during mouth preparation. This permits the use of a circumferential clasp arm originating well below the occlusal surface in a satisfactory manner. If the tooth is to be modified during mouth preparations, it should be indicated on the diagnostic cast with a colored pencil.
When the retentive area is located objectionably high on the abutment tooth or the undercut is too severe, interference may also exist on tooth surfaces that support retentive clasps. Such areas of extreme or high convexity must be considered as areas of interference and should be reduced accordingly. These areas are likewise indicated on the diagnostic cast for reduction during mouth preparations.
Esthetics The path of placement thus established must still be consid-ered from the standpoint of esthetics, as to both the location of clasps and the arrangement of artificial teeth.
Clasp designs that will provide satisfactory esthetics for any given path of placement usually may be selected. In some instances, gingivally placed bar clasp arms may be used to advantage; in others, circumferential clasp arms located cervi-cally may be used. This is especially true when other abutment teeth located more posteriorly may bear the major responsi-bility for retention. In still other instances, a tapered wrought-wire retentive clasp arm may be placed to better esthetic advantage than a cast clasp arm. The placement of clasp arms for esthetic reasons does not ordinarily justify altering the path of placement at the expense of mechanical factors. How-ever, it should be considered concurrently with other factors, and if a choice between two paths of insertion of equal merit www.konkur.in 138 Part I General Concepts/Treatment Planning permits a more esthetic placement of clasp arms by one path than the other, that path should be given preference.
When anterior replacements are involved, the choice of path is limited to a more vertical one for reasons previously stated. In this instance alone, esthetics must be given pri-mary consideration, even at the expense of altering the path of placement and making all other factors conform. This factor should be remembered when the other three factors are considered so that compromises can be made at the time other factors are being considered.
FINAL PATH OF PLACEMENT The final path of placement will be the anteroposterior and lateral position of the cast, in relation to the vertical arm of the surveyor that best satisfies all factors: guiding planes, retention, interference, esthetics, and implant long axis.
All proposed mouth changes should be indicated on the diagnostic cast in red pencil, with the exception of restora-tions to be done. These may also be indicated on an accom-panying chart if desired. Extractions and surgery are given priority to allow time for healing. The remaining red marks represent actual modifications of the teeth that remain to be done, which consist of the preparation of proximal surfaces, the reduction of buccal and lingual surfaces, and the prepa-ration of rest seats. Except when they are placed in the wax pattern for a cast restoration, preparation of rest seats should always be deferred until all other mouth preparations have been completed.
The actual locations of rests will be determined by the proposed design of the denture framework. Therefore the tentative design should be sketched on the diagnostic cast in pencil after the path of placement has been decided. This is done not only to locate rest areas but also to record graphi-cally the plan of treatment before mouth preparations. In the intervening time between patient visits, other partial denture restorations may have been considered. The dentist should have the plan of treatment readily available at each succeeding appointment to avoid confusion and to keep a reminder about that which is to be done and the sequence that will be required.
The plan for treatment should include (1) the diagnostic cast with the mouth preparations and the denture design marked on it; (2) a chart showing the proposed design and the planned treatment for each abutment; (3) a working chart showing the total treatment involved that will permit a quick review and a check-off of each step as the work progresses; and (4) a record of the fee quoted for each phase of treatment that can be checked off as it is recorded on the patient’s per-manent record.
Red pencil marks on the diagnostic cast are used to indicate the locations of areas to be modified as well as the locations of rests (Figure 11-14). Although it is not nec-essary that rest areas be prepared on the diagnostic cast, it is advisable for the beginning student to have done this before proceeding to alter the abutment teeth. This applies equally to crown and inlay preparations on abutment teeth. It is advisable, however, for even the most experienced den-tist to have trimmed the stone teeth with the surveyor blade wherever tooth reduction is to be done. This identifies not only the amount to be removed in a given area but also the plane in which the tooth is to be prepared. For example, a proximal surface may need to be recontoured in only the upper third or the middle third to establish a guiding plane that will be parallel to the path of placement. This is not usually parallel to the long axis of the tooth, and if the rotary instrument is laid against the side of the tooth, the existing surface angle will be maintained, thus avoiding the need to establish a new plane that is parallel to the path of placement.
The surveyor blade, which represents the path of place-ment, may be used to advantage to trim the surface of the abutment tooth whenever a red mark appears. The resulting surface represents the amount of tooth to be removed in the mouth and indicates the angle at which the handpiece must be held. The cut surface on the stone tooth is not marked A B Figure 11-14 Diagnostic casts can serve as a visual guide for tooth preparation. A, Surveyed cast shows areas requiring tooth reduc-tion in red (mesio-occlusal rest and distal guide plane #28, cingulum rest #27), as well as path of insertion tripod marks. B, This mesially tipped molar has been diagnosed to have a ring clasp. Red markings show the necessary mesio-occlusal and disto-occlusal rests re-quired, as well as the mesial guide plane. Also shown is the reduction necessary to lower the lingual height of contour at the mesiolingual line angle. All required axial contour adjustments are determined through the appropriate use of a surveyor.
www.konkur.in 139 Chapter 11 Surveying with red pencil again, but it is outlined in red pencil to posi-tively locate the area that is to be prepared.
RECORDING RELATION OF CAST TO SURVEYOR Some method of recording the relation of the cast to the ver-tical arm of the surveyor must be used so that the cast may be returned to the surveyor for future reference, especially during mouth preparations. The same applies to the need for returning any working cast to the surveyor for shaping wax patterns, trimming blockout on the master cast, or locating clasp arms in relation to undercut areas.
Obviously the trimmed base will vary with each cast; therefore recording the position of the surveyor table is of no value. If it were, calibrations could be incorporated on the surveyor table that would allow the same position to be reestablished. Instead, the position of each cast must be established separately, and any positional record applies only to that cast.
Of several methods, two seem to be the most convenient and accurate. One method is to place three widely diver-gent dots on the tissue surface of the cast with the tip of a carbon marker with the vertical arm of the surveyor in a locked position. Preferably these dots should not be placed on areas of the cast involved in the framework design. The dots should be encircled with a colored pencil for easy iden-tification. When the cast is returned to the surveyor, it may be tilted until the tip of the surveyor blade or diagnostic stylus again contacts the three dots in the same plane. This approach, which will produce the original position of the cast and therefore the original path of placement, is known as tripoding the cast (Figure 11-15). Some dentists prefer to make tiny pits in the cast at the location of the tripoding dots to preserve the orientation of the cast and to transfer this relationship to the refractory cast.
A second method is to score two sides and the dorsal aspect of the base of the cast with a sharp instrument held against the surveyor blade (see Figure 11-15). When the cast is tilted until all three lines are again parallel to the surveyor blade, the original cast position can be reestablished. Fortu-nately, the scratch lines will be reproduced in any duplica-tion, thereby permitting any duplicate cast to be related to the surveyor in a similar manner. Whereas a diagnostic cast and a master cast cannot be made to be interchangeable, a refractory cast, which is a duplicate of the master cast, can be repositioned on the surveyor at any time. The techni-cian must be cautioned not to trim the sides of the cast on the cast trimmer and thereby lose the reference marks for repositioning.
Carbon rod substituted to mark height of contour A D C B Figure 11-15 A and B, The path of placement is determined, and the base of the cast is scored to record its relation to the surveyor for future repositioning. C, An alternate method of recording the relation of the cast to the surveyor is known as tripoding. A carbon marker is placed in the vertical arm of the surveyor, and the arm is adjusted to the height by which the cast can be contacted in three divergent locations. The vertical arm is locked in position, and the cast is brought into contact with the tip of the carbon marker. Three resultant marks are encircled with colored lead pencil for ease of identification. Reorientation of the cast to the surveyor is accomplished by tilting the cast until the plane created by three marks is at a right angle to the vertical arm of the surveyor. D, Height of contour is then delineated by a carbon marker.
www.konkur.in 140 Part I General Concepts/Treatment Planning It must be remembered that repositioning a cast on a sur-veyor at any time can involve a certain amount of human error. It has been estimated that an error of 0.2 mm can be anticipated when a cast with three reference points on its base is reoriented. This reorientation error can influence the placement of appropriate blockout wax and may result in ineffective placement of direct retainers into prescribed undercuts and improper contacts of minor connectors with guiding planes. Therefore reorientation of the cast to the sur-veyor by any method must be done with great care.
SURVEYING THE MASTER CAST The master cast must be surveyed as a new cast, but the prepared proximal guiding plane surfaces will indicate the correct anteroposterior tilt. Some compromises may be nec-essary, but the amount of guiding plane surface remaining after blockout should be the maximum for each tooth. Areas above the point of contact with the surveyor blade are not considered as part of the guiding plane area, and neither are gingival undercut areas, which will be blocked out.
The lateral tilt will be the position that provides equal retentive areas on all principal abutments in relation to the planned clasp design. Factors of flexibility, including the need for extra flexibility on distal extension abutments, must be considered when one is deciding what will provide equal retention on all abutment teeth. For example, cast circum-ferential or cast bar retention on the tooth-supported side of a Class II design should be balanced against the 18-gauge wrought-wire retention on a distal abutment only if the more rigid cast clasp engages a lesser undercut than the wrought-wire clasp arm. Therefore the degree of undercut alone does not ensure relatively equal retention unless clasp arms of equal length, diameter, form, and material are used.
Gross interference will have been eliminated during mouth preparation. Thus for a given path of placement that is pro-viding guiding planes and balanced retention, any remain-ing interference must be eliminated with blockout. If mouth preparations have been adequately planned and executed, the undercuts remaining to be blocked out should be minimal.
The base of the cast is now scored, or the cast is tripoded as described previously. The surveyor blade or diagnostic stylus then may be replaced with a carbon marker, and the height of convexity of each abutment tooth and soft tissue contours may be delineated. Similarly, any areas of interfer-ence to the rigid parts of the framework during seating and removal should be indicated with the carbon marker so areas to be blocked out or relieved can be located.
Carbon markers that become the slightest bit worn from use should be discarded. A worn (tapered) carbon marker will indicate heights of contour more occlusally located than those that actually exist. The carbon marker must be parallel to the vertical spindle of the surveyor (Figure 11-16). The diagnostic stylus should always be checked to ensure that it is not bent or distorted.
MEASURING RETENTION The surveyor is used with the master cast for two purposes: (1) to delineate the height of contour of the abutment teeth both to locate clasp arms and to identify the location and magnitude of retentive undercuts; and (2) to trim blockout of any remaining interference to placement and removal of the denture. The areas involved are those that will be crossed by rigid parts of the denture framework.
The exact undercut that retentive clasp terminals will occupy must be measured and marked on the master cast (Figure 11-17). Undercuts may be measured with an under-cut gauge, such as those provided with the Ney and Jelenko surveyors. The amount of undercut is measured in hun-dredths of an inch, with the gauges allowing measurements up to 0.03 inch. Theoretically the amount of undercut used may vary with the clasp to be used, up to a full 0.03 inch. However, undercuts of 0.01 inch are often adequate for retention by cast retainers. Tapered wrought-wire retention may safely use up to 0.02 inch without inducing undesirable torque on the abutment tooth, provided the wire retentive arm is long enough (at least 8 mm). The use of 0.03 inch is rarely, if ever, justified with any clasp. When greater retention is required, such as when abutment teeth remain on only one side of the arch, multiple abutments should be used, rather than increased retention on any one tooth.
When a source of light is directed toward the tooth being surveyed, a triangle of light is visible. This triangle is bounded by the surface of the abutment tooth on one side and the blade of the surveyor on the other, the apex being the point of contact at the height of convexity and the base of the triangle being the gingival tissues (Figure 11-18). Retention will be determined by (1) the magnitude of the angle of cervical con-vergence below the point of convexity; (2) the depth at which the clasp terminal is placed in the angle; and (3) the flexibility of the clasp arm. The intelligent application of various clasp designs with their relative flexibility is of greater importance than the ability to measure an undercut with precise accuracy.
Figure 11-16 A worn carbon marker (left) should be discard-ed because it will invariably misleadingly mark the height of con-tour for a given orientation of the cast to the vertical spindle of the surveyor. Unworn carbon (right) with an angled end is prefer-able for marking heights of contour on abutment teeth and per-forming surveys of soft tissue areas.
www.konkur.in 141 Chapter 11 Surveying The final design may now be drawn on the master cast with a fine crayon pencil, preferably one that will not come off during duplication. Graphite is usually lifted in duplica-tion, but some crayon pencil marks will withstand dupli-cation without blurring or transfer. Sizing or spraying the master cast to protect such pencil marks is usually not advis-able unless it is done with extreme care to avoid obliterating the surface detail.
BLOCKING OUT THE MASTER CAST After the path of placement and the location of undercut areas have been established on the master cast, any under-cut areas that will be crossed by rigid parts of the denture (which is every part of the denture framework but the retentive clasp terminals) must be eliminated by blockout.
In the broader sense of the term, blockout includes not only the areas crossed by the denture framework during seating and removal but also (1) those areas not involved that are blocked out for convenience; (2) ledges on which clasp patterns are to be placed; (3) relief beneath connec-tors to avoid tissue impingement; and (4) relief to provide for attachment of the denture base to the framework.
Ledges or shelves (shaped blockout) for locating clasp patterns may or may not be used (Figure 11-19). How-ever, this should not be confused with the actual block-ing out of undercut areas that would offer interference to the placement of the denture framework. Only the latter is made on the surveyor, with the surveyor blade or diag-nostic stylus used as a paralleling device.
A B Figure 11-17 A, Undercut gauge measures the depth of undercut below the height of contour. I-bar direct retainer will contact the tooth from the point of the undercut to the height of contour. The depth to which the retentive clasp arm can be placed depends not only on its length, taper, and diameter and the alloy from which it is made, but also on the type of clasp. A circumferential clasp arm is more flexible than a bar clasp arm of the same length (see Chapter 7). B, Specific measurement of the undercut gingiva to the height of contour may be ascertained with the use of an undercut gauge attached to the surveyor. Simultaneous contact of the shank of the undercut gauge at the height of contour and of the lip of a specific undercut gauge on a tooth in the infrabulge area establishes de-finitively the degree and location of undercut. Therefore the tip of the retentive arm of the direct retainer may be placed at the planned depth of the undercut.
Figure 11-18 Tooth undercut is best viewed against a good source of light passing through the triangle bounded by the sur-face of the abutment tooth, the surveyor blade, and the gingival tissues.
www.konkur.in 142 Part I General Concepts/Treatment Planning Hard inlay wax may be used satisfactorily as a blockout material. It is easily applied and is easily trimmed with the surveyor blade. Trimming is facilitated by slight warming of the surveyor blade with an alcohol torch. Whereas it is true that any wax will melt more readily than a wax-clay mixture if the temperature of the duplicating material is too high, it should be presumed that the duplicating mate-rial will not be used at such an elevated temperature. If the temperature of the duplicating material is high enough to damage a wax blockout, other distortions resulting in an inaccurate duplication will likely occur.
Paralleled blockout is necessary for areas that are cervical to guiding-plane surfaces and over all undercut areas that will be crossed by major or minor connectors. Other areas that are to be blocked out for convenience and for avoidance of difficulties in duplication should be blocked out with hard baseplate wax or oil-base modeling clay (artist’s modeling clay). Such areas include the labial surfaces and labial undercuts not involved in the denture design and the sublingual and distolingual areas beyond the limits of the denture design. These are blocked out arbitrarily with hard baseplate wax or clay, but because they have no relation to the path of placement, they do not require the use of the surveyor. Modeling clay that is water soluble should not be used when duplication proce-dures are involved.
Areas to be crossed by rigid connectors, on the other hand, should be trimmed with the surveyor blade or some other surveyor tool parallel to the path of placement (Fig-ure 11-20). This imposes a considerable responsibility on the technician. If the blockout is not sufficiently trimmed to expose guiding-plane surfaces, the effects of these guiding planes, which were carefully established by the dentist, will be nullified. If, on the other hand, the techni-cian is overzealous in paralleling the blockout, the stone cast may be abraded by heavy contact with the surveyor blade. Although the resulting cast framework would seat back onto the master cast without interference, interfer-ence to placement in the mouth would result. This would necessitate relieving the casting at the chair, which is not only an embarrassing and time-consuming operation but also one that may have the effect of obliterating guiding plane surfaces.
RELIEVING THE MASTER CAST Tissue undercuts that must be blocked out are paral-leled in much the same manner as tooth undercuts. The difference between blockout and relief must be clearly understood (Figures 11-21 and 11-22). For example, tis-sue undercuts that would offer interference to the seating of a lingual bar connector are blocked out with blockout wax and trimmed parallel to the path of placement. This does not in itself necessarily afford relief to avoid tissue impingement. In addition to such blockout, a relief of varying thickness must sometimes be used, depending on the location of the connector, the relative slope of the alveolar ridge, and the predictable effect of denture rota-tion. It must be assumed that indirect retainers, as such, or indirect retention is provided in the design of the den-ture to prevent rotation of the lingual bar inferiorly. A vertical downward rotation of the denture bases around B A Figure 11-19 The wax ledge on the buccal surface of the molar abutment will be duplicated in a refractory cast for exact placement of the clasp pattern. Note that the ledge has been carved slightly below the penciled outline of the clasp arm. This allows the gingival edge of the clasp arm to be polished and still remain in its planned relationship to the tooth when the denture is seated. It should also be noted that the wax ledge definitively establishes planned placement of the direct retainer tip into the measured undercut.
Figure 11-20 All guiding-plane areas must be parallel to the path of placement, and all other areas that will be contacted by rigid parts of the denture framework must be made free of the un-dercut by parallel blockout. Relief must also be provided for the gingival crevice and gingival margin. Black regions designate par-allel blockout at proximal guide-plane surfaces and relief along the palatal marginal gingiva.
www.konkur.in 143 Chapter 11 Surveying A B Figure 11-21 Parallel blockout (A, labial surfaces of all teeth and gingivae to the retentive clasp on tooth #2) and relief (B, marginal gingivae of palatal surfaces of teeth and at distal minor connector) in preparation for framework casting. These spaces allow planned seating of the framework without tissue trauma while accommodating the addition of acrylic-resin beneath the distal extension minor connector for base support without metal contact.
Figure 11-22 Relief and blockout of the master cast before duplication. All undercuts involved in the denture design have been blocked out parallel to the path of placement, except the retentive tips of the retainer clasps. Residual ridges have been provided 20-gauge relief for denture base material.
Figure 11-23 Sagittal section of the cast and denture frame-work. The lingual alveolar ridge slopes inferiorly and posteriorly (upper figure). When the force is directed to displace the denture base downward, the lingual bar rotates forward and upward but does not impinge on the soft tissue of the alveolar ridge (lower figure). Therefore in such instances, adequate relief to avoid im-pingement is gained when the tissue side of the lingual bar is highly polished during the finishing process.
posterior abutments places the bar increasingly farther from the lingual aspect of the alveolar ridge when this surface slopes inferiorly and posteriorly (Figure 11-23). Adequate relief of soft tissues adjacent to the lingual bar is obtained by the initial finishing and polishing of the framework in these instances. However, excessive upward vertical rotation of a lingual bar will impinge on lingual tissues if the alveolar ridge is nearly vertical or undercut to the path of placement (Figure 11-24). The region of the cast involving proposed placement of the lingual bar should, in this situation, be relieved first by parallel block-out and then by a 32-gauge wax strip. Low-fusing casting wax (such as, Kerr’s green casting wax) should not be used for this purpose; it is too easily thinned during adapting and may be affected by the temperature of the duplicating material. Pink casting wax should be used, even though it is difficult to adapt uniformly. A pressure-sensitive, adhesive-coated casting wax is preferred because it adapts www.konkur.in 144 Part I General Concepts/Treatment Planning Table 11-1 Differentiation between Parallel Blockout, Shaped Blockout, Arbitrary Blockout, and Relief Site Material Thickness Parallel Blockout Proximal tooth surfaces to be used as guiding planes Hard baseplate wax or blockout material Only undercut remaining below contact of the surveyor blade with tooth surface Beneath all minor connectors Hard baseplate wax or blockout material Only undercut remaining below contact of the surveyor blade with tooth surface Tissue undercuts to be crossed by rigid connectors Hard baseplate wax or blockout material Only undercut remaining below contact of the surveyor blade with surface of the cast Tissue undercuts to be crossed by the origin of bar clasps Hard baseplate wax or blockout material Only undercut remaining below contact of the surveyor blade with surface of the cast Deep interproximal spaces to be covered by minor connectors or linguoplates Hard baseplate wax or blockout material Only undercut remaining below contact of the surveyor blade with surface of the cast Beneath bar clasp arms to gingival crevice Hard baseplate wax or blockout material Only undercut area involved in attachment of the clasp arm to the minor connector Figure 11-24 An undercut alveolar ridge was blocked out par-allel to the path of placement in fabricating the lingual bar (up-per figure). Application of vertical force to cause rotation of the lingual bar upward can cause impingement of lingual tissue on the alveolar ridge (lower figure). To avoid impingement in these instances, not only should the master cast be blocked out paral-lel to the path of placement, but an additional relief of 32-gauge sheet wax should be used to block out the cast in such undercut areas.
readily and adheres to the cast surface. Any wax, even the adhesive type, should be sealed all around its borders with a hot spatula to prevent its lifting when the cast is moist-ened before or during duplication.
Horizontal rotational tendencies of mandibular distal extension removable partial dentures account for many of the tissue irritations seen adjacent to a lingual mandibular major connector. These irritations can usually be avoided by blocking out all undercuts adjacent to the bar paral-lel to the path of placement and then including adequate stabilizing components in the design of the framework to resist horizontal rotation. Judicious relief of the tissue side of the lingual bar with rubber wheels at the site of the irri-tation most often will correct the discrepancy. Under no circumstances should the rigidity of the major connector be jeopardized by grinding any portion of it.
Still other areas requiring relief are the areas where component parts cross the gingiva and gingival crev-ices. All gingival areas bridged by the denture frame-work should be protected from possible impingement resulting from rotation of the denture framework. Hard inlay wax may be used to block out gingival crevices (see Figure 11-21).
PARALLELED BLOCKOUT, SHAPED BLOCKOUT, ARBITRARY BLOCKOUT, AND RELIEF Table 11-1 differentiates between paralleled blockout, shaped blockout, arbitrary blockout, and relief. The same factors apply to both maxillary and mandibular arches, except that relief is ordinarily not used beneath palatal major connectors, as it is with mandibular lingual bar connectors, except when maxillary tori cannot be cir-cumvented, or when resistive median palatal raphes are encountered.
www.konkur.in 145 Chapter 11 Surveying Table 11-1 Differentiation between Parallel Blockout, Shaped Blockout, Arbitrary Blockout, and Relief—cont’d Site Material Thickness Shaped Blockout On buccal and lingual surfaces to locate plastic or wax patterns for clasp arms Hard baseplate wax Ledges for location of reciprocal clasp arms to follow height or convexity so that they may be placed as cervi-cal as possible without becoming retentive Ledges for location of retentive clasp arms to be placed as cervical as tooth contour permits; point of origin of clasp to be occlusal or incisal to height of the convexity, cross-ing the survey line at fourth terminal, and to include un-dercut area previously selected in keeping with flexibility of the clasp type being used Arbitrary Blockout All gingival crevices Hard baseplate wax Enough to just eliminate gingival crevice Gross tissue undercuts situated below areas involved in the design of denture framework Hard baseplate wax or oil-based clay Leveled arbitrarily with a wax spatula Tissue undercuts distal to the cast frame-work Hard baseplate wax or oil-based clay Smoothed arbitrarily with a wax spatula Labial and buccal tooth and tissue under-cuts not involved in denture design Hard baseplate wax or oil-based clay Filled and tapered with spatula to within the upper third or crown Relief Beneath lingual bar connectors or the bar portion of the linguoplates when indicated (see text) Adhesive wax sealed to the cast; should be wider than the major connector to be placed on it 32-gauge wax if the slope of the lingual alveolar ridge is parallel to the path of placement; 32-gauge wax after parallel blockout of undercuts if the slope of the lingual alveolar ridge is undercut to the path of placement Areas in which major connectors will con-tact thin tissue, such as hard areas so frequently found on lingual or mandibu-lar ridges and elevated palatal raphes Hard baseplate wax Thin layer flowed on with hot wax spatula; however, if the maxillary torus must be covered, the thickness of the relief must represent the difference in the degree of displacement of the tissues covering the torus and the tissues covering the residual ridges Beneath framework extensions onto ridge areas for attachment of resin bases Adhesive wax, well adapted to and sealed to the cast beyond the involved area 20-gauge wax www.konkur.in CHAPTER 12 Considerations for the Use of Dental Implants with Removable Partial Dentures CHAPTER OUTLINE Physiologic Distinction Between Prostheses Replacing Anatomy and Functional Ability Strategically Placed Implants for Removable Partial Denture Stability and Improved Patient Accommodation Movement Control With Selective Implant Placement Placement for Support Versus Retention Anatomic Concerns for Class I and II Arches Benefits of Implant Versus Surveyed Crown Strategically Lost Tooth and Implant Use Use of Abutment for Support and/or Retention Influence of Opposing Occlusion on Implant Location and Implant Abutment Design Treatment Planning Survey Considerations: Path of Insertion for Teeth and Implants Location Influenced by Anatomic and Opposing Occlusal Factors Clasp Assembly Requirements if an Implant Is Used Adjacent to Tooth Clinical Examples Summary Dental implants are increasingly being used in a variety of ways in the replacement of teeth. Providing implants to sup-port all teeth needing replacement is often preferable if indi-cated and if the patient can afford to do so. If the patient is unable to pursue an implant-only supported prosthesis, this should not keep him or her from considering an implant, because the patient still may benefit from a carefully selected implant used for critical clinical performance advantage when removable partial dentures (RPDs) are pursued. Addi-tionally, implants can be used for RPDs while allowing for future implant-only treatment options.
This chapter presents basic considerations in the selection of implants to improve prosthesis performance by increas-ing functional stability. Prosthesis instability, a commonly stated problem with removable prostheses, is presented as a potentially confounding and unique component of oral sen-sory-motor function at issue when patients are challenged with removable prostheses. When all the relevant informa-tion necessary to adequately address how best to manage a patient’s tooth loss is considered, one aspect that should be kept in mind is that among the various prostheses from which we choose (conventional fixed partial denture, implant-supported partial denture, and removable partial denture), the removable partial denture is the most unique. Exactly how we describe this uniqueness to our patients is critical to their appropriate prosthesis selection and provides a clue to us as to how dental implants can best be used in conjunction with RPDs.
Patient factors—clinical factors (dentition and residual ridge condi-tions), functional factors, psychological factors; prosthesis factors— implant, fixed, or removable partial dentures.
www.konkur.in 147 Chapter 12 Considerations for the Use of Dental Implants with Removable Partial Dentures PHYSIOLOGIC DISTINCTION BETWEEN PROSTHESES Although physiologic aspects of replacement prostheses have not been adequately researched, the fact that RPDs are less like teeth than fixed prostheses strongly suggests that we should delineate this to our patients. The most obvious difference between treatment options is the man-ner in which they are supported—conventional fixed partial dentures using teeth with periodontal ligaments versus implant prostheses using osseointegrated implants versus RPDs using teeth, or teeth and mucosa. An addi-tional difference is the rigidity associated with the pros-thesis-tooth interface connection—cemented prostheses rigidly connected to teeth versus screw-retained implant prostheses rigidly connected between implants versus clasp/attachment-retained RPD prostheses.
Given this support and rigid connection difference, it is easy to recognize that one distinguishing difference between fixed prostheses (tooth or implant supported) and tooth-tissue–supported RPDs is the potential for movement under function. The impact that such movement will have on related oral sensory and oral functional expectations of patients is potentially critical to understand. These patient-specific perspectives are variable because some of our patients are better at distinguishing oral sensory input than others, and some are more functionally capable than oth-ers. The importance of sensory-functional considerations is evident when it is realized that the oral cavity is unique from a sensory perception standpoint, because the sensory input from the mouth regions involved in the task of mastication represents one of the largest collections of sensory informa-tion processed for a single functional act in the body, rivaled closely by the multifunctional hand (Figure 12-1).
REPLACING ANATOMY AND FUNCTIONAL ABILITY With this relative importance of oral sensory function asso-ciated with mastication in mind, it would be helpful to con-sider what exactly we are attempting to replace when we discuss with our patients the management of missing teeth. This involves both the physical tools for mastication and the oral ability for precise and stable neuromuscular functions to manipulate a bolus of food. Analysis of chewing studies shows that the oral receptor feedback that guides movement of the mandible comes from various sources. The most sensi-tive input (i.e., the input that provides the most refined and precisely controlled movement) comes from the periodontal mechanoreceptors (PMRs), with additional input coming from the gingiva, mucosa, periosteum/bone, and temporo-mandibular joint (TMJ) complex. Loss of teeth leads to loss of guidance precision, which may be caused by the deficit in PMRs. No artificial replacements reestablish the same ability inherent in the PMR guidance. Yet prostheses may vary in their ability to affect the deficit on the basis of sensory-related dynamics. This notion of variable prosthesis impact suggests that either the prostheses are not the same or patients may vary in terms of neuromuscular ability.
What is meant by the term neuromuscular ability? This is the task-specific sensory-motor performance “ability” that a specific patient possesses for the task/act of masti-cation. It involves multiple factors, including the selection of a prosthesis that may vary in qualitative neuromuscular ability compared with the gold standard—natural teeth. This prosthesis influence can be favorable when replace-ments are more like teeth (in their neuromuscular sen-sory input nature), or it can be unfavorable if they are unlike teeth and potentially cause confounding sensory input that diminishes function (e.g., implant fixed partial denture versus denture base). This therapeutic variability suggests that prostheses differ in the influence they have on oral sensory receptors; therefore, this may affect our patient’s perspective on how much we have improved the rehabilitation of oral function (i.e., neurophysiologic abil-ity rehabilitation).
Stability under function is critical for limiting con-founding sensory input because instability is a significant confounder of normal function. Whether a prosthesis is stable or unstable influences whether it provides a posi-tive or a negative contribution to an already reduced peripheral receptor influence. This is compounded by the inherent neuromuscular variability seen between patients. In this discussion, neurophysiologic and neuromuscular both refer to the relationship between oral sensory input and the resultant oral func-tional movement.
Foot Hip Genitalia Toes Leg Trunk Head Lips Tongue Pharynx Intra-abdominal Upper lip Face Nose Eye Thumb Index Middle Ring Hand Wrist Forearm Elbow Arm Little Lower lip Teeth, gums, and jaw Shoulder Neck Figure 12-1 Somatosensory cortex.
www.konkur.in 148 Part I General Concepts/Treatment Planning Consequently, in some of our patients, replacement teeth may have less of a “margin” for sensory contribution with-out creating functional impediments.
STRATEGICALLY PLACED IMPLANTS FOR REMOVABLE PARTIAL DENTURE STABILITY AND IMPROVED PATIENT ACCOMMODATION To decide which treatment option is best from a patient accommodation perspective, it must be determined whether this specific patient possesses the ability to meet the chal-lenge posed by the neuromuscular deficit. If the answer is yes, then the patient may be an appropriate candidate for an RPD. However, if the answer is no and for good reasons an RPD is chosen, the use of strategically located implants to assist the patient in meeting the challenge should be discussed.
Stated differently, the success of any prosthesis may be compromised by a significant level of sensory loss. In addi-tion, if the patient also possesses an inherently poor neu-rophysiologic ability, the choice of a prosthesis that creates an additional sensory “burden” may make the situation too challenging for the patient. The greatest contributing factor to the challenge is instability, and stability is most efficiently improved by the use of dental implants.
MOVEMENT CONTROL WITH SELECTIVE IMPLANT PLACEMENT The most beneficial use of an implant with an RPD is to reduce the negative impact of any sensory input resulting from pros-thesis movement. The greatest potential for movement is seen with Kennedy Class I and II prostheses. Consequently, the greatest benefits in the use of implants with Class I and II RPDs are to gain stability, control prosthesis movement, and reduce unnatural sensations that are movement-related.
The clinician who is deciding exactly where to place an implant for the greatest benefit should take many factors into consideration. The ultimate decision will likely be influenced by a combination of important clinical issues. Basic consid-erations include factors associated with support, stability, and retention; residual ridge anatomy; strategic tooth loss; oppos-ing occlusal influence; and the condition of a terminal tooth.
Placement for Support Versus Retention When considering where to place an implant to assist an RPD, we should ask, “What will best serve the patient over time?” Certainly, we want the prostheses to be adequately retained, but as we consider what the retentive need for our RPD is, we should also ask, “Is it the most important issue that patients face?” Support for prostheses is the characteristic that resists the greatest functional forces—those associated with chewing. A well-supported prosthesis is likely to impart less displace-ment to the soft tissue; less displacement likely is associated with a more comfortable prosthesis. If the major concern is comfort, then support is more often the best treatment target for an implant. The most effective location for resisting force would be the posterior occlusal contact position—that posi-tion where the forces of occlusion are likely the greatest.
Anatomic Concerns for Class I and II Arches It is clear that placement of implants within the alveolar ridge requires adequate bone volume. Tooth loss commonly leads to residual ridge resorption; the longer from the time of extraction that a patient is seen, the more alveolar resorption is expected.
When this occurs in the posterior jaw, vertical loss of vol-ume at the ridge crest is complicated by the maxillary sinus and inferior alveolar nerve at the apical location. These anatomic boundaries require careful consideration of surgical manipula-tion and must be considered as potentially more burdensome to the patient (i.e., they increase surgical risk). It is unclear at this time whether shortened implants (≤ 8 mm) in posterior edentulous regions are adequate for the necessary resistance. Certainly, the opposing occlusion contributes to this decision and natural teeth would be expected to generate more forceful occlusal function demanding greater implant resistance, com-pared with an opposing complete denture prosthesis.
Benefits of Implant Versus Surveyed Crown Surveyed crowns have long been used to maximize sup-port, stability, and retention for RPDs. The ability to con-trol expected movement in the removable prosthesis can be improved by the use of such a carefully contoured crown. The same can be said for an implant, which presents a logi-cal comparison as to which is a better choice for patients.
As an example, if a patient is deciding between a crown on tooth #6 versus an implant at site #5, what should he or she be considering? Both can be used to provide support, stability, and retention. If the crown is designed to accom-modate an attachment, it can also be used without a visible clasp. This is a more technically demanding intervention and carries with it more challenging maintenance require-ments. Also, the typical consideration regarding caries risk should be discussed. This suggests the need to think about long-term needs when deciding between options.
The implant at site #5 can reduce the need for a clasp, while at the same time providing a reasonably simple maintenance requirement for managing attachment needs. The implant attachment can be easily accommodated within a simple RPD framework that accommodates the #5 replacement tooth and available prosthetic space for attachment needs. Variable reten-tion is available based on the type of attachment selected—an advantage over the use of a clasp, which carries the risk of frac-ture with repeated adjustment over time.
Strategically Lost Tooth and Implant Use Loss of a canine or molar abutment in an existing removable prosthesis can have a significantly negative impact on the pros-thesis. Resistance to movement toward the tissue and across the tissue (aspects of support and stability) is greatly affected. Pre-vious interventions for strategic tooth loss, such as the Swing-Lock prosthesis, have been based on recognition of the need www.konkur.in 149 Chapter 12 Considerations for the Use of Dental Implants with Removable Partial Dentures to use more teeth to meet the demands for comfortable, stable function. Use of an implant in this situation can directly address the loss of a strategic component that controls functional move-ment and may be able to utilize the existing prosthesis.
Use of Abutment for Support and/or Retention As an implant is placed more distally (in a distal extension), it serves more of a support function than a retention func-tion. In a situation where support and retention are desired, the connecting design (the actual attachment male-female components) must allow adequate resistance to supporting forces without premature deformation of the retentive com-ponent and reduction of the retaining properties.
Generally, a component designed to primarily provide retention may not serve a supportive role for the desired period of time before replacement. Use of a more resistant retentive element to better resist supportive forces may pro-vide retention in excess of the patient’s ability to comfortably remove the prosthesis for daily hygiene.
Influence of Opposing Occlusion on Implant Location and Implant Abutment Design Distal extension RPDs are significantly influenced by the type and plane of orientation of the opposing dentition. Opposing complete dentures, which exhibit a regular occlu-sal plane, encourage equal force distribution associated with occlusion, allowing an optimum scenario for comfort.
Irregular occlusal planes created by super-erupted oppos-ing natural teeth make control of functional forces more chal-lenging at the residual alveolar ridge level. Use of implants for such oral conditions may improve prosthesis comfort, espe-cially if support is fully utilized (i.e., no retainer function). In such a scenario, implant abutment design should serve to control movement toward and across the denture bearing area. This places a premium on nonresilient and nonreten-tive abutment designs. Care should be taken to closely follow such designs during the early postinsertion phase because the potential for excessive occlusal loads could have detrimental effects on abutment connections and implant stability.
TREATMENT PLANNING Survey Considerations: Path of Insertion for Teeth and Implants Engagement of implant abutments requires consideration of the prosthesis path of insertion (POI). An implant placed coincident with the POI allows greater prosthetic flexibility in abutment selection and a more favorable RPD connection (Figure 12-2).
Location Influenced by Anatomic and Opposing Occlusal Factors Implant use requires interocclusal space sufficient for prosthetic materials: denture base with framework minor connector, attachment housing, and denture tooth. There-fore, the opposing occlusion should be evaluated preopera-tively. This becomes even more critical when less residual ridge resorption occurs.
With excessive maxillary residual ridge resorption, the implant location may be more palatal/lingual, requiring con-sideration of the impact that such a location may have on palatal or lingual contours.
Clasp Assembly Requirements if an Implant Is Used Adjacent to Tooth Engagement of an implant abutment to provide horizontal and vertical resistance supplements the resistance derived from tooth engagement. Consequently, less tooth cover-age may be necessary for regions of the arch that have an implant.
If implant use is directed primarily to retention, tooth engagement for maximum resistance remains necessary to improve stability. If this is not achieved, deterioration in retention may be more rapid because of excessive forces directed to the retentive element, which could result in more rapid material deformation.
CLINICAL EXAMPLES Figures 12-3 to 12-6 present clinical examples.
Primary use: Support Figure 12-3: Kennedy Class I mandibular Figure 12-4: Kennedy Class II Combined use: Support and retention Figure 12-5: Kennedy Class III Figure 12-6: Kennedy Class II SUMMARY RPDs are unique compared with prostheses supported by natural teeth. Their uniqueness largely relates to their potentially significant negative impact on oral sensory input during function, as well as on the prosthetic bulk required. Because of this, RPDs may present a challenge to accommo-dation, of which patients should be made aware during the treatment planning stage.
Patient sensory perception and neuromuscular variabil-ity may influence prosthesis selection. The sensory deficit associated with tooth loss can be complicated by unstable prostheses; for challenging RPDs, the use of implants can significantly contribute to stability.
ACKNOWLEDGMENT The authors would like to acknowledge Dr. Tom Salinas for his helpful input during the writing of this chapter.
www.konkur.in 150 Part I General Concepts/Treatment Planning A B C D E Figure 12-2 Parallel path of insertion established with tooth guide planes (distal of the canines and mesial of the molar) and adjacent implants.
www.konkur.in A B D C Figure 12-3 Significant residual ridge resorption with a strategically missing tooth #22. The implant provided sufficient support to improve functional comfort and occlusal function.
A B C Figure 12-4 A long span modification space across the mandibular midline. The implant provided support and retention improved function and comfort while reducing functional stress on the mandibular incisors.
www.konkur.in 152 Part I General Concepts/Treatment Planning A B C Figure 12-5 Support provided by the cingulum and mesial-occlusal rests of the teeth and by the implant at site #3. Implants at sites #5 and #12 provide retention. Guide-plane surfaces augment implants for sufficient retention without clasps.
www.konkur.in 153 Chapter 12 Considerations for the Use of Dental Implants with Removable Partial Dentures A B E C D Figure 12-6 An implant-assisted Class III removable partial denture (RPD) with minimum interocclusal space in an elderly patient desiring no prosthesis clasps. Denture teeth were shaped and positioned for optimum esthetics, allowing creation of occlusal contours by acrylic-resin, attaching the teeth to the frame and attachment housings.
www.konkur.in CHAPTER 13 Diagnosis and Treatment Planning CHAPTER OUTLINE Purpose and Uniqueness of Treatment Patient Interview Shared Decision Making Clinical Examination Objectives of Prosthodontic Treatment Oral Examination Sequence of Oral Examination Diagnostic Casts Purposes of Diagnostic Casts Mounting Diagnostic Casts Sequence for Mounting Maxillary Cast to Axis-Orbital Plane Jaw Relationship Records for Diagnostic Casts Materials and Methods for Recording Centric Relation Diagnostic Findings Interpretation of Examination Data Radiographic Interpretation Periodontal Considerations Caries Risk Assessment Considerations Evaluation of the Prosthesis Foundation—Teeth and Residual Ridge Surgical Preparation Analysis of Occlusal Factors Fixed Restorations Orthodontic Treatment Need for Determining Type of Mandibular Major Connector Need for Reshaping Remaining Teeth Infection Control Differential Diagnosis: Fixed or Removable Partial Dentures Indications for Use of Fixed Restorations Indications for Removable Partial Dentures Choice Between Complete Dentures and Removable Partial Dentures PURPOSE AND UNIQUENESS OF TREATMENT The purpose of dental treatment is to respond to a patient’s needs, both the needs perceived by the patient and those demonstrated through a clinical examination and patient interview. Although similarities have been noted between partially edentulous patients (such as classification designa-tions), significant differences exist, making each patient, and the ultimate treatment, unique.
The delineation of each patient’s uniqueness occurs through the patient interview and diagnostic clinical examination pro-cess. This includes four distinct processes: (1) understanding the patient’s desires or chief concerns/complaints regarding his or her condition (including its history) through a systematic interview process; (2) ascertaining the patient’s dental needs through a diagnostic clinical examination; (3) developing a treatment plan that reflects the best management of desires and needs (with influences unique to the medical condition or oral environment); and (4) executing appropriately sequenced treatment with planned follow-up. The ultimate treatment is individualized to address disease management and the coor-dinated restorative and prosthetic needs that are unique to the patient. Provision of the best care for a patient may involve no treatment, limited treatment, or extensive treatment, and the dentist must be prepared to help patients decide the best treat-ment option given his or her individual circumstances.
PATIENT INTERVIEW Although oral health is an important aspect of overall health, it is an elective health pursuit for most individuals. Clinical Factors Related to Metal Alloys used for Removable Partial Denture Frameworks Comparative Physical Properties of Gold and Chromium-Cobalt Wrought Wire: Selection and Quality Control Summary www.konkur.in 156 Part II Clinical and Laboratory Consequently, the patient presents for professional evalua-tion (1) to address some perception of an abnormality that requires correction; or (2) to maintain optimum oral health. In either situation, but especially for the patient presenting with some chief complaint (often with an important history related to that complaint), it is mandatory that the dentist clearly understand what brings the patient to this evalua-tion. Failure to do so leads to the chance that the patient will be unhappy with the treatment result, because it might not address the very reason he or she came for help. With expe-rience, this subtle point becomes a major component of a clinician’s management focus.
A fundamental objective of the patient interview, which accompanies the diagnostic examination, is to gain a clear understanding of why the patient is presenting for evaluation; this involves having the patient describe the history related to the chief complaint. For complicated clinical problems, the interview and diagnostic examination require two appoint-ments to allow complete gathering of all diagnostic informa-tion needed to formulate a complete plan of treatment.
The interview, an opportunity to develop rapport with the patient, involves listening to and understanding the patient’s chief complaint or concern about his or her oral health. This can include clinical symptoms of pain (provoked or unpro-voked), difficulty with function, concern about appearance, problems with an existing prosthesis, or any combination of symptoms related to the teeth, periodontium, jaws, or pre-vious dental treatment. It is important to listen carefully to what the patient has stated is the reason for presenting for evaluation; this is because all subsequent information gath-ered will be used to discuss these concerns and to relate whether the proposed treatment will affect the patient in any way. Such a discussion at the outset of patient care helps to outline realistic expectations.
Although formats for sequencing the patient interview (and clinical examination) vary, to ensure thoroughness the dentist should follow a sequence that includes: 1. Chief complaint and its history 2. Medical history review 3. Dental history review, especially related to previous pros-thetic experience(s) 4. Patient expectations It is from the aforementioned interaction that patient uniqueness, as mentioned earlier, is best defined. The expec-tations described by the patient are critical to an understand-ing of whether a removable partial denture will satisfy the stated treatment goal(s). The fact that removable partial dentures by necessity require material bulk and often use oral soft tissues for support may be hard to comprehend by patients with no such prosthetic history. Helping the patient understand the normal phase of accommodation to such a prosthesis is an important discussion point in selection of a prosthesis. For those patients with a negative past prosthesis experience, it is necessary to determine before treatment is started whether the design, fit, occlusion, or lack of mainte-nance of the prosthesis can be improved to provide a more positive experience.
SHARED DECISION MAKING When helping patients understand their oral health status, com-prising both disease and deficit considerations, and the means to address both, we should carefully consider what it is they need to hear from us. For most partially edentulous patients, the discussion may involve fairly complex rehabilitation options for addressing their missing teeth. Because of this complexity, our responsibility is to help them sort through the options in an attempt to help them come to the best decision for them. Using a communication model termed shared decision making gives structure to a process where the provider and the patient iden-tify together the best course of care. This process recognizes that there may be complex “trade-offs” in care choice, and it addresses the need to fully inform patients about risks and benefits of care options, as well as ensuring that patient values and preferences play a prominent role in the process. Although it is clear that not all patients desire to participate equally in care decisions, because the options can vary significantly (some are more inva-sive, have greater risks, are accompanied by higher treatment burden than others; there are often varying maintenance needs between options), we should actively engage them in the pro-cess. This is more important given the fact that the tooth replace-ment is often an elective pursuit, and because of this, there is seldom great urgency involved in making a decision.
CLINICAL EXAMINATION OBJECTIVES OF PROSTHODONTIC TREATMENT The objectives of any prosthodontic treatment may be stated as follows: (1) the elimination of disease; (2) the preservation, restoration, and maintenance of the health of the remaining teeth and oral tissues (which will enhance the removable par-tial denture design); and (3) the selected replacement of lost teeth for the purpose of (4) restoration of function in a man-ner that ensures optimum stability and comfort in an estheti-cally pleasing manner. Preservation is a principle that protects from decisions that place too high a premium on cosmetic concerns, and it is the dentist’s obligation to emphasize the importance of restoring the total mouth to a state of health and of preserving the remaining teeth and surrounding tissues.
Diagnosis and treatment planning for oral rehabilitation of partially edentulous mouths must take into consideration the following: control of caries and periodontal disease, res-toration of individual teeth, provision of harmonious occlu-sal relationships, and the replacement of missing teeth by fixed (using natural teeth and/or implants) or removable prostheses. Because these procedures are integrally related, the appropriate selection and sequencing of treatment should precede all irreversible procedures.
The treatment plan for the removable partial denture, which is often the final step in a lengthy sequence of treat-ment, should precede all but emergency treatment. This allows abutment teeth and other areas in the mouth to be www.konkur.in 157 Chapter 13 Diagnosis and Treatment Planning properly prepared to support, stabilize, and retain the remov-able partial denture. This means that diagnostic casts, for designing and planning removable partial denture treatment, must be made before definitive treatment is undertaken. After the major factors that create functional forces are evaluated and those that resist it are understood, the removable partial denture design is drawn on the diagnostic cast, along with a detailed chart of mouth conditions and proposed treatment. This becomes the master plan for the mouth preparations and the design of the removable partial denture to follow.
As pointed out in Chapter 1, failure of removable partial dentures can usually be attributed to factors that result in poor stability. These can result from inadequate diagnosis and failure to properly evaluate the conditions present. This results in failure to prepare the patient and the oral tissues properly before the master cast is fabricated. The importance of the examination, the consideration of favorable and unfa-vorable aspects relative to movement control, and the impor-tance of planning the elimination of unfavorable influences cannot be overemphasized (see Chapter 2).
As mentioned earlier, for complex treatment, two appointments are often required. The first will likely include a preliminary oral examination (to determine the need for management of acute needs), a prophylaxis, full-mouth radiographs, diagnostic casts, and mounting records if base-plates are not required. The follow-up appointment includes mounting of the diagnostic casts (when baseplates and occlu-sion rims are needed), a definitive oral examination, review of the radiographs to augment and correlate with clinical findings, and arrangement of additional consultations when required. Following collection and synthesis of all patient and clinical information, including surveying of the casts, a treatment plan (often with options) is presented.
ORAL EXAMINATION A complete oral examination should precede any treatment decisions. It should include visual and digital examination of the teeth and surrounding tissues with a mouth mirror, explorer, and periodontal probe, vitality tests of critical teeth, and examination of casts correctly oriented on a suitable articulator. Clinical findings are augmented by and corre-lated with a complete intraoral radiographic survey.
During the examination, the objective to be kept fore-most in mind should be the consideration of possibilities for restoring and maintaining the remaining oral structures in a state of health for the longest period of time. This is best accomplished by an evaluation of factors that generate func-tional forces and those that resist them. The stability of tooth and prosthesis position is the goal of such an evaluation. The following sequence of examination allows attention to be paid to aspects of each of these critical features of evaluation for removable partial denture service.
Sequence of Oral Examination An oral examination should be accomplished in the follow-ing sequence: visual examination, pain relief and temporary restorations, radiographs, oral prophylaxis, evaluation of teeth and periodontium, vitality tests of individual teeth, determination of the floor of the mouth position, and impressions of each arch. 1. Relief of pain and discomfort and caries control by placement of temporary restorations. A preliminary examination is conducted to determine the need for management of acute needs and whether a prophylaxis is required to conduct a thorough oral examination. It is advisable not only to relieve discomfort arising from tooth defects but also to determine as early as possible the extent of caries, and to arrest further caries activity until definitive treatment can be instituted.
If tooth contours are restored with temporary restorations, the impression will not be torn on removal from the mouth, and a more accurate diagnostic cast may be obtained. 2. A thorough and complete oral prophylaxis. An adequate examination can be accomplished best with the teeth free of accumulated calculus and debris. Also, accurate diag-nostic casts of the dental arches can be obtained only if the teeth are clean; otherwise the teeth reproduced on the diagnostic casts are not a true representation of tooth and gingival contours. Cursory examination may precede an oral prophylaxis, but a complete oral examination should be deferred until the teeth have been thoroughly cleaned. 3. Complete intraoral radiographic survey (Figure 13-1). The objectives of a radiographic examination are (a) to locate areas of infection and other pathosis that may be pres-ent; (b) to reveal the presence of root fragments, foreign objects, bone spicules, and irregular ridge formations; (c) to reveal the presence and extent of caries and the rela-tion of carious lesions to the pulp and periodontal attach-ment; (d) to permit evaluation of existing restorations as to evidence of recurrent caries, marginal leakage, and overhanging gingival margins; (e) to reveal the presence of root canal fillings and to permit their evaluation as to future prognosis (the design of the removable partial denture may hinge on the decision to retain or extract an endodontically treated tooth); (f) to permit evaluation of periodontal conditions present and to establish the need and possibilities for treatment; and (g) to evaluate the al-veolar support of abutment teeth, their number, the sup-porting length and morphology of their roots, the relative amount of alveolar bone loss suffered through pathogenic processes, and the amount of alveolar support remaining.
Figure 13-1 Complete intraoral radiographic survey of remaining teeth and adjacent edentulous areas reveals much in-formation vital to effective diagnosis and treatment planning. The response of bone to previous stress is of particular value in estab-lishing the prognosis of teeth that are to be used as abutments.
www.konkur.in 158 Part II Clinical and Laboratory 4. Impressions for making accurate diagnostic casts to be mounted for occlusal examination. The casts preferably will be articulated on a suitable instrument. The impor-tance of accurate diagnostic casts and their use will be discussed later in this chapter. 5. Examination of teeth, investing structures, and residual ridges. The teeth, periodontium, and residual ridges can be explored by instrumentation and visual means. Re-cording of relevant patient history and clinical data on diagnosis charts is important to document features im-portant to clinical presentation. These can be recorded on electronic or paper charts for future reference (Figures 13-2 and 13-3).
Visual examination will reveal many of the signs of dental disease. Consideration of caries susceptibility is of primary importance. The number of restored teeth present, signs of recurrent caries, and evidence of decalcification should be noted. Only those patients with demonstrated good oral hygiene habits and low caries susceptibility should be considered good risks without resorting to pro-phylactic measures, such as the restoration of abutment teeth. At the time of the initial examination, periodontal disease, gingival inflammation, the degree of gingival reces-sion, and mucogingival relationships should be observed. Such an examination will not provide sufficient informa-tion to allow a definitive diagnosis and treatment plan. For this purpose, complete periodontal charting that includes pocket depths, assessment of attachment levels, furcation involvement, mucogingival problems, and tooth mobility should be performed. The extent of periodontal destruction REMOVABLE PARTIAL PROSTHODONTICS PATIENT NAME PATIENT NUMBER TREATMENT PLAN COLOR CODE: BLUE:---- CAST METAL RED: ----- RESIN BASE AND WROUGHT WIRE GREEN:-- AREAS TO BE CONTOURED LABORATORY INSTRUCTIONS DESIGN SPECIFICATIONS: 1.
RESTS 2.
RETENTION 3.
RECIPROCATION 4.
MAJOR CONNECTOR 5.
INDIRECT RETENTION 6.
GUIDE PLANES 7.
BASE RETENTION 8.
AREAS TO BE MODIFIED OR CONTOURED INSTRUCTOR: APPROVAL TO SEND TO LABORATORY: DATE: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 R L R L A Figure 13-2 A, Diagnosis record for recording pertinent data.
www.konkur.in 159 Chapter 13 Diagnosis and Treatment Planning B Figure 13-2, cont’d B, Treatment record chart for recording treatment plan and treatment progress.
www.konkur.in 160 Part II Clinical and Laboratory must be determined with appropriate radiographs and use of the periodontal probe.
The number of teeth remaining, the location of the edentulous areas, and the quality of the residual ridge will have a definite bearing on the proportionate amount of support that the removable partial denture will receive from the teeth and edentulous ridges. Tissue contours may appear to present a well-formed edentulous residual ridge; however, palpation often indicates that supporting bone has been resorbed and has been replaced by displaceable, fibrous connective tissue. Such a situation is common in maxillary tuberosity regions. The removable partial denture cannot be supported adequately by tissues that are easily displaced. When the mouth is prepared, this tissue should be recontoured or removed surgically, unless otherwise contraindicated. A small but stable residual ridge is prefer-able to a larger unstable ridge for providing support for the denture. The presence of tori or other bony exostoses must be detected and their presence in relation to framework design must be evaluated. Failure to palpate the tissue over the median palatal raphe to ascertain the difference in its displaceability as compared with the displaceability of the soft tissues covering the residual ridges can lead to a rocking, unstable, uncomfortable denture and to an unsat-isfied patient. Adequate relief of the palatal major connec-tors must be planned, and the amount of relief required is directly proportionate to the difference in displaceability of the tissues over the midline of the palate and the tissues covering the residual ridges.
During the examination, not only each arch but also its occlusal relationship with the opposing arch must be con-sidered separately. A situation that looks simple when the teeth are apart may be complicated when the teeth are in occlusion. For example, an extreme vertical overlap may complicate the attachment of anterior teeth to a maxillary denture. Extrusion of a tooth or teeth into an opposing edentulous area may complicate the replacement of teeth in the edentulous area or may create occlusal interference, which will complicate the location and design of clasp retainers and occlusal rests. Such findings subsequently will be evaluated further by careful analysis of mounted diag-nostic casts.
A breakdown of the fee may be recorded on the back of this chart for easy reference if adjustments or substitutions become necessary because of changes in diagnosis as the work progresses.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Right Left Name: Summary Plan: Procedures: John Doe Cr Cr Red line each completed unit Follow-up care July 1, 1999 Date: Porcelain fused to metal abutment Crowns #18, #29 Maxillary Conventional Complete denture Mandibular Class II mod. 1 removable partial denture Tissue conditioning maxillary arch Primary impression-both arches; make impression trays Preparations. Contour and rest seats #21, #28. Abutment crown preparations #18, #29 Try-in abutment crowns; pick-up for cast for framework casting Fluid wax functional impression: Make altered cast Jaw relation records; shade and mold selection Tooth arrangement Try-in abutment crowns and framework Try-in trial set-up; verify jaw relations Placement: Mandibular crowns, partial denture, maxillary complete denture Figure 13-3 Simple Working Chart. Restorations for individual teeth, crowns, and fixed partial dentures to be made may be marked on the chart and checked off as completed during mouth preparations.
www.konkur.in 161 Chapter 13 Diagnosis and Treatment Planning 6. Vitality tests of remaining teeth. Vitality tests should be given particularly to teeth to be used as abutments and those having deep restorations or deep carious lesions.
This should be done through both thermal and electronic means. 7. Determination of the height of the floor of the mouth to locate inferior borders of lingual mandibular major con-nectors. Mouth preparation procedures are influenced by the choice of major connectors (see Figure 5-6). This determination must precede altering contours of abut-ment teeth.
The fee for examination, which should include the cost of the radiographic survey and the examination of artic-ulated diagnostic casts, should be established before the examination is performed and should not be related to the cost of treatment. It should be understood that the fee for examination is based on the time involved and the service rendered, and that the material value of the radiograph and diagnostic casts is incidental to the effectiveness of the examination.
The examination record should always be available in the office for future consultation. If consultation with another dentist is requested, respect for the hazards of unnecessary radiation justifies loaning the dentist the radiograph for this purpose. However, duplicate films should be retained in the dentist’s files.
DIAGNOSTIC CASTS A diagnostic cast should be an accurate reproduction of all the potential features that aid diagnosis. These include the teeth locations, contours, and occlusal plane relation-ship; the residual ridge contour, size, and mucosal con-sistency; and the oral anatomy delineating the prosthesis extensions (vestibules, retromolar pads, pterygomaxillary notch, hard/soft palatal junction, floor of the mouth, and frena). Additional information provided by appropriate cast mounting includes occlusal plane orientation and the impact on the opposing arch, tooth-to-palatal soft tissue relationship, and tooth-to-ridge relationships both verti-cally and horizontally.
A diagnostic cast is usually made of dental stone because of its strength and the fact that it is less easily abraded than is dental plaster. Generally the improved dental stones (die stones) are not used for diagnostic casts because of their cost. Their greater resistance to abrasion does, however, justify their use for master casts.
The impression for the diagnostic cast is usually made with an irreversible hydrocolloid (alginate) in a stock (perforated or rim lock) impression tray. The size of the arch will determine the size of the tray to be used. The tray should be sufficiently oversized to ensure an opti-mum thickness of impression material to avoid distortion or tearing on removal from the mouth. The technique for making impressions is covered in more detail in Chapter 16.
Purposes of Diagnostic Casts Diagnostic casts serve several purposes as an aid to diagnosis and treatment planning. Some of these are as follows: 1. Diagnostic casts are used to supplement the oral examina-tion by permitting a view of the occlusion from the lingual, as well as from the buccal, aspect. Analysis of the existing occlusion is made possible when opposing casts are occluded, as is a study of the possibilities for improve-ment by occlusal adjustment, occlusal reconstruction, or both. The degree of overclosure, the amount of interoc-clusal space available, and the possibilities of interfer-ence with the location of rests may also be determined.
As stated previously, opportunities for improvement of the occlusal scheme, by occlusal adjustment or occlu-sal reconstruction, are best evaluated by analysis and modification of mounted diagnostic casts. Such proce-dures often include diagnostic waxing to determine the possibility of enhancing the occlusion before definitive treatment is begun (Figure 13-4). In other words, diag-nostic casts permit the dentist to plan ahead to avoid undesirable compromises in the treatment being given a patient. 2. Diagnostic casts are used to permit a topographic sur-vey of the dental arch that is to be restored by means of a removable partial denture. The cast of the arch in question may be surveyed individually with a cast sur-veyor to determine the parallelism or lack of parallelism of tooth surfaces involved and to establish their influ-ence on the design of the removable partial denture. The principal considerations in studying the parallelism of tooth and tissue surfaces of each dental arch are to deter-mine the need for mouth preparation: (a) proximal tooth surfaces, which can be made parallel to serve as guiding planes; (b) retentive and nonretentive areas of the abut-ment teeth; (c) areas of interference with placement and removal; and (d) esthetic effects of the selected path of insertion. From such a survey, a path of placement may be selected that will satisfy requirements for parallelism and retention to the best mechanical, functional, and esthetic advantage. Then mouth preparations may be planned accordingly. 3. Diagnostic casts are used to permit a logical and com-prehensive presentation to the patient of present and fu-ture restorative needs, as well as of the hazards of future neglect. Occluded and individual diagnostic casts can be used to point out to the patient (a) evidence of tooth migration and the existing results of such migration; (b) effects of further tooth migration; (c) loss of occlusal sup-port and its consequences; (d) hazards of traumatic occlu-sal contacts; and (e) cariogenic and periodontal implica-tions of further neglect. Treatment planning actually may be accomplished with the patient present so that economic considerations may be discussed. Such use of diagnostic casts permits justification of the proposed fee through the patient’s understanding of the problems involved and of the treatment needed. Inasmuch as mouth rehabilitation www.konkur.in 162 Part II Clinical and Laboratory procedures are frequently lengthy and often irreversible, there must be complete accord between dentist and patient before extensive treatment is begun, and financial arrange-ments must be consummated during the planning phase. 4. Individual impression trays may be fabricated on the di-agnostic casts, or the diagnostic cast may be used in se-lecting and fitting a stock impression tray for the final im-pression. If wax blockout is to be used in the fabrication of individual trays, a duplicate cast made from an irrevers-ible hydrocolloid (alginate) impression of the diagnostic cast should be used for this purpose. The diagnostic cast is too valuable for purposes of future reference to risk dam-age resulting from the making of an impression tray. On the other hand, if oil-based clay blockout is used, the di-agnostic cast may be used without fear of damage. 5. Diagnostic casts may be used as a constant reference as the work progresses. Penciled marks indicating the type of restoration, the areas of tooth surfaces to be modified, the location of rests, and the design of the removable partial denture framework, as well as the path of placement and removal, all may be recorded on the diagnostic cast for future reference (Figure 13-5). Then these steps may be checked off the worksheet as they are completed. Areas of abutment teeth to be modified may first be changed on the duplicate diagnostic cast by trimming the stone cast with the surveyor blade. A record is thus made of the location and degree of modification to be done in the mouth. This must be done in relation to a definite path of placement. Any mouth preparations to be accomplished with new restorations require that restored teeth be shaped in accordance with a previously determined path of placement. Even so, the shaping of abutment teeth on the duplicate diagnostic cast serves as a guide to the form of the abutment. This is particularly true if the contouring of wax patterns is to be delegated to the technician, as it may be in a busy practice. 6. Unaltered diagnostic casts should become a permanent part of the patient’s record because records of conditions A B C Figure 13-4 A, Following mounting of the diagnostic casts, tooth arrangement for the mandibular occlusal plane requirements can be accomplished. B, Following placement of the maxillary anterior teeth in an ideal position, diagnostic arrangement of occlusion results in a space posterior to surveyed crown #27. If such a finding were objectionable, alternative arrangements could be investigated. This is not possible unless a diagnostic workup is completed. C, Occlusion of the mandibular removable partial denture will be enhanced by improving the maxillary posterior occlusal plane of the super-erupted molars.
www.konkur.in 163 Chapter 13 Diagnosis and Treatment Planning existing before treatment are just as important as preop-erative radiographs. Therefore diagnostic casts should be duplicated, with one cast serving as a permanent record and the duplicate cast used in situations that may require alterations to it.
Mounting Diagnostic Casts For diagnostic purposes, casts should be related on an anatomically appropriate articulator to best understand the role occlusion may have in the design and functional stability of the removable partial denture. This becomes increasingly important as the prosthesis replaces more teeth. If the patient presents with a harmonious occlusion and the edentulous span is a tooth-bound space, simple hand articulation is generally all that is required. How-ever, when the natural dentition is not harmonious and/ or when the replacement teeth must be positioned within the normal movement patterns of the jaws, the diagnostic casts must be related in an anatomically appropriate man-ner for diagnosis. This means placement of the maxillary cast in a position relative to the opening axis on the articu-lator, which is similar to the position of the maxilla in rela-tion to the temporomandibular joint (TMJ) of the patient (Figure 13-6). The mandibular cast is then placed beneath the maxillary cast in a horizontal position dictated by mandibular rotation without tooth contact, at a minimal vertical opening.
The Glossary of Prosthodontic Terms describes an articu-lator as a mechanical device that represents the TMJs and The Glossary of Prosthodontic Terms, J Prosthet Dent. 94(1):10-92, 2005. Available at: jaw members to which maxillary and mandibular casts may be attached. Because the dominant influence on mandibu-lar movement in a partially edentulous mouth is the occlu-sal plane and the cusps of the remaining teeth, an anatomic reproduction of condylar paths is probably not necessary. Still, movement of the casts in relation to one another as influenced by the occlusal plane and the cusps of the remain-ing teeth, when mounted at a reasonably accurate distance from the axis of condylar rotation, permits a relatively valid analysis of occlusal relations. This is more anatomically accu-rate than a simple hinge mounting.
It is better that the casts be mounted in relation to the axis-orbital plane to permit better interpretation of the plane of occlusion in relation to the horizontal plane. Although it is true that an axis-orbital mounting has no functional value on a nonarcon instrument because that plane ceases to exist when opposing casts are separated, the value of such a mounting lies in the orientation of the casts in occlu-sion. (An arcon articulator is one in which the condyles are attached to the lower member as they are in nature, the term being a derivation coined by Bergström from the words articulation and condyle. Many of the more widely used articulators such as the Hanau model H series, Den-tatus, and improved Gysi have the condyles attached to the upper member and are therefore nonarcon instruments.) Figure 13-5 Proposed mouth changes and design of the re-movable partial denture framework are indicated in pencil on the diagnostic cast in relation to the previously determined path of placement. This serves as a means of communicating with the patient and as a chair-side guide to tooth modification.
Figure 13-6 Use of the facebow makes possible the recording of the spatial relationship of the maxillae to some anatomic refer-ence points and transference of this relationship to an articulator.
www.konkur.in 164 Part II Clinical and Laboratory Sequence for Mounting Maxillary Cast to Axis-Orbital Plane The initial steps allow recording of the maxilla-TMJ relationship: 1. Identify the anterior and posterior reference points for the facebow (e.g., external auditory meatus, orbitale).
2. Prepare the bite fork and occlusion rim.
3. Place the bite fork centered to the arch, indexing it to the teeth with wax or elastomer.
4. Place the facebow over the bite fork rod anteriorly.
5. Place the bow evenly into the ears posteriorly.
6. Secure the bow anteriorly.
7. Position the bow anteriorly to the third point of refer-ence (establishes the horizontal plane).
8. Secure the bite fork vertical rod, then the horizontal rod (holding the bow securely to prevent torque).
9. Release the bow anteriorly to allow spread, and disen-gage from the ears. 10. Remove the fork downward and out of the mouth with the attached bow. 11. Carefully check the security of the attachments.
The next steps allow transfer of the recorded relationship to the articulator: 1. Position the posterior reference points on the articulator (usually a posterior attachment point). 2. Secure the posterior points by securing the bow anteri-orly. 3. Vertically relate the secured bow to the articulator ante-rior reference point. 4. Seat the maxillary cast into the bite fork registration (wax or elastomer). 5. Close the articulator and check clearance for mounting plaster (modify the cast as needed). 6. Mount with low-expansion plaster.
The facebow is a relatively simple device used to obtain a transfer record for orienting a maxillary cast on an articulating instrument. Originally, the facebow was used only to transfer a radius from condyle reference points so that a given point on the cast would be the same distance from the condyle as it is on the patient. The addition of an adjustable infraorbital pointer to the facebow and the addition of an orbital plane indicator to the articulator make possible the transfer of the elevation of the cast in relation to the axis-orbital plane. This permits the maxil-lary cast to be correctly oriented in the articulator space comparable with the relationship of the maxilla to the axis-orbital plane on the patient. To accommodate this orientation of the maxillary cast and still have room for the mandibular cast, the posts of the conventional artic-ulator must be lengthened. The older Hanau model H articulator usually will not permit a facebow transfer with an infraorbital pointer.
A facebow may be used to transfer a comparable radius from arbitrary reference points, or it may be designed so that the transfer can be made from hinge axis points. The latter type of transfer requires that a hinge-bow attached to the mandible should be used initially to deter-mine the hinge axis points, to which the facebow is then adjusted for making the hinge axis transfer.
A facebow transfer of the maxillary cast, which is ori-ented to the axis-orbital plane in a suitable articulator, is an uncomplicated procedure. The Hanau series Wide-Vue Arcon 183-2, all 96H2-0 models, the Whip Mix articulator (Whip Mix Corporation, Louisville, KY), and the Denta-tus model ARH (Dentatus USA, Ltd., New York, NY) will accept this transfer. The Hanau earpiece facebow models 153 and 158, the Hanau fascia facebow 132-2SM, and the Dentatus facebow type AEB incorporate the infraorbital plane to the articulator. None of these are hinge axis bows; they are used instead at an arbitrary point.
The location of the arbitrary point or axis has long been the subject of controversy. Gysi and others have placed it 11 to 13 mm anterior to the upper third of the tragus of the ear on a line extending from the upper mar-gin of the external auditory meatus to the outer canthus of the eye. Others have placed it 13 mm anterior to the posterior margin of the center of the tragus of the ear on a line extending to the corner of the eye. Bergström has located the arbitrary axis 10 mm anterior to the center of a spherical insert for the external auditory meatus and 7 mm below the Frankfort horizontal plane.
In a series of experiments reported by Beck, it was shown that the arbitrary axis suggested by Bergström falls consistently closer to the kinematic axis than do the other two. It is desirable that an arbitrary axis is placed as close as possible to the kinematic axis. Although most authori-ties agree that any of the three axes will permit transfer of the maxillary cast with reasonable accuracy, it would seem that the Bergström point compares most favorably with the kinematic axis.
The lowest point on the inferior orbital margin is taken as the third point of reference for establishing the axis-orbital plane. Some authorities use the point on the lower margin of the bony orbit in line with the center of the pupil of the eye. For the sake of consistency, the right infraorbital point is generally used and the facebow assembled in this relationship. All three points (right and left axes and infraorbital point) are marked on the face with an ink dot before the transfer is made.
Casts are prepared for mounting on an articulator by placing three index grooves in the base of the casts. Two V-shaped grooves are placed in the posterior section of the cast and one groove in the anterior portion (Figure 13-7).
An occlusion rim properly oriented on a well-fitting record base should be used in facebow procedures involv-ing the transfer of casts representative of the Class I and II partially edentulous situations. Without occlusion rims, such casts cannot be located accurately in the imprints of the wax covering the facebow fork. Tissues covering the residual ridges may be displaced grossly when the patient closes into the wax on the facebow fork. Therefore the wax www.konkur.in 165 Chapter 13 Diagnosis and Treatment Planning imprints of the soft tissues will not be true negatives of the edentulous regions of the diagnostic casts.
For purposes of illustration, a facebow using the exter-nal auditory meatus as the posterior reference point, the Whip Mix Facebow technique (DB 2000, Whip Mix Cor-poration, Louisville, KY), will be shown. The facebow fork is covered with a polyether, polyvinyl siloxane, or a roll of softened baseplate wax with the material distributed equally on the top and on the underneath side of the face-bow fork. Then the fork should be pressed lightly on the diagnostic casts with the midline of the facebow fork cor-responding to the midline of the central incisors (Figure 13-8). This will leave imprints of the occlusal and incisal surfaces of the maxillary casts and occlusion rim on the softened baseplate wax and is an aid in correctly orient-ing the facebow fork in the patient’s mouth. The facebow fork is placed in position in the mouth, and the patient is asked to close the lower teeth into the wax to stabilize it in position. It is removed from the mouth and chilled in cold water and then replaced in position in the patient’s mouth. An alternative method of stabilizing the facebow fork and recording bases is to enlist the assistance of the patient.
If an earpiece facebow is to be used, the patient should be reminded that the plastic earpieces in the auditory canals will greatly amplify noise. With the facebow fork in position, the facebow toggle is slipped over the anterior projection of the facebow fork (Figure 13-9). The patient can assist in guiding the plastic earpieces into the external auditory meatus. The patient can then hold the arms of the facebow in place with firm pressure while the operator secures the bite fork to the facebow. This accomplishes the radius aspect of the facebow transfer.
If an infraorbital pointer is used, it is placed on the extreme right side of the facebow and angled toward the infraorbital point previously identified with an ink dot. It is then locked into position with its tip lightly touching the skin at the dot. This establishes the elevation of the facebow in relation to the axis-orbital plane. Extreme care must be taken to avoid any slip that might injure the patient’s eye.
Figure 13-7 The base of the cast has been prepared for mounting by placing three triangular grooves to allow indexing when mounted. The grooves are prepared with a 3-inch stone mounted in a laboratory lathe.
Figure 13-8 Orienting the facebow fork to the maxillary cast and occlusion rims avoids displacing the occlusion rim in the mouth through patient closure or another uneven force. Polyvinyl siloxane material has been evenly distributed around the facebow fork, and care is exercised to position the fork to be centered at the mid-incisal position without any fork extension posterior to the record base, which could cause discomfort.
1 4 2 3 Figure 13-9 The horizontal toggle clamp of the Whip Mix ear-piece facebow (1) is slid onto the shaft of the facebow fork pro-truding from the patient’s mouth. The patient then helps guide plastic earpieces into the external auditory meatus and holds them in place while the operator tightens three thumb screws (2) and centers the plastic nosepiece (3) securely on the nasion. The horizontal toggle clamp is positioned and secured near (but not touching) the lip. The T screw (4) on the vertical bar is tightened. Note: Extreme care should be exercised not to tilt the facebow out of position when tightening.
www.konkur.in 166 Part II Clinical and Laboratory With all elements tightened securely, the patient is asked to open, and the entire assembly is removed intact, rinsed with cold water, and set aside. The facebow records not only the radius from the condyles to the incisal con-tacts of the upper central incisors, but also the angular relationship of the occlusal plane to the axis-orbital plane.
The facebow must be positioned on the articulator in the same axis-orbital relation as on the patient. If an arbi-trary-type facebow is used, the calibrated condyle rods of the facebow ordinarily will not fit the condyle shafts of the articulator unless the width between the condyles just happens to be the same. With a Hanau model 132-25M facebow, the calibrations must be reequalized when in position on the articulator. For example, they have read 74 (mm) on each side of the patient but must be adjusted to read 69 (mm) on each side of the articulator. Some later model articulators have adjustable condyle rods and may be adjusted to fit the facebow. It is necessary that the facebow be centered in either case. Some facebows are self-centering, as is the Hanau Spring-Bow (Whip Mix Corporation, Louisville, KY).
The third point of reference is the orbital plane indica-tor, which must be swung to the right so that it will be above the tip of the infraorbital pointer. The entire face-bow with maxillary cast in place must be raised until the tip of the pointer contacts the orbital plane indicated. The elevation having thus been established, for all practical purposes the orbital plane indicator and the pointer may now be removed because they may interfere with placing the mounting stone.
An auxiliary device called a cast support is available; it is used to support the facebow fork and the maxillary cast during the mounting operation (Figure 13-10). With this device, the weight of the cast and the mounting stone are supported separately from the facebow, thus prevent-ing possible downward movement resulting from their combined weight. The cast support is raised to supporting contact with the facebow fork after the facebow height has been adjusted to the level of the orbital plane. Use of some type of cast support is highly recommended as an adjunct to facebow mounting.
The keyed and lubricated maxillary cast is now attached to the upper arm of the articulator with the mounting stone, thus completing the facebow transfer (Figure 13-11). Not only will the facebow have permitted the upper cast to be mounted with reasonable accuracy, it also will have served as a convenient means of supporting the cast dur-ing mounting. Once mastered, its use becomes a great con-venience rather than a time-consuming nuisance.
It is preferable that the maxillary cast be mounted while the patient is still present, thus eliminating a pos-sible reappointment if the facebow record is unaccept-able for some reason. Not too infrequently, the facebow record has to be redone with the offset-type facebow fork repositioned to avoid interference with some part of the articulator.
Jaw Relationship Records for Diagnostic Casts One of the first critical decisions that must be made in a removable partial denture service involves selection of the horizontal jaw relationship to which the removable partial denture will be fabricated (centric relation or the maximum intercuspal position). All mouth-preparation procedures depend on this analysis. Failure to make this decision cor-rectly may result in poor prosthesis stability, discomfort, and deterioration of the residual ridges and supporting teeth.
Figure 13-10 Facebow fork support used to maintain the fork and cast in position while mounting.
Figure 13-11 Facebow mounting is complete. The relation-ship of the maxillary cast to the articulator condylar components is anatomically similar to that between the patient’s maxilla and the bilateral temporomandibular joint (TMJ) complex. Any sub-sequent tooth arrangement and occlusal contact development represent the mouth more accurately than more arbitrary mount-ings. The benefits of the anatomic similarity are seen in more accurate occlusion for the finalized prosthesis (i.e., less intraoral adjustment required).
www.konkur.in 167 Chapter 13 Diagnosis and Treatment Planning It is recommended that deflective occlusal contacts in the maximum intercuspal and eccentric positions be cor-rected as a preventive measure. Not all dentists agree that centric relation and the maximum intercuspal position must be harmonious in the natural dentition. Many denti-tions function satisfactorily with the opposing teeth maxi-mally intercusped in an eccentric position without either diagnosable or subjective indications of TMJ dysfunction, muscle dysfunction, or disease of the supporting structures of the teeth. In many such situations, no attempt should be made to alter the occlusion. It is not a requirement to inter-fere with an occlusion simply, because it does not com-pletely conform to a relationship that is considered ideal.
If most natural posterior teeth remain and if no evidence of TMJ disturbances, neuromuscular dysfunction, or peri-odontal disturbances related to occlusal factors exists, the proposed restorations may be fabricated safely with maxi-mum intercuspation of remaining teeth. However, when most natural centric stops are missing, the proposed pros-thesis should be fabricated so that the maximum intercus-pal position is in harmony with centric relation. By far, the greater majority of removable partial dentures should be fabricated in the horizontal jaw relationship of centric relation. In most instances in which edentulous spaces have not been restored, the remaining posterior teeth will have assumed malaligned positions through drifting, tip-ping, or extrusion. Correction of the remaining natural occlusion to create a coincidence of centric relation and the maximum intercuspal position is indicated in such situations.
Regardless of the method used in creating a harmonious functional occlusion, an evaluation of the existing relation-ships of the opposing natural teeth must be made and is accomplished with a diagnostic mounting. This evaluation is performed in addition to, and in conjunction with, other diagnostic procedures that contribute to an adequate diag-nosis and treatment plan.
Diagnostic casts provide an opportunity to evaluate the relationships of remaining oral structures when correctly mounted on a semiadjustable articulator with use of a face-bow transfer and interocclusal records. Diagnostic casts are mounted in centric relation (most retruded relation of the mandible to the maxillae) so that deflective occlusal contacts can be correlated with those observed in the mouth. Deflec-tive contacts of opposing teeth are usually destructive to the supporting structures involved and should be eliminated. Diagnostic casts demonstrate the presence and location of such interfering tooth contacts and permit visualization of the treatment that would be necessary for their correction. Necessary alteration of teeth to harmonize the occlusion can be performed initially on duplicates of the mounted diag-nostic casts to act as guides for similar necessary corrections in the mouth. In many instances the degree of alteration required will indicate the need for crowns or onlays to be fabricated, or for recontouring, repositioning, or elimination of extruded teeth.
As mentioned previously, the maxillary cast is correctly oriented to the opening axis of the articulator by means of the facebow transfer and becomes spatially related to the upper member of the articulator in the same relationship that the maxilla has to the hinge axis and the Frankfort plane. Simi-larly, when a centric relation record is made at an established vertical dimension, the mandible is in its most retruded rela-tion to the maxilla. Therefore when the maxillary cast is cor-rectly oriented to the axis of the articulator, the mandibular cast automatically becomes correctly oriented to the opening axis, when attached to and mounted with an accurate centric relation record.
Unlike recording the fixed relationship of the maxilla to the mandibular opening axis (using the facebow transfer record), the mandibular position is recorded in space and is not a fixed point. Consequently, it is necessary to prove that the relationship of the mounted casts is correct. This can be done simply by making another interocclusal record, at cen-tric relation, fitting the casts into the record, and checking to see that the condylar elements of the articulator are snug against the condylar housings. If this is not seen, another record is made until duplicate records are produced. Because centric relation is the only jaw position that can be repeated by the patient, mountings in this position can be replicated and verified for correctness.
A straightforward protrusive record is made to adjust the horizontal condylar inclines on the articulator. Lateral eccentric records are made so that the lateral condylar incli-nations can be properly adjusted. All interocclusal records should be made as near the vertical relation of occlusion as possible. Opposing teeth or occlusion rims must not be allowed to contact when the records are made. A contact of the inclined planes of opposing teeth will invalidate an inter-occlusal record.
In some instances, a mounting of a duplicate diagnos-tic cast in the maximum intercuspal position may also be desirable to definitively study this relationship on the artic-ulator. Because articulators simulate only jaw movements, it is not unreasonable to assume that the relationship of the casts mounted in centric relation may differ minutely from the maximum intercuspal position seen on the articula-tor and observed in the mouth. When diagnostic casts are hand related by maximum intercuspation for purposes of mounting on an articulator, it is essential that three (prefer-ably four) positive contacts of opposing posterior teeth are present, having widespread molar contacts on each side of the arch. If occlusion rims are necessary to correctly ori-ent casts on an articulator, a centric relation should usually be the horizontal jaw relationship to which the removable partial denture will be constructed.
Materials and Methods for Recording Centric Relation Materials available for recording centric relation are numer-ous. Current materials include polyether materials, polyvinyl silicone materials, dimethacrylate-based materials, metallic www.konkur.in 168 Part II Clinical and Laboratory oxide paste, and wax. Of these, wax is likely to be least sat-isfactory unless carefully handled. If not uniformly softened when introduced into the mouth, it can record a position with unequal tissue placement. Also, it does not remain rigid and dimensionally stable after removal unless carefully chilled and handled upon removal (Figure 13-12).
Metallic oxide bite registration paste is quite accurate yet is not strong enough to be used alone. To improve strength, it can be supported by a gauze mesh attached to a metal frame or it may be used with occlusion rims.
Elastomeric materials are excellent for recording interoc-clusal relationships (Figure 13-13). Some are specially for-mulated for this purpose and have the qualities of extremely low viscosity, minimal resistance to closure, rapid set, low rebound, lack of distortion, and stability after removal from the mouth. Care should be exercised to ensure that no elastic rebound results when the record is used to relate the casts during the mounting procedure.
The mandibular cast should be mounted on the lower arm of the articulator, with the articulator inverted (Figure 13-14, A). The articulator is first locked in centric posi-tion, and the incisal pin is adjusted so that the anterior distance between the upper and lower arms of the articu-lator will be increased 2 to 3 mm greater than the normal parallel relationship of the arms. This is done to compen-sate for the thickness of the interocclusal record so that the arms of the articulator will again be nearly parallel when the interocclusal record is removed and the opposing casts come into contact.
The base of the cast should be keyed and lightly lubricated for future removal. With the diagnostic casts accurately seated and secured in the occlusal record, the mandibular cast is affixed with stone to the lower member of the inverted articulator.
An articulator mounting thus made will have related the casts in centric relation (see Figure 13-14, B). The dentist then can proceed to make an occlusal analysis by observing the influence of cusps in relation to one another after the articulator has been adjusted by using eccentric interocclusal records.
After an occlusal analysis has been made, the casts may be removed from their mounting for the purpose of survey-ing them individually and for other purposes as outlined previously. The indexed mounting ring record also should be retained throughout the course of treatment in the event that further study should be needed. It is advisable that the mounting be identified with the articulator that is used, so that it may always be placed back onto the same articulator.
A B Figure 13-12 A wax interocclusal record made on a cast framework. A, The modification spaces first had baseplate wax added; these were adjusted intraorally to provide space at the occlusal vertical dimension for recording wax, the wax was softened using a wax spatula and a hot water bath, the framework was placed in the mouth, and care was exercised to guide the patient to close into a previously verified (and practiced) interocclusal position deemed appropriate (in this instance, centric relation position). The record was recovered from the mouth, excess wax was removed with a warm scalpel, and the wax was chilled and replaced in the mouth to verify the record. If not verified, the wax was resoftened (with additional wax added as needed) and the procedure was repeated. B, Immediately after verifica-tion, the framework with interocclusal registration was replaced on the mandibular cast and inverted on the maxillary cast for mounting.
Figure 13-13 Elastomeric interocclusal registration material used to record mandibular position.
www.konkur.in 169 Chapter 13 Diagnosis and Treatment Planning DIAGNOSTIC FINDINGS The information gathered in the patient interview and clinical examination provides the basis for establishing whether treat-ment is indicated, and if so, what specific treatment should be considered. More than one treatment option can be consid-ered, and financial implications need to be considered against long-term expectations if the best decision is to be reached. Provision of a removable partial denture does not often pre-clude future consideration for other treatments, a fact that is not often the case for alternative treatments. The patient interview can reveal medical considerations that influence the decision to provide any prosthesis. When it is felt that general medical health is being neglected, patients should be strongly encouraged to seek a general medical examination. Alternatively, patients who regularly see their physician may be found to take multiple medications that can contribute to a dry mouth and, potentially, an altered oral microflora with some increased risk for plaque-induced disease. Although such a condition can influence any prosthodontic care, given the unique features of removable partial denture service rela-tive to the need for increased hygiene awareness and care, any factor that places an additional risk for plaque-induced dis-ease should be emphasized with the patient and corrected if possible. Health conditions that negatively affect oral mucosal health (e.g., diabetes mellitus, Sjögren syndrome, lupus, atro-phic changes) may pose a risk for patient comfort for a tissue-supported prosthesis and factor into a treatment decision.
For the patient who has had previous experience with some form of prosthesis, the patient interview provides addi-tional information that can influence treatment decisions. Identifying possible reasons (or more importantly, a lack of any reason) for both positive and negative past prosthesis experiences is important for determining whether a patient can predictably be helped. Although the clinical examina-tion will point out the oral tissue responses to such therapy, the interview will highlight the subjective patient response to therapy and provide significant information that should be pursued. As mentioned previously, a patient complaint regarding the prosthesis needs to be confirmed through evaluation. The patient generally expresses concern about a symptom that can be related to support, stability, retention, and/or appearance. Confirmation of a design feature or oral condition that can explain the symptom is required if one is to have a chance to correct it with a similar prosthesis. If examination does not confirm any such relationship, it would be difficult to proceed without some concern for repeating the patient response to therapy unless a different form of therapy is selected (e.g., replacing a problematic removable partial denture with an implant-supported prosthesis).
INTERPRETATION OF EXAMINATION DATA As a result of the oral examination, several diagnoses are made that are related to the various tissues, conditions, and clinical information gathered. The integration of these diag-noses serves as the basis for decisions that will ultimately identify the suggested treatment. The treatment decision reflects a confluence of several aspects of the patient’s past, present, and potential oral health status.
It is helpful to consider how the various diagnoses are integrated; consequently, a suggested framework is provided that highlights aspects of disease management, followed by reconstruction considerations for (1) prosthesis support; and (2) prosthesis design-specific aspects.
Disease management takes into account findings from the radiographic examination, periodontal disease and caries assessments, and pathology requiring endodontic A B Figure 13-14 A, Mandibular cast inverted on the mounted maxillary cast, making sure that the cast is fully seated into the interocclu-sal record and stabilized to the opposing cast. It is important to check the posterior occlusion rim contact to ensure that no interfering contact has altered the record. Space should be observed between the opposing record bases (or record base and opposing occlusion). B, Mounted casts demonstrating the occlusal plane as found in the mouth. The Frankfort plane of the patient is oriented parallel to the articulator base and the floor. Also, inspection of the posterior rims demonstrates space between the rims, which ensures that the re-corded position was registered without influence from rigid contacting components, only from softened wax.
www.konkur.in 170 Part II Clinical and Laboratory considerations. Reconstruction considerations include diag-noses relative to prosthesis support (teeth and residual ridges) and prosthesis-specific design elements. Prosthesis support related to the remaining teeth requires radiographic exami-nation of alveolar support and root morphology, endodon-tic evaluation, analysis of occlusal factors, assessment of the benefit for fixed prostheses or orthodontics, and evaluation of the need for extraction. Residual ridge support involves radiographic examination of ridge contours and height, and evaluation of the need for pre-prosthetic surgical interven-tion. Prosthesis-specific design considerations include deter-mination of anatomic relationships related to mandibular major connector design, the need for tooth modification to facilitate prosthesis function, and analysis of the occlusion. Each of these is considered in the following sections.
Radiographic Interpretation Many of the reasons for radiographic interpretation during oral examination are outlined herein and are considered in greater detail in other texts. Aspects of such interpretation that are most pertinent to removable partial denture con-struction are those relative to the prognosis of remaining teeth that may be used as abutments.
Disease Validation It is important to verify by clinical examination disease found through radiographic interpretation. Also, if the clini-cal examination reveals dental caries and/or periodontal disease, its severity can be confirmed by radiographic inter-pretation. It would be important to delineate caries severity, in terms of numbers of lesions and dentin/pulpal involve-ment, to gain insight as to level of disease risk associated with the patient, as well as to identify what therapy is required to maintain teeth. The same is true for periodontal disease risk and severity, as such a diagnosis affects both current and future tooth prognosis for prosthesis support.
Radiographic interpretation allows diagnosis of bone lesions associated with both the jaws and the teeth. The impli-cations for tooth stability and ridge support are important to factor into prosthesis prognosis. Surgical and postoperative management of such lesions can vary significantly with diag-nosis (benign versus malignant), and definitive prosthesis treatment is often complicated by resective procedures.
Tooth Support The quality of the alveolar support of an abutment tooth is of primary importance because the tooth will have to with-stand greater stress loads when supporting a dental pros-thesis. Abutment teeth providing total abutment support to the prosthesis, whether fixed or removable, will have to withstand a greater load and especially greater horizontal forces. The latter may be minimized by establishing a har-monious occlusion and by distributing the horizontal forces among several teeth through the use of rigid connectors. Bilateral stabilization against horizontal forces is one of the attributes of a properly designed tooth-supported removable prosthesis. In many instances, abutment teeth may be aided more than weakened by the presence of a bilaterally rigid removable partial denture.
In contrast, abutment teeth adjacent to distal extension bases are subjected not only to vertical and horizontal forces but to torque as well because of movement of the tissue-supported base. Vertical support and stabilization against horizontal movement with rigid connectors are just as important as they are with a tooth-supported prosthesis, and the removable partial denture must be designed accordingly. In addition, the abutment tooth adjacent to the extension base will be subjected to torque in proportion to the design of the retainers, the size of the denture base, the tissue sup-port received by the base, and the total occlusal load applied. With this in mind, each abutment tooth must be evaluated carefully as to the alveolar bone support present and the past reaction of that bone to occlusal stress.
It is important to judge whether the teeth and their respec-tive periodontium can favorably respond to the demands of a prosthesis. Can radiographic interpretation provide clues to predicting tooth response to increased loading from prosthe-ses? Assessment of regions within the mouth that have been subjected to increased loading can provide some clues as to the predictability of future similar response. An understand-ing of bone density, index areas, and lamina dura response is helpful for these judgments.
Bone Density The quality and quantity of bone in any part of the body are often evaluated by radiographic means. A detailed trea-tise concerning bone support of the abutment tooth should include many considerations not possible to include in this text because of space limitations. The reader should realize that subclinical variations in bone may exist but may not be observed because of the limitations inherent in technical methods and equipment.
Of importance to the dentist in evaluating the quality and quantity of the alveolar bone are the height and the quality of remaining bone. In estimating bone height, care must be taken to avoid interpretive errors resulting from angulation factors. Technically, when a radiographic exposure is made, the central ray should be directed at right angles to both the tooth and the film. The short-cone technique does not follow this principle; instead the ray is directed through the root of the tooth at a predetermined angle. This technique invariably causes the buccal bone to be projected higher on the crown than the lingual or palatal bone. Therefore in interpreting bone height, it is imperative to follow the line of the lamina dura from the apex toward the crown of the tooth until the opacity of the lamina materially decreases. At this point of opacity change, a less dense bone extends farther toward the tooth crown. This additional amount of bone represents false bone height. Thus, the true height of the bone is ordinar-ily where the lamina shows a marked decrease in opacity. At this point, the trabecular pattern of bone superimposed on the tooth root is lost. The portion of the root between the www.konkur.in 171 Chapter 13 Diagnosis and Treatment Planning cementoenamel junction and the true bone height has the appearance of being bare or devoid of covering.
Radiographic evaluation of bone quality is hazardous but is often necessary. It is essential to emphasize that changes in bone mineralization up to 25% often cannot be recog-nized by ordinary radiographic means. Optimum bone qualities are ordinarily expressed by normal-sized inter-dental trabecular spaces that tend to decrease slightly in size as examination of the bone proceeds from the root apex toward the coronal portion. The normal interproxi-mal crest is ordinarily shown by a relatively thin white line crossing from the lamina dura of one tooth to the lamina dura of the adjacent tooth. Considerable variation in the size of trabecular spaces may exist within normal limits, and the radiographic appearance of crestal alveolar bone may vary considerably, depending on its shape and the direction that the ray takes as it passes through the bone.
Normal bone usually responds favorably to ordinary stresses. Abnormal stresses, however, may create a reduc-tion in the size of the trabecular pattern, particularly in that area of bone directly adjacent to the lamina dura of the affected tooth. This decrease in size of the trabecular pat-tern (i.e., so-called bone condensation) is often regarded as a favorable bone response, indicative of an improvement in bone quality. This is not necessarily an accurate inter-pretation. Such bone changes usually indicate stresses that should be relieved, because if the resistance of the patient decreases, the bone may exhibit a progressively less favor-able response on future radiographs.
Increased thickness of the periodontal space ordinarily suggests varying degrees of tooth mobility. This should be evaluated clinically. Radiographic evidence coupled with clinical findings may suggest to the dentist the inadvis-ability of using such a tooth as an abutment. Furthermore, an irregular intercrestal bone surface should make the dentist suspicious of active bone deterioration.
It is essential that the dentist realize that radiographic evidence shows the result of changes that have taken place and may not necessarily represent the present condition. For example, periodontal disease may have progressed beyond the stage visibly demonstrated on the radiograph. As pointed out earlier, radiographic changes are not observed until approximately 25% of the mineral content has been depleted. On the other hand, bone condensation probably does represent the current situation.
Radiographic findings should serve the dentist as an adjunct to clinical observations. Too often the radio-graphic appearance alone is used to arrive at a diagnosis; therefore radiographic findings should always be con-firmed by clinical examination. Radiographic interpreta-tion also serves an important function if used periodically after the prosthesis has been placed. Future bone changes of any type suggest traumatic interference from some source. The nature of such interference should be deter-mined and corrective measures taken.
Index Areas Index areas are those areas of alveolar support that disclose the reaction of bone to additional stress. Favorable reac-tion to such stress may be taken as an indication of future reaction to an added stress load. Teeth that have been sub-jected to abnormal loading because of the loss of adjacent teeth or that have withstood tipping forces in addition to occlusal loading may be better risks as abutment teeth than those that have not been called on to carry an extra occlu-sal load (Figures 13-15 and 13-16). If occlusal harmony can be improved and unfavorable forces minimized by the reshaping of occlusal surfaces and the favorable distribution of occlusal loading, such teeth may be expected to support the prosthesis without difficulty. At the same time, other teeth, although not at present carrying an extra load, may be expected to react favorably because of the favorable reac-tion of alveolar bone to abnormal loading elsewhere in the same arch.
Other index areas are those around teeth that have been subjected to abnormal occlusal loading; that have been subjected to diagonal occlusal loading caused by tooth migration; and that have reacted to additional loading, such as around existing fixed partial denture abutments. The reaction of the bone to additional stresses in these areas may be either positive or negative, with evidence of a supporting trabecular pattern, a heavy cortical layer, and a dense lamina dura, or the reverse response. With the former, the patient is said to have a positive bone fac-tor, which means the ability to build additional support wherever needed. With the latter, the patient is said to have a negative bone factor, which means the inability to respond favorably to stress.
Alveolar Lamina Dura The alveolar lamina dura is also considered in a radio-graphic interpretation of abutment teeth. The lamina dura is the thin layer of hard cortical bone that normally lines the sockets of all teeth. It affords attachment for the fibers of the periodontal membrane, and as with all cortical bone, its function is to withstand mechanical strain. In a roentgeno-gram, the lamina dura is shown as a radiopaque white line around the radiolucent dark line that represents the peri-odontal membrane.
When a tooth is in the process of being tipped, the center of rotation is not at the apex of the root but in the apical third. Resorption of bone occurs where there is pressure, and apposition occurs where there is tension. Therefore during the active tipping process, the lamina dura is uneven, with evidence of both pressure and tension on the same side of the root. For example, in a mesially tipping lower molar, the lam-ina dura will be thinner on the coronal mesial and apicodis-tal aspects and thicker on the apicomesial and coronal distal aspects because the axis of rotation is not at the root apex but is above it. When the tooth has been tipped into an edentu-lous space by some change in the occlusion and becomes set www.konkur.in 172 Part II Clinical and Laboratory in its new position, the effects of leverage are discontinued. The lamina dura on the side to which the tooth is sloping becomes uniformly heavier, which is nature’s reinforcement against abnormal stresses. The bone trabeculations are most often arranged at right angles to the heavier lamina dura.
Thus, it is possible to say that for a given individual, nature is able to build support where it is needed, and on this basis to predict future reactions elsewhere in the arch to addi-tional loading of teeth used as abutments. However, because bone is approximately 30% organic and mostly protein and B A Figure 13-16 A, The canine has provided support for the distal extension removable partial denture for 10 years. There has obviously been positive bone response (arrow) to increased stress generated by the removable partial denture. B, The mandibular first premolar has provided support for the distal extension denture for 3 years. Bone response (arrow) to past additional stress has been unfavorable.
Figure 13-15 The reaction of bone adjacent to teeth that have been subjected to abnormal stress serves as an indication of probable reactions of that bone when such teeth are used as abutments for fixed or removable restorations. Such areas are called index areas.
www.konkur.in 173 Chapter 13 Diagnosis and Treatment Planning because the body is not able to store a protein reserve in large amounts, any change in body health may be reflected in the patient’s ability to maintain this support permanently. When systemic disease is associated with faulty protein metabolism and when the ability to repair is diminished, bone is resorbed and the lamina dura is disturbed. Therefore the loading of any abutment tooth must be kept to a minimum inasmuch as the patient’s future health status and the eventualities of aging are unpredictable.
Root Morphology The morphologic characteristics of the roots determine to a great extent the ability of prospective abutment teeth to resist successfully additional rotational forces that may be placed on them. Teeth with multiple and divergent roots will resist stresses better than teeth with fused and conical roots, because the resultant forces are distributed through a greater number of periodontal fibers to a larger amount of support-ing bone (Figure 13-17).
Periodontal Considerations An assessment of the periodontium in general and abutment teeth in particular must be made before prosthetic restora-tion. One must evaluate the condition of the gingiva, looking for adequate zones of attached gingiva and the presence or absence of periodontal pockets. The ideal periodontal condi-tion is a disease-free periodontium with adequate attached mucosa in regions at or adjacent to removable partial den-ture component parts that cross the gingival margins to best resist the mechanical challenges posed as the result of func-tion and use. The condition of the supporting bone must be evaluated, with specific attention to reduced bone support and mobility patterns recorded. If mucogingival involve-ments, osseous defects, or mobility patterns are recorded, the causes and potential treatment must be determined.
Oral hygiene habits of the patient must be determined, and efforts made to educate the patient relative to plaque control. The most decisive evidence of oral hygiene habits is the condition of the mouth before the initial prophylaxis. Good or bad oral hygiene is basic to the patient’s nature, and although it may be influenced somewhat by patient educa-tion, the long-range view must be taken. It is reasonably fair to assume that the patient will do little more in the long-term future than he has done in the past. In making decisions regarding the method of treatment based on oral hygiene, the future in years, rather than in weeks and months, must be considered. It is probably best not to give the patient the benefit of any doubt as to future oral hygiene habits. Rather, the benefit should come from protective measures where any doubt exists about future oral hygiene habits. Therefore for patients at greatest risk, an oral prophylaxis with contin-ued oral hygiene instructions should be scheduled for 3- to 4-month intervals. In addition, the patient must be advised of the importance of regular maintenance appointments for tissue-supported prostheses to maintain occlusal relation-ships. When these ongoing observations and prophylactic requirements are described, the patient is faced with the real-ization that he or she must be willing to share responsibility for maintaining the health of the mouth after restorative and prosthodontic treatment.
The remaining teeth and prosthesis require meticu-lous plaque control after placement of a removable partial denture. Because of the nature of material coverage of oral tissues, the oral microflora can change with the use of a removable prosthesis. Coupled with this microbial change is the potential for a mechanical challenge to tissue integrity if the appropriate relationship of the prosthesis and soft tissues of the residual ridge, as well as the marginal gingival, is not maintained.
Caries Risk Assessment Considerations Caries activity in the mouth, past and present, and the need for protective restorations must be considered. The decision to use full coverage is based on a need to reshape abutment teeth to accommodate the components of the removable par-tial denture, prevention of restoration breakdown when abut-ments have large direct restorations, or evidence of recurrent caries risk. Occasionally, three-quarter crowns may be used where buccal or lingual surfaces are completely sound, but intracoronal restorations (inlays) are seldom indicated in any mouth with evidence of past extensive caries or precarious areas of decalcification, erosion, or exposed cementum.
A B Figure 13-17 A, The prognosis for abutment service is more favorable for a molar with divergent roots (shaded) than for the same tooth if its roots were fused and conical. B, Evidence that prospective abutment has conical and fused roots indicates the necessity for formulating a framework design that will minimize additional stresses placed on the tooth by the abutment service.
www.konkur.in 174 Part II Clinical and Laboratory Frequent consumption of sugars can lead to carious involvement of roots, caries around restorations, or caries associated with clasps of removable partial dentures. Intel-ligent consumption of sweets (smaller amounts and less fre-quent consumption) and frequent plaque removal are the recommended countermeasures. Excellent protection from caries can be provided by fluoride applications via tooth-pastes, mouth rinses, or (in extreme cases, such as postra-diation xerostomia) 1% sodium fluoride (NaF) gels applied daily with plastic trays.
Xerostomia, caused by degeneration of salivary glands (Sjögren syndrome) or various medications, enhances the occurrence and severity of caries, as well as contributes to irritation of the oral mucosa. A possible way to alleviate xerostomia is the use of synthetic saliva, with a carboxy-methylcellulose base, which can be enriched with fluoride in an effort to counteract caries. Frequent use provides an excellent means of maintaining high fluoride intraorally for long periods of time, thus enhancing the remineraliza-tion of incipient caries. Although providing instructions for improvement of oral hygiene is a duty of the dental team, suspected problems of dietary deficiencies should be referred to a nutritionist.
Evaluation of the Prosthesis Foundation—Teeth and Residual Ridge An evaluation of the prosthesis foundation is required to ensure that an appropriately stable base of sound teeth and/ or residual ridge(s) is provided to maximize prosthesis func-tion and patient comfort. To that end, the evaluation focuses on the identification of conditions that are inconsistent with sound support and predictably stable function.
Surgical Preparation The need for pre-prosthetic surgery or extractions must be evaluated. The same criteria apply to surgical intervention in the partially edentulous arch as in the completely eden-tulous arch. Grossly displaceable soft tissues covering basal seat areas and hyperplastic tissue should be removed to pro-vide a firm denture foundation. Mandibular tori should be removed if they will interfere with the optimum location of a lingual bar connector or a favorable path of placement. Any other areas of bone prominence that will interfere with the path of placement should be removed also. The path of place-ment will be dictated primarily by the guiding plane of the abutment teeth. Therefore some areas may present interfer-ence to the path of placement of the removable partial den-ture by reason of the fact that other unalterable factors such as retention and esthetics must take precedence in selecting that path.
Clinical research in pre-prosthetic surgical concepts has contributed significant developments to management of the compromised partially edentulous patient. Bone augmenta-tion and guided bone regeneration procedures have been used with varying degrees of success as an alternative method of improving ridge support for the denture base areas. Skill and judgment must be exercised in patient selection, pro-cedural planning, and surgical and prosthetic management to optimize clinical results. Use of osseointegrated implants can provide a foundation for developing suitable abutment support for removable partial dentures. As in any surgical procedure, results depend on careful treatment planning and cautious surgical management.
Extraction of teeth may be indicated for one of the follow-ing three reasons: 1. If the tooth cannot be restored to a state of health, extrac-tion may be unavoidable. Modern advancements in the treatment of periodontal disease and in restorative pro-cedures, including endodontic therapy, have resulted in the saving of teeth that were once considered untreatable.
All reasonable avenues of treatment should be considered from both prognostic and economic standpoints before extraction is recommended. 2. A tooth may be removed if its absence will permit a more serviceable and less complicated removable partial denture design. Teeth in extreme malposition (lingually inclined mandibular teeth, buccally inclined maxillary teeth, and mesially inclined teeth posterior to an eden-tulous space) may be removed if an adjacent tooth is in good alignment and if good support is available for use as an abutment. Justification for extraction lies in the decision that a suitable restoration, which will provide satisfactory contour and support, cannot be fabricated, or that orthodontic treatment to realign the tooth is not feasible. An exception to the arbitrary removal of a malposed tooth occurs when a distal extension remov-able partial denture base would have to be made rather than using the more desirable tooth-supported base of the tooth in question. If alveolar support is adequate, a posterior abutment should be retained if at all pos-sible in preference to a tissue-supported extension base.
Teeth deemed to have insufficient alveolar support may be extracted if their prognosis is poor and if other ad-jacent teeth may be used to better advantage as abut-ments. The decision to extract such a tooth should be based on the degree of mobility and other periodontal considerations and on the number, length, and shape of the roots contributing to its support. 3. A tooth may be extracted if it is so unaesthetically located as to justify its removal to improve appearance. In this situation, a veneer crown should be considered in pref-erence to removal if the crown can satisfy the esthetic needs. If removal is advisable because of unesthetic tooth position, the biomechanical problems involved in replac-ing anterior teeth with a removable partial denture must be weighed against the problems involved in making an esthetically acceptable fixed restoration. Admittedly, the removable replacement is commonly the more esthetic of the two, despite modern advancements in retainers and pontics. However, the mechanical disadvantage of the re-movable restoration often makes the fixed replacement of missing anterior teeth preferable.
www.konkur.in 175 Chapter 13 Diagnosis and Treatment Planning Another consideration for pre-prosthetic surgery involves the decision between use of a removable partial denture and an implant-supported prosthesis. The following categories of tooth loss are presented with comparative comments ger-mane to such decisions.
Short Modification Spaces For short spans (≤3 missing teeth), natural tooth– and implant-supported fixed prostheses as well as removable partial dentures can generally be considered. Implant place-ment requires the decision that ample bone volume exists, or can be provided with minimal morbidity, to adequately house sufficient implants to support prosthetic teeth. Implant prostheses have the advantage of not requiring the use of teeth for support, stability, and retention require-ments, and consequently do not increase the functional burden on the natural dentition. Although the predict-ability of contemporary implant procedures (surgery and prosthodontics) makes them a consideration for short span prostheses, the main advantage is the opportunity to pro-vide replacement teeth without involving adjacent teeth in the reconstruction. Therefore, when the adjacent teeth are in need of restoration, a conventional prosthesis should be considered.
Longer Modification Spaces Longer span modification spaces (≥4 missing teeth) pres-ent a greater challenge for natural tooth–supported fixed prostheses. Consequently, options for treatment include the removable partial denture and the implant-supported prosthesis. An implant prosthesis has the same bone vol-ume requirements as stated earlier and will likely require additional implants for an increased span. Because resid-ual ridge resorption can be greater with longer spans, the need for augmentation may also be greater. Both of these characteristics of longer spans cause implant use to be more costly and can significantly increase the cost dif-ference between treatment options. The increased mor-bidity associated with augmentation procedures can also limit universal application. Because the removable partial denture remains largely tooth-supported (unless the span includes anterior and posterior segments that may cause it to function similar to a distal extension), the functional stability requirements should be efficiently met through the tooth support.
Distal Extension Spaces Without tooth support at each end of the missing teeth, the removable partial denture and the implant-supported pros-thesis are the primary treatment considerations (double-abutted cantilevered fixed prostheses opposing maxillary complete dentures have been suggested to be a reasonable option for some patients). It then becomes obvious that when anatomic limitations to implant placement exist and surgical measures cannot be taken to correct this, the remov-able partial denture is the only option (unless no treatment is elected). Current surgical options are available to correct most anatomic limitations, yet frequently implant therapy is not elected because of patient medical factors, concerns for the risk of surgical morbidity, increased time required for treatment, and costs. It is important to note that a com-parison of long-term maintenance requirements between these two options may demonstrate little cost difference over time. This is related to the effects of continued residual ridge resorption acting on the removable prosthesis and not the implant prosthesis.
Endodontic Treatment Abutments for removable partial dentures are required to withstand various forces depending on the classification. The requirement for a distal extension abutment is different than that of a tooth-supported prosthesis in that torsional forces exist in the distal extension situation. For this reason, an abutment for a distal extension that is endodontically treated carries a greater risk for complications than a similar tooth not involved in removable partial denture function.
Because tooth support helps control prosthesis move-ment, the need for endodontic treatment should include assessment of overdenture abutments for removable par-tial dentures, especially to control movement of distal extensions.
Analysis of Occlusal Factors From the occlusal analysis made by evaluating the mounted diagnostic casts, the dentist must decide whether it is best to accept and maintain the existing occlusion or to attempt to improve on it by means of occlusal adjustment and/or restoration of occlusal surfaces. It must be remem-bered that the removable partial denture can supplement the occlusion that exists only at the time the prosthesis is constructed. The dominant force that dictates the occlu-sal pattern will be the cuspal harmony or disharmony of the remaining teeth and their proprioceptive influence on mandibular movement. The goal of artificial tooth place-ment is to harmonize with the functional parameters of the existing occlusion providing bilateral, simultaneous functional contact.
Chapter 18 identifies schemes of occlusion recommended for partially edentulous configurations. A review of these recommendations will provide a guide for modifying the existing occlusion or developing the appropriate occlusal scheme for each partially edentulous configuration.
Improvements in the natural occlusion must be accom-plished before the prosthesis is fabricated, not subsequent to its fabrication. The objective of occlusal reconstruction by any means should be occlusal harmony of the restored dentition in relation to the natural forces already present or established. Therefore one of the earliest decisions in plan-ning reconstructive treatment must be whether to accept or reject the existing vertical dimension of occlusion and the occlusal contact relationships in centric and eccentric posi-tions. If occlusal adjustment is indicated, cuspal analysis www.konkur.in 176 Part II Clinical and Laboratory always should precede any corrective procedures in the mouth by selective grinding. On the other hand, if recon-struction is to be the means of correction, the manner and sequence should be outlined as part of the overall treatment plan.
Fixed Restorations There may be a need to restore modification spaces with fixed restorations rather than include them in the removable par-tial denture, especially when dealing with isolated abutment teeth. The advantage of splinting must be weighed against the total cost, with the weight of experience always in favor of using fixed restorations for tooth-bounded spaces unless the space will facilitate simplification of the removable par-tial denture design without jeopardizing the abutment teeth. One of the least successful removable partial denture designs is seen when multiple tooth-bounded areas are replaced with removable partial dentures in conjunction with isolated abutment teeth and distal extension bases. Biomechanical considerations and the future health of the remaining teeth should be given preference over economic considerations when such a choice is possible.
Orthodontic Treatment Occasionally, orthodontic movement of malposed teeth fol-lowed by retention through the use of fixed partial dentures makes possible a better removable partial denture design mechanically and esthetically than could otherwise be used. Although adequate anchorage for tooth movement can be a major limitation in partially edentulous arches, carefully placed implants that subsequently can be used for prosthesis support have been used to expand orthodontic applications for this patient group.
Need for Determining Type of Mandibular Major Connector As discussed in Chapter 5, one of the criteria used to deter-mine the use of the lingual bar or linguoplate is the height of the floor of the patient’s mouth when the tongue is elevated. Because the inferior borders of the lingual bar and the lin-guoplate are placed at the same vertical level, and because subsequent mouth preparations depend in part on the design of the mandibular major connector, determination of the type of major connector must be made during the oral examination. This determination is facilitated by measur-ing the height of the elevated floor of the patient’s mouth in relation to the lingual gingiva with a periodontal probe and recording the measurement for later transfer to diagnostic and master casts. It is most difficult to make a determination of the type of mandibular major connector to be used solely from a stone cast that may or may not accurately indicate the active range of movement of the floor of the patient’s mouth. Too many mandibular major connectors are ruined or made flexible because subsequent grinding of the inferior border is necessary to relieve impingement of the sensitive tissues of the floor of the mouth.
Need for Reshaping Remaining Teeth The clinical crown shapes of anterior and posterior teeth are not capable of supporting a removable partial denture framework without appropriate modification. Without the required modifications, the prosthesis does not adequately benefit from the support and stability offered by the teeth and consequently will not be comfortable to the patient. Many failures of removable partial dentures can be attrib-uted to the fact that the teeth were not reshaped properly to establish guiding planes or to receive clasp arms and occlusal rests before the impression for the master cast was made. Of particular importance are the paralleling of proximal tooth surfaces to act as guiding planes, the preparation of adequate rest areas, and the reduction of unfavorable tooth contours (Figure 13-18). To neglect planning such mouth prepara-tions in advance is inexcusable and leads to unsuccessful removable prosthesis service.
The design of clasps is dependent on the location of the retentive, stabilizing, reciprocal, and supporting areas in relation to a definite path of placement and removal. Failure to reshape unfavorably inclined tooth surfaces and, if nec-essary, to place restorations with suitable contours not only complicates the design and location of clasp retainers but also often leads to failure of the removable partial denture because of poor clasp design.
A malaligned tooth or one that is inclined unfavorably may make it necessary to place certain parts of the clasp so that they interfere with the opposing teeth. Unparallel proxi-mal tooth surfaces not only will fail to provide needed guid-ing planes during placement and removal but also will result in excessive blockout. This inevitably results in placement of the connectors so far out of contact with tooth surfaces that food traps are created. To pass lingually-inclined lower teeth, clearance for a lingual bar major connector may have to be so great that a food trap will result when the restora-tion is fully seated. Such a lingual bar will be located so that it will interfere with tongue comfort and function. These are only some of the objectionable consequences of inadequate mouth preparations.
The amount of reduction of tooth contours should be kept to a minimum, and all modified tooth surfaces not only should be repolished after reduction but also should be sub-jected to fluoride treatment to lessen the incidence of caries. If it is not possible to produce the contour desired without perforating the enamel, then the teeth should be recon-toured with an acceptable restorative material. The age of the patient, caries activity evidenced elsewhere in the mouth, and apparent oral hygiene habits must be taken into consid-eration when one is deciding between reducing the enamel and modifying tooth contours with protective restorations.
Some of the areas that frequently need correction are the lingual surfaces of mandibular premolars, the mesial and lingual surfaces of mandibular molars, the distobuccal line angle of maxillary premolars, and the mesiobuccal line angle of maxillary molars. The actual degree of inclination of teeth in relation to the path of placement and the location www.konkur.in 177 Chapter 13 Diagnosis and Treatment Planning of retentive and supportive areas are not readily interpreta-ble during visual examination. These are established during comprehensive analysis of the diagnostic cast with a sur-veyor, which should follow the visual examination.
INFECTION CONTROL The American Dental Association (ADA) follows the Centers for Disease Control (CDC) recommended infection control procedures for dentistry. The most recent recommendations were made in 2003 and included updates from the previous 1993 guidelines. Most of the updates will be familiar to prac-titioners and are already largely practiced routinely. They are designed to prevent or reduce the potential for disease trans-mission from patient to dental health care worker (DHCW), from DHCW to patient, and from patient to patient. The document emphasizes the use of “standard precautions” (which replaces the term “universal precautions”) for the prevention of exposure to and transmission of not only bloodborne pathogens but also other pathogens encountered in oral health care settings.
Major updates and additions include application of stan-dard precautions rather than universal precautions, work restrictions for health care personnel infected with, or occu-pationally exposed to, infectious diseases; management of occupational exposure to bloodborne pathogens, including postexposure prophylaxis for work exposures to hepatitis B virus (HBV), hepatitis C virus (HCV), and human immuno-deficiency virus (HIV); selection and use of devices with fea-tures designed to prevent sharps injury, contact dermatitis, and latex hypersensitivity; hand hygiene; dental unit water-lines; and biofilm and water quality; special considerations include dental handpieces and other devices attached to air lines and waterlines, saliva ejectors, radiology, parenteral medications, single-use or disposable devices, preprocedural mouth rinses, oral surgical procedures, handling of biopsy specimens and extracted teeth, laser/electrosurgery plumes, Mycobacterium tuberculosis, Creutzfeldt-Jakob disease and A B C Figure 13-18 A, Unmodified buccal surface of the mandibular premolar illustrates a typical height of contour location natural for this tooth (middle and occlusal thirds of the tooth). B, Proximal surface modification is required (hatched region) to produce a guide-plane surface. C, Buccal surface modification is needed to position the height of contour for favorable clasp location. The tooth modification is a continuation of the proximal surface modification onto the buccal surface, and generally requires less than 0.5 mm of tooth removal.
www.konkur.in 178 Part II Clinical and Laboratory other prion diseases, program evaluation, and research con-siderations. The recommendations provide guidance for measures to be taken that will reduce the risks of disease transmission, among both DHCWs and their patients.
Dental patients and DHCWs potentially may be exposed to a variety of microorganisms. Exposure can occur via blood and/or oral or respiratory secretions. The microorgan-isms may include viruses and bacteria that infect the upper respiratory tract in general, as well as cytomegalovirus, HBV, HCV, herpes simplex virus types 1 and 2, HIV, Myco-bacterium tuberculosis, staphylococci, and streptococci. The transmission of infection in the dental operatory can occur through several routes. These include direct contact (blood, oral fluids, or other secretions), indirect contact (contami-nated instruments, operatory equipment, or environmental surfaces), and contact with airborne contaminants present in droplet spatter or in aerosols of oral and respiratory flu-ids. For infection to occur via any of these routes, the “chain of infection” must be present. This includes a susceptible host, a pathogen with sufficient infectivity and numbers to cause infection, and a portal through which the pathogen may enter the host. For infection control procedures to be effective, one or more of these “links” in the chain must be broken.
Studies from the CDC report that clothing exposed to the acquired immunodeficiency syndrome (AIDS) virus may be safely used after a normal laundry cycle. A high-temperature (140° F to 160° F, 60° C to 70 °C) wash cycle with normal bleach concentrations, followed by machine drying (212° F, 100° C, or higher), is preferable if clothing is visibly soiled with blood or other body fluids. Dry cleaning and steam pressing will also kill the AIDS virus, according to these studies. Patients with oral lesions suggestive of infec-tious disease and patients with a known history of hepatitis B, AIDS, AIDS-related complex, or other infectious diseases should be referred for appropriate medical care. In addition to environmental surface and equipment disinfection, all instruments, stones, burs, and other reusable items should be disinfected in 2% glutaraldehyde for 10 minutes, cleaned of debris, rinsed, and patted dry before the sterilizing process is initiated. Heat-sensitive items can be sterilized with the use of ethylene oxide (gas).
For items that have been used in the mouth, including laboratory materials (e.g., impressions, bite registrations, fixed and removable prostheses, orthodontic appliances), cleaning and disinfection are required before they are manip-ulated in the laboratory (whether on site or at a remote loca-tion). Any item manipulated in the laboratory should also be cleaned and disinfected before placement in the patient’s mouth. Fresh pumice with iodophor should be used for each polishing procedure, and the pumice pan should be washed, rinsed, and dried after each procedure. Because materials are constantly evolving, DHCWs are advised to follow manu-facturers’ suggested procedures for specific materials rela-tive to disinfection procedures. As a guide, use of a chemical germicide that has at least an intermediate level of activity (i.e., “tuberculocidal hospital disinfectant”) is appropriate for such disinfection. Careful communication between dental office and dental laboratory regarding the specific protocol for handling and decontamination of supplies and materials is important to prevent any cross contamination.
DIFFERENTIAL DIAGNOSIS: FIXED OR REMOVABLE PARTIAL DENTURES Total oral rehabilitation (disease management, defective tooth restoration, and tooth replacement) is an objective in treating the partially edentulous patient. Although replace-ment of missing teeth by means of fixed partial dentures, either tooth or implant supported, is generally the method of choice, there are many reasons why a removable partial denture may be the better method of treatment for a specific patient.
The dentist must follow the best procedure for the wel-fare of the patient, who is always free to seek more than one opinion. Ultimately, the choice of treatment must meet the economic limitations and personal desires of the patient. The exception to this guideline is the Class III arch with a modi-fication space on the opposite side of the arch, which will provide better cross-arch stabilization and a simpler design for the removable partial denture (Figure 13-19).
Although uncommon, unilateral tooth loss is sometimes inappropriately treated with a unilateral removable partial denture in place of a fixed partial denture. This type of pros-thesis is not enhanced by cross-arch stabilization and places excessive stress on abutment teeth. Possibly more important, the risk for aspiration is significant if such a prosthesis is A A Figure 13-19 Class III, modification 2 arch, in which modifi-cation spaces on the patient’s left (spaces designated at A) will be included in the design of the removable partial denture rather than restored with a long-span fixed partial denture. The design for a removable restoration is greatly simplified, resulting in sig-nificantly enhanced stability.
www.konkur.in 179 Chapter 13 Diagnosis and Treatment Planning dislodged during use. For these reasons, use of the unilateral removable partial denture is strongly discouraged.
Indications for Use of Fixed Restorations Tooth-Bounded Edentulous Regions Generally any unilateral edentulous space bounded by teeth suitable for use as abutments should be restored with a fixed partial denture cemented to one or more abutment teeth at either end. The length of the span and the periodontal sup-port of the abutment teeth will determine the number of abutments required. As mentioned earlier, such a span could be managed with the use of dental implants if deemed fea-sible and elected by the patient. The fact that implant support does not place additional functional demands on adjacent teeth likely contributes to their preservation, although this has not been universally demonstrated.
For conventional fixed prostheses, lack of parallelism of the abutment teeth may be counteracted with copings or locking connectors to provide parallel sectional placement. Sound abutment teeth make possible the use of more con-servative retainers, such as partial-veneer crowns, or resin-bonded-to-metal restorations, rather than full crowns. The age of the patient, evidence of caries activity, oral hygiene habits, and the soundness of remaining tooth structure must be considered in any decision to use less than full coverage for abutment teeth.
Two specific contraindications for the use of unilateral fixed restorations are known. One is a long edentulous span with abutment teeth that would not be able to withstand the trauma of nonaxial occlusal forces. The other is abutment teeth, which exhibit reduced periodontal support due to periodontal disease, which would benefit from cross-arch stabilization. In either situation, a bilateral removable res-toration can be used more effectively to replace the missing teeth.
Modification Spaces A removable partial denture for a Class III arch is better supported and stabilized when a modification area on the opposite side of the arch is present. A fixed partial den-ture need not be used to restore such an edentulous area because its inclusion may simplify the design of the remov-able partial denture. However, when a modification space is bound by a lone-standing single-rooted abutment, it is better restored by means of a fixed partial denture. This acts to stabilize the at-risk tooth, and the denture is made less complicated by not having to include other abutment teeth for the support and retention of an additional edentulous space or spaces.
When an edentulous space that is a modification of a Class I or Class II arch exists anterior to a lone-standing abutment tooth, this tooth is subjected to trauma by the movements of a distal extension removable partial denture far in excess of its ability to withstand such stresses. The splinting of the lone abutment to the nearest tooth is mandatory. The abutment crowns should be contoured for support and retention of the removable partial denture; in addition, a means of support-ing a stabilizing component on the anterior abutment of the fixed partial denture or on the occlusal surface of the pontic usually should be provided.
Anterior Modification Spaces Usually any missing anterior teeth in a partially edentulous arch, except in a Kennedy Class IV arch in which only ante-rior teeth are missing, are best replaced by means of a fixed restoration. There are exceptions. Sometimes a better esthetic result is obtainable when the anterior replacements are sup-plied by a removable partial denture, at other times treat-ment is simplified by inclusion of an anterior modification space into the removable partial denture (Figure 13-20). This is also true when excessive tissue and bone resorption neces-sitates placement of the pontics in a fixed partial denture too far palatally for good esthetics or for an acceptable relation with the opposing teeth. However, in most instances, from mechanical and biological standpoints, anterior replace-ments are best accomplished with fixed restorations. The replacement of missing posterior teeth with a removable par-tial denture is then made much less complicated and gives more satisfactory results.
Replacement of Unilaterally Missing Molars (Shortened Dental Arch) Often the decision must be made to replace unilaterally missing molars (Figure 13-21). The decision must balance the impact of the treatment on the remaining oral struc-tures with the potential benefit to the patient long term. To restore the missing molars with a fixed partial denture would require a cantilever prosthesis or the use of dental implants. A cantilever-fixed prosthesis is most applicable if the second molar is to be ignored, then only first molar occlusion need be supplied with the use of a cantilever-type fixed partial denture. Occlusion need be only minimal to maintain occlu-sal relations between the natural first molar in the one arch and the prosthetic molar in the opposite arch. The cantile-vered pontic should be narrow buccolingually and need not occlude with more than one half to two thirds of the oppos-ing tooth. Often such a restoration is the preferred method of treatment. However, at least two abutments should be used to support a cantilevered molar opposed by a natural molar.
To replace unilaterally missing molars with a remov-able partial denture necessitates the use of a distal exten-sion prosthesis. This involves the major connector joining the edentulous side to retentive and stabilizing components located on the non-edentulous side of the arch. Leverage factors are frequently unfavorable, and the retainers used on the non-edentulous side are often unsatisfactory. Two factors important to consider in making the decision to provide a unilateral, distal extension removable partial den-ture include the opposing teeth and the future effect of the maxillary tuberosity.
First, the opposing teeth must be considered if it is con-sidered important to prevent extrusion and migration. This www.konkur.in 180 Part II Clinical and Laboratory A B C Figure 13-20 A, Diagnostic waxing of this complex case revealed the best means to manage replacement of tooth #6 and tooth #7 was with a fixed prosthesis, especially because the ridge defect was not severe and the adjacent teeth offered good retainer support. B, In contrast, this complex situation requires the maxillary anterior to be repositioned palatally to address an esthetic concern caused by the condition of the maxillary canines and the need to replace the posterior teeth as well. C, The anterior teeth will be more easily managed as part of the removable partial denture. (Courtesy of Dr. M. Alfaro, Columbus, OH.) A B Figure 13-21 A, Unilaterally missing molars. If the patient exhibits opposing contacts to the remaining six posterior teeth (bilateral premolars, right first and second molars), functional gain attained by replacing the left molars may be minimal. B, By contrast, the func-tional gain resulting from replacement of the posterior occlusion in this patient is likely significant.
www.konkur.in 181 Chapter 13 Diagnosis and Treatment Planning influences replacement of the missing molars far more than any improvement in masticating efficiency that might result. Replacement of missing molars on one side is seldom neces-sary for reasons of mastication alone.
Second, the future effect of a maxillary tuberosity must be considered if concern exists for tuberosity enlargement. Often when left uncovered, the tuberosity increases in size, making future occlusal treatment difficult. However, cover-ing the tuberosity with a removable partial denture base, in combination with the stimulating effect of the intermittent occlusion, helps maintain tuberosity size and position. In such an instance, it may be better to make a removable par-tial denture with cross-arch stabilization and retention than to leave a maxillary tuberosity uncovered.
Indications for Removable Partial Dentures Although a removable partial denture should be considered only when a fixed restoration is contraindicated, there are several specific indications for the use of a removable res-toration.
Distal Extension Situations Replacement of missing posterior teeth is often best accom-plished with a removable partial denture (see Figure 13-22, B), especially when implant treatment is not feasible for the patient. The exception to this includes situations in which the replacement of missing second (and third) molars is inadvisable or unnecessary, or in which unilateral replace-ment of a missing first molar can be accomplished by means of a multiple-abutment cantilevered fixed restoration or an implant-supported prosthesis. The most common partially edentulous situations are the Kennedy Class I and Class II. With the latter, an edentulous space on the opposite side of the arch is often conveniently present to aid in required retention and stabilization of the removable partial den-ture. If no space is present, selected abutment teeth can be modified to accommodate appropriate clasp assemblies, or intracoronal retainers can be used. As stated previously, all other edentulous areas are best replaced with fixed partial dentures.
After Recent Extractions The replacement of teeth after recent extractions often can-not be accomplished satisfactorily with a fixed restoration. When relining will be required later or when a fixed restora-tion using natural teeth or implants will be constructed later, a temporary removable partial denture can be used. If an all-resin denture is used rather than a cast framework removable partial denture, the immediate cost to the patient is much less, and the resin denture lends itself best to future tempo-rary modifications, including those required after implant placement and before restoration.
Tissue changes are inevitable following extractions. Tooth-bounded edentulous areas (as a result of extractions) are best initially restored with removable partial dentures. Relining of a tooth-supported resin denture base is then pos-sible. This is usually done to improve esthetics, oral clean-liness, or patient comfort. Support for such a restoration is supplied by occlusal rests on the abutment teeth at each end of the edentulous space.
Long Span A long span may be totally tooth-supported if the abutments and the means of transferring the support to the denture are adequate and if the denture framework is rigid. There is little if any difference between the support afforded a remov-able partial denture and that afforded a fixed restoration by the adjacent abutment teeth. However, in the absence of A B Figure 13-22 A, Occlusal view of the anterior ridge defect (Kennedy Class IV) shows the palatal position of the ridge crest. Incisal edges of opposing dentition require a more labial position, which would create a difficult pontic form. B, Labial view of the same cast shows the significance of the vertical bone loss. Replacement of the teeth and ridge anatomy is best accomplished with a removable partial denture.
www.konkur.in 182 Part II Clinical and Laboratory cross-arch stabilization, the torque and leverage on the two abutment teeth would be excessive. Instead, a removable denture that derives retention, support, and stabilization from abutment teeth on the opposite side of the arch is indi-cated as the logical means of replacing the missing teeth.
Need for Effect of Bilateral Stabilization In a mouth weakened by periodontal disease, a fixed resto-ration may jeopardize the future of the involved abutment teeth unless the splinting effect of multiple abutments is used. The removable partial denture, on the other hand, may act as a periodontal splint through its effective cross-arch stabilizing of teeth weakened by periodontal disease. When abutment teeth throughout the arch are properly prepared and restored, the beneficial effect of a removable partial den-ture can be far greater than that of a unilateral fixed partial denture.
Excessive Loss of Residual Bone The pontic of a fixed partial denture must be correctly related to the residual ridge and in such a manner that the contact with the mucosa is minimal. Whenever excessive resorp-tion has occurred, teeth supported by a denture base may be arranged in a more acceptable buccolingual position than is possible with a fixed partial denture (Figure 13-22).
Unlike a fixed partial denture, the artificial teeth sup-ported by a denture base can be located without regard for the crest of the residual ridge and more nearly in the position of the natural dentition for normal tongue and cheek con-tacts. This is particularly true of a maxillary denture.
Anteriorly, loss of residual bone occurs from the labial aspect. Often the incisive papilla lies at the crest of the residual ridge. Because the central incisors are normally located anterior to this landmark, any other location of artificial central incisors is unnatural. An anterior fixed partial denture made for such a mouth will have pontics contacting the labial aspect of this resorbed ridge and will be too far lingual to provide desirable lip support. Often the only way the incisal edges of the pontics can be made to occlude with the opposing lower anterior teeth is to use a labial inclination that is excessive and unnatural, and both esthetics and lip support suffer. Because the same condition exists with a removable partial denture in which the ante-rior teeth are abutted on the residual ridge, a labial flange must be used to permit the teeth to be located closer to their natural position.
The same method of treatment applies to the replace-ment of missing mandibular anterior teeth. Sometimes a mandibular anterior fixed partial denture is made six or more units in length, in which the remaining space neces-sitates leaving out one anterior tooth or using the original number of teeth but with all of them too narrow for esthet-ics. In either instance, the denture is nearly in a straight line because the pontics follow the form of the resorbed ridge. A removable partial denture will permit the location of the replaced teeth in a favorable relation to the lip and opposing dentition regardless of the shape of the residual ridge. When such a removable prosthesis is made, how-ever, positive support must be obtained from the adjacent abutments.
Unusually Sound Abutment Teeth Sometimes the reasoning for making a removable restora-tion is the desire to see sound teeth preserved in their natu-ral state and not prepared for restorations. As mentioned previously, if this decision is made because it is felt that no tooth modification is necessary for removable partial den-tures, then the prosthesis will lack tooth-derived stability and support.
When this condition exists, the dentist should not hesitate to reshape and modify existing enamel surfaces to provide proximal guiding planes, occlusal rest areas, optimum reten-tive areas, and surfaces on which nonretentive stabilizing components may be placed. Continued durability of the nat-ural teeth is best ensured if the modifications that optimize prosthesis function are provided. This is due to the fact that such modifications also ensure the most harmonious use of the natural dentition.
Abutments with Guarded Prognoses If the prognosis of an abutment tooth is questionable or if it becomes unfavorable while under treatment, it might be possible to compensate for its impending loss by a change in denture design. The questionable or condemned tooth or teeth may then be included in the original design and, if subsequently lost, the removable partial denture can be modified or remade (Figure 13-23). Most removable partial denture designs do not lend themselves well to later addi-tions, although this eventuality should be considered in the design of the denture.
When the tooth in question will be used as an abut-ment, every diagnostic aid should be used to determine its prognosis as a prospective abutment. It is usually not as difficult to add a tooth or teeth to a removable partial denture as it is to add a retaining unit when the original abutment is lost and the next adjacent tooth must be used for that purpose.
It is sometimes possible to design a removable partial denture so that a single posterior abutment, about which there is some doubt, can be retained and used at one end of the tooth-supported base. Then if the posterior abutment is lost, it could be replaced by adding an extension base to the existing denture framework. Such an original design must include provisions for future indirect retention, flexible clasping of the future abutment, and provisions for establish-ing tissue support. Anterior abutments that are considered poor risks may not be so freely used because of the prob-lems involved in adding a new abutment retainer when the original one is lost. It is rational that such questionable teeth should be condemned in favor of more suitable abutments, even though the original treatment plan must be modified accordingly.
www.konkur.in 183 Chapter 13 Diagnosis and Treatment Planning Economic Considerations Economics should not be the sole criterion used to arrive at a method of treatment. When, for economic reasons, complete treatment is out of the question and yet replacement of miss-ing teeth is indicated, the restorative procedures dictated by these considerations must be described clearly to the patient as a compromise and not as representative of the best that modern dentistry has to offer. A prosthesis that is made to satisfy economic considerations alone may provide only lim-ited success and result in more costly treatment in the future.
CHOICE BETWEEN COMPLETE DENTURES AND REMOVABLE PARTIAL DENTURES One of the more difficult decisions to make for the partially edentulous patient involves making the choice of a complete denture over a removable partial denture. Many factors need to be considered when one is making such a decision; these generally fall under the categories of tooth-related factors, factors of comparative functional expectations between pros-theses, and patient-specific factors. Because the difference between a tooth-tissue–born prosthesis and a tissue-born prosthesis can be significant, especially because it is difficult for the partially edentulous patient to conceptualize the tis-sue-born situation, such an irreversible decision is not trivial.
An evaluation of the remaining teeth determines whether any caries or periodontal disease exists. The decision as to whether a tooth is useful for inclusion in a prosthetic treat-ment plan can be made on the basis of an understanding that with appropriate disease management, the tooth pro-vides a reasonable 5-year prognosis for survival. This takes into account the added functional demand by the prosthe-sis and a risk assessment for recurrent disease. Because this scenario concerns teeth with disease, the expectation is that tooth structure and/or support is compromised. The added functional burden, along with a potentially increased risk for disease, is an important concern when one is determining the long-term benefit for retaining teeth with a removable partial denture.
If the teeth can be maintained with a reasonable progno-sis, the next questions to ask are: “Do they require restoration with surveyed crowns?” and “How much improvement to the prosthesis support, stability, and retention do they provide?” If the expected prognosis for a given tooth is questionable, the costs associated with restoration high, and the added benefit to the prosthesis low, the tooth should likely not be maintained unless the patient strongly desires to maintain all teeth. However, if the same scenario exists and the long-term impact on the support, stability, and retention of the pros-thesis is great, the decision strongly favors keeping the tooth.
The question of whether retained teeth offer a signifi-cant advantage to the prosthesis from a support, stability, and retention standpoint requires comparative evaluation of potential denture-bearing foundations. If the expectation is that an edentulous arch would have unfavorable physical fea-tures (poor ridge form, poor arch configuration, displaceable mucosa, high frena attachments, minimum denture bearing area, and/or an unfavorable jaw relationship), then retention of teeth is likely to provide a more significant benefit. If reten-tion of teeth can help to prevent or delay age-related denture-bearing foundation changes seen with complete denture use, then retention of teeth can be of significant benefit.
C A B Figure 13-23 Kennedy Class II, modification 1, where the molar abutment has a guarded prognosis. A, Anterior abutment of the modification space has a clasp assembly that accommo-dates for potential future loss of the distal molar while currently providing adequate support, stability, and retention. B, Premolar clasp assembly comprises a mesial rest, a distal guide plane, and a wrought-wire retainer design, which will accommodate future distal extension movement. C, Buccal view shows guide-plane contact and a wrought-wire location that is appropriate for a distal extension.
www.konkur.in 184 Part II Clinical and Laboratory When evaluation demonstrates that the remaining teeth have no active disease, then the often-negative impact of dis-ease management on prognosis is not a concern. The decision to maintain teeth is again based on risk assessment, costs for use of the teeth, added benefit to prosthesis functional sta-bility, and comparative functional expectations between a mucosal-borne denture and a removable partial denture that uses teeth for some support, stability, and retention.
The remaining tooth location and distribution can also affect the decision to maintain teeth. It makes a difference whether the remaining teeth are located on only one side of the arch. Having bilateral teeth remaining, especially if they are in similar locations (canines-canines, canines/ premolars-canines/premolars), offers advantages for pros-thesis design and occlusal development compared with asymmetrical tooth locations. Some teeth may not serve well as a stabilizing component for a removable partial denture and should not be maintained. If the remaining terminal tooth adjacent to a distal extension base is an incisor, the likelihood of long-term support, stability, and retention is poor.
An additional factor for consideration when one is decid-ing between a complete denture and a removable partial denture is whether there is a strong patient desire to main-tain teeth. As mentioned previously, because the change to a complete denture is a significant transformation, sufficient discussion must take place before this decision is made. The dentist must be very clear that the patient understands the functional differences between a mucosal-borne prosthesis for all aspects of function (i.e., chewing, talking, and so on) and the natural dentition or a removable partial denture.
The uniqueness of the patient is again appreciated as these issues are discussed with the patient. One patient may prefer complete dentures rather than complete oral rehabilitation, regardless of ability to pay. Another may be so determined to keep his own teeth that he will make great financial sacrifice if given a reasonable assurance of success of oral rehabilita-tion. Listening to the patient during the examination and the diagnostic procedures pays off significantly when the treatment options differ so vastly as complete and removable partial dentures often do. During the presentation of per-tinent facts, time should be allowed for patients to express themselves freely as to their desires in retaining and restor-ing their natural teeth. At this time, a treatment plan may be influenced or even drastically changed to conform to the expressed and implied wishes of the patient. For example, there may be a reasonable possibility of saving teeth in both arches through the use of removable partial dentures. With only anterior teeth remaining, a removable partial denture can be made to replace the posterior teeth with the use of good abutment support and, in the maxillary arch, use of full palatal coverage for retention and stability. If patients express a desire to retain their anterior teeth at any cost, and if the remaining teeth are esthetically acceptable and functionally sound, the dentist should make every effort to provide suc-cessful treatment. If patients prefer a mandibular removable partial denture because of fear of difficulty in wearing a man-dibular complete denture, then (all factors being acceptable) their wishes should be respected and treatment should be planned accordingly. The professional obligation to present the facts and then do the best that can be done in accordance with the patients’ expressed desires still applies.
Other patients may wish to retain remaining teeth for an indefinite but relatively short period of time, with eventual complete dentures a foregone conclusion. In this instance, the professional obligation may be to recommend interim removable partial dentures without extensive mouth prepa-ration. Such dentures will aid in mastication and will provide esthetic replacements, at the same time serving as condi-tioning restorations, which will make the later transition to complete dentures somewhat easier. Such removable partial dentures should be designed and fabricated with care, but the total cost of removable partial denture service should be considerably less.
An expressed desire on the part of patients to retain only six mandibular anterior teeth must be considered carefully before this is agreed to as the planned treatment. The advantages for patients are obvious: (1) they may retain six esthetically accept-able teeth; (2) they do not become totally edentulous; and (3) they have the advantages of direct retention for the remov-able partial denture that would not be possible if they were completely edentulous. Retaining even the mandibular canine teeth would accomplish the latter two objectives. Potential dis-advantages relate directly to the patient keeping up with pros-thesis maintenance procedures. The disadvantages relate to the poor response of the anterior maxilla to functional stress concentrated from the opposing natural dentition. If the func-tional forces of occlusion are not well distributed, the natural anterior can concentrate stress to the anterior maxillary arch. The possible result of such poorly distributed functional force includes the loss of residual maxillary bone, loosening of the maxillary denture caused by the tripping influence of the natu-ral mandibular teeth, and loss of the basal foundation for the support of future prostheses. However, if the maxillary ante-rior teeth are arranged to contact in balanced eccentric posi-tions and if patients comply with periodic recall to maintain these relationships, these problems are minimized. Prevention of this sequence of events lies in the maintenance of positive occlusal support posteriorly and the continual elimination of traumatic influence from the remaining anterior teeth. Such support is sometimes impossible to maintain without frequent relining or remaking of the lower removable partial denture base. The presence of inflamed hyperplastic tissue is a frequent sequela to continued loss of support and denture movement.
Although some patients are able to successfully func-tion with a lower removable partial denture supported only by anterior teeth against a complete maxillary denture, it is likely that undesirable consequences will result unless the patient faithfully follows the instructions of the dentist. In no other situation in treatment planning are the general health of the patient and the quality of residual alveolar bone as critical as they are in this situation.
www.konkur.in 185 Chapter 13 Diagnosis and Treatment Planning CLINICAL FACTORS RELATED TO METAL ALLOYS USED FOR REMOVABLE PARTIAL DENTURE FRAMEWORKS The cast framework offers significant advantages over the all acrylic-resin removable partial denture. In general, the ability to predictably utilize the remaining teeth for support, stabil-ity, and retention over time is best assured when the inter-face between prosthesis and teeth consists of a cast structure and not a polymer. Although the utility of all acrylic-resin prostheses can be extended if wire “rests” are provided, typi-cal polymer properties do not allow for a durable interface, which is required if one is to take advantage of the stabiliz-ing effects of tooth contact. Expectations of how the metal framework improves functional performance are related to the properties of the metal alloy. Various alloys can be con-sidered for use. Following is a discussion of the most com-mon framework alloys in use today.
Practically all cast frameworks for removable partial den-tures are made from a chromium-cobalt (Cr-Co) alloy. The popularity of chromium-cobalt alloys has been attributed to their low density (weight), high modulus of elasticity (stiff-ness), low material cost, and resistance to tarnish. The term stellite alloy historically has referred to this class of alloy. Today the more common alloys contain 60% to 63% Co, 29% to 31.5% Cr, and 5% to 6% molybdenum (Mo), with the balance including silicon (Si), manganese (Mn), iron (Fe), nitrogen (N), and carbon (C). The addition of controlled amounts of nitrogen (<0.5%) is reported to improve physi-cal properties. Titanium is also used as a removable partial denture frame material; however, production difficulties continue to hinder its widespread use. The dentist should become familiar with the alloy used by her/his laboratory and should closely monitor fit, density, and rigidity.
The following are comparable characteristics of gold alloys and chromium-cobalt alloys: (1) each is well tolerated by oral tissues; (2) they are equally acceptable esthetically; (3) enamel abrasion by either alloy is insignificant on vertical tooth surfaces; (4) a low-fusing chrome-cobalt alloy or gold alloy can be cast to wrought wire, and wrought-wire compo-nents may be soldered to either gold or chrome-cobalt alloys (these characteristics are important in overcoming the objec-tion by some dentists to the increased stiffness of chromium-cobalt alloys for the portions of direct retainers that must engage an undercut of the abutment tooth); (5) the accu-racy obtainable in casting either alloy is clinically acceptable under strictly controlled investing and casting procedures; and (6) soldering procedures for the repair of frameworks can be performed on each alloy.
Comparative Physical Properties of Gold and Chromium-Cobalt Chromium-cobalt alloys generally have less yield strength when compared with gold alloys used for removable partial dentures. Yield strength is the greatest amount of stress an alloy will withstand and still return to its original shape in an unweakened condition. Possessing a lower proportional limit, the chromium-cobalt alloys will deform permanently at lower loads than gold alloys. Therefore the dentist must design the chromium-cobalt framework so that the degree of deforma-tion expected in a direct retainer is less than a comparable degree of deformation for a gold component. The modulus of elasticity refers to the stiffness of an alloy. Gold alloys have a modulus of elasticity approximately one-half that for chro-mium-cobalt alloys for similar uses. The greater stiffness of the chromium-cobalt alloy is advantageous but at the same time offers disadvantages. Greater rigidity can be obtained with the chromium-cobalt alloy in reduced sections in which cross-arch stabilization is required, thereby eliminating an appreciable bulk of the framework. Its greater rigidity is also an advantage when the greatest undercut that can be found on an abutment tooth is in the nature of 0.05 inch. A gold retentive element would not be as efficient in retaining the restoration under such conditions as would the chromium-cobalt clasp arm.
High yield strength and low modulus of elasticity produce greater flexibility. The gold alloys are approximately twice as flexible as the chromium-cobalt alloys; in many instances, this provides a distinct advantage in the optimum location of retentive elements of the framework. The greater flexibility of the gold alloys usually permits location of the tips of retainer arms in the gingival third of the abutment tooth. The stiff-ness of chromium-cobalt alloys can be overcome by includ-ing wrought-wire retentive elements in the framework.
The bulk of a retentive clasp arm for a removable partial denture is often reduced for greater flexibility when chro-mium-cobalt alloys are used as opposed to gold alloys. This, however, is inadvisable because the grain size of chromium-cobalt alloys is usually larger and is associated with a lower proportional limit, and so a decrease in the bulk of chro-mium-cobalt cast clasps increases the likelihood of fracture or permanent deformation. The retentive clasp arms for both alloys should be approximately the same size, but the depth of undercut used for retention must be reduced by one half when chromium-cobalt is the choice of alloys. Chromium-cobalt alloys are reported to work/harden more rapidly than gold alloys, and this, associated with coarse grain size, may lead to failure in service. When adjustments by bending are necessary, they must be executed with extreme caution and limited optimism.
Chromium-cobalt alloys have a lower density (weight) than gold alloys in comparable sections and therefore are about one half as heavy as gold alloys. The weight of the alloy in most instances is not a valid criterion for selection of one metal over another because after placement of a removable partial denture, the patient seldom notices the weight of the restoration. The comparable lightness of chromium-cobalt alloys, however, is an advantage when full palatal coverage is indicated for the bilateral distal extension removable par-tial denture. Weight is a factor that must be considered when the force of gravity must be overcome, so that usually passive direct retainers will not be activated constantly to the detri-ment of abutment teeth.
www.konkur.in 186 Part II Clinical and Laboratory The hardness of chromium-cobalt alloys presents a dis-advantage when a component of the framework, such as a rest, is opposed by a natural tooth or by one that has been restored. We have observed more wear of natural teeth opposed by some of the various chromium-cobalt alloys as contrasted with type IV gold alloys.
It has been observed that gold frameworks for removable partial dentures are more prone to produce uncomfortable galvanic shock to abutment teeth restored with silver amal-gam than are frameworks made of chromium-cobalt alloy. This may not be a valid criterion for the selection of a par-ticular alloy when the dentist has complete control over the choice of restorative materials.
Commercially pure (CP) titanium and titanium in alloys containing aluminum and vanadium, or palladium (Ti-O Pd), should be considered potential future materials for removable partial denture frameworks. Their versatility and well-known biocompatibility are promising; however, long-term clinical tri-als are needed to validate their potential usefulness. Currently, when CP titanium is cast under dental conditions, the material properties change dramatically. During the casting procedure, the high affinity of the liquid metal for elements such as oxygen, nitrogen, and hydrogen results in their incorporation from the atmosphere. As interstitial alloying elements, their deleterious effect on mechanical properties is a problem. Also, reactions between molten titanium metal and the investment refractory produce gases, which cause porosity. With alpha-beta alloys, such as Ti-6Al-4V , a surface skin of alpha titanium can form (alpha-case zone), which has a tremendous effect on electro-chemical behavior and mechanical properties. This could be important for small thin structures, such as clasp assemblies and major and minor connectors. The CP grades of titanium have yield strengths that are too low for clinical use as clasps (450 MPa minimum), although the ductility is high. The much higher yield strengths of the Ti-6Al-4V alloys are the same as that of a typical bench-cooled cobalt-chromium alloy but with far superior ductility. The typical Young’s modulus of elastic-ity of titanium alloy is half that of cobalt-chromium and just slightly higher than that of type IV gold alloys. This would require a different approach to clasp design than is used with cobalt-chromium alloys and would present some advantages. Wrought titanium alloy wires are also flexible because of the same low elastic modulus. Beta alloys, which are used in ortho-dontics, have two-thirds the elastic modulus of CP titanium and Ti-6Al-4V . The joining of titanium by brazing is a problem because like-casting inert atmospheres must be used. The cor-rosion and fatigue behavior of brazed joints has yet to be tested for long-term corrosion resistance and clinical efficacy. Clinical use has demonstrated reasonable short-term results, but labora-tory fabrication difficulties need to be addressed, and long-term advantages over existing alloys must be demonstrated before titanium will gain broad clinical use.
Wrought Wire: Selection and Quality Control Wrought-wire direct retainer arms may be attached to the restoration by embedding a portion of the wire in a resin denture base, by soldering to the fabricated framework, or by casting the framework to a wire embedded in the wax pattern (Figure 13-24). The physical (mechanical) prop-erties of available wrought wires are most important con-siderations when a proper wire for the desired method of attachment is selected. These properties include yield strength or proportional limit, percentage elongation, tensile strength, and fusion temperature. After the wire is selected, the procedures to which the wire is subjected in fabricating the restoration become critical. Improper laboratory procedures can diminish certain desirable physical properties of the wrought structure, rendering it relatively useless for its intended purpose. For example, when wrought wire is heated (such as in a cast-to or solder-ing procedure), its physical properties and microstructure may be considerably altered, depending on temperature, heating time, and cooling operation. All manufacturers of wrought forms for dental applications furnish charts listing their products and the physical properties of each product. The percentage of noble metals is given. In addition, most manufacturers designate wires that may be used in a cast-to procedure. ADA Specification No. 7 addresses itself to wrought gold wire in terms of both content and minimum physical properties (Table 13-1).
The tensile strength of the wrought structure is approxi-mately 25% greater than that of the cast alloy from which it was made. The wrought structure’s hardness and strength are also greater. This means that a wrought structure that has a smaller cross section than a cast structure may be used as a retainer arm (retentive) to perform the same function. It has been suggested that minimum yield strength of 60,000 psi is required for the retentive element of a direct retainer. A percentage elongation of less than 6% is indicative that a wrought wire may not be amenable to contouring without attendant undesirable changes in microstructure.
Figure 13-24 The wrought-wire retainer arm has been con-toured to design and incorporated into the wax pattern of this frame where it will become an integral part of the framework. The wire is contoured in two planes and will be mechanically retained in the casting.
www.konkur.in 187 Chapter 13 Diagnosis and Treatment Planning Regardless of the method of attaching the wrought-wire retainer that is used (that is, embedding, soldering, or cast-to), tapering the wrought arm seems most rational. A retainer arm is in essence a cantilever that can be made more service-able and efficient by tapering. Tapering to 0.8 mm permits more uniform distribution of service stresses throughout the length of the arm, being readily demonstrated by photoelas-tic stress analysis. Uniform tapering of an 18-gauge, round wire arm can be accomplished by rapidly rotating the wire in angled contact with an abrasive disk in the dental lathe. It is then polished by rotating the wire in angled contact with a mildly abrasive rubber disk in the dental lathe. The appropri-ate taper is shown in Figure 13-25.
SUMMARY In selecting materials, it must be remembered that fun-damentals do not change. These are inviolable. It is only methods, procedures, and substances—by which the den-tist affects the best possible end result—that change. The responsibility of the decision still rests with the dentist, who must evaluate all factors in relation to the results desired. In any instance therefore, the dentist must weigh the problems involved, compare and evaluate the charac-teristics of different potential materials, and then make a decision that leads to delivery of the greatest possible ser-vice to the patient.
Table 13-1 Comparative Specifications Contained in American Dental Association Specification No. 7 Type I Type II Content of metals of the gold, platinum group (minimum) 75% 65% Minimum fusion temperature 1742° F 1898° F Minimum yield point value (hardened or oven cooled) 125,000 psi 95,000 psi Minimum elongation (hardened) 4% 2% Minimum elongation (softened) 15% 15% psi, Pounds per square inch.
D ½ D Figure 13-25 Round, 18-gauge wrought wire for the retentive component of the direct retainer assembly (clasp) is uniformly tapered to 0.8 mm from its full diameter to its terminus. Tapering should precede contouring of the wire for the retainer arm.
www.konkur.in CHAPTER 14 Preparation of the Mouth for Removable Partial Dentures CHAPTER OUTLINE Pre-Prosthetic Considerations in Partially Edentulous Mouths Extractions Impacted Teeth Malposed Teeth Cysts and Odontogenic Tumors Exostoses and Tori Hyperplastic Tissue Muscle Attachments and Frena Bony Spines and Knife-Edge Ridges Polyps, Papillomas, and Traumatic Hemangiomas Hyperkeratoses, Erythroplasia, and Ulcerations Dentofacial Deformity Dental Implants Augmentation of Alveolar Bone Periodontal Preparation Objectives of Periodontal Therapy Periodontal Diagnosis and Treatment Planning Initial Disease Control Therapy (Phase 1) Definitive Periodontal Surgery (Phase 2) Recall Maintenance (Phase 3) Advantages of Periodontal Therapy Optimization of the Foundation for Fitting and Function of the Prosthesis Conditioning of Abused and Irritated Tissues Use of Tissue Conditioning Materials Abutment Restorations Contouring Wax Patterns Rest Seats The preparation of the mouth is fundamental to a success-ful removable partial denture service. Mouth preparation, perhaps more than any other single factor, contributes to the philosophy that the prescribed prosthesis not only must replace what is missing but also must preserve the remaining tissues and structures that will enhance the removable partial denture. Because of the significant and favorable impact dental implants can have on preserva-tion of oral tissues and structures when used with remov-able partial dentures, it is common to include discussion of dental implants when considering mouth preparation for removable prostheses.
Mouth preparation follows the preliminary diagnosis and the development of a tentative treatment plan. Final treatment planning may be deferred until the response to the preparatory procedures can be ascertained. In gen-eral, mouth preparation includes procedures that address conditions that put comfortable prosthetic function at risk and include tooth alteration that are required to for proper tooth stabilization and support of the prosthesis. The objectives of the procedures involved are to cre-ate optimum health and eliminate or alter any condition that would be detrimental to the functional success of the removable partial denture.
Naturally, mouth preparation must be accomplished before the impression procedures are performed that will produce the master cast on which the removable partial denture will be fabricated. Oral surgical and periodontal procedures should precede abutment tooth preparation and should be completed far enough in advance to allow the nec-essary healing period. If at all possible, at least 6 weeks (and preferably 3 to 6 months) should be provided between surgi-cal and restorative dentistry procedures. This depends on the extent of the surgery and its impact on the overall support, stability, and retention of the proposed prosthesis.
www.konkur.in 189 Chapter 14 Preparation of the Mouth for Removable Partial Dentures PRE-PROSTHETIC CONSIDERATIONS IN PARTIALLY EDENTULOUS MOUTHS As a rule, all pre-prosthetic surgical treatment for the remov-able partial denture patient should be completed as early as possible. When possible, necessary endodontic surgery, periodontal surgery, and oral surgery should be planned so that they can be completed during the same time frame. The longer the interval between the surgery and the impression procedure, the more complete the healing and consequently the more stable the denture-bearing areas.
A variety of oral surgical techniques can prove beneficial to the clinician in preparing the patient for prosthetic replace-ments. However, it is not the purpose of this section to present the details of surgical correction. Rather, attention is called to some of the more common oral conditions or changes in which surgical intervention is indicated as an aid to remov-able partial denture design and fabrication and as an aid to the successful function of the restoration. Additional information regarding the techniques used is available in oral surgery texts and journal publications. It is important to emphasize, how-ever, that the dentist who is providing the removable partial denture treatment bears the responsibility for ensuring that the necessary surgical procedures are accomplished in accor-dance with the treatment plan. Measures to control apprehen-sion, including the use of intravenous and inhalation agents, have made the most extensive surgery acceptable to patients. Whether the dentist chooses to perform these procedures or elects to refer the patient to someone more qualified is imma-terial. The important consideration is that the patient should not be deprived of any treatment that would enhance the suc-cess of the removable partial denture.
Extractions Planned surgical procedures should occur early in the treat-ment regimen but not before a careful and thorough evaluation of each remaining tooth in the dental arch is completed (Figure 14-1). Regardless of its condition, each tooth must be evaluated in terms of its strategic importance and its potential contribu-tion to the success of the removable partial denture. Extraction of nonstrategic teeth that would present complications or those that may be detrimental to the design of the removable partial denture is a necessary part of the overall treatment plan.
Impacted Teeth All impacted teeth, including those in edentulous areas, as well as those adjacent to abutment teeth, should be consid-ered for removal. The periodontal implications of impacted teeth adjacent to abutments are similar to those for retained roots. These teeth are often neglected until serious periodon-tal implications arise.
Malposed Teeth The loss of individual teeth or groups of teeth may lead to extrusion, drifting, or combinations of malpositioning of remaining teeth (Figure 14-2). In most instances, the alveolar bone supporting extruded teeth will be carried occlusally as the teeth continue to erupt. Orthodontics may be use-ful in correcting many occlusal discrepancies, but for some patients, such treatment may not be practical because of lack of teeth for anchorage of the orthodontic appliances, cost, time to accomplish the treatment, or for other reasons.
Cysts and Odontogenic Tumors Panoramic radiographs of the jaws are recommended to sur-vey the jaws for unsuspected pathologic conditions. When A B Figure 14-1 Diagnostic mounting allows confirmation of the need for extraction after clinical examination. A, Anterior tooth position and chronic periodontal disease status require extrac-tion to address the patient’s concern of malpositioned and painful teeth. B, Root tips require immediate extraction to allow ridge healing to begin. The status of the molar (#15) requires ad-ditional workup to determine pulpal involvement of the carious lesion and the extent of occlusal reduction required to optimize the occlusal plane. The decision to maintain this tooth, although potentially costly, must consider the stabilizing effect it will have on the posterior left functional occlusion.
www.konkur.in 190 Part II Clinical and Laboratory a suspicious area appears on the survey film, a periapical radiograph should be taken to confirm or deny the presence of a lesion. All radiolucencies or radiopacities observed in the jaws should be investigated. Although the diagnosis may appear obvious from clinical and roentgenographic exami-nations, the dentist should confirm the diagnosis through appropriate consultation and, if necessary, perform a biopsy of the area and submit the specimens to a pathologist for microscopic study. The patient should be informed of the diagnosis and provided with various options for resolution of the abnormality as confirmed by the pathologist’s report.
Exostoses and Tori The existence of abnormal bony enlargements should not be allowed to compromise the design of the removable partial denture (Figure 14-3). Although modification of denture design can, at times, accommodate for exostoses, more fre-quently this results in additional stress to the supporting ele-ments and compromised function. The removal of exostoses and tori is not a complex procedure, and the advantages to be realized from such removal are great in contrast to the delete-rious effects that their continued presence can create. Ordinar-ily, the mucosa covering bony protuberances is extremely thin and friable. Removable partial denture components in prox-imity to this type of tissue may cause irritation and chronic ulceration. Also, exostoses approximating gingival margins may complicate the maintenance of periodontal health and lead to the eventual loss of strategic abutment teeth.
Hyperplastic Tissue Hyperplastic tissues are seen in the form of fibrous tuber-osities, soft flabby ridges, folds of redundant tissue in the vestibule or floor of the mouth, and palatal papillomatosis (Figure 14-4). All these forms of excess tissue should A B Figure 14-2 A, Malpositioned maxillary dentition due to loss of posterior occlusion and excessive wear of opposing mandibular anterior teeth. B, Restored dentition made possible by a combina-tion of endodontics, periodontics, and fixed and removable par-tial prosthodontics. (Courtesy of Dr. M. Alfaro, Columbus, OH.) A B Figure 14-3 Tori and exostoses.
Figure 14-4 Hyperplasia fibrous tuberosities.
www.konkur.in 191 Chapter 14 Preparation of the Mouth for Removable Partial Dentures be removed to provide a firm base for the denture. This removal will produce a more stable denture, will reduce stress and strain on the supporting teeth and tissues, and will provide a more favorable orientation of the occlusal plane and arch form for the arrangement of the artificial teeth in many instances. Appropriate surgical approaches should not reduce vestibular depth. Hyperplastic tissue can be removed with any preferred combination of scalpel, curette, electrosurgery, or laser. Some form of surgical stent should always be considered for these patients so that the period of healing is more comfortable. An old removable partial denture that is properly modified can serve as a sur-gical stent.
Muscle Attachments and Frena As a result of the loss of bone height, muscle attachments may insert on or near the residual ridge crest. The mylohy-oid, buccinator, mentalis, and genioglossus muscles are most likely to introduce problems of this nature. In addition to the problem of the attachments of the muscles themselves, the mentalis and genioglossus muscles occasionally pro-duce bony protuberances at their attachments that may also interfere with removable partial denture design. Appropriate ridge extension procedures can reposition attachments and remove bony spines, which will enhance the comfort and function of the removable partial denture.
The maxillary labial and mandibular lingual frena are the most common sources of frenum interference with denture design. These can be modified easily through any of several surgical procedures. Under no circumstances should a fre-num be allowed to compromise the design or comfort of a removable partial denture.
Bony Spines and Knife-Edge Ridges Sharp bony spicules should be removed and knifelike crests gently rounded. These procedures should be carried out with minimum bone loss. However, use of a dental implant can effectively enhance functional comfort in such an instance.
Polyps, Papillomas, and Traumatic Hemangiomas All abnormal soft tissue lesions should be excised and sub-mitted for pathologic examination before a removable partial denture is fabricated. Even though the patient may relate a history of the condition having been present for an indefinite period, its removal is indicated. New or additional stimula-tion to the area introduced by the prosthesis may produce discomfort or undesirable changes in the tumor.
Hyperkeratoses, Erythroplasia, and Ulcerations All abnormal white, red, or ulcerative lesions should be investigated, regardless of their relationship to the proposed denture base or framework. A biopsy of areas larger than 5 mm should be completed; and if the lesions are large (over 2 cm in diameter), multiple biopsies should be taken. The biopsy report determines whether the margins of the tissue to be excised can be wide or narrow. The lesions should be removed and healing accomplished before the removable partial denture is fabricated. On occasion, such as after irra-diation treatment or the excoriation of erosive lichen planus, the removable partial denture design will have to be radically modified to avoid areas of possible sensitivity.
Dentofacial Deformity Patients with a dentofacial deformity often have multi-ple missing teeth as part of their problem. Correction of the jaw deformity can simplify the dental rehabilitation. Before specific problems with the dentition can be cor-rected, the patient’s overall problem must be evaluated thoroughly. Several dental professionals (prosthodontist, oral surgeon, periodontist, orthodontist, and/or general dentist) may play a role in the patient’s treatment. These individuals must be involved in producing the diagnostic database and in planning treatment for the patient. Infor-mation obtained from a general patient evaluation done to determine the patient’s health status, a clinical evaluation directed toward facial esthetics and the status of the teeth and oral soft tissues, and analysis of appropriate diagnos-tic records can be used to produce a database. From this database, the patient’s problems can be enumerated, with the most severe problem being placed at the top of the list. Other identified problems would follow in order of their severity. It is only after this step that input from several dentists can provide a correctly sequenced final treatment plan for the patient.
Surgical correction of a jaw deformity can be made in horizontal, sagittal, or frontal planes. The mandible and max-illae may be positioned anteriorly or posteriorly, and their relationship to the facial planes may be surgically altered to achieve improved appearance. Replacement of missing teeth and development of a harmonious occlusion are almost always major problems in treating these patients.
Dental Implants A number of implant devices to support the replacement of teeth have been introduced to the dental profession. These devices offer a significant stabilizing effect on dental pros-theses through a rigid connection to living bone. The sys-tem that pioneered clinical prosthodontic applications with the use of commercially pure (CP) titanium endosseous implants is that of Brånemark and colleagues (Figure 14-5). This titanium implant was designed to provide a direct titanium-to-bone interface (osseointegrated), with basic laboratory and clinical results supporting the value of this procedure.
Implants are carefully placed using controlled surgi-cal procedures and, in general, bone healing to the device is allowed to occur before a dental prosthesis is fabricated. Long-term clinical research has demonstrated good results for the treatment of complete and partially edentulous dental patients using dental implants. Although research on implant applications with removable partial dentures has been very limited, the inclusion of strategically placed implants can www.konkur.in 192 Part II Clinical and Laboratory significantly control prosthesis movement (see Chapter 12; Figures 14-6 through 14-8).
Augmentation of Alveolar Bone Considerable attention has been devoted to ridge augmen-tation with the use of autogenous and alloplastic materials, especially in preparation for implant placement. Larger ridge volume gains necessitate consideration of autogenous grafts; however, these procedures are accompanied by con-cerns for surgical morbidity. Alloplastic materials have displayed short-term success; however, no randomized con-trolled trials have been conducted to provide evidence of B 1 2 3 4 5 6 7 A C Figure 14-5 A, Brånemark system components. From lower to upper: implant, cover screws, abutment, abutment screw, gold cylinder, and gold screw. B, Basic procedures in second-stage surgery: (1) exploration to locate cover screw; (2) removal of soft tissue; (3) removal of bony tissue; (4) removal of cover screw; (5) use of depth gauge to measure the amount of soft tissue; (6) abutment connection; and (7) placement of healing cap. C, Diagram of freestanding three-unit fixed partial denture supported by two osseointegrated implants that restore the extension base area, which would have been restored with a Class II removable partial denture if implants had not been used. (A and C, Redrawn from Hobo S, Ichida E, Garcia LT: Osseointegration and occlusal rehabilitation, Tokyo, 1989, Quintessence Publishing Company.) www.konkur.in 193 Chapter 14 Preparation of the Mouth for Removable Partial Dentures long-term increases in ridge width and height for removable prostheses.
Clinical results depend on careful evaluation of the need for augmentation, the projected volume of required mate-rial, and the site and method of placement. Considerable emphasis must be placed on a sound clinical understanding that some of the alloplastic materials can migrate or be dis-placed under occlusal loads if not appropriately supported by underlying bone and contained by buttressing soft tissues. Careful clinical judgment with sound surgical and prosthetic principles must be exercised.
PERIODONTAL PREPARATION Periodontal preparation of the mouth usually follows any oral surgical procedure and is performed simultaneously with tissue conditioning procedures. Ordinarily, tooth extraction and removal of impacted teeth and retained Edited by Vanchit John, DDS, Associate Professor and Chair, Depart-ment of Periodontics and Allied Dental Programs, Indiana University School of Dentistry, Indianapolis, Indiana.
A B C Figure 14-6 A, Implant bar and natural tooth copings used to support and retain this maxillary prosthesis. B, Tissue side of prosthesis showing the implant bar space, which when fitted will derive both support and stability from the implants while reten-tion is gained through resilient O-rings on the natural tooth cop-ings. C, Maxillary prosthesis seated and in occlusion. (Courtesy of Dr. N. Van Roekel, Monterey, CA.) C A B Figure 14-7 A, An anterior implant-supported bar demon-strating excellent access for hygiene and a parallel relationship to opposing occlusion. B, Prosthesis with implant bar space ( housing three retentive male components for retention and a flat surface for bar contact and support) and bilateral poste-rior embrasure clasps. C, Prosthesis seated and in occlusion. ( Courtesy of Dr. N. Van Roekel, Monterey, CA.) www.konkur.in 194 Part II Clinical and Laboratory roots or fragments are accomplished before definitive peri-odontal therapy is provided. However, it is strongly rec-ommended that a gross débridement be performed before tooth extraction when patients present with significant calculus accumulation. This helps limit the possibility of accidentally dislodging a piece of calculus into the extrac-tion socket, which could lead to an infection. Elimination of exostoses, tori, hyperplastic tissue, muscle attachments, and frena, on the other hand, can be incorporated with periodontal surgical techniques. In any situation, peri-odontal therapy should be completed before restorative dentistry procedures are begun for any dental patient. This is particularly true when a removable partial denture is contemplated, because the ultimate success of this restora-tion depends directly on the health and integrity of the sup-porting structures of the remaining teeth. The periodontal health of the remaining teeth, especially those to be used as abutments, must be evaluated carefully by the dentist and corrective measures instituted before a removable partial denture is fabricated. It has been demonstrated that fol-lowing periodontal therapy and with a good recall and oral hygiene program, properly designed removable partial den-tures will not adversely affect the progression of periodon-tal disease or carious lesions.
This discussion attempts to demonstrate how periodon-tal procedures affect diagnosis and treatment planning in a removable partial denture service rather than how the pro-cedures are actually accomplished. For technical details, the reader is referred to any of several excellent textbooks on periodontics.
Objectives of Periodontal Therapy The objective of periodontal therapy is the return to health of supporting structures of the teeth, creating an environment C A B Figure 14-8 A, A Class II, modification 1, maxillary arch with a posterior implant at the distal location of the extension base. B, Maxil-lary gold framework with broad palatal coverage, maximum stabilization through palatal contacts of multiple maxillary teeth, and implant position at the distal extension base. A single implant should be protected from excessive occlusal forces; consequently the broad palatal coverage and maximum bracing are important features of the overall design. The ball attachment abutment was used for retentive pur-poses. C, Occlusal view of the prosthesis with implant (see A), which provides improved retention to the distal extension base. (Courtesy of Dr. James Taylor, Ottawa, Ontario.) www.konkur.in 195 Chapter 14 Preparation of the Mouth for Removable Partial Dentures in which the periodontium may be maintained. The specific criteria for satisfying this objective are as follows: 1. Removal and control of all etiologic factors contributing to periodontal disease along with reduction or elimina-tion of bleeding on probing 2. Elimination of, or reduction in, the pocket depth of all pockets with the establishment of healthy gingival sulci whenever possible 3. Establishment of functional atraumatic occlusal relation-ships and tooth stability 4. Development of a personalized plaque control program and a definitive maintenance schedule Complete periodontal charting that includes the record-ing of pocket depths, assessment of attachment levels, and recording of furcation involvements, mucogingival prob-lems, and tooth mobility should be performed. Determining the severity of periodontal disease should also include the use of appropriate radiographs. The dentist who is consid-ering removable partial denture fabrication must be certain that these criteria have been satisfied before continuing with impression procedures for the master cast.
Periodontal Diagnosis and Treatment Planning Diagnosis The diagnosis of periodontal diseases is based on a system-atic and carefully accomplished examination of the peri-odontium. It follows the procurement of the health history of the patient and is performed with direct vision, palpation, a periodontal probe, a mouth mirror, and other auxiliary aids, such as curved explorers, furcation probes, diagnostic casts, and appropriate radiographs.
In the examination procedure, nothing is as important as careful exploration of the gingival sulcus and recording of the probing pocket depth and sites that bleed on probing with a suitably designed periodontal probe. Under no circumstances should removable partial denture fabrication begin without an accurate appraisal of sulcus/pocket depth and health. The probe is positioned as close to parallel to the long axis of the tooth as possible and is inserted gently between the gingival margin and the tooth surface, and the depth of the sulcus/ pocket is determined circumferentially around each tooth. At least six probing depth readings are recorded on the patient’s chart for each tooth. Usually depths are recorded for the distobuccal, buccal, mesiobuccal, distolingual, lingual, and mesiolingual aspects of each tooth. Sulcular health can also be assessed by the presence or absence of bleeding upon probing.
Dental radiographs can be used to supplement the clini-cal examination but should not be used as a substitute for it. A critical evaluation of the following factors should be made: (1) type, location, and severity of bone loss; (2) loca-tion, severity, and distribution of furcation involvements; (3) alterations of the periodontal ligament space; (4) alterations of the lamina dura; (5) the presence of calcified deposits; (6) the location and conformity of restorative margins; (7) eval-uation of crown and root morphologies; (8) root proximity; (9) caries; and (10) evaluation of other associated anatomic features, such as the mandibular canal or sinus proxim-ity. This information serves to substantiate the impression gained from the clinical examination.
Each tooth should be evaluated carefully for mobility. Unfortunately, there is no universally accepted standard for mobility. In general, mobility is graded according to the ease and extent of tooth movement. Normal mobility is in the order of 0.05 to 0.10 mm. Grade I mobility is present when less than 1 mm of movement occurs in a bucco-lingual direc-tion; grade II is present when mobility in the bucco-lingual direction is between 1 and 2 mm; grade III is present when greater than 2 mm of mobility occurs in the bucco-lingual direction and/or the tooth is vertically depressible.
Tooth mobility is an indication of the condition of the sup-porting structures, namely, the periodontium, and usually is caused by inflammatory changes in the periodontal ligament, traumatic occlusion, loss of attachment, or a combination of the three factors. The degree of mobility present, coupled with a determination of the causative factors responsible, provides additional information that is invaluable in plan-ning for the removable partial denture. If the causative factor can be removed, many grade I and grade II mobile teeth can become stable and may be used successfully to help support, stabilize, and retain the removable partial denture. Mobility in itself is not an indication for extraction unless the mobile tooth cannot aid in support or stability of the removable par-tial denture, or mobility cannot be reduced. (Grade III usually cannot be reversed and will not provide support or stability.) Treatment Planning Depending on the extent and severity of the periodontal changes present, a variety of therapeutic procedures ranging from simple to relatively complex may be indicated. As was the situation with the previously discussed oral surgical pro-cedures, it is the responsibility of the dentist rendering the removable partial denture treatment to see that the required periodontal care is accomplished for the patient. Periodontal treatment planning can usually be divided into three phases. The first phase is considered disease control or initial therapy because the objective is to essentially eliminate or reduce local causative factors before any periodontal surgical procedures are accomplished. The procedures that are accomplished as part of the initial preparation phase include oral hygiene instruction, scaling, and root planing and polishing, as well as endodontics, occlusal adjustment, and temporary splinting, if indicated. In many instances, carefully performed scaling and root planing combined with excellent patient compliance may negate the need for periodontal surgery.
During the second, or periodontal, surgical phase, any needed periodontal surgery, such as free gingival grafts, osseous grafts, or pocket reduction, is accomplished. It is advisable to discuss the possible need for these treat-ment procedures with the patient at the initial examination appointment or during the initial phase of therapy, because www.konkur.in 196 Part II Clinical and Laboratory this will likely involve referral of the patient to a periodon-tist. The maintenance of periodontal health is accomplished in the third phase and is always ongoing. A definitive recall schedule should be established with the patient and is usually kept at 3- to 4-month intervals.
Initial Disease Control Therapy (Phase 1) Oral Hygiene Instruction Ordinarily, dental treatment should be introduced to the patient through instruction provided in a carefully devised oral hygiene regimen. The cooperation wit-nessed by the patient’s acceptance and compliance with the prescribed procedure, as evidenced by improved oral hygiene, provides the dentist with a valuable means of evaluating that patient’s interest and the long-term prog-nosis of treatment.
For the oral hygiene routine to be successful, the patient must be convinced to follow the prescribed procedure reg-ularly and conscientiously. The most effective motivation techniques require a good understanding by the patient of his/her periodontal condition. Only then can the benefits of routine treatment become evident. Hence, an explana-tion of dental/periodontal disease, including its causes, initiation, and progression, is an important component of oral hygiene instruction. After this discussion, the patient should be instructed on the use of disclosing wafers/tab-lets, a soft/medium-bristle toothbrush, and unwaxed/ waxed dental floss. At subsequent appointments, oral hygiene can be evaluated carefully, and other oral hygiene aids (such as, an interdental and or sulcular brushes) can be incorporated as needed. Further treatment should be withheld until a satisfactory level of plaque control has been achieved. This is a particularly critical point for the patient who requires extensive restorative dentistry or a removable partial denture. Without good oral hygiene, any dental procedure, regardless of how well it is performed, is ultimately doomed to failure. The informed dentist insists that acceptable oral hygiene is demonstrated and main-tained before embarking on an extensive restorative den-tistry treatment plan.
Scaling and Root Planing One of the most important services rendered to the patient is the removal of calculus and plaque deposits from the coronal and root surfaces of the teeth. Careful scaling and root plan-ing are fundamental to the reestablishment of periodontal health. Without meticulous removal of calculus, plaque, and toxic material in the cementum, other forms of periodontal therapy cannot be successful.
The use of ultrasonic instrumentation for calculus removal followed by root planing with sharp periodontal curettes is recommended. The curette is designed specifically for root planing and, when used correctly in combination with ultrasonic instrumentation, results in calculus removal and root surface decontamination. Thorough scaling and root planing should precede definitive surgical periodontal procedures that may be indicated before removable partial denture fabrication.
Elimination of Local Irritating Factors Other Than Calculus Overhanging restoration margins and open contacts that allow food impaction should be corrected before defini-tive prosthetic treatment is begun. Although periodon-tal health predisposes to a much better environment for restorative procedures, it is not always possible or prudent to delay all restorative procedures until complete peri-odontal therapy and healing have occurred. This is espe-cially true for patients with severe carious lesions in which pulpal involvement is likely. Excavation of these areas and placement of adequate restorations must be incor-porated early in treatment. The placement of temporary or treatment fillings must not, in itself, become a local causative factor.
Elimination of Gross Occlusal Interferences Bacterial plaque accumulations and calculus deposits are the primary factors involved in the initiation and progression of inflammatory periodontal disease. However, poor restorative dentistry can contribute to damage of the periodontium, and poor occlusal relationships may act as another factor that contributes to more rapid loss of periodontal attachment. Although occlusal interferences may be eliminated through a variety of techniques, at this stage of treatment, selective grinding is the procedure generally applied. Particular atten-tion is directed to the occlusal relationships of mobile teeth. Traumatic cuspal interferences are removed by a selective grinding procedure. An attempt is made to establish a posi-tive planned intercuspal position that coincides with centric relation. Deflective contacts in the centric path of closure are removed, eliminating mandibular displacement from the closing pattern. After this, the relationship of the teeth in various excursive movements of the mandible is observed, with special attention to cuspal contact, wear, mobility, and roentgenographic changes in the periodontium. The pres-ence of working and nonworking interferences should be evaluated; if present, they should be removed.
The mere presence of occlusal abnormalities, in the absence of demonstrable pathologic change associated with the occlusion, does not necessarily constitute an indication for selective grinding. The indication for occlusal adjustment is based on the presence of a pathologic condition rather than on a preconceived articulation pattern. In the natural dentition, attempts to create bilateral balance, in the pros-thetic sense, have no place in the occlusal adjustment pro-cedure. Bilateral balanced occlusion not only is difficult to obtain in a natural dentition but also is apparently unneces-sary in view of its absence in most normal healthy mouths. Occlusion on natural teeth needs to be perfected only to the point at which cuspal interference within the patient’s func-tional range of contact is eliminated and normal physiologic function can occur.
www.konkur.in 197 Chapter 14 Preparation of the Mouth for Removable Partial Dentures Guide to Occlusal Adjustment Schuyler has provided the following guide to occlusal adjustment by selective grinding: In the study or evaluation of occlusal disharmony of the natural dentition, accurately mounted diagnostic casts are extremely helpful, if not essential, in determining static cusp-to-fossa contacts of opposing teeth and as a guide in the correction of occlusal anomalies in both centric and eccen-tric functional relations. Occlusion can be coordinated only by selective spot grinding. Ground tooth surfaces should be subsequently smoothed and polished. 1. A static coordinated occlusal contact of the maximum number of teeth (maximum intercuspal position) when the mandible is in centric relation to the maxilla should be the first objective.
a. A prematurely contacting cusp should be reduced only if the cusp point is in premature contact in both centric and eccentric relations. If a cusp point is in premature contact in centric relation only, the opposing sulcus should be deepened.
b. When anterior teeth are in premature contact in cen-tric relation, or in both centric and eccentric relations, corrections should be made by grinding the incisal edges of the mandibular teeth. If premature contact occurs only in the eccentric relation, correction must be made by grinding the lingual inclines of the maxil-lary teeth.
c. Usually, premature contacts in centric relation are relieved by grinding the buccal cusps of mandibular teeth, the lingual cusps of maxillary teeth, and the inci-sal edges of mandibular anterior teeth. Deepening the fossa of a posterior tooth or the lingual contact area in the centric relation of a maxillary anterior tooth changes and increases the steepness of the eccentric guiding inclines of the tooth; although this relieves trauma in centric relation, it may predispose the tooth to trauma in eccentric relations. 2. After establishing a static, even distribution of stress over the maximum number of teeth in centric relation, we are ready to evaluate opposing tooth contact or lack of contact in eccentric functional relations. Our attention is directed first to balancing side contacts. In extreme cases of patho-logic balancing contacts, relief may be needed even before corrective procedures in centric relation are performed.
Where balancing contacts exist, it is extremely difficult to differentiate the harmless from the destructive because we cannot visualize the influence of these fulcrum con-tacts on the functional movements of the condyle in the articular fossa. Subluxation, pain, lack of normal func-tional movement of the joint, or loss of alveolar support of the teeth involved may be evidence of excessive balancing contacts. Balancing side contacts receive less frictional wear than working side contacts, and premature contacts may develop progressively with wear. A reduction in the Courtesy of Dr. C. H. Schuyler, Montclair, NJ.
steepness of guiding tooth inclines on the working side will increase the proximity of the teeth on the balancing side and may contribute to destructive prematurities. In all corrective grinding to relieve premature or excessive contacts in eccentric relations, care must be exercised to avoid the loss of a static supporting contact in centric rela-tion. This static support in centric relation may exist with the mandibular buccal cusp fitting into the central fossa of the maxillary tooth or with the maxillary lingual cusp fitting into the central fossa of the mandibular tooth, or it may exist in both situations. Although both the max-illary lingual cusp and the mandibular buccal cusp may sometimes have a static centric contact in the sulcus of the opposing tooth, often only one of these cusps has this static contact. In such instances, the contacting cusp must be left untouched to maintain this essential support in the planned intercuspal position, and all corrective grinding to relieve premature contacts in eccentric positions would be done on the opposing tooth inclines. The mandibular buccal cusp is in a static central contact in the maxillary sulcus more often than the maxillary lingual cusp is in a static contact in its opposing mandibular sulcus. There-fore, corrective grinding to relieve premature balanc-ing contacts is more often done on the maxillary lingual cusps. 3. To obtain maximum function and distribution of func-tional stress in eccentric positions on the working side, necessary grinding must be done on the lingual surfaces of the maxillary anterior teeth. Corrective grinding on the posterior teeth at this time should always be done on the buccal cusp of the maxillary premolars and mo-lars and on the lingual cusp of the mandibular premolars and molars. The grinding of mandibular buccal cusps or maxillary lingual cusps at this time would rob these cusps of their static contact in the opposing central sulci in centric relation. 4. Corrective grinding to relieve premature protrusive con-tacts of one or more anterior teeth should be accom-plished by grinding the lingual surface of the maxillary anterior teeth. Anterior teeth should never be ground to bring the posterior teeth into contact in either protrusive position or on the balancing side. In the elimination of premature protrusive contacts of posterior teeth, neither the maxillary lingual cusps nor the mandibular buccal cusps should be ground. Corrective grinding should be done on the surfaces of the opposing teeth on which these cusps function in the eccentric position, leaving the cen-tric contact undisturbed. 5. Any sharp edges left by grinding should be rounded off.
Temporary Splinting Teeth that are mobile at the time of the initial examination frequently present a diagnostic problem for the dentist. The cause of the mobility must be determined and then a decision made for elimination of the causative factors. The response of these teeth to temporary immobilization followed by www.konkur.in 198 Part II Clinical and Laboratory appropriate treatment may be helpful in establishing a prognosis for them and may lead to a rational decision as to whether they should be retained or sacrificed. Second-ary mobility resulting from the presence of an inflamma-tory lesion may be reversible if the disease process has not destroyed too much of the attachment apparatus. Primary mobility caused by occlusal interference also may disappear after selective grinding. In instances of angular types of osse-ous defects, one should consider guided tissue regeneration (GTR) as a means of increasing attachment levels. In some situations, however, the teeth must be stabilized because of loss of supporting structure from the periodontal process.
Teeth may be immobilized during periodontal treatment by acid etching teeth with composite resin, with fiber-rein-forced resins, with cast removable splints, or with intra-coronal attachments. The latter, an example of which is the A-splint, necessitates cutting tooth surfaces and embedding a ridge connector between adjacent teeth.
After periodontal treatment is performed, splinting may be accomplished with cast removable restorations or cast cemented restorations. The preferred form of perma-nent splinting uses two or more cast restorations soldered or cast together. They may be cemented with permanent (zinc oxyphosphate or resin) cements or temporary (zinc oxide–eugenol) cements. A properly designed removable partial denture can also stabilize mobile teeth if provi-sion for such immobilization is planned as the denture is designed.
Use of a Nightguard The removable acrylic-resin splint, originally designed as an aid in eliminating the deleterious effects of nocturnal clenching and grinding, has been used to advantage for the removable partial denture patient. The nightguard may prove helpful as a form of temporary splinting if worn at night when the removable partial denture has been removed. The flat occlusal surface prevents intercuspation of the teeth, which eliminates lateral occlusal forces (Figure 14-9).
The nightguard is particularly useful before fabrication of a removable partial denture when one of the abutment teeth has been unopposed for an extended period. The peri-odontal ligament of a tooth without an antagonist undergoes changes characterized by loss of orientation of periodontal ligament fibers, loss of supporting bone, and narrowing of the periodontal ligament space. If such a tooth is suddenly returned to full function when it is carrying an increased burden, pain and prolonged sensitivity may result. However, if a nightguard is used to return some functional stimulation to the tooth, the periodontal ligament changes are reversed and an uneventful course can be experienced when the tooth is returned to full function.
Minor Tooth Movement The increased use of orthodontic procedures in conjunction with restorative and prosthetic dentistry has contributed to the success of many restorations by altering the periodontal environment in which they are placed. Malposed teeth that were once doomed to extraction should be considered now for repositioning and retention. The additional stability pro-vided for a removable partial denture by uprighting a tilted or drifted tooth may mean much in terms of comfort to the patient. The techniques employed are not difficult to master, and the rewards in terms of a better restorative dentistry ser-vice are great.
Definitive Periodontal Surgery (Phase 2) Periodontal Surgery After initial therapy is completed, the patient is reevaluated for the surgical phase. If oral hygiene is at an optimum level, yet pockets with inflammation and osseous defects are still present, a variety of periodontal surgical techniques should be considered to improve periodontal health. The proce-dures selected should have the potential to enhance the results obtained during Phase 1 therapy.
Pocket reduction or elimination may be achieved by root planing when the cause of pocket depth is edema caused by gingival inflammation. Apically positioned flap surgery or occasionally a gingivectomy may be considered for reduc-tion of suprabony pockets. Osseous resection or regen-eration using a flap approach is a form of surgical therapy that is commonly employed to help with treatment of the diseased periodontium. It must be noted that elimination of the inflammatory disease process and restoration of the periodontal attachment apparatus are the major objectives of periodontal therapy.
Periodontal Flaps. Today, use of one of the various flap procedures is the surgical approach that offers the great-est versatility. Periodontal flap surgery involves the eleva-tion of either mucosa alone or both the mucosa and the periosteum. Although there are several indications for flap elevation, the most important goal of flap elevation is to allow access to the bone and the root surfaces for complete Figure 14-9 The removable acrylic-resin splint with a flat oc-clusal plane can be used effectively as a form of temporary sta-bilization and as a means of eliminating excessive lateral forces created by clenching and grinding habits.
www.konkur.in 199 Chapter 14 Preparation of the Mouth for Removable Partial Dentures instrumentation. Other goals of the flap approach include access for pocket elimination, caries control, crown length-ening to allow for optimum restorative dental treatment, root amputation or hemisection, as required and access to the furcation of the tooth.
A decision is made before surgery is performed if the aim is resection of osseous tissue to allow for a more physi-ologic osseous anatomy and subsequently gingival contour, or to regenerate some of the lost periodontal attachment apparatus. However, sometimes changes have to be made during surgery based on the anatomy of defects following the removal of diseased granulation tissue. Osseous resec-tion involves the use of both osteoplasty and ostectomy pro-cedures. Osteoplasty refers to reshaping the bone without removing tooth-supporting bone; ostectomy includes the removal of tooth-supporting bone. Consequently, the flap is widely applied in the treatment of periodontal disease.
Guided Tissue Regeneration. GTR has been defined as those procedures that attempt regeneration of lost periodon-tal structures through differing tissue responses. The ratio-nale for GTR is based on the physiologic healing response of the tissues after periodontal surgery. After periodontal sur-gery, a race to repopulate the root surface begins among the four tissue types of the periodontium, namely, epithelium, connective tissue, periodontal ligament, and bone. Epithe-lium, which migrates at a rate of 0.5 mm per day, typically migrates first along the root surface, preventing new attach-ment. Therefore, to allow the undifferentiated mesenchymal cells from the periodontal ligament and the endosteum of bone to repopulate the root against surfaces, the epithe-lial cells and the gingival connective tissue cells should be isolated. This isolation during initial healing enables peri-odontal structures to become reestablished and may lead to better long-term health of the tooth. The GTR procedure commonly involves the use of an osseous graft along with a resorbable membrane (Figure 14-10). This technique has the potential to lead to substantial improvement of the peri-odontal condition when used around carefully selected two- and three-walled osseous defects and mandibular furcation involvements.
Periodontal Plastic Surgery. Periodontal plastic surgery, which was previously referred to as mucogingival surgery, is applied to those procedures used to resolve problems involv-ing the interrelationship between the gingiva and the alveo-lar mucosa. Mucogingival surgery consists of plastic surgical procedures that are used for correction of gingiva–mucous membrane relationships that complicate periodontal disease and may interfere with the success of periodontal treatment. The objectives of periodontal plastic surgery are several and include elimination of pockets that transverse the muco-gingival junction, creation of an adequate zone of attached gingiva, correction of gingival recession by root coverage techniques, relief of the pull of frena and muscle attachments on the gingival margin, and correction of deformities of edentulous ridges, done to permit access to the underlying alveolar process and correction of osseous deformities when there is sufficient or insufficient attached gingiva, to deepen a shallow vestibule, and to assist in orthodontic therapy. Com-monly used periodontal plastic surgical procedures include lateral sliding flaps, free gingival grafts, pedicle grafts, coro-nally positioned grafts, double papilla flaps, semilunar coro-nally positioned flaps, subepithelial connective tissue grafts, and edentulous ridge augmentation using one of the above techniques. In addition, GTR has been used for periodon-tal plastic surgical procedures. Recently, use of the commer-cially-available acellular dermal graft has gained popularity. However, the most commonly used procedure is the subepi-thelial connective tissue graft (Figure 14-11).
These plastic surgical procedures should be consid-ered whenever an abutment tooth lacks adequate attached keratinized gingiva and requires root coverage to facilitate removable partial denture construction and maintenance.
Recall Maintenance (Phase 3) Several longitudinal studies have now demonstrated the increasing importance of maintenance for all patients who have undergone any periodontal therapy. This includes not only reinforcement of plaque control measures but also thor-ough débridement of all root surfaces of supragingival and subgingival calculus and plaque by the dentist or an auxiliary.
The frequency of recall appointments should be custom-ized for the patient, depending on the susceptibility and severity of periodontal disease. It is now understood that patients with a history of moderate to severe periodontitis should be placed on a 3- to 4-month recall system to main-tain results achieved by nonsurgical and surgical therapy.
Advantages of Periodontal Therapy Periodontal therapy done before a removable prosthesis is fabricated has several advantages. First, the elimination of periodontal disease removes a primary causative fac-tor in tooth loss. The long-term success of dental treatment depends on the maintenance of the remaining oral struc-tures, and periodontal health is mandatory if further loss is to be avoided. Second, a periodontium free of disease presents a much better environment for restorative correc-tion. Elimination of periodontal pockets with the associ-ated return of a physiologic architectural pattern establishes a normal gingival contour at a stable position on the tooth surface. Thus, the optimum position for gingival margins of individual restorations can be established with accuracy. The coronal contours of these restorations can also be developed in correct relationships to the gingival margin, ensuring the proper degree of protection and functional stimulation to gingival tissues. Third, the response of strategic but ques-tionable teeth to periodontal therapy provides an important opportunity for reevaluating their prognosis before the final decision is made to include (or exclude) them in the remov-able partial denture design. And last, the overall reaction of the patient to periodontal procedures provides the dentist with an excellent indication of the degree of cooperation to be expected in the future.
www.konkur.in 200 Part II Clinical and Laboratory Even in the absence of periodontal disease, certain periodon-tal procedures may be an invaluable aid in removable partial denture construction. Through periodontal surgical techniques, the environment of potential abutment teeth may be altered to the point of making an otherwise unacceptable tooth a most satisfactory retainer for a removable partial denture.
OPTIMIZATION OF THE FOUNDATION FOR FITTING AND FUNCTION OF THE PROSTHESIS Conditioning of Abused and Irritated Tissues Many removable partial denture patients require some con-ditioning of supporting tissues in edentulous areas before A C B D E Figure 14-10 Guided tissue regeneration (GTR) procedure performed to address a furcation involvement. A, Tooth #30 presented with a Class II furcation involvement with the probe entering 3 mm in a horizontal direction. A GTR procedure using a combination of a bone graft and a nonresorbable membrane was planned. B, Following hand and ultrasonic instrumentation, decalcified freeze-dried bone allograft was grafted around the furcation. C, A nonresorbable membrane was placed over the bone graft. D, The flap was then sutured with a nonresorbable expanded polytetraethylene suture. E, Two months following surgery, the membrane was removed. Note the presence of red rubbery tissue filling the previously exposed furcation site. This tissue has the potential to form osseous tissue and close the access to the furcation entrance.
www.konkur.in 201 Chapter 14 Preparation of the Mouth for Removable Partial Dentures the final impression phase of treatment begins. Patients who require conditioning treatment often demonstrate the fol-lowing symptoms: • Inflammation and irritation of the mucosa covering den-ture-bearing areas (Figure 14-12) • Distortion of normal anatomic structures, such as incisive papillae, rugae, and retromolar pads • A burning sensation in residual ridge areas, the tongue, and the cheeks and lips These conditions are usually associated with ill-fitting or poorly occluding removable partial dentures. However, nutritional deficiencies, endocrine imbalances, severe health problems (diabetes or blood dyscrasias), and bruxism must be considered in a differential diagnosis.
If the use of a new removable partial denture or the relin-ing of a present denture is attempted without first correcting these conditions, the chances for successful treatment will be compromised because the same old problems will be perpet-uated. The patient must be made to realize that fabrication of a new prosthesis should be delayed until the oral tissues can be returned to a healthy state. If there are unresolved systemic problems, removable partial denture treatment will usually result in failure or limited success.
The first treatment procedure should consist of immedi-ate institution of a good home care program. A suggested home care program includes rinsing the mouth three times a day with a prescribed saline solution; massaging the residual ridge areas, palate, and tongue with a soft toothbrush; remov-ing the prosthesis at night; and using a prescribed therapeu-tic multiple vitamin along with a prescribed high-protein, low-carbohydrate diet. Some inflammatory oral conditions caused by ill-fitting dentures can be resolved by removing the dentures for extended periods. However, few patients are willing to undergo such inconveniences.
A B Figure 14-12 A, Inflamed and distorted denture bearing mucosa due to an ill-fitting prosthesis that is worn 24 hours a day. B, After the tissue abuse is treated via modification of the denture base with a tissue conditioning resilient liner material, the prosthesis is removed for portions of the day, and the abused tissue is massaged, the denture bearing foundation is healthy again.
A B Figure 14-11 Gingival recession addressed with subepithelial connective tissue graft procedure. A, The patient presents with evi-dence of severe gingival recession associated with teeth #6, #7, and #8. This was an esthetic problem. The patient also complained of hypersensitivity associated with these teeth. A subepithelial connective tissue graft was planned to help correct the gingival recession. B, Clinical appearance 6 months following treatment with a subepithelial connective tissue graft on teeth #6, #7, and #8. The patient was very satisfied with the postoperative appearance, and clinically the symptom of hypersensitivity was no longer significant.
www.konkur.in 202 Part II Clinical and Laboratory Use of Tissue Conditioning Materials The tissue conditioning materials are elastopolymers that continue to flow for an extended period, permitting dis-torted tissues to rebound and assume their normal form. These soft materials apparently have a massaging effect on irritated mucosa, and because they are soft, occlusal forces are probably more evenly distributed.
Maximum benefit from using tissue conditioning materi-als may be obtained by (1) eliminating deflective or interfering occlusal contacts of old dentures (by remounting in an articu-lator if necessary); (2) extending denture bases to proper form to enhance support, retention, and stability (Figure 14-13); (3) relieving the tissue side of denture bases sufficiently (2 mm) to provide space for even thickness and distribution of con-ditioning material; (4) applying the material in amounts suf-ficient to provide support and a cushioning effect (Figure 14-14); and (5) following the manufacturer’s directions for manipulation and placement of the conditioning material.
The conditioning procedure should be repeated until the supporting tissues display an undistorted and healthy appear-ance. Many dentists find that intervals of 4 to 7 days between changes of the conditioning material are clinically accept-able. Improvement in irritated and distorted tissues is usually noted within a few visits, and in some patients a dramatic improvement will be seen. Usually three or four changes of the conditioning material are adequate, but in some instances additional changes are required. If positive results are not seen within 3 to 4 weeks, one should suspect more serious health problems and request a consultation from a physician.
Abutment Restorations Equipped with the diagnostic casts on which a tentative removable partial denture design has been drawn, the den-tist is able to accomplish preparation of abutment teeth with accuracy. The information at hand should include the pro-posed path of placement, the areas of teeth to be altered and tooth contours to be changed, and the locations of rest seats and guiding planes.
During examination and subsequent treatment planning, in conjunction with a survey of diagnostic casts, each abut-ment tooth is considered individually as to what type of resto-ration is indicated. Abutment teeth presenting sound enamel surfaces in a mouth in which good oral hygiene habits are evident may be considered a fair risk for use as removable partial denture abutments. One should not be misled, how-ever, by a patient’s promise to do better as far as oral hygiene habits are concerned. Good or bad oral hygiene is a habit of long standing and is not likely to be changed appreciably because a removable partial denture is being worn. There-fore, one must be conservative in evaluating the oral hygiene habits of the patient in the future. Remember that clasps as such do not cause teeth to decay; and if the individual will keep the teeth and the removable partial denture clean, one need not condemn clasps from a cariogenic standpoint. On the other hand, more removable partial dentures have been condemned as cariogenic because the dentist did not provide for the protection of abutment teeth rather than because of inadequate care on the part of the patient.
Esthetic veneer types of crowns should be used when a canine or premolar abutment is to be restored or protected. Less frequently, the molar will have to be treated in such a manner, and except for maxillary first molars, the full cast crown is usually acceptable.
A B Figure 14-13 A, Mandibular removable partial denture with underextended bases, which contributed to tissue irritation. B, Denture bases properly extended to enhance support, stability, and retention.
Figure 14-14 Tissue conditioning should be of sufficient thickness to be resilient and not place undue stress on the soft tissue.
www.konkur.in 203 Chapter 14 Preparation of the Mouth for Removable Partial Dentures When there is proximal caries on abutment teeth with sound buccal and lingual enamel surfaces, in a mouth exhib-iting average oral hygiene and low caries activity, a gold inlay may be indicated. However, silver amalgam or composite for the restoration of those teeth with proximal caries should not be condemned, although one must admit that an inlay cast of a hard type of gold will provide the best possible support for occlusal rests, at the same time giving an esthetically pleas-ing restoration. However, an amalgam restoration, properly condensed, is capable of supporting an occlusal rest without appreciable flow over a long period.
The most vulnerable area on the abutment tooth is the proximal gingival area, which lies beneath the minor con-nector of the removable partial denture framework and is therefore subject to accumulation of debris in an area most susceptible to caries. Even when the removable partial den-ture is removed, these areas are often missed by the tooth-brush, which allows bacterial plaque and debris to remain for long periods. Because of this unique removable partial denture concern, special attention should be paid to these areas during patient education and follow-up. Even when a complete crown restoration is placed in this most vulnerable area, recurrent caries can occur. Caries risk is best managed through effective home care and professional follow-up pro-cedures, rather than through the placement of restorations.
All proximal abutment surfaces that are to serve as guid-ing planes for the removable partial denture should be pre-pared so that they will be made as nearly parallel as possible to the path of placement. Preparations may include modify-ing the contour of existing ceramic restorations, if necessary. This may be accomplished with abrasive stones or diamond finishing stones. A polished surface for the altered ceramic restoration may be restored by using any of several polishing kits supplied by manufacturers.
When preparing abutments that will receive surveyed crowns, it is important to plan for the tooth reduction neces-sary to allow placement of sufficient restorative material for durability, contour, and esthetics, as well as the contours pre-scribed for the desired clasp assembly (Figure 14-15). This can be accomplished by first modifying the axial contours A B C D Figure 14-15 A, Diagnostic cast at an orientation best for all abutments considered. The buccal survey line is too close to the mar-ginal gingival and the distal surface does not lend itself to guide-plane preparation. A surveyed crown is indicated. B, Abutment contours appropriate to clasp design (distal guide plane and mid-buccal 0.01 inch undercut) are produced in wax. C, Cast of abutment preparation provides buccal surface reduction adequate to replace with metal ceramic material at the required contour. Without careful consideration of survey line placement needs before and during preparation, it is easy to reproduce incorrect contours in finished crowns. D, Cast of a seated surveyed crown demonstrates desired contours for the clasp design chosen.
www.konkur.in 204 Part II Clinical and Laboratory of the abutments to those required of the completed crown, then starting controlled tooth reduction (preparation) to accommodate the thickness of the materials for durability, contour, and esthetics. This ensures that the wax patterns and resultant crowns can be restored to the desired form.
Contouring Wax Patterns Modern indirect techniques permit the contouring of wax patterns on the master cast with the aid of the surveyor blade. All abutment teeth to be restored with castings can be prepared at one time and an impression made that will provide an accurate stone replica of the prepared arch. Wax patterns may then be refined on separated individual dies or removable dies. All abutment surfaces facing edentulous areas should be made parallel to the path of placement by the use of the surveyor blade (Figure 14-16). This technique will provide proximal surfaces that will be parallel without any further alteration in the mouth, will permit the most positive A B C D E Figure 14-16 A, Occlusal view of full contour wax patterns, which will be splinted between crowns and across the midline with a 13-gauge splint bar. Rests are evident on the lingual surfaces of abutment wax patterns. B, Wax patterns showing labial cut-back for porcelain. Bilateral guide-plane surfaces will be reproduced in metal and are parallel to the path of insertion. C, An abutment veneered crown with an appropriate height of contour and a 0.02-inch undercut for the anticipated wrought-wire retainer. D, Completed prosthesis splinted between retainer crowns and across the midline. Splint bar with added vertical support provides indirect retention. E, Prosthesis inserted intraorally.
www.konkur.in 205 Chapter 14 Preparation of the Mouth for Removable Partial Dentures seating of the removable partial denture along the path of placement, and will provide the least amount of undesirable space beneath minor connectors for the lodgment of debris.
Rest Seats After the proximal surfaces of the wax patterns have been made parallel, and buccal and lingual contours have been established to satisfy the requirements of stability and reten-tion with the best possible esthetic placement of clasp arms, the occlusal rest seats should be prepared in the wax pat-tern rather than in the finished restoration. The placement of occlusal rests should be considered at the time the teeth are prepared to receive cast restorations so that there will be sufficient clearance beneath the floor of the occlusal rest seat. Too many times, a completed cast restoration is cemented in the mouth for a removable partial denture abutment without any provision for the occlusal rest having been made in the wax pattern. The dentist then proceeds to prepare an occlusal rest seat in the cast restoration, while ever conscious of the fact that he or she may perforate the casting during the pro-cess of forming the rest seat. The unfortunate result is usually a poorly formed rest seat that is too shallow.
If tooth structure has been removed to provide placement of the occlusal rest seat, it may be ideally placed in the wax pattern by using a No. 8 round bur to lower the marginal ridge and establish the outline form of the rest and then using a No. 6 round bur to slightly deepen the floor of the rest seat inside this lowered marginal ridge. This approach provides an occlusal rest that best satisfies the requirements that it be placed so that any occlusal force will be directed axially and that there will be the least possible interference to occlusion with the opposing teeth.
Perhaps the most important function of a rest is the divi-sion of stress loads from the removable partial denture to provide the greatest efficiency with the least damaging effect to the supporting abutment teeth. For a distal extension removable partial denture, the rest must be able to transmit occlusal forces to the abutment teeth in a vertical direction only, thereby permitting the least possible lateral stresses to be transmitted to the abutment teeth.
For this reason, the floor of the rest seat should incline toward the center of the tooth so that the occlusal forces, inso-far as possible, are centered over the root apex. Any other form but that of a spoon shape can permit locking of the occlusal rest and the transmission of tipping forces to the abutment tooth. A ball-and-socket type of relationship between occlu-sal rest and abutment tooth is the most desirable. At the same time, the marginal ridge must be lowered so that the angle formed by the occlusal rest and the minor connector will stand above the occlusal surface of the abutment tooth as little as possible and avoid interference with the opposing teeth. Simultaneously, sufficient bulk must be provided to prevent weakness in the occlusal rest at the marginal ridge. The mar-ginal ridge must be lowered and yet not be the deepest part of the rest preparation. To permit occlusal stresses to be directed toward the center of the abutment tooth, the angle formed by the floor of the occlusal rest with the minor connector should be less than 90 degrees. In other words, the floor of the occlu-sal rest should incline slightly from the lowered marginal ridge toward the center of the tooth.
This proper form can be readily accomplished in the wax pattern, if care is taken during crown or inlay preparation to provide the location of the rest. If direct restorations are used, sufficient bulk must be present in this area to allow proper occlusal rest seat form without weakening the restoration. There is insufficient evidence to show that direct restorations used as rest seats perform equally to enamel. When the rest seat is placed in sound enamel, this is best accomplished by the use of round carbide burs (No. 4, 6, and 8 sizes) that leave a smooth enamel surface.
Rest seat preparations in sound enamel (or in existing res-torations that are not to be replaced) should always follow the recontouring of proximal tooth surfaces. The preparation of proximal tooth surfaces should be done first because if the occlusal portion of the rest seat is placed first and the proxi-mal tooth surface is altered later, the outline form of the rest seat is sometimes irreparably altered.
Following proximal surface recontouring (guided plane preparation), the larger round bur is used to lower the mar-ginal ridge 1.5 to 2.0 mm while at the same time creating the relative outline form of the rest seat. The result is a rest seat preparation with marginal ridge lowered and gross outline form established but without sufficient deepening of the rest seat preparation toward the center of the tooth. A smaller round bur (a No. 4 or 6) may then be used to deepen the floor of the rest seat to a gradual incline toward the center of the tooth. Enamel rods are then smoothed by the planing action of a round bur revolving with little pressure. Abrasive rubber points are sufficient to complete the polishing of the rest seat preparation.
The success or failure of a removable partial denture depends on how well the mouth preparations were accom-plished. It is only through intelligent planning and compe-tent execution of mouth preparations that the denture can satisfactorily restore lost dental functions and contribute to the health of the remaining oral tissues.
www.konkur.in CHAPTER 15 Preparation of Abutment Teeth CHAPTER OUTLINE Classification of Abutment Teeth Sequence of Abutment Preparations on Sound Enamel or Existing Restorations Abutment Preparations Using Conservative Restorations Abutment Preparations Using Crowns Ledges on Abutment Crowns Spark Erosion Veneer Crowns for Support of Clasp Arms Splinting of Abutment Teeth Use of Isolated Teeth as Abutments Missing Anterior Teeth Temporary Crowns when a Removable Partial Denture is Being Worn Cementation of Temporary Crowns Fabricating Restorations to Fit Existing Denture Retainers After surgery, periodontal treatment, endodontic treatment, and tissue conditioning of the arch involved, the abutment teeth may be prepared to provide support, stabilization, reciprocation, and retention for the removable partial den-ture. Rarely, if ever, is the situation encountered in which alterations of the abutment are not indicated because teeth do not develop with guiding planes, rests, and contours to accommodate clasp assemblies.
A favorable response to any deep restorations, endodon-tic therapy, and the results of periodontal treatment should be established before the removable partial denture is fabri-cated. If the prognosis of a tooth under treatment becomes unfavorable, its loss can be compensated for by a change in the removable partial denture design. If teeth are lost after the removable partial denture is fabricated, then the remov-able partial denture must be added to or replaced. Most removable partial denture designs do not lend themselves well to later additions, although this possibility should be considered in the original design of a denture. Every diag-nostic aid should be used to determine which teeth are to be used as abutments or are potential abutments for future designs. When an original abutment is lost, it is extremely difficult to effectively modify the removable partial denture to use the next adjacent tooth as a retaining unit.
It is sometimes possible to design a removable partial denture so that a single posterior abutment that is question-able can be retained and used to support one end of a tooth-supported base. Then, if that posterior abutment was lost, it could be replaced with a distal extension base (see Figure 13-25). Such a design must include provision for future indi-rect retention, flexible clasping on the remaining terminal abutment, and provision for establishing tissue support by a secondary impression. Anterior abutments, which are con-sidered poor risks, may not be so freely used because of the problems involved in adding a new abutment retainer when the original one is lost. Such questionable teeth should be treatment planned for extraction in favor of a better abut-ment in the original treatment plan.
www.konkur.in 207 Chapter 15 Preparation of Abutment Teeth CLASSIFICATION OF ABUTMENT TEETH All abutment teeth must have contour modifications that are customized to the planned prosthesis design. The subject of abutment preparations may be grouped as follows: (1) those abutment teeth that require only minor modifications to their coronal portions; (2) those that are to have restorations other than complete coverage crowns; and (3) those that are to have crowns (complete coverage).
Abutment teeth that require only minor modifications include teeth with sound enamel, those with small restora-tions not involved in the removable partial denture design, those with acceptable restorations that will be involved in the removable partial denture design, and those that have exist-ing crown restorations requiring minor modification that will not jeopardize the integrity of the crown. The latter may exist as an individual crown or as the abutment of a fixed partial denture.
The use of unprotected abutments has been discussed previously. Although complete coverage of all abutments may be desirable, it is not always possible or practical. The decision to use unprotected abutments involves certain risks of which the patient must be advised and includes responsi-bility for maintaining oral hygiene and caries control. Mak-ing crown restorations fit existing denture clasps is a difficult task; however, the fact that it is possible to do may influence the decision to use uncrowned but otherwise sound teeth as abutments.
Complete coverage restorations provide the best possible support for occlusal rests. If the patient’s economic status or other factors beyond the control of the dentist prevent the use of complete coverage restorations, then an amalgam alloy restoration, if properly condensed, is capable of sup-porting an occlusal rest without appreciable flow for a long period. Any existing silver amalgam alloy restoration about which there is any doubt should be replaced with new amal-gam restorations. This should be done before guiding planes and occlusal rest seats are prepared, to allow the restoration to reach maximum strength and be polished.
Continued improvement in dimensional stability, strength, and wear resistance of composite resin restora-tions will add another dimension to the preparation and modification of abutment teeth for removable partial den-tures that should be less invasive than placement of com-plete coverage restorations and more economical.
SEQUENCE OF ABUTMENT PREPARATIONS ON SOUND ENAMEL OR EXISTING RESTORATIONS Abutment preparations on sound enamel or on existing res-torations that have been judged as acceptable should be done in the following order: 1. Proximal surfaces parallel to the path of placement should be prepared to provide guiding planes (Figure 15-1, A). 2. Tooth contours should be modified (see Figure 15-1, B and C), lowering the height of contour so that (a) the origin of circumferential clasp arms may be placed well below the occlusal surface, preferably at the junction of the middle and gingival thirds; (b) retentive clasp termi-nals may be placed in the gingival third of the crown for better esthetics and better mechanical advantage; and (c) reciprocal clasp arms may be placed on and above a height of contour that is no higher than the cervical portion of the middle third of the crown of the abutment tooth. 3. After alterations of axial contours are accomplished and before rest seat preparations are instituted, an impression of the arch should be made in irreversible hydrocolloid and a cast formed in a fast-setting stone. This cast can be returned to the surveyor to determine the adequacy of axial alterations before proceeding with rest seat prepara-tions. If axial surfaces require additional axial recontour-ing, this can be performed during the same appointment and without compromise. 4. Occlusal rest areas should be prepared that will direct oc-clusal forces along the long axis of the abutment tooth (see Figure 15-1, D). Mouth preparation should follow the re-movable partial denture design that was outlined on the diagnostic cast at the time the cast was surveyed and the treatment plan confirmed. Proposed changes to abutment teeth should be made on the diagnostic cast and outlined in colored pencil to indicate the area, amount, and angu-lation of the modification to be done (see Chapter 13). Al-though occlusal rest seats may also be prepared on the di-agnostic cast, indication of their location in colored pencil is usually sufficient for the experienced dentist because rest preparations follow a definite pattern (see Chapter 6).
ABUTMENT PREPARATIONS USING CONSERVATIVE RESTORATIONS Conventional inlay preparations are permissible on the prox-imal surface of a tooth not to be contacted by a minor con-nector of the removable partial denture. On the other hand, proximal and occlusal surfaces that support minor connec-tors and occlusal rests require somewhat different treatment. The extent of occlusal coverage (i.e., whether cusps are cov-ered) will be governed by the usual factors, such as the extent of caries, the presence of unsupported enamel walls, and the extent of occlusal abrasion and attrition.
When an inlay is the restoration of choice for an abutment tooth, certain modifications of the outline form are neces-sary. To prevent the buccal and lingual proximal margins from lying at or near the minor connector or the occlusal rest, these margins must be extended well beyond the line angles of the tooth. This additional extension may be accom-plished by widening the conventional box preparation. How-ever, the margin of a cast restoration produced for such a preparation may be quite thin and may be damaged by the clasp during placement or removal of the removable partial denture. This hazard may be avoided by extending the out-line of the box beyond the line angle, thus producing a strong restoration-to-tooth junction.
www.konkur.in 208 Part II Clinical and Laboratory In this type of preparation, the pulp is particularly vul-nerable unless the axial wall is curved to conform to the external proximal curvature of the tooth. When caries is of minimal depth, the gingival seat should have an axial depth at all points about the width of a No. 559 fissure bur. It is of utmost importance that the gingival seat be placed where it can be easily accessed to maintain good oral hygiene. The proximal contour necessary to produce the proper guiding plane surface and the close proximity of the minor connec-tor render this area particularly vulnerable to future caries. Every effort should be made to provide the restoration with maximum resistance and retention, as well as with clinically imperceptible margins. The first requisite can be satisfied by preparing opposing cavity walls 5 degrees or less from paral-lel and producing flat floors and sharp, clean line angles.
It is sometimes necessary to use an inlay on a mandibular first premolar for the support of an indirect retainer. The nar-row occlusal width bucco-lingually and the lingual inclina-tion of the occlusal surface of such a tooth often complicate the two-surface inlay preparation. Even the most exacting occlusal cavity preparation often leaves a thin and weak lin-gual cusp remaining.
A B C D Figure 15-1 Abutment contours should be altered during mouth preparations in the following sequence. A, The proximal surface is prepared parallel to the path of placement to create a guiding plane. B, Height of contour on the buccal and lingual surfaces is lowered when necessary to permit the retentive clasp terminus to be located within the gingival third of the crown, bracing part of the retentive arm at the junction of the middle and gingival thirds of the crown, and the reciprocal clasp arm on the opposite side of tooth to be placed no higher than the cervical portion of the middle third of the crown. C, The area of the tooth at which the retentive clasp arm originates should be altered if necessary to permit a more direct approach to the gingival third of the tooth: (1) incorrect position of retentive clasp arm; (2) area of tooth modified to accommodate better position of retentive clasp arm; (3) more ideal position of retentive clasp arm. D, Occlusal rest preparation that will direct occlusal forces along the long axis of the tooth should be the final step in mouth preparations.
www.konkur.in 209 Chapter 15 Preparation of Abutment Teeth ABUTMENT PREPARATIONS USING CROWNS When multiple crowns are to be restored as removable partial denture abutments, it is best that all wax patterns be made at the same time. A cast of the arch with removable dies may be used if they are stable and sufficiently keyed for accuracy. If preferred, contouring wax patterns and making them par-allel may be done on a solid cast of the arch (Figure 15-2), with individual dies used to refine margins. Modern impres-sion materials and indirect techniques make either method equally satisfactory.
The same sequence for preparing teeth in the mouth applies to the contouring of wax patterns. After the cast has been placed on the surveyor to conform to the selected path of placement and after the wax patterns have been prelimi-narily carved for occlusion and contact, proximal surfaces that are to act as guiding planes are carved parallel to the path of placement with a surveyor blade. Guiding planes are extended from the marginal ridge to the junction of the middle and gingival thirds of the tooth surface involved. One must be careful not to extend the guiding plane to the gin-gival margin because the minor connector must be relieved when it crosses the gingivae. A guiding plane that includes the occlusal two thirds or even one third of the proximal area is usually adequate without endangering gingival tissues.
After the guiding planes are parallel and any other con-touring to accommodate the removable partial denture design is accomplished, occlusal rest seats are carved in the wax pattern. This method has been outlined in Chapter 6.
It should be emphasized that critical areas prepared in wax should not be destroyed by careless spruing or polish-ing. The wax pattern should be sprued to preserve paralleled surfaces and rest areas. Polishing should consist of little more than burnishing. Rest seat areas should need only refining with round finishing burs. If some interference by spruing is unavoidable, the casting must be returned to the surveyor for proximal surface refinement. This can be done accurately with the aid of a handpiece holder attached to the vertical spindle of the surveyor or some similar machining device.
One of the advantages of making cast restorations for abutment teeth is that mouth preparations that would other-wise have to be done in the mouth may be done on the sur-veyor with far greater accuracy. It is generally impossible to make several proximal surfaces parallel to one another when preparing them intraorally. The opportunity for contour-ing wax patterns and making them parallel on the surveyor in relation to a path of placement should be used to its full advantage whenever cast restorations are being made.
The ideal crown restoration for a removable partial den-ture abutment is the complete coverage crown, which can be carved, cast, and finished to ideally satisfy all requirements for support, stabilization, and retention without compromise for cosmetic reasons (Figure 15-3). Porcelain veneer crowns can be made equally satisfactory but only by the added step of contouring the veneered surface on the surveyor before the final glaze. If this is not done, retentive contours may be excessive or inadequate.
The three-quarter crown does not permit creation of retentive areas as does the complete coverage crown. How-ever, if buccal or labial surfaces are sound and retentive areas are acceptable or can be made so by slight modification of tooth surfaces, the three-quarter crown is a conservative restoration of merit. The same criteria apply in the deci-sion to leave a portion of an abutment unprotected, as in the decision to leave any tooth unprotected that is to serve as a removable partial denture abutment.
Regardless of the type of crown used, preparation should be made to provide the appropriate depth for the occlusal rest seat. This is best accomplished by altering the axial con-tours of the tooth to the ideal before preparing the tooth and creating a depression in the prepared tooth at the occlusal rest area (Figure 15-4). Because the location of occlusal rests is established during treatment planning, this information will be known in advance of any tooth preparations. If, for example, double occlusal rests are to be used, this will be Figure 15-2 Solid cast of multiple abutment crowns for a re-movable partial denture. Wax patterns for crown #21, #28, #30, and #31 can be completed at the same time using the identical cast orientation. This allows control of the path of insertion fea-tures on all fitting surfaces of the removable prostheses.
Figure 15-3 Metal ceramic crowns for teeth #4 and #5 dem-onstrating occlusal rests in metal and evidence of palatal finish-ing procedures. The distal surface of #4 provides a guide-plane surface that is continued onto a portion of the lingual surface for maximum stabilization.
www.konkur.in 210 Part II Clinical and Laboratory known so that the tooth can be prepared to accommodate the depth of both rests. It is inexcusable when waxing a pat-tern to find that a rest seat has to be made shallower than is desirable because of post-treatment planning. It can also create serious problems when a rest seat has to be made shal-low in an existing crown or inlay because its thickness is not known. The opportunity for creating an ideal rest seat (if it has been properly treatment planned) depends only on the few seconds it takes to create a space for it.
Ledges on Abutment Crowns In addition to providing abutment protection, more ideal retentive contours, definite guiding planes, and optimum occlusal rest support, complete coverage restorations on teeth used as removable partial denture abutments offer still another advantage not obtainable on natural teeth. This is the crown ledge or shoulder, which provides effec-tive stabilization and reciprocation.
The functions of the reciprocal clasp arm have been stated in Chapter 6. Briefly, these are reciprocation, sta-bilization, and auxiliary indirect retention. Any rigid reciprocal arm may provide horizontal stabilization if it is located on axial surfaces parallel to the path of placement. To a large extent, because it is placed at the height of convexity, a rigid reciprocal arm may also act as an auxiliary indirect retainer. However, its function as a reciprocating arm against the action of the reten-tive clasp arm is limited to stabilization against possible orthodontic movement when the denture framework is in its terminal position. Such reciprocation is needed when the retentive clasp produces an active orthodon-tic force because of accidental distortion or improper design. Reciprocation, to prevent transient horizontal forces that may be detrimental to abutment stability, is most needed when the restoration is placed or when a dislodging force is applied. Perhaps the term orthodon-tic force is incorrect, because the term signifies a slight but continuous influence that would logically reach equilibrium when the tooth is orthodontically moved. Instead, the transient forces of placement and removal are intermittent but forceful, which can lead to peri-odontal destruction and eventual instability rather than to orthodontic movement.
True reciprocation is not possible with a clasp arm that is placed on an occlusally inclined tooth surface, because it does not become effective until the prosthesis is fully seated. When a dislodging force is applied, the recipro-cal clasp arm, along with the occlusal rest, breaks contact with the supporting tooth surfaces, and they are no longer effective. Thus, as the retentive clasp flexes over the height of contour and exerts a horizontal force on the abutment, reciprocation is nonexistent just when it is needed most (Figure 15-5).
True reciprocation can be obtained only by creating a path of placement for the reciprocal clasp arm that is par-allel to other guiding planes. In this manner, the inferior border of the reciprocal clasp makes contact with its guid-ing surface before the retentive clasp on the other side of the tooth begins to flex (Figure 15-6). Thus, reciprocation exists during the entire path of placement and removal. A ledge on the abutment crown acts as a terminal stop for the reciprocal clasp arm. It also augments the occlusal rest and provides indirect retention for a distal extension removable partial denture.
A ledge on an abutment crown has still another advan-tage. The usual reciprocal clasp arm is half-round, and therefore convex, and is superimposed on and increases Figure 15-4 Metal-ceramic crown preparation on tooth #21 shows mesial-occlusal rest space provided in the crown prepara-tion at the mesial. The inset picture gives a perspective of the vertical height this provides for the rest to be prepared in the wax pattern.
Path of placement Height of contour Height of contour A B Figure 15-5 A, Incorrect relationship of retentive and recipro-cal clasp arms to each other when the removable partial den-ture framework is fully seated. As the retentive clasp arm flexes over the height of contour during placement and removal, the reciprocal clasp arm cannot be effective because it is not in con-tact with the tooth until the denture framework is fully seated. B, Horizontal forces applied to the abutment tooth as the reten-tive clasp flexes over the height of contour during placement and removal. Open circles at the top and bottom illustrate that the retentive clasp is passive only at its first contact with the tooth during placement and when in its terminal position with the den-ture fully seated. During placement and removal, a rigid clasp arm placed on the opposite side of the tooth cannot provide resistance against these horizontal forces. See Figure 15-6 for a method to ensure true reciprocation.
www.konkur.in 211 Chapter 15 Preparation of Abutment Teeth the bulk of an already convex surface. A reciprocal clasp arm built on a crown ledge is actually inlayed into the crown and reproduces more normal crown contours (see Figure 15-6). The patient’s tongue then contacts a contin-uously convex surface rather than the projection of a clasp arm. Unfortunately, the enamel is not thick enough nor the tooth so shaped that an effective ledge can be created on an unrestored tooth. Narrow enamel shoulders are sometimes used as rest seats on anterior teeth, but these do not provide the parallelism that is essential to recipro-cation during placement and removal.
The crown ledge may be used on any complete or three-quarter crown restored surface that is opposite the retentive side of an abutment tooth. It is used most fre-quently on premolars and molars but also may be used on canine restorations. It is not ordinarily used on buc-cal surfaces for reciprocation against lingual retention because of the excessive display of metal, but it may be used just as effectively on posterior abutments when esthetics is not a factor.
The fact that a crown ledge is to be used should be known in advance of crown preparation to ensure suffi-cient removal of tooth structure in this area. Although a shoulder or ledge is not included in the preparation itself, adequate space must be provided so that the ledge may be made sufficiently wide and the surface above it made parallel to the path of placement. The ledge should be placed at the junction of the gingival and middle thirds of the tooth, curving slightly to follow the curvature of the gingival tissues. On the side of the tooth where the clasp arm will originate, the ledge must be kept low enough to allow the origin of the clasp arm to be wide enough for sufficient strength and rigidity.
In forming the crown ledge, which is usually located on the lingual surface, the wax pattern of the crown is completed except for refinement of the margins before the ledge is carved. After the proximal guiding planes and the occlusal rests and retentive contours are formed, the ledge is carved with the surveyor blade so that the surface above is parallel to the path of placement. Thus, a continuous guiding plane surface will exist from the proximal surface around the lingual surface.
The full effectiveness of the crown ledge can be achieved only when the crown is returned to the surveyor for refinement after casting. To afford true reciprocation, the crown casting must have a surface above the ledge that is parallel to the path of placement. This can be accom-plished with precision only by machining the casting parallel to the path of placement with a handpiece holder in the surveyor or some other suitable machining device (Figure 15-7).
Similarly, the parallelism of proximal guiding planes needs to be perfected after casting and polishing. Although it is possible to approximate parallelism and, Lingual view Surveyor blade Veneered area A B C D E Figure 15-6 A, Preparation of the ledge in a wax pattern with a surveyor blade parallel to the path of placement. B, Refinement of the ledge on casting, using a suitable stone or milling device in a handpiece attached to the dental surveyor or a specialized drill press for the same purpose. C, Approximate width and depth of the ledge formed on the abutment crown, which will permit the reciprocal clasp arm to be inlaid within the normal contours of the tooth. D, True reciprocation throughout the full path of placement and removal is possible when the reciprocal clasp arm is inlaid onto the ledge on the abutment crown. E, Direct retainer assembly is fully seated. The reciprocal arm restores the lingual contour of the abutment.
www.konkur.in 212 Part II Clinical and Laboratory at the same time, form the crown ledge on the wax pat-tern with a surveyor blade, some of its accuracy is lost in casting and polishing. The use of suitable burs such as No. 557, 558, and 559 fissure burs and true cylindrical carborundum stones in the handpiece holder permits the paralleling of all guiding planes on the finished casting with the accuracy necessary for the effectiveness of those guiding plane surfaces.
The reciprocal clasp arm is ultimately waxed on the investment cast so that it is continuous with the ledge inferiorly and contoured superiorly to restore the crown contour, including the tip of the cusp. It is obvious that polishing must be controlled so as not to destroy the form of the shoulder that was prepared in wax or the parallel-ism of the guiding plane surface. It is equally vital that the removable partial denture casting be finished with great care so that the accuracy of the counterpart is not destroyed. Modern investments, casting alloys, and pol-ishing techniques make this degree of accuracy possible.
Spark Erosion Spark erosion technology is a highly advanced system for producing the ultimate in precision fit of the reciprocal arm to the ledge on the casting. This technology uses a tool system that permits repositioning of the casting with great accuracy and an electric discharge machine that is programmed to erode minute metal particles through periodic spark intervals.
Regardless of the method or technique used, it is imperative that the predetermined cast orientation be maintained to ensure that ledges and proximal guide planes remain parallel.
Veneer Crowns for Support of Clasp Arms For cosmetic reasons, resin and porcelain veneer crowns are used on abutment teeth that would otherwise display an objectionable amount of metal. They may be present in the form of porcelain veneers retained by pins and cemented to the crown, porcelain fused directly to a cast metal substructure, porcelain fused to a machined coping, cast porcelain, pressed ceramic crowns, computer-assisted designed and machined ceramic restorations, or acrylic-resin processed directly to a cast crown. The development of abrasion-resistant composites offers materials suitable for veneering that can withstand clasp contact, thereby eliminating an undesirable display of metal.
Veneer crowns must be contoured to provide suitable retention. This means that the veneer must be slightly over-contoured and then shaped to provide the desired under-cut for the location of the retentive clasp arm (Figure 15-8). A B Figure 15-7 A milling machine used to prepare parallel surfaces, internal rest seats, lingual grooves, and ledges in cast restorations. Such a device permits more precise milling than is possible with a dental handpiece attached to the dental surveyor. To be effective, the cast must be positioned on the drill in such a manner that the previously established path of placement is maintained. A movable stage or base therefore should be adjustable until the relation of the cast to the axis of the drill has been made the same as that obtained when the cast was on the dental surveyor.
www.konkur.in 213 Chapter 15 Preparation of Abutment Teeth If the veneer is of porcelain, this procedure must precede glazing, and if it is of resin, it must precede final polish-ing. If this important step in making veneered abutments is neglected or omitted, excessive or inadequate retentive contours may result.
In limited clinical trials, porcelain laminates demon-strated resistance to wear of 5-year equivalence. The porce-lain, however, resulted in slight wear on the clasps.
The flat underside of the cast clasp makes sufficient con-tact with the surface of the veneer so that abrasion of a resin veneer may result. Although the underside of the clasp may be polished (with some loss in accuracy of fit), abrasion results from the trapping and holding of food debris against the tooth surface as the clasp moves during function. There-fore, unless the retentive clasp terminal rests on metal, glazed porcelain should be used to ensure the future retentiveness of the veneered surface. Present-day acrylic-resins, which are cross-linked copolymers, will withstand abrasion for consid-erable time, but not nearly to the same degree as porcelain. Therefore acrylic-resin veneers are best used in conjunction with metal that supports the half-round clasp terminal.
SPLINTING OF ABUTMENT TEETH Often, a tooth is considered too weak to use alone as a removable partial denture abutment because of the short length or excessive taper of a single root, or because of bone loss resulting in an unfavorable crown-to-root ratio. In such instances, splinting to the adjacent tooth or teeth can be used as a means of improving abutment support. Thus, two single-rooted teeth serve as a multi-rooted abutment.
Splinting should not be used to retain a tooth that would otherwise be condemned for periodontal reasons. When the length of service of a restoration depends on the service-ability of an abutment, any periodontally questionable tooth should be condemned in favor of using an adjacent healthy tooth as the abutment, even though the span is increased one tooth by doing so.
The most common application of the use of multiple abut-ments is the splinting of two premolars or a first premolar and a canine (Figure 15-9). Mandibular premolars generally have round and tapered roots, which are easily loosened by rotational, as well as by tipping, forces. They are the weakest of the posterior abutments. Maxillary premolars also often have tapered roots, which may make them poor risks as abutments, particularly when they will be called on to resist the leverage of a distal extension base. Such teeth are best splinted by casting or soldering two crowns together. When a first premolar to be used as an abutment has poor root form or support, it is best that it be splinted to the stronger canine.
Anterior teeth on which lingual rests are to be placed often must be splinted together to avoid orthodontic movement of individual teeth. Mandibular anterior teeth are seldom used for support, but if they are, splinting of the teeth involved is advisable. When splinting is impossible, individual lingual rests on cast restorations may be slightly inclined apically to avoid possible tooth displacement, or lingual rests may be used in conjunction with incisal rests, slightly engaging the labial surface of the teeth.
Lingual rests should always be placed as low on the cingu-lum as possible, and single anterior teeth, other than canines, should not be used for occlusal support. Where lingual rests are used on central and lateral incisors, as many teeth as pos-sible should be included to distribute the load, thereby mini-mizing the force on any one tooth. Even so, some movement of individual teeth is likely to occur, particularly when they are subjected to the forces of indirect retention or when bone support is compromised. This is best avoided by splinting several teeth with united cast restorations. The condition of the teeth and cosmetic considerations will dictate whether complete crowns, three-quarter crowns, pin ledge inlays, Figure 15-8 A porcelain veneer crown is resurveyed follow-ing adjustment, glazing, and polishing. It is important to survey crowns returned from the laboratory before cementation. The best time to ensure control of all abutment contours for a remov-able partial denture is when surveyed crowns are used and they are resurveyed before permanent placement.
Figure 15-9 First premolars and canines have been splinted in this Class I, modification 1, partially edentulous arch. The splint bar was added to provide cross-arch stabilization for splinted abutments and to support and retain the anterior seg-ment of the removable restoration. The prospective longevity of the abutments has been enhanced.
www.konkur.in 214 Part II Clinical and Laboratory resin-bonded retainers, or composite restorations will be used for this purpose.
Splinting of molar teeth for multiple abutment support is less frequently used because they are generally multi-rooted. A two- or three-rooted tooth that is not strong enough alone is probably a poor abutment risk. However, there may be notable exceptions when a molar abutment would benefit from the effect of splinting, as in a hemi-sected molar root (Figure 15-10).
USE OF ISOLATED TEETH AS ABUTMENTS The average abutment tooth is subjected to some distal tip-ping, rotation, torquing, and horizontal movement, all of which must be held to a minimum by the quality of tissue support and the design of the removable partial denture. The isolated abutment tooth, however, is subjected also to mesial tipping caused by lack of proximal contact. Despite indirect retention, some lifting of the distal extension base is inevi-table, causing torque to the abutment.
In a tooth-supported prosthesis, an isolated tooth may be used as an abutment by including a fifth abutment for additional support. Thus, rotational and horizontal forces are resisted by the additional stabilization obtained from the fifth abutment. When two such isolated abutments exist, a sixth abutment should be included for the same reason. Thus, the two canines, the two isolated premolars, and two posterior teeth are used as abutments.
In contrast, an isolated anterior abutment adjacent to a distal extension base usually should be splinted to the near-est tooth by means of a fixed partial denture. The effect is twofold: (1) the anterior edentulous segment is eliminated, thereby creating an intact dental arch anterior to the edentu-lous space; and (2) the isolated tooth is splinted to the other abutment of the fixed partial denture, thereby providing multiple abutment support. Splinting should be used here only to gain multiple abutment support rather than to sup-port an otherwise weak abutment tooth.
Although splinting is advocated for abutment teeth that are considered too weak to risk being used alone, a single abutment standing alone in the dental arch anterior to a dis-tal extension basal seat generally requires the splinting effect of a fixed partial denture (Figures 15-11 and 15-12). Even though the form and length of the root and the supporting bone seem to be adequate for an ordinary abutment, the fact that the tooth lacks proximal contact endangers the tooth when it is used to support a distal extension base removable partial denture.
A second factor that may influence the decision to use an isolated tooth as an abutment is an esthetic consideration. However, neither esthetics nor economics should deter the dentist from recommending to the patient that an isolated C A B Figure 15-10 A, Periodontal disease required removal of #30 distal and #31 mesial roots. B, The first premolar and hemi-sected roots were splinted using a five-unit fixed partial denture. C, Fixed prosthesis provided cross-arch support, stability, and retention to a Kennedy Class II removable partial denture.
www.konkur.in 215 Chapter 15 Preparation of Abutment Teeth tooth to be used as a terminal abutment should be given the advantage of splinting by means of a fixed partial denture. If compromises are necessary, the patient must assume respon-sibility for use of the isolated tooth as an abutment.
The economic aspect of the use of fixed restorations as part of the mouth preparations for a removable partial den-ture is essentially the same as that for any other splinting procedure—the best design of the fixed partial denture that will ensure the longevity of its service makes the additional procedure and expense necessary. Although it must be rec-ognized that economic considerations, combined with a particularly favorable prognosis of an isolated tooth, may influence the decision to forego the advantages of using a fixed partial denture, the original treatment plan should include this provision, even though the alternative method may be accepted for economic reasons.
MISSING ANTERIOR TEETH When a removable partial denture is used to replace missing posterior teeth, especially in the absence of distal abutments, any additional missing anterior teeth are best replaced by means of fixed restorations rather than included in the removable partial denture. In any distal extension situation, some anteroposterior rotational action results from the addi-tion of an anterior segment to the denture. The ideal treat-ment plan, which considers the anterior edentulous space separately, may result in conflict with economic and esthetic realities. Each situation must be treated according to its own merits. Often the best esthetic result can be obtained by replacing missing anterior teeth and tissues with the remov-able partial denture rather than with a fixed restoration. From a biomechanical standpoint, however, it is generally advisable that a removable partial denture should replace the missing posterior teeth only after the remainder of the ante-rior arch has been made intact by fixed restorations.
Although the need for compromise is recognized, the decision to include an anterior segment on the denture depends largely on the support available for that part of the removable partial denture. The greater the number of natu-ral anterior teeth remaining, the better is the available sup-port for the edentulous segment. If definite rest seats can be prepared on multiple abutments, the anterior segment may be treated as any other tooth-bounded modification space. Sound principles of rest support apply just as much as elsewhere in the arch. Inclined tooth surfaces should not be used for occlusal support, nor should rests be placed on unprepared lingual surfaces. The best possible support for an anterior segment is multiple support extending, if possible, posteriorly across prepared lingual rest seats on the canine teeth to mesio-occlusal rest seats on the first premolars. Such support would permit the missing anterior teeth to be included in the removable partial denture, often with some cosmetic advantages over fixed restorations.
In some instances, the replacement of anterior teeth by means of a removable partial denture cannot be avoided. However, without adequate tooth support, any such pros-thesis lacks the stability that results from replacing only the posterior teeth with the removable partial denture and the anterior teeth with fixed restorations. When anterior teeth have been lost through accident or have been missing for some time, resorption of the anterior residual ridge may have progressed to the point that neither fixed nor remov-able pontics may be butted to the residual ridge. In such Figure 15-11 Lone-standing premolar should be splinted to the canine with a fixed partial denture. Not only will the design of the removable partial denture be simplified, but the longevity of abutment service by the premolar will be greatly extended.
A B Figure 15-12 A, Isolated abutments have been splinted using splint bars. B, The removable partial denture is more adequately sup-ported by the splinting mechanism shown in A than could be realized with isolated abutments.
www.konkur.in 216 Part II Clinical and Laboratory instances, for reasons of esthetics and orofacial tissue sup-port, the missing teeth must be replaced with a denture base supporting teeth that are more nearly in their original posi-tion, considerably forward from the residual ridge. Although such teeth may be positioned to better cosmetic advantage, the contouring and coloring of a denture base to be estheti-cally pleasing require the maximum artistic effort of both the dentist and the technician. Such a removable partial denture, both from an esthetic and a biomechanical standpoint, is one of the most difficult of all prosthetic restorations. However, a splint bar, connected by abutments on both sides of the edentulous space, will provide much-needed support and retention to the anterior segment of the removable partial denture. Because the splint bar will provide vertical support, rest seats on abutments adjacent to the edentulous area need not be prepared, thus simplifying an anterior restoration to some extent.
The concept of a dual path of placement to enhance the esthetic replacement of missing anterior teeth with a remov-able partial denture is recognized. Sources of information on this concept are made available in the Selected Reading Resources under “Partial Denture Design.” TEMPORARY CROWNS WHEN A REMOVABLE PARTIAL DENTURE IS BEING WORN Occasionally, an existing removable partial denture must remain serviceable while the mouth is being prepared for a new prosthesis. In such situations, temporary crowns must be made that will support the old removable partial den-ture and will not interfere with its placement and removal. Acrylic-resin temporary crowns that duplicate the original form of the abutment teeth must be made.
The technique for making temporary crowns to fit direct retainers is similar to that used for other types of acrylic-resin temporary crowns. The principal difference is that an impression, made with an elastic impression material, must be made of the entire arch with the existing removable par-tial denture in place. It is necessary that the removable partial denture remain in the impression when it is removed from the mouth. If it remains in the mouth, it must be removed and inserted into the impression in its designated position. The impression with the removable partial denture in place is disinfected, wrapped in a wet paper towel (if irreversible hydrocolloid was used as the impression material) or placed in a plastic bag, and set aside while the tooth or teeth are being prepared for new crowns.
After the preparations are completed and the impressions and jaw relation records have been made, the prepared teeth are dried and lubricated. The original impression is trimmed to eliminate any excess, undercuts, and interproximal pro-jections that would interfere with the replacement of the impression in the mouth.
The methyl methacrylate acrylic-resins, composites, copo-lymers, and fiber-reinforced resins may serve as excellent materials for temporary crowns in conjunction with removable partial dentures. Making temporary crowns requires a small mixing cup or dappen dish; a cement spatula; and a small, dis-posable, plastic syringe. Autopolymerizing acrylic-resin of the appropriate tooth color is placed in the cup or dappen dish, and monomer is added to make a slightly viscous mix. The volume should be slightly in excess of the amount estimated to fabri-cate the temporary restorations. The mix should be spatulated to a smooth consistency and the mix immediately poured into the barrel of the disposable syringe. A small amount of the mix should be injected over and around the margins of the pre-pared teeth. The remaining material should be injected into the impression of the prepared teeth. The impression is seated into the mouth, where the dentist holds it in place until suf-ficient time has elapsed for it to reach a stiff, rubbery stage, or a consistency recommended by the manufacturer. This again must be based on experience with the particular resin used. At this time, the impression is removed. The crowns may remain in the impression. If so, they are stripped out of the impres-sion, all excess is trimmed away with scissors, and the crowns are reseated on the prepared abutments. The removable partial denture is then removed from the impression and reseated in the mouth onto the temporary crowns, which should be in a stiff-rubbery state. The patient may bring the teeth into occlu-sion to reestablish the former position and occlusal relation-ship of the existing removable partial denture.
After the resin crown or crowns have polymerized, the removable partial denture is removed and the crowns remain on the teeth. These are then carefully removed, contoured to accommodate oral hygiene access, trimmed, polished, and temporarily cemented. The result is a temporary crown that restores the original abutment contours and allows the removable partial denture to be placed and removed without interference while temporarily providing the same support to the denture that existed before the teeth were prepared.
Cementation of Temporary Crowns Cementation of temporary crowns may require slight relief of the internal surface of the crowns to accommodate the temporary cement and to facilitate removal. The temporary cement should be thin and applied only to the inside gingival margin of the crowns to ensure complete seating. As soon as the temporary cement has hardened, the occlusion should be checked and adjusted accordingly. Regardless of the type of temporary cement used, any excess that might irritate the gingivae should be removed.
FABRICATING RESTORATIONS TO FIT EXISTING DENTURE RETAINERS It is often necessary that an abutment tooth be restored with a complete crown (or other restoration) that fits the inside of the clasp of an otherwise serviceable removable partial denture. One technique for doing so is simple enough, but it requires that an indirect-direct pattern be made and therefore justifies a fee for service above that required for the usual restoration.
www.konkur.in 217 Chapter 15 Preparation of Abutment Teeth The technique for making a crown to fit the inside of a clasp is as follows: An irreversible hydrocolloid impres-sion of the mouth is made with the removable partial denture in place. This impression, which is used to make the temporary crown, is wrapped in a wet paper towel or placed in a plastic bag and set aside while the tooth is being prepared. Even though several abutment teeth are to be restored, it is usually necessary that each temporary restoration be completed before the next one is begun. This is necessary so that the original support and occlu-sal relationship of the removable partial denture can be maintained as each new temporary crown is being made. During preparation of the abutment tooth, the remov-able partial denture is replaced frequently to ascertain that sufficient tooth structure has been removed to allow for the thickness of the casting. When the preparation is completed, an individual impression of the tooth is obtained from which a stone die is made. A temporary crown is then made in the original irreversible hydrocol-loid impression, as outlined in the preceding paragraphs. It is trimmed, polished, and temporarily cemented, and the removable partial denture is returned to the mouth. The patient is dismissed after the excess cement has been removed.
On the stone die made from the individual impres-sion, a thin, autopolymerizing resin coping will be formed with a brush technique. The stone die should first be trimmed to the finishing line of the prepara-tion, which is then delineated with a pencil, and the die painted with a tinfoil substitute. A separating material, such as a tinfoil substitute, should be used and will form a thin film on a cold, dry surface. Not all tinfoil substi-tutes are suitable for this purpose. With autopolymer-izing resin powder and liquid in separate dappen dishes and a fine brush, a coping of resin of uniform thickness is painted onto the die. This should extend not quite to the pencil line representing the limit of the crown preparation. After hardening, the resin coping may be removed, inspected, and trimmed if necessary. The thin film of foil substitute should be removed before the cop-ing is reseated onto the die.
The wax pattern buildup on the resin coping is usually not begun until the patient returns. A sequence using a functional chew-in technique for occlusion would be fol-lowed establishing proximal contact and contours appro-priate for the clasp assembly as outlined in the following paragraphs.
First, the occlusal portion of the wax pattern is established by having the patient close into maximum intercuspation, followed by excursive movements (Figure 15-13, A). The wax pattern is returned to the cast, and additions are made to dull areas as required. The process is repeated until a smooth occlusal registration has been obtained. Except for narrowing of the occlusal surface and carving of grooves and spillways, this will be the occlusal anatomy of the finished restoration.
The second step is the addition of sufficient wax to establish contact relations with the adjacent tooth. At this time, the occlusal relation of the marginal ridges also must be established. Next, wax is added to buccal and lingual surfaces where the clasp arms will contact the crown, and the wax pattern is again reseated in the mouth. The clasp arms, minor connectors, and occlusal rests involved in the removable partial denture are carefully warmed with a needlepoint flame, carefully avoiding any adjacent resin, and the removable partial denture is positioned in the mouth and onto the wax pattern (see Figure 15-13, B). Several attempts may be necessary until the removable partial denture is fully seated and the components of the clasp are clearly recorded in the wax pattern. Each time the removable partial denture is removed, the pattern will draw with it and must be teased out of the clasp.
When contact with the clasp arms and the occlu-sal relation of the removable partial denture have been established satisfactorily, the temporary crown may be replaced and the patient dismissed. The crown pattern is completed on the die by narrowing the occlusal sur-face bucco-lingually, adding grooves and spillways, and refining the margins. Any wax ledge remaining below the reciprocal clasp arm may be left to provide some of the advantages of a crown ledge that were described earlier in this chapter. Excess wax remaining below the retentive clasp arm, however, must be removed to per-mit the adding of a retentive undercut later (see Figure 15-13, C).
If a veneer material is to be added, the veneer space must now be carved in the wax pattern. In such situations, the contour of the veneer may be recorded by making a stone matrix of the buccal surface, which can be reposi-tioned on the completed casting to ensure the proper con-tour of the composite veneer.
The wax pattern must be sprued with care so that essen-tial areas on the pattern are not destroyed. After casting, the crown should be subjected to a minimum of polishing because the exact form of the axial and occlusal surfaces must be maintained.
Because it is impossible to withdraw a clasp arm from a retentive undercut on the wax pattern, the casting must be made without any provision for clasp retention. After the crown has been tried in the mouth with the denture in place, the location of the retentive clasp terminal is identified by scoring the crown with a sharp instrument. Then the crown may be ground and polished slightly in this area to create a retentive undercut. The clasp terminal then may be carefully adapted into this undercut, thereby creating clasp retention on the new crown.
An alternate method for making crowns to fit existing retainers uses mounted casts with the removable prosthe-sis adapted to the working cast to develop occlusal sur-faces for the involved crowns.
www.konkur.in 218 Part II Clinical and Laboratory Ideally, all abutment teeth would best be protected with complete crowns before the removable partial denture is fabricated. Except for the possibility of recurrent caries caused by defective crown margins or gingival recession, abutment teeth so protected may be expected to give many years of satisfactory service in support, stabilization, and retention of the remov-able partial denture. Economically, a policy of insisting on complete coverage for all abutment teeth may well be justified from the long-term viewpoint. It must be recognized, however, that in practice, complete cover-age of all abutment teeth is not always possible at the time of treatment planning. Many factors influence the future health status of an abutment tooth, some of which cannot be foreseen. It is necessary that the den-tist be able to treat abutment teeth that later become defective so that their service as abutments may be restored and the serviceability of the removable partial denture maintained. Although not part of the original mouth preparations, this service accomplishes much the same objective by providing support, stability, and retention, and the dentist must be technically capable of providing this removable partial denture service when it becomes necessary.
A B C Resin coping Inlay wax Inlay wax Lingual ledge Resin coping Figure 15-13 Making of the cast crown to fit an existing removable partial denture clasp. A, Thin acrylic-resin coping is made first on the individual die of a prepared tooth. Inlay wax is then added and coping placed onto the prepared tooth where occlusal surfaces and contact relations are established directly in the mouth. The clasp assembly is warmed with a needlepoint flame only enough to soften the inlay wax, the removable partial denture is placed into the mouth, and it is guided gently into place by the opposing occlusion. This step must be repeated several times and excess wax removed or wax added until full supporting contact with the underside of the clasp assembly has been established, with the denture fully seated. Usually, the wax pattern withdraws with the denture and must be gently teased out of the clasp each time. B, The wax pattern is then placed back onto the individual die to complete the occlusal anatomy and refine the margins. Excess wax remaining below the impression of the retentive clasp arm must be removed, but the wax ledge may be left below the reciprocal clasp arm. C, Finished casting in the mouth. The terminus of the retentive clasp is then readapted to engage the undercut. It is frequently necessary to remove some interference from casting, as indicated by articulating paper placed between the clasp and the crown, until the clasp is fully seated.
www.konkur.in CHAPTER 16 Impression Materials and Procedures for Removable Partial Dentures CHAPTER OUTLINE Elastic Materials Reversible Hydrocolloids Irreversible Hydrocolloids Mercaptan Rubber–Base Impression Materials Polyether Impression Materials Silicone Impression Materials Rigid Materials Plaster of Paris Metallic Oxide Paste Thermoplastic Materials Modeling Plastic Impression Waxes and Natural Resins Impressions of the Partially Edentulous Arch Important Precautions to Be Observed in the Handling of Hydrocolloid Impressions Step-by-Step Procedure for Making a Hydrocolloid Impression Step-by-Step Procedure for Making a Stone Cast from a Hydrocolloid Impression Possible Causes of an Inaccurate and/or a Weak Cast of a Dental Arch Individual Impression Trays Technique for Making Individual Acrylic-Resin Impression Trays Impression materials used in the various phases of partial denture fabrication may be classified as rigid, thermoplastic, or elastic substances. Rigid impression materials are those that set to a rigid consistency. Thermoplastic impression materials are those that become plastic at higher tempera-tures and resume their original form when cooled. Elastic impression materials are those that remain in an elastic or flexible state after they have set and have been removed from the mouth.
Although rigid impression materials may be capable of recording tooth and tissue details accurately, they cannot be removed from the mouth without fracture and reassembly. Thermoplastic materials cannot record minute details accu-rately because they undergo permanent distortion during withdrawal from tooth and tissue undercuts. Elastic materi-als are the only ones that can be withdrawn from tooth and tissue undercuts without permanent deformation and there-fore are used generally for making impressions for remov-able partial dentures, immediate dentures, crowns, and fixed partial dentures when tooth and tissue undercuts and surface detail must be recorded with accuracy.
ELASTIC MATERIALS Reversible Hydrocolloids Reversible (agar-agar) hydrocolloids, which are fluid at higher temperatures and gel on reduction in temperature, are used primarily as impression materials for fixed restorations. They demonstrate acceptable accuracy when properly used; how-ever, the reversible hydrocolloid impression materials offer few advantages over the irreversible (alginate) hydrocolloids Some of the historical parts of this discussion have been quoted or paraphrased from McCracken WL: Impression materials in prosthetic dentistry, Dent Clin North Am 2:671-684, 1958.
www.konkur.in 220 Part II Clinical and Laboratory when used as a removable partial denture impression material. Present-day irreversible hydrocolloids are sufficiently accurate for making master casts for removable partial dentures. How-ever, control of the border width and length of impressions made with these materials is difficult.
Irreversible Hydrocolloids Irreversible hydrocolloids are used for making diagnostic casts, orthodontic treatment casts, and master casts for removable partial denture procedures. Because they are made of colloid materials, neither reversible nor irrevers-ible hydrocolloid impressions can be stored for any length of time but must be poured immediately.
These materials have low tear strength, provide less surface detail than other materials (e.g., mercaptan rub-ber base) and are not as dimensionally stable as other materials. They can, however, be used in the presence of moisture (saliva); are hydrophilic; pour well with stone; have a pleasant taste and odor; and are nontoxic, non-staining, and inexpensive. The combination reversible-irreversible hydrocolloids have demonstrated a tendency to separate and should be used with that understanding. The hydrocolloids can be acceptably disinfected with a spray solution of 2% acid glutaraldehyde, stored in 100% humidity, and poured within 1 hour.
Mercaptan Rubber–Base Impression Materials The mercaptan rubber–base (Thiokol) impression materials can also be used for removable partial denture impressions and especially for secondary corrected or altered cast impressions. To be accurate, the impression must have a uniform thickness that does not exceed 3 mm (18 inch). This necessitates the use of a carefully made individual impression tray of acrylic-resin or some other material possessing adequate rigidity and stabil-ity. Those materials that are highly cross-linked (medium and heavy body) do not recover well from deformation and should not be used when large or multiple undercuts are present. For example, when large numbers of teeth with natural tooth con-tours that display multiple undercuts remain, these materials will be subjected to clinically significant distortion upon with-drawal. The long-term dimensional stability of these materials is poor because of water loss after setting. The material must be held still during the impression-making procedure because it does not have a snap set; it should be allowed to rebound for 7 to 15 minutes after it is removed from the mouth and should then be poured immediately. Many of these materials have an unpleasant odor and can stain clothes. These materials are moderately inexpensive, have high tear strength and long working and setting times (8 to 10 minutes), and can be dis-infected in liquid, cold-sterilizing solutions. The accuracy of mercaptan rubber base is acceptable for making impressions for removable partial dentures; however, as with hydrocol-loid impression materials, certain precautions must be taken to avoid distortion of the impression. Mercaptan rubber–base impression materials do have an advantage over hydrocolloid materials in that the surface of an artificial stone poured against them is of a smoother texture and therefore appears to be smoother and harder than one poured against a hydrocol-loid material. This is probably so because the rubber material does not have the ability to retard or etch the surface of the setting stone. Despite their accuracy, this has always been a disadvantage of all hydrocolloid impression materials. The fact that a smoother surface results does not, however, preclude the possibility of a grossly inaccurate impression and stone cast resulting from other causes. Rubber-base impression materials possess a longer setting time than the irreversible hydrocol-loid materials and lend themselves better to border molding in adequate supporting trays.
Polyether Impression Materials Polyether impression material is an elastic-type material, as are the polysulfide and silicone materials. These materi-als have demonstrated good accuracy in clinical evaluations and are thixotropic, which provides good surface detail and makes them useful as a border molding material. It should be noted, however, that these materials are not compatible with the addition reaction silicone impression materials and should not be used to border mold custom trays when the silicone impression materials are to be used as the final impression material. The polyethers are also hydrophilic, which produces good wettability for easy cast forming.
The polyethers have low to moderate tear strength and much shorter working and setting times, which can limit the usefulness of the material. The flow characteristics and flexibility of the polyether materials are the lowest of any of the elastic materials. These characteristics can limit the use of polyethers in removable partial denture impression proce-dures. The stiffness of the material can result in cast breakage when removal of the cast from a custom tray is attempted. These materials have a higher permanent deformation than the addition reaction silicones. Some have an unpleasant taste, and because the material will absorb moisture, it can-not be immersed in disinfecting solutions or stored in high humidity for any extended period of time. The materials should be poured within 2 hours; however, manufacturers claim that if the impression is kept dry, clinically acceptable casts can be poured for up to 7 days.
Silicone Impression Materials The silicone impression materials are more accurate and easier to use than the other elastic impression materials. The condensation silicones have a moderate (5 to 7 minutes) working time that can be altered by adjusting the amount of the accelerator. They have a pleasant odor, moderately high tear strength, and excellent recovery from deformation. These materials can be used with a compatible putty mate-rial to form fit a custom tray. Silicone impression materials are hydrophobic, which can make cast formation a problem. These materials can be disinfected in any of the disinfecting solutions with no alteration in accuracy. Ideally, these mate-rials should be poured within 1 hour.
www.konkur.in 221 Chapter 16 Impression Materials and Procedures for Removable Partial Dentures The addition reaction silicones are the most accurate of the elastic impression materials. They have less polymeriza-tion shrinkage, low distortion, fast recovery from deforma-tion, and moderately high tear strength. These materials have a working time of 3 to 5 minutes, which can be easily modified with the use of retardants and temperature con-trols. They are available in both hydrophilic and hydrophobic forms, have no smell or taste, and also come in putty form, to assist in form fitting the impression tray at chairside. Most of the addition reaction silicones are available in automixing devices, can be poured up to 1 week after impression making with acceptable clinical results, and are stable in most steril-izing solutions. Sulfur in latex gloves and in ferric and alumi-num sulfate retraction solution may inhibit polymerization. Many of the hydrophobic types are difficult to pour with stone, and adhesion to acrylic-resin trays is not good. The putties for these materials have a relatively short shelf life, and they are more expensive than the other elastic impres-sion materials.
RIGID MATERIALS Plaster of Paris One type of rigid impression material is plaster of Paris, which has been used in dentistry for over 200 years. Although all plaster of Paris impression materials are handled in approximately the same manner, the setting and flow characteristics of each manufacturer’s product will vary. Some are pure and finely ground with only an accelerator added to expedite setting within reasonable working limits. Others are modified impression plasters to which binders and plasticizers have been added to per-mit limited border manipulation while the material is set-ting. These do not set as hard or fracture as cleanly as pure plaster of Paris and therefore cannot be reassembled with as much accuracy if fracture occurs.
Plaster of Paris was once the only material available for removable partial denture impressions, but now elastic materials have completely replaced the impression plas-ters in this phase of prosthetic dentistry. It can be used for making accurate transfers of abutment castings or cop-ings in the fabrication of fixed restorations and internal attachment dentures and for making rigid indexes and matrices for various purposes in prosthetic dentistry. Modified impression plasters may be used to record max-illomandibular relationships.
Metallic Oxide Paste A second type of rigid impression material is metallic oxide paste, which is usually some form of a zinc oxide–eugenol Some of the historical parts of this discussion have been quoted or paraphrased from McCracken WL: Impression materials in prosthetic dentistry, Dent Clin North Am 2:671-684, 1958.
combination. A number of these pastes are available; how-ever, they are not used as primary impression materials and should never be used for impressions that include remaining natural teeth. They also are not to be used in stock impres-sion trays.
Metallic oxide pastes are manufactured with a wide varia-tion of consistencies and setting characteristics. For con-venience, most of them are dispensed from two tubes; this enables the dentist to dispense and mix the correct proportion from each tube on a mixing slab. The previously prepared tray for the edentulous ridge segments is loaded and positioned in the mouth with or without any attempt at border molding. Border molding with metallic oxide impression pastes is not advisable because wrinkles will occur if movement is permit-ted at the time the material reaches its setting state.
As with all impression techniques, the accuracy of the primary impression and of the impression tray has a great influence on the final impression. Some metallic oxide pastes remain fluid for a longer period than do others, and some manufacturers claim that border molding is possible. In gen-eral, however, all metallic oxide pastes have one thing in com-mon with plaster of Paris impression materials: they all have a setting time during which they should not be disturbed and after which no further border molding is effective.
Metallic oxide pastes, which are rigid substances, can be used as secondary impression materials for complete den-tures and for extension base edentulous ridge areas of a removable partial denture if a custom impression tray has been properly designed and attached to the removable par-tial denture framework (see Chapter 17).
Metallic oxide pastes can also be used as an impression material for relining distal extension denture bases and may be used successfully for this purpose if the original denture base has been relieved sufficiently to allow the material to flow without displacement of the denture or the underlying tissues.
THERMOPLASTIC MATERIALS Modeling Plastic Like plaster of Paris, modeling plastic is among the oldest impression materials used in prosthetic dentistry. This mate-rial is most often used for border correction (border mold-ing) of custom impression trays for Kennedy Class I and Class II removable partial denture bases. It is manufactured in several different colors, each color being an indication of the temperature range at which the material is plastic and workable. A common error in the use of modeling plas-tic is that it is often subjected to higher temperatures than intended by the manufacturer. It then becomes too soft and loses some of its favorable working characteristics. If a Some of the historical parts of this discussion have been quoted or paraphrased from McCracken WL: Impression materials in prosthetic dentistry, Dent Clin North Am 2:671-684, 1958.
www.konkur.in 222 Part II Clinical and Laboratory temperature-controlled water bath is not used, a thermom-eter should be used to maintain the water temperature. If modeling plastic is softened at a temperature above that intended by the manufacturer, the material becomes brittle and unpredictable. Also, there is the ever-present danger of burning the patient when the temperature used in softening the modeling plastic is too high.
The most commonly used modeling plastic for corrected impressions of extension base areas is the red (red-brown) material, in cake form, that softens at about 132° F. It should never be softened at temperatures much above this. Neither it nor any other modeling plastic should be immersed in the water bath for an indefinite period. It should be dipped and kneaded until soft and subjected to no more heat than neces-sary before the tray is loaded and it is placed in the mouth. Then it may be flamed with an alcohol torch for the purpose of border molding, but it should always be tempered by being dipped back into the water bath before its return to the mouth to avoid burning the patient. The modeling plastic then may be chilled using a water spray before removal from the mouth, although this is not necessary if care is used in removing the impression. During sectional flaming and border molding, the modeling plastic should be chilled in ice water after each removal from the mouth; then it may be trimmed with a sharp knife without danger of fracture or distortion.
Red, gray, and green modeling plastics are obtainable in stick form for use in border molding an impression or an impression tray. The green material is the lowest fusing of the modeling plastics. The red and gray sticks have a higher and broader working range than do the cakes of like color so they may be flamed without harming the material. The gray mate-rial in stick form is preferred by some dentists for border molding because of its contrasting lighter color. The choice between the use of green and gray sticks is purely optional and entirely up to the dentist.
Some dentists still prefer to use modeling plastic as a sec-ondary impression material to record edentulous ridges in removable partial denture fabrication. When this is done, it is generally used only as a means of building up the under-side of the denture before the tissues are recorded with some secondary impression material (see Chapter 17).
Impression Waxes and Natural Resins A second group of thermoplastic impression materials con-sists of those impression waxes and resins commonly spo-ken of as mouth-temperature waxes. The most familiar of these have been the Iowa wax (Kerr Co., Romulus, MI) and the Korecta waxes (D-R Miner Dental, Concord, CA), all of which were developed for specific techniques.
Knowledge of the characteristics of mouth-temperature waxes is important if they are to be used correctly.
The Iowa wax was developed for use in recording the functional or supporting form of an edentulous ridge. It may be used as a secondary impression material or as an impression material for relining the finished removable par-tial denture to obtain support from the underlying tissues. The mouth-temperature waxes lend themselves well to all relining techniques because they will flow sufficiently in the mouth to avoid displacement of tissues. As with any relining technique, it is necessary that sufficient relief and venting be provided to give the material the opportunity to flow.
The difference between impression wax and modeling plastic is that impression waxes have the ability to flow as long as they are in the mouth and thereby permit equaliza-tion of pressure and prevent displacement. The modeling plastics flow only in proportion to the amount of flaming and tempering that can be done outside of the mouth; this does not continue after the plastic has approached mouth temperature. The principal advantage of mouth-temperature waxes is that, given sufficient time, they permit a rebound of those tissues that may have been forcibly displaced.
The impression waxes also may be used to correct the borders of impressions made of more rigid materials, thereby establish-ing optimum contact at the border of the denture. All mouth-temperature wax impressions have the ability to record border detail accurately and include the correct width of the denture border. They also have the advantage of being correctable.
Mouth-temperature waxes vary in their working charac-teristics. They are designed primarily for impression tech-niques that attempt to record the tissues under an occlusal load. In such techniques, the occlusion rim or the arrange-ment of artificial teeth is completed first. Mouth- temperature wax is then applied to the tissue side of the denture base, and the final impression is made under functional loading by using various movements to simulate functional activity. These mouth-temperature materials also may be used suc-cessfully in open-mouth impression techniques. Iowa wax will not distort after removal from the mouth at ordinary room temperatures, but the more resinous waxes must be stored at much lower temperatures to avoid flow when they are out of the mouth. Resinous waxes are not ordinarily used in removable partial denture impression techniques except for secondary impressions.
IMPRESSIONS OF THE PARTIALLY EDENTULOUS ARCH An impression of the partially edentulous arch must record accurately the anatomic form of the teeth and surround-ing tissues. This is necessary so that the prosthesis may be designed to follow a definite path of placement and removal and so that support, stability, and retention derived from the abutment teeth may be precise and accurate.
Materials that could be permanently deformed by removal from tooth or tissue undercuts should not be used. The ther-moplastic impression materials and metallic oxide pastes are therefore excluded for recording the anatomic form of the dental arch. Rubber-base materials that are highly cross-linked should not be used when large or multiple under-cuts are present because these materials will be subjected to considerable distortion upon withdrawal. The introduction of hydrocolloids as impression materials was a giant step www.konkur.in 223 Chapter 16 Impression Materials and Procedures for Removable Partial Dentures forward in dentistry. For the first time, impressions could be made of undercut areas with a material that was elastic enough to be withdrawn from those undercuts without per-manent distortion. It permitted the making of a one-piece impression, which did not require the use of a separating medium, and was and still is an acceptably accurate material when handled properly. Irreversible hydrocolloid becomes a gel via a chemical reaction as a result of mixing alginate pow-der with water. This physical change is irreversible.
All hydrocolloids are dimensionally stable only during a brief period after removal from the mouth. If exposed to the air, they rapidly lose water content, with resulting shrinkage and other dimensional changes. If immersed in water, they imbibe water, with accompanying swelling and dimensional changes. All hydrocolloid impressions should be poured immediately, but if they must be stored for a brief period, they should be in a saturated atmosphere rather than immersed in water. This can be accomplished simply by wrapping the impression in a damp paper towel or sealing it in a plastic bag.
Hydrocolloids also exhibit a phenomenon known as syn-eresis, which is associated with the giving off of a mucinous exudate. This mucinous exudate has a retarding effect on any gypsum material, which results in a soft or chalky cast sur-face. Sometimes this is detected only by close examination of the impression after removal from the cast. Nevertheless, such a cast surface is inaccurate and ultimately will result in an inaccurate removable partial denture framework. Pouring the cast immediately and using some chemical accelerator, such as potassium sulfate, to counteract the retarding effect of the hydrocolloid can prevent this inaccuracy. All modern irreversible hydrocolloid impression materials have an accel-erator incorporated into the powder and no longer need to be treated with a fixing solution.
However, some disadvantages are associated with the use of irreversible hydrocolloid. This material gels by means of a chemical reaction that is accelerated by the warmth of the tis-sues. In the irreversible hydrocolloid, gelation first takes place next to the tissues, and any movement of the tray during gela-tion will result in internal stresses that are released on removal of the impression from the mouth. A distorted and therefore inaccurate impression results from an irreversible hydrocol-loid impression that is not held immobile during gelation.
Another disadvantage of irreversible hydrocolloid is that it must be introduced into the mouth at approximately 70° F, which results in an immediate increase in the viscosity and surface tension of the material. Air bubbles are therefore harder to dispel.
Important Precautions to Be Observed in the Handling of Hydrocolloid Impressions Some important precautions to be observed in the handling of hydrocolloid are as follows: 1. The impression should not be exposed to air because some dehydration will inevitably occur and result in shrinkage. 2. The impression should not be immersed in water or dis-infectants because some imbibition will inevitably result, with an accompanying expansion. 3. The impression should be protected from dehydration by placing it in a humid atmosphere or wrapping it in a damp paper towel until a cast can be poured. To avoid volume change, this should be done within 15 minutes after removal of the impression from the mouth. 4. Exudate from the hydrocolloid has a retarding effect on the chemical reaction of gypsum products and results in a chalky cast surface. This can be prevented by pouring the cast immediately or by first immersing the impression in a solution of accelerator if an accelerator is not included in the formula.
Step-by-Step Procedure for Making a Hydrocolloid Impression The step-by-step procedure and important points to observe in the making of a hydrocolloid impression are as follows: 1. Select a suitable, sterilized, perforated or rim-lock im-pression tray that is large enough to provide a 4- to 5-mm thickness of the impression material between the teeth and tissues and the tray.
2. Build up the palatal portion of the maxillary impres-sion tray with wax or modeling plastic to ensure even distribution of the impression material and to prevent the material from slumping away from the palatal sur-face (Figure 16-1, A). At this time, it is also helpful to pack the palate with gauze that has been sprayed with a topical anesthetic. This serves to anesthetize the mi-nor salivary glands and mucous glands of the palate and thus prevent secretions as a response to smell or taste or to the physical presence of the impression material.
If gelation occurs next to the tissues while the deeper portion is still fluid, a distorted impression of the pal-ate may result, which cannot be detected in the finished A B Figure 16-1 A, Maxillary impression tray with palatal portion built up with baseplate wax to prevent impression material from sagging away from palatal surface. B, Mandibular impression tray with periphery wax added to lingual flanges to prevent tis-sues of the floor of the mouth from rising inside the tray. The posterior end of the tray is extended with periphery wax to cover the retromolar pad regions.
www.konkur.in 224 Part II Clinical and Laboratory impression. This may result in the major connector of the finished casting not being in contact with the un-derlying tissues. The maxillary tray frequently has to be extended posteriorly to include the tuberosities and the vibrating line region of the palate. Such an extension also aids in correctly orienting the tray in the patient’s mouth when the impression is made.
3. The lingual flange of the mandibular tray may need to be lengthened with wax in the retromylohyoid area or to be extended posteriorly, but it rarely ever needs to be length-ened elsewhere. Wax may need to be added inside the dis-tolingual flange to prevent the tissues of the floor of the mouth from rising inside the tray (see Figure 16-1, B).
4. Place the patient in an upright position, with the arch to be impressed nearly parallel to the floor.
5. When irreversible hydrocolloid is used, place the mea-sured amount of water (at 70° F) in a clean, dry, rubber mixing bowl (600-mL capacity). Add the correct mea-sure of powder. Spatulate rapidly against the side of the bowl with a short, stiff spatula. This should be accom-plished in less than 1 minute. The patient should rinse his or her mouth with cool water to eliminate excess saliva while the impression material is being mixed and the tray is being loaded.
6. In placing the material in the tray, avoid entrapping air.
Have the first layer of material lock through the perfo-rations of the tray or rim-lock to prevent any possible dislodgment after gelation.
7. After loading the tray, remove the gauze with the topi-cal anesthetic and quickly place (rub) some of the im-pression material on any critical areas using your finger (areas such as rest preparations and abutment teeth). If a maxillary impression is being made, place the material in the highest aspect of the palate and over the rugae.
8. Use a mouth mirror or index finger to retract the cheek on the side away from you as the tray is rotated into the mouth from the near side.
9. Seat the tray first on the side away from you, next on the anterior area, while reflecting the lip, and then on the near side, with the mouth mirror or finger for cheek re-traction. Finally, make sure that the lip is draping natu-rally over the tray. 10. Be careful not to seat the tray too deeply, leaving room for a thickness of material over the occlusal and incisal surfaces. 11. Hold the tray immobile for 3 minutes with light finger pressure over the left and right premolar areas. To avoid internal stresses in the finished impression, do not allow the tray to move during gelation. Any movement of the tray during gelation will produce an inaccurate impres-sion. If, for example, you allow the patient or the assis-tant to hold the tray in position at any time during the impression procedure, some movement of the tray will be inevitable during the transfer, and the impression will probably be inaccurate. Do not remove the impression from the mouth until the impression material has com-pletely set. 12. After releasing the surface tension, remove the impres-sion quickly in line with the long axis of the teeth to avoid tearing or other distortion. 13. Rinse the impression free of saliva with slurry water, or dust it with plaster, and rinse gently; then examine it critically. Spray the impression thoroughly with a suit-able disinfectant, and cover it immediately with a damp paper towel.
A cast should be poured immediately into a disinfected hydrocolloid impression to avoid dimensional changes and syneresis. Circumstances often necessitate some delay, but this time lapse should be kept to a minimum. A delay of 15 minutes will satisfy the disinfection requirements and should not be deleterious if the impression is kept in a humid atmo-sphere.
Step-by-Step Procedure for Making a Stone Cast from a Hydrocolloid Impression The step-by-step procedure for making a stone cast from the impression is as follows: 1. A more abrasive-resistant type IV stone should be used to form removable partial denture casts. Have the measured dental stone at hand, along with the designated quan-tity of room temperature water, as recommended by the manufacturer. A clean 600-mL rubber mixing bowl, a stiff spatula, and a vibrator complete the preparations. A No. 7 spatula also should be within reach. 2. First, pour the measure of water into the mixing bowl, and then add the measure of stone. Spatulate thoroughly for 1 minute, remembering that a weak and porous stone cast may result from insufficient spatulation. Mechanical spatulation under vacuum is preferred. After any spatula-tion other than in a vacuum, place the mixing bowl on the vibrator and knead the material to permit the escape of any trapped air. 3. After removing the impression from the damp towel, gently shake out surplus moisture and hold the impres-sion over the vibrator, impression side up, with only the handle of the tray contacting the vibrator. The impression material must not be placed in contact with the vibrator because of possible distortion of the impression. 4. With a small spatula, add the first cast material to the distal area away from you. Allow this first material to be vibrated around the arch from tooth to tooth toward the anterior part of the impression (Figure 16-2). Continue to add small increments of material at this same distal area, with each portion of added stone pushing the mass ahead of it. This avoids the entrapment of air. The weight of the material causes any excess water to be pushed around the arch and to be expelled ultimately at the opposite end of the impression. Discard this fluid material. When the impressions of all teeth have been filled, continue to add artificial stone in larger portions until the impression is completely filled. 5. The filled impression should be placed so that its weight does not distort the hydrocolloid impression material.
www.konkur.in 225 Chapter 16 Impression Materials and Procedures for Removable Partial Dentures The base of the cast can be completed with the same mix of stone. The base of the cast should be 16 to 18 mm (2⁄3 to 3⁄4 inch) at its thinnest portion and should be extended beyond the borders of the impression so that buccal, labial, and lin-gual borders will be recorded correctly in the finished cast. A distorted cast may result from an inverted impression. 6. As soon as the cast material has developed sufficient body, trim the excess from the sides of the cast. Wrap the im-pression and cast in a wet paper towel, or place it in a hu-midor, until the initial set of the stone has taken place. The impression is thus prevented from losing water by evapo-ration, which might deprive the cast material of sufficient water for crystallization. Chalky cast surfaces around the teeth are often the result of the hydrocolloid’s acting as a sponge and robbing the cast material of its necessary water for crystallization. 7. After the cast and impression have been in the humid at-mosphere for 30 minutes, separate the impression from the cast. Thirty minutes is sufficient for initial setting. Any stone that interferes with separation from the tray must be trimmed away with a knife. 8. Clean the impression tray immediately while the used im-pression material is still elastic. 9. Trimming of the cast should be deferred until final setting has occurred. The sides of the cast then may be trimmed to be parallel, and any blebs or defects resulting from air bubbles in the impression may be removed. If this is a cast for a permanent record, it may be trimmed to orthodontic specification to present a neat appearance for demonstra-tion purposes. Master casts and other working casts are ordinarily trimmed only to remove excess stone.
Possible Causes of an Inaccurate and/or a Weak Cast of a Dental Arch The possible causes of an inaccurate cast are as follows: 1. Distortion of the hydrocolloid impression (a) by use of an impression tray that is not rigid; (b) by partial dislodg-ment from the tray; (c) by shrinkage caused by dehydra-tion; (d) by expansion caused by imbibition (this will be toward the teeth and will result in an undersized rather than oversized cast); and (e) by attempting to pour the cast with stone that has already begun to set. 2. A ratio of water to powder that is too high. Although this may not cause volumetric changes in the size of the cast, it will result in a weak cast. 3. Improper mixing. This also results in a weak cast or one with a chalky surface. 4. Trapping of air, either in the mix or in pouring, because of insufficient vibration. 5. Soft or chalky cast surface that results from the retarding action of the hydrocolloid or the absorption of necessary water for crystallization by the dehydrating hydrocolloid. 6. Premature separation of the cast from the impression. 7. Failure to separate the cast from the impression for an ex-tended period.
INDIVIDUAL IMPRESSION TRAYS This chapter has previously dealt with making an impression in a rigid stock tray of the anatomic form of a dental arch for making a diagnostic cast, a working cast for restorations, or a master cast. There are times, however, when a stock tray is not suitable for making the final anatomic impression of the A B Figure 16-2 A, Stone is introduced at one posterior region of the impression, with care taken to trace the stone moving into each tooth as it rounds the arch. B, Additional stone is not needed until the stone has reached the opposite-most posterior tooth.
www.konkur.in 226 Part II Clinical and Laboratory dental arch. Most tooth-supported removable partial den-tures may be made on a master cast from such an impression. Some maxillary distal extension removable partial dentures with broad palatal coverage, particularly those for a Kennedy Class I arch, may also be made on an anatomic cast, but usu-ally these necessitate the use of an individually made tray.
A stock tray must be sufficiently rigid to avoid distor-tion during the impression and cast forming procedures and should fit the mouth with about 4 to 5 mm clearance for the impression material without interfering with teeth or bor-dering tissues; otherwise an individual tray made of some acrylic-resin tray material should be used for the final ana-tomic impression.
Most stock or disposable removable partial denture trays are of the rim-lock or perforated varieties. Both are made in a limited selection of sizes and shapes. Wide selections of trays are available that can be used for partially edentu-lous patients, including trays for both bilateral and unilateral edentulous areas.
All of these trays have reinforced borders. Although a complete denture impression tray is, or should be, made of material that permits trimming and shaping to fit the mouth, the existence of a beaded border and the rigidity of a stock removable partial denture tray allow no trimming and little shaping. The resulting impression is often a record of border tissues distorted by an ill-fitting tray rather than an impres-sion of tissues draping naturally over a slightly underex-tended impression tray.
An individual acrylic-resin tray, on the other hand, can be made with sufficient clearance for the impression material and can be trimmed just short of the vestibular reflections to allow the tissues to drape naturally without distortion. The remov-able partial denture borders may then be made as accurately as complete denture borders with equal advantages.
Although techniques have been proposed for making individual impression trays that incorporate plastic tubing for water-cooling reversible hydrocolloid impressions, the final anatomic impression usually will be made with irrevers-ible hydrocolloid, mercaptan rubber, or silicone impression materials.
Technique for Making Individual Acrylic-Resin Impression Trays The diagnostic cast is often adequate for preparation of the individual tray. However, if extensive surgery or extractions were performed after the diagnostic cast was made, a new impression in a rigid stock tray and a new cast must be made. The procedures for making the new cast are identical to those described previously.
A duplicate of the diagnostic cast, on which the individ-ual tray can be fabricated, should be made because the cast on which an individual tray is made is often damaged or must be mutilated to separate the tray from the cast. Obvi-ously the original diagnostic cast must be retained as a per-manent record in the patient’s file. Several techniques may be used to make individual impression trays. One technique for making an individual maxillary tray is described in Figures 16-3 and 16-4. This format could be used for both autopolymerizing acrylic-resin and visible light-cured (VLC) acrylic-resin. The VLC custom tray materials are premixed sheet materials that, when polymerized, provide a highly stable, distortion-free custom impression tray that is ready to use in minutes. These materials are provided by the manufacturers in sheet forms of various sizes, thick-nesses, and colors.
A technique for making an individual maxillary tray with light-polymerized resin is as follows: 1. Outline the extent of the tray on the cast with a pencil.
The tray must include all teeth and tissues that will be involved in the removable partial denture.
2. Adapt one layer of baseplate wax over the tissue surfaces and two layers over the teeth of the cast to serve as a spacer for impression material. The wax spacer should be trimmed 2 to 3 mm short of the outline drawn on the diagnostic cast. Wax covering the posterior palatal seal area should be removed so that intimate contact of the tray and tissue in this region may serve as an aid in cor-rectly orienting the tray when the impression is made.
Expose portions of the incisal edges of the central in-cisors to serve as anterior stops when placing the tray in the mouth. Bevel the wax so that the completed tray has a guiding incline that helps position the tray on the anterior stop. Other cast undercuts should be blocked out with wax or modeling compound. Note: Adapt an additional layer of baseplate wax over the teeth if the im-pression is to be made in irreversible hydrocolloid. This step is not necessary if the choice of impression material is a rubber-base or silicone type of material.
3. Paint the exposed surfaces of the cast that may be con-tacted by the light-polymerized resin tray material with a model release agent (MRA) to facilitate separation of the polymerized tray from the cast.
4. Remove the VLC tray material from the light-proof pouch and carefully cut the desired length with a knife or scalpel. Adapt the VLC material to the cast, and trim it with a knife. Be sure not to thin the material over the teeth or the posterior border area.
5. Attach a handle by molding excess VLC material into the desired shape, and blend it into the tray material in the cast. With some materials, a paper clip or similar wire may be shaped and used to reinforce the handle. Alter-natively, some manufacturers make prefabricated metal custom tray handles that may be easily adapted.
6. Place the cast with the adapted tray in the light polymer-izing unit and process according to the manufacturer’s directions—usually a maximum of 1 minute.
7. Remove the cast from the unit, and gently remove the tray from the cast. Peel the softened wax out of the tray while the wax is still warm.
8. Paint the entire impression tray with the manufacturer’s air barrier coating material, and return the tray to the unit turnstile for additional polymerizing, tissue side up.
www.konkur.in 227 Chapter 16 Impression Materials and Procedures for Removable Partial Dentures 9. When the polymerizing cycle is completed, remove the tray from the unit, and clean it with a brush and water. 10. Perfect the borders of the tray with rotary instruments (vulcanite burs, acrylic-resin trimmers, and so on), and slightly polish the external surface of the tray. 11. Place perforations (No. 8 bur size) in the VLC resin tray at 5-mm (3⁄16-inch) intervals, with the exception of the alveo-lar groove areas, if an irreversible hydrocolloid impression material is to be used (see Figures 16-3 and 16-4). 12. The finished tray must be sanitized and tried in the mouth so that any necessary corrections to the tray can be accomplished before the impression is made.
The technique for making an individual mandibular VLC resin tray follows the same procedures. The buccal shelf regions on the mandibular cast are not covered by the wax spacer because these areas provide the primary support for the mandibular removable partial denture and serve as poste-rior stops in orienting the tray in the patient’s mouth. During impression making, these areas permit selective placement of tissues in the mandibular stress-bearing areas.
If mercaptan rubber is to be used, perforations are not usually necessary to lock the material in the tray, because the adhesive provided by the manufacturer provides reli-able retention, and some confinement of these materials is desirable. However, a series of perforations are necessary in the median palatal raphe and incisive papilla areas of the maxillary tray so that excess impression material will escape through them, thus providing relief of the tissues in this area. For the same reasons, perforations are placed in the alveo-lar groove of the mandibular tray. With the use of adhesives, the impression material is not easily removed from the tray should a faulty impression have to be remade, but this is an inconvenience common to all newer elastic materials and does not prevent reuse of the impression tray. Opaque elastic impression materials and adhesives can prevent the detec-tion of undesirable pressure areas when an impression is evaluated.
Master casts made from impressions in individual acrylic-resin trays are generally more accurate than those made in rigid stock trays. The use of individual trays should A B C D Figure 16-3 A, Desired outline of the tray is drawn on the diagnostic cast. The tray must include all teeth and tissues that will be in-volved in the removable partial denture. B, One thickness of baseplate wax is adapted to the cast and is trimmed to the penciled outline, which is 2 to 3 mm short of the desired border. The posterior palatal seal region is not covered by wax but will be included in finished tray. Two thicknesses of baseplate wax cover the teeth. A window is created in the wax spacer over the incisal edges. C, A model release agent (MRA) is painted on the stone surfaces of the cast that will be contacted by the resin. D, The visible light-cured (VLC) resin tray material is removed from the light-proof wrap and shaped to the desired outline in a uniform manner. Continued www.konkur.in 228 Part II Clinical and Laboratory be considered a necessary step in making the majority of removable partial dentures when a secondary impression technique is not to be used. Reasons and methods for mak-ing a secondary impression will be considered in Chapter 17.
Final impressions for maxillary tooth–supported remov-able partial dentures often may be made in carefully selected and recontoured rigid stock impression trays. However, an individual acrylic-resin tray is preferred in those situations in the mandibular arch when the floor of the mouth closely approximates the lingual gingiva of the remaining anterior teeth. Recording the floor of the mouth at the elevation it assumes when the lips are licked is important in selecting the type of major connector to be used (see Chapter 5). Modification of the borders of an individual tray to fulfill the requirements of an adequate tray is much easier than is the modification of a metal stock tray.
E F G Figure 16-3, cont’d E, A handle is added to provide a means to place and remove the tray, as well as to pass the tray from assistant to dentist. Its form should consider the lip length and need to manipulate the perioral region. F, Before the tray is placed in the curing oven, an air barrier coating is painted on the surface. The tray is then polymerized as per manufacturers’ recommendations. G, As soon as the tray material has hardened, the tray is removed from the cast, and the wax spacer is removed from the rough tray. An acrylic-resin trimmer in the lathe is used to rough finish the tray. Holes are drilled through the tray, spaced approximately 4.5 mm apart. These holes serve to lock the impression material in the tray. In addition, excess impression material is forced out of the holes when the impression is made, thereby minimally displacement of soft oral tissues. These two features assist in correctly orienting the individualized impression tray in the mouth.
www.konkur.in 229 Chapter 16 Impression Materials and Procedures for Removable Partial Dentures A B C D E Figure 16-4 A similar technique to the one used for fabrication of the maxillary tray in Figure 16-3 is used for the mandibular tray. A, Outline of the tray is penciled on a duplicate mandibular diagnostic cast. B, A single sheet of baseplate wax is adapted to the outline of the tray, and another sheet of baseplate wax is adapted over the teeth. A window is cut in the spacer to expose the incisal edges of the lower central incisors to serve as a stop in seating the tray. C, A model release agent (MRA) is painted on regions of the cast to be in contact with the resin. D, A visible light-cured (VLC) tray material wafer is adapted over the cast and spacer. E, A handle is formed with excess tray material, as previously described. Continued www.konkur.in 230 Part II Clinical and Laboratory F G Figure 16-4, cont’d F, An air barrier coating is painted on the tray material and it is processed as described in Figure 16-3. G, Following processing, multiple holes are placed throughout the tray.
www.konkur.in CHAPTER 17 Support for the Distal Extension Denture Base CHAPTER OUTLINE Distal Extension Removable Partial Denture Factors Influencing the Support of a Distal Extension Base Contour and Quality of the Residual Ridge Extent of Residual Ridge Coverage by the Denture Base Type and Accuracy of the Impression Registration Accuracy of the Fit of the Denture Base Design of the Removable Partial Denture Framework Total Occlusal Load Applied Anatomic form Impression Methods for Obtaining Functional Support for the Distal Extension Base Selective Tissue Placement Impression Method Framework Fitting Functional Impression Technique In a tooth-supported removable partial denture, a metal base or the framework that supports an acrylic-resin base is connected to and is part of a rigid framework that permits the direct transfer of occlusal forces to the abutment teeth through the occlusal rests. Even though the denture base of the modification space(s) in a Kennedy Class III remov-able partial denture provides support for the supplied teeth, the residual ridge beneath the base is not called on to aid in the support of the removable partial denture. Therefore the resiliency of the ridge tissues, the ridge configuration, and the type of bone that supports these tissues are not factors in denture support. Regardless of the length of the edentu-lous spans, if the framework is rigid, the abutment teeth are sound enough to carry the additional load, and the occlusal rests are properly formed, support comes entirely from the abutment teeth at either end of that span. Support may be augmented by splinting and by the use of additional abut-ments, but in any event the abutments are the sole support of the removable restoration. When a dental implant is used to augment residual ridge support, the implant provides the same role of support as a distal tooth supporting a tooth-borne posterior segment of a removable partial denture.
An impression (and resultant stone cast) records the ana-tomic form of the teeth (and/or implant) and their surrounding structures and is needed to make a tooth-supported remov-able partial denture. The impression should also record the moving tissues that will border the denture in an unstrained position so that the relationship of the denture base to those tissues may be as accurate as possible. Although underexten-sion of the denture base in a tooth-supported prosthesis is the lesser of two evils, an underextended base may lead to food entrapment and inadequate facial contours, particularly on the buccal and labial sides. To accurately record the moving tissues of the floor of the mouth, an individual impression tray should be used, rather than an ill-fitting or overextended stock tray. This has been discussed at length in Chapters 5 and 16.
www.konkur.in 232 Part II Clinical and Laboratory DISTAL EXTENSION REMOVABLE PARTIAL DENTURE The distal extension removable partial denture does not have the advantage of total tooth support because one or more bases are extensions covering the residual ridge distal to the last abutment. It therefore is dependent on the residual ridge for a portion of its support.
The distal extension removable partial denture must depend on the residual ridge for some support, stability, and retention. Indirect retention, to prevent the denture from lifting away from the residual ridge, should also be incorporated in the design. The tooth-supported base is secured at either end by the action of a direct retainer and is supported at either end by a rest, whereas this degree of support and direct retention is lacking in the distal extension prosthesis. For this reason, a distal abutment should be preserved whenever possible. In the event of the loss or absence of a distal abutment tooth, the patient must be made aware of the movements to be expected with a distal extension removable partial denture and the limitations imposed on the dentist when the residual ridge must be used for support, stabil-ity, and retention for that part of the prosthesis.
FACTORS INFLUENCING THE SUPPORT OF A DISTAL EXTENSION BASE Because one of the stated objectives of prosthodontic treat-ment is the restoration of function and comfort in an estheti-cally pleasing manner, maintenance of occlusal contact in distal extension removable partial dentures demands an understanding of the factors that influence residual ridge support. Support from the residual ridge becomes more important as the distance from the last abutment increases and will depend on the following several factors: • Contour and quality of the residual ridge • Extent of residual ridge coverage by the denture base • Type and accuracy of the impression registration • Accuracy of the fit of the denture base • Design of the removable partial denture framework • Total occlusal load applied Even when ideal residual ridge characteristics that favor support are present, the ridge is less capable of resisting func-tional loads than either a tooth or implant. Because of this, the impact of a selective implant on the anticipated func-tional comfort and efficiency of a distal extension removable partial denture should be discussed and considered when indicated based on patient and recipient site factors.
Contour and Quality of the Residual Ridge The ideal residual ridge to support a denture base would con-sist of cortical bone that covers relatively dense cancellous bone, with a broad rounded crest with high vertical slopes, and is covered by firm, dense, fibrous connective tissue. Such a residual ridge would optimally support vertical and hori-zontal stresses placed on it by denture bases. Unfortunately this ideal is seldom encountered.
Easily displaceable tissue will not adequately support a denture base, and tissues that are interposed between a sharp, bony residual ridge and a denture base will not remain in a healthy state. Not only must the nature of the bone of the residual ridge be considered in developing optimum support for the denture base but also its positional relationship to the direction of forces that will be placed on it.
The crest of the bony mandibular residual ridge is most often cancellous. Because lining mucosa restricts both the buccal and lingual mucosae adjacent to teeth in the mandi-ble, loss of firm mucosa overlying the residual ridge is com-mon following tooth extraction in the posterior mandible. Pressures placed on tissues overlying the crest of the man-dibular residual ridge usually result in irritation of these tis-sues, accompanied by the sequelae of chronic inflammation. Therefore the crest of the mandibular residual ridge cannot be a primary stress-bearing region. The buccal shelf region (bounded by the external oblique line and the crest of the alveolar ridge) seems to be better suited for a primary stress-bearing role because it is covered by relatively firm, dense, fibrous connective tissue supported by cortical bone. In most instances this region bears more of a horizontal relationship to vertical forces than do other regions of the residual ridge (Figure 17-1). The slopes of the residual ridge then would become the primary stress-bearing region for resistance of horizontal and off-vertical forces.
The immediate crest of the bone of the maxillary residual ridge may consist primarily of cancellous bone. Unlike in the mandible, oral tissues that overlie the maxillary residual alve-olar bone are usually of a firm, dense nature (similar to the Figure 17-1 The dotted portion outlines the crest of the re-sidual ridge, which should be recorded in its anatomic form in impression procedures. Similarly, retromolar pads should not be displaced by impression. Buccal shelf regions are outlined by a her-ringbone pattern, and selected additional pressures may be placed on these regions for vertical support of the denture base. Lingual slopes of the residual ridge (cross-hatched) may furnish some verti-cal support to the restoration; however, these regions principally resist the horizontal rotational tendencies of the denture base and should be recorded by the impression in undisplaced form.
www.konkur.in 233 Chapter 17 Support for the Distal Extension Denture Base mucosa of the hard palate) or can be surgically prepared to support a denture base. The topography of a partially eden-tulous maxillary arch imposes a restriction on selection of a primary stress-bearing area. In spite of impression procedures, the crestal area of the residual ridge will become the primary stress-bearing area for vertically directed forces. Some resis-tance to these forces may be obtained by the immediate buc-cal and lingual slopes of the ridge. Palatal tissues between the medial palatal raphe and the lingual slope of the posterior edentulous ridge are readily displaceable and cannot be con-sidered as primary stress-bearing sites (Figure 17-2). The tis-sues covering the crest of the maxillary residual ridge must be less displaceable than the tissues that cover palatal areas, or relief of palatal tissues must be provided in the denture bases or for palatal major connectors.
Extent of Residual Ridge Coverage by the Denture Base The broader the residual ridge coverage, the greater is the distribution of the load, which results in less load per unit area (Figure 17-3). A denture base should cover as much of the residual ridge as possible and should be extended the maximum amount within the physiologic tolerance of the limiting border structures or tissues. Knowledge of these border tissues and the structures that influence their move-ment is paramount to the development of broad coverage denture bases. In a series of experiments, Kaires has shown From Kaires AK: Effect of partial denture design on bilateral force dis-tribution, J Prosthet Dent 6:373-389, 1956.
that “maximum coverage of denture-bearing areas with large, wide denture bases is of the utmost importance in withstand-ing both vertical and horizontal stresses.” It is not within the scope of this text to review the ana-tomic considerations related to denture bases. The student is referred to several articles listed in the “Selected Reading Resources” regarding this subject.
Type and Accuracy of the Impression Registration The residual ridge may be said to have two forms: the ana-tomic form and the functional form (Figure 17-4). The ana-tomic form is the surface contour of the ridge when it is not supporting an occlusal load. The functional form of the residual ridge is the surface contour of the ridge when it is supporting a functional load.
The anatomic form is recorded by a soft impression mate-rial, such as a metallic oxide impression paste, if the entire impression tray is uniformly relieved. Depending on the viscosity of the particular impression material used and the rigidity of the impression tray, it is also the form that can be recorded by mercaptan rubber, silicone, and hydrocolloid impression materials. Distortion and tissue displacement by pressure may result from confinement of the impression mate-rial within the tray and from insufficient thickness of impres-sion material between the tray and the tissues, as well as from the viscosity of the impression material; however, none of these factors is selective or physiologic in its action. These acciden-tal distortions of the tissues occur because of faulty technique. Use of the anatomic form of the residual ridge in fabricating complete dentures is quite common because of a belief that this is the most physiologic form for support of the dentures.
However, many other dentists believe that certain regions of the residual ridge(s) in a partially edentulous patient are more capable of supporting dentures than other regions. Their impression methods are directed to place more stress on pri-mary stress-bearing regions with specially constructed indi-vidual trays and at the same time record the anatomic form of other basal seat tissues, which cannot assume a stress-bearing role. The form of the residual ridge recorded under some loading, whether by occlusal loading, finger loading, specially designed individual trays, or the consistency of the recording medium, is called the functional form. This is the surface con-tour of the ridge when it is supporting a functional load. How much it differs from the anatomic form depends on the thick-ness and structural characteristics of the soft tissues overlying the residual bone. It will also differ from the anatomic form in proportion to the total load applied to the denture base. Of the two philosophies, the latter seems to be more logical.
McLean and others recognized the need to record the tis-sues that support a distal extension removable partial denture base in their functional form, or supporting state, and then relate them to the remainder of the arch by means of a second-ary impression. This was called a functional impression because it recorded the ridge relation under simulated function.
Any method, whether it records the functional relation-ship of the ridge to the remainder of the arch, or the functional Figure 17-2 The crest of the maxillary residual ridge (herring-bone pattern) is the primary supporting region for the maxillary distal extension denture base. Buccal and lingual slopes may furnish lim-ited vertical support to the denture base. It seems logical that their primary role is to counteract the horizontal rotational tendencies of the denture base. The dotted portion outlines the incisive papilla and the median palatal raphe. Relief must be provided for these regions, especially if tissues covering the palatal raphe are less displaceable than those covering the crest of the residual ridge.
www.konkur.in 234 Part II Clinical and Laboratory form of the ridge itself, may provide acceptable support for the removable partial denture. On the other hand, those who use the anatomic ridge form for the removable partial den-ture should seriously consider the need for some mechanical stress-breaker to avoid the possible cantilever action of the distal extension base against the abutment teeth.
Steffel has classified advocates of the various methods for treating the distal extension removable partial denture as follows: 1. Those who believe that ridge and tooth supports can best be equalized by the use of stress-breakers or resilient equalizers. 2. Those who insist on bringing about the equalization of ridge and tooth support by physiologic basing, which is accomplished by a pressure impression or by relining of the denture under functional stresses. 3. Those who uphold the idea of extensive stress distribution for stress reduction at any one point.
It would seem that there is little difference in the philoso-phy behind methods 2 and 3 as given by Steffel, for both the equalization of tooth and tissue support and stress distribu-tion over the greatest area are objectives of the functional type of impression. Many of the requirements and advan-tages that are associated with the distributed stress denture apply equally well to the functionally or physiologically based denture. Some of these requirements are (1) posi-tive occlusal rests; (2) an all-rigid, nonflexible framework; (3) indirect retainers to add stability; and (4) well-adapted, broad coverage bases.
Those who do not accept the theory of physiologic basing, for one reason or another, should use some form of stress-breaker between the abutment and the distal extension base. A B Figure 17-4 Comparison of anatomic and functional ridge forms. A, Original master cast with the edentulous area recorded in its anatomic form, using elastic impression material. B, Same cast after the edentulous area has been repoured to its functional form as recorded by the secondary impression.
A B Figure 17-3 Comparison of two removable partial dentures for the same patient. A, This is the replacement denture, with properly extended bases. B, This denture has severely underextended bases. Occlusal forces are more readily distributed to denture-bearing areas by the replacement denture (A).
www.konkur.in 235 Chapter 17 Support for the Distal Extension Denture Base The advantages and disadvantages of doing so have been given in Chapter 9.
Accuracy of the Fit of the Denture Base Support of the distal extension base is enhanced by intimacy of contact of the tissue surface of the base and the tissues that cover the residual ridge. The tissue surface of the denture base must optimally represent a true negative of the basal seat regions of the master cast. Denture bases have been dis-cussed in Chapter 9.
In addition, the denture base must be related to the removable partial denture framework in the same manner as the basal seat tissues were related to the abutment teeth when the impression was made. Every precaution must be taken to ensure this relationship when the altered cast technique of making a master cast is used.
Design of the Removable Partial Denture Framework Some rotation movement of a distal extension base at the distal abutment is inevitable under functional loading. It must be remembered that the extent to which abutments are subjected to rotational and torquing forces that result from masticatory function is directly related to the position and resistance of the food bolus. The greatest movement takes place at the most posterior extent of the denture base. The retromolar pad region of the mandibular residual ridge and the tuberosity region of the maxillary residual ridge there-fore are subjected to the greatest movement of the denture base (Figure 17-5). Steffel and Kratochvil have suggested that as the rotational axis is moved from a disto-occlusal rest to a more anterior location, more of the residual ridge receives vertically directed occlusal forces to support the denture base (Figure 17-6). They have suggested that occlu-sal rests may be moved anteriorly to better use the resid-ual ridge for support without jeopardizing either vertical or horizontal support of the denture by occlusal rests and guiding planes (Figure 17-7).
It is possible, however, that the proximal plate minor connector adjacent to the edentulous area will not disen-gage from or break contact with the guiding plane. When one considers impression and cast formation variables; A C B Figure 17-5 Acute dip of the short denture base is com-pared with that of the long one in the top drawing. In the bot-tom drawing, when the point of rotation is changed from C to B by the loss of more teeth, it can be seen that a proportionally greater area of the residual ridge is used to support the denture base than occurs when the fulcrum line passes through C. The amount of movement is directly related to the quality of tis-sue support. Line AC represents the length of the denture base. (See also Figure 10-3.) Figure 17-6 If rotation of the distal extension base occurs around the nearest rest, as the rest is moved anteriorly more of the residual ridge will be used to resist rotation. Compare the vertical arcs of the long-dash broken line with the arcs of the solid line. (See also Figure 10-4.) Figure 17-7 The occlusal rest is placed on the mesio-occlusal surface of the lower first premolar; this will move the point of rotation anterior to the conventionally placed disto-occlusal rest if contact of the proximal minor connector on the distal guiding plane is designed to release under function. The occlusal rest is connected to the lingual bar by a minor connector, which con-tacts the small mesiolingual prepared guiding plane. Note the vertical extension of the denture base minor connector contact-ing the distal guiding-plane surface. The lingual guiding plane is prepared to extend from the occlusal surface inferiorly to ap-proximately one third of the height of the lingual surface and is as broad as the contacting minor connector. The distal guiding plane extends from the distal marginal ridge gingivally about two thirds of the height of the distal surface. Such preparations must be designed not to lock the tooth in a viselike grip when the den-ture base moves toward the residual ridge.
www.konkur.in 236 Part II Clinical and Laboratory waxing, investing, and casting discrepancies; and finishing and polishing procedures, it may be somewhat philosophi-cal to assume that the minor connector proximal plate will have contact, and if it does, will disengage its contact with the guiding plane, especially if the tissues that support the extension denture base are healthy and demonstrate favor-able contour. It is more likely that the abutment tooth will move physiologically to contact the minor connector and that the disengagement will depend on the amount of tis-sue displacement. Geometrically, the actual amount of tissue displacement in the extension base area under occlusal load may not be enough to cause the minor connector to break contact with the guiding plane.
Total Occlusal Load Applied Patients with distal extension removable partial dentures generally orient the food bolus over natural teeth rather than prosthetic teeth. This is likely due to the more stable nature of the natural dentition, the proprioceptive feedback it pro-vides for chewing, and the possible nociceptive feedback from the supporting mucosa. This has an effect on the direc-tion and magnitude of the occlusal load to the removable partial denture and, thus, on the load transferred to the abut-ments. Given this, the support from the residual ridge should be optimized and shared appropriately with the remaining natural dentition. When the opposing natural teeth are super-erupted, control of eccentric functional contacts and the resultant horizontal loads makes chronic discomfort a clinical challenge. This is best managed by reducing the occlusal plane irregularity and enhancing extension base sta-bility through the use of a dental implant.
The number of artificial teeth, the width of their occlu-sal surfaces, and their occlusal efficiency influence the total occlusal load applied to the removable partial denture. Kaires conducted an investigation under laboratory con-ditions and concluded that “the reduction of the size of the occlusal table reduces the vertical and horizontal forces that act on the removable partial dentures and lessens the stress on the abutment teeth and supporting tissues.” ANATOMIC FORM IMPRESSION The anatomic form impression is a one-stage impression method using an elastic impression material that will pro-duce a cast that does not represent a functional relationship between the various supporting structures of the partially edentulous mouth. It will represent only the hard and soft tissues at rest. With the removable partial denture in posi-tion in the dental arch, the occlusal rest(s) will fit the rest seat(s) of the abutment teeth, whereas the denture base(s) will fit the surface of the mucosa at rest. When a mastica-tory load is applied to the extension base(s) with a food From Kaires AK: Effect of partial denture design on bilateral force dis-tribution, J Prosthet Dent 6:373-389, 1956.
bolus, the rest(s) will act as a definite stop, which will limit the part of the base near the abutment tooth from trans-mitting the load to the underlying anatomic structures. The distal end of the base(s), however, that is able to move more freely, will transmit more of the masticatory load to the underlying extension base tissues and will transmit more torque to the abutment teeth through the rigid removable partial denture framework.
It is obvious that the soft tissues that cover the ridge can-not by themselves carry any load applied to them. They act as a protective padding for the bone, which in the final analysis is the structure that receives and resists the masticatory load. Distribution of this load over a maximum area of bone is a prime requisite in preventing trauma both to the tissues of the extension base areas and to the abutment teeth.
A removable partial denture fabricated from a one-stage impression, which records only the anatomic form of basal seat tissues, places more of the masticatory load on the abut-ment teeth and that part of the bone that underlies the distal end of the extension base. The balance of the bony ridge will not function in carrying its share of the load. The result will be a traumatic load to the bone underlying the distal end of the base and to the abutment tooth, which in turn can result in bone loss and loosening of the abutment tooth. A properly prepared, individualized impression tray can be used to record the primary stress-bearing areas in a functional form and the non–stress-bearing areas in an anatomic form, just as is often accomplished in making impressions for complete dentures.
Some dentists believe that every removable partial denture should be relined before its final placement in the mouth. Some believe that tissue can be evenly displaced and use impression materials of heavy consistency. This latter practice can intro-duce traumatic stresses to the underlying tissues. Some den-tists use free-flowing pastes that produce an impression of the soft tissues at rest. A removable partial denture made accord-ing to this technique will be similar to a removable partial denture fabricated from a one-piece impression. The occlusal rest will act as a stop and prevent an even distribution of the masticatory load by the base to the edentulous ridge.
METHODS FOR OBTAINING FUNCTIONAL SUPPORT FOR THE DISTAL EXTENSION BASE The objective of any functional impression technique is to provide maximum support for the removable partial den-ture bases. This allows for the maintenance of occlusal con-tact between natural and artificial dentition and, at the same time, minimal movement of the base, which would create leverage on the abutment teeth. Although some tissue-ward movement of the distal extension base is unavoidable and is dependent on the six factors listed previously, it can be mini-mized by providing the best possible support for the denture base.
A thorough understanding of the characteristics of each of the impression materials and impression methods leads to the conclusion that no single material can record the www.konkur.in 237 Chapter 17 Support for the Distal Extension Denture Base anatomic form of the teeth and tissues in the dental arch and, at the same time, the functional form of the residual ridge. Therefore, some secondary or corrected impression method must be used.
The method selected is greatly influenced by determina-tion of the support potential of the residual ridge mucosa. Mucosa that is firm and minimally displaceable provides a different support potential than mucosa that is more eas-ily displaced. Methods for obtaining functional support for either should satisfy the two requirements for providing adequate support to the distal extension removable partial denture base. These are (1) that it records and relates the sup-porting soft tissue under some loading; and (2) that it dis-tributes the load over as large an area as possible.
Selective Tissue Placement Impression Method Soft tissues that cover basal seat areas may be placed, dis-placed, or recorded in their resting or anatomic form. Placed and displaced tissues differ in the degree of alteration from their resting form and in their physiologic reaction to the amount of displacement. For example, the palatal tissues in the vicinity of the vibrating line can be slightly displaced to develop a posterior palatal seal for the maxillary complete denture and will remain in a healthy state for extended periods. On the other hand, these tissues develop an imme-diate inflammatory response when they have been overly displaced in developing the posterior palatal seal.
Oral tissues that have been overly displaced or distorted attempt to regain their anatomic form. When they are not permitted to do this by the denture bases, the tissues become inflamed and their physiologic functions become impaired, accompanied by bone resorption. Tissues that are minimally displaced (placed) by impression procedures for definitive border control respond favorably to the additional pressures placed on them by the resultant denture bases if these pres-sures are intermittent rather than continuous.
The selective tissue placement impression method is based on these clinical observations, the histologic nature of tissues that cover the residual alveolar bone, the nature of the residual ridge bone, and its positional relationship to the direction of stresses that will be placed on it. It is further believed that by use of specially designed individual trays for impressions, den-ture bases can be developed that will use those portions of the residual ridge that can withstand additional stress and at the same time relieve the tissues of the residual ridge that cannot withstand functional loading and remain healthy.
There should be no philosophical difference between the requirement of support and coverage by bases of distal exten-sion removable partial dentures and that of complete den-tures, either maxillary or mandibular, because the objective of maximum support is the same. The tray is unquestionably the most important part of an impression. However, a tray must be so formed and modified that the impression phi-losophy of the dentist can be carried out. Making individual-ized acrylic-resin impression trays is described in Chapter 16, and a method of attaching custom trays to a removable partial denture framework is illustrated in Figure 17-8.
Framework Fitting Because the goal is to maximize soft tissue support while using the teeth to their supportive advantage, the framework fitted to the teeth while soft tissue support is registered pro-vides a means of coordinating both. This means that before the trays are attached, the framework must be fitted in the mouth as illustrated in Figure 17-9. Fitting the framework involves the following steps: 1. Use of disclosing media to identify interferences to com-pletely seating the removable partial denture framework. 2. Use of disclosing media to identify the appropriate contact(s) of the component parts of the framework dur-ing seating of the framework and when the framework is completely seated in its designated terminal position.
A B Figure 17-8 A secondary impression for the distal extension mandibular removable partial denture is made in individual trays at-tached to the denture framework. A, The framework has been tried in the mouth and fits the mouth and master cast as planned. B, The outline of acrylic-resin trays is penciled on the cast. Continued www.konkur.in 238 Part II Clinical and Laboratory E F C D G Figure 17-8, cont’d C, One thickness of baseplate wax is adapted to outlines to act as spacers so that room for the impression material exists in the finished trays. Windows are cut in the wax spacers corresponding to regions on the cast contacted by minor connec-tors (tissue stops) for denture bases. D, The framework is warmed and pressed to position on the relieved master cast. All regions of the cast that will be contacted by autopolymerizing acrylic-resin or visible light-cured (VLC) resin are painted with tinfoil substitute (Alcote) or model release agent (MRA). E, Acrylic-resin material is adapted to the cast and over the framework with finger pressure as described in Chapter 16. Excess material over the borders of the cast is removed with a sharp knife while the material is still soft. F, Polymerized acrylic-resin trays and framework are removed from the cast, and trays are trimmed to outline the wax spacer. G, Borders of the trays will be adjusted to extend 2 mm short of the tissue reflections. Holes will be placed in the trays corresponding to the crest of the residual ridge and retromolar pads to allow escape of excess impression material when an impression is made.
www.konkur.in B C D E A Figure 17-9 The framework must be evaluated to assure complete seating, full contact with the remaining dentition for stabiliza-tion, support, and retention as planned, and to allow full natural tooth contact. A, Several types of disclosing media may be used, such as stencil correction fluid, rouge, and chloroform, and disclosing fluids, pastes, and waxes. Here, a spray disclosing medium has been applied and the framework is placed with mild pressure. Incomplete seating is seen when the framework binds. It is imperative that the framework not be forced to place at this initial seating. B, A portion of the proximal plate is preventing complete seating. C, The framework is carefully adjusted as over-adjustment can result in a poorly adapted framework. D, The framework seats completely after adequate adjustment. This may require repeated disclosing and careful adjustment; however, if improvement is not seen with each framework modification, there should be concern regarding frame accuracy. E, Following complete seating and verification of appropriate tooth contacts by component parts (i.e., rests, proximal plates, stabilizing components), the occlusion must be checked and the frame-work adjusted until natural tooth contacts that exist without the framework seated are achieved with the framework in place. All adjusted regions can be carefully polished with rotary rubber points.
www.konkur.in 240 Part II Clinical and Laboratory 3. Adjusting the seated framework to the opposing occlusion.
If there are opposing frameworks, the maxillary frame-work is removed from the mouth and the mandibular frame-work is adjusted to the natural maxillary dentition; then the maxillary framework is replaced and it is adjusted to the man-dibular dentition with its framework in place. It is important to remember that the metal frameworks must allow all of the natural dentition to maintain the same designed contact relationship with the opposing arch as when the frameworks are out of the mouth. After the framework has been fitted and the custom trays have been attached, selective tissue placement impression and cast formation can be accom-plished as described in Figure 17-10.
The altered cast method of impression making is most commonly used for the mandibular distal extension par-tially edentulous arch (Kennedy Class I and Class II arch forms). A common clinical finding in these situations is greater variation in tissue mobility and tissue distortion or displaceability, which requires some selective tissue place-ment to obtain the desired support from these tissues. This variability in tissue mobility is probably related to the pat-tern of mandibular residual ridge resorption. Altered cast C D A B E Figure 17-10 Selective tissue placement impression technique. A, Tray attached at the frame try-in, which in B is seen incompletely seated. C, Completed border molding, which defines the primary bearing areas of the buccal shelves bilaterally. This bearing area and the lingual extension are seen in the final impression (D), which can be seen to be in contact with use of the pressure-indicating paste (E).
www.konkur.in 241 Chapter 17 Support for the Distal Extension Denture Base impression methods are seldom used in the maxillary arch because of the nature of the masticatory mucosa and the amount of firm palatal tissue present to provide soft tissue support. These tissues seldom require placement to provide the required support. If excessive tissue mobility is present, it is often best managed by surgical resection, as this is a primary supporting area.
Support is obtained from the primary support areas in the manner by which the flow of the impression material is con-trolled during the impression-making procedure. Restricting the flow of the material in the primary stress-bearing areas (by minimizing the amount of relief over the area when the custom tray was made) causes greater pressure to be exerted on the tissues in this area (compared with other areas of unrestricted flow where a greater amount of relief or venting of the impression tray was provided). This is often referred to as the “selected pressure” or “dynamic” impression method. By controlling the flow of the impression material with wax relief and venting, one can place or displace the soft tissues over the primary support areas so that they are the primary areas to provide support to the denture base when a remov-able partial denture is functionally loaded.
An impression for a mandibular distal extension partially edentulous arch may also be adequately made in an individu-alized, complete arch tray. To do so, not only must the tray be formed to provide proper space for the particular impres-sion materials, but provision must be made so that the func-tional form of primary stress-bearing areas can be recorded. Such an impression procedure, properly executed, can be used when metal bases are to be incorporated in the design of the restoration. There is little difference, if any, between recording the basal seats in the partially edentulous arch and recording like areas for complete dentures on an edentulous arch. A secondary impression made in custom trays attached to the framework only makes definitive border control and tissue placement a bit easier, compared with the individual-ized complete arch tray.
Functional Impression Technique When the residual ridge mucosa demonstrates a uniformly firm consistency, an impression technique that involves cap-turing the tissue form while the patient is in occlusion can be considered. Such a technique records the mucosal position and shape under the influence of a static closure force, simi-lar to functional masticatory forces. The more the mucosa displaces under function, the more rebound there is likely to be. Because the prosthesis will be under occlusal load for only a portion of a day, minimal rebound is desired so as to maintain the clasp assembly–tooth relationship. When such a technique is applied to firm, minimally displaceable mucosa, a minimal rebound effect is seen on the prosthesis position. The selective pressure technique (described earlier) can be applied to all varieties of residual ridges because it is customized to mucosal conditions, whereas the functional impression technique has limited application to a uniformly firm ridge consistency.
www.konkur.in CHAPTER 18 Occlusal Relationships for Removable Partial Dentures CHAPTER OUTLINE Desirable Occlusal Contact Relationships for Removable Partial Dentures Methods for Establishing Occlusal Relationships Direct Apposition of Casts Interocclusal Records with Posterior Teeth Remaining Occlusal Relations Using Occlusion Rims on Record Bases Jaw Relation Records Made Entirely on Occlusion Rims Establishing Occlusion by the Recording of Occlusal Pathways Materials for Artificial Posterior Teeth Arranging Teeth to an Occluding Template Establishing Jaw Relations for a Mandibular Removable Partial Denture Opposing a Maxillary Complete Denture The fourth phase in the treatment of patients with remov-able partial dentures is the establishment of a functional and harmonious occlusion. Occlusal harmony between a removable partial denture and the remaining natural teeth is a major factor in the preservation of the health of their sur-rounding structures. In the treatment of patients with com-plete dentures, the inclination of the condyle path is the only factor not within the control of the dentist. All other factors may be altered to obtain occlusal balance and harmony in eccentric positions to conform to a particular concept and philosophy of occlusion.
Balanced occlusion is desirable with complete dentures because unbalanced occlusal stresses may cause instability of the dentures and trauma to the supporting structures. These stresses can reach a point that causes movement of the den-ture bases. In removable partial dentures, however, because of the attachment of the removable partial denture to the abutment teeth, occlusal stresses can be transmitted directly to the abutment teeth and other supporting structures, resulting in sustained stresses that may be more damaging than those transient stresses found in complete dentures. Failure to provide and maintain adequate occlusion on the removable partial denture is primarily a result of (1) lack of support for the denture base; (2) the fallacy of establishing occlusion to a single static jaw relation record; and (3) an unacceptable occlusal plane.
In establishing occlusion on a removable partial denture, the influence of the remaining natural teeth is usually such that the occlusal forms of the teeth on the removable partial denture must be made to conform to an already established occlusal pattern. Occlusal adjustment or restoration may have altered this pattern. However, the pattern present at See Chapter 2, under discussion on the phases of removable partial denture service.
www.konkur.in 243 Chapter 18 Occlusal Relationships for Removable Partial Dentures the time the removable partial denture is made dictates the occlusion on the removable partial denture. The only excep-tions are those in which an opposing complete denture can be made to function harmoniously with the removable par-tial denture, or in which only anterior teeth remain in both arches and the incisal relationship can be made so that tooth contacts do not disturb denture stability or retention. In these situations, jaw relation records and the arrangement of the teeth may proceed in the same manner as with complete dentures, and the same general principles apply.
With all other types of removable partial dentures, the remaining teeth dictate the occlusion. The dentist should strive for planned contacts in centric occlusion and no interferences in lateral excursions. Although a functional relationship of the removable partial denture to the natural dentition sometimes may be adjusted satisfactorily in the mouth, extraoral adjustment is often easier for both dentist and patient, is more accurate, and can be accomplished in a more comprehensive manner.
Establishment of a satisfactory occlusion for the removable partial denture patient should include the following: (1) analy-sis of the existing occlusion; (2) correction of existing occlusal disharmony; (3) recording of centric relation or an adjusted centric occlusion; (4) harmonizing of eccentric jaw move-ments for a functional eccentric occlusion; and (5) correction of occlusal discrepancies created by the fit of the framework and during processing of the removable partial denture.
DESIRABLE OCCLUSAL CONTACT RELATIONSHIPS FOR REMOVABLE PARTIAL DENTURES The following occlusal arrangements are recommended to develop a harmonious occlusal relationship among remov-able partial dentures and to enhance stability of the remov-able partial dentures: 1. Simultaneous bilateral contacts of opposing posterior teeth must occur in centric occlusion. 2. Occlusion for tooth-supported removable partial den-tures may be arranged similarly to the occlusion seen in a harmonious natural dentition, because stability of such removable partial dentures results from the effects of di-rect retainers at both ends of the denture base. 3. Bilateral balanced occlusion in eccentric positions should be formulated when a maxillary complete den-ture (Figure 18-1) opposes the removable partial denture.
This is accomplished primarily to promote the stability of the complete denture. However, simultaneous contacts in a protrusive relationship do not receive priority over appearance, phonetics, and/or a favorable occlusal plane. 4. Working side contacts should be obtained for the man-dibular distal extension denture (Figure 18-2). These contacts should occur simultaneously with working side contacts of the natural teeth to distribute the stress over the greatest possible area. Masticatory function of the denture is improved by such an arrangement. 5. Simultaneous working and balancing contacts should be formulated for the maxillary bilateral distal extension re-movable partial denture whenever possible (Figure 18-3).
Such an arrangement will compensate in part for the un-favorable position that the maxillary artificial teeth must occupy in relation to the residual ridge, which is usually lateral to the crest of the ridge. However, this desirable relationship often must be compromised when the pa-tient’s anterior teeth have an excessively steep vertical overlap with little or no horizontal overlap. Even in this situation, working side contacts can be obtained without resorting to excessively steep cuspal inclinations. 6. Only working contacts need to be formulated for the max-illary or mandibular unilateral distal extension removable partial denture (Figure 18-4). Balancing side contacts would not enhance the stability of the denture because it is entirely tooth supported by the framework on the bal-ancing side.
Figure 18-1 Posterior occlusion of a maxillary complete den-ture opposing a Class I mandibular removable partial denture. The stability of the maxillary complete denture can be promoted by developing balanced occlusion as shown.
Figure 18-2 Bilateral distal extension mandibular removable partial denture opposed by natural dentition in the maxillary arch. Working contacts are achieved, balancing contacts are purpose-fully avoided because they would not enhance the stability of the restoration, and protrusive balance is avoided in favor of an acceptable appearance and a favorable occlusal plane.
www.konkur.in 244 Part II Clinical and Laboratory 7. In the Kennedy Class IV removable partial denture configuration, contact of opposing anterior teeth in the planned intercuspal position is desired to prevent con-tinuous eruption of the opposing natural incisors, unless they are otherwise prevented from extrusion by means of a lingual plate or auxiliary bar, or by splinting. Contact of the opposing anterior teeth in eccentric positions can be developed to enhance incisive function but should be arranged to permit balanced occlusion without excursive interferences. 8. Artificial posterior teeth should not be arranged farther distally than the beginning of a sharp upward incline of the mandibular residual ridge or over the retromolar pad (Figure 18-5). To do so would have the effect of shunting the denture anteriorly.
A harmonious relationship of opposing occlusal and incisal surfaces alone is not adequate to ensure stability of distal extension removable partial dentures. In addition, the relationship of the teeth to the residual ridges must be con-sidered. Bilateral eccentric contact of the mandibular distal extension removable partial denture need not be formulated to stabilize the denture. The buccal cusps, however, must be favorably placed to direct stress toward the buccal shelf, which is the primary support area in the mandibular arch. In such positions, the denture is not subjected to excessive tilting forces (Figure 18-6). On the other hand, the artificial teeth of the bilateral, distal extension, maxillary removable partial denture often must be placed lateral to the crest of the residual ridge (Figure 18-7). Such an unfavorable position can cause tipping of the denture, which is restrained only by direct retainer action on the balancing side. To enhance the stability of the denture, it seems logical to provide simulta-neous working and balancing contacts in these situations if possible.
METHODS FOR ESTABLISHING OCCLUSAL RELATIONSHIPS Five methods of establishing interocclusal relations for removable partial dentures will be briefly described. Before any of these is described, it is necessary that the use of a face-bow mounting of the maxillary cast and pertinent factors in A B Figure 18-3 Opposing Class I partially edentulous arches arranged to allow working side contacts of opposing posterior teeth (A) with balancing contact (B) arranged to minimize tipping of the maxillary removable partial denture and to broadly distrib-ute forces accruing to its supporting structures (abutments and residual ridges).
Working contacts Figure 18-4 When occlusion is developed for a Class II re-movable partial denture (maxillary or mandibular), only working side contacts are necessary, because the cross-arch framework stability gained from tooth engagement provides resistance to movement. Balancing side contacts do not enhance stability be-yond that provided by the contralateral teeth.
Figure 18-5 Mandibular posterior teeth should not be ar-ranged distal to the upward incline (ascending ramus) of the residual ridge. The molar tooth has been placed just anterior to a mark on the cast land area designating the beginning incline.
www.konkur.in 245 Chapter 18 Occlusal Relationships for Removable Partial Dentures removable partial denture occlusion be considered. The tech-nique for applying the facebow has been described briefly in Chapter 13.
Although a hinge axis mounting may be desirable for complete oral rehabilitation procedures, any of the common types of facebow will facilitate mounting of the maxillary cast in relation to the condylar axis in the articulating instrument with reasonable accuracy and are acceptable for a removable partial denture. As was suggested in Chapter 13, it is still bet-ter that the plane of occlusion be related to the axis-orbital plane. Because the dominant factor in removable partial denture occlusion is the remaining natural teeth and their proprioceptive influence on occlusion, a comparable radius at the oriented plane of occlusion in an acceptable instru-ment will allow reasonably valid mandibular movements to be reproduced.
Semiadjustable articulators can simulate but not duplicate jaw movement. Realization of the limitations of a specific instrument and knowledge of the procedures that can over-come these limitations are necessary if an adequate occlusion is to be created.
The recording of occlusal relationships for the partially edentulous arch may vary from the simple apposition of opposing casts (by occluding sufficient remaining natural teeth) to the recording of jaw relations in the same man-ner as for a completely edentulous patient. As long as some natural teeth remain in contact, however, the cus-pal influence that those teeth will have on functional jaw movements dictates the placement of the artificial teeth and the occlusal scheme.
The horizontal jaw relation (planned intercuspal posi-tion or centric relation) in which the restoration is to be fabricated should have been determined during diagnosis and treatment planning. Mouth preparations also should have been accomplished in keeping with this determina-tion, including occlusal adjustment of the natural denti-tion, if such was indicated. Therefore one of the following conditions should exist: (1) centric relation and planned intercuspal position coincide with no evidence of occlusal pathologic conditions; therefore the decision should be to A B Figure 18-7 A, Maxillary molar occluded in a normal horizontal relationship to the opposing molar. B, The resultant position is lateral to the supporting crest of the residual ridge. This position is functionally unfavorable because of the potentially unstable leverage effects; however, stability can be improved by arranging simultaneous working and balancing contacts in the occlusal scheme Figure 18-6 The posterior teeth in this distal extension with a narrower buccal-lingual width than the original teeth have been selected, and they are placed relative to the primary support (buc-cal shelf) to distribute the functional load to the most anatomi-cally favorable location in a manner that reduces leverage effects.
www.konkur.in 246 Part II Clinical and Laboratory fabricate the restoration in centric relation; (2) centric rela-tion and the planned intercuspal position do not coincide, but the planned intercuspal position is clearly defined and the decision has been made to fabricate the restoration in the planned intercuspal position; (3) centric relation and the planned intercuspal position do not coincide and the inter-cuspal position is not clearly defined; therefore the decision should be made to fabricate the restoration in centric relation; and (4) posterior teeth are not present in one or both arches, and the denture will be fabricated in centric relation.
Occlusal relationships may be established by using the most appropriate of the following methods to fit a particular partially edentulous situation.
Direct Apposition of Casts The first method is used when sufficient opposing teeth remain in contact to make the existing jaw relationship obvious, or when only a few teeth are to be replaced on short denture bases and no evidence of occlusal abnor-malities is found. With this method, opposing casts may be occluded by hand. The occluded casts should be held in apposition with rigid supports attached with sticky wax to the bases of the casts until they are securely mounted in the articulator.
At best, this method can only perpetuate the existing occlusal vertical dimension and any existing occlusal dishar-mony present between the natural dentition. Occlusal analy-sis and the correction of any existing occlusal disharmony should precede the acceptance of such a jaw relation record. The limitations of such a method should be obvious. Yet, such a jaw relation record is better than an inaccurate inter-occlusal record between the remaining natural teeth. Unless a record is made that does not influence the closing path of the mandible because of its bulk and/or the consistency of the recording medium, direct apposition of opposing casts at least eliminates the possibility that the patient may have a faulty jaw relationship.
Interocclusal Records with Posterior Teeth Remaining A second method, which is a modification of the first, is used when sufficient natural teeth remain to support the remov-able partial denture (Kennedy Class III or IV) but the rela-tion of opposing natural teeth does not permit the occluding of casts by hand. In such situations, jaw relations must be established as for fixed restorations with some type of inter-occlusal record.
The least accurate of these methods is the interocclusal wax record. The bulk, consistency, and accuracy of the wax will influence the successful recording of centric relation with an interocclusal wax record after chilling. Excess wax that contacts the mucosal surfaces may distort soft tissue, thereby preventing accurate seating of the wax record onto the stone casts. Distortion of wax during or after removal from the mouth may also interfere with accurate seating. Therefore, a definite procedure for making interocclusal wax records is given as follows: • A uniformly softened, metal-reinforced wafer of baseplate or set-up wax is placed between the teeth, and the patient is guided to close in centric relation (Figure 18-8). Cor-rect closure should have been rehearsed before placement of the wax so that the patient will not hesitate or devi-ate in closing. The wax then is removed and immediately chilled thoroughly in room-temperature water. It should be replaced a second time to correct the distortion that results from chilling and then again chilled after removal. • All excess wax should now be removed with a sharp knife. It is most important at this time that all wax that contacts mucosal surfaces be trimmed free of contact. The chilled wax record again should be replaced to make sure that no contact with soft tissue occurs.
A wax record can be further corrected with a freely flowing occlusal registration material, such as a metallic oxide paste, which is used as the final recording medium. In making such a corrected wax record, the opposing teeth (and also the patient’s lips and the dentist’s gloves) should first be lightly coated with petroleum jelly or a sili-cone preparation. The occlusal registration material then is mixed and applied to both sides of the metal-reinforced wax record. It is quickly placed, and the patient is assisted with closing in the rehearsed path, which will be guided by the previous wax record. After the occlusal registra-tion material has set, the corrected wax record is removed and inspected for accuracy. Any excess projecting beyond the wax matrix should be removed with a sharp knife until only the registration of the cusp tips remains. Such a record should seat on accurate casts without discrepancy or interference and will provide an accurate interocclusal record. When an intact opposing arch is present, use of an opposing cast may not be necessary. Instead, a hard stone may be poured directly into the occlusal registra-tion material record to serve as an opposing cast. How-ever, although this may be an acceptable procedure in the Figure 18-8 A wax wafer softened in a water bath is used to register interocclusal position. Once removed and cooled, it is corrected with a more rigid and accurate registration material. Be-fore it is used for mounting, the registration is trimmed to allow complete seating of the cast without distortion of the material.
www.konkur.in 247 Chapter 18 Occlusal Relationships for Removable Partial Dentures fabrication of a unilateral fixed partial denture, the advan-tages of having casts properly oriented in a suitable artic-ulator contraindicate the practice. The only exception to this occurs if the maxillary cast on which the removable partial denture is to be fabricated has been mounted pre-viously with the aid of a facebow. In such an instance, an intact mandibular arch may be reproduced in stone by pouring a cast directly into the interocclusal record.
Some of the advantages of using a metallic oxide paste over wax as a recording medium for occlusal records include (1) uniformity of consistency; (2) ease of displacement on closure; (3) accuracy of occlusal surface reproduction; (4) dimensional stability; (5) the possibility of some modifica-tion in occlusal relationship after closure, if it is made before the material sets; and (6) reduced likelihood of distortion during mounting procedures.
Three important details to be observed when one uses such a material are as follows: 1. Make sure that the occlusion is satisfactory before making the interocclusal record. 2. Be sure that the casts are accurate reproductions of the teeth being recorded. 3. Trim the record with a sharp knife wherever it engages undercuts, soft tissue, or deep grooves.
Occlusal Relations Using Occlusion Rims on Record Bases A third method is used when one or more distal exten-sion areas are present, when a tooth-supported edentulous space is large, or when opposing teeth do not meet. In these instances, occlusion rims on accurate jaw relation record bases must be used. Simple wax records of edentulous areas are never acceptable. Any wax, however soft, will displace soft tissue. It is impossible to accurately seat such a wax record on a stone cast of the arch.
With this method, the recording proceeds much the same as with the second method, except that occlusion rims are substituted for missing teeth (Figure 18-9). It is essential that accurate bases be used to help support the occlusal relation-ship. Visible light-cured (VLC) bases may be adapted to the casts through the technique described in Chapter 16 for mak-ing impression trays. Utilizing the master cast, block out unde-sirable undercuts. However, do not add any wax for spacing. Paint a thin layer of the model release agent on the cast and the relief wax to aid in removal of the base from the cast after the designated time of polymerization recommended by the man-ufacturer. Carefully, adapt the VLC base material to the cast. Do not thin the material, and do not adapt the VLC base mate-rial over the remaining teeth. Process the base in the polym-erization to set the material, and then separate the record base from the cast and remove any remaining blockout wax. Clean the base, apply the air barrier coating over the entire record base, and process (tissue side up) as directed.
Record bases may also be made entirely of autopolymer-izing acrylic-resin. Those materials used in dough form lack sufficient accuracy for this purpose unless they are corrected by relining. An acrylic-resin base may be formed by sprin-kling monomer and polymer into a shallow matrix of wax or clay after any undercuts are blocked out. If the matrix and blockout have been formed with care, interference with removal will not occur, and little trimming will be necessary. When the sprinkling method is used and sufficient time is allowed for progressive polymerization to occur, such bases are stable and accurate. Other record bases include the use of cast metal and compression molded or processed acrylic-resin bases for jaw relation records.
Relative to the third method, some mention must be made of the ridge on which the record bases are formed. If the prosthesis is to be tooth-supported or if a distal extension base is to be made on the anatomic ridge form, the bases will A B Figure 18-9 Relationship and distribution of remaining teeth for this patient requires that record bases and occlusion rims be used for accurate mounting of casts. A, Acrylic-resin record bases and hard baseplate wax occlusion rims for an edentulous maxilla and Ken-nedy Class I mandibular arch. These record bases are stable and were formed by sprinkling autopolymerizing acrylic-resin. B, Occlusion rims substitute for missing posterior teeth and provide an opportunity for posterior support when interocclusal records are made; this is most critical for the longer edentulous span on the mandibular right.
www.konkur.in 248 Part II Clinical and Laboratory be made to fit that form of the residual ridge. But if a distal extension base is to be supported by the functional form of the residual ridge, it is necessary that the recording of jaw relations be deferred until the master cast has been corrected to that functional form.
Record bases must be as nearly identical as possible to the bases of the finished prosthesis. Jaw relation record bases are useless unless they are made on the same cast or a duplicate cast on which the denture will be processed, or are them-selves the final denture bases. The latter may be a cast alloy or a processed acrylic-resin base.
Jaw relation records made by this method accomplish essentially the same purpose as the two previous methods. The fact that record bases are used to support edentulous areas does not alter the effect. In any method, the skill and care used by the dentist in making occlusal adjustments on the finished prosthesis will govern the accuracy of the result-ing occlusion.
Methods for Recording Centric Relation on Record Bases Centric relation may be recorded in many ways when record bases are used. The least accurate is the use of softened wax occlusion rims. Modeling plastic occlusion rims, on the other hand, may be uniformly softened by flaming and tem-pering, resulting in a generally acceptable occlusal record. This method is time proved, and when competently done, it is equal in accuracy to any other method.
When wax occlusion rims are used, they should be reduced in height until just out of occlusal contact at the desired vertical dimension of occlusion. A single stop is then added to maintain their terminal position as a jaw relation record is made in some uniformly soft material, which sets to a hard state. Quick-setting impression plaster, bite regis-tration paste, or autopolymerizing resin may be used. With any of these materials, opposing teeth must be lubricated to facilitate easy separation. Whatever the recording medium, it must permit normal closure into centric relation without resistance and must be transferable with accuracy to the casts for mounting purposes.
Jaw Relation Records Made Entirely on Occlusion Rims The fourth method is used when no occlusal contact occurs between the remaining natural teeth, such as when an opposing maxillary complete denture is to be made concur-rently with a mandibular removable partial denture. It may also be used in those rare situations in which the few remain-ing teeth do not occlude and will not influence eccentric jaw movements. Jaw relation records are made entirely on occlu-sion rims when either arch has only anterior teeth present (Figure 18-10).
In any of these situations, jaw relation records are made entirely on occlusion rims. The occlusion rims must be sup-ported by accurate jaw relation record bases. Here, the choice of method for recording jaw relations is much the same as that for complete dentures. Either some direct interocclusal method or a stylus tracing may be used. As with complete denture fabrication, the use of a facebow, the choice of artic-ulator, the choice of method for recording jaw relations, and the use of eccentric positional records are optional, based on the training, ability, and desires of the individual dentist.
Establishing Occlusion by the Recording of Occlusal Pathways The fifth method of establishing occlusion on the remov-able partial denture is the registration of occlusal path-ways and the use of an occluding template rather than a cast of the opposing arch. When a static jaw relation record is used, with or without eccentric articulatory movements, the prosthetically supplied teeth are arranged to occlude according to a specific concept of occlusion. On the other hand, when a functional occlusal record is used, the teeth are modified to accept every recorded eccentric jaw movement.
These movements are made more complicated by the influence of the remaining natural teeth. Occlusal har-mony on complete dentures and in complete mouth reha-bilitation may be obtained by the use of several different instruments and techniques. Schuyler has emphasized the importance of establishing first the anterior tooth relation and incisal guidance before proceeding with any complete oral rehabilitation. Others have shown the advantages of establishing canine guidance as a key to functional occlu-sion before proceeding with any functional registration against an opposing prosthetically restored arch. This is done on the theory that the canine teeth serve to guide the mandible during eccentric movements when the oppos-ing teeth come into functional contact. It also has been pointed out that the canine teeth transmit periodontal proprioceptor impulses to the muscles of mastication and thus have an influence on mandibular movement even Figure 18-10 Opposing Kennedy Class I dental arches with remaining anterior teeth only. Recording of maxillomandibular relations was accomplished by using stable record bases and oc-clusion rims.
www.konkur.in 249 Chapter 18 Occlusal Relationships for Removable Partial Dentures without actual contact guidance. However, as long as the occlusal surfaces of unrestored natural teeth remain in contact, as in many a partially dentulous mouth, these teeth will always provide the primary influence on man-dibular movement. The degree of occlusal harmony that can be obtained on a fixed or removable restoration will depend on the occlusal harmony that exists between these teeth.
Regarding occlusion, Thompson has written, “Observ-ing the occlusion with the teeth in static relations and then moving the mandible into various eccentric positions is not sufficient. A dynamic concept is necessary to produce an occlusion that is in functional harmony with the facial skeleton, the musculature, and the temporomandibular joints.” By adding “and with the remaining natural teeth,” the requirements for removable partial denture occlusion are more completely defined.
Some of the methods described previously may be applied to the fabrication of removable partial dentures in both arches simultaneously, whereas the registra-tion of occlusal pathways necessitates that an opposing arch be intact or restored to the extent of planned treat-ment. A diagnostic wax-up of both maxillary and man-dibular arches will facilitate visualization of the proposed mouth preparation and restorative procedures required to accommodate the planned occlusal scheme, correct orientation of the occlusal plane, correct arch form, and complete tooth modifications to accommodate the removable partial denture design—all at the desired ver-tical dimension of occlusion. If removable partial den-tures are planned for both arches, a decision is necessary as to which denture is to be made first and which is to bear a functional occlusal relation to the opposing arch. Generally, the mandibular arch is restored first and the maxillary removable partial denture is occluded to that restored arch. If the maxillary arch is to be restored with a complete denture or a fixed partial denture or crowns, a full diagnostic wax-up must be done before occlusion is established on the opposing removable partial denture. If opposing fixed partial dentures or opposing occluding crowns are to be fabricated, it may be advantageous to develop the occlusion and fabricate them simultaneously to ensure optimal positioning, cuspal relationship, and functional integrity.
Regardless of the method used for recording jaw rela-tions, when one arch is completely restored first, that arch is treated as an intact arch even though it is wholly or partially restored by prosthetic means. The dentist must consider at the time of treatment planning the possible advantages of establishing the final occlusion to an intact arch.
From Thompson JR: Temporomandibular disorders: diagnosis and dental treatment in the temporomandibular joint. In Sarnat B, editor: The temporomandibular joint, ed 2, Springfield, IL, 1964, Bernard G. Sarnat and Charles C. Thomas, pp 146–184.
Step-by-Step Procedure for Registering Occlusal Pathways After the framework has been adjusted to fit the mouth, the technique used for the registration of occlusal path-ways is as follows: 1. Support the wax occlusion rim with a denture base that has the same degree of accuracy and stability as the finished denture base. Ideally, this would be the final denture base, which is one of the advantages of mak-ing the denture with a metal base. Otherwise, make a temporary base of VLC resin or sprinkled autopoly-merizing acrylic-resin (see Figures 19-34 through 19-38), either of which is essentially identical to the final acrylic-resin base. In any distal extension removable partial denture, make this base on a cast that has been corrected to desired functional or supporting forms of the edentulous ridge. Place a film of hard sticky wax on the base before the wax occlusion rim is secured to it. The wax used for the occlusion rim should be hard enough to support biting stress and should be tough enough to resist fracture. Hard inlay wax has proved to be suitable for most patients. However, some indi-viduals with weak musculature or tender mouths may have difficulty in reducing this wax. In such situations, use a slightly less hard wax. Make the occlusion rim wide enough to record all extremes of mandibular movement.
2. Inform the patient that the occlusion rim must be worn for 24 hours or longer. It should be worn con-stantly, including at nighttime, except for removal during meals. With wearing and biting into a hard wax occlusion rim, a record is made of all extremes of jaw movement. The wax occlusion rim must main-tain positive contact with the opposing dentition in all excursions and must be left high enough to ensure that a record of the functional path of each cusp will be carved in wax. This record should include not only voluntary excursive movements but also involuntary movements and changes in jaw movement caused by changes in posture. Extreme jaw positions and habitu-al movements during sleep should also be recorded.
The occlusal paths, thus recorded, will represent each tooth in its three-dimensional aspect. Although the cast poured against this will resemble the opposing teeth, it will be wider than the teeth that carved it be-cause it represents those teeth in all extremes of move-ment. The recording of occlusal paths in this manner eliminates entirely the need to reproduce mandibular movement on an instrument.
Instruct the patient in the removal and placement of the removable partial denture that supports the oc-clusion rim, and explain that by chewing in all direc-tions, the wax will be carved by the opposing teeth. The opposing teeth must be cleaned occasionally of accumulated wax particles. It is necessary that the patient comprehend what is to be accomplished and www.konkur.in 250 Part II Clinical and Laboratory understand that both voluntary and involuntary move-ments must be recorded.
Before dismissing the patient, add or remove wax where indicated to provide continuous contact throughout the chewing range. To accomplish this, repeatedly warm the wax with a hot spatula and have the patient close and chew into the warmed wax rim with the opposing dentition, each time adding to any area that is deficient. Any area left unsupported by its flow under occlusal forces must be reinforced with ad-ditional wax. It is important that the wax rim be ab-solutely dry and free of saliva before additional wax is applied. Each addition of wax must be made homo-geneous with the larger mass to prevent separation or fracture of the occlusion rim during the time it is being worn. Leave the wax occlusion rim from 1 to 3 mm high, depending on whether the occlusal vertical di-mension is to be increased.
3. After 24 hours, the occlusal surface of the wax rim should show a continuous gloss, which indicates func-tional contact with the opposing teeth in all extremes of movement. Any areas deficient in contact should be added to at this time. The reasons for maintain-ing positive occlusal contact throughout the time the occlusion rim is being worn are that (a) all opposing teeth may be placed in function; (b) an opposing den-ture, if present, will become fully seated; and (c) verti-cal dimension of occlusion in the molar region will be increased; thus, the head of the mandibular condyle will be repositioned and temporomandibular tissue can return to a normal relationship.
If during this period the wax occlusion rim has not been reduced to natural tooth contact, warm it by di-recting air from the air syringe through a flame onto the surface of the wax. If the wax rim is held with the fingers during warming, a gradual softening process will result, rather than a melting of the surfaces already established. Repeatedly warm the occlusion rim and replace it in the mouth until the occlusal height has been reduced and lateral excursions have been recorded. At this time, use additional wax to support those areas left unsupported by the flow of wax to the buccal or lingual surfaces. Trim the areas obviously not involved, thus narrowing the occlusion rim as much as possible. Remove those areas projecting above the occlusal surface, which by their presence might limit functional movement.
When seating of the denture and changes in man-dibular position have been accomplished by the previ-ous period of wear, it is possible to complete the occlusal registration as an operatory procedure. However, if all involuntary movements and those caused by changes in posture are to be recorded, the patient should again wear the occlusion rim for a period of time.
4. After a second 24- to 48-hour period of wear, the reg-istration should be complete and acceptable. The re-maining teeth that serve as vertical stops should be in contact, and the occlusion rim should show an intact glossy surface representing each cusp in all extremes of movement.
Natural teeth formerly in contact will not necessarily be in contact on completion of the occlusal registration. Those teeth that have been depressed over a period of years, and those that have been moved to accommodate overclosure or mandibular rotation may not be in contact upon reestab-lishment of mandibular equilibrium. Such teeth may pos-sibly return to occlusal contact in the future or may have to be restored to occlusal contact after initial placement of the denture. Because the mandibular position may have been changed during the process of occlusal registration, the cuspal relation of some of the natural teeth may be different from before. This fact must be recognized in determining the correct restored vertical dimension of occlusion.
Occlusion thus established on the removable par-tial denture will have more complete harmony with the opposing natural or artificial teeth than can be obtained by adjustments in the mouth alone, because occlusal adjustment to accommodate voluntary movement does not necessarily prevent occlusal disharmony in all pos-tural positions or during periods of stress. Furthermore, occlusal adjustment in the mouth without occlusal analy-sis is limited by the dentist’s ability to correctly interpret occlusal markings made intraorally, whether by articulat-ing ribbon or by other means.
The registration of occlusal pathways has additional advantages. It makes obtaining jaw relations possible under actual working conditions, with the denture framework in its terminal position, the opposing teeth in function, and an opposing denture, if present, fully seated. In some instances, it also makes possible the recovery of the lost vertical dimension of occlusion, unilaterally or bilaterally, when overclosure or mandibular rotation has occurred.
The completed registration is now ready for conversion to an occluding template. This is usually done by boxing the occlusal registration with modeling clay after it has been reseated and secured onto the master or processing cast. Only the wax registration and areas for vertical stops are left exposed. It is then filled with a hard die stone to form an occluding template (see Chapter 19).
It is necessary that stone stops are used to maintain the vertical relation rather than relying on some adjust-able part of the articulating instrument, which might be changed accidentally. Also, if stone stops are used and both the denture cast and the template are mounted before they are separated, a simple hinge instrument may be used.
MATERIALS FOR ARTIFICIAL POSTERIOR TEETH Contemporary acrylic-resin teeth are generally preferred to porcelain teeth, because they are more readily modified and are thought to more nearly resemble enamel in their abra-sion potential against opposing teeth. Acrylic-resin teeth www.konkur.in 251 Chapter 18 Occlusal Relationships for Removable Partial Dentures with gold occlusal surfaces are preferably used in opposition to natural teeth restored with gold occlusal surfaces, whereas porcelain teeth are generally used in opposition to other por-celain teeth.
Acrylic-resin tooth surfaces, however, may in time become impregnated with abrasive particles, thereby becom-ing an abrasive substance themselves. This may explain why acrylic-resin teeth are sometimes capable of wearing oppos-ing gold surfaces. An evaluation of occlusal contact or lack of contact, however, should be meticulously accomplished at each 6-month recall appointment, regardless of the choice of material for posterior tooth forms.
Although some controversy may continue with regard to the use of porcelain or acrylic-resin artificial teeth, there is broad agreement that narrow (reduced bucco-lingual) occlusal surfaces are desirable. Posterior teeth that will sat-isfy this requirement should be selected, and the use of tooth forms that have excessive bucco-lingual dimension should be avoided.
Acrylic-resin teeth are easily modified and readily lend themselves to construction of cast gold surfaces on their occlusal portions. A simple procedure for fabricating gold occlusal surfaces and attaching them to acrylic-resin teeth is described in Chapter 19 under posterior tooth forms.
Arranging Teeth to an Occluding Template The occlusal surface of the artificial teeth, porcelain or resin, must be modified to occlude with the template. With this method, they are actually only raw materials from which an occlusal surface is developed that are in harmony with an existing occlusal pattern. Therefore the teeth must be occluded too high and then modified to fit the template at the established vertical dimension of occlusion.
Teeth arranged to an occluding template ordinarily should be placed in the center of the functional range. Whenever possible, the teeth should be arranged bucco-lingually in the center of the template. When natural teeth have registered the functional occlusion, this may be considered the normal physiologic position of the artificial dentition, regardless of its relation to the resid-ual ridge. On the other hand, if some artificial occlusion in the opposing arch has been recorded, such as that of an opposing denture, the teeth should be arranged in a favorable relation to their foundation, even if this means arranging them slightly buccally or lingually from the center of the template.
The teeth are usually arranged for intercuspation with the opposing teeth in a normal cuspal relationship. When-ever possible, the mesiobuccal cusp of the maxillary first molar should be located in relation to the buccal groove of the mandibular first molar and all other teeth arranged accordingly. With a functionally generated occlusion, however, it is not absolutely necessary that normal oppos-ing tooth relationships be reestablished. In the first place, the opposing teeth in a dental arch that is not contiguous may not be in normal alignment, and intercuspation may be difficult to accomplish. In the second place, the occlusal surfaces will need to be modified so that they will func-tion favorably regardless of their anteroposterior position. Because cusps modified to fit an occlusal template will be in harmony with the opposing dentition, it is not neces-sary that the teeth themselves be arranged to conform to the usual concept of what constitutes a normal anteropos-terior relationship.
ESTABLISHING JAW RELATIONS FOR A MANDIBULAR REMOVABLE PARTIAL DENTURE OPPOSING A MAXILLARY COMPLETE DENTURE It is common for a mandibular removable partial denture to be made to occlude with an opposing maxillary complete denture. The maxillary denture may already be present, or it may be made concurrently with the opposing removable partial denture. In any event, the establishment of jaw rela-tions in this situation may be accomplished by one of several previously outlined methods.
If an existing maxillary complete denture is satisfactory and the occlusal plane is oriented to an acceptable anatomic, functional, and esthetic position (which rarely occurs), then the complete denture need not be replaced and the opposing arch is treated as an intact arch as though natural teeth were present. A facebow transfer is made of that arch, and the cast is mounted on the articulator. To accomplish this, a facebow record is made with the complete denture in place. After the facebow apparatus has been removed from the patient, the complete denture is removed and an irreversible hydrocol-loid impression of the denture is made. When the impres-sion material has set, the denture is removed, cleaned, and returned to the patient. A cast is formed in the impression and then is mounted on the articulator with the facebow record. Maxillomandibular relations may be recorded on accurate record bases attached to the mandibular removable partial denture framework with the use of a suitable record-ing medium. Centric relation is recorded and is transferred to the articulator. Eccentric records can then be made to pro-gram the articulator.
In rare instances, when the mandibular removable par-tial denture replaces all posterior teeth and the anterior teeth are noninterfering, a central bearing point tracer may be mounted in the palate of the maxillary denture, and centric relation recorded by means of an intraoral stylus tracing against a stable mandibular base. If a stylus jaw rela-tion recording method is used, the stylus must be carefully removed from the denture and attached to the same palatal location of the stone cast that was transferred to the articula-tor via the facebow. The mandibular cast can then be oriented by way of the horizontal jaw relation record and attached to the articulator.
When an existing complete denture is opposing the arch on which a removable partial denture is fabricated, www.konkur.in 252 Part II Clinical and Laboratory a cast of the complete denture may be used during the fab-rication procedures. However, when the occlusion is cor-rected after processing and the removable partial denture is finalized during initial placement, the complete denture should be retrieved and mounted on the articulator with a centric relation record at the desired vertical dimension of occlusion. This will ensure a more accurate cuspal rela-tionship and will prevent abrasion of the cusp contacts that would occur if a stone cast of the denture is used. This procedure is completed when the patient is in the office so as not to deprive the patient of use of the existing complete denture.
If the relationship of the posterior teeth on the maxillary denture to the mandibular ridge is favorable and the com-plete denture is stable, jaw relations may be established by recording occlusal pathways in the mandibular arch just as for any opposing intact arch. The success of this method depends on the stability of the denture bases, the quality of tissue support, the relation of the opposing teeth to the man-dibular ridge, and the interrelation of existing artificial and natural teeth.
More often than not, the existing maxillary complete denture will not be acceptable to use because of poor tooth position. The denture will have been made to occlude with malpositioned mandibular teeth, which have since been lost, or the teeth will have been arranged without consideration for the future occlusal relation with a mandibular removable partial denture. Too often, one sees a maxillary denture with posterior teeth arranged close to the residual ridge without regard for the interarch relationship and with an occlusal plane that is too low. Usually, however, a new maxillary den-ture must be made concurrently with the mandibular remov-able partial denture, and jaw relations may be established in one of two ways.
If the mandibular removable partial denture will be tooth supported (a Kennedy Class III arch accommodating a bilat-eral removable prosthesis), the mandibular arch is restored first. The same applies to a mandibular arch that is restored with fixed partial dentures. In either situation, the mandib-ular arch is completely restored first, and jaw relations are established, as they would be to a full complement of oppos-ing teeth. Thus, the maxillary complete denture is made to occlude with an intact arch.
On the other hand, as is more often the situation, the man-dibular removable partial denture may have one or more dis-tal extension bases. The situation then necessitates that the occlusion be established on both dentures simultaneously.
All mouth preparations and restorative procedures required to correctly orient the occlusal plane, correct the arch form, accommodate the desired occlusal scheme, and accommodate the removable partial denture design must be accomplished on the remaining natural teeth. In addi-tion, all supporting tissue must be in an acceptable state of health before the final impression is made. After final impressions, which include the alter cast impression or the corrected cast impression, are made, the maxillary occlu-sion rim is contoured, occlusal vertical relation with the remaining lower teeth is established, and a facebow transfer of the maxillary arch is made. The maxillomandibular rela-tions may be recorded by any of the several methods pre-viously outlined and the articulator mounting completed. Occlusion may be established as for complete dentures, with care taken to establish a favorable tooth-to-ridge rela-tionship in both arches, an optimum occlusal plane, and cuspal harmony between all occluding teeth.
After try-in, several methods may be used to complete the restorations. Both dentures may be processed concurrently and remounted for occlusal correction, or the removable partial denture may be processed first. After the dentures are completed and remounted, the teeth—still in wax on the complete denture—are adjusted to any discrepancies that are occurring.
Occlusal discrepancies created during processing must be corrected before the patient is permitted to use the denture(s). Methods by which these discrepancies may be corrected are discussed in Chapter 19.
www.konkur.in CHAPTER 19 Laboratory Procedures CHAPTER OUTLINE Duplicating a Stone Cast Duplicating Materials and Flasks Duplicating Procedure Waxing the Removable Partial Denture Framework Forming the Wax Pattern for a Mandibular Class II Removable Partial Denture Framework Attaching Wrought-Wire Retainer Arms by Soldering Waxing Metal Bases Spruing, Investing, Burnout, Casting, and Finishing of the Removable Partial Denture Framework Spruing Investing the Sprued Pattern Burnout Casting Removing the Casting from the Investment Finishing and Polishing Making Record Bases Technique for Making a Sprinkled Acrylic-Resin Record Base Occlusion Rims Making a Stone Occlusal Template from a Functional Occlusal Record Arranging Posterior Teeth to an Opposing Cast or Template Posterior Tooth Forms Arranging Teeth to an Occluding Surface Types of Anterior Teeth Waxing and Investing the Removable Partial Denture Before Processing Acrylic-Resin Bases Waxing the Removable Partial Denture Base Investing the Removable Partial Denture Processing the Denture Remounting and Occlusal Correction to an Occlusal Template Precautions to Be Taken in Remounting Polishing the Denture Denture Borders Facial Surfaces This chapter covers only those phases of dental laboratory procedures that are directly related to removable partial denture fabrication. Familiarity with laboratory procedures relative to making fixed restorations and complete dentures is presumed. Such information is available in numerous textbooks on those subjects and is not duplicated here. For example, the principles and techniques involved in the wax-ing, casting, and finishing of single inlays, crowns, and fixed partial dentures are adequately covered in lecture material and textbooks and in manuals available to the dental student, the dental laboratory technician, and the practicing dentist. Similarly, knowledge of the principles and techniques for mounting casts, articulating teeth, and waxing, processing, and polishing complete dentures is presumed as a necessary background for the laboratory phases of removable partial denture fabrication. Therefore this chapter is directed spe-cifically toward the laboratory procedures involved in the fabrication of a removable partial denture.
DUPLICATING A STONE CAST A stone cast may be duplicated for several purposes. One is the duplication in stone of the original or corrected master cast to preserve the original. One use of such a cast is for fit-ting a removable partial denture framework without danger of fracture or abrading the surface of the original master cast. Another use might be for processing a temporary prosthesis where wax relief and blockout on the original cast allow pro-duction of a duplicate, which following processing will make insertion of the prosthesis easier.
Another purpose of duplicating a cast is to allow an investment cast to be formed for framework fabrication. Careful preparation of the master cast for production of this investment cast involves consideration of the defined path of insertion, heights of contour, and retentive and stabilization areas designed into the mouth preparations. The framework produced should be carefully evaluated on the cast for fit, just as a fixed partial denture is evaluated on a working cast with dies (Figure 19-1).
www.konkur.in 254 Part II Clinical and Laboratory Blockout should be accomplished on the master cast before an investment cast is made. On this investment cast, the wax or plastic pattern is formed. The use of preformed plastic patterns eliminates some of the danger of damaging the surface of the investment cast in the process of forming the pattern. With freehand waxing, considerable care must be taken not to score or abrade the investment cast. The metal framework is ultimately cast against its surface, and the finished casting and the original cast are then returned to the dentist after all fitting has been completed on the dupli-cate cast.
Duplicating Materials and Flasks Duplicating materials include both colloidal and silicone materials. Colloidal materials are made fluid by heating and return to a gel while cooling. The cast to be duplicated must be placed in the bottom of a suitable flask, called a duplicating flask. A duplicating flask should contain the fluid material to facilitate its cooling, to support the mold while it is being filled with the cast material, and to facili-tate removal of the cast from the mold without permanent deformation or damage to the mold. Numerous duplicat-ing flasks are available on the market.
The technique for duplicating is the same for any cast, whether or not blockout is present. However, if wax or clay blockout is present, the temperature of the duplicat-ing material must not be any higher than that recom-mended by the manufacturer, to prevent melting and distortion of the blockout material.
Ordinary baseplate wax may be used for paralleled blockout and ledges, but care must be taken that the temperature of the duplicating material is not too high because it will then melt the wax. The use of prepared blockout material, such as Ney blockout wax or Wills undercut material, may be preferred.
Duplicating Procedure The equipment needed and the steps required for dupli-cation using a silicone material are described in Figures 19-2 through 19-6.
WAXING THE REMOVABLE PARTIAL DENTURE FRAMEWORK When preformed plastic patterns are used (Figure 19-7), parts of the denture framework must be waxed freehand to prevent excessive bulk and to create the desired contours for a satisfactory custom-made removable partial denture framework (Figure 19-8).
Although most dentists do not fabricate their own remov-able partial denture castings, it is essential that they have an understanding of the dental laboratory procedures involved. A B Figure 19-1 A, Framework returned from the laboratory with the master cast. B, Replacement on the cast reveals areas of contact that can erode the cast surface. Careful fitting to the cast allows determination of potential framework adjustment regions before fitting in the mouth. If the framework is indiscriminately finished in the laboratory or before intraoral placement, critical tooth contact regions may be lost, resulting in a poorly retained and stabilized prosthesis. Most often, such an overly adjusted framework is poorly stabilized, not poorly retained.
Figure 19-2 Duplicating material with a vacuum curing unit in the background, which is used to ensure a dense mold.
www.konkur.in 255 Chapter 19 Laboratory Procedures This enables them to design the removable partial denture framework, complete a laboratory work authorization that communicates the desired design and authorizes its fabrica-tion, and evaluate the quality of the framework (Figure 19-9). Understanding and evaluation of the key features required in a completed removable partial denture framework ensure the patient of a chance to function comfortably with the fin-ished product (Figure 19-10). The contrary is also true.
Forming the Wax Pattern for a Mandibular Class II Removable Partial Denture Framework Examples that illustrate many of the essentials of waxing a removable partial denture framework are presented in Figures 19-11 through 19-20. This exercise includes the waxing of three types of direct retainers (circumferential, combination, and bar type), a mandibular lingual bar major connector, a maxillary anterior-posterior palatal strap major connector, both distal extension and tooth-bound modification spaces, and adaptation of round, 18-gauge, wrought-wire retentive arms for combination-type direct retainers.
Attaching Wrought-Wire Retainer Arms by Soldering Wrought-wire retainers may be attached to a removable par-tial denture framework after it has been cast and finished (Figures 19-21 and 19-22). The soldering procedure may be accomplished by electric soldering or by a direct-heat method with an oxygen-gas flame. With either method, care must be taken to use compatible alloys, appropriate solder, and flux in conjunction with the careful application of controlled heat.
Students are encouraged to review the Chapter 13 discus-sion of the selection of metal alloys to enhance their under-standing of the properties of solder, flux, the effects of heat on metal alloys, and the necessity for quality control in sol-dering procedures.
Figure 19-3 Duplicating mold following removal of the re-lieved and blockout cast.
Figure 19-4 Refractory cast (right) following removal from the duplication mold (left). This cast incorporates all the features of the relieved and blockout master cast generated from the clini-cal framework impression.
A B Figure 19-5 Close up of the refractory cast (A) with initial wax added to rests and the posterior bead line (B).
www.konkur.in 256 Part II Clinical and Laboratory Waxing Metal Bases A technique for forming the retentive framework for the attachment of acrylic-resin bases has been given. Two basic types of metal bases may be used instead of the resin base. The advantages of cast metal bases in preference to acrylic-resin bases have been discussed in Chapter 9. The type of base to be used must be determined before block-out and duplication so that the relief over each edentulous ridge may be provided or eliminated as required.
For an acrylic-resin base, relief for the retentive frame must be provided. No relief over the residual ridge is used for a complete metal base. For a partial metal base, the junction between metal and acrylic-resin must be clearly defined by trimming the relief along a definite, previously determined line.
One type of metal base is the complete base with a metal border to which tube teeth, cast copings, or an acrylic-resin superstructure may be attached. If a porcelain or plastic tube or grooved teeth are used, they must be posi-tioned first and the pattern waxed around them to form a coping. The teeth are then attached to the metal base by cementation or, with the use of resin teeth, are attached with additional acrylic-resin under pressure, a so-called pressed-on method of attaching acrylic-resin teeth to a metal base. Another method of attaching teeth is to wax the base to form a coping for each tooth, either by carving recesses in the wax or by waxing around artificial teeth. Rather than attaching a stock tooth, the full tooth may be waxed into occlusion, the base invested, and the wax pat-terns replaced with heat-polymerized acrylic-resin. This method permits some variation in the dimension and form of the supplied teeth that is not possible with stock teeth. It is particularly applicable to abnormally long or short spaces or when a stock tooth of desired width is not available. With modern cross-linked acrylic-resins, such processed teeth are fairly durable; however, the addition of gold occlusal surfaces is an option.
When artificial teeth are to be arranged to occlude with an opposing cast or an opposing template, the metal base must be formed for attachment of the tissue-colored den-ture resin supporting the teeth. This is the most common method of attaching teeth to a metal base. The wax pattern for the base is formed from one thickness of 24-gauge cast-ing wax, which is then reinforced at the border and forms the retention of the acrylic-resin superstructure. Because metal borders are more difficult to adjust than acrylic-resin, they are usually made somewhat short of the area normally covered with an acrylic-resin base. Also, because A B Figure 19-6 Example of a completed maxillary waxed framework on a refractory cast with the sprue access positioned. A, From occlusal. B, From anterior.
Figure 19-7 Preformed plastic patterns are available in many shapes and sizes for clasps, for minor connectors connecting to teeth, or for distal extension bases, bars, mesh, and palatal cov-erage. Because they are made of soft plastic material, they tend to stretch on removal from their backing. Therefore, care must be exercised when one is removing patterns. Their use generally requires that tacky liquid be applied first to the investment cast at their area of placement.
www.konkur.in 257 Chapter 19 Laboratory Procedures border thickness adds objectionable weight to the denture, it is made with only a slight border roll. This is one dis-advantage of the complete metal base, because the border accuracy of the impression registration cannot be used to fullest advantage, and contouring of facial and lingual sur-faces cannot be as effective as with an acrylic-resin base, in which added bulk can sometimes be used to advantage.
The border is first penciled lightly on the investment cast, and then the 24-gauge sheet casting wax is smoothly adapted. Considerable care must be taken not to stretch and thin the sheet wax in adapting it to the cast. To pre-vent wrinkling, the wax should be adapted in at least two longitudinal pieces and joined and sealed together at the ridge crest. The wax is then trimmed along the penciled outline with a warm, dull instrument to prevent scoring of the investment cast.
A single piece of 14-gauge round wax is now adapted around the border over the sheet wax. With a hot spatula, the round wax must be sealed to the cast along its outer border. Sufficient wax is flowed onto the round wax to blend it smoothly into the sheet wax, thus completing a border roll. The inner half of the round wax form remains untouched. Wax is added when needed to facilitate carving without trimming the original 24-gauge thick-ness. The result should be a rounded border that blends smoothly into the sheet wax.
The boxing for the resin, which will in turn support the artificial teeth, is now added, again with 14-gauge round wax. The proposed outline for the boxing is identified by lightly scoring the sheet wax. On this scored line, the 14-gauge round wax is adapted, thus forming the outline of the boxing.
A B C Figure 19-8 Three steps are involved in making a denture framework by using relief, blockout ledges, and ready-made pattern forms. A, Master cast with relief, blockout of undercuts at posterior right and anterior lingual regions, and shaped blockout ledges for location of retentive and nonretentive clasp arms. B, Completed pattern using lingual bar major connector pattern, plastic clasp forms resting on investment ledges, wrought wire, and open retention mesh. C, Finished casting returned to master cast.
www.konkur.in 258 Part II Clinical and Laboratory With additional wax, the ditch between the sheet wax and the outer border of the round wax is filled in and blended smoothly onto the sheet wax. This is done in the same manner as the border, with sufficient wax added to allow for smoothing and carving. As was mentioned pre-viously, the pattern should not be flamed or polished with a cloth. Instead the pattern must be smoothed by carving.
The result thus far should be a pattern reinforced at the border and at the boxing and slightly concave in between, with some of the original sheet wax exposed. The inside of the boxing is not sealed to the sheet wax, thus leaving a slight undercut for the attachment of the acrylic-resin. With a sharp blade, the margins of the boxing are then carved to a knife-edge finishing line. With the backside of the large end of the No. 7 wax spatula, this margin may be lifted slightly, further deepening the undercut beneath the finishing line.
In addition to the undercut finishing line, retention spurs, loops, or nailheads are added for retention of the acrylic-resin to be added later. Spurs are usually made of 18-gauge or smaller round wax attached at one end only at random acute angles to the sheet wax. Loops are small-gauge round (wax, resin, or metal) circles attached vertically or horizontally with space beneath for the acrylic-resin attachment. Nailheads are made of short pieces of 18-gauge round wax attached vertically to the sheet wax, with the head flattened by a slightly warmed spatula. Any method of providing retention is acceptable if it permits positive attachment of the acrylic-resin and will not interfere with the placement of artificial teeth.
A metal base waxed as described provides optimum contours with a minimum of bulk and weight and with adequate provision for the attachment of artificial teeth to the metal base. If properly designed, the more visible por-tions of the metal base will be covered with the acrylic-resin supporting the supplied teeth.
SPRUING, INVESTING, BURNOUT, CASTING, AND FINISHING OF THE REMOVABLE PARTIAL DENTURE FRAMEWORK Brumfield has listed factors that influence the excellence of a dental casting (Box 19-1).
Figure 19-9 Evaluation of the framework by the clinician is as important for removable partial denture frameworks as it is for implant and/or fixed partial denture castings. Careful scru-tiny of attention to detail in following the design specifications is necessary, as is evaluation of the fit of the framework to the cast. On this cast, initial seating of the framework reveals interference with complete seating bilaterally from the embrasure clasps. Only minor adjustment should be required to completely seat the framework. If major adjustment is needed, or if after complete seating of the framework the clasp assemblies do not contact the teeth as prescribed, the framework should be remade.
A B Figure 19-10 A, Design of a mandibular removable partial denture framework is outlined on the master cast for the tech-nician to follow in waxing and casting the framework. B, Cast framework as returned from the laboratory is evaluated intraoral-ly, revealing sufficiently accurate adaptation and design identical to the embrasure clasp and mesio-occlusal rest #28, as shown on the design cast in A.
www.konkur.in 259 Chapter 19 Laboratory Procedures Spruing Brumfield described the function of sprues as follows: The sprue channel is the opening that leads from the crucible to the cavity in which the framework is to be cast. Sprues have the purpose of leading the molten metal from the cru-cible into the mold cavity. For this purpose, they should be large enough to accommodate the entering stream and of the proper shape to lead the metal into the mold cavity as quickly as possible but with the least amount of turbulence. Sprues have the further purpose of providing a reservoir of molten metal from which the casting may draw during solidification, thus preventing porosity caused by shrinkage.
Spruing of the cast may be roughly summarized in the following three general rules: 1. Sprues should be large enough that the molten metal in them will not solidify until after the metal in the casting proper has frozen (8- to 12-gauge round wax is usually used for multiple spruing of removable partial denture castings).
Figure 19-11 Occlusal view of mandibular Kennedy Class I, modification 1 wax pattern on a refractory cast. The lingual bar ma-jor connector joins three clasp assemblies (rest, proximal plate, and Akers [RPA] clasp, wrought wire, and cast circumferential).
A Figure 19-12 Buccal view of left side of the pattern shown in Figure 19-11. A tapered retainer clasp pattern is placed on the refractory cast ledge produced by duplication of shaped blockout ledges. The cast illustrates relief provided beneath the minor con-nector, which retains the denture base resin. The tissue index at the gingival region of the proximal plate will provide an easily identifiable finish line for resin finishing and future trimming of altered cast impression.
Figure 19-13 Buccal view of modification space of the pat-tern shown in Figure 19-11. A wrought-wire clasp is contoured at the anterior extent of the modification space. Such a retainer will not overly stress the tooth if opposite denture base movement causes rotation across the fulcrum (see Figure 19-11, occlusal rests at #21 and #31). A tapered cast circumferential clasp fol-lows the shaped ledge to a distal buccal 0.01-inch undercut. Both proximal plates have been waxed with sufficient bulk to provide a complete casting. If thickness presents a problem with tooth placement, some finishing can be accomplished at a later time.
Figure 19-14 Lingual view of the pattern shown in Figure 19-11. The lingual bar major connector rigidly connects components cross-arch. The minor connector for resin retention on the left has buccal and lingual struts to allow unimpeded tooth place-ment. The tapered proximal plate on the right (tooth #28) is thicker on the lingual and thinner on the buccal, to allow close placement of the buccal surface of the denture tooth to #28. The finish line of the modification space is positioned below adjacent tooth gingival margins to allow normal lingual contouring of the resin matrix.
www.konkur.in 260 Part II Clinical and Laboratory 2. Sprues should lead into the mold cavity as directly as possible and still permit a configuration that will in-duce a minimal amount of turbulence in the stream of molten metal.
3. Sprues should leave the crucible from a common point and be attached to the wax pattern at its bulkier sec-tions, that is, no thin sections of casting should inter-vene between two bulky, unsprued portions.
The configuration of the sprues, from their point of attachment at the crucible until they reach the mold cav-ity, may be influential in reducing turbulence. One of the more important sources of difficulty in casting is the Figure 19-15 Another example of a wax pattern for a mandi-bular framework. A wrought wire has been adapted to tooth #27. The clasp demonstrates an appropriate circlet shape, allowing maximum length. The circlet shape also allows the undercut to be approached from below; this improves retention—push reten-tion is more efficient than pull retention. Ridge relief is evident beneath the resin minor connector, and the proximal plate hous-ing the wire has been waxed with bulk to facilitate casting. The plate may require adjustment at a later time.
Figure 19-16 Same pattern as shown in Figure 19-15. In-cludes a lingual plate major connector design. The external finishing line at the anterior modification space is lingual to the anticipated tooth and resin position, allowing a natural con-tour to be developed in the completed prosthesis. The posterior denture base minor connector patterns have been reinforced with additional wax to add stiffness at the junction with the proximal plate and major connector. Because of the potential for repeated flexure in this region under function, such rein-forcement is critical to long-term success with patterns such as this. Because the wrought-wire retainers are not cast, it is im-portant that they be contoured to contact the teeth as accurately as possible, as shown.
Figure 19-17 Buccal view of the pattern shown in Figure 19-16. Double wrought wires are contoured to follow the ledges, the tissue stop at the posterior of the ridge relief is shown, and an occlusal rest distal of #20 will require finishing following comple-tion of casting.
Figure 19-18 Occlusal view of the maxillary refractory cast with a wax pattern. An anatomic replica has been used to develop an anterior-posterior palatal strap major connector with beading evident at the anterior, posterior, and internal edges. Bilateral, ta-pered I-bar retainers have been waxed as extensions of the resin-retaining minor connectors, and double rests were incorporated into the design on each side.
www.konkur.in 261 Chapter 19 Laboratory Procedures entrapment of gases in the mold cavity before they have a chance to escape. If the sprue channels contain sharp right-angle turns, great turbulence is induced, which is calculated to entrap such gases and so lead to faulty cast-ings. Sprue channels should make long radii and easy turns, and should enter the mold cavity from a direction designed to prevent splashing at this point.
As was previously stated, the sprues should be attached to the bulky points of the mold pattern. If two bulky points exist with a thin section between them, each of the bulky spots should be sprued. The points of attachment should be flared out and local constrictions avoided. If this practice is followed, the sprue, which is bulky enough to freeze after the case framework has frozen, will continue to feed molten metal to the framework until it has entirely solidified. The intent is to provide sound metal in the casting proper with all shrinkage porosity forced into the sprue rod, which is later discarded.
The laboratory procedure for multiple spruing is essen-tially the same for all mandibular and maxillary castings, except those with a palatal plate. Typical examples are illustrated in Figures 19-23 and 19-24.
Two basic types of sprues are used: multiple and single. Most removable partial denture castings require multiple spruing, with the use of 8- to 12-gauge round wax shapes for main sprues and 12- to 18-gauge round wax shapes for auxiliary sprues. Occasionally, however, a single sprue is preferred for cast palates and cast metal bases for the mandibular arch when these are used as complete den-ture bases. With removable partial dentures, the use of a single sprue is limited to those maxillary frameworks in which—because of the presence of a palatal plate—it is impossible to locate multiple sprues centrally. In such situations, the single sprue may be used advantageously. A single sprue must be attached to the wax pattern so that the direction of flow of the molten metal will be paral-lel to the long axis of the single sprue. In some instances, the investment cast may have to be cut away anteriorly to make room for the attachment of the sprue; in others, the sprue may be attached posteriorly. One disadvantage of using a single sprue for large castings is that an extra-long investment ring must be used.
Figure 19-19 Buccal view of right I-bar shown in Figure 19-18. The tissue index is demonstrated gingival to the proximal plate, tissue relief is provided beneath the denture base minor connec-tor, and the I-bar shows a taper as it approaches the mid-buccal 0.01-inch undercut.
Figure 19-20 Buccal view of I-bar opposite to that shown in Figure 19-19. Many of the same features seen in Figure 19-19 are evident in this example. The tissue stop distal to the ridge relief is shown and the palatal external finishing line continues distal to the junction of the hard and soft palate (which is the terminus of the posterior palatal strap).
Figure 19-21 Wrought wire can be soldered to cast frameworks as an alternative method to incorporation in the wax pattern and casting procedure. Electric soldering is a common method for sol-dering and does not require heating of the entire framework. Follow-ing preparation of the minor connector where the wire will be sol-dered, an 18-gauge, round, wrought-wire clasp is carefully adapted and then is secured to the framework and the duplicate stone cast with the use of refractory investment. In this situation, a distolin-gual molar undercut will be used bilaterally for retention. The flux is placed at the proximal plate, followed by placement of sufficient solder, and the electric heat element tip is placed into contact with the solder while the frame is grounded at another location.
www.konkur.in 262 Part II Clinical and Laboratory Some important points to remember in multiple spruing are as follows: 1. Use a few sprues of large diameter rather than several smaller sprues.
2. Keep all sprues as short and direct as possible.
3. Avoid abrupt changes in direction; avoid T-shaped junctions as much as possible.
4. Reinforce all junctions with additional wax to prevent constrictions in the sprue channel and to prevent V-shaped sections of investment that might break away and be carried into the casting.
Investing the Sprued Pattern The investment for a removable partial denture casting consists of two parts: the investment cast on which the pat-tern is formed, and the outer investment surrounding the cast and pattern (see Figure 19-41). The latter is confined within a metal ring, which may or may not be removed after the outer investment has set. If the metal ring is not to be removed, it must be lined with a layer of cellulose, asbestos substitute, or ceramic fiber paper to allow for both setting and thermal expansion of the mold in all directions.
The investment must conform accurately to the shape of the pattern and must preserve the configuration of the pattern as a cavity after the pattern itself has been elimi-nated through vaporization and oxidation. Brumfield has listed the purposes of the investment as follows: 1. It provides the strength necessary to hold the forces ex-erted by the entering stream of molten metal until this metal has solidified into the form of the pattern.
2. It provides a smooth surface for the mold cavity so that the final casting will require as little finishing as possible; in some situations, a deoxidizing agent is used to keep surfaces bright.
A B Figure 19-22 A, An electric soldering technique has been used to solder the wrought wire. The solder has flowed into the space between the framework and the wire as well as around the wire to ensure its secure attachment to the framework. B, Fin-ished and polished wrought-wire retainer.
Box 19-1 FACTORS THAT INFLUENCE THE EXCELLENCE OF A DENTAL CASTING 1. Care and accuracy with which the cast is reproduced 2. Intelligence with which the framework is designed and proportioned 3. Care and cleanliness in waxing up the cast 4. Consideration of the expansion of the wax caused by temperature 5. Size, length, configuration, points of attachment, and manner of attachment of the sprues 6. Choice of investment 7. Location of the pattern in the mold 8. Mixing water: Amount, temperature, and impurities 9. Spatulation of the investment during mixing 10. Restraint offered to expansion of the investment caused by the investment ring 11. Setting time 12. Burnout temperature and time 13. Method of casting 14. Gases: Adhered, entrapped, and absorbed 15. Force used in throwing the metal into the mold 16. Shrinkage on cooling 17. Removal from the investment after casting 18. Scrubbing, pickling, and so on 19. Poli1.shing and finishing 20. Heat handling Figure 19-23 Mandibular cast framework with sprue intact, showing three 8-gauge sprues attached to the lingual bar. If one is concerned about completely filling distal extension base minor con-nectors, 12-gauge sprues can be attached to the minor connector.
www.konkur.in 263 Chapter 19 Laboratory Procedures 3. It provides an avenue of escape for most of the gases entrapped in the mold cavity by the entering stream of molten metal.
4. It, together with other factors, provides necessary compensation for the dimensional changes of the al-loy from the molten to the solid, cold state.
The investment for casting gold alloys is a plaster-bound silica material; compounded in such a way that the total mold expansion will offset the casting shrinkage of the alloy, which varies from 1% to 1.74% (the highest figure being the shrinkage of pure gold). Generally, the higher the percentage of gold in the alloy, the greater is the con-traction of the casting on solidifying.
Only one chromium-cobalt alloy has a sufficiently low melt-ing temperature to be cast into a plaster-silica investment mold. According to Peyton, for the other alloys that have a higher melting temperature, an investment that contains quartz powder held together by an ethyl silicate or sodium silicate binder is generally used. Expansion to offset casting shrinkage for chromium-cobalt alloys is accomplished pri-marily through thermal expansion of the mold and must be sufficient to offset their greater casting shrinkage, which is on the order of 2.3%. For this reason, the casting ring is usu-ally removed after the mold has hardened to allow for the greater mold expansion necessary with these alloys. Because the investments for chromium-cobalt alloys are generally less Note the substitution of the word alloy, as the same principles apply whether the metal is a precious metal alloy or a chromium-cobalt alloy. In some of the latter alloys, the cobalt is partially replaced by nickel; such alloys are sometimes described as Stellite alloys.
porous, there is greater danger that gases will be entrapped in the mold cavity by the molten metal. Spruing must be done with greater care, and in some instances, provision for vent-ing the mold is necessary to prevent defective castings.
Step-by-Step Procedure The technique for applying the outer investment is usually referred to as investing the pattern. Actually, the cast on which the pattern is formed is part of the investment also. This investment technique is presented in Figures 19-25 through 19-27.
Figure 19-24 For a maxillary framework, multiple 8-gauge sprues or a single main sprue located posteriorly can be used. When a single main sprue is used, additional sprues can feed critical regions for casting completeness.
Figure 19-25 Wax pattern sprued and ready to invest.
Figure 19-26 Applying the investment with a soft moistened brush to ensure complete adaptation of the investment. This helps reduce the number of bubbles and makes for a smoother cast surface.
www.konkur.in 264 Part II Clinical and Laboratory Burnout The burnout operation serves three purposes: it drives off moisture in the mold; it vaporizes and thus eliminates the pattern, leaving a cavity in the mold; and it expands the mold to compensate for contraction of the metal on cooling.
For the investment to heat uniformly, it should be moist at the start of the burnout cycle. Steam will then carry the heat into the investment during the early stages of the burnout. Therefore if the investment is not burned out on the same day that it is poured, it should be soaked in water for a few minutes before it is placed in the burnout furnace.
Just before it is placed in the furnace, the mold should be placed in the casting machine to balance the weight against the weight of the mold. At this time, the mold should be properly oriented to the machine and its cru-cible, and a scratch line should be made at the top for later repositioning of the hot mold.
The mold should be placed in the oven with the sprue hole down and the orientation mark forward. Burnout should be started with a cold oven, or nearly so. Then the temperature of the oven should be increased slowly to a temperature recommended by the manufacturer. This temperature should be maintained for the period rec-ommended by the manufacturer to ensure uniform heat penetration. More time must be allowed for plastic pat-terns, particularly palatal anatomic replica patterns. It is important that the peak temperature recommended by the manufacturer not be exceeded during the burnout period. (When a high-heat investment is used, follow the manufacturer’s instructions as to burnout temperature.) Casting The method of casting varies widely with the alloy and equipment used. All methods use force to quickly inject the molten metal into the mold cavity. This force may be cen-trifugal or air pressure; the former is more commonly used. In any case, too much or too little force is undesirable. If too little force is used, the mold is not completely filled before the metal begins to freeze. If too much force is used, exces-sive turbulence may result in the entrapment of gases in the casting. With centrifugal casting machines, this is regulated by the number of turns put on the actuating spring.
The metal may be melted with a gas-oxygen blow-torch or by an electric muffle surrounding the metal. In some commercial casting procedures and in some dental laboratories, the induction method may be used (Figure 19-28); this provides a rapid and accurate method of melting the metal. Currently available cast-ing machines include those that are electrically con-trolled to heat alloys to a specified temperature and to release the molten metal at precisely the correct casting temperature. These machines are relatively expensive and are primarily located in commercial laboratories or institutions where casting volume is high.
Removing the Casting from the Investment Chromium-cobalt alloys are usually allowed to cool in the mold, are divested (Figure 19-29), and are not cleaned by pickling. Finishing (Figure 19-30) and polishing (Figure 19-31), which are done with special high-speed equip-ment, require technical skill in the use of bench lathes. Just before they are polished (high-shine), chromium-cobalt castings are electropolished; this is a controlled deplating process (Figure 19-32).
Finishing and Polishing Some authorities hold that the sprues should not be removed from the casting until most of the polishing is completed. Although it is true that this policy may pre-vent accidental distortion, it is difficult to adhere to and is therefore somewhat impractical. Instead, reasonable care should be exercised to prevent any distortion result-ing from careless handling. A specific precaution is that a cast clasp arm should not be indiscriminately polished and then bent into place. The waxing should have been done in such a manner that a minimum of finishing is necessary and the intended relationship of the clasp to the abutment is maintained (Figure 19-33).
Actual polishing procedures may vary widely accord-ing to personal preference for certain abrasive shapes and sizes. However, the following several rules for finishing the casting are important: 1. High speeds are preferable to low speeds. Not only are they effective but in experienced hands there is less danger of the casting being caught and thrown out of the hands by the rotating instrument.
2. The wheels or points and the speed of their rotation should do the cutting. Excessive pressure heats the work, crushes the abrasive particles, causes the wheels to clog and glaze, and slows the cutting.
Figure 19-27 Investment mold has been trimmed to prepare the sprue end for burnout and subsequent complete adaptation to the casting machine.
www.konkur.in 265 Chapter 19 Laboratory Procedures A B Figure 19-28 A, An invested pattern is removed from the burnout oven following complete elimination of the wax pattern. B, An investment mold is placed into the casting machine. The induction casting process provides consistency to the casting procedure.
A B Figure 19-29 A, Divesting of the framework involves bulk removal of the investment. B, The framework is divested with aluminum oxide.
3. A definite sequence for finishing should be adopted and followed for every framework.
4. Clean polishing wheels should be used. If contaminated wheels are used, foreign particles may become embed-ded in the surface, which will lead to later discoloration.
5. Be sure that each finishing operation completely removes all scratches left by the preceding one. Remember that each successive finishing step uses a finer abrasive and therefore cuts more slowly and requires more time to accomplish.
www.konkur.in 266 Part II Clinical and Laboratory MAKING RECORD BASES Bases for jaw relation records should be made of materi-als that possess accuracy or those that can be relined to provide such accuracy. Relining may be accomplished by seating the previously-adapted base onto the tinfoil or lubricant in the cast with an intervening mix of zinc oxide–eugenol paste or with autopolymerizing acrylic-resin. Some use has been made of mercaptan and silicone impression materials for this purpose, but the wisdom of using an elastic lining material for jaw relation record bases is questionable. When rigid setting materials are used for this purpose, any undercuts on the cast must be blocked out with wax or clay to facilitate their removal without damage to the cast.
The ideal jaw relation record base is one that is processed (or cast) to the form of the master cast, becoming the perma-nent base of the completed prosthesis. Cast metal bases for complete or removable partial dentures offer this advantage, as do acrylic-resin bases that are processed directly to the mas-ter cast, thus becoming the permanent denture base. When undercuts are present, the master cast will be destroyed during removal of the base. Then existing undercuts must be blocked out inside the denture base before dental stone is poured into it to make a cast for articulator mounting. A second cast, which includes the undercuts, must be poured against the entire base to support it when the overlying acrylic-resin is processed. With both the processed base and the overlying acrylic-resin, some care must be taken to prevent visible junction lines between the original acrylic-resin base and the acrylic-resin that supports the teeth and establishes facial contours.
Some autopolymerizing acrylic-resin materials are suffi-ciently accurate for use as jaw relation bases. These are used with a sprinkling technique, which, when properly done, permits formation of a base that compares favorably with a Figure 19-30 Gross finishing is accomplished with metal finishing burs, abrasive stones, or sintered diamonds (except at contact areas).
A B Figure 19-31 The framework is finished with a rubber polishing point before final “high-shine” polish in a region adjacent to the teeth (A) and at the periphery of a major connector (B).
Figure 19-32 Electropolishing unit used for the final surface polish of frameworks in a heated polishing liquid.
www.konkur.in 267 Chapter 19 Laboratory Procedures processed base or a visible light polymerized base. A mate-rial must be selected that will polymerize in a reasonable time (usually a 12-minute monomer) and will retain its form during the sprinkling process. Because polymerization with typical shrinkage toward the cast begins immediately, alter-nate addition of monomer and polymer in small increments results in reduced overall shrinkage and greater accuracy. Another option is to use visible light-cured (VLC) denture base materials, which employ a technique similar to that described in Chapter 16 for making custom impression trays.
Technique for Making a Sprinkled Acrylic-Resin Record Base The technique for sprinkling a record base is described here because the techniques used for the VLC impression tray and record bases were presented in Chapter 16 and elsewhere in this chapter. Some blockout and lubrication of the cast are necessary (Figure 19-34). Relief of undercut areas on the cast is best accomplished with a water-soluble modeling clay or baseplate wax. Modeling clay is easily formed and shaped on the cast and is easily removed from the cast or the base with a natural-bristle toothbrush under warm running water. Wax must be flushed off the cast with hot water and possibly removed from the inside of the base before use.
Bases for jaw relation records must have maximum con-tact with the supporting tissue. The accuracy of the base will proportionate to the contact provided to the total area of inti-mate tissue. Those areas are most often undercut and require blockout of the distolingual and retromylohyoid areas of the mandibular cast, the distobuccal and labial aspects of the A B Figure 19-33 Attention to detail in forming the wax pattern not only ensures the quality of the framework, it also saves time in finish-ing and polishing the resultant casting. A, Wax pattern of the mandibular framework, which incorporates three wrought-wire retainers. B, The same framework following casting shows a smooth casting surface and attention to detail regarding casting completeness and clasp form.
A B Figure 19-34 A, Preparation of Kennedy Class I, modification 1 distal extension for sprinkle-on record base technique. Cast under-cuts are blocked out with wax, and the cast is coated with separator. B, Peripheral extent of the base (the same as peripheral extent of the prosthesis captured during border molding and an impression-making procedure for this altered cast) is outlined with rope wax to contain resin.
www.konkur.in 268 Part II Clinical and Laboratory maxillary cast, and, frequently, small multiple undercuts in the palatal rugae. These areas and any others are blocked out with a minimum of clay or wax, to obliterate as little of the surface of the cast as possible. A close-fitting base may then be made that will have the necessary accuracy and stability and yet may be lifted from and returned to the master cast without abrading it.
The cast and the blockout or relief are coated with a sepa-rating medium. Following this, the cast is wet with the mono-mer from a dropper bottle. After the surface has been wetted with the monomer, the polymer is sprinkled or dusted onto the wet surface until all the monomer has been absorbed. Sprinkling is best accomplished with a large-mouthed bot-tle with a single hole in the lid near the rim. This facilitates placement of the polymer without excess in any one area. A flexible bottle with a suitable applicator tip also may be used. The objective should be to apply polymer uniformly over the entire ridge rather than allowing excess to accumulate at the border to be trimmed later. An autopolymerizing acrylic-resin material must be used that will retain its form during the sprinkling procedure without objectionable flow into low areas (Figures 19-35 and 19-36).
Once the polymer has been sprinkled in slight excess, the monomer is again added. Flooding must be prevented; therefore the monomer must be directed over the entire surface gradually until the polymer has just absorbed the monomer. A few seconds delay before the addition of excess monomer allows the mass to reach a tacky consistency and prevents it from flowing when more monomer is added. Then the monomer may be added in excess and is immedi-ately absorbed by the application of more polymer, as before. This process is repeated selectively until a uniform layer has been built up that is just thick enough so that none of the underlying cast or relief may be seen.
The final step in sprinkling is the addition of monomer sufficient to leave a wet surface. Immediately, the cast should be placed in a covered glass dish or covered with an inverted bowl. This permits final polymerization in a saturated atmo-sphere of monomer and prevents evaporation of surface monomer. The cast should not be placed in water nor any attempt made to accelerate polymerization. Slow polymer-ization is necessary so that shrinkage toward the cast will occur. Only then will overall shrinkage be negligible and accuracy of fit ensured. This is of little consequence in mak-ing an impression tray, but it is most essential in making a sprinkled acrylic-resin base.
Although polymerization will be about 90% complete within an hour and an impression tray may even be lifted within a half-hour, it is critical that a sprinkled denture base should be left overnight before it is separated from the cast. It should be lifted dry or under lukewarm tap water. It should not be immersed in hot water because some warp-ing may occur.
A sprinkled acrylic-resin base made with the precautions outlined previously will retain its accuracy for days, or even for an indefinite period, similar to a heat-polymerized resin base (Figure 19-37) or a VLC base.
OCCLUSION RIMS It has been explained that jaw relation records for remov-able partial dentures always should be made on accurate bases that may be part of the denture casting itself or may be attached to it in exactly the same relation as the final den-ture base will be. Further, it has been stated that although use of the final denture base is best for jaw relation records, a sprinkled or corrected acrylic-resin base may be used satis-factorily. In any case, accuracy of the base supporting a max-illomandibular record must be ensured before the function of occlusion rims is considered.
Figure 19-35 The cast is wetted with monomer, and polymer resin is added in increments to uniformly control thickness. Use of a typical eyedropper may be difficult if the monomer addition can-not be controlled. Use of peripheral rope wax helps in this regard.
Figure 19-36 The record base is complete when a uniform thickness is created that provides strength and accuracy. The completed unpolymerized resin is covered to ensure polymeriza-tion without loss of surface monomer.
www.konkur.in 269 Chapter 19 Laboratory Procedures Occlusion rims may be made of several materials. The material that is most commonly used to establish static occlusal relationships is the hard baseplate wax rim. How-ever, use of a wax occlusion rim can be inaccurate when the occlusal portion of the rim is mishandled. When some soft material that sets to a rigid state, such as impression plaster or bite registration paste, is used in conjunction with wax rims to record static occlusal relations, many of the errors common to wax rims are eliminated—provided some space for the material exists between the occlusion rims, the opposing teeth, or both, at the desired vertical dimension to be recorded. Registration made on wax occlusion rims with the use of a wax registration material must be handled care-fully and mounted immediately.
Occlusion rims for static jaw relation records should be so shaped that they represent the lost teeth and their support-ing structures (Figure 19-38). An occlusion rim that is too broad and is extended beyond where prosthetic teeth will be located is inexcusable. Such rims will substantially alter the shape of the palatal vault and the arch form of the mandibu-lar arch, will crowd the patient’s tongue, will have an unwel-come effect on the patient, and will offer more resistance to jaw relation recording media than will a correctly-shaped occlusion rim.
Modeling plastic (compound) has several advantages and may be used rather than wax for occlusion rims. It may be softened uniformly by flaming, yet when chilled it becomes rigid and sufficiently accurate. It may be trimmed with a sharp knife to expose the tips of the opposing cusps to recheck or position an opposing cast into the record rim. Opposing occlusion rims of modeling plastic may be keyed with greater accuracy than opposing wax rims. Preferably, however, even those should be trimmed short of contact at the vertical dimension of occlusion, and bite registration paste should be interposed for the final record. As with wax rims, an adjustable frame may be used to support the final record.
Occlusion rims made of extra hard baseplate wax or mod-eling plastic may be used to support intraoral central bear-ing devices, intraoral tracing devices, or both. Because of its greater stability, modeling plastic is preferable for this pur-pose when the edentulous situation permits the use of flat plane tracings. An example of such a situation occurs when an opposing complete denture is made concurrently with the removable partial denture. In such a situation, model-ing plastic occlusion rims provide greater stability than wax rims, with corresponding improvement in the predictable accuracy of such a jaw relation record. Although sealing opposing occlusion rims or using clips for complete denture jaw relation records may be acceptable, particularly for an initial articulator mounting, the existence of a removable partial denture framework makes this practice hazardous. The framework with attached base should be seated accu-rately on its cast before the opposing cast is repositioned in occlusion to it, because it is necessary for the dentist to be able to see the removable partial denture framework is in its designed terminal relation to the supporting teeth before articulating the casts.
Occlusion rims for recording functional, or dynamic, occlusion must be made of a hard wax that can be carved by the opposing dentition. This method, outlined in Chapter 18, presumes that the opposing arch is intact or Figure 19-37 Once completely polymerized, the record base and the framework can be removed from the cast, finished, and prepared for addition of the record base. The tissue side of the record base (intaglio) should possess similar accuracy and stabil-ity as are seen with the completed prosthesis. Such a record base provides a significant advantage for jaw relation records when minimal teeth remain and ridge configurations along with exten-sive base areas place a premium on base accuracy and stability.
Figure 19-38 Occlusion rims are added to allow recording of jaw relation records. Placement of the wax record is dictated by the opposing tooth position and the supporting ridge character. When possible, the occlusion rim should allow recording of the jaw position within the primary bearing area of the ridge.
www.konkur.in 270 Part II Clinical and Laboratory has been restored. Functional occlusion records cannot be made by this method when both arches are restored simultaneously. Rather, an opposing arch must be as intact as the treatment plan calls for, or it must be restored by whatever prosthetic means the situation dictates. Oppos-ing removable partial dentures or an opposing complete denture may be carried concurrently up to the final occlusal record. One denture is then completed and the functional record is made in opposition to it. Often, this necessitates that all opposing teeth are articulated first in wax to establish optimum ridge relations and the correct occlusal plane. One denture is then carried to comple-tion, and the teeth that remain in wax on the opposing denture are removed while the functional occlusal record is being made.
The laboratory steps required for establishing an occlusion with the use of functional occlusal records were described in Chapter 18; however, more detail is provided in the following section. Some inlay waxes are used for this purpose because they can be carved by the oppos-ing dentition, and because most of them are hard enough to support occlusion over a period of hours or days. A wax for recording functional crown and bridge occlusion, because it is established entirely in the dental office, is selected on the basis of how well it may be carved by the opposing dentition in a relatively short time. Therefore, a softer wax may be used than is required for the recording of occlusal paths over 24 hours or longer. For this latter purpose, hard inlay wax seems to satisfy best the require-ments for a wax that is durable yet capable of recording a functional occlusal pattern. This wax is packaged in the form of sticks. A layer of hard, sticky wax is first flowed onto the surface of the denture base. Two sticks of the inlay wax are then laid parallel along the longitudinal center of the denture base and are secured to it with a hot spatula. This is the only preparation needed before the dental appointment. Because neither the height nor the width of the occlusion rim can be known in advance, and because deep warming of a chilled wax rim is difficult, the rim is not completed before the appointment.
With the patient in the chair, a hot spatula is inserted into the crevice between the two sticks of wax, making the center portion fluid between two supporting walls. Some transfer of heat to the supporting walls occurs, resulting uniform softening of the occlusion rims. The patient is asked to close into this wax rim until the natural teeth are in contact; this establishes both the height and the width of the occlusion rim. Wax is added or carved away as indi-cated, and the patient is asked to make lateral excursions. Any excess wax is then removed, and any unsupported wax is supported by addition. Finally, wax is added to increase the occlusal vertical dimension sufficiently to allow for (1) denture settling, (2) changes in jaw rela-tions brought about by the reestablishment of posterior support, and (3) carving in all mandibular excursions. When sufficient height and width have been established to accommodate all excursive movements, the patient is given instructions for chewing in the functional record and is then dismissed.
Although this discussion has been included in the chapter on laboratory procedures, the entire procedure of establishing occlusion rims is for the purpose of recording functional occlusion and should be considered a chairside procedure rather than a laboratory procedure. It is neces-sary that the purpose of a functional occlusal record be clearly understood so that subsequent laboratory steps may be accomplished in such a manner that the effect of the occlusal record may be reproduced on the finished denture.
MAKING A STONE OCCLUSAL TEMPLATE FROM A FUNCTIONAL OCCLUSAL RECORD After final acceptance of the occlusal record developed by the patient, the effectiveness of this method for estab-lishing functional occlusion on the removable partial denture depends on how accurately the following pro-cedures are carried out. For this reason, it will be given as a step-by-step procedure. If the base of the master cast (or processing cast) has not been keyed previously, do this before proceeding. Reduce the thickness and width of the base if it is so large that difficulty will be encountered in flasking. The base may not be reduced after removal from the articulator because the mounting record would be lost.
Keying may be done in several ways, but a method whereby the keyed portions are visible on the articula-tor mounting eliminates some possibility of remount-ing error. According to the preferred method, form a 45-degree bevel on the base of the cast by hand or with the model trimmer, and then add three V-shaped grooves on the anterior and posterior aspects of the base of the cast at the bevel. The bevel serves to facilitate reseating of the cast on the articulator mounting, and the mount-ing surfaces are made still more definite by the triangular grooves. Once placed at the beveled margin, the triangu-lar grooves are visible at all times, and any discrepancy may be clearly seen.
Inspect the underside of the cast framework and the denture bases, removing any particles of wax or other debris. Similarly, inspect and clean the master cast of any particles of stone, wax, or blockout material, or any other debris that might prevent the casting from being seated accurately on it.
Now seat the denture framework on the cast in its orig-inal terminal position. This is the position that is main-tained by securing it with sticky wax with all occlusal rests seated while the trial denture base is made. It is also the position that the casting assumed in the mouth while the occlusal record was being made and that must be dupli-cated on return of the denture framework to the mas-ter cast. While holding the framework in this terminal www.konkur.in 271 Chapter 19 Laboratory Procedures position, secure it again with sticky wax. (If a processing cast is used in place of the master cast, the denture base will have been made on that cast, and the same precau-tions used in returning the framework to its original posi-tion apply.) With the denture framework and the occlusal record in position, form a matrix of clay around the occlusal record to confine the hard stone, which will form the stone occlusal template. The clay matrix is the same for a metal or electroplated surface as for the wax record. The clay matrix should rise at a 45-degree angle from the buc-cal and lingual limits of the occlusal registration. Then arch the clay or a sheet of wax across from one side to the other, forming a vault that will permit lingual access when the teeth are articulated.
Leave the occlusal surfaces of a processing cast exposed so that they may act as vertical stops. This will serve to maintain the vertical relation in the articulator. Unless such stone-to-stone stops are used, the technician may alter the vertical relation in the articulator, accidentally or otherwise. Any change in vertical relations is incom-patible with a concept of dynamic conclusion because the occlusal pattern is directly related to the degree of jaw separation. Although it may be true that occlusal vertical dimension may be changed when casts are mounted in relation to the opening axis of the mandible, as long as natural cusps remain to influence mandibular movement, the occlusal vertical relation established with a functional occlusal record must not be changed in the articulator.
Cover the surfaces of the adjacent abutment teeth left exposed with sodium silicate, microfilm, or some other separating medium to ensure separation of the stone ver-tical stops. If the wax record has not been electroplated, use a hard dental stone to form the opposing template. This may be an improved stone, but the use of a stone die material is preferred. Only the occluding surface needs to be poured in the harder stone, with a less costly laboratory stone used to back it up. If this is done, add the second layer to the first before the former takes its initial set to prevent any possibility of accidental separation between the two materials.
Vibrate the stone only into the wax registration and against the stone stops. Pile on the rest of the stone, and leave it uneven to facilitate attachment to the mounting stone. Attach the occlusal template to the articulator with-out provision for removal or remounting because only the working cast needs to be keyed for remounting.
After the stone template has set, attach the occluded casts to both arms of the articulator before separating the casts. The type of articulating instrument used is of little importance because all eccentric positions are recorded on the template, and whatever instrument is used acts purely as a simple hinge or a tripod. Therefore any labora-tory articulator or tripod may be used.
Casts should be attached to the articulating instrument with stone rather than with plaster. Mounting stones are available that have been especially formulated and pre-pared to minimize the setting expansion inherent in most gypsum products. The least amount of setting expansion of the mounting medium is most desirable to maintain the intended relationship of the opposing casts.
One must remember which arch will be movable as a working cast, and the articulator mounting should be made accordingly. For example, for a mandibular denture, the template is attached to the upper arm of the articu-lator. For a maxillary denture, the template is mounted upside down on the lower arm. The keyed base of the working cast attached to the opposing arm must be coated with a light coat of microfilm, mineral oil, or petroleum jelly to facilitate its separation from the mounting stone.
After the mounting has been completed, separate the casts and remove the clay. The template, with its mount-ing, may be removed from the articulator if a mounting ring or mounting stud permits; otherwise, trimming must be done on the articulator. With a pencil, outline the limits of the occlusal registration and any excess stone around its borders. With a knife, trim the vertical stops to a sharp edge on the buccal surface where they contact the working cast.
Remove any overhanging stone, leaving the occluding template and vertical stops clearly visible and accessible. Remove the wax registration before arranging artificial teeth to the occluding template.
ARRANGING POSTERIOR TEETH TO AN OPPOSING CAST OR TEMPLATE Whether posterior teeth are to be arranged to occlude with an opposing cast or with an occlusal template, the denture base on which the jaw relation record has been made must first be removed and discarded unless metal bases are part of the denture framework, or heat-polymerized acrylic-resin bases were used. This statement is based on the assumption that where an adjustable articulator has been used to develop the occlusion, the trial dentures have been evaluated, the articulator mounting has been proved, and the articu-lator has been programmed for eccentric positions. Because record bases that are entirely tissue supported have no place in recording occlusal relations for remov-able partial dentures, the bases must be attached to the denture framework. Metal bases that are part of the prosthesis present no problem. The teeth may be arranged in wax or replaced on the metal base, depend-ing on the type of posterior tooth used, and these must be occluded directly to the opposing cast or template.
Unless occlusal relations are recorded on final acrylic-resin bases, autopolymerizing acrylic-resin bases attained by the sprinkling method or VLC bases are the most accurate and stable of bases that may be used for this purpose. (An alternate method is relining of the original impression bases, thus accomplishing the same purpose.) www.konkur.in 272 Part II Clinical and Laboratory Although static relations may be recorded successfully on corrected bases, functional registrations are best accom-plished on new acrylic-resin bases made for that purpose. In either situation, the denture cannot be completed on these bases; neither can the bases be removed conve-niently from the retentive framework during the boil out after flasking. Therefore the metal framework must be lifted from the cast, and the original record base removed by flaming its underside. Care must be taken not to allow the acrylic-resin to burn, or the cast framework will become discolored with carbon. The framework is repol-ished and is then returned to its original position on the master cast and secured there with sticky wax before the artificial teeth are arranged.
Posterior Tooth Forms Posterior tooth forms for removable partial dentures should not be selected arbitrarily. One should bear in mind that the objective in removable partial denture occlusion is harmony between natural and artificial dentition. Whether the teeth are arranged to occlude with an opposing cast or to an occlu-sal template, they should be modified to harmonize with the existing dentition. In this respect, removable partial denture occlusion may differ from complete denture occlusion. In the latter case, posterior teeth may be selected and articulated according to the dentist’s concept of what constitutes the most favorable complete denture occlusion, whereas remov-able partial denture occlusion must be made to harmonize with an existing occlusal pattern. Thus, the occlusal surfaces on the finished removable partial denture may bear little resemblance to the original occlusal surfaces of the teeth as manufactured.
Artificial tooth forms should be selected to restore the space and fulfill the esthetic demands of the missing den-tition. Manufactured tooth forms usually require modifica-tion to satisfactorily articulate with an opposing dentition. The original occlusal form, therefore, is of little importance in forming the posterior occlusion for the removable partial denture.
The posterior teeth are most often made of resin (includ-ing all resin forms—composite, interpolymer network, cross-link, double cross-link, and so on). An advantage of resin teeth is that they are more easily modified and subse-quently reshaped for masticating efficiency by the addition of grooves and spillways. Resin teeth are also more easily nar-rowed bucco-lingually to reduce the size of the occlusal table without sacrificing strength or esthetics. They may be more easily ground to fit minor connectors and irregular spaces and to avoid retentive elements of the removable partial den-ture framework. However, when resin teeth are used, the occlusion must be evaluated periodically to make sure that the occlusal surfaces have not worn out of contact. It seems that the best combinations of opposing occlusal surfaces used to maintain the established occlusion and to prevent deleterious abrasion are porcelain-to-porcelain surfaces, gold surfaces to natural or restored natural teeth, and gold surfaces to gold surfaces.
Arranging Teeth to an Occluding Surface The procedure for arranging teeth to a static relationship with an opposing cast is essentially the same as for arranging teeth to an occluding template. On the other hand, articu-lation of artificial teeth on an adjustable instrument, which reproduces to some extent mandibular movement, will fol-low more closely the customary pattern for complete denture occlusion. The procedure for arranging posterior teeth to an occluding template was presented in Chapter 18.
TYPES OF ANTERIOR TEETH Anterior teeth on removable partial dentures are concerned primarily with esthetics and the function of incising. These are best arranged when the patient is present because an added appointment for try-in would be necessary anyway. They may be arranged arbitrarily on the cast and then tried in, but a stone index of their labial surfaces should be made on the master cast after the final arrangement has been estab-lished to preserve the arrangement that the patient saw and approved.
From a purely mechanical standpoint, all missing anterior teeth are best replaced with fixed restorations rather than with the removable partial denture. However, for cosmetic or economic reasons, or in situations in which several missing anterior teeth are involved—such as in a Class IV partially edentulous arch—their replacement with the removable par-tial denture may be unavoidable.
Some types of anterior teeth used in removable partial dentures are as follows: 1. Porcelain or resin teeth, attached to the framework with acrylic-resin.
2. Ready-made resin teeth processed directly to retentive elements on the metal framework with a matching resin. This is called a pressed-on method and has the advan-tage of permitting previous selection and evaluation of the anterior teeth, plus the advantage of the use of ready-made resin teeth for labial surfaces. These are then hollowed out on the lingual surface to facilitate their permanent attachment to the denture framework with the resin of the same shade.
3. Resin teeth processed to a metal framework in the labora-tory. Tooth forms of wax may be carved on the remov-able partial denture framework and tried in the mouth, adjusted for esthetics and occlusion, and then processed in an acrylic-resin of a suitable shade. There is some question as to whether the shade and durability of such teeth are comparable with those of manufactured resin teeth, but improvements in materials have led to im-proved quality and appearance of laboratory-made teeth. Moreover, such teeth may often be shaped and charac-terized to better blend with adjacent natural teeth.
www.konkur.in 273 Chapter 19 Laboratory Procedures 4. Porcelain or acrylic-resin facings cemented to the den-ture framework. These may be tried in the mouth on a baseplate wax base and adjusted for esthetics. Ready-made plastic backings may be used, which become part of the pattern for the removable partial denture framework. The teeth are then ultimately cemented to the framework. Esthetically, these are less satisfac-tory than other types of anterior teeth, but because the plastic backing is cast as part of the removable partial denture framework, they have the advantage of greater strength and are replaced easily. A record of the mold and shade of each tooth should be kept, and only the ridge lap of the replacement teeth needs to be ground to fit. When replaceability is the main reason for its use, the stock facing should not be beveled or difficulty will be encountered in replacing it. Replacement also may be accomplished by waxing and processing a resin tooth that directly faces the metal backing. Stock tube or side-groove teeth are not ordinarily used for ante-rior teeth on removable partial dentures because of the horizontal forces that tend to dislodge them.
5. Anterior resin denture teeth can be modified to be used as resin veneers, the same as for veneer crowns and veneer pontics on fixed partial dentures. This is most applicable when the removable partial denture framework is to be cast in an alloy. Then labial surfaces may be waxed and the final carving for esthetics done in the mouth.
A modification of this method is the waxing of the ve-neer coping on a previously cast metal base. These then are cast separately and are attached to the framework by soldering. Esthetically, the result is comparable with that obtained with resin veneer crowns. This method is par-ticularly applicable when there is a desire to make the re-placed teeth match adjacent veneered abutment crowns.
WAXING AND INVESTING THE REMOVABLE PARTIAL DENTURE BEFORE PROCESSING ACRYLIC-RESIN BASES Waxing the Removable Partial Denture Base Waxing the removable partial denture base before invest-ing differs little from waxing a complete denture. The only difference is the waxing of and around exposed parts of the metal framework. At the framework–denture base junction, undercut finishing lines should be provided whenever pos-sible. Then the waxing is merely butted to the finishing line with a little excess to allow for finishing. Otherwise, small voids in the wax may become filled with investing plaster, or fine edges of the investment may break off during boil out and packing. In either situation, small pieces of investment may become embedded in the acrylic-resin at the finishing lines. This is prevented by slightly over waxing and then fin-ishing the acrylic-resin back to the metal finishing line with finishing burs. Abrasive wheels and disks should not be used for this purpose, because they will cut into the metal and may burn the acrylic-resin. Pumice and a rag wheel should be used sparingly for polishing, because they will abrade the acrylic-resin more rapidly than the metal and will leave the finishing line elevated above the adjacent acrylic-resin.
When waxing to polished metal parts that do not possess a finishing line is done, it must be remembered that no attach-ment will exist, and that over a period of time, some seepage, separation, and discoloration of the acrylic-resin in this area are inevitable. This may be prevented to some extent by rough-ening the metal whenever possible to effect some mechanical attachment by silicoating the attachment or by using one of the resin adhesives. The wax should be left 1.5 to 2.0 mm thick so that the acrylic-resin will have some bulk at its junction with the polished metal. Thin films of acrylic-resin over metal should be avoided. In finishing, these should be cut back to an area of bulk with finishing burs. Otherwise, any thin acrylic-resin film eventually will separate and become discolored and unclean as a result of marginal seepage.
Gingival forms should be waxed in accordance with modern concepts of esthetics and should be made to pre-vent entrapment of food particles. Dental students should become familiar with normal gingival architecture as found on diagnostic casts of natural dentitions, beginning in basic technique exercises with the casts of each other’s mouths. In this manner, they may acquire a better concept of gingival contours to be reproduced on prosthetic restorations.
Artificial teeth should be uncovered fully to expose the entire anatomic crown and beyond when gingival recession is simulated. Adjacent or contralateral tooth-gingival rela-tionships should be used as a guide to facilitate the harmoni-ous esthetic presentation of the gingival contours. Relatively few prosthodontic patients are in an age bracket in which some gingival recession and exposed cementum would not normally be present; this should be simulated on prosth-odontic restorations in proportion to the patient’s age. With removable partial dentures, gingival contours around the remaining natural teeth should be used as a guide to the gin-gival contours to be reproduced on the prosthesis. However, interproximal spaces are almost always filled, particularly between posterior artificial teeth.
Frush has listed the following rules for varying the height of the gingival tissue at the cervical portion of the teeth: 1. Slightly below the high lip line at the central incisors.
2. Lower than the central incisor gingival margin at the lateral incisors.
3. Higher than the central or lateral incisor gingival mar-gin at the canine.
4. Slightly lower than the canine at the premolar and variable for both premolars and molars. The correctly formed papilla should be shaped so that it will be self-cleansing. It should be carved so that it is in harmony with the interpretation of age and will be the From Frush JP: Dentogenic restorations and dynesthetics, Los Ange-les, 1957, Swissdent Foundation.
www.konkur.in 274 Part II Clinical and Laboratory deciding factor in the visible outline form of the tooth. As Frush has pointed out, even a drop of wax properly placed can change the appearance of a square tooth to a tapering or ovoid appearance. A properly formed papilla further enhances the natural appearance by increasing the color in this area.
The rules for forming the papilla were given by Frush as follows: 1. The papilla must extend to the point of tooth contact for cleanliness.
2. The papillae must be of various lengths.
3. The papilla must be convex in all directions.
4. The papilla must be shaped according to the age of the patient.
5. The papilla must end near the labial face of the tooth and must never slope inward to terminate toward the lingual portion of the interproximal surface.
The denture should be waxed and carved as for a cast res-toration, which it actually is, regardless of the material or the method of processing used. The fact that a split-mold tech-nique is used for processing does not alter the fact that the form of the denture base is to be reproduced by a casting procedure. Therefore, the denture pattern should be waxed with care in the same form as that desired for the finished restoration, rather than attempting to shape facial contours on the prosthesis during the polishing phase (Figure 19-39). Polishing should consist primarily of trimming away the flash, stippling polished surfaces when desired, and polish-ing lightly with brush wheels and pumice, followed by final polishing with a soft brush wheel and a nonabrasive shin-ing agent, such as whiting. Gross trimming and polishing with pumice should not be necessary if the denture has been properly waxed before investing.
Because the polished surfaces of any denture play an important part in both retention and control of the food bolus, buccal and lingual contours generally should be made concave. In most situations, the border thickness of the den-ture should be left as recorded in the impression. The only exceptions are the distolingual aspect of the mandibular denture base, to prevent interference with the tongue, and the distobuccal aspect of the maxillary denture base, to pre-vent interference with the coronoid process of the mandible. These are the only areas that cannot ordinarily be waxed to final contour before investing and may need to be thinned by the dentist at the time of final polishing.
Investing the Removable Partial Denture In investing a removable partial denture for processing an acrylic-resin base, it must be remembered that the denture cast must be recovered from the flask intact for remount-ing. The practice of cutting the teeth off the cast to expose the connectors and retainers, which are then embedded in the upper half of the flask, is permissible only when an existing denture base is being relined and no provision has been made for remounting. (In such a situation, it seems that this practice offers no advantage over investing the denture that is being so relined upside down in the lower half of the flask.) Because in the past some increase in occlusal verti-cal dimension has been inevitable in any split-mold process-ing technique, this method results in raising the removable partial denture framework from the supporting teeth by the amount of increase. Occlusal adjustment in the mouth may temporarily reestablish a harmonious occlusal relation with the opposing teeth, whereas the removable partial denture framework will then settle into supporting contact with the abutment teeth at the expense of the underlying ridge.
Changes in occlusal vertical dimension may be held to a minimum by the use of acrylic denture resins that can be placed in the mold in a fluid rather than in a doughy state or those that may be injected in a fluid state into a closed mold. Dimensional changes that occur during relining may also be held to a minimum by the use of autopolymeriz-ing acrylic-resins for this purpose, thus preventing thermal expansion of a mold subjected to elevated temperatures.
When two opposing removable partial dentures are being made concurrently, one is sometimes processed and placed first, and then the final occlusion is established on the second denture to a fully restored arch. In such a situation, when no natural teeth are in opposition, it is not necessary for the first denture to be remounted after pro-cessing. In all other situations, remounting to correct for Figure 19-39 Waxing of the removable partial denture base should reproduce anatomic contours normal for the specific pa-tient’s characteristics. This is especially important for regions where the junction of denture tooth and resin will be visible. Healthy interdental papillae are convex and extend to contact points of the teeth. Root prominence can be carved into wax to create a natural appearance. Stippling is typically accomplished after processing done with the use of an eccentric round bur. All of these features should be considered against the specific oral environment in which the prosthesis will be placed. A finished and polished prosthesis that demonstrates esthetic features not found in the patient’s mouth may be considered objectionable.
www.konkur.in 275 Chapter 19 Laboratory Procedures errors in occlusion is absolutely necessary. Flasking must be accomplished so that the cast may be recovered from the flask undamaged.
Minute voids in the base of the cast will have been reproduced in the stone mounting, and although the obvious larger blebs may be trimmed away, smaller blebs will remain. If the voids in the cast become filled with investing material, the effect is that two particles are try-ing to occupy the same space. Covering the base with tin-foil substitute before investing can prevent this. Coating the base and sides of the cast with petroleum jelly not only keeps the base of the cast isolated from investing material but also may allow the cast to be more easily recovered from the surrounding investment.
The entire cast, except for the wax and teeth, may then be invested in the lower half of the flask (Figure 19-40). As with a complete denture, only the supplied teeth and wax are left exposed to be invested in the upper half. Also, as with a complete denture, the investment in the lower half must be smooth and free of undercuts and must be coated with a separator to facilitate separation of the two halves of the flask.
An alternate and preferred procedure is to invest the cast only to the top of the tinfoil on the base, smooth-ing the investment and applying a reliable separator. Then a second layer of investment placed around the ana-tomic portion of the cast covers the natural teeth and the exposed parts of the denture framework. This is likewise smoothed and made free of undercuts and coated with a separator before the top half of the flask is poured. Recov-ery of the cast is thus made easier by having a shell of investment over the anatomic portion of the cast, which may be removed separately.
When the denture base is to be characterized by applying tinted acrylic-resins to the mold, care should be taken not to embed the wax border in the lower half of the flask. Bennett has pointed out the need for investing only to the border of the wax, leaving the entire surface that is to be tinted reproduced in the upper half of the flask. With this precaution, tinting may be carried all the way to the border, and later removal of the flask will not mar the tinted surface. If tinting is not to be done or is to be done only at the cervical margins of the teeth and the papillae, the wax border should be embedded in the lower half, where it may be faithfully reproduced and preserved during polishing.
The use of acrylic-resin materials that require trial packing is complicated by the presence of the retentive framework of the removable partial denture. With their use, trial packing must be done with two sheets of cello-phane between two layers of the resin dough; otherwise, the flask could not be opened without pulling the resin away from the teeth in one half of the flask or the metal framework in the other. Acrylic-resin dough is placed in A B C D Figure 19-40 Prostheses invested in processing flasks. A, Mandibular Class I master cast with exposed minor connectors of the distal extension bases. B, Denture teeth for framework shown in A. C, Maxillary distal extension minor connector embedded in investing stone. D, Opposing flask unit with denture teeth for distal extension shown in C.
www.konkur.in 276 Part II Clinical and Laboratory each half of the flask, the sheets of cellophane are placed between them, and the flask is closed for trial packing. The flask is then opened, the cellophane is removed, and the excess flash is trimmed away. Trial packing should be repeated until no excess is visible. Final closure is accom-plished without the intervening sheets of cellophane.
Acrylic-resin materials have been developed that require no trial packing. These are mixed as usual but may be poured into the mold or placed into the mold in a soft state. They offer little or no resistance to closure of the flask, yet the finished product is comparable with acrylic-resin materials packed in a doughy state. They must be used in some excess, with the excess escaping between the halves of the flask. Although they are soft enough to allow the escape of gross excess, the use of a land space is advisable to prevent a thin film from forming on the land area. Any film that exists on the land area after deflask-ing may be interpreted as an opening of the flask by that amount, hence the need for some provision for an inter-vening space to accommodate the excess and to facilitate its escape as the flask is closed.
To provide such a land space, the land area on the lower half of the flask may be painted with melted baseplate wax before the top half is poured. After wax elimination, a space then remains to accommodate any excess acrylic-resin remaining after the flask has been completely closed. It is necessary that no plaster or wax is allowed to remain on the rim of the flask, and that the flask makes metal-to-metal contact before the second half is poured. Only in this way is it possible to see that the flask is completely closed before it is placed in the curing unit.
Pouring of the top half of the flask follows the same procedure as that used with a complete denture. Although it is not absolutely necessary for the entire top half to be poured in stone, it is necessary for a stone cap of some type to be used to prevent tooth movement in an occlusal direction. This is so because of the inability of plaster to withstand closing pressures. All plaster remaining on the occlusal surfaces of the teeth should be removed and a separator added before the stone cap is poured to facilitate its removal during deflasking. If the use of stone invest-ment is preferred, a shell of improved stone or die stone may be painted or applied with the fingers onto the wax and teeth and allowed to harden before the remainder of the flask is filled with plaster. If a full stone investment is preferred, some provision should be made for easy separation during deflasking. Not only should a separa-tor stone cap be used, but metal separators or knife cuts radiating out to the walls of the flask should be placed on the partially set stone. Deflasking then is easily accom-plished by removing the stone cap and inserting a knife blade between the sections of stone.
Boil out should be deferred until the investing material has set for several hours or preferably overnight. Boil out must effectively eliminate all waxy residue; an adequate source of clean hot water must be available. Immersion of a flask containing the invested denture in boiling water for 5 minutes will adequately soften the wax supporting the artificial teeth so that flask halves may be separated and the remaining wax flushed out. After wax elimination with boiling water, the invested denture should be flushed with a solution of grease-dissolving detergent and again with clean boiling water.
Immediately after boil out, the warm mold should be painted with a thin film of tinfoil substitute, with care taken to not allow it to collect around the cervical por-tions of the teeth. No tinfoil substitute should be used on any part of the denture teeth. A second coat should be applied after the first coat has reasonably dried, and pack-ing of the mold should proceed immediately after this film has dried to the touch.
When the master cast for a distal extension remov-able partial denture has been repoured from a secondary impression, the supporting foot on the retention frame may not necessarily be in contact with the cast. Closing pressure within the flask may distort the unsupported extension of the metal framework, with subsequent rebound on deflasking. The finished denture base will then lack contact with the supporting tissue, resulting in denture rotation about the fulcrum line, similar to that which occurs after tissue resorption. To provide support for the distal extension of the metal framework during flask closure, autopolymerizing acrylic-resin should be sprinkled or painted around the tissue stop at the distal end of the framework and allowed to harden before the denture resin continues to be packed (see Figure 5-38).
PROCESSING THE DENTURE Processing follows the same procedure as that used for a complete denture. Denture base characterization tinting may be added just before packing. This is most desirable if denture base material will be visible when in the mouth. Posterior acrylic-resin bases alone ordinarily do not require characterization, but the dentist should select a denture base material that closely resembles the color of the surrounding tissue. The ideal acrylic-resin base material for removable partial dentures therefore is one that (1) may be used without trial packing; (2) possesses a shade that is compatible with surrounding tissue; (3) is dimensionally stable and accurate; (4) is dense and lends itself to polishing; and (5) polymerizes completely.
There has never been any question concerning the merits of placing tinfoil over the denture before investing; this results in a tinfoil-lined matrix and eliminates the need for a separating film. The fact remains, however, that the use of a tinfoil substitute has become universal.
At best, any tinfoil substitute creates an undesir-able film at the gingival margins of the teeth, resulting in microscopic separation between the teeth and the www.konkur.in 277 Chapter 19 Laboratory Procedures surrounding acrylic-resin. This may be shown by section-ing a finished denture and by observing the marginal dis-coloration around the cervical portions of the teeth after several months in the mouth. To some extent, injection molding obviates this objection to the use of a tinfoil sub-stitute; this is one of the principal advantages of injection molding over compression molding.
Because the use of compression molding is widespread and is likely to continue, methods are needed that elimi-nate the use of a tinfoil substitute. The layered silicone rubber method results in more complete adaptation of the acrylic-resin around the cervical portions of porcelain teeth and more complete bonding to acrylic-resin teeth. In addition, denture base tints may be applied directly to the mold without first applying a separating film.
A room-temperature polymerizing silicone rubber that has sufficient body and toughness for the purpose is applied to the wax surface of the denture and over the teeth. To prevent movement of the teeth during processing, the occlusal surfaces should be exposed before the upper half of the flask is poured. The manufacturer’s instructions must be followed as to mixing and time elapsed before the outer stone investment is added, to ensure polymerization and bonding to the overlying investment. Boil out then is completed in the usual way.
A further advantage of the layered silicone rubber method is the ease of accomplishing deflasking. If the wax carving of the denture has been completed with care before flasking, denture tints remain unaltered by unnec-essary trimming and polishing of the processed denture.
All resin base materials available up to the present time exhibit some dimensional change, both during process-ing and in the mouth. The fit of the denture is therefore dependent on the accuracy of the denture base material because impression and cast materials in use today are themselves reasonably accurate. In an attempt to mini-mize dimensional changes in the denture base, materials and techniques are being improved constantly.
Denture base materials that may be poured into the mold or placed into the mold in a soft state are also popu-lar. This technique eliminates trial packing and excessive pressures, which lead to open flasks and altered occlu-sal vertical dimension as is sometimes experienced with compression molding of acrylic-resin base materials. Activated, or autopolymerizing, acrylic-resins are com-monly used to prevent mold expansion at higher tem-peratures. Materials other than acrylic-resins are used with various techniques, some of which include styrene, vinyl, and experimentally epoxy resins. The main objec-tive behind the development of newer techniques and materials is greater dimensional accuracy and stability, combined with improved strength and better appearance.
The use of injection molding or poured materials to process removable partial denture base materials com-bines accuracy and efficiency to help create a well-fitting denture base. The Success Injection System (Dentsply Trubyte, York, PA) combines the accuracy of injection molding with Lucitone 199 (Dentsply Trubyte). The hard-ware consists of the injection unit, aluminum alloy flasks, and associated system flask components. The flasks are numbered; specifically, they are mated halves that need to be matched for more accurate results. The investment and processing techniques are as follows: 1. The cast with the completed wax-up is embedded in the designated side “1” of the flask in the usual manner, placing the cast as close as possible to the back of the flask. After any undercuts on the cast are eliminated, flattened wax sticks (approximately 7 mm in diameter) are used to build the injection sprues. For maxillary re-movable partial dentures, attach the sprue to the poste-rior border, ensuring that the sprue is sufficiently wide. For mandibular removable partial dentures, position one sprue for each base extension. Apply a separator to the investment, and place the top half of the flask on the bottom half, ensuring complete intimate met-al contact and closure of the halves. Secure the flask brackets to the flask and tighten. Place the flask on the leveler with side “2” up, and complete the investing procedure in the usual manner.
2. When the investment has set, loosen the bolts and re-move the metal flask brackets. Place the flask in boiling water (8 to 10 minutes) and complete the boil out pro-cedure. Place the metal injection insert into the back of the flask, and slide the plastic injection socket into the metal insert as far as possible. The plastic injection socket lip should rest flush against the trim of the met-al injection insert. Close the flask, position the metal flask brackets, and tighten the bolts.
3. Use the powder/liquid vials to measure sufficient resin for the removable partial denture. Note that the maxi-mum powder and/or liquid that the injection cartridge can hold is 38 g (56 mL) powder and/or 17.5 mL liq-uid. Stir the powder and/or liquid for approximately 15 seconds. Do not excessively mix. Cover the mixing jar until the material reaches the “soft pack” stage (ap-proximately 6 minutes). Do not allow the material to reach the “snap set” stage.
4. Place the resin material into the plastic injection cartridge and insert the blue plastic cartridge plug into the large open end, ribbed side out. Push the blue cartridge plug in as far as possible to compress the material, and insert the cartridge nozzle into the plastic injection socket until it seats on the injection socket’s lip. Place the metal pro-tective sleeve over the cartridge, and place the flask in the injection unit, ensuring that the bolts and the flask brackets face to the operator’s right side. Position the open slots on the cartridge sleeve facing out, and then push the sleeve up toward the unit’s cross-head, secur-ing the sleeve around the blue rubber O-ring. Tighten the unit’s hand wheel to secure the flask.
5. Complete the injection process, ensuring that the mold is completely filled by viewing the blue cartridge plug www.konkur.in 278 Part II Clinical and Laboratory through the sleeve slots until the plug stops moving. When completed, remove the flask from the unit, re-move the cartridge sleeve, and pull the plastic cartridge out of the flask with a slight twist. Keep the injection socket in place inside the metal injection insert. Fit the small, blue plastic piston cap onto the end of the press-ing device piston. Place the piston of the processing device into the plastic injection socket at the back of the flask, and screw the pressing device onto the metal injection insert until the etched groove is visible on the pin at the top of the pressing device.
6. Allow the flask to sit for 30 minutes before heat-po-lymerizing to ensure a good bond between the denture resin and the teeth. Submerge the closed flask in water at 163 ± 2° F for 112 hours. Follow with an additional 30-minute boil. An alternate polymerization method is 9 hours in a water bath of 163 ± 2° F with no boil.
Remove the flask from the polymerization tank and al-low to air cool for approximately 30 minutes. Place the flask in a lukewarm water bath to cool completely.
7. Unscrew the pressing device and loosen the bolts on the flask. Remove the flask brackets and separate the flask.
Remove the investment and divest the removable par-tial denture and cast. Cut off the injection sprue(s), and finish and polish the removable partial denture in a conventional manner.
The use of VLC denture base materials is claimed to save considerable time in providing a processed base. Manufacturers recommend the use of a light-colored cast to enhance the polymerization of the VLC material. A stone matrix must be fabricated so that the denture teeth can be positively relocated in the same position dur-ing the polymerizing process. Once the stone matrix has been completed and verified, the wax and teeth can be removed from the framework, and they can be cleaned and silicoated or coated with the resin bonding agent.
The ideal thickness for the VLC material requires a 1.5-mm space between the edentulous ridge and the retentive component of the framework. The ridge lap of the denture teeth should also be 1.5 mm above the retentive component of the framework, and the tissue side finishing line of the framework should be even with or slightly higher than the palatal side finishing line. If VLC denture base material is to be used, these requirements must be considered at the diagnosis and treatment planning stages.
Without the framework in place, a thin coat of model release agent (MRA) is applied to the denture base areas of the cast, and the VLC material is adapted to the eden-tulous denture base area and is trimmed to the general outline with a sharp blade. The framework is then seated firmly by being embedded into the uncured VLC mate-rial. Make certain that the rests, tissue stops, and other components of the removable partial denture framework are correctly positioned in their designated terminal posi-tions on the cast. Remove any excess material that may interfere with articulation of the casts or positioning of the teeth, and check to ensure that the VLC material is adapted into the tissue side finishing line of the frame-work on the cast. This will necessitate returning the cast with the framework to the articulator. Once this has been verified, remove the cast from the articulator and process it in the light-polymerizing unit for 2 minutes.
The denture teeth can then be secured to the stone matrix and related to the cast with the framework and the removable partial denture base. The denture teeth may have to be adjusted before the removable partial denture base can be fitted. To do this, trim the ridge lap areas of the teeth enough to provide the required 1.5 mm of space between the tooth and the denture base material. Remem-ber that thin areas should be avoided between the frame-work and the cast (minimum of 1.5 mm) and between the framework and the denture teeth (minimum of 1.5 mm).
The denture tooth surfaces that are to be bonded to the VLC denture base must be lightly ground and cleaned. The teeth may then be placed in modeling compound or putty to hold them while the bonding agent is applied to all of the designated surfaces to be bonded to the VLC denture base. These coated surfaces should be allowed to sit for 2 minutes and then are processed in the light polymerization unit for 1 minute.
To secure the teeth into their designated position on the VLC denture base, apply a small piece of VLC mate-rial to each tooth with the modeling tool provided by the manufacturer. With the aid of the stone matrix and by using a high-intensity light, tack each individual tooth into position on the VLC denture base.
After tacking the teeth to the VLC denture base, paint a narrow band of bonding agent on the junction line between the teeth and the denture base material. The bonding agent acts as a sealer to prevent leakage and aids bonding of the teeth to the denture base material. Allow the bonding agent to set for 2 minutes, and then process in the light unit for 1 minute. You can now complete the buccal and lingual contouring of the denture base with additional VLC mate-rial. Paint all of the exposed surfaces of the VLC denture base material with the air barrier coating material, and pro-cess in the light polymerization unit for 2 minutes. After processing, carefully remove the removable partial denture from the cast. Do not attempt to pry it off. Paint the tissue surfaces of the edentulous areas with the air barrier coat-ing material, and place in the light polymerization unit for 6 minutes with the tissue side up. Clean the removable par-tial denture with water and a brush to remove all traces of the air barrier coating. Trim and polish in a routine manner to complete the removable partial denture.
The study of the history of denture base materials is a most interesting one that has been covered elsewhere in dental literature. The future of denture base materials promises to be just as fascinating a study, but such a dis-cussion cannot be included within the scope of this book. With newer materials, the future of methyl methacrylate as www.konkur.in 279 Chapter 19 Laboratory Procedures a denture base material is uncertain despite its acceptance as the best material available since its introduction in 1937. Although it has made possible the simulation of natural tis-sue color and contours combined with ease of manipula-tion, the fact remains that it leaves much to be desired as far as accuracy and dimensional stability are concerned. Whether other and newer materials will eventually sup-plant methyl methacrylate as a denture base material remains to be seen. The fact is that the denture base of the future (1) must be capable of accurately reproducing natural tissue tones faithfully through the use of character-izing stains and customizing procedures; and (2) must not require elaborate processing procedures and equipment, which would make the cost prohibitive for general use.
REMOUNTING AND OCCLUSAL CORRECTION TO AN OCCLUSAL TEMPLATE Even with improved denture base materials and processing techniques, some movement of artificial teeth will still occur because of the dimensional instability of the wax in which the artificial teeth were arranged. Until sources of error can be eliminated, remounting will continue to be neces-sary. How well the occlusion may be perfected by remount-ing will depend on the manner in which jaw relations were transferred to an instrument and how closely the instrument is capable of reproducing functional occlusion. But even though the articulator is capable of reproducing only a static centric relation, that relation at least should be reestablished before the denture is placed.
Although it is admitted that there are limitations to the perfection of eccentric occlusion in the mouth, some believe that it can be done with more accuracy than on an instrument that is incapable of reproducing eccentric posi-tions. Correction for errors in centric occlusion, however, should not be included in a concept that presumes that cen-tric occlusion may be established satisfactorily by intraoral adjustment, followed then by perfecting of eccentric occlu-sion. Because of denture instability and the inaccessibility of the occlusion for analysis, accurate intraoral corrections are not possible. Practically, even the occlusal adjustment of natural dentition in which each tooth has its own support can best be done when preceded by an analysis of articulated diagnostic casts.
One cardinal premise must be accepted if prosthetic den-tistry is to be anything more than a haphazard procedure. It is possible to transfer centric jaw relation to an instru-ment with accuracy and to maintain this relation through-out fabrication of the prosthesis. If this is true, then centric occlusion, coinciding with centric jaw relation, with centric occlusion of the remaining natural teeth, or with both, must have been established before initial placement of the prosthe-sis. This means that occlusal correction to reestablish centric relation by remounting after final processing is an absolute necessity for the success of the restoration. Remounting after processing is accomplished by returning the cast to a keyed relationship with the articulator mounting.
Precautions to Be Taken in Remounting The following precautions should be taken to ensure the accuracy of remounting to make final occlusal adjustment before polishing and initial placement of the denture. These apply to all types of occlusal relationship records but are directed particularly at remounting an occlusal template when stone vertical stops are used: 1. Make sure that the base of the cast has been reduced to fit the flask before keying and mounting so that it will not have to be altered later. 2. Bevel the margins of the base of the cast so that it will seat in a definite boxlike manner in the articulator mounting. 3. Notch the posterior and anterior aspects of the base to ensure its return to its original position. Notches at the margins are preferable to depressions within the base be-cause the former permit a visual check of the accuracy of the remounting. 4. Lubricate lightly the base and sides of the cast before it is mounted to facilitate its easy removal from the mounting stone. 5. Add tinfoil or lightly lubricate the base and sides of the cast before flasking it so that traces of investment will not be present to interfere with remounting. 6. When remounting the cast, secure it in the articulator with sticky wax, a hot glue gun, or modeling plastic, fol-lowed by stone over both the mounting and the sides of the cast. 7. Before adjusting the occlusion, make certain that no trac-es of investment remain on the vertical stops. 8. Take care to not abrade the opposing occlusal surface dur-ing occlusal adjustment. The use of marking tape or inked ribbon is preferable to articulating paper. The artificial tooth is less likely to cut through and mar the opposing surface, and ink or dye will not build up a false opposing surface, as will wax from articulating paper. 9. Occlusal readjustment to an occlusal template is com-plete when the stone vertical stops are again in contact.
With other types of articulator mountings, readjustment is complete when the vertical pin is again in contact and any valid horizontal excursions are freed of interference.
Occlusal readjustment, as with the original articula-tion, is done at the expense of the original tooth anatomy. Occlusal surfaces should be reshaped by adding grooves and spillways and by reducing the area of the occlusal table, thus improving the masticating efficiency of the artificial tooth. Although this may be done immediately after occlusal read-justment and before initial placement of the denture, it may be deferred until completion of the final adjustment. In any event, it is a necessary step in the completion of any remov-able prosthesis.
Porcelain teeth may be reshaped with abrasive or dia-mond-mounted points. Resin teeth lend themselves better www.konkur.in 280 Part II Clinical and Laboratory to reshaping with small burs to restore functional anatomy. Either type should be repolished judiciously to prevent reduction of cuspal contacts. Although cusps may be nar-rowed, spillways added, and the total area of contact reduced to improve masticating efficiency, critical areas of contact, both vertical and horizontal, must always be preserved.
The term remounting is also applied to the mounting of a completed prosthetic restoration back into an instrument by using some kind of interocclusal records. Discrepancies in occlusion resulting from processing of tooth-supported dentures may be corrected by reattaching the indexed pro-cessing cast and denture to the same instrument on which the occlusion was formulated. However, because of some instability inherent in distal extension removable partial dentures, such dentures should be retrieved from process-ing investment, finished, and polished and prepared for performance of occlusal corrections with the use of new intraoral records. The dentist must make a remounting cast before occlusal corrections can be accomplished. This is done simply first by placing the denture in the mouth and making an irreversible hydrocolloid (alginate) impression of the denture and the remaining teeth in the arch (Figure 19-41). When the impression is removed, the denture usu-ally remains in the impression or can be accurately replaced.
Undercuts in the denture bases are blocked out, the retentive elements of the framework are covered with a thin layer of molten wax, and a remounting cast is poured in the impres-sion. The remounting casts are then oriented in the articula-tor by the same type of interocclusal records that were used to orient the casts to formulate the occlusion. These proce-dures are covered in Chapter 21 as an integral part of the initial placement appointment.
Occlusal harmony must exist before the patient is given possession of the dentures. Delaying the correction of occlu-sal discrepancies until the dentures have had a chance to settle is not justifiable.
POLISHING THE DENTURE Areas to be considered in the polishing of a removable par-tial denture are (1) the borders of the denture bases; (2) the facial surfaces; and (3) the teeth and adjacent areas.
The borders on complete metal bases will have been established previously. On partial metal bases and complete acrylic-resin bases, the accuracy with which the border may be finished depends on the accuracy of the impression record and how well this was preserved on the stone cast. Edentulous areas recorded from impressions in stock trays generally lack the accuracy at the borders that is found on casts made from impressions in individualized trays and by secondary impression methods. Border accuracy is deter-mined also by whether or not the impression recorded a functional or a static relationship of the bordering tissue attachments.
Denture Borders The principal objectives to be considered in making an impression of edentulous areas of a partially edentulous A B Figure 19-41 A, A stock, perforated tray is used to make an irreversible hydrocolloid (alginate) impression of the denture and dental arch. Blockout of undercuts in the denture base and of tips of direct retainers is necessary, so the denture can be readily removed and replaced on the resultant remounting cast, as illustrated in B. B, Remounting cast poured in stone with the prosthesis in place and inter-occlusal registration for mounting against a maxillary cast. Wax placed at the retainer tip (A) allows the prosthesis to be readily removed and replaced on the cast for occlusal correction procedures that use an articulator.
www.konkur.in 281 Chapter 19 Laboratory Procedures arch include (1) maximal support for the edentulous remov-able partial denture base; and (2) extension of the borders to obtain maximum coverage compatible with moving tis-sue. Although this second objective may be met with an adequate individualized impression tray, it is best accom-plished with a secondary impression method. Not only the extent of the border, but also its width, should be recorded accurately. Both extent and width as recorded should be preserved on the stone cast. With the exception of certain areas that are arbitrarily thinned by polishing (mentioned previously in this chapter), finishing and polishing the den-ture borders should consist only of removing any flash and artifact blebs. Otherwise, borders should be left as recorded in the impression.
When the impression is made in a stock tray, the tray itself will have influenced both the extent and the width of the border. Some areas will be left short of the total area available for denture support, whereas others will be extended beyond functional limits by overextension of the tray. Unfortunately, the technician may attempt to interpret the anatomy of the mouth and arbitrarily trim the denture borders. This presumes that the technician has an intimate knowledge of the anatomy of the mouth of the patient for whom the restoration is being made, which is most unlikely. The dentist must correct any overextension remaining after arbitrarily trimming the border from cast landmarks in the mouth. It is preferable for the dentist to finish the borders of dentures, having painstakingly developed them during impression procedures.
Facial Surfaces The facial surfaces of the denture base are those polished surfaces lying between the buccal borders and the supplied teeth. Methods have been proposed for making sectional impression records of buccal contours, thereby permit-ting the denture base to be made to conform to the facial musculature. These have never received wide acceptance and may be considered impractical in removable partial prosthodontics.
Facial surfaces may be established in wax or may be carved into the denture base after processing. Generally, it is desir-able that it be done in wax as part of the wax pattern, because it is easier to do so and because contours can best be estab-lished at a time when modifications can be made if desired. Buccal surfaces should be contoured to aid in retention of the denture by border molding, to preserve the border roll and thereby prevent food impaction, and to facilitate return of the food bolus back onto the masticating table. Lingual surfaces should be made concave to provide tongue room and to aid in retention of the denture. Polishing of concave surfaces is always more difficult than polishing of flat and convex surfaces. If such contours are established previously in wax, not only is finishing more easily accomplished, but border and gingival areas are less likely to be inadvertently altered.
Finishing Gingival and Interproximal Areas The contouring of gingival and interproximal areas after pro-cessing is difficult and generally unsatisfactory (Figure 19-42).
Modern cosmetic considerations demand that gingival carving be done around each tooth individually, with varia-tions in the height of the gingival curve and in the length of the papillae. Interproximally, the papillae should be con-vex rather than concave. The gingival attachment should be free of grooves and ditches that would accumulate debris and stain and should be as free for cleansing as possible. All this precludes gross shaping and trimming of gingival areas after processing. Gingival carving should be done in wax, and investing should be done with care to prevent blebs and artifacts. Finishing should consist only of trimming around the teeth and the papillae with small round burs to create a more natural simulation of living tissue, plus light stippling with an off-center round bur for the same reason. Polishing should consist only of light buffing with brush wheels and pumice, and finally with a soft brush wheel and a nonabra-sive polishing agent specially made for this purpose.
Pumicing of gingival areas can only serve to polish the high spots; although it may be done lightly, its use should be limited to light buffing of areas already made as smooth as in the waxing phase. Heavy pumicing of the denture resin not only creates a typical denture look but also alters the delicately carved wax surface and any plastic teeth present. If pumicing must be done, plastic teeth should be protected with adhesive masking tape during the process.
Any polishing operation on a removable partial denture done on a lathe is made hazardous by the presence of direct retainers, which can easily become caught in the polishing A B Figure 19-42 Attention to adjacent tooth-tissue contours can facilitate the production of natural-appearing prostheses. The in-terproximal papilla at A demonstrates both vertical and horizon-tal components. In general, the horizontal component increases with age. The prosthetic interproximal papilla at B exhibits only a vertical component and appears artificial. The anterior border () contour is blunted and provides an obvious and abrupt contour change. Contouring the border to bevel into the interproximal re-gion reduces its objectionable appearance.
www.konkur.in 282 Part II Clinical and Laboratory wheel. Although the least damage that might occur is the distortion of a clasp arm, there is a greater possibility that the denture may be thrown forcibly into the lathe pan, with serious damage to the framework or other parts of the den-ture. The technician must ever be conscious of this possibil-ity and must always cover any projecting clasp with the finger while it is near the polishing wheel. In addition, it is wise to keep a pumice pan well filled with wet pumice to cushion the shock should an accident occur. Any other lathe pan used in polishing should be lined with a towel or with a resilient material for the same reason. The potential hazard of retain-ers catching gloved fingers or hands that hold a removable partial denture is always present. Serious injury can result, and extreme caution must be taken to prevent such an occur-rence. The risk of infection is increased if the removable par-tial denture has been in place intraorally.
www.konkur.in CHAPTER 20 Work Authorizations for Removable Partial Dentures CHAPTER OUTLINE Work Authorization Content Function Characteristics Definitive Instructions by Work Authorizations Legal Aspects of Work Authorizations Delineation of Responsibilities by Work Authorizations WORK AUTHORIZATION A work authorization contains the written directions for lab-oratory procedures to be performed for fabrication of dental restorations. The responsibility of a dentist to the public and to the dental profession to safeguard the quality of prosth-odontic services is controlled in part through meaningful work authorizations. If work authorizations are properly completed, they provide a means for increased professional quality assurance and satisfaction in a removable partial den-ture service.
A work authorization completed by a dentist is similar to granting power of attorney. It grants authority for others to act on the dentist’s behalf and specifically prescribes what is authorized.
When properly executed, work authorizations are effec-tive channels of communication. They enhance the quality of the completed restorations by providing instructions for individually and scientifically considered prostheses.
Content Information contained in a work authorization should include the following: (1) the name and address of the den-tal laboratory; (2) the name and address of the dentist who initiates the work authorization; (3) the identification of the patient; (4) the date of work authorization; (5) the desired completion date of the request; (6) specific instructions; (7) the signature of the dentist; and (8) the registered license number of the dentist. All these requirements can be accom-modated in a simply designed form (Figure 20-1).
Function The following four important functions are performed by a work authorization: 1. It furnishes definite instructions for laboratory proce-dures to be accomplished and implies an expectation of a level of acceptable quality for the services rendered.
www.konkur.in 284 Part II Clinical and Laboratory 2. It provides a means of protecting the public from the il-legal practice of dentistry. 3. It is a protective legal document for both the dentist and the dental laboratory technician. 4. It completely delineates the responsibilities of the dentist and the dental laboratory technician.
Characteristics A work authorization must be legible, clear, concise, and readily understood. It is unreasonable to assume that labora-tory technicians are decoding experts. Sufficient information must be included in a work authorization to enable the tech-nician to understand and execute the request. Many dentists are overly presumptive in assuming that a request can be acceptably fulfilled without proper directions.
It is sound practice to provide the dental laboratory technician with adequate written instructions for each lab-oratory service required in the fabrication of a restoration. Therefore a new work authorization should accompany the material returned to the laboratory for continuing progress to complete the restoration. In a modern dental practice, it is highly improbable that a one-trip laboratory service 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 R R L L Copyright Attachments International, Inc.
Treatment plan 1. Rests 2. Retention 3. Reciprocation 4. Major connector 5. Indirect retention 6. Guide planes 7. Base retention 8. Areas to be modified or contoured Instructor: Approval to send to laboratory: Date: Color code: Blue: Cast metal Red: Resin base and wrought wire Green: Areas to be contoured Removable Partial Prosthodontics Laboratory instructions Design specifications: Patient name: Student name: Patient number: Student number: Figure 20-1 Work authorization form, designed specifically for removable partial dentures, to furnish detailed information to the laboratory technician. Form used to specifically plan framework design and to designate mouth alterations and preparations.
www.konkur.in 285 Chapter 20 Work Authorizations for Removable Partial Dentures will be adequate to provide a truly professional removable restoration.
No single work authorization form is adequate to furnish detailed instructions for accomplishing the laboratory phases in the fabrication of removable partial dentures, crowns, and fixed partial dentures, or complete dentures, or for accomplish-ing orthodontic laboratory procedures. Inherent differences in the many types of restorations themselves and differences in the laboratory phases necessary for their fabrication estab-lish a requirement for individual work authorization forms.
DEFINITIVE INSTRUCTIONS BY WORK AUTHORIZATIONS Work authorization forms may be designed so that only a minimum of writing is necessary to provide thorough instructions (Figure 20-2). The form can contain printed listings of materials and specifications that require either a checkmark or a fill-in for authorization of their use.
Another effective means for communicating, which pro-vides anatomical detail not available when drawing on paper, includes drawing the design on a diagnostic cast. Exact location 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 R R L L Copyright Attachments International, Inc.
John Doe Joe Smith 158390 Cingulum rests #22, #27 MO occlusal rest #20 MO occlusal rest #28 18 ga. loops, 12 ga. inferior border Metal tooth to replace #21 Distal #20, #22, #27 Mesial #20 Cingulum rest #22, #27 Lingual plate #22, #27, #28 Lingual bar I-bar clasp #28 I-bar clasp #22 1234 12 ga 18 ga loops 18 ga loops Cast I-bar Cast I-bar 2/15/09 Treatment plan Design specifications: 1. Rests 2. Retention 3. Reciprocation 4. Major connector 5. Indirect retention 6. Guide planes 7. Base retention 8. Areas to be modified or contoured Patient name: Student name: Instructor: Approval to send to laboratory: Date: Patient number: Student number: Color code: Blue: Cast metal Red: Resin base and wrought wire Green: Areas to be contoured Removable Partial Prosthodontics Laboratory instructions Figure 20-2 This work authorization accompanies a master cast on which the dentist has designed and drawn an outline for a remov-able partial denture framework. It is simple and is not time-consuming to execute, yet it furnishes detailed information so the request can be properly fulfilled.
www.konkur.in 286 Part II Clinical and Laboratory of rests, proximal plates, major/minor connectors, and clasps augment the written work authorization request (Figure 20-3).
A reminder space is included to designate the choice of metal for the framework. Frameworks for removable par-tial dentures are usually cast in type IV gold, chromium-cobalt alloy, or a titanium alloy. The nature of the material of the denture base may also be indicated by a checkmark. It is difficult to elicit this information from the markings on master casts.
Space is reserved on the work authorization form to fur-nish the technician with information on the dentist’s selec-tion of teeth. The responsibility for tooth selection must remain with the dentist. The success of the removable partial denture depends in part on the consideration given to the size, number, and placement of the artificial teeth and to the material from which they are made.
A display of courtesy deserved by and a demonstra-tion of respect for the laboratory technician are indicated. Indicate characterizations Pontic design (circle) Modified ridge lap Conical Hygienic Metal design (circle) Metal coping All porcelain coverage Metal occlusal excluding buccal cusp Metal occlusal including buccal cusp 1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Patient name Date submitted Laboratory name Laboratory address Metal type Denture identification Date required Guide Shade Signature License number A WHITE – Department YELLOW – Lab PINK – Patient Record Figure 20-3 A. This work authorization will accompany design and master casts. The dentist will submit a design cast with an outline for the removable partial denture frame. It is simple and is not time consuming, yet it furnishes detailed information so the request can be properly fulfilled. www.konkur.in 287 Chapter 20 Work Authorizations for Removable Partial Dentures The general request is prefaced by “please” and the specific instructions are ended with “thank you.” Do any other three words promote better relations?
A good work authorization form not only ensures clar-ity but also simplifies correct execution. Figures can be pro-vided on which diagrams may be drawn to enhance written descriptions when necessary. These diagrams may show the occlusal and lingual surfaces of the posterior teeth and the lingual surfaces of the anterior teeth. The palatal region of the maxillary arch and the lingual slopes of the mandibular alveolar ridge can also be included. These features allow a clear, diagrammatic representation of the locations of major connectors, which will complement the outline of the frame-work on the master cast.
A color-code index can be used to explain the markings on the master cast when it is submitted to the laboratory for the fabrication of a framework. For example, a green pen-cil can be used to outline the framework; red to designate the desired location of finishing lines on the framework; and black lines to denote the height of contour on teeth and soft tissue created during the survey of the cast. The color code eliminates confusion in interpreting the markings on the master cast.
Specifications for waxing the framework components for gold, chromium-cobalt, or titanium alloy castings must be furnished for the technician and are an integral part of the work authorization form. Specifications that are adequate for most removable partial denture frameworks may be listed. This feature alone saves time and effort in preparing the work authorization and serves as a handy reference for the labora-tory technician. The listing of average specifications does not preclude altering a specification when the situation necessi-tates other characteristics in a given component.
The specific instructions provided in a work authorization must be so constructed that they will be a constant source of direction and supervision for the laboratory phases of a removable partial denture service. Instructions should leave no doubt about the dentist’s requirements in a request for laboratory services. It is foolish to use undercut dimensions of 0.01 or 0.02 inch when a master cast is surveyed, unless written directions are included to incorporate these dimen-sions into the finished framework.
B Figure 20-3 Cont’d B. Occlusal view of the design cast, with the outline of the removable partial denture framework drawn. The path of placement is indicated on the cast, as well as the height of contour and location of retentive clasp undercuts, and location of guide planes, rests, and clasp locations.
www.konkur.in 288 Part II Clinical and Laboratory Work authorization blanks should be available in such a manner that a duplicate can be conveniently made and thus a copy can be supplied for both the dentist and the dental laboratory technician. The original may be a different color than the duplicate for ready identification.
LEGAL ASPECTS OF WORK AUTHORIZATIONS Although the National Association of Dental Laboratories (NADL) provides guidelines for statutory regulation, it is the inherent right of each state to implement its own regulation. Fortunately, all states exercise this control. Interpretations of acts that constitute the practice of dentistry are moderately uniform. However, statutory restrictions on dental labora-tory operations vary from state to state. Properly executed work authorizations serve to document communication and protect the professional relationship between the dentist and the dental laboratory.
Many states require that work authorizations be made in duplicate and that both the dentist and the dental labora-tory technician retain a copy for a specified period from the date of work authorization. Thus, documents are available to substantiate or refute claims and counterclaims that concern the illegal practice of dentistry, or to aid in the settlement of misunderstandings between a dentist and a dental laboratory technician.
DELINEATION OF RESPONSIBILITIES BY WORK AUTHORIZATIONS The dentist is responsible for all phases of a removable partial denture service in the strict sense of the word, although the dental laboratory technician may be requested to perform certain technical phases of the service. However, the labora-tory technician is responsible only to the dentist and never to the patient. A dentist who relegates the design of a removable partial denture to a less qualified individual accepts the pos-sibility of and responsibility for an inferior removable partial denture service.
A dentist who imposes on auxiliary personnel those responsibilities that legally and morally belong with the dentist does a great injustice to the patients, the technicians, and the dental profession. There is little doubt that the ille-gal practice of dentistry and the presently existing impasse between dentist and some dental laboratory technicians are in part a result of many individual dentists who impose unrealistic responsibility on laboratory technicians. Further-more, this unwelcome relationship may have been caused by the submission of poor impressions, casts, records, and instructions to the laboratory technician, with the demand of impossible quality in the returned restoration under threat of economic boycott.
Most dental laboratory technicians are ethical and ear-nestly desire to contribute their talents to the dental profes-sion. The dental profession is vitally interested in increasing the number of serious-minded dental auxiliary personnel to share in providing oral health care. However, until the dental profession elevates itself in the eyes of laboratory technicians and also elevates the stature of dental laboratory technology, greater availability of responsible auxiliary personnel is more fancied than real.
The dental laboratory technician is a member of a team whose objectives are the prevention of oral disease and the maintenance of oral health as adjuncts to the physical and mental well-being of the public. A good dental laboratory technician is a valuable team member working with the den-tist and contributes much to the team effort in providing oral health care for patients. The degree and quality of the team effort are the responsibility of the dentist and depend on the dentist’s knowledge, experience, technical skill, administra-tive ability, integrity, and ability to communicate effectively.
A dentist may delegate much of the laboratory phase of a removable partial denture service. Work authorizations help fulfill the moral obligation to supervise and direct those technical phases that can be accomplished by dental labora-tory technicians. Substantial indications suggest that many members of the dental profession either are not cognizant of the rewards of writing good work authorizations or are not proficient in their execution. It is not a secret that some dentists do not submit instructions when availing themselves of commercial dental laboratory services.
If the practice of prosthodontics is to remain in the con-trol of dentists, each member of the dental profession must avoid delegating responsibility to those who are less qualified to accept the responsibility. Movements to allow denturism are seemingly becoming more prevalent and possibly are related to poor laboratory communication regarding remov-able prosthodontics.
www.konkur.in CHAPTER 21 Initial Placement, Adjustment, and Servicing of the Removable Partial Denture CHAPTER OUTLINE Adjustments to Bearing Surfaces of Denture Bases Occlusal Interference from Denture Framework Adjustment of Occlusion in Harmony with Natural and Artificial Dentition Instructions to the Patient Follow-Up Services Initial placement of the completed removable partial denture, which is the fifth of six essential phases of removable partial denture service mentioned in Chapter 2, should be a rou-tinely scheduled appointment. All too often the prosthesis is quickly placed and the patient dismissed with instructions to return when soreness or discomfort develops. Patients should not be given possession of removable prostheses until denture bases have been initially adjusted as required, occlu-sal discrepancies have been eliminated, and patient educa-tion procedures have been continued.
Although it is true that some accommodation is a neces-sary part of adjusting to new dentures, many other factors are also pertinent. Among these are how well the patient has been informed of the mechanical and biological problems involved in the fabrication and wearing of a removable prosthetic res-toration, and how much confidence the patient has acquired in the excellence of the finished product. Knowing in advance that every step has been carefully planned and executed with skill and having acquired confidence in both the dentist and the excellence of the prosthesis, the patient is better able to accept the adjustment period as a necessary but transient step in learning to wear the prosthesis. This confidence could be lost if the dentist does not approach the insertion and postinsertion phases as equally important for the success of the treatment.
The term adjustment has two connotations, each of which must be considered separately. The first is adjustment of the denture bearing and occlusal surfaces of the denture made by the dentist at the time of initial placement and thereafter. The second is the adjustment or accommodation by the patient, both psychologically and biologically, to the new prosthesis.
After the resin bases have been processed and before dentures are separated from the casts, the occluding teeth must be altered to perfect the occlusal relationship between www.konkur.in 290 Part II Clinical and Laboratory opposing artificial dentition or between artificial dentition and an opposing cast or template. Denture bases must be finished to eliminate excess and perfect the contours of pol-ished surfaces for the best functional and esthetic results. This is made necessary by the inadequacies of casting proce-dures, because both the metal and resin parts of a prosthetic restoration are produced by casting methods. Unfortunately, such procedures in the laboratory rarely eliminate the need for final adjustment in the mouth to perfect the fit of the res-toration to the oral tissue.
Included in this final step in a long sequence of finishing procedures necessary to produce a biologically acceptable prosthetic restoration are the following: (1) adjustment of the bearing surfaces of the denture bases to be in harmony with the supporting soft tissue; (2) adjustment of the occlusion to accommodate the occlusal rests and other metal parts of the denture; and (3) final adjustment of the occlusion on the artificial dentition to harmonize with natural occlusion in all mandibular positions.
ADJUSTMENTS TO BEARING SURFACES OF DENTURE BASES Altering bearing surfaces to perfect the fit of the denture to the supporting tissue should be accomplished with the use of some kind of indicator paste (Figure 21-1). The paste A B C Figure 21-1 A, Tissue side of finished bases of a Kennedy Class I modification 1 removable partial denture, where paste indicates that pressure has been applied. Paste was applied following careful inspection of the tissue surface for irregularities or sharp projections, which must be eliminated before fitting in the mouth. The entire tissue surface of the bases was dried before it was coated with a thin coat of pressure indicator paste using a stiff-bristle brush. Brush marks are evident, and it is the change in the pattern of brush marks that guides adjustment. It is important to avoid thick application of indicator paste, which can hide the presence of significant pressure. B, The prosthesis can be dipped in cold water or sprayed with a provided release agent before placement in the patient’s mouth, to prevent paste from sticking to oral tissues. After careful seating of the denture, the patient can close firmly on cotton rolls for a few seconds, or the dentist can alternately apply a tissue-ward pressure over the bases to simulate functional movement. The presence of tissue contact is evident in the pattern of the paste, which is different from the brushed pattern. There is no suggestion of excessive pressure in this tissue contact pattern. However, it is not uncommon to relieve the area adjacent to the abutment sparingly. Several placements of the denture with indicator paste are usually necessary for evaluation of the accuracy of the bases. C, A different denture base recovered from the mouth after manipulation simulating function. The tissue contact reveals excessive pressure at the region lingual to the retromolar pad.
www.konkur.in 291 Chapter 21 Initial Placement, Adjustment, and Servicing of the Removable Partial Denture must be one that will be readily displaced by positive tissue contact and that will not adhere to the tissue of the mouth. Several pressure indicator pastes are commercially available. However, equal parts of a vegetable shortening and USP zinc oxide powder can be combined to make an acceptable paste. The components must be thoroughly spatulated to a homo-geneous mixture. A quantity sufficient to fill several small ointment jars may be mixed at one time.
Rather than dismissing the patient with instructions to return when soreness develops and then over-relieving the denture for a traumatized area to restore patient comfort, use a pressure indicator paste with any tissue bearing prosthetic restoration. The paste should be applied by the dentist in a thin layer over the bearing surfaces. The material should be rinsed in water so that it will not stick to the soft tissue, and then digital pressure should be applied to the denture in a tissue-ward direction. The patient cannot be expected to apply a heavy enough force to the new denture bases to regis-ter all of the pressure areas present. The dentist should apply both vertical and horizontal forces with the fingers in excess of what might be expected of the patient. The denture is then removed and inspected. Any areas where pressure has been heavy enough to displace a thin film of indicator paste should be relieved and the procedure repeated with a new film of indicator until excessive pressure areas have been eliminated. This is particularly difficult to interpret when patients exhibit xerostomia. An area of the denture base that shows through the film of indicator paste may be erroneously interpreted as a pressure spot, when actually the paste had adhered to the tissue in that area. Therefore, only those areas that show through an intact film of indicator paste should be inter-preted as pressure areas and relieved accordingly. The deci-sion to relieve an area of pressure must consider whether the pressure is in a primary, secondary, or nonsupportive den-ture bearing area. The primary denture bearing areas should be expected to show greater contact than other areas.
Pressure areas most commonly encountered are as fol-lows: in the mandibular arch—(1) the lingual slope of the mandibular ridge in the premolar area; (2) the mylohyoid ridge; (3) the border extension into the retromylohyoid space; and (4) the distobuccal border in the vicinity of the ascending ramus and the external oblique ridge; in the max-illary arch—(1) the inside of the buccal flange of the denture over the tuberosities; (2) the border of the denture lying at the malar prominence; and (3) the point at the pterygomax-illary notch where the denture may impinge on the ptery-gomandibular raphe or the pterygoid hamulus. In addition, bony spicules or irregularities in the denture base that will require specific relief may be found in either arch.
The amount of relief necessary depends on the accuracy of the impression, the master cast, and the denture base. Despite the accuracy of modern impression and cast materi-als, many denture base materials leave much to be desired in this regard, and the element of technical error is always present. It is therefore essential that discrepancies in the den-ture base are detected and corrected before the tissues of the mouth are subjected to the stress of supporting a prosthetic restoration. One of our major responsibilities to the patient is that trauma should always be held to a minimum. Therefore the appointment time for initial placement of the denture must be adequate to permit such adjustment.
OCCLUSAL INTERFERENCE FROM DENTURE FRAMEWORK Any occlusal interference from occlusal rests and other parts of the denture framework should have been eliminated before or during the establishment of occlusal relations. The denture framework should have been tried in the mouth before a final jaw relation is established, and any such inter-ference should have been detected and eliminated. Much of this need not occur if mouth preparations and the design of the removable partial denture framework are carried out with a specific treatment plan in mind. In any event, occlu-sal interference from the framework should not ordinarily require further adjustment at the time the finished denture is initially placed. For the dentist to have sent an impression or casts of the patient’s mouth to the laboratory and to receive a finished removable partial denture prosthesis without hav-ing tried the cast framework in the mouth is a dereliction of responsibility to the patient and the profession.
ADJUSTMENT OF OCCLUSION IN HARMONY WITH NATURAL AND ARTIFICIAL DENTITION The final step in the adjustment of the removable partial denture at the time of initial placement is adjustment of the occlusion to harmonize with the natural occlusion in all mandibular excur-sions. When opposing removable partial dentures are placed concurrently, the adjustment of the occlusion will parallel to some extent the adjustment of occlusion on complete dentures. This is particularly true when the few remaining natural teeth are out of occlusion. But where one or more natural teeth may occlude in any mandibular position, those teeth will influence mandibular movement to some extent. It is necessary, therefore that the artificial dentition on the removable partial denture be made to harmonize with whatever natural occlusion remains.
Occlusal adjustment of tooth-supported removable partial dentures may be performed accurately by any of several intra-oral methods. Occlusal adjustment of distal extension remov-able partial dentures is accomplished more accurately with the use of an articulator than by any intraoral method. Because distal extension denture bases will exhibit some movement under a closing force, intraoral indications of occlusal discrep-ancies, whether produced by articulating paper or disclosing waxes, are difficult to interpret. Distal extension dentures positioned on remounting casts can conveniently be related in the articulator with new, nonpressure interocclusal records, and the occlusion can be adjusted accurately at the appoint-ment for initial placement of the dentures (Figure 21-2).
The methods by which occlusal relations may be estab-lished and recorded have been discussed in Chapter 18. www.konkur.in 292 Part II Clinical and Laboratory In this chapter, the advantages of establishing a functional occlusal relationship with an intact opposing arch have been discussed, along with the limitations that exist to perfecting harmonious occlusion on the finished prosthesis by intraoral adjustment alone. Even when the occlusion on two opposing removable partial dentures is adjusted, it is best that one arch be considered an intact arch and the other one adjusted to it. This is accomplished by first eliminating any occlusal inter-ference with mandibular movement imposed by one denture and adjusting any opposing natural dentition to accommo-date the prosthetically supplied teeth. Then the opposing removable partial denture is placed, and occlusal adjust-ments are made to harmonize with both the natural denti-tion and the opposing denture, which is now considered part of an intact dental arch. Which removable partial denture is adjusted first, and which is made to occlude with it, is some-what arbitrary, with the following exceptions: If one remov-able partial denture is entirely tooth-supported and the other has a tissue-supported base, the tooth-supported denture is adjusted to final occlusion with any opposing natural teeth. That arch is then treated as an intact arch and the opposing denture adjusted to occlude with it. If both removable partial dentures are entirely tooth-supported, the one that occludes with the most natural teeth is adjusted first and the second denture is then adjusted to occlude with an intact arch. Tooth-supported segments of a tooth- and tissue-supported removable partial denture are likewise adjusted first to harmonize with any opposing natural dentition. The final adjustment of occlusion on opposing tissue-supported bases is usually done on the mandibular removable partial den-ture because this is the moving member, and the occlusion is made to harmonize with the maxillary removable partial denture, which is treated as part of an intact arch.
Intraoral occlusal adjustment is accomplished with the use of some kind of indicator and suitable mounted points and burs. Diamond or other abrasive points must be used to reduce enamel, porcelain, and metal contacts. These also may be used to reduce plastic tooth surfaces, but burs may be used for plastic with greater effectiveness. Articulation paper may be used as an indicator if one recognizes that heavy interocclusal contacts may become perforated, leaving only a light mark. Secondary contacts, which are lighter and fre-quently sliding, may make a heavier mark. Although articu-lation ribbon does not become perforated, it is not easy to use in the mouth, and differentiation between primary and secondary contacts is difficult, if not impossible, to ascertain.
In general, occlusal adjustment of multiple contacts between natural and artificial dentition when tooth-sup-ported removable partial dentures are involved follows the same principles as those for natural dentition alone. This is because removable partial dentures are retained by devices attached to the abutment teeth, whereas no mechanical A B Figure 21-2 Sequence of laboratory and clinical procedures performed for correction of occlusal discrepancies caused by process-ing of removable partial dentures. The arch with the prosthesis will require a new cast and record to provide occlusal correction. If it is a maxillary prosthesis, this involves preserving the facebow record by replacing the processed maxillary removable partial denture and cast on the articulator and indexing the occlusal surfaces using a remount jig, or making another facebow record at the try-in appoint-ment. If it is a mandibular prosthesis, the opposing arch can be prepared before insertion. A, In this example, the opposing arch is a complete denture that is not to be altered. To produce a cast for use in correcting the occlusion on the articulator, a pick-up impres-sion of the mandibular prosthesis is made. The prosthesis stays within the irreversible hydrocolloid impression; the clasps, proximal plates, and any undercut or parallel surface are carefully blocked out with wax before the remount cast is formed. B, The remount cast that is formed is then inverted and positioned with the use of an interocclusal record. www.konkur.in 293 Chapter 21 Initial Placement, Adjustment, and Servicing of the Removable Partial Denture C D E Figure 21-2, cont’d C, The mounted mandibular cast and the interocclusal record showing that the record was made without tooth contact. This allows the position recorded to not be influenced by the teeth, which could alter the closure path and introduce error. D, This example shows a maxillary complete denture that was mounted before the patient visit with the use of a remount index (preserved facebow by indexing before recovery from the processed cast) and the mandibular remount cast and interocclusal record (as in C). The record is removed and occlusal correction accomplished to control the post-process occlusion. Use of the completed prostheses pro-vides the best chance to obtain an accurate and reliable interocclusal record, given the fact that the bases are very accurate and stable. E, The goal of the remounting procedure is to provide the occlusal position prescribed by the arrangement of prosthesis teeth. It would be inappropriate to allow the patient to attempt to accommodate to a new prosthesis in which the occlusion is not optimized.
retainers are present with complete dentures. The use of more than one color of articulation paper or ribbon to record and differentiate between centric and eccentric contacts is just as helpful in adjusting removable partial denture occlu-sion as natural occlusion, and this method may be used for the initial adjustment.
For final adjustment, because one denture will be adjusted to occlude with a predetermined arch, the use of an occlusal wax may be necessary to establish points of excessive con-tact and interference. This cannot be done by articulation paper alone. An occlusal indicating wax (Figure 21-3) that is adhesive on one side, or strips of 28-gauge casting wax or other similar soft wax, may be used. This should always be done bilaterally, with two strips folded together at the mid-line. Thus, the patient is not as likely to deviate to one side as when wax is introduced unilaterally.
For centric contacts, the patient is guided to tap into the wax. Then the wax is removed and inspected for perforations under transillumination. Premature contacts or excessive contacts appear as perforated areas and must be adjusted. One of two methods may be used to locate specific areas to be relieved. Articulation ribbon may be used to mark the occlu-sion; those marks that represent areas of excessive contact are identified by referring to the wax record and are relieved accordingly. A second method is to introduce the wax strips a second time, this time adapting them to the buccal and www.konkur.in 294 Part II Clinical and Laboratory lingual surfaces for retention. After the patient has tapped into the wax, perforated areas are marked with a waterproof pencil. The wax is then stripped off and the penciled areas are relieved.
Whichever method is used, it must be repeated until occlusal balance in the planned intercuspal position has been established and uniform contacts without perforations are evident from a final interocclusal wax record. After adjust-ment has been completed, any remaining areas of interfer-ence are reduced, thus ensuring that there is no interference during the chewing stroke. Adjustments to relieve interfer-ence during the chewing stroke should be confined to buccal surfaces of mandibular teeth and lingual surfaces of maxil-lary teeth. This serves to narrow the cusps so that they will go all the way into the opposing sulci without wedging as they travel into the planned intercuspal contact. Skinner proposed giving a small bite of soft banana to chew rather than expecting the patient to chew without food actually being present. The small bolus of banana promotes normal functional activity of the chewing mechanism, yet by its soft consistency does not in itself cause indentations in the soft wax. Any interfering contacts encountered during the chew-ing stroke are thus detected as perforations in the wax, which are marked with pencil and relieved accordingly.
After adjustment of the occlusion, the anatomy of the artificial teeth should be restored to maximal efficiency by restoring grooves and spillways (food escapeways) and by narrowing the teeth bucco-lingually to increase the sharp-ness of the cusps and reduce the width of the food table. Mandibular buccal and maxillary lingual surfaces in particu-lar should be narrowed to ensure that these areas will not interfere with closure into the opposing sulci, because arti-ficial teeth used on removable partial dentures that oppose natural or restored dentition should always be considered material out of which a harmonious occlusal surface is cre-ated. Final adjustment of the occlusion should always be followed by meticulous restoration of the most functional occlusal anatomy possible. Although this may be done after a subsequent occlusal adjustment is made at a later date, the possibility that the patient may fail to return on sched-ule is always present; in the meantime, broad and inefficient occlusal surfaces may cause overloading of the supporting structures, which would be traumatogenic. Therefore, resto-ration of an efficient occlusal anatomy is an essential part of the denture adjustment at the time of placement. Again, this necessitates that sufficient time is allotted for initial place-ment of the removable partial denture to permit accomplish-ment of all necessary occlusal corrections.
Adjustments to occlusion should be repeated at a rea-sonable interval after the dentures have reached a point of equilibrium and the musculature has become adjusted to the changes brought about by restoration of occlusal contacts. This second occlusal adjustment usually may be considered sufficient until such time as tissue-supported denture bases no longer support the occlusion, and corrective measures— either reoccluding the teeth or relining the denture—must be used. However, a periodic recheck of occlusion at intervals of 6 months is advisable to prevent traumatic interference result-ing from changes in denture support or tooth migration.
INSTRUCTIONS TO THE PATIENT Finally, before the patient is dismissed, he or she should be reminded of the chronic nature of the missing tooth condi-tion and of the fact that treatment solutions, such as a remov-able partial denture, require monitoring to ensure that they continue to provide optimum function without harming the mouth.
Patients should be instructed in the proper placement and removal of the removable partial denture. They should dem-onstrate that they can place and remove the prosthesis them-selves. Clasp breakage can be avoided by instructing patients to remove the removable partial denture by the bases and not by repeated lifting of the clasp arms away from the teeth with the fingernails.
Patients should be advised that some discomfort or minor annoyance might be experienced initially. To some extent, this may be caused by the bulk of the prosthesis to which the tongue must become accustomed.
Patients must be advised of the possibility of the develop-ment of soreness despite every attempt on the part of the den-tist to prevent its occurrence. Because patients vary widely in their ability to tolerate discomfort, it is best to advise every patient that needed adjustments will be made. On the other hand, the dentist should be aware that some patients are unable to accommodate the presence of a removable prosthe-sis. Fortunately, these are few in any practice. However, the dentist must avoid any statements that might be interpreted or construed by the patient to be positive assurance tantamount Figure 21-3 Occlusal indicator wax. (Courtesy of Kerr Corp., Orange, CA.) www.konkur.in 295 Chapter 21 Initial Placement, Adjustment, and Servicing of the Removable Partial Denture to a guarantee that the patient will be able to use the pros-thesis with comfort and acceptance. Too much depends on the patient’s ability to accept a foreign object and to tolerate reasonable pressure to make such assurance possible.
Discussing phonetics with the patient in regard to the new dentures may indicate that this is a unique problem to be overcome because of the influence of the prosthesis on speech. With few exceptions, which usually result from excessive and preventable bulk in the denture design, con-tour of denture bases, or improper placement of teeth, the average patient will experience little difficulty in wearing the removable partial denture. Most hindrances to normal speech will disappear in a few days.
Similarly, perhaps little or nothing should be said to the patient about the possibility of gagging or the tongue’s reac-tion to a foreign object. Most patients will experience little or no difficulty in this regard, and the tongue will normally accept smooth, nonbulky contours without objection. Con-tours that are too thick, too bulky, or improperly placed should be avoided in the construction of the denture, but if present, these should be detected and eliminated at the time of placement of the denture. The dentist should pal-pate the prosthesis in the mouth and reduce excessive bulk accordingly before the patient has an opportunity to object to it. The area that most often needs thinning is the disto-lingual flange of the mandibular denture. Here, the denture flange should always be thinned during finishing and pol-ishing of the denture base. Sublingually, the denture flange should be reproduced as recorded in the impression, but distal to the second molar the flange should be trimmed somewhat thinner. Then, when the denture is placed, the dentist should palpate this area to ascertain that a mini-mum of bulk exists that might be encountered by the side and base of the tongue. If this needs further reduction, it should be done and the denture repolished before the patient is dismissed.
The patient should be advised of the need to keep the dentures and the abutment teeth meticulously clean. If car-iogenic processes are to be prevented, the accumulation of debris should be avoided as much as possible, particularly around abutment teeth and beneath minor connectors. Fur-thermore, inflammation of gingival tissue is prevented by removing accumulated debris and substituting toothbrush massage for the normal stimulation of tongue and food con-tact with areas that will be covered by the denture framework.
The mouth and the removable partial denture should be cleaned after eating and before retiring. Brushing before breakfast also may be effective in reducing the bacterial count, which may help to lessen acid formation after eating in the caries-susceptible individual. A removable partial den-ture may be effectively cleaned with the use of a small, soft-bristle brush. Debris may be effectively removed through the use of nonabrasive dentifrices, because they contain the essential elements for cleaning. Household cleaners and toothpastes should not be used, because they are too abrasive for use on acrylic-resin surfaces. The patient—the elderly or handicapped patient in particular—should be advised to clean the denture over a basin partially filled with water so that denture impact will be less if the denture is dropped accidentally during cleaning.
Along with brushing with a dentifrice, additional cleaning may be accomplished with the use of a proprietary denture cleaning solution. The patient should be advised to soak the dentures in the solution for 15 minutes once daily, followed by thorough brushing with a dentifrice. Although hypochlo-rite solutions are effective denture cleansers, they have a ten-dency to tarnish chromium-cobalt frameworks and should be avoided.
In some mouths, the precipitation of salivary calculus on the removable partial denture necessitates taking extra mea-sures for its removal. Thorough daily brushing of the denture will prevent deposits of calculus for many patients. However, any buildup of calculus noted by the patient between sched-uled recall appointments should be removed in the dental office. This can be quickly and readily accomplished with an ultrasonic cleaner.
Because many patients may dine away from home, the informed patient should provide some means of carrying out midday oral hygiene. Simply rinsing the removable partial denture and the mouth with water after eating is beneficial if brushing is not possible.
Opinion is divided on the question of whether or not a removable partial denture should be worn during sleep. Conditions should determine the advice given the patient, although generally the tissue should be allowed to rest by removal of the denture at night. The denture should be placed in a container and covered with water to prevent its dehydration and subsequent dimensional change. About the only situation that possibly justifies wearing removable par-tial dentures at night is when stresses generated by bruxism would be more destructive because they would be concen-trated on fewer teeth. Broader distribution of the stress load, plus the splinting effect of the removable partial denture, may make wearing the denture at night advisable. However, an individual mouth protector should be worn at night until the cause of the bruxism is eliminated.
Often the question arises whether an opposing complete denture should be worn when a removable partial denture in the other arch is out of the mouth. The answer is that if the removable partial denture is to be removed at night, the opposing complete denture should not be left in the mouth. There is no more certain way of destroying the alveolar ridge, which supports a maxillary complete denture, than to have it occlude with a few remaining anterior mandibular teeth.
The patient with a removable partial denture should not be dismissed as completed without at least one sub-sequent appointment for evaluation of the response of oral structures to the restorations and minor adjustment if needed. This should be made at an interval of 24 hours after initial placement of the denture. It need not be a lengthy appointment but should be made as a definite www.konkur.in 296 Part II Clinical and Laboratory rather than a drop-in appointment. This not only gives the patient assurance that any necessary adjustments will be made and provides the dentist with an opportunity to check on the patient’s acceptance of the prosthesis but also avoids giving the patient any idea that the dentist’s sched-ule may be interrupted at will and serves to give notice that an appointment is necessary for future adjustments.
FOLLOW-UP SERVICES The patient must understand the sixth and final phase of removable partial denture service (periodic recall) and its rationale. Patients need to understand that the support for a prosthesis (Kennedy Class I and II) may change with time. Patients may experience only limited success with the treat-ment and prostheses so meticulously accomplished by the dentist, unless they return for periodic oral evaluations.
After all necessary adjustments have been made to the removable partial denture and the patient has been instructed on proper care of the denture, the patient must also be advised as to future care of the mouth to ensure health and longev-ity of the remaining structures. How often the dentist should examine the mouth and the denture depends on the oral and physical condition of the patient. Patients who are caries- susceptible or who have tendencies toward periodontal disease or alveolar atrophy should be examined more often. Every 6 months should be the rule if conditions are normal.
The need to increase retention on clasp arms to make the denture more secure depends on the type of clasp that has been used. Increasing retention should be accomplished by contouring the clasp arm to engage a deeper part of the retentive undercut rather than by forcing the clasp in toward the tooth. The latter creates only frictional retention, which violates the principle of clasp retention. As an active force, such retention contributes to tooth or restoration movement, or both, in a horizontal direction, disappearing only when the tooth has been moved or the clasp arm has returned to a passive relationship with the abutment tooth. Unfortunately, this is almost the only adjustment that can be made to a half-round cast clasp arm. On the other hand, the round wrought-wire clasp arm may be cervically adjusted and brought into a deeper part of the retentive undercut. Thus, the passivity of the clasp arm in its terminal posi-tion is maintained but retention is increased, because it is forced to flex more to withdraw from the deeper undercut. The patient should be advised that the abutment tooth and the clasp will serve longer if the retention is held minimally, which is only that amount necessary to resist reasonable dislodging forces.
The future development of denture rocking or looseness may be the result of a change in the form of the supporting ridges rather than lack of retention. This should be detected as early as possible after it occurs and corrected by relining or rebasing. The loss of tissue support is usually so gradual that the patient may be unable to detect the need for relining. This usually must be determined by the dentist at subsequent examinations as evidenced by rotation of the distal extension denture about the fulcrum line. If the removable partial den-ture is opposed by natural dentition, the loss of base support causes loss of occlusal contact, which may be detected by having the patient close on wax or Mylar strips placed bilat-erally. If, however, a complete denture or a distal extension removable partial denture opposes the removable partial denture, the interocclusal wax test is not dependable because posterior closure, changes in the temporomandibular joint, or migration of the opposing denture may have maintained occlusal contact. In such cases, evidence of loss of ridge sup-port is determined solely by the indirect retainer leaving its seat as the distal extension denture rotates about the fulcrum.
No assurance can be given to the patient that crowned or uncrowned abutment teeth will not decay at some future time. The patient can be assured, however, that prophylactic measures in the form of meticulous oral hygiene, coupled with routine care by the dentist, will be rewarded by greater health and longevity of the remaining teeth.
The patient should be advised that maximal service may be expected from the removable partial denture if the follow-ing rules are observed: 1. Avoid careless handling of the denture, which may lead to distortion or breakage. Damage to the removable partial denture occurs while it is out of the mouth as a result of dropping it during cleaning or an accident that occurs when the denture is not worn. Fractured teeth and den-ture bases and broken clasp arms can be repaired, but a distorted framework can rarely if ever be satisfactorily re-adapted or repaired. 2. Protect teeth from caries with proper oral hygiene, proper diet, and frequent dental care. The teeth will be no less sus-ceptible to caries when a removable partial denture is be-ing worn but may be more so because of the retention of debris. At the same time, the remaining teeth have become all the more important because of oral rehabilitation, and abutment teeth have become even more valuable because of their importance to the success of the removable partial denture. Therefore, a rigid regimen of oral hygiene, diet control, and periodic clinical observation and treatment is essential for the future health of the entire mouth. Also, the patient must be more conscientious about returning periodically for examination and necessary treatment at intervals stated by the dentist. 3. Prevent periodontal damage to the abutment teeth by maintaining tissue support of any distal extension bases. As a result of periodic evaluation, this can be detected and corrected by relining or whatever procedure is indicated. 4. Accept removable partial denture treatment as something that cannot be considered permanent, but partial den-tures must receive regular and continuous care by both the patient and the dentist. The obligations for maintain-ing caries control and for returning at stated intervals for treatment must be clearly understood, along with the fact that regular charges will be made by the dentist for whatever treatment is rendered.
www.konkur.in CHAPTER 22 Relining and Rebasing the Removable Partial Denture CHAPTER OUTLINE Relining Tooth-Supported and Tooth Implant–Supported Denture Bases Relining Distal Extension Denture Bases Methods of Reestablishing Occlusion on a Relined Removable Partial Denture Differentiation between relining and rebasing the remov-able partial denture has been discussed previously in Chapter 1. Briefly, relining is the resurfacing of the tis-sue of a denture base with new material to make it fit the underlying tissue more accurately. Rebasing is the replacement of the entire denture base with new mate-rial. The artificial teeth may also need to be replaced in a rebase procedure. Relining removable partial dentures is a common occurrence in many dental practices; however, rebasing is not done as often.
In either situation, a new impression is necessary and uses the existing denture base after modifications (Figure 22-1) as an impression tray for a closed-mouth or an open-mouth impression procedure. One of several types of impression materials may be used, such as metallic oxide impression paste, rubber-base or silicone elastomers, tissue conditioning materials, or mouth-temperature wax. With a tooth-supported prosthesis, the impression method (open- or closed-mouth) is not as critical. In deciding between a closed-mouth and an open-mouth impression method for relining a distal extension removable partial denture, a major consideration is the resiliency of the mucosa cover-ing the residual ridge. As with secondary impression tech-niques, a firm mucosal foundation can likely accommodate a closed-mouth functional impression technique or an open-mouth selective pressure technique. However, when the mucosa is easily displaced, the open-mouth selective pressure technique is preferable. Both techniques should guard against framework movement during the impression procedure.
Before relining or rebasing is undertaken, the oral tissue must be returned to an acceptable state of health (Figure 22-2). For more information, refer to the Chapter 14 discussion about conditioning abused and irritated tissue.
www.konkur.in 299 Chapter 22 Relining and Rebasing the Removable Partial Denture RELINING TOOTH-SUPPORTED AND TOOTH IMPLANT–SUPPORTED DENTURE BASES When support is available at each end of a modification space and support for that restoration is derived entirely from the abutment teeth at each end of each edentulous span, reline or rebase is not as routinely needed to maintain occlusion as with a distal extension base. This support may be effective through the use of occlusal rests, boxlike internal rests, attachments, supporting ledges on abutment restora-tions, or devices within the acrylic resin base that engage an implant abutment. Except for intrusion of abutment teeth under functional stress, the supporting abutments prevent settling of the restoration toward the tissue of the residual ridge. Tissue changes that occur beneath tooth-supported denture bases do not affect the support of the denture; therefore relining or rebasing is usually done for reasons that include (1) unhygienic conditions and the trapping of debris between the denture base and the residual ridge; (2) an unsightly condition that results from the space that has developed; or (3) patient discomfort associated with lack of tissue contact that arises from open spaces between the den-ture base and the tissue. Anteriorly, loss of support beneath a denture base may lead to some denture movement, despite occlusal support and direct retainers located posteriorly. Rebasing would be the treatment of choice if the artificial teeth are to be replaced or rearranged, or if the denture base needs to be replaced for esthetic reasons, or because it has become defective.
To accomplish relining or rebasing, the original denture base must have been made of a resin material that can be relined or replaced. Tooth-supported removable partial den-ture bases made of metal as part of the cast framework gener-ally cannot be satisfactorily relined.
Because the tooth-supported denture base cannot be depressed beyond its terminal position with the occlusal rests seated and the teeth in occlusion, and because it cannot rotate about a fulcrum, a closed-mouth impression method is used. Virtually any impression material may be used, pro-vided sufficient space is allowed beneath the denture base to permit the excess material to flow to the borders—where it may be turned by the bordering tissue, or, as in the pal-ate, may be allowed to escape through venting holes without undue displacement of the underlying tissue. The qualities of each type of impression material must be kept in mind when the material to be used is selected. Ordinarily, an impression material is used that will record the anatomic form of the oral tissue.
A word of caution should be mentioned when a tooth-supported resin base is relined with autopolymerizing resin as an intraoral procedure. When one or more relatively short spans are to be relined, making an impression for relining purposes necessitates that the denture be flasked and pro-cessed. The possibilities that the vertical dimension of occlu-sion may be increased and that the denture may be distorted during laboratory procedures must be weighed against the disadvantages of using a direct-reline material. Fortunately, these materials are constantly improved with greater pre-dictability and color stability. The possibility that the original denture base will become crazed or distorted by the action of the activated monomer is minimal when the base is made of modern cross-linked resin. However, caution should be exer-cised to ensure that the older types of resin bases are compat-ible when one is relining with direct-reline resins.
Figure 22-1 Use of an existing Kennedy Class I removable partial denture base as a tray during a reline impression. The selective pressure impression philosophy requires space for the impression material that is greater over the ridge crest (second-ary stress bearing area) than at the buccal shelf region (primary stress bearing area). A pear-shaped laboratory bur is used to provide general relief (0.5 to 1.0 mm) of the denture base, with additional relief (1.0 mm) obtained over the ridge crest with a #8 round straight shank laboratory bur. Care must be taken to ensure that the tissue surface is relieved of all undercuts that could cause cast fracture when one is recovering the cast from the impression.
Figure 22-2 Kennedy Class I modification 1 arch with a re-movable partial denture that requires relining. Tissue abuse evident at the left buccal shelf region must be corrected before the reline impression is made. Management requires a period of function without the prosthesis or relief of the prosthesis in the affected region along with placement of a tissue-resilient liner in an effort to reduce the traumatic effects of pressure.
www.konkur.in 300 Part III Maintenance When relining in the mouth with a resin reline material is done with an appropriate technique, the results can be highly satisfactory, with complete bonding to the exist-ing denture base, good color stability, permanence, and accuracy. The procedure for applying a direct reline of an existing resin base is as follows: 1. Generously relieve the tissue side of the denture base.
Lightly relieve the borders. This not only provides space for an adequate thickness of new material but also eliminates the possibility of tissue impingement caused by confinement of the material.
2. Apply lubricant or tape over the polished surfaces from the relieved border to the occlusal surfaces of the teeth to prevent new resin from adhering to the preserved bases and teeth.
3. Mix the powder and the liquid in a suitable container according to the proportions recommended by the manufacturer.
4. While the material is reaching the desired consistency, have the patient rinse the mouth with cold water. At the same time, wipe the fresh surfaces of the dried denture base with a cotton pellet or small brush saturated with some of the reline resin monomer. This facilitates bonding and ensures that the surface is free of any contamination.
5. When the material has first begun to thicken, but while it is still quite fluid, apply it to the tissue side of the denture base and over the borders. Immediately place the removable partial denture in the mouth in its ter-minal position, and have the patient lightly close into occlusion. Be sure that no material flows over the oc-clusal surfaces or alters the established vertical dimen-sion of occlusion. Then, with the patient’s mouth open, manipulate the cheeks to turn the excess at the border and establish harmony with bordering attachments. If a mandibular removable partial denture is being relined, have the patient move the tongue into each cheek and then against the anterior teeth to establish a functional lingual border. It is necessary that the direct retainers be effective to prevent displacement of the denture while molding of the borders is accomplished. Otherwise, the denture must be held in its terminal position with finger pressure on the occlusal surfaces while border molding is in progress.
6. Immediately remove the denture from the mouth and, with fine curved iris scissors, trim away gross excess material and any material that has flowed onto proximal tooth surfaces and other components of the removable partial denture framework. While doing this, have the patient again rinse the mouth with cold water. Then replace the denture in its terminal posi-tion to bring the teeth into occlusion. Repeat the bor-der movements with the patient’s mouth open. By this time, or soon thereafter, the material will have become firm enough to maintain its form out of the mouth.
7. Remove the denture, quickly rinse it in water, and dry the relined surface with compressed air. Apply a generous coat of glycerin with a brush or cotton pellet to prevent frosting of the surface caused by evapora-tion of the monomer. Allow the material to polymer-ize in a container of cold water. This will eliminate any patient discomfort and tissue damage that could have resulted from exothermic heat or prolonged contact of the tissue with unreacted monomer. Although it is preferable for 20 to 30 minutes to elapse before trim-ming and polishing, it may be done as soon as the ma-terial hardens. Polymerization may be expedited and made denser by placing the denture in warm water in a pressure pot for 15 minutes at 20 psi. The masking tape must be removed before trimming is done but should be replaced over the teeth and polished surfaces below the junction of the new and old materials to protect those surfaces during final polishing.
When properly done, a direct reline is entirely accept-able for most tooth-supported removable partial denture bases made of a resin material, except when some tissue support may be obtained for long spans between abut-ment teeth. In the latter situation, a reline impression in tissue conditioning material or other suitable elastic impression material may be accomplished. The denture may then be flasked, and a processed reline may be added for optimal tissue contact and support.
RELINING DISTAL EXTENSION DENTURE BASES A distal extension removable partial denture, which derives its major support from the tissue of the residual ridge, requires relining much more often than does a tooth-supported den-ture. Because of this, distal extension bases are usually made of a resin material that can be relined to compensate for loss of support caused by tissue changes. Although tooth-sup-ported areas are relined for other reasons, the primary rea-son for relining a distal extension base is to reestablish tissue support for that base.
The need for relining a distal extension base is deter-mined by evaluating the stability and occlusion at reasonable intervals after initial placement of the denture. Before initial placement of the denture, the patient must be advised that (1) periodic examination and also relining, when it becomes necessary, are imperative; (2) the success of the removable partial denture and the health of the remaining tissue and abutment teeth depend on periodic examination and ser-vicing of both the denture and the abutment teeth; and (3) a charge will be made for these visits in proportion to the required treatment.
There are two indications of the need for relining a dis-tal extension removable partial denture base. First, loss of occlusal contact between opposing dentures or between the denture and the opposing natural dentition may be evident (see Figures 9-13 and 9-14). This is determined by having the patient close on two strips of 28-gauge, soft green or blue (casting) wax or Mylar matrix strips. If occlusal contact www.konkur.in 301 Chapter 22 Relining and Rebasing the Removable Partial Denture between artificial dentition is weak or lacking while the remaining natural teeth in opposition are making firm con-tact, the distal extension denture needs to have occlusion reestablished on the present base by altering the occlusion, thereby reestablishing the original position of the denture framework and base, or sometimes both. In most instances, reestablishing the original relationship of the denture is nec-essary, and the occlusion will automatically be reestablished.
Second, loss of tissue support that causes rotation and set-tling of the distal extension base or bases is obvious when alternate finger pressure is applied on either side of the ful-crum line. Although checking for occlusal contact alone may be misleading, such rotation is positive proof that relining is necessary. If occlusal inadequacy is detected without any evidence of denture rotation toward the residual ridge, all that needs to be done is to reestablish occlusal contact by rearranging the teeth or by adding to the occlusal surfaces with resin or cast gold onlays. On the other hand, if occlu-sal contact is adequate but denture rotation can be demon-strated, this is usually the result of migration or extrusion of opposing teeth or a shift in position of an opposing maxillary denture, thus maintaining occlusal contact at the expense of the stability and tissue support of that denture. This is often the situation when a maxillary complete denture opposes a removable partial denture. It is not unusual for a patient to complain of looseness of the maxillary complete denture and to request relining of that denture when actually it is the removable partial denture that needs relining. Relining and thus repositioning of the removable partial denture result in repositioning of the maxillary complete denture with a return of stability and retention in that denture. Therefore, evidence of rotation of a distal extension removable partial denture about the fulcrum line must be the deciding factor as to whether relining needs to be done.
Rotation tissue-ward about the fulcrum line always results in lifting of the indirect retainer(s). The framework of any distal extension removable partial denture must be in its original terminal position with indirect retainers fully seated during and at the end of any relining procedure. Any pos-sibility of rotation about the fulcrum line caused by occlusal influence must be prevented, and therefore the framework must be held in its original terminal position during the time the impression is being made. This all but eliminates the use of a closed-mouth impression procedure when unilateral or bilateral distal extension bases are relined.
The best way to ensure framework orientation throughout the impression procedure for a distal extension removable partial denture is with an open-mouth procedure done in exactly the same manner as the original secondary impres-sion (see Figure 17-10, D). The denture to be relined first is relieved generously on the tissue side (see Figure 22-1) and then is treated in the same way as the original impression base for a functional impression. The step-by-step procedure is the same, with the dentist’s three fingers placed on the two principal occlusal rests and at a third point between, prefer-ably at an indirect retainer farthest from the axis of rotation. The framework thus is returned to its original terminal posi-tion, with all tooth-supported components fully seated. The tissue beneath the distal extension base then is registered in a relationship to the original position of the denture that will ensure that (1) the denture framework will be returned to its intended relationship with the supporting teeth; (2) opti-mum tissue support will be reestablished for the distal exten-sion base; and (3) the original occlusal relationship with the opposing teeth will be restored.
Although it is true that the teeth are not allowed to come into occlusion during an open-mouth impression procedure, the original position of the denture is positively determined by its relationship with the supporting abutment teeth. Because this is the relationship on which the original occlu-sion was established, returning the denture to this position should bring about a return to the original occlusal rela-tionship if two conditions are satisfied. The first condition is that laboratory procedures during relining must be done accurately without any increase in the vertical dimension of occlusion. This is essential to any reline procedure, but it is a particular necessity with a removable partial denture because any change in occlusal vertical dimension will pre-vent occlusal rests from seating and will result in overloading and trauma to the underlying tissue. The second condition is that the opposing teeth have not extruded or migrated, or that the position of an opposing denture has not become altered irreversibly. In the latter situation, some adjustment of the occlusion will be necessary, but this should be deferred until the opposing teeth or denture and the structures asso-ciated with the temporomandibular joint have had a chance to return to their original position before denture settling occurs. One of the greatest satisfactions of a job well done is seen in the execution of an open-mouth reline procedure as described in the previous paragraph, which results in the restoration not only of the original denture relationship and of tissue support but also of the original occlusal relationship (Figure 22-3).
METHODS OF REESTABLISHING OCCLUSION ON A RELINED REMOVABLE PARTIAL DENTURE Occlusion on a relined removable partial denture may be reestablished by several methods, depending on whether the relining results in an increase in the vertical dimension of occlusion or in lack of opposing occlusal contacts. In either instance, it is usually necessary to make a remounting cast for the relined removable partial denture so the denture can be correctly related to an opposing cast or prosthesis in an articulator (Figure 22-4).
In rare instances, after a distal extension removable partial denture has been relined by the method previously described, the occlusion is found to be negative rather than positive, or the same as it was before relining. This may be the result of wear of occlusal surfaces over a period of time, a high position of the original occlusion with resulting depres-sion of opposing teeth, or other reasons. In such a situation, www.konkur.in 302 Part III Maintenance occlusion on the denture must be restored to reestablish an even distribution of occlusal loading over both natural and artificial dentition. Otherwise, the natural dentition must carry the burden of mastication unaided, and the denture becomes only a space-filling or cosmetic device.
If the artificial teeth to be corrected are resin, the occlu-sion can be reestablished by adding autopolymerizing or light-activated resin to occlusal surfaces, or by fabricating gold occlusal surfaces, which can be attached to the original replaced teeth. The original teeth may also be removed from the denture base and replaced by new teeth arranged to har-monize with the opposing occlusal surfaces. Baseplate wax may be used to support the teeth as they are arranged. This wax should be carved to restore the lingual anatomy of the teeth and the portion of the denture base that was eliminated when the original teeth were removed. A stone matrix is made that covers the occlusal and lingual surfaces of the teeth and denture flange. Then wax may be removed from the denture base and teeth and the tissue-bearing surface coated with a bonding agent. Those areas on the stone matrix intimate to the new resin to be added should be painted with a tinfoil substitute or an air barrier coating material if a vis-ible light-cured (VLC) material is to be used. The new teeth are placed in the stone matrix, and the matrix is accurately attached to the denture base with sticky wax or a hot glue gun. VLC material or an autopolymerizing resin is then used to attach the teeth. If an autopolymerizing material is used, it can be conveniently sprinkled on by a buccal approach. The buccal surface of the denture base adjacent to the teeth should be slightly overfilled so that the correct shape may A B Figure 22-4 A, An irreversible hydrocolloid pick-up impression made following reline of the mandibular removable partial denture. All regions in the framework that are below the height of contour will be coated with a thin layer of baseplate wax to allow recovery of the prosthesis from the remount cast without damage to the cast. This allows replacement of the prosthesis in the mouth for an interocclusal registration. B, Mounted mandibular prosthesis against a maxillary complete denture. The maxillary prosthesis was mounted using a remount cast made after finishing and polishing of the prosthesis. It is not uncommon to see a significant occlusal problem when one is relining a mandibular distal extension removable partial denture when the residual ridge resorption has been left unchecked for a long time. The maxillary prosthesis can often assume a more inferior position, and when the mandibular prosthesis is reoriented to the natu-ral dentition, this raises the mandibular occlusal plane, resulting in an occlusal relationship as shown. The best resolution of this requires addressing the maxillary arch at the same time and requires a repositioning of the maxillary occlusal plane.
A B Figure 22-3 A, Mandibular reline impression accomplished with an open-mouth impression technique. B, Following processing of the reline and occlusal correction with a clinical remount procedure (see Chapter 21), the occlusion is restored to the original relationship.
www.konkur.in 303 Chapter 22 Relining and Rebasing the Removable Partial Denture be restored to this portion of the base during finishing and polishing procedures. Occlusal discrepancies caused by this procedure should be corrected in the articulator by new jaw relation records if the denture has a distal extension base.
A second method involves removing the original teeth and replacing them with a hard inlay wax occlusion rim on which a functional registration of occlusal pathways is then established (see Chapter 18). The original teeth or new teeth may then be arranged to occlude with the template thus obtained and subsequently attached to the denture base with processed VLC material or autopolymerizing resin. If the lat-ter is used, the need for flasking may be eliminated by secur-ing the teeth to a stone matrix while the resin attachment is applied with a brush technique. Regardless of the method used for reattaching the teeth, the occlusion thus established should require little adjustment in the mouth and should exhibit the occlusal harmony that is possible to attain by this method.
www.konkur.in CHAPTER 23 Repairs and Additions to Removable Partial Dentures CHAPTER OUTLINE Broken Clasp Arms Fractured Occlusal Rests Distortion or Breakage of Other Components—Major and Minor Connectors Loss of a Tooth or Teeth not Involved in Support or Retention of the Restoration Loss of an Abutment Tooth Necessitating its Replacement and Making a New Direct Retainer Other Types of Repairs Repair by Soldering The need for repairing or adding to a removable partial denture will occasionally arise. However, the frequency of this occurrence should be held to a minimum by careful diagnosis, intelligent treatment planning, adequate mouth preparations, and the carrying out of an effective removable partial denture design with proper fabrication of all compo-nent parts. Any need for repairs or additions will then be the result of unforeseen complications that arise in abutment or other teeth in the arch, breakage or distortion of the denture through accident, or careless handling by the patient, rather than faulty design or fabrication.
It is important that the patient is instructed in proper placement and removal of the prosthesis so that undue strain is not placed on clasp arms, on other parts of the denture, or on contacted abutment teeth. The patient also should be advised that care must be given to the prosthesis when it is out of the mouth, and that any distortion may be irrepara-ble. It should be made clear that there can be no guarantee against breakage or distortion from causes other than obvi-ous structural defects.
BROKEN CLASP ARMS The following are several reasons for breakage of clasp arms: 1. Breakage may result from repeated flexure into and out of too severe an undercut. If the periodontal support is greater than the fatigue limit of the clasp arm, failure of the metal occurs first. Otherwise, the abutment tooth is loosened and eventually is lost because of the persistent strain that is placed on it. Locating clasp arms only where an acceptable minimum of retention exists, as determined by an accurate survey of the master cast, can prevent this type of breakage. 2. Breakage may occur as a result of structural failure of the clasp arm itself. A cast clasp arm that is not properly formed or is subject to careless finishing and polishing www.konkur.in 305 Chapter 23 Repairs and Additions to Removable Partial Dentures eventually will break at its weakest point. This can be prevented by providing the appropriate taper to flex-ible retentive clasp arms and uniform bulk to all rigid nonretentive clasp arms. Wrought-wire clasp arms may eventually fail because of repeated flexure at the region where it exits from the resin base (Figure 23-1), or at a point at which a nick or constriction occurred as a re-sult of careless use of contouring pliers. They also may break at the point of origin from the casting as a result of excessive manipulation during initial adaptation to the tooth or subsequent readaptation. Clasp breakage can best be prevented by cautioning the patient against removing the removable partial denture by sliding the clasp arm away from the tooth with the fingernails. A wrought-wire clasp arm can normally be adjusted sev-eral times over a period of years without failure. It is only when the number of adjustments is excessive that breakage is likely to occur. Wrought-wire clasp arms also may break at the point of origin because of re-crystallization of the metal. This can be prevented by proper selection of wrought wire, avoidance of burnout temperatures exceeding 1300° F, and avoidance of ex-cessive casting temperatures when a cast-to method is used. When wrought wire is attached to the framework by soldering, the soldering technique must avoid recrys-tallization of the wire. For this reason, it is best that sol-dering be done electrically to prevent the wrought wire from overheating. A low-fusing (1420° F to 1500° F), triple-thick, color-matching gold solder should be used rather than a solder that possesses a higher fusing temperature. 3. Breakage may occur because of careless handling by the patient. Any clasp arm will become distorted or will break if subjected to excessive abuse by the patient. The most common cause of failure of a cast clasp arm is dis-tortion caused by accidental dropping of the removable partial denture. A broken retentive clasp arm, regardless of its type, may be replaced with a wrought-wire reten-tive arm embedded in a resin base (see Figure 23-1, C and D) or attached to a metal base by electric soldering.
Often, this avoids the necessity of fabricating an entirely new clasp arm.
A B C D Figure 23-1 Fractured direct retainer on canine abutment. The reason for breakage is likely the long-term repeated flexure from movement associated with this 8-year-old distal extension prosthesis. The denture must be evaluated for prospective serviceability if the retainer arm is repaired. Often, the patient will best be served by replacing the denture with a new restoration. A, The cast produced from an irreversible hydrocolloid pick-up impression. The height of contour is shown in pencil, with a red line illustrating to the laboratory the location of repair wire (18-gauge). B, Clasp adapted to the designated line on the canine and fitted into the resin trough distal to the canine and palatal to the first and second premolars. Note the curvature placed at the end of the wire to prevent movement within the polymerized resin. C, Finished and polished wire repair from the buccal. D, Palatal view.
www.konkur.in 306 Part III Maintenance FRACTURED OCCLUSAL RESTS Breakage of an occlusal rest almost always occurs where it crosses the marginal ridge. Improperly prepared occlusal rest seats are the usual cause of such weakness. An occlusal rest that crosses a marginal ridge that was not lowered sufficiently during mouth preparations may be made too thin or may be thinned by adjustment in the mouth to prevent occlusal interference. Failure of an occlusal rest rarely results from a structural defect in the metal and rarely if ever is caused by accidental distortion. Therefore, the blame for such failure must often be assumed by the dentist for not having provided sufficient space for the rest during mouth preparations.
Soldering may repair broken occlusal rests. In prepara-tion for the repair, it may be necessary to alter the rest seat of the broken rest or to relieve occlusal interferences. With the removable partial denture in its terminal position, an impression is made in irreversible hydrocolloid and then is removed, with the removable partial denture remaining in the impression. The dental stone is poured into the impres-sion and is allowed to set. The removable partial denture is removed from the cast, and platinum foil is adapted to the rest seat and the marginal ridge and overlaps the guiding plane. The removable partial denture is returned to the cast and, with a fluoride flux, gold solder is electrically fused to the platinum foil and the minor connector in sufficient bulk to form an occlusal rest. An alternative solder to use is an industrial brazing alloy, which is higher fusing but responds excellently to electric soldering and does not tarnish.
DISTORTION OR BREAKAGE OF OTHER COMPONENTS—MAJOR AND MINOR CONNECTORS Assuming that major and minor connectors were originally made with adequate bulk, distortion usually occurs from abuse by the patient (Figure 23-2). All such components should be designed and fabricated with sufficient bulk to ensure their rigidity and permanence of form under normal circumstances.
Major and minor connectors occasionally become weak-ened by adjustment to prevent or eliminate tissue impinge-ment. Such adjustment at the time of initial placement may result from inadequate survey of the master cast or from faulty design or fabrication of the casting. This is inexcusable A B C D Figure 23-2 A, The maxillary juncture between major and minor connectors at the distalmost posterior molar has fractured. Thin platinum foil has been adapted to the cast beneath the fracture, the clasp assembly has been stabilized on the cast with fast-set plaster, the remainder of the prosthesis has been positioned on the cast in full contact with teeth and tissues, and the solder has been positioned for the electric tip to be placed. B, The electric soldering tip and ground are in place. C, Immediately following solder flow, the fracture has been eliminated by the solder connecting the two segments. D, The polished solder repair is ready to be cleaned and returned to the patient. The patient is told that this repair is not as strong as the original, and although it is difficult to know how long it could serve the patient, careful handling of the prosthesis is mandatory.
www.konkur.in 307 Chapter 23 Repairs and Additions to Removable Partial Dentures and reflects on the dentist. Such a restoration should be remade instead of further weakening the restoration by attempting to compensate for its inadequacies by relieving the metal. Similarly, tissue impingement that arises from inadequately relieved components results from faulty plan-ning, and the casting should be remade with enough relief to prevent impingement. Failure of any component that was weakened by adjustment at the time of initial placement is the responsibility of the dentist. However, adjustment made necessary by settling of the restoration because abutment teeth have become intruded under functional loading may be unavoidable. Subsequent failure that results from the weakening effect of such an adjustment may necessitate making a new restoration as a consequence of tissue changes. Commonly, repeated adjustment to a major or minor con-nector results in loss of rigidity to the point that the connec-tor can no longer function effectively. In such situations, a new restoration must be made, or that part must be replaced by casting a new section and then reassembling the denture by soldering. This occasionally requires disassembly of den-ture bases and artificial teeth. The cost and probable success must then be weighed against the cost of a new restoration. Generally, the new restoration is advisable.
LOSS OF A TOOTH OR TEETH NOT INVOLVED IN SUPPORT OR RETENTION OF THE RESTORATION Additions to a removable partial denture are usually simply made when the bases are made of resin (Figure 23-3). The addition of teeth to metal bases is more complex and neces-sitates casting a new component and attaching it by soldering or creating retentive elements for the attachment of a resin extension. In most instances when a distal extension denture base is extended, the need should be considered for subse-quent relining of the entire base. After the denture base has been extended, a relining procedure for both the new and the old base should be carried out to provide optimum tissue support for the restoration.
LOSS OF AN ABUTMENT TOOTH NECESSITATING ITS REPLACEMENT AND MAKING A NEW DIRECT RETAINER In the event of a lost abutment, the next adjacent tooth is usually selected as a retaining abutment, and it generally will require modification or a restoration. Any new restoration should be made to conform to the original path of placement, with proximal guiding plane, rest seat, and suitable retentive area. Otherwise, modifications to the existing tooth should be done the same as during any other mouth preparations, with proximal recontouring, preparation of an adequate occlusal rest seat, and any reduction in tooth contours neces-sary to accommodate retentive and stabilizing components. A new clasp assembly may be cast for this tooth and the den-ture reassembled with the new replacement tooth added.
OTHER TYPES OF REPAIRS Other types of repairs may include the replacement of a broken or lost prosthetic tooth, the repair of a broken resin base, or the reat-tachment of a loosened resin base to the metal framework. Break-age is sometimes the result of poor design, faulty fabrication, or use of the wrong material for a given situation. Other times, it results from an accident that will not necessarily repeat itself. If the latter occurs, repair or replacement usually suffices. On the other hand, if fracture has occurred because of structural defects, or if it occurs a second time after the denture has been repaired once before, then some change in the design—by modification of the original denture or with a new denture—may be necessary.
REPAIR BY SOLDERING Approximately 80% of all soldering in dentistry can be done electrically. Electric soldering units are available for this pur-pose, and most dental laboratories are so equipped. Electric soldering permits soldering close to a resin base without removing that base because of rapid localization of heat at the electrode. The resin base needs only to be protected with a wet casting ring liner during soldering.
Color-matching gold solder may be used for soldering both gold and chromium-cobalt alloys. A solder for gold alloys that melts at between 1420° F and 1500° F is entirely adequate for soldering gold alloys to chromium-cobalt alloys, thereby lessening the chance of recrystallizing gold wrought wire via excessive and prolonged heat. For electric soldering, triple-thick solder should be used so the addi-tional bulk of the solder will retard melting momentarily, while the carbon electrode conducts heat to the area to be soldered. For soldering chromium-cobalt alloys, a color-matching white 19K gold solder—which melts at about 1676° F—is used. An application of flux is essential for the success of any soldering operation to prevent oxidation of the parts to be joined and the solder itself. A borax-type flux is used when gold alloys are soldered. Fluoride-type flux must be used when chromium-cobalt alloys are sol-dered. When a gold alloy is to be soldered to a chromium-cobalt alloy, a fluoride-type flux should be chosen.
The following is a procedure for electric soldering: 1. Roughen both sections to be joined.
2. Adapt platinum foil to the master cast beneath the framework to serve as a backing on which the solder will flow. Lift the edges of the foil to form a trough to con-fine the flow of the solder.
3. Seat the pieces to be soldered onto the master cast and secure them temporarily with sticky wax. Over each piece, add enough soldering investment to secure them after the sticky wax has been eliminated, but leave as much metal exposed as possible.
4. After flushing off the sticky wax with hot water, secure the cast to the soldering stand. Cut sufficient solder and place conveniently nearby.
www.konkur.in 308 Part III Maintenance 5. Flux both sections. Put sufficient triple-thick solder on or in the joint to complete the soldering in one opera-tion, always starting with enough solder to complete the job.
6. Wet the carbon tip with water to aid conduction of the current, and then touch the carbon tip to the solder (be sure the solder is held firmly in place). Place the other electrode on any portion of the framework to complete the electric circuit and heat the carbon electrode. Do not push the solder with the carbon tip, but let the heat alone make the solder flow. Do not remove the carbon tip from the solder while the soldering operation is in progress; this will cause surface pitting due to arcing. After the solder has flowed, remove the electrodes, re-moving the carbon tip electrode last, and proceed to remove the work from the cast for finishing.
Torch soldering requires an entirely different approach. It is used when the solder joint is long or unusually bulky and when a larger quantity of solder has to be used. Torch soldering cannot be undertaken to repair a removable A B D C E Retention groove #10 Figure 23-3 This patient presented with an asymptomatic fractured lateral incisor. A, Clinical presentation of the fractured tooth and prosthesis. Evaluation of the prosthesis revealed it to be adequately fitting, stable, and retentive. B, Pick-up impression of the prosthesis. C, Cast formed from the pick-up impression, showing a fully seated prosthesis. D, Preparation of the prosthesis included a mechanical means for retention (which was provided by creating a recess in the resin adjacent to the missing tooth) and creating a trough at the external finishing line to repair an area of marginal breakdown. E, The finished repair, which will be taken to the mouth and checked for occlusal clearance lingual to the maxillary anterior teeth.
www.konkur.in 309 Chapter 23 Repairs and Additions to Removable Partial Dentures partial denture framework that has resin denture bases or artificial teeth supported by resin. The procedure for torch soldering is as follows: 1. Roughen both sections to be joined.
2. Adapt platinum foil to the master cast so it extends un-der both sections.
3. Seat the sections on the master cast in the correct rela-tionship, and secure them temporarily with sticky wax. Also flow sticky wax into the joint to be soldered.
4. Attach a dental bur or nail over the two sections with a liberal amount of sticky wax. Attach a second and even a third nail or bur across other areas to lend additional support. Never use pieces of wood for this purpose be-cause the wood will swell if it gets wet, thus distorting the relationship of the two sections.
5. Carefully remove the assembled casting from the mas-ter cast. Adapt a stock of utility wax directly under each section on either side of the platinum foil. After boil out is done, investment will remain in the center to support the platinum foil.
6. Invest the casting in sufficient soldering investment to secure it, and expose as much of the area to be soldered as possible. When the investment has set, boil out the sticky and utility waxes. Then place the investment in a drying oven at a temperature not exceeding 200° F until the contained moisture has been eliminated. Do not dry or preheat the investment with a torch because oxides then will be formed that will interfere with the flow of the solder.
7. Use the reducing part of the flame, which is the feath-ery part just outside the blue inner cone. Flux the joint thoroughly and dry out the flux with the outer part of the flame until it has a powdery appearance. Heat the casting until it is dull red, and then, while holding a strip of solder in the soldering tweezers, dip it into the flux and feed it into the joint, while the casting is being held at a dull-red heat with the torch. Once the solder-ing operation has begun, do not remove the flame, be-cause any cooling will cause oxides to form. The heat from the casting should be sufficient to melt the solder; therefore, do not put the flame directly on the solder be-cause it will become overheated and pitting will result.
8. After the soldering has been completed, allow the investment to cool slowly before quenching and pro-ceeding with finishing. Remember that any soldering operation that heats the entire casting is in effect a soft-ening heat-treating operation, and heat-hardening of a repaired gold alloy casting is desirable to restore its optimal physical properties.
www.konkur.in CHAPTER 24 Interim Removable Partial Dentures CHAPTER OUTLINE Appearance Space Maintenance Reestablishing Occlusal Relationships Conditioning Teeth and Residual Ridges Interim Restoration During Treatment Conditioning the Patient for Wearing a Prosthesis Clinical Procedure for Placement Tooth replacement is required for a variety of reasons. Some-times, replacements may be necessary for shorter periods of time that serve alternative purposes than permanent replace-ment, such as while tissue is healing or related treatment is being provided. When such applications require the tempo-rary use of removable partial dentures, their fabrication and use must be incorporated into a total prosthodontic treat-ment plan.
These various uses of interim prostheses for the partially edentulous mouth strive to achieve temporary goals with minimum time and expense. These prostheses are typi-cally resin with wire retention and may include components to provide tooth support. The difficulty in achieving and maintaining strategic tooth support and stability with such prostheses makes it important that patients be made aware that these prostheses are temporary and may jeopardize the integrity of adjacent teeth and the health of supporting tissue if worn for extended periods without supportive care.
Interim prostheses may be indicated as a part of total treatment for the following: 1. Sake of appearance 2. Maintenance of a space 3. Reestablishment of occlusal relationships 4. Conditioning of teeth and residual ridges 5. Interim restoration during treatment 6. Conditioning the patient for wearing a prosthesis APPEARANCE For the sake of appearance, an interim removable partial denture may replace one or more missing anterior teeth, or it may replace several teeth, both anterior and poste-rior. Such a restoration is usually made of resin, which may be produced by a sprinkling method, by the visible light-cured (VLC) method, or by waxing, flasking, and processing with autopolymerizing or heat polymerizing resin (Figure 24-1). It may be retained by circumferential wrought-wire clasps, Crozat-type clasps, interproximal spurs, or wire loops.
www.konkur.in 311 Chapter 24 Interim Removable Partial Dentures SPACE MAINTENANCE When a space results from recent extractions or traumatic loss of teeth, it is usually prudent to maintain the space while the tissue heals. In younger patients, the space should be maintained until the adjacent teeth have reached sufficient maturity to be used as abutments for fixed restorations or so that an implant can be placed. In adult patients, maintenance of the space can prevent undesirable migration and extru-sion of adjacent or opposing teeth until definitive treatment can be accomplished (Figure 24-2).
REESTABLISHING OCCLUSAL RELATIONSHIPS Interim removable partial dentures are used for the follow-ing reasons: (1) to establish a new occlusal relationship or occlusal vertical dimension; and (2) to condition teeth and ridge tissue for optimum support of the definitive removable partial denture that will follow.
Interim removable partial dentures may be used as occlu-sal splints in much the same manner as cast or resin occlusal splints are used on natural teeth. When total tooth support is available, there is little difference between a fixed and a removable occlusal splint, except that a removable splint is likely to be left out of the mouth unless the patient is actu-ally made more comfortable by its presence. This is usually true when the wearing of an occlusal splint alleviates a tem-poromandibular joint condition. In other situations, it may be advisable to cement the removable restoration to the teeth until such time as the patient has become accustomed to, and dependent on, the jaw relationship provided by the splint.
Both fixed and removable tooth-supported occlusal splints have much in common. Either of them may be elimi-nated in sections as restorative treatment is being done, thus maintaining the established jaw relation until all restorative treatment has been completed. The dentist decides whether these are to be fixed or removable, and whether they are made of a cast alloy, a composite, or a resin material.
When one or more distal extension bases exist on an occlu-sal splint, a different situation occurs. The establishment of a new occlusal relation depends on the quality of support and stability the splint receives from the denture support. Both broad coverage and functional basing of tissue-supported bases are desirable, along with some type of occlusal rest on the nearest abutments. Any tissue-supported occlusal splint should be at least relined in the mouth with an autopolymer-izing reline resin to afford optimal coverage and support for the distal extension base.
CONDITIONING TEETH AND RESIDUAL RIDGES O. C. Applegate, in an article on the choice of partial or complete denture treatment, emphasized the advantages of conditioning edentulous areas to provide stable sup-port for distal extension removable partial dentures. This is accomplished by having the patient wear an interim removable partial denture for a period of time before the final base is fabricated. In the absence of opposing occlu-sion, stimulating the underlying tissue by applying inter-mittent finger pressure to the denture base is advised.
Whether the stimulation results from occlusal or finger pressure, there seems to be little doubt that the tissues of the residual ridge become more capable of support-ing a distal extension removable partial denture when they have been previously conditioned by wearing of a restoration.
Abutment teeth also benefit from wearing of an interim restoration when such a restoration applies an occlusal load to those teeth, either through occlusal cov-erage or through occlusal rests. Commonly, a tooth that A B Figure 24-1 A, Although loss of mandibular teeth does not always have a significant esthetic impact, this interim removable partial denture was needed because of the visibility of the mandibular incisors. Provision of the mandibular left molars allowed early accom-modation of the residual ridge during the temporary prosthesis period. B, Tissue surface of the interim prosthesis revealing a rounded lingual flange and a tapered labial flange. The latter was needed to improve lip movement and reduce the feeling of bulk, both of which enhance normal lip activity.
www.konkur.in 312 Part III Maintenance is to be used as an abutment for a removable partial den-ture has been out of occlusion for some time. Immedi-ately on applying an occlusal load to that tooth sufficient to support any type of removable prosthesis, some intru-sion of the tooth will occur. If such intrusion is allowed to occur after initial placement of the final prosthesis, the occlusal relationship of the prosthesis and its relation to the adjacent gingival tissue will be altered. Perhaps this is one reason for gingival impingement, which occurs after the prosthesis has been worn for some time, even though seemingly adequate relief was provided initially. When an interim removable partial denture is worn, such abutment teeth have an opportunity to become stabilized under the loading of the temporary restoration, and intrusion will have occurred before the impression for the master cast is made. There is sufficient reason to believe that both abut-ment teeth and supporting ridge tissue are more capable of providing continued support for the removable partial denture when they have been previously conditioned by the wearing of a temporary interim restoration.
INTERIM RESTORATION DURING TREATMENT In some instances, an existing removable partial denture can be used with modifications as an interim removable partial denture. Such modifications may include relining and adding teeth and clasps to an existing denture. In other instances, an existing removable partial denture may be converted to a transitional complete denture for immediate placement while the tissue heals and an opposing arch is prepared to receive a removable partial denture. Sometimes a tempo-rary interim removable partial denture must be made to replace missing anterior teeth in a partially edentulous arch, which are ultimately to be replaced with fixed restorations. On occasion, the anterior portion of the restoration is cut away when the fixed restorations are placed; this leaves the remainder of the denture to be worn while posterior abut-ment teeth are prepared. Still another type of interim remov-able partial denture is one in which missing posterior teeth are replaced temporarily with a resin occlusion rim rather than with occluding teeth.
CONDITIONING THE PATIENT FOR WEARING A PROSTHESIS A temporary restoration may be made to aid the patient in making a transition to complete dentures when the total loss of teeth is inevitable. Such a removable partial denture also may be considered a valid part of the treatment, because the patient is at the same time being conditioned to wear a removable prosthesis. It should be considered strictly a temporary measure that provides the patient with a restora-tion for the remaining life of the natural teeth when further A B Figure 24-2 A, Malpositioned maxillary anterior teeth require extraction. Following a period of healing, the patient will decide on a definitive treatment option, which may include a removable partial denture or an implant-supported prosthesis. Because the length of time until definitive care will be provided is not known, a temporary interim prosthesis not only replaces esthetically important teeth but provides stabilization of adjacent and opposing dentition as well. B, Occlusal view of the interim prosthesis showing clasp placement at the most posterior locations without crossing of the occlusion and anterior positions bilaterally. Full palatal coverage allows less stress to the remaining maxillary dentition and may prevent prosthesis-induced gingival trauma as well as tooth movement.
www.konkur.in 313 Chapter 24 Interim Removable Partial Dentures restorative treatment of those teeth is impractical or eco-nomically or technically impossible.
This type of a removable partial denture may be worn for prolonged periods, in the meantime undergoing revi-sion, modification to include additional teeth lost, or relin-ing when such becomes necessary or advisable. The dentist should agree to provide such a removable partial denture only under the following conditions: (1) that a definite fee for the treatment is appropriate and that the fee will depend on the servicing necessary; and (2) that when further wear-ing of the transitional denture is unwise and jeopardizes the health of remaining tissue, the transition to complete den-tures will proceed.
It is imperative that a distinction is made between tem-porary restorations and a true removable partial denture service, and that the patient is advised of the purposes and limitations of such restorations.
CLINICAL PROCEDURE FOR PLACEMENT It is important to consider proper fitting of the prosthesis to ensure comfortable use during the temporary phase of treatment. Careful attention to planned use of the teeth for support, stability, and retention without undue stress from gingival tissue contact or improper occlusal loading will ensure more comfortable use.
To ensure proper use of the remaining natural teeth, the prosthesis must be completely seated in the arch. Common areas requiring adjustment to ensure complete seating include interproximal extensions, regions where clasps exit from the acrylic-resin base, tissue undercuts (labial undercuts from recent extractions or the lingual/retromylohyoid region), and any portion of the prosthesis that lies inferior to the height of contour, especially if bilaterally opposed (Figure 24-3).
Once seated, it is important to check that no undue pres-sure to the marginal gingival region is present. To facilitate this step, and to help with the complete seating requirement, it is possible to have the laboratory block out the marginal gingival region and infrabulge regions to reduce seating problems (Figure 24-4). Infrabulge regions may include lingual and palatal tooth surfaces, as well as modification space regions. Because temporary prostheses are generally fabricated on unprepared teeth, these regions often require correction. However, if the blockout is not accomplished carefully, the prosthesis may seat easily, but it may not be as stable as possible because of insufficient tooth contact. Stability and retention are improved when it is possible to have the prosthesis contact portions of the teeth superior to the height of contour because of tooth-dictated control of movement.
It is possible to create occlusal imbalance if the lingual/ palatal portion of the prosthesis is too bulky. Consequently, an opposing cast should be provided to allow placement of clasps and acrylic-resin contours that do not cause occlusal interference (Figure 24-5). Once fully seated and relieved appropriately, the occlusion should contribute to the remain-ing natural dentition (as in a definitive prosthesis) and har-monize with natural tooth–dictated function. Typically, the prosthesis should not be the sole source of occlusal contact. In such situations, the functional forces are concentrated at the acrylic resin–to-tooth junction, and predictably a change in orientation occurs, allowing tissue-ward movement and a change in occlusion with an increase in soft tissue contact.
A B B C A B Figure 24-3 A, Maxillary interim removable partial denture. Areas that require adjustment commonly include interproximal extensions (A), the region where the clasp exits from the resin (B), and tissue undercuts of prosthesis extensions (C). B, Interproximal tooth extensions and regions where marginal gingival is crossed by the prosthesis should be carefully adjusted (outlined in red).
www.konkur.in 314 Part III Maintenance A B Figure 24-4 A, Maxillary cast with an outline of the prosthesis design and with the region at the marginal gingival crossing outlined. If the cast is not relieved at the marginal gingival, the prosthesis should be corrected before insertion. B, Cast showing gingival regions relieved with baseplate wax. A duplicate of this cast will provide the necessary relief for use as a processing cast. If interproximal regions have undercuts, these can also be blocked out to allow easier insertion of a temporary prosthesis.
A B Figure 24-5 A, Evaluation of mounted casts allows determination of the occlusal impact on maxillary temporary prosthesis clasp selection and placement. Interocclusal space allows placement of a ball clasp distal to the canine and a circumferential clasp from the distal of the second molar, both without occlusal interference if the clasps are carefully contoured. B, Anterior cast modification in anticipation of postextraction ridge contours for this immediate temporary prosthesis. Diagnosis of the limited space at the anterior region before surgery allows presurgical discussion of the possible need for mandibular incisor modification to improve the interim and final prosthesis occlusion.
www.konkur.in CHAPTER 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics CHAPTER OUTLINE Maxillofacial Prosthetics Maxillofacial Classification Timing of Dental and Maxillofacial Prosthetic Care for Acquired Defects Preoperative and Intraoperative Care Interim Care Potential Complications Defect and Oral Hygiene Definitive Care Intraoral Prostheses: Design Considerations Surgical Preservation for Prosthesis Benefit Maxillary Defects Mandibular Defects Mandibular Reconstruction—Bone Grafts Maxillary Prostheses Obturator Prostheses Speech Aid Prostheses Palatal Lift Prostheses Palatal Augmentation Prostheses Mandibular Prostheses Evolution of Mandibular Surgical Resection Type I Resection Type II Resection Type III Resection Type IV Resection Type V Resection Mandibular Guide Flange Prosthesis Jaw Relation Records for Mandibular Resection Patients Summary MAXILLOFACIAL PROSTHETICS The preceding chapters have dealt with prosthesis consider-ations for partially edentulous individuals. In these patients, the extent of loss includes teeth and a varying degree of residual ridge bone, yet the remaining anatomy of the jaws and adjacent regions is functionally and physically intact. For these patients, the major distinguishing feature that affects removable partial denture design is whether the prosthesis will be tooth-supported or tooth- and tissue-supported.
The maxillofacial patient can experience unique altera-tions in the normal oral/craniofacial environment; these result from surgical resections (Figure 25-1), maxillofacial trauma, congenital defects, developmental anomalies, or neu-romuscular disease. In contrast to the above, when remov-able partial dentures are considered for these individuals, not only are tooth and tissue support considerations important, but the design must also take into account what impact the altered environment will have on prosthesis support, stabil-ity, and retention. In general, environmental changes reduce the capacity for residual teeth and tissue to provide optimum cross-arch support, stability, and retention.
As a subspecialty of prosthodontics, maxillofacial pros-thetics is concerned with the restoration and/or replacement of the stomatognathic system and associated facial structures with prostheses that may or may not be removed on a regu-lar or elective basis. This chapter discusses important back-ground information related to maxillofacial prostheses and the principals involved in removable partial denture design for the maxillofacial patient.
Maxillofacial Classification Patients can be categorized by maxillofacial defects that are acquired, congenital, or developmental. Acquired defects www.konkur.in 316 Part III Maintenance include those that are the result of trauma or of disease and its treatment. These may include a soft and/or hard palate defect resulting from removal of a squamous cell carcinoma of the region (Figure 25-2). Congenital defects are typically craniofacial defects that are present from birth. The most common of these include cleft defects of the palate that may include the premaxillary alveolus. Developmental defects are those defects that occur because of some genetic predis-position that is expressed during growth and development (Figure 25-3). Such a classification order is helpful as patients within each category share similar characteristics (beyond those features specifically related to the prosthesis design), which become part of the total management plan. For exam-ple, prosthetic management of an adult who has experienced a maxillectomy procedure can be quite different from man-agement of a patient with an unrepaired cleft palate.
Another helpful way to classify maxillofacial patients is by the type of prosthesis under consideration. Conse-quently, prostheses are said to be extraoral (cranial or facial replacement) or intraoral (involving the oral cavity); interim (for short periods of time, often perioperative) or definitive (more permanent); and treatment (used as a component of management, such as a splint or stent) prostheses.
TIMING OF DENTAL AND MAXILLOFACIAL PROSTHETIC CARE FOR ACQUIRED DEFECTS Acquired defects are the most common maxillofacial defects managed by using removable prostheses. A conceptual frame-work for the timing of dental/oral care that best emphasizes the initial important surgical requirements, followed later by the important prosthetic requirements, is helpful to consider regarding the coordination of care for patients with acquired defects. Such a framework considers preoperative and intraop-erative interim care and definitive care. Although it may seem unrelated initially, it is included in this discussion of removable partial dentures for maxillofacial applications because of the important impact that decisions made at all stages of manage-ment can have on prosthesis function and patient outcomes.
Preoperative and Intraoperative Care The planning of prosthetic treatment for acquired oral defects should begin before surgery. For the patient facing head and neck surgery, consideration should be given to dental needs that will improve the immediate postopera-tive course. Consequently, the prosthodontist who will help with management of the patient’s care should see the patient before surgery (Figure 25-4). The dental objectives of the preoperative and intraoperative care stages are to remove potential dental postoperative complications, to plan for the subsequent prosthetic treatment, and to make recommen-dations for surgical site preparation that improve structural integrity. Important patient benefits of such a preoperative consultation include the opportunity to develop the patient-clinician relationship, to discuss the functional deficits associated with the anticipated surgical procedure, and to describe how and to what extent the stages of prosthetic Figure 25-2 Large squamous cell carcinoma involving the maxillary tuberosity region, which will result in an acquired defect following surgical removal.
A B Figure 25-1 A unilateral arrangement of maxillary teeth (A), no remaining horizontal hard palate, and a surgical defect, which includes nasal and sinus cavities (B). This unique environment, which is the result of a surgical resection, requires careful application of removable prosthodontic principles modified for maxillofacial needs.
www.konkur.in 317 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics management will address them. The benefit from a prosthe-sis standpoint is that strategically important teeth, for defini-tive and/or interim prosthesis use, can be discussed with the surgical team and treatment planned for preservation.
The immediate postoperative period will be significantly challenging to the patient. If pre-existing dental disease is severe enough to potentially create symptoms during the immediate postoperative period, treatment should be pro-vided to remove such a complication. Large carious lesions, which could create pain, can be temporarily restored by end-odontic therapy if they offer some advantage for postopera-tive prosthetic function. Teeth exhibiting acute periodontal disease (such as acute necrotizing ulcerative gingivitis) should be treated, as should any periodontal condition that could potentially cause postoperative pain because of exces-sive mobility or oral infection. Any tooth deemed nonre-storable because of advanced caries or periodontal disease, and not critical for temporary use during the interim care period following temporary treatment, should be removed before, or at the time of, surgical resection. Teeth that may appear to have a limited long-term prognosis may signifi-cantly enhance prosthetic service during the initial postsur-gical period and should be maintained until the initiation of definitive care.
Impressions are made of the maxillary and mandibular arches to provide a record of existing conditions and occlusion to allow fabrication of immediate or interim prostheses (Fig-ure 25-5) and to assess the need for immediate and delayed modification of the teeth or adjacent structures to optimize prosthetic care. It is important at this stage to begin planning for the definitive prosthesis because the greatest impact on the success of the maxillofacial prosthesis stems from the integ-rity of the remaining teeth and surrounding structures.
Interim Care The major emphasis during this stage of care is the surgi-cal (and adjunctive) management needs of the patient. In today’s environment of appropriately aggressive mandibular surgical reconstruction, mandibular discontinuity defects are seldom a surgical outcome. When discontinuity defects in the mandible result following surgery, interim prosthetic care is not indicated and the discussion will be directed to the maxillary defect.
The typical maxillary acquired defect results in oral communication with the nose and/or maxillary sinus, although the composition of the surgical defect may vary widely (Figure 25-6). This creates physiologic and func-tional deficiencies in mastication, deglutition, and speech. Such defects have a negative impact on the psychological disposition of patients, especially if the defect also affects cosmetic appearance. The major deficiencies directly A B Figure 25-3 A functional jaw position developed because of a combination of tooth loss and growth discrepancy. This developmen-tal defect is illustrated by a protruded and over-closed mandibular position (A), which has created a significantly irregular maxillary occlusal plane (B).
Figure 25-4 Presurgical presentation of a patient with a maxillary malignant melanoma. The benefits of having such a visit before surgery are both psychological and functional. The psychological benefits include the chance to discuss functional deficits associated with the anticipated surgical procedure and to describe how and to what extent the stages of prosthetic manage-ment will address them. The functional benefit from a prosthe-sis standpoint is that strategically important teeth, for definitive and/or interim prosthesis use, can be discussed with the surgical team and treatment planned for preservation.
www.konkur.in 318 Part III Maintenance A B Figure 25-5 A, A maxillary cast of the presurgical oral condition, which allows consultation with the surgeon regarding resection mar-gins and the benefits of preservation of teeth. B, Another maxillary cast altered, following consultation with the head and neck surgeon, to allow fabrication of a surgical stent. Perforations are made to allow fixation to the remaining teeth and to superior anatomic regions with the use of wires.
A B C Figure 25-6 Maxillary defects. A, A resection that resulted in a small communication with the sinus, with some hard palate remaining, and adjacent mucosa typical of the oral cavity. B, A resection that did not follow classic maxillectomy technique; however, the midline resection was made through the socket of tooth #9, preserving its alveolar housing. C, A resection along the palatal midline that did not preserve oral mucosa at the resection margin, which allows chronic ulceration at this point of prosthesis fulcrum. Notice the split-thickness skin graft in the superior-lateral region. Engagement of this region can provide support to the obturator extension, minimizing movement with function.
www.konkur.in 319 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics addressed by prosthetic management at this interim care time are deglutition and speech. This immediate postsurgi-cal time is very challenging for patients, and it is impor-tant that they have been mentally prepared for it during the preoperative period. However, even with preliminary dis-cussion, the impact of the surgery is often very distressing. An initial focus on improvement in swallowing and speech with the interim prosthesis can help boost the rehabilitation process significantly.
The patient is counseled that chewing on the defect side is not allowed because of its effect on prosthesis movement. The objective of this interim obturator prosthesis is to sepa-rate the oral and nasal cavities by obturating the communi-cation. Such obturator prostheses most commonly refer to obturation of a hard palatal defect but conceptually can be considered the same for soft palatal defects at this stage of management, because both attempt to artificially block the free transfer of speech sounds and foods/liquids between the oral and nasal cavities. The advantages of having the ability to take nourishment by mouth without nasal reflux (allow-ing for nasogastric tube removal) and to communicate with family members are a significant component of early pros-thetic management. How immediately such care should be provided depends on a number of factors.
A prosthesis can be provided at surgery (see Figure 25-5, B). Such a surgical obturator prosthesis is placed at the time of surgical access closure and serves to control the surgical dressing and split-thickness skin graft during the immedi-ate postsurgical period. Such prostheses are best stabilized by appropriate wiring to remaining teeth or alveolar bone, or they may be suspended from superior skeletal struc-tures. For some patients who have teeth remaining, such an immediate surgical prosthesis could be retained by wires in the prosthesis that engage undercuts on the teeth and would be removable; however, the ability to control the surgical dressing may be less predictable with such an approach. Immediate placement of a prosthesis has been suggested to improve patient acceptance of the surgical defect, although no measure of this psychological impact has been shown; this method offers greater assurance of adequate nourishment by mouth—potentially precluding the use of a nasogastric tube.
It may be preferable to stabilize the surgical dressing by suturing a sponge bolster to provide stabilization to the split-thickness skin graft. Following the primary healing stage, the sponge with packing (or the immediate prosthesis if used) is removed by the surgeon and an interim obturator prosthesis can be placed (Figure 25-7). For the patient who has been provided with bolster obturation, the presurgical prosthodontic evaluation is very important to ensure that the patient is prepared for the transition from bolster to pros-thesis, and to ensure that plans for the prosthesis are made, especially if an interim prosthesis is to be fabricated. Interim prostheses are wire-retained resin prostheses that generally do not have teeth initially but may be modified with the addition of teeth after an initial period of accommodation (Figure 25-8).
Figure 25-7 An interim obturator prosthesis fabricated of resin, retained by wires, and provided following surgical pack removal.
A B Figure 25-8 A-B, An interim obturator prosthesis fabricated of resin, retained by wires, and including artificial teeth for cosmesis during an extended period of recovery. The superior and lateral surfaces may need modification to improve stability and retention as the surgical site matures and allows more aggressive engagement.
www.konkur.in 320 Part III Maintenance When surgical defects become large, as in a near-total maxillectomy defect, prosthesis support, stability, and reten-tion are not likely to be satisfactory unless extension into the defect can be accomplished. When teeth remain, the impact of the defect size is somewhat lessened. But when the remain-ing teeth are few or are located unilaterally in a straight line (see Figure 25-1), the mechanical advantage for prosthesis stability is less. The ability of the defect tissue to offer the needed mechanical characteristics to the interim prosthesis is unpredictable at best. It is this patient who benefits the most from a well-planned surgery that preserves oral and defective anatomy to the advantage of the prosthesis.
Potential Complications The interim phase of prosthetic management can last for 3 months or more. The primary objective is to allow the patient to pass from an active surgical (and adjunctive treat-ment) phase to an observational phase of management with minimal complications. During the transition, the patient recovers from the systemic effects of the treatment, deals with the psychological impact of the defect using his or her own coping strategy, and becomes more aware of the func-tional deficits associated with the surgical defect(s). Mini-mizing potential complications during the transition, which includes preparing the patient for those anticipated to occur, facilitates the process for the patient and family. Common interim prosthetic complications are related to tissue trauma and the associated discomfort; inadequate retention (loose-ness) of the maxillary prosthesis; incomplete obturation with leakage of air, food, and liquid around the obturator portion of the prosthesis; and the tissue effects of chemotherapy and radiation therapy.
Discomfort related to the use of interim prostheses can be due to surgical wound healing dynamics, defect conditions, mucosal effects of adjunctive treatment, and/or prosthetic fit. Common areas of surgical wound pain include junctions of the oral and lip/cheek mucosa, especially at the anterior alveolar region for maxillectomy patients. The lateral scar band produced when the skin graft heals to the oral mucosa can be the site of discomfort in some patients. When a split-thickness skin graft is not placed, discomfort caused by the prosthesis fit within the defect can be a consistent and long-term problem. The hard palate surgical margin when not covered with surgically reflected oral mucosa most often will be covered by nasal epithelium, which is also very prone to discomfort. Alveolar bone cuts that have not been rounded will perforate the oral mucosa and will be painful whether or not a prosthesis is worn. This is most frequently a finding for mandibular resection superior alveolar margins when the reconstruction has restored the lower and labial contour to the mandible, but the intraoral mucosa at the superior sur-face is under tension because of a difference in height.
The prosthesis can create discomfort via excessive static pressure from the internal surfaces or from overextension into the vestibular tissue. The prosthesis can also create dis-comfort caused by functional movement associated with swallowing and speech. As was discussed previously, pros-thesis movement is dependent on the quality of the support-ing structures. Teeth offer the best support, followed by firm edentulous ridges, and lastly, surgical defect structures. The tongue, opposing dentition, and cheek/lips place force on the prosthesis that must be resisted over a large area to prevent movement. Because the defect is least likely to be able to resist movement, the relative size and structural integrity of the defect compared with the remaining teeth and/or eden-tulous ridge determine the potential prosthesis movement and most affect the discomfort related to such movement.
When teeth are available (and especially if located both close to and far away from the defect), retention is enhanced by engaging them with prosthetic clasps. Clasp retention is the most efficient means of effectively resisting dislodgment. The clasps will require periodic adjustment to maintain their effectiveness, as the movement of the prosthesis flexes the clasps beyond their elastic recovery capacity. For edentu-lous patients, because the surgical defect allows communi-cation between cavities, the fitting surface of the prosthesis can no longer create a closed environment to develop a seal for resisting dislodgment. Consequently, during the interim phase, when complete engagement of the defect is not pos-sible because of tissue sensitivity, the careful use of denture adhesives is required to facilitate retention. The patient should be instructed that adhesives can alter the prosthesis fit and disrupt the close adaptation of the prosthesis to the remaining tissue. Used adhesive must be removed before new adhesive is reapplied, to maintain fit and hygiene. Also related to retention is the inability to completely place the prosthesis, which for maxillectomy patients can be due to contracture of the scar band. When the maxillary resection leaves the cheek unsupported by bone, the prosthesis pro-vides the necessary support for wound maturation. If the patient removes the interim obturator prosthesis for a period sufficient to allow contraction, the prosthesis will be more difficult to place. Once placed, however, the scar band will relax and subsequent removal and placement will be more easily accomplished. The discomfort associated with this phenomenon is mostly due to patient anxiety and can be effectively addressed by reassuring the patient that this is an easily handled complication.
During the immediate postoperative healing stage, the surgical defect will undergo a change in dimension that affects the prosthesis fit and seal. If space is created with the change, speech will be altered (increase in nasality) and nasal reflux with swallowing will occur. The interim prosthesis is made of easily adjustable material to allow accommoda-tion for such changes. The most common manner of adjust-ment is through the use of temporary resilient denture lining materials, which offer the ability to mold to the tissue directly and reduce the mechanical effects of movement by virtue of their viscoelastic nature. Leakage can occur quite easily when swallowing unless the patient follows certain instruc-tions. Because the prosthesis cannot offer a watertight seal that matches the presurgical state, patients will be instructed www.konkur.in 321 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics not to swallow large quantities at one time, and to hold their heads horizontal when swallowing. When the head posture is forward, as when one is taking soup from a spoon, leak-age easily occurs around the obturator component of a pros-thesis. Another difficult condition that presents difficulty in controlling leakage on swallowing is the midline soft palatal resection. The functional movement of the remaining soft palate is often very difficult to retain with a prosthesis. It is also difficult to provide an adequate seal during the interim prosthesis stage.
When combination treatment is prescribed for the patient, it is commonly provided during the postsurgi-cal phase, when the patient is using an interim prosthe-sis. The major intraoral complication associated with both radiation therapy and chemotherapy, which affects interim prosthetic service, is mucositis. A careful balance between comfort and adequate fit for speech and swallowing needs must be determined with input from the patient. If pros-thesis adjustment can offer relief to ensure completion of treatment and the patient understands the impact adjust-ment may have on speech and swallowing, then it should be accomplished.
The long-term effects of radiation therapy, especially radiation-induced xerostomia and capillary bed changes (obliterative endarteritis) within the mandible, present a potentially significant threat to any remaining dentition and to the development of osteoradionecrosis. During the interim prosthesis stage, the patient will begin to notice the xerostomic effects, which include development of thick, ropy saliva that makes swallowing more difficult, and an increase in discomfort associated with removable prostheses.
Defect and Oral Hygiene Following surgical pack removal, the defect site will mature with time and exposure to the external environment. Initial loss of incompletely consolidated skin graft, mucous secre-tions mixed with blood, and residual food debris within the cavity are common oral findings for the patient with a maxillary defect. These cause concern for patients who are unprepared and unfamiliar with these new oral findings. As they become more familiar with the surgical defect, patients should be encouraged to clean the defect of food debris and mucous secretions routinely. Defect hygiene will allow quicker healing and will improve the ability to adequately fit a prosthesis. Common defect hygiene practices include (1) lavage procedures, which include rinsing of the defect during normal showering, (2) rinsing of the defect using a bulb syringe or a modified oral irrigating device (modi-fied to provide a multiple orifice “shower” effect), and (3) manual cleaning procedures, such as the use of a sponge- handled cleaning aid. Frequently, dried mucous secretions are difficult to remove and require adequate hydration before mechanical removal.
Following surgical pack removal, the patient may be reluc-tant to begin oral hygiene practices because of oral discom-fort. As patients use the interim prosthesis, which requires daily removal and cleaning at a minimum, they will realize the need for and benefit of normal oral hygiene practices because of improved prosthesis fit and tolerance. When teeth are remaining, it is important to the success of long-term prosthesis care to maintain a high level of oral hygiene. This is more critical for patients who exhibit xerostomia and have increased risk of caries. For these patients, daily application of fluoride in custom-formed carriers is prescribed along with frequent professional cleanings. The successful use of maxillofacial prostheses is enhanced greatly by the support provided by natural teeth. Consequently, during the interim prosthetic period, periodontal management procedures are begun in anticipation of the definitive treatment to allow a smooth transition from the interim to definitive prosthetic stages.
Definitive Care Surgical reconstruction for maxillofacial defects is an evolv-ing practice for head and neck cancer care. Prosthodontic management for mandibular resection patients most com-monly involves implant prostheses supported by a microvas-cular free flap. The evolution of similar techniques for the surgical reconstruction of maxillary defects resulting in an implant-supported prosthesis is gaining in application.
When the active treatment phase has been completed and no surgical reconstruction has been provided, defini-tive prosthetic management can be initiated for as long as it takes the defect tissue to mature sufficiently to tolerate more aggressive manipulation and obturation. This phase can be considered a transition for the patient–physician rela-tionship, in which the primary emphasis shifts from active treatment to observation. The primary emphasis from the patient’s standpoint shifts to prosthetic management, and the goals and design of the prosthesis differ from those of the interim prosthesis (Figure 25-9). However, for some patients, more definitive prostheses are delayed because of Figure 25-9 A definitive (left) and interim (right) obturator prosthesis, contrasting the materials used and the obturator bulb contour. Clasp retention is more stabilizing with the definitive prosthesis because of the cast half-round clasp configuration, the use of embrasure clasps, and the opportunity for guide-plane use. Also, because the surgical site is more mature, prosthesis extension into this region to augment support, stability, and re-tention when necessary is possible.
www.konkur.in 322 Part III Maintenance general health concerns, questionable tumor prognosis or control, or failure of the patient to reach a level of oral and/ or defect hygiene that warrants more sophisticated treat-ment. Although this phase of management can be consid-ered elective, without definitive prostheses, patients are not afforded the opportunity for complete rehabilitation. It is the extended use of temporary prostheses beyond their serviceable life span that has given a poor impression of prosthetic service to many surgeons and patients. Every opportunity should be provided to the patient for the most complete rehabilitation possible, and this necessitates con-sideration of more definitive prostheses.
From the previous discussion regarding removable prosthetic physiology, the inability of these static artificial replacements to mimic their natural counterparts results in less than ideal functional measures. Factors related to the structural integrity of the surgical defect and associated reconstructions as they affect this already compromised functional capacity are important considerations, espe-cially when few teeth remain. As was stated previously, the fact that control of removable maxillofacial prostheses has a large skilled performance requirement of patients suggests that oral and defect structures adjacent to the prostheses are important for successful performance. This is crucial to an understanding of the impact that postsurgical defect charac-teristics and soft tissue reconstructions have on maxillofacial prosthesis management. The reasons for this are twofold: (1) the opportunity for maximal prosthetic benefit necessitates consideration of surgical site characteristics that are separate from classic tumor control approaches; and (2) the ability of the patient to biomechanically control large removable pros-theses following surgery may be notably hindered by surgi-cal closure/reconstruction options. Surgical outcomes that can improve prosthetic function without adversely affecting tumor control should be considered and will be described for the more common surgical defects and associated prostheses.
INTRAORAL PROSTHESES: DESIGN CONSIDERATIONS Maxillofacial prosthetics is largely a removable prosthetic discipline, with the exception of dental implant–retained prostheses for some applications. For maxillofacial recon-struction with removable partial denture prostheses, typical goals of treatment consist of a well-supported, stable, reten-tive prosthesis that is acceptable in appearance and exhib-its minimal movement under function, thereby preserving the maximum amount of supporting tissue. A strategy for achieving these goals includes maximum coverage of the edentulous ridge within the movement capacity of the mus-cular attachments, maximum engagement of the remaining teeth to help control retention and movement under func-tion, and placement of artificial teeth to facilitate mainte-nance of this prescribed tooth-tissue contact during normal functional contacts. Maintaining these basic concepts within an otherwise normal anatomic environment (relative to food control and deglutition) has provided reasonable success for patients requiring replacement of missing teeth. The chal-lenges faced in doing so for removable maxillofacial prosthe-ses are quite different.
Normal resistance to functional loads is achieved by the highly sophisticated periodontal attachment of the natural dentition, which provides support and stability to teeth. When the dentition is partially depleted and is replaced by prostheses that are tooth-supported, the support and stabil-ity of the replacement teeth remain to be provided by the natural attachment. When tooth loss includes several poste-rior teeth, replacements are placed over the residual edentu-lous ridge, and the prosthesis receives support and stability from both teeth and mucosa. When all teeth are lost, support and stability are totally provided by the mucosa covering the residual edentulous ridges. Finally, when surgical removal of tumors results in tooth and supporting structure loss, sup-port and stability are provided by combinations of remain-ing teeth and/or residual ridges and areas within the surgical defect. For partial and complete tissue-supported prostheses, the mechanism of functional load support—as provided by the mucosa—is unsuited to the task from a biological stand-point. Given this understanding, when a maxillofacial pros-thesis is required to involve a surgical defect for support and stability, it is obvious that the environment within the surgi-cal defect is even less suited to the task.
SURGICAL PRESERVATION FOR PROSTHESIS BENEFIT Maxillary Defects Surgical outcomes that influence prosthesis success can be considered as those that determine the number of maxil-lary structures removed (Figure 25-10) and/or those that affect the structural integrity and quality of the defect. For surgical defects of the hard and/or soft palate, the primary prosthetic objectives include restoration of physical separa-tion of the oral and nasal cavities in a manner that restores mastication, deglutition, speech, and facial contour to as near a normal state as possible. Typical prostheses used to achieve these objectives include the obturator prosthesis (Figure 25-11, A and B), typically referring to prostheses that obturate defects within the bony palate, and the speech aid prosthesis (see Figure 25-11, C and D), which typically refers to prostheses that restore palatopharyngeal function for defects of the soft palate.
Current preoperative diagnostic procedures have improved the ability to discern the location and regional bone involvement of tumors of the maxilla and associ-ated paranasal sinuses. Relative to prosthetically important surgical modifications, if it can be determined that tumor control does not require a classic radical maxillectomy approach or that the inferior sinus floor, hard palate, and alveolus are uninvolved, preservation of as much hard palate and alveolar bone and as many teeth as possible should be considered. Tooth preservation has the greatest impact on success because of its stabilizing effect on prosthetic move-ment. When teeth can be retained in the premaxilla for more www.konkur.in 323 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics posterior tumors or in the posterior molar region for more anterior tumors, control of prosthesis movement is more easily accomplished and prosthetic success can be consider-ably improved (see Figure 25-10). Because the classic mid-line maxillectomy defect is significantly more debilitating for the average patient than a defect whereby preservation of the premaxillary component was accomplished, inclu-sion of the anterior premaxillary component should be an individual decision based on tumor control, classic resection technique.
A B Figure 25-10 A, A maxillary defect where a tooth distal to the resection was maintained. The tooth will significantly stabilize the prosthesis by preventing movement of the obturator bulb into the defect at the distal resection margin. B, A maxillary defect that dem-onstrates preservation of the anterior arch curvature, providing enhanced stability through a tripod effect. Also evident is the use of a split-thickness skin graft in the superior-lateral region, which improves the opportunity for useful support.
A B D C Figure 25-11 A, Superior view of an obturator prosthesis demonstrating the cast framework, three posterior cast half-round clasps and an anterior I-bar clasp, and a superior obturator surface contoured to encourage secretions to flow posteriorly. B, The same prosthe-sis seated intraorally. C, A speech aid prosthesis with posterior retention and anterior indirect retention, and a resin speech bulb. D, The same prosthesis showing bilateral embrasure clasps and obturation of the palatopharyngeal defect.
www.konkur.in 324 Part III Maintenance For resections in patients with teeth, the tooth adjacent to the defect is subjected to significant force from prosthe-sis movement. When the surgical alveolar ostectomy cut is planned, the resection should be made through the extrac-tion site of the adjacent tooth to provide the most favorable prognosis for this supportive tooth (see Figure 25-6). This procedure ensures adequate alveolar support for the adjacent tooth, which is a critical tooth for prosthesis success, and improves the tooth’s prognosis for long-term survival. The midline of the hard palate is a common area of removable prosthesis pressure because of movement of the prosthesis into the defect under functional forces of swallowing and mastication. To provide the best surgical preparation for this area, when the hard palate is resected, the vertical surface of the bone cut should be covered with an advancement flap of palatal mucosa, to provide a firm and resistant mucosal covering to this region, where the prosthesis can notably act as a fulcrum.
The soft palate owes its normal function to the bilateral sling nature of the musculature, which provides the shape and movement capacity specific for speech and degluti-tion requirements. When this is altered because of surgery, there appears to be a variable response in the ability to con-tinue to provide palatopharyngeal competence based on the amount of continuous band of posterior tissue remain-ing. Often, an insufficient band of palatal tissue fails to pro-vide palatopharyngeal competence and hinders prosthetic management of the problem. To serve as a guide for deci-sion making in surgery, it has been suggested that if the required resection leaves less than one third of the poste-rior aspect of the soft palate, the entire soft palate should be removed. The exception to this would be the edentulous patient who is undergoing a radical maxillectomy. Without teeth to provide the necessary retention for one side of a prosthesis, the patient benefits from the ability to place the prosthesis above the posterior soft tissue band for retention (Figure 25-12).
Preparation of the maxillary surgical site can improve prosthesis tolerance through the use of a split-thickness skin graft (see Figure 25-1). Lining the reflected cheek flap and posterior denuded structures with a graft improves tissue response by decreasing the pain associated with functional contact often seen when this surface is left to heal second-arily. If the posterior structures, pterygoid plates, or anterior temporal bone can provide a firm supportive base for the prosthesis, a skin graft covering is extremely helpful. Later-ally, the junction of the skin and oral mucosa creates a scar contracture, which provides a natural retentive region for the obturator portion of the prosthesis. Careful attention is given to this region in fabricating a prosthesis, to maximize sup-port, stability, and retention of the prosthesis.
In general, the need to extend a prosthesis into the defect is greater for edentulous patients than patients with teeth. When teeth remain, they are used to a greater extent to sta-bilize and retain the obturator component of the prosthesis, and the defect region is not required for such objectives. However, all patients with maxillary defects should have sufficient access to the lateral-posterior region of the defect to seal the defect at a minimum. In the edentulous patient, for maximum ability to obturate a maxillary defect, access to the regions superior to the defect opening is required. Nasal turbinates and mucosal connections that do not allow full extension into the necessary retentive and supportive areas of the defect compromise function. The function of turbinates in the newly externalized environment is not beneficial for breathed air humidification or warming and consequently may not warrant preservation.
Surgical reconstruction of maxillary defects should be undertaken when restoration of the functional goals of speech, deglutition, and mastication is better served by such procedures. Surgical reconstruction of a maxillary hard palatal defect in a manner that provides separation of the oral and nasal cavities without consideration for oral space requirements for speech, or for the supportive requirements of replacement teeth, is not only incom-plete management but can preclude subsequent prosthetic management. When surgical defects measure 3 cm or less and can be reconstructed to normal contours without compromising adjacent tissue function, surgical manage-ment is an appropriate consideration. Larger defects are very difficult to surgically reconstruct and, without care-ful planning for subsequent functional needs, could cre-ate an environment incapable of supporting a prosthesis. For partial soft palatal reconstructions, it is very difficult to provide functional tissue replacement without compro-mising palatal function. In light of this unpredictability, predictable prosthetic management of such defects is most often the treatment of choice.
Mandibular Defects The functions of mastication, deglutition, speech, and oral competence (saliva control) are made possible through Figure 25-12 A maxillary obturator prosthesis demonstrat-ing a distal extension, which engages a soft palatal remnant for added retention.
www.konkur.in 325 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics coordinated efforts of separate anatomic regions, which include the oral sphincter, alveololingual and buccal sulci, alveolar ridges, floor of the mouth, mobile tongue, base of the tongue, tonsillar pillars, soft palate, hard palate, and buc-cal mucosa. The more regions that are involved in a surgical procedure, the greater is the demand on surgical reconstruc-tive efforts. When the mandible is also involved, the com-plexity of the reconstructive procedure is dependent on the location and amount of mandible to be included in the resec-tion and the decision to maintain continuity with normal mandibular position and contour (Figure 25-13). For disease involving the functional anatomy around the mandible, sur-gical outcomes that influence prosthesis success are based on decisions to take mandibular portions or segments and decisions regarding reconstruction. The primary prosthetic objectives for mandibular defects are to restore mastication and cosmesis by the replacement of teeth. Achieving the mastication goal requires an understanding that regardless of the manner of prosthesis support (natural teeth, recon-structed soft tissue, or implants), the impact of the prosthetic device on success is dependent on appropriate surgical man-agement of both soft tissue and bone.
Disease involving soft tissue structures adjacent to and enveloping the mandible necessitates consideration of a mandibular resection to ensure control. When the soft tis-sue disease is clearly separate from the mandible and does not require bone removal, surgical defects involving these structures should be surgically reconstructed and therefore do not require prosthetic management. The exception to this is the large tongue resection that may require augmentation of the palatal contours to facilitate speech production. Such a palatal augmentation prosthesis is most beneficial when coordinated speech therapy can guide the optimal prosthe-sis configuration. Other resections may appear to require palatal augmentation for speech, yet the functional problem is tongue immobility secondary to tension created by the reconstructive tissue. Consideration should be given to soft tissue reconstructions that are of sufficient size and mobil-ity, are less prone to contracture tension, and can produce a more normal alveololingual sulcus because these char-acteristics have been shown to have a significant influence on tongue mobility. Other desirable characteristics, such as sensation and lubrication, are also possible but necessi-tate a choice of which one is most required given the goals desired.
When tumors are primary to the mandible, as an amelo-blastoma is, or when they involve the mandible from adja-cent regions, surgical resection of segments of the mandible is required for tumor control. It may be difficult to always predict the functional deficit and the exact plan of recon-struction because the surgeon determines the extent of the resection based on presurgical and surgical findings. How-ever, common anatomically based mandibular resections include the lateral mandibular resection, the anterior man-dibular resection, and the hemimandibular resection. From the standpoint of the surviving mandibular resection patient, the most significant decision regarding his or her manage-ment is the decision to maintain mandibular continuity, which allows maintenance of position for adjacent intraoral and extraoral soft tissue.
Surgical evolution of procedures that maintain continu-ity for the mandible has significantly improved the oppor-tunity for functional restoration of mastication, deglutition, and speech. The debilitating effects of the discontinuity defect include a significant cosmetic deformity to the lower third of the face, decreased masticatory function secondary to unilateral closure, compromised coordination of tongue and teeth, altered speech ability, and impaired deglutition (Figure 25-14). Given an appreciation of the decreased performance seen with conventional mucosa-borne den-ture prostheses, it should be obvious that masticatory reha-bilitation for the resection patient without mandibular A B Figure 25-13 A, Marginal (left) and segmental (right) resec-tions of the mandible. When a segmental resection is not sta-bilized with a reconstruction bar or bone graft, the continuity of the mandible is lost. Such a defect is a discontinuity defect of the mandible. B, When not stabilized, the discontinuous mandible deviates toward the defect and presents significant problems with mastication restoration.
www.konkur.in 326 Part III Maintenance continuity is unpredictable at best and is never achieved for most patients. Even for patients with remaining teeth, the altered mandibular position created in time presents a significant functional and cosmetic handicap. From a prosthetic rehabilitation standpoint, the most significantly handicapped postsurgical head and neck condition is the discontinuous mandible. Consequently, such a postsurgical condition should be the rare exception (typically because of reconstruction plate failure) and should not be the planned surgical outcome.
The cosmetic deformity associated with mandibular resec-tion is improved through the use of reconstruction plates to maintain the presurgical contour to the lower jaw. This form of mandibular contour and position maintenance should be considered the minimum standard of care for mandibu-lar resection patients from a functional standpoint. Use of reconstruction plates can maintain cosmetic appearance and preserve the bilateral nature of mandibular movement. However, the use of reconstruction plates alone precludes replacement of teeth in the region of resection. Prosthetic replacement of teeth cannot be provided for regions supe-rior to the reconstruction bar because of the potential for mucosal perforation and exposure of the bar from functional loading of the soft tissue. From a masticatory function stand-point, this may not be a significant negative impact for some patients because of the maintenance of sufficient numbers of occlusal contacts postsurgically.
Mandibular Reconstruction—Bone Grafts The evolution of head and neck reconstructive surgery has been dramatic over the past three decades. The vascular-ized tissue options of the forehead and deltopectoral regions gave way to the more popular pedicled myocutaneous flaps from the 1960s to the 1970s. By the 1980s, numerous osteo-myocutaneous free-flap donor sites had been identified and were being used for mandibular reconstruction and particu-late cancellous bone marrow in formed allogeneic frames. Equally important to the functional outcome of mastication was the development of the science and the clinical applica-tion of the osseointegration phenomenon in the area of den-tal implants.
The ideal prosthetic characteristics of the replacement mandible include a stable union to proximal and distal seg-ments, restoration of contour to the lower third of the face, a rounded ridge contour with attached mucosa of 2 to 3 mm thickness, and adjacent sulci providing free movement of buccal and lingual soft tissues for food control. Regardless of the type of prosthesis to be used, the appropriate place-ment of the bone relative to the opposing arch is vital to the intended functional use. If a removable prosthesis is planned and is expected to cover the bone reconstruction, the con-tour of the developed ridge should provide a surface covered with firm, thin soft tissue, and a rounded superior contour with buccal and lingual slopes approaching parallel to each other and with sufficient vestibular depth to provide hori-zontal stability. Such a ridge condition is the surgical analog of a minimally resorbed edentulous ridge. With adequate cheek and tongue movement, this should provide a reason-able prognosis for prosthetic success, provided sufficient numbers of teeth remain on the nonresected side. For the optimum chance of prosthetic function, dental implants should be considered, and with sufficient bulk of bone and the same characteristics listed for the removable prosthe-sis, the prognosis for success is the greatest. To reiterate, the major determining factor for improved function will be the quality of the soft tissue reconstruction.
The major complications seen with mandibular recon-struction are related to the bulk of the soft tissue component and lack of mobility of the tongue. When these factors are controlled for, complications are caused most often by bone placement and size. The common use of free flaps, includ-ing bone from other regions of the body that do not possess the native mandibular shape, presents a significant degree of technical difficulty associated with the procedure. The fibula, A B Figure 25-14 A, A deviated mandibular position following segmental resection without reconstruction. The mandibular midline is left of the maxillary midline by two teeth. B, With mandibular and maxillary prostheses in place, the patient closes to a functional position that is unique to the unilateral closure pattern.
www.konkur.in 327 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics which is a popular choice for mandibular replacement, pres-ents some challenges in meeting the ideal requirements mentioned previously. Because of the straight nature of the bone, it is easy to err in both horizontal and vertical position-ing, especially for reconstructions that span to the midline. Lingual positioning requires prosthetic placement at a posi-tion that may become functionally unstable over time. Such a location requires implant positions that create a mechanical cantilever that can be detrimental to the long-term success of the implant-supported prosthesis. Posteriorly, the inability to recreate the natural ascending curve of the mandible can restrict placement of teeth and preclude restoration of com-plete occlusion on the resected side. It is common to have a mismatch in height at the anterior junction of the graft with the resident mandible. For implant-supported prostheses, this area can present significant challenges in terms of ade-quate hygiene of the implants, and over time, this can com-promise implant health. For removable prostheses, this can become a source of irritation if fulcrum-like action occurs with movement.
MAXILLARY PROSTHESES Obturator Prostheses The defining characteristic of an obturator prosthesis is that it serves to restore separation of the oral and adjacent cavities following surgical resection of tumors of the nasal and para-nasal regions (Figure 25-15). Aramany developed a classifi-cation for partially edentulous maxillectomy dental arches (Figure 25-16). The various defects resulting from resection contain and are bounded by anatomic structures and an epi-thelial lining (either transplanted skin and/or native mucosa) that are quite different from normal partially edentulous arch anatomic features. The expectation for this altered region to contribute significantly to prosthesis support, stability, or retention is infrequently met. Consequently, prosthesis A B C D Vertical displacement Less Greater Long radius sweep Short radius sweep Given horizontal flexure Figure 25-15 A, Coronal view of proposed maxillary resection. Bold lines designate typical area to be resected. B, Demonstrates the value of lateral wall height in the design of a removable partial denture obturator. As the defect side of the prosthesis is displaced, the lateral wall of the obturator will engage the scar band and aid in retaining the prosthesis. C, Coronal section with surgical obturator in place. With the prosthesis in place, the relation of the scar band (arrow) to the lateral portion of the obturator can be seen. A buccal scar band will develop at the height of the previous vestibule where the buccal mucosa and the skin graft in the surgical defect join. D, Axial view of the resected area illustrates the defect. Dotted lines indicate areas available for intraoral retention.
www.konkur.in 328 Part III Maintenance support and stabilization are largely dependent on the ability to aggressively engage the remaining teeth and residual ridge structures.
In comparison with partially edentulous arches, the movement potential for the prosthesis extension into the defect can be significant. When engagement of the disto-buccal temporal bone is possible, upward movement of the obturator bulb can be greatly minimized. Movement poten-tial increases as the remaining tooth number decreases and their arrangement becomes more linear (Figure 25-17). This illustrates the importance of maintaining teeth when possible, which allows for greater prosthesis stabilization through direct tooth engagement and through cross-arch stabilization that increases with nonlinear tooth configura-tions (Figure 25-18).
To help control potential movement, various sugges-tions have been made relative to prosthesis design. The basic principle of placing support, stabilization, and retention I II III IV V VI Figure 25-16 Aramany’s classification for partially edentulous maxillectomy dental arches: Class I—midline resection. Class II— unilateral resection. Class III—central resection. Class IV—bilateral anteroposterior resection. Class V—posterior resection. Class VI— anterior resection.
A B Figure 25-17 A, A maxillary obturator prosthesis in which remaining teeth provide significant stabilization to the obturator extension because of their number and location, which allows cross-arch prosthesis engagement. B, Another obturator prosthesis, which benefits from teeth in a linear arrangement and therefore does not have any cross-arch tooth stabilization. Obturator movement in B is likely to be significantly greater than in A. The requirement for using the defect to provide support where possible is therefore greater in B than in A.
www.konkur.in 329 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics immediately adjacent to and as far from the defect as pos-sible acts to distribute the tooth effect on prosthesis perfor-mance to the greatest mechanical advantage. Because the teeth adjacent to the anterior resection margin are often incisors, it may be necessary to consider splinting them to improve the long-term prognosis. This region is critical for prosthesis performance, and the requirement for a cingulum rest and labial retention is often difficult to optimize with-out crowns. Distally, it is often necessary to incorporate an embrasure clasp to provide maximum retention and stabili-zation. Such a clasp assembly must have sufficient room for occlusal clearance, and it is not uncommon for the opposing occlusion to need adjustment to accommodate such a rest complex. When possible, the palatal surfaces of the maxillary teeth should be surveyed to determine whether guide-plane surfaces can be produced to impart a stabilizing effect. When accomplished, the prosthesis benefits from improved move-ment resistance, and it does so with more teeth contributing to the effect, thereby distributing the stress more appropri-ately. Brown described how the vertical height of the lateral portion of the obturator above the buccal scar band can con-tribute to prosthesis movement control by helping to prevent vertical displacement (see Figure 25-15).
Speech Aid Prostheses The defining characteristics of speech aid prostheses are that they are functionally shaped to the palatopharyngeal mus-culature to restore or compensate for areas of the soft palate that are deficient because of surgery or congenital anomaly (see Figure 25-11). Such a prosthesis consists of a palatal component, which contacts the teeth to provide stability and anchorage for retention; a palatal extension, which crosses the residual soft palate; and a pharyngeal component, which fills the palatopharyngeal port during muscular function, serving to restore the speech valve of the palatopharyngeal region.
Because the typical speech aid prosthesis does not provide tooth replacement, the patient should expect only minimum functional movement. Movement of the pharyngeal exten-sion imposed by the residual palatopharyngeal muscula-ture is generally undesired and is a sign that modification is required. Common reasons for such movement include a low position, causing tongue encroachment; superior exten-sion that does not account for head flexure; or impression procedures that do not accurately record residual soft palatal position or movement.
A pediatric speech aid is a temporary prosthesis used to improve voice quality during the growing years. It is made of materials that are easily modified as growth or orthodon-tic treatment progresses. Because a speech aid has a signifi-cant posterior extension into the pharyngeal region, torque is evident from the long moment arm. A basic principle of posterior retention with anterior indirect retention must be applied to the design of such a maxillary prosthesis. Poste-rior retention is gained by the use of wrought-wire clasps around the most distal maxillary molars, whereas the ante-rior extension of the prosthesis onto the hard palate provides indirect retention. If clinical crown length and undercut are adequate to provide retention, orthodontic bands with buc-cal tie wings can be used in conjunction with the wrought wires. This design facilitates the maintenance of the pharyn-geal part of the pediatric speech aid in the proper position in the palatopharyngeal opening.
In the adult whose palatopharyngeal insufficiency is the result of a cleft palate or palatal surgery, an adult speech aid prosthesis can be constructed of more definitive materials because growth changes will not have to be accommodated. If teeth are missing, the prosthesis will incorporate a reten-tive partial denture framework. The basic design should include posterior retention and anterior indirect retention.
Palatal Lift Prostheses The defining characteristic of a palatal lift is that it positions a flaccid soft palate posteriorly and superiorly to narrow the palatopharyngeal opening for the purpose of improving A B Figure 25-18 A, Tooth arrangement that offers cross-arch stability (as in Figure 25-17, A) because of the arch curvature of the remain-ing tooth distribution and the tripod effect it allows. B, More linear arrangement of teeth does not provide cross-arch stability and places greater demand on the defect integrity for prosthesis performance.
www.konkur.in 330 Part III Maintenance oral air pressure and therefore speech. Patients who exhibit a structurally normal soft palate and pharyngeal port can demonstrate hypernasal speech caused by paralysis of the regional musculature. This condition is referred to as palato-pharyngeal incompetence because the failure lies in function, not in anatomic deficiency. The paralysis can result from a variety of neuromuscular conditions (flaccid paralysis of the soft palate from closed head injuries, cerebral palsy, muscular dystrophy, or myasthenia gravis) that have varying clinical courses. The palatal lift prosthesis must physically position the soft palate to redirect air pressure orally. In placing the soft palate, any tissue resistance met acts as a dislodging force on the prosthesis. This dislodging force must be resisted by adequate direct and indirect retention.
To efficiently maintain prosthesis position, the dislodging force is best resisted by bilateral direct retainers placed close to the posterior lift and anteriorly placed indirect retention. Success with a palatal lift prosthesis depends upon the pres-ence of a number of maxillary posterior teeth, which can provide retention for the prosthesis, coupled with an easily placed flaccid soft palate.
Palatal Augmentation Prostheses When surgical resection involving the tongue and/or floor of the mouth limits tongue mobility, it affects both speech and deglutition. With tongue mobility limitations, the contour of the palate can be augmented by a prosthesis to modify the space of Donder to allow food manipulation to be more eas-ily transferred posteriorly into the oropharynx.
Prosthesis movement potential is low because the func-tional forces involved are those imparted by the tongue dur-ing deglutition and speech, neither of which creates force similar to mastication. It is common to use a diagnostic resin augmentation prosthesis retained with wire clasps to plan the necessary contour needs. Once the appropriate palatal con-tour has been determined, a definitive augmentation pros-thesis can be constructed of cast metal with appropriately placed minor connectors for attaching the resin augmenta-tion. Bilateral rests and direct retainers should be positioned to facilitate the design for the acrylic retention because sta-bility needs related to functional force are not a significant design concern.
MANDIBULAR PROSTHESES Evolution of Mandibular Surgical Resection When mandibular continuity is preserved, as in a marginal resection (type I mandibular defect, see Figure 25-19), func-tion is least affected and the major prosthesis concern is related to the soft tissue potential for support. With good remaining dental support, near-normal function can often be achieved with prosthodontic rehabilitation.
Although it is not as common an outcome as in the past, when continuity of the mandible is lost because of a seg-mental resection that was not reconstructed, the bilateral joint complex no longer controls the remaining mandibular segment. Consequently, the function of the remaining mandibular segment is severely compromised because of loss of coordinated bilateral muscular action functioning across a bilateral joint. The resulting segmental movement is an uncoordinated action dictated by the remaining uni-lateral muscular activity within a surgical environment that changes with healing dynamics and patient rehabilitation efforts. Successful removable prosthodontic intervention for these situations necessitates a combination of clinician knowledge of the functional movements of the remaining residual mandible and concerted effort and persistence of the patient.
Historically, mandibular stabilization by bone grafts or reconstruction bars was not always a surgical goal. The major exception was the anterior defect (type V), which was recognized to pose significant airway risks if not managed. Currently, most lateral segmental man-dibulectomies are also reconstructed surgically. When the mandible is not stabilized following resection and a dis-continuity defect results, a mandibular resection prosthe-sis should be provided to restore mastication within the unique movement capabilities of the residual functioning mandible.
I II V III IV Figure 25-19 Cantor and Curtis classification of partial man-dibulectomy. (Redrawn from Cantor R, Curtis TA: Prosthetic man-agement of edentulous mandibulectomy patients. I. Anatomic, physiologic, and psychologic considerations, J Prosthet Dent 25:446-457, 1971.) www.konkur.in 331 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics Resection prostheses are those prostheses provided to patients who have acquired mandibular defects that result in loss of teeth and significant portions of the mandible. Mandibular resection results in defects that may pre-serve mandibular continuity or may result in discontinu-ity defects. These are further subclassified by Cantor and Curtis (Figure 25-19) and provide a meaningful foun-dation for a discussion of removable prosthesis design considerations.
The following discussion highlights design consid-erations for the major defect classifications outlined. A common feature among all removable resection prosthe-ses is that all framework designs should be dictated by basic prosthodontic principles of design. These include broad stress distribution, cross-arch stabilization with use of a rigid major connector, stabilizing and retaining components at locations within the arch to best mini-mize dislodging functional forces, and replacement tooth positions that optimize prosthesis stability and func-tional needs. Modifications to these principles are deter-mined on an individual basis and are greatly influenced by unique residual tissue characteristics and mandibular movement dynamics.
Type I Resection In a type I resection of the mandible, the inferior bor-der is intact and normal movements can be expected to occur. The major difference between this situation and a typical edentulous span is the nature of the soft tissue foundation. For type I resections, the denture-bearing area may be compromised by closure of the defect with the use of adjacent lining mucosa (which can reduce the bucco-lingual width), or by the presence of a split-thick-ness skin graft.
Ideally, one would like to see a firm, non-movable tissue bed with normal buccal and lingual vestibular extension. If the defect is unilateral and posterior, the framework would be typical of a Kennedy Class II design, taking into account whatever modification spaces may be present. When the marginal resection is in the anterior area, the design may be more typical of a Kennedy Class IV design (Figure 25-20).
B A C D Figure 25-20 A type I resection of the anterior mandible. A, Bilateral molars remain to stabilize an anterior extension removable partial denture. A split-thickness skin graft has been used to reconstruct the denture bearing area. B, The prosthesis showing cast clasps and the anterior extension base. C, The prosthesis in place and covering the skin graft with a configuration produced through a corrective cast impression technique. D, The resection prosthesis in occlusion. It is critical to have the remaining natural teeth occlude at the same vertical dimension as the prosthetic teeth to ensure comfortable function.
www.konkur.in 332 Part III Maintenance Anterior marginal resections sometimes include part of the anterior tongue and floor of the mouth. With loss of normal tongue function, the remaining teeth are no longer retained in a neutral zone, and as a result, they often collapse lingually because of lip pressure. If this occurs, the use of a labial bar major connector may be necessary.
Corrected cast impression procedures provide a major advantage for fabrication of removable partial dentures in partial mandibulectomy patients. Capture of the unique buccal, lingual, and labial functional contours in the final prosthesis can contribute significantly to sta-bilization of the prosthesis, especially in discontinuity defects.
Type II Resection In the type II resection, the mandible is often resected in the region of the second premolar and first molar. If no other teeth in the arch are missing, a prosthesis usually is not indicated. In some situations, however, a prosthe-sis may have to be fabricated to support the buccal tis-sue and to help fill the space between the tongue and the cheek to prevent food and saliva from collecting in the region.
Framework design should be similar to a Kennedy Class II design, with extension into the vestibular areas of the resection. This area would be considered nonfunc-tional and should not be required to support mastication. It must be remembered that extension into the defect area can place significant stress on the remaining abutment teeth; therefore, occlusal rests should be placed near the defect, and an attempt should be made to gain tripod sup-port from remaining teeth and tissue where possible.
An example of a framework design for a type II man-dibular resection with missing molars on the nonsurgi-cal side is illustrated in Figure 25-21. The choice of major connector depends on the height of the floor of the mouth as it relates to the position of the attached gingival mar-gins during function. An extension base with artificial teeth can be used on the surgical side if space is avail-able. The extent of this base is determined by a functional impression, and determination should be cautious of the potential for bone exposure at the superior margin of the resection.
B A C Figure 25-21 Type II resection and prosthesis. A, Clinical presentation of the mandibular right resection and missing mandibular left molars. B, Resection prosthesis with a cast lingual plate major connector and wrought-wire clasps. C, Resection prosthesis in place demonstrating the two-tooth extension on the defect side (patient’s right). (Courtesy of Dr. Ron Desjardins, Rochester, MN.) www.konkur.in 333 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics Retention can be achieved through the use of various types of clasp assemblies on the distal abutments. Indirect retention can be derived from rests prepared in the mesial fossae of the first premolars and/or the lingual surfaces of the canines. Unlike the result in Figure 25-21, use of an infrabulge retainer on the surgical side may be difficult if a shallow vestibule results from surgical closure. The locations of minor connectors should be physiologically determined to minimize stress on the abutment teeth and to enhance resistance to reasonable dislodging forces. Wrought-wire circumferential retainers are acceptable alternatives.
In a type II mandibular resection, where posterior and anterior teeth are missing on the defect side, the remain-ing teeth on the intact side of the arch are often present in a straight-line configuration. Embrasure clasps may be used on the posterior teeth, with an infrabulge retainer on the anterior abutment. In some situations, a rotational path design may be used to engage the natural undercuts on the mesial proximal surfaces of the anterior abut-ments. Lingual retention with buccal reciprocation on the remaining posterior teeth should be considered. The longitudinal axis of rotation in this design should be con-sidered to be a straight line through the remaining teeth. Depression of the prosthesis on the edentulous side will have less of a chance to dislodge the prosthesis if retention is on the lingual surfaces than if on the buccal. Suggested framework designs for this patient group are illustrated in Figure 25-22.
Physiologic relief of minor connectors is always rec-ommended. When the remaining teeth are in a straight line, a Swing-Lock major connector design (Swing-Lock, Inc., Milford, TX) may be used to take advantage of as many buccal and/or labial undercuts as possible. Because elderly patients often complain of difficulty manipulat-ing Swing-Lock mechanisms, in straight-line situations it A B C D Figure 25-22 A, Frame design for type II resection, no teeth missing on the nonresected side. Note the provision for extension into the resection space between tongue and cheek. B, Type II design with missing posterior teeth on the nonresected side. C, Type II design with missing anterior teeth. D, Type II design with missing anterior and posterior teeth.
www.konkur.in 334 Part III Maintenance may be possible to use alternate buccal and lingual reten-tion effectively (Figure 25-23).
In the type II resection with anterior and posterior missing teeth on the resected side and posterior missing teeth on the nonresected side, the prosthesis will have three denture base regions. This prosthesis may have a straight-line longitudinal axis of rotation, as previously discussed. Rests should be placed on as many teeth as possible, minor connectors should be placed to enhance stability, and wrought-wire retainers represent an accept-able alternative to the bar clasps.
Type III Resection A type III resection (see Figure 25-19) produces a defect to the midline or farther toward the intact side, leaving half or less of the mandible remaining.
The importance of retaining as many teeth as possible in this situation cannot be overemphasized. The design of a framework for this situation would be similar to the type II resection. The longitudinal axis of rotation is again con-sidered to be a straight line through the remaining teeth. This resection provides a greater chance of prosthesis dis-lodgment caused by lack of support under the anterior extension.
Alternating buccal and lingual retention in a rigid design or the Swing-Lock design should be considered.
Type IV Resection A type IV resection (see Figure 25-19) would use the same design concepts as type II or III resections with the corresponding edentulous areas.
If the graft does not provide an articulation and the soft tissue covering the graft is not firmly attached to the bone graft, the movement potential will be dictated by functional forces of movement coupled with soft tissue supportive capacities.
If a type IV resection extends to the midline with the extension of a graft into the defect area, but does not include temporomandibular joint reconstruction on the surgical side, the design will be similar to the type III resection with an extension base on the surgi-cal side.
If the type IV resection extends beyond the midline, with less than half of the mandible remaining, the design will be similar to the type II resection that has an exten-sion base into the surgical defect area.
Type V Resection In the type V resected mandible, when the anterior or posterior denture-bearing area of the mandible has been surgically reconstructed, the removable partial denture design is similar to the type I resection design.
The principal difference between a type V resected mandible and the intact mandible with the same tooth loss pattern lies in the management of soft tissue at the graft site. For design purposes, one should consider the residual mandibles of the type I and V resections to be similar to nonsurgical mandibles with the same tooth-loss pattern.
Mandibular Guide Flange Prosthesis As was mentioned earlier, in a discontinuity defect, the movement of the residual mandibular segment is an uncoordinated action dictated by two features unique to the specific defect and patient. The first is the remaining unilateral muscular activity that will be specific to the sur-gical resection and that will have a characteristic resting posture to the defect side with a diagonal movement on “closure.” The second is that the surgical environment will change as healing progresses, and patient efforts to train movement during this healing period will help maintain both position and movement range. To facilitate training of the mandibular segment to maintain a more midline closure pattern, clinicians have used a guide flange pros-thesis.
The mandibular guide flange prosthesis is used pri-marily as an interim training device. When no miss-ing teeth are supplied, it may be considered a training appliance rather than a prosthesis. This appliance is used in dentulous patients with non-reconstructed lat-eral discontinuity defects who have severe deviation of the mandible toward the surgical side and are unable to achieve unassisted intercuspation on the nonsurgical side (Figure 25-24). Generally these patients have had a significant amount of soft tissue removed along with the resected mandibular segment and have attained surgical closure by suturing of the lateral surface of the tongue to the buccal mucosa, which causes a deviation toward the defect side. Scarring also occurs and is worse for patients who have not been placed on an exercise program dur-ing the healing period. The guide flange prosthesis is designed to restrict the patient to vertical opening and A R R R R Figure 25-23 Conventional clasping with the use of alternat-ing buccal and lingual retention (arrows).
www.konkur.in 335 Chapter 25 Removable Partial Denture Considerations in Maxillofacial Prosthetics closing movements into maximum occlusal contacts. Over time, this guided function should promote scar relaxation, allowing the patient to make unassisted mas-ticatory contact.
The components of the guide flange prosthesis include the major and minor connectors needed to support, sta-bilize, and retain the prosthesis and the guiding mecha-nism. This may include a cast buccal guide bar and guide flange, or simply a resin flange, which engages the oppos-ing arch buccal tooth surfaces. In either case, the oppos-ing arch must provide a stable foundation to resist any forces needed to guide the deviated mandibular segment into maximum occlusal contact.
The buccal guide bar is placed as close as possible to the buccal occlusal line angle of the remaining natural teeth to allow maximum opening. The lateral position of the bar must be adequate to prevent the guide from contacting the buccal mucosa of the maxillary alveolus. The length of the bar should overlie the premolars and the first molar where possible. Retention of the maxillary frame should not be problematic because the force directed on the bar is in a palatal direction. The guide flange is attached to the mandibular major connector by two generous inter-proximal minor connectors. As with the maxillary frame, significant interproximal tooth structure must be cleared to provide the necessary bulk for the minor connec-tors. The height of the guide flange is determined by the depth of the buccal vestibule. A small hook is placed at the middle of the top of the guide to prevent disengage-ment on wide opening. Because the mandibular segment has a constant medial force, the flange acts as a powerful lever with a strong lateral force on the teeth. Therefore, extra rests and additional stabilization and retention on multiple teeth must be considered to prevent overstress-ing of individual teeth. Retention on the tooth adjacent to the defect is critical for resistance to lifting of the frame. Lingual retention in the premolar area may be considered as an aid in resistance to displacement. When necessary, missing teeth can be added to a guide flange prosthesis. Flange prostheses can be provisionally designed for mod-ification into definitive removable partial dentures after guidance is no longer necessary. This is accomplished by removal of the buccal flange and buccal guide bar com-ponents after the patient is able to make occlusal contacts B A C D Figure 25-24 A mandibular guide flange prosthesis. A, Flange extension is incorporated into a mandibular type II resection prosthe-sis using a resin extension. B, Resection prosthesis inserted. C, Opposing maxillary prosthesis designed to engage palatal surfaces of all remaining teeth for maximal stability against flange-induced forces. D, Flange extending to the buccal region of the opposing prosthesis and teeth. Upon closure, the flange will guide the mandible to maximum intercuspation, at which time the flange extension will reside in the maxillary left buccal vestibule. (Courtesy of Dr. Ron Desjardins, Rochester, MN.) www.konkur.in 336 Part III Maintenance without use of the guide. However, many patients with mandibular resections have difficulty making repeated occlusal contacts—a fact described in several studies. Occlusal considerations in mandibulectomy patients have been discussed extensively by Desjardins.
Palatal occlusal ramps have been used to guide those patients with less severe deviation than those who require a guide flange into a more stable intercuspal contact position. These prostheses incorporate a palatal ramp that simulates the function of the guide flange prosthe-sis. This inclination of the palatal ramp is determined by the severity of the deviation of the remaining man-dible. Some patients have the ability to move laterally into occlusion but have a tendency to close medially and palatally rather than close into an acceptable cuspal rela-tionship. These patients can benefit from a palatal ramp, which can be functionally generated in wax at the try-in stage. This provides a platform for occlusal contact in the entire bucco-lingual range of movement. A supplemental row of prosthetic teeth may be arranged, then removed at the boil out stage, and processed in pink acrylic resin for esthetics. Patients who have experienced both smooth and tooth-form ramps usually prefer the tooth form if the width is adequate.
JAW RELATION RECORDS FOR MANDIBULAR RESECTION PATIENTS Interocclusal records must be made using verbal guidance only for resection patients with discontinuity defects. A hands-on approach, like that used for conventional eden-tulous jaw relation records, will lead to unnatural rota-tion of the mandible and an inaccurate record. The patient should be instructed to move the mandible toward the nonsurgical side and close into a nonresistant recording medium at the preestablished occlusal vertical dimension, which will be the occlusal contact position. If the surgical side is significantly deficient, an occlusion rim may have to be extended into the defect area to support the record-ing medium. Head position is of extreme importance during registration of jaw relation records. If the patient is in a semirecumbent position in the dental chair during the recording procedure, the mandible may be retracted and deviated toward the surgical side, preventing move-ment toward the intact side. To minimize this problem, the recording should be made with the patient seated in a normal upright postural position.
Most patients with lateral discontinuity defects can make lateral movements toward the nonsurgical side, even without the presence of a lateral pterygoid muscle functioning on the balancing (surgical) side. This is pos-sible because of the compensatory effects of the hori-zontal fibers of the temporalis and the lateral pterygoid muscle on the normal side, causing a rotational effect on the remaining condyle.
SUMMARY Maxillofacial prosthetic treatment of the patient with an oral defect is among the most challenging treatments in den-tistry. Defects are highly individual and require the clini-cian to call upon all knowledge and experience to fabricate a functional prosthesis. The basic principles and concepts described throughout this text will help the clinician to suc-cessfully design maxillofacial removable partial dentures. The interested reader is encouraged to pursue maxillofacial texts for more information regarding prosthesis design for this patient group.
www.konkur.in abutment A tooth, a portion of a tooth, or a portion of an implant that serves to support and/or retain a prosthesis.
acrylic Formed from acrylic acid (e.g., acrylic resin).
acrylic resin 1: Pertaining to polymers of acrylic acid, methacrylic acid, or acrylonitrile; for example, acrylic fibers or acrylic resins. 2: Any of a group of thermoplastic resins made by polymerizing.
anatomic ridge form Surface form of the edentulous ridge when at rest or when not supporting a functional load; often recorded in a soft impression material, such as hydrocolloid or metallic oxide impression paste, and results when an impression tray is uniformly relieved.
angle of cervical convergence Angle viewed between a vertical rod contacting an abutment tooth and the axial surface of the abutment cervical to the height of contour.
appliances Devices (such as splints, orthodontic appliances, and space maintainers) worn by the patient in the course of treatment.
balanced occlusion The simultaneous contacting of maxillary and mandibular teeth on the right and left in the anterior and posterior occlusal areas in centric or any eccentric position within the functional range.
bar clasp Type of extracoronal retainer that originates from the denture base or framework, traverses soft tissue, and approaches the tooth undercut area from a gingival direction.
basal seat Oral tissues and structures of the residual ridge supporting a denture base. See also denture foundation area.
cast An accurate and positive reproduction of a maxillary or mandibular dental arch made from an impression of that arch; further designated according to the purpose for which it is made, such as diagnostic cast, master cast, or investment cast; also may be used as an infinitive (to cast) or as an adjective (cast framework, or cast metal base).
casting A metal object shaped by being poured into a mold to harden; used primarily to designate the cast metal framework of a partial denture but also may be used to describe a molded metal denture base that is actually cast into a mold.
centric jaw relation See centric relation.
centric occlusion The occlusion of opposing teeth when the mandible is in centric relation. This may or may not coincide with the maximal intercuspal position.
centric relation 1: The maxillomandibular relationship in which the condyles articulate with the thinnest avascular portion of their respective disks with the complex in the anterior-superior position against the shapes of the articular eminences. This position is independent of tooth contact. This position is clinically discernible when the mandible is directed superior and anteriorly. It is restricted to a purely rotary movement about the transverse horizontal axis (GPT-5). 2: The most retruded physiologic relation of the mandible to the maxillae to and from which the individual can make lateral movements. It is a condition that can exist at various degrees of jaw separation. It occurs around the terminal hinge axis (GPT-3). 3: The most retruded relation of the mandible to the maxillae when the condyles are in the most posterior unstrained position in the glenoid fossae from which lateral movement can be made at any given degree of jaw separation (GPT-1). 4: The most posterior relation of the lower to the upper jaw from which lateral movements can be made at a given vertical dimension (Boucher, 1953). 5: A maxilla-to-mandible relationship in which the condyles and disks are thought to be in the midmost, uppermost position. The position has been difficult to define anatomically but is determined clinically by assessing when the jaw can hinge on a fixed terminal axis (up to 25 mm). It is a clinically determined relationship of the mandible to the maxilla when the condyle disk assemblies are positioned in their most superior position APPENDIX A Glossary Certain terms in this glossary have been adapted from The Glossary of Prosthodontic Terms. Additional terminology can be reviewed in: The glossary of prosthodontic terms, ed. 8. From The Journal of Prosthetic Dentistry 94(1):10-92, 2005. Available online:
com/science/article/pii/S0022391305001757.
www.konkur.in 338 Appendix A Glossary in the mandibular fossae and against the distal slope of the articular eminence (Ash, 1993). 6: The relation of the mandible to the maxillae when the condyles are in the uppermost and rearmost position in the glenoid fossae. This position may not be able to be recorded in the presence of dysfunction of the masticatory system. 7: A clinically determined position of the mandible placing both condyles into their anterior uppermost position. This can be determined in patients without pain or derangement in the temporomandibular joint (TMJ) (Ramsfjord, 1993).
circumferential clasp Term used to designate a clasp arm that originates above the height of contour and approaches the tooth undercut from an occlusal direction.
clasp (or direct retainer) Component of the clasp assembly that engages a portion of the tooth surface and either enters an undercut for retention or remains entirely above the height of contour to act as a reciprocating element; generally used to stabilize or retain a removable prosthesis.
clasp assembly Part of a removable partial denture that acts as a direct retainer and/or stabilizer for a prosthesis by partially encompassing or contacting an abutment.
complete denture Dental prosthesis that replaces all of the natural dentition and associated structures of the maxilla or mandible. It is entirely supported by tissues (mucous membrane, connective tissues, and underlying bone).
dental cast surveyor Instrument used to determine the relative parallelism of two or more axial surfaces of teeth or other parts of a cast of a dental arch; also used to locate and delineate the contours and relative positions of abutment teeth and associated structures.
dental stones Used to form an artificial stone reproduction from an impression, and used as an investment or for mounting purposes; all dental stones are gypsum products.
denture base Part of a denture (whether it is metal or is made of a resinous material) that rests on the residual bone covered by soft tissue and to which the teeth are attached.
denture foundation area See basal seat.
direct retainer Component of a removable partial denture used to retain or prevent dislodgment; consists of a clasp assembly or precision attachment.
edentulous ridge See residual ridge.
functional impression Impression and resulting cast of the supporting form of the edentulous ridge; artificially created by means of a specially molded (individualized) impression tray or an impression material, or both, that displaces those tissues that can be readily displaced and that would be incapable of rendering support to the denture base when it is supporting functional load.
functional occlusal registration Used to designate a dynamic registration of opposing dentition rather than the recording of a static relationship of one jaw to another.
functional ridge form See functional impression.
guiding planes Two or more vertically parallel surfaces of abutment teeth shaped to direct a prosthesis during placement and removal; surfaces are parallel to the path of the placement and parallel to each other; preferably these surfaces are made parallel to the long axes of abutment teeth.
height of contour Line encircling a tooth, designating its greatest circumference at a selected position determined by a dental surveyor.
incisal rest A rest placed on the incisal edge of an anterior abutment tooth.
indirect retainer Part of a removable partial denture that assists direct retainers in preventing displacement of distal extension denture bases by resisting lever action from the opposite side of the fulcrum line.
interim denture Dental prosthesis used for a short time for reasons of esthetics, mastication, occlusal support, or convenience, or for conditioning the patient to accept an artificial substitute for missing natural teeth until a more definite prosthetic dental treatment can be provided.
internal attachment See precision attachment.
investment cast Cast compounded to withstand high temperatures without disintegrating and, incidentally, to perform certain functions relative to burnout and expansion of the mold.
lingual bar connector Component of the partial denture framework located lingual to the dental arch and above the moving tissues of the floor of the mouth but as far below the gingival tissues as possible.
lingual rest A rest that occupies a position on the lingual surface of an anterior tooth.
linguoplate Designation given when the lingual bar major connector is attached to a thin, contoured apron adjacent to the lingual surfaces of the anterior teeth in the mandible.
major connector Part of a removable partial denture that connects the components on one side of the arch to the components on the opposite side of the arch.
mold Word used to indicate either the cavity into which a casting is made or the shape of an artificial tooth; use of the term is incorrect when referring to a reproduction of a dental arch or a portion thereof.
occlusal rest A rest placed on the occlusal surface of a posterior tooth.
palatal bar Thin, broad palatal coverage with a width of less than 8 mm used as a major connector.
palatal major connector Any thin, broad palatal coverage that is used as a major connector.
palatal strap Proportionally thinner and broader than a palatal bar, although differentiation is somewhat objective.
plastic Refers to any of various substances that harden and retain their shape after being molded.
precision attachment Refers to an interlocking device, one component of which is fixed to an abutment or abutments, Boucher CO: Occlusion in prosthodontics, J Prosthet Dent 3:633-656, 1953; Ash MM: Personal communication, July 1993; Lang BR, Kelsey CC: International prosthodontic workshop on complete denture occlu-sion, Ann Arbor, 1973, The University of Michigan School of Dentistry; Ramsfjord SP: Personal communication, July 1993.
www.konkur.in 339 Appendix A Glossary while the other is integrated into a removable prosthesis to stabilize and/or retain it.
prosthesis A denture, an obturator, a fixed partial denture, or a crown.
provisional denture See interim denture.
rebasing Process that goes beyond relining and involves refitting a denture by replacing the entire denture base with new material without changing the occlusal relations of the teeth.
refractory cast See investment cast.
refractory investment Investment material that can with-stand the high temperatures of casting or soldering; plaster of Paris and artificial stone may be considered in-vestment if either is used to invest any part of a dental restoration for processing.
relining Resurfacing of a denture base with new material to make it fit the underlying tissue.
removable partial denture (RPD) Prosthesis that replaces some teeth in a partially dentate arch and can be removed from the mouth and replaced at will.
residual ridge Residual bone with its soft tissue that covers the underlying area of the denture base; the exact character of the soft tissue covering may vary, but it includes the mucous membrane and underlying fibrous connective tissue.
resin Used broadly for substances named according to their chemical composition, physical structure, and means for activation or curing, such as acrylic resin.
rest Any component of the partial denture that is placed on an abutment tooth, ideally in a prepared rest seat, so that it limits movement of the denture in a gingival direction and transmits functional forces to the tooth.
retainer Any type of clasp, attachment, device, and so on used for fixation, stabilization, or retention of a prosthesis; may be intracoronal or extracoronal and can be used as a means of retaining a removable or a fixed restoration.
retention Quality inherent in the denture that resists the vertical forces of dislodgment (e.g., the force of gravity, the adhesiveness of foods, the forces associated with opening of the jaws).
semiprecision rest Rigid metallic extension of a fixed or removable partial denture that fits into an intracoronal preparation in a cast restoration.
stability Quality of a prosthesis of being firm, stable, or constant and resisting displacement by functional, horizontal, or rotational stresses.
static form See anatomic ridge form.
stone Use of the word in dentistry should be limited to those gypsum materials that are employed for their hardness, accuracy, or abrasion resistance.
support Foundation on which a dental prosthesis rests, or to hold up and serve as a foundation.
undercut When used in reference to an abutment tooth, that portion of a tooth that lies between the height of contour and the gingiva; when used in reference to other oral structures, the contour or cross section of a residual ridge or dental arch that would prevent placement of a denture.
wax pattern Converted to a casting via elimination of the pattern by heat, leaving a mold into which the molten metal is forced by centrifugal force or other means.
www.konkur.in A textbook is rarely found to be all-inclusive in subject mat-ter related to a dental clinical discipline or subdiscipline. Therefore this section, which lists other textbooks and arti-cles from dental periodical literature, may assist in broad-ening a student’s perspective on principles and concepts of removable partial prosthodontics.
Some of the articles have historic significance and are con-sidered classics. Contemporary selections are also included. The serious student of dentistry may extract from this sec-tion background information and details related to the prog-ress of removable partial prosthodontics over the years.
We do not imply that sources have been exhausted in com-piling these lists of textbooks and articles. We have attempted to correctly classify listed articles for ready reference; how-ever, the length of the “Miscellaneous” section attests to the difficulties encountered.
TEXTBOOKS Alberktsson T, Zarb GA: The Brånemark osseointegrated implant, Chicago, 1989, Quintessence.
Anusavice KJ: Phillips’ science of dental materials, ed 11, St Louis, 2003, Saunders.
Applegate OC: Essentials of removable partial denture prosthesis, ed 3, Philadelphia, 1965, WB Saunders.
Babbush CA, et al.: Dental implants: the art and science, ed 2, St Louis, 2011, Saunders.
Beumer J, Curtis TA, Firtell DN: Maxillofacial prosthetics, St Louis, 1979, Mosby.
Block MS: Color atlas of dental implant surgery, ed 3, St Louis, 2011, Saunders.
Brand RW, Isselhard DE: Anatomy of orofacial structures, ed 7, St Louis, 2003, Mosby.
Brewer AA, Morrow RM: Overdentures, ed 2, St Louis, 1980, Mosby.
Brunette DM: Critical thinking: understanding and evaluating dental research, ed 2, Chicago, 2007, Quintessence.
Dawson PE: Functional occlusion: from TMJ to smile design, St Louis, 2007, Mosby.
Dolder EJ, Durrer GT: The bar-joint denture, Chicago, 1978, Quintessence.
Dubrul EL: Sicher and Dubrul’s oral anatomy, ed 8, St Louis/Tokyo, 1988, Ishiyaku EuroAmerica.
Dykema RW, Cunningham DM, Johnston JF: Modern practice in removable partial prosthodontics, Philadelphia, 1969, WB Saunders.
Fonseca RJ, Davis WH: Reconstructive preprosthetic oral and maxillofacial surgery, ed 2, Philadelphia, 1995, WB Saunders.
Graber G: Removable partial dentures, Stuttgart, Germany, 1988, Thieme Medical.
Graber DA, Goldstein RE, Feinman RA: Porcelain laminate veneers, Chicago, 1988, Quintessence.
Grasso JE, Miller EL: Partial prosthodontics, ed 3, St Louis, 1991, Mosby.
Hoag PM, Pawlak EA: Essentials of periodontics, ed 4, St Louis, 1990, Mosby.
Johnson DL, Stratton RJ: Fundamentals of removable prosthodontics, Chicago, 1980, Quintessence.
Johnston JF, et al.: Modern practice in crown and bridge prosthodontics, ed 4, Philadelphia, 1986, WB Saunders.
Jordon RE: Esthetic composite bonding, ed 2, St Louis, 1993, Mosby.
Kratochvil FJ: Partial removable prosthodontics, Philadelphia, 1988, WB Saunders.
Krol AJ: Removable partial denture design outline syllabus, ed 5, San Francisco, 1999, University of the Pacific School of Dentistry.
Laney WR, et al.: Maxillofacial prosthetics, Littleton, MA, 1979, PSG.
Laney WR, Gibilisco JA: Diagnosis and treatment in prosthodontics, Philadelphia, 1983, Lea & Febiger.
Little JW, et al.: Dental management of the medically compromised patient, ed 7, St Louis, 2008, Mosby.
Malamed SF: Medical emergencies in the dental office, ed 6, St Louis, 2007, Mosby.
Malone WPF, Koth DC: Tylman’s theory and practice of fixed prosthodontics, ed 8, Tokyo, 1989, IEA.
Miller CH, Palenik CJ: Infection control and management of hazardous materials for the dental team, ed 4, St Louis, 2010, Mosby.
Morrow RM, Rudd KD, Rhoads JE: Dental laboratory procedures: complete dentures, vol. 1. St Louis, 1985, Mosby.
Mosby’s dental dictionary, ed 3, St Louis, 2013, Mosby/Elsevier.
Nelson SJ: Wheeler’s dental anatomy, physiology, and occlusion, ed 9, St Louis, 2010, Saunders.
Nevins M, Mellonig JT: Periodontal therapy: clinical approaches and evidence of success, vol. 1. Chicago, 1998, Quintessence.
O’Brien WJ: Dental materials and their selection, ed 4, Chicago, 2009, Quintessence.
Okeson JP: Management of temporomandibular disorders and occlusion, ed 6, St Louis, 2008, Mosby.
Okeson JP: Orofacial pain: guidelines for assessment, diagnosis, and management, Chicago, 1996, Quintessence.
Osborne J: Osborne and Lammie’s partial dentures, ed 5, Oxford, 1986, Blackwell Scientific.
APPENDIX B Selected Reading Resources www.konkur.in 341 Appendix B Selected Reading Resources Phoenix RD, Cagna DR, DeFreest CF: Stewart’s clinical removable partial prosthodontics, ed 4, Chicago, 2008, Quintessence.
Powers JM, Sakaguchi RL: Craig’s restorative dental materials, ed 12, St Louis, 2006, Mosby.
Powers JM, Wataha JC: Dental materials: properties and manipulation, ed 9, St Louis, 2008, Mosby.
Preiskel HW: Precision attachments in prosthodontics, Chicago, 1996, Quintessence.
Rahn AO, Ivanhoe JR, Plummer KD: Textbook of complete dentures, ed 6, Shelton, CT, 2009, People’s Medical Publishing House.
Ramfjord SP, Ash Jr MM: Occlusion, ed 4, Philadelphia, 1995, WB Saunders.
Renner RP, Boucher LJ: Removable partial prosthodontics, Chicago, 1987, Quintessence.
Rosenstiel SF, Land MF, Fujimoto J: Contemporary fixed prosthodontics, ed 4, St Louis, 2006, Mosby.
Rudd KD, Rhoads JE, Morrow RM: Dental laboratory procedures, ed 2, vol. 3. St Louis, 1986, Mosby.
Sarnat BG, Laskin DM: The temporomandibular joint: a biological basis for clinical practice, ed 4, Philadelphia, 1992, WB Saunders.
Shillingberg HT, et al.: Fundamentals of fixed prosthodontics, ed 3, Chicago, 1997, Quintessence.
Singh P, Cranin AN: Atlas of oral implantology, ed 3, St Louis, 2010, Mosby.
Stratton RP, Wiebelt FJ: An atlas of removable partial denture design, Chicago, 1988, Quintessence.
Watt DM, MacGregor AR: Designing partial dentures, Littleton, MA, 1985, PSG.
Winkler S: Essentials of complete denture prosthodontics, ed 2, Littleton, MA, 1988, PSG.
Wood NK: Differential diagnosis of oral and maxillofacial lesions, ed 5, St Louis, 1997, Mosby.
Wood NK: Review of diagnosis, oral medicine, radiology, and treatment planning, ed 4, St Louis, 1999, Mosby.
Yalisove IL, Dietz Jr JB: Telescopic prosthetic therapy, Philadelphia, 1979, George F Stickley.
Zarb GA: Temporomandibular joint and masticatory muscle disorders, St Louis, 1995, Mosby.
Zarb GA, et al.: Prosthodontic treatment for edentulous patients: complete dentures and implant-supported prostheses, ed 12, St Louis, 2004, Mosby.
ABUTMENT RETAINERS: EXTERNAL AND INTERNAL ATTACHMENTS Adisman IK: The internal clip attachment in fixed removable partial denture prosthesis, N Y J Dent 32:125–129, 1962.
Ainamo J: Precision removable partial dentures with pontic abutments, J Prosthet Dent 23:289–295, 1970.
Augsburger RH: The Gilmore attachment, J Prosthet Dent 16:1090–1102, 1966.
Becker CM, Campbell MC, Williams DL: The Thompson dowel-rest system modified for chrome-cobalt removable partial denture frameworks, J Prosthet Dent 39:384–391, 1978.
Ben-Ur Z, Aviv I, Cardash HS: A modified direct retainer design for distal extension removable partial dentures, J Prosthet Dent 70:342–344, 1988.
Benson D, Spolsky VW: A clinical evaluation of removable partial dentures with I-bar retainers, Part I J Prosthet Dent 41:246, 1979.
Berg Jr T: I-bar: myth and counter myth, Dent Clin North Am 23(1):65– 75, 1979.
Blatterfein L: Design and positional arrangement of clasps for partial dentures, N Y J Dent 22:305–306, 1952.
Blatterfein L: Study of partial denture clasping, J Am Dent Assoc 43:169–185, 1951.
Brodbelt RHW: A simple paralleling template for precision attachments, J Prosthet Dent 27:285–288, 1972.
Brudvik JS, Morris HF: Stress-relaxation testing. Part III: influence of wire alloys, gauges, and lengths on clasp behavior, J Prosthet Dent 46:374, 1981.
Brudvik JS, Wormley JH: Construction techniques for wrought-wire retentive clasp arms as related to clasp flexibility, J Prosthet Dent 30:769–774, 1973.
Chandler JA, Brudvik JS: Clinical evaluation of patients eight to nine years after placement of removable partial dentures, J Prosthet Dent 51:736, 1984.
Chou TM, et al.: Photoelastic analysis and comparison of force-transmission characteristics of intracoronal attachments with clasp distal-extension removable partial dentures, J Prosthet Dent 62:313–319, 1989.
Chou TM, et al.: Stereophotogrammetric analysis of abutment tooth movement in distal-extension removable partial dentures with intracoronal attachments and clasps, J Prosthet Dent 66:343–349, 1991.
Clayton JA: A stable base precision attachment removable partial denture (PARPD): theories and principles, Dent Clin North Am 24:3–29, 1980.
Cooper H: Practice management related to precision attachment prostheses, Dent Clin North Am 24:45–61, 1980.
DeVan MM: Preserving natural teeth through the use of clasps, J Prosthet Dent 5:208–214, 1955.
Dietz WH: Modified abutments for removable and fixed prosthodontics, J Prosthet Dent 11:1112–1116, 1961.
Dixon DL, et al.: Wear of I-bar clasps and porcelain laminate restorations, Int J Prosthet 5:28–33, 1992.
Dolder EJ: The bar joint mandibular denture, J Prosthet Dent 11:689–707, 1961.
Eliason CM: RPA clasp design for distal-extension removable partial dentures, J Prosthet Dent 49:25–27, 1983.
Frank RP, Brudvik JS, Nicholls JI: A comparison of the flexibility of wrought-wire and cast circumferential clasps, J Prosthet Dent 49: 471–476, 1983.
Getz II : Making a full-coverage restoration on an abutment to fit an existing removable partial denture, J Prosthet Dent 54:335–336, 1985.
Gilson TD: A fixable-removable prosthetic attachment, J Prosthet Dent 9:247–255, 1959.
Gindea AE: A retentive device for removable dentures, J Prosthet Dent 27:501–508, 1972.
Grasso JE: A new removable partial denture clasp assembly, J Prosthet Dent 43:618–621, 1980.
Green JH: The hinge-lock abutment attachment, J Am Dent Assoc 47:175–180, 1953.
Hebel KS, Graser GN, Featherstone JD: Abrasion of enamel and composite resin by removable partial denture clasps, J Prosthet Dent 52:389, 1984.
Highton R, Caputo AA, Matyas J: Retention and stress characteristics for a magnetically retained partial denture, J Dent Res (IADR abstract 279) 62(entire issue), 1982.
Isaacson GO: Telescope crown retainers for removable partial dentures, J Prosthet Dent 22:436–448, 1969.
Ivanhoe JR: Alternative cingulum rest seat, J Prosthet Dent 54:395–396, 1985.
James AG: Self-locking posterior attachment for removable tooth-supported partial dentures, J Prosthet Dent 5:200–205, 1955.
Johnson DL, Stratton RJ, Duncanson Jr MG: The effect of single plane curvature on half-round cast clasps, J Dent Res 62:833–836, 1983.
Johnson JF: The application and construction of the pinledge retainer, J Prosthet Dent 3:559–567, 1953.
Kapur KK, et al.: A randomized clinical trial of two basic removable partial denture designs. Part I: comparisons of five-year success rates and periodontal health, J Prosthet Dent 72:268–282, 1994.
Knodle JM: Experimental overlay and pin partial denture, J Prosthet Dent 17:472–478, 1967.
Knowles LE: A dowel attachment removable partial denture, J Prosthet Dent 13:679–687, 1963.
www.konkur.in 342 Appendix B Selected Reading Resources Koper A: Retainer for removable partial dentures: the Thompson dowel, J Prosthet Dent 30:759–768, 1973.
Kotowicz WE: Clinical procedures in precision attachment removable partial denture construction, Dent Clin North Am 24:143–164, 1980.
Kotowicz WE, et al.: The combination clasp and the distal extension removable partial denture, Dent Clin North Am 17:651–660, 1973.
Kratochvil FJ, Davidson PN, Tandarts JG: Five-year study of treatment with removable partial dentures, Part I J Prosthet Dent 48:237, 1982.
Krol AJ: Clasp design for extension base removable partial dentures, J Prosthet Dent 29:408–415, 1973.
Krol AJ: RPI clasp retainer and its modifications, Dent Clin North Am 17:631–649, 1973.
Langer A: Combinations of diverse retainers in removable partial dentures, J Prosthet Dent 40:378–384, 1978.
LaVere AM: Analysis of facial surface undercuts to determine use of RPI or RPA clasps, J Prosthet Dent 56:741–743, 1986.
Leupold RJ, Faraone KL: Etched castings as an adjunct to mouth preparation for removable partial dentures, J Prosthet Dent 53: 655–658, 1985.
Lubovich RP, Peterson T: The fabrication of a ceramic-metal crown to fit an existing removable partial denture clasp, J Prosthet Dent 37:610–614, 1977.
Marinello CP, et al.: Resin-bonded etched castings with extracoronal attachments for removable partial dentures, J Prosthet Dent 66:52– 55, 1991.
Maroso DJ, Schmidt JR, Blustein R: A preliminary study of wear of porcelain when subjected to functional movements of retentive clasp arms, J Prosthet Dent 45:14, 1981.
McLeod NS: Improved design for the Thompson dowel rest semiprecision intracoronal retainer, J Prosthet Dent 40:513–516, 1978.
McLeod NS: A theoretical analysis of the mechanics of the Thompson dowel semiprecision intracoronal retainer, J Prosthet Dent 37:19–27, 1977.
Mensor Jr MC: Attachment fixation for overdentures, Part I, J Prosthet Dent 37:366–373, 1977.
Mensor Jr MC: Attachment fixation of the overdentures, Part II, J Prosthet Dent 39:16–20, 1978.
Morris HF, et al.: Stress distribution within circumferential clasp arms, J Oral Rehabil 3:387–391, 1976.
Morris HF, et al.: Stress-relaxation testing. Part II. Comparison of bending profiles, microstructures, microhardness, and surface characteristics of several wrought-wires, J Prosthet Dent 46:256, 1981.
Morris HF, et al.: Stress-relaxation testing. Part IV . Clasp pattern dimensions and their influence on clasp behavior, J Prosthet Dent 50:319, 1983.
Morrison ML: Internal precision attachment retainers for partial dentures, J Am Dent Assoc 64:209–215, 1962.
Morrow RM: Tooth-supported complete dentures: an approach to preventive prosthodontics, J Prosthet Dent 21:513–522, 1969.
Oddo Jr VJ: The movable-arm clasp for complete passivity in partial denture construction, J Am Dent Assoc 74:1009–1015, 1967.
Plotnik IJ: Internal attachment for fixed removable partial dentures, J Prosthet Dent 8:85–93, 1958.
Pound E: Cross-arch splinting vs. premature extractions, J Prosthet Dent 16:1058–1068, 1966.
Preiskel H: Precision attachments for free-end saddle prostheses, Br Dent J 127(462):468, 1969.
Preiskel H: Screw retained telescopic prosthesis, Br Dent J 130:107–112, 1971.
Prince IB: Conservation of the supportive mechanism, J Prosthet Dent 15:327–338, 1965.
Sato Y, et al.: Effect of friction coefficient on Akers clasp retention, J Prosthet Dent 78:22–27, 1997.
Seto BG, Avera S, Kagawa T: Resin bonded etched cast cingulum rest retainers for removable partial dentures, Quintessence Int 16:757– 760, 1985.
Singer F: Improvements in precision: attached removable partial dentures, J Prosthet Dent 17:69–72, 1967.
Smith RA, Rymarz FP: Cast clasp transitional removable partial dentures, J Prosthet Dent 22:381–385, 1969.
Snyder HA, Duncanson MG, Johnson D: Effect of clasp flexure on a 4-meta adhered light-polymerized composite resin, Int J Prosthodont 4:364–370, 1991.
Soderfeldt B, et al.: A multilevel analysis of factors affecting the longevity of fixed partial dentures, retainers and abutments, J Oral Rehabil 25:245–252, 1998.
Spielberger MC, et al.: Effect of retentive clasp design on gingival health: a feasibility study, J Prosthet Dent 52:397, 1984.
Stankewitz CG, Gardner FM, Butler GV: Adjustment of cast clasps for direct retention, J Prosthet Dent 45:344, 1981.
Stansbury BE: A retentive attachment for overdentures, J Prosthet Dent 35:228–230, 1976.
Stern MA, Brudvik JS, Frank RP: Clinical evaluation of removable partial denture rest seat adaptation, J Prosthet Dent 53:658–662, 1985.
Stewart BL, Edwards RO: Retention and wear of precision-type attachments, J Prosthet Dent 49:28–34, 1983.
Strohaver RA, Trovillion HM: Removable partial overdentures, J Prosthet Dent 35:624–629, 1976.
Symposium on semiprecision attachments in removable partial dentures, Dent Clin North Am 29:1–237, 1985.
Tautin FS: Abutment stabilization using a nonresilient gingival bar connector, J Am Dent Assoc 99:988–989, 1979.
Tietge JD, et al.: Wear of composite resins and cast direct retainers, Int J Prosthet 5:145–153, 1992.
Vig RG: Splinting bars and maxillary indirect retainers for removable partial dentures, J Prosthet Dent 13:125–129, 1963.
Walter JD: Anchor attachments used as locking devices in two-part removable prostheses, J Prosthet Dent 33:628–632, 1975.
Waltz ME: Ceka extracoronal attachments, J Prosthet Dent 29:167–171, 1973.
White JT: Visualization of stress and strain related to removable partial denture abutments, J Prosthet Dent 40:143–151, 1978.
Wiebelt FJ, Shillingburg Jr HT: Abutment preparation modifications for removable partial denture rest seats, Quintessence Dent Technol 9:449–451, 1985.
Williams AG: Technique for provisional splint with attachment, J Prosthet Dent 21:555–559, 1969.
Willis LM, Swoope CC: Precision attachment partial dentures. In Clark JW, editor: Clinical dentistry, vol. 5. New York, 1976, Harper & Row.
Wright SM: Use of spring-loaded attachments for retention of removable partial dentures, J Prosthet Dent 51:605–610, 1984.
Zakler JM: Intracoronal precision attachments, Dent Clin North Am 24:131–141, 1980.
Zinner ID, Miller RD, Panno FV: Clinical management of abutments with intracoronal attachments, J Prosthet Dent 67:761–767, 1992.
Zinner ID, Miller RD, Panno FV: Precision attachments, Dent Clin North Am 31:395–416, 1987.
Zinner ID, Miller RD, Panno FV: Semiprecision rest system for distal extension removable partial denture design, J Prosthet Dent 42:131– 134, 1979.
ANATOMY Bennett NG: A contribution to the study of the movements of the mandible, J Prosthet Dent 8:41–54, 1958.
Boucher CO: Complete denture impressions based upon the anatomy of the mouth, J Am Dent Assoc 31:1174–1181, 1944.
Brodie AG: Anatomy and physiology of head and neck musculature, Am J Orthod 36:831–844, 1950.
Casey DM: Palatopharyngeal anatomy and physiology, J Prosthet Dent 49:371–378, 1983.
Craddock FW: Retromolar region of the mandible, J Am Dent Assoc 47:453–455, 1953.
Haines RW, Barnett SG: The structure of the mouth in the mandibular molar region, J Prosthet Dent 9:962–974, 1959.
www.konkur.in 343 Appendix B Selected Reading Resources Martone AL, et al.: Anatomy of the mouth and related structures: I, J Prosthet Dent 11:1009–1018, 1961; II, 12:4–27, 1962; III, 12:206–219, 1962; IV , 12:409–419, 1962; V , 12:629–636, 1962; VI, 12:817–834, 1962; VII, 13:4–33, 1963; VIII, 13:204–228, 1963.
Merkeley HJ: The labial and buccal accessory muscles of mastication, J Prosthet Dent 4:327–334, 1954.
Merkeley HJ: Mandibular rearmament. I. Anatomic considerations, J Prosthet Dent 9:559–566, 1959.
Monteith BD: Management of loading forces on the mandibular extension prosthesis. Part II: Classification for matching modalities to clinical situations, J Prosthet Dent 52:832–835, 1984.
Pendleton EC: Anatomy of the face and mouth from the standpoint of the denture prosthetist, J Am Dent Assoc 33:219–234, 1946.
Pendleton EC: Changes in the denture supporting tissues, J Am Dent Assoc 42:1–15, 1951.
Pietrokovski J: The bony residual ridge in man, J Prosthet Dent 34:456–462, 1975.
Pietrokovski J, Sorin S, Zvia H: The residual ridge in partially edentulous patients, J Prosthet Dent 36:150–158, 1976.
Preti G, Bruscagin C, Fava C: Anatomic and statistical study to determine the inclination of the condylar long axis, J Prosthet Dent 49:572–575, 1983.
Roche AF: Functional anatomy of the muscles of mastication, J Prosthet Dent 13:548–570, 1963.
Silverman SI: Denture prosthesis and the functional anatomy of the maxillofacial structures, J Prosthet Dent 6:305–331, 1956.
BIOMECHANICS Asher ML: Application of the rotational path design concept to a removable partial denture with a distal-extension base, J Prosthet Dent 68:641–643, 1992.
Augthun M, et al.: The influence of spruing technique on the development of tension in a cast partial denture framework, Int J Prosthodont 7:72–76, 1994.
Avant WE: Factors that influence retention of removable partial dentures, J Prosthet Dent 25:265–270, 1971.
Avant WE: Fulcrum and retention lines in planning removable partial dentures, J Prosthet Dent 25:162–166, 1971.
Aviv I, Ben-Ur Z, Cardash HS: An analysis of rotational movement of asymmetrical distal-extension removable partial dentures, J Prosthet Dent 61:211–214, 1989.
Aydinlik E, Akay HU: Effect of a resilient layer in a removable partial denture base on stress distribution to the mandible, J Prosthet Dent 44:17–20, 1980.
Ben-Ur Z, et al.: Designing clasps for the asymmetric distal extension removable partial denture, Int J Prosthodont 9:374–378, 1996.
Berg TE, Caputo AA: Comparison of load transfer by maxillary distal extension removable partial dentures with a spring-loaded plunger attachment and I-bar retainer, J Prosthet Dent 68:492–499, 1992.
Berg TE, Caputo AA: Load transfer by a maxillary distal-extension removable partial denture with extracoronal attachments, J Prosthet Dent 68:784–789, 1992.
Bezzon OL, et al.: Surveying removable partial dentures: the importance of guiding planes and path of insertion for stability, J Prosthet Dent 78:412–418, 1997.
Bridgeman JT, et al.: Comparison of titanium and cobalt-chromium removable partial denture clasps, J Prosthet Dent 78:187–193, 1997.
Browning JD, Eick JD, McGarrah HE: Abutment tooth movement measured in vivo by using stereophotogrammetry, J Prosthet Dent 57:323–328, 1987.
Brudvik JS, Morris HF: Stress-relaxation testing. Part III: Influence of wire alloys, gauges, and lengths on clasp behavior, J Prosthet Dent 46:374–379, 1981.
Byron Jr R, Frazer RQ, Herren MC: Rotational path removable partial denture: an esthetic alternative, Gen Dent 55:245–250, 2007; quizzes 251, 264.
Cecconi BT: Effect of rest design on transmission of forces to abutment teeth, J Prosthet Dent 32:141–151, 1974.
Cecconi BT, Asgar K, Dootz E: Clasp assembly modifications and their effect on abutment tooth movement, J Prosthet Dent 27:160–167, 1972.
Cecconi BT, Asgar K, Dootz E: The effect of partial denture clasp design on abutment tooth movement, J Prosthet Dent 25:44–56, 1971.
Cecconi BT, Asgar K, Dootz E: Removable partial denture abutment tooth movement as affected by inclination of residual ridges and types of loading, J Prosthet Dent 25:375–381, 1971.
Chou TM, et al.: Photoelastic analysis and comparison of force-transmission characteristics of intracoronal attachments with clasp distal-extension removable partial dentures, J Prosthet Dent 62:313–319, 1989.
Chou TM, et al.: Stereophotogrammetric analysis of abutment tooth movement in distal-extension removable partial dentures with intracoronal attachments and clasps, J Prosthet Dent 66:343–349, 1991.
Clayton JA, Jaslow C: A measurement of clasp forces on teeth, J Prosthet Dent 25:21–43, 1971.
Craig RG, Farah JW: Stresses from loading distal extension removable partial dentures, J Prosthet Dent 39:274–277, 1978.
DeVan MM: The nature of the partial denture foundation: suggestions for its preservation, J Prosthet Dent 2:210–218, 1952.
el Charkawi HG, et al.: The effect of the resilient-layer distal-extension partial denture on movement of the abutment teeth: a new methodology, J Prosthet Dent 60:622–630, 1988.
Fisher RL: Factors that influence the base stability of mandibular distal-extension removable partial dentures: a longitudinal study, J Prosthet Dent 50:167–171, 1983.
Frank RP, Nicholls JI: A study of the flexibility of wrought-wire clasps, J Prosthet Dent 45:259–267, 1981.
Frechette AR: The influence of partial denture design on distribution of force to abutment teeth, J Prosthet Dent 6:195–212, 1956.
Goodkind RJ: The effects of removable partial dentures on abutment tooth mobility, J Prosthet Dent 30:139–146, 1973.
Goodman JJ, Goodman HW: Balance of force in precision free-end restorations, J Prosthet Dent 13:302–308, 1963.
Hall WA: Variations in registering interarch transfers in removable partial denture construction, J Prosthet Dent 30:548–553, 1973.
Harrop J, Javid N: Reciprocal arms of direct retainers in removable partial dentures, J Can Dent Assoc 4:208–211, 1976.
Henderson D, Seward TE: Design and force distribution with removable partial dentures: a progress report, J Prosthet Dent 17:350–364, 1967.
Henriques GE, et al.: Soldering and remelting influence on fatigue strength of cobalt-chromium alloy, J Prosthet Dent 78:146–152, 1997.
Hindels GW: Stress analysis in distal extension partial dentures, J Prosthet Dent 7:197–205, 1957.
Iwama CY, et al.: Cobalt-chromium-titanium alloy for removable partial dentures, Int J Prosthodont 10:309–317, 1997.
Johnson DL, Stratton RJ, Duncanson MGJ: The effect of single plane curvature on half-round cast clasps, J Dent Res 62:833–836, 1983.
Kaires AK: Partial denture design and its relation to force distribution and masticatory performance, J Prosthet Dent 6:672–683, 1956.
Knowles LE: The biomechanics of removable partial dentures and its relationship to fixed prosthesis, J Prosthet Dent 8:426–430, 1958.
Kratochvil FJ: Influence of occlusal rest position and clasp design on movement of abutment teeth, J Prosthet Dent 13:114–124, 1963.
Kratochvil FJ, Caputo AA: Photoelastic analysis of pressure on teeth and bone supporting removable partial dentures, J Prosthet Dent 3:52, 1975.
Kratochvil FJ, Thompson WD, Caputo AA: Photoelastic analysis of stress patterns on teeth and bone with attachment retainers for removable partial dentures, J Prosthet Dent 46:21–28, 1981.
Lofbers PG, Ericson G, Eliasson S: A clinical and radiographic evaluation of removable partial dentures retained by attachments to alveolar bars, J Prosthet Dent 47:126–132, 1982.
Lowe RD, Kydd WL, Smith DE: Swallowing and resting forces related to lingual flange thickness in removable partial dentures, J Prosthet Dent 23:279–288, 1970.
www.konkur.in 344 Appendix B Selected Reading Resources MacGregor AR, Miller TPG, Farah JW: Stress analysis of partial dentures, J Dent 6:125–132, 1978.
Marei MK: Measurement (in vitro) of the amount of force required to dislodge specific clasps from different depths of undercut, J Prosthet Dent 74:258–263, 1995.
Maroso DJ, Schmidt JR, Blustein R: A preliminary study of wear of porcelain when subjected to functional movements of retentive clasp arms, J Prosthet Dent 45:14–17, 1981.
Matheson GR, Brudvik JS, Nicholls JI: Behavior of wrought-wire clasps after repeated permanent deformation, J Prosthet Dent 55:226–231, 1986.
Maxfield JB, Nicholls JI, Smith DE: The measurement of forces transmitted to abutment teeth of removable partial dentures, J Prosthet Dent 41:134, 1979.
McCartney JW: Motion vector analysis of an abutment for a distal-extension removable partial denture, J Prosthet Dent 43:15–21, 1980.
McDowell GC: Force transmission by indirect retainers during unilateral loading, J Prosthet Dent 39:616–621, 1978.
McDowell GC, Fisher RL: Force transmission by indirect retainers when a unilateral dislodging force is applied, J Prosthet Dent 47:360–365, 1982.
McLeod NS: An analysis of the rotational axes of semiprecision and precision distal-extension removable partial dentures, J Prosthet Dent 48:130–134, 1982.
Morris HF, Asgar K, Tillitson E: Stress-relaxation testing. I. A new approach to the testing of removable partial denture alloys, wrought-wires, and clasp behavior, J Prosthet Dent 46:133–141, 1981.
Morris HF, Brudvik JS: Influence of polishing on cast clasp properties, J Prosthet Dent 55:75–77, 1986.
Morris HF, et al.: Stress-relaxation testing. IV. Clasp pattern dimensions and their influence on clasp behavior, J Prosthet Dent 50:319–326, 1983.
NaBadalung DP, et al.: Comparison of bond strengths of denture base resins to nickel-chromium-beryllium removable partial denture alloy, J Prosthet Dent 78:566–573, 1997.
NaBadalung DP, et al.: Frictional resistance of removable partial dentures with retrofitted resin composite guide planes, Int J Prosthodont 10:116–122, 1997.
NaBadalung DP, et al.: Laser welding of a cobalt-chromium removable partial denture alloy, J Prosthet Dent 79:285–290, 1998.
Ogata K, Shimigu K: Longitudinal study of forces transmitted from denture base to retainers of free-end saddle dentures with Akers clasps, J Oral Rehabil 18:471–480, 1991.
Plotnick IJ, Beresin VE, Simkins AB: The effects of variations in the opposing dentition on changes in the partially edentulous mandible, J Prosthet Dent I 33:278–286, 1975; II, 33:403–406, 1975; III, 33:529– 534, 1975.
Sansom BP, et al.: Rest seat designs for inclined posterior abutments: a photoelastic comparison, J Prosthet Dent 58:57–62, 1987.
Shohet H: Relative magnitudes of stress on abutment teeth with different retainers, J Prosthet Dent 21:267–282, 1969.
Smith BH: Changes in occlusal face height with removable partial dentures, J Prosthet Dent 34:278–285, 1975.
Smith BJ, Turner CH: The use of crowns to modify abutment teeth of removable partial dentures, J Dent 7:52–56, 1979.
Smyd ES: Biomechanics of prosthetic dentistry, J Prosthet Dent 4:368–383, 1954.
Stern WJ: Guiding planes in clasp reciprocation and retention, J Prosthet Dent 34:408–414, 1975.
Swoope CC, Frank RP: Stress control and design. In Clark JW, editor: Clinical dentistry, vol. 5. New York, 1976, Harper & Row.
Taylor DT, Pflushoeft FA, McGivney GP: Effect of two clasping assemblies on arch integrity as modified by base adaptation, J Prosthet Dent 47:120–125, 1982.
Tebrock OC, et al.: The effect of various clasping systems on the mobility of abutment teeth for distal-extension removable partial dentures, J Prosthet Dent 41:511, 1979.
Thompson WD, Kratochvil FJ, Caputo AA: Evaluation of photoelastic stress patterns produced by various designs of bilateral distal-extension removable partial dentures, J Prosthet Dent 38:261, 1977.
Toth RW, et al.: Shear strength of lingual rest seats prepared in bonded composite, J Prosthet Dent 56:99–104, 1986.
Vallittu PK: Comparison of the in vitro fatigue resistance of an acrylic resin removable partial denture reinforced with continuous glass fibers or metal wires, J Prosthodont 5:115–121, 1996.
Vallittu PK: Deflection fatigue of cobalt-chromium, titanium, and gold alloy cast denture clasp, J Prosthet Dent 74:412–419, 1995.
Waldmeier MD, et al.: Bend testing of wrought-wire removable partial denture alloys, J Prosthet Dent 76:559–565, 1996.
Wills DJ, Manderson RD: Biomechanical aspects of the support of partial dentures, J Dent 5:310–318, 1977.
Yurkstas A, Fridley HH, Manly RS: A functional evaluation of fixed and removable bridgework, J Prosthet Dent 1:570–577, 1951.
Zoeller GN, Kelly Jr WJ: Block form stability in removable partial prosthodontics, J Prosthet Dent 25:515–519, 1971.
CAD/CAM, RAPID PROTOTYPING OF PARTIAL DENTURES Eggbeer D, Bibb R, Williams R: The computer-aided design and rapid prototyping fabrication of removable partial denture frameworks, Proc Inst Mech Eng H219(3):195–202, 2005.
Lang LA, Tulunoglu I: A critically appraised topic review of computer-aided design/computer-aided machining of removable partial denture frameworks, Dent Clin North Am 58(1):247–255, 2014,
Sun J, Zhang FQ: The application of rapid prototyping in prostho-dontics, J Prosthodont 21(8):641–644, 2012, j.1532-849X.2012.00888.x.
CLASSIFICATION Applegate OC: The rationale of partial denture choice, J Prosthet Dent 10:891–907, 1960.
Avant WE: A universal classification for removable partial denture situations, J Prosthet Dent 16:533–539, 1966.
Bailyn M: Tissue support in partial denture construction, Dent Cosmos 70:988–997, 1928.
Beckett LS: The influence of saddle classification on the design of partial removable restoration, J Prosthet Dent 3:506–516, 1953.
Costa E: A simplified system for identifying partially edentulous arches, J Prosthet Dent 32:639–645, 1974.
Cummer WE: Partial denture service. In Anthony LP, editor: American textbook of prosthetic dentistry, Philadelphia, 1942, Lea & Febiger.
Friedman J: The ABC classification of partial denture segments, J Prosthet Dent 3:517–524, 1953.
Godfrey RJ: Classification of removable partial dentures, J Am Coll Dent 18:5–13, 1951.
Kennedy E: Partial denture construction, Dental Items of Interest3–8, 1928.
Mensor Jr MC: Classification and selection of attachments, J Prosthet Dent 29:494–497, 1973.
Miller EL: Systems for classifying partially dentulous arches, J Prosthet Dent 24:25–40, 1970.
Skinner CN: A classification of removable partial dentures based upon the principles of anatomy and physiology, J Prosthet Dent 9:240–246, 1959.
CLEFT PALATE Aram A, Subtelny JD: Velopharyngeal function and cleft palate prostheses, J Prosthet Dent 9:149–158, 1959.
Baden E: Fundamental principles of orofacial prosthetic therapy in congenital cleft palate, J Prosthet Dent 4:420–433, 1954.
Bixler D: Heritability of clefts of the lips and palate, J Prosthet Dent 33:100–108, 1975.
www.konkur.in 345 Appendix B Selected Reading Resources Buckner H: Construction of a denture with hollow obturator, lid and soft acrylic lining, J Prosthet Dent 31:95–99, 1974.
Calvan J: The error of Gustan Passavant, Plast Reconstr Surg 13:275– 289, 1954.
Cooper HK: Integration of service in the treatment of cleft lip and cleft palate, J Am Dent Assoc 47:27–32, 1953.
Dalston RM: Prosthodontic management of the cleft palate patient: a speech pathologist’s view, J Prosthet Dent 37:327–329, 1978.
Ettinger RL: Use of teeth with a poor prognosis in cleft palate prosthodontics, J Am Dent Assoc 94:910–914, 1977.
Fox A: Prosthetic correction of a severe acquired cleft palate, J Prosthet Dent 8:542–546, 1958.
Gibbons P, Bloomer H: A supportive-type prosthetic speech aid, J Prosthet Dent 8:362–369, 1958.
Graber TM: Oral and nasal structures in cleft palate speech, J Am Dent Assoc 53:693–706, 1956.
Harkins CS: Modern concepts in the prosthetic rehabilitation of cleft palate patients, J Oral Surg 10:298–312, 1952.
Harkins CS, Ivy RH: Surgery and prosthesis in the rehabilitation of cleft palate patients, J South Calif Dent Assoc 19:16–24, 1951.
Immekus JE, Aramany MA: A fixed-removable partial denture for cleft palate patients, J Prosthet Dent 34:286–291, 1975.
Landa JS: The prosthodontist views the rehabilitation of the cleft palate patient, J Prosthet Dent 6:421–427, 1956.
Lavelle WE, Zach GE: The tissue bar and Ceka anchor as aids in cleft palate rehabilitation, J Prosthet Dent 30:321–325, 1973.
Lloyd RS, Pruzansky S, Subtelny JD: Prosthetic rehabilitation of a cleft palate patient subsequent to multiple surgical and prosthetic failures, J Prosthet Dent 7:216–230, 1957.
Merkeley HJ: Cleft palate prosthesis, J Prosthet Dent 9:506–513, 1959.
Minsley GE, Warren DW, Hairfield WM: The effect of cleft palate speech aid prostheses on the nasopharyngeal airway and breathing, J Prosthet Dent 65:122–126, 1991.
Nidiffer TJ, Shipmon TH: The hollow-bulb obturator for acquired palatal openings, J Prosthet Dent 7:126–134, 1957.
Olinger NA: Cleft palate prosthesis rehabilitation, J Prosthet Dent 2:117–135, 1952.
Rosen MS: Prosthetics for the cleft palate patient, J Am Dent Assoc 60:715–721, 1960.
Rothenberg LIA: Overlay dentures for the cleft-palate patient, J Prosthet Dent 37:190–195, 1977.
Schneiderman CR, Maun MB: Air flow and intelligibility of speech of normal speakers and speakers with a prosthodontically repaired cleft palate, J Prosthet Dent 39:193–199, 1978.
Sharry JJ: The meatus obturator in cleft palate prosthesis, Oral Surg 7:852–855, 1954.
Sharry JJ: Meatus obturator in particular and pharyngeal impressions in general, J Prosthet Dent 8:893–896, 1958.
Tautin FS, Schaaf NA: Superiorly based obturator, J Prosthet Dent 33:96–99, 1975.
Walter JD: Palatopharyngeal activity in cleft palate subjects, J Prosthet Dent 63:187–192, 1990.
COMPLETE MOUTH AND OCCLUSAL REHABILITATION Brewer AA, Fenton AH: The overdenture, Dent Clin North Am 17:723– 746, 1973.
Bronstein BR: Rationale and technique of biomechanical occlusal rehabilitation, J Prosthet Dent 4:352–367, 1954.
Cohn LA: Occluso-rehabilitation, principles of diagnosis and treatment planning, Dent Clin North Am 6:281, 1962.
Curtis SR: Integrating fixed and removable provisional restorations, J Prosthet Dent 70:374–377, 1993.
Dubin NA: Advances in functional occlusal rehabilitation, J Prosthet Dent 6:252–258, 1956.
Ferencz JL: Splinting, Dent Clin North Am 31:383–393, 1987.
Kazis H: Functional aspects of complete mouth rehabilitation, J Prosthet Dent 4:833–841, 1954.
Kornfeld M: The problem of function in restorative dentistry, J Prosthet Dent 5:670–676, 1955.
Landa JS: An analysis of current practices in mouth rehabilitation, J Prosthet Dent 5:527–537, 1955.
Lang BR: Complete denture occlusion, Dent Clin North Am 40:85–101, 1996.
Mann AW, Pankey LD: Oral rehabilitation. I. Use of the P-M instrument in treatment planning and restoring the lower posterior teeth, J Prosthet Dent 10:135–150, 1960.
Mann AW, Pankey LD: Oral rehabilitation. II. Reconstruction of the upper teeth using a functionally generated path technique, J Prosthet Dent 10:151–162, 1960.
Mann AW, Pankey LD: Oral rehabilitation utilizing the Pankey-Mann instrument and a functional bite technique, Dent Clin North Am March215–230, 1959.
McCartney JW: Occlusal reconstruction and rebase procedure for distal extension removable partial dentures, J Prosthet Dent 43:695– 698, 1980.
Schuyler CH: An evaluation of incisal guidance and its influence on restorative dentistry, J Prosthet Dent 9:374–378, 1959.
Schweitzer JM: Open bite from the prosthetic point of view, Dent Clin North Am 1:269–283, 1957.
CROWNS AND FIXED PARTIAL DENTURES Alexander PC: Analysis of the cuspid protective occlusion, J Prosthet Dent 13:309–317, 1963.
Bader JD, et al.: Effect of crown margins on periodontal conditions in regularly attending patients, J Prosthet Dent 65:75–79, 1991.
Beeson PE: The use of acrylic resins as an aid in the development of patterns for two types of crowns, J Prosthet Dent 13:493–498, 1963.
Binkley TK, Binkley C: Porcelain-fused-to-metal crowns as replacements for denture teeth in removable partial denture construction, J Prosthet Dent 58:124–125, 1987.
Blackman R, Baeg R, Barghi N: Marginal accuracy and geometry of cast titanium copings, J Prosthet Dent 67:435–440, 1992.
Budtz-Jorgenson E, Isidor F: A five-year longitudinal study of cantilever fixed partial dentures compared with removable partial dentures in a geriatric population, J Prosthet Dent 64:42–47, 1990.
Caplan J: Maintenance of full coverage fixed-abutment bridges, J Prosthet Dent 5:852–854, 1955.
Cheug SP, Dimmer A: Management of worn dentition with resin-bonded cast metal lingual veneering, J Prosthet Dent 63:122–123, 1990.
Coelho DH: Criteria for the use of fixed prosthesis, Dent Clin North Am 1:299–311, 1957.
Cooper TM, et al.: Effect of venting on cast gold full crowns, J Prosthet Dent 26:621–626, 1971.
Cowgen GT: Retention, resistance and esthetics of the anterior three-quarter crown, J Am Dent Assoc 62:167–171, 1961.
Culpepper WD, Moulton PS: Considerations in fixed prosthodontics, Dent Clin North Am 23:21–35, 1979.
Dental technology standards, J Dent Technol 14:26–31, 1997.
Ekfeldt A, et al.: Changes of masticatory movement characteristics after prosthodontic rehabilitation of individuals with extensive tooth wear, Int J Prosthodont 9(6):539–546, 1996.
Elledge DA, Schorr BL: A provisional and new crown to fit with a clasp of an existing removable partial denture, J Prosthet Dent 63:541–544, 1990.
Felton DA, et al.: Effect of in vivo crown margin discrepancies on periodontal health, J Prosthet Dent 65:357–364, 1991.
Glantz PO, et al.: The devitalized tooth as an abutment in dentitions with reduced but healthy periodontium, Periodontol 2000 4:52–57, 1994.
Goldberg A, Jones RD: Constructing cast crowns to fit existing removable partial denture clasps, J Prosthet Dent 36:382–386, 1976.
www.konkur.in 346 Appendix B Selected Reading Resources Goodacre CJ, et al.: The prosthodontic management of endodontically treated teeth: a literature review. Part I. Success and failure data, treatment concepts, J Prosthodont 3:243–250, 1994.
Guyer SE: Nonrigid subocclusal connector for fixed partial dentures, J Prosthet Dent 26:433–436, 1971.
Hansen CA, Cook PA, Nelson DF: Pin-modified facial inlay to enhance retentive contours on a removable partial denture abutment, J Prosthet Dent 55:480–481, 1986.
Henderson D, et al.: The cantilever type of posterior fixed partial dentures: a laboratory study, J Prosthet Dent 24:47–67, 1970.
Johnson Jr EA: Combination of fixed and removable partial dentures, J Prosthet Dent 14:1099–1106, 1964.
Johnston JF, et al.: Construction and assembly of porcelain veneer gold crowns and pontics, J Prosthet Dent 12:1125–1137, 1962.
Kapur KK, et al.: Veterans Administration Cooperative Dental Implant Study: Comparisons between fixed partial dentures supported by blade-vent implants and removable partial dentures. Part II. Comparison of success rates and periodontal health between two treatment modalities, J Prosthet Dent 62:685–703, 1992.
Kapur KK, et al.: Veterans Administration Cooperation Dental Implant Study: Comparisons between fixed partial dentures supported by blade-vent implants and removable partial dentures. Part IV. Comparisons of patient satisfaction between two treatment modalities, J Prosthet Dent 66:517–530, 1991.
Kunisch WH, Dodd J: A conversion alternative to ceramics in a crown-and-sleeve coping prosthesis, J Prosthet Dent 49:581–582, 1983.
Leff A: New concepts in the preparation of teeth for full coverage, J Prosthet Dent 5:392–400, 1955.
Leff A: Reproduction of tooth anatomy and positional relationship in full cast or veneer crowns, J Prosthet Dent 6:550–557, 1956.
Libby G, et al.: Longevity of fixed partial dentures, J Prosthet Dent 78:127–131, 1997.
Malson TS: Anatomic cast crown reproduction, J Prosthet Dent 9: 106–112, 1959.
Marinello CP, Scharer P: Resin-bonded etched cast extracoronal attachments for removable partial dentures: clinical experiences, Int J Periodont Res Dent 7:36–49, 1987.
McArthur DR: Fabrication of full coverage restorations for existing removable partial dentures, J Prosthet Dent 51:574–576, 1984.
Mojon P, et al.: Relationship between prosthodontic status, caries and periodontal disease in a geriatric population, Int J Prosthodont 26:564–571, 1995.
Morris HF, et al.: Department of Veterans Affairs Cooperative Studies Project No. 242: Quantitative and qualitative evaluation of the marginal fit of cast ceramic, porcelain-shoulder, and cast metal full crown margins, J Prosthet Dent 67:198–204, 1992.
Moulding MB, Holland GA, Sulik WD: An alternative orientation of nonrigid connectors in fixed partial dentures, J Prosthet Dent 6:236– 238, 1992.
Mueninghoff LA, Johnson MH: Fixed-removable partial dentures, J Prosthet Dent 48:547–550, 1982.
Palmquist S, et al.: Multivariate analyses of factors influencing the longevity of fixed partial dentures, retainers and abutments, J Prosthet Dent 71:245–250, 1994.
Patur B: The role of occlusion and the periodontium in restorative procedures, J Prosthet Dent 21:371–379, 1969.
Pezzoli M, et al.: Magnetizable abutment crowns for distal-extension removable partial dentures, J Prosthet Dent 55:475–480, 1986.
Phillips RW, Biggs DH: Distortion of wax patterns as influenced by storage time, storage temperature, and temperature of wax manipulation, J Am Dent Assoc 41:28–37, 1950.
Phillips RW, Price RR: Some factors which influence the surface of stone dies poured in alginate impressions, J Prosthet Dent 5:72–79, 1955.
Phillips RW, Swartz ML: A study of adaptation of veneers to cast gold crowns, J Prosthet Dent 7:817–822, 1957.
Pound E: The problem of the lower anterior bridge, J Prosthet Dent 5:543–545, 1955.
Preston JD: Preventing ceramic failures when integrating fixed and removable prostheses, Dent Clin North Am 23:37–52, 1979.
Pruden KC: A hydrocolloid technique for pinledge bridge abutments, J Prosthet Dent 6:65–71, 1956.
Pruden WH: Full coverage, partial coverage, and the role of pins, J Prosthet Dent 26:302–306, 1971.
Rhoads JE: The fixed-removable partial denture, J Prosthet Dent 48: 122–129, 1982.
Rubin MK: Full coverage: the provisional and final restorations made easier, J Prosthet Dent 8:664–672, 1958.
Schorr BL, Peregrina AM, Elledge DA: Alternatives to posterior complete crowns: integrating foundations with cuspal protection, J Prosthet Dent 69:165–170, 1993.
Seals Jr RR, Stratton RJ: Surveyed crowns: a key for integrating fixed and removable prosthodontics, Quintessence Dent Technol 11:43–49, 1987.
Sheets CE: Dowel and core foundations, J Prosthet Dent 23:58–65, 1970.
Shooshan ED: The reverse pin-porcelain facing, J Prosthet Dent 9:284–301, 1959.
Smith GP: The marginal fit of the full cast shoulderless crown, J Prosthet Dent 7:231–243, 1957.
Smith GP: Objectives of a fixed partial denture, J Prosthet Dent 11:463–473, 1961.
Staffanou RS, Thayer KE: Reverse pin-porcelain veneer and pontic technique, J Prosthet Dent 12(1138):1145, 1962.
Thurgood BW, Thayer KE, Lee RE: Complete crowns constructed for an existing partial denture, J Prosthet Dent 29:507–512, 1973.
Treppo KW, Smith FW: A technique for restoring abutments for removable partial dentures, J Prosthet Dent 40:398–401, 1978.
Troxell RR: The polishing of gold castings, J Prosthet Dent 9:668–675, 1959.
Turner KA, Messirlian DM: Restoration of the extremely worn dentition, J Prosthet Dent 52:464–474, 1984.
Wagman SS: Tissue management for full cast veneer crowns, J Prosthet Dent 15:106–117, 1965.
Wagner AW, Burkhart JW, Fayle Jr HE: Contouring abutment teeth with cast gold inlays for removable partial dentures, J Prosthet Dent 201:330–334, 1968.
Wallace FH: Resin transfer copings, J Prosthet Dent 8:289–292, 1958.
Wang CJ, Millstein PL, Nathanson D: Effects of cement, cement space, marginal design, seating aid materials, and seating force on crown cementation, J Prosthet Dent 67:786–790, 1992.
Welsh SL: Complete crown construction for a clasp-bearing abutment, J Prosthet Dent 34:320–323, 1975.
Wheeler RC: Complete crown form and the periodontium, J Prosthet Dent 11:722–734, 1961.
Yalisove IL: Crown and sleeve-coping retainers for removable partial prostheses, J Prosthet Dent 16:1069–1085, 1966.
DENTAL LABORATORY PROCEDURES ADA Council on Scientific Affairs and ADA Council on Dental Practice: Infection control recommendations for the dental office and the dental laboratory, J Am Dent Assoc 11:395–399, 1996.
Asgar K, Peyton FA: Casting dental alloys to embedded wires, J Prosthet Dent 15:312–321, 1965.
Becker CM, Smith EE, Nicholls JI: The comparison of denture-base processing techniques. I. Material characteristics, J Prosthet Dent 37:330–338, 1977.
Berg E, et al.: Mechanical properties of laser-welded cast and wrought titanium, J Prosthet Dent 74:250–257, 1995.
Blanchard CH: Filling undercuts on refractory casts with investment, J Prosthet Dent 3:417–418, 1953.
Bolouri A, Hilger TC, Gowrylok MD: Modified flasking technique for removable partial dentures, J Prosthet Dent 34:221–223, 1975.
www.konkur.in 347 Appendix B Selected Reading Resources Brudvik JS, Nicholls JI: Soldering of removable partial dentures, J Prosthet Dent 49:762–765, 1983.
Burnett CA, et al.: Sprue design in removable partial denture casting, J Dent 24:99–103, 1996.
Calverley MJ, Moergeli Jr JR: Effect on the fit of removable partial denture frameworks when master casts are treated with cyanoacrylate resin, J Prosthet Dent 58:327–329, 1987.
Casey DM, Crowther DS, Lauciello FR: Strengthening abutment or isolated teeth on removable partial denture master casts, J Prosthet Dent 46:105–106, 1981.
Dirksen LC, Campagna SJ: Mat surface and rugae reproduction for upper partial denture castings, J Prosthet Dent 4:67–72, 1954.
Dootz ER, Craig RG, Peyton FA: Influence of investments and duplicating procedures on the accuracy of partial denture castings, J Prosthet Dent 15:679–690, 1965.
Dootz ER, Craig RG, Peyton FA: Simplification of the chrome-cobalt partial denture casting procedure, J Prosthet Dent 17:464–471, 1967.
Elbert CA, Ryge G: The effect of heat treatment on hardness of a chrome-cobalt alloy, J Prosthet Dent 15:873–879, 1965.
Elliott RW: The effects of heat on gold partial denture castings, J Prosthet Dent 13:688–698, 1963.
Enright CM: Dentist-dental laboratory harmony, J Prosthet Dent 11:393–394, 1961.
Fiebiger GE, Parr GR, Goldman BM: Remount casts for removable partial dentures, J Prosthet Dent 48:106–107, 1982.
Firtell DN, Muncheryan AM, Green AJ: Laboratory accuracy in casting removable partial denture frameworks, J Prosthet Dent 54:856–862, 1985.
Fowler Jr JA, Kuebker WA, Escobedo JJ: Laboratory procedures for the maintenance of a removable partial overdenture, J Prosthet Dent 50:121–126, 1983.
Garver DG: Updated laboratory procedure for the subpontic clasping system, J Prosthet Dent 48:734–735, 1982.
Gay WD: Laboratory procedures for fitting removable partial denture frameworks, J Prosthet Dent 40:227–229, 1978.
Gilson TD, Asgar K, Peyton FA: The quality of union formed in casting gold to embedded attachment metals, J Prosthet Dent 15:464–473, 1965.
Grunewald AH, Paffenbarger GC, Dickson G: Dentist, dental laboratory, and the patient, J Prosthet Dent 8:55–60, 1958.
Grunewald AH, Paffenbarger GC, Dickson G: The effect of molding processes on some properties of denture resins, J Am Dent Assoc 44:269–284, 1952.
Grunewald AH, Paffenbarger GC, Dickson G: The role of the dental technician in a prosthetic service, Dent Clin North Am 4:359–370, 1960.
Hanson JG, et al.: Effect on dimensional accuracy when reattaching fractured lone standing teeth of a cast, J Prosthet Dent 47:488–492, 1982.
Johnson HB: Technique for packing and staining complete or partial denture bases, J Prosthet Dent 6:154–159, 1956.
Jones DW: Thermal analysis and stability of refractory investments, J Prosthet Dent 18:234–241, 1967.
Jordan RD, Turner KA, Taylor TD: Multiple crowns fabricated for an existing removable partial denture, J Prosthet Dent 48:102–105, 1982.
Kazanoglu A, Smith EH: Replacement technique for a broken occlusal rest, J Prosthet Dent 48:621–623, 1982.
Krand M, et al.: Study on the surface of resins that burn without residues in the lost-wax procedure, J Prosthodont 5:259–265, Dec 1996.
Lanier BR, Rudd KD, Strunk RR: Making chromium-cobalt removable partial dentures: a modified technique, J Prosthet Dent 25:197–205, 1971.
Lauciello FR: Technique for remounting removable partial dentures opposing maxillary complete dentures, J Prosthet Dent 45:336–340, 1981.
Mahler DB, Ady AB: The influence of various factors on the effective setting expansion of casting investments, J Prosthet Dent 13:365–373, 1963.
Maxson BB, et al.: Quality assurance for the laboratory aspects of prosthodontic treatment, J Prosthodont 6:204–209, 1997.
May KB, Razzoog ME: Silane to enhance the bond between polymethyl methacrylate and titanium, J Prosthet Dent 73:428–431, 1995.
McCartney JW: The acrylic resin base maxillary removable partial denture: technical considerations, J Prosthet Dent 43:467–468, 1980.
Mohammed H, et al.: Button versus buttonless castings for removable partial denture frameworks, J Prosthet Dent 72:433–444, 1994.
Moreno de Delgado M, Garcia LT, Rudd KD: Camouflaging partial denture clasps, J Prosthet Dent 55:656–660, 1986.
Mori T, et al.: Titanium for removable dentures. I. Laboratory procedures, J Oral Rehabil 24:238–341, 1997.
Morris HF, et al.: The influence of heat treatments on several types of base-metal removable partial denture alloys, J Prosthet Dent 41:388– 395, 1979.
NaBadalung DP, et al.: Comparison of bond strengths of denture base resins to nickel-chromium-beryllium removable partial denture alloy, J Prosthet Dent 78:566–573, 1997.
NaBadalung DP, et al.: Effectiveness of adhesive systems for Co-Cr removable partial denture alloy, J Prosthet Dent 7:17–25, Mar 1998.
Nelson DR, et al.: Expediting the fabrication of a nickel-chromium casting, J Prosthet Dent 55:400–403, 1986.
Nelson DR, von Gonten AS, Kelly Jr TW: The cast round RPA clasp, J Prosthet Dent 54:307–309, 1985.
Palmer BL, Coffey KW: Investing and packing removable partial denture bases to minimize vertical processing error, J Prosthet Dent 56:123–124, 1986.
Parr FR, Gardner LK: The removable partial denture design template, Compendium 8 594(596):598–600, 1987.
Perry CK: Transfer base for removable partial dentures, J Prosthet Dent 31:582–584, 1974.
Peyton FA, Anthony DH: Evaluation of dentures processed by different techniques, J Prosthet Dent 13:269–281, 1963.
Quinlivan JT: Fabrication of a simple ball-socket attachment, J Prosthet Dent 32:222–225, 1974.
Radue JT, Unser JW: Constructing stable record bases for removable partial dentures, J Prosthet Dent 46:463, 1981.
Rantanen T, Eerikainen E: Accuracy of the palatal plate of removable partial dentures, and influence of laboratory handling of the investment on the accuracy, Dent Mater 2:28–31, 1986.
Raskin ER: An indirect technique for fabricating a crown under an existing clasp, J Prosthet Dent 50:580–581, 1983.
Ring M: Rest seats in existing crowns, Dent Lab Rev 60:24–25, 1985.
Ryge G, Kozak SF, Fairhurst CW: Porosities in dental gold castings, J Am Dent Assoc 54:746–754, 1957.
Sarnat AE, Klugman RS: A method to record the path of insertion of a removable partial denture, J Prosthet Dent 46:222–223, 1981.
Scandrett FR, Hanson JG, Unsicker RL: Layered silicone rubber technique for flasking removable partial dentures, J Prosthet Dent 40:349–350, 1978.
Schmidt AH: Repairing chrome-cobalt castings, J Prosthet Dent 5:385–387, 1955.
Schmitt SM, Chance DA, Cronin RJ: Refining cast implant-retained restorations by electrical discharge machining, J Prosthet Dent 73:280–283, 1995.
Schneider R: Metals used to fabricate removable partial denture frameworks, J Dent Technol 13:35–42, 1996.
Schneider RL: Adapting ceramometal restorations to existing removable partial dentures, J Prosthet Dent 49:279–281, 1983.
Schneider RL: Custom metal occlusal surfaces for acrylic resin denture teeth, J Prosthet Dent 46:98–101, 1981.
Schwalm CA, LaSpina FY: Fabricating swinglock removable partial denture frameworks, J Prosthet Dent 45:216–220, 1981.
Schwedhelm ER, et al.: Fracture strength of type IV and type V die stone as a function of time, J Prosthet Dent 78:554–559, 1997.
Shay JS, Mattingly SL: Technique for the immediate repair of removable partial denture facings, J Prosthet Dent 47:104–106, 1982.
www.konkur.in 348 Appendix B Selected Reading Resources Smith GP: The responsibility of the dentist toward laboratory procedures in fixed and removable partial denture prostheses, J Prosthet Dent 13:295–301, 1963.
Smith RA: Clasp repair for removable partial dentures, J Prosthet Dent 29:231–234, 1973.
Stade EH, et al.: Influence of fabrication technique on wrought-wire clasp flexibility, J Prosthet Dent 54:538–543, 1985.
Stankewitz CG: Acrylic resin blockout for interim removable partial dentures, J Prosthet Dent 40:470–471, 1978.
Swoope CC, Frank RP: Fabrication procedures. In Clark JW, editor: Clinical dentistry, vol. 5. New York, 1976, Harper & Row.
Sykora O: A new tripoding technique, J Prosthet Dent 44:463–464, 1980.
Sykora O: Removable partial denture design by Canadian laboratories: a retrospective study, J Can Dent Assoc 61:615–621, 1995.
Tambasco J, et al.: Laser welding in the dental laboratory: an alternative to soldering, J Dent Technol 13:23–31, May 1996.
Teppo KW, Smith FW: A method of immediate clasp repair, J Prosthet Dent 30:77–80, 1975.
Tran CD, Sherraden DR, Curtis TA: A review of techniques of crown fabrication for existing removable partial dentures, J Prosthet Dent 55:671–673, 1986.
Tuccillo JJ, Nielsen JP: Compatibility of alginate impression materials and dental stones, J Prosthet Dent 25:556–566, 1971.
Ulmer FC, Ward JE: Simplified technique for production of a distal-extension removable partial denture remounting cast, J Prosthet Dent 41:473–474, 1979.
von Gonten AS, Nelson DR: Laboratory pitfalls that contribute to embrasure clasp failure, J Prosthet Dent 53:136–138, 1985.
Williams HN, Falkler Jr WA, Hasler JF: Acinetobacter contamination of laboratory dental pumice, J Dent Res 62:1073–1075, 1983.
Zalkind M, Avital R, Rehany A: Fabrication of a replacement for a broken attachment, J Prosthet Dent 51:714–716, 1984.
DENTURE ESTHETICS: TOOTH SELECTION AND ARRANGEMENT Askinas SW: Facings in removable partial dentures, J Prosthet Dent 33:633–636, 1975.
Culpepper WD: A comparative study of shade-matching procedures, J Prosthet Dent 24:166–173, 1971.
DeVan MM: The appearance phase of denture construction, Dent Clin North Am 1:255–268, 1957.
Engelmeier RL: Complete-denture esthetics, Dent Clin North Am 40:71–84, 1996.
Fields Jr H, Birtles JT, Shay J: Combination prosthesis for optimum esthetic appearance, J Am Dent Assoc 101:276–279, 1980.
French FA: The selection and arrangement of the anterior teeth in prosthetic dentures, J Prosthet Dent 1:587–593, 1951.
Frush JP, Fisher RD: How dentogenic restorations interpret the sex factor, J Prosthet Dent 6:160–172, 1956.
Frush JP, Fisher RD: How dentogenics interprets the personality factor, J Prosthet Dent 6:441–449, 1956.
Frush JP, Fisher RD: Introduction to dentogenic restorations, J Prosthet Dent 5:586–595, 1955.
Hughes GA: Facial types and tooth arrangement, J Prosthet Dent 1:82–95, 1951.
Krajicek DD: Natural appearance for the individual denture patient, J Prosthet Dent 10:205–214, 1960.
Lang BR: Complete denture occlusion, Dent Clin North Am 40:85–101, 1996.
Levin EI: Dental esthetics and the golden proportion, J Prosthet Dent 40:244–252, 1978.
Lombardi RE: Factors mediating against excellence in dental esthetics, J Prosthet Dent 38:243–248, 1977.
Myerson RL: The use of porcelain and plastic teeth in opposing complete dentures, J Prosthet Dent 7:625–633, 1957.
Payne AGL: Factors influencing the position of artificial upper anterior teeth, J Prosthet Dent 26:26–32, 1971.
Pound E: Applying harmony in selecting and arranging teeth, Dent Clin North Am 6:241–258, 1962.
Pound E: Lost—fine arts in the fallacy of the ridges, J Prosthet Dent 4:6–16, 1954.
Pound E: Recapturing esthetic tooth position in the edentulous patient, J Am Dent Assoc 55:181–191, 1957.
Roraff AR: Instant photographs for developing esthetics, J Prosthet Dent 26:21–25, 1971.
Smith BJ: Esthetic factors in removable partial prosthodontics, Dent Clin North Am 23:53–63, 1979.
Sykora O: Fabrication of a posterior shade guide for removable partial dentures, J Prosthet Dent 50:287–288, 1983.
Tillman EJ: Molding and staining acrylic resin anterior teeth, J Prosthet Dent 5:497–507, 1955. Dent Abstr 1:111, 1956.
Van Victor A: The mold guide cast: its significance in denture esthetics, J Prosthet Dent 13:406–415, 1963.
Van Victor A: Positive duplication of anterior teeth for immediate dentures, J Prosthet Dent 3:165–177, 1953.
Vig RG: The denture look, J Prosthet Dent 11:9–15, 1961.
Wallace DH: The use of gold occlusal surfaces in complete and partial dentures, J Prosthet Dent 14:326–333, 1964.
Weiner S, Krause AS, Nicholas W: Esthetic modification of removable partial denture teeth with light-cured composites, J Prosthet Dent 57:381–384, 1987.
Wolfson E: Staining and characterization of acrylic teeth, Dent Abstr 1:41, 1956.
Young HA: Denture esthetics, J Prosthet Dent 6:748–755, 1956.
Zarb GA, MacKay HF: Cosmetics and removable partial dentures: the Class IV partially edentulous patient, J Prosthet Dent 46:360–368, 1981.
DIAGNOSIS AND TREATMENT PLANNING Academy of Prosthodontics: Principles, concepts and practices in prosthodontics, J Prosthet Dent 73:73–94, 1995.
Applegate OC: Evaluating oral structures for removable partial dentures, J Prosthet Dent 11:882–885, 1961.
Bartels JC: Diagnosis and treatment planning, J Prosthet Dent 7:657–662, 1957.
Beaumont AJ: An overview of esthetics with removable partial dentures, Quintessence Int 33:745–755, 2002.
Bezzon OL, et al.: Surveying removable partial dentures: the importance of guiding planes and path of insertion for stability, J Prosthet Dent 78:412–418, 1997.
Blatterfein L, Kaufman EG: Prevention of problems with removable partial dentures: Council on Dental Materials, Instruments, and Equipment, J Am Dent Assoc 100:919–921, 1980.
Bolender CL, Swenson RD, Yamane C: Evaluation of treatment of inflammatory papillary hyperplasia of the palate, J Prosthet Dent 15:1013–1022, 1965.
Budtz-Jorgensen E: Restoration of the partially edentulous mouth: a comparison of overdentures, removable partial dentures, fixed partial dentures and implant treatment, J Dent 24:237–244, July 1996.
Casey DM, Lauciello FR: A review of the submerged-root concept, J Prosthet Dent 43:128–132, 1980.
Contino RM, Stallard H: Instruments essential for obtaining data needed in making a functional diagnosis of the human mouth, J Prosthet Dent 7:66–77, 1957.
Dreizen S: Nutritional changes in the oral cavity, J Prosthet Dent 16: 1144–1150, 1966.
Dummer PMH, Cidden J: The upper anterior sectional denture, J Prosthet Dent 41:146–152, 1979.
Dunn BW: Treatment planning for removable partial dentures, J Prosthet Dent 11:247–255, 1961.
www.konkur.in 349 Appendix B Selected Reading Resources Faine MP: Dietary factors related to preservation of oral and skeletal bone mass in women, J Prosthet Dent 73:65–72, 1995.
Foster TD: The use of the face-bow in making permanent study casts, J Prosthet Dent 9:717–721, 1959.
Frechette AR: Partial denture planning with special reference to stress distribution, J Prosthet Dent 1:700–707, (disc, 208-209), 1951.
Friedman S: Effective use of diagnostic data, J Prosthet Dent 9:729–737, 1959.
Garver DC, et al.: Vital root retention in humans: a preliminary report, J Prosthet Dent 40:23–28, 1978.
Garver DC, Fenster RK: Vital root retention in humans: a final report, J Prosthet Dent 43:368–373, 1980.
Guyer SE: Selectively retained vital roots for partial support of overdentures: a patient report, J Prosthet Dent 33:258–263, 1975.
Harvey WL: A transitional prosthetic appliance, J Prosthet Dent 14: 60–70, 1964.
Heintz WD: Treatment planning and design: prevention of errors of omission and commission, Dent Clin North Am 23:3–12, 1979.
Henderson D, Hickey JC, Wehner PJ: Prevention and preservation: the challenge of removable partial denture service, Dent Clin North Am 9:459–473, 1965.
House MM: The relationship of oral examination to dental diagnosis, J Prosthet Dent 8:208–219, 1958.
Kabcenell JL: Planning for individualized prosthetic treatment, J Prosthet Dent 34:405–407, 1975.
Kaldahl WB, Becher CM: Prosthetic contingencies for future tooth loss, J Prosthet Dent 54:1–6, 1985.
Kanno T, Carlsson GE: A review of the shortened dental arch concept focusing on the work by the Käyser/Nijmegen group, J Oral Rehabil 33:850–862, 2006.
Kayser AF: Limited treatment goals: shortened dental arches, Periodontol 2000, 4:7–14, 1994.
Killebrew RF: Crown construction and splinting of mobile partial denture abutments, J Am Dent Assoc 70:334–338, 1965.
Krikos AA: Preparing guide planes for removable partial dentures, J Prosthet Dent 34:152–155, 1975.
Lambson GO: Papillary hyperplasia of the palate, J Prosthet Dent 16:636–645, 1966.
Langer Y, et al.: Modalities of treatment for the combination syndrome, J Prosthodont 4:76–81, June 1995.
Lopes I, Norlau LA: Specific mechanics for abutment uprighting, Aust Dent J 25:273–278, 1980.
McCracken WL: Differential diagnosis: fixed or removable partial dentures, J Am Dent Assoc 63:767–775, 1961.
McGill WJ: Acquiring space for partial dentures, J Prosthet Dent 17:163–165, 1967.
Miller EL: Critical factors in selecting removable prosthesis, J Prosthet Dent 34:486–490, 1975.
Miller EL: Planning partial denture construction, Dent Clin North Am 17:571–584, 1973.
Mopsik ER, et al.: Surgical intervention to reestablish adequate intermaxillary space before fixed or removable prosthodontics, J Am Dent Assoc 95:957–960, 1977.
Moulton GH: The importance of centric occlusion in diagnosis and treatment planning, J Prosthet Dent 10:921–926, 1960.
Nassif J, Blumenfeld WL: Joint consultation services by the periodontist and prosthodontist, J Prosthet Dent 29:55–60, 1973.
Nassif J, Blumenfeld WL, Tarsitano JT: Dialogue—a treatment modality, J Prosthet Dent 33:696–700, 1975.
Payne SH: Diagnostic factors which influence the choice of posterior occlusion, Dent Clin North Am 1:203–213, 1957.
Rudd KD, Dunn BW: Accurate removable partial dentures, J Prosthet Dent 18:559–570, 1967.
Saunders TR, Gillis RE, Desjardins RP: The maxillary complete denture opposing the mandibular bilateral distal-extension partial denture: treatment considerations, J Prosthet Dent 41:124–128, 1979.
Sauser CW: Pretreatment evaluation of partially edentulous arches, J Prosthet Dent 11:886–893, 1961.
Seiden A: Occlusal rests and rest seats, J Prosthet Dent 8:431–440, 1958.
Silverman SI: Differential diagnosis: fixed or removable prosthesis, Dent Clin North Am 31:347–362, 1987.
Swoope CC, Frank RP: Removable partial dentures indications and planning. In Clark JE, editor: Clinical dentistry, vol. 5. New York, 1976, Harper & Row.
Turner CE, Shaffer FW: Planning the treatment of the complex prosthodontic case, J Am Dent Assoc 97:992–993, 1978.
Uccellani EL: Evaluating the mucous membranes of the edentulous mouth, J Prosthet Dent 15:295–303, 1965.
Vahidi F: The provisional restoration, Dent Clin North Am 31:363–381, 1987.
Wagner AG: Instructions for the use and care of removable partial dentures, J Prosthet Dent 26:481–490, 1971.
Waldron CA: Oral leukoplakia, carcinoma, and the prosthodontist, J Prosthet Dent 15:367–376, 1965.
Welker WA, Kramer DC: Claspless chrome-cobalt transitional removable partial dentures, J Am Dent Assoc 96:814–818, 1978.
Wöstmann B, et al.: Indications for removable partial dentures: a literature review, Int J Prosthodont 18:139–145, 2005.
Wright P, Hellyer PH: Gingival recession related to removable partial dentures in older patients, J Prosthet Dent 74:602–607, 1995.
Young HA: Diagnostic survey of edentulous patients, J Prosthet Dent 5:5–14, 1955.
IMPLANTS AND REMOVABLE PARTIAL DENTURES Grossmann Y, Levin L, Sadan A: A retrospective case series of implants used to restore partially edentulous patients with implant-supported removable partial dentures: 31-month mean follow-up results, Quintessence Int 39:665–671, 2008.
Grossmann Y, Nissan J, Levin L: Clinical effectiveness of implant-supported removable partial dentures: a review of the literature and retrospective case evaluation, J Oral Maxillofac Surg 67:1941– 1946, 2009.
Kaufmann R, et al.: Removable dentures with implant support in strategic positions followed for up to 8 years, Int J Prosthodont 22: 233–241, 2009.
Mijiritsky E: Implants in conjunction with removable partial dentures: a literature review, Implant Dent 16:146–154, 2007.
Mijiritsky E, et al.: Implant tooth-supported removable partial denture with at least 15-year long-term follow-up, Clin Implant Dent Relat Res, 2013.
Strassburger C, Kerschbaum T, Heydecke G: Influence of implant and conventional prostheses on satisfaction and quality of life: a literature review. Part 2. Qualitative analysis and evaluation of the studies, Int J Prosthodont 19:339–348, 2006.
IMPRESSION MATERIALS AND METHODS: THE PARTIAL DENTURE BASE Akerly WB: A combination impression and occlusal registration technique for extension-base removable partial dentures, J Prosthet Dent 39:226–229, 1978.
Appleby DC, et al.: The combined reversible hydrocolloid/irreversible hydrocolloid impression system: clinical application, J Prosthet Dent 46:48–58, 1981.
Applegate OC: An evaluation of the support for the removable partial denture, J Prosthet Dent 10:112–123, 1960.
Applegate OC: The partial denture base, J Prosthet Dent 5:636–648, 1955.
Bailey LR: Rubber base impression techniques, Dent Clin North Am 1:156–166, 1957.
Bauman R, DeBoer J: A modification of the altered cast technique, J Prosthet Dent 47:212–213, 1982.
www.konkur.in 350 Appendix B Selected Reading Resources Beaumont AJ: Sectional impression for maxillary Class I removable partial dentures and maxillary immediate dentures, J Prosthet Dent 49:438–441, 1983.
Berkey D, Berg R: Geriatric oral health issues in the United States, Int Dent J 51:254–264, 2001.
Beyerle MP, et al.: Immersion disinfection of irreversible hydrocolloid impressions. Part I. Microbiology, Int J Prosthodont 7:234–238, May 1994.
Birnbach S: Impression technique for maxillary removable partial dentures, J Prosthet Dent 51:286, 1984.
Blatterfein L, Klein IE, Miglino JC: A loading impression technique for semiprecision and precision removable partial dentures, J Prosthet Dent 43:9–14, 1980.
Boretti G, Bickel M, Geering AH: A review of masticatory ability and efficiency, J Prosthet Dent 74:400–403, 1995.
Carlsson GE: Masticatory efficiency: the effect of age, the loss of teeth and prosthetic rehabilitation, Int Dent J 34:93–97, 1984.
Chaffee NR, et al.: Dimensional accuracy of improved dental stone and epoxy resin die materials. Part I. Single die, J Prosthet Dent 77:131–135, 1997.
Chaffee NR, et al.: Dimensional accuracy of improved dental stone and epoxy resin die materials. Part II. Complete arch form, J Prosthet Dent 77:235–238, 1997.
Chai J, et al.: Clinically relevant mechanical properties of elastomeric impression materials, Int J Prosthodont 11:219–223, 1998.
Chase WW: Adaptation of rubber-base impression materials to removable denture prosthetics, J Prosthet Dent 10:1043–1050, 1960.
Chau VB, et al.: In-depth disinfection of acrylic resin, J Prosthet Dent 74:309–313, 1995.
Chen MS, et al.: An altered-cast impression technique that eliminates conventional cast dissecting and impression boxing, J Prosthet Dent 57:471–474, 1987.
Cho GC, et al.: Tensile bond strength of polyvinyl siloxane impressions bonded to a custom tray as a function of drying time, Part I, J Prosthet Dent 73:419–423, 1995.
Chong MP, et al.: The tear test as a means of evaluating the resistance to rupture of alginate impression materials, Aust Dent J 16:145–151, 1971.
Clark RJ, Phillips RW: Flow studies of certain dental impression materials, J Prosthet Dent 7:259–266, 1957.
Cohen BI, et al.: Dimensional accuracy of three different alginate impression materials, J Prosthodont 4:195–199, 1995.
Corso M, et al.: The effect of temperature changes on the dimensional stability of polyvinyl siloxane and polyether impression materials, J Prosthet Dent 79:626–631, 1998.
Cserna A, et al.: Irreversible hydrocolloids: a comparison of antimicrobial efficacy, J Prosthet Dent 71:387–389, 1994.
Davidson CL, Boere G: Liquid-supported dentures. Part I. Theoretical and technical considerations, J Prosthet Dent 68:303–306, 1990.
Davidson CL, Boere G: Liquid-supported dentures. Part II. Clinical study: a preliminary report, J Prosthet Dent 68:434–436, 1990.
Davis BA, et al.: Effect of immersion disinfection on properties of impression materials, J Prosthodont 3:31–34, 1994.
DeFreitas JF: Potential toxicants in alginate powders, Aust Dent J 25:224–228, 1980.
Dixon DL, Breeding LC, Ekstrand KG: Linear dimensional variability of three denture base resins after processing and in water storage, J Prosthet Dent 68:196–200, 1992.
Dixon DL, Ekstrand KG, Breeding LC: The transverse strengths of three denture base resins, J Prosthet Dent 66:510–513, 1991.
Dootz ER: Fabricating non-precious metal bases, Dent Clin North Am 24:113–122, 1980.
Dootz ER, Craig RG: Comparison of the physical properties of eleven soft denture liners, J Prosthet Dent 67:707–712, 1992.
Douglas CW, Shih A, Ostry L: Will there be a need for complete dentures in the United States in 2020? J Prosthet Dent 87:5–8, 2002.
Douglas CW, Watson AJ: Future needs for fixed and removable partial dentures in the United States, J Prosthet Dent 87:9–14, 2002.
Drennon DG, Johnson GH: The effect of immersion disinfection of elastomeric impressions on the surface detail reproduction of improved gypsum casts, J Prosthet Dent 63:233–241, 1990.
Fitzloff RA: Functional impressions with thermoplastic materials for reline procedures, J Prosthet Dent 52:25–27, 1984.
Frank RP: Analysis of pressures produced during maxillary edentulous impression procedures, J Prosthet Dent 22:400–403, 1969.
Fusayama T, Nakazato M: The design of stock trays and the retention of irreversible hydrocolloid impressions, J Prosthet Dent 21:136–142, 1969.
Gelbard S, et al.: Effect of impression materials and techniques on the marginal fit of metal castings, J Prosthet Dent 71:1–6, 1994.
Gilmore WH, Schnell RJ, Phillips RW: Factors influencing the accuracy of silicone impression materials, J Prosthet Dent 9:304–314, 1959.
Hans S, Gunne J: Masticatory efficiency and dental state: a comparison between two methods, Acta Odont Scand 43:139–146, 1985.
Harris Jr WT: Water temperature and accuracy of alginate impressions, J Prosthet Dent 21:613–617, 1969.
Harrison JD: Prevention of failures in making impressions and dies, Dent Clin North Am 23:13–20, 1979.
Heartwell Jr CM, et al.: Comparison of impressions made in perforated and nonperforated rimlock trays, J Prosthet Dent 27:494–500, 1972.
Helkimo E, Carlsson GE, Helkimo M: Chewing efficiency and state of the dentition, Acta Odont Scand 36:33–41, 1978.
Herfort TW, et al.: Viscosity of elastomeric impression materials, J Prosthet Dent 38:396–404, 1977.
Hesby RM, et al.: Effects of radiofrequency glow discharge on impression material surface wettability, J Prosthet Dent 77:414–422, 1997.
Holmes JB: Influence of impression procedures and occlusal loading on partial denture movement, J Prosthet Dent 15:474–481, 1965.
Hondrum SO, et al.: Effects of long-term storage on properties of an alginate impression material, J Prosthet Dent 77:601–606, 1997.
Hudson WC: Clinical uses of rubber impression materials and electroforming of casts and dies in pure silver, J Prosthet Dent 8:107–114, 1958.
Huggett R, et al.: Dimensional accuracy and stability of acrylic resin denture bases, J Prosthet Dent 68:634–640, 1992.
Iglesias A, et al.: Accuracy of wax, autopolymerized, and light-polymerized resin pattern materials, J Prosthodont 5:193–200, 1996.
Ivanovski S, et al.: Disinfection of dental stone casts: antimicrobial effects and physical property alterations, Dent Mater 11:19–23, 1995.
James JS: A simplified alternative to the altered-cast impression technique for removable partial dentures, J Prosthet Dent 53:598, 1985.
Jasim FA, Brudvik JS, Nicholls JI: Impression distortion from abutment tooth inclination in removable partial dentures, J Prosthet Dent 54:532–538, 1985.
Johnson GH, et al.: Dimensional stability and detail reproduction of irreversible hydrocolloid and elastomeric impressions disinfected by immersion, J Prosthet Dent 79:446–453, 1998.
Johnston JF, Cunningham DM, Bogan RG: The dentist, the patient, and ridge preservation, J Prosthet Dent 10:288–295, 1960.
Jones RH, et al.: Effect of provisional luting agents on polyvinyl siloxane impression materials, J Prosthet Dent 75:360–363, 1996.
Kawamura Y: Recent concepts of the physiology of mastication, Adv Oral Biol 1:77–109, 1964.
Kawano F, et al.: Comparison of bond strength of six soft denture liners to denture base resin, J Prosthet Dent 68:368–371, 1992.
Koran III A: Impression materials for recording the denture bearing mucosa, Dent Clin North Am 24:97–111, 1980.
Kramer HM: Impression technique for removable partial dentures, J Prosthet Dent 11:84–92, 1961.
Landesman HM, Wright WE: A technique for making impressions on patients requiring complete and removable partial dentures, CDA J 14:20–24, 1986.
www.konkur.in 351 Appendix B Selected Reading Resources Langenwalter EM, Aquilino SA: The dimensional stability of elastomeric impression materials following disinfection, J Prosthet Dent 63:270– 276, 1990.
Leach CD, Donovan TE: Impression technique for maxillary removable partial dentures, J Prosthet Dent 50:283–286, 1983.
Leake JL, Hawkins R, Locker D: Social and functional impact of reduced posterior dental units in older adults, J Oral Rehab 21:1–10, 1994.
Lee IK, et al.: Evaluation of factors affecting the accuracy of impressions using quantitative surface analysis, Oper Dent 20:246–252, 1995.
Lee RE: Mucostatics, Dent Clin North Am 24:81–96, 1980.
Lepe X, et al.: Accuracy of polyether and addition silicone after long-term immersion disinfection, J Prosthet Dent 78:245–249, 1997.
Leupold RJ: A comparative study of impression procedures for distal extension removable partial dentures, J Prosthet Dent 16:708–720, 1966.
Leupold RJ, Flinton RJ, Pfeifer DL: Comparison of vertical movement occurring during loading of distal-extension removable partial denture bases made by three impression techniques, J Prosthet Dent 68:290–293, 1992.
Leupold RJ, Kratochvil FJ: An altered-cast procedure to improve support for removable partial dentures, J Prosthet Dent 15:672–678, 1965.
Liedberg B, Spiechowicz E, Owall B: Mastication with and without removable partial dentures: an intraindividual study, Dysphagia 10:107–112, 1995.
Loh PL, et al.: An evaluation of microwave-polymerized resin bases for removable partial dentures, J Prosthet Dent 79:389–392, 1998.
Lucas W, Luke H: The processes of selection and breakage in mastication, Arch Oral Biol 28:813–818, 1983.
Lund PS, Aquilino SA: Prefabricated custom impression trays for the altered cast technique, J Prosthet Dent 66:782–783, 1991.
Manly RS, Vinton P: A survey of the chewing ability of denture wearers, J Dent Res 30:314–321, 1951.
Matis BA, et al.: The effect of the use of dental gloves on mixing vinyl polysiloxane putties, J Prosthodont 6:189–192, 1997.
Millar BJ, et al.: The effect of surface wetting agent on void formation in impressions, J Prosthet Dent 77:54–56, 1997.
Millar BJ, et al.: In vitro study of the number of surface defects in monophase and two-phase addition silicone impressions, J Prosthet Dent 80:32–35, 1998.
Mitchell JV, Damele JJ: Influence of tray design upon elastic impression materials, J Prosthet Dent 23:51–57, 1970.
Mitchener RW, Omori MD: Putty materials for stable removable partial denture bases, J Prosthet Dent 53:435–436, 1985.
Morrow RM, et al.: Compatibility of alginate impression materials and dental stones, J Prosthet Dent 25:556–566, 1971.
Myers GE: Electroformed die technique for rubber base impressions, J Prosthet Dent 8:531–535, 1958.
Myers GE, Wepfer GG, Peyton FA: The Thiokol rubber base impression materials, J Prosthet Dent 8:330–339, 1958.
Nishigawa G, et al.: Efficacy of tray adhesives for the adhesion of elastomer rubber impression materials to impression modeling plastics for border molding, J Prosthet Dent 79:140–144, 1998.
O’Brien WJ: Base retention, Dent Clin North Am 24:123–130, 1980.
Olin PS, et al.: The effects of sterilization on addition silicone impressions in custom and stock metal trays, J Prosthet Dent 71:625–630, 1994.
Oosterhaven SP, et al.: Social and psychological implications of missing teeth for chewing ability, Comm Dent Oral Epid 16:79–82, 1988.
Parker MH, et al.: Comparison of occlusal contacts in maximum intercuspation for two impression techniques, J Prosthet Dent 78:255–259, 1997.
Pfeiffer KA: Clinical problems in the use of alginate hydrocolloid, Dent Abstr 2:82, 1957.
Phillips RW: Factors affecting the surface of stone dies poured in hydrocolloid impressions, J Prosthet Dent 2:390–400, 1952.
Phillips RW: Factors influencing the accuracy of reversible hydrocolloid impressions, J Am Dent Assoc 43:1–17, 1951.
Phillips RW: Physical properties and manipulation of rubber impression materials, J Am Dent Assoc 59:454–458, 1959.
Pratten DH, Covey DA, Sheats RD: Effect of disinfectant solutions on the wettability of elastomeric impression materials, J Prosthet Dent 63:223–227, 1990.
Prieskel HW: Impression techniques for attachment retained distal extension removable partial dentures, J Prosthet Dent 25:620–628, 1971.
Prinz JF, Lucas PW: Swallow thresholds in human mastication, Arch Oral Biol 40:401–403, 1995.
Rapuano JA: Single tray dual-impression technique for distal extension partial dentures, J Prosthet Dent 24:41–46, 1970.
Redford M, et al.: Denture use and the technical quality of dental prostheses among persons 18-74 years of age: United States, 1988-1991, J Dent Res 75:714–725, 1996.
Render PJ: An impression technique to make a new master cast for an existing removable partial denture, J Prosthet Dent 67:488–490, 1992.
Rudd KD, et al.: Comparison of effects of tap water and slurry water on gypsum casts, J Prosthet Dent 24:563–570, 1970.
Rudd KD, Morrow RM, Bange AA: Accurate casts, J Prosthet Dent 21:545–554, 1969.
Rudd KD, Morrow RM, Strunk RR: Accurate alginate impressions, J Prosthet Dent 22:294–300, 1969.
Samadzadeh A, et al.: Fracture strength of provisional restorations reinforced with plasma-treated woven polyethylene fiber, J Prosthet Dent 5:447–450, 1997.
Scott GK, et al.: Check bite impressions using irreversible alginate/ reversible hydrocolloid combination, J Prosthet Dent 77:83–85, 1997.
Sherfudhin H, et al.: Preparation of void-free casts from vinyl polysiloxane impressions, J Dent 24:95–98, 1996.
Silver M: Impressions and silver-plated dies from a rubber impression material, J Prosthet Dent 6:543–549, 1956.
Smith RA: Secondary palatal impressions for major connector adaptation, J Prosthet Dent 24:108–110, 1970.
Steffel VL: Relining removable partial dentures for fit and function, J Prosthet Dent 4:496–509, 1954; J Tenn Dent Assoc 36:35–43, 1956.
Taylor TD, Morton Jr TJ: Ulcerative lesions of the palate associated with removable partial denture castings, J Prosthet Dent 66:213– 221, 1991.
Thompson GA, et al.: Effects of disinfection of custom tray materials on adhesive properties of several impression material systems, J Prosthet Dent 72:651–656, 1994.
Tjan AH, et al.: Marginal fidelity of crowns fabricated from six proprietary provisional materials, J Prosthet Dent 77:482–485, 1997.
Vahidi F: Vertical displacement of distal-extension ridges by different impression techniques, J Prosthet Dent 40:374–377, 1978.
van Waas M, et al.: Relationship between wearing a removable partial denture and satisfaction in the elderly, Comm Dent Oral Epid 22:315– 318, 1994.
Vandewalle KS, et al.: Immersion disinfection of irreversible hydrocolloid impressions with sodium hypochlorite. Part II. Effect on gypsum, Int J Prosthodont 7:315–322, 1994.
Verran J, et al.: Microbiological study of selected risk areas in dental technology laboratories, J Dent 24:77–80, 1996.
Wang HY, et al.: Vertical distortion on distal extension ridges and palatal area of casts made by different techniques, J Prosthet Dent 75:302–308, 1996.
Wang RR, Nguyen T, Boyle AM: The effect of tray material and surface condition on the shear bond strength of impression materials, J Prosthet Dent 74:449–454, 1995.
Wilson JH: Partial dentures: relining the saddle supported by the mucosa and alveolar bone, J Prosthet Dent 3:807–813, 1953.
Young JM: Surface characteristics of dental stone: impression orientation, J Prosthet Dent 33:336–341, 1975.
Yurkstas AA: The masticatory act, J Prosthet Dent 15:248–260, 1965.
www.konkur.in 352 Appendix B Selected Reading Resources Zinner ID: Impression procedures for the removable component of a combination fixed and removable prosthesis, Dent Clin North Am 31:417–440, 1987.
MAXILLOFACIAL PROSTHESIS Ackerman AJ: Maxillofacial prosthesis, Oral Surg 6:176–200, 1953.
Ackerman AJ: The prosthetic management of oral and facial defects following cancer surgery, J Prosthet Dent 5:413–432, 1955.
Adams D: A cantilevered swinglock removable partial denture design for the treatment of the partial mandibulectomy patient, J Oral Rehabil 12:113–118, 1985.
Brown KE: Fabrication of a hollow-bulb obturator, J Prosthet Dent 21:97–103, 1969.
Brown KE: Reconstruction considerations for severe dental attrition, J Prosthet Dent 44:384–388, 1980.
Cantor R, et al.: Methods for evaluating prosthetic facial materials, J Prosthet Dent 21:324–332, 1969.
Curtis TA, Cantor R: The forgotten patient in maxillofacial prosthetics, J Prosthet Dent 31:662–680, 1974.
Desjardins RP: Prosthodontic management of the cleft palate patient, J Prosthet Dent 33:655–665, 1975.
Firtell DN, Curtis TA: Removable partial denture design for the mandibular resection patient, J Prosthet Dent 48:437–443, 1982.
Firtell DN, Grisius RJ: Retention of obturator: removable partial dentures: a comparison of buccal and lingual retention, J Prosthet Dent 43:211–217, 1980.
Gay WD, King GE: Applying basic prosthodontic principles in the dentulous maxillectomy patient, J Prosthet Dent 43:433–435, 1980.
Goll G: Design for maximal retention of obturator prosthesis for hemimaxillectomy patients (letter), J Prosthet Dent 48:108–109, 1982.
Immekus JE, Aramy M: Adverse effects of resilient denture liners in overlay dentures, J Prosthet Dent 32:178–181, 1974.
Kelley EK: Partial denture design applicable to the maxillofacial patient, J Prosthet Dent 15:168–173, 1965.
King GE, Martin JW: Cast circumferential and wire clasps for obturator retention, J Prosthet Dent 49:799–802, 1983.
Metz HH: Mandibular staple implant for an atrophic mandibular ridge: solving retention difficulties of a denture, J Prosthet Dent 32:572–578, 1974.
Monteith GG: The partially edentulous patient with special problems, Dent Clin North Am 23:107–115, 1979.
Moore DJ: Cervical esophagus prosthesis, J Prosthet Dent 30:442–445, 1973.
Myers RE, Mitchell DL: A photoelastic study of stress induced by framework design in a maxillary resection, J Prosthet Dent 61:590–594, 1989.
Nethery WJ, Delclos L: Prosthetic stent for gold-grain implant to the floor of the mouth, J Prosthet Dent 23:81–87, 1970.
Shifman A, Lepley JB: Prosthodontic management of postsurgical soft tissue deformities associated with marginal mandibulectomy. Part I. Loss of the vestibule, J Prosthet Dent 48:178–183, 1982.
Smith Jr EH: Prosthetic treatment of maxillofacial injuries, J Prosthet Dent 5:112–128, 1955.
Strain JC: A mechanical device for duplicating a mirror image of a cast or moulage in three dimensions, J Prosthet Dent 5:129–132, 1955.
Toremalm NG: A disposable obturator for maxillary defects, J Prosthet Dent 29:94–96, 1973.
Weintraub GS, Yalisove IL: Prosthodontic therapy for cleidocranial dysostosis: report of case, J Am Dent Assoc 96:301–305, 1978.
Wright SM, Pullen-Warner EA, LeTissier DR: Design for maximal retention of obturator prosthesis for hemimaxillectomy patients, J Prosthet Dent 47:88–91, 1982.
Young JM: The prosthodontist’s role in total treatment of patients, J Prosthet Dent 27:399–412, 1972.
MISCELLANEOUS Abere DJ: Post-placement care of complete and removable partial dentures, Dent Clin North Am 23:143–151, 1979.
Academy of Denture Prosthetics: Principles, concepts and practices in prosthodontics, J Prosthet Dent 61:88–109, 1989.
Adisman IK: What a prosthodontist should know, J Prosthet Dent 21:409–416, 1969.
American Association of Dental Schools: Curricular guidelines for removable prosthodontics, J Dent Educ 44:343–346, 1980.
Applegate OC: Conditions which may influence the choice of partial or complete denture service, J Prosthet Dent 7:182–196, 1957.
Applegate OC: Factors to be considered in choosing an alloy, Dent Clin North Am 4:583–590, 1960.
Asgar K, Techow BO, Jacobson JM: A new alloy for partial dentures, J Prosthet Dent 23:36–43, 1970.
Atwood DA: Practice of prosthodontics: past, present, and future, J Prosthet Dent 21:393–401, 1970.
Augsburger RH: Evaluating removable partial dentures by mathematical equations, J Prosthet Dent 22:528–543, 1969.
Backenstose WM, Wells JG: Side effects of immersion-type cleansers on the metal components of dentures, J Prosthet Dent 37:615–621, 1977.
Baker CR: Difficulties in evaluating removable partial dentures, J Prosthet Dent 17:60–62, 1967.
Baker CR: Occlusal reactive prosthodontics, J Prosthet Dent 17:566– 569, 1967.
Barrett DA, Pilling LO: The restoration of carious clasp-bearing teeth, J Prosthet Dent 15:309–311, 1965.
Bates JF: Studies related to fracture of partial dentures, Br Dent J 120:79–83, 1966.
Beck HO: Alloys for removable partial dentures, Dent Clin North Am 4:591–596, 1960.
Beck HO: A clinical evaluation of the arcon concept of articulation, J Prosthet Dent 9:409–421, 1959.
Beck HO, Morrison WE: Investigation of an arcon articulator, J Prosthet Dent 6:359–372, 1956.
Becker CM, Bolender CL: Designing swinglock partial dentures, J Prosthet Dent 46:126–132, 1981.
Bergman B, Hugoson A, Olsson CO: Caries, periodontal and prosthetic findings in patients with removable partial dentures: a ten-year longitudinal study, J Prosthet Dent 48:506–514, 1982.
Blanco-Dalmau L: The nickel problem, J Prosthet Dent 48:99–101, 1982.
Blatterfein L, Pearce RL, Jackota JT: Minimum acceptable procedures for satisfactory removable partial denture service, J Prosthet Dent 27:84–87, 1972.
Bolender CL, Becker CM: Swinglock removable partial dentures: where and when, J Prosthet Dent 45:4–10, 1981.
Boucher CO: Writing as a means for learning, J Prosthet Dent 27: 229–234, 1972.
Budtz-Jorgensen E, Isidor F: Cantilever bridges or removable partial dentures in geriatric patients: a two-year study, J Oral Rehabil 14:239–249, 1987.
Cavalaris CJ: Pathologic considerations associated with partial dentures, Dent Clin North Am 17:585–600, 1973.
Chandler JA, Brudvik JS: Clinical evaluation of patients eight to nine years after placement of removable partial dentures, J Prosthet Dent 51:736–743, 1984.
Chen MS, et al.: Simplicity in interim tooth-supported removable partial denture construction, J Prosthet Dent 54:740–744, 1985.
Cotmore JM, et al.: Removable partial denture survey: clinical practice today, J Prosthet Dent 49:321–327, 1983.
Coy RE, Arnold PD: Survey and design of diagnostic casts for removable partial dentures, J Prosthet Dent 32:103–106, 1974.
Cunningham DM: Comparison of base metal alloys and type IV gold alloys for removable partial denture frameworks, Dent Clin North Am 17:719–722, 1973.
www.konkur.in 353 Appendix B Selected Reading Resources Diaz-Arnold AM, Langenwalter EM, Hatch LK: Cast restorations made to existing removable partial dentures, J Prosthet Dent 61: 414–417, 1989.
Dukes BS, Fields Jr H: Comparison of disclosing media used for adjustment of removable partial denture frameworks, J Prosthet Dent 45:380–382, 1981.
Elliott RW: The effects of heat on gold partial denture castings, J Prosthet Dent 13:688–698, 1963.
Ettinger RL: The acrylic removable partial denture, J Am Dent Assoc 95:945–949, 1977.
Ettinger RL, Beck JD, Jakobsen J: Removable prosthodontic treatment needs: a survey, J Prosthet Dent 51:419–427, 1984.
Ewing JE: The construction of accurate full crown restorations for an existing clasp by using a direct metal pattern technique, J Prosthet Dent 15:889–899, 1965.
Farah JW, MacGregor AR, Miller TPG: Stress analysis of disjunct removable partial dentures, J Prosthet Dent 42:271–275, 1979.
Federation of Prosthodontic Organizations: Guidelines for evaluation of completed prosthodontic treatment for removable partial dentures, J Prosthet Dent 27:326–328, 1972.
Fenton AH, Zarb GA, MacKay HF: Overdenture oversights, Dent Clin North Am 23:117–130, 1979.
Fields H, Campfield RW: Removable partial prosthesis partially supported by an endosseous blade implant, J Prosthet Dent 31:273–278, 1974.
Firtell DN, Kouyoumdjian JH, Holmes JB: Attitudes toward abutment preparation for removable partial dentures, J Prosthet Dent 55: 131–133, 1986.
Fish SF: Partial dentures, Br Dent J 128:243–246, 289–293, 339–344, 398–402, 446–453, 495–502, 547–551, 590–592, 1970.
Fisher R: Relation of removable partial denture base stability to sex, age, and other factors, J Dent Res (IADR abstract 613) 59:entire issue, 1980.
Frank RP: Evaluating refractory cast wax-ups for removable partial dentures, J Prosthet Dent 35:388–392, 1976.
Girardot RL: The physiologic aspects of partial denture restorations, J Prosthet Dent 3:689–698, 1953.
Gordon SR: Measurement of oral status and treatment need among subjects with dental prostheses: are the measures less reliable than the prostheses? Part I. Oral status in removable prosthodontics, J Prosthet Dent 65:664–668, 1991.
Harrison WM, Stansbury BE: The effect of joint surface contours on the transverse strength of repaired acrylic resin, J Prosthet Dent 23:464–472, 1970.
Heintz WD: Principles, planning, and practice for prevention, Dent Clin North Am 17:705–718, 1973.
Helel KS, Graser GN, Featherstone JD: Abrasion of enamel and composite resin by removable partial denture clasps, J Prosthet Dent 52:389–397, 1984.
Henderson CW, et al.: Evaluation of the barrier system: an infection control system for the dental laboratory, J Prosthet Dent 58:517–521, 1987.
Izikowitz L: A long-term prognosis for the free-end saddle-bridge, J Oral Rehabil 12:247–262, 1985.
Jankelson BH: Adjustment of dentures at time of insertion and alterations to compensate for tissue changes, J Am Dent Assoc 64:521–531, 1962.
Jones RR: The lower partial denture, J Prosthet Dent 2:219–229, 1952.
Kaaber S: Twelve year changes in mandibular bone level in free end saddle denture wearers, J Dent Res (IADR abstract 1367) 60:entire issue, 1981.
Kaires AK: A study of partial denture design and masticatory pressures in a mandibular bilateral distal extension case, J Prosthet Dent 8:340– 350, 1958.
Kelly E: Changes caused by a mandibular removable partial denture opposing a maxillary complete denture, J Prosthet Dent 27:140–150, 1972.
Kelly E: Fatigue failure in denture base polymers, J Prosthet Dent 21:257–266, 1969.
Kelly EK: The physiologic approach to partial denture design, J Prosthet Dent 3:699–710, 1953.
Kessler B: An analysis of the tongue factor and its functioning areas in dental prosthesis, J Prosthet Dent 5:629–635, 1955.
Klein IE, Blatterfein L, Kaufman EG: Minimum clinical procedures for satisfactory complete denture, removable partial denture, and fixed partial denture services, J Prosthet Dent 22:4–10, 1969.
Kratochvil FJ: Maintaining supporting structures with a removable partial prosthesis, J Prosthet Dent 25:167–174, 1971.
Kratochvil FJ, Caputo AA: Photoelastic analysis of pressure on teeth and bone supporting removable partial dentures, J Prosthet Dent 32:52–61, 1974.
Kratochvil FJ, Davidson PN, Guijt J: Five-year survey of treatment with removable partial dentures, Part I, J Prosthet Dent 48:237–244, 1982.
Landa JS: The troublesome transition from a partial lower to a complete lower denture, J Prosthet Dent 4:42–51, 1954.
Lanser A: Tooth-supported telescope restorations, J Prosthet Dent 45:515–520, 1981.
Lechner SK: A longitudinal survey of removable partial dentures. I. Patient assessment of dentures, Aust Dent J 30:104–111, 1985.
Lechner SK: A longitudinal survey of removable partial dentures. II. Clinical evaluation of dentures, Aust Dent J 30:194–197, 1985.
Lechner SK: A longitudinal survey of removable partial dentures. III. Tissue reactions to various denture components, Aust Dent J 30: 291–295, 1985.
Lee MW, et al.: O-ring coping attachments for removable partial dentures, J Prosthet Dent 74:235–241, 1995.
Lewis AJ: Failure of removable partial denture castings during service, J Prosthet Dent 39:147–149, 1978.
Lewis AJ: Radiographic evaluation of porosities in removable partial denture castings, J Prosthet Dent 39:278–281, 1978.
Lopuck SE, Reitz PV, Altadonna J: Hinge for a unilateral maxillary arch prosthesis, J Prosthet Dent 45:446–448, 1981.
Lorton L: A method of stabilizing removable partial denture castings during clinical laboratory procedures, J Prosthet Dent 39:344–345, 1978.
MacEntee MI, Hawbolt EB, Zahel JI: The tensile and shear strength of a base metal weld joint used in dentistry, J Dent Res 60:154–158, 1981.
Maetani T, et al.: Effect of TFE coating on plaque accumulation on dental castings, J Dent Res (IADR abstract 1359) 60:entire issue, 1981.
Maison WG: Instructions to denture patients, J Prosthet Dent 9:825–831, 1959.
Makrauer FL, Davis JS: Gastroscopic removal of a partial denture, J Am Dent Assoc 94:904–906, 1977.
Marcus SE, et al.: The retention and tooth loss in the permanent dentition of adults: United States, 1988-1991, J Dent Res 75:684–695, 1996.
Martone AL: The challenge of the partially edentulous mouth, J Prosthet Dent 8:942–954, 1958.
Martone AL: The fallacy of saving time at the chair, J Prosthet Dent 7:416–419, 1957.
Massler M: Geriatric nutrition: the role of taste and smell in appetite, J Prosthet Dent 32:247–250, 1980.
McCracken WL: Auxiliary uses of cold-curing acrylic resins in prosthetic dentistry, J Am Dent Assoc 47:298–304, 1953.
McCracken WL: A comparison of tooth-borne and tooth-tissue-borne removable partial dentures, J Prosthet Dent 3:375–381, 1953.
McCracken WL: A philosophy of partial denture treatment, J Prosthet Dent 13:889–900, 1963.
Means CR, Flenniken IE: Gagging: a problem in prosthetic dentistry, J Prosthet Dent 23:614–620, 1970.
Mehringer EJ: The saliva as it is related to the wearing of dentures, J Prosthet Dent 4:312–318, 1954.
Michell DL, Wilke ND: Articulators through the years. I. Up to 1940, J Prosthet Dent 39:330–338, 1978; II, from 1940, 39:451–458, 1978.
Miller EL: Clinical management of denture-induced inflammations, J Prosthet Dent 38:362–365, 1977.
www.konkur.in 354 Appendix B Selected Reading Resources Mohamed SE, Schmidt JR, Harrison JD: Articulators in dental education and practice, J Prosthet Dent 36:319–325, 1976.
Morris HF, Asgar K: Physical properties and microstructure of four new commercial partial denture alloys, J Prosthet Dent 33:36–46, 1975.
Neufeld JO: Changes in the trabecular pattern of the mandible following the loss of teeth, J Prosthet Dent 8:685–697, 1958.
Oatlund SG: Saliva and denture retention, J Prosthet Dent 10:658–663, 1960.
Ogle RE, Sorensen SE, Lewis EA: A new visible light-cured resin system applied to removable prosthodontics, J Prosthet Dent 56:497–506, 1986.
Osborne J, Lammie GA: The bilateral free-end saddle lower denture, J Prosthet Dent 4:640–652, 1954.
Overton RG, Bramblett RM: Prosthodontic services: a study of need and availability in the United States, J Prosthet Dent 27:329–339, 1972.
Pascoe DF, Wimmer J: A radiographic technique for the detection of internal defects in dental castings, J Prosthet Dent 39:150–157, 1978.
Phillips RW, Leonard LJ: A study of enamel abrasion as related to partial denture clasps, J Prosthet Dent 6:657–671, 1956.
Plainfield S: Communication distortion: the language of patients and practitioners of dentistry, J Prosthet Dent 22:11–19, 1969.
Prieskel HW: The distal extension prosthesis reappraised, J Dent 5:217– 230, 1977.
Ramsey WO: The relation of emotional factors to prosthodontic service, J Prosthet Dent 23:4–10, 1970.
Raybin NH: The polished surface of complete dentures, J Prosthet Dent 13:236–239, 1963.
Removable prosthodontics, Dent Clin North Am 28(entire issue), 1984.
Renggli HH, Allet B, Spanauf AJ: Splinting of teeth with fixed bridges: biological effect, J Oral Rehabil 11:535–537, 1984.
Reynolds JM: Crown construction for abutments of existing removable partial dentures, J Am Dent Assoc 69:423–426, 1964.
Rissen L, et al.: Effect of fixed and removable partial dentures on the alveolar bone of abutment teeth, J Dent Res (IADR abstract 1368) 60:entire issue, 1981.
Rissen L, et al.: Six-year report of the periodontal health of fixed and removable partial denture abutment teeth, J Prosthet Dent 54:461–467, 1985.
Rothman R: Phonetic considerations in denture prosthesis, J Prosthet Dent 11:214–223, 1961.
Rudd KD, Dunn BW: Accurate removable partial dentures, J Prosthet Dent 18:559–570, 1967.
Rushford CB: A technique for precision removable partial denture construction, J Prosthet Dent 31:377–383, 1974.
Ruyter IE, Svendsen SA: Flexural properties of denture base polymers, J Prosthet Dent 43:95–104, 1980.
Sadig W, Fahmi F: The modified swing-lock: a new approach, J Prosthet Dent 74:428–431, 1995.
Savage RD, MacGregor AR: Behavior therapy in prosthodontics, J Prosthet Dent 24:126–132, 1970.
Schabel RW: Dentist-patient communication: a major factor in treatment prognosis, J Prosthet Dent 21:3–5, 1969.
Schabel RW: The psychology of aging, J Prosthet Dent 27:569–573, 1972.
Schmitt SM: Combination syndrome: a treatment approach, J Prosthet Dent 54:307–309, 1985.
Schole ML: Management of the gagging patient, J Prosthet Dent 9: 578–583, 1959.
Schopper AF: Loss of vertical dimension: causes and effects: diagnosis and various recommended treatments, J Prosthet Dent 9:428–431, 1959.
Schopper AF: Removable appliances for the preservation of the teeth, J Prosthet Dent 4:634–639, 1954.
Schulte JK, Smith DE: Clinical evaluation of swinglock removable partial dentures, J Prosthet Dent 44:595–603, 1980.
Schuyler CH: Stress distribution as the prime requisite to the success of a partial denture, J Am Dent Assoc 20:2148–2154, 1963.
Schwarz WD, Barsby MJ: Design of partial dentures in dental practice, J Dent 6:166–170, 1978.
Sears VH: Comprehensive denture service, J Am Dent Assoc 64:531–552, 1962.
Skinner EW, Gordon CC: Some experiments on the surface hardness of dental stones, J Prosthet Dent 6:94–100, 1956.
Skinner EW, Jones PM: Dimensional stability of self-curing denture base acrylic resin, J Am Dent Assoc 51:426–431, 1955.
Smith DE: Removable prosthodontics research—quo vadis? J Prosthet Dent 62:707–711, 1989.
Smith FW, Applegate OC: Roentgenographic study of bone changes during exercise stimulation of edentulous areas, J Prosthet Dent 11:1086–1097, 1961.
Stendahl CG, Grob DJ: Detection of binding areas on removable partial denture frameworks, Dent Clin North Am 23:101–106, 1979.
Swoope CC, Frank RP: Insertion and post-insertion care. In Clark J, editor: Clinical dentistry, vol. 5. New York, 1976, Harper & Row.
Sykora O: Definitive immediate cast removable partial dentures, Can Dent Assoc J 51:767–769, 1985.
Sykora O: Extracoronal removable partial denture service in Canada, J Prosthet Dent 39:37–41, 1978.
Tallgren A: Alveolar bone loss in denture wearers as related to facial morphology, Acta Odontol Scand 28:251–270, 1970.
Taylor TD, et al.: Prosthodontic survey. I. Removable prosthodontic laboratory survey, J Prosthet Dent 52:598–601, 1984.
Taylor TD, et al.: Prosthodontic survey. II. Removable prosthodontic curriculum survey, J Prosthet Dent 52:747–749, 1984.
Teppo KW, Smith FW: A method of immediate clasp repair, J Prosthet Dent 34:77–80, 1975.
Trainor JE, Elliott Jr RW: Removable partial dentures designed by dentists before and after graduate level instruction: a comparative study, J Prosthet Dent 27:509–514, 1972.
von Gonten AS, Nelson DR: Laboratory pitfalls that contribute to embrasure clasp failure, J Prosthet Dent 53:136–138, 1985.
von Gonten AS, Palik JF: Tooth preparation guide for embrasure clasp designs, J Prosthet Dent 53:281–282, 1985.
Wagner AG: Maintenance of the partially edentulous mouth and care of the denture, Dent Clin North Am 17:755–768, 1973.
Wagner AG, Forgue EG: A study of four methods of recording the path of insertion of removable partial dentures, J Prosthet Dent 35:267–272, 1976.
Wallace DH: The use of gold occlusal surfaces in complete and partial dentures, J Prosthet Dent 14:326–333, 1964.
Walter JD: Partial denture technique. I. Introduction, Br Dent J 147:241–243, 1979; II. The purpose of the denture: choice of material, 147:302–304, 1979; III. Supporting the denture, 148:13–16, 1980; IV. Guide planes, 148:70–72, 1980.
Weaver RE, Goebel WM: Reactions to acrylic resin dental prostheses, J Prosthet Dent 43:138–142, 1980.
Whitsitt JA, Battle LW, Jarosz CJ: Enhanced retention for the distal extension-base removable partial denture using a heat-cured resilient soft liner, J Prosthet Dent 52:447–448, 1984.
Williams EO, Hartman GE: Instructional aid for teaching removable partial denture design, J Prosthet Dent 48:222, 1982.
Wise HB, Kaiser DA: A radiographic technique for examination of internal defects in metal frameworks, J Prosthet Dent 42:594–595, 1979.
Young HA: Factors contributory to success in prosthodontic practice, J Prosthet Dent 5:354–360, 1955.
Young Jr L: Try-in of the removable partial denture framework, J Prosthet Dent 46:579–580, 1981.
Zach GA: Advantages of mesial rests for removable partial dentures, J Prosthet Dent 33:32–35, 1975.
Zerosi C: A new type of removable splint: its indications and function, Dent Abstr 1:451–452, 1956.
www.konkur.in 355 Appendix B Selected Reading Resources Zurasky JE, Duke ES: Improved adhesion of denture acrylic resins to base metal alloys, J Prosthet Dent 57:520–524, 1987.
MOUTH PREPARATIONS Alexander JM, Van Sickels JE: Posterior maxillary osteotomies: an aid for a difficult prosthodontic problem, J Prosthet Dent 41:614–617, 1979.
Atwood DA: Reduction of residual ridges in the partially edentulous patient, Dent Clin North Am 17:745–754, 1973.
Axinn S: Preparation of retentive areas for clasps in enamel, J Prosthet Dent 34:405–407, 1975.
Belinfante LS, Abney Jr JM: A teamwork approach to correct a severe prosthodontic problem, J Am Dent Assoc 91:357–359, 1975.
Dixon DL, Breeding LC, Swift EJ: Use of a partial coverage porcelain laminate to enhance clasp retention, J Prosthet Dent 63:55–58, 1990.
Glann GW, Appleby RC: Mouth preparations for removable partial dentures, J Prosthet Dent 10:698–706, 1960.
Johnston JF: Preparation of mouths for fixed and removable partial dentures, J Prosthet Dent 11:456–462, 1961.
Jones RM: Dentin exposed and decay incidence in removable partial denture rest seats, Int J Prosthodont 5:227–236, 1992.
Kahn AE: Partial versus full coverage, J Prosthet Dent 10:167–178, 1960.
Kapur KK, et al.: A randomized clinical trial of two basic removable partial denture designs. Part I. Comparisons of five-year success rates and periodontal health, J Prosthet Dent 72:268–282, 1994.
Laney WR, Desjardins RP: Comparison of base metal alloys and type IV gold alloys for removable partial denture framework, Dent Clin North Am 17:611–630, 1973.
Lorey RE: Abutment considerations, Dent Clin North Am 24:63–79, 1980.
Marquardt GL: Dolder bar joint mandibular overdenture: a technique for nonparallel abutment teeth, J Prosthet Dent 36:101–111, 1976.
McArthur DR, Turvey TA: Maxillary segmental osteotomies for mandibular removable partial denture patients, J Prosthet Dent 41:381–387, 1979.
McCarthy JA, Moser JB: Mechanical properties of tissue conditioners. I. Theoretical considerations, behavioral characteristics and tensile properties, J Prosthet Dent 40:89–97, 1978.
McCarthy JA, Moser JB: Mechanical properties of tissue conditioners. II. Creep characteristics, J Prosthet Dent 40:334–342, 1978.
McCracken WL: Mouth preparations for partial dentures, J Prosthet Dent 6:39–52, 1956.
Mills M: Mouth preparation for removable partial denture, J Am Dent Assoc 60:154–159, 1960.
Mopsik ER, et al.: Surgical intervention to reestablish adequate intermaxillary space before fixed or removable prosthodontics, J Am Dent Assoc 95:957–960, 1977.
Nishimura RD: Etched metal cingulum rest retainer, J Am Dent Assoc 112:177–179, 1986.
Phillips Jr RJ: Design sequence and mouth preparation for the removable partial denture, J Calif Dent Assoc 25:363–370, 1997.
Phillips RW: Report of the Committee on Scientific Investigation of the Academy of Restorative Dentistry, J Prosthet Dent 13:515–535, 1963.
Schorr L, Clayman LH: Reshaping abutment teeth for reception of partial denture clasps, J Prosthet Dent 4:625–633, 1954.
Stamps JT, Tanquist RA: Restoration of removable partial denture rest seats using dental amalgam, J Prosthet Dent 41:224–227, 1979.
Stern WJ: Guiding planes in clasp reciprocation and retention, J Prosthet Dent 34:408–414, 1975.
Swoope CC, Frank RP: Mouth preparation. In Clark JW, editor: Clinical dentistry, vol. 5. New York, 1976, Harper & Row.
Tiege JD, et al.: In vitro investigation of the wear of resin composite materials and cast direct retainers during removable partial denture placement and removal, Int J Prosthodont 5:145–153, 1992.
Tucker KM, Heget HS: The incidence of inflammatory papillary hyperplasia, J Am Dent Assoc 93:610–613, 1976.
Wong R, Nicholls JI, Smith DE: Evaluation of prefabricated lingual rest seats for removable partial dentures, J Prosthet Dent 48:521–526, 1982.
OCCLUSION, JAW RELATION RECORDS, AND TRANSFER METHODS Applegate OC: Loss of posterior occlusion, J Prosthet Dent 4:197–199, 1954.
Baraban DJ: Establishing centric relation and vertical dimension in occlusal rehabilitation, J Prosthet Dent 12:1157–1165, 1962.
Bauman R: Minimizing postinsertion problems: a procedure for removable partial denture placement, J Prosthet Dent 42:381–385, 1979.
Beck HO: Choosing the articulator, J Am Dent Assoc 64:468–475, 1962.
Beck HO: A clinical evaluation of the arcon concept of articulation, J Prosthet Dent 9:409–421, 1959.
Beck HO: Selection of an articulator and jaw registration, J Prosthet Dent 10:878–886, 1960.
Beckett LS: Accurate occlusal relations in partial denture construction, J Prosthet Dent 4:487–495, 1954.
Berke JD, Moleres I: A removable appliance for the correction of maxillomandibular disproportion, J Prosthet Dent 17:172–177, 1967.
Berman MH: Accurate interocclusal records, J Prosthet Dent 10: 620–630, 1960.
Beyron HL: Characteristics of functionally optimal occlusion and principles of occlusal rehabilitation, J Am Dent Assoc 48:648–656, 1954.
Beyron HL: Occlusal changes in adult dentition, J Am Dent Assoc 48:674–686, 1954.
Beyron HL: Occlusal relationship, Int Dent J 2:467–496, 1952.
Block LS: Preparing and conditioning the patient for intermaxillary relations, J Prosthet Dent 2:599–603, 1952.
Block LS: Tensions and intermaxillary relations, J Prosthet Dent 4:204–207, 1954.
Boos RH: Basic anatomic factors of jaw position, J Prosthet Dent 4:200– 203, 1954.
Boos RH: Maxillomandibular relations, occlusion, and the temporo-mandibular joint, Dent Clin North Am 6:19–35, 1962.
Boos RH: Occlusion from rest position, J Prosthet Dent 2:575–588, 1952.
Borgh O, Posselt U: Hinge axis registration: experiments on the articulator, J Prosthet Dent 8:35–40, 1958.
Boucher CO: Occlusion in prosthodontics, J Prosthet Dent 3:633–656, 1953.
Braly BV: Occlusal analysis and treatment planning for restorative dentistry, J Prosthet Dent 27:168–171, 1972.
Breeding LC, et al.: Accuracy of three interocclusal recording materials used to mount a working cast, J Prosthet Dent 71:265–270, 1994.
Cerveris AR: Vibracentric equilibration of centric occlusion, J Am Dent Assoc 63:476–483, 1961.
Christensen PB: Accurate casts and positional relation records, J Prosthet Dent 8:475–482, 1958.
Clayton JA, Kotowicz WE, Zahler JM: Pantographic tracings of mandibular movements and occlusion, J Prosthet Dent 25:389–396, 1971.
Cohn LA: Factors of dental occlusion pertinent to the restorative and prosthetic problem, J Prosthet Dent 9:256–257, 1959.
Collett HA: Balancing the occlusion of partial dentures, J Am Dent Assoc 42:162–168, 1951.
Colman AJ: Occlusal requirements for removable partial dentures, J Prosthet Dent 17:155–162, 1967.
D’Amico A: Functional occlusion of the natural teeth of man, J Prosthet Dent 11:899–915, 1961.
Draper DH: Forward trends in occlusion, J Prosthet Dent 13:724–731, 1963.
www.konkur.in 356 Appendix B Selected Reading Resources Emmert JH: A method for registering occlusion in semiedentulous mouths, J Prosthet Dent 8:94–99, 1958.
Farmer JB, Connelly ME: Treatment of open occlusions with onlay and overlay removable partial dentures, J Prosthet Dent 51:300–303, 1984.
Fedi PF: Cardinal differences in occlusion of natural teeth and that of artificial teeth, J Am Dent Assoc 62:482–485, 1926.
Fountain HW: Seating the condyles for centric relation records, J Prosthet Dent 11:1050–1058, 1961.
Freilich MA, Altieri JW, Wahle JJ: Principles of selecting interocclusal records for articulation of dentate and partially dentate casts, J Prosthet Dent 68:361–367, 1992.
Gilson TD: Theory of centric correction in natural teeth, J Prosthet Dent 8:468–474, 1958.
Granger ER: The articulator and the patient, Dent Clin North Am 4:527–539, 1960.
Hansen CA, et al.: Simplified procedure for making gold occlusal surfaces on denture teeth, J Prosthet Dent 71:413–416, 1994.
Hausman M: Interceptive and pivotal occlusal contacts, J Am Dent Assoc 66:165–171, 1963.
Henderson D: Occlusion in removable partial prosthodontics, J Prosthet Dent 27:151–159, 1971.
Hindels GW: Occlusion in removable partial denture prosthesis, Dent Clin North Am 6:137–146, 1962.
Hughes GA, Regli CP: What is centric relation? J Prosthet Dent 11:16–22, 1961.
Ivanhoe JR, Vaught RD: Occlusion in the combination fixed removable prosthodontic patient, Dent Clin North Am 31:305–322, 1987.
Jankelson B: Considerations of occlusion on fixed partial dentures, Dent Clin North Am 3:187–203, 1959.
Jeffreys FE, Platner RL: Occlusion in removable partial dentures, J Prosthet Dent 10:912–920, 1960.
Kapur KK, et al.: A randomized clinical trial of two basic removable partial denture designs. Part I. Comparisons of five-year success rates and periodontal health, J Prosthet Dent 72:268–282, 1994.
Lang BR: Complete denture occlusion, Dent Clin North Am 40:85–101, 1996.
Lauritzen AG, Bodner GH: Variations in location of arbitrary and true hinge axis points, J Prosthet Dent 11:224–229, 1961.
Lay LS, et al.: Making the framework try-in, altered cast impression and occlusal registration in one appointment, J Prosthet Dent 75:446–448, Apr 1996.
Lindblom G: Balanced occlusion with partial reconstructions, Int Dent J 1:84–98, 1951.
Lindblom G: The value of bite analysis, J Am Dent Assoc 48:657–664, 1954.
Long Jr JH: Location of the terminal hinge axis by intraoral means, J Prosthet Dent 23:11–24, 1970.
Lucia VO: Centric relation theory and practice, J Prosthet Dent 10:849– 956, 1960.
Lucia VO: The gnathological concept of articulation, Dent Clin North Am 6:183–197, 1962.
Lundquist DO, Fiebiger GE: Registration for relating the mandibular cast to the maxillary cast based on Kennedy’s classification system, J Prosthet Dent 35:371–375, 1976.
Mann AW, Pankey LD: The PM philosophy of occlusal rehabilitation, Dent Clin North Am 7:621–636, 1963.
McCollum BB: The mandibular hinge axis and a method of locating it, J Prosthet Dent 10:428–435, 1960.
McCracken WL: Functional occlusion in removable partial denture construction, J Prosthet Dent 8:955–963, 1958.
McCracken WL: Occlusion in partial denture prosthesis, Dent Clin North Am 6:109–119, 1962.
Mehta JD, Joglekar AP: Vertical jaw relations as a factor in partial dentures, J Prosthet Dent 21:618–625, 1969.
Meyer FS: The generated path technique in reconstruction dentistry. I and II, J Prosthet Dent 9:354–366, 432–440, 1959.
Millstein PL, Kronman JH, Clark RE: Determination of the accuracy of wax interocclusal registrations, J Prosthet Dent 25:189–196, 1971.
Moore AW: Ideal versus adequate dental occlusion, J Am Dent Assoc 55:51–56, 1957.
Moulton GH: The importance of centric occlusion in diagnosis and treatment planning, J Prosthet Dent 10:921–926, 1960.
Nayyar A, Bill Jr JA, Twiggs SW: Comparison of interocclusal recording materials for mounting a working cast, J Dent Res (IADR abstract 1216) 60:entire issue, 1981.
Nuttall EB: Establishing posterior functional occlusion for fixed partial dentures, J Am Dent Assoc 66:341–348, 1963.
O’Leary TJ, Shanley DB, Drake RB: Tooth mobility in cuspid-protected and group-function occlusions, J Prosthet Dent 27:21–25, 1972.
Olsson A, Posselt U: Relationship of various skull reference lines, J Prosthet Dent 11:1045–1049, 1961.
Reitz PV: Technique for mounting removable partial dentures on an articulator, J Prosthet Dent 22:490–494, 1969.
Reynolds JM: Occlusal wear facets, J Prosthet Dent 24:367–372, 1970.
Ricketts RM: Occlusion: the medium of dentistry, J Prosthet Dent 21:39–60, 1969.
Robinson MJ: Centric position, J Prosthet Dent 1:384–386, 1951.
Scaife Jr RR, Holt JE: Natural occurrence of cuspid guidance, J Prosthet Dent 22:225–229, 1969.
Scandrett FR, Hanson JC: Technique for attaching the master cast to its split mounting index, J Prosthet Dent 40:467–469, 1978.
Schireson S: Grinding teeth for masticatory efficiency and gingival health, J Prosthet Dent 13:337–345, 1963.
Schuyler CH: An evaluation of incisal guidance and its influence in restorative dentistry, J Prosthet Dent 9:374–378, 1959.
Schuyler CH: Factors contributing to traumatic occlusion, J Prosthet Dent 11:708–715, 1961.
Schuyler CH: Factors of occlusion applicable to restorative dentistry, J Prosthet Dent 3:772–782, 1953.
Schuyler CH: Fundamental principles in the correction of occlusal disharmony: natural and artificial (grinding), J Am Dent Assoc 22:1193–1202, 1935.
Sears VH: Centric and eccentric occlusions, J Prosthet Dent 10:1029–1036, 1960.
Sears VH: Occlusal pivots, J Prosthet Dent 6:332–338, 1956.
Sears VH: Mandibular equilibration, J Am Dent Assoc 65:45–55, 1962.
Sears VH: Occlusion: the common meeting ground in dentistry, J Prosthet Dent 2:15–21, 1952.
Shanahan TEJ, Leff A: Interocclusal records, J Prosthet Dent 10:842– 848, 1960.
Silverman MM: Determination of vertical dimension by phonetics, J Prosthet Dent 6:465–471, 1956. Dent Abstr 2:221, 1957.
Skurnik H: Accurate interocclusal records, J Prosthet Dent 21:154–165, 1969.
Stuart CE: Accuracy in measuring functional dimensions and relations in oral prosthesis, J Prosthet Dent 9:220–236, 1959.
Teteruck WR, Lundeen HC: The accuracy of an ear face-bow, J Prosthet Dent 16:1039–1046, 1966.
Trushkowsky RD, Guiv B: Restoration of occlusal vertical dimension by means of a silica-coated onlay removable partial denture in conjunction with dentin bonding: a clinical report, J Prosthet Dent 66:283–286, 1991.
Wagner AG: A technique to record jaw relations for distally edentulous dental arches, J Prosthet Dent 29:405–407, 1973.
Weinberg LA: Arcon principle in the condylar mechanism of adjustable articulators, J Prosthet Dent 13:263–268, 1963.
Weinberg LA: An evaluation of basic articulators and their concepts. I and II, J Prosthet Dent 13:622–863, 1963.
Weinberg LA: An evaluation of the face-bow mounting, J Prosthet Dent 11:32–42, 1961.
Weinberg LA: The transverse hinge axis: real or imaginary, J Prosthet Dent 9:775–787, 1959.
www.konkur.in 357 Appendix B Selected Reading Resources PARTIAL DENTURE DESIGN Akagawa Y, et al.: A new telescopic crown system using a soldered horizontal pin for removable partial dentures, J Prosthet Dent 69:228–231, 1993.
Antos Jr EW, Tenner RP, Foerth D: The swinglock partial denture: an alternative approach to conventional removable partial denture service, J Prosthet Dent 40:257–262, 1978.
Avant EW: Indirect retention in partial denture design, J Prosthet Dent 16:1103–1110, 1966.
Axinn S, O’Connor Jr RP, Kopp EN: Immediate removable partial denture frameworks, J Am Dent Assoc 95:583–585, 1977.
Beaumont Jr AJ, Bianco HJ: Microcomputer-aided removable partial denture design: the next evolution, J Prosthet Dent 62:551–556, 1989.
Becker CW, Bolender CL: Designing swinglock partial dentures, J Prosthet Dent 46:126–132, 1981.
Becker CM, Kaiser DA, Goldfogel MH: Evolution of removable partial denture design, J Prosthodont 3:158–166, 1994.
Ben-Ur Z, et al.: Designing clasps for the asymmetric distal extension removable partial denture, Int J Prosthodont 9:374–378, July 1996.
Ben-Ur Z, et al.: Rigidity of major connectors when subjected to bending and torsion forces, J Prosthet Dent 62:557–562, 1989.
Berg E: Periodontal problems associated with use of distal extension removable partial dentures: a matter of construction? J Oral Rehabil 12:369–379, 1985.
Berg Jr T: I-bar: myth and counter-myth, Dent Clin North Am 23:65–75, 1979.
Berg Jr T, Caputo AA: Anterior rests for maxillary removable partial dentures, J Prosthet Dent 39:139–146, 1978.
Blatterfein L: A systematic method of designing upper partial denture bases, J Am Dent Assoc 46:510–525, 1953.
Blatterfein L: The use of the semiprecision rest in removable partial dentures, J Prosthet Dent 22:301–306, 1969.
Bolouri A: Removable partial denture design for a few remaining natural teeth, J Prosthet Dent 39:346–348, 1978.
Breeding L, Dixon DL: Prosthetic restoration of the anterior edentulous space, J Prosthet Dent 67:144–148, 1992.
Bridgeman JT, et al.: Comparison of titanium and cobalt-chromium removable partial denture clasps, J Prosthet Dent 78:187–193, 1997.
Brown DT, Desjardins RP, Chao EY: Fatigue failure in acrylic resin retaining minor connectors, J Prosthet Dent 58:329–335, 1987.
Browning JD, et al.: Effect of positional loading of three removable partial denture clasp assemblies on movement of abutment teeth, J Prosthet Dent 55:347–351, 1986.
Browning JD, Meadors LW, Eick JD: Movement of three removable partial denture clasp assemblies under occlusal loading, J Prosthet Dent 55:69–74, 1986.
Brudvik J, Reimers D: The tooth-removable partial denture interface, J Prosthet Dent 68:924–927, 1992.
Budtz-Jorgensen E, et al.: Alternate framework designs for removable partial dentures, J Prosthet Dent 80:58–66, July 1998.
Burns DR, Ward JE, Nance GL: Removable partial denture design and fabrication survey of the prosthodontic specialist, J Prosthet Dent 62:303–307, 1989.
Campbell LD: Subjective reactions to major connector designs for removable partial dentures, J Prosthet Dent 36:507–516, 1977.
Campbell SD, Weiner H: The hinged-clasp assembly removable partial denture, J Prosthet Dent 63:59–61, 1990.
Casey DM, Lauciello FR: A method for marking the functional depth of the floor of the mouth, J Prosthet Dent 43:108–111, 1980.
Cecconi BT: Lingual bar design, J Prosthet Dent 28:635–639, 1973.
Cecconi BT, et al.: The component partial: a new RPD construction system, J Calif Dent Assoc 25:363–370, 1997.
Cowles KR: Partial denture design: a simple teaching aid, J Prosthet Dent 47:219, 1982.
Davenport JC, et al.: The acquisition and validation of removable partial denture design knowledge. I. Methodology and overview, J Oral Rehabil 23:152–157, 1996.
Davenport JC, et al.: The acquisition and validation of removable partial denture design knowledge. II. Design rules and expert reaction, J Oral Rehabil 23:811–824, 1996.
Demer WJ: An analysis of mesial rest-I-bar clasps designs, J Prosthet Dent 36:243–253, 1976.
Dunny JA, King GE: Minor connector designs for anterior acrylic resin bases: a preliminary study, J Prosthet Dent 34:496–497, 1975.
Eick JD, et al.: Abutment tooth movement related to fit of a removable partial denture, J Prosthet Dent 57:66–72, 1987.
Ettinger RL: The acrylic removable partial denture, J Am Dent Assoc 85:945–949, 1977.
Farmer JB, et al.: Interim removable partial dentures: a modified technique, Quintessence Dent Technol 8:511–516, 1985.
Feingold FM, Grant AA, Johnson W: The effect of partial denture design on abutment tooth and saddle movement, J Oral Rehabil 13:549–557, 1986.
Firtell DN: Effect of clasp design upon retention of removable partial dentures, J Prosthet Dent 20:43–52, 1968.
Firtell DN, Grisius RJ, Muncheryan AM: Reaction of the anterior abutment of a Kennedy Class II removable partial denture to various clasp arm designs: an in vitro study, J Prosthet Dent 53:77– 82, 1985.
Fisher RL, Jaslow C: The efficiency of an indirect retainer, J Prosthet Dent 33:24–30, 1975.
Fisher RL, McDowell GC: Removable partial denture design and potential stress to the periodontium, Int J Periodont Restor Dent 4:34–47, 1984.
Frank RP: Direct retainers for distal-extension removable partial dentures, J Prosthet Dent 56:562–567, 1986.
Frank RP: An investigation of the effectiveness of indirect retainers, J Prosthet Dent 38:494–506, 1977.
Frantz WR: Variations in a removable maxillary partial denture design by dentists, J Prosthet Dent 34:625–633, 1975.
Ghamrawy E: Oral ecologic response caused by removable partial dentures, J Dent Res (IADR abstract 2898) 61:entire issue, 1982.
Ghamrawy E: Plaque formation and crevicular temperature relation to minor connector position, J Dent Res (IADR abstract 387) 61:entire issue, 1982.
Giradot RL: History and development of partial denture design, J Am Dent Assoc 28:1399–1408, 1941.
Hansen CA: Metal minibases in removable prosthodontics, J Prosthet Dent 54:442–446, 1985.
Hansen CA, Campbell DJ: Clinical comparison of two mandibular major connector designs: the sublingual bar and the lingual plate, J Prosthet Dent 54:805–809, 1985.
Henderson D: Major connectors for mandibular removable partial dentures, J Prosthet Dent 30:532–548, 1973.
Henderson D: Major connectors: united it stands, Dent Clin North Am 17:661–668, 1973.
Hero H, et al.: Ductility and structure of some cobalt-base dental casting alloys, Biomaterials 5:201–208, 1984.
Highton R, Caputo AA, Rhodes S: Force transmission and retentive capabilities utilizing labial and palatal I-bar partial dentures, J Dent Res (IADR abstract 1214) 60:entire issue, 1981.
Iwama CY, et al.: Cobalt-chromium-titanium alloy for removable partial dentures, Int J Prosthodont 10:309–317, 1997.
Jacobson TE: Rotational path partial denture design: a 10-year clinical follow-up. Parts I and II, J Prosthet Dent 71:271–282, 1994.
Jacobson TE, Krol AJ: Rotational path removable partial denture design, J Prosthet Dent 48:370–376, 1982.
Jordan LG: Designing removable partial dentures with external attachments (clasps), J Prosthet Dent 2:716–722, 1952.
www.konkur.in 358 Appendix B Selected Reading Resources Kapur KK, et al.: A randomized clinical trial of two basic removable partial denture designs. Part I. Comparisons of five-year success rates and periodontal health, J Prosthet Dent 72:268–282, 1994.
Kelly EK: The physiologic approach to partial denture design, J Prosthet Dent 3:699–710, 1953.
King GE: Dual-path design for removable partial dentures, J Prosthet Dent 39:392–395, 1978.
King GE, Barco MT, Olson RJ: Inconspicuous retention for removable partial dentures, J Prosthet Dent 39:505–507, 1978.
Knodle JM: Experimental overlay and pin partial denture, J Prosthet Dent 17:472–478, 1967.
Ko SH, McDowell GC, Kotowicz WE: Photoelastic stress analysis of mandibular removable partial dentures with mesial and distal occlusal rests, J Prosthet Dent 56:454–460, 1986.
Krikos AA: Artificial undercuts for teeth which have unfavorable shapes for clasping, J Prosthet Dent 22:301–306, 1969.
Lanser A: Telescope retainers for removable partial dentures, J Prosthet Dent 45:37–43, 1981.
Latta GH, et al.: Wear of visible light-cured restorative materials and removable partial denture direct retainers, J Prosthodont 6:104–109, 1997.
LaVere AM, Freda AL: A simplified procedure for survey and design of diagnostic casts, J Prosthet Dent 37:680–683, 1977.
LaVere AM, Krol AJ: Selection of a major connector for the extension base removable partial denture, J Prosthet Dent 30:102–105, 1973.
LaVere AM, Smith RC, Serka RJ: Cross-arch bar splint, J Prosthet Dent 67:82–84, 1992.
Lindquist TJ, et al.: Effectiveness of computer-aided partial denture design, J Prosthodont 6:122–127, 1997.
Lorencki SF: Planning precision attachment restorations, J Prosthet Dent 21:506–508, 1969.
Luk K, et al.: Unilateral rotational path removable partial dentures for tilted mandibular molars, J Prosthet Dent 78:102–105, 1997.
Marxkors R: Mastering the removable partial denture. Part I. Basic reflections about construction, J Dent Technol 14:34–39, 1997.
Marxkors R: Mastering the removable partial denture. Part II. Connection of partial dentures to the abutment teeth, J Dent Technol 14:24–30, 1997.
Maxfield JB, Nicholls JE, Smith DE: The measurement of forces transmitted to abutment teeth of removable partial dentures, J Prosthet Dent 41:134–142, 1979.
McCartney JW: Lingual plating for reciprocation, J Prosthet Dent 42:624–625, 1979.
McCracken WL: Contemporary partial denture designs, J Prosthet Dent 8:71–84, 1958.
McCracken WL: Survey of partial denture designs by commercial dental laboratories, J Prosthet Dent 12:1089–1110, 1962.
McHenry KR, Johansson DE, Christensson LA: The effect of removable partial denture framework design on gingival inflammation: a clinical model, J Prosthet Dent 68:799–803, 1992.
Meeuwissen R, Keltjens HM, Battistugzi PG: Cingulum bar as a major connector for mandibular removable partial dentures, J Prosthet Dent 66:221–223, 1991.
Monteith BD: Management of loading forces on mandibular distal-extension prostheses. I. Evaluation of concepts for design, J Prosthet Dent 52:673–681, 1984.
Monteith BD: Management of loading forces on mandibular distal-extension prostheses. II. Classification for matching modalities to clinical situations, J Prosthet Dent 52:832–836, 1984.
Myers RE, et al.: A photoelastic study of rests on solitary abutments for distal-extension removable partial dentures, J Prosthet Dent 56:702– 707, 1986.
NaBadalung DP, et al.: Frictional resistance of removable partial dentures with retrofitted resin composite guide planes, Int J Prosthodont 10:116–122, 1997.
Naim RI: The problem of free-end denture bases, J Prosthet Dent 16:522–532, 1966.
Navas MTR, del Campo ML: A new free-end removable partial denture design, J Prosthet Dent 70:176–179, 1993.
Pardo-Mindan S, Ruiz-Villandiego JC: A flexible lingual clasp as an esthetic alternative: a clinical report, J Prosthet Dent 69:245–246, 1993.
Perry C: Philosophy of partial denture design, J Prosthet Dent 6:775– 784, 1956.
Pienkos TE, et al.: The strength of multiple major connector designs under simulated functional loading, J Prosthet Dent 97:299–304, 2007.
Pipko DJ: Combinations in fixed-removable prostheses, J Prosthet Dent 26:481–490, 1971.
Potter RB, Appleby RC, Adams CD: Removable partial denture design: a review and a challenge, J Prosthet Dent 17:63–68, 1967.
Radford DR, Walter JD: A variation in minor connector design for partial denture, Int J Prosthet 6:50–53, 1993.
Russell MD, Tumer P: A three-part sectional design for an upper removable partial denture with an anterior modification, Br Dent J 162:24–26, 1987.
Rybeck Jr SA: Simplicity in a distal extension partial denture, J Prosthet Dent 4:87–92, 1954.
Schmidt AH: Planning and designing removable partial dentures, J Prosthet Dent 3:783–806, 1953.
Schuyler CH: The partial denture as a means of stabilizing abutment teeth, J Am Dent Assoc 28:1121–1125, 1941.
Schwartz RS, Murchison DG: Design variations of the rotational path removable partial denture, J Prosthet Dent 58:336–338, 1987.
Seals Jr RR, Schwartz IS: Successful integration of fixed and removable prosthodontics, J Prosthet Dent 53:763–766, 1985.
Shifman A: Use of an Adam’s clasp for a cast unilateral removable partial denture, J Prosthet Dent 61:703–705, 1989.
Shohet H: Relative magnitudes of stress on abutment teeth with different retainers, J Prosthet Dent 21:267–282, 1969.
Steffel VL: Fundamental principles involved in partial denture design, J Am Dent Assoc 42:534–544, 1951.
Steffel VL: Simplified clasp partial dentures designed for maximum function, J Am Dent Assoc 32:1093–1100, 1945.
Sykora O: Removable partial denture design by Canadian dental laboratories: a retrospective study, J Can Dent Assoc 61:615–621, 1995.
Sykora O, Calikkocaoglu S: Maxillary removable partial denture designs by commercial dental laboratories, J Prosthet Dent 22:633–640, 1970.
Tautin FS: Abutment stabilization using a nonresilient gingival bar connector, J Am Dent Assoc 99:988–998, 1979.
Thompson WD, Kratochvil FJ, Caputo AA: Evaluation of photoelastic stress patterns produced by various designs of bilateral distal-extension removable partial dentures, J Prosthet Dent 38:261–273, 1977.
Tsao DH: Designing occlusal rests using mathematical principles, J Prosthet Dent 23:154–163, 1970.
Unger JW, Badr SE: Esthetic placement of bar-clasp direct retainers, J Prosthet Dent 56:381–382, 1986.
Vallittu PK: Comparison of the in vitro fatigue resistance of an acrylic resin removable partial denture reinforced with continuous fibers or metal wires, J Prosthodont 5:115–121, 1996.
Vofa M, Kotowicz WE: Plaque retention with lingual bar and lingual plate major connectors, J Dent Res (AADR abstract 609) 59:entire issue, 1980.
Wagner AC, Traweek FC: Comparison of major connectors for removable partial dentures, J Prosthet Dent 47:242–245, 1982.
Waller NI: The root rest and the removable partial denture, J Prosthet Dent 33:16–23, 1975.
Walter JD: Alternative major connectors for mandibular partial dentures, Restorative Dent 2 80:82–84, 1986.
Warren AB, Caputo AA: Load transfer to alveolar bone as influenced by abutment design for tooth supported dentures, J Prosthet Dent 33:137–148, 1975.
www.konkur.in 359 Appendix B Selected Reading Resources Weinberg LA: Lateral force in relation to the denture base and clasp design, J Prosthet Dent 6:785–800, 1956.
Williams RJ, et al.: Use of a cast flexible plate as a hinge substitute in a hinge-lock design removable partial denture, J Prosthet Dent 80:220–223, 1998.
Zach GA: Advantages of mesial rests for removable partial dentures, J Prosthet Dent 33:32–35, 1975.
Zoller GN, et al.: Technique to improve surveying in confined areas, J Prosthet Dent 73:223–224, 1995.
PERIODONTAL CONSIDERATIONS Amsterdam M, Fox L: Provisional splinting: principles and technics, Dent Clin North Am 3:73–99, 1959.
App GR: Periodontal treatment for the removable partial prosthesis patient: another half century, Dent Clin North Am 17:601–610, 1973.
Applegate OC: The interdependence of periodontics and removable partial denture prosthesis, J Prosthet Dent 8:269–281, 1958.
Aydinlik E, Dayangac B, Celik E: Effect of splintings on abutment tooth movement, J Prosthet Dent 49:477–480, 1983.
Bates JF, Addy M: Partial dentures and plaque accumulation, J Dent 6:285–293, 1978.
Bazirgan MK, Bates JF: Effect of clasp design on gingival health, J Oral Rehabil 14:271–281, 1987.
Becker CM, Kaldahl WB: Using removable partial dentures to stabilize teeth with secondary occlusal traumatism, J Prosthet Dent 47:587– 594, 1982.
Berg TE, Caputo AA: Maxillary distal extension removable partial denture abutments with reduced periodontal support, J Prosthet Dent 70:245–250, 1993.
Bergman B: Periodontal reactions related to removable partial dentures: a literature review, J Prosthet Dent 58:454–458, 1987.
Bergman B, Ericson G: Cross-sectional study of the periodontal status of removable partial denture patients, J Prosthet Dent 61:208–211, 1989.
Brill N, et al.: Ecologic changes in the oral cavity caused by removable partial dentures, J Prosthet Dent 38:138–148, 1977.
Clarke NG: Treatment planning for fixed and removable partial dentures: a periodontal view, J Prosthet Dent 36:44–50, 1976.
Dello Russo NM: Gingival autografts as an adjunct to removable partial dentures, J Am Dent Assoc 104:179–181, 1982.
Erperstein H: The role of the prosthodontist in the treatment of periodontal disease, Int Dent J 36:18–29, 1986.
Fisher RL, McDowell GC: Removable partial denture design and potential stress to the periodontium, Int J Periodont Res Dent 4:34–47, 1984.
Garfield RE: A prosthetic solution to the periodontally compromised/ furcation involved abutment tooth, I, Quintessence Int 15:805–813, 1984.
Gilson CM: Periodontal considerations, Dent Clin North Am 24:31–44, 1980.
Gomes BC, et al.: A clinical study of the periodontal status of abutment teeth supporting swinglock removable partial dentures: a pilot study, J Prosthet Dent 46:7–13, 1981.
Gomes BC, Renner RP, Bauer PN: Periodontal considerations in removable partial dentures, J Am Dent Assoc 101:496–498, 1980.
Hall WB: Periodontal preparation of the mouth for restoration, Dent Clin North Am 24:195–213, 1980.
Hirschfeld Z, et al.: New sustained release dosage form of chlorhexidine for dental use: use for plaque control in partial denture wearers, J Oral Rehabil 11:477–482, 1984.
Isidor F, Budtz-Jorgensen E: Periodontal conditions following treatment with cantilever bridges or removable partial dentures in geriatric patients: a 2-year study, Gerodontics 3:117–121, 1987.
Ivancie GP: Interrelationship between restorative dentistry and periodontics, J Prosthet Dent 8:819–830, 1958.
Jacobson TE: Periodontal considerations in removable partial denture design, Compendium 8: 530–534,536–539, 1987.
Jordan LG: Treatment of advanced periodontal disease by prosthodontic procedures, J Prosthet Dent 10:908–911, 1960.
Kimball HD: The role of periodontia in prosthetic dentistry, J Prosthet Dent 1:286–294, 1951.
Krogh-Poulsen W: Partial denture design in relation to occlusal trauma in periodontal breakdown, Int Dent J 4:847–867, 1954; also Acad Rev 3:18–23, 1955.
McKenzie JS: Mutual problems of the periodontist and prosthodontist, J Prosthet Dent 5:37–42, 1955.
Morris ML: Artificial crown contours and gingival health, J Prosthet Dent 12:1146–1155, 1962.
Nevin RB: Periodontal aspects of partial denture prosthesis, J Prosthet Dent 5:215–219, 1955.
Orban BS: Biologic principles in correction of occlusal disharmonies, J Prosthet Dent 6:637–641, 1956.
Overby GE: Esthetic splinting of mobile periodontally involved teeth by vertical pinning, J Prosthet Dent 11:112–118, 1961.
Perel ML: Periodontal consideration of crown contours, J Prosthet Dent 26:627–630, 1971.
Picton DCA, Wills DJ: Viscoelastic properties of the periodontal ligament and mucous membrane, J Prosthet Dent 40:263–272, 1978.
Rissin L, et al.: Effect of age and removable partial dentures on gingivitis and periodontal disease, J Prosthet Dent 42:217–223, 1979.
Rudd KD, O’Leary TJ: Stabilizing periodontally weakened teeth by using guide plane removable partial dentures: a preliminary report, J Prosthet Dent 16:721–727, 1966.
Schuyler CH: The partial denture and a means of stabilizing abutment teeth, J Am Dent Assoc 28:1121–1125, 1941.
Schwalm CA, Smith DE, Erickson JD: A clinical study of patients 1 to 2 years after placement of removable partial dentures, J Prosthet Dent 38:380–391, 1977.
Seibert JS, Cohen DW: Periodontal considerations in preparation for fixed and removable prosthodontics, Dent Clin North Am 31:529– 555, 1987.
Spiekermann H: Prosthetic and periodontal considerations of free-end removable partial dentures, Int J Periodont Restor Dent 6:148–163, 1986.
Sternlicht HC: Prosthetic treatment planning for the periodontal patient, Dent Abstr 2:81–82, 1957.
Stipho HDK, Murphy WM, Adams D: Effect of oral prostheses on plaque accumulation, Br Dent J 145:47–50, 1978.
Talkov L: Survey for complete periodontal prosthesis, J Prosthet Dent 11:124–131, 1961.
Tebrock OC, et al.: The effect of various clasping systems on the mobility of abutment teeth for distal-extension removable partial dentures, J Prosthet Dent 41:511–516, 1979.
Thayer HH, Kratochvil FJ: Periodontal considerations with removable partial dentures, Dent Clin North Am 24:195–213, 1980.
Thomas BOA, Gallager JW: Practical management of occlusal dysfunctions in periodontal therapy, J Am Dent Assoc 46:18–31, 1953.
Trapozzano VR, Winter CR: Periodontal aspects of partial denture design, J Prosthet Dent 2:101–107, 1952.
Waerhaug J: Justification for splinting in periodontal therapy, J Prosthet Dent 22:201–208, 1969.
Ward HL, Weinberg LA: An evaluation of periodontal splinting, J Am Dent Assoc 63:48–54, 1961.
PHYSIOLOGY: MANDIBULAR MOVEMENT Brekke CA: Jaw function. I. Hinge rotation, J Prosthet Dent 9:600–606, 1959. II. Hinge axis, hinge axes, 9:936–940, 1959; III. Condylar placement and condylar retrusion, 10:78–85, 1960.
Brotman DN: Contemporary concepts of articulation, J Prosthet Dent 10:221–230, 1960.
Budtz-Jorgensen E: Restoration of the occlusal face height by removable partial dentures in elderly patients, Gerodontics 2:67–71, 1986.
www.konkur.in 360 Appendix B Selected Reading Resources Emig GE: The physiology of the muscles of mastication, J Prosthet Dent 1:700–707, 1951.
Fountain HW: The temporomandibular joints: a fulcrum, J Prosthet Dent 25:78–84, 1971.
Gibbs CH, et al.: Functional movements of the mandible, J Prosthet Dent 26:604–620, 1971.
Jankelson B: Physiology of human dental occlusion, J Am Dent Assoc 50:664–680, 1955.
Jemt T, Hedegard B, Wickberg K: Chewing patterns before and after treat-ment with complete maxillary and bilateral distal-extension mandibular removable partial dentures, J Prosthet Dent 50:566–569, 1983.
Kurth LE: Centric relation and mandibular movement, J Am Dent Assoc 50:309–315, 1955.
Kurth LE: Mandibular movement and articulator occlusion, J Am Dent Assoc 39:37–46, 1949.
McMillen LB: Border movements of the human mandible, J Prosthet Dent 27:524–532, 1972.
Messerman T: A concept of jaw function with a related clinical application, J Prosthet Dent 13:130–140, 1963.
Naylor JG: Role of the external pterygoid muscles in temporomandibular articulation, J Prosthet Dent 10:1037–1042, 1960.
Plotnick IJ, Beresin VE, Simkins AB: The effects of variations in the opposing dentition on changes in the partially edentulous mandible. I. Bone changes observed in serial radiographs, J Prosthet Dent 33:278–286, 1975.
Plotnick IJ, Beresin VE, Simkins AB: The effects of variations in the opposing dentition on changes in the partially edentulous mandible. III. Tooth mobility and chewing efficiency with various maxillary dentitions, J Prosthet Dent 33:529–534, 1975.
Posselt U: Movement areas of the mandible, J Prosthet Dent 7:375–385, 1957.
Posselt U: Studies in the mobility of the human mandible, Acta Odontol Scand 10:19–160, 1952.
Posselt U: Terminal hinge movement of the mandible, J Prosthet Dent 7:787–797, 1957.
Saizar P: Centric relation and condylar movement, J Prosthet Dent 26:581–591, 1971.
Schweitzer JM: Masticatory function in man, J Prosthet Dent 11:625– 647, 1961.
Shanahan TEJ: Dental physiology for dentures: the direct application of the masticatory cycle to denture occlusion, J Prosthet Dent 2:3, 1952.
Shore NA: Educational program for patients with temporomandibular joint dysfunction (ligaments), J Prosthet Dent 23:691–695, 1970.
Sicher H: Positions and movements of the mandible, J Am Dent Assoc 48:620–625, 1954.
Skinner CN: Physiology of the occlusal coordination of natural teeth, complete dentures, and partial dentures, J Prosthet Dent 17:559–565, 1967.
Sostenbo HR: CE Luce’s recordings of mandibular movement, J Prosthet Dent 11:1068–1073, 1961.
Tallgren A, Mizutani H, Tryda G: A two-year kinesiograph, study of mandibular movement patterns in denture wearers, J Prosthet Dent 62:594–600, 1989.
Ulrich J: The human temporomandibular joint: kinematics and actions of the masticatory muscles, J Prosthet Dent 9:399–406, 1959.
Vaughan HC: The external pterygoid mechanism, J Prosthet Dent 5:80– 92, 1955.
REBASING AND RELINING Blatterfein L: Rebasing procedures for removable partial dentures, J Prosthet Dent 8:441–467, 1958.
Bolouri A, et al.: A procedure for relining a complete or removable partial denture without the use of wax, J Prosthet Dent 79:604–606, May 1998.
Breeding LC, Dixon DL, Lund TS: Dimensional changes of processed denture bases after relining with three resins, J Prosthet Dent 66:650– 656, 1991.
Grady RD: Objective criteria for relining distal-extension removable partial dentures: a preliminary report, J Prosthet Dent 49:178–181, 1983.
McGivney GP: A reline technique for extension base removable partial dentures. In Lefkowitz W, editor: Proceedings of the Second International Prosthodontic Congress, St Louis, 1979, Mosby.
Steffel VL: Relining removable partial dentures for fit and function, J Prosthet Dent 4:496–509, 1954.
Turck MD, Richards MW: Microwave processing for denture relines, repairs, and rebases, J Prosthet Dent 69:340–343, 1993.
Wilson JH: Partial dentures: relining the saddle supported by the mucosa and alveolar bone, J Prosthet Dent 3:807–813, 1953.
Yasuda N, et al.: New adhesive resin to metal in removable prosthodontics field, J Dent Res (IADR abstract 213) 59:entire issue, 1980.
STRESS-BREAKER DESIGNS Bartlett AA: Duplication of precision attachment partial dentures, J Prosthet Dent 16:1111–1115, 1966.
Bickley RW: Combined splint-stress breaker removable partial denture, J Prosthet Dent 21:509–512, 1969.
Cecconi BT, Kaiser C, Rahe A: Stress-breakers and the removable partial denture, J Prosthet Dent 34:145–151, 1975.
Hansen CA, Singer MT: The segmented framework removable partial denture, J Prosthet Dent 47:765–768, 1987.
Hirschtritt E: Removable partial dentures with stress-broken extension bases, J Prosthet Dent 7:318–324, 1957.
James AC: Stress-breakers which automatically return the saddle to rest position following displacement: mandibular distal extension partial dentures, J Prosthet Dent 4:73–81, 1954.
Kabcenell JL: Stress-breaking for partial dentures, J Am Dent Assoc 63:593–602, 1961.
Kane BE: Buoyant stress equalizer, J Prosthet Dent 14:698–704, 1964.
Kane BE: Improved buoyant stress equalizer, J Prosthet Dent 17:365– 371, 1967.
Levin B: Stress-breakers: a practical approach, Dent Clin North Am 23:77–86, 1979.
Levitch HC: Physiologic stress-equalizer, J Prosthet Dent 3:232–238, 1953.
MacGregor AR: Stress-breaking in partial dentures, Aust Prosthodont Soc Bull 16:65–70, 1986.
Marris FN: The precision dowel rest attachment, J Prosthet Dent 5:43– 48, 1955.
Neill DJ: The problem of the lower free-end removable partial denture, J Prosthet Dent 8:623–634, 1958.
Plotnik IJ: Stress regulator for complete and partial dentures, J Prosthet Dent 17:166–171, 1967.
Reitz PV, Caputo AA: A photoelastic study of stress distribution by a mandibular split major connector, J Prosthet Dent 54:220–225, 1985.
Reitz PV, Sanders JL, Caputo AA: A photoelastic study of a split palatal major connector, J Prosthet Dent 51:19–23, 1984.
Simpson DH: Considerations for abutments, J Prosthet Dent 5:375–384, 1955.
Terrell WH: Split bar technique applicable to both precision attachment and clasp cases, J South Calif Dent Assoc 9:10–14, 1942.
Zinner ID: A modification of the Thompson Dowel rest for distal-extension removable partial dentures, J Prosthet Dent 61:374–378, 1989.
SURVEYING Applegate OC: Use of paralleling surveyor in modern partial denture construction, J Am Dent Assoc 27:1317–1407, 1940.
Atkinson HF: Partial denture problems: surveyors and surveying, Aust J Dent 59:28–31, 1955.
Bezzon OL, et al.: Surveying removable partial dentures: the importance of guiding planes and path of insertion for stability, J Prosthet Dent 78:412–418, 1997.
www.konkur.in 361 Appendix B Selected Reading Resources Chestner SC: A methodical approach to the analysis of study cases, J Prosthet Dent 4:622–624, 1954.
Hanson JC: Surveying, J Am Dent Assoc 91:826–828, 1975.
Katulski EM, Appleyard WN: Biological concepts of the use of the mechanical cast surveyor, J Prosthet Dent 7:627–634, 1959.
Knapp JC, Shotwell JL, Kotowicz WE: Technique for recording dental cast-surveyor relations, J Prosthet Dent 41:352–354, 1979.
McCarthy MF: An intraoral surveyor, J Prosthet Dent 61:462–464, 1989.
Solle W: An improved dental surveyor, J Am Dent Assoc 60:727–731, 1960.
Wagner AC, Forque EC: A study of four methods of recording the path of insertion of removable partial dentures, J Prosthet Dent 35:267–272, 1976.
Yilmaz C: Optical surveying of casts for removable partial dentures, J Prosthet Dent 34:292–296, 1975.
Zoller GN, et al.: Technique to improve surveying in confined areas, J Prosthet Dent 73:223–224, Feb 1995.
WORK AUTHORIZATIONS Brown ET: The dentist, the laboratory technician, and the prescription law, J Prosthet Dent 15:1132–1138, 1965.
Dutton DA: Standard abbreviations (and definitions) for use in dental laboratory work authorizations, J Prosthet Dent 27:94–95, 1972.
Gehl DH: Investment in the future, J Prosthet Dent 18:190–201, 1968.
Henderson D: Writing work authorizations for removable partial dentures, J Prosthet Dent 16:696–707, 1966.
Henderson D, Frazier Q: Communicating with dental laboratory technicians, Dent Clin North Am 14:603–615, 1970.
Leeper SH: Dentist and laboratory: a love-hate relationship, Dent Clin North Am 23:87–99, 1979.
Quinn L: Status of the dental laboratory work authorization, J Am Dent Assoc 79:1189–1190, 1969.
Travaglini EA, Jannetto LB: A work authorization format for removable partial dentures, J Am Dent Assoc 6:429–431, 1978.
www.konkur.in |
188758 | https://courses.lumenlearning.com/tulsacc-collegealgebrapclc/chapter/determinants-and-cramers-rule/ | Determinants and Cramer’s Rule
Determinants of 2-by-2 Square Matrices
The determinant of a 2×2 square matrix is a mathematical construct used in problem solving that is found by a special formula.
Learning Objectives
Practice finding the determinant of a 2×2 matrix
Key Takeaways
Key Points
The determinant of a 2×2 matrix [abcd] is defined to be ad−bc.
A matrix is often used to represent the coefficients in a system of linear equations, and the determinant can be used to solve those equations.
Any matrix has a unique inverse if its determinant is nonzero.
Key Terms
determinant: The unique scalar function over square matrices which is distributive over matrix multiplication, multilinear in the rows and columns, and takes the value of 1 for the unit matrix. Its abbreviation is “det“.
What is a determinant?
A matrix is often used to represent the coefficients in a system of linear equations, and the determinant can be used to solve those equations. The use of determinants in calculus includes the Jacobian determinant in the change of variables rule for integrals of functions of several variables. Determinants are also used to define the characteristic polynomial of a matrix, which is essential for eigenvalue problems in linear algebra. In analytical geometry, determinants express the signed n-dimensional volumes of n-dimensional parallelepipeds. Sometimes, determinants are used merely as a compact notation for expressions that would otherwise be unwieldy to write down.
It can be proven that any matrix has a unique inverse if its determinant is nonzero. Various other theorems can be proved as well, including that the determinant of a product of matrices is always equal to the product of determinants; and, the determinant of a Hermitian matrix is always real.
The determinant of a matrix [A] is denoted det(A), det A, or |A|. In the case where the matrix entries are written out in full, the determinant is denoted by surrounding the matrix entries by vertical bars instead of the brackets or parentheses of the matrix.
For instance, the determinant of the matrix [abde] is written ∣∣∣abde∣∣∣.
Determinant of a 2-by-2 Matrix
In linear algebra, the determinant is a value associated with a square matrix. It can be computed from the entries of the matrix by a specific arithmetic expression, shown below:
For a 2×2 matrix, [abcd],
the determinant ∣∣∣abcd∣∣∣ is defined to be ad−bc.
Example 1: Find the determinant of the following matrix:
[4−275]
The determinant ∣∣∣4−275∣∣∣ is:
(4⋅5)−(−2⋅7)=20−(−14)=34
Cofactors, Minors, and Further Determinants
The cofactor of an entry (i,j) of a matrix A is the signed minor of that matrix.
Learning Objectives
Explain how to use minor and cofactor matrices to calculate determinants
Key Takeaways
Key Points
Let A be an m×n matrix and k an integer with [latex]0
The first minor of a matrix Mij is formed by removing the ith row and jth column of the matrix, and retrieving the determinant of the smaller matrix.
The cofactor of an element aij of a matrix A, written as Cij is defined as (−1)i+jMij.
Key Terms
cofactor: The signed minor of an entry of a matrix.
minor: The determinant of some smaller square matrix, cut down from matrix A by removing one or more of its rows or columns.
Cofactor and Minor: Definitions
Cofactor
In linear algebra, the cofactor (sometimes called adjunct) describes a particular construction that is useful for calculating both the determinant and inverse of square matrices. Specifically the cofactor of the (i,j) entry of a matrix, also known as the (i,j) cofactor of that matrix, is the signed minor of that entry.
The cofactor of aij entry of a matrix is defined as:
Cij=(−1)i+jMij
Minor
To know what the signed minor is, we need to know what the minor of a matrix is. In linear algebra, a minor of a matrix A is the determinant of some smaller square matrix, cut down from A by removing one or more of its rows or columns. Minors obtained by removing just one row and one column from square matrices (first minors) are required for calculating matrix cofactors.
Let A be an m×n matrix and k an integer with [latex]0Calculate the Determinant
The determinant of any matrix can be found using its signed minors. The determinant is the sum of the signed minors of any row or column of the matrix scaled by the elements in that row or column.
Calculating the Minors
The following steps are used to find the determinant of a given minor of a matrix A:
Choose an entry aij from the matrix.
Cross out the entries that lie in the corresponding row i and column j.
Rewrite the matrix without the marked entries.
Obtain the determinant of this new matrix.
Mij is termed the minor for entry aij.
Note: If i+j is an even number, the cofactor coincides with its minor: Cij=Mij. Otherwise, it is equal to the additive inverse of its minor: Cij=−Mij
Calculating the Determinant
We will find the determinant of the following matrix A by calculating the determinants of its cofactors for the third, rightmost column and then multiplying them by the elements of that column.
⎡⎢⎣147305−1911⎤⎥⎦
As an example, we will calculate the determinant of the minor M23, which is the determinant of the 2×2 matrix formed by removing the 2nd row and 3rd column. A black dot represents an element we are removing.
∣∣
∣∣14∙∙∙∙−19∙∣∣
∣∣=∣∣∣14−19∣∣∣=(9−(−4))=13
Since i+j=5 is an odd number, the cofactor is the additive inverse of its minor: −(13)=−13
We multiply this number by a23=5, giving −65.
The same process is carried out to find the determinants of C13 and C33, which are then multiplied by a13 and a33, respectively. The determinant is then found by summing all of these:
detA=a13detC13+a23detC23+a33detC33=7⋅27−5⋅13+11⋅−12=−8
Cramer’s Rule
Cramer’s Rule uses determinants to solve for a solution to the equation Ax=b, when A is a square matrix.
Learning Objectives
Use Cramer’s Rule to solve for a single variable in a system of linear equations
Key Takeaways
Key Points
Cramer’s Rule only works on square matrices that have a non-zero determinant and a unique solution.
Consider the linear system {ax+by=ecx+dy=f, which in matrix format is [abcd][xy]=[ef]. Assume the determinant is non-zero. Then, x and y and be found by Cramer’s rule: x=∣∣∣ebfd∣∣∣∣∣∣abcd∣∣∣=ed−bfad−bc and y=∣∣∣aecf∣∣∣∣∣∣abcd∣∣∣=af−ecad−bc.
Cramer’s Rule is efficient for solving small systems and can be calculated quite quickly; however, as the system grows, calculating the new determinants can be tedious.
Key Terms
determinant: The unique scalar function over square matrices which is distributive over matrix multiplication, multilinear in the rows and columns, and takes the value of 1 for the unit matrix. Its abbreviation is “det“.
square matrix: A matrix having the same number of rows as columns.
“Cramer’s Rule” is another way to solve a system of linear equations with matrices. It uses a formula to calculate the solution to the system utilizing the definition of determinants.
Cramer’s Rule: Definition
Cramer’s Rule is an explicit formula for the solution of a system of linear equations with as many equations as unknowns, i.e. a square matrix, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the vector of right hand sides of the equations.
Cramer’s Rule: Formula
Rules for a 2×2 Matrix
Consider the linear system:
[abcd][xy]=[ef]
Assume the determinant is non-zero. Then, x and y can be found by Cramer’s rule:
x=∣∣∣ebfd∣∣∣∣∣∣abcd∣∣∣=ed−bfad−bc
And:
y=∣∣∣aecf∣∣∣∣∣∣abcd∣∣∣=af−ecad−bc
Rules for a 3×3 Matrix
Given:
⎡⎢⎣abcdefghi⎤⎥⎦⎡⎢⎣xyz⎤⎥⎦=⎡⎢⎣jkl⎤⎥⎦
Then the values of x, y and z can be found as follows:
x=∣∣
∣∣jbckeflhi∣∣
∣∣∣∣
∣∣abcdefghi∣∣
∣∣y=∣∣
∣∣ajcdkfgli∣∣
∣∣∣∣
∣∣abcdefghi∣∣
∣∣z=∣∣
∣∣abjdekghl∣∣
∣∣∣∣
∣∣abcdefghi∣∣
∣∣
Using Cramer’s Rule
Example 1: Solve the system using Cramer’s Rule:
{3x+2y=10−6x+4y=4
In matrix format:
[32−64][xy]=
x=∣∣∣ebfd∣∣∣∣∣∣abcd∣∣∣=ed−bfad−bc
x=∣∣∣10244∣∣∣∣∣∣32−64∣∣∣=10⋅4−2⋅4(3⋅4)−[2⋅(−6)]=3224=43
y=∣∣∣aecf∣∣∣∣∣∣abcd∣∣∣=af−ecad−bc
y=∣∣∣310−64∣∣∣∣∣∣32−64∣∣∣=(3⋅4)−10⋅(−6)−[2⋅(−6)]=7224=3
The solution to the system is (43,3). |
188759 | https://www.youtube.com/watch?v=jrFLb9ZoZH0 | Multiplicity of zeros of polynomials | Polynomial graphs | Algebra 2 | Khan Academy
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The polynomial p(x)=(x-1)(x-3)² is a 3rd degree polynomial, but it has only 2 distinct zeros. This is because the zero x=3, which is related to the factor (x-3)², repeats twice. This is called multiplicity. It means that x=3 is a zero of multiplicity 2, and x=1 is a zero of multiplicity 1. Multiplicity is a fascinating concept, and it is directly related to graphical behavior of the polynomial around the zero.
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Transcript:
so what we have here are two different polynomials p1 and p2 and they have been expressed in factored form and you can also see their graphs this is the graph of y is equal to p1 of x in blue and the graph of y is equal to p2x in white what we're going to do in this video is continue our study of zeros but we're going to look at a special case when something interesting happens with their with the zeros so let's just first look at p1's zeros so i'll set up a little table here because it'll be useful so the first column let's just make it the zeros the x values at which at which our polynomial is equal to zero and that's pretty easy to figure out from factored form when x is equal to one the whole thing is going to be equal to zero because zero times anything is zero when x is equal to two by the same argument and when x is equal to three and we can see it here on the graph when x equals one the graph of y is equal to p1 intersects the x-axis it does it again at the next zero x equals two and at the next zero x equals three we can also see the property that between consecutive zeros our function our polynomial maintains the same sign so between these first two or actually before this first zero it's negative then between these first two it's positive then the next two it's negative and then after that it is positive now what about p2 well p2 is interesting because if you were to multiply this out it would have the same degree as p1 in either case you would have an x to the third term you would have a third degree polynomial but how many zeros how many distinct unique zeros does p2 have pause this video and think about that well let's just list them out so our zeros well once again if x equals one this whole expression is going to be equal to zero so we have zero at x equals one and we can see that our white graph also intersects the x-axis at x equals one and then if x is equal to three this whole thing's going to be equal to zero and we can see that it intersects the x-axis at x equals three and then notice this next this next part of the expression would say oh well we have zero at x equals three but we already said that so we actually have two zeros for a third degree polynomial so something very interesting is happening in some ways you could say that hey it's trying to reinforce that we have a zero at x minus three and this notion of having multiple parts of our factored form that would all point to the same zero that is the idea of multiplicity so let me write this word down so multiplicity multiple i'll write it out there and i will write it over here multiplicity and so for each of these zeros we have a multiplicity of one they're only they're only deduced one time when you look at it at factored form only one of the factors points to each of those zeros so they all have a multiplicity of one for p2 the first zero has a multiple of one only one of the expressions points to a zero of one or would become zero if x would be equal to one but notice out of our factors when we have it in factored form out of our factored expressions or our expression factors i should say two of them become zero when x is equal to three this one and this one are going to become zero and so here we have a multiplicity of two and i encourage you to pause this video again and look at the behavior of graphs and see if you can see a difference between the behavior of the graph when we have a multiplicity of one versus when we have a multiplicity of two all right now let's look through it together we can look at p1 where all of the zeros have a multiplicity of one and you can see every time we have a zero we are crossing the x-axis not only are we intersecting it but we are crossing it we're crossing the x-axis there we're crossing it again and we're crossing it again so at all of these we have a sign change around that zero but what happens here well on the first zero that has a multiplicity of one that only makes one of the factors equal zero we have a sign change just like we saw with p1 but what happens at x equals 3 where we have a multiplicity of 2. well there we intersect the x axis still p of 3 is 0 but notice we don't have a sign change we were positive before and we are positive after we touch the x-axis right there but then we go back up and the general idea and i encourage you to test this out and think about why this is true is that if you have an odd multiplicity and let me write this down if the multiplicity is odd so if it's 1 3 5 7 etc then you're going to have a sign change sign change while if it is even as the case of two or four or six you're going to have no sign change no sign no sign change one way to think about it in an example where you have a multiplicity of two so let's just use this zero here where x is equal to three when x is less than three both of these are going to be negative and a negative times a negative is a positive and when x is greater than 3 both of them are going to be positive and so in either case you have a positive so notice you saw no sign change another thing to appreciate is thinking about the number of zeros relative to the degree of the polynomial and what you see is is that the number of zeros number of zeros is at most equal to the degree of the polynomial so it is going to be less than or equal to the degree of the polynomial and why is that the case well you might not all your zeros might have a multiplicity of one in which case the number of zeros is equal is going to be equal to the degree of the polynomial but if you have a zero that has a higher than one multiplicity well then you're going to have fewer distinct zeros another way to think about it is if you were to add all the multiplicities that that is going to be equal to the degree of your polynomial |
188760 | https://encrypt.a41.io/primitives/modular-arithmetic/modular-reduction/barrett-reduction | Barrett Reduction
,The primary goal of Barrett reduction is to perform modular reduction efficiently. The standard calculation, x−⌊x/n⌋⋅n, involves a division (x/n), which can be slow. Barrett reduction aims to replace this division.
The Core Idea
Barrett reduction uses multiplications, subtractions, and bit shifts (fast division by powers of 2) instead of a general division. It achieves this using a precomputed value based on the modulus n to approximate the quotient value ⌊x/n⌋.
How it works
Setup and Precomputation
Suppose we are working with integers base b. (Typically b=2 in computers). Consider an integer k such that bk>n. Often, k is chosen such that bk is roughly n2 (e.g., if n fits in w bits, k=2w is common). Now, we precompute the value m:
m
⌊
b
k
/
n
⌋
which acts as a scaled approximation of division by n.
Approximating the Quotient
The true remainder is r=x−q⋅n, where the true quotient is q=⌊x/n⌋. So in Barrett's method we would like to estimate q without dividing x by n.
Consider the product x⋅m. Substituting the definition of m, we get:
x
⋅
m
x
⋅
⌊
b
k
/
n
⌋
≤
x
⋅
(
b
k
/
n
)
Dividing both sides by bk (which is a right shift by k if b=2):
(
x
⋅
m
)
/
b
k
≤
(
x
⋅
b
k
/
n
)
/
b
k
x
/
n
This gives us the approximation to q:
⌊(
x
⋅
m
)
/
b
k
⌋
≤
⌊
x
/
n
⌋
=
q
Let's denote this approximation as q^:
q
^
=
⌊(
x
⋅
m
)
/
b
k
⌋
≤
q
How Close is the Approximation?
Notice that we can write bk=⌊bk/n⌋⋅n+r′ for some remainder r′<n. Solving for m, we havem=(bk−r′)/n. Replacing m for the q^ equation, we get:
q
^
=
⌊(
x
⋅
m
)
/
b
k
⌋
=
⌊
x
⋅
b
k
(
b
k
−
r
′
)
/
n
⌋
⌊(
x
/
n
)
−
(
x
⋅
r
′
/
(
n
⋅
b
k
))⌋
Comparing q^ to q=⌊x/n⌋, the difference depends on the term (x⋅r′/(n⋅bk)), which is small if bk is large enough relative to x.
For example, if bk>n2, then (x⋅r′/(n⋅bk))<1 which gives us a nice tight bound q−1≤q^ so we need only 1 subtraction at the worst case. If we choose a smaller k such that bk>n, q^ can get as low as q−2, but typically we choose a large bk>n2.
Final Reduction Algorithm
Precomputation (once for fixed n):
Choose k (e.g., k=2⋅bitlength(n)). We assume b=2.
Calculate m=⌊2k/n⌋.
Reduction (for each x):
Calculate q^=⌊(x⋅m)/2k⌋ (intermediate product x⋅m might need k+bitlength(x/n) bits; often only the higher bits are needed since we do a k bit-shift afterwards).
Calculate r^=x−q^⋅n.
Correction: while r^≥n, r^=r^−n
Cost Analysis of Modular Multiplication
Single-Precision case
Suppose the modulus n has bit width of w<word_size. Then, the cost looks like:
Computing x=a⋅b: multiplication of bitwidth w×w→2w .
Computing x⋅m: multiplication of bitwidth 2w×(k−w)→w+k.
Computing q^⋅n: multiplication of bitwidth w×w→2w.
TODO(BATZORIG ZORIGOO): add multi-precision case
Written by BATZORIG ZORIGOO from A41
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Modular Reduction
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Montgomery Reduction
Last updated |
188761 | https://artofproblemsolving.com/wiki/index.php/Pascal%27s_Identity?srsltid=AfmBOopKP5IhMVXx1sSsRgg3WhPCTFDY-_riNb8cWL4QQHLhOpWYBDSI | Art of Problem Solving
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Pascal's Identity
Pascal's Identity is a useful theorem of combinatorics dealing with combinations (also known as binomial coefficients). It can often be used to simplify complicated expressions involving binomial coefficients.
Pascal's Identity is also known as Pascal's Rule, Pascal's Formula, and occasionally Pascal's Theorem.
Contents
[hide]
1 Theorem
2 Proof
3 Alternate Proofs
4 History
5 See Also
6 External Links
Theorem
Pascal's Identity states that
for any positive integers and . Here, is the binomial coefficient .
This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things.
Proof
If then and so the result is trivial. So assume . Then
Alternate Proofs
Here, we prove this using committee forming.
Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways.
Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways.
But we already know they can be picked in ways, so
Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, .
History
Pascal's identity was probably first derived by Blaise Pascal, a 17th century French mathematician, whom the theorem is named after.
Pascal also did extensive other work on combinatorics, including work on Pascal's triangle, which bears his name. He discovered many patterns in this triangle, and it can be used to prove this identity. The method of proof using that is called block walking.
See Also
Combinatorics
Combination
Committee forming
Combinatorial_identity#Hockey-Stick_Identity
External Links
Pascal's Identity at Planet Math
Pascal's Identity at Wolfram's Math World
Retrieved from "
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Combinatorics
Theorems
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188762 | https://www.quora.com/Whats-the-difference-between-disk-method-washer-method-and-shell-method-in-calculus | Something went wrong. Wait a moment and try again.
Solids of Revolution
Integration Process
Washer Method
Disk (mathematics)
Shell Method
Calculus Computations
Volume of Revolution
Calculus 1
5
What's the difference between disk method, washer method and shell method in calculus?
Razin Sazzad
Lives in Bozeman, MT
·
6y
They are all methods of finding out the volume of solids of revolution using a definite integral.
They are all methods of finding out the volume of solids of revolution using a definite integral.
Related questions
Which is the better method, the shell method or disk integration?
What are the differences between the shell, washer, and disks methods in calculus, and what makes them better for different problems?
How do I figure which type of volume problem is necessary for disk method, shell method, and washer method? Also give examples of problems and solution along with rectangles.
When should I use The Washer Method or the Disk Method in Calculus?
Can we replace all problems using the shell method with the washer method?
Brian Smith
I've been editing and writing math ed materials for 15 yrs
·
Upvoted by
Musunuri Bhavana
, BS-MS Mathematics, Indian Institute of Science Education and Research, Pune (2024) · Author has 1.5K answers and 1.4M answer views
·
6y
The disk method uses an infinitesimally thick slice of the area beneath a curve and rotates it around an axis to create a circle. That’s why you’ll see πr2 in the formula.
The washer method uses the disk method twice, once to find the exterior volume, and again to subtract the vacated interior volume. That’s why you see a subtraction problem in the formula.
Shells uses the area beneath the curve, but it rotates it to create a cylinder instead of a circle. That’s why they’re called shells, and why you will see 2π times the radius (to get the circumference) multiplied by the height (to ge
The disk method uses an infinitesimally thick slice of the area beneath a curve and rotates it around an axis to create a circle. That’s why you’ll see πr2 in the formula.
The washer method uses the disk method twice, once to find the exterior volume, and again to subtract the vacated interior volume. That’s why you see a subtraction problem in the formula.
Shells uses the area beneath the curve, but it rotates it to create a cylinder instead of a circle. That’s why they’re called shells, and why you will see 2π times the radius (to get the circumference) multiplied by the height (to get the surface area of the cylinder).
Nick Henderson
6y
Originally Answered: What's the difference between washer method and shell method in calculus?
·
The main difference between the washer and shell methods in calculus is the orientation to the axis of rotation. The washer method you use a dx if you rotate around the x axis. The shell method, you use dy for rotation around the x axis.
The washer method is used between two curves. It creates a disc with a hole in it essentially, so there is some thickness to the washer. The shell method has only one curve and a distance from the axis. This yields a thin layer of the surface. The washer method gives you a disc with a smaller disc taken out from the middle with thickness given by the difference
The main difference between the washer and shell methods in calculus is the orientation to the axis of rotation. The washer method you use a dx if you rotate around the x axis. The shell method, you use dy for rotation around the x axis.
The washer method is used between two curves. It creates a disc with a hole in it essentially, so there is some thickness to the washer. The shell method has only one curve and a distance from the axis. This yields a thin layer of the surface. The washer method gives you a disc with a smaller disc taken out from the middle with thickness given by the difference of the two discs’ radii. The shell method is a hollow disc with thickness of 1 point.
David Kaplan
Ph.D. in Mathematics, University of South Florida
·
Author has 556 answers and 368.1K answer views
·
6y
Disks slice a volume of rotation around the x-axis with circular disks centered on the x-axis, with area pi[f(x)]^2 and volume pi[f(x)]^2dx. With dx infinitesimally narrow the integral sums these disks to give the exact area.
Washers use the same method as disks but computer the volume when the figure has a hollow parrallel along the x-axis. The calculation is outer volume minis inner volume to give the volume between.
The shell method divides a volume into concentric cans of infintessilimal thickness. Think of each can as having the volume of a Folger's coffee can without the coffee, the top
Disks slice a volume of rotation around the x-axis with circular disks centered on the x-axis, with area pi[f(x)]^2 and volume pi[f(x)]^2dx. With dx infinitesimally narrow the integral sums these disks to give the exact area.
Washers use the same method as disks but computer the volume when the figure has a hollow parrallel along the x-axis. The calculation is outer volume minis inner volume to give the volume between.
The shell method divides a volume into concentric cans of infintessilimal thickness. Think of each can as having the volume of a Folger's coffee can without the coffee, the top, or the bottom. Snip it and lay it flat. It's volume is it's width times height times thickness. The width is 2piradius, with the radius being “x" when it's a revolution about the y-axis. The height is f(x) and the thickness is dx. The volume is obtained by summing / integrating the infinitely many concentric can volumes, each one 2pixf(x). When the revolution is about x = a instead of the y-axis, the radius changes from x to (a — x).
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Related questions
What is the difference between the methods for calculating volume in calculus like disk, washer, cross section, shell, etc.?
When is disk or washer method of calculus used in real life?
How do you deal with problems in life?
Can cylinder shell method replace disk and washer methods?
When do you use the shell method vs the washer method?
Mike Saleh
J.S.D in Bullshitology & More Bullshitology, Lalloopanjoo University (Graduated 1900)
·
Author has 152 answers and 171.2K answer views
·
6y
The disk method is typically easier when evaluating revolutions around the x-axis, whereas the shell method is easier for revolutions around the y-axis---especially for which the final solid will have a hole in it (hence shell).
The disk method is:
V=π∫(r(x))^2dx (limits of integral a to b)
The shell method is:
V=2π∫xf(x)dx (limits of integral a to b)
While the disk method is about stacking disks of varying radii and shape (defined by the revolution of r(x) along the x-axis at each x , the shell method is about vertically layering rings (defined by 2πx, where x is the radius of the ring) of varying
The disk method is typically easier when evaluating revolutions around the x-axis, whereas the shell method is easier for revolutions around the y-axis---especially for which the final solid will have a hole in it (hence shell).
The disk method is:
V=π∫(r(x))^2dx (limits of integral a to b)
The shell method is:
V=2π∫xf(x)dx (limits of integral a to b)
While the disk method is about stacking disks of varying radii and shape (defined by the revolution of r(x) along the x-axis at each x , the shell method is about vertically layering rings (defined by 2πx, where x is the radius of the ring) of varying thickness and shape f(x).
ka kahath ho baboova? Lallan taap bahut marat ho!
Joshua Love
Studied Physics (college major) & Mathematics (Graduated 2020)
·
Author has 253 answers and 501.2K answer views
·
6y
Without being too technical they all achieve the same goal of calculating volumes of rotated solids. However, they are derived from different geometric interpretations of volume in mind. Firstly, the disk and washer method are basically the same. The washer method is a generalization of the disk method such that when g(x)=0 you get the formula for the disk method. More so, volume can be thought of as stacking up cross sections of area to some height h or a volume can be thought of the tight meshing of various surface areas or tightly packing various sized shells within one another sort of like
Without being too technical they all achieve the same goal of calculating volumes of rotated solids. However, they are derived from different geometric interpretations of volume in mind. Firstly, the disk and washer method are basically the same. The washer method is a generalization of the disk method such that when g(x)=0 you get the formula for the disk method. More so, volume can be thought of as stacking up cross sections of area to some height h or a volume can be thought of the tight meshing of various surface areas or tightly packing various sized shells within one another sort of like Russian dolls. The disk and washer method conceive of volume as the summation of circular cross sections or stacking up disk for some length and the shell method conceives of volume as the tightly packing shells together of various sizes. For example, to futher illustrate what I mean by packing surface area you can think of the volume of sphere as the encapsulation of hollow spheres or surface areas with radius lengths that range from 0 to R. 0 being the center of the circle R being the radius of the sphere and the largest outer most shell or hollow sphere. Think Russian dolls and that's how the shell method works.
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Mangalam Gopal
Ph.D in Mathematics, University of Michigan (Graduated 1967)
·
Author has 283 answers and 290.4K answer views
·
6y
These are used in computing the volumes of solids generated when a plane region is rotated about the coordinate axes or horizontal or vertical lines.
For example, consider the plane region R in the first quadrant bounded below by the X-axis and above by the curve y = f(x), for x in [a, b].
Suppose R is revolved about the X-axis. By dividing the region using vertical strips, the elementary volume will be the volume of a DISK with radius y and thickness dx: Pi y^2 dx.
On the other hand, if R is rotated about the Y-axis, then the elementary volume will be the volume of a cylindrical SHELL with radiu
These are used in computing the volumes of solids generated when a plane region is rotated about the coordinate axes or horizontal or vertical lines.
For example, consider the plane region R in the first quadrant bounded below by the X-axis and above by the curve y = f(x), for x in [a, b].
Suppose R is revolved about the X-axis. By dividing the region using vertical strips, the elementary volume will be the volume of a DISK with radius y and thickness dx: Pi y^2 dx.
On the other hand, if R is rotated about the Y-axis, then the elementary volume will be the volume of a cylindrical SHELL with radius x, height y and thickness dx: 2 Pi x y dx.
Finally, if R is rotated about the line y = -2, then the elementary volume is the volume of a WASHER with outer radius y+2 , inner radius 2 and thickness dx: Pi [(2+ y)^2 - 4] dx.
Murray Eisenberg
Former Professor (now Emeritus) at University of Massachusetts, Amherst (1965–present)
·
Author has 193 answers and 154.4K answer views
·
6y
A partial answer: The so-called disk method and washer method are just special instances of finding the volume of a solid in case the solid is a solid of revolution. Both are just special cases of the more general notion that volume is obtained by integrating the cross-sectional area formed by slicing parallel planes through the solid.
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Marisa Manno
Studied at University of Pittsburgh (Graduated 2022)
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6y
Originally Answered: When do you know whether to use shell method or washer method in calculus?
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Either method works for every problem. Which one is easier depends on the specific problem. When you look at the diagram you can decide if it would be easier if the region is split up into vertical or horizontal sections. Deciding that comes easier with doing more problems and drawing out the region and splitting it up
Jeff Suzuki
Mathematician and math historian
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Author has 2K answers and 2.7M answer views
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3y
Basically the shape of the representative volume.
The disk method uses short, stubby cylinders (AKA disks); the shell method uses slabs wound into a circle (AKA shells).
The washer method is not really a separate method. Rather it’s the mathematical answer to “How do I carve an elephant from a block of marble?”
(The answer is “Remove everything that isn’t the elephant”)
Charles G. Allensworth
I've taken up to Cal II Differential Equations per Coursework.
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Author has 117 answers and 288.6K answer views
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9y
Related
When should I use The Washer Method or the Disk Method in Calculus?
The disk method and the washer method are very similar. Probably the best way to think of it as trying to calculate the volume for a roll of quarters. The quarters do not have holes (at least American quarters... coins for non-US folks).
So you would calculate the volume of the cylinder made up of those quarters by finding the diameter of that quarter, and dividing it by two for the radius. You would then, of course, calculate the thickness of the quarters. As the volume of a cylinder is height times pi times the radius-squared (or one-half the diameter squared) you would have the approxima
The disk method and the washer method are very similar. Probably the best way to think of it as trying to calculate the volume for a roll of quarters. The quarters do not have holes (at least American quarters... coins for non-US folks).
So you would calculate the volume of the cylinder made up of those quarters by finding the diameter of that quarter, and dividing it by two for the radius. You would then, of course, calculate the thickness of the quarters. As the volume of a cylinder is height times pi times the radius-squared (or one-half the diameter squared) you would have the approximate volume of a single quarter. Once you have the volume of your squashy little cylinder, multiply by the number of quarters in that roll.
Now for the fun part: what if ALL the quarters have the same sized hole in the center? You would calculate the radius for that hole, so that the volume of the quarter minus the volume for that hole equals the volume of your punctured quarters.
As you are probably aware of, in calculus, when you integrate a function, the resulting function can be used to calculate the area under a curve between two given points a and b.
fig. 1 from wikipedia, smurf-blue is the area under the curve.
This is great for finding a 2-d area on it's size (an infinite number of wires standing on end next to each other)... but we need volume.
And of course, the volume of a cylinder is h • pi • r-squared.
For a cylinder, there are smooth sides and the radius is fixed and constant.... but what about for a jar? What about a vase whose curvy outside looks like the shape above? The radius, of course, varies from the outer edge to the center.
Now, realize that the area of a circle is simply 2•pi•r integrated to get surface area
=> (2•pi•r^2)/2 = pi•r^2 and, again, times the height gives the volume of a cylinder.
As our radius varies, and the radius is defined by a function (f(x) = y-value = radius at a given point), if we integrate pi(fx)^2... we get a volume. In this case a wobbly sided vase filled with cement, composed of an infinite number of 2d disks stacked on top of one another.
=>V=pi∙∫ba (f(x))2dx
The same works in case you have an f(y), swap x for y, and you have the volume for that cement-filled vase.
For the washer method, such as there were holes in those quarters, there is another function that defines that hole. Imagine you pulled out the cement, (carefully), out of that vase so that there wasn't a spot of cement left in the vase and the cement was entirely intact. The volume of the cement is the same as the volume of empty space within the vase. So, want to find out the volume for the ceramic that composes the vase?
=>Voutervaseedge=pi∙∫ba (f(xvase))2dx
=>Vcement=pi∙∫ba (f(xcement))2dx
=>Voutervaseedge−Vcement=Vvase′s−ceramic
=> Volume of outer edge of a washer minus the volume of the washer's hole.
=> Volume of glass composing a cup => Volume created by outer edge - volume of orange juice capacity (to the brim!)
Most people, though, just notate it as integrate f(outer edge radius)^2 - f(inner edge)^2 dx or dy if the function is f(y) for those outer and inner edges:
=>Vcylindrical−shape=pi(∫ba (f(xouter))2−(f(xinner))2)dx)
fig2. From the University of Bakersfield.
While this is for equations rotated about the x-axis it also works for rotated about the y-axis. Usually for those, you're integrating f(y). Sometimes, if given an f(x), finding an f(y) is done by, for example:
f(x)=x2 => y = x2=>√y=x=f(y)
One thing to remember, as in the above image, if you were to rotate about the y-axis, you would need to probably find the relative maxima for the two intervals, and figure out the volume in two steps. Your washers shouldn't have more than one hole. Sometimes it's easier to use what's called the "shell" method aka the "onion" method.
I apologize for being verbose but I hope this helps!
Phil Scovis
I play guitar. And sing in the car.
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Author has 6.9K answers and 13.3M answer views
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9y
Related
How does one know whether to use the disk, washer, or shell methods?
The disk method is used when the curve y=f(x) is revolved around the x-axis. The shell method is used when the curve y=f(x) is revolved around the y-axis.
If the curve is x=f(y), use the shell method for revolving around the x-axis, and the disk method for revolving around the y-axis.
If either the disk or shell method produce an integral that is too difficult, you can find the inverse of the function, and switch to the other method.
The “washer” method is just the disk method applied to two curves, and subtract the results. It is applied when you would normally use the disk method, but there is
The disk method is used when the curve y=f(x) is revolved around the x-axis. The shell method is used when the curve y=f(x) is revolved around the y-axis.
If the curve is x=f(y), use the shell method for revolving around the x-axis, and the disk method for revolving around the y-axis.
If either the disk or shell method produce an integral that is too difficult, you can find the inverse of the function, and switch to the other method.
The “washer” method is just the disk method applied to two curves, and subtract the results. It is applied when you would normally use the disk method, but there is a hollow created by a second curve.
Hope that helps.
Alex Vaegle
Chemistry/ Calculus Tutor (sometimes)
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Author has 150 answers and 403.9K answer views
·
7y
Related
How does one know whether to use the disk, washer, or shell methods?
This was actually one of the most challenging topics for me to learn about in calculus. Probably because none of the textbooks or online sources didn’t just come out and say what to do in plain English:
When everything is in terms of x, integrate using dx and when everything is in terms of y, integrate using dy.
This was actually one of the most challenging topics for me to learn about in calculus. Probably because none of the textbooks or online sources didn’t just come out and say what to do in plain English:
When everything is in terms of x, integrate using dx and when everything is in terms of y, integrate using dy.
Related questions
Which is the better method, the shell method or disk integration?
What are the differences between the shell, washer, and disks methods in calculus, and what makes them better for different problems?
How do I figure which type of volume problem is necessary for disk method, shell method, and washer method? Also give examples of problems and solution along with rectangles.
When should I use The Washer Method or the Disk Method in Calculus?
Can we replace all problems using the shell method with the washer method?
What is the difference between the methods for calculating volume in calculus like disk, washer, cross section, shell, etc.?
When is disk or washer method of calculus used in real life?
How do you deal with problems in life?
Can cylinder shell method replace disk and washer methods?
When do you use the shell method vs the washer method?
When should you use the cylindrical shell method vs the disk and washer method?
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188763 | https://m1maths.com/P2-3-venn-diagrams.pdf | M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 1 M1Maths.com homepage Other probability modules Probability: P2-3 Venn Diagrams • using Venn diagrams to determine numbers and probabilities Summary Learn Solve Revise Answers Summary A Venn diagram shows which of two or three groups certain items or people belong to. Venn diagrams can be used to work out probabilities. Learn Venn Diagrams with Dots A Venn diagram is a visual way of showing which of two or three groups certain items or people belong to. For example, a school might have 12 teachers. Suppose 4 of them teach English, 3 teach maths and 5 teach neither. We can show this visually on a Venn diagram like this: The way the diagram works is that those who teach English are represented by dots in the English circle; those who teach Maths are represented by dots in the maths circle and those who teach neither are represented by dots in neither circle. But suppose one of the Maths teachers was given an English class as well. That teacher then needs to be in both circles. Giving her two dots might be misleading because we would have 13 dots and so might think that there are 13 teachers. Instead we make the circles overlap, so that one dot can be in both circles, like this: English Maths M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 2 This Venn diagram now has 5 dots in the English circle, showing that 5 of the teachers teach English. It has 3 dots in the Maths circle, showing that 3 of the teachers teach Maths. And it has 5 teachers not in either circle, showing that 5 teachers don’t teach either. And there are still just 12 dots showing that there are just 12 teachers. Practice Q1 This Venn diagram shows the students in Class 8C who have a cat or a dog at home. (a) How many have a cat and a dog? (b) How many have a cat? (c) How many don’t have a cat? (d) How many have a cat, but not a dog? (e) How many have a dog? (f) How many don’t have a dog? (g) How many have a dog but not a cat? (h) How many have neither? (i) How many students in the class? Make sure you can get all parts of Q1 right before you go on. Venn Diagrams with Numbers Now, using dots is ok when there are just a few people. But suppose the Venn diagram for English and Maths teachers above was for a big school where 12 teachers teach just English, 14 teach just maths, 5 teachers teach both English and Maths and 49 teachers teach neither. Using dots would be messy and anyone who wants to read the diagram will have to do a lot of counting. So instead, we tend to just write the numbers in each region of the Cat Dog English Maths M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 3 diagram instead of using dots. In fact, we tend to use numbers even when the numbers are small. So the Venn diagram for the big school would look like this: Practice Q2 This Venn diagram shows the students that play guitar and volleyball at Cabbage Tree High School. (a) How many play just guitar? (b) How many play guitar? (c) How many don’t play guitar? (d) How many play volleyball? (e) How many play volleyball but not guitar? (f) How many don’t play volleyball? (g) How many play guitar and volleyball? (h) How many play neither? (i) How many students in the school? Drawing Venn Diagrams We can draw Venn diagrams from information about how many people are in each group. For example, if we are told that there were 22 people in a club, that 5 of them were over 50, 13 of them were female and 6 of them were neither, we can draw the diagram like this: English Maths 12 14 5 49 Guitar Volleyball 1 2 3 2 8 106 Over 50 Female 1 1 2 3 6 M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 4 Note that we have to work out that there are 2 in the overlap by seeing that 5 + 13 + 6 = 24 and thus that 2 people must have been counted twice. Once we know this, we can work out that there must be 3 in the rest of the Over 50 circle and 11 in the rest of the female section. Practice Q3 Draw a Venn diagram to show the following information. There are 17 people in the swimming club. 5 race freestyle, 4 race breast stroke, 10 race neither. Q4 Use your Venn diagram from Q3 to find out: (a) How many race freestyle and breast stroke. (b) How many race freestyle but not breast stroke. (c) How many race breast stroke but not freestyle. (d) How many race freestyle or breast stroke or both. Q5 Draw a Venn diagram to show the following. Of the 30 people in Rudolf’s class, 14 have a pool at home, 21 have a barbeque and 10 have both. Q6 Use your Venn diagram from Q5 to find out: (a) How many have neither. (b) How many have a pool but not a barbeque. (c) How many have a barbeque but not a pool. (d) How many don’t have a pool. Q7 In a group of 12 sailors, 4 were sea sick and 2 of those were drunk as well. If only one was neither, draw a suitable Venn diagram to show this information. Venn Diagrams with 3 Circles Venn diagrams can be drawn with 3 circles. You may not need to be able to deal with these, but if you can master them, the 2-circle ones will seem easy. The Venn diagram to the right shows the students in a class who study Music, French and Art. Music French 3 1 2 4 6 1 1 3 Art M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 5 Note that, to find out how many people do Music, we add up all the numbers inside the Music circle. There are 7. Likewise, we can see that 11 do art and 6 do French. Of the students who do French, 2 do Art. One students does all three subjects. And so on. The good news is that it’s not geometrically possible to have a Venn diagram with more than 3 circles that still shows all the possible regions. Practice Q8 Use this 3-circle Venn diagram for Class 8A to answer the questions below. (a) How many students in the class? (b) How many girls in the class? (c) How many boys? (Assume all students are either a boy or a girl.) (d) How many blonde girls are there? (e) How many blonde girls over 165 cm? (f) How many blonde boys over 165 cm? (g) How many students are <165 cm? (h) How many girls are >165 cm? (i) How many blonde girls <165 cm? (j) How many blonde boys < 165 cm? (k) How many dark-haired boys <165 cm? (Assume anyone not blonde is dark-haired.) Q9 Of the 15 people on a bus, 7 had dogs, 4 had cats and 5 had birds. 3 had both dogs and cats and one of those had birds as well. One had just cats and 2 had just birds. Present this information as a Venn diagram. Q10 Of the 18 people in a restaurant, all but 6 are having potatoes. 6 are having beef and potatoes. 4 are having fish and potatoes and 3 are having beef and fish, one of whom is also having potatoes. More are having beef than fish. Another 3 are having none of those. Present this information on a Venn diagram. Blonde >165 cm tall 2 3 5 6 6 2 1 4 Female M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 6 Probability Consider this Venn diagram of people on a tour bus who have a cold and who have a fungal infection on their feet. If you got on the bus and sat next to someone at random, what is the probability that they would have a cold? It would be the number of people who have a cold divided by the total number of people on the bus. From the diagram, we can see that the number with a cold is 11 and that the total number on the bus is 22. So the probability that the person you sit next to has a cold is 11/22 or 50%. Note that this is the same question as ‘What fraction of the people on the bus have a cold?’ Other probabilities can be worked out in the same way. Suppose you got on the bus and sat next to someone with a cold. What is the probability that they would have a fungal infection too? Well, there are 11 people with a cold. 5 of those also have a fungal infection. So the probability would be 5/11. Practice Q11 Use the Cold-Fungal infection Venn diagram above to answer these questions. (a) What is the probability that a random person on the bus would have a fungal infection? (b) What is the probability that a random person on the bus would have a cold and a fungal infection? (c) What is the probability that a random person on the bus would have neither? (d) What is the probability that a random person on the bus would have a cold, but not a fungal infection? (e) If you got on the bus and sat next to someone with a cold, what is the probability that they will have a fungal infection too? (f) If you got on the bus and sat next to someone with a fungal infection, what is the probability that they will have a cold too? (g) If you got on the bus and sat next to someone with a cold, what is the probability that they won’t have a fungal infection? Cold Fungal infection 2 1 5 6 9 M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 7 (h) If you got on the bus and sat next to someone without a cold, what is the probability that they will have a fungal infection? (i) If you got on the bus and sat next to someone with a fungal infection, what is the probability that they won’t have a cold? Below is a Venn diagram showing the numbers of people in a class who are studying French, Chemistry and Music. If we took one of the music students at random, what is the probability that they would be studying chemistry as well? As you can probably tell, to work this out, we look at how many music students there are – 11. Then we look at how many of these study chemistry – 3. Then the probability is 3/11. Practice Q12 Use the French – Chemistry – Music Venn diagram above to answer these questions. (a) What is the probability that a chemistry student picked at random would be studying music? (b) What fraction of the French students do chemistry? (c) What fraction of the chemistry students don’t do French? (d) If you picked a student who doesn’t do music, what is the probability that they would do French? (e) If you picked a student who doesn’t do French, what is the probability that they wouldn’t do any of these three subjects? (f) If you picked a student who doesn’t do chemistry, what is the probability that they don’t do music? (g) If you pick a student who does music, what is the probability that they do French, but not chemistry? French Chemistry 9 4 4 7 5 2 1 3 Music M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 8 (h) What fraction of those doing chemistry do French or music, but not both? Solve Q51 Harry has 22 pets. 11 are tortoises, 5 are guinea pigs, 7 weigh more than 500 g and this include 4 of the tortoises. 3 fit none of these categories. Present this information as a Venn diagram. Q52 If you picked a whole number at random, what is the probability that it would be rational number? Q53 If you picked a rational number at random, what is the probability that it would be an integer? Q54 If you picked an integer at random, what is the probability that it would be a whole number? Revise Revision Set 1 Q61 This Venn diagram shows how many people who live in Caylee Street like rock music and how many like Rap. (a) How many like rock and rap? (b) How many like rock but not rap? (c) How many like neither? (d) If you picked someone at random from the street, what is the probability that they would like rap? (e) What fraction of people in the street like rap? (f) If you picked someone at random from the street, what is the probability that they would like rap but not rock? (g) If you picked someone at random from the street who likes rock, what is the probability that they would also like rap? (h) If you picked someone at random from the street who doesn’t like rock, what is the probability that they would like rap? (i) If you picked someone at random from the street who doesn’t like rap, what is the probability that they wouldn’t like rock either? Rock Rap 2 1 5 6 9 M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 9 Q62 This Venn diagram is of the people who live in Arundel Street. It shows how many have a barbeque, how many have an indoor fire and how many smoke. (a) How many have an indoor fire? (b) How many have a barbeque but don’t smoke? (c) How many smoke, but don’t have an indoor fire or a barbeque? (d) If you picked someone from the street at random, what if the probability that they would smoke? (e) If you picked a smoker from the street, what is the probability that they would have a fire? (f) If you picked a smoker with a fire, what is the probability that they would have a barbeque? Q63 Draw a Venn diagram to show the following information. 5 of my 12 friends own a cat and 8 own a dog. 3 own neither. Q64 Use your Venn diagram from the last question to answer the following (a) How many own both? (b) If you picked a dog owner at random, what would be the probability that they would own a cat? Q65 Draw a Venn diagram to show the following information: Of the 32 people in a life boat, 12 are sea sick, 5 have a head ache and 11 are children. There are 2 sea sick children with headaches. 5 of the children are well and 6 of them are sea sick. There are 3 sea sick adults without headaches. Q66 Use your Venn diagram from Q73 (check it’s right first) to answer these questions. (a) What fraction of the children have headaches? (b) What fraction of the adults are well? (c) If you picked a sea sick passenger at random, what is the probability that they would have a head ache? (d) I you picked an adult at random, what is the probability that they would not be sea sick? Answers Q1 (a) 2 (b) 7 (c) 7 (d) 5 (e) 6 (f) 8 (g) 4 (h) 3 (i) 14 Q2 (a) 1 (b) 9 (c) 109 (d) 11 (e) 3 (f) 107 (g) 8 (h) 106 (i) 118 Barbeque Fire 1 4 5 7 2 1 0 2 Smoke M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 10 Q3 Q4 (a) 2 (b) 3 (c) 2 (d) 7 Q5 Q6 (a) 5 (b) 4 (c) 11 (d) 16 Q7 Q8 (a) 29 (b) 13 (c) 16 (d) 6 (e) 2 (f) 3 (g) 21 (h) 3 (i) 4 (j) 5 (k) 6 Q9 Q10 Q11 (a) 7/22 (b) 5/22 (c) 9/22 (d) 6/22 (e) 5/11 (f) 5/7 (g) 6/11 (h) 2/11 (i) 2/7 Q12 (a) 3/16 (b) 6/13 (c) 10/16 (d) 8/24 (e) 7/22 (f) 11/19 (g) 3/11 (h) 5/16 Freestyle Breast stroke 2 1 2 3 10 Pool Barbeque 11 1 10 4 5 Drunk Sea sick 2 2 7 1 Dogs Cats 1 2 2 5 2 1 0 2 Birds Beef Fish 0 2 1 3 3 1 3 5 Potatoes M1Maths.com Other Probability Modules P2-3 Venn Diagrams Page 11 Q51 Q52 1. All whole numbers are rational Q53 0. There are an infinite number of rational numbers between each pair of integers. Q54 ½ The 0 will become insignificant among the infinite number of integers and whole numbers. Q61 (a) 5 (b) 6 (c) 9 (d) 7/22 (e) 7/22 (f) 2/22 (g) 5/11 (h) 2/11 (i) 9/15 Q62 (a) 6 (b) 9 (c) 2 (d) 5/22 (e) 6/22 (f) 1 Q63 Q64 (a) 2 (b) 4/8 Q65 Q66 (a) 2/11 (b) 15/21 (c) 5/12 (d) 15/21 Tortoises Guinea Pigs 5 0 7 3 3 0 0 4 > 500 g Sea sick Head ache 0 3 3 15 5 2 0 4 Children Cat Dog 4 4 1 3 |
188764 | https://pubmed.ncbi.nlm.nih.gov/15034230/ | Effects of zolpidem and zaleplon on sleep, respiratory patterns and performance at a simulated altitude of 4,000 m - PubMed
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. 2004;49(3):154-62.
doi: 10.1159/000076723.
Effects of zolpidem and zaleplon on sleep, respiratory patterns and performance at a simulated altitude of 4,000 m
Maurice Beaumont1,Denise Batéjat,Olivier Coste,Pascal Van Beers,Anthony Colas,Jean-Michel Clère,Christophe Piérard
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1 Department of Integrative Physiology, Institut de Médecine Aérospatiale du Service de Santé des Armées (IMASSA), Brétigny-sur-Orge, France. mbeaumont@imassa.fr
PMID: 15034230
DOI: 10.1159/000076723
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Clinical Trial
Effects of zolpidem and zaleplon on sleep, respiratory patterns and performance at a simulated altitude of 4,000 m
Maurice Beaumont et al. Neuropsychobiology.2004.
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. 2004;49(3):154-62.
doi: 10.1159/000076723.
Authors
Maurice Beaumont1,Denise Batéjat,Olivier Coste,Pascal Van Beers,Anthony Colas,Jean-Michel Clère,Christophe Piérard
Affiliation
1 Department of Integrative Physiology, Institut de Médecine Aérospatiale du Service de Santé des Armées (IMASSA), Brétigny-sur-Orge, France. mbeaumont@imassa.fr
PMID: 15034230
DOI: 10.1159/000076723
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The effects of zolpidem or zaleplon on sleep architecture, respiratory patterns and performance were assessed at a simulated altitude of 4,000 m. Twelve male healthy subjects spent 4 nights in a decompression chamber, 1 at sea level (baseline), 3 at 4,000 m to test zolpidem (10 mg), zaleplon (10 mg) and placebo, given 15 min before switching the lights off. Sleep and respiratory patterns were analysed using polysomnography. Cognitive and physical performance was examined the next morning at sea level conditions. The study demonstrates that both zolpidem and zaleplon improved slow wave sleep at altitude, with zolpidem showing more marked effects than zaleplon. Both agents did not adversely affect respiration at altitude during the night, or cognitive or physical performance the next morning at the dosages used in this study. Thus, climbers may safely use both hypnotic agents.
Copyright 2004 S. Karger AG, Basel
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188765 | https://www.geeksforgeeks.org/maths/how-to-calculate-dice-probabilities/ | How to calculate Dice Probabilities?
Last Updated :
23 Jul, 2025
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Probability is the chance or likelihood of an event happening. It is represented by a number between 0 and 1. The higher the probability, the greater the chances of the event occurring.
A probability of 0 means the event is impossible and cannot happen.
A probability of 1 means the event is certain to happen.
The probability value can never exceed 1.
[Probability in Maths]
Probability Formula
The Formula for Probability is:
P(A)=Total number of possible outcomesNumber of favorable outcomes
➣ Check Probability Formulas in detail: [Learn Here]
Here, we will discuss how to calculate the probabilities for rolling one die and two dice.
DICE
Dice are small, throwable objects used in games of chance, usually with six faces numbered 1 to 6. In probability, dice are used to calculate the chances of getting specific outcomes, like a certain number or a particular sum when rolling one or more dice.
One Die Roll
The simplest case of dice probability involves calculating the chance of rolling a specific number on a fair die. A standard die has six faces, numbered from 1 to 6, with each outcome being equally likely.
The basic formula is:
P(A) = {Number of favourable outcomes / Total number of possible outcomes}
For example, the probability of rolling a specific number, say a 1, is:
P(1) = 1/6
This probability is the same for each of the numbers 1, 2, 3, 4, 5, or 6, since each face of the die is equally likely to land face up. Therefore, the probability of rolling any particular number between 1 and 6 is always:
P(n) = 1/6
Note:-
To express a probability as a percentage, multiply it by 100. For example, the probability of rolling a 6 on a fair six-sided die is 1/6 or approximately 0.167. When multiplied by 100, this gives about 16.7%.
Two or More Dice
When rolling two dice, the probability calculation becomes slightly more complex. For example, the probability of rolling two sixes involves multiplying the individual probabilities.
When rolling two standard dice, the total number of possible outcomes becomes 36. The total possible outcome sample space is given through the table below:
| Dice B ⇨ Dice A ⇩ | 1 | 2 | 3 | 4 | 5 | 6 |
--- --- ---
| 1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| 2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| 3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| 4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| 5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| 6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
Unconventional probabilities have the rule that one must multiply the individual probabilities jointly to attain the outcome. Therefore, the formula for this is,
Probability of both = Probability of result one × Probability of result two
For Example P(Two 6s) = P(6 on first die) x P(on second die) = 1/6 x 1/6 = 1/36
Total Score from Two or More Dice
If you want to find the probability of getting a specific total(e.g, sum of 4) when rolling two dice, you need to count all possible combinations that give the total:
For example: to get sum as 4, the following combination are possible:
(1,3)
(2,2)
(3,1)
So, there are 3 favourable outcomes out of 36 total possible outcomes(since each die has 6 faces:6 x 6 = 36).
(Sum of 4) = 3/36 = 1/12
➣ Check Similar Articles:
How to calculate Card Probabilities?
How to calculate Coin Probabilities?
Similar Problems on Dice Probabilities
Question 1: Find the probability of retrieving a sum of 8 on throwing two dice.
Solution:
There are 36 total likely results on throwing two dice i.e., 6² = 6 × 6 = 36.
There are 5 total possibility of retrieving a sum of 8 on throwing two dice i.e., (2, 6), (3, 5), (4, 4), (5, 3), (6, 2).
Hence, the probability of retrieve a sum of 8 on throwing two dice is 5/36.
Question 2: Shawn tosses a die 400 times, and he documents the score of getting 6 30 times. What could be the probability of
a) retrieving a score of 6?
b) retrieving a score under 6?
Solution:
a) P (getting a score of 6)
= Number of times getting 6/total times
= 30/400
= 3/40
b) P (getting a score under 6)
= number of times getting under 6/total times
= 370/400
= 37/40
Question 3: What is the probability of retrieving a sum of 6 if two dice are thrown?
Solution:
When two dice are rolled, n(S) = 36. Let, A be the event of getting a sum of 6. Then,
A = {(3, 3), (2, 4), (4, 2), (1, 5), (5, 1)}
n(A) = 5
Hence, the required probability will be, P(A) = n(A)/n(S) = 5/36.
Question 4: Find the probability of throwing two dice and retrieving a sum of 4.
Solution:
The set of possible outcomes when we roll a die are {1, 2, 3, 4, 5, 6}
So, when two dice are rolled, there are 6 × 6 = 36 chances.
When we roll two dice, the probability of retrieving number 4 is (1, 3), (2, 2), and (3, 1).
So, the number of favorable outcomes = 3
Total number of possibilities = 36
Probability = {Number of likely affair } ⁄ {Total number of affair} = 3 / 36 = 1/12.
Thus, 1/12 is the probability of rolling two dice and retrieving a sum of 4.
➣ Also check
Probability Questions (Basic)
Probability Questions (Medium)
Probability Questions (Hard)
How to calculate Dice Probabilities?
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作为一个高中生。。上次闭门造车思考了 i i i^{i} 之后，发现获得的褒贬不一。于是这次准备来写点正常的东西......
高中刚学函数的时候，我们应该都学过最经典的两个函数，指数函数与对数函数。有一次本人碰到了这样一题：
已 知 l o g x 2≤2,求 x 的 取 值 范 围。已知log_{x}2\leq2,求x的取值范围。
聪明的我（划掉）立刻就想到要分类讨论，就是x大于1或小于1两种情况。具体做法不写了，反正答案是：
x∈(0,1)∪[2,+∞)x\in(0,1)\cup[\sqrt{2},+\infty)
作为一个数形结合的粉丝，觉得这种题目还要讨论实在是浪费时间，于是乎，就想研究一下自变量在底数的对数函数的性质与图像。即：
y=l o g x a,(a∈(0,+∞))y=log_{x}a,(a\in(0,+\infty))
研究一个函数，首先要看其定义域。毫无疑问，其定义域即底数的取值范围。
D：x∈(0,1)∪(1,+∞)D：x\in(0,1)\cup(1,+\infty)
然后是单调性和值域。这俩都不是很明显，于是只能看其导函数了。问题是，这个函数并不属于初等基本函数，这可咋办？对了，没学过这个函数的导数但我知道对数函数的导数以及导数的倒数运算（真拗口）法则啊。即：
f(x)=l o g a x,则 f′(x)=1 x l n a f(x)=log_{a}x,则f'(x)=\frac{1}{xlna} ，
以及
g(x)=1 f(x),且 f(x)的 导 数 记 为 f′(x),则 g′(x)=−f′(x)f 2(x)g(x)=\frac{1}{f(x)},且f(x)的导数记为f'(x),则g'(x)=-\frac{f'(x)}{f^{2}(x)} .
又因为我们的新函数： y=l o g x a=1 l o g a x y=log_{x}a=\frac{1}{log_{a}x} ,那么直接带公式就可以了。
y′=−1 x l n a l o g x 2 a=−1 x⋅l n a⋅l o g x 2 a y'=-\frac{\frac{1}{xlna}}{log_{x}^{2}a}=-\frac{1}{x\cdot lna\cdot log_{x}^{2}a}
由于 l o g x 2 a log_{x}^{2}a 一项恒正，因此只要判断 l n a lna 的正负即可。
由此可得单调性：
1、当 0<a<1 0<a<1 时， y′>0 y'>0 ,因此其单调区间为：在 （0，1），（1，+∞）（0，1），（1，+\infty） 上单调递增。
2、当 a>1 a>1 时， y′<0 y'<0 ,因此其单调区间为：在 （0，1），（1，+∞）（0，1），（1，+\infty） 上单调递减。
随后是值域。
我们知道 y=l o g a x y=log_{a}x 的值域为R，因此新函数y=l o g x a=1 l o g a x y=log_{x}a=\frac{1}{log_{a}x} 的值域自然是：
C：y∈(−∞,0)∪(0,+∞)C：y\in(-\infty,0)\cup(0,+\infty) 了。
随后是图像。
我们知道，当a取值不同，对数函数的图像也不相同。
1、当 0<a<1 时，
因为\lim_{x \rightarrow 0}{log_{a}x}=+\infty ,因此 \lim_{x \rightarrow 0}{\frac{1}{log_{a}x}}=0
因为 \lim_{x \rightarrow 1}{log_{a}x}=0 ，因此 \lim_{x \rightarrow 1}{\frac{1}{log_{a}x}}=\infty
由于当x小于1时， {log_{a}x} 大于零，因此在1处的左极限为 \lim_{x \rightarrow 1^{+}}{\frac{1}{log_{a}x}}=+\infty ，同理 右极限为\lim_{x \rightarrow 1^{-}}{\frac{1}{log_{a}x}}=-\infty 。
因为 \lim_{x \rightarrow +\infty}{log_{a}x}=-\infty ，因此 \lim_{x \rightarrow +\infty}{\frac{1}{log_{a}x}}=0 。
其渐近线为 x=1
由此我们可以画出当 0<a<1 时的函数图像。如下图：
这里我取 a=\frac{1}{e} .
2、 a>1 时，同上。
图像如下：
这里我取 a=e 。
好啦，图像也画完了，下面来看看此函数的运算性质。
f(x)+f(y)=\frac{1}{log_{a}x}+\frac{1}{log_{a}y}=\frac{log_{a}xy}{log_{a}x\cdot log_{a}y}=\frac{log_{x}a\cdot log_{y}a}{log_{xy}a}=\frac{f(x)\cdot f(y)}{f(xy)}
即： f(xy)[f(x)+f(y)]=f(x)\cdot f(y)
减法同理。
那么到此为止，此函数的基本性质大概就到此为止啦。虽然没啥用，但数学这种东西，玩的不就是没用与好玩吗？况且这玩意指不定对某题真有些作用呢。这篇文章先写到这里，日后我也会不断分享自己在数学方面的个人小研究的~
软广告 上回文章:关于i^i 381711?utm_source=qq&utm_medium=social
P.S.我觉得没啥人会看qwq
编辑于 2017-09-17 22:58
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王二小
这个。。。。 至于费那么大劲吗， log_{a}(x) = ln(x)/ ln(a)
所以。 log_{x}(a) = ln(a)/ ln(x)
简单的说， log_{a}(x) 就是 ln(x) 的倒数称上个常量
2017-09-17
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法会因由
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王二小
感谢您的建议~
2017-09-19
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法会因由
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幻想乡的观测者
不好意思刚看到回复。。然而我比您想象的辣鸡的多。。因为并不知道x^x与欧拉定理的关系。。我去研究一下。。分圆多项式和欧拉定理虽然知道但很了解的很浅。非常感谢您给的指导~(以及前面并没有真的指责剧透😂只是打个趣，本来只是想写写其基本性质以及初等应用，比如判断a^x=x^b的解的情况之类的比较初等的玩意。毕竟还是高中生😂)
2017-09-19
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LION Skywalker
2024-07-17
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meiyddd
感谢，学习了！希望能继续看见你的更多的这种思考！
2019-01-01
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柯芳
喜欢你的认真劲。
我们同届。
2017-10-03
回复2
法会因由
作者
谢谢😁
2017-10-03
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根号二
不是有传说中的换底公式吗。。。
2017-10-02
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法会因由
作者
是，但是换了底还是对数和反比例的复合
2017-10-03
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黑择明
感谢喔，我好像知道怎么对一个东西进行研究了。。。。
2017-10-01
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法会因由
作者
有帮助就好~
2017-10-01
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一苇
思路清晰，逻辑性强，很好，帮到我了，我高中没学好这个函数，
2017-09-20
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橙海雪莲
傻孩子你是不知道换底公式么。
2017-09-20
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法会因由
作者
知道啊 但问题是换来换去始终是对数与反比例的复合的形式 我只是觉得好玩而已😂不用太在意
2017-09-20
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暮紫骏
对数函数的倒数
基本初等函数经过有限次四则运算或者复合得到的函数叫做初等函数。
我们一般不研究那么多奇怪的初等函数的性质
2017-09-18
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法会因由
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嗯嗯，我只是觉得好玩而已😂
2017-09-18
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currim卅
可以换底吗
2017-09-17
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可以啊
2017-09-17
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188767 | https://optimization-online.org/wp-content/uploads/2002/10/554.pdf | The Thirteen Spheres: A New Proof Kurt M. Anstreicher∗ October, 2002 Abstract. The “thirteen spheres problem,” also know as the “Gregory-Newton problem” is to determine the maximum number of three-dimensional spheres that can simultaneously touch a given sphere, where all the spheres have the same radius. The history of the problem goes back to a disagreement between Isaac Newton and David Gregory in 1694. Using a combination of harmonic analysis and linear programming it can be shown that the maximum cannot exceed 13, but in fact 13 is impossible. The argument that the maximum is 12 uses an ad-hoc construction that does not appear to extend to higher dimensions. In this paper we describe a new proof that uses linear programming bounds and properties of spherical Delaunay triangulations.
Keywords: Kissing problem, spherical codes, Voronoi decomposition, Delaunay triangula-tion ∗Dept. of Management Sciences, University Of Iowa, Iowa City, Ia 52242, kurt-anstreicher@uiowa.edu 1 1 Introduction For n ≥3 let Sn−1 = {x ∈ℜn : xT x = 1}, and −1 < z < 1. A finite set C = {xi}M i=1 ⊂Sn−1 is called a spherical z-code if xT i xj ≤z for all i ̸= j. For z = 1 2 , {xi} correspond to contact points between Sn−1 and M non-overlapping spheres of radius one that are all incident to Sn−1. Maximizing the number M of such spheres is called the kissing problem in dimension n, and the maximal M is called the kissing number. The kissing problem in dimension 3 has a long history, going back to a celebrated disagreement between Isaac Newton and David Gregory in 1694. Newton believed that for n = 3 the kissing number was 12, while Gregory thought that 13 might be possible. In fact M = 13 is not possible, but the proof was quite long in coming. Several German papers in 1874/75 described approaches to the problem (see [2, First edition] and for references), but the first proof accepted as being complete is due to Sch¨ utte and van der Waerden , in 1953. A subsequent proof by Leech in 1956 is now the widely-cited standard. Versions of Leech’s proof also appear in the more recent references [2, First edition] and .
For general n and z there are several approaches that provide upper bounds on the size M of a z-code C ⊂Sn−1; see for example [5, 16]. In low dimensions the best known bounds are obtained using a combination of harmonic analysis and linear programming [5, 6, 7, 10, 16].
For z = 1 2 this approach leads to a complete characterization of maximal codes in dimensions n = 8 and 24 [4, 5], but for n = 3 the result is a bound of 13. The proof that for n = 3 the kissing number is 12 is not based on a general methodology, but instead uses an ad-hoc construction to obtain a contradiction if M = 13. The idea of the proof is simple and elegant, but there are many details that require verification. In fact the authors of [2, First edition] decided that the version published there was sufficiently incomplete that the entire chapter was removed in the second edition. A more serious drawback of the standard proof is that it seems impossible to extend it to higher dimensions, for two reasons. First, the proof is based on a construction that is tailored to S2, and second, the final contradiction is obtained by showing that a certain graph is not planar. The kissing number is unknown in dimensions other than 3, 8, and 24; for example for n = 4 a 1 2 -code with M = 24 is known, but the linear programming bound is 25.
2 In this paper we describe a new proof that the kissing number is 12 for n = 3. Our approach is based on linear programming bounds and properties of the spherical Delaunay decomposition associated with C. We do not claim that the proof here is “simpler” than the standard one . However, because the new proof is based on structure associated with C that is not dependent on n = 3, we believe that it has a much better chance of being extended to higher dimensions, for example to n = 4.
An outline of the paper is as follows. In Section 2 we describe the linear programming bounds on M that apply to a z-code C ⊂Sn−1. (We will eventually utilize bounds based on z > 1 2 .) In Section 3 we consider the spherical Voronoi and Delaunay decompositions associated with C. We describe an approach for obtaining lower bounds on the surface areas of spherical Delaunay triangles that gives another proof that M ≤13. In Section 4 we use information from linear programming bounds to reduce the range of possible included angles that can occur in a spherical Delaunay triangle.
This reduction provides a considerable improvement in the lower bound for the total surface area of a spherical Delaunay star (the spherical triangles containing a given point xi ∈C) not consisting of 5 spherical triangles.
The resulting lower bounds on the areas of Delaunay stars are sufficient to prove that M = 13 is impossible.
The kissing number problem is closely related to problems of packing spheres in ℜn [5, 16].
There has recently been considerable progress by T. Hales in settling the Kepler conjecture regarding the maximal density of sphere packings in dimension 3. See for a detailed discussion, and references.
2 Linear programming bounds Let C = {xi}M i=1 be a spherical z-code in ℜn, n ≥3. In this section we describe a well-known linear programming bound for the size M of such a code. The distance distribution of the code is the function λ(·) : [−1, 1] →ℜ+ defined as λ(s) = |{(i, j) : xT i xj = s}| M .
(1) 3 It is then easy to see that λ(1) = 1, (2a) −1≤s≤z λ(s) = M −1, (2b) λ(s) ≥ 0, −1 ≤s ≤z.
(2c) Let Φk(·), k = 0, 1, . . . denote the Gegenbauer, or ultraspherical, polynomials Φk(t) = P (β,β) k (t) k+β k , (3) where P (β,β) k is the Jacobi polynomial with β = (n −3)/2 . The normalization in (3) is chosen so that Φk(1) = 1 for all k. Using techniques from harmonic analysis it can be shown (, [5, Chapter 9, 13], [16, Chapter 8]), that 1 + −1≤s≤z λ(s)Φk(s) ≥0, k = 1, 2, . . . .
(4) From (2) and (4), using k = 1, . . . , K, a bound on M can be obtained via the semi-infinite linear programming problem LP : max −1≤s≤z λ(s) s.t.
−1≤s≤z λ(s)Φk(s) ≥−1, k = 1, . . . , K, λ(s) ≥0, −1 ≤s ≤z.
The dual of LP is the problem LD : min K k=1 fk s.t.
K k=1 fkΦk(s) ≤−1, −1 ≤s ≤z, (5) fk ≥0, k = 1, . . . , K.
In practice it is impossible to solve LD exactly due to the infinite number of constraints.
However a solution that is approximately feasible for LD can easily be adjusted to provide a valid bound on M, as described in the followng lemma.
4 Lemma 1 [5, p.339] Suppose that 0 ≤ϵ < 1, and for a given n and z, f ≥0 satisfies the constraints of LD with the right-hand side of the constraints (5) relaxed to −(1 −ϵ). Then M ≤1 + 1 1−ϵ(K i=1 fk). for any z-code in ℜn.
Proof: As described above M ≤1 + vLD, where vLD denotes the optimal objective value in LD. But under the assumptions of the lemma 1 1−ϵf is feasible in LD, and therefore vLD ≤ 1 1−ϵ( K i=1 fk). 2 In practice Lemma 1 can be used to obtain a valid bound on M by discretizing [−1, z] using N + 1 points si = −1 + iδ, i = 0, . . . , N, where δ = 1+z N . LD is then solved using the constraints corresponding to {si}N i=0, resulting in a solution f ≥0. The value ϵ required in the lemma is easily obtained by bounding the derivative of K k=1 fkΦk(·) on the intervals [si−1, si], i = 1, . . . , N. For z = 1 2 this approach obtains the best known bounds on M for dimensions 4 ≤n ≤24 , and leads to a complete characterization of the maximal 1 2 -codes in dimensions 8 and 24 [4, 5]. For n = 3, using K = 16 obtains a bound on M of about 13.16. This bound can be reduced somewhat by imposing additional valid inequalities on the distance distribution λ(·), but to our knowledge a linear programming bound below 13 has never been obtained. (Better linear programming bounds on M = |C| can be obtained for special cases, for example the case that C is antipodal .) 3 Spherical Delaunay triangulations Let C = {xi}M i=1 ⊂S2, M ≥3. The spherical Voronoi cell associated with xi is the set Vi = {x ∈S2 : ∥x, xi∥g ≤∥x, xj∥g ∀j ̸= i}, where ∥x, v∥g = cos−1(xTv) denotes geodesic distance on S2. Each Voronoi cell is a spherical polygon whose sides are arcs of great circles, and ∪M i=1Vi = S2. The spherical Delaunay1 decomposition, which is dual to the decomposition of S2 into Voronoi cells, is obtained by connecting each xi and xj such that Vi and Vj share a nontrivial boundary with a geodesic arc. If the points of C are in general position (no three points on one great circle) then the 1The alternative translation Delone appears often in the literature.
5 Figure 1: Spherical Voronoi cells and Delaunay triangulation Delaunay cells are all spherical triangles, and in any case spherical triangles can be obtained by adding additional arcs if necessary. We refer to a triangulation of S2 so obtained as a spherical Delaunay triangulation (SDT). The following proposition describes the well-known Delaunay property that holds for a SDT.
Proposition 2 Let {xi, xj, xk} be the vertices of a spherical triangle in a SDT. The there is a spherical cap {x : ∥x, v∥g ≤R} that circumscribes {xi, xj, xk} such that ∥xl, v∥g ≥R for all xl ∈C.
The spherical Voronoi cells and Delaunay triangles for a typical C are depicted in Figure 1. The spherical caps circumscribing the Delaunay triangles for the same C are depicted in Figure 2. Figures 1 and 2 were produced by the ModeMap applet of D. Watson .
Suppose now that C = {xi}M i=1 is a maximal 1 2 -code. For an arbitrary spherical Delaunay triangle it is possible that the circumscribing spherical cap described in Proposition 2 has spherical radius R ≥π 2. However this is only possible if all of the points in C lie on one side of a hyperplane through the origin. This is clearly impossible if C is a maximal 1 2 -code, so henceforth we assume that R < π 2 for any spherical cap circumscribing a spherical Delaunay triangle. Note that this also implies that all edges in the SDT have length less than π.
6 Figure 2: Spherical Delaunay triangulation and circumscribing caps Consider a given spherical triangle in a SDT associated with C. Let a, b, c denote the geodesic lengths of the sides, and α, β, γ denote the corresponding opposite angles. We begin by collecting some useful formulas from spherical trigonometry [11, 13]. The law of cosines for spherical triangles is cos c = cos a cos b + sin a sin b cos γ.
(6) The area is given by the spherical excess E = α + β + γ −π, (7) where α + β can be obtained from a, b and γ using Gauss’ formula cos 1 2 (α + β) = sin 1 2 γ cos 1 2 (a + b) cos 1 2 c .
(8) When a = b it is straightforward to show that the spherical radius of a circumscribing spherical cap is given by tan R = tan 1 2 a cos 1 2 γ , (9) and in the general case [11, p.78] tan R = tan 1 2 c sin(γ −1 2 E).
(10) 7 In what follows we will consider γ to be fixed, and choose a and b to minimize the area of the triangle, subject to valid constraints. We write c(a, b | γ), E(a, b | γ) and R(a, b | γ) to denote c, E and R as functions of a and b for a given γ. Since C is a 1 2 -code we certainly have a ≥π 3, b ≥π 3, c(a, b | γ) ≥π 3. In addition we must have R(a, b | γ) ≤π 3, since otherwise we could add another point at the center of the circumscribing cap and still have a 1 2 -code, which is impossible if C is maximal. To minimize the area of a spherical Delaunay triangle with included angle γ, associated with a maximal 1 2 -code C, we are thus led to the problem Pγ: min E(a, b | γ) s.t.
c(a, b | γ) ≥π 3, R(a, b | γ) ≤π 3, a ≥b ≥π 3.
In the next two lemmas we give a complete characterization of the solution of Pγ.
Let γ0 = cos−1( 1 3).
Lemma 3 Pγ is infeasible for γ > 2γ0. For γ0 ≤γ ≤2γ0 the solution of Pγ is a = b = π 3.
Proof: It is obvious that the area E(a, b | γ) is monotonically increasing in a and b, and therefore the minimum possible value corresponds to a = b = π 3. Using (6) and (9) it is straightforward to compute that c( π 3, π 3 | γ) ≥π 3 for γ ≥γ0, and R( π 3, π 3 | γ) ≤π 3 for γ ≤2γ0.
Moreover for γ ≥π 2 it is easy to see that R(a, b | γ) is monotonically increasing in a and b, and therefore Pγ is infeasible for γ > 2γ0. 2 Lemma 4 Pγ is infeasible for γ < 1 2 γ0. For 1 2 γ0 ≤γ ≤γ0 the solution of Pγ has b = c = π 3.
Proof: Consider Pγ with the constraint on R ignored. We will show that the solution of this problem has b = c(a, b | γ) = π 3, and then consider the effect of the constraint on R. To start we will argue that b = π 3. It is straightforward to compute that ∂cos c(a, b | γ) ∂a + ∂cos c(a, b | γ) ∂b = (cos γ −1) sin(a + b), (11a) ∂cos c(a, b | γ) ∂a −∂cos c(a, b | γ) ∂b = −(cos γ + 1) sin(a −b), (11b) 8 Suppose that we have a ≥b > π 3 with c(a, b | γ) ≥ π 3.
From (11a), if a + b ≥π then simultaneously decreasing a and b will increase c(a, b | γ) while also decreasing E(a, b | γ).
This process could be continued until either b = π 3 or a + b = π. In the first case we are done, so consider now the case where a ≥b > π 3, a + b ≤π. From (11b) simultaneously increasing a and decreasing b will increase c(a, b | γ). Moreover, from (8) this same operation will decrease α + β, and therefore also E(a, b | γ), until we reach b = π 3.
We have now shown that the solution of Pγ, with the constraint on R ignored, has b = π 3.
But decreasing a will decrease E(a, b | γ), and c( π 3, π 3 | γ) < π 3 for γ < γ0. We must therefore have c(a, b | γ) = π 3 in the solution, for γ ≤γ0. This completes the analysis of the problem with the constraint on R ignored.
Next we consider the effect of the constraint on R. For b = c = π 3, (6) implies that cos a = 1 −3 cos2 γ 1 + 3 cos2 γ .
(12) The analog of (9) gives tan R = 1 √ 3 cos 1 2α, so R ≤π 3 is equivalent to cos 1 2 α ≥1 3, or cos α ≥−7 9 . From the analog of (6) we obtain cos a = 1 4 + 3 4 cos α, (13) so cos α ≥−7 9 is equivalent to cos a ≥−1 3 .
(14) Comparing (12) and (14) we conclude that the solution with b = c = π 3 results in R ≤π 3 for γ ≥cos−1 2 3 = 1 2 γ0. However, for all γ ≤γ0 the solution with b = c = π 3 simultaneously minimizes the numerator of the right-hand side in (10) while maximizing the denominator, and therefore Pγ is infeasible for γ < 1 2 γ0. 2 For 1 2 γ0 ≤γ ≤2γ0 we can easily obtain the minimal area associated with the solution of Pγ by computing α and β. In the case of γ0 ≤γ ≤2γ0 we have a = b = π 3, from Lemma 3. Therefore α = β, and cos c = 1 4 + 3 4 cos γ, from (6). Using (8) and ordinary trigonometric identities we obtain cos α = cos β = 1 −cos γ 5 + 3 cos γ .
(15) 9 0.550 0.575 0.600 0.625 0.650 0.675 0.700 0 35 70 105 140 175 γ (degrees) Area Figure 3: Minimal area of Delaunay triangle with included angle γ For 1 2 γ0 ≤γ ≤γ0 we have b = c = π 3, from Lemma 4. Therefore β = γ, and (12) and (13) together imply that cos α = 1 −5 cos2 γ 1 + 3 cos2 γ .
(16) From (15) and (16) we can easily compute the solution value of Pγ as a function of γ, using (7). In Figure 3 we plot the resulting minimal area. (The vertical lines in the figure will be explained in the next section and can be ignored for the moment.) The minimal value of 3γ0 −π ≈.55129 occurs at γ0, 1 2 γ0, and 2γ0. Note that since the total surface area of S2 is 4π ≈12.5664, this lower bound implies that there can be at most 22 spherical Delaunay triangles in a SDT associated with C. Let E and F denote the number of edges and faces (triangles) in this SDT. From Euler’s theorem we have F + M = E + 2, and 3F = 2E for a triangulation. Then M = 1 2 F + 2, so F ≤22 gives another proof that M ≤13.
Let ni denote the number of edges incident to xi in the SDT (equal to the number of Voronoi cells sharing a nontrivial border with Vi if the points of C are in general position).
Note that if M = 13, then 13 i=1 ni = 2E = 66, (17) implying that ni ̸= 5 for at least one i. For each xi the spherical Delaunay star at xi is the union of the spherical Delaunay triangles in the SDT that have xi as a vertex. Our proof is 10 based on examining the total surface area of spherical Delaunay stars, in particular the star at an xi with ni ̸= 5. Unfortunately the analysis of Pγ above is not sufficient to produce a contradiction based on the total surface area of such a star. For each spherical triangle in the star at a point xi we take γ to be the angle between the 2 edges incident to xi. For ni = 5 we can then take 4 spherical triangles with γ = γ0, and one with γ = 2π −4γ0, to get a lower bound of the total surface area of the star equal to about 2.7965. For ni = 6 we can use 2 triangles with γ = 1 2 γ0 in place of one of the triangles with γ = γ0 to get a lower bound of about 3.3478. Summing the bounds on the areas of the stars for 12 points xi with ni = 5, and one with ni = 6, and dividing by 3 then obtains a lower bound on the total area of the triangles in the SDT of about 12.3019. This is greater than 22(.55129) ≈12.1284 but still less than 4π. In fact the same lower bound can be achieved with any combination of ni having 3 ≤ni ≤9, i = 1, . . . , 13 satisfying (17).
In the next section we show how the analysis of a SDT can be improved by incorporating information from the LP bounds described in Section 2.
4 The kissing number for n = 3 As described in the previous two sections, bounds based on linear programming and an analysis of spherical Delaunay triangulations both individually suffice to prove that M = |C| ≤13 for a 1 2 -code C in S2. In this section we show how these two methodologies can be combined to give a new proof that M ≤12. Our approach is based on strengthening the constraint on R used in the formulation of the problem Pγ of the previous section. Recall that the logic of this constraint is based on the fact that if C is a maximal 1 2 -code, then it is impossible to have R ≥π 3 for a spherical Delaunay triangle since otherwise C could be augmented to obtain a larger 1 2 -code. We now consider the possibility of adding a point to C such that the augmented code C+ is a z-code, with z > 1 2 . The following lemma is useful for obtaining valid constraints on the distance distribution of the augmented code C+.
Lemma 5 Suppose that a spherical triangle has c ≥π 3, 0 ≤za ≤cos a ≤cos b ≤zb < 1.
11 Then cos γ ≤max 1 2 −z2 a 1 −z2 a , 1 2 −zazb (1 −z2 a)(1 −z2 b) .
Moreover if za < cos a then the above inequality is also strict.
Proof: Consider the problem to minimize γ subject to the given constraints on a, b, and c.
By assumption b ≤a < π 2, so (11a) implies that for any γ, a and b can be simultaneously increased resulting in an increase in c. Thus we may assume that cos a = za. Using (6) and ordinary trigonometric identities we conclude that cos2 γ ≤ ( 1 2 −za cos b)2 (1 −z2 a)(1 −cos2 b).
(18) The numerator of the right-hand side of (18) is convex in cos b, while the denominator is positive and concave in cos b. Therefore the ratio is a quasiconvex function of cos b, implying that the maximum must occur at one of the bounds zb or za. This proves the weak inequality of the lemma. The strict inequality is an immediate consequence of (11a); if cos a > za then a and b can both be increased, resulting in a strict increase in c, allowing a strict decrease in γ. 2 Note that for za = 1 7, zb = 1 2 Lemma 5 implies that if π 3 ≤b ≤a < cos−1( 1 7) and c ≥π 3, then γ > π 6. This fact is an important element of Leech’s proof .
Lemma 6 Suppose that C+ = {xi}14 i=1 is a z-code, .5 < z ≤.66, where C = {xi}13 i=1 is a 1 2 -code. Let z5 = sin2 π 5 −1 4 sin π 5 ≈.5257 .
Then the distance distribution for C+ satisfies the constraints: 1.
.2<s≤z λ(s) ≤5, 2.
.5 z5.
Proof: For part 1 we apply Lemma 5 with za = .2, zb = .66, and obtain γ > π 3. It follows that for each xi, i = 1, . . . , 14 there can be at most five xj having .2 < xT i xj ≤.66. Part 12 2 follows from this same observation, and the fact that C is a 1 2 -code. For part 3 we use za = z5, zb = .66 and obtain γ > 2π 5 , so there can be at most four xi, i = 1, . . . , 13 with z5 < xT i x14 ≤.66. 2 Theorem 7 Suppose that C = {xi}13 i=1 is a 1 2 -code. Then cos R ≥.6595 for any Delaunay triangle in a SDT associated with C.
Proof: We consider the problem LP from Section 2 with .5 < z ≤.66, K = 10, and the additional constraints on λ(·) from Lemma 6. Forming the dual problem, discretizing the constraints, and adjusting the solution value as described in Lemma 1, we obtain a bound2 strictly less than 14 for .5 < z ≤.6595. It follows from the Delaunay property that R ≤cos−1(.6595) for any Delaunay triangle associated with C, since otherwise we could construct an augmented code C+, |C+| = 14 whose distance distribution was feasible for all the constraints of LP. 2 Let Pγ,z denote the problem Pγ, from the previous section, but with the constraint on R replaced by cos R(a, b | γ) ≥z.
Lemmas 3 and 4 can easily be generalized to the case of z > 1 2 , as described below.
Lemma 8 Suppose that 1 2 ≤z ≤ 2 3. Let γ+ = 2 cos−1 z2 3(1−z2). Then Pγ,z is infeasible for γ > γ+. For γ0 ≤γ ≤γ+ the solution of Pγ,z is a = b = π 3.
Proof: The proof follows that of Lemma 3, except that cos R( π 3, π 3 | γ) ≤z for γ ≤γ+. For γ < π 2 it is not obvious that R(a, b | γ) is monotone in a and b. However it is easy to show that a = b = π 3 minimizes c(a, b | γ) while also minimizing E(a, b | γ). For γ ≤π 2 this implies that a = b = π 3 simultaneously minimizes the numerator of the right-hand-side of (10) while maximizing the denominator, again implying that Pγ is infeasible for γ > γ+. 2 Lemma 9 Suppose that 1 2 ≤z ≤ 2 3. Let γ−= cos−1 1 −4 3z2. Then Pγ,z is infeasible for γ < γ−. For γ−≤γ ≤γ0 the solution of Pγ,z has b = c = π 3.
2Full details of the solutions of the linear programming problems are available from the author. A bound strictly less than 14 is actually obtained for z up to approximately .65976.
13 Proof: The proof is identical to that of Lemma 4, except that b = c = π 3 results in cos R ≥z for γ ≥γ−. 2 In Figure 3 the two vertical lines correspond to the values γ+ and γ−from Lemmas 8 and 9, respectively, using z = .6595. It is obvious from the figure that the use of z = .6595, as opposed to the original z = .5, considerably reduces the range of possible values of the included angle γ. This reduction has a substantial effect on the lower bound for the area of a spherical Delaunay star at a point xi having ni ̸= 5 triangles. From the figure it is also apparent that the area bounds for γ = γ+ and γ = γ−coincide. In fact it is easy to show that this is the case for any 1 2 ≤z ≤ 2 3.
Consider now a SDT associated with a 1 2 -code C = {xi}13 i=1. Using Lemma 8 and (15), the minimal area of a spherical Delaunay triangle in this SDT, with included angle γ0 ≤γ ≤γ+, is bounded from below by f +(γ) = γ + 2 cos−1 1 −cos γ 5 + 3 cos γ −π.
From Lemma 9 and (16), the minimal area of a spherical Delaunay triangle in the SDT, with included angle γ−≤γ ≤γ0, is bounded from below by f −(γ) = 2γ + cos−1 1 −5 cos2 γ 1 + 3 cos2 γ −π.
Since f +(·) and f −(·) are both concave on their domains of definition, the minimum possible area for a spherical Delaunay star at xi containing ni triangles corresponds to all but one of the triangles in the star having included angle γ equal to either γ0, γ+, or γ−, with the last chosen to make the sum of the angles equal to 2π. Enumerating the very small number of possibilities, continuing to use z = .6595, we obtain the bounds in Table 1. (Note that 4 ≤ni ≤7 follows from γ+ < 2π 3 , γ−> π 4.) In the table we also report the number of triangles with included angles equal to γ0, γ+, and γ−that achieve the minimum total area, and the minimal average area (total area divided by number of triangles). The area and average area are rounded down to give valid lower bounds.
Theorem 10 Let C = {xi}M i=1 be a 1 2 -code in S2. Then M ≤12.
Proof: Assume that M = 13.
Recall that the minimum possible area for a spherical Delaunay triangle in a SDT associated with C is 3γ0 −π > .55128. If there is a j with 14 Table 1: Bounds for areas of Delaunay stars Triangles γ0 γ+ γ− Area Average 4 2 1 0 2.44178 .61044 5 4 0 0 2.79648 .55929 6 2 0 3 3.66747 .61124 7 0 0 6 4.66593 .66656 nj = 7, then using the bound from Table 1 the total area of the triangles in the SDT would be at least 4.66593 + 15(.55128) = 12.93513 > 4π. Therefore there can be no j with nj = 7.
Next suppose that there is a j with nj = 4. Then there must be at least 2 points xi with ni = 6. Summing over the Delaunay stars, the total area is bounded from below by 2.44178 + 2(3.66747) + 50(.55929) 3 ≈12.58 > 4π, so there can be no j with nj = 4. It must then be the case that nj = 6 for one j, and ni = 5 for all i ̸= j. For convenience suppose that j = 1, and {xi}7 i=2 are the six other nodes in the star at x1. The area of the star at x1 is bounded from below by 3.66747. For i = 2, . . . , 7 the Delaunay star at xi contains 2 spherical triangles in common with the star at x1, and 3 other triangles. Summing the areas of these 6 stars we get twice the area of the star at x1, plus the areas of 18 spherical triangles. Finally for i = 8, . . . , 13 the area of the star at xi is bounded from below by 2.79648. Summing over all the stars, the total area is bounded from below by 3(3.66747) + 18(.55128) + 6(2.79648) 3 ≈12.56811 > 4π, and therefore M = 13 is impossible. 2 Although the area bounds in Table 1 are sufficient to complete the proof of Theorem 10, it is worth noting that the bound for the area of a star with ni = 6 from Table 1 clearly cannot be attained. This bound corresponds to three spherical triangles with γ = γ−, b = π 3, a ≈1.6862, two with γ = γ0, a = b = π 3, and one with γ slightly less that γ0, b = π 3, and a slightly greater than π 3. It is obvious that it is impossible to assemble 6 such spherical triangles into a spherical star. It is also interesting to note that the last case considered in the proof, nj = 6 for one j, and ni = 5 for all i ̸= j, could alternatively be excluded 15 using a graph-theoretic argument. The dual of this SDT would be a cubic, planar graph that divides S2 into one hexagon and 12 pentagons. It is easy to show that no such graph can exist. Leech’s proof also uses arguments based on area to eliminate all but one case, which corresponds to an unrealizable planar graph. However the area arguments here are completely different, as is the graph in the final case.
16 References M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards Appl. Math Series 55, U.S. Dept. of Commerce, Washington DC, 1972.
M. Aigner and G. Ziegler, Proofs from THE BOOK. Springer, Berlin, 1998; Second edition, Springer, Berlin 2001.
K.M. Anstreicher, Improved linear programming bounds for antipodal spherical codes.
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E. Bannai and N.J.A. Sloane, Uniqueness of certain spherical codes. Canadian J. Math.
33 (1981), 437-449.
J.H. Conway and N.J.A. Sloane, Sphere Packings, Lattices and Groups. Third edition, Springer, New York, 1999.
P. Delsarte, J.-M. Goethals, and J.J. Seidel, Spherical codes and designs. Geom. Dedicat.
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G.A. Kabatiansky and V.I. Levenshtein, Bounds for packings on a sphere and in space.
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J. Leech, The problem of the thirteen spheres. Math. Gazette 40 (1956), 22-23.
J.C. Lagarias, Bounds for local density of sphere packings and the Kepler conjecture.
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V.I. Levenshtein, Universal bounds for codes and designs. In Handbook of Coding The-ory, Volume 1, V.S. Pless and W.C. Huffman, eds., North-Holland, Amsterdam, 1998, 499-648.
E. Miller, Plane and Spherical Trigonometry. Third edition, Leach, Shewell and San-born, Boston, 1894.
H.-S. Na, C.-N. Lee and O. Cheong, Voronoi diagrams on the sphere. Dept. of Mathe-matics, Pohang University of Science and Technology, Korea (2001).
H.L. Pearson, Formulas from Algebra, Trigonometry, and Analytic Geometry. In Hand-book of Applied Mathematics, Second Edition, C.E. Pearson, ed., Van Nostrand Reinhold, New York, 1983, Chapter 1.
K. Sch¨ utte and B.L. van der Waerdan, Das Problem der dreizehn Kugeln. Math. Annalen 53 (1953), 325-334.
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188768 | https://eichrom.com/wp-content/uploads/2018/02/Neutron-Attenuation-White-Paper-by-D-M-rev-2-1.pdf | Neutron Shielding Materials Daniel R. McAlister, Ph.D. PG Research Foundation, Inc. 1955 University Lane Lisle, IL 60532, USA Revision 2.1 June 22, 2018 INTRODUCTION Understanding the fundamentals of the interactions of neutrons with matter is an important step in working safely in areas where neutron radiation may be encountered. The following monograph will cover the basic principles and terminology of neutron sources and shielding. A more detailed treatment of each topic may be found in references [1-5]. DEFINITION OF COMMON TERMS Barn – The standard unit for neutron cross sections, equal to 10-24 cm2, denoted by the symbol b. Typical cross section values range between 0.001 and 1000 barns(b). Compound Nucleus – An unstable nucleus formed during neutron capture. The neutron combines with the target nucleus, adding its kinetic energy and binding energy, forming a new nucleus in an excited state. If sufficient energy is present, one or more nucleons from the compound nucleus may overcome nuclear binding forces and escape the nucleus in a process called evaporation. Excess energy may also be released in the form of gamma radiation. The typical lifetime of a compound nucleus is 10-14 to 10-20 seconds. Cross Section – A value which describes the probability that a nuclear reaction will occur. For neutrons, the cross section is related to the geometric cross section of the target nucleus. Therefore, cross sections are typically expressed in cm2 or barns(b) = 10-24 cm2. The cross section for a given nuclear interaction is also dependent on other factors, such as the speed of the neutron, the type of interaction (scattering, capture, etc..), and the stability of the target nucleus. Cross sections are typically defined for specific nuclear reactions and for the overall probability of nuclear reaction with a target nucleus. The term cross section may refer to the microscopic cross section, or the interaction of neutron(s) with a single target nucleus, or macroscopic cross section interaction of neutron(s) with a thick layer of material. Cross sections may be further divided into cross section values for individual types of interaction (scattering, absorption). Elastic Scattering – A scattering interaction in which the neutron-target system has essentially the same kinetic energy before and after interaction. Elastic scattering of neutrons may alter the direction and speed of the neutron, but will not alter the identity of the neutron or the target or cause nuclear excitation in the target. Relatively small kinetic energy losses in the neutron-target system may occur through atomic or molecular excitations. Epithermal Neutrons – Neutrons of higher energy than thermal neutrons, typically ~0.1eV and 1keV. Some resources may characterize epithermal neutrons with slightly different energy ranges. Fast Neutrons – Neutrons with energy >0.1 MeV. Some references may characterize fast neutrons with slightly different energy ranges. Fission – A nuclear reaction or decay process (spontaneous fission) in which the nucleus of an atom splits into lighter parts. Fission may produce lighter nuclides as well as additional neutrons, photons and large amounts of energy. Fission is one possible outcome from neutron capture reactions. Inelastic Scattering – A scattering interaction in which a neutron transfers energy to a target nucleus causing nuclear excitation. The excited target nucleus then returns to a non-excited state through the emission of gamma radiation. The identity of the neutron and target nucleus are not altered. However, there is a net loss in kinetic energy of the neutron-target system. Macroscopic Cross Section – A value (), with units of cm-1, that describes the probability of interaction of a neutron with a thick layer of target material. Mean Free Path – The average distance traveled by a moving particle (neutron) in a target medium between interactions with the target material. Microscopic Cross Section – A value (), typically in cm-2 or barns, which describes the probability of the interaction of a neutron with a single target nucleus. Moderating Power – A value relating the effectiveness of a material in the lowering of neutron energy through scattering reactions (moderation/thermalization). The moderating power is defined as the product of macroscopic scattering cross section () and the average logarithmic energy loss upon scattering (). As moderating power increases, less material is required to achieve the same degree of moderation. Moderating Ratio – A value relating the effectiveness of a material in lowering the neutron energy through scattering reactions, which also takes into account neutron capture cross sections. The moderating ratio is defined as the moderating power divided by the macroscopic neutron capture cross section for the material. Materials with large moderating ratios are good neutron moderators and poor neutron absorbers. Nucleon – A particle that makes up an atomic nucleus. (neutrons or protons) Neutron Capture – A nuclear reaction in which a neutron and target nucleus collide and merge, forming a heavier nucleus (compound nucleus). Depending on factors, including the energy of the incident neutron and the nuclear properties of the target nucleus, the neutron capture may be followed by the emission of gamma radiation or atomic particles and/or in the fission of the compound nucleus. Neutron Excitation Function – A plot of cross section vs neutron energy for a given neutron-target system. Neutron Fluence – The neutron flux integrated over a period of time with units of neutrons/cm2. Neutron Flux – A measure of the intensity of neutron radiation, expressed in neutrons/cm2/sec, corresponding to the rate of flow of neutrons. Neutron Moderation – See Neutron Thermalization. Neutron Scattering – An interaction between a neutron and matter which results in a change in velocity of the neutron. Scattering may be elastic or inelastic. Neutron Thermalization – The process of neutron energy reduction (moderation) to thermal values (~0.025eV) through scattering reactions. Resonance Neutrons – Neutrons which are strongly captured in the resonance of U-238 and some commonly used detector materials (In, Au), typcially 1-300eV. Resonance Peaks – Sharp peaks in the plot of cross section vs neutron energy (neutron excitation function) for a given neutron-target system. The peaks correspond to nuclear energy level spacings in the target material. Neutrons at resonance energies exhibit increased probability of interaction with the target (higher cross section). Thermal Neutrons – Neutrons in thermal equilibrium with their surroundings, typically ~0.025eV. NEUTRON SOURCES Neutron emission is typically associated with the fission of uranium or plutonium fuel in a nuclear reactor. However, there are many other potential sources of neutrons that may be encountered. A list of some common neutron sources is presented in Table 1. Neutron Interactions with Matter Like gamma radiation, neutrons undergo scattering and absorption interactions with matter. These interactions form the basis for methods used to shield and measure neutron radiation. However, unlike gamma radiation, which interacts primarily with the electrons in matter, neutrons interact primarily with the nucleus. Consequently, the types of materials favored for neutron shielding are quite different than the dense, high atomic number absorbers which are most effective in the attenuation of gamma radiation. Additionally, whereas isotopes of an element will have essentially identical gamma attenuation properties, isotopes of an element often have significantly different neutron attenuation properties. In general, for fast (high energy) neutrons, scattering interactions are more likely than capture interactions. As the energy of neutrons is reduced through scattering interactions (neutron thermalization/moderation), all neutron interactions increase in probability and neutron capture interactions become more important. Scattering interactions for neutrons can be divided into elastic (kinetic energy of neutron-target system conserved) and inelastic scattering (kinetic energy of the Type Examples Comment 87Br→87Kr+ b -→86Kr + n 11Li→11Be + b -→9Be + 2n Mixture of alpha emitter and low atomic number element.
~30 neutrons per one million alpha emissions.
Neutron energy from 0.5-4MeV depending on alpha energy and target.
Major sources are spontaneous fission of 244Cm and 242Cm and (a, n) reactions on oxygen Gamma radiation in excess of nuclear binding energy causes neutron emission from target Light Ion D + T → n + 4He Deuterium or tritium ions are accelerated into deuterium or Accelerators D + D → n + 3He tritium hydride targets, producing neutrons from fusion.
ISIS neutron source High energy protons impacting on depleted uranium, SNS at ORNL tugsten or tantalum target strip (spall) neutrons Table 1. Neutron Sources PuBe, AmBe, AmLi 88Y-9Be, 124Sb-9Be Spallation Sources (g,n) Sources (a,n) Sources Induced Fission Spontaneous Fission Delayed Neutron Emission Spent Nuclear Fuel neutron bombardment of 235U Yields average of 2.5 neutrons with an average energy of 2MeV.
Neutron-rich nuclides in excited states may emit neutrons. Important for fission products.
252Cf 3.1% branch of 252Cf decay. Yields 3.7 neutrons per fission, average energy 2.3MeV.
neutron-target system is lowered through excitation of the target nucleus and subsequent gamma emission). In neutron capture, a neutron and target nucleus collide and merge, forming a heavier nucleus (compound nucleus). Depending on the energy of the incident neutron and the nuclear properties of the target nucleus, the neutron capture may be followed by the emission of gamma radiation, atomic particles, and/or in the fission of the compound nucleus. Neutron capture reactions are denoted by X(n,a)Y, where X is the target nucleus, n is the incident neutron, a is the ejected particle(s) or gamma ray, and Y is the nucleus after absorption of the neutron and particle or gamma emission. Table 2 lists several types of neutron capture reactions. Neutron Cross Sections The probability that neutron-target interactions will occur is expressed using cross sections, denoted by the symbol . Neutron cross sections are related to the geometric cross section of the target nucleus. Therefore, cross sections are typically expressed in cm2 or barns(b) = 10-24 cm2. The cross section for a given nuclear interaction is also dependent on other factors, such as the speed of the neutron, the type of interaction, and the stability of the target nucleus. Cross sections are typically defined for specific nuclear reactions and for the overall probability of nuclear reaction (t) with a target nucleus. t = ei + i + c + f + …… el = elastic scattering cross section i = inelastic scattering cross section c = capture cross section (may be split into individual capture reactions) f = fission cross section Cross section values typically range from 0.0001 to 1000 barns. Plots of cross section vs neutron energy for a given target are called neutron excitation functions. A typical neutron excitation function is depicted in Figure 1. In general, the cross section decreases with increasing neutron energy. However, at some neutron energies sharp increases in cross section may be observed. These sharp peaks are known as resonance peaks and correspond to nuclear energy level spacings in the target material. Reaction Neutron Energy (n,g) 0-500 keV (n,p) 0.5-50 MeV (n,a) 0.5-50 MeV (n,2n) 1-50 MeV (n,np) 1-50 MeV (n,2p) 1-50 MeV Fission Thermal to Fast Table 2. Common Neutron Capture Reactions The cross section discussion to this point has focused on the interaction of neutron(s) with a single target nucleus, or the microscopic cross section (). The microscopic cross section is useful for understanding the fundamental interaction processes for neutrons and matter. However, for the attenuation of neutrons in shielding material, it is better to discuss the interaction of neutron(s) with a thick layer of material, or the macroscopic cross section (). The total macroscopic cross section (t) can be defined as: t = Nt where N = the atom density of the target material t is the total microscopic cross section. The intensity of a neutron beam, I(x), passing through a target material of thickness x can be expressed as: I(x) = Ioe-Ntx -or- I(x) = Ioe-x This function allows the calculation of the fraction of neutrons at a given energy that will pass through a thickness (x) of a given target or shielding material without undergoing any type of scattering or capture interaction. Calculations for composite materials can be performed by using the sum of the macroscopic cross sections of each individual element: = 1 + 2 + 3 ….. where the atom density of each element (Ni) is: Ni = NAni/M where is the density of the composite material, M is the molecular or unit weight of the composite material, NA is Avagadro’s number, and ni is the number of atoms of the element in the molecule or composite unit. The accurate calculation of total dose reduction for a given thickness of shielding is much more difficult due to the possibility of multiple scattering and capture reactions and the production of secondary radiation (gamma emission, etc.). The calculation of dose reduction in neutron systems must take into account contributions from scattered neutrons with lower energy and secondary radiation, and therefore requires more sophisticated calculation techniques, such as Monte Carlo N-Particle Transport Code (MCNP) , employing extensive libraries of cross sections. Appendix I. contains neutron attenuation data for a wide range of materials calculated using MCPN66. Figure 1. Neutron Excitation Function Common Neutron Shielding Materials Neutron shielding materials are typically constructed from low atomic number elements (hydrogen, carbon, and oxygen) with high scattering cross sections that can effectively moderate or thermalize incident neutrons. Shielding for small sources is often constructed from polyethylene or paraffin, while shielding for larger sources is made from concrete or large pools/tanks of water. Elements with high capture cross sections for thermal neutrons (boron, cadmium, and gadolinium) are often dispersed in the shielding material to capture moderated/thermalized neutrons. Borated polyethylene, layers of B4C and aluminum, boron-aluminum alloys, and boric acid in water are examples of materials incorporating boron. Adding boron to neutron shielding materials reduces the dose from secondary gamma production from radiative capture (n,g). Boron, specifically the ~20% naturally abundant boron-10, has a very high (n,a) capture reaction, which yields much lower energy gamma radiation than the radiative capture reactions (n,g) of hydrogen, oxygen, or carbon. Neutron shielding may also incorporate high atomic weight elements or layers of higher atomic weight shielding material to reduce dose from gamma radiation produced from neutron capture interactions (n, g). Lead, bismuth, and tungsten are often used due to their high density, good gamma attenuation characteristics, and relatively benign activation products. However, the relatively high (n,g) cross-section of tungsten should be considered when choosing shielding for areas with high neutron fields, due to the significant secondary gamma radiation dose that can be produced from the capture of neutrons escaping the primary neutron shield. Lead or bismuth may be better choices for shielding (n,g), due to their much lower (n,g) cross-sections. Resonance Peaks 1/ region Moderating power and moderating ratio are two measures commonly used to describe the moderating effectiveness of materials. Materials with high moderating power typically have high scattering cross sections and induce large energy losses in the neutron for each scattering interaction. Moderating ratio, also takes into account capture cross section for the material. Materials with high moderating ratios have high moderating power and low probability of neutron capture interactions. Provided below (Tables 3, 4 and 5) are brief descriptions of the attenuation characteristics and physical properties of some materials commonly used in neutron shielding. Selection of the appropriate shielding material requires consideration of many factors including neutron moderation and capture properties, production of secondary radiation (n,g), potential neutron activation reactions which can induce radioactivity in the shielding material, and chemical and physical compatibility of the shielding material with the environment in which it will be used. References 1) Thomas E. Johnson and Brian K. Birky, Health Physics and Radiological Health, 4th ed., Wolters Kluwer/Lippincott Williams and Wilkins, 2012. 2) J. Kenneth Shultis and Richard E. Faw, “Radiation Shielding,” American Nuclear Society, Inc. La Grange Park, IL, 2000. 3) William D. Ehmann and Diane E. Vance, Radiochemistry and Nuclear Methods of Analysis, John Wiley and Sons, Inc., New York, 1991. 4) Phillip M. Rinard, “Neutron Interactions with Matter,” 5) O.W. Herman and C.W. Alexander, “A Review of Spent-Fuel Photon and Neutron Source Spectra,” ORNL/CS/TM-205, DE86-006764. 6) “MCNP6 User’s Manual,” version 1.0, May 2013, LA-CP-13-00634, Rev. 0 7) Neutron Scattering Lengths and Cross Sections, NIST Center for Neutron Research, From reference . Material Half Thickness, cm Parrafin 6.6 Water 5.4 12% Boric Acid-Water 5.3 Aluminum 7.8 Steel 4.9 Lead 6.8 Table 3. Half Thickness for Po-Be neutrons (4MeV) Average Capture Atomic Thermal Thermal Gamma Element Number Scattering Capture Emission (keV) Notes High scattering cross section.
Energetic gamma upon capture.
Relatively high capture cross section.
Energetic gamma upon capture.
B 5 5.24 0.103 7005 High capture cross section and relatively low B (n,a) 5 5.24 767 478 energy gamma emmision for (n,a) reaction.
C 6 5.551 0.0035 4945 O 8 4.232 0.00019 1469 Al 13 1.503 0.231 3737 Fe 26 11.62 2.56 4620 Difficult to shield activation products.
Very high capture cross section. Toxic metal.
Difficult to shield activation products.
Gd 64 180 49700 1440 Very high capture cross section. Expensive.
Good activation characteristics.
Adds good gamma attenuation. Expensive.
Adds good gamma attenuation. Low activation for pure grades.
Good activation characteristics Adds good gamma attenuation. Table compiled from data in references and .
High fission cross section.
Moderate capture cross section. Very low fission cross section. Good gamma attenuation.
Very high fission cross section.
Very high fission cross section.
Very high fission cross section.
U-234 92 19.3 100 (fission = 67) 574 (fission = 531) 12.9 92 U-233 681 (fission = 585) 14 92 U-235 2.68 (fission = 1.68E-05) 8.87 92 U-238 (depleted) 1017 (fission = 747) 7.7 94 Pu-239 cross section (b) Table 5. Neutron Attenuation Properties of Selected Materials H 1 82.02 0.3326 2223 Cd 48 6.5 2520 1522 Li 3 1.37 70.5 2094 Pb 82 11.118 0.171 7336 W 74 4.6 18.3 1758 Bi 83 9.156 0.0338 4118 Appendix I. MCNP Data 0 20 40 60 80 100 120 1E-3 0.01 0.1 1 10 100 TFlex (R)-W Concrete (2.2g/cm 3) Concrete (3.2g/cm 3) TFlex (R)-Bi Steel Borated Poly (5%) Lead Water % Transmission of Fission Neutron Dose cm shielding MCNP Calculation Figure 1. Attenuation of Neutron Dose. Plotted as transmission for clarity. Watt Fission Spectrum. 0 20 40 60 80 100 120 1E-3 0.01 0.1 1 10 100 TFlex (R)-W Concrete (2.2g/cm 3) Concrete (3.2g/cm 3) TFlex (R)-Bi Steel Borated Poly (5%) Lead Water % Gamma Dose from all (n,g) reactions cm shielding MCNP Calculation Figure 2. Secondary Gamma Dose (% of unshielded neutron dose) from all (n,g) reactions, including radiative capture and scatter of Watt Fission Neutrons. |
188769 | https://magoosh.com/gre/mastering-quantitative-comparison-questions-on-the-gre/ | GRE®Blog
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Mastering Quantitative Comparison Questions on the GRE
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GRE Math Question Types
The Quantitative Reasoning section of the GRE is all about flexing your math muscles and solving problems. One unique—and sometimes tricky—type of question you’ll encounter here is the Quantitative Comparison (QC). These questions require you to compare two quantities and determine their relationship, rather than providing a single numerical answer. In this article, we will explore strategies and tips to help you tackle QC questions with confidence and accuracy.
What are quantitative comparison questions?
In QC questions, you’re given two quantities—creatively named Quantity A and Quantity B. Your task is to compare these quantities and determine their relationship using one of the following options:
(A) Quantity A is greater.(B) Quantity B is greater.(C) The two quantities are equal.(D) The relationship cannot be determined from the information given.
You don’t need to find the exact values of Quantity A and Quantity B. Instead, focus on their relationship using the info the question gives you.
Example question
x^3=27
Quantity A: x
Quantity B: The smallest odd prime number
To solve the given equation x^3=27, take the cube root of both sides to find x=3.
A prime number is divisible by only two factors, 1 and itself. Thus, 1 is not a prime number, 2 is the smallest even prime number, and three is the smallest odd prime number. So, if x=3, and Quantity B = 3, the correct answer is (C) The two quantities are equal.
Tips and tricks for tackling QC questions
Keep It Simple, Smarty: Before diving into complex calculations, try simplifying both quantities. Cancel out terms, look for common factors, or convert expressions into something easier to handle. This could make comparing them a whole lot simpler (and faster!).
Estimate Like a Pro: Many QC questions don’t require super precise calculations. Instead, try using approximations or estimations to quickly compare the sizes of the quantities. This could save you time and help avoid calculation errors.
Plug in numbers: If the quantities involve variables, try plugging in specific numbers. Test different scenarios to figure out the relationship between Quantity A and Quantity B. This can be especially helpful with complex algebraic expressions.
Watch out for restrictions: Keep an eye out for any restrictions or limitations in the question. Certain conditions or constraints could impact the relationship between the quantities. So pay attention to the details!
Quantitative Comparison questions on the GRE might seem intimidating at first, but with a solid game plan and plenty of practice, you can definitely conquer them. Remember to simplify where you can, use approximations, and always stay mindful of any restrictions. By honing these skills, you’ll gain the confidence you need to ace the Quantitative Reasoning section and get the GRE score you’re aiming for.
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16: Vibrations with One Degree of Freedom
Mechanics Map (Moore, 2nd Edition)
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Sun, 18 Aug 2024 17:46:22 GMT
16.2: Viscous Damped Free Vibrations
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16.2: Viscous Damped Free Vibrations
Last updated Aug 18, 2024
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16.1: Undamped Free Vibrations
16.3: Friction (Coulomb) Damped Free Vibrations
Page ID 111433
Jacob Moore & Contributors
Pennsylvania State University Mont Alto via Mechanics Map
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Four Viscous Damping Cases:
1. 1. ζ=0: Undamped
2. 2. ζ>1: Overdamped
3. 3. ζ=1: Critically damped
4. 4. ζ<1: Underdamped
Comparison of Viscous Damping Cases:
Question 1:
Solution:
Question 2:
Solution:
Question 3:
Solution:
Viscous damping is damping that is proportional to the velocity of the system. That is, the faster the mass is moving, the more damping force is resisting that motion. Fluids like air or water generate viscous drag forces.
Figure 16.2.1: A diagram showing the basic mechanism in a viscous damper. As the system (mass) attached to the loop at the top vibrates up and down, the damper will resist motion in both directions due to the piston passing through the fluid. Image by Egmason, CC-BY SA.
We will only consider linear viscous dampers, that is where the damping force is linearly proportional to velocity. The equation for the force or moment produced by the damper, in either x or θ, is:
(16.2.1)F→c=cx˙→,
(16.2.2)M→c=cθ˙→,
where c is the damping constant. This is a physical property of the damper based on the type of fluid, size of the piston, etc. Note that the units of c change depending on whether it is damping linear motion (N-s/m) or rotational motion (N-m s/rad).
Figure 16.2.2: Diagram of a hanging mass-spring system, with a linear viscous damper, in equilibrium position. The spring is stretched from its natural length.
When the system is at rest in the equilibrium position, the damper produced no force on the system (no velocity), while the spring can produce force on the system, such as in the hanging mass shown above. Recall that this is the equilibrium position, but the spring is NOT at its unstretched length, as the static mass produces an extention of the spring.
Figure 16.2.3: Free body diagram of the system in equilibrium position. The spring is at its equilibrium position, but it is stretched and does produce a force.
If we perturb the system (applying an initial displacement, an initial velocity, or both), the system will tend to move back to its equilibrium position. What that movement looks like will depend on the system parameters (m, c, and k).
Figure 16.2.4: The system in a perturbed position. The spring is stretched further and the damper is extended, compared to their equilibrium positions.
To determine the equation of motion of the system, we draw a free body diagram of the system with perturbation and apply Newton's Second Law.
Figure 16.2.5: Free body diagram of the system with perturbation.
The process for finding the equation of motion of the system is again:
Sketch the system with a small positive perturbation (x or θ).
Draw the free body diagram of the perturbed system. Ensure that the spring force and the damper force have directions opposing the perturbation.
Find the one equation of motion for the system in the perturbed coordinate using Newton's Second Law. Keep the same positive direction for position, and assign positive acceleration in the same direction.
Move all terms of the equation to one side, and check that all terms are positive. If all terms are not positive, there is an error in the direction of displacement, acceleration, and/or spring or damper force.
For the example system above, with mass m, spring constant k and damping constant c, we derive the following:
(16.2.3)∑F x=ma x=mx¨
(16.2.4)−F k−F c=mx¨
(16.2.5)−kx−c(x)˙=mx¨
(16.2.6)mx¨+c(x)˙+kx=0
This gives us a differential equation that describes the motion of the system. We can rewrite it in normal form:
(16.2.7)mx¨+cx˙+kx=0
(16.2.8)⇒x¨+c mx˙+k mx=0
(16.2.9)⇒x¨+2ζω nx˙+ω n 2x=0
As before, the term ω n is called the angular natural frequency of the system, and has units of rad/s.
(16.2.10)ω n 2=k m;ω n=k m
ζ (zeta) is called the damping ratio. It is a dimensionless term that indicates the level of damping, and therefore the type of motion of the damped system.
(16.2.11)ζ=c c c=actual damping critical damping
The expression for critical damping comes from the solution of the differential equation. The solution to the system differential equation is of the form
(16.2.12)x(t)=ae rt,
where a is constant and the value(s) of r can be can be obtained by differentiating this general form of the solution and substituting into the equation of motion.
(16.2.13)mr 2e rt+cre rt+ke rt=0
(16.2.14)⇒(mr 2+cr+k)e rt=0
Because the exponential term is never zero, we can divide both sides by that term and get:
(16.2.15)mr 2+cr+k=0.
Using the quadratic formula, we can find the roots of the equation:
(16.2.16)r 1,2=−c±c 2−4mk 2m
Critical damping occurs when the term under the square root sign equals zero:
(16.2.17)c c 2=4mk
(16.2.18)c c=2mk=2mω n
Four Viscous Damping Cases:
There are four basic cases for the damping ratio. For the solutions that follow in each case, we will assume that the initial perturbation displacement of the system is x 0 and the initial perturbation velocity of the system is v 0.
1. ζ=0: Undamped
(16.2.19)c=0
This is the case covered in the previous section. Undamped systems oscillate about the equilibrium position continuously, unless some other force is applied.
Figure 16.2.6: Response of an undamped system.
2. ζ>1: Overdamped
(16.2.20)c 2>4mk
Roots are both real and negative, but not equal to each other. Overdamped systems move slowly toward equilibrium without oscillating.
Figure 16.2.7: Response of an overdamped system.
The response for an overdamped system is:
(16.2.21)x(t)=a 1e(−c+c 2−4mk 2m)t+a 2e(−c−c 2−4mk 2m)t,
(16.2.22)where a 1=−v 0+r 2x 0 r 2−r 1 and a 2=v 0+r 1x 0 r 2−r 1.
3. ζ=1: Critically damped
(16.2.23)c 2=4mk(=c c 2)
Roots are real and both equal to −ω n. Critically-damped systems will allow the fastest return to equilibrium without oscillation.
Figure 16.2.8: Response of an critically-damped system.
The solution for a critically-damped system is:
(16.2.24)x(t)=(A+Bt)e−ω nt,
(16.2.25)where A=x 0 and B=v 0+x 0ω n.
4. ζ<1: Underdamped
(16.2.26)c 2<4mk
The roots are complex numbers. Underdamped systems do oscillate around the equilibrium point; unlike undamped systems, the amplitude of the oscillations diminishes until the system eventually stops moving at the equilibrium position.
Figure 16.2.9: Response of an underdamped system.
The solution for an underdamped system is:
(16.2.27)x(t)=[C 1sin(ω dt)+C 2cos(ω dt)]e−ω nζt,
(16.2.28)where C 1=v 0+ω nζx 0 ω d,C 2=x 0,and ζ=c 2mω n.
This can alternately be expressed as:
(16.2.29)x(t)=Asin(ω dt+ϕ)e−ω nζt
Where:
(16.2.30)A=(v 0+ω nζx 0)2+(x 0ω d)2 ω d 2 ϕ=tan−1(x 0ω d v 0+ω nζx 0)ζ=c 2mω n
ω d is called the damped natural frequency of the system. It is always less than ω n:
(16.2.31)ω d=ω n1−ζ 2.
The period of the underdamped response differs from the undamped response as well.
(16.2.32)Undamped:τ n=2π ω n
(16.2.33)Underdamped:τ d=2π ω d
Comparison of Viscous Damping Cases:
Figure 16.2.10: Responses for all four types of system (or values of damping ratio) in viscous damping. All four systems have the same mass and spring values, and have been given the same initial perturbations (initial position and initial velocity); this is apparent because they start at the same y-intercept and have the same slope at x=0.
In the figure above, we can see that the critically-damped response results in the system returning to equilibrium the fastest. Also, we can see that the underdamped system amplitude is quite attenuated compared to the undamped case.
Question 1:
A 15 kg block on a frictionless surface is attached to a spring (k = 300 N/m). Find the damping constant, c, that will make the system critically damped.
Solution:
Question 2:
A 20 kg block on a frictionless surface is attached to a spring (k = 700 N/m) and a damper (c = 35 N-s/m). If the initial perturbation, x 0, is 0.2 m (v 0 = 0), how many half-cycles will it take for the amplitude of the oscillation to peak at half the original displacement or less (i.e. |x peak|<=|x 0/2|)?
Solution:
Question 3:
A bar of length 1.5m mass of 2kg is pinned to the ceiling. A spring, k=50N/m, is attached to the bottom of the bar and a damper, c=10Ns/m, is attached halfway down. Given a small angle displacement, find the damped frequency and the roots.
Solution:
This page titled 16.2: Viscous Damped Free Vibrations is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jacob Moore & Contributors (Mechanics Map) via source content that was edited to the style and standards of the LibreTexts platform.
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188772 | https://www.math.univ-paris13.fr/~schmid/personal/schmid_33t.pdf | AN APPLICATION OF CODING THEORY TO ESTIMATING DAVENPORT CONSTANTS ALAIN PLAGNE AND WOLFGANG A. SCHMID Abstract. We investigate a certain well-established generalization of the Dav-enport constant. For j a positive integer (the case j = 1, is the classical one) and a finite Abelian group (G, +, 0), the invariant Dj(G) is defined as the smallest ℓsuch that each sequence over G of length at least ℓhas j disjoint non-empty zero-sum subsequences. We investigate these quantities for elemen-tary 2-groups of large rank (relative to j). Using tools from coding theory, we give fairly precise estimates for these quantities. We use our results to give improved bounds for the classical Davenport constant of certain groups.
1. Introduction For a given Abelian group (G, +, 0), the Davenport constant, denoted D(G), is defined as the smallest integer ℓsuch that each sequence over G of length at least ℓ has a non-empty zero-sum subsequence, i.e. the sum of its terms is 0. Equivalently, D(G) is the maximal length of a minimal zero-sum sequence over G, i.e.
the maximal length of a sequence of elements of G summing to 0 and with no proper subsequence summing to 0.
It is considered as a central object in combinatorial number theory since Dav-enport popularized it in the 60’s (as reported in ), notably for its link with algebraic number theory, see e.g. or . In fact it seems that the first paper that deals with this invariant was written by Rogers , who himself attributes the paternity of the problem to Sudler.
This invariant has become the prototype of algebraic invariants of combinatorial flavour. Since the 60’s, the theory of these invariants has highly developed in several directions; see for instance the survey article or Chapters 5, 6, and 7 of .
Let G be written, as is always possible, as a direct sum of cyclic groups G ∼ = Cn1 ⊕· · · ⊕Cnr with integers 1 < n1 | · · · | nr (r denotes the rank of G, and nr the exponent, except for r = 0 where the exponent is 1). Then, the basic lower bound for Davenport constant is (1) D(G) ≥1 + r X i=1 (ni −1); to see this, note that a sequence containing only, for each cyclic component Cni (1 ≤i ≤n), one generating element ni −1 times, has no non-empty zero-sum subsequence.
It is known that for groups of rank at most two and for p-groups (p, a prime), inequality (1) is in fact an equality; this was obtained independently in and 2010 MSC: 11B30, 11B75, 11P70, 94B05, 94B65 W.S. is supported by the Austrian Science Fund (FWF): J 2907-N18.
1 2 A. PLAGNE, W. A. SCHMID [25, 26]. For groups of rank at least four, equality is definitely not the rule (see [1, 11, 18]). In the case of groups of rank three, it is conjectured that equality always holds but this conjecture is wide open (see ).
Concerning upper bounds, the best general result is the following D(G) ≤exp(G) 1 + log |G| exp(G) proved in [12, 23].
In view of the depicted situation, it appears that it is generically very difficult to determine the Davenport constant of a group (of rank at least three). In particular, despite various works related to the Davenport constant over the years, its actual value was only determined for a few additional – beyond the ones known since the end of the 60’s – families of groups; see for a recent contribution.
The j-wise Davenport constants are defined depending on a positive integer j.
We define Dj(G) to be the smallest ℓsuch that each sequence over G of length at least ℓhas j disjoint non-empty zero-sum subsequences. Equivalently, the maximal length of a zero-sum sequence over G that cannot be decomposed into j + 1 non-empty zero-sum sequences. Evidently D1(G) = D(G) and for any positive j one has Dj(G) ≤jD(G).
This variant of the Davenport constant was introduced by Halter-Koch in order to determine the order of magnitude of the counting function of the set of algebraic integers of some number field that are not divisible by a product of j + 1 irreducible algebraic integers of this number field. It is x log x(log log x)Dj(G)−1 where G denotes the ideal class group of the number field considered. Moreover, knowledge of these invariants is useful in investigations of the Davenport constant itself (cf. below for details and recall e.g. that a considerable part of , determining D(C2 3 ⊕C3n), is devoted to determining Dj(C3 3) and closely related problems).
The case of cyclic groups is the simplest one since then it is easy to see that jn −1 < Dj(Cn) ≤jD(Cn) = jn (for the lower bound, simply take a generating element repeated that number of times). The case of groups of rank two is also known , as well as the case of certain closely related groups [17, Section 6.1].
But in general computing (even bounding) Dj(G) is quite more complicated than for D(G), in particular for (elementary) p-groups. For example, Dj(G) for all j, is only known for the following elementary p-groups of rank greater than two: C3 2, C4 2, C5 2, and C3 3 (see [8, 14, 3]).
The main difference is that it seems that the types of arguments used to de-termine D(G) for p-groups cannot be applied. At first this might seem surprising, however recall that the same phenomenon is encountered for the invariant η(G) (cf. Section 2) and other closely related invariants such as the Erd˝ os–Ginzburg–Ziv constant. In particular, the difficulty of the problem seems to increase with the rank of the groups considered (to be precise, the key quantity is the size of the rank relative to the exponent), as then D(G) and η(G) are far apart (cf. Section 2 for the relevance of this fact).
In this paper and from now on, we focus on the case of elementary 2-groups with large rank. The reason for this is two-fold. On the one hand, it is an interesting case; the rank is ‘maximal’ relative to the exponent. On the other hand, the special CODING THEORY AND DAVENPORT CONSTANTS 3 nature of the group allows certain arguments that fail in more general situations; for instance recall that to show η(Cr 2) = 2r is almost trivial, yet the problem of determining η(Cr p) for any (fixed) odd prime p and arbitrary r is wide open (it is even open, which order of magnitude is to be expected).
While it is easy to determine that D1(Cr 2) = r + 1 – we effectively consider a vector space over a field with a unique non-zero element, and having no non-empty zero-sum subsequences is thus equivalent to linear independence – the situation becomes more complicated for larger values of j. We observe that r < D1(Cr 2) ≤Dj(Cr 2) ≤jD1(Cr 2) ≤j(r + 1).
Thus, for a fixed positive integer j, the sequence (Dj(Cr 2))r∈N has to grow linearly.
Our main aim here is to make this statement more precise and to study the following relevant quantity Dj(Cr 2) r , of which we examine the asymptotic behaviour when r becomes large. The quan-tities αj = lim inf r→+∞ Dj(Cr 2) r and βj = lim sup r→+∞ Dj(Cr 2) r will be considered and investigated. By the above crude reasoning we only get that 1 ≤αj ≤βj ≤j. Our first theorem, which gives explicit estimates for small values of j, improves on this. It is a generalization of results by Koml´ os (lower bound, quoted in [24, 5]) and by Katona and Srivastava (upper bound) for the case j = 2 that we recall for the sake of completeness in the statement of the theorem.
Theorem 1. For each sufficiently large integer r we have 1.261 r ≤ D2(Cr 2) ≤ 1.396 r, 1.500 r ≤ D3(Cr 2) ≤ 1.771 r, 1.723 r ≤ D4(Cr 2) ≤ 2.131 r, 1.934 r ≤ D5(Cr 2) ≤ 2.478 r, 2.137 r ≤ D6(Cr 2) ≤ 2.815 r, 2.333 r ≤ D7(Cr 2) ≤ 3.143 r, 2.523 r ≤ D8(Cr 2) ≤ 3.464 r, 2.709 r ≤ D9(Cr 2) ≤ 3.778 r, 2.890 r ≤ D10(Cr 2) ≤ 4.087 r.
As will be apparent from our arguments, having our method at hand, to expand this list further is merely a computational effort.
Notice that the question whether αj = βj (independently of the value that this constant would take) seems not obvious.
After the study of small values for j, we turn our attention to the case of large j’s.
Although we are unable to prove that αj and βj are equal, which seems conceivable, we show that they at least grow at the same speed and more precisely determine their order of magnitude.
Theorem 2. When j tends to infinity, we have the following: log 2 j log j ≲lim inf r→+∞ Dj(Cr 2) r ≤lim sup r→+∞ Dj(Cr 2) r ≲2 log 2 j log j .
4 A. PLAGNE, W. A. SCHMID We believe that the bound for the lim inf is closer to the actual value. A heuristic suggests that the lim sup in Theorem 2 is in fact close to log 2 (j/ log j) as well.
More precisely, we formulate the following conjecture.
Conjecture 3. For any positive integer j, the limit γj = lim r→+∞ Dj(Cr 2) r exists and one has γj ∼log 2 j log j as j tends to infinity.
For results in the converse scenario, that is fixed but arbitrary r, and j goes to infinity, see .
One of the main reasons for studying j-wise Davenport constants is the fact that they are important in obtaining results on the Davenport constant itself. The connection is encoded in the following inequality, due to Delorme, Ordaz, and Quiroz . For a finite Abelian group G and a subgroup H one has (2) D(G) ≤DD(H)(G/H).
Among others, this inequality encodes the classical form of the inductive method, originally introduced to determine the Davenport constant for groups of rank two (see [28, 11, 26]).
We can apply our results on the j-wise Davenport constants to obtain improved bounds on the Davenport constant for certain types of groups (the 2-rank has to be ‘large’ relative to the order). We only formulate it explicitly for a quite special type of group, which however, due to its extremal nature, is of relevance in this context.
Corollary 4. When n tends to infinity, we have lim sup r→+∞ D(Cr−1 2 ⊕C2n) r ≲2 log 2 n log n.
For comparison, the general bound mentioned above yields only lim sup r→+∞ D(Cr−1 2 ⊕C2n) r ≲2 log 2 n.
We immediately give the short proof of this result.
Proof. By (2), we get that D(Cr−1 2 ⊕C2n) ≤DD(Cn)(Cr 2) = Dn(Cr 2).
By Theorem 2, the claim follows.
□ We finish this Introduction with outlining the plan of the present article. In Section 2, we explain the methods and the prerequisites we need in the course of this article. In Section 3, we derive the lower bounds of our two results while in Section 4 the upper bounds are proved. Finally, in Section 5, we discuss the heuristic leading to Conjecture 3. If true, this heuristic would establish at least the asymptotic equivalence of αj and βj and imply the second part of Conjecture 3.
CODING THEORY AND DAVENPORT CONSTANTS 5 2. The methods We outline the methods we use to establish our results.
2.1. Zero-sum subsequences of bounded length. Let G be an Abelian group.
For x a real number, let s≤x(G) denote the smallest element ℓ∈N ∪{+∞} such that each sequence of length at least ℓhas a zero-sum subsequence of length at most x.
Evidently, s≤x(G) = s≤⌊x⌋(G), yet for technical reasons it is useful to define s≤x(G) for non-integral x as well.
A prominent special case of this definition is x = exp(G), the resulting invariant is typically denoted by η(G); note that for x < exp(G), one has s≤x(G) = +∞.
Also, note that for x ≥D(G) we have s≤x(G) = D(G).
For results on s≤x(G), for generic x, mainly for elementary 2- and 3-groups, see e.g. [8, 3], and for a very closely related problem (cf. below). For recent results on η(G), focusing on lower bounds, see [10, 9].
To determine η(G) seems to be a very difficult problem in general. For example, η(C6 3) was determined only recently , despite the fact that the problem of deter-mining η(Cr 3) is fairly popular (see for a detailed outline of several problems, and their respective history, that are equivalent to determining η(Cr 3)).
Yet, in the case of elementary 2-groups, to determine both η(Cr 2) and D(Cr 2) is almost trivial. However, for other values of x even for elementary 2-groups the problem of determining s≤x(Cr 2) is not at all trivial, namely, it is equivalent to a central problem of coding theory (cf. below).
It is known, in particular by the work of Delorme, Ordaz, and Quiroz , that the invariants s≤x(G) can be used to derive upper bounds for Dj(G). More specifically, we have (this is Lemma 2.4 in ) (3) Dj+1(G) ≤min i∈N max{Dj(G) + i, s≤i(G) −1}.
Thus, knowing D1(G) = D(G) and the constants s≤i(G), or bounds for these con-stants, one can obtain, recursively applying estimate (3), bounds for Dj(G). Notice however that even exact knowledge of D(G) and s≤i(G) for all i can be insufficient to determine Dj(G) exactly via this method, which in general is not optimal.
2.2. Coding theory enters the picture. We recall that the link between coding theory and combinatorial number theory is not new. For instance, Cohen, Litsyn, and Z´ emor used coding theoretic bounds in the Sidon problem. The general paper by two of these authors provides a worthwhile introduction to the links between the two problematics. Also, Freeze used coding theory in the present context.
One of the reasons for this connection is the following folkloric lemma .
Lemma 5. The minimal distance of a binary linear code C is equal to the minimal length of a zero-sum subsequence of columns of a parity check matrix of C.
In the case of present interest, Cohen and Z´ emor pointed out a connection between s≤i(Cr 2) and coding theory. As the situation at hand is slightly different from the one in that paper, and this connection is central to our investigations, we recall and slightly expand it in some detail.
First, we give a technically useful definition.
In the present context, we call a function f : [0, 1] →[0, 1] upper-bounding if it is decreasing (not necessarily 6 A. PLAGNE, W. A. SCHMID strictly), continuous and each [n, k, d] code satisfies k n ≤f d n .
In other words, upper-bounding functions are the functions intervening in the upper bounds of the rate of a code by a function of its normalized minimal distance.
We call a function asymptotically upper-bounding if it has the same properties as an upper-bounding function, except that the inequality only has to hold for [n, k, d] codes with sufficiently large n.
Notice that in both cases, the assumptions on decreasingness and continuity are not restrictive at all since these assumptions are usually fulfilled.
Lemma 6. Let f be an upper-bounding function. Let d, n, and r be three positive integers satisfying 2 ≤d ≤n −1 and n −r n > f d + 1 n , then s≤d(Cr 2) ≤n.
Moreover, the same assertion holds true for f an asymptotically upper-bounding function if we impose that n is sufficiently large (depending on f).
Proof. Let S = g1, . . . , gn be an arbitrary finite sequence over Cr 2. We shall prove that it contains a zero-sum subsequence of length at most d.
It is immediate that S has a zero-sum subsequence of length 1 if and only if 0 occurs in S, and that S has a zero-sum subsequence of length 2 if and only if some element occurs at least twice in S. Thus, since d ≥2, we may assume that S does neither contain 0 nor an element more than once i.e. we effectively have to study the case of subsets of Cr 2 \ {0} (and not the one of general sequences in Cr 2).
We assert that we may assume that the elements appearing in S generate Cr 2. To see this, note that if g is an element of Cr 2 not contained in the subgroup generated by the elements of S, and if T denotes the sequence obtained by appending g to S, then each zero-sum subsequence of T is in fact a zero-sum subsequence of S. Thus, if S has no zero-sum subsequence of length at most d and the elements of S do not generate Cr 2, then the longer sequence T, defined as above, neither has a zero-sum subsequence of length at most d. So, it suffices to establish an upper bound on the length of sequences S such that the elements appearing in S generate Cr 2.
We choose some basis of Cn 2 . We consider the binary linear code C ⊂Cn 2 of length n whose parity check matrix is A = [g1 | · · · | gn] ∈Mr,n (identify the gi’s with their coordinate vectors with respect to some basis of Cr 2, and consider them as column vectors, and use the just chosen basis of Cn 2 ). Notice that the rank of this matrix is equal to r in view of our assumption that the gi’s generate Cr 2. Let m be the minimal distance of C. By definition, the code C is an [n, n −r, m] binary linear code. But by assumption since f is upper-bounding and n −r n > f d + 1 n , an [n, n −r, d + 1] code cannot exist. This implies that m < d + 1, or equivalently d ≥m.
CODING THEORY AND DAVENPORT CONSTANTS 7 We conclude by applying Lemma 5 which shows that S possesses a zero-sum subsequence of length m.
The additional claim for asymptotically upper-bounding functions, is immediate in view of the just given argument.
□ 3. Lower bounds In this section, we establish the lower bounds for Dj(Cr 2), for large r, contained in Theorems 1 and 2. Specifically, we prove the following asymptotic lower bound in r, which immediately yields both lower bounds (replacing log(j + 1) by log j is asymptotically, in j, irrelevant).
Proposition 7. Let j be a positive integer. Then Dj(Cr 2) ≥log 2 j log(j + 1) r as r tends to infinity.
Notice that the case j = 1 is essentially trivial, while the case j = 2 of this result, formulated in the context of coding theory, is attributed to Koml´ os in and .
Indeed, our proof will generalize Koml´ os’ approach (in the form given in ). This proof can be seen as probabilistic, yet we prefer to present it via a direct counting argument. Notice that it is non-constructive.
We need the following well-known lemma (see e.g. ).
Lemma 8. Let n and k be two positive integers, n ≥k. In an n-dimensional vector space over a field with 2 elements, the number of k-dimensional subspaces is equal to the 2-ary binomial coefficient defined as n k = (2n −1) · · · (2n−k+1 −1) (2k −1) · · · (2 −1) .
Moreover, the number of k-dimensional subspaces containing a fixed j-dimensional subspace, k ≥j, is equal to n −j k −j .
We can now prove Proposition 7.
Proof of Proposition 7. For the entire proof, we fix an arbitrary positive integer j > 1. As D1(Cr 2) = D(Cr 2) = r + 1, we can ignore the case j = 1.
We shall now prove that for each integer n larger than or equal to r + j (this condition is technically convenient later on) and less than (j log 2/ log(j + 1)) r (notice that, for r large enough, since j ≥2, such n’s always exist), one can find a sequence of cardinality n which does not contain j disjoint zero-sum subsequences.
This will prove our result.
To each sequence S = g1, . . . , gn over Cr 2, with n ≥r + j, having the properties that S does neither contain 0 nor an element at least twice, we associate, as de-scribed above, an [n, n −r] code (contained in Cn 2 , and we fix some basis). This linear code, automatically, has a minimal distance of at least 3. Conversely, any [n, n −r, d] code with d ≥3 can be obtained in this way (cf. ).
The following remark is central: the condition that S has j disjoint zero-sum subsequences translates to the condition that the associated code contains j non-zero codewords c1, . . . , cj such that intersection of the support (the set of indices 8 A. PLAGNE, W. A. SCHMID of non-zero coordinates) of cu and cv is empty for all distinct u, v ∈{1, . . . , j}. A code having this property will be called j-inadmissible, otherwise it will be called j-admissible.
We first produce an upper bound on the total number of [n, n −r] codes that are j-inadmissible.
By definition any j-inadmissible code contains c1, . . . , cj with the above men-tioned property. These ci’s generate a j-dimensional vector space since the ci’s are certainly independent: the non-zero coordinates of each ci are unique to that element.
Let V denote the set of all subsets {d1, . . . , dj} ⊂Cn 2 \ {0} such that the inter-section of the support of du and dv is empty for all distinct u, v ∈{1, . . . , j}; thus, in particular, all the di’s are distinct.
We note that a code C is j-inadmissible if and only if V ⊂C for some V ∈V (this V is not necessarily unique). Moreover Lemma 8 implies that for each V ∈V there are n−j n−r−j codes containing V ; note that if V ⊂C then C also contains the vector space generated by V , which is j-dimensional, and apply Lemma 8. It follows that the total number of j-inadmissible codes cannot exceed |V| n −j n −r −j .
In order to estimate |V| we make the following remark: each element of {1, . . . , n} has to belong to either the support of exactly one of the di’s or to none (obviously, the information which element of {1, . . . , n} belongs to each of the di’s uniquely determines the element of V). Thus, for each element of {1, . . . , n} there are (at most) j + 1 possibilities. This readily gives |V| ≤(j + 1)n. (We ignore the slight improvements that could be obtained from the fact that the ordering of the di’s is irrelevant and the supports are non-empty, as they would not affect our estimate).
We therefore infer that the total number of j-inadmissible [n, n −r] codes is bounded above by (j + 1)n n −j n −r −j .
Again by Lemma 8, it follows that the ratio of the total number of j-inadmissible [n, n −r] codes divided by the total number of [n, n −r] codes is bounded above by (j + 1)n n−j n−r−j n n−r = (j + 1)n n Y k=n−j+1 2k−r −1 2k −1 ≤ (j + 1)n n Y k=n−j+1 2k−r 2k = (j + 1)n2−rj = 2n log2(j+1)−rj.
Here, log2 refers to the logarithm in basis 2.
Thus, it follows that as soon as (n log2(j + 1) −rj) is negative, that is n r < j log2(j + 1), the existence of at least one admissible code is guaranteed.
CODING THEORY AND DAVENPORT CONSTANTS 9 From this we deduce (the condition n ≥r + j becomes irrelevant) Dj(Cr 2) ≥log 2 j log(j + 1) r as r tends to infinity.
□ 4. Upper bounds 4.1. The crucial lemma. The following lemma is central for our investigations.
Lemma 9. Let f be an asymptotic upper-bounding function. Let j be a positive integer and p be a real number such that one has Dj(Cr 2) ≤pr for each sufficiently large r. Let finally c denote a solution to the inequality p + c −1 p + c > f c p + c .
Then for each sufficiently large integer r, we have Dj+1(Cr 2) ≤(p + c)r.
Proof. By assumption, we have that (p + c)r −r (p + c)r = p + c −1 p + c > f c p + c = f cr (p + c)r ≥f cr + 1 (p + c)r , f being decreasing. If we substitute ⌊(p + c)r⌋for n and ⌊cr⌋for d, we obtain (for r sufficiently large and by the continuity of f) n −r n > f d + 1 n , an equation of the type given in Lemma 6 and we may therefore deduce that s≤⌊cr⌋(Cr 2) ≤n = ⌊(p + c)r⌋≤(p + c)r.
This now implies, by (3) and using our assumption, that Dj+1(Cr 2) ≤ min i∈N max{pr + i, s≤i(Cr 2) −1} ≤ max{pr + ⌊cr⌋, s≤⌊cr⌋(Cr 2) −1} ≤ max{(p + c)r, (p + c)r −1} = (p + c)r, as wanted.
□ 4.2. The upper bounds in Theorem 1. To obtain a proof of these upper bounds, we use the approach described in Section 2 in combination with a bound on the parameters of linear codes originally due to McEliece, Rodemich, Rumsey, and Welch , that we recall here (see e.g. ): Let us define h to be the binary entropy function, that is (for 0 ≤u ≤1), h(u) = −u log2 u −(1 −u) log2(1 −u) and g(u) = h((1 −√1 −u)/2). Then the function f equal to f1(δ) = ( min0≤u≤1−2δ 1 + g(u2) −g(u2 + 2δu + 2δ) if δ ≤1/2 0 otherwise is an asymptotically upper-bounding function and we may thus apply Lemma 9.
10 A. PLAGNE, W. A. SCHMID We define a sequence (uj)j∈N recursively. We set u1 = 1 and, for j ≥1, let uj+1 be defined as the solution (if it exists, which is always the case in practice, this solution has to be unique) to the equation 1 − 1 Uj + uj+1 = f1 uj+1 Uj + uj+1 where we define Uj = u1 + · · · + uj, the sum of the j first values of the sequence (uj)j∈N. The sequence (Uj)j∈N corresponds to the coefficient in the upper bound of Theorem 1 (U1 = 1, U2 = 1.395 . . . , U3 = 1.770 . . . , . . . ), which therefore follows by a repeated application of Lemma 9.
We point out that for our problem it is actually useful to use this bound, as opposed to bounds whose numerical evaluation is simpler, since for our problem we encounter δ in a fairly wide range. A simpler strategy, regarding computations, which is – in view of classical results on these bounds – obviously worse, though only slightly so, would be to use another bound proved in (see also ), that is the function f2(δ) = ( h 1/2 − p δ(1 −δ) if δ ≤1/2 0 otherwise for the first few values, namely 2, 3, 4 (for 2 this yields the identical bound, yet a slightly weaker one for 3, 4) where δ is still fairly large, and then to switch to using a bound that is better for small δ such as the Elias–Bassalygo bound (see and ), that is f3(δ) = ( 1 −h (1 − √ 1 −2δ)/2 if δ ≤1/2 0 otherwise with h as above (this is the case u = 0 of (4.2)). Using this approach, we would for instance get the values 1.776 for j = 3, 2.147 for j = 4, 2.512 for j = 5 and 4.172 for j = 10 that is, slightly but noticeably weaker bounds.
4.3. The upper bound in Theorem 2. For the asymptotics we use the method described in Section 2 in combination with the Hamming bound, that is the function f4 defined by f4(δ) = 1 −h δ 2 .
We use this bound as the resulting analytic expressions and asymptotic calcula-tions are simpler than for stronger bounds. On the other hand, asymptotically, using say Elias–Bassalygo would not yield a better result. Interestingly, for this particular choice of upper-bounding function, a very elementary proof (cf. below) of the conclusion of Lemma 9 can be obtained with additive means, avoiding the coding theoretic argument that this function is an upper-bounding function. This makes our proof of Theorem 2 essentially self-contained.
Proof. We start with a sequence S = g1, . . . , gn in Cr 2, where n = ⌊(p + c)r⌋. The set {1, . . . , n} has ⌊cr/2⌋ X j=0 ⌊(p + c)r⌋ j > ⌊(p + c)r⌋ ⌊cr/2⌋ CODING THEORY AND DAVENPORT CONSTANTS 11 subsets of size at most cr/2. Since by assumption (p + c) h c 2(p + c) > 1 and since, as r tends to infinity, αr βr ∼c 2α h(β/α)r (for some constant c depending on α and β), it follows that, for sufficiently large r, the number of such sets exceeds 2r. This implies that there exist two distinct subsets I, J of {1, . . . , n} such that X i∈I gi = X i∈J gi.
This yields X i∈I△J gi = 0, – where I△J denotes the symmetric difference of I and J –, that is a non-empty zero-sum sequence of length at most cr.
□ Using the upper-bounding function f4, we build a sequence, let us call it (vj)j∈N such that v1 = 1 and (4) 1 Vj + vj+1 = h vj+1 2(Vj + vj+1) , where Vj = v1 + · · · + vj for any integer j ≥1. It can be checked easily that such a sequence is well-defined and that for any j ≥1, one has vj ≤1. Then, rewriting (4) as 1 − 1 Vj + vj+1 = f4 vj+1 Vj + vj+1 , we may repeatedly apply Lemma 9 and obtain the inequality Dj(Cr 2) ≤Vj r for each integer j and all sufficiently large r (relative to j).
By the lower bounds in Theorem 2 proved in the preceding Section, we already know that Vj tends to infinity when j tends to infinity since (5) Vj ≳Dj(Cr 2) r ≫ j log j as r tends to infinity ; while vj remains bounded by 1.
We can now develop (4) to obtain the desired asymptotics. At the first order when j tends to infinity we obtain 1 Vj + vj+1 = − vj+1 2(Vj + vj+1) log2 vj+1 2(Vj + vj+1) + O vj+1 Vj + vj+1 or equivalently 2 log 2 vj+1 = −log vj+1 2(Vj + vj+1) + O(1) = log(Vj + vj+1) −log vj+1 + O(1) = log Vj −log vj+1 + O(1), 12 A. PLAGNE, W. A. SCHMID and therefore 2 log 2 vj+1 + log vj+1 = log Vj + O(1).
It follows that, as j tends to infinity, vj+1 ∼2 log 2 log Vj .
By (5), we obtain Vj+1 −Vj = vj+1 ≲2 log 2 log j and therefore, summing all these estimates yields Vj ≲2 log 2 j−1 X k=1 1 log k ∼2 log 2 j log j from which the upper bound of Theorem 2 follows.
5. Heuristics In this section, we discuss the quality of the bound in Theorem 1. In investi-gations on intersecting codes, which are equivalent to determining D2(Cr 2), Cohen and Lempel put forward the heuristic that one can expect that the rate of an intersecting code will not exceed the Gilbert–Varshamov bound 1 −h(δ).
Extrapolating this heuristic to the investigation of Dj(Cr 2), which might be too op-timistic as the restrictions imposed on the code corresponding to extremal example associated to Dj(Cr 2) get weaker as j increases, this suggests to use the function 1 −h(δ) as if it were an upper-bounding function.
An argument similar to the one in the proof of Theorem 2 thus yields an opti-mistic heuristic bound βj ≲log 2 j log j .
Regarding heuristic numerical values we get, for example, 1.294 for j = 2, 1.550 for j = 3, 1.784 for j = 4, 2.003 for j = 5, and 2.984 for j = 10.
Acknowledgment The authors would like to thank G. Cohen for helpful discussions.
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Alain Plagne, Wolfgang A. Schmid Centre de Math´ ematiques Laurent Schwartz UMR 7640 du CNRS ´ Ecole polytechnique 91128 Palaiseau cedex France E-mail address: plagne@math.polytechnique.fr, schmid@math.polytechnique.fr |
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1 Dipolar coupling and NMR spectroscopy
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Magnetic dipole–dipole interaction
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Direct interaction between two magnetic dipoles
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Magnetic dipole–dipole interaction, also called dipolar coupling, refers to the direct interaction between two magnetic dipoles. Roughly speaking, the magnetic field of a dipole goes as the inverse cube of the distance, and the force of its magnetic field on another dipole goes as the first derivative of the magnetic field. It follows that the dipole-dipole interaction goes as the inverse fourth power of the distance.
Suppose m1 and m2 are two magnetic dipole moments that are far enough apart that they can be treated as point dipoles in calculating their interaction energy. The potential energy H of the interaction is then given by:
where μ0 is the magnetic constant, r̂ is a unit vector parallel to the line joining the centers of the two dipoles, and |r| is the distance between the centers of m1 and m2. Last term with -function vanishes everywhere but the origin, and is necessary to ensure that vanishes everywhere. Alternatively, suppose γ1 and γ2 are gyromagnetic ratios of two particles with spin quanta S1 and S2. (Each such quantum is some integral multiple of 1/2.) Then:
where is a unit vector in the direction of the line joining the two spins, and |r| is the distance between them.
Finally, the interaction energy can be expressed as the dot product of the moment of either dipole into the field from the other dipole:
where B2(r1) is the field that dipole 2 produces at dipole 1, and B1(r2) is the field that dipole 1 produces at dipole 2. It is not the sum of these terms.
The force F arising from the interaction between m1 and m2 is given by:
The Fourier transform of H can be calculated from the fact that
and is given by[citation needed]
Dipolar coupling and NMR spectroscopy
[edit]
The direct dipole-dipole coupling is very useful for molecular structural studies, since it depends only on known physical constants and the inverse cube of internuclear distance. Estimation of this coupling provides a direct spectroscopic route to the distance between nuclei and hence the geometrical form of the molecule, or additionally also on intermolecular distances in the solid state leading to NMR crystallography notably in amorphous materials.
For example, in water, NMR spectra of hydrogen atoms of water molecules are narrow lines because dipole coupling is averaged due to chaotic molecular motion. In solids, where water molecules are fixed in their positions and do not participate in the diffusion mobility, the corresponding NMR spectra have the form of the Pake doublet. In solids with vacant positions, dipole coupling is averaged partially due to water diffusion which proceeds according to the symmetry of the solids and the probability distribution of molecules between the vacancies.
Although internuclear magnetic dipole couplings contain a great deal of structural information, in isotropic solution, they average to zero as a result of diffusion. However, their effect on nuclear spin relaxation results in measurable nuclear Overhauser effects (NOEs).
The residual dipolar coupling (RDC) occurs if the molecules in solution exhibit a partial alignment leading to an incomplete averaging of spatially anisotropic magnetic interactions i.e. dipolar couplings. RDC measurement provides information on the global folding of the protein-long distance structural information. It also provides information about "slow" dynamics in molecules.
See also
[edit]
J-coupling
Magic angle
Residual dipolar coupling
Nuclear Overhauser effect
Magnetic moment
Zero field splitting
References
[edit]
Malcolm H. Levitt, Spin Dynamics: Basics of Nuclear Magnetic Resonance. ISBN 0-471-48922-0.
^ Abragam, A. (1961) The Principles of Nuclear Magnetism. Oxford University Press, Oxford.
^ Gabuda, S.P.; Lundin, A.G.(1969) Diffusion of Water Molecules in Hydrates and NMR Spectra. JETP, 28 (3), 555. Archived 2018-05-17 at the Wayback Machine
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Expressing ζ k+ζ−k ζ k+ζ−k as a polynomial in ζ+ζ−1 ζ+ζ−1.
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Let ζ ζ be an n n-th root of unity and let χ:=ζ+ζ−1 χ:=ζ+ζ−1. Then ζ k+ζ−k=P k(χ)ζ k+ζ−k=P k(χ) where P k∈Z[X]P k∈Z[X] is a polynomial not depending on n n. For example we have
ζ 2+ζ−2 ζ 3+ζ−3 ζ 4+ζ−4 ζ 5+ζ−5 ζ 6+ζ−6=====χ 2−2,χ 3−3 χ,χ 4−4 χ 2+2,χ 5−5 χ 3+5 χ,χ 6−6 χ 4+9 χ 2+18,so so so so so P 2 P 3 P 4 P 5 P 6=====X 2−2,X 3−3 X,X 4−4 X 2+2,X 5−5 X 3+5,X 6−6 X 4+9 X 2+18.ζ 2+ζ−2=χ 2−2,so P 2=X 2−2,ζ 3+ζ−3=χ 3−3 χ,so P 3=X 3−3 X,ζ 4+ζ−4=χ 4−4 χ 2+2,so P 4=X 4−4 X 2+2,ζ 5+ζ−5=χ 5−5 χ 3+5 χ,so P 5=X 5−5 X 3+5,ζ 6+ζ−6=χ 6−6 χ 4+9 χ 2+18,so P 6=X 6−6 X 4+9 X 2+18.
It isn't hard to see that P a b=P a∘P b=P b∘P a P a b=P a∘P b=P b∘P a for all positive integers a a and b b, and that we have a recurrence relation
P a=X a−∑i=1(a i)P a−2 i,P a=X a−∑i=1(a i)P a−2 i,
where we take the convention that P k=0 P k=0 for all k<0 k<0, and P 0=1 P 0=1. My question is:
Is there a simple explicit expression for P k P k?
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edited Aug 22, 2021 at 21:09
ServaesServaes
asked Nov 15, 2018 at 16:38
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2 This may help you: en.wikipedia.org/wiki/Chebyshev_polynomialsSeewoo Lee –Seewoo Lee 2018-11-15 16:40:57 +00:00 Commented Nov 15, 2018 at 16:40
@SeewooLee Can you turn that comment into an answer?Pedro –Pedro♦ 2018-11-15 16:43:59 +00:00 Commented Nov 15, 2018 at 16:43
Anyone care to explain the downvote?Servaes –Servaes 2020-09-25 21:29:18 +00:00 Commented Sep 25, 2020 at 21:29
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As I said, Chebyshev polynomials are exactly what you said.
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answered Nov 15, 2018 at 17:55
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188776 | https://mathcracker.com/discriminant-formula-calculator | MathCracker.com
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Discriminant Formula Calculator
Instructions:
Use this calculator to find the discriminant of a quadratic equation, showing all the steps. Please type in a
valid quadratic equation in the form box below.
Enter a valid quadratic equation (Ex: 2x^2 + 3x - 2 = 0, etc.)
Discriminant Formula
This calculator will use the discriminant formula showing all the steps for a quadratic equation that you provide.
You need to provide a valid quadratic equation, something like 2x²+x-1=0, which comes already simplified, or you
can provide something that is a valid quadratic expression, but needs further simplification like
2x²+3x-1 = 3/4x - 4/5.
Once a valid quadratic equation is provided, all you need to do is to click the "Calculate" button, and all the steps of the calculation will be
provided to you.
The simplified quadratic equation in the form ax² + bx + c = 0 will be used for the calculation of the discriminant, which will indicate right away the
nature of the roots: Two real roots, one real root, or two complex roots.
The discriminant formula
How to find the discriminant of a quadratic equation? Once you have the quadratic equation in the form ax² + bx + c = 0, you can apply directly the discriminant formula:
Discriminant meaning
Once you have applied the above formula, and you get a value (\Delta) for the discriminant, what is its meaning?
What is the meaning of two conjugate complex roots? Graphically, it is simply a parabola that does not cross
the x-axis.
On the other hand, two different real roots implies graphically that the parabola crosses the x-axis at two points. A discriminant equal to zero indicates
that the parabola is tangent to the x-axis.
Why would care about the discriminant?
The discriminant provides you with an easy form to assess the types of root a quadratic equation has, without actually solving the equation.
Naturally, we can see that the discriminant literally appears in the quadratic formula, so it is obviously linked
with the process of calculating quadratic roots.
Example: Calculating discriminant
Find the discriminant of the following equation: (x^2+ 3x + 10 = 0)
Solution: We need to solve the following given quadratic equation (\displaystyle x^2+3x+10=0).
For a quadratic equation of the form (a x^2 + bx + c = 0), the discriminant is computed using the following formula:
In this case, we have that the equation we need to solve is (\displaystyle x^2+3x+10 = 0), which implies that corresponding coefficients are:
Plugging these values into the formula we get:
Therefore, the discriminant for the given quadratic equation is (\Delta = \displaystyle -31 < 0),
which is negative, and that indicates that the given equation (\displaystyle x^2+3x+10=0) has two different conjugate complex roots.
This concludes the calculation of the determinant.
Example: Discriminant calculation
Find the discriminant of the following equation: (3x^2 - 2x + 4 = 0)
Solution: In this case, since the quadratic equation we need to solve is (\displaystyle x^2+3x+10 = 0), which
is in its simplified form, the corresponding coefficients are:
Plugging these values into the above formula we find that:
So then, the discriminant for the given quadratic equation is (\Delta = \displaystyle -44 < 0), which is negative. Therefore, the given equation
(3x^2 - 2x + 4 = 0) has two different conjugate complex roots.
This concludes the calculation.
Example: Discriminant Meaning
Without solving the equation (2x^2 - 3x - 10 = 0), indicate the nature of its roots.
Solution: In this case, we need to solve is (2x^2 - 3x + 1 = 0), so then the corresponding coefficients are:
Plugging these values into the determinant formula we find that:
So then, the discriminant for the given quadratic equation is (\Delta = 89 > 0), which is positive. Therefore, without solving the equation, we know that
the given equation
(2x^2 - 3x - 10 = 0) has two different real roots.
More quadratic calculators
Dealing with quadratic functions and equations is very common in Algebra. Computing roots of quadratic equations is tightly
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Geometrically, the discriminant will indicate the type of disposition of a the parabola that represents the quadratic function and the x-axis.
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Cylinder of Greatest Volume Inscribed in a Cube with its Axis as the Diagonal
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$\begingroup$
Given a cube of side $a$, what is the volume of the greatest cylinder that can be inscribed in it, such that its axis coincides with the longest diagonal of the cube?
It is quite obvious that for obtaining maximum volume, we would have to stretch the cylinder until it touches the cubes faces at six distint points, but is that even possible? Ill explain what I mean by converting this (sort of) to a $2D$ problem.
Lets say I have a rectangle $ABCD$, and inside the rectange is another (smaller) rectangle with its axis being the diagonal $AC$. (axis here refers to the line passing through the center and parallel to any one pair of sides). Now, intuitively or even by drawing a picture its evident that it shouldnt be possible to have all four corners of the smaller rectangle touching the bigger rectangle, leaving one corner hanging in the air. Is there a way to prove/disprove this? Its quite possible in the case of a square, though.
What I did here was take a side-view of the original configuration, where the bigger rectangle is determined by the farthest edges (of length $a$) of the cube and the shorter diagonals (length $\sqrt 2 a$).
Im looking for is way to relate the height and radius of the cylinder, after which the task is trivial using calculus.
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edited Apr 20, 2020 at 22:07
VishuVishu
asked Apr 17, 2020 at 21:21
VishuVishu
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$\endgroup$
6
$\begingroup$ If the inscribed cylinders axis is a long diagonal, then it will touch at six points, not four: three faces converge symmetrically at each vertex. That symmetry is key: if you the faces that converge at either end of the diagonal, you get an equilateral triangle. $\endgroup$
amd
– amd
2020-04-17 22:21:13 +00:00
Commented Apr 17, 2020 at 22:21
$\begingroup$ @amd Yes, my bad. Can you explain your last statement? $\endgroup$
Vishu
– Vishu
2020-04-17 22:29:57 +00:00
Commented Apr 17, 2020 at 22:29
$\begingroup$ Left out a bunch of words. Sorry. It should read if you intersect the faces that converge at either end of the diagonal with a plane perpendicular to the diagonal, you get an equilateral triangle. $\endgroup$
amd
– amd
2020-04-17 22:34:18 +00:00
Commented Apr 17, 2020 at 22:34
$\begingroup$ @amd and how would that help? $\endgroup$
Vishu
– Vishu
2020-04-17 22:48:44 +00:00
Commented Apr 17, 2020 at 22:48
$\begingroup$ The circle inscribed in that triangle is an end cap of the cylinder. It also shows that each end of the cylinder contacts the cube at three points. $\endgroup$
amd
– amd
2020-04-17 22:49:59 +00:00
Commented Apr 17, 2020 at 22:49
| Show 1 more comment
2 Answers 2
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+50
$\begingroup$
Position the cube so that a vertex is at the origin and the cube lies in the first octant. The long diagonal has length $a\sqrt3$, so if the cylinders height is $h$, its near cap lies at a distance of $\frac12\left(a\sqrt3-h\right)$ from the origin. The normal to this caps plane is $(1,1,1)$, so an equation of this plane is $$x+y+z=\frac{\sqrt3}2\left(a\sqrt3-h\right).$$ The expression on the right-hand side is the axis-intercept of this plane with all three coordinate axes. The end cap touches the $x$-$y$ plane at the midpoint of the $x$- and $y$-intercepts, so using the Pythagorean theorem, we can obtain the square of its radius, namely $$\frac38\left(a\sqrt3-h\right)^2-\frac14(a\sqrt3-h)^2 = \frac18\left(a\sqrt3-h\right)^2.$$ The rest of the task, as you say, is a trivial calculus exercise.
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answered Apr 20, 2020 at 23:46
amdamd
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$\begingroup$
Following amds comment consider a plane containing the cup circle (of radius $r$) of the cylinder. The plane cuts a pyramid (of height $h$) from the cube. Let $V$ be the volume of the pyramid and $S$ be the area of its base, which is an equilateral triangle with a side $s$. Then $V=\frac 13 Sh=\frac {\sqrt 3}{12}s^2h$. On the other hand, $V=\frac {d^3}6$, where $d$ is the length of the part of an edge of the cube cut by the plane. Also we have $s^2=2d^2$. It follows $V=\frac {\sqrt 3}{12}s^2h=\frac {s^3}{12\sqrt{2}}$ and so $s=h\sqrt{6}$. The circle inscribed into the triangle has radius $r=\frac s{2\sqrt{3}}=\frac h{\sqrt 2}$. The cyliner has height $H=a\sqrt{3}-2h$ and the volume $$\pi r^2H=\pi\frac {h^2}{2}\left(a\sqrt{3}-2h \right).$$
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answered Apr 20, 2020 at 20:02
Alex RavskyAlex Ravsky
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$\endgroup$
4
$\begingroup$ How is $V=\frac{d^3}{6}$? Is there some formula Im not seeing? $\endgroup$
Vishu
– Vishu
2020-04-20 20:37:07 +00:00
Commented Apr 20, 2020 at 20:37
$\begingroup$ We have $V=\frac {d^3}6$ because the edges of the pyramid incident to the vertex $A$ of the cube are mutually perpendicular and have length $d$. So an area of any face incident to $A$ equals $\frac {d^2}2$ and the height of the pyramid perpendicular to the face has length $d$. $\endgroup$
Alex Ravsky
– Alex Ravsky
2020-04-21 01:34:31 +00:00
Commented Apr 21, 2020 at 1:34
$\begingroup$ Nice. Can any face of a pyramid be taken as its base for volume calculations? $\endgroup$
Vishu
– Vishu
2020-04-21 07:32:23 +00:00
Commented Apr 21, 2020 at 7:32
$\begingroup$ @Tavish Yes, if all the faces of the pyramid are triangles. $\endgroup$
Alex Ravsky
– Alex Ravsky
2020-04-21 07:58:19 +00:00
Commented Apr 21, 2020 at 7:58
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🏎️Engineering Mechanics – Dynamics Unit 7 Review
7.1 Work-energy principle for rigid bodies
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Engineering Mechanics – Dynamics
Unit 7 – Energy and Momentum in Rigid Body Dynamics
Topic: 7.1
🏎️Engineering Mechanics – Dynamics Unit 7 Review
7.1 Work-energy principle for rigid bodies
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
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🏎️Engineering Mechanics – Dynamics
Unit & Topic Study Guides
Particle Kinematics
Kinetics of Particles in Engineering Mechanics
Work, Energy, and Power in Dynamics
Impulse and Momentum in Dynamics
Rigid Body Kinematics
Kinetics of Rigid Bodies in Dynamics
Energy and Momentum in Rigid Body Dynamics
7.1 Work-energy principle for rigid bodies
7.2 Conservation of energy for rigid bodies
7.3 Impulse-momentum principle for rigid bodies
7.4 Conservation of angular momentum
7.5 Rigid body collisions
3D Dynamics in Engineering Mechanics
Vibrations
Gyroscopic Motion in Engineering Dynamics
Orbital Mechanics in Engineering Dynamics
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The work-energy principle for rigid bodies is a powerful tool in dynamics, connecting work done on a system to changes in its energy. It offers an alternative to Newton's laws, simplifying analysis of complex motions in both linear and rotational systems.
This principle encompasses kinetic energy (translational and rotational), work by external forces, potential energy, and power. It's particularly useful for solving problems involving variable forces, constraints, and multi-body systems, often reducing vector problems to scalar equations.
Work-energy principle overview
Fundamental concept in Engineering Mechanics – Dynamics linking work done on a system to changes in its energy
Provides alternative approach to Newton's laws for analyzing rigid body motion
Applies to both linear and rotational motion of rigid bodies in dynamic systems
Kinetic energy of rigid bodies
Translational kinetic energy
Energy associated with linear motion of a rigid body's center of mass
Calculated using the formula K E t r a n s=1 2 m v 2 KE_{trans} = \frac{1}{2}mv^2 K E t r an s=2 1m v 2
Depends on total mass of the body and velocity of its center of mass
Scalar quantity independent of direction of motion
Rotational kinetic energy
Energy due to rotational motion of a rigid body about an axis
Expressed as K E r o t=1 2 I ω 2 KE_{rot} = \frac{1}{2}I\omega^2 K E ro t=2 1I ω 2
Relies on moment of inertia (I) and angular velocity (ω)
Varies with axis of rotation and mass distribution within the body
Total kinetic energy
Sum of translational and rotational kinetic energies
Represented by K E t o t a l=K E t r a n s+K E r o t KE_{total} = KE_{trans} + KE_{rot}K E t o t a l=K E t r an s+K E ro t
Crucial for analyzing complex motions (rolling, tumbling)
Conserved in absence of external forces or torques
Work done on rigid bodies
Work by external forces
Product of force and displacement in direction of force
Calculated using W=∫F⋅d r W = \int \mathbf{F} \cdot d\mathbf{r}W=∫F⋅d r
Includes work done by gravity, friction, and applied forces
Can be positive (energy added) or negative (energy removed)
Work by internal forces
Forces acting between particles within the rigid body
Sum to zero for perfectly rigid bodies due to equal and opposite reactions
Negligible in most engineering applications involving rigid body analysis
Becomes significant in deformable body mechanics
Net work calculation
Algebraic sum of work done by all external forces and torques
Accounts for both linear and angular displacements
Expressed as W n e t=∑W e x t e r n a l W_{net} = \sum W_{external}W n e t=∑W e x t er na l
Determines overall energy change in the system
Potential energy in rigid bodies
Gravitational potential energy
Energy stored due to object's position in gravitational field
Calculated using P E g=m g h PE_g = mgh P E g=m g h for uniform gravitational fields
Reference height (h) can be chosen arbitrarily
Significant in problems involving vertical motion or inclined planes
Elastic potential energy
Energy stored in deformed elastic objects (springs, flexible beams)
Computed using P E e=1 2 k x 2 PE_e = \frac{1}{2}kx^2 P E e=2 1k x 2 for linear springs
Depends on spring constant (k) and displacement from equilibrium (x)
Applicable in vibration analysis and impact problems
Conservation of energy
Conservative vs non-conservative forces
Conservative forces (gravity, spring force) conserve mechanical energy
Work done by conservative forces independent of path taken
Non-conservative forces (friction, air resistance) dissipate energy
Presence of non-conservative forces requires consideration of energy loss
Energy conservation in closed systems
Total energy remains constant in absence of external work
Expressed as E i n i t i a l=E f i n a l E_{initial} = E_{final}E ini t ia l=E f ina l or Δ K E+Δ P E=0\Delta KE + \Delta PE = 0 Δ K E+Δ PE=0
Allows prediction of motion without detailed force analysis
Simplifies problem-solving for complex mechanical systems
Work-energy theorem for rigid bodies
Derivation from Newton's laws
Starts with Newton's second law for translation and rotation
Integrates force and torque equations over displacement and angle
Results in W n e t=Δ K E t o t a l W_{net} = \Delta KE_{total}W n e t=Δ K E t o t a l
Connects work done to change in kinetic energy of the rigid body
Applications to planar motion
Analyzes combined translation and rotation in a plane
Useful for problems involving rolling without slipping
Accounts for both linear and angular velocities
Simplifies analysis of complex motions (wheels, gears, pendulums)
Power in rigid body dynamics
Instantaneous power
Rate of energy transfer or work done at a specific moment
Calculated using P=F⋅v P = \mathbf{F} \cdot \mathbf{v}P=F⋅v for linear motion
For rotational motion, P=τ ω P = \tau \omega P=τ ω
Crucial for analyzing time-dependent energy transfer
Average power
Work done divided by time interval
Expressed as P a v g=W t P_{avg} = \frac{W}{t}P a vg=t W
Useful for assessing overall energy efficiency of systems
Applied in design of motors, engines, and mechanical devices
Energy methods in problem-solving
Work-energy vs Newton's laws
Work-energy principle often simplifies complex dynamic problems
Avoids need for detailed force and acceleration analysis
Particularly useful when only initial and final states matter
Can handle problems with variable forces more easily
Advantages of energy approach
Reduces vector problems to scalar equations
Eliminates need for coordinate systems in many cases
Provides insight into energy transfer and conservation
Effective for problems involving constraints or complex geometries
Collision analysis using work-energy
Elastic vs inelastic collisions
Elastic collisions conserve both momentum and kinetic energy
Inelastic collisions conserve momentum but not kinetic energy
Perfectly inelastic collisions result in objects sticking together
Analysis involves comparing pre-collision and post-collision energies
Coefficient of restitution
Measure of "bounciness" in collisions
Defined as ratio of relative velocities after and before impact
Ranges from 0 (perfectly inelastic) to 1 (perfectly elastic)
Used to predict post-collision velocities in impact problems
Energy in systems of rigid bodies
Energy transfer between bodies
Occurs through work done by contact forces or distance forces
Involves conversion between kinetic and potential energies
Governed by conservation of energy for the entire system
Crucial in analyzing multi-body dynamics (gear trains, linkages)
Work-energy in multi-body systems
Applies work-energy principle to system as a whole
Accounts for internal work between bodies and external work on system
Simplifies analysis of complex mechanical assemblies
Useful in studying energy flow in machines and mechanisms
Numerical methods for work-energy
Energy-based simulations
Utilize energy conservation principles for time-stepping algorithms
Often more stable than force-based methods for long-term simulations
Effective for systems with many degrees of freedom
Applied in computer graphics, robotics, and virtual reality
Computational approaches
Finite element analysis for deformable bodies and energy distribution
Particle-based methods for granular materials and fluid-structure interaction
Optimization algorithms based on minimum energy principles
Machine learning techniques for predicting energy-efficient designs
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Questions tagged [circles]
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For elementary questions concerning circles (or disks). A circle is the locus of points in a plane that are at a fixed distance from a fixed point. Use this tag alongside [geometry], [Euclidean geometry], or something similar. Do not use this tag for more advanced topics, such as complex analysis or topology.
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188780 | https://en.wikipedia.org/wiki/Constitutive_equation | Jump to content
Constitutive equation
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From Wikipedia, the free encyclopedia
Relation between two physical quantities which is specific to a substance
In physics and engineering, a constitutive equation or constitutive relation is a relation between two or more physical quantities (especially kinetic quantities as related to kinematic quantities) that is specific to a material or substance or field, and approximates its response to external stimuli, usually as applied fields or forces. They are combined with other equations governing physical laws to solve physical problems; for example in fluid mechanics the flow of a fluid in a pipe, in solid state physics the response of a crystal to an electric field, or in structural analysis, the connection between applied stresses or loads to strains or deformations.
Some constitutive equations are simply phenomenological; others are derived from first principles. A common approximate constitutive equation frequently is expressed as a simple proportionality using a parameter taken to be a property of the material, such as electrical conductivity or a spring constant. However, it is often necessary to account for the directional dependence of the material, and the scalar parameter is generalized to a tensor. Constitutive relations are also modified to account for the rate of response of materials and their non-linear behavior. See the article Linear response function.
Mechanical properties of matter
[edit]
The first constitutive equation (constitutive law) was developed by Robert Hooke and is known as Hooke's law. It deals with the case of linear elastic materials. Following this discovery, this type of equation, often called a "stress-strain relation" in this example, but also called a "constitutive assumption" or an "equation of state" was commonly used. Walter Noll advanced the use of constitutive equations, clarifying their classification and the role of invariance requirements, constraints, and definitions of terms
like "material", "isotropic", "aeolotropic", etc. The class of "constitutive relations" of the form stress rate = f (velocity gradient, stress, density) was the subject of Walter Noll's dissertation in 1954 under Clifford Truesdell.
In modern condensed matter physics, the constitutive equation plays a major role. See Linear constitutive equations and Nonlinear correlation functions.
Definitions
[edit]
| Quantity (common name/s) | (Common) symbol/s | Defining equation | SI units | Dimension |
--- ---
| General stress, pressure | P, σ | F is the perpendicular component of the force applied to area A | Pa = N⋅m−2 | [M][L]−1[T]−2 |
| General strain | ε | D, dimension (length, area, volume) ΔD, change in dimension of material | 1 | Dimensionless |
| General elastic modulus | Emod | | Pa = N⋅m−2 | [M][L]−1[T]−2 |
| Young's modulus | E, Y | | Pa = N⋅m−2 | [M][L]−1[T] −2 |
| Shear modulus | G | | Pa = N⋅m−2 | [M][L]−1[T]−2 |
| Bulk modulus | K, B | | Pa = N⋅m−2 | [M][L]−1[T]−2 |
| Compressibility | C | | Pa−1 = m2⋅N−1 | [M]−1[L][T]2 |
Deformation of solids
[edit]
Friction
[edit]
Friction is a complicated phenomenon. Macroscopically, the friction force F at the interface of two materials can be modelled as proportional to the reaction force R at a point of contact between two interfaces through a dimensionless coefficient of friction μf, which depends on the pair of materials:
This can be applied to static friction (friction preventing two stationary objects from slipping on their own), kinetic friction (friction between two objects scraping/sliding past each other), or rolling (frictional force which prevents slipping but causes a torque to exert on a round object).
Stress and strain
[edit]
The stress-strain constitutive relation for linear materials is commonly known as Hooke's law. In its simplest form, the law defines the spring constant (or elasticity constant) k in a scalar equation, stating the tensile/compressive force is proportional to the extended (or contracted) displacement x:
meaning the material responds linearly. Equivalently, in terms of the stress σ, Young's modulus E, and strain ε (dimensionless):
In general, forces which deform solids can be normal to a surface of the material (normal forces), or tangential (shear forces), this can be described mathematically using the stress tensor:
where C is the elasticity tensor and S is the compliance tensor.
Solid-state deformation
[edit]
Several classes of deformation in elastic materials are the following:
Plastic
: The applied force induces non-recoverable deformation in the material when the stress (or elastic strain) reaches a critical magnitude, called the yield point.
Elastic
: The material recovers its initial shape after deformation.
Viscoelastic
: If the time-dependent resistive contributions are large, and cannot be neglected. Rubbers and plastics have this property, and certainly do not satisfy Hooke's law. In fact, elastic hysteresis occurs.
Anelastic
: If the material is close to elastic, but the applied force induces additional time-dependent resistive forces (i.e. depend on rate of change of extension/compression, in addition to the extension/compression). Metals and ceramics have this characteristic, but it is usually negligible, although not so much when heating due to friction occurs (such as vibrations or shear stresses in machines).
Hyperelastic
: The applied force induces displacements in the material following a strain energy density function.
Collisions
[edit]
The relative speed of separation vseparation of an object A after a collision with another object B is related to the relative speed of approach vapproach by the coefficient of restitution, defined by Newton's experimental impact law:
which depends on the materials A and B are made from, since the collision involves interactions at the surfaces of A and B. Usually 0 ≤ e ≤ 1, in which e = 1 for completely elastic collisions, and e = 0 for completely inelastic collisions. It is possible for e ≥ 1 to occur – for superelastic (or explosive) collisions.
Deformation of fluids
[edit]
The drag equation gives the drag force D on an object of cross-section area A moving through a fluid of density ρ at velocity v (relative to the fluid)
where the drag coefficient (dimensionless) cd depends on the geometry of the object and the drag forces at the interface between the fluid and object.
For a Newtonian fluid of viscosity μ, the shear stress τ is linearly related to the strain rate (transverse flow velocity gradient) ∂u/∂y (units s−1). In a uniform shear flow:
with u(y) the variation of the flow velocity u in the cross-flow (transverse) direction y. In general, for a Newtonian fluid, the relationship between the elements τij of the shear stress tensor and the deformation of the fluid is given by
: with and
where vi are the components of the flow velocity vector in the corresponding xi coordinate directions, eij are the components of the strain rate tensor, Δ is the volumetric strain rate (or dilatation rate) and δij is the Kronecker delta.
The ideal gas law is a constitutive relation in the sense the pressure p and volume V are related to the temperature T, via the number of moles n of gas:
where R is the gas constant (J⋅K−1⋅mol−1).
Electromagnetism
[edit]
Constitutive equations in electromagnetism and related areas
[edit]
See also: Permittivity, Permeability (electromagnetism), and Electrical conductivity
In both classical and quantum physics, the precise dynamics of a system form a set of coupled differential equations, which are almost always too complicated to be solved exactly, even at the level of statistical mechanics. In the context of electromagnetism, this remark applies to not only the dynamics of free charges and currents (which enter Maxwell's equations directly), but also the dynamics of bound charges and currents (which enter Maxwell's equations through the constitutive relations). As a result, various approximation schemes are typically used.
For example, in real materials, complex transport equations must be solved to determine the time and spatial response of charges, for example, the Boltzmann equation or the Fokker–Planck equation or the Navier–Stokes equations. For example, see magnetohydrodynamics, fluid dynamics, electrohydrodynamics, superconductivity, plasma modeling. An entire physical apparatus for dealing with these matters has developed. See for example, linear response theory, Green–Kubo relations and Green's function (many-body theory).
These complex theories provide detailed formulas for the constitutive relations describing the electrical response of various materials, such as permittivities, permeabilities, conductivities and so forth.
It is necessary to specify the relations between displacement field D and E, and the magnetic H-field H and B, before doing calculations in electromagnetism (i.e. applying Maxwell's macroscopic equations). These equations specify the response of bound charge and current to the applied fields and are called constitutive relations.
Determining the constitutive relationship between the auxiliary fields D and H and the E and B fields starts with the definition of the auxiliary fields themselves:
where P is the polarization field and M is the magnetization field which are defined in terms of microscopic bound charges and bound current respectively. Before getting to how to calculate M and P it is useful to examine the following special cases.
Without magnetic or dielectric materials
[edit]
In the absence of magnetic or dielectric materials, the constitutive relations are simple:
where ε0 and μ0 are two universal constants, called the permittivity of free space and permeability of free space, respectively.
Isotropic linear materials
[edit]
In an (isotropic) linear material, where P is proportional to E, and M is proportional to B, the constitutive relations are also straightforward. In terms of the polarization P and the magnetization M they are:
where χe and χm are the electric and magnetic susceptibilities of a given material respectively. In terms of D and H the constitutive relations are:
where ε and μ are constants (which depend on the material), called the permittivity and permeability, respectively, of the material. These are related to the susceptibilities by:
General case
[edit]
For real-world materials, the constitutive relations are not linear, except approximately. Calculating the constitutive relations from first principles involves determining how P and M are created from a given E and B.[note 1] These relations may be empirical (based directly upon measurements), or theoretical (based upon statistical mechanics, transport theory or other tools of condensed matter physics). The detail employed may be macroscopic or microscopic, depending upon the level necessary to the problem under scrutiny.
In general, the constitutive relations can usually still be written:
but ε and μ are not, in general, simple constants, but rather functions of E, B, position and time, and tensorial in nature. Examples are:
Dispersion and absorption where ε and μ are functions of frequency. (Causality does not permit materials to be nondispersive; see, for example, Kramers–Kronig relations.) Neither do the fields need to be in phase, which leads to ε and μ being complex. This also leads to absorption.
Nonlinearity where ε and μ are functions of E and B.
Anisotropy (such as birefringence or dichroism) which occurs when ε and μ are second-rank tensors,
Dependence of P and M on E and B at other locations and times. This could be due to spatial inhomogeneity; for example in a domained structure, heterostructure or a liquid crystal, or most commonly in the situation where there are simply multiple materials occupying different regions of space. Or it could be due to a time varying medium or due to hysteresis. In such cases P and M can be calculated as:
in which the permittivity and permeability functions are replaced by integrals over the more general electric and magnetic susceptibilities. In homogeneous materials, dependence on other locations is known as spatial dispersion.
As a variation of these examples, in general materials are bianisotropic where D and B depend on both E and H, through the additional coupling constants ξ and ζ:
In practice, some materials properties have a negligible impact in particular circumstances, permitting neglect of small effects. For example: optical nonlinearities can be neglected for low field strengths; material dispersion is unimportant when frequency is limited to a narrow bandwidth; material absorption can be neglected for wavelengths for which a material is transparent; and metals with finite conductivity often are approximated at microwave or longer wavelengths as perfect metals with infinite conductivity (forming hard barriers with zero skin depth of field penetration).
Some man-made materials such as metamaterials and photonic crystals are designed to have customized permittivity and permeability.
Calculation of constitutive relations
[edit]
See also: Computational electromagnetics
The theoretical calculation of a material's constitutive equations is a common, important, and sometimes difficult task in theoretical condensed-matter physics and materials science. In general, the constitutive equations are theoretically determined by calculating how a molecule responds to the local fields through the Lorentz force. Other forces may need to be modeled as well such as lattice vibrations in crystals or bond forces. Including all of the forces leads to changes in the molecule which are used to calculate P and M as a function of the local fields.
The local fields differ from the applied fields due to the fields produced by the polarization and magnetization of nearby material; an effect which also needs to be modeled. Further, real materials are not continuous media; the local fields of real materials vary wildly on the atomic scale. The fields need to be averaged over a suitable volume to form a continuum approximation.
These continuum approximations often require some type of quantum mechanical analysis such as quantum field theory as applied to condensed matter physics. See, for example, density functional theory, Green–Kubo relations and Green's function.
A different set of homogenization methods (evolving from a tradition in treating materials such as conglomerates and laminates) are based upon approximation of an inhomogeneous material by a homogeneous effective medium (valid for excitations with wavelengths much larger than the scale of the inhomogeneity).
The theoretical modeling of the continuum-approximation properties of many real materials often rely upon experimental measurement as well. For example, ε of an insulator at low frequencies can be measured by making it into a parallel-plate capacitor, and ε at optical-light frequencies is often measured by ellipsometry.
Thermoelectric and electromagnetic properties of matter
[edit]
These constitutive equations are often used in crystallography, a field of solid-state physics.
Electromagnetic properties of solids
| Property/effect | Stimuli/response parameters of system | Constitutive tensor of system | Equation |
| Hall effect | E, electric field strength (N⋅C−1) J, electric current density (A⋅m−2) H, magnetic field intensity (A⋅m−1) | ρ, electrical resistivity (Ω⋅m) | |
| Direct Piezoelectric Effect | σ, Stress (Pa) P, (dielectric) polarization (C⋅m−2) | d, direct piezoelectric coefficient (C⋅N−1) | |
| Converse Piezoelectric Effect | ε, Strain (dimensionless) E, electric field strength (N⋅C−1) | d, direct piezoelectric coefficient (C⋅N−1) | |
| Piezomagnetic effect | σ, Stress (Pa) M, magnetization (A⋅m−1) | q, piezomagnetic coefficient (A⋅N−1⋅m) | |
Thermoelectric properties of solids
| Property/effect | Stimuli/response parameters of system | Constitutive tensor of system | Equation |
| Pyroelectricity | P, (dielectric) polarization (C⋅m−2) T, temperature (K) | p, pyroelectric coefficient (C⋅m−2⋅K−1) | |
| Electrocaloric effect | S, entropy (J⋅K−1) E, electric field strength (N⋅C−1) | p, pyroelectric coefficient (C⋅m−2⋅K−1) | |
| Seebeck effect | E, electric field strength (N⋅C−1 = V⋅m−1) T, temperature (K) x, displacement (m) | β, thermopower (V⋅K−1) | |
| Peltier effect | E, electric field strength (N⋅C−1) J, electric current density (A⋅m−2) q, heat flux (W⋅m−2) | Π, Peltier coefficient (W⋅A−1) | |
Photonics
[edit]
Refractive index
[edit]
The (absolute) refractive index of a medium n (dimensionless) is an inherently important property of geometric and physical optics defined as the ratio of the luminal speed in vacuum c0 to that in the medium c:
where ε is the permittivity and εr the relative permittivity of the medium, likewise μ is the permeability and μr are the relative permeability of the medium. The vacuum permittivity is ε0 and vacuum permeability is μ0. In general, n (also εr) are complex numbers.
The relative refractive index is defined as the ratio of the two refractive indices. Absolute is for one material, relative applies to every possible pair of interfaces;
Speed of light in matter
[edit]
As a consequence of the definition, the speed of light in matter is
for special case of vacuum; ε = ε0 and μ = μ0,
Piezooptic effect
[edit]
The piezooptic effect relates the stresses in solids σ to the dielectric impermeability a, which are coupled by a fourth-rank tensor called the piezooptic coefficient Π (units K−1):
Transport phenomena
[edit]
Definitions
[edit]
Definitions (thermal properties of matter)
| Quantity (common name/s) | (Common) symbol/s | Defining equation | SI units | Dimension |
| General heat capacity | C, heat capacity of substance | | J⋅K−1 | [M][L]2[T]−2[Θ]−1 |
| linear thermal expansion coefficient | L, length of material (m) α, coefficient linear thermal expansion (dimensionless) ε, strain tensor (dimensionless) | | K−1 | [Θ]−1 |
| Volumetric thermal expansion coefficient | β, γ V, volume of object (m3) p, constant pressure of surroundings | | K−1 | [Θ]−1 |
| Thermal conductivity | κ, K, λ, q, heat flux through material (W/m2) ∇T, temperature gradient in material (K⋅m−1) | | W⋅m−1⋅K−1 | [M][L][T]−3[Θ]−1 |
| Thermal conductance | U | | W⋅m−2⋅K−1 | [M][T]−3[Θ]−1 |
| Thermal resistance | R Δx, displacement of heat transfer (m) | | m2⋅K⋅W−1 | [M]−1[L][T]3[Θ] |
Definitions (electrical/magnetic properties of matter)
| Quantity (common name/s) | (Common) symbol/s | Defining equation | SI units | Dimension |
| Electrical resistance | R | | Ω, V⋅A−1 = J⋅s⋅C−2 | [M][L]2[T]−3[I]−2 |
| Resistivity | ρ | | Ω⋅m | [M]2[L]2[T]−3[I]−2 |
| Resistivity temperature coefficient, linear temperature dependence | α | | K−1 | [Θ]−1 |
| Electrical conductance | G | | S = Ω−1 | [M]−1[L]−2[T]3[I]2 |
| Electrical conductivity | σ | | Ω−1⋅m−1 | [M]−2[L]−2[T]3[I]2 |
| Magnetic reluctance | R, Rm, | | A⋅Wb−1 = H−1 | [M]−1[L]−2[T]2 |
| Magnetic permeance | P, Pm, Λ, | | Wb⋅A−1 = H | [M][L]2[T]−2 |
Definitive laws
[edit]
There are several laws which describe the transport of matter, or properties of it, in an almost identical way. In every case, in words they read:
: Flux (density) is proportional to a gradient, the constant of proportionality is the characteristic of the material.
In general the constant must be replaced by a 2nd rank tensor, to account for directional dependences of the material.
| Property/effect | Nomenclature | Equation |
---
| Fick's law of diffusion, defines diffusion coefficient D | D, mass diffusion coefficient (m2⋅s−1) J, diffusion flux of substance (mol⋅m−2⋅s−1) ∂C/∂x, (1d)concentration gradient of substance (mol⋅dm−4) | |
| Darcy's law for fluid flow in porous media, defines permeability κ | κ, permeability of medium (m2) μ, fluid viscosity (Pa⋅s) q, discharge flux of substance (m⋅s−1) ∂P/∂x, (1d) pressure gradient of system (Pa⋅m−1) | |
| Ohm's law of electric conduction, defines electric conductivity (and hence resistivity and resistance) | V, potential difference in material (V) I, electric current through material (A) R, resistance of material (Ω) ∂V/∂x, potential gradient (electric field) through material (V⋅m−1) J, electric current density through material (A⋅m−2) σ, electric conductivity of material (Ω−1⋅m−1) ρ, electrical resistivity of material (Ω⋅m) | Simplest form is: More general forms are: |
| Fourier's law of thermal conduction, defines thermal conductivity λ | λ, thermal conductivity of material (W⋅m−1⋅K−1 ) q, heat flux through material (W⋅m−2) ∂T/∂x, temperature gradient in material (K⋅m−1) | |
| Stefan–Boltzmann law of black-body radiation, defines emmisivity ε | I, radiant intensity (W⋅m−2) σ, Stefan–Boltzmann constant (W⋅m−2⋅K−4) Tsys, temperature of radiating system (K) Text, temperature of external surroundings (K) ε, emissivity (dimensionless) | For a single radiator: For a temperature difference 0 ≤ ε ≤ 1; 0 for perfect reflector, 1 for perfect absorber (true black body) |
See also
[edit]
Defining equation (physical chemistry)
Governing equation
Principle of material objectivity
Rheology
Notes
[edit]
^ The free charges and currents respond to the fields through the Lorentz force law and this response is calculated at a fundamental level using mechanics. The response of bound charges and currents is dealt with using grosser methods subsumed under the notions of magnetization and polarization. Depending upon the problem, one may choose to have no free charges whatsoever.
References
[edit]
^ Truesdell, Clifford; Noll, Walter (2004). Antman, Stuart S. (ed.). The Non-linear Field Theories of Mechanics. Springer. p. 4. ISBN 3-540-02779-3.
^ See Truesdell's account in Truesdell The naturalization and apotheosis of Walter Noll. See also Noll's account and the classic treatise by both authors: Truesdell, Clifford; Noll, Walter (2004). "Preface" (Originally published as Volume III/3 of the 1965 Encyclopedia of Physics). In Antman, Stuart S. (ed.). The Non-linear Field Theories of Mechanics (3rd ed.). Springer. p. xiii. ISBN 3-540-02779-3.
^ Jørgen Rammer (2007). Quantum Field Theory of Nonequilibrium States. Cambridge University Press. ISBN 978-0-521-87499-1.
^ Encyclopaedia of Physics (2nd Edition), R.G. Lerner, G.L. Trigg, VHC publishers, 1991, ISBN (Verlagsgesellschaft) 3-527-26954-1, ISBN (VHC Inc.) 0-89573-752-3
^ Essential Principles of Physics, P.M. Whelan, M.J. Hodgeson, 2nd Edition, 1978, John Murray, ISBN 0 7195 3382 1
^ Kay, J.M. (1985). Fluid Mechanics and Transfer Processes. Cambridge University Press. pp. 10 & 122–124. ISBN 9780521316248.
^ The generalization to non-isotropic materials is straight forward; simply replace the constants with tensor quantities.
^ Halevi, Peter (1992). Spatial dispersion in solids and plasmas. Amsterdam: North-Holland. ISBN 978-0-444-87405-4.
^ Jackson, John David (1999). Classical Electrodynamics (3rd ed.). New York: Wiley. ISBN 0-471-30932-X.
^ Note that the 'magnetic susceptibility' term used here is in terms of B and is different from the standard definition in terms of H.
^ TG Mackay; A Lakhtakia (2010). Electromagnetic Anisotropy and Bianisotropy: A Field Guide. World Scientific. Archived from the original on 2010-10-13. Retrieved 2012-05-22.
^ Aspnes, D.E., "Local-field effects and effective-medium theory: A microscopic perspective", Am. J. Phys. 50, pp. 704–709 (1982).
^
Habib Ammari; Hyeonbae Kang (2006). Inverse problems, multi-scale analysis and effective medium theory : workshop in Seoul, Inverse problems, multi-scale analysis, and homogenization, June 22–24, 2005, Seoul National University, Seoul, Korea. Providence RI: American Mathematical Society. p. 282. ISBN 0-8218-3968-3.
^
O. C. Zienkiewicz; Robert Leroy Taylor; J. Z. Zhu; Perumal Nithiarasu (2005). The Finite Element Method (Sixth ed.). Oxford UK: Butterworth-Heinemann. p. 550 ff. ISBN 0-7506-6321-9.
^ N. Bakhvalov and G. Panasenko, Homogenization: Averaging Processes in Periodic Media (Kluwer: Dordrecht, 1989); V. V. Jikov, S. M. Kozlov and O. A. Oleinik, Homogenization of Differential Operators and Integral Functionals (Springer: Berlin, 1994).
^
Vitaliy Lomakin; Steinberg BZ; Heyman E; Felsen LB (2003). "Multiresolution Homogenization of Field and Network Formulations for Multiscale Laminate Dielectric Slabs" (PDF). IEEE Transactions on Antennas and Propagation. 51 (10): 2761 ff. Bibcode:2003ITAP...51.2761L. doi:10.1109/TAP.2003.816356. Archived from the original (PDF) on 2012-05-14.
^
Gilbert, Anna C. (May 2000). Coifman, Ronald R. (ed.). Topics in Analysis and Its Applications: Selected Theses. Singapore: World Scientific Publishing Company. p. 155. ISBN 981-02-4094-5.
^
Edward D. Palik; Ghosh G (1998). Handbook of Optical Constants of Solids. London UK: Academic Press. p. 1114. ISBN 0-12-544422-2.
^ "2. Physical Properties as Tensors". www.mx.iucr.org. Archived from the original on 19 April 2018. Retrieved 19 April 2018.
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高斯消元法(Gauss Elimination)【超详解&模板】
高斯消元法，是线性代数中的一个算法，可用来求解线性方程组，并可以求出矩阵的秩，以及求出可逆方阵的逆矩阵。高斯消元法的原理是：若用初等行变换将增广矩阵 化为 ，则AX = B与CX = D是同解方程组。
1、线性方程组
1）构造增广矩阵，即系数矩阵A增加上常数向量b（A|b）
2）通过以交换行、某行乘以非负常数和两行相加这三种初等变化将原系统转化为更简单的三角形式（triangular form）
注：这里的初等变化可以通过系数矩阵A乘上初等矩阵E来实现
3）从而得到简化的三角方阵组，注意它更容易解
4）这时可以使用向后替换算法（Algorithm for Back Substitution）求解得
z=-4/-4=1, y=4-2z=4-2=2, x= (1-y-z)/2=(1-2-1)/2=-1
总结上面过程，高斯消元法其实就是下面非常简单的过程
原线性方程组 ——> 高斯消元法 ——> 下三角或上三角形式的线性方程组 ——> 前向替换算法求解（对于上三角形式，采用后向替换算法）
补充1：
高斯－若尔当消元法（Gauss-Jordan Elimination）
相对于高斯消元法，高斯－若尔当消元法最后的得到线性方程组更容易求解，它得到的是简化行列式。其转化后的增高矩阵形式如下，因此它可以直接求出方程的解，而无需使用替换算法。但是，此算法的效率较低。
例子如下：
解为
个人感觉区别就是对每行进行了归一化处理
补充2：
介绍了最基本的高斯消元法，现在看看应用于实际问题的实用算法
因为实际应用中，我们总是利用计算机来分析线性系统，而计算机中以有限的数来近似无限的实数，因此产生舍入误差（roundoff error），进而对解线性系统产生很多影响。
一个t位（即精度为t）以为基的浮点数的表达形式为：，。对于一个实数x，其浮点近似值为最接近x的浮点数，必要时进行近似。
例1：对2位以10为基的浮点算法，。
例2：同样考虑，。
以下面系统为例，看看在高斯消元中采用浮点算法会产生什么效果。
当以精确解法时，通过将第一行乘以m=89/47，并从第二行中减去得到，进而利用后向替换算法得x=1，y=-1。
当以3位以10为基的浮点算法时，乘子变为，因为，因此第一步高斯消元后得
。此时，因为不能将第2行第1列位置变为0，所以不能将其三角化。从而，我们只能接受将这个位置值赋为0，而不管其实际浮点值。因此，3位浮点高斯消元的结果为，后向算法计算结果为。
尽管无法消除近似误差的影响，可以采用一些技术来尽量减小这类机器误差。部分主元消元法在高斯消元的每一步，都选择列上最大值为轴（通过行变换将其移动）。
下面给出列主元消去的代码（所谓列主元消去法是在系数矩阵中按列选取元素绝对值得最大值作为主元素，然后交换所在行与主元素所在行的位置，再按顺序消去法进行消元。）
1 列选主元消元法 2/ 3Gauss.cpp 4功能：列选主元消元法 5/ 6 7 "Gass.h " 8 9// 高斯消元法（列选主元） 10 void double int n) 11{ 12 int i,j; 13// 列选主元并消元成上三角 14// 回代求解 15 " 上三角的结果\n "); 16 3); 17 for 1) 18 { 19 for 1) 20 1 1]; 21 1a[i][i]; 22 } 23 return ; 24} 25// 选择列主元并进行消元 26 void double int n) 27{ 28 int i,j,k,maxRowE; 29 double// 用于记录消元时的因数 30 for 1) 31 { 32j; 33 for) 34 iffabs(a[maxRowE][j])) 35 i; 36 ifj) 37// 与最大主元所在行交换 38// 消元 39 for 1) 40 { 41a[j][j]; 42 for 1) 43temp; 44 } 45 " 第%d列消元后：\n ",j); 46 3); 47 } 48 return; 49} 50 void double int int int n) 51{ 52 int k; 53 double temp; 54 for 1) 55 { 56 a[m][k]; 57 a[maxRowE][k]; 58 temp; 59 } 60 return ; 61} 62 63// 测试函数 64 int main() 65{ 66 double 4][MAXNUM]; 67 68 int i,n,j; 69 70 1 10.001 1 22.000 1 33.000 1 41.000; 71 2 11.000 2 23.712 2 34.623 2 42.000; 72 3 12.000 3 21.070 3 35.643 3 43.000; 73 3); 74 for 1 3) 75 "X%d=%f\n " 4]); 76 return 0; 77} 78// 输出矩阵 79 void double int n) 80{ 81 int i,j; 82 for 1) 83 { 84 for 1 1) 85 { 86 "%f , ",a[i][j]); 87 } 88 " \n "); 89 } 90
测试结果：
2、逆矩阵
下面来说说高斯消元法在编程中的应用。
首先，先介绍程序中高斯消元法的步骤：(我们设方程组中方程的个数为equ，变元的个数为var，注意：一般情况下是n个方程，n个变元，但是有些题目就故意让方程数与变元数不同)
把方程组转换成增广矩阵。
利用初等行变换来把增广矩阵转换成行阶梯阵。枚举k从0到equ – 1，当前处理的列为col(初始为0) ，每次找第k行以下(包括第k行)，col列中元素绝对值最大的列与第k行交换。如果col列中的元素全为0，那么则处理col + 1列，k不变。
转换为行阶梯阵，判断解的情况。
① 无解当方程中出现(0, 0, …, 0, a)的形式，且a != 0时，说明是无解的。
② 唯一解条件是k = equ，即行阶梯阵形成了严格的上三角阵。利用回代逐一求出解集。
③ 无穷解。条件是k < equ，即不能形成严格的上三角形，自由变元的个数即为equ – k，但有些题目要求判断哪些变元是不缺定的。 这里单独介绍下这种解法：首先，自由变元有var - k个，即不确定的变元至少有var - k个。我们先把所有的变元视为不确定的。在每个方程中判断不确定变元的个数，如果大于1个，则该方程无法求解。如果只有1个变元，那么该变元即可求出，即为确定变元。以上介绍的是求解整数线性方程组的求法，复杂度是O(n3)。浮点数线性方程组的求法类似，但是要在判断是否为0时，加入EPS，以消除精度问题。
实例代码：
有多组测试数据。每组测试数据先输入一个整数n，表示方阵的阶。然后下面输入n阶方阵。输出其逆矩阵。若无逆矩阵，则输出No inverse matrix。
1 2 3 4 5 using namespace std; 6 7 const double 6; 8 9 bool const double// 用于判断有无逆矩阵 10{ 11 return eps; 12} 13 14 void double const int n ) 15{ 16 new double [n]; 17 for int 0i ) 18 new double[n]; 19} 20 21 void double const int n ) 22{ 23 for int 0i ) 24 { 25 for int 0 j ) 26 matrix[i][j]; 27 } 28} 29 30 bool double double const int n ) 31{ 32 int i, j; 33 for 0// 初始化一个单位矩阵 34 { 35 for 0 j ) 36 { 37 if j ) 38 1; 39 else 40 0; 41 } 42 } 43 for 0i ) 44 { 45 int i; 46 for 1j ) 47 { 48 if matrix1[i][rowmaxpos] ) 49 j; 50 } 51 for// 按从大到小的顺序排列矩阵 52 { 53 swap( matrix1[j][rowmaxpos], matrix1[j][i]); 54 swap( matrix2[j][rowmaxpos], matrix2[j][i]); 55 } 56 ifis_zero(matrix1[i][i]) ) 57 { 58 int matrix1[i][i]; 59 for// 归一化矩阵 60 { 61 divisor; 62 divisor; 63 } 64 for 1// 高斯消元法处理行列式 65 { 66 int matrix1[j][i]; 67 for int k ) 68 { 69 multiple; 70 multiple; 71 } 72 } 73 } 74 else 75 return false; 76 } 77 return true; 78} 79 80 void double const int n ) 81{ 82 for int 0i ) 83 { 84 for int 0 j ) 85 ' '; 86endl; 87 } 88} 89 90 void double const int n ) 91{ 92 for int 0 i ) 93 delete [] matrix[i]; 94 delete [] matrix; 95} 96 97 int main() 98{ 99 int n; 100 double matrix1; 101 double matrix2; 102 while n ) 103 { 104 create( matrix1, n ); 105 create( matrix2, n ); 106 input( matrix1, n); 107 if ( inverse(matrix1, matrix2, n) ) 108 output( matrix2, n ); 109 else 110 " No inverse matrix " endl; 111 destroy( matrix1, n ); 112 destroy( matrix2, n ); 113 } 114 return 0; 115
扩展1：
利用矩阵的初等行变换也可以判断一个矩阵是否可逆，即分块矩阵(A︱E)经过初等行变换，原来A的位置不能变换为单位阵E，那么A不可逆。
扩展2：
利用矩阵初等行变换解矩阵方程
对于一般的矩阵方程，我们可以先求自变量系数矩阵的逆，然后乘以结果矩阵即可得到自变量矩阵
ps：最后来点福利：
下面是几道OJ上的高斯消元法求解线性方程组的题目：
POJ 1222 EXTENDED LIGHTS OUT 1681 Painter's Problem 1753 Flip Game 1830 开关问题
POJ 3185 The Water Bowls
2947 Widget Factory 9]之间。POJ 1166 The Clocks
POJ 2065 SETI
POJ 1487 Single-Player Games + 递归的做法分解的。首先用栈的思想求出该结点的孩子数，然后递归分别求解各个孩子。这题解方程组也与众不同...首先是求解浮点数方程组，要注意精度问题，然后又询问不确定的变元，按前面说的方法求解。一顿折腾后，这题居然写了6000+B...而且囧的是巨人C++ WA，G++ AC，可能还是精度的问题吧...看这题目，看这代码，就没有改的欲望...hdu OJ 2449
fze OJ 1704
Sgu 275 To xor or not to xor
pps：对矩阵内涵的思考
如果不熟悉线性代数的概念，要去学习自然科学，现在看来就和文盲差不多。”，然而“按照现行的国际标准，线性代数是通过公理化来表述的，它是第二代数学模型，这就带来了教学上的困难。” 矩阵究竟是什么东西？向量可以被认为是具有n个相互独立的性质（维度）的对象的表示，矩阵又是什么呢？我们如果认为矩阵是一组列（行）向量组成的新的复合向量的展开式，那么为什么这种展开式具有如此广泛的应用？特别是，为什么偏偏二维的展开式如此有用？如果矩阵中每一个元素又是一个向量，那么我们再展开一次，变成三维的立方阵，是不是更有用？ 矩阵的乘法规则究竟为什么这样规定？为什么这样一种怪异的乘法规则却能够在实践中发挥如此巨大的功效？很多看上去似乎是完全不相关的问题，最后竟然都归结到矩阵的乘法，这难道不是很奇妙的事情？难道在矩阵乘法那看上去莫名其妙的规则下面，包含着世界的某些本质规律？如果是的话，这些本质规律是什么？ 行列式究竟是一个什么东西？为什么会有如此怪异的计算规则？行列式与其对应方阵本质上是什么关系？为什么只有方阵才有对应的行列式，而一般矩阵就没有（不要觉得这个问题很蠢，如果必要，针对m x n矩阵定义行列式不是做不到的，之所以不做，是因为没有这个必要，但是为什么没有这个必要）？而且，行列式的计算规则，看上去跟矩阵的任何计算规则都没有直观的联系，为什么又在很多方面决定了矩阵的性质？难道这一切仅是巧合？ 矩阵为什么可以分块计算？分块计算这件事情看上去是那么随意，为什么竟是可行的？ 对于矩阵转置运算AT，有(AB)T = BTAT，对于矩阵求逆运算A-1，有(AB)-1 = B-1A-1。两个看上去完全没有什么关系的运算，为什么有着类似的性质？这仅仅是巧合吗？ 为什么说P-1AP得到的矩阵与A矩阵“相似”？这里的“相似”是什么意思？ 特征值和特征向量的本质是什么？它们定义就让人很惊讶，因为Ax =λx，一个诺大的矩阵的效应，竟然不过相当于一个小小的数λ，确实有点奇妙。但何至于用“特征”甚至“本征”来界定？它们刻划的究竟是什么？今天先谈谈对线形空间和矩阵的几个核心概念的理解。首先说说空间(space)，这个概念是现代数学的命根子之一，从拓扑空间开始，一步步往上加定义，可以形成很多空间。线形空间其实还是比较初级的，如果在里面定义了范数，就成了赋范线性空间。赋范线性空间满足完备性，就成了巴那赫空间；赋范线性空间中定义角度，就有了内积空间，内积空间再满足完备性，就得到希尔伯特空间。总之，空间有很多种。你要是去看某种空间的数学定义，大致都是“存在一个集合，在这个集合上定义某某概念，然后满足某些性质”，就可以被称为空间。这未免有点奇怪，为什么要用“空间”来称呼一些这样的集合呢？大家将会看到，其实这是很有道理的。我们一般人最熟悉的空间，毫无疑问就是我们生活在其中的（按照牛顿的绝对时空观）的三维空间，从数学上说，这是一个三维的欧几里德空间，我们先不管那么多，先看看我们熟悉的这样一个空间有些什么最基本的特点。仔细想想我们就会知道，这个三维的空间：1. 由很多（实际上是无穷多个）位置点组成；2. 这些点之间存在相对的关系；3. 可以在空间中定义长度、角度；4. 这个空间可以容纳运动，这里我们所说的运动是从一个点到另一个点的移动（变换），而不是微积分意义上的“连续”性的运动，事实上，不管是什么空间，都必须容纳和支持在其中发生的符合规则的运动（变换）。你会发现，在某种空间中往往会存在一种相对应的变换，比如拓扑空间中有拓扑变换，线性空间中有线性变换，仿射空间中有仿射变换，其实这些变换都只不过是对应空间中允许的运动形式而已。因此只要知道，“空间”是容纳运动的一个对象集合，而变换则规定了对应空间的运动。下面我们来看看线性空间。线性空间的定义任何一本书上都有，但是既然我们承认线性空间是个空间，那么有两个最基本的问题必须首先得到解决，那就是：1. 空间是一个对象集合，线性空间也是空间，所以也是一个对象集合。那么线性空间是什么样的对象的集合？或者说，线性空间中的对象有什么共同点吗？2. 线性空间中的运动如何表述的？也就是，线性变换是如何表示的？我们先来回答第一个问题，回答这个问题的时候其实是不用拐弯抹角的，可以直截了当的给出答案。线性空间中的任何一个对象，通过选取基和坐标的办法，都可以表达为向量的形式。通常的向量空间我就不说了，举两个不那么平凡的例子：L1. 最高次项不大于n次的多项式的全体构成一个线性空间，也就是说，这个线性空间中的每一个对象是一个多项式。如果我们以x0, x1, ..., xn为基，那么任何一个这样的多项式都可以表达为一组n+1维向量，其中的每一个分量ai其实就是多项式中x(i-1)项的系数。值得说明的是，基的选取有多种办法，只要所选取的那一组基线性无关就可以。这要用到后面提到的概念了，所以这里先不说，提一下而已。L2. 闭区间[a, b]上的n阶连续可微函数的全体，构成一个线性空间。也就是说，这个线性空间的每一个对象是一个连续函数。对于其中任何一个连续函数，根据魏尔斯特拉斯定理，一定可以找到最高次项不大于n的多项式函数，使之与该连续函数的差为0，也就是说，完全相等。这样就把问题归结为L1了。后面就不用再重复了。所以说，向量是很厉害的，只要你找到合适的基，用向量可以表示线性空间里任何一个对象。这里头大有文章，因为向量表面上只是一列数，但是其实由于它的有序性，所以除了这些数本身携带的信息之外，还可以在每个数的对应位置上携带信息。为什么在程序设计中数组最简单，却又威力无穷呢？根本原因就在于此。这是另一个问题了，这里就不说了。下面来回答第二个问题，这个问题的回答会涉及到线性代数的一个最根本的问题。线性空间中的运动，被称为线性变换。也就是说，你从线性空间中的一个点运动到任意的另外一个点，都可以通过一个线性变化来完成。那么，线性变换如何表示呢？很有意思，在线性空间中，当你选定一组基之后，不仅可以用一个向量来描述空间中的任何一个对象，而且可以用矩阵来描述该空间中的任何一个运动（变换）。而使某个对象发生对应运动的方法，就是用代表那个运动的矩阵，乘以代表那个对象的向量。简而言之，在线性空间中选定基之后，向量刻画对象，矩阵刻画对象的运动，用矩阵与向量的乘法施加运动。是的，矩阵的本质是运动的描述。如果以后有人问你矩阵是什么，那么你就可以响亮地告诉他，矩阵的本质是运动的描述可是多么有意思啊，向量本身不是也可以看成是n x 1矩阵吗？这实在是很奇妙，一个空间中的对象和运动竟然可以用相类同的方式表示。能说这是巧合吗？如果是巧合的话，那可真是幸运的巧合！可以说，线性代数中大多数奇妙的性质，均与这个巧合有直接的关系。 “矩阵是运动的描述”，到现在为止，好像大家都还没什么意见。但是我相信早晚会有数学系出身的网友来拍板转。因为运动这个概念，在数学和物理里是跟微积分联系在一起的。我们学习微积分的时候，总会有人照本宣科地告诉你，初等数学是研究常量的数学，是研究静态的数学，高等数学是变量的数学，是研究运动的数学。大家口口相传，差不多人人都知道这句话。但是真知道这句话说的是什么意思的人，好像也不多。简而言之，在我们人类的经验里，运动是一个连续过程，从A点到B点，就算走得最快的光，也是需要一个时间来逐点地经过AB之间的路径，这就带来了连续性的概念。而连续这个事情，如果不定义极限的概念，根本就解释不了。古希腊人的数学非常强，但就是缺乏极限观念，所以解释不了运动，被芝诺的那些著名悖论（飞箭不动、飞毛腿阿喀琉斯跑不过乌龟等四个悖论）搞得死去活来。因为这篇文章不是讲微积分的，所以我就不多说了。有兴趣的读者可以去看看齐民友教授写的《重温微积分》。我就是读了这本书开头的部分，才明白“高等数学是研究运动的数学”这句话的道理。不过在我这个《理解矩阵》的文章里，“运动”的概念不是微积分中的连续性的运动，而是瞬间发生的变化。比如这个时刻在A点，经过一个“运动”，一下子就“跃迁”到了B点，其中不需要经过A点与B点之间的任何一个点。这样的“运动”，或者说“跃迁”，是违反我们日常的经验的。不过了解一点量子物理常识的人，就会立刻指出，量子（例如电子）在不同的能量级轨道上跳跃，就是瞬间发生的，具有这样一种跃迁行为。所以说，自然界中并不是没有这种运动现象，只不过宏观上我们观察不到。但是不管怎么说，“运动”这个词用在这里，还是容易产生歧义的，说得更确切些，应该是“跃迁”。因此这句话可以改成：“矩阵是线性空间里跃迁的描述”。可是这样说又太物理，也就是说太具体，而不够数学，也就是说不够抽象。因此我们最后换用一个正牌的数学术语——变换，来描述这个事情。这样一说，大家就应该明白了，所谓变换，其实就是空间里从一个点（元素/对象）到另一个点（元素/对象）的跃迁。比如说，拓扑变换，就是在拓扑空间里从一个点到另一个点的跃迁。再比如说，仿射变换，就是在仿射空间里从一个点到另一个点的跃迁。附带说一下，这个仿射空间跟向量空间是亲兄弟。做计算机图形学的朋友都知道，尽管描述一个三维对象只需要三维向量，但所有的计算机图形学变换矩阵都是4 x 4的。说其原因，很多书上都写着“为了使用中方便”，这在我看来简直就是企图蒙混过关。真正的原因，是因为在计算机图形学里应用的图形变换，实际上是在仿射空间而不是向量空间中进行的。想想看，在向量空间里相一个向量平行移动以后仍是相同的那个向量，而现实世界等长的两个平行线段当然不能被认为同一个东西，所以计算机图形学的生存空间实际上是仿射空间。而仿射变换的矩阵表示根本就是4 x 4的。又扯远了，有兴趣的读者可以去看《计算机图形学——几何工具算法详解》。一旦我们理解了“变换”这个概念，矩阵的定义就变成：“矩阵是线性空间里的变换的描述。”到这里为止，我们终于得到了一个看上去比较数学的定义。不过还要多说几句。教材上一般是这么说的，在一个线性空间V里的一个线性变换T，当选定一组基之后，就可以表示为矩阵。因此我们还要说清楚到底什么是线性变换，什么是基，什么叫选定一组基。线性变换的定义是很简单的，设有一种变换T，使得对于线性空间V中间任何两个不相同的对象x和y，以及任意实数a和b，有：T(ax + by) = aT(x) + bT(y)，那么就称T为线性变换。定义都是这么写的，但是光看定义还得不到直觉的理解。线性变换究竟是一种什么样的变换？我们刚才说了，变换是从空间的一个点跃迁到另一个点，而线性变换，就是从一个线性空间V的某一个点跃迁到另一个线性空间W的另一个点的运动。这句话里蕴含着一层意思，就是说一个点不仅可以变换到同一个线性空间中的另一个点，而且可以变换到另一个线性空间中的另一个点去。不管你怎么变，只要变换前后都是线性空间中的对象，这个变换就一定是线性变换，也就一定可以用一个非奇异矩阵来描述。而你用一个非奇异矩阵去描述的一个变换，一定是一个线性变换。有的人可能要问，这里为什么要强调非奇异矩阵？所谓非奇异，只对方阵有意义，那么非方阵的情况怎么样？这个说起来就会比较冗长了，最后要把线性变换作为一种映射，并且讨论其映射性质，以及线性变换的核与像等概念才能彻底讲清楚。我觉得这个不算是重点，如果确实有时间的话，以后写一点。以下我们只探讨最常用、最有用的一种变换，就是在同一个线性空间之内的线性变换。也就是说，下面所说的矩阵，不作说明的话，就是方阵，而且是非奇异方阵。学习一门学问，最重要的是把握主干内容，迅速建立对于这门学问的整体概念，不必一开始就考虑所有的细枝末节和特殊情况，自乱阵脚。接着往下说，什么是基呢？这个问题在后面还要大讲一番，这里只要把基看成是线性空间里的坐标系就可以了。注意是坐标系，不是坐标值，这两者可是一个“对立矛盾统一体”。这样一来，“选定一组基”就是说在线性空间里选定一个坐标系。就这意思。好，最后我们把矩阵的定义完善如下：“矩阵是线性空间中的线性变换的一个描述。在一个线性空间中，只要我们选定一组基，那么对于任何一个线性变换，都能够用一个确定的矩阵来加以描述。”理解这句话的关键，在于把“线性变换”与“线性变换的一个描述”区别开。一个是那个对象，一个是对那个对象的表述。就好像我们熟悉的面向对象编程中，一个对象可以有多个引用，每个引用可以叫不同的名字，但都是指的同一个对象。如果还不形象，那就干脆来个很俗的类比。比如有一头猪，你打算给它拍照片，只要你给照相机选定了一个镜头位置，那么就可以给这头猪拍一张照片。这个照片可以看成是这头猪的一个描述，但只是一个片面的的描述，因为换一个镜头位置给这头猪拍照，能得到一张不同的照片，也是这头猪的另一个片面的描述。所有这样照出来的照片都是这同一头猪的描述，但是又都不是这头猪本身。同样的，对于一个线性变换，只要你选定一组基，那么就可以找到一个矩阵来描述这个线性变换。换一组基，就得到一个不同的矩阵。所有这些矩阵都是这同一个线性变换的描述，但又都不是线性变换本身。但是这样的话，问题就来了如果你给我两张猪的照片，我怎么知道这两张照片上的是同一头猪呢？同样的，你给我两个矩阵，我怎么知道这两个矩阵是描述的同一个线性变换呢？如果是同一个线性变换的不同的矩阵描述，那就是本家兄弟了，见面不认识，岂不成了笑话。好在，我们可以找到同一个线性变换的矩阵兄弟们的一个性质，那就是：若矩阵A与B是同一个线性变换的两个不同的描述（之所以会不同，是因为选定了不同的基，也就是选定了不同的坐标系），则一定能找到一个非奇异矩阵P，使得A、B之间满足这样的关系：A = P-1BP线性代数稍微熟一点的读者一下就看出来，这就是相似矩阵的定义。没错，所谓相似矩阵，就是同一个线性变换的不同的描述矩阵。按照这个定义，同一头猪的不同角度的照片也可以成为相似照片。俗了一点，不过能让人明白。而在上面式子里那个矩阵P，其实就是A矩阵所基于的基与B矩阵所基于的基这两组基之间的一个变换关系。关于这个结论，可以用一种非常直觉的方法来证明（而不是一般教科书上那种形式上的证明），如果有时间的话，我以后在blog里补充这个证明。这个发现太重要了。原来一族相似矩阵都是同一个线性变换的描述啊！难怪这么重要！工科研究生课程中有矩阵论、矩阵分析等课程，其中讲了各种各样的相似变换，比如什么相似标准型，对角化之类的内容，都要求变换以后得到的那个矩阵与先前的那个矩阵式相似的，为什么这么要求？因为只有这样要求，才能保证变换前后的两个矩阵是描述同一个线性变换的。当然，同一个线性变换的不同矩阵描述，从实际运算性质来看并不是不分好环的。有些描述矩阵就比其他的矩阵性质好得多。这很容易理解，同一头猪的照片也有美丑之分嘛。所以矩阵的相似变换可以把一个比较丑的矩阵变成一个比较美的矩阵，而保证这两个矩阵都是描述了同一个线性变换。这样一来，矩阵作为线性变换描述的一面，基本上说清楚了。但是，事情没有那么简单，或者说，线性代数还有比这更奇妙的性质，那就是，矩阵不仅可以作为线性变换的描述，而且可以作为一组基的描述。而作为变换的矩阵，不但可以把线性空间中的一个点给变换到另一个点去，而且也能够把线性空间中的一个坐标系（基）表换到另一个坐标系（基）去。而且，变换点与变换坐标系，具有异曲同工的效果。线性代数里最有趣的奥妙，就蕴含在其中。理解了这些内容，线性代数里很多定理和规则会变得更加清晰、直觉。首先来总结一下前面两部分的一些主要结论：1. 首先有空间，空间可以容纳对象运动的。一种空间对应一类对象。2. 有一种空间叫线性空间，线性空间是容纳向量对象运动的。3. 运动是瞬时的，因此也被称为变换。4. 矩阵是线性空间中运动（变换）的描述。5. 矩阵与向量相乘，就是实施运动（变换）的过程。6. 同一个变换，在不同的坐标系下表现为不同的矩阵，但是它们的本质是一样的，所以本征值相同。 下面让我们把视力集中到一点以改变我们以往看待矩阵的方式。我们知道，线性空间里的基本对象是向量，而向量是这么表示的： [a1, a2, a3, ..., an] 矩阵呢？矩阵是这么表示的： a11, a12, a13, ..., a1n a21, a22, a23, ..., a2n ... an1, an2, an3, ..., ann 不用太聪明，我们就能看出来，矩阵是一组向量组成的。特别的，n维线性空间里的方阵是由n个n维向量组成的。我们在这里只讨论这个n阶的、非奇异的方阵，如果一组向量是彼此线性无关的话，那么它们就可以成为度量这个线性空间的一组基，从而事实上成为一个坐标系体系，其中每一个向量都躺在一根坐标轴上，并且成为那根坐标轴上的基本度量单位（长度1）。现在到了关键的一步。看上去矩阵就是由一组向量组成的，而且如果矩阵非奇异的话（我说了，只考虑这种情况），那么组成这个矩阵的那一组向量也就是线性无关的了，也就可以成为度量线性空间的一个坐标系。结论：矩阵描述了一个坐标系。之所以矩阵又是运动，又是坐标系，那是因为——“运动等价于坐标系变换”。对不起，这话其实不准确，我只是想让你印象深刻。准确的说法是：“对象的变换等价于坐标系的变换”。或者：“固定坐标系下一个对象的变换等价于固定对象所处的坐标系变换。” 说白了就是： “运动是相对的。” 让我们想想，达成同一个变换的结果，比如把点(1, 1)变到点(2, 3)去，你可以有两种做法。第一，坐标系不动，点动，把(1, 1)点挪到(2, 3)去。第二，点不动，变坐标系，让x轴的度量（单位向量）变成原来的1/2，让y轴的度量（单位向量）变成原先的1/3，这样点还是那个点，可是点的坐标就变成(2, 3)了。方式不同，结果一样。从第一个方式来看，那就是我在《理解矩阵》1/2中说的，把矩阵看成是运动描述，矩阵与向量相乘就是使向量（点）运动的过程。在这个方式下， Ma = b的意思是： “向量a经过矩阵M所描述的变换，变成了向量b。”而从第二个方式来看，矩阵M描述了一个坐标系，姑且也称之为M。那么： Ma = b的意思是： “有一个向量，它在坐标系M的度量下得到的度量结果向量为a，那么它在坐标系I的度量下，这个向量的度量结果是b。” 这里的I是指单位矩阵，就是主对角线是1，其他为零的矩阵。而这两个方式本质上是等价的。我希望你务必理解这一点，因为这是本篇的关键。正因为是关键，所以我得再解释一下。在M为坐标系的意义下，如果把M放在一个向量a的前面，形成Ma的样式，我们可以认为这是对向量a的一个环境声明。它相当于是说： “注意了！这里有一个向量，它在坐标系M中度量，得到的度量结果可以表达为a。可是它在别的坐标系里度量的话，就会得到不同的结果。为了明确，我把M放在前面，让你明白，这是该向量在坐标系M中度量的结果。” 那么我们再看孤零零的向量b： b 多看几遍，你没看出来吗？它其实不是b，它是： Ib 也就是说：“在单位坐标系，也就是我们通常说的直角坐标系I中，有一个向量，度量的结果是b。” 而 Ma = Ib的意思就是说： “在M坐标系里量出来的向量a，跟在I坐标系里量出来的向量b，其实根本就是一个向量啊！”这哪里是什么乘法计算，根本就是身份识别嘛。从这个意义上我们重新理解一下向量。向量这个东西客观存在，但是要把它表示出来，就要把它放在一个坐标系中去度量它，然后把度量的结果（向量在各个坐标轴上的投影值）按一定顺序列在一起，就成了我们平时所见的向量表示形式。你选择的坐标系（基）不同，得出来的向量的表示就不同。向量还是那个向量，选择的坐标系不同，其表示方式就不同。因此，按道理来说，每写出一个向量的表示，都应该声明一下这个表示是在哪个坐标系中度量出来的。表示的方式，就是 Ma，也就是说，有一个向量，在M矩阵表示的坐标系中度量出来的结果为a。我们平时说一个向量是[2 3 5 7]T，隐含着是说，这个向量在 I 坐标系中的度量结果是[2 3 5 7]T，因此，这个形式反而是一种简化了的特殊情况。注意到，M矩阵表示出来的那个坐标系，由一组基组成，而那组基也是由向量组成的，同样存在这组向量是在哪个坐标系下度量而成的问题。也就是说，表述一个矩阵的一般方法，也应该要指明其所处的基准坐标系。所谓M，其实是 IM，也就是说，M中那组基的度量是在 I 坐标系中得出的。从这个视角来看，M×N也不是什么矩阵乘法了，而是声明了一个在M坐标系中量出的另一个坐标系N，其中M本身是在I坐标系中度量出来的。回过头来说变换的问题。我刚才说，“固定坐标系下一个对象的变换等价于固定对象所处的坐标系变换”，那个“固定对象”我们找到了，就是那个向量。但是坐标系的变换呢？我怎么没看见？请看：Ma = Ib我现在要变M为I，怎么变？对了，再前面乘以个M-1，也就是M的逆矩阵。换句话说，你不是有一个坐标系M吗，现在我让它乘以个M-1，变成I，这样一来的话，原来M坐标系中的a在I中一量，就得到b了。我建议你此时此刻拿起纸笔，画画图，求得对这件事情的理解。比如，你画一个坐标系，x轴上的衡量单位是2，y轴上的衡量单位是3，在这样一个坐标系里，坐标为(1，1)的那一点，实际上就是笛卡尔坐标系里的点(2, 3)。而让它原形毕露的办法，就是把原来那个坐标系:2 00 3的x方向度量缩小为原来的1/2，而y方向度量缩小为原来的1/3，这样一来坐标系就变成单位坐标系I了。保持点不变，那个向量现在就变成了(2, 3)了。怎么能够让“x方向度量缩小为原来的1/2，而y方向度量缩小为原来的1/3”呢？就是让原坐标系：2 00 3被矩阵：1/2 00 1/3左乘。而这个矩阵就是原矩阵的逆矩阵。下面我们得出一个重要的结论：“对坐标系施加变换的方法，就是让表示那个坐标系的矩阵与表示那个变化的矩阵相乘。”再一次的，矩阵的乘法变成了运动的施加。只不过，被施加运动的不再是向量，而是另一个坐标系。如果你觉得你还搞得清楚，请再想一下刚才已经提到的结论，矩阵MxN，一方面表明坐标系N在运动M下的变换结果，另一方面，把M当成N的前缀，当成N的环境描述，那么就是说，在M坐标系度量下，有另一个坐标系N。这个坐标系N如果放在I坐标系中度量，其结果为坐标系MxN。在这里，我实际上已经回答了一般人在学习线性代数是最困惑的一个问题，那就是为什么矩阵的乘法要规定成这样。简单地说，是因为：1. 从变换的观点看，对坐标系N施加M变换，就是把组成坐标系N的每一个向量施加M变换。2. 从坐标系的观点看，在M坐标系中表现为N的另一个坐标系，这也归结为，对N坐标系基的每一个向量，把它在I坐标系中的坐标找出来，然后汇成一个新的矩阵。3. 至于矩阵乘以向量为什么要那样规定，那是因为一个在M中度量为a的向量，如果想要恢复在I中的真像，就必须分别与M中的每一个向量进行内积运算。我把这个结论的推导留给感兴趣的朋友吧。应该说，其实到了这一步，已经很容易了。综合以上1/2/3，矩阵的乘法就得那么规定，一切有根有据，绝不是哪个神经病胡思乱想出来的。我已经无法说得更多了。矩阵又是坐标系，又是变换。到底是坐标系，还是变换，已经说不清楚了，运动与实体在这里统一了，物质与意识的界限已经消失了，一切归于无法言说，无法定义了。道可道，非常道，名可名，非常名。矩阵是在是不可道之道，不可名之名的东西。到了这个时候，我们不得不承认，我们伟大的线性代数课本上说的矩阵定义，是无比正确的：“矩阵就是由m行n列数放在一起组成的数学对象。”好了，这基本上就是我想说的全部了。还留下一个行列式的问题。矩阵M的行列式实际上是组成M的各个向量按照平行四边形法则搭成一个n维立方体的体积。对于这一点，我只能感叹于其精妙，却无法揭开其中奥秘了。也许我掌握的数学工具不够，我希望有人能够给我们大家讲解其中的道理了。我不知道是否讲得足够清楚了，反正这一部分需要您花些功夫去推敲。此外，请大家不必等待这个系列的后续部分。以我的工作情况而言，近期内很难保证继续投入脑力到这个领域中，尽管我仍然对此兴致浓厚。不过如果还有（四）的话，可能是一些站在应用层面的考虑，比如对计算机图形学相关算法的理解。但是我不承诺这些讨论近期内会出现了。
“分”的反义字是“和”，是我们熟悉的字。比如：2+3=5，从左往右运算，我们叫求和。那么“分”呢，既然是反义字，就把上面的等式反过来：5=2+3。把一个对象表示成两个以至更多的对象的和，这个过程叫分析。通常来说，分析对象应当与被分析对象一致。是数就都是数，是函数就都是函数，是向量就都是向量，是矩阵就都是矩阵。求和是数学里最基本的运算，减、乘、除是从求和中衍生出来的。而更高级的幂、指、对、三角、微积分等，也是一层一层建立起来的， 最根本的还是这个求和。求和最简单，最容易计算，性质也最简单。所以成了分析的基本出发点。分析的妙处在于，通过分析可以将较复杂的对象划分为较简单的对象。比如2和3就比5简单。单独研究2的性质，再单独研究3的性质，再通过简单的求和，就可以把握5的性质。把复杂的东西划分成若干简单对象的和，对各简单对象搞各个击破，再加起来，复杂的东西也就被掌握了。分析是西方思想中一个根本性的东西。西方人认为，事物总是有因果的，看到了结果，要分析原因。所谓分析原因，就是找出一堆因素，说明这堆因素合起来导致了结果。西方人认为，事物总是可以分析的。看到了整体，就要把那些合成这个整体的局部一一分析出来。 现代科学很大一部分就是这么回事。大学数学里，有很多内容就是在讲分析。数学里的分析还要把含义拓展，就是把一个数学对象合理地表示成若干更简单对象与实数系数之积的和。但微积分和线性代数各有侧重。微积分研究的是无穷项求和。无穷项之和与有穷项之和是本质不同的。但是无穷项之和是无法运算的，至少不实际。所以要想办法通过一种办法用有穷项之和来近似的代替，这就是逼近。逼近成立的条件是收敛，就是说，只有从一个收敛的无穷项的开头截出一部分来求和，才能被认为是逼近。华人数学家项武义说，微积分就逼近这一板斧，但是无往而不利。微积分主要研究函数，连续函数的因变量y会由于自变量x的变化而变化。这种变化也是要分析的。当x从 x0变成x1时，y是怎样从y0变到y1 的？按照上面的说法，“y的变化（y1-y0)”这一个数学对象，要用一系列比较简单的“变化”相加来表示。数学家找到了一个收敛的“变化”对象的序列，排在头一位的是一个线性的变化量，它的系数就是导数，它本身就是微分dy。数学家又发现，当x的变化量无穷小时，从这个无穷的、收敛的“变化”对象序列中，只要截出第一项，也就是微分dy，就无论如何可以精确描述y的变化了。曾在一本书上见过这样的说法，泰勒公式是数学分析的顶峰。不知道是不是有道理。我自己觉得是这么回事。有了泰勒公式，我们可以任意精确地算一个函数在某一点上的值。毕竟只是实数求和嘛。但是为了表示泰勒公式，我们却用了一个挺复杂的连加代数式。代数式不能象实数那样简单加起来得到一个对象，它只能表示成和的形式。这是我们意识到，在这个连加式中各对象存在某些特别的不同，使它们没法简单地加到一起。 因此我们有必要讨论，把一些性质不同的东西加到一起所形成的这个对象有什么性质。 这就是向量。微积分研究如何把一个对象分解为无穷项同质对象之和，线性代数研究“有限项异质对象之和”这个新对象的性质。一方面，上面说过，微积分到最后还是要化无穷为有穷，化精确为逼近；另一方面，异质对象经过某种处理可以转化为同质对象。比如不同次的幂函数是异质对象，但是一旦代入具体数值则都可以转化为实数，变成了同质对象。因此线性代数研究的问题对微积分很重要。故我认为大学里应先讲线性代数，后讲微积分。我们的微积分教学，将重点过分倾注在微分和积分的运算上了，其实实践中更为重要的是我们称为“级数”的那部分内容。即研究如何将一个量表达为一个数项级数，如何将一个函数表达为一个函数项级数。线性代数把异质对象之和（向量）作为研究的基础，研究这些新定义的对象加起来又可以表示什么。其结论是，有限数量的向量连加起来，有可能具有这样的能力，即同维的全部向量都可以表示成这些向量的和。这样的一组具有充分表现能力的向量，是线性无关的向量，组成了一个向量空间，而它们自己构成了这个向量空间里的一组基。回到分析的概念上，一个向量总可以表示为若干个同阶向量之和，这就是向量的分析。但是并不是所有的这些分析都具有相同的价值。在某种运算中，某种特别的分析能够提供特别优越的性，从而大大简化运算。比如在大多数情况下，将一个向量表示成一组单位正交基向量的和，就能够在计算中获得特别的便利。面对某个问题，寻找一个最优越的分析形式，把要研究的对象合理地表示成具有特殊性质的基对象与实数系数之积的和，这是分析的重要步骤，也是成功的关键。在这种表示式中，系数称为坐标。经典的方法都是以找到一组性质优良的基为开端的，例如：傅立叶分析以正交函数系为基，因此具有优良性质，自1904年以来取代幂函数系，成为分析主流。在曲线和曲面拟合中，正交多项式集构成了最佳基函数。 拉格朗日插值多项式具有一个特别的性质，即在本结点上为1，在其他结点上为0。有限元中的形函数类似拉氏插值多项式。结构动力学中的主振型迭加法，也是以相互正交的主振型为基，对多质点体系位移进行分析的。
举两个例子：说到采样，大家的第一反应肯定是一个词“2倍”（采样定理）。学得比较扎实的，可能还会把为什么是2倍解释清楚。但我对采样的理解是：采样实际上是在进行正交分解，采样值不过是在一组正交基下分解的系数。如果原信号属于该组正交基所张成的线性子空间，那么该信号就能无失真的恢复（满足采样定理）。学过信号处理的朋友，你知道这组正交基是什么吗？:)第二个例子是关于为什么傅里叶变换在线性系统理论中如此重要？答案可能五花八门，但我认为我的理解是比较深入的：原因是傅里叶基是所有线性时不变算子的特征向量（和本文联系起来了）。这句话解释起来比较费工夫，但是傅里叶变换能和特征向量联系起来，大家一定感觉很有趣吧。
特征向量确实有很明确的几何意义，矩阵(既然讨论特征向量的问题，当然是方阵，这里不讨论广义特征向量的概念，就是一般的特征向量)乘以一个向量的结果仍是同维数的一个向量，因此，矩阵乘法对应了一个变换，把一个向量变成同维数的另一个向量，那么变换的效果是什么呢？这当然与方阵的构造有密切关系，比如可以取适当的二维方阵，使得这个变换的效果就是将平面上的二维向量逆时针旋转30度，这时我们可以问一个问题，有没有向量在这个变换下不改变方向呢？可以想一下，除了零向量，没有其他向量可以在平面上旋转30度而不改变方向的，所以这个变换对应的矩阵(或者说这个变换自身)没有特征向量(注意：特征向量不能是零向量)，所以一个变换的特征向量是这样一种向量，它经过这种特定的变换后保持方向不变，只是进行长度上的伸缩而已(再想想特征向量的原始定义Ax=cx,你就恍然大悟了,看到了吗？cx是方阵A对向量x进行变换后的结果，但显然cx和x的方向相同)，而且x是特征向量的话，ax也是特征向量(a是标量且不为零)，所以所谓的特征向量不是一个向量而是一个向量族，另外，特征值只不过反映了特征向量在变换时的伸缩倍数而已，对一个变换而言，特征向量指明的方向才是很重要的，特征值不是那么重要，虽然我们求这两个量时先求出特征值，但特征向量才是更本质的东西！
比如平面上的一个变换，把一个向量关于横轴做镜像对称变换，即保持一个向量的横坐标不变，但纵坐标取相反数，把这个变换表示为矩阵就是[1 0;0 -1]，其中分号表示换行，显然[1 0;0 -1][a b]'=[a -b]',其中上标'表示取转置，这正是我们想要的效果，那么现在可以猜一下了，这个矩阵的特征向量是什么?想想什么向量在这个变换下保持方向不变，显然，横轴上的向量在这个变换下保持方向不变(记住这个变换是镜像对称变换，那镜子表面上(横轴上)的向量当然不会变化)，所以可以直接猜测其特征向量是 [a 0]'(a不为0)，还有其他的吗？有，那就是纵轴上的向量，这时经过变换后，其方向反向，但仍在同一条轴上，所以也被认为是方向没有变化，所以[0 b]'(b不为0)也是其特征向量，去求求矩阵[1 0;0 -1]的特征向量就知道对不对了!
//最近一直在做一个数论专题，后期有待整理，先将有关资料收藏下，在学习高斯消元的时候看了czyuan大牛的此文获益匪浅，czyuan的此份模板可以解决大多高斯问题，当然并不是万能的，其中建立矩阵是难点，需要自己琢磨，并且对于方程组是否有解、解的个数以及自由元等问题也需要自己做题慢慢思考，自己做了两三道题前前后后在建矩阵以及对一些解的问题在Gauss函数中改了几十次，逐渐摸索，还不算掌握的好，有待再慢慢练。
高斯消元法，是线性代数中的一个算法，可用来求解线性方程组，并可以求出矩阵的秩，以及求出可逆方阵的逆矩阵。高斯消元法的原理是：若用初等行变换将增广矩阵 化为 ，则AX = B与CX = D是同解方程组。所以我们可以用初等行变换把增广矩阵转换为行阶梯阵，然后回代求出方程的解。
以上是线性代数课的回顾，下面来说说高斯消元法在编程中的应用。
首先，先介绍程序中高斯消元法的步骤：(我们设方程组中方程的个数为equ，变元的个数为var，注意：一般情况下是n个方程，n个变元，但是有些题目就故意让方程数与变元数不同)
把方程组转换成增广矩阵。
利用初等行变换来把增广矩阵转换成行阶梯阵。枚举k从0到equ – 1，当前处理的列为col(初始为0) ，每次找第k行以下(包括第k行)，col列中元素绝对值最大的列与第k行交换。如果col列中的元素全为0，那么则处理col + 1列，k不变。
转换为行阶梯阵，判断解的情况。
① 无解当方程中出现(0, 0, …, 0, a)的形式，且a != 0时，说明是无解的。
② 唯一解条件是k = equ，即行阶梯阵形成了严格的上三角阵。利用回代逐一求出解集。
③ 无穷解。条件是k < equ，即不能形成严格的上三角形，自由变元的个数即为equ – k，但有些题目要求判断哪些变元是不缺定的。 这里单独介绍下这种解法：首先，自由变元有var - k个，即不确定的变元至少有var - k个。我们先把所有的变元视为不确定的。在每个方程中判断不确定变元的个数，如果大于1个，则该方程无法求解。如果只有1个变元，那么该变元即可求出，即为确定变元。以上介绍的是求解整数线性方程组的求法，复杂度是O(n3)。浮点数线性方程组的求法类似，但是要在判断是否为0时，加入EPS，以消除精度问题。
下面讲解几道OJ上的高斯消元法求解线性方程组的题目：
POJ 1222 EXTENDED LIGHTS OUT 1681 Painter's Problem 1753 Flip Game 1830 开关问题
POJ 3185 The Water Bowls
2947 Widget Factory 9]之间。POJ 1166 The Clocks
POJ 2065 SETI
POJ 1487 Single-Player Games + 递归的做法分解的。首先用栈的思想求出该结点的孩子数，然后递归分别求解各个孩子。这题解方程组也与众不同...首先是求解浮点数方程组，要注意精度问题，然后又询问不确定的变元，按前面说的方法求解。一顿折腾后，这题居然写了6000+B...而且囧的是巨人C++ WA，G++ AC，可能还是精度的问题吧...看这题目，看这代码，就没有改的欲望...hdu OJ 2449
fze OJ 1704
Sgu 275 To xor or not to xor
1/ 用于求整数解得方程组. / 2 3 4 string 5 6 using namespace std; 7 8 const int 105; 9 10 int var// 有equ个方程，var个变元。增广阵行数为equ, 分别为0到equ - 1，列数为var + 1，分别为0到var. 11 int a[maxn][maxn]; 12 int// 解集. 13 bool// 判断是否是不确定的变元. 14 int free_num; 15 16 void void) 17{ 18 int i, j; 19 for 0) 20 { 21 for 0 var 1) 22 { 23 " "; 24 } 25 endl; 26 } 27 endl; 28} 29 30 int int int b) 31{ 32 int t; 33 while 0) 34 { 35 b; 36 b; 37 t; 38 } 39 return a; 40} 41 42 int int int b) 43{ 44 return gcd(a, b); 45} 46 47// 高斯消元法解方程组(Gauss-Jordan elimination).(-2表示有浮点数解，但无整数解，-1表示无解，0表示唯一解，大于0表示无穷解，并返回自由变元的个数) 48 int void) 49{ 50 int i, j, k; 51 int// 当前这列绝对值最大的行. 52 int// 当前处理的列. 53 int ta, tb; 54 int LCM; 55 int temp; 56 int free_x_num; 57 int free_index; 58// 转换为阶梯阵. 59 0// 当前处理的列. 60 for 0 var) 61// 枚举当前处理的行. 62// 找到该col列元素绝对值最大的那行与第k行交换.(为了在除法时减小误差) 63 k; 64 for 1) 65 { 66 if i; 67 } 68 if k) 69// 与第k行交换. 70 for var 1) swap(a[k][j], a[max_r][j]); 71 } 72 if 0) 73// 说明该col列第k行以下全是0了，则处理当前行的下一列. 74 continue; 75 } 76 for 1) 77// 枚举要删去的行. 78 if 0) 79 { 80 lcm(abs(a[i][col]), abs(a[k][col])); 81 abs(a[k][col]); 82 if 0// 异号的情况是两个数相加. 83 for var 1) 84 { 85 tb; 86 } 87 } 88 } 89 } 90 Debug(); 91// 1. 无解的情况: 化简的增广阵中存在(0, 0, ..., a)这样的行(a != 0). 92 for) 93// 对于无穷解来说，如果要判断哪些是自由变元，那么初等行变换中的交换就会影响，则要记录交换. 94 if 0 return 1; 95 } 96// 2. 无穷解的情况: 在var (var + 1)的增广阵中出现(0, 0, ..., 0)这样的行，即说明没有形成严格的上三角阵. 97// 且出现的行数即为自由变元的个数. 98 if var) 99 { 100// 首先，自由变元有var - k个，即不确定的变元至少有var - k个. 101 for 1 0) 102 { 103// 第i行一定不会是(0, 0, ..., 0)的情况，因为这样的行是在第k行到第equ行. 104// 同样，第i行一定不会是(0, 0, ..., a), a != 0的情况，这样的无解的. 105 0// 用于判断该行中的不确定的变元的个数，如果超过1个，则无法求解，它们仍然为不确定的变元. 106 for 0 var) 107 { 108 if 0 j; 109 } 110 if 1 continue// 无法求解出确定的变元. 111// 说明就只有一个不确定的变元free_index，那么可以求解出该变元，且该变元是确定的. 112 var]; 113 for 0 var) 114 { 115 if 0 x[j]; 116 } 117// 求出该变元. 118 0// 该变元是确定的. 119 } 120 return var// 自由变元有var - k个. 121 } 122// 3. 唯一解的情况: 在var (var + 1)的增广阵中形成严格的上三角阵. 123// 计算出Xn-1, Xn-2 ... X0. 124 for var 1 0) 125 { 126 var]; 127 for 1 var) 128 { 129 if 0 x[j]; 130 } 131 if 0 return 2// 说明有浮点数解，但无整数解. 132 a[i][i]; 133 } 134 return 0; 135} 136 137 int void) 138{ 139 "Input.txt " " r ", stdin); 140 int i, j; 141 while " %d %d " var EOF) 142 { 143 0 sizeof(a)); 144 0 sizeof(x)); 145 1 sizeof// 一开始全是不确定的变元. 146 for 0) 147 { 148 for 0 var 1) 149 { 150 " %d "a[i][j]); 151 } 152 } 153// Debug(); 154 Gauss(); 155 if 1 "无解!\n "); 156 else if 2 "有浮点数解，无整数解!\n "); 157 else if 0) 158 { 159 "无穷多解! 自由变元个数为%d\n ", free_num); 160 for 0 var) 161 { 162 if " x%d 是不确定的\n " 1); 163 else "x%d: %d\n " 1, x[i]); 164 } 165 } 166 else 167 { 168 for 0 var) 169 { 170 "x%d: %d\n " 1, x[i]); 171 } 172 } 173 " \n "); 174 } 175 return 0; 176
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Laurent Series : A Key to Understanding Singularities
Last Updated : 23 Jul, 2025
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Laurent series is a powerful tool in complex analysis, providing a way to represent complex functions as a series that includes both positive and negative powers of the variable. This series is particularly useful for functions that have singularities—points where they are not analytic.
In simple terms, the Laurent series allows us to express a complex function f(z) around a point z 0 as a combination of terms that can extend to both positive and negative degrees. This makes it more versatile than the Taylor series, which only includes non-negative powers.
In this article we will briefly discuss about Laurent Series, its definition, formula and solved example of Laurent Series also we will discuss the key difference between Laurent and Taylor Series
What is Laurent’s Series?
Laurent series is a representation of a complex function that generalizes theTaylor seriesto include terms with negative exponents. It allows for the expression of functions that have singularities, particularly useful for functions that are not analytic at certain points. Unlike the Taylor series, which is valid only in a region where the function is analytic, Laurent's series can represent functions in annular regions (ring-shaped areas) around singularities.
Laurent Series consists of two parts: the principal part (terms with negative powers) and the regular part (terms with non-negative powers). Laurent's series is instrumental in complex analysis, particularly in the study of residues and the evaluation of complex integrals.
Table of Content
What is Laurent’s Series?
Formula for Laurent Series
Convergence of Laurent Series
Radius of Convergence
Difference between Laurent Series and Taylor Series
Solved Examples of Laurent Series
Applications of Laurent Series
Solved Examples on Laurent Series
Practice Problems on Laurent Series
Formula for Laurent Series
Formally, for a function f(z) defined on an annulus A = {z ∈ C : r < |z - z 0| < R}, the Laurent series expansion of f(z) about a point z 0 is given by:
f(z)=∑n=−∞∞a n(z−z 0)n f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n f(z)=∑n=−∞∞a n(z−z 0)n
Where,
a n are the coefficients determined by:
a n=1 2 π i∫C f(w)(w−z 0)n+1 d w a_n = \frac{1}{2\pi i} \int_{C} \frac{f(w)}{(w - z_0)^{n+1}} \, dw a n=2 πi 1∫C(w−z 0)n+1 f(w)d w
C is a closed contour around z 0 within the region of analyticity.
The series can be split into two parts:
Principal Part: The terms with negative powers of
∑n=−1−∞a n(z−z 0)n\sum_{n=-1}^{-\infty} a_n (z - z_0)^n∑n=−1−∞a n(z−z 0)n
Regular Part: The terms with non-negative powers of
∑n=0∞a n(z−z 0)n\sum_{n=0}^{\infty} a_n (z - z_0)^n∑n=0∞a n(z−z 0)n
Convergence of Laurent Series
Convergence of the Laurent series occurs on an annulus; defined as {z: r 1< | z – z 0 | < r 2}.
For a Laurent series to converge, the positive and negative degree terms of the power series must converge. This convergence is uniform on compact sets within the annulus. As a result, the series defines a holomorphic function on this region.
Radius of Convergence
Radius of convergence of a power series is the distance within which the series converges to a finite value.
The convergence of a Laurent series depends on the distance from the point z 0 and can be divided into three regions:
Interior of the Inner Radius (r1): The series converges for |z - z 0| < r 1 only if a n = 0 for all n < 0 , reducing it to a Taylor series.
Annulus ( r1< |z - z0| < r2): The series converges in this annular region. This is the most general case for Laurent series, where both positive and negative powers of (z - z 0) are present.
Exterior of the Outer Radius ( r2):The series converges for |z - z_0| > r_2 if a_n = 0 for all n \geq 0 , turning it into a series in negative powers of (z - z_0).
Convergence Criteria
Some of the common criteria for convergence of series are:
Cauchy-Hadamard Theorem: For a series ∑n=−∞∞a n(z−z 0)n\sum_{n=-\infty}^{\infty} a_n (z - z_0)^n∑n=−∞∞a n(z−z 0)n, define:
1 r 1=lim supn→∞∣a−n∣1/n \frac{1}{r_1} = \limsup_{n \to \infty} |a_{-n}|^{1/n} r 11=lim sup n→∞∣a−n∣1/n
1 r 2=lim supn→∞∣a n∣1/n \frac{1}{r_2} = \limsup_{n \to \infty} |a_n|^{1/n} r 21=lim sup n→∞∣a n∣1/n
The series converges in the annulus r1< |z - z0| < r2 .
Absolute Convergence: If the Laurent series converges at some point z 1 , then it converges absolutely at every point z such that |z - z 0| = |z 1 - z 0|.
Uniform Convergence:The series converges uniformly on compact subsets within the annulus r1< |z - z0| < r2.
Difference between Laurent Series and Taylor Series
Some of the key differences between laurent and taylor series are listed in the following table:
| Aspect | Laurent Series | Taylor Series |
---
| Definition | Represents a function as a series with both positive and negative powers of (z - z 0) | Represents a function as a series with only non-negative powers of (z - z 0). |
| General Form | _f_(_z_)=∑∞_n_=−∞_a_ _n_(_z_−_z_ 0)_n_ | f(_z_)=∑∞_n_=0 _a_ _n_(_z_−_z_ 0)_n_ |
| Region of Convergence | Annulus( r 1 < |z - z 0| < r 2 ) | z - z 0 |
| Components | Contains both a principal part (negative powers) and an analytic part (non-negative powers). | Contains only the analytic part (non-negative powers). |
| Singularities | Can handle isolated singularities within the annulus | Cannot handle singularities; requires function to be analytic |
| Existence | a n=1 2 π i∫γ f(ζ)(ζ−z 0)n+1 d ζ a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta a n=2 πi 1∫γ(ζ−z 0)n+1 f(ζ)d ζ where γ is a contour around z 0. | a n=f(n)(z 0)n!a_n = \frac{f^{(n)}(z_0)}{n!}a n=n!f(n)(z 0) where f n(z 0) is the n~~th~~ derivative at z 0. |
| Analytic Requirement | Always exists for functions with isolated singularities, providing a unique representation. | Exists only for analytic functions within the radius of convergence. |
Solved Examples of Laurent Series
Example 1: Find the Laurent series for f(z)= z+1/z around z0= 0.
Solution:
The given function can be expressed as:
f(z)=z z+1 z=1+1 z f(z) = \frac{z}{z} + \frac{1}{z} = 1 + \frac{1}{z}f(z)=z z+z 1=1+z 1
Hence, the Laurent series is:
f(z)=1+1 z f(z) = 1 + \frac{1}{z}f(z)=1+z 1
This series is valid in the region 0<∣z∣<∞.
Example 2: Find the Laurent series for f(z)= z/z2+ 1 around z0=i.
Solution:
Using partial fractions, the function can be decomposed as:
f(z)=1 2(1 z−i)+1 2(1 z+i)f(z) = \frac{1}{2} \left( \frac{1}{z - i} \right) + \frac{1}{2} \left( \frac{1}{z + i} \right)f(z)=2 1(z−i 1)+2 1(z+i 1)
The term 1/z + i is analytic at z=i and can be expanded using a geometric series:
1 z+i=1 2 i∑n=0∞(−z−i 2 i)n\frac{1}{z + i} = \frac{1}{2i} \sum_{n = 0}^{\infty} \left( -\frac{z - i}{2i} \right)^n z+i 1=2 i 1∑n=0∞(−2 i z−i)n
Therefore, the Laurent series is:
The region of convergence is 0<∣z−i∣<2.
Applications of Laurent Series
The applications of the Laurent Series are as follows:
Complex analysis: Helps in studying the behavior of complex functions near singularities.
Residue calculus: Used for evaluating complex integrals via the residue theorem.
Engineering: Applied in signal processing and control theory for stability analysis.
Physics: Utilized in quantum mechanics and electrodynamics for potential expansions.
Mathematical modeling: Assists in solving differential equations and in the analysis of stability and bifurcation.
Read More,
Taylor Series
Maclurian Series
Residue Theorem
Power Series
Couchy Theorem
Solved Examples on Laurent Series
Example 1: Find the Laurent series for f(z)= z+1/z around 𝑧0=v0and determine the region of convergence.
Solution:
The given function is:
f(z)=z+1 z=1+1 z f(z) = \frac{z + 1}{z} = 1 + \frac{1}{z}f(z)=z z+1=1+z 1
Here, the function is already in the form of a Laurent series. We can write:
f(z)=1+1 z f(z) = 1 + \frac{1}{z}f(z)=1+z 1
The term 1 represents the analytic part with non-negative powers of z.
The term 1/𝑧 represents the principal part with negative powers of z.
Region of Convergence:
The series is valid for 0 < ∣z∣ < ∞.
Example 2: Find the Laurent series for f(z)= z/ z2+ 1 around z0=i. Identify the region where your answer is valid and the singular part.
Solution:
First, use partial fractions to decompose f(z):
f(z)=z z 2+1=1 2(1 z−i+1 z+i)f(z) = \frac{z}{z^2 + 1} = \frac{1}{2} \left( \frac{1}{z - i} + \frac{1}{z + i} \right)f(z)=z 2+1 z=2 1(z−i 1+z+i 1)
Expanding 1/z+i around z 0=i:
1 z+i=1 2 i⋅1 1+z−i 2 i=1 2 i∑n=0∞(−z−i 2 i)n\frac{1}{z + i} = \frac{1}{2i} \cdot \frac{1}{1 + \frac{z - i}{2i}} = \frac{1}{2i} \sum_{n=0}^{\infty} \left( -\frac{z - i}{2i} \right)^n z+i 1=2 i 1⋅1+2 i z−i1=2 i 1∑n=0∞(−2 i z−i)n
The Laurent series is:
f(z)=1 2⋅1 z−i+1 4 i∑n=0∞(−z−i 2 i)n f(z) = \frac{1}{2} \cdot \frac{1}{z - i} + \frac{1}{4i} \sum_{n=0}^{\infty} \left( -\frac{z - i}{2i} \right)^n f(z)=2 1⋅z−i 1+4 i 1∑n=0∞(−2 i z−i)n
Singular Part:
The term 1/z − i represents the principal part.
Region of Convergence:
The series is valid for 0<∣z−i∣<2.
Practice Problems on Laurent Series
Problem 1: Determine the Laurent series for f(z)= 1/z(z−1) around z 0=0.
Problem 2:Compute the Laurent series for f(z)= 1/z 2 + 4 around z 0=2i.
Problem 3:Find the Laurent series for f(z)= e 2/z 3 around z 0 =0.
Conclusion
Laurent series is a powerful tool in complex analysis, extending the capabilities of the Taylor series to functions with singularities. It provides a comprehensive method for representing and analyzing complex functions in annular regions, making it indispensable for both theoretical and applied mathematics. Its applications span various fields, highlighting its versatility and importance in mathematical sciences.
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188783 | https://philosophy.stackexchange.com/questions/36351/examples-of-falsifiability | Skip to main content
Examples of Falsifiability
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I came across the notion of falsifiability quite recently.
The wikipedia article on the same states that:
Falsifiability or
refutability of a statement, hypothesis, or theory is the inherent
possibility that it can be proven false. A statement is called
falsifiable if it is possible to conceive of an observation or an
argument which negates the statement in question. In this sense,
falsify is synonymous with nullify, meaning to invalidate or "show to
be false".
For a statement to be questioned using observation, it needs to be at
least theoretically possible that it can come into conflict with
observation.
While I can understand the general concept - I would like to have a deeper understanding of the same. Popper mentions that this notion differentiates science from pseudo - science.
Can someone please give me some examples for the same? - So that I might understand the idea more intuitively.
Specifically if you could provide what would be the falsifiability arguments/observations would be for:
Newton's theory of gravitation.
Heliocentralism
Theorem of calculus.
Probability theory.
Basically two popular theories from the realm of physics and two popular theories from mathematics (which I might possibly be familiar with), would do. Need not be just these four.
philosophy-of-science
falsifiability
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edited Jul 2, 2016 at 14:57
E...
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asked Jul 2, 2016 at 12:49
user12196user12196
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Newton's theory: a free apple "falling" from the floor to the ceiling.
Mauro ALLEGRANZA
–
Mauro ALLEGRANZA
2016-07-02 15:29:38 +00:00
Commented
Jul 2, 2016 at 15:29
Right - if we observe that theory of gravitation would be falsified.
user12196
–
user12196
2016-07-02 15:43:37 +00:00
Commented
Jul 2, 2016 at 15:43
Of course, the more complex is the theory, more difficult is to found "simple" falsifying conditions like that. When many "factors" are involved, a falsifying experiment must "manage" all of them. Consider the well-known discovery of Neptune by Urbain Le Verrier : a potential "falsifier" has been transformed into a brilliant "verification".
Mauro ALLEGRANZA
–
Mauro ALLEGRANZA
2016-07-02 15:56:26 +00:00
Commented
Jul 2, 2016 at 15:56
2
For mathematical theories, it is not so clear if Popper's criteria applies. In principle, we can say that the only way to "falsify" a math theory is proving his inconsistency.
Mauro ALLEGRANZA
–
Mauro ALLEGRANZA
2016-07-02 15:57:23 +00:00
Commented
Jul 2, 2016 at 15:57
@MauroALLEGRANZA Can you give me some sources to study up Urbain Le Verrier's discovery - specifically how it relates to falsifiability.
user12196
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user12196
2016-07-02 17:25:19 +00:00
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Jul 2, 2016 at 17:25
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The best way to understand Popper is to read Popper. There are a few commentators who have correctly understood his ideas, but the vast bulk of commentary on Popper is not even able to state his ideas correctly. Lakatos, Feyerabend and Kuhn are especially bad and should be avoided.
To understand falsification properly, you need to understand Popper's theory of knowledge more broadly. Most philosophers of science who take science seriously and think it is good are inductivists: they believe in a process called induction. Induction supposedly involves (1) taking observations, (2) using them to make theories, and then (3) showing those theories are true or probably true by more observations. People have looked at many phenomena such as the night sky, biology, medicine and so on, without learning much for thousands of years. So just observing stuff doesn't do much good. If you don't know what to look for, just observing will not produce progress, so step (1) is impossible. In addition, explanations don't follow from observations. The theory of stars has implications for many events we will never observe, e.g. - supernovae that took place before there were human observers, and those events don't follow from observations without a theory of how stars change. So steps (2) and (3) are also impossible.
So if we don't get theories from observation how do we get them? We guess. You look for a problem: some issue that is not explained by current ideas. You guess solutions to that problem. You then criticise the proposed solutions. This criticism may involve experiments, but many theories can be eliminated without doing experiments, e.g. - inconsistent theories.
An experiment involves looking for a situation in which two or more different ideas about how the world works make different predictions. You then either set up that situation or look for an existing system that realises that situation. Newton's theory of gravity and Einstein's general theory of relativity made different predictions about Mercury, and Newton's theory was refuted.
Some philosophers make a lot of fuss about the possibility that you might do an experiment wrong or misinterpret the results. But as Popper pointed out in Logic of Scientific Discovery, Chapter V (especially Section 29), this problem is solved by his epistemology. If an experiment contradicts an existing theory, that's a problem. This problem can be solved by any guess that explains the difference and is not eliminated by some criticism. The discovery of Neptune was taken as an example above, so let's look at it. An unsolved problem was found in explaining the orbits of some planets. Urbain Le Verrier guessed that there might be another planet. He worked out some constraints on where the planet could be to produce such effects, Johann Gottfried Galle looked for it and found it. If Galle had not found the planet that problem would have remained unsolved. Perhaps some other explanation could have been found to reconcile Newtonian mechanics with observation, perhaps not. Popper recommended that a proposed solution to a scientific problem should be rejected if it was ad hoc: if it had no implications beyond the problem it was invented to solve.
I'm going to skip the heliocentric theory because it is fairly similar to Newtonian mechanics. If you want a long list of examples, see the introduction to "Realism and the Aim of Science" by Popper.
Mathematical theories are about abstractions. They can be critically discussed, but not experimentally tested. 1+1 = 2 even though it is possible to think of examples of putting two objects together and only getting one object as a result. If you move two piles of sand together, you may only get one pile. So you have to think carefully about what systems you take as models of mathematical operations such as addition. For a discussion see "Realism and the Aim of Science" by Popper Chapter III, Section 24.
As far as probability is concerned, the best existing explanations have been provided by David Deutsch, see
For explanations of Popper's positions, see "Objective Knowledge" by Popper, Chapter 1, "Realism and the Aim of Science" by Popper, "Logic of Scientific Discovery" by Popper, "The Fabric of Reality" by David Deutsch, Chapters 3 and 7, and "The Beginning of Infinity" by David Deutsch, Chapters 1,2,4 and 13.
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answered Jul 4, 2016 at 10:29
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In the comments to Cort Ammon's answer you say:
"So we can't falsify mathematical theories? I thought Popper's method was a way of distinguishing the scientific from the non scientific - does that imply mathematical constructs are not scientific or is there something wrong with Popper's method."
Exactly, mathematical theories are not scientific theories. Mathematics is about abstract mathematical objects, Science is about empirically observables phenomena. The truth of mathematical statements are prove using logic and reason alone, while the truth of statements in physics, chemistry, biology, etc...are proven by experiment and observation. This was best described by David Hume, with his distinction known as Hume's Fork:
"All the objects of human reason or enquiry may naturally be divided into two kinds, to wit, Relations of Ideas, and Matters of fact. Of the first kind are the sciences of Geometry, Algebra, and Arithmetic ... [which are] discoverable by the mere operation of thought ... Matters of fact, which are the second object of human reason, are not ascertained in the same manner; nor is our evidence of their truth, however great, of a like nature with the foregoing."
- An Enquiry Concerning Human Understanding
So things like the fundamental theorem of calculus and probability theory can't be falsified because they don't correspond to anything observable. They, like all mathematical truths are proved solely using the rules and axioms of logic.
This is the whole point of falsification, one has to attempt to show that they empirically observe a phenomena that contradicts their theory. So the Newton's theory of gravity says that apples should fall every time we let go of them in midair. Pre Popper's falsificationism, Newton's theory is falsified if someone raises an apple lets go of it, and instead of it falling it hovers in the air or goes upwards.
Similarly per Popper, heliocentrism will be falsified the day that Venus or Mars, or one of the other planets is observed in a different orbit then the one predicted by the theory.
This points to an interesting problem with Popper's theory, that of auxiliary hypotheses (also called the Duhem-Quine thesis, or they idea that all observations are theory laden): Consider that at the beginning of the 19th century the orbit of Uranus was different than what was predicted by Newtoninan mechanics and heliocentrism. But astronomers, instead of abandoning the theory, concluded that there was an unknown planet modifying the orbit of Uranus, which they later confirmed and called Neptune. So the dilemma is: When observation contradicts theory, is the theory falsified? or is there missing data that can explain the mismatch between theory and predictions?
The issue of how to solve the problem of auxiallry hypotheses is still debated, and hasn't been solved yet. See the ideas of W.V.O Quine, Thomas Kuhn, Imre Lakatos and Paul Feyerabend, all in response to Popper's concept of falsification.
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answered Jul 3, 2016 at 18:44
Alexander S KingAlexander S King
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The thing is that the wiki article also says arguments could be used to falsify-so I thought it might be applicable for mathematics constructs as well.
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Falsification is an excellent and easy to understand system in principle, but much more nuanced in implementation. The easiest to falsify hypotheses are those famous ones such as "all swans are white," which can be falsified by observing a black swan. Of course, this assumes we all agree on what is a swan is. Hypotheses get murkier from there.
When it comes to real meaningful scientific hypotheses, falsification is typically more of an extended process rather than an instantaneous event. A scientific theory which is falsifiable is one where some results could cast substantial doubt on the hypothesis, and that doubt can be compounded by future tests.
For example, if one believed the hypothesis that light acts as a wave, one would be surprised to see particle like behavior. The photoelectric effect is one such effect that we now know exhibits particle like behavior. The first time one observes particle like behavior from an experiment, one might assume the results were a measurement error. Doing it a second time would begin casting doubt on the theory that light always behaves like a wave. Having dozens of scientists all run such experiments multiple times, and each discovering particle like effects would eventually "falsify" the hypothesis.
This process is even more complicated due to statistics. If I claim there is a gaussian error term on my results, you can never truly prove that my theory is wrong, because there is always a non-zero chance that you simply observed random luck. However, in practice, once the probability of such chance events is low enough, we declare a theory "falsified." How high one has to go is discipline dependent. In sociology, we regularly see error terms permitting 10% or even 20% due to unexplained factors. In particle physics, a hypothesis is not declared "confirmed" until those unexplained factors account for no more than 0.00001% of the total observed effects. This is because subatomic particles behave quite regularly, and we're able to generate as many results as needed to attain such high degrees of confidence. In sociology, it is much harder to repeat experiments and there is a great deal of variance between individuals, so the best we can do is lower degrees of confidence.
As for your list of examples:
Newton's theory of gravitation.
It is generally accepted that the motion of planets is almost completely governed by gravitational interactions. If we were to observe the motion of the planets, and find substantial deviations not explained by his theory of gravity, this would either falsify his theory or show that there are other forces at work. I believe we actually do see results which would falsify his work: you have to account for relativity to explain some movements (particularly in cases near a black hole)
Heliocentralism
Heliocentralism actually cannot be proven nor disproven because it is merely a model. Its more akin to a coordinate system transform than a theory. However, if one assumes geocentralism, one is forced to admit many strange forces which account for all of the movement we see in the planets. If one assumes heliocentralism, the movement can be explained entirely with simple conservative gravity models like Newton's theory of gravity. It is the simplicity of the heliocentric model that made it so effective.
Theorem of calculus.
Consider if we couldn't go anywhere, because of Xeno's paradox. This would demonstrate that the assumptions we make regarding limits are false. That being said, calculus is a mathematical construct. All that we can really falsify is its usefulness in describing the world around us.
Probability theory.
Once again this is a mathematical construct, making it difficult to falsify. However, one could argue that it is "falsified" by demonstrating that it does not effectively model reality. One major assumption in much of probability is IID: the idea that observations are (I)ndependent (I)denticaly (D)istributed. If there was a reason to argue that this assumption is invalid in the real world, then much of probability would not apply. This actually does occur when exploring the human mind. In many cases the assumption of IID is very poorly founded, so many simplifications that probability would permit are simply invalid when discussing the behavior of the mind.
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answered Jul 3, 2016 at 5:52
Cort AmmonCort Ammon
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So we can't falsify mathematical theories? I thought Popper's method was a way of distinguishing the scientific from the non scientific - does that imply mathematical constructs are not scientific or is there something wrong with Popper's method. Theories in physics use many mathematical constructs as well - so if they are not falsifiable how come the concepts on physics are?
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@novice I cannot answer the question "does that imply mathematical constructs are not scientific" directly, because it would involve a detailed discussion of exactly what "not scientific" means to you. That would be better suited for a chat room, rather than comments. However, I will try my best to answer obliquely. The validity of a mathematical construct is based in the validity of its assumptions and the validity of the rules of inference associated with it, not physical reality. It is only as one seeks to apply said constructs to the real world that the concept of falsification...
Cort Ammon
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... becomes meaningful. Without that application to physical reality, mathematical theories are subject to far more stringent validity requirements than falsification would ever ask from them. However, when applying such constructs to science, one does make the assumption that those constructs are indeed valid. If the real numbers do not, indeed, form a field, calculus falls apart. On the other hand, it is totally valid to have mathematical constructs which do not have an immediately obvious connection to reality. The concept of complex numbers, for instance, was a mathematical...
Cort Ammon
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Cort Ammon
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... curiosity until Euler's function connected them to cyclic motion. Now they are used constantly in science. If you are interested in this question of the validity of mathematical construct, I recommend a beautiful video by Vsauce, How to Count past Infinity. He does an excellent job of explaining the subtle distinction between scientific theories and mathematical ones.
Cort Ammon
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Cort Ammon
2016-07-03 16:44:02 +00:00
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Jul 3, 2016 at 16:44
Thanks for your comments as well as the resource you have provided. I confess I am not sure how Popper gives arguments/justifications in favor of the notion of falsifiability to distinguish science from pseudo-science. I read the part about observation and argument - and thought that an argument for the falsifiability of mathematical constructs must be possible - instead of an observation. Let me go into this more deeply and come back with more questions.
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I'm not sure just how useful falsification is in explaining the actual progress of science, in the sense of the revision of its basic concepts; its not for example a source of new ideas, but of pruning out what is given on the basis of what is known. Its a minor mode of progress, and not its major mode. A major mode would tell us how to find new ideas, unfortunately such a philosphical stone is illusionary.
For example, calculus can be explained by attempting to give meaning to 0/0; this, in terms of the usual arithmetic operations, is nonsensical; however mathematicians like to 'close up' operations; 0/0 can in fact be given meaning by thinking of it as dx/dy; of course this opens up the whole new world of calculus.
Similarly, no meaning could be given to the square root of -1; eventually one was found that was useful: i -the imaginary; and it again opened up a whole new world of complex geometry.
Probability is a concept with intuitive appeal; yet Quantum Mechanics relies on the notion of the square root of probability, and in fact a great deal of the bizarre beahviour can be explained on the basis of this new concept which still hasn't found a properly ontological basis in the same way that the infinitesimal or the imaginary has.
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answered Jul 4, 2016 at 10:54
Mozibur UllahMozibur Ullah
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-1 This answer does not even attempt to address the question.
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Hypercalcemia …
Douglas J Lanska MD MS MSPH
General Neurology
Updated
02.06.2025
Released
11.03.1993
Expires For CME
02.06.2028
Hypercalcemia
Author
: Douglas J Lanska MD MS MSPH
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Overview
Hypercalcemia is associated with a broad range of neurologic manifestations that have been ascribed to both central nervous system and peripheral nervous system dysfunction. Reported neurologic manifestations can include weakness, fatigue, confusion, posterior reversible leukoencephalopathy syndrome, a Creutzfeldt-Jakob–like syndrome due to hypercalcemic encephalopathy, stupor, and coma. It remains unclear if the weakness associated with hypercalcemia is primarily due to the central nervous system and peripheral nervous system effects of hypercalcemia. In this article, the author reviews the clinical spectrum of neurologic dysfunction associated with hypercalcemia, as well as the evaluation and management of hypercalcemia.
Key points
| | |
--- |
| | • Depending on the severity and rate of development, hypercalcemia can produce varying degrees of a generalized encephalopathy ranging from mild impairment of attention to coma. |
| | • Primary hyperparathyroidism and malignancy-associated hypercalcemia are the most common causes of hypercalcemia, together accounting for more than 90% of cases. |
| | • Hypercalcemia in the setting of malignancy is a common oncologic emergency and develops in 20% to 30% of patients with cancer. |
| | • For patients with severe hypercalcemia (greater than 13.5 mg/dL) or moderate hypercalcemia and significant clinical manifestations, the initial management entails strategies that directly lower the calcium concentration, independent of the underlying cause. |
Historical note and terminology
The parathyroid glands were discovered (but not named) in 1852 by comparative anatomist and paleontologist (later Sir) Richard Owen (1804-1892) in the necropsy of an Indian rhinoceros that died at the London Zoo (218; 61; 43; 124; 96). In his description, Owen referred to the glands as "a small compact yellow glandular body attached to the thyroid at the point where the veins emerged" (218; 61). The significance of this report was only evident in retrospect, and Owen was much more famously recognized for naming the Dinosauria (ie, dinosaurs) in the 1830s and infamously as an antievolutionist opposed to Charles Darwin and his proponent Thomas Huxley after publication of Darwin’s The Origin of Species by Means of Natural Selection (1859) (183; 110; 111; 99; 227). Although Owen was notorious for usurping the work of others and passing it off as his own, in this case Owen was apparently responsible for the observation, although he had no idea of its significance.
British surgeon and anatomist Richard Owen
British surgeon and anatomist Richard Owen (1804-1892) c1856 with the skull of a crocodile. In 1852, Owen discovered the parathyroid glands in the necropsy of an Indian rhinoceros. However, he neither named the glands, nor underst...
Unaware of Owen’s earlier work, published as it was in what was then a relatively obscure society proceeding, the parathyroid glands were identified decades later in humans in 1880 by Ivar Sandstrom (1852-1889), a 25-year-old medical student working as a praelector (lecturer) in anatomy at the University of Uppsala, Sweden (235; 236; 48; 58; 144; 145). In his classic paper, On a New Gland in Man and Fellow Animals (in translation), he described what he called the “glandulae parathyroidae” (parathyroid glands) in dogs, cats, rabbits, oxen, horses, and man (gross and micro). Sandstrom’s principal interest was the organ in man, and he examined 50 individuals and found in most of them two parathyroid glands on each side. Unfortunately, Sandstrom's report was not well received, and he later committed suicide at age 37 years.
Swedish physician and anatomist Ivar Sandstrom (1852-1889)
In 1880, Sandstrom identified the parathyroid glands in man and various other mammals. (Courtesy of Wikipedia and Wellcome Library. Public domain.)
Parathyroid glands (gross appearance)
Gross appearance of the parathyroid glands (marked as “gl. pth.” in the Figure) in man as illustrated by Swedish physician and anatomist Ivar Sandstrom in 1880. (Public domain.)
Parathyroid glands (microscopic appearance)
Microscopic appearance of the parathyroid glands in man as illustrated by Swedish physician and anatomist Ivar Sandstrom in 1880. (Public domain.)
The clinical importance of the parathyroid glands was not appreciated until 1891, when French physiologist Eugène Émile Gley (1857-1930) observed that tetany and death following experimental thyroidectomy in dogs occurred only if the excised material included the glandulae parathyroidae described by Sandström (104). To this point, tetany in association with thyroidectomy had been misattributed to the removal of the thyroid gland. Because of Gley’s discovery, parathyroid glands have sometimes been referred to as "Gley's glands."
French physiologist Eugène Émile Gley (1857-1930)
Gley observed that tetany and death following experimental thyroidectomy in dogs occurred only if the excised material included the glandulae parathyroidae (parathyroid glands) described by Ivar Sandstrom in 1880. (Courtesy of the...
From 1903 to 1908, American pathologist William G MacCallum (1874-1944) and Swiss-U.S. pharmacologist (and later the first head of the U.S. National Cancer Institute from 1938-1943) Carl Voegtlin (1879-1960), both working at Johns Hopkins, demonstrated that tetany following parathyroidectomy was the result of the hypocalcemia (182; 181; 142; 186; 86). Not only was there a “marked reduction in the calcium content of the tissues especially of the blood and brain, during tetany” following parathyroidectomy, but the “injection of a solution of a salt of calcium into the circulation of an animal in tetany promptly checks all the symptoms and restores the animal to an apparently normal condition.” MacCallum and Voegtlin also showed that variable production of tetany following parathyroidectomy in animal experiments depended on the presence of residual parathyroid tissue, a result that was not infrequent because of the variable number and location of the parathyroid glands. In 1909, William B Berkeley and S P Beebe, at Cornell University Medical College in New York, described correction of hypocalcemic tetany with parathyroid extract in man (38).
American pathologist William G MacCallum (1874-1944)
From 1903 to 1908, MacCallum studied the effects of parathyroidectomy. In 1908, MacCallum and colleague Carl Voegtlin, both working at Johns Hopkins, demonstrated that tetany following parathyroidectomy was the result of the hypoc...
Swiss-American pharmacologist Carl Voegtlin
Swiss-American pharmacologist (and later the first head of the U.S. National Cancer Institute) Carl Voegtlin (1879–1960). In 1908, Voegtlin and colleague William G MacCallum, both working at Johns Hopkins, demonstrated that tetany...
In 1891, German pathologist Friedrich Daniel von Recklinghausen (1833-1910) described osteitis fibrosa cystica, which is characterized by a loss of bone mass, a weakening of the bones as their calcified supporting structures are replaced with fibrous tissue (peritrabecular fibrosis), and the formation of cyst-like brown tumors in and around the bone. This is also known as osteitis fibrosa, osteodystrophia fibrosa, and Recklinghausen disease of bone (which should not be confused with Recklinghausen disease, neurofibromatosis type I). By 1914 Austrian pathologist Jacob Erdheim (1874-1937), working in Vienna, suggested that parathyroid pathology may cause skeletal abnormalities, and this was documented the following year by Z Schlagenhaufer from the observation that in patients with osteitis fibrosa cystica, only one parathyroid gland is typically enlarged (ie, a parathyroid adenoma). If the parathyroid enlargement had been somehow due to or in response to the bony changes, then all of the parathyroid glands routinely should have been similarly enlarged. This ultimately led to the use of parathyroidectomy as a treatment for osteitis fibrosa cystica beginning in 1925.
German pathologist Friedrich Daniel von Recklinghausen
German pathologist Friedrich Daniel von Recklinghausen (1833-1910) is most known for his monograph in 1882 (Über die multiplen Fibrome der Haut und ihre Beziehung zu den multiplen Neuromen) in which he reviewed previous literature...
Osteitis fibrosa cystica (x-ray)
Osteitis fibrosa cystica is characterized by a loss of bone mass, a weakening of the bones as their calcified supporting structures are replaced with fibrous tissue (peritrabecular fibrosis), and formation of cyst-like brown tumor...
Austrian pathologist Jacob Erdheim (1874-1937)
By 1914 Erdheim, working in Vienna, suggested that parathyroid pathology may cause skeletal abnormalities. Photograph reproduced with permission from Erdheim’s great niece, Claudia Erdheim. The image is the private property of Cla...
Parathyroid surgery began before that, though. British surgeon Sir John Bland-Sutton (1855-1936) had described a postmortem specimen of a parathyroid tumor in 1886, had surgically removed a parathyroid cyst in 1909, and had performed a parathyroidectomy for a parathyroid tumor some time before 1917 (79). In 1907, Herbert M Evans, working with American surgeon William Stewart Halsted (1852-1922) at Johns Hopkins in Baltimore, described the vascular supply of the parathyroid glands in man, and in the same paper Halsted discussed preservation of the parathyroid glands with thyroid surgery (117). Evans careful drawing of the parathyroid glands was later used by anatomist Henry Gray in his Anatomy of the Human Body (1918). In 1909, Halsted attempted both iso- and auto-transplantation of parathyroid tissue by transplanting canine parathyroid glands into thyroid tissue and under the skin (116). In 1925, Viennese surgeon Felix Mandl (1892-1957), at the Hochenegg Clinic, performed a successful parathyroidectomy as a means of alleviating the bone disease of hyperparathyroidism; his patient was a 34-year-old tram-car conductor with severe osteitis fibrosa cystica (187).
British surgeon Sir John Bland-Sutton (1855-1936)
Bland-Sutton had described a postmortem specimen of a parathyroid tumor in 1886, had surgically removed a parathyroid cyst in 1909, and had performed a parathyroidectomy for a parathyroid tumor some time before 1917. (Courtesy of ...
American surgeon William Stewart Halsted (1852-1922)
In 1909, Halstead, working at Johns Hopkins in Baltimore, attempted both iso- and auto-transplantation of parathyroid tissue, by transplanting canine parathyroid glands into thyroid tissue and under the skin. (Courtesy of the U.S....
Parathyroid glands (drawing)
Drawing of the parathyroid glands by Herbert M Evans, working with American surgeon William Stewart Halsted (1852-1922). Evans careful drawing of the parathyroid glands was subsequently used in a later edition (1918) of Anatomy of...
In 1923 Adolph M Hanson (1880-1959), and two years later Canadian biochemist James B Collip (1892-1965) independently isolated parathyroid hormone from crude glandular extracts (123; 92; 119; 120; 121; 122; 70; 70; 212; 272; 49; 30; 176; 177; 148). The purification of parathyroid hormone greatly accelerated experimental studies to determine the effect of the hormone on bone and kidneys. In addition, American medical physicist Rosalyn Sussman Yalow (1921-2011) successfully developed radioimmunoassays for peptide hormones, including parathyroid hormone (42; 39; 39; 40; 41; 250; 274). Yalow was awarded a Nobel Prize for Physiology or Medicine in 1977.
Canadian biochemist James B Collip (1892-1965)
In 1923, following the earlier work of Adolph M Hanson, Collip isolated parathyroid hormone from crude glandular extracts. (Courtesy of Wikipedia. Public domain.)
American medical physicist and Nobel laureate Rosalyn Sussman Yalow (1921-2011)
Yalow successfully developed radioimmunoassays for peptide hormones, including parathyroid hormone. (Courtesy of Wikipedia. Public domain.)
From the late 1920s until 1956 (when he suffered a career-ending postoperative complication of chemopallidectomy for early-onset Parkinson disease, ie, intracranial hemorrhage with resulting akinetic mutism), American endocrinologist Fuller Albright (1900-1969) and associates at the Massachusetts General Hospital in Boston studied numerous aspects of disordered parathyroid gland function and conducted landmark metabolic balance studies that clearly defined several of the diseases associated with parathyroid dysfunction, as well as related disorders of calcium and phosphorus metabolism (34; 06; 04; 05; 07; 09; 15; 14; 16; 02; 18; 134; 03; 03; 68; 50; 11; 12; 194; 72; 22; 26; 32; 33; 127; 131; 167; 258; 132; 89; 220; 155; 154; 97; 188). In 1929, Albright colleague Read McLane Ellsworth (1899-1970) diagnosed the first case of idiopathic hypoparathyroidism (10). Albright and colleagues noted that most patients treated with parathyroidectomy for primary hyperparathyroidism and osteitis fibrosa cystica also had nephrolithiasis or nephrocalcinosis (06), established the concept of secondary hyperparathyroidism (06), described hyperparathyroidism due to adrenal hyperplasia (06), described vitamin D-resistant rickets and effective treatment with high doses of vitamin D (08), established a primary effect of vitamin D is to increase intestinal absorption of calcium (16), and described postmenopausal osteoporosis (02), hypercalcemia with disuse osteoporosis (02), pseudohypoparathyroidism (18), the milk-alkali syndrome (17), pseudo-pseudohypoparathyroidism (11), and idiopathic hypercalciuria (12).
The subsequent assay, sequencing, and cloning of parathyroid hormone led to the further elaboration of the multiple actions of the hormone and of the abnormalities associated with dysfunction of the parathyroid glands.
In 1957, Walter T St Goar, at the College of Physicians and Surgeons of Columbia University in New York, emphasized the abdominal manifestations of hyperparathyroidism and proposed a mnemonic triad for recognizing the disorder as a “disease of stones, bones, and abdominal groans” (253). St Goar had been influenced to pursue studies in this area by Fuller Albright while St Goar and his wife were interns and residents at Massachusetts General Hospital. As St Goar elaborated (253):
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| | Gastrointestinal symptoms appear to represent a clue to the earlier recognition of some cases of hyperparathyroidism…Unexplained episodes of nausea and vomiting, unexplained anorexia and weight loss, peptic ulcers which do not respond in the usual way to therapy, [marked constipation,] and a variety of unexplained abdominal pains should all lead to a consideration of hyperparathyroidism as a possible diagnosis. Hyperparathyroidism, which has been popularly thought of by medical men as a ‘disease of stones and bones,’ might be recognized both earlier and more frequently if it were widely regarded as a ‘disease of stones, bones and abdominal groans. |
St Goar recognized that the abdominal manifestations of hyperparathyroidism are nonspecific, but he hoped that greater recognition of their prominence in this disorder might speed clinical recognition and treatment (253):
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| | These gastrointestinal symptoms are meaningless in themselves. An awareness of their occurrence in hyperparathyroidism, however, may prove helpful in recognizing other nonspecific manifestations of a potentially reversible disease and thus lead to its earlier diagnosis. |
In 1961, William C Mieher Jr., Yvan Thibaudeau, and Boy Frame, at Henry Ford Hospital in Detroit, emphasized the neuropsychiatric features of hyperparathyroidism, which can include apathy, agitated depression, psychosis with hallucinations or delusions, paranoia, and dementia. They modified St Goar’s mnemonic triad into a mnemonic quadrad by adding “psychic moans” to reflect the neuropsychiatric manifestations: “we wish to add a postscript to St Goar's description and emphasize that hyperparathyroidism is a disease of stones, bones, abdominal groans, and psychic moans” (198).
In 1965 Charles E Boonstra and Charles E Jackson, at the Caylor-Nickel Clinic in Bluffton, Indiana, emphasized the chronic fatigue and nonspecific irritability seen in many patients with hyperparathyroidism (46). Boonstra further modified the existing mnemonic for hyperparathyroidism from a quadrad into a pentad by adding “fatigue overtones” to reflect the fatigue and “nervous irritability” often seen in patients with hyperparathyroidism, even when more specific findings are either absent or not clinically manifest (46):
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| | The majority of patients with hyperparathyroidism manifested nonspecific fatigue and nervous irritability that were alleviated by excision of the parathyroid adenoma. The tiredness noted by many patients…was often present in our patients on arising even though it became worse with activity and was partially relieved by resting. St. Goar (1957) proposed that hyperparathyroidism be thought of as a disease of ‘stones, bones and abdominal groans’ to which Mieher, Thibaudeau, and Frame (1961) added ‘and psychic moans.’ Perhaps this statement should be amplified to ‘stones, bones, abdominal groans, and psychic moans with fatigue overtones. |
Clinical manifestations
Presentation and course
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| | • Hypercalcemia is often asymptomatic, and more than half of cases are identified by an unanticipated elevation in serum calcium on biochemical screening studies, whereas other cases present with symptoms and signs of their underlying diseases (eg, bone pain from osseous metastases or kidney stones in hyperparathyroidism). |
| | • Symptoms and signs of hypercalcemia per se depend on the severity and rate of development of the hypercalcemia. |
| | • Patients with serum calcium levels less than 11 mg/dl are rarely symptomatic from hypercalcemia, whereas those with levels between 11 mg/dl and 14 mg/dl may be symptomatic, and those with levels above 14 mg/dl are uniformly symptomatic and at risk of severe organ damage. |
| | • Ectopic soft-tissue calcification is increasingly likely as serum calcium levels rise above 13 mg/dl. |
| | • Symptoms tend to be more likely and more severe if hypercalcemia develops rapidly. |
| | • Hypercalcemia can cause a broad range of nonspecific systemic manifestations, including dehydration, fatigue, weight loss, anorexia, constipation, nausea, vomiting, abdominal pain, and pruritus. |
| | • The most specific sign of chronic hypercalcemia is band keratopathy, a condition in which metastatic calcification occurs in the medial and lateral margins of the cornea adjacent to the scleral limbus. |
| | • Depending on the severity and rate of development, hypercalcemia can produce varying degrees of a generalized encephalopathy, ranging from mild impairment of attention to coma. |
| | • Patients with hypercalcemic encephalopathy can also present with a posterior reversible leukoencephalopathy syndrome, or a subacute Creutzfeldt-Jakob-like syndrome of progressive dementia. |
| | • Weakness is a common symptom of hypercalcemia. |
| | • Hypercalcemia in the setting of malignancy is a common oncologic emergency and ultimately develops in 10% to 30% of patients with cancer. |
Hypercalcemia is often asymptomatic, and more than half of the cases are identified by an unanticipated elevation in serum calcium on biochemical screening studies. Other cases present with symptoms and signs of their underlying diseases (eg, bone pain from osseous metastases or kidney stones in hyperparathyroidism). Symptoms and signs of hypercalcemia per se depend on the severity and rate of development of the hypercalcemia (271). Patients with serum calcium levels less than 11 mg/dl are rarely symptomatic from hypercalcemia, whereas those with levels between 11 mg/dl and 14 mg/dl may be symptomatic, and those with levels above 14 mg/dl (severe hypercalcemia) are uniformly symptomatic and at risk of severe organ damage. Ectopic soft-tissue calcification is increasingly likely as serum calcium levels rise above 13 mg/dl. Symptoms tend to be more likely and more severe if hypercalcemia develops rapidly.
Hypercalcemia can cause a broad range of nonspecific systemic manifestations, including dehydration, fatigue, weight loss, anorexia, constipation, nausea, vomiting, abdominal pain, and pruritus. Most of these can be traced to dysfunction of the renal, cardiovascular, and gastrointestinal systems (88; 63; 135; 44), and in the setting of malignancy-associated hypercalcemia may be difficult to differentiate from tumor- or treatment-related symptoms (171; 137). Hypercalcemia impairs the ability of the renal tubules to respond to antidiuretic hormone (vasopressin), causing a form of nephrogenic diabetes insipidus that decreases the ability of the kidneys to concentrate the urine by removing free water, which in turn produces polyuria and contributes to dehydration (as do nausea and vomiting). Other renal effects include azotemia, renal stones, and nephrocalcinosis (ie, deposition of calcium salts in the renal parenchyma). Cardiovascular manifestations include atrial or ventricular arrhythmias, and the arrhythmogenic effects of hypercalcemia may be potentiated by concomitant therapy with digitalis or by underlying cardiac disease (269). Anorexia, nausea, vomiting, constipation, and ileus are common gastrointestinal symptoms; pancreatitis and peptic ulcer disease are less commonly associated conditions involving the gastrointestinal system.
Primary hyperparathyroidism is the most common cause of hypercalcemia and is most often identified in postmenopausal women with hypercalcemia and parathyroid hormone (PTH) levels that are either elevated or inappropriately normal (248). The clinical presentation of primary hyperparathyroidism includes three types: (1) target organ involvement of the renal and skeletal systems; (2) mild asymptomatic hypercalcemia; and (3) high PTH levels in the context of persistently normal albumin-corrected and ionized serum calcium values. Factors that influence the clinical presentation include the extent to which biochemical screening is employed, the prevalence of vitamin D deficiency, and whether PTH levels are a routine part of the evaluation of osteopenia and osteoporosis. When biochemical screening is common, asymptomatic primary hyperparathyroidism is the most likely presentation. If vitamin D deficiency is prevalent and biochemical screening is routinely employed, symptomatic disease with skeletal abnormalities is the likely presentation. Finally, when PTH levels are used in the evaluation of low bone mineral density, the normocalcemic presentation is likely.
Osteoclastic bone resorption in hyperparathyroidism causes loss of bone mineral density and, in advanced cases, peritrabecular fibrosis and cyst-like brown tumors in and around the bone, producing "osteitis fibrosa cystica." These advanced changes produce softening of the bones, fragility-related bone fractures, and a moth-eaten appearance of the bones on x-ray studies. Osteitis fibrosa cystica was first described by Gerhard Engel in 1864 and Friedrich Daniel von Recklinghausen in 1890 but is now infrequently seen. Less advanced loss of bone mineral density in hyperparathyroidism still increases the risk of fragility-related bone fractures.
Hand x-rays in hyperparathyroidism
X-ray of the hands showing brown tumors in a patient with hyperparathyroidism. (Contributed by Ahmed Haroud. Courtesy of Radiopaedia.org and Wikipedia.)
German pathologist Friedrich Daniel von Recklinghausen
German pathologist Friedrich Daniel von Recklinghausen (1833-1910) is most known for his monograph in 1882 (Über die multiplen Fibrome der Haut und ihre Beziehung zu den multiplen Neuromen) in which he reviewed previous literature...
The most specific sign of chronic hypercalcemia is band keratopathy, a condition in which metastatic calcification occurs in the medial and lateral margins of the cornea adjacent to the scleral limbus. Band keratopathy differs in appearance from arcus senilis, which begins superiorly and inferiorly and eventually extends around the margins of the cornea (annulus senilis) (280).
Band keratopathy
Band keratopathy in a patient with aphakia (due to congenital cataracts) and a history of partial iridectomy. Note the white calcium deposits along the medial and lateral margins of the cornea adjacent to the scleral limbus. (Phot...
Arcus senilis
Arcus senilis (for comparison with band keratopathy). (Source Loren A Zech Jr and Jeffery M Hoeg 2008.)
Depending on the severity and rate of development, hypercalcemia can produce varying degrees of a generalized encephalopathy, ranging from mild impairment of attention to coma. Between these two extremes, patients commonly manifest an acute confusional state associated with apathy, depression, or lethargy (219). Personality changes and psychosis have also been described (262; 63; 283; 217), although primary hyperparathyroidism may present with various severe psychiatric symptoms, even in mild hypercalcemia (217). Rarely, other neurologic manifestations may occur, including seizures (60) and parkinsonism (215).
Several case reports have highlighted patients with hypercalcemia who presented with a posterior reversible leukoencephalopathy syndrome (151; 65; 211). Hypercalcemia associated with posterior reversible leukoencephalopathy syndrome was due to a variety of causes, including excessive oral intake of calcium, plasmacytoma, carcinoma, and AIDS-associated mycobacterial infection. Although immune suppression and arterial hypertension are well known to predispose patients to the development of posterior reversible leukoencephalopathy syndrome, isolated hypercalcemia in normotensive immunocompetent patients is a relatively recently described etiology (151; 65; 211).
Patients with hypercalcemic encephalopathy can also present with a subacute Creutzfeldt-Jakob-like syndrome of progressive dementia (229). EEG may show bursts of 1.5 to 2 Hz intermittent rhythmic delta activity superimposed on low-voltage background activity. Unlike in Creutzfeldt-Jakob disease, myoclonic jerks and periodic discharges are rare in hypercalcemic encephalopathy. Clinical and EEG abnormalities may resolve after normalization of serum calcium levels.
Weakness is a common symptom of hypercalcemia, and one that may have multiple etiologies. Although the pathophysiology of objective weakness due to hypercalcemia is not always clear, it is thought to be primarily a manifestation of CNS dysfunction, based on the presence of hyperreflexia and extensor plantar responses in some well-detailed case reports, and the lack of convincing evidence of dysfunction within the motor unit (166). At times, weakness and muscle wasting occur in combination with brisk muscle stretch reflexes, and the clinical picture can mimic amyotrophic lateral sclerosis. Some have postulated a neurogenic source of weakness in such cases. However, the proposal that hyperparathyroidism can cause a form of motor neuron disease remains highly contentious (221; 166; 138). The presence of paraparesis or quadriparesis and upper motor neuron signs, with or without a sensory level, should also prompt consideration of spinal cord compression due to a brown tumor arising in a vertebral body, a rare complication of primary or secondary hyperparathyroidism (262). In addition, in the setting of primary hyperparathyroidism, instances of true myopathy have been documented with support from electrophysiologic and histologic studies, but such cases appear to be uncommon (265; 165; 149).
Hypercalcemia in the setting of malignancy is a common oncologic emergency and ultimately develops in 10% to 30% of patients with cancer (254; 171; 230; 105). Hypercalcemia due to malignancy is the most common cause of hypercalcemia in hospitalized patients (19). Overall, the point prevalence of hypercalcemia among cancer patients is only about 2% to 3% (102), but the prevalence is greater in patients with advanced cancer. The point prevalence of hypercalcemia of malignancy varies by tumor type and is highest for patients with multiple myeloma (8% to 10%) and lowest among colorectal and prostate cancer patients (102). Patients with malignancy-related hypercalcemia can present with a prominent encephalopathy, dehydration, and generalized weakness. Hypercalcemia is the most common life-threatening metabolic disorder in patients with advanced-stage cancers and is a poor prognostic indicator (24). Cancer therapy can also be a cause of hypercalcemia; in particular, immune checkpoint inhibitor-related hypercalcemia is becoming increasingly recognized (137).
Prognosis and complications
Malignancy-associated hypercalcemia has a dire prognosis: the mean survival in patients with cancer once hypercalcemia supervenes is approximately 30 days. The prognosis of hypercalcemia is otherwise excellent, provided that the underlying disease is identified and appropriately treated.
Severe hypercalcemia itself is not generally immediately life-threatening (113). A retrospective observational study over a 5-year period of patients admitted to the adult emergency department of a large tertiary hospital found no cases of immediately life-threatening cardiac arrhythmias or neurologic complications associated with hypercalcemia above 4 mmol/l (16 mg/dl) (113).
Clinical vignette
A 55-year-old woman was brought to the emergency room by her family because of increasing confusion. Her family reported that she had been depressed and withdrawn for many months, but she had recently become less responsive and more somnolent. She was diagnosed with breast cancer 6 months earlier and was treated with a modified radical mastectomy and radiation. Review of systems was remarkable for complaints of headache, polyuria, and lightheadedness on standing. Neurologic examination showed a decreased level of arousal, poor attention, and an inability to follow complex commands. Cranial nerve, motor, sensory, coordination, and muscle stretch reflex examinations were unremarkable. Plantar responses were downgoing. Serum calcium was 14 mg/dL (3.5 mmol/L). A bone survey revealed multiple osteolytic metastatic lesions. Initial treatment included volume expansion with normal saline, calcitonin, and zoledronic acid.
Biological basis
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| | • Calcium is transported through the bloodstream as dissolved divalent cations (Ca2+) or bound to proteins (eg, serum albumin). |
| | • Parathyroid hormone, secreted by the parathyroid gland, regulates (1) the resorption of calcium from bone (the major calcium storage site in the body); (2) reabsorption in the kidney back into circulation; and (3) increases in the activation of vitamin D3 to calcitriol. |
| | • Calcitriol, the active form of vitamin D3, promotes absorption of calcium from the intestines and the mobilization of calcium ions from bone matrix. |
| | • Hyperparathyroidism and malignancy-associated hypercalcemia are by far the most common causes of hypercalcemia, together accounting for more than 90% of cases. |
| | • Primary hyperparathyroidism is most often caused by a parathyroid adenoma. |
| | • Homeostasis of calcium concentration depends primarily on the parathyroid’s capacity to regulate calcium fluxes between the extracellular fluid space, the intestine, the kidneys, and the other bones. |
Etiology and pathogenesis
Calcium regulation and balance in the human body. Calcium is transported through the bloodstream as dissolved divalent cations (Ca2+) or bound to proteins (eg, serum albumin). Calcium levels are tightly regulated.
Calcium regulation
Calcium is transported through the bloodstream as dissolved divalent cations (Ca2+) or bound to proteins (eg, serum albumin). Calcium levels are tightly regulated. Parathyroid hormone, secreted by the parathyroid gland, regulat...
Calcium balance in the human body
The movement of calcium ions from the plasma (central square) to and from various body compartments. The sizes of the red arrows are roughly in proportion to the quantities moved in and out of the plasma per day. The influence of ...
The parathyroid gland. Although the number varies, there are usually four parathyroid glands, two on each side. The parathyroid glands are usually located on the posterior surface of the thyroid gland. The position of the parathyroid glands varies considerably. Most (71%) are located at the level of the middle or lower third of the thyroid gland, some (6%) are located at the level of the upper third of the thyroid gland, and the rest are located away from the thyroid gland, even into the mediastinum (125).
Parathyroid glands typical location (anterior-posterior view)
The parathyroid glands are located usually on the posterior surface of the thyroid gland. Although the number varies, there are usually four parathyroid glands, two on each side. (Courtesy of Wikimedia Commons. Creative Commons...
Parathyroid glands typical location (lateral view)
The parathyroid glands are located usually on the posterior surface of the thyroid gland. Although the number varies, there are usually four parathyroid glands, two on each side. (Courtesy of Wikimedia Commons. Creative Commons...
Parathyroid glands to thyroid gland relationship
The parathyroid glands, and their relationship to the thyroid gland, the three laryngeal nerves, and the inferior thyroid artery (from behind). The original image has been modified by coloring the parathyroid glands purple for eas...
Parathyroid glands (range in position)
This shows a composite of the positions of the parathyroid glands found in 25 cadavers in relation to the thyroid gland and trachea. The thyroid gland is illustrated as transparent so the parathyroid gland positions can be seen. T...
Like all endocrine glands, the parathyroid gland has a capsule and is well vascularized. There are two main cell types: principal (chief) cells and oxyphil cells. Principal (chief) cells are the predominant cell type. Principal cells have clear-cut cell outlines and an eosinophilic cytoplasm containing lipofuscin pigment granules and moderate amounts of glycogen. Principal cells produce parathyroid hormone. Less common are oxyphil cells, which are larger than principal cells, and have light hyperchromatic to eosinophilic cytoplasmic staining and an abundant amount of cytoplasm. Oxyphil cells also secrete parathyroid hormone. Adipose tissue is frequently found in this gland in older individuals.
Parathyroid gland histology (low magnification)
Low-magnification micrograph of the parathyroid gland. Hematoxylin and eosin stain. The predominant cell type is the chief cell, with its intense hyperchromatic to eosinophilic cytoplasmic staining; chief cells manufacture parathy...
Parathyroid gland histology (intermediate magnification)
Intermediate-magnification micrograph of the parathyroid gland. Hematoxylin and eosin stain. (Courtesy of Wikimedia Commons. Creative Commons Attribution-Share Alike 3.0 Unported.)
Parathyroid gland histology (high magnification)
High-magnification micrograph of the parathyroid gland. Hematoxylin and eosin stain. (Courtesy of Wikimedia Commons. Creative Commons Attribution-Share Alike 3.0 Unported.)
Disorders associated with hypercalcemia. A broad range of disorders may lead to hypercalcemia (240; 63; 139; 204; 237). Primary hyperparathyroidism and malignancy-associated hypercalcemia are by far the most common causes of hypercalcemia, together accounting for more than 90% of cases (204; 251).
Parathyroid adenoma (low magnification)
Low-magnification micrograph of the parathyroid adenoma (left side). Hematoxylin and eosin stain. Parathyroid adenomas as small, solitary, clearly demarcated nodules within the parathyroid gland. They consist of a single cell popu...
Parathyroid adenoma (intermediate magnification)
Intermediate-magnification micrograph of the parathyroid adenoma (left side). Hematoxylin and eosin stain. In this case, normal parathyroid gland with prominent adipose tissue is seen on the right side of the image. (Courtesy of W...
Parathyroid adenoma (high magnification)
High-magnification micrograph of the parathyroid adenoma (left side). Hematoxylin and eosin stain. In this case, normal parathyroid gland with prominent adipose tissue is seen on the right side of the image. (Courtesy of Wikimedia...
Parathyroid hyperplasia (low magnification)
(Courtesy of Wikimedia Commons. Creative Commons Attribution-Share Alike 3.0 Unported.)
Parathyroid hyperplasia (intermediate magnification)
(Courtesy of Wikimedia Commons. Creative Commons Attribution-Share Alike 3.0 Unported.)
Parathyroid hyperplasia (high magnification)
(Courtesy of Wikimedia Commons. Creative Commons Attribution-Share Alike 3.0 Unported.)
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| Primary hyperparathyroidism (63; 190; 44) | | |
| | • Primary hyperparathyroidism is most common in patients older than 40 years, with an average age of about 55 years, and women are two to three times more likely to be affected than men. Primary hyperparathyroidism is rare in childhood. | |
| | • Primary hyperparathyroidism is most often caused by a parathyroid adenoma, a benign tumor in a single parathyroid gland (80% to 85% of cases), or parathyroid hyperplasia, an enlargement of all four parathyroid glands (15% to 20% of cases), and rarely by parathyroid carcinoma (less than 0.1% of cases) or a multiple endocrine neoplasia syndrome. | |
| | • Parathyroid hyperplasia may be sporadic or occur in the multiple endocrine neoplasia syndromes with pheochromocytoma, islet cell tumors, pituitary tumors, or medullary carcinoma of the thyroid. | |
| | • A significant proportion of patients with hyperparathyroidism do not undergo appropriate evaluation and surgical referral. In a large series of 10,432 patients with hypercalcemia, only 31% had parathyroid hormone levels measured, and 28% had a documented diagnosis of hypercalcemia in the medical record; of those with classic hyperparathyroidism only 22% had a surgical referral (27). | |
| | • End-organ damage develops before or within 5 years of diagnosis for about two thirds (62% in one study) of patients (25). | |
| | • Hyperparathyroidism-induced hypercalcemic crisis is a rare presentation of primary hyperparathyroidism (178). | |
| Malignancy-associated hypercalcemia (207; 254; 67; 171; 230; 210; 243; 157; 102; 105; 162; 200; 203; 21). | | |
| | • Malignancy-associated hypercalcemia is the most common cause of non-parathyroid hypercalcemia (106). | |
| | • Malignancy-associated hypercalcemia is a common complication, occurring in 10% to 30% of patients with cancer (254; 171; 230; 105). | |
| | • Patients with malignancy-associated hypercalcemia almost always have advanced clinically evident disease. | |
| | • Production of humoral factors by the primary tumor is collectively known as humoral hypercalcemia of malignancy and is the mechanism responsible for 80% of cases of malignancy-associated hypercalcemia (67). | |
| | • The malignancies that are most frequently associated with humoral hypercalcemia of malignancy are hematological malignancies (eg, multiple myeloma, non-Hodgkin lymphoma, leukemias) and solid cancers (particularly renal, breast, ovarian, and endometrial carcinomas as well as squamous cell carcinomas of the lung, head and neck, or esophagus) (254; 24). In rare cases, lymphoma may present with hypercalcemia, and CT is less sensitive for lymphoma than for most solid tumors, so the diagnosis may be missed or delayed (21). | |
| | • Humoral hypercalcemia of malignancy is usually due to secretion of parathyroid hormone-related protein by malignant tumors, but other humoral factors may be involved in rare cases (eg, tumor production of 1,25(OH)2D or parathyroid hormone) (257; 67; 210; 243; 105; 162; 203; 107). Parathyroid hormone-related protein is encoded by a separate gene (cytogenetic location: 12p11.22), but it shares sequence homology with the amino-terminal domain of parathyroid hormone (cytogenetic location: 11p15.3); this allows parathyroid hormone-related protein to cross-react at a common G protein receptor, the type 1 PTH/PTHrP receptor (PTHR1), resulting in similar skeletal effects and effects on calcium and phosphorus metabolism to that produced by parathyroid hormone (106). Hypercalcemia with concomitant elevation of both serum parathyroid hormone and parathyroid-related protein levels has been reported in a patient with advanced gastric carcinoma and multiple liver metastases (210). Severe hypercalcemia with concomitant elevation of both PTHrP and 1,25(OH)2D has been reported in a patient with non-Hodgkin lymphoma that expressed both PTHrP and CYP27B1 (this enzyme carries out the second of two reactions to convert vitamin D to its active form, 1,25-dihydroxyvitamin D3, which is also known as calcitriol) (107). | |
| | • Local osteolytic hypercalcemia is due to skeletal invasion by malignant cells, as in multiple myeloma (which makes osteoclast activating factor that stimulates osteoclasts to resorb bone), breast or lung cancer metastatic to bone, lymphoma, and occasionally other malignancies (eg, acute lymphoblastic leukemia) (157). This is the mechanism responsible for 20% of cases of malignancy-associated hypercalcemia (67). | |
| | • Absorptive hypercalcemia is due to excess 1,25(OH)2D production by malignancies (eg, lymphoma) (230; 200). | |
| | • Hypercalcemia in patients with malignancies can also occur to nonmalignancy-related causes (105). | |
| Granulomatous disorders (245; 153; 57; 238; 81) | | |
| | • A wide variety of granulomatous diseases cause hypercalcemia (247) due to enhanced extra-renal conversion of 25-hydroxy vitamin D to 1,25(OH)2D by activated macrophages within the granulomas. | |
| | • Sarcoidosis is the granulomatous disorder most commonly associated with hypercalcemia (205; 23; 129; 199). The incidence of hypercalcemia in sarcoidosis ranges from 10% to 20%, but over 40% have hypercalciuria, often with nephrolithiasis. It may present as hypercalcemia without evident systemic manifestations. | |
| | • Other granulomatous causes of hypercalcemia include berylliosis, mycobacteria infections (tuberculosis and granulomatous leprosy), fungal infections (histoplasmosis, blastomycosis, and coccidiomycosis), Wegener granulomatosis, foreign body granulomas, and eosinophilic granuloma. | |
| | • Rare cases of hypercalcemia in granulomatous disorders are not explained by elevated levels of vitamin D or its metabolites (247). | |
| Endocrine disorders (other than hyperparathyroidism) | | |
| | • Hyperthyroidism, which is associated with both increased bone resorption and increased bone turnover. | |
| | • Addisonian crisis due to primary adrenal insufficiency is a rare cause of hypercalcemia that resolves with glucocorticoid treatment (206; 251) | |
| | • Pheochromocytoma may be part of a multiple endocrine neoplasia syndrome, or hypercalcemia may result from production of parathyroid hormone-related protein by the pheochromocytoma (256; 251). | |
| | • Pancreatic islet tumors that secrete vasoactive intestinal polypeptide (251). These tumors are associated with the watery diarrhea hypokalemia achlorhydria syndrome, also called pancreatic cholera (51). | |
| | • Acromegaly may be a cause of 1,25(OH)2D-dependent hypercalcemia (242). | |
| | • Familial hypocalciuric hypercalcemia ("familial benign hypocalciuric hypercalcemia") resembles primary hyperparathyroidism but is relatively benign in comparison: the hypercalcemia is usually asymptomatic (190; 179; 268; 90; 246; 202; 189; 259; 94). Familial hypocalciuric hypercalcemia is generally transmitted as an autosomal dominant disorder (179; 202). Most cases are caused by a loss of function mutations in the CaSR gene, which encodes a calcium-sensing receptor that is expressed in parathyroid and kidney tissue (94). The perceived lack of calcium levels by the parathyroid glands results in high levels of parathyroid hormone secretion and, therefore, hypercalcemia. The diagnosis can be confirmed by genetic testing for a mutation in the gene encoding the calcium-sensing receptor (90). Homozygous mutations in CaSR may lead to life-threatening forms of neonatal severe hyperparathyroidism. Transient neonatal hyperparathyroidism may occur in affected neonates if the mutation is paternally inherited (130). | |
| | • The mitochondrial enzyme 1,25-dihydroxyvitamin D3 24-hydroxylase, a member of the cytochrome P450 superfamily of enzymes, is encoded by the CYP24A1 gene. Mutations in CYP24A1 may cause failure to metabolize 1,25-dihydroxyvitamin D, with resultant chronic hypercalcemia, hypercalciuria, or nephrolithiasis (140; 55; 228; 225). Pregnant women with a CYP24A1 gene mutation are at increased risk of hypercalcemia (due to upregulation of calcitriol) and fetal demise (55; 228; 225). Pathogenic mutations of CYP24A1 should be considered in the differential diagnosis of hypercalcemia with low parathyroid hormone concentrations, particularly if there is a reduced ratio of 24,25-dihydroxyvitamin D to 25-hydroxyvitamin D (225). Diagnosis is confirmed by genetic analyses (225). Monoallelic carriers have significant rates of nephrolithiasis (19%), nephrocalcinosis (5%), and symptomatic hypercalcemia (6%) (55). Strictly avoiding vitamin D supplementation may be effective in preventing or reducing the degree of hypercalcemia in affected individuals (228; 225). | |
| | • Malignancies and systemic lupus erythematosus can be associated with elevated parathyroid hormone-related protein (257; 67; 210; 243; 105; 162; 281; 203). | |
| Medications | | |
| | | • Recombinant human PTH (168). Transient hypercalcemia can occur in patients with hypoparathyroidism receiving recombinant human PTH because of overtreatment, usually during acute illness (168). |
| | • Vitamin A intoxication (267; 233; 47), usually with doses over 50,000 units daily. Vitamin A is provided in supplements and animal sources (animal liver, fish liver oil, dairy, and eggs). | |
| | • Vitamin D intoxication in health faddists or as a complication of treatment, usually with high-dose ergocalciferol that is stored in fat depots for months (51; 112; 184; 223; 261; 73; 152; 100; 168; 238; 53). Vitamin-D toxicity is the second most common cause of hypercalcemia after primary hyperparathyroidism (152). In a systematic review of vitamin D toxicity from overcorrection of vitamin D deficiency, patients presented with serum 25-hydroxy vitamin D concentrations ranging between 150 and 1220 ng/mL and serum calcium concentrations between 11.1 and 23.1 mg/dL (100). Most of the reported patients showed symptoms of vitamin D toxicity, including vomiting, dehydration, pain, and loss of appetite (100). The underlying causes of overcorrection of vitamin D deficiency included manufacturing errors and overdosing by patients or prescribers (100). In the elderly population with hypervitaminosis D and vitamin D intoxication, most patients were normocalcemic, but severe hypercalcemia is also reported, which can be life-threatening and result in death (53). | |
| | • Thiazide diuretics (52; 80; 168). Thiazide-induced hypercalcemia is usually attributed to enhanced renal calcium reabsorption by increasing Na/Ca exchange in the distal convoluted tubule (eg, changing pre-existent asymptomatic normocalcemic or intermittently hypercalcemic hyperparathyroidism into the classic hypercalcemic hyperparathyroidism) (168). The hypercalcemic effect of thiazides also occurs in anephric patients, so nonrenal mechanisms contribute to the development of thiazide-associated hypercalcemia (158). | |
| | • Lithium carbonate (172; 195). Lithium causes hypercalcemia mainly by drug-induced hyperparathyroidism (168). | |
| | • Theophylline preparations (usually seen following "bolus" administration of aminophylline in emergency room settings) (192) | |
| | • Estrogen and antiestrogen use in women with breast cancer and skeletal metastases (170) | |
| | • Corticosteroid therapy (215) | |
| | • Foscarnet (103; 28; 226; 106) | |
| | • Milk-alkali syndrome is characterized by the triad of hypercalcemia, renal insufficiency, and metabolic alkalosis that results from overconsumption of calcium-containing products (35; 156; 128; 209). During pregnancy, there is a physiological increase in calcium absorption, and in this setting milk-alkali syndrome can be life-threatening (156). | |
| Acute and chronic renal failure | | |
| | • Causes of hypercalcemia in acute and chronic renal failure include vitamin D therapy, calcium carbonate use, aluminum intoxication, and severe secondary or tertiary hyperparathyroidism (52; 190; 93). Hypercalcemia may occur rarely during the recovery phase of acute renal failure (237). | |
| | • In patients with end-stage renal disease, neither uncorrected nor albumin-corrected total serum calcium levels are reliable indicators or ionized calcium levels (214). Most patients with end-stage renal disease and elevated ionized calcium levels are falsely categorized as normocalcemic using conventional total serum calcium assays (214). Such “hidden hypercalcemia” in patients with end-stage renal disease is associated with a significantly higher risk of death (214). | |
| Advanced chronic liver disease (161) | | |
| | • This may be transient and may not require treatment (161). | |
| Immobilization(255; 64; 169; 54; 128; 197) | | |
| | • Severe hypercalcemia may result from immobilization in the setting of paraplegia, quadriplegia, extensive skeletal fractures, or prolonged enforced bedrest. | |
| | • Immobilization hypercalcemia usually occurs with immobilization in the setting of increased skeletal turnover (eg, children or adolescents, Paget disease of the bone, primary or secondary hyperparathyroidism, cancer) (255; 64; 169; 128). | |
| Protein abnormalities | | |
| | • Hyperalbuminemia. Approximately 50% of circulating calcium is bound to albumin. Disorders that raise or lower albumin concentrations cause corresponding alterations in total but not ionized serum calcium. This is a formula to correct for changes in albumin concentration: corrected calcium = serum calcium + 0.8 (normal albumin – patient albumin). | |
| | • Calcium-binding immunoglobulins in patients with multiple myeloma. Rarely, immunoglobulins are produced that specifically bind calcium and lead to asymptomatic elevation in total but not ionized serum calcium (196). | |
| Infantile hypercalcemia | | |
| | • Williams syndrome (or Williams-Beuren syndrome, OMIM #194050). Infantile hypercalcemia, usually mild and transient, is associated with multiple congenital developmental defects, including mental retardation, gregariousness, profound visuo-spatial impairment, “elfin” facies, a low nasal bridge, and supravalvular aortic stenosis (240; 237). Other findings can include chronic serous otitis media, hyperacusis, and obstructive sleep apnea (201). It is caused by a deletion of about 26 genes from the long arm of chromosome 7 (7q11.23). The syndrome was described in 1961 by New Zealand cardiologist John C P Williams (b 1922). The hypercalcemia associated with Williams syndrome typically, but not invariably, resolves by the first year (126). The cause of the hypercalcemia in Williams syndrome is still unclear (74; 101; 231; 159; 224; 175). | |
| | • Idiopathic infantile hypercalcemia (OMIM #143880). Idiopathic infantile hypercalcemia (oddly, the "idiopathic" persists even though responsible mutations in several genes have been identified) is a rare inborn form of severe hypersensitivity to vitamin D, which tends to abate by 1 year of age (193; 278; 234; 239; 238; 78; 59; 84; 98; 252; 146; 85; 264; 66; 174; 173; 282). Infantile hypercalcemia1 (HCINF1) is caused by a CYP24A1 mutation (cytochrome p450, family 24, subfamily a, polypeptide 1) on chromosome 20q13.2, resulting in slow inactivation of 1,25-dihydroxy-vitamin D3 with subsequent hypercalcemia. Common signs include lethargy, psychomotor retardation, failure to thrive, muscle hypotonia, dehydration, constipation, and nephrocalcinosis. Laboratory investigations show hypercalcemia, elevated vitamin D levels, hypercalciuria, and low parathyroid hormone levels (282). Infantile hypercalcemia-2 (HCINF2) is caused by a homozygous or compound heterozygous mutation in the SLC34A1 gene (solute carrier family 34; type II sodium/phosphate cotransporter; OMIM 182309) on chromosome 5q35. Infantile hypercalcemia with nephrocalcinosis can also be caused by a heterozygous mutation in the SLC34A3 gene (solute carrier family 34; sodium/phosphate cotransporter) on chromosome 9q34.3 (174). Biallelic mutations in the SLC34A3 gene cause hereditary hypophosphatemic rickets with hypercalciuria and also, rarely, with intermittent hypercalcemia (77; 260). The clinical presentation of mild so-called "idiopathic" infantile hypercalcemia is variable, as is the underlying genetic cause (174; 173). Dietary calcium and vitamin D restriction does not consistently normalize elevated 1,25(OH)2D concentrations or uniformly prevent worsening of renal calcification (173). The milder form of idiopathic infantile hypercalcemia has a distinctive vitamin D metabolite profile and is primarily associated with heterozygous SLC34A1 and SLC34A3 variants, whereas biallelic variants in the CYP24A1 or SLC34A1 genes are associated with severe idiopathic infantile hypercalcemia (174). | |
| | • Benign familial hypercalciuric hypercalcemia (252). At least three types of benign familial hypercalciuric hypercalcemia have been identified: type I with autosomal dominant inheritance due to heterozygous loss-of-function mutations in the CASR gene (601199), which encodes the calcium-sensing receptor, on chromosome 3q13.3-q21.1, OMIM #145980; type II with autosomal dominant inheritance due to heterozygous mutations in the GNA11 gene (139313) on chromosome 19p13.3, OMIM #145981; and type III with autosomal dominant inheritance due to heterozygous mutations in the AP2S1 gene (602242) on chromosome 19p13.32, OMIM # 600740. | |
| | • Iatrogenic hypervitaminosis D (147; 252). | |
| | • Subcutaneous fat necrosis (29; 273; 276; 20; 82; 45; 180; 284; 01; 252). | |
| | | |
Genetic changes associated with familial hypocalciuric hypercalcemia
In familial hypocalciuric hypercalcemia (FHH) types 1-3, inactivating mutations in CaSR, GNA11, or AP2S1 lead to loss of function of signaling through the CaSR pathway and, therefore, require higher extracellular calcium concen...
Primary hyperparathyroidism is most often caused by a parathyroid adenoma.
Parathyroid adenoma
Parathyroid glands in a patient with hyperparathyroidism caused by a parathyroid adenoma. (Source Bruce Blaum.)
Homeostasis of calcium concentration depends primarily on the parathyroid’s capacity to regulate calcium fluxes between the extracellular fluid space, the intestine, the kidneys, and the other bones (135). Parathyroid hormone induces renal calcium reabsorption in the distal tubule, activates osteoclastic bone resorption, and stimulates the conversion of vitamin D precursors to the active form of vitamin D [1,25(OH)2D], which in turn increases intestinal calcium absorption. In hypercalcemia due to hyperparathyroidism, the problem is an excess secretion of parathyroid hormone with otherwise intact homeostatic control, resulting in an elevated but relatively stable concentration of calcium (so-called equilibrium hypercalcemia) (88). In hypercalcemia due to other causes, compensatory mechanisms ultimately fail, despite appropriate suppression of parathyroid hormone. Typically, one of the central features of this decompensation, occurring in the more advanced stages of primary hyperparathyroidism as well, relates to the hypercalciuria-induced diuresis that leads to volume depletion and decreased glomerular filtration rate. The resulting decreased delivery of calcium and increased tubular reabsorption (which may already be increased if the parathyroid hormone or parathyroid-related protein is elevated) diminishes the kidney’s capacity to counteract the underlying problem (88; 52).
Calcium ions from the serum diffuse easily across the blood-brain barrier, and levels can directly correlate with calcium levels in both cerebrospinal fluid and brain parenchyma (143). Nevertheless, the precise mechanisms by which hypercalcemia causes neurologic symptoms remain unclear.
Epidemiology
The epidemiology of hypercalcemia depends on the underlying disorder.
Prevention
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| | • The methods for prevention of hypercalcemia depend on the underlying disorder. |
| | • Only some causes of hypercalcemia are preventable (eg, vitamin D and A intoxication). |
| | • Intravenous bisphosphonates are effective in reducing the incidence of hypercalcemia in patients with breast cancer and known skeletal metastases. |
The methods for preventing hypercalcemia depend on the underlying disorder. Some causes are not preventable (eg, primary hyperparathyroidism, cancer, sarcoidosis), whereas others clearly are (eg, vitamin D and A intoxication). Intravenous bisphosphonates have been shown to be effective in reducing the incidence of hypercalcemia in patients with breast cancer and known skeletal metastases (222).
Differential diagnosis
Confusing conditions
The symptoms of hypercalcemia are nonspecific, and especially in the setting of malignancy-associated hypercalcemia the symptoms caused by hypercalcemia may be difficult to differentiate from tumor- or treatment-related symptoms (171).
Hypercalcemia should be considered in the differential diagnosis of encephalopathy, generalized weakness, and posterior reversible leukoencephalopathy syndrome. The likelihood of hypercalcemia being present increases if commonly associated symptoms are present, such as dehydration, fatigue, weight loss, anorexia, constipation, nausea, vomiting, or abdominal pain. Note that hypermagnesemia may closely mimic the effects of hypercalcemia.
An important consideration in the differential diagnosis of patients with a malignancy and a clinical picture of encephalopathy is tumor lysis syndrome. This oncologic emergency occurs most commonly after the initiation of cytotoxic chemotherapy in patients with high-grade lymphomas or other hematologic malignancies and rarely with solid tumors. Massive cell lysis results in multiple metabolic derangements, including elevated serum uric acid 8.0 mg/dL or greater, hyperkalemia 6.0 mmol/L or greater, hyperphosphatemia 6.5 mg/dL or greater, and hypocalcemia 7.0 mg/dL or less. Clinical presentation can include encephalopathy, nausea, vomiting, anorexia, heart failure, cardiac dysrhythmias, seizures, tetany, and death (69). In contrast to malignancy-associated hypercalcemia, patients with tumor lysis syndrome are more likely to suffer diarrhea than constipation.
Associated or underlying disorders
The most common causes of hypercalcemia are primary hyperparathyroidism and malignancy, with primary hyperparathyroidism being the most common cause in ambulatory patients, and malignancy-related hypercalcemia being the most common cause among hospitalized patients. A history of hypercalcemia for over a year, in the absence of known malignancy, weight loss, and other systemic signs and symptoms, generally excludes malignancy-associated hypercalcemia. Less common causes of hypercalcemia include, for example, endocrine disorders (the most common cause in this group being hyperthyroidism), granulomatous disorders (the most common cause in this group being sarcoidosis), medications (especially thiazine diuretics, vitamin D, vitamin A, calcium carbonate, and lithium), acute and chronic renal failure, and immobilization (240).
The differential diagnosis of infantile hypercalcemia includes idiopathic infantile hypercalcemia (OMIM #143880), benign familial hypercalciuric hypercalcemia (types I-III), Williams syndrome (or Williams-Beuren syndrome, OMIM #194050), iatrogenic hypervitaminosis D, neonatal hyperparathyroidism, primary hyperparathyroidism, malignancy, granulomatous disease, Jansen metaphyseal dysplasia, and subcutaneous fat necrosis (252).
Diagnostic workup
| | |
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| | • In most clinical circumstances, total serum calcium levels accurately reflect ionized calcium levels. |
| | • Hypoalbuminemia and end-stage renal disease are two common situations where total serum calcium levels may be misleading. |
| | • Recognizing that all calcium in the body is ionized, the term “ionized calcium” refers to the free ionic fraction that is physiologically active in blood. |
| | • About half of the calcium circulating in the blood is protein-bound. Ninety percent of protein-bound calcium is linked to albumin, whereas the remaining 10% is bound to a variety of globulins. |
| | • In patients with end-stage renal disease, neither uncorrected nor albumin-corrected total serum calcium levels are reliable indicators of ionized calcium levels. |
| | • When the cause of hypercalcemia is not immediately obvious, careful history taking, with particular attention to medication use, will sometimes suggest the underlying problem. |
| | • Appropriate initial laboratory testing for hypercalcemia in adults generally includes serum total and ionized calcium, magnesium, albumin, phosphate, alkaline phosphatase, creatinine, and parathyroid hormone. |
| | • The principal ECG finding in hypercalcemia is a short QT interval, but EKG changes may mimic an acute myocardial infarction, and in severe hypercalcemia, an Osborn wave (or J wave, ie, notching of the terminal QRS complex) may be present, mimicking the EKG findings in hypothermia. |
| | • In hyperparathyroidism, x-rays may show evidence of a moth-eaten appearance to the bones, so-called brown tumors, and fragility fractures. |
| | • EEG abnormalities associated with hypercalcemic encephalopathy include excess theta activity, delta and theta slowing, bursts of 1.5 to 2 Hz intermittent rhythmic delta activity superimposed on low-voltage background activity, and diffuse but predominantly occipital spike-slow-wave complexes. |
Laboratory testing. In most clinical circumstances, total serum calcium levels accurately reflect ionized calcium levels. However, hypoalbuminemia and end-stage renal disease are two common situations where total serum calcium levels may be misleading (214).
The term “ionized calcium” refers to the free ionic fraction that is physiologically active in blood. Ionized calcium competes with hydrogen ions to bind to negatively charged sites on albumin and other calcium-binding proteins. About half of the calcium circulating in the blood is protein-bound. Ninety percent of protein-bound calcium is linked to albumin, whereas the remaining 10% is bound to a variety of globulins. There are 12 potential binding sites on each albumin molecule, but only about 10% to 15% are utilized under normal conditions. Therefore, each 1 g/dL reduction in the serum albumin concentration lowers total calcium concentration by approximately 0.8 mg/dL (0.2 mmol/L) without affecting the ionized calcium concentration and, consequently, without producing any symptoms or signs of hypocalcemia.
Elevated calcium levels can be classified by severity (271; 232). Note that there are minor variations in the thresholds for these categories in different authoritative reports.
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| | • Mild hypercalcemia is defined as an elevated total calcium less than 12 mg/dL (< 3 mmol/L) or elevated ionized calcium less than 8.0 mg/dL (< 2.0 mmol/L). |
| | • Moderate hypercalcemia is defined as total calcium 12 to 13.9 mg/dl (3.0 to 3.5 mmol/L) or ionized calcium of 8.0 to 9.9 mg/dL (2.0 to 2.5 mmol/L) |
| | • Severe hypercalcemia is defined as total calcium 14 mg/dL or greater (≥ 3.5 mmol/L) or ionized calcium 10 mg/dL or greater (≥ 2.5 mmol/L) |
In patients with end-stage renal disease, neither uncorrected nor albumin-corrected total serum calcium levels are reliable indicators or ionized calcium levels (214). Most patients with end-stage renal disease and elevated ionized calcium levels are falsely categorized as normocalcemic using conventional total serum calcium assays (214). Such “hidden hypercalcemia” in patients with end-stage renal disease is associated with a significantly higher risk of death (214).
The most important initial test to evaluate hypercalcemia in a person without known metastatic cancer is a serum intact parathyroid hormone (PTH), which distinguishes PTH-dependent from PTH-independent causes (271).
Elevated PTH may occur with primary, secondary, or tertiary forms of hyperparathyroidism (Table 1). Normally, PTH release is triggered by low calcium or high phosphate levels. PTH has several actions: (1) it stimulates bone resorption through the action of osteoclasts, increasing calcium and phosphate levels; (2) it acts on the kidney to increase calcium resorption from the distal convoluted tubule and decrease phosphate resorption; and (3) it acts on an enzyme in the kidneys (1-alpha-hydroxylase) to convert 25-hydroxyvitamin D to its active form, 1,25-dihydroxyvitamin D, which then targets the gut to increase calcium and phosphate absorption.
Control of parathyroid hormone synthesis or secretion
Parathyroid hormone (PTH) synthesis or secretion can be decreased in the chief cell of the parathyroid gland by different mechanisms: (1) increased extracellular calcium concentrations by activation of the calcium-sensing recep...
Physiology of parathyroid hormone
Abbreviations: 1,25-DOH-Vit-D: 1,25 dihydroxy hydroxycholecalciferol; 25-OH-Vit-D: 25 monohydroxy hydroxycholecalciferol; CKD: chronic kidney disease; FGF: fibroblast factor; PTH: parathyroid hormone; Vit D: vitamin D. (Source:...
The net effect of PTH is to increase calcium levels and decrease phosphate levels. Secondary and tertiary forms of hyperparathyroidism are most common with chronic kidney disease. Secondary hyperparathyroidism (SHPT) is an increased secretion of PTH due to parathyroid hyperplasia caused by triggers, such as hypocalcemia, hyperphosphatemia, or decreased active vitamin D (249; 208). In secondary hyperparathyroidism due to advanced chronic kidney disease, the kidney has trouble getting rid of phosphates and difficulty converting vitamin D to its active form (115).
Pathophysiology of secondary hyperparathyroidism
Abbreviations: 1,25DHD: 1,25 dihydroxy vitamin D; Ca-R: calcium receptor; CKD: chronic kidney disease; FGF-23: fibroblast growth factor-23; FGFR: fibroblast growth factor receptor; iPh: inorganic phosphate; PTH: parathyroid hor...
The resultant rising serum phosphate binds serum calcium, so serum calcium goes down, stimulating the parathyroid gland to secrete PTH. PTH then stimulates the release of calcium (and phosphate) from bones in an indirect process through osteoclasts; it also stimulates the kidneys to activate vitamin D to facilitate gastrointestinal absorption of calcium (although chronic kidney disease also reduces the synthesis of activated vitamin D). Another cause of secondary hyperparathyroidism is vitamin D deficiency. Diagnosis of secondary hyperparathyroidism can be made with high confidence by documenting an increased serum PTH level with an elevated phosphate level, a low or normal calcium level, and an underlying renal failure or vitamin D deficiency (277). Given the pathophysiology of secondary hyperparathyroidism, the treatment involves optimizing chronic kidney disease, a low-phosphate diet, phosphate binders (which bind phosphate in the gut and thereby decrease phosphate absorption), and ergocalciferol (vitamin D2) (Vestergaard and Thomsen 2011). Tertiary hyperparathyroidism develops as a consequence of longstanding secondary hyperparathyroidism; eventually, the parathyroid gland undergoes hyperplasia and begins secreting PTH irrespective of serum calcium levels (Vestergaard and Thomsen 2011; 141). The treatment of tertiary hyperparathyroidism is similar to primary hyperparathyroidism and typically involves parathyroidectomy.
Table 1. Lab Studies in Primary, Secondary, and Tertiary Hyperparathyroidism
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| Hyperparathyroidism | Serum calcium | Serum phosphate | PTH |
| Primary | H (or high N) | L | H |
| Secondary | L or maybe N | H (failure of renal excretion) | H |
| Tertiary | H (or high N) | H or N (eg, with phosphate binders) | H |
| Symbols: H, high; L, low; N, normal | | | |
| PTH promotes renal excretion of phosphates (assuming normal renal function) | | | |
Rarely, endogenous antibodies can interfere with 25-hydroxy vitamin D immunoassays (36). In one case with a myeloma-related IgG monoclonal gammopathy, hypercalcemia with low parathyroid hormone levels and apparently very high 25-hydroxyvitamin D levels mimicked vitamin D intoxication (36). Alternative vitamin D assays are available, which can give correct results in these rare circumstances.
The clinical setting often dictates the extent of the diagnostic workup for hypercalcemia, as the likely etiology may be evident at the time of presentation (eg, malignancy or renal failure). When the cause is not immediately obvious, careful history taking, with particular attention to medication use, will sometimes suggest the underlying problem. Appropriate initial laboratory testing in adults generally includes serum total and ionized calcium, magnesium, albumin, phosphate, alkaline phosphatase, creatinine, and parathyroid hormone (93). The highly sensitive and specific immunoassay for parathyroid hormone offers a straightforward approach to the diagnosis of hyperparathyroidism. Assays measuring intact parathyroid hormone should be used because the results of these assays are independent of renal function. The finding of an elevated parathyroid hormone-related peptide is helpful in confirming a diagnosis of humoral malignancy-associated hypercalcemia: parathyroid hormone-related peptide acts via parathyroid hormone receptors but is not detected by parathyroid hormone assays. When these latter two tests fail to provide a diagnosis, measurement of plasma 1,25(OH)2D may be useful; an elevated level in this setting would raise the possibility, for example, of a granulomatous disorder. Diagnosis of one of the less frequent causes of hypercalcemia (eg, sarcoidosis) may require a more focused and intensive search (240).
Different causes of infantile hypercalcemia can be distinguished by blood and urine studies, and then targeted genetic studies in appropriate circumstances (252). The following blood studies are recommended: calcium, phosphate, parathyroid hormone, 25(OH)D, and 1,25(OH)2D. Urine calcium and phosphate are also recommended.
Electrocardiography. The principal ECG finding in hypercalcemia is a short QT interval, but EKG changes may mimic an acute myocardial infarction, and in severe hypercalcemia an Osborn wave (or J wave, ie, notching of the terminal QRS complex) may be present, mimicking the EKG findings in hypothermia (216).
Osborn wave
Osborn wave (J wave, ie, notching of the terminal QRS complex) on EKG, which may be seen in hypothermia and severe hypercalcemia. (Source J E Roediger. Courtesy of Wikipedia.)
Bone x-rays. In hyperparathyroidism, x-rays may show evidence of a moth-eaten appearance to the bones, so-called brown tumors, and fragility fractures.
Hand x-rays in hyperparathyroidism
X-ray of the hands showing brown tumors in a patient with hyperparathyroidism. (Contributed by Ahmed Haroud. Courtesy of Radiopaedia.org and Wikipedia.)
Parathyroid imaging. Primary hyperparathyroidism causing hypercalcemia or end-organ damage (eg, kidney stones or osteoporosis) should be treated whenever possible by parathyroidectomy. Accurate preoperative location of parathyroid adenomas is crucial for surgery planning, particularly for minimally invasive surgery. Ultrasonography is usually performed as an initial approach to localize parathyroid adenomas, followed by 99mTc-sestamibi scintigraphy with SPECT/CT or 4D-CT where feasible (163). 18F-fluorocholine positron emission tomography/computed tomography (18F-FCH PET/CT) is the most sensitive method for parathyroid adenoma detection, and it can be combined with 4D-CT to increase its diagnostic performance. Parathyroid imaging is not routinely used in secondary hyperparathyroidism because parathyroidectomy is not usually part of the management of these patients. Parathyroid imaging is used for tertiary hyperparathyroidism because total or subtotal parathyroidectomy is often performed. Because 18F-FCH PET/CT is the most sensitive modality in multigland disease, it is the preferred imaging technique in tertiary hyperparathyroidism; however, cost and availability limit its use.
Electroencephalography. EEG abnormalities associated with hypercalcemic encephalopathy include excess theta activity, delta and theta slowing, bursts of 1.5 to 2 Hz intermittent rhythmic delta activity superimposed on low-voltage background activity, and diffuse but predominantly occipital spike-slow-wave complexes (150; 229). Periodic discharges are rare in hypercalcemic encephalopathy (229). EEG abnormalities may resolve after normalization of serum calcium levels (229).
Management
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| | • The management of hypercalcemia depends on its severity and the nature of the underlying disorder. |
| | • In cases of primary hyperparathyroidism, surgery is generally indicated for all symptomatic patients and for patients with asymptomatic, mild hypercalcemia who meet certain clinical or laboratory criteria. |
| | • In patients with advanced incurable cancer, withholding treatment of hypercalcemia may be the most appropriate course of action. |
| | • Patients with mild hypercalcemia (less than 12 mg/dl) do not require emergent intervention. |
| | • Treatment of patients with moderate hypercalcemia (12 to 13.5 mg/dl) is guided by the clinical status of the patient. |
| | • In the setting of severe hypercalcemia, treatment with intravenous bisphosphonates is recommended because bisphosphonates are well tolerated and have a more prolonged effect than most other agents. |
| | • Calcitonin can be a useful initial adjunct for severe hypercalcemia because of its rapid onset of action. |
| | • Dietary calcium restriction (less than 400 mg/day) is indicated in patients with vitamin D intoxication, granulomatous disorders, milk-alkali syndrome, severe hyperparathyroidism, and in patients with lymphomas that produce 1,25(OH)2D. |
| | • Dialysis using a low-calcium dialysate is an effective means of lowering the serum calcium and is indicated particularly in selected patients with congestive heart failure or acute reversible renal failure. |
The management of hypercalcemia depends on its severity and the nature of the underlying disorder (88; 63; 52; 136; 254; 263).
Treatment or correction of the underlying cause. Treatment or correction of the underlying cause, when possible, may be the only necessary intervention when hypercalcemia is mild. In cases of primary hyperparathyroidism, surgery is generally indicated for all symptomatic patients and for patients with asymptomatic, mild hypercalcemia who meet certain clinical or laboratory criteria (52; 266; 44; 95). Approximately 1.6% of patients with sporadic primary hyperparathyroidism eventually develop recurrence following parathyroidectomy (279). In patients with advanced incurable cancer, withholding treatment of hypercalcemia may be the most appropriate course of action.
Mild hypercalcemia (less than 12 mg/dl). Patients with mild hypercalcemia (less than 12 mg/dl) do not require emergent intervention. Instead, they should be counseled to avoid possible exacerbating factors, including excess calcium intake (more than 1000 mg/day), and alternative medications should be prescribed in place of those that can lead to hypercalcemia (ie, lithium carbonate and thiazide diuretics). Patients should also be instructed to avoid immobilization, avoid salt restriction, and maintain adequate fluid intake.
In patients older than 50 years with serum calcium levels less than 1 mg above the upper limit of normal and no evidence of skeletal or kidney disease, and no obvious known cause, observation may be appropriate (271). If mild hypercalcemia is due to primary hyperparathyroidism, parathyroidectomy may be considered depending on age, serum calcium level, and kidney or skeletal involvement (271).
Moderate hypercalcemia (12 to 13.9 mg/dl). Treatment of patients with moderate hypercalcemia (12 to 13.9 mg/dl) is guided by the clinical status of the patient. If the patient is asymptomatic, a conservative approach may be sufficient. However, if the patient is symptomatic, especially with signs of encephalopathy, then a more aggressive approach is required. The goal of therapy is to alleviate symptoms rather than to rapidly normalize serum calcium.
Severe hypercalcemia (greater than 14 mg/dL). For patients with severe hypercalcemia or moderate hypercalcemia and significant clinical manifestations, the initial management entails strategies that directly lower the calcium concentration independent of the underlying cause, including measures that either restore extracellular fluid volume or increase renal calcium excretion of inhibit bone resorption.
Restore extracellular fluid volume. Because almost all patients in this setting are dehydrated, volume repletion is the essential first step toward reestablishing calcium homeostasis. Volume repletion requires careful monitoring for fluid overload, especially in this patient population that is at risk for renal and cardiac dysfunction. Hydration alone is rarely sufficient to normalize calcium levels, and additional pharmacotherapy is typically required.
Enhancing renal calcium excretion. Once extracellular fluid volume has been restored, saline diuresis with normal saline infusion facilitates calcium excretion. Furosemide adds little to the effect of saline diuresis and may interfere with the restoration and maintenance of adequate extracellular fluid volume.
Antiresorptive therapies. Agents that inhibit bone resorption include bisphosphonates (eg, zoledronate and pamidronate), denosumab, calcitonin, plicamycin, and gallium nitrate. Of these, bisphosphonates are the most widely used because of their relative safety and efficacy profiles (285).
In the setting of severe hypercalcemia, treatment with intravenous bisphosphonates is recommended because bisphosphonates are well tolerated and have a more prolonged effect than most other agents. An intravenous infusion of bisphosphonate pamidronate disodium (60 to 90 mg) will become effective within 48 hours, and the hypocalcemic effect peaks in 1 week and can last up to 2 weeks (213). Pamidronate is less effective in patients with humoral-related hypercalcemia (114). For this reason, zoledronic acid is often used as a first-line bisphosphonate in malignancy-associated hypercalcemia (62). It has a higher response rate in normalizing serum calcium and a more prolonged effect of up to 32 days (185). Side effects can include fever, hypocalcemia, hypophosphatemia, nausea, pruritus, and acute renal failure. Osteonecrosis, particularly of the jaw, is an infrequent but important side effect of bisphosphonate therapy that is most common in patients with malignancy-associated hypercalcemia (87). Although there may be an increased risk of atypical femoral neck fractures with prolonged use of bisphosphonates in patients with osteoporosis, this is of little relevance in the setting of malignancy-associated hypercalcemia, and the overall benefit in these patients far outweighs the minimal risk of atypical fracture (244). Available data are conflicting regarding whether chronic use of bisphosphonates is associated with increased risk of esophageal cancer, but again, with the time course involved in treating malignancy-associated hypercalcemia, there is little relevance in this setting (56; 109).
Subcutaneous denosumab is used for bisphosphonate-refractory hypercalcemia (eg, as a second-line therapy for hypercalcemia of malignancy) and in patients with renal failure (62; 160; 31; 263). Denosumab is a human monoclonal antibody that acts as a RANKL inhibitor, a critical mediator of bone resorption and overall bone density through its ability to stimulate osteoclast formation and activity. Denosumab binds RANKL, preventing RANKL from activating RANK, its receptor on the osteoclast surface. Osteoclast formation, function, and survival are inhibited with reduced RANK–RANKL binding, and consequently, bone resorption decreases and bone mass increases (118).
Calcitonin can be a useful initial adjunct for severe hypercalcemia because of its rapid onset of action. The addition of calcitonin will produce a rapid but mild reduction in serum calcium levels (1 to 2 mg/dl) (164); calcitonin acts primarily by decreasing bone reabsorption, but it also increases renal excretion of calcium. It is effective for up to 48 hours, but tachyphylaxis eventually develops.
Gallium nitrate and plicamycin are additional therapeutic options, though both have significant side effects (285; 75).
Treatments with restricted indications
Dietary calcium restriction. Dietary calcium restriction (less than 400 mg/day) is indicated in patients with vitamin D intoxication, granulomatous disorders, milk-alkali syndrome, severe hyperparathyroidism, and in patients with lymphomas that produce 1,25(OH)2D.
Glucocorticoids. Glucocorticoids may be useful in a select group of patients, namely those with myeloma, other hematologic malignancies, sarcoidosis, and vitamin D intoxication. Glucocorticoids act to lower serum calcium by several mechanisms, including inhibiting cytokine release, producing direct cytotoxic effects on some tumor cells, decreasing intestinal calcium absorption, and increasing urinary calcium excretion.
Dialysis. Dialysis using a low-calcium dialysate is an effective means of lowering the serum calcium and is indicated in selected patients (37). It is particularly effective in patients with congestive heart failure or acute reversible renal failure.
Special circumstances
Hypercalcemia of malignancy. As indicated earlier with general considerations on the management of hypercalcemia, the treatment of hypercalcemia of malignancy consists of enhancing renal calcium excretion (mostly through hydration with isotonic fluids) and the use of antiresorptive therapies. Intravenous zoledronic acid has often been considered the first-line treatment, with subcutaneous denosumab used for bisphosphonate-refractory hypercalcemia and in patients with renal failure (62; 133). According to Endocrine Society 2023 guidelines, in adults with hypercalcemia of malignancy, intravenous hydration and intravenous bisphosphonate or denosumab is recommended (strong recommendation) with the guidelines committee conditionally favoring denosumab over bisphosphonates, based on low-quality evidence (and the inference that denosumab is associated with greater suppression of bone turnover compared with zoledronic acid) (83; 91; 191; 241). Dialysis may be needed in some cases (37).
Conditional suggestions of the Endocrine Society based on low-quality evidence include the following (83; 91):
| | |
--- |
| | • In adults with severe hypercalcemia of malignancy (serum calcium level > 14 mg/ dL), consider combination therapy with calcitonin and an intravenous bisphosphonate or denosumab. |
| | • For refractory or recurrent hypercalcemia of malignancy despite treatment with intravenous bisphosphonate, consider the addition of denosumab. |
| | • For hypercalcemia due to high calcitriol levels with symptomatic or severe hypercalcemia despite glucocorticoid therapy, consider the addition of intravenous bisphosphonate or denosumab. |
| | • Patients with parathyroid carcinoma can be treated with calcimimetic and/or antiresorptive therapy, depending on the severity and results from initial treatment. |
Hypophosphatemia. Oral phosphorus replacement should be used only in patients with phosphorus values less than 3.0 mg/dL and normal renal function to minimize the risk of ectopic soft-tissue calcification.
Immobilized patients. Mobilization is critical in reversing hypercalcemia in immobilized patients. This must include weight bearing. Passive range-of-motion exercises are inadequate.
Special considerations
Pregnancy
Although none of the common forms of hypercalcemia are epidemiologically linked with pregnancy, hypercalcemia due to any of the possible underlying disorders may first come to light during pregnancy.
Primary hyperparathyroidism. Hypercalcemic disorders in pregnant women are usually due to primary hyperparathyroidism (76). In general, in patients with primary hyperparathyroidism, parathyroidectomy is delayed until after parturition, but severe cases may warrant second-trimester parathyroidectomy.
Loss-of-function mutations of the CYP24A1 gene. Loss-of-function mutations of the CYP24A1 gene cause a deficiency of the CYP24A1 enzyme, which is involved in the catabolism of 1,25-dihydroxyvitamin D3 (calcitriol). Because pregnancy is associated with an upregulation of the active vitamin D hormone calcitriol, loss-of-function mutations of the CYP24A1 gene are likely to trigger symptomatic hypercalcemia in affected women during pregnancy (55; 225). Pathogenic mutations of CYP24A1 should be considered in the differential diagnosis of hypercalcemia with low parathyroid hormone concentrations, particularly if there is a reduced ratio of 24,25-dihydroxyvitamin D to 25-hydroxyvitamin D, or if the 1,25-dihydroxyvitamin D3 level is elevated while the 25-hydroxyvitamin D level is still within the reference range (228; 225). Diagnosis is confirmed by genetic analyses (225). In affected women, pregnancy is associated with high rates of obstetric complications (55). Strictly avoiding vitamin D supplementation may be effective in preventing or reducing the degree of hypercalcemia (228; 225). Calcium and 25-hydroxyvitamin D levels should be monitored in routine blood tests during pregnancy (228). Hypercalcemia in a newborn should be carefully evaluated and treated as hypercalciuria can lead to nephrocalcinosis (228).
Anesthesia
General anesthesia is best avoided in patients with hypercalcemia, except in emergencies, because of the risk of cardiac arrhythmias.
Media
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PMID 3564966
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PMID 15698586
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Age range of presentation
: 0 month to 65+ years
Sex preponderance
: male=female
Heredity
: heredity may be a factor
autosomal dominant
Population groups selectively affected
: none selectively affected
Occupation groups selectively affected
: none selectively affected
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ICD-11
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OMIM
: Hypercalcemia, familial benign: #145980, %145981
Hypercalcemia, infantile (Williams syndrome, or Williams–Beuren syndrome): #194050
Hypercalcemia, Idiopathic infantile: #143880
Hypercalciuric hypercalcemia, benign familial: Type I: #145980
Hypercalciuric hypercalcemia, benign familial: Type II: #145981
Hypercalciuric hypercalcemia, benign familial: Type III: # 600740
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188786 | https://math.stackexchange.com/questions/1810510/chinese-remainder-theorem-when-gcd-is-not-1 | systems of equations - Chinese Remainder Theorem When GCD is not 1 - Mathematics Stack Exchange
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Chinese Remainder Theorem When GCD is not 1
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I've got this system of equations that I'm trying to solve. I'm pretty sure I have to use the CRT, but to my understanding, it can only be used when GCD of all the mods is 1. I don't want an answer because this is a homework problem. I just have a question. The system is
x≡1 mod 2 x≡1 mod 2
x≡2 mod 3 x≡2 mod 3
x≡3 mod 4 x≡3 mod 4
x≡4 mod 5 x≡4 mod 5
x≡5 mod 6 x≡5 mod 6
x≡0 mod 7 x≡0 mod 7
gcd(2,4) ≠≠ 1, gcd(3,6) ≠≠ 1, gcd(2,6) ≠≠ 1, gcd(4,6) ≠≠ 1. So I can't use the Chinese Remainder Theorem here. My question is, can I simplify it as follows
If I list out the first congruence class of mod 2, then the numbers are all either 1 or 3 mod 4
If I list out the second congruence class of mod 3, then the numbers are all either 2 or 5 mod 6.
With the above observations, I simplify the system to
x≡3 mod 4 x≡3 mod 4
x≡4 mod 5 x≡4 mod 5
x≡5 mod 6 x≡5 mod 6
x≡0 mod 7 x≡0 mod 7
Now I observe that when I write out the thrid congruence class in mod 4, then numbers are all either 3, 1, or 5 mod 6. So I simplify the above to
x≡4 mod 5 x≡4 mod 5
x≡5 mod 6 x≡5 mod 6
x≡0 mod 7 x≡0 mod 7
Can I simplify the system to the above three equations?
Any help would be appreciated Thanks
systems-of-equations
gcd-and-lcm
chinese-remainder-theorem
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edited Aug 2, 2018 at 1:36
Steven Alexis Gregory
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asked Jun 3, 2016 at 2:53
The_QuestionerThe_Questioner
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Note that if x≡3(mod 4)x≡3(mod 4) then necessarily x≡1(mod 2)x≡1(mod 2). Simiarly, if x≡5(mod 6)x≡5(mod 6) then x≡2(mod 3)x≡2(mod 3). So you can actually remove the equations x≡1(mod 2)x≡1(mod 2) and x≡2(mod 3)x≡2(mod 3) because they are implied by the other equations.
Now 4 4 and 6 6 are still not coprime. You should be able to show that x≡3 mod 4 x≡3 mod 4 and x≡5 mod 6 x≡5 mod 6 if and only if x≡11(mod 12)x≡11(mod 12). So those two equations can be replaced by the one equation x≡11(mod 12)x≡11(mod 12). Now you only have three equations, and gcd(12,5,7)=1 gcd(12,5,7)=1, so you can apply the Chinese remainder theorem.
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answered Jun 3, 2016 at 3:01
kccukccu
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Thanks for that. I just tried out my method as well as yours and they both give the same answer. Is my method correct as well, or would it not work in other cases?The_Questioner –The_Questioner 2016-06-03 03:10:09 +00:00 Commented Jun 3, 2016 at 3:10
I don't understand what you mean by "Now I observe that when I write out the thrid congruence class in mod 4, then numbers are all either 3, 1, or 5 mod 6." The three equations you are left with do not give a correct answer. By my calculations, they give x≡119(mod 210)x≡119(mod 210). Yet x=329 x=329 satisfies this but not x≡3(mod 4)x≡3(mod 4).kccu –kccu 2016-06-03 17:00:27 +00:00 Commented Jun 3, 2016 at 17:00
The only time you can get rid of an equation is if it is implied by the other equations, e.g. x≡3(mod 4)⇒x≡1(mod 2)x≡3(mod 4)⇒x≡1(mod 2). In my answer I combined two equations into one, which is okay because they are equivalent: x≡3(mod 4)and x≡5(mod 6)⇔x≡11(mod 12).x≡3(mod 4)and x≡5(mod 6)⇔x≡11(mod 12). You removed the equation x≡3(mod 4)x≡3(mod 4), but this equation is not implied by the others, as I showed with the example x=329 x=329.kccu –kccu 2016-06-03 17:02:28 +00:00 Commented Jun 3, 2016 at 17:02
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A shortcut first. Note that
x≡1 mod 2 x≡1 mod 2
x≡2 mod 3 x≡2 mod 3
x≡3 mod 4 x≡3 mod 4
x≡4 mod 5 x≡4 mod 5
x≡5 mod 6 x≡5 mod 6
is equivalent to
x≡−1 mod 2 x≡−1 mod 2
x≡−1 mod 3 x≡−1 mod 3
x≡−1 mod 4 x≡−1 mod 4
x≡−1 mod 5 x≡−1 mod 5
x≡−1 mod 6 x≡−1 mod 6
which is equivalent to
x≡−1 mod 60 x≡−1 mod 60
where 60=l c m(2,3,4,5,6)60=l c m(2,3,4,5,6)$
Then you only need to solve
x≡−1 mod 60 x≡−1 mod 60
x≡0 mod 7 x≡0 mod 7
In more general cases, I find it easiest to decompose each equivalence into equivalent prime-power equivalences. It sucks up a bit more space but makes it easier to find what is essential.
Note x≡3 mod 4 x≡3 mod 4 implies x≡1 mod 2 x≡1 mod 2. So we can remove x≡1 mod 2 x≡1 mod 2.
Note x≡5 mod 6 x≡5 mod 6 implies
x≡1 mod 2 x≡2 mod 3 x≡1 mod 2 x≡2 mod 3
We have already removed x≡1 mod 2 x≡1 mod 2. So, if we leave x≡2 mod 3 x≡2 mod 3, then we can remove x≡5 mod 6 x≡5 mod 6, which is the more complicated congruence.
x≡1 mod 2 x≡1 mod 2
x≡2 mod 3 x≡2 mod 3
x≡3 mod 4 x≡3 mod 4
x≡4 mod 5 x≡4 mod 5
x≡5 mod 6 x≡5 mod 6
x≡0 mod 7 x≡0 mod 7
Note that the remaining congruences are amenable to the CRT.
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answered Aug 2, 2018 at 2:01
Steven Alexis GregorySteven Alexis Gregory
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188787 | https://www.cuemath.com/questions/how-many-yards-in-a-mile/ | How many yards in a mile?
Mile is also used to describe or translate a wide range of units that were derived from the Roman mile, nautical mile.
Answer: 1 mile is 1760 yards.
Let's check.
Explanation:
Miles and yards are the units of measurements commonly used while calculating and measuring the area of a given place.
The conversion factor for converting miles to yards is given as, 1 mile = 1760 yards.
You can also use this easy and accurate metric conversion calculator to convert the given quantity from yards to miles in seconds.
Therefore, we can say 1 mile is accurately equal to 1760 yards. |
188788 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3372784/ | Better Prognosis in Newborns with Trisomy 13 Who Received Intensive Treatments: A Retrospective Study of 16 Patients - PMC
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Better Prognosis in Newborns with Trisomy 13 Who Received Intensive Treatments: A Retrospective Study of 16 Patients
Keiko Tsukada
Keiko Tsukada
1 Department of Pediatrics, Dokkyo Medical University School of Medicine, Kitakobayashi 880, Mibu, Tochigi 321-0293 Japan
Find articles by Keiko Tsukada
1, George Imataka
George Imataka
1 Department of Pediatrics, Dokkyo Medical University School of Medicine, Kitakobayashi 880, Mibu, Tochigi 321-0293 Japan
Find articles by George Imataka
1,✉, Hiroshi Suzumura
Hiroshi Suzumura
1 Department of Pediatrics, Dokkyo Medical University School of Medicine, Kitakobayashi 880, Mibu, Tochigi 321-0293 Japan
Find articles by Hiroshi Suzumura
1, Osamu Arisaka
Osamu Arisaka
1 Department of Pediatrics, Dokkyo Medical University School of Medicine, Kitakobayashi 880, Mibu, Tochigi 321-0293 Japan
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1
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1 Department of Pediatrics, Dokkyo Medical University School of Medicine, Kitakobayashi 880, Mibu, Tochigi 321-0293 Japan
✉
Corresponding author.
Issue date 2012.
© The Author(s) 2012
This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.
PMC Copyright notice
PMCID: PMC3372784 PMID: 22487910
Abstract
Intensive treatment for newborns with trisomy 13 is controversial because of their lethal prognosis. We report the better life prognosis of patients with trisomy 13 who received intensive treatment. At our hospital, we provided an intensive management to such patients including resuscitation and surgical procedures as required. Herein, we present the results of a retrospective study (1989–2010) of 16 trisomy 13 cases who received an intensive treatment. None was diagnosed to have trisomy 13 before birth; 9 were delivered by C-section and oxygen was administered to all patients during postpartum resuscitation. Mechanical ventilation was used in 9 patients after tracheal intubation and tracheotomy was performed in 2 patients when withdrawing of extubation was difficult. Regarding prognosis, 9 patients died, 3 were referred to another hospital, and 4 were discharged from the hospital. Four and 7 patients died within 7 and 30 days after birth, respectively. Nine patients survived for >1 month, 7 for >180 days, and 5 for >3 years. Median survival for 16 patients was 733 days. The patients who received intensive treatments survived longer compared to the previous data. This study provides useful information concerning genetic counseling, especially from an ethical point of view, before providing intensive management to newborns with trisomy 13.
Keywords: Trisomy 13, Prognosis, Intensive treatment, Survival, Natural history
Introduction
Trisomy 13, first reported by Patau et al . , is the third common trisomy in live born infants. It is clinically associated with various anomalies, found in 80%, of cardiac and circulatory systems, such as mainly ventricular septal defect, patent ductus arteriosus, auricular septal defect and 60–80% with cleft lip and/or palate of mouth, and 50% or more with the following: central nervous system with holoprosencephaly, hearing system with apparent deafness, cranium with microcephaly with sloping forehead, eyes with microphthalmia and colobomata of iris, retinal dysplasia, abnormal auricle helices, skin with capillary hemangioma of forehead, distal palmar axial triadii, simian crease, hypoconvex narrow fingernails, flexion of fingers with or without overlapping and camptodactyly, polydacyly of hand, skeletal of thin posterior ribs and hypoplasia of pelvis, males with cryptorchidism/abnormal scrotum and females with bicornuate uterus and another malformation accompanied by severe psychomotor disorder [2, 3]. The prevalence studies from Denmark and United Kingdom reported birth prevalence in live born of trisomy 13 as ~1 in 20,000 to 1 in 29,000 [4, 5]. The Hawaiian study reported prevalence at birth of trisomy 13 as 1 in 12,048 in live born infants . Jones described that the incidence was ~1 in 5,000 births . Generally, natural prevalence of delivery at birth of trisomy 13 is between 1 in 5,000 and 1 in 29,000, which is the third most common autosomal trisomy in newborns following trisomy 21 and trisomy 18.
The approach to the management policy of a third-trimester fetus and infant with trisomy 13 and 18 is quite complicated, and the existing literature is controversial. The principal reason for the complexity surrounding decision making in the care of infants with trisomy 13 relates to high neonatal and infant mortality in both conditions [7, 8]. Avoidance of delivery by cesarean section when a fetus is known to have trisomy 13 appears to be the trend in Denmark as shown by obstetrics literature. There was strong consensus among physicians working in perinatal medicine toward abortion for social reasons or because of abnormal prenatal diagnostic results that abortion is acceptable until week 21 in the case of trisomy 13 . Moreover, there tends to be extremely negative information about survival of trisomy 13. It is important to acknowledge that the prognosis is lethal. Approximately, 50% of infants with trisomy 13 will live longer than a week and 5–10% of infants will live past 1 year [7, 8]. Tradition, as reflected in the pediatric literature, also indicates a nonintervention approach in the newborn management of trisomy 13. Bos et al. summarized these issues, arguing that early diagnosis was very important so that surgery could be withheld. The important issues that emerge in the counseling and management of newborn infants with trisomy 13 are high infant mortality. Survivors have severe mental retardation, often seizures, and fail to thrive. So that surgical or orthopedic corrective procedures should be withheld in early infancy to await the outcome of the first few months .
In Japan, the policies for the treatment of chromosomal anomalies including trisomy 13 and 18 with poor long-term prognoses are determined based on either of two controversial concepts: (i) the provision of thorough love and care while avoiding excessive intensive treatment; or (ii) the provision of active intensive treatment including resuscitation and surgery according to the clinical conditions of an infant and in accordance with the wishes of the infant’s parents [11, 12]. The most important problems in this argument were that there was no evidence about the improvement of prognosis of trisomy 13 treated at the Neonatal Intensive Care Unit (NICU) of Dokkyo Medical University Hospital. In the same context, Kosho et al. [11, 13] investigated the prognosis of intensive treatments in 24 Japanese patients with trisomy 18 at NICU. This report argued better prognosis of trisomy 18 through treatment at NICU based on improved survival period, and the data provide helpful guidelines for clinicians by offering the best information on treatment options for families of patients with trisomy 18. However, there is still a lack of precise data concerning the clinical details and prognosis regarding intensive treatments of patient with trisomy13.
The NICU at the Dokkyo Medical University Hospital provides the intensive treatments including resuscitation of intratracheal intubation, respiratory support, and some kind of surgery for trisomy 13 and 18, if informed consent is obtained from the family after counseling. At NICU, the five tentative treatment policies for trisomy 13 and 18 with poor long-term prognoses of survival are as follows: (1) The treatment policy is determined without relation with trisomy 13 and 18, and is determined after the pathology, treatment options and risks involved have been explained to the parents; (2) The actual initial treatment with resuscitation and medical therapy after delivery are provided to neonates who have ordinary cases, and surgical treatment is also not limited; (3) With respect to surgical treatment and resuscitation, even in case of long-term survival; (4) Psychological care for family is regularly provided by clinical psychotherapists; and (5) Management at home is actively discussed if the family wishes for it.
In this study, in order to determine the survival period of trisomy 13 patients who received intensive treatment, we retrospectively analyzed the detailed clinical data of 16 patients with trisomy 13, admitted to NICU at the Dokkyo Medical University Hospital from 1989 to 2010.
Patients and Methods
This study comprised 6,230 newborn infants who were hospitalized at the neonatal care unit of Dokkyo University Hospital, Japan during the period of 22 years from 1989 to 2010. Newborn infants presenting with external malformations or organ malformations that were observed during the clinical examinations underwent chromosomal analysis, and 183 patients (2.94%) were diagnosed with chromosomal anomalies. Of these, 138 patients (2.22%) were diagnosed with numerical autosomal aberration. A breakdown of these patients of numerical autosomal aberration shows trisomy 21 in 83 patients (1.33%), trisomy 18 in 39 patients (0.63%), and trisomy 13 in 16 patients (0.26%). Other chromosomal abnormalities were 45 patients (0.72%) including 2 patients with trisomy 8 mosaic and 2 patients with 4p-syndrome and others. In this study, we retrospectively investigated the NICU medical records at our hospital regarding the clinical details of 16 patients with trisomy 13 who received intensive treatment.
Results
Family History, Ages of the Parents, History of Pregnancy, and Delivery (Table1)
Table 1.
Family history, age of parents, pregnancy, and delivery regarding 16 patients with trisomy 13
| Patients | Age of parents (father/mother) | History of previous delivery | Gender | Prenatal ultrasonographic findings | Cesarean section | Apgar score (1 min/5 min) | Gestational age (weeks) | Birth weight (g) |
:--- :--- :--- :---
| 1 | 31/29 | – | M | Hydrocephalus IUGR | – | 3/4 | 35 | 1,582 |
| 2 | 32/32 | – | F | Holoprosencephaly umbilical hernia severe IUGR | + | 1/2 | 35 | 1,756 |
| 3 | 32/31 | – | M | IUGR | + | 3/6 | 40 | 2,162 |
| 4 | 35/42 | + | M | Hydronephrosis | + | 1/3 | 33 | 3,378 |
| 5 | 41/40 | + | M | IUGR oligoamnios | + | 2/5 | 37 | 2,254 |
| 6 | 22/20 | – | F | IUGR | – | 6/7 | 37 | 2,237 |
| 7 | 33/33 | – | F | Hydrocephalus oligoamnios | + | 8/9 | 35 | 2,540 |
| 8 | 30/35 | + | M | – | – | 8/9 | 36 | 2,380 |
| 9 | 34/32 | – | F | Holoprosencephaly IUGR | + | 1/3 | 36 | 2,784 |
| 10 | 22/21 | – | M | IUGR | – | 8/8 | 37 | 2,342 |
| 11 | 36/35 | – | F | Hydrocephalus IUGR polydactyly origoamnios | – | 4/9 | 35 | 1,960 |
| 12 | 29/33 | – | M | Severe IUGR | – | 3/7 | 34 | 1,602 |
| 13 | 39/37 | + | M | Origoamnios | | | | |
| 14 | 36/36 | – | F | Severe IUGR | – | 8/10 | 37 | 1,746 |
| 15 | 47/34 | + | M | IUGR | – | 4/8 | 39 | 3,366 |
| 16 | 41/37 | – | M | IUGR | + | 3/6 | 35 | 1,950 |
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M male, F female, IUGR intrauterine growth retardation
None of the 16 patients with trisomy 13 had chromosomal abnormalities in their family histories. The age of the mothers ranged from 20 to 42 years (average: 32.9 years) including mothers who were over 35-years-old for 7 patients and over 40-years-old for 2 patients. The age of the fathers ranged from 22 to 47 years (average: 33.8 years) including fathers who were over 35-years-old for 7 patients and over 40-years-old for 3 patients. Regarding the history of previous delivery, 5 patients had parous mothers. In terms of gender, there were 10 males and 6 females. A definitive prenatal diagnosis of amniotic fluid test for trisomy 13 was not carried out for any of the 16 patients. However, in 15 patients diagnosed with trisomy 13, some type of abnormality was indicated in the ultrasonography test during the course of pregnancy, such as intrauterine growth retardation in 11 patients and brain malformations in 5 patients. The term of delivery ranged from 33 to 40 gestational weeks (average: 35.9 gestational weeks). Regarding the method of delivery, 9 patients were delivered via a cesarean section and 7 patients were delivered via a spontaneous cephalic delivery. The delivery term for the performance of cesarean section ranged from 33 to 40 gestational weeks (average: 35.8 gestational weeks).
The average of Apgar Score at the time of birth was 4.25 points for 1 min scores and 6.5 points for 5 min scores. The birth weight ranged from 1,582 to 3,378 g, with an average weight of 2,243 g. The breakdown includes 7 patients weighing 1,500 to 2,000 g, 5 patients weighing 2,000 to 2,500 g, 2 patients weighing 2,500 to 3,000 g, and 2 patients weighting at least 3,000 g.
Chromosomal Analysis, External Malformations, and Organ Malformations (Table2)
Table 2.
Chromosomal analysis, external malformations, organ malformations with 16 patients of trisomy 13
| Patients | Karyotype | (Major external malformations) | (Organ malformations) |
:--- :--- |
| Head | Face | Abdomen | Extremities others | Brain malformation | Congenital heart disease | Respiratory complications |
| 1 | Full trisomy | – | Cleft lip and palate lack of nose | Umbilcal hernia intestinal fistula | Polydactyly | Holoprosencephaly (alober type) | VSD ASD PDA PH | Unseparate superior and inferior with hypoplastic lung |
| 2 | Full trisomy | – | Cleft lip and palate | Umbilcal hernia | – | Holoprosencephaly (alober type) | VSD ASD DORV | Respiratory failure |
| 3 | Full trisomy | Scalp defect hemangioma | Cleft lip and palate | – | Polydactyly cryptorchidism | – | PDA ECD DORV PS | Tracheal stenosis tracheoesophageal fistula respiratory failure |
| 4 | Full trisomy | Scalp defect | – | Prune belly syndrome umbilical hernia | – | – | VSD DORV | Hypoplastic lung tracheal stenosis respiratory failure |
| 5 | Full trisomy | Low set ears | Cleft lip and palate | Umbilical hernia | Finger apposition anomaly | – | ECD PS | Respiratory failure |
| 6 | Robertson type trisomy | – | – | Necrotizing enterocolitis | – | – | VSD CoA | PPHN respiratory failure |
| 7 | Full trisomy | – | Cleft lip and palate | – | – | Holoprosencephaly (alober type) | VSD PDA | Respiratory failure |
| 8 | Full trisomy | Scalp defect low set ears | – | Umbilical hernia | Polydactyly narrow fingernails | – | TOF MAPCA | |
| 9 | Full trisomy | – | Cleft lip and palate lack of nose | – | – | Holoprosencephaly (semi-lober type) | VSD ASD CoA | Respiratory failure |
| 10 | Full trisomy | – | Cleft lip and palate | Inguinal hernia | – | – | TOF | Respiratory failure |
| 11 | Full trisomy | – | Cleft lip and palate | – | Polydactyly | – | PDA CoA | PPHN respiratory failure |
| 12 | Full trisomy | – | – | Umbilical hernia | – | – | ASD PDA | Vocal cord anomaly |
| 13 | Full trisomy | Scalp defect low set ears | Cleft lip and palate | Umbilical hernia | Polydactyly | – | TOF PA | Respiratory failure |
| 14 | Full trisomy | – | Cleft lip and palate | – | – | Dandy–Walker malformation | ASD | Respiratory failure |
| 15 | Mosaic type trisomy | Scalp defect hemangioma | – | Bowel malrotation inguinal hernia | Narrow fingernails micropenis buried penis | Septum pellucidum fenestration olfactory aplasia | ASD PDA | Respiratory failure |
| 16 | Full trisomy | Low set ears | – | Undescended testis | Polydactyly | – | ASD PDA | Respiratory failure |
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VSD ventriculoseptal defect, ASD atrial septal defect, PDA patent ductus arteriosus, DORV double-outlet right ventricle, ECD endcardial cusion defect, PS: pulmonary stenosis, CoA coarctation of the aorta, TOF tetralogy of Fallot, PPHN persistent pulmonary hypertension of newborn, MAPCA major aortopulmonary collateral arteries
The chromosomal karyotypes of all 16 patients were diagnosed using a G-band method. The r karyotypes included full trisomy 13 in 14 patients, mosaic type of trisomy 13 in 1 patient, and Robertsonian type of translocation in 1 patient.
The major external craniofacial malformations included scalp defects in 5 patients for the head and cleft lip and palate in 10 patients for the face; malformation of abdomen included umbilical hernia in 8 patients. Polydactyly was noted in 6 patients for the extremities. Among the major organ malformations, 3 patients had alobar and 1 patient had semi-lobar type of holoprosencephaly. Dandy-Walker malformation was observed 1 patient and olfactory aplasia and fenestration of the septum pellucidum in 1 patient for the brain. As for the heart, some type of congenital cardiac disorder was observed in all patients. The breakdown includes atrial septal defect in 7 patients, ventricular septal defect in 6 patients, patent ductus arteriosus in 7 patients, coarctation of the aorta in 3 patients, endocardial cushion defect in 2 patients, tetralogy of Fallot (TOF) in 3 patients, and double-outlet right ventricle (DORV) in 3 patients. Combined respiratory malformations included congenital hypoplastic lung in 2 cases and congenital tracheal stenosis in 2 cases.
Administration of Oxygen and the Use of Mechanical Ventilation and Surgical Treatment (Table3)
Table 3.
Administration of oxygen and the use of mechanical ventilation and surgical treatment, others with 16 patients of trisomy 13
| Patients | Resuscitation of oxygen | Mechanical ventilation | Mode of ventilation | Surgical treatment | Others |
:--- :--- :--- |
| 1 | + | – | – | – | – |
| 2 | + | + | IMV | – | Seizures |
| 3 | + | – | – | – | – |
| 4 | + | + | IMV, HFO | Operation for paracentesis of hydronephrosis | – |
| 5 | + | – | – | – | – |
| 6 | + | – | – | – | – |
| 7 | + | – | – | – | – |
| 8 | + | + | IMV | Operation for umbilical hernia | |
| 9 | + | + | IMV | – | Seizures |
| 10 | + | – | – | – | HOT |
| 11 | + | – | – | Plastic operation for cleft lip and palate operation for PDA ligation subcravian method for CoA | – |
| 12 | + | + | IMV | – | Seizures |
| 13 | + | + | IMV | Blalock–Taussig operation for TOF | – |
| 14 | + | + | IMV | – | – |
| 15 | + | + | IMV | Tracheotomy operation for inguinal hernia operation for buried penis | Seizures |
| 16 | + | + | IMV | Tracheotomy | HOT |
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IMV intermittent mandatory ventilation, HFO high frequency oscillation, HOT home oxygen therapy, PDA patent ductus arteriosus, CoA coarctation of the aorta, TOF tetralogy of Fallot
Signs of acute respiratory failure were observed in 13 patients after birth. To resuscitate, oxygen was administered to all 16 patients. Mechanical ventilators were used for 9 patients to control acute respiratory failure after tracheal intubation. The tracheal intubation was unsuccessful in 1 patient due to severe tracheal stenosis.
The various types of congenital malformations were observed in all 16 patients with trisomy 13; consultations with specialized surgeons were conducted to determine whether surgical treatment would be possible. Due to the difficulty of tube withdrawal after prolonged intubation, tracheotomy was performed for 2 patients. After tracheotomy, respiratory conditions in both patients were stabilized and they were able to survive for long-time period over 7 years. Abdominal operation for umbilical hernia was performed in 1 patient and for inguinal hernia in 1 patient. Plastic operation of the cleft lip and palate was performed in 1 patient in order to improve the cosmetic problem at 210 days of birth before discharge. Cardiac surgery was performed in 2 patients with trisomy 13. Ligation for patent ductus arteriosus and subclavian methods for coarctation of the aorta were carried out for 1 patient but she was died at 592 days of birth because of respiratory failure. Cardiac operation using Blalock–Taussig method was performed for TOF in 1 patient at 330 days of birth and she has survived for more than 3 years and 3 months now.
Outcomes and Main Cause of Death (Table4)
Table 4.
Outcomes and life prognosis with 16 patients of trisomy 13
| Patients | Discharge(days) | Transfer another hospital (days) | Home care | Survival time (days) | Main cause of death |
:--- :--- :--- |
| 1 | – | – | – | 12 h | Respiratory failure due to hypoplastic lung |
| 2 | – | – | – | 1 | DORV |
| 3 | – | – | – | 1 | DORV tracheal stenosis |
| 4 | – | – | – | 2 | DORV |
| 5 | – | – | – | 11 | Heart failure |
| 6 | – | – | – | 14 | PPHN respiratory failure |
| 7 | – | – | – | 21 | Heart failure respiratory failure |
| 8 | – | – | – | 39 | Heart failure |
| 9 | – | – | – | 50 | Heart failure respiratory failure |
| 10 | 204 | – | + | 325 | Bacterial infection |
| 11 | 255 | – | + | 592 (live) | – |
| 12 | 63 | 63 | + | 1186 | Respiratory failure |
| 13 | 349 | 349 | + | 1219 (live) | – |
| 14 | 251 | 251 | + | 1842 (live) | – |
| 15 | 336 | – | + | 2,705 | s/o central apnea |
| 16 | 331 | – | + | 3,713 (live) | – |
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DORV double-outlet right ventricle, PPHN persistent pulmonary hypertension of newborn, s/o suspect of
In-hospital death at NICU occurred in case of 9 patients, whereas 7 patients were discharged with the condition of homecare. Among them, 3 patients were referred by our hospital to another specialized institution close to the patients’ homes for training purpose of homecare nursing at days 349, 251, and 63 after birth. The ages of 4 patients discharged from our hospital to their homes while they were alive were 336, 331, 256, and 204 days after birth. The average of median hospital stay regarding 7 patients who could be discharged including stay in another specialized institute was 256 days after birth. The survival period of the16 patients with trisomy 13 ranged from death at day 1 (within 24 h of birth) to 10 years and 1 month (this patient lives until now).
The breakdown includes death at 1st day of birth, within 24 h, for 1 patient (6.25%), death within 7 days of birth for 4 patients (25.00%), and death within 1 month of birth for 7 patients (43.75%). On the other hand, 9 patients survived for more than 1 month (56.25%), 6 patients survived for more than 1 year (37.50%), and 3 patients survived for more than 5 years (18.75%). Therefore, the survival rates at ages of 1 day, 1 week, 1 month, 6 months, 1 year, 3 years and 5 years were 93.75, 75.00, 56.25, 43.75, 37.50, 31.25, and 18.75%, respectively. Median survival time for all 16 patients, both males and females, were 733, 887, and 534 days, respectively.
We also reviewed the data in detail to analyze the main cause of death. Some characteristic trend was recognized as follows: death occurred in 4 patients within 1 week of birth due to a major organ anomaly; especially, 3 patients had DORV and in 2 of 3, intubation with ventilation management was performed. On the other hand, 4 of 6 patients from long-survival group (over 365 days, now living) had some kind of surgery that included tracheotomy in 2 patients and cardiac surgery in other 2 patients. Moreover, 5 of 6 patients from long-survival group (over 365 days) had a history of ventilation therapy.
The longest survival time was set at 365 days (Fig.1) and 1,825 days equal 5 years (Fig.2) in the Kaplan–Meier survival curves prepared from the mortality data of all 16 patients with trisomy 13. These Kaplan–Meier survival curves showed mainly two groups: one group that survived for about 2 months and the other group survived for over 365 days.
Fig.1.
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Kaplan–Meier survival curves. Kaplan–Meier survival curves are shown for patients with trisomy 13 who received treatment at the neonatal intensive care unit (NICU) of Dokkyo Medical University Hospital. The data represent 16 trisomy 13 patients. The vertical and horizontal axes represent % survival and survival time (days), respectively, setting the longest survival time at 365 days
Fig.2.
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Kaplan–Meier survival curves. Kaplan–Meier survival curves are shown for patients with trisomy 13 who received treatment at the neonatal intensive care unit (NICU) of Dokkyo Medical University Hospital. The data represent 16 trisomy 13 patients. The vertical and horizontal axes represent % survival and survival time (days), respectively, setting the longest survival time at 1,825 days
Discussion
Among autosomal chromosomal abnormalities, birth is generally believed to be possible in patients of trisomies 21, 18, and 13. Of these, trisomies 18 and 13 are also referred to as fatal chromosomal abnormalities that lead to the development of many severe malformations . Most patients with trisomy 18 and 13 present with severe psychomotor delay, even if they survive for a long period [4, 15, 16]. Historically since the 1980s, resuscitation and surgical treatments for trisomy 18 and 13 have been discussed both from medical and ethical standpoints, taking into consideration the quality of life for the patients and families, medical and economical issues, and the medical standards in different countries. In Japan, the concept of “no treatments beyond current treatments” as proposed by Nishida and Sakamoto of the Tokyo Women’s Medical University has influenced many medical institutions with regard to medical care and management of newborn infants. However, recently, there are few reports of patients with trisomy 13 in Japan who have survived for long periods . Notably, the current problem in Japan is, the primary doctors who tend to patients with trisomy 13 or trisomy 18 provide healthcare by their own medical environments and policies, and no discrete guidelines in this concern have been established. Regarding the medical policy of negative intensive treatment performed without the explanation of detailed medical information, a part of support group members with trisomy 18 or trisomy 13 in Japan have expressed a desire for more active intensive treatment after informed consent.
As for the prognoses of trisomy 13, there is one known report published in 2003 by the American Research Group regarding a large-scale survey of 5,515 cases. According to this report, the average age at the time of death was 10 days, but 5.6% of the cases survived for more than 1 year . Besides, other long-term survival cases were reported including a 38-year-old adult case of trisomy 13 . There have been several other reports about long-term survivors with trisomy 13 [4, 5, 8, 16, 19]. As summarized in Table5, our data represent significantly longer median survival periods for the patients with trisomy 13 treated intensively at the NICU.
Table 5.
Compared survival rates of trisomy 13 in our study with previously reported
| Survival time (age) | Denmark n=76 % surviving | England n=16 % surviving | Scotland n=84 % surviving | Atlanta, USA n=114 % surviving | Taiwan n=28 % surviving | Japan (our study) n=16 % survival |
:--- :--- :---
| 1 day | 77 | 69 | 75 | 86 | 89 | 93.75 |
| 7 days | 40 | 38 | 50 | 56 | 61 | 75 |
| 1 month | 23 | 13 | 28 | 30 | 29 | 56.25 |
| 6 months | 10 | 0 | ND | 11 | 7 | 43.75 |
| 1 year | ND | 0 | 3 | 9 | 4 | 37.5 |
| 3 years | ND | 0 | ND | ND | 4 | 31.25 |
| 5 years | ND | 0 | ND | ND | ND | 18.75 |
| Median survival (days) | 2.5 | 4 | 8.5 | 7 | 9 | 733 |
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ND no data
Of note, two reports have compared between active intensive treatment and non-active treatment with regard to prognosis of trisomy 18 [11, 14] but no such reports are available regarding trisomy 13. Herein, we present for the first time a study of patients with trisomy 13 who received active intensive treatment and discuss the clinical picture and prognosis. Our data suggest that the patients with trisomy 13 who have survived over 60 days after birth may have a high probability of long-term survival. In the previous report on survival of trisomy 18 patients, those who survived longer than 7 months after birth might have a high probability of long-term survival prognosis . In this study, we should not only consider 9/16 (56.25%) of patients with trisomy 13 who died in the hospital without discharge but also those 7/16 (43.75%) of patients who were discharged or transferred for purpose of homecare. As our data reveal, one of the major causal factors involved in long-term survival of patients with trisomy 13 is whether the provision of active resuscitation should follow immediately after birth, and then the various surgical treatments should be performed in survivors such as tracheotomy and cardiac surgery. Two previous studies reported intensive cardiac management in patients with trisomy 13 or trisomy 18 [20, 21]. These studies summarized intensive cardiac management including pharmacological intervention, performed to improve the survival periods in patients with trisomy 13 or trisomy 18. In our study, two patients underwent cardiac surgery. One surgical procedure involved ligation of patent ductus arteriosus and subclavian methods for coarctation of the aorta, the other was Blalock–Taussig procedure for TOF; and both survived for long periods: 592 days and more than 3 years and 3 months (living until date), respectively. In addition, all 3 patients with DORV died within 2 days of birth, despite that fact that ventilation management was performed in 2 cases.
In conclusion, although our clinical data represent a small number of trisomy 13 patients, we did observe significantly better survival prognosis in the patients who received intensive treatment management. Our clinical results underscore the need to consider obtaining informed consent and counseling from patients’ families for the future determinations of treatment policy for trisomy 13 patients.
Acknowledgments
This work was supported by three grants: one from Japanese Ministry of Education, Culture, Sports, Science and Technology (MEXT) (Ref. KAKENHI 18790733), second Grant-in-Aid for Young Scientists (B), and the Investigator-Initiated Research Grant from Dokkyo Medical University (No. 2005-01-5/No. 2009).
Open Access
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21.Kaneko Y, Kobayashi J, Yamamoto Y, et al. Intensive cardiac management in patients with trisomy 13 or trisomy 18. American Journal of Medical Genetics. 2008;146A:1372–1380. doi: 10.1002/ajmg.a.32311. [DOI] [PubMed] [Google Scholar]
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188789 | https://www.statisticshowto.com/hypergeometric-distribution-examples/ | Skip to content
Hypergeometric Distribution: Examples and Formula
< List of probability distributions < Hypergeometric Distribution
Contents:
What is the Hypergeometric Distribution?
Formula
Worked Examples
Real Life Uses
What is the Hypergeometric Distribution?
The hypergeometric distribution is a discrete probability distribution that calculates the probability an event happens k times in n trials when sampling from a small population without replacement. In other words, it describes the probability of getting a certain number of “successes” in a sample of a given size, drawn from a population of a known size, without putting any of the “successes” back into the population.
The hypergeometric distribution is very similar to the binomial distribution. In fact, the binomial distribution is a very good approximation of the hypergeometric distribution as long as you are sampling 5% or less of the population. Therefore, in order to understand the hypergeometric distribution, you should be familiar with the binomial distribution. Plus, you should be comfortable with using the combinations formula.
Hypergeometric Distribution Formula
The (somewhat formal) definition for the hypergeometric distribution, where X is a random variable, is: Where:
K is the number of successes in the population,
k is the number of observed successes,
N is the population size,
n is the number of draws.
Hypergeometric Distribution Example 1
A deck of cards contains 20 cards: 6 red cards and 14 black cards. 5 cards are drawn randomly without replacement. What is the probability that exactly 4 red cards are drawn? The probability of choosing exactly 4 red cards is: P(4 red cards) = # samples with 4 red cards and 1 black card / # of possible 4 card samples.
Using the combinations formula, the problem becomes: In shorthand, the above formula can be written as:
(6C414C1)/20C5
where
6C4 means that out of 6 possible red cards, we are choosing 4.
14C1 means that out of a possible 14 black cards, we’re choosing 1.
Solution = (6C414C1)/20C5 = 1514/15504 = 0.0135 The binomial distribution doesn’t apply here, because the cards are not replaced once they are drawn. In other words, the trials are not independent events. For example, for 1 red card, the probability is 6/20 on the first draw. If that card is red, the probability of choosing another red card falls to 5/19.
Hypergeometric Distribution Example 2
A small voting district has 101 female voters and 95 male voters. A random sample of 10 voters is drawn. What is the probability exactly 7 of the voters will be female? Solution: 101C795C3/(196C10)= (17199613200138415)/18257282924056176 = 0.130 Where:
101C7 is the number of ways of choosing 7 females from 101 and
95C3 is the number of ways of choosing 3 male voters from 95
196C10 is the total voters (196) of which we are choosing 10.
That’s because if 7/10 voters are female, then 3/10 voters must be male. Check out our YouTube channel for hundreds of statistics help videos!
Application of Hypergeometric Distribution in Real Life: Examples
The hypergeometric distribution describes the number of successes in a sequence of n trials from a finite population without replacement. At first glance, it might seem that this is a purely academic distribution, but there are actually many different applications of the hypergeometric distribution in real life. One of the most common applications of the hypergeometric distribution is in industrial quality control, such as calculating probabilities for defective parts produced in a factory. Let’s say a factory line produces 1% defective parts that are boxed at the end of the line; The hypergeometric distribution can be used to model the occurrence of defective parts for quality control purposes, as long as the items being sampled from a box are not replaced.
In electrochemistry, the hypergeometric distribution can predict the effect of surface deterioration on electrode behavior for any electrode process with two competing reactions. This gives valuable information about the effectiveness of electrode-electrolyte interfaces and improves interpretation of the measurements of surface properties .
If you play poker, the hypergeometric distribution can tell you the probability of getting 3 of the same suit in a 5 card hand (or any number of other card/hand combinations).
The PowerBall lottery game is a televised, two part drawing. In the first stage, five white balls are drawn randomly from a bowl of 49 balls. In the second stage, one red ball (the PowerBall) is drawn randomly from a bowl of 42 balls. The probability of success changes from one draw to the next (the balls are not replaced), so the probabilities for the game can be modeled with the hypergeometric distribution .
Binomial vs. hypergeometric distribution
The hypergeometric distribution is very similar to the binomial distribution. In fact, the binomial distribution is a good approximation of the hypergeometric distribution if you are sampling 5% or less of the population.
A key difference between the hypergeometric distribution and the binomial distribution is that the former is a discrete probability distribution, while the latter is a continuous probability distribution. This means that the hypergeometric distribution deals with situations where the number of possible outcomes is finite and countable, while the binomial distribution deals with situations where the number of possible outcomes is infinite and uncountable. For example, if you roll a die 10 times, the number of possible outcomes is finite and countable (1, 2, 3, 4, 5, or 6), so we would use the hypergeometric distribution to determine the probability of getting a specific number of 6’s. On the other hand, if you measure the length of a piece of wire to 10 decimal places, the number of possible outcomes is infinite and uncountable, so we would use the binomial distribution to model the probability of getting a certain range of values.
In addition to its applications in sampling without replacement, the hypergeometric distribution has other important applications in various fields such as genetics, ecology, and epidemiology. For example, it can be used to calculate the probability of getting a certain number of disease cases in a population of a given size, without assuming that the number of cases follows a normal distribution.
One important point to note is that the hypergeometric distribution assumes that the population size is much larger than the sample size. In other words, the proportion of “successes” in the population doesn’t change significantly after each draw. If the sample size is large relative to the population size, then the hypergeometric distribution is not an appropriate model and we should use the binomial distribution instead.
In conclusion, probability distributions are powerful tools for analyzing and interpreting data in various fields. While the binomial distribution is widely known and used, it’s important to understand that there are other probability distributions that are just as essential. The hypergeometric distribution is one such distribution, which is used to model situations where we are sampling without replacement. By understanding the key differences between the hypergeometric and binomial distributions, we can choose the appropriate distribution for our analysis and avoid common mistakes.
References
Fahidy, T. (2012). An application of hypergeometric distribution theory to competitive processes at deteriorating electrode surfaces. In Electrochemistry Communications, 282-284
Anderson, J. & Schmidt, J. (2002). Playing Powerball?. Proceedings. Annual Conference on Taxation and Minutes of the Annual Meeting of the National Tax Association. pp. 377-382.
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188790 | https://artofproblemsolving.com/wiki/index.php/Modular_arithmetic?srsltid=AfmBOooQFxzw3Zo8xErId3KhPgJZ1mcSVZINt3fB9WJ-IgW-m_MNHydt | Art of Problem Solving
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Modular arithmetic
Modular arithmetic is a special type of arithmetic that involves only integers. Since modular arithmetic is such a broadly useful tool in number theory, we divide its explanations into several levels:
Introduction to modular arithmetic
Intermediate modular arithmetic
Olympiad modular arithmetic
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188791 | https://www.youtube.com/watch?v=NlpykbGDzF8 | Simultaneous Equations Three Variables Using Elimination - Math lesson
tecmath
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Posted: 2 Sep 2020
Simultaneous equations three variables using elimination - easy - eliminate down from three equations with three variables, to two equations with two variables...to your answer!
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Transcript:
good day welcome to the techmath channel I'm Josh in this video we're going to be using the elimination method to solve simultaneous equations that involve three variables so ones like you can see here we have three equations and they have three variables x y and Zed and the way we're going to do this is as follows we're going to by combining reduce these equations down to two equations that involve two variables and that will then allow us to use the elimination method on two variable two equations to reduce down to one variable and go back from there so as you see this is how this works so the way that we're going to do this is as follows the first thing we're going to do is we are going to combine two of these equations here so the ones I'm going to combine is equation one and equation two and you can see that by doing this if you look at the Zed variable we have negative Z here and positive Z here these guys are going to cancel each other out and we're going to be left with an equation with only two variables the X and Y so let's do that so if we were to combine these guys 3x and 2x gives us 5x and pos2 Y and -3 y gives us y the Zs cancel each other out negative Z and positive Z cancel each other out and this is equal to 11 + 7 which is equal to 18 so we have our first equation which has only two variables X and Y the next thing we're going to do is we're going to combine two other equations but the one we have to combine it has to have this third equation in it in in order for this to work so we're going to combine equation 2 and equation 3 here as you can see if we consider the Zed variable you can see that this is positive Z and this is -2 Z so what we're going to have to do is we are going to have to multiply this entire equation by two so we to do that this is what we would get so we would get 2x 2 which is equal 4x - 3 y 2 which is - 6 y and POS 2 Z this is going to be equal to 7 2 which is 14 cool so equations now that we're actually going to be combining is the modified version of this second equation that we have down here so let's go through and let's do that so what do we get when we do that I'm going to write the answer underneath so 5x + 4x gives us 9x positive y- 6 y gives us -5 y -2 Z and pos2 Z well they cancel each other out and this is going to be equal to 12 + 14 which is 26 so now as you can see we have two equations here the ones that are in blue and both of them havly have two variables they have the X variable and the Y variable as you can see so now what we can do is we can combine these two equations and what we can do is reduce down to only one variable so let's go through and do that so if we have a look at our two equations I'm going to eliminate the Y variable okay leave only leaving us with the X here so the way that I'm going to do that if you have a look we have - 5 y here and Y here so if I multiply this entire equation by5 I'll end up with a positive 5 Y which will cancel out down here so I'm going to give the result of that down here so let's do that uh what do we have we have 5x -5 is going to be -25 5x - y -5 is POS 5 y we also have 18 -5 which is going to give us 90 cool all right now that means we can go through and eliminate because as you can see these guys are going to cancel each other out what do we end up with we end up with 9x - 25x which is equal to -6x and this is equal to 26 - 90 which is equal to -64 okay easy to solve 16 well -16 goes into both of them so we end up with x = 4 and that's the very very first variable that we managed to solve there now what we do is we're going to substituting this into one of our equations that has two variables the X and the Y we're going to find out what Y is and then we're going to use our answers there and substitute into one of these equations and then we'll have all our answers so let's go through and do that so first off let's substitute x = 4 into one of these equations let's put it into this one up here okay so we have x = 4 this is going to be 5 4 - y = 18 cool so what do we end up with we end up with 20 - y = 18 and therefore when we solve this we take 20 off both sides we end up with -2 so Y is = to 2 20 - 2 is = 18 so I'm going to actually put that in a different color we have our second variable here that is y is equal to 2 all right cool now what we do is we have x = 4 we have Y = 2 we substitute these into one of our original equations and we're going to find out what Z equals so let's go through and do that I'm just going to write out this entire equation again so we have 3x + 2 y this is equation 1 - Z = 11 all right so what do we do when we work this one out uh 3x well X is equal to 4 so 3 4 is 12 plus 2 Y 2 2 is = 4 - Z = 11 cool 12 + 4 is = 16 - Z = 11 I reckon you'll be able to see what Z equals so 16 - 5 is equal to 11 so Z is equal to 5 we have our three variables here we have X is equal to 4 we have y is equal to 2 and Z is equal to 5 you go through and sub these in just to check you go okay 2 4 is 8 - 6 there 2 - 6 is 2 + 5 is 7 so it's correct there uh 20 + 2 is 22 - uh 2 5 which is 10 uh 22 - 10 is equal to 12 so go through check make sure it's okay and that's how you solve it we reduce from three equations with three variables down to two equations with two variables and then down to one equation one variable anyway that's how you solve simultaneous equations using three variables and the elimination method thank you for watching we'll see you next time bye |
188792 | https://scienceready.com.au/pages/forces-on-an-inclined-surface?srsltid=AfmBOopRAHLLb9vTF3JxSEU09PvmiwpjcHcy8aHbccwcA35VEvDOrIYY | Forces on an Inclined Surface: Friction vs. No Friction –HSC Physics – Science Ready
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Forces On An Inclined Surface
This topic is part of the HSC Physics course under the section Forces, Acceleration and Energy.
Physics textbooks
HSC Physics Syllabus
explore the concept of net force and equilibrium in one-dimensional and simple two-dimensional contexts using: (ACSPH050)
– algebraic addition
– vector addition
– vector addition by resolution into components
solve problems or make quantitative predictions about resultant and component forces by applying the following relationships:
– F A B=−F B A F A B=-F B A
Physics textbooks
– F x=F cos θ F x=F cos θ, F y=F sin θ F y=F sin θ
conduct a practical investigation to explain and predict the motion of objects on inclined planes (ACSPH098)
Forces on an Inclined Surface WITHOUT Friction
Forces on an Inclined Surface WITH Friction
Forces on an inclined surface is a crucial topic to grasp in HSC Physics as it forms the basis for understanding various real-world applications such as ramps, inclined planes, and more.
Physics textbooks
Forces on an Inclined Surface
When an object is placed on an inclined surface, it experiences various forces acting upon it.
The two primary forces at play are:
Gravitational force (F g F g): The force exerted by Earth on the object, which acts vertically downwards. This is also known as the weight force.
Normal force (F N F N): The perpendicular force exerted by the surface on the object, acting at a 90-degree angle to the surface.
Resolving the Weight Force
To analyse the forces acting on the object, we need to resolve the gravitational force into two components:
Perpendicular to the surface (F g⊥F g⊥): The component of gravitational force that acts in the opposite direction of the normal force.
Parallel to the surface (F g∥F g‖): The component of gravitational force that acts along the incline.
Using trigonometry, we can derive the following relationships:
F g⊥=F g×cos θ F g⊥=F g×cosθ
F g∥=F g×sin θ F g‖=F g×sinθ
Where θ is the angle between the inclined surface and the horizontal plane, and F g=m g F g=m g, with m representing the mass of the object and g being the acceleration due to gravity (approximately 9.81 m/s²).
No Friction Scenario
In the absence of friction, the only force acting on the object along the incline is F g∥F g‖. This force causes the object to accelerate down the incline, and the acceleration can be calculated using Newton's second law:
F n e t=m a F n e t=m a
F g∥=m a F g‖=m a
m g×sin θ=m a m g×sinθ=m a
a=g×sin θ a=g×sinθ
In this case, the acceleration of the object is independent of its mass and solely depends on the angle of the incline and the acceleration due to gravity. In the absence of friction, the parallel component of weight force is never balanced by any other forces, so the mass will also undergo motion with acceleration.
With Friction Scenario
When friction is present, we need to consider the force of friction (F f F f) acting on the object.
The force of friction opposes the motion of the object and is proportional to the normal force acting on it:
F f=μ×F n F f=μ×F n
Where μ is the coefficient of friction (a dimensionless value).
In this scenario, the net force acting on the object along the incline is given by:
F n e t=F g∥−F f F n e t=F g‖−F f
Using Newton's second law, we can determine the acceleration:
m a=F g∥−F f m a=F g‖−F f
Since F n=F g⊥F n=F g⊥, we can rewrite the frictional force as:
F f=μ×F g⊥F f=μ×F g⊥
F f=μ t i m e s m g cos θ F f=μ t i m e s m g cosθ
Now, we can find the acceleration:
a=F g∥−F f m a=F g‖−F f m
a=m g sin θ−μ m g cos θ m a=m g sinθ−μ m g cosθ m
a=g(sin θ−μ cos θ)a=g(sinθ−μ cosθ)
In this case, the acceleration of the object is dependent on the angle of the incline, the acceleration due to gravity, and the coefficient of friction.
When the parallel component of weight force acting down the incline is balanced by friction acting up the incline, the net force acting on the mass would equal zero. As a result, the object is in equilibrium (either at rest or moving at constant velocity).
When a mass is at rest on an inclined surface, the parallel component of weight force is balanced by static friction.
When a mass is moving at constant velocity, the parallel component of weight force is balanced by kinetic friction.
When the parallel component of weight force is greater than the kinetic friction, the mass will accelerate down the incline.
RETURN TO MODULE 2: DYNAMICS
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188793 | https://goldbook.iupac.org/terms/view/RT06783/plain | Title: redox potential
Long Title: IUPAC Gold Book - redox potential
DOI: 10.1351/goldbook.RT06783
Status: current
Definition
Any oxidation-reduction (redox) reaction can be divided into two half-reactions: one in which a chemical species undergoes oxidation and one in which another chemical species undergoes reduction. If a half-reaction is written as a reduction, the driving force is the reduction potential. If the half-reaction is written as oxidation, the driving force is the oxidation potential related to the reduction potential by a sign change. So the redox potential is the reduction/oxidation potential of a compound measured under standards conditions against a standard reference half-cell. In biological systems the standard redox potential is defined at pH – 7.0 versus the hydrogen electrode and partial pressure of hydrogen = 1 bar.
Related Terms
- bar:
- electrode potential:
- oxidation:
- partial pressure:
Source
- PAC, 1997, 69, 1251. 'Glossary of terms used in bioinorganic chemistry (IUPAC Recommendations 1997)' on page 1294 (
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- html:
- json:
- xml:
Citation: Citation: 'redox potential' in IUPAC Compendium of Chemical Terminology, 5th ed. International Union of Pure and Applied Chemistry; 2025. Online version 5.0.0, 2025. 10.1351/goldbook.RT06783
License: The IUPAC Gold Book is licensed under Creative Commons Attribution-ShareAlike CC BY-SA 4.0 International ( for individual terms.
Collection: If you are interested in licensing the Gold Book for commercial use, please contact the IUPAC Executive Director at executivedirector@iupac.org .
Disclaimer: The International Union of Pure and Applied Chemistry (IUPAC) is continuously reviewing and, where needed, updating terms in the Compendium of Chemical Terminology (the IUPAC Gold Book). Users of these terms are encouraged to include the version of a term with its use and to check regularly for updates to term definitions that you are using.
Accessed: 2025-09-03T12:51:59+00:00 |
188794 | https://math.stackexchange.com/questions/206158/solving-recurrence-relation-in-2-variables | Solving recurrence relation in 2 variables - Mathematics Stack Exchange
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Solving recurrence relation in 2 variables
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We already know how to solve a homogeneous recurrence relation in one variable using characteristic equation. Does a similar technique exists for solving a homogeneous recurrence relation in 2 variables. More formally, How can we solve a homogeneous recurrence relation in 2 variables? For example,
F(n,m) = F(n-1,m) + F(n,m-1)
Given some initial conditions, how can we solve the above recurrence relation?
recurrence-relations
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asked Oct 2, 2012 at 19:27
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1 You might be intrested in cellular automatons and number triangles.mick –mick 2012-10-02 21:20:30 +00:00 Commented Oct 2, 2012 at 21:20
If im not mistaken if your recursion contains no minus , division , root or logaritm then F(n,n) is usually expressible in closed form. If not then by adding the concept of superfunctions it increases the probability alot.mick –mick 2012-10-02 21:23:20 +00:00 Commented Oct 2, 2012 at 21:23
@mick For the current question we can safely assume that the recursion is a simple linear recursion with no constants.gibraltar –gibraltar 2012-10-03 05:06:57 +00:00 Commented Oct 3, 2012 at 5:06
1 You might be intrested in en.wikipedia.org/wiki/Master_theoremmick –mick 2012-10-03 12:37:16 +00:00 Commented Oct 3, 2012 at 12:37
1 Can you apply the master theorem to multi-variable recurrences?Joey Eremondi –Joey Eremondi 2013-08-14 21:58:19 +00:00 Commented Aug 14, 2013 at 21:58
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You can use generating functions, as we did in the single variable case.
Let G(x,y)=∑m,n≥0 F(n,m)x n y m G(x,y)=∑m,n≥0 F(n,m)x n y m. We'll express G G in a nice form from which one can recover F(n,m)F(n,m).
As you didn't specify initial conditions, let
H 1(x)=∑n≥0 F(n,0)x n,H 2(y)=∑m≥0 F(0,m)y m,c=F(0,0)H 1(x)=∑n≥0 F(n,0)x n,H 2(y)=∑m≥0 F(0,m)y m,c=F(0,0)
By the recurrence of G G, if we multiply it by 1−x−y 1−x−y, most of the terms will cancel. I'll elaborate on that.
I choose 1−x−y 1−x−y in a similar manner to that of constructing the characteristic polynomial in one variable: 1 1 corresponds to F(n,m)F(n,m), x x to F(n−1,m)F(n−1,m) and y y to F(n,m−1)F(n,m−1), i.e. F(n−a,m−b)F(n−a,m−b) is replaced by x a y b x a y b.
G(x,y)(1−x−y)=∑m,n≥0 F(n,m)(x n y m−x n+1 y m−x n y m+1)=G(x,y)(1−x−y)=∑m,n≥0 F(n,m)(x n y m−x n+1 y m−x n y m+1)=
We'll group coefficients of the same monomial:
∑m,n≥1(F(n,m)−F(n−1,m)−F(n,m−1))x n y m+∑m,n≥1(F(n,m)−F(n−1,m)−F(n,m−1))x n y m+
∑n≥1(F(n,0)−F(n−1,0))x n+∑m≥1(F(0,m)−F(0,m−1))y m+F(0,0)=∑n≥1(F(n,0)−F(n−1,0))x n+∑m≥1(F(0,m)−F(0,m−1))y m+F(0,0)=
H 1(x)(1−x)+H 2(y)(1−y)−c H 1(x)(1−x)+H 2(y)(1−y)−c
So, finally,
G(x,y)=H 1(x)(1−x)+H 2(y)(1−y)−c 1−x−y G(x,y)=H 1(x)(1−x)+H 2(y)(1−y)−c 1−x−y
(Compare this to the relation F i b(x)=x 1−x−x 2 F i b(x)=x 1−x−x 2 where F i b F i b is the generating function of the Fibonacci sequence.)
How do we recover F F? We use the formal identity 1 1−x−y=∑i≥0(x+y)i 1 1−x−y=∑i≥0(x+y)i. Let S(x,y)=H 1(x)(1−x)+H 2(y)(1−y)−c=∑n,m s n,m x n y m S(x,y)=H 1(x)(1−x)+H 2(y)(1−y)−c=∑n,m s n,m x n y m. It gives us:
G(x,y)=∑i≥0 S(x,y)(x+y)i=∑n,m≥0(∑a,b≥0 s a,b(n+m−a−b n−a))x n y m G(x,y)=∑i≥0 S(x,y)(x+y)i=∑n,m≥0(∑a,b≥0 s a,b(n+m−a−b n−a))x n y m
So F(n,m)=∑a,b≥0 s a,b(n+m−a−b n−a)F(n,m)=∑a,b≥0 s a,b(n+m−a−b n−a). I have an hidden assumption - that S S is a polynomial! Otherwise convergence becomes an issue.
I guess that your initial conditions are F(n,0)=1,F(0,m)=δ m,0 F(n,0)=1,F(0,m)=δ m,0, which give S(x,y)=1 S(x,y)=1, so F(n,m)=(n+m n)F(n,m)=(n+m n).
EDIT: In the general case, where F(n,m)=∑a,b c a,b F(n−a,m−b)F(n,m)=∑a,b c a,b F(n−a,m−b) where the sum is over finitely many tuples in N 2−∖{(0,0)}N 2−∖{(0,0)}, the generating function will be of the form H(x,y)1−∑a,b c a,b x a y b H(x,y)1−∑a,b c a,b x a y b where H H depends on the initial conditions.
When we had one variable, we wrote q(x)1−∑a i x i=∑q i(x)1−r i x q(x)1−∑a i x i=∑q i(x)1−r i x where r−1 i r i−1 is a root of 1−∑a i x i 1−∑a i x i and used 1 1−c x=∑c i x i 1 1−c x=∑c i x i.
With 2 variables, this is not always possible, but we can write 1 1−∑a,b c a,b x a y b=∑i≥0(∑a,b c a,b x a y b)i 1 1−∑a,b c a,b x a y b=∑i≥0(∑a,b c a,b x a y b)i and use the binomial theorem to expand. We can also use complex analysis methods to derive asymptotics of F(n,m)F(n,m) from the generating functions.
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edited Jan 5, 2013 at 21:59
answered Jan 5, 2013 at 17:08
OfirOfir
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3 Is any CAS software able to solve it automatically?skan –skan 2017-12-10 20:29:14 +00:00 Commented Dec 10, 2017 at 20:29
2 Can you recommend a textbook, technical note or paper where this method is explained in more detail. What materials did you use when learning this?nilo de roock –nilo de roock 2021-01-25 13:18:13 +00:00 Commented Jan 25, 2021 at 13:18
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Use generating functions like the one variable case, but with a bit of extra care. Define:
G(x,y)=∑r,s≥0 F(r,s)x r y s G(x,y)=∑r,s≥0 F(r,s)x r y s
Write your recurrence so there aren't subtractions in indices:
F(r+1,s+1)=F(r+1,s)+F(r,s+1)F(r+1,s+1)=F(r+1,s)+F(r,s+1)
Multiply by x r y s x r y s, sum over r≥0 r≥0 and s≥0 s≥0. Recognize e.g.:
∑r,s≥0 F(r+1,s)x r y s∑r,s≥0 F(r+1,s+1)x r y s=1 x(G(x,y)−∑s≥0 F(0,s)y s)=G(x,y)−G(0,y)x=1 x(G(x,y)−∑s≥0 F(0,s)y s−∑r≥0 F(r,0)x s+F(0,0))=G(x,y)−G(0,y)−G(x,0)+F(0,0)x y∑r,s≥0 F(r+1,s)x r y s=1 x(G(x,y)−∑s≥0 F(0,s)y s)=G(x,y)−G(0,y)x∑r,s≥0 F(r+1,s+1)x r y s=1 x(G(x,y)−∑s≥0 F(0,s)y s−∑r≥0 F(r,0)x s+F(0,0))=G(x,y)−G(0,y)−G(x,0)+F(0,0)x y
Here G(0,y)G(0,y) and G(x,0)G(x,0) are boundary conditions. If you are lucky, the resulting equation can be solved for G(x,y)G(x,y).
In the specific case of binomial coefficients, you have F(r,0)=F(0,r)=1 F(r,0)=F(0,r)=1, so that G(x,0)=1 1−x G(x,0)=1 1−x and G(0,y)=1 1−y G(0,y)=1 1−y:
G(x,y)−1/(1−y)−1/(1−x)+1 x y=G(x,y)−1/(1−y)x+G(x,y)−1/(1−x)y G(x,y)−1/(1−y)−1/(1−x)+1 x y=G(x,y)−1/(1−y)x+G(x,y)−1/(1−x)y
The result is:
G(x,y)=1 1−x−y=∑n≥0(x+y)n G(x,y)=1 1−x−y=∑n≥0(x+y)n
This is:
[x r y s]G(x,y)=(r+s r)=(r+s s)[x r y s]G(x,y)=(r+s r)=(r+s s)
as expected.
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answered Apr 15, 2014 at 22:14
vonbrandvonbrand
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3 That's a great answer. Did you figure out yourself or it's on the books?skan –skan 2017-03-11 19:10:23 +00:00 Commented Mar 11, 2017 at 19:10
1 My own derivartion.vonbrand –vonbrand 2018-05-30 13:42:16 +00:00 Commented May 30, 2018 at 13:42
2 @skan That means he refuses to share his sources.nilo de roock –nilo de roock 2021-01-25 13:20:19 +00:00 Commented Jan 25, 2021 at 13:20
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You will need to specify F(0,r)F(0,r) and F(s,0)F(s,0) as initial conditions. Your recurrence is precisely that for Pascal's triangle. If you specify F(0,r)=F(s,0)=1 F(0,r)=F(s,0)=1 you will have F(n,m)=(n+m n)F(n,m)=(n+m n). You can use linearity to turn it into a sum over initial conditions and binomial coefficients. If your initial condition is F(1,0)=1,F(r,0)=F(0,s)=0 F(1,0)=1,F(r,0)=F(0,s)=0 you will get a Pascal's triangle shifted down to the left by one slot, so F(m,n)=(m+n−1 m−1)F(m,n)=(m+n−1 m−1)
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answered Oct 2, 2012 at 20:29
Ross MillikanRoss Millikan
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does recurrence relation in more than one variable also increases the initial conditions required to solve the recurrence relation? For example if recurrence relation has three variables and the order is one, can it be solved by an initial condition like F(n=0,m=0,k=0)=1 or more conditions are required? I am asking because this is a non separable initial condition.Userhanu –Userhanu 2025-09-23 10:43:26 +00:00 Commented Sep 23 at 10:43
1 @Userhanu:you typically need initial conditions over one less dimension than variables. With one variable you need a number of points. In this answer with two variables we need two rays. With three you would need a few planes.Ross Millikan –Ross Millikan 2025-09-23 13:32:09 +00:00 Commented Sep 23 at 13:32
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You can express it as linear equation systems on a grid and solve it with your favourite Linear Algebra method.
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answered Jan 27, 2023 at 22:21
mathreadlermathreadler
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188795 | https://www.sciencedirect.com/science/article/abs/pii/S0016236118311608 | Analysis of dry, wet and superwet in situ combustion using a novel conical cell experiment - ScienceDirect
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Volume 234, 15 December 2018, Pages 482-491
Full Length Article
Analysis of dry, wet and superwet in situ combustion using a novel conical cell experiment
Author links open overlay panel Alireza Alamatsaz a, Gordon R.Moore b, Sudarshan A.Mehta b, Matthew G.Ursenbach b
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Abstract
A conical combustion cell was built and utilized in combustion experiments at the In Situ Combustion Research Group at the University of Calgary to investigate the effect of continuous in situ air flux drop on the dynamics of the combustion process and to identify the characteristics of the process most importantly the extinction air flux. Nine set of top-down (gravity stable) dry, wet and superwet combustion experiments were conducted at different injection rates in sand packs representing typical Athabasca heavy oil reservoirs. The temperature profiles, the produced combustion gases and liquids as well as the unpacked core were examined. Minimum injected air flux at the termination of the experiments was calculated; gas phase combustion parameters were analyzed. Some of the features specific to combustion experiments in conical cells were investigated.
It is believed that the formation of fuel rich condition especially in presence of water at the location of the combustion front at certain locations of the core at certain circumstances tends to prevent energy generation at the combustion front even at high air injection rates. Also, it was shown that the counter current convection/diffusion of fuel gases from the combustion front to upstream locations of the core also known as the roll cell effect adversely influences the advancement of the high temperature combustion front which in turn impacts the oil mobilization and recovery.
Combustion performance analysis using oil recovered/volume burned calculations for dry conical tests and oil recovered vs. volume steamed and oxygen oil ratio vs. volume burned as the crucial economical parameters of the design of wet combustion processes were calculated. This study greatly enhances the understanding of the complexity of the in situ combustion process and illustrates its characteristic behavior at or close to its exhaustion.
Graphical abstract
A conical combustion cell was conceived and utilized in dry, wet and super wet combustion experiments at the In Situ Combustion Research Group at the University of Calgary to investigate the effect of continuous in situ air flux drop on the dynamics of the combustion process and to identify the characteristics of the process most importantly the extinction air flux. Minimum injected air flux at the termination of the experiments was calculated; gas phase combustion parameters were analyzed. Some of the features specific to combustion experiments in conical cells were investigated. Combustion performance analysis using oil recover/volume burned calculations for dry conical tests and oil recovered vs. volume steamed and oxygen oil ratio vs. volume burned as the crucial economical parameters of the design of wet combustion processes were calculated. This study greatly enhances the understanding of the complexity of the combustion process and illustrates its characteristic behavior at or close to its exhaustion.
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Introduction
Measurement of the air flux that is requisite for the generation, sustenance and propagation of combustion front is necessary to calculate the minimum required capacity of the air injection facility and match this capacity to the volume of reservoir which is to be swept by the thermal zone . It is essential to know the injection air flux at which the fire front extinguishes.
The combustion literature annals various experimental tools introduced to investigate not only the physical and chemical aspects of a fireflood process during air injection based enhanced oil recovery processes at different conditions but also the strength and weakness of each experimental tool. Pressurized differential scanning calorimeters (PDSC) are traditionally used to see the effect of temperature and time on heat generation rate under oxidizing atmosphere, to identify temperature ranges for significant mass transfer effects, and to obtain activation energies corresponding to different heat generation events. However, heat flow traces coming off of PDSC’s are known to be sensitive to heating rate, sample size, and gas injection rate. Additionally, with PDSC’s, one cannot replicate gas flux or oil core matrix contact . Thermo Gravimetric Analysis (TGA) tools are another class of experiments that primarily measure the mass loss of fire-flooded cores with time and yield compositional analysis . In conjunction with PDSC, TGAs can also yield kinetics parameters such as activation energies. Yet, similar to PDSC, TGA is sensitive to the heating rate, sample size and gas injection rate and on top of all, can neither capture the effect of (high) operating pressures nor replicate gas flux on oil/core samples , , . Accelerated Rate Calorimetric (ARC) tools yield kinetic parameters . Yet, same problem as PDSC and TGA makes it impossible to quantify the actual air flux at the interface of core and injected air , . Shortcomings of the above tools in small sample-based kinetic studies led to the development of a laboratory tool to measure kinetic parameters of combustion at low air fluxes. Ramped Temperature Oxidation (RTO) is an experimental setup developed at the In Situ Combustion Research Group at the University of Calgary in the 1990’s where the heating of re-combined cores are controlled; a heating ramp rate is imposed to trigger a forced combustion and allow obtaining information on the temperature at which significant changes in the energy generation or consumption rates occur. However, RTO experiments require significant number of costly tests to enable post-testing residual oil composition. Among various research tools, one-dimensional combustion tube has gained much attention , , as it yields important combustion parameters used for designing field scale air injection projects i.e. air and fuel requirements, field scale flue gas composition, burn stability and so on. However, when one dimensional (1-D) combustion tube tests are performed at low air injection rates, due to the high heat capacity of this laboratory equipment –that is primarily designed for elevated temperature and pressure operation– oxygen addition reactions are promoted by the heat transfer through the core holder walls. Therefore, when operated at elevated pressures, the combustion tubes are unable to operate at the low air fluxes required to establish the minimum possible air injection flux while maintaining the combustion reactions in an effective mode.
The authors previously proposed a conical combustion cell which overcomes the above-explained shortcoming of other experimental tools and enables the experimental study of combustion process at a large range of air fluxes in a single experiment without having to adjust the injected air rate to change the air flux manually , . The primary aim of the development of the conical cell is to obtain the minimum air flux below which the combustion process could no longer sustain itself at bond scission reaction mode. The Appendix gives the necessary details of the conical cell shown in Fig. 1, the configuration of the experimental setup and its operation including the heating process and heat loss control during an experiment and the configuration of and the methodology to analyze the basic results, etc. This article discusses the collective findings and advanced analyses for six dry, two wet and one superwet combustion experiments that were conducted using the conical combustion cell.
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Experimental
As will be elaborated, the medium of the test or the sand pack had more or less the same formulation in all nine runs. The bitumen, water and gas saturation were 0.40–0.43, 0.17–0.20 and 0.37–0.40, respectively except for Test 1 where these saturation values were 0.33, 0.38 and 0.28, respectively. Sand, water and bitumen physical properties were the same as those described in Appendix and elsewhere , . Air injection pressure for all the tests was constant at 3550 kPa, typical of
Conclusions
Ignition temperature levels of 350 °C are believed to result in large air requirements during the early stages of the experiments. It is expected that lower ignition temperatures result in less air requirement in early periods of combustion processes.
In order to identify the location of the fire front in similar combustion processes, the temperature level of 400 °C can be picked to be associated with the combustion front in both dry and wet combustion experiments.
Using the conical combustion
Recommendations
Persistent transition from stoichiometric to what appeared to be fuel rich condition while the combustion front traveled through the intersection between the cylindrical and the conical parts of the core suggests a design modification with a conical cell cross section change with smoother build up i.e. acute angle with core main axis would yield lower fuel gas influx into the reaction zone at the intersection of the cylindrical and conical sections.
The only parameter changed in the experiments
Acknowledgement
Authors acknowledge National Science and Engineering Research Council (NSERC), Edward Wichert Graduate Scholarship and the Department of Chemical and Petroleum Engineering at the University of Calgary for financial support and the members of In Situ Combustion Research Group at the University of Calgary for technical assistance throughout this research.
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A parametric study on in-situ hydrogen production from hydrocarbon reservoirs – Effect of reservoir and well properties
2024, International Journal of Hydrogen Energy Citation Excerpt :
The composition of hydrocarbon within the reservoir pore space also affects the hydrogen yield of the in-situ hydrogen production process[26,27]. The nature of the combustion front, the amount of heat released, and the overall temperature increase during combustion are all dependent on the hydrocarbon composition in the reservoir [18,28–30]. Hydrocarbon properties can be classified into several categories based on API gravity (density), viscosity, or chemical composition. Show abstract Energy transition is a key driver to combat climate change and achieve zero carbon future. Sustainable and cost-effective hydrogen production will provide valuable addition to the renewable energy mix and help minimize greenhouse gas emissions. This study investigates the performance of in-situ hydrogen production (IHP) process, using a full-field compositional model as a precursor to experimental validation The reservoir model was simulated as one geological unit with a single point uniform porosity value of 0.13 and a five-point connection type between cell to minimize computational cost. Twenty-one hydrogen forming reactions were modelled based on the reservoir fluid composition selected for this study. The thermodynamic and kinetic parameters for the reactions were obtained from published experiments due to the absence of experimental data specific to the reservoir. A total of fifty-four simulation runs were conducted using CMG STARS software for 5478 days and cumulative hydrogen produced for each run was recorded. Results generated were then used to build a proxy model using Box-Behnken design of experiment method and Support Vector Machine with RBF kernel. To ascertain accuracy of the proxy models, analysis of variance (ANOVA) was conducted on the variables. The average absolute percentage error between the proxy model and numerical simulation was calculated to be 10.82%. Optimization of the proxy model was performed using genetic algorithm to maximize cumulative hydrogen produced. Based on this optimized model, the influence of porosity, permeability, well location, injection rate, and injection pressure were studied. Key results from this study reveals that lower permeability and porosity reservoirs supports more hydrogen yield, injection pressure had a negligible effect on hydrogen yield, and increase in oxygen injection rate corelated strongly with hydrogen production until a threshold value beyond which hydrogen yield decreased. The framework developed in the study could be used as tool to assess candidate reservoirs for in-situ hydrogen production.
### Effect of secondary water body on the in-situ combustion behaviors
2022, Fuel Show abstract In-situ combustion (ISC) is an effective alternative method for the development of heavy oil reservoirs, especially in the later stage of steam injection and a large majority of high water-saturated channels have been formed during this phase. The water-saturated channels, also known as secondary water body (SWB), is regarded as a sensitive factor to the ISC performance. The effect of the SWB with different scales and locations was investigated by using combustion tube experiments. Results show that the existence of SWB is conducive to improving the propagation speed of combustion front combined with the reduction of combustion stability during the ISC process. As combustion front propagates to the SWB, the temperature is quickly decreased accompanied by an unstable propagation process, which inevitably induces the significant reduction of the O 2 availability and CO x emission concentration. Additionally, oversized SWB would lead to a worse combustion process with much lower CO x concentration and O 2 availability or even fall into an extinguishment process. Better combustion performance and oil recovery factor are obtained when SWB has large spacing to the gas injection well or exists in the lower layer. Whereas the premature O 2 breakthrough and unstable fire front propagation are generated if the SWB is close to the gas injection well or forms in the upper layer. The findings of this study are significant for better understanding the ISC performance with the consideration of SWB in heavy oil reservoirs.
### Influence of steam on the coking characteristics of heavy oil during in situ combustion
2020, Fuel Citation Excerpt :
Lapene et al. found that steam slowed down heavy oil LTO reactions and reduced the oxygen consumption rate under 0.5–0.7 MPa . Alamatsaz et al. and Yang et al. found that ISC was carried out more easily with steam according to a combustion tube test. Yang believed that the main mechanism was the promotion of asphaltene displacement by steam. Show abstract When in situ combustion is applied to reservoirs that have been exploited by steam recovery, the influence of high water saturation on the thermal conversion of heavy oil into coke needs to be evaluated. A Xinjiang heavy oil was used as the sample. A thermogravimetric analyzer and a pressurized reactor were used to study the influence of steam on the pyrolysis and oxidation characteristics of heavy oil under atmospheric pressure (0.1 MPa) and reservoir pressure (4 MPa), respectively. Under atmospheric pressure, the presence of steam showed little effect on the pyrolysis and low temperature oxidation characteristics of heavy oil but accelerated the rate of coke oxidation. Under reservoir pressure in an inert atmosphere, steam showed little influence on the coke yield but increased the proportion of alkanes and hydrogen in gas products due to the hydrogen donation of steam. Additionally, the oxidation activity of coke produced with steam decreased. The main reason was a decrease in the number of active sites of coke oxidation according to kinetic analysis. Under reservoir pressure in an oxidizing atmosphere, the coke yield decreased significantly because of the existence of steam, but the coke oxidation activity remained unchanged. The main reason for this coke yield decrease was that steam accelerated coke oxidation and increased the oxidation consumption of the produced coke in the coking temperature range of 250–400°C. In addition, steam promoted the cleavage of oxygen-containing functional groups in heavy oil conversion intermediates, thus inhibiting the formation of coke precursors.
### High-temperature drying behavior and kinetics of lignite tested by the micro fluidization analytical method
2019, Fuel Citation Excerpt :
Generally, for the direct utilization of lignite, the process of water evaporation accounts for the energy of 20−25% in raw coal [7–9]. Moreover, for most of the high-temperature thermal conversion processes, such as pyrolysis, gasification, and combustion, coal drying or water evaporation is always considered as an initial step, limiting the total conversion rate and energy utilization efficiency . So, examining the drying behavior of lignite and calculating its kinetics at high temperature, become very necessary and essential to understand the thermal-chemical conversion process and guide the design of a novel reactor. Show abstract To simulate the drying behavior of lignite in the high-temperature thermal conversion process, a micro fluidized bed reactor was adopted to measure the drying characteristics of moist char from 773 K to 1173 K and calculate the drying kinetics. Both temperature and the initial moisture strongly affected the drying behavior by changing the curve shape and drying time. The higher the initial moisture, the more obvious the influence of drying temperature. Compared to the conventional low-temperature lignite drying, the rate curve tested by MFBRA only had the growing-rate zone and falling-rate zone, but did not include the common constant-rate zone. Under the experimental condition, the Lewis drying model can describe the drying characteristics of wet char in MFBRA very well. The apparent diffusion coefficient and activation energy ranged from 1.765 E−10 to 4.756 E−10 m 2/s, and 12.83–14.97 kJ/mol, respectively. The activation energy was much lower than those in literature.
### Workflow of the In Situ Combustion EOR Method in Venezuela: Challenges and Opportunities
2023, ACS Omega
### The Geoscience of In-Situ Combustion and High-Pressure Air Injection
2022, Geosciences Switzerland
View all citing articles on Scopus
View full text
© 2018 Elsevier Ltd. All rights reserved.
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第一课时 等差数列的前n项和公式及相关性质
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项和公式第一课时 差数列的前
n
n
理解等差数列的通项公式与前
n
n
项和公式及相关性质的过程中，发展数学运算和逻辑推理素养
.
n
项和公式
1.
＋
＝？当其他同学忙于把
100
岁的高斯却用下面的方法迅速算出了正确答案：
＋
＋
＋
＋…＋
＋
＝
5 050.
n
＝
n
，那么高斯的计算方法可以表示为
(
a
1
＋
a
100
)
＋
(
a
2
＋
a
99
)
＋
…
＋
(
a
50
＋
a
51
)
＝
101
×
50
＝
5 050.
可以发现，高斯在计算中利用了
a
1
＋
a
100
＝
a
2
＋
a
99
＝
…
＝
a
50
＋
a
51
这一特殊关系，这就是上一节我们学过的性质，它使不同数的求和问题转化了相同数
(
即
101)
的求和，从而简化了运算
.2.
问题
结合问题
1
，对于一般的等差数列，如何求其前
n
项和
S
n
？设其首项为
a
1
，公差为
d
.
提示
倒序相加法
⇒
2
S
n
＝
(
a
1
＋
a
n
)
＋
(
a
2
＋
a
n
－
1
)
＋
…
＋
(
a
n
＋
a
1
)
＝
n
(
a
1
＋
a
n
)
，即
S
n
＝，上述过程实际上用到了等差数列性质里面的首末
“
等距离
”
的两项的和相等
.3.
填空
(1)
等差数列的前
n
项和公式已知量 首项、末项与项数 首项、公差与项数
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求和公式
S
n
＝
S
n
＝
na
1
＋
(2)
两个公式的关系：把
a
n
＝
a
1
＋
(
n
－
1)
d
代入
S
n
＝中，就可以得到
S
n
＝
na
1
＋
d
.
温馨提醒
S
n
＝与
S
n
＝
na
1
＋
d
均为等差数列前
n
项和公式，注意灵活选择、应用
.
当已知
a
1
，
a
n
时，多用
S
n
＝；当已知
a
1
，
d
时，多用
S
n
＝
na
1
＋
d
.4.
做一做
(1)
在等差数列
{
a
n
}
中，
S
10
＝
120
，那么
a
1
＋
a
10
＝
(
)A.12 B.24 C.36 D.48 (2)
在等差数列
{
a
n
}
中，
d
＝
2
，
a
n
＝
11
，
S
n
＝
35
，则
a
1
＝
(
)A.5
或
7 B.3
或
5C.7
或－
1 D.3
或－
15.
思考辨析
正确的在后面的括号内打“√”，错误的打“×”.
(1)
设等差数列
{
a
n
}
的前
n
项和为
S
n
，则
S
n
与
a
n
不可能相等
.(
) (2)
等差数列
{
a
n
}
的前
n
项和
S
n
＝
.(
)(3)
数列的前
n
项和就是指从数列的第
1
项
a
1
起，一直到第
n
项
a
n
所有项的和
.(
)
二、等差数列前
n
项和的性质
1.
思考
等差数列的前
n
项和为
S
n
，你能发现
S
n
与
S
2
n
的关系吗？
提示
S
2
n
＝
a
1
＋
a
2
＋
…
＋
a
n
＋
a
n
＋
1
＋
…
＋
a
2
n
＝
S
n
＋
(
a
1
＋
nd
)
＋
(
a
2
＋
nd
)
＋
…
＋
(
a
n
＋
nd
)
＝
2
S
n
＋
n
2
d
，同样我们发现
S
3
n
＝
3
S
n
＋
3
n
2
d
，这里出现了一个有意思的数列
S
n
，
S
2
n
－
S
n
＝
S
n
＋
n
2
d
，
S
3
n
－
S
2
n
＝
S
n
＋
2
n
2
d
，
…
，是一个公差为
n
2
d
的等差数列
.2.
思考
在等差数列
{
a
n
}
中，如果项数为
2
n
，那么
S
偶
与
S
奇
之间存在哪些关系？
提示
∵
S
偶
＝＝＝
na
n
＋
1
，
S
奇
＝＝＝
na
n
，
∴
S
偶
－
S
奇
＝
n
(
a
n
＋
1
－
a
n
)
＝
nd
，＝
.3.
填空
(1)
若数列
{
a
n
}
是公差为
d
的等差数列，则数列也是等差数列，且公差为
.(2)
若
S
m
，
S
2
m
，
S
3
m
分别为等差数列
{
a
n
}
的前
m
项，前
2
m
项，前
3
m
项的和，则
S
m
，
S
2
m
－
S
m
，
S
3
m
－
S
2
m
也成等差数列，公差为
m
2
d
.(3)
设两个等差数列
{
a
n
}
，
{
b
n
}
的前
n
项和分别为
S
n
，
T
n
，则＝
.(4)
若等差数列的项数为
2
n
，则
S
2
n
＝
n
(
a
n
＋
a
n
＋
1
)
，
S
偶
－
S
奇
＝
nd
，＝
(
S
奇
≠
0).
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(5)
若等差数列的项数为
2
n
＋
1
，则
S
2
n
＋
1
＝
(2
n
＋
1)
a
n
＋
1
(
a
n
＋
1
是数列的中间项
)
，
S
偶
－
S
奇
＝－
a
n
＋
1
，＝
(
S
奇
≠
0).(6)
在等差数列中，若
S
n
＝
m
，
S
m
＝
n
，则
S
m
＋
n
＝－
(
m
＋
n
).4.
做一做
等差数列
{
a
n
}
中，
S
2
＝
4
，
S
4
＝
9
，则
S
6
＝
________.
题型一 等差数列前
n
项和公式的基本运算例
1
在等差数列
{
a
n
}
中：
(1)
已知
a
5
＋
a
10
＝
58
，
a
4
＋
a
9
＝
50
，求
S
10
；
(2)
已知
S
7
＝
42
，
S
n
＝
510
，
a
n
－
3
＝
45
，求
n
.
思维升华
等差数列中基本计算的两个技巧
(1)
利用基本量求值
.(2)
利用等差数列的性质解题
.
训练
1 (1)
设
S
n
是等差数列
{
a
n
}
的前
n
项和
.
若
a
1
＝－
2 021
，
S
6
－
2
S
3
＝
18
，则
S
2 023
＝
(
)A.
－
2 021 B.2 021
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C.2 022 D.2 023(2)(
多选
)
设等差数列
{
a
n
}
的前
n
项和为
S
n
(
n
∈
N
)
，当首项
a
1
和公差
d
变化时，若
a
1
＋
a
8
＋
a
15
是定值，则下列各项中为定值的是
(
)A.
a
7
B.
a
8
C.
S
15
D.
S
16
题型二 等差数列前
n
项和性质及应用例
2 (1)
在等差数列
{
a
n
}
中，
a
1
＝
1
，其前
n
项和为
S
n
，若－＝
2
，则
S
10
等于
(
)A.10 B.100 C.110 D.120 (2)
等差数列
{
a
n
}
的前
m
项和为
30
，前
2
m
项和为
100
，求数列
{
a
n
}
的前
3
m
项的和
S
3
m
. (3)
两个等差数列
{
a
n
}
，
{
b
n
}
的前
n
项和分别为
S
n
和
T
n
，已知＝，求的值
.
思维升华
等差数列前
n
项和运算的几种思维方法
(1)
整体思路：利用公式
S
n
＝，设法求出整体
a
1
＋
a
n
，再代入求解
.(2)
待定系数法：利用当公差
d
≠
0
时
S
n
是关于
n
的二次函数，设
S
n
＝
An
2
＋
Bn
(
A
≠
0)
，列出方程组求出
A
，
B
即可，或利用是关于
n
的一次函数，设＝
an
＋
b
(
a
≠
0)
进行计算
.
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188798 | https://en.wikipedia.org/wiki/Expectiminimax | Published Time: 2004-11-10T14:41:55Z
Expectiminimax - Wikipedia
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1 Pseudocode
2 Expectimax search
3 Alpha-beta pruning
4 See also
5 References
Expectiminimax
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From Wikipedia, the free encyclopedia
Variation of the minimax algorithm
Expectiminimax| Class | Search algorithm |
| Worst-caseperformance | O(b m n m){\displaystyle {\mathcal {O}}(b^{m}n^{m})}, where n{\displaystyle n} is the number of distinct dice throws |
| Best-caseperformance | O(b m){\displaystyle {\mathcal {O}}(b^{m})}, in case all dice throws are known in advance |
The expectiminimax algorithm is a variation of the minimax algorithm, for use in artificial intelligence systems that play two-player zero-sum games, such as backgammon, in which the outcome depends on a combination of the player's skill and chance elements such as dice rolls. In addition to "min" and "max" nodes of the traditional minimax tree, this variant has "chance" ("move by nature") nodes, which take the expected value of a random event occurring. In game theory terms, an expectiminimax tree is the game tree of an extensive-form game of perfect, but incomplete information.
In the traditional minimax method, the levels of the tree alternate from max to min until the depth limit of the tree has been reached. In an expectiminimax tree, the "chance" nodes are interleaved with the max and min nodes. Instead of taking the max or min of the utility values of their children, chance nodes take a weighted average, with the weight being the probability that child is reached.
The interleaving depends on the game. Each "turn" of the game is evaluated as a "max" node (representing the AI player's turn), a "min" node (representing a potentially-optimal opponent's turn), or a "chance" node (representing a random effect or player).
For example, consider a game in which each round consists of a single die throw, and then decisions made by first the AI player, and then another intelligent opponent. The order of nodes in this game would alternate between "chance", "max" and then "min".
Pseudocode
[edit]
The expectiminimax algorithm is a variant of the minimax algorithm and was firstly proposed by Donald Michie in 1966. Its pseudocode is given below.
function expectiminimax(node, depth)
if node is a terminal node or depth = 0
return the heuristic value of node
if the adversary is to play at node
// Return value of minimum-valued child node
let α := +∞
foreach child of node
α := min(α, expectiminimax(child, depth-1))
else if we are to play at node
// Return value of maximum-valued child node
let α := -∞
foreach child of node
α := max(α, expectiminimax(child, depth-1))
else if random event at node
// Return weighted average of all child nodes' values
let α := 0
foreach child of node
α := α + (Probability[child] × expectiminimax(child, depth-1))
return α
Note that for random nodes, there must be a known probability of reaching each child. (For most games of chance, child nodes will be equally-weighted, which means the return value can simply be the average of all child values.)
Expectimax search
[edit]
Expectimax search is a variant described in Universal Artificial Intelligence: Sequential Decisions Based on Algorithmic Probability (2005) by Tom Everitt and Marcus Hutter.
Alpha-beta pruning
[edit]
Bruce Ballard was the first to develop a technique, called -minimax, that enables alpha-beta pruning in expectiminimax trees. The problem with integrating alpha-beta pruning into the expectiminimax algorithm is that the scores of a chance node's children may exceed the alpha or beta bound of its parent, even if the weighted value of each child does not. However, it is possible to bound the scores of a chance node's children, and therefore bound the score of the CHANCE node.
If a standard iterative search is about to score the i{\displaystyle i}th child of a chance node with N{\displaystyle N} equally likely children, that search has computed scores v 1,v 2,…,v i−1{\displaystyle v_{1},v_{2},\ldots ,v_{i-1}} for child nodes 1 through i−1{\displaystyle i-1}. Assuming a lowest possible score L{\displaystyle L} and a highest possible score U{\displaystyle U} for each unsearched child, the bounds of the chance node's score is as follows:
score≤1 n((v 1+…+v i−1)+v i+U×(n−i)){\displaystyle {\text{score}}\leq {\frac {1}{n}}\left((v_{1}+\ldots +v_{i-1})+v_{i}+U\times (n-i)\right)}
score≥1 n((v 1+…+v i−1)+v i+L×(n−i)){\displaystyle {\text{score}}\geq {\frac {1}{n}}\left((v_{1}+\ldots +v_{i-1})+v_{i}+L\times (n-i)\right)}
If an alpha and/or beta bound are given in scoring the chance node, these bounds can be used to cut off the search of the i{\displaystyle i}th child. The above equations can be rearranged to find a new alpha & beta value that will cut off the search if it would cause the chance node to exceed its own alpha and beta bounds:
α i=N×α−(v 1+…+v i−1)+U×(n−i){\displaystyle \alpha {i}=N\times \alpha -\left(v{1}+\ldots +v_{i-1}\right)+U\times (n-i)}
β i=N×β−(v 1+…+v i−1)+L×(n−i){\displaystyle \beta {i}=N\times \beta -\left(v{1}+\ldots +v_{i-1}\right)+L\times (n-i)}
The pseudocode for extending expectiminimax with fail-hard alpha-beta pruning in this manner is as follows:
function -minimax(node, depth, α, β)
if node is a terminal node or depth = 0
return the heuristic value of node
if node is a max or min node
return the minimax value of the node
let N = numSuccessors(node)
// Compute α, β for children
let A = N (α - U) + U
let B = N (β - L) + L
let sum = 0
foreach child of node
// Limit child α, β to a valid range
let AX = max(A, L)
let BX = min(B, U)
// Search the child with new cutoff values
let score = -minimax(child, depth - 1, AX, BX)
// Check for α, β cutoff conditions
if score <= A
return α
if score >= B
return β
sum += score
// Adjust α, β for the next child
A += U - v
B += L - v
// No cutoff occurred, return score
return sum / N
This technique is one of a family of variants of algorithms which can bound the search of a CHANCE node and its children based on collecting lower and upper bounds of the children during search. Other techniques which can offer performance benefits include probing each child with a heuristic to establish a min or max before performing a full search on each child, etc.
See also
[edit]
Minimax
Alpha–beta pruning
Negamax
Expected value
References
[edit]
^ Jump up to: abcdRussell, Stuart Jonathan; Norvig, Peter; Davis, Ernest (2010). Artificial Intelligence: A Modern Approach. Prentice Hall. pp.177–178. ISBN978-0-13-604259-4.
^Michie, D. (1966). "Game-Playing and Game-Learning Automata". Advances in Programming and Non-Numerical Computation. pp.183–200. doi:10.1016/B978-0-08-011356-2.50011-2. ISBN978-0-08-011356-2.
^Ballard, Bruce W. (September 1983). "The -minimax search procedure for trees containing chance nodes". Artificial Intelligence. 21 (3): 327–350. doi:10.1016/S0004-3702(83)80015-0.
^Hauk, Thomas; Buro, Michael; Schaeffer, Jonathan (2006). "Rediscovering -Minimax Search". Computers and Games. Lecture Notes in Computer Science. Vol.3846. pp.35–50. doi:10.1007/11674399_3. ISBN978-3-540-32488-1.
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188799 | https://chem-textbook.ucalgary.ca/chapter-9-main/stoichiometry-of-gaseous-substances-mixtures-and-reactions-introduction/ | Stoichiometry of Gaseous Substances, Mixtures, and Reactions - Introduction - UCalgary Chemistry Textbook
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